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Engineering Mathematics
P. Sivaramakrishna Das
Professor of Mathematics
and
Head of the P.G. Department of Mathematics (Retired)
Ramakrishna Mission Vivekananda College
Mylapore, Chennai
Presently
Professor of Mathematics
and
Head of the Department of Science and Humanities
K.C.G College of Technology
(a unit of Hindustan Group of Institutions
Karapakkam, Chennai)
C. Vijayakumari
Professor of Mathematics (Retired)
Queen Mary’s College (Autonomous)
Mylapore, Chennai
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Copyright © 2017 Pearson India Education Services Pvt. Ltd
Published by Pearson India Education Services Pvt. Ltd, CIN: U72200TN2005PTC057128, formerly
known as TutorVista Global Pvt. Ltd, licensee of Pearson Education in South Asia.
No part of this eBook may be used or reproduced in any manner whatsoever without the publisher’s
prior written consent.
This eBook may or may not include all assets that were part of the print version. The publisher
reserves the right to remove any material in this eBook at any time.
ISBN 978-93-325-1912-1
eISBN 978-93-325-8776-2
Head Office: 15th Floor, Tower-B, World Trade Tower, Plot No. 1, Block-C, Sector 16,
Noida 201 301, Uttar Pradesh, India.
Registered office: 4th Floor, Software Block, Elnet Software City, TS-140, Block 2 & 9, Rajiv
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Fax: 080-30461003, Phone: 080-30461060
www.pearson.co.in, Email: companysecretary.india@pearson.com
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Dedicated
to
Our Beloved Parents
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Brief Contents
Prefacexxxi
About the Authors
xxxiii
A. ALGEBRA
1. Matrices
1.1
2. Sequences and Series
2.1
B. CALCULUS
3. Differential Calculus
3.1
4. Applications of Differential Calculus
4.1
5. Differential Calculus of Several Variables
5.1
6. Integral Calculus
6.1
7. Improper Integrals
7.1
8. Multiple Integrals
8.1
9. Vector Calculus
9.1
C. DIFFERENTIAL EQUATIONS
10. Ordinary First Order Differential Equations
10.1
11. Ordinary Second and Higher Order Differential Equations
11.1
12. Applications of Ordinary Differential Equations
12.1
13. Series Solution of Ordinary Differential Equations and Special Functions
13.1
14. Partial Differential Equations
14.1
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Brief Contents
D. COMPLEX ANALYSIS
15. Analytic Functions
15.1
16. Complex Integration
16.1
E. SERIES AND TRANSFORMS
17. Fourier Series
17.1
18. Fourier Transforms
18.1
19. Laplace Transforms
19.1
F. APPLICATIONS
20. Applications of Partial Differential Equations
Index A01_ENGINEERING_MATHEMATICS-I _FM - (Reprint).indd 8
20.1
I.1
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Contents
Prefacexxxi
About the Authors
xxxiii
1.
Matrices
1.1
1.0 Introduction
1.1
1.1 Basic Concepts
1.1
1.1.1 Basic Operations on Matrices
1.4
1.1.2 Properties of Addition, Scalar Multiplication and Multiplication
1.5
1.2 Complex Matrices
1.7
Worked Examples
1.10
Exercise 1.1
1.13
Answers to Exercise 1.1
1.14
1.3 Rank of a Matrix
1.14
Worked Examples
1.16
Exercise 1.21.23
Answers to Exercise 1.21.24
1.4 Solution of System of Linear Equations
1.24
1.4.1 Non-homogeneous System of Equations
1.24
1.4.2 Homogeneous System of Equations
1.25
1.4.3 Type 1: Solution of Non-homogeneous System of Equations
1.26
Worked Examples1.26
1.4.4 Type 2: Solution of Non-homogeneous Linear Equations Involving
Arbitrary Constants
1.34
Worked Examples1.34
1.4.5 Type 3: Solution of the System of Homogeneous Equations
1.38
Worked Examples1.38
1.4.6 Type 4: Solution of Homogeneous System of Equation Containing
Arbitrary Constants
1.41
Worked Examples1.41
Exercise 1.31.44
Answers to Exercise 1.31.45
1.5 Matrix Inverse by Gauss–Jordan method
1.46
Worked Examples1.47
Exercise 1.41.53
Answers to Exercise 1.41.53
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1.6 Eigen Values and Eigen Vectors
1.54
1.6.0 Introduction
1.54
1.6.1 Vector
1.54
Worked Examples
1.55
1.6.2 Eigen Values and Eigen Vectors
1.56
1.6.3 Properties of Eigen Vectors
1.57
Worked Examples
1.58
1.6.4 Properties of Eigen Values
1.67
Worked Examples
1.70
Exercise 1.5
1.72
Answers to Exercise 1.5
1.73
1.6.5 Cayley-Hamilton Theorem
1.73
Worked Examples
1.75
Exercise 1.6
1.82
Answers to Exercise 1.6
1.83
1.7 Similarity Transformation and Orthogonal Transformation
1.83
1.7.1 Similar Matrices
1.83
1.7.2 Diagonalisation of a Square Matrix
1.84
1.7.3 Computation of the Powers of a Square Matrix
1.85
1.7.4 Orthogonal matrix
1.86
1.7.5 Properties of orthogonal matrix
1.86
1.7.6 Symmetric Matrix
1.87
1.7.7 Properties of Symmetric Matrices
1.88
1.7.8Diagonalisation by Orthogonal Transformation or Orthogonal Reduction 1.89
Worked Examples
1.90
1.8 Real Quadratic Form. Reduction to Canonical Form
1.96
Worked Examples
1.99
Exercise 1.7
1.111
Answers to Exercise 1.7
1.112
Short Answer Questions1.113
Objective Type Questions1.114
Answers1.116
2.
Sequences and Series
2.0 Introduction
2.1 Sequence
2.1.1 Infinite Sequence
2.1.2 Finite sequence
2.1.3 Limit of a sequence
2.1.4 Convergent sequence
2.1.5 Oscillating sequence
2.1.6 Bounded sequence
2.1.7 Monotonic Sequence
Worked Examples
Exercise 2.1
Answers to Exercise 2.1
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2.1
2.1
2.1
2.1
2.2
2.2
2.2
2.2
2.2
2.3
2.3
2.9
2.9
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2.2 Series
2.9
2.2.1 Convergent Series
2.9
2.2.2 Divergent series2.10
2.2.3 Oscillatory series2.10
2.2.4 General properties of series2.10
2.3 Series of Positive Terms
2.10
2.3.1 Necessary Condition for Convergence of a Series
2.10
2.3.2 Test for convergence of positive term series2.11
2.3.3 Comparison tests2.11
Worked Examples2.13
Exercise 2.22.17
Answers to Exercise 2.22.18
2.3.4 De’ Alembert’s Ratio Test
2.18
Worked Examples2.21
Exercise 2.32.25
Answers to Exercise 2.32.26
2.3.5 Cauchy’s Root Test
2.27
Worked Examples2.28
2.3.6 Cauchy’s Integral Test
2.30
Worked Examples2.32
Exercise 2.42.36
Answers to Exercise 2.42.36
2.3.7 Raabe’s Test
2.36
Worked Examples2.37
Exercise 2.52.41
Answers to Exercise 2.52.42
2.3.8 Logarithmic Test
2.42
Worked Examples2.44
2.4 Alternating Series
2.46
2.4.1 Leibnitz’s Test
2.46
Worked Examples2.47
2.5 Series of Positive and Negative Terms
2.50
2.5.1 Absolute Convergence and Conditional Convergence
2.50
2.5.2 Tests for absolute convergence2.50
Worked Examples2.51
Exercise 2.62.55
Answers to Exercise 2.62.55
2.6 Convergence of Binomial Series
2.56
2.7 Convergence of the Exponential Series
2.57
2.8 Convergence of the Logarithmic Series
2.57
2.9 Power Series
2.58
2.9.1 Hadmard’s Formula
2.59
2.9.2 Properties of power series2.60
Worked Examples2.60
Exercise 2.72.66
Answers to Exercise 2.72.67
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Short Answer Questions
Objective Type Questions
Answers
2.67
2.69
2.70
3.
Differential Calculus
3.1
3.0 Introduction
3.1
3.1 Successive Differentiation
3.2
Worked Examples
3.3
Exercise 3.1
3.6
3.1.1 The nth Derivative of Standard Functions
3.7
Worked Examples
3.11
Exercise 3.2
3.16
Answers to Exercise 3.2
3.17
Worked Examples
3.18
Exercise 3.33.24
3.2 Applications of Derivative
3.25
3.2.1 Geometrical Interpretation of Derivative
3.25
3.2.2 Equation of the Tangent and the Normal to the Curve y = f(x)3.25
Worked Examples3.26
Exercise 3.43.33
Answers to Exercise 3.43.34
3.2.3 Length of the Tangent, the Sub-Tangent, the Normal and the Sub-normal
3.34
Worked Examples3.36
Exercise 3.53.38
Answers to Exercise 3.53.38
3.2.4 Angle between the Two Curves
3.38
Worked Examples3.39
Exercise 3.63.42
Answers to Exercise 3.63.43
3.3 Mean-value Theorems of Derivatives
3.43
3.3.1 Rolle’s Theorem
3.43
Worked Examples3.44
3.3.2 Lagrange’s Mean Value Theorem
3.47
Worked Examples3.49
3.3.3 Cauchy’s Mean Value Theorem
3.53
Worked Examples3.54
Exercise 3.73.56
Answers to Exercise 3.73.58
3.4 Monotonic Functions
3.58
3.4.1 Increasing and Decreasing Functions
3.58
3.4.2 Piece-wise Monotonic Function
3.58
3.4.3 Test for increasing or decreasing functions3.59
Worked Examples3.60
Exercise 3.83.65
Answers to Exercise 3.83.66
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3.5 Generalised Mean Value Theorem
3.66
3.5.1 Taylor’s Theorem with Lagrange’s Form of Remainder
3.66
3.5.2 Taylor’s series
3.68
3.5.3 Maclaurin’s theorem with Lagrange’s form of remainder
3.68
3.5.4 Maclaurin’s series
3.68
Worked Examples
3.69
Exercise 3.9
3.74
Answers to Exercise 3.9
3.74
3.5.5 Expansion by Using Maclaurin’s Series of Some Standard Functions
3.75
Worked Examples
3.75
3.5.6 Expansion of Certain Functions Using Differential Equations
3.78
Worked Examples
3.78
Exercise 3.10
3.81
Answers to Exercise 3.10
3.82
3.6 Indeterminate Forms
3.82
0
3.6.1 General L’Hopital’s Rule for
Form
3.84
0
Worked Examples
3.85
Exercise 3.11
3.94
Answers to Exercise 3.11
3.94
3.7 Maxima and Minima of a Function of One Variable
3.94
3.7.1 Geometrical Meaning
3.96
3.7.2 Tests for Maxima and Minima
3.96
Summary
3.97
Worked Examples
3.97
Exercise 3.12
3.103
Answers to Exercise 3.123.104
3.8 Asymptotes
3.104
Worked Examples3.105
3.8.1 A General Method
3.108
3.8.2 Asymptotes Parallel to the Coordinates Axes
3.110
Worked Examples
3.110
3.8.3 Another Method for Finding the Asymptotes
3.113
Worked Examples
3.114
3.8.4 Asymptotes by Inspection
3.115
Worked Examples
3.116
3.8.5 Intersection of a Curve and Its Asymptotes
3.116
Worked Examples
3.116
Exercise 3.13
3.121
Answers to Exercise 3.13 3.122
3.9 Concavity
3.122
Worked Examples3.124
Exercise 3.143.127
Answers to Exercise 3.143.128
3.10 Curve Tracing
3.128
3.10.1 Procedure for Tracing the Curve Given by the Cartesian
Equation f(x, y) = 0.
3.128
Worked Examples3.129
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3.10.2 Procedure for Tracing of Curve Given by Parametric Equations
x = f(t), y = g(t)3.137
Worked Examples3.137
3.10.3 Procedure for Tracing of Curve Given by Equation in Polar
Coordinates f(r, u) = 0
3.141
Worked Examples3.142
Exercise 3.153.146
Answers to Exercise 3.153.146
Short Answer Questions3.148
Objective Type Questions3.149
Answers3.152
4.
Applications of Differential Calculus
4.1 Curvature in Cartesian Coordinates
4.1.0 Introduction
4.1.1 Measure of Curvature
4.1.2 Radius of Curvature for Cartesian Equation of a Given Curve
4.1.3 Radius of Curvature for Parametric Equations
Worked Examples
4.1.4 Centre of Curvature and Circle of Curvature
4.1.5 Coordinates of the Centre of Curvature
Worked Examples
Exercise 4.1
Answers to Exercise 4.1
4.1.6 Radius of Curvature in Polar Coordinates
Worked Examples
4.1.7 Radius of Curvature at the Origin
Worked Examples
4.1.8 Pedal Equation or p - r Equation of a Curve
Worked Examples
4.1.9 Radius of Curvature Using the p - r Equation of a Curve
Worked Examples
Exercise 4.2
Answers to Exercise 4.2
4.2 Evolute
4.2.1 Properties of Evolute
4.2.2 Procedure to Find the Evolute
Worked Examples
Exercise 4.3
Answers to Exercise 4.3
4.3 Envelope
4.3.1Method of Finding Envelope of Single Parameter Family of Curves
Worked Examples
4.3.2 Envelope of Two Parameter Family of Curves
Worked Examples
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4.1
4.1
4.1
4.1
4.2
4.4
4.4
4.11
4.12
4.13
4.15
4.16
4.17
4.19
4.22
4.23
4.25
4.26
4.28
4.29
4.30
4.31
4.31
4.31
4.34
4.34
4.41
4.41
4.42
4.42
4.43
4.45
4.45
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4.3.3 Evolute as the Envelope of Normals
4.48
Worked Examples4.49
Exercise 4.44.52
Answers to Exercise 4.44.53
Short Answer Questions4.54
Objective Type Questions4.54
Answers4.56
5.
Differential Calculus of Several Variables
5.1
5.0 Introduction
5.1
5.1 Limit and Continuity
5.1
Worked Examples
5.4
Exercise 5.1
5.6
Answers to Exercise 5.1
5.6
5.2 Partial Derivatives
5.6
∂z ∂z
5.7
5.2.1 Geometrical Meaning of
,
∂x ∂y
5.2.2 Partial Derivatives of Higher Order
5.8
5.2.3 Homogeneous Functions and Euler’s Theorem
5.8
Worked Examples
5.9
5.2.4 Total Derivatives
5.15
Worked Examples
5.17
Exercise 5.25.24
Answers to Exercise 5.25.26
5.3 Jacobians
5.26
5.3.1 Properties of Jacobians
5.27
Worked Examples5.29
5.3.2 Jacobian of Implicit Functions
5.35
Worked Examples5.35
Exercise 5.35.37
Answers to Exercise 5.35.38
5.4Taylor’s Series Expansion for Function of Two Variables
5.38
Worked Examples5.39
Exercise 5.45.44
Answers to Exercise 5.45.44
5.5Maxima and Minima for Functions of Two Variables
5.45
5.5.1 Necessary Conditions for Maximum or Minimum
5.46
5.5.2 Sufficient Conditions for Extreme Values of f (x, y )
5.46
5.5.3 Working Rule to Find Maxima and Minima of f (x, y )
5.46
Worked Examples5.47
5.5.4 Constrained Maxima and Minima
5.51
5.5.5 Lagrange’s Method of (undetermined) Multiplier
5.51
5.5.6 Method to Decide Maxima or Minima
5.52
Worked Examples5.56
Exercise 5.55.65
Answers to Exercise 5.55.66
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5.6 Errors and Approximations
Worked Examples
Exercise 5.6
Answers to Exercise 5.6
Short Answer Questions
Objective Type Questions
Answers
5.67
5.68
5.72
5.73
5.73
5.74
5.76
6.
Integral Calculus
6.1
6.0 Introduction
6.1
6.1 Indefinite Integral
6.1
6.1.1 Properties of Indefinite Integral
6.1
6.1.2 Integration by Parts
6.3
6.1.3 Bernoulli’s Formula
6.3
6.1.4 Special Integrals
6.3
Worked Examples
6.4
Exercise 6.1
6.9
Answers to Exercise 6.1
6.9
6.2 Definite Integral (Newton–Leibnitz formula)
6.10
6.2.1 Properties of Definite Integral
6.10
Worked Examples6.15
Exercise 6.26.27
Answers to Exercise 6.26.27
b
6.3 Definite Integral
∫ f ( x) dx
as Limit of a Sum
6.28
a
6.3.1 Working Rule
6.28
Worked Examples6.29
Exercise 6.36.32
Answers to Exercise 6.36.33
6.4 Reduction Formulae
6.33
n
n
6.4.1 The Reduction Formula for (a) ∫ sin x dx and (b) ∫ cos x dx 6.33
∫ tan x dx and (b) ∫ cot x dx 6.36
The Reduction Formula for (a) ∫ sec x dx and (b) ∫ cosec x dx 6.37
6.4.2 The Reduction Formula for (a)
n
n
n
n
6.4.3
Worked Examples6.38
6.4.4The Reduction Formula for ∫ sin m x cos n x dx , Where m, n are
Non-negative Integers6.45
Worked Examples6.47
6.4.5The Reduction Formula for (a) xm (log x) ndx, (b) xn sin mx dx,
(c) xn cos mx dx6.49
ax
m
ax
n
6.4.6 The Reduction Formula for (a) ∫ e sin x dx and (b) ∫ e cos x dx 6.51
∫
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6.4.7 The Reduction Formula for (a) ∫ cos m x sin nx dx and (b) ∫ cos m x cos nx dx 6.52
Exercise 6.4
6.55
Answers to Exercise 6.4
6.55
6.5 Application of Integral Calculus
6.55
6.5.1 Area of Plane Curves
6.56
6.5.1 (a) Area of Plane Curves in Cartesian Coordinates
6.56
Worked Examples
6.57
Exercise 6.5
6.66
Answers to Exercise 6.5
6.67
6.5.1 (b) Area in Polar Coordinates
6.67
Worked Examples
6.68
Exercise 6.6
6.72
Answers to Exercise 6.6
6.72
6.5.2 Length of the Arc of a Curve
6.72
6.5.2 (a) Length of the Arc in Cartesian Coordinates
6.72
Worked Examples
6.73
Exercise 6.7
6.78
Answers to Exercise 6.7
6.79
6.5.2 (b) Length of the Arc in Polar Coordinates
6.79
Worked Examples
6.80
Exercise 6.8
6.81
Answers to Exercise 6.8
6.81
6.5.3 Volume of Solid of Revolution
6.82
6.5.3(a) Volume in Cartesian Coordinates
6.82
Worked Examples
6.83
Exercise 6.9
6.89
Answers to Exercise 6.9
6.90
6.5.3 (b) Volume in Polar Coordinates
6.91
Worked Examples
6.91
Exercise 6.10
6.93
Answers to Exercise 6.10
6.93
6.5.4 Surface Area of Revolution
6.93
6.5.4(a) Surface Area of Revolution in Cartesian Coordinates
6.93
Worked Examples
6.94
Exercise 6.11
6.99
Answers to Exercise 6.11
6.99
6.5.4 (b) Surface Area in Polar Coordinates
6.100
Worked Examples6.100
Exercise 6.126.102
Answers to Exercise 6.126.103
Short Answer Questions6.103
Objective Type Questions6.103
Answers6.106
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7.
Improper Integrals
7.1
7.1 Improper Integrals
7.1
7.1.1 Kinds of Improper Integrals and Their Convergence
7.1
Worked Examples
7.4
Exercise 7.17.13
Answers to Exercise 7.17.13
7.1.2 Tests of Convergence of Improper Integrals
7.14
Worked Examples
7.15
Exercise 7.27.27
Answers to Exercise 7.27.27
7.2 Evaluation of Integral by Leibnitz’s Rule
7.27
7.2.1 Leibnitz’s Rule—Differentiation Under Integral Sign for Variable Limits
7.28
Worked Examples7.28
Exercise 7.3 7.47
Answers to Exercise 7.37.47
7.3 Beta and Gamma functions
7.47
7.3.1 Beta Function 7.47
7.3.2 Symmetric property of beta function7.48
7.3.3 Different forms of beta function7.48
7.4 The Gamma Function
7.49
7.4.1 Properties of Gamma Function
7.50
7.4.2 Relation between Beta and Gamma Functions
7.51
Worked Examples7.55
Exercise 7.47.69
Answers to Exercise 7.47.69
7.5 The Error Function
7.70
7.5.1 Properties of Error Functions
7.70
7.5.2 Series expansion for error function7.71
7.5.3 Complementary error function7.71
Worked Examples7.72
Exercise 7.57.76
Answers to Exercise 7.57.76
Short Answer Questions7.76
Objective Type Questions7.77
Answers7.78
8.
Multiple Integrals
8.1 Double Integration
8.1.1 Double Integrals in Cartesian Coordinates
8.1.2 Evaluation of Double Integrals
Worked Examples
Exercise 8.1
Answers to Exercise 8.1
8.1.3 Change of Order of Integration
Worked Examples
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8.1
8.1
8.1
8.2
8.3
8.6
8.7
8.7
8.8
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Exercise 8.28.15
Answers to Exercise 8.28.15
8.1.4 Double Integral in Polar Coordinates
8.16
Worked Examples8.16
8.1.5 Change of Variables in Double Integral
8.19
Worked Examples8.19
Exercise 8.38.26
Answers to Exercise 8.38.27
8.1.6 Area as Double Integral
8.27
Worked Examples8.28
Exercise 8.48.31
Answers to Exercise 8.48.31
Worked Examples8.32
Exercise 8.58.37
Answers to Exercise 8.58.37
8.2 Area of a Curved Surface
8.37
8.2.1 Surface Area of a Curved Surface
8.38
8.2.2 Derivation of the Formula for Surface Area
8.38
8.2.3 Parametric Representation of a Surface
8.41
Worked Examples8.41
Exercise 8.68.49
Answers to Exercise 8.68.49
8.3 Triple Integral in Cartesian Coordinates
8.49
Worked Examples8.50
Exercise 8.78.55
Answers to Exercise 8.78.56
8.3.1 Volume as Triple Integral
8.56
Worked Examples8.56
Exercise 8.88.63
Answers to Exercise 8.88.64
Short Answer Questions8.64
Objective Type Questions8.64
Answers8.66
9.
Vector Calculus
9.0 Introduction
9.1 Scalar and Vector Point Functions
9.1.1 Geometrical Meaning of Derivative
9.2 Differentiation Formulae
9.3 Level Surfaces
9.4Gradient of a Scalar Point Function or Gradient of a Scalar Field
9.4.1 Vector Differential Operator
9.4.2 Geometrical Meaning of ∇φ
9.4.3 Directional Derivative
9.4.4 Equation of Tangent Plane and Normal to the Surface
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9.1
9.1
9.1
9.2
9.3
9.4
9.4
9.4
9.4
9.5
9.5
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9.4.5 Angle between Two Surfaces at a Common Point
9.6
9.4.6 Properties of gradients
9.6
Worked Examples
9.8
Exercise 9.1
9.20
Answers to Exercise 9.1
9.21
9.5Divergence of a Vector Point Function or Divergence of a Vector Field
9.22
9.5.1 Physical Interpretation of Divergence
9.22
9.6Curl of a Vector Point Function or Curl
9.23
1 of a Vector Field
9.6.1 Physical Meaning of Curl wF = curlv
9.23
2
Worked Examples
9.24
Exercise 9.2
9.30
Answers to Exercise 9.2
9.31
9.7 Vector Identities
9.31
Worked Examples
9.37
9.8 Integration of Vector Functions
9.39
9.8.1 Line Integral
9.40
Worked Examples
9.40
Exercise 9.3
9.46
Answers to Exercise 9.3
9.47
9.9 Green’s Theorem in a Plane
9.47
9.9.1 Vector Form of Green’s Theorem
9.50
Worked Examples
9.50
9.10 Surface Integrals
9.56
9.10.1 Evaluation of Surface Integral
9.57
9.11 Volume Integral
9.58
Worked Examples
9.58
9.12 Gauss Divergence Theorem
9.62
9.12.1 Results Derived from Gauss Divergence Theorem
9.64
Worked Examples
9.68
9.13 Stoke’s Theorem
9.81
Worked Examples
9.83
Exercise 9.4
9.97
Answers to Exercise 9.49.100
Short Answer Questions9.100
Objective Type Questions9.101
Answers9.102
10. Ordinary First Order Differential Equations
10.0 Introduction
10.1 Formation of Differential Equations
Worked Examples
Exercise 10.1
Answers to Exercise 10.1
10.2 First Order and First Degree Differential Equations
10.2.1 Type I
Variable Separable Equations
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10.1
10.1
10.2
10.2
10.5
10.6
10.6
10.6
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Worked Example
10.6
Exercise 10.2
10.9
Answers to Exercise 10.2
10.9
10.2.2 Type II
Homogeneous Equation
10.10
Worked Examples10.10
Exercise 10.310.13
Answers to Exercise 10.310.14
10. 2.3 Type III
Non-Homogenous Differential Equations of the First Degree
10.14
Worked Examples10.16
Exercise 10.410.21
Answers to Exercise 10.410.21
10.2.4 Type IV
Linear Differential Equation
10.22
Worked Examples10.23
Exercise 10.510.27
Answers to Exercise 10.510.27
10.2.5 Type V Bernoulli’s Equation
10.28
Worked Examples10.28
Exercise 10.610.31
Answers to Exercise 10.610.31
10.2.6 Type VI Riccati Equation
10.31
Worked Examples10.33
Exercise 10.710.36
Answers to Exercise 10.710.36
10.2.7 Type VII First Order Exact Differential Equations
10.37
Worked Examples10.39
Exercise 10.810.41
Answers to Exercise 10.810.42
10.3 Integrating Factors
10.42
Worked Examples10.43
10.3.1Rules for Finding the Integrating Factor for Non-Exact Differential
Equation Mdx + Ndy = 0
10.45
Worked Examples10.46
Exercise 10.910.56
Answers to Exercise 10.910.56
10.4Ordinary Differential Equations of the First Order but of Degree Higher
than One
10.56
10.4.1 Type 1 Equations Solvable for p 10.57
Worked Examples10.57
Exercise 10.1010.59
Answers to Exercise 10.1010.60
10.4.2 Type 2
Equations Solvable for y10.60
Worked Examples10.61
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10.4.3 Type 3
Equations Solvable for x10.64
Worked Examples10.65
Exercise 10.1110.67
Answers to Exercise 10.1110.67
10.4.4 Type 4 Clairaut’s Equation
10.67
Worked Examples10.68
Exercise 10.1210.71
Answers to Exercise 10.1210.71
Short Answer Questions10.71
Objective Type Questions10.72
Answers10.74
11. Ordinary Second and Higher Order Differential Equations
11.1
11.0 Introduction
11.1
11.1Linear Differential Equation with Constant Coefficients
11.1
11.1.1 Complementary Function
11.1
11.1.2 Particular Integral
11.2
Worked Examples
11.3
Exercise 11.111.19
Answers to Exercise 11.111.19
11.2Linear Differential Equations with Variable Coefficients
11.21
11.2.1 Cauchy’s Homogeneous Linear Differential Equations
11.21
Worked Examples11.22
11.2.2 Legendre’s Linear Differential Equation
11.29
Worked Examples11.30
Exercise 11.211.32
Answers to Exercise 11.211.33
11.3Simultaneous Linear Differential Equations with
Constant Coefficients
11.34
Worked Examples11.34
Exercise 11.311.43
Answers to Exercise 11.311.44
11.4 Method of Variation of Parameters
11.44
11.4.1 Working rule11.45
Worked Examples11.45
Exercise 11.411.51
Answers to Exercise 11.411.52
11.5 Method of Undetermined Coefficients
11.52
Worked Examples11.54
Exercise 11.511.60
Answers to Exercise 11.511.60
Short Answers Questions11.60
Objective Type Questions11.61
Answers 11.63
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12. Applications of Ordinary Differential Equations
12.1
12.0 Introduction
12.1
12.1 Applications of Ordinary Differential Equations of First Order
12.1
12.1.1 Law of Growth and Decay
12.1
12.1.2 Newton’s Law of Cooling of Bodies
12.2
Worked Examples
12.2
Exercise 12.1
12.7
Answers To Exercise 12.1
12.8
12.1.3 Chemical Reaction and Solutions
12.8
Worked Examples
12.9
Exercise 12.212.12
Answers to Exercise 12.212.13
12.1.4 Simple Electric Circuit
12.13
Worked Examples12.14
Exercise 12.312.19
Answers to Exercise 12.312.19
12.1.5 Geometrical Applications
12.20
12.1.5 (a) Orthogonal Trajectories in Casterian Coordinates
12.20
Worked Examples12.21
12.1.5 (b) Orthogonal Trajectories in Polar Coordinates
12.23
Worked Examples12.24
Exercise 12.412.26
Answers to Exercise 12.412.27
12.2 Applications of Second Order Differential Equations
12.27
12.2.1 Bending of Beams
12.27
Worked Examples12.29
12.2.2 Electric Circuits
12.34
Worked Examples12.34
Exercise 12.512.38
Answers to Exercise 12.512.39
12.2.3 Simple Harmonic Motion (S.H.M)
12.40
Worked Examples12.41
Exercise 12.612.43
Answers to Exercise 12.612.44
Objective Type Questions12.44
Answers12.45
13. Series Solution of Ordinary Differential Equations and Special Functions
13.0 Introduction
13.1 Power Series Method
13.1.1 Analytic Function
13.1.2 Regular Point
13.1.3 Singular Point
13.1.4 Regular and Irregular Singular Points
Worked Examples
Exercise 13.1
Answers to Exercise 13.1
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13.1
13.1
13.1
13.1
13.2
13.2
13.2
13.3
13.9
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13.2
13.3
13.4
Frobenius Method
13.9
Worked Examples13.11
Exercise 13.213.33
Answers to Exercise 13.213.33
Special Functions
13.34
Bessel Functions
13.34
13.4.1 Series Solution of Bessel’s Equation
13.34
13.4.2 Bessel’s Functions of the First Kind
13.37
Worked Examples13.39
13.4.3 Some Special Series
13.40
13.4.4 Recurrence Formula for Jn (x)13.41
13.4.5 Generating Function for Jn (x) of Integral Order
13.44
Worked Examples13.46
13.4.6 Integral Formula for Bessel’s Function Jn (x) 13.49
Worked Examples13.53
13.4.7 Orthogonality of Bessel’s Functions
13.56
13.4.8 Fourier–Bessel Expansion of a Function f(x)13.59
Worked Examples13.60
13.4.9 Equations Reducible to Bessel’s Equation
13.62
Worked Examples13.62
Exercise 13.313.65
Answers to Exercise 13.313.66
13.5 Legendre Functions
13.66
13.5.1 Series Solution of Legendre’s Differential Equation
13.66
13.5.2 Legendre Polynomials
13.71
13.5.3 Rodrigue’s Formula
13.71
Worked Examples13.73
13.5.4 Generating Function for Legendre Polynomials
13.74
Worked Examples13.75
13.5.5 Orthogonality of Legendre Polynomials in [-1, 1]
13.77
Worked Examples13.80
13.5.6 Fourier–Legendre Expansion of f(x) in a Series of Legendre
Polynomials13.83
Worked Examples13.83
Exercise 13.413.85
Answers to Exercise 13.413.85
14. Partial Differential Equations
14.1
14.0 Introduction
14.1
14.1 Order and Degree of Partial Differential Equations
14.1
14.2 Linear and Non-linear Partial Differential Equations
14.1
14.3 Formation of Partial Differential Equations
14.2
Worked Examples
14.2
Exercise 14.114.15
Answers to Exercise 14.114.15
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14.4 Solutions of Partial Differential Equations
14.16
14.4.1Procedure to find general integral and singular
integral for a first order partial differential equation14.17
Worked Examples14.17
Exercise 14.214.20
Answers to Exercise 14.214.20
14.4.2 First Order Non-linear Partial Differential Equation of Standard Types
14.20
Worked Examples14.21
Exercise 14.314.25
Answers to Exercise 14.314.25
Worked Examples14.26
14.4.3 Equations Reducible to Standard Forms
14.33
Worked Examples14.35
Exercise 14.414.38
Answers to Exercise 14.414.38
14.5 Lagrange’s Linear Equation 14.39
Worked Examples14.41
Exercise 14.514.48
Answers to Exercise 14.514.48
14.6 Homogeneous Linear Partial Differential Equations of the Second
and Higher Order with Constant Coefficients
14.49
14.6.1 Working Procedure to Find Complementary Function
14.50
14.6.2 Working Procedure to Find Particular Integral
14.51
Worked Examples14.53
Exercise 14.614.66
Answers to Exercise 14.614.67
14.7 Non-homogeneous Linear Partial Differential Equations of the
Second and Higher Order with Constant Coefficients
14.68
Worked Examples14.69
Exercise 14.714.73
Answers to Exercise 14.714.73
Short Answer Questions14.74
Objective Type Questions14.74
Answers 14.76
15. Analytic Functions
15.0 Preliminaries
15.1 Function of a Complex Variable
15.1.1 Geometrical Representation of Complex Function or Mapping
15.1.2 Extended complex number system
15.1.3 Neighbourhood of a point and region
15.2 Limit of a Function
15.2.1 Continuity of a function
15.2.2 Derivative of f(z)
15.2.3 Differentiation formulae
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15.1
15.1
15.2
15.3
15.4
15.5
15.5
15.6
15.6
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15.3 Analytic Function
15.8
15.3.1Necessary and Sufficient Condition for f(z)to be Analytic
15.8
15.3.2 C-R equations in polar form15.10
Worked Examples15.11
Exercise 15.115.20
Answers to Exercise 15.115.21
15.4 Harmonic Functions and Properties of Analytic Function
15.21
15.4.1Construction of an Analytic Function Whose Real or
Imaginary Part is Given Milne-Thomson Method
15.23
Worked Examples15.25
Exercise 15.215.32
Answers to Exercise 15.215.33
15.5 Conformal Mapping
15.33
15.5.1 Angle of rotation15.34
15.5.2 Mapping by elementary functions15.36
Worked Examples15.37
Exercise 15.315.72
Answers to Exercise 15.315.74
15.5.3 Bilinear Transformation
15.79
Worked Examples15.82
Exercise 15.415.89
Answers to Exercise 15.415.90
Short Answer Questions15.90
Objective Type Questions15.91
Answers15.92
16. Complex Integration
16.1
16.0 Introduction
16.1
16.1 Contour Integral
16.1
16.1.1 Properties of Contour Integrals
16.1
Worked Examples
16.2
16.1.2 Simply Connected and Multiply Connected Domains
16.4
16.2 Cauchy’s Integral Theorem or Cauchy’s Fundamental Theorem
16.4
16.2.1 Cauchy-Goursat Integral Theorem
16.5
16.3 Cauchy’s Integral Formula
16.6
16.3.1 Cauchy’s Integral Formula for Derivatives
16.7
Worked Examples
16.7
Exercise 16.116.12
Answers to Exercise 16.116.13
16.4 Taylor’s Series and Laurent’s Series
16.14
16.4.1 Taylor’s Series
16.14
16.4.2 Laurent’s series16.15
Worked Examples16.16
Exercise 16.216.22
Answers to Exercise 16.216.23
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16.5 Classification of Singularities
16.24
16.6 Residue
16.26
16.6.1 Methods of Finding Residue
16.26
16.7 Cauchy’s Residue Theorem
16.27
Worked Examples16.28
Exercise 16.316.34
Answers to Exercise 16.316.36
16.8 Application of Residue Theorem to Evaluate Real Integrals
16.36
16.8.1 Type 1
16.36
Worked Examples16.37
16.8.2 Type 2. Improper Integrals of Rational Functions
16.44
Worked Examples16.46
16.8.3 Type 3
16.50
Worked Examples16.50
Exercise 16.416.55
Answers to Exercise 16.416.56
Short Answer Questions16.56
Objective Type Questions16.58
Answers16.60
17
Fourier Series
17.1
17.0 Introduction
17.1
17.1 Fourier series
17.2
17.1.1 Dirichlet’s Conditions
17.2
17.1.2 Convergence of Fourier Series
17.3
Worked Examples
17.5
17.2 Even and Odd Functions
17.15
17.2.1 Sine and Cosine Series
17.15
Worked Examples17.16
Exercise 17.117.23
Answers to Exercise 17.117.25
17.3 Half-Range Series
17.26
17.3.1 Half-range Sine Series
17.27
17.3.2 Half-range cosine series17.27
Worked Examples17.28
Exercise 17.217.36
Answers to Exercise 17.217.37
17.4 Change of Interval
17.38
Worked Examples17.39
17.5 Parseval’s Identity
17.47
Worked Examples17.47
Exercise 17.317.50
Answers to Exercise 17.317.52
17.6 Complex Form of Fourier Series
17.53
Worked Examples17.55
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Exercise 17.417.59
Answers to Exercise 17.417.59
17.7 Harmonic Analysis
17.60
17.7.1 Trapezoidal Rule
17.60
Worked Examples17.62
Exercise 17.517.68
Answers to Exercise 17.517.69
Short Answer Questions17.69
Objective Type Questions17.70
Answers17.72
18. Fourier Transforms
18.1
18.0 Introduction
18.1
18.1 Fourier Integral Theorem
18.1
18.1.1 Fourier Cosine and Sine Integrals
18.2
Worked Examples
18.2
18.1.2 Complex Form of Fourier Integral
18.6
18.2 Fourier Transform Pair
18.7
18.2.1 Properties of Fourier transforms
18.8
Worked Examples18.12
Exercise 18.118.21
Answers to Exercise 18.118.22
18.3 Fourier Sine and Cosine Transforms
18.23
18.3.1 Properties of Fourier Sine and Cosine Transforms
18.24
Worked Examples18.29
Exercise 18.218.39
Answers to Exercise 18.218.39
18.4 Convolution Theorem
18.40
18.4.1 Definition: Convolution of Two Functions
18.40
18.4.2 Theorem 18.1: Convolution theorem or Faltung theorem18.41
18.4.3 Theorem 18.2 : Parseval’s identity for Fourier transforms or
Energy theorem18.41
Worked Examples18.43
Exercise 18.318.51
Answers to Exercise 18.318.52
Short Answer Questions 18.52
Objective Type Questions18.53
Answers18.54
19. Laplace Transforms
19.0 Introduction
19.1Condition for Existence of Laplace Transform
19.2Laplace Transform of Some Elementary Functions
19.3 Some Properties of Laplace Transform
Worked Examples
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19.1
19.1
19.1
19.2
19.4
19.5
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Exercise 19.1
19.9
Answers to Exercise 19.119.10
19.4Differentiation and Integration of Transforms
19.11
Worked Examples19.12
Exercise 19.219.20
Answers to Exercise 19.219.20
19.5Laplace Transform of Derivatives and Integrals
19.21
Worked Examples19.23
19.5.1Evaluation of Improper Integrals using Laplace Transform
19.25
Worked Examples19.25
19.6Laplace Transform of Periodic Functions and Other Special Type of Functions 19.27
Worked Examples19.29
19.6.1 Laplace Transform of Unit Step Function
19.36
19.6.2 Unit impulse function19.37
19.6.3 Dirac-delta function19.37
19.6.4 Laplace transform of delta function19.37
Worked Examples19.38
Exercise 19.319.39
Answers to Exercise 19.319.41
19.7 Inverse Laplace Transforms
19.41
19.7.1 Type 1 – Direct and shifting methods19.43
Worked Examples19.43
19.7.2 Type 2 – Partial Fraction Method
19.44
Worked Examples19.44
19.7.3 Type 3 – 1. Multiplication by s and 2. Division by s19.48
Worked Examples19.48
19.7.4Type 4 – Inverse Laplace Transform of Logarithmic and
Trigonometric Functions
19.50
Worked Examples19.50
Exercise 19.419.53
Answers to Exercise 19.419.54
19.7.5 Type 5 – Method of Convolution
19.55
Worked Examples19.57
Exercise 19.519.60
Answers to Exercise 19.519.61
19.7.6 Type 6: Inverse Laplace Transform as Contour Integral
19.61
Worked Examples19.62
Exercise 19.619.64
Answers to Exercise 19.619.65
19.8Application of Laplace Transform to the Solution of Ordinary
Differential Equations
19.65
19.8.1 First Order Linear Differential Equations with Constant Coefficients
19.65
Worked Examples19.65
19.8.2Ordinary Second and Higher Order Linear Differential Equations
with Constant Coefficients
19.68
Worked Examples19.68
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19.8.3 Ordinary Second Order Differential Equations with Variable Coefficients 19.72
Worked Examples19.72
Exercise 19.719.75
Answers to Exercise 19.719.76
19.8.4 Simultaneous Differential Equations
19.77
Worked Examples19.77
19.8.5 Integral–Differential Equation
19.83
Worked Examples19.83
Exercise 19.819.85
Answers to Exercise 19.819.86
Short Answer Questions19.86
Objective Type Questions19.86
Answers19.88
20. Applications of Partial Differential Equations
20.1
20.0 Introduction
20.1
20.1 One Dimensional Wave Equation – Equation of Vibrating String
20.2
20.1.1 Derivation of Wave Equation
20.2
20.1.2Solution of One-Dimensional Wave Equation by the Method of
Separation of Variables (or the Fourier Method)
20.3
Worked Examples
20.5
Exercise 20.120.34
Answers to Exercise 20.120.35
20.1.3 Classification of Partial Differential Equation of Second Order
20.36
Worked Examples20.37
Exercise 20.220.38
Answers to Exercise 20.220.38
20.2 One-Dimensional Equation of Heat Conduction (In a Rod)
20.39
20.2.1 Derivation of Heat Equation
20.39
20.2.2 Solution of Heat Equation by Variable Separable Method
20.40
Worked Examples20.42
Exercise 20.320.62
Answers to Exercise 20.320.63
Worked Examples20.64
Exercise 20.420.68
Answers to Exercise 20.420.69
20.3 Two Dimensional Heat Equation in Steady State
20.69
20.3.1 Solution of Two Dimensional Heat Equation
20.70
Worked Examples20.71
Exercise 20.520.83
Answers to Exercise 20.520.84
Short Answer Questions20.85
Objective Type Questions20.86
Answers20.88
IndexI.1
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Preface
This book Engineering Mathematics is written to cover the topics that are common to the syllabi of
various universities in India. Although this book is designed primarily for engineering courses, it
is also suitable for Mathematics courses and for various competitive examinations. The aim of the
book is to provide a sound understanding of Mathematics. The experiences of both the authors in
teaching undergraduate and postgraduate students from diverse backgrounds for over four decades
have helped to present the subject as simple as possible with clarity and rigour in a step-by-step
approach.
This book has many distinguishing features. The topics are well organized to create
self-confidence and interest among the readers to study and apply the mathematical tools in
engineering and science disciplines. The subject is presented with a lot of standard worked examples
and exercises that will help the readers to develop maturity in Mathematics.
This book is organized into 20 chapters. At the end of each chapter, short answer questions and
objective questions are given to enhance the understanding of the topics.
Chapter 1 focuses on the applications of matrices to the consistency of simultaneous linear equations
and Eigen value problems.
Chapter 2 discusses convergence of sequence and series.
Chapter 3 deals with differentiation and applications of derivative, Rolle’s Theorem, mean value
theorems, asymptotes and curve tracing.
Chapter 4 deals with the geometrical application of derivative in radius of curvature, centre of
curvature, evolute and envelope.
Chapter 5 elaborates calculus of several variables.
Chapter 6 deals with integral calculus and applications of integral calculus.
Chapter 7 discusses improper integrals, and beta and gamma functions.
Chapter 8 focuses on multiple integrals.
Chapter 9 deals with vector calculus.
Chapter 10 discusses solution of various types of first order differential equations.
Chapter 11 is concerned with the solution of second order and higher order linear differential
equations.
Chapter 12 deals with some applications of ordinary differential equations.
Chapter 13 conforms to series solution of ordinary differential equations and special functions.
Chapter 14 focuses on solution of partial differential equations.
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Preface
Chapter 15 examines analytic functions.
Chapter 16 focuses on complex integration.
Chapter 17 deals with Fourier series.
Chapter 18 pertains to Fourier transforms.
Chapter 19 discusses Laplace transforms.
Chapter 20 is concerned with applications of partial differential equations.
Mathematics is a subject that can be mastered only through hard work and practice. Follow the
maximum, Mathematics without practice is blind and practice without understanding is futile.
“Tell me and I will forget
Show me and I will remember
Involve me and I will understand”
—Confucius
We hope that this book is student-friendly and that it will be well received by students and teachers.
We heartily welcome valuable comments and suggestions from our readers for the improvement of
this book, which may be addressed to profpsdas@yahoo.com.
ACKNOWLEDGEMENTS
P. Sivaramakrishna Das: I express my gratitude to our chairperson, Dr Elizabeth Varghese, and
the directors of K.C.G. College of Technology for giving me an opportunity to write this book.
I am obliged to my department colleagues for their encouragement.
The inspiration to write this book came from my wife, Prof. C. Vijayakumari, who is also the
co-author of this book.
P. Sivaramakrishna Das and C. Vijayakumari: We are grateful to the members of our family for
lending us their support for the successful completion of this book.
We are obliged to Sojan Jose, R. Dheepika and C. Purushothaman of Pearson India Education
Services Pvt. Ltd, for their diligence in bringing this work out to fruition.
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P. Sivaramakrishna Das
C. Vijayakumari
3/2/2017 6:17:56 PM
About the Authors
Prof. Dr P. Sivaramakrishna Das started his career in 1967 as assistant
professor of Mathematics at Ramakrishna Mission Vivekananda College,
Chennai, his alma mater and retired as Head of the P.G. Department of
Mathematics from the same college after an illustrious career spanning
36 years.
Currently, he is professor of Mathematics and Head of the Department
of Science and Humanities, K.C.G. College of Technology, Chennai
(a unit of Hindustan Group of Institutions).
P. Sivaramakrishna Das has done pioneering research work in the
field of “Fuzzy Algebra” and possess a Ph.D. in this field. His paper
on fuzzy groups and level subgroups was a fundamental paper on fuzzy
algebra with over 600 citations and it was the first paper from India. With a teaching experience
spanning over 49 years, he is an accomplished teacher of Mathematics at undergraduate and
postgraduate levels of Arts and Science and Engineering colleges in Chennai. He has guided several
students to obtain their M.Phil. degree from the University of Madras, Chennai.
He was the most popular and sought-after teacher of Mathematics in Chennai during 1980s for
coaching students for IIT-JEE. He has produced all India 1st rank and several other ranks in IIT-JEE.
He was also a visiting professor at a few leading IIT-JEE training centres in Andhra Pradesh.
Along with his wife C. Vijayakumari, he has written 10 books covering various topics of
Engineering Mathematics catering to the syllabus of Anna University, Chennai, and has also written
“Numerical Analysis”, an all India book, catering to the syllabi of all major universities in India.
Prof. Dr C. Vijayakumari began her career in 1970 as assistant professor
of Mathematics at Government Arts College for Women, Thanjavur, and has
taught at various Government Arts and Science colleges across Tamil Nadu
before retiring as professor of Mathematics from Queen Mary’s College
(Autonomous), Chennai after an illustrious career of spanning 36 years.
As a visiting professor of Mathematics, she has taught the students at
two engineering colleges in Chennai. With a teaching experience spanning
over 40 years, she is an accomplished teacher of Mathematics and Statistics
at both undergraduate and postgraduate levels. She has guided many
students to obtain their M.Phil. degree from the University of Madras,
Chennai and Bharathiar University, Coimbatore.
Along with her husband P. Sivaramakrishna Das, she has co-authored several books on Engineering
Mathematics catering to the syllabus of Anna University, Chennai and has also co-authored “Numerical
Analysis”, an all India book, catering to the syllabi of all major universities in India.
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1
Matrices
1.0
INTRODUCTION
The concept of matrices and their basic operations were introduced by the British mathematician
Arthur Cayley in the year 1858. He wondered whether this part of mathematics will ever be used.
However, after 67 years, in 1925, the German physicist Heisenberg used the algebra of matrices in his
revolutionary theory of quantum mechanics. Over the years, the theory of matrices have been found as
an elegant and powerful tool in almost all branches of Science and Engineering like electrical networks,
graph theory, optimisation techniques, system of differential equations, stochastic processes, computer
graphics, etc. Because of the digital computers, usage of matrix methods have become greatly fruitful.
In this chapter, we review some of the basic concepts of matrices. We shall discuss two important
applications of matrices, namely consistency of system of linear equations and the eigen value problems.
1.1
BASIC CONCEPTS
Definition 1.1 Matrix
A rectangular array of mn numbers (real or complex) arranged in m rows (horizontal lines) and
n columns (vertical lines) and enclosed in brackets [ ] is called an m × n matrix.
The numbers in the matrix are called entries or elements of the matrix.
Usually an m × n matrix is written as
⎡ a11
⎢
⎢ a21
⎢A
A= ⎢
⎢a
⎢ i1
⎢A
⎢
⎢⎣ am1
a12
a13
… a1 j
a22
a23
… a2 j
A
A
ai 2
ai 3
A
am 2
A
am 3 … amj
A
…
aij
… a1n ⎤
⎥
… a2 n ⎥
A ⎥
⎥
… ain ⎥
⎥
⎥
⎥
… amn ⎥⎦
where aij is the element lying in the ith row and jth column, the first suffix refers to row and the second
suffix refers to column.
The matrix A is briefly written as
A = [aij]m × n, i = 1, 2, 3, …, m, j = 1, 2, 3, …, n
If all the entries are real, then the matrix A is called a real matrix.
Definition 1.2 Square Matrix
In a matrix, if the number of rows = number of columns = n, then it is called a square matrix of order n.
If A is a square matrix of order n, then A = [aij]n × n, i = 1, 2, 3, …, n; j = 1, 2, 3, …, n.
Definition 1.3 Row Matrix
A matrix with only one row is called a row matrix.
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EXAMPLE 1.1
Let A = [a11 a12 a13 … a1n]. It is a row matrix with n columns. So, it is of type 1 × n.
EXAMPLE 1.2
Let A = [1, 2, 3, 4]. It is a row matrix with 4 columns. So, it is a row matrix of type 1 × 4.
Definition1.4 Column Matrix
A matrix with only one column is called a column matrix.
EXAMPLE 1.3
⎡ a11 ⎤
⎢a ⎥
⎢ 21 ⎥
A = ⎢a31 ⎥
⎢ ⎥
⎢:⎥
⎢⎣an1 ⎥⎦
Let
It is a column matrix with n rows. So, it is of type n × 1.
EXAMPLE 1.4
⎡1⎤
⎢0⎥
⎢ ⎥
Let
A = ⎢ −2⎥
⎢ ⎥
⎢1⎥
⎢⎣ 3 ⎥⎦
It is a column matrix with 5 rows. So, it is of type 5 × 1.
Definition 1.5 Diagonal Matrix
A square matrix A = [aij] with all entries aij = 0 when i ≠ j is is called a diagonal matrix.
In other words a square matrix in which all the off diagonal elements are zero is called a diagonal
matrix.
EXAMPLE 1.5
0 … 0⎤
0 … 0 ⎥⎥
is a diagonal matrix of order n.
⎥
:
⎥
0 … ann ⎦
⎡a11 0
⎢0 a
22
(1) A = ⎢
⎢:
:
⎢
0
⎣0
⎡2 0 0 ⎤
(2) A = ⎢⎢0 3 0 ⎥⎥ is a diagonal matrix of order 3.
⎢⎣0 0 −4 ⎥⎦
⎡ −1
⎢0
(3) A = ⎢
⎢0
⎢
⎣0
0
2
0
0
0
0
3
0
0⎤
0 ⎥⎥
is a diagonal matrix of order 4.
0⎥
⎥
0⎦
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1.3
Definition 1.6 Scalar Matrix
In a diagonal matrix if all the diagonal elements are equal to a non-zero scalar a, then it is called a
scalar matrix.
EXAMPLE 1.6
⎡a 0 0 ⎤
A = ⎢⎢ 0 a 0 ⎥⎥ is a scalar matrix.
⎢⎣ 0 0 a ⎥⎦
Definition 1.7 Unit Matrix or Identity Matrix
In a diagonal matrix, if all the diagonal elements are equal to 1, then it is called a Unit matrix or
identity matrix.
EXAMPLE 1.7
⎡1 0 0⎤
⎡1 0⎤ ⎢
[1], ⎢
, ⎢0 1 0 ⎥⎥ are identity matrices of orders 1, 2, 3 respectively. They are denoted by I1, I2, I3.
⎥
⎣0 1⎦ ⎢
⎣0 0 1 ⎥⎦
In general, In is the identity matrix of order n.
Definition 1.8 Zero Matrix or Null Matrix
In a matrix (rectangular or square), if all the entries are equal to 0, then it is called a zero matrix or
null matrix.
EXAMPLE 1.8
⎡0 0 0 0⎤
⎡0 0 ⎤
A =⎢
⎥ , B = ⎢0 0 0 0 ⎥ are zero matrices of types 2 × 2 and 2 × 4.
0
0
⎦
⎣
⎦
⎣
Definition 1.9 Triangular matrix
A square matrix A = [aij] is said to be an upper triangular matrix if all the entries below the main
diagonal are zero.
That is aij = 0 if i > j
A square matrix A = [aij] is said to be a lower triangular matrix if all the entries above the main
diagonal are zero.
That is aij = 0 if i < j
EXAMPLE 1.9
⎡4
⎡ −1 2 3 ⎤
⎢0
(1) The matrices A = ⎢⎢ 0 1 4 ⎥⎥ and B = ⎢
⎢0
⎢⎣ 0 0 5 ⎥⎦
⎢
⎣0
1
2
0
0
0 2⎤
3 1 ⎥⎥
are upper triangular matrices.
0 −2⎥
⎥
0 5⎦
⎡1 0 0 ⎤
⎡ 2 0⎤
⎢
⎥
(2) The matrices A = ⎢
⎥ and B = ⎢ 2 −1 0 ⎥ are lower triangular matrices.
⎣ −1 0 ⎦
⎢⎣0 2 1 ⎥⎦
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1.1.1 Basic Operations on Matrices
Definition 1.10 Equality of Matrices
Two matrices A = [aij] and B = [bij] of the same type m × n are said to be equal if aij = bij for all i, j and
is written as A = B.
Definition 1.11 Addition of Matrices
Let A = [aij] and B = [bij] of the same type m × n. Then A + B = [cij], where cij = aij + bij for all i and j
and A + B is of type m × n.
EXAMPLE 1.10
⎡ −1 2 3⎤
⎡1 2 3 ⎤
⎡ −1 + 1 2 + 2 3 + 3 ⎤ ⎡0 4 6 ⎤
If A = ⎢
⎥ and B = ⎢1 0 −2⎥ , then A + B = ⎢ 0 + 1 1 + 0 5 − 2⎥ = ⎢1 1 3⎥
0
1
5
⎦
⎣
⎣
⎦
⎣
⎦ ⎣
⎦
We see that A and B are of type 2 × 3 and A + B is also of type 2 × 3.
Definition 1.12 Scalar Multiplication of a Matrix
Let A = [aij] be an m × n matrix and k be a scalar, then kA = [kaij].
EXAMPLE 1.11
a
⎡a
If A = ⎢ 11 12
a
a
⎣ 21 22
a13 ⎤
⎡ ka
, then kA = ⎢ 11
a23 ⎥⎦
⎣ ka21
ka12
ka22
ka13 ⎤
.
ka23 ⎥⎦
−a12 −a13 ⎤
⎡ −a
In particular if k = −1, then − A = ⎢ 11
⎥.
⎣ −a21 −a22 −a23 ⎦
Multiplication of Matrices
If A and B are two matrices such that the number of columns of A is equal to the number of rows of
B, then the product AB is defined. Two such matrices are said to be conformable for multiplication.
In the product AB, A is known as pre-factor and B is known as post-factor.
Definition 1.13 Let A = [aij] be an m × p matrix and B = [bij] be an p × n matrix, then AB is defined and
p
AB = [cij] is an m × n matrix, where cij = ∑ aik b kj .
k =1
That is cij is the sum of the products of the corresponding elements of the ith row of A and the jth
column of B.
EXAMPLE 1.12
⎡ 1 1 2⎤
⎡ 1 2⎤
⎢ 1 3⎥ and B = ⎢ 3 1 ⎥
⎢
⎥
⎢
⎥
⎢⎣ 2 2 1 ⎥⎦
⎢⎣ 2 1 ⎥⎦
Since A is of type 3 × 3 and B is of type 3 × 2, AB is defined and AB is of type 3 × 2.
Let A = 0
⎡ 1 1 2⎤ ⎡ 1 2⎤ ⎡1⋅1 + 1⋅ 3 + 2 ⋅ 2 1⋅ 2 + 1⋅1 + 2 ⋅1 ⎤ ⎡ 8 5 ⎤
A B = ⎢⎢0 1 3⎥⎥ ⎢⎢ 3 1 ⎥⎥ = ⎢⎢ 0 ⋅1 + 1⋅ 3 + 3 ⋅ 2 0 ⋅ 2 + 1⋅1 + 3 ⋅1⎥⎥ = ⎢⎢ 9 4 ⎥⎥
⎢⎣ 2 2 1 ⎥⎦ ⎢⎣ 2 1 ⎥⎦ ⎢⎣ 2 ⋅1 + 2 ⋅ 3 + 1⋅ 2 2 ⋅ 2 + 2 ⋅1 + 1⋅1⎥⎦ ⎢⎣10 7 ⎥⎦
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Matrices ■
1.5
Note If A and B are square matrices of order n, then both AB and BA are defined, but not necessarily
equal. That is, AB ≠ BA, in general.
So, matrix multiplication is not commutative.
1.1.2 Properties of addition, scalar multiplication and multiplication
1. If A, B, C are matrices of the same type, then
(i) A + B = B + A
(ii) A + (B + C) = (A + B) + C
(iii) A + 0 = A
(iv) A + (−A) = 0
(v) a (A + B) = a A + a B
(vi) (a + b)A = a A + b A
(vii) a (bA) = (a b)A for any scalars a, b.
2. If A, B, C are conformable for multiplication, then
(i) a (AB) = (a A)B = A(a B)
(ii) A(BC) = (AB)C
(iii) (A + B)C = AC + BC, where A and B are of type m × p and C is of type p × n.
(iv) If A is a square matrix, then
A2 = A × A, A3 = A2 × A, …, An = An − 1 × A
Definition 1.14 Transpose of a Matrix
Let A = [aij] be an m × n matrix. The transpose of A is obtained by interchanging the rows and
columns of A and it is denoted by AT.
∴ A T = [a ji ] is a n × m matrix.
Properties:
(i) (AT)T = A
(iii) (AB)T = BT AT
(ii) (A + B)T = AT + BT
(iv) (aA)T = aAT
Definition 1.15 Symmetric Matrix
A square matrix A = [aij] of order n is said to be symmetric if AT = A.
This means [aji] = [aij] ⇒ aji = aij for i, j = 1, 2, …n
Thus, in a symmetric matrix elements equidistant from the main diagonal are the same.
EXAMPLE 1.13
⎡1
⎡a h g ⎤
⎢
−2
⎢
⎥
A = ⎢ h b f ⎥ and B = ⎢⎢
3
⎢
⎢⎣ g f c ⎥⎦
⎢⎣ 4
−2
3
0
5
5
2
7 8
4 ⎤
⎥
7⎥
are symmetric matrices of orders 3 and 4.
8 ⎥
⎥
4 ⎥⎦
Definition 1.16 Skew-Symmetric Matrix
A square matrix A = [aij] of order n is said to be skew-symmetric if AT = −A.
This means
[aji] = −[aij] ⇒
aji = − aij for all i, j = 1, 2, …, n
In particular, put j = i, then aii = − aii
⇒ 2aii = 0 ⇒ aii = 0 for all i = 1, 2, …, n
So, in a skew-symmetric matrix, the diagonal elements are all zero and elements equidistant from
the main diagonal are equal in magnitude, but opposite in sign.
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EXAMPLE 1.14
2 −3⎤
⎡0
⎡ 0 1⎤
⎢
4 ⎥⎥ are skew-symmetric matrices of orders 2 and 3.
A =⎢
⎥ and B = ⎢ −2 0
1
0
−
⎦
⎣
⎢⎣ 3 −4 0 ⎥⎦
Definition 1.17 Non-Singular Matrix
A Square matrix A is said to be non-singular if A ≠ 0 ( A means determinant of A).
If A = 0, then A is singular.
Definition 1.18 Minor and Cofactor of an Element
Let A = [aij] be a square matrix of order n. If we delete the row and column of the element aij, we get
a square submatrix of order (n − 1).
The determinant of this submatrix is called the minor of the element aij and is denoted by Mij.
The cofactor of aij in A is A ij = ( −1)i + j M ij
EXAMPLE 1.15
⎡1 6 2⎤
A = ⎢⎢0 −2 4 ⎥⎥
⎢⎣ 3 1 2 ⎥⎦
The cofactor of a11 = 1 is A 11 = ( −1)1+1
−2 4
= −4 −4
1 2
The cofactor of a12 = 6 is A 12 = ( −1)1+ 2
0 4
= − (−12) = 12
3 2
= −8
1 2
= − (4 −0) = −4
0 4
Similarly, we can determine the cofactors of other elements.
The cofactor of a32 = 1 is A 32 = ( −1)3+ 2
Definition 1.19 Adjoint of a Matrix
Let A = [aij] be a square matrix. The adjoint of A is defined as the transpose of the matrix of cofactors
of the elements of A and it is denoted by adj A.
⎡ A 11
⎢A
Thus, adj A = ⎢ 21
⎢ :
⎢
⎣ A n1
A 12
A 21
:
A n2
… A 1n ⎤T
… A 2n ⎥
⎥
: ⎥
… A nn ⎥⎦
Properties: If A and B are square matrices of order n, then
(ii) (adj A) A = A (adj A) = A In.
(i)
adj AT = (adj A)T
(iii) adj(AB) = (adj A) (adj B)
Using property (ii), we define inverse.
Definition 1.20 Inverse of a Matrix
If A is a non-singular matrix, then the inverse of A is defined as
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adj A
and it is denoted by A−1.
A
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Matrices ■
A −1 =
∴
1.7
adj A
A
EXAMPLE 1.16
⎡1 6 2⎤
Find the inverse of A = ⎢0 −2 4 ⎥ .
⎢
⎥
⎢⎣ 3 1 2 ⎥⎦
Solution.
⎡1 6 2⎤
A = ⎢⎢0 −2 4 ⎥⎥
Given
⎣⎢ 3 1 2 ⎥⎦
1 6 2
∴
A = 0 −2 4 = 1(−4 −4) −6(0 − 12) + 2(0 + 6) = −8 + 72 + 12 = 76 ≠ 0
3 1 2
adj A
.
Since A ≠ 0, A is non-singular and hence A−1 exists and A −1 =
A
We shall find the cofactors of the elements of A
−2 4
0 4
A 11 = ( −1)1+1
= ( −4 − 4) = −8,
A 12 = ( −1)1+ 2
= −(0 − 12) = 12
1 2
3 2
A 13 = (−
−1)1+ 3
0 −2
= (0 + 6) = 6,
3 1
A 21 = ( −1) 2 +1
6 2
= −(12 − 2) = −10
1 2
A 22 = ( −1) 2 + 2
1 2
= ( 2 − 6) = −4,
3 2
A 23 = ( −1) 2 + 3
1 6
= −(1 − 18) = 17
3 1
A 31 = ( −1)3+1
6 2
= ( 24 + 4) = 28,
−2 4
A 32 = ( −1)3+ 2
1 2
= −( 4 − 0) = −4
0 4
A 33 = ( −1)3+ 3
1 6
= ( −2 − 0) = −2
0 −2
⎡ −8 12 6 ⎤
⎡ −8 −10 28 ⎤
⎢
⎥
adj A = ⎢ −10 −4 17 ⎥ = ⎢⎢12 −4 −4 ⎥⎥
⎢⎣ 28 −4 −2⎥⎦
⎢⎣ 6 17 −2 ⎥⎦
T
∴
∴
1.2
A
−1
⎡ −8 −10 28 ⎤
1 ⎢
12 −4 −4 ⎥⎥
=
76 ⎢
⎢⎣ 6 17 −2 ⎥⎦
COMPLEX MATRICES
A matrix with at least one element as complex number is called a complex matrix.
Let A = [aij] be a complex matrix.
The conjugate matrix of A is denoted by A and A = ⎡⎣aij ⎤⎦ .
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EXAMPLE 1.17
⎡ 2i
A =⎢
⎣3 − 2i
−i
0
2⎤
is a complex matrix.
3⎥⎦
⎡ 2i
The conjugate of A is A = ⎢
⎢⎣3 − 2i
−i
0
2⎤ ⎡ −2i i 2⎤
⎥=⎢
⎥
3⎥⎦ ⎣3 + 2i 0 3⎦
[{ conjugate of a + ib = a − ib]
We denote ( A ) by A*.
∴ A* is the transpose of the conjugate of A.
In the above example
T
⎡ −2i 3 + 2i ⎤
A = ⎢⎢ i
0 ⎥⎥
⎢⎣ 2
3 ⎥⎦
*
Note We have
∴ If
⎡⎣ A T ⎤⎦ = ⎡⎣ A T ⎤⎦
∴ A ∗ = ⎡⎣ A T ⎤⎦
A = [a ji ], then A T = [a ji ], ⎡⎣ A T ⎤⎦ =[a ji ]
∴
A ∗ = [a ji ]
Definition 1.21 Hermitian Matrix
A complex square matrix A is said to be a Hermitian matrix if A* = A and Skew-Hermitian matrix if
A* = −A.
A Hermitian matix is also denoted by AH.
If A = [aij], then A * = [a ji ] ∴ A* = A ⇒ a ji = aij for all i and j
Put j = i, then aii = aii ⇒ aii are real numbers.
So, the diagonal elements of a Hermitian matrix are real numbers.
The elements equidistant from the main diagonal are conjugates.
A* = −A ⇒ a ji = −aij for all i and j
Put j = i, then aii = −aii
If aii = a + ib, then aii = a − ib
∴ a − ib = −(a + ib) ⇒ 2a = 0 ⇒ a = 0
∴ aii = ib, which is purely imaginary if b ≠ 0 and 0 if b = 0.
∴ the diagonal elements of a Skew-Hermitian matrix are all purely imaginary or 0 and the elements
equidistant from the main diagonal are conjugates with opposite sign.
Properties: If A and B are complex matrices, then
1. ( A) = A,
2. A + B = A + B
3. aA = a A
4. A B = A B
7. (aA).* = aA *
5. (A*).* = A
8. (AB).* = B*A*
6. (A + B).* = A* + B*
Definition 1.22 Unitary Matrix
A complex square matrix is said to be unitary if AA* = A*A = I
From the definition it is obvious that A* is the inverse of A.
∴
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Matrices ■
1.9
EXAMPLE 1.18
⎡ 21
Show that A 5 ⎢
⎣3 1 i
Solution.
Given
32 i ⎤
is a Hermitian matrix.
2 ⎥⎦
⎡ −1 3 − i ⎤
A=⎢
2 ⎥⎦
⎣3 + i
T
∴
T
⎡ −1 3 − i ⎤
⎡ −1 3 − i ⎤
⎡ −1 3 + i ⎤
T
A * = (A ) = ⎢
=
=A
⎥
⎢3 − i
⎥ = ⎢3 + i
2
2 ⎥⎦
2 ⎥⎦
⎣
⎦
⎣
⎢⎣3 + i
∴ A is Hermitian matrix.
EXAMPLE 1.19
32i 312 i ⎤
⎡ 1
⎢
Show that B 5 ⎢ 31i
0
223i ⎥⎥ is a Hermitian matrix.
⎢⎣322 i 213i 22 ⎥⎦
Solution.
Since the diagonal elements are real and elements equidistant from the main diagonal are conjugates,
B is a Hermitian matrix.
EXAMPLE 1.20
11 i ⎤
⎡ 2i
Show that A 5 ⎢
is a Skew-Hermitian matrix.
0 ⎥⎦
⎣2(1 2 i )
Solution.
1+ i⎤
⎡ 2i
A =⎢
Given
⎥
⎣ −(1 − i ) 0 ⎦
Since the diagonal elements are purely imaginary or zero and (1 + i) and − (1 − i) are conjugates with
opposite sign, A is Skew-Hermitian matrix.
EXAMPLE 1.21
11 i
2 25i ⎤
⎡ 2i
⎢
Show that B 5 ⎢ 2(1 2 i )
0
2 1 3i ⎥⎥ is skew-Hermitian.
⎢⎣2( 2 15i ) 2( 2 2 3i )
3i ⎥⎦
Solution.
1+ i
2 − 5i ⎤
⎡ 2i
Given B = ⎢ −(1 − i )
0
2 + 3i ⎥⎥
⎢
⎢⎣ −( 2 + 5i ) −( 2 − 3i )
3i ⎥⎦
In B, the diagonal elements are purely imaginary or zero and the elements equidistant from the main
diagonal are conjugates with opposite sign. So, B is skew-Hermitian matrix.
Note If A is a real matrix, then the definition of unitary
⇒
AAT = ATA = I.
In this case A is called an orthogonal matrix. So, if A is an orthogonal matrix, then AT = A−1.
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WORKED EXAMPLES
EXAMPLE 1
⎡ 2 1i
If A 5 ⎢
⎣ 25
Solution.
Given
3 21 13i ⎤
, then show that AA* is a Hermitian matrix.
i 4 22 i ⎥⎦
⎡ 2 + i 3 −1 + 3i ⎤
A =⎢
⎥
⎣ −5 i 4 − 2i ⎦
⎡ 2−i
T
−1 − 3i ⎤
∴
A* = [ A ]
= ⎢⎢ 3
4 + 2i ⎥⎦
⎢⎣ −1 − 3i
We have to prove AA* is a Hermitian matrix.
That is to prove (AA*)* = AA*
T
Now
⎡2 − i
=⎢
⎣ −5
3
−i
⎡ 2−i
⎡ 2 + i 3 −1 + 3i ⎤ ⎢
AA* = ⎢
⎥⎢ 3
⎣ −5 i 4 − 2i ⎦ ⎢
⎣ −1 − 3i
−5 ⎤
−i ⎥⎥
4 + 2i ⎥⎦
−5 ⎤
−i ⎥⎥
4 + 2i ⎥⎦
⎡( 2 + i )( 2 − i ) + 3 ⋅ 3 + ( −1 + 3i )( −1 − 3i ) ( 2 + i )( −5) + 3( −i ) + ( −1 + 3i ) ( 4 + 2i ) ⎤
=⎢
( −5)( −5) + i ( −i ) + ( 4 − 2i )( 4 + 2i ) ⎥⎦
⎣ −5( 2 − i ) + i ⋅ 3 + ( 4 − 2i )( −1 − 3i )
⎡
22 + 1 + 9 + 1 + 32
−10 − 5i − 3i − 4 + 10i + 6i 2 ⎤
=⎢
⎥
2
25 − i 2 + 4 2 + 22
⎣ −10 + 5i + 3i − 4 − 10i + 6i
⎦
24
−20 + 2i ⎤
−14 + 2i − 6 ⎤ ⎡ 24
⎡
=⎢
=⎢
⎥
46
46 ⎥⎦
⎣ −14 − 2i − 6
⎦ ⎣ −20 − 2i
∴
⎡ 24
( A A *)* = ⎢
⎢⎣ −20 − 2i
[{ i 2 = −1]
T
−20 + 2i ⎤
⎥
46 ⎥⎦
−20 − 2i ⎤
⎡ 24
⎡ 24
=⎢
=⎢
⎥
46 ⎦
⎣ −20 + 2i
⎣ −20 − 2i
⇒
(AA*)* = AA*
Hence, AA* is a Hermitian matrix.
T
−20 + 2i ⎤
= AA*
46 ⎥⎦
EXAMPLE 2
Show that every square complex matrix can be expressed uniquely as P + iQ, where P and Q are
Hermitian matrices.
Solution.
Let A be any square complex matrix.
We shall rewrite A as
A=
M01_ENGINEERING_MATHEMATICS-I _CH01_Part A.indd 10
1
⎡1
⎤
[A + A *] + i ⎢ ( A − A *) ⎥
2
2
i
⎣
⎦
5/30/2016 4:34:54 PM
Matrices ■
1
1
( A + A *),
Q = ( A − A *),
2
2i
We shall now prove P and Q are Hermitian.
Put P =
1.11
then A = P + iQ.
*
Now,
⎡1
⎤
⎡1
⎤ 1
P* = ⎢ ( A + A *) ⎥ = ⎢ ( A * + ( A *) *⎥ = ( A * + A ) = P
⎣2
⎦
⎣2
⎦ 2
∴ P is Hermitian.
*
and
1
1
1
⎡1
⎤
Q* = ⎢ ( A − A*) ⎥ =
( A * −( A*) *] = − [ A * − A] = ( A − A*) = Q
[
2i
2i
2i
⎣ 2i
⎦
∴ Q is Hermitian.
We shall now prove the uniqueness of the expression A = P + iQ.
If possible, let
A = R + iS
where R and S are Hermitian matrices.
∴
R* = R and S* = S
Now,
A* = (R + iS)* = R* + (iS)* = R* − iS* = R − iS
(1) + (2) ⇒
A + A* = 2R ⇒ R =
(1)
[by property]
(2)
1
( A + A *) = P
2
1
( A − A *) = Q
2i
∴ the expression A = P + iQ is unique.
(1) − (2) ⇒
A − A* = 2iS ⇒ S =
EXAMPLE 3
If A is any square complex matrix, prove that (1) A 1 A* is Hermitian and (ii) A 5 B 1 C, where
B is Hermitian and C is Skew-Hermitian.
Solution.
Given A is a square complex matrix.
(i) Let P = A + A*
∴
P* = (A + A*)* = A* + (A*)* = A* + A = A + A* = P
∴
P is Hermitian
Hence, A + A* is Hermitian.
To prove (ii): Since A is square complex matrix, we can write A as
1
1
A = ( A + A*) + ( A − A*) = B + C
2
2
1
1
where B = ( A + A *) is Hermitian by part (i) and C = ( A − A *)
2
2
[by property]
*
1
1
1
⎡1
⎤
⇒ C * = ⎢ ( A − A*) ⎥ = [( A * −( A*) *] = [ A * − A] = − [ A − A*] = −C
2
2
2
⎣2
⎦
∴ C is Skew- Hermitian.
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EXAMPLE 4
1 12 i ⎤
⎡ 0
21
If A 5 ⎢
⎥ , then show that (I 2 A) (I 1 A) is a unitary matrix.
2
1
1
2
i
0
⎣
⎦
Solution.
Given
∴
∴
1 + 2i ⎤ ⎡ 0
1 + 2i ⎤
⎡1 0 ⎤
⎡ 0
A =⎢
=⎢
. Let I = ⎢
⎥
⎥.
⎥
0 ⎦ ⎣ −(1 − 2i )
0 ⎦
⎣ −1 + 2i
⎣0 1 ⎦
1 + 2i ⎤
1 + 2i ⎤
⎡1 0 ⎤ ⎡ 0
⎡ 1
I + A= ⎢
=⎢
⎥ + ⎢ −(1 − 2i )
⎥
0
1
−
−
0
(
1
2
i
)
1 ⎥⎦
⎣
⎦ ⎣
⎦
⎣
1
1 + 2i
I +A =
= 1 + (1 − 2i )(1 + 2i ) = 1 + 1 + 4 = 6 ≠ 0
−(1 − 2i )
1
Inverse of I + A exists and (I + A ) −1 =
adj (I + A )
I +A
1 − 2i ⎤
⎡ 1
⎡ 1
=⎢
adj(I + A ) = ⎢
⎥
1 ⎦
⎣ −(1 + 2i )
⎣1 − 2i
T
∴
∴
(I + A ) −1 =
1⎡ 1
6 ⎢⎣1 − 2i
−(1 + 2i ) ⎤
1 ⎥⎦
−(1 + 2i ) ⎤
1 ⎥⎦
1 + 2i ⎤ ⎡ 1
⎡1 0 ⎤ ⎡ 0
=
I −A = ⎢
⎥ − ⎢ −1 + 2i
0
1
0 ⎥⎦ ⎢⎣1 − 2i
⎣
⎦ ⎣
∴ (I − A )(I + A ) −1 =
1⎡ 1
6 ⎢⎣1 − 2i
−(1 + 2i ) ⎤
1 ⎥⎦
⎡ 1
⎢1 − 2i
⎣
−(1 + 2i ) ⎤
1 ⎥⎦
−(1 + 2i ) ⎤
1 ⎥⎦
=
1 ⎡1 − (1 + 2i )(1 − 2i ) −(1 + 2i ) − (1 + 2i ) ⎤
6 ⎢⎣ (1 − 2i ) + (1 − 2i ) −(1 − 2i )(1 + 2i ) + 1⎥⎦
=
−2(1 + 2i ) ⎤
1 ⎡1 − (1 + 4) −2(1 + 2i ) ⎤ 1 ⎡ −4
= ⎢
= B , say
⎢
⎥
−4 ⎥⎦
6 ⎣ 2(1 − 2i ) −(1 + 4) + 1⎦ 6 ⎣ 2(1 − 2i )
T
Now,
B* =
− 2(1 + 2i ) ⎤
1 ⎡ −4
⎢
⎥
6 ⎢⎣ 2(1 − 2i )
− 4 ⎥⎦
2(1 + 2i ) ⎤
−2(1 − 2i ) ⎤
1 ⎡ −4
1 ⎡ −4
= ⎢
= ⎢
⎥
−4 ⎥⎦
−4 ⎦
6 ⎣ 2(1 + 2i )
6 ⎣ −2(1 − 2i )
T
To prove
Now,
B = (I − A) (I + A)−1 is unitary, verify BB* = I
BB * =
=
−2(1 + 2i ) ⎤ ⎡ −4
2(1 + 2i ) ⎤
1 ⎡ −4
⎢
⎥
⎢
−4 ⎦ ⎣ −2(1 − 2i )
−4 ⎥⎦
36 ⎣ 2(1 − 2i )
1 ⎡16 + 4(1 + 2i )(1 − 2i ) −8(1 + 2i ) + 8(1 + 2i ) ⎤
36 ⎢⎣ −8(1 − 2i ) + 8(1 − 2i ) 4(1 + 2i))(1 − 2i ) + 16 ⎥⎦
M01_ENGINEERING_MATHEMATICS-I _CH01_Part A.indd 12
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Matrices ■
=
1.13
0
⎤ 1 ⎡36 0 ⎤ ⎡1 0 ⎤
1 ⎡16 + 4(1 + 4)
⎢
⎥ = 36 ⎢ 0 36 ⎥ = ⎢0 1 ⎥ = I .
0
4
(
1
+
4
)
+
6
36 ⎣
⎦
⎣
⎦ ⎣
⎦
∴ B is unitary.
Hence, (I − A)(I + A)−1 is unitary.
Note Another method: To prove B is unitary, verify B* = B−1
EXERCISE 1.1
⎡3
⎡7 0⎤
1. If A + B = ⎢
⎥ , A − B = ⎢0
2
5
⎣
⎦
⎣
2. Find x, y, z and w if
6 ⎤ ⎡ 4
⎡x y ⎤ ⎡ x
3⎢
⎥ = ⎢ −1 2w ⎥ + ⎢ z + w
z
w
⎣
⎦ ⎣
⎦ ⎣
0⎤
, find A and B
3⎥⎦
x + y⎤
3 ⎥⎦
3. If matrix A has x rows and x +5 columns and B has y rows and 11 − y columns such that both AB and BA
exist, then find x and y.
⎡ 2 3 4⎤
⎡ 1 3 0⎤
4. If A = ⎢⎢ 1 2 3⎥⎥ and B = ⎢ −1 2 1 ⎥ , then find AB and BA and test their equality.
⎢
⎥
⎢⎣ −1 1 2⎥⎦
⎢⎣ 0 0 2⎥⎦
⎡
⎢ 0
5. If A = ⎢
⎢ tan a
⎢⎣
2
a⎤
− tan ⎥
2
⎥ , show that I + A = [I − A ] ⎡cos a − sin a ⎤
⎢ sin a cos a ⎥
⎣
⎦
0 ⎥⎥
⎦
⎡ cos a sin a ⎤
T
6. If A = ⎢
⎥ , then verify that AA = I2.
⎣ − sin a cos a ⎦
7. If A is a square matrix, then show that A can be expressed as A = P + Q, where P is symmetric and Q is
skew-symmetric.
⎡
A + AT
A − AT ⎤
,Q =
⎢ Hint: Take P =
⎥
2
2 ⎦
⎣
⎡2 0 1⎤
⎢
⎥
8. If A = ⎢ 2 1 3⎥ and f(x) = x2 − 5x + 6, then find f(A).
⎢⎣1 −1 0 ⎥⎦
⎡ 1 −1 1 ⎤
⎢
3 0 ⎥⎥, then prove that A(adj A) =
9. If A = 2
⎢
⎢⎣18 2 10 ⎥⎦
⎡0 0 0⎤
⎢0 0 0⎥ .
⎢
⎥
⎢⎣0 0 0 ⎥⎦
⎡1 0 0 ⎤
10. Find the inverse of A = ⎢3 3 0 ⎥ in terms of adj A.
⎢
⎥
⎢⎣5 2 −1⎥⎦
1 ⎡1 + i −1 + i ⎤
is unitary.
11. Show that A = ⎢
2 ⎣1 + i 1 − i ⎥⎦
12. If A and B are orthogonal matrices of the same order, prove that AB is orthogonal.
[Hint: AAT = I, BBT = I. Compute AB(AB)T = A(BBT)AT = AAT = I].
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13. If A and B are Hermitian matrices of the same order, prove that
(i) A + B is Hermitian
(ii) AB + BA is Hermitian
(iii) iA is Skew-Hermitian
(iv) AB − BA is Skew-Hermitian
14. Find the inverse of the following matrices.
⎡ 2 −1 4 ⎤
3 3⎤
⎡4
(i) ⎢ −3 0 1 ⎥
(ii) ⎢ −1 0 −1⎥
⎥
⎢
⎢
⎥
⎢⎣ −1 1 2⎥⎦
⎢⎣ −4 −4 −3⎥⎦
⎡ 3 −3 4 ⎤
15. If A = ⎢⎢ 2 −3 4 ⎥⎥ , then show that A3 = A−1.
⎢⎣0 −1 1 ⎥⎦
ANSWERS TO EXERCISE 1.1
⎡2 0⎤
⎡5 0 ⎤
1. A = ⎢
⎥ , B = ⎢1 1 ⎥
1
4
⎦
⎣
⎦
⎣
2. x = 2, y = 4, z = 1, w = 3
4. AB ≠ BA
8.
14.
1.3
⎡ 1 −1 −3 ⎤
f ( A) = ⎢⎢ −1 −1 −10 ⎥⎥
⎢⎣ −5 4
4 ⎥⎦
3 3⎤
⎡4
⎡ 1 −6 1 ⎤
(i) A−1 = ⎢⎢ 5 8 −14 ⎥⎥ (ii) A−1 = ⎢⎢ −1 0 −1⎥⎥
⎢⎣ −4 −4 3 ⎥⎦
⎢⎣ −3 −1 −3 ⎥⎦
3. x = 3, y = 8
⎡1
⎢
10. A−1 = ⎢ −1
⎢
⎢
⎢3
⎣
0
1
3
2
3
0⎤
⎥
0⎥
⎥
⎥
−1⎥
⎦
RANK OF A MATRIX
Let A = [aij] be an m × n matrix. A matrix obtained by omitting some rows and columns of A is called
a submatrix of A.
The determinant of a square submatrix of order r is called a minor of order r of A.
EXAMPLE 1.22
⎡ 2 3 4 −1⎤
Consider A = ⎢⎢0 3 4 0 ⎥⎥
⎢⎣ 3 −2 −1 2 ⎥⎦
2 3 4
⎡2 3 4 ⎤
⎢
⎥
Omitting the fourth column, we get the submatrix A 1 = ⎢0 3 4 ⎥ and A 1 = 0 3 4 is a
minor of order 3.
⎢⎣ 3 −2 −1⎥⎦
3 −2 −1
⎡3 −1⎤
Omitting the first and third columns and the third row, we get the submatrix A 2 = ⎢
⎥ and
⎣3 0 ⎦
3 −1
is a minor of order 2. Since A 2 = 3 ≠ 0, it is called a non-vanishing minor of order 2.
A2 =
3 0
But
3 4
= 0 , so it is called a vanishing minor of order 2.
3 4
M01_ENGINEERING_MATHEMATICS-I _CH01_Part A.indd 14
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Matrices ■
1.15
Definition 1.23 Rank of a Matrix
Let A be an m × n matrix. A is said to be of rank r if (1) at least one minor of A of order r is not zero
and (ii) every minor of A of order (r + 1) (and higher order) is zero.
The rank of A is denoted by r(A) or r(A).
Note
(1) The definition says rank of A is the order of the largest non-vanishing minor of A.
(2) The Rank of Zero matrix is zero.
(3) All non-zero matrices have rank ≥ 1.
(4) The rank of an m × n matrix is less than or equal to the min {m, n}.
(5) r(A) = r(AT)
(6) If In is the unit matrix of order n, then I n = 1 ≠ 0 and so, r(In) = n.
To find the rank of a matrix A, we have to identify the largest non-vanishing minor. This process
involves a lot of computations and so it is tedious for matrices of large type.
To reduce the computations, we apply elementary transformations and transform the given matrix
to a convenient form, namely Echelon form or normal form.
Elementary transformations
1. Interchange of any two rows (or columns)
2. Multiplication of elements of any row (or column) by a non-zero number k.
3. Addition to the elements of a row (column), the corresponding elements of another row (column)
multiplied by a fixed number.
Note When an elementary transformation is applied to a row, it is called a row transformation and
when it is applied to a column, it is called a column transformation.
Notation: The following symbols will be used to denote the elementary row operations.
(i) Ri ↔ Rj means ith row and jth row are interchanged.
(ii) Ri → kRi means the elements of ith row is multiplied by k (≠0)
(iii) Ri → Ri + kRj means the jth row is multiplied by k and added to the ith row.
Similarly we indicate the column transformations by Ci ↔ Cj, Ci → kCi, Ci → Ci + kCj
Definition 1.24 Equivalent Matrices
Two matrices A and B of the same type are said to be equivalent if one matrix can be obtained from the
other by a sequence of elementary row (column) transformations. Then we write A ~ B.
Results:
1. The Rank of a matrix is unaffected by elementary transformations.
2. Equivalent matrices have the same rank.
Definition 1.25 Echelon Matrix
A matrix is called a row-echelon matrix if (1) all zero rows (i.e., rows with zero elements only), if any,
are on the bottom of the matrix and (ii) each leading non-zero element is to the right of the leading
non-zero element in the preceding row.
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EXAMPLE 1.23
⎡2
⎡1 2 3⎤
⎢0
A = ⎢⎢0 1 2⎥⎥ , B = ⎢
⎢0
⎢⎣0 0 0 ⎥⎦
⎢
⎣0
⎡1 −1 0 4 5⎤
⎢0 −1 2 1 3⎥
⎥
C=⎢
⎢0 0 0 6 1 ⎥
⎢
⎥
⎣0 0 0 0 0 ⎦
1 −1
0 −1
0 0
0 0
0⎤
2⎥⎥
,
0⎥
⎥
0⎦
⎡1 2 3⎤
D = ⎢⎢0 2 1 ⎥⎥ are row echelon matrices.
⎢⎣0 0 2⎥⎦
and
Note
Triangular matrix is a special case of an echelon matrix.
Result: If a matrix A is equivalent to a row echelon matrix B, then r(A) = the number of non-zero rows
of B.
In the above examples, r(A) = 2, r(B) = 2, r(C) = 3, r(D) = 3.
WORKED EXAMPLES
⎡1
⎢2
Find the rank of the matrix A 5 ⎢
⎢3
⎢
⎣6
Solution.
⎡1 2 3 0 ⎤ ⎡1 2
⎢ 2 4 3 2⎥ ⎢
⎥ ∼ ⎢0 0
Given A = ⎢
⎢ 3 2 1 3⎥ ⎢0 −4
⎥ ⎢
⎢
⎣6 8 7 5⎦ ⎣0 −4
EXAMPLE 1
2
4
2
8
3
−3
−8
−11
3
⎡1 2
⎢0 0 −3
∼⎢
⎢0 −4 −8
⎢
⎣0 0 −3
⎡1 2 3
⎢0 −4 −8
∼⎢
⎢0 0 −3
⎢
⎣0 0 −3
3
3
1
7
0⎤
2 ⎥⎥
, by reducing to an echelon matrix.
3⎥
⎥
5⎦
0⎤
2⎥⎥ R 2 → R 2 + ( −2)R1
3⎥ R 3 → R 3 + ( −3)R1
⎥
5⎦ R 4 → R 4 + ( −6)R1
0⎤
2⎥⎥
3⎥
⎥
2⎦ R 4 → R 4 − R 3
0⎤
3⎥⎥ R 2 ↔ R 3
2⎥
⎥
2⎦
⎡1 2 3 0⎤
⎢0 −4 −8 3⎥
⎥
∼⎢
⎢0 0 −3 2⎥
⎥
⎢
0 0⎦ R 4 → R 4 − R 3
⎣0 0
∴
= B, which is a row echelon matrix.
r(A) = the number of non-zero rows in B = 3
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Matrices ■
1.17
EXAMPLE 2
⎡1 2 3 21⎤
Determine the rank of the matrix A 5 ⎢⎢ 3 6 9 23 ⎥⎥ , by reducing to an echelon matrix.
⎢⎣ 2 4 6 22 ⎥⎦
Solution.
⎡1 2 3 −1⎤ ⎡1 2 3 −1⎤
Given A = ⎢⎢ 3 6 9 −3⎥⎥ ∼ ⎢0 0 0 0 ⎥ R 2 → R 2 + ( −3)R1
⎢
⎥
⎢⎣ 2 4 6 −2⎥⎦ ⎢⎣0 0 0 0 ⎥⎦ R 3 → R 3 + ( −2)R1
= B, which is a row echelon matrix.
r(A) = the number of non-zero rows in B = 1
∴
EXAMPLE 3
⎡6
⎢5
Find the value of k if the rank of the matrix ⎢
⎢3
⎢
⎣2
Solution.
Let
⎡6
⎢5
A =⎢
⎢3
⎢
⎣2
3 5
2 3
1 2
1 1
9⎤ ⎡
1
6 ⎥⎥ ⎢
⎢
3 ⎥ ∼ ⎢5
⎥ ⎢
k ⎦ ⎢3
⎢⎣ 2
1
2
2
1
1
1
⎡
⎢1 2
⎢
⎢0 − 1
⎢
2
∼⎢
1
⎢0 −
⎢
2
⎢
⎢0 0
⎢⎣
1
⎡
⎢1 2
⎢
⎢0 − 1
⎢
2
∼⎢
⎢0 0
⎢
⎢
⎢0 0
⎢⎣
M01_ENGINEERING_MATHEMATICS-I _CH01_Part A.indd 17
5
6
3
2
1
3
2
1
1
5
3
2
1
9⎤
6 ⎥⎥
is 3.
3⎥
⎥
k⎦
3⎤
2⎥
⎥
6⎥
3⎥
⎥
k ⎥⎦
5
6
7
−
6
3
−
6
4
−
6
⎤
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
k − 3⎥
⎥⎦
5
6
7
−
6
4
6
4
−
6
⎤
⎥
⎥
⎥
⎥
⎥
0 ⎥
⎥
⎥
k − 3⎥
⎥⎦
R1 →
3
2
3
−
2
3
−
2
1
R1
6
R 2 → R 2 + ( −5)R1
R 3 → R 3 + ( −3)R1
R 4 → R 4 + ( −2)R1
3
2
3
−
2
R3 → R3 − R 2
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1
⎡
⎢1 2
⎢
⎢0 − 1
∼⎢
2
⎢
⎢0 0
⎢
⎢0 0
⎣
5
6
7
−
6
4
6
0
⎤
⎥
⎥
⎥
⎥
⎥
0 ⎥
⎥
k − 3⎥⎦ R 4 → R 4 + R 3
3
2
3
−
2
=B
Given r(A) = 3. So, the number of non-zero rows of B should be 3.
∴
k−3=0 ⇒ k=3
Definition 1.26 Elementary Matrix
A matrix obtained from a unit matrix by performing a single elementary row (column) transformation
is called an elementary matrix.
Since unit matrices are non-singular square matrices, elementary matrices are also non-singular.
EXAMPLE 1.24
⎡1 0 0 ⎤ ⎡1 0 0⎤
I 3 = ⎢⎢0 1 0 ⎥⎥ z ⎢⎢0 1 1 ⎥⎥
⎢⎣0 0 1 ⎥⎦ ⎢⎣0 0 1 ⎥⎦ C 3 → C 3 + C 2
This is an elementary matrix.
⎡1 0 0 ⎤
Similarly, ⎢⎢0 3 0 ⎥⎥ got by R2 → 3R2 is an elementary matrix.
⎢⎣0 0 1 ⎥⎦
Definition 1.27 Normal form of a Matrix
Any non-zero matrix A of rank r can be reduced by a sequence of elementary transformations to the
⎡I r 0 ⎤
form ⎢
⎥ , where Ir is a unit matrix of order r.
⎣0 0⎦
This form is called a normal form of A.
⎡I ⎤
Other normal forms are Ir, ⎢ r ⎥ , [Ir, 0].
⎣0⎦
Theorem 1.1
Let A be an m × n matrix of rank r. Then there exist non-singular matrices P and Q of orders m and n
⎡I r 0 ⎤
respectively such that PAQ = ⎢
⎥
⎣0 0⎦
Note Each elementary row transformation of A is equivalent to pre multiplying A by the corresponding
elementary matrix. Each elementary column transformation is equivalent to post multiplying A
by the corresponding elementary matrix. So, there exists elementary matrices P1, P2, …, Pk and
Q1, Q2, …, Qt such that
⎡I r 0 ⎤
P1 P2 … Pk A Q1 Q2 … Qt = ⎢
⎥
⎣0 0⎦
M01_ENGINEERING_MATHEMATICS-I _CH01_Part A.indd 18
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Matrices ■
1.19
⎡I r 0 ⎤
PAQ = ⎢
⎥
⎣0 0⎦
P = P1 P2 … Pk, Q = Q1 Q2 … Qt
⇒
where
Working rule to find normal form and P, Q:
Let A be a non-zero m × n matrix write A = ImAIn(which is obviously true). Reduce A on the L. H. S to
normal form by applying elementary row and column transformations on A.
Each elementary row transformation of A will be applied to Im on R. H. S and each elementary
column transformation of A will be applied to In on R. H. S.
⎡I r 0 ⎤
= PAQ.
After a sequence of suitable applications of elementary transformations, we get ⎢
0 0 ⎥⎦
⎣
Then the rank of A is the rank of I = r
r
WORKED EXAMPLES
EXAMPLE 1
⎡0 1 2 1 ⎤
Reduce the matrix ⎢⎢ 1 2 3 1⎥⎥ to normal form and hence find the rank.
⎢⎣ 3 1 1 3 ⎥⎦
Solution.
Let
⎡ 0 1 2 1 ⎤ ⎡ 1 0 2 1 ⎤ C1 ↔ C 2
A = ⎢⎢1 2 3 2⎥⎥ ∼ ⎢ 2 1 3 2⎥
⎢
⎥
⎢⎣ 3 1 1 3⎥⎦ ⎢⎣1 3 1 3⎥⎦
⎡1 0 0 0⎤
∼ ⎢⎢ 2 1 −1 0 ⎥⎥ C 3 → C 3 + ( −2)C1
⎢⎣ 1 3 −1 2⎥⎦ C 4 → C 4 − C1
⎡1 0 0 0⎤
∼ ⎢⎢0 1 −1 0 ⎥⎥ R 2 → R 2 + ( −2)R1
⎢⎣0 3 −1 2⎥⎦ R 3 → R 3 − R1
⎡1 0 0 0⎤
∼ ⎢⎢0 1 0 0 ⎥⎥
⎢⎣0 3 2 2⎥⎦ C 3 → C 3 + C 2
⎡1
∼ ⎢⎢0
⎢⎣0
⎡1
∼ ⎢⎢0
⎢⎣0
M01_ENGINEERING_MATHEMATICS-I _CH01_Part A.indd 19
0 0 0⎤
1 0 0 ⎥⎥
0 2 2⎥⎦ R 3 → R 3 + ( −3)R 2
0 0 0⎤
1 0 0 ⎥⎥
1
0 1 1 ⎥⎦ R 3 → R 3
2
5/30/2016 4:35:14 PM
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Engineering Mathematics
⎡1 0 0 0⎤
∼ ⎢⎢0 1 0 0 ⎥⎥
⎢⎣0 0 1 0 ⎥⎦ C 4 → C 4 − C 3
= [I 3 : 0 ]
This is the normal form of A and so the r(A) = 3
EXAMPLE 2
⎡1 21 21⎤
Let A 5 ⎢⎢1 1
1 ⎥⎥ . Find matrices P and Q such that PAQ is in the normal form. Also find rank
⎢⎣3 1
1 ⎥⎦
of A.
Solution.
Given
⎡1 −1 −1⎤
A = ⎢⎢1 1 1 ⎥⎥
⎢⎣3 1 1 ⎥⎦ 3× 3
Consider
A = I3AI3
⎡1 −1 −1⎤ ⎡1 0 0 ⎤ ⎡1 0 0 ⎤
⎢1 1 1 ⎥ = ⎢0 1 0 ⎥ A ⎢0 1 0 ⎥
⇒
⎢
⎥ ⎢
⎥ ⎢
⎥
⎣⎢3 1 1 ⎥⎦ ⎢⎣0 0 1 ⎥⎦ ⎢⎣0 0 1 ⎥⎦
Our aim is to reduce the LHS matrix to normal form.
Also row operations to be applied to pre factor and column operations to be applied to post factor.
Apply, C 2 → C 2 + C1 ⎡1 0 0 ⎤ ⎡1 0 0 ⎤ ⎡1 1 1 ⎤
⎢1 2 2 ⎥ = ⎢0 1 0 ⎥ A ⎢0 1 0 ⎥
C 3 → C 3 + C1
⎢
⎥ ⎢
⎥ ⎢
⎥
⎢⎣3 4 4 ⎥⎦ ⎢⎣0 0 1 ⎥⎦ ⎢⎣0 0 1 ⎥⎦
and to post factor
R 2 → R 2 − R1
R 3 → R 3 + ( −3)R1
⎡1 0 0 ⎤ ⎡ 1 0 0 ⎤ ⎡1 1 1⎤
⎢0 2 2⎥ = ⎢ −1 1 0 ⎥ A ⎢0 1 0 ⎥
⎢
⎥ ⎢
⎥ ⎢
⎥
⎢⎣0 4 4 ⎥⎦ ⎢⎣ −3 0 1 ⎥⎦ ⎢⎣0 0 1 ⎥⎦
and to pre factor
R 3 → R 3 + ( −2)R 2
and to pre factor
1
R2 → R2
2
and to pre factor
⎡1 0 0⎤ ⎡ 1 0 0⎤ ⎡1 1 1⎤
⎢0 2 2⎥ = ⎢ −1 1 0 ⎥ A ⎢ 0 1 0 ⎥
⎢
⎥ ⎢
⎥ ⎢
⎥
⎢⎣0 0 0 ⎥⎦ ⎢⎣ −1 −2 1 ⎥⎦ ⎢⎣ 0 0 1 ⎥⎦
⎡ 1
⎡1 0 0⎤ ⎢
⎢0 1 1⎥ = ⎢ − 1
⎢
⎥ ⎢ 2
⎢⎣0 0 0 ⎥⎦ ⎢
⎣ −1
M01_ENGINEERING_MATHEMATICS-I _CH01_Part A.indd 20
0 0⎤
⎥ ⎡1 1 1⎤
1
0 ⎥ A ⎢⎢0 1 0 ⎥⎥
2
⎥
⎢0 0 1 ⎥⎦
−2 1 ⎥⎦ ⎣
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Matrices ■
C3 → C3 − C 2
and to post factor
⎡ 1
⎡1 0 0⎤ ⎢
⎢0 1 0⎥ = ⎢ − 1
⎢
⎥ ⎢ 2
⎢⎣0 0 0 ⎥⎦ ⎢
⎣ −1
⎡I 2
⎢0
⎣
⇒
1.21
0 0⎤
⎥ ⎡1 1 0 ⎤
1
0 ⎥ A ⎢⎢0 1 −1⎥⎥
2
⎥
⎢0 0 1 ⎥⎦
−2 1 ⎥⎦ ⎣
0⎤
= PA Q
0 ⎥⎦
This shows r(A) = 2, and
⎡ 1
⎢ 1
P = ⎢−
⎢ 2
⎢ −1
⎣
0 0⎤
⎡1 1 0 ⎤
⎥
1
0 ⎥ , Q = ⎢⎢0 1 −1⎥⎥
2
⎥
⎢⎣0 0 1 ⎥⎦
−2 1 ⎥⎦
EXAMPLE 3
⎡ 1 21 21 2 ⎤
Let A 5 ⎢⎢ 4 2
2 21⎥⎥ . Find the non-singular matrices P and Q, such that PAQ is in the
⎢⎣ 2 2
0 22 ⎥⎦
normal form. Also find the rank of A.
Solution.
Given
⎡ 1 −1 −1 2 ⎤
A = ⎢⎢ 4 2 2 −1⎥⎥
⎢⎣ 2 2 0 −2⎥⎦ 3× 4
Consider
A = I3AI4
⇒
⎡1
⎡ 1 −1 −1 2 ⎤ ⎡ 1 0 0 ⎤
⎢
⎢ 4 2 2 −1⎥ = ⎢ 0 1 0 ⎥ A ⎢0
⎢
⎥ ⎢
⎥
⎢0
⎢⎣ 2 2 0 −2⎥⎦ ⎢⎣ 0 0 1 ⎥⎦
⎢
⎣0
0 0 0⎤
1 0 0 ⎥⎥
0 1 0⎥
⎥
0 0 1⎦
Our aim is to reduce the LHS matrix to normal form. Also row operations to be applied to pre factor
and column operations to be applied to post factor.
C 2 → C 2 + C1
C 3 → C 3 + C1
C 4 → C 4 + ( −2)C1
and to post factor
⎡1
⎡1 0 0 0 ⎤ ⎡1 0 0 ⎤
⎢
⎢ 4 6 6 −9 ⎥ = ⎢0 1 0 ⎥ A ⎢0
⎢
⎥ ⎢
⎥
⎢0
⎢⎣ 2 4 2 −6 ⎥⎦ ⎢⎣0 0 1 ⎥⎦
⎢
⎣0
R 2 → R 2 + ( −4)R1
R 3 → R 3 + ( −2)R1
and to pre factor
⎡1
⎡1 0 0 0 ⎤ ⎡ 1 0 0 ⎤
⎢
⎢0 6 6 −9 ⎥ = ⎢ −4 1 0 ⎥ A ⎢0
⎢
⎥ ⎢
⎥
⎢0
⎢⎣0 4 2 −6 ⎥⎦ ⎢⎣ −2 0 1 ⎥⎦
⎢
⎣0
M01_ENGINEERING_MATHEMATICS-I _CH01_Part A.indd 21
1 −2⎤
0 0 ⎥⎥
1 0⎥
⎥
0 1⎦
1
1
0
0
1
1
0
0
1 −2⎤
0 0 ⎥⎥
1 0⎥
⎥
0 1⎦
5/30/2016 4:35:18 PM
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■
Engineering Mathematics
1
R2 → R2
3
1
R3 → R3
2
and to pre factor
3
C4 → C4 + C2
2
and to post factor
C3 → C3 − C 2
and to post factor
R3 → R3 − R 2
and to pre factor
1
R2 → R2
2
R 3 → ( −1)R 3
and to pre factor
⇒
where
⎡
⎢ 1
⎡1 0 0 0 ⎤ ⎢
⎢0 2 2 −3⎥ = ⎢ − 4
⎢
⎥ ⎢ 3
⎢⎣0 2 1 −3⎥⎦ ⎢
⎢ −1
⎣
⎡
⎢ 1
⎡1 0 0 0⎤ ⎢
⎢0 2 2 0 ⎥ = ⎢ − 4
⎢
⎥ ⎢ 3
⎢⎣0 2 1 0 ⎥⎦ ⎢
⎢ −1
⎣
0
1
3
0
0
1
3
0
⎡
⎢ 1
⎡1 0 0 0⎤ ⎢
⎢0 2 0 0⎥ = ⎢ − 4
⎢
⎥ ⎢ 3
⎢⎣0 2 −1 0 ⎥⎦ ⎢
⎢ −1
⎣
⎡
⎢ 1
⎡1 0 0 0 ⎤ ⎢
⎢0 2 0 0 ⎥ = ⎢ − 4
⎢
⎥ ⎢ 3
⎢⎣0 0 −1 0 ⎥⎦ ⎢
⎢ 1
⎢⎣ 3
⎡
⎢ 1
⎡1 0 0 0⎤ ⎢
⎢0 1 0 0⎥ = ⎢ − 2
⎢
⎥ ⎢ 3
⎢⎣0 0 1 0 ⎥⎦ ⎢
⎢− 1
⎢⎣ 3
0
1
3
0
⎤
⎡1
0⎥
⎥
⎢0
0⎥ A ⎢
⎥
⎢0
⎢
1 ⎥⎥
⎣0
2⎦
1
1
0
0
⎡
⎤
⎢1 1
0⎥
⎢
⎥
⎢
⎥
0 A ⎢0 1
⎥
⎢
1 ⎥⎥
⎢0 0
⎢⎣0 0
2⎦
⎡
⎤
⎢1 1
0⎥
⎢
⎥
⎢
⎥
0 A ⎢0 1
⎥
⎢
1 ⎥⎥
⎢0 0
⎢⎣0 0
2⎦
1 −2⎤
0 0 ⎥⎥
1 0⎥
⎥
0 1⎦
1⎤
1 − ⎥
2
⎥
3 ⎥
0
2 ⎥
⎥
1 0 ⎥
0 1 ⎥⎦
1⎤
0 − ⎥
2
⎥
3 ⎥
−1
2 ⎥
⎥
1
0 ⎥
0
1 ⎥⎦
1
6
1
3
1⎤
⎡
⎤
⎢1 1 0 − 2 ⎥
0⎥
⎥
⎢
⎥
3 ⎥
⎢
0 ⎥ A ⎢ 0 1 −1
⎥
2 ⎥
⎥
⎢
⎥
0 ⎥
1⎥
⎢0 0 1
⎢⎣0 0 0
1 ⎥⎦
2 ⎥⎦
1⎤
⎡
⎤
⎢1 1 0 − 2 ⎥
0 ⎥
⎥
⎢
⎥
⎢ 0 1 −1 3 ⎥
⎥
0 A ⎢
⎥
2 ⎥
⎥
⎢
⎥
0 0 1
0 ⎥
1⎥
⎢
−
⎢⎣0 0 0
1 ⎥⎦
2 ⎥⎦
0
1
6
1
3
1⎤
⎡
⎤
⎢1 1 0 − 2 ⎥
0 ⎥
⎢
⎥
⎥
⎢ 0 1 −1 3 ⎥
⎥
0 , Q=⎢
⎥
2 ⎥
⎢
⎥
⎥
0 0 1
0 ⎥
1⎥
⎢
−
⎢⎣0 0 0
1 ⎥⎦
2 ⎥⎦
0
1
3
1
−
3
0
[I3 : 0] = PAQ,
⎡
⎢ 1
⎢
2
P = ⎢−
⎢ 3
⎢ 1
⎢−
⎢⎣ 3
and the rank of A = 3
Remark: To find the rank of a matrix, the simplest method is to reduce to row echelon form.
M01_ENGINEERING_MATHEMATICS-I _CH01_Part A.indd 22
5/30/2016 4:35:19 PM
Matrices ■
1.23
EXERCISE 1.2
Find the rank of the following matrices reducing to echelon form.
⎡ 2 3 −1
⎢1 −1 −2
1. ⎢
⎢3 1 3
⎢
⎣6 3 0
−1⎤
−4 ⎥⎥
−2⎥
⎥
−7⎦
3 0 −2⎤
⎡4
4. ⎢ 3 4 −1 3 ⎥
⎢
⎥
⎢⎣ −7 −7 1 5 ⎥⎦
⎡0
⎢1
2. ⎢
⎢3
⎢
⎣1
1 −3 −1⎤
0 1 1 ⎥⎥
1 0 2⎥
⎥
1 −2 0 ⎦
⎡ 1 2 −1
⎢4 1 2
5. ⎢
⎢ 3 −1 1
⎢
⎣1 2 0
⎡1
⎢2
3. ⎢
⎢3
⎢
⎣6
2
4
2
8
3
3
1
7
0⎤
2⎥⎥
3⎥
⎥
5⎦
⎡1 1 2 1⎤
⎢ −1 −1 −2 1 ⎥
⎢
⎥
6. ⎢ 1 2 1 −1⎥
⎢
⎥
⎢ 1 3 0 −3⎥
⎢⎣ 1 1 2 3 ⎥⎦
3⎤
1 ⎥⎥
2⎥
⎥
1⎦
⎡ 1 2 −2 3 ⎤
⎡3 1 −5 −1⎤
⎢ 2 5 −4 6 ⎥
⎥
8. ⎢
9. ⎢1 −2 1 −5⎥
⎢
⎥
⎢ −1 −3 2 −2⎥
⎢⎣1 5 −7 2 ⎥⎦
⎢
⎥
⎣ 2 4 −4 6 ⎦
⎡1 3 4 3⎤
10. Find the rank of the matrix A = ⎢⎢3 9 12 3⎥⎥ , by reducing to an echelon matrix.
⎢⎣1 3 4 1⎥⎦
⎡1
⎢1
7. ⎢
⎢2
⎢
⎣3
1 1
3 −2
0 −3
3 −3
1⎤
1 ⎥⎥
2⎥
⎥
3⎦
⎡ −2 −1 −1⎤
11. Find the rank of the matrix A = ⎢⎢12 8 6 ⎥⎥ .
⎢⎣10 5 6 ⎥⎦
⎡ 2 1 −3 −6 ⎤
12. Reduce the matrix ⎢ 3 −3 1
2 ⎥⎥ to normal form and hence find the rank.
⎢
⎢⎣1 1 1
2 ⎥⎦
⎡ 4 4 −3
⎢ 1 1 −1
13. Find the values of k if the rank of ⎢
⎢k 2 2
⎢
⎣9 9 k
1⎤
0 ⎥⎥
is 3.
2⎥
⎥
3⎦
⎡ 2 1 −1 3⎤
14. Find the values of a and b if the matrix ⎢1 −1 2 4 ⎥ is of rank 2.
⎢
⎥
⎢⎣ 7 −1 a b ⎥⎦
⎡1 −2 3 1 ⎤
15. Find the values of a and b if the matrix ⎢ 2 1 −1 2⎥
⎢
⎥
⎢⎣6 −2 a b ⎥⎦
⎡ 1 −1 2
⎢4 1 0
16. Reduce to normal form and find the rank of ⎢
⎢0 3 1
⎢
⎣0 1 0
M01_ENGINEERING_MATHEMATICS-I _CH01_Part A.indd 23
is of rank 2.
−3⎤
2 ⎥⎥
.
4⎥
⎥
2⎦
5/30/2016 4:35:23 PM
1.24
■
Engineering Mathematics
⎡0 1 2 1 ⎤
17. Reduce to normal form and find the rank of ⎢1 2 3 2⎥ .
⎢
⎥
⎢⎣ 3 1 1 3⎥⎦
⎡1 1 2 ⎤
18. If A = ⎢⎢1 2 3 ⎥⎥ , then find non-singular matrices P and Q such that PAQ is in normal form and find its
⎣⎢0 −1 −1⎦⎥
rank.
ANSWERS TO EXERCISE 1.2
1.
6.
11.
16.
3
2. 2
3. 3
3
7. 3
8. 3
3
12. 3
13. k = 2
[I4, 0], rank = 4
17. [I3, 0], rank = 3
⎡ 1 0 0⎤
⎡1 −1 −1⎤
⎢
⎥
18. P = ⎢ −1 1 0 ⎥ , Q = ⎢⎢0 1 −1⎥⎥ and rank = 2.
⎢⎣ −1 1 1 ⎥⎦
⎢⎣0 0 1 ⎥⎦
1.4
4. 2
9. 3
14. a = 4, b = 18
5. 3
10. 2
15. a = 4, b = 6
SOLUTION OF SYSTEM OF LINEAR EQUATIONS
There are many problems in science and engineering whose solution often depends upon a system of
linear equations.
The equation a1x1 + a2x2 + … + anxn = b is called a non-homogeneous linear equation in n variables
x1, x2, …, xn where b ≠ 0 and at least one ai ≠ 0.
If b = 0, then the equation a1x1 + a2x2 + … + anxn = 0 is called a homogenous linear equation in
x1, x2, …, xn.
1.4.1 Non-homogeneous System of Equations
Consider the system of m linear equations in n variables x1, x2, …, xn
a11x1 + a12x2 + … + a1nxn = b1
a21x1 + a22x2 + … + a2nxn = b2
:
am1x1 + am2x2 + … + amnxn = bm, where at least one bi ≠ 0
⎡ b1 ⎤
⎡ x1 ⎤
⎡ a11 a12 … a1n ⎤
⎢a
⎥
⎢
⎥
⎢x ⎥
a22 … a2 n ⎥
b
If A = ⎢ 21
, B = ⎢ 2 ⎥, X = ⎢ 2⎥,
⎢ :
⎢:⎥
⎢:⎥
:
: ⎥
⎢
⎥
⎢ ⎥
⎢ ⎥
…
a
a
a
b
⎣ m1 m 2
⎣ m⎦
⎣x n ⎦
mn ⎦
then the system of equations can be written as a single matrix equation AX = B.
The matrix A is called the coefficient matrix.
A solution of the system is a set of values of x1, x2, …, xn which satisfy the m equations.
The system of equations is said to be consistent if it has at least one solution.
If the system has no solution, then the system of equations is said to be inconsistent.
The condition for the consistency of the system is given by Rouche’s theorem.
We shall state the theorem without proof.
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Theorem 1.2 Rouche’s Theorem The system of linear equations AX = B is consistent if and only if the
coefficient matrix A and the augmented matrix [A, B] have the same rank. That is., r(A) = r([A, B])
Working rule:
Let AX = B represent a system of m equations in n variables.
1.
2.
3.
4.
Write down the coefficient matrix A and the augmented matrix [A, B]. Find r(A), r([A, B])
If r(A) ≠ r([A, B]), then the system is inconsistent. That is it has no solution.
If r(A) = r([A, B]) = n, the number of variables, then the system is consistent with unique solution.
If r(A) = r([A, B]) < n, the number of variables, then the system is consistent with infinite number
of solutions.
If the rank is r, then in this case the solution set will contain n − r parameters or arbitrary constants.
To get the solutions we assign arbitrary values to n − r variables and write down the solutions in
terms of them.
For example, the system x + y + z = 1, 2x − y + 3z = −1, 2x + 5y + z = 5 is consistent with infinite number of
solutions.
Here r = 2, n = 3.
∴ the solution set will contain n − r = 3 − 2 = 1 parameter.
We assign an arbitrary value to one variable, say y.
Put y = k and solve for x and z in terms of k.
The solution set is
x = 4 + 2k, y = k, z = −3 − 3k,
where k is any real number.
Note
If m = n, then A is a square matrix and the system of equations AX = B has unique solution if A is
non-singular. That is., A ≠ 0, then r(A) = number of variables n.
The unique solution is X = A−1B.
1.4.2 Homogeneous System of Equations
Consider the homogeneous system
a11x1 + a12x2 + … + a1nxn = 0
a21x1 + a22x2 + … + a2nxn = 0
:
am1x1 + am2x2 + … + amnxn = 0
⎡ a11 a12
⎢a
a22
If A = ⎢ 21
⎢ :
:
⎢
⎣am 1 am 2
… a1n ⎤
⎡ x1 ⎤
⎢ ⎥
… a2 n ⎥
⎥ , X = ⎢ x 2 ⎥ , then the matrix equation is AX = 0.
⎢:⎥
: ⎥
⎢ ⎥
⎥
… amn ⎦
⎣x n ⎦
For this system x1 = 0, x2 = 0, …, xn = 0 is always a solution. This is called the trivial solution.
If A ≠ 0, the r(A) = n and the only solution is the trivial solution. So, the condition for
non-trivial solution is A = 0 (or r(A) < n).
In solving equations we use only row operations.
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1.4.3 Type 1: Solution of Non-homogeneous System of Equations
WORKED EXAMPLES
(A) Non-homogeneous system with unique solution
EXAMPLE 1
Test for consistency and solve 2x 2 y 1 z 5 7, 3x 1 y 2 5z 5 13, x 1 y 1 z 5 5.
Solution.
The given equations are
x+y+z=5
2x − y + z = 7
3x + y − 5z = 13
We have rearranged the equation for convenience in reducing to row echelon form.
The coefficient matrix is
and the augmented matrix is
⎡1 1 1 ⎤
A = ⎢⎢ 2 −1 1 ⎥⎥
⎢⎣ 3 1 −5⎥⎦
⎡1 1 1 : 5 ⎤
⎢
[A , B ] = ⎢ 2 −1 1 : 7 ⎥⎥ ∼
⎢⎣ 3 1 −5 : 13⎥⎦
⎡1 1 1 : 5 ⎤
⎢0 −3 −1 : −3⎥ R → R − 2R
2
2
1
⎢
⎥
⎢⎣0 −2 −8 : −2⎥⎦ R 3 → R 3 − 3R1
⎡1 1 1 : 5 ⎤
⎢
⎥
1
1
:1 ⎥ R2 → − R2
∼ ⎢0 1
3
3
⎢
⎥
⎢0 −1 −4 : −1⎥
⎣
⎦ R → 1R
3
3
2
⎤
⎡
⎢1 1
1
: 5⎥
⎢
⎥
1
∼ ⎢0 1
: 1⎥
⎢
⎥
3
⎢
⎥
11
⎢0 0 −
: 0⎥ R 3 → R 3 + R 2
⎢⎣
⎥⎦
3
From the last matrix, we find
M01_ENGINEERING_MATHEMATICS-I _CH01_Part A.indd 26
⎡
⎤
⎢1 1
1 ⎥
⎢
⎥
1 ⎥
A ∼ ⎢0 1
⎢
3 ⎥
⎢
⎥
⎢ 0 0 − 11⎥
⎢⎣
3 ⎥⎦
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Matrices ■
1.27
The number of non-zero rows in the equivalent matrices of A and [A, B] are 3.
∴
r(A) = 3, r([A, B]) = 3
⇒
r(A) = r([A, B]) = 3, the number of variables.
So, the equations are consistent with unique solution.
From the reduced matrix [A, B], we find the given equations are equivalent to
x + y + z = 5,
1
z =1
3
x+1+0=5
y+
∴
y = 1 and
So, the unique solution is x = 4, y = 1, z = 0.
and
⇒
11
z =0
3
x = 4.
−
⇒
z=0
EXAMPLE 2
Test for the consistency and solve x 1 2y 1 z 5 3, 2x 1 3y 1 2z 5 5, 3x 2 5y 1 5z 5 2,
3x 1 9y 2 z 5 4.
Solution.
The given equations are
The coefficient matrix is
x + 2y + z = 3
2x + 3y + 2z = 5
3x − 5y + 5z = 2
3x + 9y − z = 4.
⎡1 2 1 ⎤
⎢2 3 2 ⎥
⎥
A =⎢
⎢ 3 −5 5 ⎥
⎢
⎥
⎣ 3 9 −1⎦
The augmented matrix is
⎡1 2 1
⎢2 3 2
[ A, B] = ⎢
⎢ 3 −5 5
⎢
⎣ 3 9 −1
: 3⎤ ⎡1 2
: 5 ⎥⎥ ⎢⎢0 −1
∼
: 2⎥ ⎢0 −11
⎥ ⎢
: 4⎦ ⎣0 3
⎡1 2
⎢0 −1
∼⎢
⎢0 0
⎢
⎣0 0
⎡1 2
⎢0 −1
∼⎢
⎢0 0
⎢
⎣0 0
M01_ENGINEERING_MATHEMATICS-I _CH01_Part A.indd 27
1 :3 ⎤
0 : −1⎥⎥ R2 → R2 − 2 R1
2 : −7⎥ R3 → R3 − 3R1
⎥
−4 : −5⎦ R4 → R4 − 3R1
1 :3 ⎤
0 : −1⎥⎥
2 : 4 ⎥ R3 → R3 − 11R2
⎥
−4 : −8⎦ R4 → R4 + 3R2
1 :3 ⎤
0 : −1⎥⎥
1 :2 ⎥
⎥
1 :2 ⎦
1
R3
2
1
R4 → − R4
4
R3 →
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⎡1 2
⎢0 −1
[A , B ] ∼ ⎢
⎢0 0
⎢
⎣0 0
⇒
From this last matrix we find
⎡1 2
⎢0 −1
A ∼⎢
⎢0 0
⎢
⎣0 0
1 :3 ⎤
0 : −1⎥⎥
1 :2 ⎥
⎥
0 : 0 ⎦ R 4 → R4 − R3
1⎤
0 ⎥⎥
1⎥
⎥
0⎦
The number of non-zero rows in the equivalent matrices of A and [A B] are 3.
∴
⇒
r(A) = 3, r([A, B]) = 3
r(A) = r([A, B]) = 3, the number of variables.
So, the equations are consistent with unique solution
From the reduced matrix [A, B], we find the given equations are equivalent to
x + 2y + z = 3, −y = −1 and z = 2
∴
y = 1, z = 2 and so x + 2 ⋅ 1 + 2 = 3 ⇒ x = −1
So, the unique solution is x = −1, y = 1, z = 2.
EXAMPLE 3
Solve x2yz 5 e, xy2z3 5 e, x3y2z 5 e using matrices.
Solution.
The given equations are
x2yz = e
xy2z3 = e
x3y2z = e
(1)
(2)
(3)
Taking logarithm to the base e on both sides of (1), (2) and (3), we get
log e x 2 yz = log e e ⇒
log e x 2 + log e y + log e z = 1
⇒
2 log e x + log e y + log e z = 1
log e xy z = log e e ⇒ log e x + 2 log e y + 3 log e z = 1
2 3
and
log x 3 y 2 z = log e e ⇒ 3 log e x + 2 log e y + log e z = 1
For simplicity, put x1 = logex, y1 = logey, z1 = logez
∴ the equations are
The coefficient matrix is
M01_ENGINEERING_MATHEMATICS-I _CH01_Part A.indd 28
2x1 + y1 + z1 = 1
x1 + 2y1 + 3z1 = 1
3x1 + 2y1 + z1 = 1
⎡ 2 1 1⎤
A = ⎢⎢1 2 3⎥⎥
⎢⎣ 3 2 1⎥⎦
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1.29
The augmented matrix is
⎡ 2 1 1 : 1⎤ ⎡1
[A , B ] = ⎢⎢1 2 3 : 1⎥⎥ ∼ ⎢⎢ 2
⎢⎣ 3 2 1 : 1⎥⎦ ⎢⎣ 3
⎡1
∼ ⎢⎢0
⎢⎣0
2 3 : 1⎤ R1 ↔ R 2
1 1 : 1⎥⎥
2 1 : 1⎥⎦
2
3
:1 ⎤
−3 −5 : −1⎥⎥ R 2 → R 2 − 2R1
−4 −8 : −2⎥⎦ R 3 → R 3 − 3R1
⎤
⎡
⎢1 2
3
:1 ⎥
⎥
⎢
[A , B ] ∼ ⎢0 −3 −5 : −1 ⎥
⎢
4
4
2⎥
⎢0 0 −
: − ⎥ R3 → R3 − R 2
3
3
3⎦
⎣
⇒
3 ⎤
⎡1 2
⎢0 −3 −5 ⎥
⎥
A ∼⎢
From the last matrix, we find
⎢
4⎥
⎢0 0 − ⎥
3⎦
⎣
The number of non-zero rows in the equivalent matrices of A and [A, B] are 3.
∴
r(A) = 3, r([A, B]) = 3
⇒
r(A) = r([A, B]) = 3, the number of variables.
So, the equations are consistent with unique solution.
From the reduced matrix [A, B], we find that the given equations are equivalent to
x1 + 2y1 + 3z1 = 1
(4)
(5)
−3y1 − 5z1 = −1
4
2
− z1 = −
⇒
3
3
and
z1 =
1
2
Substituting in (5), we get
−3 y1 − 5 ⋅
1
= −1 ⇒
2
Substituting in (4) we get
⇒
1
⎛ 1⎞
x1 + 2 ⎜ − ⎟ + 3 ⋅ = 1
⎝ 2⎠
2
∴
3 y1 = 1 −
⇒
x1 = 1 −
1
2
⇒
x = e2
⇒
y=e
x1 +
1
2
⇒
log e x =
y1 = −
1
⇒
2
z1 =
1
2
M01_ENGINEERING_MATHEMATICS-I _CH01_Part A.indd 29
⇒
y1 = −
1
=1
2
⇒
x1 =
5
3
=− ⇒
2
2
log e y = −
log e z =
1
2
1
2
⇒
1 1
=
2 2
1
−
z=e
1
2
= e
1
2
1
2
=
1
e
= e
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So, the unique solution is
x = e,
1
y=
e
, z= e.
(B) Non-homogeneous system with infinite number of solutions
EXAMPLE 4
By investigating the rank of relevant matrices, show that the following equations possess a one
parameter family of solutions: 2x 2 y 2 z 5 2, x 12y 1 z 5 2, 4x 2 7y 2 5z 5 2.
Solution.
The given equations are
The coefficient matrix is
x +2y + z = 2
2x − y − z = 2
4x − 7y − 5z = 2
⎡1 2 1 ⎤
A = ⎢⎢ 2 −1 −1⎥⎥
⎢⎣ 4 −7 −5⎥⎦
The augmented matrix is
⎡ 1 2 1 : 2⎤ ⎡ 1 2
[A , B ] = ⎢⎢ 2 −1 −1 : 2⎥⎥ ∼ ⎢⎢0 −5
⎢⎣ 4 −7 −5 : 2⎥⎦ ⎢⎣0 −15
⎡1 2
⎢
∼ ⎢0 −5
⎢⎣0 0
1 :2 ⎤
−3 : −2⎥⎥ R 2 → R 2 − 2R1
−9 : −6 ⎥⎦ R 3 → R 3 − 4 R1
1 :2 ⎤
−3 : −2⎥⎥
0 : 0 ⎥⎦ R 3 → R 3 − 3R 2
⎡1 2 1 ⎤
From the last matrix we find
A ∼ ⎢⎢0 −5 −3⎥⎥
⎢⎣0 0 0 ⎥⎦
The number of non-zero rows of equivalent matrices of A and [A, B] are 2
∴
⇒
r(A) = 2,
r([A, B]) = 2
r(A) = r([A, B]) = 2 < the number of variables 3.
So, the equations are consistent with infinite number of solutions involving one parameter,
since n − r = 3 − 2 = 1.
From the reduced matrix [A, B], we find that the given equations are equivalent to
x + 2y + z = 2
− 5y − 3z = − 2
(1)
⇒ 5y + 3z = 2
(2)
Assign arbitrary value to one of the variables.
Put z = k in (2) ∴
M01_ENGINEERING_MATHEMATICS-I _CH01_Part A.indd 30
5y + 3k = 2
⇒
y =
2 − 3k
5
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1.31
Substituting in (1), we get,
x+
2( 2 − 3k )
+k = 2
5
⇒
x = 2−
2( 2 − 3k )
10 − 4 + 6k − 5k k + 6
−k =
=
5
5
5
1
1
∴ the solution set is x = ( k + 6), y = ( 2 − 3k ), z = k , where k is any real number.
5
5
EXAMPLE 5
Solve, if the equations are consistent: x 2 y 1 2z 5 1, 3x 1 y 1 z 5 4, x 1 3y 23z 5 2,
5x 2 y 1 5z 5 6.
Solution.
The given equations are
The coefficient matrix is
x − y + 2z = 1
3x + y + z = 4
x + 3y − 3z = 2
5x − y + 5z = 6
⎡1 −1 2 ⎤
⎢3 1 1 ⎥
⎥
A =⎢
⎢1 3 −3⎥
⎢
⎥
⎣ 5 −1 5 ⎦
The augmented matrix is
⎡1 −1 2
⎢3 1 1
[A , B ] = ⎢
⎢1 3 −3
⎢
⎣5 −1 5
From the last matrix, we find
: 1 ⎤ ⎡1 −1 2
: 4 ⎥⎥ ⎢⎢0 4 −5
∼
: 2⎥ ⎢0 4 −5
⎥ ⎢
: 6 ⎦ ⎣ 0 4 −5
: 1⎤
: 1⎥⎥ R 2 → R 2 − 3R1
: 1⎥ R 3 → R 3 − R1
⎥
: 1⎦ R 4 → R 4 − 5R1
⎡ 1 −1 2
⎢ 0 4 −5
∼⎢
⎢0 0 0
⎢
⎣0 0 0
: 1⎤
: 1 ⎥⎥
: 0⎥ R 3 → R 3 − R 2
⎥
: 0⎦ R 4 → R 4 − R 2
⎡1 −1 2 ⎤
⎢0 4 −5⎥
⎥
A ∼⎢
⎢0 0 0 ⎥
⎢
⎥
⎣0 0 0 ⎦
The number of non-zero rows of the equivalent matrices of A and [A, B] are 2.
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∴
r(A) = 2,
r([A, B]) = 2
⇒
r(A) = r([A, B]) = 2 < 3, the number of variables.
So, the equations are consistent with infinite number of solutions involving one parameter,
since n − r = 3 − 2 = 1.
From reduced matrix [A, B], we find that the given equations are equivalent to
x − y + 2z = 1
4y − 5z = 1
Put z = k, in (2) then
(1)
(2)
4y − 5k = 1 ⇒
y =
1 + 5k
4
Substituting in (1), we get
1 + 5k
1 + 5k
4 + 1 + 5k − 8k 5 − 3k
x−
+ 2k = 1
⇒
x = 1+
− 2k =
=
4
4
4
4
∴ the solution set is
5 − 3k
1 + 5k
x=
,y =
, z = k , where k is any real number.
4
4
EXAMPLE 6
Test the consistency of the system of equations and solve, if consistent: x1 1 2x2 2 x3 2 5x4 5 4,
x1 1 3x2 2 2x3 2 7x4 5 5, 2x1 2 x2 1 3x3 5 3.
Solution.
The given equations are
x1 + 2x2 − x3 − 5x4 = 4
x1 + 3x2 − 2x3 − 7x4 = 5
2x1 − x2 + 3x3 + 0x4= 3
The coefficient matrix is
⎡1 2 −1 −5⎤
A = ⎢⎢1 3 −2 −7⎥⎥
⎢⎣ 2 −1 3 0 ⎥⎦
The augmented matrix is
⎡1 2 −1 −5 : 4 ⎤ ⎡1
[A , B ] = ⎢⎢1 3 −2 −7 : 5 ⎥⎥ ∼ ⎢⎢0
⎢⎣ 2 −1 3 0 : 3⎥⎦ ⎢⎣0
⎡1
∼ ⎢⎢0
⎢⎣0
From the last matrix, we find
⎡1
A ∼ ⎢⎢0
⎢⎣0
M01_ENGINEERING_MATHEMATICS-I _CH01_Part A.indd 32
2 −1 −5 : 4 ⎤
1 −1 −2 : 1 ⎥⎥ R 2 → R 2 − R1
−5 5 10 : −5⎥⎦ R 3 → R 3 − 2R1
2 −1 −5 : 4 ⎤
1 −1 −2 : 1 ⎥⎥
0 0 0 : 0 ⎥⎦ R 3 → R 3 + 5R 2
2 −1 −5⎤
1 −1 −2⎥⎥
0 0 0 ⎥⎦
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Matrices ■
1.33
The number of non-zero rows of the equivalent matrices of A and [A, B] are 2.
∴
r(A) = 2,
r([A, B]) = 2
⇒
r(A) = r([A, B]) = 2 < 4, the number of variables.
So, the equations are consistent with infinite number of solutions containing two parameters,
since n − r = 4 − 2 = 2.
From the reduced matrix of [A, B] we find that the given equations are equivalent to
x1 + 2x2 − x3 − 5x4 = 4
(1)
and
x2 − x3 − 2x4 = 1
(2)
Put x3 = k1, x4 = k2, then (2) ⇒
Substituting in (1), we get
x2 − k1 − 2k2 = 1
⇒
x2 = 1 + k1 + 2k2
⇒
x1 = 2 − k1 + k2
x1 + 2(1 + k1 + 2k2) − k1 − 5k2 = 4
⇒
x1 + 2 + 2k1 + 4k2 − k1 − 5k2 = 4
⇒
x1 + k1 − k2 = 2
∴ the solution set is
x1 = 2 − k1 + k2,
x2 = 1 + k1 + 2k2,
x3 = k1,
x4 = k2, where k1, k2 are any real numbers.
(C) Non-homogeneous system with no solution
EXAMPLE 7
Examine for the consistency of the following equations 2x 1 6y 1 11 5 0, 6x 1 20y 2 6z 1 3 5 0,
6y 2 18z 1 1 5 0.
Solution.
The given equations are
2x + 6y + 0z = −11
6x + 20y − 6z = −3
0x + 6y − 18z = −1
The coefficient matrix is
0 ⎤
⎡2 6
⎢
A = ⎢6 20 −6 ⎥⎥
⎢⎣0 6 −18⎥⎦
The augmented matrix is
11⎤
1
⎡
1 3
0 : − ⎥ R1 → R1
0 : −11⎤ ⎢
⎡2 6
2
2
⎢
⎥
[A , B ] = ⎢⎢6 20 −6 : −3 ⎥⎥ ∼ ⎢6 20 −6 : −3 ⎥
⎢⎣0 6 −18 : −1 ⎥⎦ ⎢⎣0 6 −18 : −1 ⎥⎦
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11⎤
⎡
⎢1 3 0 : − 2 ⎥
⎢
⎥
∼ ⎢0 2 −6 : 30 ⎥ R 2 → R 2 − 6 R1
⎢ 0 6 −18 : −1 ⎥
⎣
⎦
11⎤
⎡
⎢1 3 0 : − 2 ⎥
⎢
⎥
[A , B ] ∼ ⎢ 0 2 −6 : 30 ⎥
⎢ 0 0 0 : −91 ⎥ R → R − 3R
⎣
⎦
3
3
2
⇒
⎡1 3 0 ⎤
From the last matrix, we find
A ∼ ⎢⎢0 2 −6 ⎥⎥
⎢⎣0 0 0 ⎥⎦
The number of non-zero rows in the equivalent matrices of A and [A, B] are 2 and 3 respectively.
∴
r(A) = 2, r([A, B]) = 3
⇒
r(A) ≠ r([A, B]).
Hence, the equations are inconsistent and the system has no solution.
1.4.4 Type 2: Solution of Non-homogeneous Linear Equations Involving Arbitrary
Constants
WORKED EXAMPLES
EXAMPLE 1
Show that the system of equations 3x 2 y 1 4z 5 3, x 1 2y 2 3z 5 22, 6x 1 5y 1 lz 5 23 has
at least one solution for any real number l. Find the set of solutions when l 5 25.
Solution.
The given equations are
x + 2y − 3z = −2,
3x − y + 4z = 3
6x + 5y + lz = −3
⎡1 2 −3⎤
The coefficient matrix is A = ⎢⎢ 3 −1 4 ⎥⎥
⎢⎣6 5 l ⎥⎦
and the augmented matrix is
: −2⎤
−3
⎡1 2 −3 : −2⎤ ⎡1 2
[A , B ] = ⎢⎢ 3 −1 4 : 3 ⎥⎥ ∼ ⎢⎢0 −7
: 9 ⎥⎥ R 2 → R 2 − 3R1
13
⎢⎣6 5 l : −3⎥⎦ ⎢⎣0 −7 l + 18 : 9 ⎥⎦ R 3 → R 3 − 6 R1
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−3 : −2⎤
⎡1 2
[A , B ] ∼ ⎢⎢0 −7 13
: 9 ⎥⎥
⎢⎣0 0 l + 5 : 0 ⎥⎦ R 3 → R 3 − R 2
⇒
−3 ⎤
⎡1 2
A ∼ ⎢⎢0 −7 13 ⎥⎥
⎢⎣0 0 l + 5⎥⎦
Case (i): If l + 5 ≠ 0 ⇒ l ≠ −5, the number of non-zero rows in the equivalent matrices of [A, B]
and A are 3.
From the last matrix we find
∴
⇒
r(A) = 3, r([A, B]) = 3
r(A) = r([A, B]) = 3, the number of variables.
So, the equations are consistent with unique solution.
Case (ii): If l + 5 = 0 ⇒ l = −5, then we get
⎡1 2 −3 : −2⎤
[A , B ] ∼ ⎢⎢0 −7 13 : 9 ⎥⎥
⎢⎣0 0 0 : 0 ⎥⎦
and
⎡1 2 −3⎤
⎢
A ∼ ⎢0 −7 13 ⎥⎥
⎢⎣0 0 0 ⎥⎦
The number of non-zero rows of equivalent matrices of [A, B] and A are 2.
∴
r(A) = 2, r([A, B]) = 2
⇒
r(A) = r([A, B]) = 2 < 3, the number of variables.
So, the equations are consistent with infinite number of solutions involving one parameter
since n − r = 3 − 2 = 1.
From cases (i) and (ii), we find that the equations are consistent for all values of l.
We shall now find the solution when l = −5. The solutions will contain one parameter.
In this case from the last matrix [A, B], we find the equations are equivalent to
Put z 5 k in (2), then
Substituting in (1) we get
x + 2y − 3z = −2
(1)
− 7y + 13z = 9
(2)
−7y + 13k = 9
⇒ 7 y = 13k − 9 ⇒ y =
1
(13k − 9)
7
1
2
x + 2 ⋅ (13k − 9) − 3k = −2 ⇒ x = −2 − (13k − 9) + 3k
7
7
1
1
= [ −14 − 26 k + 18 + 21k ] = [4 − 5k ]
7
7
∴ the solutions are
1
1
x = [4 − 5k ], y = [13k − 9], z = k ,
7
7
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EXAMPLE 2
Find the values of a and b if the equations x 1 y 1 2z 5 2, 2x 2 y 1 3z 5 2, 5x 2 y 1 az 5 b have
(i) no solutions, (ii) unique solution and (iii) infinite number of solutions.
Solution.
The given equations are
x + y + 2z = 2
2x − y +3z = 2
5x − y + az = b
⎡1 1 2⎤
The coefficient matrix is A = ⎢⎢ 2 −1 3⎥⎥
⎢⎣ 5 −1 a ⎥⎦
The augmented matrix is
2
⎡1 1 2 : 2⎤ ⎡1 1
⎢
⎢
⎥
−1
[A , B ] = ⎢ 2 −1 3 : 2⎥ ∼ ⎢0 −3
⎢
⎢⎣ 5 −1 a : b ⎥⎦ ⎣0 −6 a − 10
2
⎡1 1
⎢
∼ ⎢0 −3 −1
⎢⎣0 0 a − 8
From this matrix, we find
:
2⎤
: − 2 ⎥⎥ R 2 → R 2 − 2R1
: b − 10 ⎥⎦ R 3 → R 3 − 5R1
:
2⎤
: − 2 ⎥⎥
: b − 6 ⎥⎦ R 3 → R 3 − 2R 2
2 ⎤
⎡1 1
A ∼ ⎢⎢0 −3 −1 ⎥⎥
⎢⎣0 0 a − 8⎥⎦
Case (i): The equations have no solution
⇒
r(A) ≠ r([A, B])
This is possible, if r(A) = 2 and r([A, B]) = 3
⇒
a − 8 = 0 and b − 6 ≠ 0
⇒
a = 8 and b ≠ 6.
Case (ii): The equations have unique solution ⇒ r(A) = r([A, B]) = 3.
∴
a − 8 ≠ 0 and b is any real number.
∴
a ≠ 8 and b is any real number.
Case (iii): The equations have infinite number of solutions
⇒
r(A) = r([A, B]) = 2 < 3, the number of variables.
This is possible, if a − 8 = 0 and b − 6 = 0 ⇒ a = 8, b = 6
Thus, no solution
⇒ a = 8, b ≠ 6
Unique solution
⇒ a ≠ 8, b is any real number
Infinite number of solutions ⇒ a = 8, b = 6.
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EXAMPLE 3
For what values of k, the equations x 1 y 1 z 5 1, 2x 1 y 1 4z 5 k and 4x 1 y 1 10z 5 k2 have
(i) a unique solution, (ii) infinite number of solutions, (iii) no solution and solve them completely
in each case of consistency.
Solution.
The given equations are
x+y+z=1
2x + y + 4z = k
4x + y + 10z = k2
⎡1 1 1 ⎤
A = ⎢⎢ 2 1 4 ⎥⎥
⎢⎣ 4 1 10 ⎥⎦
The coefficient matrix is
The augmented matrix is
:1 ⎤
⎡1 1 1 : 1 ⎤ ⎡1 1 1
⎢
⎢
⎥
[A , B ] = ⎢ 2 1 4 : k ⎥ ∼ ⎢ 0 −1 2 : k − 2 ⎥⎥ R 2 → R 2 − 2R1
⎢⎣ 4 1 10 : k 2 ⎥⎦ ⎢⎣0 −3 6 : k 2 − 4 ⎥⎦ R 3 → R 3 − 4 R1
:1
⎡1 1 1
⎤
⎢
⎥
∼ ⎢0 −1 2
:k −2
⎥
⎢⎣0 0 0 : k 2 − 4 − 3( k − 2) ⎥⎦ R 3 → R 3 − 3R 2
:1
⎤
⎡1 1 1
⎢
[ A , B ] ∼ ⎢ 0 −1 2
: k − 2 ⎥⎥
⎢⎣0 0 0 : k 2 − 3k + 2⎥⎦
⇒
From the last matrix, we find
⎡1 1 1⎤
A ∼ ⎢⎢0 −1 2⎥⎥
⎢⎣0 0 0 ⎥⎦
Since the number of non-zero rows is 2, r(A) = 2
(i) If
k2 − 3k + 2 = 0 ⇒
(k − 2)(k − 1) = 0 ⇒ k = 1, k = 2
:1 ⎤
⎡1 1 1
⎢
[A , B ] ∼ ⎢0 −1 2 : k − 2⎥⎥
⎢⎣0 0 0
: 0 ⎥⎦
∴
∴
r[ A , B ] = 2
So, r(A) = r([A,B]) = 2 < 3, the number of variables.
∴
the system of equations is consistent with infinite number of solutions if k = 1 or k = 2.
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(ii) If k2 − 3k + 2 ≠ 0
⇒
k ≠ 1 and k ≠ 2, then r([A, B]) = 3
∴
r(A) ≠ r([A, B])
So, the system is inconsistent and has no solution if k ≠ 1 and k ≠ 2.
Now we shall find the solutions if k = 1 and k = 2.
⎡1 1 1 : 1 ⎤
If k = 1, then [A , B ] = ⎢⎢0 −1 2 : −1⎥⎥
⎢⎣0 0 0 : 0 ⎥⎦
So, the equivalent equations are
Put z = k1, then
∴
x + y + z = 1 and −y + 2z = −1
−y + 2k1 = −1 ⇒
y = 1 + 2k1
x + 1 + 2k1 + k1 = 1
⇒
x = −3k1
∴ the solutions are x = 3k1,
y = 1 + 2k1, z = k1, where k1 is any real number
⎡1 1 1 : 1 ⎤
If k = 2, then [A , B ] = ⎢⎢0 −1 2 : 0 ⎥⎥
⎢⎣0 0 0 : 0 ⎥⎦
So, the equivalent equations are x + y + z = 1 and −y + 2z = 0 ⇒ y = 2z.
Put z = k2, then y = 2k2
∴
x + 2k2 + k2 = 1
∴ the solutions are x = 1 − 3k2,
y = 2k2,
⇒
x = 1 − 3k2
z = k2, where k2 is any real number.
1.4.5 Type 3: Solution of the System of Homogeneous Equations
WORKED EXAMPLES
EXAMPLE 1
Find all the non-trivial solutions of 7x 1 y 2 2z 5 0, x 1 5y 2 4z 5 0, 3x 2 2y 1 z 5 0.
Solution.
The given equations are
7x + y − 2z = 0
x + 5y − 4z = 0
3x − 2y + z = 0
⎡7 1 −2 ⎤
The coefficient matrix is A = ⎢⎢1 5 −4 ⎥⎥
⎢⎣ 3 −2 1 ⎥⎦
Since R. H. S of the equations is zero it is enough, we consider A instead of augmented matrix
⎡7 1 −2 : 0 ⎤
[A , B ] = ⎢⎢1 5 −4 : 0 ⎥⎥ , because r(A) = r([A, B]) always.
⎣⎢ 3 −2 1 : 0 ⎦⎥
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⎡1 5
A ∼ ⎢⎢7 1
⎢⎣ 3 −2
⎡1 5
∼ ⎢⎢0 −34
⎣⎢0 −17
1.39
−4 ⎤
−2 ⎥⎥ R1 ↔ R 2
1 ⎥⎦
−4 ⎤
26 ⎥⎥ R 2 → R 2 − 7R1
13 ⎥⎦ R 3 → R 3 − 3R1
−4 ⎤
⎡1 5
1
⎢
∼ ⎢0 17 −13⎥⎥ R 2 → − R 2
2
⎢⎣0 −17 13 ⎥⎦
⎡1 5 −4 ⎤
∼ ⎢⎢0 17 −13⎥⎥
⎢⎣0 0
0 ⎥⎦ R 3 → R 3 + R 2
The number of non-zero rows is 2.
∴ r(A) = 2 < the number of variables.
∴ the number of solutions is infinite containing n − r = 3 − 2 = 1 parameter.
From the last equivalent matrix, the given equations are equivalent to
x + 5y − 4z = 0
y
z
and
17 y − 13z = 0 ⇒
=
=k
13 17
∴
y = 13k, z = 17k.
and (1) ⇒
x + 5 × 13k − 4 × 17k = 0 ⇒ x = 3k
∴ the solutions are x = 3k, y = 13k, z = 17k, where k is any real number.
(1)
Aliter: Since the number of equations is same as the number of variables, the coefficient matrix A is a
square matrix.
∴
7 1 −2
A = 1 5 −4 = 7 ⋅ (5 − 8) − 1⋅ (1 + 12) − 2( −2 − 15) = −21 − 13 + 34 = 0.
3 −2 1
∴ r(A) < 3, the number of variables.
∴ the homogeneous system has non-trivial solutions.
The solutions can be obtained by the rule of cross multiplication, using first and second equations
x
y
z
=
=
x
y
z
−4 + 10 −2 + 28 35 − 1
−2
1
7
1
x
y
z
⇒
=
=
6 26 34
−4
5
1
5
x
y
z
⇒
=
=
= k ,say
3 13 17
∴ the solutions are
x = 3k , y = 13k , z = 17k , where k is any reaal number.
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EXAMPLE 2
Solve completely the homogeneous system 3x 1 4y 2z 2 6w 5 0, 2x 1 3y 1 2z 2 3w 5 0,
2x 1 y 2 14z 2 9w 5 0, x 1 3y 1 13z 1 3w 5 0.
Solution.
The given equations are
The coefficient matrix is
⎡3
⎢2
A =⎢
⎢2
⎢
⎣1
3x + 4y − z − 6w = 0
2x + 3y + 2z − 3w = 0
2x + y − 14z − 9w = 0
x + 3y + 13z + 3w = 0
4 −1 −6 ⎤ ⎡1
3 2
−3⎥⎥ ⎢⎢ 2
∼
1 −14 −9 ⎥ ⎢ 2
⎥ ⎢
3 13
3 ⎦ ⎣3
⎡1
⎢0
∼⎢
⎢0
⎢
⎣0
⎡1
⎢0
∼⎢
⎢0
⎢
⎣0
⎡1
⎢0
⎢
∼ ⎢0
⎢
⎣0
⇒
⎡1
⎢0
A ∼⎢
⎢0
⎢
⎣0
3
3
1
4
3
−3
−5
13
2
−14
−1
13
−24
−40
−5 −40
3 13
−3 −24
−5 −40
0
0
3 ⎤ R1 ↔ R 4
−3⎥⎥
−9 ⎥
⎥
−6 ⎦
3 ⎤
−9 ⎥⎥ R 2 → R 2 − 2R1
−15⎥ R 3 → R 3 − 2R1
⎥
−15⎦ R 4 → R 4 − 3R1
3 ⎤
−9 ⎥⎥
−15⎥
⎥
0 ⎦ R 4 → R 4 − R3
3 13 3⎤
1
R2 → − R2
1 8 3⎥⎥
3
1 8 3⎥
1
⎥ R3 → − R3
5
0 0 0⎦
3 13 3⎤
1 8 3⎥⎥
0 0 0⎥ R 3 → R 3 − R 2
⎥
0 0 0⎦
The last equivalent matrix has 2 non-zero rows.
∴ r(A) = 2 < 4, the number of variables.
∴ the equations have infinite number of solutions and will contain n − r = 4 − 2 = 2 parameters.
The given equations are equivalent to x + 3y + 13z + 3w = 0
and
y + 8z + 3w = 0
y = −8k1 − 3k2
Put z = k1 and w = k2, then
and
x + 3(−8k1 − 3k2) + 13k1 + 3k2 = 0
⇒
x − 24k1 − 9k2 + 13k1 + 3k2 = 0
⇒
x − 11k1 − 6k2 = 0 ⇒ x = 11k1 + 6k2
∴ the solution set is x = 11k1 + 6k2, y = −8k1 − 3k2,
z = k1,
w = k2,
where k1, k2 are any real numbers.
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EXAMPLE 3
Find all the non-trivial solutions of x 2 y 1 z 5 0, 2x 1 y 2 z 5 0, x 1 5y 2 5z 5 0.
Solution.
The given equations are
x−y+z=0
2x + y − z = 0
x + 5y − 5z = 0
⎡1 −1 1 ⎤
The coefficient matrix is A = ⎢⎢ 2 1 −1⎥⎥
⎢⎣1 5 −5⎥⎦
1 −1 1
A = 2 1 −1
1 5 −5
∴
= 1(−5 + 5) − (−1)(−10 + 1) + (10 − 1) = 0 − 9 + 9 = 0
∴ r(A) < 3, the number of variables.
∴ the homogeneous system has non-trivial solutions.
The solutions can be obtained by the rule of cross multiplication using first and second equations.
x
y
z
x
y
z
=
=
+1 − 1 2 + 1 1 + 2
1
−1
1
−1
x y z
−1
⇒
= =
1
2
1
0 3 3
x y z
⇒
= = =k
0 1 1
∴
x = 0, y = k, z = k
∴ the solution set is x = 0, y = k, z = k, where k is any real number.
1.4.6 Type 4: Solution of Homogeneous System of Equation Containing Arbitrary
Constants
WORKED EXAMPLES
EXAMPLE 1
If x 5
a
b
c
,y 5
,z 5
, then prove that 1+ xy + yz + zx = 0.
b 2c
c 2a
a 2b
Solution.
The given equations are
a
⎫
⇒ x(b − c) = a ⇒ a − bx + cx = 0⎪
b−c
⎪
b
⎪
y=
⇒ y(c − a) = b ⇒ ay + b − cy = 0⎬
c−a
⎪
c
⎪
z=
⇒ z ( a − b) = c ⇒ az − bz − c = 0⎪
a−b
⎭
x=
and
M01_ENGINEERING_MATHEMATICS-I _CH01_Part A.indd 41
(1)
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It is a system of homogeneous equations in a, b, c.
From the given equations, it is obvious that a, b, c cannot be simultaneously zero or no two of them
a
b
c
equal [For if all a, b, c are zero, then x =
, y=
, z=
are indeterminate, do not exist.]
b−c
c−a
a−b
∴ the system of equations (1) has non-trivial solutions
∴
A =0
1
⇒ y
+z
−x x
1 −y = 0
− z −1
⇒
1( −1 − yz ) − ( − x )( − y + yz ) + x( − yz − z ) = 0
⇒
− 1 − yz − xy + xyz − xyz − xz = 0
⇒
− 1 − xy − yz − zx = 0
⇒
1 + xy + yz + zx = 0
EXAMPLE 2
Find the values of l for which the equations
( l 2 1) x 1 (3l 1 1) y 1 2 lz 5 0
(l 2 1) x 1 (4 l 2 2) y 1 ( l 1 3) z 5 0
2 x 1 (3l 1 1) y 1 3( l21 )z 5 0
are consistent and find x : y : z, when l has the smallest of these values. What happens when l has
the greater of these values?
Solution.
The given equations are
(l − 1)x + (3l + 1) y + 2lz = 0
(l − 1)x + ( 4l − 2) y + (l + 3)z = 0
2x + (3l + 1) y + 3(l − 1)z = 0
Since the system is homogeneous it is always consistent with at least the trivial solution x = 0, y = 0
and z = 0. It will have non-trivial solution if A = 0
⇒
⇒
l − 1 3l + 1
2l
l − 1 4l − 2 l + 3 = 0
2
3l + 1 3(l − 1)
6l 3l + 1
2l
6l 4l − 2 l + 3 = 0
6l 3l + 1 3(l − 1)
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1 3l + 1
2l
6l 1 4l − 2 l + 3 = 0
1 3l + 1 3(l − 1)
⇒
1 3l + 1
2l
6l 0 l − 3 −l + 3 = 0 R2 → R2 − R1
0
0
l −3
R3 → R3 − R1
⇒
⇒
⇒
6l (l − 3) 2 = 0
l = 0 or (l − 3) 2 = 0 ⇒ l = 3
∴
l = 0 or 3.
∴ the least value is l = 0
When l = 0, the equations become
−x + y = 0 ⇒
x −y =0 ⇒ x = y
− x − 2 y + 3z = 0 ⇒ − y − 2 y + 3z = 0 ⇒ y = z
2x + y − 3z = 0 ⇒ 2 y + y − 3z = 0 ⇒ y = z
and
∴
x = y = z ∴ x : y : z = 1:1:1
When l = 3, then the equations become
2 x + 10 y + 6 z = 0
⇒
x + 5 y + 3z = 0
2 x + 10 y + 6 z = 0
⇒
x + 5 y + 3z = 0
2 x + 10 y + 6 z = 0
⇒
x + 5 y + 3z = 0
∴ the equations coincide with
x + 5y + 3z = 0
∴ the solution set is a two parameter family, since n = 3, r = 1 ∴ n − r = 2.
Put y = k1, z = k2 in (1), then
x + 5k1 + 3k2 = 0 ⇒ x = −5k1 − 3k2
So, the solutions are x = −5k1 − 3k2,
y = k1,
(1)
z = k2, where k1, k2 are any real numbers.
EXAMPLE 3
If the system of equations x 5 cy 1 bz, y 5 az 1 cx, z 5 bx 1 ay have non-trivial solutions, prove
that a2 1 b2 1 c2 1 2abc 5 1 and the solutions are
x : y : z 5 12 a 2 : 12 b 2 : 12 c 2 .
Solution.
The given equations are
x − cy − bz = 0
cx − y + az = 0
bx + ay − z = 0
It is a homogeneous system in x, y, z
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The coefficient matrix is
⎡1 −c −b ⎤
A = ⎢⎢ c −1 a ⎥⎥
⎢⎣b a −1⎥⎦
Given that this homogeneous system has non-trivial solution
∴
1 −c −b
A = 0 ⇒ c −1 a = 0
b a −1
⇒
1(1 − a 2 ) − ( −c)( −c − ab) + ( −b)( ac + b) = 0
⇒
1 − a 2 − c 2 − abc − abc − b 2 = 0
⇒
1 − ( a 2 + b 2 + c 2 − 2abc) = 0
⇒
a2 + b 2 + c 2 + 2abc = 1
(1)
The non-trivial solutions are obtained by the rule of cross multiplication from first and second
equations.
x
y
z
=
=
−ac − b −bc − a −1 + c 2
x
y
z
⇒
=
=
ac + b bc + a 1 − c 2
x
⇒
( ac + b)
2
x
⇒
a c + b + 2abc
2 2
2
x
⇒
a c +1− a − c
2 2
2
2
x
⇒
=
=
=
=
y
(bc + a)
2
z
=
(1 − c 2 ) 2
y
b c + a + 2abc
2 2
2
y
b c +1− b − c
2 2
2
2
y
(1 − a )(1 − c )
(1 − b )(1 − c )
x
y
z
⇒
=
=
2
2
1− a
1− b
1 − c2
2
⇒
∴
2
2
2
=
=
=
z
(1 − c 2 ) 2
z
(1 − c 2 ) 2
using (1)
z
(1 − c 2 ) 2
x : y : z = 1 − a2 : 1 − b2 : 1 − c2
EXERCISE 1.3
1. Test the consistency of the following system of equation and solve, if consistent.
(i) 2x − 3y +7z = 5,
(ii) 2x + 5y + 3z = 1,
(iii) 3x + y + 2z = 3,
3x + y − 3z = 13,
2x + 19y − 47z = 32
−x + 2y + z = 2 , x + y + z = 0
2x − 3y − z = −3,
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(iv) x + y + z = 3, x + 2y + 2z = 4, x + 4y + 9z = 6
(v) 2x − y + z = 7,
(vi) x + 2y − z = 3,
3x + y − 5z = 13,
3x − y + 2z = 1,
(vii) x1 − x2 + x3 + x4 = 2,
(viii) x1 + 2x2 − x3 = 3,
(ix) 4x − 2y + 6z = 8,
(x) 2x + y + 5z = 4,
x+y+z=5
2x − 2y + 3z = 2,
x1 + x2 − x3 + x4 = −4,
3x1 − x2 − 2x3 = 1,
x + y − 3z = −1,
x − y + z = −1
x1 + x2 + x3 − x4 = 4,
2x1 − 2x2 + 3x3 = 2,
x1 + x2 + x3 + x4 = 0
x1 − x2 + x3 = −1
15x − 3y + 9z = 21
3x − 2y + 2z = 2, 5x− 8y − 4z = 1
2. Find all the values of a and b for which the equations.
x + y + z = 3, x + 2y + 2z = 6, x + ay + 3z = b have (1) no solution, (ii) unique solution and (iii) infinite number
of solutions.
3. Discuss the solutions of ax + 2y + z = 1, x + 2ay + z = 6, x + 2y + az = 1.
4. Determine the values of k such that the equation kx + y + z = 1, x + ky + z = 1, x + y + kz = 1 have (i) unique
solution, (ii) infinite number of solutions and (iii) no solution.
5. Investigate for what values of l and m the equations x + y +z = 6, x + 2y +3z = 10, x + 2y + lz = m have
(a) no solution, (b) unique solution and (c) infinite number of solutions.
6. Find all non-trivial solutions of x1 + 2x2 + x3 = 0, 3x1 + x2 − x3 = 0.
7. Find the value of l, if the equations 3x1 + x2 − lx3 = 0, 4x1 − 2x2 − 3x3 = 0, 2lx1 − 4x2 + lx3 = 0 have
non-trivial solution. Hence, find the solutions.
8. Find the non-trivial solution of the equations x + 5y + 3z = 0, 5x + y − l = 0, x + 2y + lz = 0 and the values
of l.
9. Determine the values of l for which the system of equations lx1 − 2x2 + x3 = 0, lx1 + (1 − l)x2 + x3 = 0,
2x1 − x2 + 2lx3 = 0 has non-trivial solutions and find the solution in each case.
10. Determine the values of l for which the system of equations 2x1 − 2x2 + x3 = lx1, 2x1 − 3x2 + 2x3 = lx2,
−x1 + 2x2 = lx3 has non-trivial solutions and find the solutions.
ANSWERS TO EXERCISE 1.3
(i) In consistent.
(ii) x = −1, y = 0, z = 1;
(iii) x = 1, y = 2, z = −1
(iv) x = 2, y = 1, z = 0;
(v) x = 4, y = 1, z = 0
(vi) x = −1, y = 4, z = 4;
(vii) x1 = 1, x2 = 2, x3 = 2, x4 = −2w
(viii) x1 = −1, x2 = 4, x3 = 4
(ix) x = 1, y = 3k − 2, z = k for all k.
(x) In consistent. Hence, no solution.
2. (i) No solution if a = 3, b ≠ 9,
(ii) Unique solution if a = 3, b is any real number.
(iii) Infinite number of solutions if a = 3, b = 9.
3. (i) Unique solution if a ≠ 1, b ≠ −2,
(ii) No solution if a = 1, b ≠ 1 and a = −2, b ≠ −2
(iii) Infinite number of solutions if a = 1, b = 1 and a = −2, b = −2.
1.
4.
(i) Unique solution if k ≠ 2, 1.
(ii) Infinite number of solutions if k = 1
and (iii) No solution if k = − 2.
5. (a) No solution if l = 3 and m ≠ 10
(b) Unique solution if l ≠ 3 and m is any real number.
(c) Infinite number of solution if l = 3, m =10
6. x1 = 3k,
x2 = −4k,
x3 = 5k for all k.
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7. When l = 9;
When l = −1;
8. When l = 1;
9. When l = 3;
When l = 1;
When l = −1;
10. When l = 1;
When l = −3;
1.5
x1 = 21k,
x1 = k,
x = 3k,
x1 = −11k,
x1 = k,
x1 = k,
x1 = 2k1 − k2,
x1 = −k,
x2 = 27k,
x2 = −13k,
y = −2k,
x2 = −16k,
x2 = 0,
x2 = 0,
x2 = k1,
x2 = −2k,
x3 = 10k for all k.
x3 = 10k for all k.
z = 3k
for all k.
x3 = k
for all k.
x3 = −k
for all k.
x3 = k
for all k.
x3 = k2
for all k1, k2.
x3 = k
for all k.
MATRIX INVERSE BY GAUSS–JORDAN METHOD
We shall explain the method for 3 × 3 matrix.
Let
⎡ a11 a12
A = ⎢⎢a21 a22
⎢⎣a31 a32
a13 ⎤
a23 ⎥⎥
a33 ⎥⎦
⎡ x 11 x 12
If A is non-singular, then there exists a 3 × 3 matrix X = ⎢⎢ x 21 x 22
⎢⎣ x 31 x 32
such that AX = I
⎡a11 a12 a13 ⎤ ⎡ x 11 x 12 x 13 ⎤ ⎡1 0
⎢a
⎥ ⎢
⎥ ⎢
⇒
⎢ 21 a22 a23 ⎥ ⎢ x 21 x 22 x 23 ⎥ = ⎢0 1
⎢⎣a31 a32 a33 ⎥⎦ ⎢⎣ x 31 x 32 x 33 ⎥⎦ ⎢⎣0 0
x 13 ⎤
x 23 ⎥⎥
x 33 ⎥⎦
0⎤
0 ⎥⎥
1 ⎥⎦
This equation is equivalent to the three equations below:
⎡a11 a12 a13 ⎤ ⎡ x 11 ⎤ ⎡1 ⎤
⎢a
⎥ ⎢ ⎥ ⎢ ⎥
⎢ 21 a22 a23 ⎥ ⎢ x 21 ⎥ = ⎢0 ⎥
⎢⎣a31 a32 a33 ⎥⎦ ⎢⎣ x 31 ⎥⎦ ⎢⎣0 ⎥⎦
and
(1)
⎡a11 a12
⎢a
⎢ 21 a22
⎢⎣a31 a32
a13 ⎤
a23 ⎥⎥
a33 ⎥⎦
⎡ x 12 ⎤ ⎡0 ⎤
⎢ x ⎥ = ⎢1 ⎥
⎢ 22 ⎥ ⎢ ⎥
⎢⎣ x 32 ⎥⎦ ⎢⎣0 ⎥⎦
(2)
⎡a11 a12
⎢a
⎢ 21 a22
⎢⎣a31 a32
a13 ⎤
a23 ⎥⎥
a33 ⎥⎦
⎡ x 13 ⎤ ⎡0 ⎤
⎢ x ⎥ = ⎢0⎥
⎢ 23 ⎥ ⎢ ⎥
⎢⎣ x 33 ⎥⎦ ⎢⎣1 ⎥⎦
(3)
Equation (1) is a system of linear equations. Solving by Jordan’s method (or by Gauss elimination
⎡ x 11 ⎤
method), we get x 11 , x 21 , x 31 and so the vector ⎢⎢ x 21 ⎥⎥ is known. Similarly solving (2) and (3), we get the
⎢⎣ x 31 ⎥⎦
other columns of X, and hence, X is known. This matrix X is the inverse of A.
M01_ENGINEERING_MATHEMATICS-I _CH01_Part A.indd 46
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Matrices ■
1.47
⎡1 ⎤
Now to solve equation (1), we start with the augmented matrix [ A , I 1 ] where I 1 = ⎢⎢0 ⎥⎥ and trans⎢⎣0 ⎥⎦
form by row operations so that A is reduced to unit matrix in Jordan’s method, then we write the solution for x 11 , x 21 , x 31 directly.
The same procedure is applied to solve (2) and (3) by writing [ A , I 2 ] and [ A , I 3 ]. In practice, we
will not do this individually and convert A into a unit matrix, but we start with ⎡⎣ A⏐I 1I 2 I 3 ⎤⎦ = ( A , I ) and
convert A into unit matrix by row operations and find X.
Working rule
Consider the augmented matrix [A, I ], where I is the identity matrix of the same order as A.
By row operations, reduce A into a unit matrix, then correspondingly I will be changed into a
matrix X. This matrix X is the inverse of A. It is advisable to change the pivot element to 1 before
applying row operations at each step.
WORKED EXAMPLES
1
3⎤
⎡ 1
Using Gauss–Jordan method, find the inverse of the matrix A 5 ⎢⎢ 1
3 23 ⎥⎥ .
⎢⎣22 24
4 ⎥⎦
Solution.
EXAMPLE 1
1 3⎤
⎡ 1
⎢
Given A = ⎢ 1 3 −3⎥⎥ . To find A–1
⎢⎣ −2 −4 4 ⎥⎦
Consider the augmented matrix
1 3 : 1 0 0⎤ ⎡ 1
1 3 : 1 0 0⎤
⎡ 1
⎢
⎥
⎢
[ A , I ] = ⎢ 1 3 −3 : 0 1 0⎥ ∼ ⎢0 2 −6 : −1 1 0⎥⎥ R 2 → R 2 − R1
⎢⎣ −2 −4 4 : 0 0 1⎥⎦ ⎢⎣0 −2 10 : 2 0 1⎥⎦ R 3 → R 3 + 2R1
⎡1
⎢
⎢
0
∼⎢
⎢
⎢
⎢0
⎣
⎡
⎢1
⎢
∼ ⎢0
⎢
⎢
⎢0
⎢⎣
M01_ENGINEERING_MATHEMATICS-I _CH01_Part A.indd 47
1
3 :
−1
5 :
1 0
1 1
1 −3 : −
2 2
3
2
1
1 −3 : −
2
1
0 2 :
2
0
6 :
1
0
−
1
2
1
2
1
2
0⎤
⎥
1
⎥
0 ⎥ R 2 → R 2 [The pivot 2 in R2
2
is reduced to 1]
⎥
1
1⎥
R3 → R3
2
2⎥
⎦
⎤
0 ⎥ R1 → R1 + ( −1) R 2
⎥
0⎥
⎥
⎥
1⎥
R3 → R3 + R 2
2 ⎥⎦
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3
⎡
⎢1 0 6 :
2
⎢
1
∼ ⎢0 1 −3 : −
⎢
2
⎢
1
⎢0 0
1 :
⎢⎣
4
−
⎡
⎢ 1 0 0 : 0 −2
⎢
1
5
∼ ⎢0 1 0 :
⎢
4
4
⎢
1
⎢0 0 1 : 1
⎢⎣
4
4
∴ the inverse matrix of A is A −1
1
2
1
2
1
4
−3 ⎤
2⎥
⎥
3⎥
4⎥
⎥
1⎥
4 ⎥⎦
⎤
0⎥
⎥
0⎥
⎥
⎥
1⎥
4 ⎥⎦
R3 →
1
R 3 (The pivot 2 in R3 is
2
reduced to 1)
R1 → R1 + ( −6 ) R3
R2 → R2 + 3R3
3⎤
⎡
⎢ 0 −2 − 2 ⎥
⎥
⎢
1
5
3⎥
⎢
=
⎢4
4
4⎥
⎥
⎢
1
1⎥
⎢1
⎢⎣ 4
4
4 ⎥⎦
EXAMPLE 2
1
2⎤
⎡4
Find the inverse of the matrix A 5 ⎢⎢ 2
3 21⎥⎥ by Gauss–Jordan method.
⎢⎣ 1 22
2 ⎥⎦
Solution.
1 2⎤
⎡4
Given A = ⎢⎢ 2
3 −1⎥⎥ . To find A–1
⎢⎣ 1 −2 2⎥⎦
Consider the augmented matrix
[A , I ]
1 2 : 1 0 0 ⎤ ⎡ 1 1 1 : 1 0 0 ⎤ R1 → 1 R1 (The pivot 4 in R1
⎡4
⎢
⎥
4 2
4
4
is reduced to 1)
= ⎢⎢ 2
3 −1 : 0 1 0 ⎥⎥ ⎢
⎥
∼ ⎢2
3 −1 : 0 1 0 ⎥
⎢⎣ 1 −2 2 : 0 0 1⎥⎦ ⎢
⎥
⎣ 1 −2 2 : 0 0 1⎦
1
⎡
⎢1
4
⎢
5
∼ ⎢0
⎢
2
⎢
9
⎢0 −
⎢⎣
4
M01_ENGINEERING_MATHEMATICS-I _CH01_Part A.indd 48
1
:
2
1
4
1
−2 : −
2
3
1
: −
2
4
⎤
0 0⎥
⎥
1 0 ⎥ R 2 → R 2 + ( −2) R1
⎥
⎥
0 1⎥ R 3 → R 3 − R1
⎥⎦
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Matrices ■
1
1
1
⎡
:
⎢1 4
2
4
⎢
−4
−1
∼ ⎢0
1
:
⎢
5
5
⎢
9
3
1
−
⎢0
: −
4
2
4
⎣⎢
⎤
0 0⎥
⎥
2
2
0⎥ R 2 → R 2
⎥
5
5
⎥
0 1⎥
⎥⎦
1.49
5
⎡
⎤
⎢The pivot 2 in R 2 is reduced to 1⎥
⎣
⎦
7
3 −1 ⎤
⎡
:
0⎥ R → R + ⎛ − 1 ⎞ R
⎢1 0
⎜⎝
⎟ 2
1
1
10
10 10
4⎠
⎥
⎢
−
4
−
1
2
∼ ⎢0 1
:
0⎥
⎥
⎢
5
5
5
⎥
⎢
9
9
⎢0 0 − 3 : − 7
1⎥ R 3 → R 3 + R 2
⎢⎣
⎥
10
10 10
4
⎦
7
3 −1
⎡
⎢ 1 0 10 : 10 10
⎢
−4
−1
2
∼ ⎢0 1
:
⎢
5
5
5
⎢
7
⎢0 0
1 :
−3
⎢⎣
3
−4
⎡
⎢1 0 0 : 3
⎢
5
∼ ⎢0 1 0 :
⎢
3
⎢
7
⎢0 0 1 :
⎢⎣
3
2
−2
−3
∴ the inverse of A is A −1
⎤
0⎥
⎥
0⎥
⎥
⎥
10
−10 ⎥ R 3 → − R 3
3
3 ⎥⎦
⎛ −7 ⎞
R1 → R1 + ⎜ ⎟ R 3
⎝ 10 ⎠
7⎤
3⎥
⎥
8
− ⎥
3⎥
⎥
−10 ⎥
3 ⎥⎦
⎡ −4
⎢ 3
⎢
5
=⎢
⎢ 3
⎢
⎢ 7
⎢⎣ 3
3
⎡
⎤
⎢The pivot − 10 in R 3 is reduced to 1⎥
⎣
⎦
R2 → R2 +
2
−2
−3
4
R3
5
7⎤
3⎥
⎥
8⎥
−
3⎥
⎥
−10 ⎥
3 ⎥⎦
EXAMPLE 3
Solve the system of equations x 1 y 1 3 z 5 4 ; x 1 3 y 23 z 5 2; 22 x 2 4 y 2 4 z 58 by finding the
matrix inverse by Gauss–Jordan method.
Solution.
The given system of equations is
M01_ENGINEERING_MATHEMATICS-I _CH01_Part A.indd 49
x + y + 3z = 4
x + 3y − 3z = 2
−2x − 4 y − 4 z = 8
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The coefficient matrix is
1 3⎤
⎡ 1
⎡4⎤
⎡x ⎤
A = ⎢⎢ 1 3 −3⎥⎥ , B = ⎢⎢ 2 ⎥⎥ and X = ⎢⎢ y ⎥⎥
⎢⎣ −2 −4 −4 ⎥⎦
⎢⎣8 ⎥⎦
⎢⎣ z ⎦⎥
∴ the system of equations is A X = B ⇒ X = A −1B .
We find A −1 by the method of matrix inverse by Gauss–Jordan method.
Consider the augmented matrix
1 3 : 1 0 0⎤
⎡ 1
⎢
[ A , I ] = ⎢ 1 3 −3 : 0 1 0⎥⎥
⎢⎣ −2 −4 −4 : 0 0 1⎥⎦
1 3 : 1 0 0⎤
⎡1
⎢
∼ ⎢0
2 −6 : −1 1 0 ⎥⎥ R 2 → R 2 − R1
⎢⎣0 −2
2 : 2 0 1⎥⎦ R 3 → R 3 + 2R1
⎡1 1 3 :
1 0
⎢
⎢
1 1
0
1 −3 : −
∼⎢
2 2
⎢
⎢
1 0
⎢0 −1 1 :
⎣
0⎤
⎥
⎥
1
0⎥ R 2 → R 2
2
⎥
1⎥
1
R3 → R3
2⎥
2
⎦
⎡
2 0
⎢1 0 4 :
⎢
1 1
∼ ⎢0
1 −3 : −
⎢
2 2
⎢
⎢ 0 −1 1 :
1 0
⎢⎣
1⎤ R → R + R
1
1
3
2⎥
⎥
0⎥
⎥
⎥
1⎥
2 ⎥⎦
1⎤
2⎥
⎥
0⎥
⎥
⎥
1⎥
2 ⎥⎦ R 3 → R 3 + R 2
1⎤
⎡
2
0
⎢1 0 4 :
2⎥
⎥
⎢
1
1
0⎥
∼ ⎢0 1 −3 : −
⎥
⎢
2
2
⎥
⎢
1
1
1
⎢0 0
1 : −
−
− ⎥ R → ⎛ − 1⎞ R
⎜⎝
⎟ 3
3
⎢⎣
4
4
4 ⎥⎦
2⎠
⎡
4 :
2 0
⎢1 0
⎢
1 1
∼ ⎢ 0 1 −3 : −
⎢
2 2
⎢
1
1
⎢ 0 0 −2 :
⎢⎣
2 2
M01_ENGINEERING_MATHEMATICS-I _CH01_Part A.indd 50
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Matrices ■
∴
the inverse of A is A −1
∴
⇒
⎡
⎢1 0
⎢
∼ ⎢0 1
⎢
⎢
⎢0 0
⎢⎣
⎡
⎢ 3
⎢
5
= ⎢−
⎢ 4
⎢
⎢− 1
⎢⎣ 4
0 :
3
5
4
1
: −
4
3⎤
2⎥
⎥
3
− ⎥
4⎥
⎥
1⎥
−
4 ⎥⎦
0 : −
1
1
1
4
1
−
4
−
1
1
4
1
−
4
−
1.51
3 ⎤ R → R − 4R
1
1
3
2⎥
⎥
3
− ⎥ R 2 → R 2 + 3R 3
4⎥
⎥
1
− ⎥
4 ⎥⎦
X = A −1B
3 ⎤ ⎡4⎤
⎡x ⎤ ⎡ 3
1
⎢
⎢ ⎥
2⎥ ⎢ ⎥
⎥
⎢ ⎥ ⎢ 5
1
3⎥ ⎢ ⎥
⎢y ⎥ = ⎢−
⎢ 2⎥
−
−
4
4⎥ ⎢ ⎥
⎢ ⎥ ⎢ 4
⎥
⎢ ⎥ ⎢ 1
1
1⎥ ⎢ ⎥
⎢
−
−
−
⎢⎣ z ⎥⎦
⎢ ⎥
⎢⎣ 4
4
4 ⎥⎦ ⎣ 8 ⎦
⎡12 + 12 + 12⎤ ⎡ 26 ⎤
⎥
⎢
⎥ ⎢
⎢
⎥ ⎢ 23 ⎥
1
= ⎢ −5 − − 6 ⎥ = ⎢ − ⎥
2
⎢
⎥ ⎢ 2⎥
1
⎢
⎥ ⎢ 7⎥
⎢ −1 − 2 − 2 ⎥ ⎢ − 2 ⎥
⎦
⎣
⎦ ⎣
∴ the solution is x = 26, y = −
23
7
, z =−
2
2
EXAMPLE 4
Solve the system of equations 2 x 1 y 1 2 z 510 ; 2 x 1 2 y 1 z 5 9; x 1 2 y 1 2 z 511 by finding the
inverse by Gauss–Jordan method.
Solution.
The given system of equations is
The coefficient matrix is
2x + y + 2z = 10
2x + 2 y + z = 9
x + 2 y + 2z = 11
⎡ 2 1 2⎤
⎡10 ⎤
⎢
⎥
A = ⎢ 2 2 1 ⎥ , B = ⎢⎢9 ⎥⎥
⎢⎣1 2 2⎥⎦
⎢⎣11 ⎥⎦
M01_ENGINEERING_MATHEMATICS-I _CH01_Part A.indd 51
⎡x ⎤
and X = ⎢⎢ y ⎥⎥
⎢⎣ z ⎥⎦
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∴ the system of equation is A X = B ⇒ X = A −1B .
We find A −1 by the method of matrix inverse by Gauss–Jordan method.
Consider the augmented matrix
⎡2 1 2 : 1 0 0⎤
[ A , I ] = ⎢⎢ 2 2 1 : 0 1 0⎥⎥
⎢⎣1 2 2 : 0 0 1 ⎥⎦
⎡
⎢1
⎢
∼ ⎢2
⎢
⎢1
⎣
⎡
⎢1
⎢
∼ ⎢0
⎢
⎢0
⎣
1
2
2
⎤ R → 1R
1
1
1
0 0⎥
2
2
⎥
: 0 1 0⎥
⎥
: 0 0 1⎥
⎦
1
⎤
:
0 0⎥
2
⎥
: −1 1 0 ⎥ R 2 → R 2 − 2R1
⎥
1
: −
0 1⎥ R 3 → R 3 − R1
2
⎦
1 :
1
2 2
1
1
2
1 −1
3
1
2
3
1
1
⎡
⎤
⎢ 1 0 2 : 1 − 2 0 ⎥ R1 → R1 − 2 R 2
⎢
⎥
∼ ⎢0 1 −1 : −1
1 0⎥
⎢
⎥
5
3
3
⎢0 0
: 1 −
1⎥ R 3 → R 3 − R 2
2
2
2
⎣
⎦
3
1
⎡
⎤
⎢ 1 0 2 : 1 − 2 0⎥
⎢
⎥
∼ ⎢0 1 −1 : −1
1 0⎥
⎢
2
3 2⎥
2
⎢0 0
⎥ R3 → R3
−
1 :
5
5 5⎦
5
⎣
2
⎡
⎢1 0 0 :
5
⎢
3
⎢
∼ 0 1 0 : −
⎢
5
⎢
2
⎢0 0 1 :
⎢⎣
5
∴ the inverse of A is
A −1
⎡ 2
⎢ 5
⎢
3
= ⎢−
⎢ 5
⎢
⎢ 2
⎢⎣ 5
M01_ENGINEERING_MATHEMATICS-I _CH01_Part A.indd 52
2
5
2
5
3
−
5
2
5
2
5
3
−
5
3
3⎤
− ⎥ R1 → R1 − R 3
2
5
⎥
2⎥
R 2 → R 2 + R3
5⎥
⎥
2⎥
5 ⎥⎦
3⎤
− ⎥
5
⎥
2⎥
5⎥
⎥
2⎥
5 ⎥⎦
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Matrices ■
∴
1.53
X = A −1B
⎡x ⎤ ⎡ 2
⎢ ⎥ ⎢ 5
⎢ ⎥ ⎢ 3
⎢y ⎥ = ⎢−
⎢ ⎥ ⎢ 5
⎢ ⎥ ⎢ 2
⎢⎣ z ⎥⎦ ⎢
⎢⎣ 5
⇒
2
5
2
5
3
−
5
3⎤
− ⎥ ⎡10 ⎤
5 ⎢ ⎥
⎥
2⎥ ⎢ ⎥
⎢9⎥
5⎥ ⎢ ⎥
⎥
2⎥ ⎢ ⎥
⎢ ⎥
5 ⎥⎦ ⎣11⎦
⎡ 20 18 33 ⎤ ⎡ 38 − 33 ⎤ ⎡1 ⎤
⎢ 5 + 5 − 5 ⎥ ⎢ 5 ⎥ ⎢ ⎥
⎥
⎢
⎥ ⎢
30 18 22 ⎥ ⎢ 40 − 30 ⎥ ⎢ ⎥
⎢
= ⎢ 2⎥
= − + +
=
⎢ 5 5
5 ⎥ ⎢ 5 ⎥ ⎢ ⎥
⎥
⎢
⎢
⎥
⎢ ⎥
⎢ 20 − 27 + 22 ⎥ ⎢ 42 − 27 ⎥ ⎢ ⎥
3
⎢⎣ 5
5
5 ⎥⎦ ⎢⎣ 5 ⎥⎦ ⎣ ⎦
∴ the solution is x = 1, y = 2, z = 3 .
EXERCISE 1.4
Find the matrix inverse by Gauss–Jordan method
1 3⎤
⎡ 1
1. ⎢ 1 3 −3⎥
⎢
⎥
⎢⎣ −2 −4 −4 ⎥⎦
⎡ 2 2 6⎤
2. ⎢ 2 6 −6 ⎥
⎢
⎥
⎣⎢ 4 −8 −8⎥⎦
3.
⎡ 2 1 2⎤
⎢ 2 2 1⎥
⎢
⎥
⎢⎣ 1 2 2⎥⎦
0⎤
⎡ 8 −4
⎡ 2 −2 4 ⎤
⎡ 2 1 1⎤
⎥
⎥
5. ⎢ 2
6. ⎢ 3 2 3⎥
4. ⎢ −4
−
4
3
2
8
⎢
⎥
⎢
⎥
⎢
⎥
⎢⎣ 0 −4
⎢⎣ −1 1 1⎥⎦
⎢⎣ 1 4 9⎥⎦
8⎥⎦
7. Solve the system of linear equations x + y + z = 9, 2x − 3y + 4 z = 13, 3x + 4 y + 5z = 40, finding the
inverse matrix by Gauss–Jordan method.
8. Solve the system of equations 2 x + y + z = 10, 3 x + 2 y + 3 z = 18, x + 4 y + 9 z = 16, finding the inverse
matrix by Gauss–Jordan method.
ANSWERS TO EXERCISE 1.4
3⎤
⎡
1
⎢ 3
2⎥
⎢
⎥
5
1
3
1. A −1 = ⎢⎢ −
− ⎥⎥
4
4
4
⎢
⎥
1
1
1
⎢−
−
− ⎥
⎢⎣ 4
4
4 ⎥⎦
4. A −1 =
⎡3 2 1 ⎤
1 ⎢
2 4 2⎥⎥
16 ⎢
⎢⎣1 2 3 ⎥⎦
7. x = 1, y = 3, z = 5
2. A −1 =
5. A −1 =
⎡12 4 6 ⎤
1 ⎢
1 5 −3⎥⎥
56 ⎢
⎢⎣ 5 −3 −1⎥⎦
⎡ −5 2 −16
1 ⎢
0 2
4
10 ⎢
⎢⎣ 5 0 10
8. x = 7, y = −9, z = 5
M01_ENGINEERING_MATHEMATICS-I _CH01_Part A.indd 53
⎤
⎥
⎥
⎥⎦
⎡ 2 2 −3 ⎤
1⎢
−2 2 2 ⎥⎥
5⎢
⎢⎣ 2 −3 2 ⎥⎦
3.
A −1 =
6.
5 1 ⎤
⎡
−
⎢ −3
2 2 ⎥
⎢
⎥
17
3 ⎥
A −1 = ⎢⎢12 −
2
2 ⎥
⎢
⎥
7
1
⎢ −5
⎥
−
⎢⎣
2 2 ⎥⎦
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1.6 EIGEN VALUES AND EIGEN VECTORS
1.6.0 Introduction
In this section we study certain numbers associated with a square matrix, called eigen values and
certain vectors associated with them, called eigen vectors. The problem of determining eigen values
and eigen vectors of a square matrix is called an eigen value problem. The eigen value problem arise in
a wide range of physical and engineering applications such as mechanical system, electrical networks,
Markov processes, elastic deformations, etc.
Eigen value problems are used in diagonalisation of a matrix of a quadratic form.
1.6.1 Vector
The vector 2i + 3 j − 4 k can be regarded as the triplet (2, 3, −4).
Definition 1.28 An ordered n-tuple (x1, x2, …, xn) of numbers x1, x2, …, xn is called an n-dimensional
vector.
For example the triplet (2, 3, −4) is a 3-dimensional vector. (1, 0, −2, 3) is a 4-dimensional vector.
A row matrix is also called a row vector and a column matrix is called a column vector.
Definition 1.29 If X1 = (a1, a2, …, an), X2 = (b1, b2, …, bn) be two n-dimensional vectors, then their
sum and scalar multiplications are
X1 + X2 = (a1 + b1, a2 + b2, …, an + bn),
aX1 = (aa1, aa2, …, aan), which are n-dimensional vectors.
X1 = X2 if and only if a1 = b1, a2 = b2, …, an = bn
Definition 1.30 Linear Combination
If X1, X2, …, Xr are r vectors of n-dimension and if a1, a2, …, ar are numbers, then the vector
a1X1 + a2X2 + … + arXr is called a linear combination of the vectors X1, X2, …, Xr.
Definition 1.31 Linearly dependent and independent vectors
(a) The set of vectors X1, X2, …, Xr is said to be linearly dependent if there exist numbers
a1, a2, …, ar, not all zero, such that a1X1 + a2X2 + … + arXr = 0
(b) The set of vectors X1, X2, …, Xr is said to be linearly independent if any relation of the form
a1X1 + a2X2 + … + arXr = 0 ⇒ a1 = 0, a2 = 0, …, ar = 0
Note
(i) If X1, X2, …, Xr are linearly dependent, then some vector is a linear combination of others.
(ii) In a plane or 2-dimensional space, non-collinear vectors are linearly independent vectors whereas
collinear vectors are dependent vectors.
In 3-dimesional space, non-coplanar vectors are linearly independent vectors whereas coplanar
vectors are dependent vectors.
Example: i = (1, 0, 0 ), j = (0, 1, 0 ), k = (0 , 0, 1) are linearly independent vectors.
(iii) Any set of vectors containing zero vector 0 is a linearly dependent set.
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1.55
(iv) Rank of an m × n matrix A is equal to the maximum number of independent column vectors or
row vectors of A.
(v) A useful result to test linear independence:
Let X1, X2, …, Xn be n vectors of n-dimensional space. Let A be the matrix having these n-vectors as
columns (or rows). Then A is a square matrix of order n. If A ≠ 0 , then X1, X2, …, Xn are linearly
independent.
If A = 0 , then X1, X2, …, Xn are linearly dependent.
WORKED EXAMPLES
EXAMPLE 1
Show that the vectors (1, 2, 3), (3, 22, 1), (1, 26, 25) are linearly dependent.
Solution.
1⎤
⎡1 3
Let A = ⎢⎢ 2 −2 −6 ⎥⎥ with the vectors as columns.
⎢⎣ 3
1 −5⎥⎦
1 3
1
Then A = 2 −2 −6 = 1⋅ (10 + 6) − 3( −10 + 18) + 1⋅ ( 2 + 6) = 16 − 24 + 8 = 0
3
1 −5
∴
the vectors (1, 2, 3), (3, −2, 1) and (1, −6, −5) are linearly dependent.
EXAMPLE 2
Show that the vectors X1 = (1, 2, 23, 4), X2 = (3, 21, 2, 1), X3 = (1, 25, 8, 27) are linearly
dependent and find the relation between them.
Solution.
⎡ 1 2 −3 4 ⎤
Let A = ⎢⎢3 −1 2
1⎥⎥ with the vectors as rows.
⎢⎣ 1 −5 8 −7⎥⎦
We shall use elementary row operations.
4⎤
⎡ 1 2 −3
⎢
∴
A ∼ ⎢0 −7 11 −11⎥⎥ R 2 → R 2 − 3R1 = R 2′
R 3 → R 3 − R1 = R 3′
⎣⎢0 −7 11 −11⎥⎦
4⎤
⎡ 1 2 −3
⎢
∼ ⎢0 −7 11 −11⎥⎥
⎢⎣0
0 0
0 ⎥⎦ R 3 → R 3′ − R 2′ = R 3′′
Since the maximum number of non-zero rows is 2, which is less than the number of vectors, the given
vectors are linearly dependent.
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The relation between them is obtained as below.
R 3′′ = 0 ⇒
R 3′ − R 2′ = 0
⇒ R 3 − R1 − ( R 2 − 3R1 ) = 0
⇒
R 3 − R 2 + 2R1 = 0
Since the rows are vectors, we get X3 − X2 + 2X1 = 0 which is the relation between the vectors.
Note The rows of the matrix are the given vectors. So, only row operations must be used to find the
relationship between the vectors.
1.6.2 Eigen Values and Eigen Vectors
Definition 1.32 Let A be a square matrix of order n. A number l is called an eigen value of A if there
exists a non-zero column matrix X such that AX = lX. Then X is called an eigen vector of A corresponding to l.
⎡ x1 ⎤
⎢x ⎥
If
A = ⎡⎣aij ⎤⎦
and X = ⎢ 2 ⎥ , then AX = lX ⇒ (A − lI)X = 0.
n×n
⎢:⎥
⎢ ⎥
⎣x n ⎦
This will represent a system of linear homogeneous equations in x1, x2, …, xn.
Since X ≠ 0 at least one of the xi ≠ 0.
Hence, the homogeneous system has nontrivial solutions.
∴ the determinant of coefficients A − lI = 0.
This equation is called the characteristic equation of A.
The determinant A − lI , on expansion, will be a nth degree polynomial in l and is known as the
characteristic polynomial of A.
The roots of the characteristic equation are the eigen values of A.
Definition 1.33 Characteristic Equation and Characteristic Polynomial
If l is a characteristic root of a square matrix A, then A − lI = 0 is called the characteristic equation
of A.
The polynomial A − lI in l is called the characteristic polynomial of A.
Note
(1) The word ‘eigen’ is German, which means ‘characteristic’ or ‘proper’. So, an eigen value is also
known as characteristic root or proper value. Sometimes it is also known as latent root.
⎡a
(2) If A = ⎢ 11
⎣ a21
a12 ⎤
⎡1 0 ⎤
, I=⎢
⎥
⎥ , then the characteristic equation of A is
a22 ⎦
⎣0 1⎦
A − lI = 0
⇒
a11 − l
a12
=0
a21
a22 − l
⇒
(a11 − l )(a22 − l ) − a21a12 = 0
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Matrices ■
⇒
a11 ⋅ a22 − (a11 + a22 )l + l 2 − a21a12 = 0
⇒
l 2 − (a11 + a22 )l + (a11a22 − a21a12 ) = 0
⇒
l 2 − S 1l + S 2 = 0
where
1.57
S1 = a11 + a22 = sum of the diagonal elements of A.
S2 = a11 ⋅ a22 − a21 ⋅ a12 = A
⎡ a11 a12
(3) If A = ⎢a21 a22
⎢
⎢⎣a31 a32
⇒
a13 ⎤
⎡1 0 0 ⎤
a23 ⎥⎥ , I = ⎢⎢0 1 0 ⎥⎥ , then the characteristic equation of A is
⎢⎣0 0 1 ⎥⎦
a33 ⎥⎦
A − lI = 0
a11 − l
a12
a13
a21
a22 − l
a23 = 0
a31
a32
a33 − l
Expanding this determinant we will get
l3 − S1l2 + S2l − S3= 0,
S1 = sum of the diagonal elements of A
S2 = sum of the minors of elements of the main diagonal
S3 = A
We will use this formula in problems.
where
Definition 1.34 The set of all distinct eigen values of the square matrix A is called the spectrum of A.
The largest of the absolute values of the eigen values of A is called the spectral radius of A. The set
of all eigen vectors corresponding to an eigen value l of A, together with zero vector, forms a vector
space which is called the eigenspace of A corresponding to l.
1.6.3 Properties of Eigen Vectors
Theorem 1.3
(1) Eigen vector corresponding to an eigen value is not unique.
(2) Eigen vectors corresponding to different eigen values are linearly independent.
Proof
(1) Let l be an eigen value of a square matrix A of order n.
Let X be an eigen vector corresponding to l.
Then
AX = lX
Multiply by a constant C
∴
C(AX) = C(lX) ⇒ A(CX) = l(CX)
Since
C ≠ 0, X ≠ 0 we have CX ≠ 0
∴ CX is an eigen vector corresponding to l for any C ≠ 0. Hence, eigen vector is not unique
for the eigen value l.
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(2) Let l1, l2 be two different eigen values of A.
Let X1, X2 be the corresponding eigen vectors.
∴
and
AX1 = l1X1
AX2 = l2X2
(1)
(2)
We have to prove X1 and X2 are linearly independent.
Suppose
a1X1 + a2X2 = 0
then
A(a1X1 + a2X2) = 0
a1 ( AX1 ) + a 2 ( AX 2 ) = 0
⇒
a1 (l1X1 ) + a 2 (l 2 X 2 ) = 0
⇒
(3)
(a1l1 )X1 + (a 2 l 2 )X 2 = 0
⇒
(4)
Multiply (3) by l1, we get
l1 (a1X1 ) + l1 (a 2 X 2 ) = 0
⇒
(4) − (5)
(a1l1 )X1 + (a 2 l1 )X 2 = 0
(5)
a 2 ( l 2 − l1 ) X 2 = 0
(6)
⇒
l1 ≠ l 2
Since
( l 2 − l1 ) X 2 ≠ 0
∴
∴
(6) ⇒ a 2 = 0
Thus,
∴
⇒ l 2 − l1 ≠ 0 and X 2 ≠ 0
∴
(3) ⇒ a1X1 = 0
a1X1 + a 2 X 2 = 0
⇒
⇒
a1 = 0, since X1 ≠ 0.
a1 = 0 and a 2 = 0
X1 and X2 are linearly independent.
Note
(1) If all the n eigen values l1, l2, …, ln of A are different, then the corresponding eigen vectors
X1, X2, …, Xn are linearly independent.
(2) A given eigen vector of A corresponds to only one eigen value of A.
(3) Eigen vectors corresponding to equal eigen values may be linearly independent or dependent.
WORKED EXAMPLES
EXAMPLE 1
⎡4 1 ⎤
Find the eigen values and eigen vectors of the matrix ⎢
⎥.
⎣3 2 ⎦
Solution.
⎡4 1⎤
Let A = ⎢
⎥.
⎣ 3 2⎦
The characteristic equation of A is A − lI = 0
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4−l
1
=0
3
2−l
⇒
⇒
1.59
l 2 − S1l + S2 = 0
where S1 = sum of the diagonal elements of A = 4 + 2 = 6
S2 = A =
∴ the characteristic equation is l2 − 6l + 5 = 0
which are the eigen values of A.
4 1
= 8−3= 5
3 2
⇒
(l − 1) (l − 5) = 0
⇒ l = 1, 5
To find eigen vectors:
⎡ x1 ⎤
Let X = ⎢ ⎥ be an eigen vector of A corresponding to l.
⎣x 2 ⎦
Then
( A − lI)X = 0
1 ⎤
⎡4 − l
⎢ 3
2 − l ⎥⎦
⎣
⇒
⎡ x 1 ⎤ ⎡0 ⎤
⎢ x ⎥ = ⎢0 ⎥
⎣ 2⎦ ⎣ ⎦
( 4 − l )x 1 + x 2 = 0 ⎫
⎬
3x 1 + ( 2 − l )x 2 = 0 ⎭
⇒
Case (i) If l = 1, then equations (I) become
3x1 + x2 = 0 and 3x1 + x2 = 0
Choosing x1 = 1, we get x2 = −3.
∴
eigen vector is
Case (ii) If l = 5, then equations (I) become
−x1 + x2 = 0 and 3x1 − 3x2 = 0
Choosing x1 = 1, we get x2 = 1
(I)
∴ x2 = −3x1
⎡ 1⎤
X1 = ⎢ ⎥
⎣ −3⎦
∴ x1 = x2
∴ eigen vector is
⎡1⎤
X2 = ⎢ ⎥
⎣1⎦
⎡ 1⎤ ⎡1⎤
Thus, eigen values of A are 1, 5 and the corresponding eigen vectors are ⎢ ⎥ , ⎢ ⎥
⎣ −3⎦ ⎣1⎦
Note In case (i) we have only one equation 3x1 + x2 = 0 to solve for x1 and x2. So, we have infinite
number of solutions x1 = k, x2 = −3k, for any k ≠ 0. We have chosen the simplest solution.
⎡ k⎤
⎡ 1⎤
Infact ⎢
= k ⎢ ⎥ is an eigen vector for l = 1 for any k ≠ 0. So, for l = 1 there are many eigen
⎥
⎣ −3k ⎦
⎣ −3⎦
vectors. This verifies property 1.
EXAMPLE 2
Show that the real matrix ⎡ a
⎢⎣2b
b ⎤ has two eigen vectors
a ⎥⎦
⎡1⎤
⎡ 1⎤
⎢i ⎥ and ⎢ −i ⎥ , where b ≠ 0.
⎣ ⎦
⎣ ⎦
Solution.
⎡ a b⎤
Let A = ⎢
⎥.
⎣ −b a⎦
The characteristic equation of A is A − lI = 0
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a−l
b
=0
−b
a−l
⇒
⇒
l 2 − S1l + S2 = 0
where S1 = sum of the diagonal elements of A = a + a = 2a
S2 = A =
a b
= a2 + b 2
−b a
∴ the characteristic equation is
l2 − 2al + (a2 + b2) = 0 ⇒ l = 2a ± 4a2 − 4(a2 + b 2 )
=
2a ± −4b 2 2a ± 2ib
=
= a + ib or a − ib
2
2
which are the eigen values of A.
To find eigen vectors:
⎡x ⎤
Let X = ⎢ 1 ⎥ be an eigen vector of A corresponding to l.
⎣x 2 ⎦
b ⎤ ⎡ x 1 ⎤ ⎡0⎤
⎡a − l
( A − lI)X = 0 ⇒ ⎢
Then
⎥ ⎢ ⎥=⎢ ⎥
⎣ −b a − l ⎦ ⎣ x 2 ⎦ ⎣0 ⎦
(a − l )x 1 + bx 2 = 0⎫
⎬
−bx 1 + (a − l )x 2 = 0⎭
Case (i) If l = a + ib, then the equations (I) become
⇒
(I)
(a − (a + ib ))x 1 + bx 2 = 0
−bx 1 + (a − a − ib )x 2 = 0
and
⇒
−ibx 1 + bx 2 = 0
⇒
x 2 = ix 1
and
−bx 1 − ibx 2 = 0
⇒
− x 1 = ix 2
⇒
x 2 = ix 1
⇒
i x 1 = ix 2
2
So, we have only one equation x 2 = ix 1
Choosing x1 = 1, we get x2 = i ∴
an eigen vector is
Case (ii) If l = a − ib, then the equations (I) become
⎡1⎤
X1 = ⎢ ⎥
⎣i ⎦
(a − a + ib )x 1 + bx 2 = 0
−bx 1 + (a − a + ib )x 2 = 0
and
ibx 1 + bx 2 = 0 ⇒ x 2 = −ix 1
⇒
−bx 1 + ibx 2 = 0 ⇒ x 2 = −ix 1
Choosing x1 = 1, we get x2 = −i
⎡ 1⎤
∴ an eigen vector is X 2 = ⎢ ⎥
⎣ −i ⎦
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1.61
Thus, the eigen values of A are a + ib, a − ib and the corresponding eigen vectors are
⎡1⎤
⎡1⎤
X1 = ⎢ ⎥ and X 2 = ⎢ ⎥
⎣i ⎦
⎣ −i ⎦
EXAMPLE 3
⎡3 24 4 ⎤
⎢
Find the eigen values and eigen vectors of the matrix ⎢1 22 4 ⎥⎥ .
⎢⎣1 21 3 ⎥⎦
Solution.
⎡3 −4 4 ⎤
Let A = ⎢⎢1 −2 4 ⎥⎥
⎣⎢1 −1 3 ⎦⎥
The characteristic equation of A is A − lI = 0
⇒
where
3−l
−4
4
1
−2 − l
4 =0
1
−1
3−l
⇒
l 3 − S1l 2 + S2 l − S3 = 0
S1 = sum of main diagonal elements of A = 3 + (−2) + 3 = 4
S2 = sum of minors of diagonal elements of A
3 4
3 −4
−2 4
=
+
+
= −6 + 4 + 9 − 4 + ( −6) + 4 = −2 + 5 + ( −2) = 1
1 3
1 −2
−1 3
and
S3 = A = 3( −6 + 4) + 4(3 − 4) + 4( −1 + 2) = −6 − 4 + 4 = −6
∴ the characteristic equation is l3 − 4l2 + l + 6 = 0
We choose integer factors of constant term 6 for trial solution. We find l = −1 is a root. To find the
other roots we perform synthetic division
Other roots are given by
⇒
⇒
⇒
l 2 − 5l + 6 = 0
(l − 2)(l − 3) = 0
l = 2 or 3
∴ the eigen values are l = −1, 2, 3
−1
1 −4 1 6
0 −1 5 −6
1 −5 6
0
[different roots]
To find eigen vectors:
⎡ x1 ⎤
Let X = ⎢⎢ x 2 ⎥⎥ be an eigen vector corresponding to the eigen value l.
⎢⎣ x 3 ⎥⎦
Then
4 ⎤
⎡3 − l −4
4 ⎥⎥
( A − lI)X = 0 ⇒ ⎢⎢ 1 −2 − l
⎢⎣ 1
3 − l ⎥⎦
−1
M01_ENGINEERING_MATHEMATICS-I _CH01_Part B.indd 61
⎡ x 1 ⎤ ⎡0 ⎤
⎢ x ⎥ = ⎢0 ⎥
⎢ 2⎥ ⎢ ⎥
⎢⎣ x 3 ⎥⎦ ⎢⎣0 ⎥⎦
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(3 − l ) x 1 − 4 x 2 + 4 x 3 = 0 ⎫
⎪
x 1 − ( 2 + l )x 2 + 4 x 3 = 0⎬
x 1 − x 2 + (3 − l )x 3 = 0⎪⎭
⇒
(I)
Case (i) If l = −1, then the equations (I) become
4x1 − 4x2 + 4x3 = 0
The different equations are
x1 − x2 + x3 = 0
x1 − x2 + 4x3 = 0
and x1 − x2 + 4x3 = 0
x1 − x2 + x3 = 0
and x1 − x2 + 4x3 = 0
x1
By rule of cross multiplication, we get
⇒
⇒
x3
x1
x
= 2 =
−4 + 1 1 − 4 −1 + 1
x
x1
x
= 2 = 3 ⇒
−3 −3
0
x2
x3
−1
1
1
−1
−1
4
1
−1
x1 x2 x3
=
=
1
1
0
⎡1 ⎤
Choosing x1 = 1, x2 = 1, x3 = 0, we get an eigen vector X1 = ⎢1 ⎥
⎢⎣0 ⎥⎦
Case (ii) If l = 2, then equations (I) become
x1 − 4x2 + 4x3 = 0, x1 − 4x2 + 4x3 = 0
∴ the different equations are
x1 − 4x2 + 4x3 = 0
By the rule of cross multiplication, we get
⇒
x3
x1
x
= 2 =
−4 + 4 4 − 1 −1 + 4
x
x1 x 2
=
= 3 ⇒
0
3
3
and x1 − x2 + x3 = 0
and x1 − x2 + x3 = 0
x1
x1 x 2 x 3
=
=
0
1
1
x2
x3
−4
4
1
−4
−1
1
1
−1
⎡0 ⎤
Choosing x1 = 0, x2 = 1, x3 = 1, we get an eigen vector X 2 = ⎢1 ⎥
⎢⎣1 ⎥⎦
Case (iii) If l = 3, then equations (I) become
0x1 − 4x 2 + 4x 3 = 0 ⇒ 0x1 − x 2 + x 3 = 0
x 1 − 5x 2 + 4 x 3 = 0 and x 1 − x 2 + 0 x 3 = 0
The equations are different, but only two of them are independent. So, we can choose any two of them
to solve. From the first two equations, we get
x1
x2
x3
x
x1
x
= 2 = 3
1
−1
0
−1
−4 + 5 1 − 0 0 + 1
x
x1 x 2
⇒
4
−5
1
−5
=
= 3
1
1
1
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⎡1⎤
Choosing x1 = 1, x2 = 1, x3 = 1, we get an eigen vector X 3 = ⎢1⎥
⎢⎣1⎥⎦
Thus, the eigen values of A are −1, 2, 3 and corresponding eigen vectors are
⎡1 ⎤
⎡0 ⎤
⎡1⎤
X1 = ⎢1 ⎥ , X 2 = ⎢1 ⎥ , X 3 = ⎢1⎥
⎢⎣0 ⎥⎦
⎢⎣1 ⎥⎦
⎢⎣1⎥⎦
Note
(1) We are using the following integer root theorem for trial solution. “For the equation of the form
xn + an − 1 xn − 1 + an − 2 xn − 2 + … + a1x + a0 = 0 with integer coefficients ai, any rational root is an
integer and is a factor of the constant term a0”. So, it is enough we try factors of the constant term
for integer solutions. If there is no integer solution, then real roots should be irrational.
(2) In the above problem the eigen values −1, 2, 3 are different. So, by property (2) the eigen vectors
are linearly independent. We shall verify this:
∴
⎡1 1 0⎤
Consider B = ⎢0 1 1 ⎥ with the eigen vectors as rows.
⎥
⎢
⎢⎣1 1 1 ⎥⎦
Then B = 1⋅ 0 − 1( −1) + 0 = 1 ≠ 0
X1, X2, X3 are linearly independent.
EXAMPLE 4
⎡2
Find the eigen values and eigen vectors of ⎢ 1
⎢⎣ 1
Solution.
⎡2 2 1⎤
Let A = ⎢1 3 1 ⎥
⎢⎣1 2 2⎥⎦
1⎤
1⎥ .
2 ⎥⎦
2
3
2
The characteristic equation of A is A − l I = 0
⇒
2−l
2
1
1
3−l
1 =0
1
2
2−l
⇒
l 3 − S1l 2 + S2 l − S3 = 0
where S1 = sum of the diagonal elements of A = 2 + 3 + 2 = 7
S2 = sum of minors of the diagonal elements of determinant A
3 1
2 1
2 2
=
+
+
= 6 − 2 + 4 − 1 + 6 − 2 = 11
2 2
1 2
1 3
S3 = A = 2(6 − 2) − 2( 2 − 1) + 1( 2 − 3) = 8 − 2 − 1 = 5
∴ the characteristic equation is l3 − 7l2 + 11l − 5 = 0
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Choose the integer factors of constant term −5 for trial.
The integer factors of −5 are −5, 1, or −1, 5.
We find l = 1 is a root.
Other roots are given by l2 − 6l + 5 = 0
1 1 −7 11 −5
0
1 −6
5
⇒
(l − 1) (l − 5) = 0
⇒
l = 1, 5
∴ the eigen values are l = 1, 1, 5 (Two equal eigen values)
1 −6
5
0
To find eigen vectors:
⎡ x1 ⎤
Let X = ⎢⎢ x2 ⎥⎥ be an eigen vector of A corresponding to the eigen value l.
⎢⎣ x3 ⎥⎦
⎡2 − l
Then ( A − lI)X = 0 ⇒ ⎢ 1
⎢⎣ 1
⇒
⎡ x 1 ⎤ ⎡0 ⎤
⎢ x 2 ⎥ = ⎢0 ⎥
⎢ x ⎥ ⎣⎢0 ⎥⎦
⎣ 3⎦
( 2 − l ) x 1 + 2x 2 + x 3 = 0 ⎫
⎪
x 1 + (3 − l ) x 2 + x 3 = 0 ⎬
x 1 + 2x 2 + ( 2 − l )x 3 = 0 ⎪⎭
2
3−l
2
1 ⎤
1 ⎥
2 − l ⎥⎦
(I)
Case (i) If l = 5, then the equations (I) become
−3x1 + 2x2 + x3 = 0, x1 − 2x2 + x3 = 0 and x1 + 2x2 − 3x3 = 0
These 3 equations are different, but only 2 of them are independent.
So, we can choose any two of them to solve for x1, x2, x3.
From last two equations, by the rule of cross multiplication, we get
⇒
⇒
x
x1
x
= 2 = 3
6 − 2 1+ 3
2+ 2
x3
x1
x
= 2 =
4
4
4
x3
x1
x2
=
=
1
1
1
x2
x1
x3
−2
1
1
−2
2
−3
1
2
⎡1⎤
Choosing x1 = 1, x2 = 1, x3 = 1, we get an eigen vector X1 = ⎢⎢1⎥⎥
⎣⎢1⎦⎥
Case (ii) If l = 1, then the equations (I) become
x1 + 2x2 + x3 = 0, x1 + 2x2 + x3 = 0 and x1 + 2x2 + x3 = 0
We have only one equation x1 + 2x2 + x3 = 0 to solve for x1, x2, x3. Assign arbitrary values for two variables and solve for the third.
Choose x3 = 0, then x1 + 2x2 = 0 ⇒ x1 = −2x2
⎡ −2⎤
Choose x2 = 1, ∴ x1 = −2, we get an eigen vector X 2 = ⎢⎢ 1⎥⎥
⎢⎣ 0 ⎥⎦
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We shall find one more solution from x1 + 2x2 + x3 = 0
Choose x2 = 0 then x1 + x3 = 0 ⇒ x3 = − x1
Choose x1 = 1 ∴ x3 = −1
⎡ 1⎤
⎢
∴ another eigen vector corresponding to l = 1 is X 3 = ⎢ 0 ⎥⎥
⎢⎣ −1⎥⎦
Thus, eigen values of A are 5, 1, 1 and the corresponding eigen vectors are
⎡ −2⎤
⎡ 1⎤
⎡1⎤
⎢
⎥
⎢
⎥
X1 = ⎢1⎥ , X 2 = ⎢ 1⎥ and X 3 = ⎢⎢ 0 ⎥⎥
⎣⎢1⎦⎥
⎣⎢ 0 ⎦⎥
⎣⎢ −1⎦⎥
Note Though the eigen values are not different, we could find independent eigen vectors.
⎡ 1 1 1⎤
For, consider B = ⎢⎢ −2 1 0 ⎥⎥ with the vectors as rows
⎣⎢ 1 0 −1⎦⎥
B = 1( −1 − 0) − 1( 2 − 0) + 1(0 − 1) = −1 − 2 − 1 = −4 ≠ 0
Then
∴ X1, X2, X3 are linearly independent.
EXAMPLE 5
⎡ 6
⎢
Find the eigen values and eigen vectors of the matrix ⎢ 14
⎢⎣ 7
Solution.
⎡ 6 −6 5⎤
Let A = ⎢⎢14 −13 10 ⎥⎥
⎢⎣ 7 −6 4 ⎥⎦
−6 5 ⎤
−13 10 ⎥⎥ .
−6 4 ⎥⎦
The characteristic equation of A is A − lI = 0
6−l
−6
5
3
2
14
−13 − l 10 = 0 ⇒ l − S1l + S2 l − S3 = 0
7
−6
4−l
where S1 = sum of the diagonal elements of A = 6 + (−13) + 4 = −3
S2 = sum of minors of elements of the diagonal of A
−13 10
6 5
6 −6
=
+
+
−6 4
7 4
14 −13
⇒
= ( −52 + 60) + ( 24 − 35) + ( −78 + 84) = 8 − 11 + 6 = 3
S3 = A = 6( −52 + 60) + 6(56 − 70) + 5( −84 + 91)
= 48 + 6( −14) + 5(7) = 48 − 84 + 35 = −1
∴ the characteristic equation is l3 + 3l2 + 3l + 1 = 0
⇒
(l + 1)3 = 0
Three equal eigen values.
M01_ENGINEERING_MATHEMATICS-I _CH01_Part B.indd 65
⇒
l = −1, −1, −1
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To find eigen vectors:
⎡ x1 ⎤
Let X = ⎢⎢ x2 ⎥⎥ be an eigen vector corresponding to the eigen value l.
⎢⎣ x3 ⎥⎦
Then
( A − lI)X = 0
−6
5 ⎤
⎡6 − l
⎢ 14
−13 − l 10 ⎥⎥
⎢
⎢⎣ 7
−6
4 − l ⎥⎦
⎡ x 1 ⎤ ⎡0⎤
⎢ x ⎥ = ⎢0⎥
⎢ 2⎥ ⎢ ⎥
⎢⎣ x 3 ⎥⎦ ⎢⎣ 0 ⎥⎦
( 6 − l ) x 1 − 6 x 2 + 5x 3 = 0 ⎫
⎪
14 x 1 − (13 + l )x 2 + 10 x 3 = 0 ⎬
7x 1 − 6 x 2 + ( 4 − l )x 3 = 0 ⎪⎭
⇒
⇒
(I)
If l = −1, then the equations (I) become
7 x1 − 6 x2 + 5 x3 = 0
14 x1 − 12 x2 + 10 x3 = 0 ⇒ 7 x1 − 6 x2 + 5 x3 = 0
7 x1 − 6 x2 + 5 x3 = 0
We have only one equation 7x1 − 6x2 + 5x3 = 0
Assign arbitrary values to two variables and find the third.
We shall find 3 vectors.
x
x
Putting x1 = 0, we get −6x2 + 5x3 = 0 ⇒ 6x2 = 5x3 ⇒ 2 = 3
5
6
⎡0 ⎤
Choosing x2 = 5, x3 = 6, we get an eigen vector X1 = ⎢⎢ 5⎥⎥
⎣⎢6 ⎦⎥
x
x
Putting x2 = 0, we get 7x1 + 5x3 = 0 ⇒ 7x1 = −5x3 ⇒ 1 = 3
5 −7
⎡ 5⎤
⎢
Choosing x1 = 5, x3 = −7, we get the second eigen vector X 2 = ⎢ 0 ⎥⎥
Again, putting x3 = 0, we get
⎢⎣ −7⎥⎦
x1 x2
=
6
7
⎡6 ⎤
Choosing x1 = 6, x2 = 7, we get the third eigen vector X 3 = ⎢⎢ 7⎥⎥
⎢⎣0 ⎥⎦
Thus, eigen values of A are −1, −1, −1 and the corresponding eigen vectors are
⎡ 5⎤
⎡0 ⎤
⎡6 ⎤
⎢
⎥
⎢
⎥
⎢
and
X1 = ⎢ 5 ⎥ , X 2 = ⎢ 0 ⎥
X 3 = ⎢ 7⎥⎥ ,
⎢⎣6 ⎥⎦
⎢⎣ −7⎥⎦
⎢⎣0 ⎥⎦
7x1 − 6x2 = 0
⇒
7x1 = 6x2
⇒
which are different.
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5 6⎤
⎡0
⎢
Note If B = ⎢ 5 0 7⎥⎥ with the eigen vectors as rows, then
⎢⎣6 −7 0 ⎥⎦
B = 0 − 5(0 − 42) + 6( −35 − 0) = 210 − 210 = 0
∴ the vectors X1, X2, X3 are linearly dependent. However, any two of them are linearly independent.
Geometrically, it means that all the vectors are coplanar, but any two of them are non-collinear.
In this example we have seen −1 is the only eigen value of 3 × 3 matrix and two linearly
independent eigen vectors.
1.6.4 Properties of Eigen Values
1. A square matrix A and its transpose AT have the same eigen values.
Proof Eigen values of A are the roots of its characteristic equation
A − lI = 0
We know
( A − l I)T = A T − (l I)T
(1)
[{ ( A + B)T = A T + BT ]
= A T − l IT = A T − l I
∴
( A − l I)T = A T − l I
For any square matrix B,
∴
(2)
BT = B
( A − l I)T = A − l I
From (2) and (3),
[{ I T = I]
(3)
A − lI = AT − lI .
This shows that the characteristic polynomial of A and AT are the same.
Hence, the characteristic equations of A and AT is (1).
■
∴
A and AT have the same eigen values.
2. Sum of the eigen values of a square matrix A is equal to the sum of the elements on its main
diagonal.
Proof Let A be a square matrix of order n.
A − lI = 0
Then the characteristic equation of A is
⇒
l n − S1l n −1 + S2 l n − 2 − ... + ( −1) n Sn = 0
(1)
where S1 = sum of the diagonal elements of A
If l1, l2, …, ln are the roots of (1), then l1, l2, …, ln are the eigen values of A.
From theory of equations,
−coefficient of l n −1
sum of the roots of (1) is =
coefficient of l n
⇒
∴
l1+ l2 + … + ln = −(−S1) = S1
the sum of the eigen values = l1+ l2 + … + ln = S1
= sum of the diagonal elements of the matrix A.
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Note Sum of the diagonal elements of A is called the trace of A.
∴
Sum of the eigen values = trace of A
3. Product of the eigen values of a square matrix A is equal to A .
Proof Let A be a square matrix of order n.
Then its characteristic equation is
A − lI = 0
l n − S1l n −1 + S2 l n − 2 − ... + ( −1) n Sn = 0
⇒
(1)
where Sn = A .
If l1, l2, …, ln are the n roots of (1), then from theory of equations,
the product of roots = ( −1) n
constant term
coefficient of l n
⇒
l1 l 2 … l n = ( −1) n ( −1) n Sn = ( −1) 2 n Sn = Sn = A
∴
the product of the eigen values = l1 l 2 … l n = Sn = A .
[{ (−1)2n = 1]
■
Note If at least one eigen value is 0, then A = 0 ∴ A is a singular matrix.
If all the eigen values are non-zero, then A ≠ 0
∴ A is a non-singular if all the eigen values are non-zero.
4. If l1, l2, …, ln are non-zero eigen values of square matrix of order n, then
21
1 1
1
are
,
, …,
l1 l 2
ln
eigen values of A .
Proof Let l be any non-zero eigen value of A, then there exists a non-zero column matrix X such that
AX = lX. Since all the eigen values are non-zero, A is non-singular.
−1
∴ A exists.
−1
−1
∴
A (AX) = A (lX)
⇒
(A A)X = l(A X)
⇒
IX = l(A X)
−1
−1
−1
−1
X = l( A X ) ⇒
⇒
−1
−1
1
1
X = A X ⇒ A X = X.
l
l
[{ l ≠ 0]
−1
1
is an eigen value of A .
l
This is true for all the eigen values of A.
So,
∴
−1
1 1
1
,
, …, are the eigen values of A .
l1 l 2
ln
−1
Note that the eigen vector for A corresponding to
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1
is also X.
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5. If l1, l2, ???, ln are the eigen values of A, then
(i) cl1, cl2, ???, cln are the eigen values of cA, where c ≠ 0
(ii) l1m , l 2m , …, l mn are the eigen values of Am, where m is a positive integer.
Proof
Let l be any eigen value of A, then there exists a non-zero column matrix X such that
AX = lX
(i) Multiply by c ≠ 0 then
c(AX) = c(lX) ⇒ (cA) X = (cl) X
∴ cl is an eigen value of cA.
This is true for all eigen values of A.
∴ cl1, cl2, …, cln are the eigen values of cA.
(ii) Now
A2X = A(AX) = A(lX) = l(AX) = l(lX) = l2X
∴
AX=lX
2
⇒
2
1.69
(1)
[using (1)]
l is an eigen value of A .
2
2
Similarly, A X = A(A X) = A(l2X) = l2(AX) = l2(lX) = l3X
3
2
A3X = l3X ⇒ l3 is an eigen value of A3.
Proceeding in this way, we have AmX = l mX for any positive integer m.
This is true for all eigen values.
∴ l1m , l 2m , …, l mn are the eigen values of Am.
■
6. If l1, l2, …, ln are the eigen values of A, then
(i) l1 2 K, l2 2 K, …, ln 2 K are the eigen values of A 2 KI.
2
2
2
(ii) a 0 l1 1 a1 l1 1 a 2 , a 0 l 2 1 a1 l 2 1 a 2 , …, a 0 l n 1 a1 l n 1 a 2 are the eigen values of
a 0 A 2 1 a1 A 1 a 2 I.
Proof
Let l be any eigen value of A. Then
where X ≠ 0 is a column matrix.
∴
AX = lX
(1)
AX − KX = lX − KX
⇒
(A − KI)X = (l − K)X
∴ l − K is an eigen value of A − KI.
This is true for all eigen values of A.
∴ l1 − K, l2 − K, …, ln − K are the eigen values of A − KI.
(ii) We have AX = lX and A2X = l2X.
∴ a0 (A2X) = a0 (l2X) and a1 (AX) = a1 (lX)
∴
a0 (A2X) + a1 (AX) = a0 (l2X) + a1 (lX)
Adding a2X on both sides, we get
a0(A2X) + a1(AX) + a2X = a0(l2X) + a1(lX) + a2X
⇒
(a0A2 + a1A + a2I)X = (a0l2 + a1l + a2)X
This means a0l2 + a1l + a2 is an eigen value of a0A2 + a1A + a2I.
This is true for all eigen values of A.
∴ a 0 l 12 + a1l 1+ a 2 , a 0 l 22 + a1l 2 + a 2 , …, a 0 l 2n + a1l n + a 2 are the eigen values of
a 0 A 2 + a1A + a 2 I.
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Note
⎡1 0 0 ⎤
⎥
⎢
1. The eigen values of the unit matrix ⎢0 1 0 ⎥ are 1, 1, 1 and the corresponding eigen vectors are
⎢⎣0 0 1 ⎥⎦
⎡1 ⎤ ⎡0 ⎤ ⎡0 ⎤
⎢0 ⎥ , ⎢1 ⎥ , ⎢0 ⎥ ,
⎢ ⎥ ⎢ ⎥ ⎢ ⎥ which are independent.
⎢⎣0 ⎥⎦ ⎢⎣0 ⎥⎦ ⎢⎣1 ⎥⎦
⎡l1 a12 a13 ⎤
⎢
⎥
2. The eigen values of a triangular matrix ⎢ 0 l 2 a23 ⎥ are the main diagonal elements l1, l2, l3.
⎢⎣ 0
0 l 3 ⎥⎦
3. If l is an eigen value of A then AX = lX. We have seen
A2X = l2X, …, AmX = lmX.
Thus, the eigen values of A, A2, …, Am are l, l2, …, lm which are all different. But they all have
the same eigen vector X.
Similarly, l and a0l2 + a1l + a2 are eigen values of A and
a0A2 + a1A + a2I. But they have the same eigen vector X.
WORKED EXAMPLES
EXAMPLE 1
⎡ 1 2 −2 ⎤
⎢
3 ⎥⎥ .
Find the sum and product of the eigen values of the matrix ⎢ 1 0
⎢⎣ −2 −1 −3 ⎥⎦
Solution.
⎡ 1 2 −2 ⎤
3 ⎥⎥
A = ⎢⎢ 1 0
⎢⎣ −2 −1 −3 ⎥⎦
Let
Sum of the eigen values = Sum of the elements on the main diagonal
= 1 + 0 + (−3) = −2
1 2 −2
Product of the eigen values = A = 1 0
3
−2 −1 −3
= 1(0 + 3) − 2 (−3 + 6) − 2(−1 − 0) = 3 − 6 + 2 = −1
EXAMPLE 2
5⎤
⎡ 3 10
−1
⎢
⎥
If 2 and 3 are eigen values of A 5 ⎢22 23 24 ⎥ , find the eigen values of A and A3.
⎢⎣ 3
5
7 ⎥⎦
Solution.
5⎤
⎡ 3 10
Given A = ⎢ −2 −3 − 4 ⎥
⎢
⎥
⎢⎣ 3 5
7⎥⎦
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Also given 2 and 3 are two eigen values of A.
Let l be the 3rd eigen value.
We know, sum of the eigen values = sum of the diagonal elements.
⇒
2 + 3 + l = 3 + (−3) + 7 ⇒ l = 2
So, eigen values of A are 2, 2, 3
−1
1 1 1
∴ the eigen values of A are , , and the eigen values of A3 are 23, 23, 33 ⇒
2 2 3
1.71
8, 8, 27.
EXAMPLE 3
⎡5 4 ⎤
⎡4 ⎤
If ⎢ ⎥ is an eigen vector of the matrix ⎢
⎥ , then find the corresponding eigen value.
⎣1 2⎦
⎣1 ⎦
Solution.
⎡4⎤
⎡5 4 ⎤
and X = ⎢ ⎥ . If l is the eigen value corresponding to an eigen vector X, then
Let A = ⎢
⎥
⎣1 ⎦
⎣1 2 ⎦
(A − lI)X = 0
4 ⎤ ⎡ 4⎤ ⎡0⎤
⎡5 − l
=
⇒ (5 − l ) ⋅ 4 + 4 = 0 ⇒ l = 6
⎢ 1
2 − l ⎥⎦ ⎢⎣1 ⎥⎦ ⎢⎣0 ⎥⎦
⎣
⇒
EXAMPLE 4
⎡3
If A 5 ⎢0
⎢⎣0
1
2
0
4⎤
6 ⎥ , then find the eigen values of A2 2 2A + I.
5 ⎥⎦
⎡3 1 4⎤
A = ⎢0 2 6 ⎥
⎢⎣0 0 5 ⎥⎦
Since A is a triangular matrix, the eigen values are the diagonal elements.
Solution.
Given
∴
3, 2, 5 are the eigen values of A.
∴ the eigen values of A − 2A + I are 32 − 2 ⋅ 3 + 1, 22 − 2 ⋅ 2 + 1, 52 − 2 ⋅ 5 + 1
i.e., the eigen values of A2 − 2A + I are 4, 1, 16.
2
EXAMPLE 5
⎡ 6
The product of two eigen values of the matrix A 5 ⎢22
⎢⎣ 2
value.
Solution.
Let l1, l2, l3 be the eigen values of A.
Given
l1 ⋅ l2 = 16
We know that
l1 ⋅ l2 ⋅ l3 = A
⇒
⇒
22
3
21
2⎤
21⎥ is 16. Find the third eigen
3 ⎥⎦
6 −2 2
16l 3 = −2
3 −1 = 6(9 − 1) + 2( −6 + 2) + 2( 2 − 6)
2 −1 3
16l 3 = 48 − 8 − 8 = 32 ⇒ l 3 = 2
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EXAMPLE 6
Find the eigen values of the matrix ⎡ 1
⎢⎣25
1
and 21.
6
22 ⎤ . Hence, find the matrix whose eigen values are
4 ⎥⎦
Solution.
1 −2⎤
.
Let A = ⎡⎢
4 ⎥⎦
⎣ −5
The characteristic equation of A is A − lI = 0
1− l
−2
= 0 ⇒ l 2 − S1l + S2 = 0
−5 4 − l
⇒
∴
where S1 = 1 + 4 = 5 and S2 = A = 4 − 10 = −6
the characteristic equation is l2 − 5l − 6 = 0 ⇒
(l − 6) (l + 1) = 0 ⇒ l = 6, −1
−1
1
Since 6, −1 are the eigen values of A, by property (4), , − 1 are the eigen values of A .
6
−1
So, the required matrix is A =
1
1 ⎡ 4 5⎤
adj A = − ⎢
6 ⎣ 2 1⎥⎦
A
T
1 ⎡ 4 2⎤
=− ⎢
6 ⎣ 5 1 ⎥⎦
EXAMPLE 7
⎡ 3 21⎤
If a, b are the eigen values of ⎢
form the matrix whose eigen values are a3, b3.
5 ⎥⎦
⎣21
Solution.
⎡ 3 −1⎤
Let A = ⎢
⎥.
⎣ −1 5⎦
Since a, b are the eigen values of A, by property 5(ii), a3, b 3 are the eigen values of A3.
⎡ 3 −1⎤ ⎡ 3 −1⎤ ⎡10 −8⎤
A2 = A ⋅ A = ⎢
⎥ ⎢
⎥=⎢
⎥
⎣ −1 5⎦ ⎣ −1 5⎦ ⎣ −8 26 ⎦
⎡ 10 −8⎤ ⎡ 3 −1⎤ ⎡ 38 −50 ⎤
A3 = A 2 A = ⎢
⎥ ⎢
⎥=⎢
⎥
⎣ −8 26 ⎦ ⎣ −1 5⎦ ⎣ −50 138⎦
Now
and
EXERCISE 1.5
Find eigen values and eigen vectors of the following matrices.
⎡ 2
1. ⎢ 2
⎢⎣ −7
2
1
2
⎡ 6
5. ⎢ −2
⎢⎣ 2
−2
3
−1
0⎤
1⎥
−3⎥⎦
2⎤
−1⎥
3⎥⎦
⎡ 4 −20 −10 ⎤
⎢
4 ⎥⎥
2. ⎢ −2 10
⎢⎣ 6 −30 −13⎥⎦
⎡ −2
3. ⎢ 2
⎢⎣ −1
2
1
−2
−3⎤
−6 ⎥
0 ⎥⎦
⎡ 7 −2 0 ⎤
4. ⎢ −2 6 −2⎥
⎢
⎥
⎢⎣ 0 −2
5⎥⎦
⎡1 1 3⎤
⎢
⎥
6. ⎢1 5 1⎥
⎢⎣3 1 1⎥⎦
⎡ 2
7. ⎢ −1
⎢⎣ 1
−1
2
−1
1⎤
−1⎥
2⎥⎦
5⎤
⎡ 3 10
⎢
⎥
8. ⎢ −2 −3 −4 ⎥
⎢⎣ 3 5
7⎥⎦
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⎡ −5 −5 −9⎤
9 18⎥⎥
⎢⎣ −2 −3 −7⎥⎦
9. ⎢⎢ 8
1.73
⎡ 2 1 1⎤
10. ⎢⎢1 2 1⎥⎥
⎢⎣0 0 1⎥⎦
ANSWERS TO EXERCISE 1.5
⎡ 2⎤ ⎡ 2⎤ ⎡ 1⎤
1. l = 1, 3, −4; eigen vectors ⎢⎢ −1⎥⎥ , ⎢⎢ 1⎥⎥ , ⎢⎢ −3⎥⎥
⎣⎢ −4 ⎥⎦ ⎢⎣ −2⎥⎦ ⎢⎣ 13⎥⎦
⎡ 5 ⎤ ⎡ 2⎤ ⎡ 0 ⎤
⎢ ⎥ ⎢ ⎥ ⎢ ⎥
2. l = 0, −1, 2; eigen vectors ⎢1 ⎥ , ⎢0 ⎥ , ⎢ 1⎥
⎢⎣0 ⎥⎦ ⎢⎣1 ⎥⎦ ⎢⎣ −2⎥⎦
⎡ 2⎤ ⎡ 3⎤ ⎡ −1⎤
3. l = −3, −3, 5; eigen vectors ⎢ −1⎥ , ⎢0 ⎥ , ⎢ −2⎥
⎢⎣ 0 ⎥⎦ ⎢⎣1 ⎥⎦ ⎢⎣ 1⎥⎦
⎡ 1 ⎤ ⎡ 2⎤ ⎡ 2⎤
4. l = 3, 6, 9; eigen vectors ⎢ 2⎥ , ⎢ 1⎥ , ⎢ −2⎥
⎢⎣ 2⎥⎦ ⎢⎣ −2⎥⎦ ⎢⎣ 1⎥⎦
⎡ 1⎤ ⎡1 ⎤ ⎡ 2⎤
⎢ ⎥ ⎢ ⎥ ⎢ ⎥
5. l = 2, 2, 8; eigen vectors ⎢ 0 ⎥ , ⎢ 2⎥ , ⎢ −1⎥
⎢⎣ −2⎥⎦ ⎢⎣0 ⎥⎦ ⎢⎣ 1⎥⎦
⎡ −1⎤ ⎡ 1⎤ ⎡1 ⎤
6. l = −2, 3, 6; eigen vectors ⎢⎢ 0 ⎥⎥ , ⎢⎢ −1⎥⎥ , ⎢⎢ 2⎥⎥
⎢⎣ 1⎥⎦ ⎢⎣ 1⎥⎦ ⎢⎣1 ⎥⎦
⎡1 ⎤ ⎡0 ⎤ ⎡ 1⎤
⎢ ⎥ ⎢ ⎥ ⎢ ⎥
7. l = 1, 1, 4; eigen vectors ⎢1 ⎥ , ⎢1 ⎥ , ⎢ −1⎥
⎢⎣0 ⎥⎦ ⎢⎣1 ⎥⎦ ⎢⎣ 1⎥⎦
⎡ 1⎤
⎡ 5⎤
8. l = 3, 2, 2; eigen vectors X1 = ⎢⎢ 1⎥⎥ and X 2 = X 3 = ⎢⎢ 2⎥⎥
⎢⎣ −2⎥⎦
⎢⎣ −5⎥⎦
⎡ 3⎤
⎢ ⎥
=
9. l = −1, −1, −1; eigen vectors X1 = X2 = X3 ⎢ −6 ⎥
⎣⎢ 2⎦⎥
⎡ 1⎤
⎡ 0⎤
⎡1 ⎤
10. l = 1, 1, 3; eigen vectors X1 = ⎢ 0 ⎥ , X 2 = ⎢ 1⎥ , X 3 = ⎢1 ⎥
⎢ ⎥
⎢ ⎥
⎢ ⎥
⎢⎣ −1⎥⎦
⎢⎣ −1⎥⎦
⎢⎣0 ⎥⎦
1.6.5
Cayley-Hamilton Theorem
Theorem 1.4 Every square matrix satisfies its characteristic equation
Proof
Let A = [ aij ]n × n be a square matrix of order n.
Then the Characteristic polynomial is
a11 − l
A − lI =
a21
:
an1
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a12
a13 …
a22 − l a23 …
an 2
a1n
a2 n
an3 … ann − l
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The coefficient of ln is (−1)n from the product of (a11 − l) (a22 − l)... (ann − l).
A − lI = ( −1) n ⎡⎣l n + a1l n −1 + a2 l n − 2 + … + an ⎤⎦
∴ let
(1)
Since the elements of A − lI are at most first degree in l, the elements of adj (A − lI) are ordinary
polynomials in l of degree at most (n − 1).
∴ adj (A − lI) can be written as a matrix polynomial in l of degree n − 1.
Let
adj( A − lI ) = B 0 l n −1 + B1l n − 2 + B 2 l n − 3 + … B n n −1
(2)
where B0, B1,…, Bn − 1 are n × n matrices.
The elements of these matrices are function of aij.
We know that if A is a n × n matrix, then A (adj A ) = A I = (adj A) A
where I is n × n identity matrix.
∴ we have ( A − lI )adj( A − lI ) = A − lI = I
Substituting from (1) and (2), we have
( A − lI )[B 0 l n −1 + B1l n − 2 + B 2 l n − 3 + … + B n − 2 l + B n −1 ]
= ( −1) n (l n + a1l n −1 + a2 l n − 2 + …an −1l + an )
Equating the coefficients of lm, ln − 1, ln − 2… l and term independent of l, we get
−IB0 = (−1)n,
AB1 − IB2 = (−1)n a2I, … ,
AB0 − IB0 = (−1)n a1I,
ABn − 2 − IBn − 1 = (−1) an − 1 I,
ABn − 1= (−1)n anI
Pre-multiplying the above equation by An, An − 1, An − 2, …, A, I, we get
− An B0 = ( −1) n An
An B0 − An −1 B1 = ( −1) n a1 An −1
An −1 B1 − An − 2 B2 = ( −1) n a2 An − 2
:
A Bn − 2 − ABn −1 = ( −1) n an −1 A
2
ABn −1 = ( −1) n an I
Adding we get,
( −1) n [A n + a1A n −1 + a2 A n − 2 + ... an I ] = 0
⇒
A n + a1A n −1 + a2 A n − 2 + ... an I = 0
This means A satisfies the equation l n + a1l n −1 + a2 l n − 2 + … + an = 0 ,
which is the characteristic equation of A.
Hence, the theorem.
Properties: Cayley–Hamilton Theorem has the following two important properties:
1. To find the inverse of a non-singular matrix A
2. To find higher integral power of A
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1.75
WORKED EXAMPLES
EXAMPLE 1
2⎤
⎡1
4
Verify that A 5 ⎢
⎥ satisfies its characteristic equation and hence find A .
⎣ 2 21⎦
Solution.
⎡ 1 2⎤
Given
A=⎢
⎥
⎣ 2 −1⎦
A − lI = 0
The characteristic equation of A is
⇒
l 2 − S1l + S2 = 0
where S1 = 1 + ( −1) = 0 and S2 = A = −1 − 4 = −5
∴ the characteristic equation is
l2 − 5 = 0
(1)
By Cayley-Hamilton theorem, A satisfies (1). That is A − 5I = 0
We shall now verify this by direct computation.
2
(2)
⎡1 2 ⎤ ⎡1 2 ⎤ ⎡ 5 0 ⎤
A2 = A ⋅ A = ⎢
⎥
⎥=⎢
⎥⎢
⎣ 2 −1⎦ ⎣ 2 −1⎦ ⎣0 5⎦
⎡1 0 ⎤
I=⎢
⎥
⎣0 1 ⎦
∴
⎡1 0⎤ ⎡ 5 0⎤ ⎡ 5 0⎤ ⎡0 0⎤
⎡5 0⎤
A 2 − 5I = ⎢
− 5⎢
⎥
⎥=⎢
⎥=⎢
⎥−⎢
⎥
⎣ 0 1 ⎦ ⎣ 0 5⎦ ⎣ 0 5⎦ ⎣ 0 0 ⎦
⎣0 5⎦
2
A − 5I = 0.
⇒
Hence, A satisfies its characteristic equation.
2
To find A4: We have
A = 5I
∴
⎡ 5 0 ⎤ ⎡ 25 0 ⎤
A4 = 5A2 = 5 ⎢
⎥.
⎥=⎢
⎣0 5⎦ ⎣ 0 25⎦
[from (2)]
EXAMPLE 2
⎡1 4 ⎤
Verify Cayley-Hamilton theorem for the matrix A 5 ⎢
⎥ and find its inverse. Also express
⎣2 3 ⎦
A5 2 4A4 2 7A3 1 11A2 2 A 2 10I as a linear polynomial in A.
Solution.
⎡1 4 ⎤
A=⎢
⎥
⎣ 2 3⎦
Given
The characteristic equation of A is
where
A − lI = 0
⇒
l 2 − S1l + S2 = 0
S1 = 1 + 3 = 4, S2 = A = 3 − 8 = −5
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∴ the characteristic equation is l2 − 4l − 5 = 0
By Cayley-Hamilton theorem, A satisfies (1).
(1)
∴
(2)
A2 − 4A − 5I = 0
We shall now verify this by direct computations.
⎡1 4 ⎤ ⎡1 4 ⎤ ⎡ 9 16 ⎤
A2 = A ⋅ A = ⎢
⎥
⎥=⎢
⎥⎢
⎣ 2 3 ⎦ ⎣ 2 3⎦ ⎣8 17⎦
⎡1 4 ⎤
⎡1 0 ⎤
⎡9 16 ⎤
−4⎢
−5⎢
A 2 − 4A − 5I = ⎢
⎥
⎥
⎥
⎣8 17⎦
⎣ 2 3⎦
⎣0 1 ⎦
⎡9 16 ⎤ ⎡ 4 16 ⎤ ⎡ 5 0 ⎤ ⎡9 − 4 − 5 16 − 16 − 0 ⎤ ⎡0 0 ⎤
=⎢
⎥−⎢
⎥−⎢
⎥=⎢
⎥=⎢
⎥
⎣8 17⎦ ⎣ 8 12⎦ ⎣ 0 5⎦ ⎣ 8 − 8 − 0 17 − 12 − 5 ⎦ ⎣0 0 ⎦
A2 − 4A − 5I = 0.
⇒
Hence, the theorem is verified.
21
To find A : We have
−1
Multiply by A , we get
5I = A2 − 4A
−1
−1
−1
5A = A A2 − 4 A A
⎡1 0⎤
⎡1 4 ⎤
5 A−1 = A − 4I = ⎢
⎥ − 4 ⎢0 1⎥
2
3
⎦
⎣
⎦
⎣
⎡1 4 ⎤ ⎡ 4 0 ⎤ ⎡1 − 4 4 − 0 ⎤ ⎡ −3 4 ⎤
=⎢
⎥=⎢
⎥−⎢
⎥
⎥=⎢
⎣ 2 3 ⎦ ⎣ 0 4 ⎦ ⎣ 2 − 0 3 − 4 ⎦ ⎣ 2 −1⎦
1 ⎡ −3 4 ⎤
5 ⎢⎣ 2 −1⎥⎦
Finally, to find A5 − 4A4 − 7A3 + 11A2 − A − 10I:
Consider the polynomial l5 − 4l4 − 7l3 + 11l2 − l − 10
Divide the polynomial (3) by l2 − 4l − 5.
Division is indicated below.
l 3 − 2l + 3
∴
A −1 =
(3)
l 2 − 4l − 5 l 5 − 4l 4 − 7l 3 + 11l 2 − l − 10
l 5 − 4l 4 − 5l 3
− 2l 3 + 11l 2 − l
−2l 3 + 8l 2 + 10l
3l 2 − 11l − 10
3l 2 − 12l − 15
l+5
∴ We get the quotient l3 − 2l + 3 and remainder l + 5.
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∴
1.77
l5 − 4l4 − 7l3 + 11l2 − l − 10 = (l2 − 4l − 5)(l3 − 2l + 3) + l + 5
Replace l by A, we get
A5 − 4A4 − 7A3 + 11A2 − A − 10I = (A2 − 4A − 5I) (A3 − 2A + 3I) + A + 5I
= 0 + A + 5I = A + 5I
[using (2)]
which is a linear polynomial in A.
EXAMPLE 3
1⎤
⎡ 2 21
⎢
Find the characteristic equation of the matrix A given A 5 ⎢21
2 21⎥⎥ . Hence, find A21 and A4.
2 ⎥⎦
⎣⎢ 1 21
Solution.
⎡ 2 −1 1⎤
Given
A = ⎢⎢ −1 2 −1⎥⎥
⎢⎣ 1 −1 2⎥⎦
The characteristic equation of A is A − lI = 0
2−l
−1
1
−1 2 − l
−1 = 0
1
−1 2 − l
⇒
⇒
l 3 − S1l 2 + S2 l − S3 = 0
where S1 = sum of the diagonal elements of A = 2 + 2 + 2 = 6
S2 = sum of the minors of the diagonal elements of A
=
2 −1
2 1
2 −1
+
+
= 4 −1+ 4 −1+ 4 −1 = 9
1 2
−1 2
−1 2
S3 = A = 2( 4 − 1) + ( −2 + 1) + (1 − 2) = 6 − 1 − 1 = 4
∴
the characteristic equation is l3 − 6l2 + 9l − 4 = 0
By Cayley-Hamilton theorem, A satisfies its characteristic equation
∴
A3 − 6A2 + 9A − 4I = 0
⇒
4I = A3 − 6A2 + 9A
4IA−1 = A3A−1 − 6A2 ⋅ A−1 + 9A A−1
Multiply by A−1,
⇒
But
(1)
4A−1 = A2 − 6A + 9I
⎡ 2 −1 1⎤ ⎡ 2 −1 1⎤
A 2 = ⎢⎢ −1 2 −1⎥⎥ ⎢⎢ −1 2 −1⎥⎥
⎢⎣ 1 −1 2⎥⎦ ⎢⎣ 1 −1 2⎥⎦
2 + 1 + 2⎤ ⎡ 6 −5 5⎤
⎡ 4 + 1 + 1 −2 − 2 − 1
⎢
= ⎢ −2 − 2 − 1
1 + 4 + 1 −1 − 2 − 2⎥⎥ = ⎢⎢ −55 6 −5⎥⎥
1 + 1 + 4 ⎥⎦ ⎢⎣ 5 −5 6 ⎥⎦
⎣⎢ 2 + 1 + 2 −1 − 2 − 2
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⎡1 0 0⎤
⎡ 2 −1 1⎤
⎡ 6 −5 5⎤
⎥
⎥
⎢
⎢
4 A = ⎢ −5 6 −5⎥ − 6 ⎢ −1 2 −1⎥ + 9 ⎢⎢ 0 1 0 ⎥⎥
⎢⎣ 0 0 1 ⎥⎦
⎢⎣ 1 −1 2⎥⎦
⎢⎣ 5 −5 6 ⎥⎦
−5 + 6
5 − 6 ⎤ ⎡ 3 1 −1⎤
⎡6 − 12 + 9
⎢
= ⎢ −5 + 6 6 − 12 + 9
−5 + 6 ⎥⎥ = ⎢⎢ 1 3 1⎥⎥
⎢⎣
−5 + 6 6 − 12 + 9⎥⎦ ⎢⎣ −1 1 3⎥⎦
5−6
−1
∴
∴
⎡ 3 1 −1⎤
1⎢
A = ⎢ 1 3 1⎥⎥
4
⎢⎣ −1 1 3⎥⎦
(1) ⇒
A 3 = 6 A 2 − 9A + 4I
∴
A 4 = 6 A 3 − 9A 2 + 4 A
−1
[Multiplying by A]
= 6[6 A − 9A + 4I] − 9A + 4A
2
2
= 27A 2 − 50A + 24I
⎡ 2 −1 1⎤
⎡1 0 0⎤
⎡ 6 −5 5⎤
⎥
⎢
⎥
⎢
= 27 ⎢ −5 6 −5⎥ − 50 ⎢ −1 2 −1⎥ + 24 ⎢⎢ 0 1 0 ⎥⎥
⎢⎣ 1 −1 2⎥⎦
⎢⎣ 0 0 1 ⎥⎦
⎢⎣ 5 −5 6 ⎥⎦
−135 + 50
135 − 50 ⎤ ⎡ 86 −85 85⎤
⎡162 − 100 + 24
= ⎢⎢ −135 + 50
−135 + 50 ⎥⎥ = ⎢⎢ −85 86 −85⎥⎥
162 − 100 + 24
⎢⎣ 135 − 50
−135 + 50
162 − 100 + 24 ⎥⎦ ⎢⎣ 85 −85 86 ⎥⎦
EXAMPLE 4
Use Cayley-Hamilton theorem to find the matrix
⎡2 1 1 ⎤
⎢
A 2 5A 1 7A 2 3A 1 8A 2 5A 1 8A 2 2A 1 I if the matrix A 5 ⎢0 1 0 ⎥⎥ .
⎢⎣1 1 2 ⎥⎦
Solution.
⎡2 1 1⎤
Given
A = ⎢⎢0 1 0 ⎥⎥
⎢⎣1 1 2⎥⎦
8
7
6
5
4
The characteristic equation is
⇒
3
2
A − lI = 0
2−l
1
1
0
1− l
0 = 0 ⇒ l 3 − S1l 2 + S2 l − S3 = 0
1
1
2−l
where S1 = 2 + 1 + 2 = 5
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S2 =
1.79
1 0
2 1
2 1
+
+
= 2 + 4 −1+ 2 = 7
1 2
1 2
0 1
S3 = A = 2 ⋅ 2 − 1⋅ 0 + 1( −1) = 4 − 1 = 3
∴ the characteristic equation is
By Cayley-Hamilton theorem, we get
l3 − 5l2 + 7l − 3 = 0
A3 − 5A2 + 7A − 3I = 0
(1)
We have to find the matrix
A8 − 5A7 + 7A6 − 3A5 + 8A4 − 5A3 + 8A2 − 2A + I = f(A), say
We shall rewrite this matrix polynomial in terms of
A3 − 5A2 + 7A − 3I
∴
the polynomial
But
∴
f(A) = A5(A3 − 5A2 + 7A − 3I) + 8A4 − 5A3 + 8A2 − 2A + I
= 8A4 − 5A3 + 8A2 − 2A + I
= 8A(A3 − 5A2 + 7A − 3I) + 35A3 − 48A2 + 22A + I
= 35A3 − 48A2 + 22A + I
= 35(A3 − 5A2 + 7A − 3I) + 127A2 − 223A + 106I
= 127A2 − 223A + 106I
⎡2 1 1⎤ ⎡2 1 1⎤ ⎡5 4 4⎤
2
A = ⎢⎢0 1 0 ⎥⎥ ⎢⎢0 1 0 ⎥⎥ = ⎢⎢ 0 1 0 ⎥⎥
⎣⎢1 1 2⎦⎥ ⎣⎢1 1 2⎦⎥ ⎣⎢ 4 4 5 ⎦⎥
[Using (1)]
[Using (1)]
[Using (1)]
⎡1 0 0⎤
⎡2 1 1⎤
⎡5 4 4⎤
⎥
⎥
⎢
⎢
f ( A ) = 127 ⎢ 0 1 0 ⎥ − 223 ⎢0 1 0 ⎥ + 106 ⎢⎢0 1 0 ⎥⎥
⎢⎣0 0 1 ⎥⎦
⎢⎣1 1 2⎥⎦
⎢⎣ 4 4 5 ⎥⎦
508 − 223
508 − 223 ⎤ ⎡ 295 285 285⎤
⎡635 − 446 + 106
⎥ = ⎢ 0 10
⎢
=⎢
0
127 − 223 + 106
0
0 ⎥⎥
⎥ ⎢
⎢⎣ 508 − 223
508 − 223
635 − 446 + 106 ⎥⎦ ⎢⎣ 285 285 295⎥⎦
Note: Otherwise divide l8 − 5l7 + 7l6 − 3l5+ 8l4− 5l3 + 8l2 − 2l + 1 by l3 − 5l2 + 7l − 3 and
proceed as in example 2.
EXAMPLE 5
⎡1 0 0 ⎤
If A 5 ⎢⎢1 0 1 ⎥⎥ , then show that An 5 An 2 2 1 A2 2 I for n $ 3. Hence, find A50.
⎢⎣0 1 0 ⎥⎦
Solution.
⎡1 0 0 ⎤
Given
A = ⎢⎢1 0 1 ⎥⎥ .
⎢⎣0 1 0 ⎥⎦
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The characteristic equation of A is A − lI = 0
⇒
where
l 3 − S1l 2 + S2 l − S3 = 0
S1 = 1, S2 =
0 1
1 0
1 0
+
+
= −1 + 0 + 0 = −1
1 0
0 0
1 0
S3 = A = −1
and
∴ the characteristic equation is l − l − l + 1 = 0
By Cayley-Hamilton theorem, A statisfies (1)
(1)
∴
A3 − A2 − A + I = 0
⇒
A3 − A2 = A − I
Multiplying (2) by A, A2, …, An − 3, we get the equations
(2)
3
2
A 4 − A3 = A 2 − A
A5 − A 4 = A3 − A 2
A6 − A5 = A 4 − A3
:
:
A n − A n −1 = A n − 2 − A n − 3
Adding all these equations, we get
An − A2 = An−2 − I
⇒
A n = A n − 2 + ( A 2 − I) for all n ≥ 3
A
A
n−2
n−4
=A
n−4
=A
n −6
:
(i)
+A −I
2
+ A2 − I
:
:
∴
A n = ( A n − 4 + A 2 − I) + ( A 2 − I)
⇒
A n = A n − 4 + 2( A 2 − I)
(ii)
=A
n −6
+ A − I + 2( A − I)
⇒
A =A
n −6
+ 3( A 2 − I)
(iii)
⇒
A =A
n −8
+ 4( A − I)
(iv)
n
n
A
If n is even, then
2
2
2
A
⎛ n − 2⎞ 2
A n = A n −( n − 2) + ⎜
( A − I)
⎝ 2 ⎟⎠
⎡
2
4
2
⎢observe the coefficients of A − I in (i), (ii), (iii) … and index of A. We see 2 = 1 in (i), 2 = 2 in
⎣
6
n−2
⎤
8
(ii), = 3 in (iii), = 4 in (iv) and so on
in the last one ⎥
2
2
2
⎦
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⎛ n − 2⎞ 2 n − 2
An = A2 + ⎜
A −
I
⎝ 2 ⎟⎠
2
∴
An =
⇒
1.81
n 2 ⎛ n − 2⎞
A −⎜
I
⎝ 2 ⎟⎠
2
Putting n = 50, we get A50 = 25A2 − 24I
But
∴
⎡1 0 0⎤
A = A ⋅ A = ⎢⎢1 0 1 ⎥⎥
⎢⎣0 1 0 ⎥⎦
2
A
50
⎡1 0 0 ⎤ ⎡1 0 0 ⎤
⎢1 0 1 ⎥ = ⎢1 1 0 ⎥
⎢
⎥ ⎢
⎥
⎢⎣0 1 0 ⎥⎦ ⎢⎣1 0 1 ⎥⎦
⎡1 0 0 ⎤
⎡1 0 0⎤ ⎡ 1 0 0⎤
⎢
⎥
= 25 ⎢1 1 0 ⎥ − 24 ⎢⎢0 1 0 ⎥⎥ = ⎢⎢ 25 1 0 ⎥⎥
⎢⎣1 0 1 ⎥⎦
⎢⎣0 0 1 ⎥⎦ ⎢⎣ 25 0 1 ⎥⎦
EXAMPLE 6
2⎤
⎡1
n
If A 5 ⎢
⎥ , then find A in terms of A and I.
2
2
1
⎣
⎦
Solution.
⎡ 1 2⎤
A=⎢
⎥
⎣ 2 −1⎦
Given
The characteristic equation is
l2 − 5 = 0
[see example 17]
By Cayley-Hamiltons theorem
A2 − 5I = 0
(1)
To find An, consider the polynomial ln
Dividing ln by l2 − 5, we get ln = (l2 − 5) f(l) + al + b
(2)
where f(l) is the quotient and al + b is the remainder.
We shall now find the values of a and b.
l = 5, − 5
The eigen values of A are
( 5)
Substitute l = 5 in (2) then
n
= 0+a 5+b
⇒
a 5 + b = ( 5)
n
(3)
Substitute l = − 5 in (2), then ( − 5 ) = 0 + a ( − 5 ) + b
n
−a 5 + b = ( − 5 )
⇒
n
(4)
2b = ( 5 ) + ( − 5 )
n
(3) + (4) ⇒
( 5 ) + (− 5 )
n
b=
∴
(3) − (4) ⇒ 2 5a = ( 5 ) − ( − 5 )
n
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n
2
n
= ( 5)
n
(1 + ( −1) n )
2
n
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( 5 ) − (− 5 )
n
∴
a=
Replacing l by A in (2), we get
2 5
n
= ( 5)
n
(1 − ( −1) n )
2
A n = ( A 2 − 5I)f( A ) + aA + bI = 0 + aA + bI
∴
n ⎛ 1 − ( −1) n ⎞
n ⎛ 1 + ( −1) n ⎞
An = ( 5) ⎜
A + ( 5) ⎜
⎟
⎟⎠ I.
⎠
⎝
⎝
2
2
EXERCISE 1.6
Verify Cayley-Hamilton theorem for the following matrices and hence find their inverses.
⎡ 1 3 7⎤
1. ⎢ 4 2 3⎥
⎢
⎥
⎢⎣ 1 2 1⎥⎦
3⎤
⎡1 0
2. ⎢ 2 1 −1⎥
⎢
⎥
⎢⎣ 1 −1 1⎥⎦
3⎤
⎡ −1 0
⎢ 8 1 −7⎥
3. ⎢
⎥
⎢⎣ −3 0
8⎥⎦
⎡ 7 2 −2⎤
⎢
⎥
4. ⎢ −6 −1 2⎥
⎢⎣ 6 2 −1⎥⎦
⎡ 2 −1 2⎤
5. Verify that the matrix A = ⎢⎢ −1 2 −1⎥⎥ satisfies its characteristic equation and hence find A4.
⎢⎣ 1 −1 2⎥⎦
⎡ 7 3⎤
n
3
6. A = ⎢
⎥ , find A in terms of A and I using Cayley-Hamilton theorem and hence find A .
⎣2 6⎦
⎡ 1 2 3⎤
4
7. Find A using Cayley-Hamilton theorem for the matrix A = ⎢⎢ 2 −1 4 ⎥⎥ . Find A4 + A3 − 18A2 − 39A + 2I
⎢⎣ 3 1 −1⎥⎦
8. Find the eigen values and eigen vectors of the system of equations 10x1 + 2x2 + x3 = lx1, 2x1 + 10x2 + x3 =
lx2, 2x1 + x2 + 10x3 = lx3
[Hint: Equations can be rewritten as (10 − l)x1 + 2x2 + x3 = 0, 2x1 + (10 − l)x2 + x3 = 0,
2x1 + x2 + (10 − l)x3 = 0
⎡10 2 1 ⎤
⎡ x1 ⎤
If A = ⎢⎢ 2 10 1 ⎥⎥ and X = ⎢⎢ x 2 ⎥⎥ then these equations in matrix form is (A − lI)X = 0 and so A − lI = 0
⎢⎣ 2 1 10 ⎥⎦
⎢⎣ x 3 ⎥⎦
is the characteristic equation of A and l = 8, 9, 13. Eigen vectors are given by (I)]
A
is an eigen value of the matrix adj A.
l
⎡
A ⎤
(
X
⇒ (adj A )X =
AX = lX ⇒ (adj A) (AX) = (adj A) (lX) A X = l(adj A )X
⎢Hint:
l ⎥⎦
⎣
9. If l is an eigen value of a non-singular matrix A, show that
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1.83
⎡ 1 2⎤
10. Verify Cayley-Hamilton theorem for the matrix A = ⎢
⎥ and find its inverse and also find
⎣ −1 3⎦
A6 − 4A5 + 8A4 − 12A3 + 14A2.
⎡ 13 −3 5⎤
⎢
⎥
11. Verify Cayley-Hamilton theorem and find the inverse of A = ⎢ 0 4 0 ⎥
⎢⎣ −15 9 −7⎥⎦
ANSWERS TO EXERCISE 1.6
1.
⎡ −4 11 −5⎤
1 ⎢
−1 −6
25⎥⎥
35 ⎢
⎢⎣ 6
1 −10 ⎥⎦
⎡ −3 −2 2⎤
1⎢
4.
6
5 −2⎥⎥
3⎢
⎢⎣ −6 −2
5⎥⎦
6. A n =
9n − 4 n
5
1 ⎡3 −2⎤ ⎡ 1 −8⎤
10. A = ⎢
,
5 ⎣1 −1⎥⎦ ⎢⎣ 4 −7⎥⎦
⎡0 3 3⎤
1⎢
3 2 −7⎥⎥
9⎢
⎢⎣ 3 −1 −1⎥⎦
3.
⎡ 8 0 −3⎤
⎢ −43 1 17⎥
⎢
⎥
⎢⎣ 3 0 −1⎥⎦
⎡ 124 −123 162⎤
⎢
96 −123⎥⎥
5. A = ⎢ −95
⎢⎣ 95 −95 124 ⎥⎦
4
⎡ 7 3⎤ 9 ⋅ 4 n − 4 ⋅ 9 n
⎢2 6⎥ +
5
⎣
⎦
⎡ 248 101 218⎤
⎢
⎥
7. A 4 = ⎢ 272 109 50 ⎥
⎢⎣104 98 204 ⎥⎦
−1
2.
⎡ 463 399⎤
⎡1 0 ⎤
3
⎢0 1 ⎥ ; A = ⎢ 266 330 ⎥
⎣
⎣
⎦
⎦
⎡ 3⎤
⎢ ⎥
8. ⎢ −2⎥ ,
⎢⎣ 2⎥⎦
⎡ 1⎤
⎢ 1⎥ ,
⎢ ⎥
⎢⎣ −3⎥⎦
⎡1⎤
⎢1⎥
⎢⎥
⎢⎣1⎥⎦
5⎤
⎡ 7 −6
1 ⎢
11. A =
0
4
0 ⎥⎥
64 ⎢
⎢⎣15 18 −13⎥⎦
−1
1.7 SIMILARITY TRANSFORMATION AND ORTHOGONAL TRANSFORMATION
1.7.1 Similar Matrices
Definition 1.35 Let A and B be square matrices of order n. A is said to be similar to B if there exists a
non-singular matrix P of order n such that
A = P−1BP
(1)
The transformation (1) which transforms B into A is called a similarity transformation. The matrix P
is called a similarity matrix.
Note We shall now see that if A is similar to B then B is similar to A.
A = P−1BP ⇒ PA P−1 = B
(Premultiplying by P and postmultiplying by P−1)
⇒
(P−1)−1 A(P−1) = B
⇒
Q−1AQ = B
(2)
where
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The relation (2) means B is similar to A. Thus, if A is similar to B, then B is similar to A.
Hence, we simply say A and B are similar matrices.
An important property of similarity transformations is that they preserve eigen values, which is
proved in the next theorem.
Theorem 1.5 Similar matrices have the same eigen values.
Proof Let A and B be two similar matrices of order n.
Then B = P−1 AP, by definition.
∴ the characteristic polynomial of B is B − lI
Now
⇒
B − lI = P −1AP − lI = P −1AP − lP −1IP
= P −1 ( A − lI)P
= P −1 A − lI P
[{
AB = A B ]
= A − lI P −1 P = A − lI P −1P = A − lI I
⇒
B − lI = A − lI
[{
I = 1]
∴ A and B have the same characteristic polynomial and hence have the same characteristic equation.
So, A and B have the same eigen values.
Note Similar matrices A and B have the same determinant value i.e., A = B .
For
1.7.2
B = P −1AP ⇒
B = P −1AP = P −1 A P = A P −1P = A I = A
Diagonalisation of a Square Matrix
Definition 1.36 A square matrix A is said to be diagonalisable if there exists a non-singular matrix P
such that P−1 AP = D, where D is a diagonal matrix. The matrix P is called a modal matrix of A.
The next theorem provides us with a method of diagonalisation.
Theorem 1.6 If A is a square matrix of order n, having n linearly independent eigen vectors and M
is the matrix whose columns are the eigen vectors of A, then M−1 AM = D, where D is the diagonal
matrix whose diagonal elements are the eigen values of A.
Proof Let X1, X2, …, Xn be n linearly independent eigen vectors of A corresponding to the eigen
values l1, l2, …, ln of A.
∴
AXi = liXi, i = 1, 2, 3, …, n.
Let M = [X1 X2 … Xn] be the matrix formed with the eigen vectors as columns.
Then
AM = [AX1 AX2 AX3 … AXn]
= [l1X1 l2X2 l3X3 … lnXn]
0 0
0⎤
⎡ l1
⎢ 0 l
0 …⎥⎥
2
⎢
0 l 3 …⎥
= [ X1 X 2 … X n ] ⎢ 0
⎢
⎥
: :
: ⎥
⎢:
⎢ 0
0 0 l n ⎥⎦
⎣
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⇒
AM = MD,
1.85
0
0⎤
⎡ l1
⎢ 0 l
0 ⎥⎥
2
where D = ⎢
⎢: :
:⎥
⎥
⎢
⎣ 0 … ln ⎦
M −1AM = D
⇒
The matrix M which diagonalises A is called the modal matrix of A and the resulting diagonal matrix
D whose elements are eigen values of A is called the spectral matrix of A.
1.7.3
Computation of the Powers of a Square Matrix
Diagonalisation of a square matrix A is very useful to find powers of A, say Ar.
By the theorem 1.6,
D = M−1AM
∴
D 2 = ( M −1AM ) ( M −1AM )
= M −1A( MM −1 )AM = M −1AIAM = M −1A 2 M
Similarly,
D3 = D 2 D
= ( M −1A 2 M ) ( M −1AM )
= M −1A 2 ( MM −1 )AM = M −1A 2 IAM = M −1A 3 M
Proceeding in this way, we can find
D r = M −1A r M
∴
⎡l1r
0
⎢
r
0
l
2
A r = MD r M −1 , where D r = ⎢
⎢… …
⎢
0
⎢⎣ 0
… 0⎤
⎥
… 0⎥
… …⎥
⎥
… l rn ⎥⎦
Note
(1) If the eigen values l1, l2, …, ln of A are different then the corresponding eigen vectors X1,
X2, …, Xn are linearly independent by theorem 1.2.1 (2). So, A can be diagonalised.
(2) Even if 2 or more eigen values are equal, if we can find independent eigen vectors corresponding
to them (see worked example 6), then A can be diagonalised.
Thus, independence of eigen vectors is the condition for diagonalisation.
Working rule to diagonalise a n 3 n matrix A by similarity transformation:
Step 1: Find the eigen values l1, l2, …, ln
Step 2: Find linearly independent eigen vectors X1, X2, …, Xn
Step 3: Form the modal matrix M = [X1 X2 … Xn]
Step 4: Find M−1 and AM.
0 0 … 0⎤
⎡ l1
⎢ 0 l
0 … 0 ⎥⎥
2
Step 5: Compute M−1AM = D = ⎢
⎢ :
: : :
:⎥
⎢
⎥
0 … … ln ⎦
⎣ 0
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1.7.4
Orthogonal Matrix
Engineering Mathematics
Definition 1.37 A real square matrix A is said to be an orthogonal matrix if AAT = ATA = I, where I
is the unit matrix of the same order as A.
From this definition it is clear that AT = A−1.
So, an orthogonal matrix is also defined as below.
Definition 1.38 A real square matrix A is orthogonal if AT = A−1.
EXAMPLE 1.25
⎡ cos u sin u ⎤
Prove that A 5 ⎢
⎥ is orthogonal.
⎣2sin u cos u⎦
Solution.
Given
⎡ cos u sin u ⎤
A=⎢
⎥
⎣ − sin u cos u⎦
∴
⎡ cos u sin u ⎤
2
2
A =⎢
⎥ = cos u + sin u = 1
⎣ − sin u cos u⎦
∴
A is non-singular.
−1
A =
Hence,
T
adj A ⎡ cos u sin u ⎤
T
=⎢
⎥ =A
sin
u
cos
u
−
A
⎦
⎣
[{
A = 1]
∴ A is orthogonal.
1.7.5
Properties of Orthogonal Matrix
1. If A is orthogonal, then AT is orthogonal.
Proof Given A is orthogonal.
∴ AAT = ATA = I
Reversing the roles of A and AT, we see ATA = AAT = I
⇒
AT is orthogonal.
Note Since AT = A−1, it follows A−1 is orthogonal.
2. If A is an orthogonal matrix, then A 561.
Proof Given A is orthogonal. Then AAT = 1
⇒
But we know
∴
AA T = 1
⇒
A AT = 1
AT = A
A A =1 ⇒
A = 1 ⇒ A = ±1
2
3. If l is an eigen value of an orthogonal matrix A, then
Proof Given A is orthogonal and l is an eigen value of A.
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is also an eigen value of A.
l
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1.87
Then AT = A−1. By property (4) of eigen values,
1
is an eigen value of A−1 and so an eigen value of AT.
l
By property (1), A and AT have same eigen values.
1
1
∴
is an eigen of A and hence l, are eigen values of orthogonal matrix A.
l
l
4. If A and B are orthogonal matrices, then AB is orthogonal.
Proof Given A and B are orthogonal matrices.
∴
AT = A−1 and BT = B−1
(AB)T = BT AT = B−1 A−1 = (AB)−1
Now
∴ AB is orthogonal.
5. Eigen values of an orthogonal matrix are of magnitude 1.
Proof Let A be an orthogonal matrix and let l be an eigen value of A.
Then AX = lX, where X ≠ 0
Taking complex conjugate, we get A X = l X
(1)
∴ A=A
But A is real matrix
AX = l X
Hence,
( AX )T = (l X )T
Taking transpose,
T
⇒
Multiplying (2) and (1) we get
X AT = l X
T −1
T
T −1
⇒
X A =lX
T
[{ A T = A −1 ] (2)
T
( X A )( AX ) = (l X )(lX )
T
−1
T
X ( A A )X = ll X X
⇒
T
2
T
2
⇒
T
X X= l X X
⇒
1= l
⇒
[{ X X ≠ 0 as X ≠ 0]
l =1
This is true for all eigen values of A.
Hence, eigen values of A are of absolute value 1.
1.7.6
■
Symmetric Matrix
Definition 1.39 Real Square Matrix
The matrix A = [aij]n × n is said to be symmetric if AT = A.
⎡ 1 −1 3⎤
⎡ 1 −1⎤ ⎢
⎥
Example: ⎢
,
⎥ ⎢ −1 0 4 ⎥ are symmetric matrices.
−
1
0
⎦ ⎢
⎣
⎣ 3 4 2⎥⎦
Note that the elements equidistant from the main diagonal are the same.
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1.7.7
Properties of Symmetric Matrices
Engineering Mathematics
1. Eigen values of a symmetric matrix are real.
Proof Let A be a symmetric matrix of order n and l be an eigen value of A.
Then there exists X ≠ 0 such that AX = lX
Taking complex conjugate,
AX=lX
⇒
AX = l X
Taking transpose, ( AX )T = (l X )T
⇒
(1)
[{ A is real A = A ]
T
( X )T A T = l ( X )T
⇒ X A=lX
T
(2)
[{ A T = A ]
Post multiplying by X,
T
( X A)X = (l X)T X
⇒
T
T
T
T
X (AX ) = l (X X )
⇒
X (lX ) = l ( X X )
⇒
l( X X ) = l( X X )
⇒
l=l
T
T
[{ X X ≠ 0 as X ≠ 0]
T
∴ l iss real
This is true for all eigen values.
∴ eigen values of a symmetric matrix are real.
■
2. Eigen vectors corresponding to different eigen values of a symmetric matrix are orthogonal
vectors.
Proof Let A be a symmetric matrix of order n. ∴ AT = A.
Let l1, l2 be two different eigen values of A. Then l1, l2 are real, by property (1).
∴ there exist X1 ≠ 0, X2 ≠ 0 such that
AX1 = l1X1
AX2 = l2X2
and
T
2
Premultiplying (1) by X , we get
⇒
Premultiplying (2) by X1T , we get
⇒
Taking transpose of (3), we get
⇒
⇒
From (4) and (5) we get,
M01_ENGINEERING_MATHEMATICS-I _CH01_Part B.indd 88
(1)
(2)
X ( AX1 ) = X l1X1
T
2
T
2
X T2 AX1 = l1 ( X T2 X1 )
(3)
X1T (AX 2 ) = X1T l 2 X 2
X1T AX 2 = l 2 ( X1T X 2 )
( X AX1 ) = (l1X X1 )
T
2
T
T
2
(4)
T
X1T A T X 2 = l1X1T X 2
X1T AX 2 = l1 (X1T X 2 )
(5)
l1 ( X X 2 ) = l 2 ( X X 2 )
T
1
T
1
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Matrices ■
⇒
1.89
(l1 − l 2 )( X1T X 2 ) = 0
Since l1 ≠ l 2 , l1 − l 2 ≠ 0,
∴ X1T X 2 = 0
⇒ X1 and X2 are orthogonal.
■
Remark: If X1 = (a1, b1, c1) and X2 = (a2, b2, c2) be two 3-dimensional vectors, they are orthogonal if
their dot product is 0 ⇒ a1a2 + b1b2 + c1c2 = 0
⎡ a2 ⎤
⎡ a1 ⎤
⎥
⎢
If we treat them as column matrices, X1 = ⎢ b1 ⎥ X 2 = ⎢⎢ b2 ⎥⎥ , then the matrix product
⎣⎢ c2 ⎦⎥
⎣⎢ c1 ⎦⎥
⎡ a2 ⎤
X X 2 = [a1 b1 c1 ] ⎢⎢ b2 ⎥⎥ = a1a2 + b1b 2 + c1c 2
⎢⎣ c2 ⎥⎦
T
1
So, X1 and X2 are orthogonal if X1T X 2 = 0 or X T2 X1 = 0 .
Thus, we can treat column matrices as vectors and verify dot product = 0.
2. The unit vector in X1 is
X1
and it is called a normalised vector.
a 1 b 21 c 2
Note For any square matrix eigen vectors corresponding to different eigen values are linearly
independent, but for a symmetric matrix, they are orthogonal, pairwise.
1.7.8
2
Diagonalisation by Orthogonal Transformation or Orthogonal Reduction
Definition 1.40 A Square Matrix
A is said to be orthogonally diagonalisable if there exists an orthogonal matrix N such that
N−1 AN = D ⇒ NT AN = D [ { NT = N−1]
This transformation which transforms A into D is called an orthogonal transformation.
The next theorem gives a method of orthogonal reduction.
Theorem 1.7 Let A be a symmetric matrix of order n. Let X1, X2, …, Xn be eigen vectors of A which are
pairwise orthogonal. Let N be the matrix formed with the normalised eigen vectors of A as columns.
Then N is an orthogonal matrix such that N−1 AN = D ⇒ NT AN = D.
N is called normalised modal matrix of A or normal modal matrix of A.
Working rule for orthogonal reduction of a n 3 n symmetric matrix.
Step 1: Find the eigen values l1, l2, …, ln
Step 2: Find the eigen vectors X1, X2, …, Xn which are pairwise orthogonal.
Step 3: Form the normalised modal matrix N with the normalised eigen vectors as columns.
Step 4: Find NT and AN.
0
0⎤
⎡ l1
⎢ 0 l
0 ⎥⎥
2
Step 5: Compute N T AN = D = ⎢
⎢ : … …⎥
⎢
⎥
⎣ 0 … ln ⎦
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WORKED EXAMPLES
EXAMPLE 1
1
1⎤
⎡3
Diagonalise the matrix A = ⎢1
3 21⎥⎥ by means of an orthogonal transformation.
⎢
⎢⎣1 21
3 ⎥⎦
1
1⎤
⎡3
⎢
3 −1⎥⎥
Solution. Given A = ⎢1
3 ⎦⎥
⎣⎢1 −1
A is a symmetric matrix, since the elements equidistant from the main diagonal are the same.
The characteristic equation of A is A − l I = 0
⇒
3−l 1
1
1
3 − l −1 = 0
1
−1
3−l
where
S1 = 3 + 3 + 3 = 9
S2 =
⇒ l3 – S1l2 + S2l – S3 = 0
3 −1
3 1
3 1
+
+
= (9 − 1) + (9 − 1) + (9 − 1) = 24
−1
3
1 3
1 3
3 1 1
S 3 = 1 3 −1 = 3(9 − 1) − 1(3 + 1) + ( −1 − 3) = 24 − 4 − 4 = 16
1 −1 3
∴ The characteristic equation is l3 − 9l2 + 24l − 16 = 0
By trial we find l = 1 is a root.
Other roots are given by l2 – 8l + 16 = 0
⇒
(l − 4)2 = 0
⇒
l = 4, 4
∴ the eigen values are
l = 1, 4, 4.
1
1 −9 24 −16
0
1 −8 16
1
− 8 16
0
To find eigen vectors:
⎡x1 ⎤
⎢ ⎥
Let X = ⎢ x 2 ⎥ be an eigen vector corresponding to l.
⎢⎣ x 3 ⎥⎦
1
1 ⎤ ⎡ x 1 ⎤ ⎡0 ⎤
⎡3 − l
⎢ 1
3−l
−1 ⎥⎥ ⎢⎢ x 2 ⎥⎥ = ⎢⎢0 ⎥⎥
Then (A − lI)X = 0 ⇒ ⎢
−1 3 − l ⎦⎥ ⎣⎢ x 3 ⎦⎥ ⎣⎢0 ⎥⎦
⎣⎢ 1
⇒
M01_ENGINEERING_MATHEMATICS-I _CH01_Part B.indd 90
(3 − l ) x 1 + x 2 + x 3 = 0 ⎫
⎪
x 1 + (3 − l ) x 2 − x 3 = 0 ⎬
x 1 − x 2 + (3 − l )x 3 = 0 ⎪⎭
(I)
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Matrices ■
1.91
Case (i) If l = 1 then equations (I) become
2x 1 + x 2 + x 3 = 0, x 1 + 2x 2 − x 3 = 0 and x 1 − x 2 + 2x 3 = 0
Choosing the first two equations, we have
x2
x1
⇒
x
x1
x
= 2 = 3
−1 − 2 1 + 2 4 − 1
x
x1 x 2
=
= 3 ⇒
−3 3
3
x1 x 2
=
−1 1
=
x3
1
x3
1
1
2
1
2
−1
1
2
⎡ −1⎤
⎢
Choosing x1 = −1, x2 = 1, x3 = 1, we get an eigen vector X1 = 1⎥
⎢ ⎥
⎢⎣ 1⎥⎦
Case (ii) If l = 4, then the equations (I) become
−x1 + x 2 + x 3 = 0 ⇒ x1 − x 2 − x 3 = 0
x 1 − x 2 − x 3 = 0 and x 1 − x 2 − x 3 = 0
We get only one equation x1 – x2 – x3 = 0
(1)
To solve for x1, x2, x3, we can assign arbitrary values for two of the variables and we shall find 2
orthogonal vectors.
⎡1 ⎤
⎢
Put x3 = 0, x2 = 1, then x1 = 1, we get an eigen vector X 2 = ⎢1 ⎥⎥ .
⎢⎣0 ⎥⎦
⎡a⎤
Let X 3 = ⎢⎢ b ⎥⎥ be ⊥ to X2. Then a + b = 0 ⇒ b = −a and X3 should satisfy (1) ∴ a – b – c = 0
⎢⎣ c ⎥⎦
⎡ 1⎤
⎢
⎥
Choosing a = 1, we get b = −1 and c = 2, ∴ X 3 = ⎢ −1⎥
⎣⎢ 2⎦⎥
Thus, the eigen values are l = 1, 4, 4, and the corresponding eigen vectors are
⎡ −1⎤
⎡ 1⎤
⎡ 1⎤
⎢
⎥
⎢
⎥
⎢
X1 = ⎢ 1⎥ , X 2 = ⎢ 1⎥ , X 3 = ⎢ −1⎥⎥
⎢⎣ 1⎥⎦
⎢⎣0 ⎥⎦
⎢⎣ 2⎥⎦
⎡
⎢−
⎢
The normalised eigen vectors are ⎢
⎢
⎢
⎢
⎢
⎣
M01_ENGINEERING_MATHEMATICS-I _CH01_Part B.indd 91
1 ⎤
⎥
3⎥
1 ⎥
⎥,
3⎥
1 ⎥
⎥
3⎦
⎡
⎢
⎢
⎢
⎢
⎢
⎢⎣
1 ⎤
⎥
2⎥
1 ⎥
⎥,
2⎥
0 ⎥⎦
⎡ 1 ⎤
⎢ ⎥
⎢ 6⎥
⎢ −1 ⎥
⎢ ⎥
⎢ 6⎥
⎢ 2 ⎥
⎢ ⎥
⎣ 6⎦
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⎡ −1
⎢
⎢ 3
⎢ 1
So, the normalised modal matrix N = ⎢
⎢ 3
⎢ 1
⎢
⎣ 3
1 ⎤
⎥
6⎥
−1 ⎥
⎥
6⎥
2 ⎥
⎥
6⎦
1
2
1
2
0
1⎤
⎡3 1
⎢
AN = ⎢1 3 −1 ⎥⎥
⎢⎣1 −1
3⎥⎦
⎡
⎢−
⎢
⎢
T
N AN = ⎢
⎢
⎢
⎢
⎣
∴
1
1
3
1
3
1
2
1
2
1
6
−
6
⎡
⎢
⎢
⎢
∴ NT = ⎢
⎢
⎢
⎢
⎣
⎡
⎢−
⎢
⎢
⎢
⎢
⎢
⎢
⎣
1
1
3
1
2
1
3
1
2
1 ⎤
⎥
3⎥
⎥
0⎥
⎥
2 ⎥
⎥
6⎦
⎡
⎢−
⎢
⎢
⎢
⎢
⎢
⎢
⎣
1
3
1
3
1
2
1
2
−1
6
6
1 ⎤ ⎡ −1
⎥ ⎢
6⎥ ⎢ 3
−1 ⎥ ⎢ 1
⎥=⎢
6⎥ ⎢ 3
2 ⎥ ⎢ 1
⎥ ⎢
6⎦ ⎣ 3
0
3
−1
1
4
3
1
2
4
3
1
2
3
0
1 ⎤
⎥
3⎥
⎥
0⎥
⎥
2 ⎥
⎥
6⎦
4
2
4
2
0
4 ⎤
⎥
6⎥
−4 ⎥
⎥
6⎥
8 ⎥
⎥
6⎦
4 ⎤
⎥
6⎥
−4 ⎥
⎥
6⎥
8 ⎥
⎥
6⎦
⎡ 1 0 0⎤
= ⎢⎢0 4 0 ⎥⎥ , which is a diagonal matrix.
⎢⎣0 0 4 ⎥⎦
EXAMPLE 2
The eigen vectors of a 3 3 3 real symmetric matrix A corresponding to the eigen values 2, 3, 6
are [1, 0, 21]T, [1, 1, 1]T, [21, 2, 21]T respectively, find the matrix A.
Solution.
Given A is symmetric and the eigen values are different.
So, the eigen vectors are orthogonal pairwise.
The normalised eigen vectors are
⎡
⎢
⎢
⎢
⎢
⎢−
⎢⎣
1 ⎤
⎥
2⎥
0⎥ ,
⎥
1 ⎥
2 ⎥⎦
M01_ENGINEERING_MATHEMATICS-I _CH01_Part B.indd 92
⎡
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎣
1 ⎤
⎥
3⎥
1 ⎥
⎥,
3⎥
1 ⎥
⎥
3⎦
⎡
⎢−
⎢
⎢
⎢
⎢
⎢
⎢
⎣
1 ⎤
⎥
6⎥
2 ⎥
⎥
6⎥
−1 ⎥
⎥
6⎦
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Matrices ■
⎡
⎢
⎢
⎢
So, the normalised modal matrix is N = ⎢
⎢
⎢
⎢−
⎣
1
1
2
0
1
2
3
1
3
1
3
1.93
1 ⎤
⎥
6⎥
2 ⎥
⎥
6⎥
1 ⎥
−
⎥
6⎦
−
Then by orthogonal reduction theorem 1.7, page 1.89
⎡2 0 0⎤
⎢
N AN = D = ⎢0 3 0 ⎥⎥ , since 2, 3, 6 are the eigen values.
⎢⎣0 0 6 ⎥⎦
T
But NT = N−1
∴ N−1AN = D ⇒ A = ND N−1 = NDNT
⎡ 1
⎢
⎢ 2
⎢
A=⎢ 0
⎢
⎢ −1
⎢
⎣ 2
⇒
⎡ 2
⎢
⎢
=⎢ 0
⎢
⎢
⎢
⎢⎣ − 2
1
3
1
3
1
3
3
3
3
1 ⎤
⎥
6⎥
2 ⎥
⎥
6⎥
1 ⎥
−
⎥
6⎦
−
− 6⎤
⎥
⎥
⎥
2 6⎥
⎥
⎥
− 6 ⎥⎦
⎡ 3 −1 1⎤
= ⎢⎢ −1 5 −1⎥⎥ ,
⎢⎣ 1 −1 3⎥⎦
⎡2 0 0⎤ ⎡
⎥ ⎢
⎢
⎥ ⎢
⎢
⎢0 3 0 ⎥ ⎢
⎥ ⎢
⎢
⎥ ⎢
⎢
⎢⎣0 0 6 ⎥⎦ ⎢
⎢−
⎣
⎡
⎢
⎢
⎢
⎢
⎢
⎢
⎢−
⎣
1
1
1 ⎤
⎥
2⎥
1 ⎥
⎥
3⎥
1 ⎥
−
⎥
6⎦
0 −
2
1
1
3
1
3
2
6
6
1 ⎤
⎥
2⎥
1 ⎥
⎥
3⎥
1 ⎥
−
⎥
6⎦
0 −
2
1
1
3
1
3
2
6
6
which is the required matrix.
EXAMPLE 3
⎡ 2 0 1⎤
Diagonalise the matrix A = ⎢⎢0 3 0 ⎥⎥ . Hence, find A3.
⎢⎣ 1 0 2 ⎥⎦
Solution.
⎡ 2 0 1⎤
Given A = ⎢0 3 0 ⎥ which is a symmetric matrix.
⎢
⎥
⎢⎣ 1 0 2⎥⎦
So, we shall diagonalise by orthogonal transformation.
M01_ENGINEERING_MATHEMATICS-I _CH01_Part B.indd 93
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The characteristic equation of A is A − lI = 0
2−l
0
1
0
3−l
0 = 0 ⇒ l 3 − S1l 2 + S2 l − S3 = 0
1
0
2−l
⇒
where
S1 = 2 + 3 + 2 = 7
S2 =
3 0
2 1
2 0
+
+
= 6 + 3 + 6 = 15
0 2
1 2
0 3
S3 = A = 2(6) − 0 + 1( −3) = 12 − 3 = 9
∴ the characteristic equation is l 3 − 7l 2 + 15l − 9 = 0
By trial l = 1 is a root.
Other roots are given by l 2 − 6l + 9 = 0
1
⇒
(l − 3) 2 = 0 ⇒ l = 3, 3
∴ the eigen values are l = 1, 3, 3
1 −7 15 −9
0
1 −6
9
1 −6
9
0
To find eigen vectors:
⎡ x1 ⎤
Let X = ⎢⎢ x 2 ⎥⎥ be an eigen vector corresponding to eigen value l.
⎢⎣ x 3 ⎥⎦
0
1 ⎤ ⎡ x 1 ⎤ ⎡0 ⎤
⎡2 − l
⎢
( A − lI)X = 0 ⇒ ⎢ 0
3−l
0 ⎥⎥ ⎢⎢ x 2 ⎥⎥ = ⎢⎢0 ⎥⎥
Then
⎢⎣ 1
0
2 − l ⎥⎦ ⎢⎣ x 3 ⎥⎦ ⎣⎢0 ⎥⎦
⇒
( 2 − l )x 1 + 0x 2 + x 3 = 0 ⎫
⎪
0 x 1 + (3 − l ) x 2 + 0 x 3 = 0 ⎬
x 1 + 0 x 2 + ( 2 − l )x 3 = 0 ⎪⎭
(I)
Case (i) If l = 1, then equations (I) become
x1 + x3 = 0 ⇒ x3 = −x1, 2x2 = 0 ⇒ x2 = 0 and x1 + x3 = 0 ⇒ x3 = −x1
⎡ 1⎤
⎢
Choose x1 = 1. ∴ x3 = −1
∴ an eigen vector is X1 = ⎢ 0 ⎥⎥
⎢⎣ −1 ⎥⎦
Case (ii) If l = 3, then equations (I) become
− x 1 + x 3 = 0, 0 x 2 = 0 and x 1 − x 3 = 0 ⇒ x 1 = x 3
and x2 can take any value
Choosing x1 = 1, we get x3 = 1 and choose x2 = 0
M01_ENGINEERING_MATHEMATICS-I _CH01_Part B.indd 94
(2)
⎡1 ⎤
∴ an eigen vector is X 2 = ⎢⎢0 ⎥⎥
⎢⎣1 ⎥⎦
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Matrices ■
1.95
⎡a⎤
We shall now choose X 3 = ⎢⎢b ⎥⎥ orthogonal to X2
⎣⎢c ⎦⎥
∴ dot product = 0 ⇒ a + c = 0 and X3 should satisfy equations (2)
∴
a – c = 0 and 0b = 0
⎡0 ⎤
⎢
Solving, we get a = c = 0 and choose b = 1 ∴ X 3 = 1 ⎥
⎢ ⎥
⎢⎣0 ⎥⎦
⎡ 1⎤
⎢ 0⎥ ,
Thus, the eigen values are 1, 3, 3 and the corresponding eigen vectors are ⎢ ⎥
⎢⎣ −1 ⎥⎦
Clearly they are pairwise orthogonal vectors.
1 ⎤
⎥
2⎥
0⎥ ,
⎥
1 ⎥
2 ⎥⎦
⎡
⎢
⎢
∴ the normalised modal matrix N = ⎢
⎢
⎢−
⎢⎣
⎡
⎢
⎢
⎢
⎢
⎢
⎢⎣
∴
2
∴
∴
N−1AN = D
2
1
2
0
⎡
⎢
⎢
⎢
T
∴ N =⎢
⎢
⎢⎣
1
2
0
1
1
⎡0 ⎤
⎢1⎥
⎢ ⎥
⎢⎣0 ⎥⎦
⎤
0⎥
2
⎥
0 1⎥
⎥
1
0⎥
⎥⎦
2
1
⎡2 0 1⎤
AN = ⎢⎢0 3 0 ⎥⎥
⎢⎣1 0 2⎥⎦
⎡
⎢
⎢
N T AN = ⎢
⎢
⎢
⎢⎣
1 ⎤
⎥
2⎥
0 ⎥,
⎥
1 ⎥
2 ⎥⎦
0 −
0
1
⎡
⎢
⎢
⎢
⎢
⎢−
⎢⎣
1
1 ⎤
⎥
2⎥
1 ⎥
⎥
2⎥
0 ⎥⎦
2
⎡
⎢
⎢
⎢
⎢
⎢−
⎢⎣
1
2
1
2
0
⎤ ⎡
0⎥ ⎢
2
⎥ ⎢
0 1⎥ = ⎢
⎥ ⎢
1
0⎥ ⎢ −
⎥⎦ ⎢⎣
2
1
2
0
1
1
2
0
1
2
1
1 ⎤
⎥
2⎥
1 ⎥
⎥
2⎥
0 ⎥⎦
1
3
0 −
0
2
0
1
2
⎤
0⎥
2
⎥
0 3⎥
⎥
3
0⎥
2
⎦⎥
⎤
0⎥
⎡1 0 0 ⎤
2
⎥ ⎢
0 3⎥ = ⎢0 3 0 ⎥⎥ = D
⎥ ⎢
3
0 0 3⎥⎦
0⎥ ⎣
⎥⎦
2
3
⇒ A = ND N−1
A3 = ND3N−1 = ND3NT
M01_ENGINEERING_MATHEMATICS-I _CH01_Part B.indd 95
⎡0 ⎤
⎢1⎥
⎢ ⎥
⎢⎣0 ⎥⎦
∴
⎡
⎢
⎢
The normalised eigen vectors are ⎢
⎢
⎢−
⎢⎣
⎡1 ⎤
⎢0⎥ ,
⎢ ⎥
⎢⎣1 ⎥⎦
[
NT = N−1]
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⇒
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Engineering Mathematics
⎡
⎢
⎢
A3 = ⎢
⎢
⎢−
⎢⎣
⎡
⎢
⎢
=⎢
⎢
⎢−
⎢⎣
1.8
1
⎤
0⎥
2
⎥
0 1⎥
⎥
1
0⎥
⎥⎦
2
⎡ 1 0 0⎤
⎥
⎢
⎥
⎢
⎢ 0 27 0⎥
⎥
⎢
⎥
⎢
⎢⎣ 0 0 27⎦⎥
⎤
0⎥
2
⎥
0 1⎥
⎥
1
0⎥
2
⎦⎥
⎡ 1
0
⎢
2
⎢
⎢ 27
0
⎢
⎢ 2
⎢⎣ 0 27
1
2
0
1
2
1
1
2
0
1
2
⎡
⎢
⎢
⎢
⎢
⎢
⎢⎣
1
2
1
2
0
0
0
1
−1 ⎤
⎥
2⎥
1 ⎥
⎥
2⎥
0 ⎥⎦
−1 ⎤ ⎡14 0 13⎤
⎥
⎥ ⎢
2⎥ ⎢
⎥
27 ⎥ = ⎢ 0 27 0 ⎥
⎥
⎥ ⎢
2⎥ ⎢
⎥
0 ⎥⎦ ⎣⎢13 0 14 ⎥⎦
REAL QUADRATIC FORM. REDUCTION TO CANONICAL FORM
Definition 1.41 A homogeneous polynomial of second degree in any number of variables is called a
quadratic form.
For example
(i) x 2 + 4 xy + 4 y 2
(ii) ax 2 + by 2 + cz 2 + 2hxy + 2gyz + 2fzx
(iii) x 12 + x 22 + 3x 32 + x 42
are quadratic forms in 2, 3 and 4 variables respectively.
n
Definition 1.42 The general quadratic form in n variables x1, x2, …, xn is
n
∑∑a x x
j =1 i =1
ij
i
j
, where aij are
real numbers such that aij = aji for all i, j = 1, 2, 3, …, n.
n
n
Usually the quadratic form is denoted by Q and Q = ∑ ∑ aij xi x j .
j =1 i =1
1. Matrix form of Q
⎡ x1 ⎤
⎡ a11 a12
⎢x ⎥
⎢a
a22
2
21
If X = ⎢ ⎥ , A = ⎢
⎢ …⎥
⎢… …
⎢ ⎥
⎢
⎣x n ⎦
⎣an1 an2
…
…
…
…
a1n ⎤
a2n ⎥⎥
, where aij = aji, then A is a symmetric matrix and the quadratic
…⎥
⎥
ann ⎦
form
n
n
Q = ∑ ∑ aij x ij can be written as Q = XT AX.
j =1 i =1
A is called the matrix of the quadratic form.
⎡1 2⎤ ⎡ x ⎤
For example the quadratic form x2 + 4xy + 4y2 can be written in the matrix form [x y ] ⎢
⎥ ⎢ ⎥.
⎣2 4⎦ ⎣ y ⎦
⎡1 2⎤
⎡x ⎤
Here X = ⎢ ⎥ and A = ⎢
⎥.
⎣2 4⎦
⎣y ⎦
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Note Since the quadratic form is Q = XT AX, it is obvious that the characteristics or properties of Q
depend on the characteristics of A.
2. Canonical form of Q
Definition 1.43 A quadratic form Q which contains only the square terms of the variables is said to
be in canonical form.
2
2
2
2
For example x2 + y2, x2 – y2, x2 + y2 – 4z2 and x 1 + x 2 + 2x 3 + x 4 are in canonical forms because they
contain only square terms of the variables.
3. Reduction of Q to canonical form by orthogonal transformation
Let Q = XT AX be a quadratic form in n variables x1, x2, …, xn and A = [aij] be the symmetric matrix
of order n of the quadratic form. We will reduce A to diagonal form by an orthogonal transformation
X = NY, where N is the normalised modal matrix of A. Then NT AN = D, where D is the diagonal
matrix containing the eigen values of A.
If l1, l2, …, ln are the eigen values of A, then
0 0 … 0⎤
⎡ l1
⎢ 0 l
0 … 0 ⎥⎥
2
⎢
D=
⎢ … … … … …⎥
⎢
⎥
0 0 … ln ⎦
⎣ 0
Q = X T AX = (NY)T A(NY) = Y T (N T AN)Y = Y T DY
∴
If
0
⎡ l1
⎡ y1 ⎤
⎢ 0 l
⎢y ⎥
2
Y = ⎢ 2 ⎥ , then Q = [ y 1 y 2 … y n ] ⎢
⎢… …
⎢…⎥
⎢
⎢ ⎥
0
⎣ 0
⎣y n ⎦
… 0⎤ ⎡ y 1 ⎤
… 0 ⎥⎥ ⎢⎢ y 2 ⎥⎥
… …⎥ ⎢ … ⎥
⎥⎢ ⎥
… ln ⎦ ⎣y n ⎦
⎡ y1 ⎤
⎢y ⎥
= [l1 y 1 l 2 y 2 … l n y n ] ⎢ 2 ⎥ = l1 y 12 + l 2 y 22 + … + l n y n2
⎢…⎥
⎢ ⎥
⎣y n ⎦
This is the required quadratic form.
Note In the canonical form the coefficients are the eigen values of A. Since A is a symmetric matrix,
the eigen values of A are all real. So, the eigen values may be positive, negative or zero. Hence, the
terms of the canonical form may be positive, negative or zero. By using the canonical form or the
eigen values, we can characterise the quadratic form.
Definition 1.44 If A is the matrix of the quadratic form Q in the variables x1, x2, …, xn, then the rank
of Q is equal to the rank of A.
If rank of A < n, where n is the number of variables or order of A, then A = 0 and Q is called a
singular form.
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4. Index, signature and rank of quadratic form
Definition 1.45 Let Q = XT AX be a quadratic form in n variables x1, x2, …, xn.
X = [x1, x2, …, xn]T and A is the matrix of the quadratic form.
(i) Index of the quadratic form is the number of positive eigen values of A.
(ii) Signature of the quadratic form is the difference between the number of positive and
negative eigen values of A.
(iii) Rank of the quadratic form is the number of positive and negative eigen values of A.
Usually index is denoted by p, signature by s and rank by r.
5. Definite and indefinite quadratic forms
Definition 1.46 Let Q = XT AX be a quadratic form in n variables x1, x2, …, xw.
i.e., X = [x1 … xn]T and A is the matrix of the quadratic form.
(i) Q is said to be positive definite if all the n eigen values of A are positive.
i.e., if r = n and p = n
e.g., y 12 + y 22 + … + y n2 is positive definite.
(ii) Q is said to be negative definite if all the n eigen values of A are negative.
i.e., if r = n, p = 0
e.g., − y 12 − y 22 − … − y n2 is negative definite.
(iii) Q is said to be positive semidefinite if all the n eigen values of A are ≥ 0 with at least one
eigen value = 0.
i.e., if r < n and p = r
e.g., y 12 + y 22 + … + y r2 , where r < n, is positive semi-definite.
(iv) Q is said to be negative semidefinite if all the n eigen values of A are ≤ 0 with at least one
value = 0.
i.e., if r < n and p = 0
e.g., − y 12 − y 22 − … − y r2 , where r < n, is negative semi definite.
(v) Q is said to be indefinite if A has positive and negative eigen values.
e.g., y 12 + y 22 − y 32 − y 42 + … + y n2 is indefinite.
6. We can also find the nature of a quadratic form without finding the eigen values of A or without
reducing to canonical form but by using the principal minors of A as below.
Definition 1.47 Let Q = XT AX be a quadratic form in n variables x1, x2, …, xn and let the matrix of
the quadratic form be
⎡ a11 a12
⎢a
a
A = ⎢ 21 22
⎢… …
⎢
⎣an1 an2
Let D1 = a11 = a11,
Finally Dn = A .
D2 =
… a1n ⎤
… a2n ⎥
⎥
… …⎥
⎥
… ann ⎦
a11 a12
a11 a12
=
a
a22
D
, 3
21
a21 a22
a31 a32
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a13
a23 and so on.
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The determinants D1, D2, D3, …, Dn are called the principal minors of A.
The quadratic form Q is said to be
(i) positive definite if Di > 0 for all i = 1, 2, …, n
(ii) negative definite if (−1)i Di > 0 for all i = 1, 2, …, n
i.e., D1, D3, D5, … are negative and D2, D4, D6 … are positive.
(iii) positive semi-definite if Di ≥ 0 for all i = 1, 2, 3, …, n and at least one Di = 0.
(iv) negative semi-definite if (−1)i Di ≥ 0 for all i = 1, 2, 3, …, n and at least one Di = 0.
(v) indefinite in all other cases.
7. Law of intertia of a quadratic form
In the reduction of a quadratic form to canonical form the number of positive and negative terms
are independent of the choice of the transformation.
In other words, the signature of a real quadratic form is invariant under a real non-singular
transformation. This property is called the law of inertia of the quadratic form.
WORKED EXAMPLES
EXAMPLE 1
Write down the matrix of the quadratic form 2 x 12 2 2 x 22 1 4 x 32 1 2 x 1 x 2 2 6 x 1 x 3 1 6 x 2 x 3 .
Solution.
The Q.F is 2x 12 − 2x 22 + 4 x 32 + 2x 1x 2 − 6 x 1x 3 + 6 x 2 x 3
It has 3 variables x1, x2, x3.
So, the matrix of the quadratic form is a 3 × 3 symmetric matrix.
⎡ a11 a12
A = ⎢⎢a21 a22
⎢⎣a31 a32
a13 ⎤
a23 ⎥⎥
a33 ⎥⎦
Here
a11 = coefficient of x 12 = 2
a12 = a21 =
1
1
(coefficient of x1x2) = ( 2) = 1
2
2
a22 = coefficient of x 22 = −2
a13 = a31 =
1
1
(coefficient of x1x3) = ( −6) = −3
2
2
a33 = coefficient of x 32 = 4
a23 = a32 =
1
1
(coefficient of x2x3) = (6) = 3
2
2
∴
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1 −3⎤
⎡ 2
⎢
3⎥⎥
A = ⎢ 1 −2
⎢⎣ −3
3 4 ⎥⎦
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EXAMPLE 2
⎡2 4 5⎤
⎢
⎥
Write down the quadratic form corresponding to the matrix ⎢ 4 3 1 ⎥ .
⎢⎣ 5 1 1 ⎥⎦
Solution.
⎡ 2 4 5⎤
⎢
⎥
Let A = ⎢ 4 3 1⎥ , which is a 3 × 3 symmetric matrix.
⎣⎢ 5 1 1⎦⎥
So, the quadratic form has 3 variables x1, x2, x3.
⎡ x1 ⎤
Let X = ⎢⎢ x 2 ⎥⎥ , then the quadratic form is
⎢⎣ x 3 ⎥⎦
⎡ 2 4 5⎤
Q = X AX = [ x1 x2 x3 ] ⎢⎢ 4 3 1⎥⎥
⎢⎣ 5 1 1⎥⎦
T
⎡ x1 ⎤
⎢x ⎥
⎢ 2⎥
⎢⎣ x3 ⎥⎦
⎡ x1 ⎤
5 x1 + x2 + x3 ] ⎢ x2 ⎥
⎢x ⎥
⎣ 3⎦
= ( 2 x1 + 4 x2 + 5 x3 ) x1 + ( 4 x1 + 3 x2 + x3 ) x2 + (5 x1 + x2 + x3 ) x3
= [2 x1 + 4 x2 + 5 x3
4 x1 + 3 x2 + x3
= 2 x12 + 4 x1 x2 + 5 x3 x1 + 4 x1 x2 + 3 x22 + x3 x2 + 5 x1 x3 + x2 x3 + x32
= 2 x12 + 3 x22 + x32 + 8 x1 x2 + 10 x1 x3 + 2 x2 x3
⎡ a11 a12 a13 ⎤ ⎡ 2 4 5⎤
Aliter Given
A = ⎢⎢a21 a22 a23 ⎥⎥ = ⎢⎢ 4 3 1⎥⎥
⎢⎣a31 a32 a33 ⎥⎦ ⎢⎣ 5 1 1⎦⎥
Then the quadratic form in 3 variables x1, x2, x3 is
Q = X T AX = a11x 12 + a22 x 22 + a33 x 32 + 2(a12 )x 1x 2 + 2(a13 )x 1x 3 + 2(a23 )x 2 x 3
= 2x 12 + 3x 22 + x 32 + 2( 4)x 1x 2 + 2(5)x 1x 3 + 2(1)x 2 x 3
= 2x 12 + 3x 22 + x 32 + 8x 1x 2 + 10x 1x 3 + 2x 2 x 3
EXAMPLE 3
Discuss the nature of the following quadratic forms.
(i) 6x2 + 3y2 + 3z2 – 4xy 2 2yz + 4zx
(ii) 6 x 12 1 3 x 22 1 14 x 32 1 4 x 2 x 3 1 18 x 1 x 3 1 4 x 1 x 2
(iii) xy + yz + zx
(iv) 10x2 + 2y2 + 5z2 + 6yz 2 10zx 2 4xy.
Solution.
(i) The Q.F is 6x2 + 3y2 + 3z2 – 4xy − 2yz + 4zx, having 3 variables x, y, z.
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The matrix of the quadratic form is
⎡ 6 −2 2⎤
A = ⎢⎢ −2
3 −1⎥⎥ .
⎢⎣ 2 −1 3⎥⎦
The principal minors are
D1 = | 6 | = 6 > 0; D 2 =
6 −2
= 18 − 4 = 14 > 0
−2
3
6 −2 2
D3 = A = −2
3 −1 = 6(9 − 1) + 2( −6 + 2) + 2( 2 − 6) = 48 − 8 − 8 = 32 > 0
2 −1 3
Since D1, D2, D3 are positive, the quadratic form is positive definite.
(ii) The quadratic form is 6 x 12 + 3x 22 + 14 x 32 + 4 x 2 x 3 + 18x 1x 3 + 4 x 1x 2 , having 3 variables x1, x2, x3.
The matrix of the Q.F is
⎡6 2 9 ⎤
A = ⎢⎢ 2 3 2 ⎥⎥ .
⎢⎣ 9 2 14 ⎥⎦
D2 =
D1 = 6 > 0;
The principal minors are
6 2
= 18 − 4 = 14 > 0
2 3
6 2 9
and D3 = A = 2 3 2
9 2 14
= 6(42 − 4) – 2(28 − 18) + 9(4 − 27) = 6(38) – 20 + 9(−23) = 228 – 20 – 207 = 1 > 0
Since D1, D2, D3 are positive, the quadratic form is positive definite.
(iii) The quadratic form is xy + yz + zx in 3 variables x, y, z
1 1⎤
⎡
⎢0 2 2⎥
⎢
⎥
1
1⎥
The matrix of the quadratic form is A = ⎢
0
⎢2
2⎥
⎢
⎥
⎢1 1 0⎥
⎢⎣ 2 2
⎥⎦
0
The principal minors are D1 = 0, D2 =
0
and
D3 =
1
2
1
2
1
2
0
1
2
1
2
1
2
0
= −
1
<0
4
1
2
0 1 1
1
1
1
1
1
= 1 0 1 = [ 0 − 1(0 − 1) + (1 − 0) ] = × 2 = > 0
2
8
8
8
4
1 1 0
0
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Since D1 = 0, D2 < 0, D3 > 0, the quadratic form is indefinite.
(iv) The Q.F 10 x 2 + 2 y 2 + 5z 2 + 6 yz − 10 zx − 4 xy is in three variables x, y, z.
⎡ 10 −2 −5⎤
⎢
The matrix of the quadratic form A = ⎢ −2
2
3⎥⎥
⎢⎣ −5
3 5⎥⎦
10 −2
= 20 − 4 = 16 > 0
The principal minors are D1 = 10 > 0, D 2 =
−2
2
10 −2 −5
D3 = −2 2
3 = 10(10 − 9) + 2 (−10 + 15) – 5(−6 + 10) = 10 +10 – 20 = 0
−5
3 5
Since D1 > 0, D2 > 0 and D3 = 0, the quadratic form is positive semi-definite.
EXAMPLE 4
Determine l so that l(x2 1 y2 1 z2) 1 2xy 2 2yz 1 2zx is positive definite.
⎡l
The matrix of the quadratic form is A = ⎢⎢ 1
⎢⎣ 1
Solution.
The principal minors are
1⎤
−1⎥⎥
l ⎥⎦
1
l
−1
D1 = l, D 2 =
l 1
= l 2 − 1;
1 l
D3 = A = l (l 2 − 1) − (l + 1) + ( −1 − l )
= (l + 1)(l(l + 1) − 1 − 1)
= (l
l + 1)(l 2 − l − 2 = (l + 1) 2 (l − 2)
Given the quadratic form is positive definite.
∴
D1 > 0, D2 > 0 and D3 > 0
l > 0, l2 – 1 > 0
⇒
(l + 1)(l − 1) > 0 ⇒
and
(l + 1) (l − 2) > 0 ⇒ l − 2 > 0
∴ the common values of l are l > 2
2
l>1
( l > 0)
⇒ l>2
[∴ (l + 1)2 > 0]
∴
⇒
EXAMPLE 5
Show that the quadratic form ax 12 2 2bx 1 x 2 1 cx 22 is positive definite if a . 0 and ac 2 b2 . 0.
Solution.
⎡ a −b ⎤
The matrix of the quadratic form is A = ⎢
c ⎥⎦
⎣ −b
a −b
= ac − b 2 .
−b
c
Given a > 0 and ac – b2 > 0. ∴ D1 > 0 and D2 > 0.
Hence, the Q.F is positive definite.
The principal minors are D1 = a, D 2 = A =
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EXAMPLE 6
Reduce 6 x 2 1 3 y 2 1 3 z 2 2 4 xy 2 2 yz 1 4 xz into a canonical form by an orthogonal
reduction and find the rank, signature, index and the nature of the quadratic form.
Solution.
Given quadratic form is 6 x 2 + 3y 2 + 3z 2 − 4 xy − 2 yz + 4 xz
⎡ 6 −2 2⎤
3 −1⎥⎥
The matrix of the Q.F is A = ⎢⎢ −2
⎢⎣ 2 −1 3⎥⎦
The characteristic equation is A − lI = 0
6 − l −2
2
−2 3 − l
−1 = 0
2
−1 3 − l
⇒
where
⇒
l 3 − S1l 2 + S2 l − S3 = 0
S1 = 6 + 3 + 3 = 12
S2 =
3 −1
6 2
6 −2
+
+
= 8 + 14 + 14 = 36
−1 3
2 3
−2
3
S3 = A = 6(9 − 1) + 2( −6 + 2) + 2( 2 − 6) = 48 − 8 − 8 = 32
∴ the characteristic equation is l 3 − 12l 2 + 36l − 32 = 0
By trial, l = 2 is a root.
Other roots are given by l2 − 10l + 16 = 0
⇒
(l − 2)(l − 8) = 0 ⇒
l = 2, 8
2
1 −12
36 −32
0
2 −20
32
1 −10
16
0
∴ the eigen values are l = 2, 2, 8
To find eigen vectors:
⎡ x1 ⎤
⎢ x ⎥ be an eigen vector corresponding to eigen value l.
X
=
Let
⎢ 2⎥
⎢⎣ x 3 ⎥⎦
2 ⎤
⎡6 − l −2
⎢
Then ( A − lI)X = 0 ⇒ ⎢ −2 3 − l
−1 ⎥⎥
⎢⎣ 2
−1 3 − l ⎥⎦
⎡ x 1 ⎤ ⎡0 ⎤
⎢ x ⎥ = ⎢0 ⎥
⎢ 2⎥ ⎢ ⎥
⎢⎣ x 3 ⎥⎦ ⎢⎣0 ⎥⎦
(6 − l ) x 1 − 2 x 2 + 2 x 3 = 0 ⎫
⎪
−2x 1 + (3 − l )x 2 − x 3 = 0 ⎬
2x 1 − x 2 + (3 − l )x 3 = 0 ⎪⎭
⇒
(I)
Case (i) If l = 8, then equations (I) become
−2 x1 − 2 x2 + 2 x3 = 0 ⇒ x1 + x2 − x3 = 0
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−2 x1 − 5 x2 − x3 = 0 ⇒ 2 x1 + 5 x2 + x3 = 0
2 x1 − x2 − 5 x3 = 0
From the first two equations we get
x
x1
x2
=
= 3
1 + 5 −2 − 1 5 − 2
x
x1 x 2
⇒
=
= 3
6 −3
3
x3
x1 x 2
=
=
⇒
2 −1
1
x1
x2
x3
1
−1
1
1
5
1
2
5
⎡ 2⎤
Choosing x1 = 2, x2 = −1, x3 = 1, we get an eigen vector X1 = ⎢⎢ −1⎥⎥
⎢⎣ 1⎥⎦
Case (ii) If l = 2, then equations (I) become
4 x 1 − 2x 2 + 2x 3 = 0 ⇒ 2x 1 − x 2 + x 3 = 0
−2x 1 + x 2 − x 3 = 0 ⇒ 2x 1 − x 2 + x 3 = 0
2x 1 − x 2 + x 3 = 0
So, we get only one equation 2x1 − x2 + x3 = 0
Choosing x3 = 0, we get 2x1 – 2x2 = 0 ⇒ x2 = 2x1
(1)
⎡1 ⎤
⎢
Choosing x1 = 1, we get x2 = 2 and x3 = 0 ∴ an eigen vector is X 2 = ⎢ 2⎥⎥
⎢⎣0 ⎥⎦
⎡a⎤
⎢
⎥
We shall find another eigen vector X 3 = ⎢b ⎥ orthogonal to X2
⎢⎣c ⎥⎦
∴
Also X3 satisfies (1)
∴
a + 2b = 0 ⇒ a = −2b
2a – b + c = 0 ⇒
−4b – b + c = 0 ⇒ c = 5b
⎡ −2⎤
Choosing b = 1, we get c = 5 and a = −2 and eigen vector X 3 = ⎢ 1 ⎥
⎢ ⎥
⎣⎢ 5⎦⎥
Thus, the eigen values are 8, 2, 2 and the corresponding eigen vectors are
⎡ −2⎤
⎡ 2⎤
⎡1⎤
⎢
⎥
⎢
⎥
X1 = ⎢ −1⎥ , X 2 = ⎢ 2⎥ , X 3 = ⎢⎢ 1⎥⎥
⎢⎣ 1⎥⎦
⎢⎣ 0 ⎥⎦
⎢⎣ 5⎥⎦
Clearly X3 is orthogonal to X1 and X2. Also X1, X2 are orthogonal.
⎡ 2 ⎤
⎢ ⎥
⎢ 6⎥
⎢ −1 ⎥
The normalised eigen vectors are ⎢ ⎥ ,
⎢ 6⎥
⎢ 1 ⎥
⎢ ⎥
⎣ 6⎦
M01_ENGINEERING_MATHEMATICS-I _CH01_Part B.indd 104
⎡
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎣
1 ⎤
⎥
5⎥
2 ⎥
⎥,
5⎥
⎥
0 ⎥
⎦
⎡
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎣
−2 ⎤
⎥
30 ⎥
1 ⎥
⎥
30 ⎥
5 ⎥
⎥
30 ⎦
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⎡ 2
⎢
⎢ 6
⎢ −1
The normalised modal matrix N = ⎢
⎢ 6
⎢ 1
⎢
⎣ 6
Then
1
5
2
5
0
1.105
−2 ⎤
⎥
30 ⎥
1 ⎥
⎥
30 ⎥
5 ⎥
⎥
30 ⎦
⎡8 0 0 ⎤
N AN = D = ⎢⎢0 2 0 ⎥⎥
⎢⎣0 0 2⎥⎦
T
⎡ y1 ⎤
The orthogonal transformation X = NY, where Y = ⎢⎢ y 2 ⎥⎥ reduces the given quadratic form to
⎢⎣ y 3 ⎥⎦
⎡8 0 0 ⎤ ⎡ y 1 ⎤
⎥⎢ ⎥
⎢
YT DY = [y1 y2 y3] ⎢0 2 0 ⎥ ⎢ y 2 ⎥
⎢⎣0 0 2⎥⎦ ⎢⎣ y 3 ⎥⎦
= 8y 12 + 2y 22 + 2 y 32 , which is the canonical form.
∴ rank of the Q.F = 3, index = 3, signature = 3
The Q.F is positive definite, since all the eigen values are positive.
EXAMPLE 7
Find out the type of conic represented by 17x2 2 30xy 1 17y2 5 128 after reducing the quadratic
form 17x2 2 30xy 1 17y2 to canonical form by an orthogonal transformation.
Solution.
Given quadratic form is 17x2 – 30xy + 17y2
⎡ 17 −15⎤
A=⎢
⎥
⎣ −15 17⎦
The matrix of the Q.F is
The characteristic equation of A is A − lI = 0
⇒
⇒
17 − l −15
=0
−15 17 − l
(17 − l ) 2 − 152 = 0 ⇒ (17 − l ) 2 = 152 ⇒ 17 − l = ±15 ⇒ l = 2 or 32.
To find eigen vectors:
⎡x ⎤
If X = ⎢ 1 ⎥ be an eigen vector corresponding to eigen value l.
⎣x 2 ⎦
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Then
⎡17 − l −15 ⎤ ⎡ x1 ⎤ ⎡0 ⎤
( A − lI)X = 0 ⇒ ⎢
⎥ ⎢ ⎥=⎢ ⎥
⎣ −15 17 − l ⎦ ⎣ x2 ⎦ ⎣0 ⎦
Engineering Mathematics
(17 − l )x 1 − 15x 2 = 0 ⎫
⎬
−15x 1 + (17 − l )x 2 = 0 ⎭
⇒
(I)
Case (i) If l = 2, then equations (I) become
15x 1 − 15x 2 = 0, − 15x 1 + 15x 2 = 0 ⇒ x 1 = x 2
Choosing x1 = 1, we get x2 = 1
⎡1⎤
∴ an eigen vector is X1 = ⎢ ⎥
⎣1⎦
Case (ii) If l = 32, then equation (I) become
−15 x1 − 15 x2 = 0 and
− 15 x1 − 15 x2 = 0 ⇒ x2 = − x1
⎡ 1⎤
Choose x1 = 1, we get x2 = −1 ∴ an eigen vector is X 2 = ⎢ ⎥
⎣ −1⎦
∴
the normalised eigen vectors are
The normalised modal matrix
⎡ 1
⎢
2
N= ⎢
⎢ 1
−
⎢
⎣ 2
⇒
⎡
⎢
N T AN = ⎢
⎢
⎢
⎣
1
2
1
2
⎡
⎢
⎢
⎢
⎢
⎣
1 ⎤
⎥
2⎥
,
1 ⎥
⎥
2⎦
⎡
⎢
⎢
⎢
⎢−
⎣
1 ⎤
⎥
2⎥
1 ⎥
⎥
2⎦
1 ⎤
⎥
2⎥
1 ⎥
⎥
2⎦
1 ⎤
⎥
2⎥
1 ⎥
−
⎥
2⎦
⎡ 17
⎢
⎢
⎢⎣ −15
−15⎤
⎥
⎥
17 ⎥⎦
⎡
⎢
⎢
⎢
⎢
⎣
1
2
1
2
1 ⎤
⎥
2 ⎥ = ⎡2 0⎤ = D
⎥
⎢
−1 ⎥ ⎣0 32⎦
⎥
2⎦
⎡y ⎤
The transformation X = NY, where Y = ⎢ 1 ⎥ , reduces the given quadratic form to
⎣ y2 ⎦
⎡ 2 0 ⎤ ⎡ y1 ⎤
2
2
Y T DY = [ y1 y2 ] ⎢
⎥ ⎢ ⎥ = 2 y1 + 32 y2 , which is the canonical form.
⎣0 32⎦ ⎣ y2 ⎦
But the given quadratic from = 128
∴
2 y12 + 32 y22 = 128 ⇒
y12 y22
+
= 1, which represents an ellipse.
64 4
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EXAMPLE 8
Reduce the quadratic form 8x2 1 7y2 1 3z2 2 12xy 1 4xz 2 8yz to the canonical form by an
orthogonal transformation. Find one set of values of x, y, z (not all zero) which will make the
quadratic form zero.
Solution.
Given quadratic form is 8x2 + 7y2 + 3z2 – 12xy + 4xz – 8yz
2⎤
⎡ 8 −6
⎢
The matrix of the quadratic form is A = ⎢ −6
7 −4 ⎥⎥
⎢⎣ 2 −4
3⎥⎦
The characteristic equation is A − lI = 0
8−l
−6
2
−6 7 − l
−4 = 0 ⇒ l 3 − S1l 2 + S2 l − S3 = 0
2
−4 3 − l
⇒
where S1 = 8 + 7 + 3 = 18
S2 =
7 −4
8 2
8 −6
+
+
−4
3
2 3
−6
7
= ( 21 − 16) + ( 24 − 4) + (56 − 36) = 5 + 20 + 20 = 45
S3 = A = 8( 21 − 16) + 6( −18 + 8) + 2( 24 − 14) = 40 − 60 + 20 = 0
∴ the characteristic equation is l 3 − 18l 2 + 45l = 0
⇒
l (l 2 − 18l + 45) = 0
⇒
l(l − 3)(l − 15) = 0 ⇒ l = 0, 3,15.
To find eigen vectors:
⎡ x1 ⎤
If X = ⎢⎢ x 2 ⎥⎥ be an eigen vector of the eigen value l of A, then
⎢⎣ x 3 ⎥⎦
2 ⎤ ⎡ x1 ⎤
⎡8 − l −6
( A − lI)X = 0 ⇒ ⎢⎢ −6 7 − l −4 ⎥⎥ ⎢⎢ x 2 ⎥⎥ = 0
⎢⎣ 2
−4 3 − l ⎥⎦ ⎢⎣ x 3 ⎥⎦
⇒
M01_ENGINEERING_MATHEMATICS-I _CH01_Part B.indd 107
(8 − l )x 1 − 6 x 2 + 2x 3 = 0 ⎫
⎪
−6 x 1 + (7 − l )x 2 − 4 x 3 = 0 ⎬
2x 1 − 4 x 2 + (3 − l )x 3 = 0 ⎪⎭
(I)
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Case (i) If l = 0, then the equations (I) become
8 x1 − 6 x2 + 2 x3 = 0
⇒
4 x1 − 3 x2 + x3 = 0
−6 x1 + 7 x2 − 4 x3 = 0 and 2 x1 − 4 x2 + 3 x3 = 0
From the first and third equations, we get
x3
x1
x2
=
=
−9 + 4 2 − 12 −16 + 6
x
x1
x
= 2
= 3
−5 −10
−10
x3
x1 x 2
=
=
1
2
2
⇒
⇒
x1
x2
x3
−3
1
4
−3
−4
3
2
−4
⎡1 ⎤
Choosing x1 = 1, x2 = 2, x3 = 2, we get an eigen vector X1 = ⎢ 2⎥
⎢ ⎥
⎢⎣ 2⎥⎦
⎡ 2⎤
⎡ 2⎤
⎢
⎥
Similarly, we can find for l = 3, X 2 = ⎢ 1⎥ and for l = 15, X 3 = ⎢⎢ −2⎥⎥
⎢⎣ 1⎥⎦
⎢⎣ −2⎥⎦
∴ the normalised modal matrix
⎡1
⎢3
⎢
N = ⎢2
⎢3
⎢
⎢2
⎢⎣ 3
2
3
1
3
−2
3
2⎤
3⎥
2⎤
⎡1 2
⎥
−2 ⎥ = 1 ⎢ 2
1 −2⎥⎥
3⎢
3⎥
⎢⎣ 2 −2
1⎥⎦
⎥
1⎥
3 ⎥⎦
⎡0 0 0 ⎤
⎢0 3 0 ⎥
T
=
=
N
AN
D
∴
⎢
⎥
⎢⎣0 0 15⎥⎦
⎡ y1 ⎤
⎢ ⎥
The transformation X = NY, where Y = ⎢ y 2 ⎥ , reduces the quadratic form to the canonical form
⎢⎣ y 3 ⎥⎦
⎡0 0 0 ⎤
Y DY = [ y 1 y 2 y 3 ] ⎢⎢0 3 0 ⎥⎥
⎢⎣0 0 15⎥⎦
T
2⎤
⎡1 2
⎡x ⎤
1
1 −2⎥⎥
The transformation is ⎢⎢ y ⎥⎥ = ⎢⎢ 2
3
⎢⎣ 2 −2
⎢⎣ z ⎥⎦
1⎥⎦
M01_ENGINEERING_MATHEMATICS-I _CH01_Part B.indd 108
⎡ y1 ⎤
⎢y ⎥
2
2
2
⎢ 2 ⎥ = 0 y 1 + 3y 2 + 15y 2
⎢⎣ y 3 ⎥⎦
⎡ y1 ⎤
⎢y ⎥
⎢ 2⎥
⎢⎣ y 3 ⎥⎦
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Matrices ■
1.109
1
1
1
x = ( y1 + 2 y2 + 2 y3 ), y = ( 2 y1 + y2 − 2 y3 ) and z = ( 2 y1 − 2 y2 + y3 )
3
3
3
∴
Quadratic form = 0
⇒
0 y 12 + 3y 22 + 15y 32 = 0
⇒
3y 22 + 15y 32 = 0 ⇒ y 2 = 0 and y 3 = 0
and y1 can take any value, we shall choose y1 = 3
∴
x = 1, y = 2, z = 2
Hence, this set of values will make the quadratic form = 0
EXAMPLE 9
Reduce the quadratic form x 12 1 2 x 22 1 x 32 2 2 x 1 x 2 1 2 x 2 x 3 to the canonical form through an
orthogonal transformation and hence show that it is positive semi-definite. Also give a non-zero
set of values (x1, x2, x3) which makes the quadratic form zero.
Solution.
Given the quadratic form is x 12 + 2x 22 + x 32 − 2x 1x 2 + 2x 2 x 3
The matrix of the quadratic form is
⎡ 1 −1 0 ⎤
A = ⎢⎢ −1 2 1⎥⎥
⎢⎣ 0
1 1⎥⎦
The characteristic equation of A is A − lI = 0
1− l
−1
0
−1 2 − l
1 = 0 ⇒ l 3 − S1l 2 + S2 l − S3 = 0
0
1
1− l
⇒
where
S1 = 1 + 2 + 1 = 4
2 1
1 0
1 −1
+
+
= 1+1+1 = 3
S2 =
1 1
0 1
−1 2
S3 = A = 2 − 1 + 1( −1) = 0
∴ the characteristic equation is l 3 − 4l 2 + 3l = 0
⇒
l(l 2 − 4l + 3) = 0 ⇒ l(l − 1)(l − 3) = 0 ⇒ l = 0, 1, 3
To find eigen vectors:
⎡ x1 ⎤
If X = ⎢⎢ x 2 ⎥⎥ be an eigen vector corresponding to an eigen value l of A, then
⎢⎣ x 3 ⎥⎦
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(A − lI)X = 0 ⇒
−1
0 ⎤
⎡1 − l
⎢ −1 2 − l
1 ⎥⎥
⎢
⎢⎣ 0
1
1 − l ⎥⎦
⎡ x 1 ⎤ ⎡0 ⎤
⎢ x ⎥ = ⎢0 ⎥
⎢ 2⎥ ⎢ ⎥
⎢⎣ x 3 ⎥⎦ ⎢⎣0 ⎥⎦
(1 − l ) x1 − x2 + 0 x3 = 0
⇒
− x1 + ( 2 − l ) x2 + x3 = 0
(I)
0 x1 + x2 + (1 − l ) x3 = 0
Case (i) If l = 0, then the equations (I) become
x1 − x2 = 0 ⇒
x1 = x2 , − x1 + 2 x2 + x3 = 0 and x2 + x3 = 0 ⇒
x3 = − x2
⎡ 1⎤
Take x2 = 1, then x1 = 1, x3 = −1 ∴ an eigen vector is X1 = ⎢⎢ 1⎥⎥
⎢⎣ −1⎥⎦
Case (ii) If l = 1, the equations (I) become
0x1 − x 2 = 0 ⇒ x 2 = 0
−x1 + x 2 + x 3 = 0 ⇒ x1 = x 3
and
⎡ 1⎤
Take x3 = 1, then x1 = 1
∴ an eigen vector is X 2 = ⎢⎢0 ⎥⎥
⎢⎣ 1⎥⎦
Case (iii) If l = 3, then equations (I) become
−2 x1 − x2 = 0 ⇒
x2 = −2 x1 , − x1 − x2 + x3 = 0 and x2 − 2 x3 = 0 ⇒
x2 = 2 x3
⎡ −1⎤
⎢
Take x2 = 2, then x1 = −1, x3 = 1 ∴ an eigen vector is X 3 = ⎢ 2⎥⎥
⎢⎣ 1⎥⎦
Thus, the eigen values are l = 0, 1, 3 and the eigen vectors are
⎡ 1⎤
⎡ 1⎤
⎡ −1⎤
⎢
⎥
⎢
⎥
X1 = ⎢ 1⎥ , X 2 = ⎢0 ⎥ , X 3 = ⎢⎢ 2⎥⎥
⎢⎣ −1⎥⎦
⎢⎣ 1⎥⎦
⎢⎣ 1⎥⎦
⎡ 1 ⎤
⎢ ⎥
⎢ 3⎥
⎢ 1 ⎥
∴ the normalized eigen vectors are ⎢ ⎥ ,
⎢ 3⎥
⎢ −1 ⎥
⎢ ⎥
⎣ 3⎦
M01_ENGINEERING_MATHEMATICS-I _CH01_Part B.indd 110
⎡
⎢
⎢
⎢
⎢
⎢
⎢⎣
1 ⎤
⎥
2⎥
0 ⎥,
⎥
1 ⎥
2 ⎦⎥
⎡
⎢−
⎢
⎢
⎢
⎢
⎢
⎢
⎣
1 ⎤
⎥
6⎥
2 ⎥
⎥
6⎥
1 ⎥
⎥
6⎦
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Matrices ■
⎡ 1
⎢
⎢ 3
⎢ 1
∴ the normalised modal matrix is N = ⎢
⎢ 3
⎢ −1
⎢
⎣ 3
1
−
2
0
1
2
1.111
1 ⎤
⎥
6⎥
2 ⎥
⎥
6⎥
1 ⎥
⎥
6⎦
⎡0 0 0 ⎤
∴ the diagonal matrix is N AN = D = ⎢⎢0 1 0 ⎥⎥
⎢⎣0 0 3⎥⎦
The orthogonal transformation is X = NY
⎡ y1⎤
where Y = ⎢⎢ y 2 ⎥⎥ and the canonical form is Y T DY = y 22 + 3y 32
⎢⎣ y 3 ⎥⎦
T
∴ the quadratic form is positive semi-definite.
⎡ 1
⎢
3
⎡ x1 ⎤ ⎢
⎢ ⎥ ⎢ 1
The transformation X = NY ⇒ ⎢ x 2 ⎥ = ⎢
3
⎢⎣ x 3 ⎥⎦ ⎢
⎢ −1
⎢
⎣ 3
∴
x1 =
1
3
y1 +
1
2
y2 −
1
6
y3 , x2 =
1
3
y1 +
1
−
2
0
1
2
2
6
1 ⎤
⎥
6⎥
⎡ y1⎤
2 ⎥⎢ ⎥
⎥ y2
6⎥⎢ ⎥
⎢y ⎥
1 ⎥ ⎣ 3⎦
⎥
6⎦
y3 and x3 = −
1
3
y1 +
1
2
y2 +
1
6
y3
These equation make the quadratic form = 0
⇒
y 22 + 3y 32 = 0 ⇒ y 2 = 0 and y 3 = 0
[Since sum of squares of real numbers = 0 ⇒ each number = 0] and y1 can take any real value.
∴
x1 =
1
3
y1, x 2 =
1
3
y1, x3 = −
1
3
y1
Choosing y 1 = 3 , one set of values of x1, x2, x3 is x1 = 1, x2 = 1, x3 = −1.
EXERCISE 1.7
Diagonlaise the following matrices by orthogonal transformation.
⎡ 6 −2 2⎤
3 −1⎥⎥
1. A = ⎢⎢ −2
⎢⎣ 2 −1 3⎥⎦
⎡ 10 −2 −5⎤
2. A = ⎢⎢ −2 2
3⎥⎥
⎢⎣ −5
3 5⎥⎦
2⎤
⎡ 8 −6
⎢
3. A = ⎢ −6
7 −4 ⎥⎥
⎢⎣ 2 −4
3⎥⎦
⎡ 2 −1 1⎤
4. A = ⎢⎢ −1 2 −1⎥⎥
⎢⎣ 1 −1 2⎥⎦
⎡ 7 −2 0 ⎤
5. A = ⎢⎢ −2 6 −2⎥⎥
⎢⎣ 0 −2
5⎥⎦
1 −1⎤
⎡ 2
⎢
6. A = ⎢ 1 1 −2⎥⎥
⎢⎣ −1 −2
1⎥⎦
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7. Find the symmetric matrix A whose eigen values and eigen vectors are given
⎡ 1⎤
(i) eigen values are 0, 2, and eigen vectors ⎢ ⎥ ,
⎣ −1⎦
⎡1⎤
⎢1⎥ .
⎣⎦
⎡ 1⎤
(ii) eigen values are 1, 2, 3 and eigen vectors ⎢⎢ −1⎥⎥ ,
⎢⎣ 0 ⎥⎦
⎡0 ⎤
⎢0 ⎥ ,
⎢ ⎥
⎢⎣ 1⎥⎦
⎡ 1⎤
⎢ 1⎥ .
⎢ ⎥
⎢⎣0 ⎥⎦
8. Reduce the quadratic form 8x 12 + 7x 22 + 3x 32 − 12x 1x 2 − 8x 2 x 3 + 4 x 3x 1 to the canonical form through an
orthogonal transformation and hence show that it is positive semi-definite.
9. Reduce the quadratic form 2x 12 + 6 x 22 + 2x 32 + 8x 1x 3 to canonical form by orthogonal reduction.
10. Reduce the quadratic form 3x 12 + 5x 22 + 3x 32 − 2x 2 x 3 + 2x 3x 1 − 2x 1x 2 to the canonical form by orthogonal
reduction.
11. Find the nature, index, signature and rank of the following Q.F, without reducing to canonical form.
(i) 3x 12 + 5x 22 + 3x 32 − 2x 2 x 3 + 2x 3x 1 − 2x 1x 2 .
(ii) 10 x 2 + 2 y 2 + 5z 2 + 6 yz − 10 zx − 4 xy .
(iii) 3x 2 − 2 y 2 − z 2 − 4 xy + 8xz + 12 yz .
12. Determine the nature of the following quadratic form f(x1, x2, x3) = x 12 + 2x 22 .
13. Find the nature of the quadratic form 2x2 + 2xy + 3y2.
14. Find the index, signature and rank of the Q.F in 3 variables x 12 + 2x 22 − 3x 32 .
15. Reduce the quadratic form x 12 + 5x 22 + x 32 + 2x 1x 2 + 2x 2 x 3 + 6 x 3x 1 to canonical form through an orthogonal
transformation.
16. Reduce the quadratic form 2x 12 + 6 x 22 + 2x 32 + 8x 1x 3 to canonical form.
17. Reduce the given quadratic form Q to its canonical form using orthogonal transformation
Q = x2 + 3y2 + 3z2 − 2yz.
ANSWERS TO EXERCISE 1.7
1. l = 2, 2, 8; eigen vectors [0 1 1]T, [1 1 −1]T, [2 –1 1]T
2. l = 0, 3, 14; eigen vectors [1 −5 4]T, [1 1 1]T, [−3 1 2]T
3. l = 0, 3, 15; eigen vectors [1 2 2]T, [2 1 −2]T, [2 −2 1]T
⎡ 22 −21 21⎤
⎢
⎥
4. l = 1, 1, 4; eigen vectors [1 1 0] , [−1 1 2] , [1 −1 1] , A = ⎢ −21 22 −21⎥
⎢⎣ 21 −21 22⎥⎦
T
T
T
3
5. l = 3, 6, 9; eigen vectors [1 2 2]T, [2 1 −2]T, [2 −2 1]T
⎡ −1 0 0 ⎤
6. D = ⎢⎢ 0 1 0 ⎥⎥
⎢⎣ 0 0 4 ⎥⎦
⎡1 1⎤
7. (i) A = ⎢
⎥
⎣1 1⎦
M01_ENGINEERING_MATHEMATICS-I _CH01_Part B.indd 112
⎡2 1 0⎤
(ii) ⎢ 1 2 0 ⎥
⎢
⎥
⎢⎣0 0 2⎥⎦
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Matrices ■
8. 3y 22 + 15y 32
11.
9. −2 y 12 + 6 y 22 + 6 y 32
1.113
10. 2 y 12 + 3y 22 + 6 y 32
(i) Eigen values are 2, 3, 6; positive definite, index = 3, signature = 3, rank = 3.
(ii) Eigen values are 0, 3, 14; positive semi-definite, index = 2, signature = 2, rank = 2.
(iii) Eigen values are 3, 6, −9; indefinite, index = 2, signature = 1, rank = 3.
12. It is a positive semi-definite form
13. The quadratic form is positive definite.
14. Eigen value = 1, 2, −3, Index = 2, Signature = 1, Rank = 3
15. −2 y 12 + 3y 22 + 6 y 32
16. −2 y 12 + 6 y 22 + 6 y 32
17. y 12 + 2 y 22 + 4 y 32
SHORT ANSWER QUESTIONS
⎡5 4 ⎤
2
1. If A 5 ⎢
⎥ , find the eigen values of A 2 2A 1 I.
⎣1 2⎦
1 2⎤
2. Prove that A 5 ⎡
and 23A21 have the same eigen values.
⎢ 2 1⎥
⎣
⎦
⎡ 2 2 1⎤
3. Two eigen values of the matrix A 5 ⎢⎢ 1 3 1⎥⎥ are equal to 1 each, find the eigen values of A21.
⎣⎢ 1 2 2 ⎦⎥
⎡ 2 0 1⎤
⎢
⎥
4. Find the sum and product of the eigen value of the matrix A 5 ⎢0 2 0 ⎥ .
⎢⎣ 1 0 2 ⎥⎦
6⎤
6
⎡ 4
3
2 ⎥⎥ are equal and they are double the third. Find the eigen values
5. Two eigen values of A 5 ⎢⎢ 1
⎢⎣21 25 22 ⎥⎦
of A2.
6. If l1, l2, ..., ln are the eigen values of an n 3 n
values of A3.
⎡ 1 1 3⎤
7. The matrix A 5 ⎢⎢ 1 5 1⎥⎥ has an eigen vector
⎣⎢3 1 1⎦⎥
matrix A, then show that l13 , l 32 ,…, l 3n are the eigen
⎡21⎤
⎢ 0 ⎥ , find the corresponding eigen value of A.
⎢ ⎥
⎢⎣ 1⎥⎦
⎡ 1 4⎤
8. Using Cayley-Hamilton theorem find the inverse of ⎢
⎥
⎣2 3⎦
9. If 2, 3 are eigen values of a square matrix A of order 2, express A2 in terms of A and I.
10. If A is an orthogonal matrix, show that A21 is also orthogonal.
11. For a given matrix A of order 3, A 5 32 and two of its eigen values are 8 and 2. Find the sum of the
eigen values
⎡ cos u sin u 0⎤
12. Check whether the matrix B = ⎢⎢2sin u cos u 0⎥⎥ is orthogonal? Justify.
⎢⎣ 0
0
1⎥⎦
⎡1 2 ⎤
13. Use Cayley-Hamilton theorem to find A4 2 4A3 2 5A2 1 A 1 2I when A 5 ⎢
⎥.
⎣4 3 ⎦
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14. True or false: If A and B are two invertible matrices then AB and BA have the same eigen values.
⎡2 2 1 ⎤
15. Find the eigen vector corresponding to the eigen value 1 of the matrix A 5 ⎢⎢1 3 1 ⎥⎥ .
⎢⎣1 2 2 ⎥⎦
2 22 ⎤
⎡ 1
⎢
0
3 ⎥⎥ .
16. Find the sum and product of the eigen values of the matrix A 5 ⎢ 1
⎢⎣22 21 23 ⎥⎦
17. A is a singular matrix of order three, 2 and 3 are the eigen values. Find the third eigen value.
18. Find the nature of the quadratic form 2x2 + 2xy + 3y2.
19. If the quadratic form ax2 + 2bxy + cy2 is positive definite (or negative definite) then prove that the
quadratic equation ax2 + 2bx + c = 0 has imaginary roots.
20. Find the index and signature of the quadratic form x 12 1 2 x 22 2 5 x 32 .
OBJECTIVE TYPE QUESTIONS
A. Fill up the blanks
⎡1 9⎤ ⎡ x
1. If ⎢
⎥=⎢
⎣2 0⎦ ⎣z
y 2⎤
⎥ , y < 0, then x − y + z + w = ________.
w⎦
⎡1 5⎤
⎡ −2 5⎤
2. If A − 2B = ⎢
⎥ , 2A − 3B = ⎢ 0 7⎥ , then B = ________.
3
7
⎣
⎦
⎣
⎦
⎡ cos 2 a
cos a sin a ⎤
p
, then M(a) ⋅ M(b) = ________.
3. If M (a) = ⎢
⎥ , if a and b differ by
sin 2 a ⎦
2
⎣cos a sin a
⎡ 1 0 2⎤
4. A = ⎢⎢0 2 1 ⎥⎥ satisfies the equation x3 − 6x2 + 7x + 2 = 0, then A−1 = ________.
⎢⎣ 2 0 3⎥⎦
⎡1 3 4 3⎤
⎡5 4 ⎤
6. Eigen values of ⎢
5. The rank of the matrix ⎢ 3 9 12 9 ⎥ is ________.
⎥ are ________.
⎢
⎥
⎣1 2 ⎦
⎢⎣ −1 −3 −4 −3⎥⎦
⎡ −2 −1⎤
7. An Eigen vector corresponding to the Eigen value −1 of ⎢
⎥ is ________.
⎣5 4⎦
⎡ 2 −2 2 ⎤
⎢
⎥
8. If 2 is Eigen value of ⎢1 1 1 ⎥ , then the other Eigen values are ________.
⎢⎣1 3 −1⎥⎦
⎡ 1 7 5⎤
9. The sum of the square of the Eigen values of the matrix ⎢⎢0 2 9⎥⎥ is ________.
⎢⎣0 0 5⎥⎦
⎡a 4 ⎤
10. The Eigen values of the matrix ⎢
⎥ are −2 and 3, then the values of a and b are ________.
⎣1 b ⎦
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Matrices ■
1.115
11. The nature of the quadratic form 2x 12 − x 22 is ________.
12. The nature of the quadratic form 2xy + 2yz + 2zx is ________.
13. The matrix of the quadratic form 3x 12 + 3x 22 − 5x 32 − 2x 1x 2 − 6 x 1x 3 − 6 x 2 x 3 is ________.
⎡ cos u sin u ⎤
14. The matrix A = ⎢
⎥ is ________.
⎣ − sin u cos u⎦
15. If A is an orthogonal matrix, then A−1 is ________.
B. Choose the correct answer
⎡2
⎢x
⎢
1. If the matrix ⎢ 1
⎢
⎢1
⎣
(a) ±2
−1
x
1
x
⎤
2 ⎥
⎥
2x 2 ⎥ is singular, then the value of x is
⎥
2 ⎥
⎦
(b) ±1
(c) ±3
⎡1
⎢
2. If v is a complex cube root of unity, then the matrix ⎢v2
⎢v
⎣
(a) Singular
(b) non-singular
3. If A2 − A + I = 0, then the inverse of A is
(a) A − I
(b) I − A
v2
v
1
(d) None of these
v⎤
⎥
1 ⎥ is
2⎥
v ⎦
(c) symmetric
(d) skew-symmetric
(c) A + I
(d) A
⎡ 5 5a a ⎤
4. Let A = ⎢⎢0 a 5a ⎥⎥ . If A 2 = 25, then a is equal to
⎢⎣0 0
5 ⎥⎦
(a) 25
(b) 1
(c)
1
5
(d) 5
5. If a, b, c are in A.P, then the system of equations 3x + 4y + 5z = a, 4x + 5y + 6z = b, 5x + 6y + 7z = c is
(a) consistent
(b) consistent with unique solution
(c) consistent with infinite number solutions
(d) consistent with finite number solutions
6. If A is non-singular, then the equation AB = 0 implies
(a) B is non-singular
(b) B is singular
(c) B = 0
(d) None of these
7. The system of equations x + 2y − z = 6, 3x − y − 2z = 3, 4x + 3y + z = 9 is
(a) consistent with unique solution
(b) consistent with infinite number of solutions
(c) inconsistent
(d) None of these
⎡ 3 −1 1 ⎤
8. Two Eigen values of the matrix A = ⎢⎢ −1 5 −1⎥⎥ are 3 and 6. Then the Eigen values of A−1 are
⎢⎣ 1 −1 3 ⎥⎦
1 1
(a) 1, ,
3 6
(b)
1 1
−1, ,
3 6
M01_ENGINEERING_MATHEMATICS-I _CH01_Part B.indd 115
(c)
1 1 1
− , ,
2 3 6
(d)
1 1 1
, ,
2 3 6
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Engineering Mathematics
9. If the Eigen values of a matrix A of order 3 are 2,3, and 4, then the Eigen values of adj A are
1 1 1
1 1 1
(a) 12, 8, 6
(b)
, ,
(c)
, ,
(d) None of these
2 3 4
12 8 6
10. If the sum of two Eigen values of 3 × 3 matrix A is equal to its trace, then the value of A is equal to
(b) −1
(a) 1
(c) 0
(d) None of these
⎡1 0 0 ⎤
11. The product of two Eigen values of the matrix A = ⎢0 3 −1⎥ is 2, then the third Eigen values is
⎥
⎢
⎢⎣0 −1 3 ⎥⎦
(a) 2
(b) 4
(c) 5
(d) 6
12. If A and B are invertible matrices of the same order, such that AB = BA then A and B are
(a) similar
(b) dissimilar
(c) have different Eigen values
(d) None of these
13. The index and signature of the quadrature for x 12 + 2x 22 − 5x 32 are
(a) 1, 2
(b)1, −2
(c) 2, 1
(d) 2, −2
14. The matrix of the quadratic form 3x2 + 2y2 − 4xy is
(a)
⎡ 3 −2⎤
⎢ −2 2 ⎥
⎣
⎦
⎡ 1 2⎤
(b) ⎢
⎥
⎣ −2 1 ⎦
⎡ −3 2 ⎤
⎢ 2 −2⎥
⎣
⎦
(c)
15. The nature of the quadratic form 2xy + 2yz + 2zx is
(a) indefinite
(b) definite
⎡ −1 −2⎤
(d) ⎢
⎥
⎣ 2 −1⎦
(c) positive definite
(d) negative definite
ANSWERS
A. Fill up the blanks
⎡ −3
⎢
=⎢1
⎢
⎢2
⎣
1. 6
⎡ −4 −5⎤
2. ⎢
⎥
⎣ −6 −7⎦
3. 0
4. A
5. rank = 1
6. 1, 6
⎡1⎤
7. ⎢ ⎥
⎣ −1⎦
8. −2, 2
−1
10. a = 2, b = −1 or a = 1, b = 2
11. indefinite
12. indefinite
⎡ 3 −1 3 ⎤
⎢
⎥
13. ⎢ −1 3 −3⎥
⎢⎣ −3 −3 −5⎥⎦
14. orthogonal
15. A −1 = A T
B. Choose the correct answer
1. (b)
2. (a) or (e)
3. (b)
10. (c)
11. (b)
12. (a) 13. (c)
M01_ENGINEERING_MATHEMATICS-I _CH01_Part B.indd 116
4. (c)
14. (a)
5. (c)
15. (a)
6. (c)
0
1
2
0
2⎤
1 ⎥⎥
2⎥
−1⎥⎦
9. 30
7. (a)
8. (d)
9. (a)
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Sequences and Series
2.0
2
INTRODUCTION
Many of the functions which are encountered in mathematical applications are represented by an
infinite series. The sum of an infinite series may or may not exist.
1 1 1 …
+ + + ∞ is equal to 2, whereas
2 4 8
the sum of the infinite geometric series 1 + 2 + 4 + 8 + … ∞ is ∞, which is not a real number and so
the sum does not exist. The usage of an infinite series, whose sum does not exist, will lead to absurd
conclusions in scientific investigations. Thus, an infinite series must be tested for the existence of its
sum. This aspect is the study of convergence of the infinite series and it is of vital importance to the
students of engineering and science.
In common usage of the English language, the words ‘sequence’ and ‘series’ are used in the
same sense to suggest a succession of things or events arranged in some order. But in Mathematics
‘sequence’ and ‘series’ are different concepts.
For example, the sum of the infinite geometric series 1 +
2.1 SEQUENCE
2.1.1 Infinite Sequence
Definition 2.1 If for every positive integer n there is associated a unique real number sn, then the
ordered set of numbers s1, s2, …, sn, … or {s1 , s 2 , …, s n , …} is called an infinite sequence. sn is called
the nth term or general term of the sequence.
The sequence is briefly written as {sn }∞n =1 or {sn }.
Precisely a sequence of real numbers is a function s: N → R, where N is the set of natural numbers
(or positive integers) and R is the set of real numbers.
The image of n ∈ N is the real number sn.
Examples of Infinite Sequences
(1) {1, 2, 4, 8, …, 2 n , …}
(2) {1, − 1, 1, − 1, …, ( −1) n , …}
(3)
{}
(4)
{2 + ( −1) } ,
(5)
{
1
, n = 1, 2, 3, …
n
n
n = 1, 2, 3, …
}
1
2 3
4
, − 2 , 2 , − 4 ,…
3 3 3
3
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Engineering Mathematics
2.1.2 Finite Sequence
Definition 2.2 If the domain of a sequence is a finite set of integers {1, 2, 3, …,n}, then the sequence
{s1, s2, …, sn} is called a finite sequence.
That is if the sequence has finite number of terms, then it is called a finite sequence.
For example: {1, 3, 5, 7, …, 199} is a finite sequence.
2.1.3 Limit of a Sequence
Definition 2.3 Let {sn} be a sequence of real numbers. A real number l is said to be the limit of the
sequence {sn}, if for every given e > 0, there exists a positive integer n0 (depending on e) such that
sn − l < e for all n ≥ n0 .
Symbolically, we write lim s n = l or s n → l as n → ∞.
n →∞
Note that when limit exists, it is unique.
2.1.4 Convergent Sequence
Definition 2.4 If a sequence of real numbers {sn } has the limit l, then the sequence is said to be
convergent and it converges to l.
If the sequence does not have a limit l, then it is said to be divergent.
That is if lim s n = ∞ or − ∞, then the sequence is divergent.
n →∞
For instance, the sequence {1, 2, 3, …, n, …} is divergent.
Examples
(1) The sequence {1, 1, 1, …} converges to 1.
1 1
1
1
(2) The sequence 1, , , …, , … converges to 0, since lim = 0.
n→∞
n
2 3
n
{
}
(3) The sequence {1, 2, 3, …} is divergent, since lim n = ∞.
n→ 0
(4) The sequence {−1, 1, −1, 1, −1, …} is divergent, since the limit does not exist.
Note A sequence {sn} is called a null sequence if it converges to zero.
2.1.5 Oscillating Sequence
Definition 2.5 If the sequence of real numbers {sn } diverges, but does not diverge to ` or −`, then the
sequence is said to be an oscillating sequence.
For example, the sequence {−1, 1, −1, 1, …} oscillates between −1 and 1.
2.1.6 Bounded Sequence
Definition 2.6 A sequence of real numbers {sn } is said to be bounded above if there exists a number
M such that
s n ≤ M ∀ n = 1, 2, 3, …
and bounded below if there exists a number m such that
m ≤ sn
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Sequences and Series ■
2.3
M is called an upper bound and m is called a lower bound of {sn }.
In other words, the sequence {sn } is bounded if there exists numbers m and M such that
m ≤ s n ≤ M ∀ n = 1, 2, 3, …
2.1.7 Monotonic sequence
Definition 2.7 A sequence {sn } is said to be a (monotonically) increasing sequence if s n ≤ s n +1 ∀ n ∈ N
and (monotonically) decreasing sequence if s n ≥ s n +1 ∀ n ∈ N
A sequence which is either increasing or decreasing is called a monotonic sequence.
A sequence {sn} is strictly increasing if sn < sn + 1 ∀ n ∈ N and strictly decreasing if sn , sn + 1 ∀ n ∈ N.
A sequence which is either strictly increasing or strictly decreasing is called strictly monotonic
We now state some important results:
1.
2.
3.
4.
Every convergent sequence is bounded.
An increasing sequence bounded above is convergent.
A decreasing sequence bounded below is convergent.
An unbounded sequence is not convergent. However bounded sequence need not be convergent.
For example: The sequence {1, −1, 1, −1, …} is bounded by −1 and 1, but not convergent. It is an
oscillating sequence.
5. A monotonic sequence is convergent if and only if it is bounded.
WORKED EXAMPLES
EXAMPLE 1
Test the convergence of the following infinite sequences:
⎧ 3n ⎫
(i) ⎨
2 ⎬
⎩ n 1 7n ⎭
(ii)
⎧⎪⎛ n ⎞ 2 ⎫⎪
(iii) ⎨⎜
⎟ ⎬
⎩⎪⎝ n 21 ⎠ ⎭⎪
{
n 11 2 n
}
⎧ 2 n31 7 n ⎫
(iv) ⎨ 3
.
2 ⎬
⎩ 5 n 1 3n ⎭
Solution.
(i) The given sequence is
{
3n
n + 7n 2
}
3n
3
=
1 + 7n
n + 7n 2
3
1
lim s n = lim
= = 0.
n →∞
n →∞ 1 + 7n
∞
∴ the nth term is s n =
∴
Hence, the sequence is convergent and converges to 0.
(ii) The given sequence is { n + 1 − n }
∴ the nth term is sn = n + 1 − n =
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(
n + 1 − n )( n + 1 + n )
(
n +1 + n)
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2.4
Engineering Mathematics
=
n +1− n
n +1 + n
=
1
n +1 + n
1
1
= = 0.
n +1 + n ∞
Hence, the sequence is convergent and converges to 0.
∴
lim sn = lim
n →∞
n →∞
⎧ n ⎞ 2⎫
(iii) The given sequence is ⎨⎛⎜
⎟ ⎬
⎩⎝ n −1⎠ ⎭
2
2
1
⎛ n ⎞
⎛ 1 ⎞
∴ the nth term is sn = ⎜
=
=
2
⎜
⎝ n −1⎟⎠
1⎟
⎛ 1⎞
⎜⎝ 1 − ⎟⎠
−
1
⎜⎝
⎟
n
n⎠
1
⎡ 1
⎤
lim sn = lim
=1
∴
2
⎢⎣{ n → 0, as n → ∞ ⎥⎦
n →∞
n →∞
⎛ 1⎞
⎜⎝1 − ⎟⎠
n
Hence, the sequence is convergent and converges to 1.
3
(iv) The given sequence is ⎧⎨ 2n + 7n ⎫⎬
3
⎩ 5n + 3n2 ⎭
7⎞
⎛
7
n3 ⎜ 2 + 2 ⎟ 2 + 2
3
+
2
n
7
n
⎝
⎠
n
n
∴ the nth term is s n =
=
=
3
5n 3 + 3n 2
3⎞
3⎛
5+
n ⎜5 + ⎟
⎝
⎠
n
n
7
2+ 2
⎡ 1 1
⎤
n =2
lim s n = lim
{ , 2 → 0 as n → ∞ ⎥
∴
⎢
n →∞
n →∞
3
5
⎣ n n
⎦
5+
n
2
Hence, the sequence is convergent and converges to .
5
EXAMPLE 2
Test the convergence of the following sequences:
n22 n
(i) s n 5 2
(ii) s n 5 2 1 (21) n (iii)
2n 1 n
{
}
1
2 3
4
,2 2 , 3 ,2 4 ,… .
3 3 3
3
Solution.
⎛ 1⎞
1
n 2 ⎜1 − ⎟ 1 −
n2 − n
⎝ n⎠
n
(i) The given sequence is s n = 2
=
=
2n + n
1
1⎞
2 ⎛
n ⎜2+ ⎟ 2+
⎝
n
n⎠
∴
1
1
n
lim s n = lim
= .
n →∞
n →∞
1 2
2+
n
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Sequences and Series ■
2.5
1
Hence, the sequence is convergent and converges to .
2
⎧2 + 1 = 3 if n is even
(ii) The given sequence is s n = 2 + ( −1) n = ⎨
⎩2 − 1 = 1 if n is odd
∴
lim sn = lim 3 = 3 if n is even
and
lim s n = lim 1 = 1 if n is odd
n →∞
n →∞
n →∞
n →∞
Since the limit is not unique, the sequence is not convergent. But it oscillates between 1 and 3.
Hence, the sequence is an oscillating sequence.
(iii) The given sequence is ⎧⎨ 1 , − 2 , 3 , − 4 , …⎫⎬ .
2
3
34
⎩3 3 3
⎭
⎧n
⎪⎪ 3n
n
∴ the nth term is s n = ( −1) n −1 ⋅ n = ⎨
3
⎪− n
⎪⎩ 3n
if n is odd
if n is even
n
=0
3n
Now, if n is odd
lim sn = lim
and if n is even
⎛ n⎞
lim sn = lim ⎜ − n ⎟ = 0
n →∞
n →∞ ⎝
3 ⎠
∴
lim sn = 0
n →∞
n →∞
[by L - Hopital’s rule]
[by L - Hopital’s rule]
n →∞
∴ the sequence is convergent and converges to 0.
EXAMPLE 3
⎧ 2n 2 7 ⎫
Show that the sequence ⎨
⎬ is monotonic increasing. Hence or otherwise prove that it is
⎩ 3n 1 2 ⎭
convergent.
Solution.
Let the given sequence be {s n } =
{ }
2n − 7
3n + 2
∴
sn =
2n − 7
3n + 2
To prove it is monotonic increasing sequence, we have to prove sn ≤ sn +1
Now
sn+1 =
∀ n ∈N
2( n + 1) − 7 2n − 5
=
3( n + 1) + 2 3n + 5
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Engineering Mathematics
2n − 7 2n − 5
−
3n + 2 3n + 5
(3n + 5)( 2n − 7) − (3n + 2)( 2n − 5)
=
(3n + 2)(33n + 5)
s n − s n +1 =
∴
=
⇒
6 n 2 − 11n − 35 − (6 n 2 − 11n − 10)
25
=−
(3n + 2)(3n + 5)
(3n + 2)(3n + 5)
s n − s n +1 < 0
∀ n ∈Ν
⇒
sn < sn + 1 ∀ n ∈Ν
∴ the sequence {sn } is monotonic increasing.
We know that monotonic increasing sequence, bounded above is convergent. So, we have to prove {sn }
is bounded above.
The given sequence is ⎧⎨ − 7 , − 3 , − 1 , …⎫⎬
⎩ 5 7 11 ⎭
Consider 12sn
2n − 7 3n + 2 − 2n + 7 n + 9
=
=
> 0 ∀ n ∈N
3n + 2
3n + 2
3n + 2
∀ n ∈ N ⇒ 1 > sn
∀ n ∈ N ⇒ sn < 1 ∀ n ∈ N
1 − sn > 0
∴
1 − sn = 1 −
⇒
∴ the sequence is bounded above.
Hence, the sequence is monotonically increasing and bounded above.
∴ the sequence is convergent.
Aliter: To prove the sequence is convergent
Given
∴
7⎞
7⎞
⎛
⎛
2n ⎜ 1 − ⎟ 2 ⎜ 1 − ⎟
⎝ 2n ⎠
⎝ 2n ⎠
=
2⎞
2⎞
⎛
⎛
3n ⎜ 1 + ⎟ 3 ⎜1 + ⎟
⎝ 3n ⎠
⎝ 3n ⎠
7⎞
⎛
1−
2 ⎜⎝ 2n ⎟⎠ 2
=
lim sn = lim
n →∞
n →∞ 3 ⎛
2⎞ 3
⎜⎝1 + ⎟⎠
3n
2n − 7
sn =
=
3n + 2
Hence, the sequence is convergent.
EXAMPLE 4
Show that the sequence whose nth term is
1
1
1
1
1…1
n 11 n 1 2
n1n
; nPN is convergent.
Solution.
Let the given sequence be {sn }.
∴ the nth term is
sn =
1
1
1
+
+ …+
n +1 n + 2
n+n
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Sequences and Series ■
s n+1 =
and
2.7
1
1
1
1
1
+
+ …+
+
+
n+2 n+3
n + n 2n + 1 2n + 2
To prove the sequence is monotonically increasing.
That is to prove s n − s n +1 ≤ 0 ∀ n ∈ N
Now s n − s n +1 =
1
1
1
1
1
1
1 ⎤
⎡ 1
+
+ …+
−
+
+ …+
+
+
n +1 n + 2
n + n ⎢⎣ n + 2 n + 3
n + n 2n + 1 2n + 2 ⎥⎦
1
1
1
−
−
n + 1 2n + 1 2n + 2
1
1
1
1
1
=
−
−
=
−
n + 1 2n + 1 2( n + 1) 2( n + 1) 2n + 1
2n + 1 − 2(nn + 1)
1
=
=−
< 0 ∀ n ∈N
2( n + 1)( 2n + 1)
2( n + 1)( 2n + 1)
=
⇒
s n − s n +1 < 0 ∀ n ∈ N
⇒
s n < s n +1
∀ n ∈N
∴ the sequence is monotonically increasing.
A monotonically increasing sequence bounded above is convergent.
∴ we have to prove that the sequence is bounded above.
That is to prove sn ≤ M for n ∈ N
Now
sn =
1
1
1
+
+ …+
n +1 n + 2
n+n
1
1
<
n +1 n
we know
n +1 > n ⇒
Similarly,
1
1
1
1
< , …,
<
n+2 n
n+n n
∴
sn <
1 1 … 1 n
+ + + = =1
n n
n n
⇒
sn < 1
∴ the sequence {sn } is bounded above.
Hence, the given sequence is monotonically increasing and bounded above.
∴ the sequence is convergent.
EXAMPLE 5
⎧ 1 ⎫
Show that the sequence ⎨
⎬ is monotonically decreasing and convergent.
⎩ 3n 1 5 ⎭
Solution.
1
Let the given sequence be {sn } =
.
3n + 5
{ }
∴ the nth term is
sn =
1
3n + 5
M02_ENGINEERING_MATHEMATICS-I _XXXX_CH02.indd 7
and
sn+1 =
1
1
=
3( n + 1) + 5 3n + 8
5/12/2016 10:53:18 AM
2.8
■
Engineering Mathematics
To prove the sequence is monotonically decreasing.
That is to prove sn − sn +1 ≥ 0 ∀ n ∈ N
1
1
3n + 8 − (3n + 5)
3
−
=
=
> 0 ∀ n ∈N
3n + 5 3n + 8 (3n + 5)(3n + 8) (3n + 5)(3n + 8)
Now
s n − s n +1 =
⇒
s n − s n +1 > 0 ∀ n ∈ N
⇒
s n > s n +1
∀ n ∈N
∴ the given sequence is decreasing sequence.
A decreasing sequence bounded below is convergent.
∴ we have to prove the sequence is bounded below.
That is to prove m ≤ sn , n ∈ N.
1 1 1
, , ,…
8 11 14
∴ it is bounded below by 0.
Hence, the sequence is convergent.
The given sequence is
EXAMPLE 6
n
⎧
⎫
Show that the sequence ⎨ 2
⎬ is decreasing and convergent.
n
n
1
21
⎩
⎭
Solution.
n
Let the given sequence be {sn } = 2
n + n −1
{
∴ the nth term is
and
}
n
n + n −1
n +1
n +1
n +1
sn+1 =
= 2
= 2
2
( n + 1) + n + 1 − 1 n + 2n + 1 + n n + 3n + 1
sn =
2
We have to prove that the sequence is decreasing.
That is to prove
Now
⇒
sn ≥ sn +1
s n − s n +1 =
∀ n ∈N
n
n +1
− 2
n + n − 1 n + 3n + 1
2
=
n ( n 2 + 3n + 1) − ( n + 1)( n 2 + n − 1)
( n 2 + n − 1)( n 2 + 3n + 1)
=
n 3 + 3n 2 + n − ( n 3 + n 2 − n + n 2 + n − 1)
n 2 +11
=
( n 2 + n − 1)( n 2 + 3n + 1)
( n 2 + n − 1)( n 2 + 3n + 1)
s n − s n +1 > 0
∀ n ∈N
⇒
s n > s n +1
∀ n ∈N
∴ the sequence is a decreasing sequence.
M02_ENGINEERING_MATHEMATICS-I _XXXX_CH02.indd 8
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Sequences and Series ■
2.9
2 3
The given sequence is 1, , ,…
5 11
∴ the sequence is a bounded below by 0 [and above by 1].
Hence, the sequence is convergent
1. Show that the sequence
EXERCISE 2.1
{ }
n converges to 1.
n +1
⎧ n2 ⎫
1
2. Show that the sequence ⎨ 2 ⎬ converges to .
2
⎩ 2 n + 1⎭
2
⎧
⎫
3. Show that the sequence ⎨ 3n + n ⎬ converges to 3.
2
⎩ n + 3n + 1⎭
{ }
{ }
4. Test the convergence of the sequence n − 2 .
n+2
5. Test the convergence of the sequence 3n + 4 .
2n + 1
⎧ 3 + ( −1) n ⎫
6. Test the convergence of the sequence ⎨
⎬.
⎩ n2 ⎭
ANSWERS TO EXERCISE 2.1
4. Converges
2.2
5. Converges
6. Converges
SERIES
Definition 2.8 If {un } be a sequence of real numbers, then the expression u1 + u2 + u3 + u4 + … + un + …
∞
is called an infinite series and it is denoted by ∑ un or
n =1
∑u .
n
th
un is called the n term of the series.
2.2.1 Convergent Series
Definition 2.9 Let u1 + u 2 + u3 + … + u n + … be an infinite series.
If s n = u1 + u 2 + … + u n , then s n is called the nth partial sum of the series. If the sequence of partial sums
∞
∞
n=1
n=1
{sn } converges to l, then we say that the series ∑ un converges to l and it is written as ∑ un = l .
Then l is called the sum of the series.
M02_ENGINEERING_MATHEMATICS-I _XXXX_CH02.indd 9
5/12/2016 10:53:28 AM
■
2.10
Engineering Mathematics
2.2.2 Divergent Series
∞
Definition 2.10 If the sequence of partial sums {sn } of the infinite series diverges, then the series ∑ un
n=1
diverges.
That is, if lim s n = ∞, then
n →∞
∞
∑u
n =1
∞
n
diverges to ` and if lim s n = −∞, then ∑ un diverges to −`.
n →∞
n=1
2.2.3 Oscillatory Series
Definition 2.11 If the sequence of partial sums {sn } of the infinite series diverges, but does not diverge
to + ` or −`, then the sequence {sn } is said to oscillate. Then we say that the series
∞
∑u
n
is an
n=1
oscillatory series.
Examples
(1) 1 +
1
1 1 1
(2) 1 − + 2 − 3 + … + ( −1) n n + …
3 3 3
3
1 1
1
1
+ 2 + 3 + …+ n + …
2 2
2
2
∞
(3) 1 − 1 + 1 − 1 + …
(4)
∞
2n
∑ n! n
(5)
n
n =1
n=1
x n −1
∑ (2n − 1)
p
2.2.4 General Properties of Series
1. The convergence or divergence of an infinite series is unaffected by addition or removal of finite
number of terms.
2. The convergence or divergence of an infinite series is unaffected when each term of the series is
multiplied by a non-zero number.
∞
∞
n=1
n=1
3. If ∑ un and ∑ vn are convergent series with sums a and b respectively, then for any pair of real
numbers l and m, the series
∞
∑ [lu
n =1
2.3
n
± mv n ] converges with sum la ± mb .
SERIES OF POSITIVE TERMS
The discussion of the convergence of any type of series of real numbers depend upon the series of
positive terms. So, we shall discuss in detail the series with positive terms.
∞
Definition 2.12 A series ∑ un , where un > 0 ∀ n ∈ N, is called a series of positive terms.
n =1
A series of positive terms can either converge or diverge to `. It can never oscillate.
2.3.1 Necessary Condition for Convergence of a Series
Theorem 2.1 If the series of positive terms ∑ un is convergent, then lim un = 0
n →∞
M02_ENGINEERING_MATHEMATICS-I _XXXX_CH02.indd 10
5/12/2016 10:53:35 AM
Sequences and Series ■
Proof
Let
2.11
Sn = u1 + u2 + … + un−1 + un
Sn−1 = u1 + u2 + … + un−1
then
∴
Sn − Sn−1 = un
Since ∑ un is convergent, lim s n = l , where l is finite.
n →∞
lim u n = lim(s n − s n −1 ) = lim s n − lim s n −1 = l − l = 0
∴
n →∞
n →∞
n →∞
n →∞
This is only a necessary condition, but not sufficient.
That is if lim u n = 0, we cannot say the series is convergent.
n →∞
For example: the series
1+
1
1 1 … 1 …
+ + + + is divergent, but lim u n = lim = 0.
n
→∞
n
→∞
n
2 3
n
∞
* If lim u n ≠ 0 , then the series ∑ u n is not convergent.
n→∞
■
n51
2.3.2 Test for Convergence of Positive Term Series
The definition of convergence of a series depends on the limit of the sequence of partial sum {sn }.
But in practice it will be difficult to find sn in many cases.
So, it is necessary to device methods by which we can decide the convergence or divergence of a
series without finding the partial sum sn .
A standard technique used in studying convergence of positive term series is comparison test.
The given series ∑ un is compared with a known series ∑ vn , which is known as auxiliary series.
2.3.3 Comparison Tests
1. If ∑ un and ∑ vn are positive term series such that un ≤ c vn ∀ n ∈ N, for some positive
constant c, then ∑ un is convergent if ∑ vn is convergent.
If un ≥ c vn ∀ n ∈N and if ∑ vn is divergent, then ∑ un is divergent.
2. Limit form: Let ∑ un and ∑ vn be two positive term series such that lim
n →∞
Then ∑ un and ∑ vn behave alike.
un
= l (l ≠ 0).
vn
If ∑ vn converges, then ∑ un converges and if ∑ vn diverges, then ∑ un diverges.
Here we compare ∑ un with ∑ vn .
Proof
Given ∑ un and ∑v n are series of positive terms and lim
n →∞
Since
un
> 0 for all n = 1, 2, 3, …, we have l > 0.
vn
M02_ENGINEERING_MATHEMATICS-I _XXXX_CH02.indd 11
un
= l (l ≠ 0).
vn
5/12/2016 10:53:45 AM
2.12
■
Engineering Mathematics
Choose ε > 0, such that l − ε > 0.
Then by the definition of limit, there exists a positive integer n0 such that
un
− l < ε for all n ≥ n0
vn
∴
−ε <
un
−l < ε
vn
⇒
l−ε <
un
<l+ε
vn
⇒
[{ x − a < ε ⇒ a − ε < x < a + ε]
( l − ε ) v n < un < ( l + ε ) v n
∀ n ≥ n0
[{ vn ≥ 0 ∀ n ∈ N ]
Consider un < (l + ε )vn
Case (i): If ∑ vn is convergent, then ∑ vn is a finite number.
∴
(l + ε ) ∑ vn is finite number
But
∑ un < ( l + ε ) ∑ v n
∴
∑ un < a finite number ∀ n ≥ n0
⇒
Hence, ∑ un is convergent.
∑ un < a finite number as n → ∞
Case (ii): If ∑ vn is divergent, then ∑ vn → ∞ as n → ∞
Consider
∴
( l − ε ) v n < un
∀ n ≥ n0
( l − ε ) ∑ v n < ∑ un
⇒
∴ ∑ un is divergent.
∑ un > ( l − ε ) ∑ v n
⇒
∑ un → ∞ as n → ∞
■
Note
1. If l = 0, then ∑ un is convergent if ∑ vn is convergent.
2. If l = ∞, then ∑ un is divergent if ∑ vn is divergent.
3. In order to discuss the convergence of ∑ un by comparison test, we consider ∑ vn whose convergence
is known already.
Two standard series used for comparison are the following.
(i) The geometric series with positive terms a + ar + ar 2 + … , where a > 0 and r > 0.
It converges if 0 < r < 1 and diverges if r ≥ 1.
∞
1 1
1
1
1
+ p + p + … + p + … = ∑ p , where p > 0.
p
1
2
3
n
n
n =1
It converges if p > 1 and diverges if p ≤ 1.
The p-series is also known as harmonic series of order p.
In many problems, the auxiliary series is chosen as the p-series for particular values of p.
(ii) The p-series is
M02_ENGINEERING_MATHEMATICS-I _XXXX_CH02.indd 12
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Sequences and Series ■
2.13
⎛ 1⎞
For choosing the auxiliary series we write un in the form f ⎜ ⎟ , then decide vn.
⎝ n⎠
1
1 ⎛ 1⎞
For example if un = p f ⎜ ⎟ , Then we take v n = p , p > 0.
⎝ n⎠
n
n
WORKED EXAMPLES
EXAMPLE 1
Test the convergence of the series
1
3
5
1
1
1….
1⋅ 2 ⋅ 3 2 ⋅ 3 ⋅ 4 3 ⋅ 4 ⋅ 6
Solution.
Let the given series be ∑ un .
1
3
5
∴
∑ un =
+
+
+…
1⋅ 2 ⋅ 3 2 ⋅ 3 ⋅ 4 3 ⋅ 4 ⋅ 6
The numerators 1, 3, 5, … are in A.P. So, the nth term is 1 + ( n − 1)2 = 2n − 1]. In the denominator, first
factors are 1, 2, 3,… and the nth term is n, the second factors are 2, 3, 4,… and the nth term is (n + 1)
and the third factors are 3, 4, 5,… and the nth term is n + 2.
1⎞
1⎞
⎛
⎛
2n ⎜ 1 − ⎟
⎜⎝1 − ⎟⎠
⎝
⎠
−
2
1
2
n
2
n
2
n
=
= 2
Then the nth term is u n =
n ( n + 1)( n + 2)
⎛ 1⎞ ⎛ 2⎞ n ⎛ 1⎞ ⎛ 2⎞
n 3 ⎜1 + ⎟ ⎜1 + ⎟
⎜⎝1 + ⎟⎠ ⎜⎝1 + ⎟⎠
⎝ n⎠ ⎝ n⎠
n
n
1
Take nn 5 2
n
1⎞
1⎞
⎛
⎛
2 ⎜1 − ⎟
⎜⎝1 − ⎟⎠
⎝
un
2
2n ⎠
2n
∴
= 2
× n2 =
vn n ⎛ 1 ⎞ ⎛ 2 ⎞
⎛ 1⎞ ⎛ 2⎞
⎜⎝1 + ⎟⎠ ⎜⎝1 + ⎟⎠
⎜⎝1 + ⎟⎠ ⎜⎝1 + ⎟⎠
n
n
n
n
∴
⎡
⎢⎣
∴
1⎞
⎛
2 ⎜1 − ⎟
⎝ 2n ⎠
un
= 2 ( ≠ 0)
lim = lim
n →∞ v
n →∞ ⎛
1⎞ ⎛ 2⎞
n
⎜⎝1 + ⎟⎠ ⎜⎝1 + ⎟⎠
n
n
1
⎤
→ 0 as n → ∞ ⎥
⎦
n
∴ by comparison test ∑ un and ∑ vn behave alike.
But ∑ vn = ∑
1
is convergent, since p = 2 > 1 in p-series
n2
Hence, ∑ un is convergent.
EXAMPLE 2
∞
Test the convergence of
⎛ ( n11)( n12) ⎞
⎟⎠ .
n2 n
n51
∑ ⎜⎝
Solution.
Let the given series be ∑ un .
M02_ENGINEERING_MATHEMATICS-I _XXXX_CH02.indd 13
∴
∞
⎛ ( n + 1)( n + 2) ⎞
∑ un = ∑ ⎜
⎟⎠
n2 n
n =1 ⎝
5/12/2016 10:54:04 AM
2.14
■
Engineering Mathematics
un =
Then the nth term is
Take nn 5
( n + 1)( n + 2)
n2 n
=
⎛ 1⎞ ⎛ 2⎞
n 2 ⎜1 + ⎟ ⎜1 + ⎟
⎝ n⎠ ⎝ n⎠
n2 n
=
⎛ 1⎞ ⎛ 2⎞
⎜⎝1 + ⎟⎠ ⎜⎝1 + ⎟⎠
n
n
n
1
n
⎛ 1⎞ ⎛ 2⎞
1+
1+
un ⎜⎝ n ⎟⎠ ⎜⎝ n ⎟⎠
⎛ 1⎞ ⎛ 2⎞
=
× n = ⎜1 + ⎟ ⎜1 + ⎟
⎝ n⎠ ⎝ n⎠
vn
n
∴
∴
lim
n →∞
But ∑ v n = ∑
un
⎛ 1⎞ ⎛ 2⎞
= lim ⎜1 + ⎟ ⎜1 + ⎟ = (1 + 0)(1 + 0) = 1 ( ≠ 0)
⎝ n⎠ ⎝ n⎠
n
→∞
vn
⎡ 1
⎤
⎢⎣{ n → 0 as n → ∞ ⎥⎦
1
1
is divergent, since p = < 1 in p-series.
1/ 2
2
n
Hence, ∑ un is divergent.
EXAMPLE 3
∑(
∞
Discuss the convergence of
3
)
11 n 3 2 n .
n 51
Solution.
Let the given series be ∑ un .
∴
∑ un = ∑ ( 3 1 + n3 − n )
∞
n =1
1/ 3
⎡⎛
⎤
1
1⎞
Then the n term is u n = 1 + n − n = n 1 + 3 − n = n ⎢⎜1 + 3 ⎟ − 1⎥
n
⎣⎝ n ⎠
⎦
3
th
3
3
1
< 1, so expanding by binomial series, we get,
n3
For large values of n,
1⎞
⎛
⎜⎝1 + 3 ⎟⎠
n
1/ 3
1⎛1 ⎞
−1
1 1 3 ⎜⎝ 3 ⎟⎠
= 1+ ⋅ 3 +
2!
3 n
1⎛1 ⎞ ⎛1 ⎞
2
⎜ − 1⎟ ⎜ − 2⎟
⎛ 1⎞
3⎝3 ⎠ ⎝3 ⎠
⎜⎝ 3 ⎟⎠ +
3!
n
3
⎛ 1⎞ …
⎜⎝ 3 ⎟⎠ +
n
1 1 1 1 …
= 1+ ⋅ 3 −
+
3 n
9 n6
⇒
∴
1⎞
⎛
⎜⎝1 + 3 ⎟⎠
n
1/ 3
1 1 1 1
−1 = ⋅ 3 − ⋅ 6 + …
3 n
9 n
1/ 3
⎡⎛
⎤
1⎞
⎛1 1 1 1
⎞
n ⎢⎜1 + 3 ⎟ − 1⎥ = n ⎜ ⋅ 3 − ⋅ 6 + …⎟
⎝
⎠
⎝
⎠
3
9
n
n
n
⎣
⎦
1 1 1 1
1
= ⋅ 2 − ⋅ 5 +… = 2
3 n
9 n
n
M02_ENGINEERING_MATHEMATICS-I _XXXX_CH02.indd 14
⎛ 1 1 1 …⎞
⎜⎝ − ⋅ 3 + ⎟⎠
3 9 n
5/12/2016 10:54:09 AM
Sequences and Series ■
un =
∴
Take nn 5
1
n2
∴
∴
∴
lim
2.15
⎛ 1 1 1 …⎞
⎜⎝ − ⋅ 3 + ⎟⎠
3 9 n
1
n2
un
1 ⎛ 1 1 1 …⎞
1 1 1 …
2
=
+
⎜ − ⋅ + ⎟⎠ × n = −
v n n 2 ⎝ 3 9 n3
3 9 n3
un
⎛1 1 1
⎞ 1
= lim ⎜ − ⋅ 3 + …⎟ =
( ≠ 0)
⎠ 3
n →∞ v
n →∞ ⎝ 3
9
n
n
⎡ 1
⎤
⎢⎣{ n 3 → 0 as n → ∞ ⎥⎦
∴ by comparison test, ∑ un and ∑ vn behave alike.
1
But ∑ vn = ∑ 2 is convergent, since p = 2 > 1 in p-series.
n
Hence, ∑ un is convergent.
EXAMPLE 4
∑(
∞
Test the convergence of
)
n 4 11 2 n 4 21 .
n 51
Solution.
Let the given series be ∑ un .
∑ un = ∑ ( n 4 + 1 − n 4 − 1 )
∞
∴
n =1
Then the nth term is u n = n 4 + 1 − n 4 − 1
=
(
n4 + 1 + n4 −1
n + 1 − ( n − 1)
4
=
∴
∴
1
n2
un
=
vn
4
n4 + 1 + n4 −1
2
=
Take nn 5
n 4 + 1 − n 4 + 1) ( n 4 + 1 + n 4 − 1)
n +1 + n −1
4
4
=
2
⎛
⎛
1 ⎞
1 ⎞
n2 ⎜ 1+ 4 ⎟ + n2 ⎜ 1 − 4 ⎟
⎝
⎠
⎝
n
n ⎠
2 × n2
⎛
1
1⎞
n2 ⎜ 1 + 4 + 1 − 4 ⎟
⎝
n
n ⎠
lim
n →∞
un
= lim
vn n →∞
M02_ENGINEERING_MATHEMATICS-I _XXXX_CH02.indd 15
1+
1
1
+ 1− 4
4
n
n
2
⎛
1
1 ⎞
n2 ⎜ 1+ 4 + 1 − 4 ⎟
⎝
n
n ⎠
2
=
2
=
1+
=
1
1
+ 1− 4
4
n
n
2
= 1 ( ≠ 0)
1+1
1
⎡
⎤
⎢⎣{ n 4 → 0 as n → ∞ ⎥⎦
5/12/2016 10:54:15 AM
2.16
■
Engineering Mathematics
∴ by comparison test, ∑ un and ∑ vn behave alike.
1
is convergent, since p = 2 > 1 in p-series.
n2
Hence, ∑ un is convergent.
But ∑ vn = ∑
EXAMPLE 5
⎛ n 11 2 n ⎞
⎟⎠ .
np
n 51
∞
∑ ⎜⎝
Discuss the convergence of
Solution.
∞ ⎛
⎞
∴ ∑ un = ∑ ⎜ n + 1p− n ⎟
⎝
⎠
Let the given series be ∑ un .
n
n =1
Then the nth term is un =
=
=
=
(
n +1 − n
np
n + 1 − n )( n + 1 + n )
np ( n +1 + n )
n +1− n
np ( n +1 + n )
1
n
p
(
n +1 + n )
=
1
⎛
1 ⎞
n p ⋅ n1/ 2 ⎜ 1 + + 1⎟
⎝
n ⎠
=
1
⎛
1 ⎞
n p +1/ 2 ⎜ 1 + + 1⎟
⎝
n ⎠
1
Take nn 5
n
p1
1
2
un
=
vn
∴
∴
lim
n →∞
1
n
1
p+
2
⎛
1 ⎞
⎜⎝ 1 + + 1⎟⎠
n
×n
p+
1
2
=
1
⎛
1 ⎞
⎜⎝ 1 + + 1⎟⎠
n
un
1
1
1
⎛
⎞
= lim
=
( ≠ 0)
=
⎟
v n n →∞ ⎜
1
1
2
+
1
⎜ 1 + + 1⎟
⎝
⎠
n
⎤
⎡ 1
⎢⎣{ n → 0 as n → ∞ ⎥⎦
∴ by comparison test ∑ un and ∑ vn behave alike.
1
But ∑ vn = ∑
n
p+
1
2
is convergent if p +
Hence, ∑ un is convergent if p >
1
>1 ⇒
2
p>
1
1
and divergent if p + ≤ 1 ⇒
2
2
p≤
1
2
1
1
and divergent if p ≤
2
2
M02_ENGINEERING_MATHEMATICS-I _XXXX_CH02.indd 16
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Sequences and Series ■
2.17
EXAMPLE 6
∞
Discuss the convergence of
3 n 21
.
2 n 11
∑
n51
Solution.
Let the given series be ∑ un .
∴
∞
∑ un = ∑
n =1
n
3n − 1
=
2n + 1
Then the nth term is un =
3n − 1
2n + 1
1
1
n
1− n
3n = ⎛ 3 ⎞ 2
3
⎜ ⎟
1 ⎝ 2⎠ 1+ 1
1+ n
2n
2
32 1 −
n
22
n
⎛ 3⎞ 2
Take nn 5 ⎜ ⎟
⎝ 2⎠
1
3n
1
1
1+ n
1− n
2 =
3
n
1
1+ n
⎛ 3⎞ 2
2
⎜⎝ ⎟⎠
2
1−
n
⎛ 3⎞ 2
⎜⎝ ⎟⎠
2
un
=
vn
∴
u
lim n = lim
n →∞ v
n →∞
n
1
3n = 1 ( ≠ 0 )
1
1+ n
2
1−
⎡ 1 1
⎤
⎢⎣{ 3n , 2 n → 0 as n → ∞ ⎥⎦
∴ by comparison test, ∑ un and ∑ vn behave alike.
n
n
∞ ⎛
3 2
3⎞
But ∑ vn = ∑ ⎛⎜ ⎞⎟ = ∑ ⎜
is a geometric series with common ratio r =
⎝ ⎠
⎝ 2 ⎟⎠
n =1 2
n =1
∞
3
>1
2
∴ ∑ vn is divergent.
Hence, ∑ un is divergent.
EXERCISE 2.2
Test the convergence of the following series:
1.
1
1
1 …
+
+
+
1⋅ 2 2 ⋅ 3 3 ⋅ 4
∞
3.
∑ (1 + n )
n =1
1
, p > 0, q > 0
( 2 + n )q
p
M02_ENGINEERING_MATHEMATICS-I _XXXX_CH02.indd 17
∞
2.
∑
n =1
4.
2 n3 + 3
5n3 + 7
1
2
3
+
+
+…
1 + 2 −1 1 + 2 −2 1 + 2 −3
5/12/2016 10:54:27 AM
■
2.18
5.
1⋅ 2
3⋅ 4
5⋅6
+ 2 2 + 2 2 +…
2
2
3 ⋅4
5 ⋅6
7 ⋅8
6. 1 +
22 − 12 32 − 22 … n 2 − ( n − 1) 2 … ⎡
1⎤
+
+ + 2
+
⎢⎣Take v n = n ⎥⎦
22 + 1 32 + 22
n + ( n − 1) 2
∞
7.
Engineering Mathematics
∑
n =1
9. 1 +
∞
np
n +1 − n
2
8.
1
∑ n log n
n= 2
3
1 2
3
+ 3 + 4 +…
2
2
3
4
10.
3
5
7
+ 2 2 + 2 2 +…
2
1 ⋅2
2 ⋅3 3 ⋅ 4
2
∞
1
⎞
11. Test the convergence of the series ∑ ⎛
⎜⎝
⎟.
n + n + 1⎠
n =1
∑(
∞
12. Test the convergence of
)
n2 + 1 − n .
n =1
13. Test the convergence of the series
2 −1
3 −1
4 −1 …
+
+
+ ∞.
33 − 1 43 − 1 53 − 1
14. Test the convergence of the series
2 3 4 …
+ + + .
1p 2 p 3p
∞
1
1
15. Test the convergence of ∑ sin .
n
n=1 n
ANSWERS TO EXERCISE 2.2
1. Convergent
2. Divergent
3. Convergent if p + q > 1 and divergent if p + q ≤ 1
4. Divergent
5. Convergent
3
3
7. Convergent if p < − and divergent if p ≥ −
2
2
10. Convergent
9. Divergent
12. Divergent
13. Convergent
14. Convergent if p > 2 and divergent if p ≤ 2
6. Divergent
8. Divergent
11. Divergent
15. Convergent
2.3.4 De’ Alembert’s Ratio Test
∞
Let
∑u
n
be a series of positive terms such that lim
n=1
n →∞
un
= l . Then the series ∑ un is convergent if
un +1
l > 1, divergent if l < 1 and the test fails to give a definte result if l = 1.
That is ∑ un may converge or diverge if l = 1.
M02_ENGINEERING_MATHEMATICS-I _XXXX_CH02.indd 18
5/12/2016 10:54:35 AM
Sequences and Series ■
2.19
Proof
Given ∑ un is a series of positive terms and lim
n →∞
Since un > 0 ∀ n ∈ N,
Since lim
n →∞
un
= l.
un +1
un
> 0 ⇒ l > 0.
un +1
un
= l , by definition of limit, given ε > 0, there exists a positive integer n0 such that
un +1
un
− l < ε ∀ n ≥ n0
u n +1
−ε <
⇒
⇒
un
− l < ε ∀ n ≥ n0
u n +1
l−ε <
(i) Let l > 1:
un
< l + ε ∀ n ≥ n0
u n +1
Choose ε > 0 such that l − ε > 1, then
l−ε <
un
< l + ε ∀ n ≥ m [for this ε, n0 is m]
u n +1
un
u n +1
Consider
l−ε <
∀ n≥m
⇒
un
> l−ε ∀ n≥ m
u n +1
Replace n by m, m + 1, m + 2, …, n − 1; we get
um
u
u
u
> l − ε, m +1 > l − ε, m + 2 > l − ε, …, n −1 > l − ε
um+1
um + 2
um + 3
un
Multiplying all these inequalities, we get
u m u m +1 u m + 2 … u n −1 > (l − ε)(l − ε ) …(l − ε)
⋅
⋅
u m +1 u m + 2 u m + 3
un
⇒
um
n−m
> (l − ε )
un
⇒
um ( l − ε ) n
>
un ( l − ε ) m
M02_ENGINEERING_MATHEMATICS-I _XXXX_CH02.indd 19
[n − m factors]
∀ n ≥ m +1 ⇒
∀ n ≥ m +1
5/12/2016 10:54:41 AM
2.20
■
Engineering Mathematics
⇒
um
(l − ε) m
> un
(l − ε) n
∀ n ≥ m +1⇒
1
(l − ε) n
⇒
un < um ( l − ε ) m ⋅
⇒
∑ un < um ( l − ε ) m ∑
1
(l − ε) n
But ∑
1
1
is an infinite geometric series with r =
<1
n
l
−
ε
(l − ε)
∴ ∑
1
converges as n → ∞
(l − ε) n
(ii) Let l , 1:
∴
∀ n ≥ m +1
∀ n ≥ m +1
⎡
⎢⎣{ l − ε > 1 ⇒
1
⎤
<1
l − ε ⎥⎦
∑u n is convergent if l > 1
Choose ε > 0 such that l + ε < 1.
Then there exists a positive integer k such that
l−ε <
Consider
un
<l+ε ∀ n≥k
u n +1
[for this ε, n0 is k]
un
<l+ε ∀ n≥k .
u n +1
Replacing n by k, k + 1, k + 2, …, n − 1, we get
uk
< l + ε,
u k +1
u k +1
un −1
< l + ε, …,
<l+ε
uk + 2
un
Multiplying all these inequalities, we get
uk uk +1 … un −1
⋅
< (l + ε )(l + ε ) …(l + ε)
u k +1 u k + 2
un
uk
(l + ε) n
< (l + ε) n − k =
un
(l + ε) k
⇒
⇒
[(n−k) factors]
uk ( l + ε ) k ⋅
1
< un
(l + ε) n
1
(l + ε) n
⇒
un > uk ( l + ε ) k ⋅
⇒
∑ un > uk ( l + ε ) k ∑
M02_ENGINEERING_MATHEMATICS-I _XXXX_CH02.indd 20
1
.
(l + ε) n
5/12/2016 10:54:46 AM
Sequences and Series ■
But ∑
1
1
is a geometric series with common ratio r =
>1
l+ε
(l + ε) n
∴ the series ∑
2.21
1
⎡
⎤
⎢{ l + ε < 1, l + ε > 1⎥
⎣
⎦
1
is divergent.
(l + ε) n
Hence, ∑ un is divergent if l < 1.
(iii) Let l = 1. Then ∑ un may converge or diverge.
We have lim
n →∞
But ∑
1
Consider ∑ un = ∑ .
n
un
n +1
⎛ 1⎞
= lim
= lim ⎜1 + ⎟ = 1
⎝ n⎠
n
→
∞
n
→∞
n
un +1
1
is divergent, since p = 1 in p-series.
n
Now consider the series ∑ un = ∑
1
n2
2
un
( n + 1) 2
⎛ 1⎞
= lim
2
=
lim
1
+
=1
⎜
⎟
n →∞ u
n →∞
n
n→∞ ⎝
n +1
n⎠
∴
lim
But ∑
1
is convergent, since p = 2 > 1 in p-series.
n2
So, when lim
n→∞
un
= 1, the test fails to give definite answer as ∑ un may be convergent or divergent.
un +1
■
Note Sometimes this test is stated as below.
If lim
n→∞
un +1
= l , then ∑ un is convergent if l < 1 and divergent if l > 1. The test fails if l = 1.
un
WORKED EXAMPLES
EXAMPLE 1
∞
Test the convergence of the series
Solution.
Let the given series be ∑ un .
n ! 2n
.
n
n51 n
∑
∴
Then
M02_ENGINEERING_MATHEMATICS-I _XXXX_CH02.indd 21
∞
n !2 n
n
n =1 n
∑ un = ∑
un =
n!2 n
( n + 1)! 2 n +1
and un +1 =
n
n
( n + 1) n +1
5/12/2016 10:54:54 AM
2.22
■
Engineering Mathematics
un
n ! 2 n ( n + 1) n +1
= n ⋅
u n +1
( n + 1)! 2 n +1
n
∴
n
=
∴
⎛ 1⎞
n
nn ⎜1 + ⎟
⎝ n⎠
1 ⎛ 1⎞
=
=
+
1
⎜
⎟
2n n
2 ⎝ n⎠
( n +1)( n + 1) n
2n n ( n + 1)
n
un
e
1 ⎛ 1⎞
= lim ⎜1 + ⎟ = > 1
⎠
n →∞ u
n→∞ 2 ⎝
n
2
n +1
[{ 2 < e < 3]
lim
∴ by De’ Alembert’s ratio test ∑ un is convergent.
EXAMPLE 2
Test the convergence or divergence of the series
x
x2
x3
1
1
1 … ∞,
1⋅ 2 2 ⋅3 3 ⋅ 4
x > 0.
Solution.
Let the given series be ∑ un .
x
x2
x3 …
+
+
+
1⋅ 2 2 ⋅ 3 3 ⋅ 4
xn
x n +1
un =
and u n +1 =
( n + 1)( n + 2)
n ( n + 1)
∴
∑ un =
Then
un
xn
( n + 1)( n + 2) n + 2 1 ⎡ 2 ⎤ 1
=
⋅
=
⋅ = 1+ ⋅
u n +1 n ( n + 1)
n x ⎢⎣ n ⎥⎦ x
x n +1
∴
∴
lim
n →∞
un
⎛ 2⎞ 1 1
= lim ⎜1 + ⎟ ⋅ = , x > 0
⎝ n⎠ x x
n
→
∞
un +1
∴ by De’Alemberts’ ratio test, ∑ un is convergent if
1
>1 ⇒
x
x < 1 and divergent if
1
<1 ⇒
x
x > 1.
If x = 1, then the test fails to give a conclusion.
In this case, the series becomes
∴
un =
1
=
n( n + 1)
∴
un
=
vn
1
⎛ 1⎞
n2 ⎜1 + ⎟
⎝ n⎠
1
1
1 …
+
+
+ ∞
1⋅ 2 2 ⋅ 3 3 ⋅ 4
1
⎛ 1⎞
n2 ⎜1 + ⎟
⎝ n⎠
× n2 =
M02_ENGINEERING_MATHEMATICS-I _XXXX_CH02.indd 22
.
Take vn =
1
n2
1
1+
1
n
5/12/2016 10:55:04 AM
Sequences and Series ■
∴
lim
n →∞
2.23
un
1
= lim
= 1 ( ≠ 0)
n
→∞
1
vn
1+
n
∴ ∑ un and ∑ vn behave alike by comparison test.
But ∑ vn = ∑
1
is convergent, since p = 2 > 1 in p-series.
n2
∴ ∑ un is convergent if x = 1.
Hence, the given series is convergent if 0 < x ≤ 1 and divergent if x . 1.
EXAMPLE 3
Test for the convergence of the series
1
x2
x4
x6
1
1
1
1 … ∞.
2 1 3 2
4 3 5 4
Solution.
Let the given series be ∑ un
∴
∑ un =
un =
Then
2 1
+
x2
3 2
x 2n−2
( n + 1) n
+
x4
4 3
and
+
x6
5 4
+ …∞
u n +1 =
x 2( n +1) −2
( n + 1 + 1) n + 1
=
x 2n
( n + 2) n + 1
⎛ 2⎞
1+
un
( n + 2) n + 1 n + 2 n + 1 1 ⎜⎝ n ⎟⎠
x 2n −2
1 1
1+ ⋅ 2 .
=
⋅
=
⋅
=
1
u n +1 ( n + 1) n
n +1
n x2
n x
x 2n
1+
n
∴
∴
1
⎛ 2⎞
⎜⎝1 + ⎟⎠
un
n 1+ 1 ⋅ 1 = 1 , x ≠ 0
lim
= lim
n →∞ u
n→∞ ⎛
1⎞
n x2 x2
n +1
⎜⎝1 + ⎟⎠
n
∴ by De’ Alembert’s ratio test, ∑ un is convergent if
1
2
> 1 ⇒ x < 1 ⇒ −1 < x < 1, x ≠ 0
x2
1
< 1 ⇒ x 2 > 1 ⇒ x < −1 or x > 1.
x2
x = 0, the series is convergent, trivially.
and divergent if
If
If
1
=1 ⇒
x2
x2 = 1 ⇒
x = ±1, the test fails to give a conclusion.
M02_ENGINEERING_MATHEMATICS-I _XXXX_CH02.indd 23
5/12/2016 10:55:13 AM
2.24
■
Engineering Mathematics
1
In this case, the series is
2 1
∴
lim
n →∞
1
=
( n + 1) n
1
3 2
+
1
4 3
+…
1
.
⎡ 1⎤
n ⎢1 + ⎥
⎣ n⎦
3
un
n2
⎛ 1⎞
= 3
= ⎜1 + ⎟
vn
⎛ 1⎞ ⎝ n⎠
n 2 ⎜1 + ⎟
⎝ n⎠
un =
∴
∴
+
Take vn =
3
2
1
3
n2
un
⎛ 1⎞
= lim ⎜1 + ⎟ = 1 ( ≠ 0 )
⎝ n⎠
n
→
∞
vn
So, by comparison test ∑u n and ∑ vn behave alike.
But ∑ vn = ∑
1
3
2
is convergent, since p =
3
> 1 in p-series
2
n
∴ ∑ un is convergent if x 2 = 1. When x = 0 the series is trivially convergent
Hence, the given series is convergent if −1 ≤ x ≤ 1 and is divergent if x < −1 or x > 1.
EXAMPLE 4
Discuss the convergence of the series
1
1
1
1
1
1
1
1 … ∞, for positive values of x.
1 1 x 1 1 2 x 2 1 1 3x 3 1 1 4 x 4
Solution.
Let the given series be ∑ un
∴
Then
∴
⇒
∑ un =
un =
1
1
1
1
+
+
+
+…
1 + x 1 + 2x 2 1 + 3x 3 1 + 4x 4
1
1+ nx n
and un +1 =
1
1 + ( n + 1) x n +1
un
1 + ( n + 1)x n +1
=
u n +1
1 + nx n
un
u n +1
M02_ENGINEERING_MATHEMATICS-I _XXXX_CH02.indd 24
1 + ( n + 1)x n +1
nx n
⎡⎣ Dividing Nr and Dr by nx n ⎤⎦
=
1 + nx n
nx n
1
1
⎛ n + 1⎞
⎛ 1⎞
x
+⎜
+ ⎜1 + ⎟ x
n
n
⎝ n ⎟⎠
⎝ n⎠
nx
nx
=
=
1
1
1+ n
+1
n
nx
nx
5/12/2016 10:55:26 AM
Sequences and Series ■
1
⎛ 1⎞
+ ⎜1 + ⎟ x
n
⎝ n⎠
un
lim
= lim nx
,
n →∞ u
n→∞
1
n +1
1+ n
nx
∴
If x > 1, then
2.25
x > 0.
1
1
→ 0 as n → ∞ and → 0 as n → ∞
nx n
n
∴
lim
n →∞
un
= x >1
u n +1
∴ by De’ Alembert’s ratio test, the series ∑ un is convergent if x > 1
If x = 1, then the series becomes
1 1 1 …
+ + + , which is divergent, since p = 1, in p-series
2 3 4
If 0 < x < 1, then
∴
xn < 1 ∀ n ≥ 1
nx n < n
∀ n ≥1
1
1
>
1+ x 2
1
1
1 + 2x2 < 1 + 2 = 3 ⇒
>
2
1 + 2x
3
1
1
1 + 3x 2 < 1 + 3 = 4 ⇒
> and so on.
1 + 3x 2 4
1
1
∴
,…
>
n
1+ n
1 + nx
1
1
1
1 1 1
∴
+
+
+ … > + + + …,
2
2
1 + x 1 + 2x
2 3 4
1 + 3x
which is divergent.
∴ ∑ un is divergent if 0 < x < 1.
Thus, the given series is convergent if x > 1 and divergent if 0 < x ≤ 1.
∴
1+ x < 2
⇒
EXERCISE 2.3
Test for the convergence or divergence of the following series.
1. 1 +
3.
2 p 3p 4 p …
+ ∞,
+ +
2 ! 3! 4 !
p>0
∞
2.
n!3n
∑
n
n=1 n
1
2
3
+
+
+…
1 + 2 1 + 22 1 + 32
n
2
3
4. 1 + 2x + 6 x + 14 x + … + 2 − 2 x n −1 + … ∞, x > 0, n ≥ 2
5
9
17
2n + 1
∞
5.
∑
n =1
n n
x , x >0
n +1
M02_ENGINEERING_MATHEMATICS-I _XXXX_CH02.indd 25
∞
6.
∑
n =1
n
n +1
2
xn, x > 0
5/12/2016 10:55:36 AM
■
2.26
Engineering Mathematics
∞
7.
xn
, x >0
∑
2n
n =1 1 + x
9.
4
4 ⋅12
4 ⋅12 ⋅ 20 …
+
+
+
18 18 ⋅ 27 18 ⋅ 27 ⋅ 36
8. 1 +
10.
11.
x 1 x 3 1⋅ 3 x 5 1⋅ 3 ⋅ 5 x 7 …
+ ⋅ +
+
+ , x >0
1 2 3 2⋅ 4 5 2⋅ 4⋅6 7
12.
2 −1
3 −1
4 −1 …
+ 3
+ 3
+
3 −1 4 −1 5 −1
∞
14.
13.
3
n
∑ (n + 1)(n + 2)x
n
12 ⋅ 22 12 ⋅ 22 ⋅ 32 …
+
+
1⋅ 3 ⋅ 5 1⋅ 3 ⋅ 5 ⋅ 7 ⋅ 9
1⋅ x 22 x 2 33 x 3 4 4 x 4 …
+
+
+
+ , x >0
1!
2!
3!
4!
1 1⋅ 3 1⋅ 3 ⋅ 5
1⋅ 3 ⋅ 5 ⋅ 7 …
+
+
+
+
4 4 ⋅ 7 4 ⋅ 7 ⋅10 4 ⋅ 7 ⋅10 ⋅13
,x >0
n =1
15. Prove that the series 1 +
a + 1 ( a + 1)( 2a + 1) ( a + 1)( 2a + 1)(3a + 1) …
+
+
+ ∞
b + 1 (b + 1)( 2b + 1)
(b + 1)(3b + 1)
converges if b > a > 0 and diverges if a ≥ b > 0.
n3 + 1
.
n
n =1 2 + 1
∞
16. Discuss the convergence of the series ∑
17. Test the convergence of the series shown below
x n −1
1 x x2 …
+ … ∞, p > 0 and x > 0.
+ p+ p + +
p
( 2 n − 1) p
1 3
5
18. Test the convergence of the series
1 22 32 4 2 52 …
+ + + + + by ratio test.
3 32 33 34 35
19. Test for the convergence of the series
x
x2
x3 …
+
+
+ , x > 0 and x ≠ 1.
1⋅ 2 3 ⋅ 4 5 ⋅ 6
ANSWERS TO EXERCISE 2.3
Convergent for all p > 0
2. Divergent
3. Convergent
Convergent if 0 < x < 1; divergent if x ≥ 1
5. Convergent if 0 < x < 1; divergent if x ≥ 1
Convergent if 0 < x < 1; divergent if x ≥ 1
Convergent if 0 < x < 1; and x > 1; divergent if x = 0
8. Divergent
1
1
9. Convergent
10. Convergent if 0 < x < ; divergent if x ≥
e
e
11. Convergent if 0 < x ≤ 1; divergent if x > 1
12. Convergent
13. Convergent
14. Convergent if 0 < x < 1 and divergent if x ≥ 1 16. Convergent
1.
4.
6.
7.
17. Convergent if 0 < x < 1, p > 0 and if x = 1, p > 1. It is divergent if x > 1, p > 0 and if x = 1, p ≤ 1.
18. Convergent
M02_ENGINEERING_MATHEMATICS-I _XXXX_CH02.indd 26
19. The series ∑ un is convergent if x <1 and divergent if x > 1.
5/12/2016 10:56:07 AM
Sequences and Series ■
2.27
2.3.5 Cauchy’s Root Test
1
∞
If ∑ un is a series of positive terms and lim u nn = l , then ∑ un converges if l < 1 and diverges if l > 1.
n→∞
n=1
The test fails to give a definite conclusion if l = 1.
Proof
1
∞
Given ∑ un is a series of positive terms and lim u nn = l
n →∞
n=1
∴ by the definition of limit, given ε > 0, there exists an integer n0 such that
1
n
n
1
n
n
u − l < ε for n $ n0
⇒ l − ε < u < l + ε for n ≥ n0
(i) Let l < 1. Choose ε > 0 such that l + ε < 1.
Then there exists a positive m such that
1
l − ε < unn < l + ε ∀ n ≥ m
⇒
( l − ε ) n < un < ( l + ε ) n
un < ( l + ε ) n
Consider
∀ n≥m
⇒ ∑ un < ∑ ( l + ε ) n
But ∑(l + ε ) n is a geometric series with common ratio r = l + ε < 1
∴ the series ∑(l + ε ) n is convergent.
Hence, ∑ un is convergent by comparison test
(ii) Let l > 1. Choose ε > 0 such that l − ε > 1.
Then there exists a positive integer k such that
1
l − ε < unn < l + ε for all n ≥ k
1
⇒
(l − ε) n < un < (l + ε ) n for all n ≥ k
Consider
( l − ε ) n < un
∴
⇒ un > ( l − ε ) n
∀ n≥k
∑ un > ∑ ( l − ε ) n
But ∑(l − ε) n is a geometric series with common ratio r = l – ε > 1
∴
∑(l − ε) n is divergent.
Hence, ∑ un is divergent by comparison test.
M02_ENGINEERING_MATHEMATICS-I _XXXX_CH02.indd 27
5/12/2016 10:56:14 AM
2.28
■
Engineering Mathematics
(iii) If l = 1, then the test fails to give a definite conclusion.
1
1
1
1
⎛ 1⎞ n
⎛ 1 ⎞n
1
1
n
Consider ∑ and ∑ 2 , we find that lim u n = lim ⎜ ⎟ = 1 and lim u nn = lim ⎜ 2 ⎟ = 1
⎝
⎠
n →∞
n→∞ n
n →∞
n→∞ ⎝ n ⎠
n
n
But ∑
1
1
is divergent and ∑ 2 is convergent.
n
n
∴ ∑ un may be convergent or divergent if l = 1
■
Note
1. Root test is more general or stronger than the ratio test, because there are cases where the ratio
test fails but root test gives definite conclusion.
2. The root test is used when the general term un contains index interms of n.
WORKED EXAMPLES
EXAMPLE 1
∞
1⎞
⎛
Test the convergence of ∑ ⎜1 1 ⎟
⎝
⎠
n
n 51
2 n2
.
Solution.
Let the given series be ∑ un .
Then
∴
∞
⎛ 1⎞
∑ un = ∑ ⎜ 1 + ⎟
⎝ n⎠
n =1
∴
⎛ 1⎞
u n = ⎜1 + ⎟
⎝ n⎠
− n2
1
1
⎡⎛ 1 ⎞ − n ⎤ n ⎡ 1 ⎤ − n
(u n ) = ⎢⎜1 + ⎟ ⎥ = ⎢1 + ⎥
⎣ n⎦
⎣⎝ n ⎠ ⎦
2
1
n
⇒
⎛ 1⎞
lim unn = lim ⎜1 + ⎟
n→∞
n→∞ ⎝
n⎠
− n2
−n
= e −1 =
1
<1
e
[{ 2 < e < 3]
∴ by Cauchy’s root test, ∑ un is convergent.
EXAMPLE 2
3
∞
1 ⎞
Test the convergence of ∑ ⎛⎜1 1
⎟
n⎠
n 51 ⎝
2n 2
.
Solution.
3
Let the given series be ∑ un .
M02_ENGINEERING_MATHEMATICS-I _XXXX_CH02.indd 28
∴
∞
1 ⎞
⎛
∑ un = ∑ ⎜ 1 +
⎟
n⎠
n =1 ⎝
−n2
5/12/2016 10:56:20 AM
Sequences and Series ■
2.29
3
−n2
1 ⎞
⎛
un = ⎜ 1 +
⎟
⎝
n⎠
Then
1
1
1 ⎞
⎛
⎜⎝1 +
⎟
n⎠
∴
1
n
n
lim u = lim
n →∞
n →∞
n2
⎤n
⎥
⎥
⎥
⎥⎦
1
⎡
[ un ] = ⎢
3
n2
⎢⎛
1 ⎞
⎢ ⎜1 +
⎟
n⎠
⎢⎣ ⎝
=
3
1 ⎞
⎛
⎜⎝1 +
⎟
n⎠
1
n
⇒
1
=
1
=
3
1
n2 ⋅
1 ⎞
⎛
⎜⎝1 +
⎟
n⎠
n
1
1 ⎞
⎛
⎜⎝1 +
⎟
n⎠
=
n
1
n2
=
1
1 ⎞
⎛
⎜⎝1 +
⎟
n⎠
n
1
<1
e
∴ by Cauchy’s root test, ∑ un is convergent.
EXAMPLE 3
Test the convergence of the series given below
⎛ 22
2⎞
⎜⎝ 2 2 ⎟⎠
1
1
21
Solution.
Let the given series be ∑ un.
∴
⎛ 22 2 ⎞
∑ un = ⎜ 2 − ⎟
⎝1
1⎠
−1
⎛ 33 3 ⎞
+⎜ 3 − ⎟
⎝ 2 2⎠
⎡ ( n + 1) n +1 n + 1⎤
−
Then un = ⎢
⎥
n ⎦
⎣ n n +1
−2
⎛ 33
3⎞
1⎜ 3 2 ⎟
⎝2
2⎠
⎛ 44 4 ⎞
+⎜ 4 − ⎟
⎝3
3⎠
22
⎛ 44 4 ⎞
1 ⎜ 41 ⎟
⎝3
3⎠
23
1… ∞.
−3
+…
−n
⎡⎛ n + 1⎞ n +1 n + 1⎤
= ⎢⎜
⎥
⎟ −
n ⎦
⎣⎝ n ⎠
−n
⎡ ⎛ 1 ⎞ n +1 ⎛ 1 ⎞ ⎤
= ⎢⎜1 + ⎟ − ⎜1 + ⎟ ⎥
⎝ n⎠ ⎦
⎣⎝ n ⎠
−n
⎡⎛ 1 ⎞ n ⎛ 1 ⎞ ⎛ 1 ⎞ ⎤
= ⎢⎜1 + ⎟ ⋅ ⎜1 + ⎟ − ⎜1 + ⎟ ⎥
⎣⎝ n ⎠ ⎝ n ⎠ ⎝ n ⎠ ⎦
−n
1
∴
−n n
−1
⎧⎛ 1 ⎞ n ⎛ 1 ⎞ ⎛ 1 ⎞ ⎫
⎪⎧ ⎡⎛ 1 ⎞ n ⎛ 1 ⎞ ⎛ 1 ⎞ ⎤ ⎪⎫
⎨
⎬
u = ⎢⎜1 + ⎟ ⎜1 + ⎟ − ⎜1 + ⎟ ⎥
= ⎨⎜ 1 + ⎟ ⎜ 1 + ⎟ − ⎜ 1 + ⎟ ⎬
⎩⎝ n ⎠ ⎝ n ⎠ ⎝ n ⎠ ⎭
⎩⎪ ⎣⎝ n ⎠ ⎝ n ⎠ ⎝ n ⎠ ⎦ ⎭⎪
1
n
n
1
n
⎡
1
1
1 ⎤
∴ lim u nn = lim ⎢⎛⎜1 + ⎞⎟ ⎛⎜1 + ⎞⎟ − ⎛⎜1 + ⎞⎟ ⎥
n →∞
n →∞ ⎣ ⎝
n⎠ ⎝ n⎠ ⎝ n⎠ ⎦
−1
= [e ⋅1 − 1]−1 =
1
<1
e −1
[{ 2 < e < 3, e − 1 > 1]
∴ by Cauchy’s root test, ∑ un is convergent.
M02_ENGINEERING_MATHEMATICS-I _XXXX_CH02.indd 29
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■
2.30
Engineering Mathematics
EXAMPLE 4
Discuss the nature of the series
2
3
1 2
⎛ 3⎞
⎛ 4⎞
+ x + ⎜ ⎟ x 2 + ⎜ ⎟ x 3 + … ∞ for x > 0 .
⎝
⎠
⎝ 5⎠
2 3
4
Solution.
Let ∑ un be the series, omitting the first term.
2
2
4
⎛ 3⎞
∑ un = x + ⎜ ⎟ x 2 + x 3 + …
⎝
⎠
3
4
5
∴
⎛ n +1 ⎞ n ⎡⎛ n + 1 ⎞
un = ⎜
x = ⎢⎜
⎟
⎝ n + 2 ⎟⎠
⎣⎝ n + 2 ⎠
n
Then
⎤
x⎥
⎦
n
1
1
1+
⎧⎪ ⎡⎛ n + 1 ⎞ ⎤ n ⎫⎪ n ⎛ n + 1 ⎞
nx
u = ⎨ ⎢⎜
⎟ x ⎥ ⎬ = ⎜⎝
⎟x =
2
n + 2⎠
⎩⎪ ⎣⎝ n + 2 ⎠ ⎦ ⎭⎪
1+
n
1
n
n
∴
1⎞
⎛
1+
⎜
⎟
n
lim u = lim ⎜
x=x
n →∞
n →∞
2⎟
⎜1+ ⎟
⎝
n⎠
1
n
n
∴
∴ by Cauchy’s root test, ∑ un converges if x , 1 and diverges if x . 1.
If x = 1, the test fails to give a definite conclusion.
1 ⎤
⎡
n
⎢⎣1 + n + 1⎥⎦
1
1
1
⎛ n +1 ⎞
=
=
=
In this case, u n = ⎜
=
n
n
n +1
n
⎝ n + 2 ⎟⎠
1 ⎞
1 ⎤
⎛
⎛ n + 1 + 1⎞
⎡
⎛ n + 2⎞
1
+
1
+
⎜⎝
⎟
⎜⎝
⎟
⎜⎝
⎟
⎢⎣ n + 1⎥⎦
n +1 ⎠
n + 1⎠
n +1⎠
lim un = lim
∴
n →∞
n →∞
1 ⎞
⎛
⎜⎝1 +
⎟
n + 1⎠
1 ⎞
⎛
⎜⎝ 1 +
⎟
n + 1⎠
n +1
=
n +1
⎡
⎤
1 ⎤
⎡
1
+
= e⎥
⎢{ lim
⎥
n→∞ ⎢
⎣ n + 1⎦
⎢⎣
⎦⎥
1
≠0
e
∴ when x = 1, the series is divergent by theorem 2.1, page 2.10.
Hence, the series is convergent if x , 1 and divergent if x $ 1.
2.3.6 Cauchy’s Integral Test
If u(x) is positive, decreasing and integrable function in [1, `) such that u( n) = un
∞
∞
n=1
1
∀ n ∈ N, then the
series ∑ un and ∫ u ( x )dx converge or diverge together.
∞
∞
If ∫ u ( x )dx converges (i.e., the value of the integral exists), then ∑ un converges.
1
M02_ENGINEERING_MATHEMATICS-I _XXXX_CH02.indd 30
n=1
5/12/2016 10:56:30 AM
Sequences and Series ■
∞
∞
1
n=1
2.31
If ∫ u ( x )dx diverges (i.e., the value does not exist), then ∑ un diverges.
Proof
∞
Let {sn} be the sequence of partial seem of the series ∑ un .
n=1
sn = u1 + u2 + u3 + … + un
∴
(1)
n
I n = ∫ u ( x )dx , n = 2, 3, 4, …
Let
1
Since u(x) is monotonic decreasing in [1, `), we have
u( n) ≥ u( x ) ≥ u( n + 1) for n ≤ x ≤ n + 1, n ∈ N
Also u(x) is positive and integrable on [1, `)
n +1
∫
∴
u( n)dx ≥
n +1
∫
n
u( x )dx ≥
n
u( n) ≥
⇒
n +1
∫ u(n + 1)dx
n
n +1
∫ u( x) dx ≥ u(n + 1)
∀ n ∈N
n
Since u(n) = un, n ∈ N; we have
un ≥
n +1
∫ u(x ) dx ≥ u
n +1
n
Putting n = 1, 2, 3, …, n − 1 and adding, we get
2
3
1
2
u1 + u 2 + … + u n −1 ≥ ∫ u ( x )dx + ∫ u ( x )dx + … +
≥ u 2 + u3 + … + u n
n
∫ u(x )dx
n −1
n
sn −1 ≥ ∫ u( x )dx ≥ sn − u1
⇒
∀ n ∈N
[From (1)]
1
sn −1 ≥ I n ≥ sn − u1
⇒
I n ≥ sn − u1
Taking
⇒
sn − u1 ≤ I n
∀ n ∈N
∀ n ∈N
∀ n ∈N
sn ≤ I n + u1
∴
lim sn ≤ lim I n + lim u1
n →∞
n →∞
n →∞
⇒
∞
∞
lim sn ≤ ∫ u( x )dx + u1
n →∞
1
If ∫ u ( x )dx is convergent, then lim sn is finite
1
n →∞
∴
M02_ENGINEERING_MATHEMATICS-I _XXXX_CH02.indd 31
∑ un is convergent.
5/12/2016 10:56:35 AM
■
2.32
Engineering Mathematics
sn −1 ≥ I n
Taking
⇒
∞
lim sn −1 ≥ lim I n
n→∞
n →∞
⇒
lim sn −1 ≥ ∫ u( x )dx.
n→∞
1
∞
If ∫ u ( x )dx does not exist, then lim sn −1 does not exist.
n →∞
1
∴
∑u
Thus,
n
is divergent.
∞
∑ un and ∫ u(x )dx behave alike.
■
1
WORKED EXAMPLES
EXAMPLE 1
∞
Discuss the convergence of the p-series
1
∑n
p
, p > 0.
n=1
Solution.
Let the given series be ∑ un .
∴
∞
1
p
n =1 n
∑ un = ∑
1
1
Consider u ( x ) = p , x ∈[1, ∞)
.
p
x
n
u(x) is positive and decreasing function of x for x > 1.
∴ by Cauchy’s integral test,
un =
Then
∞
∑ un and ∫ u ( x )dx behave alike.
1
∞
∞
1
1
∞
∞
1
∫ u( x)dx = ∫ x
Now
p
dx.
1
If p 5 1, then ∫ u ( x )dx = ∫ dx = [log x ]1∞ = ∞
x
1
1
∴ ∑ un diverges if p = 1
So, the integral diverges.
∞
∞
1
1
If p > 1, then ∫ u ( x )dx = ∫
∞
− p +1
1
⎤
dx = ⎡ x
p
⎢
⎥
x
⎣ − p + 1⎦1
∞
1 ⎡ 1 ⎤
1 ⎛1 ⎞
1
1
[{ p − 1 > 0]
=
=
=
⎜ − 1⎟ = −
1 − p ⎢⎣ x p −1 ⎥⎦1 1 − p ⎝ ∞ ⎠
1− p p −1
Since the integral exists, by Cauchy’s integral test, ∑ un is convergent if p > 1.
M02_ENGINEERING_MATHEMATICS-I _XXXX_CH02.indd 32
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Sequences and Series ■
∞
2.33
∞
∞
⎡ x − p +1 ⎤
1
1
(∞ − 1) = ∞
If p < 1, then ∫ u ( x )dx = ∫ p dx = ⎢
⎥ =
⎣ − p + 1⎦1 1 − p
1
1 x
∴ the integral does not exist and hence, divergent.
So, ∑ un is divergent if p < 1.
[{ 1 − p > 0]
Thus, ∑ un is convergent if p > 1 and divergent if p ≤ 1.
EXAMPLE 2
Test the convergence of the series 1 1 1 1 1 1 … 1 1 1 … ∞.
2 5 10
1 1 n2
Solution.
Let the given series be ∑ un
∴
∑ un =
un =
Then
1 1 1 …
1
+…
+ + + +
2 5 10
1+ n2
1
.
1 + n2
Consider u ( x ) =
1
,
1+ x 2
x ≥1
∴ u( x ) is positive and decreasing function of x.
∞
By cauchy’s integral test ∑ un and ∫ u ( x )dx behave alike.
1
∞
Now
∞
1
∫ u( x)dx = ∫ 1 + x
1
dx = [ tan −1 x ]1 = tan −1 ∞ − tan −1 1 =
∞
2
1
p p p
− = .
2 4 4
∞
∴ ∫ u ( x )dx is convergent.
1
Hence, ∑u n is convergent.
EXAMPLE 3
Test the convergence of the series
1
1
1
1 … ∞, p > 0.
1
1
p
p
2 (log e 2 )
3(1og e 3)
4(log e 4)p
Solution.
∞
Let the given series be ∑ un
n=2
∞
∴
∑u
n=2
n
=
1
1
1
+
+ … ∞,
+
p
p
2(log e 2)
3(log e 3)
4(log e 4) p
p>0
∞
1
, p>0
p
n = 2 n(log e n)
=∑
M02_ENGINEERING_MATHEMATICS-I _XXXX_CH02.indd 33
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■
2.34
Engineering Mathematics
un =
Then
1
n (loge n ) p
Consider u ( x ) =
1
, p > 0, x ≥ 2
x (loge x ) p
Clearly u(x) is positive and decreasing function of x for x ≥ 2
∞
∑ un and ∫ u ( x )dx behave alike.
∴ by Cauchy’s integral test,
2
∞
∞
1
∫ u( x)dx = ∫ x(log
Now
2
2
When x = 2,
x)
Put t = log x ∴ dt =
dx.
1
dx
x
t = log 2 and when x = `, t = log ` = `
∞
∫ u(x )dx =
∴
e
p
2
∞
If p > 1, then ∫ u( x )dx =
2
∞
1
dt
p
t
loge 2
∫
∞
∞
⎡ t − p +1 ⎤
t − p dt = ⎢
⎥
∫
⎣ − p + 1⎦ loge 2
loge 2
∞
=
1 ⎛ 1 ⎞
⎜
⎟
1 − p ⎝ t p −1 ⎠ loge 2
=
1 ⎡1
1
1
1
1
1
⎤
−
=−
⋅
=
⋅
p −1 ⎥
p −1
⎢
(1 − p ) (loge 2)
( p − 1) (loge 2) p−1
1 − p ⎣ ∞ (loge 2) ⎦
[{ p − 1 > 0]
∞
∞
2
n= 2
∴ the integral ∫ u ( x )dx exists and hence, ∑ un is convergent for p > 1
∞
∴
∞
∞
⎡ t − p +1 ⎤
1
1
∫1 u(x )dx = log∫ 2 t p dt = ⎢⎣ − p + 1⎥⎦ = 1 − p (∞ − loge 2) = ∞
loge 2
e
If p < 1, then
[{ 1 − p > 0]
∞
∫ u(x )dx
does not exist and hence, divergent
2
∴
∞
∑u
n
is divergent if p < 1
n= 2
∞
If p 5 1, then ∫ u ( x )dx =
2
∴
∞
1
dt = (log t )∞log 2 = log ∞ − log log 2 = ∞
t
loge 2
∫
∞
∫ u( x)dx does not exist and hence, divergent
2
∴
∞
∑u
n=2
n
is divergent if p = 1
Thus, the given series ∑ un is convergent if p > 1 and divergent if 0 < p ≤ 1.
M02_ENGINEERING_MATHEMATICS-I _XXXX_CH02.indd 34
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Sequences and Series ■
2.35
EXAMPLE 4
Using integral test, determine whether the series
Solution.
Let the given series be ∑ un
un =
Then
ln n
.
n2
ln n
is convergent or divergent.
2
n51 n
∑
∞
ln n
[ln n = log en ]
2
n =1 n
∑ un = ∑
∴
Consider u ( x ) =
ln x
,x ≥2
x2
(1) [{ u ( x ) = 0 if x = 1]
u ( x ) > 0 ∀ x ≥ 2.
Clearly
To prove u(x) is monotonic function of x, test the sign of u ′( x ).
Differentiating (1) w.r. to x, we get
1
x 2 ⋅ − ln x ⋅ 2 x
x − 2 x ln x x(1 − 2 ln x ) 1 − 2 ln x
x
u ′( x ) =
=
=
=
4
x
x4
x4
x3
If x ≥ 2, then ln x ≥ ln 2
[{ logarithm is an increasing function]
∴
2 ln x ≥ 2 ln 2
∴
u ′( x ) < 0 ∀ x ≥ 2
⇒
−2 ln x ≤ −2 ln 2
⇒ 1 − 2 ln x ≤ 1 − ln 4 < 0
[{ ln 4 = 1.39]
Hence, u ( x ) is decreasing in [2, `)
∞
∴ by Cauchy’s integral test, ∑ un and
n= 2
∞
∫
But
ln 2
∞
∫ u( x)dx behave alike
ln 2
∞
ln x
dx .
x2
ln 2
u( x )dx = ∫
Put ln x = t ∴ dt =
1
dx
x
When x = 2, t = ln 2 and when x = ∞, t = ln ∞ = ∞
∞
∴
∫
ln 2
u( x )dx =
∞
∞
t
−t
∫ln 2 e t dt = ln∫2 te dt
[{ ln x = t ⇒ x = e t ]
∞
⎡ e−t
e−t ⎤
= ⎢t
− 1⋅
⎥
( −1) 2 ⎦ ln 2
⎣ −1
[ by Bernoulli’s formula ]
∞
= ⎡⎣ −e − t (t + 1) ⎤⎦ ln 2 = ⎡⎣0 + e − ln 2 (ln 2 + 1) ⎤⎦ = e
∞
∞
2
n= 2
ln
1
2
(ln 2 + 1) =
1
(ln 2 + 1)
2
∴ the integral ∫ u( x )dx is convergent and hence, ∑ un is convergent.
M02_ENGINEERING_MATHEMATICS-I _XXXX_CH02.indd 35
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■
2.36
Engineering Mathematics
EXERCISE 2.4
Test the convergence of the following:
1. 1 +
∞
3.
x x2 x3 …
+
+ + ,
2 32 43
xn
∑n
n =1
n
,
2.
⎡ ( n + 1)x n ⎤
∑
⎢
⎥,
n n +1 ⎦
n =1 ⎣
4.
∑ (n
∞
x >0
∞
x>e
n − 1)
x >0
∞
n
5.
n =1
1
⎤
⎥
n= 2 ⎣ n (log n) ⎦
⎡
∑⎢
ANSWERS TO EXERCISE 2.4
2. Convergent if x < 1 and divergent if x ≥ 1
4. Convergent
5. Divergent
1. Convergent
3. Convergent
2.3.7 Raabe’s Test
⎛ u
⎞
Let ∑ un be a series of positive terms such that lim n ⎜ n − 1⎟ = l . Then the series ∑ un is convergent
n→∞ ⎝ u
⎠
n +1
if l > 1 and divergent if l < 1.
The test fails to give a definite result if l = 1.
We first state the theorem of comparison of ratios of two series.
un +1 vn +1
for all values of n ∈ N, then
<
un
vn
u
v
∑ un is convergent if ∑ vn is convergent, whereas if n +1 > n +1 for all values of n, then ∑ un is
un
vn
divergent if ∑ vn is divergent.
If ∑ un and ∑ vn are two series of positive terms and if
Proof
1
Given ∑ un is a series of positive terms, we compare with ∑ vn , where vn = p .
n
We know that ∑ vn is convergent if p > 1 and hence, ∑ un will be convergent if ∑ vn is convergent
u n +1 v n +1
and if
<
∀ n ∈ N and if p > 1
un
vn
⇒
un
v
> n
un +1 vn +1
But
vn
( n + 1) p ⎛ n + 1⎞
⎛ 1⎞
=
=⎜
⎟⎠ = ⎜⎝1 + ⎟⎠
p
⎝
v n +1
n
n
n
∀ n ∈ N and if p > 1
p
= 1+
∴
⇒
p
p 1 p ( p − 1) 1 …
⋅ +
⋅ 2+
1! n
2!
n
un
p 1 p ( p − 1) 1 …
> 1+ ⋅ +
⋅ 2+
u n +1
1! n
2!
n
⎛ un
⎞ p 1 p ( p − 1) 1 …
⎜⎝ u − 1⎟⎠ > 1! ⋅ n + 2 ! ⋅ n 2 +
n +1
M02_ENGINEERING_MATHEMATICS-I _XXXX_CH02.indd 36
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Sequences and Series ■
2.37
p ( p − 1) 1 …
⎛ u
⎞
n ⎜ n − 1⎟ > p +
⋅ 2+
2!
n
⎝ u n +1 ⎠
⇒
⎛ u
⎞
lim n ⎜ n − 1⎟ > p .
n →∞ ⎝ u
⎠
n +1
∴
∑ v n is convergent if p > 1.
But
⎛ u
⎞
⎛ u
⎞
∴ ∑u n is convergent if lim n ⎜ n − 1⎟ > p > 1 and divergent if lim n ⎜ n − 1⎟ < 1.
n
→∞
n →∞ ⎝ u
⎝ un +1 ⎠
⎠
n +1
⎛ u
⎞
The test fails if lim n ⎜ n − 1⎟ = 1
n →∞ ⎝ u
⎠
n +1
WORKED EXAMPLES
EXAMPLE 1
Test the convergence of the series
x1
1⋅3 ⋅5 x7 …
1 x3
1⋅ 3 x5
1 ∞, x > 0 .
1
1
2 ⋅4 ⋅6 7
2 3
2⋅4 5
Solution.
Let ∑ un be the series, omitting the first term,
∴
Then
and
1 x 3 1⋅ 3 x 5 1⋅ 3 ⋅ 5 x 7 …
+
+
+ ∞, x > 0
2 3 2⋅ 4 5 2⋅ 4⋅6 7
∑ un =
un =
u n +1 =
1⋅ 3 ⋅ 5…( 2n − 1) x 2 n +1
2 ⋅ 4 ⋅ 6 … 2n ( 2n + 1)
1⋅ 3 ⋅ 5…( 2n − 1)( 2n + 1) x 2 n + 3
⋅
2 ⋅ 4 ⋅ 6 … 2n ( 2n + 2) ( 2n + 3)
u n 1⋅ 3 ⋅ 5…( 2n − 1) x 2 n +1 2 ⋅ 4 ⋅ 6 … 2n ( 2n + 2)( 2n + 3) 1
⋅
=
⋅
u n +1
2 ⋅ 4 ⋅ 6 … 2n 2n + 1 1⋅ 3 ⋅ 5…( 2n − 1)( 2n + 1) x 2 n + 3
⇒
un
( 2n + 2)( 2n + 3) 1
=
u n +1
( 2n + 1) 2
x2
⇒
un
=
u n +1
M02_ENGINEERING_MATHEMATICS-I _XXXX_CH02.indd 37
2⎞ ⎛
3⎞
⎛
n2 ⎜ 2 + ⎟ ⎜ 2 + ⎟
⎝
n⎠ ⎝
n⎠
1⎞
⎛
n2 ⎜ 2 + ⎟
⎝
n⎠
2
(1)
2⎞ ⎛
3⎞
⎛
⎜⎝ 2 + ⎟⎠ ⎜⎝ 2 + ⎟⎠ 1
1
n
n
⋅ 2
⋅ 2 =
2
x
x
1⎞
⎛
⎜⎝ 2 + ⎟⎠
n
(2)
5/12/2016 10:57:13 AM
2.38
■
Engineering Mathematics
2⎞ ⎛
3⎞
⎛
⎜⎝ 2 + ⎟⎠ ⎜⎝ 2 + ⎟⎠ 1 2 ⋅ 2 1
un
1
n
n ⋅ =
lim
= lim
⋅ 2 = 2
2
2
2
n →∞ u
n →∞
x
x
x
2
1
⎛
⎞
n +1
⎜⎝ 2 + ⎟⎠
n
∴
∴ by the De’ Alembert’s ratio test,
∑ un is convergent if
and divergent if
1
>1 ⇒
x2
x2 < 1 ⇒
x2 −1 < 0 ⇒ 0 < x < 1
[{ x > 0]
1
<1 ⇒ x2 >1 ⇒ x >1
x2
1
=1 ⇒ x2 =1 ⇒ x =1
x2
If x = 1, the test fails to give a definite conclusion.
In this case, we use Raabe’s test.
When x = 1, the series is
1 1 1⋅ 3 1 1⋅ 3 ⋅ 5 1 …
∑ un = ⋅ +
⋅ +
⋅ +
2 3 2⋅ 4 5 2⋅ 4⋅6 7
[{ x > 0]
Now
un
( 2n + 2)( 2n + 3)
=
u n +1
( 2n + 1) 2
∴
∴
∴
[{ x > 0]
[from (1)]
un
( 2n + 2)( 2n + 3)
−1 =
−1
( 2n + 1) 2
un+1
=
( 2n + 2)( 2n + 3) − ( 2n + 1) 2
( 2n + 1) 2
=
4n2 + 10n + 6 − ( 4n2 + 4n + 1)
6n + 5
=
( 2n + 1) 2
( 2n + 1) 2
⎛ u
⎞ n(6 n + 5)
=
n ⎜ n − 1⎟ =
⎝ un+1 ⎠ ( 2n + 1) 2
5⎞
⎛
n2 ⎜ 6 + ⎟
⎝
n⎠
5
n
=
2
2
1⎞
1⎞
⎛
2⎛
4n ⎜1 + ⎟
4 ⎜1 + ⎟
⎝ 2n ⎠
⎝ 2n ⎠
6+
5
6+
6 3
⎛ un
⎞
n
lim n ⎜
− 1⎟ = lim
= = >1
2
n →∞ ⎝ u
n
→∞
4 2
⎠
1⎞
⎛
n +1
4 ⎜1 + ⎟
⎝ 2n ⎠
∴by Raabe’s test ∑ un is convergent if x = 1.
Hence, the given series is convergent if 0 < x ≤ 1 and divergent if x > 1.
M02_ENGINEERING_MATHEMATICS-I _XXXX_CH02.indd 38
5/12/2016 10:57:17 AM
Sequences and Series ■
2.39
EXAMPLE 2
Test the convergence or divergence of
x21
22 4
22 ⋅ 42
22 ⋅ 42 ⋅ 62
x 8 1 ….
x 1
x6 1
3⋅4
3⋅4 ⋅5⋅6
3⋅ 4 ⋅5 ⋅6 ⋅7 ⋅8
Solution.
Omitting the first term, let the given series be ∑ un
∴
∑ un =
Then
un =
22 4
22 ⋅ 4 2 6
22 ⋅ 4 2 ⋅ 6 2
x +
x +
x8 +…
3⋅ 4
3⋅ 4 ⋅ 5⋅ 6
3⋅ 4 ⋅ 5⋅ 6 ⋅ 7 ⋅8
2 2 ⋅ 4 2 …( 2n ) 2
x 2n+ 2
3 ⋅ 4 ⋅ 5 ⋅ 6 …( 2n + 1)( 2n + 2)
2 2 ⋅ 4 2 … ( 2 n ) 2 ⋅ ( 2 n + 2) 2
x 2n+ 4
3 ⋅ 4 ⋅ 5 ⋅ 6 …( 2n + 1)( 2n + 2)( 2n + 3)( 2n + 4)
and
u n+1 =
∴
un
( 2n + 3)( 2n + 4) 1
=
⋅ 2
( 2n + 2) 2
un+1
x
(1)
3⎞ ⎛
4⎞
3⎞ ⎛
4⎞
⎛
⎛
n2 ⎜ 2 + ⎟ ⎜ 2 + ⎟
2+ ⎟ ⎜2+ ⎟
⎜
⎝
⎠
⎝
⎝
⎠
⎝
⎠
n
n⎠ ⋅ 1 , x ≠ 0
n
n ⋅ 1 =
=
2
2
2
x
x2
2⎞
2⎞
⎛
⎛
2+ ⎟
n2 ⎜ 2 + ⎟
⎜
⎝
⎝
n⎠
n⎠
∴
3⎞ ⎛
4⎞
⎛
⎜⎝ 2 + ⎟⎠ ⎜⎝ 2 + ⎟⎠ 1 2 ⋅ 2 1
un
1
n
n ⋅ =
lim
= lim
⋅ 2 = 2
2
2
2
n →∞ u
n →∞
x
x
2 x
2⎞
⎛
n +1
⎜⎝ 2 + ⎟⎠
n
∴ by De’ Alembert’s ratio test,
1
∑ un is convergent if 2 > 1 ⇒
x
x 2 < 1 ⇒ −1 < x < 1, x ≠ 0.
1
< 1 ⇒ x 2 > 1 ⇒ x < −1 or x > 1.
2
x
If x2 = 1, the test fails, so we use Raabe’s test.
un
( 2n + 3)( 2n + 4)
In this case,
=
u n +1
( 2n + 2) 2
and divergent if
∴
[From (1)]
un
( 2n + 3)( 2n + 4)
−1
−1 =
u n +1
( 2 n + 2) 2
=
( 2n + 3)( 2n + 4) − ( 2n + 2) 2
( 2 n + 2) 2
=
6n + 8
4 n 2 + 14 n + 12 − ( 4 n 2 + 8n + 4)
=
( 2n + 2) 2
( 2 n + 2) 2
M02_ENGINEERING_MATHEMATICS-I _XXXX_CH02.indd 39
5/12/2016 10:57:22 AM
2.40
■
∴
∴
Engineering Mathematics
⎛ u
⎞ n (6 n + 8)
n ⎜ n − 1⎟ =
=
⎝ u n +1 ⎠ ( 2n + 2) 2
8⎞
⎛
n2 ⎜ 6 + ⎟
⎝
n⎠
8
n
=
2
2
2⎞
2⎞
⎛
2 ⎛
n ⎜2+ ⎟
2+ ⎟
⎜
⎝
⎝
n⎠
n⎠
6+
8
6+
⎛ un
⎞
n = 6 = 3 >1
lim n ⎜
− 1⎟ = lim
2
n →∞ ⎝ u
n
→∞
4 2
⎠
2⎞
⎛
n +1
⎜⎝ 2 + ⎟⎠
n
∴ by Raabe’s test ∑ un is convergent if x 2 = 1 ⇒ x = ±1.
∴ the given series is convergent if −1 ≤ x ≤ 1 [{ when x = 0; it is trivially convergent]
and divergent if x < −1 or x > 1.
EXAMPLE 3
∞
Test the convergence of the series
∑ ( n !) 2
n =1
Solution.
The given series be ∑ un
∴
x 2n
.
( 2 n )!
∞
∑ un = ∑ ( n!) 2
n =1
un =
Then
x 2n
( 2n)!
( n!) 2 x 2 n
( 2n)!
[( n + 1)!]2 x 2 n + 2 [( n + 1)!]2 x 2 n + 2
=
[2( n + 1)]!
( 2n + 2)!
and
u n +1 =
∴
un
( n !) 2 x 2 n
( 2n + 2)!
=
⋅
u n +1
( 2n)! [( n + 1)!]2 x 2 n + 2
⇒
=
( n!) 2 ( 2n)!(( 2n + 1)( 2n + 2) 1
⋅ 2
( 2n)!
( n !) 2 ( n + 1) 2
x
=
( 2n + 1)2( n + 1) 1
⋅ 2, x≠0
( n + 1) 2
x
un
2( 2n + 1) 1
=
⋅
un+1
n + 1 x2
(1)
1⎞
1⎞
⎛
⎛
2n ⎜ 2 + ⎟
2⎜ 2 + ⎟
⎝
⎠ 1
⎝
n⎠ ⋅ 1
n
=
⋅ =
1
x2
⎛ 1⎞ x2
1+
n ⎜1 + ⎟
⎝ n⎠
n
M02_ENGINEERING_MATHEMATICS-I _XXXX_CH02.indd 40
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Sequences and Series ■
2.41
1⎞
⎛
2⎜2 + ⎟
⎝
un
n⎠ ⋅ 1 = 2⋅2⋅ 1 = 4
lim
= lim
n →∞ u
n →∞
1
x2
x2 x2
n +1
1+
n
∴
∴ by ratio test, ∑ un is convergent if
4
>1 ⇒ x2 < 4 ⇒ x2 − 4 < 0
x2
⇒
−2 < x < 2, x ≠ 0
When x = 0, trivially the series is convergent and divergent if x < −2 or x > 2
When
4
=1 ⇒
x2
x 2 = 4, the test fails. So, we use Raabe’s test.
In this case, from (1), we get
un
2( 2n + 1) 1 ( 2n + 1)
=
⋅ =
un+1
n + 1 4 2( n + 1)
un
2n + 1 − 2( n + 1)
1
2n + 1
−1 =
−1 =
=−
u n +1
2( n + 1)
2( n + 1)
2( n + 1)
∴
1
n
⎛ u
⎞
n ⎜ n − 1⎟ = −
=−
2( n + 1)
⎛ 1⎞
⎝ un+1 ⎠
2 ⎜1 + ⎟
⎝ n⎠
∴
−1
1
⎛ u
⎞
lim n ⎜ n − 1⎟ = lim
= − <1
n →∞ ⎝ u
n
→∞
1
2
⎛
⎞
⎠
n +1
2 ⎜1 + ⎟
⎝ n⎠
∴
∴ by Raabe’s test, ∑ un is divergent if x 2 = 4 ⇒ x = ±2
∴ the given series is convergent if −2 < x < 2 and divergent if x ≤ −2 or x ≥ 2
EXERCISE 2.5
Test the convergence of the following series.
1 x 3 1⋅ 3 ⋅ 5 ⋅ x 4 1⋅ 3 ⋅ 5 ⋅ 7 ⋅ 9 x 6 …
1. 1 + ⋅ +
+
+ ∞, x > 0
2 4 2 ⋅ 4 ⋅ 6 ⋅ 8 2 ⋅ 4 ⋅ 6 ⋅ 8 ⋅10 12
2
2…
2
∞
2. ∑ 1 ⋅ 2 ( n + 1)
…( 2n + 1)( 2n + 3)
n =1 1 ⋅ 3
2
3.
3
3
⎛ 5⎞
⎛ 4⎞
x + ⎜ ⎟ x 2 + ⎜ ⎟ x 3 + … ∞, x > 0
⎝ 4⎠
⎠
⎝
4
5
M02_ENGINEERING_MATHEMATICS-I _XXXX_CH02.indd 41
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■
2.42
Engineering Mathematics
2
6
14
2 n − 2 n −1 …
4. 1 + x + x 2 + x 3 + … + n
x + ∞,
5
9
17
2 +1
⎛ xn ⎞
,
2n ⎟
⎠
n =1
∞
5.
∑ ⎜⎝ 1 + x
x >0
2
2
n
6. 1 + x + x + x + … + x + … ∞,
2 5 10
n2 + 1
∞
7.
⎛
3 ⋅ 6 ⋅ 9 …3n
∑ ⎜⎝ 7 ⋅10 ⋅13…(3n + 4) ⋅ x
n
n =1
8. 1 +
x >0
⎞
⎟⎠ ,
x >0
x >0
x 1⋅ 3 2 1⋅ 3 ⋅ 5 3 …
+
x +
x + ,
2 2⋅ 4
2⋅ 4⋅6
x >0
⎛ 4 ⋅ 7 ⋅ …(3n + 1) n ⎞
⋅ x ⎟ , x > 0.
⎠
1⋅ 2 ⋅ 3 ⋅ 4 … n
n =1
∞
9. Test the convergence of
∑ ⎜⎝
∞
10. Test the convergence of
n +1
∑ (n + 2)(n + 3) x
n
, x > 0.
n =1
ANSWERS TO EXERCISE 2.5
Convergent if 0 < x ≤ 1 and divergent if x > 1
Convergent if 0 < x < 1 and divergent if x ≥ 1
Convergent if 0 < x < 1 and divergent if x ≥ 1
Convergent if 0 < x < 1 or x > 1 and divergent if x = 1
Convergent if 0 < x ≤ 1 and divergent if x > 1
Convergent if 0 < x ≤ 1 and divergent if x > 1
Convergent if 0 < x < 1 and divergent if x ≥ 1
1
1
and divergent if x ≥ .
9. Convergent if 0 < x <
3
3
10. Convergent if x < 1 and divergent if x ≥ 1.
1.
3.
4.
5.
6.
7.
8.
2. Divergent
2.3.8 Logarithmic Test
⎛ u ⎞
Let ∑ un be a series of positive terms such that lim n loge ⎜ n ⎟ = l .
n →∞
⎝ u n +1 ⎠
Then the series ∑ un is convergent if l > 1 and divergent if l < 1. The test fails to give a definite
result if l = 1.
Proof
⎛ u ⎞
Given ∑ un is a series of positive terms such that lim n loge ⎜ n ⎟ = l .
n →∞
⎝ u n +1 ⎠
M02_ENGINEERING_MATHEMATICS-I _XXXX_CH02.indd 42
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Sequences and Series ■
We compare ∑ un with ∑ vn where vn =
2.43
1
np
We know that ∑ vn is convergent if p > 1.
∴ ∑ un is convergent if
u n +1 v n +1
<
and p > 1 ⇒
un
vn
But
∴
vn
( n + 1) p ⎛ 1 ⎞
=
= ⎜1 + ⎟
⎝ n⎠
vn +1
np
⎛ v ⎞
⎛ 1⎞
log e ⎜ n ⎟ = log e ⎜1 + ⎟
⎝ n⎠
⎝ v n +1 ⎠
un
v
> n and p > 1
un+1 vn +1
p
p
1
1
1
⎛ 1⎞
⎛1
⎞
= p log e ⎜1 + ⎟ = p ⎜ − 2 + 3 − 4 + …⎟
⎝ n⎠
⎝ n 2n 3n 4 n
⎠
∴
⎛ v ⎞
1
1
1
⎛1
⎞
n loge ⎜ n ⎟ = n p ⎜ − 2 + 3 − 4 + …⎟
⎝
⎠
v
n
2
3
4
n
n
n
⎝ n +1 ⎠
1
1
1 …⎞
⎛
= p ⎜1 −
+
−
+ ⎟
⎝ 2n 3n 2 4 n 3
⎠
⇒
⎛ v ⎞
1
1
1
⎛
⎞
+ 2 − 3 + …⎟
n loge ⎜ n ⎟ = p ⎜1 −
⎝
⎠
v
2
n
3
4
n
n
⎝ n +1 ⎠
∴
⎛ v ⎞
lim n loge ⎜ n ⎟ = p
n →∞
⎝ v n +1 ⎠
∴
⎛ u ⎞
⎛ v ⎞
lim n loge ⎜ n ⎟ > lim n loge ⎜ n ⎟ = p
n →∞
n
→∞
u
⎝ n +1 ⎠
⎝ v n +1 ⎠
But ∑ vn is convergent if p > 1.
⎛ u ⎞
∴ ∑ un will be convergent if lim n loge ⎜ n ⎟ > p > 1
n →∞
⎝ u n +1 ⎠
⎛ u ⎞
Similarly, we can prove ∑ un is divergent if lim n loge ⎜ n ⎟ < 1.
n →∞
⎝ u n +1 ⎠
⎛ u ⎞
The test fails if lim n loge ⎜ n ⎟ = 1.
n →∞
⎝ u n +1 ⎠
M02_ENGINEERING_MATHEMATICS-I _XXXX_CH02.indd 43
■
5/12/2016 10:57:41 AM
2.44
■
Engineering Mathematics
WORKED EXAMPLES
EXAMPLE 1
Test the convergence and divergence of the series
2
32
43 3 54 4 …
11 x 1 x 2 1
x 1 x 1 , x > 0.
2!
3!
4!
5!
Solution.
Let the given series be ∑ un .
∴
∑ un = 1 +
n n −1 n −1
x
n!
un =
Then
2
32
43
54
x + x 2 + x 3 + x 4 + …, x > 0
2!
3!
4!
5!
and u n +1 =
( n + 1) n n
x
( n + 1)!
∴
un
n n −1 n −1 ( n + 1)!
n n −1
1
x ⋅
=
=
⋅
u n +1
n!
( n + 1) n x n ( n + 1) n −1 x
⇒
un
=
u n +1
∴
lim
n →∞
un
un +1
⎛ 1⎞
1+
1
1
1 ⎜⎝ n ⎟⎠ 1
⋅
=
⋅
=
⋅
n −1
x ⎛ 1 ⎞ n −1 x ⎛ 1 ⎞ n x
1⎞
n −1 ⎛
n ⎜1 + ⎟
⎜⎝1 + ⎟⎠
⎜⎝1 + ⎟⎠
⎝ n⎠
n
n
⎛ 1⎞
⎜⎝1 + ⎟⎠ 1 1
n ⋅ =
= lim
n
n →∞
⎛ 1 ⎞ x ex
+
1
⎜⎝
⎟
n⎠
n n −1
1
1
1
> 1 ⇒ x < and diverges if
<1 ⇒
ex
e
ex
1
When x = , the test fails. So, we use logarithmic test.
e
un
1
∴
=
⋅e
u n +1 ⎛ 1 ⎞ n −1
1
+
⎜⎝
⎟
n⎠
∴ by ratio test, ∑ un is convergent if
∴
⎛ u ⎞
⎛ 1⎞
loge ⎜ n ⎟ = loge e − loge ⎜1 + ⎟
⎝ n⎠
⎝ u n +1 ⎠
(1)
x>
1
e
[from (1)]
n −1
⎛ 1⎞
= 1 − ( n − 1) loge ⎜1 + ⎟
⎝ n⎠
1
1
1
⎛1
⎞
= 1 − ( n − 1) ⎜ − 2 + 3 − 4 + …⎟
⎝ n 2n
⎠
3n
4n
1 1
1
1 …⎞
3
5 …
⎛
= 1 − ⎜1 −
− +
+
+ ⎟=
−
+
⎝ 2n n 3n 2 2n 2
⎠ 2n 6 n 2
M02_ENGINEERING_MATHEMATICS-I _XXXX_CH02.indd 44
5/12/2016 10:57:46 AM
Sequences and Series ■
⎛ u ⎞
5 …⎞ 3 5 …
⎛ 3
−
+ ⎟= −
+
n log ⎜ n ⎟ = n ⎜
⎝ 2n 6 n 2
⎠ 2 6n
⎝ u n +1 ⎠
∴
∴
2.45
⎛ u ⎞
⎛ 3 5 …⎞ 3
+ ⎟ = >1
lim n log ⎜ n ⎟ = lim ⎜ −
⎝ 2 6n
⎠ 2
n →∞
n
→∞
⎝ u n +1 ⎠
1
∴ by logarithmic test, ∑ un is convergent if x = .
e
1
1
∴ the given series is convergent if 0 < x ≤ and divergent if x > .
e
e
EXAMPLE 2
Test the convergence of 1 1
(1!) 2
( 2 !) 2 2 ( 3 !) 2 3 …
x1
x 1
x 1 , x > 0.
2!
4!
6!
Solution.
Let ∑ un be the series, omitting the first term
∞
∑u
n
=
n=1
Then
un =
(1!) 2
( 2 !) 2 2 (3!) 2 3 …
x+
x +
x + ,
2!
4!
6!
( n!) 2 n
x
( 2n)!
and
un +1 =
x >0
[( n + 1)!]2 n +1
x
( 2n + 2)!
un
( n!) 2 n ( 2n + 2)! 1
x ⋅
=
u n +1 ( 2 n ) !
[( n + 1)!]2 x n +1
∴
=
⇒
un
=
un+1
∴
( 2n + 1)( 2n + 2) 1 2( 2n + 1)( n + 1) 1 2( 2n + 1) 1
⋅ =
⋅ =
⋅
x
x
n +1 x
( n + 1) 2
( n + 1) 2
1⎞
1⎞
⎛
⎛
4n ⎜1 + ⎟
4 ⎜1 + ⎟
⎝ 2n ⎠ 1
⎝ 2n ⎠ 1
⋅ =
⋅
⎛ 1⎞ x
⎛ 1⎞ x
n ⎜1 + ⎟
1+ ⎟
⎜
⎝ n⎠
⎝ n⎠
(1)
1⎞
⎛
4 ⎜1 + ⎟
⎝ 2n ⎠ 1 4
un
lim
= lim
⋅ =
n →∞ u
n →∞ ⎛
1⎞ x x
n +1
1
+
⎜⎝
⎟
n⎠
4
4
> 1 ⇒ x < 4 and is divergent if < 1 ⇒
x
x
When x = 4, the test fails. So, we use logarithmic test.
∴ by ratio test, ∑ un is convergent if
∴ If x = 4, then
1⎞
⎛
1
4 ⎜1 + ⎟
⎝ 2n ⎠ 1 1 + 2n
un
=
⋅ =
1
1
u n +1
4
1+
1+
n
n
M02_ENGINEERING_MATHEMATICS-I _XXXX_CH02.indd 45
x > 4.
[from (1)]
5/12/2016 10:57:53 AM
■
2.46
Engineering Mathematics
∴
un
1⎞
⎛
⎛ 1⎞
= loge ⎜1 + ⎟ − loge ⎜1 + ⎟
⎝
⎠
⎝ n⎠
u n +1
2n
loge
2
=
∴
n log e
∴
lim n loge
1 7 1 …
1 1 ⎛ 1 ⎞ … ⎛ 1 1 1 …⎞
+ ⋅ −
− ⎜ ⎟ + −⎜ − ⋅ 2 + ⎟ = −
⎝
⎠
⎝
⎠
2n 8 n 2
n 2 n
2n 2 2n
un
1 7 1
⎛ 1 7 1
⎞
= n ⎜ − + ⋅ 2 − …⎟ = − + ⋅ 2 − …
⎝
⎠
2n 8 n
2 8 n
u n +1
un
1
⎛ 1 7 1
⎞
= lim ⎜ − + ⋅ 2 − …⎟ = − < 1
n
→∞
⎝
⎠
2 8 n
2
u n +1
n →∞
∴ by logarithmic test, ∑ un is divergent if x = 4.
∴ the given series is convergent if 0 < x < 4 and divergent if x ≥ 4.
2.4
ALTERNATING SERIES
Definition 2.13 A series of the form u1 − u 2 + u3 − u 4 + … ∞, where un > 0 ∀ n ∈ N is called an alternating series.
Examples
1 1
1
+ 2 − 3 + … is an alternating series.
2 2
2
2
∞
n
(2) ∑ ( −1) n 3
is an alternating series.
n +1
n =1
(1) 1 −
That is −
12
22
32
+
−
+ … is an alternating series, because it is
13 + 1 23 + 1 33 + 1
⎛ 12
⎞
22
32
−⎜ 3
− 3
+ 3
+ …⎟
⎝1 +1 2 +1 3 +1
⎠
2.4.1 Leibnitz’s Test
Statement: If the alternating series u1 − u 2 + u3 − u 4 + … is such that (i) u n +1 ≤ u n
and (ii) lim un = 0, then the series is convergent.
∀n
n →∞
Proof
Given un > 0 ∀ n
and (i) u1 ≥ u 2 ≥ u3 ≥ u 4 ≥ …u n ≥ u n +1 ≥ … ,
and
(ii)
lim un = 0
n →∞
Consider the even partial sum s2n.
∴
Since
s 2n = (u1 − u 2 ) + (u3 − u 4 ) + … + (u 2 n −1 − u 2 n ) = s 2n − 2 + (u 2 n −1 − u 2 n )
u2 n −1 ≥ u2 n , u2 n −1 − u2 n ≥ 0 ∀ n ,
M02_ENGINEERING_MATHEMATICS-I _XXXX_CH02.indd 46
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Sequences and Series ■
∴
s2 n ≥ s2 n − 2
2.47
∀ n.
0 ≤ s 2 ≤ s 4 ≤ s6 ≤ … ≤ s 2 n ≤ …
Hence,
Also we can write
s2n = u1 − (u2 − u3 ) − (u4 − u5 ) − (u2 n − 2 − u2 n −1 ) − u2 n
Every difference in the brackets are non-negative and u2 n > 0
∴
s2 n ≤ u1
∀ n
So, the sequence of partial seems {s2n} is increasing and bounded above.
Hence, it is convergent.
∴
lim s2 n = l .
n →∞
Now compute the limit of odd partial sum.
s 2n+1 = s2 n + u2 n +1
Now
∴
lim s2 n +1 = lim( s2n + u2n +1 )
n →∞
n →∞
lim un = 0 and lim s2n = l , lim sn = l , we have lim s2 n +1 = l + 0 = l .
Since
n→∞
n →∞
n →∞
n →∞
Since both even and odd partial sums converge to l, we have lim sn = l , and so the series is convergent.
n →∞
When lim u n ≠ 0, lim s 2 n ≠ lim s 2 n +1
n →∞
n →∞
n →∞
∴ the given series is oscillatory.
Thus, in an alternating series if the terms are decreasing with lim un = 0, then it is convergent.
■
WORKED EXAMPLES
EXAMPLE 1
Test the convergence of the series 1 2
1
2 2
1
1
3 3
2
1
4 4
1 ….
Solution.
The given series 1 −
1
2 2
+
1
3 3
−
1
4 4
un =
+ … is an alternating series with
1
and
n n
We know that
( n + 1) n + 1 > n n
⇒
1
1
<
( n + 1) n + 1 n n
M02_ENGINEERING_MATHEMATICS-I _XXXX_CH02.indd 47
un+1 =
⇒
1
( n + 1) n + 1
un +1 < un
∀ n ≥ N.
5/12/2016 10:58:06 AM
2.48
■
Engineering Mathematics
i.e., the terms are decreasing.
lim un = lim
Now
n →∞
n →∞
1
n n
=
1
=0
∞
Hence, by Leibnitz’s test, the given alternating series is convergent.
EXAMPLE 2
⎛ (21) n 2 1 n ⎞
.
2 n 2 1 ⎟⎠
n 51
∞
Test the convergence of the series
Solution.
∞
The given series ∑ ( −1) n −1
n =1
∑ ⎜⎝
2 3 4
n
= 1 − + − + … is an alternating series with
2n − 1
3 5 7
un =
n
2n − 1
and
u n +1 =
n +1
n +1
=
2( n + 1) − 1 2n + 1
n +1
n
−
2n + 1 2n − 1
( n + 1)( 2n − 1) − n ( 2n + 1)
=
( 2n + 1)( 2n − 1)
u n +1 − u n =
∴
=
∴
2n 2 + n − 1 − ( 2n 2 + n )
1
=− 2
< 0 ∀ n ∈N
2
4n − 1
4n − 1
un +1 < un
∀ n ∈ N.
That is the terms are decreasing
lim un = lim
and
n →∞
n →∞
n
1
1
= lim
= ≠ 0.
n
→∞
1⎞ 2
2n − 1
⎛
⎜⎝ 2 − ⎟⎠
n
∴ the given series is not convergent, but it is oscillatory.
EXAMPLE 3
Discuss the convergence of the series
x
x2
x3
x4
2
1
2
1 … ∞, 0 < x < 1 .
11 x 11 x2 11 x3 11 x4
Solution.
The given series
x
x2
x3
x4
−
+
−
+…
1+ x 1+ x 2 1+ x 3 1+ x 4
is an alternating series with un =
∴
u n +1 − u n =
x n +1
xn
and un +1 =
n
1 + x n +1
1+ x
x n +1
xn
x n +1 (1 + x n ) − x n (1 + x n +1 )
−
=
n +1
n
(1 + x n +1 )(1 + x n )
1+ x
1+ x
M02_ENGINEERING_MATHEMATICS-I _XXXX_CH02.indd 48
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Sequences and Series ■
∴
=
x n +1 + x n +1 ⋅ x n − x n − x n ⋅ x n +1
(1 + x n +1 )(1 + x n )
=
x n +1 − x n
x n ( x − 1)
=
<0
(1 + x n +1 )(1 + x n ) (1 + x n +1 )(1 + x n )
2.49
[{ 0 < x < 1 ⇒ x − 1 < 0]
un +1 < un
∀ n ∈N
That is the terms are decreasing
xn
=0
n →∞
n →∞ 1 + x n
∴ by Leibnitz’s test, the given series is convergent.
lim un = lim
and
[{ 0 < x < 1, x n → 0 as n → ∞]
EXAMPLE 4
2
⎛
n (1 1 n ) ⎞
(
2
1
)
.
∑⎜
(1 1 n3 ) ⎟⎠
n 51 ⎝
∞
Test the convergence of
Solution.
⎛ 1 + 12 1 + 22 1 + 32
⎞
1 + 12 1 + 22 1 + 32
+
−
+ ... = − ⎜
−
+
− ...⎟
The given series is −
3
3
3
⎝ 1 + 13 1 + 23 1 + 33
⎠
1+1 1+ 2 1+ 3
Consider, the series is
1 + 12 1 + 22 1 + 32 1 + 4 2 1 + 52 …
−
+
−
+
+
1 + 13 1 + 23 1 + 33 1 + 43 1 + 53
It is an alternating series with
un =
∴
1 + ( n + 1) 2
1 + n2
n 2 + 2n + 2
and un+1 =
3 = 3
3
1 + ( n + 1)
1+ n
n + 3n2 + 3n + 2
1 + n2
n 2 + 2n + 2
−
2
n + 3n + 3n + 2 1 + n3
(1 + n3 )( n 2 + 2n + 2) − (1 + n2 )(( n3 + 3n2 + 3n + 2)
=
(1 + n3 )( n3 + 3n2 + 3n + 2)
u n +1 − u n =
=
3
n3 + 2n + 2n + n5 + 2n4 + 2n3 − ( n3 + 3n2 + 3n + 2 + n5 + 3n4 + 3n3 + 2n2 )
(1 + n3 )( n3 + 3n2 + 3n + 2)
=−
∴
u n +1 < u n
( n4 + 2n3 + 4 n2 + n)
<0
(1 + n3 )( n3 + 3n2 + 3n + 2)
[{ n is positive]
∀ n ≥ 1.
So, the terms are decreasing.
1⎞
⎛
1
n2 ⎜1 + 2 ⎟
1+ 2
⎝ n ⎠
1
n
= =0
= lim
and
n
→∞
1
1
⎛
⎞ ∞
⎛
⎞
n ⎜1 + 3 ⎟
n3 ⎜ 1 + 3 ⎟
⎝ n ⎠
⎝ n ⎠
∴ by Leibnitz’s test, the given series is convergent.
⎛ 1 + n2 ⎞
lim un = lim ⎜
⎟ = lim
n →∞
n →∞ ⎝ 1 + n3 ⎠
n →∞
M02_ENGINEERING_MATHEMATICS-I _XXXX_CH02.indd 49
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■
2.50
2.5
Engineering Mathematics
SERIES OF POSITIVE AND NEGATIVE TERMS
Let u1 + u 2 + u3 + … + u n + … be a series containing both positive and negative terms.
That is ui’s may be positive or negative.
Alternating series is a special case of series with positive and negative terms.
Example
1−
1 1 1 1 1 1 …
− + − − + − is a series with positive and negative terms.
2 4 3 6 8 5
2.5.1 Absolute Convergence and Conditional Convergence
Definition 2.14 Let ∑ un be a series of positive and negative terms. The series ∑ un is said to be absolutely convergent if the series ∑ un is convergent, where un is the absolute value of un.
The series ∑ un is said to be conditionally convergent if it is convergent but not absolutely
convergent.
EXAMPLE 1
The series 1 −
1 1 1 …
is absolutely convergent because the series of absolute terms,
+ − +
23 33 43
∑ un = 1 +
1 1 1 …
is convergent, by p-series ({ p = 3 > 1)
+ + +
23 33 43
EXAMPLE 2
1 1 1 …
+ − + ∞ is convergent by Leibnitz’s test. But the series of absolute values,
2 3 4
1 1 1
∑ u n = 1 + + + + … ∞ is divergent by p-series ({ p = 1).
2 3 4
1 1 1
So, 1 − + − + … ∞ is conditionally convergent.
2 3 4
The series 1 −
Results:
1. A series ∑ un which is absolutely convergent is itself convergent, but the converse is not true.
2. In an absolute convergent series, the series formed by the positive terms alone is convergent and
the series formed by negative terms alone is convergent.
3. If the terms of an absolutely convergent series is rearranged the series remains convergent and its
sum is unaltered.
2.5.2 Tests for Absolute Convergence
The test for series of positive terms is used for testing absolute convergence because ∑ un is a series
of positive terms.
1. Comparison test: If lim
n →∞
un
= l ( ≠ 0) then ∑ un and ∑ vn behave alike.
vn
where ∑ vn is an auxiliary series of positive terms.
M02_ENGINEERING_MATHEMATICS-I _XXXX_CH02.indd 50
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Sequences and Series ■
2.51
If ∑ vn is convergent, then ∑ un is convergent.
If ∑ vn is divergent, then ∑ un is divergent.
2. De’ Alembert’s ratio test
If lim
n →∞
un
= l , then ∑ un is convergent if l > 1 and divergent if l < 1. The test fails if l = 1.
un +1
3. Cauchy’s root test: If lim un
1
n
n →∞
= l , then ∑ un is convergent if l < 1 and divergent if l > 1.
The test fails if l = 1
Most cases can be tackled with comparison test and ratio test.
WORKED EXAMPLES
EXAMPLE 1
Prove that the series
sin x sin 2 x sin 3 x …
converges absolutely.
2
1
2
13
23
33
Solution.
Let the given series be ∑ un
∴
sin x
sin 2x
sin 3x …
+
+
+
3
3
1
2
33
The series of absolute terms is ∑ u n =
sin nx ≤ 1 ∀ n ∈ N
We know that
⇒
sin nx
1
≤ 3
3
n
n
∀ n ∈N
∞
1
.
3
n =1 n
∑ un ≤ ∑
∴
∞
But
sin x sin 2x sin 3x …
−
+ 3 − ,
13
23
3
∑ un =
1
∑n
n =1
3
=
1 1 1 …
+ + + is convergent by p-series, since p = 3 > 1.
13 23 33
∴ ∑ un is convergent by comparison test.
∴ the given series is absolutely convergent.
EXAMPLE 2
For what values of x, the series x 2
x2
2
1
x3
3
2
x4
4
1 … ∞ is convergent.
Solution.
Let the given series be ∑ un .
∴
∑ un = x −
x2
2
+
x3
3
M02_ENGINEERING_MATHEMATICS-I _XXXX_CH02.indd 51
−
x4
4
+ … is a series of positive and negative terms.
5/12/2016 10:58:57 AM
2.52
■
Engineering Mathematics
The series of absolute terms is
x2
∑ un = x +
∴
un =
un
∴
u n +1
∴
lim
n →∞
=
2
xn
n
x
x
x4
+
4
=
n +1
+…
x n +1
u n +1 =
n +1
⋅
3
3
and
n
xn
+
n +1
n +1 1
1 1
⋅ = 1+ ⋅
n
x
n x
⎛
un
1 1⎞ 1
= lim ⎜ 1 + ⋅ ⎟ =
un +1 n→∞ ⎝
n x⎠ x
1
> 1 ⇒ x < 1 ⇒ − 1 < x < 1, x ≠ 0
x
∴ by ratio test, ∑ un is convergent if
when x = 0, trivially convegent
∴ the given series is absolutely convergent and hence, convergent if −1 < x < 1.
When x = 1, the series is
∑ un = 1 −
un =
Here
Clearly, n + 1 > n
⇒
1
2
+
1
3
−
1
and un +1 =
n
n +1 > n
⇒
1
4
+ …, which is an alternating series.
1
n +1
1
1
<
n +1
n
u n +1 < u n
⇒
So, the terms of the series are decreasing and lim un = lim
n →∞
n →∞
∀ n ∈N
∀ n ∈N
1
=0
n
Hence, by Leibnitz’s test, the series is convergent.
∴ the given series ∑ un is convergent if −1 < x ≤ 1.
EXAMPLE 3
∑ {( 2 1)
∞
Show that the series
n
n 51
}
⎡ n 2 1 1 2 n ⎤ is conditionally convergent.
⎣
⎦
Solution.
The given series is
∑ {( −1)
∞
n =1
n
}
⎡⎣ n2 + 1 − n⎤⎦ = − ( 12 + 1 − 1) + ( 22 + 1 − 2) − ( 32 + 1 − 3) + ( 4 2 + 1 − 4 ) − …,
=−
{(
M02_ENGINEERING_MATHEMATICS-I _XXXX_CH02.indd 52
}
12 + 1 − 1) − ( 22 + 1 − 2) + ( 32 + 1 − 3) − ( 4 2 + 1 − 4 ) + …
5/12/2016 10:59:04 AM
Sequences and Series ■
2.53
which is an alternating series with
(
un = n 2 + 1 − n =
⇒
un =
∴
un+1 =
n2 + 1 − n2
n +1 + n
2
n2 + 1 − n) ( n 2 + 1 + n)
(
n2 + 1 + n)
1
=
(1)
n +1 + n
2
1
( n + 1) + 1 + ( n + 1)
2
[{ n ≥ 1]
( n + 1) 2 + 1 + ( n + 1) > n 2 + 1 + n
It is obvious
1
⇒
( n + 1) + 1 + ( n + 1)
2
<
1
n +1 + n
2
⇒
un +1 < un
∀ n ∈N
So, the terms of the series are decreasing
1
lim un = lim
Now
=
1
=0
∞
n +1 + n
∴ by Leibnitz’s test, the series is convergent.
To test the conditional convergence of the given series ∑ un , we test the convergence of the series
∑ un of absolute terms
Now
n →∞
n →∞
2
∑ u n = ∑ ( −1) n { n 2 + 1 − n} = ∑ ( n 2 + 1 − n )
∞
∞
n =1
n =1
1
1
∴
un = n 2 + 1 − n =
Take
vn =
1
n
∴
un
1
1
×n =
⎧
⎫
1
1
1+ 2 +1
n ⎨ 1 + 2 + 1⎬
n
n
⎩
⎭
∴
vn
lim
n →∞
=
n +1 + n
2
=
1
n 1+ 2 + n
n
⎡⎣{ | ( −1) n | = 1 and n2 + 1 − n > 0 ⎤⎦
=
1
⎧
⎫
1
n ⎨ 1 + 2 + 1⎬
n
⎩
⎭
[from (1)]
un
1
1
= lim
=
( ≠ 0)
2
vn n→∞ 1 + n + 1 2
But ∑ vn = ∑
1
is divergent by p-series, since p = 1
n
∴ by comparison test, ∑ un is divergent.
Thus, ∑ un is convergent and ∑ un is divergent.
Hence, the given series ∑ un is conditionally convergent.
M02_ENGINEERING_MATHEMATICS-I _XXXX_CH02.indd 53
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2.54
■
Engineering Mathematics
EXAMPLE 4
Test whether the series is conditionally convergent or not.
1
11 2 11 2 1 3 11 2 1 3 1 4 …
2 3 1
2
1 ∞.
3
2
3
43
53
Solution.
The given series is
1 1+ 2 1+ 2 + 3 …
− 3 +
− ,
23
3
43
which is an alternating series with
1 + 2 + 3 + … + n n ( n + 1)
n
un =
=
=
2( n + 1) 3 2( n + 1) 2
( n + 1)3
and
∴
∴
un+1 =
un +1 − un =
n +1
n +1
=
2
2( n + 1 + 1)
2( n + 2) 2
(1)
(2)
n +1
n
−
2
2( n + 2)
2( n + 1) 2
=
( n + 1)3 − n ( n + 2) 2
2( n + 1) 2 ( n + 2) 2
=
n 3 + 3n 2 + 3n + 1 − n ( n 2 + 4 n + 4)
2( n + 1) 2 ( n + 2) 2
=
−n 2 − n + 1
−( n 2 + n − 1)
=
< 0 ∀ n ∈N
2
2
2( n + 1) ( n + 2)
2( n + 1) 2 ( n + 2) 2
un +1 < un
∀ n ∈N
∴ the terms of the given series ∑ un are decreasing
and
lim u n = lim
n →∞
n →∞
n
= lim
2( n + 1) 2 n →∞
1
⎛ 1⎞
2n ⎜ 1 + ⎟
⎝ n⎠
2
=
1
=0
∞
∴ by Leibnitz’s test, the series ∑ un is convergent.
To test the conditional convergence of the given series ∑ un, we test the convergence of the series ∑ un
of absolute terms.
1 1+ 2 1+ 2 + 3 …
+ 3 +
+
23
3
43
n
n
1
un =
=
=
2
2
2
2( n + 1)
⎛ 1⎞
⎛ 1⎞
2n 2 ⎜ 1 + ⎟
2n ⎜ 1 + ⎟
⎝ n⎠
⎝ n⎠
∑ un =
Take
∴
vn =
un
=
vn
[from(1)]
1
n
1
⎛ 1⎞
2n ⎜ 1 + ⎟
⎝ n⎠
2
M02_ENGINEERING_MATHEMATICS-I _XXXX_CH02.indd 54
⋅n =
1
⎛ 1⎞
2 ⎜1 + ⎟
⎝ n⎠
5/12/2016 10:59:16 AM
Sequences and Series ■
∴
lim
n →∞
2.55
un
1
1
= lim
=
( ≠ 0)
vn n→∞ ⎛ 1 ⎞ 2
2 ⎜1 + ⎟
⎝ n⎠
∴ by comparison test, ∑ un and ∑ vn behave alike.
But ∑ vn = ∑
1
is divergent by p-series, since p = 1
n
∑ un is divergent.
∴
Thus, the series ∑ un is convergent and ∑ un is divergent.
Hence, the given series ∑ un is conditionally convergent.
EXERCISE 2.6
Test the nature of convergence of the following series
∞
1.
∑ ( −1)
n
2.
n
n =1
( −1) n
n!
n =1
∑
7.
∑ ( −1)
∞
n
n =1
∞
10.
n =1
3.
∞
( n + 1)
n
∑
( −1) n −1 n
n = 0 2n − 1
6.
8.
∑
( −1) n x n
n!
n =1
9. x −
11.
n
( −1) n x n
n +1
n=0
∑
x3 x5 x7 …
+
−
+
3
5
7
5 7 9 11 13 …
− + − + −
2 4 6 8 10
12. Discuss the convergence of the series
1
1
1
1 …
−
+
−
+ .
1⋅ 2 3 ⋅ 4 5 ⋅ 6 7 ⋅ 8
13. Discuss the convergence of the series 1 − 2x + 3x 2 − 4 x 3 + … ∞, 0 < x <
∞
14. Test the series
∑ ( −1)
n =1
n
n
n!
∞
∞
n +1
∑ ( −1)
n =1
∞
( −2) n
∑ n3
( −2) n
n2
n =1
5.
∞
4.
∑
∞
1
1
.
2
n
for absolute convergence and conditional convergence.
n +4
2
ANSWERS TO EXERCISE 2.6
1.
3.
5.
7.
9.
11.
13.
Conditionally convergent
Convergent
Oscillatory
Not convergent
Converges if −1 ≤ x ≤ 1
Not convergent
Convergent
M02_ENGINEERING_MATHEMATICS-I _XXXX_CH02.indd 55
2.
4.
6.
8.
10.
12.
14.
Divergent
Convergent
Absolutely convergent for −1 < x < 1
Converges for all x
Absolutely convergence
Convergent
Condionally convergent
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We shall now test the convergence of some important series.
1. Binomial series
2. Exponential series
3. Logarithmic series
2.6
CONVERGENCE OF BINOMIAL SERIES
The binomial series 1 +
n
n ( n − 1) 2 n ( n − 1)( n − 2) 3 … n ( n − 1)( n − 2) …( n − r + 1) r …
x+
x +
x + +
x + ∞
r!
1!
2!
3!
is absolutely convergent if | x | < 1.
Proof
∞
Let the given series be denoted by ∑ ur
[omitting the first term].
r=1
n ( n − 1)( n − 2) …( n − r + 1) r
x
r!
n ( n − 1)( n − 2) …( n − r + 1)( n − r ) r +1
u r +1 =
x
( r + 1)!
ur =
Then
and
ur
r +1 1
=
⋅
ur+1 n − r x
∴
(1)
1
ur
r +1 1 1+ r 1
=
⋅ =
⋅
n
ur+1
n−r x
x
−1
r
∴
1
1+ 1
ur
1
1
r
lim
= lim
= −1 =
r →∞ u
r →∞ n
x
x
x
r +1
−1
r
∴ by the ratio test, the series is convergent if
1
>1 ⇒
x
x < 1 and divergent if
1
<1 ⇒
x
x > 1.
When x = 1, the test fails.
When x = 1,
1
ur
r +1 1+ r
=
=
ur+1 n − r n
−1
r
1
1+
u
r
r = 1 = −1 < 1
∴
lim
= lim
r →∞ u
r →∞ n
−1
r +1
−1
r
∴ by ratio test, the series is divergent when x = 1.
∴ the binomial series is absolutely convergent if x < 1.
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[using(1)]
■
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Sequences and Series ■
2.57
Note The sum of the binomial series is (1 + x)n.
∴
2.7
nx n( n − 1) 2 n( n − 1)( n + 2) 3
+
x +
x
1!
2!
3
n( n − 1)( n − 2) …( n − r + 1) r …
x +
+ …+
r!
(1 + x ) n = 1 +
if x < 1
CONVERGENCE OF THE EXPONENTIAL SERIES
The exponential series 1 +
x x2 x3 … xn …
+
+ + +
+ ∞ converges absolutely for all values of x.
n!
1! 2 ! 3!
Proof
∞
Let the given series be denoted by ∑ un
[omitting the first term]
n=1
un =
Then
xn
n!
and u n +1 =
x n +1
( n + 1)!
∴
un +1
x n +1 n!
x
=
⋅ n =
( n + 1)! x
un
n +1
∴
un +1
x
x
=
=
un
n +1 n +1
∴
lim
n →∞
un +1
x
= lim
= 0 ∀ x ∈R
n →∞ n + 1
un
Here l < 1 and so the series ∑ un is absolutely convergent for all x ∈R.
Hence, the exponential series is convergent for all values of x ∈R.
■
Note The sum of the exponential series is ex.
∴
2.8
ex = 1+
x x2 x3 … xn …
+
+ + +
+ ∞ ∀ x ∈R
1! 2 ! 3!
n!
CONVERGENCE OF THE LOGARITHMIC SERIES
x 2 x 3 x 4 … ( −1) n x n …
The logarithmic series x −
is convergent for all values of x in
+
−
+ +
+
2
3
4
n
−1 < x ≤ 1.
Proof
Let the given series be ∑ un .
∴
∑ un = x −
x2 x3 x4 …
+
−
+
2
3
4
Now the series of absolute terms is
∑ un = x +
M02_ENGINEERING_MATHEMATICS-I _XXXX_CH02.indd 57
x2
x3
x4 …
+
+
+
2
3
4
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x
un =
n
∴
n
and
u n +1 =
x n +1
n +1
n
u
un
x n +1 n +1 1
= n =
⋅
=
⋅
un +1
un +1
n x n +1
n x
∴
∴
lim
n →∞
un
⎧⎛ n + 1⎞ 1 ⎫
⎧⎛ 1 ⎞ 1 ⎫ 1
= lim ⎨⎜
⎟ ⋅ ⎬ = lim ⎨⎜1 + ⎟ ⋅ ⎬ =
un +1 n →∞ ⎩⎝ n ⎠ x ⎭ n →∞ ⎩⎝ n ⎠ x ⎭ x
∴ by ratio test, the series ∑ un is convergent if
1
>1 ⇒
x
x < 1 and divergent if
1
< 1 ⇒ x > 1.
x
∴ the series ∑ un is absolutely convergent if x < 1 ⇒ − 1 < x < 1.
Hence, the series is convergent if −1 < x < 1.
1 1 1 …
+ − +
2 3 4
1
1
It is an alternating series with un = and un +1 =
n
n +1
1
1
<
∀n> N
n +1 > n ⇒
Now
n +1 n
1
So, the terms of the series are decreasing and lim un = lim = 0
n →∞
n →∞ n
∴ by Leibnitz’s test the series (1) is convergent.
∴ the logarithmic series
When x = 1, the series is 1 −
x−
(1)
⇒
un +1 < un
∀ n ∈ N.
x2 x3 x4 …
+
−
+ ∞ is convergent if −1 < x ≤ 1.
2
3
4
■
Note The sum of the series is log e (1+ x )
∴
log e (1 + x ) = x −
and when x = 1,
2.9
log e 2 = 1 −
x2 x3 x4 …
+
−
+ ∞, − 1 < x ≤ 1
2
3
4
1 1 1 …
+ − +
2 3 4
POWER SERIES
Definition 2.15 Real Power Series A series of the form a0 + a1x + a2 x 2 + … + an x n + … is called a real
power series, where a0 , a1 , a2 , …, an , … are real coefficients independent of x and x is a real variable.
∞
The power series is written as ∑ an x n
n= 0
Binomial series, exponential series and logarithmic series are few special power series.
A more general form of the power series is
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Sequences and Series ■
∞
∑ a ( x − a)
n=0
n
n
2.59
2
n
= a0 + a1 ( x − a) + a2 ( x − a) + … + an ( x − a) + …
It is called a power series in (x − a) [or about the point a], where a0, a1, a2, , … are real numbers and
x is a real variable.
Definition 2.16 Radius of Convergence of a Power Series
∞
The power series
∑a x
n
n
is said to have radius of convergence R if the series converges for all x
n= 0
satisfying x < R (i.e., − R < x < R ).
This interval is called the interval of convergence of the power series.
2.9.1 Hadmard’s Formula
∞
Consider the power series ∑ an x n . If lim an
Theorem 2.2
1
n
n →∞
n=0
=
1
, then the power series will converge
R
absolutely if x < R and diverge if x > R
lim an
n →∞
1
n
=
1
is known as Hadmard’s formula.
R
Proof
∞
Let the given power series be ∑ un = ∑ an x n . Then un = an x n
n=0
un = an x n = an x n
∴
⇒
un
∴
lim u n
1
n
1
n
n →∞
1
= an n ⋅ x n
= lim an
n →∞
1
n
1
n
1
= an n ⋅ x
x =
⎡
⎢⎣{ lim an
1
x.
R
x
<1 ⇒
R
x
divergent if > 1 ⇒
R
∴ by Cauchy’s root test, the series ∑ un is convergent if
x <R
and
x > R.
∴ ∑ un is absolutely convergent if x < R
1
n
=
1⎤
R ⎥⎦
∞
∴ the power series ∑ an x n converges absolutely if x < R.
n= 0
That is, the power series converges in the interval − R < x < R.
■
Note
1. The interval (−R, R) is called the interval of convergence of the power series.
At the end points x = −R and x = R, the power series may or may not converge.
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2. If lim an
Engineering Mathematics
1
n
n →∞
1
=0 ⇒ R=∞
R
= 0, then
∞
So, the power series ∑ an x n converges absolutely for all real values of x.
n=0
3. If lim an
n →∞
1
n
1
=∞ ⇒ R=0
R
= ∞, then
∞
∴ the power series ∑ an x n will converge only at the point x = 0
n=0
1
n
4. We know that lim an
n →∞
= lim
n →∞
an +1
if the right hand side limit exists.
an
a
1
= lim n +1
n
→∞
R
an
an
(1)
=R
an +1
So, the radius of convergence of a power series can also be obtained by using (1) whenever this
limit could be easily evaluated.
∴
⇒
lim
n →∞
2.9.2 Properties of Power Series
The power series is very useful and convenient to deal with, because within the interval of convergence
it can be treated as a polynomial.
It has the following properties.
∞
1. A power series
∑ an x n may be differentiated term by term. The resulting series
n= 0
∞
∑ na x
n =1
n
n −1
and
the given series will have the same radius of convergence.
∞
2. A power series ∑ an x n can be integrated term by term.
n= 0
∞
3. Two power series
∑a x
n=0
n
∞
n
and ∑ bn x n may be added, subtracted and multiplied. The resultant
n=0
series converges in the common interval of convergence.
4. We can divide one power series by another power series if the denominator series is not zero at
x = 0.
∞
If f ( x ) = ∑ an x n , x ∈( − R , R )
n=0
∞
and
g ( y ) = ∑ bn y n , g ( y ) ∈( − R , R ),
n=0
then we can substitute g(y) for x. That is we can substitute one power series in another.
5. The power series expansion of a function is unique.
WORKED EXAMPLES
EXAMPLE 1
Find the radius of convergence and interval of convergence of the series
∞
n=0
M02_ENGINEERING_MATHEMATICS-I _XXXX_CH02.indd 60
n!
∑n
n
xn.
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Sequences and Series ■
2.61
Solution.
∞
Let the given power series be ∑ an x n
∴
∞
n! n
x
n
n
n=0
∑ an x n = ∑
n=0
n= 0
Here
∞
n!
( n + 1)!
an = n and an +1 =
n
( n + 1) n +1
an
n ! ( n + 1) n +1 ( n + 1)( n + 1) n ⎛ n + 1⎞
⎛ 1⎞
= n⋅
=
=⎜
⎟⎠ = ⎜⎝1 + ⎟⎠
n
⎝
n
an +1 n ( n + 1)!
n
n ( n + 1)
n
∴
n
n
∴ the radius of convergence is R = lim
n →∞
an
⎛ 1⎞
= lim ⎜1 + ⎟ = e
⎝ n⎠
n
→∞
an +1
So, the interval of convergence is −e < x < e.
EXAMPLE 2
Find the region of convergence of the series
x2
x5
x2
x3
x4
1 2 2 2 1 2 1 … ∞.
2
2
3
4
5
Solution.
∞
Let the given power series be ∑ an x n
n=1
∴
∞
∑a x
n =1
n
n
x x2 x3 x4 x5 …
−
+
−
+
− ∞
12 22 32 4 2 5 2
=
an =
Here
( −1) n −1
( −1) n
and an +1 =
2
n
( n + 1) 2
an
( − ) n −1 ( n + 1) 2
⎡ n + 1⎤
=
⋅
= −⎢
n
2
( −1)
an +1
n
⎣ n ⎥⎦
∴
2
an
⎡ 1⎤
⎛ n + 1⎞
⎡ n + 1⎤
= ⎢1 + ⎥
= −⎜
⎟⎠ = ⎢
⎥
⎝
an+1
n
⎣ n⎦
⎣ n ⎦
2
2
2
2
a
⎛ 1⎞
∴ the radius of convergence is R = lim n = lim ⎜1 + ⎟ = 1
n →∞ a
n →∞ ⎝
n⎠
n+ 2
∴ the series converges in −1 < x < 1
Now we test the convergence of the series at the end points.
1 1 1 …
+ − +
22 32 4 2
1
It is an alternating series with un = 2 and the terms are decreasing
n
When x = 1, the series is 1 −
Now
lim un = lim
n →∞
M02_ENGINEERING_MATHEMATICS-I _XXXX_CH02.indd 61
n →∞
1
=0
n2
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■
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∴ by Leibnitz’s test, the series is convergent.
When x = −1, the series is
−1 −
1 1 1 …
1 1 1
⎛
⎞
− 2 − 2 − = − ⎜ 1 + 2 + 2 + 2 + …⎟
2
⎝
⎠
2 3 4
2 3 4
This series is convergent by p-series, since p = 2 > 1
Hence, the region of convergence for the given power series is −1 ≤ x ≤ 1.
EXAMPLE 3
⎛ (21) n 3 n ⎞
⎜⎝ n x ⎟⎠ and the interval of convergence.
∑
8
n=0
∞
Find the radius of convergence of the series
Solution.
( −1) n x 3n
x3 x6 x9 …
=
1
−
+
−
+
8 82 83
8n
n=0
∞
The given series is
∑
Put y = x 3 (to reduce to usual form)
∴ the series is 1 −
y y2 y3 …
+
−
+
8 82 83
which is a power series in y with an =
( −1) n
8n
and
an +1 =
∴
an
( −1) n
8 n +1
= n ×
= −8
an +1
8
( −1) n +1
∴
an
= −8 = 8
an+1
( −1) n +1
8 n +1
∴ the radius of convergence of the series in y is
R = lim
n→∞
an
= lim 8 = 8
an+1 n→∞
∴ the power series in y converges in the interval −8 < y < 8
Now we test the convergence of the series at the end points
When y = 8, the series in y becomes
1 − 1 + 1 − 1 + 1 − … ∞, which is not convergent (it oscillates between −1 and 1)
When y = −8, the series in y becomes
1 + 1 + 1 + 1 + … ∞ , which is not convergent.
Hence, the power series in y is not convergent at the end points
∴ the interval of convergence for the power series in y is
−8 < y < 8 ⇒
−8 < x 3 < 8
⇒
−2 < x < 2
[{ y = x 3 ]
∴ the interval of convergence of the given series is −2 < x < 2
and the radius of convergence is 2.
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Sequences and Series ■
EXAMPLE 4
∞
Find the interval of convergence of the series
∑
( x 1 2) n 21
n
n 51
2.63
.
Solution.
The given series is
∞
∑
( x + 2) n −1
n =1
n
= 1+
x +2
2
+
( x + 2) 2
3
( x + 2)3
+
4
+ …+
( x + 2) n
n +1
+…
2
3
n
Put y = x + 2, then the series becomes 1 + y + y + y + … + y
+…
2
3
4
n +1
This is a power series in y.
Here
an =
∴
an
=
an+1
∴
an
=
an+1
1
n +1
1
n +1
1
and an +1 =
n+2
n+2
=
n +1
× n+2 =
2
n
1
1+
n
1+
2
2
1+
n =
n
1
1
1+
1+
n
n
1+
∴ the radius of convergence of the series in y is
2
1+
an
n =1
R = lim
= lim
n →∞ a
n →∞
1
n +1
1+
n
∴ the power series in y converges in the interval −1 < y < 1
Now, we shall test the convergence of the series at the end points.
When y = 1, the series becomes 1 +
1
2
This is divergent by p-series, since p =
+
1
3
+ …+
1
n +1
+…
1
<1
2
When y = −1, the series becomes
1−
This is an alternating series with un =
M02_ENGINEERING_MATHEMATICS-I _XXXX_CH02.indd 63
1
2
+
1
3
−
1
4
+ … ∞.
1
and the terms are decreasing.
n
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■
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lim u n = lim
Now
n →∞
1
n →∞
n
=
1
=0
∞
∴ by Leibnitz’s test the series is convergent, if y = −1.
∴ the interval of convergence of the power series in y is
−1 ≤ y < 1 ⇒
−1 ≤ x + 2 < 1 ⇒ − 3 ≤ x < −1
∴ the interval of convergence of the given series is −3 ≤ x < −1 and the radius of convergence = 1,
since the centre is −2
EXAMPLE 5
⎛ (22 ) n ( 2 x 1 1) n ⎞
⎟⎠ into a power series and then find (i) the radius of
∑ ⎜⎝
n2
n 51
∞
Convert the series
convergence and (ii) the interval of convergence.
Solution.
The given series is
∞
∑ ( −2)n
n =1
( 2x + 1) n
( 2x + 1) 2 8( 2x + 1)3 …
x
+
=
−
2
(
2
+
1
)
+
4
−
n2
32
22
Put y = 2 x + 1, then the series is
∞
∑ ( −2)n
n =1
yn
4 y 2 8y 3 …
y
=
−
2
+
− 2 +
n2
22
3
This is a power series in y.
Here
an =
( −2) n
n2
and
an +1 =
( −2) n +1
( n + 1)2
∴
an
( −2) n ( n + 1) 2
⎡ 1⎤ 1
⎛ n + 1⎞ 1
⋅ = − ⎢1 + ⎥ ⋅
=
×
= −⎜
⎝ n ⎟⎠ 2
an +1
( −2) n +1
n2
⎣ n⎦ 2
∴
an
⎛ 1⎞ ⎛ 1⎞ ⎛ 1 ⎞ 1
= − ⎜1 + ⎟ ⋅ ⎜ ⎟ = ⎜1 + ⎟ ⋅
⎝ n ⎠ ⎝ 2⎠ ⎝ n ⎠ 2
an+1
2
2
2
2
∴ the radius of convergence of the series in y is
2
R = lim
n →∞
an
⎛ 1⎞ 1 1
= lim ⎜1 + ⎟ ⋅ =
an +1 n →∞ ⎝ n ⎠ 2 2
1
1
∴ the power series in y converges in − < y <
2
2
Now we test the convergence of the series at the end points.
1
When y = , the series becomes
2
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Sequences and Series ■
2.65
∞
( −2) n 1
( −1) n
1 1 1
⋅
=
= −1 + 2 − 2 + 2 − …
∑
∑
2
n
2
2
2 3 4
n =1 n
n =1 n
∞
This is an alternating series with u n =
1
and the terms are decreasing
n2
lim u n = lim
Now
n →∞
n →∞
1
1
= =0
2
∞
n
∴ by Leibnitz’s test the series is convergent.
1
When y = − , the series becomes
2
∞
∞
( −2) n
1
1
1 1 1
⋅
= ∑ 2 = 1+ 2 + 2 + 2 + …
∑
2
n
(
−
2
)
2
3 4
n
n
n =1
n =1
which is convergent by p-series, since p = 2 > 1.
∴ the interval of convergence of the power series in y is
⇒
1
1
− ≤ y≤
2
2
⇒
1 1
1 1
− − ≤x≤ −
2 4
4 2
⇒
1
1
− ≤ 2x + 1 ≤
2
2
3
1
− ≤x≤−
4
4
⇒
−
1
1 1
≤ x+ ≤
4
2 4
[dividing by 2]
3
1
≤x≤−
4
4
1
1
and the radius of convergence is , since the centre is −
4
2
∴ the interval of convergence of the given series is −
EXAMPLE 6
For what values of x, the series
1
1
1
1
1
1
1
1 … ∞ converges.
1 2 x 2 (1 2 x ) 2
3(1 2 x ) 3
4 (1 2 x ) 4
Solution.
The given series is
1
1
1
1
+
+
+
+ …∞
2
3
1 − x 2(1 − x ) 3(1 − x ) 4(1 − x ) 4
It is defined if x ≠ 1.
1
, then the given series is
Put y =
1− x
y+
y2 y3 y4 …
+
+
+
2
3
4
This is a power series in y.
Here
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an =
1
n
and an +1 =
1
n +1
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∴
an
n +1
1
=
= 1+
an +1
n
n
∴
an
1
1
= 1+ = 1+
an+1
n
n
∴
lim
n →∞
an
⎛ 1⎞
= lim ⎜1 + ⎟ = 1.
an +1 n →∞ ⎝ n ⎠
∴ the radius of convergence is R = lim
n→∞
an
=1
an+1
∴ the power series in y converges in −1 < y < 1.
Now, we test the convergence of the series at the end points.
1 1 1
When y = 1, the series becomes 1 + + + + …, which is divergent;
2 3 4
1 1 1
When y = −1, the series becomes −1 + − + − …
2 3 4
⎡ 1 1 1
⎤
= − ⎢1 − + − + …⎥ = − loge 2
⎣ 2 3 4
⎦
∴ the series is convergent if y = −1
∴ the interval of convergence of the power series in y is
−1 ≤ y < 1
⇒
−1 ≤
If x < 1, then 1 – x > 0.
⇒
x −1 ≤ 1 < 1− x
x − 1 ≤ 1 and 1 < 1 − x
⇒
x ≤ 2 and x < 0 ⇒ x < 0
If x > 1, then 1 − x < 0.
∴ (1)⇒
( −1)(1 − x ) ≥ 1 > 1 − x
⇒
x −1 ≥ 1 > 1− x
x − 1 ≥ 1 and 1 > 1 − x
∴
(1)
( −1)(1 − x ) ≤ 1 < 1 − x
∴ (1) ⇒
∴
1
<1
1− x
⇒
x≥2
and x > 0 ⇒ x ≥ 2.
∴ the values of x for which the given series converges is x < 0 or x ≥ 2.
EXERCISE 2.7
Find the radius of convergence and interval convergence of the following series (1 to 7)
∞
1.
∞
x 2n
∑
n=0 n !
∞
∞
⎛ 2⎞
2n
⎜⎝ ⎟⎠ ⋅ x
∑
3
n=0
3.
∑
5.
( x + 2) n
∑
n2
n =1
6.
( x − 2) n
∑
3n
n =1
n2
1
4. ∑ ⎛⎜1 + ⎞⎟ ⋅ ( x − 1) 2 n
⎝
n⎠
n =1
n
2.
∞
M02_ENGINEERING_MATHEMATICS-I _XXXX_CH02.indd 66
( 2n )! n
⋅x
n
n=0 2
∞
∞
7.
n
∑ ( x 2 + 1) 2 ⋅
n= 0
2n
3n + n3
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Sequences and Series ■
2.67
8. Find the radius of convergence and the interval of convergence and the behaviour at the end
x x2
x3
x4
points of the interval of convergence of the power series 1 − + 2 − 3 + 4 − …, a > 0.
a 2a 3a 4a
9. Find the region of convergence of the series x +
x2
2
+
x3
3
+
x4
4
+ … ∞.
( x − 3) 2 n
.
3n
n= 0
∞
10. Find the radius of convergence and the interval of convergence of the series ∑
n
∞ ⎡⎛ ( 2
x − 1) ⎞ ⎤
11. Find the interval of convergence of the series ∑ ⎢⎜
⎥ by changing it into a power series.
⎝ 8n ⎟⎠ ⎦
n= 0 ⎣
( x + 2) n
.
n ⋅ 3n
n =1
∞
12. Find the region of convergence of the series ∑
ANSWERS TO EXERCISE 2.7
3
3
;−
<x<
2
2
1. R = ∞; − ∞ < x < ∞
2. R =
1 1
1
3. R = ; − < x <
4 4
4
4. R =
5. R = 1; − 3 ≤ x ≤ −1
6. R = 3; − 1 < x < 5
7. R =
5
5
5
<x<
;−
2
2
2
3
2
1
1
1
;1−
< x < 1+
e
e
e
8. R = a, −a < x < a
9. R = 1, −1 ≤ x < 1
10. R = 3 , 3 − 3 < x < 3 + 3 , since 3 is the centre.
11. R = 3, −3 < x < 3
12. −5 ≤ x < 1, R = 3
SHORT ANSWER QUESTIONS
1. Examine the series 1 1
1 1 1…
for convergence.
1 1
2 4 8
2. The decimal representation of a number 0 ⋅ d 1d 2d 3d 4 … is that
0 ⋅ d 1d 2d 3d 4 … 5
d
d1
d
d
1 2 1 3 1 4 1…
10 10 2 10 3 10 4
where di is one of the numbers 0, 1, 2, 3, … 9. Show that the series on the right hand side of equation is
always convergent.
1
3. Test the convergence of an infinite series whose nth term is sin 2 .
n
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∞
4. Test the convergence of
∑n
n 51
3
n
.
11
5. Test the convergence of the series
∞
6. Let
∑u
n 51
n
1
2
3
1
1
1 ….
1⋅ 3 3 ⋅ 5 5 ⋅ 7
be a series of positive terms with the property un+1 < un ; n ≥ 1. By an application of ratio
∞
test, can we conclude that
∑u
n
converges? Justify your answer.
n 51
7. Prove that the series
2
3
4
1 1 1… is convergent.
3! 4 ! 5 !
8. Test the convergence of
n3 2n 1 1
.
n!
n 51
∑
9. Test the convergence of the series 1 1
10. Is the series 1 2
2p
3p
4p …
1
1
1 .
2 ! 3! 4 !
1 1 1 …
absolutely convergent? or conditionally convergent?
1 2 1
2 3 4
3 4 5 …
1 2 1 .
2 3 4
11. Discuss the convergence of the series 2 2
12. Test the convergence of the series
2
3
4
5
2 2 1 2 2 2 1 … ∞.
12
2
3
4
13. Test the convergence of the series 1 2
2 3 4 …
1 2 1 .
3 5 7
14. Discuss the convergence of the series
∑ ⎜⎝ (21)
∞
n 51
⎛
n 21
n ⎞
.
n 2 1 1 ⎟⎠
15. Discuss the convergence of the series 1 2
1
1
1
1 2 1 ….
2! 4! 6!
16. Discuss the convergence of the series 1 2
1
1
1
1 2 2 1….
2 2 32
4
17. Discuss the convergence of the series
1
2
3
4 …
2 1
2
1 .
6 11 16 21
18. Test the convergence of the series 5 2 4 2 1 1 5 2 4 2 1 1 5 2 4 2 1 1 … ∞.
19. The series 2 2 5 1 3 1 2 2 5 1 3 1 2 2 5 1 3 1 … ∞ is convergent or not.
20. Test the convergence of the series 1 1
1
1
1
1
1
1
1
2 2 2 2 1 2 1 2 2 2 2 2 1… ∞.
2
2
3
4
5
6
7
8
21. Test the convergence of the sequence
3 1 4 3 2 1 4 2 33 1 4 3 …
,
,
, .
4 1 5 4 2 1 5 2 4 3 1 53
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Sequences and Series ■
∞
22. Test the convergence of the series
1
∑ (log n)
n52
n
2.69
.
23. Test the convergence of the series 1 2 1 1 1 2 1 1… ∞.
2 2 3 3 4 4
OBJECTIVE TYPE QUESTIONS
A. Fill up the blanks
1 1 1
1. The series 1 + + +
+ … is _____________.
3 9 27
2. The series 1 + 3 + 9 + 27 + … is _____________
3. If lim un ≠ 0, then the series of positive terms ∑ un is _____________.
n →∞
4. The series
∞
5.
1
2
3
+
+
+ … is _____________.
1.2 3.4 5.6
1
∑
n + n +1
n =1
is _____________.
∞
6. If ∑ un is a convergent series of positive terms, then lim un = _____________.
n →∞
n=1
7. The series
1 1 1 1
− + − + … is _____________.
2 3 4 5
8. The series 1 −
9. The series
1 1 1
+ − + … is _____________.
2! 4 ! 6 !
1
1
1
1
−
+
−
+ … is _____________.
1.2 3.4 5.6 7.8
⎧ n 2 − 2n ⎫
10. The sequence ⎨ 2
⎬ converges to _____________.
⎩ n +n ⎭
B. Choose the correct answer
1. The sequence {sn}, where sn = 1 +
(a) 1
2. The sequence
(b)
{
n +1 − n
(a) 0
3. The series 1 +
(a) convergent
}
∞
n =1
1 1 … 1
+ + + n +1 converges to
2 22
2
1
2
2
+
1
3
+
(d) 3
converges to
(b) 1
1
(c) 2
(c)
1
2
(d) −1
1
+ … is
4
(b) divergent
M02_ENGINEERING_MATHEMATICS-I _XXXX_CH02.indd 69
(c) oscillates
(d) None of these
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1
1
1
1
+
+
+
+ … is
2 2.22 3.23 4.24
(a) convergent
(c) oscillates
1
1
1
1 …
+
−
+
− is
5. The series 1 −
3
5
7
9
(a) absolutely convergent
(c) not convergent
4. The series
(b) divergent
(d) conditionally convergent
(b) conditionally convergent
(d) divergent
1 1 1 1 …
+ − + − is
2 4 8 16
(a) absolutely convergent
(c) divergent
(b) conditionally convergent
(d) None of these
2 3
4
5
− + − + … is
12 22 32 4 2
(a) conditionally convergent
(c) not convergent
(b) absolutely convergent
(d) None of these
6. The series 1 −
7. The series
∞
8. The series ∑ ( −1) n sin
n =1
1
is
n
(a) absolutely convergent
(c) not convergent
9. The series
∑ ( −1)
n −1
.
(b) not absolutely convergent
(d) None of these
n
is
2n − 1
(b) divergent
(d) conditionally convergent
(a) convergent
(c) absolutely convergent
10. The series
1 1 1 …
1
+ + + +
+ … is
2 5 10
1 + n2
(b) divergent
(d) None of these
(a) convergent
(c) conditionally convergent
ANSWERS
A. Fill up the blanks
1. convergent
2. divergent
3. divergent
4. divergent
5. divergent
6. 0
7. convergent
8. convergent
9. convergent
B. Choose the correct answer
1. (a)
2. (a)
3. (b)
4. (a)
M02_ENGINEERING_MATHEMATICS-I _XXXX_CH02.indd 70
5. (b)
6. (a)
7. (b)
8. (b)
10. 1
9. (b)
10. (a)
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Differential Calculus
3.0
3
INTRODUCTION
Calculus is one of the remarkable achievements of the human intellect. It is a collection of fascinating
and exciting ideas rather than a technical tool.
Calculus has two main divisions namely differential calculus and integral calculus. They serve
to solve a variety of priblems that arise in science, engineering and other fields including social
sciences.
Differential calculus had its origin from the problem of tangent to a curve and integral calculus had
its origin from the problem of finding plane area. The concept of derivative which measure the rate of
change of a function is the central idea in differential calculus.
We assume that the reader is familiar with the basic concepts of limit, continuity and differentiability.
In this chapter we deal with successive differentiation, equation of tangent and normal, length of the
tangent, length of the normal, length of the sub-tangent and length of the sub-normal, Rolle’s theorem,
Mean value theorems, Taylor’s series, Maclaurin’s series, Indeterminate forms, Maxima and minima
and curve tracing.
Basic rules of differentiation
If u and v are differentiable functions of x, then
(1)
(2)
(3)
(4)
(5)
(6)
d
du
( ku ) = k
, where k is constant
dx
dx
d
du dv
(u ± v ) =
±
dx
dx dx
d
dv
du
(u ⋅ v ) = u + v
dx
dx
dx
d ⎛ u⎞
⎜ ⎟=
dx ⎝ v ⎠
v
du
dv
−u
dx
dx
v2
[product rule]
[quotient rule]
d
d
dv
[u (v ( x ))] =
[u (v ( x ))] ⋅
dx
dv
dx
[composite function rule]
d
( k ) = 0, where k is a constant.
dx
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Now we state the basic formulae in differential calculus.
Function
y = f ( x)
Derivative
dy
= f ′( x )
dx
12.
1
xn
−
1
x
13.
sin −1 x
ex
ex
14.
cos −1 x
−
4.
sin x
cos x
15.
tan −1 x
1
1+ x2
5.
cos x
− sin x
16.
( ax + b) n
n( ax + b) n−1 ⋅ a
6.
tan x
sec 2 x
17.
e ax + b
e ax + b ⋅ a
7.
sec x
sec x tan x
18.
loge (ax + b )
1
⋅a
ax + b
8.
cos ec x
− cosec x cot x
19.
cos( ax + b)
− sin( ax + b) ⋅ a
9.
cot x
− cosec 2 x
20.
sin( ax + b)
cos( ax + b) ⋅ a
10.
x
1
2 x
21.
tan( ax + b)
sec 2 ( ax + b) ⋅ a
1
x2
22.
ax
a x ⋅ log e a
Function
y = f ( x)
Derivative
dy
= f ′( x )
dx
1.
xn
nx n−1
2.
log e x
3.
11.
3.1
1
x
−
n
x n+1
1
1 − x2
1
1 − x2
SUCCESSIVE DIFFERENTIATION
dy
Let y = f ( x ) be a differentiable function of x. The derivative
is called the first derivative of y
dx
w.r.to x.
In general, it is function of x.
dy
d2 y
The derivative of
is called the second derivative of y w.r.to x and it is denoted as
dx
dx 2
2
d y d dy
Thus, 2 = ⎛⎜ ⎞⎟ .
dx
dx ⎝ dx ⎠
d 3y
d3 y
Similarly, the derivative of
is called the third derivative of y w.r.to x and it is denoted as 3
3
dx
dx
d3 y d ⎛ d2 y ⎞
and 3 = ⎜ 2 ⎟ and so on.
dx
dx ⎝ dx ⎠
M03_ENGINEERING_MATHEMATICS-I _CH03_Part A.indd 2
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Differential Calculus ■
The nth differential coefficient of y w.r.to x is denoted by
Thus, the successive derivative of y are
3.3
d n y d ⎛ d n−1 y ⎞
.
=
dx n dx ⎜⎝ dx n−1 ⎟⎠
dy d 2 y d 3 y
dn y
,
, …,
.
,
2
3
dx dx
dx
dx n
These derivatives are also denoted by
(1) y1 , y2 ,… yn … or (2) Dy , D 2 y , D 3 y , …, D n y , …
2
n
where D = d , D 2 = d 2 , …, D n = d n , …
dx
dx
dx
(3) f ′( x ), f ″( x ), f ″′( x ),…, f ( n ) ( x ) or (4) y ′, y ″, y ″′,…, y ( n ) .
The process of finding second and higher order derivative is called successive differentiation.
For example: If y = x 7 , then
dy
= 7x 6 ,
dx
d 3y
d ⎛ d 2y ⎞
= 42.5x 4 = 210 x 4 ,
=
dx 3 dx ⎜⎝ dx 2 ⎟⎠
d 5y
= 840 ⋅ 3x 2 = 2520 x 2 ,
dx 5
d 7y
= 5040,
dx 7
d 2y
d ⎛ dy ⎞
5
5
=
⎜ ⎟ = 7.6 ⋅ x = 42x
2
dx ⎝ dx ⎠
dx
d 4y
= 210 ⋅ 4 x 3 = 840 x 3
dx 4
d6y
= 2520 × 2x = 5040 x
dx 6
d8y
=0
dx 8
and other higher derivatives vanish.
Note Let y = x 3
∴
dy
d 2y
d 3y
= 3x 2 ,
= 6x ,
=6
2
dx
dx
dx 3
2
Now
⎛ dy ⎞
2 2
4
⎜⎝ ⎟⎠ = (3x ) = 9x
dx
3
⎛ dy ⎞
and ⎜ ⎟ = (3x 2 )3 = 27x 6 .
⎝ dx ⎠
Hence, we find that,
d 2 y ⎛ dy ⎞
≠⎜ ⎟
dx 2 ⎝ dx ⎠
2
and
d 3y
⎛ dy ⎞
≠⎜ ⎟
⎝ dx ⎠
dx 3
3
n
In general,
d n y ⎛ dy ⎞
≠⎜ ⎟ .
dx n ⎝ dx ⎠
WORKED EXAMPLES
EXAMPLE 1
Find the first two differential coefficients w.r.to x of x 2 cos x.
Solution.
Let
y = x 2 cos x
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Differentiating w.r.to x,
dy
= x 2 [ − sin x ] + cos x ⋅ 2 x = − x 2 sin x + 2 x cos x.
dx
Again differentiating w.r.to x,
d 2y
= −[x 2 cos x + sin x ⋅ 2x ] + 2[x ( − sin x ) + cos x ⋅1]
dx 2
= − x 2 cos x − 2x sin x − 2x sin x + 2 cos x = ( 2 − x 2 ) cos x − 4 x sin x .
EXAMPLE 2
If y 5 e ax sin bx , then show that
d 2y
dy
22 a
1 ( a 2 1b 2 ) y 5 0.
2
dx
dx
Solution.
Given
y = e ax sin bx
Differentiating w.r.to x,
dy
= e ax cos bx ⋅ b + sin bxe ax ⋅ a = e ax [b cos bx + a sin bx ].
dx
Again differentiating w.r.to x,
d 2y
= e ax [b ( − sin bx )b + a cos bx ⋅ b ] + [b cos bx + a sin bx ]e ax ⋅ a
dx 2
= e ax [ −b 2 sin bx + ab cos bx + ab cos bx + a2 sin bx ]
= e ax [(a2 − b 2 ) sin bx + 2ab cos bx ].
d 2y
dy
− 2a
+ (a 2 + b 2 ) y
dx
dx 2
= e ax [(a2 − b 2 ) sin bx + 2ab cos bx ] − 2a[e ax {b cos bx + a sin bx }] + (a2 + b 2 )e ax sin bx
L.H.S =
= e ax [a2 sin bx − b 2 sin bx + 2ab cos bx − 2ab cos bx − 2a2 sin bx + a2 sin bx + b 2 sin bx ]
= e ax × 0 = 0 = R.H.S..
EXAMPLE 3
If y 5 cos 21 x , then show that (12x 2 ) y2 2 xy1 5 0.
Solution.
Given
Differentiating w.r.to x,
y = cos −1 x
dy
1
=−
.
dx
1 − x2
2
1
⎛ dy ⎞
⎜⎝ ⎟⎠ =
dx
1− x 2
Again differentiating w.r.to x,
Squaring,
(1 − x 2 )2 y1 y2 + y12 ( −2 x ) = 0
M03_ENGINEERING_MATHEMATICS-I _CH03_Part A.indd 4
⇒
⇒
y 12 =
1
1− x 2
⇒
(1 − x 2 ) y 2 − xy 1 = 0.
(1 − x 2 ) y 12 = 1.
[Dividing by 2y1]
5/30/2016 7:07:02 PM
Differential Calculus ■
3.5
EXAMPLE 4
If y 5
2
3
dy
log x
2 d y
3 d y
,
then
show
that
x
1
6
x
1 4x
2 4 y 5 0.
2
3
2
dx
dx
dx
x
Solution.
loge x
.
x2
Differentiating w.r.to x,
Given
y =
dy
=
dx
1
x 2 ⋅ − log e x ⋅ 2 x x − 2 x ⋅ log x x[1 − 2 log x ] 1 − 2 log x
x
e
e
e
=
=
=
.
x4
x4
x4
x3
Again differentiating w.r.to x,
⎛ 2⎞
x 3 ⎜ − ⎟ − (1 − 2 loge x )3x 2
⎝ x⎠
x6
2
2
−2x − 3x (1 − 2 loge x ) x 2 [ −5 + 6 loge x ] 6 loge x − 5
=
=
=
x6
x6
x4
Again differentiating w.r.to x,
d 2y
=
dx 2
6
− (6 loge x − 5)4 x 3
x
x8
3
3
6 x − 24 x loge x + 20 x 3 x 3 [26 − 24 loge x ] 26 − 24 loge x
=
=
=
x8
x8
x5
d 3y
=
dx 3
∴
x4 ⋅
L.H.S = x 3
d3 y
d2 y
dy
+ 6x2 2 + 4x − 4 y
3
dx
dx
dx
x 3 [26 − 24 loge x ] 6 x 2 [6 loge x − 5] 4x [1 − 2 loge x ]
log x
+
+
− 4 e2
5
4
3
x
x
x
x
26 − 24 loge x 6(6 loge x − 5) 4(1 − 2 loge x )
loge x
=
+
+
−4
x2
x2
x2
x2
1
= 2 [26 − 24 loge x + 36 loge x − 30 + 4 − 8 loge x − 4 loge x ]
x
30 − 30 + 36 loge x − 36 loge x
=
= 0 = R.H.S.
x2
=
EXAMPLE 5
If x 5 a cos 3 u, y 5 b sin3 u then find
d 2y
.
dx 2
Solution.
Given x = a cos3 u and y = b sin 3 u , which are parametric equations.
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Differentiating w.r.to u, we get,
dx
= 3a cos 2 u( − sin u) = −3a cos 2 u sin u
du
∴
∴
and
dy
= 3b sin 2 u cos u.
du
dy dy/d u
3b sin 2 u cos u
b
=
=
= − tan u.
2
dx dx/d u −3a cos u sin u
a
d 2 y d ⎛ dy ⎞
d ⎛ b
b
1
b
⎞ du
=
= − sec 2 u ⋅
= 2 sec 4 u cosecu.
⎜⎝ ⎟⎠ =
⎜⎝ − tan u⎟⎠
2
2
dx dx
du
a
dx
a
dx
−3a cos u sin u 3a
EXAMPLE 6
If x 5 sin t and y 5 sin pt, then prove that (1 2 x2 )y2 2 xy1 1 p2y 5 0.
Solution.
x = sin t
Given
(1)
and
y = sin pt
(2)
Differentiating (1) and (2) w.r.to t, we get
dx
dy
= cos t ,
and
= p cos pt
dt
dt
p cos pt
dy dy/dt p cos pt
.
⇒ y1 =
=
=
cos t
dx dx/dt
cos t
∴
y12 =
Squaring,
⇒
2
2
2
2
p 2 cos 2 pt p (1 − sin pt ) p (1 − y )
=
=
1 − sin 2 pt
1− x 2
cos 2 t
(1 − x ) 2 y12 = p 2 (1 − y 2 ).
Differentiating w.r.to x,
(1 − x 2 )2 y1 y2 + y12 ( −2 x ) = −2 p 2 yy1
Dividing by 2y1, we get
(1 − x 2 ) y2 − xy1 = − p 2 y
⇒
(1 − x 2 ) y2 − xy1 + p 2 y = 0.
EXERCISE 3.1
4
1. If y = e − x cos x , then prove that d y + 4 y = 0.
dx 4
2
2. If y = sin(sin x ), then prove that d y + tan x dy + y cos 2 x = 0 .
dx
dx 2
2
d
y
2
= n( n + 1) y .
3. If y = Ax n+1 + Bx − n , then prove that x
dx 2
d 2y
dy
2
2
2
+ 2x
+ n 2 y = 0 , where n = p + k .
dt
dt 2
d2 y
dy
5. If y = a cos(log x ) + b sin(log x ), then show that x 2 2 + x + y = 0.
dx
dx
ax 2 + bx + c
6. If y =
, then show that (1 − x ) y 3 = 3y 2 .
1− x
4. If y = Ae − kt cos( pt + e), then show that
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Differential Calculus ■
3.7
d 2y
2t 3
1 ⎛ 1⎞
⎛ 1⎞
.
=
7. If x = ⎜ t + ⎟ , y = ⎜ t − ⎟ , then prove that
⎝ t⎠
dx 2 (1 − t 2 )3
2⎝ t⎠
d2 y
dy
− x − 2 = 0.
2
dx
dx
2
−1
2
9. If y = (tan x ) , then prove that ( x + 1) y 2 + 2( x 2 + 1) y 1 − 2 = 0.
8. If y = (sin −1 x ) 2 , then show that (1 − x 2 )
10. If xy = ae x + be − x , then prove that x
d 2y
dy
+2
− xy = 0 .
2
dx
dx
2
2
11. If y 2 = ax 2 + 2bx + c , where a, b, c are constants, then show that d x = b − ac .
dy 2 ( ax + b)3
2a3 xy
3
3
2
.
12. If x + y = 3axy , prove that D y =
( ax − y 2 )3
13. If y = e3x(ax + b), then prove that
d2 y
dy
− 6 + 9y = 0 .
2
dx
dx
d2x
dx
+ 2n + n 2 x = 0 .
2
dt
dt
2
d
y
dy
15. If y = x n −1 log x , show that x 2 − ( n − 2)
− ( n − 1)x n − 2 = 0 .
dx
dx
d 2y
dy
16. If xy = ae x + be − x , then prove that x 2 + 2
− xy = 0.
dx
dx
14. If x = (a + bt)⋅e − nt, then show that
17. If x = 2 cos t − cos 2t, y = 2 sin t − sin 2t, then prove that the value of
p
d 2y
when t = is −3.
2
dx
2
3.1.1 The nth Derivative of Standard Functions
1. Find the nth derivative of eax.
Solution.
Let
y = eax
Differentiating w.r.to x, successively we get,
(1)
y1 = ae ax ,
y2 = a ⋅ ae ax = a 2 e ax
y3 = a 2 ⋅ ae ax = a3 e ax ,...
yn = a n −1 ⋅ ae ax = a n e ax
(2)
Note
(1) Putting a = 1 in (1) and (2), we get, y = ex and
(2) Since a = e loge a , a x = e x loge a
So, if y = ax, then
∴
yn = ex
dn x
log a
(a ) = (loge a) n e x e = (loge a) n ax
dx n
yn = (loge a)n ax.
2. Find the nth derivative of (ax 1 b)m.
Solution.
Let
y = (ax + b)m
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3.8
■
Engineering Mathematics
Differentiating w.r.to x, successively,
y 1 = m (ax + b ) m −1 ⋅ a = am (ax + b ) m −1
y 2 = a ⋅ m ( m − 1)(ax + b ) m − 2 ⋅ a = a2 m ( m − 1)(ax + b ) m − 2
y3 = a3 m( m − 1)( m − 2)( ax + b) m −3
:
n
y n = a m ( m − 1)( m − 2) …( m − n + 1)(ax + b ) m − n
This result is true for all values of m.
Particular cases
Case 1: If m is a positive integer and equal to n, then
yn = a n ⋅ n( n − 1)( n − 2) ⋅⋅⋅ ( n − n + 1)( ax + b) n− n
= a n n( n − 1)( n − 2) ⋅⋅⋅ 2 ⋅1 = n! a n
∴
dn
( ax + b) n = n! a n .
dx n
Since
yn = n! a n is a constant, yn+1 = 0, y n+2 = 0 and so on.
That is (n + 1)th, (n + 2)th and all higher order derivatives of (ax + b)n are zero.
Case 2: If m is a integer and m < n, then the nth derivative of (ax + b)m = 0.
Case 3: If a = 1, b = 0, then
y = xm
yn = m( m − 1) ⋅⋅⋅ ( m − n + 1) x m − n
If m is a positive integer and m = n then
yn = n( n − 1)( n − 2) ⋅⋅⋅ ( n − n + 1) x n − n = n( n − 1)( n − 2) ⋅⋅⋅ 2 ⋅1 = n !
3. Find the nth derivative of
1
.
ax 1 b
Solution.
1
= ( ax + b) −1
ax + b
Differentiating w.r.to x, successively,
Let
y=
y1 = ( −1)( ax + b) −2 ⋅ a
=
( −1)a
( ax + b) 2
y2 = a( −1)( −2)( ax + b) −3 ⋅ a
=
( −1) 2 2 ! a 2
( ax + b)3
y3 = a 2 ( −1)( −2)( −3)( ax + b) −4 ⋅ a =
( −1)3 3! a3
( ax + b) 4
:
yn = a n−1 ( −1)( −2)( −3) ⋅⋅⋅ ( n − 1)( n)( ax + b) − ( n+1) ⋅ a
⇒
yn =
( −1) n n !an
(ax + b ) n +1
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Differential Calculus ■
3.9
Particular case
If a = 1, b = 0, then y =
1
x
yn =
∴
( − ) n n!
, where n is a positive integer.
x n+1
4. Find the nth derivative of log (ax 1 b).
Solution.
Let
Differentiating w.r.to x,
y = loge (ax + b)
1
⋅a
ax + b
Again differentiating w.r.to x, we get
= a( ax + b) −1
y1 =
( −1)a 2
( ax + b) 2
( −1) 2 2! a3
y3 = a 2 ( −1)( −2)( ax + b) −3 ⋅ a =
( ax + b)3
:
an
.
yn = ( −1) n−1 ( n − 1)!
( ax + b) n
y2 = a ⋅ ( −1)( ax + b) −2 ⋅ a
=
5. Find the nth derivative of sin( ax 1 b ).
Solution.
Let
y = sin( ax + b)
Differentiating w.r.to x successively, we get
p
y1 = cos( ax + b) ⋅ a = a cos( ax + b) = a sin ⎛⎜ ax + b + ⎞⎟
⎝
2⎠
p⎞
⎛
y2 = a cos ⎜ ax + b + ⎟ ⋅ a
⎝
2⎠
p p⎞
⎛
= a 2 sin ⎜ ax + b + + ⎟
⎝
2 2⎠
p⎞
⎛
= a 2 sin ⎜ ax + b + 2 ⋅ ⎟
⎝
2⎠
p⎞
p p⎞
p⎞
⎛
⎛
⎛
y3 = a 2 cos ⎜ ax + b + 2 ⋅ ⎟ ⋅ a = a3 sin ⎜ ax + b + 2 ⋅ + ⎟ = a3 sin ⎜ ax + b + 3 ⎟
⎝
⎝
⎝
2⎠
2 2⎠
2⎠
:
p⎞
⎛
yn = a n sin ⎜ ax + b + n ⋅ ⎟
⎝
2⎠
Particular case
If a = 1, b = 0 , then y = sin x
∴
p⎞
⎛
yn = sin ⎜ x + n ⎟ .
⎝
2⎠
6. Find the nth derivative of cos( ax 1 b).
Solution.
Let
y = cos( ax + b)
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■
3.10
Engineering Mathematics
Differentiating w.r.to x successively, we get
p
y1 = − sin( ax + b) ⋅ a = − a sin (ax + b) = a cos ⎛⎜ ax + b + ⎞⎟
⎝
2⎠
p⎞
p⎞
p p⎞
⎛
⎛
⎛
y2 = − a sin ⎜ ax + b + ⎟ ⋅ a = a 2 cos ⎜ ax + b + + ⎟ = a 2 cos ⎜ ax + b + 2 ⋅ ⎟
⎝
⎝
⎝
2⎠
2⎠
2 2⎠
:
p⎞
⎛
yn = a n cos ⎜ ax + b + n ⋅ ⎟
⎝
2⎠
Particular case:
If a = 1, b = 0 , then y = cos x.
∴
p⎞
⎛
yn = cos ⎜ x + n ⋅ ⎟ .
⎝
2⎠
7. Find the nth derivative of e ax sin( bx 1 c ).
Solution.
Let
y = e ax sin(bx + c)
Differentiating w.r.to x, successively we get
ax
y1 = e ax cos(bx + c) ⋅ b + sin(bx + c). e ax ⋅ a = e [b cos(bx + c) + a sin(bx + c)]
a = r cos u, b = r sin u, then
If
( a 2 + b 2 ) = r 2 cos 2 u + r 2 sin 2 u = r 2 (cos 2 u + r 2 sin 2 u) = r 2
and
∴
tan u =
b,
a
∴
⇒
r = ( a 2 + b 2 )1 2
⎛ b⎞
u = tan −1 ⎜ ⎟ .
⎝ a⎠
y1 = e ax [r sin u cos(bx + c) + r cos u sin( bx + c)]
= re ax [sin u cos(bx + c) + cos u sin( bx + c)] = re ax sin(bx + c + u)
∴
y2 = r[e ax cos(bx + c + u) ⋅ b + sin(bx + c + u)e ax ⋅ a]
= re ax [r sin u cos(bx + c + u) + r cos u sin( bx + c + u)]
= r 2 e ax sin(bx + c + u + u) = r 2e ax sin(bx + c + 2u).
Similarly,
y3 = r 3e ax sin(bx + c + 3u)
:
yn = r n e ax sin(bx + c + nu)
⇒
b⎞
⎛
yn = ( a 2 + b 2 ) n 2 ⋅ e ax sin ⎜ bx + c + n tan −1 ⎟ .
⎝
a⎠
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Differential Calculus ■
3.11
8. Find the nth derivative of e ax cos( bx 1 c ).
Solution.
Let
y = e ax cos( bx + c)
Differentiating w.r.to x, successively we get
y1 = e ax ( − sin(bx + c) ⋅ b) + cos(bx + c) ⋅ e ax ⋅ a
⇒
y1 = e ax [a cos(bx + c) − b sin(bx + c)]
Put
a = r cos u, b = r sin u , then r 2 = a 2 + b 2 ⇒ r = ( a 2 + b 2 )1 2
and
tan u =
∴
b
a
⇒
⎛ b⎞
u = tan −1 ⎜ ⎟ .
⎝ a⎠
y1 = e ax [r cos u cos(bx + c ) − r sin u sin( bx + c)]
= re ax [cos u cos(bx + c ) − sin u sin( bx + c)] = re ax cos( bx + c + u)
y2 = r[e ax ( − sin( bx + c + u) ⋅ b) + cos( bx + c + u)e ax ⋅ a]
= re ax [a cos(bx + c + u) − b sin(bx + c + u)]
= re ax [r cos u cos( bx + c + u) − r sin u sin(bx + c + u)]
= r 2 e ax [cos u cos(bx + c + u) − sin u sin( bx + c + u)]
⇒
y 2 = r 2e ax cos(bx + c + u + u) = r 2e ax cos(bx + c + 2u).
Similarly, y3 = r 3e ax cos(bx + c + 3u)
:
y n = r n e ax cos(bx + c + nu)
b⎞
⎛
yn = ( a 2 + b 2 ) n 2 ⋅ e ax cos ⎜ bx + c + n tan −1 ⎟ .
⎝
a⎠
⇒
WORKED EXAMPLES
1. The n th derivative using partial fractions
EXAMPLE 1
Find the nth derivative of (i)
Solution.
x .
(i)
( x − 2) 2
x
2x
(ii)
.
( x 22) 2
(2 x11)( x 21)
Let
y=
x
( x − 2) 2
1
⎛
⎞
Split the R.H.S into partial fractions. ⎜ to reduce to the form
m⎟
( ax + b) ⎠
⎝
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Engineering Mathematics
Let
x
B
= A +
2
( x − 2)
( x − 2) ( x − 2) 2
⇒
x = A( x − 2) + B.
Put x = 2, then
2 = A× 0 + B
Equating the coefficients of x, A = 1.
∴
∴
2x
.
( 2 x + 1)( x − 1)
B = 2.
x
1
2
=
+
2
( x − 2)
( x − 2) ( x − 2) 2
1
2
y=
+
( x − 2) ( x − 2) 2
then
(ii)
⇒
Let
yn =
( −1) n n!
( −1) n ( n + 1)! .
+ 2⋅
n +1
( x − 2)
( x − 2) n+ 2
y=
2x
( 2 x + 1)( x − 1)
[Refer formula 2]
Split the R.H.S into partial fractions.
2x
Let
= A + B
( 2 x + 1)( x − 1) 2 x + 1 x − 1
⇒
2 x = A( x − 1) + B( 2 x + 1)
1
⎛ 1⎞
⎛ 1 ⎞
Put x = − , then 2 ⎜ − ⎟ = A ⎜ − − 1⎟
⎝ 2⎠
⎝ 2 ⎠
2
3
2
− A = −1 ⇒ A =
2
3
2
2 ⋅1 = B( 2 ⋅1 + 1) ⇒
3B = 2
⇒ B=
Put x = 1, then
3
2
x
2
1
2
1
∴
= ⋅
+ ⋅
( 2 x + 1)( x − 1) 3 2 x + 1 3 x − 1
∴
y=
Hence,
yn =
=
EXAMPLE 2
Find the nth derivative of
Solution.
Let
⇒
2
1
2 1
⋅
+ ⋅
3 2x + 1 3 x −1
2 ( −1) n n ! 2n 2 ( −1) n n ! .
⋅
+
3 ( 2 x + 1) n +1 3 ( x − 1) n +1
[Refer formula 3]
( −1) n n ! 2n +1 2( −1) n n ! .
+
3( 2 x + 1) n +1 3( x − 1) n +1
x4
.
( x 21)( x 22)
y =
x4
( x − 1)( − 2)
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Differential Calculus ■
3.13
Split the R.H.S into partial fractions.
Since the degree of the Nr. is greater than the degree of Dr., it is an improper fraction.
We divide x 4 by ( x − 1)( x − 2) = x 2 − 3x + 2 and then split the proper fraction into partial
fractions.
Dividing
∴
x4
15 x − 14
= x 2 + 3x + 7 +
( x − 1)( x − 2)
( x − 1) ( x − 2)
Let
15 x − 14
A
B
=
+
( x − 1)( x − 2) x − 1 x − 2
⇒
x 2 + 3x + 7
x 2 − 3x + 2 x 4
x 4 − 3x 3 + 2 x 2
⎯⎯⎯⎯⎯⎯⎯
15x − 14 = A ( x − 2) + B ( x − 1)
Put x = 1, then 15 ⋅1 − 14 = A(1 − 2) + B ⋅ 0
Put x = 2, then 15 ⋅ 2 − 14 = B( 2 − 1)
⇒
3x 3 − 2 x 2
A = −1
⇒ B = 16
15 x − 14
1
16
=−
+
( x − 1)( x − 2)
( x − 1) x − 2
∴
3x 3 − 9 x 2 + 6 x
⎯⎯
⎯⎯⎯⎯⎯⎯⎯⎯⎯
7x2 − 6x
7 x 2 − 21x + 14
⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯
15 x − 14
4
∴
x
1
16
= x 2 + 3x + 7 −
+
( x − 1)( x − 2)
x −1 x − 2
∴
y = x 2 + 3x + 7 −
Hence,
y n = −1⋅
1
16
+
x −1 x − 2
( −1) n n !
( −1) n n !
( −1) n +1 n ! 16( −1) n n !
, n>2
=
+
16
+
( x − 1) n +1
( x − 2) n +1
( x − 1) n +1 ( x − 2) n +1
2. The n th derivatives of trigonometric functions
If y is a simple power of sine or cosine or product of sine and cosine, then they can be expressed as a
sum of sines and cosines of multiple angles and nth derivative can be found.
EXAMPLE 3
Find the nth derivative of (i) sin2x (ii) sin3x.
Solution.
(i) sin2x
Let
∴
y = sin 2 x =
1 − cos 2x 1 cos 2x
= −
2
2
2
p⎞
1
p⎞
⎛
⎛
y n = − 2n cos ⎜ 2x + n ⎟ = −2n −1 cos ⎜ 2x + n ⎟ .
⎝
⎝
2
2⎠
2⎠
M03_ENGINEERING_MATHEMATICS-I _CH03_Part A.indd 13
[Refer the formula 5]
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3.14
■
Engineering Mathematics
(ii) sin3x
Let
y = sin 3 x ⇒ =
Hence, y n =
1
3
1
[3 sin x − sin 3 x ] = sin x − sin 3x [{ sin 3x = 3 sin x − 4 sin 3 x ]
4
4
4
3 ⎛
p⎞ 1
p⎞
⎛
sin ⎜ x + n ⎟ − 3n sin ⎜ 3x + n ⎟
⎝
⎝
4
2⎠ 4
2⎠
[Refer the formula 5]
EXAMPLE 4
Find the nth derivative of (i) cos4x (ii) sinx sin2x sin3x.
Solution.
(i) cos4x
⎡1 + cos 2x ⎤
⎥
2
⎣
⎦
2
2
2
y = cos 4 x = (cos x ) = ⎢
Let
⇒
=
1
[1 + 2 cos 2x + cos 2 2x ]
4
=
1 1
1 ⎡1 + cos 4 x ⎤ 1 1 1
1
+ cos 2x + ⎢
⎥ = 4 + 8 + 2 cos 2x + 8 cos 4 x
4 2
4⎣
2
⎦
y =
yn =
Hence,
1
3 1
+ cos 2x + cos 4 x
8 2
8
1 n
np ⎞ 1 n
np ⎞
⎛
⎛
⋅ 2 cos ⎜ 2x +
⎟⎠ + 4 cos ⎜⎝ 4 x +
⎟.
⎝
2
2
8
2 ⎠
(ii) sinx sin2 x sin3 x
Let y = sin x sin 2 x sin 3x
1
2
1
2
We know that sin Asin B = ⎡⎣cos ( A − B ) − cos ( A + B )⎤⎦ and sin A cos B = ⎡⎣sin ( A + B ) + sin ( A − B )⎤⎦
∴
y=
=
1
1
sin x [ cos x − cos 5 x ] = {sin x cos x − sin x cos 5 x}
2
2
1
1 ⎧1
1
⎫ 1
⎨ sin 2 x − [sin 6 x − sin 4 x ]⎬ = sin 2 x − (sin 6 x − sin 4 x )
4
2 ⎩2
2
⎭ 4
Hence, yn =
⇒
1 n
np ⎞ 1 n
np ⎞ 1 n
np ⎞
⎛
⎛
⎛
2 sin ⎜ 2 x +
− 6 sin ⎜ 6 x +
+ 4 sin ⎜ 4 x +
⎟
⎟
⎟
⎝
⎝
⎝
4
2 ⎠ 4
2 ⎠ 4
2 ⎠
[Refer formula 5]
np ⎞
np ⎞
np ⎞
⎛
⎛
⎛
yn = 2n − 2 sin ⎜ 2 x +
− 3n ⋅ 2n − 2 sin ⎜ 6 x +
+ 4 n −1 sin ⎜ 4 x +
⎟
⎟
⎟
⎝
⎝
⎝
2 ⎠
2 ⎠
2 ⎠
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Differential Calculus ■
3.15
EXAMPLE 5
3
5
Find the n th differential coefficient of sin u cos u.
Solution.
let x = cos u + i sin u = e iu
Let
y = sin 3 u cos5 u
∴
1
1
1
=
= iu = e iu = cos u − i sin u
x cos u + i sin u e
∴
1
= 2 cos u
x
x+
and
and
x−
1
= 2i sin u .
x
By De−Moivre’s theorem,
x n = cos nu + i sin nu
xn +
1
= 2cos nu
xn
and
and
xn −
5
∴
1⎞
⎛
25 cos5 u = ⎜ x + ⎟
⎝
x⎠
∴
1⎞ ⎛
1⎞
⎛
23 i 3 sin 3 u ⋅ 25 cos5 u = ⎜ x − ⎟ ⎜ x + ⎟
⎝
x⎠ ⎝
x⎠
⇒
1
= 2i sin nu
xn
1⎞
⎛
23 i 3 sin 3 u = ⎜ x − ⎟
⎝
x⎠
and
3
1
= cos nu − i sin nu
xn
3
5
3
⎡⎛
1⎞ ⎛
1⎞ ⎤ ⎛
1⎞ ⎛
1⎞ ⎛
1⎞
−i 28 sin 3 u cos5 u = ⎢⎜ x − ⎟ ⎜ x + ⎟ ⎥ ⎜ x + ⎟ = ⎜ x 2 − 2 ⎟ ⎜ x + ⎟
⎝
⎠
⎝
⎠
⎝
⎠
⎝
⎠
⎝
x
x
x
x
x⎠
⎣
⎦
2
3
2
3
1 ⎞⎛
1⎞
⎛
= ⎜ x 6 − 3x 2 + 2 − 6 ⎟ ⎜ x 2 + 2 + 2 ⎟
⎝
x
x ⎠⎝
x ⎠
= x 8 + 2 x 6 + x 4 − 3 x 4 − 6 x 2 − 3 +3 +
6
3
1
2 1
+ 4− 4− 6− 8
2
x
x
x
x
x
1⎞
1⎞
1⎞
1⎞
⎛
⎛
⎛
⎛
= ⎜ x8 − 8 ⎟ + 2 ⎜ x6 − 6 ⎟ + 2 ⎜ x 4 − 4 ⎟ − 6 ⎜ x 2 − 2 ⎟
⎝
⎝
⎝
⎝
x ⎠
x ⎠
x ⎠
x ⎠
−i 28 sin 3 u cos5 u = 2i sin 8u + 2 ⋅ 2i sin 6u − 2 ⋅ 2i sin 4u − 6 ⋅ 2i sin 2u ⋅
⇒
∴
−27 sin 3 u cos5 u = sin 8u + 2 sin 6u − 2 sin 4u − 6 sin 2u ⋅
sin 3 u cos5 u = −
=
[dividing by 2i]
1
[sin 8u + 2 sin 6u − 2 sin 4u − 6 sin 2u]
27
1
[6 sin 2u + 2 sin 4u − 2 sin 6u − sin 8u]
27
M03_ENGINEERING_MATHEMATICS-I _CH03_Part A.indd 15
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■
3.16
∴
y=
∴
yn =
Engineering Mathematics
1
[6 sin 2u + 2 sin 4u − 2 sin 6u − sin 8u]
27
1
27
⎡
np ⎞
np ⎞
np ⎞
np ⎞ ⎤ .
⎛
⎛
⎛
n
n
n
n⎛
⎢6 ⋅ 2 sin ⎜⎝ 2u + 2 ⎟⎠ + 2 ⋅ 4 sin ⎜⎝ 4u + 2 ⎟⎠ − 2 ⋅ 6 ⎜⎝ 6u + 2 ⎟⎠ − 8 sin ⎜⎝ 8u + 2 ⎟⎠ ⎥
⎣
⎦
EXAMPLE 6
If y 5
log x
(21) n n! ⎡
1 1
1⎤
, then prove that yn 5
log e x212 2 ⋅⋅⋅ 2 ⎥ .
x
2 3
n⎦
x n11 ⎢⎣
Solution.
log e x
y=
x
Differentiating w.r.to x, we get
Given
(1)
1 1
1
1
( −1)1!
⎛ 1⎞
⋅ + log e x ⋅ ⎜ − 2 ⎟ = 2 − 2 log e x =
[log e x − 1] .
⎝
⎠
x x
x
x
x
x2
Again differentiating w.r.to x, we get
y1 =
⎧1 1
⎛ 2 ⎞⎫
y2 = ( −1) ⎨ 2 ⋅ + (log e x − 1) ⋅ ⎜ − 3 ⎟ ⎬
⎝ x ⎠⎭
⎩x x
2
⎧1
⎫
= ( −1) ⎨ 3 − 3 (log e x − 1)⎬
x
⎩x
⎭
=
( −1) 2 2 ⎧
1 ⎫ ( −1) 2 2 ! ⎧
1⎫
log
−
1
−
x
⎨ e
⎬=
⎨log e x − 1 − ⎬
3
3
2⎭
2⎭
x ⎩
x
⎩
Again differentiating w.r.to x, we get
⎡1 1 ⎛
1⎞ ⎛ 3 ⎞ ⎤
y3 = ( −1) 2 2 ! ⎢ 3 ⋅ + ⎜ log e x − 1 − ⎟ ⎜ − 4 ⎟ ⎥
⎝
x
2⎠ ⎝ x ⎠ ⎦
x
⎣
⎧1
3
= ( −1) 2 2 ! ⎨ 4 − 4
x
⎩x
1⎞ ⎫
⎛
⎜⎝ loge x − 1 − ⎟⎠ ⎬
2 ⎭
( −1)3 2 ! × 3 ⎧
1 1⎫
( −1)3 3! ⎧
1 1⎫
⎨log e x − 1 − − ⎬ =
⎨log e x − 1 − − ⎬ .
4
4
2
3
2 3⎭
x
x
⎩
⎭
⎩
Proceeding in this way or by induction, we get
=
yn =
( −1) n n ! ⎡
1 1
1⎤
log e x − 1 − − ⋅⋅⋅ − ⎥ .
n +1
⎢
2 3
n⎦
x
⎣
EXERCISE 3.2
1. Find the nth derivative of (i) x 2 sin 3 x (ii) x 2 cos 4 x (iii) sin 3 x sin 2 x.
2. Find the nth derivative of x 4 + loge (3x 2 + 5x − 2) .
M03_ENGINEERING_MATHEMATICS-I _CH03_Part A.indd 16
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Differential Calculus ■
3. Find the nth derivative of
3.17
1
x2
. 4. Find the nth derivative of 2
.
x − 6x + 8
2x + 7x + 6
2
5. Find the nth derivative of sin 3 x cos 2 x .
6. Find the nth derivative of e 3x cos 6 x + log 4 x .
ANSWERS TO EXERCISE 3.2
⎛ p
⎞
1. (i) x2 ⋅ 3n sin ⎜ n + 3x ⎟ + 2nx ⋅ 3n−1 sin
⎝ 2
⎠
⎛ p
⎞
(ii) x2 ⋅ 4n cos ⎜ n + 4 x ⎟ + 2nx4n−1 cos
⎝ 2
⎠
(iii)
p
⎡
⎤
n−2
⎢( n − 1) 2 + 3x ⎥ + n(n − 1) 3 sin
⎣
⎦
p
⎡
⎤
n−2
⎢( n − 1) 2 + 4x ⎥ + n (n − 1)⋅ 4 cos
⎣
⎦
1
(−1)n . n!
2
5. −
p
⎡
⎤
⎢ ( n − 2) 2 + 4 x ⎥
⎣
⎦
1
⎛ p
⎞ 1
⎛ np
⎞
cos ⎜ n + x ⎟ − 5n cos ⎜
+ 5x ⎟
⎝ 2
⎠ 2
⎝ 2
⎠
2
⎡ 3n
1 ⎤
2. (−1)n−1 (n −1)! ⎢
+
⎥
n
3
1
2) n ⎦
(
x
−
)
(
x
+
⎣
4.
p
⎡
⎤
⎢( n − 2) 2 + 3x ⎥
⎣
⎦
⎡
⎤
2n −1
4
3. (−1)n n! ⎢9.
−
n +1
n +1 ⎥
(
2
3
)
(
2
)
x
+
x
+
⎣
⎦
⎡
⎤
1
1
+
⎢−
n +1
n +1 ⎥
( x − 4) ⎦
⎣ ( x − 2)
1 ⎡ n
⎛ p
⎞
⎛ np
⎞
⎛ np
⎞⎤
5 sin ⎜ n + 5u⎟ − 3n sin ⎜
+ 3u⎟ − 2 sin ⎜
+ u⎟ ⎥
⎢
⎝ 2
⎠
⎝ 2
⎠
⎝ 2
⎠⎦
24 ⎣
n
1
⎛ p
⎞
6. ( 45) 2 . e 3x cos ⎜ n + 6 x ⎟ + ( −1) n −1 ( n − 1)! n
⎝ 2
⎠
x
Theorem 3.1 Leibnitz’s theorem
Statement: If y = uv is the product of two differentiable functions u and v of x, then the nth derivative of y is
yn = uvn + nC1u1vn −1 + nC2 u2 vn − 2 + … + nCr ur vn − r + … + un v,
where ur and vr are the r th derivatives of u and v respectively.
Proof
Let
Given y = uv
P(n): y n = uv n + nC1u1v n −1 + nC 2u 2v n − 2 + … + nC r u r v n − r + … + u n v
(1)
We prove the theorem by induction on n.
Basic step:
p (1): y 1 = uv 1 + u1v
Which is true from product rule.
M03_ENGINEERING_MATHEMATICS-I _CH03_Part A.indd 17
∴
p(1) is true
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3.18
Engineering Mathematics
Inductive step: We assume that the theorem is true for n = k (>1).
y k = uv k + kC1u1v k −1 + … + kC k u k v is true
i.e., p(k) is true ⇒
(2)
To prove p(k + 1) is true.
That is to prove
y k +1 = uv k +1 + k +1 C1u1v k + k +1 C 2u 2v k −1 + … + k +1 C k +1u k +1v .
Differentiating (2) w.r.to x, we get
y k +1 = uv k +1 + u1v k + kC1 [u1v k + u 2v k −1 ] + kC 2 [u 2v k −1 + u3v k − 2 ] + … + kC k −1 [u k −1v 2 + u k v 1 ]
+ kC k [u k v 1 + u k +1v ]
⇒
y k +1 = uv k +1 + ( kC1 + 1) u1v k + ( kC 2 + kC1 ) u 2v k −1 + … + ( kC k + kC k −1 ) u k v 1 + u k +1v
⇒
y k +1 = uv k +1 + k +1 C1u1v k + k +1 C 2u 2v k −1 + … + k +1 C k u k v 1 + u k +1v
∴
P(k + 1) is true.
Thus,
P(k) is true
⇒
[{ n+1Cr = nCr + nCr −1 ]
p(k + 1) is true.
∴ By induction p(n) is true for all values of n ∈ N.
Hence, the theorem is true for all values of n ∈ N.
y n = uv n + nC1u1v n −1 + nC 2u 2v n − 2 + … + nC r u r v n − r + … + u n v .
∴
WORKED EXAMPLES
EXAMPLE 1
If y 5 cos( m sin21 x ), then prove that (1 2 x 2 ) y 2 2 xy 1 1 m 2 y 5 0. Hence prove that
(1 2 x 2 ) y n 1 2 2(2 n11) xy n 111 ( m 2 2n 2 ) y n 5 0 .
Solution.
y = cos( m sin −1 x )
Given
⇒
cos −1 y = m sin −1 x
(1)
Differentiating (1) w.r.to x, we get
−1
1− y
Squaring,
2
⋅
dy
1
= m⋅
dx
1 − x2
y12
m2
=
2
1 − y 1 − x2
⇒
⇒
−1
1− y
2
⋅ y1 =
m
1 − x2
(1 − x 2 ) y12 = m 2 (1 − y 2 ) .
Differentiating w.r.to x, we get
(1 − x 2 ) 2 y1 y2 − 2 xy12 = −2m 2 yy1
M03_ENGINEERING_MATHEMATICS-I _CH03_Part A.indd 18
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Differential Calculus ■
3.19
Dividing by 2 y1 on both sides,
(1 − x 2 ) y2 − xy1 = − m 2 y
⇒
(1 − x 2 ) y2 − xy1 + m 2 y = 0
(2)
Differentiating (2) w.r.to x, n times by Leibnitz’s theorem, we get,
(1 − x 2 ) yn + 2 + nC1 ( −2 x ) yn +1 + nC2 ( −2) yn − [ xyn +1 + nC1 ⋅1⋅ yn ] + m 2 yn = 0
⇒
n( n − 1)
yn − xyn +1 − nyn + m 2 yn = 0
1⋅ 2
(1 − x 2 ) y n + 2 − ( 2n + 1) xy n +1 + [m 2 − n ( n − 1) − n ] y n = 0
⇒
(1 − x 2 ) y n + 2 − ( 2n + 1) xy n +1 + ( m 2 − n 2 + n − n ) y n = 0
⇒
(1 − x 2 ) y n + 2 − ( 2n + 1) xy n +1 + ( m 2 − n 2 ) y n = 0.
⇒
(1 − x 2 ) yn+ 2 − 2nx yn+1 − 2 ⋅
EXAMPLE 2
2
2
If y 5 a cos(log x ) 1 b sin (log x ) , then show that x y n 1 2 1(2 n 1 1) xy n 111 ( n 1 1) y n 5 0 .
Solution.
Given
y = a cos(log x ) + b sin(log x )
(1)
Differentiating (1) w.r.to x, we get
1
1 1
y1 = a( − sin(log x )) ⋅ + b cos(log x ) ⋅ = [ − a sin(log x ) + b cos(log x )]
x
x x
⇒
xy 1 = −a sin(log x ) + b cos(log x ).
Again differentiating w.r.to x, we get
1
1
xy2 + y1 = − a cos(log x ) ⋅ + b( − sin(log x ))
x
x
1
y
= − [a cos(log x ) + b cos(log x ) = −
x
x
⇒
x 2 y2 + xy1 = − y
⇒
x 2 y2 + xy1 + y = 0
(2)
Differentiating (2) w.r.to x, n times by Leibnitz’s theorem, we get,
x 2 y n + 2 + nC1 2xy n +1 + nC 2 2 y n + xy n +1 + nC1 ⋅1⋅ y n + y n = 0
⇒
x 2 yn+ 2 + 2nxyn+1 + 2 ⋅
n( n − 1)
yn + xyn+1 + nyn + yn = 0
1⋅ 2
⇒
x 2 yn+ 2 + ( 2n + 1) xyn+1 + [n( n − 1) + n + 1] yn = 0
⇒
x 2 yn+ 2 + ( 2n + 1) xyn+1 + ( n2 − n + n + 1) yn = 0
⇒
M03_ENGINEERING_MATHEMATICS-I _CH03_Part A.indd 19
x 2 yn+ 2 + ( 2n + 1) xyn+1 + ( n2 + 1) yn = 0.
5/30/2016 7:08:04 PM
■
3.20
Engineering Mathematics
EXAMPLE 3
(
If y 5 x 1 1 1x 2
) , then prove that (11x ) y
y = (x + 1 + x )
m
2
Solution.
Given
Differentiating (1) w.r.to x, we get
2
(
)
(
)
(
)
= m x + 1+ x 2
= m x + 1+ x 2
m2 y 2
1+ x2
Again differentiating w.r.to x,
y 12 =
m −1
⎛
⎞
1
⋅ 2x ⎟
⎜1 +
2
⎝ 2 1+ x
⎠
m −1
⎛
x ⎞
⎜1 +
⎟
⎝
1+ x 2 ⎠
m −1
(1)
(
m x + 1+ x 2
⎛ x + 1+ x 2 ⎞
⎜
⎟ =
1+ x 2
1+ x 2 ⎠
⎝
)
m
my
=
1+ x 2
(1 + x 2 ) y 12 = m 2 y 2
⇒
(1 + x 2 )2 y1 y2 + 2 xy12 = 2m 2 yy1
⇒
1 (2 n 11) xy n 111( n 2 2m 2 ) y n 50 .
m
y1 = m x + 1+ x 2
Squaring,
n12
⇒
(1 + x 2 ) y2 + xy1 = m 2 y
[dividing by 2 y1]
(1 + x 2 ) y2 + xy1 − m 2 y = 0 .
(2)
Differentiating (2) w.r.to x, n times by Leibnitz’s theorem, we get,
(1 + x 2 ) yn+ 2 + nC1 2 xyn+1 + nC2 2 yn + xyn+1 + nC1 ⋅1⋅ yn − m 2 yn = 0
⇒
(1 + x 2 ) y n + 2 + 2nx y n +1 + 2 ⋅
n ( n − 1)
y n + xy n +1 + ny n − m 2 y n = 0
1⋅ 2
⇒
(1 + x 2 ) yn+ 2 + ( 2n + 1) xyn+1 + [n( n − 1) + n − m 2 ] yn = 0
⇒
(1 + x 2 ) yn+ 2 + ( 2n + 1) xyn+1 + ( n2 − n + n − m 2 ) yn = 0
⇒
(1 + x 2 ) yn+ 2 + ( 2n + 1) xyn+1 + ( n2 − m 2 ) yn = 0 .
Problems to find y n (0)
EXAMPLE 4
If y 5 (sin21 x ) 2 , then prove that (1 2x 2 ) y 2 2xy 1 22 5 0 .
(1 2x 2 ) y n 1 2 2(2 n 1 1) xy n 11 2n 2 y n 50 and find y n (0).
Solution.
Given
y = (sin −1 x ) 2
M03_ENGINEERING_MATHEMATICS-I _CH03_Part A.indd 20
Hence
show
that
(1)
5/30/2016 7:08:08 PM
Differential Calculus ■
3.21
Differentiating (1) w.r.to x, we get
y1 =
Squaring,
y 12 =
2 sin −1 x
(2)
1− x 2
4(sin −1 x ) 2
⇒
( 1 − x 2 )2
(1 − x 2 ) y12 = 4 y
(3)
Differentiating (3) w.r.to x, we get
(1 − x 2 )2 y1 y2 − 2 xy12 = 4 y1
⇒
(1 − x 2 ) y2 − xy1 = 2.
[dividing by 2 y1]
(4)
Differentiating (4) w.r.to x, n times using Leibnitz’s theorem, we get
(1 − x 2 ) yn+ 2 + nC1 ( −2 x ) yn+1 + nC2 ( −2) yn − [ xyn +1 + nC1 ⋅1⋅ yn ] = 0
⇒
(1 − x 2 ) yn + 2 − 2nxyn +1 − 2 ⋅
n( n − 1)
yn − xyn +1 − nyn = 0
1⋅ 2
⇒
(1 − x 2 ) y n + 2 − ( 2n + 1)xy n +1 − [n( n − 1) + n ]y n = 0
⇒
(1 − x 2 ) y n + 2 − ( 2n + 1)xy n +1 − [n 2 − n + n ]y n = 0
⇒
(1 − x 2 ) y n + 2 − ( 2n + 1)xy n +1 − n 2 y n = 0
(5)
To find y n (0)
Put x = 0 in (5), we get
yn + 2 ( 0 ) − n 2 yn ( 0 ) = 0 .
Put x = 0 in (2), we get
y 1 ( 0) =
Put x = 0 in (4), we get
We have
2 sin −1 0
1− 0
= 0.
y2 ( 0 ) − 2 = 0
⇒
yn + 2 ( 0 ) − n 2 yn ( 0 ) = 0
⇒
y2 ( 0 ) = 2 .
yn + 2 ( 0 ) = n 2 yn ( 0 ) .
If n is odd, n = 1, 3, 5, 7, …, then we have
y 3 ( 0) = y 1 ( 0) = 0 ,
y5 (0) = 32 y3 (0) = 0,
y7 (0) = 52 y5 (0) = 0 and so on.
∴ If n is odd, yn (0) = 0.
If n is even, n = 2, 4, 6, 8, …, then we have
y2 ( 0 ) = 2
y 4 (0) = 22 y 2 (0) = 22 ⋅ 2,
y6 (0) = 4 2 y4 (0) = 2 ⋅ 22 ⋅ 4 2 , y8 (0) = 6 2 y6 (0) = 2 ⋅ 22 ⋅ 4 2 ⋅ 6 2 and so on.
∴ If n is even,
yn (0) = 2 ⋅ 22 ⋅ 4 2 ⋅ 6 2 ⋅⋅⋅ ( n − 2) 2
∴
⎧2 ⋅ 22 ⋅ 4 2 ⋅ 6 2 ⋅⋅⋅ ( n − 2) 2 ,
y n ( 0) = ⎨
⎩0,
M03_ENGINEERING_MATHEMATICS-I _CH03_Part A.indd 21
if n is even
if n is odd.
5/19/2016 2:15:20 PM
3.22
■
Engineering Mathematics
EXAMPLE 5
21
Find the value of the n th derivate of e a sin
x
for x 5 0.
Solution.
−1
Let
y = e a sin x
Differentiating w.r. to x, we get
y1 = e a sin −1 x
(1)
a
1− x
2
=
ay
1− x 2
⇒
(1 − x2)y 12 = a2 y2
Again differentiating w.r. to x, we get
(1 − x2)2y1y2 − 2xy 12 = a2 . 2yy1
⇒
(1 − x2)y2 − xy1 = a2y
⇒
(1 − x2) y2 − xy1 − a2y = 0
Differentiating (3) w.r. to x, n times using Leibnitz’s theorem, we get
(2)
(3)
(1− x2) yn + 2 + n C1(−2x)yn + 1 + n C 2(−2) yn − [xyn+1 + n C1. 1. yn] − a2 yn = 0
⇒
n( n − 1)
yn − xyn+1 − nyn − a2 yn = 0
1.2
(1 − x2) yn + 2 − (2n + 1) x yn+1 − [n (n − 1) + n + a2] yn = 0
(1 − x2) yn + 2 − (2n + 1) xyn+1 − (n2 + a2) yn = 0
(1 − x2) yn + 2 − 2nx yn + 1 − 2
⇒
⇒
Put x = 0 in (4), then
(4)
yn + 2 ( 0 ) − ( n 2 + a 2 ) yn ( 0 ) = 0
⇒
yn + 2 ( 0 ) = ( n 2 + a 2 ) yn ( 0 )
Put
x = 0 in (1), then
y(0) = 1
y1 (0) = a
y2 ( 0 ) = a 2 y ( 0 )
Putting n = 1, 3, 5, 7, … in (5) we get
Put
Put
(5)
x = 0 in (2), then
x = 0 in (3), then
⇒
y2 ( 0 ) = a 2 ⋅ 1 = a 2
y3 (0) = (12 + a 2 ) ⋅ y1 (0) = (12 + a 2 ) ⋅ a = a(12 + a 2 )
y5 (0) = (32 + a 2 ) ⋅ y3 (0) = a(12 + a 2 )(32 + a 2 )
y7 (0) = (52 + a 2 ) ⋅ y5 (0) = a(12 + a 2 )(32 + a 2 )(52 + a 2 )
and so on.
If n is odd,
yn (0) = a(12 + a 2 )(32 + a 2 ) ⋅⋅⋅ [( n − 2) 2 + a 2 ]
Putting n = 2, 4, 6, 8, … in (5), we get
y4 ( 0 ) = ( 2 2 + a 2 ) y2 ( 0 ) = a 2 ( 2 2 + a 2 )
y6 (0) = ( 4 2 + a 2 ) y4 (0) = a 2 ( 22 + a 2 )( 4 2 + a 2 )
y8 (0) = (6 2 + a 2 ) y6 (0) = a 2 ( 22 + a 2 )( 4 2 + a 2 )(6 2 + a 2 )
and so on.
M03_ENGINEERING_MATHEMATICS-I _CH03_Part A.indd 22
5/19/2016 2:15:25 PM
Differential Calculus ■
3.23
If n is even, yn (0) = a 2 ( 22 + a 2 )( 4 2 + a 2 )(6 2 + a 2 ) ⋅⋅⋅ [( n − 2) 2 + a 2 ].
2
2
2
2
2
2
2
⎪⎧a ( 2 + a )( 4 + a ) ⋅⋅⋅ [( n − 2) + a ],
yn ( 0 ) = ⎨ 2
2
2
2
2
2
⎩⎪a(1 + a )(3 + a ) ⋅⋅⋅ [(n − 2) + a ],
if n is even
if n is odd
EXAMPLE 6
Considering x 2 n 5 x n ⋅ x n and using Leibnitz’s theorem prove that
11
n 2 n 2 ( n 21) 2 n 2 ( n 21) 2 ( n 2 2) 2
(2 n )!
.
1
1
1 ⋅⋅⋅ 5
12
12 ⋅ 2 2
12 ⋅ 2 2 ⋅ 3 2
( n!) 2
Solution.
y = x 2n
Let
Differentiating w.r.to x, n times we get
yn = 2n( 2n − 1)( 2n − 2) ⋅⋅⋅ ( 2n − n + 1) x 2 n− n
= 2n( 2n − 1)( 2n − 2) ⋅⋅⋅ ( n + 1) x n
2n( 2n − 1)( 2n − 2) ⋅⋅⋅ ( n + 1)n ⋅⋅⋅ 2 ⋅1 n
x
1⋅ 2 ⋅ 3 ⋅⋅⋅ n
( 2n)! n
yn =
x
( n!)
=
⇒
Also
y = x 2n = x n ⋅ x n .
Let u = x n
(1)
and v = x n
∴
y = uv
By Leibnitz’s theorem,
yn = u vn + nc1u1vn−1 + nc2 u2 vn− 2 + ⋅⋅⋅ + ncn−1un−1v1 + un v
u1 = nx n−1 ,
u2 = n( n − 1) x n− 2
u3 = n ( n − 1)( n − 2)x n − 3 , ⋅⋅⋅
u n − 2 = n ( n − 1)( n − 2) ⋅⋅⋅ [n − ( n − 3)]x n − ( n − 2 )
= n ( n − 1)( n − 2) ⋅⋅⋅ 3 ⋅ x 2 = n ( n − 1)( n − 2) ⋅⋅⋅ 4 ⋅ 3 ⋅ 2 ⋅1 x 2 = n ! x 2
1⋅ 2
1⋅ 2
n − ( n −1)
=
n
(
n
−
1
)(
n
−
2
)
⋅⋅⋅
2x = n ! x
u n −1 = n ( n − 1)( n − 2) ⋅⋅⋅ [n − ( n − 2)] ⋅ x
∴
un = n !
Similarly, if v = x n , then vn = n!
∴
yn = x n n! + nC1nx n−1 ⋅ n! x + nC2 ⋅ n( n − 1) x n− 2 ⋅
+ nC 3 n ( n − 1)( n − 2)x n − 3
M03_ENGINEERING_MATHEMATICS-I _CH03_Part A.indd 23
n! 2
x
1⋅ 2
n! 3
x + ⋅⋅⋅ + n ! x n
1⋅ 2 ⋅ 3
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= x n n! + n ⋅ n ⋅ n! x n +
+
n( n − 1)
n! n
n( n − 1)
x
1⋅ 2
1⋅ 2
n( n − 1)( n − 2)
n! n
n( n − 1)( n − 2)
x + ⋅⋅⋅ + n! x n
1⋅ 2 ⋅ 3
1⋅ 2 ⋅ 3
⎤
⎡ n2 n2 ( n − 1) 2 n2 ( n − 1) 2 ( n − 2) 2
= n! x n ⎢1 + 2 + 2 2 +
+ ⋅⋅⋅ + 1⎥
2
2
2
1
1
⋅
2
1
⋅
2
⋅
3
⎣
⎦
(2)
From (1) and (2), we get
⎤ ( 2n)! n
⎡ n2 n2 ( n − 1) 2 n2 ( n − 1) 2 ( n − 2) 2
n! x n ⎢1 + 2 + 2 2 +
+ ⋅⋅⋅ + 1⎥ =
x
2
2
2
1 ⋅2
1 ⋅ 2 ⋅3
n!
⎦
⎣ 1
⇒
1+
( 2n)!.
n2 n2 ( n − 1) 2 n2 ( n − 1) 2 ( n − 2) 2
+ 2 2 +
+ ⋅⋅⋅ + 1 =
2
2
2
2
1
1 ⋅2
1 ⋅ 2 ⋅3
( n!) 2
EXERCISE 3.3
1. If y = ( x + 1 + x 2 ) , then prove that (1 + x 2 ) y n + 2 + ( 2n + 1)xy n +1 + ( n 2 − m 2 ) y n = 0 .
m
2. If y = (1 − x ) − a e − ax , then prove that (1− x )
dy
= axy and (1 − x ) yn+1 − ( n + ax ) yn − nayn−1 = 0 .
dx
3. If y = tan −1 x , then prove that (1 + x 2 ) yn+ 2 + ( 2n + 1) xyn+1 + n( n + 1) yn = 0 .
4. If y = (sin −1 x ) 2 , then prove that (1 − x 2 ) yn+ 2 − ( 2n + 1) xyn+1 − n2 yn = 0 .
5. If y = (cos −1 x ) 2 , then prove that (1 − x 2 ) yn+ 2 − ( 2n + 1) xyn+1 − n2 yn = 0 .
6. If y = sin −1 x then prove that (i) (1 − x 2 ) y 2 − xy 1 = 0 (ii) (1 − x 2 ) y n + 2 − ( 2n + 1)xy n +1 − n 2 y n = 0
n
⎛y⎞
⎛x⎞
7. If cos −1 ⎜ ⎟ = log ⎜ ⎟ , then prove that x 2 y n + 2 + ( 2n + 1)xy n +1 + 2n 2 y n = 0.
⎝b⎠
⎝ n⎠
(
)
2
8. If y = ⎡⎢log x + 1 + x 2 ⎤⎥ , then prove that (1 + x 2 ) y 2 + xy 1 = 2 and hence show that
⎣
⎦
(1 + x 2 ) yn + 2 + ( 2n + 1) xyn +1 + n2 yn = 0 .
9. If y 1 m + y −1 m = 2x , then prove that ( x 2 − 1) y 2 + xy 1 − m 2 y = 0 and hence show that
( x 2 − 1) y n + 2 + ( 2n + 1)xy n +1 + ( n 2 − m 2 ) y n = 0 .
10. If x = sin t , y = sin pt , then prove that (i) (1 − x 2 ) y 2 − xy 1 + p 2 y = 0
(ii) (1 − x 2 ) y n + 2 − ( 2n + 1)xy n +1 − ( n 2 − p 2 ) y n = 0
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Differential Calculus ■
3.2
3.25
APPLICATIONS OF DERIVATIVE
3.2.1 Geometrical Interpretation of Derivative
Let f be a differentiable function on [a, b].
The graph of f is the set {( x , y ) y = f ( x ), x ∈[a, b ]}.
That is y = f ( x ) is the equation of the graph of f,
Let c, c + h ∈[a, b] .
y
So, that P (c, f (c)) and Q(c + h, f (c + h)) be the
corresponding points of the curve y = f ( x ).
Then the slope of the chord
PQ =
Q
P
f (c + h) − f (c) f (c + h) − f (c)
=
c+h−c
h
θ
T
o
Suppose the point Q moves along the curve towards
P, then the chord PQ approaches to a definite line PT
in the limit as Q → P .
This line PT is called the tangent line to the curve at P.
∴
lim(slope of chord PQ ) = lim
Q→ p
h→ 0
C
C+n
x
Fig. 3.1
f (c + h) − f (c)
h
= f ′(c) , if the limit exists.
So, when f ′(c) exists, it is the slope of the tangent PT at P.
∴ the equation of the tangent at P is.
y − f (c) = f ′(c)( x − c).
3.2.2 Equation of the Tangent and the Normal to the Curve y = f(x)
1. The equation of the tangent at ( x 1 , y 1 )
The given curve is y = f ( x ).
Let P ( x1 , y1 ) be any point on the curve.
Let m be the slope of the tangent at ( x1 , y1 ).
∴ m=
dy
dx
( x1 , y 1 )
∴ the equation of the tangent at P ( x1 , y1 ) is y − y1 = m( x − x1 ) .
2. The equation of the normal at ( x 1 , y 1 )
The normal at P ( x1 , y1 ) is a straight line through P and perpendicular to the tangent at P .
∴ If m1 is the slope of the normal at P , then m ⋅ m1 = −1 ⇒ m1 = −
That is the slope of the normal = −
M03_ENGINEERING_MATHEMATICS-I _CH03_Part A.indd 25
1
.
m
1
.
the slope of the tangent
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∴ the equation of the normal at P ( x1 , y1 ) is
y − y1 = −
1
( x − x1 ) .
m
Note The derivative f ′(c ) of f at c is defined as a real number. However, for geometrical convenience,
we extend the definition to include ±∞ .
We define f ′(c ) = ∞, if f ′(c − ) = ∞ and f ′(c + ) = ∞ and f ′(c) = − ∞, if f ′(c − ) = −∞ and
f ′(c + ) = −∞ .
Thus, if dy = 0 , the tangent is parallel to the x−axis.
dx
If
⎛
⎞
dy
dx
= ∞ ⎜ or
= 0⎟ , the tangent is parallel to the y−axis.
dx
⎝
dy
⎠
WORKED EXAMPLES
EXAMPLE 1
Find the equation of the tangent and the normal to the curve x 2 2y 2 5144 at the point P(13, 5) .
If the normal at the point P meets the x−axis at G, then find the coordinates of the mid point of
PG.
Solution.
The given curve is
x 2 − y 2 = 144
Differentiating w.r.to x, we get
2x − 2 y
dy
=0
dx
(1)
⇒
y
dy
=x
dx
⇒
dy x .
=
dx y
dy 13 .
=
dx 5
At the point (13, 5):
13
.
5
∴ the equation of the tangent at the point P(13, 5) is
∴ the slope of the tangent at the point P(13, 5) is m =
y −5=
13
( x − 13)
5
⇒
⇒
5( y − 5) = 13( x − 13)
5 y − 25 = 13 x − 169
The slope of the normal at the point P(13, 5) is m1 = −
⇒ 13x − 5 y − 144 = 0
(2)
1
5
=− .
m
13
∴ the equation of the normal at the point P(13, 5) is
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Differential Calculus ■
y −5= −
⇒
5
( x − 13)
13
13 y − 65 = −5 x + 65
3.27
⇒ 13( y − 5) = −5( x − 13)
⇒
5 x + 13 y − 130 = 0 .
(3)
y
The normal at P(13, 5) meets the x−axis at G as in Fig 3.2.
Let M be the midpoint of PG.
To find the coordinates of M:
Since the normal at P meets the x-axis at G, y = 0.
∴ putting y = 0 in (3), we get
5 x − 130 = 0 ⇒ x = 130 = 26 .
5
∴ G is ( 26, 0) and P is (13, 5).
(13, 5)P
M
G(28, 0)
x
Fig. 3.2
∴ the coordinates of the midpoint M is ⎡13 + 26 , 5 + 0 ⎤ = ⎡ 39 , 5 ⎤ .
⎢ 2
2 ⎥⎦ ⎢⎣ 2 2 ⎥⎦
⎣
EXAMPLE 2
Find the points on the curve y 5 x 4 26 x 3 113 x 2 210 x 15 where the tangents are parallel to the
line y 5 2 x and prove that two of the points have the same tangent.
Solution.
The given curve is
y = x 4 − 6 x 3 + 13 x 2 − 10 x + 5
Let the tangent at the point ( x1 , y1 ) on the curve be parallel to the line
(1)
y = 2x.
∴ the slope of the tangent at the point ( x1 , y1 ) = The slope of the line y = 2 x .
Since the slope of the line = 2, the slope of the tangent is 2.
Differentiating (1) w.r.to x, we have
dy
= 4 x 3 − 18 x 2 + 26 x − 10
dx
∴ the slope of the tangent at the point ( x1 , y1 ) is m = 4 x 13 − 18x 12 + 26 x 1 − 10
∴
4 x 13 − 18x 12 + 26 x 1 − 10 = 2
⇒
4 x 13 − 18x 12 + 26 x 1 − 12 = 0
⇒
2x 13 − 9x 12 + 13x 1 − 6 = 0 .
By inspection x1 = 1 is a root. The other roots are given by
2x 12 − 7x 1 + 6 = 0
1 2 − 9 13 − 6
⇒
( 2 x1 − 3)( x1 − 2) = 0
0
2 −7
6
⇒
2 x1 − 3 = 0, x1 − 2 = 0
2 −7
6
0
3
∴
x1 = , x1 = 2.
2
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∴ the roots are x1 = 1, 2,
3
2
When x1 = 1, y1 = 1 − 6 + 13 − 10 + 5 = 3
When x1 = 2, y1 = 24 − 6 × 23 + 13 × 22 − 10 × 2 + 5 = 16 − 48 + 52 − 20 + 5 = 5.
4
When x 1 =
3
2
3
3
1
15
⎛ 3⎞
⎛ 3⎞
⎛ 3⎞
, y 1 = ⎜ ⎟ − 6 × ⎜ ⎟ + 13 × ⎜ ⎟ − 10 × + 5 = [81 − 324 + 468 − 240 + 80] = .
⎝ 2⎠
⎝ 2⎠
⎝ 2⎠
2
2
16
16
⎛ 3 15 ⎞
∴ the points are (1, 3),( 2, 5), ⎜ , ⎟ .
⎝ 2 16 ⎠
At these points the tangents to the curve are parallel to the line y = 2 x .
Now we have to prove that at two of these points, the tangents are equal.
At the point (1, 3)
The equation of the tangent is
y − 3 = 2( x − 1) ⇒
At the point ( 2 , 5 )
The equation of the tangent is
y − 5 = 2( x − 2) ⇒
y − 3 = 2x − 2
⇒
2x − y + 1 = 0
(2)
y − 5 = 2x − 4
⇒
2x − y + 1 = 0
(3)
The equations (2) and (3) are the same.
∴ the tangents at the points (1, 3) , ( 2, 5) are the same.
EXAMPLE 3
Find the equations of the tangents from the origin to the curve y 5 4 x 3 22 x 5 .
Solution.
The given curve is y = 4 x 3 − 2 x 5 .
Let a tangent from the origin to the curve touch it at the point P( x 1 , y 1 ) .
∴ OP is a tangent to the curve and y1 = 4 x 13 − 2 x 51
The slope of OP = y1 ,
x1
But the slope of OP =
dy
dx
(2)
[since ( x1 , y1 ) ≠ (0, 0)]
.
( x1 , y 1 )
Differentiating (1) w.r.to x, we get
12x 12 − 10 x 14 =
dy
= 12 x 2 − 10 x 4 .
dx
dy
= 12x 12 − 10 x 14 .
dx
At the point P ( x 1 , y 1 ) :
∴
(1)
y1
x1
M03_ENGINEERING_MATHEMATICS-I _CH03_Part A.indd 28
⇒
12x 13 − 10 x 15 = y 1
(3)
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Differential Calculus ■
3.29
Solving (2) and (3), we find ( x 1 , y 1 ) .
From (2) and (3), we get
12x 13 − 10 x 15 = 4 x 13 − 2x 15
When x1 = 1,
⇒
8x 15 = 8x 13 ⇒ x 12 = 1 ⇒ x1 = ±1. ({ x 1 ≠ 0, dividing by 8x 13 )
y1 = 4 − 2 = 2 .
When x1 = −1, y1 = 4( −1) − 2( −1) = −4 + 2 = −2 .
∴ the points of contact of the tangents from the origin to the curve are (1, 2) and ( −1, − 2).
At the point (1, 2 ) :
The slope of the tangent is m =
y1 2
= = 2.
x1 1
y
P
∴ the equation of the tangent is
y − 2 = 2( x − 1) ⇒ y = 2 x.
(1, 2)
− 2
At the point ( −1, −2) :
y1 −2
=
= 2.
x1 −1
∴ the equation of the tangent is
The slope of the tangent is m =
o
2 x
Q
(−1, −2)
Fig. 3.3
y − ( −2) = 2( x − ( −1))
⇒
y + 2 = 2( x + 1) ⇒ y = 2 x.
So, from the origin, same tangent is drawn to the curve. See Fig. 3.3.
EXAMPLE 4
Find the equations of the tangent at any point P on y 2 5 x 3 . If the tangent at P meets the curve
again at Q and the line OP and OQ (O is the origin) make angles a,b with the x 2 axis , then
prove that tan a 522 tan b.
Solution.
The given curve is
y 2 = x3
2 3
Let P (t , t ) be any point on the curve.
Differentiating (1) w.r.to x, we get
2y
dy
= 3x 2
dx
⇒
(1)
dy 3 x 2 .
=
dx 2 y
2 2
At the point P (t 2 , t 3 ): dy = 3(t ) = 3 t .
dx
2t 3
2
3
∴ the slope of the tangent at P is m = t .
2
M03_ENGINEERING_MATHEMATICS-I _CH03_Part A.indd 29
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∴ the equation of the tangent at P is
3
y − t 3 = t( x − t 2 )
2
⇒
⇒
y
2( y − t 3 ) = 3t ( x − t 2 )
P(t 2, t 3)
2 y − 2t 3 = 3tx − 3t 3
⇒
2 y = 3tx − t 3
(2)
o
x
Q
It meets the curve (1) in Q.
To find Q : Solve (1) and (2).
Squaring (2), we get 4 y 2 = (3tx − t 3 ) 2
Fig. 3.4
∴
4 ⋅ x 3 = 9t 2 x 2 − 6t 4 x + t 6 ⇒ 4 x 3 − 9t 2 x 2 + 6t 4 x − t 6 = 0
We see x = t 2 is a root of (3).
Since the line (2) is a tangent at P to the curve, two points of intersection coincide at P .
So, x = t 2 is a repeated root of (3),
We remove these factors by synthetic division.
t 2 4 − 9t 2 6t 4 − t 6
∴ the third root is given by
0 4t 2 − 5t 4 t 6
2
t
4x − t 2 = 0 ⇒ x =
t 2 4 − 5t 2
0
t4
4
2
4
0 4t − t
Substituting in (2), we get
2
3
t
t
2 y = 3⋅ t ⋅ − t3 = −
4
4
⇒
t3
y =− .
8
4 −t2
(3)
0
2
3
∴ Q is ⎛ t , − t ⎞ .
⎜⎝ 4
8 ⎟⎠
Given that OP and OQ make angles a and b with the x−axis.
∴ the slope of OP = tan a
and the slope of OQ = tanb
∴
⇒
t3 − 0
= tan a ⇒ t = tan a
t2 − 0
t3
−0
t
= tanb ⇒ − = tanb ⇒ t = −2 tanb
⇒ 28
t
2
−0
4
−
tan a = −2 tan b .
EXAMPLE 5
If the tangent at ( x 1 , y 1 ) to the curve x 3 1 y 3 5 a 3 meets the curve in ( h, k ) , then show that
h k
1 11 5 0 .
x1 y 1
M03_ENGINEERING_MATHEMATICS-I _CH03_Part A.indd 30
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Differential Calculus ■
3.31
Solution.
x 3 + y 3 = a3
(1)
Also given ( x 1 , y 1 ) is a point on (1). ∴ x 13 + y 13 = a3
(2)
Given curve is
Differentiating (1) w.r.to x, we get
3x 2 + 3 y 2
At the point P ( x1 , y1 ):
dy
=0
dx
⇒
y2
dy
= − x2
dx
dy − x 2 .
= 2
dx
y
⇒
x2
dy
= − 12 .
dx
y1
∴ the slope of the tangent at the point ( x 1 , y 1 ) is m = −
x 12
.
y 12
∴ the equation of the tangent at the point ( x1 , y1 ) is
y − y1 = −
x12
( x − x1 )
y12
⇒
⇒
y12 y − y13 = − x12 x + x13
x12 x + y12 y = x13 + y13
This tangent meets the curve in ( h , k ).
( h , k ) is a point on the curve
(3) − ( 4) ⇒
⇒
( 2) − (3) ⇒
⇒
∴ (5) ⇒
(6 )
⇒
∴
∴
⇒ x12 x + y12 y = a3
[using (2)]
x12 h + y12 k = a3
h3 + k 3 = a 3
(3)
(4)
x 12 h + y 12 k − h 3 − k 3 = 0
h ( x 12 − h 2 ) + k ( y 12 − k 2 ) = 0
⇒
h ( x 12 − h 2 ) = − k ( y 12 − k 2 )
(5)
⇒
x 12 ( x 1 − h ) = − y 12 ( y 1 − k )
(6)
x 13 + y 13 − x 12 h 2 − y 12 k = 0
x 12 ( x 1 − h ) + y 12 ( y 1 − k ) = 0
h ( x 12 − h 2 ) − k ( y 12 − k 2 )
=
x 12 ( x 1 − h ) − y 12 ( y 1 − k )
h( x 1 + h ) k ( y 1 + k )
=
x 12
y 12
⇒
h k h2 k 2
− +
−
=0
x 1 y 1 x 12 y 12
⇒
⎛h k⎞ ⎡
h k⎤
⎜⎝ x − y ⎟⎠ ⋅ ⎢1 + x + y ⎥ = 0
1
1
1
1⎦
⎣
M03_ENGINEERING_MATHEMATICS-I _CH03_Part A.indd 31
⇒
h h2
k k2
+ 2 =
+
x 1 x 1 y 1 y 12
⇒
h k ⎛ h k ⎞⎛ h k ⎞
− +
−
+
=0
x1 y1 ⎜⎝ x1 y1 ⎟⎠ ⎜⎝ x1 y1 ⎟⎠
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Engineering Mathematics
⇒
1+
⎤
h k
h k =0 ⎡
+
−
≠ 0 as ( h , k ) and ( x 1 , y 1 ) are different points ⎥
⎢{
x1 y1
x
y
1
1
⎣
⎦
h k
+ +1 = 0
x1 y1
∴
EXAMPLE 6
t2
t3
, y5
are concurrent, Prove
If the tangent at three points t1, t2, t3 on the curve x 5
3
11 t
11 t 3
that t 1 1 t 2 1 t 3 5 0 .
Solution.
The equation of the curve is given in parametric form
t2
t3
(1)
and
y
=
1+ t 3
1+ t 3
First we find the equation of the tangent at any point ‘t’ on the curve.
Differentiating (1) and (2) w.r.to t, we get
x=
(2)
dx (1 + t 3 )2t − t 2 ⋅ 3t 2 2t + 2t 4 − 3t 4
2t − t 4
=
=
=
.
dt
(1 + t 3 ) 2
(1 + t 3 ) 2
(1 + t 3 ) 2
dy (1 + t 3 )3t 2 − t 3 ⋅ 3t 2 3t 2 + 3t 5 − 3t 5
3t 2
=
=
=
3 2
3 2
dt
(1 + t )
(1 + t )
(1 + t 3 ) 2
3t 2
dy dy /dt (1 + t 3 ) 2
3t 2
3t
=
=
=
=
.
∴
dx dx /dt
2t − t 4 2 − t 3
2t − t 4
(1 + t 3 ) 2
3t
∴ slope of the tangent at ‘t’ is m =
2 − t 3.
∴ the equation of the tangent at ‘t’ is
y−
⇒
⇒
t3
3t
=
3
1+ t
2 −t3
⎛
t2 ⎞
x
−
⎜⎝
1 + t 3 ⎟⎠
(1 + t 3 ) y − t 3
3t ⎡ (1 + t 3 )x − t 2 ⎤
=
⎢
⎥
3
1+ t
2 − t 3 ⎣ 1+ t 3
⎦
(1 + t 3 ) y − t 3 =
3t
[(1 + t 3 )x − t 2 ]
2 −t3
⇒ (1 + t 3 )( 2 − t 3 ) y − t 3 ( 2 − t 3 ) = 3t (1 + t 3 )x − 3t 3
⇒
(1 + t 3 )( 2 − t 3 ) y = 3t (1 + t 3 )x + 2t 3 − t 6 − 3t 3
⇒
(1 + t 3 )( 2 − t 3 ) y = 3t (1 + t 3 )x − t 3 (1 + t 3 )
M03_ENGINEERING_MATHEMATICS-I _CH03_Part A.indd 32
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Differential Calculus ■
⇒
( 2 − t 3 ) y = 3tx − t 3
3.33
[Dividing by (1 + t3)]
⇒
3tx + (t 3 − 2) y − t 3 = 0 ⇒ ( y − 1)t 3 + 3tx − 2 y = 0.
∴ the equation of the tangent at t1, t2, t3 are
( y − 1)t13 + 3xt1 − 2 y = 0, ( y − 1)t 23 + 3xt 2 − 2 y = 0, ( y − 1)t 33 + 3xt 3 − 2 y = 0.
Given that the three tangents at t1, t2, t3 are concurrent.
Let (x1, y1) be the point of concurrency.
∴ the point (x1, y1) will satisfy the equations of the tangents at t1, t2, t3.
∴
( y 1 − 1)t13 + 3x 1t1 − 2 y 1 = 0, ( y 1 − 1)t 23 + 3x 1t 2 − 2 y 1 = 0 and ( y 1 − 1)t 33 + 3x 1t 3 − 2 y 1 = 0.
These three equations imply that t1, t2, t3 are the roots of the equation
( y 1 − 1)t 3 + 3x 1t − 2 y 1 = 0
∴
sum of the roots = −
coeff t 2
coeff t 3
⇒
t1 + t 2 + t 3 =
0
= 0.
y1 −1
EXERCISE 3.4
Find the equation of tangents to the curve y = ( x 3 − 1)( x − 2) at the point where it meets the x-axis.
Find the equation of the tangents from the origin to the curve y = 2 x 2 + 1.
Find the equation of normal to the curve 3x 2 − y 2 = 8 which is parallel to the line x + 3y = 4.
Find the equation of the tangent to the curve y = 7 x − x 2 at the point (3, 12) on it. Also find the
equation of the normal at the point.
5. Find the points on the curve y 2 = 2 x(3 − x 2 ) at which the tangents are parallel to the x-axis.
1.
2.
3.
4.
6. Find the points on the curve y 3 = x 2 ( 2a − x ), where the tangents are parallel to y-axis.
7. If the tangent at (x1, y1) on the curve y3 + x3 – 9xy + 1 = 0 is parallel to the x-axis, prove that at
d2 y
18
the point 2 =
.
dx
27 − x13
t −1
t +1
at the point t = 2.
8. Find the equation of the tangent to the curve x =
,y =
t +1
t −1
9. Find the abscissa of the point on the curve ay2 = x3 at which the normal cuts off equal intercepts
on the coordinate axes.
[Hint: A line makes equal intercepts if its slope = 1 or −1].
10. If the tangent at (1, 3) on the parabola y = 4x − x2 cuts the parabola y = x2 − 6x + k at two different
points, find the values of k.
11. Find the equation of the tangent to the curve x2 + 4y2 =16 at the point which is such that it is the
mid point of the portion of the tangent intercepted between the coordinate axes.
12. Find the points on x2 = y3 at which the normal pass through (0, 4).
13. The curve y = ax2 + bx + c passes through the points (−1, 0) and (0, −2). The tangent to the curve
at the latter point makes 135° with the x-axis.
14. If p, q are the lengths of the perpendiculars from the origin to the tangent and normal at a point
on x 2/3 + y 2/3 = a 2/3 , then prove that 4 p 2 + q 2 = a 2 .
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15. Show that the portion of the tangent at any point on x m y n = a m+ n intercepted between the
coordinate axes is divided by the point in the ratio m : n.
16. Find the equation of the straight line which is tangent at one point and normal at another point of
the curve x = 3t, y = 2t3.
17. Find all the tangents to the curve y = cos (x + y), −2p ≤ x ≤ 2p , that are parallel to the line
x + 2y = 0.
18. Prove that the condition for the line x cos a + y sin a = p to touch the curve x m y n = a m+ n is
p m+ n ⋅ m m nn = ( m + n) m+ n a m + n cos m a sin n a.
x3
at the point (2, −2).
19. Find the equation of the tangent and the normal to the curve y 2 =
4−x
20. Find the equation of the tangent and the normal to the curve y ( x − 2)( x − 3) − x + 7 = 0 at the
point where it cuts the x - axis.
x y
21. Prove that the line + = 1 touches the curve y = b e − x a at the point where the curve crosses
a b
the y - axis.
n ( n −1)
n ( n −1)
⎛x⎞
⎛y⎞
22. If x cos a + y sin a = p touches the curve ⎜ ⎟
+⎜ ⎟
= 1, then prove that
⎝ a⎠
⎝b⎠
(a cos a) n + (b sin a) n = p n .
ANSWERS TO EXERCISE 3.4
y + 3x =3; y − 7x + 14 = 0
y = ±2 2 x
x + 3y + 8 = 0
y = x + 9; y = x + 15
(1, 2); (1, −2)
(0, 0); (2a, 0)
9x + y − 6 = 0
4a
9.
9
10. k < 17
1.
2.
3.
4.
5.
6.
8.
8 4⎞
⎛
,
12. ⎜ ±
⎝ 3 2 3 ⎟⎠
13. a = 1, b = 1, c = −2. At (2, 0) the equation of
the tangent is y = 3 x − 6
16. y = ± 2( x − 2)
1⎛
p⎞
1⎛
p⎞
17. y = − ⎜ x − ⎟ ; y = − ⎜ x + 3 ⎟ .
2⎝
2⎠
2⎝
2⎠
19. 2x + y − 2 = 0; x − 2 y − 6 = 0.
20. x − 20 y − 7 = 0; 20 x + y − 140 = 0
11. x + 2 y = 4 2 , x − 2 y = 4 2 ,
x + 2 y = −4 2 , x − 2 y = −4 2 .
3.2.3 Length of the Tangent, the Sub-Tangent, the Normal and the Sub-normal
Proof: Let y = f ( x ) be the equation of the given curve. Let P(x, y) be any point on the curve.
Let the tangent at the point P(x, y) meet the x-axis at T. Let the normal at the point P meet the
x−axis at N.
Draw PM perpendicular to the x-axis.
∴ TM is the projection of PT on the x-axis and MN is the projection of PN on the x-axis.
(i) PT is the length of the tangent at the point P(x, y)
(ii) PN is the length of the normal at the point P(x, y)
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Differential Calculus ■
(iii) TM is the length of the sub−tangent at P
(iv) MN is the length of the sub−normal at P.
y
Tangent
Let the tangent at P make an angle c with the x-axis.
That is PTN = c.
Since angle between two lines is equal to the angle
their perpendiculars, MPN = c.
∴ from Δ TPM, sin c =
PM
PT
⇒
PT =
P (x, y)
ψ
Normal
y
ψ
o T
M
N
PM
sin c
Fig. 3.5
= ycosec c = y 1 + cot 2 c
⎛ dy ⎞
We know that tan c = ⎜ ⎟ = y1
⎝ dx ⎠ P
∴ Length of the tangent =
From Δ PMN, cos c =
PM
PN
3.35
cot c =
∴
1
y1
∴
PT = y 1 +
1
y
=
1 + y 12
2
y1
y1
y
1 + y12 .
y1
⇒
PN =
PM
= y sec c = y 1 + tan 2 c = y 1 + y 12
cos c
Length of the normal = y 1 + y12 .
∴
Length of the sub−tangent = TM
Length of the sub−normal = MN
From Δ PTM,
∴
tan c =
PM
TM
Length of the sub-tangent =
From ΔPMN ,
tan C =
⇒
TM =
PM
y
=
tan c y 1
y
y1
MN
PM
⇒
MN = PM tan C = yy1
∴ Length of the sub-normal = yy1
Length of the tangent is y 11y 12 and Length of the normal is y 1 1y 12
y1
Length of the sub-tangent is y and Length of the sub-normal is yy 1
y1
⎛ dy ⎞
where y 5 f ( x ), y 1 5 ⎜ ⎟
.
⎝ dx ⎠ atP ( x , y )
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WORKED EXAMPLES
EXAMPLE 1
x2 y 2
1 2 5 1 varies as the abscissa of the
a2
b
point. Also find the length of the sub-tangent, tangent and normal at that point.
Prove that the sub-normal at any point of the ellipse
Solution.
x2 y2
+
=1
a2 b2
Let P ( x, y ) be any point on the curve.
Differentiating (1) w.r.to x, we get
(1)
The given curve is the ellipse
2 x 2 y dy
+
=0
a 2 b 2 dx
y1 = −
At the point P ( x , y ) :
dy
b2 x
=− 2 .
dx
a y
⇒
b2 x
⋅ .
a2 y
⎛ b2 x ⎞
b2
b2
Length of the sub-normal = yy1 = y ⋅ ⎜ − 2 ⎟ = − 2 x = k x , where k = + 2
a
a
⎝ a y⎠
∴ the sub-normal varies as the abscissa of the point.
y
=
Length of the sub-tangent =
y1
Length of the tangent =
Now
a2 y 2
y
=
.
b2 x b2 x
− 2
a y
y
1 + y 12
y1
1 + y 12 = 1 +
∴ Length of the tangent =
b4 x 2 b2
=
y
a4 y 2
b2
y
.
−b 2 x y
a2 y
x2 y 2
+
a4 b 4
x2 y 2
x2 y 2
2
+
=
a
+ .y
a4 b 4
a4 b 4
and Length of the normal = y 1 + y 12 = y .
b2
y
x2 y 2
x2 y 2
+ 4 = b2 4 + 4
4
a
b
a
b
EXAMPLE 2
Find the length of the tangent, normal, sub-tangent and sub-normal for the cycloid
x 5 a (t 1 sin t ), y 5 a(1 2 cos t ).
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Differential Calculus ■
3.37
Solution.
The given curve is the cycloid
x = a(t + sin t )
(1)
y = a(1 − cos t )
and
(2)
Differentiating (1) and (2) w.r.to t, we get
dx
t
dy
t
t
= a(1 + cos t ) = 2a cos 2
= a sin t = 2a sin cos
and
dt
2
dt
2
2
dy dy /dt
=
=
dx dx /dt
∴
Length of the tangent
t
t
2a sin cos
2
2 = tan t
t
2
2a cos 2
2
∴
t
y 1 = tan .
2
y
a(1 − cos t )
t
1 + y12 =
1 + tan 2
t
2
= y1
tan
2
=
t
t
2a sin 2
t
2 ⋅ sec =
2 ⋅ cos t ⋅ sec t = 2a sin t ,
t
t
2
2
2
2
tan
sin
2
2
2a sin 2
y
a(1 − cos t )
=
=
Length of the sub-tangent =
t
y1
tan
2
=
2a sin 2
tan
({ 0 ≤ t ≤ 2p)
t
2
t
2
t
2 ⋅ cos t = 2a sin t cos t = a sin t .
t
2
2
2
sin
2
2a sin 2
Length of the normal = y 1 + y 12 = a(1 − cos t ) 1 + tan 2
t
2
t
2 = 2a tan t ⋅ sin t
t
2
2
cos
2
t
t
t
Length of sub-normal = yy1 = a(1 − cos t ) ⋅ tan = 2a sin 2 tan .
2
2
2
t
t
= 2a sin ⋅ sec = 2a
2
2
sin 2
2
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EXERCISE 3.5
1. Show that the length of the sub-normal at any point on the exponential curve y = be x / a varies as
the square of the ordinate.
u
⎡
⎤
2. For the curve y = a sin u, x = a ⎢log cot + cos u⎥ . Find the length of the tangent, normal,
2
⎣
⎦
p
sub-tangent and sub-normal at u = .
4
3. Show that in the curve by2 = (x + a)3, the square of the sub-tangent varies as the sub-normal.
4. Show that in the parabola y2 = 4ax, the sub-tangent at any point is double the abscissa and the sub
normal is constant.
5. Find the value of n for which the length of the subnormal of the curve xyn = an+1 is constant.
x
6. Prove that the sub-tangent for any point on the curve y = be a is of constant length.
7. Find the length of the sub-tangent and sub-normal at the point ‘t’on the curve x = a (cost + t sint)
and y = a (sint − t cost).
ANSWERS TO EXERCISE 3.5
a 13 5a a
,
,
5
2 5 2
7. a (sint − t cost)cost, a (sint − t cost) tant.
2. a 13,
5. n = −2
3.2.4 Angle between the Two Curves
The angle between two curves at a point of intersection is defined as the angle between the tangents
to the curves at the point.
Let the two curves C1, C2 intersect at the point P. Let m1, m2 be the slopes of the tangents at the
point P of the two curves
m − m2
If u is the angle between the two tangents, then tan u = 1
.
1 + m1m2
Note
(1) This formula always gives the acute angle between the two curves.
⎛ m − m2 ⎞
.
(2) When this formula is used for further algebraic computations, we use it as tan u = ± ⎜ 1
⎝ 1 + m1m 2 ⎟⎠
If u = 90°, then curves intersect orthogonally at the point P.
In this case, tan u = tan 90° = ∞ ⇒ 1 + m1m2 = 0 ⇒ m1m2 = −1
Conversely, if m1m2 = −1, then
tan u = ∞ ⇒
u = 90°.
∴ the condition for two curves to intersect orthogonally at the point P is m1 m 2 521 .
(3) If u = 0, then tanu = 0 ⇒ m1 = m2.
Conversely, if m1 = m2, then u = 0.
∴ the condition for two curves to touch at the point P is m1 = m2.
(4) If f (x, y) = 0 is a rational algebraic equation passing through the origin (0, 0), then the equation
of the tangent at the origin is obtained by equating the lowest degree terms to zero.
For example, the tangent at the origin to the parabola y2 = 4x is x = 0.
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Differential Calculus ■
3.39
f
dy
(5) If the equation of the curve is in implicit form f(x, y) = 0, then
= − x , where fx, fy are partial
dx
fy
differential coefficients of f w.r.to x and y respectively.
WORKED EXAMPLES
EXAMPLE 1
Find the angle of intersection of the curves x2 5 2y and x2 1 y2 5 8.
Solution:
The given curves are
x2 = 2y
(1)
x2 + y2 = 8
and
(2)
To find the point of intersection solve (1) and (2).
∴
2y + y2 = 8
⇒
y2 + 2y − 8 = 0 ⇒
(y + 4) (y − 2) = 0
⇒
y = −4, 2.
When y = 2,
x = 4 ⇒ x = ± 2.
When y = 4,
x2 = −8 < 0. ∴ x is imaginary.
∴ the points of intersection are P(2, 2) and Q(−2, 2).
Differentiating (1) w.r.to x, we get
2
2
dy
= 2x
dx
⇒
dy
= x = m1
dx
⇒
dy
y
= − = m2
dx
x
Differentiating (2) w.r.to x, we get
2x + 2 y
dy
=0
dx
m1 = 2 and m2 = −
At the point (2, 2):
2
= −1
2
Let u1 be the angle between the curves at the point (2, 2).
∴
tan u1 =
m1 − m2
2 +1
3
=
=
=3 ⇒
1 + m1m2
1 + 2( −1)
−1
u1 = tan −1 3.
At the point Q(22, 2)
m1 = −2 and m2 = −
Let u 2 be the angle between the curves (−2, 2)
( −2)
=1
2
m1 − m2
−2 − 1
−3
−1
=
=
= 3. ⇒ u 2 = tan 3.
1 + m1m2
1 + ( −2) ⋅1 −1
∴ the angle of intersection of the curves at the points P(2, 2) and Q(−2, 2) are the same.
∴
tan u 2 =
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EXAMPLE 2
Show that the ellipse 4 x 2 1 9 y 2 572 cuts the hyperbola x 2 2y 2 5 5 orthogonally.
Solution.
The given curves are
4 x 2 + 9 y 2 = 72
(1) and x 2 − y 2 = 5
The two curves are symmetric with respect to both the axes.
Let ( x 1 , y 1 ) be any point of intersection.
Let m1 , m 2 be the slopes of the tangents at the point ( x 1 , y 1 ) .
Differentiating (1) w.r.to x, we get
dy
⇒ dy = −8x = −4 x
8 x + 18 y
=0
dx
dx 18y
9y
Differentiating (2) w.r.to x,we get
2x − 2 y
At the point ( x 1 , y 1 ):
dy
=0
dx
m1 =
Now
m1 × m2 =
−4 x1
9 y1
⇒
dy x .
=
dx y
and
m2 =
(2)
x1 .
y1
−4 x1 x1
4x 2
⋅ = − 12 .
9 y1 y1
9 y1
(3)
Since, ( x 1 , y 1 ) is a point on both the curves.
4 x 12 + 9 y 12 = 72
(5) × 4 ⇒
(4)
and
x 12 − y 12 = 5
4 x 12 − 4 y 12 = 20
(6)
Subtracting, (4) − (6) ⇒ 13 y12 = 52
⇒
∴
⇒
x 12 − 4 = 5
(5)
52 = 4.
13
x 12 = 9
y 12 =
4 9
m1m 2 = − ⋅ = −1 .
9 4
Hence, the curves cut orthogonally at the point ( x 1 , y 1 ) .
Since ( x 1 , y 1 ) is an arbitrary point of intersection, it follows that the two curves cut orthogonally
at all the points of intersection.
∴
EXAMPLE 3
Show that the condition for ax 2 1 by 2 51 and a ′x 2 1 b ′y 2 51 to cut orthogonally is 1 2 1 5 1 2 1 .
a b a′ b ′
Solution.
The given curves are
ax 2 + by 2 = 1
(1)
and
a ′x 2 + b ′y 2 = 1
(2)
Let ( x 1 , y 1 ) be any point of intersection of (1) and (2).
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Differential Calculus ■
3.41
Let m1 , m 2 be the slope of the tangents to the curves at the point ( x 1 , y 1 ) .
Differentiating (1) w.r.to x, we get
2ax + 2by
dy
=0
dx
⇒
dy
ax
=− .
dx
by
dy
=0
dx
⇒
dy
a′ x .
=−
dx
b′ y
Differentiating (2) w.r.to x, we get
2a ′x + 2b ′y
At the point ( x 1 , y 1 ):
m1 = −
ax1
by1
and
m2 = −
a′ x1
.
b′ y1
The condition for orthogonality is
m1m2 = −1 = −
ax 1 ⎛ a ′x 1 ⎞
⋅ −
= −1
by 1 ⎜⎝ b ′y 1 ⎟⎠
⇒
aa ′x 12
= −1
bb ′y 12
(3)
Since ( x 1 , y 1 ) is a point on the two curves, we get
ax 12 + by 12 = 1 and a ′x 12 + b ′y 12 = 1
∴
ax 12 + by 12 = a ′x 12 + b ′y 12
⇒
( a − a ′ ) x12 = −(b − b ′ ) y12
Substituting in (3), we get
aa ′ ⎡ (b − b ′ ) ⎤
−
= −1
bb ′ ⎢⎣ a − a ′ ⎥⎦
⇒
b − b ′ a − a′
=
bb ′
aa ′
⇒
aa ′ ⎡ b − b ′ ⎤
⋅
=1
bb ′ ⎢⎣ a − a ′ ⎥⎦
⇒
1 1 1 1
− = −
b ′ b a′ a
⇒
⇒
x12
(b − b ′)
=−
2
( a − a ′)
y1
1 1 1 1.
− = −
a b a′ b ′
EXAMPLE 4
x12
touch at the point (2, 4). Also find the
x 21
equation of the common tangent at the point (2, 4).
Show that the curves y 5 6 1 x 2 x 2 and y 5
Solution.
The given curves are
x+2
x −1
When x = 2, from (1),
y = 6 + 2 − 4 = 4. ∴ (2, 4) is a point on (1).
2+2
When x = 2, from (2),
∴ (2, 4) is a point on (2).
y=
= 4.
2 −1
∴ the point (2, 4) satisfies the equations of the two curves and it is a common point.
Now to prove that the two curves touch each other at the point (2, 4).
That is to prove the slope of the tangents to the curves are equal at the point (2, 4).
Let m1 and m2 be the slopes of the tangents to the curves at the point (2, 4)
y = 6 + x − x2
M03_ENGINEERING_MATHEMATICS-I _CH03_Part A.indd 41
(1)
and
y=
(2)
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To prove m1 5 m 2
Differentiating (1) w.r.to x, we get
dy
= 1 − 2 x.
dx
Differentiating (2) w.r.to x, we get
dy ( x − 1) ⋅1 − ( x + 2) ⋅1
−3 .
=
=
dx
( x − 1) 2
( x − 1) 2
At the point (2, 4):
m1 = 1 − 2 × 2 = 1 − 4 = −3 and m2 = −
∴ at the point (2, 4), m1 = m2.
Hence, the two curves touch each other at the point (2, 4).
The equation of the common tangent at the point (2, 4) is
y − 4 = −3( x − 2) = −3x + 6 ⇒ 3x + y − 10 = 0.
3
= −3.
( 2 − 1) 2
EXERCISE 3.6
I. Find the angle of intersection of the following pairs of curves.
x 2 − y 2 = a 2and x 2 + y 2 = 2a 2
2. xy = 10 and x2 + y2 = 21
2
3
y = 2x and y (5 − x) = x
4. y2 = 4x and x2 = 4y
2
2
2
xy = a and x + y =2x
x2 = ay and x3 + y3 = 3axy, at the point other than the origin.
y = 2x and x3 + y3 = 6xy
8. 2y2 = ax and x2 = 4ay 2
x
2
2
2
y = 4x and 8x + y − 6y = 0 at the point (0, 0) 10. y = 4 − x and y = 4 −
2
1
11. Find the angle of intersection of the curves y = x 3 and y = 6 − x 2 at the point (2, 2).
4
2
x +3
x − 7x + 11
cut each other at the point (2, 1) at an angle
12. Show that the curves y = 2
and y =
x −1
x +1
of 45°.
13. Find the angle of intersection of the parabolas y 2 = 4ax and x 2 = 4ay .
1.
3.
5.
6.
7.
9.
II. Show that the curves touch at the indicated points and find the equation of the common
tangent.
1. y = x 3 − 3 x 2 − 8 x − 4 and y = 3 x 2 + 7 x + 4 at the point (−1, 0).
2. x2 = ay and x3 + y3 = 3axy at the point (0, 0)
3x 2 p
p
− at the point ⎛⎜ , 0⎞⎟
2p 6
⎝3 ⎠
y = x3 + x +1 and 2y = x3 + 5x at the point (1, 3)
Show that the following pairs of curves cut orthogonally.
x2 – y2 = 8 and xy = 3
2. x3 – 3xy2 = –2 and y3 – 3x2y = –2 at the point (1, 1)
3
3
x + y + 2y + x = 0 and xy + 2x = y at the origin (0, 0).
3. y = 2 sin x − 3 and y =
4.
III.
1.
3.
4. y = x 2 and 6 y = 7 − x 2 at the point (1, 1). 5. y 2 = 4( x + 1) and y 2 = 36(9 − x ).
M03_ENGINEERING_MATHEMATICS-I _CH03_Part A.indd 42
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3.43
ANSWERS TO EXERCISE 3.6
I.
1.
p
4
⎛ 1⎞
⎛ 1⎞
3. tan −1 ⎜ ⎟ and tan −1 ⎜ ⎟
⎝ 3⎠
⎝ 2⎠
21
2. tan −1 ⎛⎜ ⎞⎟
⎝ 20 ⎠
3
p
4. tan −1 ⎛⎜ ⎞⎟ and
⎝ 4⎠
2
5. 0, the curves touch each other. 6. tan −1 ( 21/3 )
6
7. tan −1 ⎛⎜ ⎞⎟ , tan −1 ( 2)
⎝ 13 ⎠
8.
10.
p
⎛ 3⎞
, tan −1 ⎜ ⎟
⎝ 5⎠
2
9.
⎛ 7⎞
11. tan −1 ⎜ ⎟
⎝ 11⎠
p
⎛ 1⎞
, tan −1 ⎜ ⎟
⎝ 3⎠
4
⎛ 3⎞
13. tan −1 ⎜ ⎟
⎝ 4⎠
II.
1. y = x + 1
3.3
2. y = 0
p
2
3. y − x +
p
=0
3
4. 4x − y = 1.
MEAN-VALUE THEOREMS OF DERIVATIVES
The mean−value theorems for derivatives play an important role in calculus because many basic
properties of functions can be deduced from it. However, mean−value theorems are obtained from a
special case due to the French Mathematician Michael Rolle.
3.3.1 Rolle’s Theorem
Let f be a real function defined on the closed interval [a, b] such that
(i) f is continuous on [a, b]
(ii) f is derivable in the open interval (a, b) and
(iii) f (a) = f(b), then there exists at least one point c ∈( a, b) such that f ′( c ) 5 0 .
Geometrical Meaning of Rolle’s Theorem
If y = f ( x ) be a continuous curve with end points A and B, having tangent at every point between
A and B and the ordinates of A and B are equal, then there exists at least one point P on the curve
between A and B such that the tangent at P is parallel to the x-axis.
The geometrical meaning is clear from the diagrams.
y
y
P1
P
A
o
o
a
b
Fig. 3.6
M03_ENGINEERING_MATHEMATICS-I _CH03_Part A.indd 43
B
A
B
x
a
b x
P2
Fig. 3.7
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Algebraic Meaning
If a and b are two consecutive roots of the polynomial equation f ( x ) = 0, then there is at least one root
of f ′( x ) = 0, between a and b.
Physical Meaning
The instantaneous rate of change of f at some point c between a and b is zero.
WORKED EXAMPLES
EXAMPLE 1
Test the application of Rolle’s theorem for the following functions.
1
(i) x on[21, 1]
(ii) cos on[21, 1]
x
(iii) x 2 on[2, 3]
(iv) tan x on[0, p].
Solution.
Then f is continuous on [ −1, 1]
(i) Let f ( x ) = x , − 1 ≤ x ≤ 1.
and f ( −1) = −1 = 1, f (1) = 1 = 1
∴ f ( −1) = f (1).
⎧ − x if − 1 ≤ x < 0
We have f ( x ) = ⎨
0 ≤ x ≤1
⎩ x if
⎧ −1 if
∴ f ′( x ) = ⎨
⎩ 1 if
∴ f ′( 0 − ) ≠ f ′( 0 + )
∴ f ′(0) does not exist.
−1 < x < 0
0 < x <1
Hence, the second condition is not satisfied.
So, Rolle’s theorem can not be applied.
1
(ii) Let f ( x ) = cos , x ∈[ −1, 1]
x
f ( −1) = cos( −1) = cos 1, f (1) = cos 1
∴ f ( −1) = f (1)
But f is not continuous at x = 0.
∴ Rolle’s theorem can not be applied
So, f ′(0) does not exist.
(iii) Let f ( x ) = x 2 , x ∈[2, 3] and f is continuous and differentiable on [2, 3]
But f ( 2) = 22 = 4
and
f (3) = 32 = 9 ∴ f (2) ≠ f (3).
Hence, Rolle’s theorem can not be applied.
(iv) Let f ( x ) = tan x, 0 ≤ x ≤ p
p
p
p
, tan x = tan = ∞ and so, the function f ( x ) is discontinuous at .
2
2
2
Hence, first condition is not satisfied.
When x =
But f (0) = 0 and f (p) = 0
∴ f (0) = f (p).
Hence, Rolle’s theorem can not be applied.
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3.45
EXAMPLE 2
a0
a
a
a
1 1 1 2 1…1 n21 1 a n 5 0, then there
n 11 n n 21
2
n
n21
n22
exists at least one x in (0, 1) such that a0 x 1a1 x 1 a 2 x 1…1 a n21 x 1 a n 50.
If a0, a1, a2, … an are real numbers such that
Solution.
Consider the function
f ( x ) = a0
Then
f (0) = 0
and
f (1) =
∴
x n +1
xn
x n −1 …
x2
+ a1
+ a2
+ + an −1
+ an x , x ∈[ 0, 1]
2
n +1
n
n −1
a0
a
a
a
+ 1 + 2 + … + n −1 + an = 0
n +1 n n −1
2
[Given]
f (0) = f (1).
Since f(x) is a polynomial, it is continuous in [0,1] and differentiable in (0, 1).
So, all the conditions of Rolle’s theorem are satisfied.
Hence, by Rolle’s theorem, there exists at least one x ∈(0, 1) such that f ′(x) = 0.
( n + 1) x n
( n − 1) x n− 2 …
2x
nx n−1
+ a1
+ a2
+ + an−1
+ an
2
n +1
n
n −1
= a0 x n + a1x n −1 + a2 x n − 2 + … + an −1x + an
But
f ′( x ) = a0
∴
a0 x n + a1x n −1 + … + an −1x + an = 0
for at least one x ∈ (0, 1).
EXAMPLE 3
Using Rolle’s theorem, prove that there is no real a for which the equation x 2 23 x 1 a 5 0 has
two different roots in [21, 1].
Solution.
Suppose there is a real a for which x 2 − 3 x + a = 0 has two different roots a, b in [−1, 1].
Then the interval [a, b] is contained in the interval [−1, 1]
[a, b] ⊆ [−1, 1].
i.e.,
Consider f ( x ) = x 2 − 3x + a. x ∈[a, b].
Since f ( x ) is a polynomial in x, it is continuous and differentiable in (a, b ).
Given a and b are the roots of f(x) = 0.
∴
f (a) = 0 and f (b) = 0 ⇒ f (a) = f (b).
So, the condition of Rolle’s theorem are satisfied by f (x) in [ a, b ].
∴ by Rolle’s theorem, there exists at least one c ∈[a, b] such that f ′(c ) = 0 .
Now
f ′( x ) = 2x − 3
M03_ENGINEERING_MATHEMATICS-I _CH03_Part B.indd 45
∴ f ′(c) = 2c − 3
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f ′(c ) = 0
⇒
2c − 3 = 0
⇒
c=
3
.
2
3
∉[ −1,1] and so, c ∉[a, b] .
2
This is a contradiction. So, our assumption is wrong
i.e., there is no a for which x 2 − 3 x + a = 0 has two different roots in [−1, 1].
But
c=
EXAMPLE 4
Prove that between any two roots of e x sin x 5 1 , there is at least one root of e x cos x 1 1 5 0 .
Solution.
Let a and b be any two roots of e x sin x = 1.
e a sin a = 1 ⇒ sin a = e − a
∴
e − a − sin a = 0
⇒
e b sin b = 1 ⇒ sin b = e −b
and
and
e −b − sin b = 0.
Consider the function
f ( x ) = e − x − sin x , x ∈[a, b]
Then
f (a) = e − a − sin a = 0 and f (b) = e −b − sin b = 0
∴
f (a) = f (b).
(1)
Since e − x and sinx are continuous in [a, b] and differentiable in ( a, b ),
f ( x )is continuous in [a, b] and differentiable in ( a, b ).
Hence, all the conditions of Rolle’s theorem are satisfied.
∴ by Rolle’s theorem, there is a c ∈(a, b) such that f ′(c ) = 0
But
f ′( x ) = −e − x − cos x
∴ f ′(c) = −e − c − cos c
∴
f ′(c ) = 0 ⇒ − e − c − cos c = 0 ⇒ − 1 − e c cos c = 0 ⇒ e c cos c + 1 = 0.
∴ c is a root of e x cos x + 1 = 0 and c ∈ (a, b)
Thus, between any two roots of e x sin x = 1 , there is a root of e x cos x + 1 = 0.
EXAMPLE 5
sin x sin a sinb
p
f
(
x
)
5
cos x cos a cos b , 0 < a < b < . Show that f ′( x ) 5 0 has a root between a and b.
If
2
tan x tan a tanb
Solution.
Consider f (x) in [a, b].
∴
M03_ENGINEERING_MATHEMATICS-I _CH03_Part B.indd 46
sin a sin a sin b
f (a) = cos a cos a cos b = 0
tan a tan a tan b
[{ C1 = C2]
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Differential Calculus ■
and
sin b sin a sin b
f (b) = cos b cos a cos b = 0
tan b tan a tan b
∴
f (a) = f (b)
3.47
[{ C1 = C3]
⎛ p⎞
⎛ p⎞
Since sinx, cosx, tanx are continuous and differentiable in ⎜ 0, ⎟ and [a, b] ⊂ ⎜ 0, ⎟ , it follows
⎝ 2⎠
⎝ 2⎠
that f (x) is continuous and differentiable in (a, b) .
So, all the three conditions of Rolle’s theorem are satisfied by f (x) in [a, b] .
∴ by Rolle’s theorem, there is a c ∈(a, b) such that f ′(c ) = 0 .
f ′( x ) = 0 has a root c between α and β.
⇒
3.3.2 Lagrange’s Mean Value Theorem
Let f be a real function defined on the closed interval [a, b ] such that
(i) f is continuous on [a, b ] and (ii) f is derivable in the open interval (a, b).
Then there exists at least one point c ∈(a, b ) such that
f ′( c ) 5
Proof
f (b ) 2 f ( a )
b 2a
Consider F ( x ) = f ( x ) + A x , x ∈[a, b ] , where A is chosen such that
F ( a) = F ( b)
Since f and x are continuous on [a, b] and derivable in (a, b), F is continuous on [a, b] and derivable
in (a, b).
Further
F ( a) = F ( b)
⇒
⇒
F ( a) + Aa = F (b) + Ab
A( a − b) = f (b) − f ( a)
⇒
A= −
[ f ( b) − f ( a) ] .
b−a
(1)
So, F satisfies all the conditions of Rolle’s theorem on [a, b].
∴ by Rolle’s theorem, there exists at least one c ∈(a, b ) such that F ′(c ) = 0 .
But
F ′( x ) = f ′( x ) + A ∴ F ′(c ) = f ′(c ) + A
∴
F ′(c ) = 0
⇒
f ′(c ) =
⇒
f ′(c ) + A = 0
f (b ) − f (a)
b −a
⇒
f ′(c ) = − A
[using (1)]
Geometrical meaning of Lagrange’s Mean Value Theorem
If y = f ( x ) is a continuous curve with A and B as end points and at each point between A and B, the
curve has a tangent, then there is atleast one point P on the curve between A and B at which the tangent
is parallel to the chord AB.
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f ( b) − f ( a)
.
b−a
A is (a, f(a)) and B is (b, f (b). So, slope of chord AB =
Slope of the tangent at the point P (c, f (c)) is f ′(c)
y
y
P
B
B
A
o
A
a
c
b
o
x
a
Fig. 3.8
b
x
Fig. 3.9
∴
f ′(c) =
f ( b) − f ( a)
.
b−a
Physical meaning of Lagrange’s Mean Value Theorem
The instantaneous rate of change of f at some point between a and b is equal to the average rate of
change of f on [a, b].
Another form of Lagrange’s Mean Value Theorem
If b − a = h , then b = a + h, h > 0 , then c ∈ (a, b ) ⇒ c = a + uh , 0 < u < 1.
a
c
a+h
Lagrange’s mean value theorem is
f ′ ( a + uh ) =
f ( a + h ) − f ( a)
h
f (a + h ) = f (a) + hf ′(a + uh ), 0 < u < 1.
⇒
Deductions of Lagrange’s Mean Value Theorem
(1) If f is a continuous on a closed interval [a, b] and derivable in (a, b) such that f ′( x ) = 0 ∀x ∈ (a, b ),
then f is constant on [a, b].
i.e.,
f (x ) = k
∀x ∈[a, b ]
In particular,
f ( x ) = f (a) ∀x ∈[a, b ]
(2) If f and g are real functions which are continuous on [a, b] and derivable in (a, b) such that
f ′( x ) = g ′( x ) ∀x ∈ (a, b ), then
f (x ) − g (x ) = k
∀x ∈[a, b ], where k is a constant.
(3) If f is continuous at c and if lim f ′( x ) = l and lim f ′( x ) = l , then f ′(c ) exists and f ′(c ) = l .
x →c −
Proof
∴
x→c +
Given lim f ′( x ) = l
x→c
lim f ′( x ) = l
x→c −
M03_ENGINEERING_MATHEMATICS-I _CH03_Part B.indd 48
and
lim f ′( x ) = l .
x→c +
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Differential Calculus ■
Consider lim f ′( x ) = l .
c
x→c +
x
3.49
c+h
So, there exists an interval (c , c + h ), h > 0 where f ′( x ) exists for every x ∈ (c , c + h ).
∴ f is continuous in [c , c + h ] .
If x is a point in this interval., by Lagrange’s mean value theorem
lim
x→c +
⇒
Similarly, we can prove that
f ( x ) − f (c )
= f ′(t ), c < t < x .
x −c
f ( x ) − f (c)
= lim f ′(t ) = lim+ f ′( x ) = l
x→c +
x→c
x−c
f ′(c + ) = l
f ′( c − ) = l
∴ f ′( c ) = l .
WORKED EXAMPLES
EXAMPLE 1
Find c of Lagrange’s mean value theorem for the following functions
(i) x ( x 2 1)( x 2 2) in ⎡0,
⎢
⎣
1⎤
2 ⎥⎦
(ii)
x 3 1 x in [1, 2].
Solution.
1
f (x) = x(x − 1) (x − 2), x ∈ ⎡⎢0, ⎤⎥ .
⎣ 2⎦
(i) Let
(1)
1
1
Since f (x) is a polynomial function, it is continuous on ⎢⎡0, ⎤⎥ and differentiable in ⎛⎜ 0, ⎞⎟ .
⎝
2⎠
⎣ 2⎦
So, the conditions of Lagrange’s mean value theorem are satisfied.
1
∴ by Lagrange’s mean value theorem, there exists c ∈ ⎛⎜ 0, ⎞⎟ such that
⎝ 2⎠
f ′(c) =
f ( b) − f ( a)
b−a
(2)
1
Here a = 0, b = . Now f(a) = f(0) = 0
2
1⎛1 ⎞⎛1
⎞ 1 ⎛ 1⎞ ⎛ 3⎞ 3
⎛ 1⎞
f(b) = f ⎜ ⎟ = ⎜⎝ − 1⎟⎠ ⎜⎝ − 2⎟⎠ = ⎜ − ⎟ ⎜ − ⎟ = .
⎝ 2⎠
2 ⎝ 2⎠ ⎝ 2⎠ 8
2 2
2
and
Differentiating (1) w.r.to x, we get
f ′( x ) = x{( x − 1) ⋅1 + ( x − 2) ⋅1} + ( x − 1)( x − 2) ⋅1
= x 2 − x + x 2 − 2 x + x 2 − 3x + 2 = 3x 2 − 6 x + 2
∴
f ′(c) = 3c 2 − 6c + 2
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∴ (1) ⇒
∴
3
−0
3
3c 2 − 6c + 2 = 8
=
1
4
−0
2
c=
⇒
6 ± 36 − 4 ⋅ 3 ⋅
6
⎛ 1⎞
1 + 0.76 = 1.76 ∉ ⎜ 0, ⎟
⎝ 2⎠
Clearly,
3c 2 − 6c +
5
=0
4
5
4 = 6 ± 36 − 15 = 6 ± 21 = 1 ± 21 = 1 ± 0.76.
6
6
6
⎛ 1⎞
and 1 − 0.76 = 0.24 ∈ ⎜ 0, ⎟
⎝ 2⎠
∴
c = 0.24
(ii) Let f ( x ) = x 3 + x , x ∈[1, 2] .
Since f (x) is a polynomial, it is continuous on [1, 2] and differentiable in (1, 2).
So, the conditions of Lagrange’s mean value theorem are satisfied.
∴ by Lagrange’s mean value theorem, there exists a c ∈(1, 2) such that
f ′(c ) =
Here a =1, b = 2.
Now
f (b ) − f (a)
b −a
f (1) = 1 + 1 = 2
and f (2) = 23 + 2 = 10
f ′( x ) = 3 x 2 + 1
∴ (1) ⇒
3c 2 + 1 =
10 − 2
=8
2 −1
⇒
(1)
∴
f ′(c) = 3c 2 + 1
3c 2 − 7 = 0
But c = −1.53 ∉ (1, 2) and c = 1.53 ∈(1, 2)
⇒
∴c=
c2 =
7
7
⇒ c=±
= ±1.53 .
3
3
7
3
EXAMPLE 2
Using Lagrange’s mean value theorem prove that tan21b 2 tan21 a < b 2 a, where b > a.
Solution.
⎛ p p⎞
Consider f ( x ) = tan −1 x , x ∈[a, b] . Clearly [a, b] ⊆ ⎜ − , ⎟
⎝ 2 2⎠
1
, f ( x ) is continuous on [a, b] and
Since tan −1 x is continuous on [a, b] and f ′( x ) =
1+ x 2
differentiable in (a, b).
So, the condition of Lagrange’s mean value theorem are satisfied.
∴ by Lagrange’s mean value theorem, there exists c ∈(a, b) such that
f ′(c ) =
M03_ENGINEERING_MATHEMATICS-I _CH03_Part B.indd 50
f (b ) − f (a)
.
b −a
(1)
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Differential Calculus ■
3.51
Here a = a, b = b.
∴
f (a) = tan −1 a, f (b) = tan −1 b and f ′(c) =
∴ (1) ⇒
1
tan −1 b − tan −1 a
=
b−a
1 + c2
Since
0<
⇒
1
1 + c2
tan −1 b − tan −1 a =
b−a
1 + c2
b−a
1
< b − a.
< 1 and b − a > 0, we have 0 <
2
1 + c2
1+ c
tan −1 b − tan −1 a < b − a if b > a.
∴
Note In particular, if a = 0, b = x, then tan −1 x − tan −1 0 < x − 0,
EXAMPLE 3
If 0 < a < b, then prove that
(i)
⇒
tan −1 x < x, if x > 0.
b2a
b2a
< tan21 b 2 tan21 a <
. Deduce that
2
11 b
11 a 2
p
4 p 1
3
1
< tan21 < 1
3 4 6
4 25
(ii)
p 1
p 1
1 < tan21 2 < 1 .
4 5
4 2
Solution.
Consider the function f ( x ) = tan −1 x , x ∈[a, b ] . Then f ′( x ) =
1
, x ∈ (a, b ).
1+ x 2
Clearly f ( x ) is continuous on [a, b ] and differential in (a, b ).
So, the conditions of Lagrange’s mean value theorem are satisfied.
∴ by Lagrange’s mean value theorem, there exists c ∈( a, b), such that
f ′(c) =
where f ( a) = tan −1 a, f (b) = tan −1 b
∴ (1) ⇒
and
f ( b) − f ( a)
b−a
f ′(c ) =
1
tan −1 b − tan −1 a
=
2
1+ c
b−a
⇒
(1)
1
1+ c2
b −a
= tan −1 b − tan −1 a
1+ c2
(2)
Since a < c < b and a, b, c are positive numbers, a 2 < c 2 < b 2 .
∴
1 + a2 < 1 + c2 < 1 + b2
We have b > a. ∴ b − a > 0.
M03_ENGINEERING_MATHEMATICS-I _CH03_Part B.indd 51
⇒
1
1
1
>
>
.
2
2
1+ a
1+ c
1 + b2
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b−a b−a b−a
<
<
1 + b2 1 + c2 1 + a2
∴
b−a b−a b−a
>
>
1 + a2 1 + c2 1 + b2
∴
b −a
b −a
< tan −1 b − tan −1 a <
2
1+ b
1 + a2
⇒
[using (2)](3)
4
in (3)
3
To deduce (i): Put a = 1 and b =
4
4
−1
−1
−1 4
−1
3
< tan
− tan 1 < 3
16
3
1+1
1+
9
∴
3
4 p 1
< tan −1 − <
25
3 4 6
⇒
p 3
4 p 1
+
< tan −1 < + .
4 25
3 4 6
⇒
To deduce (ii): Put a = 1 and b = 2 in (3)
2 −1
2 −1
< tan −1 2 − tan −1 1 <
1+ 4
1+1
∴
1
p 1
< tan −1 2 − <
5
4 2
⇒
⇒
p 1
p 1
+ < tan −1 2 < + .
4 5
4 2
EXAMPLE 4
If f : [0, 4] → R
for a , b ∈ (0, 4) .
is
differentiable,
then
prove
that
[ f (4)] 2 2 [ f (0)] 2 5 8f ′( a ) f (b )
Solution.
Given f is a real function defined on [0, 4] and differentiable and so it is continuous on [0, 4] .
So, the conditions of Lagrange’s mean value theorem are satisfied.
∴ by Lagrange’s mean value theorem, there exists a ∈(0, 4) , such that
f ′(a) =
f ( 4) − f ( 0) .
4−0
⇒
f ( 4) − f (0) = 4f ′(a) .
(1)
Since f is continuous on [0, 4], it takes all values between its maximum and minimum values.
f ( 4) + f ( 0)
In particular, there is b ∈(0, 4) such that
= f ( b) .
2
⇒
f ( 4) + f (0) = 2f (b )
(1) × ( 2) ⇒
(2)
[f ( 4) − f (0)][f ( 4) + f (0)] = 4f ′(a)2f (b )
⇒
M03_ENGINEERING_MATHEMATICS-I _CH03_Part B.indd 52
[f ( 4)]2 − [f (0)]2 = 8f ′(a)f (b ),
a, b ∈ (0, 4) .
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3.53
3.3.3 Cauchy’s Mean Value Theorem
Let f and g be two functions defined on [a, b] such that
(i) f and g are continuous on [a, b]
and (iii) g ′( x ) ≠ 0 ∀x ∈ (a, b ) .
(ii) f and g are differentiable in (a, b)
f ′( c ) f ( b ) 2 f ( a )
.
5
g ′( c ) g ( b ) 2 g ( a )
F ( x ) = f ( x ) + A g ( x ), x ∈[a, b ]
Then, there exists at least one c ∈(a, b ) such that
Proof
Consider the function
F ( a) = F ( b)
where A is chosen such that
f (a) + A g (a) = f (b ) + A g (b )
⇒
⇒
A [g (a) − g (b )] = f (b ) − f (a)
⇒
A=
f (b ) − f (a)
[f (b ) − f (a)]
⇒ A =−
g (a) − g (b )
g (b ) − g (a)
(1)
Since f and g are continuous on [a, b] and differentiable in (a, b), F is continuous on [a, b] and
differentiable in (a, b) and F(a) = F(b).
So, F satisfies the conditions of Rolle’s theorem.
∴ by Rolle’s theorem, there exists c ∈(a, b ) such that F ′(c) = 0.
But
F ′( x ) = f ′( x ) + Ag ′( x ) ⇒ F ′(c) = f ′(c) + Ag ′(c)
∴
F ′(c) = 0
∴
Corollary:
⇒ f ′(c ) + A g ′(c ) = 0 ⇒ f ′(c) = − Ag ′(c) ⇒
f ( b) − f ( a)
f ′( c)
=
g ( b) − g ( a)
g ′( c)
f ′( c )
= −A
g ′( c )
[using (1)].
If f (a) = 0, g(a) = 0, then
f ′(c ) f (b )
=
,
g ′(c ) g (b )
a < c < b.
Note The corollary can be algebraically interprepted as below:
If the polynomial equations f (x) = 0, g(x) = 0 have a common root a, then
f ( x ) f ′( c )
=
g ( x ) g ′( c )
for some c ∈(a, x ) .
Formula:
f1 ( x ) f 2 ( x ) f 3 ( x )
1. If F ( x ) = g1 ( x ) g 2 ( x ) g3 ( x )
h1 ( x ) h2 ( x ) h3 ( x )
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where fi , gi , hi , i = 1, 2, 3 are differentiable functions of x, then
f 1′( x ) f 2′( x ) f 3′( x ) f 1 ( x ) f 2 ( x ) f 3 ( x ) f 1 ( x ) f 2 ( x ) f 3 ( x )
F ′( x ) = g1 ( x ) g 2 ( x ) g 3 ( x ) + g1′( x ) g 2′ ( x ) g 3′ ( x ) + g1 ( x ) g 2 ( x ) g 3 ( x )
h1 ( x ) h2 ( x ) h3 ( x ) h1 ( x ) h2 ( x ) h3 ( x ) h1′( x ) h2′ ( x ) h3′ ( x )
2. If F ( x ) =
f1 ( x )
f2 ( x)
g1 ( x ) g 2 ( x )
, then F ′( x ) =
f 1′( x ) f 2′( x ) f 1 ( x ) f 2 ( x )
+
.
g1 ( x ) g 2 ( x ) g1′( x ) g 2′ ( x )
The formula can be used column wise also.
WORKED EXAMPLES
EXAMPLE 1
Verify Cauchy’s mean value theorem for f(x) 5 x3, g(x) 5 x2 in [1, 2].
Solution.
Given
f (x) = x3
and
g(x) = x2
in [1, 2].
f (x), g(x) being polynomials, they are continuous on [1, 2] and differentiable in (1, 2)
and g ′( x ) = 2 x ≠ 0 for ∀x ∈(1, 2)
So, all the conditions of Cauchy’s mean value theorem are satisfied.
∴by Cauchy’s mean value theorem, there exists c ∈(1, 2) such that
Here a = 1, b = 2.
∴
f ′( c ) f ( b) − f ( a)
=
.
g ′( c ) g ( b) − g ( a)
f (a) = 1
f (b) = 23 = 8,
and
f ′( x ) = 3 x 2 and
∴(1) ⇒
3c 2 8 − 1 7
=
=
2c 4 − 1 3
(1)
g(a) = 1
g ′( x ) = 2 x ∴ f ′(c) = 3c 2
⇒
3
7
c=
2
3
⇒
c=
and
g(b) = 22 = 4
and g ′(c) = 2c
14
∈(1, 2)
9
Hence, the theorem is verified.
EXAMPLE 2
Verify Cauchy’s mean value theorem and find c, if f ( x ) 5 e x , g ( x ) 5 e 2x , x ∈[ a, b ].
Solution.
Given
f (x ) = e x
and
g ( x ) = e − x , x ∈[a, b ] .
Both f and g are continuous on [a, b] and differentiable in (a, b) and g ′( x ) = −e − x ≠ 0∀x ∈( a, b) .
So, all the conditions of Cauchy’s mean value theorem are satisfied.
∴ by Cauchy’s mean value theorem, there exists c in (a, b) such that
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f ′( c ) f ( b) − f ( a)
.
=
g ′( c ) g ( b) − g ( a)
But
∴
(1) ⇒
∴
f(a) = ea
and
f(b) = eb,
f ′( x ) = e x
and
g ′( x ) = −e − x
ec
eb − ea
=
−e − c e − b − e − a
2c = a + b
∴
(1)
g ( a) = e − a
and
g ( b) = e − b
f ′(c) = e c
and g ′(c ) = −e − c
eb − ea
eb − ea a b
a +b
= a
⋅ e ⋅ e = −e
b
1 1
e −e
−
eb ea
a+b
c=
∈( a, b)
2
⇒
−e 2c =
⇒
3.55
⇒
e 2c = e a+ b
Hence, the theorem is verified.
EXAMPLE 3
sin a 2 sin b
p
, using Cauchy’s mean value theorem, prove that
5 cotu for
cos b 2 cos a
2
some u ∈( a, b).
If 0 < a < b <
Solution.
⎛ p⎞
Consider the functions f(x) = sin x and g(x) = cos x, x ∈[ a, b], where [ a, b] ⊆ ⎜ 0, ⎟
⎝ 2⎠
Both the functions f and g are continuous on [ a, b] and differentiable in ( a, b) and g ′( x ) = − sin x ≠ 0
in ( a, b) .
So, all the conditions of Cauchy’s mean value theorem are satisfied.
∴ by Cauchy’s mean value theorem there exists u ∈( a, b) such that
f ′(u) f (b) − f ( a)
=
g ′(u) g (b) − g ( a)
But
f (a) = sin a
f ′( x ) = cos x
(1)
f (b) = sin b ;
and
and g ′( x ) = − sin x . ∴
∴ (1) ⇒
cos u
sin b − sin a
=
− sin u cos b − cos a
⇒
− cot u =
sin b − sin a
cos b − cos a
⇒
g (a) = cos a and
g (b) = cos b
f ′(u) = cos u
g ′(u) = − sin u
sin a − sin b
= cot u , u ∈[a, b] .
cos b − cos a
EXAMPLE 4
Using mean value theorem, show that x > log e (1 + x ) > x −
Solution.
Consider the function
and
x2
if x > 0.
2
f(x) = loge(1 + x), x ≥ 0
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1
1
,
f ′′( x ) = −
, x>0
1+ x
(1 + x ) 2
We know that
f(x) = f(0) + xf ′(ux), 0 < u < 1
1
∴
f(0) = log(1 + 0) = 0, f ′(ux ) =
1+ ux
1
x
∴
loge(1 + x) = 0 + x
=
1 + ux 1 + ux
1
x
<1 ⇒
<x
Since u > 0, x > 0, 1 + ux > 1 ⇒
1 + ux
1 + ux
∴
loge(1 + x) < x ⇒ x > loge(1 + x)
∴
f ′( x ) =
Again f ( x ) = f (0) + xf ′(0) +
But
∴
⇒
(1) ⇒
∴
f ′(0) = 1,
(2)
2
x
f ′′(ux ) , up to second derivative 0 < u < 1
2
f ″(ux) = −
1
(1 + ux ) 2
loge (1 + x ) = 0 + x ⋅1 +
loge (1 + x) − x = −
loge(1 + x) – x > −
x2
2
⎛
⎞
x2
1
=
x
−
,0 < u < 1
⎜−
⎟
⎜⎝ (1 + ux )2 ⎟⎠
2(1 + ux ) 2
x2
2(1 + ux ) 2
x
x2
<x⇒
< x2
1 + ux
(1 + ux ) 2
From (2) and (3), we get
(1)
⇒−
x2
x2
>
−
2
2(1 + ux ) 2
x2
x2
⇒ loge (1 + x ) > x −
2
2
x > loge (1 + x ) > x −
(3)
x2
if x > 0.
2
EXERCISE 3.7
I. Verify Rolle’s theorem for the following functions.
1⎤
⎡
3
2
1. f ( x ) = 2x + x − 4 x − 2, x ∈ ⎢ − 2 , − ⎥ .
2⎦
⎣
⎡ p p⎤
x
2. f ( x ) = e cos x , x ∈ ⎢ − , ⎥
⎣ 2 2⎦
2
3. f ( x ) = 4 − x , x ∈[ −2, 2]
2
4. f ( x ) = sin x , x ∈[0, p]
⎧x 2 + 1, 0 ≤ x ≤ 1
6. f ( x ) = ⎨
⎩3 − x , 1 < x ≤ 2
m
n
7. f ( x ) = ( x − a) ( x − b ) where m and n are positive integers.
5
4
5. f ( x ) = 1 − x , x ∈[ −1, 1]
8. f ( x ) = x 2 n −1 (a − x ) 2 n in [0, a].
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9. f ( x ) = loge
x 2 + ab
in [a, b], a, b > 0.
x (a + b )
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⎡ p 5p ⎤
II. 1. Find c of Rolle’s theorem for the function f ( x ) = e x (sin x − cos x ), x ∈ ⎢ , ⎥
⎣4 4 ⎦
2. Considering the function f ( x ) = ( x − 2) loge x , show that the equation x log e x = 2 − x has a root
between 1 and 2.
p
3. Apply Rolle’s theorem for sin x cos 2x and find x such that 0 < x < .
4
III. Verify Lagrange’s mean value theorem for the following functions.
1. f ( x ) = ( x − 1)( x − 2)( x − 4), x ∈[0, 4]
23
2. f ( x ) = x , x ∈[ −1, 1]
1
⎧
⎪x cos , x ≠ 0
in [ −1, 1]
3. f ( x ) = ⎨
x
⎪⎩0,
x =0
4. f (x) = loge x in [1, e].
IV.
x
1. Verify Lagrange’s mean value theorem and find the point on the curve y =
between the
1− x
⎛ −5 ⎞
points A(2, −2) and B ⎜ 5, ⎟ at which the tangent is parallel to the chord AB.
⎝ 4⎠
2. Prove that
p
1
3 p 1
−
> cos −1 > − by Lagrange’s mean value theorem.
3 5 3
5 3 8
3. For any two real numbers a and b (a < b), prove that a2 + ab + b2 = 3c2, for some c ∈ (a, b) using
Lagrange’s mean value theorem.
4. For the quardratic function f ( x ) = lx 2 + mx + n , x ∈ (a, b ), find u of Lagrange’s mean value
theorem.
5. If f (x) and g(x) are continuous on a ≤ x ≤ b and derivable in a < x < b, then prove that
f ( a)
g ( a)
f ( b)
f ( a)
= ( b − a)
g ( b)
g ( a)
f ′( c )
.
g ′( c )
[Hint: Consider f(x) =
f ( a)
g ( a)
f ( x)
, x ∈ [a, b].
g( x)
x =0
⎧3,
⎪ 2
6. For what values of a, m, b does the function f ( x ) = ⎨ − x + 3x + a, 0 < x < 1
⎪mx + b ,
1≤ x ≤ 2
⎩
satisfy the hypothesis of Lagrange’s mean value theorem on the interval [0, 2].
2
7. Find u of Lagrange’s mean value theorem for f ( x ) = 3x − 2x + 4, x ∈[2, 3]
V. Using Lagrange’s mean value theorem, prove the following
1.
x
< loge (1 + x ) < x , x > 0
1+ x
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2. 0 <
1
1
− < 1, x > 0
loge (1 + x ) x
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1
ex −1
loge
< 1, x > 0
x
x
x
[Hint: Take f ( x ) = e in [0, x]]
1
4. If f ( x ) = cos x, applying Lagrange’s mean value theorem in [0, h ] prove that lim u = .
x →0+
2
[Hint: f ( h) = f (0) + hf ′(uh), 0 < u < 1 ]
ANSWERS TO EXERCISE 3.7
II. 1. c = p,
IV. 1.
3.4
3⎞
⎛
⎜⎝ 3, − ⎟⎠
2
p
6
1
4. u =
2
3. x =
6. a = 3, b = 4, m = 1
7. u =
1
2
MONOTONIC FUNCTIONS
Monotonic functions form an important class of functions in Mathematics. Because most of the
functions that occur in various fields, in practice, are monotonic functions or sum of monotonic
functions.
It will be seen that all bounded monotonic functions are integrable. In this section, we use mean
value theorems to deduce the properties of monotonic functions, using sign of derivative.
3.4.1 Increasing and Decreasing Functions
Definition 3.1 Let f be a function defined on [a, b ] .
If for every pair of points x 1 , x 2 ∈[a, b ]
(i) x 1 < x 2 ⇒ f ( x 1 ) ≤ f ( x 2 ), then f is increasing (or non−decreasing) on [a, b].
(ii) x 1 < x 2 ⇒ f ( x 1 ) < f ( x 2 ), then f is strictly increasing on [a, b].
(iii) x1 < x2 ⇒ f ( x1 ) ≥ f ( x2 ) , then f is decreasing (or non−increasing) on [a, b].
(iv) x1 < x2 ⇒ f ( x1 ) > f ( x2 ) , then f is strictly decreasing on [a, b].
A function f which is either increasing or decreasing on [a, b] is called a monotonic function on [a, b].
If f is strictly increasing or strictly decreasing on [a, b], then f is called strictly monotonic.
3.4.2 Piece−wise Monotonic Function
Definition 3.2 A function f is said to be piece−wise monotonic on [a, b] if the interval [a, b] can
be partitioned into finite number of sub−intervals such that in each of the open sub−intervals f is
monotonic
That is f is piece−wise monotonic on [a, b] if its graph consists of a finite number of monotonic
pieces.
Note
(1) A characteristic property of monotonic function is that it has it finite left and right limits at each
interior point.
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y
o
y
a
b
x
o
x
Increasing
Strictly increasing
Fig. 3.10
Fig. 3.11
y
o
y
a
D
Strictly increasing
x
o
a
b
Decreasing
x
Fig. 3.13
Fig. 3.12
y
o
3.59
y
a
b
b
x
a
o
Strictly decreasing
d
c
e
x
Piece-wise monotonic
Fig. 3.14
Fig. 3.15
If f is increasing on [a, b] and c ∈(a, b ), then f (c − ) and f (c + ) exist and
f (c −) ≤ f (c ) ≤ f (c + )
Further
f (a) ≤ f (a+ )
and
f (b −) ≤ f (b ) .
(2) If f is strictly increasing and continuous on [a, b], then f : [a, b ] → [f (a), f (b )] is bijective and
f −1 : [f (a), f (b )] → [a, b ] exists and f −1 is continuous and strictly increasing on [ f (a), f (b )].
Similarly, we can state (1) and (2) for decreasing functions.
3.4.3 Test for Increasing or Decreasing Functions
Theorem 3.2 Let f be continuous on the closed interval [ a , b ] and let f 9(x) exist for each point
x ∈( a , b ) .
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(i) If f ′( x ).0 ; x e( a , b ) , then f is strictly increasing on [a, b].
(ii) If f ′( x ) ≥ 0 ; x e( a , b ) , then f is increasing on [a, b].
(iii) If f ′( x ) < 0 ; x e( a , b ) , then f is strictly decreasing on [a, b].
(iv) If f ′( x ) ≤ 0 ; x e( a , b ) , then f is decreasing on [a, b].
(v) If f ′( x ) 5 0 ; x e( a , b ), then f is constant on [a, b].
Proof
Given f is continuous on [a, b] and f ′( x ) exists for each x ∈(a, b ).
Let x 1 , x 2 ∈[a, b ] be any two points with x 1 < x 2.
Applying Lagrange’s mean value theorem for f on [x 1 , x 2 ], there is c ∈( x 1 , x 2 ) such that
f ′(c ) =
⇒
f (x 2 ) − f (x1 )
x 2 − x1
f ( x 2 ) − f ( x 1 ) = ( x 2 − x 1 ) f ′(c )
(1)
(i) Let f ′(c ) > 0
Since f ′(c ) > 0 and x 2 − x 1 > 0 , we have from (1)
f (x 2 ) − f (x1 ) > 0
Thus, x 1 < x 2
⇒ f (x 2 ) > f (x1 )
⇒
f (x1 ) < f (x 2 )
⇒ f ( x 1 ) < f ( x 2 ) So, by definition, f is strictly increasing on [a, b]
(ii) Let f ′(c ) ≥ 0
Since f ′(c ) ≥ 0 and x 1 − x 2 > 0 , we have from (1)
f (x 2 ) − f (x1 ) ≥ 0 ⇒ f (x 2 ) ≥ f (x1 )
Thus,
⇒
f (x1 ) ≤ f (x 2 )
x 1 < x 2 ⇒ f ( x 1 ) ≤ f ( x 2 ) By definition, f is increasing on [a, b].
Similarly, we can prove (iii), (iv) and (v).
Corollary:
If f and g are continuous functions on [a, b] such that f ′( x ) = g ′( x )∀x ∈ ( a, b), then
f ( x ) − g ( x ) = k ∀ x ∈[a, b ], where k is a constant.
WORKED EXAMPLES
EXAMPLE 1
Determine the interval of monotonicity of the following functions
(i) 2 x 3 2 9 x 2 2 24 x 1 7
(ii) 2 x 2 2 lnx.
Solution.
(i) Let
∴
f ( x ) = 2x 3 − 9x 2 − 24 x + 7
f ′( x ) = 6 x 2 − 18x − 24 = 6( x 2 − 3x − 4)
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∴
f ′( x ) > 0
if
( x − 4)( x + 1) > 0
x < −1
⇒
or
3.61
x > 4.
∴ the function is strictly increasing in the interval ( −∞, −1] and [4, ∞).
⇒ f ′( x ) < 0
if
( x − 4)( x + 1) < 0
⇒
−1 < x < 4
−∞
−1
x
4
∞
∴ the function is strictly decresing in the interval −1 ≤ x ≤ 4.
(ii) Let g ( x ) = 2 x 2 − 1n x, x > 0
and
g ′( x ) > 0
if
4x −
∴ g ′( x ) = 4 x −
1
>0
x
⇒
1
x
4x 2 − 1 > 0
Since x > 0, 2x + 1 > 0 and so, 2x − 1 > 0
[{ 1nx = loge x ]
⇒
⇒ x>
( 2x + 1)( 2x − 1) > 0, x > 0 .
1
.
2
⎡1 ⎞
∴ the function is strictly increasing in the interval ⎢ , ∞⎟ .
⎣2 ⎠
1
Now
g ′( x ) < 0 if 4 x − < 0 ⇒ 4 x 2 − 1 < 0 ⇒
x
Since x > 0, 2x + 1 > 0 and so, 2 x − 1 < 0 ⇒ x <
( 2 x + 1)( 2 x − 1) < 0
1
.
2
1
∴ the function is strictly decreasing in the interval 0 < x ≤ .
2
EXAMPLE 2
1
⎛ p⎞
Prove that x sin x 1 cos x 1 cos 2 x is strictly increasing in ⎜ 0, ⎟ .
⎝ 2⎠
2
Solution.
p
1
Let f ( x ) = x sin x + cos x + cos 2 x , 0 ≤ x ≤ .
2
2
∴
1
f ′( x ) = x cos x + sin x ⋅1 − sin x + ⋅ 2 cos x ( − sin x ) = x cos x − cos x sin x
2
⇒
f ′( x ) = cos x ( x − sin x ),
0<x <
p
.
2
p
, cos x > 0 and to decide the sign of x − sin x , we consider the function
2
p
x
p
∴ g ′( x ) = 1 − cos x = 2 sin 2 > 0 in 0 < x < .
g ( x ) = x − sin x in 0 ≤ x ≤
2
2
2
p
∴ g ( x ) is strictly increasing in 0 ≤ x ≤ .
2
For 0 < x <
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⎛ p⎞
g ( x ) > g (0) for x > 0, x ∈⎜ 0, ⎟ .
⎝ 2⎠
∴
p
2
p
⎛ p⎞
f ′( x ) > 0 in 0 < x < . ∴ f ( x ) is strictly increasing in ⎜ 0, ⎟ .
⎝ 2⎠
2
But g(0) = 0 − sin 0 = 0
Hence,
∴
g ( x ) > 0 for x > 0, in 0 < x <
EXAMPLE 3
⎧ x
,
⎪
If f ( x ) 5 ⎨ sin x
⎪1,
⎩
0<x≤
p
2
x 50
x
p
⎡ p⎤
then prove that (i) f is strictly increasing in ⎢0, ⎥ and (ii) 1 <
< .
2
sin
x
2
⎣
⎦
Solution.
p
⎧ x
, 0<x ≤
⎪
Given f ( x ) = ⎨ sin x
2
⎪⎩1,
x =0
lim f ( x ) = lim
x →0+
x →0+
x
=1
sin x
and
f (0) = 1 ⇒ lim f ( x ) = f (0)
x →0+
⎡ p⎤
∴ the function is continuous at x = 0 from the right and hence continuous in the interval ⎢0, ⎥ .
⎣ 2⎦
f ′( x ) =
Now
For 0 < x <
sin x ⋅1 − x ⋅ cos x
sin 2 x
=
cos x
[tan x − x ] ,
sin 2 x
p
.
2
0<x <
p cos x
,
> 0 and to decide the sign of tan x − x , we consider the function
2 sin 2 x
g ( x ) = tan x − x in 0 ≤ x <
∴ g ( x ) is strictly increasing in 0 ≤ x <
∴ g ′( x ) = sec 2 x − 1 = tan 2 x > 0
p
.
2
∴ g (x ) > 0
But g(0) = tan0 − 0,
Hence, f ′( x ) > 0 for 0 < x <
p
2
p
. ∴
2
M03_ENGINEERING_MATHEMATICS-I _CH03_Part B.indd 62
in
0<x <
p
.
2
So, g ( x ) > g (0) for x > 0.
⇒
tan x − x > 0 for 0 < x <
f ( x ) is strictly increasing in 0 ≤ x <
p
.
2
p
.
2
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Differential Calculus ■
p
Hence, 0 < x <
2
⇒
⎛ p⎞
f ( 0) < f ( x ) < f ⎜ ⎟
⎝ 2⎠
⇒
x
1<
<
sin x
p
2
p
sin
2
⇒
1<
3.63
x
p
< .
sin x 2
EXAMPLE 4
Use the function f ( x ) 5 x 1/ x , x > 0, to determine the bigger of the two numbers
(i) e p and pe
(ii) (202) 303 and (303) 202 .
Solution.
Given
f ( x ) = x 1 x , x > 0.
Taking logarithm to the base e on both sides, we get
∴ Differentiating w.r.to x,
1
loge x .
x
1
1 1
⎛ 1⎞
f ′( x ) = ⋅ + loge x ⎜ − 2 ⎟
⎝ x ⎠
f (x )
x x
f ′( x ) = f ( x ) ⋅
⇒
f ′( x ) < 0
f ′( x ) 1
=
[1 − loge x ]
f (x ) x 2
1
x 1/x
1
[
−
log
x
]
=
[1 − loge x ].
e
x2
x2
x 1/x
[1 − loge x ] < 0 ⇒
x2
∴ f(x) is strictly decreasing for all x ≥ e .
∴
⇒
if
1 − loge x < 0
⇒
loge x > 1 ⇒ x > e .
( e > 1)
∴
loge f ( x ) = loge x 1 x =
(i) Since p > e, we have
f (p) < f (e )
⇒
p1/p < e 1/e
⇒
pe < e p
Raising to the power pe on both sides we have
(p1/p )pe < (e 1/e )pe
∴ e p is the bigger number.
(ii) Since 303 > 202, we have
f (303) < f (202)
⇒
[(303)1/303 ]( 303)( 202 ) < [( 202)1/202 ]( 303)( 202 )
1/303
< ( 202)1/202
⇒ (303)
⇒
(303) 202 < ( 202)303
∴ ( 202)303 is the bigger number.
EXAMPLE 5
Prove that
tan x 2 x 2
p
>
for 0 < x 1 < x 2 < .
tan x 1 x 1
2
Solution.
⎛ p⎞
We know that for x 1 , x 2 ∈ ⎜ 0, ⎟ , tanx1, tanx2 are positive.
⎝ 2⎠
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tan x 2 x 2
>
tan x 1 x 1
∴
⇒
tan x 2 tan x 1
>
x2
x1
To prove this inequality, consider the function
tan x
p
, 0<x < .
x
2
x sec 2 x − tan x ⋅1
f ′( x ) =
x2
f (x ) =
∴
1
p
sec 2 x ⎡
sin x
sec 2 x ⎡
⎤
2 ⎤
x
x
x − sin 2x ⎥ , 0 < x < .
−
cos
=
2
2
⎢
⎥
⎢
2
2
cos x
x ⎣
x ⎣
⎦
⎦
p sec 2 x
1
For 0 < x < ,
> 0 and to decide the sign of x − sin 2x, we consider the function
2
2
x2
=
1
p
g ( x ) = x − sin 2x , 0 ≤ x ≤ .
2
2
1
⎛ p⎞
g ′( x ) = 1 − ⋅ 2 cos 2x = 1 − cos 2x = 2 sin 2 x > 0 ∀x ∈ ⎜ 0, ⎟ .
⎝ 2⎠
2
p
∴ g(x) is strictly increasing in 0 ≤ x ≤ .
∴ g(x) > g(0) for x > 0
2
1
1
⎡ p⎤
But g(0) = 0 − sin 0 = 0.
∴ g ( x ) > 0 for x > 0 ⇒ x − sin 2x > 0 for x > 0 in ⎢0, ⎥
2
2
⎣ 2⎦
∴
Hence, f ′( x ) > 0 for 0 < x <
∴
x1 < x 2
⇒
tan x 1 tan x 2
<
x1
x2
p
p
. ∴ f (x) is strictly increasing in 0 < x < .
2
2
⎛ p⎞
⇒ f ( x 1 ) < f ( x 2 ) in ⎜ 0, ⎟
⎝ 2⎠
⇒
EXAMPLE 6
Show that x > loge (1 1 x) >
Solution.
First we shall prove
Consider
x 2 tan x 2
<
x 1 tan x 1
⇒
tan x2 x2
p
>
for 0 < x1 < x2 <
tan x1 x1
2.
x
for x > 0.
11 x
x > loge(1 + x), x > 0.
f ( x ) = x − loge (1 + x ), x ≥ 0
1
1+ x −1
x
=
=
> 0 for x > 0.
1+ x
1+ x
1+ x
∴ f(x) is strictly increasing for x ≥ 0.
∴
f ′( x ) = 1 −
Hence, f (x) > f (0) for x > 0. But f (0) = 0 − loge (1 + 0) = 0.
∴
f ( x ) > 0 for x > 0
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Differential Calculus ■
∴
x − loge (1 + x ) > 0 for x > 0 ⇒ x > loge (1 + x ) for x > 0
x
Now we shall prove loge (1 + x ) >
for x > 0.
1+ x
Consider g ( x ) = loge (1 + x ) −
∴
g ′( x ) =
3.65
(1)
x
, x ≥ 0.
1+ x
1
(1 + x ) ⋅1 − x .1
1
1
1+ x −1
x
−
=
−
=
=
> 0 for x > 0
2
2
2
1+ x
1 + x (1 + x )
(1 + x )
(1 + x )
(1 + x ) 2
∴ g(x) is strictly increasing for x ≥ 0.
∴
g ( x ) > g (0) for x > 0. But
∴
g(x) > 0 for x > 0 ⇒
⇒
g(0) = loge(1 + 0) − 0 = 0.
log e (1 + x ) −
x
> 0 for x > 0
1+ x
log e (1 + x ) >
x
1+ x
for x > 0
x > log e (1 + x ) >
x
1+ x
for x > 0.
(2)
From (1) and (2), we get
EXERCISE 3.8
1. Using sign of derivative, prove the following
1
x
<
if 0 < x < 1
(a) x < log
1− x 1− x
(b)
2 sin x
⎛ p⎞
<
< 1 in ⎜ 0, ⎟
⎝ 2⎠
p
x
(c) x −
x3
< sin x < x for x > 0
6
(e) x −
x2 x3
x2
x3
for x > 0
+
> loge (1 + x ) > x −
+
2
3
2 3(1 + x )
(d) 1 + x < x < 1 + xe x for x > 0
2. Determine the intervals of increase and decrease for the following functions
(a) f ( x ) = x 3 + 2x − 5
(b) f ( x ) = log(1 − x 2 )
(c) f ( x ) = cos x − x
3. Prove that (a + b ) n = an + b n if 0 ≤ n ≤ 1 and a > 0, b > 0.
⎡
n
n
⎢ Hint: Take f ( x ) = 1 + x − (1 + x ) , x ≥ 0, then put x =
⎣
(
a⎤
.
b ⎥⎦
)
4. Show that 1+ x In x + x 2 + 1 ≥ 1 + x 2 ∀x ≥ 0.
5. Prove the inequality tan x > x +
p
x3
if 0 < x < .
2
3
⎛ p p⎞
6. Find the behaviour of the function f ( x ) = 2 sin x + tan x − 3x in ⎜ − , ⎟ .
⎝ 2 2⎠
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ANSWERS TO EXERCISE 3.8
2. (a) Increases ∀x ∈ R
(b) Increases in (−1, 0) and decreases in (0, 1)
(c) Increases ∀x ∈ R
3.5
6. Decreases
GENERALISED MEAN VALUE THEOREM
In many applications it is useful to approximate a continuous function by a polynomial function
which is the simplest continuous function. Taylor’s and Maclaurin’s theorems are important tools
which provide such an approximation for real functions. Mean value theorems relate the value of the
functions and its first order derivative, where as, Taylor’s and Maclaurin’s theorems generalise this
relation to higher order derivatives. Hence these theorems can be considered as “Generalised mean
value theorems”.
3.5.1 Taylor’s Theorem with Lagrange’s form of Remainder
If f is a real function defined on [a, a + h ] such that
(i) The ( n −1)th derivative f
and (ii) The n th derivative f
that
(n)
( n −1)
is continuous on [a, a + h ]
exists in [a, a + h ], then there exists a number u between 0 and 1 such
f ( a + h ) = f ( a) +
h
h2
h3
f ′(a) + f ′′(a) + f ′′′(a)
1!
2!
3!
n −1
hn
h
+ …+
f ( n −1) (a) + f ( n ) (a + uh ), 0 < u < 1
( n − 1)!
n!
Proof
Given f is a real function defined on [a, a + h ] such that the (n – 1)th derivative f
so, f , f ′, f ′′, …f ( n −1) are continuous.
( n −1)
is continuous and
Consider the function
f( x ) = f ( x ) + (a + h − x )f ′( x ) +
+…+
(a + h − x ) 2
(a + h − x ) 3
f ′′( x ) +
f ′′′( x )
2!
3!
( a + h − x ) n −1 ( n −1)
(a + h − x)n
( x) +
f
A
( n − 1)!
n!
where A is a constant to be determined such that
⇒
f ( a) +
f(a) = f(a + h ) .
n −1
h
h
h
hn
f ′( a) +
f ′′( a) + … +
f ( n −1) ( a) +
A = f ( a + h)
1!
2!
( n − 1)!
n!
Since f , f ′, f ′′, …f
2
(1)
(2)
( n −1)
are continuous on [a, a + h ] and a + h − x, ( a + h − x ) 2 , …, ( a + h − x ) n are
continuous, we get f is continuous on [a, a + h ] .
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Differential Calculus ■
3.67
Further f , f ′, f ′′, …f ( n −1) and a + h − x, (a + h − x ) 2 , …(a + h − x ) n are derivable in (a, a + h ) and
also we get f is derivable in (a, a + h ).
Further f(a) = f(a + h ).
So, f satisfies the conditions of Rolle’s theorem.
∴ by Rolle’s theorem, there exists a c ∈ (a, a + h ) such that f′(c ) = 0.
Since a < c < a + h , we can write c = a + uh , 0 < u < 1.
∴
Now
f ′ ( a + u h ) = 0,
f′( x ) = f ′( x ) − f ′( x ) + (a + h − x )f ′′( x ) − (a + h − x )f ′′( x ) + …
=
+
(a + h − x ) n −2
f
( n − 2)!
+
(a + h − x ) n −1
f
( n − 1) !
(a + h − x ) n −1
f
( n − 1)!
∴
⇒
0 < u < 1.
(n)
(x ) −
( n −1)
(n)
(x ) −
(x ) −
(a + h − x ) n − 2
f
( n − 2)!
( n −1)
(x )
n (a + h − x ) n −1
A
n!
(a + h − x ) n −1
A
( n − 1)!
f ′ ( a + uh ) = 0
( a + h − a − uh )
( n − 1)!
n −1
f
(n)
(a + h − a − uh ) n −1 A
=0
( n − 1)!
( h − uh ) n −1 ( n )
( h − uh ) n −1 A
f ( a + uh ) =
( n − 1)!
( n − 1)!
( a + uh ) −
⇒
⇒
Substituting in (2), we get
f ( a + h ) = f ( a) +
A =f
(n)
(a + uh ).
h
h2
h3
f ′(a) + f ′′(a) + f ′′′(a) + …
1!
2!
3!
n −1
hn
h
+
f ( n −1) (a) + f ( n ) (a + uh ),
( n − 1)!
n!
0 < u <1
This is called Taylor’s theorem.
Note
hn (n)
f (a + uh ), 0 < u < 1, is called the Lagrange’s Remainder after n terms in
n!
the Taylor’s series expansion of f (a + h ) and it is denoted by R n .
(2) If n = 1, then we get
f (a + h ) = f (a) + hf ′(a + uh ),
0 < u <1
(1) The ( n +1)th term
which is Lagrange’s mean value theorem.
(3) If we put a + h = x or h = x − a, that is if the interval is taken as [a, x ], then Taylor’s theorem takes
another form
M03_ENGINEERING_MATHEMATICS-I _CH03_Part B.indd 67
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( x − a)
( x − a) 2
( x − a)3
f ′( a) +
f ′′( a) +
f ′′′( a) + …
1!
2!
3!
( x − a) n ( n )
( x − a) n −1 ( n −1)
f
( a) +
f ( a + u( x − a))
0 < u <1
+
( n − 1)!
n!
f ( x ) = f ( a) +
This is called Taylor’s formula about the point a.
If the reminder R n → 0 as n → ∞, we get Taylor’s series.
3.5.2 Taylor’s Series
If f is a real function defined on [a, a + h ] such that
(i) f has derivatives of all orders in the interval [a, a + h ] and
(ii) the remainder after n terms
hn
f
n!
Rn =
then
(n)
f ( a + h ) = f ( a) +
(a + uh ) → 0 as n → ∞,
h
h2
hn
f ′(a) + f ′′(a) + … + f
n!
1!
2!
(n)
( a) + … ∞
This infinite series is called Taylor’s series.
Taylor’s series about the point a is
f ( x ) = f ( a) +
( x − a) 2
( x − a) n
( x − a)
f ′ ( a) +
f ′′(a) + … +
f
n!
1!
2!
(n)
( a) + … ∞
Instead of the interval [a, a + h ] , if the interval is [0, x], then we get Maclaurin’s theorem and
Maclaurin’s series.
3.5.3 Maclaurin’s Theorem with Lagrange’s Form of Remainder
If f is a real function defined on [0, x] such that
(i) The ( n −1) th derivative f ( n −1) is continuous on [0, x ] and (ii) The nth derivative f
then there exists a number u between 0 and 1 such that
f ( x ) = f ( 0) +
x
x2
x n −1
f ′(0) + f ′′(0) + … +
f
1!
2!
( n − 1)!
( n −1)
( 0) +
xn
f
n!
(n)
(n)
exists in [0, x ],
(ux ), 0 < u < 1.
If the reminder term R n → 0 as n → ∞, we get Maclaurin’s series.
3.5.4 Maclaurin’s Series
If f is a real function defined on [0, x ] such that (i) f has derivatives of all orders
and (ii) the reminder term
Rn =
xn
f
n!
M03_ENGINEERING_MATHEMATICS-I _CH03_Part B.indd 68
(n)
(uh ) → 0 as n → ∞,
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Differential Calculus ■
f ( x ) = f ( 0) +
then
x
x2
xn
f ′(0) + f ′′(0) + ⋅⋅⋅ + f
1!
2!
n!
(n)
3.69
(0) + ⋅⋅⋅
This infinite series is called Maclaurin’s series.
It is also known as the Maclaurin’s series expansion for f ( x ) in powers of x.
Remark: (1) It should be noted that technically there is no distinction between Taylor’s and Maclaurin’s
series. Each of which seeks to express the value of the function at any point (in an interval) interms
of the value of the various derivatives of the function at another point and the distance between the
two points.
(2) In order to expand a given function as an infinite series using Taylor’s series or Maclaurin’s
series, it is necessary to verify R n → 0 as n → ∞. However in practical situations we will be dealing
with functions involving trigonometric, exponential, logarithmic or algebraic functions which satisfy
this condition. Hence, we obtain the expansion formally assuming this condition.
WORKED EXAMPLES
EXAMPLE 1
Expand sin x as a finite series in powers of x, with remainder in Lagrange’s form. Hence find
the series for sin x.
Solution.
Let
f ( x ) = sin x .
∴
⎛p
⎞
f ′( x ) = cos x = sin ⎜ + x ⎟ ,
⎝2
⎠
⎛ 2p
⎞
f ′′( x ) = − sin x = sin ⎜
+ x⎟
⎝ 2
⎠
⎛ 3p
⎞
f ′′′( x ) = − cos x = sin ⎜
+ x ⎟ , …, f
⎝ 2
⎠
(n)
⎛ np
⎞
( x ) = sin ⎜
+ x⎟ .
⎝ 2
⎠
By Maclaurin’s theorem with Lagrange’s reminder, we have
f ( x ) = f ( 0) +
⇒
x
x2
x n −1
f ′(0) + f ′′(0) + ⋅⋅⋅ +
f
1!
2!
( n − 1)!
sin x = sin(0) +
Here
Rn =
∴
Rn =
( 0) +
xn
f
n!
(n)
(ux ), 0 < u < 1
2
3
x
⎛ 3p
⎞
⎛p ⎞ x
⎛ 2p ⎞ x
+ 0⎟ + …
sin ⎜ + 0⎟ + sin ⎜
+ 0⎟ + sin ⎜
⎝ 2
⎠
⎠ 2!
⎝ 2
⎠ 3!
1! ⎝ 2
+
=x−
( n −1)
n
x n −1
⎛ ( n − 1)p
⎞ x
⎛ np
⎞
sin ⎜
+ 0⎟ + sin ⎜
+ ux ⎟ , 0 < u < 1
⎠ n!
⎝ 2
⎠
( n − 1)! ⎝
2
p⎞ x n
x 3 x 5 … x n −1
⎞
⎛
⎛ np
sin ⎜ ( n − 1) ⎟ + sin ⎜
+
− +
+ ux ⎟ , 0 < u < 1.
⎠
⎝ 2
3! 5!
( n − 1)! ⎝
2 ⎠ n!
xn
⎛ np
⎞
sin ⎜
+ ux ⎟ and f is in [0, x ].
⎝ 2
⎠
n!
n
xn
⎛ np
⎞ x
sin ⎜
+ ux ⎟ ≤ ,
⎝ 2
⎠
n!
n!
M03_ENGINEERING_MATHEMATICS-I _CH03_Part B.indd 69
⎛ np
⎞
since sin ⎜
+ ux ⎟ ≤ 1.
⎝ 2
⎠
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■
3.70
Let u n =
∴
Engineering Mathematics
xn
x n +1
, then u n +1 =
.
n!
( n + 1)!
x
u n +1
x n +1 n !
=
⋅
=
un
( n + 1)! x n
( n + 1)
∴
lim
n →∞
x
u n +1
= lim
=0 ∴
n →∞ n + 1
un
R n → 0 as n → ∞ ∴
Hence,
lim u n = 0.
n →∞
sin x = x −
Result: If {u n } is a sequence such that lim
n →∞
[by the result quoted below]
x3 x5 x7 …
+
−
+
3! 5! 7 !
u n +1
= l , 0 < l < 1, then lim u n = 0.
n →∞
un
EXAMPLE 2
Find the Taylor’s series expansion of cos x about x 5
p
.
4
Solution.
Let f(x) = cosx
p
The Taylor’s series expansion of f(x) about x =
4
⎛ p⎞
is f(x) = f ⎜ ⎟
⎝ 4⎠
1⎛
p⎞
⎜⎝ x − ⎟⎠
1!
4
+
⎛ p⎞
f ′⎜ ⎟
⎝ 4⎠
2
+
1⎛
p⎞
⎛ p⎞
⎜⎝ x − ⎟⎠ f ″ ⎜⎝ ⎟⎠
2!
4
4
3
+
1⎛
p⎞
⎛ p⎞
⎜⎝ x − ⎟⎠ f ′″ ⎜⎝ ⎟⎠
3!
4
4
4
1⎛
p ⎞ ( 4) ⎛ p ⎞
⎜ x − ⎟⎠ f ⎜⎝ ⎟⎠ + …
4! ⎝
4
4
we have
f(x) = cosx ,
+
f (x) = −sin x,
f ″( x ) = −cosx,
f ″′( x ) = sinx,
f(4) (x) = cosx,
∴ Taylor’s series is
1
⎛ p⎞
⎛ p⎞
f ′ ⎜ ⎟ = cos ⎜ ⎟ =
⎝ 4⎠
⎝ 4⎠
2
1
⎛ p⎞
⎛ p⎞
f ′ ⎜ ⎟ = −sin ⎜ ⎟ = −
⎝ 4⎠
⎝ 4⎠
2
1
⎛ p⎞
⎛ p⎞
f ″ ⎜ ⎟ = − cos ⎜ ⎟ = −
⎝ 4⎠
⎝ 4⎠
2
1
⎛ p⎞
⎛ p⎞
f ″′ ⎜⎝ 4 ⎟⎠ = sin ⎜⎝ 4 ⎟⎠ =
2
1
⎛ p⎞
⎛ p⎞
f(4) ⎜ ⎟ = cos ⎜ ⎟ =
and so on
⎝ 4⎠
⎝ 4⎠
2
2
p⎞
⎛
x− ⎟
3
4
p⎞ ⎛ 1 ⎞
1
p ⎞ ⎛ 1 ⎞ ⎜⎝
p⎞ ⎛ 1 ⎞ 1 ⎛
4⎠ ⎛ 1 ⎞ 1 ⎛
⎛
+
x
−
...
f (x ) =
+
x
+ ⎜x − ⎟ ⎜−
−
+
−
⎜
⎟
⎜
⎟
⎟
⎜⎝
⎟
4 ⎠ ⎜⎝ 2 ⎟⎠
4⎠ ⎝
2!
4 ⎠ ⎜⎝ 2 ⎟⎠ 4 ! ⎝
2 ⎝
2⎠
2 ⎠ 3! ⎝
⇒
sinx =
2
3
4
⎤
1 ⎡ ⎛
p⎞ 1 ⎛
p⎞
p⎞
p⎞
1⎛
1⎛
⎢1 − ⎜ x − ⎟ − ⎜ x − ⎟ + ⎜ x − ⎟ + ⎜ x − ⎟ − ...⎥
4 ⎠ 2! ⎝
4⎠
3! ⎝
4⎠
4! ⎝
4⎠
2 ⎢⎣ ⎝
⎦⎥
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Differential Calculus ■
3.71
EXAMPLE 3
Using Taylor’s theorem, prove that x 2
x3
x3
x5
< sinx < x 2 1
for x > 0.
6
6
120
Solution.
x3
x3 x5
for x > 0
< sin x < x − +
6
6 120
For proving this inequality in terms of x, we consider Maclaurin’s formula upto third and fifth
degree terms with Lagrange’s reminder.
Maclaurin’s series upto the third term with Lagrange’s reminder is
To prove x −
Let f ( x ) = sin x.
f ( x ) = f ( 0) +
f ( x ) = sin x ,
f (0) = sin 0 = 0
f ′( x ) = cos x,
f ′(0) = cos 0 = 1
f ′′( x ) = − sin x ,
f ′′(0) = − sin 0 = 0
f ′′′( x ) = − cos x ,
f ′′′(0) = − cos 0 = −1
We have
∴
and
x
x2
x3
f ′(0) + f ′′(0) + f ′′′(u1x ) ,
1!
2!
3!
f
( 4)
( x ) = sin x ,
f
( 5)
f
(6)
∴
f
( 4)
(0) = sin 0 = 0
( x ) = cos x ,
f
( 5)
(0) = cos 0 = 1
( x ) = − sin x ,
f
(6)
(0) = − sin 0 = 0
sin x = 0 +
We have for x > 0,
−
0 < u1 < 1
x
x2
x3
x3
⋅1 + ⋅ 0 + ( − cos u1x ) = x − cos u1 ( x ) .
6
1!
2!
3!
x3
x3
cos u1x > −
3!
6
[whether cos u1x is +ve or –ve]
x3
x3
x3
cos u1x > x −
⇒ sin x ≥ x −
6
6
6
Also Maclaurin’s series up to fifth degree term with Lagrange’s reminder is
⇒
x−
f ( x ) = f ( 0) +
⇒
x
x2
x3
x4
f ′( 0) +
f ′′(0) +
f ′′′(0) + f
4!
1!
2!
3!
sin x = 0 + x ( +1) +
=x−
( 4)
( 0) +
x5
f
5!
( 5)
(1)
(u 2 x ), 0 < u 2 < 1
x2
x3
x4
x5
(0) + ( −1) + (0) + (cos u 2 x ), 0 < u 2 < 1
2!
3!
4!
5!
x3 x5
cos u 2 x , 0 < u 2 < 1
+
6 120
We have −1 ≤ cos u 2 x ≤ 1 and x > 0.
M03_ENGINEERING_MATHEMATICS-I _CH03_Part B.indd 71
∴
x5
x5
cos u 2 x ≤
120
120
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∴
Engineering Mathematics
x−
x3 x5
x3 x5
cos u 2 x ≤ x − +
+
6 120
6 120
From (1) and (2), we get
x−
But equality holds only when x = 0 .
∴
x3 x5
+
6 120
sin x ≤ x −
(2)
x3
x3 x5
.
≤ sin x ≤ x − +
6
6 120
∴
x−
x3
x3 x5
< sin x < x − +
6
6 120
for
x > 0.
EXAMPLE 4
x x2 x4 …
ex
1 x x3 …
Show that log e (11e x ) 5 log e 2 1 1 2
1 and hence deduce that
5
1 2 1
2 8 192
11e x 2 4 48
Solution.
Let f ( x ) = loge (1 + e x )
Maclaurin’s series for f ( x ) is
x
x2
x3
x4
f ′(0) + f ′(0) + f ′′′(0) + f
1!
2!
3!
4!
f ( x ) = loge (1 + e x )
f ( x ) = f ( 0) +
we have
( 4)
( 0) + …
1
⋅ex
1+ ex
(1 + e x )e x − e x ⋅ e x
e x + e 2x − e 2x
ex
f ′′( x ) =
.
=
=
(1 + e x ) 2
(1 + e x ) 2
(1 + e x ) 2
∴
f ′( x ) =
f ′′′( x ) =
(1 + e x ) 2 ⋅ e x − e x 2(1 + e x ) ⋅ e x
(1 + e x )e x [1 + e x − 2e x ]
=
(1 + e x ) 4
(1 + e x ) 4
⇒
f ′′′( x ) =
e x (1 − e x ) e x − e 2 x
=
(1 + e x )3
(1 + e x )3
and
f ( 4) ( x) =
(1 + e x )3 [e x − 2e 2 x ] − (e x − e 2 x ) ⋅ 3(1 + e x ) 2 ⋅ e x
(1 + e x )6
=
(1 + e x ) 2 [(1 + e x )(e x − 2e 2 x ) − 3e x (e x − e 2 x )]
(1 + e x )6
⇒
∴
(1)
f
( 4)
(x ) =
(1 + e x )(e x − 2e 2 x ) − 3e x (e x − e 2 x ) .
(1 + e x ) 4
f (0) = loge (1 + 1) = loge 2 , f ′(0) =
f ′′′(0) =
1−1
= 0,
(1 + 1)3
M03_ENGINEERING_MATHEMATICS-I _CH03_Part B.indd 72
f ( 4 ) ( 0) =
1
1
= , f ′′(0) = 1 = 1
1+1 2
(1 + 1) 2 4
(1 + 1)[1 − 2] − 3 ⋅1(1 − 1)
2
1
=− =− .
4
16
8
(1 + 1)
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Differential Calculus ■
3.73
∴ (1) becomes,
f ( x ) = loge 2 +
⇒
loge (1 + e x ) = loge 2 +
To deduce the expansion of
x 1 x2 1 x3
x 4 ⎛ 1⎞
⋅ + ⋅ + × 0 + ⎜ − ⎟ +…
1! 2 2! 4 3!
4! ⎝ 8⎠
x x2 x4 …
+
−
+
2 8 192
ex
, differentiate (2) w.r.to x,
1+ e x
ex
1 2 x 4 x3 …
=
+
−
+
1 + e x 2 8 192
∴
(2)
⇒
ex
1 x x3 …
=
+ − + .
1 + e x 2 4 48
EXAMPLE 5
d
(sin21 x ) 5 (1 2x 2 )21/ 2 and the binomial series, obtain the first four non−zero
dx
terms of the Taylor’s series for sin21 x and hence obtain the first five non−zero terms of the
Using the fact
Taylor’s series for cos 21 x.
Solution.
d
(sin −1 x ) = (1 − x 2 ) −1 2
dx
Given
1⎛1 ⎞
1⎛1 ⎞⎛1
⎞
+2
+1
+1
1 2 2 ⎜⎝ 2 ⎟⎠ 2 2 2 ⎜⎝ 2 ⎟⎠ ⎜⎝ 2 ⎟⎠ 2 3 …
= 1+ x +
(x ) +
(x ) + , x < 1
2
2!
3!
d
x 2 1⋅ 3 4 1⋅ 3 ⋅ 5 6 …
(sin −1 x ) = 1 +
x +
x +
+
dx
2 2⋅ 4
2⋅ 4⋅6
⇒
(1)
Integrating (1) w.r.to x, we get
sin −1 x = c + x +
When x = 0, sin −1 x = 0
∴c = 0
sin −1 x = x +
∴
We know that sin −1 x + cos −1 x =
⇒
cos −1 x =
x 3 1⋅ 3 x 5 1⋅ 3 ⋅ 5 x 7 …
+
+
+
2⋅3 2⋅ 4 5 2⋅ 4⋅6 7
1 x 3 1⋅ 3 x 5 1⋅ 3 ⋅ 5 x 7 …
+
+
+
2 3 2⋅ 4 5 2⋅ 4⋅6 7
p
.
2
p
p ⎡
1 x 3 1⋅ 3 x 5 1⋅ 3 ⋅ 5 x 7 …⎤
− sin −1 x = − ⎢ x +
+
+
+ ⎥
2
2 ⎣
2 3 2⋅ 4 5 2⋅ 4⋅6 7
⎦
=
p
1 x 3 1⋅ 3 x 5 1⋅ 3 ⋅ 5 x 7 …
−x −
−
−
−
2
2 3 2⋅ 4 5 2⋅ 4⋅6 7
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EXERCISE 3.9
1. Find the Taylor’s series expansion of loge sin x about x = 3 [or in powers of (x − 3].
p
2. Find the Taylor’s series expansion of sin x about x = .
4
p
⎛
⎞
3. Write down the Taylor’s series up to x4 for tan ⎜ x + ⎟ .
⎝
4⎠
4. Expand f ( x ) = sin 2 x − x 2e − x by Taylor’s formula up to x4.
x2
x2
≤ e x ≤ 1 + x + e x for x ≥ 0.
2
2
6. If f ( x ) = loge (1 + x ), x > 0 using Maclaurin’s theorem, then show that for 0 < u < 1.
5. Prove that 1 + x +
x2
x3
+
.
2 3(1 + ux )3
x2
x3
Deduce that loge (1 + x ) < x −
+
for x ≥ 0.
2 3(1 + ux )3
log e (1 + x ) = x −
7. Write down Maclaurin’s formula for the function f ( x ) = 1 + x with Lagrange’s reminder R 3.
x x2
Estimate the error in the approximation 1 + x ~1 + −
when x = 0.2.
2 8
8. Using Maclaurin’s series expand tan x up to the term containing x 5 .
52
9. Write Taylor’s series for f ( x ) = (1 − x ) with Lagrange’s form of remainder up to 3 terms in the
interval [0, 1]
10. Apply Taylor’s theorem to express x about the point x = 1, up to third degree.
11. Expand loge x as a Taylor’s series in powers of ( x −1) and hence evaluate loge 1.1 to 4 places of
decimals.
12. Calculate the approximate value of 10 to four decimal places using Taylor’s series.
ANSWERS TO EXERCISE 3.9
cos ec 2 3
cos ec 2 3 cot 3
( x − 3) 2 +
( x − 3)3
1. logesinx = logesin3 + cot3(x – 3) −
3
3
2
2
cos ec 3(1 + 3 cot 3)
−
( x − 3) 4 + …
12
2. sinx =
2
3
4
⎤
1 ⎡ ⎛
p⎞ 1 ⎛
p⎞
p⎞
p⎞
1⎛
1⎛
⎢1 + ⎜ x − ⎟ − ⎜ x − ⎟ − ⎜ x − ⎟ + ⎜ x − ⎟ + …⎥
4 ⎠ 2! ⎝
4⎠
3! ⎝
4⎠
4! ⎝
4⎠
2 ⎢⎣ ⎝
⎥⎦
8
10
p⎤
⎡
3. tanx ⎢ x + ⎥ = 1 + 2x + 2x2 + x3 + x4 +…
3
3
4⎦
⎣
5
4. f ( x ) = x 3 − x 4 + …
6
M03_ENGINEERING_MATHEMATICS-I _CH03_Part B.indd 74
6. loge (1 + x ) = x −
x2 x3
for x ≥ 0.
+
2
3
5/19/2016 5:09:43 PM
Differential Calculus ■
7. the error is less than
9. (1 − x )5 2 = 1 −
1
2.103
8. tan x = x +
x = 1+
x3 2 x 2
+
3 15
5
15 x 2 15
1
x3
x+ ⋅ −
, 0 < u <1
2
4 2 ! 8 (1 − ux )1 2 3!
4
1
1 ( x − 1) 3 ( x − 1) 15
−7 2 ( x − 1)
, 0 < u < 1.
+
− [1 + u( x − 1) ] ⋅
( x − 1) −
4!
2
4 2!
8 3!
16
2
10.
3.75
3
11. 0.0953
12. 3.1623
3.5.5 Expansion by Using Maclaurin’s Series of Some Standard Functions
Sometimes for a given function f obtaining the derivatives f ′, f ′′, f ′′′, … would be difficult for writing
the Maclaurin’s series. In such cases, we use Maclaurin’s series expansion of some standard functions,
when we require few terms of the resulting series.
We list the Maclaurin’s series expension of some of the standard functions.
1. (1 + x ) n = 1 + nx +
2. e x = 1 +
n ( n − 1) 2 n ( n − 1)( n − 2) 3 …
x +
x + , x <1
2!
3!
x x2 x3 …
+
+ + , x ∈R
1! 2 ! 3!
3. loge (1 + x ) = x −
4. sin x = x −
x2 x3 x4 …
+
−
+ , x <1
2
3
4
x3 x5 x7 …
+
−
+
3! 5! 7 !
6. sinh x = x +
5. cos x = 1 −
x3 x5 …
+
+
3! 5!
7. cosh x = 1 +
8. tan −1 x = x −
x3 x5 …
+
+
3! 5!
9. sin −1 x = x +
1 x 3 1⋅ 3 x 5 1⋅ 3 ⋅ 5 x 7 …
+
+
+ ,
2 3 2⋅ 4 5 2⋅ 4⋅6 7
10. cos −1 x =
x2 x4 x6 …
+
−
+
2 4! 6!
x2 x4 …
+
+
2! 4 !
−1 ≤ x ≤ 1
p
1 x 3 1⋅ 3 x 5 1⋅ 3 ⋅ 5 x 7 …
−x −
−
−
− , −1 ≤ x ≤ 1
2
2 3 2⋅ 4 5 2⋅ 4⋅6 7
WORKED EXAMPLES
EXAMPLE 1
Expand log e (1 1 sinx ) 5 x 2
x2 x3 x4 …
1 2 1 .
2
6 12
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Solution.
We know that
loge (1 + x ) = x −
x2 x3 x4 x5 …
+
−
+ + x <1
2
3
4
5
∴ loge (1 + sin x ) = sin x −
(sin x ) 2 (sin x )3 (sin x ) 4 (sin)5 …
+
−
+
+
2
3
4
5
2
1⎛
x3 x5 … 1 ⎛
x 2 x 5 …⎞
x 2 x 5 …⎞
=x− +
− − ⎜x −
+
− ⎟ + ⎜x −
+
− ⎟
3! 5 !
2⎝
3! 5 !
3⎝
3! 5!
⎠
⎠
⎡⎣{ sin x ≤ 1⎤⎦
3
4
1⎛
x 2 x 5 …⎞ …
− ⎜x −
+
− ⎟ +
4⎝
3! 5!
⎠
= x−
⎞ 1
1
x3 x5 … 1 ⎛ 2
x4
+
− − ⎜ x − 2 ⋅ + …⎟ + ( x 3 − …) − ( x 4 − …)
6 120
2⎝
6
4
⎠ 3
= x−
x 2 x3 x3 x 4 x 4 …
x 2 x3 x 4
+ − +
− + = x − + − +…
2
3 6
6
4
2
6 12
EXAMPLE 2
⎛ 1 1 x 2 21 ⎞ 1 ⎡
x 3 x 5 x 7 …⎤
2
1 ⎥.
Show that tan21 ⎜
⎟ 5 ⎢x 2 1
x
3
5
7
⎦
⎝
⎠ 2⎣
Solution.
⎛ 1 + x 2 − 1⎞
Let
f ( x ) = tan −1 ⎜
⎟
x
⎝
⎠
Put
∴
Now
x = tan u. ∴
1 + x 2 = 1 + tan 2 u = sec 2 u = sec u
⎛ sec u − 1⎞
f ( x ) = tan −1 ⎜
⎝ tan u ⎟⎠
1
u
u
−1
sin
2 sin 2
sec u − 1 cos u
1 − cos u
2
2 = tan u
=
=
=
=
sin u
tan u
u
u
u
sin u
2
cos
2 sin cos
cos u
2
2
2
∴
u⎞ u 1
⎛
f ( x ) = tan −1 ⎜ tan ⎟ = = tan −1 x
⎝
2⎠ 2 2
⇒
f (x ) =
⎤
1⎡
x3 x5 x7
x
−
+
−
+ …⎥ .
⎢
2⎣
3
5
7
⎦
M03_ENGINEERING_MATHEMATICS-I _CH03_Part B.indd 76
⎛ 1 + x 2 − 1⎞ 1 ⎡
⎤
x3 x5 x7
=
x
−
+
−
+ …⎥ .
⇒ tan −1 ⎜
⎟
⎢
x
3
5
7
⎦
⎝
⎠ 2⎣
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Differential Calculus ■
3.77
EXAMPLE 3
Show that
e x cos x 5 1 1 21/ 2 x ⋅ cos
p x2
2 p x 3 3/ 2
3p … x n n / 2
⎛ n p⎞ …
1
2 ⋅ cos
1
2 cos
1 1
2 cos ⎜
1 .
⎝ 4 ⎟⎠
4
2!
4
3!
4
n!
Solution.
⎡ e ix + e − ix ⎤ 1 x (1+ i )
f ( x ) = e x cos x = e x ⎢
+ e x (1− i ) ]
⎥ = [e
2
2
⎣
⎦
Let
1
[(1 + i )e x (1+ i ) + (1 − i )e x (1− i ) ]
2
1
f ′′( x ) = [(1 + i ) 2 e x (1+ i ) + (1 − i ) 2 e x (1− i ) ]
2
f 9( x ) =
1
[(1 + i )3 e x (1+ i ) + (1 − i )3 e x (1− i ) ]
2
:
:
f ′′′( x ) =
1
[(1 + i ) n e x (1+ i ) + (1 − i ) n e x (1− i ) ]
2
1
f ( n ) (0) = [(1 + i ) n + (1 − i ) n ]
2
f
∴
(n)
(x ) =
p
p⎤
np
np ⎤
⎡
⎡
1 + i = 2 ⎢cos + i sin ⎥ and (1 + i ) n = 2n 2 ⎢cos
+ i sin
4
4
4
4 ⎥⎦
⎣
⎦
⎣
But
For, let 1+ i = r (cos u + i sin u) . ∴ r cos u = 1 and r sin u = 1
⇒
r 2 cos 2 u + r 2 sin 2 u = 2 ⇒ r 2 (cos 2 u + sin 2 u) = 2
and
tan u = 1 ⇒ u =
∴
f
(n)
( 0) =
=
⇒
r2 = 2 ⇒ r = 2
p
4
np
np ⎤
⎡
and (1 + i ) n = 2n 2 ⎢cos
+ i sin
4
4 ⎥⎦
⎣
np
np ⎤
⎡
(1 − i ) n = 2n 2 ⎢cos
− i sin
4
4 ⎥⎦
⎣
p
p⎤
⎡
∴ 1 + i = 2 ⎢cos + i sin ⎥
4
4⎦
⎣
Similarly,
⇒
f
(n)
[by De−movire’s theorem]
np
np ⎫ n/2 ⎧
np
np ⎫ ⎤
1 ⎡ n/2 ⎧
+ i sin
− i sin
⎢ 2 ⎨cos
⎬ + 2 ⎨cos
⎬⎥
2 ⎢⎣
4
4 ⎭
4
4 ⎭ ⎥⎦
⎩
⎩
np
np
np
np ⎫ 1 n / 2
np
1 n/2 ⎧
⋅ 2 ⎨cos
+ i sin
+ cos
− i sin
⎬ = ⋅ 2 ⋅ 2 cos
4
2
4
4
4
4 ⎭ 2
⎩
(0) = 2n / 2 cos
M03_ENGINEERING_MATHEMATICS-I _CH03_Part B.indd 77
np
4
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Maclaurin’s series is
f ( x ) = f ( 0) +
x
x2
x3
xn
f ′(0) + f ′′(0) + f ′′′(0) + ⋅⋅⋅ + f n (0) + ⋅⋅⋅⋅
1!
2!
3!
n!
But f(0) = 1.
∴
∴
f ′(0) = 21/2 cos
e x cos x = 1 +
p
p
3p
, f ′′(0) = 2 cos , f ′′′(0) = 23/2 cos
, ⋅⋅⋅, f
4
4
4
(n)
(0) = 2n / 2 cos
np
4
p x2
np …
x 1/ 2
2p x 3 3/2
3p … x n n/2
+
2 cos + 2 cos
+ 2 cos
+ +
2 cos
4
1!
4 2!
4
3!
4
n!
3.5.6 Expansion of Certain Functions Using differential Equations
WORKED EXAMPLES
EXAMPLE 1
Expand cos( m sin21 x ) as a power series.
Solution.
Let
Differentiating w.r.to x, we get
y = cos( m sin −1 x )
y 1 = − sin( m sin −1 x ).
(1)
m
1− x 2
1 − x 2 y 1 = − m sin( m sin −1 x ).
⇒
Squaring,
(1 − x 2 ) y 12 = ( − m ) 2 sin 2 ( m sin −1 x ) = m 2 [1 − cos 2 ( m sin −1 x )]
⇒
(1 − x 2 ) y 12 = m 2 (1 − y 2 )
(2)
Differentiating w.r.to x, we get
(1 − x 2 )2 y 1 y 2 + y 12 ( −2x ) = m 2 ( −2 yy 1 )
⇒
2(1 − x 2 ) y 1 y 2 − 2xy 12 = −2m 2 yy 1
Dividing by 2y1, we get
(1 − x 2 ) y 2 − xy 1 = − m 2 y
⇒ (1 − x 2 ) y 2 − xy 1 + m 2 y = 0
(3)
Differentiating n times using Leibnitz’s theorem, we get
(1 − x 2 ) y n + 2 + nC1 ( −2x ) y n +1 + nC 2 ( −2) y n − {xy n +1 + nC1 ⋅1⋅ y n } + m 2 y n = 0
⇒
(1 − x 2 ) y n + 2 + 2nxy n +1 − 2
M03_ENGINEERING_MATHEMATICS-I _CH03_Part B.indd 78
n ( n − 1)
y n − xy n +1 − ny n + m 2 y n = 0
1⋅ 2
5/19/2016 5:10:39 PM
Differential Calculus ■
⇒
(1 − x 2 ) y n + 2 − ( 2n + 1)xy n +1 + ( − n 2 − n + n + m 2 ) y n = 0
⇒
(1 − x 2 ) y n + 2 − ( 2n + 1)xy n +1 − ( n 2 − m 2 ) y n = 0
Putting x = 0, we get y(0) = cos (0) = 1 ∴
y 1 ( 0) = − m
sin 0
1− 0
(4)
=0
(1 − 0) y 2 (0) − 0 + m 2 y (0) = 0
⇒
y 2 ( 0) = − m 2 y ( 0) = − m 2
From (4), we get (1 − 0) y n + 2 (0) − 0 − ( n 2 − m 2 ) y n (0) = 0
⇒
y n + 2 ( 0) = ( n 2 − m 2 ) y n ( 0)
From (3), we get
3.79
(5)
Putting n = 1, 2, 3, 4, … in (5), we get
y 3 (0) = (1 − m 2 ) y 1 (0) = 0
y 4 (0) = ( 22 − m 2 ) y 2 (0) = ( 22 − m 2 )( − m 2 ) = m 2 ( m 2 − 22 )
y 5 (0) = (32 − m 2 ) y 3 (0) = 0
y 6 ( 0) = ( 4 2 − m 2 ) y 4 ( 0)
= ( 4 2 − m 2 )m 2 ( m 2 − 22 ) = − m 2 ( m 2 − 22 )( m 2 − 4 2 ) and so on.
∴ Maclaurin’s series for y = f(x) is
f ( x ) = f ( 0) +
⇒
x
x2
x3
x4
f ′(0) + f ′′(0) + f ′′′(0) + f
1!
2!
3!
4!
cos( m sin −1 x ) = y (0) +
= 1+
( 4)
( 0) +
x5
f
5!
( 5)
( 0) +
x6
f
6!
(6)
(0) + ⋅⋅⋅
x6
x
x2
x3
x4
x5
y 6 (0) + ⋅⋅⋅
y 1 ( 0) +
y 2 ( 0) + y 3 ( 0) +
y 4 ( 0) + y 5 ( 0) +
5!
6!
1!
2!
3!
4!
x
x2
x3
x4 2 2
x5
x6
× 0 + (−m 2 ) + × 0 +
m ( m − 22 ) +
× 0 + [ − m 2 ( m 2 − 22 )( m 2 − 4 2 )] + ⋅⋅⋅
1!
2!
3!
4!
5!
6!
cos( m sin −1 x ) = 1 − m 2
⇒
x2
x4
x6
+ m 2 ( m 2 − 22 ) − m 2 ( m 2 − 22 )( m 2 − 4 2 ) + ⋅⋅⋅ ⋅
2!
4!
6!
Note Deduce the series for cos mu
Put u = sin −1 x ⇒ x = sin u
∴
cos ( m u) = 1 −
m2
m 2 ( m 2 − 22 ) 4
m 2 ( m 2 − 22 )( m 2 − 4 2 ) 6
sin 2 u +
sin u −
sin u + ⋅⋅⋅
2!
4!
6!
EXAMPLE 2
Expand sin[ln ( x 2 1 2 x 11)] as a power series using Maclaurin’s series up to x5.
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Solution.
Let
y = sin[ln( x 2 + 2x + 1)]
⇒
y = sin[ln( x + 1) 2 ] = sin[2 ln( x + 1)]
(1)
Differentiating w.r.to x,
⇒
y 1 = cos[2 ln( x + 1)]⋅
2
x +1
(2)
( x + 1) y 1 = 2 cos[ 2 ln( x + 1)]
⇒
( x + 1) 2 y 12 = 4 cos 2 [2 ln( x + 1)] = 4[1 − sin 2 ( 2 ln( x + 1))]
Squaring,
( x + 1) 2 y 12 = 4[1 − y 2 ].
Differentiating w.r.to x, we get
( x + 1) 2 2 y 1 y 2 + y 12 ⋅ 2( x + 1) = 4( −2 yy 1 )
⇒
⇒
⇒
2( x + 1) 2 y 1 y 2 + 2( x + 1) y 12 = −8yy 1
( x + 1) 2 y 2 + ( x + 1) y 1 = −4 y
[dividing by 2y1]
( x + 1) 2 y 2 + ( x + 1) y 1 + 4 y = 0
(3)
Differentiating n times, using Leibnitz’s formula, we get
( x + 1) 2 y n + 2 + nC1 2( x + 1) y n +1 + nC 2 2 y n + ( x + 1) y n +1 + nC1 ⋅1⋅ y n + 4 y n = 0
⇒
( x + 1) 2 y n + 2 + 2n ( x + 1) y n +1 + 2 ⋅
n ( n − 1)
y n + ( x + 1) y n +1 + ny n + 4 y n = 0
1⋅ 2
⇒
( x + 1) 2 y n + 2 + ( 2n + 1)( x + 1) y n +1 + ( n ( n − 1) + n + 4) y n = 0
⇒
( x + 1) 2 yn + 2 + ( 2n + 1)( x + 1) yn +1 + ( n2 + 4) yn = 0
⇒
( x + 1) 2 y n + 2 = −( 2n + 1)( x + 1) y n +1 − ( n 2 + 4) y n
Put x = 0, then (1)
(2) ⇒
(3) ⇒
(4) ⇒
⇒
(4)
y (0) = sin[ln(1)] = sin 0 = 0.
y 1 (0) = cos[2 ln(1)] ⋅
y 2 ( 0) + y 1 ( 0) + 4 y ( 0) = 0
⇒
2
= 2 cos 0 = 2
0 +1
y 2 ( 0) + 2 + 0 = 0
y n + 2 (0) = −( 2n + 1) y n +1 (0) − ( n 2 + 4) y n (0)
M03_ENGINEERING_MATHEMATICS-I _CH03_Part B.indd 80
⇒
y2 (0) = −2
(5)
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Differential Calculus ■
3.81
Putting n = 1, 2, 3, … in (5), we get
y 3 (0) = −( 2 + 1) y 2 (0) − (12 + 4) y 1 (0) = −3( −2) − (12 + 4)2 = 6 − 10 = −4
y 4 (0) = −( 4 + 1) y 3 (0) − ( 22 + 4) y 2 (0) = −5( −4) − 8( −2) = 20 + 16 = 36
y 5 (0) = −(6 + 1) y 4 (0) − (32 + 4) y 3 (0) = −7(36) − 13( −4) = −252 + 52 = −200
and so on.
∴ Maclaurin’s series for y = f(x) is
f ( x ) = f ( 0) +
= y ( 0) +
x
x2
x3
x4
f ′(0) + f ′′(0) + f ′′′(0) + f
1!
2!
3!
4!
( 4)
( 0) +
x5
f
5!
( 5)
( 0) + …
x
x2
x3
x4
x5
y 1 ( 0) +
y 2 ( 0) + y 3 ( 0) +
y 4 ( 0) + y 5 ( 0) + …
1!
2!
3!
4!
5!
x
x2
x3
x4
x5
× 2 + ( −2) + ( −4) +
× 36 + ( −200) + …
1!
2!
3!
4!
5!
4
36
200
∴ sin[ln( x 2 + x + 1)] = 2x − x 2 − x 3 + x 4 −
x5 +…
3!
4!
5!
2 3 3
5
= 2x − x 2 − x + x 4 − x 5 + …
3
2
3
= 0+
EXERCISE 3.10
I. Using Maclaurin’s series expand the following functions in powers of x.
(a) loge (1+ e x )
(e)
x
sin x
(b) e sin x
(c) e x sec x
(d) e x sin x
(f) e x cos x
(g) loge sec x
(h)
cos x sinh x
II. Forming differential equations prove the following.
(a) tan −1 x = x −
x3 x5 …
+
−
3
5
x2
x4
x6
x8
+ 2 ⋅ 22
+ 2.22 ⋅ 4 2
+ 2 ⋅ 22 ⋅ 4 2 ⋅ 6 2
+ ⋅⋅⋅
2!
4!
6!
8!
x2
x3
x5
x6
(c) e x loge (1 + x ) = x +
+ 2 + 9 − 35 + ⋅⋅⋅
2!
3!
5!
6!
(b) (sin −1 x ) 2 = 2 ⋅
(d) e x cos a cos( x sin a) = 1 + x cos a +
x2
x3
cos 2a + cos 3a + ⋅⋅⋅
2!
3!
1 x 3 1⋅ 3 x 5 1⋅ 3 ⋅ 5 x 7
(e) sin −1 x = x + ⋅ +
+
+ ⋅⋅⋅
2 3 2⋅ 2 5 2⋅ 4⋅6 7
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3.82
Engineering Mathematics
a2 2 a(a2 + 1) 3 a2 (a2 + 22 ) 4 ...
x +
x +
x +
2!
3!
4!
−1
(g) If e tan x = a0 + a1 x + a2 x 2 + … + an x n + …, then prove that ( n + 2)an + 2 + nan = an +1 .
(h) If an is the coefficient of xn in the expansion of ex sin x, then show that
(f) e a sin
−1
x
= 1 + ax +
np
sin
an −1 an − 2 an − 3
2 .
an −
+
−
+ ⋅⋅⋅ =
1!
2!
3!
n!
ANSWERS TO EXERCISE 3.10
I. (a) log e 2 +
(c) 1 + x + 2
(e) 1 +
1
1 x2 1 x4
x+
−
+ ⋅⋅⋅
2
4 2! 8 4 !
(b) 1 + x +
x2
x3
x4
x5
+ 4 ⋅ + 12 ⋅ + 36 ⋅ + …
2!
3!
4!
5!
(d) x + 2
x2
x4
+7
+…
6
360
x2
x3
x5
+ 2⋅ − 4⋅ +…
2!
3!
5!
(f) 1 + x − 2
x2 x4 x6
(g)
+
+
+…
2 12 45
3.6
x2 3 4
− x +…
2! 4 !
(h) x −
x3
x4
− 22
+ ⋅⋅⋅
3!
4!
x3 x5 …
+
−
3 30
INDETERMINATE FORMS
Let f and g be functions defined in a neighbour hood of a, except possibly at a.
If lim f ( x ) and lim g ( x ) exist and lim g ( x ) ≠ 0, then lim
x →a
x →a
x →a
lim
x→a
x →a
f (x )
exists and
g (x )
f ( x)
f ( x ) lim
5 x→a
.
g ( x ) lim g ( x )
x→a
If lim g ( x ) = 0 and lim f ( x ) ≠ 0, then the limit does not exist.
x →a
x →a
f (x )
is of the
However, if both the limits are zero. i.e., lim f ( x ) = 0 and lim g ( x ) = 0, then lim
x →a
x →a
x →a g ( x )
0
form , which is a meaningless expression, known as “indeterminate form”.
0
f (x )
This does not mean that lim
does not exist.
x →a g ( x )
0
sin x
is of the form , but we know that it has finite limit 1.
For example, lim
x →0
x
0
Other indeterminate forms are
∞
,
∞
M03_ENGINEERING_MATHEMATICS-I _CH03_Part B.indd 82
0 × ∞,
∞ − ∞, 0°, 1∞ and ∞ 0.
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Differential Calculus ■
3.83
These indeterminate forms cannot be evaluated by ordinary methods of limits.
A special method was devised by the French mathematician “L’Hopital”, a student of the famous
mathematician Johann Bernoulli.
0
form
0
Let f and g be the two functions defined in a neighbourhood (a − d, a + d) of a, except possibly at a,
d > 0, such that
Theorem 3.3 L’Hopital’s rule for
(i) lim f ( x ) = 0, lim g ( x ) = 0
x →a
x →a
(ii) f and g are differentiable in (a − d, a + d) except possibly at a.
i.e., f ′( x ) and g ′( x ) exist and g ′( x ) ≠ 0, for every x ∈ (a − d, a + d) , except at x = a
and
f ′( x )
f (x )
f ′( x )
exists, then lim
= lim
.
x → a g ′( x )
x →a g ( x )
x → a g ′( x )
(iii) lim
Proof
Given f and g are defined and differentiable in a neighbourhood (a − d, a + d) of a, except
possibly at x = a . We shall define F and G in (a − d, a + d) such that
⎧ f ( x ) ∀x ∈ ( a − d, a + d) and x ≠ a
F ( x) = ⎨
when x = a
⎩0,
⎧ g ( x ) ∀x ∈ ( a − d, a + d) and x ≠ a
G( x) = ⎨
when x = a
⎩ 0,
Clearly F and G are continuous and derivable on (a − d, a + d) except possibly at a.
Now
lim F ( x ) = lim f ( x ) = F (a)
and
lim G ( x ) = lim g ( x ) = G (a)
x →a
x →a
x →a
a−δ
x →a
a
x
a+δ
∴ F and G are continuous at a also.
If x > a, then in the interval [a, x ] , F and G satisfy the conditions of Cauchy’s mean value theorem.
∴
F ( x ) − F (a) F ′(c )
=
for some c ∈(a, x )
G ( x ) − G (a) G ′(c )
Given F (a) = 0 and G (a) = 0.
F ( x ) F ′(c )
=
,
G ( x ) G ′(c )
∴
We have
F(x ) = f (x )
∴
F ′(c ) = f ′(c ) and
M03_ENGINEERING_MATHEMATICS-I _CH03_Part B.indd 83
and
G (x ) = g (x )
G ′(c ) = g ′(c )
a<c<x
∀x ∈ (a, a + d)
[{
c ∈ (a, x ) ]
5/19/2016 5:11:50 PM
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3.84
Engineering Mathematics
f ( x ) f ′(c )
=
.
g ( x ) g ′(c )
∴ we get
As x → a, c → a
∴
f ′( x )
exists.
x → a + g ′( x )
Given lim
∴
lim
x →a +
f (x )
f ′( x )
= lim
x
→
a
+
g (x )
g ′( x )
f (x )
exists.
x →a + g ( x )
lim
f (x )
f ′( x )
= lim
.
x →a − g ( x )
x → a − g ′( x )
If x < a , similarly, we can prove
lim
Thus, L’Hopital’s rule is
lim
x →a
f (x )
f ′( x )
= lim
.
g ( x ) x → a g ′( x )
Note Suppose lim f ′( x ) = 0, lim g ′( x ) = 0 and f ″( x ), g ″( x ) exists, g ″( x ) ≠ 0 we have
x →a
x →a
f ′( x )
f ″( x )
= lim
.
x → a g ′( x )
x →a g ″( x )
0
form, then
0
f (x )
f ″( x )
= lim
.
x →a g ( x )
x →a g ″( x )
∴ lim
lim
We continue this till the value is obtained.
3.6.1
General L’Hopital’s Rule for
0
0
form
Theorem 3.4 If f and g are functions defined in a neighbourhood (a − d, a + d) of a except possibly
at a, d > 0, such that
(i) lim f ( x ) = 0,
lim f ′( x ) = 0,
lim f ″( x ) = 0, ...,
lim f (n −1) ( x ) = 0
and lim g ( x ) = 0,
lim g ′( x ) = 0,
lim g ″( x ) = 0, ...,
lim g (n −1) ( x ) = 0
x→a
x→a
(ii) f
(n)
(iii) lim
x →a
x→a
x→a
x→a
x→a
x→a
x→a
( x ), g ( n ) ( x ) exist and g ( n ) ( x ) ≠ 0∀x ∈ (a − d, a + d)x ≠ a and
f ( n ) (x )
f (x )
f (n ) ( x )
exists, then lim
= lim (n )
.
(n)
x →a g ( x )
x →a
g (x )
g (x )
f (x ) 0
f (x )
f (n ) ( x )
is form, then also L’Hopital’s rule is valid. That is lim
= lim (n ) .
x →∞ g ( x )
x →∞ g ( x )
x →∞
0
g (x )
Note If lim
Theorem 3.5
Indeterminate form
(i) lim f ( x ) = ∞, lim g ( x ) = ∞
x →a
∞
∞
x →a
(ii) f ′( x ), g ′( x ) exist and g ′( x ) ≠ 0 for every x ∈ (a − d, a + d) , except possibly at x = a and
(iii) lim
x →a
f ′( x )
f (x )
f ′( x )
exists, then lim
= lim
.
x →a g ( x )
x → a g ′( x )
g ′( x )
M03_ENGINEERING_MATHEMATICS-I _CH03_Part C.indd 84
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Differential Calculus ■
Note
(1) Clearly f ( x ), g ( x ) are positive functions.
So, we can use L’Hopital’s rule for
∴
0
form.
0
1
f (x )
g (x )
lim
= lim
x →a g ( x )
x →a
1
f (x )
3.85
⎡0
⎤
⎢ 0 form ⎥
⎣
⎦
0
∞
form and can be evaluated using
All other indeterminate forms can be rewritten as form or
0
∞
L’Hopital’s rule.
(2) While evaluating some of the limits the usage of standard limits such as
x
sinx
lim
51 ,
x →0
x
1⎞
⎛
lim ⎜ 11 ⎟ = e ,
x→∞ ⎝
x⎠
lim tanx 5 1 ,
x →0
lim(1 + x )1 x 5 e
x →0
and the series expansions of e , (1 1x ) , log(1 1x ), sinx , cos x , etc. may be used.
x
n
WORKED EXAMPLES
0 ∞
Type I: Problems of the type , , ∞ − ∞, 0 × ∞
0 ∞
EXAMPLE 1
e ax 2e bx
.
x →0
x
Evaluate lim
Solution.
e ax − e bx
.
x →0
x
Here f ( x ) = e ax − e bx
and
The given limit is lim
∴
g (x ) = x .
lim f ( x ) = lim(e ax − e bx ) = 1 − 1 = 0 and lim g ( x ) = lim x = 0
x →0
∴
lim
x →0
x →0
x →a
x →0
f (x ) 0
is form.
g (x ) 0
By L’Hopital’s rule, we have
f (x )
f ′( x )
ae ax − be bx
= lim
= ae 0 − be 0 = a − b
lim
= lim
x →0 g (x )
x → 0 g ′( x )
x →0
1
EXAMPLE 2
⎡ xe x 2 log e (1 1 x ) ⎤
Evaluate lim ⎢
⎥.
x →0
x2
⎦
⎣
Solution.
⎡ xe x − loge (1 + x ) ⎤
The given limit is lim ⎢
⎥.
x →0
x2
⎣
⎦
Here f ( x ) = xe x − loge (1 + x )
lim f ( x ) = lim[ xe x − loge (1 + x )] = 0.1 − loge 1 = 0
x →0
x →0
M03_ENGINEERING_MATHEMATICS-I _CH03_Part C.indd 85
g ( x ) = x 2.
and
and lim g ( x ) = lim x 2 = 0
x →0
x →0
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■
3.86
∴ lim
x →0
Engineering Mathematics
f (x ) 0
is form.
g (x ) 0
By L’Hopital’s rule,
x ⋅ e x + e x ⋅1 −
f (x )
f ′( x )
= lim
lim
= lim
x →0
x →0 g (x )
x → 0 g ′( x )
= lim
1
x +1
2x
( x + 1)e x −
x →0
1
x +1
⎡0
⎤
⎢ 0 form ⎥
⎣
⎦
2x
( x + 1)e x + e x ⋅1 −
= lim
x →0
( −1)
( x + 1) 2
2
( x + 2)e x +
= lim
x→0
1
( x + 1) 2
2
=
,
by L’Hopital’s rule
2 +1 3
= .
2
2
EXAMPLE 3
Evaluate lim
x →0
tan x 2 x
.
x 2 tan x
Solution.
The given limit is lim
x →0
tan x − x
.
x 2 tan x
f ( x ) = tan x − x
Here
g ( x ) = x 2 tan x .
and
lim f ( x ) = lim (tan x − x ) = 0 and
x →0
∴
lim
x →0
lim g ( x ) = lim x 2 tan x = 0
x →0
x →0
x →0
f (x ) 0
is form.
g (x ) 0
By L’Hopital’s rule,
f (x )
f ′( x )
sec 2 x − 1
lim
= lim
= lim 2
x →0 g (x )
x → 0 g ′( x )
x → 0 x sec 2 x + 2x tan x
1
⎡
⎢ 1 − sec 2 x
= lim ⎢
x →0
⎢ x 2 + 2x tan x
⎢⎣
sec 2 x
⎤
⎥
⎥
⎥
⎥⎦
[Dividing Nr. and Dr. by sec 2 x ]
⎤
⎡
(1 − cos 2 x )
= lim ⎢ 2
⎥
x → 0 x + 2x sin x cos x
⎦
⎣
⎡
⎤
sin 2 x
= lim ⎢ 2
⎥
x → 0 x + x sin 2x
⎦
⎣
M03_ENGINEERING_MATHEMATICS-I _CH03_Part C.indd 86
⎛0
⎞
⎜⎝ form⎟⎠
0
5/19/2016 1:06:57 PM
Differential Calculus ■
= lim
x →0
2 sin x cos x
,
2x + 2x cos 2x + sin 2x ⋅1
sin 2x
2x + 2x cos 2x + sin 2x
sin 2x
2x
= lim
x →0
⎛ sin 2x ⎞
1 + cos 2x + ⎜
⎝ 2x ⎟⎠
3.87
[by L’Hopital’s rule]
= lim
x →0
=
[Dividing Nr. and Dr. by 2x ]
sin 2x
⎡
⎤
= 1⎥
⎢{ lim
x →0
2x
⎣
⎦
1
1
= .
1+1+1 3
EXAMPLE 4
Evaluate lim
x →∞
xn
.
ex
Solution.
xn
.
x →∞ e x
The given limit is lim
f (x ) = x n
Here
and
lim f ( x ) = x ∞ = ∞
∴
x →∞
∴
lim
x →∞
f (x )
g (x )
is
and
g (x ) = e x .
lim g ( x ) = e ∞ = ∞
x →∞
∞
form.
∞
By L’Hopital’s rule,
f (x )
f ′( x )
nx n −1
= lim
= lim x
x →∞ g ( x )
x →∞ g ′ ( x )
x →∞ e
lim
Now applying L’Hopital’s rule ( n −1) times, we get f
(n)
⎤
⎡∞
⎢ ∞ form ⎥
⎣
⎦
( x ) = n ! and g ( n ) ( x ) = e x .
f (x )
f ( n ) (x )
n! n! n!
= lim ( n )
= lim x = ∞ = = 0 .
x →∞ g ( x )
x →∞ g
x →∞ e
∞
e
(x )
∴
lim
EXAMPLE 5
Evaluate lim log tan 2 x tan 3 x.
x →0
Solution.
The given limit is lim log tan 2 x tan 3x .
x →0
We know that
∴
logb a =
log tan 2 x tan 3x =
loge a
, using change of base rule in quotient form.
loge b
loge tan 3x
loge tan 2x
M03_ENGINEERING_MATHEMATICS-I _CH03_Part C.indd 87
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3.88
∴
■
Engineering Mathematics
loge tan 3x
x → 0 log tan 2x
e
lim log tan 2 x tan 3x = lim
x →0
1
⋅ sec 2 3x ⋅ 3
3
x
tan
= lim
x →0
1
⋅ sec 2 2x ⋅ 2
tan 2x
⎤
⎡∞
⎢ ∞ form ⎥
⎣
⎦
[by L’Hopital’s rule]
sin 2x
cos 2 2x
3
tan 2x sec 2 3x 3
2
cos
x
= lim
= lim
2 x → 0 tan 3x sec 2 2x 2 x → 0 sin 3x
cos 2 3x
cos 3x
⇒
lim log tan 2 x tan 3 x =
x→0
=
3
3
2 sin 2 x cos 2 x
sin 4 x
lim
= lim
x
→
0
x
→
0
2
2
2 sin 3 x cos 3 x
sin 6 x
3
4 cos 4 x 3 4
lim
= × = 1.
x
→
0
2
6 cos 6 x 2 6
⎡0
⎤
⎢ 0 form ⎥
⎣
⎦
[by L’Hopital’s rule]
EXAMPLE 6
Evaluate lim log e (1 2 x ) cot
x →1
p
x.
2
Solution.
p
The given limit is lim loge (1 − x ) cot x .
x →1
2
p
p
Now
lim loge (1 − x ) cot x = loge 0 × cot = − ∞ × 0 form.
x →1
2
2
0
∞
So, we have to rewrite as or form.
0
∞
∴
⎡
⎤
⎢ loge (1 − x ) ⎥
p
lim loge (1 − x ) cot x = lim ⎢
⎥
x →1
x →1
2
⎢ tan p x ⎥
⎣
⎦
2
M03_ENGINEERING_MATHEMATICS-I _CH03_Part C.indd 88
⎤
⎡∞
⎢ ∞ form ⎥
⎣
⎦
−1
⎡
⎤
⎢ (1 − x ) ⎥
⎥,
= lim ⎢
x →1
⎢ sec 2 p x ⋅ p ⎥
⎢⎣
2
2 ⎥⎦
[by L’Hopital’s rule]
⎡ 2p ⎤
⎢ cos 2 x ⎥
2
= − lim ⎢
⎥
p x →1 ⎢ 1 − x ⎥
⎣
⎦
⎛0
⎞
⎜⎝ form⎟⎠
0
5/19/2016 1:07:09 PM
Differential Calculus ■
⎡
p ⎛
p ⎞ p⎤
⎢ 2 cos 2 x ⎜⎝ − sin 2 x ⎟⎠ ⋅ 2 ⎥
2
⎥
= − lim ⎢
p x →1 ⎢
−1
⎥
⎢
⎥
⎣
⎦
=
3.89
[by L’Hopital’s rule]
2
p⎛
p⎞ p
× 2 cos ⎜ − sin ⎟ ⋅ = 0
⎝
p
2
2⎠ 2
EXAMPLE 7
1 ⎞
⎛ 1
Evaluate lim ⎜ 2 2
⎟.
x →0 ⎝ x
sin 2 x ⎠
Solution.
1 ⎞
⎛ 1
The given limit is lim ⎜ 2 − 2 ⎟
x →0 ⎝ x
sin x ⎠
1 ⎞
⎛ 1
Now lim ⎜ 2 − 2 ⎟ = ∞ − ∞ form.
x →0 ⎝ x
sin x ⎠
∴
So, we have to rewrite as
0
∞
or
form.
0
∞
1 ⎤
sin 2 x − x 2
⎡1
lim ⎢ 2 − 2 ⎥ = lim 2 2
x →0 x
sin x ⎦ x → 0 x sin x
⎣
sin 2 x
−1
2
= lim x 2
x →0
sin x
⎛0
⎞
⎜⎝ form⎟⎠
0
⎛0
⎞
⎜⎝ form⎟⎠
0
[dividing Nr. and Dr. by x 2 ]
x 2 2 sin x cos x − sin 2 x ⋅ 2x
x4
= lim
x →0
2 sin x cos x
= lim
2x sin x [x cos x − sin x ]
2 sin x cos x × x 4
= lim
x cos x − sin x
x − tan x
= lim
3
x
→
0
x cos x
x3
x →0
x →0
[by L’Hopital’s rule]
⎡0
⎤
⎢ 0 form ⎥
⎣
⎦
1 − sec 2 x
x →0
3x 2
1
1−
2
cos 2 x = lim cos x − 1
= lim
2
2
x →0
x → 0 3x cos 2 x
3x
= lim
[by L’Hopital’s rule]
⎛
1
1
sin 2 x ⎞
−1
⎛ tan x ⎞
= lim ⎜ − 2
= − lim ⎜
=
⋅1 = −
⎟
2
⎟
x →0 ⎝
x
→
0
⎝ x ⎠
3
3
3
3x cos x ⎠
2
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3.90
Engineering Mathematics
Type II: Problems of the type 08, ∞ 8,1∞
If A = lim [f ( x ) ]
g (x )
x →a
is one of these forms, we take logarithm and rewrite loge A = lim g ( x ) ⋅ loge f ( x )
x →a
0
∞
and evaluate.
in the form or
0
∞
EXAMPLE 8
1/ x
⎛ sinx ⎞
Evaluate lim ⎜
⎟ .
x → 01 ⎝ x ⎠
Solution.
1x
⎛ sin x ⎞
The given limit is lim ⎜
⎟ .
x →0+ ⎝ x ⎠
Let
∴
⎛ sin x ⎞
A = lim ⎜
⎟
x →0+ ⎝ x ⎠
loge A = lim
x →0+
1x
(1∞ form)
1
⎛ sin x ⎞
loge ⎜
⎝ x ⎟⎠
x
(∞ ⋅ 0 form )
⎛ sin x ⎞
1
0
∞
loge ⎜
= loge 1 = 0 and lim = ∞ , were write in the form or .
Since lim
⎝ x ⎟⎠
x →0
x →0 x
0
∞
∴
⎛ sin x ⎞
loge ⎜
⎝ x ⎟⎠
loge A = lim
x →0+
x
1
sin x ⎡ x cos x − sin x ⋅1⎤
⎥⎦
x ⎢⎣
x2
,
= lim
x →0+
1
⎡0
⎤
⎢⎣ 0 form ⎥⎦
[by L’Hopital’s rule]
= lim
x cos x − sin x
x sin x
⎡0
⎤
⎢ 0 form ⎥
⎣
⎦
= lim
x ( − sin x ) + cos x ⋅1 − cos x
,
x cos x + sin x ⋅1
[by L’Hopital’s rule]
x →0+
x →0+
− x sin x
x → 0 + x cos x + sin x
= lim
= lim
x →0+
−[x cos x + sin x ⋅1]
,
x ( − sin x ) + cos x ⋅1 + cos x
⎡0
⎤
⎢ 0 form ⎥
⎣
⎦
[by L’Hopital’s rule]
−[ x cos x + sin x ] 0
= = 0.
x → 0 + − x sin x + 2 cos x
2
= lim
∴
log e A = 0 ⇒ A = e 0 = 1 .
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Differential Calculus ■
3.91
EXAMPLE 9
Evaluate lim (sec x ) cot x.
x →p/2
Solution.
cot x
The given limit is lim (sec x ) .
x →p 2
∴
(∞
A = lim (sec x )
cot x
Let
x →p 2
loge A = lim loge (sec x )
cot x
= lim cot x loge sec x
x →p 2
x →p 2
= lim
x →p 2
log sec x
tan x
1
⋅ sec x tan x
= lim sec x 2
x →p 2
sec x
= lim
x →p 2
form
)
[0 ⋅ ∞ form]
⎤
⎡∞
⎢⎣ ∞ form ⎥⎦
[by L’Hopital’s rule]
tan x
sin x
= lim
⋅ cos 2 x = lim sin x cos x = 1⋅ 0 = 0
2
x
→
p
2
x →p 2
cos x
sec x
log e A = 0 ⇒
∴
0
A = e0 = 1
EXAMPLE 10
1/ x
⎡ a x 1b x 1 c x 1 d x ⎤
Evaluate lim ⎢
⎥ , where a, b, c, d are positive numbers.
x →0
4
⎣
⎦
Solution.
1/x
⎡ ax + b x + c x + d x ⎤
The given limit is lim ⎢
⎥
x →0
4
⎣
⎦
1/x
Let
⎡ ax + b x + c x + d x ⎤
A = lim ⎢
⎥
x →0
4
⎣
⎦
[1∞ form]
1/x
⎡ ax + b x + c x + d x ⎤
∴ l oge A = lim loge ⎢
⎥
x →0
4
⎣
⎦
⎡ ax + bx + cx + d x ⎤
log e ⎢
⎥
4
⎣
⎦
⇒ loge A = lim
x →0
x
⎡ ax + b x + c x + d x ⎤
1
= lim loge ⎢
⎥
x →0 x
4
⎣
⎦
[∞ ⋅ 0 form]
⎡0
⎤
⎢ 0 form ⎥
⎣
⎦
1
1
[ax loge a + b x loge b + c x loge c + d x loge d ] ×
x
x
4
a +b +c +d
4
= lim
x →0
1
x
x
M03_ENGINEERING_MATHEMATICS-I _CH03_Part C.indd 91
[by L’Hospital’s
rule]
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3.92
Engineering Mathematics
1
[a 0 log e a + b 0 log e b + c 0 log e c + d 0 log e d ]
a + b + c0 + d 0
1
1
= [log e a + logg e b + log e c + log e d ] = log e ( abcd ) = log e ( abcd )1/4
4
4
=
0
0
⇒ l oge A = loge (abcd )1/ 4
⇒ A = (abcd )1/ 4.
EXAMPLE 11
Evaluate lim(12 x 2 )1/loge (1 2 x ) .
x →1
Solution.
The given limit is lim (1 − x 2 )1/ loge (1− x ) .
x →1−
Let
A = lim (1 − x 2 )1/ loge (1− x ) .
∴
loge A = lim[loge (1 − x 2 )1/ loge (1− x ) ]
[00 form]
x →1−
x →1−
⎤
⎡
1
= lim ⎢
loge (1 − x 2 ) ⎥
x →1− log (1 − x )
⎣ e
⎦
⎡∞
⎤
⎢ ∞ form ⎥
⎣
⎦
⎡ log (1 − x 2 ) ⎤
= lim ⎢ e
⎥
x →1−
⎣ loge (1 − x ) ⎦
⎡∞
⎤
⎢ ∞ form ⎥
⎣
⎦
⎡⎡ 1
⎤⎤
( −2x ) ⎥ ⎥
⎢⎢
2
(1 − x )
⎦⎥,
= lim ⎢ ⎣
x →1− ⎢
⎥
1
⎢ 1 − x ( −1) ⎥
⎣
⎦
[by L’ Hopital’s rule]
⎡ 2 x(1 − x ) ⎤
⎡ 2 x ⎤ 2 ⋅1
= lim ⎢
=
=1
⎥ = xlim
x →1− (1 + x )(1 − x )
→1− ⎢1 + x ⎥
⎣
⎦ 1+1
⎣
⎦
∴
loge A = 1
⇒ A = e1 = e
Type III: Problems to find the limit using the expansions of sin x, cos x, tan x, ex, loge(1 1 x)
We know
(1) sin x = x −
x3 x5 …
+ −
3! 5!
(2) cos x = 1 −
(3) tan x = x +
x3 2 5 …
+ x +
3 15
(4) e x = 1 +
(5) log e (1 + x ) = x −
x2 x4 …
+
−
2! 4 !
x x 2 x3 x 4 …
+
+ +
+
1! 2 ! 3! 4 !
x 2 x3 x 4 …
+ − +
2
3
4
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Differential Calculus ■
3.93
EXAMPLE 12
Determine a and b so that lim
x →0
a sin 2 x 2b sin x
51 .
x3
Solution.
Given
lim
x →0
a sin 2x − b sin x
= 1.
x3
⎡
⎤
⎡
⎤
( 2x )3 ( 2x )5
x3 x5
a ⎢ 2x −
+
− …⎥ − b ⎢ x − +
− …⎥
3!
5!
3! 5!
⎦
⎣
⎦ =1
lim ⎣
x →0
x3
⇒
⇒
lim
x →0
1
x3
⎡
⎤
⎛ −8a b ⎞ 3 ⎛ 32a b ⎞ 5
⎢( 2a − b )x + ⎜⎝ 3! + 3!⎟⎠ x + ⎜⎝ 5! − 5!⎟⎠ x + …⎥ = 1
⎦
⎣
⎡ 2a − b ⎛ b − 8a ⎞ ⎛ 32a − b ⎞ 2
⎤
lim ⎢ 2 + ⎜
⎟⎠ + ⎜⎝
⎟⎠ x − …⎥ = 1
x →0
⎝
6
120
⎣ x
⎦
⇒
2a − b
→ ∞ and hence the limit is infinite, which contradicts the
x2
hypothesis that the limit is finite and equal to1.
Suppose 2a − b ≠ 0 , then as x → 0,
∴
2a − b = 0 ⇒ b = 2a
−8a + b
=1
6
and
∴
(1)
⇒ − 8a + 2a = 6 ⇒ −6a = 6 ⇒ a = −1
b = 2a = 2( −1) = −2 .
(2)
So, a = −1, b = −2
EXAMPLE 13
If lim
x →0
sin 2 x 1 a sin x
is finite, then find the value of a and the limit.
x3
Solution.
The given limit is lim
x →0
sin 2x + a sin x
and it is finite, say k.
x3
∴
⇒
⇒
⇒
lim
x →0
1
x →0 x 3
lim
sin 2x + a sin x
=k
x3
⎡
( 2x )3 ( 2x )5 ... ⎛
x 3 x 5 ...⎞ ⎤
2
−
+
+
+
−
+
+ ⎟⎥ = k
x
a
x
⎢
⎜⎝
3!
5!
3! 5!
⎠⎦
⎣
lim
x →0
⎤
1 ⎡
⎛ 8 a⎞
⎛ 32 a ⎞
( 2 + a)x − ⎜ + ⎟ x 3 + ⎜ + ⎟ x 5 + ...⎥ = k
3 ⎢
⎝
⎠
⎝
⎠
3! 3!
5! 5!
x ⎣
⎦
⎡ ( 2 + a) ⎛ 8 + a ⎞ ⎛ 32 + a ⎞ 2 ...⎤
+
x + ⎥=k
lim ⎢ 2 − ⎜
x →0
⎝ 6 ⎟⎠ ⎜⎝ 120 ⎟⎠
⎣ x
⎦
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Engineering Mathematics
2+a
Suppose 2 + a ≠ 0, then as x → 0, 2 → ∞ and hence the limit is infinite, which contradicts the
x
fact that the limit is finite.
⎛ 8 + a⎞
⎛ 8 − 2⎞
∴
2+a = 0
⇒ a = −2 and − ⎜
=k
⇒ −⎜
= k ⇒ k = −1
⎟
⎝ 6 ⎠
⎝ 6 ⎟⎠
∴
a = −2 and the limit is −1.
EXERCISE 3.11
I. Evaluate the following limits.
log e (1 + x )
x→0
3x − 1
2. lim
e x − e sin x
x → 0 x − sin x
1 ⎞
⎛1
6. lim ⎜ −
(sec x − tan x )
⎟ 7. xlim
x →0 ⎝ x
→p 2
sin x ⎠
1. lim
5. lim
9. lim x
tan
px
2
loge (1 − x 2 )
x → 0 log cos x
e
tan x − x
x →0
x3
10. lim(cos 2x )1/x
x →1
2
x →0
⎛ tan x ⎞
13. lim ⎜
⎟
x →0 ⎝
x ⎠
1x
x →0
2
x
sin x sin −1 x − x 2
x →0
x4
8. lim
e x − e − x − 2x
x →0
x − sin x
x − tan x
x →0
x3
12. lim
15. lim(tan x ) tan 2 x
⎛ sin x ⎞
16. lim ⎜
⎟
x →0 ⎝ x ⎠
x→
1
(1 + x )1/x − e + ex
2 .
17. lim
x →0
x2
4. lim
11. lim
2
14. lim(cos x )cot
loge x 2
x →0 log cot 2 x
3. lim
p
4
(1 + x )1/ x − e
x →0
x
18. lim
19. lim
x →0
sin x /( x − sin x )
x cos x − sin x
.
x 2 sin x
⎧ x (1 + a cos x ) − b sin x ⎫
20. Find the value of a and b if lim ⎨
⎬ = 1.
x →0
x3
⎩
⎭
ANSWERS TO EXERCISE 3.11
1. log3 e
−3
40
1
15.
e
8.
3.7
1
3
3. 2
4. −1
9. e −2 p
10. e −2
11. −
2.
16.
1
e
17.
11e
24
1
3
e
18. −
2
5. 1
6. 0
7. 0
12. 2
13. e1/3
14. e −1/2
19. −
1
3
20. a = −
5
3
and b = −
2
2
MAXIMA AND MINIMA OF A FUNCTION OF ONE VARIABLE
We have already seen the applications of derivative in problems of tangent and normal, in deciding
increasing and decreasing nature of a function in an interval. We shall now use it to locate maxima
and minima of a function.
In calculus the term “maximum” is used in two senses “absolute” maximum and “relative” maximum.
Similarly, absolute minimum and relative minimum.
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Differential Calculus ■
3.95
Definition 3.3 Let f be the function defined on [a, b ] and let c ∈(a, b ) .
Then (i) f is said to have a relative maximum (or local maximum) at c, if there is a neighbourhood
(c − d, c + d) of c such that f ( x ) < f (c )∀x ∈ (c − d, c + d) , x ≠ c .
That is f (c ) is the greatest value in a neighbourhood of c.
(ii) f is said to have a relative minimum (or local minimum) at c, if there is a neighbourhood
(c − d, c + d) of c such that f ( x ) > f (c )∀x ∈ (c − d, c + d) , x ≠ c .
That is f (c ) is the least value in a neighbourhood of c.
Note (1) If f (c ) is a relative maximum or relative minimum, then f (c ) is called an extreme value
of f at c or extremum of f at c.
Definition 3.4 Let f be defined on [a, b ]. f is said to have an absolute maximum (or global maximum)
on [a, b ] if there is at least one point c ∈[a, b ] such that f ( x ) ≤ f (c )∀x ∈[a, b ].
Inotherwords, the largest value of f on [a, b ] is called the absolute maximum
Definition 3.5 Let f be defined on [a, b ]. f is said to have an absolute minimum (or global minimum)
on [a, b ] if there is at least one point c ∈[a, b ] such that f ( x ) ≥ f (c )∀x ∈[a, b ].
That is the least value of f on [a, b ] is called the absolute minimum (or global minimum).
Note
1. Given a function f defined on [a, b ], the absolute maximum and the absolute minimum need not
exist.
⎧1
if 0 < x ≤ 1
⎪
For example: If f ( x ) = ⎨ x
y
⎪⎩0 if x = 0
Then f has no absolute maximum on [a, b ].
However, if f is continuous on a closed and
bounded interval [a, b ], then absolute maximum
and absolute minimum exist.
2. From the definition of maximum and minimum
it is obvious that f (c ) is an extreme value of f at
c, iff f ( x ) − f (c ) keeps the same sign for all x,
other than c, in some neighbourhood of c.
o
a
( )
C1
( )
C2
( )
Ca
b
x
Fig. 3.16
Theorem 3.6 A necessary condition for the existence of an extremum at an interior point
Let f be a function defined on the interval [a, b ] and c ∈(a, b ) . If f(c) is an extremum at c and if f ′(c )
exists, then f ′(c ) = 0 .
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■
Engineering Mathematics
3.7.1 Geometrical Meaning
Let y = f ( x ) be the graph of f on [a, b ], then P(c , f (c )) is a point on the curve y = f ( x ).
If f (c ) is a maximum value, then in (c − d, c ] the curve is
increasing and so f ′( x ) > 0 in (c − d, c ) and decreasing in
(c , c + d) .
y
P
That is f ′( x ) < 0 in (c , c + d) .
If f ′(c ) exists, then f ′(c ) must be zero.
That is the tangent is parallel to the x-axis , because it is
increasing up to the point P and momentarily at rest at P and then
decreasing.
Similarly, if f (c ) is a minimum value, then f ′(c ) = 0 .
c−δ
Note
1. The points where f ′( x ) = 0 are called stationary points of f.
2. The converse of the above theorem is not true.
Similarly, if f ′(c ) = 0, then f (c ) is not an extremum.
x
c+δ
c
Fig. 3.17
y
−δ
δ
o
x
For example: Consider f ( x ) = x 3.
f ′( x ) = 3x 2 ∴ f ′(0) = 0.
But f (0) is neither a maximum nor a minimum because there is no
neighbourhood of 0 in which f ( x ) − f (0) keeps the same sign for x ≠ 0.
For, f ( x ) = x 3 < 0
if x < 0 and
f (x ) = x 3 > 0
if
Fig. 3.18
x > 0.
3. It is possible that f (c ) is an extreme value of f
even if f ′(c ) does not exist.
For example: Consider f ( x ) = x .
y
y = −x
y=x
We know that f ′(0) does not exist.
But f (0) is a minimum value of f ( x ) .
In fact, f (0) is the absolute minimum.
o
x
Fig. 3.19
Definition 3.6 Critical Points
Let f be a function defined on [a, b ]. The points x ∈[a, b ] at which f ′( x ) = 0 or f ′( x ) does not exist
are called critical points of f.
For f ( x ) = x , x = 0 is a critical point but not a stationary point.
3.7.2
Tests for Maxima and Minima
(1) Second Derivative Test
Let f be a function defined on [a, b ] and let f be twice differentiable in a neighbourhood (c − d, c + d)
of c ∈( a, b) and f ′(c ) = 0. Suppose f ″(c ) ≠ 0 , then
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1.
f (c ) is a maximum if f ″(c ) < 0
and
3.97
2. f (c ) is a minimum if f ″(c ) > 0 .
Note If f ″(c ) = 0 , then the second derivative test cannot be applied.
In this case, we use the following general test involving higher derivatives or the first derivative
test.
(2) General Test
Let f be differentiable n times and f ′(c) = 0, f ″(c) = 0, ..., f (n −1) (c) and f (n) (c) ≠ 0 .
If n is even, then
1. f (c ) is a maximum if f
(n)
(c ) < 0
2. f (c ) is a minimum if f
(n)
(c ) > 0.
If n is odd, then f (c ) is neither a maximum nor a minimum.
(3) First Derivative Test
Let f be defined on [a,b] and c ∈ (a,b).
Let f be differentiable in a neighbourhood (c − d, c + d) of c, except possibly at c.
(i) If f ′( x ) > 0 for x < c and f ′( x ) < 0 for x > c in the neighbourhood of c, then f (c ) is a
maximum value.
That is, f ′( x ) changes from positive to negative in the neighbourhood of c as x increases.
(ii) If f ′( x ) < 0 for x < c and f ′( x ) > 0 for x > c in the neighbourhood of c, then f (c ) is a
minimum value.
That is, f ′( x ) changes from negative to positive in the neighbourhood of c as x increases.
SUMMARY
To find the maximum and minimum values of a function f on [a, b ].
(i) Find the critical points. That is, find the points where f ′( x ) = 0 or f ′( x ) does not exist.
(ii) Use the second derivative test or the first derivative test and decide the maximum and minimum.
(iii) Absolute maximum will occur at a relative maximum or at the end points.
Absolute minimum will occur at a relative minimum or at the end points.
WORKED EXAMPLES
EXAMPLE 1
Find the maxima and minima of the function 10 x 6 2 24 x 5 115 x 4 2 40 x 3 1108 .
Solution.
Let f ( x ) = 10 x 6 − 24 x 5 + 15x 4 − 40 x 3 + 108.
∴ f ′( x ) = 60 x 5 − 120 x 4 + 60 x 3 − 120 x 2 = 60( x 5 − 2x 4 + x 3 − 2x 2 )
f ″( x ) = 60[5x 4 − 8x 3 + 3x 2 − 4 x ] .
For maxima or minima
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∴
60( x 5 − 2x 4 + x 3 − 2x 2 ) = 0
⇒
⇒
x 2 [x 3 − 2x 2 + x − 2] = 0
⇒
⇒
x 2 ( x − 2)( x 2 + 1) = 0
⇒
x 5 − 2x 4 + x 3 − 2x 2 = 0
x 2 [x 2 ( x − 2) + ( x − 2)] = 0
x = 0 or 2
[{ x 2 + 1 ≠ 0]
When x = 2, f ″( 2) = 60 (5 ⋅ 24 − 8 ⋅ 23 + 3 ⋅ 22 − 4 ⋅ 2) = 60 × 20 > 0.
When x = 2, the function is minimum.
f ( 2) = 10 ⋅ 26 − 24 ⋅ 25 + 15 ⋅ 24 − 40 ⋅ 23 + 108
The minimum value is
= 640 − 324 + 240 − 320 + 108 = 344 .
When x = 0, f ′′( x ) = 0, the test fails.
Note f ′′′( x ) = 60 [20 x 3 − 24 x 2 + 6 x − 4]
∴ f ′′′(0) = −240 ≠ 0 .
So, f (0) is neither a maximum nor a minimum.
EXAMPLE 2
Find the maximum and minimum values of f ( x ) 5 4 2 x 2 , x ∈ [24 , 4 ]. Also find the absolute
maximum and absolute minimum, if they exist.
Solution.
Let f ( x ) = 4 − x 2 , x ∈[ −4, 4] .
y
We know 4 − x = 4 − x if 4 − x ≥ 0
2
⇒
2
2
x −4≤ 0 ⇒ −2≤ x ≤ 2
2
x = −4
o
and 4 − x 2 = −( 4 − x 2 ), if 4 − x 2 < 0 .
⇒ x 2 − 4 > 0 ⇒ x < −2 or x > 2
∴
⎧⎪4 − x 2 if − 2 ≤ x ≤ 2
f (x ) = ⎨ 2
⎩⎪x − 4 if x < −2 or x > 2
∴
⎧ −2x if − 2 < x < 2
f ′( x ) = ⎨
⎩2x if x < −2 or x > 2
(0, 4)
(−2, 0)
(0, −4)
x=4
x
(2, 0)
Fig. 3.20
⎧ −2 if − 2 < x < 2
f ″( x ) = ⎨
⎩2 if x < −2 or x > 2
∴
f ′( x ) = 0 ⇒ x = 0 ∈ ( −2, 2).
At x = −2, 2 , f ′( x ) does not exist. [Since, f is continuous at x = 2 and f ′( 2 −) = −4 , f ′( 2+ ) = 4
∴ f ′( 2) does not exist. Similarly, f ′( −2) does not exist]
The critical points are x = 0, − 2, 2 .
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When x = 0, f ″( x ) = −2 < 0 .
∴ f ( x ) has a maximum at x = 0 and the maximum value = 4.
Since f ′( x ) does not exist at x = −2, x = 2, we use the first derivative test.
In a neighbourhood of −2,
f ′( x ) < 0 if x < −2 and f ′( x ) > 0 if x > −2.
So, f ( −2) is a minimum.
Similarly, in a neighbourhood of 2,
f ′( x ) < 0 if x < 2 and f ′( x ) > 0 if x > 2.
So, f ( 2) is a minimum and the minimum value is zero.
The least value of f ( x ) in [ −4, 4] is 0. ∴ the absolute minimum = 0.
Though f (0) = 4 is a relative maximum, it is not the absolute maximum.
Absolute maximum occur at the end points x = 4 or x = −4 and the value is
f ( 4) = 4 − 4 2 = 4 − 16 = −12 = 12.
EXAMPLE 3
⎛ 1⎞
In a submarine cable the speed of signalling varies as x 2 log e ⎜ ⎟ , where x is the ratio of the
⎝x⎠
radius of the core to that of the covering. Find the values of x for which the speed of signaling is
maximum.
Solution.
Let S be the speed of signalling in a submarine cable.
⎛ 1⎞
S = kx 2 log e ⎜ ⎟ = − kx 2 log e x,
⎝ x⎠
Then
∴
dS
= −k
dx
where x > 0, k > 0
⎡ 2 1
⎤
⎢ x ⋅ x + loge x ⋅ 2x ⎥ = − kx [1 + 2 loge x ].
⎣
⎦
dS
=0
dx
− kx[1 + 2 log e x ] = 0
For maximum or minimum,
⇒
⇒
1 + 2 loge x = 0 ⇒ loge x = −
Now
When x =
∴ when x =
1
e
,
1
e
,
1
−
1
1
⇒x =e 2 =
2
e
[{ kx ≠ 0]
d 2S
= −k
dx 2
1
⎡
⎤
⎢ x ⋅ 2 ⋅ x + (1 + 2 loge x ) ⋅1⎥ = − k [2 + 1 + 2 loge x ] = − k [3 + 2 loge x ]
⎣
⎦
d 2S
= −k
dx 2
⎡
1 ⎤
⎢3 + 2 loge
⎥ < 0.
e⎦
⎣
S is maximum.
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EXAMPLE 4
A factory D is to be connected by a road with a straight railway line on which a town A is
situated. The distance DB of the factory to the railway line is 5 3 km. Length AB of the railway
line is 20km. Freight charges on the road are twice the charges on the railway. At what point
P ( PA > PB ) on the railway line should the road DP be connected so as to ensure minimum
freight charges from the factory to the town.
Solution.
D is the factory, A is the town and AB is the straight railway line.
Given DB = 5 3 km.
Let BP = x , then PA = 20 − x
and
DP = x 2 + 75 is the road.
Let k be the freight charge on railway. Then 2k is the freight charge on road.
If F is the total freight charges, then F = 2k x 2 + 75 + k ( 20 − x ), 0 ≤ x ≤ 20.
∴
dF
1
= 2k ⋅
⋅ 2x + k ( −1)
2
dx
2 x + 75
B
⎤
⎡ 2x
=k⎢
− 1⎥ .
2
⎣ x + 75 ⎦
5 3
dF
=0
For maximum or minimum,
dx
⎤
⎡ 2x
⇒ k⎢
− 1⎥ = 0
2
⎣ x + 75 ⎦
⇒
Now
When x = 5,
x
P
20 − x
A
90°
x 2 + 75
D
⇒
2x
x 2 + 75
2x = x 2 + 75 ⇒ 4 x 2 = x 2 + 75
−1 = 0
⇒
Fig. 3.21
3x 2 = 75 ⇒ x 2 = 25 ⇒ x = 5 [{ x > 0]
⎡
⎤
2x
k ⎢ x 2 + 75 ⋅ 2 − 2x ⋅
⎥
k ⎡ 2( x 2 + 75) − 2x 2 ⎤⎦
150 k
d F
2 x 2 + 75 ⎦ = ⎣
.
= 2
⎣
=
2
2
( x + 75)3/ 2
2
( x + 75) x + 75
dx 2
2
( x + 75 )
2
d 2F
150 k
150 k
=
=
>0
dx 2 ( 25 + 75)3/ 2
103
∴ when x = 5, F is minimum.
So, the freight charge will be minimum if the road is connected to the railway line at a distance 5 km
from B or 15 km from the town A.
EXAMPLE 5
A rectangular sheet of metal has four equal square portions removed at the corners and the sides
are then turned up so as to form an open rectangular box. Show that when the volume contained
1
in the box is a maximum, the depth will be ⎡a 1b 2 a 2 2ab 2b 2 ⎤ , where a and b the sides of
⎦
b⎣
the original rectangle.
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Solution.
Given a and b be the sides of the rectangular sheet of metal.
Let x be the side of square cut off from the corners.
∴ the dimensions of the box formed by folding up the sides are a − 2x , b − 2x and x.
The volume of the box
V = x (a − 2x )(b − 2x )
⇒
V = x [ab − 2(a + b )x + 4 x 2 ]
⇒
V = abx − 2(a + b )x 2 + 4 x 3
∴
dV
= ab − 4(a + b )x + 12x 2 .
dx
For maximum or minimum,
dV
=0
dx
x
x
x
b − 2x
x
a − 2x
x
Fig. 3.22
12x 2 − 4(a + b )x + ab = 0
⇒
x=
4(a + b ) ± 16(a + b ) 2 − 48ab
24
=
4(a + b ) ± 4 (a + b ) 2 − 3ab
24
=
(a + b ) ± a2 + b 2 + 2ab − 3ab
6
a + b ± a2 + b 2 − ab
6
1⎡
= a + b + a2 + b 2 − ab ⎤
⎦
6⎣
=
Now
When x =
(
or
1⎡
a + b − a2 + b 2 − ab ⎤ .
⎦
6⎣
d 2V
= 24 x − 4(a + b ) = 4[6 x − (a + b )].
dx 2
)
1
a + b − a2 + b 2 − ab ,
6
⎡ ⎛ a + b − a 2 + b 2 − ab ⎞
⎤
d 2V
⎢6 ⎜
=
4
⎟ − ( a + b) ⎥
2
6
dx
⎢⎣ ⎝
⎥⎦
⎠
= 4 ⎡ a + b − a 2 + b 2 − ab − (aa + b) ⎤ = −4 a 2 + b 2 − ab < 0.
⎣
⎦
∴ when the depth is
1⎡
a + b − a2 + b 2 − ab ⎤ , V is maximum.
⎦
6⎣
EXAMPLE 6
A cone circumscribed a sphere of radius r. Prove that when the volume of the cone is minimum,
⎛ 1⎞
its height is 4r and the semi−vertical angle is sin21 ⎜ ⎟ .
⎝ 3⎠
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Solution.
Let the sphere be of radius r. Let the cone be of radius R and height h.
Let the semi−vertical angle of the cone be u. Let V be the volume of the cone.
A
1
V = pR 2 h.
3
Let O be centre of the sphere. D be the centre of the
base circle of the cone.
Then Volume
∴
OD = r
∴
OA = h − r
From Δ A BD ,
BD
= tan u
AD
From Δ AEO,
EO
= tan u.
EA
But
∴
∴
∴
Now
h
90°
r
O
E
AD = h
and
r
B
⇒ BD = h tan u
R
D
C
Fig. 3.23
EA = OA2 − r 2 = ( h − r ) 2 − r 2 = h2 + r 2 − 2rh − r 2 = h2 − 2hr
tan u =
r
∴ BD = h ⋅
h − 2hr
2
r
h − 2hr
2
=
hr
h − 2hr
2
.
h3
h2
1 2
1
1
h2r 2
= pr 2
V = p 2
.h = pr
h( h − 2r ) 3
( h − 2r )
3
3 h − 2hr
dV pr 2 [( h − 2r )2h − h 2 ⋅1]
pr 2 h[2h − 4 r − h ] pr 2 h ( h − 4 r )
=
.
=
=
dh
3
3( h − 2r ) 2
3( h − 2r ) 2
( h − 2r ) 2
For maximum or minimum
⇒
θ
dV
=0
dh
pr 2 h ( h − 4 r )
=0
3( h − 2r ) 2
⇒
h( h − 4r ) = 0
⇒
h − 4r = 0
⇒
h = 4r
[{h ≠ 0]
d 2V pr 2 [( h − 2r ) 2 ⋅ ( 2h − 4 r ) − ( h 2 − 4 rh )2( h − 2r )]
=
3
( h − 2r ) 4
dh 2
=
pr 2 ( h − 2r )[( h − 2r )2( h − 2r ) − 2( h 2 − 4 rh )]
3
( h − 2r ) 4
⇒
d 2V pr 2 [2( h − 2r ) 2 − 2h( h − 4 r )]
=
.
3
dh2
( h − 2r )3
When h = 4 r ,
d 2V pr 2 2( 4 r − 2r ) 2 − 2h × 0 pr 2 2 pr
=
=
⋅ =
>0
3 2r
3
3
dh2
( 4 r − 2r )3
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∴ when h = 4 r , V is minimum.
Now
sinu =
OE
r
r
1
=
=
=
OA h − r 4 r − r 3
⎛ 1⎞
u = sin −1 ⎜ ⎟ .
⎝ 3⎠
⇒
EXERCISE 3.12
1. Find the maximum and minimum values of x 5 − 5x 4 + 5x 3 − 1.
2. Find the maximum and minimum values of
x 2 + x +1
.
x 2 − x +1
3. Find the maximum and minimum values of a2 sin 2 x + b 2 cos 2 x .
4. If xy = 4, find the maximum and minimum values of 4 x + 9 y .
5. Find the maximum and minimum values of x − x 2 on [ −2, 2].
x2 y 2
+
= 1. Find the equation
6. An isosceles triangle with vertex (0, 2) is inscribed in the ellipse
9
4
of the base if the area of the triangle is a maximum.
7. A manufacturer plans to construct a cylindrical can to hold one cubic meter of liquid. If the cost
of constructing the top and bottom of the can is twice the cost of constructing the side, what are
the dimensions of the most economical can?
8. Find the rectangle of maximum area with sides parallel to the coordinate axes which can be
inscribed in the figure bounded by two parabolas 3y = 12 − x 2 and 6 y = x 2 − 12.
9. The cost of fuel in running an engine is proportional to the square of the speed and is Rs. 48
per hour for speed of 16 km/hr and other costs per hour amount Rs. 300. What is the most
economical speed and the cost of a journey of 400 km.
p
10. Show that the function sin x (1+ cos x ) is a maximum when x = . Does this function have
3
minimum at x = 0 or p ?
⎧⎪ x − 2 + a2 − 9a − 9
11. Let f ( x ) = ⎨
⎪⎩2x − 3,
if x < 2
if x ≥ 2.
Find the values of a if f ( x ) has a local minimum at x = 2.
[Hint: f ( x ) = 2 − x + a 2 − 9a − 9 if r < 2
= 2x − 3
if x ≥ 2
and
lim f ( x ) ≥ f ( 2)]
x→2−
12. Find the maximum possible slope for a tangent line to the curve y =
8
.
1 + 3e − x
13. Find the area of the largest rectangle with lower base on the x-axis and the upper vertices on
y = 12 − x 2.
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ANSWERS TO EXERCISE 3.12
1. Maximum at x = 2, Maximum value = −9
2. Maximum at x = 1, Maximum value = 3
Minimum at x = 3, Minimum value = −28
1
Minimum at x = −1, Minimum value =
No extremum at x = 0
3
p
3. Maximum at x = , Maximum value = a
4. Maximum at x = 3, Maximum value = −24
2
Minimum at x = 3, Minimum value = 24
Minimum at x = p, Minimum value = b
1
5. Maximum at x = −2, x = and x = 2, Maximum values are 6, 4 and 2
2
Minimum at x = 0 and x = 1, Minimum values are 0, 0
6. y + 1 = 0,
7. h = 4 r
8. 16
9. 40 km per hour and Rs. 6000.
10. x = p, f ″(p) = − sin p( 4 cos p + 1) = 0
11. a ∈ ( −∞, − 1] ∪ [10, ∞).
Hence, f ( x ) is neither a maximum nor a minimum at x = p.
12. So, the greatest possible slope is when x = loge 3 and greatest slope = 2.
13. Maximum area = 32 sq.unit.
3.8
ASYMPTOTES
The study of asymptotes is yet another aspect of characterizing the shape of a curve. In this section we
study rectilinear asymptote. Roughly, an asymptote to an infinite curve is a straight line touching the
curve at an infinite distance from the origin.
In order that a curve to have asymptote it should extend up to infinity. Closed curves like circle and
ellipse will not have asymptotes. But every curve extending up to infinity need not have asymptotes
for example parabola y 2 = 4ax extends up to infinity, yet it has no asymptote.
We shall now formally define an asymptote.
Definition 3.7 A point P( x , y ) on an infinite curve is said to tend to infinity (i.e., P → ∞) along the
curve as either x or y or both tend to ∞ or −∞ as P moves along the curve.
Definition 3.8 Asymptote
A straight line at a finite distance from the origin is called an asymptote of an infinite curve, if when a
point P on the curve tends to ∞ along the curve, the perpendicular distance from P to the line tends to 0.
An asymptote parallel to the x-axis is called a horizontal asymptote and an asymptote parallel to
the y-axis is called a vertical asymptote.
An asymptote which is not parallel to either axis will be called an oblique asymptote.
Theorem 3.7 If y = mx + c (where m and c are finite) is an asymptote of an infinite curve, then
⎛y⎞
m = lim ⎜ ⎟ and c = lim( y − mx ),
x →∞ ⎝ x ⎠
x →∞
where P( x , y ) is any point on the infinite curve.
Proof
Given P( x , y ) be any point on the infinite curve.
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The perpendicular distance from P( x , y ) to the line y − mx − c = 0
d=
is
y − mx − c
1 + m2
3.105
(1)
(2)
If the line (1) is an asymptote to the curve, then d → 0 as P → ∞. i.e., as x → ∞(or −∞).
∴
lim( y − mx − c ) = 0
⇒
x →∞
lim( y − mx ) = c.
x →∞
y
1
− m = ( y − mx )
x
x
Also
y
1
⎛y
⎞
lim ⎜ − m ⎟ = lim( y − mx ) lim = c × 0 = 0 ⇒ lim = m
x
→∞
x
→∞
x
→∞
⎝x
⎠
x
x
∴
x →∞
⎛y⎞
lim ⎜ ⎟ = m and c = lim( y − mx ).
x →∞
⎝x⎠
Conversely, if these two limits exists as P → ∞, then y − mx − c → 0 as x → ∞
and hence, d → 0 as P → ∞
∴
y = mx + c is an asymptote.
Hence,
x →∞
Note
(1) In the theorem m and c are finite. If m = 0, then the asymptote is parallel to x-axis.
(2) The above theorem gives a method of finding asymptotes not parallel to y-axis.
Working rule:
Given a curve f ( x , y ) = 0.
⎛y⎞
(i) Find lim ⎜ ⎟ , where y = f( x ).
x →∞ ⎝ x ⎠
For different branches of the curve, we may get different values for this limit.
(ii) If m is one such value, then find lim( y − mx ).
x →∞
Let this limit be c, then y = mx + c is an asymptote to the curve.
Note
The above method will give all asymptotes not parallel to y-axis.
To find asymptotes not parallel to x-axis, we start with x = my + d and x = f( y ),
⎛x⎞
where m = lim ⎜ ⎟ and d = lim( x − my ).
y →∞ ⎝ y ⎠
y →∞
WORKED EXAMPLES
EXAMPLE 1
Find the asymptotes of the curve y 5
3x
.
x 22
Solution.
3x
The given curve is y =
.
x −2
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When x = 2, y → ∞.
∴ x = 2 is a vertical asymptote.
2y
.
Also rewriting the equation as x interms of y, x =
y −3
When y = 3, x → ∞.
So, y = 3 is a horizontal asymptote.
Note: Determination of asymptotes parallel to the axes
Let x = k be an asymptote parallel to the y-axis. Then
lim (x − k ) = 0
y →∞
⇒
lim x = k .
y →∞
Find the values of x, for which y → ∞. For each value of x we get a vertical asymptote x = k .
Similarly, to find the asymptote parallel to the x-axis, find the values of y for which x → ∞.
For each value of y, we get a horizontal asymptote y = k .
EXAMPLE 2
Find the vertical and horizontal asymptotes of the curve y 5
3x 2
.
x 2 1 2 x 215
Solution.
The curve is y =
3x 2
3x 2
⇒
( x − 3)( x + 5)
x 2 + 2 x − 15
When x = −5 and x = 3, y → ∞
∴ x = −5 and x = 3 are vertical asymptotes.
⎡
⎤
⎡
⎤
⎢
⎥
⎢
⎥
⎡
⎤
3x 2
3
3x 2
⎢
⎥ = lim ⎢
Now lim y = lim ⎢ 2
⎥=3
⎥ = xlim
x →∞
x →∞ x + 2x − 15
→∞ ⎢
x
→∞
2
15
2 15 ⎞ ⎥
2 ⎛
⎣
⎦
⎢1 + − ⎥
x
1
+
−
⎢ ⎜⎝ x x 2 ⎟⎠ ⎥
⎣ x x2 ⎦
⎣
⎦
∴ y = 3 is the horizontal asymptote.
EXAMPLE 3
Find the vertical and horizontal asymptotes of the graph of the function f ( x ) 5
x 2 29
.
x 2 1 3x
Solution.
Let the equation of the given curve be
y =
When x = 0, y → ∞.
Now
x2 −9
( x + 3)( x − 3) x − 3
=
=
.
2
x ( x + 3)
x
x + 3x
y
∴ x = 0 is a vertical asymptote.
⎛ x − 3⎞
⎛ 3⎞
lim y = lim ⎜
= lim ⎜1 − ⎟ = 1.
⎟
x →∞
x →∞ ⎝
x ⎠ x →∞ ⎝ x ⎠
y=1
∴ y = 1 is the horizontal asymptote.
0
x=3
x
Note
The graph has a break at x = 0 i.e., discontinuous at x = 0
and continuous for all other values of x.
The y-axis x = 0 and y = 1 are the asymptotes.
Fig. 3.24
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Differential Calculus ■
EXAMPLE 4
Find the asymptote of the curve y 5
3.107
2x
1 5x .
x 23
Solution.
y =
The given curve is
When x = 3, y → ∞ .
2x
+ 5x .
x −3
∴ x = 3 is a vertical asymptote.
⎛
⎞
⎜ 2
⎟
⎛ 2x
⎞
lim y = lim ⎜
+ 5 x ⎟ = lim ⎜
+ 5 x ⎟ → ∞.
x →∞
x →∞ ⎝ x − 3
3
⎠ x →∞
⎜1−
⎟
⎝
⎠
x
Now
∴ there is no horizontal asymptote.
To find the oblique asymptote
⎛y⎞
⎛ 2
⎞ 2
+ 5⎟ = + 5 = 5
We know m = lim ⎜ ⎟ = lim ⎜
x →∞ ⎝ x ⎠
x →∞ ⎝ x − 3
⎠ ∞
And
⎛
⎞
2
⎜ 2 ⎟
⎛ 2x
⎞
⎛ 2x ⎞
c = lim( y − mx ) = lim ⎜
+ 5x − 5x ⎟ = lim ⎜
= li m
=
= 2.
x →∞
x →∞ ⎝ x − 3
3 ⎟ 1− 0
⎠ x →∞ ⎝ x − 3 ⎟⎠ x →∞ ⎜
⎜1− ⎟
⎝
x⎠
∴
y = 5x + 2 is an oblique asymyptote.
EXAMPLE 5
Find the asymptotes if any, of the curve y 5 xe 1/ x .
Solution.
The given curve is
When x → 0 + ,
y = xe 1/ x .
1
y = lim xe1/ x
→ ∞. ∴ e1/ x → ∞ and xlim
x→0+
→0+
x
e1/ x
x → 0 + 1/ x
= lim
e 1/ x (1/x 2 )
,
x →0+
1
− 2
x
= lim e 1/ x = ∞.
= lim
(0 ⋅ ∞ form)
⎛∞
⎞
⎜⎝ form⎟⎠
∞
[ by L’hopital’s rule]
x →0+
∴ as x → 0 +, y → ∞. Hence, x = 0 is a vertical asymptote.
It can be seen that as x → ∞, y → ∞ and so, there is no horizontal asymptote.
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Engineering Mathematics
To find the oblique asymptote
We know
y
xe1/ x
m = lim = lim
= lim e1/ x = e 0 = 1.
x →∞ x
x →∞
x →∞
x
and
c = lim( y − mx ) = lim( xe1/ x − x ) = lim x (e 1/ x − 1)
x →∞
x →∞
(∞ ⋅ 0 form)
x →∞
(e1/ x − 1)
e1/ x ( −1/ x 2 )
= lim
= lim e1/x = e 0 = 1
x →∞
x →∞
x →∞
1/ x
−1/ x 2
⇒
c = lim
∴
y = x +1 is the oblique asymptote.
⎛0
⎞
⎜⎝ form⎟⎠
0
3.8.1 A General Method
Find the asymptotes of the rational algebraic curve f ( x , y ) 5 0
Consider the general algebraic curve of nth degree in x and y
a0 x n + a1x n −1 y + a2 x n − 2 y 2 + … + an −1xy n −1 + an y n
+b1x n −1 + b 2 x n − 2 y + … + b n y n −1 + c 2 x n − 2 + c3 x n − 3 y + … + c n y n − 2 + …
+( rn −1x + rn y ) + s n = 0
It can be rewritten as
n −1
⎡
y
y2
yn ⎤
⎛y⎞
x n ⎢a0 + a1 + a2 2 + … + an −1 ⎜ ⎟ + an n ⎥
⎝x⎠
x
x
x ⎥⎦
⎢⎣
2
n −1
⎡
⎡
y
y
y ⎤
y
y n−2 ⎤
+ x n −1 ⎢b1 + b 2 + b3 2 + … + b n n −1 ⎥ + x n − 2 ⎢c 2 + c3 + … + c n n − 2 ⎥ + …
x
x
x
x ⎦
x ⎦
⎣
⎣
(1)
y⎞
⎛
+ x ⎜ rn −1 + rn ⎟ + s n = 0
⎝
x⎠
It is of the form
⎛y⎞
⎛y⎞
⎛y⎞
⎛y⎞
⎛y⎞
x n fn ⎜ ⎟ + x n −1fn −1 ⎜ ⎟ + x n − 2fn − 2 ⎜ ⎟ + … + x f1 ⎜ ⎟ + f0 ⎜ ⎟ = 0.
⎝x⎠
⎝x⎠
⎝x⎠
⎝x⎠
⎝x⎠
(2)
y
⎛y⎞
Where fr ⎜ ⎟ is a polynomial of degree r in .
⎝x⎠
x
y
c
= m + in (2).
x
x
c
c
c
⎞
⎛
⎞
⎛
⎞
⎛
∴
x n fn ⎜ m + ⎟ + x n −1fn −1 ⎜ m + ⎟ + x n − 2fn − 2 ⎜ m + ⎟ + … = 0.
⎝
⎝
⎝
x⎠
x⎠
x⎠
Expanding by Taylor’s theorem, we get
To find the point of intersection of the line y = mx + c with (2), put
⎤
⎡
c
1 c2
x n ⎢fn ( m ) + fn′ ( m ) +
fn′′( m ) + …⎥
2
x
2! x
⎣
⎦
⎤
⎡
1 c2
c
fn′′−1 ( m ) + …⎥
+ x n −1 ⎢fn −1 ( m ) + fn′−1 ( m ) +
2
2
!
x
x
⎣
⎦
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3.109
⎤
⎡
1 c2
c
fn′′− 2 ( m ) + …⎥ + … = 0.
+ x n − 2 ⎢fn − 2 ( m ) + fn′− 2 ( m ) +
2
2
!
x
x
⎣
⎦
⇒
x n fn ( m ) + x n −1 [cfn′ ( m ) + fn −1 ( m )]
⎤
⎡c2
+ x n − 2 ⎢ fn′′ ( m ) + cfn′−1 ( m ) + fn − 2 ( m ) ⎥ + … = 0.
⎣ 2!
⎦
n
Dividing by x , we get
1
fn ( m ) + [cfn′ ( m ) + fn −1 ( m )]
x
+
Also from (2), we get
⎤
1 ⎡c2
fn′′ ( m ) + cfn′−1 ( m ) + fn − 2 ( m ) ⎥ + … = 0
2 ⎢
x ⎣ 2!
⎦
(3)
⎛y⎞ 1
⎛y⎞ 1
⎛y⎞
fn ⎜ ⎟ + fn −1 ⎜ ⎟ + 2 fn − 2 ⎜ ⎟ + … = 0.
⎝x⎠ x
⎝x⎠ x
⎝x⎠
(4)
⎛y⎞
y = mx + c is an asymptote if lim ⎜ ⎟ = m .
x →∞ ⎝ x ⎠
Hence, from (4), we get fn ( m ) = 0.
(5)
The real values of m are the slopes of the asymptotes.
Substituting (5) in (3), we get
1
1
[cfn′ ( m ) + fn −1 ( m )] + 2
x
x
⎤
⎡c2
⎢ fn′′( m ) + cfn′−1 ( m ) + fn − 2 ( m ) ⎥ + … = 0.
⎣ 2!
⎦
Multiplying by x and taking limit as x → ∞, we get
cfn′ ( m ) + fn −1 ( m ) = 0
fn −1 ( m )
if fn′ ( m ) ≠ 0.
(6)
fn′ ( m )
If m1, m2, …, mr are the real roots of fn ( m ) = 0, then the corresponding values of c from (6)
are c1, c2, …, cr
∴ the asymptotes are
⇒
c=−
y = m1x + c1 ,
y = m 2 x + c 2 , …,
y = m r x + cr .
Note
(1) Suppose fn′ ( m ) = 0 and fn −1 ( m ) ≠ 0 then c is infinite and hence, there is no asymptote to the
curve, in this case.
(2) Suppose fn′ ( m ) = 0 and fn ( m ) = 0 then cfn ( m ) + fn −1 ( m ) = 0 is an identity.
If fn′ ( m ) = 0, then fn ( m ) = 0 has repeated roots.
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Let the repeated roots be m1, m1, then c is given by
c2
fn′′( m) + cfn′−1 ( m) + fn − 2 ( m) = 0 if fn′′( m ) ≠ 0.
2
If c1, c2 are the roots, then y = m1x + c1 and y = m1 x + c2 are parallel asymptotes.
Working Rule to find oblique asymptotes of algebraic rational function f(x, y) = 0
(1) Put x = 1, y = m in the highest degree terms.
That is in the nth degree terms and find fn ( m ).Solve fn ( m ) = 0 to find the real roots m1 , m 2 ,..., m r .
(2) Put x = 1, y = m in the next highest degree terms. That is in the (n − 1)th degree terms and get
fn −1 ( m ).
Then find c = −
fn −1 ( m )
if fn′ ( m ) ≠ 0.
fn′ ( m )
Find c1, c2, …, cr corresponding to m1, m2, … mr.
Then the asymptotes are
y = m1 x + c1 ,
y = m2 x + c2 , ...,
y = mr x + cr .
(3) If fn′ ( m ) = 0 and fn′−1 ( m ) = 0 and two roots of fn ( m ) = 0 are equal say m1, m1, then the values of
c are given by
c2
fn′′( m ) + cfn′−1 ( m ) + fn′− 2 ( m ) = 0 if fn′′( m ) ≠ 0.
2
If c1, c2 are the roots, then we get parallel asymptotes y = m1x + c1 and y = m1x + c 2 .
3.8.2 Asymptotes parallel to the coordinates axes
Let f(x, y) = 0 be the rational algebraic equation of the given curve.
(1) To find the asymptotes parallel to the x-axis, equate to zero the coefficients of highest power
of x.
The linear factors of this equation are the asymptotes parallel to the x-axis.
If the highest coefficient is constant or if the linear factors are imaginary, then there is no
horizontal asymptotes.
(2) To find the asymptotes parallel to the y-axis, equate to zero the coefficients of the highest power
of y.
The real linear factors of this equation are the asymptotes parallel to the y-axis.
If the highest coefficient is constant or if the linear factors are imaginary, then there is no vertical
asymptotes.
WORKED EXAMPLES
EXAMPLE 1
Find the asymptotes of the curve
x 3 1 2 x 2 y 2 xy 2 2 2 y 3 1 4 y 2 1 2 xy 1 y 215 0.
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Differential Calculus ■
3.111
Solution.
The given curve is
x 3 + 2x 2 y − xy 2 − 2 y 3 + 4 y 2 + 2xy + y − 1 = 0.
It is a third degree equation. The third degree terms are
x 3 + 2x 2 y − xy 2 − 2 y 3 .
Put x = 1, y = m, we get
Solve
f3 ( m ) = 1 + 2m − m 2 − 2m 3
(1)
f3 ( m ) = 0 ⇒ 1 + 2m − m 2 − 2m 3 = 0
⇒
1 + 2m − m 2 (1 − 2m ) = 0
1
(1 + 2m )(1 − m )(1 + m ) = 0 ⇒ m = − , 1, − 1
2
Now put x =1, y = m in the second degree terms 4y2 + 2xy.
We get
f2 ( m ) = 4 m 2 + 2m = 2m ( 2m + 1)
⇒
(1 + 2m )(1 − m 2 ) = 0
Now
c=−
⇒
fn −1 ( m )
f (m )
=− 2
fn′ ( m )
f3′( m )
But
f3 ( m ) = 1 + 2m − m 2 − 2m 3
∴
f3′ ( m ) = 2 − 2m − 6 m 2 = −2(3m 2 + m − 1)
−2m ( 2m + 1)
m ( 2m + 1)
.
=
2
−2(3m + m − 1) 3m 2 + m − 1
∴
c=
1
When m = − ,
2
c=0
When m = −1
c=
( −1)( −2 + 1)
1
=
= 1.
3( −1) 2 + ( −1) − 1 3 − 1 − 1
When m = 1
c=
1( 2 ⋅1 + 1)
3
= =1
2
3 ⋅1 + 1 − 1 3
1
∴ the asymptotes are y = − x , y = − x + 1, y = x + 1.
2
Note Since the coefficient of x3 and y3 are constants, there is no asymptotes parallel to x-axis and
y-axis.
EXAMPLE 2
Find the asymptotes of the curve y 3 1 x 2 y 1 2 xy 2 2 y 1 1 5 0.
Solution.
The given curve is
y 3 + x 2 y + 2xy 2 − y + 1 = 0.
It is cubic equation.
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Since coefficient of y 3 is 1, a constant, there is no asymptotes parallel to the y-axis.
The highest degree term in x is x 2 and the coefficient of x 2 is y, equating the coefficient of x 2 to
zero we get y = 0 is the asymptote, which is the x−axis.
To find the other asymptotes
Put x = 1, y = m in the cubic terms y 3 + x 2 y + 2xy 2 .
∴
f3 ( m ) = m 3 + m + 2m 2
∴
f3′( m ) = 3m 2 + 4 m + 1.
There is no second degree terms.
∴
f2 ( m ) = 0
and
c= −
f2 ( m )
if f3′( m ) ≠ 0
f3′( m )
⇒
m ( m + 1) 2 = 0
Solving f3 ( m ) = 0
⇒
m 3 + 2m 3 + m = 0
⇒
m ( m 2 + 2m + 1) = 0
When m = 0, c = 0.
⇒
(1)
m = 0 or m = −1, − 1.
∴ the asymptote isy = 0.
But when m = −1, we can’t find c using (1), because f3′( −1) = 0.
∴ we can find c using
c2
f3′′( m ) + cf2′( m ) + f1 ( m ) = 0
2
f3 ″( m ) = 6 m + 4,
Now
∴
c2
(6 m + 4 ) + 0 − m = 0
2
When m = −1,
f2 ′( m ) = 0,
f1 ( m ) = −1
⇒ c 2 (3m + 4) − m = 0
c 2 (3( −1) + 4) − 1 = 0
⇒
c 2 − 1 = 0 ⇒ c = ±1
∴ there are two parallel asymptotes
y = − x + 1 and y = − x − 1.
∴ the three asymptotes are
y = 0, y = − x + 1, y = − x − 1.
EXAMPLE 3
Find the asymptotes of the curve x 5
t2
t2 12
,
.
y
5
11t
11t 3
Solution.
The equation of the curve is given in parametric form
x=
M03_ENGINEERING_MATHEMATICS-I _CH03_Part D.indd 112
t2
t2 + 2
and y =
.
3
1+ t
1+ t
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Differential Calculus ■
3.113
When t = −1, x → ∞ and y → ∞, we get an asymptote.
We know that y = mx + c is an asymptote if m = lim
x →∞
y
and c = lim( y − mx )
x →∞
x
where (x, y) is a point on the curve.
⎡t2 + 2⎤
⎢
⎥
⎛y⎞
We have m = lim ⎜ ⎟ = lim ⎢ 1 +2 t ⎥
t →−1
x →∞ ⎝ x ⎠
⎢ t ⎥
⎢⎣ t 3 + 1 ⎥⎦
({ as x → ∞, t → −1).
⎡ (t 2 + 2)(t 2 − t + 1) ⎤ (( −1) 2 + 2)[( −1) 2 − ( −1) + 1]
= lim ⎢
= (1 + 2)(1 + 1 + 1) = 9.
⎥=
t →−1
t2
( −1) 2
⎣
⎦
and
c = lim( y − mx ) = lim( y − 9x )
x →∞
x →∞
⎡t 2 + 2
t2 ⎤
−9
= lim ⎢
⎥
t →−1 1 + t
1+ t 3 ⎦
⎣
(t 2 + 2)(1 + t 3 ) − 9t 2 (t + 1)
t →−1
(1 + t )(1 + t 3 )
= lim
[(t 2 + 2)(t 2 − t + 1) − 9t 2 ](t + 1)
t →−1
(1 + t )(1 + t 3 )
= lim
(t 2 + 2)(t 2 − t + 1) − 9t 2
t →−1
(1 + t 3 )
= lim
t 4 − t 3 + t 2 + 2t 2 − 2t + 2 − 9t 2
t →−1
(1 + t 3 )
= lim
⎡0
⎤
⎢ 0 form ⎥
⎣
⎦
t 4 − t 3 + 6t 2 − 2t + 2
t →−1
1+ t 3
= lim
4t 3 − 3t 2 − 12t − 2
t →−1
3t 2
= lim
=
[by L’Hopital’s rule]
4( −1)3 − 3( −1) 2 − 12( −1) − 2 −4 − 3 + 12 − 2 12 − 9
=
=
= 1.
3
3
3( −1) 2
∴ the asymptote is y = 9x + 1.
3.8.3 Another Method for Finding the Asymptotes
Let the equation of the curve be nth degree in x and y.
Suppose the nth degree curve can be put in the form (ax + by + c )Pn −1 + Fn −1 = 0, where Pn −1 and Fn −1
denote polynomials of degree ( n −1) in x and y.
Any line parallel to ax + by + c = 0 that cut the curve in two points at infinity is an asymptote and
⎛F ⎞
it is given by ax + by + c + lim ⎜ n −1 ⎟ = 0, if the limit is finite.
a
y = − x →∞ ⎝ Pn −1 ⎠
b
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Suppose ax + by + c is a factor of Pn−1, then the equation of the curve takes the form
( ax + by + c) 2 Pn − 2 + Fn − 2 = 0 and the parallel asymptotes are given by
1/ 2
⎡ ⎛ F ⎞⎤
ax + by + c = ± ⎢lim ⎜ − n − 2 ⎟ ⎥
⎢⎣ ⎝ Pn − 2 ⎠ ⎥⎦
a
, when x , y → ∞ along y = − b x .
If the equation is (ax + by + c)Pn−1 + Fn−2 = 0, then ax + by + c = 0 is an asymptote.
WORKED EXAMPLES
EXAMPLE 1
Find the asymptotes of x 3 1 y 3 5 3axy .
Solution.
The equation of the given curve is
⇒
x 3 + y 3 = 3axy .
( x + y )( x 2 − xy + y 2 ) − 3axy = 0.
This is of the form ( x + y )Pn −1 + Fn −1 = 0.
∴ the asymptotes parallel to x + y = 0is
⎡ −3axy ⎤
x + y + lim ⎢ 2
⎥=0
y = − x →∞ x − xy + y 2
⎣
⎦
⇒
−3ax ( − x )
=0
x →∞ x − x ( − x ) + ( − x ) 2
⇒
x + y + lim
2
3ax 2
= 0 ⇒ x + y + lim a = 0
x →∞ 3x 2
x →∞
There is no asymptote parallel to the axes. It has only one asymptote.
⇒
x + y + lim
⇒
x + y + a = 0.
2⎞
2⎞
⎛
⎛
8 x ⎜1 + ⎟
8 ⎜1 + ⎟
⎝ 8x ⎠
⎝ 8x ⎠
= lim
=8
x →∞
2
⎛ 2⎞
1−
x ⎜1 − ⎟
⎝ x⎠
x
⎡ 2
⎤
⎢{ x → 0 ⎥
⎣
⎦
EXAMPLE 2
Find the asymptotes of ( x 1 y ) 2 ( x 1 2 y 1 2) 5 x 1 9 y 22.
Solution.
The equation of the given curve is ( x + y ) 2 ( x + 2 y + 2) = x + 9 y − 2.
This is of the form ( x + y ) 2 Pn − 2 + Fn − 2 = 0.
The asymptotes parallel to x + y = 0 are
( x + y ) 2 = lim
y = − x →∞
x + 9y − 2
x − 9x − 2
= lim
x + 2 y + 2 x →∞ x − 2 x + 2
8x + 2
= lim
= lim
x →∞ x − 2
x →∞
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Differential Calculus ■
∴
3.115
x + y = ±2 2 are two asymptotes.
Now the equation is of the form
( x + 2 y + 2) Pn −1 + Fn − 2 = 0
∴ x + 2 y + 2 = 0 ia an asymptote.
Hence, x + y = ±2 2 , x + 2 y + 2 = 0 are the three asymptotes. [Work out this by the general
method]
EXAMPLE 3
Find the asymptotes of x 3 22 x 2 y 1xy 2 1x 2 2xy 12 50.
Solution.
It is a third degree equation in x and y.
Since the coefficient of x 3 is constant there is no asymptote parallel to the x-axis.
Since the coefficient of y 3 is x, asymptote parallel to the y-axis is x = 0.
That is the y-axis itself.
Factorising the third degree terms
⇒
x ( x 2 − 2xy + y 2 ) + x ( x − y ) + 2 = 0
⇒
x ( y − x )2 − x ( y − x ) + 2 = 0
⇒
x (x − y )2 + x (x − y ) + 2 = 0
( y − x )2 − ( y − x ) +
Asymptote parallel to y − x = 0 is given by ( y − x ) 2 − ( y − x ) + lim
y = x →∞
⇒
⇒
2
= 0.
x
2
=0
x
( y − x )2 − ( y − x ) = 0
( y − x )[( y − x ) − 1] = 0
∴ the asymptotes are x = 0,
y − x = 0,
⇒
y − x = 0, y − x − 1 = 0
y − x − 1 = 0.
3.8.4 Asymptotes by Inspection
In certain cases, we can find the asymptotes of an rational algebraic equation without any calculations.
If the equation can be rewritten in the form Fn + Fn − 2 = 0, where Fn is a polynomial of degree n in x
and y and Fn − 2 is a polynomial of degree almost n − 2.
If Fn can be factored into linear factors so that no two of them represent parallel straight lines, then
Fn = 0 gives all the asymptotes.
x2 y 2
For example: The equation of the hyperbola 2 − 2 = 1is of the Fn + Fn − 2 = 0, where
a
b
x2 y 2 ⎛ x y ⎞ ⎛ x y ⎞
−
=⎜ − ⎟⎜ + ⎟.
a2 b 2 ⎝ a b ⎠ ⎝ a b ⎠
x y
x y
So, the asymptotes are − = 0 and + = 0.
a b
a b
Fn =
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WORKED EXAMPLES
EXAMPLE 4
Find the asymptotes of ( x 1 y )( x 2 y )( x 2 2 y 2 4) 5 3 x 1 7 y 2 6.
Solution.
The given curve is
( x + y )( x − y )( x − 2 y − 4) − (3x + 7 y − 6) = 0.
This is of the form Fn + Fn −2 = 0, where Fn = F3 = ( x + y )( x − y )( x − 2 y − 4) and Fn is the product of
linear factors, which do not represent parallel lines.
∴ the asymptotes are given by
Fn = 0
⇒
( x + y )( x − y )( x − 2 y − 4) = 0
∴ The asymptotes of the given curve are x + y = 0,
x − y = 0,
x − 2y − 4 = 0 .
3.8.5 Intersection of a Curve and Its Asymptotes
Any asymptote of an algebraic curve of nth degree cuts the curve in two points at infinity and in ( n − 2)
other points. So, the n asymptotes of the curve cut it in atmost n ( n − 2) points.
If the equation of the curve is written in the form Fn + Fn − 2 = 0, where Fn is of nth degree and is a
product of n linear factors and Fn −2 is of degree atmost n − 2, then, the equation of the asymptote is
given by Fn = 0.
So, the point of intersection of the curve and the asymptote are obtained by solving Fn = 0,
Fn + Fn −2 = 0 and hence such points lie on the curve Fn − 2 = 0.
Note
If C is the equation of the curve and A is the combined equation of the asymptotes, then the curve on
which the points intersection of the asymptotes lie is C − A = 0.
WORKED EXAMPLES
EXAMPLE 5
Show that the asymptotes of the cubic x 2 y 2 xy 2 1 xy 1 y 2 1 x 2 y 5 0 cut the curve again in
three points which lie on the line x 1 y 5 0 .
Solution.
The given curve is
x 2 y − xy 2 + xy + y 2 + x − y = 0
Since the coefficient of x 2 is y, the asymptote parallel to the x-axis is y = 0.
Since the coefficient of y 2 is 1− x , the equation of the asymptote parallel to the y-axis is
1 − x = 0 ⇒ x = 1.
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Now the equation can be rewritten as
xy ( x − y ) + ( xy + y 2 + x − y ) = 0
This is of the form ( x − y )Pn −1 + Fn −1 = 0.
∴ the asymptote parallel to x − y = 0 is
Fn −1
=0
x = y →∞ P
n −1
x − y + lim
⇒
⎡x2 + x2 + x − x ⎤
x − y + lim ⎢
⎥=0
x →∞
x2
⎣
⎦
⎡ xy + y 2 + x − y ⎤
x − y + lim ⎢
⎥=0
x = y →∞
xy
⎣
⎦
⇒
⎛ 2x 2 ⎞
⇒ x − y + lim ⎜ 2 ⎟ = 0
x →∞ ⎝ x
⎠
⇒
x − y + 2 = 0.
∴ the asymptotes are y = 0, x − 1 = 0, x − y + 2 = 0.
The curve cannot have more than 3 asymptotes.
∴ their combined equation is
y ( x − 1)( x − y + 2) = 0 ⇒
⇒
x 2 y − xy 2 + 2xy − xy + y 2 − 2 y = 0
∴
A ≡ x 2 y − xy 2 + xy + y 2 − 2 y = 0.
⇒
( xy − y )( x − y + 2) = 0
x 2 y − xy 2 + xy + y 2 − 2 y = 0.
The curve is
C ≡ x 2 y − xy 2 + xy + y 2 + x − y = 0.
The point of intersection of the asymptotes lie on the curve.
C − A = 0.
⇒ x + y = 0,
which is a straight line and the number of points of intersection is 3(3 – 2) = 3.
EXAMPLE 6
Show that the four asymptotes of the curve
( x 2 2 y 2 )( y 2 2 4 x 2 ) 1 6 x 3 25 x 2 y 23 xy 2 12 y 2 2 x 2 1 3 xy 21 50
cut the curve again in eight points which lie on a conic.
Solution.
The given curve is ( x 2 − y 2 )( y 2 − 4 x 2 ) + 6 x 3 − 5x 2 y − 3xy 2 + 2 y 2 − x 2 + 3xy − 1 = 0.
Put x = 1, y = m in the fourth degree terms, we get
f4 ( m ) = (1 − m 2 )( m 2 − 4).
Put x = 1, y = m in the third degree terms, we get
f3 ( m ) = 6 − 5m − 3m 2 .
∴
∴
f4′ ( m ) = (1 − m 2 )( 2m ) + ( m 2 − 4)( −2m )
= 2m [1 − m 2 − m 2 + 4] = 2m [5 − 2m 2 ]
c=−
M03_ENGINEERING_MATHEMATICS-I _CH03_Part D.indd 117
⎡ 6 − 5m − 3m 2 ⎤
f3 ( m )
= −⎢
.
2 ⎥
f4′ ( m )
⎣ 2 m (5 − 2 m ) ⎦
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Solving f4 ( m ) = 0, we get
(1 − m 2 )( m 2 − 4) = 0
⇒
1 − m 2 = 0 or m 2 − 4 = 0
⇒
When m = −1,
⎡ 6 − 5( −1) − 3( −1) 2 ⎤
4
6+5−3
8
= .
c = −⎢
=−
=
2 ⎥
( −2) ⋅ 3 2 × 3 3
⎣ 2( −1)(5 − 2( −1) ) ⎦
∴ asymptote is
y = −x +
When m = 1,
c=−
∴ the asymptote is
y =x+
When m = −2,
c=−
∴ the asymptote is
y = −2x −
When m = 2,
c=−
∴ the asymptote is
y = 2x −
4
3
⇒
y +x −
m = ±1 or m = ±2.
4
= 0.
3
(6 − 5 ⋅1 − 3 ⋅12 )
( −2) 1
=−
= .
2
2⋅3 3
2 ⋅1(5 ⋅1 − 2 ⋅1 )
1
3
⇒
y −x −
1
= 0.
3
[6 − 5( −2) − 3( −2) 2 ]
4
1
[6 + 10 − 12]
=− .
=−
=−
2
4×3
3
−4[5 − 8]
2( −2)[5 − 2( −2) ]
1
⇒
3
y + 2x +
1
= 0.
3
[6 − 5 × 2 − 3 ⋅ 22 ]
[6 − 10 − 12]
16
4
=−
=−
=− .
4(5 − 8)
4×3
3
2 ⋅ 2(5 − 2 ⋅ 22 )
4
3
⇒
y − 2x +
4
= 0.
3
The fourth degree equation has 4 asymptotes.
∴ the combined equation of the asymptotes is
4⎞ ⎛
1⎞ ⎛
1⎞ ⎛
4⎞
⎛
⎜⎝ y + x − ⎟⎠ ⎜⎝ y − x − ⎟⎠ ⎜⎝ y + 2x + ⎟⎠ ⎜⎝ y − 2x + ⎟⎠ = 0
3
3
3
3
⇒
1
4
4⎤ ⎡ 2
4
1
4⎤
⎡ 2
2
2
⎢ y − x − 3 ( y + x ) − 3 ( y − x ) + 9 ⎥ ⎢ y − 4 x + 3 ( y + 2x ) + 3 ( y − 2x ) + 9 ⎥ = 0
⎣
⎦⎣
⎦
⇒
4
1
4
( y 2 − x 2 )( y 2 − 4 x 2 ) + ( y 2 − x 2 )( y + 2x ) + ( y 2 − x 2 )( y − 2x ) + ( y 2 − x 2 )
3
3
9
1
4
1
4
− ( y + x )( y 2 − 4 x 2 ) − ( y + x )( y + 2x ) − ( y + x )( y − 2x ) − ( y + x )
3
9
9
27
4
16
4
16
− ( y − x )( y 2 − 4 x 2 ) − ( y − x )( y + 2x ) − ( y − x )( y − 2x ) − ( y − x )
3
9
9
27
4
16
4
16
+ ( y 2 − 4 x 2 ) + ( y + 2 x) + ( y − 2 x) +
=0
9
27
27
81
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3.119
⇒
( x 2 − y 2 )( y 2 − 4 x 2 ) − 3xy 2 + 6 x 3 − 5x 2 y +
17 2 2 2 5
4
16
y + x + xy − x − = 0
9
9
3
3
81
∴
A ≡ ( x 2 − y 2 )( y 2 − 4 x 2 ) − 3xy 2 + 6 x 3 − 5x 2 y +
17 2 2 2 5
4
16
y + x + xy − x − = 0
9
9
3
3
81
C ≡ ( x 2 − y 2 )( y 2 − 4 x 2 ) + 6 x 3 − 5x 2 y − 3xy 2 + 2 y 2 − x 2 + 3xy − 1 = 0.
The points of intersection lie on the curve
2y 2 −
5
4
16
17 2
2
y − x 2 − x 2 + 3xy − xy + x − 1 +
=0
9
9
3
3
81
1 2 11 2 4 xy 4
65
y − x +
+ x−
=0
9
9
3
3
81
⇒
⇒
y 2 − 11x 2 + 12xy + 12x −
65
= 0.
9
which is a hyperbola.
∴
⇒
C − A = 0.
[
h2 − ab > 0]
EXAMPLE 7
Determine the asymptotes of the curve
4 ( x 4 1 y 4 ) 217 x 2 y 2 2 4 x ( 4 y 2 2 x 2 ) 1 2 ( x 2 2 2 ) 5 0 and show that they pass through the points
of intersection of the curve with the ellipse x 2 1 4 y 2 5 4 .
Solution.
The given curve is
4( x 4 + y 4 ) − 17x 2 y 2 − 4 x ( 4 y 2 − x 2 ) + 2( x 2 − 2) = 0
Put x = 1, y = m in the fourth degree terms, we get
f 4 ( m ) = 4(1 + m 4 ) − 17m 2
∴
f ′4 ( m ) = 16 m 3 − 34 m
Put x = 1, y = m in the third degree terms, we get f3 ( m ) = −4( 4m 2 − 1) .
∴
c=−
f 3 (m )
f ′4 ( m )
=
4( 4m 2 − 1)
2( 4 m 2 − 1)
=
3
16m − 34m 8m 3 − 17m
Solving, f 4 ( m ) = 0 , we get
⇒
4(1 + m 4 ) − 17m 2 = 0
⇒
4 m 4 − 17m 2 + 4 = 0
4 m 4 − 16 m 2 − m 2 + 4 = 0
⇒
4 m 2 ( m 2 − 4) − 1( m 2 − 4) = 0
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⇒ ( 4 m 2 − 1)( m 2 − 4) = 0
⇒
4m 2 − 1 = 0
or
m2 − 4 = 0
⇒
m=±
⎡ ⎛ 1⎞
⎤
⎡ 1 ⎤
2 ⎢ 4 ⋅ ⎜ − ⎟ − 1⎥
2 ⎢ 4 ⋅ − 1⎥
⎝
⎠
2
⎢⎣
⎥⎦
4 ⎦
= ⎣
= 0.
c=
3
17
⎛ 1⎞
⎛ 1⎞
−1 +
8 ⎜ − ⎟ − 17 ⎜ − ⎟
2
⎝ 2⎠
⎝ 2⎠
2
1
When m = − ,
2
∴ the asymptote is y = −
1
When m = ,
2
When m = −2,
x
2
or
m = ± 2.
⇒ 2y + x = 0 .
⎡ ⎛ 1⎞ ⎤
2 ⎢ 4 ⋅ ⎜ ⎟ − 1⎥
⎝ 4⎠ ⎦
= 0 . ∴ the asymptote is
c= ⎣
1
⎛ 1⎞
⎛ ⎞
8 ⎜ 3 ⎟ − 17 ⎜ ⎟
⎝ 2⎠
⎝2 ⎠
c=
1
2
y =
x
⇒ 2y − x = 0 .
2
2 × 15
30
2( 4 ⋅ 4 − 1)
=
=
= −1.
3
8( −2) − 17( −2) −64 + 34 −30
∴ the asymptote is
y = −2x − 1 ⇒ y + 2x + 1 = 0.
When m = 2,
c=
∴ the asymptote is
y = 2x + 1 ⇒ y − ( 2x + 1) = 0.
2 × 15 30
2( 4 ⋅ 4 − 1)
=
=
= 1.
3
8 × 2 − 17 × 2 64 − 34 30
The 4 th degree equation has 4 asymptotes.
Now the combined equation of the asymptotes is
( 2 y + x )( 2 y − x )[ y + ( 2x + 1)][ y − ( 2x + 1)] = 0
⇒
( 4 y 2 − x 2 )[ y 2 − ( 2x + 1) 2 ] = 0
⇒
( 4 y 2 − x 2 )[ y 2 − ( 4 x 2 + 4 x + 1)] = 0
⇒
( 4 y 2 − x 2 )( y 2 − 4 x 2 − 4 x − 1) = 0
⇒
( 4 y 2 − x 2 )( y 2 − 4 x 2 ) − 4 x ( 4 y 2 − x 2 ) − ( 4 y 2 − x 2 ) = 0
⇒
4 y 4 − 17x 2 y 2 + 4 x 4 − 16 xy 2 + 4 x 3 − 4 y 2 + x 2 = 0
⇒
4( x 4 + y 4 ) − 17x 2 y 2 − 16 xy 2 + 4 x 3 − 4 y 2 + x 2 = 0
∴
A ≡ 4( x 4 + y 4 ) − 17x 2 y 2 − 16 xy 2 + 4 x 3 − 4 y 2 + x 2 = 0
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The curve is
3.121
C ≡ 4( x 4 + y 4 ) − 17x 2 y 2 − 16 xy 2 + 4 x 3 + 2x 2 − 4 = 0 .
The four asymptotes intersect the curve in 4( 4 − 2) = 8 points and they lie on the curve
C − A = 0.
⇒
4y 2 + x 2 − 4 = 0
⇒
x 2 + 4y 2 = 4.
which is an ellipse.
EXERCISE 3.13
I. Obtain the horizontal and the vertical asymptotes, if any, of the following curves.
1. y =
x
x −2
4. x 2 + 5y 2 = 1
x
x +1
10. y = sec x
7. y =
2
2. y =
x2
1+ x
5. y = loge x , x > 0
3x − 1
x +2
11. y = tan x
8. y =
x +2
x −3
II. Find the asymptotes of the following curves.
13. y = e x
14. y =
3. y =
x
x −1
2
6. y = e − x
2
x2 + 2
x 2 −1
12. xy = loge x , x > 0
9. y =
15. y =
2x 2
x +3
1. x 2 y + xy 2 + xy + y 2 + 3x = 0
2. ( x + y )( x − y )( 2x − y ) − 4 x ( x − 2 y ) + 4 x = 0
3. 2x 3 − x 2 y − 2xy 2 + y 3 − 4 x 2 + 8xy − 4 x + 1 = 0
4. x 2 y 2 − x 2 y − xy 2 + x + y + 1 = 0
5. ( x + y ) 2 ( x + 2 y + 2) − ( x + 9 y − 2) = 0
6. y 3 − 2xy 2 − x 2 y + 2x 3 + 3y 2 − 7xy + 2x 2 + 2 y + 2x + 1 = 0
7. y 3 + x 2 y + 2xy 2 − y + 1 = 0
8. x 3 + 2x 2 y − 4 xy 2 − 8y 3 − 4 x + 8y = 1
9. y 2 = x 2 ( x − y )
10. 8x 2 + 10 xy − 3y 2 − 2x + 4 y − 2 = 0
11. ( x 2 − y 2 )( x + 2 y + 1) + x + y + 1 = 0
12. x 3 − 2x 2 y + xy 2 + x 2 − xy + 2 = 0
13. ( y + x − 1)( y + 2x + 1)( y + 3x − 2)( y − x ) + x 2 − y 2 + 5 = 0
III. Show that the asymptotes of the cubic x 3 − 2 y 3 + xy ( 2x − y ) + y ( x − y ) + 1 = 0 cuts the curve in
three points which lie on the straight line x − y + 1 = 0.
IV. Show that the four asymptotes of the curve
( x 2 − y 2 )( y 2 − 4 x 2 ) + 6 x 3 − 5x 2 y − 3xy 2 + 2 y 3 − x 2 + 3xy − 1 = 0 cuts in 8 points which lie on a
circle.
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ANSWERS TO EXERCISE 3.13
I.
1. x = 2, y = 1
2. x = −1
3. x = ±1, y = 0
4. No, it is an ellipse which is a finite curve.
5. x = 0
6. y = 0
8. x = −2, y = 3
7. y = 0
9. x = ±1, y = 1
11. x = ( 2n + 1)
10. x = ( 2n + 1)
p
, n = 0, 1, 2, 3, …
2
p
, n = 0, 1, 2, 3, …
2
13. y = 0
12. y = 0
14. x = 3 and y = 1
15. x = 3, y = 2x − 6
II.
1. x = −1, y = 0, y = − x
2. y = x + 9, y = 2x − 4, y = − x + 2
3. y = − x + 2, y = x + 2, y = 2x − 4
4. x = 0, y = 0, x = 1, y = 1
5. x + 2 y + 2 = 0, x + y = ±2 2
6. y = x − 1, y = − x − 2, y = 2x
7. y = 0, x + y = 1, x + y = −1
9. y = x − 1
11. x + 2y + 1 = 0; y = x; y = −x
13. y + x − 1 = 0,
3.9
y + 2x + 1 = 0,
x
x
x
, y = − + 1, y = − − 1
2
2
2
10. 3y = −2x + 1
8. y =
12. x = 0, y = x, y = x + 1
y + 3x − 2 = 0
and
y−x=0
CONCAVITY
In Section 3.4, we have seen that the sign of first derivative of a function tells us where the function is
increasing or decreasing. Critical points are the points where the first derivative is zero or the points
where the first derivative does not exist. At these points, local maximum or local minimum occurs.
We shall now discuss another aspect of the shape of a curve called concavity. All these concepts are
needed to draw the graph of a function.
Definition 3.9 Let f be a differentiable function in the interval (a, b). The graph of f, viz, the curve
given by the equation y = f ( x ) is said to be concave up in (a, b) if the curve lies above every tangent
to the curve in (a, b)
The curve is said to be concave down in (a, b) if the curve lies below every tangent to the curve
in (a, b)
Note
Concave up is sometimes referred as convex down and concave down is referred as convex up.
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x=a
x=b
x=a
3.123
x=b
Concave up
Concave down
Fig. 3.25
Fig. 3.26
Theorem 3.8 Criterion for Concavity
Let f be defined on [a, b] and let f ″ exist in (a, b).
1. If f ″ (x) < 0 ∀ x ∈(a, b ), then the graph of f, viz, the curve y = f ( x ) is concave down in (a, b ).
2. If f ″ (x) > 0 ∀ x ∈(a, b ), then the graph of f, viz, the curve y = f ( x ) is concave up in (a, b ).
Definition 3.10 Point of Inflexion
A point P on the curve y = f ( x ) is said to be a point of inflexion, if the curve has a tangent at the point
P and the curve changes from concave up to concave down or vice versa at the point P.
Criteria for point of inflexion (or inflection)
1. If f be a function such that f ″(c) = 0 and f ″′(c) ≠ 0 then the point (c, f (c)) is a point of inflexion
on the curve y = f ( x ).
2. Let f be a function such that f ″(x) changes sign in a neighbourhood (c − d, c + d) of c as x
increases, then the point (c, f (c)) is a point of inflexion on the curve y = f ( x ).(even if f ″(c) = 0
or f ″(c) does not exist).
Note
1. The position of the point of inflexion on a curve is independent of the position of x and y axes.
Therefore, the point of inflexion is unaffected by the
y
dy
= ∞, we may
interchange of these x and y axes. When
dx
2
dx d x
use
,
to determine the point of inflexion.
dy dy 2
2. At a point of inflexion, the curve crosses the tangent at the
point.
For example, for the curve y = x 3, x = 0 is a point of
inflexion, the tangent at the point is x-axis
y = x3
O
x
Fig. 3.27
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WORKED EXAMPLES
EXAMPLE 1
Test the concavity of the curve y 5 log e x .
y
Solution.
The given curve is y = loge x
y = loge x
(1)
Since the domain of loge x is x > 0, we test
O
x
(1, 0)
the concavity in the interval (0, ∞).
Differentiating (1) w.r.t. x, we get
dy 1
d2 y
1
= and
= − 2 < 0∀ x > 0
2
dx x
dx
x
Therefore, the entire curve is concave down in the
interval (0, ∞)
Fig. 3.28
EXAMPLE 2
Find the ranges of values of x for which the curve y 5 x 4 2 6 x 3 1 12 x 2 1 4 x 1 10 is concave up
or down. Further, find the points of inflexion.
Solution.
The given curve is
y = x 4 − 6 x 3 + 12 x 2 + 4 x + 10,
x ∈ ( −∞, ∞)
dy
= 4 x 3 − 18 x + 24 x + 4
dx
∴
d2 y
= 12 x 2 − 36 x + 24 = 12( x 2 − 3 x + 2) = 12( x − 1)( x − 2)
dx 2
and
When 1 < x < 2, ( x − 1)( x − 2) < 0
d 2y
<0
dx 2
The curve is concave down in (1, 2).
Similarly, if x < 1 or x > 2, then ( x − 1)( x − 2) > 0
∴ if 1 < x < 2, then
∴
if x < 1 or x > 2, then
−∞
1
2
∞
2
d y
>0
dx 2
The curve is concave up in ( −∞, 1) and (2, ∞).
To find the point of inflexion, we have
M03_ENGINEERING_MATHEMATICS-I _CH03_Part D.indd 124
d2 y
=0
dx 2
and
d3 y
≠0
dx 3
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Differential Calculus ■
Now,
d2 y
=0 ⇒
dx 2
and
d3 y
= 24 x − 36
dx 3
When x = 1,
d3 y
= 24 ⋅1 − 36 = −12 ≠ 0
dx 3
When x = 2,
d3 y
= 24 ⋅ 2 − 36 = 12 ≠ 0
dx 3
12( x − 1)( x − 2) = 0
3.125
⇒ x = 1 or 2
When x = 1 and x = 2, the curve has the points of inflexion.
When x = 1,
y = 1 − 6 × 1 + 12 × 1 + 4 × 1 + 10 = 21 .
When x = 2,
y = 24 − 6 × 23 + 12 × 22 + 4 × 2 + 10 = 34
The points of inflexion on the curve are (1, 21) and (2, 34).
EXAMPLE 3
Test the concavity of the curve x 2 y 1 a 2 ( x 1 y ) 5a 3 and show that the points of inflexion lie on
the line x 1 4 y 53a.
Solution.
The given curve is
x 2 y + a 2 ( x + y ) = a3
⇒
y ( x 2 + a 2 ) = a3 − a 2 x
⇒
y =
a 2 (a − x )
x 2 + a2
(1)
Differentiating (1) w.r. to x
dy a 2 [( x 2 + a 2 )( −1) − ( a − x )2 x ]
=
dx
( x 2 + a2 )2
=
and
a 2 [ − x 2 − a 2 − 2ax + 2 x 2 ] a 2 [ x 2 − 2ax − a 2 ]
=
( x 2 + a2 )2
( x 2 + a2 )2
d 2 y a 2 [( x 2 + a 2 ) 2 ( 2 x − 2a) − ( x 2 − 2ax − a 2 )2.( x 2 + a 2 ) ⋅ 2 x ]
=
dx 2
( x 2 + a2 )4
=
a 2 ( x 2 + a 2 )[2( x 2 + a 2 )( x − a) − 4 x( x 2 − 2ax − a 2 )]
( x 2 + a2 )4
=
2a 2 [ x 3 − ax 2 + a 2 x − a3 − 2 x 3 + 4 ax + 2a 2 x ]
( x 2 + a 2 )3
=
2a 2 [ − x 3 − a3 + 3ax 2 + 3a 2 x ]
( x 2 + a 2 )3
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=
2a2 [ −( x 3 + a3 ) + 3ax ( x + a)]
( x 2 + a 2 )3
=−
d2 y
=0
dx 2
∴
⇒ ( x + a)[ x 2 − 4 ax + a 2 ] = 0 ⇒
x + a= 0 ⇒
Now,
and
2a2 ( x + a) 2
2a2 ( x + a) 2
2
⎡
⎡ x − 4ax + a2 ⎤⎦
⎤
−
+
−
3
ax
=
−
x
ax
a
⎦
( x 2 + a 2 )3 ⎣
( x 2 + a 2 )3 ⎣
x = −a
−∞
x 2 − 4a x + a2 = 0
x=
⇒
x + a = 0 or x 2 − 4a x + a2 = 0
−a
(2 − 3) a (2 + 3) a
∞
4 a + 2 3a
4a + 16a2 − 4a2
=
= ( 2 ± 3 )a
2
2
x = − a, x = ( 2 − 3 )a, x = ( 2 + 3 )a
∴
2
d y
2a 2
=
−
( x + a)[ x − ( 2 − 3 )a][ x − ( 2 + 3 )a]
dx 2
( x 2 + a 2 )3
and
If x < −a, all the three factors are negative.
⇒
( x + a)[x − ( 2 − 3 )a] [x − ( 2 + 3 )a] < 0
and
−
2a2
< 0 always
( x 2 + a 2 )3
2
d y
>0
dx 2
∴
If
−a < x < ( 2 − 3 )a,
then
If
( 2 − 3 )a < x < ( 2 + 3 )a,
then
If
x > ( 2 + 3 )a,
then
d 2y
<0
dx 2
d 2y
>0
dx 2
d 2y
<0
dx 2
∴ the curve is concave up in the intervals ( −∞, − a), [( 2 − 3 )a, ( 2 + 3 )a] and concave down in the
intervals [ −a, ( 2 − 3 )a] and [( 2 + 3 )a, ∞)
The points of inflexion are at x = −a, ( 2 − 3 )a, ( 2 + 3 )a.
When x = − a,
y=
When x = ( 2 − 3 )a,
y=
⇒
y=
a 2 ( a + a)
=a
a2 + a2
a 2 [ a − ( 2 − 3 ) a]
(2 − 3) a + a
2
a( 3 − 1)
4[2 − 3 ]
M03_ENGINEERING_MATHEMATICS-I _CH03_Part D.indd 126
2
=
2
=
a3 [ 3 − 1]
a [4 − 4 3 + 3 + 1]
2
a( 3 − 1)
2( 3 − 1)
2
=
a( 3 − 1)
8−4 3
[∴ ( 3 − 1) 2 = 4 − 2 3 = 2( 2 − 3 )]
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Differential Calculus ■
a
=
When x = ( 2 + 3 )a,
y=
⇒
y=
=
2( 3 − 1)
=
a( 3 + 1)
2( 3 − 1)( 3 + 1)
a 2 [ a − ( 2 + 3 ) a]
(2 + 3)2 a2 + a2
− a( 3 + 1)
4( 2 + 3 )
−a
2( 3 + 1)
=
=
=
=
3.127
a( 3 + 1)
4
a3 [ − 3 − 1]
a 2 ( 4 + 3 + 4 3 + 1)
− a( 3 + 1)
[∴ ( 3 + 1) 2 = 3 + 1 + 2 3 = 2( 2 + 3 )
2( 3 + 1) 2
−a( 3 − 1)
2( 3 + 1)( 3 − 1)
=
−a( 3 − 1)
4
a
⎛
⎞
B ⎜⎝ ( 2 − 3 ) a, ( 3 + 1)⎟⎠
4
Thus, the points of inflexion are the points A(−a, a),
a
⎛
⎞
and C = ⎜⎝ ( 2 + 3 ) a, − ( 3 − 1)⎟⎠
4
To prove the points A, B, and C are collinear, we have to prove the slope of AB = the slope BC.
a
a
3 +1− a
( 3 + 1 − 4)
1 ⎡ 3 − 3⎤
1
4
Now, the slope of AB =
= 4
= ⎢
⎥=−
4
( 2 − 3 )a + a
a( 2 − 3 + 1) 4 ⎣ 3 − 3 ⎦
a
a
a
− ( 3 − 1) − ( 3 + 1) − ⎡⎣ 3 − 1 + 3 + 1⎤⎦
1
12 3
4
and slope of BC = 4
= 4
=−
=−
42 3
4
( 2 + 3 )a − ( 2 − 3 )a
a ⎡⎣ 2 + 3 − 2 − 3 ⎤⎦
∴
slope of AB = slope of BC
Therefore, the points A, B, C are collinear.
The equation of the line in which the points of inflexion is
1
y − a = − ( x + a)
4
⇒
4 y − 4a = − x − a
⇒
x + 4 y = 3a
EXERCISE 3.14
1. Show that y = x 4 is concave upwards at the origin.
2. Find the intervals in which the curve y = 3 x 5 − 40 x 3 + 3 x − 20 is concave upwards or
downwards.
3. Find the intervals in which the curve y = 3 x 2 − 2 x 3 is concave upwards or concave downwards.
4. Show that the curve y =
6x
has three points of inflexion and they are collinear.
x +3
2
5. Find the points of inflexion of the curve y 2 = x( x + 1) 2 .
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ANSWERS TO EXERCISE 3.14
2. Concave up in (−2, 0), ( 2, ∞) and concave down in ( −∞, − 2) and (0, 2).
1⎞
⎛1 ⎞
⎛
3. Concave up in ⎜ −∞, − ⎟ and concave down in ⎜ , ∞⎟ .
⎝2 ⎠
⎝
2⎠
⎛1 4 ⎞ ⎛1
4 ⎞
5. ⎜ ,
, ⎜ ,−
⎟
⎟
⎝ 3 3 3⎠ ⎝ 3
3 3⎠
3.10
CURVE TRACING
In dealing with the problems of finding the area of curves, length of arc, volume of solids of
revolution, surface area of revolution etc, it is necessary to know the shape of the curve represented
by the equation.
It is not always possible to draw the curve by plotting few of the points. We can only draw the curve
with the knowledge of the important characteristics of the curve like increasing, decreasing nature,
maxima and minima, special points on the curve, concavity and convexity, asymptotes of the curve
etc.
We shall now give the general procedure for tracing the graph of y = f ( x ). The equation may be
given in cartesian form, parametric form or polar form.
3.10.1 procedure for Tracing the Curve Given by the Cartesian Equation f(x, y) 5 0.
1. Symmetry
The curve is symmetrical
(i) about the x-axis if the equation is even degree in y .
(ii) about the y-axis if the equation is even degree in x .
(iii) about the origin O , when ( x , y ) is replaced by ( − x , − y ) , the equation is unaltered
(iv) about the line y = x if the equation is unaltered when x and y are interchanged. i.e., ( x , y ) is
replaced by ( y , x )
(v) about the line y = − x if the equation is unaltered when( x , y ) is replaced by ( − y , − x ).
2. Special points on the curve
Intersection with the axes and the origin, points of inflection etc.
3. Tangents at the origin
It is obtained by equating the lowest degree terms to zero, if it is a polynomial equation in x and y
passing through the origin.
4. Asymptotes
Find the vertical, horizontal and oblique asymptotes.
5. Region
Identify the domain or region of the plane in which the graph exists.
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Differential Calculus ■
3.129
dy
dx
Determine the intervals of increasing, decreasing, Critical points etc.
6. Sign of
d 2y
dx 2
Intervals of concavity upwards and downwards and point of inflexion.
8. Loop
If the curve intersects the line of symmetry at two points A and B , then there is a loop
between A and B .
7. Sign of
Note
However, the order of the steps can be interchanged depending on the nature of the equation of the curve.
WORKED EXAMPLES
EXAMPLE 1
Trace the curve y 2 5 x 3.
[It is called the semi-cubical parabola]
Solution.
The given curve is y 2 = x 3 .
1. Symmetry
The given equation is even degree in y, so the curve is symmetrical about the x-axis.
2. Origin: it is a point on the curve.
3. Tangent at the origin:
The tangent at the origin is got by equating the lowest degree terms to zero.
That is y 2 = 0 ⇒ y = 0
∴ the x-axis is the tangent at the origin.
4. Region
y 2 ≥ 0 ⇒ x 3 ≥ 0 ⇒ x ≥ 0. ∴ the curve lies on the right side of y-axis.
5. Sign of
dy
dx
y 2 = x 3 ⇒ 2y
dy
= 3x 2
dx
⇒
dy 3x 2
=
dx
2y
dy
> 0 That is, the curve is increasing for all x ≥ 0 and y > 0.
dx
So, the curve is increasing in the first quadrant.
dy
Also if y < 0,
< 0 i.e., the curve is decreasing for all x ≥ 0 and y < 0.
dx
So, the curve is decreasing in the 4 th quadrant.
∴ if
y > 0,
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6. Sign of
d 2y
dx 2
Engineering Mathematics
dy
⎡
y ⋅ 2x − x 2 ⋅
d 2y 3 ⎢
dx
= ⎢
dx 2 2 ⎢
y2
⎣
⎤
⎡
2 dy
⎥ 3 ⎢ 2xy − x ⋅ dx
⎥= ⎢
y2
⎥ 2⎢
⎦
⎣
=
∴
⎤
⎥
⎥
⎥
⎦
3 x
3 x
3 x4 3 x4
3 x4 1 3 x
⎡ 4 y 2 − 3x 3 ⎤⎦ = ⋅ 3 ⎡⎣ 4 x 3 − 3x 3 ⎤⎦ =
=
=
⋅ =
.
3
2
3 ⎣
2y
4 y
4y
4 yy
4 y x3 4 y
d 2y
d 2y
>
0
if
y
>
0
(as
already
x
>
0)
and
< 0 if y < 0.
dx 2
dx 2
∴ the curve is concave up if y > 0 and concave
down if y < 0.
So, the curve is concave up in the first quadrant and the
curve is concave down in the fourth quadrant.
y
o
x
7. Asymptotes
It has no asymptotes.
With these information we shall draw the curve.
The curve is as shown in Fig. 3.29.
Fig. 3.29
EXAMPLE 2
Trace the curve y 2 ( 2 a 2 x ) 5 x 3. [This curve is called the Cissoid of Diocles]
Solution.
The given equation of the curve is y 2 =
x3
.
2a − x
(1)
1.Symmetry
The equation is even degree in y, so the curve is symmetrical about the x-axis.
2. Origin: It is a point on the curve.
The tangent at the origin is given by y 2 = 0 ⇒ y = 0 That is the x-axis is the tangent at the origin.
3. Region:
y2 ≥ 0 ⇒
x3
x3
≥0 ⇒
≤0
x − 2a
2a − x
⇒
x3
≤ 0 ⇒ 0 ≤ x < 2a .
x − 2a
∴ the curve lies between the lines x = 0 and x = 2a .
4. Asymptote
When x → 2a , y → ∞ ∴ x = 2a is a vertical asymptote.
dy
dx
Differentiating (1) with respect to x, we get
5. Sign of
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Differential Calculus ■
2y
3.131
y
dy ( 2a − x ) ⋅ 3x 2 − x 3 ( −1) 6ax 2 − 2x 3
=
=
dx
( 2a − x ) 2
( 2a − x ) 2
dy 2x 2 (3a − x ) x 2 (3a − x )
=
=
dx 2 y ( 2a − x ) 2 y ( 2a − x ) 2
⇒
o
Since 0 ≤ x < 2a, 3a − x > 0 and ( 2a − x ) 2 > 0.
∴
dy
>0
dx
if
y >0
and
dy
<0
dx
if
y <0
x
x = 2a
Fig. 3.30
∴ the curve is increasing in the first quadrant and the curve is decreasing in the 4 quadrant
Also it touches x = 2a at infinity
With these informations, we can draw the curve.
The curve is as shown in Fig. 3.30.
th
EXAMPLE 3
Trace the curve y 2 5
x 2 (a2 2 x 2 )
. [This curve is called Lemniscate of Bernoulli]
a2 1 x 2
Solution.
The given equation of the curve is y 2 =
x 2 (a 2 − x 2 )
a2 + x 2
(1)
1. Symmetry
The equation is even degree in x and y .
So, the curve is symmetrical about the x-axis as well as the y-axis.
When ( x , y ) is replaced by ( − x , − y ) equation is unaltered.
∴ the curve is symmetrical about the origin.
2. Origin
Origin is a point on the curve.
3. Tangent at the origin
We have
y2 =
x 2 (a 2 − x 2 )
a2 + x 2
⇒
y 2 (a 2 + x 2 ) = x 2 a 2 − x 4
⇒
a2 ( x 2 − y 2 ) = x 2 y 2 + x 4
Tangent at the origin is got by equating the lowest degree terms to zero.
i.e.,
x2 − y 2 = 0
⇒ y = ±x
∴ y = x and y = − x are the tangents at the origin.
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Engineering Mathematics
4. Special points
To find the point of intersection with the x-axis, put y = 0 in (1)
∴
x 2 (a 2 − x 2 ) = 0
⇒
x2 = 0
or
a2 − x 2 = 0
⇒
x = 0, 0
or
x = ±a
So, the curve passes through the origin twice and the points ( −a, 0), (a, 0) .
To find the intersection with the y-axis, put x = 0 in (1).
∴
y =0
So, it meets the y-axis only at the origin.
5. Region
x 2 (a 2 − x 2 )
a2 + x 2
y 2 ≥ 0 ⇒ x 2 (a2 − x 2 ) ≥ 0
y
y2 =
⇒
⇒ x 2 − a2 ≤ 0 ⇒
y = −x
y=x
a2 − x 2 ≥ 0
−a≤ x ≤ a.
B
(−a, 0)
O
∴ the curve lies between x = −a and x = a.
6. Loop
Since the curve meets the line of symmetry, the
x-axis, at O(0, 0), A( a, 0), B( − a, 0) there is a loop
between O and A and a loop between O and B.
With these informations, we shall draw the curve.
The curve is as shown in Fig. 3.31.
A
(a, 0)
x
Fig. 3.31
EXAMPLE 4
Trace the curve x 3 1 y 3 5 3axy , a > 0. [This curve is called the Folium of Descartes]
Solution.
The given equation of the curve is x 3 + y 3 = 3axy , a > 0
1. Symmetry
The equation is unaltered if x and y are interchanged.
So, the curve is symmetric about the line y = x .
2. Origin
Origin lies on the curve.
3. Tangent at the origin
Tangents at the origin are got by equating the lowest degree terms to zero
∴
xy = 0 ⇒
(1)
x = 0, y = 0
So, the y-axis and the x-axis are the tangents at the origin.
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Differential Calculus ■
3.133
4. Special points
To find the point of intersection with y = x, put y = x in (1)
∴
x 3 + x 3 = 3ax 2
⇒
2x 3 − 3ax 2 = 0
⇒
x 2 ( 2x − 3a) = 0
⇒
x 2 = 0 or 2x − 3a = 0
⇒
x = 0, 0 or x =
3a
2
When x = 0, y = 0
When x =
3a
3a
,y =
2
2
⎛ 3a 3a ⎞
∴ the point of intersections are O(0, 0) and A ⎜ , ⎟
⎝ 2 2⎠
The curve meets the axes only at the origin, twice.
5. Loop
Since the curve intersects the line of symmetry
y = x at O and A, there is a loop between O and A.
6. Asymptotes
The coefficients of x3 and y3 are constants and so
there is no vertical or horizontal asymptotes.
To find the oblique asymptotes of
y
y=
x=0
x 3 − y 3 − 3axy = 0
O
A
x
3a , 3a
2 2
y=0
x
Put x = 1, y = m in the highest degree terms x 3 + y 3
∴
f3 ( m ) = 1 + m 3 , f′3 ( m ) = 3m 2
x+y+a=0
Now put x = 1, y = m in −3axy
∴
f2 ( m ) = −3am .
Solve,
Now
f3 ( m ) = 0
c=
⇒
Fig. 3.32
1+m =0
3
⇒
m = −1
−f2 ( m )
−(3am ) a
=−
=
f3′( m )
m
3m 2
a
= −a ∴ asymptote is y = −x − a
⇒
x+y+a=0
−1
With these informations, we can draw the curve. The curve is as shown in Fig. 3.32.
When m = −1, c =
EXAMPLE 5
Trace the curve y 2 5( x 21)( x 2 2 )( x 23).
Solution.
The equation of the given curve is y 2 = ( x − 1)( x − 2)( x − 3)
1. Symmetry
The equation is even degree in y and so the curve is symmetrical about the x-axis.
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2. Special points
To find the point of intersection with the x-axis, put y = 0
∴
( x − 1)( x − 2)( x − 3) = 0
⇒
x = 1, 2, 3
It does not intersect the y-axis, because when x = 0, y 2 = −6 < 0. So, y is imaginary
3. Region
−∞
If x < 1, then x − 1 < 0, x − 2 < 0, x − 3 < 0
∴
( x − 1)( x − 2)( x − 3) < 0
⇒
1
2
∞
3
y 2 < 0 ∴ y is imaginary
So, the curve does not exist if x < 1
But if 1 ≤ x ≤ 2
and
x ≥ 3, y 2 ≥ 0
So, the curve lies in between x = 1 and x = 2 and x ≥ 3.
4. Loop
The curve lies between the points A (1, 0)and B( 2, 0) and symmetric about the x-axis and so there is
a loop between A and B
dy
5. Sign of
dx
y 2 = ( x − 1)( x − 2)( x − 3) = x 3 − 6 x 2 + 11x − 6
dy 3x 2 − 12x + 11
dy
2
⇒
=
2y
= 3x − 12x + 11
dx
2y
dx
dy
dy
< 0, when y < 0
If x > 3,
> 0, when y > 0 and
dx
dx
So, for all x ≥ 3, the curve is strictly increasing above the x-axis and strictly decreasing below the
x-axis.
With these information, we shall draw the graph of the curve.
The curve is as shown in Fig. 3.33.
y
O
(1, 0)
x
(2, 0) (3, 0)
x=1
x=2
x=3
Fig. 3.33
EXAMPLE 6
Trace the curve whose equation is y 5
M03_ENGINEERING_MATHEMATICS-I _CH03_Part E.indd 134
x2 11
.
x 2 21
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Differential Calculus ■
Solution.
The equation of the given curve is y =
3.135
x 2 +1
x 2 −1
(1)
1. Symmetry
Since the equation is even degree in x, the curve is symmetrical about the y-axis.
2. Asymptotes
When x = −1 and x = 1, y → ∞
x = −1 and x = 1 are vertical asymptotes.
∴
x 2 +1
lim y = lim 2
= lim
x →∞
x →∞ x − 1
x →∞
1
x2 =1
1
1− 2
x
1+
∴ y = 1 is the horizontal asymptote.
3. Region
y =
x 2 +1
x 2 −1
Series x 2 ≥ 0 ⇒
⇒
y ( x 2 − 1) = x 2 + 1
⇒
x 2 ( y − 1) = y + 1
⇒
x2 =
y +1
y −1
y +1
≥ 0 ⇒ y ≤ −1 or y ≥ 1
y −1
The curve lies in the part of y ≤ −1 and y > 1
i.e., the curve lies above the line y = 1 and below the line y = −1 for all x ≠ ±1
4. Sign of
dy
dx
x2 + 1
x2 −1
dy ( x 2 − 1) ⋅ 2 x − ( x 2 + 1)( 2 x ) 2 x[ x 2 − 1 − x 2 − 1]
4x
=
=
=− 2
2
2
2
2
dx
( x − 1)
( x − 1)
( x − 1) 2
y=
∴
∴
dy
< 0 if x > 0
dx
and
dy
> 0 if x < 0
dx
So, the curve is increasing if x < 0 and is decreasing if x > 0
When x = 0, y = −1. In the interval (−1, 1) the curve increases upto the point (0, −1) and then
decreases.
If −1 < x < 1, then x 2 − 1 < 0. ∴ y < 0
So, in this part, the curve lies below the x-axis.
dy
< 0 and so, the curve is decreasing.
If x > 1, then
dx
dy
If x < −1,then
> 0 and so, the curve is increasing.
dx
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3.136
■
5. Sign of
d 2y
dx 2
We have
∴
Engineering Mathematics
dy
4x
=− 2
dx
( x − 1) 2
2
2
2
d 2 y −4 ⎡⎣( x − 1) ⋅1 − x ⋅ 2( x − 1)( 2 x ) ⎤⎦
=
dx 2
( x 2 − 1) 4
= −4
If x > 1,
If x < −1,
( x 2 − 1)[ x 2 − 1 − 4 x 2 ] 4(1 + 3 x 2 )
=
( x 2 − 1) 4
( x 2 − 1)3
d 2y
> 0 ∴ the curve is concave up
dx 2
[∴ x2 − 1 > 0]
d 2y
> 0 ∴ the curve is concave up
dx 2
6. Assymptotes
x = −1, x = 1 are the vertical asymptotes. y = 1 is the horizontal asymptote.
The curve lies in the region y < −1 and y > 1 and decreasing if x > 1 and concave up.
Increasing if x < −1 and concave up
We draw the curve.
The curve is as shown in Fig. 3.34.
y
y=1
(0, 1)
x
O
(0, −1)
y = −1
x = −1
x=1
Fig. 3.34
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Differential Calculus ■
3.137
3.10.2 Procedure for Tracing of Curve Given by Parametric Equations x 5 f(t), y 5 g(t)
If the current coordinates (x, y) of the curve are expressed interms of another variable t, then t is called
the parameter and the equations x = f(t), y = g(t) are called the parametric equations of the curve.
1. Symmetry
(i) If f(−t) = f(t) and g(−t) = −g(t), then the curve is symmetrical about the x-axis.
(ii) If f(−t) = −f(t) and g(−t) = g(t), then the curve is symmetrical about the y-axis.
(iii) If f(−t) = f(t) and g(−t) = g(t), then the curve is symmetrical in the opposite quadrants.
2. Special points
To find the points of intersection with x-axis, put y = 0 ⇒ g(t) = 0
To find the points of intersection with the y-axis, put x = 0 ⇒ f(t) = 0.
3. Region
Determine the limits of x and y and hence the limt of t
dy
4. Sign of derivative
dx
dx
dy
Find
and the values of t for which x and y are increasing or decreasing.
and
dt
dx
Find the tangent parallel to the axes.
dy
= 0 or ∞.
dx
d 2y
Also check for concavity. i.e.,
> 0 or < 0
dx 2
5. Period
If x and y are periodic functions of t with a common period, study the curve in this period.
i.e.,
Note If it is possible to eliminate the parameter t and get the Cartesian form, then we can trace the
curve by the first method.
WORKED EXAMPLES
EXAMPLE 7
Trace the curve x 5 a cos t 1
Solution
Given
a
t
log e tan 2 ; y 5 a sin t .
2
2
a
t
x = a cos t + loge tan 2
2
2
x = f (t)
Let
and
y = a sin t .
and
y = g(t).
[This curve is called the tractrix]
1. Symmetry
a
a
t
⎛ t⎞
f ( −t ) = a cos( −t ) + loge tan 2 ⎜ − ⎟ = a cos t + loge tan 2 = f (t )
⎝ 2⎠
2
2
2
and
g ( −t ) = a sin( −t ) = −a sin t = − g (t ).
∴ the curve is symmetrical about the x-axis.
2. Intersection with the axes
To find the intersection with the x-axis, put y = 0
M03_ENGINEERING_MATHEMATICS-I _CH03_Part E.indd 137
∴
a sin t = 0 ⇒ t = 0, p, 2p, ...
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When t = 0, x → ∞ and y = 0
∴ x-axis is an asymptote to the curve
p
When t = ,
x = 0 and y = a
2
∴ the curve intersects the y-axis at (0, a).
3. Sign of derivative
dx
a
= −a sin t + ⋅
dt
2
t
t 1
⋅ 2 tan ⋅ sec 2 ⋅
t
2
2 2
tan 2
2
1
a 1
⋅
= −a sin t + ⋅
t
t
2
tan cos 2
2
2
a
a(1 − sin 2 t ) a cos 2 t
= −a sin t +
=
=
sin t
sin t
sin t
and
1
y
π
t=
2
dy
= a cos t .
dt
t=0
dy
a cos t
dy
sin t
∴
= dt =
=
= tan t
2
dx
dx
a cos t cos t
dt
sin t
t=π
O
p dy
,
= ∞ and the point is (0, a)
2 dx
∴y-axis is tangent at (0, a)
0<t <
p
, then
2
x
Fig. 3.35
y
When t =
If
(0, a)
(0, a)
dx
dy
> 0,
>0
dt
dt
∴ x increases from −∞ to 0 and y increase
from 0 to a
i.e., (x, y) varies from (−∞, 0) to (0, a)
x′
O
x
p
dx
dy
< t < p, then
> 0,
<0
2
dt
dt
(0, −a)
∴ x increases from 0 to ∞
y′
and y decreases from a to 0
Fig. 3.36
i.e., (x, y) varies from (0, a) to (∞, 0).
∴ the graph above the x-axis is as in Fig. 3.35
Since the curve is symmetric about the x-axis, taking reflection about the x-axis, we get the graph of
the given equation as in Fig. 3.36.
⎛ a − a2 − y 2 ⎞
a
Note The cartesian equation of the tractix is x = a2 − y 2 + log ⎜
⎟
2
2
2
⎝a+ a −y ⎠
If
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Differential Calculus ■
3.139
EXAMPLE 8
Trace the curve x 5 a ( u 2 sin u), y 5 a (1 2 cos u).
[This curve is called a cycloid]
Cycloid is a curve traced out by a fixed point on the circumference of a circle when it rolls on a
fixed straight line without slipping.
This fixed line is called the base of the curve.
Solution.
Given parametric equations are x = a(u − sin u) and y = a(1 − cos u)
Let x = f (u) and y = g (u)
1. Symmetry
f ( −u) = a( −u − sin( −u)) = a( −u + sin u) = −a(u − sin u) = −f (u)
g ( −u) = a(1 − cos( −u)) = a(1 − cos u) = g (u)
∴ the curve is symmetric about the y-axis.
We shall consider the graph for u ≥ 0.
2. To find the point of intersection with the x-axis
Put y = 0 ⇒ a(1 − cos u) = 0 ⇒ cos u = 1
⇒ u = 0, 2p, 4p, …
When u = 0, x = 0 and y = 0
∴ the origin corresponds to u = 0 and the origin lies on the curve.
Since −1 ≤ cos u ≤ 1, 0 ≤ y ≤ 2a.
When u = 2p , x = a( 2p − sin 2p) = 2pa and y = a(1 − 1) = 0 .
∴ the points of intersections with x-axis are (0, 0) and ( 2pa, 0) .
When u = p, x = a(p − sin p) = ap and y = a(1 − cos p) = 2a , which is the maximum value of y.
3. Sign of the derivative
dx
u
dy
= a(1 − cos u) = 2a sin 2 and
= a sin u
du
2
du
If 0 < u < p, then
dx
dy
> 0 and
>0
du
du
∴ x increases from 0 to 2ap and y increases from
0 to 2a.
We shall draw one arch of the curve between
u = 0 and u = 2p .
i.e., between the points (0, 0) and ( 2ap, 0) on the
curve.
We see that the curve increases from (0, 0)
to ( ap, 2a) and decreases from ( ap, 2a) to
( 2ap, 0) as in Fig. 3.37.
M03_ENGINEERING_MATHEMATICS-I _CH03_Part E.indd 139
(aπ, 2a)
y = 2a
(2aπ, 0)
O
x
Fig. 3.37
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y
y = 2a
−6aπ
−4aπ
−2aπ
4aπ
2aπ
O
6aπ
x
Fig. 3.38
Similarly another arch is from u = 2p and u = 4p .
Because of the symmetry about the y-axis, we can reflect about the y-axis and get the full graph as
in Fig. 3.38.
The x-axis is the base line on which the circle rolls. Then the points 0, 2ap, 4ap, 6ap, are the points
where the fixed point of the circle touches the base line.
Note
There are different forms of cycloids with base line the x-axis, y = 0 or the line y = 2a .
1. The parametric equations are x = a(u + sin u), y = a(1 + cos u)
When u = 0, x = 0 and y = 2a
So, the point corresponding to u = 0 is (0, 2a)
The curve meets the x-axis, y = 0
⇒
1 + cos u = 0
⇒
y
⇒ cos u = −1
u = p, 3p, …, − p, …
(0, 2a)
So, one arch of the curve is between
u = −p and p
−3aπ
When u = −p , x = −ap and y = 0.
−aπ O
aπ
3aπ
x
When u = p , x = ap and y = 0
Fig. 3.39
So, the graph is as shown in the Fig. 3.39.
2. The
parametric
equations
x 5 a(u1sinu), y 5 a(12cosu)
When u = 0, x = 0, y = 0.
The curve meets the x-axis, y = 0
⇒
are
⇒
a(1 − cos u) = 0
cos u = 1
⇒
u = 0, 2p, 4p, …, −2p, − 4p, …
When u = 2p , x = 2ap, y = 0 and when u = 4p, x = 4ap, y = 0
When u = −2p, x = −2ap, y = 0 and when u = −4p, x = −4ap, y = 0
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Differential Calculus ■
3.141
dx
dy
= a(1 + cosu),
= a sinu
du
du
u
u
2a sin .cos
dy
a sin u
2
2 = tan u
=
=
dx a(1+ cosu)
2
2 u
2a cos
2
dy
=0
at u = 0, 2p, 4p, ......
dx
dy
and
=∞
at u = p, 3p, 5p, ......
dx
When u varies from 0 to p, x increases from 0 to ap and y increases from 0 to 2a
When u varies from p to 2p, x increases from ap to 2ap and y decreases from 2a to 0.
So, the graph is as shown in Fig. 3.40.
y
(aπ, 2a)
y = 2a
(0, 2a)
2a
(−3aπ, 0)
(−aπ , 0)
(3aπ, 0) x
O (0, 0) (aπ , 0)
Fig. 3.40
3.10.3 Procedure for Tracing of Curve given by Equation in Polar Coordinates f(r, u) 5 0
Let the polar equation of the curve be r = f (u).
Relation between Cartesian and polar is x = r cos u and
y = r sin u
1. Symmetry
(i) When u is replaced by −u and if the equation is
unaltered, then the curve is symmetrical about the
initial line u = 0 (i.e., the x-axis).
(ii) When u is replaced by p − u and if the equation is
unaltered, then the curve is symmetrical about the
p
line u = (i.e., the y-axis).
2
(iii) When u is replaced by p + u and if the equation is
unaltered, then the curve is symmetrical about the
pole 0.
p
(iv) When u is replaced by − u and if the equation
2
is unaltered, then the curve is symmetrical about
p
the line u = (i.e., the line y = x).
4
M03_ENGINEERING_MATHEMATICS-I _CH03_Part E.indd 141
y
θ=
π
2
P (r, θ )
r
θ
0
x
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(v) When u is replaced by p and r is replaced by –r and if the equation is unaltered, then the
curve is symmetrical about the pole.
2. Pole
If r = 0 for u = a, then the curve passes through the pole and the line u = a is tangent at the
pole.
3. Region
If r is imaginary for values of u lying between u = u1 , and u = u 2 , then the curve does not lie
between the lines u = u1 , and u = u 2 .
4. Points of intersection
p
p
3p
Determine the points where the curve meets the lines u = 0, u = , u = , u = p, u =
4
2
2
5. Tangent line
du
Find the values of f with the formula tan f = r
where f is the angle between the tangent at
dr
the point p ( r , u) and the radius vector OP.
6. Loop
If a curve meets a line u = a at A and B and the curve is symmetrical about the line, then a loop
of the curve exists between A and B.
WORKED EXAMPLES
EXAMPLE 1
Trace the curve r 5 a (1 1 cos u). [This curve is called a cardioid]
Solution.
The given curve is r = a(1 + cos u)
1. Symmetry
If u is replaced by −u, then r = a(1 + cos( −u)) = a(1+ cos u)
Therefore, the equation is unaltered. Hence, the curve is symmetrical about the initial line u = 0.
2. r = 0 ⇒ 1 + cos u = 0 ⇒ cos u = −1 ⇒ u = p
∴ the tangent at the pole is the line u = p
3. When u = 0, r = 2a, which is the maximum value of r.
When u varies from 0 to p, r decreases from 2a to 0.
4. We know that tan f = r
Therefore,
∴
du
.
dr
dr
= a( − sin u)
du
⇒
du
1
=−
dr
a sin u
⎛ −1 ⎞
tan f = a(1 + cos u) ⎜
=−
⎝ a sin u ⎟⎠
f=
2 cos 2
u
2
u
u
2 sin cos
2
2
= − cot
u
⎛ p u⎞
= tan ⎜ + ⎟
⎝ 2 2⎠
2
p u
+
2 2
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Differential Calculus ■
When u = 0, f =
3.143
p
2
∴ the tangent at the point (2a, 0) is perpendicular to the initial line u = 0
u:
r:
points:
p
3
3a
2a
2
A B
0
p
2
2p
3
a
2
D
a
C
p
0
O
The curve is as in figure A to O
By symmetry about u = 0, by reflecting the point ABCDO about u = 0,
we get the full curve as in Fig. 3.41.
y
θ = 2π
3
θ=
B
C
π
3
D
a
a
2
θ=π
3a
2
A θ=0 x
(2a, 0)
2a
O
Fig. 3.41
EXAMPLE 2
Trace the curve r 5 a sin 3u. [This curve is called 3 leaved rose]
Solution.
The given curve is r = a sin 3u.
1. Symmetry
When u is replaced by p − u, r = a sin 3(p − u) = a sin(3p − 3u) = a sin 3u
So, the equation is unaltered.
Hence, the curve is symmetrical about the line u =
p
(i.e., y-axis)
2
2. The maximum value of r is a, when sin 3u is maximum,
That is, when sin 3u = 1.
⇒ 3u =
p 5p 9p
,
,
,.....
2 2 2
⇒
M03_ENGINEERING_MATHEMATICS-I _CH03_Part E.indd 143
u=
p 5p 9p
,
,
, ……
6 6 6
⇒
u = 30°, 150°, 270°, ...
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So, the curve lies within circle r = a, and r varies from –a to 0.
We get the third loop OBO.
3. r = 0 ⇒ sin 3u = 0
⇒
3u = 0, p, 2p, 3p, 4p, 5p
⇒
u = 0,
p 2p
4p 5p
,
, p,
,
3 3
3 3
p
2p
4p
5p
,u=
, u = p, u =
,u=
are tangents at the ori3
3
3
3
gin and the curve passes through the pole.
Therefore,
u = 0, u =
4. Loop
p
As u varies from 0 to , r varies from 0 to a. In other words, the curve is from O to A.
6
p p
As u varies from to , r varies from a to 0. In other words, the curve is from A to O.
6
3
Therefore, a loop OAO is formed.
p
p
, reflecting about u = , we get the second loop in the second
2
2
varies from
By the symmetry about u =
quadrant. As u
4p 3p
to
, r varies from 0 to –a and
3
2
3p 5p
to
, r varies
as u varies from
2
3
from −a to O. we get the third loop
y
5π
θ=
6
Limacon of Pascal
The polar curve r = a + b cos u, where
a, b > 0 is called Limacon of Pascal.
2π
3
θ=
π
3
θ=
C
OBO.
Note
1. More generally, the curve is of the
form r = a sin nu or r = a cos nu.
When n is odd, it is called n-leaved
rose
2. When n is even, it is a 2n-leaved
rose.
For example, r = a sin 2u will have
4 leaves and the curve lies within
the circle r = a.
θ=
θ=π
A
O
θ=0
B
4π
θ=
3
π
6
θ=
θ=
x
5π
3
3π
2
Fig. 3.42
When a = b, it becomes the cardioid
a
r = a(1 + cos u), which is discussed in worked example 1. When < 1, that is, a < b, it is called a
b
a
Limacon of Pascal with an inner loop. When 1 < < 2, it is called a dimpled Limacon and when
b
a
≥ 2 , it is called a Convex Limacon.
b
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Differential Calculus ■
3.145
EXAMPLE 3
Trace the curve r 5 1 1 2 cos u.
Solution.
∴ a<b
The given curve is r = 1 + 2 cos u. Here a = 1, b = 2.
Therefore, it is a Limacon with an inner loop.
1. Symmetry
When u is replaced by −u, we get r = 1 + 2 cos( −u) = 1 + 2 cos u
Therefore, the equation is unaltered. The curve is symmetrical about the initial line u = 0
2. Pole
1
2p 4p
⇒u=
,
2
3 3
2p
4p
,u=
are the tangents at the pole.
Inotherwords, the curve passes through the pole and u =
3
3
3. When u = 0, r 5 3 is the farthest point (3, 0).
r = 0 ⇒ cos u = −
When u =
p
, r = 1. There is no asymptote, since r is finite for every value of u.
2
Since the curve is symmetric about u = 0, we shall find the variation of r as u varies from 0 to p.
u= 0
and
r= 3
The points are
A
p
6
p
3
p
2
2p
3
1+ 3
B
2
C
1
D
0
O
5p
6
p
1− 3
E
−1
F
The curve is ABCDO and OEF.
By symmetry about u = 0, we get the full curve as in Fig. 3.43.
y
θ=
θ=
2π
3
D
θ=
C
5π
6
θ=π
π
3
B
F
O
θ=
π
6
(3, 0)
A
θ=0
x
E
Fig. 3.43
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EXERCISE 3.15
Trace the following curves
1. y 2 =
x3
2a − x
2. a 2 y 2 = x 2 ( a 2 − x 2 )
2x − 1
x2 −1
x3
7. y 2 =
x −1
2
4. y 2 =
10. r = a(1 − cos u)
2
3. y 2 =
2
x 2 (a − x)
a+ x
5. x 3 + y 3 = a 3
6. y 2 x 2 = x 2 − a 2
8. x 2 ( x 2 + y 2 ) = a 2 ( x 2 − y 2 )
9. r = a sin 2u
11. r = 1 − 2sin u
12. x =
3at
3at 2
,y =
,a>0
3
1+ t
1+ t 3
[Hint: Folium of Descartes]
14. xy 2 = a2 (a − x ), a > 0
13. y (a − x ) = x (a + x ), a > 0.
2
2
15. x = a cos3 u, y = a sin 3u.
ANSWERS TO EXERCISE 3.15
y
1.
y
2.
y = −x
x
O
20
y=x
(−a, 0)
(a, 0)
x
O
x = 2a
3.
4.
y
y
(0, 1)
y=x
x
(−1, 0)
( 1 , 0) (1, 0)
2
(0, −1)
y = −x
x=1
x = −1
x = −a
M03_ENGINEERING_MATHEMATICS-I _CH03_Part E.indd 146
x
x=
1
2
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Differential Calculus ■
5.
6.
y
3.147
y
y=1
(0, a)
x
x
(−a, 0)
(a, 0)
(0, −a)
y = −1
x=a
y
7.
8.
y
y=x
y = x −1
2
x=1
(−a, 0)
x
(− 1 , 0)
2
(a, 0)
O
x
y = −x
y = x +1
2
y
9.
θ=
3π
4
θ=
10.
π
2
θ=
y
π
4
θ=
π
2
a
θ=π
2a
O
θ=0
x
x
θ=π
M03_ENGINEERING_MATHEMATICS-I _CH03_Part E.indd 147
θ=0
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■
Engineering Mathematics
11.
y
θ=
13.
π
2
y
y = −x
θ=0
x
B
(−a, 0)
A
(a, 0)
O
x
x=a
y=x
3π
θ=
2
14.
y
15.
y
(0, a)
x=a
O
(a, 0)
x
(−a, 0)
O
(a, 0) x
x=0
(0, −a)
SHORT ANSWER QUESTIONS
1. Find the first two differential coefficients of y 5 e 2 x ? cos 3 x .
2. Find
d 3y
dx 3
if y 5 x 3log x .
3. If y 5 a cos mx 1 b sin mx, then prove that
d 2y
dx 2
1 m 2 y 5 0.
d2 y
dy
26 19y 50 .
dx
dx 2
5. Show that the length of the sub-tangent at any point of the curve xm yn 5 am 1 n varies as the abscissa.
4. If y 5 e 3 x ( ax 1 b ), then prove that
y2
x
.
, prove that the length of the normal is
c
c
x3
.
7. Find the equation of the tangent at the point (2, 22) on the curve y 2 5
42x
8. Show that the curves y 5 x2 and 6y 5 7 2 x3 cut orthogonally at the point (1, 1).
6. For the catenary y 5 c cos h
d 2y
dy
2x
5 0.
dx
dx 2
10. Find ‘c’ of Lagrange’s mean value theorem for f(x) 5 ln x in [1, e].
9. If y 5 sin21 x, then prove that (1 2 x2)
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Differential Calculus ■
3.149
11. Find the value of ‘a’ if x3 2 ax2 1 3x 1 1 is strictly increasing ; x [ R.
12. Using Taylor’s theorem, express the polynomial 2x3 1 7x2 1 x 1 6 in powers of (x 2 1).
⎛ x2 ⎞
1 x3
x
f 9( x ) 1
f 0( x ) 1 …
13. Prove that f ⎜
⎟ 5 f (x ) 2
2 ( x 1 1) 2
x 11
⎝ x 11 ⎠
14. Expand esinx by Maclaurin series up to terms containing x4.
15. Evaluate lim
x →0
tanx 2 sinx
x
e ax 2e bx
.
x →0
x
.
3
16. Evaluate lim
17. Prove that the curve y 5 x4 is concave upwards at the origin.
18. Find the asymptotes of the curve y 5
3x
which are the parallel to the x-axis.
x 22
19. Find the vertical asymptotes of the curve y 5
3x 2
x 2 1 2x 1 5
.
1
20. If the function f(x) 5 sinx 2 asin2x 2 sin 3 x 1 2ax is increasing for all x [ R, then find the value of a.
3
OBJECTIVE TYPE QUESTIONS
A. Fill up the blanks
1. The equation of tangent at the point where the curve y = b e
−x
a
cuts the y-axis is __________
2. The value of ‘t’ for which the tangent is perpendicular to x-axis is __________
3. The length of the sub-tangent at any point on the curve y = b e
−x
a
is __________
x2 y 2
−
= 1 is __________
a2 b 2
5. The length of the sub-tangent at any point on the parabola y2 = 4ax is __________
4. The length of the sub-normal at any point on the hyperbola
6. The maximum value of
1
2
(sin x − cos x ) is __________
x 2 + 2x − 1
is __________
x +2
8. The curve y = x3 − 3x2 − 9x + 9 has a point of inflexion at x = __________
7. The vertical asymptote of the curve y =
9. The equations of the tangents at the origin are __________
10. The interval in which f ( x ) =
x
1+ x2
is strictly increasing is __________
1
11. Using the function f(x) = x x ; x > 0, the bigger of the two numbers pe and ep is __________
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3.150
12. lim
■
x cos x − sin x
x 2 sin x
x →0
13. lim
x →∞
14. lim
Engineering Mathematics
x4
ex
is __________
2
cos 3x − 1
x2
x →0
= __________
is __________
15. The curve y = x4 at the origin is concave __________
B. Choose the correct answer
1
1. x x , x > 0, is a decreasing function if
(a) x < e
(c) x >
(b) x > e
1
e
(d) x <
1
e
2. The value of lim (tan x)tan2x is
x→
(a)
p
4
1
e
(b) e
e
(c)
(d)
1
e
x
are
x−2
(b) x = 2, y = −1
(c) x = 2, y = −1
3. The vertical and horizontal asymptotes of y =
(a) x = 2, y = 1
(d) x = 2, y = 2
4. The equation of the normal to the curve x = 4y passing through the point (1, 2) is
2
(a) x + y + 3 = 0
(b) x − y − 3 = 0
(c) x + y − 3 = 0
5. The two curves x3 − 3xy2 + 2 = 0 and 3x2y − y3 − 2 = 0
(a) cut at right angles
(b) touch each other
p
(c) cut at an angle
(d) None of these
4
f ( 4) − f ( 0)
= f ′(c) is
6. If f(x) = x3 + 3x, then the value of c ∈ (0, 4) such that
4−0
(a)
19
3
4
(b)
(c)
3
4
3
(d) None of these
(d) None of these
7. If f(x) = x3 − 6x2 − 36x + 7, then f(x) is strictly increasing for the value of x.
(a) x < −1 and x > 5
(b) x < −2 and x > 6
(c) x < −3 and x > 3
(d) x < −4 and x > 1
8. Maximum value of f(x) =
(a) −
1
4
x
x2 + x + 4
(b) −
1
3
on [−1, 1] is
(c)
1
6
(d)
1
5
9. The curve y = 3x5 − 40x3 + 3x − 20 is concave up in the intervals
(a) (−2, 0) and (2, ∞)
(b) (−3, 0) and (1, ∞)
(c) (0, 2) and (2, ∞)
(d) None of these
M03_ENGINEERING_MATHEMATICS-I _CH03_Part E.indd 150
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Differential Calculus ■
3.151
10. The point of inflexion of the curve y2 = x(x + 1)2 is
⎛ 2 −4 ⎞
(b) ⎜ ,
⎝ 3 3 3 ⎟⎠
⎛1 4 ⎞
(a) ⎜ ,
⎝ 3 3 3 ⎟⎠
⎛ 2 5 2⎞
(c) ⎜ ,
(d) None of these
⎟
⎝ 3 3 3⎠
11. The maximum value of the function y = x(x − 1)2, 0 ≤ x ≤ 2, is
(a) 0
(b)
4
27
(c) −4
(d) None of these
12. If a < 0, then the function f(x) = eax + e−ax is decreasing for all values of x, where
(a) x > 0
(b) x < 0
(c) x > 1
(d) x < 1
13. Which of the following functions satisfies the conditions of Rolle’s theorem?
x −1
,0≤x≤1
(a) f ( x ) =
(b) f ( x ) = x ( x − 1) , 0 ≤ x ≤ 1
x
(c)
f ( x) =
tan x
,0≤x≤p
x
(d)
1 −1
1
f ( x ) = sin ,
≤x≤
x p
p
14. If 2a + 3b + c = 0, then at least one root of the equation ax2 + bx + c = 0 lies in the interval
(a) (0, 1)
(b) (1, 2)
(c) (2, 3)
(d) (1, 3)
15. If the line y = 2x + k is a tangent to the curve x2 = 4y, then k is equal to
1
(a) 4
(b)
(c) −4
2
(d) −
1
2
16. The radius of curvature at any point on the curve 2x2 + 2y2 + 5x − 2y + 1 = 0 is
(a)
4
21
(b)
21
4
(c)
5
2
(d) None of these
17. The family of straight lines 2y − 4x + l = 0 has envelope
(a) x2 + y2 = 2
(b) x2 − y2 = 2
(c) x =
1
y
2
(d) No envelope
18. Envelope of the family of straight lines x cosu + ysinu = a, where a is constant, is
(a) x2 − 2xy = 0
(b) x2 − y2 = a2
(c) x2 + y2 = a2
(d) None of these
19. The centre of curvature at any point on the curve x2 + y2 − 2x + 4y + 2 = 0 is
(a) (−2, 4)
(b) (1, −2)
(c) (−1, 2)
(d) None of these
20. The centre of curvature of y = x2 at origin is
1 1
(a) ⎛⎜ , ⎞⎟
⎝ 2 2⎠
1
(b) ⎛⎜ 0, ⎞⎟
⎝ 2⎠
21. Asymptote parallel to x-axis and y-axis are
(a) y = 1 and x = 1
(b) y = 1 and x = 2
1
(c) ⎛⎜ , 0⎞⎟
⎝2 ⎠
1 1
(d) ⎛⎜ , ⎞⎟
⎝ 4 4⎠
(c) y = 2 and x = 0
(d) y = 2 and x = 1
22. Number of oblique asymptotes of the curve x y + xy + xy + y + 3x = 0 is
(a) 0
(b) 1
(c) 2
(d) 3
23. The number of asymptotes of the curve y (x − y ) − 2ay + 2a x = 0 is
(a) 1
(b) 2
(c) 3
(d) 4
2
2
M03_ENGINEERING_MATHEMATICS-I _CH03_Part E.indd 151
2
2
2
2
3
3
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3.152
■
Engineering Mathematics
24. The equation of the asymptotes of x3 + y3 = 3axy, a > 0, is
(a) x − y − a = 0
(b) x + y + a = 0
(c) x + y − a = 0
(d) x − y + a = 0
25. The asymptotes of the curve (y − x) (y − 2x) − (y − x)(y − 6x) + x − 5y + 3 = 0 parallel to the line y − x = 0
are
(a) y − x − 1 = 0 and y − x + 1 = 0
(b) y − x − 2 = 0 and y − x + 2 = 0
(c) y − x − 1 = 0 and y − x − 4 = 0
(d) None of these
2
ANSWERS
A. Fill up the blanks
1.
x x
+ =1
a b
2. t =
1
2
7. x = −2
6. 1
12. −
11. ep
1
3
3. a
2
4. b x
2
a
8. x = 1
9. y = ±x
13. 0
14. −
9
2
5. 2 x
10. [−1, 1]
15.
up
B. Choose the correct answer
1. (b)
11. (b)
21. (c)
2. (a)
12. (a)
22. (b)
3. (a)
13. (b)
23. (c)
4. (c)
14. (a)
24. (b)
M03_ENGINEERING_MATHEMATICS-I _CH03_Part E.indd 152
5. (a)
15. (b)
25. (c)
6. (b)
16. (b)
7. (b)
17. (d)
8. (a)
18. (c)
9. (c)
19. (b)
10. (a)
20. (b)
5/19/2016 8:49:15 PM
Applications of Differential
Calculus
4
4.1 CURVATURE IN CARTESIAN COORDINATES
4.1.0 Introduction
To characterize a curve completely we have seen various aspects of the curve such as increasing and
decreasing nature, maximum and minimum points, concavity and convexity, symmetry and special
points such as points of inflexion etc. Another aspect to characterize the shape of a curve is the degree
of its bending or curvature.
In many practical problems we are concerned with the bending of a curve at different points or
the bending of two curves such as rail tracks. The concept of curvature is considered while laying rail
tracks and designing highways. The curvature at a point is a numerical measure of the rate of bending
of a curve.
4.1.1 Measure of Curvature
Definition 4.1 Let Γ be a curve that does not intersect itself and having tangents at each point. Let A
be a fixed point on the curve from which arc length is measured. Let P and Q be neighbouring points on the curve so
Y
that AP = s and AQ = s + Δs.
∴ length of arc PQ = Δs
Let the tangents at P and Q make angles c and c + Δc
respectively with the positive direction of x-axis.
∴ Δc is the angle between the tangents at P and Q.
Precisely, Δc is the angle through which the tangent
turns from P to Q as P moves along the arc through the
distance Δs.
Q
Δψ
Δs
P
s
A
ψ
ψ + Δψ
O
X
1. The angle Δc is called the angle of contigence of the
arc PQ or the total curvature of the arc PQ.
Fig. 4.1
2. The ratio Δc is called the average curvature of the arc PQ.
Δs
Δc d c
=
and it is denoted by the greek letter
3. The curvature of the curve at P is defined as lim
Δs →0 Δs
ds
dc
k (kappa). Thus, k =
.
ds
Note
1. s and c are called the intrinsic coordinates of P and f(s, c) = 0 is called the intrinsic equation of
the curve.
dc
2. Since the difference in angles and difference in arc lengths are Δc and Δs , we have k =
.
ds
So, curvature is a positive quantity.
M04_ENGINEERING_MATHEMATICS-I _CH04_Part-A.indd 1
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4.2
■
Engineering Mathematics
Theorem 4.1 The curvature of a circle at any point is a constant and is equal to the reciprocal of the
radius of the circle.
Proof Consider a circle with centre C and radius r. Let P and Q be two neighbouring points on the
circle.
Let the angles which the tangents at P, Q make with x-axis
Y
ˆ = Δc
be c, c + Δc
∴ PCQ
Arc AP = s and arc AQ = s + Δs so that arc PQ = Δs
But we know arc Δs = rΔc [From trigonometry]
Δc 1
C Δψ
=
Δs r
r
A
Δc 1
P
∴
lim
=
[{ the radius r is constant ]
Δs → 0 Δs
r
dc 1
ψ
⇒
=
ds r
O
1
∴ curvature at the point P is a constant =
r
1
Fig. 4.2
Hence, curvature at any point of the circle is a constant =
r
= reciprocal of its radius
∴
Q
ψ + Δψ
X
Note If r → ∞, the curvature tends to zero. i.e., when radius r → ∞, the circle approaches a straight
line.
Hence, the curvature of a straight line is zero at any of its points. In otherwords, the straight line
does not bend at any point.
Definition 4.2 Radius of Curvature
If the curvature at a point P on a curve is k, then 1 is called the radius of curvature at P (if k ≠ 0).
k
Radius of curvature is denoted by Greek letter r.
Thus, r =
1 ds
=
.
k dc
Note From the definition of curvature it is obvious that we should know the intrinsic equation of the
curve. This is not easy in many cases. Generally, equation of a curve is given in Cartesian or polar
coordinates. So, we shall derive formula for radius of curvature for Cartesian equation of a given curve.
4.1.2 Radius of Curvature for Cartesian Equation of a Given Curve
Let y = f(x) be the equation of a curve, then we know that at the point (x, y),
dy
= tan c, where c is the
dx
angle made by the tangent at (x, y) with the positive direction of the x-axis.
d2 y
dc
= sec 2 c
∴
dx 2
dx
d c ds
= (1 + tan 2 c)
ds dx
M04_ENGINEERING_MATHEMATICS-I _CH04_Part-A.indd 2
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■
Applications of Differential Calculus
Y
⎡ ⎛ dy ⎞ 2 ⎤ 1 ds
= ⎢1 + ⎜ ⎟ ⎥ ⋅
⎣ ⎝ dx ⎠ ⎦ r dx
∴
y = f(x)
⎡ ⎛ dy ⎞ 2 ⎤
⎢1 + ⎜⎝ ⎟⎠ ⎥
dx ⎦ ds
r= ⎣
⋅
d2 y
dx
2
dx
P
But we know that
ψ
( ds) 2 = ( dx ) 2 + ( dy ) 2
2
∴
⎛ dy ⎞
⎛ ds ⎞
⎜⎝ ⎟⎠ = 1 + ⎜⎝ ⎟⎠
dx
dx
4.3
O
X
2
Fig. 4.3
1/ 2
⇒
∴
⇒
2
ds ⎡ ⎛ dy ⎞ ⎤
= ⎢1 + ⎜ ⎟ ⎥
dx ⎣ ⎝ dx ⎠ ⎦
⎡ ⎛ dy ⎞ 2 ⎤
⎢1 + ⎜⎝ ⎟⎠ ⎥
dx ⎦
r= ⎣
d2 y
dx 2
(1 + y12 )3/ 2
r=
y2
where y1 =
3/ 2
(1)
dy
d2 y
and y2 = 2
dx
dx
Note
1. When calculating r only positive value should be taken i.e., numerical value of r is taken as
radius of curvature, since it cannot be negative. If y2 > 0 , the curve is concave up and if y2 < 0
then it is concave down or convex up at the point.
2. At a point of inflexion i.e., when y2 = 0 , the curvature is defined as zero.
3/ 2
⎡ ⎛ dx ⎞ 2 ⎤
⎢1 + ⎜ ⎟ ⎥
(1 + x12 )3/ 2
⎣ ⎝ dy ⎠ ⎦
(2)
3. If the equation of curve is x = f(y), then r =
if x2 ≠ 0
=
d2x
x2
dy 2
d2x
dx
x1 =
and x2 = 2
where
dy
dy
dy
4. If at a point
= ∞ formula (1) cannot be used. i.e., if the tangent is parallel to y-axis,
dx
dx
= 0 . So, we use formula (2) in such cases.
then
dy
M04_ENGINEERING_MATHEMATICS-I _CH04_Part-A.indd 3
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4.4
4.1.3
■
Engineering Mathematics
Radius of Curvature for Parametric Equations
If the equation of curve is given by parametric equations x = f (t ), y = g (t ), then we find
Let
∴
∴
dx
dy
= x ′,
= y′
dt
dt
dy
dy dt
y′
=
=
dx dx x ′
dt
2
d y
d ⎛ dy ⎞ d ⎛ y ′ ⎞ dt
=
⎜ ⎟ = ⎜ ⎟⋅
dx 2 dx ⎝ dx ⎠ dt ⎝ x ′ ⎠ dx
x ′y ′′ − y ′x ′′ 1 x ′y ′′ − y ′x ′′
=
⋅ =
x′
( x ′)2
( x ′ )3
dx dy
,
dt dt
3/ 2
∴
⎡ ⎛ y ′⎞ 2 ⎤
⎢1 + ⎜⎝ ⎟⎠ ⎥
x′ ⎦
the radius of curvature r = ⎣
x ′y ′′ − y ′x ′′
( x ′ )3
=
( x ′ 2 + y ′ 2 ) 3/ 2
x ′y ′′ − y ′x ′′
[in magnitude]
Note Radius of curvature for parametric equations can be obtained by using formula (1).
WORKED EXAMPLES
EXAMPLE 1
1
Find the radius of curvature at the point ⎛⎜ ,
⎝4
1⎞
⎟ on
4⎠
x 1 y 5 1.
Solution.
The given curve is
x + y =1
Differentiating w.r.to x, we get
1
⇒
2 x
+
1
(1)
dy
=0
2 y dx
⋅
y
dy
=−
dx
x
=
x −1
= 1−
x
1
x
[From (1) y = 1 − x ]
= 1 − x −1/ 2
d y
1
⎛ 1
⎞
= − ⎜ − ⋅ x −3/ 2 ⎟ = 3/ 2
⎝ 2
⎠ 2x
dx 2
dy
1
= 1−
= 1 − 2 = −1
dx
1/4
2
⎛ 1 1⎞
At the point ⎜ , ⎟ ,
⎝ 4 4⎠
and
M04_ENGINEERING_MATHEMATICS-I _CH04_Part-A.indd 4
d2 y
=
dx 2
1
⎛ 1⎞
2⋅⎜ ⎟
⎝ 4⎠
3/ 2
=
4 3/ 2 4 ⋅ 2
=
=4
2
2
5/12/2016 10:08:39 AM
Applications of Differential Calculus
4.5
y1 = −1 and y2 = 4
∴
∴
■
the radius of curvature r =
(1 + y 12 )3/ 2 (1 + 1)3/ 2 2 2
1
=
=
=
y2
y2
4
2
EXAMPLE 2
3a 3a
Find the radius of curvature at the point ⎛⎜ , ⎞⎟ on the curve x 3 1 y 3 53 axy.
⎝ 2 2⎠
Solution.
The given curve is x 3 + y 3 = 3axy .
Differentiating w.r.to x, we get
(1)
dy
⎤
⎡ dy
= 3a ⎢ x + y ⋅1⎥
dx
⎣ dx
⎦
dy 2
dy
ay − x 2
2
[ y − ax ] = ay − x ⇒
= 2
dx
dx y − ax
3x 2 + 3 y 2
⇒
(2)
Differentiating (2) w.r.to x, we get
⎛ dy
⎞
⎛ dy
⎞
( y 2 − ax ) ⎜ a − 2 x ⎟ − ( ay − x 2 ) ⎜ 2 y − a⎟
⎝
⎠
⎝ dx
⎠
d y
dx
=
2
2
2
dx
( y − ax )
2
3a 9a 2
−
dy
4 = −1
= 22
3a
dx 9a
− a⋅
4
2
a⋅
⎛ 3a 3a ⎞
At the point ⎜
,
,
⎝ 2 2 ⎟⎠
and
⎛ 9a 2
⎛ 3a 2 9a 2 ⎞
3a ⎞
− a ⋅ ⎟ ( − a − 3a) − ⎜
−
⎟ ( −3a − a)
⎜
⎝ 2
d y ⎝ 4
2⎠
4 ⎠
=
2
2
dx
⎛ 9a 2 3a 2 ⎞
−
⎜⎝
⎟
4
2 ⎠
⎡ 3a2 ⎛ 3a2 ⎞ ⎤
⎛ 3a2 ⎞
( −4a) ⎢
−⎜−
4
2
−
⋅
⋅
a
⎟
⎜⎝
⎟
⎥
4 ⎠ = −8a = − 32
⎣ 4 ⎝ 4 ⎠⎦ =
=
2
2
3a
3a2
⎛ 3a2 ⎞
⎛ 3a2 ⎞
⎜⎝
⎟⎠
⎜⎝
⎟⎠
4
4
4
∴
2
y1 = −1 and y2 = −
32
3a
3/ 2
(1 + y 12 )3/ 2 = (1 + 1) = −2 2 × 3a = − 3a
∴ the radius of curvature r =
32
32
8 2
y2
−
3a
3a
Since r is positive,
r=
8 2
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4.6
■
Engineering Mathematics
EXAMPLE 3
xy 2 5 a 3 2 x 3 at ( a, 0).
Find the radius of curvature of the curve
Solution.
The given curve is xy 2 = a3 − x 3
Differentiating w.r.to x, we get
dy
+ y 2 ⋅1 = −3x 2
dx
x ⋅ 2y
At the point (a, 0),
dy
=∞
dx
So, we use the formula,
∴
r=
dy
−(3x 2 + y 2 )
=
dx
2xy
⇒
dx
=0 ⇒
dy
x1 = 0
(1 + x12 )3/ 2
x2
dx
2xy
=− 2
dy
3x + y 2
Now,
Differentiating w.r.to y
⎧
dx ⎞
dx
⎛
⎛
⎞⎫
−2 ⎨(3 x 2 + y 2 ) ⎜ x ⋅1 + y ⎟ − xy ⎜ 3 ⋅ 2 x + 2 y ⎟ ⎬
dy
dy
⎝
⎠
⎝
⎠⎭
d x
⎩
=
2
2
2 2
dy
(3 x + y )
2
At the point(a, 0),
∴
3a 2 ⋅ a
2
d2x
= −2
=−
2
2 2
(3a )
3a
dy
the radius of curvature r =
Since r is positive,
r=
∴
x1 = 0 and x2 = −
2
3a
(1 + x 12 )3/ 2 (1 + 0)3/ 2
3a
=
=−
x2
2
−2/3a
3a
2
EXAMPLE 4
x2
y2
1
5 1, show that the radius of curvature at an end of the major axis is equal
a2
b2
to the semi-latus rectum of the ellipse.
In an ellipse
Solution.
x2 y 2
+
=1
a2 b 2
The given curve is
(1)
An end of the major axis is (a, 0)
Differentiating (1) w.r.to x, we get
1
1
dy
dy
b2 x
⋅ 2x + 2 2 y
=0⇒
=− 2
2
dx
dx
a
b
a y
M04_ENGINEERING_MATHEMATICS-I _CH04_Part-A.indd 6
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Applications of Differential Calculus
dx
dy
=0 ⇒
=∞ ∴
dy
dx
(1 + x12 )3/ 2
r=
x2
At the point (a, 0),
So, we use the formula
■
4.7
x1 = 0
dx
a2 y
=− 2⋅
dy
b x
Now
Differentiating w.r.to y, we get
dx ⎤
⎡
x ⋅1 − y ⋅ ⎥
⎢
dy ⎦
d x
a
=− 2 ⋅⎣
dy 2
b
x2
2
d2x
a2 a
a
=− 2 2 =− 2
2
dy
b a
b
At the point (a, 0),
∴ the radius of curvature
Since r is positive, r =
EXAMPLE 5
For the curve y 5
⎛ 2r ⎞
⎜⎝ ⎟⎠
a
2/3
2
2
r=
∴ x1 = 0 and
x2 = −
a
b2
b2
(1 + 0)3/ 2
=−
a
a
− 2
b
b2
, which is the length of the semi-latus rectum of the ellipse.
a
ax
, if r is the radius of curvature at any point (x, y), show that
a1 x
2
⎛ y⎞
⎛ x⎞
5⎜ ⎟ 1 ⎜ ⎟ .
⎝ x⎠
⎝ y⎠
Solution.
y=
The given curve is
ax
a+ x
(1)
Differentiating w.r.to x, we get
dy a[(a + x ) ⋅1 − x ⋅1]
a2
=
=
2
dx
(a + x )
(a + x ) 2
and
d2 y
2a 2
=
−
dx 2
( a + x )3
At the point (x, y),
∴ the radius of curvature
M04_ENGINEERING_MATHEMATICS-I _CH04_Part-A.indd 7
y1 =
r=
a2
2a 2
y
and
=
−
2
(a + x)2
( a + x )3
(1 + y 12 )3/ 2
y2
⎡
a4 ⎤
⎢1 + (a + x ) 4 ⎥
⎦
=⎣
2a2
−
(a + x ) 3
3/ 2
5/12/2016 10:09:07 AM
4.8
■
Engineering Mathematics
Since r is positive,
3/ 2
r=
⇒
2r ⎡⎣( a + x ) 4 + a 4 ⎤⎦
=
a
a3 ( a + x )3
⎛ 2r ⎞
⎜⎝ ⎟⎠
a
⇒
We have y =
⎡⎣(a + x ) 4 + a4 ⎤⎦
⎡⎣(a + x ) 4 + a4 ⎤⎦
=
3/ 2
2a2
2aa2 ⋅ (a + x )3
⎡⎣(a + x ) 4 ⎤⎦
(a + x ) 3
ax
a+ x
⇒
⎛ 2r ⎞
⎜⎝ ⎟⎠
a
∴
y
a
=
x a+ x
2/3
3/ 2
⎡(a + x ) 4 + a4 ⎤⎦
=⎣
2/3
⎡⎣a3 (a + x )3 ⎤⎦
(a + x ) 4 + a 4 (a + x ) 2
a2
= 2
=
+
2
2
a (a + x )
a
(a + x ) 2
2/3
2
2⎤
⎡
wer ⎥
⎢⎣ raising to the pow
3⎦
[dividing term by term]
x a+ x
=
y
a
and
⎛ x⎞ ⎛ y⎞
=⎜ ⎟ +⎜ ⎟
⎝ y⎠ ⎝ x⎠
3/ 2
2
EXAMPLE 6
If r1, r2 be the radii of curvatures at the ends of a focal chord of a parabola whose latus rectum
is 2l, prove that (r1 )22/3 1 (r2 )22/3 5 l 22/3.
Solution.
Let the parabola be
y 2 = 4ax
We know that 4a is the length of the latus rectum.
Given 2l is the latus rectum.
l
∴
4 a = 2l ⇒ a =
2
Let S be the focus (a, 0).
(1)
P(at 2, 2at)
Let PQ be the focal chord.
Let P be ( at 2 , 2at ) and Q be ( at12 , 2at1 ).
y − y 2 2at1 − 2at
= 2
The Slope of PQ = 1
x1 − x 2
at1 − at 2
=
Slope of SP
2a(t1 − t )
a(t1 + t )(t1 − t )
=
2
t1 + t
=
2at − 0
2at
2t
=
=
at 2 − a a(t 2 − 1) t 2 − 1
S
O
(a, 0)
Q(at 12, 2at 1)
[{ t1 ≠ t ]
Since slope of PQ = Slope of SP
M04_ENGINEERING_MATHEMATICS-I _CH04_Part-A.indd 8
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Applications of Differential Calculus
⇒
2
2t
= 2
t1 + t t − 1
∴
t 2 − 1 = t (t1 + t ) ⇒ t 2 − 1 = tt1 + t 2
⇒ t1t = −1
⇒ t1 = −
1
t
■
4.9
(2)
The radius of curvature at ( at 2 , 2at ) is r = 2a(1 + t 2 )3/ 2
∴ At the point P( at 2 , 2 at ), r 1= 2a(1 + t 2 )3/ 2
2
−
⇒
−
2
⇒ r1 3 =
r 1 3 = ( 2a) −2 / 3 (1 + t 2 ) −1
( 2a) −2 / 3
1+ t 2
At the point Q (at 12 , 2 at 1 ), r 2 = 2a(1 + t12 )3/ 2
2
r2 3 = ( 2a) −2 / 3 (1 + t12 )
−
⇒
−1
−1
= ( 2a)
∴
∴
r1
r1
−
2
3
−
2
3
+ r2
+ r2
−
2
3
−
2
3
−2 / 3
2 −1
t2
1⎞
⎛
−2 / 3 (1 + t )
[Using (2)]
= ( 2a) −2 / 3 ⋅
⎜⎝1 + 2 ⎟⎠ = ( 2a)
−2
t
t
1+ t 2
1
t2
+ ( 2a) −2 / 3 ⋅
2
1+ t
1+ t 2
⎛ 1
t2 ⎞
= ( 2a) −2 / 3 ⎜
+
⎟
2
⎝1+ t
1+ t 2 ⎠
1+ t 2
= ( 2a) −2 / 3
= ( 2a) −2 / 3 = l −2 / 3
1+ t 2
= ( 2a) −2 / 3
[{ 2a = l]
= l −2 / 3
EXAMPLE 7
Prove that the radius of curvature at any point of the cycloid x 5 a(u1sin u) , y 5 a( 1 2 cos u) is
u
4 a cos .
2
Solution.
The given curve is
∴
∴
∴
x = a(u + sin u)
and
y = a(1 − cos u)
dx
u
dy
= a(1 + cos u) = 2a cos 2
= a sin u
and
du
2
du
dy
u
u
2a sin cos
dy d u
a sin u
2
2 = tan u
=
=
=
u
u
dx dx
2
2a cos 2
2a cos 2
du
2
2
u 1
1
d 2 y d ⎛ dy ⎞ d u
= sec 2 ⋅ ⋅
=
=
⎜ ⎟⋅
u
2 2
dx 2 d u ⎝ dx ⎠ dx
2a cos 2
2
M04_ENGINEERING_MATHEMATICS-I _CH04_Part-A.indd 9
sec 4
u
2
4a
5/12/2016 10:09:20 AM
4.10
■
Engineering Mathematics
u
y1 = tan
2
∴
and
y2 =
sec 4
u
2
4a
3
∴
the radius of curvature r =
(1 + y 12 )3/ 2
y2
2
⎛
u
2 u⎞
4asec3
⎜⎝1 + tan ⎟⎠
2 =
2 = 4 acos u
=
u
u
2
sec 4
sec 4
2
2
4a
EXAMPLE 8
If r1 , r2 be the radii of curvature at the points P and Q on the cycloid x 5 a(u 1 sin u),
y 5 a( 1 2 cos u), where the tangents are at right angles, then r12 1 r22 5 16a 2 .
Solution.
Given x = a(cos u + sin u), y = a(1 − cos u)
From Example 7, the radius of curvature at any point u is
u
r = 4 a cos
2
Let the parameters of the points P be u1 and Q be u2, then
u
u
r1 = 4 a cos 1 , r2 = 4 a cos 2
2
2
u
u1
But the slope of tangent at P is tan and the slope of the tangent at Q is tan 2
2
2
Since tangents are perpendicular to each other, we get
tan
u1
u
tan 2 = −1 ⇒
2
2
tan
∴
∴
⇒
tan
u2
⎛p u ⎞
= tan ⎜ + 1 ⎟
⎝2 2⎠
2
u2
=−
2
⇒
u
1
= − cot 1
u1
2
tan
2
u 2 p u1
= +
2
2 2
u
⎛p u ⎞
r2 = 4 a cos ⎜ + 1 ⎟ = −4 a sin 1
⎝ 2 2⎠
2
u1
u
+ 16 a 2 sin 2 1
2
2
u
u⎞
⎛
2
2
2
r12 + r22 = 16 a 2 ⎜ cos 2 1 + sin 2 1 ⎟ ⇒ r1 + r2 = 16 a
⎝
2
2⎠
r12 + r22 = 16 a 2 cos 2
EXAMPLE 9
Find the points on the parabola y 2 5 4 x at which the radius of curvature is 4 2.
Solution.
Given
y2 = 4x
M04_ENGINEERING_MATHEMATICS-I _CH04_Part-A.indd 10
(1)
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Applications of Differential Calculus
■
4.11
Differentiating w.r.to x,
2y
dy 2
dy
=
=4 ⇒
dx y
dx
d2 y
2 dy
4
=− 2⋅ =− 3
dx 2
y dx
y
and
Let (a, b) be the point on the curve at which the radius of curvature is 4 2.
Since (a, b) is on
y2 = 4x,
b2 = 4a
At the point (a, b),
d2 y
4
=− 3
2
dx
b
dy 2
= ,
dx b
r=
(1 + y )
2 3/ 2
1
y2
r=
Given
r=4 2
∴
(b 2 + 4)3/ 2
=4 2 ⇒
4
⇒
43 (a + 1)3 = 16 2 × 2
(a + 1)3 =
4⎞
⎛
⎜⎝1 + 2 ⎟⎠
b
=
−4
b3
y1 =
2
b
=−
(b 2 + 4)3 / 2
(b 2 + 4)3/ 2
=−
3
4
4b
y2 = −
and
4
b3
3/ 2
( b 2 + 4 ) 3/ 2
4
Since r is positive,
⇒
∴
(2)
( 4a + 4)3/ 2 = 16 2
43/ 2 (a + 1)3/ 2 = 16 2
[squaring both sides]
16 2 × 2
= 8 = 23
43
⇒
b 2 = 4a
⇒
(2) is
⇒
a +1 = 2 ⇒ a = 1
b 2 = 4 ⇒ b = ±2
∴ the points are (1, 2) and (1, −2).
4.1.4
Centre of Curvature and Circle of Curvature
Let Γ be a simple curve having tangent at each point. At any point P on this curve we can draw a circle
having the same curvature at P as the curve Γ.
This circle is called the circle of curvature and its centre is called the centre of curvature and its
radius is the radius of curvature of Γ at P.
How to draw the circle of curvature is given in the next definition.
Definition 4.3 Let Γ be a simple curve and let P be a point of Γ.
Draw the normal at P to the curve Γ in the direction of the concavity of the curve (which is the positive
direction of the normal) and cut off a segment PC = r, the radius of curvature of Γ at P.
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4.12
■
Engineering Mathematics
The point C is called the centre of curvature of the given curve at P.
The circle with centre C and radius r (passing through P) is
called the circle of curvature of the given curve at P.
tangent at P
P
C
Note
1. From the definition of circle of curvature it follows that at the
given point, the curvature of the curve and curvature of the
circle are the same.
2. It is quite possible that the circle of curvature at a point crosses
the curve as in Fig. 4.4, just as a tangent line crosses the curve
at the point of inflexion.
4.1.5
Fig. 4.4
Y
Coordinates of the Centre of Curvature
Let P (x, y) be the point on y = f ( x ). Let C ( x , y ) be the centre
of curvature at P.
Then PC = r.
Let the tangent at P make an angle c with the x-axis.
=c
Then
[{ angle between two lines = angle
between their perpendiculars]
NP
From the right angle ΔCNP, sin c =
⇒ NP = r sin c
r
Now
x = OM = OL − ML = OL − NP = x − r sin c
and
y = MC = MN + MC = y + r cos c
Since
circle of
curvature
ρ
dy
= y1,
dx
y1
sin c =
,
1 + y 12
C(x, y)
ρ
ψ
N
P(x, y)
ψ
O
M
L
X
Fig. 4.5
1 + y 12
y1
tan c =
cos c =
ψ
1
1 + y 12
1
Fig. 4.6
(1 + y12 )3/ 2
y2
and
r=
∴
x = x−
(1 + y12 )3/ 2
y1
⋅
y2
1 + y12
⇒
x = x−
y1 (1 + y12 )
y2
(1)
and
y = y+
(1 + y12 )3/ 2
1
⋅
y2
1 + y12
⇒
y = y+
(1 + y12 )
y2
(2)
Thus, the centre of curvature ( x , y ) is given by (1) and (2).
∴ the equation of the circle of curvature at P is ( x − x ) 2 + ( y − y ) 2 = r 2
Note The centre of the curvature formula holds if y2 > 0 or < 0.
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Applications of Differential Calculus
■
4.13
WORKED EXAMPLES
EXAMPLE 1
Find the circle of curvature at (3, 4) on xy 5 12.
Solution.
The given curve is
xy = 12
∴
dy
12
=− 2
dx
x
and
d 2y
24
=+ 3
2
dx
x
At the point (3, 4)
dy
12
4
=− =−
dx
9
3
and
d 2 y 24 8
=
=
dx 2 27 9
∴
y1 = −
4
3
y2 =
y=
⇒
and
12
x
8
9
The centre of curvature ( x , y ) is given by
⎛ 16 ⎞
⎜1 + ⎟⎠
y 1 (1 + y 12 )
4 25 43
⎛ −4 ⎞ ⎝
9
= 3+ ⋅
=
x=x−
= 3−⎜ ⎟
⎝
⎠
8
y2
3
3 8
6
9
16
1+
(11 + y 12 )
9 = 4 + 25 = 57
= 4+
y =y+
8
y2
8
8
9
⎛ 43 57 ⎞
( x, y) = ⎜ , ⎟
⎝ 6 8⎠
∴ the centre of Curvature is
and the radius of curvature
r=
(1 + y )
y2
2 3/ 2
1
⎛ 4⎞
⎜⎝1 + ⎟⎠
9
=
8
9
3/ 2
=
125
( 25)3/ 2
=
8 24
9×3×
9
∴ the equation of circle of curvature at (3, 4) is
2
( x − x )2 + ( y − y )2 = r 2
⇒
EXAMPLE 2
2
2
43 ⎞
57 ⎞
⎛ 125 ⎞
⎛
⎛
⎟ .
⎜⎝ x − ⎟⎠ + ⎜⎝ y − ⎟⎠ = ⎜⎝
6
8
24 ⎠
2
Show that the circle of curvature of
Solution.
The given curve is
2
a a
a2
3a
3a
x 1 y 5 a at ⎛⎜ , ⎞⎟ is ⎛⎜ x 2 ⎞⎟ 1 ⎛⎜ y 2 ⎞⎟ 5 .
⎝ 4 4⎠ ⎝
⎝
4⎠
4⎠
2
x+ y = a
⇒
y = a− x
(1)
Differentiating w.r.to x, we get
1
dy
1
=−
dx
2 y
2 x
⋅
M04_ENGINEERING_MATHEMATICS-I _CH04_Part-A.indd 13
⇒
y
dy
=−
dx
x
(2)
5/12/2016 10:09:43 AM
4.14
■
Engineering Mathematics
( a− x)
dy
a
= 1−
= 1 − a ⋅ x −1/ 2
=−
dx
x
x
Now
d2 y
a −3/ 2
x
= − a ( −1/2 x −3/ 2 ) =
2
2
dx
∴
⎛ a a⎞
At the point ⎜ , ⎟
⎝ 4 4⎠
a/4
dy
=−
= −1
dx
a/4
d 2y
a ⎛ a⎞
=
⋅⎜ ⎟
2
2 ⎝ 4⎠
dx
and
−3 / 2
the radius of curvature r =
a 4⋅2 4
⋅
=
2 a a a
=
y 1 = −1 and y 2 =
∴
∴
[from (1)]
4
a
(1 + y12 )3/ 2 (1 + 1)3/ 2 a 2
a
=
=
=
y2
4/a
2
2
a a
At the point ⎛⎜ , ⎞⎟ , the coordinates of the centre of curvature ( x , y ) is given by
⎝ 4 4⎠
y (1 + y 12 ) a ( −1)(1 + 1) a a 3a
x=x− 1
= −
= + =
4
4/a
4 2 4
y2
y =y+
(1 + y 12 ) a 1 + 1 a a 3a
= + =
= +
4
4
4 2 4
y2
a
3a 3a
∴ the centre of curvature is ( x , y ) = ⎛⎜ , ⎞⎟
⎝ 4 4⎠
a a
∴ the circle of curvature at ⎛⎜ , ⎞⎟ is
⎝ 4 4⎠
2
( x − x )2 + ( y − y )2 = r 2
⇒
2
3a ⎞ ⎛
3a ⎞
a2
⎛
⎜⎝ x − ⎟⎠ + ⎜⎝ y − ⎟⎠ = .
4
4
2
EXAMPLE 3
Find the centre of curvature and equation of the circle of curvature at the point P on the curve
y 5 e x where the curve crosses the y-axis.
Solution.
The given curve is
∴
y = ex
dy
= ex
dx
(1)
and
d2 y
= ex
dx 2
Also given P is the point on the y-axis where the curve crosses it.
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Applications of Differential Calculus
■
4.15
Equation of y-axis is x = 0.
∴
y = e0 = 1
∴ P is (0, 1)
∴ At the point P,
dy
= 1,
dx
and
(1 + y )
r=
2 3/ 2
1
y2
d 2y
=1
dx 2
=
∴
y1 = 1 and y2 = 1
(1 + 1)3/ 2
=2 2
1
At the point P, the coordinates of the centre of curvature ( x , y ) is given by
y 1 (1 + y 12 )
x =x−
y2
y =y+
and
∴ the centre of curvature is
= 0−
1(1 + 1)
= −2
1
1 + y 12
(1 + 1)
= 1+
=3
y2
1
( x , y ) = ( −2, 3)
∴ the equation of the circle of curvature is
( x − x )2 + ( y − y )2 = r 2
2
2
⇒ ( x + 2) + ( y − 3) = 8
EXERCISE 4.1
3
1. Find the radius of curvature at (−2, 0) on the curve y = x + 8.
x
2. Find r for the curve y = c cosh at the point (0, c).
c
x2 y2
3. Find the radius of curvature at any point ( a cos u, b sin u) on the ellipse 2 + 2 = 1.
a
b
(
)
x
−
a ax
e +e a .
2
2
5. Find the radius of curvature of x = 2c( y − c) where it crosses the y-axis.
4. Find the radius of curvature at any point (x, y) on the curve y =
6. Show that the radius of curvature of the curve x = 3a cos u − a cos 3u, y = 3a sin u − a sin 3u is
3a sin u.
p
7. Find the radius of curvature of the curve x = a cos u, y = sin u at u = .
4
8. Find the radius of curvature at x = c on the curve xy = c 2.
9. Find the radius of curvature at x = 1 on y = log e x.
a( x 2 + y 2 )
10. Find the radius of curvature at ( −2a, 2a) on y =
.
x2
t
11. Find the radius of curvature of the curve x = e t cos t , y = e sin t at (1, 0).
12. Find the radius of curvature of the parabola x = at 2 , y = 2at at t.
13. Find the radius of curvature at any point ‘u’ on x = a log(sec u + tan u), y = a sec u.
M04_ENGINEERING_MATHEMATICS-I _CH04_Part-A.indd 15
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4.16
■
Engineering Mathematics
2
14. Find the radius of curvature at y = 2a on the curve y = 4 ax.
p
5 5
15. Prove that at the point x = of the curve y = 4 sin x − sin 2 x, r =
.
2
4
1/ 2
1/ 2
2( ax + by )3/ 2
x
y
.
16. Prove that the radius of curvature at any point (x, y) on ⎛⎜ ⎞⎟ + ⎛⎜ ⎞⎟ = 1 is r =
ab
⎝ b⎠
⎝ a⎠
2/3
2/3
2/3
17. Show that the radius of curvature at the point ( a cos3 u, a sin 3 u) on the curve x + y = a is
3a sin u cos u.
18. Prove that the radius of curvature at any point of the astroid x 2 / 3 + y 2 / 3 = a 2 is three times the
length of the perpendicular from the origin to the tangent at that point.
x
19. Find the radius of curvature and the curvature at the point (0, c) on the curve y = c cosh .
c
20. If r is the radius of curvature at any point P on the parabola y 2 = 4 ax and S is its focus, show
that r2 varies as (SP)3.
21. Find r for the curve x = a(cos t + t sin t ), y = a(sin t − t cos t ).
u
22. Show that the radius of curvature at any point of the curve x = ae (sin u − cos u),
y = ae u (sin u + cos u) is twice the perpendicular distance from the origin to the tangent at the point.
23. Show that the measure of curvature of the curve
ab
x
y
.
+
= 1 at any point (x, y) on it is
2( ax + by )3/ 2
a
b
x2 y2
24. Find the centre of curvature of the hyperbola 2 − 2 = 1 at the point (a sec u, b tan u) .
a
b
25. Find the centre of curvature of y 2 = 4ax at an end of the latus rectum.
26. Find the centre of curvature at (c, c) on xy = c 2.
⎛ 1 1⎞
27. Find the circle of curvature at ⎜ , ⎟ on x + y = 1.
⎝ 4 4⎠
28. Find the circle of curvature for the curve x 3 + y 3 = 3 xy at the point ⎛⎜ 3 , 3 ⎞⎟ on it.
⎝ 2 2⎠
4
4
29. Find the coordinates of the centre of curvature of the curve x + y = 2a 2 xy at (a, a).
x2 y2
30. If the centre of curvature of the ellipse 2 + 2 = 1 at one end of the minor axis lies at the other
a
b
end. Find the eccentricity of the ellipse.
[Hint At (0, b), the centre of curvature is (0, −b) ∴ r = 2b compute r and find e]
2
31. Find the equation of the circle of curvature at (3, 6) on y = 12 x.
32. Find the equation of the circle of curvature at (c, c) on xy = c2.
x
33. Find the radius of curvature and centre of curvature at any point (x, y) on the curve y = c log sec .
c
34. Find the radius of curvature and centre of curvature of the curve x 4 + y 4 = 2 at the point (1, 1).
ANSWERS TO EXERCISE 4.1
1
( a sin 2 u + b 2 cos 2 u)3/ 2
ab
1. 6
2. c
3.
5. c
7. a
8. c 2
M04_ENGINEERING_MATHEMATICS-I _CH04_Part-A.indd 16
4. a cosh 2
9.
x
a
2 2
3
5/12/2016 10:10:10 AM
Applications of Differential Calculus
10. 2a
11.
14. r = 4a 2 .
18. r = 3p
22. r = 2p
12. 2 a (1 + t 2 )
2
5 5
4
16. r =
2
17. 3asinθ cosθ
21. at
25. (5a, − 2a)
2
3
3
1
27. ⎛⎜ x − ⎞⎟ + ⎛⎜ y − ⎞⎟ =
⎝
⎠
⎝
⎠
4
4
2
26. ( 2c, 2c)
28. x 2 + y 2 −
6a 6a
29. ⎛⎜ , ⎞⎟
⎝ 7 7⎠
21
432
( x + y) +
=0
8
128
30.
31. ( x − 15) 2 + ( y + 6) 2 = 288
32. ( x − 2c) 2 + ( y − 2c) 2 = 2c 2
x
x ⎛
⎞
33. c sec , ⎜ x − c tan , y + c ⎟
⎠
c ⎝
c
34.
4.1.6
4.17
13. asec 2u
2
(ax + by)3/2 .
ab
4
1
2
3
19. Curvature =
20. r = (SP)
a
c
a2 + b2
−( a 2 + b 2 ) 3
24. x =
sec3u; y =
tan u
a
b
15. r =
■
2 ⎛2
,⎜ ,
3 ⎝3
1
2
2⎞
⎟
3⎠
Radius of Curvature in Polar Coordinates
Let the equation of the curve in polar coordinates be r = f(u)
Let P(r, u) be any point on the curve
Let r be the radius of curvature at the point P.
Let O be the pole and OA be the initial line. Draw the tangent at P and it meets OA at the point B.
Let PB make an angle c with OA.
Let C be a fixed point on the curve from which the arc length is measured.
Let CP = s and CQ = s + Δs. so that the arc PQ = Δs.
Let AOP = u and f the angle between the tangent at P and the radius vector OP
ds
∴ the radius of curvature is
r=
Q
dc
r+
dc
df
= 1+
du
du
∴
We know
Δs
c=u+f
and from Fig. 4.7
[ From Δ OPB]
tanf = r
Differentiating with respect to u, we get
du
r
=
dr
dr
du
P
Δr
s
r
φ
C
ψ
θ
O
B
A
Fig. 4.7
2
d r
dr dr
d 2 r ⎛ dr ⎞
⋅
− r 2 ⎜⎝ ⎟⎠ − r 2
df d u d u
du
du
2
d
u
sec f
=
=
2
2
du
⎛ dr ⎞
⎛ dr ⎞
⎜⎝ ⎟⎠
⎜⎝ ⎟⎠
du
du
M04_ENGINEERING_MATHEMATICS-I _CH04_Part-A.indd 17
2
5/12/2016 10:10:19 AM
4.18
■
Engineering Mathematics
⎡ ⎛ dr ⎞ 2
d 2r
⎢⎜ ⎟ − r 2
1 ⎢ ⎝ du ⎠
df
du
=
2
d u sec 2 f ⎢
dr
⎛ ⎞
⎢
⎜⎝ ⎟⎠
du
⎣
⇒
⎤
⎥
⎥
⎥
⎥
⎦
⎡ ⎛ dr ⎞ 2
d 2r
⎢⎜ ⎟ − r 2
1
du
⎢ ⎝ du ⎠
=
2
2
⎢
1 + tan f
⎛ dr ⎞
⎢
⎜⎝ ⎟⎠
du
⎣
⎡ ⎛ dr ⎞ 2
d 2r
⎢⎜ ⎟ − r 2
1
du
⎢ ⎝ du ⎠
=
2
r2 ⎢
⎛ dr ⎞
1+
2 ⎢
⎜⎝ ⎟⎠
du
⎛ drr ⎞ ⎣
⎜⎝ ⎟⎠
du
⎤
⎥
⎥
⎥
⎥
⎦
⎤
⎥
⎥
⎥
⎥
⎦
⎡ ⎛ dr ⎞ 2
d 2r
⎢⎜ ⎟ − r 2
du
⎢ ⎝ du ⎠
=
2
2
⎢
⎛ dr ⎞
⎛ dr ⎞
r2 + ⎜ ⎟ ⎢
⎜⎝ ⎟⎠
⎝ du⎠ ⎣
du
⎛ dr ⎞
⎜⎝ ⎟⎠
du
2
⎤ ⎛ dr ⎞ 2
d 2r
⎥ ⎜ ⎟ −r 2
du
⎥ = ⎝ du ⎠
2
⎥
⎛ dr ⎞
r2 + ⎜ ⎟
⎥
⎝ du ⎠
⎦
2
∴
dc
df
= 1+
= 1+
du
du
d 2r
⎛ dr ⎞
⎜⎝ ⎟⎠ − r 2
du
du
⎛ dr ⎞
r2 + ⎜ ⎟
⎝ du ⎠
2
2
2
2
d 2r
⎛ dr ⎞ ⎛ dr ⎞
r +⎜ ⎟ +⎜ ⎟ −r 2
⎝ du ⎠ ⎝ du ⎠
du
=
⎛ dr ⎞
r +⎜ ⎟
⎝ du ⎠
2
2
We know
∴ the radius curvature
ds
⎛ dr ⎞
= r2 + ⎜ ⎟
⎝ du ⎠
du
r = ds ⋅ d u =
du d c
d 2r
⎛ dr ⎞
r + 2⎜ ⎟ − r 2
⎝ du ⎠
du
2
2
=
⎛ dr ⎞
r2 + ⎜ ⎟
⎝ du ⎠
2
2
⎡ 2 ⎛ dr ⎞ 2 ⎤
⎢r + ⎜ ⎟ ⎥
⎝ d u ⎠ ⎥⎦
⎢⎣
2
d 2r
⎛ dr ⎞
r2 + 2 ⎜ ⎟ − r 2
⎝ du ⎠
du
⎛ dr ⎞
r2 + ⎜ ⎟
⎝ du ⎠
2
3/ 2
⇒
⎡ 2 ⎛ dr ⎞ 2 ⎤
⎢r + ⎜ ⎟ ⎥
⎝ d u ⎠ ⎥⎦
⎢
r= ⎣
2
d 2r
⎛ dr ⎞
r2 + 2 ⎜ ⎟ − r 2
⎝ du ⎠
du
M04_ENGINEERING_MATHEMATICS-I _CH04_Part-A.indd 18
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Applications of Differential Calculus
■
4.19
3/ 2
⎡ r 2 + r12 ⎤⎦
r = ⎣2
r + 2r12 − rr2
⇒
where r1 =
dr
du
and r2 =
d 2r
.
du2
WORKED EXAMPLES
EXAMPLE 1
Find the radius of curvature of the cardioid r = a (1 + cos u) and show that
Solution.
The given equation is
r2
is a constant.
r
r = a (1 + cos u)
r=
The radius curvature is
(r
2
+ r12
)
(1)
3
2
r 2 + 2r12 − rr2
Differentiating (1) with respect to u, we get
r1 =
dr
= a( − sin u) = − a sin u
du
r2 =
and
d 2r
= − a cos u
du2
r 2 + r12 = r 2 + a2 sin 2 u
Now
= a2 (1 + cos u) 2 + a2 sin 2 u
= a2 ⎡⎣1 + cos 2 u + 2 cos u + sin 2 u⎤⎦
= a2 ⎡⎣1 + cos 2 u + sin 2 u + 2 cos u⎤⎦ = 2a2 [1 + cos u] = 2ar
3
∴
and
3
⎡⎣ r 2 + r12 ⎤⎦ 2 = [ 2ar ] 2
r 2 + 2r12 − rr2 = r 2 + 2a 2 sin 2 u − r ( − a cos u)
= a 2 (1 + cos u) 2 + 2a 2 sin 2 u + a ⋅ a (1 + cos u) cos u
= a 2 ⎡⎣1 + cos 2 u + 2cos u + 2 sin 2 u + coos u + cos 2 u⎤⎦
= a 2 ⎡⎣1 + 2(cos 2 u + sin 2 u) + 3cos u⎤⎦
[{ cos 2 u + sin 2 u = 1]
= 3a 2 (1 + cos u) = 3ar
3
∴ the radius of curvature is
⇒
[ 2ar ]2
r=
r2 =
M04_ENGINEERING_MATHEMATICS-I _CH04_Part-A.indd 19
3ar
( 2ar )3
(3ar )
2
=
8
ar
9
⇒
r2 8
= a,
9
r
constant
5/12/2016 10:10:26 AM
4.20
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Engineering Mathematics
EXAMPLE 2
Find the radius of curvature at any point on the curve rn = ansin nu, where a is a constant.
Solution.
The given curve is
rn = ansin nu
Taking log on both sides, we get
log e r n = log e a n + log e sin nu
⇒
n log e r = n log e a + log e sin nu
(1)
The radius of curvature at any point is
3
( r 2 + r12 ) 2
r= 2
r + 2r12 − rr2
(2)
Differentiating (1) with respect to u, we get
n⋅
⇒
Now
1 dr
1
cos nu ⋅ n
=
r d u sin nu
dr
= r cot nu ⇒ r1 = r cot nu
du
dr
d 2r
= − nr cos ec 2 nu + r cot 2 nu
r2 =
= r ⎡⎣ − cos ec 2 nu ⋅ n⎤⎦ + cot nu
du
d u2
r 2 + r12 = r 2 + r 2 cot 2 nu = r 2 (1 + cot 2 nu) = r 2 cosec 2 nu
⎡⎣ r 2 + r12 ⎤⎦
∴
3/ 2
3
= ⎡⎣ r 2 cosec 2 nu⎤⎦ 2 = r 3 cosec3 nu
r 2 + 2r12 − r r2 = r 2 + 2r 2 cot 2 nu − r ( − nr cosec 2 nu + r cot 2 nu)
and
= r 2 + r 2 cot 2 nu + nr 2 cosec 2 nu
= r 2 (1 + cot 2 nu) + nr 2 cosec 2 nu
= r 2 cosec2 nu + nr 2 cosec 2 nu = ( n + 1) r 2 cosec 2 nu
Substituting in (2), we get
rcosec nu
r 3 cosec3 nu
r=
=
2
2
n +1
( n + 1)r cosec nu
=
1
r
r a n a n r − n +1
⋅
=
⋅ =
n + 1 sin nu n + 1 r n
n +1
[Using (1)]
Note
1. When n = 1, the given curve is r = a sinu
which is a circle and the radius of curvature is
M04_ENGINEERING_MATHEMATICS-I _CH04_Part-A.indd 20
r=a
r −1+1 a
=
1+1 2
5/12/2016 10:10:29 AM
Applications of Differential Calculus
■
4.21
1
2. When n = , the given curve is
2
u
u a
⇒ r = a sin 2 = (1 − cos u)
2
2 2
which is a cardioid and the radius of curvature is
1
1
r 2 = a 2 sin
1
1
− +1
2
2
a2 r 2
ar
r=
= a1/ 2 r1/ 2 =
1
3
3
+1
2
1
3. When n = − , the given curve is
2
1
1
−
−
u
⎛ u⎞
r 2 = a 2 sin ⎜ − ⎟ = − a −1/ 2 sin
⎝ 2⎠
2
1
⇒
r
1/ 2
=−
u
1
sin
1/ 2
2
a
1 1 2u
= si n
2
r a
⇒
⇒
a 1 − cos u
=
r
2
⇒
2a
= 1 − cos u
r
which is a parabola and the radius of curvature is
−
1
⎛ 1⎞
− ⎜ − ⎟ +1
a 2 r ⎝ 2⎠
r=
1
− +1
2
3
= −2a
3
−1/ 2 2
r2
r = −2 1/ 2
a
3
r2
Since r is positive, r = 2 1/ 2
a
4. When n = 2, the given curve is
r2 = a2 sin2u
which is lemniscate of Bernoulli and radius of curvature is
r=
a 2 r −2 +1 a 2
=
2 +1
3r
5. When n = −2, the given curve is
r −2 = a −2 sin( −2u) = − a −2 sin 2u
1
sin 2u
=− 2
r2
a
⇒
which is a rectangular hyperbola and the radius of curvature is
r=
Since r is positive, r =
a −2 r − ( −2 ) +1
r3
= −a −2 r 3 = − 2
−2 + 1
a
r3
a2
M04_ENGINEERING_MATHEMATICS-I _CH04_Part-A.indd 21
5/12/2016 10:10:34 AM
4.22
■
Engineering Mathematics
4.1.7 Radius of Curvature at the Origin
Newton’s Method
1. If the curve y = f(x) passes through the origin and the x-axis is tangent at the origin O(0,0), then
x2
at the point O(0,0), the radius of curvature
r = lim
x→0 2 y
Proof
Let y = f(x) be the equation of the curve.
Since the x-axis is a tangent at the origin O(0, 0) to the curve, we have
dy
=0
⇒ y1 = 0
dx
3
The radius of curvature
r=
(1 + y12 )
1
=
y2
y2
∴ at the origin O(0, 0),
Now consider
r=
1
y2
2
(1)
x2
2x
= lim
x →0 2y
x →0 2y
1
lim
x2 1
=
x→0 2 y
y2
⇒
[By L’ Hopital’s rule]
lim
(2)
⎛x ⎞
∴ from (1) and (2), we get at the origin O(0, 0), r = lim ⎜ ⎟
x →0 ⎝ 2y ⎠
2
2. If the curve y = f(x) passes through the origin and the y-axis is tangent at the origin O(0, 0), then
y2
at O(0, 0), r = lim .
y→0 2 x
Proof
The curve is y = f(x)
Since the y-axis is tangent at the origin O(0, 0) to the curve
dy
dx
=∞ ⇒
= 0 ⇒ x1 = 0
dx
dy
3
But, the radius of curvature
(1 + x 12 ) 2
1
r=
=
x2
x2
∴ at the origin O(0, 0),
r=
1
x2
(1)
Now, consider
⎛y2⎞
2y
lim ⎜ ⎟ = lim
y → 0 ⎝ 2x ⎠
y →0
dx
2⋅
dy
= lim
y →0
M04_ENGINEERING_MATHEMATICS-I _CH04_Part-A.indd 22
⎡1⎤ 1
y
= lim ⎢ ⎥ =
x1 y →0 ⎣ x 2 ⎦ x 2
[ By L’Hopital’s rule ]
[ By L’Hopital’s rule ]
5/12/2016 10:10:38 AM
Applications of Differential Calculus
■
4.23
y2
.
y→0 2 x
From (1) and (2), we get, at the origin O(0, 0), r = lim
3. If the curve y = f(x) passes through the origin O(0, 0), but neither the x-axis nor the y-axis is a
3
⎡1 + p 2 ⎤⎦ 2
, where p = f ′(0), q = f ′′(0).
tangent at the origin, then at the origin O(0, 0), r = ⎣
q
Proof
In this case we use Maclaurin’s series expansion for f(x)
∴
y = f ( x ) = f ( 0) +
x
x2
f ′(0) + f ′′(0) + ⋅⋅⋅
1!
2!
Putting p = f ′(0), q = f ′′(0) , we get
y = px +
q 2
x + ⋅⋅⋅
2
[{ f(0) = 0]
3
Thus, at the origin O(0, 0),
⎡1 + p 2 ⎤⎦ 2
.
r= ⎣
q
Note
1. In this case we may not be able to find y1, y at the origin O(0, 0) in the usual way.
So, we substitute
q 2
x + ⋅⋅⋅ in the given equation and equate the like coefficients of p and q.
2
2. If the curve passes through the origin, then the equation of the tangent at the origin is obtained by
equating the lowest degree terms to zero.
y = px +
WORKED EXAMPLES
EXAMPLE 1
Find the radius of curvature at the origin for x 31 y 3 22 x 2 16 y 5 0.
Solution.
The given curve is
x3 + y3 − 2 x 2 + 6 y = 0
(1)
Since there is no constant term in the equation, it passes through the origin (0, 0).
The tangent at the origin O(0, 0) is obtained by equating the lower degree term to zero.
In (1) the lowest degree term is 6y
∴ y = 0 is the tangent at the origin. i.e., the x-axis the tangent at the origin.
∴ the radius of curvature
M04_ENGINEERING_MATHEMATICS-I _CH04_Part-A.indd 23
⎛ x2 ⎞
r = lim ⎜ ⎟ ⇒
x →0 ⎝ 2y ⎠
⎛ x2 ⎞
2r = lim ⎜ ⎟
x→0 ⎝ y ⎠
5/12/2016 10:10:43 AM
4.24
■
Engineering Mathematics
Divide the equation (1) by y.
∴
x3
x2
+ y2 − 2 + 6 = 0
y
y
⇒
⎛ x2 ⎞
⎛ x2 ⎞
x ⎜ ⎟ + y2 − 2 ⎜ ⎟ + 6 = 0
⎝ y⎠
⎝ y⎠
(2)
As x → 0, we have y → 0.
∴ from the equation (2), we get 0.2r + 0 − 2.2r + 6 = 0 ⇒
4r = 6 ⇒ r =
3
2
EXAMPLE 2
Find the radius of curvature at the origin for 2 x 4 1 3 y 4 1 4 x 2 y 1 xy 2 y 2 1 2 x 5 0.
Solution.
The given curve is
2 x 4 + 3 y 4 + 4 x 2 y + xy − y 2 + 2 x = 0
(1)
There is no constant term in the equation (1). Therefore it passes through the origin.
The tangent at the origin is obtained by equating the lowest degree term in the equation (1) to
zero.
The lowest degree term in equation (1) is 2x.
∴ x = 0, is the tangent at the origin. That is, the y-axis is the tangent at the origin.
⎛y2⎞
⎛ 2⎞
r = lim ⎜ ⎟ ⇒ 2r = lim y
y → 0 ⎝ 2x ⎠
y →0 ⎜
⎝ x ⎟⎠
∴ the radius of curvature
Dividing (1) by x, we get
2x 3 + 3 ⋅
⇒
y4
y2
+ 4 xy + y −
+2=0
x
x
⎛ y2 ⎞
⎛ y2 ⎞
2 x 3 + 3 y 2 ⋅ ⎜ ⎟ + 4 xy + y − ⎜ ⎟ + 2 = 0
⎝ x⎠
⎝ x⎠
As x → 0, y → 0.
∴from the equation( 2), we get
(2)
0 + 0.2r + 0 + −2r + 2 = 0 ⇒ 2r = 2 ⇒ r = 1
EXAMPLE 3
Find the radius of curvature at the origin for the curve y 2 ( a 2 x ) 5 x 2 ( a + x ) .
Solution.
The given curve is
y 2 (a − x) = x 2 (a + x)
(1)
It passes through the origin. The equation of the tangent at the origin is obtained by equating the
lowest degree terms to zero.
M04_ENGINEERING_MATHEMATICS-I _CH04_Part-A.indd 24
5/12/2016 10:10:49 AM
Applications of Differential Calculus
■
4.25
The lowest degree term in equation (1) is ay 2 − ax 2
∴the tangent at the origin is
ay 2 − ax 2 = 0 ⇒
y 2 = x2
⇒ y = ±x .
Which are neither parallel to the x-axis nor to the y-axis.
q
y = px + x 2 + … in equation (1), we get
2
∴ substituting
q 2
⎛
⎞
⎜⎝ px + x + …⎟⎠
2
2
(a − x ) = x 2 (a + x )
⇒
(p x
⇒
ap 2 x 2 + apqx 3 − p 2 x 3 + … = ax 2 + x 3
2
2
)
+ pqx 3 + … (a − x ) = ax 2 + x 3
Equating the coefficients of x2 and x3 on both sides, we get
ap 2 = a
⇒
p2 = 1
⇒
p = ±1 and apq − p 2 = 1
aq − 1 = 1
⇒
aq = 2
⇒
− aq − 1 = 1
⇒
aq = −2 ⇒
2
a
2
q=−
a
When p = 1,
When p = −1,
q=
(1 + p )
r=
3
2 2
At the origin (0, 0):
When p = 1, q =
2
,
a
q
r=
(1 + 1) 2 2a 2
=
=a 2
2
2
a
3
2
When p = −1, q = − ,
a
(1 + 1) 2
r=
= −a 2
2
−
a
Since r is positive, in both cases r = a 2
4.1.8 Pedal Equation or p – r Equation of a Curve
Let O be the pole and OA be the initial line. Let P be the point (r, u) on the curve r = f(u).
Let PT be the tangent at the point P.
Let OP = r and the angle between OP and PT be f. Let OM be the perpendicular drawn from the
pole to the tangent at the point P.
Let OM = p.
From Δ OPM,
OM
= sin (180 − f) ⇒ ΟΜ = OP sin f ⇒ p = r sin f
OP
M04_ENGINEERING_MATHEMATICS-I _CH04_Part-A.indd 25
5/12/2016 10:10:56 AM
4.26
⇒
■
Engineering Mathematics
1
1
= 2 2
2
p
r sin f
M
(
1
1
= 2 cosec 2f = 2 1 + cot 2 f
r
r
2
1 ⎡ ⎛ 1 dr ⎞ ⎤
= 2 ⎢1 + ⎜
⎥
⎟
r ⎢⎣ ⎝ r d u ⎠ ⎥⎦
∴
1
1
1 ⎛ dr ⎞
5 1 ⎜ ⎟
p2 r 2 r 4 ⎝ d u ⎠
0
)
18
−
φ
P
φ
p
r
ψ
O
2
TA
Fig. 4.8
which is called the p – r equation or pedal equation of a curve, r = f(u ).
WORKED EXAMPLES
EXAMPLE 1
Find the p – r equation of the cardioid r 5 a(12cosu).
Solution.
The given equation is
r = a(1 − cos u)
(1)
The pedal equation of the curve (1) is
1
1 1
= 2+ 4
2
p
r
r
⎛ dr ⎞
⎜⎝ ⎟⎠
du
2
Differentiating (1) with respect to u, we get
dr
= a [ −( − sin u) ] = a sin u
du
∴ the pedal equation is
r 2 + a 2 sin 2 u
1
1 1 2 2
=
=
+
a
sin
u
r4
p2 r2 r4
2
a 2 (1 − cos u ) + a 2 sin 2 u
=
r4
a 2 ⎡1 + cos 2 u − 2 cos u + sin 2 u⎤⎦
= ⎣
r4
a 2 ⎡1 + cos 2 u + sin 2 u − 2 cos u ⎤⎦ 2a 2
= ⎣
= 4 (1 − cos u)
r4
r
1
2a
2a
= 4 ⋅r = 3 ⇒
2
p
r
r
M04_ENGINEERING_MATHEMATICS-I _CH04_Part-A.indd 26
p2 =
r3
.
2a
5/12/2016 10:11:01 AM
Applications of Differential Calculus
■
4.27
EXAMPLE 2
Find the p – r equation of the curve r m 5 a m cos mu.
Solution.
The given equation is
The p – r equation is
r m = a m cos mu
(1)
1
1 1 ⎛ dr ⎞
= + ⎜ ⎟
p 2 r 2 r 4 ⎝ du ⎠
2
(2)
Differentiating (1) with respect to u, we get
m r m −1
dr
am
dr
= − ma m sin mu ⇒
= − m −1 sin mu
du
du
r
2
a2m
⎛ dr ⎞
a2 m
a2m
=
sin 2 mu = 2 m − 2 (1 − cos 2 m u) = 2 m − 2
⎜⎝ ⎟⎠
2m− 2
du
r
r
r
⇒
⎛ r 2m ⎞
⎜⎝1 − a 2 m ⎟⎠
Substituting in Equation(2), we get
1
1 1 ⎡ a2m
= 2 + 4 ⎢ 2m− 2
2
p
r
r ⎣r
1
a2m
=
p2 r 2m+ 2
⇒
⇒
⎛ r 2m ⎞ ⎤ 1
1
a2m
1
−
=
+
− 2
⎥
m
2
2
2m+ 2
⎜⎝ a ⎟⎠
r
r
⎦ r
p 2 a 2 m = r 2 m + 2 = r 2( m +1) ⇒ pa m = r m +1
Note
1. When m = 1, the curve is r = a cos u, which is a circle.
The pedal equation is pa = r 2
2. When m = 2, the curve is r 2 = a2 cos 2u, which is lemniscate of Bernoulli.
The (p – r) equation is pa 2 = r 3
1
3. When m = , the curve is
2
1
1
u
u a
⇒ r = a cos 2 = (1 + cos u)
r 2 = a 2 cos
2
2 2
Which is a cardioid and the p – r equation is
1
1
pa 2 = r 2
+1
3
= r2
⇒
p2 =
r3
a
1
1
1
−
−
1
u
⎛ u⎞
4. When m = − , the curve is r 2 = a 2 cos ⎜ − ⎟ = a 2 cos
⎝ 2⎠
2
2
⇒
1
r
Squaring,
1
2
=
1
1
2
cos
u
2
a
1 1
u
1 + cos u
= cos 2 =
2
2a
r a
M04_ENGINEERING_MATHEMATICS-I _CH04_Part-A.indd 27
⇒
2a
= 1 + cos u
r
5/12/2016 10:11:10 AM
4.28
■
Engineering Mathematics
Which is a parabola and the p – r equation is
pa
−
1
2
=r
1
− +1
2
1
= r2
⇒
p 2 a −1 = r
⇒
p 2 = ar
5. When m = −1, the curve is
r −1 = a −1 cos( −u) ⇒
1 1
= cos u ⇒
r a
a
= cos u, a straight line.
r
and the p – r equation is
pa −1 = r −1+1 = 1
⇒
p=a
When m = −2, the curve is
r −2 = a −2 cos( −2u)
1
1
1
= cos 2u = 2
⇒
⇒
r 2 a2
a sec 2u
which is a rectangular hyperbola and the p – r equation is
pa −2 = r −2 +1 = r −1
⇒
p 1
=
a2 r
⇒
r 2 = a2 sec 2u
pr = a 2
4.1.9 Radius of Curvature Using the p 2 r Equation of a Curve
Let r = f (u) be the equation of the curve.
We know that the radius of curvature at any point ( r , u) is
3/ 2
⎡ 2 ⎛ dr ⎞ 2 ⎤
3/ 2
⎢r + ⎜ ⎟ ⎥
⎝ d u ⎠ ⎥⎦
[r 2 + r12 ]
⎢⎣
= 2
r=
2
2
d 2 r r + 2r1 − rr2
⎛ dr ⎞
2
r + 2⎜ ⎟ − r 2
⎝ du ⎠
du
For some curves, it is not easy to find r using the above formula. In such cases, we find r using the
(p – r) equation of the curve.
Prove that the radius curvature r5r
dr
dp
Proof
Let O be the pole and OA be the intial line. Let P( r , u) be any point on the curve r = f (u)
Let r be the radius of curvature at P.
Let PT be the tangent at the point P and it meets OA at T.
Draw OM perpendicular to PT
Let OP = r and OM = p
∠AOP = u . Let ∠ OPT = f and ∠PTA = c
∴
c = u+f
M04_ENGINEERING_MATHEMATICS-I _CH04_Part-A.indd 28
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■
Applications of Differential Calculus
4.29
From ΔOPM,
OM
= sin f ⇒ p = r sin f
OP
We know
sin f = r
du
dr
and cosf =
ds
ds
P
du
1
= r⋅
dr
dr
du
We have
p = r sin f
Differentiating with respect to r, we get
∴
φ
r
tan u = r
ψ
θ
O
T
A
p
M
dp
df
= r cos f
+ sin f
Fig. 4.9
dr
dr
dr d f
du
=r
+r
ds dr
ds
df
du
d
dc
⎡ df d u ⎤
=r
+r
= r⎢
+ ⎥ = r (u + f) = r
ds
ds
ds
ds
⎣ ds ds ⎦
ds
dr
=r
dc
dP
⇒ r=r
⎡
ds ⎤
⎢{ r = d c ⎥
⎣
⎦
dr
dP
∴ the radius of curvature at the point P ( r , u) on the curve is r = r
dr
dP
WORKED EXAMPLES
EXAMPLE 3
Find the radius curvature at any point on the cardioid r 5 a( 1 2 cosu) using the p – r equation
of the curve.
Solution.
The given curve is
The p – r equation is
r = a(1 − cos u)
p2 =
r3
2a
(1)
[Refer worked example 1, page 4.26]
Let r be the radius curvature at any point P( r , u) on the curve,
∴
Now
r=r
p2 =
M04_ENGINEERING_MATHEMATICS-I _CH04_Part-A.indd 29
dr
dp
r3
2a
⇒
r 3 = 2ap 2
5/12/2016 10:11:30 AM
4.30
■
Engineering Mathematics
Differentiating w.r.to p, we get
3
∴
∴
dr
dr 4 ap 4 a r 2
2
3r 2
= 2a ⋅ 2 p ⇒ r
=
=
⋅
=
2ar
1
dp
dp
r
r
3
( 2a) 2
2
2ar
the radius of curvature r =
3
[using (1)]
EXAMPLE 4
Find the (p – r) equation of the curve x 2 1 y 2 5 ax and hence, deduce its radius of curvature.
Solution.
The given curvature is x 2 + y 2 = ax.
Put x = r cos u, y = r sin u.
∴ Its polar equation is
r 2 cos 2 u + r 2 sin 2 u = ar cos u
⇒
r 2 (cos 2 u + sin 2 u) = ar cos u ⇒
r 2 = ar cos u ⇒ r = a cos u
The (p – r) equation is
1
1 1 ⎛ dr ⎞
= + ⎜ ⎟
p 2 r 2 r 4 ⎝ du ⎠
r = a cos u
We have
∴
∴
2
dr
= −a sin u
du
1
1 1
1
= 2 + 4 a2 sin 2 u = 4 [r 2 + a2 sin 2 u]
2
p
r
r
r
1 2 2
a2 [cos 2 u + sin 2 u] a2
2
2
= 4
[
a
cos
u
+
a
sin
u
]
=
r4
r
r4
r 4 = a2 p 2 ⇒ r 2 = ap
=
⇒
The radius of curvature is
r=r
(1)
dr
dp
Differentiating (1) w.r to p, we get
2r
∴ the radius of curvature
dr
=a ⇒
dp
a
r = r⋅ =
2r
dr
a
=
dp 2r
a
2
Note that r = a cos u is a circle of diameter a. So, the radius =
a
a
. Hence, r = .
2
2
EXERCISE 4.2
1. Find the radius of curvature of the rectangular hyperbola r 2 = a 2 sec 2u.
2. Find the radius of curvature at the point ( r , u) on the curve r n = a n cos nu.
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Applications of Differential Calculus
■
4.31
3. Find the radius of curvature at a point on the cardioid r = a(1 − cos u).
a2
4. Show that the radius of curvature of r 2 = a2 cos 2u is .
3r
5. Find the radius of curvature at the origin for the following curves.
(1) y 4 + x 3 + a( x 2 + y 2 ) − a2 y = 0
( 2) a( y 2 − x 2 ) = x 3
(3) x 3 − x 2 y − 4 y 3 + 5x 2 − 2xy + 3y 2 − 8y = 0
( 4) y − x = x 2 + 2xy + y 2
6. Find the radius of curvature using p−r equation for the following curves.
(1) r = a sin u ( 2) r m = a m sin mu
(3) r = a e u cot a ( 4)
2a
= 1 − cos u
r
ANSWERS TO EXERCISE 4.2
r − n +1
n +1
a
6. (1) r 2 = ap, r =
2
1.
r3
a2
2. a n
3. 2 2ar
3
5. (1)
a
2
( 2) 2 3a
( 2) pa m = r m +1, r = a m
(3) 45
( 4)
1
2 2
− m +1
r
m +1
3
r
(3) p = r sin a, r =
sin a
4.2
( 4) p = ar , r =
2
2r 2
a
EVOLUTE
Definition 4.4 The locus of centre of curvature of a given curve Γ is called the evolute of the curve.
The given curve Γ is called an involute of the evolute. In fact, for the evolute there are many
involutes.
4.2.1
Properties of Evolute
The evolute has some interesting properties.
Property I
The normal at any point P to a given Curve is a tangent to the evolute at the centre of curvature of P.
Proof
Let P(x, y) be any point on the curve y = f(x). Let ( x, y) be the centre of curvature at P to the given
curve.
Centre of curvature is given by
x = ON = OM − NM = x − BP
Let r be the radius of curvature at the point P (x, y)
BP
= sin c
∴
r
⇒
∴
BP = r sin c
x = x − r sin c
M04_ENGINEERING_MATHEMATICS-I _CH04_Part-A.indd 31
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4.32
and
But
■
Engineering Mathematics
y = CN = BN + BC = y + BC
BC
= cos c
r
y
(x, y )
C
⇒ BC = r cos c
ψ ρ
y = y + r cos c
PC is normal to the curve at
−1
∴ slope of PC =
tan c
B
We now show that the slope of the evolute at the
point ( x, y) = slope of PC.
Since the evolute is the locus of centre of
curvature, the point ( x, y) is any point on the evolute.
∴ the slope of evolute
But
∴
dy
dy
=
= ds ,
dx dx
ds
ψ
O
M
x
where s is arc length.
dx dx ⎧
dc
dr ⎫
=
− ⎨r cos c
+ sin c
⎬
ds ds ⎩
ds
ds ⎭
and
dx
= cos c
ds
dx
dr
1
= cos c − r cos c. − sin c
ds
r
ds
dr
= cos c − cos c − sin c
ds
dx
dr
= − sin c
ds
ds
Now
d y dy
dy ⎤
dr
⎡
=
+ r ⎢ − cos c ⎥ + cos c
ds ds
ds ⎦
ds
⎣
⇒
N
Fig. 4.10
⇒
∴
P(x, y)
(1)
and
dy
= sin c
ds
dy
dr
1
= sin c − r sin c + cos c
ds
r
ds
dr
= sin c − sin c + cos c
ds
dy
dr
= cos c
ds
ds
(2)
dr
ds = − cot c = − 1 = slope of pC
∴
dr
tan c
− sin c
ds
∴ the normal at P is tangent to the evolute at its centre of curvature.
dy
=
dx
cos c
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■
Applications of Differential Calculus
4.33
Property 2
The length of an arc of the evolute is equal to the difference between the radii of curvature of the given
curve which are tangent to the arc at its extremities, provided that along the arc of the given curve, r
increases or decreases.
Proof
We know the differential arc length of any curve is
( ds)2 = ( dx )2 + ( dy )2
Let C1 C2 be an arc of the evolute, C1 C2 are the centres of curvature of P1 P2 respectively.
For the evolute if s′ is the arc length of evolute measured from the fixed point A and ( x, y) is anypoint on the evolute.
∴
( ds ′ )2 = ( dx )2 + ( d y )2
(3)
dx
dr
dy
dr
= − sin c
and
= cos c
ds
ds
ds
ds
But
[from property (1)]
Squaring and adding, we get
y
2
2
⎛ dx ⎞
⎛ dy ⎞
⎛ dr ⎞
⎜⎝ ⎟⎠ + ⎜⎝ ⎟⎠ = ⎜⎝ ⎟⎠
ds
ds
ds
But
2
2
(4)
2
⎛ ds ′ ⎞
⎛ dx ⎞ ⎛ d y ⎞
⎜⎝
⎟ = ⎜⎝ ⎟⎠ + ⎜⎝ ⎟⎠
ds ⎠
ds
ds
⎛ dr ⎞
=⎜ ⎟
⎝ ds ⎠
⇒
∴ integrating,
2
P2
Using (4)
⇒
C1
[Dividing (3) by ds]
2
ds ′
dr
=±
ds
ds
⇒
C2
ρ2
s1′
ρ1
A
P1
O
x
Fig. 4.11
ds ′ = ± d r
s ′ = ±r + c
If arc AC1 = s1′ and arc AC2 = s 2′ , then arc C1 C2 = s 2′ − s1′
If r1 and r2 are radii of curvatures at the points to P1, P2 on the given curve, then
s1′ = ± r1 + c and s 2′ = ± r2 + c
∴
∴
s2′ − s1′ = ± (r2 − r1 )
arc C1C 2 = r2 − r1
Hence, the result.
Note
For a given curve Γ there is only one evolute C. But there are many curves for which C is the
evolute. So, there are many involutes.
M04_ENGINEERING_MATHEMATICS-I _CH04_Part-A.indd 33
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4.34
■
Engineering Mathematics
Let P1 P2 P3 P4 …, be the given curve
C1 C2 C3 C4 … be its evolute, where Ci is the
centre of curvature at Pi
If the evolute is taken as the given curve
P1 P2 P3 P4 … is an evlute. P1 C1 P2 C2 … are
normal at P1, P2, …, respectively touching the
evolute at C1,C2 … respectively.
Now construct the curve P′1 P′2 P′3 … such
that P1P′1 = P2P′2 = P3P′3 = …
Then the curve P′1 P′2 P′3 … is an involute.
Similarly, we can construct many involutes
for the given curve C1 C2 C3 …
C4
C1 C2
P1
P 1′
P2
P3
P4
P ′2
P ′3
P ′4
4.2.2 Procedure to Find the Evolute
Let y = f(x) (1) be the equation of the given
curve. If ( x , y ) is the centre of curvature at
any point P (x, y) on (1), then
y 1 (1 + y 12 )
x =x−
( 2)
y2
C3
Fig. 4.12
(1 + y )
y =y+
2
1
and
y2
(3)
Eliminating x, y using (1), (2) and (3), we get a relation in x , y . Replacing x by x and y by y, we get
the equation of locus of ( x , y ), which is the evolute of the given curve.
The process of elimination of x and y would become simpler if the point (x, y) is taken in terms of
a parameter t.
WORKED EXAMPLES
EXAMPLE 1
Find the equation of the evolute of the parabola y 2 5 4 ax.
Solution.
The given curve is
y 2 = 4 ax.
(1)
2
Let P ( at , 2at ) be any point on the parabola.
dy
dy 2a
Differentiating w.r.to x,
2y
= 4a ⇒
=
dx
dx
y
2
2
d y
2a dy
4a
and
=− 2⋅
=− 3
dx 2
y dx
y
2
At the point (at , 2at),
∴
dy
2a 1
=
=
and
dx 2at t
y1 =
M04_ENGINEERING_MATHEMATICS-I _CH04_Part-A.indd 34
1
and
t
d 2y
4a2
1
=
−
=−
dx 2
2at 3
( 2at )3
y2 = −
1
2at 3
5/12/2016 10:11:59 AM
Applications of Differential Calculus
■
4.35
The centre of curvature ( x , y ) at P is given by
1⎛
1⎞
⎜⎝1 + 2 ⎟⎠
y 1 (1 + y 12 )
t
t = at 2 + 2a (1 + t 2 )
x =x−
= at 2 −
= 3at 2 + 2a
1
y2
−
2at 3
3at 2 = x − 2a
∴
and
1 + y12
y = y+
y2
⇒
y = −2at 3
t2 =
⇒
x − 2a
3a
x − 2a ⎞
⇒ t = ⎛⎜
⎝ 3a ⎟⎠
1/ 2
(2)
1
2
t
= 2at +
= 2at − 2at (1 + t 2 ) = −2at 3
1
−
2at 3
1+
(3)
Eliminating t from (2) and (3) we get,
⎛ x − 2a ⎞
y = −2a ⋅ ⎜
⎝ 3a ⎟⎠
Squaring both sides,
∴
( y )2 =
4a 2
( x − 2a)3
27a3
3/ 2
⇒
27a( y ) 2 = 4( x − 2a)3
the locus of ( x , y ) is 27ay 2 = 4( x − 2a)3 ,
which is the equation of the evolute of the parabola.
EXAMPLE 2
Find the equation of the evolute of the ellipse
x2 y2
1 5 1.
a 2 b2
Solution.
x2 y2
+
=1
a2 b2
The given curve is
(1)
Let P (a cos u, b sin u) be any point on the ellipse
Differentiating w.r.to x, we get
∴
At the point (a cos u, b sin u)
M04_ENGINEERING_MATHEMATICS-I _CH04_Part-B.indd 35
2
x 2 y dy
+
=0
a 2 b 2 dx
⇒
dy
b2 x
=− 2
dx
a y
dy ⎤
⎡
y ⋅1 − x ⋅ ⎥
d2 y
b2 ⎢
dx
=− 2⎢
⎥
dx 2
a ⎣
y2
⎦
b cos u
dy
b 2 a cos u
=−
=− 2⋅
a sin u
dx
a b sin u
5/12/2016 10:26:41 AM
4.36
■
Engineering Mathematics
⎡
⎛ −b cos u ⎞ ⎤
b sin u − a cos u ⎜
⎝ a sin u ⎟⎠ ⎥
d2 y
b2 ⎢
⎢
⎥
=
−
b 2 sin 2 u
dx 2
a2 ⎣
⎦
b 1
−b 2 [b sin 2 u + b cos 2 u]
=− 2
= 2
a
b 2 sin 3 u
a sin 3 u
b cos u
b
∴
y1 = −
and
y2 = − 2 3
a sin u
a sin u
The centre of curvature ( x , y ) at the point P is given by
and
2
2
⎛ − b cos u ⎞ ⎛ b cos u ⎞
1+ 2 2 ⎟
⎜
⎟
⎜
⎝ a sin u ⎠ ⎝
y (1 + y )
a sin u ⎠
x = x− 1
= a cos u −
b 1
y2
− 2
a sin 3 u
2
1
⎛ b 2 cos 2 u ⎞
= a cos u − a cos u sin 2 u ⎜1 + 2 2 ⎟
⎝
a sin u ⎠
= a cos u − a cos u sin 2 u −
= a cos u (1 − sin 2 u) −
= a cos3 u −
⇒
and
⇒
x=
b2
cos3 u
a
b2
cos3 u
a
b2
cos3 u
a
=
a2 − b2
cos3 u
a
a2 − b2
cos3 u
a
(1)
b 2 cos 2 u
1+ y
a 2 sin 2 u
y = y+
= b sin u +
b
y2
− 2 3
a sin u
a 2 sin 3 u ⎛ b 2 cos 2 u ⎞
= b sin u −
⎜⎝1 + 2
⎟
b
a sin 2 u ⎠
a2
= b sin u − sin 3 u − b sin u cos 2 u
b
a2
= b sin u (1 − cos 2 u) − sin 3 u
b
2
a
b 2 − a2
= b sin 3 u − sin 3 u =
sin 3 u
b
b
2
1
y=−
M04_ENGINEERING_MATHEMATICS-I _CH04_Part-B.indd 36
1+
a2 − b2 3
sin u
b
(2)
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Applications of Differential Calculus
■
4.37
Eliminate u from (1) and (2).
ax
= cos3 u
2
a − b2
From (1), we get
by ⎞
⎛
sin u = ⎜ − 2
⎝ a − b 2 ⎟⎠
Similarly, from (2)
⎡ ( ax ) ⎤
⎢⎣ a 2 − b 2 ⎥⎦
2/3
⎡ −by ⎤
+⎢ 2
⎣ a − b 2 ⎥⎦
1/ 3
2/3
=1
( ax ) 2/ 3
(by ) 2/ 3
+ 2
=1 ⇒
2
2 2/3
(a − b )
( a − b 2 ) 2/3
⇒
1/ 3
cos 2 u + sin 2 u = 1
We know that
⇒
⎛ ax ⎞
cos u = ⎜ 2
⎝ a − b 2 ⎟⎠
⇒
( ax ) 2/ 3 + (by ) 2/ 3 = ( a 2 − b 2 ) 2/ 3
∴ the locus of ( x , y ) is ( ax ) 2/ 3 + (by ) 2/ 3 = ( a 2 − b 2 ) 2/ 3 ,
which is the equation of the evolute of the ellipse.
EXAMPLE 3
Find the evolute of the rectangular hyperbola xy 5 c 2 .
Solution.
The given curve is
xy = c 2 .
c
Let P ⎛⎜ ct , ⎞⎟ be any point on (1)
⎝ t⎠
y=
(1) ⇒
c2
x
dy
c2
=− 2
dx
x
∴
At the point ⎛⎜ ct ,
⎝
(1)
c⎞
⎟,
t⎠
dy
c2
1
=− 2 2 =− 2
dx
ct
t
1
y1 = − 2
t
∴
and
d 2 y 2c 2
= 3
dx 2
x
and
d 2 y 2c 2
2
=
=
dx 2 c 3t 3 ct 3
and
y2 =
2
ct 3
The centre of curvature ( x , y ) at the point P is given by
y (1 + y 12 )
x =x− 1
= ct −
y2
⇒
x=
c
(3t 4 + 1)
2t 3
M04_ENGINEERING_MATHEMATICS-I _CH04_Part-B.indd 37
−
1⎛
1⎞
1+ 4 ⎟
2 ⎜
⎝
t
t ⎠ = ct + ct ⎛ 1 + 1 ⎞ = 3ct + c
⎜
⎟
2
2 2t 3
2 ⎝ t4 ⎠
3
ct
(2)
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4.38
■
Engineering Mathematics
1
1 + y 12 c 1 + t 4
y =y+
= +
2
y2
t
ct 3
and
=
⇒
y=
∴
x +y =
=
c c 4
3c ct 4
c
+ (t + 1) =
+
= (3 + t 4 )
t 2t
2t 2t
2t
c
(3t 2 + t 6 )
2t 3
c
[3t 4 + 1 + 3t 2 + t 6 ]
2t 3
c
c
[1 + 3t 2 + 3t 4 + t 6 ] = 3 (1 + t 2 )3
3
2t
2t
c ⎛ t 2 + 1⎞
x+y= ⎜
⎟
2⎝ t ⎠
⇒
1/ 3
⎛ c⎞
( x + y )1/ 3 = ⎜ ⎟
⎝ 2⎠
⇒
x −y =
Also
=
3
⎛ t 2 + 1⎞
⎜⎝
⎟
t ⎠
(4)
c
[3t 4 + 1 − 3t 2 − t 6 ]
2t 3
c
c
c ⎡1 − t 2 ⎤
[1 − 3t 2 + 3t 4 − t 6 ] = 3 (1 − t 2 )3 = ⎢
⎥
3
2⎣ t ⎦
2t
2t
⎛ c⎞
( x − y )1/ 3 = ⎜ ⎟
⎝ 2⎠
⇒
(3)
1/ 3
⎡1 − t 2 ⎤
⎢ t ⎥
⎣
⎦
3
(5)
Eliminating t from (4) and (5), we get the equation of the evolute.
Now,
⎛ c⎞
(x + y )2/ 3 − (x − y )2/ 3 = ⎜ ⎟
⎝ 2⎠
2/3
⎛ c⎞
=⎜ ⎟
⎝ 2⎠
2/3
⎛ c⎞
=⎜ ⎟
⎝ 2⎠
⇒
⎡⎛ 1 + t 2 ⎞ 2 ⎛ 1 − t 2 ⎞ 2 ⎤
⎢⎜
⎟ −⎜
⎟ ⎥
⎣⎝ t ⎠ ⎝ t ⎠ ⎦
⎡ (1 + t 2 ) 2 − (1 − t 2 ) 2 ⎤
⎢
⎥
t2
⎣
⎦
2/3
⋅
2
1
1−
4t 2 c 2 / 3
3
3
=
⋅
=
c
= ( 4c ) 2 / 3
4
4
t2
22 / 3
( x + y ) 2 / 3 − ( x − y ) 2 / 3 = ( 4c ) 2 / 3
∴ the locus of ( x , y ) is
( x + y ) 2 / 3 − ( x − y ) 2 / 3 = ( 4c ) 2 / 3 ,
which is the equation of the evolute.
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Applications of Differential Calculus
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4.39
EXAMPLE 4
Show that the evolute of the cycloid x 5 a( u 2 sin u), y 5 a(12 cos u) is another cycloid.
Solution.
Let P ‘u’ be any point on the cycloid.
Given the parametric equation of the cycloid.
x = a(u − sin u)
and
∴
∴
and
∴
y = a(1 − cos u)
dx
u
dy
u
u
= a(1 − cos u) = 2a sin 2 and
= a sin u = 2a sin cos
du
2
du
2
2
dy
u
u
u
2a sin cos
cos
dy d u
2
2
2 = cot u
=
=
=
u
u
dx dx
2
2a sin 2
sin
du
2
2
u
cosec 4
d2 y
d ⎛
u ⎞ du
u
1
1
2
=
= −cosec 2 ⋅ ⋅
=−
⎜ cot ⎟⎠
2 dx
2 2
4a
dx 2 d u ⎝
2 u
2a sin
2
4 u
cosec
u
2
and y2 = −
y1 = cot
2
4a
The centre of curvature ( x , y ) at u is given by
u⎞
⎛
1 + cot 2 ⎟
y 1 (1 + y 12 )
u ⎜⎝
2⎠
x =x−
= a(u − sin u) − cot
u
2
y2
− cosec 4
2
4a
u
u
4 a cos cosec 2
2⋅
2
= a(u − sin u) +
u
4 u
sin
cosec
2
2
u
u
= a(u − sin u) + 4 a cos ⋅ sin
2
2
= a(u − sin u) + 2a ⋅ sin u = au + a sin u
⇒
and
x = a(u + sin u)
(1)
u
1 + cot 2
1 + y 12
2
y =y+
= a(1 − cos u) +
y2
4 u
−cosec
2
4a
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4.40
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Engineering Mathematics
u
2
= a(1 − cos u) − 4a
4 u
cosec
2
u
u
u
2 u
= a(1 − cos u) − 4a sin
= a2 sin 2 − 4a sin 2 = −2a sin 2
2
2
2
2
y = − a(1 − cos u)
cosecc 2
⇒
(2)
Elimination of u from (1) and (2) is very difficult.
∴ the locus of ( x , y ) is given by the parametric equations
x = a(u + sin u) and y = − a(1 − cos u),
which is another cycloid.
EXAMPLE 5
t
Show that the evolute of the tractrix x 5 a ⎛⎜ cos t 1 log tan ⎞⎟ , y 5 a sin t is the catenary
⎝
2⎠
x
y 5 a cosh .
a
Solution.
Let t be any point on the given curve
t⎞
⎛
x = a ⎜ cos t + log tan ⎟
⎝
2⎠
∴
∴
dx
t 1⎤
1
⎡
= a ⎢ − sin t +
⋅ sec 2 ⋅ ⎥
t
dt
2 2
tan
⎢
⎥
⎣
2
⎦
t
⎤
⎡
cos
⎢
⎥
2
= a ⎢ − sin t +
⎥
t
t
⎢
2 sin cos 2 ⎥
⎣
2⎦
2
y = a sin t
and
dy
= a cos t
dt
⎡1 − sin 2 t ⎤ a cos 2 t
1
1 ⎤
⎡
⎤
⎡
= a ⎢ − sin t +
=
a
−
sin
t
+
=
a
⎢
⎥=
⎢⎣
t
t⎥
sin t
sin t ⎥⎦
⎣ sin t ⎦
2 sin cos ⎥
⎢
⎣
2
2⎦
dy
a cos t
sin t
dy dt
=
=
=
= tan t
2
cos t cos t
dx dx
a
dt
sin t
d 2y d
=
dx 2 dt
∴
and
sin t
sin t
⎛ dy ⎞ dt
= sec 2 t ⋅
=
⎜⎝ ⎟⎠
2
dx dx
a cos t a cos 4 t
y1 = tan t and y2 =
sin t
a cos 4 t
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Applications of Differential Calculus
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4.41
The centre of curvature ( x , y ) is given by
x =x−
y 1 (1 + y 12 )
t ⎞ tan t (1 + tan 2 t )
⎛
= a ⎜ cos t + loge tan ⎟ −
⎝
sin t
2⎠
y2
a coss 4 t
t⎤
sin t sec 2 t cos 4 t
⎡
⋅
= a ⎢cos t + loge tan ⎥ − a
⎣
2⎦
cos t
sin t
t⎤
⎡
= a ⎢ cos t + loge tan ⎥ − a cos t
⎣
2⎦
⇒
x = a log e tan
y =y+
and
t
⇒
2
x
t
= log e tan
a
2
1 + y 12
1 + tan 2 t
= a sin t +
sin t
y2
a cos 4 t
= a sin t +
a sec 2 t cos 4 t
sin t
a
(sin 2 t + cos 2 t )
=a
=
= a⋅
sin t
sin t
x
e a = tan
(1)
(1)
x
−
t
⇒ e a =
2
t
a⎛ 1
t⎞
2 ⇒ y=
+ tan
⎜
t
2⎟
2
t
⎜⎝ tan
⎟⎠
2 tan
2
2
1 + tan 2
(2)
1
tan
t
2
x
x
x
−
a ax
a x
(e + e a ) = (e a + e a ) = a cosh
a
2
2
x
∴ the locus of ( x , y ) is y = a cosh , which is a catenary.
a
(2) ⇒
y=
EXERCISE 4.3
1. Find the evolute of the curve x = a(cos u + u sin u), y = a(sin u − u cos u).
x2 y2
−
= 1.
a2 b2
3. Find the evolute of the parabola x 2 = 4 y .
4. Show that the evolute of the cycloid x = a(t + sin t ), y = a(1 − cos t ) is given by x = a(t − sin t ),
y − 2a = a(1 + cos t ).
2. Find the evolute of the hyperbola
ANSWERS TO EXERCISE 4.3
1. x + y = a
2
2
2
2. ( ax ) 2/ 3 − (by ) 2/ 3 = ( a 2 + b 2 ) 2/ 3
M04_ENGINEERING_MATHEMATICS-I _CH04_Part-B.indd 41
3. 27 x 2 = 4( y − 2)3
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4.42
4.3
■
Engineering Mathematics
ENVELOPE
1
(1) where m is a parameter. For different values
m
of m, we have different straight lines and so (1) represents a family of straight lines. Each member of
this family touches the curve y 2 = 4 x. So, these lines cover the curve y 2 = 4 x. This curve is called the
envelope of the family of lines. We shall now define envelope.
Consider the system of straight lines y = mx +
Definition 4.5 Let f ( x , y , a) = 0 be a single parameter family of curves, where a is the parameter.
The envelope of this family of curves is a curve which touches every member of the family.
4.3.1 Method of Finding Envelope of Single Parameter Family of Curves
1. Given the curves f ( x, y, a) = 0
(1)
∂
(2)
f ( x, y, a) = 0
∂a
Eliminate a from (1) and (2). The eliminant, if exists, is an equation in x and y. It is the envelope
of the family (1).
2. It is impossible to eliminate a from (1) and (2), then solve for x and y in terms of a. It will give
the parametric representation of the envelope.
3. If the equation of the family of curves (1) can be written in the form of a quadratic in the parameter
a as
A a2 + B a + C = 0
(3)
where A, B, C are functions of x and y,
Then the envelope is B2 − 4AC = 0
−B
For, differentiating (3) w.r.to a,we get 2A a + B = 0 ⇒ a =
2
A
Substituting in (3), we get
Find
A⋅
B2
B2
B2
−
+
C
=
0
⇒
−
+C= 0
4A
4A 2 2A
⇒ B2 − 4AC = 0
⇒
B2 − 4AC = 0
which is the equation of the envelope.
Note
(1) A point P (a, b) is a singular point of a curve f ( x, y, a) = 0 (a fixed)
∂f
∂f
and
if it satisfies (1)
=0
( 2)
=0
∂x
∂y
(1)
(3)
P is said to be an ordinary point if atleast one of (2) and (3) is not satisfied.
(2) The characteristic points of the family of curves f ( x, y, a) = 0 are those ordinary points of the
family where the equations
∂f ( x , y , a )
f ( x, y, a) = 0,
= 0 simultaneously hold.
∂a
Characteristic points are isolated on each curve.
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Applications of Differential Calculus
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4.43
Infact, the envelope of a family of curves f ( x, y, a) = 0, a is a parameter, is the locus of their
isolated characteristic points.
(3) Not every single parameter family has envelope. For example, the family of concentric circles
x 2 + y 2 = a 2 has no envelope, as there is no characteristic point.
WORKED EXAMPLES
EXAMPLE 1
Find the envelope of the family lines y = mx ± a 2 m2 − b 2, where m is the parameter.
Solution.
Given family is
y = mx ± a 2 m2 − b 2 , m is the parameter
⇒
y − mx = ± a 2 m 2 − b 2
⇒
( y − mx ) 2 = a 2 m 2 − b 2
⇒
y 2 − 2mxy + m 2 x 2 = a 2 m 2 − b 2
m 2 ( x 2 − a 2 ) − 2mxy + ( y 2 + b 2 ) = 0
⇒
This is quadratic in m
Here A = x 2 − a 2 ,
B = −2 xy,
C = y 2 + b2
∴
the envelope is B2 − 4 AC = 0
⇒
4 x 2 y 2 − 4( x 2 − a 2 )( y 2 + b 2 ) = 0
⇒
x 2 y 2 − ( x 2 y 2 + b2 x 2 − a2 y 2 − a2 b2 ) = 0
⇒
b2 x 2 − a2 y 2 − a2b2 = 0
[÷ by 4]
2 2
2 2
2 2
⇒ b x −a y = a b
which is a hyperbola.
2
2
⇒ x 2 − y2 = 1
a
b
EXAMPLE 2
Find the envelope of the straight lines represented by x cos a 1 y sin a 5 a sec a, where a is the
parameter.
Solution.
Given family is x cos a + y sin a = a sec a , where a is the parameter
Dividing by cos a, x + y tan a = a sec 2 a = a(1 + tan 2 a)
⇒
a tan 2 a − y tan a + ( a − x ) = 0 ,
which is a quadratic in tan a
Here A = a, B = − y, C = a − x
∴ the envelope is B2 − 4 AC = 0
⇒
y 2 − 4 a( a − x ) = 0 ⇒
y 2 = 4 a( a − x )
which is a parabola.
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Engineering Mathematics
EXAMPLE 3
Find the envelope of the lines x cos 3 a 1 y sin3 a 5 a , where a is a parameter.
Solution.
Given x cos3 a + y sin 3 a = a, where a is the parameter.
Differentiating w.r.to a, we get
(1)
x ⋅ 3 cos 2 a( − sin a) + y ⋅ 3 sin 2 a ⋅ cos a = 0
x cos 2 a sin a = y sin 2 a cos a
⇒
⇒
tana =
∴
x
sin a =
Substituting in (1), x ⋅
x +y
2
,
2
cos a =
x2 + y2
x
y
x
α
y
x +y
2
2
y3
x3
+ y⋅ 2
=a
2 3/ 2
(x + y )
( x + y 2 ) 3/ 2
y
Fig. 4.13
2
xy ( x 2 + y 2 )
xy
=a ⇒
=a
( x 2 + y 2 )3 / 2
( x 2 + y 2 )1/ 2
⇒
xy = a x 2 + y 2 ⇒ x 2 y 2 = a 2 ( x 2 + y 2 ),
⇒
which is the envelope.
EXAMPLE 4
Find the envelope of the family of straight lines y cos a 2 x sin a 5 a cos 2a , a being the
parameter.
Solution
Given
y cos a − x sin a = a cos 2a
where a is a parameter
Differentiating partially w.r.to a, we get
− y sin a − x cos a = −2a sin 2a
⇒
y sin a + x cos a = 2a sin 2a
(1)
(2)
(1)sin a − ( 2) cos a ,
⇒
− x sin 2 a − x cos 2 a = a cos 2a sin a − 2a sin 2a ⋅ cos a
⇒
− x(sin 2 a + cos 2 a) = a(sin a cos 2a − 2 sin 2a cos a)
⇒
x = a[2 sin 2a cos a − sin a cos 2a
= a ⎡⎣ 4 sin a cos 2 a − sin a (cos 2 a − sin 2 a) ⎤⎦
⇒
x = a ⎡⎣3 sin a cos 2 a + sin 3 a ⎤⎦
M04_ENGINEERING_MATHEMATICS-I _CH04_Part-B.indd 44
(3)
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Applications of Differential Calculus
■
4.45
(1) cos a + ( 2)sin a
y(cos 2 a + sin 2 a) = a [cos 2a cos a + 2 sin 2a sin a]
⇒
y = a ⎡⎣(cos 2 a − sin 2 a) cos a + 4 sin 2 a cos a ⎤⎦
⇒
= a[cos3 a + 3 sin 2 a cos a]
(4)
x − y = a[sin 3 a + 3 sin a cos 2 a − cos3 a − 3 sin 2 a cos a]
∴
= a[sin a − cos a]3
⎛ x − y⎞
⎜⎝
⎟
a ⎠
⇒
1/ 3
= sin a − cos a
(5)
x + y = a ⎡⎣sin 3 a + 3 sin a cos 2 a + 3 sin 2 a cos a + sin 3 a ⎤⎦
= a(sin a + cos a)3
and
⎛ x + y⎞
⎜⎝
⎟
a ⎠
∴
1/ 3
= sin a + cos a
(6)
Squaring and adding (5) and (6), we get
( x − y)2/ 3 ( x + y)2/ 3
+
= (sin a − cos a) 2 + (sin a + cos a) 2
2/3
2/3
a
a
= sin 2 a + cos 2 a − 2 sin a cos a + sin 2 a cos 2 a + 2 sin a cos a
= 2(cos 2 a + sin 2 a) = 2
⇒
( x − y)) 2 / 3 + ( x + y ) 2 / 3 = 2a 2 / 3
which is the envelope.
4.3.2 Envelope of Two Parameter Family of Curves
1. Let
f ( x , y , a, b) = 0
be a 2-parameter family of curves, where a, b are the parameters such that
f (a, b) = 0
(1)
(2)
Find b in terms of a from (2) and substitute in (1) and thus the problem is reduced to one
parameter family and proceed as above.
2. The following method is more convenient in many cases. For a fixed point (x, y) of the envelope
db
treating b as a function of a differentiate (1) and (2) w.r.to a and eliminate
from these equations.
da
Using this relation with (1) and (2), eliminate a and b. The eliminant, if exists, gives the envelope.
WORKED EXAMPLES
EXAMPLE 1
Find the envelope of the famly of straight lines
c is a constant.
M04_ENGINEERING_MATHEMATICS-I _CH04_Part-B.indd 45
x
y
2
1 51, where ab 5 c and a , b are parameters,
a b
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4.46
■
Engineering Mathematics
Solution.
x y
+ =1
a b
Substituting in (1), we get
(1)
Given
ab = c2
and
x y
+
=1
a c2
a
⇒
x 1
+ ay = 1
a c2
⇒
b=
c2
a
(2)
2
2
2
⇒ c x + a y = ac
ya 2 − c 2 a + c 2 x = 0
⇒
This is a quadratic in the parameter a.
Here A = y, B = − c 2 , C = c 2 x
c4 − 4 y ⋅ c2 x = 0
∴ the envelope is B2 − 4AC = 0 ⇒
which is a rectangular hyperbola.
⇒
4 xy = c 2
EXAMPLE 2
x y
1 51 where the parameters a and b are
a b
related by the equation a n 1 b n 5 c n , c being a constant.
Find the envelope of the family of straight lines
Solution.
x y
+ =1
a b
Given
(1)
an + b n = c n
and
(2)
Differentiating (1) and (2) with respect to a, treating b as a function of a, we get
−
x
y db
−
=0 ⇒
a 2 b 2 da
na n −1 + nb n −1 ⋅
and
db
=0
da
⇒
b2 x
db
=− 2
da
a y
(3)
db
a n −1
= − n −1
da
b
(4)
2
n −1
From (3) and (4) we get, b x = a
⇒ x = y
2
n −1
a y b
a n+1 b n+1
x
y
x y
+
a = b = a b = 1
an bn an + bn cn
∴
⎡ using (1) and ( 2)
⎤
⎢
⎥
⎢and since each ratio = sum of Nrs ⎥
sum of Drs ⎥⎦
⎢⎣
x
y
1
=
=
a n +1 b n +1 c n
∴
Now
⇒
x
y
a = b
an
bn
x
a n+1
=
1
cn
1
⇒
a = (c n x ) n +1
M04_ENGINEERING_MATHEMATICS-I _CH04_Part-B.indd 46
and
y
b
n +1
=
1
cn
⇒
1
b = (c n y ) n +1
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Applications of Differential Calculus
■
4.47
Substitute in (1), then we get
c
n / n +1
x
y
+
=1 ⇒
⋅ x1 / n + 1 c n / n + 1 ⋅ y 1 / n + 1
n
n
n
x n +1 + y n +1 = c n +1
which is the required envelope.
Note
Some of the important particular cases are
x y
1. For n = 1, we get + = 1, a + b = c. Then the envelope is x + y = c
a b
x y
2. For n = 2, we get + = 1, a 2 + b 2 = c 2 . Then the envelope is x 2/ 3 + y 2/ 3 = c 2/ 3 , which is the
a b
astroid.
x y
3. For n = 3, we get + = 1, a3 + b3 = c 3 . Then the envelope is x 3/ 4 + y 3/ 4 = c 3/ 4
a b
x y
4. For n = 1 and c = 1, we get + = 1, a + b = 1. Then the envelope is x + y = 1
a b
EXAMPLE 3
Prove that the envelope of the system of lines
by
l m
1 5 1 is the curve
a b
x
y
1
51.
a
b
x
y
1 5 1, where the parameters l and m are connected
l
m
Solution.
x y
(1)
and
+ =1
l m
Treating m as a function of l, differentiate (1) and (2) w.r.to l.
Given
∴
−
and
l m
+ =1
a b
(2)
x
y dm
− 2⋅
=0
2
l
m dl
⇒
dm
m2 x
=− 2
dl
l y
(3)
1 1 dm
+ ⋅
=0
a b dl
⇒
dm
b
=−
dl
a
(4)
From (3) and (4) we get,
m 2x
b
− 2 =−
⇒
a
l y
x
y
x y
+
1
l = m = l m
=
bl
am bl + am bl + am
∴
Now
x
y
=
⇒
2
bl
am 2
l m
+ =1
a b
M04_ENGINEERING_MATHEMATICS-I _CH04_Part-B.indd 47
⇒
y
x
l = m
bm
bl
[using (1)]
bl + am = ab.
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4.48
■
Engineering Mathematics
x
y
1
=
=
2
2
bl
am
ab
∴
x
1
=
2
ab
bl
⇒
⇒ l 2 = ax
⇒
l = ax
m = by
Similarly,
by
ax
+
=1
a
b
which is the required envelope.
Substituting in (2),
⇒
x
y
+
=1
a
b
EXAMPLE 4
Find the envelope of a system of concentric and coaxial ellipses of constant area.
Solution.
Let the common axes of the system of ellipses by the coordinate axes and the common centre be the
origin
x2 y2
(1)
∴ the equation of the family of ellipses is 2 + 2 = 1
a
b
where a and b are the parameters.
Given the area of the ellipse is constant. Let it be A.
But we know that the area of the ellipse is p ab.
∴
A = pab ⇒ b =
A
pa
x 2 p2 a2 2
x2
y2
⇒
+ 2 y =1
+
=1
2
2
a2
A
A
a
2 2
pa
which is the given family of ellipses with parameter a.
∴ differentiating (2) w.r.to a, we get
p2 2
x2
p2
ay = a −3 x 2 = 3
−2a −3 x 2 + 2 2 ay 2 = 0 ⇒
2
A
a
A
∴ (1) becomes
A2 x 2
p2
⇒
a4 y 2 =
∴ (2) becomes
p2 A x 2
x2
+ 2⋅
⋅ y =1
Ax A py
py
⇒
⇒ a4 =
p
p
p
xy + xy = 1 ⇒ 2 xy = 1
A
A
A
where c 2 =
A
2p
A2 x 2
p2 y 2
⇒
⇒ a2 =
xy =
A
2p
(2)
Ax
py
⇒
xy = c 2
∴ envelope is a rectangular hyperbola.
4.3.3 Evolute as the Envelope of Normals
A property of evolute is that the normal at a point P on the given curve G is a tangent to the evolute
with the centre of curvature as point of contact. Hence, the envelope of normals to G is the same as the
locus of the centre of curvature and hence, it is the evolute.
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Applications of Differential Calculus
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4.49
Procedure to find the evolute as the envelope of normals
1. Take any point on the given curve in terms of a parameter, if possible.
2. Find the equation of the normal at that point.
3. Find the envelope of the normal.
4. This envelope is the evolute of the given curve.
WORKED EXAMPLES
EXAMPLE 1
Find the evolute of
x2 y2
1 51 as envelope of normals.
a 2 b2
Solution.
x2 y2
+
=1
a2 b2
First we shall find the equation of the normal at P( a cos u, b sin u) .
Let P ( a cos u, b sin u) be any point on
2 x 2 y dy
+
=0
a 2 b 2 dx
Differentiating w.r.to x,
At the point ‘P’,
∴
∴
⇒
dy
b2 x
=− 2
dx
a y
b
dy
b 2 a cos u
= − cot u = slope of the tangent at P.
=− 2⋅
dx
a b sin u
a
slope of the normal at P = −
the equation of the normal is
y − b sin u =
1
a
a sin u
= tan u =
b
b
b cos u
− cot u
a
a sin u
( x − a cos u)
b cos u
⇒
b cos u ⋅ y − b 2 sin u cos u = a sin u ⋅ x − a 2 sin u cos u
⇒
a sin u ⋅ x − b cos uy = ( a 2 − b 2 )sin u cos u
⇒
ax
by
−
= (a2 − b2 )
cos u sin u
(1)
where u is the parameter.
Differentiating (1) w.r.to u partially, we get
ax
by
⇒
−
( − sin u) + 2 cos u = 0
2
cos u
sin u
sin u
by
ax
= − 2 cos u
⇒
2
cos u
sin u
⇒
ax
cos3 u
M04_ENGINEERING_MATHEMATICS-I _CH04_Part-B.indd 49
=−
by
sin 3 u
⇒
ax
by
−
cos u = sin u
cos 2 u sin 2 u
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Engineering Mathematics
ax
⎛ by ⎞
+ −
cos u ⎜⎝ sin u ⎟⎠
ax
by
−
= cos u sin u = a 2 − b 2
=
1
cos 2 u + sin 2 u
∴ each ratio
ax
∴
cos3 u
−
and
= a −b
2
by
3
sin u
⎛ ax ⎞
⇒ cos u = ⎜ 2
⎝ a − b 2 ⎟⎠
ax
3
= a2 − b2
a2 − b2
⇒ sin3 u = −
by ⎞
⎛
⇒ sin u ⎜ − 2
⎝ a − b 2 ⎟⎠
by
a −b
2
1
2
3
1
3
cos2u + sin2u = 1
we know that
⎛ ax ⎞
⎜⎝ 2
⎟
a − b2 ⎠
∴
⇒ cos u =
2
[Using (1)
2
( ax )
⇒
2
2
by ⎞
⎛
+⎜− 2
⎝ a − b 2 ⎟⎠
2
3
(a − b )
2
3
2
(by )
+
3
3
3
(a − b )
2
2
2
2
=1
= 1 ⇒ ( ax )
2
3
+ (by )
2
3
= (a2 − b2 )
2
3
3
which is the evolute of the given curve.
EXAMPLE 2
Find the evolute of the parabola y 2 54 x as the envelope of normals.
Solution.
Let P (t 2 , 2t ) be any point on the parabola y 2 = 4 x . First we shall find the equation of the normal at P.
Differentiating w.r.to x, we get
2y
At the point P,
dy
=4 ⇒
dx
dy 2
=
dx y
dy 2 1
= = = slope of the tangent at P
dx 2t t
∴
slope of the normal at P = −t
∴
the equation of the normal at P is
y − 2t = −t ( x − t 2 )
⇒
y + tx = 2t + t 3
(1)
where t is the parameter.
Differentiating (1) w.r.to t, we get
x = 2 + 3t 2
We have
⇒ t2 =
x−2
⎛ x − 2⎞
⇒ t=⎜
⎝ 3 ⎟⎠
3
1/ 2
(2)
y + tx = 2t + t 3
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Applications of Differential Calculus
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4.51
y = t [ − x + 2 + t 2 ] = t [t 2 − ( x − 2)]
⇒
⎛ x − 2⎞
=⎜
⎝ 3 ⎟⎠
⇒
⎛ x − 2⎞
y =⎜
⎝ 3 ⎟⎠
Squaring,
⎛ x − 2⎞ ⎡ 4
⎤
y2 = ⎜
( x − 2) 2 ⎥
⎝ 3 ⎟⎠ ⎢⎣ 9
⎦
1/ 2
⎡x − 2
⎤
⎢⎣ 3 − ( x − 2) ⎥⎦
[Using (2)]
⎡ 2
⎤
⎢⎣ − 3 ( x − 2) ⎥⎦
4
( x − 2) 3
27
which is the evolute of the parabola.
⇒
1/ 2
y2 =
⇒ 27 y 2 = 4( x − 2)3 ,
EXAMPLE 3
Find the evolute of the hyperbola
x2 y2
− 51 as the envelope of its normals.
a 2 b2
Solution.
x2 y2
−
=1
a2 b2
We have x = a sec u, y = b tan u, as parametric equation of the hyperbola.
We shall find the equation of the normal at the point “u”
dx
dy
= a sec u tan u and
= b sec 2 u.
du
du
The equation of the hyperbola is
∴
dy
dy d u
b sec 2 u
=
=
dx dx a sec u tan u
du
b sec u b 1 cos u
b
=
=
=
= slope of the tangent.
a tan u a cos u sin u a sin u
∴ the slope of the normal = −
a sin u
b
∴ the equation of the normal at the point ‘u’ is
a sin u
( x − a sec u)
b
by − b 2 tan u = − a sin ux + a 2 tan u
y − b tan u = −
⇒
⇒
⇒
a sin u ⋅ x + by = ( a 2 + b 2 ) tan u
ax cos u + by cot u = ( a 2 + b 2 )
(1)
where u is the parameter.
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Differentiating (1) partially w.r.to u, we get
− ax sin u − by cosec 2u = 0
⇒
− ax sin u = by
1
sin 2 u
θ
by
sin u = −
ax
3
⇒
⇒
∴
⎛ by ⎞
sin u = ⎜ −
⎝ ax ⎟⎠
cos u =
13
=−
Fig. 4.14
(by )1/ 3
( ax )1/ 3
( ax ) 2/ 3 − (by ) 2/ 3
( ax )1/ 3
Substituting in (1) we get
ax ⋅
( ax ) 2/ 3 − (by ) 2/ 3 by ⋅ ( ax ) 2/ 3 − (by ) 2/ 3
+
= a2 + b2
( ax )1/ 3
−(by )1/ 3
⇒
( ax ) 2/ 3 ( ax ) 2/ 3 − (by ) 2/ 3 − (by ) 2/ 3 ( ax ) 2/ 3 − (by ) 2/ 3 = a 2 + b 2
⇒
⎡⎣( ax ) 2/ 3 − (by ) 2/ 3 ⎤⎦ ( ax ) 2/ 3 − (by ) 2/ 3 = a 2 + b 2
⎡⎣( ax ) 2/ 3 − (by ) 2/ 3 ⎤⎦
⇒
3/ 2
= a2 + b2
( ax ) 2/ 3 − (by ) 2/ 3 = ( a 2 + b 2 ) 2/ 3 ,
⇒
which is the evolute of the hyperbola.
EXERCISE 4.4
x
+ yt = 2c, t being a parameter.
t
x
y
2. Find the envelope of the family of lines cos u + sin u = 1, u being the parameter.
a
b
1. Find the envelope of the family of lines
3. Find the envelope of the family of lines y = mx − 2am − am 3, where m is the parameter.
4. Find the envelope of x cos u + y sin u = r , where u is the parameter.
5. Find the envelope of ( x − a) 2 + y 2 = 4a, a is the parameter.
6. Find the envelope of y = mx + am 2 , m being the parameter.
7. Find the envelope of the family of ellipses
M04_ENGINEERING_MATHEMATICS-I _CH04_Part-B.indd 52
x2 y2
+
= 1 for which a + b = c, c is a constant.
a2 b2
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Applications of Differential Calculus
8. Find the envelope of the family of parabolas y = x tan a −
■
4.53
gx 2
, a being the parameter.
2u 2 cos 2 a
9. Find the envelope of ( x − a) 2 + ( y − a) 2 = 2a.
10. Find the envelope of the family of curves
a2
b2
cos u − sin u = c for different values of u.
x
y
11. Find the envelope of the family of straight lines y = mx ± a 1 + m 2 where m is the parameter.
12. Find the envelope of the family of straight lines y = mx +
13. Find the envelope of
1
.
m
x cos a y sin a
+
= 1, a is the parameter.
a
b
14. Find the envelope of the family of curves x 2 ( x − a) + ( x + a)( y − m ) 2 = 0, where m is a parameter
and a is a constant.
x2 y2
15. Find the envelope of 2 + 2 = 1, where a n + b n = c n , a and b are the parameters and c is a
a
b
constant.
x y
m n
m+ n
+ = 1, where a, b are the parameters and are related by a b = c .
a b
x y
17. Find the envelope of the family of lines + = 1 subject to the condition a + b = 1.
a b
16. Find the envelope of
18. Find the evolute of the parabola x2 = 4ay, treating it as the envelope of normals.
ANSWERS TO EXERCISE 4.4
x2 y2
+
=1
a2 b2
1. xy = c 2
2.
3. 27ay 2 = 4( x − 2a)3
4. x 2 + y 2 = r 2
5. y 2 = 4(1 + x )
6. x 2 + 4 ay = 0
7. x 2/ 3 + y 2/ 3 = c 2/ 3
2
2
8. y = u − gx
2 g 2u 2
9. ( x + y + 1) 2 = 2( x 2 + y 2 )
10. a 4 y 2 + b 4 x 2 = c 2
11. x 2 + y 2 = a 2
12. y 2 = 4 x
x2 y 2
+
=1
a2 b 2
14. x = 0, x = a
15. x n + 2 + y n + 2 = c n + 2
13.
16. x m y n =
cm +n ⋅ m m ⋅ nn
(m + n)m + n
M04_ENGINEERING_MATHEMATICS-I _CH04_Part-B.indd 53
17.
x + y =1
2n
2n
2n
18. 27ax2 = 4 (y − 2a)3
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SHORT ANSWER QUESTIONS
1. Find the curvature of x 1 y 24 x 2 6 y 1 10 5 0 at any point on it.
2. Define curvature of a plane curve and what is the curvature of a straight line?
2
2
3. What is the radius of curvature at (4, 3) on the curve x 2 1 y 2 5 25 ?
4. Find the curvature of y 2 54 x at its vertex.
5. Find the curvature of the curve y 5 c logsec
x
at any point.
c
p
on y 5 4 sin x.
2
7. True or false. When the tangent at a point on a curve is parallel to the x-axis, then the curvature at that
point is same as the second derivative at that point.
6. Find the radius of curvature at x 5
8. Find the radius of curvature of the curve given by x 5 3 12 cos u, y 5 4 1 2 sin u.
9. Find the radius of curvature of the curve x 5 acosu, y 5 bsinu at any point ‘u’.
10. Find the centre of curvature of y 5 x 2 at the origin.
11. Define the circle of curvature at a point P(x1, y1) on the curve y 5 f(x).
12. Write down the equation of the circle of curvature of a curve at a given point.
c
⎛c
⎞
cos 3 t , sin3 t ⎟ , find the evolute of the curve.
⎝a
⎠
a
13. If the centre of curvature of curve is ⎜
14. Show that the family of straight lines 2 y 2 4 x 1 l 5 0, has no envelope, where l is a parameter.
15. Find the envelope of the family of straight lines y 5 mx 6 m2 21 where m is a parameter.
16. Find the envelope of x 21 y 2 2ax cos u2by sin u50, where u is the parameter.
17. Find the envelope of the family of circles ( x 2 a) 21 y 2 54 a , where a is the parameter.
a
18. Find the envelope of the curve y 5 mx 1 where m is a parameter.
m
gx 2
, a being the parameter.
19. Find the envelope y 5 x tan a 2 2
2 u cos 2 a
20. Find the envelope of the family of lines
x
1 yt 5 2 c , where t is the parameter.
t
OBJECTIVE TYPE QUESTIONS
A.
1.
2.
3.
4.
Fill up the blanks
The curvature of the curve 2x2 + 2y2 + 5x − 2y + 1 = 0 at any point is ____________.
The radius of curvature at (4, 3) on x2 + y2 = 25 is __________.
For the curve x2 = 2c (y − c), the radius of curvature at (0, c) is ___________.
The radius of curvature of the curve y = ex at the point (0, 1) is __________.
5. The radius of curvature at (0, p ) on r = a cosu is ____________.
2
6. The curvature at any point on 3x + 4y = 8 is ___________.
7. The center of curvature at any point of 2x2 + 2y2 + 5x − 2y + 1 = 0 is ___________.
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Applications of Differential Calculus
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4.55
8. For the p-r equation of a curve, the radius of curvature is given by ___________.
9. The evolute of the curve x2 + y2 + 8x − 6y + 7 = 0 is ___________.
10. The envelope of the family of lines y = mx + am2, where m is the parameter is __________.
B. Choose the correct answer
1. For a curve y = f(x) if
(a)
(1 + y 12 )3 2
y2
dy
= ∞ at a point, then the radius of curvature at the point is given by the formula
dx
(b)
(1 + x 12 )3 2
x2
2. The curvature at the point (c, c) on xy = c2 is
1
(a)
2c
(b)
c
(c)
y2
(1 + y 12 )
1
(c)
2c
x2
(1 + x 12 )
(d)
(d) none of these
3. The radius of curvature at the point (1, 1) on x4 + y4 = 2 is
(a)
1
(b)
2
3
2
3
(c)
2
x
4. Find the value of p at (0, c) on y = c cosh
c
(a) c
(b) 2c
5. The curvature at any point on x2 + y2 = 9 is
(c)
1
c
1
(c) 2
3
6. The radius of curvature at the origin on the curve y2 (a − x) = x2 (a + x) is
(a) 3
(b)
(a) a
(b) 2a
7. Center of curvature of y = x2 at (0, 0) is
1
(a) ⎛⎜ ,
⎝2
1⎞
⎟
2⎠
(b) ⎛⎜ 0,
⎝
1⎞
⎟
2⎠
3
(b) 2
1
2c
(d)
(d) none of these
(c) a 2
(d) none of these
1
(c) ⎛⎜ , 0⎞⎟
⎝2 ⎠
(d) none of these
8. If a plane curve has a constant radius of curvature at any point, then the curve is
(a) a straight line
(b) a parabola
(c) a circle
(d)
9. The radius of curvature of x3 − y3 − 2x2 + 6y = 0 at (0, 0) is
3
1
(a) 1
(b)
(c)
(d)
2
2
10. The locus of centre of curvature is called
(a) Envelope
(b) Evolute
(c) circle of curvature
(d)
11. The centre of curvature at the point (2, 2) on the curve xy = a2 is
(a) (4, 1)
(b) (4, 4)
(c) (2, 2)
(d)
12. The value of ‘a’ for which the radius of curvature of the curve x2 = 2ay at (0, 0) is
(a) 1
1
(d)
(c) 3
an ellipse
2
none of these
(3, 3)
(d)
13. The radius of curvature at any point on the curve whose p-r equation is p2 = ar varies as
(a) r
3
1
(b) r 2
(c) r 2
14. The envelope of the lines x cosu + y sinu = a where u is the parameter is
(a) x2 + y2 = a2
(b) x2 − y2 = a2
(c) x + y = a
M04_ENGINEERING_MATHEMATICS-I _CH04_Part-B.indd 55
1
2
(d) r
−
1
2
(d) x − y = a
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15. The envelope of the family of lines y = mx + 1 + x 2 , where m is a parameter, is
(a) x2 + y2 = 1
(b) x2 − y2 = 1
(c) x + y = 1
(d) x − y = 1
ANSWERS
A. Fill up the blanks
4
1.
21
a
5.
2
9. (−4, 3)
2. 5
3. c
4. 2 5
6. 0
7. ⎛ −5 , 1 ⎞
8. r
⎜⎝ 4 2 ⎟⎠
dr
dp
10. x2 + 4ay = 0
B. Choose the correct answer
1. b
2. c
3. c
4. a
11. b
12. c
13. b
14. a
M04_ENGINEERING_MATHEMATICS-I _CH04_Part-B.indd 56
5. b
15. a
6. c
7. b
8. c
9. b
10. b
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Differential Calculus of
Several Variables
5.0
5
INTRODUCTION
There are many practical situations in which a quantity of interest depends on the values of two or
more variables.
For example
(i) The volume of a circular cylinder is V = pr2h, where r is the radius of the base circle and h is the
height of the cylinder. So, V is a function of two variables r and h.
(ii) The volume of a rectangular parallelopiped is V = lbh, where l, b, h are the length, breadth and
height. Here V is a function of three variables l, b, h.
Similarly we can have functions of more than two or three variables. But, for simplicity, we shall deal
with functions of two variables and the arguments and results can be extended to more than two variables.
5.1
LIMIT AND CONTINUITY
Definition 5.1 Function of two variables
Let S be a subset of R2. A function f: S → R is a rule which assigns to every (x, y) ∈ S a unique real
number in R, denoted by f(x, y).
We say f(x, y) is a function of two independent variables x and y.
S is called the domain of the function f and the range is a subset of R.
EXAMPLE 1
If f ( x , y ) 5
x 2 1 3x
, find the domain and f(1, 3).
x 2y
Solution.
Domain of f is the set of all points in the plane at which f(x, y) exists. f(x, y) is defined for all x ≠ y
{
So, domain D = ( x , y ) ∈ R 2 x ≠ y
}
Geometrically, D is the xy-plane, except the line y = x.
f (1, 3) =
1 + 3 ⋅ 3 10
=
= −5
1− 3
−2
Neighbourhood of a point in the plane
Definition 5.2 The d-neighbourhood of the point
(a,b)is the disc
⇒
δ
( x, y ) − ( a, b) < d
( x − a) 2 + ( y − b) 2 < d
A neighbourhood may also be taken as a square
0 < ⏐x − a⏐ < d and 0 < ⏐y − b⏐ < d
M05_ENGINEERING_MATHEMATICS-I _CH05_Part A.indd 1
(a, b)
Fig. 5.1
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Engineering Mathematics
Limit of a function
Definition 5.3 Let f be a function defined on S ⊂ R2. The function f is said to tend to the limit l as
(x, y) → (a, b) if to every e > 0, ∃ d > 0, such that f ( x, y ) − l < e, for all (x, y) satisfying ( x, y ) − ( a, b) < d
Then we write symbolically, lim f ( x, y ) = l or
( x , y )→( a, b )
lim f ( x, y ) = l or f ( x, y ) → l as ( x, y ) → ( a, b)
x→a
y→b
This limit is called the double limit or simultaneous limit of f(x, y)
Note
(1) If for every (x, y) ∈ S ⊂ R2, there is a unique z assigned by f, then z = f(x, y). Geometrically this
represents a surface.
(2) If
lim f ( x, y ) = l and if y = f(x) is a function such that f(x) → b as x → a,
( x , y )→( a, b )
then f(x, f(x)) → l as x → a.
That is lim f ( x, f( x )) = l .
x→a
(3) To test limit f(x, y) does not exist.
Find any two paths y = f1(x), y = f2(x) in the domain of f such that lim f ( x, f1 ( x )) = l1 and
x→a
lim f ( x, f2 ( x )) = l2
x→a
If l1≠ l2, then the limit of the function does not exist.
EXAMPLE 2
Show that the
⎡ x2 ⎤
⎢
⎥ does not exist.
( x , y ) →( 0 , 0 ) x 4 1 y 2
⎣
⎦
lim
Solution.
Given f ( x, y ) =
x2 y
x4 + y2
Choose two paths y = mx and y = mx2 and test.
Along y 5 mx
lim
⎡ x 2 ⋅ mx ⎤
0
mx
=
=0
f ( x, y ) = lim ⎢ 4
⎥ = lim
x→ a ⎣ x + m2 x 2 ⎦
x→0 x 2 + m2
0 + m2
lim
⎡ x2 y ⎤
x 2 ⋅ mx 2
m
m
f ( x, y ) = lim ⎢ 4
=
lim
= lim
=
2⎥
x→0
x→0 x 4 + m2 x 4
x→0 1 + m2
x
+
y
1
+
m2
⎦
y = mx 2 ⎣
( x , y )→( 0, 0 )
Along y 5 mx2
( x , y )→( 0, 0 )
This depends on m and so for different values of m, we will get different limit values.
Hence, the limits along different paths are different.
∴ the limit does not exist.
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Differential Calculus of Several Variables ■
5.3
Note The existence of lim f ( x, f( x )) does not imply the existence of lim f ( x, y ).
x→a
x→a
y→b
Repeated limits or iterated limits
Definition 5.4 If f(x, y) is defined in a neighbourhood of (a, b) and if lim f ( x, y ) exists, then the limit
x→a
is a function of y and the limit as y → b is written as lim lim f ( x, y ). This limit is called repeated limit
y→b x→a
of f(x, y) as x → a first and then as y → b.
Similarly, we can define the repeated limit lim lim f ( x, y ). The two repeated limits may or may not
x→a y→b
exist and when they exist, they may or may not be equal. Even if the repeated limits have the same
value, the double limit may not exsist.
If the double limit lim f ( x, y ) exists, then we cannot say repeated limits exist. But if the
Remark
x→a
y→b
repeated limits exist and are not equal, then the double limit cannot exist.
(2) If the double limit exist and repeated limits exist, then they are equal.
That is
lim
( x , y )→( a, b )
f ( x, y ) = lim lim f ( x, y )
x→a y→b
EXAMPLE 3
If f ( x , y ) 5
x2 2 y 2
, where ( x , y ) ≠ ( 0 , 0 ), find the repeated limits and double limit, if they
x2 1 y 2
exist.
Solution.
f ( x, y ) =
Given
x2 − y2
x2 + y2
∴
lim lim f ( x, y ) = lim lim
x2 − y2
x2
lim
=
= lim 1 = 1
x 2 + y 2 x→0 x 2 x→0
and
lim lim f ( x, y ) = lim lim
x2 − y2
− y2
= lim 2 = lim ( −1) = −1
2
2
y→0 y
y→0
x +y
x→0
y→0
y→0
x→0
x→0
y→0
y→0
x→0
Since the repeated limits are unequal, double limit does not exist.
Continuity of a function
Definition 5.5 A function f(x, y) defined in SCR2 is said to be continuous at the point (a, b) if
lim
f ( x , y ) 5 f ( a, b)
( x , y )→( a , b )
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5.4
■
Engineering Mathematics
WORKED EXAMPLES
EXAMPLE 1
Text the continuity of the function
xy
⎧
⎪ 2
f ( x , y ) 5 ⎨ x 1 y2
⎪
0
⎩
if ( x , y ) ≠ ( 0 , 0 )
if ( x , y ) 5 ( 0 , 0 )
at the origin.
Solution.
xy
⎧
⎪ 2
f ( x, y ) = ⎨ x + y 2
⎪
0
⎩
Given
lim f ( x, y ) =
∴
( x , y )→( 0, 0 )
if ( x, y ) ≠ (0, 0)
if ( x, y ) = (0, 0)
xy
lim
( x , y )→( 0, 0 )
x + y2
2
We shall verify
lim
( x , y )→( 0, 0 )
f ( x, y ) = f (0, 0) by ∈ − d definition
Let ∈ > 0 be given.
f ( x , y ) − f ( 0, 0 ) =
x +y
2
−0 =
x y
x y < x2 + y2
∴
f ( x, y ) − f (0, 0) < ∈
⇒
⇒
[
x2 + y 2
x < x 2 + y 2 and
But
∴
xy
2
∴
Then
f(0,0) = 0]
y < x2 + y2
x y
x +y
2
2
< x2 + y2
x2 + y2 < ∈
Take d = ∈, then ( x − 0) 2 + ( y − 0) 2 < d
f ( x, y ) − f (0, 0) < ∈ if
Thus,
∴
by definition
lim
( x , y )→( 0, 0 )
( x − 0) 2 + ( y − 0) 2 < d
f ( x , y ) = f ( 0, 0 ) = 0.
Hence, f ( x, y ) is continuous at (0, 0).
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Differential Calculus of Several Variables ■
5.5
EXAMPLE 2
If f ( x , y ) 5
2 xy
, then test lim f ( x , y ) exists or not.
x →0
x2 1 y2
y →0
Solution.
f ( x, y ) =
Given
lim f ( x, y ) = lim
∴
x→0
y→0
x→0
y→0
2 xy
x + y2
2
y = mx
Take a path
∴
2 xy
x2 + y2
lim f ( x, y ) = lim
x→0
y→0
x→0
2 x ⋅ mx
2m
2m
= lim
=
2 2
2
x
→
0
1+ m
1 + m2
x +m x
2
The limit depends on m and so for different values of m, we get different limits. So, the limit is not
unique.
Hence, the limit does not exist.
EXAMPLE 3
Find the limit and test for continuity of the function.
⎧ x3 2 y3
⎪
f ( x , y )5 ⎨ x 1 y
⎪
0
⎩
if x 1 y ≠ 0
if x 1 y 5 0
at the point (0, 0).
Solution.
⎧ x3 − y3
⎪
f ( x, y ) = ⎨ x + y
⎪ 0
⎩
Given
if x + y ≠ 0
if x + y = 0
By the definition of the function f(0, 0) = 0
Now
lim
( x , y )→( 0, 0 )
f ( x, y ) =
lim
( x , y )→( 0, 0 )
x3 − y3
x+ y
Take a path y = mx3 − x, m ≠ 0
∴
lim
( x , y )→( 0, 0 )
f ( x, y ) = lim
x→0
x 3 − ( mx 3 − x )3
x + mx 3 − x
x 3 − x 3 ( mx 2 − 1)3
x →0
mx 3
1 − ( mx 2 − 1)3 1 − ( −1)3 2
= ,m ≠0
= lim
=
x →0
m
m
m
= lim
Since the limit depends on m, for different values of m, we will have different limit values.
So, the limit is not unique. Hence, limit does not exist.
∴ the function is not continuous at (0, 0).
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5.6
■
Engineering Mathematics
EXERCISE 5.1
1. Evaluate the following limits, if they exist.
(i) lim
x →0
y →0
xy
x +y2
2
(ii) lim
x→0
y→0
(iii) lim x ( y − 2)
x → 2 y ( x − 2)
y →2
xy 3
x + y6
2
2. Test continuity of the following
⎧ x 2 + 4 y if ( x, y ) ≠ (1, 2)
(i) f ( x, y ) = ⎨
if ( x, y ) = (1, 2)
⎩ 0
⎧ xy 2
(ii) f ( x, y ) = ⎪⎨ x 2 + y 2
⎪ 0
⎩
if x ≠ 0, y ≠ 0
if x = 0, y = 0
⎧ x2 − y2
(iii) f ( x, y ) = ⎪⎨ x 2 + y 2
⎪ 0
⎩
if ( x, y ) ≠ (0, 0)
if ( x, y ) = (0, 0)
ANSWERS TO EXERCISE 5.1
1. (i) does not exist. (ii) does not exist (iii) does not exist.
2. (i) not continuous
(ii) not continuous
[Hint limit does not exist. Choose paths y = x, x = y3]
(iii) not continuous
5.2
PARTIAL DERIVATIVES
Functions of two or more independent variables appear in many practical problems more often than
functions of one independent variable. The concept of derivative of a single variable function f(x)
is extended to functions of two or more variables. Suppose f(x, y) is a function of two independent
variables x and y, we treat y as constant and find the derivative of f(x, y) w.r.to x, then the derivative is
called a partial derivative. Partial derivatives find applications in a wide variety of fields like fluid
dynamics, electricity, physical sciences, econometrics, probability theory etc.
Definition 5.6 Let z = f(x, y) be a real function of two independent variables x and y. Let (x0, y0) be a
point in the domain of f. The partial derivative of f(x, y) w.r.to x at (x0, y0) is the limit
f ( x0 1 h, y0 ) 2 f ( x0 , y0 )
, if the limit exists.
h→ 0
h
∂f
∂z
or ⎛ ⎞
or fx(x0, y0).
Then it is denoted by ⎛⎜ ⎞⎟
⎝ ∂x ⎠ ( x , y ) ⎜⎝ ∂x ⎟⎠ ( x , y )
0
0
0
0
lim
Similarly, partial derivative of f(x, y) w.r.to y at (x0, y0) is the limit
lim
k→ 0
It is denoted by ⎛ ∂f ⎞
⎜⎝ ∂y ⎟⎠
(x
0,
y0 )
f ( x0 , y0 + k ) − f ( x0 , y0 )
, if the limit exists.
k
⎛ ∂z ⎞
or ⎜ ⎟
⎝ ∂y ⎠ ( x
M05_ENGINEERING_MATHEMATICS-I _CH05_Part A.indd 6
or f y ( x0 , y0 )
0,
y0 )
5/12/2016 10:24:19 AM
Differential Calculus of Several Variables ■
5.7
If the partial derivatives of z = f(x, y) exist at any point in its domain, then the partial derivatives
∂z
∂f
or
or fx, assuming the point as (x, y).
w.r.to x is simply written as
∂x
∂x
Similarly, the partial derivative w.r.to y is written as ∂z or ∂f or fy
∂y
∂y
5.2.1 Geometrical Meaning of
∂z ∂ z
,
∂x ∂y
Let z = f(x, y) be a real function of two independent variables x, y. For every point (x, y) in its domain
R in the xy plane there is a real number z, where f(x, y) = z. The set of all points (x, y, z), where
z = f(x, y), in space determine a surface S. This surface is called the graph of the function f.
Thus, z = f(x, y) represents a surface in space.
z = f(x, y0)
P
Z
The equation y = y0 represents a vertical plane (parallel to
xoz plane) intersecting the surface in a curve C : z = f(x, y0).
(x0, y0, z0)
∂z
at (x0, y0) represents the slope of the
The partial derivative
∂x
tangent to this curve at the point (x0, y0, z0), where z0 = f(x0, y0).
Similarly, the partial derivative ∂z at (x0, y0) is the slope of
O
Y
∂y
the tangent to the curve z = f(x0, y) at the point (x0, y0, z0), where
z0 = f(x0, y0).
X
Fig. 5.2
Note
(1) If z = f(x, y) then f(x, y) = k for all points in the domain of f, where k is a constant, is called a level
curve of function f.
(2) If u = f(x, y, z) be a function of three independent variables x, y, z then the graph of f is a
4-dimensional surface.
f(x, y, z) = c, where c is a constant, is called a level surface. For different c, we have different level
surfaces. No two level surfaces intersect.
(3) A function f(x, y) may not be continuous at a point, but still it can have partial derivatives with
respect to x and y at that point.
⎧0 if xy ≠ 0
.
⎩1 if xy = 0
For example, consider f(x, y) = ⎨
We shall find the limit (x, y) → (0, 0) along y = x, where f(x, y) = 0, except at (0, 0).
lim
( x , y )→( 0,0 )
f (x , y ) =
lim
( x , y )→( 0,0 )
0=0
But f(0, 0) = 1. Hence, f is not continuous at (0, 0).
Now, fx(0, 0) = lim
h →0
f ( 0 + h , 0 ) − f ( 0, 0 )
0−0
= lim
=0
h →0 h
h
f ( 0, 0 + k ) − f ( 0, 0 )
0−0
= lim
=0
k
→
0
k
k
Thus, the partial derivatives exist at the origin (0, 0), but f is not continuous at the origin (0, 0).
This is different from functions of single variable where the existence of derivatives implies
continuity.
and f y (0, 0) = lim
k →0
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5.8
■
Engineering Mathematics
5.2.2 Partial Derivatives of Higher Order
Let z = f(x, y) be a function of two independent variables. The derivatives ∂f , ∂f are called partial
∂x ∂y
derivatives of first order, which are again functions of x, y and can be differentiated partially w.r.to
x, y. These are called partial derivatives of second order and denoted by
∂ ⎛ ∂f ⎞
⎜ ⎟
∂x ⎝ ∂x ⎠
∂ ⎛ ∂f ⎞
∂x ⎜⎝ ∂y ⎟⎠
∂ ⎛ ∂f ⎞
∂y ⎜⎝ ∂y ⎟⎠
∂ ⎛ ∂f ⎞
⎜ ⎟
∂y ⎝ ∂x ⎠
∂ 2f
∂x 2
∂ 2f
=
∂x ∂y
∂ 2f
= 2
∂y
∂ 2f
=
∂y ∂x
=
or
f xx
or
f yx
or
f yy
or
f xy
It can be shown that if fx, fy and fxy are continuous, then fxy = fyx. In fact the elementary functions that we
come across satisfy these conditions. In many practical applications also these conditions are satisfied.
So, we shall assume this in our discussions.
Differentiating the second order derivatives partially w.r.to x, y, we get third order derivatives.
5.2.3 Homogeneous Functions and Euler’s Theorem
Definition 5.7 A function f(x, y) is said to be homogeneous of degree (or order) n if f(tx, ty) = tn f(x, y)
for any positive t.
x6 + y6
For example (1): f ( x, y ) = 4
is homogeneous of degree 2,
x − y4
since
Note that
(2)
since
f (tx, ty ) =
t 6 x6 + t 6 y6 t 6 ( x6 + y6 ) t 2 ( x6 + y6 )
= t 2 f ( x, y )
= 4 4
=
4 4
4 4
4
4
4
t x −t y
t (x − y )
x −y
⎛ ⎛ y⎞6 ⎞
⎜ 1 + ⎜⎝ x ⎟⎠ ⎟
⎛ y⎞
⎟ = x2F ⎜ ⎟ .
f ( x, y ) = x 2 ⎜
4
⎝ x⎠
⎜ ⎛ y⎞ ⎟
1
−
⎜
⎟
⎜⎝ ⎝ ⎠ ⎟⎠
x
y
f ( x, y ) = tan −1 is homogeneous functions of degree 0,
x
ty
y
f (tx, ty ) = tan −1 = tan −1 = t 0 f ( x, y )
tx
x
Theorem 5.1 Euler’s theorem
If f(x, y) is a homogeneous function of degree n in x and y having continuous partial derivatives, then
∂f
∂f
x
1 y
5 nf ( x , y ).
∂x
∂y
Proof
Given f(x, y) is a homogeneous function in x and y of degree n, we can write
M05_ENGINEERING_MATHEMATICS-I _CH05_Part A.indd 8
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Differential Calculus of Several Variables ■
5.9
⎛ y⎞
f ( x, y ) = x n F ⎜ ⎟
⎝ x⎠
Let
∂f
⎛ y⎞
⎛ y⎞ ⎛ −y⎞
= x n ⋅ F ′ ⎜ ⎟ ⎜ 2 ⎟ + nx n −1F ⎜ ⎟
⎝ x⎠
⎝ x⎠ ⎝ x ⎠
∂x
∴
∴
x
∂f
⎛ y⎞
⎛ y⎞
= − x n −1 y F ′ ⎜ ⎟ + nx n F ⎜ ⎟
⎝ x⎠
⎝ x⎠
∂x
(1)
∂f
⎛ y⎞
⎛ y⎞ 1
= x n ⋅ F ′ ⎜ ⎟ ⋅ = x n −1 F ′ ⎜ ⎟
⎝ x⎠
⎝
⎠
∂y
x x
∴
y
∂f
⎛ y⎞
= x n −1 y F ′ ⎜ ⎟
⎝ x⎠
∂y
(2)
∂f
∂f
⎛ y⎞
+y
= nx n F ⎜ ⎟ = n f ( x, y )
⎝ x⎠
∂x
∂y
This theorem can be extended to homogeneous function of any number of variables.
If f(x, y, z) is a homogeneous function of degree n in three independent variables x, y, z and differentiable then x ∂f + y ∂f + z ∂f = nf .
■
∂x
∂y
∂z
(1) + (2) ⇒
x
WORKED EXAMPLES
EXAMPLE 1
If u 5 log (x3 1 y3 1 z3 2 3xyz), then prove that
3
(i) ∂u 1 ∂u 1 ∂u 5
∂x ∂y ∂z x 1 y 1 z
Solution.
Given
2
9
(ii) ⎛ ∂ 1 ∂ 1 ∂ ⎞ u 52
⎜⎝ ∂x ∂y ∂z ⎟⎠
( x 1 y 1 z )2
u = log (x3 + y3 + z3 − 3xyz)
∴
∂u
1
[3 x 2 − 3 yz ]
= 3
3
∂x x + y + z 3 − 3 xyz
Similarly,
∂u
3[ y 2 − zx ]
= 3
∂y x + y 3 + z 3 − 3 xyz
∴
and
=
3[ x 2 − yz ]
x 3 + y 3 + z 3 − 3 xyz
∂u
3[ z 2 − xy ]
= 3
∂z x + y 3 + z 3 − 3 xyz
∂u ∂u ∂u 3[ x 2 1 y 2 1 z 2 2 yz 2 zx 2 xy ]
1
1
5
∂x ∂y ∂z
x 3 1 y 3 1 z 3 23 xyz
We know that
x 3 + y 3 + z 3 − 3 xyz = ( x 2 + y 2 + z 2 − xy − yz − zx )( x + y + z )
∴
∂u ∂u ∂u
3( x 2 + y 2 + z 2 − xy − yz − zx )
3
+
+
=
=
∂x ∂y ∂z ( x + y + z )( x 2 + y 2 + z 2 − xy − yz − zx ) x + y + z
M05_ENGINEERING_MATHEMATICS-I _CH05_Part A.indd 9
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■
5.10
Engineering Mathematics
⎛ ∂
∂
∂⎞
3
⎜⎝ ∂x + ∂y + ∂z ⎟⎠ u = x + y + z
(ii) ⇒
(1)
∂
∂
∂
Operating ⎛⎜ + + ⎞⎟ on both sides, we get
⎝ ∂x ∂y ∂z ⎠
2
⎛ ∂
⎛ ∂
⎞
∂
∂⎞
∂
∂ ⎞⎛
3
⎜⎝ ∂x + ∂y + ∂z ⎟⎠ u = ⎜⎝ ∂x + ∂y + ∂z ⎟⎠ ⎜⎝ x + y + z ⎟⎠
=
⎞ ∂ ⎛
⎞ ∂ ⎛
⎞
3
3
∂ ⎛
3
+
+
⎜
⎟
⎜
⎟
⎜
∂x ⎝ x + y + z ⎠ ∂y ⎝ x + y + z ⎠ ∂z ⎝ x + y + z ⎟⎠
=3
∂
∂
∂
( x + y + z ) −1 + 3 ( x + y + z ) −1 + 3 ( x + y + z ) −1
∂x
∂y
∂z
= − 3( x + y + z ) −2 ⋅1 − 3( x + y + z ) −2 ⋅1 − 3( x + y + z ) −2 ⋅1
3
3
3
9
=−
−
−
=−
2
2
2
(x + y + z )2
(x + y + z ) (x + y + z ) (x + y + z )
EXAMPLE 2
If u 5 (x 2 y) (y 2 z) (z 2 x), then prove that (i)
∂u
∂u
∂u
∂u ∂u ∂u
1 1 5 0 (ii) x 1 y 1 z 5 3u
∂
x
∂
y
∂z
∂x ∂y ∂z
Solution.
Given u = (x − y) (y − z) (z − x)
∴
∂u
= ( y − z )[( x − y )( −1) + ( z − x ) ⋅1]
∂x
= ( y − z )( − x + y + z − x ) = ( y − z )( y + z ) − 2( y − z ) x = y 2 − z 2 − 2 yx + 2 zx
Similarly,
(i)
∂u
= z 2 − x 2 − 2 zy + 2 xy and
∂y
∂u
= x 2 − y 2 − 2 xz + 2 yz
∂z
∂u ∂u ∂u
+ + = y 2 − z 2 − 2 yx + 2 zx + z 2 − x 2 − 2 zy + x 2 − y 2 − 2 xz + 2 yz = 0
∂x ∂y ∂z
(ii) u is a homogeneous function of degree 3, since
u ( xt , yt , zt ) = (tx − ty ) (ty − tz ) (tz − tx ) = t 3 ( x − y ) ( y − z ) ( z − x )
= t 3 ( x − y ) ( y − z ) ( z − x ) = t 3 u( x , y , z )
So, by Euler’s theorem, we get
x
∂u
∂u
∂u
+y +z
= 3u
∂x
∂y
∂z
M05_ENGINEERING_MATHEMATICS-I _CH05_Part A.indd 10
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Differential Calculus of Several Variables ■
5.11
EXAMPLE 3
∂u
∂u
∂u
⎛ x y z⎞
If u 5 f ⎜ , , ⎟ then prove that x 1 y 1 z 5 0.
∂x
∂y
∂z
⎝ y z x⎠
Solution.
Given u is a function of x, y, z
⎛ x y z⎞
u ( x, y, z ) = f ⎜ , , ⎟
⎝ y z x⎠
and
⎛ x y z⎞
⎛ tx ty tz ⎞
u (tx, ty, tz ) = f ⎜ , , ⎟ = f ⎜ , , ⎟ = u( x, y, z )
⎝ y z x⎠
⎝ ty tz tx ⎠
∴
∴ u(x, y, z) is a homogeneous function of degree 0 in x, y, z
By Euler’s theorem, we get x
∂u
∂u
∂u
+y +z
= 0.
∂x
∂y
∂z
EXAMPLE 4
If u 5 f(x 2 y, y 2 z, z 2 x) show that
∂u ∂u ∂u
1 1 5 0.
∂x ∂y ∂z
Solution.
Given
u = f(x − y, y − z, z − x)
Put
x1 = x − y, y1 = y − z, z1 = z − x, then
u = f(x1, y1, z1), where x1, y1, z1 are functions of x, y, z.
We know du =
∂u
∂u
∂u
⋅ dx1 +
⋅ dy1 +
⋅ dz1
∂x1
∂y1
∂z1
∴
∂u ∂u ∂x1 ∂u ∂y1 ∂u ∂z1
=
⋅
+
⋅
+
⋅
∂x ∂x1 ∂x ∂y1 ∂x ∂z1 ∂x
Now
x1 = x − y
∴
y1 = y − z
∴
and
z1 = z − x
∴
∴
∂u ∂u ∂x1 ∂u ∂y1 ∂u ∂z1
=
+
+
∂x ∂x1 ∂x ∂y1 ∂x ∂z1 ∂x
⇒
∂u ∂u ∂u
=
−
∂x ∂x1 ∂z1
(1)
Similarly,
∂u ∂u ∂x1 ∂u ∂y1 ∂u ∂z1
=
⋅
+
⋅
+
⋅
∂y ∂x1 ∂y ∂y1 ∂y ∂z1 ∂y
⇒
∂u
∂u ∂u
=−
+
∂y
∂x1 ∂y1
(2)
M05_ENGINEERING_MATHEMATICS-I _CH05_Part A.indd 11
∂x1
∂x1
∂x1
= 1,
= −1,
=0
∂x
∂y
∂z
∂y1
∂y1
∂y1
= 0,
= 1,
= −1
∂x
∂y
∂z
∂z1
= −1,
∂x
∂z1
= 0,
∂y
∂z1
=1
∂z
5/12/2016 10:24:40 AM
5.12
■
and
Engineering Mathematics
∂u ∂u ∂x1 ∂u ∂y1 ∂u ∂z1
+
⋅
+
⋅
=
⋅
∂z ∂x1 ∂z ∂y1 ∂z ∂z1 ∂z
⇒
∂u
∂u ∂u
+
=−
∂z
∂y1 ∂z1
(3)
∂u ∂u ∂u ∂u ∂u ∂u ∂u ∂u ∂u
+
+
=
−
−
+
−
+
=0
∂x ∂y ∂z ∂x1 ∂z1 ∂x1 ∂y1 ∂y1 ∂z1
(1) + (2) + (3) ⇒
EXAMPLE 5
If u 5 sin21
∂u
∂u 1
⎛ x1 y ⎞
, then prove that x 1 y 5 tan u .
⎜⎝ x 1 y ⎟⎠
∂x
∂y 2
Solution.
Given
u = sin −1
⎛ x+ y ⎞
⎜⎝ x + y ⎟⎠
x+ y
Let f ( x, y ) =
tx + ty
tx + ty
= t 1/ 2
⎛ x + y ⎞ 1/ 2
= t f ( x, y )
⎜⎝ x + y ⎟⎠
∴ f is a homogeneous function of degree
By Euler’s theorem, we get x
⇒
⇒
⇒
x
x+ y
x+ y
∴ f ( x, y ) = sin u
x+ y
∴ f (tx, ty ) =
sin u =
⇒
1
2
∂f
∂f 1
+y
= f
∂x
∂y 2
∂
∂
1
(sin u ) + y (sin u ) = sin u
∂x
∂y
2
x cos u
[since f = sin u]
∂u
∂u 1
+ y cos u
= sin u
∂x
∂y 2
x
∂u
∂u 1 sin u
+y
=
∂x
∂y 2 cos u
⇒
x
∂u
∂u 1
+y
= tan u
∂x
∂y 2
Another result on homogeneous functions which follow from Euler’s theorem is given below.
Theorem 5.2
If u(x, y) is homogeneous function of degree n in x and y with all first and second derivatives
continuous, then
2
∂2 u
∂2 u
2 ∂ u
1
2
1
5 n( n 2 1)u
xy
y
∂x ∂y
∂x 2
∂y 2
Proof Given u(x, y) is a homogeneous function of x and y of degree n.
∂u
∂u
So, by Euler’s theorem x
+y
= nu
∂x
∂y
x2
M05_ENGINEERING_MATHEMATICS-I _CH05_Part A.indd 12
(1)
5/12/2016 10:24:45 AM
Differential Calculus of Several Variables ■
5.13
Differentiating (1) partially w.r.to x, we get
x
∂ 2 u ∂u
∂ 2u
∂u
+
+y
=n
2
∂x
∂x ∂y
∂x
∂x
x
∂ 2u
∂ 2u
∂u
+
= ( n − 1)
y
2
∂x ∂y
∂x
∂x
x2
∂2u
∂2u
∂u
+
xy
= ( n − 1) x
2
∂x ∂y
∂x
∂x
⇒
Multiplying by x,
(2)
Now differentiating (1) w.r.to y, we get
∂2u
∂ 2 u ∂u
∂u
x⋅
+ y⋅ 2 +
=n
∂y ∂x
∂y
∂y
∂y
⇒
x
Multiplying by y,
(2) + (3) ⇒
⇒
xy
x2
∂2u
∂2u
∂u
+ y 2 = ( n − 1)
∂x ∂y
∂y
∂y
⎡
∂2u
∂2u ⎤
=
assu
u
ming
⎥
⎢
∂x ∂y ∂y ∂x ⎦
⎣
∂2u
∂2u
∂u
+ y 2 2 = ( n − 1) y
∂x ∂y
∂y
∂y
2
∂2u
∂2u
∂u ⎤
⎡ ∂u
2 ∂ u
2
+
xy
+
y
= ( n − 1) ⎢ x
+ y ⎥ = ( n − 1)nu
2
2
∂x ∂y
∂y ⎦
∂x
∂y
⎣ ∂x
(3)
[Using (1)]
x 2 uxx + 2 xyuxy + y 2 uyy = n( n − 1)u
WORKED EXAMPLES
EXAMPLE 6
∂2 z
∂2 z
∂2 z
⎛ y⎞
⎛ y⎞
1 y 2 2 5 0.
If z 5 xf ⎜ ⎟ 1 g ⎜ ⎟ , then show that x 2 2 1 2 xy
⎝ x⎠
⎝ x⎠
∂x
∂x ∂y
∂y
Solution.
y
y
Given z = xf ⎛⎜ ⎞⎟ + g ⎛⎜ ⎞⎟
⎝ x⎠
⎝ x⎠
Let
⎛ y⎞
⎛ y⎞
u = xf ⎜ ⎟ , v = g ⎜ ⎟
⎝ x⎠
⎝ x⎠
∴ z = u+v
(1)
⎡
⎛ y⎞ ⎤
⎛ ty ⎞
u(tx, ty ) = tx f ⎜ ⎟ = t ⎢ x f ⎜ ⎟ ⎥ = tu
⎝ x⎠⎦
⎝ tx ⎠
⎣
∴ u is homogeneous of degree 1.
By theorem 5.2
∂2 u
∂2 u
∂2 u
x 2 2 1 2 xy
1 y 2 2 5 n( n 2 1)u 5 0
∂x ∂y
∂x
∂y
Now
∴
[{ n = 1]
(2)
⎛ y⎞
v( x, y ) = g ⎜ ⎟
⎝ x⎠
⎛ y⎞
⎛ ty ⎞
v(tx, ty ) = g ⎜ ⎟ = g ⎜ ⎟ = v( x, y )
⎝ x⎠
⎝ tx ⎠
M05_ENGINEERING_MATHEMATICS-I _CH05_Part A.indd 13
5/12/2016 10:24:49 AM
■
5.14
Engineering Mathematics
∴ v is homogeneous of degree 0.
∂2 v
∂2 v
∂2 v
x 2 2 + 2 xy
+ y 2 2 = n( n − 1)v = 0
∴
∂x ∂y
∂x
∂y
(2) + (3) ⇒
x2
2
∂2
∂2
2 ∂
u
+
v
+
2
xy
u
+
v
+
y
(u + v ) = 0
(
)
(
)
∂x ∂y
∂x 2
∂y 2
⇒
x2
EXAMPLE 7
If u 5 x 2 tan21
[{ n = 0] (3)
∂2 z
∂2 z
∂2 z
+ 2 xy
+ y2 2 = 0
2
∂x ∂y
∂x
∂y
y
x
∂2u
∂2u
∂2u
2 y 2 tan21 , then find the value of x 2 2 1 2 xy
+ y2 2 .
x
y
∂x ∂y
∂y
∂x
Solution.
Given u( x, y ) = x 2 tan −1
∴
x
y
− y 2 tan −1
x
y
⎛ tx ⎞
⎛ ty ⎞
u(tx, ty ) = t 2 x 2 tan −1 ⎜ ⎟ − t 2 y 2 tan −1 ⎜ ⎟
⎝ tx ⎠
⎝ ty ⎠
y
x⎤
⎡
= t 2 ⎢ x 2 tan −1 − y 2 tan −1 ⎥ = t 2 u( x, y )
x
y⎦
⎣
∴ u(x, y) is homogeneous of degree 2 and it is differentiable twice, partially.
∴ by theorem 5.2,
∂2u
∂2u
∂2u
x 2 2 + 2 xy
+ y 2 2 = 2( 2 − 1)u = 2u
∂x ∂y
∂x
∂y
EXAMPLE 8
If u 5 xy, then show that (i) uxy 5 uyx
(ii) uxxy 5 uxyx
Solution.
Given u = xy
∴
ux = yxy − 1
and
uy = xy logex
Differentiating (1) again w.r.to x, uxx = y(y − 1)xy − 2
Differentiating again w.r.to y, we get
[treating y as constant] (1)
[treating x as constant] (2)
uxxy = y( y − 1)x y −2 log e x + x y −2 [ y ⋅1 + (y − 1) ⋅1]
= x y −2 [ y( y − 1) log e x + 2y − 1]
(3)
Differentiating (1) w.r.to y, we get
uxy = y ⋅ x y −1 log e x + x y −1 ⋅1
⇒ uxy = x y −1 [1 + y log e x ]
(4)
1
uyx = x y ⋅ + log e x ⋅ y x y −1
x
⇒ uyx = x y −1 [1 + y log e x ]
(5)
Differentiating (2) w.r.to x, we get
M05_ENGINEERING_MATHEMATICS-I _CH05_Part A.indd 14
5/12/2016 10:24:52 AM
Differential Calculus of Several Variables ■
5.15
From (4) and (5), we get uxy = uyx, which is (i)
Again differentiating (4) w.r.to x, we get
⎡ 1⎤
uxyx = x y −1 ⎢ y ⋅ ⎥ + (1 + y log e x ) ⋅ ( y − 1) x y − 2
⎣ x⎦
y−2
= x ⋅ y + ( y − 1)(1 + y loog e x ) x y − 2
⇒
uxyx = x y − 2 [ y + ( y − 1)(1 + y log e x )] = x y − 2 [ y( y − 1) log e x + 2 y − 1]
(6)
From (3) and (6), we get uxxy = uxyx, which is (ii)
EXAMPLE 9
If r 2 5 x 2 1 y 2 1 z 2 , then prove that
∂2 r ∂2 r ∂2 r 2
1
1
5 .
∂x 2 ∂y 2 ∂z 2 r
Solution.
Given
∴
∂r
∂r x
= 2x ⇒
=
∂x
∂x r
∂r
x
r ⋅1 − x
r − x⋅
2
2
∂2 r
∂x =
r = r −x
=
∂x 2
r2
r2
r3
r 2 = x 2 + y 2 + z 2 ∴ 2r
Similarly,
∂2 r r 2 − y 2
=
∂y 2
r3
∴
∂2 r ∂2 r ∂2 r r 2 − x 2 + r 2 − y 2 + r 2 − z 2
+
+
=
∂x 2 ∂y 2 ∂z 2
r3
and
=
∂2 r r 2 − z 2
=
∂z 2
r3
3r 2 − ( x 2 + y 2 + z 2 ) 3r 2 − r 2 2r 2 2
=
= 3 =
r
r3
r3
r
5.2.4 Total Derivatives
Let u = f(x, y) be a function of 2 variables x, y. If x and y are continuous functions of t then z will be
ultimately a function of t only or z is a composite function of t. Then we can find the ordinary
du
which is called the total derivative of u to distinguish it from the partial derivatives ∂u , ∂u .
derivative
dt
∂x ∂y
du ∂u dx ∂u dy
We have
=
+
dt ∂x dt ∂y dt
This is also know as chain rule for one independent variable.
Proof u = f(x, y), x = F(t), y = G(t).
Giving increment Δt to t will result in increments Δx, Δy and Δu in x, y and u.
∴
Δu = f ( x + Δx, y + Δy ) − f ( x, y )
= f ( x + Δx, y + Δy ) − f ( x, y + Δy ) + f ( x, y + Δy ) − f (xx, y )
∴
Δu f ( x + Δx, y + Δy ) − f ( x, y + Δy ) f ( x, y + Δy ) − f ( x, y )
=
+
Δt
Δt
Δt
=
f ( x + Δx, y + Δy ) − f ( x, y + Δy ) Δx f ( x, y + Δy ) − f ( x, y ) Δy
+
Δx
Δt
Δy
Δt
M05_ENGINEERING_MATHEMATICS-I _CH05_Part A.indd 15
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5.16
■
Engineering Mathematics
∴
du
Δu
= lim
,
Δ
t
→
0
dt
Δt
∴
du ∂f dx ∂f dy
=
+
dt ∂x dt ∂y dt
as Δt → 0, Δx → 0, Δy → 0
or
du ∂u dx ∂u dy
=
+
dt ∂x dt ∂y dt
[{ u = f ] (1)
Cor (1) In differential form the result (1) can be written as
df =
∂f
∂f
dx + dy
∂x
∂y
or
du =
∂u
∂u
dx +
dy
∂x
∂y
du is called the total differential of u.
Similarly, if u = f(x, y, z) of 3 independent variable x, y, z, then the total differential is
du =
∂u
∂u
∂u
dx +
dy +
dz
∂x
∂y
∂z
Note: The function z = f(x, y) is differentiable at the point (x0, y0) if the first partial derivatives fx, fy
exist at (x0, y0) and are continuous at (x0, y0) and
dz = f x ( x 0 , y 0 )dx + f y ( x 0 , y 0 )dy .
From this result, it follows that if f (x, y) is differentiable at (x0, y0), then f is continuous at
(x0, y0).
Cor (2) If u = f(x, y) where x and y are function of t1, t2, then u is ultimately a function of t1, t2 and so
z is a composite function of t1, t2.
Then we have partial derivatives of u w.r.to t1, t2
∴
∂u ∂u ∂x ∂u ∂y
=
+ ⋅
∂t1 ∂x ∂t1 ∂y ∂t1
∂u ∂u ∂x ∂u ∂y
=
+ ⋅
∂t 2 ∂x ∂t 2 ∂y ∂t 2
and
These are chain rules for two independent variables.
Cor (3) Differentiation of implicit functions
The equation f(x, y) = 0 defines y implicitly as a function of x. Suppose the function f(x, y) is differentiable, then the total differential
df = 0
⇒
df
df
dx +
dy = 0
dx
dy
⇒
∂f ∂f dy
+
=0
∂x ∂y dx
⇒
∂f
−
dy
= ∂x
∂f
dx
∂y
M05_ENGINEERING_MATHEMATICS-I _CH05_Part A.indd 16
⇒
f
dy
= − x if f y ≠ 0
dx
fy
5/12/2016 10:25:07 AM
Differential Calculus of Several Variables ■
5.17
WORKED EXAMPLES
EXAMPLE 1
If u 5 x2 y3, x 5 log t, y 5 et, then find
Solution.
Given u = x2 y3,
We know du =
x = log t,
du
.
dt
y = et.
∂u
∂u
dx +
dy
∂x
∂y
So, u is ultimately a function of t
du ∂u dx ∂u dy .
=
+
dt ∂x dt ∂y dt
∴
∂u
∂u
dx 1 dy
= 2 xy 3 ,
= x2 ⋅ 3y2 ,
= ,
= et
∂x
∂y
dt t dt
du
1
= 2 xy 3 ⋅ + 3 x 2 y 2 e t
dt
t
e 3t log t
1
2
[2 + 3t log t ]
= 2 log t ⋅ e 3t ⋅ + 3(log t ) 2 ⋅ e 2t ⋅ e t = (log t )ee 3t + 3(log t ) 2 e 3t =
t
t
t
But
∴
EXAMPLE 2
If u 5 sin21 ( x 2 y ), where x = 3t, y = 4t3, then show that
Solution.
Given
We know
But
du
3
5
, −1 < t < 1.
dt
12 t2
u = sin−1(x − y), where x = 3t, y = 4t3
∂u
∂u
du =
dx +
dy
∴ du = ∂u dx + ∂u dy
∂x
∂y
dt ∂x dt ∂y dt
∂u
1
∂u
1
=
=
,
( −1) and dx = 3, dy = 12t 2
∂x
∂y
1 − ( x − y)2
1 − ( x − y)2
dt
dt
du
1
3(1 − 4t 2 )
=
(3 − 12t 2 ) =
dt
1 − ( x − y)2
1 − ( x − y)2
∴
Now 1 − (x − y)2 = 1 − (3t − 4t3)2 = 1 − (9t2 + 16t6 − 24t4) = 1 − 9t2 + 24t4 − 16t6
Since sum of the coefficients of R. H. S = 0, t2 = 1 will satisfy the polynomial 1 − 9t2 + 24t4 − 16t6
∴
1 − ( x − y ) 2 = (t 2 − 1)( −16t 4 + 8t 2 − 1)
1 −16
24 −9
1
= −(t 2 − 1)(16t 4 − 8t 2 + 1)
= −(t − 1)( 4t − 1) = (1 − t )(1 − 4t )
2
du
=
dt
∴
2
2
3(1 − 4t 2 )
(1 − t )(1 − 4t )
2
2 2
=
2
3
1− t
2
2 2
0
− 16
−16
8
8 −1
−1
0
[since 1 − 4t 2 > 0]
EXAMPLE 3
Find
du
if u 5 cos (x2 1 y2) and a2x2 1 b2y2 5 c2.
dx
Solution.
Given
u = cos (x2 + y2) and a2x2 + b2y2 = c2
M05_ENGINEERING_MATHEMATICS-I _CH05_Part A.indd 17
5/12/2016 10:25:12 AM
5.18
■
Engineering Mathematics
We know du =
∂u
∂u
dx +
dy
∂x
∂y
But a2x2 + b2y2 = c2, then
Here
∴
Now
∴
∴
du ∂u ∂u dy
=
+
dx ∂x ∂y dx
f
dy
=− x
dx
fy
f = a2x2 + b2y2 − c2
dy −2a 2 x
a2 x
=
=− 2
2
dx
2b y
b y
∴
fx = 2a2x
fy = 2b2y
and
∂u
∂u
= − sin( x 2 + y 2 ) ⋅ 2 x and
= − sin( x 2 + y 2 ) ⋅ 2 y
∂x
∂y
⎛ a2 x ⎞
du
= −2 x sin( x 2 + y 2 ) − 2 y sin( x 2 + y 2 ) ⋅ ⎜ − 2 ⎟
dx
⎝ b y⎠
⎧ a2 ⎫
⎛ a2 ⎞ 2 x
= −2 x sin( x 2 + y 2 ) ⎨1 − 2 ⎬ = 2 x sin( x 2 + y 2 ) ⋅ ⎜ 2 − 1⎟ = 2 ( a 2 − b 2 ) sin( x 2 + y 2 )
⎝b
⎠ b
⎩ b ⎭
EXAMPLE 4
2
If u 5 f(x, y), x 5 r cos u, y 5 r sin u, then show that ⎛⎜ ∂u ⎞⎟ 1 ⎛ ∂u ⎞ 5 ⎛⎜ ∂u ⎞⎟ 1 1 ⎛⎜ ∂u ⎞⎟ .
⎜⎝ ∂y ⎟⎠
⎝ ∂r ⎠
⎝ ∂x ⎠
r 2 ⎝ ∂u ⎠
Solution.
Since x and y are functions of r and u, u is a composite function of r and u. So, we have partial
derivatives of u w.r.to r, u
∂u ∂u ∂x ∂u ∂y ∂u ∂u ∂x ∂u ∂y
∂u
∂u
=
+
,
=
+
du =
dx +
dy ∴
We know that
∂r ∂x ∂r ∂y ∂r
∂u ∂ x ∂u ∂y ∂u
∂x
∂y
2
2
x = r cos u and y = r sin u
∂x
∂y
= cos u
and
= sin u
∂r
∂r
∂x
∂y
= − r sin u and
= r cos u
∂u
∂u
∂u
∂u ∂u ∂x ∂u ∂y ∂u
cos u + sin u
=
⋅ + ⋅ =
∂y
∂r ∂x ∂r ∂y ∂r ∂x
Since
∴
∂ u ∂ u ∂ x ∂u ∂y ∂u
∂u
=
⋅ + ⋅
=
( − r sin u) + ( r cos u)
∂ u ∂ x ∂ u ∂y ∂u ∂x
∂y
and
1 ∂u
∂u
∂u
= − sin u + cos u
r ∂u
∂x
∂y
⇒
∂u
∂u
⎛ ∂u ⎞ ⎛ 1 ∂u ⎞
⎞
⎛ ∂u
⎞ ⎛ ∂u
⎜⎝ ⎟⎠ + ⎜⎝
⎟⎠ = ⎜ cos u + sin u⎟ + ⎜ − sinn u + cos u⎟
∂r
r ∂u
∂y
∂y
⎠
⎝ ∂x
⎠ ⎝ ∂x
2
∴
2
2
2
2
∂u ∂u
⎛ ∂u ⎞
⎛ ∂u ⎞
= ⎜ ⎟ cos 2 u + ⎜ ⎟ sin 2 u + 2
cos u sin u
⎝ ∂x ⎠
∂x ∂y
⎝ ∂y ⎠
2
2
∂u ∂u
⎛ ∂u ⎞
⎛ ∂u ⎞
+ ⎜ ⎟ sin 2 u + ⎜ ⎟ cos 2 u − 2 ⋅ sin u cos u
⎝ ∂x ⎠
⎝ ∂y ⎠
∂x ∂y
2
M05_ENGINEERING_MATHEMATICS-I _CH05_Part A.indd 18
2
5/12/2016 10:25:17 AM
Differential Calculus of Several Variables ■
5.19
⎛ ∂u ⎞
⎛ ∂u ⎞
= ⎜ ⎟ [cos 2 u + sin 2 u] + ⎜ ⎟ [sin 2 u + cos 2 u]
⎝ ∂x ⎠
⎝ ∂y ⎠
2
2
1 ⎛ ∂u ⎞
⎛ ∂u ⎞
⎛ ∂u ⎞ ⎛ ∂u ⎞
⎜⎝ ⎟⎠ + 2 ⎜⎝ ⎟⎠ = ⎜⎝ ⎟⎠ + ⎜ ⎟
∂r
r ∂u
∂x
⎝ ∂y ⎠
2
⇒
2
2
2
EXAMPLE 5
∂u
∂u
∂u
⎛ y 2 x z 2 x⎞
1y2
1z2
5 0.
If u 5 f ⎜
,
, then prove that x 2
⎟
zx ⎠
⎝ xy
∂x
∂y
∂z
Solution.
⎛ y − x z − x⎞
,
u= f⎜
⎝ xy
zx ⎟⎠
Given
r=
Put
x− y 1 1
= −
xy
y x
s=
and
z−x 1 1
= −
zx
x z
∴ u is a function of r, s and r and s are functions of x, y, z
We know du =
∴
∂u
∂u
dr +
ds
∂r
∂s
∂u ∂u ∂r ∂u ∂s
∂u ∂u ∂r ∂u ∂s
=
⋅ + ⋅
,
=
⋅ + ⋅
,
∂x ∂r ∂x ∂s ∂x
∂y ∂r ∂y ∂s ∂y
But
r=
1 1
−
y x
∴
and
s=
1 1
−
x z
∴
∴
and
∂u ∂u ∂ r ∂ u ∂ s
=
⋅ + ⋅
∂z ∂ r ∂ z ∂ s ∂ z
∂r
∂r
∂r
1
1
=+ 2 ,
=− 2 ,
=0
∂x
∂y
∂z
x
y
∂s
∂s 1
∂s
1
=0
=
,
=− 2 ,
∂y
∂z z 2
∂x
x
∂u ∂u 1 ∂u ⎛ 1 ⎞
=
⋅ + ⎜− ⎟
∂x ∂r x 2 ∂s ⎝ x 2 ⎠
⇒
x2
∂u ∂u ∂u
=
−
∂x ∂r ∂s
(1)
∂u ∂u ⎛ 1 ⎞ ∂u
+ ⋅0
=
⋅ −
∂y ∂r ⎜⎝ y 2 ⎟⎠ ∂s
⇒
y2
∂u
∂u
=−
∂y
∂r
(2)
∂u ⎛ ∂u
∂u 1 ⎞
= ⎜ ⋅0 + ⋅ 2 ⎟
∂z ⎝ ∂r
∂s z ⎠
⇒
z2
∂u ∂u
=
∂z ∂s
(3)
(1) + (2) + (3) ⇒ x 2 ∂u + y 2 ∂u + z 2 ∂u = ∂u − ∂u − ∂u + ∂u = 0
∂x
∂y
∂y ∂r ∂s ∂r ∂s
M05_ENGINEERING_MATHEMATICS-I _CH05_Part A.indd 19
5/12/2016 10:25:22 AM
5.20
■
Engineering Mathematics
EXAMPLE 6
If z is a function of x and y and x 5 u cos a 2 v sin a, y 5 u sin a 1 v cos a, then show that
∂2 z ∂2 z ∂2 z ∂2 z
1
5
1
.
∂x 2 ∂y 2 ∂u 2 ∂v 2
Solution.
Given z is a composite function of u and v and x = u cos a − v sin a, y = u sin a + v cos a
∂z
∂z
dx + dy
∂x
∂y
We have
dz =
∴
∂z ∂z ∂x ∂z ∂y
=
+
∂u ∂x ∂u ∂y ∂u
(1)
and
∂z ∂z ∂x ∂z ∂y
=
+
∂v ∂x ∂v ∂y ∂v
(2)
∂x
∂x
= cos a,
= − sin a ,
∂u
∂v
∂z ∂z
∂z
=
cos a + sin a
∂u ∂x
∂y
But,
∴
operator
∂
∂
∂
= cos a + sin a
∂u
∂x
∂y
(3)
∂z ∂z
∂z
∂
∂⎞
⎛
( − sin a) + cos a = ⎜ − sin a + cos a ⎟ z
=
⎝
∂v ∂x
∂y
∂x
∂y ⎠
and
∴
∂y
= cos a
∂v
∂z
∂z
∂z ⎛
∂
∂⎞
= cos a + sin a
= cos a + sin a ⎟ z
∂u
∂x
∂y ⎜⎝
∂x
∂y ⎠
⇒
∴
∂y
= sin a,
∂u
operator
∂ ⎛
∂
∂⎞
= ⎜ − sin a + cos a ⎟
∂v ⎝
∂x
∂y ⎠
(4)
Now
∂2 z
∂ ⎛ ∂z ⎞ ⎛
∂
∂ ⎞⎛
∂z
∂z ⎞
=
⎜⎝ ⎟⎠ = ⎜ cos a + sin a ⎟ ⎜ cos a + sin a ⎟
2
⎝
⎠
∂u ∂u
∂x
∂y ⎝
∂x
∂y ⎠
∂u
= cos 2 a
∂2 z
∂2 z
∂2 z
∂2 z
+ cos a sin a
+ sin a cos a
+ sin 2 a 2
2
∂x ∂y
∂y ∂x
∂y
∂x
∂2 z
∂2 z
∂2 z
∂2 z
2
2
=
cos
a
+
2
sin
a
cos
a
+
sin
a
∂x ∂y
∂u 2
∂x 2
∂y 2
and
⎡
∂2 z ⎤
∂2 z
=
⎢assuming
⎥
∂x ∂y ∂y ∂x ⎦
⎣
(5)
∂2 z
∂ ⎛ ∂z ⎞ ⎛
∂
∂ ⎞⎛
∂z
∂z ⎞
=
⎜ ⎟ = − sin a + cos a ⎟ ⎜ − sin a + cos a ⎟
∂y ⎠
∂x
∂y ⎠ ⎝
∂x
∂v 2 ∂v ⎝ ∂v ⎠ ⎜⎝
= sin 2 a
M05_ENGINEERING_MATHEMATICS-I _CH05_Part A.indd 20
∂2 z
∂2 z
∂2 z
∂2 z
2
−
sin
cos
−
sin
cos
+
cos
a
a
a
a
a
∂x ∂y
∂y ∂x
∂x 2
∂y 2
5/12/2016 10:25:27 AM
Differential Calculus of Several Variables ■
∂2 z
∂2 z
∂2 z
∂2 z
2
2
=
sin
a
−
2
sin
a
cos
a
+
cos
a
∂x ∂y
∂v 2
∂x 2
∂y 2
5.21
(6)
∂2 z ∂2 z
∂2 z
∂2 z
2
2
2
2
+
=
(cos
a
+
sin
a
)
+
(sin
a
+
cos
a
)
∂u 2 ∂v 2
∂x 2
∂y 2
(5) + (6) ⇒
=
∂2 z ∂2 z
+
.
∂ x 2 ∂y 2
EXAMPLE 7
If z 5 f ( x , y ) where x 5 u 2 2 v 2 , y 5 2uv, prove that
Solution.
Given
z = f ( x, y )
Since z is a function of x, y,we have dz =
⎛ ∂2 z ∂2 z ⎞
∂2 z ∂2 z
1 2 5 4( u2 1 v 2 ) ⎜ 2 1 2 ⎟ .
2
∂u
∂v
∂y ⎠
⎝ ∂x
where x = u 2 − v 2, y = 2uv
∂z
∂z
dx + dy
∂x
∂y
(1)
But x and y are function of u and v.
So, z is ultimately a function of u and v.
∂z ∂z ∂x ∂z ∂y
=
+
∂u ∂x ∂u ∂y ∂u
∴
But
∴
∴
⇒
x = u2 − v2
and
y = 2uv
∂x
∂y
= 2u
and
= 2v
∂u
∂u
∂z ∂z
∂z
∂z
∂z ⎛
∂
∂⎞
=
⋅ 2u + ⋅ 2v = 2u + 2v
= ⎜ 2u + 2v ⎟ z
∂y ⎠
∂u ∂x
∂y
∂x
∂y ⎝ ∂x
∂
∂
∂
= 2u + 2v
∂u
∂x
∂y
Now
∂2 z
∂ ⎛ ∂z ⎞
∂ ⎛ ∂z
∂z ⎞
∂ ⎛ ∂z ⎞
∂ ⎛ ∂z ⎞
=
2u + 2v ⎟ = 2 ⎜ u ⎟ + 2 ⎜ v ⎟
⎜ ⎟=
∂y ⎠
∂u ⎝ ∂y ⎠
∂u ⎝ ∂x ⎠
∂u 2 ∂u ⎝ ∂u ⎠ ∂u ⎜⎝ ∂x
⇒
⎧ ∂z
⎧ ∂z
∂2 z
∂ ⎛ ∂z ⎞ ⎫
∂ ⎛ ∂z ⎞ ⎫
= 2 ⎨ ⋅1 + u ⋅ ⎜ ⎟ ⎬ + 2 ⎨ ⋅ 0 + v ⎜ ⎟ ⎬
2
⎝
⎠
∂u ∂x ⎭
∂u ⎝ ∂y ⎠ ⎭
∂u
⎩ ∂x
⎩ ∂y
=2
∂z
∂ ⎛ ∂z ⎞
∂ ⎛ ∂z ⎞
+ 2u ⎜ ⎟ + 2v ⎜ ⎟
∂x
∂u ⎝ ∂x ⎠
∂u ⎝ ∂y ⎠
=2
⎡ ∂
⎡ ∂
∂z
∂ ⎤ ⎛ ∂z ⎞
∂ ⎤ ⎛ ∂z ⎞
+ 2u ⎢ 2u + 2v ⎥ ⎜ ⎟ + 2v ⎢ 2u + 2v ⎥ ⎜ ⎟
∂x
∂y ⎦ ⎝ ∂x ⎠
∂y ⎦ ⎝ ∂y ⎠
⎣ ∂x
⎣ ∂x
=2
⎧ ∂ ⎛ ∂z ⎞
⎧ ∂ ⎛ ∂z ⎞
∂z
∂ ⎛ ∂z ⎞ ⎫
∂ ⎛ ∂z ⎞ ⎫
+ 2u ⎨2u ⎜ ⎟ + 2v ⎜ ⎟ ⎬ + 2v ⎨2u ⎜ ⎟ + 2v ⎜ ⎟ ⎬
∂x
∂y ⎝ ∂y ⎠ ⎭
∂y ⎝ ∂x ⎠ ⎭
⎩ ∂x ⎝ ∂x ⎠
⎩ ∂x ⎝ ∂y ⎠
M05_ENGINEERING_MATHEMATICS-I _CH05_Part A.indd 21
5/12/2016 10:25:32 AM
■
5.22
∴
Engineering Mathematics
2
2
∂2 z
∂z
∂2 z
∂2 z
2 ∂ z
2 ∂ z
2
4
u
4
uv
4
uv
4
v
=
+
+
+
+
∂x
∂y ∂x
∂x ∂y
∂u 2
∂x 2
∂y 2
Also ∂z = ∂z ⋅ ∂x + ∂z ⋅ ∂y
∂v ∂x ∂v ∂y ∂v
∴
and
∂x
= −2v,
∂v
(1)
∂y
= 2u
∂v
∂z ∂z
∂z
∂z
∂z ⎛
∂
∂⎞
=
( −2v ) + 2u = −2v + 2u
= ⎜ −2v + 2u ⎟ z
∂v ∂x
∂y
∂x
∂y ⎝
∂x
∂y ⎠
∂
∂
∂
= −2v + 2u
∂v
∂x
∂y
∂2 z
∂ ⎛ ∂z ⎞ ∂ ⎛
∂ ⎛ ∂z ⎞
∂z
∂z ⎞
∂ ⎛ ∂z ⎞
Now
=
−2v + 2u ⎟ = −2 ⎜ v ⎟ + 2 ⎜ u ⎟
⎜⎝ ⎟⎠ =
2
⎜
∂v ∂v
∂v ⎝ ∂y ⎠
∂v ⎝
∂x
∂y ⎠
∂v ⎝ ∂x ⎠
∂v
∴
⇒
⇒
⎧ ∂z
⎧ ∂z
∂2 z
∂ ⎛ ∂z ⎞ ⎫
∂ ⎛ ∂z ⎞ ⎫
= −2 ⎨ ⋅1 + v ⋅ ⎜ ⎟ ⎬ + 2 ⎨ ⋅ 0 + u ⎜ ⎟ ⎬
2
∂v ⎝ ∂x ⎠ ⎭
∂v ⎝ ∂y ⎠ ⎭
∂v
⎩ ∂x
⎩ ∂y
= −2
∂ ⎛ ∂z ⎞
∂z
∂ ⎛ ∂z ⎞
− 2v ⎜ ⎟ + 2u ⎜ ⎟
∂v ⎝ ∂x ⎠
∂x
∂v ⎝ ∂y ⎠
= −2
⎧
⎧
∂ ⎫ ⎛ ∂z ⎞
∂
∂
∂ ⎫ ⎛ ∂z ⎞
∂z
− 2v ⎨ −2v + 2u ⎬ ⎜ ⎟ + 2u ⎨ −2v + 2u ⎬ ⎜ ⎟
∂x
∂y ⎭ ⎝ ∂x ⎠
∂x
∂y ⎭ ⎝ ∂y ⎠
∂x
⎩
⎩
= −2
∂ ⎛ ∂z ⎞
∂z
∂ ⎛ ∂z ⎞
∂ ⎛ ∂z ⎞
∂ ⎛ ∂z ⎞
+ 4v 2 ⎜ ⎟ − 4uv ⎜ ⎟ − 4uv ⎜ ⎟ + 4u 2 ⎜ ⎟
∂y ⎝ ∂y ⎠
∂x
∂x ⎝ ∂x ⎠
∂y ⎝ ∂x ⎠
∂x ⎝ ∂y ⎠
2
2
∂2 z
∂z
∂2 z
∂2 z
2 ∂ z
2 ∂ z
=
−
+
−
−
+
2
4
4
4
4
v
uv
uv
u
∂x
∂y ∂x
∂x ∂y
∂v 2
∂x 2
∂y 2
(2)
Adding (1) and (2), we get
2
2
∂2 z ∂2 z
2
2 ∂ z
2
2 ∂ z
(
u
v
)
u
v
+
=
+
+
+
4
4
(
4
4
)
∂u 2 ∂v 2
∂x 2
∂y 2
⎛ ∂2 z ∂2 z ⎞
= 4(u 2 + v 2 ) ⎜ 2 + 2 ⎟
∂y ⎠
⎝ ∂x
Note The same problem is asked as below also. If g ( x, y ) = c(u, v ) where u = x 2 − y 2 and v = 2xy,
2
2
2
⎛ 2
⎞
then prove that ∂ g + ∂ g = 4( x 2 + y 2 ) ∂ c + ∂ c .
2
2 ⎟
2
2
⎜
∂x
∂y
∂v ⎠
⎝ ∂u
EXAMPLE 8
Transform the equation zxx 1 2 zxy 1 zyy 5 0 by changing the independent variables using
u 5 x 2 y and v 5 x 1 y.
Solution.
Given
M05_ENGINEERING_MATHEMATICS-I _CH05_Part A.indd 22
zxx + 2zxy + zyy = 0
(1)
5/12/2016 10:25:37 AM
Differential Calculus of Several Variables ■
5.23
and
u = x − y, v = x + y
In the given equation independent variables are x and y. We have to change them to u and v.
So, we treat z as a function of u and v
∴
∴
Now
∂z
∂z
∂z ∂z ∂u ∂z ∂v
∂u + dv ⇒
=
⋅ + ⋅
∂u
∂v
∂x ∂u ∂x ∂v ∂x
∂z ∂z ∂z ⎛ ∂
∂⎞
∂
∂
∂
=
+
=⎜ + ⎟z ⇒
=
+
⎝
⎠
∂x ∂u ∂v
∂u ∂v
∂x ∂u ∂v
dz =
∂ ⎛ ∂z ⎞ ∂ ⎛ ∂z ⎞ ∂ ⎛ ∂z ⎞ ∂ ⎛ ∂z ⎞
⎜ ⎟+ ⎜ ⎟+ ⎜ ⎟+ ⎜ ⎟
∂u ⎝ ∂u ⎠ ∂u ⎝ ∂v ⎠ ∂v ⎝ ∂u ⎠ ∂v ⎝ ∂v ⎠
∂2 z ∂2 z ∂2 z
∂2 z ∂2 z
=
+
+
+
∂ x 2 ∂u 2 ∂u ∂v ∂v ∂u ∂v 2
(2)
∂u
= −1,
∂y
∂u
=1
∂y
Now
∂z ∂z ∂u ∂z ∂v
=
+
∂y ∂u ∂y ∂v ∂y
∴
∂z
∂z ∂z ⎛ ∂
∂⎞
∂
∂
∂
=− +
= ⎜− + ⎟ z ⇒
=− +
∂y
∂u ∂v ⎝ ∂u ∂v ⎠
∂y
∂u ∂v
∴
and
∂ ⎛ ∂z ⎞ ∂ ⎛ ∂z ⎞ ∂ ⎛ ∂z ⎞ ∂ ⎛ ∂z ⎞
⎜− ⎟ − ⎜ ⎟ + ⎜− ⎟ + ⎜ ⎟
∂u ⎝ ∂u ⎠ ∂u ⎝ ∂v ⎠ ∂v ⎝ ∂ u ⎠ ∂ v ⎝ ∂ v ⎠
∂2 z ∂2 z ∂2 z
∂2 z ∂2 z
+
=
−
−
∂y 2 ∂u 2 ∂u ∂v ∂v ∂u ∂v 2
(3)
∂2 z
∂ ⎛ ∂z ⎞ ⎛ ∂
∂ ⎞ ⎛ ∂z ∂z ⎞
=
= ⎜ + ⎟ ⎜−
+ ⎟
⎜
⎟
⎝
∂x ∂y ∂x ⎝ ∂y ⎠
∂u ∂v ⎠ ⎝ ∂u ∂v ⎠
=
⇒
and
∂2 z
∂ ⎛ ∂z ⎞ ⎛ ∂
∂ ⎞ ⎛ ∂z ∂z ⎞
=
= ⎜− + ⎟ ⎜− + ⎟
2
⎜
⎟
∂y ⎝ ∂y ⎠ ⎝ ∂u ∂v ⎠ ⎝ ∂u ∂v ⎠
∂y
=−
⇒
∂u
∂v
= 1,
=1
∂x
∂x
∂2 z
∂ ⎛ ∂z ⎞ ⎛ ∂
∂ ⎞ ⎛ ∂z ∂z ⎞
=
⎜⎝ ⎟⎠ = ⎜⎝ + ⎟⎠ ⎜⎝ + ⎟⎠
2
∂x ∂x
∂u ∂v ∂u ∂v
∂x
=
⇒
and
∂ ⎛ ∂z ⎞ ∂ ⎛ ∂z ⎞ ∂
⎜− ⎟ +
⎜ ⎟+
∂u ⎝ ∂u ⎠ ∂u ⎝ ∂v ⎠ ∂v
⎛ ∂z ⎞ ∂ ⎛ ∂z ⎞
⎜⎝ − ⎟⎠ +
⎜ ⎟
∂u
∂v ⎝ ∂v ⎠
∂2 z
∂2 z ∂2 z
∂2 z ∂2 z
=− 2+
−
+
∂x ∂y
∂u ∂v ∂v ∂u ∂v 2
∂u
(4)
Substituting (2), (3) and (4) in (1), we get
⎛ ∂2 z
∂2 z
∂2 z
∂2 z
∂2 z
∂2 z
∂2 z
∂2 z ⎞
−
+ 2⎟
+
+
+ 2 + 2⎜ − 2 +
2
∂u ∂v ∂v ∂u ∂v
∂ u ∂ v ∂ v ∂ u ∂v ⎠
∂u
⎝ ∂u
+
M05_ENGINEERING_MATHEMATICS-I _CH05_Part A.indd 23
∂2 z
∂2 z
∂2 z
∂2 z
−
−
+
=0
∂u 2 ∂u ∂v ∂v ∂u ∂v 2
5/12/2016 10:25:43 AM
5.24
■
Engineering Mathematics
⇒
4
∂2 z
∂2 z
∂2 z
2
2
+
−
=0
∂v 2
∂u∂v
∂v∂u
∂2 z ∂2 z
∂2 z
+
−
= 0 ⇒ 2 zvv + zvu − zuv = 0
∂v 2 ∂u ∂v ∂v ∂u
which is the transformed equation.
⇒
2
Note In general z uv ≠ z vu . If z uv = z vu , then z vv = 0
EXERCISE 5.2
2
2
1. If z = x3 + y3 − 3axy, then show that ∂ z = ∂ z .
∂x ∂y ∂y ∂x
2. If u = x 2 tan −1
y
x
∂2u
x2 − y2
− y 2 tan −1 , then show that
= 2
.
x
y
∂x ∂y x + y 2
x2 + y2
∂z ∂z
∂z ∂z
, then show that ⎛⎜ − ⎞⎟ = 4 ⎛⎜1 − − ⎞⎟ .
x+ y
⎝ ∂x ∂y ⎠
⎝ ∂x ∂y ⎠
2
3. If z =
4. If u = f(r), where r =
x 2 + y 2 , prove that
∂2u ∂2u
1
+ 2 = f ′′( r ) + f ′( r ).
2
r
∂x
∂y
5. If z = f(x + ct) + g (x − ct), where c is a constant, prove that
6. If z = sin −1
∂z
∂z
x
y
+ tan −1 , then prove that x + y
= 0.
y
x
∂x
∂y
∂2 z
∂2 z
= c2 2 .
2
∂t
∂x
7. If z =
∂z
∂z
1
= −2 z .
, then prove that x + y
2
∂x
∂y
x + y + xy
8. If u =
x 3 y − xy 3
∂2u
∂2u
∂2u
, then prove that x 2 2 + 2 xy
+ y 2 2 = 6 xy( x − y ).
x+ y
∂x ∂y
∂x
∂y
2
⎛ x3 + y3 ⎞
9. If u = log e ⎜
, then prove that x ∂u + y ∂u = 2.
⎝ x − y ⎟⎠
∂x
∂y
⎛ x2 + y2 ⎞
10. If u = sin −1 ⎜
, then prove that x ∂u + y ∂u = tan u.
⎝ x + y ⎟⎠
∂x
∂y
11. If u = (x − y)4 + (y − z)4 + (z − x)4, then find the value of ∂u + ∂u + ∂u .
∂x ∂y ∂z
x2 y2
∂u
∂u
12. If sin u =
, then show that x + y
= 3tan u.
∂x
∂y
x+ y
13. If u =
x y z
∂u
∂u
∂u
+ + , then show that x + y + z
= 0.
y z x
∂x
∂y
∂z
M05_ENGINEERING_MATHEMATICS-I _CH05_Part A.indd 24
5/12/2016 10:25:52 AM
Differential Calculus of Several Variables ■
5.25
⎛ x3 + y 3 ⎞
∂u
∂u
= sin 2u.
, then prove that x + y
14. If u = tan −1 ⎜
∂x
∂y
⎝ x − y ⎟⎠
y
15. If u = log( x 2 + y 2 ) + tan −1 , then prove that uxx + uyy = 0.
x
∂ 2u
x2 − y2
2
−1 y
2
−1 x
− y tan
16. If u = x tan
, then prove that
.
= 2
x
y
∂x ∂y x + y 2
17. If u = x2 + y2 + z2 and x = e2t, y = e2t cos 3t, z = e2t sin 3t, then find
du
as a total derivative.
dt
du
.
dt
If u = f(r, s) where r = x + at, s = y + bt and x, y, t are independent variables, then show that
∂u
∂u
∂u
= a +b .
∂t
∂x
∂y
du
.
If u = x2 + y2 + z2 and x = et, y = et sin t, z = et cos t, then find
dt
∂z
∂z
∂z
∂z
= x + 2y .
If z = f(x, y), where x = u + v, y = uv, then prove that u + v
∂u
∂v
∂x
∂y
If u = f(x2 + 2yz, y2 + 2zx), then prove that
18. If u = x3 + y3, x = a cos t, y = b sin t, then find
19.
20.
21.
22.
( y 2 − zx )
∂u
∂u
∂u
+ ( x 2 − yz ) + ( z 2 − xy )
= 0.
∂x
∂y
∂z
1 ∂u 1 ∂u 1 ∂u
+
+
= 0.
2 ∂x 3 ∂y 4 ∂z
2
2
∂z
If z = log (u2 + v) where u = e x + y , v = x 2 + y , then find
and ∂z .
∂x
∂y
dy
y
(
y
−
x
log
y
)
x
y
, using partial derivative method.
If y = x , then show that
=
dx x( x − y log x )
If u = f(x2 − y2, y2 − z2, z2 − x2), then prove that 1 ∂u + 1 ∂u + 1 ∂u = 0 .
x ∂x y ∂y z ∂z
du
2
2
t
Find
when
u
=
x
y,
x
=
t
,
y
=
e
,
dt
du
1
If u = xy + yz + zx where x = , y = e t , and z = e − t, then find .
dt
t
23. If u = f(2x − 3y, 3y − 4z, 4z − 2x), then prove that
24.
25.
26.
27.
28.
29. Find
x
du
t
2
, when u = sin , x = e , y = t .
y
dt
∂z ∂z
∂z
∂z
−
=x
−y
.
∂u ∂v
∂x
∂y
∂z
∂z
∂z
31. If z = f(x, y), where x = eu cos v, y = eu sin v, show that y
.
+x
= e 2u
∂u
∂v
∂y
∂v
∂u
32. If u2 + 2v2 = 1 − x2 + y2 and u2 + v2 = x2 + y2 − 2, then find
and .
∂x
∂x
dz
2
2
3
3
2
33. If z = x + y and x + y + 3axy = 5a , then find the value of
when x = y = a.
dx
30. If z is a function of x and y and x = eu 1 e2v, y 5 e2u 2 ev, then show that
M05_ENGINEERING_MATHEMATICS-I _CH05_Part A.indd 25
5/12/2016 10:26:02 AM
■
5.26
Engineering Mathematics
ANSWERS TO EXERCISE 5.2
28. 22 [t sinh t − cosh t ]
t
⎛ et ⎞
et
29. 3 cos ⎜ 2 ⎟ [t − 2]
⎝t ⎠
t
11. 0
17. 8e4t
18. 3 sin t cos t (b3 sin t − a3 cos t)
20. 4e2t
32.
24. 2 x( 2u + 1) , 4 yu + 1
u2 + v
u2 + v
27. t3(4 + t) et
2
5.3
2
∂u 3 x ∂v −2 x
= ,
=
∂x u ∂x
v
33. 0
JACOBIANS
Jacobians have many important applications such as functional dependence, transformation of variable in multiple integrals, problems in partial differentiation and in the study of existence of implicit
functions determined by a system of functional equations.
Definition 5.8
(1) If u and v are continuous functions of two independent variables x and y, having first order partial
∂u ∂u
derivatives, then the determinant ∂x ∂y is called the Jacobian determinant or Jacobian of u
∂v ∂v
∂x ∂y
u, v ⎞
and v with respect to x and y and is denoted by ∂ (u, v ) or J ⎛⎜
or J.
⎝ x, y ⎟⎠
∂ ( x, y )
Thus,
∂u
∂ ( u , v ) ∂x
=
∂ ( x , y ) ∂v
∂x
∂u
∂y
∂v
∂y
(2) If u, v, w are continuous functions of three independent variables x, y, z having first order partial
derivatives then the Jacobian of u, v, w with respect to x, y, z is defined as
∂u
∂x
∂( u , v , w )
∂v
=
∂( x , y , z )
∂x
∂w
∂x
∂u
∂y
∂v
∂y
∂w
∂y
∂u
∂z
∂v
∂z
∂w
∂z
Similarly, we can define Jacobians for functions of 4 or more variables.
M05_ENGINEERING_MATHEMATICS-I _CH05_Part A.indd 26
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Differential Calculus of Several Variables ■
5.27
5.3.1 Properties of Jacobians
For simplicity we shall prove the properties of Jacobians for two variables. However, they can be
extended to any number of variables.
∂( u , v ) ∂( x , y )
⋅
=1
Property 1 If u and v are functions of x and y, then
∂( x , y ) ∂( u , v )
Proof Let u = f1(x, y) and v = f2(x, y) be continuous functions of two independent variables x and y
∂( u , v )
having first order partial derivatives then J =
∂( x , y )
The condition for these equations to be solvable for x and y is J ≠ 0.
If x = g1(u, v) and y = g2(u, v) and J ′ =
u = f1 ( x, y )
We have
∴
∂( x , y )
, then to prove JJ′ = 1
∂( u , v )
du =
∂u
∂u
dx +
dy
∂x
∂y
(1)
and
dv =
∂v
∂v
dx + dy
∂x
∂y
(2)
Since u and v are independent variables from differentials (1) and (2) we get
∂u
∂u
∂v
∂v
= 1,
= 0,
= 1,
=0
∂u
∂v
∂v
∂u
∴
∂u ∂x ∂u ∂y
∂u ∂x ∂u ∂y
+
= 1 and
+
=0
∂x ∂u ∂y ∂u
∂x ∂v ∂y ∂v
∂v ∂x ∂v ∂y
+
= 1 and
∂x ∂v ∂y ∂v
Now
∂u
∂x
JJ′ =
∂v
∂x
∂u
∂y
∂v
∂y
∂v ∂x ∂v ∂y
+
=0
∂x ∂u ∂y ∂u
∂x
∂u
∂y
∂u
∂x
∂v
∂y
∂v
∂u ∂x ∂u ∂y
⋅ +
∂x ∂u ∂y ∂u
=
∂v ∂x ∂v ∂y
⋅ + ⋅
∂x ∂u ∂y ∂u
∂u ∂x ∂u ∂y
⋅ +
1 0
∂ x ∂v ∂y ∂v
=
=1
∂v ∂x ∂v ∂y
0 1
⋅ +
∂x ∂v ∂y ∂v
■
Property 2 Jacobians of composite functions or chain rule
If u and v are functions of p and q, where p and q are functions of x and y, then ∂(u, v ) = ∂(u, v ) ⋅ ∂( p, q)
∂ ( x , y ) ∂ ( p, q ) ∂ ( x , y )
Proof If u and v are continuous functions of p and q and p and q are functions of x and y, then
du =
∂u
∂u
∂v
∂v
dp + ⋅ dq and dv =
dp + ⋅ dq
∂p
∂q
∂p
∂q
M05_ENGINEERING_MATHEMATICS-I _CH05_Part A.indd 27
5/12/2016 9:35:46 AM
5.28
■
Engineering Mathematics
∂u ∂u ∂p ∂u ∂q
∂v ∂v ∂p ∂v ∂q
=
⋅ + ⋅
and
=
⋅ + ⋅
∂x ∂p ∂x ∂q ∂x
∂x ∂p ∂x ∂q ∂x
∴
∂u ∂u ∂p ∂u ∂q
∂v ∂v ∂p ∂v ∂q
=
⋅ + ⋅
and
=
⋅ + ⋅
∂y ∂p ∂y ∂q ∂y
∂y ∂p ∂y ∂q ∂y
∂u
∂(u, v ) ∂( p, q) ∂p
∴
⋅
=
∂( p, q) ∂( x, y ) ∂v
∂p
∂u
∂q
∂v
∂q
∂p
∂x
∂q
∂x
∂p
∂y
∂q
∂y
∂u ∂p ∂u ∂q
⋅ + ⋅
∂p ∂x ∂q ∂x
=
∂v ∂p ∂v ∂q
⋅ + ⋅
∂p ∂x ∂q ∂x
∂u
∂u ∂p ∂u ∂q
+
∂x
∂p ∂y ∂q ∂y
=
∂v
∂v ∂p ∂v ∂q
+
∂p ∂y ∂q ∂y
∂x
∂u
∂y
∂( u , v )
=
∂v ∂( x , y )
∂y
∂ ( u , v ) ∂ ( u , v ) ∂ ( p, q )
=
⋅
∂ ( x , y ) ∂ ( p, q ) ∂ ( x , y )
∴
■
Note Extension to three variables
(1) ∂(u, v, w ) ⋅ ∂( x, y, z ) = 1
∂( x , y , z ) ∂( u , v , w )
and
(2) ∂(u, v, w ) = ∂(u, v, w ) ⋅ ∂( p, q, r )
∂( x, y, z ) ∂( p, q, r ) ∂( x, y, z )
If u and v are functions of two independent variables x and y and u and v are functionally
∂( u , v )
= 0.
dependent [i.e., f(u, v) = 0], then
∂( x , y )
Proof If u and v are not independent, then there is a relation between u and v. Let f(u, v) = 0 be the
relation between u and v.
Differentiating with respect to x and y we have,
Property 3
∂f
∂f
∂u + ∂v = 0
∂u
∂v
∴
∂f ∂u ∂f ∂v
+ ⋅ =0
∂u ∂x ∂v ∂x
Eliminating
(1)
∂f ∂f
from (1) and (2), we get
,
∂u ∂v
∂u ∂u
∂x ∂y
=0
∂v ∂v
∂x ∂y
and
∂f ∂u ∂f ∂v
⋅ + ⋅ =0
∂u ∂y ∂v ∂y
⇒ ∂( u , v ) = 0
∂( x , y )
(2)
■
Note
(1) The converse of property 3 is also true.
(2) The property can be extended to functions of more than two variables.
M05_ENGINEERING_MATHEMATICS-I _CH05_Part A.indd 28
5/12/2016 9:35:51 AM
Differential Calculus of Several Variables ■
If x = f(u, v), y = g(u, v) and h(x, y) = F(u, v), then
Property 4
∫∫ h( x, y) dx dy = ∫∫ F (u, v) J
∂( x , y )
is the Jacobian of transformation.
∂( u , v )
Similarly, if x, y, z are functions of u, v, w, then the Jacobian of transformation J =
∫∫∫ f ( x, y, z ) dx dy dz = ∫∫∫ F(u, v, w ) J
R
du dv ,
R′
R
where dx dy = J du dv and J =
5.29
∂( x , y , z )
and
∂( u , v , w )
du dv dw
R
WORKED EXAMPLES
EXAMPLE 1
If u 5 x2 1 1, v 5 y2 2 2, then find ∂( u, v ) .
∂( x, y )
Solution.
∂u ∂u
∂x ∂y
∂
(
u
,
v
)
We have
=
∂ ( x , y ) ∂v ∂v
∂x ∂y
Given
u = x2 + 1
∴
∂u
= 2 x and
∂x
∂u
=0
∂y
and
v = y2 − 2
∴
∂v
= 0 and
∂x
∂v
= 2y
∂y
∂ ( u, v ) 2 x 0
=
= 4 xy.
0 2y
∂ ( x, y )
∴
EXAMPLE 2
If x 5 r cos u, y 5 r sin u, then find the Jacobian of x and y with respect to r and u.
Solution.
∂x
(
x
,
y
)
∂
∂r
The Jacobian of x and y with respect to r and u is
=
∂ ( r , u) ∂y
∂r
Given
x = r cos u
∴
and
y = r sin u
∴
∂x
∂u
∂y
∂u
∂x
∂x
= cos u and
= − r sin u
∂r
∂u
∂y
∂y
= sin u and
= r cos u
∂r
∂u
cos u − r sin u
∴ ∂( x , y ) =
= r cos 2 u + r sin 2 u = r (cos 2 u + sin 2 u) = r
∂( r , u) sin u r cos u
M05_ENGINEERING_MATHEMATICS-I _CH05_Part A.indd 29
5/12/2016 9:35:58 AM
5.30
■
Engineering Mathematics
Note x = r cos u, y = r sin u transforms cartesian coordinates into polar coordinates.
dx dy = J dr d u = r dr d u
∫∫ dx dy = ∫∫ r dr d u
R
by property 4.
R
EXAMPLE 3
If x 5 u (1 2 v), y 5 uv, then compute J and J′ and prove that JJ′ 51.
Solution.
∂( x , y )
,
∂( u , v )
We know
J=
Given
x = u(1 − v)
and
∴
J′ =
∂( u , v )
∂( x , y )
To prove JJ′ = 1
∂x
∂x
= 1 − v and
= −u
∂u
∂v
∂y
∂y
=u
=v
and
∴
y = uv
∂v
∂u
∂x ∂x
∂( x , y ) ∂u ∂v 1 − v − u
J=
=
=
= u(1 − v ) + uv = u − uv + uv = u
v
u
∂( u , v ) ∂y ∂y
∂u ∂v
∴
∂u ∂u
∂( u , v ) ∂x ∂y
=
Now to find J ′ =
∂( x , y ) ∂v ∂v
∂x ∂y
We have to find u and v in terms of x and y.
We have x = u(1 – v) = u – uv and y = uv
∂u
∴
x=u–y ⇒ u=x+y
∴
= 1 and
∂x
y
y
and
y = uv
⇒ v= =
u x+ y
∂v
y
∴
=−
=1
∂x
( x + y)2
∂v ( x + y ) ⋅1 − y ⋅1
x
=
=
and
2
∂y
( x + y)
( x + y)2
∂u ∂u
∂x ∂y
∂( u , v )
=
∴
J′ =
∂( x , y ) ∂v ∂v
∂x ∂y
=
∴
1
−y
( x + y)2
JJ′ = u ⋅
∂u
=1
∂y
⎡ d ⎛ 1 ⎞
1 ⎤
⎢{ dx ⎜⎝ x + a ⎟⎠ = −
⎥
( x + a) 2 ⎦
⎣
1
x
1
1
y
x+ y
=
+
=
=
=
x
2
2
2
x
y
u
+
x
+
y
(
x
y
)
(
x
y
)
(
)
+
+
( x + y)2
1
= 1.
u
M05_ENGINEERING_MATHEMATICS-I _CH05_Part A.indd 30
5/12/2016 9:36:05 AM
Differential Calculus of Several Variables ■
5.31
EXAMPLE 4
If u 5 2xy, v 5 x2 2 y2, x 5 r cos u, y 5 r sin u, evaluate ∂(u, v ) without actual substitution.
∂( r , u)
Solution.
Given u, v are functions of x and y and x and y are functions of r and u.
So, by property 2 of composite function
∂(u , v ) ∂(u , v ) ∂( x , y )
=
⋅
∂( r , u) ∂( x , y ) ∂( r , u)
Given
u = 2 xy
∴
∂u
= 2 y and
∂x
and
v = x2 − y2
∴
∂v
= 2x
∂x
∂u
∂( u , v ) ∂x
∴
=
∂( x , y ) ∂v
∂x
Since x = r cos u,
∂u
= 2x
∂y
∂v
= −2 y
∂y
and
∂u
∂y
2 y 2x
=
= −4 y 2 − 4 x 2 = −4( x 2 + y 2 )
∂v
2 x −2 y
∂y
y = r sin u
⇒
x2 + y2 = r2 cos2 u + r2 sin2 u = r2
∂( u , v )
= −4 r 2
∂( x , y )
∂( x , y )
From example 2, we have
=r
∂( r , u)
∴
∂( u , v )
= −4 r 2 ⋅ r = −4 r 3
∂( r , u)
∴
EXAMPLE 5
For the transformation x 5 r sin u cos f, y 5 r sin u sin f, z 5 r cos u, compute the Jacobian of
x, y, z with respect to r, u, f.
Solution.
The Jacobian of transformation is
∂x
∂r
∂( x , y , z )
∂y
J=
=
∂r
∂( r , u, f)
∂z
∂r
Given x = r sin u cos f
∂x
∴
= sin u cos f,
∂r
Z
∂x
∂u
∂y
∂u
∂z
∂u
∂x
∂f
∂y
∂f
∂z
∂f
∂x
= r cos u cos f,
∂u
M05_ENGINEERING_MATHEMATICS-I _CH05_Part A.indd 31
θ
r
P (x, y, z)
(r, θ, φ)
r sinθ
Y
φ
M
X
Fig. 5.3
∂x
= − r sin u sin f
∂f
5/12/2016 9:36:15 AM
■
5.32
y = r sin u sin f
and
∴
Engineering Mathematics
∂y
= sin u sin f,
∂r
z = r cos u
∂z
= cos u,
∂r
∂y
= r cos u sin f,
∂u
∂y
= r sin u cos f
∂f
∂z
= − r sin u,
∂u
∂z
=0
∂f
sin u cos f r cos u cos f − r sin u sin f
∴
J = sin u sin f r cos u sin f r sin u cos f
cos u
−r sin u
0
Expanding using third row, we get
J = cos u [r 2 cos u sin u cos 2 f + r 2 cos u sin u sin 2 f] + r sin u[r sin 2 u cos 2 f + r sin 2 u sin 2 f]
= r 2 sin u[cos 2 u(cos 2 f + sin 2 f] + r 2 sin u[sin 2 u(cos2 f + sin 2 f)]
= r 2 sin u[cos 2 u + sin 2 u] = r 2 sin u
Note This is the transformation of cartesian coordinates to spherical polar coordinates (r, u, f)
dx dy dz = J dr d u df = r 2 sin u ⋅ dr d u df
EXAMPLE 6
In cylindrical polar coordinates x 5 r cos f, y 5 r sin f, z 5 z, show that
Solution.
We have
Z
∂x
∂r
∂( x , y , z ) ∂y
=
∂(r, f, z ) ∂r
∂z
∂r
Given
and
∴
∴
∂( x , y , z )
5 r.
∂( r, f, z )
∂x
∂f
∂y
∂f
∂z
∂f
x = r cos f
∂x
= cos f,
∂r
∂x
∂z
∂y
∂z
∂z
∂z
P (x, y, z)
z (ρ, φ, z)
φ
X
∂x
= −r sin f,
∂f
∂x
=0
∂z
∂y
= r cos f,
∂f
∂y
=0
∂z
∂z
= 0,
∂f
∂z
=1
∂z
Y
ρ
M
Fig. 5.4
y = r sin f
∂y
= sin f,
∂r
z=z
∂z
= 0,
∂r
M05_ENGINEERING_MATHEMATICS-I _CH05_Part A.indd 32
5/12/2016 9:36:35 AM
Differential Calculus of Several Variables ■
∴
5.33
cos f −r sin f 0
∂( x , y , z )
= sin f r cos f 0
∂(r, f, z )
0
0
1
= r cos 2 f + r sin 2 f = r(cos 2 f + sin 2 f) = r
Note This is the transformation of cartesian coordinate to cylindrical coordinates (r, f, z).
dx dy dz = J d r d f dz = r d r d f dz
EXAMPLE 7
If u 5
∂( u, v , w)
yz
zx
xy
5 4.
,v5 ,w5
, then show that
x
y
z
∂( x , y , z)
Solution.
∂u
∂x
∂( u , v , w )
∂v
We have,
=
∂( x , y , z )
∂x
∂w
∂x
Given
∴
∂u
∂y
∂v
∂y
∂w
∂y
∂u
∂z
∂v
∂z
∂w
∂z
u=
yz
x
∴
∂u − yz
= 2 ,
∂x
x
∂u z
=
∂y x
and
∂u y
=
∂z x
v=
zx
y
∴
∂v z
= ,
∂x y
∂v
zx
= − 2 and
∂y
y
∂v x
=
∂z y
w=
xy
z
∴
∂w y
= ,
∂x z
∂w x
=
∂y z
∂w
xy
=− 2
∂z
z
− yz
x2
∂( u , v , w )
z
=
∂( x , y , z )
y
y
z
z
x
− zx
y2
x
z
and
y
x
x
y
− xy
z2
1
from I row
x2
1
1
from II row and 2 from III row]
2
y
z
[Take
1
= 2 2 2
x y z
− yz
zy
zy
M05_ENGINEERING_MATHEMATICS-I _CH05_Part A.indd 33
xz
xy
− zx xy
xz − xy
5/12/2016 9:36:41 AM
■
5.34
Engineering Mathematics
x2 y2 z2
= 2 2 2
x y z
−1
1
1
1 −1
1
1
1 −1
[Take yz from I column
zx from II column and xy from III column]
= −1( +1 − 1) − 1( −1 − 1) + 1(1 + 1) = 0 + 2 + 2 = 4
EXAMPLE 8
If u 5 x 1 y 1 z, uv 5 y 1 z, uvw 5 z, then find ∂( x , y , z) .
∂( u, v , w)
Solution.
Given
u = x + y + z,
uv = y + z,
uvw = z
∴
u = x + uv
⇒ x = u − uv
y = uv − z = uv − uvw
and
z = uvw
∂x
∂x
Now
x = u − uv
∴
= 1 − v,
= −u
∂u
∂v
∂y
∂y
= v − vw ,
= u − uw
y = uv − uvw ∴
∂u
∂v
∂z
∂z
and
∴
= vw ,
= uw ,
z = uvw
∂u
∂v
∂x
∂u
∂( x , y , z )
∂y
∴
=
∂( u , v , w )
∂u
∂z
∂u
∂x
∂v
∂y
∂v
∂z
∂v
∂x
∂w
1− v
−u
∂y
= v − vw u − uw
∂w
vw
uw
∂z
∂w
and
and
and
∂x
=0
∂w
∂y
= − uv
∂w
∂z
= uv
∂w
0
−uv
uv
1− v
−u
= uv v − vw u − uw
vw
uw
1 − v −u 0
= uv v
u 0
vw uw 1
0
−1
1
[Taking out uv from C3 ]
R 2 → R 2 + R3
= uv[(1 − v )u + uv ] = uv(u − uv + uv ) = u 2 v
[Expanding by C3 ]
EXAMPLE 9
If u 5
x1 y
, v 5 tan21 x 1 tan21 y find ∂( u, v ) .
12xy
∂( x, y )
Solution.
Given
u=
x+ y
1 − xy
M05_ENGINEERING_MATHEMATICS-I _CH05_Part A.indd 34
and
v = tan −1 x + tan −1 y
5/12/2016 9:36:47 AM
Differential Calculus of Several Variables ■
tan v = tan(tan −1 x + tan −1 y ) =
Now
5.35
x+ y
tan(tan −1 x ) + tan(tan −1 y )
=
=u
−1
−1
1 − tan(tan x ) ⋅ tan(tan y ) 1 − xy
∴ u and v are not independent, That is u and v are functionally dependent.
∴ by property 3,
∂( u , v )
= 0.
∂( x , y )
5.3.2 Jacobian of Implicit Functions
If y1, y2, y3, … yn are implicitly given as functions of x1, x2, …, xn by the functional equations
fi (x1, x2, … xn, y1, y2, y3, … yn) = 0 for i = 1, 2, … n, then
∂( f1 , f 2 , …, f n )
∂( f1 , f 2 , …, f n ) ∂( y1 , y2 , …, yn )
= ( −1) n
⋅
∂( x1 , x2 , …, xn )
∂( y1 , y2 , …, yn ) ∂( x1 , x2 , …, xn )
WORKED EXAMPLES
EXAMPLE 10
∂( F, G, H)
∂( x, y, z )
If F 5 xu 1 v 2 y, G 5 u2 1 vy 1 w, H 5 zu 2 v 1 uw compute (1)
( 2)
∂(u, v, w )
∂(u, v, w )
∂( F, G, H)
( 3)
.
∂( x, y, z )
Solution.
F = xu + v − y,
G = u2 + vy + w
and
These equations implicitly define x, y, z interms of u, v, w.
∴ by the Jacobian of implicit functions, we have
H = zu − v + vw
∂( F, G, H)
∂( F, G, H) ∂( x, y, z )
= ( −1)3
⋅
∂( u , v , w )
∂( x , y , z ) ∂( u , v , w )
⇒
Given
∴
and
F = xu + v − y
1
∂( x , y , z )
∂( F, G, H )
= ( −1)3
×
∂( u , v , w )
∂(u, v, w ) ∂( F, G, H)
∂( x , y , z )
∂F
∂F
∂F
= u,
= −1,
=0
∂x
∂y
∂z
G = u2 + vy + w
∂G
∂G
∂G
= 0,
= v,
=0
∂x
∂y
∂z
and
∂F
= x,
∂u
∂F
= 1,
∂v
∂F
=0
∂w
and
∂G
= 2u ,
∂u
∂G
= y,
∂v
∂G
=1
∂w
and
∂H
= z,
∂u
∂H
= −1 + w ,
∂v
∂H
=v
∂w
H = zu − v + vw
∂H
∂H
= 0,
= 0,
∂x
∂y
∂H
=u
∂z
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Engineering Mathematics
x
1
0
∂( F, G, H)
∴
= 2u
y
1
= x{vy − ( −1 + w )} − 1( 2uv − z ) = x(1 − w + vy ) + z − 2uv
∂( u , v , w )
z −1 + w v
∂F
∂x
∂( F, G, H) ∂G
∴
=
∂( x , y , z )
∂x
∂H
∂x
∂F
∂y
∂G
∂y
∂H
∂y
∂F
∂z
u −1 0
∂G
= 0 v 0 = u( vu ) = u 2 v
∂z
0 0 u
∂H
∂z
3
∴ ∂( x, y, z ) = ( −1) [ x( vy + 1 − w ) − 2uv + z ] = x[w − 1 − vy ] + 2uv − z
∂( u , v , w )
u2v
u2v
EXAMPLE 11
If x 1 y 1 z 2 u 5 0, y 1 z 1 uv 5 0, z 2 uvw 5 0 then find
∂( x, y, z )
.
∂(u, v, w )
Solution.
Given
x + y +z − u = 0,
y + z − uv = 0,
z − uvw = 0
These equations implicitly define x, y, z interms of u, v, w
Let f1 = x + y + z − u,
f2 = y + z − uv
and
f3 = z − uvw
To find ∂( x, y, z ) .
∂( u , v , w )
By the Jacobian of implicit functions, we have
∂( f1 , f 2 , f 3 )
∂( f1 , f 2 , f 3 ) ∂( x, y, z )
= ( −1)3
⋅
∂( u , v , w )
∂ ( x , y , z ) ∂( u , v , w )
∂( x , y , z )
= ( −1)3
∂( u , v , w )
∴
We have
∴
∂( f1 , f 2 , f 3 )
∂( u , v , w )
∂( f1 , f 2 , f 3 )
∂( x , y , z )
f1 = x + y + z − u
∂f1
∂f1
∂f
= 1,
= 1, 1 = 1
∂x
∂y
∂z
and
∂f1
∂f1
∂f1
= −1,
= 0,
=0
∂u
∂v
∂w
and
∂f 2
∂f
∂f
= − v , 2 = − u, 2 = 0
∂u
∂v
∂w
and
∂f 3
∂f
∂f
= − vw , 3 = − uw , 3 = − uv
∂u
∂v
∂w
f2 = y + z − uv
∂f 2
∂f
∂f
= 0, 2 = 1, 2 = 1
∂x
∂y
∂z
and
∴
f3 = z − uvw
∂f 3
∂f
∂f
= 0, 3 = 0, 3 = 1
∂x
∂y
∂z
M05_ENGINEERING_MATHEMATICS-I _CH05_Part A.indd 36
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Differential Calculus of Several Variables ■
∴
∂f1
∂x
∂( f1 , f 2 , f3 ) ∂f 2
=
∂( x , y , z )
∂x
∂f 3
∂x
−1
∂( f1, f 2 , f 3 )
= −v
∂( u , v , w )
− vw
∴
∂f1
∂y
∂f 2
∂y
∂f 3
∂y
0
−u
−uw
5.37
∂f1
∂z
1 1 1
∂f 2
= 0 1 1 =1
∂z
0 0 1
∂f 3
∂z
0
0 = ( −1)[( −u )( −uv )] = −u 2 v
−uv
∂( x , y , z )
( −u 2 v )
= ( −1)3 ⋅
= u2v
∂( u , v , w )
1
EXERCISE 5.3
1. If x = sin u cos v, y = sin u sin v, then find ∂( x, y ) .
∂( u , v )
2. If x = er sec u, y = er tan u, then show that
∂( x , y )
∂( x , y ) ∂( r , u)
= e 2 r sec u . Verify
⋅
= 1.
∂( r , u)
∂( r , u) ∂( x , y )
∂( x , y ) ∂(u , v )
3. If x = u(1 + v), y = v (1 + u), then find ∂( x, y ) and show that
⋅
= 1.
∂(u , v ) ∂( x , y )
∂( u , v )
∂( x , y ) ∂( u , v )
4. If x = u + v, y = u − v, then find ∂( x, y ) and prove that
⋅
= 1.
∂( u , v ) ∂( x , y )
∂( u , v )
5. If x = e2u cos v, y = e2u sin v, then find ∂( x, y ) and ∂(u, v ) .
∂( u , v )
∂( x , y )
6. If x = u − v2, y = u + v2, then find ∂( x, y ) .
∂( u , v )
7. If u = x2 + y2 + z2, v = x + y + z, w = xy + yz + zx, then show that
∂( u , v , w )
= 0. Is u, v, w
∂( x , y , z )
functionally related? If so find the relation between them.
8. If u = y + z, v = x +2z2, w = x − 4yz − zy2, then find ∂(u, v, w ) .
∂( x , y , z )
y2
x2 + y2
∂
(
u
,
v
)
9. If u =
, then find
.
,v=
2x
2x
∂( x , y )
2
2
10. If u3 + v3 = x + y, u2 + v2 = x3 + y3, then prove that ∂(u, v ) = 1 ( y − x ) .
∂( x, y ) 2 uv(u − v )
11. If x = v2 + w2, y = w2 + u2, z = u2 + v2, then prove that ∂( x, y, z ) = 0 .
∂( u , v , w )
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12. If u = x + y + z, v = xy + yz + zx, w = x3 + y3 +z3 − 3xyz, then prove that ∂(u, v, w ) = 0 .
∂( x , y , z )
13. If u =
1
x2
,v=
, w = x + y + zy 2 , then find ∂(u, v, w ) . Hence, find ∂( x, y, z ) .
x
y
∂( x , y , z )
∂( u , v , w )
∂( u , v , w )
= x2 y .
∂( x , y , z )
15. If x = a(u + v), y = b(u − v) and u = r2 cos 2u, v = r2 sin 2u, then find ∂( x , y ) .
∂( r , u)
y
16. If u = x + y, v =
, then find ∂(u, v ) .
x+ y
∂( x, y )
14. If u = x(1 − y), v = xy(1 − z), w = xyz, then prove that
17. If x = uv, y =
u +v
, then find ∂ (u, v ) .
u −v
∂ ( x, y )
18. If u = xyz, v = x2 + y2 + z2, w = x + y + z, then find ∂( x, y, z ) .
∂(u, v, w )
19. If u = x2 − 2y, v = x + y + z, w = x − 2y + 3z, then find ∂(u, v, w )
∂( x, y, z )
ANSWERS TO EXERCISE 5.3
1. sin u cos u
5. 2e4u, 4v
y
9.
2x
16.
5.4
1
u
2. e2r sec u
6. 4v
2
2
10. y − x
2uv(u − v )
2
17. (u − v )
4uv
3. 1 + u + v
7. v2 = u + 2w
13. 1
18.
1
2( x − y )( y − z )( z − x )
4. −2
y( y 2 − 2)
8.
2x2
15. 8ab r3
19. 10 x + 4
TAYLOR’S SERIES EXPANSION FOR FUNCTION OF TWO VARIABLES
The Taylor’s series expansion of a single variable function f(x) in a neighbourhood of a point a is
f ( a + h ) = f ( a) +
h
h2
f ′(a) + f ′′(a) + …
1!
2!
which is an infinite power series in h.
x2
f ′′(0) + …
2!
These ideas are extended to a function f(x, y) of two independent variables x, y. We state the theorem.
Maclaurin’s series is f ( x ) = f (0) + xf ′(0) +
Theorem 5.3 Taylor’s theorem
Let f(x, y) be a function of two independent variables x, y defined in a region R of the xy-plane and let
(a, b) be a point in R. Suppose f(x, y) has all its partial derivatives in a neighbourhood of (a, b), then
M05_ENGINEERING_MATHEMATICS-I _CH05_Part A.indd 38
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Differential Calculus of Several Variables ■
5.39
∂⎞
⎛ ∂
f (a + h , b + k ) = f (a, b ) + ⎜ h
+k
f (a, b )
∂y ⎟⎠
⎝ ∂x
1⎛ ∂
∂⎞
∂⎞
1⎛ ∂
+k
h
f (a, b ) + ⎜ h
+k
f (a, b ) + …
⎜
⎟
2 ! ⎝ ∂x
∂y ⎠
∂y ⎟⎠
3! ⎝ ∂x
1
i.e., f (a + h , b + k ) = f (a, b ) + ⎡⎣ h f x (a, b ) + k f y (a, b ) ⎤⎦ + [h 2 f xx (a, b ) + 2hk f xy (a, b ) + k 2 f yy (a, b )]
2!
1 3
+ [h f xxx (a, b ) + 3h 2 k f xxy (a, b ) + 3hk 2 f xyy (a, b ) + k 3f yyy (a, b )] + …
3!
2
3
+
Modified forms
1. Put x = a + h, y = b + k, then h = x − a, k = y − b
∴ the Taylor’s series can be written as
f ( x , y ) = f (a, b ) + {( x − a)f x (a, b ) + ( y − b )f y (a, b )}
+
1
{( x − a) 2 f xx (a, b ) + 2( x − a)( y − b )f xy (a, b ) + ( y − b ) 2 f yy (a, b )} + …
2!
(1)
This series is known as the Taylor’s series expansion of f(x, y) in the neighbourhood of (a, b) or
about the point (a, b).
2. Putting a = 0, b = 0, we get the expansion of f(x, y) in the neighbourhood of (0, 0)
f ( x , y ) = f (0, 0) + [xf x (0, 0) + yf y (0, 0)]
1 2
[x f xx (0, 0) + 2xy f xy (0, 0) + y 2 f yy (0, 0)] + …
2!
This is called Maclaurin’s series for f(x, y) in powers of x and y.
+
Note Taylor’s formula gives polynomial approximation to a function of two variables about a given
point.
WORKED EXAMPLES
EXAMPLE 1
Expand tan21
y
about (1, 1) upto the second degree terms.
x
Solution.
We know the expansion of f(x, y) about the point (a, b) as Taylor’s series is
f ( x , y ) = f (a, b ) + [( x − a)f x (a, b ) + ( y − b )f y (a, b )]
+
1
[( x − a) 2 f xx (a, b ) + 2( x − a)( y − b )f xy (a, b ) + ( y − b ) 2 f yy (a, b )] + …
2!
Here (a, b) 5 (1, 1)
∴
f ( x , y ) = f (1, 1) + [( x − 1)f x (1, 1) + ( y − 1)f y (1, 1)]
+
1
[( x − 1) 2 f xx (1, 1) + 2( x − 1)( y − 1)f xy (1, 1) + ( y − 1) 2 f yy (1, 1)] + …
2!
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■
Given
Engineering Mathematics
f ( x, y ) = tan −1
y
,
x
1
y2
1+ 2
x
1
fy =
y2
1+ 2
x
∴
fx =
f (1, 1) = tan −1 1 =
−y
⎛ −y ⎞
⎜⎝ 2 ⎟⎠ = 2
x
x +y2
⋅
1
x
=
x x2 + y 2
⎧( x2 + y2 ) ⋅ 0 − y ⋅ 2x ⎫
2 xy
f xx = − ⎨
⎬= 2
2
2 2
(
x
+
y
)
(
x
+ y 2 )2
⎩
⎭
f xy =
=
∂ ⎛ −y ⎞
,
∂y ⎜⎝ x 2 + y 2 ⎟⎠
f x (1, 1) =
−1
1
=−
1+1
2
f y (1, 1) =
1
1
=
1+1 2
f xx (1, 1) =
2 ⋅1⋅1 1
=
(1 + 1) 2 2
f xy (1, 1) =
1−1
=0
(1 + 1) 2
( x 2 + y 2 )( −1) − ( − y )22 y
y2 − x2
= 2
2
2 2
(x + y )
( x + y 2 )2
f yy =
=
∴
p
4
f (x , y ) =
∂ ⎛ x ⎞
,
∂y ⎜⎝ x 2 + y 2 ⎟⎠
f yy (1, 1) =
−2
1
=−
2
2
2
(x2 + y2 ) ⋅ 0 − x ⋅ 2 y
−2 xy
= 2
( x 2 + y 2 )2
( x + y 2 )2
p ⎡
1⎤
⎛ 1⎞
+ ⎢( x − 1) ⎜ − ⎟ + ( y − 1) ⋅ ⎥
⎝
⎠
4 ⎣
2
2⎦
1⎡
1
⎛ −1⎞ ⎤
+ ⎢( x − 1) 2 ⋅ + ( x − 1)( y − 1) ⋅ 0 + ( y − 1) 2 ⎜ ⎟ ⎥ + …
⎝ 2 ⎠⎦
2⎣
2
⇒
tan −1
y p 1
1
1
1
= − ( x − 1) + ( y − 1) + ( x − 1) 2 − ( y − 1) 2
x 4 2
2
4
4
EXAMPLE 2
⎛ p⎞
Expand ex cos y near the point ⎜1, ⎟ by Taylor’s series as far as quadratic terms.
⎝ 4⎠
Solution.
We know Taylor’s series about the point (a, b) is
f ( x , y ) = f (a, b ) + [( x − a)f x (a, b ) + ( y − b )f y (a, b )]
+
1
[( x − a) 2 f xx (a, b ) + 2( x − a)( y − b )f xy (a, b ) + ( y − b ) 2 f yy (a, b )] + …
2!
⎛ p⎞
Here (a, b) 5 ⎜1, ⎟
⎝ 4⎠
M05_ENGINEERING_MATHEMATICS-I _CH05_Part A.indd 40
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Differential Calculus of Several Variables ■
∴
5.41
p⎞ ⎛ p⎞ ⎤
⎛ p⎞ ⎡
⎛ p⎞ ⎛
f ( x , y ) = f ⎜1, ⎟ + ⎢( x − 1)f x ⎜1, ⎟ + ⎜ y − ⎟ f y ⎜1, ⎟ ⎥
⎝ 4⎠ ⎣
⎝ 4⎠ ⎝
4⎠ ⎝ 4⎠⎦
2
p⎞ ⎛ p⎞ ⎛
p⎞
1⎡
⎛ p⎞ ⎤
⎛ p⎞
⎛
+ ⎢( x − 1) 2 f xx ⎜1, ⎟ + 2( x − 1) ⎜ y − ⎟ f xy ⎜1, ⎟ + ⎜ y − ⎟ f yy ⎜1, ⎟ ⎥ + …
⎝ 4⎠⎦
⎝ 4⎠
⎝
4⎠
2⎣
4⎠ ⎝ 4⎠ ⎝
e
p
⎛ p⎞
f ⎜ 1, ⎟ = e ⋅ cos =
⎝ 4⎠
4
2
p
e
⎛ p⎞
f x ⎜1, ⎟ = e ⋅ cos =
,
⎝ 4⎠
4
2
f ( x, y ) = e x cos y,
Given
f x = e x ⋅ cos y,
p
e
⎛ p⎞
f y ⎜1, ⎟ = −e ⋅ sin = −
⎝ 4⎠
4
2
e
p
⎛ p⎞
f xx ⎜1, ⎟ = e ⋅ cos =
,
⎝ 4⎠
4
2
f y = −e x sin y,
f xx = e x cos y,
f yy = −e x ⋅ cos y
f xy = −e x sin y,
∴
e x cos y =
e
2
+ ( x − 1)
e
2
p
e
⎛ p⎞
f xy ⎜1, ⎟ = −e ⋅ssin = −
⎝ 4⎠
4
2
p
e
⎛ p⎞
f yy ⎜1, ⎟ = −e ⋅ cos = −
⎝ 4⎠
4
2
p⎞ ⎛ e ⎞
⎛
+⎜y − ⎟ ⎜−
⎟
⎝
4⎠ ⎝
2⎠
2
1⎡
e
p⎞ ⎛ e ⎞ ⎛
p⎞ ⎛ e ⎞ ⎤ …
⎛
y
+ ⎢( x − 1) 2 ⋅
+ 2( x − 1) ⎜ y − ⎟ ⎜ −
+
−
−
⎜
⎟
⎟⎥+
⎟
⎝
2⎣
4⎠ ⎝
4 ⎠ ⎜⎝
2
2⎠ ⎝
2⎠⎦
2
e ⎡
p⎞ 1
p⎞ 1 ⎛
p⎞ ⎤
⎛
⎛
2
=
⎢1 + ( x − 1) − ⎜⎝ y − ⎟⎠ + ( x − 1) − ( x − 1) ⎜⎝ y − ⎟⎠ − ⎜⎝ y − ⎟⎠ ⎥
4
2
4
2
4 ⎦
2⎣
EXAMPLE 3
Expand ex loge (1 1 y) in powers of x and y upto terms of third degree.
Solution.
Required the expansion in powers of x and y and so Maclaurin’s series is to be used.
1
We know f ( x, y ) = f (0, 0) + [ x f x (0, 0) + y f y (0, 0)] + [ x 2 f xx (0, 0) + 2 xy f xy (0, 0) + y 2 f yy (0, 0)]
2
+
Here (a,b) 5 (0, 0)
Given
1 3
[ x f xxx (0, 0) + 3x 2 y f xxy (0, 0) + 3xy 2 f xyy (0, 0) + y 3 f xyy (0, 0)] + …
3!
f(x, y) = ex log (1 + y),
M05_ENGINEERING_MATHEMATICS-I _CH05_Part A.indd 41
f(0, 0) = e0 log (1 + 0) = 0,
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5.42
■
∴
Engineering Mathematics
fx = e x log(1 + y ),
fy = e x ⋅
1
,
1+ y
fy(0, 0) = 1
fxx = e x log(1 + y ),
fxy(0, 0) = 1,
ex
,
(1 + y ) 2
fyy(0, 0) = −1
f xxx = e x log (1 + y ),
fxxx(0, 0) = 0
1
= ex ⋅
,
1+ y
fxxy(0, 0) = 1,
f xxy
f xyy = −
f yyy =
ex
,
(1 + y ) 2
⇒
fxyy(0, 0) = −1,
x
2e
,
(1 + y )3
fyyy(0, 0) = 2
f ( x, y ) = 0 + x ⋅ 0 + y ⋅1 +
∴
fxx(0, 0) = 0
1
,
1+ y
fxy = e x ⋅
f yy = −
fx(0, 0) = 0
1 2
[ x ⋅ 0 + 2 xy ⋅1 + y 2 ⋅ ( −1)]
2
1
+ [ x 3 ⋅ 0 + 3 x 2 y ⋅1 + 3 xy 2 ( −1) + y 3 ⋅ 2]
6
e x log(1 + y ) = y + xy −
1 2 1 2
1
1
y + x y − xy 2 + y 3
2
2
2
3
EXAMPLE 4
Expand x2y 1 3y 2 2 in powers of x 2 1 and y 1 2 using Taylor’s theorem.
Solution.
We know
f ( x, y ) = f ( a, b) + [( x − a) f x ( a, b) + ( y − b) f y ( a, b)]
1
[( x − a) 2 f xx ( a, b) + 2( x − a)( y − b) f xy ( a, b) + ( y − b) 2 f yy ( a, b)]
2!
1
+ [( x − a)3 f xxx ( a, b) + 3( x − a) 2 ( y − b) f xxy ( a, b)
3!
+ 3( x − a)( y − b) 2 f xyy ( a, b) + ( y − b)3 f yyy ( a, b)] + …
+
Here (a, b) 5 (1, 22)
Given
f(x, y) = x2 y + 3y − 2,
f(1, −2) = −2 − 6 − 2 = −10
∴
f x = 2 xy,
f x (1, − 2) = 2 ⋅1⋅ ( −2) = −4
f y = x + 3,
f y (1, − 2) = 1 + 3 = 4
2
f xx = 2 y,
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f xx (1, − 2) = 2( −2) = −4
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Differential Calculus of Several Variables ■
∴
f yy = 0,
f yy (1, − 2) = 0
f xy = 2 x,
f xy (1, − 2) = 2 ⋅1 = 2
f xxx = 0,
f xxx (1, − 2) = 0
f xxy = 2
f xxy (1, − 2) = 2
f xyy = 0
f xyy (1, − 2) = 0
f yyy = 0
f yyy (1, − 2) = 0
5.43
x 2 y + 3y − 2 = −10 + [( x − 1)( −4) + ( y + 2) ⋅ 4]
1
+ [( x − 1) 2 ( −4) + 2( x − 1)( y + 2) ⋅ 2 + ( y + 2) 2 ⋅ 0]
2
1
+ [( x − 1)3 ⋅ 0 + 3( x − 1) 2 ( y + 2) ⋅ 2 + 3( x − 1)( y + 2) 2 ⋅ 0 + 0] + …
6
= −10 − 4( x − 1) + 4( y + 2) − 2( x − 1) 2 + 2( x − 1)( y + 2) + ( x − 1) 2 ( y + 2)
Note Since the given function is 3rd degree in x, y, the expansion terminates with 3rd degree terms.
EXAMPLE 5
If f(x, y) 5 tan21(xy) compute an approximate value of f(0.9, 21.2).
Solution.
We shall use Taylor’s series to find the approximate value. The point (0.9, −1.2) is close to the point
(1, −1). So, we shall find the Taylor’s series about (1, −1).
1
[( x − 1) 2 f xx (1, − 1)
2!
+ 2( x − 1)( y + 1)f xy (1, − 1) + ( y + 1) 2 f yy (1, − 1)] + …
f ( x , y ) = f (1, − 1) + [( x − 1)f x (1, − 1) + ( y + 1)f y (1, − 1)] +
Here
(a, b) 5 (1, 21)
Given
f ( x, y ) = tan −1 xy,
1
⋅ y,
1+ x2 y2
1
fy =
⋅ x,
1 + x2 y2
(1 + x 2 y 2 ) ⋅ 0 − 2 xy 2 ⋅ y
−2 xy 3
f xx =
=
2 2 2
(1 + x y )
(1 + x 2 y 2 ) 2
fx =
f xy =
=
∂ ⎛
y ⎞
⎜
∂y ⎝ 1 + x 2 y 2 ⎟⎠
f (1, − 1) = tan −1 ( −1) =
−1
2
1
f y (1, − 1) =
2
2
1
f xx (1, − 1) = − = −
4
2
−p
4
f x (1, − 1) =
f xy (1, − 1) = 0
(1 + x 2 y 2 ) ⋅1 − y ⋅ 2 x 2 y
1 − x2 y2
=
(1 + x 2 y 2 ) 2
(1 + x 2 y 2 ) 2
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f yy =
∴
(1 + x 2 y 2 ) ⋅ 0 − x ⋅ 2 x 2 y
−2 x 3 y
=
(1 + x 2 y 2 ) 2
(1 + x 2 y 2 ) 2
f yy (1, − 1) =
2 1
=
4 2
−p
1 1⎡
1
1⎤
⎛ −1⎞
+ ( x − 1) ⎜ ⎟ + ( y + 1) + ⎢( x − 1) 2 ⋅ + 0 + ( y + 1) 2 ⋅ ⎥
⎝
⎠
4
2
2 2⎣
2
2⎦
−p 1
1
f ( x, y ) = tan −1 xy =
+ [ −( x − 1) + ( y + 1)] + [( x − 1) 2 + ( y + 1) 2 ]
4 2
4
tan −1 xy =
Put x 5 0.9, y 5 21.2
∴
p 1
1
+ (0.1 − 0.2) + (0.01 + 0.04)
4
4 2
−p 0.1 0.05
=
−
+
= −0.7854 − 0.05 + 0.0125 = −0.8229
4
2
4
f (0.9, − 1.2) = −
Note We have approximated tan−1 xy by a second degree polynomial in x and y. Using this polynomial
we have found the approximate value of f(0.9, −1.2) = −0.8229
But by direct computation f(0.9, −1.2) = tan−1(−1.08) = tan−1 1.08 = −0.8238
correct upto 4 decimal places.
The error is only 0.0009, which is negligible.
EXERCISE 5.4
( x + y ) 2 ( x + y )3 …
+
−
2
3
( x + y )3 ( x + y )5 …
Using Taylor’s series, verify that tan −1 ( x + y ) = x + y −
+
−
3
5
1
by Taylor’s series upto second degree terms.
Expand
1+ x − y
Find the Taylor’s series expansion of sin x sin y as a polynomial in x and y upto second degree.
p
Expand ex sin y about the point ⎛⎜ −1, ⎞⎟ upto third degree terms using Taylor’s series.
⎝
4⎠
p⎞
⎛
Expand sin (xy) in powers of (x − 1) and ⎜⎝ y − ⎟⎠ upto the second degree terms.
2
Expand ex cos y in powers of x and y at (0, 0) upto third degree term, by Taylor’s theorem.
1. Using Taylor’s series, verify that log (1 + x + y ) = x + y −
2.
3.
4.
5.
6.
7.
8. Expand exy in powers of (x − 1) and (y − 1) upto third degree terms, by Taylor’s series.
ANSWERS TO EXERCISE 5.4
3. 1 − x + y + x2 − 2xy + y2
5.
4. xy
1 ⎡
p⎞ 1
p⎞ 1 ⎛
p⎞
⎛
⎛
2
⎢1 + ( x + 1) + ⎜⎝ y − 4 ⎟⎠ + 2 ( x + 1) + ( x + 1) ⎜⎝ y − 4 ⎟⎠ − 2 ⎜⎝ y − 4 ⎟⎠
e 2⎣
2
p⎞ 1
p⎞
p⎞
1
1
1⎛
⎛
⎛
+ ( x + 1)3 + ( x − 1) 2 ⎜ y − ⎟ − ( x + 1) ⎜ y − ⎟ − ⎜ y − ⎟
⎝
⎠
⎝
⎠
⎝
6
2
4
2
4
6
4⎠
M05_ENGINEERING_MATHEMATICS-I _CH05_Part A.indd 44
2
3
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Differential Calculus of Several Variables ■
6. 1 −
p2
p
p⎞ 1 ⎛
p⎞
⎛
( x − 1) 2 − ( x − 1) ⎜ y − ⎟ − ⎜ y − ⎟
⎝
⎠
⎝
8
2
2
2
2⎠
5.45
2
7. 1 + x + 1 ( x 2 − y 2 ) + 1 ( x 3 − 3xy 2 ) + …
2
6
1
2
2
8. e {1 + ( x − 1) + ( y − 1) + [( x − 1) + 4( x − 1)( y − 1) + ( y − 1) ]
2
1
+ [( x − 1)3 + 9( x − 1) 2 ( y − 1) + 9( x − 1)( y − 1) 2 + ( y − 1)3 ] + …}
6
5.5
MAXIMA AND MINIMA FOR FUNCTIONS OF TWO VARIABLES
You have learned maxima and minima of a function f(x) of a single variable in x. We shall extend these
ideas to a function f(x, y) of two variables in x and y. We shall derive the conditions of maxima and
minima as an application of quadratic form.
Definition 5.9 Let f(x, y) be a continuous function defined in a closed and bounded domain D of the
xy-plane and let (a, b) be an interior point of D.
(i) f(a, b) is said to be a local maximum or relative maximum value of f(x, y) at the point (a, b), if
there exists a neighbourhood N of (a, b) such that f(x, y) < f(a, b) for all points (x, y) in N, other
than the point (a, b). And,
(ii) f(a, b) is said to be a local minimum or relative minimum if f(x, y) > f(a, b) for all points (x, y)
in N, other than the point (a, b)
Note
(1) A common name for relative maximum or relative minimum is extreme value.
A relative maximum or relative minimum is simply referred to as maximum or minimum.
(2) In contrast, the greatest value of f(x, y) over the entire domain including the boundary is called
the global maximum or the absolute maximum value of f(x, y) on D and smallest value of
f(x, y) over the entire domain D is called the global minimum or absolute minimum.
Z
Z
z = f(x, y)
z = f(x, y)
Y
Y
X
(a, b)
(x, y)
Maximum
Fig. 5.5
M05_ENGINEERING_MATHEMATICS-I _CH05_Part A.indd 45
(a, b)
X
(x, y)
Minimum
Fig. 5.6
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Definition 5.10 Stationary point of f (x , y )
A point (a, b) satisfying fx = 0 and fy = 0 is called a stationary point of f(x, y).
5.5.1
Necessary Conditions for Maximum or Minimum
If f(a, b) is an extreme value of f(x, y) at (a, b), then (a, b) is a stationary point of f(x, y) if fx
and fy exist at (a, b) and fx(a, b) = 0, fy(a, b) = 0
Note
(1) But the converse is not true i.e., if (a, b) is a stationary point of f(x, y), then (a, b) need not be an
extreme point.
For example, consider the function f(x, y) defined by
Z
⎧0 if x = 0 or y = 0
f ( x, y ) = ⎨
⎩1, otherwise
P (a, b, z1)
then fx(0, 0) = 0, fy(0, 0) = 0.
But f(0, 0) is not an extreme value.
i.e., (0, 0) is not an extreme point.
Some more conditions are needed to ensure the extreme
Y
value. They are the sufficient conditions.
(2) The graph of the function f(x, y) is the surface
D
(a, b)
z = f(x, y). Corresponding to the point (a, b) in D,
P (a, b, z1), where z1 = f(a, b), is a point on the surface.
X
If (a, b) is stationary point of the function f(x, y), then
Fig. 5.7
P (a, b, z1) is a stationary point on the surface.
If the tangent plane exists at the stationary point on the surface, then it will be parallel to the
xy-plane z = 0, f(a, b) is called a stationary value of f(x, y).
(3) Stationary points on a surface are usually classified into three categories: maxima, minima and
saddle points. If the surface is regarded as a mountain landscape we can visualise these categories as
mountain tops, bottoms of valleys and mountain passe (saddle is the seat for a rider on horse back).
5.5.2
Sufficient Conditions for Extreme Values of f (x, y ).
Let (a, b) be a stationary point of the differentiable function f(x, y).
i.e., fx(a, b) = 0, fy(a, b) = 0.
Let fxx(a, b) = r, fxy(a, b) = s, fyy(a, b) = t.
(i)
(ii)
(iii)
(iv)
If rt − s2 > 0 and r < 0, then f(a, b) is a maximum value.
If rt − s2 > 0 and r > 0, then f(a, b) is a minimum value.
If rt − s2 < 0, then f(a, b) is not an extreme value, but (a, b) is a saddle point of f(x, y).
If rt − s2 = 0 then no conclusion is possible and further investigation is required.
5.5.3
Working Rule to find Maxima and Minima of f (x, y )
Step 1: Find f x =
∂f
,
∂x
fy =
∂f
and solve fx = 0, fy = 0 as simultaneous equations in x and y.
∂y
Let (a, b), (a1, b1),… be the solutions, which are stationary points of f(x, y).
M05_ENGINEERING_MATHEMATICS-I _CH05_Part B.indd 46
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Differential Calculus of Several Variables ■
Step 2: Find r = f xx =
5.47
∂2 f
∂2 f
∂2 f
,
,
.
s
f
t
f
=
=
=
=
xy
yy
∂x ∂y
∂x 2
∂y 2
Step 3: Evaluate r, s, t at each stationary point.
At the stationary point (a, b)
(i)
(ii)
(iii)
(iv)
If rt − s2 > 0 and r < 0, then f(a, b) is a maximum value of f(x, y).
If rt − s2 > 0 and r > 0, then f(a, b) is a minimum value of f(x, y).
If rt − s2 < 0, then (a, b) is a saddle point.
If rt − s2 = 0, no conclusion can be made; further investigation is required.
Note
(i) Instead of r, s, t, we can also use the symbols A, B, C respectively.
(ii) The expression rt − s2 enables us to discriminate the stationary points and so it is called the
discriminant of the function f(x, y).
Definition 5.11 Critical Point
A point (a, b) is a critical point of f(x, y) if fx = 0 and fy = 0 at (a, b) or fx and fy do not exist at (a, b).
Maxima or minima occur at a critical point.
Note Generally, in this book we deal with differentiable functions f(x, y). So, critical points are all
stationary points.
WORKED EXAMPLES
EXAMPLE 1
Examine f(x, y) 5 x3 1 y3 2 12x 2 3y 1 20 for its extreme values.
Solution.
Given
f(x, y) = x3 + y3 − 12x − 3y + 20
∴
f x = 3 x 2 − 12,
r = f xx = 6 x,
fy = 3 y2 − 3
s = f xy = 0 and t = f yy = 6 y
To find the stationary points, solve fx = 0 and fy = 0
∴
3 x 2 − 12 = 0 ⇒
x2 = 4 ⇒
x = ±2
and
3y2 − 3 = 0 ⇒
y2 = 1 ⇒
y = ±1
The points are (2, 1) (2, −1), (−2, 1), (−2, −1)
At the point (2, 1)
r = 6 ⋅ 2 = 12 > 0,
∴
rt − s2 = 12 × 6 = 72 > 0
s=0
and t = 6 ⋅ 1 = 6
and r > 0
∴ (2, 1) is a minimum point.
Minimum value = f(2, 1) = 23 + 1 − 12 × 2 − 3 ⋅ 1 + 20 = 8 + 1 − 24 − 3 + 20 = 2
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At the point (22, 1)
r = 6 × (−2) = −12 < 0, s = 0
∴
rt − s2 = −12 × 6 − 0 = −72 < 0
∴ (−2, 1) is a saddle point.
t=6⋅1=6
and
At the point (2, 21)
r = 6 ⋅ 2 = 12 > 0, s = 0
∴
rt − s2 = 12(−6) − 0 = −72 < 0
∴ (2, −1) is a saddle point.
t = 6(−1) = −6
and
At the point (22, 21)
r = 6(−2) = −12 < 0, s = 0,
∴
t = 6(−1) = −6
and
rt − s = (−12)(−6) − 0 = 72 > 0 and r < 0
2
∴ (−2, −1) is a maximum point.
Maximum value = f(−2, −1) = (−2)3 + (−1)3 − 12(−2) − 3(−1) + 20 = −8 − 1 + 24 + 3 + 20 = 38
EXAMPLE 2
Discuss the maxima and minima of f(x, y) 5 x3y2(1 2 x 2 y).
Solution.
Given
f(x, y) = x3y2(1 − x − y) = x 3 y 2 − x 4 y 2 − x 3 y 3
∴
f x = 3x 2 y 2 − 4 x 3 y 2 − 3x 2 y 3 ,
f y = 2 x 3 y − 2 x 4 y − 3x 3 y 2
r = f xx = 6 xy 2 − 12 x 2 y 2 − 6 xy 3 ,
s = f xy = 6 x 2 y − 8 x 3 y − 9 x 2 y 2
t = f yy = 2 x 3 − 2 x 4 − 6 x 3 y
To find the stationary points, solve fx = 0 and fy = 0
∴
⇒
3x2y2 − 4x3y2 − 3x2y3 = 0
and
2 x 3 y − 2 x 4 y − 3x 3 y 2 = 0
⇒
x 2 y 2 (3 − 4x − 3y ) = 0
(1)
x = 0, y = 0 or 3 − 4x − 3y = 0
∴
⇒
x 3 y[2 − 2 x − 3 y ] = 0
(2)
x = 0, y = 0 or 2 − 2x − 3y = 0
We find that (0, 0) satisfies the equations (1) and (2)
3 − 4x − 3y = 0 ⇒ 4x + 3y = 3
2 − 2x − 3y = 0 ⇒ 2x + 3y = 2
Solving
and
(3) − (4) ⇒
When x =
1
,
2
When x = 0,
1
2. + 3 y = 2
2
1
2
1
⇒y=
3
2.0 + 3y = 2
⇒ y=
2x = 1
( 4) ⇒
(3) ⇒
y=1
M05_ENGINEERING_MATHEMATICS-I _CH05_Part B.indd 48
and
(4)
⇒
(3)
(4)
⇒ x=
2
3
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Differential Calculus of Several Variables ■
When y = 0,
⇒
(3)
x=
3
and
4
⇒
(4)
1 1
∴ the stationary points are (0, 0), ⎛⎜ , ⎞⎟ , ⎛⎜ 0,
⎝ 2 3⎠ ⎝
5.49
x=1
2⎞
⎛3 ⎞
⎟⎠ , (0, 1), ⎜⎝ , 0⎟⎠ , (1, 0)
3
4
2
3
At the points (0, 0), ⎛⎜ 0 , ⎞⎟ , ( 0 , 1), ⎛⎜ , 0 ⎞⎟ and (1, 0 )
⎝4 ⎠
⎝ 3⎠
r = 0, s = 0, t = 0 ∴ rt − s2 = 0
∴ we cannot say maximum or minimum. Further investigation is required.
1
At the point ⎛⎜ ,
⎝2
1⎞
⎟
3⎠
2
1 ⎛ 1⎞
1 1
1 1 1 1 1
1
= − − =− <0
r = 6 ⋅ ⎜ ⎟ − 12 ⋅ ⋅ − 6 ⋅ ⋅
2 ⎝ 3⎠
4 9
2 27 3 3 9
9
1 1
1 1 1 1 1
1
1 1
s = 6 ⋅ ⋅ − 8⋅ ⋅ − 9⋅ ⋅ = − − = −
4 3
8 3
4 9 2 3 4
12
1
1 1 2 1 2
1
1
t = 2⋅ − 2⋅ − 6⋅ ⋅ = − − = −
8
16
8 3 8 8 8
8
∴
1
1
1
⎛ 1 ⎞ ⎛ 1 ⎞ ⎛ −1⎞
−
rt − s 2 = ⎜ − ⎟ ⎜ − ⎟ − ⎜ ⎟ =
=
>0
⎝ 9 ⎠ ⎝ 8 ⎠ ⎝ 12 ⎠
72 144 144
∴
rt − s2 > 0 and r < 0
2
1 1
∴ the point ⎛⎜ , ⎞⎟ is a maximum point.
⎝ 2 3⎠
1 1
1 1
1 1
1
The maximum value = ⋅ ⎛⎜1 − − ⎞⎟ =
⋅ =
⎝
⎠
8 9
2 3
72 6 432
EXAMPLE 3
Find the maximum and minimum values of sin x sin y sin (x 1 y), 0 < x, y < p.
Solution.
Given
∴
f(x, y) = sin x sin y sin (x 1 y)
f x = sin y[sin x cos( x + y ) + sin( x + y ) cos x ]
= sin y sin( x + x + y ) = sin y sin(2
2 x + y)
f y = sin x[sin y cos( x + y ) + sin( x + y ) cos y ] = sin x sin( x + 2 y )
r = f xx = sin y cos( 2 x + y ) ⋅ 2 = 2 sin y cos( 2 x + y )
⇒
s = f xy = sin y cos( 2 x + y ) + sin( 2 x + y )(cos y ) = sin( 2 x + y + y )
⇒
s = sin( 2 x + 2 y )
t = f yy = 2 sin x cos( x + 2 y )
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To find the stationary points, solve fx = 0 and fy = 0
∴
sin y sin( 2 x + y ) = 0
(1)
and
sin x sin( x + y ) = 0
(2)
Since x, y ≠ 0 and sin x ≠ 0, sin y ≠ 0
∴ (1) ⇒
and (2) ⇒
Since 0 < x,
sin (2x + y) = 0
sin (x + 2y) = 0
y < p,
0 < 2x < 2p
and 0 < y < p
Adding
0 < 2x + y < 3p
Similarly,
0 < x + 2y < 3p
∴
sin (2x + y) = 0
and
sin (x + 2y) = 0
If 2x + y = p and x + 2y = p, then x − y = 0 ⇒ x = y
∴
3x = p ⇒
x=
p
3
∴
y =
⇒ 2x + y = p or 2p
⇒ x + 2y = p or 2p
p
3
p p
∴ one point is ⎛⎜ , ⎞⎟
⎝ 3 3⎠
If 2x + y = p and x + 2y = 2p, then x − y = −p
⇒
∴
⇒ 3 y = 3p ⇒ y = p which is
2 (y − p) + y = p
x = y −p,
not admissible since y < p
Similarly, 2x + y = 2p and x + 2y = p are also not admissible.
Now, take 2x + y = 2p
∴
x + 2 y = 2p , then
and
3 x = 2p ⇒ x =
2p
3
x−y=0
∴
y=
⇒
x=y
2p
3
2p 2p ⎞
∴ another point is ⎛⎜
,
⎝ 3 3 ⎟⎠
p p
2p 2p ⎞
So, the stationary points are ⎛⎜ , ⎞⎟ and ⎛⎜
,
⎝ 3 3⎠
⎝ 3 3 ⎟⎠
p p
At the point ⎛⎜ , ⎞⎟
⎝ 3 3⎠
r = 2 sin
∴
3
p
cos p = −2
< 0,
3
2
s = sin
rt − s 2 = 3 −
M05_ENGINEERING_MATHEMATICS-I _CH05_Part B.indd 50
4p
3
p⎞
p
⎛
= sin ⎜ p + ⎟ = −
and t = 2 sin ⋅ cos p = − 3
⎝
⎠
3
3
2
3
3 9
= >0
4 4
and r < 0
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Differential Calculus of Several Variables ■
5.51
p p
∴ the point ⎛⎜ , ⎞⎟ is a maximum point
⎝ 3 3⎠
p
p
2p
3 3 3 3 3
⎛ p p⎞
=
⋅
=
The maximum value = f ⎜ , ⎟ = sin sin sin
⎝ 3 3⎠
3
3
3
2 2 2
8
2p 2p ⎞
At the point ⎛⎜
,
⎟
⎝ 3
3 ⎠
r = 2 sin
2 3
8p
p
3
2p
2p
cos 2p =
= 3 > 0, s = sin
= sin =
and t = 2 sin
cos 2p = 3
3
2
3
3
2
3
∴
rt − s 2 = 3 ⋅ 3 −
3 9
= > 0 and r > 0
4 4
⎛ 2p 2p ⎞
,
is a minimum point.
∴ the point ⎜
⎝ 3 3 ⎟⎠
2p 2p ⎞
2p
2p
4p
3 3⎛
3⎞
3 3
The minimum value = f ⎛⎜
,
=
sin
sin
sin
=
−
⎟
⎜
⎟=−
⎝ 3 3 ⎠
3
3
3
2 2 ⎝ 2 ⎠
8
5.5.4
Constrained Maxima and Minima
In many practical problems on maxima and minima we have to find the extreme values of a function
of two or more variables which are not independent but are connected by some relation. For example,
suppose we want to find the
maximum value of
x2 + y2 + z2
(1)
subject to the condition
2x + 3y + 5z = 4
(2)
One method is to find z from (2) and substitute in (1), then it reduces to a function u (x, y) of
two independent variables x and y. As above, we can find the maximum or minimum value of
u (x, y).
Suppose the relation between the variables is complicated, then finding z interms of x and y will be
difficult or impossible. In such cases we use the versatile Lagrange’s multiplier method, in which the
introduction of a multiplier enables us to solve the constrained extreme problems without solving the
constrained equation for one variable in terms of others.
5.5.5
Lagrange’s Method of (undetermined) Multiplier
Let f(x, y, z) be the function whose extreme values are to be found subject to the restriction
f (x, y, z) = 0
Between the variables x, y, z construct the auxiliary function
(1) .
F(x, y, z) = f(x, y, z) + lf(x, y, z)
where l is an undetermined parameter independent of x, y, z.
l is called Lagrange’s multiplier.
Any relative extremum of f(x, y, z) subject to (1) must occur at a stationary point of F(x, y, z).
The stationary points of F are given by
∂F
= 0,
∂x
M05_ENGINEERING_MATHEMATICS-I _CH05_Part B.indd 51
∂F
= 0,
∂y
∂F
= 0,
∂z
∂F
=0
∂l
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Engineering Mathematics
⇒
f x + lfx = 0, f y + lfy = 0, f z + lfz = 0 and f( x , y , z ) = 0
⇒
fy
fx
f
=
= z = −l and f( x, y, z ) = 0
fx fy fz
Solving these equations, we find the values of x, y, z, which are the stationary points of F, giving the
maximum and minimum values of f(x, y, z).
Note This method does not specify the extreme value obtained is a maximum or minimum. It is
usually decided from the physical and geometrical considerations of the problem.
A method on the basis of quadratic form is given below to decide maxima or minima at the
stationary point for constrained maxima and minima.
5.5.6
Method to Decide Maxima or Minima
We shall see sufficient conditions given by quadratic form of the differentials.
1. For unconstrained functions.
Let u = f(x, y) be a function of two variables.
∴ the total differential du = fx dx + fy dy
Necessary conditions for maxima or minima of u = f(x, y) is
du = 0 ⇒
f x dx + f y dy = 0
⇒ fx = 0, fy = 0, since dx, dy may take any value.
The sufficient condition for minimum is d 2u > 0 and maximum is d 2u < 0.
Thus, du = 0 and d 2u > 0 are the necessary and sufficient conditions for minimum.
Similarly, du = 0 and d 2u < 0 are the necessary and sufficient conditions for maximum.
Now, d 2u = d(fx) dx + d(fy) dy
∂f y ⎞
⎛ ∂f y
∂f
⎛ ∂f
⎞
= ⎜ x dx + x dy⎟ dx + ⎜
dx +
dy dy
⎝ ∂x
∂y ⎠
⎝ ∂x
∂y ⎟⎠
= ( f xx dx + f xy dy )dx + ( f xy dx + f yy dy )dy
= f xx ( dx ) + f xy dx dy + f xy dxdy + f yy ( dy )
2
[Assuming f xy = f yx ]
2
= f xx ( dx ) 2 + 2 f xy dx dy + f yy ( dy ) 2
Thus, d 2u is a quadratic form in dx, dy.
The matrix of the quadratic form is called the Hessian matrix.
⎡ f xx f xy ⎤
H= ⎢
⎥
⎢⎣ f xy f yy ⎥⎦
Its principal minors are D1 = fxx = r
D2 =
f xx
f xy
f xy
f yy
M05_ENGINEERING_MATHEMATICS-I _CH05_Part B.indd 52
= f xx f yy − ( f xy ) 2 = rt − s 2
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Differential Calculus of Several Variables ■
5.53
For minimum d 2u > 0
i.e., the quadratic form is positive definite.
∴
D1 > 0, D2 > 0 ⇒ r > 0 and rt − s2 > 0
For maximum,
d 2u < 0
⇒
D1 < 0, D2 > 0 ⇒ r < 0 and rt − s2 > 0
This can be extended to three or more variables.
The necessary and sufficient conditions are (i) for maximum du = 0 and d 2u < 0
and (ii) for minimum du = 0 and d 2u > 0.
2. If u = f(x, y, z), then du = fx dx + fy dy + fz dz and
d 2 u = f xx ( dx ) 2 + f yy ( dy ) 2 + f zz ( dz ) 2 + 2 f yz dydz + 2 f zx dxdz + 2 f xy dxdy
du = 0 ⇒ f x = 0,
f y = 0,
fz = 0
which gives the stationary points.
The matrix of the quadratic form in dx, dy, dz is the
⎡ f xx
⎢
Hessian H = ⎢ f yx
⎢f
⎣ zx
The principal minors are D1 = fxx, D 2 =
f xx
f xy
f yx
f yy
f xy
f yy
f zy
f xz ⎤
⎥
f yz ⎥
f zz ⎥⎦
and D3 = H
At a stationary point (a, b, c), if D1 > 0, D2 > 0, and D3 > 0, then u is minimum.
If D1 < 0, D2 > 0, and D3 < 0 then u is maximum.
In the same way we can extend to function of n variables f(x1, x2, …, xn)
3. We shall now see how the Hessian changes in the discussion of constrained maxima and
minima.
For example, consider the quadratic form in two variables.
ax
Q = ax2 +2hxy + by2 with linear constraint ax + by = 0 ⇒ y = −
b
2
∴
Since
x2
⎛ ax ⎞
⎛ ax ⎞
Q = ax 2 + 2hx ⎜ − ⎟ + b ⎜ − ⎟ = 2 [ab2 − 2hab + ba 2 ]
⎝ b⎠
⎝ b⎠
b
x2
> 0, Q > 0 or < 0 if ab2 − 2hab + ba2 > 0 or < 0
b2
0 a b
We can easily see that − [ab2 − 2hab + ba2] = a a h
b h b
∴
0 a b
Q > 0 if a a h < 0
b h b
M05_ENGINEERING_MATHEMATICS-I _CH05_Part B.indd 53
and
0 a b
Q < 0 if a a h > 0
b h b
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Engineering Mathematics
0 a b
The determinant a a h is made up of the matrix of coefficients of the quadratic form Q
b h b
which is bordered by coefficients of the linear constraint. So, this is called a bordered determinant.
Thus, the corresponding matrix is called the bordered Hessian matrix.
⎡ 0 a b⎤
H = ⎢⎢a a h ⎥⎥ , bordered by the linear constraint.
⎢⎣b h b ⎥⎦
0 a b
0 a
The bordered principle minors are D1 =
= −a 2 and D 2 = a a h
a a
b h b
2
Since D1 = −a is always negative,
Q > 0,
if D2 < 0 and Q < 0,
if D2 > 0
4. We shall now consider quadratic form in three variables.
Q = a11 x12 + a22 x22 + a33 x32 + 2a12 x1 x2 + 2a23 x2 x3 + 2a31 x3 x1 subject to the linear constraint
b1x1 + b2x2 + b3x3 = 0.
Then the corresponding bordered Hessian is
⎡0
⎢b
H= ⎢ 1
⎢b2
⎢
⎣ b3
b1
a11
a21
a31
b2
a12
a22
a32
b3 ⎤
a13 ⎥⎥
a23 ⎥
⎥
a33 ⎦
0
b1
, D 2 = b1
a11
b2
b1
a11
b2
a12 and D3 = H
a21
a22
The bordered principle minors are
0
D1 =
b1
Since D1 = −b is always negative, Q > 0 if D2 < 0 and D3 < 0 and Q < 0 if D2 > 0 and D3 < 0.
Similarly, we can discuss conditions for more than three variables.
5. Extremum with general constraints
Extreme values of u = f(x, y) subject to f (x, y) = 0
Stationary points are given by du = 0, f = 0
At a stationary point, u is maximum if d 2u < 0 and df = 0
and
u is minimum if d 2u > 0 and df = 0
2
2
where d u = fxx(dx) + fyy(dy)2 + 2fxy dx dy is a quadratic form in dx, dy and
df = 0 ⇒ fx dx + fy dy = 0, which is linear in dx, dy.
So, the corresponding bordered Hessian matrix is
2
1
⎡0
⎢
H = ⎢fx
⎢f
⎣ y
M05_ENGINEERING_MATHEMATICS-I _CH05_Part B.indd 54
fx
f xx
f yx
fy ⎤
⎥
f xy ⎥
f yy ⎥⎦
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Differential Calculus of Several Variables ■
0
fx
and D 2 = fx
fx
fy
0
The bordered principal minors are D1 =
fx
fx
fy
f xx
f xy .
f yx
f yy
5.55
D1 = −f < 0 always
∴ u is maximum (i.e., d 2u < 0), if D2 > 0
and u is minimum (i.e., d 2u > 0), if D2 < 0
2
x
6. Extreme values of u 5 f(x, y, z) subject to f (x, y, z) 5 0.
The bordered Hessian matrix is
⎡0
⎢
⎢fx
H=⎢
f
⎢ y
⎢fz
⎣
The bordered principal minors are
0
D1 =
fx
fx
fy
f xx
f xy
f yx
f yy
f zx
f zy
0
fx
, D 2 = fx
f xx
fy
fz ⎤
⎥
f xz ⎥
f yz ⎥
⎥
f zz ⎥⎦
fx
fy
f xx
f xy , D3 = H
f yx
f yy
Since D1 = −fx2 < 0 always,
u is maximum (i.e., d 2u < 0), if D2 > 0 and D3 < 0
and u is minimum (i.e., d 2u > 0), if D2 < 0 and D3 < 0
7. Sufficient conditions for Lagrange’s method
Extreme values of f(x, y, z) subject to f(x, y, z) = 0
Form the auxiliary function F(x, y, z) = f(x, y, z) + lf(x, y, z), where l is the Lagrange’s
multiplier.
Stationary points are given by Fx = 0, Fy = 0, Fz = 0, f = 0
At the stationary point we have maximum if d2F < 0, df = 0
⇒ fxdx + fydy + fzdz = 0 and minimum if d2F > 0, df = 0
The corresponding bordered Hessian is
⎡0
⎢
⎢fx
H=⎢
f
⎢ y
⎢fz
⎣
The bordered principal minor D1 =
0
fx
fx
fy
Fxx
Fxy
Fyx
Fyy
Fzx
Fzy
fz ⎤
⎥
Fxz ⎥
Fyz ⎥
⎥
Fzz ⎥⎦
fx
= −fx2 which is always negative.
Fxx
∴ At a stationary point f(x, y, z) is minimum if D2 < 0 and D3 < 0 and maximum if D2 > 0 and D3 < 0
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WORKED EXAMPLES
EXAMPLE 1
A rectangular box, open at the top, is to have a volume of 32 cc. Find dimensions of the box which
requires least amount of material for its construction.
Solution.
Let x, y, z be the length, breadth and height of the box. Given volume of the box is 32 cc.
⇒
xyz = 32,
x, y, z > 0
(1)
We want to minimize the amount of material for its construction. i.e., surface area of the box is to be
minimized.
Surface area S = xy + 2xz + 2yz
[{ top is open]
(2)
We shall solve by two methods.
z=
32
xy
S = xy + 2( x + y )
32
xy
Method 1 xyz = 32
∴
⇒
∂S
64
= y− 2
∂x
x
∴
⎛ 1 1⎞
⇒ S = xy + 64 ⎜ + ⎟
⎝ x y⎠
∂S
64
= x− 2
∂y
y
and
To find stationary points, solve
∂S
∂S
= 0 and
=0
∂x
∂y
⇒
64
=0
x2
y−
and x −
64
=0
y2
⇒
x 2 y = 64
and
xy 2 = 64
∴
x 2 y = xy 2
⇒
x=y
x = 64
⇒
x=4 ∴ y=4
∴ (3) ⇒
∴ stationary point is (4, 4)
3
(3)
[{ x > 0, y > 0]
∂2 S
∂ 2 S 128
= 1 and t = 2 = 3
∂x ∂y
∂y
y
Now
r=
∂ 2 S 128
= 3 ,
∂x 2
x
s=
∴ at the point (4, 4),
r=
128
= 2 > 0,
43
s = 1 and t =
128
=2
43
∴
rt − s2 = 2 ⋅ 2 − 1 = 3 > 0
2
Since r > 0 and rt − s > 0, it is a minimum point
When x = 4, y = 4, z =
32
=2
16
∴ dimensions of the box are x = 4 cms, y = 4 cms and z = 2 cms.
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Differential Calculus of Several Variables ■
5.57
Method 2 Lagrange’s method
We have to minimise S = xy + 2xz + 2yz
Subject to
xyz = 32 ⇒ xyz − 32 = 0
Form the auxiliary function
(1)
(2)
F(x, y, z) = xy + 2xz + 2yz + l(xyz − 32)
where l is the Lagrange’s multiplier.
∴
Fx =
∂F
∂F
∂F
= y + 2 z + lyz , Fy =
= x + 2 z + lxz , Fz =
= 2 x + 2 y + lxy and Fl = f
∂x
∂y
∂z
To find stationary points, solve Fx = 0, Fy = 0,
Fx = 0 ⇒ y + 2z + lyz = 0
Fz = 0,
⇒
⇒ xy + 2zx = −lxyz
y + 2z = −lyz
f=0
[multiplying by x] (3)
Fy = 0 ⇒ x + 2z + lxz = 0
⇒
x + 2z = − lxz ⇒
and
[multiplying by y] (4)
2xz + 2yz = −lxyz
[multiplying by z] (5)
Fz = 0 ⇒ 2x + 2y + lxy = 0
⇒
2x + 2y = −lxy
From (3), (4) and (5)
⇒
and
⇒
xy + 2 zx = xy + 2 zy = 2 xz + 2 yz
⇒ 2 zx = 2 zy
xy + 2 zx = xy + 2 zy
⇒ x= y
xy + 2 zx = 2 xz + 2 yz
⇒ x = 2z
∴
Substituting in (2), we get
⇒
xy = 2 yz
x = y = 2z
2 z ⋅ 2 z ⋅ z = 32
⇒
∴ (6) ⇒ x = 4, y = 4
∴ the stationary point is (4, 4, 2)
So, the dimensions are 4 cms, 4 cms, 2 cms.
Remark
xy + 2zy = −lxyz
(6)
4 z 3 = 32
⇒
z3 = 8
⇒ z=2
We have not justified S is minimum at (4, 4, 2). We shall use the bordered Hessian to decide.
⎡0
⎢
⎢fx
H= ⎢
f
⎢ y
⎢fz
⎣
When x = 4, y = 4, z = 2,
fx
fy
Fxx
Fxy
Fyx
Fyy
Fzx
Fzy
fz ⎤
⎥
Fxz ⎥
Fyz ⎥
⎥
Fzz ⎥⎦
Fx = 0 ⇒ 4 +4 + 8l = 0 ⇒ l = −1
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∴
Fx = y +2z − yz,
Fy = x +2z − xz and Fz = 2x +2z − xy
∴
Fxx = 0,
Fyy = 0,
Fzz = 0 and Fyx = Fxy = 1 − z;
Fxx = 0,
Fyy = 0,
Fzz = 0
Fxz = Fzx = 2 − y
At (4, 4, 2)
Fxy = 1 − 2 = −1, Fxz = 2 − 4 = −2,
fx = 8,
∴
fy = 8,
Fyz = 2 − 4 = −2
fz = 16
8 16 ⎤
⎡0 8
⎢ 8 0 −1 −2⎥
⎥
H= ⎢
⎢ 8 −1 0 −2⎥
⎢
⎥
⎣16 −2 −2 0 ⎦
The bordered principal minors are D1 =
0 8
= −64 < 0;
8 0
0 8 8
0 1 1
D 2 = 8 0 −1 = 64 1 0 −1 = 64( −2) = −128 < 0
8 −1 0
1 −1 0
0
8
D3 =
8
8
8 16
0 −1 −2
= 8.8
−1 0 −2
16 −2 −2 0
0 1
1
2
1 0 −1 −2
1 −1 0 −2
2 −2 −2 0
0 1 1
1 0 −1
= 64 × ( +2)( +2)
1 −1 0
1 −1 −1
0 1 0
1 0 −1
= 256
1 −1 +1
1 −1 0
0
−1
0
+1
1
−1
−1
0
C3 → C3 − C 2
C4 → C4 − C2
1 −1 −1
= 256( −1) 1 1
0 , expanding by R1
1 0
1
1 0 0
= −256 1 2 1
1 1 2
C2 → C2 + C2
C3 → C3 + C1
= −256( 4 − 1) = −256 × 3 = −768 < 0
M05_ENGINEERING_MATHEMATICS-I _CH05_Part B.indd 58
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Differential Calculus of Several Variables ■
5.59
Since D1 < 0, D2 < 0, D3 < 0
d 2F is positive definite and hence minimum.
∴ the function S is minimum when x = 4, y = 4 and z = 2.
EXAMPLE 2
Find the shortest and longest distance from the point (1, 2, −1) to the sphere x2 1 y2 1 z2 5 24,
using Lagrange’s method of constrained maxima and minima.
Solution.
Let P(x, y, z) be a point on the sphere
x2 + y2 + z2 = 24
and A be (1, 2, −1).
The distance AP = ( x − 1) 2 + ( y − 2) 2 + ( z + 1) 2
Let f(x, y, z) = (x − 1)2 + (y − 2)2 + (z + 1)2
AP is minimum or maximum if f(x, y, z) is minimum or maximum.
So, we minimize or maximize f(x, y, z) subject to x2 + y2 + z2 = 24
Let f(x, y, z) = x2 + y2 + z2 − 24 = 0
Form the auxillary function
(1)
(2)
F(x, y, z) = f(x, y, z) + lf(x, y, z)
where l is the Lagrange’s multiplier.
F(x, y, z) = (x − 1)2 + (y − 2)2 + (z + 1)2 + l (x2 + y2 + z2 − 24)
∴
Fx = 2( x − 1) + 2lx,
Fy = 2( y − 2) + 2ly,
To find the stationary points solve Fx = 0,
Fy = 0,
Fz = 2( z + 1) + 2lz and Fl = f
Fz = 0,
f=0
x −1
1
−l =
= 1−
x
x
∴
Fx = 0 ⇒ 2( x − 1) + 2lx = 0
⇒
x − 1 = −lx
⇒
⇒
Fy = 0 ⇒ 2( y − 2) + 2ly = 0 ⇒
y − 2 = − ly
⇒
−l =
y −2
2
= 1−
y
y
and
Fz = 0 ⇒ 2( z + 1) + 2lz = 0
z + 1 = − lz
⇒
−l =
z +1
1
= 1+
z
z
⇒
∴
1−
1
2
1
= 1− = 1+
x
y
z
Now
1−
1
2
= 1−
x
y
⇒
−
1
2
=−
x
y
⇒
y = 2x
and
1−
1
1
= 1+
x
z
⇒
−
1 1
=
x z
⇒
z = −x
∴
2 x = y = −2 z ⇒
We have
⇒
x2 + y2 + z2 = 24
x 2 + 4 x 2 + x 2 = 24
⇒
x=
y
= −z
2
6 x 2 = 24 ⇒ x 2 = 4 ⇒ x = ±2
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When x = 2, y = 4, z = −2 and when x = −2, y = −4, z = +2
The stationary points are P1 (2, 4, −2) and P2 (−2, −4, 2)
AP1 = 1 + 4 + 1 = 6 and AP2 = 9 + 36 + 9 = 3 6
∴
the shortest distance = 6 and the longest distance = 3 6
EXAMPLE 3
The temperature T at any point (x, y, z) in space is T 5 400 xyz2. Find the highest temperature on
the surface of the unit sphere x2 1 y2 1 z2 5 1.
Solution.
We want to maximize
T = 400 xyz2
Subject to
f(x, y, z) = x2 + y2 + z2 − 1 = 0
Auxiliary function is
F(x, y, z) = T + lf(x, y, z)
⇒
F(x, y, z) = 400 xyz2 + l( x2 + y2 + z2 − 1)
(1)
where l is the Lagrange’s multiplier.
∴
Fx = 400 yz 2 + 2lx,
Fy = 400 xz 2 + 2ly,
To find stationary points, solve Fx = 0,
∴
Fz = 800 xyz + 2lz and Fl = f
Fy = 0,
Fz = 0,
f=0
Fx = 0 ⇒ 400 yz 2 + 2lx = 0
⇒
400 yz 2 = −2lx
Fy = 0 ⇒ 400 xz 2 + 2ly = 0
⇒
200 xz 2
= −l
y
(3)
Fz = 0 ⇒ 800 xyz + 2lz = 0
⇒
400xy = − l
(4)
⇒
200 yz 2
= −l
x
(2)
From (2), (3) and (4), we get
200 yz 2 200 xz 2
=
= 400 xy
x
y
Now
200 yz 2 200 xz 2
=
x
y
⇒
y2 = x2
⇒
y = ±x
(5)
and
200 xz 2
= 400 xy
y
⇒
z2 = 2 y2
⇒ z = ± 2⋅y
(6)
Substituting in (1) we get
⇒ 4 y2 = 1
y2 + y2 + 2 y2 = 1
∴
M05_ENGINEERING_MATHEMATICS-I _CH05_Part B.indd 60
x=±
1
2
and
⇒ y2 =
1
4
⇒
y=±
1
2
1
⎛ 1⎞
z = ± 2 ⋅⎜± ⎟ = ±
⎝ 2⎠
2
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Differential Calculus of Several Variables ■
5.61
1
The stationary points are given by x = ± ,
2
1
1
y=± , z=±
2
2
These give 8 stationary points. We want the maximum value of T = 400 xyz2, and so we must have
xy positive. This will occur at 4 of the points.
1 1 1 ⎞ ⎛ 1 1 −1 ⎞ ⎛ 1 1 1 ⎞ ⎛ 1 1
1 ⎞
i.e., at the points ⎛⎜ , ,
, , ,
, − ,− ,
, − ,− ,−
⎟
⎝ 2 2 2 ⎟⎠ ⎜⎝ 2 2 2 ⎟⎠ ⎜⎝ 2 2 2 ⎟⎠ ⎜⎝ 2 2
2⎠
1 1 1
∴ maximum T = 400 × ⋅ ⋅ = 50°C
2 2 2
EXAMPLE 4
Find the volume of the largest rectangular parallelopiped that can be inscribed in the ellipsoid
x2
y2 z2
1
1 51.
a2
b2 c 2
Solution.
Given the ellipsoid
x2 y2 z2
+
+
=1
a2 b2 c2
By the symmetry of the ellipsoid, for the largest parallelopiped, the edges must be parallel to the
coordinate axes and the centre coincides with the centre (0, 0, 0) of the ellipsoid.
Let P (x, y, z) be the coordinates of a vertex on the
ellipsiod, then the dimensions of the rectangular
Z
parallelopiped (or cuboid) are 2x, 2y, 2z respectively.
∴
volume V = 2x ⋅ 2y ⋅ 2z
= 8xyz
Let f( x , y , z ) =
P
x2 y 2 z 2
+
+
−1 = 0
a2 b 2 c 2
(1)
O
We want to maximize V subject to f(x, y, z) = 0
Form the auxiliary equation F(x, y, z) = V + lf (x, y, z),
where l is the Lagrange’s multiplier.
⎛ x2 y 2 z 2 ⎞
⇒ F( x , y , z ) = 8xyz + l ⎜ 2 + 2 + 2 − 1⎟
⎝a
⎠
b
c
Y
X
Fig. 5.8
lx
ly
lz
Fy = 8 zx + 2 2 ,
Fz = 8 xy + 2 2 and
Fl = f
2 ,
a
b
c
To find stationary points solve Fx = 0,
Fy = 0,
Fz = 0,
f=0
∴
∴
Fx = 8yz + 2
Fx = 0 ⇒ 8yz + 2
lx
=0
a2
⇒
M05_ENGINEERING_MATHEMATICS-I _CH05_Part B.indd 61
⇒
4 yz = −
lx
a2
⇒
4 xyz = −l
−
4 xyz x 2
= 2
l
a
x2
a2
[multiplying by x]
(2)
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y
=0
b2
⇒
−
4 xyz y 2
= 2
l
b
[multiplying by y]
(3)
z
=0
a2
∴ From (2), (3) and (4), we get
⇒
−
4 xyz z 2
= 2
l
b
[multiplying by z]
(4)
Fy = 0 ⇒ 8xz + 2l
Fz = 0 ⇒ 8xy + 2l
x2 y2 z2
=
=
a2 b2 c2
We have
x2 y2 z2
+
+
=1
a2 b2 c2
∴
x2 x2 x2
+
+
=1 ⇒
a2 a2 a2
x2
=1 ⇒
a2
x2 =
a2
3
⇒
x=±
a
3
c
and z = ±
3
3
So, the stationary points are given by
Similarly, y = ±
∴
3
b
x=±
a
3
,y=±
b
3
,z=±
c
3
∴ there are 8 stationary points.
Since we want maximum value of V, choose the points with the product of xyz positive.
This will occur at 4 of the points. They are
⎛ a b c ⎞ ⎛ a −b −c ⎞ ⎛ −a −b c ⎞ ⎛ −a b −c ⎞
,
,
,
,
,
,
,
,
⎟,⎜
⎟
⎜⎝
⎟,⎜
⎟, ⎜
3
3 3⎠ ⎝ 3 3 3⎠ ⎝ 3 3 3⎠ ⎝ 3 3 3⎠
∴
maximum V =
8abc
3 3
EXAMPLE 5
Divide the number 24 into three parts such that the continued product of the first, square of the
second and the cube of the third may be maximum.
Solution.
Let 24 be divided into 3 parts x, y, z, so that
x + y + z = 24
∴
x + y + z − 24 = 0
2
where x, y, z > 0
(1)
3
and the product is xy z
We have to maximize this product subject to (1)
Let f(x, y, z) = xy2 z3 and f (x, y, z) = x2 + y2 + z2 − 24
Form the auxiliary function
F(x, y, z) = f(x, y, z) + lf(x, y, z)
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Differential Calculus of Several Variables ■
5.63
where l is the Lagrange’s multiplier.
⇒ F( x , y , z ) = xy 2 z 3 + l ( x + y + z − 24)
∴
Fx = y 2 z 3 + l,
Fy = 2 xyz 3 + l ,
To find stationary points, solve Fx = 0,
∴
Fz = 3 xy 2 z 2 + l and Fl = f
Fy = 0,
Fz = 0,
f=0
Fx = 0 ⇒
y 2z 3 + l = 0
⇒
y 2 z 3 = −l
(2)
Fy = 0 ⇒
2xyz 3 + l = 0
⇒
2xyz 3 = −l
(3)
Fz = 0 ⇒ 3xy 2 z 2 + l = 0
⇒
3xy 2 z 2 = −l
(4)
y 2 z 3 = 2xyz 3 = 3xy 2 z 2
From (2), (3), (4)
∴
y 2 z 3 = 2xyz 3 ⇒
y = 2x
and
xy 2 z 3 = 3 xy 2 z 2 ⇒
z = 3x
Subsituting in (1), we get
x + 2 x + 3 x = 24
∴
y=8
⇒
and
6 x = 24 ⇒ x = 4
z = 12
∴ the product is maximum if the parts are 4, 8, 12.
Note We shall test it is indeed maximum.
⎡0
⎢
⎢fx
The bordered Hessian matrix is H = ⎢
f
⎢ y
⎢fz
⎣
Now
Fx = y2z3 + l
⇒
⇒
⇒
fx
fy
Fxx
Fxy
Fyx
Fyy
Fzx
Fzy
fz ⎤
⎥
Fxz ⎥
Fyz ⎥
⎥
Fzz ⎥⎦
Fxx = 0,
Fxy = 2 yz 3 ,
Fxz = 3 y 2 z 2
Fy = 2 xy 2 z 3 + l ⇒
Fyy = 2 xz 3 ,
Fyx = 2 yz 3 ,
Fyz = 6 xyz 2
Fz = 3 xy 2 z 2 + l ⇒
Fzz = 6 xy 2 z ,
Fzx = 3 y 2 z 2 ,
Fzy = 6 xyz 2
When x = 4, y = 8, z = 12
Fxx = 0, Fxy = Fyx = 2 ⋅ 8 123 = 24 ⋅123 , Fyy = 2 ⋅ 4 ⋅123 = 23 ⋅123
Fyz = Fzy = 6 ⋅ 4 ⋅ 8 122 = 24 ⋅123 , Fzx = Fxz = 3.82 ⋅122 = 24 ⋅123
Fzz = 6 ⋅ 4 ⋅ 82 ⋅12 = 27 ⋅122
Now f(x, y, z) = x + y + z − 24
⇒
M05_ENGINEERING_MATHEMATICS-I _CH05_Part B.indd 63
fx = 1,
fy = 1,
fz = 1
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∴ the bordered Hessian matrix is
1
⎡0
⎢1
0
H= ⎢
⎢1 24 ⋅123
⎢
4
3
⎣1 2 ⋅12
1
2 ⋅123
23 ⋅123
24 ⋅123
4
1 ⎤
2 ⋅123 ⎥⎥
24 ⋅123 ⎥
⎥
27 ⋅122 ⎦
4
⎡0 1 ⎤
The principal bordered minors are D1 = ⎢
⎥ = −1 < 0
⎣1 0⎦
0
1
D2 = 1
0
4
1 2 ⋅123
1
24 ⋅123
23 ⋅123
= −1[23 ⋅123 − 24 ⋅123 ] + 1⋅ [24 ⋅123 ]
= −8 ⋅123 + 16 ⋅123 + 16 ⋅123 = 24 ⋅123 > 0
0
1
1
0
D3 =
4
1 2 ⋅123
1 24 ⋅123
1
2 ⋅123
23 ⋅123
24 ⋅123
1
2 ⋅123
24 ⋅123
27 ⋅122
0
1
1
0
=
4
1 2 ⋅123
1 24 ⋅123
0
24 ⋅123
−8 ⋅123
0
0
24 ⋅123
0
6
−2 ⋅122
4
1 24 ⋅123
= −1 1 −8 ⋅123
1
0
4
24 ⋅123
0
−26 ⋅122
C3 → C3 − C 2
C4 → C4 − C2
expanding by R1
1 2 12
= ( −1)2 ⋅12 ⋅ 2 ⋅12 1 −1 0
1 0 4
3
3
4
[Taking out 23⋅123 from c2 and 24⋅122 from c3]
2
= ( −1)27 125 [1⋅12 − 4( −1 − 2)]
[expanding by R 3 ]
= ( −1)2 ⋅12 [12 + 12] = ( −1)2 ⋅12 ⋅ 24 < 0
7
5
7
5
Since D1 < 0, D2 < 0 and D3 < 0, f(x, y, z) is maximum at (4, 8, 12).
EXAMPLE 6
Find the maximum value of xm yn zp subject to x 1 y 1 z 5 a.
Solution.
Let f(x, y, z) = xm yn zp
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(1)
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Differential Calculus of Several Variables ■
Maximize (1) subject to
Form the auxiliary function
f(x, y, z) = x + y + z − a = 0
5.65
(2)
F(x, y, z) = f(x, y, z) + lf(x, y, z)
where l is the Lagrange’s multiplier.
⇒
∴
F( x, y, z ) = x m y n z p + l( x + y + z − a)
Fy = nx m y n −1 z p + l, Fz = px m y n z p −1 + l and Fl = f
Fx = mx n −1 y n z p + l ,
To find stationary points, solve Fx = 0,
Similarly,
Fy = 0, Fz = 0,
f=0
mx n −1 y n z p + l = 0 ⇒ mx n −1 y n z p = −l
Fx = 0
⇒
Fy = 0
⇒
nx m y n −1z p = −l
(3)
Fz = 0
⇒
px m y n z p−1 = −l
(4)
(2)
∴ From (2), (3) and (4), we get
mx m −1 y n z p = nx m y n −1 z p = px m y n z p −1
⇒
m n p
= =
x y z
⇒
x y z
x+ y+z
a
= = =
=
m n p m+n+ p m+n+ p
∴
x=
[dividing by x m y n z p ]
am
,
m+n+ p
y=
an
ap
, z=
m+n+ p
m+n+ p
am
an
ap ⎞
Stationary point is ⎛⎜
,
,
⎝ m + n + p m + n + p m + n + p ⎟⎠
m
n
am ⎞ ⎛
an ⎞ ⎛
ap ⎞
Maximum value of f = ⎛⎜
⎟
⎜
⎟
⎜
⎝ m + n + p ⎠ ⎝ m + n + p ⎠ ⎝ m + n + p ⎟⎠
=
p
a m + n + p ⋅ m m nn p p
( m + n + p) m + n + p
EXERCISE 5.5
Find the extreme values of the function f(x, y) = x3 + y3 − 3x − 12y + 20.
Find the maximum and minimum values of x2 − xy + y2 − 2x + y.
Find the maximum and minimum values of x3 + 3xy2 − 15y2 + 72x.
Find the maxima and minima of the function x3y2(12 − x − y).
1 1
5. Find the extreme values of the function x 2 + xy + y 2 + + .
x y
1.
2.
3.
4.
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6. Find the extreme values of the function y4 − x4 + 2(x2 −y2).
7. If the perimeter of a triangle is a given constant, show that the area is maximum when the triangle
is equilateral.
8. Locate the stationary points of x4 + y4 − 2x2 + 4xy − 2y2 and determine their nature.
9. Examine maximum and minimum values of sinx + siny + sin (x + y), 0 < x < p, 0 < y <p.
10. Find the dimensions of the rectangular box open at the top, of maximum capacity whose surface
is 432 sq.cm.
11. Examine f(x, y) = x3 + y3 − 3axy for maximum and minimum values.
12. Find the extreme values, if any, of the function f(x, y) = x4 + y2 + x2y.
13. A flat circular plate is heated so that the temperature at any point (x, y) is u(x, y) = x2 + 2y2 − x.
Find the coldest point on the plate.
14. In a plane triangle, find the maximum value of cos A cos B cos C.
15. Find the minimum value of x2 + y2 + z2, given that xyz = a3.
16. Find the maximum and minimum distances of the point (3, 4, 12) from the sphere x2 + y2 + z2 = 1.
17. Find the maximum and minimum distances from the origin to the curve 5x2 + 6xy+ 5y2 − 8 = 0.
18. Find the minimum values of the function x2 + y2 + z2 subject to the condition
ax + by + cz = a + b + c.
19. Find the minimum values of the function x2 + y2 + z2 subject to the condition xy + yz + zx = 3a2.
20. A thin closed rectangular box is to have one edge equal to twice the other and a constant volume
72 m3. Find the least surface area of the box.
21. Show that the area of the greatest rectangle that can be inscribed in an ellipse 4x2 + 9y2 = 36, having its sides parallel to the axes is 12 sq. units.
22. Find the shortest distance from the origin to the curve x2 + 8xy + 7y2 = 225.
23. Using Lagrange’s method of multipliers, show that the stationary value of a3x2 + b3y2 + c3z2, where
a+b+c
a+b+c
a+b+c
1 1 1
,y =
, and z =
.
+ + = 1, occurs at x =
a
b
c
x y z
24. If u = x2 + y2 − z2 where ax + by + cz − p = 0, then find the stationary value of u.
25. Find the minimum value of x2yz3 subject to 2x + y + 3z = a.
5xyz
26. If xyz = 8, find the values of x, y, z for which u =
is a maximum.
x + 2y + 4z
27. Test for the extreme of the function f(x, y) = x4 + y4 − x2 − y2 − 1.
ANSWERS TO EXERCISE 5.5
1. (i) Minimum at (1, 2); Minimum value = 2
(ii) Maximum at (−1, −2); Maximum value = 38
2. Minimum at (1, 0); Minimum value = −1
3. (i) Maximum at (4, 0); Maximum value = 112
(ii) Minimum at (6, 0); Minimum value = 108
5. Minimum at ⎛ 1 , 1 ⎞ ; Minimum value = 34 / 3
⎜⎝ 3
⎟
3 3 3⎠
4. Maximum at (6, 4); Maximum value = 6912
6. (i) Minimum at (0, ±1); Minimum value = −1 and (ii) Maximum at (±1, 0); Maximum value = 1
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Differential Calculus of Several Variables ■
5.67
8. (i) Stationary points are (0, 0) ( 2 , − 2 ) and ( − 2 , 2 ) and Minimum at ( 2 , − 2 ) and
(− 2, 2 )
p p
3 3
9. Maximum at ⎛⎜ , ⎞⎟ ; and maximum value
⎝ 3 3⎠
2
10. Dimensions 12, 12, 6 cms.
11. Maximum value = −a3 if a < 0 and minimum value = −a3 if a > 0
1
14. 1
12. Minimum value 0
13. ⎛⎜ , 0⎞⎟
15. Minimum value = 3a2
⎝2 ⎠
8
16. Maximum distance = 14, minimum distance = 12
17. Maximum distance = 4, and minimum distance = 1
2
18. Minimum value = ( a + b + c)
19. Minimum value = 3a2
2
2
2
a
+
b
+
c
20. 108 m2
22. shortest distance = 5
P2
a+b+c
a+b+c
a+b+c
23. x =
24. u = 2
,y=
,z=
(a + b 2 + c 2 )
a
b
c
a
25. ⎛⎜ ⎞⎟
⎝ 6⎠
6
26. x = 4, y = 2, z = 1
1 ⎞
3
⎛ 1
and value = −
,±
27. Maximum at (0, 0) and value = −1 and minimum at ⎜ ±
⎟
⎝
2
2
2⎠
5.6
ERRORS AND APPROXIMATIONS
All physical quantities, when measured however precisely, are subjected to small errors in their
observed values, resulting in a cumulative error in the dependent variable. Our aim is to estimate such
error.
If u = f(x, y), then the total differential relation is,
∂f
∂f
du =
dx +
dy , which is an exact relation.
∂x
∂y
An approximate error relation is obtained by replacing differentials by increments Δx, Δy and Δu.
∂f
∂f
∴Δu =
Δx + Δy , giving the total error Δu in u approximately in terms of the errors Δx and
∂x
∂y
Δy in x and y, respectively.
Δx, Δy and Δu are called the absolute errors in x, y and u, respectively.
Δx Δy
Δu
,
and
are called the relative errors in x, y and u, respectively.
x y
u
Δx
Δy
Δu
× 100,
× 100 and
× 100 are called the percentage errors in x, y and u, respectively.
x
y
u
If relative error or percentage errors are given in problems, it is advantages to take logarithm and
differentiate.
If y = f(x), then dy = f ′(x) dx
∴
Δy = f ′(x) Δx approximately
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WORKED EXAMPLES
EXAMPLE 1
The dimensions of a cone are given by radius 5 4 cm, and altitude 5 6 cm. What is the error in
calculation of its volume, if there is a shortage of 0.01 cm in the measures used.
Solution.
Let r be the radius and h be the height of the cone in centimetre.
1
The volume of the cone V = pr 2 h
3
∴
(1)
The differential relation is
dV =
∂v
∂V
dr +
dh
∂r
∂h
ΔV =
∂V
∂V
Δr +
Δh
∂r
∂h
The error relation is
∂V 1
= ph2r and
∂r 3
From equation (1), we get
∴
ΔV =
∂V 1 2
= pr
∂h 3
1
1
2phr Δr + pr 2 Δh
3
3
Given r = 4 cm and h = 6 cm,
Δr = Δh = −0.01 = −
∴
ΔV =
∴
(2)
1
100
2p
⎛ 1 ⎞ p 2⎛ 1 ⎞
×6 × 4⎜−
+ 4 −
⎝ 100 ⎟⎠ 3 ⎜⎝ 100 ⎟⎠
3
16p 3
p
64p
=−
=−
cm
[48 + 16] = −
75
300
300
the volume is decreased by
16p 3
cm .
75
EXAMPLE 2
8pIl
. If l is decreased
t 2r 4
by 2%, r is increased by 2% and t is increased by 1.5%, then show that the value of N is diminished
by 13% approximately.
The torsional rigidity of length of a wire is obtained from the formula N 5
Solution.
8pIl
t 2r 4
Taking logarithm on both sides of equation (1), we get
Given N =
(1)
log e N = log e 8pI + log e l − 2 log e t − 4 log e r
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Differential Calculus of Several Variables ■
5.69
Taking differentials, we get
∴
∴ Error relation is
dN
dl
dt
dr
= 0+ −2 −4
N
l
t
r
ΔN Δl
Δt
Δr
=
−2 −4
N
l
t
r
ΔN
Δl
Δt
Δr
× 100 =
× 100 − 2 × 100 − 4
× 100
N
l
t
r
∴
Given: l is decreased by 2%.
∴
r is increased by 2%
∴
t is increased by 1.5%.
∴
Δl
× 100 = −2
l
Δr
× 100 = 2
r
Δt
× 100 = 1.5
t
ΔN
× 100 = −2 − 2(1.5) − 4 × 2 = −2 − 3 − 8 = −13
N
∴ the value of N is diminished by 13%.
∴
EXAMPLE 3
In estimating the cost of a pile of bricks measured as 6 m 3 50 m 3 4 m, the tape is stretched
1% beyond the standard length. If the count is 12 bricks in 1 m3 and bricks cost `100 per 1000,
then find the approximate error in cost.
Solution.
Let l, b and h be the length, breadth and height of the rectangular pile, respectively, in meters. Hence,
its volume
V=lbh
(1)
Taking loge on both sides, we get
log e V = log e l + log e b + log e h
Taking differentials, we get
1
1
1
1
dV = dl + db + dh
V
l
b
h
∴ Error relation is
M05_ENGINEERING_MATHEMATICS-I _CH05_Part B.indd 69
ΔV
Δl Δb Δh
=
+
+
V
l
b
h
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Given V = 6 × 50 × 4 = 1200 m3
Δl
Δb
Δh
× 100 =
× 100 =
× 100 = 1
l
b
h
and
Δl Δb Δh
1
=
=
=
l
b
h 100
⇒
ΔV
3
3
3
=
⇒ ΔV =
×V =
× 1200 = 36 m3
V
100
100
100
Given the number of bricks in 1 cubic meter is 12.
∴
∴ number of bricks in ΔV = 36 × 12 = 432
Cost of 1000 bricks is `100
∴
Cost of 432 bricks is = 432 ×
⇒
100
= 43.2
1000
Error in cost is = `43.20
EXAMPLE 4
1 1 2
2 5 . If equal errors d are made in
v
u f
Δf
the determinations of u and v, then show that the relative error
in focal length is given by
f
⎛1 1⎞
d⎜ 1 ⎟ .
⎝u v ⎠
The focal length of a mirror is given by the formula
Solution.
Given
1 1 2
− =
v u f
⇒ 2 f −1 = v −1 − u −1
Taking differential,
−
∴
2
1
1
df = − 2 dv + 2 du
2
f
v
u
Error relation is
−2
Δu = Δv = d
Given
∴
⇒
Δf
Δv Δu
=− 2 + 2
v
u
f2
d d
2 Δf
− ⋅
=− 2+ 2
f f
v
u
−
2 Δf
1⎞
⎛ 1 1⎞ ⎛ 1 1⎞
⎛ 1 1⎞ ⎛ 2 ⎞
⎛ 1
⋅
= d⎜ 2 − 2 ⎟ = d⎜ + ⎟ ⎜ − ⎟ = d⎜ + ⎟ ⋅⎜ − ⎟
⎝ u v⎠ ⎝ u v⎠
⎝ u v⎠ ⎝ f ⎠
⎝u
f f
v ⎠
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Differential Calculus of Several Variables ■
Δf
⎛ 1 1⎞
= d⎜ + ⎟
⎝u v ⎠
f
⇒
∴
5.71
⎛ 1 1⎞
the relative error in f is d ⎜ + ⎟ .
⎝u v ⎠
EXAMPLE 5
If a triangle be slightly disturbed so as to remain inscribed in the same circle, then prove that
Δa
Δb
Δc
1
1
5 0.
cos A
cos B cos C
A
Solution.
We know from trigonometry that
A
c
b
R
a
b
c
=
=
= 2R
sin A sin B sin C
B
⇒
a = 2 R sin A , b = 2 R sin B
and
c = 2R sin C
B
C
a
(1)
C
Fig. 5.9
When the triangle inscribed in the circle of radius R is slightly disturbed, the sides and angles will be
slightly changed. Here R is constant, since the circle is not changed.
Taking differentials to the relation (1), we get
da = 2 R cos A ⋅ dA,
db = 2 R cos B ⋅ dB
and
dc = 2 R cos C ⋅ dC
∴ the error relations are
Δa = 2 R cos A ⋅ ΔA,
Δb = 2 R cos B ⋅ ΔB
and
Δa
Δb
= 2 R ΔA,
= 2 R ΔB
and
cos A
cos B
∴
∴
Δc = 2 R cos C ⋅ ΔC
Δc
= 2 R ΔC
cos C
Δa
Δb
Δc
+
+
= 2R [ ΔA + ΔB + ΔC ]
cos A cos B cos C
But
∴
Hence,
A + B + C = p,
dA + dB + + dC = 0
Δa
Δb
Δc
+
+
= 2R ⋅ 0 = 0
cos A cos B cos C
and
ΔA + ΔB + ΔC = 0
EXAMPLE 6
The angles of a triangle ABC are calculated from the sides a, b, and c. If small errors
da, db, and dc are made in the measurements of the sides, then show that the error in the angle
a
A is approximately dA 5
( da 2 db cos c 2 dc cos c ).
2Δ
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Solution.
From trigonometry, we know that the relation between sides and angle A of a triangle is
a2 = b 2 + c 2 − 2bc cos A.
Taking differentials, we get
⇒
2ada = 2bdb + 2cdc − 2[cos A(bdc + cdb) − (bc sin AdA)]
ada = bdb + cdc − b cos Adc − c cos Adb + bc sin AdA
bc sin AdA = ada − bdb − cdc + b cos Adc + c cos Adb
Since area of ΔABC is
Δ=
1
bc sin A ⇒ bc sin A = 2 Δ
2
∴
2Δ ⋅ dA = ada − (b − c cos A )db − (c − b cos A )dc
Given that the errors in a, b and c are da, db and dc respectively.
So, the error in A is dA
∴
2ΔdA = ada − (b − c cos A )db − (c − b cos A )dc
We know by projection formula that
and
∴
⇒
b = c cos A + a cos C
⇒ b − c cos A = a cos C
c = a cos B + b cos A
⇒
c − b cos A = a cos B
2ΔdA = ada − a cos C db − a cos Bdc = a(da − cos C db − cos Bdc)
dA =
a
[da − cos C ⋅ db − cos B ⋅ dc ]
2Δ
EXERCISE 5.6
1. The pressure p and the volume v of a gas are connected by pv1.4 = c, a constant. Find the percent1
age increase in pressure corresponding to % diminish in the volume.
2
2. If q is calculated from q = Kr 2 h , where K is a constant, then show that a small % error in r is
four times as serious as the same % error in h.
3. Find the percentage error in the area of an ellipse if one per cent error is made in measuring the
major and minor axes.
1
E
4. The resistance of a circuit was found by using the formula C = . If there be possible errors of
10
R
1
amperes in C and
volt in E, what is the possible error in R, given C = 18 amp, E = 100 volts.
20
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Differential Calculus of Several Variables ■
5.73
5. The range R of a projectile which starts with a velocity v at an angle of elevation a is given by
1
v2
R=
sin 2a. Find the percentage error in R due to an error of 1% in v and an error of % in a.
2
g
l
6. The time T of a complete oscillation of a simple pendulum of length l is given by T = 2p
,
g
where g is a constant. Find the approximate error in the calculated value of T corresponding to an
error of 2% in the value of l.
7. The work that must be done to propel a ship of displacement D for distance s in time t is
3
s 2D 2
proportional to
. Estimate roughly the percentage increase of work necessary where the
t2
distance is increased by 1%, the time is diminished by 1% and the displacement of the ship is
diminished by 3%.
1
8. If the area of a triangle is calculated from the formula Δ = bc sin A , and if b and c are measured
2
correctly, but A is taken as 60° with possible error of 5′, then calculate the possible percentage of
error is Δ.
ANSWERS TO EXERCISE 5.6
1. 0.7%
3. 2%
4. 0.028 ohm
5. ( 2 + a cot 2a)%
6. 1%
1
7. − %
2
8. 0.084%
SHORT ANSWER QUESTIONS
⎛x 1 y ⎞
∂z
∂z
, then show that x
1. If u 5 log ⎜
1 y 5 3.
∂x
∂y
⎝ x 1 y ⎟⎠
4
2. If u 5 sin21
4
x
y
∂u
∂u
1 tan21 , then prove that x
1y
5 0.
y
x
∂x
∂y
∂u
∂u
u
⎛ x1 y ⎞
, then show that x
1y
1 cot 5 0.
⎜⎝ x 1 y ⎟⎠
∂x
∂y
2
∂z
∂z
∂z
4. If z 5 f(x, y), x 5 u 2 v, y 5 uv, then prove that (u 1 v )
5u
2v
.
∂x
∂u
∂v
du
5. If u = x2 1 y2, x 5 e2t, y 5 e2t cos 3t, then find
as a total derivative.
dt
3. If u 5 2cos21
6. If u 5 f(x, y) and x 5 r 1 s, y 5 r 2 s, then prove that
7. Find
∂u ∂u
∂u
1
52 .
∂r
∂s
∂x
x
du
if u = , x = e t ; y = log e t .
y
dt
∂u
∂u
∂u
y
z
1y
1z .
1 , then find x
∂x
∂y
∂z
x x
dy
if xy 1 yx 5 a, a is a constant and x, y > 0.
9. Find
dx
8. If u 5
10. Find
dy
if xy 5 yx.
dx
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⎛ y⎞
∂2 z
∂2 z
11. If z 5 sin ⎜ ⎟ , then show that
5
.
⎝ x⎠
∂y ∂x ∂x ∂y
∂2 z
∂2 z
12. If z 5 f(x 1 ay) 1 g(x 2 ay), where a is constant, then prove that 2 5 a 2 2 .
∂y
∂x
∂( r , u )
13. If x = r cos u, y = r sin u, then find
.
∂( x , y )
14. If x 5 u(1 1 v), y 5 v(1 1 u), then find ∂( x , y ) .
∂( u , v )
y2
x2
∂( x , y )
.
15. If u 5 , v 5 , then find
x
y
∂( u , v )
∂( u , v )
16. If u 5 x2 + y2, v 5 2xy and x 5 r cos u, y 5 r sin u, then find
.
∂( r , u )
17. If x2y 1 3y 2 2 is expanded as Maclaurin’s series, then find the value at the point (1, 0).
18. Find the Taylor’s series expansion of xy near the point (1, 1) upto first degree terms.
19. Find the stationary points of f(x, y) 5 x3 1 3xy2 2 15x2 2 15y2 1 72x.
20. Find the stationary points of f(x, y) 5 x3 1 y3 2 3x 2 12y 1 20
OBJECTIVE TYPE QUESTIONS
A. Fill up the blanks
⎛ x − y⎞
∂u
∂u
1. If u = sin −1 ⎜
⎟ then x ∂x + y ∂y = ____________.
⎝ x + y⎠
y z
∂u
∂u
+ y = _________.
2. If u = + then x
x x
∂x
∂y
3. The slope of the curve 2xy − ln xy = 2 at the point (1, 1) is ___________.
4. If x = r cosu, y = r sinu then
5. If x3 + y3 = 3axy, then
∂r ∂x
. = ___________.
∂x ∂r
dy
= _____________.
dx
6. If u = x + y, y = uv, then
∂( x , y )
= ___________.
∂( u , v )
2x − y
y
∂( x , y )
, v = , then
= ___________.
2
2
∂( u , v )
8. The necessary and sufficient conditions for f(x, y) to have a relative maximum at the point (a, b) is
___________.
9. The maximum value of f(x, y) = x2 + y2 subject to the constraint x = 1 is _________.
10. If the point (1, 1) is a stationary point of f(x, y) and if fxx = 6xy3, fxy = 9x2y, and fyy = 6x2y, then the point (1, 1)
is ___________.
7. If u =
B. Choose the correct answer
∂u
∂u ∂u
x y z
+y
+
1. If u = + + , then x
is equal to
∂x
∂y ∂z
y z x
(a) u
(b) 0
M05_ENGINEERING_MATHEMATICS-I _CH05_Part B.indd 74
(c) 2u
(d) none of these
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Differential Calculus of Several Variables ■
2. If u2 + 2v2 = 1 − x2 + y2 and u2 + v2 = x2 + y2 − 2, then
(a) x
(b) 3x
3. If u = f(y − z, z − x, x − y), then
(a) x + y + z
∂v
is equal to
∂x
3x
(c)
u
(d) −
5.75
2x
v
∂u ∂u ∂u
is equal to
+ +
∂x ∂y ∂z
(b) 1 + x + y + z
(c) 1
(d) 0
(c) sin u
(d) none of these
⎛ x+ y ⎞
∂u
∂u
4. If u = tan −1 ⎜
⎟ , then x ∂x + y ∂y is equal to
⎝ x + y⎠
(a) 2 cos2u
(b) sin2 u
5. If x = r cosu, y = r sinu, then the value of
(a) 0
∂ 2u ∂ 2u
is
+
∂x 2 ∂y 2
(b) 1
∂r
∂x
(d)
∂r
∂y
(c) 2u
(d)
−
(c)
2
2
2
x 2 + y 2 + z 2 , then ∂ u + ∂ u + ∂ u is equal to
∂x 2 ∂y 2 ∂z 2
6. If u =
(a) 4u
(b)
2
u
u
2
∂u
∂u
⎛ y⎞
+y
is equal to
7. If u = sin −1 ⎜ ⎟ , then x
⎝ x⎠
∂x
∂y
(b) u
(a) 0
(c)
u
2
(d) 2u
2
2
∂f
∂f
−1 ⎛ x + y ⎞
is equal to
8. If f = sin ⎜
, then x + y
⎝ x + y ⎟⎠
∂x
∂y
(b) 2f
(a) f
9. If z = e x sin y , where x = ln t and y = t2, then
(c) tan f
dz
is
dt
(a)
ex
(sin y − 2t 2 cos y )
t
(b)
ex
(sin y + 2t 2 cos y )
t
(c)
ex
(cos y + 2t 2 sin y )
t
(d)
ex
(cos y − 2t 2 sin y )
t
10. If x = u + v and y = u − v, then
(a) 0
11. If u =
(a) y
(d) sin f
∂( x , y )
is equal to
∂(u , v )
(b) 1
(c) 2
(d) − 2
y2
x2 + y2
∂(u , v )
and v =
, then
is equal to
2x
2x
∂( x , y )
(b) x
M05_ENGINEERING_MATHEMATICS-I _CH05_Part B.indd 75
(c)
y
2x
(d)
x
2y
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12. Expansion of ex cos y in powers of x and y upto first degree by Taylor series is
(a) 1 + x
(b) 1 + y
(c) 1 + x + y
(d) none of these
13. The Taylor series expansion of xy about the point (1, 1) upto first degree is
(a) 1 + x
(b) 1 + (x − 1)
(c) 2 + (x − 1)
(d) 1 + (y − 1)
14. The expansion of e is power of (x − 1) and (y − 1) upto first degree terms by Taylor series is
xy
(a) 1 + ( x − 1) + ( y − 1)
(b) e + e ( x − 1) + e( y − 1)
(c) 2e + e( x − 1) + e( y − 1)
(d) none of these
15. The linear approximation of f ( x, y ) = x 2 − xy +
1 2
y + 3 at (3, 2) is
2
(b) 2x − y − 2
(d) none of these
(a) 8 + 4(x − 3) − (y − 2)
(c) 3 − 2(x − 3) − (y − 2)
16. A minimum point of f(x, y) = x2 + y2 + 6x + 12 is
(a) (3, 0)
(b) (−3, 0)
(c) (0, 3)
(d) (0, −3)
17. A stationary point of f(x, y) = x2 − xy + y2 − 2x + y is
(a) (1, 1)
(b) (1, 0)
(c) (0, 1)
(d) (−1, 0)
18. The nature of the stationary point (0, −1) for the function f(x, y), if fxx = 4 − 12x2, fxy = 0, and
fyy= −4 + 12y2 is
(a) minimum point
(b) maximum point
(c) saddle point
(d) cannot decide
dw
19. If w = xy + z, x = cos t, y = sin t, and z = t, then
at t = 0 is
dt
(a) 0
(b) 1
(c) 2
(d) −2
20. The side a and angle A of a triangle ABC remain constant, where as the other element of the triangle slightly
vary, then
(a)
db
dc
+
=0
cos C cos B
(b)
db
dc
=
cos B cos C
(c)
db
dc
+
=0
cos B cos C
(d) none of these
ANSWERS
A. Fill up the blanks
1. 0
2. 0
3. − 1
6. u
7. 2
8. f x = 0, f y = 0, rt − s 2 > 0, r < 0
2
5. ay − x
2
y − ax
9. 1
4. cos2u
10. saddle point
B. Choose the correct answer
1. (b)
2. (d)
3. (d)
4. (d)
11. (c)
12. (a)
13. (b) 14. (b)
M05_ENGINEERING_MATHEMATICS-I _CH05_Part B.indd 76
5. (a)
15. (a)
6. (b)
16. (b)
7. (a)
17. (b)
8. (c)
18. (a)
9. (b)
19. (c)
10. (d)
20. (c)
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Integral Calculus
6
6.0 INTRODUCTION
Calculus is one of the remarkable achievements of human intellect. It is a collection of fascinating
and exciting ideas rather than a technical tool. Calculus has two main divisions ‘differential calculus’
and ‘integral calculus’.
Both had their origin from geometrical problems.
Integral calculus had its origin from the problem of area and differential calculus had its origin
from the problem of tangent to a curve.
The term integration means summation. Infact definite integral is the process of finding a limit of a sum.
The integration symbol ‘∫ ’ was divised by stretching the summation symbol ‘S’ conveying the
meaning of the process.
6.1 INDEFINITE INTEGRAL
We consider indefinite integral as reverse process of differentiation.
Definition 6.1
If f(x) is continuous function of x such that F′(x) 5 f(x) in [a, b], then F(x) 1 c is defined as the
indefinite integral of f(x) and is denoted by ∫ f ( x ) dx.
Thus, ∫ f ( x ) dx = F ( x ) + c
Here f(x) is called the integrand and the arbitrary constant c is called the constant of integration,
dx indicates that the variable of integration is x. f(x)dx is called an element of integration.
F(x) is called an antiderivative or primitive of f(x).
F(x) + c is referred to as the most general primitive.
Note In computing indefinite integral, no interest is shown on the interval [a, b]. It is to be understood
that ∫ f ( x )dx = F ( x ) + c is valid in some suitable sub-interval.
Leibnitz used the symbol
∫ f ( x)dx to denote general primitive of f.
6.1.1 Properties of Indefinite Integral
From the definition of indefinite integral, we have the following properties
d ⎡
f ( x )dx ⎤⎦ = f ( x )
1.
dx ⎣ ∫
2.
3.
4.
5.
∫ d ( f ( x)) = f ( x) + c
d ( ∫ f ( x )dx ) = f ( x )dx
∫ [ f ( x) ± g ( x)]dx = ∫ f ( x)dx ± ∫ g ( x)dx
∫ kf ( x)dx = k ∫ f ( x)dx, where k is a constant.
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Engineering Mathematics
Table of Integrals
Now we list the standard indefinite integrals derived from the derivatives of standard functions
x n +1
if n ≠ −1
n +1
1.
n
∫ x dx =
3.
∫ e dx = e
5.
7.
dx
= log e x + c
x
2.
∫
4.
∫ sin x dx = − cos x + c
∫ cos xdx = sin x + c
6.
∫ tan xdx = log
∫ sec x dx = log
8.
∫ cos ec x dx = − log
x
x
+c
e
sec x + tan x + c
⎛p x⎞
or loge tan ⎜ + ⎟ + c
⎝ 4 2⎠
9. ∫ cot xdx = log sin x + c
e
sec x + c
e
cos ec x + cot x + c
or loge tan
10.
∫ sec
2
x
+c
2
xdx = tan x + c
11.
∫ cos ec xdx = − cot x + c
12.
∫ sec x tan x dx = sec x + c
13.
∫ cos ec x cot x dx = − cos ec x + c
14.
∫a
15.
∫a
a+ x
dx
1
log e
+c
=
2
a−x
2a
−x
16.
∫
17.
∫
18.
∫
19.
∫
21.
n
∫ [ f ( x)] f ′( x)dx =
23.
∫
a 2 − x 2 dx =
x a2 − x 2 a2
x
+ sin −1 + c
2
2
a
24.
∫
a 2 + x 2 dx =
x x 2 + a2 a2
+ log e ( x + x 2 + a 2 ) + c
2
2
25.
∫
x 2 − a 2 dx =
x x 2 − a2 a2
− log e ( x + x 2 − a 2 ) + c
2
2
2
2
dx
a −x
dx
2
2
x 2 − a2
dx
a +x
2
2
= sin −1
x
+ c , − a < x < a, a > 0.
a
= loge x + x 2 − a2 + c , x > a > 0
= log e x + a 2 + x 2 + c
[ f ( x )]n +1
+ c if n ≠ −1
n +1
26. If ∫ f ( x )dx = F ( x ) + c, then
27.
x
∫ a dx =
x
dx
1
= tan −1 + c , a ≠ 0
2
a
a
+x
x−a
dx
1
log e
+c
=
2
2
x+a
2a
x −a
2
20.
∫x
22.
∫
dx
x −a
2
2
=
1
x
sec −1 + c
a
a
f ′( x )
dx = log e f ( x ) + c
f ( x)
1
∫ f (ax + b)dx = a F (ax + b) + c
ax
+ c , a > 0, a ≠ 1
log e a
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Integral Calculus ■
6.3
In the above formula the derivative of the R.H.S is the integrand.
In the evaluation of the integrals, three main techniques are used.
They are
1. Integration by substitution
2. Integration by partial fractions
3. Integration by parts
6.1.2 Integration by Parts
If u and v are differentiable function of x, then ∫ uv dx = uv1 − ∫ u ′ v1dx
du
, v1 = ∫ v dx
dx
Integration by parts is used when the integrand is a product of two functions.
The success of this method depends upon the proper choice of u as that function which comes first
in the word ‘ILATE’, where
I – inverse circular function,
L – logarithmic function
A – algebraic function,
T – Trigonometric function
E – exponential function
where u ′ =
6.1.3 Bernoulli’s Formula
If u and v are differentiable functions of x, then
∫ uv dx = uv
1
− u ′ v2 + u ″ v3 − u ″′ v4 + …
where primes denote differentiation and suffixes denote integration.
du
,
dx
That is
u′ =
and
v 1 = ∫ vdx ,
u″ =
d 2u
,
dx 2
v 2 = ∫ v 1dx ,
u ″′ =
d 3u …
,
dx 3
v 3 = ∫ v 2dx ,
v 4 = ∫ v 3dx , …
If u is a polynomial in x, then Bernoulli’s formula terminates.
6.1.4 Special Integrals
1.
∫ e [ f ( x) + f ′( x)] dx = e
2.
ax
∫ e cos bx dx =
e ax
[a cos bx + b sin bx ]
a2 + b2
3.
∫e
e ax
[a sin bx − b cos bx ]
a + b2
x
ax
sin bx dx =
x
f ( x) + c
2
Solution.
To prove ∫ e ax sin bx dx =
e ax
[a sin bx − b cos bx ]
a2 + b2
Let I = ∫ e ax sin bx dx.
It is a product of two functions. So, we use integration by parts to evaluate the integral.
Taking
u = e ax ,
u ′ = ae ax .
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■
Engineering Mathematics
v 1 = ∫ vdx = ∫ sin bx dx = −
v = sin bx ,
and
cos bx
b
I = uv1 − ∫ u ′ v1dx
[
⎡ cos bx ⎤
ax ⎡ cos bx ⎤
= e ax ⎢ −
⎥ − ∫ ae ⎢ − b ⎥ dx
b
⎣
⎦
⎣
⎦
e ax coss bx a ax
=−
+ ∫ e cos bx dx
b
b
ax
e cos bx a ⎧ e ax sin bx
sin bx ⎫
=−
+ ⎨
− ∫ ae ax
⎬ dx
b
b⎩
b
b ⎭
I=−
⇒
I=
[Again integrating by parts]
a2
e ax cos bx a ax
+ 2 e sin bx − 2 ∫ e ax sin bx dx
b
b
b
a2
e ax
[
a
sin
bx
−
b
cos
bx
]
−
I
b2
b2
⇒
⎛ a2 ⎞
e ax
1
+
I
=
[a sin bx − b cos bx ]
⎜⎝ b 2 ⎟⎠
b2
⇒
⎛ a2 + b2 ⎞
e ax
I
=
[a sin bx − b cos bx ]
⎜⎝ b 2 ⎟⎠
b2
[
⇒
I=
e ax
[a sin bx − b cos bx ]
a2 + b2
ax
∫ e sin bx dx =
e ax
[a sin bx − b cos bx ]
a2 + b2
ax
∫ e cos bx dx =
e ax
[a cos bx + b sin bx ]
a2 + b2
Similarly, (2) is
Remember as below:
d
⎡
⎤
⎢ a sin bx − dx (sin bx ) ⎥
⎣
⎦
ax
d
e
⎡
⎤
ax
∫ e cos bx dx = a2 + b2 ⎢⎣a cos bx − dx (cos bx)⎥⎦
∫e
ax
sin bx dx =
e ax
a + b2
2
WORKED EXAMPLES
EXAMPLE 1
Evaluate
∫e
2x
cos 2 x sin 3 x dx .
Solution.
Let I = ∫ e 2 x cos 2 x sin 3 x dx
1
We know sin A cos B = [sin( A + B) + sin( A − B)]
2
M06_ENGINEERING_MATHEMATICS-I _CH06_PART A.indd 4
5/19/2016 4:42:39 PM
Integral Calculus ■
[
[
I = ∫ e2x
6.5
1
1
sin 3x cos 2 x = [sin(3 x + 2 x ) + sin(3 x − 2 x )] = [sin 5 x + sin x ]
2
2
1 2x
1 2x
1
(sin 5 x + sin x ) dx = ∫ e sin 5 x dx + ∫ e sin x dx
2
2
2
2x
1 e
1 e2x
= ⋅
[2 sin 5 x − 5 cos 5 x ] + ⋅
[2 sin x − cos x ] + c
2 4 + 25
2 4 +1
e2x
e2 x
=
[2 sin 5 x − 5 cos 5 x ] +
[2 sin x − cos x ] + c
58
10
EXAMPLE 2
Evaluate ∫ x 2 tan21 x dx .
Solution.
Let
I = ∫ x 2 tan −1 x dx
Take
u = tan −1 x and v = x 2
Integrating by parts, we get
[
I = uv1 − ∫ u ′ v1 dx = tan −1 x ⋅
u′ =
1
1+ x2
and v1 = ∫ x 2 dx =
x3
3
x3
x3
1
dx
−∫
3
(1 + x 2 ) 3
x3
x3
1
dx
tan −1 x − ∫
3
3 1+ x2
x3
1 x(1 + x 2 ) − x
dx
= tan −1 x − ∫
3
3
1+ x2
x3
1⎧ ⎛
x ⎞ ⎫
= tan −1 x − ⎨∫ ⎜ x −
⎟ dx ⎬
3⎩ ⎝
3
1+ x2 ⎠ ⎭
=
I=
⇒
∫x
⇒
2
tan −1 x dx =
⎤
1 ⎡x2 1
x3
tan −1 x − ⎢ − loge (1 + x 2 ) ⎥ + c
3
3⎣ 2 2
⎦
x3
x2 1
tan −1 x −
+ loge (1 + x 2 ) + c
3
6 6
EXAMPLE 3
Evaluate
∫x
n
log e x dx .
Solution.
Let I = ∫ x n loge x dx
Take u = loge x and v = x n [
u′ =
Integrating by parts, we get
1
x n +1
and v1 = ∫ x n dx =
x
n +1
I = uv1 − ∫ u ′ v1 dx = loge x ⋅
M06_ENGINEERING_MATHEMATICS-I _CH06_PART A.indd 5
1 x n +1
x n +1
−∫ ⋅
dx
n +1
x n +1
5/19/2016 4:42:42 PM
6.6
■
Engineering Mathematics
x n +1 log e x
1
x n dx
−
n +1
n +1∫
x n +1 log e x
x n +1
1
=
−
+c
( n + 1) ( n + 1)
n +1
=
=
[
n
∫ x log x dx =
EXAMPLE 4
Evaluate
∫3
x
x n +1 log e x
x n +1
x n +1
c
[( n + 1) log e x − 1] + c
−
+
=
n +1
( n + 1) 2
( n + 1) 2
x n +1
[( n + 1) log x − 1] + c
( n + 1) 2
sin 2 x dx .
Solution.
Let
I = ∫ 3x sin 2 x dx
Take
u = 3x , v = sin 2 x
Integrating by parts, we get
[ u ′ = 3x log e 3
and
v1 = −
cos 2 x
2
⎛ cos 2x ⎞
⎛ cos 2x ⎞
I = 3x ⎜ −
− ∫ 3x loge 3 ⎜ −
⎟
⎟ dx
⎝
⎝
2 ⎠
2 ⎠
3x
1
cos 2x + loge 3∫ 3x cos 2x dx
2
2
x
3 cos 2x 1
sin 2x ⎤
⎡ sin 2x
dx ⎥
=−
+ loge 3 ⎢3x
− ∫ 3x loge 3
2
2
2
2
⎣
⎦
=−
3x cos 2x 1
1
2
+ (loge 3)3x sin 2x − ( loge 3) ∫ 3x sin 2x dx
2
4
4
3x cos 2x 1
1
2
I =−
+ (loge 3) ? 3x sin 2x − ( loge 3) I
2
4
4
=−
⇒
3x cos 2x 1
⎛ 1
⎞
⇒ ⎜1 + (loge 3) 2 ⎟ I = −
+ (loge 3) ? 3x sin 2x
⎝ 4
⎠
2
4
⎛ 4 + (loge 3) 2 ⎞
3x cos 2x 1
=
−
+ (loge 3) ? 3x sin 2x
⇒ ⎜
I
⎟⎠
4
2
4
⎝
3x
[
I=
[ −2 cos 2x + loge 3 sin 2x ] + c
4 + (loge 3) 2
EXAMPLE 5
Evaluate
∫x e
3 22 x
dx .
Solution.
Let
I = ∫ x 3 e −2 x dx
M06_ENGINEERING_MATHEMATICS-I _CH06_PART A.indd 6
5/19/2016 4:42:45 PM
Integral Calculus ■
6.7
By Bernoulli’s formula,
I = uv1 − u ′ v2 + u ″ v3 − u ″′ v4 + …
and v = e −2 x
u = x3
where
u ′ = 3 x 2 , u ″ = 6 x, u ″′ = 6 and v1 = ∫ e −2 x dx =
[
v2 =∫
1 e −2 x
e −2 x
e −2 x
dx =
=
,
( −2) ( −2) ( −2) 2
−2
v4 = ∫
e −2 x
e −2 x
e −2 x
dx =
=
3
3
( −2)
( −2) ( −2) ( −2) 4
I = x3
[
3 −2 x
∫ x e dx = −
[
v3 = ∫
e −2 x
−2
e −2 x
e −2 x
e −2 x
dx
=
=
( −2) 2
( −2) 2 ( −2) ( −2)3
⎡ x 3 3x 2 6 x 6 ⎤
e −2 x
e −2 x
e −2 x
e −2 x
− 3x 2
+ 6x
−6
+ c = e −2 x ⎢ − −
−
− ⎥+c
2
3
4
4
8 16 ⎦
−2
( −2)
( −2)
( −2)
⎣ 2
e −2 x
[4 x 3 + 6 x 2 + 6 x + 3] + c
8
EXAMPLE 6
Evaluate
∫x
2
sin 2 x dx .
Solution.
Let I = ∫ x 2 sin 2 x dx
By Bernoulli’s formula,
I = uv1 − u′v2 + u″v3 − u′″v4 +…
Where
u = x2
and v = sin 2x
cos 2 x
∴
u′ = 2x, u″ = 2, u′″ = 0 and v1 = ∫ sin 2 x dx = −
2
cos 2 x
sin 2 x
V2 = ∫ −
dx = −
2
4
sin 2 x
1 ⎛ cos 2 x ⎞ 1
V3 = ∫ −
=− ⎜
⎟ = cos 2 x
4
4⎝ 2 ⎠ 8
cos 2 x
⎛ sin 2 x ⎞
⎛ cos 2 x ⎞
I = x2 ⎜ −
+c
⎟ − 2 x ⎜⎝ −
⎟ +2
⎝
2 ⎠
4 ⎠
8
∴
1
1
x
= − x 2 cos 2 2 x + sin 2 x + cos 2 x + c
4
2
2
1
⎡ 2 x sin 2 x + (1 − 2 x 2 ) cos 2 x ⎤⎦ + c
4⎣
=
EXAMPLE 7
2
⎛ 12 x ⎞
Evaluate ∫ e x ⎜
dx .
⎝ 1 1 x 2 ⎟⎠
Solution.
(1 − x ) 2
⎛ 1− x ⎞
I =∫e ⎜
dx = ∫ e x
dx
2⎟
⎝1+ x ⎠
(1 + x 2 ) 2
2
Let
x
M06_ENGINEERING_MATHEMATICS-I _CH06_PART A.indd 7
5/19/2016 4:42:48 PM
6.8
■
Engineering Mathematics
= ∫ ex
If
f ( x) =
1
, then
1 + x2
(1 + x 2 − 2x )
dx = ∫ e x
(1 + x 2 ) 2
f ′( x ) = −
⎡ 1
2x ⎤
−
⎢
⎥ dx
2
1
(
1
+
x
+
x 2 )2 ⎦
⎣
2x
(1 + x 2 ) 2
I = ∫ e x [ f ( x ) + f ′( x )]dx = e x f ( x ) + c =
[
ex
+c
1 + x2
EXAMPLE 8
Evaluate ∫
e x ( x 3 1 x 1 1)
dx .
( x 2 1 1) 3 / 2
Solution.
I = ∫ ex
Let
If
⎡
⎤
( x 3 + x + 1)
[ x( x 2 + 1) + 1]
x
1
dx = ∫ e x
dx = ∫ e x ⎢ 2
+ 2
2
3/ 2
2
3/ 2
1/ 2
3/ 2 ⎥
( x + 1)
( x + 1)
( x + 1) ⎦
⎣ ( x + 1)
x
f ( x) = 2
, then f ′( x ) =
( x + 1)1 / 2
=
1
( x 2 + 1)1/ 2 ⋅1 − x ⋅ ( x 2 + 1)1 / 2 −1 ⋅ 2 x
2
2
2
⎡⎣( x + 1)1 / 2 ⎤⎦
( x 2 + 1)1 / 2 − ( x 2 + 1) −1 / 2 x 2
( x 2 + 1)
x2
1
( x + 1)1 / 2 x 2 + 1 − x 2
= 2
= 2
3/ 2
2
( x + 1)
( x + 1)3 / 2
x +1
( x 2 + 1)1 / 2 −
=
2
I = ∫ e x [ f ( x ) + f ′( x )]dx = e x f ( x ) + c = e x
[
x
+c
(1 + x 2 )1 / 2
EXAMPLE 9
Evaluate
∫e
tan21 x
?
(1 1 x 1 x 2 )
dx .
1 1 x2
Solution.
I = ∫ e tan
Let
−1
x
⋅
(1 + x + x 2 )
dx
1+ x2
dx
and x = tan t
1+ x2
Put
t = tan −1 x
[
I = ∫ e t [1 + tan 2 t + tan t ]dt = ∫ e t [sec 2 t + tan t ]dt = ∫ e t [tan t + sec 2 t ]dt
If
[
∴ dt =
f (t ) = tan t , then f ′(t ) = sec 2 t
I = ∫ e t [ f (t ) + f ′(t )]dt = e t f (t ) + c = e t tan t + c = xe tan
M06_ENGINEERING_MATHEMATICS-I _CH06_PART A.indd 8
−1
x
+c
5/19/2016 4:42:51 PM
Integral Calculus ■
6.9
EXERCISE 6.1
Evaluate the following integrals:
1.
∫x
4.
∫e
7.
∫ (log x )
10.
2
3
13.
∫e
16.
∫e
19.
∫x
x
6.
∫x
9.
∫e
sin 3 x cos 5 x dx
12.
⎛ x + 3x + 3 ⎞
⎜⎝ ( x + 2) 2 ⎟⎠ dx
∫x e
15.
∫e
18.
∫
∫x
cos 4 x dx
5.
∫ x log
dx
8.
∫x e
cos x dx
11.
2
⎛ x +1 ⎞
⎜⎝ ( x + 2) 2 ⎟⎠ dx
2
tan −1 x dx
5 2x
∫e
2x
e
x dx
dx
2
14.
∫e
cos 6 x cos 2 x dx
17.
∫ sin
sin x dx
20.
∫x
4x
4
∫x
2.
2x
∫x
3.
sin 3 x dx
x
2
−1
x dx
2
sec −1 x dx
2
sin x dx
3x
sin 5 x dx
2 −5 x
x
dx
⎛ 1 + sin x ⎞
⎜⎝
⎟ dx
1 + cos x ⎠
a 2 + x 2 dx
cos 2 x dx
ANSWERS TO EXERCISE 6.1
2
2
1
1. − x 2 cos 3 x + x sin 3 x + cos 3 x + c
3
9
9
2.
⎤
x3
1 ⎡ x2 1
tan −1 x − ⎢ − log(1 + x 2 ) ⎥ + c
3
3⎣ 2 2
⎦
3.
x3
x2 −1 ⎡1 2
⎤
sec −1 x −
( x − 1) + 2⎥ + c
3
3 ⎢⎣ 3
⎦
4.
e2x
[2 cos 4 x + 4 sin 4 x ] + c
20
5.
x2
x2
log x − + c
2
4
6. 2 x sin x + cos x( 2 − x 2 ) + c
5
5
15
15
15 ⎤
⎡
8. e 2 x ⎢ x 5 − x 4 + x 3 − x 2 + x − ⎥
2
2
4
4
8⎦
⎣
7. x(log x ) 2 − 2 x log x + 2 x + c
e3 x
[3 sin 5 x − 5 cos 5 x ] + c
34
e2x
e2x
[sin 8 x − 4 cos 8 x ] −
[sin 2 x − cos 2 x ] + c
11.
6
8
9.
⎛ x − 1⎞
13. e x ⎜
+c
⎝ x + 1⎟⎠
16.
10. x 3 sin x + 3 x 2 cos x − 6 x sin x − cos x + c
⎡ x2 2x
2 ⎤
12. e −5 x ⎢ − +
−
⎥+c
⎣ 5 25 125 ⎦
x
⎛ x +1⎞
14. e x ⎜
+c
15. e x ⋅ tan + c
⎟
⎝ x + 2⎠
2
e4 x
e4 x
[cos 8 x + 2 sin 8 x ] +
[cos 4 x + sin 4 x ] + c
40
16
17. x sin −1 x + 1 − x 2 + c
19. 4 x( x 2 − 6) sin x − ( x 4 − 12 x 2 + 24) cos x + c
M06_ENGINEERING_MATHEMATICS-I _CH06_PART A.indd 9
x a2 + x 2 a2
+ loge x + a2 + x 2 + c
2
2
1
20. ⎡⎣(1 + 2 x 2 ) sin 2 x + 2 x cos 2 x ⎤⎦ + c
4
18.
5/19/2016 4:42:58 PM
■
6.10
6.2
Engineering Mathematics
DEFINITE INTEGRAL (NEWTON–LEIBNITZ FORMULA)
Definition 6.2
If f(x) is a continuous function on [a, b] and F ′( x ) = f ( x ) on [a, b], then the definite integral
b
∫ f ( x ) dx
is defined as F(b) 2 F(a).
a
b
∫ f ( x )dx = F ( b) − F ( a )
[
a
Note
1. The difference F(b) − F(a) is written symbolically [ F ( x ) ]a.
b
b
Hence, the definite integral
∫ f ( x)dx = [ F ( x)]
b
a
= F ( b) − F ( a)
a
2. Newton–Leibnitz formula gives a practical method of computing definite integrals when an
anti derivative of the integrand is known.
b
3. In a definite integral ∫ f ( x ) dx, when substitution or transformation of variable is made, it should
a
be either an increasing function or a decreasing function in the given interval.
6.2.1 Properties of Definite Integral
If f(x) is a continuous and integrable function of x in [a, b], then the following properties are satisfied.
a
1.
b
b
2.
∫
a
3.
4.
a
c
b
a
c
f ( x ) dx = ∫ f ( x ) dx + ∫ f ( x ) dx, a < c < b
b
b
a
a
a
a
0
0
∫ f ( x) dx = ∫ f (a + b − x)dx
∫ f ( x)dx = ∫ f (a − x)dx
a
5.
b
∫ f (x )dx = −∫ f (x ) dx
∫
−a
a
f ( x )dx = 2∫ f ( x ) dx if f(x) is an even function of x
(i.e., f(−x) = f(x) ∀ x ∈[−a, a])
0
a
and
2a
6.
∫
0
∫
−a
f ( x ) dx = 0 if f(x) is an odd function of x
a
f ( x )dx = 2∫ f ( x ) dx if f(2a − x) = f(x)
0
b
7. If f ( x ) ≤ g ( x ), then
∫
a
(i.e., f(−x) = −f(x) ∀ x ∈[−a, a])
2a
and
∫
f ( x ) dx = 0 if f(2a − x) = −f(x)
0
b
f ( x ) dx ≤ ∫ g ( x ) dx
M06_ENGINEERING_MATHEMATICS-I _CH06_PART A.indd 10
a
5/19/2016 4:43:01 PM
Integral Calculus ■
b
∫
Note
b
b
a
a
6.11
f ( x ) dx = ∫ f (u ) du = ∫ f (t ) dt
a
That is the value of a definite integral is unaffected by the change of dummi variable, if the limits and
function are the same.
Periodic Function
Definition 6.3
A real function f is said to be periodic if there exists a positive number T such that
f ( x + T ) = f ( x)
∀x ∈ R.
The smallest such T is called the period of the function.
EXAMPLE
We know
cos( x + 2p) = cos x , cos( x + 4p) = cos x , cos( x + 6p) = cos x and so on.
The smallest one is 2p. So, 2p is the period of cos x.
8. If f(x) is periodic with period T, then
a + nT
(i)
∫
a
b + nT
(iii)
∫
a + nT
T
nT
f ( x ) dx = n∫ f ( x ) dx, n ∈ Z
(ii)
a
∫
0
T
f ( x ) dx = n∫ f ( x ) dx, n ∈ Z
0
b
f ( x ) dx = ∫ f ( x ) dx, n ∈ Z
a
Now we shall prove these formulae.
1.
a
b
b
a
∫ f (x ) dx 52∫ f (x ) dx
Proof
Let F ′( x ) = f ( x ) on [a, b]
By Newton–Leibnitz formula,
b
∫ f ( x) dx = [ F ( x)]
b
a
= F ( b) − F ( a)
a
a
and
∫ f ( x) dx = [ F ( x)]
a
b
= F ( a) − F ( b) = − [ F ( b) − F ( a) ]
b
[
M06_ENGINEERING_MATHEMATICS-I _CH06_PART A.indd 11
a
b
b
a
∫ f ( x) dx = − ∫ f ( x) dx
■
5/19/2016 4:43:03 PM
■
6.12
2.
Engineering Mathematics
b
c
a
a
a
c
∫ f (x ) dx 5 ∫ f (x ) dx 1 ∫ f (x ) dx
Proof
We have
b
∫ f ( x) dx = F (b) − F (a)
a
c
∫ f ( x) dx = F (c) − F (a)
a
b
∫ f ( x) dx = F (b) − F (c)
and
c
c
b
a
c
∫ f (x ) dx + ∫ f (x ) dx = F(c ) − F(a) + F(b ) − F(c )
∴
b
= F (b ) − F (a) = ∫ f ( x ) dx
a
c
a
a
a
c
∫ f ( x) dx = ∫ f ( x) dx + ∫ f ( x) dx
∴
3.
b
b
b
a
a
■
∫ f (x ) dx 5 ∫ f (a 1 b 2 x ) dx
Proof
R.H.S =
b
∫ f (a + b − x) dx
a
Put t = a + b − x
∴
dt = − dx ⇒ dx = − dt
When x = a, t = b and when x = b, t = a
a
a
b
b
R.H.S = ∫ f (t )( − dt ) = − ∫ f (t ) dt
∴
b
b
a
b
a
b
a
a
[by property 1]
= ∫ f (t ) dt = ∫ f ( x )dx = L.H.S
∴
4.
∫ f (x ) dx = ∫ f (a + b − x ) dx
a
a
0
0
■
∫ f (x ) dx 5 ∫ f (a 2 x ) dx
Proof
In 3, put a = 0, b = a, then
a+b− x=a− x
a
∴
∫
0
a
f ( x ) dx = ∫ f ( a − x ) dx.
M06_ENGINEERING_MATHEMATICS-I _CH06_PART A.indd 12
■
0
5/19/2016 4:43:07 PM
Integral Calculus ■
5.
a
a
2a
0
6.13
∫ f (x ) dx 5 2 ∫ f (x ) dx if f(x) is an even function of x and
a
∫ f (x ) dx 5 0
if f(x) is an odd function of x
2a
Proof
a
a
0
∫
f ( x ) dx =
−a
∫
−a
f ( x ) dx + ∫ f ( x ) dx
(1) [by property 2]
0
0
Let I =
∫
f ( x ) dx
−a
Put x = −t
When
∴ dx = − dt
x = −a, t = a and when x = 0, t = 0
0
I = ∫ f ( −t )( − dt )
∴
−a
0
a
a
a
0
0
= − ∫ f ( −t ) dt = ∫ f ( −t ) dt = ∫ f ( − x ) dx
If f(x) is an even function of x, then f(−x) = f(x). ∴
[by note]
0
a
a
0
∫ f (x ) dx = ∫ f (x ) dx
0
If f(x) is an odd function of x, then f(−x) = −f(x). ∴
∫
−a
a
f ( x ) dx = − ∫ f ( x ) dx
0
Substituting in (1), we get
a
∫
−a
a
and
∫
−a
6.
a
a
a
0
0
0
f ( x ) dx = ∫ f ( x ) dx + ∫ f ( x ) dx = 2∫ f ( x ) dx
a
a
0
0
f ( x ) dx = − ∫ f ( x ) dx + ∫ f ( x ) dx = 0
2a
a
0
0
∫ f (x ) dx 5 2 ∫ f (x ) dx
if f (x) is an even function of x
if f (x) is an odd function of x.
■
if f ( 2 a 2 x ) 5 f ( x ) and
2a
∫
f ( x ) dx 5 0 if
f ( 2a 2 x ) 52 f ( x )
0
Proof
2a
∫
0
a
2a
0
a
f ( x ) dx = ∫ f ( x ) dx + ∫ f ( x ) dx
(1)
2a
Let I =
∫
f ( x ) dx
a
M06_ENGINEERING_MATHEMATICS-I _CH06_PART A.indd 13
5/19/2016 4:43:10 PM
6.14
■
Engineering Mathematics
Put 2a − x = t
∴
− dx = dt ⇒ dx = − dt
When x = a, t = a and when x = 2a, t = 0
0
0
a
a
a
0
0
I = ∫ f ( 2a − t )( − dt ) = − ∫ f ( 2a − t )dt = ∫ f ( 2a − t ) dt = ∫ f ( 2a − x ) dx
∴
a
2a
a
a
0
I = ∫ f ( x ) dx = ∫ f ( 2a − x ) dx
∴
2a
∫
If f (2a−x) = f (x), then
a
2a
If f (2a−x) = − f (x) , then
∫
a
a
f ( x ) dx = ∫ f ( x ) dx .
0
a
f ( x ) dx = − ∫ f ( x ) dx .
0
Substituting in (1), we get
2a
∫
0
2a
∫
0
∫
a
a
a
0
0
0
a
a
0
0
f ( x ) dx = ∫ f ( x ) dx − ∫ f ( x ) dx = 0
a1nT
8. (i )
a
f ( x ) dx = ∫ f ( x ) dx + ∫ f ( x ) dx = 2∫ f ( x ) dx
if f ( 2a − x ) = f ( x )
if f ( 2a − x ) = − f ( x )
■
T
f ( x ) dx 5 n∫ f ( x ) dx
a
Proof
Given f(x) is periodic with period T.
f ( x + T ) = f ( x)
[
a + nT
Now
∫
f ( x ) dx =
a
a +T
∫
∀x ∈ R
f ( x ) dx +
a + 2T
∫
and
f ( x + rT ) = f ( x )
f ( x ) dx + … +
a + ( r +1)T
∫
a +T
a
(1)
∀r ∈ Z
f ( x ) dx + … +
a + rT
a + nT
∫
f ( x ) dx
a + ( n −1)T
a + ( r +1)T
Consider,
∫
I=
f ( x ) dx
a + rT
Put x = y + rT ∴ dx = dy
When x = a + rT ,
y + rT = a + rT
When x = a + ( r + 1)T ,
y + rT = a + ( r + 1)T ⇒ y = a + rT + T − rT = a + T
a + ( r +1)T
[
∫
f ( x ) dx =
a + rT
a +T
∫
⇒y =a
f ( y + rT )dy =
a
a +T
∫
f ( y ) dy =
a
a +T
∫
f ( x ) dx
[using (1)]
a
Putting r = 1, 2, 3, …, (n − 1) in I, we get
a + 2T
∫
f ( x ) dx =
a +T
M06_ENGINEERING_MATHEMATICS-I _CH06_PART A.indd 14
a +T
∫
a
a + 3T
f ( x ) dx,
∫
a + 2T
f ( x ) dx =
a +T
∫
f ( x ) dx, … ,
a
5/19/2016 4:43:14 PM
Integral Calculus ■
a + nT
∫
a +T
∫
f ( x ) dx =
a + ( n −1)T
a + nT
∫
[
6.15
f ( x ) dx
a
a +T
∫
f ( x ) dx =
a
f ( x ) dx +
a +T
∫
a 1 nT
∫
T
0
0
∫
f ( x ) dx + … +
a
a +T
∫
f ( x ) dx
a
a1T
f ( x ) dx 5 n
a
nT
a +T
a
a
⇒
f ( x ) dx +
∫
■
f ( x ) dx
a
8. (ii ) ∫ f ( x ) dx 5 n ∫ f ( x ) dx
nT
∫
a = 0 in 8(i), we get
Putting
0
b 1 nT
8. ( iii )
∫
a 1 nT
b + nT
∫
a + nT
∴
b + nT
∫
a + nT
■
0
b
f ( x ) dx 5 ∫ f ( x ) dx , n e z .
a
x = nT + y
x = a + nT,
x = b + nT,
Put
When
When
∴
T
f ( x ) dx 5 n∫ f ( x ) dx .
∴ dx = dy
a + nT = nT + y
b + nT = nT + y
then
then
b
b
a
a
⇒y=a
⇒y=b
f ( x )dx = ∫ f ( y + nT )dy = ∫ f ( y )dy since f is of period T.
b
f ( x )dx = ∫ f ( x )dx,
n ∈ z.
a
WORKED EXAMPLES
EXAMPLE 1
p
2
Show that
∫ sin
0
n
sin n x
p
dx 5 .
4
x 1 cos n x
Solution.
Let
I=
p
2
∫ sin
n
0
Also
I=
sin n x
dx
x + cos n x
p
2
⎛p
⎞
sin n ⎜ − x ⎟
⎝2
⎠
0
⎛p
⎞
⎛p
⎞
sin ⎜ − x ⎟ + cos n ⎜ − x ⎟
⎝2
⎠
⎝2
⎠
∫
n
M06_ENGINEERING_MATHEMATICS-I _CH06_PART A.indd 15
(1)
dx =
p
2
∫ cos
0
cos n x
dx
x + sin n x
n
(2)
5/19/2016 4:43:15 PM
■
6.16
Engineering Mathematics
p
p
2
2
p
sin n x + cos n x
p
2I = ∫ n
dx
=
dx = [ x ]02 =
∫
n
2
0
0 sin x + cos x
(1) + (2) ⇒
p
I=
⇒
4
∴
p
2
sin n x
p
∫0 sin n x + cosn x dx = 4
Note Since the right hand side is independent of n, this is true for all n.
p
2
Similarly,
∫ cos
0
cos n x
p
dx =
4
x + sin n x
for any n.
n
EXAMPLE 2
p
4
Prove that
p
∫ log (11 tan x ) dx = 8 log
e
e
2.
0
Solution.
p
4
I = ∫ loge (1 + tan x ) dx
Let
(1)
0
p
4
Also
Now
∴
⎡
⎛p
⎞⎤
I = ∫ loge ⎢1 + tan ⎜ − x ⎟ ⎥ dx
⎝4
⎠⎦
⎣
0
p
tan − tan x
1 − tan x
⎛p
⎞
4
tan ⎜ − x ⎟ =
=
p
⎝4
⎠
1 + tan x
1 + tan tan x
4
1 − tan x 1 + tan x + 1 − tan x
2
⎛p
⎞
1 + tan ⎜ − x ⎟ = 1 +
=
=
.
⎝4
⎠
1 + tan x
1 + tan x
1 + tan x
∴
(1) + (2) ⇒
[by property 4]
p
4
2 ⎞
⎛
I = ∫ loge ⎜
dx
⎝ 1 + tan x ⎟⎠
0
p
4
(2)
p
4
2 ⎞
⎛
2 I = ∫ log e (1 + tan x ) dx + ∫ log e ⎜
⎟ dx
⎝
1
+
tan
x⎠
0
0
p
4
⎡
2 ⎞⎤
⎛
= ∫ ⎢log e (1 + tan x ) + log e ⎜
⎟ dx
⎝
1 + tan x ⎠ ⎥⎦
0 ⎣
p
4
⎡
2 ⎞⎤
⎛
= ∫ ⎢log e (1 + tan x ) ⎜
⎟ dx
⎝
+
1
tan
x ⎠ ⎥⎦
0 ⎣
M06_ENGINEERING_MATHEMATICS-I _CH06_PART A.indd 16
5/19/2016 4:43:19 PM
Integral Calculus ■
6.17
p
4
p
⎛ p⎞
= ∫ log e 2dx = log e 2 [ x ]04 = log e 2 ⎜ ⎟
⎝ 4⎠
0
∴
I=
p
log e 2
8
EXAMPLE 3
3p
4
x
∫ 1 1 sin x dx 5 p ⎡⎣
Prove that
p
4
Solution.
I=
Let
3p
4
2 2 1⎤⎦.
x
∫ 1 + sin x dx
(1)
p
4
First remove x by using property 4.
f (x ) =
Here
∴
∴
Since
x
,
1 + sin x
a=
p
,
4
b=
3p
4
⎛ p 3p
⎞
f (a + b − x) = f ⎜ +
− x ⎟ = f (p − x )
⎝4
⎠
4
f (p − x ) =
p−x
p−x
=
1 + sin(p − x ) 1 + sin x
b
b
a
a
3p
4
p−x
( 2)
∫ f ( x)dx = ∫ f (a + b − x)dx , we have I = ∫ 1 + sin x dx
(1) + (2) ⇒
2I =
p
4
3p
4
∫
p
4
x
dx +
1 + sin x
3p
4
=p∫
p
4
3p
4
=p∫
p
4
3p
4
∫
p
4
p−x
dx =
1 + sin x
3p
4
∫
p
4
x+p− x
dx =
1 + sin x
3p
4
p
∫ 1 + sin x dx
p
4
dx
⎛p
⎞
1 + cos ⎜ − x ⎟
⎝2
⎠
dx
1⎛p
⎞
2 cos 2 ⎜ − x ⎟
⎠
2⎝ 2
M06_ENGINEERING_MATHEMATICS-I _CH06_PART A.indd 17
5/19/2016 4:43:22 PM
6.18
■
Engineering Mathematics
p
=
2
3p
4
∫ sec
p
4
2
⎛ p x⎞
⎜⎝ − ⎟⎠ dx
4 2
3p
⎡ ⎛ p x⎞ ⎤ 4
tan
−
p ⎢ ⎜⎝ 4 2 ⎟⎠ ⎥
⎥
= ⎢
1
2⎢
⎥
−
⎢
⎥p
2
⎣
⎦
4
⎡ ⎛ p 3p ⎞
⎛ p p⎞ ⎤
= −p ⎢ tan ⎜ − ⎟ − tan ⎜ − ⎟ ⎥
⎝ 4 8⎠⎦
⎣ ⎝4 8 ⎠
⎡ ⎛ p⎞
⎛ p⎞ ⎤
= −p ⎢ tan ⎜ − ⎟ − tan ⎜ ⎟ ⎥
⎝
⎠
⎝ 8⎠⎦
8
⎣
p
p⎤
⎡
= −p ⎢ − tan − tan ⎥
8
8⎦
⎣
⇒
2I = 2p tan
∴
I =p
(
p
= 2p
8
(
1°
p
⎡
⎤
⎢{ tan 8 = tan 22 2 = 2 − 1⎥
⎣
⎦
)
2 −1
)
2 −1
EXAMPLE 4
p
2
Evaluate ∫ log e sin xdx .
0
Solution.
p
2
Let
I = ∫ loge sin xdx
(1)
0
p
2
Also
(1) + (2) ⇒
⎛p
⎞
I = ∫ loge sin ⎜ − x ⎟ dx
⎝
⎠
2
0
p
2
= ∫ loge cos xdx
(2)
[by property 4]
0
p
2
p
2
p
2
0
0
0
2 I = ∫ log e sin xdx + ∫ log e cos xdx = ∫ (log e sin x + log cos x ) dx
p
2
= ∫ loge sin x cos x dx
0
M06_ENGINEERING_MATHEMATICS-I _CH06_PART A.indd 18
5/19/2016 4:43:25 PM
Integral Calculus ■
p
2
⎛ sin 2 x ⎞
dx
= ∫ log e ⎜
⎝ 2 ⎟⎠
0
6.19
[{ sin 2 x = 2 sin x cos x ]
p
2
= ∫ (loge sin 2x − loge 2)dx
0
p
2
p
2
= ∫ (log e sin 2 x dx ) − ∫ log e 2dx
0
p
2
0
2 I = I1 − log e 2 [ x ]
⇒
p
2
0
p
⎡p
⎤
= I1 − log e 2 ⎢ − 0 ⎥ = I1 − log e 2 .
2
⎣2
⎦
I1 = ∫ loge sin 2 x dx
where
0
Put
t = 2x
∴ dt = 2dx ⇒ dx =
p
, t = p.
2
When x = 0, t = 0 and when x =
p
I1 = ∫ loge sin t
∴
0
=
dt
2
p
2
1
⋅ 2 loge sin t dt
2 ∫0
p
2
= ∫ loge sin t dt
0
p
2 I = I − log e 2 ⇒
2
∴
p
1
dt
= ∫ loge sin t dt
20
2
p
2
⎡ By property 6
⎤
⎢
⎥
⎢{ f (p − t ) = log sin (p − t ) ⎥
⎢
⎥
= log sin t = f (t )⎦
⎣
= ∫ loge sin x dx = I
0
I=−
p
log e 2 .
2
EXAMPLE 5
1⎞
⎛
log e ⎜ x 1 ⎟
⎝
x⎠
dx 5 p log e 2.
Show that ∫
2
(1 1 x )
0
∞
Solution.
1⎞
⎛
log e ⎜ x + ⎟
⎝
x⎠
I =∫
dx
2
(1 + x )
0
∞
Let
Put x = tan u
∴
dx = sec 2 u d u
When x = 0, tan u = 0 ⇒ u = 0 and when x = ∞, tan u = ∞ ⇒ u =
M06_ENGINEERING_MATHEMATICS-I _CH06_PART A.indd 19
p
.
2
5/19/2016 4:43:30 PM
6.20
∴ I=
■
p
2
∫
Engineering Mathematics
1 ⎞
⎛
2
log e ⎜ tan u +
⎟ sec u
⎝
tan u ⎠
(1 + tan u)
2
0
p
2
⎛ 1 + tan 2 u ⎞
d u = ∫ log e ⎜
du
⎝ tan u ⎟⎠
0
p
2
⎛ sec 2 u ⎞
= ∫ log e ⎜
du
⎝ tan u ⎟⎠
0
p
2
cos u ⎞
⎛ 1
= ∫ log e ⎜
⋅
⎟ du
2
⎝
sin u ⎠
cos
u
0
p
2
1
⎛
⎞
= ∫ log e ⎜
du
⎝ sin u cos u ⎟⎠
0
p
2
2
⎛
⎞
= ∫ log e ⎜
⎟ du
⎝
2 sin u cos u ⎠
0
p
2
⎛ 2 ⎞
= ∫ log e ⎜
du
⎝ sin 2u ⎟⎠
0
p
2
p
2
0
0
= ∫ log e 2d u − ∫ log e sin 2ud u
p
2
p
2
0
= log e 2 [u] − ∫ log e sin 2ud u
0
p
2
p
p
= log e 2 − ∫ loge sin 2ud u = log e 2 − I1
2
2
0
p
2
where
p
p
2
1
p
I1 = ∫ loge sin 2ud u = ∫ loge sin ud u = ∫ loge sin ud u = − log e 2
20
2
0
0
[
I=
=
M06_ENGINEERING_MATHEMATICS-I _CH06_PART A.indd 20
[Refer example 4]
p
⎛ p
⎞
log e 2 − ⎜ − log e 2⎟
⎝ 2
⎠
2
p
p
log e 2 + log e 2 = p log e 2 .
2
2
5/19/2016 4:43:34 PM
Integral Calculus ■
6.21
EXAMPLE 6
p
Prove that
∫
0
⎛p
⎞
x sin 2 x sin ⎜ cos x ⎟
⎝2
⎠
8
dx 5 2 .
2x 2 p
p
Solution.
⎛p
⎞
x sin 2 x sin ⎜ cos x ⎟
⎝2
⎠
Let I = ∫
dx
2x − p
0
p
Put 2 x − p = 2t ∴ 2dx = 2dt ⇒ dx = dt
When x = 0, 2t = −p
⇒t = −
I=
[
p
p
and When x = p, 2t = 2p − p ⇒ t =
2
2
⎛p
⎛p ⎞
⎛ p ⎞⎞
⎜⎝ + t ⎟⎠ sin(p + 2t ) sin ⎜ cos ⎜⎝ + t ⎟⎠ ⎟
⎝2
⎠
2
2
p
2
∫
2t
p
−
2
dt
⎛ p
⎞
⎛p ⎞
⎜⎝ + t ⎟⎠ ( − sin 2t ) ⋅ sin ⎜⎝ − sin t ⎟⎠
2
2
dt
= ∫
t
2
p
p
2
−
2
p
=
4
I=
⇒
⎛p
⎞
sin 2t ⋅ sin ⎜ sin t ⎟
⎝2
⎠
1
dt +
∫p
t
2
p
2
−
2
⎛p
⎞
∫ sin 2t sin ⎝⎜ 2 sin t ⎟⎠ dt
−
p
2
p
1
I1 + I 2
4
2
⎛p
⎞
sin 2t ⋅ sin ⎜ sin t ⎟
⎝2
⎠
where I 1 = ∫
dt
t
p
p
2
−
p
2
2
and
I2 =
p
2
⎛p
⎞
∫ sin 2t ? sin ⎜⎝ 2 sin t ⎟⎠ dt
−
p
2
⎛p
⎞
sin 2t sin ⎜ sin t ⎟
⎝2
⎠
Let f (t ) =
t
⎛p
⎞
sin( −2t ) sin ⎜ sin( −t )⎟
⎝2
⎠
[ f ( −t ) =
−t
M06_ENGINEERING_MATHEMATICS-I _CH06_PART A.indd 21
⎛p
⎞
sin 2t sin ⎜ sin t ⎟
⎝2
⎠
=−
= − f (t )
t
[{ sin( −t ) = − sin t ]
5/19/2016 4:43:37 PM
■
6.22
Engineering Mathematics
∴ f (t ) is an odd function of t.
[ I1 =
p
2
∫
f (t )dt = 0
[by property 5]
p
−
2
⎛p
⎞
g (t ) = sin 2t sin ⎜ sin t ⎟
⎝2
⎠
Now, let
[
⎛p
⎞
⎛p
⎞
g ( −t ) = sin( −2t ) sin ⎜ sin( −t )⎟ = sin 2t sin ⎜ sin t ⎟ = g (t )
⎝2
⎠
⎝2
⎠
∴ g (t ) is an even function of t.
I2 =
[
p
2
p
2
⎛p
⎞
g (t )dt = 2 ∫ g (t ) dt = 2 ∫ sin 2t sin ⎜ sin t ⎟ dt
⎝2
⎠
p
0
0
∫
−
p
2
[by property 5]
2
p
2
⎛p
⎞
= 4 ∫ sin t cos t sin ⎜ sin t ⎟ dt
⎝
⎠
2
0
Put
u=
p
sin t ⇒
2
When t = 0, u =
sin t =
2u
p
∴ costdt =
p
p
sin 0 = 0 and when t =
2
2
p
2
[
2
du .
p
,
u=
p
p p
sin =
2
2 2
p
16 2
2u
2
I 2 = 4∫
⋅ sin u ⋅ du = 2 ∫ u sin u du .
p
p
p 0
0
Integrating by parts, we get
p
⎧
⎫
2
p
16 ⎪
⎪
2
I 2 = 2 ⎨[ u( − cos u ) ]0 − ∫ 1⋅ ( − cos u )du ⎬
p ⎪
0
⎪
⎩
⎭
p
⎧
⎫
2
16 ⎪
⎪
= 2 ⎨[0] + ∫ cos u du ⎬
p ⎪
0
⎪
⎩
⎭
=
[
I=
p
16
16
sin u ]02 = 2
2 [
p
p
16
⎡ p
⎤ 16
⎢sin 2 − sin 0 ⎥ = p 2 (1 − 0) = p 2
⎣
⎦
p
1
p
1 16
8
I1 + I 2 = × 0 + × 2 = 2
4
4
2
2 p
p
M06_ENGINEERING_MATHEMATICS-I _CH06_PART A.indd 22
5/19/2016 4:43:42 PM
Integral Calculus ■
6.23
EXAMPLE 7
1
Prove that ∫ cot21 (1 2 x 1 x 2 )dx 5
0
p
2 log e 2.
2
Solution.
1
⎛
⎞
dx
Let I = ∫ cot −1 (1 − x + x 2 )dx = ∫ tan −1 ⎜
2⎟
⎝1− x + x ⎠
0
0
1
1
1
⎛ x + (1 − x ) ⎞
dx
= ∫ tan −1 ⎜
⎝ 1 − x (1 − x ) ⎟⎠
0
1
= ∫ [tan −1 x + tan −1 (1 − x )] dx
0
1
1
= ∫ tan −1 x dx + ∫ tan −1 (1 − x ) dx
0
0
1
1
0
0
= ∫ tan −1 x dx + ∫ tan −1 x dx
[by property 4]
1
= 2∫ tan −1 x dx
0
1
⎧⎪
⎫⎪
1
1
= 2 ⎨ ⎡⎣ tan −1 x ⋅ x ⎤⎦ 0 − ∫
⋅ x dx ⎬
2
0 1+ x
⎩⎪
⎭⎪
1⎫
1
⎧
= 2 ⎨[tan −1 1 − 0] − ⎡⎣log e (1 + x 2 ) ⎤⎦ 0 ⎬
2
⎩
⎭
⎧p 1
⎫ p
= 2 ⎨ − [loge 2 − loge 1]⎬ = − loge 2
⎩4 2
⎭ 2
EXAMPLE 8
1
Prove that
∫ tan
−1
(1 − x + x 2 )dx 5 log e 2.
0
Solution.
p
p
We know that tan −1 u + cot −1 u =
⇒ tan −1 u = − cot −1 u
2
2
p
∴
tan −1 (1 − x + x 2 ) = − cot −1 (1 − x + x 2 )
2
1
1
⎡p
⎤
∴ ∫ tan −1 (1 − x + x 2 )dx = ∫ ⎢ − cot −1 (1 − x + x 2 )dx ⎥
2
⎦
0
0 ⎣
p
= ∫ dx − ∫ cot −1 (1 − x + x 2 ) dx
2
0
0
1
=
1
p 1 ⎛p
⎞ p p
[ x ]0 − ⎜ − log e 2⎟ = − + log e 2 = log e 2 [using worked example 7]
⎝2
⎠ 2 2
2
M06_ENGINEERING_MATHEMATICS-I _CH06_PART A.indd 23
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6.24
■
Engineering Mathematics
EXAMPLE 9
p
Prove that
Solution.
Let
x dx
p2
5
∫0 a 2 cos2 x 1 b2 sin2 x 2ab .
p
I =∫
0
p
I =∫
Also
0
0
(1) + (2) ⇒
p
2I = ∫
0
[
f ( x) =
(p − x )
a cos 2 x + b 2 sin 2 x
2
dx
( 2)
p
1
a 2 cos 2 x + b 2 sin 2 x
f ( 2 a − x ) = f (p − x ) =
2 I = 2p ∫
0
⇒
[by property 4]
x+p− x
1
dx = p∫ 2
dx
2
2
2
2
2
2
2
a cos x + b sin x
0 a cos x + b sin x
p
2
[
(1)
(p − x ) dx
a 2 cos 2 (p − x ) + b 2 sin 2 (p − x )
p
I =∫
⇒
Let
x dx
a 2 cos 2 x + b 2 sin 2 x
1
1
= 2
= f ( x)
2
2
2
a cos (p − x ) + b sin (p − x ) a cos x + b 2 sin 2 x
2
2
dx
a 2 cos 2 x + b 2 sin 2 x
p
2
[by property 6]
p
2
sec 2 dx
dx
I = p∫
=
p
∫0 [a2 + b2 tan 2 x]
2
2
2
2
0 cos x[ a + b tan x ]
Put t = tan x ∴ dt = sec 2 x dx .
When x = 0, t = tan 0 = 0 and when x =
∞
[
I = p∫
0
p
dt
= 2
2
2 2
b
a +b t
∞
p
p
, t = tan = ∞
2
2
∞
∫a
dt
2
0
b2
+ t2
⎡
⎤
2
p 1 ⎢ −1 t ⎥
p
p ⎡p
⎤ p
[tan −1 ∞ − tan −1 0] =
= 2 ⎢ tan
− 0⎥ =
.
⎥ =
⎢
a⎥
ab
ab ⎣ 2
b a⎢
⎦ 2ab
b⎣
b ⎦0
Note In the interval (0, p) we cannot put t = tan x as it is not increasing there and discontinuous at
p
⎛ p⎞
⎛ p⎞
x = . So, we reduced the interval (0, p) to ⎜ 0, ⎟ by property 6, so that in ⎜ 0, ⎟ , tan x is strictly
⎝
⎠
⎝ 2⎠
2
2
increasing.
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Integral Calculus ■
6.25
Integrals of Periodic Functions
EXAMPLE 10
4p
∫
Evaluate
1 2 cos 2 x dx .
0
Solution.
I=
Let
4p
∫
1 − cos 2 x dx
=
4p
∫
2 sin 2 x dx
0
0
4p
= 2 ∫ sin x dx
[{
x 2 = x = x if x ≥ 0 and x = − x if x < 0]
0
But sin x is periodic with period p. That is, T = p in property 8(1).
4p
∫
∴ we have
sin x dx = 4
0
p
∫ sin x dx
0
But sin x ≥ 0 if x ∈ (0, p)
[
p
p
0
0
∴ sin x = sin x
4∫ sin x dx = 4∫ sin x dx = 4 [ − cos x ]0 = −4[cos p − cos 0] = −4[ −1 − 1] = 8
p
∴
I = 2 ×8 = 8 2
EXAMPLE 11
p
Evaluate
∫ sin x1cos x dx .
0
Solution.
p
I = ∫ sin x + cos x dx
Let
0
1
⎡ 1
⎤
sin x + cos x = 2 ⎢
cos x ⎥
sin x +
2
2
⎣
⎦
Now
p
p⎞
⎡ p
⎤
⎛
= 2 ⎢sin ⋅ sin x + cos ⋅ cos x ⎥ = 2 cos ⎜ x − ⎟
⎝
4
4
4⎠
⎣
⎦
p⎞
⎛
sin x + cos x = 2 cos ⎜ x − ⎟ is of period p.
⎝
4⎠
But
p
p⎞
⎛
I = 2 ∫ cos ⎜ x − ⎟ dx
⎝
4⎠
0
[
Put t = x −
p
.
4
[ dt = dx
M06_ENGINEERING_MATHEMATICS-I _CH06_PART A.indd 25
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■
6.26
Engineering Mathematics
When x = 0, t = −
p
p 3p
and when x = p, t = p − =
.
4
4
4
3p
4
I= 2
[
∫
cos t dt
p
−
4
But
⎛ p p⎞
cos t ≥ 0 in ⎜ − , ⎟
⎝ 4 2⎠
and
⎛ p 3p ⎞
cos t < 0 in ⎜ , ⎟
⎝2 4 ⎠
∴
⎛ p p⎞
cos t = cos t in ⎜ − , ⎟ ,
⎝ 4 2⎠
and
⎛ p 3p ⎞
cos t = − cos t in ⎜ , ⎟
⎝2 4 ⎠
3p
⎧ p2
⎫
4
⎪
⎪
I = 2 ⎨ ∫ cos t dt + ∫ − cos t dt ⎬
p
⎪− p
⎪
2
⎩ 4
⎭
[
3p
p
⎧
⎫
= 2 ⎨[sin t ] 2p − [sin t ]p4 ⎬
−
⎩
4
2 ⎭
⎧⎡ p
1
1
p ⎤⎫
⎤
⎡
⎛ p ⎞ ⎤ ⎡ 3p
= 2 ⎨ ⎢sin − sin ⎜ − ⎟ ⎥ − ⎢sin
− sin ⎥ ⎬ = 2 ⎢1 +
−
+ 1⎥ = 2 2.
⎝
⎠
2
4
4
2
2
2 ⎦
⎦⎭
⎣
⎦ ⎣
⎩⎣
EXAMPLE 12
np 1 v
∫
Prove that
sin x dx 5 2n 1 1 2 cos v if 0 ≤ v < p and n ∈ Z .
0
Solution.
Let
I=
where I1 =
np + v
∫
0
np
sin x dx =
∫ sin x dx .
np
∫
sin x dx +
np + v
∫
sin x dx = I 1 + I 2
np
0
But sin x is periodic with period p.
0
p
[
I1 = n∫ sin x dx
[by property 8(i)]
0
p
= n∫ sin x dx
[since sin x > 0 in ( 0, p ) ]
0
= n [ − cos x ]0 = − n[cos p − cos 0] = − n[ −1 − 1] = 2n
p
and
I2 =
np + v
∫
sin x dx
np
Put x = np + y
[ dx = dy. When x = np, y = 0 and when x = np + v, y = v .
M06_ENGINEERING_MATHEMATICS-I _CH06_PART A.indd 26
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Integral Calculus ■
v
v
v
0
0
0
I 2 = ∫ sin ( np + y ) dy = ∫ sin y dy = ∫ sin y dy
[
6.27
[{ 0 ≤ y ≤ v < p, sin y > 0]
= [ − cos y ]0 = −[cos v − cos 0] = 1 − cos v.
v
I = I1 + I 2 = 2n + 1 − cos v .
[
EXERCISE 6.2
Evaluate the following integrals:
3p
4
1.
2p
dx
∫ 1 + cos x
2.
cos −1 x
∫0 x dx
5.
∫ sin
p
4
0
∞
1
4.
2
p
2
∫ 1+
8.
p
2
3
∫ u sin u du
∫
11.
0
13.
x sin x
2
0
x
tan x
dx
1 − x2
p
6.
x dx
∫ 1 + sin
2
x
0
a
9.
dx
dx
∫ x+
a2 + x 2
0
100 p
cos x − sin x dx
12.
0
∫
1 − cos 2 x dx
0
p
2
dx
log x
∫
0
dx
dx
0
p
p
3.
0
x tan x
dx
7. ∫
sec x + tan x
0
∫ 1 + cos
tan −1 x
∫ x (1 + x )
∞
10.
1
x sin 2 n x
dx, n > 0
x + cos 2 n x
2n
x dx
∫ sin x + cos x
14.
0
a
15. Prove that
∫
−a
a
f ( x )dx = ∫ [ f ( x ) + f ( − x )]dx. Hence, evaluate
0
p
2
∫ u sin u cos udu.
−
p
2
ANSWERS TO EXERCISE 6.2
1. 2
2. p 2
⎛p ⎞
7. p ⎜ − 1⎟
⎝2 ⎠
8.
13.
p2
4
14.
3. −
p
4
p
2 2
9.
⋅ loge
(
4.
p
log 2
2
p
4
10.
2p
3
)
15.
p
4
2 +1
M06_ENGINEERING_MATHEMATICS-I _CH06_PART A.indd 27
p
log 2
2
5.
p
log 2
2
11. 2 2 − 1
6.
p2
2 2
12. 200 2
5/20/2016 10:09:10 AM
6.28
■
Engineering Mathematics
b
6.3 DEFINITE INTEGRAL ∫ f ( x )dx AS LIMIT OF A SUM
a
Let f be a bounded function defined on the finite interval [a, b] .
Divide [a, b] into n sub-intervals by the points x1 , x2 , …, xn −1 , not necessarily equidistant.
Take x0 = a and xn = b .
Now [x 0 , x 1 , x 2 , …, x n ] form a partition P of [a, b] and we have n sub-intervals.
I r = [ xr −1 , xr ], r = 1, 2, 3, …, n ,
Let dr = xr − xr −1 be the length of I r .
n
Then form the sum ∑ f ( cr ) dr .
Let cr be any point in I r .
r =1
This sum is called the Riemann sum of f ( x ) on [ a, b ] for the partition P.
The length of the largest sub-interval is denoted by d and is called the norm of the partition.
n
The function f ( x ) is Reimann integrable if lim ∑ f (cr )d r exists and is independent of the choice
n →∞
d → 0 r =1
of the interval and the point cr.
This limit is called the definite integral of f ( x ) over [a, b].
b
n
We write ∫ f ( x ) dx = lim ∑ f (cr )dr
a
n →∞
d → 0 r =1
In practical applications, for convenience, we take the length of the sub-interval equal to h =
where n is the number of sub-intervals.
b−a
,
n
Then x 1 = a + h , x 2 = a + 2h , …, x r = a + rh , …, b = x n = a + nh
and c r is taken as an end point of I r .
n −1
b
∫ f ( x) dx = lim h∑ f (a + rh)
Then
h→ 0
n →∞
a
(1)
r=0
Or
b
∫
a
If a = 0, b = 1, then h =
1
n
1 n −1
∑
n →∞ n
r=0
[ lim
n
f ( x ) dx = lim h∑ f (a + rh)
n →∞
h→ 0
( 2)
r =1
⎛ r⎞
f ⎜ ⎟ = ∫ f ( x )dx
⎝ n⎠ 0
1
This formula enables us to evaluate the limit of a certain sums interms of the integrals.
6.3.1 Working Rule
1 ⎛ r⎞
f⎜ ⎟
n →∞
⎝ n⎠
r =1 n
n
1. First rewrite the given limit of a sum in the form lim ∑
M06_ENGINEERING_MATHEMATICS-I _CH06_PART A.indd 28
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Integral Calculus ■
2. Treat
6.29
r
1
⎛ r⎞
as x and as dx, then f ⎜ ⎟ as f ( x )
⎝ n⎠
n
n
3. Then the summation become integral as n → ∞
4. Limits are obtained from x =
r
1
n
. When r = 1, x = → 0 as n → ∞ and when r = n, x = = 1
n
n
n
1
5. Then the given limit is
1
∫
f ( x )dx and 6. Evaluate
0
∫ f ( x)dx by using methods of integration.
0
1 ⎛ r⎞
f ⎜ ⎟ , then the limits of integration are 1, 4 because if
n →∞
⎝ n⎠
r =1 n
4n
Note Suppose the limit is lim ∑
4
r = 1, x =
1
4n
→ 0 as n → ∞ and r = 4 n , x =
= 4 and the integral is ∫ f ( x ) dx .
n
n
0
WORKED EXAMPLES
EXAMPLE 1
n
1
Show that lim
∑
n →∞
4n 2 r 2
2
r 51
Solution.
n
p
.
6
1
The given limit is lim ∑
n →∞
5
4n − r 2
2
r =1
⎛ r⎞
Rewrite in the form f ⎜ ⎟ . So, take out n2 as common.
⎝ n⎠
n
lim ∑
[
Treat
n →∞
r =1
4n − r
2
r
1
as x and as dx.
n
n
[
n
lim ∑
n →∞
r =1
n
1
2
= lim ∑
n →∞
r =1
1
n 4−
When r = 1, x =
1
4n2 − r 2
1
=∫
0
2
r
n2
1 n
∑
n →∞ n
r =1
= lim
1
⎛ r⎞
n 4−⎜ ⎟
⎝ n⎠
2
1
n
→ 0 as n → ∞ and when r = n , x = = 1
n
n
1
x⎤
1
p
⎡
= ⎢sin −1 ⎥ = sin −1 − sin −1 0 = .
2
2 ⎦0
2
6
⎣
4−x
dx
EXAMPLE 2
⎡ 1
4
9
n2 ⎤
Find the value of lim ⎢ 3
1 3
1 3
1…1 3 ⎥ .
n→ ∞ n + 1
n + 8 n + 27
2n ⎦
⎣
M06_ENGINEERING_MATHEMATICS-I _CH06_PART A.indd 29
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■
6.30
Engineering Mathematics
Solution.
r2
3
n →∞
r =1 n + r
n
The given limit is lim ∑
3
⎛ r⎞
Rewrite in the form f ⎜ ⎟ .
⎝ n⎠
n
r2
= lim ∑
3
n →∞
n →∞
r =1
r =1 n + r
r2
n
lim ∑
[
Treat
So, take out n3 as common.
3
⎛ r ⎞
n 3 ⎜1 + 3 ⎟
⎝ n ⎠
3
⎛ r⎞
⎜⎝ ⎟⎠
n
2
1 n
∑
3
n →∞ n
r =1 ⎛
⎛ r⎞ ⎞
⎜1 + ⎜⎝ n ⎟⎠ ⎟
⎝
⎠
= lim
r
1
= x and as dx.
n
n
When r = 1, x =
1
n
→ 0 as n → ∞ and when r = n, x = = 1
n
n
[
lim ∑
1
1
r2
x2
1 3x 2
dx
=
=
∫0 1 + x 3 3 ∫0 1 + x 3 dx
3
3
n →∞
r =1 n + r
n
⎤
⎡ f ′( x )
⎢{ f ( x ) dx = log e f ( x ) ⎥
⎣
⎦
=
1
1
⎡⎣log e (1 + x 3 ) ⎤⎦
0
3
=
1
1
log e (1 + 1) − log(1 + 0) ] = log e 2
[
3
3
⎡ 1
4
9
n2 ⎤ 1
…
+
+
+
+
[ lim
⎢
⎥ = log e 2
x →∞ n3 + 1
2 n3 ⎦ 3
n3 + 8 n3 + 27
⎣
EXAMPLE 3
⎡⎛
n⎞ ⎤
1⎞ ⎛
2⎞ ⎛
Show that lim ⎢⎜ 1 + ⎟ ⎜ 1 + ⎟ … ⎜ 1 + ⎟ ⎥
n→∞ ⎝
n⎠ ⎝
n⎠ ⎝
n⎠ ⎦
⎣
1/ n
4
5 .
e
Solution.
1/ n
⎡⎛ 1 ⎞ ⎛ 2 ⎞ ⎛ n ⎞ ⎤
A = lim ⎢⎜1 + ⎟ ⎜1 + ⎟ … ⎜1 + ⎟ ⎥
n →∞ ⎝
n⎠ ⎝ n⎠ ⎝ n⎠ ⎦
⎣
To convert the product into sum, take logarithm on both sides
Let
1/ n
⎡⎛ 1 ⎞ ⎛ 2 ⎞ ⎛ n ⎞ ⎤
[{ loge is a continuous function]
loge A = lim loge ⎢⎜1 + ⎟ ⎜1 + ⎟ … ⎜1 + ⎟ ⎥
n →∞
⎣⎝ n ⎠ ⎝ n ⎠ ⎝ n ⎠ ⎦
⎡⎛ 1 ⎞ ⎛ 2 ⎞ ⎛ n ⎞ ⎤
1
= lim loge ⎢⎜1 + ⎟ ⎜1 + ⎟ … ⎜1 + ⎟ ⎥
n →∞ n
⎣⎝ n ⎠ ⎝ n ⎠ ⎝ n ⎠ ⎦
M06_ENGINEERING_MATHEMATICS-I _CH06_PART A.indd 30
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Integral Calculus ■
= lim
n →∞
6.31
1⎡
⎛ 1⎞
⎛ 2⎞
⎛ n⎞ ⎤
loge ⎜1 + ⎟ + loge ⎜1 + ⎟ + … + loge ⎜1 + ⎟ ⎥
⎝ n⎠
⎝ n⎠
⎝ n⎠ ⎦
n ⎢⎣
1 n
⎛ r⎞
loge ⎜1 + ⎟
∑
n→∞ n
⎝ n⎠
r =1
= lim
Treat
r
1
= x and as dx.
n
n
When r = 1, x =
1
n
→ 0 as n → ∞ and when r = n, x = = 1
n
n
1
[
log e A = ∫ log e (1 + x )dx
0
Integrating by parts, we get
1
log e A = [ log e (1 + x ) ? x ]0 − ∫
1
1
? xdx
1
+
x
0
1
= [ log e (1 + 1) − log e 1] − ∫
x
dx
1
x
+
0
1+ x −1
dx
1+ x
0
1
= log e 2 − ∫
1 ⎞
⎛
= log e 2 − ∫ ⎜1 −
⎟ dx
⎝
1
+
x⎠
0
1
1
1
1
dx
x
1
+
0
= log e 2 − ∫ dx + ∫
0
= log e 2 − [ x ]0 + [ log e (1 + x ) ]0
1
1
= log e 2 − (1 − 0 ) + log e 2 − log e 1 = 2 log e 2 − 1
⇒
loge A = loge 22 − loge e = loge 4 − loge e = loge
1/ n
[ lim ⎡⎛1 + 1 ⎞ ⎛1 + 2 ⎞ … ⎛1 + n ⎞ ⎤
⎟
⎜
⎟⎜
⎟ ⎜
n →∞ ⎢ ⎝
n ⎠ ⎝ n ⎠ ⎝ n ⎠ ⎥⎦
⎣
=
4
e
⇒ A=
4
e
4
e
EXAMPLE 4
n
2
2
⎡⎛
⎛
n2 ⎞ ⎤
1 ⎞⎛
22 ⎞ ⎛
32 ⎞
Show that lim ⎢⎜ 1 + 2 ⎟ ⎜ 1 + 2 ⎟ ⎜ 1 + 2 ⎟ … ⎜ 1 + 2 ⎟ ⎥
n→∞ ⎝
n ⎠⎝
n ⎠ ⎝
n ⎠
n ⎠ ⎥
⎝
⎢⎣
⎦
2 / n2
4
5 .
e
Solution.
n
2
2
⎡⎛
1 ⎞ ⎛ 22 ⎞ ⎛ 32 ⎞ … ⎛ n2 ⎞ ⎤
Let A = lim ⎢⎜1 + 2 ⎟ ⎜1 + 2 ⎟ ⎜1 + 2 ⎟
⎜⎝1 + n2 ⎟⎠ ⎥
n →∞ ⎝
n ⎠⎝ n ⎠ ⎝ n ⎠
⎢⎣
⎥⎦
M06_ENGINEERING_MATHEMATICS-I _CH06_PART A.indd 31
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5/20/2016 10:09:26 AM
6.32
■
Engineering Mathematics
To convert the product into sum, take logarithm on both sides
[
n
2
⎡⎛
⎛ n2 ⎞ ⎤
1 ⎞ ⎛ 22 ⎞
loge A = lim log e ⎢⎜1 + 2 ⎟ ⎜1 + 2 ⎟ … ⎜1 + 2 ⎟ ⎥
n →∞
⎝ n ⎠ ⎥⎦
⎢⎣⎝ n ⎠ ⎝ n ⎠
2 n2
n
2
⎡⎛
2
1 ⎞ ⎛ 22 ⎞ … ⎛ n2 ⎞ ⎤
= lim 2 log e ⎢⎜1 + 2 ⎟ ⎜1 + 2 ⎟
⎜⎝1 + n2 ⎟⎠ ⎥
n →∞ n
⎢⎣⎝ n ⎠ ⎝ n ⎠
⎦⎥
⎛ 22 ⎞ …
⎛ n2 ⎞ ⎤
2 ⎡
1⎞
⎛
log
+
+
log
+
+
+
log
1
2
1
n
⎢
⎜
⎟
e
e ⎜
e ⎜1 + 2 ⎟ ⎥
n →∞ n 2
⎝ n2 ⎠
⎝ n2 ⎟⎠
⎝ n ⎠⎦
⎣
2
n
n
⎛ r2 ⎞
⎛ r ⎞
2
1
⎛ r⎞
= lim 2 ∑ r loge ⎜1 + 2 ⎟ = lim ∑ 2 ⎜ ⎟ log e ⎜1 + 2 ⎟
n →∞ n
⎝ n ⎠
⎝ n ⎠ n→∞ n r =1 ⎝ n ⎠
r =1
= lim
r
1
= x and as dx.
n
n
1
n
When r = 1, x = → 0as n → ∞ and when r = n, x = = 1
n
n
Treat
1
[
loge A = ∫ 2x loge (1 + x 2 ) dx
0
Put 1 + x 2 = t ∴ 2 xdx = dt.
When x = 0, t = 1 and when x = 1, t = 2
2
[
log e A = ∫ log t dt
1
Integrating by parts, we get
2
1
2
log e A = [ log e t ⋅ t ]1 − ∫ ⋅ t dt
t
1
2
= 2 log e 2 − ∫ dt = log e 22 − [t ]1 = log e 4 − ( 2 − 1) = log e 4 − 1
2
1
⇒
[
log e A = log e 4 − log e e = log e
n
2
2
⎡⎛
⎛ n2 ⎞ ⎤
1 ⎞ ⎛ 22 ⎞ ⎛ 32 ⎞
lim ⎢⎜1 + 2 ⎟ ⎜ 1 + 2 ⎟ ⎜1 + 2 ⎟ … ⎜1 + 2 ⎟ ⎥
n →∞ ⎝
n ⎠⎝ n ⎠ ⎝ n ⎠
⎝ n ⎠ ⎥⎦
⎢⎣
4
e
⇒
A=
4
e
2 n2
=
4
.
e
EXERCISE 6.3
Evaluate the following limits as integrals
1
1⎤
⎡ 1
+
+ …+ ⎥.
1. lim ⎢
n →∞ n + 1
n+2
2n ⎦
⎣
M06_ENGINEERING_MATHEMATICS-I _CH06_PART A.indd 32
n −1
⎡
⎤
1
2. lim ∑ ⎢ 2
.
2 12 ⎥
n →∞
r = 0 ⎣ (n + r )
⎦
5/20/2016 10:09:30 AM
Integral Calculus ■
n ⎡
⎤
n
⎥.
3. lim ∑ ⎢
3
n →∞
r =1 ⎢ [ n + 3( r − 1)] ⎥
⎣
⎦
⎡ n +1
n+2
2n ⎤
4. lim ⎢ 3 2 + 3 2 + … + 3 2 ⎥ .
n →∞
n
n ⎦
⎣ n
∞
⎡ r3 ⎤
5. lim ∑ ⎢ 4
.
4 ⎥
n →∞
r =1 ⎣ r + n ⎦
6. lim ∑
r2
.
3
n →∞
r =1 ( n + r )
n
n ⎡
n2 − r 2
8. lim ∑ ⎢
2
n →∞
r =1 ⎢
⎣ n
n
r
.
2
n →∞
r =1 n + r
7. lim ∑
2
⎡ ( n3 + 1)( n3 + 23 ) …( n3 + n3 ) ⎤
9. lim ⎢
⎥ .
n →∞
n3
⎣
⎦
1n
n −1
11. lim ∑
n →∞
r=0
1
n2 − r 2
6.33
⎤
⎥.
⎥⎦
r4
.
5
5
n →∞
r =1 n + r
n
10. lim ∑
1n
⎡⎛
1 ⎞ ⎛ 22 ⎞ ⎛ n2 ⎞ ⎤
12. lim ⎢⎜1 + 2 ⎟ ⎜1 + 2 ⎟ … ⎜1 + 2 ⎟ ⎥ .
n →∞ ⎝
n ⎠ ⎝ n ⎠ ⎝ n ⎠⎦
⎣
.
⎤
⎡n
n
n
n
13. lim ⎢ 2 + 2 2 + 2
+ …+ 2
⎥.
2
2
n →∞ n
n +1 n + 2
n + ( n − 1) ⎥⎦
⎢⎣
ANSWERS TO EXERCISE 6.3
2.
p
2
3.
1
3
4.
6.
3
8
7.
1
log e 2
2
8.
p
4
9. 4e
11.
p
2
13.
p
4
6.4
p
12. 2e 2
−2
)
(
1. log e 2
2
2 2 −1
3
p
3
5.
1
log e 2
4
10.
1
log e 2
5
−3
REDUCTION FORMULAE
Integrals of type ∫ sin n x dx, ∫ tan n x dx, ∫ x n e ax dx cannot be evaluated directively. Applying integration
by parts, we can reduce an integral with index n > 0, called the order of the integral, to an integral
of the reduced order with a smaller index. The relation between the given integral and the reduced
integral of lower order is called the reduction formula.
We derive the reduction formula for some standard integrals
6.4.1 The Reduction Formula for (a) ∫ sinn x dx and (b) ∫ cos n x dx
(a)
∫ sin
n
x dx
Solution.
Let
I n = ∫ sin n x dx = ∫ sin n −1 x sin x dx
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Taking u = sin n −1 x ,
v = sin x and integrating by parts, we get
I n = sin n −1x ( − cos x ) − ∫ ( n − 1) sin n − 2 x cos x ( − cos x )dx
= − sin n −1x cos x + ( n − 1)∫ sin n − 2 x cos 2 x dx
= − sin n −1x cos x + ( n − 1)∫ sin n − 2 x ( 1 − sin 2 x ) dx
= − sin n −1x cos x + ( n − 1)∫ sin n − 2 x dx − ( n − 1)∫ sin n x dx
= − sin n −1 x cos x + ( n − 1)I n − 2 − ( n − 1)I n
(1 + n − 1)I n = − sin n −1x cos x + ( n − 1)I n − 2
[
nI n = − sin n −1x cos x + ( n − 1)I n − 2
⇒
In = −
[
sin n −1x cos x n − 1
+
I n−2
n
n
[ the reduction formula is ∫ sin n x dx = −
(b)
∫ cos
n
sin n −1x cos x n − 1
I n−2 .
+
n
n
x dx
Solution.
Let
Taking
I n = ∫ cos n x dx = ∫ cos n −1x cos x dx
u = cos n −1x ,
v = cos x and integrating by parts, we get
I n = cos n −1 x sin x − ∫ ( n − 1) cos n − 2 x ( − sin x ) sin x dx
= cos n −1x sin x + ( n − 1)∫ cos n − 2 x sin 2 x dx
= cos n −1x sin x + ( n − 1)∫ cos n − 2 x (1 − cos 2 x ) dx
= cos n −1x sin x + ( n − 1)∫ cos n − 2 x dx − ( n − 1) ∫ cos n x dx
⇒
I n = cos n −1x sin x + ( n − 1)I n − 2 − ( n − 1)I n
(1 + n − 1)I n = cos n −1x sin x + ( n − 1)I n − 2
⇒
∴
nI n = cos n −1x sin x + ( n − 1)I n − 2
In =
cos n −1x sin x n − 1
+
I n−2
n
n
∴ the required reduction formula for
M06_ENGINEERING_MATHEMATICS-I _CH06_PART A.indd 34
∫ cos x dx
n
=
cos n −1x sin x n − 1
I n−2 .
+
n
n
5/20/2016 10:09:40 AM
Integral Calculus ■
Deduction:
If n is a non-negative integer, then prove that
p
p
⎧n −1 n − 3 n − 5 … 3 1 p
2
2
⎪⎪ n ⋅ n − 2 ⋅ n − 4 4 ⋅ 2 ⋅ 2
n
n
sin
x
dx
=
cos
x
dx
=
⎨
∫0
∫0
⎪ n − 1 ⋅ n − 3 ⋅ n − 5 … 4 ⋅ 2 ⋅1
⎪⎩ n n − 2 n − 4 5 3
6.35
if n is even
if n is odd
Solution.
p
2
Let
I n = ∫ sin n x dx
0
By reduction formula, we have
p
⎡ sin n −1x cos x ⎤ 2 n − 1
I n = ⎢−
I n−2
⎥ +
n
n
⎦0
⎣
⇒
⇒
[
In = 0 +
I n−2 =
In =
n −1
n −1
I n−2 =
I n−2
n
n
n−3
I n−4,
n−2
I n−4 =
n−5
I n − 6 and so on.
n−4
n −1 n − 3 n − 5 …
⋅
⋅
, the last integral is I1 or I 0
n n−2 n−4
Case 1: If n is even, then
In =
n −1 n − 3 n − 5 … 3 1
⋅
⋅
⋅ ⋅ I0
n n−2 n−4 4 2
p
2
But
0
[
p
I 0 = ∫ sin 0 x dx = [ x ]02 =
In =
p
2
n −1 n − 3 n − 5 … 3 1 p
⋅
⋅
⋅ ⋅ , if n is even
n n−2 n−4 4 2 2
Case 2: if n is odd, then
In =
n −1 n − 3 n − 5 … 4 2
⋅
⋅
⋅ ⋅ I1
n n−2 n−4 5 3
p
2
But
p
p
⎡
⎤
I 1 = ∫ sin x dx = [ − cos x ]02 = − ⎢cos − cos 0 ⎥ = −1(0 − 1) = 1
2
⎣
⎦
0
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Engineering Mathematics
∴
In =
n −1 n − 3 n − 5 … 4 2
⋅
⋅
⋅ ⋅1
n n−2 n−4 5 3
⎧n −1 n − 3 n − 5 … 3 1 p
⎪⎪ n ⋅ n − 2 ⋅ n − 4 4 ⋅ 2 ⋅ 2
n
sin
x
dx
=
⎨
∫0
⎪ n − 1 ⋅ n − 3 ⋅ n − 5 … 4 ⋅ 2 ⋅1
⎪⎩ n n − 2 n − 4 5 3
p
2
∴
if n is even
if n is odd
Similarly, we get
⎧n −1 n − 3 n − 5 … 3 1 p
⎪⎪ n ⋅ n − 2 ⋅ n − 4 4 ⋅ 2 ⋅ 2
n
∫0 cos x dx = ⎨ n − 1 n − 3 n − 5 4 2
⎪
… ⋅ ⋅1
⋅
⋅
⎪⎩ n n − 2 n − 4 5 3
p
2
if n is even
if n is odd
⎧n −1 n − 3 n − 5 … 3 1 p
⎪⎪ n ⋅ n − 2 ⋅ n − 4 4 ⋅ 2 ⋅ 2
n
n
sin
x
dx
=
cos
x
dx
=
⎨
∫0
∫0
⎪ n − 1 ⋅ n − 3 ⋅ n − 5 … 4 ⋅ 2 ⋅1
⎪⎩ n n − 2 n − 4 5 3
p
2
[
p
2
if n is even
if n is odd
6.4.2 The Reduction Formula for (a) ∫ tan n x dx and (b) ∫ cot n x dx
(a)
∫ tan
n
x dx
Solution.
I n = ∫ tan n x dx = ∫ tan n − 2 x tan 2 x dx
Let
= ∫ tan n − 2 x (sec 2 x − 1)dx
= ∫ tan n − 2 x sec 2 x dx − ∫ tan n − 2 x dx
(tan x ) n − 2 +1
(tan x ) n −1
− I n−2 =
− I n−2
n − 2 +1
n −1
(tan x ) n −1
n
tan
x
dx
=
− I n−2
∫
n −1
=
⇒
(b)
∫ cot
n
x dx
Solution.
Let
I n = ∫ cot n x dx = ∫ cot n − 2 x cot 2 x dx
= ∫ cot n − 2 x( cosec 2 x − 1) dx
M06_ENGINEERING_MATHEMATICS-I _CH06_PART A.indd 36
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Integral Calculus ■
6.37
= − ∫ cot n − 2 x ( −cosec 2 x ) dx − ∫ cot n − 2 x dx
⇒
In = −
n
∫ cot x dx = −
∴
(cot x ) n − 2 +1
(cot x ) n −1
− I n−2 = −
− I n−2
n −1
n − 2 +1
(cot x ) n −1
− I n−2 .
n −1
n
n
6.4.3 The Reduction Formula for (a) ∫ sec x dx and (b) ∫ cosec x dx
(a)
∫ sec
n
x dx
Solution
Let I n = ∫ sec n x dx = ∫ sec n − 2 x sec 2 x dx
Take u = sec n − 2 x, v = sec 2 x and
integrating by parts, we get
I n = sec n − 2 x tan x − ∫ ( n − 2) sec n − 3 x sec x tan x tan x dx
= sec n − 2 x tan x − ( n − 2)∫ sec n − 2 x tan 2 x dx
= sec n − 2 x tan x − ( n − 2)∫ sec n − 2 x (sec 2 x − 1) dx
= sec n − 2 x tan x − ( n − 2)∫ sec n x dx + ( n − 2)∫ sec n − 2 x dx
I n = sec n − 2 x tan x − ( n − 2) I n + ( n − 2) I n − 2
⇒
⇒
(1 + n − 2) I n = sec n − 2 x tan x + ( n − 2) I n − 2
⇒
( n − 1) I n = sec n − 2 x tan x + ( n − 2) I n − 2
In =
[
∫ sec
[
(b)
n
x dx =
sec n − 2 x tan x ( n − 2)
+
I n−2
( n − 1)
( n − 1)
sec n − 2 x tan x ( n − 2)
+
I n−2 .
( n − 1)
( n − 1)
∫ cosec x dx
n
Solution.
Let
I n = ∫ cosec n x dx = ∫ cosec n − 2 x cosec 2 x dx
Take
u = cosec n − 2 x, v = cosec 2 x
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Integrating by parts, we get
I n = cosec n − 2 x ( − cot x ) − ∫ ( n − 2)cosec n − 3 x ( −cosecx cot x )( − cot x ) dx
= −cosec n − 2 x cot x − ( n − 2)∫ cosec n − 2 x cot 2 x dx
= −cosec n − 2 x cot x − ( n − 2)∫ cosec n − 2 x (cosec 2 x − 1)dx
= −cosec n − 2 x cot x − ( n − 2)∫ cosec n x dx + ( n − 2)∫ cosec n − 2 x dx
= −cosec n − 2 x cot x − ( n − 2) I n + ( n − 2) I n − 2
⇒
(1 + n − 2) I n = −cosec n − 2 x cot x + ( n − 2) I n − 2
⇒
( n − 1) I n = −cosec n − 2 x cot x + ( n − 2) I n − 2
In = −
cosec n − 2 x cot x n − 2
+
I n−2
n −1
n −1
x dx = −
cosec n − 2 x cot x n − 2
+
I n−2
n −1
n −1
[
∫ cot
[
n
WORKED EXAMPLES
EXAMPLE 1
2a
Evaluate
∫
0
x3
2ax − x 2
dx .
Solution.
2a
Let
I=
∫
0
Put x = 2a sin 2 u
x3
2ax − x 2
[
dx
dx = 4a sin u cos u d u
When x = 0, sin u = 0 ⇒ u = 0 and when x = 2a, sin u = 1
[
I=
p/2
∫
0
= 16 a3
( 2a sin 2 u)3 4 a sin u cos u
2a ? 2a sin 2 u − ( 2a sin 2 u) 2
p/2
∫
0
= 16 a3
d u = 32a
⇒u=
p/2
4
∫
p
2
sin 7 u cos u
2a sin u 1 − sin 2 u
du
0
sin 6 u cos u
du
cos u
p/2
∫ sin
6
u d u = 16a3
0
M06_ENGINEERING_MATHEMATICS-I _CH06_PART A.indd 38
5 3 1 p 5 3
? ? ? = pa
6 4 2 2 2
[{ n = 6 is even]
5/20/2016 10:09:54 AM
Integral Calculus ■
6.39
EXAMPLE 2
∞
Evaluate
dx
∫ (1 + x
0
2 8
)
.
Solution.
∞
dx
2 8
0 (1 + x )
Let I = ∫
Put x = tan u
[ dx = sec 2 ud u
When x = 0, tan u = 0 ⇒ u = 0 and when x = ∞, tan u = ∞ ⇒ u =
[I=
p/2
∫
0
p
2
sec 2 u
d u p / 2 sec 2 u
du
(1 + tan 2 u)8 = ∫
16
0 sec u
=
p/2
∫ cos
14
u du =
0
13 11 9 7 5 3 1 p 429p
?
?
? ? ? ? ? =
[{ n = 14 is even]
14 12 10 8 6 4 2 2 4090
EXAMPLE 3
If I n =
tn
t n +1
,
then
show
that
I
+
I
=
dt . Hence, evaluate I 6.
dt
n
+
2
n
∫ (1 + t 2 )
n +1
Solution.
tn
dt
(1 + t 2 )
In = ∫
Given
t = tan x
Put
[
In = ∫
⇒
In =
⇒
I n + I n−2 =
∴ dt = sec 2 x dx
tan n x
sec 2 x dx = ∫ tan n x dx
(1 + tan 2 x )
tan n −1 x
− I n−2
n −1
tan n −1 x
n −1
[using reduction formula 6.4.2(a)]
⇒
I n + I n−2 =
t n −1
n −1
(1)
t n + 2 −1
⇒
n + 2 −1
I n+ 2 + I n =
t n +1
n +1
( 2)
Replacing n by n + 2 in (1), we get
I n+ 2 + I n+ 2− 2 =
To evaluate I 6 .
I6 = ∫
Put n = 6 in (1), we get
I6 + I 4 =
t6
dt
1+ t 2
t5
5
M06_ENGINEERING_MATHEMATICS-I _CH06_PART A.indd 39
⇒
I6 =
t5
− I4
5
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6.40
Engineering Mathematics
=
⎤ t5 t3
t5 ⎡t3
t5 t3
− ⎢ − I2 ⎥ = − + I2 = − + t − I0.
5 ⎣3
5 3
⎦ 5 3
But
I 0 = ∫ tan 0 x dx = ∫ dx = x = tan −1 t
[
I6 =
t5 t3
− + t − tan −1 t + c
5 3
EXAMPLE 4
If I n =
p/ 4
∫
tan n xdx , then prove that (1) ( n − 1)[ I n + I n −1 ] = 1 and (2) n[ I n +1 + I n − 2 ] = 1 .
0
Hence, evaluate I 5.
Solution.
In =
Given
p/4
∫
tan n x dx
0
1. To prove ( n 2 1)[ I n 1 I n 2 1 ] 5 1
By reduction formula 6.4.2 (a), we have
p
⎡ tan −1 x ⎤ 4
1 ⎡ −1 p
−1 ⎤
In = ⎢
⎥ − I n−2 =
⎢ tan 4 − tan 0 ⎥ − I n − 2
1
n
1
n
−
−
⎣
⎦
⎣
⎦0
1
1
(1 − 0) − I n − 2 =
− I n−2
n −1
n −1
1
In =
− I n−2
n −1
=
⇒
1
n −1
2. To prove n[ I n 1 1 1 I n 2 1 ] 5 1
⇒
I n + I n −2 =
⇒
( n − 1)[ I n + I n − 2 ] = 1
(1)
( 2)
The result (1) is true for all n ≥ 2. Replacing n by ( n +1) in ( 2) , we get
( n + 1 − 1)[ I n +1 + I n +1− 2 ] = 1 ⇒ n [ I n +1 + I n −1 ] = 1
To find I 5
p
4
I 5 = ∫ tan 5 x dx
0
Put n = 5 in (1), we get
I5 =
1
1 ⎡1
1
⎤
− I 3 = − ⎢ − I1 ⎥ = − + I1
4
4 ⎣2
4
⎦
M06_ENGINEERING_MATHEMATICS-I _CH06_PART A.indd 40
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Integral Calculus ■
6.41
p
4
p
p
⎡
⎤
I1 = ∫ tan x dx = [ log sec x ]04 = ⎢log sec − log sec 0 ⎥
4
⎣
⎦
0
1
= log e 2 − log e 1 = log e 2
2
1
1 1
I 5 = log e 2 − = [ 2 log e 2 − 1]
2
4 4
But
[
EXAMPLE 5
a
If un = ∫ x n e − x dx , then prove that un − ( n + a )un −1 + a ( n − 1)un − 2 = 0 .
0
Solution.
a
un = ∫ x n e − x dx
Given
0
−x
Take u = x , v = e . Integrating by parts, we get
n
a
a
⎡ e − x ⎤ a n −1 e − x
un = ⎢ x n
dx = −[a n e − a − 0] + n∫ x n −1 e − x dx
⎥ − ∫ nx
−1
⎣ −1 ⎦ 0 0
0
un = − a n e − a + n un −1
⇒
un −1 = − a n −1e − a + ( n − 1) un − 2
[
un − ( n + a)un −1 = − a n e − a + nun −1 − nun −1 − aun −1
[
= − a n e − a − aun −1
= − a n e − a − a[ − a n −1e − a + ( n − 1) un − 2 ]
= − a n e − a + a n e − a − a( n − 1) un − 2 = − a( n − 1) un − 2
[
un − ( n + a)un −1 + a( n − 1) un − 2 = 0
EXAMPLE 6
If I n = ∫ x n e x dx , then show that I n + nI n −1 = x n e x . Hence, find I 4 .
Solution.
Given
I n = ∫ x n e x dx
Take u = x n , v = e x. Integrating by parts, we get
I n = x n e x − ∫ n ⋅ x n −1 e x dx = x n e x − n∫ x n −1 e x dx ⇒ I n = x n e x − nI n −1
⇒
(1)
I n + nI n −1 = x n e x
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Engineering Mathematics
To find I 4 = ∫ x 4 e x dx
Put n = 4 in (1) .
I 4 = x 4 e x − 4 I 3 = x 4 e x − 4[ x 3 e x − 3I 2 ]
[
= x 4 e x − 4 x 3 e x + 12 I 2
= x 4 e x − 4 x 3 e x + 12[ x 2 e x − 2 I1 ]
= x 4 e x − 4 x 3 e x + 12 x 2 e x − 24 I1
= x 4 e x − 4 x 3 e x + 12 x 2 e x − 24[ xe x − I 0 ]
= x 4e x − 4 x 3e x + 12x 2e x − 24 xe x + 24∫ e x dx
[{ I 0 = ∫ e x dx ]
= x 4e x − 4 x 3e x + 12x 2e x − 24 xe x + 24e x
= e x [x 4 − 4 x 3 + 12x 2 − 24 x + 24]
EXAMPLE 7
p
2
⎛ p⎞
If un = ∫ x n sin x dx ( n ≥ 0 ) , then prove that un + n( n − 1)un − 2 = n ⎜ ⎟
⎝ 2⎠
0
n −1
, n ≥ 0.
Hence, evaluate u2.
Solution.
p
2
Given
un = ∫ x n sin x dx
0
Take u = x n , v = sin x . Integrating by parts, we get
p
2
0
p
2
p
2
0
0
un = ⎡⎣ x n ( − cos x )⎤⎦ − ∫ nx n −1 ( − cos x ) dx = 0 + n ∫ x n −1 cos x dx
⎧
⎫
p
⎪
⎪
= n ⎨ ⎡⎣ x n −1 sin x ⎤⎦ 02 − ∫ ( n − 1) x n − 2 sin x dx ⎬
0
⎪
⎪
⎩
⎭
p
2
⎤
⎡ p2
⎡⎛ p ⎞ n −1
⎤
p
⎢ n−2
⎥
= n ⎢⎜ ⎟ sin − 0 ⎥ − n( n − 1) ⎢ ∫ x sin x dx ⎥
2
⎢⎣⎝ 2 ⎠
⎥⎦
⎢⎣ 0
⎥⎦
⇒
⇒
⎛ p⎞
un = n ⎜ ⎟
⎝ 2⎠
n −1
− n( n − 1)un − 2
⎛ p⎞
un + n( n − 1)un − 2 = n ⎜ ⎟
⎝ 2⎠
M06_ENGINEERING_MATHEMATICS-I _CH06_PART A.indd 42
n −1
(1)
5/20/2016 10:10:15 AM
Integral Calculus ■
6.43
To find u2.
Put n = 2 in (1) , then we get
⎛ p⎞
u2 + 2( 2 − 1)u0 = 2 ⋅ ⎜ ⎟
⎝ 2⎠
2 −1
⇒ u2 + 2u0 = p
p
2
p
u0 = ∫ sin x dx = − [ cos x ] 02 = −(0 − 1) = 1
But
0
u2 + 2 ⋅1 = p ⇒ u2 = p − 2
[
EXAMPLE 8
p
2
1
n −1
I n − 2 + 2 , n ≥ 2 . Hence, find I 4 .
n
n
If I n = ∫ u sin n u d u , then prove that I n =
0
Solution.
p
2
Given
p
2
I n = ∫ u sin u d u, = ∫ u sin n − 2 u sin 2 u d u
n
0
0
p
2
= ∫ u (sin u) n − 2 (1 − cos 2 u) d u
0
p
2
= ∫ u(sin u)
0
n−2
p
2
d u − ∫ u (sin u) n − 2 cos 2 u d u
0
p
2
= I n − 2 − ∫ u cos u[(sin u) n − 2 cos u] d u
0
⇒
I n = I n−2
p
p
⎧
⎫
n −1 2
2
⎤
⎡
u
(sin
)
(sin u) n −1 ⎪
⎪
d u⎬
− ⎨ ⎢u cos u
⎥ − [u( − sin u) + cos u]
n −1
n − 1 ⎦ 0 ∫0
⎪⎣
⎪
⎩
⎭
p/2
p/2
⎧⎪
⎫⎪
1
1
n
= I n − 2 − ⎨0 +
u
sin
u
d
u
−
(sin u) n −1 cos ud u⎬
∫
∫
n −1 0
n −1 0
⎭⎪
⎩⎪
p
= I n−2
1
1 ⎡ sin n u ⎤ 2
−
In +
⎥
⎢
n −1
n − 1 ⎣ n ⎦0
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6.44
Engineering Mathematics
= I n−2 −
⇒
I n = I n −2 −
1
1 ⎡ np
⎤
sin
In +
− sin n 0 ⎥
2
n −1
n( n − 1) ⎢⎣
⎦
1
1
In +
n −1
n( n − 1)
⇒
1 ⎞
1
⎛
⎜⎝1 +
⎟ I n = I n−2 +
n − 1⎠
n( n − 1)
⇒
1
⎛ n − 1 + 1⎞
⎜⎝
⎟⎠ I n = I n − 2 +
n −1
n( n − 1)
⇒
1
⎛ n ⎞
⎜⎝
⎟⎠ I n = I n − 2 +
n −1
n( n − 1)
⇒
In =
1
1
n −1
n −1
n −1
⇒ In =
I n−2 +
⋅
I n−2 + 2
n
n n( n − 1)
n
n
I4 =
3
1
I2 + 2 ,
4
4
To find I 4
If n = 4, then
p
2
I2 =
1
1
I0 + 2
2
2
p
But
⎡ u2 ⎤ 2 1 p2 p2
.
I 0 = ∫ u du = ⎢ ⎥ = ⋅
=
8
⎣ 2 ⎦0 2 4
0
[
I2 =
1 p2 1 p2 + 4
⋅
+ =
2 8 4
16
[
I4 =
3 p 2 + 4 1 3(p 2 + 4) + 4 3p 2 + 16
⋅
+
=
=
.
4 16
16
64
64
EXAMPLE 9
p
p
2
2
1
If I n 5 ∫ cot x dx , show that I n 5
2 I n − 2 , n ≥ 2 . Hence, evaluate ∫ cot 4 x dx.
n −1
p
0
n
4
Solution.
Given
p
2
I n = ∫ cot n x dx
p
4
M06_ENGINEERING_MATHEMATICS-I _CH06_PART A.indd 44
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Integral Calculus ■
6.45
By reduction formula for ∫ cot n x dx, we have
p/2
⎡ (cot x ) n −1 ⎤
−1
In = ⎢
[0 − 1] − I n − 2
⎥ − I n−2 =
n −1
⎣ n − 1 ⎦p/ 4
⇒
In =
[Refer page 6.36]
1
− I n−2 , n ≥ 2
n −1
(1)
p
2
I 4 = ∫ cot 4 x dx
To find I 4 .
p
4
I4 =
Put n = 4 in (1) , then
1
1
2
1
− I 2 = − (1 − I 0 ) = −1 + + I 0 = − + I 0.
3
3
3
3
p
2
p
I 0 = ∫ dx = [ x ]p2 =
But
p
4
4
p p p
− = .
2 4 4
2 p 3p − 8
.
I4 = − + =
3 4
12
[
6.4.4 The Reduction Formula for ∫ sin m x cos n x dx , Where m, n are Non-negative
Integers
Solution.
Let
I m , n = ∫ sin m x cos n x dx
= ∫ sin m x cos n −1 x cos x dx = ∫ cos n −1 x(sin m x cos x ) dx
Take u = cos n −1 x, v = sin m x cos n x.
I m , n = cos n −1 x
Integrating by parts, we get
sin m +1 x
sin m +1 x
− ∫ ( n − 1) cos n − 2 x ( − sin x )
dx
m +1
m +1
=
cos n −1 x sin m +1 x n − 1
sin m x sin 2 x cos n − 2 x dx
+
m +1
m +1∫
=
cos n −1 x sin m +1 x n − 1
sin m x(1 − cos 2 x ) cos n − 2 x dx
+
m +1
m +1∫
=
cos n −1 x sin m +1 x n − 1
n −1
sin m x cos n − 2 x dx −
sin m x cos n x dx
+
m +1
m +1∫
m +1∫
M06_ENGINEERING_MATHEMATICS-I _CH06_PART A.indd 45
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Engineering Mathematics
=
cos n −1 x sin m +1 x n − 1
n −1
I m,n − 2 −
I m,n
+
m +1
m +1
m +1
⇒
n −1 ⎞
sin m +1 x cos n −1 x n − 1
⎛
I
I m,n − 2
1
+
=
+
⎜⎝
⎟ m,n
m + 1⎠
m +1
m +1
⇒
sin m +1 x cos n −1 x n − 1
⎛ m + 1 + n − 1⎞
I m,n − 2
+
⎜⎝
⎟⎠ I m , n =
m +1
m +1
m +1
⇒
sin m +1 x cos n −1 x n − 1
⎛ m + n⎞
I m,n − 2
+
⎜⎝
⎟⎠ I m , n =
m +1
m +1
m +1
I m,n =
[
Deduction:
sin m +1 x cos n −1 x
n −1
+
I m,n − 2
m+n
m+n
p
2
The Reduction formula for ∫ sin m x cos n x dx , where m, n are non-negative integers.
0
Solution.
p
2
Let I m , n = ∫ sin m x cos n x dx
0
By reduction formula, we have
p
I m,n
⇒
⎡ sin m +1 x cos n −1 x ⎤ 2 n − 1
n −1
=⎢
I m,n − 2
I m,n − 2 = 0 +
⎥ +
+
+
m
m
n
m
n
+n
⎦0
⎣
I m,n =
[
I m,n − 2 =
[
I m,n =
n −1
I m,n − 2
m+n
n−3
I m,n − 4 ,
m+n−2
I m,n − 4 =
n−5
I m,n − 4
m+n−4
n −1
n−3
n−5 …
⋅
⋅
, the last integral is I m ,1 or I m , 0 .
m+n m+n−2 m+n−4
Case 1: If n is odd, reduce by n, then we get
I m,n =
n −1
n−3
n−5 … 2
⋅
⋅
I m ,1
m+n m+n−2 m+n−4 m+3
p
2
But
I m ,1
p
⎡ sin m +1 x ⎤ 2
1
= ∫ sin x cos x dx = ⎢
⎥ =
⎣ m + 1 ⎦0 m + 1
0
m
M06_ENGINEERING_MATHEMATICS-I _CH06_PART A.indd 46
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Integral Calculus ■
6.47
n −1
n−3
n−5 … 2
1
⋅
⋅
⋅
m + n m + n − 2 m + n − 4 m + 3 m +1
Similarly, if m is odd, reduce by m.
[
I m,n =
[
I=
m −1 m − 3
2
1
,
…
⋅
m + n m + n − 2 n + 3 n +1
Case 2: If n is even, reduce by n, we get
1
n −1
n−3
n−5 … 3
I m,n =
⋅
⋅
⋅
I m,0
m+n m+n−2 m+n−4 m+4 m+2
p
2
I m , 0 = ∫ sin m x dx =
But
0
=
I m,n =
[
=
m −1 m − 3 … 3 1 p
⋅
⋅ ⋅
m m −2 4 2 2
m −1 m − 3 … 2
⋅
⋅1
m m −2 3
if m is even
if m is odd
n −1
n−3 … 3
1 m −1 m − 3 … 3 1 p
⋅
⋅
⋅
⋅
⋅ ⋅ .
m+n m+n−2 m+4 m+2 m m−2 4 2 2
if m is even
n −1
n −3 … 3
1
m −1 m − 3 … 2
⋅
⋅
⋅
⋅
⋅1
m +n m +n−2 m +3 m +2 m m −2 3
if m is odd.
Note If both m and n are odd or even reduce by smaller index and use case (1) and m, n are small
numbers, we can directly integrate by substitution.
WORKED EXAMPLES
EXAMPLE 1
p
2
Evaluate
∫ sin
8
x cos 7 x dx .
0
Solution.
p
2
I 8, 7 = ∫ sin 8 x cos 7 x dx .
Let
Here m = 8, n = 7 is odd.
0
I m,n =
6 4 2 1
16
n −1
n−3 … 2
1
= ⋅ ⋅ ⋅ =
.
⋅
⋅
m + n m + n − 2 m + 3 m + 1 15 13 11 9 6435
EXAMPLE 2
a
Evaluate
∫x
2
( a 2 2x 2 ) 3/ 2 dx.
0
Solution.
Let
a
I = ∫ x 2 ( a 2 −x 2 )3/ 2 dx
0
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6.48
■
Engineering Mathematics
Let x = a sin u
∴ dx = a cos ud u
When x = 0, sin u = 0 ⇒ u = 0 and when x = a, sin u = 1 ⇒ u =
I=
[
p/2
∫a
p
2
sin 2 u(a2 − a2 sin 2 u)3 / 2 a cos ud u
2
0
= a6
p/2
∫ sin
2
u(1 − sin 2 u)3/ 2 cos ud u
2
u cos3 u cos ud u = a6
0
= a6
p/2
∫ sin
p/2
0
∫ sin
2
u cos 4 u d u
0
Here both the indices are even.
[ m = 4, n = 2.
[ we take the smaller one as n.
I = a6 ⋅
[
1
6
p/2
∫ cos
4
ud u =
0
a6 3 1 p pa6
⋅ ⋅ ⋅ =
6 4 2 2
32
EXAMPLE 3
4
Evaluate
∫x
3
4 x 2 x 2 dx.
0
Solution.
4
Let
I = ∫ x 3 4 x − x 2 dx
0
Put x = 4 sin 2 u
∴ dx = 8 sin u cos ud u
p
When x = 0, sin u = 0 ⇒ u = 0 and when x = 4, sin u = 1 ⇒ u =
2
[
I=
p/2
∫ (4 sin
2
u)3 4 ⋅ 4 sin 2 u − 16 sin 4 u ⋅
8 sin u cos ud u
0
=
p/2
∫
43 sin 6 u ⋅ 4 sin u 1 − sin 2 u ⋅ 8 sin u cos ud u
0
p/2
p/2
0
0
= 2048 ∫ sin 8 u ⋅ cos u ⋅ cos ud u = 2048 ∫ sin 8 u ⋅ cos 2 ud u
Both indices are even and n = 2 is smaller.
[
I = 2048 ⋅
M06_ENGINEERING_MATHEMATICS-I _CH06_PART A.indd 48
1
10
p/2
∫ sin
0
8
ud u = 2048 ⋅
1 7 5 3 1 p
⋅ ⋅ ⋅ ⋅ ⋅ = 28p
10 8 6 4 2 2
5/20/2016 10:10:38 AM
Integral Calculus ■
EXAMPLE 4
Evaluate
6.49
p
6
∫ cos
4
3usin 3 6ud u
0
Solution.
p
6
Let I = ∫ cos 4 3u sin 3 6ud u
0
1
dx = d u
3
p
p p
When u = 0, x = 0 and when u = , x = 3 =
6
6 2
Put x = 3u
∴ dx = 3d u ⇒
Now sin 3 6u = sin 3 2 x = ( 2 sin x cos x )3 = 8 sin 3 x cos3 x
p
2
p
1
82
8 2 1 1
I = ∫ cos 4 x 8 sin 3 x cos3 x dx = ∫ cos 7 x sin 3 xdx = ⋅ ⋅ =
3
30
3 10 8 15
0
[
Aliter:
I=
p
2
p
2
8
8
cos 7 x sin 3 xdx = ∫ cos 7 x sin x(1 − cos 2 x )dx
3 ∫0
30
=
p
2
p
2
8
cos 7 x sin xdx − ∫ cos 9 x sin xdx
3 ∫0
0
p
p
8 ⎡ cos8 x ⎤ 2 ⎡ cos10 x ⎤ 2
= ⎢−
⎥ − ⎢−
⎥
3⎣
8 ⎦0 ⎣
10 ⎦ 0
8⎡ 1
1
⎤ 8 ⎡ 1 1 ⎤ 8 ⎡10 − 8 ⎤ 8 ⎡ 2 ⎤ 1
= ⎢ − (0 − 1) + (0 − 1) ⎥ = ⎢ − ⎥ = ⎢
⎥= ⎢
⎥=
3⎣ 8
10
⎦ 3 ⎣ 8 10 ⎦ 3 ⎣ 8 × 10 ⎦ 3 ⎣ 8 × 10 ⎦ 15
6.4.5 The Reduction Formula For (a) ∫ xm(log x)ndx, (b) ∫ xn sin mx dx, (c)
(a)
∫x
m
∫ xn cos mx dx
(log e x ) n dx
Solution.
Let I m , n = ∫ x m (log e x ) n dx = ∫ (log e x ) n x m dx
Take u = (log e x ) n , v = x m. Integrating by parts, we get
[
m +1
n −1 1 x
x m +1
− ∫ n(log e x ) ⋅ ⋅
dx
m +1
x m +1
n −1
x m +1 (log e x ) n
n
=
−
(log e x ) ⋅ x m dx
∫
m +1
m +1
I m , n = (log e x ) n ⋅
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6.50
Engineering Mathematics
I m,n =
⇒
(b)
∫x
n
x m +1 (log x ) n
n
−
I m , n −1
m +1
m +1
sin mx dx
Solution.
Let I n = ∫ x n sin mx dx
Take u = x n , v = sin mx. Integrating by parts, we get
cos mx ⎞
⎛ cos mx ⎞
n −1 ⎛
In = xn ⋅ ⎜ −
⎟ − nx ⎜⎝ −
⎟ dx
⎝
m ⎠ ∫
m ⎠
x n cos mx n n −1
+ ∫ x cos mxdx
m
m
x n cos mx n ⎡ n −1 sin mx
sin mx ⎤
=−
+ ⎢x
− ∫ ( n − 1) x n − 2
dx ⎥
m
m⎣
m
m
⎦
=−
[ Aggain integrating by parts]
x n cos mx n n −1
n( n − 1) n − 2
+ 2 x sin mx −
x sin mxdx
m
m
m2 ∫
x n cos mx n n −1
n( n − 1)
+ 2 x sin mx −
In = −
I n−2
m
m
m2
=−
⇒
(c)
∫x
n
cos mx dx
Solution.
I n = ∫ x n cos mx dx
Let
Take u = x n , v = cos mx . Integrating by parts, we get
sin mx
sin mx
− ∫ nx n −1
dx
m
m
x n sin mx n
=
− ∫ x n −1 sin mxdx
m
m
n
x sin m x n ⎡ n −1 ( − cos mx )
( − cos mx ) ⎤
dx ⎥
=−
− ⎢x
− ∫ ( n − 1)x n − 2
m
m⎣
m
m
⎦
In = x n ⋅
x n sin mx
n
n ( n − 1) n − 2
x cos mxdx
+ 2 x n −1 cos mx −
m
m
m2 ∫
n ( n − 1)
x n sin mx
n
I n−2
In = −
+ 2 x n −1 cos m x −
m
m2
m
=−
⇒
M06_ENGINEERING_MATHEMATICS-I _CH06_PART A.indd 50
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Integral Calculus ■
6.51
6.4.6 The Reduction Formula for (a) ∫ e ax sin m x dx and (b) ∫ e ax cos n x dx
(a)
∫e
ax
sin n x dx
Solution.
Let I n = ∫ e ax sin n xdx = ∫ sin n xe ax dx
Take u = sin n x, v = e ax . Integrating by parts, we get
e ax
e ax
− ∫ n sin n −1x cos x
dx
a
a
e ax sin n x n
=
− ∫ sin n −1x cos xe ax dx
a
a
ax
n
e ax ⎫
e sin x n ⎧ n −1
e ax
dx ⎬
=
− ⎨sin x cos x
− ∫ ⎡⎣sin n −1x ( − sin x ) + cos x ( n − 1) sin n − 2 x cos x ⎤⎦
a
a
a⎩
a
⎭
ax
n
e sin x n ax
n
=
− 2 e sin n −1x cos x + 2 ∫ ⎡⎣ − sin n x + ( n − 1) sin n − 2 x cos 2 x ⎤⎦ e ax dx
a
a
a
e ax sin n −1x
n
a sin x − n cos x ] + 2 ∫ ⎡⎣ − sin n x + ( n − 1) sin n − 2 x (1 − sin 2 x ) ⎤⎦ e ax dx
=
[
2
a
a
e ax sin n −1x
n
n ( n − 1) ax
=
a sin x − n cos x ] + 2 ∫ ( −1 − n + 1) sin n xe ax dx +
e sin n − 2 x dx
[
2
a
a
a2 ∫
I n = sin n x
e ax sin n −1x
n 2 ax
n ( n − 1)
a
sin
x
n
cos
x
e sin n x dx +
I n−2
−
−
[
]
2
2 ∫
a
a
a2
e ax
n
n ( n − 1)
I n−2
= 2 sin n −1x [a sin x − n cos x ] − 2 I n +
a
a
a2
=
⎛ n2 ⎞
e ax sin n −1 x
n( n − 1)
I
a sin x − n cos x ] +
I n−2
1
+
=
[
n
2
⎜⎝ a 2 ⎟⎠
a
a2
⇒
⎛ n2 + a 2 ⎞
e ax sin n −1 x
n( n − 1)
I
=
a sin x − n cos x ] +
I n−2
[
n
2
2
⎜⎝ a ⎟⎠
a
a2
⇒
⇒
(b)
∫e
ax
In =
e ax sin n −1 x
n( n − 1)
[ a sin x − n cos x ] + ( n2 + a2 ) I n − 2
n2 + a 2
In =
e ax sin n −1 x
n( n − 1)
[ a sin x − n cos x ] + (n2 + a2 ) I n − 2
n2 + a 2
cos n xdx
Solution.
Let I n = ∫ e ax cos n xdx = ∫ cos n xe ax dx
M06_ENGINEERING_MATHEMATICS-I _CH06_PART A.indd 51
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Engineering Mathematics
Take u = cos n x , v = e ax. Integrating by parts, we get
e ax
e ax
− ∫ n cos n −1 x( − sin x ) x
dx
a
a
e ax cos n x n
=
+ ∫ cos n −1 x sin xe ax dx
a
a
ax
n
e ax ⎫
e cos x n ⎧
e ax
=
+ ⎨∫ cos n −1 x sin x ⋅
− ∫ ⎡⎣cos n −1 x cos x + ( n − 1) cos n − 2 x( − sin x ) sin x ⎤⎦
dx ⎬
a
a
a⎩
a
⎭
ax
n
e cos x n ax
n
=
+ 2 e cos n −1 x sin x − 2 ∫ ⎡⎣cos n x − ( n − 1) cos n − 2 x sin 2 x ⎤⎦ e ax dx
a
a
a
e ax cos n −1 x
n
=
[ a cos x + n sin x ] − a2 ∫ ⎡⎣cos n x − (n − 1) cos n − 2 x (1 − cos 2 x)⎤⎦ e ax dx
a2
e ax cos n −1 x
n
a cos x + n sin x ] − 2 ∫ ⎡⎣(1 + n − 1) cos n x − ( n − 1) cos n − 2 x ⎤⎦ e ax dx
=
[
2
a
a
e ax cos n −1 x
n2 ax
n( n − 1) ax
a
x
n
x
e cos n xdx +
e cos n − 2 xdx
cos
sin
=
+
−
[
]
2
2 ∫
a
a
a2 ∫
I n = cos n x
⇒
In =
e ax cos n −1 x
n2
n( n − 1)
cos
sin
a
x
+
n
x
−
I +
I n−2
[
]
2
2 n
a
a
a2
⇒
⎛ n2 ⎞
e ax cos n −1 x
n( n − 1)
[ a cos x + n sin x ] + a2 I n − 2
⎜⎝1 + a 2 ⎟⎠ I n =
a2
⇒
⎛ n2 + a 2 ⎞
e ax cos n −1 x
n( n − 1)
I
=
a cos x + n sin x ] +
I n−2
[
2
⎜⎝ a 2 ⎟⎠ n
a
a2
[
In =
e ax cos n −1 x
n( n − 1)
a cos x + n sin x ] + 2
I n−2
[
2
2
(n + a2 )
n +a
[
ax
n
∫ e cos xdx =
e ax cos n −1 x
n( n − 1)
a cos x + n sin x ] + 2
I n−2
[
2
2
(n + a2 )
n +a
6.4.7 The Reduction Formula for (a) ∫ cos m x sinn x dx and (b) ∫ cos m x cos nx dx
Deduce if f ( m, n) =
p/2
∫ cos
m
x cos nx dx, then prove that f ( m, n) =
0
m
f ( m − 1, n − 1) and hence
m +1
p
prove that f ( n, n) = n +1 where m, n are non-negative integers.
2
(a)
∫ cos
m
x sin nx dx
Solution.
I m , n = ∫ cos m x sin nxdx
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Integral Calculus ■
6.53
Take u = cos m x,
v = sin nx
Integrating by parts, we get
⎛ cos nx ⎞
⎛ cos nx ⎞
m −1
I m , n = cos m x ⎜ −
⎟⎠ − ∫ m cos x( − sin x ) ⎜⎝ −
⎟ dx
⎝
n
n ⎠
=−
Now
sin( n − 1) x = sin( nx − x ) = sin nx cos x − cos nx sin x
⇒
cos nx sin x = sin nx cos x − sin( n −1) x
cos m x cos nx m
− ∫ cos m −1 x [sin nx cos x − sin( n − 1) x ] dx
n
n
m
m
cos x cos nx m
=−
− ∫ cos m x sin nxdx + ∫ cos m −1 x sin( n − 1) xdx
n
n
n
[
I m,n = −
⇒
I m,n = −
⇒
[
[
∫ cos
cos m x cos nx m
m
− I m , n + I m −1, n −1
n
n
n
cos m x cos nx m
⎛ m⎞
I
1
+
=
−
+ I m −1, n −1
⎜⎝
⎟ m,n
n⎠
n
n
⇒
(b)
cos m x cos nx m
− ∫ cos m −1 x cos nx sin xdx
n
n
m+n
cos m x cos nx m
I m,n = −
+ I m −1, n −1
n
n
n
cos m x cos nx
m
+
I m,n = −
I m −1, n −1
m+n
m+n
cos m x cos nx
m
m
cos
x
sin
nxdx
=
−
+
I m −1, n −1
∫
m+n
m+n
m
x cos nxdx
Solution.
Let I m , n = ∫ cos m x cos nxdx
m
Take u = cos x ,
v = cos nx .
I m,n =
=
Integrating by parts, we get
cos m x sin nx
⎛ sin nx ⎞
− ∫ m cos m −1 x( − sin x ) ⎜
dx
⎝ n ⎟⎠
n
cos m x sin nx m
+ ∫ cos m −1 x sin nx sin xdx.
n
n
cos( n − 1) x = cos( nx − x ) = cos nx cos x + sin nx sin x
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Engineering Mathematics
[
[
⇒
sin nx sin x = cos( n − 1) x − cos nx cos x
cos m x sin nx m
+ ∫ cos m −1 x [ cos( n − 1) x − cos nx ]dx
I m,n =
n
n
m
cos m x sin nx m
=
+ ∫ cos m −1 x cos( n − 1) xdx − ∫ cos m x cos nxdx
n
n
n
cos m x sin nx m
m
+ I m −1, n −1 − I m , n
n
n
n
m
cos x sin nx m
=
+ I m −1, n −1
n
n
I m,n =
⇒
⎛ m⎞
⎜⎝1 + ⎟⎠ I m , n
n
⇒
cos m x sin nx m
⎛ m + n⎞
I
=
+ I m −1, n −1
⎜⎝
⎟ m,n
n ⎠
n
n
[
Deduction:
Given
I m,n =
f ( m, n) =
cos m x sin nx
m
+
I m −1, n −1
m+n
m+n
p/2
∫ cos
m
0
[
If m = n, then
⇒
x cos nxdx
p/2
⎡ cos m x sin nx ⎤
m
f ( m − 1, n − 1)
=⎢
⎥ +
m+n
m+n
⎦0
⎣
m
f ( m − 1, n − 1)
= 0+
m+n
m
f ( m, n) =
f ( m − 1, n − 1)
m+n
n
f ( n, n) =
f ( n − 1, n − 1)
n+n
1
f ( n, n) = f ( n − 1, n − 1)
2
1 1
= ⋅ f ( n − 2, n − 2)
2 2
1
= 2 f ( n − 2, n − 2)
2
1 1
= 2 ⋅ f ( n − 3, n − 3)
2 2
1
= 3 f ( n − 3, n − 3)
2
1
= 4 f ( n − 4, n − 4)
2
:
1
= n f (0, 0)
2
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Integral Calculus ■
f (0, 0) =
But
p/2
∫
cos 0 x cos 0 xdx
=
p/2
0
f ( n, n) =
[
∫
dx = [ x ]0
p/2
0
=
6.55
p
2
1 p
p
⋅ =
2 n 2 2 n +1
EXERCISE 6.4
1. Evaluate the following integrals
p/2
(a)
∫
p/2
sin8 x dx
(b)
0
(c)
0
∫ sin
15
x cos3 x dx
(f)
∫ (a
2
0
0
∫
p/2
sin 7 x cos5 x dx
(d)
0
∞
p/2
(e)
∫
p/2
sin 7 x dx
0
1
3. If I n = ∫ x 2 (1 − x 3 ) n dx, prove that I n =
0
∫
0
x cos 4 x dx
x3
dx
+ x 2 )5
a
p/2
6
0
2. If I n = ∫ ( a 2 − x 2 ) n dx where n is a positive integer, then prove that I n =
4. If I n =
∫ sin
2na 2
I n −1 .
2n + 1
1
n
I n −1 . Hence, find ∫ x 2 (1 − x 3 )7 dx.
n +1
0
n
⎛ p⎞
x n cos x dx, show that I n + n( n − 1) I n − 2 = ⎜ ⎟ .
⎝ 2⎠
∞
5. Obtain the reduction formula for I n = ∫ e − x sin n x dx and show that (1 + n2 ) I n = n( n − 1) I n − 2 .
0
Hence, evaluate I4.
ANSWERS TO EXERCISE 6.4
1. (a)
3.
6.5
1
24
35p
256
(b)
5.
16
35
(c)
1
120
(d)
3p
512
(e)
1
144
(f)
1
24 a6
24
85
APPLICATION OF INTEGRAL CALCULUS
In this section, we deal with some of the important applications of Integral Calculus.
They are
1.
2.
3.
4.
Area of plane curves
Length of arc of plane curves
Volume of solids of revolution
Area of surface of revolution
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Engineering Mathematics
6.5.1 Area of Plane Curves
6.5.1 (a) Area of Plane Curves in Cartesian Coordinates
1. If f ( x ) is continuous, positive and bounded in
y
b
[a, b], then
∫ f ( x) dx
y = f(x)
geometrically represents
a
the area bounded by the curve y = f ( x ), the
x-axis and the abscissae x = a and x = b
b
b
a
a
∴ area A = ∫ ydx = ∫ f ( x )dx
2. If y = f ( x ) crosses the x-axis (as in Fig 6.2)
at x = c in [a, b], then the area is given by
c
A = ∫ f ( x )dx +
a
x=a
O
x=b
x
Fig. 6.1
y
y = f(x)
b
∫ f ( x)dx ,
c
x=c
O x=a
x=b
x
b
Since
∫ f ( x)dx < 0,
for area, we take the
Fig. 6.2
c
absolute value.
y
y=d
3. If the area is bounded by the curve x = g ( y ), the
y-axis and the ordinates y = c, y = d , then the
area
d
d
c
c
A = ∫ xdy = ∫ g ( y )dy
x = g(y)
y=c
x
O
Fig. 6.3
4. Area bounded between two curves
If f ( x ) ≤ g ( x ) ∀ x ∈[a, b], then the area
bounded between the curves y = f ( x ) and
y = g ( x ) in [a, b] is
y
y = g(x)
b
A = ∫ [ g ( x ) − f ( x ) ] dx
a
Oy=a
x=b x
y = f(x)
Fig. 6.4
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Integral Calculus ■
6.57
WORKED EXAMPLES
EXAMPLE 1
x2 y2
1 51.
a 2 b2
Solution.
x2 y2
The given curve is the ellipse 2 + 2 = 1
a
b
Find the area of the ellipse
(1)
b
area A = ∫ ydx
∴
a
y
The area in the four quadrants are equal, because
the ellipse is symmetric w.r.to both the axis.
B
A′
x ′(−a, 0)
2
2
2
2
Equation (1) ⇒ y = 1 − x = a − x
2
2
2
b
a
a
When y = 0,
∴
B′
b 2
y=±
a − x2
a
⇒
x2
=1 ⇒
a2
A
(a, 0) x
O
y′
x 2 = a2
⇒
Fig. 6.5
x = ±a
area of the ellipse A = 4 × Area in the first quadrant
a
a
b 2
a − x 2 dx
a
0
= 4∫ ydx = 4∫
0
a
b ⎡ x a2 − x 2 a2
x⎤
=4 ⎢
+ sin −1 ⎥
a ⎢⎣
2
2
a ⎥⎦
0
a
⎤
b⎡
a2
p
sin −1 1 − sin −1 0 ⎥ = 2ab ⋅ = pab
= 4 ⎢0 +
a⎣
2
2
⎦0
(
)
EXAMPLE 2
Find the area of the curve x 2/ 3 + y 2/ 3 = a 2/ 3 .
Solution.
The given curve is x 2 / 3 + y 2 / 3 = a 2 / 3 .
(1)
b
∴
Area A = ∫ ydx
a
The curve is symmetric w.r.to both the axes. ∴ the area in the four quadrants are equal.
Equation (1) ⇒
y 2 / 3 = a 2 / 3 − x 2 / 3 ⇒ y = ⎡⎣ a 2 / 3 − x 2 / 3 ⎤⎦
3/ 2
When y = 0, x 2 / 3 = a 2 / 3 ⇒ x 2 = a 2 ⇒ x = ± a
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Engineering Mathematics
∴
area A = 4 × Area in the first quadrant
a
a
0
0
y
(0, a)
= 4∫ ydx = 4∫ ( a 2 / 3 − x 2 / 3 )3/ 2 dx
Put x = a sin u
∴ dx = 3a sin 2 u cos ud u
3
x ′ (−a, 0)
When x = 0, sin u = 0 ⇒ u = 0 and
p
2
When x = a, sin u = 1 ⇒ u =
x
y′
3/ 2
p/2
(a, 0)
(0, −a)
∴ area A = 4 ∫ ⎡⎣ a 2 / 3 − a 2 / 3 sin 2 u⎤⎦ 3a sin 2 u cos ud u
Fig. 6.6
0
= 12a 2
O
p/2
∫ (1 − sin
2
u)3/ 2 sin 2 u cos ud u
0
= 12a 2
p/2
∫ cos u sin
3
2
u cos ud u
2
ud u
0
= 12a 2
p/2
∫ cos u sin
4
0
= 12a 2
=
2 −1
4+2
p/2
[ Using reduction formula ]
∫ cos udu
4
0
12a 2 3 1 p 3 2
⋅ ⋅ ⋅ = pa
6 4 2 2 8
[ { n = 4 is even ]
EXAMPLE 3
Show that the area of the loop of the curve ay 2 5 x 2 ( a2x ) is
8a 2
.
15
Solution.
The given curve is ay 2 = x 2 ( a − x )
To find the loop of the curve, first trace the curve.
Since the equation is of even degree in y, it is symmetric about the x-axis.
To find the intersection with the x-axis, put y = 0 in (1)
[
x 2 (a − x) = 0 ⇒
(1)
x = 0, 0, a.
If x > a, y 2 is negative ⇒ y is imaginary. So, the curve does not exit beyond x = a.
Tangents at the origin is obtained by equating the lowest degree terms to zero.
[
ay 2 − ax 2 = 0 ⇒
y2 = x2
⇒
y = ±x
∴ the loop of the curve is a shown in Fig 6.7.
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Integral Calculus ■
6.59
Let A be the area of the loop of the curve.
A = 2 × area of the loop above the x-axis
[
a
a
= 2∫ ydx = 2∫
0
=
2
x a−x
a
0
y
dx
y=x
a
∫x
a
a − xdx
0
Put
t = a − x ∴ − dx = dt ⇒ dx = − dt
When x = 0, t = a and when
x = a, t = 0
[
A=
2
a
0
∫ (a − t ) t ( −dt ) = −
2
2
a
a
a
0
∫ (a − t )
x
y = −x
tdt
a
∫ (at
a
=
(a, 0)
O
1/ 2
− t 3/ 2 ) dt
Fig. 6.7
0
a
=
2 ⎡ at 3/ 2 t 5 / 2 ⎤
−
⎢
⎥
a ⎣ 3/ 2 5 / 2 ⎦ 0
=
4 ⎡ a3 / 2 a5 / 2 ⎤
4 ⎡ a5 / 2 a5 / 2 ⎤
4
a5 / 2 8a 2
⋅
−
=
−
=
⋅
2
⋅
=
a
⎢
⎥
⎢
⎥
15
3
5 ⎦
5 ⎦
15
a⎣
a⎣ 3
a
EXAMPLE 4
Find the area bounded by the curve y 2 ( 2a 2 x ) 5 x 3 and its asymptote.
Solution.
x3
( 2a − x )
The equation is even degree in y. So, the curve is symmetric about the x-axis.
To find the point of intersection with the x-axis, put y = 0 in (1)
The given curve is y 2 ( 2a − x ) = x 3
x3 = 0 ⇒
∴
⇒
y2 =
(1)
x = 0.
When x = 2a, y2 is infinite
∴
x = 2a is an asymptote.
Tangent at the origin is y = 0, the x-axis.
The curve will be as shown in the figure.
Let A = area bounded by the asymptote
∴ A = 2 × area above the x-axis
2a
2a
0
0
= 2 ∫ y dx = 2 ∫ x
x
dx
2a − x
M06_ENGINEERING_MATHEMATICS-I _CH06_PART A.indd 59
y
(2a, 0)
x′
O
x
x = 2a
y′
Fig. 6.8
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■
6.60
Put
Engineering Mathematics
x = 2a sin 2 u
[ dx = 4 a sin u cos u d u
When x = 0, sin u = 0 ⇒ u = 0 and when x = 2a, sin 2 u = 1 ⇒ u =
p
2
A = 2 ∫ 2a sin 2 u
[
0
p
2
p
2
2a sin 2 u
4 a sin u cos u d u
2a − 2a sin 2 u
= 16 a 2 ∫ sin 3 u ?
sin u cos u
0
1 − sin 2 u
p
2
du
p
2
cos u
3 1 p
= 16 a ∫ sin u ⋅
d u = 16 a 2 ∫ sin 4 u d u = 16 a 2 ⋅ ⋅ ⋅ = 3pa 2 .
cos
u
4
2 2
0
0
2
4
EXAMPLE 5
Compute the area bounded by the curve y 5 x 4 22 x 31x 2 13 , the x 5 axis and the ordinates
corresponding to the points of minimum of the function.
Solution.
Given y = x 4 − 2 x 3 + x 2 + 3
For maximum or minimum
4 x3 − 6 x 2 + 2 x = 0
⇒
2 x[2 x 2 − 3 x + 1] = 0
When x = 0,
dy
= 4 x3 − 6 x 2 + 2 x
dx
⇒
2 x( 2 x − 1)( x − 1) = 0
dy
=0
dx
⇒
Now
[
d2 y
= 12 x 2 − 12 x + 2
dx 2
d2 y
= 2 > 0.
dx 2
⇒
1
x = 0, , 1
2
[ y is minimum.
2
When x =
1 d2 y
1
⎛ 1⎞
, 2 = 12 ? ⎜ ⎟ − 12 ? + 2 = 3 − 6 + 2 = −1 < 0
⎝ 2⎠
2 dx
2
When x = 1,
d2 y
= 12 ? 1 − 12 ? 1 + 2 = 2 > 0
dx 2
[ y is maximun.
[ y is minimum
∴ the minimum points correspond to x = 0 and x = 1 and the curve is above the x-axis in this interval.
1
1
0
0
∴ required area is A = ∫ ydx = ∫ ( x 4 − 2 x 3 + x 2 + 3) dx
1
⎤
⎡ x5
6 − 15 + 10 + 90 91
x 4 x3
1 1 1
= ⎢ − 2 + + 3x ⎥ = − + + 3 − 0 =
=
5
2
3
5
4
3
30
30
⎦0
⎣
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Integral Calculus ■
6.61
EXAMPLE 6
The gradient of a curve at any point is x 2 24 x13 and the curve passes through ( 3, 1) . Find the
area enclosed by this curve, the x-axis and the maximum and minimum ordinates.
Solution.
Let y = f ( x ) be the equation the curve.
Given the slope of the curve at any point is x 2 − 4 x + 3.
We know that the slope of the curve at any point is the same as the slope of the tangent at that point.
dy
At any point( x, y ), the slope of the tangent is .
dx
dy
= x 2 − 4x + 3
dx
[
Integrating w.r.to x,
y = ∫ ( x 2 − 4 x + 3) dx
⇒
y=
x3
x2
− 4 + 3x + c
3
2
⇒
y=
x3
− 2 x 2 + 3x + c
3
It passes through the point (3, 1)
[
1=
33
− 2 ? 32 + 3 ? 3 + c = 9 − 18 + 9 + c ⇒ c = 1
3
∴ the equation of the curve is
x3
− 2 x 2 + 3x + 1
3
For maximum or minimum,
y=
dy
=0
dx
⇒ x2 − 4x + 3 = 0
and
d2 y
= 2x − 4
dx 2
When x =1,
d2 y
= 2 ? 1 − 4 = −2 < 0
dx 2
(1)
⇒
( x − 3)( x − 1) = 0 ⇒ x = 1, 3
∴ y is maximum at x = 1.
d2 y
= 6−4 = 2> 0
∴ y is minimum at x = 3
dx 2
∴ area bounded by the curve (1) , the x-axis and the maximum and the minimum ordinates is
When x =3,
3
3
3
⎤
⎛ x3
⎞
⎡1 x4
x3
x2
A = ∫ y dx = ∫ ⎜ − 2 x 2 + 3 x + 1⎟ dx = ⎢ ?
+3?
+ x⎥
−2?
3
3
4
3
2
⎠
⎦1
⎣
1
1⎝
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6.62
Engineering Mathematics
1 4 4 2 3 3 3 2 2
(3 − 1 ) − (3 − 1 ) + (3 − 1 ) + (3 − 1)
12
3
2
1
2
3
= × 80 − × 26 + × 8 + 2
12
3
2
20 52
20 − 52 + 42 10
− + 14 =
=
=
3
3
3
3
=
EXAMPLE 7
Find the area of the propeller shaded region enclosed by the curves x 2 y1 3 50 and x 2 y1 5 50.
Solution.
The given curves are
y
x − y1 3 = 0 ⇒ x 3 = y
(1)
x− y
( 2)
15
and
=0⇒ x = y
5
(1, 1)
To find the points of intersection solve (1) and ( 2)
(−1, 0)
∴
x 3 = x 5 ⇒ x 3 ( x 2 − 1) = 0
(1, 0) x
O
x = 0, −1, +1
⇒
When x = −1, y = −1 and when x = 1, y = 1
(−1, −1)
The curves are symmetric about the origin.
Area bounded by the curves is
A = 2[area in the I quadrant]
1
(
Fig. 6.9
)
= 2 ∫ x 3 − x 5 dx
0
⎡ x 4 x6 ⎤
⎡1 1⎤
⎡3 − 2⎤
⎛ 1⎞ 1
=2⎢ − ⎥ =2⎢ − ⎥ =2⎢
⎥ = 2 ⎜⎝ 12 ⎟⎠ = 6
4
6
4
6
12
⎣
⎦
⎣
⎦
⎦
⎣
EXAMPLE 8
Find the area between the curves y 5 x 4 1x 3 1 16 x 1 4 and y 5 x 4 1 6 x 2 1 8 x 1 4 .
Solution.
Let
f ( x ) = x 4 + x 3 + 16 x + 4
y
g( x) = x 4 + 6 x 2 + 8x + 4
Now
f ( x) − g( x) = x3 − 6 x 2 + 8x
(0, 4)
The points of intersection of the two
curves is given by f ( x ) − g ( x ) = 0
⇒
x3 − 6 x 2 + 8x = 0
O
x=2
x=4
x
Fig. 6.10
M06_ENGINEERING_MATHEMATICS-I _CH06_PART A.indd 62
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Integral Calculus ■
⇒
6.63
x( x 2 − 6 x + 8) = 0
x( x + 2)( x − 4) = 0
⇒
x = 0, 2, 4
⇒
When x = 0, y = 4.
In the interval [0, 4] , the curves intersect at x = 2.
Required area is A =
2
4
0
2
∫ ( f ( x) − g ( x))dx + ∫ ( f ( x) − g ( x))dx
2
2
0
0
∫ f ( x) − g ( x)dx = ∫ ( x
Now
3
− 6 x 2 + 8 x ) dx
2
⎡ x 4 6 x3 8x 2 ⎤
24
− 2 ? 23 + 2 ? 22 = 4 − 16 + 16 = 4
=⎢ −
+
⎥ =
4
3
4
4
⎦0
⎣
and
4
4
2
2
3
2
∫ ( f ( x) − g ( x)) dx = ∫ ( x − 6 x + 8 x) dx
4
⎡ x 4 6 x3 8x 2 ⎤
=⎢ −
+
⎥
3
4 ⎦2
⎣4
1 4
( 4 − 24 ) − 2( 43 − 23 ) + 2( 4 2 − 22 )
4
1
= ( 240) − 2(56) + 4(12) = 60 − 112 + 48 = −4
4
=
∴
Area A = 4 + −4 = 4 + 4 = 8.
EXAMPLE 9
Find the area bounded by y 5 x , x ∈[0, 1], y 5 x 2 , x ∈[1, 2] and y 52x 2 12 x14, x ∈[0, 2] .
Solution.
The given curves are y =
⇒
and
x , x ∈[0, 1]
y 2 = x, x ∈[0, 1]
(1)
y = x 2 , x ∈[1, 2]
( 2)
y = − x2 + 2x + 4
(3)
= −( x 2 − 2 x ) + 4 = −[( x − 1) 2 − 1] + 4 = −( x − 1) 2 + 5
⇒
y − 5 = −( x − 1) 2 ,
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Which is a downward parabola with vertex
(1, 5) and axis x = 1 as in Fig 6.11.
2
1
2
0
0
1
y
(1, 5)
[ area A = ∫ y( 3) dx − ∫ y(1) dx − ∫ y(2) dx
2
x2 = y
1
y = −x 2 + 2x + 4
2
= ∫ ( − x 2 + 2x + 4) dx − ∫ xdx − ∫ x 2dx
0
0
y2 = x
1
x=1 x=2
O
2
1
x
2
⎤ ⎡x ⎤ ⎡x ⎤
⎡ x
x
= ⎢− + 2 + 4 x⎥ − ⎢
⎥ −⎢ ⎥
2
⎦ 0 ⎣ 3 2 ⎦ 0 ⎣ 3 ⎦1
⎣ 3
3
2
32
3
y2 = x
23
2
1
+ 22 + 4 ⋅ 2 − 0 − [1 − 0] − [23 − 13 ]
3
3
3
8
2 7
17
−17 + 36 19
= − + 4 + 8 − − = − + 12 =
=
3
3 3
3
3
3
=−
Fig. 6.11
EXAMPLE 10
e t 1 e 2t
e t 2 e 2t
is a point on the hyperbola x 2 2 y 2 51 . Show that the
, y5
2
2
area bounded by this hyperbola and the lines joining its centre to the points corresponding to
For any real t , x 5
t1 and2 t1 is t1.
Solution.
The given equation of the hyperbola is
x2 − y2 = 1
x=
Also given
(1)
et + e − t
et − e − t
,y=
; t ∈R
2
2
( 2)
Y
⎛e +e e −e ⎞
[ ⎜
,
are the coordinates of any
2 ⎟⎠
⎝ 2
point on the rectangular hyperbola x 2 − y 2 = 1.
Centre of the hyperbola is the origin O.
t
−t
t
−t
Let P be the point on the hyperbola corresponding
to the parameter t = t1.
⎛ e t1 + e − t1 e t1 − e − t1 ⎞
,
[P=⎜
2
2 ⎟⎠
⎝
M06_ENGINEERING_MATHEMATICS-I _CH06_PART A.indd 64
P
A
O
R
t1
X
Q
Fig. 6.12
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Integral Calculus ■
6.65
Let Q be the point on the curve corresponding to t = −t1.
⎛ e − t1 + e t1 e − t1 − e t1 ⎞
,
Q=⎜
2
2 ⎟⎠
⎝
[
⎛ e t1 + e − t1
e t1 − e − t1 ⎞
,−
=⎜
2
2 ⎟⎠
⎝
[ P and Q have same x-coordinates but y-coordinates have opposite signs.
Hence, Q is the image of Pin the x-axis and so PQ is perpendicular to the x-axis
Since the points P and Q and the curve are symmetric about the x-axis, OP and OQ are symmetric
about the x-axis.
So, the area bounded by OP, OQ and the curve is = 2 (area above x-axis)
Let PQ cuts the x-axis at R.
[ the required area = 2[Area of the right angled ΔOPR − area APR ]
where area APR is the area bounded by the curve, the x-axis and the line PR.
Now
area of ΔOPR =
=
1
OR ⋅ PR
2
1 e t1 + e − t1 e t1 − e − t1 1 2 t 1
⋅
= (e − e −2t1 )
2
2
2
8
t1
the area APR = ∫ y
and
0
Since y =
dx
dt
dt
et − e − t
et + e − t
and x =
2
2
[{ t = 0 corresponds to A]
∴
dx e t − e − t
=
2
dt
[{ t = t1 corresponds to P ]
e t − e −t e t − e −t
⋅
dt
2
2
0
t1
[
area A PR = ∫
t
=
11 t
(e − e − t ) 2 dt
4 ∫0
=
1 1 2t
(e + e −2t − 2)dt
4 ∫0
t
t1
⎤
1 ⎡ e 2t e −2t
= ⎢ +
− 2t ⎥
4⎣ 2
−2
⎦0
=
∴
t
1 ⎡ e 2t1 e −2t1
⎛1 1
⎞⎤ 1
−
− 2t1 − ⎜ − − 0⎟ ⎥ = (e 2t1 − e −2t1 ) − 1
⎢
⎝ 2 2 ⎠⎦ 8
4⎣ 2
2
2
⎡1
t ⎫⎤
⎛t ⎞
⎧1
required area A = 2 ⎢ (e 2t1 − e −2t1 ) − ⎨ (e 2t1 − e −2t1 ) − 1 ⎬ ⎥ = 2 ⎜ 1 ⎟ = t1.
⎝ 2⎠
2 ⎭⎦
⎩8
⎣8
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EXERCISE 6.5
1. Find the area bounded by the curve
x + y = 1and the coordinate axes.
2. Find the area bounded by the parabola and its latus rectum.
3. Find the area bounded by the curve y = x 3 − 4 x and the x-axes.
4. Find the area of the curve y 2 = x 4 (9 − x 2 ).
5. Find the area bounded by the curve and its asymptote
x3
a2 x
(ii) y 2 =
(i) y 2 =
2− x
a−x
(iii) xy 2 = a 2 ( a − x )
6. Find the area of the loop of the curve
(i) a 2 y 2 = x 3 ( a − x )
(ii) 3ay 2 = x( x − a) 2
(iii) y 2 =
a2 (a2 − x 2 )
a2 + x 2
7. Find the area in the I quadrant bounded by y 2 = x, the x-axis and the line x − y = 2.
8. Find the area bounded by y 2 = 4 ax and x 2 = 4by.
9. Find the area bounded by the parabola y = x 2 and the line 2 x − y + 3 = 0.
10. Show that the larger of the two areas into which the circle x 2 + y 2 = 64 a 2 is divided by the parabola
16 a 2
(8p − 3 ).
y 2 = 12ax is
3
11. Find the area bounded by the parabola x = −2 y 2 , x = 1 − 3 y 2 .
12. Find the area bounded by x 2 = 4 y and y =
8
.
x2 + 4
13. Find the area of the region bounded by the parabola y = − x 2 − 2 x + 3, the tangent at the point
P( 2, −5) on the curve and the y-axis.
2
2 ⎛ a + x⎞
.
14. Find the area of the loop of the curve y = x ⎜
⎝ a − x ⎟⎠
15. Find the area of the curve y = sin x bounded by the x-axis (i) in [0, 2p] and (ii) in [−p, p].
16. Compute the area bounded by the curve by y =
x and y = x 2 .
17. Find the area bounded by the curve x 2 = 4 y and the straight line x = 4 y − 2.
18. Show that the parabola y 2 = x divides the circle x 2 + y 2 = 2 into two portions whose area are in
the ratio (9p − 2) : (3p + 2).
19. Find the area bounded by one arch of the cycloid x = a(u − sin u), y = a(1 − cos u) and its base.
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Integral Calculus ■
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ANSWERS TO EXERCISE 6.5
1.
1
6
2.
5. (i) 3p
6. (i)
7.
p 2
a
8
10
3
12. 2p −
16. 1
3
13.
31
p
4
32
3
11.
4
3
a2
(p + 4).
2
15. (i) 4, (ii) 4
(iii) pa 2
8 3 2
a
45
(ii)
4.
3. 8
(ii) pa 2
8.
4
3
8 2
a
3
(iii)
16 ab
3
9.
8
3
14.
1
(p − 2)a 2
2
17. 9
8
6.5.1 (b) Area in Polar Coordinates
Formula: The area bounded by the curve r = f (u) and the radius vectors u = a and u = b is
b
A=
1 2
r du.
2 ∫a
Proof
B
Q (r + Δr, θ + Δθ)
Given r = f (u) is the equation of the curve.
P (r, θ)
Let A and B be two points on the curve with radii
vectors u = a and u = b
β
f (u) is continuous in [a, b]
Δθ
A
α
Let P ( r , u) and Q( r + Δr , u + Δu) be
neighbouring points on the curve.
O
x
Fig. 6.13
Let ΔA be the element area of the strip OPQ.
Then
∴
The limit of
1 2
r Δu approximately.
2
1
∑ ΔA = ∑ 2 r 2 Δu
ΔA =
∑ ΔA as Δu → 0 is the area of OAB.
b
∴ area of the region OAB =
b
1 2
1 2
∫a 2 r du = 2 ∫a r du.
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WORKED EXAMPLES
EXAMPLE 1
Find the area of the cardioid r 5 a(1 1 cos u).
Solution.
The given curve is
∴
r = a(1 + cos u).
b
1
area A = ∫ r 2 d u
2a
r
The equation is unaffected if u is changed to −u,
because cos( −u) = cos u.
θ=π O
θ
θ=0
x
2θ
∴ the curve is symmetric about the initial line OX
and u varies from 0 to p.
∴
Fig. 6.14
area of the curve = 2 (area above OX)
Area A = 2 ×
p
p
1 2
r d u = ∫ a 2 (1 + cos u) d u
2 ∫0
0
p
= a 2 ∫ (1 + 2 cos u + cos 2 u) d u
0
p
1 + cos 2u ⎤
⎡
= a 2 ∫ ⎢1 + 2 cos u +
⎥ du
2
⎦
0 ⎣
p
cos 2u ⎤
⎡3
= a 2 ∫ ⎢ + 2 cos u +
du
2
2 ⎥⎦
0 ⎣
p
sin 2u ⎤
⎡3
= a 2 ⎢ u + 2 sin u +
4 ⎥⎦ 0
⎣2
2
sin 2p
⎡3
⎤ 3pa
− 0⎥ =
= a 2 ⎢ p + 2 sin p +
4
2
⎣2
⎦
EXAMPLE 2
Find the area outside the circle r 5 2a cos u and inside the cardioid r 5 a(11 cos u) .
Solution.
Given the circle r = 2a cos u
(1)
and the cardioid r = a(1 + cos u)
( 2)
O
(a, 0)
(2a, 0) x
The required area is as shown in the Fig 6.15,
since the circle lies inside the cardioid.
From (1), when u = 0, r = 2a
and when u =
Fig. 6.15
p
,r=0
2
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Integral Calculus ■
6.69
To find the point of intersection, solve (1) and ( 2)
a(1 + cos u) = 2a cos u
∴
When u =
⇒
cos u = 1 ⇒ u = 0 or 2p
p
1
, r = 2a ⋅ = a
3
2
[From (1)]
That is the circle lies inside the cardioid.
Required area A = Area of the cardioid − Area of the circle
Area of the cardioid =
3pa 2
2
[by example 1]
Area of the circle = pa 2 , since radius is a.
∴
required area =
pa 2
3pa 2
.
− pa 2 =
2
2
EXAMPLE 3
Find the area of a loop of the curve r 5 a sin 3u .
Solution.
Given the curve is r = a sin 3u
When u = 0,
r=0
y
p
p
, r = a sin = a, which is the
6
2
maximum value of r.
When u =
When u =
p
,
3
θ=
π
2
θ=
π
3
r = a sin p = 0
θ=
p
, x goes from 0 to A
6
p
p
and as u varies from
to , x comes from A
6
3
So, as u varies from 0 to
π
6
A
60°
30°
O
θ=0
x
to 0.
p
So, as u varies from 0 to , we get a loop as
3
in Fig. 6.16.
Fig. 6.16
p
13
Area of the loop = ∫ r 2 du
20
p
p
13
a2 3 ⎡1 − cos 6u ⎤
= ∫ a2 sin 2 3u d u = ∫ ⎢
⎥ du
20
2 0⎣
2
⎦
a2
=
4
p
2
sin 6u ⎤ 3 a2 ⎡ p sin 2p
⎤ pa
⎡
u
0
−
=
−
−
=
⎢
⎥ 12
6 ⎦⎥ 0
4 ⎢⎣ 3
6
⎣
⎦
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EXAMPLE 4
Show that the area between the cardioids r 5 a(11 cos u) and r 5 a(1 2 cos u) is
( 3p 2 8 ) 2
a .
2
Solution.
The given equations of the two cardioids are
r = a(1 + cos u)
(1)
r = a(1 − cos u)
( 2)
The area common to the cardioids is the two shaded regions as in Fig 6.17, which are equal in area,
because both the curves are symmetric about the initial line.
Common Area A = 2[area of the part above the line of OX ].
r = a(1 − cos θ)
r = a(1 + cos θ)
P
θ=
π
2
A′
(−2a, 0)
A
r = 2a
O
r = 2a
(2a, 0)
x
Fig. 6.17
The two cardioids interact at u =
p 3p
p 3p
,
. since a(1 − cosu) = a(1 + cosu) ⇒ cosu = 0 ⇒ u = ,
.
2 2
2 2
p
p
But area of the loop above the line OX = 2 ⋅
where r is from the cardioid ( 2).
p
2
Now
∫r
0
p
2
2
2
12 2
2
r
d
u
=
∫0 r du
2 ∫0
p
2
d u = ∫ a 2 (1 − cos u) 2 d u = a 2 ∫ (1 − 2 cos u + cos 2 u) d u
0
0
p
2
1 + cos 2u ⎞
⎛
= a 2 ∫ ⎜1 − 2 cos u +
⎟⎠ d u
⎝
2
0
=a
p
2
2
⎛3
∫ ⎜⎝ 2 − 2 cos u +
0
cos 2u ⎞
⎟ du
2 ⎠
p
sin 2u ⎞ 2
⎛3
= a ⎜ u − 2 sin u +
⎟
⎝2
4 ⎠0
2
p sin p
(3p − 8)
⎛3 p
⎞
⎛ 3p
⎞
= a 2 ⎜ ⋅ − 2 sin +
− 0⎟ = a 2 ⎜
− 2⎟ = a 2
⎝2 2
⎠
⎝ 4
⎠
2
4
4
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Integral Calculus ■
6.71
a 2 (3p − 8)
4
2
a
a2
common Area A = 2 × (3p − 8) = (3p − 8) .
4
2
∴ area of the loop above the x-axis =
∴
EXAMPLE 5
Prove that the area of the loop of the curve x 3 1 y 3 5 3axy is
3a 2
.
2
Solution.
The given curve is x 3 + y 3 = 3axy
(1)
Transform (1) to polar coordinates by putting x 5 r cos u and y 5 r sin u
∴ the equation (1) becomes r 3 cos3 u + r 3 sin 3 u = 3ar cos u sin u
⇒
⇒
r 3 (cos3 u + sin 3 u) = 3ar 2 cos u sin u
r (cos3 u + sin 3 u) = 3a cos u sin u ⇒
If r = 0, then cos u sin u = 0 ⇒
r=
3a sin u cos u
cos3 u + sin 3 u
sin 2u
= 0 ⇒ sin 2u = 0 ⇒ 2u = 0 or 2u = p
2
p
, which are the limits for u.
2
p
As u varies from 0 to , we get a loop of the curve, because r varies from 0 to 0.
2
For the figure, refer the Fig 3.32, page 3.133
⇒
u = 0 or u =
∴ area of the loop is A =
1
2
p/ 2
∫
r 2 du =
0
1
2
p/2
∫
0
9a 2
=
2
9a 2 sin 2 u cos 2 u
du
(cos3 u + sin 3 u) 2
p/2
∫
0
sin 2 u cos 2 u
9a 2
d
u
=
2
cos6 u(1 + tan 3 u) 2
p/2
∫
0
tan 2 u sec 2 u
du
(1 + tan 3 u)
1
Put t = 1 + tan u ∴ dt = 3 tan u sec ud u ⇒ tan sec u d u 5 dt .
3
p
3
3 p
When u = 0, t = 1 + tan 0 ⇒ t = 1 and when u = , t = 1 + tan
⇒ t=∞
2
2
3
∴
2
2
2
2
∞
∞
9a 2 1 dt 3a 2 ∞ −2
3a 2 ⎡ t −2 +1 ⎤
A=
t
dt
=
=
∫
2
⎥
⎢
2 1t 3
2 ∫1
2 ⎣ −2 + 1⎦1
=−
3a 2 −1 ∞
3a 2
⎡⎣t ⎤⎦ = −
1
2
2
M06_ENGINEERING_MATHEMATICS-I _CH06_PART A.indd 71
∞
3a 2 ⎡ 1 ⎤
3a 2
3a 2
⎡1 ⎤
⎢ t ⎥ = − 2 ⎢ ∞ − 1⎥ = − 2 [0 − 1] = 2
⎣ ⎦1
⎣
⎦
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EXERCISE 6.6
1. Find the area of the cardioid r = a(1 − cos u).
2. Find the area of circle r = 3 + 2sin u..
3. Find the area of the lemniscate r 2 = a 2 cos 2u.
4. Find the area common to the circles r = a 2 and r = 2a cos u.
5. Find the area of the loop of the curve r = a sin 2u.
6. Find the area of circle r = 2a cos u.
7. Show that the curve r = 3 + 2 cos u consists of a single oval and find its area.
ANSWERS TO EXERCISE 6.6
1.
3pa 2
2
6. pa 2
2. pa 2
4. a 2 (p − 1)
3. a 2
5.
pa 2
.
8
7. 11p
6.5.2 Length of the Arc of a Curve
The process of finding the length of a continuous curve is known as rectification.
A curve having arc length is said to be a rectifiable curve.
As in the case of area, we can find the arc length in Cartesian and polar coordinates.
6.5.2 (a) Length of the Arc in Cartesian Coordinates
Let y = f ( x ) be the Cartesian equation of the curve whose length is required between x = a and x = b.
y
B
Q(x + Δx, y + Δy)
P(x, y) Δs
Δy
A
Δx
F
x=a
x=b
x
Fig. 6.18
Let the arc length be measured from a fixed point F on the curve. Let the lines x = a and x = b meet
the curve at Aand Brespectively.
Let FA = s1 and FB = s2 .
Let P ( x, y ) and Q( x + Δx, y + Δy ) are neighbouring points on the curve such that FP = s and
FQ = s + Δs.
Let PQ = Δs be the element arc.
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Integral Calculus ■
The sum of such element arcs
∑ Δs gives approximately arc AB.
6.73
s2
The limit when the largest element Δs → 0, we have the length of arc AB = ∫ ds
s1
1. Since A and B on the curve correspond to x = a and x = b,
b
ds
dx
dx
a
we have arc length = ∫
ds
⎛ dy ⎞
= 1+ ⎜ ⎟
⎝ dx ⎠
dx
We know ( ds) 2 = ( dx ) 2 + ( dy ) 2 ⇒
2
2
⎛ dy ⎞
s = ∫ 1 + ⎜ ⎟ dx
⎝ dx ⎠
a
b
∴
2. If the points A and B on the curve corresponding to y = c and y = d , then the arc length
2
d
⎛ dx ⎞
ds
= ∫ dy = ∫ 1 + ⎜ ⎟ dy
dy
⎝ dy ⎠
c
c
d
3. Parametric form
If x = f (t ) and y = g (t ) be the parametric equations of the given curve y = f ( x ) and the limits of
t are t1 and t 2 , then arc length
t2
t
2
2
2
ds
⎛ dx ⎞ ⎛ dy ⎞
dt = ∫ ⎜ ⎟ + ⎜ ⎟ dt
⎝ dt ⎠ ⎝ dt ⎠
dt
t1
t1
=∫
WORKED EXAMPLES
EXAMPLE 1
Find the length of arc of the curve x 3 = y 2 from x = 0 to x = 1
Solution:
Given
x3 = y 2
a = 0, b = 1
and
(1)
2
ds
⎛ dy ⎞
dx = ∫ 1 + ⎜ ⎟ dx
⎝ dx ⎠
dx
0
a
b
Length of arc s = ∫
1
Differentiating (1) w.r.to x, we get
3x 2 = 2 y
dy
dx
⇒
dy 3 x 2
=
dx 2 y
2
∴
9x 4
9x 4
9x
⎛ dy ⎞
1+ ⎜ ⎟ = 1+ 2 = 1+ 3 = 1+
⎝ dx ⎠
4
4y
4x
∴
9x
⎛ dy ⎞
1+ ⎜ ⎟ = 1+
⎝ dx ⎠
4
2
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Engineering Mathematics
∴ s = ∫ 1+
0
1/ 2
9x
⎛ 9x ⎞
dx = ∫ ⎜1 + ⎟ dx
⎝
4
4⎠
0
1
1
⎡ ⎛ 9 ⎞ 3/ 2 ⎤
⎢ ⎜⎝1 + x ⎟⎠ ⎥
4
⎥
=⎢
⎢ 9 3 ⎥
⎢⎣ 4 × 2 ⎥⎦ 0
8 ⎡⎛ 9 ⎞
=
⎢⎜1 + ⎟
27 ⎣⎝ 4 ⎠
3/ 2
⎡
⎤
( ax + b) n +1
n
if n ≠ −1⎥
{
(
ax
+
b
)
=
⎢ ∫
a( n + 1)
⎣
⎦
⎤ 8 ⎡13 13 ⎤ 1
− 1⎥ =
− 1⎥ =
[13 13 − 8]
⎢
⎦ 27
⎦ 27 ⎣ 8
EXAMPLE 2
Find the length of one loop of the curve 3ay 2 5 x ( x 2 a )2 .
Solution.
Given
3ay 2 = x( x − a) 2
It is even degree in y and so symmetric about the
x-axis.
(1)
y
When y = 0, x ( x − a) 2 = 0 ⇒ x = 0, a, a
A (a, 0)
That is the curve meets the x-axis at x = 0 and
x = a two times
So, we get a loop between x = 0 and x = a as in
Fig 6.19.
Let A be the point (a, 0) on the x-axis
a
OA = ∫
Length of the arc
0
∴
x
O
x=a
Fig. 6.19
ds
dx
dx
length of the loop = 2 × the length of arc OA
2
ds
⎛ dy ⎞
dx = 2∫ 1 + ⎜ ⎟ dx
⎝ dx ⎠
dx
0
0
a
a
= 2∫
Differentiating (1) w.r.to x, we get
6ay
⇒
dy
= x ⋅ 2( x − a) + ( x − a) 2 ⋅1
dx
= ( x − a) + ( 2x + x − a) = ( x − a)(3x − a)
dy ( x − a)(3 x − a)
=
dx
6 ay
( x − a ) 2 (3 x − a ) 2 ( x − a ) 2 (3 x − a ) 2 (3 x − a ) 2
⎛ dy ⎞
=
=
⎜⎝ ⎟⎠ =
dx
12ax
36 a 2 y 2
12ax( x − a) 2
2
∴
(3 x − a )
12ax + (3 x − a)
⎛ dy ⎞
=
1+ ⎜ ⎟ = 1+
⎝ dx ⎠
12ax
12ax
2
∴
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Integral Calculus ■
=
12ax + 9 x 2 − 6 ax + a 2 9 x 2 + 6 ax + a 2 (3 x + a) 2
=
=
12ax
12ax
12ax
(3x + a) 2
3x + a
=
12ax
2 3a x
2
⎛ dy ⎞
1+ ⎜ ⎟ =
⎝ dx ⎠
∴
3x + a
a
∴
6.75
length of the loop = 2∫
0
2 3a x
dx
a ⎤
⎡
∫ ⎢3 x + x ⎥⎦ dx
3a 0 ⎣
1
a
=
1
a
=
3a
∫ ⎡⎣3x
1/ 2
+ ax −1/ 2 ⎤⎦ dx
0
a
x 1/ 2 ⎤
1 ⎡ x 3/ 2
+ a⋅
=
⎢3 ⋅
⎥
1/ 2 ⎦ 0
3a ⎣ 3/ 2
=
1
3a
⎡⎣ 2a3 / 2 + 2a ⋅ a1/ 2 − 0 ⎤⎦ =
1
3a
⎡⎣ 2a3 / 2 + 2a3 / 2 ⎤⎦ =
4a ⋅ a1/ 2
3a
=
4a
3
EXAMPLE 3
Find the length of the curve x 2 / 31 y 2 / 3 5a 2 / 3 .
Solution.
The given curve is x 2 / 3 + y 2 / 3 = a 2 / 3
(1)
It is symmetric w.r.to both the axes
∴ the length of the arc is the same in all four
quardrants as in Fig 6.20.
When y = 0, x 2 / 3 = a 2 / 3 ⇒ x 2 = a 2 ⇒ x ± a
When x = 0, y
2/3
=a
2/3
⇒y =a
2
2
⇒ y = ±a
∴ length of the arc AB = length of the arc BC = length
of the arc CD = length of the arc DA
∴ length of the curve = 4 × length of the arc AB
a
= 4×∫
0
ds
dx
dx
y
B
(0, a)
C
x ′ (−a, 0)
A
O
(0, −a)
(a, 0) x
D
y′
Fig. 6.20
Differentiating (1) w.r.to x, we get
2 −31 2 −31 dy
x + y
=0
3
3
dx
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y −1/ 3
⇒
dy
= − x −1/ 3 ⇒
dx
dy
x −1/ 3
y +1/ 3
= − −1/ 3 = − 1/ 3
dx
y
x
2
y 2/3
⎛ dy ⎞
=
⎜⎝ ⎟⎠
dx
x 2/3
∴
∴
y 2/3 x 2/3 + y 2/3
a2/3
⎛ dy ⎞
1 + ⎜ ⎟ = 1 + 2/3 =
=
⎝ dx ⎠
x 2/3
x
x 2/3
∴
⎛ dy ⎞
1+ ⎜ ⎟ =
⎝ dx ⎠
∴
length of the curve is s = 4 ∫ a1/ 3 x −1/ 3dx
2
2
[from (1)]
a2 / 3 a1/ 3
= 1/ 3 = a1/ 3 x −1/ 3
2/3
x
x
a
0
a
⎡
⎤
2/3 a
−1/ 3 +1
⎢
⎥
⎤
x
1/ 3
1/ 3 ⎡ x
1/ 3
2/3
= 4a ⎢
⎥ = 4a ⎢
⎥ = 6a (aa − 0) = 6a
1
⎣ 2/3 ⎦ 0
⎢ − + 1⎥
⎣ 3 ⎦0
EXAMPLE 4
Find the length of the curve x 2 ( a 2 2 x 2 ) 5 8a 2 y 2.
Solution.
Given curve is x 2 ( a 2 − x 2 ) = 8a 2 y 2
(1)
The equation of the curve is of even degree in x and y and so the curve is symmetric w.r.to both the
axes.
y
If y = 0, then x 2 (a2 − x 2 ) = 0 ⇒ x = 0, 0 or x = −a, a
That is it meets the x-axis at the arigin x = 0 twice,
x = −a and x = a.
B
A
If x = 0, y = 0 and if x = ±a, y = 0
(−a, 0)
(a, 0) x
∴ the curve passes through the origin and meets the
O
x-axis at the points A( a, 0) and B( − a, 0) .
∴we get two loops of the curve as in Fig 6.21.
∴ total length of the curve is
s = 4 × length of the arc OA
Fig. 6.21
2
ds
⎛ dy ⎞
4∫ dx = 4 × ∫ ⎜1 + ⎟ dx
⎝ dx ⎠
dx
0
0
a
a
Differentiating (1) w.r.to x, we get
8a2 2 y
dy
= x 2 ( −2x ) + (a2 − x 2 )2x
dx
= −2 x 3 + 2a 2 x − 2 x 3 = −4 x 3 + 2a 2 x = 2 x[a 2 − 2 x 2 ]
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Integral Calculus ■
∴
dy x[a 2 − 2 x 2 ]
=
dx
8a 2 y
∴
[ x( a 2 − 2 x 2 )]2
⎛ dy ⎞
=
⎜⎝ ⎟⎠
dx
(8a 2 y ) 2
6.77
2
=
(a 2 − 2 x 2 ) 2
x 2 (a 2 − 2 x 2 ) 2
x 2 (a 2 − 2 x 2 ) 2
=
=
8a2 .8a2 y 2
8a2 ⋅ x 2 (a2 − x 2 ) 8a2 (a2 − x 2 )
[from (1)]
(a 2 − 2 x 2 ) 2
⎛ dy ⎞
1+ ⎜ ⎟ = 1+ 2 2
⎝ dx ⎠
8a (a − x 2 )
2
∴
=
8a2 (a2 − x 2 ) + (a2 − 2x 2 ) 2
8a2 (a2 − x 2 )
=
8a 4 − 8a 2 x 2 + a 4 − 4 a 2 x 2 + 4 x 4
8a 2 ( a 2 − x 2 )
=
9a 4 − 12a 2 x 2 + 4 x 4 (3a 2 − 2 x 2 ) 2
= 2 2
8a 2 ( a 2 − x 2 )
8a ( a − x 2 )
2
∴
⎛ dy ⎞
1+ ⎜ ⎟ =
⎝ dx ⎠
(3a2 − 2x 2 ) 2
3a2 − 2x 2
=
2
2
2
8a (a − x ) 2a 2 a2 − x 2
3a 2 − 2 x 2
a
∴ Length of the curve s = 4
∫
0
a
= 4∫
0
=
2a 2 a 2 − x 2
2(a2 − x 2 ) + a2
2a 2 a2 − x 2
dx
dx
a
a
⎤
a2
2⎡
2
2
a
x
dx
dx ⎥
2
−
+
⎢∫
∫
a ⎣0
a2 − x 2 ⎦
0
a
⎧
a⎫
⎫⎤
2 ⎪ ⎡ ⎧⎪ x a2 − x 2 a2
−1 x ⎪
−1 x ⎤ ⎪
2 ⎡
⎢
⎥
+
a
=
2
sin
sin
+
⎨ ⎨
⎬
⎬
⎢
a ⎪ ⎢ ⎩⎪
a ⎪⎭ ⎥
a ⎥⎦ 0 ⎪
2
2
⎣
⎦0
⎩⎣
⎭
=
2
⎡0 + a 2 (sin −1 1 − sin −1 0) + a 2 (sin −1 1 − sin −1 0) ⎤⎦
a ⎣
=
2⎡ 2 p
2 2
p⎤
.a p = pa 2
a . + a2 . ⎥ =
a
a ⎢⎣ 2
2⎦
EXAMPLE 5
Find the perimeter of the loop of the curve x 5 t 2 and y 5 t −
t3
.
3
Solution.
Given
x = t 2 and y = t −
t3
3
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2
⎡ t3 ⎤
⎛ t2 ⎞
⎛ x⎞
y 2 = ⎢t − ⎥ = t 2 ⎜ 1 − ⎟ = x ⎜ 1 − ⎟
⎝ 3⎠
3⎠
⎝
⎣ 3⎦
2
2
⎛ x⎞
⎛ x⎞
y = 0, x ⎜1 − ⎟ = 0 ⇒ x = 0 or ⎜⎝1 − ⎟⎠ = 0 ⇒
3
⎝ 3⎠
2
⇒
When
2
∴ the curve meets the x-axis at the origin
and at the point (3, 0), twice.
∴ the loop of the curve is as shown in the
Fig 6.22.
Let A be the point (3, 0)
When x = 0, t = 0 and when x = 3, t = 3
Length of the loop = 2 × arc length of OA.
Since the equation of the curve is in
parametric form, the length of the loop is
t2
s = 2∫
t1
x = 0, x = 3, 3
y
A
O
x
(3, 0)
Fig. 6.22
ds
dt
dt
where t1 = 0 and t 2 = 3.
2
2
⎛ dx ⎞ ⎛ dy ⎞
s = 2 ∫ ⎜ ⎟ + ⎜ ⎟ dt
⎝ dt ⎠ ⎝ dt ⎠
0
3
∴
dx
= 2t and
dt
⇒
dy
3t 2
= 1−
= 1− t2
dt
3
x = t2 ⇒
∴
⎛ dx ⎞ ⎛ dy ⎞
2
2 2
2
2
4
4
2
2 2
⎜⎝ ⎟⎠ + ⎜⎝ ⎟⎠ = 4t + (1 − t ) = 4t + 1 − 2t + t = t + 2t + 1 = (1 + t )
dt
dt
∴
⎛ dx ⎞ ⎛ dy ⎞
2 2
2
⎜⎝ ⎟⎠ + ⎜⎝ ⎟⎠ = (1 + t ) = 1 + t
dt
dt
2
2
2
2
y =t −
t3
3
We have
3
∴
s = 2 ∫ (1 + t 2 )dt
0
⎡
⎡ t3 ⎤
3 3⎤
= 2 ⎢t + ⎥ = 2 ⎢ 3 +
⎥ = 2 ⎡⎣ 3 + 3 ⎤⎦ = 4 3
3 ⎦
⎣ 3 ⎦0
⎣
3
EXERCISE 6.7
1. Find the length of the following curves
⎛ 10 5 ⎞
⎛2 ⎞
, 2⎟ .
(i) 9x 2 = 4(1 + y 2 )3 from the point ⎜ , 0⎟ to the point ⎜
⎝3 ⎠
⎠
⎝ 3
(ii) 2 y = ( x − 1)(3 − x ) between x = 1 and x = 3.
(iii) y 2 = 4 ax cut off by the line 3 y = 8 x .
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Integral Calculus ■
6.79
2. Find the perimeter of the loop of the curves.
(i) 6 ay 2 = x( x − 2a)3
(ii) 9 xy 2 = ( x − 2a)( x − 5a)3
3. Find the length of the curve x = 2u − sin 2u, y = 2 sin 2 u as u varies from 0 to 2p.
4. Find the length of the curve x = at 2 cos t , y = at 2 sin t from the origin to the point t = 5.
p
5. Find the length of the curve x = a(cos u + u sin u), y = a(sin u − u cos u) from u = 0 to u = .
2
6. Prove that the length of parabola y2 = 4ax cut off by the latus rectum is 2a[ 2 + log(1 + 2 )]
7. Find the length of one complete arch of the cycloid x = a(u − sin u) and y = a(1 − cos u) .
ANSWERS TO EXERCISE 6.7
1. (i)
2. (i)
22
3
8a
3
(ii)
2 + log(1 + 2 )
(ii) 4 a 3
3. 8
⎛ 15
⎞
(iii) ⎜ + log 2⎟ a
⎝ 16
⎠
19a
3
4.
5.
p2 a
8
7. 8a
6.5.2 (b) Length of the Arc in Polar Coordinates
Let r = f (u) be the equation of the curve. Let A and B be two points on the curve with vectorial angles
b
ds
a and b. Then the length of the arc AB is
s=∫
du .
du
a
We know the differential arc in polars is
(ds ) 2 = r 2 (d u) 2 + (dr ) 2
2
∴
⎛ ds ⎞
⎛ dr ⎞
2
⎜⎝ ⎟⎠ = r + ⎜⎝ ⎟⎠
du
du
2
⇒
b
∴
2
2
⎛ dr ⎞
s = ∫ r 2 + ⎜ ⎟ du
⎝ du ⎠
a
r2
When the limits for r are given, the arc length is s = ∫
r1
M06_ENGINEERING_MATHEMATICS-I _CH06_PART B.indd 79
ds
⎛ dr ⎞
= r2 + ⎜ ⎟
⎝ du ⎠
du
r
2
2
ds
⎛ du ⎞
dr = ∫ 1 + r 2 ⎜ ⎟ dr.
⎝ dr ⎠
dr
r1
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WORKED EXAMPLES
EXAMPLE 1
Find the length of the cardioid r 5 a (1 1 cos u) . Also show that the upper half is bisected by
p
u5 .
3
Solution.
The equation of the given curve is r = a(1 + cos u)
(1)
The cardioid is symmetric about the initial line
Ox as shown in Fig 6.23
So, the length of the curve is
B
θ=π
2 × length of the arc OBA
p
p
π
3
O
θ=0
A
x
2
ds
⎛ dr ⎞
d u = 2∫ r 2 + ⎜ ⎟ d u
⎝ du ⎠
d
u
0
0
= 2∫
Differentiating (1) w.r.to u, we get
Fig. 6.23
dr
= a( − sin u) = − a sin u
du
2
⇒
⎛ dr ⎞
2
2
⎜⎝ ⎟⎠ = a sin u
du
2
∴
⎛ dr ⎞
r 2 + ⎜ ⎟ = a 2 (1 + cos u) 2 + a 2 sin 2 u
⎝ du ⎠
= a2 (1 + 2 cos u + cos 2 u + sin 2 u)
= a 2 (1 + 2 cos u + 1) = a 2 ( 2 + 2 cos u) = 2a 2 (1 + cos u) = 4 a 2 cos 2
u
2
2
u
u
⎛ dr ⎞
∴ r 2 + ⎜ ⎟ = 4 a 2 cos 2 = 2a cos
⎝ du ⎠
2
2
p
⎡ sin u/2 ⎤
⎡ p
⎤
u
∴
s = 2∫ 2a cos d u = 4a ⎢
⎥ = 8a ⎢sin 2 − sin 0 ⎥ = 8a(1 − 0) = 8a
/
1
2
2
⎣
⎦
⎣
⎦0
0
∴ upper half curve is of length 4a.
p
Now, length of arc AB =
p
3
ds
∫0 du du =
p
3
∫
0
2
⎛ dr ⎞
r 2 + ⎜ ⎟ du
⎝ du ⎠
p
3
u
= ∫ 2a cos d u
2
0
p
∴
⇒
⎡ sin u/2 ⎤ 3
⎡ p
⎤
⎡1
⎤
= 2a ⎢
= 4a ⎢sin − sin 0 ⎥ = 4a ⎢ − 0 ⎥ = 2a
⎥
⎣ 6
⎦
⎣2
⎦
⎣ u/2 ⎦ 0
arc AB = half of the upper half of the cardioid.
p
the line u = bisects the upper half of the cardioid.
3
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Integral Calculus ■
6.81
EXAMPLE 2
Prove that the length of the equiangular spiral r 5 ae u cot a between the points with radii vectors
r1 and r2 is r1 2 r2 sec a.
Solution.
r2
The equation of the given curve is r = aeu cota
ds
Since the limits for r are given, the length of the arc is s = ∫ dr
dr
r1
(1)
2
r2
⎛ du ⎞
s = ∫ 1 + r 2 ⎜ ⎟ dr
⎝ dr ⎠
r
∴
1
Differentiating (1) w.r.to u, we get
dr
= ae u cot a cot a
du
du
1
1
=
⇒
=
dr ae u cot a .cot a r cot a
2
∴
r
1
du
⎛ du ⎞
=
= tan a ⇒ r 2 ⎜ ⎟ = tan 2 a
⎝ dr ⎠
dr cot a
2
∴
⎛ du ⎞
1 + r 2 ⎜ ⎟ = 1 + tan 2 a = sec 2 a
⎝ dr ⎠
⇒
⎛ du ⎞
1 + r 2 ⎜ ⎟ = sec 2 a = sec a
⎝ dr ⎠
2
r2
∴
s = ∫ sec a dr = sec a [r ]rr12 ⇒ s = sec a [r2 − r1 ]
if r2 > r1
r1
Note: If r2 < r1, s = r2 − r1 sec a, since s is positive.
EXERCISE 6.8
1. Find the perimeter of the cardioid r = 5(1 + cos u) .
2. Find the length of the parabola r (1 + cos u) = 2a cut off by its latus rectum.
u
3. Find the perimeter of the curve r = a sin 3 .
3
4. Find the perimeter of the curve r = a(cos u + sin u) 0 ≤ u ≤ p.
ANSWERS TO EXERCISE 6.8
1. 40
2. 2a ⎡⎣ 2 + log(1 + 2 ) ⎤⎦
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3.
3pa
2
4.
2pa
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6.5.3 Volume of Solid of Revolution
The volume of solid of revolution is obtained by revolving a plane area about line in the plane. This
line is called the axis of revolution.
6.5.3(a) Volume in Cartesian Coordinates
Formula 1: The volume of the solid of revolution obtained by revolving the area bounded by y = f ( x ),
the x-axis, x = a and x = b about the x-axis is
b
V = ∫ py 2 dx
a
Proof Let y = f ( x ) be the equation of the curve.
Let A and B be the points on the curve with
x = a, x = b.
The area ABCD is revolved about the x-axis, a
solid of revolution is generated.
Let P(x, y) and Q ( x + Δx , y + Δy ) be two
neighbouring points on the curve.
The element area is y Δx .
An element volume is generated by the element
area y Δx , which is practically a rectangle as Δx
is small.
When y Δx is revolved about the x-axis we get a
circular disc of radius y and thickness Δx.
∴
ΔV = py 2 Δ x
⇒
y
B
Q
P
x=b
A
y
x=a
O
Δx
C
D
x
Fig. 6.24
∑ Δv = ∑ py
2
Δx .
The sum of such element volume is approximately the required volume.
∴ in the limit, as Δx → 0 we get the volume
b
V = ∫ py 2 dx
a
Formula 2:
The volume generated by revolving the area bounded by x = g(y), y = c and y = d about the
d
y-axis is
V = ∫ px 2dy
c
Formula 3: If the parametric equations of the curve are given by x = f(t) and y = g(t), then volume of
the solid obtained by revolving area about the x-axis is
t2
V = ∫ py 2
t1
dx
dt
dt
and when revolved about the y-axis
t4
V = ∫ px 2
t3
M06_ENGINEERING_MATHEMATICS-I _CH06_PART B.indd 82
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dt
dt
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Integral Calculus ■
Formula 4: If f 2 ( x ) < f1 ( x )∀x ∈[a, b] and the area
bounded by the curves y = f1 ( x ), y = f 2 ( x )
and x = a, x = b (that is area ABCD) is revolved
about the x-axis, the volume of the solid
generated is
y
y = f2(x)
A
B
x=a
x=b
b
V = ∫ p(y 22 − y 12 )dx
C
a
where y1 = f1 ( x ), y2 = f 2 ( x ).
Similarly, the area bounded by the
x = g ( y ), x = h( y ) and y = c, y = d is revolved
about the y-axis, the volume of the solid generated is
D
x=a
O
V = ∫ p(x − x ) dy
2
1
c
x=b
y = f1(x)
x
Fig. 6.25
y
y=d
d
2
2
6.83
x = g(y)
x = h(y)
where x1 = g(y) and x2 = h(y)
y=c
Formula 5: Solid of revolution about any
x
line L in the xy plane.
y
O
Let y = f ( x ) be the equation of curve.
The given line L is in the xy plane is taken as the
Fig. 6.26
x-axis.
Let A and B be two points on the curve. Draw AC and BD perpendicular to the line L.
When the area ACDB as in Fig 6.27 is revolved
y
y = f(x)
about the line L, we get the required volume of solid of
B
revolution.
L
Q
Let PQNM be the element area perpendicular to CD.
P
A
When the element area is revolved about the line L,
D
we get a circular disc of height PM and width MN.
N
M
The element volume ΔV is the volume of the circular
C
disc
O
∴
ΔV = p( PM ) 2 .( MN )
Fig. 6.27
The limit of the sum of such element volume is the
x
volume of the solid of revolution.
∴ Volume V =
OD
∫ p(PM ) d (OM ).
2
OC
WORKED EXAMPLES
EXAMPLE 1
x2 y2
Find the volume of the solid generated by revolving the ellipse 2 1 2 51, a > b be the major
a
b
axis.
Solution.
x2 y2
The equation of the ellipse is 2 + 2 = 1
a
b
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Engineering Mathematics
The x-axis is the major axis.
The ellipse meets the x-axis at x = −a, a.
y
a
∴
Volume V =
∫ py dx
2
A′
(−a, 0)
−a
2
Now
2
y
x
= 1− 2
2
b
a
⇒
y2 =
2
b
(a2 − x 2 )
a2
A
(a, 0) x
O
a
b2 2
( a − x 2 )dx
2
a
−a
V = p∫
∴
= 2p
= 2p
2 a
b
a2
∫
b2
a2
⎡ 2
x3 ⎤
b2
⎢ a x − ⎥ = 2p 2
3 ⎦0
a
⎣
Fig. 6.28
( a 2 − x 2 ) dx
[{ a 2 = x 2 is even function]
0
a
Note If revolved about the minor axis (y-axis), Volume =
⎡ 2
a2 ⎤
b 2 2a 3 4
= pab 2 .
⎢ a a − ⎥ = 2p 2 .
3
3
3
a
⎣
⎦
4 2
pa b
3
EXAMPLE 2
A sphere of radius a is divided into two parts by a plane at a distance
that the ratio of the volume of two parts is 5:27.
Solution.
A sphere of radius a is obtained by revolving the
semi-circular area of radius a as in figure about the
x-axis.
a
The sphere is cut off by a plane at a distance from
2
the centre (0,0) means the area of the semi-circle is cut
a
from the centre. Show
2
y
A2
A1
x=a
O (0, 0)
a
off by the line x =
2
Let V1 and V2 be the two volumes generated by the two
areas A1 and A2.
x=
x
a
2
Fig. 6.29
Equation of the circle is x 2 + y 2 = a 2
(1)
[ Volume V1 is generated by the area bounded the portion of the circle (1) and the lines
a
x = , x = a.
2
[ Volume
a
⎡
x3 ⎤
V 1 = ∫ py dx = p ∫ (a − x ) dx = p ⎢a2 x − ⎥
3 ⎦a 2
⎣
a2
a2
a
a
2
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⎡ ⎛
a⎞ 1 ⎛
a3 ⎞ ⎤
= p ⎢ a 2 ⎜ a − ⎟ − ⎜ a3 − ⎟ ⎥
2⎠ 3 ⎝
8 ⎠⎦
⎣ ⎝
3
⎡ a 1 7a ⎤
7⎤
5pa3
3 ⎡1
3 ⎡12 − 7 ⎤
.
= p ⎢a 2 . − .
=
⎥ = pa ⎢ − ⎥ = pa ⎢
⎥
24
⎣ 2 24 ⎦
⎣ 24 ⎦
⎣ 2 3 8 ⎦
4 3
pa .
3
4
5pa3
pa3 (32 − 5) 27pa3
Volume V2 = pa3 −
=
=
3
24
24
24
3
3
5pa 27pa
:
= 5 : 27
V1 : V 2 =
24
24
We know that the volume of the sphere of radius a is
∴
∴
EXAMPLE 3
Find the volume of a spherical cap of height h cut off from a solid sphere of radius a.
Solution.
The equation of the circle of radius a is x 2 + y 2 = a 2.
Required the volume of the sphere cap of height h
cut off from a sphere of radius a.
∴
y
OA = a − h, AB = h
C
If the area ABC is revolved about the x-axis,
then we get the spherical cap of height h.
a
∴ required volume V =
∫ py dx
2
a−h
O
a
=p
∫
y 2 dx
A h B (a, 0)
(a − h, 0)
a−h
a
=
∫ (a
2
− x 2 )dx
Fig. 6.30
a−h
a
⎡
x3 ⎤
= p ⎢a2 x − ⎥
3 ⎦a−h
⎣
1
⎡
⎤
= p ⎢a2 (a − (a − h )) − (a3 − (a − h )3 ) ⎥
3
⎣
⎦
1
⎡
⎤
= p ⎢a2 h − (a3 − a3 + 3a2 h − 3ah 2 + h 3 ) ⎥
3
⎣
⎦
⎡ 3a2 h − 3a2 h + 3ah 2 − h 3 ⎤
= p⎢
⎥
3
⎣
⎦
=
p
ph2
(3ah2 − h3 ) =
(3a − h)
3
3
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Engineering Mathematics
Note Sometimes the spherical cap formula is given in
terms of base radius of the cap and its height h.
If we assume the base radius of the spherical
cap is c. i.e., AC = c
Then OC2 = OA2 + AC2
= ( a − h) + c
2
⇒
⇒
∴
y
C
a
c
2
a−h
a 2 = a 2 − 2ah + h2 + c 2
2ah = h2 + c 2 ⇒ a =
Volume of the cap =
=
O
h2 + c 2
2h
A h B (a, 0) x
(a − h, 0)
Fig. 6.31
⎤
ph ⎡ ( h + c )
− h⎥
⎢3
3 ⎣
2h
⎦
2
2
2
ph 2
ph 2
[3h 2 + 3c 2 − 2h 2 ] =
[h + 3c 2 ]
6h
6
EXAMPLE 4
The area bounded by one arch of the cycloid x 5 a ( u 2 sinu), y 5 a (1 2 cos u) and its base is
revolved about its base. Find the volume generated.
Solution.
The parametric equations of the cycloid are x = a(u − sinu), y = a(1 − cos u)
The base is the x-axis.
The curve meets the x-axis y = 0 ∴ cosu = 1 ⇒ u = 0, 2p
The volume of the solid generated by revolving the area bounded by one arch of the given curve and
its base (x-axis) about the x-axis is
V =
2p
∫
p
py 2
0
We have
dx
dx
d u = p∫ y 2
du
du
du
0
x = a(u − sin u)
∴
dx
= a(1 − cos u)
du
2p
V = p ∫ a2 (1 − cos u) 2 a(1 − cos u) d u
∴
0
2p
2p
⎛
= pa3 ∫ (1 − cos u)3 d u = pa3 ∫ ⎜ 2 sin 2
⎝
0
0
Put t =
u
2
∴
1
du = dt
2
⇒
3
2p
u⎞
⎛ 6 u⎞
3
⎟⎠ d u = 8pa ∫ ⎜⎝ sin ⎟⎠ d u
2
2
0
d u = 2dt
When u = 0, t = 0 and when u = 2p, t = p
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Integral Calculus ■
p
p
0
0
6.87
∴ V = 8pa3 ∫ sin 6 t 2dt = 16pa3 ∫ sin 6 t dt
[ using property f (p − 0) = sin 6 (p − t )
p
2
= sin 6 t = f (t )
= 16pa3 × 2 ∫ sin 6 t dt
⎤
⎥
∴ ∫ f (t )dt = 2 ∫ f (t )dt ⎥
0
0
⎥⎦
0
p
2
p
5 3 1 p
= 32pa3 . . . . = 5pa3
6 4 2 2
EXAMPLE 5
The area bounded by y 2 5 4 x and the line x 5 4 is revolved about the line x 5 4. Find the
volume of the solid of revolution.
Solution.
Given
y2 = 4x
(1)
Let the line x = 4 meets the parabola in A and B
When
x = 4, y2 = 16 ⇒ y ± 4
∴ A is (4, 4) and B is (4, −4)
The area OAB is revolved about the line x = 4
to get the solid of revolution.
Let P(x, y) be any point on the curve.
Draw PM perpendicular to the line AB.
∴
y
P(x, y)
O
A(4, 4)
M
C
(4, 0)
N
y 2 = 4x
x
x=4
PM = 4 − ON = 4 − x
The line x = 4 is parallel to the y-axis.
∫ p( PM )
−4
Fig. 6.32
4
4
∴ required volume V =
B(4, −4)
2
dy = p ∫ ( 4 − x ) 2 dy
−4
2
4
⎛
y2⎞
= p ∫ ⎜ 4 − ⎟ dy
4⎠
−4 ⎝
2
⎛
y2⎞
= 2p∫ ⎜ 4 − ⎟ dy
4⎠
⎝
4
0
⎡
⎢{ the function
⎢⎣
2
⎤
⎛
y2⎞
⎥
4
−
is
even
⎜⎝
4 ⎟⎠
⎥⎦
4
⎛
⎞
y4
= 2p∫ ⎜16 +
− 2 y 2 ⎟ dy
16
⎠
0⎝
4
⎡
1 y5
y3⎤
= 2p ⎢16 y + . − 2. ⎥
16 5
3 ⎦0
⎣
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⎡
1 45
43 ⎤
= 2p ⎢16 × 4 + .
− 2. ⎥
16 5
3⎦
⎣
⎡ 1 2⎤
⎛ 15 + 3 − 10 ⎞
⎛ 8 ⎞ 1024p
= 2p × 43 ⎢1 + − ⎥ = 128p ⎜
⎟⎠ = 128p ⎜⎝ ⎟⎠ =
⎝
15
5
3
15
15
⎣
⎦
EXAMPLE 6
Find the volume generated when the area bounded by the parabolas y 2 5 4 2 x and y 2 5 4 2 4 x
revolves
1. about the common axis of the two curves
2. about the y-axis.
Solution.
The given parabolae are
and
y 2 = 4 − x = −( x − 4)
(1)
y 2 = 4 − 4 x = −4( x − 1)
(2)
y
B (0, 2)
For the first parabola, the x-axis is the axis and
the vertex is (4, 0).
For the second parabola, the axis is the x-axis
and the vertex is (1, 0).
∴ the common axis is the x-axis.
To find the point of intersection, solve (1) and (2).
∴
A
O
C
4 − x = 4 − 4 x ⇒ 3x = 0 ⇒ x = 0
(1, 0)
(4, 0)
x
(0, −2)
Fig. 6.33
When x = 0, y 2 = 4 ⇒ y ± 2.
∴ the points of intersection are (0, 2), (0, −2).
The common area is as shown in the Fig 6.33.
The volume of the solid generated by revolving the common area about the x-axis is the same as
the volume of the solid generated by revolving the area above the x-axis, about the x-axis.
4
1
∴ required volume V = ∫ py12 dx − ∫ py22 dx ,
0
2
2
where y 1 = 4 − x , y 2 = 4 − 4x
0
4
1
= p∫ ( 4 − x ) dx − p∫ ( 4 − 4 x )dx
0
0
4
1
⎡ (1 − x ) 2 ⎤
⎡ (4 − x)2 ⎤
= p⎢
⎥
⎥ − 4p ⎢
⎣ −2 ⎦ 0
⎣ −2 ⎦ 0
p
[( 4 − 4) 2 − ( 4 − 0) 2 ] + 2p[(1 − 1) 2 − (1 − 0) 2 ]
2
p
p
= − [0 − 16] + 2p[0 − 1] = × 16 − 2p = 8p − 2p = 6p
2
2
=−
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Integral Calculus ■
6.89
2. If the area is revolved about the y-axis, then the volume generated is
2
V=
∫ p( x
2
1
− x22 ) dy , where x1 = 4 − y 2 , x2 =
−2
1
(4 − y 2 )
4
1
⎡
⎤
= p ∫ ⎢( 4 − y 2 ) 2 − ( 4 − y 2 ) 2 ⎥ dy
16
⎣
⎦
−2
2
2
15
( 4 − y 2 ) 2 dy
16
−2
= p∫
=
15p 2
16
2
∫ (4 − y
2 2
) dy
−2
15p
( 4 − y 2 ) 2 dy
=
8 ∫0
2
15p
(16 − 8y 2 + y 4 ) dy
=
8 ∫0
2
15p ⎡
y3 y5⎤
=
⎢16 y − 8 + ⎥
8 ⎣
3
5 ⎦0
2
=
15p ⎡
8
1
⎤
16 × 2 − × 23 + × 25 ⎥
8 ⎢⎣
3
5
⎦
=
15p × 32 ⎡ 2 1 ⎤
⎢1 − 3 + 5 ⎥
8
⎣
⎦
⎡{ ( 4 − y 2 ) is an even function y,⎤
⎥
⎢2
2
⎢ ( 4 − y 2 ) 2 dy = 2 ( 4 − y 2 ) 2 dy ⎥
∫0
⎥
⎢∫
⎦
⎣ −2
8
⎡15 − 10 + 3 ⎤
= 15p × 4 ⎢
= 15p × 4 × = 32p
⎥
15
15
⎣
⎦
Remark: From the above problem, we observe that the solids generated revolving the same area about
two different axes of revolution are different. Hence, volume generated are different.
EXERCISE 6.9
1. Find the volume of the solid generated by revolving about the x-axis, the area bounded by
x1/ 2 + y1/ 2 = a1/ 2 and the coordinates axes.
x2
+ 2 and the line 5 x − 8 y + 14 = 0 is revolved about the x-axis. Find
2. The area bounded by y 2 =
4
the volume of the solid generated.
2
2
2 ⎡
⎡
⎛ x2
⎞ ⎤
1 5x
891p ⎤
⎢ Hint : V = p ∫ ⎢ ⎛⎜ + 7⎞⎟ − ⎜ + 2⎟ ⎥ dx =
⎥
⎠ ⎝ 4
16 ⎝ 2
1280 ⎥
⎠
⎢
⎥
1/ 2 ⎢
⎦
⎣
⎣
⎦
3. Find the volume if the area of the loop of y 2 = x 2 ( x + 4) is revolved about x-axis.
4. The area of the loop of y 2 (1 + x ) = x 2 (1 − x ) is revolved about the x-axis. Find the volume of the
solid of revolution.
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Engineering Mathematics
5. The area bounded by the portion of the curve y = e x sin x between x = 0 and x = p, revolves about
the x-axis. Find the volume generated.
6. Find the volume of the solid generated by revolving the area of the curve y = x 3 , y = 0 and x = 2.
ax 3 − x 4
7. Find the volume of the solid obtained by revolving the area of the curve y 2 =
about the
a2
x-axis.
8. The volume of the solid generated by revolving the area bounded by y( x 2 + a 2 ) = a3 and its
asymptote about the asymptote.
9. The area bounded by y 2 = 4 ax and x 2 = 4 ay, a > 0, revolves about the x-axis. Show that the
96pa2
.
volume of the solid formed is V =
5
10. Compute the volume of the solid generated by revolving about the y-axis, the area bounded by
y = x 2 and 8 x = y 2 .
11. Find the volume of the solid generated when the area of the loop of the curve y 2 = x( 2 x − 1) 2
resolves about the x-axis.
12. Find the volume of a right circular cone of base radius r and height h by integration.
13. When the area of the curve x 2 / 3 + y 2 / 3 = a2 / 3 in the first quadrant is revolved about the x-axis, find
the volume of the solid generated.
14. Find the volume of the solid generated by revolving the loop of the curve 3ay 2 = x ( x − a) 2 , about
the x-axis.
x
15. Find the volume of the solid generated by revolving the catenary y = a cosh about the x-axis
a
between x = 0 and x = b.
x2
between y = 0 and y = 5
16. A bowl has a shape that can be generated by revolving the graph y =
2
about the y-axis. Find the volume of bowl.
17. Find the volume of the frustrum of a right circular cone whose lower base has radius R, upper
base is of radius r and height h.
18. If the curve (a − x ) y 2 = a2 x revolved about its asymptote, find the volume formed.
19. The area bounded by y 2 = 4 x and the line x = 4 above the x-axis is revolved about the x-axis. Find
the volume of the solid generated.
20. Find the volume of the solid if the area included between the curve xy 2 = a2 (a − x ) and its
asymptote is revolved about the asymptote.
ANSWERS TO EXERCISE 6.9
1.
pa
15
2.
891p
1280
3.
64p
3
4⎤
⎡
4. p ⎢ 2 log 2 − ⎥
3⎦
⎣
6.
64p
5
7.
pa3
20
8.
p2 a2
2
9.
3
11.
p
48
16. 25p
12. pr 2 h.
17.
13.
16pa3
105
ph 2
[R + rh + r 2 ]
3
M06_ENGINEERING_MATHEMATICS-I _CH06_PART B.indd 90
5.
p 2p
⎡e − 1⎤⎦
8⎣
96pa 2
5
10.
24p
5
14.
pa 3
36
15. =
18.
p 2 a3
2
19. 32p
pa 3 2b / a
pa2b
[e
− e −2b / a ] +
8
2
20.
p 2 a3
2
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Integral Calculus ■
6.91
6.5.3 (b) Volume in Polar Coordinates
1. Revolution about the initial line
Let r = f (u) be the equation of the given curve
When arc OAB bounded by the given curve and
radii vector u = a and u = b is revolved about the
initial line, the volume of the solid generated is
y
θ=
B
r = f(θ )
π
2
A
b
V =
β
2p 3
r sin u d u
3 ∫a
α
p
2. Revolution about the line u =
2
θ=0
O
x
Fig. 6.34
When the area OAB is revolved about the line
b
u=
2p 3
p
, the volume is V =
r cos u d u.
3 ∫a
2
WORKED EXAMPLES
EXAMPLE 1
Find the volume of the solid generated by revolving the area of the cardioid r 5 a(1 2 cos u)
about the initial line.
Solution.
P
r = a(1 − cos u)
Given
(1)
Since the volume of the solid generated by revolving the
area of the cardioid about the initial line is same as the
volume generated by revolving the area OPA above the
initial line, about the initial line.
b
2p 3
V =
r sin u d u,
Required volume
3 ∫a
where a = 0 and b = p
A
O
x
Fig. 6.35
For, when r = 0, 1 − cos u = 0 ⇒ cos u = 1 ⇒ u = 0
and when r = 2a, 1 − cos u = 2 ⇒ cos u = −1 ⇒ u = p
∴
V =
p
p
2p 3
2p 3
r sin u d u =
a (1 − cos u)3 sin ud u
∫
3 0
3 ∫0
p
2pa3 ⎛
2
=
⎜⎝ 2 sin
∫
3 0
=
3
u⎞
u
u
⎟⎠ 2 sin cos d u
2
2
2
p
32pa3
u
u
sin 7 cos d u
∫
3 0
2
2
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Engineering Mathematics
p
32pa3 ⎡ sin s u/2 ⎤
=
⎥
⎢
3 ⎣ (1 / 2) × 8 ⎦ 0
=
8pa3
3
⎡
sin n +1 a u ⎤
n
{
sin
a
u
cos
a
u
d
u
=
⎢ ∫
⎥
a( n + 1) ⎦
⎣
8pa3
8pa3
⎡ 8p
8 ⎤
sin
sin
[
]
−
0
=
1
−
0
=
⎢
⎥
2
3
3
⎣
⎦
EXAMPLE 2
Show that the volume of the solid generated by revolving the lemniscate r 2 5 a 2 cos 2 u about the
pa 3
p
line u 5 is 2
.
2
8
Solution.
Given
r 2 = a 2 cos 2u
b
2p 3
Required volume is V =
r cos u d u
3 ∫a
(1)
y
θ=
π
2
θ=
p⎤
⎡
⎢since the area is revolved about the line u = 2 ⎥
⎣
⎦
If we replace r by −r, then
π
4
x
π
θ=−
4
( − r ) 2 = a 2 cos 2u ⇒ r 2 = a 2 cos 2u
∴ the equation is unaffected.
Fig. 6.36
When u is changed to −u, the equation is unaffected,
since cos(−2u) = cos 2u.
∴ the curve is symmetric about the initial line and pole respectively.
p
p
⇒ u=
2
4
cos 2u = 1 ⇒ 2u = 0 ⇒ u = 0
When
r = 0,
cos 2u = 0 ⇒ 2u =
When
r = a,
When
r = −a, cos 2u = 1 ⇒ 2u = 0 ⇒ u = 0
We get two loops of the curve as in Fig 6.36.
p
The volume of the solid generated by revolving the area of the lemniscate about the line u = is
2
equal to 2 times the volume generated by the area above Ox of one loop of the curve revolving about
the line u =
p
.
2
p
2p 4 3
∴ required volume is V = 2 ×
r cos u d u
3 ∫0
Now
r 2 = a 2 cos 2u ⇒ r = a(cos 2u)1/ 2
∴
4pa3 2
2p 3
3/ 2
(1 − 2 sin 2 u)3/ 2 cos u d u
V = 2×
a
(cos
2
u
)
cos
u
d
u
=
3 ∫0
3 ∫0
M06_ENGINEERING_MATHEMATICS-I _CH06_PART B.indd 92
p
⇒ r 3 = a3 (cos 2u)3/ 2
p
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Integral Calculus ■
Put
2 sin u = sin f
∴
2 cos u d u = cosf df ⇒ cos u d u =
When u = 0, sin f = 0 ⇒ f = 0 and when u =
V =
∴
=
=
4pa
3
p
3 2
4pa
∫ (1 − sin
2
f)3/2 ⋅
1
2
6.93
cos f df
p sin f = 2 sin p = 2 ⋅ 1 = 1 ⇒ f = p
,
4
2
2
4
cos f
0
2
df
p
3 2
3 2
4pa3
3 2
∫ cos
3
f cos f df
0
p
2
4
∫ cos f df =
0
4pa3 . 3 . 1 . p p 2 a3
=
=
3 2 4 2 2 4 2
2p 2 a3
8
EXERCISE 6.10
1. Find the volume of the solid generated by revolving the curve r = a + b cos u, a > b about the
initial line.
p
p
2. The area of the loop of r = a cos 3u lying between u = − and u = is revolved about the initial
6
6
p
⎡
⎤
6
2p 3
⎢
⎥
line. Find the volume generated. ⎢ Hint : V = ∫
r sin u d u⎥
3
0
⎢
⎥
⎣
⎦
3. Find the volume of solid generated by revolving the area of the cardioid r = a(1 + cos u) about the
initial line.
4. Find the volume of the solid formed by rotating the area of r 3 = a 2 cos u about its line of symmetry.
ANSWERS TO EXERCISE 6.10
1.
4
pa( a 2 + b 2 )
3
2.
19pa3
960
3.
8pa3
3
4.
8pa3
15
6.5.4 Surface Area of Revolution
An arc of a curve is revolved about an axis, a surface is generated. This surface is called the surface
of revolution and its area is the surface area.
We find the surface area in Cartesian and polar coordinates.
6.5.4(a) Surface Area of Revolution in Cartesian Coordinates
Let y = f ( x ) be the equation of the curve.
Let AB be an arc on the curve.
Let PQ = Δ s be an element arc in between the points A and B.
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6.94
■
Engineering Mathematics
Let the coordinates of P be (x, y) and the
coordinates of Q be ( x + Δ x, y + Δ y ).
The element arc Δ s is revolved about the x-axis,
we get the element surface as a circular ring of
radius y and width Δ s. Let Δ S be the element
surface area generated by the element arc Δ s .
y
B
P(x, y) ΔB
A
y = f(x)
Q(x + Δx, y + Δy)
M
∴ Δ S = 2py Δ s
x
O
The sum of such element surface areas
= ∑ Δ S = ∑ 2py Δ s
Fig. 6.37
s2
∴ the surface area is S = lim ∑ Δ S = lim ∑ 2py Δ s = ∫ 2py ds
Δx → 0
Δs → 0
Δx → 0
s1
with proper limits s1 and s2.
(a) If the limits for x are known, say x = a and x = b, then
b
S = ∫ 2py
a
2
ds
⎛ dy ⎞
dx = ∫ 2py 1 + ⎜ ⎟ dx
⎝ dx ⎠
dx
a
b
(b) If the limits for y are known say y = c and y = d, then
d
S = ∫ 2px
c
2
b
⎛ dx ⎞
ds
dy = ∫ 2px 1 + ⎜ ⎟ dy
dy
⎝ dy ⎠
a
(c) If the equation of the curve is given in parametric form, x = f (t ), y = g (t ) and the limits for t are
t = t1 and t = t2, then
t2
S = ∫ 2py
t1
t
2
2
2
⎛ dx ⎞ ⎛ dy ⎞
ds
dt = ∫ 2py ⎜ ⎟ + ⎜ ⎟ dt
⎝ dt ⎠ ⎝ dt ⎠
dt
t1
WORKED EXAMPLES
EXAMPLE 1
x2
3
cut off by the straight line y 5 is revolved about the y-axis.
2
2
Find the surface area of revolution.
The portion of the curve y 5
Solution.
x2
, which is a parabola with vertex (0, 0).
The given curve is y =
2
M06_ENGINEERING_MATHEMATICS-I _CH06_PART B.indd 94
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Integral Calculus ■
y
3
It is symmetric about the y-axis. Let the line y =
2
intersect the parabola at the points A and B
x
∫
0
x2
y =
2
we have
∴
2
dy 2 x
=
=x ∴
dx
2
dx 1
=
dy x
2
∴
⎛ dx ⎞
1
x2 + 1
1+ ⎜ ⎟ = 1+ 2 =
⎝ dy ⎠
x
x2
∴
S = 2p ∫ x
3/ 2
0
Fig. 6.38
⎛ dx ⎞
ds
dy = 2p ∫ x 1 + ⎜ ⎟ dy
dy
⎝ dy ⎠
0
3/ 2
2px
x
O
The surface obtained by revolving arc AOB about the
y-axis, is the same as the surface obtained by revolving
arc OA about the y-axis.
∴ the surface area generated is
3/ 2
A y = 3/2
B
3
∴ the portion of the curve Cut off by the line y = is the
2
arc AOB as in figure.
S=
6.95
2
⎛ dx ⎞
1
⇒ ⎜ ⎟ = 2
⎝ dy ⎠
x
2
∴
⎛ dx ⎞
1+ ⎜ ⎟ =
⎝ dy ⎠
x2 + 1
=
x2
x2 + 1
x
3/ 2
x2 + 1
dy = 2p ∫ 2 y + 1 dy
x
0
[{ x2 = 2y]
3/ 2
⎡ ( 2 y + 1)3/ 2 ⎤
= 2p ⎢
⎥
⎣ 2 ⋅ ( 3 / 2) ⎦ 0
=
2p ⎡⎛ 3 ⎞
⎢⎜ 2 ⋅ + 1⎟
3 ⎢⎣⎝ 2 ⎠
⎤
− ( 2 ⋅ 0 + 1)3/ 2 ⎥
⎥⎦
14p
2p 3
2p
− 1] =
[2 − 1] =
(8 − 1) =
3
3
3
=
2p 2 3/ 2
[( 2 )
3
3/ 2
EXAMPLE 2
Find the surface area formed by revolving four-cusped hypocycloid (astroid) x 2 / 3 1 y 2 / 3 5 a 2 / 3
about the x-axis.
Solution.
The given curve is
x 2/3 + y 2/3 = a2/3
(1)
The curve is symmetric w.r.to both the axes.
Let x-axis meets the curve at the points A and C and the y-axis meets the curve at B and D
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6.96
Engineering Mathematics
When y = 0, x 2 / 3 = a 2 / 3 ⇒ x 2 = a 2 ⇒ x = ± a
∴
y
A is ( a, 0), C is ( −a, 0).
B(0, a)
Similarly, B is (0, a) and D is (0, − a)
C
A
By symmetry the four arcs AB, BC, CD and DA
x ′ (−a, 0)
are equal.
[ the surface area generated by revolving the
curve about the x-axis is equal to twice the surface
area generated by the arc AB about the x-axis. x varies from 0 to a.
O
D(0, −a)
y′
2
ds
⎛ dy ⎞
dx = 4p∫ y 1 + ⎜ ⎟ dx
⎝ dx ⎠
dx
0
0
Differentiating (1) w.r.to x, we get
a
a
[ Surface area S = 2 × ∫ 2py
(a, 0) x
Fig. 6.39
2 −1/ 3 2 −1/ 3 dy
x + y
=0
3
3
dx
dy
x −1/ 3 + y −1/ 3
=0
dx
⇒
dy
x −1/ 3
y1/ 3
= − −1/ 3 = − 1/ 3
dx
y
x
⇒
2
y 2/3
⎛ dy ⎞
∴ ⎜ ⎟ = 2/3
⎝ dx ⎠
x
∴
y 2/3
x 2 / 3 + y 2 / 3 a2 / 3
⎛ dy ⎞
1 + ⎜ ⎟ = 1 + 2/3 =
= 2/3
⎝ dx ⎠
x
x 2/3
x
∴
⎡ a2/3 ⎤
⎛ dy ⎞
1 + ⎜ ⎟ = ⎢ 2/3 ⎥
⎝ dx ⎠
⎣x ⎦
2
1/ 2
2
We have
x 2/3 + y 2/3 = a2/3
a
[
S = 4p∫ (a2 / 3 − x 2 / 3 )3/ 2 ⋅
Let
t =a
0
2
2/3
[Using (1)]
a1/ 3
x1/ 3
=
⇒
y 2/3 = a2/3 − x 2/3
⇒
y = ( a 2 / 3 − x 2 / 3 )3 / 2
a
a1/ 3
1
dx = 4pa1/ 3 ∫ (a2 / 3 − x 2 / 3 )3/ 2 ⋅ 1/ 3 dx
x 1/ 3
x
0
− x 2/3
2 2 −1
2 1
[ 2tdt = − x 3 dx = −
dx
3
3 − 13
x
⇒
tdt = −
1 1
dx
3 − 13
x
⇒
1
−3tdt =
x
−
1
3
dx
When x = 0, t 2 = a 2 / 3 ⇒ t = a1/ 3 and when x = a, t = a 2 / 3 − a 2 / 3 = 0
0
[
S = 4pa1/ 3
∫
0
1/ 3
(t 2 )3/ 2 ( −3tdt ) = −12pa
∫t
4
dt
a1/ 3
a1/ 3
a1/ 3
= 12pa1/ 3
∫t
4
dt
0
a1/33
= 12pa
1/ 3
M06_ENGINEERING_MATHEMATICS-I _CH06_PART B.indd 96
1/ 3 5
⎤ 12pa1/ 3 ⋅ a5 / 3 12pa 2
⎡t5 ⎤
)
1/ 3 ⎡ ( a
=
12
p
a
−
0
=
⎢ ⎥
⎢
⎥=
5
5
⎣ 5 ⎦0
⎣ 5
⎦
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Integral Calculus ■
6.97
EXAMPLE 3
Find the surface generated by revolving the portion of the curve y 2 5 4 1 x cut off by the straight
line x 5 2, about the x-axis.
Solution.
The given curve is y 2 = 4 + x
(1)
is a parabola with vertex ( −4, 0).
It is symmetric about the x-axis. Let A be the vertex.
∴ A is ( −4, 0).
Let the straight line x = 2 intersect the parabola at
the points B , B ′.
The straight line x = 2 meets the x-axis at ( 2, 0).
The required surface area generated by revolving
the arc AB about the x-axis.
x varies from −4 to 2.
2
∴ Surface area is
S=
∫ 2py
−4
y
x′
A
(−4, 0)
B
(2, 0)
O
x
y ′ B′
Fig. 6.40
2
ds
⎛ dy ⎞
dx = 2p ∫ y 1 + ⎜ ⎟ dx
⎝ dx ⎠
dx
−4
2
Differentiating (1) w.r.to x, we get
2y
dy
=1 ⇒
dx
dy
1
=
dx 2 y
2
1
⎛ dy ⎞
⇒ ⎜ ⎟ = 2
⎝ dx ⎠
4y
1
4 y2 + 1
⎛ dy ⎞
1+ ⎜ ⎟ = 1+ 2 =
∴
⎝ dx ⎠
4y
4 y2
2
∴
2
∴
S = 2p ∫ y
−4
4y 2 + 1
dx
2y
2
⎛ dy ⎞
1+ ⎜ ⎟ =
⎝ dx ⎠
4 y2 + 1
2y
2
= p ∫ 4 y 2 + 1 dx
−4
2
= p ∫ 4( x + 4) + 1 dx
−4
2
= p ∫ 4 x + 17 dx
−4
2
⎡ ( 4 x + 17)(1/ 2 ) +1 ⎤
= p⎢
⎥
⎣ 4((1/2) + 1) ⎦ −4
2
⎡ ( 4 x + 17)3/ 2 ⎤
= p⎢
⎥
⎣ 4 × (3/2) ⎦ −4
p
[( 4 ⋅ 2 + 17)3/ 2 − ( 4( −4) + 17)3/ 2 ]
6
p
p
= [( 25)3/ 2 − 1] = [(52 )3/ 2 − 1]
6
6
p
p
62p
p 3
= [5 − 1] = [125 − 1] = × 124 =
6
6
3
6
=
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■
Engineering Mathematics
EXAMPLE 4
t
Compute the surface area generated when an arc of the curve x 5 t 2 , y 5 ( t 2 2 3) between
3
the points of intersection of the curve and the x-axis is revolved about the x-axis.
Solution.
The given curve is
t
x = t 2 , y = (t 2 − 3)
3
Which is parametric form
t2
[
S = ∫ 2py
t1
(1)
y
2
t2
2
ds
⎛ dx ⎞ ⎛ dy ⎞
dt = 2p∫ y ⎜ ⎟ + ⎜ ⎟ dt
⎝ dt ⎠ ⎝ dt ⎠
dt
t1
O
To find the limits
t 2
When y = 0,
(t − 3) = 0 ⇒ t = 0, t = ± 3
3
Fig. 6.41
When t = 0, x = 0 and when t = ± 3 , x = 3
[ We get the loop of the curve as in figure.
[ t varies from t = 0 to t = 3
[ the required surface area is
2
(3, 0) x
2
⎛ dx ⎞ ⎛ dy ⎞
S = 2p ∫ y ⎜ ⎟ + ⎜ ⎟ dt
⎝ dt ⎠ ⎝ dt ⎠
0
3
t
y = (t 2 − 3) =
3
dy 1 2
= (3t − 3)
dt 3
We have x = t 2 and
∴
dx
= 2t
dt
and
2
2
2
2
1 3
(t − 3t )
3
= (t 2 − 1)
∴
⎛ dx ⎞ ⎛ dy ⎞
2
2
2
2
4
2
4
2
2
2
⎜⎝ ⎟⎠ + ⎜⎝ ⎟⎠ = 4t + (t − 1) = 4t + t − 2t + 1 = t + 2t + 1 = (t + 1)
dt
dt
∴
⎛ dx ⎞ ⎛ dy ⎞
2
2
2
⎜⎝ ⎟⎠ + ⎜⎝ ⎟⎠ = (t + 1) = t + 1
dt
dt
⎤
⎡
⎢Since t < 3 ⇒ t 2 < 3 ⇒ t 2 − 3 < 0
⎥
⎢
⎥
2
⇒ −(t − 3) > 0 and t > 0 ⎥
⎢
⎢
⎥
t
⎢∴
⎥
y = − (t 2 − 3) > 0
3
⎣
⎦
3
3
2p
2
p
(t 3 − 3t )(t 2 + 1)dt = −
(t 3 − 3t )(t 2 + 1) dt
S=−
3 ∫0
3 ∫0
3
∴
∴
t
S = 2p ∫ − (t 2 − 3)(t 2 + 1)dt
3
0
=−
2p
3
3
∫ (t
5
− 2t 3 − 3t ) dt
0
M06_ENGINEERING_MATHEMATICS-I _CH06_PART B.indd 98
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Integral Calculus ■
=−
2p ⎡ t 6
t4
t2 ⎤
2
3
−
−
⎥
⎢
3 ⎣6
4
2 ⎦0
( ) − ( 3)
⎡
2p ⎢ 3
=−
3 ⎢ 6
⎢⎣
=−
6
2
4
6.99
3
( 3)
−3
2
2
⎤
− 0⎥
⎥
⎦⎥
2p ⎡ 33 32 32 ⎤
2p ⎡ 9 9 9 ⎤
2p ⎡ 9 ⎤
− − ⎥=−
− = 3p
⎢ − − ⎥=−
⎢
3 ⎣6 2 2⎦
3 ⎣2 2 2⎦
3 ⎢⎣ 2 ⎦⎥
EXERCISE 6.11
1. Find the surface area generated by revolving the arc of the curve 8 y 2 = x 2 (1 − x 2 ) about the x-axis.
2. Find the surface area generated by revolving the curve 3 y = x 3 between x = −2 and x = 2 about
the x-axis.
3. Find the surface area generated by revolving the loop of the curve 9 y 2 = x( x − 3) 2 about the x-axis.
4. An arc of the curve ay 2 = x 5 from x = 0 to x = 4 a is revolved about the y-axis, find the surface
area generated.
5. Find the surface area of the right circular cone of height h and base radius r.
6. A quadrant of a circle of radius 2 revolves about the tangent at one end. Show that the surface area
generated is 4p(p − x ).
7. The part of the parabola y 2 = 4 x cut off by the latus rectum revolves about the tangent at the
vertex. Find the curved surface of the real thus generated.
8. Find the area of the surface generated by revolving the cardioid x = a( 2 cos u − cos 2u),
y = a( 2 sin u + sin 2u) about the x-axis.
9. Find the area of the surface generated by revolving one arch of the cardioid x = a(t − sin t )
y = a(1 − cos t ) about the x-axis.
10. The asteroid x = a sin 3 t , y = a cos3 t is revolved about the x-axis. Find the surface area generated.
11. Find the surface area obtained by revolving a loop of the curve 9ax 2 = y(3a − y ) 2 about the y-axis.
x2 y2
12. Find the surface area of the ellipsoid formed by revolving the ellipse 2 + 2 = 1 about the x-axis.
a
b
Deduce the surface area of the sphere of radius a.
ANSWERS TO EXERCISE 6.11
1.
5.
10.
p
2
1 2
pr h
3
12pa 2
.
5
2. (34 17 − 2)
p
9
7. pa2 ⎡3 2 − log(1 + 2 ) ⎤
⎣
⎦
11. 3pa2
M06_ENGINEERING_MATHEMATICS-I _CH06_PART B.indd 99
3. 3p
8.
128 2
pa
5
4.
128
pa2 (1 + 125 10 )
1215
9.
64 2
pa
3
1
⎡
⎤
12. 2pab ⎢ 1 − e 2 + sin −1 e ⎥ , 4pa2
e
⎣
⎦
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6.100
■
Engineering Mathematics
6.5.4 (b) Surface Area in Polar Coordinates
Let r = f (u) be the equation of the curve.
Let A and B be two points on the curve with vectorial
angles a and b.
1. If the arc AB is revolved about the initial line
u = 0 (i.e., the x-axis) then the surface is generated.
y
α
b
ds
S = ∫ 2py
du
du
a
= ∫ 2pr sin u
a
A
β
O
b
r = f(θ)
π
θ=
2
The area of the surface is S = ∫ 2py ds with
limits for s.
∴
B
θ=0
x
Fig. 6.42
b
2
ds
⎛ dr ⎞
d u = ∫ 2pr sin u r 2 + ⎜ ⎟ d u
⎝ du ⎠
du
a
[{ y = r sin u]
p
2. If the arc AB is revolved about the line u = (i.e., about the y-axis, then a surface is generated)
2
The area of the surface is
S = ∫ 2pxds within suitable limits
b
= ∫ 2px
a
b
2
ds
⎛ dr ⎞
d u = ∫ 2pr cos u r 2 + ⎜ ⎟ d u
⎝ du ⎠
du
a
[{ x = r cos u]
WORKED EXAMPLES
EXAMPLE 1
Find the area of the surface formed by revolving the lemniscate r 2 5 a 2 cos 2 u about the polar
axis (polar axis is the initial line).
Solution.
Given the curve
r 2 = a 2 cos 2u ⇒ r = a cos 2u
p
p
≤ 2u ≤
2
2
p
p
− ≤u≤
4
4
cos 2u ≥ 0 ⇒ −
⇒
and
3p
5p
3p
5p
≤ 2u ≤
⇒
≤u≤
2
2
4
4
The curve is symmetric about the initial line
M06_ENGINEERING_MATHEMATICS-I _CH06_PART B.indd 100
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Integral Calculus ■
p
, r = 0, 0
2
We get the two loops of the curve as in Fig 6.43.
The required surface area is
S = area of the surface generated by revolving the two
loops about the initial line.
= 2[area of the surface generated by revolving the
arc OA]
[{ the curve is symmetric]
y
When u = 0, r = ± a and when u =
p/4
=2∫
0
p/4
θ=
6.101
π
2
θ=
B(−a, 0)
O
π
4
A(a, 0)
x
Fig. 6.43
2
ds
⎛ dr ⎞
2py
d u = 4p ∫ r sin u r 2 + ⎜ ⎟ d u
⎝ du ⎠
du
0
r = a cos 2u
We have
dr
1
a sin 2u
=a
( − sin 2u) ⋅ 2 = −
du
2 cos 2u
cos 2u
∴
2
a 2 sin 2 2u
⎛ dr ⎞
⎜⎝ ⎟⎠ =
du
cos 2u
∴
2
∴
a2 sin 2 2u
⎛ dr ⎞
r + ⎜ ⎟ = a2 cos 2u +
⎝ du ⎠
cos 2u
2
=
2
∴
⎛ dr ⎞
r2 + ⎜ ⎟ =
⎝ du ⎠
∴
2
2
2
a2 cos 2 2u + a2 sin 2 2u a ⎡⎣cos 2u + sin 2u ⎤⎦
a2
=
=
cos 2u
cos 2u
cos 2u
a
cos 2u
p/4
S = 4p ∫ a cos 2u sin u
0
= 4pa2
a
cos 2u
p/4
du
∫ sin ud u = 4pa [− cos u]
2
p/4
0
0
p
⎡
⎤
= −4pa2 ⎢cos − cos 0 ⎥
4
⎣
⎦
⎡ 2 ⎤
⎡ 2 − 2⎤
⎤
⎡ 1
2
− 1⎥ = −4pa2 ⎢
= −4pa2 ⎢
− 1⎥ = −4pa2 ⎢
⎥ = 2pa ⎡⎣ 2 − 2 ⎤⎦
⎣ 2 ⎦
⎣ 2
⎦
⎣ 2 ⎦
EXAMPLE 2
Find the area of the surface generated by revolving one branch of the lemniscates r 5 a cos 2 u
about the tangent at the origin.
M06_ENGINEERING_MATHEMATICS-I _CH06_PART B.indd 101
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6.102
Engineering Mathematics
Solution.
Given
r = a cos 2u
The curve has two loops as in Fig 6.44.
p
u=
is a tangent at the origin to the right side loop.
4
p
This loop is revolved about u = .
4
Let P ( r , u) be any point on the curve.
p
Draw PM perpendicular to the tangent u = .
4
From the right angled triangle OPM, we get
PM
⎛p
⎞
= sin ⎜ − u⎟
⎝4
⎠
OP
(1)
y
θ=
M
π
2
π
4
P(r, θ )
θ=
θ
x
O
π
θ= −
4
Fig. 6.44
⎛p
⎞
⎛p
⎞
⎛p
⎞
PM = OP sin ⎜ − u⎟ = r sin ⎜ − u⎟ = a cos 2u sin ⎜ − u⎟
⎝4
⎠
⎝4
⎠
⎝4
⎠
∴ S=
p/4
∫
2pPM
−p/ 4
ds
du
du
p/4
2
⎛p
⎞
⎛ dr ⎞
= 2p ∫ a cos 2u sin ⎜ − u⎟ r 2 + ⎜ ⎟ d u
⎝4
⎠
⎝ du ⎠
−p/ 4
= 2p
p/4
a
⎛p
⎞
du
a cos 2u sin ⎜ − u⎟
⎝4
⎠ cos 2u
−p/ 4
∫
= 2pa2
⎡
⎢ From example 1, page 6.100
⎣
2
⎤
a
⎛ dr ⎞
2
⎥
r +⎜ ⎟ =
⎝ du ⎠
cos 2u ⎥
⎦
p/4
⎛p
⎞
sin ⎜ − u⎟ d u
⎝
⎠
4
−p/ 4
∫
p/4
⎡
⎛p
⎞⎤
⎢ − cos ⎜⎝ 4 − u⎟⎠ ⎥
⎥
= 2pa2 ⎢
−1
⎢
⎥
⎢
⎥
⎣
⎦ −p/ 4
p/4
⎡ ⎛p
⎡
⎞⎤
⎛ p⎞ ⎤
= 2pa2 ⎢cos ⎜ − u⎟ ⎥
= 2pa2 ⎢cos 0 − cos ⎜ ⎟ ⎥ = 2pa2
⎝
⎠
⎝ 2⎠⎦
4
⎣
⎦ −p / 4
⎣
EXERCISE 6.12
1. Find the surface area generated by rotating the cardioid r = a(1 + cos u) about the initial line.
2. Find the surface area generated when the curve r = 4 + 2 cos u revolves about its axis.
M06_ENGINEERING_MATHEMATICS-I _CH06_PART B.indd 102
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Integral Calculus ■
3. The portion of the parabola r =
the surface area generated.
6.103
2a
cut off by the latus rectum revolves about the axis. Find
1 + cos u
4. Find the surface area generated by revolving the curve r = 2a cos u about the initial line.
5. Find the area of the surface generated by revolution of the curve r = 2a sin u about the polar axis.
ANSWERS TO EXERCISE 6.12
1.
32pa
5
37p
5
2
2.
3.
8 2
pa
3
4. 4pa2
5. 4p 2 a2
SHORT ANSWER QUESTIONS
Evaluate the following integrals
1. ∫ e −3 x sin 4 x dx
log x
dx
x2
4.
∫
7.
∫e
10.
∫
13.
∫ cosec x dx
2.
∫e
5.
∫ sin 6 x cos 2 x dx
3x
cos 2 x dx
3.
∫x
6.
∫ cos 3x sin 2 x dx
2
sin x 3 dx
3
x
sin e dx
x −1
dx
x−2
∫ cos
14.
log x dx
15.
x dx
17.
∫ sin
6
x cos x dx
20.
∫
2
0
6
x cos x dx
18.
n
2
2
n →∞
n =1 n + r
23.
∫ sin
6
x dx
∫ sin
2
x cos 4 x dx
0
x
x + 5− x
p
2
dx
21.
∫ sin
0
1
n
22. lim ∑
x dx
p
2
7
0
3
8
6
0
p
2
9
p
2
∫ sin
∫x
∫ sec
p
2
n
0
0
19.
12.
1
3
p
2
16.
(1 − x 2 ) 2
9. ∫
dx
x6
x
8. ∫
dx
25 + 4 x 2
x ex
11. ∫
dx
( x + 1) 2
x
2
dx
∫0 ( x + 1)( x 2 + 1)
24.
3
sin 3 x
dx
x + cos3 x
x −1
∫ ( x + 1)
3
dx
0
OBJECTIVE TYPE QUESTIONS
A. Fill up the blanks
p
2
ln 3
ex
dx =___________
1. ∫
1+ ex
ln 2
p
2
3.
∫
0
sin x
sin x + cos x
dx =___________
M06_ENGINEERING_MATHEMATICS-I _CH06_PART B.indd 103
2.
sin x
∫ sin x + cos x dx
=___________
0
p
4.
∫ sin
3
x cos 4 x dx =___________
−p
5/19/2016 3:17:39 PM
■
6.104
7
5.
Engineering Mathematics
x
∫
x + 9−x
2
dx =___________
p
7.
p
2
6.
ex
2
0
∫
x⎞
⎛ 2x
⎜⎝ sec + 2 tan ⎟⎠ dx =___________
2
2
n
sin 4 x
dx =___________
sin x
0
8. lim ∑
∫
n →∞
r =1
1
=___________
n+r
9. The area of the region in the first quadrant bounded by y-axis and curves y = sin x and y = cos x
is______________
10. The length of the arc of the curve 6xy = x4 + 3 from x = 1 to x = 2 is___________
11. The area of the surface of the solid generated by the revolution of the line segment y = 2x from x = 0 to x = 2
about x-axis is__________
12. The area bounded by y2 = x and x2 = y is________
p
13. The length of the arc of the curve y = logc sec x between x = 0 and x = is ________
6
14. If the area of the curve y2 = 4x bounded by y = 0 and x = 1 is rotated about the line x = 1, then the volume of
the solid generated is____________
15. The surface area of the surface generated by the revolution of the line segment y = x + 1 from x = 0 to x = 2
about the x-axis is equal to ___________
B. Choose the correct answer
1.
∫
(sin x + cos x )
1 + sin 2x
(a) sin x
dx is equal to
(b) cos x
(c) x
(d) tan x
⎛ b⎞
(c) log e ⎜ ⎟ .log e ( ab)
⎝ a⎠
(d) None of these
b
2.
log e x
dx is equal to
x
a
∫
⎛ a⎞
1
⎛ b⎞
log e ⎜ ⎟ .log e ( ab) (b) log e ⎜ ⎟ .log e ( ab)
⎝ b⎠
⎝ a⎠
2
(a)
1
3.
∫ 5x − 3 dx
is
0
(a) −
8−x
5
4.
1
2
∫
x + 8−x
3
∞
x2
∫2
x
13
10
(c)
1
2
(d)
23
10
(b)
1
2
(c)
3
2
(d) 3
dx is
(a) 1
5.
(b)
dx is equal to
0
(a) (loge2)−2
(b) 2loge2
(c) 2(loge2)−3
⎡n
⎤
n
n
n
+… 2
6. nlim
⎢ + 2 2+ 2
2
2 ⎥ is equal to
→∞ n 2
n +1 n + 2
n + ( n − 1) ⎦
⎣
3p
p
p
(a)
(b)
(c)
4
4
3
M06_ENGINEERING_MATHEMATICS-I _CH06_PART B.indd 104
(d) None of these
(d) None of these
5/19/2016 3:17:44 PM
Integral Calculus ■
6.105
p
2
7.
∫ sin
5
x cos 7 x dx is equal to
p
−
2
(b) p
(a) 0
5p
4
(c)
(d) None of these
∞
8. The value of
m n
(a)
p
2
9.
2 m+n
∫ 1+
tanx
tan x
0
(a)
x n −1
∫0 (1 + x )m + n dx is equal to
(b)
m n
m+n
m n
(c)
m + n +1
(d) None of these
dx is equal to
p
2
(b)
p
4
3p
4
(c)
(d) 0
x
1
1
10. If the function f is continuous for all x ≥ 0 and satisfies ∫ f (t )dt = − + x 2 + x sin 2x + cos 2x then the
2
2
0
⎛ p⎞
value of f ′ ⎜ ⎟ is
⎝ 4⎠
(a) p − 2
(b) p + 2
(c) 2 − p
(d) − p
p
11. The length of the curve y = log sec x between the points with abscissae 0, is equal to
3
(a) log e ( 2 + 3)
(b) log e ( 3 + 1)
(c) log e ( 2 + 1)
(d) log e ( 2 + 3
12. The area bounded by the parabola y2 = 4ax and its latus rectum is given by
a
(a)
∫ y dx
0
a
(b) 2∫ 4axdx
a
(c)
0
y2
∫0 4a dy
a
(d) 2 ∫ 4axdx
−a
13. The area of the cardiod r = a(1−cosu) is given by
(a) 3pa2
(b) 6pa2
(c) pa2
(d)
3pa2
2
14. The volume of the solid obtained by revolving the area of the parabola y2 = 4ax cut off by the latus rectum
about the tangent at the vertex is given by
(a)
pa3
5
(b)
2pa3
5
(c)
4pa3
3
(d)
2pa3
3
15. The volume of the solid generated by the revolution of r = 2a cosu about the initial line is given by
(a)
2pa3
3
(b)
4pa3
3
M06_ENGINEERING_MATHEMATICS-I _CH06_PART B.indd 105
(c)
8pa3
3
(d) None of these
5/19/2016 3:17:50 PM
■
6.106
Engineering Mathematics
ANSWERS
A. Fill up the blanks
1. loge
4
3
2.
p
4
3.
p
4
p
6. e 2
11. 8 5p
7. 0
12.
1
3
8. ln 2
13.
B. Choose the correct answer
1. (c)
2. (a)
3. (b) 4. (a)
11. (d) 12. (b)
13. (d) 14. (c)
M06_ENGINEERING_MATHEMATICS-I _CH06_PART B.indd 106
1
log e 3
2
5. (c)
15. (b)
5.
5
2
10.
17
12
4. 0
9.
14.
6. (b)
2 −1
16p
15
7. (a)
15. 8 2p
8. (b)
9. (b)
10. (c)
5/19/2016 3:17:53 PM
7
Improper Integrals
7.1
IMPROPER INTEGRALS
b
The definite integral
∫ f (x )dx
is defined as the limit of a sum under two conditions (i) the interval
a
b
[a, b] is of finite length and (ii) f is defined and bounded on [a, b]. Then ∫ f ( x )dx is called a proper
a
integral.
But there are many practical problems where f is unbounded on [a, b] or the interval is not finite.
Such integrals are known as improper integrals.
1
dx
,
x
0
For example: ∫
7.1.1
∞
∞
dx
∫ 1+ x
2
∫e
,
−x
x 2dx are improper integrals.
−∞
0
Kinds of Improper Integrals and Their Convergence
(a) Improper integrals of the first kind
∞
If f is continuous and the interval is infinite, then the infinite integrals ∫ f ( x )dx ,
a
∞
b
∫
−∞
f ( x )dx ,
∫ f (x ) dx
−∞
are called improper integrals of the first kind.
∞
b
1. We define ∫ f ( x )dx = lim ∫ f ( x )dx
b →∞
a
a
b
if (i) the proper integral ∫ f ( x ) dx exists for every b > a
a
b
(ii) the limit lim ∫ f ( x ) exists with value equal to A, where A is finite.
b →∞
a
∞
Then ∫ f ( x )dx is said to converge to A.
a
∞
A is called the value of the integral and we write ∫ f ( x )dx = A .
a
∞
Otherwise, ∫ f ( x )dx is said to diverge.
a
b
2. We define
∫
−∞
b
f ( x )dx = lim ∫ f ( x )dx
a →−∞
a
b
if (i) the proper integral ∫ f ( x )dx exists for every a < b
a
M07_ENGINEERING_MATHEMATICS-I _CH07_Part A.indd 1
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7.2
■
Engineering Mathematics
b
(ii) the limit lim ∫ f ( x )dx exists with value equal to B, where B is finite
a →−∞
a
b
Then
∫ f (x )dx is said to converge to B.
−∞
b
B is called the value of the integral and we write
∫ f (x )dx = B .
−∞
b
Otherwise,
∫ f (x )dx is said to diverge.
−∞
3. We define
∞
∫
c
f ( x )dx =
−∞
∫
−∞
∞
f ( x )dx + ∫ f ( x )dx for some c,
c
if both the integrals on the R.H.S converge.
∞
Then
∫ f ( x )dx
is said to converge to the sum of the values.
2∞
∞
The integral
∫ f (x )dx is said to diverge if at least one of the integrals on the R.H.S diverges.
−∞
b
Note The integral ∫ f ( x )dx for every b ≥ a is analogus to “partial sum” in infinite series and so it may
a
be considered as “partial integral”.
(b) Improper integrals of the second kind
b
If the interval [a, b] is finite and f is unbounded at one or more points on [a, b], then ∫ f ( x )dx is called
a
an improper integral of the second kind.
1. If f is unbounded at a only (i.e., f has an infinite discontinuity at a), then we define
b
∫ f (x )dx = lim
∈→ 0 +
a
b
∫ f (x )dx ,
0 <∈< b − a ,
a +∈
b
if (i) the proper integral
∫ f (x ) dx exists
a +∈
(ii) the limit exists and is equal to A.
b
Then the integral ∫ f ( x )dx is said to converge to A.
a
b
A is called the value of the integral and we write ∫ f ( x )dx = A .
a
b
Otherwise, ∫ f ( x )dx is said to diverge.
a
M07_ENGINEERING_MATHEMATICS-I _CH07_Part A.indd 2
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Improper Integrals ■
7.3
2. If f is unbounded at b only, then we define
b
∫ f (x )dx = lim
∈→ 0 +
a
b −∈
∫ f (x )dx ,
0 <∈< b − a.
a
b −∈
if (i) the proper integral
∫ f (x ) dx exists,
a
b −∈
(ii) the limit lim
∈→ 0 +
∫ f (x )dx exists and is equal to B.
a
b
Then ∫ f ( x )dx is said to converge to B.
a
b
B is called the value of the integral and we write ∫ f ( x )dx = B
a
b
Otherwise, ∫ f ( x )dx is said to diverge.
a
3. If f is unbounded at c only, a < c < b, then we define
b
∫ f (x )dx = lim
∈→ 0 +
a
c −∈
∫
b
f ( x )dx + lim
d→ 0 +
a
∫ f (x )dx ,
c+d
if both the limits exist.
b
Then ∫ f ( x )dx is said to converge to the sum of the limits.
a
b
If at least one of the limits fail to exist, then ∫ f ( x )dx is said to diverge.
a
Remark:
Cauchy’s Principal Value
∞
1. If
∫
−∞
a
f ( x )dx is defined as lim ∫ f ( x )dx and if the limit exists with value A, then A is called
a →∞
−a
∞
Cauchy’s principal value (CPV) of
∫ f (x )dx .
−∞
∞
Note that Cauchy’s principal value of
∫ f (x )dx may exist even if the improper integral diverges.
−∞
2. Similarly, Cauchy’s principal value of
b
⎤
⎡c −∈
f
(
x
)
dx
=
lim
f
(
x
)
dx
+
f ( x )dx ⎥ if the R.H.S limits exist.
⎢
∫a
∫
∫
∈→ 0 +
c +∈
⎦
⎣a
b
3. If the improper integral converges to A, then CPV = A.
M07_ENGINEERING_MATHEMATICS-I _CH07_Part A.indd 3
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7.4
■
Engineering Mathematics
WORKED EXAMPLES
Problems based on improper integral of the first kind
EXAMPLE 1
∞
dx
, if it exists.
1
x2
1
0
Evaluate the improper integral ∫
Solution.
∞
b
dx
dx
= lim ∫
2
b →∞ 1 + x 2
+
x
1
0
0
I =∫
Let
= lim[tan −1 x ]b0
b →∞
= lim[tan −1 b − tan −1 0] = tan −1 ∞ − tan −1 0 =
b →∞
p
p
−0 =
2
2
EXAMPLE 2
∞
Evaluate ∫
1
1
dx if it exists.
x2
Solution.
∞
I =∫
Let
1
b
1
1
dx = lim ∫ 2 dx
b →∞ x
x2
1
b
⎡ x −2+1 ⎤
= lim ⎢
⎥
b →∞ ⎣ −2 + 1 ⎦
1
b
⎡ 1⎤
⎡ 1 ⎤ ⎡ 1 ⎤
= lim ⎢ − ⎥ = lim ⎢ − + 1⎥ = ⎢ − + 1⎥ = 1
b →∞ ⎣
x ⎦1 b→∞ ⎣ b ⎦ ⎣ ∞ ⎦
EXAMPLE 3
Evaluate
∞
∫ xe
2x 2
dx if it exists.
2∞
Solution.
I=
Let
∞
0
∫
xe − x dx =
2
−∞
−∞
0
I1 =
where
∫ xe
t = x2 ∴
2
2
0
0
−x2
−∞
Put
∫
∞
xe − x dx + ∫ xe − x dx = I 1 + I 2
dt = 2xdx
dx = lim ∫ xe − x dx
2
a →−∞
⇒
a
xdx =
1
dt
2
When x = a, t = a2 and when x = 0, t = 0
∴
0
2
⎡ e −t ⎤
1
dt 1
I 1 = lim ∫ e ⋅ = lim ⎢ ⎥ = − lim [e 0 − e − a ]
a →−∞
a →− ∞ ⎣ −1 ⎦ 2
a →− ∞
2
2
2
a
a2
1
1
1
= − [1 − e − ∞ ] = − [1 − 0] = −
2
2
2
0
−t
M07_ENGINEERING_MATHEMATICS-I _CH07_Part A.indd 4
5/12/2016 9:52:47 AM
Improper Integrals ■
∞
7.5
a
I 2 = ∫ xe − x dx = lim ∫ xe − x dx
2
2
and
a→ ∞
0
0
a2
= lim ∫ e −t
a→ ∞
0
1
dt
2
a2
⎡ e −t ⎤
1
= lim ⎢ ⎥
2 a→ ∞ ⎣ −1 ⎦ 0
2
1
1
1
1
= − lim [e − a − e 0 ] = − [e − ∞ − 1] = − [0 − 1] =
2
2 a→ ∞
2
2
I=
∴
∞
∫ xe
−x2
−∞
1 1
dx = − + = 0
2 2
EXAMPLE 4
∞
∫a
Evaluate
2
0
dx
, a > 0, if it exists.
1x 2
Solution.
∞
c
dx
dx
= lim ∫ 2
2
2
→∞
c
0 a +x
0 a +x
I =∫
Let
2
c
x⎤
1⎡
= lim ⎢ tan −1 ⎥
c→ ∞ a ⎣
a ⎦0
c
1⎡
1 p p
⎤ 1
= lim ⎢ tan −1 − tan −1 0 ⎥ = [ tan −1 ∞ − 0 ] = . =
c→ ∞ a ⎣
a
a 2 2a
⎦ a
EXAMPLE 5
0
Evaluate
∫ x sin xdx , if it exists.
2∞
Solution.
0
Let
I=
∫
−∞
0
x sin xdx = lim ∫ x sin xdx
a →−∞
a
= lim [x ( − cos x ) − ( − sin x )]a0
[ by Bernoulli’s formula,
a→ − ∞
= lim [ − x cos x + sin x ]a0
Here u = x , v = sin x ]
a→ − ∞
= lim [ 0 − ( −a cos a + sin a) ] = lim [a cos a − sin a]
a→ − ∞
a→ − ∞
But sin a and cos a oscillate finitely between −1 and +1.
Since a cos a → ∞ as a → ∞, limit is −∞
0
∴
the integral I =
∫ x sin xdx
diverges to −∞.
−∞
M07_ENGINEERING_MATHEMATICS-I _CH07_Part A.indd 5
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7.6
■
Engineering Mathematics
EXAMPLE 6
∞
Evaluate
dx
∫ x (log
e
e
x )3
, if convergent.
Solution.
∞
I =∫
Let
e
Put t = loge x
∴ dt =
a
dx
dx
= lim ∫
3
3
a
→∞
x (loge x )
e x (log e x )
1
dx
x
When x = e, t = loge e = 1 and when x = a, t = loge a
loge a
∴
I = lim
a →∞
∫
1
dt
= lim
t 3 a→∞
loge a
∫
(t −3 )dt
1
loge a
⎡ t −2 ⎤
= lim ⎢ ⎥
a → ∞ ⎣ −2 ⎦
1
log a
e
1
⎡1⎤
= − lim ⎢ 2 ⎥
2 a→ ∞ ⎣ t ⎦1
1
1
1
1⎡1 ⎤
1
⎡
⎤
= − lim ⎢
− 1 = − ⎢ − 1⎥ = − [0 − 1] =
2 a→ ∞ ⎣ (loge a) 2 ⎥⎦
2 ⎣∞ ⎦
2
2
EXAMPLE 7
∞
Evaluate ∫ e 2x sin xdx , if it exists.
0
Solution.
∞
Let
b
I = ∫ e − x sin xdx = lim ∫ e − x sin xdx
0
b →∞
0
b
⎤
⎡e −x
= lim ⎢
( − sin x − cos x ) ⎥
b→ ∞ ⎣ 2
⎦0
1
= − lim [e − x (sin x + cos x )]b0
2 b→ ∞
1
= − lim [e −b (sin b + cos b ) − e 0 (0 + 1)]
2 b→ ∞
1
1
1
= − [e − ∞ − 1] = − [0 − 1] =
2
2
2
M07_ENGINEERING_MATHEMATICS-I _CH07_Part A.indd 6
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Improper Integrals ■
7.7
EXAMPLE 8
∞
dx
, a > 0 , p > 0 is convergent if p > 1 and divergent if p ≤ 1.
p
a x
Prove that ∫
Solution.
∞
b
dx
= lim ∫ x − p dx , a > 0, p > 0
p
b →∞
x
a
a
I =∫
Let
b
⎡ x − p +1 ⎤
⎡ b − p +1 − a − p +1 ⎤
= lim ⎢
= lim ⎢
⎥
⎥
b →∞ − p + 1
⎣
⎦a b → ∞ ⎣ − p + 1 ⎦
p > 1,then −p + 1 < 0. ∴ as b → ∞, b−p + 1 → 0
0 − a − p +1
a1− p
=−
∴ I=
if p > 0
1− p
− p +1
If p < 1, then −p + 1 > 0. ∴ as b → ∞, b−p + 1 → ∞
∴ I = ∞ if p < 1
If
b
b
1
I = lim ∫ dx = lim [ loge x ]a
b →∞
b →∞ x
a
= lim[log b − log a] = ∞ − log a = ∞
When p = 1,
b →∞
∴
I is convergent if p > 1 and divergent if 0 < p ≤ 1
∞
dx
is convergent if p > 1 and divergent if 0 < p ≤ 1.
p
a x
That is ∫
∞
Note
1
∫x
2
dx is convergent, since p = 2 > 1.
1
EXAMPLE 9
∞
Evaluate
x 13
∫ ( x 21)( x
2
2
11)
dx .
Solution.
∞
Let
x +3
x +3
dx
dx = lim ∫
2
2
b
→∞
2 ( x − 1)( x + 1)
2 ( x − 1)( x + 1)
b
I =∫
Let
x +3
A
Bx + C
=
+
( x − 1)( x 2 + 1) x − 1 x 2 + 1
x + 3 = A(x2 + 1) + (Bx + C)(x − 1)
⇒
Put x = 1, then
4 = A(1 + 1) ⇒
Put x = 0, then
2
Equating coefficients of x ,
2A = 4 ⇒ A = 2
3=A−C
⇒
C = A − 3 = 2 − 3 = −1
0=A+B
⇒
B = −A = −2
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∴
x +3
2
−2x − 1
2
2x
1
=
+ 2
=
− 2
− 2
2
( x − 1)( x + 1) x − 1 x + 1 x − 1 x + 1 x + 1
2x
1 ⎤
⎡ 2
I = lim ∫ ⎢
− 2
− 2 ⎥ dx
b →∞ ⎣ x − 1
x
x
+
1
+ 1⎦
2
b
∴
b
= lim ⎡⎣ 2 loge ( x − 1) − loge ( x 2 + 1) − tan −1 x ⎤⎦ 2
b →∞
b
= lim ⎡⎣loge ( x − 1) 2 − loge ( x 2 + 1) − tan −1 x ⎤⎦ 2
b →∞
b
⎡
⎤
⎡ ( x − 1) 2 ⎤
−1
= lim ⎢loge ⎢ 2
⎥ − tan x ⎥
b →∞ ⎣
⎣ x +1 ⎦
⎦2
⎡
⎤
⎛ ( 2 − 1) 2 ⎞
⎛ (b − 1) 2 ⎞
−1
= lim ⎢loge ⎜ 2
− tan −1 b − loge ⎜ 2
⎟⎠ − ( − tan 2) ⎥
⎟
⎝
b →∞
⎝
⎠
b +1
2 +1
⎣
⎦
⎡
⎢
= lim ⎢loge
b →∞ ⎢
⎢
⎣
⎡
⎢
= lim ⎢loge
b →∞ ⎢
⎢
⎣
⎤
⎛ 2 ⎛ 1⎞ 2 ⎞
⎥
⎜ b ⎜⎝1 − b ⎟⎠ ⎟
1
⎜
⎟ − tan −1 b − loge + tan −1 2⎥
⎥
5
⎜ b 2 ⎛1 + 1 ⎞ ⎟
⎟⎟
⎜⎝ ⎜⎝
⎥
b2 ⎠ ⎠
⎦
⎤
⎛ ⎛ 1⎞ 2 ⎞
⎥
⎜ ⎜⎝1 − b ⎟⎠ ⎟
1
⎜
⎟ − tan −1 b − loge + tan −1 2⎥
⎥
5
⎜ ⎛1 + 1 ⎞ ⎟
⎟⎟
⎜⎝ ⎜⎝
⎥
b2 ⎠ ⎠
⎦
1
+ tan −1 2
5
p
⎛p
⎞
= − + tan −1 2 + loge 5 = loge 5 − ⎜ − tan −1 2⎟ = loge 5 − cot −1 2
⎝
⎠
2
2
= loge 1 − tan −1 ∞ − loge
EXAMPLE 10
∞
dv
.
21
v)
0 (11v )(11 tan
Evaluate the improper integral ∫
2
Solution.
Let
∞
dv
2
+
v
+ tan −1 v )
(
1
)(
1
0
I =∫
Here
f (v ) =
1
(1 + v 2 )(1 + tan −1 v )
When v = 0,
f ( 0) =
1
=1
(1 + 0)(1 + 0)
∴ v = 0 is a point of continuity of f(v).
∴
M07_ENGINEERING_MATHEMATICS-I _CH07_Part A.indd 8
b
dv
−1
0 (1 + v )(1 + tan v )
I = lim ∫
b →∞
2
5/12/2016 9:53:07 AM
Improper Integrals ■
7.9
b
dv
2
v
(
1
+
)(
1
+ tan −1 v )
0
I1 = ∫
Let
Put
t = tan−1v
∴ dt =
1
dv
1+ v 2
When v = 0, t = tan −1 0 = 0 and when v = b , t = tan −1 b
tan −1 b
∴
I1 =
∫
0
dt
tan −1 b
= [ loge (1 + t ) ]0
1+ t
= loge (1 + tan −1 b ) − loge 1 = loge (1 + tan −1 b )
∴
p
I = lim ⎡⎣loge (1 + tan −1 b ) ⎤⎦ = loge (1 + tan −1 ∞) = loge ⎛⎜1 + ⎞⎟
b→ ∞
⎝
2⎠
Problems based on improper integral of the second kind
EXAMPLE 11
3
dx
Evaluate the improper integral ∫
9 2x 2
0
, if it exists.
Solution.
3
dx
I =∫
Let
9− x2
0
Here the integrand is
1
f (x ) =
9− x2
1
= ∞. ∴ f(x) is unbounded when x = 3.
0
∴ the integrand is unbounded when x = 3 and the interval [0, 3] is finite.
So, it is an improper integral of the second kind.
When x = 3, f (3) =
3
∴
I =∫
0
dx
9−x
2
= lim
∈→ 0 +
3 −∈
∫
0
dx
9− x2
3 −∈
x⎤
⎡
= lim ⎢sin −1 ⎥
∈→ 0 + ⎣
3 ⎦0
3
−∈
p
⎡
⎤
− sin −1 0 ⎥ = sin −1 1 − 0 =
= lim ⎢sin −1
∈→ 0 + ⎣
3
2
⎦
[as ∈→ 0, 3− ∈→ 3]
EXAMPLE 12
2
dx
, if it exists.
x
21
(
)2/3
0
Evaluate the improper integral ∫
Solution.
2
Let
dx
x
−
(
1) 2 / 3
0
I =∫
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Here the integrand is f ( x ) =
When x = 1,
f (1) =
1
( x − 1) 2 / 3
1
=∞
0
∴ f(x) is unbounded at x = 1.
∴ the integrand is unbounded when x = 1 and the interval [0, 2] is finite.
Hence, it is an improper integral of the second kind.
2
∴
1
2
dx
dx
dx
+∫
2/3 = ∫
2/3
2/3
x
−
(
1
)
0
0 ( x − 1)
1 ( x − 1)
I =∫
= lim
1−∈
∫
∈→ 0 +
= lim
0
1−∈
∫
∈→ 0 +
2
dx
dx
+ lim ∫
2/3
2/3
0
d
→
+
( x − 1)
1+ d ( x − 1)
2
( x − 1) −2 / 3 dx + lim
d→ 0 +
0
∫ (x − 1)
−2 / 3
dx
1+ d
1− ∈
2
⎡ ( x − 1)1/ 3 ⎤
⎡ ( x − 1)1/ 3 ⎤
= lim ⎢
+ lim ⎢
⎥
⎥
∈→ 0 ⎣
d→ 0 ⎣
1 / 3 ⎦0
1 / 3 ⎦1+ d
= 3 lim[(1− ∈−1)1/ 3 − ( −1)1/ 3 ] + 3 lim[11/ 3 − (1 + d − 1)1/ 3 ]
∈→ 0
d→ 0
= 3 lim[( − ∈)
1/ 3
∈→ 0
− ( −1)] + 3 lim (1 − d1/ 3 )
d→ 0
[{ ( −1)1/ 3 = −1]
= 3[0 + 1] + 3[1 − 0] = 6
EXAMPLE 13
2
Evaluate the improper integral ∫
0
dx
, if it exists.
2 x 2x 2
Solution.
2
I =∫
Let
0
Here the integrand is f ( x ) =
2
dx
dx
=
2x − x 2 ∫0 x ( 2 − x )
1
x (2 − x )
1
1
= ∞ and when x = 2, f ( 2) = = ∞
0
0
∴ f(x) is unbounded at x = 0 and x = 2
∴ the integrand is unbounded and the interval [0, 2] is finite.
Hence, it is an improper integral of the second kind.
2
1
2
dx
dx
dx
∴
I =∫
=∫
+∫
(
2
)
(
2
)
(
2
x
−
x
x
−
x
x
− x)
0
0
1
When x = 0, f (0) =
1
dx
+ lim
∈→ 0 ∫ x ( 2 − x )
d→ 0
0 +∈
= lim
M07_ENGINEERING_MATHEMATICS-I _CH07_Part A.indd 10
2−d
∫
1
dx
= I1 + I 2
x (2 − x )
5/12/2016 9:53:18 AM
Improper Integrals ■
1
A
B
= +
x (2 − x ) x 2 − x
Let
7.11
⇒ 1 = A ( 2 − x ) + Bx
A=
1
2
Putting x = 0,
1 = 2A ⇒
∴
1
1 1 1
1
= ⋅ + ⋅
x (2 − x ) 2 x 2 (2 − x )
⎛1 1
dx
and
1
Putting x = 2,
1 = 2B
⇒ B=
1
2
⎞
∫ x (2 − x ) = ∫ ⎜⎝ 2 ⋅ x + 2(2 − x ) ⎟⎠ dx
∴
=
1
1
1
⎛ x ⎞
loge x − loge ( 2 − x ) = loge ⎜
⎝ 2 − x ⎟⎠
2
2
2
1
x ⎤
dx
⎡1
= lim ⎢ loge
∈→ 0 ∫ x ( 2 − x )
∈→ 0 ⎣ 2
2
x ⎥⎦ 0 +∈
−
0 +∈
1
I 1 = lim
Now,
(1)
[using (1)]
⎧1 ⎡
1
0+ ∈ ⎤ ⎫
= lim ⎨ ⎢loge
− loge
⎬
∈→ 0 2
2
−
1
2
−
(0 + ∈) ⎥⎦ ⎭
⎩ ⎣
⎧1 ⎡
∈ ⎤⎫
= lim ⎨ ⎢loge 1 − loge
⎬
∈→ 0 ⎩ 2 ⎣
2 − ∈⎥⎦ ⎭
⎧
1
1
⎛ ∈ ⎞⎫
= − lim ⎨loge ⎜
⎟⎠ ⎬ = − loge 0 = ∞
⎝
∈→
0
2
2− ∈ ⎭
2
⎩
I 2 = lim
and
d→ 0
2−d
∫
1
2− d
dx
⎡1
⎛ x ⎞⎤
= lim log
x ( 2 − x ) d→ 0 ⎢⎣ 2 e ⎜⎝ 2 − x ⎟⎠ ⎥⎦1
=
⎡
1
1 ⎤
⎛ 2−d ⎞
− loge
lim ⎢loge ⎜
⎟
d
→
0
⎝ 2 − ( 2 − d) ⎠
2
2 − 1⎥⎦
⎣
=
⎡
⎤
1
⎛ 2 − d⎞
lim loge ⎜
− loge 1⎥
⎝ d ⎟⎠
2 d→ 0 ⎢⎣
⎦
=
1
2
1
lim loge − 0 = loge ∞ = ∞
d→
0
2
0
2
[using (1)]
I = I1 + I 2 = ∞
∴
i.e., the limit does not exist.
2
∴
the integral ∫
0
dx
is divergent.
x (2 − x )
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EXAMPLE 14
1
Evaluate the improper integral
∫ log
e
| x | dx .
21
Solution.
1
1
−1
0
I = ∫ loge x dx = 2∫ loge x dx
Let
⎡⎣loge x is even ⎤⎦
1
= 2∫ loge x dx
⎡⎣ x = x if x ≥ 0 ⎤⎦
0
Here f ( x ) = loge x
As x → 0+, log e x → ∞. So, f(x) is unbounded at x = 0.
∴ the integrand is unbounded and the interval [−1, 1] is fininte.
Hence, it is an improper integral of the second kind.
1
1
⎫⎪
⎧⎪
1
1
I = lim ∫ loge xdx = lim ⎨[ loge x ⋅ x ]∈ − ∫ ⋅ x dx ⎬
∈→ 0 +
∈→ 0 +
x
∈
∈
⎭⎪
⎩⎪
But
= lim {0 − ∈ loge ∈−[x ]1∈}
∈→ 0 +
= lim [ − ∈ loge ∈−(1− ∈)]
∈→ 0 +
= lim [ − ∈ loge ∈−1 + ∈]
∈→ 0 +
= 2 ⎡ lim [ − ∈ loge ∈] − 1 + 0 ⎤
⎣⎢ ∈→ 0 +
⎦⎥
= 2 lim [ − ∈ loge ∈] − 2
∈→ 0 +
loge ∈
1/ ∈
1/ ∈
= lim
= lim ( − ∈) = 0
∈→ 0+ −1/ ∈2
∈→ 0 +
I = 2[0] − 2 = −2
⎡∞
⎤
⎢⎣ ∞ form ⎥⎦
lim (∈ loge ∈) = lim
Now
∈→ 0 +
∴
∈→ 0 +
[ by L-Hopital’s rule]
EXAMPLE 15
1
Evaluate
∫
0
x
12 x 2
dx .
Solution.
Let
1
x
I =∫
dx
1− x 2
Hence, it is an improper integral of the second kind
0
f (x ) =
Here
When x = 1, f (1) =
1
= ∞.
0
x
1− x 2
∴ f(x) is unbounded at x = 1.
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Improper Integrals ■
7.13
∴ the integrand is unbounded and the interval is finite
∴
I = lim
∈→ 0 +
1−∈
∫
x
1− x 2
0
dx = lim
1−∈
∫
∈→ 0 +
0
1
−
1
− (1 − x 2 ) 2 ( −2x ) dx
2
1− ∈
1
⎡
2 2 ⎤
x
1
1
(
−
)
⎥
= lim ⎢ −
⎥
∈→ 0 + ⎢ 2
1
⎥
⎢
⎦0
⎣
2
1− ∈
1
⎡
⎤
= − lim ⎢⎣(1 − x 2 ) 2 ⎥⎦ 0
∈→ 0 +
{
}
1
1
⎡
⎤
= − lim ⎢⎣[1 − (1− ∈) 2 ] 2 − 1⎥⎦ = − (1 − 1) 2 − 1 = −(0 − 1) = 1
∈→ 0 +
1
∫
∴
0
x
1− x 2
dx = 1
EXERCISE 7.1
Test the convergence of the following improper integrals using definition. Find the value if convergent.
∞
∞
∞
dx
1
1. ∫ 2
2. ∫ p dx , p > 0
3. ∫ x − ( 3/ 2 )dx
1
1 x
1 x
∞
4.
x
∫−∞ 1 + x 4 dx
2
7.
∫
0
p
2
10.
2
dx
8. ∫
2
1 (1 − x )
1
∫ tan xdx
11.
∫ x sin x dx
14.
∫
−1
p
−
2
∞
13.
dx
∫−∞ x 2 + 2x + 5
6.
∞
0
∫
2
∫e
1
∞
2
dx
4−x
∞
∞
5.
9.
∫e
x
2
dx
+4
− px
dx , p > 0
0
1+ x
dx
1− x
1
12.
dx
∫x
p
,p>0
0
x dx
( x 2 − 3)3
ANSWERS TO EXERCISE 7.1
1. 1
2.
1
if p > 1, divergent if p ≤ 1
p −1
3.
4. 0
5.
p
2
6. Divergent
p
2
10. Divergent
1
12.
if p < 1, Divergent if p ≥ 1
p −1
7.
M07_ENGINEERING_MATHEMATICS-I _CH07_Part A.indd 13
8. Divergent
11. p
13. Divergent
9.
1
2
1
if p > 0
p
14. 1
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7.1.2
Tests of Convergence of Improper Integrals
Engineering Mathematics
As in the case of series of positive terms, we have tests for convergence of improper integrals with
positive integrand.
(a) Tests of convergence of improper integrals of the first kind
We state the following theorems without proof.
Theorem 7.1 Let f ( x ) > 0 ∀ x ≥ a. Then
∞
∫ f (x )dx
is convergent if and only if there exists a
a
b
constant M > 0 such that ∫ f ( x )dx ≤ M
∀ b ≥ a.
a
Comparison Test
Theorem 7.2 If f(x) and g(x) are both positive and continuous on [a, ∞) and 0 ≤ f ( x ) ≤ g ( x ) ∀ x ≥ a,
∞
∞
∞
∞
a
a
a
a
then ∫ f ( x )dx converges if ∫ g ( x )dx converges and ∫ f ( x )dx ≤ ∫ g ( x )dx .
∞
∞
a
a
Note The above result is equivalently, if ∫ f ( x )dx diverges, then ∫ g ( x )dx diverges.
Limit form of Comparison Test
Theorem 7.3 If f(x) and g(x) are both positive and continuous on [a, ∞) such that lim
∞
∞
a
a
x →∞
f (x )
= l , 0 < l < ∞,
g (x )
then ∫ f ( x )dx and ∫ g ( x )dx converge or diverge together.
That is both the integrals behave alike.
∞
∞
∞
∞
a
a
a
a
If ∫ g ( x ) dx is convergent, then ∫ f ( x )dx is convergent, and if ∫ g ( x )dx is divergent, then ∫ f ( x )dx
is divergent.
∞
∞
f (x )
= 0, then ∫ f ( x )dx is convergent if ∫ g ( x )dx is convergent.
x →∞ g ( x )
a
a
Note If lim
As in the case of series, here also we consider the following improper integrals for comparison,
∞
dx
, a > 0 and p > 0 is convergent if p > 1 and divergent if 0 < p ≤ 1.
p
a x
1. The p-integral, ∫
∞
2.
∫e
− ax
dx is convergent if a > 0 and divergent if a ≤ 0.
0
Absolute Convergence
∞
∞
a
a
The improper integral ∫ f ( x )dx is said to be absolutely convergent if ∫ f ( x ) dx is convergent.
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Improper Integrals ■
7.15
WORKED EXAMPLES
EXAMPLE 1
∞
x
dx .
1
x4
1
1
Test the convergence of ∫
Solution.
∞
x
∫ 1+ x
Given
dx
4
1
f ( x) =
Here
f (x )
=
g (x )
∴
lim
x
=
1 + x4
x
1⎞
⎛
x ⎜1 + 4 ⎟
⎝ x ⎠
4
1
1⎞
⎛
x 3 ⎜1 + 4 ⎟
⎝ x ⎠
f (x )
= lim
g (x ) x →∞
1
⋅x3 =
Take g ( x ) =
1
.
x3
1
1+
1
x4
∞
∫ f (x ) dx and
1
∞
1
1
⋅
.
3
1⎞
x ⎛
+
1
⎜⎝
⎟
x4 ⎠
= 1 ( ≠ 0)
1
1+ 4
x
∴ by the comparison test,
x →∞
=
∞
∫ g (x ) dx
behave alike.
1
∞
dx
is convergent by p-integral, since p = 3 > 1.
3
1 x
But ∫ g ( x )dx = ∫
1
∴
∞
∞
1
1
xdx
∫ f (x )dx = ∫ 1 + x
EXAMPLE 2
∞
4
is convergent.
dx
Test the convergence of ∫
1 (11 x ) x
.
Solution.
∞
Given
dx
∫ (1 + x )
1
Here
∴
f ( x) =
f (x )
=
g (x )
x
1
(1 + x ) x
x
3/ 2
.
=
1
.
⎡ 1⎤
x 3/ 2 ⎢1 + ⎥
⎣ x⎦
Take g ( x ) =
1
.
x 3/ 2
1
1
⋅ x 3/ 2 =
1
⎡ 1⎤
1+
1+
x
⎣⎢ x ⎥⎦
M07_ENGINEERING_MATHEMATICS-I _CH07_Part A.indd 15
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Engineering Mathematics
∴
lim
x →∞
f (x )
1
= lim
= 1 ( ≠ 0)
x
→
∞
1
g (x )
1+
x
∴ by the comparison test,
∞
∫ f (x ) dx and
1
∞
∞
1
1
But ∫ g ( x )dx = ∫
∴
∞
∫ g (x ) dx
behave alike.
1
dx
3
is convergent by p-integral, since p = > 1.
3/ 2
x
2
∞
∞
dx
∫1 f (x )dx = ∫1 (1 + x ) x is convergent.
EXAMPLE 3
∞
Test the convergence of ∫ e 2x dx .
2
0
Solution.
∞
∞
1
I = ∫ e − x dx = ∫ e − x dx + ∫ e − x dx = I 1 + I 2
2
Let
2
0
2
0
1
1
I 1 = ∫ e − x dx is proper integral
2
where
0
∞
I 2 = ∫ e − x dx is an improper integral of the first kind
2
and
1
we shall test the convergent of I2
f ( x ) = e − x . Consider g ( x ) =
2
Here
1
x2
2
f (x )
x2
= x 2e − x = x 2
g (x )
e
∴
∴
lim
x →∞
⎤
⎡∞
⎢ ∞ form ⎥
⎣
⎦
f ( x)
x2
= lim x 2
g ( x ) x →∞ e
= lim
x →∞
2x
e ⋅ 2x
x2
∞
= lim
x →∞
1
e
x2
=
1
=0
∞
dx
is convergent by p-integral, since p = 2 > 1.
2
1 x
But ∫
[L-Hopital’s rule]
∞
∴ I 2 = ∫ e − x dx is convergent.
2
1
I = I 1 + I 2 is convergent.
∴
∞
Hence, ∫ e − x dx is convergent.
2
0
M07_ENGINEERING_MATHEMATICS-I _CH07_Part A.indd 16
5/12/2016 9:54:20 AM
Improper Integrals ■
EXAMPLE 4
7.17
∞
Test the convergence of
sin 2 x
∫0 x 2 dx .
Solution.
∞
∞
1
sin 2 x
sin 2 x
sin 2 x
dx
=
dx
+
∫0 x 2
∫1 x 2 dx .
x2
0
I =∫
Let
sin 2 x
= 1.
x →0
x2
But
lim
∴ 0 is not a point of infinite discontinuity.
1
sin 2 x
dx is a proper-integral and it has finite value.
x2
0
∞
sin 2 x
So, we shall test the convergence of ∫ 2 dx , which is an improper integral.
x
1
Hence, ∫
We know
sin 2 x
1
≤ 2
x2
x
∀ x ≥1 ∴
∞
∞
sin 2 x
1
∫1 x 2 dx ≤ ∫1 x 2 dx .
∞
dx
is convergent by p-integral, since p = 2 > 1
2
1 x
But ∫
∞
sin 2 x
∫1 x 2 dx is convergent, by the comparison test
∴
∞
sin 2 x
dx is convergent.
x2
0
Hence, ∫
EXAMPLE 5
∞
Test the convergence of
∫
1
log x
dx .
x2
Solution.
∞
log x
dx .
2
1 x
∫
Given
log x
, x ≥ 1.
x2
Here
f (x ) =
∴
f ( x ) log x 3/ 2 log x
.
= 2 ⋅x =
g (x )
x
x
∴
lim
x →∞
f (x )
log x
= lim
g ( x ) x →∞ x
M07_ENGINEERING_MATHEMATICS-I _CH07_Part A.indd 17
Consider g ( x ) =
1
.
x 3/ 2
⎤
⎡∞
⎢⎣ ∞ form ⎥⎦
5/12/2016 9:54:29 AM
7.18
■
Engineering Mathematics
1
2
= lim x = lim
=0
x →∞
x →∞
1
x
2 x
∞
∞
1
1
But ∫ g ( x )dx = ∫
[by L’Hopital’s Rule]
dx
3
is convergent by p-integral, since p = > 1.
x 3/ 2
2
∞
∞
log x
dx is convergent.
2
1 x
∫ f (x )dx = ∫
∴
1
EXAMPLE 6
∞
Test the convergence of
∫x
1
dx
11 x 2
.
Solution.
∞
∫x
Given
1
f (x ) =
Here
f (x )
=
g (x )
∴
lim
x →∞
dx
1+ x 2
1
x 1+ x 2
.
1
x 2 1+
f (x )
= lim
g (x ) x →∞
1
=
1
x2
1
1
1+ 2
x
x 2 1+
⋅x2 =
.
1
x2
Take g ( x ) =
1
x2
1
1+
1
x2
= 1 ( ≠ 0)
∴ by the comparison test,
∞
∫ f (x ) dx and
1
∞
∞
1
1
But ∫ g ( x ) dx = ∫
∫ g (x ) dx
behave alike.
1
1
dx is convergent by p-integral, since p = 2 > 1.
x2
∞
∞
1
1
∫ f (x )dx = ∫ x
∴
EXAMPLE 7
∞
dx
1+ x 2
is convergent.
∞
dx
.
x
e
11
0
Test the convergence of ∫
M07_ENGINEERING_MATHEMATICS-I _CH07_Part A.indd 18
5/12/2016 9:54:34 AM
Improper Integrals ■
7.19
Solution.
∞
∫e
Given
0
dx
.
+1
Here
x
e x +1 ≥ e x
We know
∞
∀x ≥0
⇒
f (x ) =
1
.
ex +1
1
1
≤ x
e +1 e
x
∞
1
dx
∫0 e x + 1 dx ≤ ∫0 e x
∴
∞
∞
dx
∫e
But
= ∫ e − x dx is convergent, since a = 1 > 0.
x
0
0
∞
∫e
∴
0
dx
is convergent, by comparison test.
+1
x
EXAMPLE 8
∞
Test the convergence of
cos x
∫ 11 x
2
dx .
0
Solution.
∞
cos x
∫ 1+ x
Given
2
dx .
f (x ) =
Here
0
cos x
1
≤
2
1+ x
1+ x 2
But
∞
dx
∫ 1+ x
Now
∀x ≥0
∴
∞
= [ tan −1 x ]0 = tan −1 ∞ − tan −1 0 =
0
∞
dx
∫0 1 + x 2 is convergent.
∴
∞
Hence,
cos x
∫ 1+ x
2
2
0
∞
2
cos x
∫ 1+ x
∞
⇒
∫
0
cos x
1+ x 2
∞
1
dx
2
0 1+ x
dx ≤ ∫
p
p
−0 = .
2
2
cos x
dx is convergent.
1+ x 2
dx is absolutely convergent.
0
(b) Test of convergence of Improper Integrals of the second kind
We state the following theorems without proof.
b
Theorem 7.4 Let f(x) be positive and integrable in a < x ≤ b and f(x) is unbounded at a. Then ∫ f ( x )dx
b
will converge, if there exists a positive number M such that
∫ f (x )dx < M
a
∀ ∈, where 0 <∈< b − a.
a +∈
M07_ENGINEERING_MATHEMATICS-I _CH07_Part A.indd 19
5/12/2016 9:54:43 AM
■
7.20
Engineering Mathematics
Comparison Test
b
Theorem 7.5 If f(x) and g(x) be positive and integrable in a < x ≤ b and f(x) ≤ g(x) then
∫ g (x )dx converges.
converges if
∫ f (x )dx
a
b
a
Limit form of Comparison Test
Theorem 7.6 Let f(x) and g(x) be positive and integrable in a < x ≤ b. If lim
x →a +
b
b
∫ f (x )dx
f (x )
= c ( ≠ 0), then
g (x )
and
∫ g (x )dx
behave alike.
a
a
That is both converge or diverge.
b
b
a
a
∫ g (x )dx is convergent, then ∫ f (x )dx is convergent.
If
b
b
a
a
∫ g (x )dx is divergent, then ∫ f (x )dx is divergent.
If
Note
f (x )
= 0 and
x →a + g ( x )
1. If lim
b
b
a
a
∫ g (x )dx converges, then ∫ f (x )dx converges.
f (x )
= ∞ (or − ∞) and
x →a + g ( x )
2. If lim
b
b
a
a
∫ g (x )dx diverges, then ∫ f (x )dx
diverges.
Improper Integrals for Comparison
b
1.
dx
∫ (x − a)
l
converges if l < 1 and diverges if l ≥ 1
a
b
2.
dx
∫ (b − x )
m
converges if m < 1 and diverges if m ≥ 1.
a
WORKED EXAMPLES
EXAMPLE 1
1
Test the convergence of the improper integral ∫
0
x
1/ 3
dx
.
(11 x )
Solution.
1
Given
∫x
0
1/ 3
dx
.
(1 + x )
M07_ENGINEERING_MATHEMATICS-I _CH07_Part A.indd 20
Here
f (x ) =
1
.
x (1 + x )
1/ 3
5/12/2016 9:54:51 AM
Improper Integrals ■
1
=∞
0
1
g ( x ) = 1/ 3
x
When x = 0,
∴ f(x) is unbounded at x = 0
f ( 0) =
Take
7.21
f (x )
x 1/ 3
1
= lim 1/ 3
= lim
= 1 ( ≠ 0)
x →0 g (x )
x →0 x
x
→0
1+ x
(1 + x )
∴
lim
∴ by the comparison test,
1
∫ f (x )dx and
0
1
But
∴
1
dx
∫ g (x )dx = ∫ x
0
0
1
1
∫ f (x )dx = ∫ x
0
1/ 3
1/ 3
0
1
∫ g (x )dx
behave alike.
0
is convergent, since l =
1
<1
3
dx
is convergent
(1 + x )
EXAMPLE 2
1
Test the convergence of the improper integral ∫
0
dx
.
x (11 x ) 3
2
Solution.
1
∫x
Given
2
0
When x = 0, f (0) =
1
=∞
0
dx
.
(1 + x )3
Here
1
x2
g (x ) =
∴
f (x )
1
1
= 2
⋅x2 =
3
g ( x ) x (1 + x )
(1 + x )3
lim
x →0
1
.
x (1 + x )3
2
∴ f(x) is unbounded when x = 0
Take
∴
f ( x) =
f (x )
1
= lim
g ( x ) x → 0 (1 + x )3
= 1 ( ≠ 0)
∴ by the limits form of comparison test,
1
∫ f (x )dx and
0
But
∴
1
1
0
0
1
1
1
∫ g (x )dx = ∫ x
∫ f (x )dx = ∫
0
M07_ENGINEERING_MATHEMATICS-I _CH07_Part A.indd 21
0
2
1
∫ g (x )dx behave alike.
0
dx is divergent, since l = 2 > 1
dx
is divergent.
x (1 + x )3
2
5/12/2016 9:55:18 AM
7.22
■
Engineering Mathematics
EXAMPLE 3
Test the convergence and evaluate the improper integral
2
x
1
e
∫ log
x
dx .
Solution.
2
Given
x
∫ log
1
e x
dx .
When x = 1, f ( x ) =
f (x ) =
Here
1
=∞
loge 1
x
, x ≥1
loge x
∴ f(x) is unbounded when x = 1
1
x loge x
Take
g (x ) =
∴
f (x )
x
=
⋅ x loge x = x 3/ 2
g ( x ) loge x
∴
lim
x →1
f (x )
= lim x 3/ 2 = 1 ( ≠ 0)
g ( x ) x →1
∴ by the limits comparison test,
2
∫ f (x )dx and
1
2
2
1
1
∫ g (x )dx behave alike.
1
1
∫ g (x )dx = ∫ x log
But
2
e
x
dx
2
2
1
1/x
dx = lim ∫
dx
∈→ 0 ∫ x log x
∈→ 0
log
e
e x
1+∈
1+∈
= lim
= lim [ loge (loge x ) ]1+∈
2
∈→ 0
= lim [ loge (loge 2) − log (loge 1 + ∈) ]
∈→ 0
= loge loge 2 − log (loge 1) = loge loge 2 − log 0 = ∞
2
∴
∫ g (x )dx is divergent.
1
2
2
x
1
1
e
∫ f (x ) dx = ∫ log
∴
EXAMPLE 4
1
Test the convergence of the improper integral
x
dx is divergent.
dx
∫x
2
.
21
Solution.
1
Given
dx
∫x
2
.
−1
M07_ENGINEERING_MATHEMATICS-I _CH07_Part A.indd 22
Here
f (x ) =
1
.
x2
5/12/2016 9:55:25 AM
Improper Integrals ■
When x = 0, f ( x ) =
7.23
1
=∞
0
∴ f(x) has infinite discontinuity at x = 0. (f(x) is unbounded at x = 0).
It is improper integral of second kind.
1
0
∈
1
1
dx
dx
dx
dx
dx
lim ∫ 2 + lim ∫ 2
∫−1 x 2 = −∫1 x 2 + ∫0 x 2 = ∈→
0−
d
→
0
+
−1 x
d x
Hence,
∈
1
⎡ x −1 ⎤
⎡ x −1 ⎤
+ lim ⎢
= lim ⎢
⎥
⎥
∈→ 0 − ⎣ −1 ⎦
d → 0 + ⎣ −1 ⎦
−1
d
∈
1
⎡1⎤
⎡1⎤
= − lim ⎢ ⎥ − lim ⎢ ⎥
d→ 0 + ⎣ x ⎦
∈→ 0 − ⎣ x ⎦
d
−1
⎡ 1⎤
⎤
⎡1
= − lim ⎢ − ( −1) ⎥ − lim ⎢1 + ⎥ = ∞ + ∞ = ∞
∈→ 0 − ⎣ ∈
d
→
0
+
⎦
⎣ d⎦
∴ the given integral is divergent.
1
dx
is divergent.
2
−1 x
∴∫
EXAMPLE 5
∞
Prove that ∫ sin ( x 2 )dx converge.
0
Solution.
∞
I = ∫ sin ( x 2 )dx
Let
0
The function is bounded, but the interval is infinite.
So, it is improper integral of the first kind.
Put
t = x2
∴ t >0
When x = 0, t = 0 and when
∞
I = ∫ sin t
0
lim
t →0+
⇒ dx =
dt
dt
=
2x 2 t
x = ∞, t = ∞
∴
But
dt = 2xdx
and
sin t
t
dt
2 t
=
∞
1 sin t
dt .
2 ∫0 t
⎛ sin t ⎞
= lim t ⎜
= 0 ⋅1 = 0
⎝ t ⎟⎠
t →0+
∴ 0 is not a discontinuity.
∴
M07_ENGINEERING_MATHEMATICS-I _CH07_Part A.indd 23
I=
1
2
p/2
∫
0
sin t
t
dt +
∞
1 sin t
dt = I 1 + I 2
2 p∫/ 2 t
5/12/2016 9:55:30 AM
■
7.24
Engineering Mathematics
I1 =
where
I2 =
and
1
2
p/2
∫
sin t
0
t
dt is a proper integral.
∞
1 sin t
dt is improper integral.
2 p∫/ 2 t
=
∞
∞
1 ⎪⎧ −1/ 2
1
⎪⎫
⋅ ( − cos t ) ⎤⎦ p / 2 − ∫ − ⋅ t −3/ 2 ( − cos t )dt ⎬
⎨ ⎡⎣t
2 ⎩⎪
2
p/ 2
⎭⎪
=
∞
∞
1 ⎪⎧
1 cos t ⎪⎫
1 cos t
⎨0 − ∫ 3/ 2 dt ⎬ = − ∫ 3/ 2 dt
2 ⎪⎩
2 p/2 t
4 p/2 t
⎪⎭
∞
But
∴
1
3
cos t
1
≤ 3/ 2 and ∫ 3/ 2 dt is convergent by p-integral, since p = > 1
3/ 2
t
2
t
t
p/ 2
∞
cos t
dt is absolutely convergent and hence, convergent.
3/ 2
p/2 t
∫
I2 is convergent and I1 is proper integral.
∴ I is convergent.
∞
That is ∫ sin ( x 2 )dx is convergent.
0
∞
Note Similarly, we can prove ∫ cos ( x 2 )dx is convergent.
0
These two integrals are called Fresnel’s integrals.
They are useful in explaining the concept of light diffraction.
EXAMPLE 6
∞
Show that
sin x
dx converges.
x
0
∫
Solution.
∞
sin x
dx
x
0
I =∫
Let
sin x
= 1 ∴ 0 is a point of continuity.
x
So, it is an improper integral of the first kind.
We know
We write
lim
x →0
I=
p/2
∫
0
where
I1 =
p/2
∫
0
M07_ENGINEERING_MATHEMATICS-I _CH07_Part A.indd 24
∞
sin x
sin x
dx = I 1 + I 2
dx + ∫
x
x
p/2
sin x
dx is a proper integral having a finite value
x
5/12/2016 9:55:35 AM
Improper Integrals ■
I2 =
and
7.25
∞
sin x
dx is an improper integral.
x
p/2
∫
b
= lim
b →∞
sin x
dx
x
p/ 2
∫
b
b
⎧⎪ ⎡ 1
⎫⎪
1
⎤
= lim ⎨ ⎢ ( − cos x ) ⎥ − ∫ − 2 ( − cos x ) dx ⎬
b →∞ ⎣ x
⎦p/ 2 p/ 2 x
⎪⎩
⎪⎭
b
cos x ⎪⎫
⎪⎧⎛ 1
⎞
dx ⎬
= lim ⎨⎜ − cos b − 0⎟ − ∫
2
⎠
b →∞ ⎝
⎪⎭
p/2 x
⎩⎪ b
b
⎧⎪⎛ 1
cos x ⎫⎪
⎞
= lim ⎨⎜ − cos b ⎟ − ∫
dx ⎬
2
⎠
b →∞ ⎝
⎪⎩ b
⎪⎭
p/2 x
1
cos x
≤ 2
2
x
x
Now
b
∴ by the comparison test,
∫
p
2
∀x≥
cos x
dx and
x2
b
1
∫x
2
p
2
∴
b
∫
p
2
b
cos x
1
≤
dx
dx
∫
2
2
x
p x
2
dx behave alike.
p
2
∞
But
∴
1
dx is convergent by p-integral, since p = 2 > 1
2
p/ 2 x
∫
∞
cos x
dx is convergent.
x2
∫
p/2
∞
∴
cos x
dx is absolutely convergent and hence, convergent.
2
p/2 x
∴
I1 is proper integral and I2 is convergent.
∫
∴
I is convergent.
∞
That is
sin x
dx is convergent.
x
0
∫
EXAMPLE 7
Show that
p/ 2
2
∫ sin x log sin x dx is convergent and its value is log e .
e
0
Solution.
Let
I=
p/2
∫ sin x log sin x dx .
e
Here
f ( x ) = sin x loge sin x
0
As x → 0 +, log sin x → −∞ and so f(x) is unbounded at x = 0.
∴
I = lim
∈→ 0 +
p/2
∫ sin x log sin x dx .
e
∈
M07_ENGINEERING_MATHEMATICS-I _CH07_Part A.indd 25
5/12/2016 9:55:42 AM
7.26
■
Engineering Mathematics
Integrating by parts,
p/2
⎫⎪
⎧⎪
1
p/2
I = lim ⎨[ loge sin x ⋅ ( − cos x ) ]∈ − ∫
⋅ cos x ( − cos x )dx ⎬
∈→0 +
sin x
⎪⎩
⎪⎭
∈
p/2
⎧⎪
1 − sin 2 x ⎫⎪
p/2
= lim ⎨[ − cos x loge sin x ]∈ + ∫
dx ⎬
∈→0 +
sin x
⎩⎪
⎭⎪
∈
p
p
⎪⎧ ⎡
⎤
= lim ⎨ ⎢ − cos loge sin + cos ∈ loge sin ∈⎥ +
∈→0 + ⎣
⎦
2
2
⎩⎪
⎧⎪
= lim ⎨0 + cos ∈ loge sin ∈+
∈→ 0 +
⎪⎩
p/2
⎛ 1
⎞
⎪⎫
∫ ⎜⎝ sin x − sin x ⎟⎠ dx ⎬⎭⎪
∈
⎫⎪
p/2
∫ (cosec x − sin x ) dx ⎬⎪⎭
∈
p/2
⎧
x
∈
∈⎞ ⎡
⎛
⎤ ⎫
= lim ⎨cos ∈ log e ⎜ 2 sin cos ⎟ + ⎢log e tan + cos x ⎥ ⎬
⎝
∈→ 0 +
⎦∈ ⎭
2
2⎠ ⎣
2
⎩
⎧
∈
∈
∈⎞ ⎡
p
p
⎤⎫
⎛
= lim ⎨cos ∈ ⎜ log e 2 + log e sin + log e cos ⎟ + ⎢log e tan + cos − log e tan − cos ∈⎥ ⎬
⎝
⎠
∈→ 0 + ⎩
⎦⎭
2
2 ⎣
4
2
2
∈
⎡
⎤
sin
⎢
∈
∈
2 − cos ∈⎥
= lim ⎢log e 2 ⋅ cos ∈+ cos ∈ log e sin + cos ∈ log e cos + 0 − log e
⎥
∈→ 0 +
∈
2
2
⎢
⎥
cos
⎣
2
⎦
{
{
= lim (loge 2 − 1) cos ∈+ cos ∈ loge sin
∈→0 +
∈
∈
∈ ⎛
+ cos ∈ loge cos − ⎜ loge sin − loge cos
⎝
2
2
2
= lim (loge 2 − 1) cos ∈−(1 − cos ∈) loge sin
∈→ 0 +
But
∴
(1 − cos ∈) loge sin
lim (1 − cos ∈) loge sin
∈→ 0 +
2 loge t
= lim
t →0
1
t2
and
∴
log e cos
}
∈⎞ ⎫
⎟⎬
2⎠ ⎭
∈
∈
∈
+ cos ∈ loge cos + loge cos
2
2
2
∈
∈
∈
= 2 sin 2 loge sin
2
2
2
∈
∈
= lim 2t 2 loge t , where t = sin → 0 as ∈→ 0
2 t →0
2
2
⎛∞
⎞
t = lim ( −t 2 ) = 0
⎜⎝ form⎟⎠ = lim
t →0
2 t →0
∞
− 3
t
[L-Hopital’s rule]
∈
→ log e cos 0 = log e 1 = 0 as ∈→ 0
2
I = (log e 2 − 1) cos 0 = log e 2 − log e e = log e
M07_ENGINEERING_MATHEMATICS-I _CH07_Part A.indd 26
2
.
e
5/20/2016 11:19:14 AM
Improper Integrals ■
7.27
EXERCISE 7.2
Test the convergence of the following improper integrals.
∞
∞
1.
log x
∫1 x 3 dx
∞
4.
x +1
∫
x
1
3
2.
5.
dx
1
x
9. ∫
dx
1 1+ x x
tan
∞
11.
8.
1
16.
∫x
−1
p/2
17.
∫
0
dx
3
x
dx
14.
2
dx
(1 + e x )
cos x
1+ x 3
dx
2 + cos x
1 ⎤
⎡
>
⎢ Hint:
⎥
x
x⎦
⎣
∞
dx
∫0 x 1/ 3 (1 + x 2 )
sin x
dx
2
1 x
10.
∫
12.
∫
1
0
p/2
1
sin x
∫0 x 3 dx
∫
0
1 ⎤
⎡
⎢⎣ Hint: compare with g ( x ) = x 3/ 2 ⎥⎦
p
13.
x
6.
1 ⎤
⎡
⎢⎣ Hint: compare with g ( x ) = x 5 / 2 ⎥⎦
x ( x − 1)( x − 2)
3
2 + cos x
1
dx
∫
1
sin x
dx
3
1 x
∫
∫x
∞
∫
∞
cos mx
∫0 1 + x 2 dx
∞
3.
∞
∞
7.
∞
dx
∫1 x + sin 2 x
15.
∫
0
dx
x (1 − x )
dx
cos x
0
1
⎤
⎡
dx
dx
⎢ Hint: I = ∫ 3 + ∫ 3 both are divergent.⎥
x 0x x
−1 x
⎣
⎦
sin x
dx
xn
2
18.
∫
0
log x
2−x
dx
ANSWERS TO EXERCISE 7.2
1.
4.
7.
10.
13.
16.
7.2
convergent
divergent
convergent
convergent
convergent
divergent
2.
5.
8.
11.
14.
17.
divergent
convergent
divergent
convergent
convergent
convergent if n < 2
3.
6.
9.
12.
15.
18.
convergent
convergent
convergent
convergent
divergent
convergent
Evaluation of Integral by Leibnitz’s Rule
In engineering applications we come across integrals involving a parameter such as
1
x a −1
dx , where a ≥ 0.
e x
∫ log
0
The evaluation of such integrals is difficult by the usual methods of integration. Leibnitz’s rule changes
this integral, by differentiation, into a simpler integral which can be easily evaluated.
M07_ENGINEERING_MATHEMATICS-I _CH07_Part A.indd 27
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7.28
■
Engineering Mathematics
7.2.1 Leibnitz’s Rule—Differentiation Under Integral Sign for Variable limits
Theorem 7.7 If f(x, a), a(a) and b(a) are differentiable functions of a, where a is a parameter and
∂f
is continuous, then
∂a
b (a)
⎫⎪ b ( a ) ∂f ( x , a)
d ⎧⎪
db
da
a
f
x
dx
dx + f [b (a), a]
(
,
)
− f [a(a), a]
.
⎨
⎬= ∫
∂
a
d a ⎩⎪a(∫a )
d
a
d
a
⎭⎪ a( a )
The proof of the theorem is beyond the scope of the book.
Corollary 1: If the limits a and b are constants (independent of the parameter a), then the Leibnitz’s
b
b
d ⎪⎧
∂f ( x , a )
⎪⎫
dx
rule becomes
⎨∫ f ( x , a)dx ⎬ = ∫
d a ⎩⎪ a
⎭⎪ a ∂a
Corollary 2: If the limits are functions of a, a(a) and b(a), but f is independent of a, then the
Leibnitz’s rule becomes
b (a)
d ⎪⎧
db
da
⎪⎫
− f [a(a)]
⎨ ∫ f ( x )dx ⎬ = f [b (a)]
d a ⎪⎩a( a )
da
da
⎪⎭
Note
b (a)
1. When
∫
f ( x , a)dx is integrated and evaluated, it will be ultimately a function of a, say g(a). So,
a( a )
differentiation of the integral w.r.to a is ordinary derivative
dg
.
da
But f(x, a) is a function two variables x and a. So, derivative of f w.r.to a is partial derivative
2. When f is independent of a, it is a function of x alone and so
∂f
= 0.
∂a
∂f
.
∂a
So, corollary 2 does not contain integral on the R.H.S.
3. Leibnitz’s rule can be used even if one of the limits of integration is infinite.
∞
That is, the integral is of the form ∫ f ( x , a)dx or
a
∞
∫
f ( x , a)dx .
a( a )
WORKED EXAMPLES
EXAMPLE 1
1
Evaluate ∫
0
x a 21
dx , a ≥ 0 , using Leibnitz’s rule.
log e x
Solution.
x a −1
dx
loge x
0
1
Let
F (a) = ∫
M07_ENGINEERING_MATHEMATICS-I _CH07_Part A.indd 28
(1)
5/20/2016 11:19:45 AM
Improper Integrals ■
7.29
[When the integral is evaluated, it will be a function of a].
Differentiating w.r.to a, using corollary (1) of Leibnitz’s rule, we get
1
1
dF
1
∂ ⎛ x a −1⎞
∂ a
dx
=∫
=
∫0 loge x ⋅ ∂a (x − 1) dx
d a 0 ∂a ⎜⎝ loge x ⎟⎠
⎡ d x
⎤
x
⎢⎣{ dx [a ] = a loge a⎥⎦
1
1
⋅ x a loge x dx
log
x
e
0
=∫
1
⎡ x a +1 ⎤
1
= ∫ x dx = ⎢
⎥ =
⎣ a + 1⎦ 0 a + 1
0
1
a
dF
1
=
da a + 1
⇒
Integrating w.r.to a, we get
1
d a = log e (a + 1) + C
a +1
F (0) = log e 1 + C ⇒ F (0) = C
F (a) = ∫
∴
(2)
Put a = 0 in (1), we get
x0 − 1
1−1
dx = ∫
dx = 0 ⇒ C = 0
x
log
log
e
e x
0
0
1
1
F ( 0) = ∫
xa − 1
∫0 loge x dx = log (a + 1)
1
∴
F (a) = log e (a +1)
⇒
EXAMPLE 2
∞
Evaluate ∫ e 2ax
0
sin x
dx ,
x
∞
sin x
p
dx 5 .
2
x
0
a ≥ 0 and hence, show that ∫
Solution.
∞
F ( a ) = ∫ e − ax
Let
0
sin x
dx
x
(1)
Differentiating w.r.to a, using cor (1) of Leibnitz’s rule, we get
∞
∞
dF
∂ ⎛ − ax sin x ⎞
sin x ∂ − ax
(e ) dx
=
⎜e
⎟ dx = ∫
d a ∫0 ∂a ⎝
x ⎠
x ∂a
0
[{ the limits are constants]
∞
sin x − ax
e ( − x )dx
x
0
=∫
∞
= − ∫ e − ax sin xdx
0
M07_ENGINEERING_MATHEMATICS-I _CH07_Part A.indd 29
5/20/2016 11:19:52 AM
■
7.30
Engineering Mathematics
∞
⎡ e − ax
⎤
= −⎢ 2
( −a sin x − cos x )⎥
⎦0
⎣a +1
⎡
⎢{
⎣
∴
=
ax
∫ e sin bxdx =
⎤
e ax
[a sin bx − b cos bx ]⎥
2
2
a +b
⎦
∞
1
1
1
⎡⎣e − ax (a sin x + cos x ) ⎤⎦ 0 = 2
[0 − 1] = − 2
a +1
a +1
a +1
2
Integrating w.r.to a, we get
F (a) = − ∫
⇒
F (a) = − tan −1 a + C
∴ F (∞) = − tan −1 ∞ + C = −
a = ∞,
Put
da
a2 + 1
p
+C
2
From (1), we get,
∞
F (∞) = ∫ e −∞
0
∴
F(a) =
∞
⇒ ∫ e − ax
0
∞
sin x
dx = ∫ 0 dx = 0
x
0
∴
0=−
p
p
+C ⇒ C =
2
2
p
− tan −1 a
2
sin x
p
dx = − tan −1 a = cot−1 a
2
x
∞
sin x
dx ,
x
0
To deduce the value of ∫
p⎤
⎡
−1
−1
⎢⎣{ tan a + cot a = 2 ⎥⎦
(2)
put a = 0 in (2) .
∞
sin x
p
p
dx = − tan −1 0 =
2
2
x
0
∫
∴
EXAMPLE 3
a
⎞
d ⎛
1
21 x
dx ⎟ 5 2 a tan21 a 2 log ( a 2 11), using Leibnitz’s rule. Hence, evaluate
⎜ ∫ tan
da ⎝ 0
a ⎠
2
2
Prove that
a2
∫ tan
21
0
x
dx.
a
Solution.
a2
Let
F ( a) =
∫ tan
0
−1
x
dx
a
M07_ENGINEERING_MATHEMATICS-I _CH07_Part A.indd 30
(1)
5/20/2016 11:20:00 AM
Improper Integrals ■
7.31
Differentiating w.r.to parameter a using Leibnitz’s rule, we get
dF
=
da
a2
∂ ⎛
∫ ∂a ⎜⎝ tan
−1
0
2
x⎞
−1 ⎛ a ⎞
⎟⎠ dx + tan ⎜⎝ ⎟⎠ ⋅ 2a − 0
a
a
a2
=
1 ⎛ x⎞
−1
⎜ − ⎟ dx + 2a tan a
x 2 ⎝ a2 ⎠
0
1+ 2
a
∫
a2
= −∫
0
x
dx + 2a tan −1 a
a + x2
2
⎡
⎢{
⎣
a2
1
= − ⎡⎣loge (a2 + x 2 ) ⎤⎦ 0 + 2a tan −1 a
2
f ′( x )
∫ f (x ) dx = log
e
⎤
f (x )⎥
⎦
1
= − [loge (a2 + a4 ) − loge a2 ] + 2a tan −1 a
2
1
a2 (1 + a2 )
1
= − loge
+ 2a tan −1 a = 2a tan −1 a − loge (1 + a2 )
2
2
a2
Integrating w.r.to a, we get
1
loge (1 + a2 ) da
2∫
⎡
a2
1 a2 ⎤ 1 ⎡
1
⎤
⋅ 2a ⋅ a da⎥
= 2 ⎢ tan −1 a ⋅ − ∫
⋅ da⎥ − loge (1 + a2 ) ⋅ a − ∫
2
1 + a2 2 ⎦ 2 ⎢⎣
1 + a2
⎦
⎣
F (a) = ∫ 2a tan −1 a da −
a2
a
a2
daa − loge (1 + a2 ) + ∫
da
2
2
1 + a2
1+ a
a
= a2 tan −1 a − loge (1 + a2 ) + C
2
= a2 tan −1 a − ∫
Putting a = 0, we get F (0) = 0 + C = C
Putting a = 0 in (1), we get F(0) = 0
∴
∴ C=0
a
F (a) = a2 tan −1 a − loge (1 + a2 )
2
a2
∫ tan
⇒
0
−1
x
a
dx = a2 tan −1 a − loge (1 + a2 )
a
2
EXAMPLE 4
∞
Let I 5 ∫ e
⎛ a⎞
2x 2 2⎜ ⎟
⎝x⎠
2
dx , Prove that
0
dI
5 22 I . Hence, find the value of I.
dx
Solution.
∞
Given
I = ∫e
⎛ a⎞
−x 2 −⎜ ⎟
⎝x⎠
0
M07_ENGINEERING_MATHEMATICS-I _CH07_Part A.indd 31
2
∞
dx = ∫ e
−x2
e
−
a2
x2
dx
(1)
0
5/20/2016 11:20:11 AM
7.32
■
Engineering Mathematics
Differentiating (1) w.r.to a using Leibnitz’s rule, we get
(
∞
a2
)
dI
∂ − x2 − x2
e ⋅e
dx
=∫
da 0 ∂a
a
a
∞
∞
− 2 ⎛
2a ⎞
∂ ⎛ − x2 ⎞
− x2
− x2
⎜ e ⎟ dx = ∫ e e x ⎜⎝ − 2 ⎟⎠ dx = −2a∫ e
∂a ⎝
x
⎠
0
0
2
2
∞
= ∫e
− x2
0
a
x
⇒x =
a
y
y =
When
x = 0, y = ∞ and when x = ∞, y = 0
0
a2
∴
∞
− 2
2
2
2
dI
= 2∫ e − y e y dy = −2∫ e − x e − ( a /x ) dx = −2I
da
∞
0
⇒
⎞
⎟ dx
⎟⎠
⎛y2⎞⎛ a ⎞
⎜⎝ 2 ⎟⎠ ⎜ − 2 ⎟ dy
a ⎝ y ⎠
a2
0
2
a
dy
y2
Put
− 2
2
dI
= −2a∫ e y e − y
da
∞
∴ dx = −
⎛ − a2
⎜e x
⎜⎝ 2
x
∴
dI
= −2da
I
Integrating, we get
⇒
⇒
dI
= −2∫ da
I
−2 a
C
C
−2 a
log I = −2a + C ⇒ I = e −2a +C = e ⋅ e = Ke , where K = e
∫
Putting a = 0, then I = Ke 0 = K
Putting a = 0 in (1), then
∞
I (0) = ∫ e − x dx =
2
0
I=
∴
p
2
⇒
K=
p
2
[Result]
p −2a
e
2
EXAMPLE 5
∞
Prove that ∫
0
tan21 ax
p
dx 5 log (1 1 a ), a ≥ 0 .
2
2
x (1 1 x )
Solution.
∞
Let
F (a) = ∫
0
tan −1 ax
dx
x (1 + x 2 )
(1)
Differentiating (1) w.r.to a, using Leibnitz’s rule, we get
∞
∞
∂ ⎛ tan −1 ax ⎞
dF
∂
1
−1
=∫ ⎜
=
dx
∫0 x (1 + x 2 ) ∂a (tan ax )dx
da 0 ∂a ⎝ x (1 + x 2 ) ⎟⎠
∞
=∫
0
M07_ENGINEERING_MATHEMATICS-I _CH07_Part A.indd 32
∞
1
1
1
⋅ xdx = ∫
dx
2
x (1 + x 2 ) 1 + a2 x 2
1
+
x
1 + a2 x 2 )
(
)(
0
5/20/2016 11:20:25 AM
Improper Integrals ■
We shall put
7.33
1
into partial fractions.
(1 + x )(1 + a2 x 2 )
2
Since it is a function of x2, treating x2 as u, we write the special partial fraction
1
A
B
=
+
2
(1 + u )(1 + a u ) (1 + u ) (1 + a2u )
1 = A (1 + a2u ) + B (1 + u )
⇒
Put
u = −1,
Put
u=−
1 = A(1 − a 2 )
then
⇒ A=
1
1 − a2
⎛ a 2 − 1⎞
1⎞
a2
⎛
1 = B ⎜1 − 2 ⎟ ⇒ B ⎜ 2 ⎟ = 1 ⇒ B = −
⎝ a ⎠
⎝ a ⎠
1 − a2
1
, then
a2
∴
a2
1
1
1
1
=
⋅
−
⋅
2
2
2
(1 + u )(1 + a u ) 1 − a 1 + u 1 − a 1 + a2u
⇒
1
1
1
1
a2
=
⋅
−
⋅
2 2
2
2
(1 + x )(1 + a x ) 1 − a 1 + x
(1 − a2 ) 1 + a2 x 2
∴
2
∞
⎡ 1
dF
1
1 ⎤
a2
= ∫⎢
⋅
−
⋅
⎥ dx
2
2
da 0 ⎣1 − a 1 + x
(1 − a2 ) 1 + a2 x 2 ⎦
∞
∞
=
1
1
dx
a2
−
dx
2 ∫
2
2 ∫
1− a 0 1+ x
(1 − a ) 0 1 + a2 x 2
=
2
1
[ tan −1 x ]∞0 − a 2 ∫
2
1− a
(1 − a ) 0
=
1
[ tan −1 ∞ − tan −1 0] − 1 2 1 ⎡⎢ tan −1 x ⎤⎥
2
1/a ⎦ 0
(1 − a ) 1/a ⎣
1− a
=
a
1 ⎡p
⎤
[ tan −1 ∞ − tan −1 0]
− 0⎥ −
⎦ (1 − a2 )
1 − a2 ⎢⎣ 2
=
1 p
p
a ⎡p
⎤ p (1 − a)
=
−
− 0⎥ =
2
2 ⎢
2
⎦ 2 (1 − a ) 2(1 + a)
(1 − a ) 2 (1 − a ) ⎣ 2
∞
dx
1
⎛
⎞
a2 ⎜ 2 + x 2 ⎟
⎝a
⎠
∞
M07_ENGINEERING_MATHEMATICS-I _CH07_Part A.indd 33
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7.34
■
Engineering Mathematics
dF
p
=
da 2(1 + a)
∴
Integrating w.r.to a, we get
F ( a) =
p da
p
= log e (1 + a) + C
∫
2 1+ a 2
Put a = 0, then
F ( 0) =
p
log 1 + C = 0 + C = C
2
But from (1)
F ( 0) = ∫
∞
0
∴
F ( a) =
∞
tan −1 0
dx = 0
x(1 + x 2 )
[ C=0
p
log e (1 + a), a ≥ 0
2
tan −1 ax
p
dx = log e (1 + a), a ≥ 0
2
2
)
∫ x(1 + x
⇒
0
EXAMPLE 6
1
log e (11 ax )
log e (11 x )
p
≥
and
hence,
show
that
0
dx
,
a
dx 5 log e 2.
∫
2
2
8
11 x
11 x
0
0
a
Evaluate ∫
Solution.
log e (1 + ax )
dx
1+ x2
0
a
F ( a) = ∫
Let
(1)
Differentiating (1) w.r.to a, we get by Leibnitz’s rule
a
log e (1 + a 2 )
dF
∂ ⎛ log e (1 + ax ) ⎞
=∫ ⎜
+
⋅1 − log 1
dx
⎟
⎠
da 0 ∂a ⎝ 1 + x 2
1 + a2
log e (1 + a 2 )
1 ⎛ ∂
⎞
−0
+
ax
dx
1
log
(
+
)
⎜
⎟
e
2 ⎝
⎠
∂a
1 + a2
0 1+ x
a
=∫
log e (1 + a 2 )
1
1
⋅
xdx
⋅
+
2
1 + ax
1 + a2
0 1+ x
a
=∫
a
log e (1 + a 2 )
dF
x
dx
=∫
⋅
+
da 0 (1 + x 2 )(1 + ax )
1 + a2
⇒
To evaluate this integral, we put
Let
∴
x
into partial fractions.
(1 + x )(1 + ax )
2
x
Ax + B
C
=
+
2
1 + ax
(1 + x )(1 + ax ) 1 + x
2
x = ( A x + B )(1 + ax ) + C (1 + x 2 )
M07_ENGINEERING_MATHEMATICS-I _CH07_Part A.indd 34
5/20/2016 11:20:39 AM
Improper Integrals ■
Put
1
x = − , then
a
−
1
1⎞
⎛
= C ⎜1 + 2 ⎟
⎝ a ⎠
a
⇒ C
1
(1 + a 2 )
=−
2
a
a
⇒ C=−
7.35
a
1 + a2
Equating coefficients of x2, we get
0 = Aa + C ⇒ A = −
C
1⎛
a ⎞
1
= − ⎜−
⎟=
a
a ⎝ 1 + a2 ⎠ 1 + a2
Equating constants, we get
0 = B + C ⇒ B = −C =
a
1 + a2
a
1
x+
2
x
a
a2 − a ⋅ 1
1
1
+
+
=
2
2
(1 + x )(1 + ax )
(1 + x )
(1 + a2 ) (1 + ax )
∴
x +a
a
1
1
⋅
−
⋅
(1 + a2 ) (1 + x 2 ) (1 + a2 ) (1 + ax )
x
a
1
1
1
a
⋅
−
⋅
=
⋅
+
2
2
2
2
2
(1 + a ) (1 + x ) (1 + a ) (1 + x ) (1 + a ) (1 + ax )
=
a
∴
a
a
a
xdx
xdx
a
dx
a
dx
1
∫0 (1 + x 2 )(1 + ax ) = (1 + a2 ) ∫0 (1 + x 2 ) + (1 + a2 ) ∫0 (1 + x 2 ) − (1 + a2 ) ∫0 (1 + ax )
a
1 1
a
[ tan −1 x ]a0 − a 2 1 [loge (1 + ax )]a0
⎡⎣loge (1 + x 2 ) ⎤⎦ 0 +
2
2
(1 + a ) 2
(1 + a )
(1 + a ) a
1
a
[ tan −1 a − tan −1 0]
⎡⎣loge (1 + a2 ) − loge 1⎤⎦ +
=
2(1 + a2 )
(1 + a2 )
1
⎡loge (1 + a2 ) − loge 1⎤⎦
−
2 ⎣
(1 + a )
=
=−
∴
1
a tan −1 a
loge (1 + a2 ) +
2
2(1 + a )
(1 + a2 )
1
a tan −1 a
dF
1
2
1
=−
+
loge (1 + a2 )
+
+
log
(
)
a
e
da
2(1 + a2 )
(1 + a2 ) (1 + a2 )
=
loge (1 + a2 ) a tan −1 a
+
(1 + a2 )
2(1 + a2 )
Integrating w.r.to a, we get
F ( a) =
=
1 loge (1 + a2 )
a tan −1 a
+
da
∫ (1 + a2 ) da
2 ∫ (1 + a2 )
a tan −1 a
1 ⎡
1 ⎤
2
log
(
+
)
⋅
da
+
1
a
e
∫ (1 + a2 ) da
2 ∫ ⎢⎣
(1 + a2 ) ⎥⎦
M07_ENGINEERING_MATHEMATICS-I _CH07_Part A.indd 35
5/20/2016 11:20:45 AM
■
7.36
Engineering Mathematics
=
1⎡
1
a tan −1 a
⎤
2
−1
−1
1
2
log
(
+
a
)
⋅
tan
a
−
⋅
a
tan
a
da
+
e
∫ 1 + a2
⎥⎦ ∫ (1 + a2 ) da
2 ⎢⎣
=
a tan −1 a
a tan −1 a
1
da
loge (1 + a2 ) ⋅ tan −1 a − ∫
+
∫ 1 + a2 da
2
1 + a2
⇒
F ( a) =
1
loge (1 + a2 ) ⋅ tan −1 a + C
2
Put a = 0, then
F ( 0) =
1
loge (1 + 0) ⋅ tan −1 0 + C = C
2
But from (1), we get F (0) = 0
F ( a) =
Hence,
[ C=0
1
loge (1 + a2 ) ⋅ tan −1 a
2
loge (1 + ax )
1
dx = loge (1 + a2 ) ⋅ tan −1 a
2
2
1+ x
0
a
∫
⇒
(2)
To deduce, put a 5 1 in (2).
loge (1 + x )
p p
1
1
dx = loge 2 ⋅ tan −1 1 = loge 2 ⋅ = loge 2
2
2
2
4 8
1+ x
0
1
∫
∴
EXAMPLE 7
∞
Prove that
e2 x
(12 e2ax )dx 5 log e (11 a ), a ≥ 0.
x
0
∫
Solution.
∞
e −x
(1 − e − ax )dx
x
0
Differentiating (1) w.r.to a, we get
Let
F ( a) = ∫
(1)
∞
∞ −x
⎞
dF
e ∂
∂ ⎛ e −x
=∫ ⎜
(1 − e − ax )⎟ dx = ∫
(1 − e − ax )dx
⎝
⎠
da 0 ∂a x
x
a
∂
0
∞
e −x
( −e − ax )( − x ) dx
x
0
=∫
∞
∞
= ∫ e − x ⋅ e − ax dx = ∫ e − ( a +1) x dx
0
0
∞
⎡ e − ( a +1) x ⎤
1
1
[e −∞ − e 0 ] =
=⎢
⎥ =−
(
a
1
)
1
a
1
a
−
+
+
+
⎣
⎦0
M07_ENGINEERING_MATHEMATICS-I _CH07_Part A.indd 36
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Improper Integrals ■
7.37
Integrating w.r.to a,
da
= loge (1 + a) + C
1+ a
F (0) = loge 1 + C = C
⇒
Put
F ( a) = ∫
a = 0, then
∞
e −x
(1 − 1)dx = 0
x
0
But from (1), we get
F ( 0) = ∫
Hence,
F (a) = loge (1 + a)
[ C=0
∞
e −x
− ax
∫0 x (1 − e )dx = loge (1 + a)
⇒
EXAMPLE 8
p
Evaluate ∫ log(1 1 a cos x )dx ,
0 < a < 1.
0
Solution.
p
F (a) = ∫ log(1 + a cos x )dx
Let
(1)
0
Differentiating w.r.to a, by Leibnitz’s rule, we get
p
p
dF
∂
1
=
[log (1 + a cos x )]dx = ∫
⋅ cos x dx
d a ∫0 ∂a
a
1
+
cos x
0
∴
p
=
a cos x
1
dx
a ∫0 1 + a coos x
=
1 1 + a cos x − 1
dx
a ∫0 1 + a cos x
=
1 ⎡
1
⎤
1−
dx
a ∫0 ⎢⎣ 1 + a cos x ⎥⎦
=
1
1
1
dx − ∫
dx
a ∫0
a 0 1 + a cos x
=
p
p 1
1
1 p 1
1
dx
[x ]0 − ∫
dx = − ∫
a a 0 1 + a cos x
a
a 0 1 + a cos x
p
p
p
p
p
(2)
p
1
dx
1
+
a
cos x
0
I =∫
Let
Put t = tan
x
.
2
x 1
⋅ dx
2 2
2dt
2dt
2dt
dx =
=
=
x
x
1+ t 2
sec 2
1 + tan 2
2
2
∴ dt = sec 2
⇒
M07_ENGINEERING_MATHEMATICS-I _CH07_Part A.indd 37
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7.38
■
Engineering Mathematics
When x = 0, t = tan 0 = 0 and when x = p, t = tan
∴
p
=∞
2
⎡
1 − tan 2
⎢
⎢{ cos x =
⎢
1 + tan 2
⎣
∞
∞
2dt
1
dt
⋅
I =∫
= 2∫
2
2
2
1− t 1+ t
+ + a(1 − t 2 )
0
01 t
1+ a
2
∞
1+ t
dt
= 2∫
+ + (1 − a)t 2
01 a
x⎤
2⎥
x ⎥⎥
2⎦
∞
dt
⎡1 + a 2 ⎤
0
+t ⎥
(1 − a) ⎢
⎣1 − a
⎦
= 2∫
=
=
2
1
⋅
1− a 1+ a
1− a
2
=
(1 − a)(1 + a)
2
⋅
∞
⎡ −1 t ⎤
⎢ tan
⎥ ,
1+ a ⎥
⎢
⎢⎣
1 − a ⎥⎦ 0
0 < a <1
[ tan −1 ∞ − tan −1 0]
p
p
=
2
1 − a2
p
1− a
dF p 1
= − ⋅
d a a a 1 − a2
∴
2
Integrating w.r.to a, we get
p
da
da
da
⎡p
⎤
F (a) = ∫ ⎢ −
d a = p∫
− p∫
= p log a − p∫
⎥
2
2
a
a 1− a
a 1 − a2
⎣a a 1− a ⎦
I1 = ∫
Let
Put
da
a 1 − a2
a = sin u ∴d a = cos ud u
I1 = ∫
cos ud u
sin u 1 − sin u
2
=∫
cos ud u
du
=∫
sin u cos u
sin u
= ∫ cosec ud u
= − loge (cosec u + cot u)
cos u ⎤
⎡ 1
= − loge ⎢
+
sin
sin u ⎥⎦
u
⎣
⎡1 + 1 − a 2 ⎤
⎡1 + cos u ⎤
=
−
log
= − loge ⎢
⎥
e ⎢
⎥
⎣
a
⎦
⎣ sin u ⎦
M07_ENGINEERING_MATHEMATICS-I _CH07_Part A.indd 38
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Improper Integrals ■
7.39
⎡1 + 1 − a 2 ⎤
F (a) = p loge a + p loge ⎢
⎥ +C
⎣
a
⎦
2
⎡
⎡1 + 1 − a ⎤ ⎤
= p ⎢loge a + loge ⎢
⎥⎥ + C
a
⎣
⎦⎦
⎣
⎡ (1 + 1 − a 2 ) ⎤
= p loge ⎢a ⋅
⎥ + C = p loge (1 + 1 − a 2 ) + C
⎣
⎦
a
∴
Put a = 0, then
F (0) = p log e 2 + C
From (1), we get
F (0) = ∫ log 1dx = 0
p
0
∴
p loge 2 + C = 0 ⇒ C = −p loge 2
⎡1 + 1 − a 2 ⎤
F (a) = p loge (1 + 1 − a 2 ) − p loge 2 = p loge ⎢
⎥
⎣
⎦
2
∴
∴
⎡1 + 1 − a 2 ⎤
log
(
1
+
a
cos
x
)
dx
=
p
log
⎥
e ⎢
∫0
2
⎣
⎦
p
EXAMPLE 9
p/2
Show that
∫
0
log e (11 y sin 2 x )
dx 5 p ⎡⎣ 11 y 21⎤⎦ , y ≥ 0.
sin 2 x
Solution.
p/2
loge (1 + y sin 2 x )
dx
sin 2 x
0
Differentiating (1) w.r.to y, using Leibnitz’s rule, we get
F(y ) =
Let
dF
=
dy
p/ 2
∫
0
∫
(1)
p/2
∂ ⎡ log (1 + y sin 2 x ) ⎤
⎡ 1 ∂
2
=
dx
⎥
⎢
∫
2
⎢ sin 2 x ∂y log (1 + y sin x )
∂y ⎣
sin x
⎦
⎣
0
(
=
p/2
∫
0
=
p/2
∫
p/2
∫
0
M07_ENGINEERING_MATHEMATICS-I _CH07_Part A.indd 39
⎦
1
⎡ 1
2 ⎤
⎢ sin 2 x ⋅ (1 + y sin 2 x ) ⋅ sin x ⎥ dx
⎣
⎦
1
⎡
⎤
⎢1 + y sin 2 x ⎥ dx
⎦
0 ⎣
p/2
2
⎡
⎤
sec x
= ∫ ⎢ 2
⎥ dx
2
0 ⎣ sec x + y tan x ⎦
=
)⎤⎥ dx
[Multiplying Nr and Dr byy sec 2 x ]
p/ 2
⎡
⎤
⎤
⎡
sec 2 x
sec 2 x
⎢
⎥ dx = ∫ ⎢
⎥ dx
2
2
2
⎣1 + tan x + y tan x ⎦
0 ⎣1 + (1 + y ) tan x ⎦
5/20/2016 11:21:26 AM
7.40
■
Put
t = tan x.
Engineering Mathematics
[ dt = sec 2 xdx
When x = 0, t = tan 0 = 0 and when x =
∞
∞
dF
dt
1
=
=
dy ∫0 1 + (1 + y )t 2 (1 + y ) ∫0
∴
p
p
, t = tan = ∞
2
2
dt
1
+t2
(1 + y )
∞
⎡ −1 ⎛
t
⎞⎤
⎢ tan ⎜
⎟⎥
1
1 ⎢
⎜
⎟⎥
⎝ (1 + y ) ⎠ ⎥⎦ 0
(1 + y ) ⎢⎣
1
[ tan −1 ∞ − tan −1 0] = 1 ⎡⎢ p − 0⎤⎥ = p
=
1+ y ⎣ 2
1+ y
⎦ 2 1+ y
=
1
⋅
(1 + y )
1
Integrating w.r.to y, we get
F ( y) = ∫
p
2 1+ y
⋅ dy =
Put y = 0, then
F ( 0) = p + C
But from (1),
F ( 0) =
p/2
∫
0
log 1
dx = 0
sin 2 x
∴
p + C = 0 ⇒ C = −p
F ( y ) = p 1 + y − p = p ⎡⎣ 1 + y − 1⎤⎦
∴
∴
p
p (1 + y )1/ 2
(1 + y ) −1/ 2 dy =
+ C = p 1+ y + C
∫
2
2 1/ 2
p/2
∫
0
log (1 + y sin 2 x )
dx = p
sin 2 x
{
}
1+ y −1
EXAMPLE 10
1
1
0
0
Differentiating ∫ x m dx w.r.to m successively, evaluate ∫ x m (log e x ) n dx , where n ∈ N .
Solution.
1
Let
F ( m ) = ∫ x m dx
0
Differentiating w.r.to m, using Leibnitz’s rule, we get
dF
∂
=∫
( x m )dx = ∫ x m loge x dx
dm 0 ∂m
0
1
M07_ENGINEERING_MATHEMATICS-I _CH07_Part A.indd 40
1
d x
⎡
⎤
x
⎢⎣{ dx (a ) = a loge a⎥⎦
5/20/2016 11:21:54 AM
Improper Integrals ■
7.41
Differentiating again w.r.to m, using Leibnitz’s rule, we get
d 2F
∂
=∫
( x m loge x )dx
2
∂m
dm
0
1
1
= ∫ loge x
0
∂
( x m )dx
∂m
1
1
0
0
= ∫ loge x ⋅ x m loge x dx = ∫ x m (loge x ) 2 dx
Proceeding in the same way, differentiating n times w.r.to m, we get
1
dnF
= x m (loge x ) n dx
dm n ∫0
(1)
1
1
⎡ x m+1 ⎤
1
1
F ( m ) = ∫ x m dx = ⎢
[1 − 0] =
⎥ =
⎣ m + 1⎦ 0 m + 1
m +1
0
But
1
dF
=−
dm
( m + 1) 2
∴
1
2!
d2F
= ( −1)( −2) ⋅
= ( −1) 2 ⋅
2
3
( m + 1)
( m + 1)3
dm
d3F
( −3)
= ( −1) 2 ⋅ 2!
dm3
( m + 1) 4
and
= ( −1)3
3!
( m + 1) 4
Proceeding in the same way, differentiating n times w.r.to m, we get
dnF
( −1) n n !
=
n
dm
( m + 1) n +1
(2)
From (1) and (2), we get
1
∫x
m
(loge x ) n dx = ( −1) n
0
n!
( m + 1) n +1
EXAMPLE 11
∞
Evaluate
dx
∫ a 1x
2
2
and then using the method of differentiation under the sign of integration
∞
0
∞
expression for
∞
dx
dx
and hence, obtain an
,
2
2 2 ∫
2
2 3
0 (a 1x )
0 (a 1x )
[i.e., Leibnitz’s rule] find the value of the integrals ∫
∫ (a
2
0
dx
.
1 x 2 ) n11
Solution.
∞
Given
∫a
0
2
dx
, a>0
+ x2
M07_ENGINEERING_MATHEMATICS-I _CH07_Part A.indd 41
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7.42
■
Engineering Mathematics
∞
∞
We know
1 ⎡p
⎤
dx
1 ⎡ −1 x ⎤
1
−1
−1
∫0 a2 + x 2 = a ⎢⎣tan a ⎥⎦ 0 = a [ tan ∞ − tan 0] = a ⎢⎣ 2 − 0⎥⎦
∞
∫a
⇒
0
2
dx
p
=
2
2a
+x
(1)
Differentiating (1) w.r.to a, by Leibnitz’s rule, we get
∞
d ⎡
dx ⎤ d ⎛ p ⎞
⎜ ⎟
⎢∫ 2
⎥=
da ⎣ 0 a + x 2 ⎦ da ⎝ 2a ⎠
∞
∂ ⎛
∫ ∂a ⎜⎝ a
⇒
2
0
∞
⇒
∂
p⎛ 1 ⎞
1 ⎞
⎟ dx = ⎜⎝ − 2 ⎟⎠
2
+ x2 ⎠
a
+ x 2 ) −1 ⎤⎦ dx = −
p
2a2
+ x 2 ) −2 2a dx = −
p
2a2
∫ ∂a ⎡⎣(a
2
0
⇒
∞
∫ ( −1)(a
2
0
1
1⎤
dy
⎡
⎢⎣{ y = x ⇒ dx = − x 2 ⎥⎦
∞
1
p
dx = − 2
2
2 2
(
a
+
x
)
2
a
0
−2a∫
⇒
∞
∫ (a
⇒
2
0
1
p
dx = 3
4a
+ x 2 )2
(2)
Differentiating (2) w.r.to a, by Leibnitz’s rule, we get
∞
d
d ⎛ p ⎞
1
dx =
⎜
⎟
∫
2
2 2
da 0 (a + x )
da ⎝ 4a3 ⎠
⇒
∞
∂ ⎛
∫ ∂a ⎜⎝ (a
1
d ⎛ p −3 ⎞
⎞
dx =
⎜ a ⎟⎠
da ⎝ 4
+ x 2 ) 2 ⎟⎠
2
0
∞
⇒
∂
∫ ∂a ( a
2
+ x 2 ) −2 dx =
0
∞
⇒
∫ ( −2)(a
2
0
⇒
⇒
p
( −3)a −4
4
3
+ x 2 ) −3 2adx = − pa −4
4
∞
1
3p
dx = −
2 3
4 a4
0 (a + x )
−4a∫
2
∞
∫ (a
2
0
M07_ENGINEERING_MATHEMATICS-I _CH07_Part A.indd 42
dx
3p
=
2 3
+x )
16a5
(3)
5/20/2016 11:22:14 AM
Improper Integrals ■
7.43
Differentiating (3) w.r.to a, we get
∞
⇒
d
1
d ⎛ 3p ⎞
dx =
⎜
⎟
∫
2
2 3
da 0 (a + x )
da ⎝ 16a5 ⎠
⇒
∫ ∂a ( a
∞
∂
2
+ x 2 ) −3 dx =
0
∞
+ x 2 ) −4 2adx = −
3 ⋅ 5p −6
a
16
−6a∫ (a2 + x 2 ) −4 dx = −
3 ⋅ 5p −6
a
16
∫ ( −3)(a
⇒
d ⎛ 3p −5 ⎞
a ⎟
⎜
⎠
da ⎝ 16
2
0
∞
⇒
0
∞
1
3 ⋅ 5p 1
dx =
2 4
16 a6
0 (a + x )
6a∫
⇒
2
∞
∫ (a
⇒
dx
3 ⋅ 5p 1
=
+ x 2 ) 4 6 ⋅16 a7
2
0
(4)
We observe that
∞
dx
p
=
2
2a
+x
=
1 p
⋅
a 2
dx
p
= 3
2 2
+x )
4a
=
1 1 p
⋅ ⋅
a3 2 2
dx
3p
=
2 3
16 a5
+x )
=
1 3 1 p
⋅ ⋅ ⋅
a5 4 2 2
∫a
2
0
∞
∫ (a
2
0
∞
∫ (a
2
0
∞
∫ (a
2
0
dx
3 × 5p 1
1 5 3 1 p
=
⋅ 7 = 7⋅ ⋅ ⋅ ⋅
2 4
6 × 16 a
a 6 4 2 2
+x )
Proceeding in this way, from the pattern, we find that
∞
∫ (a
0
∞
∫ (a
and
0
2
2
dx
1 2n − 3 2n − 5 3 1 p
… ⋅ ⋅
= 2 n −1 ⋅
⋅
2 n
2n − 2 2n − 4 4 2 2
a
+x )
dx
1 2n − 1 2n − 3 3 1 p
= 2 n +1 ⋅
⋅
… ⋅ ⋅
2 n +1
2n 2n − 2 4 2 2
a
+x )
EXAMPLE 12
∞
e 2ax 2 e 2bx
dx using Leibnitz’s rule.
x
0
Evaluate ∫
M07_ENGINEERING_MATHEMATICS-I _CH07_Part A.indd 43
5/20/2016 11:22:21 AM
7.44
■
Engineering Mathematics
Solution.
∞
e − ax − e −bx
dx
x
0
F(a, b) is a function of two parameters a and b.
Differentiating (1) partially w.r.to a and b, by Leibnitz’s rule, we get
F (a, b ) = ∫
Let
(1)
∞
∞ − ax
∂F
∂ ⎛ e − ax − e −bx ⎞
e ( −x )
=∫ ⎜
dx = ∫
dx
⎟
⎠
∂a 0 ∂a ⎝
x
x
0
∞
= − ∫ e − ax dx
0
∞
⎡ e − ax ⎤
1 −∞
1
1
0
= −⎢
⎥ = [e − e ] = [0 − 1] = −
a
a
⎣ −a ⎦ 0 a
∞
∞
e −bx ( − x )
∂F
∂ ⎛ e − ax − e −bx ⎞
dx
=∫ ⎜
dx = ∫ −
⎟
⎠
x
∂b 0 ∂b ⎝
x
0
and
∞
= ∫ e −bx dx
0
∞
⎡ e −bx ⎤
1 −∞
1
1
0
=⎢
⎥ = − [e − e ] = − [0 − 1] =
b
b
b
b
−
⎦0
⎣
dF =
But
Integrating,
∂F
∂F
1
1
da +
db = − da + db
∂a
∂b
a
b
1
1
F (a, b ) = − ∫ da + ∫ db
a
b
= − loge a + loge b + C = loge
Putting a = 1, b = 1,
F (1, 1) = log 1 + C = 0 + C = C
But from (1), we get
F (1, 1) = ∫
b
+C
a
∞
e −x − e −x
dx = 0 [ C = 0
x
0
∴
F (a, b ) = loge
b
a
∞
e − ax − e −bx
b
dx = loge .
x
a
0
∫
∴
EXAMPLE 13
1
Evaluate
∫
0
x a 2x b
dx ,
x
a , b > 0 , using Leibnitz’s rule.
M07_ENGINEERING_MATHEMATICS-I _CH07_Part A.indd 44
5/20/2016 11:22:33 AM
Improper Integrals ■
7.45
Solution.
1
F (a, b ) = ∫
Let
0
xa − xb
dx
x
(1)
F(a, b) is a function of two parameters a and b.
Differentiating (1) partially w.r.to a and b, by Leibnitz’s rule, we get
1
1
x a loge x
∂F
∂ ⎛ xa − xb ⎞
=∫ ⎜
dx
=
dx
⎟
∫
∂a 0 ∂a ⎝ x ⎠
x
0
1
= ∫ loge x ⋅ x a −1dx
0
1
⎡
⎛ xa ⎞⎤ 1 1 xa
= ⎢loge x ⎜ ⎟ ⎥ − ∫ ⋅ dx
⎝ a ⎠ ⎦0 0 x a
⎣
1
=
1
1
⎡⎣loge 1⋅1a − 0 ⎤⎦ − ∫ x a −1dx
a
a0
1
1 ⎡xa ⎤
1
1
= 0 − ⎢ ⎥ = − 2 [1 − 0] = − 2
a ⎣ a ⎦0
a
a
and
1
1
x b loge x
∂F
∂ ⎛ xa − xb ⎞
dx
=∫ ⎜
dx
=
−
⎟
∫
x
∂b 0 ∂b ⎝ x ⎠
0
1
= − ∫ loge x ⋅ x b −1dx
0
1
⎧⎪ ⎡
⎛ x b ⎞ ⎤ 1 1 x b ⎫⎪
= − ⎨ ⎢loge x ⎜ ⎟ ⎥ − ∫ ⋅ dx ⎬
⎝ b ⎠ ⎦0 0 x b
⎪⎩ ⎣
⎪⎭
1
⎤ 1 ⎡xb ⎤
1⎡
1
1
1
= − ⎢[ loge 1 − 0 ] − ∫ x b −1dx ⎥ = ⎢ ⎥ = 2 (1 − 0) = 2
b⎣
b0
b
⎦ b ⎣ b ⎦0 b
1
dF =
But
∂F
∂F
1
1
da +
db = − 2 da + 2 db
∂a
∂b
a
b
Integrating, we get
F (a, b ) = − ∫
⎛ a −1 ⎞ b −1
1 1
da
db
+ ∫ 2 = −⎜ ⎟ +
+C = − +C
2
⎝ −1 ⎠ −1
a b
a
b
Putting a = 1, b = 1, F(1, 1) = C
1−1
dx = 0 [ C = 0
x
0
1
But from (1), we get
∴
F (1, 1) = ∫
F (a, b ) =
M07_ENGINEERING_MATHEMATICS-I _CH07_Part A.indd 45
1 1
−
a b
∴
1
∫
0
xa − xb
1 1
dx = − , a, b > 0
x
a b
5/20/2016 11:22:44 AM
7.46
■
Engineering Mathematics
EXAMPLE 14
1
Evaluate ∫
0
x a 2x b
dx ,
log e x
a , b > 0 , using Leibnitz’s rule.
Solution.
1
F (a, b ) = ∫
Let
0
xa − xb
dx
loge x
(1)
F(a, b) is a function of two parameters a and b.
Differentiating (1) partially w.r.to a and b, by Leibnitz’s rule, we get
1
1
x a loge x
∂F
∂ ⎛ xa − xb ⎞
=∫ ⎜
dx
=
∫0 loge x dx
∂a 0 ∂a ⎝ loge x ⎟⎠
1
1
⎡ x a +1 ⎤
1
1
[1 − 0] =
= ∫ x adx = ⎢
⎥ =
1
1
a
a
a
+
+
+1
⎣
⎦
0
0
and
1
1
x b loge x
∂F
∂ ⎛ xa − xb ⎞
=∫ ⎜
dx
=
−
∫0 loge x dx
∂b 0 ∂b ⎝ loge x ⎟⎠
1
⎡ x b +1 ⎤
1
1
[1 − 0] = −
= − ∫ x dx = − ⎢
⎥ =−
b +1
b +1
⎣ b + 1⎦ 0
0
1
b
But
dF =
∂F
∂F
1
1
da +
db =
da −
db
∂a
∂b
a +1
b +1
Integrating,
F (a, b ) = ∫
da
db
−∫
a +1
b +1
⎛ a + 1⎞
+C
= loge (a + 1) − loge (b + 1) + C = loge ⎜
⎝ b + 1⎟⎠
Putting a 5 1, b 5 1,
F (1, 1) = log 1 + C
⇒ F (1, 1) = 0 + C = C
But from (1), we get
(1 − 1)
dx = 0 [
log
e x
0
1
F (1, 1) = ∫
∴
⎛ a + 1⎞
F (a, b ) = loge ⎜
⎝ b + 1⎟⎠
M07_ENGINEERING_MATHEMATICS-I _CH07_Part A.indd 46
C=0
1
[
∫
0
xa − xb
⎛ a + 1⎞
dx = loge ⎜
⎝ b + 1⎟⎠
loge x
5/20/2016 11:22:49 AM
Improper Integrals ■
7.47
EXERCISE 7.3
p
log (1 + sin a cos x )
dx .
cos x
0
1. Evaluate using Leibnitz’s rule ∫
∞
∞
0
0
2. Differentiately ∫ e − ax dx successively by using Leibnitz’s rule evaluate ∫ x n e − ax dx .
∞
e −x
(1 − cos ax )dx using Leibnitz’s rule.
x
0
3. Evaluate ∫
p
dx
=
a + b cos x
0
4. Given ∫
p
p
a2 − b 2
p
dx
cos x
dx .
and ∫
2
(
a
+
b
cos
x
)
(
a
+
b cos x ) 2
0
0
, a > b find the values of ∫
∞
5. Evaluate ∫ e − x cos ax dx by using Leibnitz’s rule.
2
0
∞
cos ax − cos bx
dx , a > 0, b > 0.
x
0
6. Evaluate ∫
∞
⎛ a2 ⎞
7. Evaluate ∫ log ⎜1 + 2 ⎟ dx , a > 0.
⎝ x ⎠
∞
8.
0
∫e
0
−x
sin ax
dx , a ≥ 0.
x
ANSWERS TO EXERCISE 7.3
1. pa
4.
2. ( −1) n
pa
pb
,− 2
2 3/ 2
(a − b )
(a + b 2 ) 3 / 2
2
7. pa
n!
a n +1
p − a2 / 4
e
2
5.
3.
1
loge (1 + a 2 )
2
6. log
b
a
8. tan−1 a
7.3 BETA AND GAMMA FUNCTIONS
7.3.1 Beta Function
Definition 7.1
1
The definite integral ∫ x m −1 (1 − x ) n −1 dx , m > 0, n > 0 is defined as the beta function and it is denoted
0
by b( m, n ).
1
∴
b( m, n ) = ∫ x m −1 (1 − x ) n −1 dx , m > 0, n > 0
0
m, n are called the parameters of beta function.
This integral is also known as the first Eulerian integral.
M07_ENGINEERING_MATHEMATICS-I _CH07_Part A.indd 47
5/20/2016 11:22:56 AM
7.48
■
7.3.2
Symmetric Property of Beta Function
Engineering Mathematics
b( m, n ) = b( n , m )
The beta function is symmetric with respect to its parameters.
Proof
1
b( m, n ) = ∫ x m −1 (1 − x ) n −1 dx
We have
(1)
0
Put x = 1 − y in (1), then 1 − x = y and dx = −dy
When x = 0, y = 1 and when x = 1, y = 0
0
b( m, n ) = ∫ (1 − y ) m −1 y n −1 ( −dy )
∴
1
0
1
1
0
= − ∫ y n −1 (1 − y ) m −1 dy = ∫ y n −1 (1 − y ) m −1 dy = b( n , m )
7.3.3
■
Different Forms of Beta Function
(1) Beta Function is an Improper Integral
∞
That is
Proof
By definition,
x m −1
dx
m +n
0 (1 + x )
b( m , n ) = ∫
1
b( m , n ) = ∫ x m −1 (1 − x ) n −1 dx
0
y
Put x =
1+ y
(1 + y ) ⋅1 − y ⋅1
1
∴ dx =
dy =
dy
2
(1 + y )
(1 + y ) 2
y
⇒ (1+ y )x = y ⇒ x + xy = y
1+ y
When x = 0, y = 0 and when x = 1, y = ∞
Now x =
∞
∴
⎛ y ⎞
b( m, n ) = ∫ ⎜
⎝ 1 + y ⎟⎠
0
∞
=∫
0
∴
m −1
y ⎞
⎛
⎜⎝1 − 1 + y ⎟⎠
y ( x − 1) = − x
⇒
n −1
⇒
y =
x
1− x
dy
(1 + y ) 2
∞
y m −1 (1 + y − y ) n −1
y m −1
dy = ∫
dy
m −1+ n −1+ 2
m +n
(1 + y )
0 (1 + y )
∞
x m −1
dx
m +n
0 (1 + x )
b( m, n ) = ∫
■
This integral is an improper integral of the first kind.
(2) Beta Function Interms of Trignometric Function
p/2
b( m , n ) = 2 ∫ sin 2 m 21 u cos 2 n21 ud u
0
M07_ENGINEERING_MATHEMATICS-I _CH07_Part B.indd 48
5/12/2016 9:59:24 AM
Improper Integrals ■
Proof
7.49
1
b( m , n ) = ∫ x m −1 (1 − x ) n −1 dx
By definition
0
x = sin 2 u ∴ dx = 2 sin u cos ud u
Put
When x = 0, u = 0 and when x = 1, u =
b( m, n ) =
∴
p
2
p/2
∫ (sin
2
u) m −1 (1 − sin 2 u) n −1 2 sin u cos ud u
0
p/2
= 2 ∫ sin 2 m − 2 u cos 2 n − 2 u sin u cos ud u
0
p/2
b( m, n ) = 2 ∫ sin 2 m −1 u cos 2 n −1 ud u
⇒
(1)
0
Note In some practical problems we come across definite integrals involving trignometric functions
which can be evaluated interms of beta functions.
p/2
b( m, n ) = 2 ∫ sin 2 m −1 u cos 2 n −1 ud u
We have
0
p/2
∫ sin
⇒
2 m −1
u cos 2 n −1 ud u =
0
1
b( m , n )
2
If
p = 2m − 1
⇒
2m = p + 1
⇒
and
q = 2n − 1
⇒
2n = q + 1
⇒
p/2
∫ sin
∴
p
u cosq ud u =
0
7.4
p +1
2
q +1
n=
2
m=
1 ⎛ p + 1 q + 1⎞
b⎜
,
⎟
2 ⎝ 2
2 ⎠
■
THE GAMMA FUNCTION
Definition 7.2
∞
The integral ∫ e − x x n −1dx ( n > 0) is defined as the gamma function with parameter n and it is denoted
by Γn.
0
∴
∞
Gn = ∫ e − x x n −1 dx , ( n . 0)
0
This integral is also known as Euler’s integral of the second kind.
M07_ENGINEERING_MATHEMATICS-I _CH07_Part B.indd 49
5/12/2016 9:59:29 AM
7.50
■
7.4.1
Properties of Gamma Function
Engineering Mathematics
(1) Prove that Γ1 5 1 .
Proof
∞
∞
∞
⎡e −x ⎤
Γ1 = ∫ e − x x 1−1dx = ∫ e − x dx = ⎢
⎥ = −[0 − 1] = 1
⎣ −1 ⎦ 0
0
0
By definition,
■
(2) Prove that Γ ( n 1 1) 5 nΓn
Proof
∞
By definition
Γn = ∫ e − x x n −1dx
0
∞
∴
Γ( n + 1) = ∫ e − x x n +1−1dx
0
∞
= ∫ e − x x n dx
0
∞
∞
∞
−x
⎡ e −x ⎤
− x n −1
n −1 e
−
= ⎢x n
nx
dx
=
n
⎥
∫0 e x dx = nΓn
−1
⎣ −1 ⎦ 0 ∫0
∴
Γ ( n + 1) = nΓn
This is true for all positive values of n.
(3) If n is an Integer ≥ 1, then Γn 5 ( n 21)!
We have
Γ ( n + 1) = nΓn
⇒
Γn = ( n − 1)Γ ( n − 1)
= ( n − 1)( n − 2)Γ ( n − 2)
= ( n − 1)( n − 2) … 3 ⋅ 2 ⋅1⋅ Γ1
⇒
Γn = ( n − 1)( n − 2) … 3 ⋅ 2 ⋅1 = ( n − 1)!
■
⎛ 1⎞
(4) Prove that Γ ⎜ ⎟ 5 p
⎝ 2⎠
Proof
∞
∞
⎛ 1⎞
By definition,
Γ ⎜ ⎟ = ∫ e − x x 1/ 2 −1dx = ∫ e − x x −1/ 2dx
⎝ 2⎠ 0
0
Put x = y 2
∴ dx = 2 ydy
When x = 0, y = 0 and when x = ∞, y = ∞
∴
∞
∞
2
2
⎛ 1⎞
Γ ⎜ ⎟ = ∫ e − y y −1 ⋅ 2 ydy = 2∫ e − y dy
⎝ 2⎠ 0
0
M07_ENGINEERING_MATHEMATICS-I _CH07_Part B.indd 50
5/12/2016 9:59:34 AM
Improper Integrals ■
∞
∞
2
⎛ 1⎞
Γ ⎜ ⎟ = 2∫ e − y dy
⎝ 2⎠
0
Now
7.51
and
∞
2
⎛ 1⎞
Γ ⎜ ⎟ = 2∫ e − x dx
⎝ 2⎠
0
∞
2
2
⎛ 1⎞ ⎛ 1⎞
Γ ⎜ ⎟ ⋅ Γ ⎜ ⎟ = 2∫ e − x dx ⋅ 2∫ e − y dy
⎝ 2⎠ ⎝ 2⎠
0
0
∴
∞∞
2
⎡ ⎛ 1⎞ ⎤
− ( x2 + y2 )
dxdy
⎢Γ ⎜⎝ 2 ⎟⎠ ⎥ = 4∫ ∫ e
⎣
⎦
0 0
⇒
[since the limits are constants]
Put x = r cos u, and y = r sin u
∴
r 2 = x 2 + y 2 and
dxdy = rdrd u
and
y
x
[{ Jacobian value = r ]
tan u =
When x, y varies from 0 to ∞, r varies from 0 to ∞ and u varies from 0 to
2
p/ 2 ∞
⎤
⎡p / 2 ⎤ ⎡ ∞ − r2
⎡ ⎛ 1⎞ ⎤
−r2
=
4
d
u
e
r
dr
Γ
4
e
r
dr
d
u
=
⎥
⎢
⎥
⎢
∫
∫
⎜
⎟
⎢ ⎝ 2⎠ ⎥
∫0 ∫0
⎦
⎦ ⎣0
⎣0
⎣
⎦
∴
∞
= 4[u]p0 / 2 ∫ e − r r dr = 4 ⋅
2
0
Let r2 = u
∴ 2rdr = du ⇒ rdr =
p
2
∞
∞
2
p − r2
e r dr = 2p∫ e − r r dr
∫
20
0
du
2
When r = 0, u = 0 and when r = ∞, u = ∞
∴
∞
∞
2
⎡ ⎛ 1⎞ ⎤
−u
− u du
⎢Γ ⎜⎝ 2 ⎟⎠ ⎥ = 2p∫ e 2 = p∫ e du
0
⎣
⎦
0
∞
⎡ e −u ⎤
−∞
0
= p⎢
⎥ = −p[e − e ] = −p(0 − 1) = p
⎣ −1 ⎦ 0
∴
7.4.2
■
Γ(1/ 2) = p
RELATION BETWEEN BETA AND GAMMA FUNCTIONS
Prove that b( m , n ) 5
Γm ⋅ Γn
, m > 0, n > 0
Γ( m 1 n)
Proof
∞
By definition, Γm = ∫ e −t t m −1dt
0
Let
t = x 2 ∴ dt = 2 xdx
When t = 0, x = 0 and when t = ∞, x = ∞
∴
∞
∞
Γm = ∫ e − x x 2 m − 2 ⋅ 2xdx = 2∫ e − x x 2 m −1dx
2
0
M07_ENGINEERING_MATHEMATICS-I _CH07_Part B.indd 51
2
(1)
0
5/12/2016 9:59:40 AM
■
7.52
Engineering Mathematics
∞
Γn = 2∫ e − y y 2 n −1dy
2
Similarly,
(2)
0
∴ (1) × (2) ⇒
∞
∞
Γm ⋅ Γn = 2∫ e − x x 2 m −1dx ⋅ 2∫ e − y y 2 n −1dy
2
2
0
0
∞∞
Γm ⋅ Γn = 4 ∫ ∫ e − ( x
⇒
+y 2 )
2
x 2 m −1 y 2 n −1dx dy
(3)
0 0
[since the limits are constants]
Changing to polar coordinates by putting
x = r cos u, y = r sin u
r 2 = x 2 + y 2 , tan u =
we get
y
x
and dxdy = rdrd u
When x, y varies from 0 to ∞, r varies from 0 to ∞ and u varies from 0 to
∴ (3) becomes
p/2 ∞
p
2
Γm ⋅ Γn = 4 ∫ ∫ e − r ( r cos u) 2 m −1 ( r sin u) 2 n −1 rdrd u
2
0 0
p/ 2 ∞
= 4 ∫ ∫ e − r r 2 m + 2 n − 2 sin 2 n −1 u cos 2 m −1 u r dr d u
2
0 0
∞
p/ 2
= 4 ∫ sin 2 n −1 u cos 2 m −1 ud u ⋅ ∫ e − r r 2 m + 2 n −1dr
2
0
0
∞
= 2b( n , m ) ⋅ ∫ e − r r 2 m + 2 n −1dr
2
(4)
0
∞
I = ∫ e − r r 2 m + 2 n −1dr
2
Let
0
Put t = r
∴ dt = 2rdr ⇒ dr =
2
dt
dt
= 1/ 2
2r 2t
When r = 0, t = 0 and when r = ∞, t = ∞
∞
I = ∫ e −t (t 1/ 2 )
∴
0
2 m + 2 n −1
∞
dt
1
= ∫ e −t t m + n −1/ 2 ⋅ t −1/ 2dt
1/ 2
20
2t
=
∞
1
1 −t m + n −1
e t
dt = Γ( m + n )
∫
2
20
⇒
Γ(m + n)
2
Γm ⋅ Γn = b( m, n)Γ ( m + n)
∴
b( m , n ) =
∴
(4)
⇒
[ by definiition]
Γm ⋅ Γn = 2b( n , m ) ⋅
M07_ENGINEERING_MATHEMATICS-I _CH07_Part B.indd 52
[{ b( m, n) = b( n, m)]
Γm ⋅ Γn
Γ(m + n)
5/12/2016 9:59:46 AM
Improper Integrals ■
7.53
Corollary
p/2
Prove that
∫
⎛ p + 1⎞ ⎛ q + 1⎞
Γ⎜
⋅Γ
⎝ 2 ⎟⎠ ⎜⎝ 2 ⎟⎠
sin u cos ud u =
⎛ p + q + 2⎞
2Γ ⎜
⎟⎠
⎝
2
p
0
p/2
We have
∫ sin
p
q
u cosq ud u =
0
⇒
p/2
∫
⇒
p/2
∫
0
1
sin u cos ud u =
2
p
0
1 ⎛ p + 1 q + 1⎞
b⎜
,
⎟
2 ⎝ 2
2 ⎠
q
⎛ p + 1⎞ ⎛ q + 1⎞
Γ⎜
⋅Γ
⎝ 2 ⎟⎠ ⎜⎝ 2 ⎟⎠
⎛ p + 1 q + 1⎞
+
Γ⎜
⎟
⎝ 2
2 ⎠
⎛ p + 1⎞ ⎛ q + 1⎞
Γ⎜
⋅Γ
⎝ 2 ⎟⎠ ⎜⎝ 2 ⎟⎠
sin p u cosq ud u =
⎛ p + q + 2⎞
2Γ ⎜
⎟⎠
⎝
2
(5)
(1) In particular, if we put p 5 n, q 5 0 in (5), we get
p/2
∫
0
p/2
∫
Similarly,
0
If n = 2, we get,
⎛ n + 1⎞
Γ⎜
⎝ 2 ⎟⎠
p
⋅
cos u d u =
⎛ n + 2⎞ 2
Γ⎜
⎝ 2 ⎟⎠
p/2
∫
0
⎛ n + 1⎞
⎛ n + 1⎞ ⎛ 1 ⎞
Γ⎜
Γ⎜
⋅Γ
⎝ 2 ⎟⎠
⎝ 2 ⎟⎠ ⎜⎝ 2 ⎟⎠
p
=
sin u d u =
⋅
+
2
n
n
+
2
2
⎛
⎞
⎛
⎞
2Γ ⎜
Γ⎜
⎝ 2 ⎟⎠
⎝ 2 ⎟⎠
n
n
⎛ 2 + 1⎞
Γ⎜
⎝ 2 ⎟⎠
p
⋅
sin u d u =
2
2
+
2
⎛
⎞
Γ⎜
⎝ 2 ⎟⎠
2
1 ⎛ 1⎞
⎛ 3⎞
Γ⎜ ⎟
Γ
⎝ 2 ⎠ p 2 ⎜⎝ 2 ⎟⎠ p
⋅
=
⋅
=
=
Γ2
2
1!
2
p/2
∫ cos
Similarly,
2
u du =
0
p p p
=
4
4
p
4
(2) In particular, if we put p 5 0, q 5 0 in (5), then we get
p/2
∫ sin
0
u cos 0 ud u =
0
M07_ENGINEERING_MATHEMATICS-I _CH07_Part B.indd 53
Γ (1/ 2)Γ (1/ 2)
2Γ ( 2/2)
5/12/2016 9:59:50 AM
7.54
■
Engineering Mathematics
p/2
∫ du =
(Γ1/ 2) 2
2Γ1
⇒
[u]p0 / 2 =
(Γ1/ 2) 2
2
⇒
⎡ ⎛ 1⎞ ⎤
⎢Γ ⎜⎝ 2 ⎟⎠ ⎥ = p
⎣
⎦
⇒
0
⇒
p (Γ1/ 2) 2
=
2
2
⇒
⎛ 1⎞
Γ⎜ ⎟ = p
⎝ 2⎠
2
⎛ 1 1⎞ ⎛ ⎛ 1⎞⎞
(3) b ⎜⎝ , ⎟⎠ 5 ⎜ Γ ⎜ ⎟ ⎟
⎝ ⎝ 2⎠⎠
2 2
2
Proof
2
⎛ 1⎞ ⎛ 1⎞ ⎛ ⎛ 1⎞ ⎞
Γ ⎜ ⎟ Γ ⎜ ⎟ ⎜Γ ⎜ ⎟⎟
2
⎝ 2⎠ ⎝ 2⎠ ⎝ ⎝ 2⎠ ⎠
⎛ ⎛ 1⎞ ⎞
⎛ 1 1⎞
=
= ⎜Γ ⎜ ⎟⎟
b⎜ , ⎟ =
⎝ 2 2⎠
⎝ ⎝ 2⎠ ⎠
Γ1
⎛ 1 1⎞
Γ⎜ + ⎟
⎝ 2 2⎠
1
1
(4) Prove that ∫ x m (1 2 x n ) p 5
n
0
Proof
Let
⎛ m 1 1⎞
Γ ( p 1 1) Γ ⎜
⎝ n ⎟⎠
m 1 1⎞
⎛
Γ ⎜ p 111
⎟
⎝
n ⎠
1
I = ∫ x m (1 − x n ) p dx
0
Put
1− x = y
n
⇒ x = 1− y
n
⇒ x = (1 − y )1/ n
Differentiating both sides, we get
dx =
1
1− n
−1
1
1
(1 − y ) n ( −1) dy = − (1 − y ) n dy
n
n
When x = 0, y = 1 and when x = 1, y = 0
∴
0
m
1− n
⎡ 1
⎤
I = ∫ (1 − y ) n y p ⎢ − (1 − y ) n ⎥ dy
⎣ n
⎦
1
m 1− n
+
1 p
n
n
y
(
1
y
)
dy
−
∫
n1
0
=−
1
=
m 1
+ −1
1 p
n n
y
1
y
dy
(
−
)
∫
n0
=
m +1
−1
1 p +1−1
y
(1 − y ) n dy
∫
n0
a
⎤
⎡b
f
x
dx
f ( x )dx ⎥
(
)
=
−
⎢∫
∫
b
⎦
⎣a
1
M07_ENGINEERING_MATHEMATICS-I _CH07_Part B.indd 54
5/12/2016 9:59:54 AM
Improper Integrals ■
1 ⎛
m + 1⎞ 1
= b ⎜ p + 1,
⎟=
n ⎝
n ⎠ n
∴
1
1
∫0 x (1 − x ) dx = n
m
n p
7.55
⎛ m + 1⎞
Γ ( p + 1)Γ ⎜
⎝ n ⎟⎠
m + 1⎞
⎛
Γ ⎜ p +1+
⎟
⎝
n ⎠
⎛ m + 1⎞
Γ ( p + 1)Γ ⎜
⎝ n ⎟⎠
m + 1⎞
⎛
Γ ⎜ p +1+
⎟
⎝
n ⎠
WORKED EXAMPLES
EXAMPLE 1
1
Find the value of ∫ x 7 (1 2 x ) 6 dx .
0
Solution.
1
b( m , n ) = ∫ x m −1 (1 − x ) n −1 dx
We know
0
1
∫x
Now
0
1
7
(1 − x )6 dx = ∫ x 8 −1 (1 − x )7 −1 dx = b(8, 7) =
0
Γ8 ⋅ Γ 7 7 !⋅ 6 !
=
Γ (8 + 7) 14 !
EXAMPLE 2
1
Evaluate ∫ x 11 (1 2 x ) 5 dx .
0
Solution.
1
b( m , n ) = ∫ x m −1 (1 − x ) n −1 dx
We know
0
1
∫x
Now
0
1
11
(1 − x )5 dx = ∫ x 12 −1 (1 − x )6 −1 dx = b(12, 6) =
0
Γ12 ⋅ Γ 6
11!⋅ 5!
=
Γ (12 + 16)
27 !
EXAMPLE 3
p/ 2
Evaluate
∫ sin
8
u cos 7 ud u.
0
Solution.
We know
p/2
∫ sin
p
u cosq ud u =
1 ⎛ p + 1 q + 1⎞
b⎜
,
⎟
2 ⎝ 2
2 ⎠
u cos 7 ud u =
1 ⎛ 8 + 1 7 + 1⎞
b⎜
,
⎟
2 ⎝ 2
2 ⎠
0
∴
p/2
∫ sin
8
0
M07_ENGINEERING_MATHEMATICS-I _CH07_Part B.indd 55
[ p = 8, q = 7]
5/12/2016 9:59:58 AM
7.56
■
Engineering Mathematics
1 ⎛9 ⎞ 1
= b ⎜ , 4⎟ =
2 ⎝2 ⎠ 2
1
=
2
⎛ 9⎞
Γ ⎜ ⎟ Γ4
⎝ 2⎠
⎛9
⎞
Γ ⎜ + 4⎟
⎝2
⎠
⎛ 9⎞
⎛ 9⎞
Γ ⎜ ⎟ × 3!
1⋅ 2 ⋅ 3 ⋅ Γ ⎜ ⎟
⎝ 2⎠
⎝ 2⎠
16
16
=
=
=
15 13 11 9 ⎛ 9 ⎞ 15 ⋅13 ⋅11⋅ 3 6435
⎛ 17 ⎞
Γ⎜ ⎟
2⋅ ⋅ ⋅ ⋅ Γ ⎜ ⎟
⎝ 2⎠
2 2 2 2 ⎝ 2⎠
EXAMPLE 4
p/ 2
∫ sin
Evaluate
5
ud u.
0
Solution.
We know that
p/2
∫ sin
p
u cosq ud u =
0
p/2
∫
∴
sin 5 ud u =
0
1 ⎛ p + 1 q + 1⎞
b⎜
,
⎟
2 ⎝ 2
2 ⎠
p/2
∫ sin
5
u cos 0 ud u =
0
1 ⎛ 1⎞ 1
= b ⎜ 3, ⎟ =
2 ⎝ 2⎠ 2
1 ⎛ 5 + 1 0 + 1⎞
b⎜
,
⎟
2 ⎝ 2
2 ⎠
[ p = 5, q = 0]
⎛ 1⎞
⎛ 1⎞
⎛ 1⎞
Γ3 ⋅ Γ ⎜ ⎟
2 !⋅ Γ ⎜ ⎟
Γ⎜ ⎟
⎝ 2⎠
⎝ 2⎠ 1
⎝ 2⎠
8
=
=
=
5 3 1 ⎛ 1 ⎞ 15
2
⎛ 6 + 1⎞
⎛ 7⎞
Γ⎜
Γ⎜ ⎟
⋅ ⋅ Γ⎜ ⎟
⎝ 2 ⎟⎠
⎝ 2⎠
2 2 2 ⎝ 2⎠
EXAMPLE 5
p/ 2
Evaluate
∫
cos 8 ud u.
0
Solution.
p/2
We know that
∫ sin
p
u cosq ud u =
0
∴
p/2
∫ cos
8
ud u =
0
1 ⎛ p + 1 q + 1⎞
b⎜
,
⎟
2 ⎝ 2
2 ⎠
p/2
∫ sin
0
[ p = 0, q = 8]
u cos8 ud u
0
1 ⎛ 0 + 1 8 + 1⎞ 1 ⎛ 1
,
b⎜
⎟ = b⎜ ,
2 ⎠ 2 ⎝2
2 ⎝ 2
⎛ 1⎞ ⎛ 9⎞
Γ
⋅Γ
1 ⎜⎝ 2 ⎟⎠ ⎜⎝ 2 ⎟⎠
=
2
⎛ 9 1⎞
Γ⎜ + ⎟
⎝ 2 2⎠
⎛ 1⎞ 7 5 3 1 ⎛ 1⎞
⋅ ⋅ ⋅ Γ
Γ
1 ⎜⎝ 2 ⎟⎠ 2 2 2 2 ⎜⎝ 2 ⎟⎠
=
=
2
Γ5
=
M07_ENGINEERING_MATHEMATICS-I _CH07_Part B.indd 56
9⎞
⎟
2⎠
1 7 ⋅ 5 ⋅ 3p 35p 35p
= 8 =
256
25 1 ⋅ 2 ⋅ 3 ⋅ 4
2
5/12/2016 10:00:01 AM
Improper Integrals ■
7.57
EXAMPLE 6
p/ 2
∫
Evaluate
tan u d u.
0
Solution.
We know that
p/2
∫ sin
p
1 ⎛ p + 1 q + 1⎞
b⎜
,
⎟
2 ⎝ 2
2 ⎠
u cosq ud u =
0
p/2
∫
Now,
p/ 2
∫ sin
tan u d u =
1/ 2
0
u ⋅ cos −1/ 2 ud u
0
1 ⎞
⎛1
+1 − +1
1 ⎜2
⎟
= b⎜
, 2 ⎟
⎝
2
2
2 ⎠
1
1⎤
⎡
⎢⎣{ p = 2 , q = − 2 ⎥⎦
1 ⎛ 3 1⎞
b⎜ , ⎟
2 ⎝ 4 4⎠
⎛ 3⎞ ⎛ 1⎞
⎛ 1⎞ ⎛ 3⎞
Γ
Γ
⋅Γ
⋅Γ
1 ⎜⎝ 4 ⎟⎠ ⎜⎝ 4 ⎟⎠ 1 ⎜⎝ 4 ⎟⎠ ⎜⎝ 4 ⎟⎠ 1 ⎛ 1 ⎞ ⎛ 3 ⎞
= Γ⎜ ⎟ Γ⎜ ⎟ .
tan u d u =
=
2
Γ1
2 ⎝ 4⎠ ⎝ 4⎠
2
⎛ 3 1⎞
Γ⎜ + ⎟
⎝ 4 4⎠
=
p
2
∫
⇒
0
EXAMPLE 7
p/2
Evaluate
1
∫
tan u
0
d u.
Solution.
We know that
p/2
∫ sin
p
u cosq ud u =
0
p/2
∴
∫
0
1
tan u
du =
1 ⎛ p + 1 q + 1⎞
b⎜
,
⎟
2 ⎝ 2
2 ⎠
p/ 2
∫ sin
−1/ 2
u cos1/ 2 ud u
0
1 ⎞
⎛ 1
− +1 +1
1 ⎜ 2
⎟
= b⎜
,2 ⎟
2 ⎝ 2
2 ⎠
⇒
p
2
∫
0
1 ⎛1
du = b ⎜ ,
2 ⎝4
tan u
1
M07_ENGINEERING_MATHEMATICS-I _CH07_Part B.indd 57
3⎞ 1
⎟=
4⎠ 2
⎛ 1⎞ ⎛ 3⎞
⎛ 1⎞ ⎛ 3⎞
Γ
⋅Γ
Γ⎜ ⎟ Γ⎜ ⎟
⎝ 4 ⎠ ⎝ 4 ⎠ 1 ⎜⎝ 4 ⎟⎠ ⎜⎝ 4 ⎟⎠ 1 ⎛ 1 ⎞ ⎛ 3 ⎞
= Γ ⎜ ⎟ ⋅Γ ⎜ ⎟ .
=
2
Γ1
2 ⎝ 4⎠ ⎝ 4⎠
⎛ 1 3⎞
Γ⎜ + ⎟
⎝ 4 4⎠
5/12/2016 10:00:05 AM
■
7.58
Engineering Mathematics
EXAMPLE 8
p/ 2
p/ 2
∫
Evaluate
tan u d u 3
0
1
∫
tan u
0
d u.
Solution.
From Examples 6 and 7, we have
p/2
∫
p/2
1
∫
tan u d u ×
0
du =
tan u
0
1 ⎛ 1⎞ ⎛ 3⎞ 1 ⎛ 1⎞ ⎛ 3⎞ 1 ⎡ ⎛ 1⎞ ⎛ 3⎞ ⎤
Γ⎜ ⎟ Γ⎜ ⎟ × Γ⎜ ⎟ Γ⎜ ⎟ =
Γ⎜ ⎟ Γ⎜ ⎟
2 ⎝ 4 ⎠ ⎝ 4 ⎠ 2 ⎝ 4 ⎠ ⎝ 4 ⎠ 4 ⎢⎣ ⎝ 4 ⎠ ⎝ 4 ⎠ ⎥⎦
2
EXAMPLE 9
Show that
p/ 2
∫
p/ 2
sin u d u 3
0
∫
1
sin u
0
d u 5 p.
Solution.
We know that
p/2
∫ sin
p
u cosq ud u =
0
p/2
∫
∴
sin u d u =
p/2
∫ sin
1/ 2
0
0
1 ⎛ p + 1 q + 1⎞
b⎜
,
⎟
2 ⎝ 2
2 ⎠
⎛1
⎞
+1
1 ⎜2
0 + 1⎟
u cos ud u = b ⎜
,
⎟
2 ⎝ 2
2 ⎠
0
1 ⎛ 3 1⎞ 1
= b⎜ , ⎟ =
2 ⎝ 4 2⎠ 2
p
2
∫
⇒
0
p/2
∫
and
0
⎛ 3⎞ ⎛ 1⎞
⎛ 3⎞ ⎛ 1⎞
Γ ⎜ ⎟ ⋅Γ ⎜ ⎟
Γ
⋅Γ
⎝ 4 ⎠ ⎝ 2 ⎠ 1 ⎜⎝ 4 ⎟⎠ ⎜⎝ 2 ⎟⎠
=
2
⎛ 3 1⎞
⎛ 5⎞
Γ⎜ + ⎟
Γ⎜ ⎟
⎝ 4 2⎠
⎝ 4⎠
⎛ 3⎞ ⎛ 1⎞
⎛ 3⎞ ⎛ 1⎞
⎛ 3⎞ ⎛ 1⎞
⋅Γ
Γ ⎜ ⎟ ⋅Γ ⎜ ⎟
Γ
Γ ⎜ ⎟ ⋅Γ ⎜ ⎟
⎝ 4⎠ ⎝ 2⎠
⎝ 4⎠ ⎝ 2⎠
1 ⎜⎝ 4 ⎟⎠ ⎜⎝ 2 ⎟⎠
1
=2
sin u d u =
=
2 ⎛ 5 ⎞ ⎛ 5 ⎞ 2 ⎛ 1⎞ ⎛ 1⎞
⎛ 1⎞
Γ⎜ ⎟
⎜⎝ ⎟⎠ Γ ⎜⎝ ⎟⎠
⎜⎝ − 1⎟⎠ Γ ⎜⎝ − 1⎟⎠
⎝ 4⎠
4
4
4
4
1
sin u
du =
p/ 2
∫
0
⎛ 1
− +1
1
⎜
sin −1/ 2 u cos 0 ud u = b ⎜ 2 ,
2 ⎝ 2
⎞
1⎟
⎟
2⎠
⎛ 1⎞ ⎛ 1⎞
⎛ 1⎞ ⎛ 1⎞
Γ ⎜ ⎟ ⋅Γ ⎜ ⎟
Γ ⎜ ⎟ ⋅Γ ⎜ ⎟
⎝
⎠
⎝
⎠
1 ⎛ 1 1⎞ 1
4
2 = 1 ⎝ 4⎠ ⎝ 2⎠
= b⎜ , ⎟ =
2
2 ⎝ 4 2⎠ 2
⎛ 1 1⎞
⎛ 3⎞
Γ⎜ + ⎟
Γ⎜ ⎟
⎝ 4 2⎠
⎝ 4⎠
∴
p/2
∫
0
sin u d u ×
p/2
∫
0
⎛ 1⎞ ⎛ 1⎞
⎛ 3⎞ ⎛ 1⎞
Γ ⎜ ⎟ ⋅Γ ⎜ ⎟ Γ ⎜ ⎟ ⋅Γ ⎜ ⎟
⎝ 4⎠ ⎝ 2⎠ 1 ⎝ 4⎠ ⎝ 2⎠
⎛ 1⎞
⎛ 1⎞
= Γ⎜ ⎟ ×Γ⎜ ⎟ = p × p = p
du = 2
⎝
⎠
⎝ 2⎠
3
1
2
2
⎛ ⎞
⎛ ⎞
sin u
Γ⎜ ⎟
Γ⎜ ⎟
⎝ 4⎠
⎝ 4⎠
1
M07_ENGINEERING_MATHEMATICS-I _CH07_Part B.indd 58
5/12/2016 10:00:09 AM
Improper Integrals ■
7.59
EXAMPLE 10
∞
Γn
, where a and n are positive.
an
Prove that ∫ e 2ax x n 21dx 5
0
Solution.
∞
Γn = ∫ e − x x n −1dx .
We know
∞
I = ∫ e − ax x n −1dx
Let
0
Put t = ax
∴ dt = adx ⇒ dx =
0
dt
a
When x = 0, t = 0 and when x = ∞, t = ∞
∞
⎡t ⎤
I = ∫ e −t ⎢ ⎥
⎣a⎦
0
∴
∞
∫e
⇒
− ax
x n −1dx =
0
n −1
1
dt
= n
a
a
∫
∞
0
e − t t n −1dt =
Γn
an
Γn
an
EXAMPLE 11
∞
Show that
∫x
n 2a 2 x 2
e
dx 5
0
∞
∫ cos ( x
1
⎛ n 11 ⎞
Γ⎜
⎟ . Deduce that
2 a n 11 ⎝ 2 ⎠
∞
2
0
∞
∫e
2a 2 x 2
dx 5
0
p
. Hence, show that
2a
1 p
.
2 2
) dx 5 ∫ sin ( x 2 ) dx 5
0
Solution.
∞
Γn = ∫ e − x x n −1dx .
We know
∞
Let
0
Put ax = t .
∴ adx =
1
2 t
dt
⇒ dx =
I = ∫ x n e − a x dx
2 2
0
dt
2a t
When x = 0, t = 0 and when x = ∞, t = ∞
∞
t n / 2 − t dt
⋅e
n
2a t
0 a
I =∫
=
∞
∫x
∴
n
∞
∞
n 1
n +1
−1
−
1
1
1
⎛ n + 1⎞
e − t t 2 2 dt = n +1 ∫ e − t t 2 dt = n +1 Γ ⎜
n +1 ∫
⎝ 2 ⎟⎠
2a
2a 0
2a 0
1
⎛ n + 1⎞
Γ⎜
n +1
⎝ 2 ⎟⎠
2a
(1)
dx =
1 ⎛ 1⎞
p
Γ⎜ ⎟ =
⎝
⎠
2a
2
2a
(2)
a2 =
(1 − i ) 2 1 − 1 − 2i
=
= −i
2
2
e − a x dx =
2 2
0
In (1), put n = 0.
∞
∫e
∴
− a2 x 2
0
In (2), put a =
1− i
2
, then
M07_ENGINEERING_MATHEMATICS-I _CH07_Part B.indd 59
⇒
−a 2 = i
5/12/2016 10:00:17 AM
7.60
∴
(2)
■
Engineering Mathematics
∞
⇒
∫e
ix 2
dx =
0
∞
∫ (cos (x
⇒
2
p
⎛1− i ⎞
2⎜
⎝ 2 ⎟⎠
(1 + i )
p (1 + i )
=
)
)
2 (1 + i (1 − i
2 2
) + i sin ( x 2 ) ) dx = p
0
∴ equating real and imaginary parts, we get
∞
∫ cos (x
2
) dx =
∞
p
0
2 2
∞
∞
0
0
and
∫ sin (x
) dx =
0
2
2
∫ cos (x ) dx = ∫ sin (x ) dx =
∴
2
p
2 2
1 p
2 2
EXAMPLE 12
1
Show that Γn 5 ∫ ⎛⎜ log ⎞⎟
⎝
y⎠
0
1
n 21
dy , n > 0 .
Solution.
∞
Γn = ∫ e − x x n −1dx , n > 0
We have
0
y = e −x
Put
⇒ ex =
1
y
⇒ x = log
1
y
∴ dx = 1 ⎛ − 1 ⎞ dy = − 1 dy
1 ⎜⎝ y 2 ⎟⎠
y
y
When x = 0, y = 1 and when x = ∞, y = 0
1⎞
⎛
Γn = ∫ y ⎜ log ⎟
⎝
y⎠
1
0
∴
n −1
1⎞
⎛ 1 ⎞
⎛
⎜⎝ − y dy ⎟⎠ = − ∫ ⎜⎝ log y ⎟⎠
1
0
n −1
1⎞
⎛
dy = ∫ ⎜ log ⎟
⎝
y⎠
0
1
n −1
dy
EXAMPLE 13
1
Prove that ∫
0
1
x 2 dx
12 x
4
3∫
0
dx
11 x
4
p
5
4 2
.
Solution.
1
Let
I =∫
0
1
Let
I1 = ∫
0
1
x 2dx
1− x
4
x 2dx
1− x 4
×∫
0
1
dx
1+ x 4
= ∫ x 2 (1 − x 4 )
M07_ENGINEERING_MATHEMATICS-I _CH07_Part B.indd 60
−
1
2
dx .
0
5/12/2016 10:00:25 AM
Improper Integrals ■
7.61
⎛ m + 1⎞
Γ ( p + 1) Γ ⎜
⎝ n ⎟⎠
1
Property (4) is ∫ x m (1 − x n ) p dx =
m + 1⎞
n ⎛
0
Γ ⎜ p +1+
⎟
⎝
n ⎠
1
⎛ 1 ⎞ ⎛ 2 + 1⎞
Γ ⎜ − + 1⎟ Γ ⎜
1 ⎝ 2 ⎠ ⎝ 4 ⎟⎠
I 1 = ∫ x 2 (1 − x ) dx =
2 + 1⎞
4
⎛ 1
0
Γ ⎜ − +1+
⎟
⎝ 2
4 ⎠
1
∴
1
−
4
2
1
=
4
1
Let I 2 = ∫
0
⎛ 3⎞
⎛ 3⎞
⎛ 1⎞ ⎛ 3⎞
⎛ 3⎞
Γ⎜ ⎟ Γ⎜ ⎟
Γ⎜ ⎟
Γ⎜ ⎟
Γ⎜ ⎟
⎝ 4⎠ 1
⎝ 4⎠
⎝ 2⎠ ⎝ 4⎠ 1
⎝ 4⎠
p
=
= p
=
p
5
1
5
1
4
⎛ ⎞ 4
⎛ 1⎞
⎛ ⎞
⎛ ⎞
Γ⎜ ⎟
Γ⎜ ⎟
Γ⎜ ⎟
Γ⎜ ⎟
⎝ 4⎠
⎝ 4⎠
⎝ 4⎠
4 ⎝ 4⎠
dx
1+ x 4
Put x 2 = tan u ∴ 2 xdx = sec 2 ud u ⇒ dx =
When x = 0, u = 0 and when x = 1, u =
∴
p/4
I2 =
∫
0
sec 2 ud u
(
1 + tan 2 u 2 tan u
)
2u = f,
∴ 2d u = d f
When u = 0, f = 0 and when u =
sec 2 ud u
2 tan u
p
4
=
=
Let
1
[Here m = 2, n = 4, p = − ]
2
1
2
1
2
⇒
p/4
∫
0
sec 2 ud u
p/4
∫
0
sin u
cos u
sec u
1
sin u cos u
du =
du =
2
2
p/4
∫
0
du
sin 2u
df
2
p
p
,f=
4
2
∴
M07_ENGINEERING_MATHEMATICS-I _CH07_Part B.indd 61
I2 =
2
⋅
2
p/2
∫ sin
0
−1/ 2
f
2
df
=
2
4
p/2
∫ sin
−1/ 2
f df
0
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=
2 1 ⎛ −1/ 2 + 1 0 + 1⎞
b⎜
,
⎟
4 2 ⎝
2
2 ⎠
=
2 ⎛ 1 1⎞
b⎜ , ⎟
8 ⎝ 4 2⎠
1
⎡
⎤
⎢⎣{ p = − 2 , q = 0 ⎥⎦
⎛ 1⎞
⎛ 1⎞ ⎛ 1⎞
⎛ 1⎞ ⎛ 1⎞
Γ⎜ ⎟
Γ⎜ ⎟ Γ⎜ ⎟
Γ⎜ ⎟ Γ⎜ ⎟
⎝ 4⎠
p
2 ⎝ 4⎠ ⎝ 2⎠
2 ⎝ 4⎠ ⎝ 2⎠
=
=
=
⋅
8
8
⎛ 1 1⎞
⎛ 3⎞
4 2 Γ ⎛ 3⎞
Γ⎜ + ⎟
Γ⎜ ⎟
⎜⎝ ⎟⎠
⎝ 4 2⎠
⎝ 4⎠
4
1
I = I1 × I 2 = ∫
∴
0
x 2 dx
1 − x4
1
×∫
0
dx
1+ x4
⎛ 3⎞
⎛ 1⎞
Γ⎜ ⎟
Γ
⎝ 4⎠
p ⎜⎝ 4 ⎟⎠
p
= p
⋅
=
⎛ 1⎞
⎛ 3⎞
Γ⎜ ⎟ 4 2 Γ⎜ ⎟ 4 2
⎝ 4⎠
⎝ 4⎠
EXAMPLE 14
∞
xc
dx 5 Γ( c 11) / (log e c ) c 11 .
x
0 c
Show that ∫
Solution.
∞
∞
∞
xc
xc
dx = ∫ x log c dx = ∫ e − x loge c x c dx
x
e
0 c
0 e
0
I =∫
Let
Put x loge c = t
∴
loge c ⋅ dx = dt
⇒ dx =
When x = 0, t = 0 and when x = ∞, t = ∞
∴
∞
[{ e
x log e c
=e
log e c x
= cx ]
dt
loge c
∞
1
Γ (c + 1)
⎛ t ⎞ dt
I = ∫ e −t ⎜
=
e −t t c +1−1dt =
c +1 ∫
⎟
(loge c )c +1
⎝ loge c ⎠ loge c (loge c ) 0
0
c
EXAMPLE 15
1⎞
⎛
Prove that b ⎜ m , ⎟ 5 2 2 m 21 b( m , m ).
⎝
2⎠
Solution.
p/2
We know that
b( m, n ) = 2 ∫ sin 2 m −1 u cos 2 n −1 ud u
0
∴
p/2
b( m, m ) = 2 ∫ sin 2 m −1 u cos 2 m −1 ud u
0
p/2
= 2 ∫ (sin u cos u) 2 m −1 d u
0
p/2
⎛ sin 2u ⎞
=2∫ ⎜
⎝ 2 ⎟⎠
2 m −1
du
0
=
M07_ENGINEERING_MATHEMATICS-I _CH07_Part B.indd 62
2
22 m −1
p/2
∫
0
sin 2 m −1 2u d u =
1
22 m − 2
p/2
∫ sin
2 m −1
2u d u
0
5/12/2016 10:00:41 AM
Improper Integrals ■
Let 2u = f [ 2d u = d f ⇒ d u =
When u = 0,
∴
f = 0 and when u =
b( m , m ) =
=
7.63
df
2
p
,f= p
2
p
1
22 m − 2
sin 2 m −1 f d f
∫0
2
1
2
2⋅
2 m −1
p/2
∫ sin
2 m −1
f df
[{ sin (p − f) = sin f]
0
p/2
⇒
22 m −1 b( m , m ) = 2 ∫ sin 2 m −1 f d f
0
p/2
= 2 ∫ sin 2 m −1 f cos 0 fd f
0
⎛ 2m − 1 + 1 0 + 1⎞
= b⎜
,
⎟
⎝
2
2 ⎠
⇒
⎛ 1⎞
22 m −1 b( m , m ) = b ⎜ m , ⎟
⎝
2⎠
EXAMPLE 16
1⎞
p
⎛
Prove that Γm ⋅ Γ ⎜ m 1 ⎟ 5 2 m 21 Γ ( 2 m ).
⎝
2⎠ 2
Solution.
From Example 15, we have
⎛ 1⎞
22 m −1 b( m , m ) = b ⎜ m , ⎟
⎝
2⎠
⇒
⇒
⇒
Γm Γm
=
22 m −1
Γ(m + m )
22 m −1 Γm
=
Γ ( 2m )
⎛ 1⎞
Γm ⋅ Γ ⎜ ⎟
⎝ 2⎠
1⎞
⎛
Γ⎜m + ⎟
⎝
2⎠
Γm ⋅ Γn ⎤
⎡
⎢∴ b( m , n ) = Γ ( m + n ) ⎥
⎣
⎦
p
1⎞
⎛
Γ⎜m + ⎟
⎝
2⎠
1⎞
p
⎛
Γm Γ ⎜ m + ⎟ = 2 m −1 Γ ( 2m )
⎝
2⎠ 2
M07_ENGINEERING_MATHEMATICS-I _CH07_Part B.indd 63
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EXAMPLE 17
Prove that b( n, n ) 5
p ? Γn
.
1⎞
⎛
2 n 21
2
Γ ⎜ n1 ⎟
⎝
2⎠
Solution.
From Example 15, we have
⎛ 1⎞
22 m −1 b( m , m ) = b ⎜ m , ⎟
⎝
2⎠
If m = n , we get
⎛ 1⎞
22 n −1 b( n , n ) = b ⎜ n , ⎟
⎝ 2⎠
⇒
⎛ 1⎞
Γn ⋅ Γ ⎜ ⎟
⎝ 2⎠
22 n −1 b( n , n ) =
1⎞
⎛
Γ⎜n + ⎟
⎝
2⎠
⇒ b( n , n ) =
p ⋅ Γn
1⎞
⎛
22 n −1 Γ ⎜ n + ⎟
⎝
2⎠
EXAMPLE 18
Prove that
b( m 11, n ) b( m , n 11) b( m , n )
5
5
.
m
n
m1n
Solution.
We have
b( m , n ) =
Γm Γn
Γ(m + n)
Γ ( m + 1)Γn
Γ(m + 1 + n)
∴
b( m + 1, n ) =
∴
b( m + 1, n )
Γ ( m + 1) ⋅ Γn
m Γm ⋅ Γn
Γm ⋅ Γn
=
=
=
m
m Γ ( m + n + 1) m Γ ( m + n + 1) ( m + n )Γ ( m + n )
⇒
b( m + 1, n ) b( m , n )
=
m
m+n
and
b( m , n + 1) Γm ⋅ Γ ( n + 1)
=
n
nΓ ( m + n + 1)
=
(1)
Γm ⋅ n Γn
Γm ⋅ Γn
=
n ⋅ ( m + n )Γ ( m + n ) ( m + n )Γ ( m + n )
b( m , n + 1) b( m , n )
=
n
m+n
From (1) and (2), we get
⇒
(2)
b( m + 1, n ) b( m , n + 1) b( m , n )
=
=
m
n
m+n
M07_ENGINEERING_MATHEMATICS-I _CH07_Part B.indd 64
5/12/2016 10:00:53 AM
Improper Integrals ■
7.65
EXAMPLE 19
Prove that b( m 11, n ) 1 b( m , n 11) 5 b( m , n ).
Solution.
b( m , n ) =
We have
Γm Γn
Γ(m + n)
∴
b( m +1, n ) =
Γ ( m + 1) ⋅ Γn
m Γm ⋅ Γn
=
Γ ( m + n + 1) Γ ( m + n + 1)
⇒
b( m + 1, n ) =
m Γm ⋅ Γn
m
=
⋅ b( m , n )
( m + n )Γ ( m + n ) ( m + n )
and
b( m , n + 1) =
Γm ⋅ Γ ( n + 1)
Γm ⋅ n Γn
=
Γ ( m + n + 1) ( m + n )Γ ( m + n )
⇒
b( m , n + 1) =
n
Γm Γn
⋅
(m + n) Γ(m + n)
⇒
b( m , n + 1) =
n
⋅ b( m , n )
(m + n)
(1)
(2)
Adding (1) and (2), we get
m
n
b( m , n ) +
b( m , n )
m+n
m+n
m+n
=
b( m , n )
m+n
⇒
b( m + 1, n ) + b( m , n + 1) =
⇒
b( m + 1, n ) + b( m , n + 1) = b( m , n )
Aliter By example 18,
b( m + 1, n ) b( m , n + 1) b( m , n )
=
=
m
n
m+n
By rule of ratio and proportion, each ratio =
b( m + 1, n ) + b( m , n + 1)
m+n
∴
b( m + 1, n ) + b( m , n + 1) b( m , n )
=
m+n
m+n
⇒
b( m + 1, n ) + b( m , n + 1) = b( m , n )
EXAMPLE 20
1
∫x
1
m
(1 2x n ) p dx in terms of gamma function and hence, evaluate
0
M07_ENGINEERING_MATHEMATICS-I _CH07_Part B.indd 65
∫
0
dx
12 x n
.
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Engineering Mathematics
Solution.
1
1
m
n p
∫0 x (1 − x ) dx = n
we have (i)
1
∫
(ii)
0
1
dx
1− x n
⎛ m + 1⎞
Γ ( p + 1)Γ ⎜
⎝ n ⎟⎠
m + 1⎞
⎛
Γ ⎜ ( p + 1) +
⎟
⎝
n ⎠
= ∫ (1 − x n )
−
1
2
[Refer property 4 Page 7.54]
dx
0
1
2
Here m = 0, p = −
1
∫
∴
0
⎛ 1 ⎞ ⎛ 0 + 1⎞
Γ ⎜ − + 1⎟ Γ
1 ⎝ 2 ⎠ ⎜⎝ n ⎟⎠
=
1 − x n n Γ ⎡⎛ − 1 + 1⎞ + 0 + 1⎤
⎢ ⎜⎝ 2 ⎟⎠
n ⎥⎦
⎣
dx
1
=
n
⎛ 1⎞ ⎛ 1⎞
1
Γ⎜ ⎟ Γ⎜ ⎟
Γ
⎝ 2⎠ ⎝ n⎠
p
n
=
⎛ 1 1⎞
n
⎛ n + 2⎞
Γ⎜ + ⎟
Γ⎜
⎝ n 2⎠
⎝ 2n ⎟⎠
1
⎡
⎤
⎢⎣{ Γ 2 = p ⎥⎦
EXAMPLE 21
1
∫x
Evaluate
5
(12 x 3 )10 dx .
0
Solution.
1
I = ∫ x 5 (1 − x 3 )10 dx
Let
0
Here m = 5, n = 3, p = 10
1
1
I = ∫ x 5 (1 − x 3 )10 dx =
3
0
∴
=
⎛ 5 + 1⎞
Γ (10 + 1)Γ ⎜
⎝ 3 ⎟⎠
5 + 1⎞
⎛
Γ ⎜ 10 + 1 +
⎟
⎝
3 ⎠
[Using property 4 Page 7. 54]
1 Γ (11)Γ ( 2)
Γ (11) × 1
1
=
=
.
3 Γ (13)
3 × 12 × 11 Γ (11) 396
EXAMPLE 22
1
Express
∫x
7
(1 2 x 4 ) 9 dx in terms of gamma function and evaluate.
0
Solution.
1
Let
I = ∫ x 7 (1 − x 4 )9 dx
0
M07_ENGINEERING_MATHEMATICS-I _CH07_Part B.indd 66
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Improper Integrals ■
7.67
Here m = 7, n = 4, p = 9
1
I=
n
∴
⎛ m + 1⎞
Γ ( p + 1)Γ ⎜
⎝ n ⎟⎠
m + 1⎞
⎛
Γ ⎜ p +1+
⎟
⎝
n ⎠
[Using property 4 Page 7.54]
1
=
4
⎛ 7 + 1⎞
Γ (9 + 1)Γ ⎜
⎝ 4 ⎟⎠ 1 Γ (10)Γ ( 2)
1
Γ (10) × 1
=
=
=
7 + 1⎞
4 × 12 × 11 × 10Γ(10) 5280
4 Γ (13)
⎛
Γ ⎜ 9 +1+
⎟
⎝
4 ⎠
,
(2) ∫ (1 2x 3 ) 5 dx ,
EXAMPLE 23
1
1
dx
Evaluate (1) ∫
12 x
0
4
0
1
(3) ∫ x 5 (1 2 x 4 ) 7 dx ,
0
1
(4) ∫
0
dx
12 x n
.
Solution.
1
(1)
∫
0
dx
1− x 4
1
∫
dx
1
= ∫ x 0 (1 − x 4 ) −1/ 2 dx
1− x 4 0
Here m = 0, n = 4, p = −1/2
0
1
∴
∫
0
⎛ m + 1⎞
Γ ( p + 1)Γ ⎜
⎝ n ⎟⎠
1
[Using property 4 Page 7.54]
= ⋅
1 − x 4 n Γ ⎛ p + 1 + m + 1⎞
⎜⎝
⎟
n ⎠
⎛ 1⎞
⎛ 1⎞ ⎛ 1⎞
⎛ 1 ⎞ ⎛ 0 + 1⎞
Γ⎜ ⎟
Γ⎜ ⎟ Γ⎜ ⎟
Γ ⎜ − + 1⎟ Γ ⎜
⎟
⎝ 4⎠
⎝
⎠
⎝
⎠
⎝
⎠
⎠
⎝
p
1
1
2
4 =
4
2
⋅
= ⋅
= ⋅
0 + 1⎞
4
4
4
⎛ 1 1⎞
⎛ 3⎞
⎛ 1
Γ⎜ ⎟
Γ ⎜ − +1+
Γ⎜ + ⎟
⎟
⎝ 2 4⎠
⎝ 4⎠
⎝ 2
4 ⎠
dx
1
(2)
∫ (1 − x
3 5
) dx
0
1
1
0
0
3 5
0
3 5
∫ (1 − x ) dx = ∫ x (1 − x ) dx
Here m = 0, n = 3, p = 5
∴
1
1
3 5
∫0 (1 − x ) dx = n ⋅
M07_ENGINEERING_MATHEMATICS-I _CH07_Part B.indd 67
⎛ m + 1⎞
Γ ( p + 1)Γ ⎜
⎝ n ⎟⎠
m + 1⎞
⎛
Γ ⎜ p +1+
⎟
⎝
n ⎠
[Using property 4 Page 7.54]
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Engineering Mathematics
⎛ 0 + 1⎞
Γ (5 + 1)Γ ⎜
⎝ 3 ⎟⎠
1
= ⋅
0 + 1⎞
3
⎛
Γ ⎜5 +1+
⎟
⎝
3 ⎠
⎛ 1⎞
⎛ 1⎞
5!⋅ Γ ⎜ ⎟
Γ6 ⋅ Γ ⎜ ⎟
⎝ 3⎠
⎝ 3⎠ 1
1
= ⋅
= ⋅
1⎞ 3
3 ⎛
⎛ 19 ⎞
Γ⎜ ⎟
Γ ⎜6 + ⎟
⎝
⎝ 3⎠
3⎠
⎛ 1⎞
Γ⎜ ⎟
⎝ 3⎠
120
36
729
.
=
=
=
3 16 13 10 7 4 1 1 16 ⋅13 ⋅ 7 1456
. . . . . Γ
3 3 3 3 3 3 3
1
(3)
∫x
5
(1 − x 4 )7 dx
0
Here m = 5, n = 4, p = 7
1
1
5
4 7
∫0 x (1 − x ) dx = n ⋅
We know
1
(4)
∫
0
⎛ m + 1⎞
Γ ( p + 1)Γ ⎜
⎝ n ⎟⎠
m + 1⎞
⎛
Γ ⎜ p +1+
⎟
⎝
n ⎠
[Using Property 4]
1
= ⋅
4
⎛ 5 + 1⎞
Γ (7 + 1)Γ ⎜
⎝ 4 ⎟⎠
5 + 1⎞
⎛
Γ ⎜ 7 +1+
⎟
⎝
4 ⎠
1
= ⋅
4
1 ⎛ 1⎞
⎛ 3⎞
Γ8 ⋅ Γ ⎜ ⎟
7! ⋅ Γ ⎜ ⎟
⎝ 2⎠ 1
2 ⎝ 2 ⎠ = 7! p Γ ⎛ 19 ⎞
=
⎜⎝ ⎟⎠
3
8
2
⎛
⎞ 4
⎛ 19 ⎞
Γ⎜ ⎟
Γ ⎜8 + ⎟
⎝
⎝ 2⎠
2⎠
dx
1− x n
Here m = 0, n = p = −
1
We know
∫
0
1
2
⎛ 1 ⎞ ⎛ 0 + 1⎞
Γ ⎜ − + 1⎟ Γ ⎜
⎝ 2 ⎠ ⎝ n ⎟⎠
1
= ∫ x 0 (1 − x n ) −1/ 2 dx =
0 + 1⎞
n
⎛ 1
1 − xn 0
Γ ⎜ − +1+
⎟
⎝ 2
n ⎠
dx
1
1
=
n
M07_ENGINEERING_MATHEMATICS-I _CH07_Part B.indd 68
[m = 0, n, p = −1/2]
⎛ 1⎞ ⎛ 1⎞
⎛ 1⎞
Γ⎜ ⎟ Γ⎜ ⎟
Γ⎜ ⎟
⎝ n⎠
⎝ n ⎠ ⎝ 2⎠
p
=
n
⎛ 1 1⎞
⎛ 1 1⎞
Γ⎜ + ⎟
Γ⎜ + ⎟
⎝ n 2⎠
⎝ n 2⎠
5/12/2016 10:01:14 AM
Improper Integrals ■
7.69
EXERCISE 7.4
1. Prove that
p/2
3. Prove that
p/2
b( m + 1, n ) m
= .
b( m, n + 1) n
∫
cos x dx ×
2. Prove that
0
p/2
0
1 ⎛ 1⎞ ⎛ 3⎞
cot ud u = Γ ⎜ ⎟ Γ ⎜ ⎟ .
2 ⎝ 4⎠ ⎝ 4⎠
∫
∫
cos x
0
1⎤
⎡
4. Prove that ∫ ⎢log ⎥
⎦
⎣
x
0
1
dx
= p.
∞
3
5
5. Evaluate ∫ xe − x dx given that Γ ⎛⎜ ⎞⎟ = 0.902.
⎝
3⎠
0
p/2
7. Find the value of
∫
∫ sin
3
x cos5 / 2 x dx .
0
1
0
dx = Γn.
p/2
6. Find the value of
8. Prove that ∫
sin11 u cos12 u d u.
n −1
0
1 ⎛ 2 1⎞
= b⎜ , ⎟.
5 ⎝ 5 2⎠
1− x
xdx
5
b
9. Prove that ∫ ( x − a) m (b − x ) n dx = (b − a) m + n +1 b( m + 1, n + 1).
a
1
10. Prove that ∫ x m (log x ) n dx =
0
⎛
11. Show that ∫ x q −1 ⎜ log
⎝
1
0
1⎞
⎟
x⎠
( −1) n n !
, where n is positive m > −1.
( m + n ) n +1
p −1
dx =
Γp
where p > 0, q > 0.
qn
1
1
12. Evaluate (i)
7
9
∫ x (1 − x ) dx
(ii)
1
∫x
13
(1 − x )12 dx
0
0
13. Evaluate (i)
∫x
1
7
(1 − x 2 )7 dx
(ii)
0
∫x
1
3
(1 − x 5 )8 dx
(iii)
0
∫x
2
(1 − x 3 ) 4 dx
0
∞
14. Express ∫ x p −1e − kx dx ( k > 0) in terms of gamma function.
0
x 2dx
dx
p
⋅
= .
4 1/ 2 ∫
4 1/ 2
4
0 (1 − x )
0 (1 − x )
1
15. Show that ∫
1
ANSWERS TO EXERCISE 7.4
5. 0.451
7 ! 9 !⎞
12. (i) ⎛⎜
⎝ 17 ! ⎟⎠
6.
13! 12 !⎞
(ii) ⎛⎜
⎝ 24 ! ⎟⎠
8
77
13. (i)
M07_ENGINEERING_MATHEMATICS-I _CH07_Part B.indd 69
7.
12 13/ 2
25/ 2
1
490
⎛ 4⎞ ⎛ 4⎞
Γ⎜ ⎟ Γ⎜ ⎟
⎝ 5⎠ ⎝ 5⎠
(ii)
⎛ 24 ⎞
2Γ ⎜ ⎟
⎝ 5⎠
(iii)
2
15
14.
Γp
kp
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7.5
Engineering Mathematics
THE ERROR FUNCTION
The normal distribution in probability theory is the most important and useful distribution. Many
random phenomena that occur in nature, industry and scientific studies are found to follow normal
distribution.
It is also known as Gaussian distribution in honour of the german mathematician Gauss, because
Gauss obtained this distribution from the study of errors of measurements made a large number of
z
1
times. The standard normal probability integral is
2p
∫e
−
z2
2
dz
0
Definition 7.3
The integral
2
x
∫e
p
−t 2
dt is defined as the error function of x and it is denoted by er f(x).
0
∴
er f ( x ) =
2
x
∫e
p
−t 2
dt
0
Note
1. It is called error function because it is very closely related to normal distribution, which is derived
from the study of errors of measurements. It also arises in solution of partial differential equations, thermodynamics and engineering applications.
2. The error function is also known as the probability integral.
3. The error function is a “special function” because it cannot be evaluated by usual methods of
integration.
7.5.1
Properties of Error Functions
(1) Error function is an odd function
Proof
x
2
2
We have
er f ( x ) =
e −t dt
∫
p0
∴
er f ( − x ) =
2
−x
∫e
p
−t 2
dt
0
[ du = −dt
u = −t
When t = 0, u = 0 and when t = − x , u = x
Put
∴
er f ( − x ) =
2
x
−u
∫ e ( −du ) = −
p0
∴ er f(x) is an odd function of x.
(2) er f ( x ) 1er f (2 x ) 50
Proof
we have
2
2
x
∫e
p
−u 2
er f ( x ) + er f ( − x ) = er f ( x ) − er f ( x ) = 0
M07_ENGINEERING_MATHEMATICS-I _CH07_Part B.indd 70
du = −er f ( x )
■
0
[ using (1)]
■
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Improper Integrals ■
7.71
(3) er f ( 0 ) 5 0 and (4) er f ( ∞) 51
Proof
2
x
∫e
p
er f ( x ) =
we have
−t 2
dx
0
∴
2
er f (∞) =
p
∞
2
−t
∫ e dx =
2
p
0
⋅
⎡ ∞ −x2
p⎤
⎢{ ∫ e dx =
⎥
2 ⎦
⎣ 0
p
=1
2
■
This proves that the total area under error function is 1.
7.5.2
Series Expansion for Error Function
2 ⎡
1 x 3 1 x 5 1 x 7 …⎤
+
−
+ ⎥
⎢x −
⎦
p ⎣ 1! 3 2 ! 5 3! 7
er f ( x ) =
Proof
2
x
∫e
p
er f ( x ) =
We have
−t 2
■
dt
0
⎛ t2 t4 t6
⎞
⎜⎝1 − + − + …⎟⎠ dt
∫
1! 2 ! 3!
p
2
=
x
0
x
2 ⎡ 1 t 3 1 t 5 1 t 7 …⎤
=
+
−
+ ⎥
⎢t −
p ⎣ 1! 3 2 ! 5 3! 7
⎦0
2 ⎡
1 x 3 1 x 5 1 x 7 …⎤
+ ⎥
x
−
+
−
⎢
⎦
p ⎣ 1! 3 2! 5 3! 7
=
7.5.3
Complementary Error Function
Definition 7.4
∞
2
2
The complementary error function of x is defined as
e −t dt and it is denoted by er f c ( x ).
∫
px
∞
2
∫e
p
er f c ( x ) =
∴
−t 2
dt .
x
(5) Prove that er f c ( x ) 5 1 2er f ( x )
Proof
er f ( x ) =
We have
2
x
∫e
p
0
2
∞
∫e
p
and
−t 2
dt = 1
−t 2
dt = 1
−t 2
dt ,
er fc ( x ) =
2
∞
∫e
p
−t 2
dt
x
[by property(4)]
0
⇒
2
p
x
−t
∫ e dt +
2
0
2
∞
∫e
p
x
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er f ( x ) + er fc ( x ) = 1
⇒
⇒
er fc ( x ) = 1 − er f ( x )
(6) Prove that er fc ( x ) 1 er fc (2x ) 5 2
Proof
we have
■
er fc ( x ) + er fc ( − x ) = 1 − er f ( x ) + 1 − er f ( − x )
= 2 − [er f ( x ) + er f ( − x )] = 2 − 0 = 2
■
WORKED EXAMPLES
EXAMPLE 1
Prove that
d
2 a 2a 2 x 2
e
[er f ( ax )] 5
.
dx
p
Solution.
We know that, the series expansion of
∴
er f ( x ) =
2 ⎡
1 x 3 1 x 5 1 x 7 …⎤
+
+
+ ⎥
⎢x −
⎦
p ⎣ 1! 3 2 ! 5 3! 7
er f (ax ) =
2 ⎡
1 a3 x 3 1 a5 x 5 1 a7 x 7 …⎤
ax
+ ⎥
−
+
−
⎢
1! 3
2! 5
3! 7
⎦
p⎣
d
2 ⎡ a3 x 2 a5 x 4 a7 x 6 …⎤
[er f (ax )] =
+
−
+ ⎥
⎢a −
dx
1!
2!
3!
⎦
p⎣
=
2a ⎡ a2 x 2 a4 x 4 a6 x 6 …⎤
−
+ ⎥
+
⎢1 −
1!
2!
3!
⎦
p⎣
=
2a ⎡ (a2 x 2 )1 (a2 x 2 ) 2 (a2 x 2 )3 …⎤ 2a − a2 x 2
e
+
−
+ ⎥=
⎢1 −
1!
2!
3!
⎦
p⎣
p
EXAMPLE 2
Evaluate
d
[er f ( ax )] by method of differentiation under integral sign.
dx
Solution.
By definition er f ( x ) =
x
2
−t
∫ e dt
2
p
∴
er f (ax ) =
0
2
ax
∫e
p
−t 2
dt
0
By Leibnitz’s rule, page 7.28 [or theorem 7.7]
∴
d
2
[er f ( ax )] =
dx
p
=
∂
∫ ∂x (e ) dt + e
ax
−t 2
− a2 x 2
⋅a − 0
0
ax
⎤ 2a − a2 x 2
2 ⎡
− a2 x 2
e
⎢ ∫ 0dt + ae
⎥=
p ⎣0
p
⎦
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Improper Integrals ■
7.73
EXAMPLE 3
u
Evaluate ∫ er f ( x ) dx by changing the order of integration.
0
Solution.
u
Now
u
⎡ 2
∫ er f (x )dx = ∫ ⎢⎣
0
0
x
∫e
p
−t 2
0
u x
⎤
2 ⎡ −t 2 ⎤
dt ⎥ dx =
⎢ ∫ e dt ⎥ dx
∫
p 0 ⎣0
⎦
⎦
Changing the order of integration,
⎛ u ⎞ −t 2
∫ ⎜ ∫ dx ⎟ e dt
p 0⎝t ⎠
u
∫ er f (x )dx =
2
u
2
u
=
0
∫ [x ] e
p
u
t
−t 2
dt
0
2
=
p
2
=
p
= u⋅
t
u
−t
∫ (u − t ) e dt
2
t=u
(a, u)
t=x
0
u
2
u
∫e
p
−t 2
dt −
x=0
x=u
0
o
u
2
∫ te
p
0
= u er f (u ) −
−t
∫ te dt
2
p
0
2
u
2
−t
∫ ue dt −
−t 2
x
dt
0
2
I1
p
u
I 1 = ∫ te −t dt
2
where
0
Put y = t 2
∴ dy = 2tdt
⇒ tdt =
dy
2
When t = 0, y = 0 and when t = u, y = u 2
u2
∴
I1 =
−y
∫e
0
2
1
dy 1 − y u2 −1 − u2
= [ e ]0 = [ e − e 0 ] = − [ e − u − 1]
2 2
2
2
∴
2 (e −u − 1)
er
f
(
x
)
dx
=
u
er
f
(
u
)
+
∫0
2
p
⇒
∫ er f (x )dx = u er f (u ) +
2
u
u
0
M07_ENGINEERING_MATHEMATICS-I _CH07_Part B.indd 73
1
p
(e −u − 1)
2
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EXAMPLE 4
Find the relation between the cumulative distribution function of the standard normal random
variable Z and the error function.
Solution.
The probability density function of the standard normal randam variable Z
1
f( z ) =
e −z
2p
2
, −∞ < z < ∞
/2
If F is the cumulative distribution function of Z, then
z
0
z
−∞
−∞
0
∫ f( z ) dz = ∫ f( z ) dz + ∫ f( z ) dz
F ( z) = P(Z ≤ z) =
z
⇒
F(z ) =
⇒
F(z ) =
2
1
1
+
e − z / 2 dz
∫
2
2p 0
(1)
z
2
1
1
+
e −u / 2 du
∫
2
2p 0
But error function in z is
2
z
∫e
p
er f ( z ) =
−t 2
dt
0
Put t =
u
2
∴ dt =
in the error function.
du
2
When t = 0, u = 0 and when t = z , u = 2 z
∴
⇒
2
er f ( z ) =
p
2
er f ( z ) =
2p
2z
∫
e−u
2
/2
du
2
0
2z
∫
e − u / 2 du
2
(2)
0
But from (1),
F ( 2 z) =
1
1
+
2
2p
2z
∫
e − u / 2 du
2
0
1 1⎡ 2
= + ⎢
2 2 ⎢⎣ 2p
2z
∫
0
⎤ 1 1
2
e − u / 2 du ⎥ = + er f ( z )
⎥⎦ 2 2
[Using (2)]
Replacing 2z by x, we get
F(x ) =
1 1
⎛ x ⎞
+ er f ⎜
⎝ 2 ⎟⎠
2 2
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Improper Integrals ■
7.75
EXAMPLE 5
Compute er f ( 0 . 5 ) correct to 3 places of decimals.
Solution.
The series expansion of error function is
2 ⎡
1 x 3 1 x 5 1 x 7 …⎤
x
−
+
−
+ ⎥
⎢
⎦
p ⎣ 1! 3 2! 5 3! 7
er f ( x ) =
(1)
Put x = 0.5 in (1), then we get
2 ⎡
(0.5)3 (0.5)5 (0.5)7 …⎤
+
−
+ ⎥
⎢ 0. 5 −
1! ⋅ 3
2! ⋅ 5
3! ⋅ 7
p⎣
⎦
2
[0.5 − 0.041667 + 0.003125 − 0.000186 + …]
=
p
er f (0.5) =
=
0.922544
= 0.5204
1.772454
EXAMPLE 6
∞
Prove that ∫ e 2x
2
22 ax
2
p
[1 2er f ( a )]e a .
2
dx 5
0
Solution.
∞
−x
∫e
2
− 2 ax
0
∞
= ∫ e −( x
2
+ 2 ax )
0
∞
dx = ∫ e −[( x + a )
2
− a2 ]
dx
0
∞
= ∫ e −( x + a)
0
2
+ a2
∞
dx = e a ∫ e − ( x + a ) dx
2
2
0
Put u = x + a ∴ du = dx
When x = 0, u = a and when x = ∞, u = ∞
∴
∞
∫e
− ( x + a )2
0
∞
dx = ∫ e −u du
2
0
er fC ( x ) =
We know that
2
∞
∫e
p
−t 2
dt
x
er fC (a) =
∴
2
∞
∫e
p
−t 2
dt
a
∞
∫e
⇒
−t 2
a
∴
∞
∫e
− x 2 − 2 ax
dt =
p
er fC (a)
2
dx = e a
2
0
M07_ENGINEERING_MATHEMATICS-I _CH07_Part B.indd 75
p
2
p
er fC (a) =
[1 − er f (a)]e a
2
2
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■
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EXERCISE 7.5
b
1. Prove that ∫ e − x dx =
2
0
p
[er f (b ) − er f (a)].
2
t
2. Prove that ∫ er f (ax )dx = t ⋅ er f (at ) +
0
1
a p
[e
− a2t 2
− 1] .
[Hint Applying integration by parts with u = er f ( x ) and v = 1]
3. Prove that
d
2a − a2 x 2
er f c ( x ) ] = −
e .
[
dx
p
4. Compute er f (0.3) .
ANSWERS TO EXERCISE 7.5
4. 0.3248
SHORT ANSWER QUESTIONS
Problems based on improper integrals of the first kind
∞
∞
1. Test the convergence of
2. Test the convergence of ∫ e
dx
∫0 11 x 2
∞
3. Test the convergence of
∞
dx
∫0 x 5
∞
5. Test the convergence of ∫
1
dx
∫x
5
∞
x x 21
6. Test the convergence of
∫e
1
∞
2a x
dx , if a > 0.
8. Test the convergence of ∫
0
2∞
∞
9. Test the convergence of
dx
1
2
∫e
7. Test the convergence of
4. Test the convergence of
dx
∞
2x
0
x
∫ e dx
0
1
dx
13
dx
x 4 12
∞
10. Test the convergence of
x
∫e
2x
dx
0
∞
dx
11
0
Problems based on improper integrals of the second kind
11. Test the convergence of
12. Test the convergence
∫e
x
2
dx
0
( x 21) 3
∫
1
1
13. Test the convergence of ∫
0
dx
and find the value, if it is convergent.
x
1
14. Test the convergence of
dx
x
0
∫
M07_ENGINEERING_MATHEMATICS-I _CH07_Part B.indd 76
1
15. Test the convergence of
∫
21
dx .
2
x3
5/12/2016 10:02:03 AM
Improper Integrals ■
1
16. Test the convergence of
dx
∫
12 x 2
0
1
17. Prove that
1
∫x
3
7.77
.
dx has Cauchy’s principal value, but not the general value.
21
2
18. State true or false
dx
has a finite value
x2
22
∫
1
20. Evaluate the improper integral
∫
0
1
19. Test the convergence of
∫
21
dx
x
xdx
1 2x 2
21. Prove that er f(x) 1 er f(2 x) = 0
22. Evaluate
d
[er f ( ax )]
dx
OBJECTIVE TYPE QUESTIONS
A. Fill up the blanks
3
1.
∫
1
x
4−x + x
p
2
3.
∫ sin
∞
dx = __________
2.
0
9
x dx = _________
4.
e − x dx = __________
∫e
−x
dx = _________
−∞
∞
5. Cauchy’s principal value of
x
∫− ∞ 8 − x 3 dx = __________ 6.
∞
ln x
dx is _________ (convergent or divergent)
x2
1
∫
1
dx
is _________ (convergent or divergent)
x
0
∫
ln 2
8.
∫x
−2
e
−
1
x
dx is _________ (convergent or divergent)
0
∞
9.
3
0
0
7.
∫x
2 + cos x
dx is _________ (convergent or divergent) 10.
x
p
∫
∞
dx
∫
−∞
x4 +1
is _________ (convergent or divergent)
B. Choose the correct answer
p
1. ∫ sin 3 x (1 − cos x ) 2 dx is equal to
0
(a)
1
2
(b)
4
5
(c)
8
5
(d)
5
4
(b)
1
12
(c)
5
8
(d)
3
4
(c)
p
2
(d)
p
4
(c)
p
4
(d)
3p
4
2. b(3, 2) is equal to
(a)
3
2
3
3. The improper integral
∫
0
(a) 1
∞
4.
dx
∫1+ x
2
dx
converges to
9 − x2
(b) p
is equal to
1
(a) 0
(b) p
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■
7.78
Engineering Mathematics
∞
5.
ln x
dx is equal to
x3
1
(a) 1
∫
1
6.
⎛
1⎞
∫ ⎜⎝ ln x ⎟⎠
(b)
n
(c)
(b)
1
2
(c) 2
(d)
⎛ 3 1⎞
(b) b ⎜ , ⎟
⎝ 4 2⎠
⎛ 1 1⎞
(c) b ⎜ , ⎟
⎝ 4 2⎠
(d) None of these
(b) ln (m + 1)
(c) ln
(b) 1
(c) 0
n −1
∞
dx
3
(a) 1
1
∫
0
(a)
x2
1− x4
1
(d) None of these
n
is equal to
e
8.
(d)
dx , n > 0, is
(a) n!
∫ x (ln x )
1
4
(c)
0
7.
1
2
(b) 2
1
3
dx =
1 ⎛ 3 1⎞
b⎜ , ⎟
4 ⎝ 4 2⎠
x m −1
dx , m ≥ 0, is
ln x
0
1
9.
∫
(a) ln m
10. er f c ( x ) + er f c (c − x ) =
(a) 2
⎛ 1 ⎞
(d) ln ⎜
⎝ m + 1⎟⎠
1
m
(d) 4
ANSWERS
A. Fill up the blanks
1. 1
2. 2.6
3.
6. divergent
7. convergent
128
4. 1
5. 0
315
8. convergent
9. divergent
5. (c)
7. (b)
10. convergent
B. Choose the correct answer
1. (c)
2. (b)
3. (c)
4. (c)
M07_ENGINEERING_MATHEMATICS-I _CH07_Part B.indd 78
6. (b)
8. (a)
9. (a)
10. (a)
5/12/2016 10:02:21 AM
8
Multiple Integrals
8.1
DOUBLE INTEGRATION
Double integrals occur in many practical problems in science and engineering. It is used in problems
involving area, volume, mass, centre of mass. In probability theory it is used to evaluate probabilities
of two dimensional continuous random variables.
8.1.1
Double Integrals in Cartesian Coordinates
A double integral is defined as the limit of a sum. Let f(x, y) be a continuous function of two independent variables x and y defined in a simple closed region R. Sub-divide R into element areas
ΔA1 , ΔA 2 , …, ΔA n by drawing lines parallel to the coordinate axes.
y
ΔAi
(xi , yi )
R
O
x
Fig. 8.1
Let (xi, yi) be any point in ΔA i .
n
Find the sum f ( x1, y1 )DA 11 f ( x2 , y2 )DA 2 1…1 f ( xn , yn )DA n 5∑ f ( xi , yi ) DA i
i =1
Increase the number of sub-divisions indefinitely large.
i.e., n → ∞ so that each ΔA i → 0.
n
In this limit, if the sum exists, i.e.,nlim
→∞
∑ f (x , y ) ΔA
i
ΔA i → 0 i =1
i
i
exists, then it is called the double integral of
f(x, y) over the region R and it is denoted by
∫∫ f ( x, y) dxdy.
R
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Engineering Mathematics
Note The continuity of f(x, y) is a sufficient condition for the existence of double integral, but not
necessary. The double integral exists even if finite number of discontinuous points are there in R, but
f should be bounded.
8.1.2
Evaluation of Double Integrals
In practice, a double integral is computed by repeated single variable integration, integrate with
respect to one variable treating the other variable as constant.
Case 1: If the region R is a rectangle given by R = {( x, y ) a ≤ x ≤ b , c ≤ y ≤ d}
where a, b, c, d are constants, then
y
b
⎡
⎤
d
d
⎡
b
⎤
∫∫ f ( x, y) dx dy = ∫ ⎢⎣∫ f ( x, y) dy ⎥⎦ dx = ∫ ⎢⎣∫ f ( x, y) dx ⎥⎦ dy
R
a
c
c
a
y=d
R
If the limits are constants the order of integration is immaterial,
provided proper limits are taken and f(x, y) is bounded in R
Case 2: If the region R is given by
y=c
x=a
O
x=b
Fig. 8.2
R = {( x, y ) a ≤ x ≤ b , g ( x ) ≤ y ≤ h ( x )}
y
where a and b are constants, then
b h(x )
⎤
⎡
f
(
x
,
y
)
dx
dy
=
∫∫R
∫a ⎢⎢ g ∫( x ) f (x, y ) dy ⎥⎥ dx
⎣
⎦
Here the limits of x are constants and the limits of y are functions
of x, so we integrate first with respect to y and then with respect
to x.
h(x )
R
g(x )
x=a
O
x=b
R = {( x , y ) g ( y ) ≤ x ≤ h ( y ) , c ≤ y ≤ d}
h(y)
y
where c and d are constants then
y=d
⎡
⎤
f ( x , y ) dx ⎥ dy
⎥⎦
c ⎣g (y )
h(y )
∫∫ f (x , y ) dx dy = ∫ ⎢⎢ ∫
R
x
Fig. 8.3
Case 3: If the region R is given by
d
x
Since the limits of x are functions of y, we integrate first w.r.to
x and then w.r.to y.
R
y=c
g(y)
O
x
Note
Fig. 8.4
(1) When variable limits are involved we have to integrate first
w.r.to the variable having variable limits and then w.r.to the
variable having constant limits.
(2) When all the limits are constants, the order of dx, dy determine the limits of the variable.
M08_ENGINEERING_MATHEMATICS-I _CH08.indd 2
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Multiple Integrals ■
8.3
WORKED EXAMPLES
EXAMPLE 1
1 2
∫ ∫ x ( x 1 y ) dydx .
Evaluate
0 1
Solution.
1 2
1 2
⎡
⎤
Let I = ∫ ∫ x( x + y ) dy dx = ∫ ⎢ ∫ x( x + y ) dy ⎥ dx
0 1
0 ⎣1
⎦
2
⎡
y2 ⎤
= ∫ x ⎢ xy + ⎥ dx
2 ⎦1
⎣
0
1
1
⎧⎡
22 ⎤ ⎡
1 ⎤⎫
= ∫ x ⎨ ⎢ x ⋅ 2 + ⎥ − ⎢ x ⋅1 + ⎥ ⎬ dx
2⎦ ⎣
2 ⎦⎭
⎩⎣
0
1
1
⎡ x 3 3x 2 ⎤
3⎞
3x ⎞
1 3 13
⎛
⎛
= ∫ x ⎜ x + ⎟ dx = ∫ ⎜ x 2 + ⎟ dx = ⎢ +
⎥ = + =
⎝
⎠
⎝
⎠
2
2
⎣ 3 2 ⋅ 2 ⎦ 0 3 4 12
0
0
1
EXAMPLE 2
1 1
Evaluate
dxdy
∫∫
.
12 x 2 12 y 2
0 0
Solution.
1 1
1
dxdy
Let I = ∫ ∫
1− x
0 0
2
1− y
2
1
dx
=∫
1− x
0
2
⋅∫
0
dy
1 − y2
= [sin −1 x ]10 [sin −1 y ]10 = (sin −1 1 − sin −1 0) (sin −1 1 − sin −1 0) =
p p p2
⋅ =
4
2 2
Note We could write the integral in Example 2 as a product of two integrals because the limits are
constants and the functions could be factorised as x terms and y terms. This is not possible in Example 1,
even though the limits are constants.
EXAMPLE 3
a 2 2x 2
a
Evaluate
∫ ∫
0
x 2 y dxdy .
0
Solution.
a
Let I = ∫
0
a2 − x 2
∫
0
a2 − x 2
a
x y dxdy = ∫
∫
2
0
x 2 y dydx
0
⎡ y2 ⎤
=∫x ⎢ ⎥
⎣ 2 ⎦0
0
a
2
M08_ENGINEERING_MATHEMATICS-I _CH08.indd 3
a2 − x 2
dx
5/19/2016 8:24:26 PM
8.4
■
Engineering Mathematics
a
=
1 2 2
x ( a − x 2 )dx
2 ∫0
a
a
1
1 ⎡ x3 x5 ⎤
1⎡
a 3 a 5 ⎤ 1 2a 5 a 5
= ∫ ( a 2 x 2 − x 4 )dx = ⎢ a 2
− ⎥ = ⎢a2 ⋅ − ⎥ = ⋅
=
20
2⎣ 3
5 ⎦0 2 ⎣
3 5 ⎦ 2 15 15
EXAMPLE 4
Evaluate
∫∫ xy dxdy over the positive quadrant of the circle x
2
1y 2 5 a2 .
R
Solution.
Given that the region R is bounded by the coordinate axes y = 0, x = 0 and the circle x2 + y2 = a2.
So, the region of integration is the shaded region OAB as in Fig. 8.5.
To find the limits for x, consider a strip PQ parallel to x-axis, x varies from x = 0 to x = a2 − y 2 .
When we move the strip to cover the region it moves from y = 0 to y = a.
∴ limits for y are y = 0 and y = a
∴
a
a2 − y 2
0
0
∫∫ xy dxdy = ∫ ∫
R
y
xy dxdy
⎡ x2 ⎤
= ∫ y⋅⎢ ⎥
⎣ 2 ⎦0
0
a
(0, a)
B
x 2 + y 2 = a2
a2 − y 2
dy
P
Q
O
(a, 0)
A
a
=
1
y( a 2 − y 2 ) dy
2 ∫0
=
a
1
1 ⎡ y2 y4 ⎤
( a 2 y − y 3 ) dy = ⎢ a 2
− ⎥
∫
20
2⎣ 2
4 ⎦0
a
=
x
Fig. 8.5
1 ⎡ 2 a2 a4 ⎤ 1 a4 a4
⎢a ⋅ − ⎥ = ⋅ =
2⎣
2
4⎦ 2 4
8
EXAMPLE 5
Evaluate ∫∫ xy dxdy , where A is the region bounded by x 5 2a and the curve x2 5 4ay.
A
Solution.
Given that the shaded region OAB is the region of integration bounded by y = 0, x = 2a and the parabola x2 = 4ay as in Fig 8.6.
We first integrate w.r.to y and then w.r.to x.
To find the limits for y, we take a strip PQ parallel to the y-axis, its lower end P lies on y = 0 and
x2
upper end Q lies on x 2 = 4ay ⇒ y =
4a
M08_ENGINEERING_MATHEMATICS-I _CH08.indd 4
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Multiple Integrals ■
∴ the limits for y are y = 0 and y =
x2
.
4a
y
y
x 2 = 4ay
B
When the strip is moved to cover the area,
x varies from x = 0 to x = 2a.
x = 2a
Q
2
∴
x
2a 4a
2a
0 0
0
x2
∫∫ xy dxdy = ∫ ∫ xy dydx = ∫
R
8.5
⎡ y2 ⎤ 4a
x ⎢ ⎥ dx
⎣ 2 ⎦0
2a
1
x4
1
= ∫ x⋅
dx =
2 0 16 a 2
32a 2
2a
∫
0
O
P A(2a, 0) x
Fig. 8.6
2a
1 ⎡ 26 a 6 ⎤ a 4
1 ⎛ x6 ⎞
x dx =
⎜ ⎟ =
⎢
⎥=
32a 2 ⎣ 6 ⎦ 3
32a 2 ⎝ 6 ⎠ 0
5
EXAMPLE 6
Evaluate ∫∫ xy 2 y 2 dxdy, where R is the triangle with vertex (0, 0), (10, 1), (1, 1).
R
Solution.
Given that the region of integration is the triangle OAB as shown as Fig. 8.7.
Equation of OA is
y −0 x −0
=
0 −1 0 −1
⇒
y =x
y
y −0 x −0
x
⇒ y =
=
0 − 1 0 − 10
10
We first integrate w.r.to x and then w.r. to y.
To find the limits for x, take a strip PQ parallel
to the x-axis. Its left end P is on x = y and right end
Q is on x = 10y.
∴ the limits for x are x = y and x = 10y.
When the strip is moved to cover the region,
y varies from 0 to 1.
∴
∫∫
R
1 10 y
xy − y 2 dxdy = ∫
∫
0 y
1 10 y
xy − y 2 dxdy = ∫
∫
A(1, 1)
(0, 1)
Equation of OB is
P
B
(10, 1)
Q
O(0, 0)
x
Fig. 8.7
1
1
y 2 ( x − y ) 2 dxdy
0 y
1
1
1 ⎡10 y
⎤
= ∫ y 2 ⎢ ∫ ( x − y ) 2 dx ⎥ dy
⎢⎣ y
0
⎦⎥
⎡ ( x − y)
=∫y ⎢
3
⎢⎣ 2
1
0
M08_ENGINEERING_MATHEMATICS-I _CH08.indd 5
1
2
3
2
⎤
⎥
⎥⎦
10 y
dy
y
⎡
( x − a ) n +1 ⎤
n
{
(
x
−
a
)
dx
=
⎢ ∫
⎥
n +1 ⎦
⎣
5/19/2016 8:24:33 PM
■
8.6
Engineering Mathematics
=
1
3
3
⎫
2 12 ⎧
y ⎨(10 y − y ) 2 − ( y − y ) 2 ⎬ dy
∫
30 ⎩
⎭
=
1
1
3
⎡ y3 ⎤
2 12
2 3 2
2
9
3
d
y
=
18
y
(
y
)
dy
=
y
⎢ ⎥ dy = 6[1 − 0] = 6.
3 ∫0
3 ∫0
⎣ 3 ⎦0
1
EXAMPLE 7
Evaluate
∫∫ x dxdy
over the region R bounded by y2 5 x and the lines x 1 y 5 2, x 5 0, x 5 1.
R
Solution.
Given that the region of integration is the shaded
region OAB as in Fig. 8.8.
To find A, solve x + y = 2 and y2 = x
y2 = 2−y
⇒
y
A(1, 1)
P
y2 +y −2= 0
⇒
⇒
O
(y + 2)(y −1) = 0 ⇒
x = 4, 1
y2 = x
Q
B (0, 2)
x=1
x+y=2
(1, 0)
x
y = −2, 1
∴
∴ A is (1, 1) and B is (0, 2) which is the point of
Fig. 8.8
intersection of x = 0 and x + y = 2.
It is convenient to integrate with respect to y first
and hence find y limits.
Take a strip PQ parallel to y-axis. P lies on y2 = x and Q lies on x + y = 2.
∴ the limits for y are y = x and y = 2 − x .
When the strip is moved to cover the region, x varies from 0 to 1.
1 2− x
∴
∫∫ x dxdy = ∫ ∫
R
0
x
1
x dydx = ∫ x ⋅ [ y ]2 −x x dx
0
1
= ∫ x ⎡⎣ 2 − x − x ⎤⎦ dx
0
1
⎡ x 2 x 3 x 5/ 2 ⎤
1 2 15 − 5 − 6 4
=
= ∫ ( 2 x − x 2 − x 3/ 2 ) dx = ⎢ 2 − −
⎥ = 1− − =
2
3
5
/
2
3 5
15
15
⎣
⎦0
0
1
EXERCISE 8.1
∫∫ xy dxdy over the first quadrant of the circle x + y = a .
Evaluate x dxdy over the region bounded by the hyperbola xy = 6 and the lines y = 0, x = 1, x = 3.
∫∫
Evaluate
∫∫ xy − y dxdy , where R is a triangle with vertices (0, 0), (5, 1) and (1, 1).
2
1. Evaluate
2.
3.
2
2
2
2
R
4. Evaluate ∫∫ ( x + y ) 2 dxdy over the area bounded the ellipse
M08_ENGINEERING_MATHEMATICS-I _CH08.indd 6
x2 y 2
+
= 1.
a2 b 2
5/19/2016 8:24:37 PM
Multiple Integrals ■
8.7
5. Evaluate ∫∫ ( x 2 + y 2 ) , where R is the region bounded by x = 0, y = 0 and x + y = 1.
R
6. Evaluate ∫∫ e 2 x + 3 y dxdy over the triangle bounded by x = 0, y = 0 and x + y = 1.
x −y
x −y
dydx ≠ ∫ ∫
dxdy .
3
3
0 0 (x + y )
0 0 (x + y )
1 1
1 1
7. Show that ∫ ∫
8. Compute the value of ∫∫ y dxdy , where R is the region in the first quadrant bounded by the ellipse
R
x2 y 2
+
= 1.
a2 b 2
9. Evaluate ∫∫ xy dxdy , where A is the domain bounded by x-axis, ordinate x = 2a and curve x2 = 4ay.
A
a b
b a
0 0
0 0
10. Show that ∫ ∫ ( x + y ) dxdy = ∫ ∫ ( x + y ) dydx .
4 2
dxdy
11. Evaluate ∫ ∫
.
xy
2 1
12. Evaluate
1 1+ x 2
∫ ∫
0
0
dydx
.
1+ x 2 + y 2
13. Evaluate ∫∫ xy ( x + y ) dxdy over the area between y = x and y = x.
2
R
14. Evaluate ∫∫ xy dxdy , where R is the region bounded by the parabola y2 = x, the x-axis and the line
R
x + y = 2, lying on the first quadrant.
15. Evaluate ∫∫ y dxdy over the region R bounded by y = x and y = 4x − x2.
R
ANSWERS TO EXERCISE 8.1
4
a
2. 24
4
1
6. (e − 1) 2 ( 2e + 1)
6
1.
p
11. (log2) 2
8.1.3
12. 4
loge (1 + 2 )
16
9
ab 2
8.
3
3.
13.
3
56
p ab 2
(a + b 2 )
4
a4
9.
3
1
6
ab
10.
(a + b )
2
4.
14.
5.
3
8
15.
54
5
Change of Order of Integration
a h(x )
The double integral with variable limits for y and constant limits for x is ∫
∫
f ( x , y ) dydx . To evaluate
b g (x )
this integral, we integrate first w.r.to y and then w.r.to x. This may sometimes be difficult to evaluate. But change in the order of integration will change the limits of y from c to d where c and d are
d h1 ( y )
constants and the limits of x from g1(y) to h1(y). The double integral becomes ∫
∫
f ( x , y ) dxdy and
c g1 ( y )
hence the evaluation may be easy. To evaluate this integral, we integrate first w.r.to x and then w.r.to y.
M08_ENGINEERING_MATHEMATICS-I _CH08.indd 7
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8.8
■
Engineering Mathematics
This process of changing a given double integral into an equal double integral with order of integration changed is called Change of order of integration.
For doing this we have to identify the region R of integration from the limits of the given double
integral. Sometimes this region R may split into two regions R1 and R2 when we change the order of
integration and hence the given double integral ∫∫ f ( x , y ) dxdy will be the sum of two double integrals.
R
∫∫ f (x , y ) dxdy = ∫∫ f (x , y ) dxdy + ∫∫ f (x , y ) dxdy
i.e.,
R
R1
R2
WORKED EXAMPLES
EXAMPLE 1
∞∞
Evaluate ∫ ∫
0 x
e 2y
dydx by changing the order of integration.
y
Solution.
∞∞
e −y
dydx
y
0 x
I = ∫∫
Let
The region of integration is bounded by y = x, y = ∞, x = 0, x = ∞.
∴ the region is unbounded as in Fig. 8.9.
In the given integral, integration is first with respect to y and then w.r.to x.
After changing the order of integration, first integrate w.r.to x and then w.r.to y. To find the limits of x,
take a strip PQ parallel to x-axis (see Fig. 8.10) with P on the line x = 0 and Q on the line x = y respectively.
y
y=x
y
y=x
P
O
O
x
Fig. 8.9
Given order of integration
Q
x
Fig. 8.10
After the change of order of integration
∴ the limits of x are x = 0 and x = y and the limits of y are y = 0 and y = ∞
∞ y
−y
0 0
y
∴ I = ∫∫ e
∞
∞
∞ −y
∞
⎡ e− y ⎤
e− y
e
−∞
0
⋅ [ x ]0y dy = ∫
⋅ ydy = ∫ e − y dy = ⎢
⎥ = −(e − e ) = −(0 − 1) = 1
y
⎦
y
1
⎣
−
0
0
0
0
dxdy = ∫
EXAMPLE 2
Evaluate by changing the order of integration
4 a 2 ax
∫ ∫
0
M08_ENGINEERING_MATHEMATICS-I _CH08.indd 8
dydx .
x2
4a
5/19/2016 8:24:46 PM
Multiple Integrals ■
8.9
Solution.
4 a 2 ax
Let I =
∫ ∫
0
dydx
x2
4a
The region of integration is bounded by y =
x2
, y = 2 ax and x = 0, x = 4a.
4a
x2
2
⇒ x 2 = 4ay is a parabola and y = 2 ax ⇒ y = 4ax is a parabola.
4a
In the given integral, integration is first w.r.to y and then w.r.to x. After changing the order of
integration, we have to integrate first w.r.to x and then w.r. to y.
y =
x 2 = 4ay
y
x 2 = 4ay
y
y 2 = 4ax
y 2 = 4ax
A(4a, 4a)
A(4a, 4a)
P
Q
O
y =0
O
x
x
Fig. 8.12
After the change of order of integration
Fig. 8.11
Given order of integration
To find the points of intersection of the curves x2 = 4ay and y2 = 4ax, solve the two equations.
x 4 = 16a2 y 2 = 16a2 ⋅ 4ax = 64a3 x
x ( x 3 − 64a3 ) = 0
⇒
⇒
x 3 − 64 a3 = 0 ⇒
Now
When x = 0, y = 0 and when x = 4a, y =
x = 0 and x 3 − 64a3 = 0
x 3 = 64 a3 = ( 4 a)3
⇒
x = 4a
x 2 16a2
=
= 4a
4a
4a
Points of intersection are O(0, 0) and A is (4a, 4a)
Now to find the x limits, take a strip PQ parallel to the x-axis (see Fig. 8.11) where P lies on y2 = 4ax
and Q lies on x2 = 4ay.
y2
∴ the limits of x are x =
and x = 2 ay
4a
When the strip is moved to cover the region, y varies from 0 to 4a.
4a 2 a y
∴ I=
∫ ∫
a
4a
dxdy =
y2
4a
∫ [ x]
2 a y
y2 / 4 a
dy
0
4a
=
∫
0
⎡
y2 ⎤
2
a
y
−
⎢
⎥ dy
4a ⎦
⎣
M08_ENGINEERING_MATHEMATICS-I _CH08.indd 9
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■
8.10
Engineering Mathematics
4a
=
⎡
∫ ⎢⎣2a
1/ 2
y1/ 2 −
0
y2 ⎤
⎥ dy
4a ⎦
4a
⎡
y 3/ 2 1 y 3 ⎤
4 a1/ 2
1 ( 4a)3 32a 2 16a 2 16 a 2
3/ 2
(
4
)
−
= ⎢ 2a1/ 2
=
a
−
=
−
=
⎥
3/2 4a 3 ⎦ 0
3
4a 3
3
3
3
⎣
EXAMPLE 3
a
Change the order of integration in
∫
a1 a 2 2 y 2
∫
xy dxdy and then evaluate it.
0 a2 a 2 2 y 2
Solution.
a + a2 − y 2
a
Let I = ∫
∫
xy dxdy
0 a − a2 − y 2
The region of integration is bounded by the lines y = 0, y = a and the curves x = a − a 2 − y 2 , x = a + a 2 − y 2
x = a ± a2 − y 2
i.e.,
⇒
2
2
2
x − a = ± a2 − y 2 ⇒ ( x − a) + y = a
which is a circle with (a, 0) as centre and radius a.
The region of integration is the upper semi-circle OAB as in Fig. 8.14.
The original order is first integration w.r.to x and then w.r.to y. After changing the order of
integration, first integrate w.r.to y and then w.r.to x. To find the limits of y, take a strip PQ parallel to y-axis
(see Fig. 8.14), where P lies on y = 0 and Q lies on the circle ( x − a) 2 + y 2 = a2 .
y
y
A
y=a
Q
(x − a)2 + y 2 = a2
x
O
(a, 0)
O
B(2a, 0)
A
y=a
(x − a)2 + y 2 = a2
x
P (a, 0)
B(2a, 0)
Fig. 8.14
After the change of order of integration
Fig. 8.13
Given order of integration
∴ the limits of y are y = 0 and y = a 2 − ( x − a) 2 = 2ax − x 2
When the strip is moved to cover the region, x varies from 0 to 2a.
∴
I=
2 a 2 ax − x 2
∫ ∫
0
0
xydydx = ∫
2a
0
M08_ENGINEERING_MATHEMATICS-I _CH08.indd 10
⎡ y2 ⎤
x⎢ ⎥
⎣ 2 ⎦0
2 ax − x 2
dx.
5/19/2016 8:24:58 PM
Multiple Integrals ■
8.11
2a
=
1
x [2ax − x 2 ]dx
2 ∫0
=
1
( 2ax 2 − x 3 )dx
2 ∫0
=
4
4
4
2
1 ⎡ x3 x 4 ⎤
1 ⎡ ( 2a)3 ( 2a) 4 ⎤ 1 ⎡16 a 16 a ⎤ 1 16 a
=
−
=
= a4
a
−
=
a
−
2
2
⎢
⎥
⎢
⎥ 2⎣ 3
⎢
⎥
4 ⎦ 2 12
3
2⎣ 3
4 ⎦0
2⎣
3
4 ⎦
2a
2a
EXAMPLE 4
1
22 x
0
x2
∫ ∫
Change the order of integration in
xy dydx and hence evaluate.
Solution.
1 2− x
Let I = ∫
∫
xy dydx
0 x2
The region of integration is bounded by x = 0, x = 1, y = x2, y = 2 – x.
In the given integral, first integrate with respect to y and then w.r.to x. After changing the order we
have to first integrate w.r.to x, then w.r.to y.
y
y
y = x2
B
(0, 2)
P′
P
x=1
x2 = 2− x
x
y=2−x
Fig. 8.15
Given order of integration
⇒
A(1, 1)
Q
O
x
y=2−x
To find A, solve y = x 2 ,
Q′
C
A(1, 1)
O
y = x2
B
Fig. 8.16
After the change of order of integration
y = 2−x
⇒
x2 + x − 2 = 0
⇒
( x + 2)( x − 1) = 0 ⇒ x = −2, 1
Since the region of integration is OAB, x = 1 ⇒ y = 1
∴ A is (1, 1) and B is (0, 2), which is the point of intersection of y-axis x = 0 and y = 2 – x
Now to find the x limits, take a strip parallel to the x-axis. We see there are two types of strips PQ and
P′ Q′ after the change of order of integration (see Fig. 8.16) with right end points Q and Q′ are respectively
on the parabola y = x2 and the line y = 2 – x. So, the region OAB splits into two regions OAC and CAB
as in Fig. 8.16.
M08_ENGINEERING_MATHEMATICS-I _CH08.indd 11
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8.12
■
Engineering Mathematics
Hence, the given integral I is written as the sum of two integrals
In the region OAC,
x varies from 0 to y and y varies from 0 to 1
x varies from 0 to 2 – y and y varies from 1 to 2
In the region CAB,
∴
I=
∫∫ xy dxdy = ∫∫ xy dxdy + ∫∫ xy dxdy
OAB
OAC
1
=∫
y
∫
0 0
CAB
2 2− y
x y dx dy + ∫
1
∫
xy dx dy
0
2− y
y
1
2
⎡ x2 ⎤
⎡ x2 ⎤
= ∫ y ⋅ ⎢ ⎥ dy + ∫ y ⎢ ⎥
⎣ 2 ⎦0
⎣ 2 ⎦0
0
1
1
dy
2
=
1
1
y y dy + ∫ y ⋅ ( 2 − y ) 2 dy
2 ∫0
21
=
1 2
1
y dy + ∫ y( 4 − 4 y + y 2 )dy
2 ∫0
21
1
2
1
1 ⎡ y3 ⎤ 1
= ⎢ ⎥ + ∫ ( 4 y − 4 y 2 + y 3 )dy
2 ⎣ 3 ⎦0 2 1
2
2
=
1 1 ⎡ y2
y3 y 4 ⎤
+ ⎢4 − 4 + ⎥
6 2⎣ 2
3
4 ⎦1
=
1 1⎡ 2 2 4 3 3 1 4 4 ⎤
+ 2( 2 − 1 ) − ( 2 − 1 ) + ( 2 − 1 ) ⎥
6 2 ⎢⎣
3
4
⎦
=
1 1⎡
4
1
9 3
⎤ 1 1 [72 − 112 + 45] 1 5
+ ⎢6 − × 7 + × 15⎥ = + .
= +
=
=
6 2⎣
3
4
12
6 24 24 8
⎦ 6 2
EXAMPLE 5
1
Evaluate
2−x2
∫ ∫
0
x
x
x +y2
2
dydx by changing the order of integration.
Solution.
Let
1
2−x 2
0
x
I=∫
∫
x
x2 + y 2
dydx
The region of integration is bounded by x = 0, x = 1 and y = x, y = 2 − x 2 .
Now y = 2 − x2 ⇒ y2 = 2 − x2
x2 + y2 = 2, which is a circle, with centre (0, 0) and radius 2.
The region of integration is OAB as in Fig. 8.18.
To find A, solve
y = x and x 2 + y 2 = 2
M08_ENGINEERING_MATHEMATICS-I _CH08.indd 12
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Multiple Integrals ■
∴
⇒
x2 + x2 = 2
⇒
2x 2 = 2
x = ±1
Since A is in the first quadrant, x = 1 ∴
∴ A is (1, 1) and B is (0, 2 ) , which is the point
of intersection of x = 0 and x2 + y2 = 2
In the given integral, integration is w.r.to y
first and then w.r.to x. After changing the order of
integration, first integrate w.r.to x and then w.r.to y.
To find the x limits, take a strip parallel to the x-axis.
We see there are two strips PQ and P′Q′ with ends
Q, Q′ on the line y = x and circle x2 + y2 = 2 respectively.
So, the region splits into 2 regions OAC and CAB.
In the region OAC,
x varies from 0 to y and y varies from 0 to 1
In the region CAB,
x varies from 0 to 2 − y 2
∫∫
OAC
x
x +y
2
2
dxdy +
∫∫
CAB
x
x +y2
2
2 −1/ 2
x dxdy +
0 0
Fig. 8.17
Given order of integration
B
P′
(0, 1)C
P
1
1 ⎡ ( x 2 + y 2 )1 2 ⎤
1
dy +
= ∫⎢
⎥
1
20⎢
2
⎥
2
⎣
⎦0
2
∫
1
1
= ∫ [( y 2 + y 2 )1/ 2 − ( y 2 )1/ 2 ]dy +
0
0
x 2 + y2 = 2
( x 2 + y 2 ) −1/ 2 x dxdy
0
1 y
= ∫ ( 2 y − y )dy +
A (1, 1)
Q
O
∫ ∫
2
1
1
= ∫ ∫ ( x 2 + y 2 ) −1/ 2 2 x dxdy +
200
2
1
y=x
Q′
2 − y2
1
y
x=1
Fig. 8.18
After the change of order of
integration
2
= ∫ ∫ (x + y )
2
x 2 + y2 = 2
O
dxdy
1 y
A (1, 1)
x=0
and y varies from 1 to 2
∴ I=
y=x
B
y=1
8.13
2 − y2
∫ ∫
1
( x 2 + y 2 ) −1/ 2 2 x dxdy
0
⎡ ( x 2 + y 2 )1 2 ⎤
⎢
⎥
1
⎢
⎥
2
⎣
⎦0
2 − y2
dy
2
∫ [(2 − y
2
+ y 2 )1/ 2 − ( y 2 )1/ 2 ]dy
1
2
∫(
2 − y )dy
1
1
⎡
y2 ⎤ ⎡
y2 ⎤
= ⎢( 2 − 1) ⎥ + ⎢ 2 y − ⎥
2 ⎦0 ⎣
2 ⎦1
⎣
2
1
2 ⎡
1⎤
= ( 2 − 1) + 2 ⋅ 2 − − ⎢ 2 ⋅1 − ⎥
2
2 ⎣
2⎦
=
2 −1
( 2 2 − 1)
+ 2 −1−
=
2
2
M08_ENGINEERING_MATHEMATICS-I _CH08.indd 13
2 −1+ 2 − 2 2 +1 2 − 2
=
2
2
5/19/2016 8:25:12 PM
■
8.14
Engineering Mathematics
EXAMPLE 6
Show that
a
ay
0
0
∫∫
2a 2a2y
xy dxdy 1 ∫
a
∫
0
a 2 a 2x
xy dxdy 5 ∫
0
∫
xy dydx and hence evaluate.
x2
a
Solution.
a
Let I = ∫
0
ay
∫
0
2a 2a − y
xy dxdy + ∫
a
∫
xy dxdy.
0
The given integral I has same integrand defined over two region R1 and R2 given by the two double
integrals.
y
Region R1 is bounded by y = 0, y = a and x = a,
x 2 = ay
B(0, 2a)
x = ay ⇒ x 2 = ay
y = 2a
2
Q
x = a and x = ay intersect at A (a, a)
R2
A(a, a)
C
Region R2 is given by y = a and y = 2a and x = 0,
y=a
x
=
0
x = 2a − y.
R1
(2a, 0)
P
The regions R1 and R2 are as shown in Fig. 8.19.
O
x
x
R1 is the shaded region OAC
x + y = 2a
R2 is the shaded region CAB
x=a
The line x + y = 2a also passes through A and B.
Combining the two regions R1 and R2, we get the
Fig. 8.19
shaded region OAB. In the given integral, we have
to integrate first with respect to x and then w.r.to y. Changing the order integration, we first integrate
w.r.to y, then w.r.to x. To find the y limits, take a strip PQ parallel to the y-axis with P on x2 = ay ⇒
x2
and Q on x + y = 2a ⇒ y = 2a − x.
a
x2
and y = 2a − x and the limits for x are x = 0 and x = a
∴ the limits for y are y =
a
y=
a
∴ I=∫
2a 2a − y
ay
∫
0 0
xy dxdy +
∫ ∫
a
0
a 2a − x
xy dxdy = ∫
0
∫
xy dydx
x2
a
2a− x
⎡ y2 ⎤
= ∫ x ⎢ ⎥ 2 dx
⎣ 2 ⎦x
0
a
a
=
⎡
1
x4 ⎤
x ⋅ ⎢( 2a − x ) 2 − 2 ⎥ dx
∫
20 ⎣
a ⎦
=
a
1 ⎡ 2
x4 ⎤
2
4
4
x
a
−
ax
+
x
−
⎥ dx
⎢
2 ∫0 ⎣
a2 ⎦
=
a
1 ⎛ 2
x5 ⎞
2
3
−
+
−
4
4
a
x
ax
x
⎜
⎟ dx
2 ∫0 ⎝
a2 ⎠
a
a
1⎡
x2
x3 x 4
x6 ⎤
= ⎢ 4a 2
− 4a +
− 2 ⎥
2⎣
2
3
4 a ⋅ 6 ⎦0
M08_ENGINEERING_MATHEMATICS-I _CH08.indd 14
5/19/2016 8:25:15 PM
Multiple Integrals ■
8.15
=
1 ⎡ 2 2 4 a 3 a 4 a6 ⎤
a +
−
⎥
⎢ 2a ⋅ a −
2⎣
3
4 6a2 ⎦
=
1 ⎡ 4 4 a 4 a 4 a 4 ⎤ a 4 [24 − 16 + 3 − 2] a 4 9 3a 4
+
− ⎥=
=
⋅ =
⎢ 2a −
2⎣
3
4
6⎦ 2
12
2 12
8
EXERCISE 8.2
Change the order of integration in the following integrals and evaluate.
a 2a − x
1.
∫ ∫
2
a/ 2 a − y
2
∫ ∫
0
6.
∫∫ x
0 y
0 x /a
3.
x+ y
dxdy
2
+ y2
a a
2.
xy dydx
∞ y
2
log( x 2 + y 2 ) dxdy , a > 0
∫ ∫ ye
4.
− y2 / x
1 2−x
dxdy
5.
0 0
y
∫∫
xy dydx
0 x2
∫∫ xy dxdy , where R is region bounded by the line x + 2y = 2 and axes in the first quadrant.
R
2
7.
4 − y2
∫ ∫
0
2− y
3
16 − x 2
4 4
10.
∫ ∫
0
x dydx
0
9.
∫∫
dydx
1 1+ y
2
x 2 dydx
14.
0
∫ ∫
0
0
( x 2 + y 2 ) dydx
0 x / 4a
12.
dxdy
1+ x2 + y2
∫ ∫
2
0 x2 / 4
0
∫ ∫
x
∫0 ∫y x 2 + y 2 dxdy
4 2 x
11.
a 2 ax
13.
2 a 3a − x
a a
8.
y dxdy
15.
1
2 − x2
0
x
3
4− y
∫ ∫
∫ ∫
0
x
dxdy
x2 + y2
( x + y ) dxdy
1
a
2
16.
1 2− y
a2 − x 2
∫ ∫
0
y 2 dydx
17.
x
∫∫
0
3 6/ x
xy dxdy
18.
∫∫
x 2 dydx.
1 y=0
y
ANSWERS TO EXERCISE 8.2
1.
3a4
8
2.
pa a
+ log e 2
4
2
3.
1⎞
p a2 ⎛
⎜⎝ log e a − ⎟⎠
2
2
4.
1
2
6.
1
6
7.
4
3
8.
pa
4
9.
314 4
a
35
11.
64
3
13.
4a4
7
14.
17.
1
3
18. 24
15.
241
60
12. 1 −
1
2
4
16.
a
( 2 + p)
32
M08_ENGINEERING_MATHEMATICS-I _CH08.indd 15
5.
3
8
10. 10
p
log(1 + 2 )
4
5/19/2016 8:25:23 PM
8.16
■
8.1.4
Double Integral in Polar Coordinates
Engineering Mathematics
To evaluate the double integral of f(r, u) over a region R in polar coordinates, generally we integrate
u2 r = f 2 ( u )
first w.r.to r and then w.r.to u. So, the double integral is
∫ ∫
f ( r , u) drd u
u1 r = f1 ( u )
However, whenever necessary, the order of integration may be changed with suitable changes in the
limits. As in Cartesian, when we integrate w.r.to r, treat u as constant.
WORKED EXAMPLES
EXAMPLE 1
p
2 2 cos u
∫ ∫
Evaluate
2
p
2
r 2 drd u.
0
Solution.
Let I =
p
2 2 cos u
∫ ∫
p
−
2
r 2 drd u =
p
2
⎡r ⎤
3
∫ ⎢⎣ 3 ⎥⎦
p
−
2
0
2 cos u
du =
0
=
1
3
8
3
=
p
2
∫ 8 cos
3
ud u
p
−
2
p
2
∫ cos
−
p
2
p
2
3
[{ cos3 u is an even
8
ud u = ⋅ 2 ∫ cos3 ud u
3 0
16 2
32
⋅ ⋅1 =
3 3
9
function ]
[Using formula ]
Important Formulae
p/2
∫
cos n xdx =
0
∫ sin
n
xdx =
n −1 n − 3 2
⋅
… ⋅1 if n is odd and n ≥ 3
n n−2 3
xdx =
n −1 n − 3 1 p
⋅
… ⋅ if n is even
n n−2 2 2
0
p/2
∫
and
p/2
cos n xdx =
0
p/2
∫ sin
n
0
∫ [ f ( x)] f ′( x) dx =
n
[ f ( x )]n +1
if n ≠ −1
n +1
EXAMPLE 2
Evaluate ∫∫ rsin u drd u over the area of the cardioid r 5 a(1 1 cosu) above the initial line.
A
Solution.
Let
I = ∫∫ r sin udrd u
A
M08_ENGINEERING_MATHEMATICS-I _CH08.indd 16
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Multiple Integrals ■
First integrate w.r.to r.
Take a radial strip OP, its ends are on r = 0 and
r = a(1 + cosu). When it is moved to cover the area,
u varies from 0 to p
p a (1+ cos u )
∴ I=∫
0
∫
0
a (1+ cos u )
p
⎡ r2 ⎤
r sin u drd u = ∫ ⎢ ⎥
⎣ 2 ⎦0
0
=
sin ud u
8.17
P
θ=π
θ
O
p
1 2
a (1 + cos u) 2 sin ud u
2 ∫0
θ=0
x
Fig. 8.20
p
=−
a2
(1 + cos u) 2 ( − sin u)d u
2 ∫0
=−
a 2 ⎡ (1 + cos u)3 ⎤
⎢
⎥
2 ⎣
3
⎦0
=−
8a 2 4 a 2
a2
a2
⎡⎣(1 + cos p)3 − (1 + cos 0)3 ⎤⎦ = − ⎡⎣(1 − 1)3 − (1 + 1)3 ⎤⎦ =
=
6
6
6
3
⎡ d
⎤
⎢{ du (1+ cos u) = − sin u⎥
⎣
⎦
p
EXAMPLE 3
Evaluate
∫∫ r drd u, over the area bounded between the circles r 5 2 cos u and r 5 4 cos u .
3
Solution.
Let
I = ∫∫ r 3drd u,
y
A
where the region A is the area between the circles
r = 2 cos u and r = 4 cos u
The area A is the shaded area in the Fig. 8.21
We first integrate w.r.to r. So, take a radius vector OPQ,
where r varies from P to Q.
∴ r varies from 2 cos u to 4 cos u
When PQ is varied to cover the area A between
r = 2 cos u and r = 4 cos u, u varies from −
∴
p
2 4 cos u
p
2
2
2
4 cos u
π
4
Q
P
θ
O
Area
p p
to
2
2
⎡ r4 ⎤
I = ∫ ∫ r 3 drd u = ∫ ⎢ ⎥
du
− p 2 cos u
− p ⎣ 4 ⎦ 2 cos u
θ=
θ=−
x
π
2
Fig. 8.21
p
1 2
= ∫ ( 4 4 cos 4 u − 24 cos 4 u)d u
4 −p
2
=
p
2
1
( 256 − 16) cos 4 u d u
4 −∫p
2
M08_ENGINEERING_MATHEMATICS-I _CH08.indd 17
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■
8.18
Engineering Mathematics
p
p
2
240 2
4
4
=
cos
u
d
u
=
60
×
2
∫0 cos u d u
4 −∫p
[{ cos 4 u is even ]
2
3 1 p 45p
= 120 × ⋅
=
4 22
2
[Using formula ]
EXAMPLE 4
Evaluate
rdrd u
∫∫
r 1a
2
R
2
2
2
, where R is the area of one loop of the leminiscate r 5 a cos 2u .
Solution.
Let I = ∫∫
R
rdrd u
θ=
r 2 + a2
First integrate with respect to r
Take a radial strip OP, its ends are r = 0 and
P
O
When the strip covers the region, u varies
p p
from − to
4
4
I=
p
4 a cos 2 u
∫ ∫
−p
4
0
r =a
θ
r = a cos 2 u
∴
π
r
r +a
2
2
drd u =
p
4
⎡1 a
∫p ⎢⎢⎣ 2
−
cos 2 u
∫
0
θ=−
⎤
( r 2 + a 2 ) −1/ 2 2rdr ⎥ d u
⎥⎦
2θ
x
π
Fig. 8.22
4
1
=
2
=
=
−1
+1 ⎤
⎡ 2
2 2
(
r
a
)
+
⎥
⎢
∫p ⎢ −1
⎥
+1 ⎥
− ⎢
4 ⎣
⎦0
2
p
4
∫ ⎡⎣(r
2
p
−
4
p
4
∫ ⎡⎣(a
−
=
a cos 2 u
p
4
p
4
2
+ a 2 )1/ 2 ⎤⎦
a cos 2 u
0
du
du
cos 2 u + a 2 )1/ 2 − ( a 2 )1/ 2 ⎤⎦ d u
p
4
∫ {a[cos 2 u + 1]
1/ 2
− a}d u
p
−
4
=
p
4
∫ [a(2 cos
−
M08_ENGINEERING_MATHEMATICS-I _CH08.indd 18
2
u)1/ 2 − a]d u
p
4
5/19/2016 8:25:41 PM
Multiple Integrals ■
p
4
8.19
[{ 2 cos u − 1 is even function ]
= 2 ∫ a( 2 cos u − 1)d u
0
= 2a ⎡⎣ 2 sin u − u⎤⎦
p/4
0
⎧⎡
⎫
⎡⎛
⎤
1 p⎞
p p⎤
⎡ p⎤
− ⎟ − 0 ⎥ = 2a ⎢1 − ⎥
= 2a ⎨ ⎢ 2 sin − ⎥ − ( 2 sin 0 − 0)⎬ = 2a ⎢⎜ 2 ⋅
⎝
⎠
4
4
⎦
⎣ 4⎦
⎣
4
⎩
⎭
2
⎣
⎦
8.1.5
Change of Variables in Double Integral
The evaluation of a double integral, sometimes become simpler if the variables of integration are
transformed suitably into new variables.
For example, from cartesian coordinates to polar coordinates or to some variables u and v.
1. Change of variables from x, y to the variables u and v.
Let
f ( x, y ) dxdy be the given double integral.
∫∫
R
∂( x, y )
Suppose x = g(u, v), y = h(u, v) be the transformations. Then dxdy = J dudv , where J =
is
∂(u, v )
the Jacobian of the transformation.
∴
∫∫ f ( x, y) dxdy = ∫∫ F (u, v)
R
J dudv
R
2. Change of variable from Cartesian to polar coordinates
Let ∫∫ f ( x, y ) dxdy be the double integral.
R
Let x = rcosu, y = rsinu be the transformation from Cartesian to polar coordinates.
Then dxdy = J drd u
∂( x, y )
is the Jacobian of transformation.
∂( r, u)
where J =
and
∂x
∂r
J=
∂y
∂r
∂x
∂ u cos u − r sin u
=
= r cos 2 u + r sin 2 u = r (cos 2 u + sin 2 u) = r
∂y
sin u r cos u
∂u
∴
dxdy = rdrd u
and
∴
∫∫ f ( x, y) dxdy = ∫∫ F (r, u) r drdu
R
R
WORKED EXAMPLES
EXAMPLE 1
∞
∞∞
Evaluate ∫ ∫ e 2( x
0 0
2
1y 2 )
dxdy by changing to polar coordinates and hence evaluate ∫ e
M08_ENGINEERING_MATHEMATICS-I _CH08.indd 19
2x 2
dx .
0
5/19/2016 8:25:45 PM
■
8.20
Engineering Mathematics
Solution.
∞∞
I = ∫ ∫ e −( x
Let
2
+ y2 )
y
dxdy
0 0
Since x varies from 0 to ∞ and y varies from 0 to ∞, it
is clear that the region of integration is the first quadrant
as in Fig. 8.23
To change to polar coordinates, put x = rcosu, y = rsinu
∴
dxdy = rdrdu
2
and
x + y2 = r2cos2u + r2sin2u = r2(cos2 u + sin2 u) = r2
p
∴ r varies from 0 to ∞ and u varies from 0 to
2
p
2 ∞
∫∫e
− r2
∴
I=
Put
r2 = t
When
r = 0, t = 0 and when r = ∞, t = ∞
∴
P
r
θ
O
x
Fig. 8.23
rdrd u
0 0
⇒
2rdr = dt
⇒
p
p
2
rdr =
dt
2
p
∞
⎤
⎡1 ∞
1 2 ⎡ e−t ⎤
12
0
I = ∫ ⎢ ∫ e − t dt ⎥d u = ∫ ⎢ ⎥ d u = − ∫ (e −∞ − e )d u
⎦
2
1
2
⎣
−
2
0
0
0
0 ⎣
0
⎦
=−
∞∞
∴
∫∫e
− ( x2 + y2 )
dxdy =
0 0
p
2
p
2
1 p2 p
1
1
0
1
d
d
u
=
[u]0 =
(
−
)
u
=
2
4
2 ∫0
2 ∫0
p
4
To find ∫ e 2x dx
2
∞∞
Now,
∫∫e
−( x 2 + y 2 )
0 0
∞
dxdy = ∫ e
−x2
0
∞
∫e
∴
− x2
dx =
0
∞
dx ⋅ ∫ e
−y 2
⇒
dy
0
∞
p ⎡ − x2 ⎤
= ⎢ ∫ e dx ⎥
4 ⎣0
⎦
2
⎡
⎢{
⎣
∞
∫e
0
− x2
∞
⎤
2
dx = ∫ e − y dy ⎥
0
⎦
p
p
=
2
4
EXAMPLE 2
2
Evaluate
2 x 2x 2
∫ ∫
0
0
x
x 1y 2
2
dydx by changing into polar coordinates.
Solution.
2
Let
I=∫
0
2x − x 2
∫
0
x
x +y2
2
dydx
The limits for y are y = 0 and y = 2x − x 2
M08_ENGINEERING_MATHEMATICS-I _CH08.indd 20
5/19/2016 8:25:51 PM
Multiple Integrals ■
y = 2x − x 2
Now,
⇒
⇒
y 2 = 2x − x 2
x 2 + y 2 − 2x = 0
which is a circle with centre (1, 0) and radius r = 1
and x varies from 0 to 2.
∴ the region of integration is the upper semi-circle
as in Fig. 8.24
To change to polar coordinates,
put x = rcosu, y = rsinu
∴ dx dy = r dr du
∴
x2 + y2 − 2x = 0
y
8.21
( x − 1) 2 + y 2 = 1,
⇒
x 2 + y 2 − 2x = 0
or r = 2cosθ
P
x=2
r
θ
O
(2, 0) x
(1, 0)
Fig. 8.24
⇒ r 2 cos 2 u + r 2 sin 2 u − 2r cos u = 0
r 2 − 2r cos u = 0
⇒
⇒
r ( r − 2 cos u) = 0
⇒
Limits of r are r = 0 and r = 2cosu and limits of u are u = 0 and u =
I=
∴
p
2 2 cos u
∫ ∫
0
0
r cos u
rdrd u =
r
p
2 2 cos u
∫ ∫
0
r = 0, 2 cos u
p
2
r cos u drd u
0
p
2
⎡ 2 cos u ⎤
= ∫ cos u ⎢ ∫ rdr ⎥ d u
0
⎣ 0
⎦
p
2
2 cos u
⎡ r2 ⎤
= ∫ cos u ⎢ ⎥
⎣ 2 ⎦0
0
du
p
p
2
4
12
3 −1
= ∫ cos u 4 cos 2 ud u = 2 ∫ cos3 d u = 2 ⋅
⋅1 =
3
3
20
0
EXAMPLE 3
2 a 2 ax 2x 2
By changing into polar coordinates, evaluate the integral
∫ ∫
0
( x 2 1 y 2 )dydx .
0
Solution.
Let
I=
2 a 2 ax − x 2
∫ ∫
0
( x 2 + y 2 ) dydx
y
0
The limits for y are y = 0 and y = 2ax − x 2
Now,
y = 2ax − x 2
⇒ x 2 + y 2 − 2ax = 0
⇒
y 2 = 2ax − x 2
⇒
( x − a) 2 + y 2 = a2
which is a circle with centre (a, 0) and radius r = a.
∴ x varies from 0 to 2a
M08_ENGINEERING_MATHEMATICS-I _CH08.indd 21
x 2 + y 2 − 2ax = 0
or r = 2acosθ
P
x = 2a
r
θ
O
(2a, 0)x
Fig. 8.25
5/19/2016 8:25:58 PM
8.22
■
Engineering Mathematics
∴ the region of integration is the upper semi circle as in Fig. 8.25.
To change to polar coordinates, put x = rcos u and y = rsin u.
∴
dxdy = rdrdu and x 2 + y 2 = r 2 cos 2 u + r 2 sin 2 u = r 2
and x 2 + y 2 − 2ax = 0 ⇒ r 2 − 2ar cos u = 0
∴ r varies from 0 to 2a cos u
∴
I=
p
2 2 a cos u
∫ ∫
0
0
and
⇒
r ( r − 2a cos u) = 0
u varies from 0 to
⇒
r = 0, r = 2a cos u
p
2
p
2
⎤
⎡ 2 a cos u
r 2 ⋅ r drd u = ∫ ⎢ ∫ r 3 dr ⎥ d u
0 ⎣ 0
⎦
p
2
2 a cos u
⎡ r4 ⎤
= ∫⎢ ⎥
⎣ 4 ⎦0
0
du
p
2
p
cos 4 u
16 a 4 2
3a 4 p
4 3 1 p
4
= ∫ ( 2a) 4
du =
cos
d
=
4
a
⋅
⋅
⋅
=
u
u
4 2 2
4
4
4 ∫0
0
EXAMPLE 4
4a y
Evaluate
∫∫
0 y2
4a
x 2 2y 2
dxdy by changing to polar coordinates.
x 2 1y 2
Solution.
4a y
Let
I=
∫∫
0 y2
4a
x2 − y 2
dxdy
x2 + y 2
P
y2
and x = y
4a
y 2 = 4 ax and y = x
r
Given, the limits for x are x =
⇒
y=x
y
y 2 = 4ax
θ
x
And the limits for y are y = 0 and y = 4a
To find the point of intersection of y 2 = 4ax and y = x ,
solve the two equations.
Now y2 = 4ax ⇒ y2 = 4ay ⇒ y(y − 4a) = 0 ⇒ y = 0, y = 4a
Fig. 8.26
∴
x = 0, x = 4a
∴ the points are (0, 0), (4a, 4a)
∴ the region of integration is the shaded region as in Fig. 8.26 which is bounded by y2 = 4ax and y = x.
To change to polar coordinates, put x = rcosu, y = rsinu
∴
∴
and
dxdy = rdrdu and x2 + y2 = r2
x2 − y2 = r2cos2u − r2sin2u = r2(cos2u − sin2u) = r2cos2u
y2 = 4ax becomes
M08_ENGINEERING_MATHEMATICS-I _CH08.indd 22
r2sin2u = 4a ⋅ rcosu
⇒ r(rsin2u − 4acosu) = 0
5/19/2016 8:26:01 PM
Multiple Integrals ■
4a cos u
sin 2 u
4a cos u
p p
∴ limits for r are 0,
and u varies from to .
2
4
2
sin u
⇒
∴
r = 0 and rsin2u − 4acosu = 0
I=
p 4 a cos u
2 sin 2 u
∫ ∫
p
4
0
⇒
8.23
r=
[{ slope of the line is tan u = 1 ⇒ u =
p
]
4
u
p
⎡ 4 a cos
⎤
sin 2 u
2
r 2 cos 2 u
⎢
⎥
rdrd u = ∫ cos 2u ⎢ ∫ rdr ⎥ d u
r2
p
⎣ 0
⎦
4
p
2
4 a cos u
⎡ r 2 ⎤ sin2 u
du
= ∫ cos 2u ⎢ ⎥
⎣ 2 ⎦0
p
4
p
12
16 a 2 × cos 2 u
= ∫ cos 2 u
du
2p
sin 4 u
4
p
16a2 2
cos 2 u
(cos 2 u − sin 2 u) 4 d u
=
∫
2 p
sin u
4
p
2
⎛ cos 2 u ⎞ sin 2 u cos 2 u
= 8a2 ∫ ⎜ 2 − 1⎟
du
⎝
⎠
sin 4 u
p sin u
4
p
2
= 8a2 ∫ (cot 2 u − 1)
p
4
cos 2 u
du
sin 2 u
p
2
= 8a2 ∫ (cosec 2 u − 1 − 1) cot 2 ud u
p
4
= 8a
p
2
2
∫ (cosec u − 2) cot
2
2
u du
p
4
p
2
= 8a2 ∫ (cosec 2u cot 2 u − 2 cot 2 u) d u
p
4
p
⎤
⎡ p2
2
⎢
⎥
2
2
2
2
= 8a ⎢ ∫ cosec u cot ud u − 2 ∫ cot ud u⎥
p
⎢⎣ p4
⎥⎦
4
M08_ENGINEERING_MATHEMATICS-I _CH08.indd 23
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■
8.24
Engineering Mathematics
p
⎤
⎡ p2
2
⎥
2 ⎢
2
2
2
= 8a ⎢ ∫ cot u cosec ud u − 2 ∫ (cosec u − 1) d u⎥
p
⎥⎦
⎢⎣ p4
4
⎤
⎡ p2
p
⎥
2 ⎢
= 8a ⎢ ∫ − cot 2 u ( −cosec 2 u)d u − 2[ − cot u − u]p2 ⎥
4 ⎥
⎢⎣ p4
⎦
p
⎡ 1
⎧ p p ⎛
p p ⎞ ⎫⎤
= 8a 2 ⎢ − [cot 3 u]p2 + 2 ⎨cot + − ⎜ cot + ⎟ ⎬ ⎥
⎝
2 2
4 4 ⎠ ⎭⎦
⎩
⎣ 3
4
⎡ 1⎛
p
p⎞
p
p⎞ ⎤
⎛
= 8a 2 ⎢ − ⎜ cot 3 − cot 3 ⎟ + 2 ⎜ 0 + − 1 − ⎟ ⎥
⎝
⎠
⎝
2
4
2
4⎠⎦
⎣ 3
⎡ 1
p ⎤ 8a 2
4a 2
⎛ p p⎞ ⎤
⎡1
(3p − 10) =
(3p − 10)
= 8a 2 ⎢ − ( −1) − 2 + 2 ⎜ − ⎟ ⎥ = 8a 2 ⎢ − 2 + ⎥ =
⎝ 2 4⎠⎦
2⎦
6
3
⎣3
⎣ 3
EXAMPLE 5
a
a 2 2y 2
0
0
Evaluate ∫
∫
( x 2 1 y 2 ) dxdy by changing into polar coordinates.
Solution.
a
I=∫
Let
0
a2 − y 2
∫
( x 2 + y 2 ) dxdy
0
Limits for x are x = 0 and x = a2 − y 2
⇒
x = a2 − y 2
Now
x 2 = a2 − y 2
which is circle with centre (0, 0) and radius a
Limits for y are y = 0 and y = a
∴ the region of integration is as in Fig. 8.27
bounded by y = 0, y = a and x = 0, x = a2 − y 2
To change to polar coordinates,
∴
dxdy = rdrdu
x2 + y2 = a2
⇒
and
r2 = a2
p
2
⇒
r
O
r=±a
p
a
4
⎡a 3 ⎤
⎡ r4 ⎤
2
∴ I = ∫ ∫ r ⋅ rdrd u = ∫ ⎢ ∫ r dr ⎥ d u = ∫ ⎢ ⎥ d u =
⎣ 4 ⎦0
0
0 0
0 ⎣0
⎦
M08_ENGINEERING_MATHEMATICS-I _CH08.indd 24
(0, a)
x 2 + y 2 = a2
or r = a
P
θ
(a, 0) x
x2 + y2 = r2
∴ in the given region, r varies from 0 to a and
p
u varies from 0 to
2
p
2 a
x 2 + y 2 = a2
y=a
put x = rcosu, y = rsinu
∴
⇒
Fig. 8.27
p
4
p
a4
a4
pa 4
2
=
=
d
u
[
u
]
.
0
∫0 4
4
8
5/19/2016 8:26:08 PM
Multiple Integrals ■
8.25
EXAMPLE 6
Evaluate ∫∫ ydxdy, where R is the region bounded by the semi-circle x2 1 y2 5 2ax and the x-axis
R
and the lines y 5 0 and y 5 a.
Solution.
I = ∫∫ ydxdy
Let
R
The region R is as in Fig. 8.28
y
We have x2 + y2 = 2ax
⇒
x 2 − 2ax + y 2 = 0
⇒
( x − a) 2 + y 2 = a2
(2, 2)
r
θ P
O
which is a circle with centre (a, 0) and
radius a
To change to polar coordinates,
put x = rcosu, y = rsinu
∴
y= a
x 2 + y 2 = 2ax
or r = 2acosθ
x
(2a, 0)
Fig. 8.28
dxdy = rdrd u and x 2 + y 2 = r 2
Now
x 2 + y 2 = 2ax
⇒
r 2 = 2ar cos u
r 2 − 2ar cos u = 0
⇒
⇒
r ( r − 2a cos u) = 0
∴ r varies from 0 to 2a cos u and u varies from 0 to
∴ I=
p
2 2 a cos u
∫ ∫
0
0
⇒
r = 0, r = 2a cos u
p
2
p
2
2 a cos u
r sin u ⋅ rdrd u = ∫ sin u ⎡⎢ r 2 ∫
dr ⎤⎥ d u
⎦
⎣ 0
0
p
2
⎡ r3 ⎤
= ∫ sin u ⎢ ⎥
⎣3⎦
0
=
2 a cos u
du
p
2
1
sin u ( 2a)3 cos3 ud u
3 ∫0
p
8a3 2
=
cos3 u sin ud u
3 ∫0
p
2a 3 ⎡ 4 p
2a 3
2a 3
8a3 ⎡ − cos 4 u ⎤ 2
⎤
− cos 0 ⎥ = −
cos
(0 − 1) =
=
⎥ =−
⎢
⎢
⎦
3 ⎣ 4 ⎦0
3 ⎣
2
3
3
M08_ENGINEERING_MATHEMATICS-I _CH08.indd 25
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8.26
■
Engineering Mathematics
EXERCISE 8.3
Polar Coordinates
p
2 a 2
p
2 2 a cos u
∫∫
1. Evaluate
0
rdrd u.
2. Evaluate
∫ ∫
p
4
0
rdrd u.
0
3. Find the area of a loop of the curve r = a sin3u. 4. Find the area of a loop of the curve r = a cos3u.
5. Find the area common to the circles r = a 2 and r = 2acosu.
p
2 ∞
6. Find the area of the cardioid r = a(1 − cosu).
7. Evaluate
∫ ∫ (r
0 0
rdrd u
.
2
+ a2 )2
8. Evaluate ∫∫ r 3 drd u, where A is the area between the circles r = 2 sin u and r = 4 sin u.
A
p
2 a cos u
∫ ∫
9. Evaluate
0
p a (1+ cos u )
10. Evaluate ∫
r a − r drd u.
2
2
0
0
∫
r 2 cos u dr d u.
0
Change of Variables
a a
xdxdy
.
2
x
(
+ y 2 )3 / 2
y
11. Change into polar coordinates and evaluate ∫ ∫
0
∫∫ (x
12. Evaluate
+ y 2 )7 / 2dxdy by changing into polar coordinates where R is the region bounded
2
R
by the circle x2 + y2 = 1.
2 a 2 ax − x 2
13. Change into polar coordinates and evaluate
∫ ∫
0
∫∫
14. Evaluate
R
( x 2 + y 2 )dxdy .
0
xydxdy by changing into polar coordinates, where R is the region in the positive
x2 + y 2
quadrant.
15. Evaluate
∫∫
R
dxdy
x + y 2 + a2
by changing into polar coordinates, where R is the I quadrant.
2
x 2y 2
∫∫R x 2 + y 2 dxdy by changing into polar coordinates, where R is the annular region
between the circles x2 + y2 = 4 and x2 + y2 = 16.
16. Evaluate
[Hint I =
2p 4
∫ ∫r
2
cos 2 u sin 2 udrd u ]
0 2
17. Evaluate
∫∫
2
2
a 2 − x 2 − y 2 dxdy where R is the semi-circle x + y = ax in the I quadrant, changing
R
to polar coordinates.
M08_ENGINEERING_MATHEMATICS-I _CH08.indd 26
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Multiple Integrals ■
a a
18. Evaluate
∫∫
0 y
8.27
x 2dxdy by changing to polar coordinates.
x2 +y2
n
19. Evaluate
2
2
2
2
∫ ∫ xy( x + y ) 2 dxdy over the positive quadrant of x + y = 4, supposing n + 3 > 0.
2
20. Transforming to polar coordinates evaluate the integral
4 − x2
∫ ∫
0
( x 2 y + y 3 ) dxdy .
0
ANSWERS TO EXERCISE 8.3
1.
pa2
2
2.
a2
(p − 2)
4
3.
pa2
12
4.
pa2
12
6.
3pa2
2
7.
p
4a 2
8.
45p
2
9.
a3
(3p − 4)
18
10.
5pa3
8
12.
2p
9
13.
3pa4
4
14.
a3
6
15.
p
4a 2
17.
a3
(3p − 4)
18
18.
a3
log e ( 2 + 1)
3
19.
2n + 3
n+4
20.
32
5
11.
pa
4
16. 15 p
8.1.6
5. a2 (p − 1)
Area as Double Integral
(a) Area as double integral in cartesian coordinates
Double integrals are used to compute area of bounded plane regions. The area A of a plane
bounded region R in cartesian coordinates is
A5
∫∫ dxdy.
R
(i) If the region R is bounded by curves y = f1(x), y = f2(x) and lines x = a, x = b where a and b are
constants, then
b f2 (x)
⎡
⎤
A 5 ∫ ⎢ ∫ dy ⎥dx
a ⎢
⎣ f1 (x) ⎥⎦
(ii) If the region R is bounded by curves x = g1(y), x = g2(y) and line y = c, y = d where c and d are
constants, then
d g 2 (y)
⎡
⎤
A 5 ∫ ⎢ ∫ dx ⎥ dy
c ⎢
⎣ g1 (y) ⎥⎦
M08_ENGINEERING_MATHEMATICS-I _CH08.indd 27
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8.28
■
Engineering Mathematics
WORKED EXAMPLES
EXAMPLE 1
Find the area bounded by the ellipse
x2 y 2
1 51, using double integration.
a2 b 2
Solution.
x2 y 2
Equation of the ellipse is 2 + 2 = 1
a
b
y
By the symmetry of the curve,
ellipse is A = 4 × Area in the first quadrant
the
area
of
y = b 1−
the
(0, b)
x2
a2
a b 1− x 2 / a 2
= 4∫
∫
0
(a, 0)
dydx
x
0
a
= 4∫ [ y ]0b
1− x 2 / a 2
dx
0
a
= 4∫ b 1 −
0
Fig. 8.29
a
x2
4b
dx =
a 2 − x 2 dx
∫
2
a
a
0
a
4b ⎡
a2
p
x⎤
4b ⎡ x a 2 − x 2 a 2
−1 ⎤
=
+ sin −1 ⎥ =
⎢
⎢0 + sin 1⎥ = 2ab ⋅ = pab
a ⎣
2
2
a ⎣
a ⎦0
⎦
2
2
EXAMPLE 2
Using double integral find the area enclosed by the curves y 5 2x2 and y2 5 4x.
Solution.
The region of integration is the shaded region
(as in Fig. 8.30) bounded by y2 = 4x and y = 2x2
To find A, solve the equations
y2 = 4x4
⇒
4x = 4x4 ⇒
∴
y 2 = 4x
A
Q
y2 = 4x and y = 2x2
⇒
y = 2x 2
y
(1, 1)
x=1
P
O
x(x3 − 1) = 0 ⇒
x
x = 0, 1
A is (1, 1)
Required area = ∫∫ dxdy
Fig. 8.30
R
Take a strip PQ parallel to y axis with P lies on
y = 2x2, Q lies on y2 = 4x ⇒ y = 2 x
∴ the limits of y are y = 2x2 to y = 2 x and the limits of x are x = 0 to x = 1
area =
1
1
1
⎡ x 3/ 2
x3 ⎤
4 2 2
∫0 ∫2 dydx = ∫ [y ]22x2x dx = ∫ [2 x − 2x 2 ]dx = ⎢⎣2 3/ 2 − 2 3 ⎥⎦ 0 = 3 − 3 = 3
2x
0
0
1 2 x
∴
M08_ENGINEERING_MATHEMATICS-I _CH08.indd 28
6/3/2016 8:20:19 PM
Multiple Integrals ■
8.29
EXAMPLE 3
Find the smaller of the areas bounded by y 5 2 2 x and x2 1 y2 5 4 using double integral.
Solution.
The region R is the shaded part in Fig. 8.31.
Required area A = ∫∫ dxdy
R
y
To find limits for y, take a strip PQ parallel to
the y-axis, P lies on y = 2 – x and Q lies on the
circle x2 + y2 = 4
∴ y limits are y = 2 – x to y = 4 − x 2 and the
x limits are x = 0 to x = 2
∴
⎡
A = ∫⎢
⎣
0 ⎢
2
4 − x2
∫
2−x
B(0, 2)
Q
P
O
x 2 + y2 = 4
A(2, 0)
x
⎤
dy ⎥dx
⎥⎦
2
= 0∫ [ y ]2 −4x− x dx
y=2−x
Fig. 8.31
2
2
= ∫ ⎡⎣ 4 − x 2 − ( 2 − x ) ⎤⎦ dx
0
2
⎡x
4
x
x2 ⎤
p
4
4 − x 2 + sin −1 − 2 x + ⎥ = 0 + 2(siin −1 1 − sin −1 0) − 2 ⋅ 2 + = 2 ⋅ − 4 + 2 = p − 2
=⎢
2
2
2 ⎦0
2
2
⎣2
EXAMPLE 4
Find the area bounded by the parabola y2 5 4 2 x and y2 5 4 2 4x as a double integral and
evaluate it.
Solution.
Given
y2 = 4 − x = − (x − 4) is a parabola with vertex (4, 0) and towards
the negative x-axis, axis of symmetry the x-axis.
and
y2 = 4 − 4x = − 4 (x − 1) is a parabola with vertex (1, 0) and towards
the negative x-axis, axis of symmetry the x-axis.
To find the points of intersections, solve y2 = 4 − x and y2 = 4 − 4x,
∴
4 − x = 4 − 4x ⇒ 3x = 0 ⇒ x = 0
and
y2 = 4 − x ⇒ y2 = 4 ⇒ y = ±4 and the points are (0, 2), (0, −2)
Draw the graph and determine the region. The region is the shaded region as in Fig. 8.32.
Both curves are symmetric about x-axis.
∴ required area A = 2 Area above the x-axis = 2∫∫ dxdy
R
M08_ENGINEERING_MATHEMATICS-I _CH08.indd 29
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8.30
■
Engineering Mathematics
It is convenient to take strip PQ parallel to the
x-axis. P lies on y2 = 4 − 4x
and Q lies on
y
y 2 = 4 − 4x
(0, 2)
y = 4 − x.
and
Q
P
y2
4
2
y = 4 − x ⇒ x = 4 − y2
y2 = 4 − 4x ⇒ x = 1 −
Now
y 2= 4 − x
2
B
A (1, 0)
and the limits of y are y = 0, y = 2
x
(4, 0)
(0, −2)
2 ⎡4−y
2
⎤
2
∴ area A = 2∫ ⎢ ∫ dx ⎥ dy = 2∫ [ x ]4 −yy2 dy
1
−
⎥
0 ⎢ y2
0
4
⎢⎣1− 4 ⎥⎦
2
⎡
⎛
y2 ⎞ ⎤
= 2∫ ⎢ 4 − y 2 − ⎜1 − ⎟ ⎥ dy
⎝
4 ⎠⎦
0 ⎣
2
Fig. 8.32
2
⎡
3 ⎞
3 y3 ⎤
8⎤
⎛
⎡
= 2∫ ⎜ 3 − y 2 ⎟ dy = 2 ⎢3 y − ⋅ ⎥ = 2 ⎢3 × 2 − ⎥ = 2[6 − 2] = 8
⎝
⎠
⎣
4
4 3 ⎦0
4⎦
⎣
0
2
EXAMPLE 5
Using double integration find the area of the parallelogram whose vertices are A(1, 0), B(3, 1),
C(2, 2), D(0,1)
Solution.
The given points A(1, 0), B(3, 1), C(2, 2) and D(0,1) are the vertices of a parallelogram ABCD.
Required area is the area of the parallelogram
y
ABCD as in Fig 8.33.
Area of the parallelogram ABCD
C(2, 2)
= 2 (area of the triangle ABD)
We shall find the equations of AB and AD.
B
D
We know the equation of the line joining the
(3, 1)
(0, 1)
the points (x1, y1) and (x2, y2) is
Q
P
O
y − y1
x − x1
=
y 1 − y 2 x1 − x 2
A(1, 0)
x
Fig. 8.33
∴ equation of AB, the line joining (1, 0) and (3, 1)
y − 0 x −1
=
0 −1 1− 3
⇒
y =
1
( x − 1)
2
(1)
Equation of AD, the line joining (1, 0), (0, 1) is
y − 0 x −1
=
⇒ y = −x +1
0 −1 1− 0
M08_ENGINEERING_MATHEMATICS-I _CH08.indd 30
(2)
5/19/2016 8:26:48 PM
Multiple Integrals ■
Area of Δ ABD =
8.31
∫∫ dx dy
ABD
Take a strip PQ parallel to the x-axis with P is on (2) and Q is on (1).
∴ x = −y + 1 and x = 2y + 1 and y varies from 0 to 1.
1
1 2 y +1
⎤
⎡
2 y +1
area of ΔABD = ∫ ⎢ ∫ dx ⎥ dy = ∫ [ x ]− y +1 dy
0
0 ⎢
⎣ − y +1 ⎥⎦
∴
[ Note that BD is parallel to the x-axis]
1
1
1
⎡ y2 ⎤
3
= ∫ [2 y + 1 − ( − y + 1)]dy = ∫ 3 y dy = 3 ⎢ ⎥ =
⎦
⎣
2
2
0
0
0
∴ area of the parallelogram ABCD is = 2 ×
3
= 3.
2
EXERCISE 8.4
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
13.
Find the area bounded by the parabola x2 = 4y and the straight line x – 2y + 4 = 0.
Evaluate the area bounded by y = x and y = x2.
Evaluate the area bounded by y2 = 4ax and x2 = 4ay.
Evaluate the area bounded by y = 4x – x2 and y = x.
x2 y 2
x y
Evaluate the smaller area bounded by
+
= 1 and the line + = 1.
9
4
3 2
2
2
Evaluate the smaller area bounded by x + y = 4 and x + y = 2.
Evaluate the area bounded by y2 = 4x, x + y = 3 and the X-axis.
x2
Evaluate the area bound by y =
, y = ax , x = 0 and x = 4a.
4a
Find the area common to y2 = x and x2 + y2 = 4.
Find the area bounded by y2 = 4 − x, y2 = x.
Find the area of the curve a2y2 = x2(2a − x).
Find the area of a circle of radius a by double integration.
Find the area between the parabola y = 4x – x2 and the line y = x by double integration.
ANSWERS TO EXERCISE 8.4
1. 9
2.
1
6
3.
6. p − 2
7.
10
3
8.
11. 4a
12. pa2
16a2
3
4.
16a2
3
9
13.
2
(b) Area as double integral in polar coordinates
As double integral, area in polar coordinates is
9
2
9. 3 3 + 4
5.
p
3
10.
3
( p − 2)
2
16 2
3
∫∫ rdrd u
R
where R is the region for which the area is required.
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WORKED EXAMPLES
EXAMPLE 1
Find the area bounded between r 5 2cosu and r 5 4cosu.
Solution.
Area A = ∫∫ rdrdu
R
where the region R is the region between the circles
r = 2cosu and r = 4cosu
The area is the shaded region as in Fig. 8.34.
We first integrate w.r.to r and so, we take the radius
vector OPQ. When PQ is moved to cover the area A,
r varies from r = 2cosu to r = 4cosu,
p
p
and u varies from u = − to u =
2
2
∴ Area A =
p
2 4 cos u
p
2
−
2
=
Q
O
θ P
X
4 cos u
⎡ r2 ⎤
∫p 2 cos∫ u rdrdu = ∫p ⎢⎣ 2 ⎥⎦ 2 cos u du
−
r = 2cosθ
r = 4cosθ
Y
π
θ=
2
θ= −
2
1
2
Fig. 8.34
p
2
∫ (4
−
π
2
2
cos 2 u − 22 cos 2 u) d u
p
2
p
2
p/2
1 p
2
= 6 ∫ cos 2 u d u = 6 × 2 ∫ cos 2 u du
u = 2 × 6 ⋅ ⋅ = 3p [{ cos u is even ]
2
2
p
0
−
2
EXAMPLE 2
Find the area of one loop of the leminiscate r2 5 a2cos2u.
Solution.
Given
r2 = a2cos2u
Area of the loop = ∫∫ rdrdu , where R is the region as in Fig. 8.35.
R
Since the loop is symmetric about the initial line, required area is twice the area above the
initial line.
First we integrate w.r.to r
In this region, take a radial strip OP, its ends are
r = 0 and r = a cos 2u
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Multiple Integrals ■
When the strip is moved to cover the region R,
p
u varies from 0 to (above ox)
4
θ=
p
4 a cos 2 u
Required Area A = 2 ∫
0
∫
π
P
rdrd u
θ
0
x
O
p
4
8.33
a cos 2 u
⎡⎛ r ⎞ ⎤
= 2 ∫ ⎢⎜ ⎟ ⎥
⎝ 2 ⎠ ⎦0
0 ⎣
2
du
p
4
θ=−
= ∫ a 2 cos 2u d u
π
Fig. 8.35
0
p
4
p
2
2
⎡ sin 2 u ⎤ 4 a ⎛ p
⎞ a
= a ∫ cos 2u d u = a ⎢
sin
sin
0
=
−
=
⎜
⎟
⎠
2 ⎝
2
2
⎣ 2 ⎥⎦ 0
0
2
2
EXAMPLE 3
Find the area of a loop of the curve r 5 a sin3u.
Solution
Given r = asin3u
The area of the loop = ∫∫ rdrd u
θ=
R
But the loop is formed by two consecutive values of u
when r = 0.
When r = 0, asin3u = 0
p
3u = 0 or p ⇒ u = 0 or
⇒
3
and r varies from r = 0 to r = asin3u
∴
area of the loop =
p
3 a sin 3 u
∫ ∫
0
=
0
p
3
π
3
X
O
Fig. 8.36
a sin 3u
⎡ r2 ⎤
rdrd u = ∫ ⎢ ⎥
⎣ 2 ⎦0
0
du
p
3
1 2 2
a sin 3 ud u
2 ∫0
p
a 2 3 1 − cos 6 u
=
du
2 ∫0
2
a2
=
4
M08_ENGINEERING_MATHEMATICS-I _CH08.indd 33
p
sin 6u ⎤ 3 a 2 ⎡ p sin 2p − sin 0 ⎤ pa 2
⎡
u
−
=
−
⎢⎣
⎥⎦ = 12
6 ⎥⎦ 0
4 ⎢⎣ 3
6
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EXAMPLE 4
Find the area which is inside the circle r 5 3acosu and outside the cardioid r 5 a(1 1 cosu).
Solution.
Given
r = 3acosu
(1)
and
r = a(1 + cosu)
(2)
r = 3acosθ
Required area A = ∫ ∫ rdrdu
Eliminating r from (1) and (2), we get
P′
3a cos u = a(1 + cos u)
⇒
2 cos u = 1
∴
u=−
⇒
cos u =
P
θ
O
1
2
p p
or
3
3
x
r = a(1 + cosθ)
Required area is the shaded region as in Fig. 8.37.
Fig. 8.37
Since both the curves are symmetrical about the
initial line, required area is twice the area above the initial line.
In this region take a radial strip OPP′ where P lies on (2) and P′ lies on (1).
When it moves, it will cover the required area.
p
∴ r varies from a(1 + cosu) to 3acosu and u varies from 0 to .
3
p
3 r = 3 a cos u
p
3
3a cos u
⎡ r2 ⎤
Required area = 2 ∫
rdrd
u
=
2
∫
∫0 ⎢⎣ 2 ⎥⎦a(1+ cos u)d u
0 r = a (1+ cos u )
p
3
= ∫ [9a2 cos 2 u − a2 (1 + cos u) 2 ] d u
0
p
3
= a 2 ∫ [9 cos 2 u − (1 + 2 cos u + cos 2 u)] d u
0
=a
p
3
2
∫ [8 cos
2
u − 1 − 2 cos u] d u
0
p
3
{
}
⎤
⎡ 1 + cos 2 u
= a 2 ∫ ⎢8
− 1 − 2 cos u⎥ d u
2
⎦
⎣
0
p
⎡ ⎛
⎤3
sin 2 u ⎞
= a ⎢4 ⎜ u +
− u − 2 sin u ⎥
⎟
2 ⎠
⎣ ⎝
⎦0
2
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Multiple Integrals ■
8.35
⎡ ⎛
⎤
2p ⎞
⎢ ⎜ p sin 3 ⎟ p
⎥
p
= a ⎢4 ⎜ +
− − 2 sin − 0 ⎥
⎟
2 ⎠ 3
3
⎣ ⎝3
⎦
2
⎡ 4p
3 p
3⎤
⎡ 4p p ⎤
= a2 ⎢
+2
− − 2 ⎥ = a2 ⎢
− ⎥ = pa 2
2
3
2 ⎦
3⎦
⎣ 3
⎣ 3
EXAMPLE 5
Find the area common to r5a 2 and r = 2acosu.
Solution.
Given
r=a 2
(1)
r = 2acosu
and
(2)
y
(1) is a circle with centre (0, 0) and radius a 2
(2) is a circle with centre (a, 0) and radius a
r =a 2
Solve (1) and (2) to find the point of intersection.
∴
Q
a 2 = 2a cos u
1
p
cos u =
⇒ u=
4
2
⇒
r = 2a cosθ
B
C
P
π/4
O
A
x
Since the circles are symmetrical about the initial
line OX,
required area = 2 [area OABC] = 2 [area OAB
+ area OBC]
Fig. 8.38
In OAB, take a strip OP.
When OP moves it covers the area OAB. Ends of OP are, r = 0 and r = a 2
p
∴ r varies from 0 to a 2 and u varies from 0 to
4
In the area OBC, take a strip OQ. Ends of OQ are, r = 0 and r = 2acosu
When OQ moves it covers the area OBC.
p p
∴ r varies from 0 to 2acosu and u varies from to
4
2
∴
p
⎤
⎡ p4 a 2
2 2 a cos u
⎢
⎥
Required area = 2 ⎢ ∫ ∫ rdrd u + ∫ ∫ rdrd u⎥
p
0
⎢⎣ 0 0
⎥⎦
4
p
4
a 2
p
2 a cos u
2
⎡ r2 ⎤
⎡ r2 ⎤
= 2∫ ⎢ ⎥ d u + 2∫ ⎢ ⎥
⎣ 2 ⎦0
p ⎣ 2 ⎦0
0
du
4
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p
4
p
2
0
p
4
= ∫ 2a2d u + ∫ 4a2 cos 2 ud u
p
2
p
4
0
⎛ 1 + cos 2u ⎞
= 2a2 [u] + 4a2 ∫ ⎜
⎟⎠ d u
⎝
2
p
4
p
p
sin 2 u ⎤ 2
⎡
= 2a ⋅ + 2a 2 ⎢u +
2 ⎥⎦ p
4
⎣
2
4
=
⎡p p 1 ⎛
pa
p⎞ ⎤
+ 2a 2 ⎢ − + ⎜ sin p − sin ⎟ ⎥
⎝
2
2
4
2
2⎠⎦
⎣
=
2
pa 2
pa 2
⎡ p 1 ⎤ pa
+
− a 2 = a 2 (p − 1)
+ 2a 2 ⎢ − ⎥ =
⎣ 4 2⎦
2
2
2
2
EXAMPLE 6
Find the area inside the circle r 5 asinu but lying outside the cardiod r 5 a(1 2 cosu).
Solution.
Given
r = a sinu
(1)
and
r = a(1 − cosu)
Area = ∫ ∫ rdrdu
(2)
r = a(1 − cosθ)
Y
θ=
Eliminating r from (1) and (2), we get
Q
P
asinu = a(1 − cosu)
⇒
π
2
sinu + cosu = 1
O
r = a sinθ
X
Squaring, sin2u + cos2u + 2sinu cosu = 1
⇒
1 + 2sin2u = 1
Fig. 8.39
⇒ sin 2u = 0 ⇒
∴
p
2
a sin u
p
2
2u = 0, p ⇒ u = 0 or
p
2
a sin u
⎡ r2 ⎤
Area = ∫ ∫ rdrd u = ∫ ⎢ ⎥
du
⎣ 2 ⎦ a (1− cos u )
0 a (1− cos u )
0
p
12
= ∫ [a 2 sin 2 u − a 2 (1 − cos u) 2 ] d u
20
p
a2 2
=
[sin 2 u − (1 − 2 cos u + cos 2 u)] d u
2 ∫0
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Multiple Integrals ■
8.37
p
a2 2
=
{−1 + 2 cos u − (cos 2 u − sin 2 u)} d u
2 ∫0
p
⎡p
⎤
2
a2 ⎢ 2
⎥
2
2
[ −1 + 2 cos u}d u − ∫ {cos u − sin u] d u⎥
=
∫
⎢
2 ⎣0
0
⎦
p
2 2
p
2
p
2
0
0
=
a
2
=
p⎤ a ⎡ p
a
a ⎡ p
⎤ a
[ −u + 2 sin u] =
− + 2 sin ⎥ =
− + 2⎥ = [4 − p]
2
2 ⎢⎣ 2
2 ⎦ 2 ⎢⎣ 2
⎦ 4
[since ∫ cos 2 d u = ∫ sin 2 ud u]
∫ ( −1 + 2 cos u) du
0
p
2
0
2
2
2
2
EXERCISE 8.5
Find the area bounded between r = 2sinu and r = 4sinu.
Find the area of one loop of r = acos3u.
Find the area that lies inside the cardioid r = a(1 + cosu) and outside the circle r = a.
Find the area of the cardioid
(i) r = a(1 + cosu), (ii) r = 4(1 + cosu)
5. Find by double integration, the area lying inside the cardioid r = 1 + cosu and out the parabola
r(1 + cosu) = 1.
6. Calculate the area included between the curve r = a(secu + cosu) and its asymptote.
7. Find the area of the cardioid r = a(1 − cosu).
1.
2.
3.
4.
ANSWERS TO EXERCISE 8.5
1. 3p
5.
8.2
9p + 16
12
2.
pa2
12
3.
6.
5pa2
4
2
7. 3pa
2
a2
(p + 8)
4
4. (i)
3pa2
2
(ii) 24p
AREA OF A CURVED SURFACE
Introduction
Let D ⊂ R, say D = [a, b]. If f : D → R is a function, then the graph of the function f is the set of points
{(x, y) : y = f(x) ∀ x ∈ D} which is a subset of R2.
This subset of R2 is called a curve in R2 whose equation is
y = f(x) ∀ x [a, b]
In implicit form the equation of the curve is F(x, y) = 0
For example y = x2 is the equation of the parabola in explicit form, where as x2 − y = 0 is the
implicit form of the equation of the parabola.
Let D ⊂ R2
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If f : D → R is a function, then the graph of the function f is the set of points
{(x, y, z) : z = f(x, y) ∀ (x, y) ∈ D}
3
which is a subset of R .
This subset of R3 is called a surface in R3, whose equation is
z = f(x, y) ∀ (x, y) ∈ D
This explicit form is called Monge’s form of the equation of the surface.
The general form of the surface is the implicit form
F(x, y, z) = 0
Sphere, cone, cylinder are surfaces in R3.
The equation x2 + y2 + z2 = a2 is a sphere in R3 or 3-dimensional space.
The equation
x2 + y2 = a2 is a cylinder in R3 or 3-dimensional space.
The equation
x2 + y2 = 4z2 is a cone in R3 or 3-dimensional space.
Smooth surface
Definition 8.1 A surface S is said to be smooth if at each point unique normal exists and it varies
continuously as the point moves on S.
Piece-wise Smooth surface
Definition 8.2 A surface S is said to be piece-wise smooth if it can be divided into a finite number of
smooth surfaces.
For example: the surface of a cube is a piece-wise smooth surface.
8.2.1
Surface Area of a Curved Surface
In earlier classes you have seen the area of surface of revolution. That is a surface obtained by
revolving an arc of a curve about an axis.
For example the surface of a sphere is obtained by revolving the semi-circle about its bounding diameter.
This surface area is expressed as an integral of a function of a single independent variable.
2
⎛ dy ⎞
ds
We know that surface area = ∫ 2py
dx = 2p∫ y 1 + ⎜⎝ ⎟⎠ dx
dx
dx
a
a
where y = f(x).
But the general problem of finding the area of a curved surface S is found as a double integral over
the orthogonal projection D of S on one of the coordinate planes.
This is possible if any line perpendicular to the chosen coordinate plane meets the surface S in not
more than one point.
b
8.2.2
b
Derivation of the Formula for Surface Area
Let S be a surface of finite area represented by the equation F(x, y, z) = 0.
Let D be the orthogonal projection of S on the xy-plane as in Fig. 8.40
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Multiple Integrals ■
8.39
z
S P(x, y, z)
dS
y
dxdy
dx
dy
x
D
Fig. 8.40
Divide the region D into element rectangular areas by drawing lines parallel to x-axis and y-axis.
Let dS be the element area of the surface whose projection is shaded, which is a rectangle of sides dx, dy
∴ element area = dxdy
Let P(x, y, z) be any point on dS and n be the outward unit normal at P.
Then n =
∇F
∂F
∂F
∂F
, where ∇F = i
+j
+k
∇F
∂x
∂y
∂z = i Fx + j Fy + k Fz
∴
∇F = Fx 2 + Fy 2 + Fz 2
Let g be the angle between the plane of dS and the plane of dxdy.
We know that the angle between two planes is the angle between their normals.
The normal to the plane of dS is n and the normal to the plane of dxdy is k.
∴
cos g =
n⋅k
n⋅k
= n ⋅ k , Since n and k are unit vectors.
We always take the acute angle, which is given by cos g = n ⋅ k
Since dxdy is the projection of dS, we have
dxdy = cos g × dS
Integrating,
S = ∫∫
D
M08_ENGINEERING_MATHEMATICS-I _CH08.indd 39
⇒
dS =
dxdy
dxdy
=
cos g ∫∫
D n⋅k
dxdy
cos g
(1)
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Engineering Mathematics
Similarly, projecting on yz plane, we get
dydz
S = ∫∫
D1 n ⋅ i
where, D1 is the orthogonal projection of S on the yz plane.
Projecting on the zx-plane, we get
S = ∫∫
D2
(2)
dzdx
(3)
n⋅ j
Cartesian form of Surface Area
Since
∇F i Fx + j Fy + k Fz
∇F =
Fx 2 + Fy 2 + Fz 2
n=
Fz
n⋅k =
∴
Similarly,
S = ∫∫
n⋅k =
Fx + Fy + Fz
dxdy
Fx 2 + Fy 2 + Fz 2
2
= ∫∫
Fz
Fx + Fy 2 + Fz 2
2
dx dy
(4)
Fx 2 + Fy 2 + Fz 2
dydz =
dy dz
∫∫D
Fx
n ⋅i
1
(5)
Fx 2 + Fy 2 + Fz 2
dzdx =
dz dx
S = ∫∫
∫∫
Fy
D2
D2 n ⋅ j
(6)
D
S = ∫∫
D1
and
∴
2
2
n⋅k
Fz
D
Corollary If the equation of the surface is given explicitly or rewritten as z = f(x, y).
Then
f(x, y) − z = 0
Here
F(x, y, z) = f(x, y) − z
∴
∂F ∂f ∂z
=
=
∂y ∂y ∂y
∂F ∂f ∂z
,
=
=
∂x ∂x ∂x
∂F
= −1
∂z
⎛ ∂z ⎞ ⎛ ∂z ⎞
S = ∫∫ ⎜ ⎟ + ⎜ ⎟ + 1 dxdy
⎝ ∂x ⎠ ⎝ ∂y ⎠
D
2
∴ (4) ⇒
and
2
If the equation of the surface is given by x = f(y, z), then as above
(5) ⇒
⎛ ∂x ⎞ ⎛ ∂x ⎞
S = ∫∫ ⎜ ⎟ + ⎜ ⎟ + 1 dydz
⎝ ∂y ⎠ ⎝ ∂z ⎠
D1
2
2
If the equation of the surface is given by y = f2(x, z), then
(6) ⇒
⎛ ∂y ⎞ ⎛ ∂y ⎞
S = ∫∫ ⎜ ⎟ + ⎜ ⎟ + 1 dz dx
⎝ ∂x ⎠ ⎝ ∂z ⎠
D2
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2
2
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Multiple Integrals ■
8.2.3
8.41
Parametric Representation of a Surface
The parametric equations of a surface F(x, y, z) = 0 are written interms of two parameters as
x = x(u, v), y = y(u, v), z = (u, v) where (u, v) ∈ D ⊂ R2 in the u, v-plane. u and v are the parameters.
For example: The parametric equation of the equation of the sphere x2 + y2 + z2 = a2 in spherical
polar coordinates are x = a sin u cos f, y = a sin u sin f, and z = a cos u; where u and f are the
parameters.
WORKED EXAMPLES
EXAMPLE 1
Find the surface area of the sphere of radius a.
Solution.
x2 + y2 + z2 = a2
Let
(1)
be the equation of the sphere.
Since the sphere is symmetric about all the coordinate axes
the surface area S = 8 × Surface area of the sphere in the
positive octant.
The projection of the surface in the first octant is a
quadrant of the circle x2 + y2 = a2 as in Fig 8.42.
∴
S = 8∫∫
surface area
D
dxdy
∇F
∇F
n=
where
F = x2 + y2 + z2 − a2
Fx = 2x,
∴
y
O
n⋅k
But
∴
y
x
Fy = 2y, Fz = 2z
Fig. 8.41
∇F = 2 x i + 2 y j + 2 z k
y
∇F = 4 x 2 + 4 y 2 + 4 z 2 = 4a2 = 2a
∴
∴
∴
n=
2x i + 2 y j + 2z k = x i + y j + z k
a
a
a
2a
=
=
y
z ⎞
⎛x
n ⋅ k = ⎜ i + j + k ⎟ ⋅ k = z = 1 a2 − x 2 − y 2
⎝a
a
a ⎠
a
a
[ using (1)]
S = 8∫∫
D
dxdy
1 2
a − x2 − y 2
a
M08_ENGINEERING_MATHEMATICS-I _CH08.indd 41
a
= 8a∫
0
a2 − y 2
∫
0
−
+
=
dxdy
a − x2 − y 2
2
Fig. 8.42
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8.42
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Engineering Mathematics
a ⎡
= 8a∫ ⎢
⎢
0 ⎣
a2 − y 2
∫
0
⎤
⎥ dy
( a 2 − y 2 ) − x 2 ⎥⎦
dx
x
⎡
⎤
= 8a∫ ⎢sin −1
2
2 ⎥
a − y ⎦0
0 ⎣
a
a2 − y 2
dy
a⎧
⎫⎪
⎛ a2 − y 2 ⎞
⎪
= 8a∫ ⎨sin −1 ⎜
− sin −1 0⎬ dy
⎟
2
2
0⎪
⎝ a −y ⎠
⎩
⎭⎪
p
dy = 4ap[ y ]a0 = 4pa[a − 0] = 4pa2
2
0
a
a
= 8a∫ (sin −1 1 − 0) dy = 8a∫
0
Aliter: Since the sphere is symmetric in all the 8 octants. Consider the sphere in the I octant and
project it on the xy-plane. We get the quadrant of the circle x2 + y2 = a2.
⎛ ∂z ⎞ ⎛ ∂z ⎞
S = 8∫∫ ⎜ ⎟ + ⎜ ⎟ + 1 dxdy
⎝ ∂x ⎠ ⎝ ∂y ⎠
D
2
∴
Surface area
2
where D is the region of the circle in the first quadrant as in Fig.8.42
The equation of the sphere is
x2 + y2 + z2 = a2
Treating z as a function of x and y and differentiating partially w. r. to x and y respectively, we get
∂z
∂z
x
=0 ⇒
=−
∂x
∂x
z
∂z
∂z
y
2 y + 2z
=0 ⇒
=−
∂y
∂y
z
2x + 2z
and
2
∴
x 2 + y 2 + z 2 a2
x2 y 2
⎛ ∂z ⎞ ⎛ ∂z ⎞
1
1
=
= 2
+
+
=
+
+
⎜⎝ ⎟⎠ ⎜ ⎟
∂x
⎝ ∂y ⎠
z2
z
z2 z2
∴
a
⎛ ∂z ⎞ ⎛ ∂z ⎞
⎜⎝ ⎟⎠ + ⎜ ⎟ + 1 =
∂x
⎝ ∂y ⎠
z
2
2
∴
2
a
Surface area
a
S = 8∫∫ dxdy = 8a∫
z
0
D
a2 − y 2
∫
0
1
a −x −y
2
2
dxdy = 4pa2
[as above]
EXAMPLE 2
Find the surface area of the cone x 2 1 y 2 5 4 z 2 lying above the xy-plane and inside the cylinder
x 2 1 y 2 5 3ay .
Solution.
The equation of the cone is
M08_ENGINEERING_MATHEMATICS-I _CH08.indd 42
x 2 + y 2 = 4z 2
(1)
5/19/2016 8:27:31 PM
Multiple Integrals ■
The equation of the cylinder is x 2 + y 2 = 3ay
8.43
z
(2)
The surface area of (1) lying inside (2) is required.
Project this surface on the xy-plane.
x 2 + y 2 = 4z 2
2
⎛ ∂z ⎞
∂z
surface area S = ∫∫ ⎛⎜ ⎞⎟ + ⎜ ⎟ + 1 dxdy
⎝
⎠
∂x
⎝ ∂y ⎠
D
2
∴
y
x + y = 3ay
2
Differentiating (1) partially w. r. to x, y treating z as a
function of x and y, we get
∂z
= 2x
∂x
∂z
8z
= 2y
∂y
⇒
8z
and
x
Fig. 8.43
2
x2
y2
x 2 + y 2 + 16 z 2 20 z 2 5
⎛ ∂z ⎞ ⎛ ∂z ⎞
+
+1 =
=
=
⎜⎝ ⎟⎠ + ⎜ ⎟ + 1 =
2
2
∂x
⎝ ∂y ⎠
16 z 2
16 z 2 4
16 z
16 z
2
∴
⇒
∂z
x
=
∂x 4 z
∂z
y
=
∂y 4 z
2
2
⎛ ∂z ⎞
∂z
5
∴ ⎛⎜ ⎞⎟ + ⎜ ⎟ + 1 =
⎝ ∂z ⎠ ⎝ ∂y ⎠
2
2
∴
5
Surface area S =
∫∫D 2 dxdy
=
5
5
5 2
dxdy =
(area of the circle x 2 + y 2 = 3ay ) =
pr
2 ∫∫
2
2
D
where r is the radius of the circle.
2
x 2 + y 2 = 3ay
∴
S=
⇒ x 2 + y 2 − 3ay = 0
5 ⎛ 3a ⎞
p⎜ ⎟
2 ⎝ 2⎠
2
=
∴
3a
⎛ 3a ⎞
radius r = ⎜ ⎟ =
⎝ 2⎠
2
9 5 2
pa
8
EXAMPLE 3
Find the area cut from the sphere x 2 1 y 2 1 z 2 5a 2 by the cylinder x 2 1 y 2 5ax .
Solution.
We interchange the x, y axis for convenience as in Fig 8.44.
The equation of the sphere is
x 2 + y 2 + z 2 = a2
(1)
The equation of the cylinder is
x 2 + y 2 = ax
M08_ENGINEERING_MATHEMATICS-I _CH08.indd 43
(2)
5/19/2016 8:27:36 PM
8.44
■
Engineering Mathematics
Both the surfaces are symmetric about the axes.
∴ surfaces above and below the xy-plane are the
same. Since the cylinder lies on the side of the
positive x-axis, the required surface area
z
= 2 (surface area above the xy-plane)
x 2 + y 2 + z 2 = a2
Project the surface of the sphere cut off by the
cylinder onto the xy plane.
This is the circle x 2 + y 2 = ax .
O
x
x 2 + y 2 = ax
2
⎛ ∂z ⎞
∂z
∴ S = 2∫∫ ⎛⎜ ⎞⎟ + ⎜ ⎟ + 1 dxdy
⎝ ∂x ⎠ ⎝ ∂y ⎠
D
2
y
Fig. 8.44
Differentiating (1) partially w. r. to x, y, treating z
as a function of x and y, we get
and
2x + 2z
∂z
=0 ⇒
∂x
∂z
x
=−
∂x
z
2 y + 2z
∂z
=0 ⇒
∂y
∂z
y
=−
∂y
z
2
∴
x 2 + y 2 + z 2 a2
x2 y 2
⎛ ∂z ⎞ ⎛ ∂z ⎞
= 2
⎜⎝ ⎟⎠ + ⎜ ⎟ + 1 = 2 + 2 + 1 =
∂x
⎝ ∂y ⎠
z2
z
z
z
∴
a
⎛ ∂z ⎞ ⎛ ∂z ⎞
⎜⎝ ⎟⎠ + ⎜ ⎟ + 1 = =
∂x
⎝ ∂y ⎠
z
2
2
2
∴
a
a − x2 − y 2
2
S = 2∫∫
D
a
a − x2 − y 2
2
dxdy
where D is the circle in the xy plane.
The equation of this circle is
x 2 + y 2 = ax
∴
⇒
a
S = 2a∫
ax − x 2
∫
0 − ax − x 2
a
= 4a∫
0
y 2 = ax − x 2
1
(a − x 2 ) − y 2
2
ax − x 2
∫
0
⇒
1
(a − x ) − y
2
2
2
⎡
⎞⎤
⎛
y
= 4 a∫ ⎢sin −1 ⎜
⎥
2
2 ⎟
⎝ a − x ⎠ ⎥⎦ 0
0 ⎢
⎣
a
y = ± ax − x 2
dydx
dydx
⎡{ (a2 − x 2 ) − y 2 is even function of y ⎤
⎣
⎦
ax − x 2
dx
a
a
a
⎛ ax − x 2 ⎞
⎛ x( a − x ) ⎞
x
−1
= 4 a∫ sin −1 ⎜
=
=
dx
4
a
4
sin
dx
a
sin −1
dx
⎟
⎜
⎟
∫
∫
2
2
2
2
a+ x
⎝ a −x ⎠
⎝ a −x ⎠
0
0
0
M08_ENGINEERING_MATHEMATICS-I _CH08.indd 44
5/19/2016 8:27:40 PM
Multiple Integrals ■
t = sin −1
Put
x
a+x
⇒
∴
cos 2 t = 1 − sin 2 t = 1 −
⇒
a+x =
∴
⇒
sin 2 t =
x
a+x
x
a+ x− x
a
=
=
a+ x
a+ x
a+ x
a
= a sec 2 t
cos 2 t
⇒
x = a sec 2 t − a = a(sec 2 t − 1) = a tan 2 t
dx = 2a tan t sec 2 t dt
When x = 0, t = 0
and
a
∫ sin
∴
0
−1
when x = a, t = sin −1
a
∫ sin
0
−1
2
=
p
4
p
4
x
dx = ∫ t ⋅ 2a tan t sec 2 t dt = 2a ∫ t (tan t ⋅ sec 2 t ) dt
a+x
0
0
du = dt
∴
1
p
4
We integrate using integration by parts.
So, take
u=t
and
∴
x
a+x
sint =
8.45
and
dv = tan t sec 2 t dt
2
∫ dv = ∫ tan t sec t dt
⇒v=
tan 2 t
2
p
p
⎧
⎫
2
4
4
⎤
⎡
x
tan 2 t ⎪
⎪ t ⋅ tan t
dx = 2a ⎨ ⎢
dt ⎬
⎥ − ∫ 1⋅
a+x
2
⎪
⎪⎣ 2 ⎦0 0
⎩
⎭
p
⎧
⎫
4
⎪⎛ p
⎪
⎞
2 p
2
= a ⎨⎜ ⋅ tan
− 0⎟ − ∫ (sec t − 1) dt ⎬
⎠ 0
4
⎪⎝ 4
⎪
⎩
⎭
p
⎧p
⎫
= a ⎨ − [tan t − t ]04 ⎬
⎩4
⎭
⎧p ⎛
p p⎞ ⎫
p
p
⎡p ⎤ a
= a ⎨ − ⎜ tan − ⎟ ⎬ = a
−1+
= a ⎢ − 1⎥ = (p − 2)
⎝
⎠
⎣2 ⎦ 2
4
4
4
4
4
⎩
⎭
{
∴
}
a
S = 4a ⋅ (p − 2) = 2a2 (p − 2)
2
Note This problem can also be stated as below.
Find the surface area of the portion of the sphere x2 + y2 + z2 = a2 lying inside the cylinder x2 + y2 = ax.
EXAMPLE 4
Find the surface area of the part of the plane x 1 y 1 z 5 2a which lies in the first octant and is
bounded by the cylinder x2 1 y2 5 a2.
Solution.
The required surface area is the part of plane
x + y + z = 2a
M08_ENGINEERING_MATHEMATICS-I _CH08.indd 45
5/19/2016 8:27:45 PM
■
8.46
Engineering Mathematics
bounded by x = 0, y = 0, z = 0
z
and the cylinder x + y = a
2
2
2
The projection of the surface on the xy plane
is the quadrant of the circle
x2 + y2 = a2.
(0, 0, 2a)
⎛ ∂z ⎞ ⎛ ∂z ⎞
S = ∫∫ ⎜ ⎟ + ⎜ ⎟ + 1 dxdy
⎝ ∂x ⎠ ⎝ ∂y ⎠
D
2
∴
2
The surface is z = 2a − x − y
∂z
= −1,
∂x
∴
(0, 2a, 0)
∂z
= −1
∂y
(2a, 0, 0)
2
∂z
∂z
∴ ⎛⎜ ⎞⎟ + ⎛⎜ ⎞⎟ + 1 = 3
⎝ ∂x ⎠ ⎝ ∂y ⎠
2
y
x2 + y2 = a2
⎛ ∂z ⎞ ⎛ ∂z ⎞
⎜⎝ ⎟⎠ + ⎜ ⎟ + 1 = 1 + 1 + 1 = 3
∂x
⎝ ∂y ⎠
2
∴
O
2
x
Fig. 8.45
1
2
2
2
∴S =
∫∫ 3 dxdy = 3 ∫∫ dxdy = 3 × 4 area of the circle x + y = a
D
D
=
3 2
pa
4
[{ radius of the circle = a]
EXAMPLE 5
Find the surface area of the cylinder x 2 1 y 2 5 a 2 cut out by the cylinder x 2 1 z 2 5 a 2 .
Solution.
Given two right circular cylinders
with z-axis as the axis of the cylinder
and
x 2 + y 2 = a2
(1)
x 2 + z 2 = a2
(2)
with y-axis as the axis of the cylinder.
Both the cylinders are symmetric about the three axis.
∴ the surface area are the same in all octants.
Projecting the surface x 2 + y 2 = a2 on the xz-plane, we get the required surface area S.
⎛ ∂y ⎞ ⎛ ∂y ⎞
S = 8∫∫ ⎜ ⎟ + ⎜ ⎟ + 1 dzdx
⎝ ∂x ⎠ ⎝ ∂z ⎠
D1
2
∴
2
where D1 is the circle x 2 + z 2 = a2 in the xz-plane.
The surface is
x 2 + y 2 = a2
M08_ENGINEERING_MATHEMATICS-I _CH08.indd 46
5/19/2016 8:27:49 PM
Multiple Integrals ■
Differentiating, partially w. r. to x
and z, treating y as a function of x and z,
we get
2x + 2 y
and 2y
∂y
=0 ⇒
∂x
∂y
x
=−
∂x
y
∂y
=0 ⇒
∂z
∂y
=0
∂z
x2 + z2 = a2
2
=
∴
z
x2
⎛ ∂y ⎞ ⎛ ∂y ⎞
⎜⎝ ⎟⎠ + ⎜⎝ ⎟⎠ + 1 = 2 + 1
∂x
∂z
y
2
∴
∴ Surface area
O
x 2 + y 2 a2
= 2
y2
y
a
⎛ ∂y ⎞ ⎛ ∂y ⎞
⎜⎝ ⎟⎠ + ⎜⎝ ⎟⎠ + 1 = =
∂x
y
∂z
2
8.47
2
x2 + y2 = a2
a
x
a − x2
2
Fig. 8.46
⎛ ∂y ⎞ ⎛ ∂y ⎞
S = 8∫∫ ⎜ ⎟ + ⎜ ⎟ + 1 dxdz
⎝ ∂x ⎠ ⎝ ∂z ⎠
D1
2
a2 − x 2
a
= 8∫
a − x2
2
0
a ⎡
= 8a∫ ⎢
0 ⎢
⎣
a2 − x 2
∫
0
a
= 8a∫
0
a
= 8a∫
0
2
a
∫
0
y
⎤
dz ⎥ dx
⎥⎦
a2 − x 2
1
1
a −x
2
dzdx
2
1
a2 − x 2
[z ]0 a
2
−x 2
dx
a
⎡⎣ a2 − x 2 − 0 ⎤⎦ dx = 8a dx = 8a[x ]a = 8a2
0
∫
0
EXAMPLE 6
Find the surface area of the cylinder x 2 1 z 2 5 4 lying inside the cylinder x 2 1 y 2 5 4 .
Solution.
In the above example 5, putting a = 2, we get the surface area.
∴
surface area S = 8 ⋅ 22 = 32.
EXAMPLE 7
Find the surface area cut off from the cylinder x2 1 y2 5 ax by the sphere x2 1 y2 1 z2 5 a2.
Solution
The equation of the sphere is
M08_ENGINEERING_MATHEMATICS-I _CH08.indd 47
x 2 + y 2 + z 2 = a2
(1)
5/19/2016 8:27:51 PM
■
8.48
Engineering Mathematics
The equation of the cylinder is
x 2 + y 2 = ax
z
(2)
The surface area of the cylinder cut off
by the sphere is required.
Projecting the surface on the xz-plane,
we get the required surface area S.
x 2 + y 2 + z 2 = a2
⎛ ∂y ⎞ ⎛ ∂y ⎞
S = 2∫∫ ⎜ ⎟ + ⎜ ⎟ + 1 dx dz
⎝ ∂x ⎠ ⎝ ∂z ⎠
D1
2
∴
2
O
where D1 is the region obtained by
eliminating y2 from (1) and (2)
∴
z 2 + ax = a2
x 2 + y 2 = ax
(3)
y
The surface is x 2 + y 2 = ax
Differentiating partially w.r.to x and z,
treating y as function of x and z, we get
and
Fig. 8.47
2x + 2 y
∂y
=a ⇒
∂x
∂y a − 2 x
=
∂x
2y
2y
∂y
=0 ⇒
∂z
∂y
=0
∂z
(a − 2 x ) 2
⎛ ∂y ⎞ ⎛ ∂y ⎞
+1
⎜⎝ ⎟⎠ + ⎜⎝ ⎟⎠ + 1 =
∂x
∂z
4y 2
2
∴
2
=
(a − 2 x ) 2 + 4 y 2
4y 2
a2 − 4ax + 4 x 2 + 4(ax − x 2 )
a2
=
4y 2
4y 2
=
a
∴ ⎛⎜ ∂y ⎞⎟ + ⎛⎜ ∂y ⎞⎟ + 1 = a =
⎝ ∂x ⎠ ⎝ ∂z ⎠
2y
2 ax − x 2
2
[ using ( 2)]
2
[ using ( 2)]
[x and y axes are interchanged for convenience of the figure]
We have
z 2 + ax = a 2
∴
x
S = 2∫∫
D1
⇒
z 2 = a 2 − ax ⇒
a
a
2 ax − x
2
dx dz = a∫
z = ± a 2 − ax
a 2 − ax
∫
0 − a 2 − ax
1
ax − x 2
dz dx
2
1
⎤
⎡
= a∫ ⎢
[ z ] a −2 ax ⎥ dx
− a − ax
2
0 ⎣ ax − x
⎦
a
1
⎧
⎡ a 2 − ax + a 2 − ax ⎤ ⎫ dx
= a∫ ⎨
2 ⎣
⎦⎬
0 ⎩ ax − x
⎭
a
M08_ENGINEERING_MATHEMATICS-I _CH08.indd 48
5/19/2016 8:27:56 PM
Multiple Integrals ■
a
= 2a ∫
0
a
= 2a ∫
0
a 2 − ax
ax − x 2
8.49
dx
a( a − x )
dx
x( a − x )
a
a
= 2a∫
0
a
a
dx = 2a a ∫ x
x
0
1
−
2
⎡ 12 ⎤
1
x
dx = 2a a ⎢ ⎥ = 4a a (a 2 − 0) = 4a2
⎢ 1 ⎥
⎢ ⎥
⎣ 2 ⎦0
EXERCISE 8.6
1. Find the surface area of that part of the plane x + y + z = a intercepted by the coordinate planes.
2. Find the surface area of that part of the plane
x y z
+ + = 1 intercepted by coordinate planes.
a b c
3. Find the surface area of the portion of the cylinder x 2 + y 2 = 4 y lying inside the sphere
x 2 + y 2 + z 2 = 16.
4. Find the surface area of the portion of the cone x 2 + y 2 = 3z 2 lying above the xy-plane inside the
cylinder x 2 + y 2 = 4 y .
5. Find the area of the surface of the sphere x 2 + y 2 + z 2 = 9a2 cut off by the cylinder x 2 + y 2 = 3ax .
ANSWERS TO EXERCISE 8.6
1.
8.3
3 2
a
2
2.
1 2 2
b c + c 2 a 2 + a 2b 2
2
3. 64
4.
8p
3
5. 9a2 (p − 2)
TRIPLE INTEGRAL IN CARTESIAN COORDINATES
Let f(x, y, z) be a continuous function at every point in a closed and bounded region D in space.
Subdivide the region into a number of element volumes by drawing planes parallel to the
coordinate planes. Let ΔV1, ΔV2, …, ΔVn be the number of element volumes formed. Let (xi, yi, zi) be
n
any point in ΔVi, where ΔVi = Δxi Δyi Δzi. Form the sum ∑ f ( x i , y i , z i ) ΔVi . The limit of the sum as
i =1
n → ∞ and ΔVi. → 0, if it exists, is called the triple integral of f(x, y, z) over D and is denoted by
∫ ∫ ∫ f ( x , y , z ) dV
D
or
∫ ∫ ∫ f ( x , y , z ) dxdydz
(1)
D
As in the case of double integrals, the triple integral is evaluated by three successive integration of
single variable.
M08_ENGINEERING_MATHEMATICS-I _CH08.indd 49
5/19/2016 8:27:59 PM
■
8.50
Engineering Mathematics
Consider the triple integral
z 1 y1 x1
∫ ∫ ∫ f ( x , y , z ) dxdydz
z0 y0 x0
(1) If all the limits are constants, then the integration can be performed in any order with proper
limits,
z1 y1 x1
x1 y1 z1
x1 z1 y1
z0 y0 x0
x0 y0 z0
x0 z0 y0
∫ ∫ ∫ f (x , y , z ) dxdydz = ∫ ∫ ∫ f (x , y , z ) dzdydx = ∫ ∫ ∫ f (x , y , z ) dydzdx
i.e.,
(2) If x0 = f0(y, z),
x1 = f1(y, z),
y1 = g1(z),
z1 y1 x1
b y1 = g1 ( z ) x1 = f1 ( y, z )
z0 y0 x0
a y 0 = g0 ( z ) x 0 = f 0 ( y, z )
∫∫
then
y0 = g0(z),
∫ f (x , y , z ) dxdydz = ∫
∫
∫
z0 = a,
z1 = b,
f ( x , y , z ) dx dy dz
First we integrate w.r.to x, treating y and z as constants and substitute limits of x. Next integrate
the resulting function of y and z w.r.to y, treating z as constant and substitute the limits of y.
Finally we integrate the resulting function of z w.r.to z and substitute the limits of z.
WORKED EXAMPLES
EXAMPLE 1
1 2 2
∫∫∫x
Evaluate
2
yz dxdydz .
0 0 1
Solution.
1 2 2
Let I = ∫ ∫ ∫ x 2 yz dxdydz
0 0 1
1
2
2
1
2
2
2
⎛ 2⎞ ⎛ 2⎞ ⎛ 3⎞
⇒ I = ∫ zdz ∫ ydy ∫ x dx = ⎜ z ⎟ ⎜ y ⎟ ⎜ x ⎟ = 1 ⋅ 4 ⎛⎜ 8 − 1 ⎞⎟ = 7
⎝ 2 ⎠ 0 ⎝ 2 ⎠ 0 ⎝ 3 ⎠ 1 2 2 ⎝ 3 3⎠ 3
0
0
1
[{ limits are constants]
EXAMPLE 2
a b c
Evaluate ∫ ∫ ∫ ( x 2 1 y 2 1 z 2 ) dxdydz .
0 0 0
Solution.
c
a b
⎡x3
⎤
I = ∫ ∫ ∫ ( x 2 + y 2 + z 2 ) dxdydz = ∫ ∫ ⎢ + ( y 2 + z 2 )x ⎥ dydz
3
⎣
⎦0
0 0 0
0 0
a b c
Let
a b
⎡ c3
⎤
= ∫ ∫ ⎢ + ( y 2 + z 2 ) c ⎥ dydz
3
⎣
⎦
0 0
a b
⎛ c2
⎞
= c ∫ ∫ ⎜ + y 2 + z 2 ⎟ dydz
⎝ 3
⎠
0 0
M08_ENGINEERING_MATHEMATICS-I _CH08.indd 50
5/19/2016 8:28:03 PM
Multiple Integrals ■
8.51
b
⎡c2
⎤
y3
= c∫ ⎢ y +
+ z 2 y ⎥ dz
3
3
⎦0
0 ⎣
a
a
⎡ c 2b b 3
⎤
= c∫ ⎢
+ + z 2b ⎥ dz
3
3
⎦
0 ⎣
a
⎡ c2 b2
⎤
= bc ∫ ⎢ + + z 2 ⎥ dz
3 3
⎦
0 ⎣
a
⎡⎛ b 2 + c 2 ⎞
⎡⎛ c 2 b 2 ⎞
⎡ a2 + b2 + c2 ⎤
z3 ⎤
a3 ⎤
= bc ⎢⎜ + ⎟ z + ⎥ = bc ⎢⎜
a + ⎥ = abc ⎢
⎟
⎥
3 ⎦0
3⎦
3
⎣
⎦
⎣⎝ 3 ⎠
⎣⎝ 3 3 ⎠
EXAMPLE 3
log 2 x x1y
∫∫∫
Evaluate
0
0
e ( x1y1z ) dxdydz.
0
Solution.
log 2 x x + y
Let I =
∫∫∫
0
0
log 2
e ( x + y + z ) dzdydx =
0
0
log 2 x
=
∫
(2x+2 y)
− e x + y )dydz =
∫ (e
0
0
1
2
log 2
1
=
2
log 2
=
∫
∫
0
∫
log 2
x+ y
e x + y .[e z ] dydz =
∫
e x + y .(e x + y − 1)dydz
0
log 2
∫
0
x
⎫
⎪⎧ 2 x ⎡ e 2 y ⎤
x
y x⎪
⎨e . ⎢ ⎥ − e .[e ]0 ⎬ dz
⎪⎩
⎪⎭
⎣ 2 ⎦0
1
⎡⎣e 2 x (e 2 x − 1) − 2e x (e x − 1) ⎤⎦ dz =
2
log 2
∫
⎡⎣e 4 x − e 2 x − 2e 2 x + 2e x ⎤⎦ dx
0
(e 4 x − 3e 2 x + 2e x )dx
0
loge 2
=
⎤
1 ⎡ e4x
e2x
3
−
+ 2e x ⎥
⎢
⎦0
2⎣ 4
2
=
⎞ ⎛ 1 3 ⎞⎤
1 ⎡⎛ e 4 loge 2 3 2 loge 2
− e
+ 2e loge 2 ⎟ − ⎜ − + 2⎟ ⎥
⎢⎜⎝
⎠ ⎝ 4 2 ⎠⎦
2⎣
4
2
=
1 ⎡ e loge 16 3 loge 4
3 ⎤ 1 ⎡16 3 ⋅ 4
3⎤
3
3 5
− e
+ 2e loge 2 − ⎥ = ⎢ −
+ 2⋅ 2 − ⎥ = 2 − 3 + 2 − = 1− =
⎢
2⎣ 4
2
4⎦ 2 ⎣ 4
2
4⎦
8
8 8
[{ e loge x = x ]
EXAMPLE 4
42 z
Evaluate
4 z2 x 2
∫∫ ∫
0
0
dydxdz.
0
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Engineering Mathematics
Solution.
42 z
I=∫
Let
0
4z −x 2
∫ ∫
0
0
42 z
dydxdz = ∫
∫ [y ]
4z −x 2
0
0
dxdz
[Treating x , z constants]
0
42 z
=∫
∫
0
4 z − x 2 dxdz
0
2 z
x ⎤
4z
⎡x
= ∫⎢
sin −1
4z − x 2 +
⎥ dz
2
2
2 z ⎦0
0 ⎣
4
[Treating z constant ]
4
= ∫ [ z 4 z − 4 z + 2 z sin −1 1 − 0] dz
0
4
4
= ∫ 2z
0
EXAMPLE 5
4
⎡ z2 ⎤
p
p
dz = p∫ z dz = p ⎢ ⎥ = (16) = 8p
⎦
2
⎣
2
2
0
0
log 2 x x1 log 2
Evaluate
∫∫ ∫
0
0
e x1y1z dzdydx .
0
Solution.
log 2 x x + loge y
Let I =
∫∫ ∫
0
0
log 2 x
e x + y + z dzdydx =
0
∫ ∫e
0
x
⋅ e y [e z ]0x + log y dydx
0
log 2 x
=
∫ ∫e
0
⋅ e y [e x + log y − e 0 ] dydx
x
0
log 2 x
=
∫ ∫e e
x y
0
(e x ⋅ e log y − 1)dydx
0
log 2 x
=
∫ ∫e e
x y
0
[{ e loge y = y ]
(e x ⋅ y − 1) dydx
0
log 2 x
=
∫ ∫ (e
0
log 2
=
∫
0
2x
⋅ ye y − e x ⋅ e y ) dydx
0
x
⎤
⎡ 2x x y
x
y
⎢e ∫ ye dy − e ∫ e dy ⎥ dx
0
⎦
⎣ 0
log 2
=
∫ {e
2x
[ y ⋅ e y − 1⋅ e y ]0x − e x [e y ]0x }dx
2x
[ xe x − e x − (0 − 1)] − e x (e x − 1)}dx
[Using Bernoulli’s formula ]
0
log 2
=
∫ {e
0
log 2
=
∫ {( x − 1)e
3x
+ e 2 x − e 2 x + e x }dx
0
M08_ENGINEERING_MATHEMATICS-I _CH08.indd 52
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Multiple Integrals ■
8.53
log 2
∫ {( x − 1)e
=
3x
+ e x }dx
0
loge 2
⎡
⎤
e3x
e3x
= ⎢( x − 1)
− 1⋅
+ ex ⎥
3
9
⎣
⎦0
⎧1
1
⎡ 1 1 ⎤⎫
= ⎨ (log e 2 − 1)e 3 loge 2 − e 3 loge 2 + e loge 2 − ⎢ − − + 1⎥ ⎬
9
3
⎣ 3 9 ⎦⎭
⎩
3
1
5
8
= (log e 2 − 1) ⋅ 8 − + 2 −
[{ e 3 loge 2 = e loge 2 = 23 = 8 and e loge 2 = 2]
3
9
9
8 8
5 8
19 1
8
= log e 2 − − + 2 − = log e 2 − = ( 24 log 2 − 19)
3
3 9
9 3
9 9
EXAMPLE 6
a 2 2 x 2 2 y2
a 22x2
a
Evaluate
∫ ∫
0
dzdydx
∫
0
a 2 x2 2 y2 2 z2
2
0
.
Solution.
a
Let
I=∫
∫
0
a
=∫
=∫
∫
0
a − x2
=∫
dzdydx
(a − x 2 − y 2 ) − z 2
2
0
a2 − x 2
z
⎡ −1
⎤
⎢sin
2
2
2 ⎥
a − x − y ⎦0
⎣
a2 − x 2 − y 2
dydx
⎡
⎢{
⎣
∫
x⎤
= sin −1 ⎥
a⎦
a −x
dx
2
2
2
∫
0
a
∫
0
0
a
a2 − x 2 − y 2
a 2 − x2
[sin −1 1 − sin −1 0] dydx
0
a2 − x 2
∫
0
0
p
dydx
2
=
2
2
p
[ y ]0 a − x dx
∫
20
=
p
a 2 − x 2 dx
2 ∫0
a
a
a
⎤ p a2 p p2 a2
p ⎡x 2
p⎡
a2
x⎤
a2
= ⎢
a − x 2 + sin −1 ⎥ = ⎢0 + sin −1 1 − 0 ⎥ = ⋅ ⋅ =
⎦ 2 2 2
8
2 ⎣2
2
2
a ⎦0 2 ⎣
EXAMPLE 6(A)
Evaluate
1
12 x 2
12 x 2 2 y2
0
0
0
∫ ∫
∫
dxdydz
12 x 2 2 y 2 2 z 2
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Engineering Mathematics
Solution.
In example 6, putting a = 1, we get
1 1− x 2
∫ ∫
0
1− x 2 − y 2
0
∫
0
dxdydz
p 2 ⋅1 p 2
=
8
8
=
1− x 2 − y 2 − z 2
EXAMPLE 7
Evaluate
∫ ∫ ∫ xyz dxdydz
over the volume V enclosed by the three coordinate planes and the
V
plane
x y z
1 1 51.
a b c
Solution.
Let V be the volume enclosed by the plane
x y z
+ + = 1 and it meets the coordinate axes in
a b c
A(a, 0, 0), B(0, b, 0), C(0, 0, c) respectively.
The projection of V on the xy-plane is the ΔOAB
x y
bounded by
x = 0, y = 0, + = 1
a b
z
(0, 0, c)
C
⎛ x y⎞
z varies from 0 to z = c ⎜1 − − ⎟
⎝ a b⎠
A x
x
a
+
∫
0
∫
0
∫
0
0
=
1
2 ∫0
⎛ x y⎞
c ⎜ 1− − ⎟
a b⎠
∫
0
2 a
c
2
∫x
0
xyz dzdydx
⎡ z2 ⎤ ⎝
xy ⎢ ⎥
⎣ 2 ⎦0
⎛ x⎞
b ⎜ 1− ⎟
a ⎝ a⎠
=
= 1
0
⎛ x⎞
b ⎜ 1− ⎟
a ⎝ a⎠
=∫
y
b
Fig. 8.48
⎛ x⎞ ⎛ x y⎞
b ⎜ 1− ⎟ c ⎜ 1− − ⎟
a ⎝ a⎠ ⎝ a b⎠
I=∫
(a, b, 0)
y
B
O
x
y varies from 0 to b ⎛⎜1− ⎞⎟
⎝ a⎠
and x varies from 0 to a.
⎛ x
y⎞
z = C ⎜1−
−
⎜
⎝ a
b⎠
0
2
⎛ x y⎞
xyc 2 ⎜1 − − ⎟ dydx
⎝ a b⎠
⎛ x⎞
b ⎜ 1− ⎟
⎝ a⎠
∫
dydx
2
⎡⎛ x ⎞ y ⎤
y ⎢⎜1 − ⎟ − ⎥ dydx
⎣⎝ a ⎠ b ⎦
⎛ x⎞
b ⎜ 1− ⎟
a⎠
4
⎝
⎡ ⎧⎛ x ⎞ y ⎫ 3
⎧⎛ x ⎞ y ⎫ ⎤
⎢ y ⎨⎜ 1 − ⎟ − ⎬
⎨⎜⎝1 − ⎟⎠ − ⎬ ⎥
a
⎝ a⎠ b ⎭
c2
a
b⎭ ⎥
= ∫x⎢ ⎩
− 1⋅ ⎩
1
−3 ⎛ −1 ⎞ ⎥
2 0 ⎢
− ⋅3
⋅ ⎜ ⋅ 4⎟ ⎥
⎢
b
b ⎝ b ⎠ ⎦0
⎣
M08_ENGINEERING_MATHEMATICS-I _CH08.indd 54
dx
[Using Bernouli’s formula ]
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Multiple Integrals ■
=
4
a
⎡b ⎛ x ⎞
c2
b2 ⎛
⎛ x⎞ ⎞⎤
−
−
−
−
1
(
0
)
0
0
1
x
−
⎥ dx
⎢
⎜
⎟
⎜
⎟
⎜
⎝ a ⎠ ⎟⎠ ⎦
2 ∫0 ⎣ 3 ⎝ a ⎠
12 ⎝
=
c2 ⋅ b2
⎛ x⎞
x ⎜1 − ⎟ dx
24 ∫0 ⎝ a ⎠
8.55
4
a
a
6
⎡ ⎛ x⎞5
⎛ x⎞ ⎤
1− ⎟
1− ⎟ ⎥
⎜
⎜
2 2
2 2 ⎢
⎝ a⎠
b c ⎢ ⎝ a⎠
⎥ =bc
=
− 1⋅
x⋅
1
5 ⎛ −6 ⎞ ⎥
24 ⎢
24
− ⋅5
− ⋅⎜ ⎟ ⎥
⎢
⎝
⎠
a
a
a
⎦0
⎣
⎡
⎤ b2c2 a2 a2b2c2
a2
−
−
0
0
1
(
)
⋅
=
⎥=
⎢
30
24 30
720
⎣
⎦
EXERCISE 8.7
Evaluate the following integrals
1.
dxdydz
∫ ∫ ∫ ( x + y + z + 1)
3
, where D is the region bounded by x + y + z = 1 and the coordinate planes.
D
1 1− x x + y
2.
∫∫ ∫
0 0
1 1− x 2
xdzdydx.
3.
0
0
6.
∫
0
dxdydz
.
1 − x2 − y2 − z2
0
a2 − r 2
p
2 a cos u a
1 z x+z
4.
∫ ∫
1− x 2 − y 2
∫ ∫ ∫ ( x + y + z ) dydxdz.
5.
−1 0 x − z
∫ ∫ ∫
0
0
rdzdrd u.
0
∫ ∫ ∫ xyz dxdydz , where D is the region interior to the sphere x
2
+ y2 + z2 = a2 in the I octant.
D
7.
∫ ∫ ∫ xyz dxdydz
taken over the volume for which x, y, z ≥ 0 and x2 + y2 + z2 = 9.
1 1− x ( x + y )
2
z = 5 x = 6 y = 36 − x
∫ ∫
8.
∫
dydxdz .
z = 0 x = −6 y = − 36 − x
9.
∫ ∫ ∫ (x
2
+ y + z ) dxdydz.
2
2
12.
−c −b −a
1 1− x x + y
14.
∫∫ ∫
e dzdydx.
1 1− x 2
1− x 2 − y 2
0 0
17.
0
z
0
∫ ∫
0
xdzdydx.
10.
0
∫
15.
∫∫ ∫
0 0
0
3 1
xy
1 1
x
0
∫∫ ∫
e x + y + z dzdydx.
∫ ∫
0
a x x+y
c b a
11.
∫∫ ∫
0 0
2
⎛ x⎞ ⎛ x y⎞
b ⎜ 1− ⎟ c ⎜ 1− − ⎟
a ⎝ a⎠ ⎝ a b⎠
2
∫
0
1 1− x 1− x − y
13.
∫∫ ∫
0 0
16.
∫∫ ∫
0 0
xyz dxdydz.
0
1 1− x ( x + y )
xy dzdydx.
x 2 zdzdydx.
0
2
x dzdydx.
0
xyz dxdydz.
0
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Engineering Mathematics
ANSWERS TO EXERCISE 8.7
1.
1
5
log e 2 −
2
16
8. 180p
12.
8.3.1
5pa3
64
1
4
3.
9.
1
10
10.
a3bc 2
360
11.
8abc 2
(a + b2 + c2 )
3
13.
1
720
14.
1
2
1 4a
[e − 6e 2a + 8e a − 3]
8
16. 1
10
p2
8
2.
4. 0
5.
6.
a6
48
7.
243
19
2 ⎡2
⎤
(9 3 − 1) − log e 3⎥
⎢
5 ⎣5
⎦
15.
17. 1
48
Volume as Triple Integral
Triple integrals can be used to evaluate volume V of a finite bounded region D in space.
The volume V =
dxdydz .
∫∫∫
D
[Taking f(x, y, z) = 1 in (1) of 8.3, page 8.49, we get the volume]
WORKED EXAMPLES
EXAMPLE 1
Find the volume of the tetrahedron bounded by the plane
planes.
x
y
z
1 1 51 and the coordinate
a b
c
Solution.
The region of integration is the region bounded by
x y z
+ + = 1, x = 0, y = 0, z = 0.
a b c
z
C
Its projection in the xy-plane is the ΔOAB
x y
bounded by x = 0, y = 0 and + = 1
a b
∴
volume V =
O
∫ ∫ ∫ dxdydz
B
D
0
∫
0
⎛ x⎞
b ⎜ 1− ⎟
a ⎝ a⎠
=∫
0
y
A
x
⎛ x⎞ ⎛ x y⎞
b ⎜ 1− ⎟ c ⎜ 1− − ⎟
a ⎝ a⎠ ⎝ a b⎠
=∫
⎛ x
y⎞
z = C ⎜1− −
⎟
⎝ a
b⎠
∫
∫
x
dzdydx
⎛ x y⎞
c ⎜ 1− − ⎟
a b⎠
+
y
b
= 1
Fig. 8.49
0
[z ]0⎝
a
dydx
0
M08_ENGINEERING_MATHEMATICS-I _CH08.indd 56
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Multiple Integrals ■
⎛ x⎞
b ⎜ 1− ⎟
a ⎝ a⎠
=∫
∫
0
0
8.57
⎛ x y⎞
c ⎜1 − − ⎟ dydx
⎝ a b⎠
⎛ x⎞
b ⎜ 1− ⎟
a⎠
⎡⎛ x ⎞
y2⎤ ⎝
= c ∫ ⎢⎜1 − ⎟ y − ⎥
⎝ a⎠
2b ⎦ 0
0 ⎣
a
dx
2
a
⎡⎛ x ⎞ ⎛ x ⎞ 1
⎛ x⎞ ⎤
= c ∫ ⎢⎜1 − ⎟ b ⎜1 − ⎟ − b 2 ⎜1 − ⎟ ⎥ dx
⎝ a ⎠ ⎝ a ⎠ 2b ⎝ a ⎠ ⎦
0 ⎣
a
⎡⎛ x⎞3 ⎤
1−
2
a
− abc
bc ⎛ x ⎞
bc ⎢⎢ ⎜⎝ a ⎟⎠ ⎥⎥
abc
=
−
=
=
1
dx
[0 − 1] =
⎜
⎟
∫
⎝
⎠
⎥
2 0
a
2 ⎢ 1
6
6
⎢⎣ − a ⋅ 3 ⎥⎦ 0
EXAMPLE 2
Find the volume bounded by the cylinder x2 1 y2 5 4 and the planes y 1 z 5 4, z 5 0.
Solution.
Required volume of the cylinder x2 + y2 = 4, cut off between the planes z = 0 and y + z = 4 is
V = ∫ ∫ ∫ dxdydz
D
∴ z varies from z = 0 to z = 4 − y
The projection of the region in the xy plane is
2
x + y2 = 4 ⇒ y = ± 4 − x 2
∴ y varies from − 4 − x 2
x-varies from – 2 to 2
y + z= 4
∫ ∫ ∫ dzdydx
−2 − 4 − x 2
=
−2
4 − x2
∫ ∫
4−x
2
∫ ∫
x
O
0
[z ]04 − y dydx
−2 − 4 − x 2
=
and
4 − x2 4 − y
2
∴ Volume V =
to + 4 − x 2
z
x2 + y2 = 4
x
Fig. 8.50
2
( 4 − y ) dydx
−2 − 4 − x 2
⎡
y2⎤
= ∫ ⎢4y − ⎥
⎣
2 ⎦−
−2
2
2
=
∫
−2
4 − x2
{
dx
4 − x2
}
1
4 ⎡ 4 − x 2 − ( − 4 − x 2 ) ⎤ − [4 − x 2 − ( 4 − x 2 )] dx
⎣
⎦ 2
2
= 8 ∫ 4 − x 2 dx
−2
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Engineering Mathematics
2
= 8 ⋅ 2∫ 4 − x 2 dx
[{ 4 − x 2 is even ]
0
2
⎡ x 4 − x 2 4 −1 x ⎤
p
= 16 ⎢
+ sin
⎥ = 16[0 + 2 sin −1 1 − (0 + 0)] = 16 ⋅ 2 = 16p
⎣
2
2
2 ⎦0
2
EXAMPLE 3
Change to spherical polar coordinates and hence evaluate
∫∫∫ x
V
volume of the sphere x2 1 y2 1 z2 5 a2.
2
dxdydz
where V is the
1 y2 1 z2
Solution.
I = ∫∫∫
V
1
dxdydz
x +y2 +z2
2
Using spherical polar coordinates (r, u, f), x = rsinucosf, y = rsinusinf, z = rcosu
Then the Jacobian of transformation is
J=
∴
∂( x , y , z )
= r 2 sin u
∂( r , u, f)
[Ref.Chapter 5, worked example 5, Page 5.31]
dxdydz = J drd u df = r 2 sin u drd u df
x 2 + y 2 + z 2 = r 2 sin 2 cos 2 f + r sin 2 u sin 2 f + r 2 cos 2 u
= r 2 sin 2 u [cos 2 f + sin 2 f] + r 2 cos 2 u = r 2 [sin 2 u + cos 2 u] = r 2
p
2p 2 a
∴
I = 2∫
∫∫
0 0 0
r 2 sin u drd u d f
r2
p
2p 2 a
= 2∫
∫ ∫ sin u drd udf
0 0 0
2p
p
2
a
0
0
0
p
= 2 ∫ df∫ sin u∫ dr = 2[f]02p [ − cos u]02 [r ]a0 = 2 ⋅ 2p[ −0 + 1][a − 0] = 4pa
EXAMPLE 4
Find the volume of the region of the sphere x2 1 y2 1 z2 5 a2 lying inside the cylinder x2 1 y2 5 ay.
Solution.
x2 + y2 = ay
2
⇒
x 2 + y 2 − ay = 0
M08_ENGINEERING_MATHEMATICS-I _CH08.indd 58
⇒
a⎞
a2
⎛
x2 + ⎜ y − ⎟ =
⎝
2⎠
4
5/23/2016 5:55:47 PM
Multiple Integrals ■
z
a
which is a circle with centre (0, a/2) radius r = in
2
the xy-plane, z = 0
So, the cylinder has this circle as guiding curve
and generators parallel to the z-axis.
x2 + y2 + z2 = a2 is a sphere with centre (0, 0, 0)
and radius = a.
The volume inside the cylinder bounded by the
sphere is symmetric about the xy-plane. So, the
required volume = 2 (volume inside the cylinder)
above the xy-plane.
Its projection in the xy-plane is the circle
x2 + y2 = ay