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Table of contents :
Cover......Page 1
Contents......Page 8
Preface to the Revised Edition......Page 12
Symbols and Basic Formulae......Page 14
1.2 Convergence of Sequences......Page 18
1.3 The Upper and Lower Limits of a Sequence......Page 20
1.4 Cauchy’s Principle of Convergence......Page 21
1.5 Monotonic Sequence......Page 23
1.6 Theorems on Limits......Page 25
1.7 Subsequences......Page 28
1.8 Series......Page 29
1.9 Comparison Tests......Page 32
1.10 D’alemberi’s Ratio Test......Page 37
1.11 Cauchy’s Root Test......Page 42
1.12 Raabe’s Test......Page 44
1.13 Logarithmic Test......Page 48
1.14 De Morgan–Berirand Test......Page 50
1.15 Gauss’s Test......Page 51
1.16 Cauchy’s Integral Test......Page 53
1.17 Cauchy’s Condensation Test......Page 55
1.18 Kummer’s Test......Page 57
1.19 Alternating Series......Page 58
1.20 Absolute Convergence of a Series......Page 60
1.21 Convergence of the Series of the Type......Page 65
1.22 Derangement of Series......Page 67
1.23 Nature of Non-Absolutely Convergent Series......Page 68
1.24 Effect of Derangement of Non-Absolutely Convergent Series......Page 69
1.25 Uniform Convergence......Page 71
1.26 Uniform Convergence of a Series of Functions......Page 73
1.27 Properties of Uniformly Convergent Series......Page 74
1.28 Power Series......Page 75
Exercises......Page 76
2.1 Successive Differentiation......Page 80
2.2 Leibnitz’s Theorem and its Applications......Page 84
2.3 General Theorems......Page 88
2.4 Taylor’s Infinite Series and Power Series Expansion......Page 95
2.6 Expansion of Functions......Page 96
2.7 Indeterminate Forms......Page 106
Exercises......Page 115
3.1 Radius of Curvature of Intrinsic Curves......Page 120
3.2 Radius of Curvature for Cartesian Curves......Page 121
3.3 Radius of Curvature for Parametric Curves......Page 125
3.4 Radius of Curvature for Pedal Curves......Page 127
3.5 Radius of Curvature for Polar Curves......Page 128
3.6 Radius of Curvature at the Origin......Page 131
3.7 Center of Curvature......Page 133
3.9 Equation of the Circle of Curvature......Page 134
3.10 Chords of Curvature Parallel to the Coordinate Axes......Page 137
3.11 Chord of Curvature in Polar Coordinates......Page 138
3.12 Miscellaneous Examples......Page 140
Exercises......Page 145
4.1 Determination of Asymptotes When the Equation of the Curve in Cartesian form is Givens......Page 148
4.2 The Asymptotes of the General Rational Algebraic Curve......Page 149
4.3 Asymptotes Parallel to Coordinate Axes......Page 150
4.4 Working Rule for Finding Asymptotes of Rational Algebraic Curve......Page 151
4.5 Intersection of a Curve and its Asymptotes......Page 154
4.6 Asymptotes by Expansion......Page 156
4.7 Asymptotes of the Polar Curves......Page 157
4.9 Concavity, Convexity and Singular Points......Page 159
4.10 Curve Tracing (Cartesian Equations)......Page 163
4.11 Curve Tracing (Polar Equations)......Page 168
4.12 Curve Tracing (Parametric Equations)......Page 170
Exercises......Page 171
Chapter 5: Functions of Several Variables......Page 176
5.3 The Differential Coefficients......Page 177
5.5 Higher-Order Partial Derivatives......Page 178
5.6 Envelopes and Evolutes......Page 183
5.7 Homogeneous Functions and Euler’s Theorem......Page 185
5.8 Differentiation of Composite Functions......Page 190
5.9 Transformation from Cartesian to Polar Coordinates and Vice Versa......Page 194
5.10 Taylor’s Theorem for Functions of Several Variables......Page 196
5.11 Approximation of Errors......Page 199
5.12 General Formula for Errors......Page 200
5.13 Tangent Plane and Normal to a Surface......Page 203
5.15 Properties of Jacobian......Page 205
5.16 Necessary and Sufficient Conditions for Jacobian to Vanish......Page 208
5.17 Differentiation Under the Integral Sign......Page 209
5.18 Miscellaneous Examples......Page 212
5.19 Extreme Values......Page 216
5.20 Lagrange’s Method of Undetermined Multipliers......Page 223
Exercises......Page 228
6.3 Equation of the Normal at a Point of a Curve......Page 232
6.4 Lengths of Tangent, Normal, Sub-Tangent and Subnormal at any Point of a Curve......Page 236
Exercises......Page 237
7.2 Properties of Beta Function......Page 238
7.4 Properties of Gamma Function......Page 241
7.5 Relation Between Beta and Gamma Functions......Page 242
7.6 Dirichlet’s and Liouville’s Theorems......Page 247
7.7 Miscellaneous Examples......Page 249
Exercises......Page 250
8.1 Reduction Formulas for sinn x dx and cosn x dx......Page 252
8.2 Reduction Formula for sinm x cosn x dx 8.3......Page 254
8.3 Reduction Formulas for tann x dx and secn x dx 8.5......Page 256
8.4 Reduction Formulas for xn sinmx dx and xn cosmx dx......Page 257
8.5 Reduction Formulas for x n eaxdx and xm (log x)n dx......Page 258
8.7 Reduction Formula For......Page 259
Exercises......Page 260
9.1.1 Area of a Curve Given by the Cartesian Equation......Page 262
9.1.2 Area of a Curve Given by Polar Equation......Page 267
9.2.1 Length of a Curve......Page 270
Exercises......Page 275
10.1 Centre of Gravity......Page 278
10.2 Moment of Inertia......Page 281
10.3 Mean Values of a Function......Page 282
Exercises......Page 283
11.1 Volume of the Solid of Revolution (Cartesian Equations)......Page 284
11.2 Volume of the Solid of Revolution (Parametric Equations)......Page 289
11.3 Volume of the Solid of Revolution (Polar Curves)......Page 291
11.4 Surface of the Solid of Revolution (Cartesian Equations)......Page 292
11.5 Surface of the Solid of Revolution (Parametric Equations)......Page 294
11.6 Surface of the Solid of Revolution (Polar Curves)......Page 296
Exercises......Page 297
12.1 Double Integrals......Page 300
12.3 Evaluation of Double Integrals (Cartesian Coordinates)......Page 301
12.4 Evaluation of Double Integrals (Polar Coordinates)......Page 305
12.5 Change of Variables in a Double Integral......Page 307
12.6 Change of Order of Integration......Page 311
12.7 Area Enclosed by Plane Curves (Cartesian and Polar Coordinates)......Page 315
12.8 Volume and Surface Area as Double Integrals......Page 318
12.9 Triple Integrals and their Evaluation......Page 325
12.10 Change to Spherical Polar Coordinates from Cartesian Coordinates in a Triple Integral......Page 329
12.11 Volume as a Triple Integral......Page 332
12.12 Miscellaneous Examples......Page 336
Exercises......Page 338
Chapter 13: Vector Calculus......Page 344
13.1 Differentiation of a Vector......Page 354
13.2 Partial Derivatives of a Vector Function......Page 361
13.3 Gradient of a Scalar Field......Page 362
13.5 Properties of a Gradient......Page 363
13.6.1 Directional Derivatives along Coordinate Axes......Page 364
13.8 Physical Interpretation of Divergence......Page 369
13.11 The Laplacian Operator......Page 371
13.12 Properties of Divergence and Curl......Page 375
13.14 Line Integral......Page 380
13.15 Work Done by a Force......Page 384
13.16 Surface Integral......Page 386
13.17 Volume Integral......Page 390
13.18 Gauss’s Divergence Theorem......Page 392
13.19 Green’s Theorem in a Plane......Page 398
13.20 Stoke’s Theorem......Page 402
13.21 Miscellaneous Examples......Page 407
Exercises......Page 415
14.3 Direction Ratios and Direction Cosines of a Line......Page 424
14.4 Section Formulae—Internal Division of a Line by a Point on the Line......Page 425
14.4.1 External Division of a Line by a Point on the Extended Line......Page 426
14.5 Straight Line in Three Dimensions......Page 429
14.6 Angle Between Two Lines......Page 432
14.7 Shortest Distance Between Two Skew Lines......Page 434
14.8 Equation of a Plane......Page 441
14.10 Equation of a Plane Passing through Three Points......Page 442
14.12 Equation of a Plane Passing through Two Point and Parallel to a Line......Page 443
14.13 Angle Between Two Planes......Page 447
14.15 Perpendicular Distance of a Point From a Plane......Page 449
14.16 Planes Bisecting the Angles Between Two Planes......Page 450
14.18 Planes Passing through the Intersection of Two Given Planes......Page 452
14.19 Sphere......Page 454
14.20 Equation of a Sphere Whose Diameter is the Line Joining Two Given Points......Page 457
14.21 Equation of a Sphere Passing through Four Points......Page 456
14.24 Angle of Intersection of Two Spheres......Page 459
14.25 Condition of Orthogonality of Two Spheres......Page 460
14.27 Equation of a Cylinder with Given Axis and Guiding Curves......Page 463
14.28 Right Circular Cylinder......Page 464
14.31 Equation of a Cone with Given Vertex and Guiding Curve......Page 466
14.32 Right Circular Cone......Page 468
14.33 Right Circular Cone with Vertex (α, β, γ), Semi-Vertical Angle and the (l, m, n) Direction Cosines of the Axis.......Page 469
14.34 Conicoids......Page 470
14.36 Shape of the Hyperboloid of One Sheet......Page 471
14.38 Shape of the Elliptic Cone......Page 472
14.40 Tangent Plane at a Point of Central Conicoid......Page 473
14.42 Equation of Normal to the Central Conicoid at any Point (α, β, γ) on it......Page 474
14.43 Miscellaneous Examples......Page 477
Exercises......Page 479
15.1 Propositions......Page 490
15.2 Basic Logical Operations......Page 491
15.2.1 Translating from English to Symbols......Page 492
15.2.2 Truth Table for Exclusive OR......Page 493
15.3 Logical Equivalence Involving Tautologies and Contradictions......Page 495
15.4 Conditional Propositions......Page 496
Exercises......Page 507
16.1 Fuzzy Set......Page 510
16.2 Standard Operations on a Fuzzy Set......Page 512
16.3 Many Valued Logic......Page 514
Exercises......Page 515
17.1 Definitions and Basic Concepts......Page 518
17.2 Special Graphs......Page 520
17.3 Subgraphs......Page 523
17.4 Isomorphisms of Graphs......Page 525
17.5 Walks, Paths and Circuits......Page 527
17.6 Eulerian Paths and Circuits......Page 531
17.6.1 Methods for Finding Euler Circuit......Page 536
17.7 Hamiltonian Circuits......Page 538
17.7.1 Travelling Salesperson Problem......Page 543
17.8 Matrix Representation of Graphs......Page 544
17.9 Planar Graphs......Page 546
17.10 Colouring of Graph......Page 553
17.11 Directed Graphs......Page 556
17.12 Trees......Page 560
17.13 Isomorphism of Trees......Page 565
17.14 Representation of Algebraic Expressions by Binary Trees......Page 568
17.15 Spanning Tree of a Graph......Page 571
17.16.1 Dijkstra’s Shortest Path Algorithm......Page 573
17.16.2 Shortest Path if All Edges Have Length 1......Page 576
17.17.1 Prim Algorithm......Page 577
17.17.2 Kruskal’s Algorithm......Page 580
17.18 Cut Sets......Page 582
17.18.1 Relation Between Spanning Trees, Circuits and Cut Sets......Page 584
17.19 Tree Searching......Page 586
17.19.1 Procedure to Evaluate an Expression Given in Polish Form......Page 587
17.20 Transport Networks......Page 588
Exercises......Page 595
Index......Page 602

Citation preview

Engineering Mathematics Volume 1

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Engineering Mathematics Volume 1

BABU RAM Formerly Dean, Faculty of Physical Sciences, Maharshi Dayanand University, Rohtak

Delhi • Chennai • Chandigarh

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Copyright © 2012 Dorling Kindersley (India) Pvt. Ltd Licensees 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 present in this eBook at any time. ISBN 9788131784709 eISBN xxxxxxxxxxxxx Head Office: A-8(A), Sector 62, Knowledge Boulevard, 7th Floor, NOIDA 201 309, India Registered Office: 11 Local Shopping Centre, Panchsheel Park, New Delhi 110 017, India

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





In the Memory of My Parents Smt. Manohari Devi and Sri. Makhan Lal



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



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Contents Preface xi Symbols and Basic Formulae xiii 1 Sequences and Series

2 Successive Differentiation, Mean Value Theorems and Expansion of Functions 1.1

1.1 Sequences 1.1 1.2 Convergence of Sequences 1.1 1.3 The Upper and Lower Limits of a Sequence 1.3 1.4 Cauchy’s Principle of Convergence 1.4 1.5 Monotonic Sequence 1.6 1.6 Theorems on Limits 1.8 1.7 Subsequences 1.11 1.8 Series 1.12 1.9 Comparison Tests 1.15 1.10 D’alemberi’s Ratio Test 1.20 1.11 Cauchy’s Root Test 1.25 1.12 Raabe’s Test 1.27 1.13 Logarithmic Test 1.31 1.14 De Morgan–Berirand Test 1.33 1.15 Gauss’s Test 1.34 1.16 Cauchy’s Integral Test 1.36 1.17 Cauchy’s Condensation Test 1.38 1.18 Kummer’s Test 1.40 1.19 Alternating Series 1.41 1.20 Absolute Convergence of a Series 1.43 1.21 Convergence of the Series of the ∞

Type ∑ unν n 1.48 n =1

1.22 Derangement of Series 1.50 1.23 Nature of Non-Absolutely Convergent Series 1.51 1.24 Effect of Derangement of NonAbsolutely Convergent Series 1.52 1.25 Uniform Convergence 1.54 1.26 Uniform Convergence of a Series of Functions 1.56 1.27 Properties of Uniformly Convergent Series 1.57 1.28 Power Series 1.58 Exercises 1.59

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2.1

2.1 Successive Differentiation 2.1 2.2 Leibnitz’s Theorem and its Applications 2.5 2.3 General Theorems 2.9 2.4 Taylor’s Infinite Series and Power Series Expansion 2.16 2.5 Maclaurin’s Infinite Series 2.17 2.6 Expansion of Functions 2.17 2.7 Indeterminate Forms 2.27 Exercises 2.36 3 Curvature

3.1

3.1 Radius of Curvature of Intrinsic Curves 3.1 3.2 Radius of Curvature for Cartesian Curves 3.2 3.3 Radius of Curvature for Parametric Curves 3.6 3.4 Radius of Curvature for Pedal Curves 3.8 3.5 Radius of Curvature for Polar Curves 3.9 3.6 Radius of Curvature at the Origin 3.12 3.7 Center of Curvature 3.14 3.8 Evolutes and Involutes 3.15 3.9 Equation of the Circle of Curvature 3.15 3.10 Chords of Curvature Parallel to the Coordinate Axes 3.18 3.11 Chord of Curvature in Polar Coordinates 3.19 3.12 Miscellaneous Examples 3.21 Exercises 3.26 4 Asymptotes and Curve Tracing

4.1

4.1 Determination of Asymptotes When the Equation of the Curve in Cartesian form is Givens 4.1

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viii n contentS 4.2 The Asymptotes of the General Rational Algebraic Curve 4.2 4.3 Asymptotes Parallel to Coordinate Axes 4.3 4.4 Working Rule for Finding Asymptotes of Rational Algebraic Curve 4.4 4.5 Intersection of a Curve and its Asymptotes 4.7 4.6 Asymptotes by Expansion 4.9 4.7 Asymptotes of the Polar Curves 4.10 4.8 Circular Asymptotes 4.12 4.9 Concavity, Convexity and Singular Points 4.12 4.10 Curve Tracing (Cartesian Equations) 4.16 4.11 Curve Tracing (Polar Equations) 4.21 4.12 Curve Tracing (Parametric Equations) 4.23 Exercises 4.24 5 Functions of Several Variables

5.1

5.1 Continuity of a Function of two Variables 5.2 5.2 Differentiability of a Function of two Variables 5.2 5.3 The Differential Coefficients 5.2 5.4 Distinction Between Derivatives and Differential Coefficients 5.3 5.5 Higher-Order Partial Derivatives 5.3 5.6 Envelopes and Evolutes 5.8 5.7 Homogeneous Functions and Euler’s Theorem 5.10 5.8 Differentiation of Composite Functions 5.15 5.9 Transformation from Cartesian to Polar Coordinates and Vice Versa 5.19 5.10 Taylor’s Theorem for Functions of Several Variables 5.21 5.11 Approximation of Errors 5.24 5.12 General Formula for Errors 5.25 5.13 Tangent Plane and Normal to a Surface 5.28 5.14 Jacobians 5.30 5.15 Properties of Jacobian 5.30 5.16 Necessary and Sufficient Conditions for Jacobian to Vanish 5.33 5.17 Differentiation Under the Integral Sign 5.34 5.18 Miscellaneous Examples 5.37

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5.19 Extreme Values 5.41 5.20 Lagrange’s Method of Undetermined Multipliers 5.48 Exercises 5.53 6 Tangents and Normals

6.1

6.1 Introduction 6.1 6.2 Equation of the Tangent at a Point of a Curve 6.1 6.3 Equation of the Normal at a Point P(x1, y1) of a Curve 6.1 6.4 Lengths of Tangent, Normal, SubTangent and Subnormal at any Point of a Curve 6.5 Exercises 6.6 7 Beta and Gamma Functions

7.1

7.1 7.2 7.3 7.4 7.5

Beta Function 7.1 Properties of Beta Function 7.1 Gamma Function 7.4 Properties of Gamma Function 7.4 Relation Between Beta and Gamma Functions 7.5 7.6 Dirichlet’s and Liouville’s Theorems 7.10 7.7 Miscellaneous Examples 7.12 Exercises 7.13 8 Reduction Formulas

8.1

n 8.1 Reduction Formulas for ∫ sin x dx and

∫ cos

n

x dx 8.1 8.2 Reduction Formula for m n ∫ sin x cos x dx 8.3

8.3 Reduction Formulas for n ∫ sec x dx 8.5

∫ tan

8.5 Reduction Formulas for m n ∫ x (log x) dx 8.7 8.6 Reduction Formula for

∫x e

8.4 Reduction Formulas for and ∫ x n cos mx dx 8.6

I mn = ∫ cos m x sin nx dx. 8.8

8.7 Reduction Formula for ∫ Exercises 8.9

∫x

n

n

dx x ( 2+ a

x dx and

n

sin mx dx ax

dx and

2 n

)

8.8

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ix contentS n  9 Quadrature and Rectification

9.1

9.1 Quadrature 9.1 9.2 Rectification 9.9 Exercises 9.14 10 Centre of Gravity and Moment of Inertia

13 Vector Calculus 10.1

10.1 Centre of Gravity 10.1 10.2 Moment of Inertia 10.4 10.3 Mean Values of a Function 10.5 Exercises 10.6 11 Volumes and Surfaces of Solids of Revolution

11.1

11.1 Volume of the Solid of Revolution (Cartesian Equations) 11.1 11.2 Volume of the Solid of Revolution (Parametric Equations) 11.6 11.3 Volume of the Solid of Revolution (Polar Curves) 11.8 11.4 Surface of the Solid of Revolution (Cartesian Equations) 11.9 11.5 Surface of the Solid of Revolution (Parametric Equations) 11.11 11.6 Surface of the Solid of Revolution (Polar Curves) 11.13 Exercises 11.14 12 Multiple Integrals

12.1

12.1 Double Integrals 12.1 12.2 Properties of a Double Integral 12.2 12.3 Evaluation of Double Integrals (Cartesian Coordinates) 12.2 12.4 Evaluation of Double Integrals (Polar Coordinates) 12.6 12.5 Change of Variables in a Double Integral 12.8 12.6 Change of Order of Integration 12.12 12.7 Area Enclosed by Plane Curves (Cartesian and Polar Coordinates) 12.16 12.8 Volume and Surface Area as Double Integrals 12.19 12.9 Triple Integrals and their Evaluation 12.26 12.10 Change to Spherical Polar Coordinates from Cartesian Coordinates in a Triple Integral 12.30

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12.11 Volume as a Triple Integral 12.33 12.12 Miscellaneous Examples 12.37 Exercises 12.40 13.1

13.1 Differentiation of a Vector 13.11 13.2 Partial Derivatives of a Vector Function 13.18 13.3 Gradient of a Scalar Field 13.19 13.4 Geometrical Interpretation of a Gradient 13.20 13.5 Properties of a Gradient 13.20 13.6 Directional Derivatives 13.21 13.7 Divergence of a Vector-Point Function 13.26 13.8 Physical Interpretation of Divergence 13.26 13.9 Curl of a Vector-Point Function 13.28 13.10 Physical Interpretation of Curl 13.28 13.11 The Laplacian Operator 13.28 13.12 Properties of Divergence and Curl 13.32 13.13 Integration of Vector Functions 13.37 13.14 Line Integral 13.37 13.15 Work Done by a Force 13.41 13.16 Surface Integral 13.43 13.17 Volume Integral 13.47 13.18 Gauss’s Divergence Theorem 13.49 13.19 Green’s Theorem in a Plane 13.55 13.20 Stoke’s Theorem 13.59 13.21 Miscellaneous Examples 13.64 Exercises 13.72 14 Three-Dimensional Geometry

14.1

14.1 Coordinate Planes 14.1 14.2 Distance Between Two Points 14.1 14.3 Direction Ratios and Direction Cosines of a Line 14.1 14.4 Section Formulae—Internal Division o f a Line by a Point on the Line 14.2 14.5 Straight Line in Three Dimensions 14.6 14.6 Angle Between Two Lines 14.9 14.7 Shortest Distance Between Two Skew Lines 14.11 14.8 Equation of a Plane 14.18 14.9 Equation of a Plane Passing through a Given Point and Perpendicular to a Given Direction 14.19 14.10 Equation of a Plane Passing through Three Points 14.19

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x n contentS 14.11 Equation of a Plane Passing through a Point and Parallel to Two Given Vectors 14.20 14.12 Equation of a Plane Passing through Two Point and Parallel to a Line 14.20 14.13 Angle Between Two Planes 14.24 14.14 Angle Between a Line and a Plane 14.26 14.15 Perpendicular Distance of a Point From a Plane 14.26 14.16 Planes Bisecting the Angles Between Two Planes 14.27 14.17 Intersection of Planes 14.29 14.18 Planes Passing through the Intersection of Two Given Planes 14.29 14.19 Sphere 14.31 14.20 Equation of a Sphere Whose Diameter is the Line Joining Two Given Points 14.33 14.21 Equation of a Sphere Passing through Four Points 14.33 14.22 Equation of the Tangent Plane to a Spherem 14.36 14.23 Condition of Tangency 14.36 14.24 Angle of Intersection of Two Spheres 14.36 14.25 Condition of Orthogonality of Two Spheres 14.37 14.26 Cylinder 14.40 14.27 Equation of a Cylinder with Given Axis and Guiding Curves 14.40 14.28 Right Circular Cylinder 14.41 14.29 Cone 14.43 14.30 Equation of a Cone with its Vertex at the Origin 14.43 14.31 Equation of a Cone with Given Vertex and Guiding Curve 14.43 14.32 Right Circular Cone 14.45 14.33 Right Circular Cone with Vertex (a , b , g ), Semi-Vertical Angle q, and (l, m, n) the Direction Cosines of the Axis. 14.46 14.34 Conicoids 14.47 14.35 Shape of an Ellipsoid 14.48 14.36 Shape of the Hyperboloid of One Sheet 14.48 14.37 Shape of the Hyperboloid of Two Sheets 14.49 14.38 Shape of the Elliptic Cone 14.49 14.39 Intersection of a Conicoid and a Line 14.50 14.40 Tangent Plane at a Point of Central Conicoid 14.50

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14.41 Condition of Tangency 14.51 14.42 Equation of Normal to the Central Conicoid at any Point ( a , b , g ) on it 14.51 14.43 Miscellaneous Examples 14.54 Exercises 14.56 15 Logic

15.1

15.1 Propositions 15.1 15.2 Basic Logical Operations 15.2 15.3 Logical Equivalence Involving Tautologies and Contradictions 15.6 15.4 Conditional Propositions 15.7 Exercises 15.18 16 Elements of Fuzzy Logic

16.1

16.1 Fuzzy Set 16.1 16.2 Standard Operations on a Fuzzy Set 16.3 16.3 Many Valued Logic 16.5 16.4 Fuzzy Logic 16.6 16.5 Fuzzy Propositions 16.6 Exercises 16.6 17 Graphs

17.1

17.1 Definitions and Basic Concepts 17.1 17.2 Special Graphs 17.3 17.3 Subgraphs 17.6 17.4 Isomorphisms of Graphs 17.8 17.5 Walks, Paths and Circuits 17.10 17.6 Eulerian Paths and Circuits 17.14 17.7 Hamiltonian Circuits 17.21 17.8 Matrix Representation of Graphs 17.27 17.9 Planar Graphs 17.29 17.10 Colouring of Graph 17.36 17.11 Directed Graphs 17.39 17.12 Trees 17.43 17.13 Isomorphism of Trees 17.48 17.14 Representation of Algebraic Expressions by Binary Trees 17.51 17.15 Spanning Tree of a Graph 17.54 17.16 Shortest Path Problem 17.56 17.17 Minimal Spanning Tree 17.60 17.18 Cut Sets 17.65 17.19 Tree Searching 17.69 17.20 Transport Networks 17.71 Exercises 17.78 Index I.1

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Preface to the Revised Edition The book Engineering Maths I caters to the requirements of the revised syllabi of various Indian universities. Eight more chapters have been incorporated in this book to enable the students cover all the topics in the syllabus. Accordingly, the contents of the present book has also been divided into two volumes. Volume I of the book consists of six new chapters, namely, Tangents and Normals, Rectification and Quadrature, Centre of Gravity and Moments of Inertia, Logic, Elements of Fuzzy Logic, and Graphs in addition to the eleven chapters of Part I of first edition of the book. Volume II of the book consists of two new chapters entitled Calculus of Variation and Dynamics in addition to the fifteen chapters of Part II of the first edition of the book The contents of some of the previous chapters have been rearranged along with additions and refinements. A number of examples, selected generally from various university question papers, have been added in almost all chapters of this book. I would like to thank Suresh Kumar Godara for the effort he has taken in preparing the figures of the book. My wife Meena Kumari and daughter-in-law Ritu provided the moral support during the revision of this book. My son Aman Kumar, working with Goldman-Sach, Bangalore, offered, as usual, valid comments on the contents of some chapters. Special thanks are due to Thomas Mathew Rajesh, Anita Yadav and Vamanan Namboodiri at Pearson Education for their constructive support. Suggestions and feedback on the contents of the book will be gratefully acknowledged. Babu Ram

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Symbols and Basic Formulae 1

2

Greek Letters a alpha β beta γ gamma Г capita gamma δ delta Δ capital delta ε epsilon ι iota θ theta λ lambda µ mu ν nu ω omega Ω capital omega

c

φ Φ ψ Ψ ξ η ζ χ π σ ∑ τ ρ κ

phi capital phi psi capital psi xi eta zeta chi pi sigma capital sigma tau rho kapha

(c) product of the roots is equal to , a (d) b2 – 4ac = 0 ⇒ the roots are equal, (e) b2 – 4ac > 0 ⇒ the roots are real and distinct, (f) b2 – 4ac < 0 ⇒ the roots are complex, (g) if b2 – 4ac is a perfect square, the roots are rational. 3

(i) loga 1= 0, loga 0 = – ∞ for a > 1, loga a = 1. loge 2 = 0.6931, loge 10 = 2.306, log10 e = 0.4343. (ii) loga P + loga q = loga pq.

Algebraic Formulae (i) Arithmetic Progression a, a + d, a + 2d,…, nth term Tn = a + (n–1)d; Sum of n terms = n2 [2a + (n–1)d ]. (ii) Geometrical Progression: a, ar, ar2,…, nth term Tn= arn–1;

a(1 − r n ) . 1− r (iii) Arithmetic Mean of two number a and b is 12 ( a + b ) (iv) Geometric Mean of two numbers a and b is ab. (v) Harmonic Mean of two numbers a and b Sum of n terms =

2 ab

is a + b . (vi) If ax2 + bx + c = 0 is quadratic, then (a) its roots are given by

−b ± b − 4ac , 2a 2

b a

(b) the sum of the roots is equal to − ,

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Properties of Logarithm

4

p . q

(iii)

log a p − log a q = log a

(iv)

log a p q = q log a p.

(v)

log a n = log b b.log b n =

log b n . log b a

Angles Relations (i) 1 radian =

1800

π

(ii) 10 = 0.0174 radian 5

Algebraic Signs of Trigonometrical Ratios: (a) First quadrant: All trig. Ratios positive. (b) Second quadrant: sin θ and cosec θ positive, all others negative. (c) Third quadrant: tan θ and cot θ positive, all others negative. (d) Fourth quadrant: cos θ and sec θ positive, all others negative.

are are are are

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xiv n SyMbolS and baSic forMulae 6 Commonly Used Values of Triganometrical Ratios:

sin

π 2

= 1, cos

cos ec sin

π 6

π 3

π 4

6

π 3

=

cos ec 7

π

=

cos ec sin

2

=

cos ec

sin

π

π 4

π 2

= 0, tan

= 1, sec

π 2

π 2

(d) cos (A – B) = cos A cos B + sin A sin B

= ∞,

= ∞, cos

π 2

= 0,

1 π 3 π 1 , cos = , tan = , 2 6 2 6 3 = 2, sec

π 6

=

3 π , cos = 2 3 2 π = , sec 3 3 1 π , cos = 4 2 π = 2, sec 4

2 3

, cot

π 6

(i) tan 2 A =

2 tan A 1 + tan 2 A 1 − tan 2 A 1 + tan 2 A

sin 2 A 2 tan A = cos 2 A 1 − tan 2 A

(j) sin 3A = 3 sin A –4 sin3A (k) cos 3A = 4 cos3A – 3 cos A

3tan A − tan 3 A 1 − 3tan 2 A A+ B A− B cos (m) sin A + sin B = 2sin 2 2 A+ B A− B sin (n) sin A − sin B = 2 cos 2 2 A+ B A− B cos (o) cos A + cos B = 2 cos 2 2 A+ B B−A sin (p) cos A − cos B = 2sin 2 2 1 (q) sin A cos B = sin ( A + B ) + sin ( A − B ) 2 (l) tan 3 A =

Trig. Ratios of Allied Angles

± sametrig. ratio θ when n is even  ± co − ratio of θ when n is odd

( n.90 ± θ ) = 

For example: sin (4620) = sin[900(52) – 600]

3 . 2 Similarly, cos ec (2700 – θ ) = cos ec (900 (3) – θ ) = – secθ . = sin (–600) = –sin 600 = –

toc.indd 12

tan A − tan B 1 + tan A tan B

= 2 cos 2 A – 1 =

1 π , tan = 3, 2 3 π 1 = 2, cot = , 3 3 1 π , tan = 1, 4 2 π = 2, cot = 1. 4

(a) sin (A + B) = sin A cos B + cos A sin B (b) sin (A – B) = sin A cos B – cos A sin B (c) cos (A + B) = cos A cos B – sin A sin B

(f) tan ( A − B ) =

(h) cos 2A = cos 2 A – sin 2 A = 1 – 2sin 2 A

= 3,

Transformations of Products and Sums

tan A + tan B 1 − tan A tan B

(g) sin 2A = 2sin Acos A =

(a) sin(–θ ) = –sinθ , cos (–θ ) = cosθ , tan(–θ ) = –tan θ , Cosec (–θ ) = –cosecθ , sec (–θ ) = secθ , cot(–θ ) = –cotθ . (b) Any trig ratio of (n. 90 + θ ) =

8

(e) tan ( A + B ) =

(r) cos A sin B =

1 sin ( A + B ) − sin ( A − B ) 2

cos A cos B =

1  cos ( A + B ) − cos ( A − B )  2

(s)

(t) sin A sin B = 1 cos ( A − B ) − cos ( A + B ) 2

9

Expressions for sin (a) sin

A 2

=±

A A A , cos and tan 2 2 2

1 − cos A 2

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SyMbolS and baSic forMulae n xv

d x (e ) = e x dx

(b) cos

A 2

=±

1 + cos A 2

(g)

(c) tan

A 2

=±

1 − cos A 1 + cos A

(i)

d 1 (log a x) = dx x log a

(d) sin

A 2

+ cos

(k)

d ( ax + b)n = na ( ax + b)n −1 dx

(l)

dn m ax + b ) = m ( m − 1)( m − 2) n ( dx

(e) sin

A 2

− cos

A 2

= ± 1 + sin A

A 2

= ± 1 − sin A

a b c = = (sine formulae) sin A sin B sin C b2 + c2 − a 2   2bc  2 2 2 c +a −b  cos B =  cosine formulae 2ca  a 2 + b2 − c2  cos C =  2cb 

(b)

cos A =

(c) a = b cos C + c cos B 

 b = c cos A + a cos C  Projection fromulae. c = a cos B + b cos A 

11

Permutations and Combinations Formulae

Pr =

n! , (n − r )!

n

Cr =

n! = n Cn − r , r !(n − r )!

n

C0 = Cn = 1

n

12

(m)

(

)

d sin −1 x = dx

(o) d ( tan −1 x ) = dx

(q)

1 1 − x2

)

d 1 cos −1 x = − dx 1 − x2

(

)

1 1 (p) d cot −1 x = dx 1 + x2 1 + x2

d 1 sec −1 x = dx x x2 − 1

(

(n)

m−n

(

(r)

)

d ( cos ec −1 x ) = − 12 dx x x −1

(s) d ( sin h x ) = cos h x (t) d (cos h x ) = sin h x dx

dx

(u) D n ( uv ) = D n u + nc1 D n –1uDv + nc2 D n – 2uD 2 v +... + n Cr D n − r uD r v + ... + n C n uD n v

(Leibnitz’s Formulae) 13

Integration Formulae

(a) ∫ sin xdx = − cos x (b)

∫ cos xdx = sin x

(c) ∫ tan xdx = − log cos x (d)

∫ cot xdx = log sin x

(e)

∫ sec xdx = log (sec x + tan x)

(f)

∫ cosec x dx = log ( cosec x − cot x )

Differentiations Formulae

(c) d (tan x) = sec2 x (d)

d (cot x) = − cos ec 2 x dx

(g) ∫ sec 2 xdx = tan x (h) ∫ cos ec 2 xdx = − cot x

d (sec x) = sec x tan x dx

d (cos ec x) = − cos ec x cot x dx

(i)

dx

(e)

toc.indd 13

d 1 (log a x) = dx x

(j)

n

d (a) d (sin x ) = cos x (b) dx (cos x ) = − sin x dx

d (a x ) = a x log e a dx

...( m − n + 1)( ax + b )

10 Relations Between Sides and Angles of a Triangle (a)

(h)

(f)

x x ∫ e dx = e

(j)

x ∫ a dx =

ax log e a

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xvi n SyMbolS and baSic forMulae π

x n +1 , n ≠ −1 (l) ∫ x dx = n +1

1 (k) ∫ dx = log e x x (m) ∫ a (o) ∫ x (q) ∫ (s) (t)

(u)

2

2

n

dx 1 x = tan −1 − x2 a a

dx 1 x−a = log eˆ x+a − a 2 2a

dx a2 + x2

= sinh −1

x a

dx

(n) ∫ a 2 − x 2 (p) ∫

=

1 a+x 2a log eˆ a − 2 x

dx a −x 2

2

dx

(r) ∫

x2 − a2

= sin −1 = cosh

x a

−1

a 2 + x 2 dx =

x a2 + x2 a2 x + sinh −1 2 2 a

∫

x 2 − a 2 dx =

x x −a a x + cosh −1 2 2 a

∫

a 2 − x 2 dx =

x a2 − x2 a2 x + sin −1 2 2 a

∫

2

2

x a

2

2

(y) ∫ sin m x cos n x dx 0

 ( m − 1)( m − 3) …( n − 1)( n − 3) ... if m and n are not   (m + n ) ( m + n − 2)( m + n − 4) ... simultaneously even = ( m − 1)( m − 3)( n − 5)…π   ( m + n )( m + n − 2 ) (m + n − 4) … 2 if both m and n are even 

14

1

(a) β(m, n) = ∞

0

(c) Γ(n+1) = nΓ(n) and Γ(n+1)= n! if n is positive integer (d) Γ(1) = 1= Γ(2) and Γ( 12 ) = π

(f)

2

Γ ( m )Γ ( n ) Γ ( m + n)

π 2

∫ sin

p

=

x cos q xdx

0

=

π 2

(x) ∫ sin x dx = ∫ COS x dx 0

(1 − x)n −1 dx converges

− x n −1 (b) Γ(n) = ∫ e x dx converges for n > 0

e ax ( a cos bx − b sin bx ) (w) ∫ e cos bx dx = 2 a + b2 π

m −1

0

(e) β(m, n) =

ax

∫x

for m, n > 0.

eax (v) ∫ e sin bx dx = 2 ( a sin bx − b cos bx ) a + b2 ax

n

n

π

0

 ( n − 1)( n − 3)( n − 5 ) ... if n is odd   n ( n − 2 )( n − 4 ) ... =  ( n − 1)( n − 3)( n − 5 ) ...π if n is even  n ( n − 2 )( n − 4 ) ... 2 

toc.indd 14

Beta and Gamma Functions

2

(g)

∫ 0

=

1  p +1 q +1 , β  2  2 2  Γ

( )Γ( ) 2Γ ( ) p +1 2

q +1 2

p + q +1 2

π 2

−

1

tan θ dθ = ∫ sin 2θ cos 2 θ dθ Γ(

1

0

)Γ( ) 1 1 3 = 2 Γ ( 4 ) Γ ( 4 ). 2Γ (1) 3 4

1 4

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1

Sequences and Series

The intuitive concept of sequences of numbers involves not only a set of numbers but also an order in which these numbers have been placed. This suggests that for each positive integer, there is a number associated in the sequence. Thus, real sequence is a function whose domain is the set N of natural numbers and range a set of real numbers. Our aim in this chapter is to study the convergence of real sequences and then to apply the results to study the convergence behavior of various infinite series 1.1  SEQUENCES A function f : NgR whose domain is the set N of all natural numbers and range a set of real numbers is called a sequence of real number or simply a real sequence. If n ∈ N, then f(n) is generally denoted by xn, an, or un and is called the nth term of the sequence {xn}, {an} or {un}. The set of all distinct terms of a sequence is called its range. For example, consider the sequence {xn } = {( −1) n } = {−1,1, −1,1, −1…}. Its nth term is xn = (–1)n and the range is {–1,1} Similarly, for the sequence {xn } = { 1n }n∈N , the nth term is xn = 1n . All the elements in this sequence are distinct. Thus, the range for this sequence is an infinite set A sequence {xn}defined by xn = c, where c is a fixed real number, for all n ∈ N, is called a constant sequence. A sequence xn is said to be bounded above if there exists a real number k, such that xn < K for all n ∈ N. The number K is called an upper bound of the sequence {xn}.

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A sequence {xn} is said to be bounded below if there exists a real number k, such that xn > k for all n ∈ N. The number k is called a lower bound of the sequence{xn}. A sequence which is bounded both above and below is called a bounded sequence. Thus a sequence {xn} is said to be bounded if there exist two real numbers k and K such that k < xn < K for all n ∈ N. If we choose M = max {| k |,| K |}, then {xn} is bounded if | xn| < M for all n ∈ N. If there exists no real number M such that | xn| < M for all n ∈ N, then the sequence {xn} is said to be unbounded. For example, consider the sequence { 1n }n∈N . Since 0 < xn < 1 for all n ∈ N, the sequence { 1n } is bounded. Here 1 is an upper bound, while 0 is a lower bound. We note that no element of the sequence is actually equal to zero, while one element is equal to the upper bound 1. Thus for this sequence, the upper bound is attained while the lower bound is not attained. But if we consider the sequence {xn} = {2n–1}, then {xn} = {1,22,23,…}. For this sequence 1 < an, n ∈ N. But there no real number K such that xn < K for all n ∈ N. Therefore the given sequence is not bounded above. 1.2  CONVERGENCE OF SEQUENCES A sequence {xn} is said to converge to the limit l, if for every e > 0 there exists a positive integer n0(e) such that | xn – l | < e for all n > n0(e).

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1.2  n  Chapter One If the sequence {xn} converges to the limit l, xn = λ . then we write nlim →∞

A sequence {xn} is said to be divergent if lim xn is not finite, that is, if lim xn is +∞ or −∞ .

n →∞

For

{xn } =

example,

{ } 1 2n

n∈N

n →∞

consider

the

. Then lim xn = lim n →∞

Hence the sequence

1 n n →∞ 2

sequence = 0 (finite).

{ } is convergent. 1 2n

On the other hand, if we consider the sequence {xn} = {n2}, then lim n 2 = ∞ . Hence the sequence n →∞ {n2} is divergent. A sequence {xn}which neither converges to a finite number nor diverges to ∞ or −∞ , is called an oscillatory sequence. For example, consider the sequence {xn} = {(–1)n}. Here, the even elements are all +1 and so x2 n → 1 , whereas the odd elements are –l and so x2 n −1 → −1 . Hence the sequence {(–1)n} oscillates finitely between −1 and +1. On the other hand, if {xn} = {n (–1)n}, then the sequence oscillates infinitely between −∞ and +∞ . Remark 1.1. The positive integer n0 in the definition of convergence of the sequence, in general, depends on the value of e. For example, consider the sequence {an } = 1, 12 , 13 , … for which an = 1n (n = 1, 2, …). We would naturally guess that the sequence has the limit 0. To prove that 0 is the limit, we choose e > 0. Then for the convergence we must find n0 such that |an – 0| < e for all n > n0, that is, such that 1 (1) − 0 < ε for all n ≥ n0 n or such that

1 < ε for all n ≥ n0 . n Thus, if we choose n0 such that n10 < ε , then since 1 < n10 if n > n0 , (1) will hold. But n1 < ε if and n 0 only if n0 > ε1 . Thus the condition (1) is satisfied if we choose the integer n­­0 greater than ε1 . Thus, the sequence { 1n } converges to 0 if n0 > ε1 . Obviously, n­­0 depends upon e in this case.

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If we consider the sequence {an}, where an = 1(n = 1, 2, …), then the given sequence has the limit l = 1. To prove this, we note that an – l = 1– 1 = 0 so that for any e > 0, | an – l | < e for all n > 1. Thus for any e > 0, we can choose n0 = 1. This is one of the rare cases where n0 does not depend on e. EXAMPLE 1.1

Show that the sequence {an} = {n} does not have a limit. Solution. Assume, on the contrary, that the sequence {n} has the limit l. Then for a given e > 0, there must exist a positive integer n0 such that that is,

|an – l | < e for all n > n0, | n – l | < e for all n > n0,

Choose e = 1. Then, we have | n – l | < 1 for all n > n0,

or

l – 1 < n < l + 1 for all n > n0 . Thus all values of n satisfying n > n0 must lie between l – 1 and l + 1. This is absurd. Thus, we arrive at a contradiction and so the given sequence has no limit. EXAMPLE 1.2 ( −1) If an = 2 + n , find n0 such that 2

(i) | an − 2 |
n0. What conclusion can you draw regarding the convergence of {an}? Solution. (i) We have ( −1) n 1 | an − 2 | = = 2 . n2 n Hence, an − 2 < 101 implies 4

1 1 or n 2 > 104 or n > 102 . < n 2 10 4 Therefore, if we choose n0 = 102 = 100, then 1 an − 2 < 4 for all n > n0 . 10

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SequenceS and SerieS   n 1.3 (ii) The expression |an – 2 |< e implies 1 1 1 . < ε or n 2 > or n > ε n2 ε Choose a positive integer n0 > 1ε . Hence for this n0, we have |an – 2 | < e for all n < n0, Therefore, we conclude that the sequence {an} converges to 2. EXAMPLE 1.3 If an = 32nn++51 , show that lim an = 32 . n →∞

Solution. Let e > 0. Then

an −

2 2n + 1 2 = − 3 3n + 5 3

=

−7 3(3n + 5)

=

7 max (m , m ) 1 2

= e for all n > max (m 1 , m 2 ) . If we select ε = 12 | λ − µ |, then

| λ − µ | < 12 | λ − µ | . which is absurd.

Hence, the sequence {xn} cannot converge to two different limits. Theorem 1.2. Every convergent sequence is bounded. Proof: Let the sequence {xn} converge to the limit

l. Then, by the definitionof convergence, to each e > 0, there exists a positive integer m such that |xn – l| < e for all n > m, or l– e < xn < l + e for all n > m. If k and K are the least and greatest of x1, x2…, xm–1, l – e, l + e, then k < x < K for all n ∈ N . n

Hence the sequence {xn} is bounded.

implies 3n + 5 >

or n>

7 3ε

7 5 − . 9ε 3

Let n0 be a positive integer greater than 97ε − 53 . Then, 2 an − < ε for all n > n0 . 3 Hence, lim an = 32 . n →∞

Theorem 1.1. Every convergent sequence has a unique limit. Proof: Suppose, on the contrary, that a sequence {xn} converges to two distinct real numbers l and m. Then for e > 0, there exists positive integers m1and m2 such that ε | x n − λ | < for all n ≥ m1 and 2 ε | x n − µ | < for all n ≥ m2 . 2 Therefore, for n > max (m1, m2), we have |l – m| = |l – xn + xn – m | < |xn – l| + | xn – m|

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Remark 1.2.

The converse of Theorem 1.2 may not be true. There are bounded sequences which do not converge. For example, the sequence {(–1)n} is bounded but it does not converge. In fact, it does not have a unique limit. 1.3  THE UPPER AND LOWER LIMITS OF A SEQUENCE A number M is said to be the least upper bound of a sequence if (i) no element of the sequence {xn} exceeds M (ii) at least one element of the sequence {xn} exceeds M – e, where e is any positive number. Similarly, a number m is called the greatest lower bound of a sequence {xn} if (i) no element of the sequence is less than m (ii) at least one element of the sequence is less than m + e , where e is a positive number. For example, for the sequence {xn } = { 1n } = {1, 12 , …, 1n , …} , the least upper bound is 1, whereas the greatest lower bound is 0. Since one element in the sequence is equal to the least

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1.4  n  Chapter One upper bound, we say that the least upper bound is attained by this sequence. Since the greatest lower bound is not attained, there are infinite number of elements of the sequence that are less than m + e. A superior number for a given sequence is a number such that no element or only a finite number of elements of the sequence exceed this number. An inferior number for a given sequence is a number such that only a finite number of elements or no element of the sequence is less than this number. Thus, (i) if a sequence is bounded above, then superior numbers always exists (ii) if a sequence is bounded below, then inferior number always exists. The greatest lower bound of the set of superior numbers of a sequence is called the upper limit or limit superior of the sequence. It is denoted ¯ by U or lim an . The least upper bound of the set of inferior numbers of a sequence {an}is called the lower limit or limit inferior of the sequence. We denote it by L or lim an. For example, (i) In the sequence { 1n } , the element 1, 12 , 13 ,… 1 , are all superior numbers. If we take 985 only 984 number of the sequence are 1 . The greatest lower bound greater than 985 of the superior number is 0. Hence, the upper limit of the sequence is 0. On the other hand, the inferior numbers of the given sequence are 0 and all negative numbers. Hence the lower limit is zero. (ii) The superior numbers of the sequence { n+n 1} consist of 1 and all numbers greater than 1. The greatest lower bound of the set of superior number is 1. Hence, the upper n limit U for this sequence is 1. Also n+1 is an inferior number for all value of n. The least upper bound of this set of inferior numbers is 1. Thus L = U = 1. Also, upper bound M =1 and lower bound of the sequence is m = 12 . The lower bound is attained by the sequence, but not the upper bound.

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(iii) The superior numbers of the sequence (–1)n consists of 1 and all numbers greater than 1. The greatest lower bound of the set of superior numbers is 1. Hence, the upper limit U = 1. The inferior numbers of this sequence are −l, −2, …. The least upper bound of this set is −l. Thus L = −l. (iv) Consider the sequence {(−1) n (1 + 1n )} = {−2, 32 , − 34 , 54 , − 65 ,…} . Clearly, M = 32 and m = 2. Further, 32 , 54 , 76 ,… are superior numbers and −2, 34 , − 65 , … are inferior numbers.

Therefore, U = 1and L = –1.

Theorem 1.3. The necessary and sufficient condition that a sequence should be convergent is that its upper and lower limits are equal. Proof: Suppose that the sequence {xn} converges

to the limit l. Then to each e > 0, there exists a positive integer n0, such that or

| xn – l | < e for all n > n0

l – e < xn < l + e for all n > n0.

Hence l + e is a superior number. Then U, being the greatest lower bound of the set of superior numbers, we have U < l. But l – e is not a superior number. Therefore, U cannot be less than l. Hence U = l. Similarly, l – e is an inferior number for all e > 0 and the least upper bound of such number is l. Thus L = l. Hence U = L = l. Suppose conversely that U = L. By definition, only a finite number of terms of the sequence exceed U + e and only a finite number of terms are less than L – e. Thus, except for finite number of terms, all the other terms lie between L – e and U + e, that is, between U – e and U + e . Hence, U – e < xn< U + e after a xn = U and certain value of n. This implies nlim →∞ so the sequence {xn} converges. 1.4  CAUCHY’S PRINCIPLE OF CONVERGENCE The following fundamental theorem is useful for determining whether a sequence converges or not.

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SequenceS and SerieS   n 1.5 Theorem 1.4. (Cauchy’s Principle of Convergence). A necessary and sufficient condition for the convergence of a sequence {xn} is that for each e > 0 there exists a positive integer n0 such that | xm – xn| < e for all m > n > n0. Proof: Necessity: If the sequence {xn} converges

to the limit l, then to each e > 0 there exists a positive integer n0 such that ε | x n − λ | < for all n > n0 , 2 ε | x m − λ | < for all m > n0 . 2 Therefore, | xm − xn | = | xm − λ − xn + λ | = | ( x m − λ) − ( x n − λ) | ≤ | ( x m − λ) | + | ( x n − λ) |


n0 .

2 2 Sufficiency: Suppose | xm – xn | < e for all m > n > n0. Therefore, xm – e < xn < xm + e for all m > n > n0. Thus, xm + e is a superior number whereas xm – e is an inferior number. Therefore, if U and L denotes upper and lower limits, respectively, we have U < xm + e and L > xm – e , which yields U – L < 2e. Also U > L, that is, U – L > 0. Hence, 0 < U – L < 2e. Since e > 0 is arbitrary, it follows that U = L. Hence, by Theorem 1.3, the sequence {xn} converges. This completes the proof of Theorem 1.4. A sequence {xn} is called a Cauchy Sequence or a Fundamental Sequence if for each e > 0, there exists a positive integer n0 such that | xm – xn | < e for m > n > n0. Thus, Theorem 1.4 may be restated in the following way:

Theorem 1.5. A sequence of real numbers is convergent if and only if it is a Cauchy Sequence.

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EXAMPLE 1.4

Show that the sequence  cos nπ  {x n } =    n 

is a Cauchy Sequence. Solution. We have

| xn − xm | =

cos nπ cos m π − n m

m | cos nπ | + n | cos m π | nm m+n . = nm If m > n, we have n + m 2m 2 . | xn − xm | ≤ < = nm nm n ≤

Choose e > 0 and n0 > ε2 . Then | xn – xm | < e for n,m > n0. Hence {xn} is a Cauchy Sequence. EXAMPLE 1.5

Using Cauchy’s Principle of Convergence, show that the sequence {xn}, where xn = 1 + 12 + 13 + … + 1n , is not convergent. Solution. Suppose, on the contrary, that {xn} is

convergent. Then taking ε = 12 , the Cauchy’s Principal yields | x 2m − x m |


1 1 1 + + …+ (m terms) 2m 2m 2m

=

m 1 = , 2m 2

| x 2m − x m | >

1, 2

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1.6  n  Chapter One which is a contradiction to (1). Hence, the sequence is not convergent. 1.5  MONOTONIC SEQUENCE A sequence {x n} is said to be monotonic increasing or steadily increasing if xn+1 > xn for all n. A sequence {xn} is said to be monotonic decreasing or steadily decreasing if xn+1 < xn for all n. A sequence {xn} is said to be monotonic if it is either monotonic increasing or monotonic decreasing. A sequence {xn} is strictly increasing if xn+1 > xn for all n and strictly decreasing if xn+1 < xn for all n. For example, (i) 1 + 12 , 13 , 14 , … is a monotonic decreasing sequence. (ii)

{ n+n 1}

is a monotonic increasing sequence.

(iii) {n} is a monotonic increasing sequence. Theorem 1.6. A monotonic sequence always tends to a limit (finiteor infinite) (Thus, a monotonic sequence is either convergent or divergent, it cannot oscillate). Proof: Consider the monotonic increasing sequence, {xn}. Suppose that its upper bound M is finite Therefore, xn > M – e for some value of n, say n0 and, hence, for all n > n0. Therefore, M – e < xn < M for all n > n0 xn = M proving that {xn} is convergent. and so nlim →∞ If M is infinite we can find n0 such that xn is greater than any positive N for all n > n0. Therefore, lim xn = ∞ . Hence, the theorem is n →∞ proved for monotonic increasing sequence. Also, we have shown that the limit of a monotonic increasing sequence is its upper bound. Similarly, a monotonic decreasing sequence tends to its lower bound. If the lower bound is finite, say L, the monotonic decreasing sequence converges to L. If the lower bound is not finite, then the sequence converges to −∞ .

M01_Baburam_ISBN _C01.indd 6

Theorem 1.7. A necessary and sufficient condition for the convergence of a monotonic sequence is that it is bounded. Proof: Necessity: We have already proved in Theorem 1.2 that a convergent sequence is bounded. Sufficiency: Suppose that {xn} is a bounded monotonic increasing sequence. Thus range of {xn} is bounded and so, by completeness property, it has least upper bound, say M. We assert that {xn} converges to M. To prove it, let e > 0 be a positive number. Since M is supremum, there exists at least one member say xm such that xm > M – e . Since {xn} is monotonic increasing, we have xn > xm > M – e for all n > m. Also, since M is supremum, xn < M < M + e for all n. Hence M – e < xn < M + e for all n > m or |xn – M | < e for all n > m. Hence, {xn} converges to M. The case of a bounded monotonic decreasing sequence can be considered similarly. EXAMPLE 1.6

Show that the sequence {xn}, where 1 1 1 xn = 1 + + + …+ 1! 2! n! is convergent. Solution. We have 1 xn +1 − xn = > 0 for all n. (n + 1)! Hence {xn} is monotonic increasing. Further, xn = 1+

1 1 1 1 + + …+ < 1 + 1 + 1! 2! 2 n!

+ +

1 2

n−1

< 1+

1 =3 1 − 12

and so {xn} is bounded. Hence by Theorem 1.6, the given sequence converges to its upper bound, which is less than 3. In fact, xn → e = 2.71… , the base of the natural logarithm.

1/2/2012 11:33:37 AM

SequenceS and SerieS   n 1.7 EXAMPLE 1.7

y

 1 lim 1 +  y →∞  y and so the result follows from Example 1.7.

Show that the sequence {xn}, where n

is convergent. Solution. We have

 1 x n = 1 +  ,  n

EXAMPLE 1.9 n

 1 xn = 1 +  .  n

By Binomial Theorem for a positive integer, xn = 1 +

n n(n − 1) 1 n(n − 1)(n − 2) 1 + . 2+ . 3 n 2! 3! n n

+…+

= 1+1+

n(n − 1) …1 1 . n n! n

1 − 1n (1 − 1n )(1 − n2 ) + 2! 3!

+…+

(1 − 1n )(1 − n2 )…(1 − nn−1 ) n!

.

1 − n1+1 (1 − n1+1 )(1 − n2+1 ) + 2 3!

+… +

(1 – )(1 – 1 n +1

2 n + 1+ 1

)…(1 − )

( n + 1)!

n n +1

All the factors in the numerators of xn and xn+1 are positive. Further, each factor in the numerator of xn+1 is greater than the corresponding factor in the numerator of xn whereas the denominators of xn and xn+1 are same. Therefore, xn+1 > xn and so {xn} is monotonic increasing. Also, 1 1 xn < 1 + 1 + + …+ 2! n! 1 1 1 1 < 1 + 1 + + 2 + 3 + …+ n−1 < 3 2 2 2 2 and so {xn} is bounded. Hence, {xn} converges to a finite positive limit whose value is less than xn = e = 2.71… . 3. In fact the nlim →∞ EXAMPLE 1.8

2n − 5 2n − 7 − > 0 for all n. 3n + 5 3n + 2 Hence, the sequence is monotonic increasing. The sequence is bounded above and below. The upper and lower bounds being 32 and –1. Further, 2 − 7n 2 lim xn = lim = . n →∞ n →∞ 3 + 2 3 n xn +1 − xn =

Therefore, changing n to n +1, we get xn +1 = 1 + 1 +

Show that the sequence whose nth term is 2n − 7 , xn = 3n + 2 (i) is monotonic increasing, (ii) is bounded above, (iii) is bounded below, and (iv) has a limit. Solution. For the given sequence, 2(n + 1) − 7 2n − 5 2n − 7 , xn +1 = . = xn = 3n + 2 3(n + 1) + 2 3n + 5 Then

EXAMPLE 1.10

Show that lim(n)1/ n = 1 . n →∞

Solution. Let an = (n)1/n = 1 + hn, where hn > 0.

Then,

n = ann = (1 + hn ) n

1 n ( n − 1) hn2 2 1 n(n − 1) 2 . +…+ n(n − 1) … hnn > hn n! 2

= 1 + nhn +

Thus, hn2
0 be given. Then, 1/ 2

Show that lim(1 + x)1/ x = e . x→0

Solution. Substituting x = 1y , we get

M01_Baburam_ISBN _C01.indd 7

2  2    < ε if and only if n > 1 + ε 2 . n −1 If n0 is a positive integer greater than 1 + ε22 , then

1/2/2012 11:33:38 AM

1.8  n  Chapter One  2  − ε < 0 < hn <   n − 1

1/ 2

0. Since, an → a , bn → b as n → ∞,

there exists positive integers m1 and m2 such that ε | an − a | < for all n ≥ m1 2 and ε | bn − b | < for all n ≥ m . 2 2 Therefore, for m = max (m1, m2), we have ε | an − a | < for all n ≥ m 2 and ε | b n − b | < for all n ≥ m . 2 Therefore, | (an ± b n ) − (a ± b ) | = | (an − a) ± (b n − b ) |

ε

ε

+ = ε for all n ≥ m . 2 2 Hence, by definition of convergence, the sequence {(an + bn)} converges to a + b. (ii) Let e > 0. Then, as in part (i),

ε ε | an − a | < , | b n − b | < for all n ≥ m .

2 2 Also, since {an} and {bn} are convergent sequences, they are bounded. Therefore, there exist positive real numbers k and K such that |an| < k and |bn| < k for all n ∈ N . Thus, for n > m, we have

M01_Baburam_ISBN _C01.indd 8

=

2



( k + | b |) .

(iii) If b ≠ 0 , we have b − bn 1 1 . − = bn b bb n Choose n0 such that | bn – b| < e |b| | bn| for all n > n0. Therefore, 1 1 − < ε for all n > n0 . bn b Hence if bn → b as n → ∞ , then provided b ≠ 0 . Now an → a , a

1 bn

→

1 b

→ b1 , b ≠ 0

1 bn

Therefore, by part (ii), bnn → a ( b1 ) =

a b

as n → ∞ .

Theorem 1.9. If {bn} is a monotonic increasing sequence tending to infinity and bn>0 for all n, then lim

n →∞

an a − an , = lim n +1 n →∞ bn bn +1 − bn

provided, the latter limit exists. Proof: Suppose that

an+1 − an n →∞ bn+1 − bn

lim

=λ

(finite).

Therefore, to each e > 0, there exists a positive integer n0 such that

an + 1 − a n − λ < ε for all n ≥ n0 . bn +1 − bn

≤ | an − a | + | b n − b |


λ − ε − ε = λ − 2ε  bn  bn and

 b (λ + ε )  1 − m  bn

 am < λ + ε + ε = λ + 2ε . +  bn

Thus,

an < (λ + ε ) n − m am

( λ − ε)

1

1

1– mn

amn < ann < ( λ + ε)

1

amn .

Setting An = (λ − ε )1− am and Bn = (λ + ε )1− am m n

m n

1 n

1 n

and taking into account that am is finite, we have lim An = λ − ε and lim Bn = λ + ε .

n →∞

n →∞

Therefore, we can choose n0 such that An > λ − 2ε and Bn < λ + 2ε for all n ≥ n0

and so 1 n

λ − 2ε < an < λ + 2ε for all n ≥ n0 . Hence 1

λ − 2ε < Hence,

an < λ + 2ε for all n ≥ n0 . bn an =λ. n →∞ b n lim

Theorem 1.10. (Cauchy). If an is positive for all values of n, then a +1 lim a1/n n = lim n , n →∞ n →∞ a n provided, the latter limit exists. (While studying convergence of infinite series, we shall observe that this theorem of Cauchy implies that whenever D’Alembert’s Ratio Test is applicable, Cauchy’s Root Test is also applicable to test the convergence of a series of positive terms). = λ (finite). Therefore, for e > 0, we have a λ − ε < n +1 < λ + ε for all n ≥ m . an an+1 n →∞ an

Proof: Suppose that lim

M01_Baburam_ISBN _C01.indd 9

lim ann = λ.

n →∞

Theorem 1.11. (Cauchy’s First Theorem on Limits). If lim an = λ , then n →∞

 a + a2 + …+ an lim  1 n 

n →∞

  = λ. 

Proof: Write bn = an – l. Since lim an = λ , it

follows that lim bn = 0 . Also,

n →∞

n →∞

a1 + a2 + …+ an b + b + …+ bn . =λ+ 1 2 n n Therefore, to prove the theorem, it is sufficient to show that if lim bn = 0 , then n →∞

b1 + b2 + …+ bn = 0. n Since {bn}converges, it is bounded and so there exists k > 0 such that | bn | < K for all n. Since lim b n = 0, to each e > 0, there exists a n →∞ positive integer m such that |bn| < e for all n > m. lim

n →∞

1/2/2012 11:33:40 AM

1.10  n  Chapter One Therefore,

Proof: Write an = an – A. Since an → A as

b1 + b2 + …+ b n n =

b1 + b2 + …b m b m +1 + b m + 2 + …+ b n + n n

| b | + | b2 | +…+ | bm | ≤ 1 n | b m +1 | + | b m + 2 | +…+ | b n | + n mK ( n − m )ε < + for all n > m . n n mK ε < + . n 2

n → ∞ it follows that an and hence | an |→ 0 as n → ∞ . Therefore, a b + a2 bn −1 + …+ an b1 + 1 n . n

A (b1 + b2 + …+ b n ) n a1bn + a2 bn −1 + …+ an b1 + . n b n = B , by Theorem 1.11 b + b +…+ b → B Since lim n n →∞ as n → ∞ Since {bn} converges, it is bounded and, therefore, there exists K > 0 such that | bn | < K. Therefore, =

1

and so b + b + …+ b n = 0. lim 1 2 n →∞ n This proves the theorem. Remark 1.3. The converse of Theorem 1.11 is not true. For example, consider the sequence {an} = {(–1)n}. Then a1 + a2 + …+ an 0 = 1 n − n

if n is even if n is odd.

 | a | + | a2 | +…+ | an |  ≤K 1 . n   Since | an |→ 0 as n → ∞, by Theorem 1.11, | a1 | + | a2 | +…+ | an | n

a1 + a2 + …+ an = 0. n But the sequence {an} is not convergent n →∞

Theorem 1.12. (Cesaro). If the sequence {an} and {bn} tend to definite finite limits A and B, respectively, then a b + a2b n −1 + …+ an b1 = A B. lim 1 n n →∞ n

M01_Baburam_ISBN _C01.indd 10

tends to 0 as n → ∞ . Hence

a1b n + a2b n −1 + …+ an b1 = A B. n →∞ n

lim

EXAMPLE 1.11

Show that 1 lim (1 + 21/ 2 + 31/3 + …+ n1/ n ) = 1. n →∞ n Solution. Let an = n1/n. Then lim an = lim n n = 1. 1

→∞

n →∞

Therefore, by Cauchy’s First Theorem on limits, a1 + a2 + …+ an → 1 as n → ∞, n

Therefore, lim

n

a1bn + a2 bn −1 + …+ an b1 n

If m1 is a positive integer greater than 2 mK ε , ε mK then n < 2 when n > m1. Thus for n > max (m, m1), we have b1 + b2 + …+ b n ε ε < + =ε n 2 2

2

that is, 1

1

1

1 + 2 2 + 33 + …+ n n → 1 as n → ∞. n EXAMPLE 1.12

Show that 1 1 1 1 lim 1 + + +  +  = 0. n →∞ n  2 3 n

1/2/2012 11:33:41 AM

SequenceS and SerieS   n 1.11 1 1 Solution. Let an = n . Then lim an = lim n = 0. n →∞

n →∞

Therefore, by Cauchy’s First Theorem on limits, lim

n →∞

a1 + a2 +  + an = 0, n

that is, lim

n →∞

1 + 12 +  + n

1 n

= 0.

EXAMPLE 1.13

Show that

1+ 2 + + n 1 lim = . n →∞ 2 n2

Solution. Let

an = 1+ 2 + … + n and bn = n2. Then an +1 − an 1 (1 + 1n ) = bn +1 − bn n (1 + 1n )2 − 1

and so lim

n →∞

=

 1  1 + n1   1 2 n 1 + n2 + n − 1

=

1 + n1 2 + n1

an +1 − an 1 = . b n +1 − b n 2

Then, by Theorem 1.9, we have lim

n →∞

an a −a 1 = lim n +1 n = , n →∞ bn b n +1 − b n 2

that is, lim

n →∞

1 + 2 + …+ n 1 = . 2 n2

1.7  SUBSEQUENCES Let {an} be a sequence and let n1 < n2 < n3 < … < nk 0,

there exists a positive integer n0 such that | an – l | < e for all n > n0. Then taking n > nk > n0, we have | an – l | < e for all n > n0.

{ } converges to l.

Hence an

k

Remark 1.4. The converse of Theorem 1.13 need not be true. For example, consider the sequence {an}= {(–1)n}. This sequence is not convergent. But the subsequence {an} = {1,1,1,…} whose all terms are 1 converges to 1 as n → ∞ . Thus, the subsequence {a2n} of {an} is convergent but the original sequence {an} is not convergent. However, if the union of the subsequences is the whole sequence, then it seems plausible that the whole sequence will converge to the same limit. For example, the converse of Theorem 1.13 holds good if we take two subsequences—one formed by odd terms alone and the other by even terms alone—if these two sequences tend to a limit, then the whole sequence also converges to that limit. Thus we have the following theorem. Theorem 1.14. If for a sequence {an}, the subsequences {a2n–1} and {a2n} converges to a, then {an} converges to a. Proof: Since a2 n −1 → a as n → ∞, given e > 0, there exists a positive integer m1 such that

| a2n–1 – a | < e for all n > m1.

(1)

1/2/2012 11:33:42 AM

1.12  n  Chapter One Since a2n → a as n → ∞, given e > 0, there exists a positive integer m2 such that | a2n – a | < e for all n > m2.



(2)

Let m = max (m1, m2) and let n > 2m. If n is even and equal to 2p, then p > m > m2 and so by (2), | an – a | < e for all n > 2m.



(3)

On the other hand, if n is odd and equal to 2p + 1, then since n > 2m, p > m > m1, we have by (1), | an – a | < e for all n > 2m.



(4)

Hence, whether n is even or odd, we have from (3) and (4) that

Theorem 1.15. Every bounded sequence {an} of real numbers contains a convergent subsequence. Proof: Let A be the range of the bounded sequence {an}. If A is finite, then there is at least one point of A, say b, which appears infinite number of times in the sequence. Hence {b,b,b,…} is a subsequence of {an} which converges to b. If A is infinite, then it is a bounded infinite subset of real numbers. But, by Weierstrass’s Theorem, a bounded infinite subset has a limit point, say a, on the real line. Choose n1 so that | an − a |< 1. Having chosen n1, n2,…, ni–1, there is an integer ni > ni–1 such that ani − a < 1i . Hence, the subsequence {an } converges to a. 1

j

1.8  SERIES An expression of the form a1 + a2 + … + an + …, ∞

denoted by ∑an , is called an infinite series. n =1 The term an is called the nth term of the series. Further, S n = a1 + a2 +  + an ∞

is called the partial sum of the series

∑a . n

n =1

The definition of convergence or divergence of the series ∑a depends on the convergence or divergence of the sequence {S n }∞n =1 of partial sums. ∞

n

n =1

M01_Baburam_ISBN _C01.indd 12

be a series of real number with

n

n =1

partial sums

S n = a1 + a2 +…+ an .

sequence {S }

converges to l, we say that the

series ∞

∞

∑an n =1

∑an = l. n =1

∞ n n =1

If the

converges to the sum l and we write ∞

If {S n }∞n =1 diverges, then the series ∑an n =1

also diverges. In case of infinite series, the Cauchy’s criteria for the convergence takes the form |Sm – Sn| < e, m > n > n0 (e), that is, | an +1 + an + 2 +  + am |< ε , m > n > n0 (ε ).

| an – a | < e for all n > 2m Therefore, {an}converges to a.

∞

∑a

Let

If we take m = n + 1, then the last expression takes the form | an +1 | n0 (∈) or

lim an = 0.

which is a necessary condition for the convergence of the series

∞

∑a

n

.

n =1

However, lim an = 0 is not a sufficient condition for the convergence of the series of positive terms. For example, ∞ ∞ 1 ∞ 1 , ∑ 1 ∑1 n , ∑ n n = 2 n log n n =1

∞

∑a

n

n =1

are all divergent series inspite of the fact that lim an = 0. But the condition is sometimes n→∞ useful. Infact, if lim an ≠ 0, we can at once say that the series is not convergent. Remark 1.5. Two convergent series can be added (or subtracted) term by term to give a convergent series. Thus, if Sun = S and Svn = T, then the sum of n terms of the series obtained by addition is Sn + Tn. But lim( S n + Tn ) = lim S n + lim Tn = S + T by n →∞ n →∞ n →∞ Theorem 1.8. Hence, the sum series converges to S+T. (ii) We note that the sum of two non-convergent series form a convergent series. For example, let S = 1−1+1−1+1−1+ …

1/2/2012 11:33:43 AM

SequenceS and SerieS   n 1.13 and T = 0 + 1 − 1 + 1 − 1 + 1 −… We observe that S and T are both oscillatory, but S + T =1. Theorem 1.16. (Pringsheim). If the terms of the series Sun of positive terms steadily (monotonically) decrease, then it is necessary but not a sufficient condition for the convergence nun = 0. of the series that nlim →∞

Proof: The condition is necessary: Let the series Sun of positive terms be convergent. Then, by Cauchy’s Principle of Convergence, for a given e > 0 there exists a positive integer m such that

ε

for all n ≥ m. 2 Since the terms are monotonically decreasing, un > un +1 > un + 2 > … > um +1 . Hence, the above expression reduces to | um +1 + um + 2 + …+ un |
0 for all n.

M01_Baburam_ISBN _C01.indd 13

Thus Sn+1 > Sn for all n. Hence{Sn}is a monotonically increasing sequence. If {Sn} is bounded above, then {Sn} is monotonically increasing and bounded above sequence and so, by Theorem 1.7, it is convergent. Hence San is convergent. If {Sn} is not bounded above, then Sn is monotonically increasing and not bounded above and so it diverges to + ∞. Hence, Sun diverges to + ∞. Theorem 1.18. Convergence, divergence, or oscillation of a series of positive terms is not affected by the addition or omission of a finite number of its terms. Proof: Let Sun be a series of positive terms. It is

sufficient to show that the two serie

u1 + u2 + …+ um + um +1 + um + 2 + … and

um +1 + um + 2 + ……

converge or diverge together. To show it, let Sn and Tn denote the nth partial sums, respectively, of the above two series. Then S n = u1 + u2 + …+ un and Tn = um +1 + um + 2 + …+ um + n = (u1 + u2 + …+ um + n ) − (u1 + u2 + …+ um ) = Sm + n − Sm . But Sm, being the sum of a finite number of terms of Sun, is a fixed finite quantity. Therefor (i) If Sm+n tends to a finite limit, then Tn also tends to a finite limi (ii) If Sm+n tends to ∞ , then Tn also tends to ∞

lim Sm + n does not exist, the lim Tn also (iii) If ∞→ n →∞ 0 does not exist. Thus, the two sequences Sn and Tn converge or diverge together. Hence, the series u1 + u2 + … + um+1 + um+2 + … and the series um+1 + um+2 + … converge or diverge together.

1/2/2012 11:33:43 AM

1.14  n  Chapter One EXAMPLE 1.14

Using

Cauchy’s

general

principle

convergence, show that the series converge.

∞

∑ n =1

1 n

does not

Solution. We shall prove our result by contradiction. So, suppose that the given series is convergent. We choose ε = 1 . Then, by 2 Cauchy’s criterion, we have 1 1 | an − am |< for all n > m > n0   2 2 or

1 1 1 1 + +  < for all n > m > n  1  0  m +1 m + 2 n 2 2

or

1 1 1 1 1 + +  + < for all n > m > n0   . m +1 m + 2 n 2 2 We take n = 2m and so we should have 1 1 1 1 + + …+ < . m +1 m + 2 2m 2

But, 1 1 1 + ++ m +1 m + 2 2m 1 1 1 > + ++ (m terms) 2m 2m 2m m 1 = = . 2m 2 Thus, we arrive at a contradiction. Hence, the given series is not convergent. EXAMPLE 1.15

Test the convergence of the series 1 2 3 n + + + …+ +… 4 6 8 2(n + 1) Solution. The nth term of the given series is

un =

n 1 = . 2(n + 1) 2(1 + 1n )

lim un =

of

n →∞

1 2

≠ 0.

Thus necessary condition for convergence is not satisfied. Hence Sun is divergent. EXAMPLE 1.16

Discuss the convergence of the geometric series 1 + a + a 2 + a 3 + Solution: The partial sum for the given series is S n = 1 + a + a 2 +  + a n −1

=

If | a | < 1, then

an −1 , a ≠ 1. a −1

Sn =

1 − an and so 1− a

lim S n = 1−1a since, a n → 0 as n → ∞.

n →∞

Thus, if | a | < 1, then San–1 is convergent 1 and its sum is 1 − a . If a = 1, then S n = 1 + 1 + 1 (n terms) = n and so lim S n = ∞. Hence San–1 is divergent. n →∞

If a >1, then Sn > n and so lim S n = ∞. n →∞ Hence the series is divergent. If a = –1, the series becomes 1 − 1 + 1 + 1 − 1 + and then

= 1. S 2 n = 0 and 2 S n −1 Hence the series oscillates between 0 and 1. If a < –1, the successive terms increase in magnitude. Then, a 2n − 1 (+ ve) a −1 is monotonic increasing while S2n+1 is negative and numerically increasing with n. Hence the series oscillates between + ∞ and − ∞. S2 n =

Therefore,

M01_Baburam_ISBN _C01.indd 14

1/2/2012 11:33:44 AM

SequenceS and SerieS   n 1.15 EXAMPLE 1.17

Show by direct summation of n terms that the series

be the partial sums of Sun and Svn, respectively. Since un < vn for all n, we have Sn < Tn and so

1 1 1 + + + 1.2 2.3 3.4

lim Sn < lim Tn . Since Svn is convergent,

n →∞

n →∞

lim Tn is finite and so lim S n is finite. Hence,

is convergent. Solution. The nth term of the given series is

an =

1 1 1 = − . n(n + 1) n n + 1

Therefore, 1 2 1 1 a2 = − 2 3 1 1 a3 = − 3 4 1 1 an = − n n +1 Then the partial sum is

and Tn = v1 + v2 + …+ vn be partial sums of Sun and Svn, respectively. Since un > vn, it follows that Sn > Tn and so

lim Sn > lim Tn = ∞ , since Sv diverges. Thus n →∞ n

n →∞

1 . n +1

lim Sn = 1 − 0 = 1.

n →∞

Hence, the given series converges to 1. 1.9  COMPARISON TESTS The convergence or divergence behavior of a given series is generally determined by comparing its terms with the terms of another series whose convergence behavior is known. Such comparisons are called Comparison Tests. Comparison Test l(a): If Sun and Svn are two series of positive terms such that un < vn and Svn is convergent, then Sun is also convergent. S n = u1 + u2 + …+ un and Tn = v1 + v2 + …+ vn

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Comparison Test 1(b): If Sun and Svn are two series of positive terms such that un > vn and Svn is divergent, then Sun also diverges. S n = u1 + u2 + …+ un

Therefore,

Proof: Let

Sun is convergent.

Proof: Let

a1 = 1 −

S n = a1 + a2 + …+ an = 1 −

n →∞

n →∞

lim Sn = ∞ and hence, Sun diverges.

n →∞

Comparison Test II. If Sun and Svn are two series of positive terms, then u

n = l (finite and non-zero), then Sun (i) if nlim →∞ vn

and Svn both converge or diverge together u

(ii) if lim vnn = 0 and Svn converges, then Sun n→∞ also converges u

(iii) if lim vnn = ∞ and Svn diverges, then Sun also n→∞ diverges. Proof: (i). Since un and vn are positive terms, un vn

> 0 and so lim uvn ≥ 0. But lim νu = l ≠ 0. n n →∞ n

n

n→∞

Therefore, l > 0. Since lim uvn = l , for every n n →∞

e > 0, there exists a positive integer m such that that is,

un vn

− l < ε for all n > m,

l −ε
1. Hence, by Comparison Test I(a), the series Sun converges, if Svn converges. Also un > q vn, where q < 1. If Svn diverges then Sun also diverges. Similarly, we can show that Svn converges or diverges according as Sun converges or diverges. u

lim vnn = 0, given e > 0 there exists a (ii) Since n→∞ positive integer m such that

un − 0 < ε for all n > m vn

or

un < ε for all n > m vn un < ε vn for all n > m.

−ε
0, however n→∞ un large, there exists integer m such that vn > M for all n > m. Hence un > Mvn for all n > m. Since Svn diverges, it follows by Comparison Test I(b) that Sun also diverges.

Hence

u

v

u

v

(i) if unn+1 < vnn+1 and Svn converges, then Sun also converges. (ii) if unn+1 > vnn+1 and Svn diverges, then Sun also diverges. Proof: Let Sn and Tn be, respectively, the partial

sums of the series Sun and Svn . Then

 u u u u u u  S n = u1 1 + 2 + 3 . 2 + 4 . 3 . 2 +    u1 u2 u1 u3 u2 u1   v v v v v v  < u1 1 + 2 + 3 . 2 + 4 . 3 . 2 +   ,  v1 v2 v1 v3 v2 v1 

M01_Baburam_ISBN _C01.indd 16

u1 . Tn n →∞ v 1

n →∞

u1

Tn is finite and so is Since Svn converges, nlim →∞ ν lim Sn which in turn implies that Sun converges. 1

n →∞

Part (ii). This can be proved by replacing ‘’ in the above proof. EXAMPLE 1.18 ∞

Show that the series ∑ n1p , called the Harmonic n =1

Series, converges if p > 1 and diverges if p < 1. Solution. Take p > 1, first. Group the terms of the

series as follows:

1  1 1   1 1 1 1  + + + + + +  1p  2 p 3 p   4 p 5 p 6 p 7 p 

n

of positive terms, then

u1 u (v1 + v2 +  + vn ) = 1 Tn v1 v1

=

lim S n < lim

1  1 +  p + + p 15 8

n

Comparison Test III: If Sun and Svn are two series

u n +1 u n +1 < , un vn

  + 

But 1 1 1 1 2 1 + p < p + p = p = p −1 p 2 3 2 2 2 2 1 1 1 1 1 1 1 1 + p+ ρ+ p < ρ+ ρ+ p+ ρ ρ 4 5 6 7 4 4 4 4 1 1 = ρ −1 = p −1 2 , 4 (2 ) 1 1 1 1 1 + p +  + p < p −1 = p −1 3 , p 8 9 15 8 (2 ) and so on. Hence, ∞

1

∑n n =1

p

< 1+

1 1 1 + + + . 2 p −1 (2 p −1 ) 2 (2 p −1 )3

The right hand side is a Geometric progression 1 with common ratio 2 < 1, since p > 1. Hence, by Comparison Test the series is convergent. p−1

1/2/2012 11:33:46 AM

SequenceS and SerieS   n 1.17 Solution. The nth term of the given series is

If p = 1, then the series is 1+

1 1 1 + + + 2 3 4

and 1 1 1+ = 1+ , 2 2 1 1 1 1 1, + > + = 3 4 4 4 2 1 1 1 1 1 1 1 1 + + + > + + + 5 6 7 8 8 8 8 8 1 = 2 and so on. Therefore, the given series is less than the series 1 1 1 1+ + + + … 2 2 2 Leaving aside the first term, this series is a Geometric progression whose common ratio is 1. Therefore, this series is divergent and so is the given series. If 0 < p < 1, then np < n, that is, n1 < 1n . 1 But the series ∑ 1n is divergent. Hence, ∑ n p is divergent for p < 1. If p = 0, then the series becomes 1 + 1 + 1 + ..., which is divergent. p

EXAMPLE 1.19

Test the convergence of the series 1 1 1 + + +… 1.2 2.3 3.4 Solution. We note that

1 1 1 1 1 1 1 + + + < + 2 + 2 + = ∑ 2 . 1.2 2.3 3.4 1 2 3 n

But ∑ n2 is convergent (Harmonic Series for p = 2). Hence, the given series is convergent. 1

EXAMPLE 1.20

Test the convergence of the series (n + 1)(n + 2) ∑ (n2 + 1)(n2 + 2) .

M01_Baburam_ISBN _C01.indd 17

(1 + 1n )(1 + n2 ) (n + 1)(n + 2) = 2 . 2 2 (n + 1)(n + 2) n (1 + n12 )(1 + n22 )

un =

Take vn =

1 n2

. Then,

(1 + 1n )(1 + n2 ) un = lim n →∞ v n →∞ (1 + 1 )(1 + 2 ) n n2 n2 lim

= 1, finite and non-zero Therefore, Sun and Svn converge or diverge together. But the series

∑v = ∑ n

Therefore, Sun is convergent.

1 n2

converges.

EXAMPLE 1.21

Examine the convergence of the series

∑ [(n

3

1

+ 1) 3 − n].

Solution. The nth term of the series is 1

un = (n3 + 1) 3 − n 1

1

  1  3 1 3  =  n3  1 + 3   − n = n  1 + 3  − n  n    n 

(

= n  1 + n13 

)

1 3

− 1 

 1 1 1 ( 1 − 1) = n 1 + ⋅ 3 + 3 3 2!  3 n

( ) 1 n3

2

 +  − 1 

1 1 1   3 − 9 . n3 −   

=

n n3

=

1 1 1 1  − . −  . n 2  3 9 n3 

Choose vn = lim

n →∞

1 n2

. Then

un 1 1 1  = lim  − . 3 − …. →∞ n vn 3 9 n  = 13 , finite and non − zero.

Hence, Sun and Svn converge or diverge together.

But ∑vn = ∑ n12 is convergent. Therefore, Sun is also convergent.

1/2/2012 11:33:46 AM

1.18  n  Chapter One EXAMPLE 1.22

EXAMPLE 1.24

Examine the convergence of the series

Examine the convergence of the series

∑(

n +1 − n) .

Solution: The nth term of the given series is

un =

n +1 − n

n +1 − n

=

n +1 + n

1p 2q

(ii)

∑

1 n

sin 1n

Solution. (i) The nth term of the series is

un =

np np 1 . = = q q q q− p 1 q (n + 1) n (1 + n ) n (1 + 1n )

n +1 + n 1 n  1 + 1n + 1

Take vn =

1 n

p

p

Take

n +1− n

= =

[ n +1 + n]

+ 23q + 34q +…

(i)

.

lim

un 1 = lim 1 n →∞ v n →∞ 1 + n n +1 lim

1 2

Then, n →∞

. Then

=

vn =

finite and non-zero

Therefore, Sun and Svn converge or diverge together. But ∑ 1n diverges. Therefore, Sun is also divergent.

n

1 .

q− p

un 1 = lim = 1. q →∞ n vn (1 + 1n )

Therefore, Sun and Svn converge or diverge together. But Σvn = Σ nq1− p converges if q – p > 1 and diverges, if q – p < 1. Hence, Sun converges if q – p > 1 and diverges if q – p < 1. (ii) Here un =

1 1 sin n n

Since lim sin1 n = 1, it follows that sin 1n ∼ 1

EXAMPLE 1.23

Examine the convergence of the series ∞

n . ∑ 2 n =1 n + 1 Solution. The nth term of the series is

un = Take vn =

n n 1 = 2 = 3 . 1 n + 1 n (1 + n2 ) n 2 (1 + 12 ) n

un ∼

n→∞ 1 n2

n

. We, therefore, take vn =

1 n2

1 n

and so

, and then

2

sin 1 un n 1 sin = lim 1 n = 1. = lim n →∞ v n →∞ n n n →∞ n n lim

Therefore, Sun and Svn converge or diverge together. But Σvn = Σ n12 converges. Hence, Sun is also convergent, by Comparison Test.

2

1 n 32

. . Then

u 1 lim n = lim = 1, finite and non-zero. n →∞ v n →∞ 1 + 1 n n2 Therefore, Sun and Svn converge or diverge 1 together. But the series Σvn = Σ 32 is n convergent. Hence, Sun is also convergent.

M01_Baburam_ISBN _C01.indd 18

EXAMPLE 1.25

If Sun is a convergent series of positive terms, show that Σun2 is also convergent. Give an example to show that the converse need not be true. Solution. Since Sun is convergent, un → 0 as n → ∞ . Therefore, there exists a positive

integer m such that 0 < un < 1 for all n > m and so un2 ≤ un for all n > m. Hence, by Comparison

1/2/2012 11:33:47 AM

SequenceS and SerieS   n 1.19 2 Test, Σun is convergent.

2n 3 + 5 . 5 +1

∑ 4n

However, the converse need not be true. 1 For example, if we take Σun = Σ n , then

Σu = Σ . The series Σ Σ but 1n does not converge. 2 n

1 n2

1 n2

is convergent,

Solution. The nth term of the given series is

un =



=

EXAMPLE 1.26

Show that the series 1 1 1 1 + + + + 2 32 23 34 is convergent.

n Take v =

which is a geometric series with common ratio 1 2

and so, converges. Hence, by Comparison Test, the given series converges. EXAMPLE 1.27

Show that the series 1 3 5 + + + 1.2.3 2.3.4 3.4.5 is convergent. Solution. The nth term of the given series is

2n − 1 n(n + 1)(n + 2) 2 − 1n n (1 + 1n )(1 + 2

2 n

).

u 1 Taking vn = n2 , we note that lim vnn = 2 n→∞ (finite). Therefore, Sun and Svn converge or diverge together. But Σ n12 is convergent. Hence Sun is convergent.

EXAMPLE 1.28

Examine the convergence of the series

M01_Baburam_ISBN _C01.indd 19

5

5 n3 1 n5

5 n3

2

1 n5

. Then

2 + n23 1 un = lim = , finite n →∞ v n →∞ 4 + 1 2 n n5

series in less than or equal to the corresponding term of the series 1 1 1 1 + + + + , 2 2 2 23 2 4

=

( ) = (2 + ) n (4 + ) n (4 + ) . n3 2 +

lim

Solution. Since 2 < 3, each term, of the given

un =

1 n2

2n 3 + 5 4n 3 + 1

and

non-zero.

Hence, Sun and Svn converge or diverge together. But Σvn = Σ 12 converges. Therefore, Sun n converges. EXAMPLE 1.29

Test the convergence of the series 1 1 1 + + + 1.2.3 2.3.4 3.4.5 Solution. The nth term of the series is 1 1 un = = n(n + 1)(n + 2) n3 (1 + 1n )(1 + n2 ) Take vn =

1 n3



. Then, u 1 lim n = lim n →∞ v n →∞ (1 + 1 )(1 + 2 ) n n n

=1, finite and non-zero Therefore, Sun and Svn converge or diverge together. But the series Σvn = Σ n13 is convergent. Hence, the given series Sun is also convergent. EXAMPLE 1.30

Examine the convergence of the series

∑(

)

n4 + 1 − n4 − 1 .

Solution. The nth term of the series is

un =

=

n4 + 1 − n4 − 1

n4 + 1 − n4 − 1 n4 + 1 + n4 − 1

.[ n 4 + 1 + n 4 − 1]

1/2/2012 11:33:48 AM

1.20  n  Chapter One

=

n4 + 1 + n4 − 1

(

n 2  1 + n14  2 n2

lim

n →∞

) ( 1 2

)

+ 1 − n14

1 2

 

.

and Svn = Srn, r < 1 is convergent. Therefore, by Comparison Test, Sun also converges. To prove (ii), proceed with the same process as in (i) and obtain

. Then, un = lim vn n →∞ (1 +

1 1 n4

1 2

1 , finite and non-zero 2 Therefore, the series Sun and Svn converge or diverge together. But the series Σvn = Σ n22 is convergent. Hence, the given series is convergent. 1.10  D’ALEMBERI’S RATIO TEST Let Sun be a series of positive terms. Then un+1 n →∞ un un+1 n →∞ un

(i) Sun is convergent if lim (ii) Sun is divergent if lim

1

= 1, this test gives no information about (If lim the convergence or divergence of the series Sun). Proof: Suppose that

un +1 = l. n →∞ u n lim

Then to each e > 0, there exists a positive integer m such that

and so

u n +1 − l < ε for all n ≥ m un

u n +1 < l + ε for all n ≥ m . un If l < 1, choose e > 0 such that l + e = r < 1. Then, u n +1 < r for all n ≥ m un and so u n +1 r n +1 < n for all n ≥ m . un r l −ε
r = n , r > 1 for all n ≥ m . un r

1

) + (1 − n14 ) 2

=

un+1 n →∞ un

un +1 vn +1 < un vn



2

Take vn =

Take vr = rn. Then

2

=

Take vn = rn. Then, un +1 vn +1 > un vn and Svn = Srn diverges. Hence, by Comparison Test, Sun also diverges. Remark 1.7. For comparison with other tests, D’Alembert Ratio Test is generally used in its inverted form: “If Sun is a series positive terms and un u lim u n > 1 then Su converges. If lim un+1 < 1, n →∞

n

n +1

then Sun diverges.”

n →∞

EXAMPLE 1.31

Examine the convergence of the series 12 ⋅ 22 12 ⋅ .22. ⋅ 32 + + . 1⋅ 3 ⋅ 5 1⋅ 3 ⋅ 5 ⋅ 7 ⋅ 9 Solution. The nth term of the given series is 1+

un =

12. ⋅ 22 ⋅ 32  n 2 1⋅ 3 ⋅ 5 (4n − 5)(4n − 3)

Then, u n +1 = and so

12 ⋅ 22 ⋅ 32  n 2 (n + 1) 2 1⋅ 3 ⋅ 5 (4n − 5)(4n − 3)(4n − 1)(4n + 1) un (4n − 1)(4n + 1) = u n +1 (n + 1) 2

1/2/2012 11:33:49 AM

SequenceS and SerieS   n 1.21 = =

Therefore,

16n 2 − 1 (n + 1) 2 16 − n12

(1 − 1n )2

un (n + 2) n + 1 1 = lim ⋅ un +1 n→∞ (n + 1) n x 2

lim

n →∞

.

1

n(1 + n2 ) n (1 + 1n ) 2 1 1 . . 2 = 2 . = lim n →∞ n(1 + 1 ) x x n 4

Further, 16 − n12 16 un = lim = >1 n →∞ u n →∞ n +1 (1 − 1n )2 1 . lim

Hence, by D’Alembert’s Ratio Test, the series Sun converges. EXAMPLE 1.32

Examine the convergence of the series 1+

1 3 , we have n2 u 1 lim n = lim = 1 , finite and non-zero: n →∞ v n →∞ 1 + 1 n n Hence, by Comparison Test, Sun and Svn

Taking vn =

3 3.6 2 3.6.9 3 x+ x + x + 7 710 7.10.13

Solution. Neglecting the first term, we hav

3 ⋅ 6 ⋅ 9 3n x n and 7 ⋅10 ⋅13 (3n + 4) 3 ⋅ 6 ⋅ 9 (3n)(3n + 3) = x n +1 . 7 ⋅10 ⋅13 (3n + 4)(3n + 7)

un = u n +1

converge or diverge together. But Σvn = Σ

lim



un 3n + 7 1 = lim . →∞ n u n +1 3n + 3 x

Test the convergence of the following series:

Therefore, by D’Alembert’s Ratio Test, the 1 given series converges if X > 1 or if x < 1 and diverges if X1 < 1 or if x > 1 . If x = 1 , then D’Alembert’s Ratio Test gives no information. EXAMPLE 1.33

Test the convergence of the series 2 1

+

x2 3 2

+

x4 4 3

+

x6 5 4

+ , x > 0 .

un =

(n + 1) n

M01_Baburam_ISBN _C01.indd 21

and so un +1 =

x2n (n + 2) n + 1

(i)

xn

∑ n!

2 p 3 p 4ρ + + + 2! 3! 4! Solution. (i) The nth term of the series is xn un = . n! Therefore, x n +1 u n +1 = (n + 1)!

(ii) 1 +

and so

Solution. The nth term of the given series is

x2n− 2

3

EXAMPLE 1.34

n(3 + 73 ) 1 1 . = = lim n →∞ n(3 + 3 ) x x n

1

1

n2 converges. Hence Sun also converges. Therefore, the given series converges for x2 < 1 and diverges for x2 < 1.

Therefore, n →∞

Thus, by D’Alembert’s Ratio Test, the given series converges if x12 > 1, that is if, x2 < 1, and diverges if 12 < 1 , that is, if x2 > 1. When x2 = 1, x the D’Alembert’s Ratio Test fails and we have 1 1 un = = 3 . (n + 1) n n 2 (1 + 1n )

.

lim

n →∞

u n +1 x = lim n →∞ un n +1

1/2/2012 11:33:50 AM

1.22  n  Chapter One = 0, less than 1, for all finite values of x. Hence, by D’Alembert’s Ratio Test, the given series converges for all finite values of x. (ii) We have un =

np , n!

u n +1 =

(n + 1) (n + 1)!

un n +1 = lim n →∞ u n →∞ n +1 (1 + 1n ) p

Hence, the given series converges for all values of p.

(ii)

( 13 ) + ( 13⋅⋅25 ) + ( 13⋅⋅25⋅⋅37 ) + ( 13⋅⋅25⋅⋅37⋅⋅49 )

(iii)

∑

2

2

2

2

+.

n! nn

un =

n 2 (n + 1) 2 (n + 1) 2 (n + 2) 2 , un +1 = n! (n + 1)!

and so

EXAMPLE 1.35

un n 2 (n + 1) 2 (n + 1)n ! = lim . n →∞ u n →∞ n! (n + 1) 2 (n + 2) 2 n +1 lim

Examine the convergence of the series n ∑ 2 xn , x > 0 n +1

n3 (1 + 1n ) = ∞. n →∞ n 2 (1 + 2 ) 2 n

= lim

Solution. For the given series

n n2 + 1

n +1

x , u n +1 = n

(n + 1) 2 + 1

x

Hence, the given series converges by D’Alembert’s Ratio Test. (ii) For the given series

n +1

2

and so un n n 2 + 2 + 2n 1 = lim ⋅ ⋅ n →∞ u n →∞ n +1 x n2 + 1 n +1 lim

= lim

n →∞

2 2 1 1 + n2 + n 1 1 ⋅ = 1 + 1n 1 + n12 x x

Hence, by D’Alembert’s Ratio Test, Sun converges if 1x > 1 , that is, if x < 1 and diverges if 1x < 1 , that is, if x > 1. When x = 1, this test gives no information. But in that case, un =

n 2 ( n +1)2 n!

Solution. (i) For the given series

= ∞ for all values of p.



∑

(i)

lim

un =

EXAMPLE 1.36

Test for the convergence of the series p

and so

given series diverges for x = 1. Hence, the given series converges if x < 1 and diverges if x > 1.

n n +1 2

n n 2 (1 +

=

1 n2

)

=

1 n

⋅

1 1+

1 n2

.

Taking vn = 1n we note that u 1 lim n = lim = 1, finite and non-zero. n →∞ v n →∞ 1 + 12 n n

Therefore, Sun and Svn converge or diverge together. But Σvn = Σ 1n diverges. Therefore, the

M01_Baburam_ISBN _C01.indd 22

 1.2.3.4 …..n  un =   ,  3.5.7.9....(2n + 1)  2

 1.2.3.4....(n + 1)  un +1 =   .  3.5.7.9....(2n + 3)  Therefore, (2 + 1n ) 2 u (2n + 3) 2 lim n = lim = lim = 2. 2 n →∞ u n →∞ ( n + 1) n →∞ (1 + 1 ) 2 n +1 n Hence, the given series converges D’Alembert’s Ratio Test. (iii) The nth term of the given series is n! un = n . n Therefore, (n + 1)! (n + 1)n ! n! u n +1 = = = n +1 n +1 (n + 1) (n + 1) (n + 1) n

by

and so un (n + 1) n  n + 1 = lim = n →∞ u n →∞  n  nn n +1

n

lim

1/2/2012 11:33:50 AM

SequenceS and SerieS   n 1.23 = (1 + 14 ) = e = 2.7 , greater than 1. Hence, the given series converges by D’Alembert’s Ratio Test. n

EXAMPLE 1.37

Test for the convergence of the series (i) x + 2 x 2 + 3 x 3 + 4 x 4 + 

n →∞

x x 2 x3 x4 +  (iv) 1 + 2 + 5 + 10 +  + 2 n +1 Solution. (i) The given series is Snxn . Therefore, un = nx , un +1 = (n + 1) x n

n +1

un nx n = lim n →∞ u n →∞ ( n + 1) x n +1 n +1 lim



= lim

n →∞

1 1 = . 1 (1 + n ) x x

Hence, by D’Alembert’s Ratio Test, the series converges if 1x > 1 , that is, if x < 1 and diverges if 1x < 1 , that is, if x > 1. If x = 1, the Ratio Test gives no information about convergence. But for x = 1, the series becomes Sn, which is divergent. Hence, the given series converges for x < 1 and diverges for x > 1. (ii) We have un =

xn x n +1 , u n +1 = (2n)! (2n + 2)!

Therefore, un x n (2n + 2)(2n + 1)(2n)! = lim ⋅ n →∞ u n →∞ (2n)! x n +1 n +1 lim

(2n + 2)(2n + 1) n →∞ x

= lim



n 2 (2 + n2 )(2 + 1n ) = ∞ >1. n →∞ x

= lim

Hence, by D’Alembert’s Ratio Test, the given series converges.

M01_Baburam_ISBN _C01.indd 23

and so lim

xn (ii) ∑ (2n)! n (iii) ∑ 2 x n , x > 0 n +1

and so

(iii) The nth term of the series is n un = 2 xn , x > 0 . n +1 Therefore, (n + 1) u n +1 = x n +1 (n + 1) 2 + 1 un n (n + 1) 2 + 1 = lim 2 xn . →∞ n u n +1 n + 1 (n + 1) x n +1 n( n 2 + 2 + 2n) 1 . n →∞ ( n + 1)( n 2 + 1) x

= lim



= lim

n →∞

(

)

n3 1 + n22 + n2

n (1 + 3

1 n

) (1 +

1 n2

)

1 1 = . x x

Hence, by D’Alembert’s Ratio Test, the series converges if 1x > 1 or x < 1 and diverges if 1 x < 1 or x > 1 . If x = 1 , the Ratio Test gives no information. But in that case, n n 1 . un = 2 = = n + 1 n 2 1 + 12 n 1 + 12

(

n

)

(

n

)

Take vn = . Then, 1 n

u

lim vnn = 1, finite and non-zero

n→∞

Hence, Σu n and Σv n converges or diverges together. But Σvn = Σ 1n diverges. Therefore Σu n also diverges. Hence, the given series converges for x < 1 and diverges for x ≥ 1 . (iv) Neglecting the first term, the nth term of the series is un =

x n +1 xn , . u = n + 1 (n + 1) 2 + 1 n2 + 1

Therefore, un x n n 2 + 2 + 2n = lim 2 . n →∞ u n →∞ n + 1 x n +1 n +1 lim

= lim . n →∞

(

n2 1 +

(

2 n2

n 1+ 2

+ 1 n2

2 n

)

).1 = 1 . x

x

Therefore, the series converges if > 1 or x < 1 and diverges for 1x < 1 or x > 1 . For x = 1 , the D’Alembert’s Ratio Test given no information. 1 x

1/2/2012 11:33:51 AM

1.24  n  Chapter One But in such a case, un =

Taking vn =

1 n2

1 1 = n2 + 1 n2 1 +

(



)

.

, we get

un 1 = lim 1 n →∞ v n →∞ 1 + n n2 lim

1 n2

if αβ < 1 or a > b > 0 . If a = b , then lim uun = 1 n →∞ n+1 and so the Ratio Test gives no information. But then the series becomes 1 + 1 + 1 +…,

(

which is divergent. Hence, the given series converges if b > a > 0 and diverges if a > b > 0. (ii) The given series is

)

= 1, finite and non –zero Hence, Sun and Svn converge or diverge together. But Σvn = Σ n12 is convergent. Hence, Sun converges for x = 1. Thus, the given series is convergent for x < 1 and diverges for x > 1. Examine the convergence for the following series of positive term.

lim

n →∞

4

6

(α + 1)(2α + 1) … (nα + 1) . ( β + 1)(2 β + 1) … (nβ + 1)

Therefore, un +1

(α + 1)(2α + 1) … (nα + 1)[(n + 1)α + 1] = ( β + 1)(2 β + 1) … (nβ + 1)[(n + 1) β + 1]

and so lim

n →∞

un (n + 1) β + 1 = lim un +1 n →∞ (n + 1)α + 1 (1 + 1n ) β + 1n β = . n →∞ (1 + 1 )α + 1 α n n

= lim

Hence, by D’Alembert’s Ratio Test, the series is β convergent if α > 1 or b > a > 0 and divergent

M01_Baburam_ISBN _C01.indd 24

4 ⋅12 ⋅ 20 … (8n − 4)(8n + 4) 18 ⋅ 27 ⋅ 36 … (9n + 9)(9n + 18)

un 9n + 18 = lim →∞ n u n +1 8n + 4 n →∞

(iii) 1 + 1 . x + 1.3.5 . x + 1.3.5.7.9 . x +… 2 4 2.4.6 8 2.4.6.8.10 12 Solution. (i) Without taking notice of the first term, the nth term of the given series is un =

4 ⋅12 ⋅ 20 … (8n − 4) . 18 ⋅ 27 ⋅ 36 … (9n + 9)

= lim

4 4.12 4.12.20 + + +… 18 18.27 18.27.36 2

un +1 =

and so

α + 1 (α + 1)(2α + 1) + + β + 1 (β + 1)(2β + 1) (α + 1)(2α + 1)(3α + 1) + +… ( β + 1)(2 β + 1)(3β + 1)

(i) 1 +

(ii)

un =

Therefore,

EXAMPLE 1.38



4 4.12 4.12.20 + + + 18 18.27 18.27.36 whose nth term is given by

n ( 9 + 18n ) n ( 8 + n4 )

=

9 > 1. 8

Hence, by D’Alembert’s Ratio Test, the given series converges. (iii) For the series (iii), we have 1.3.5… (4n − 7) x 2 n − 2 un = . , (4n − 6) (4n − 4) un +1 =

1.3.5… (4n − 7)(4n − 5)(4n − 3) x 2 n . . 2.4.6 … (4n − 6)(4n − 4)(4n − 2) 4n

Therefore, u (4n − 2)(4n − 4) 4n 1 lim n = lim . . n →∞ u n →∞ (4 n − 5)(4 n − 3) (4 n − 4) x 2 n +1 16n 2 − 8n 1 1 . = 2 n →∞ 16n 2 − 32n + 15 x 2 x

= lim

Hence, by D’Alembert Ratio Test, the series converges if

1 x2

> 1 or x2 < 1 and diverges if x2

> 1 . If x2 = 1, the test fails. But if x2 = 1, then 1 ⋅ 3 ⋅ 5… (4n − 7) un = 2 ⋅ 4.6 … (4n − 6)(4n − 4)

1/2/2012 11:33:52 AM

SequenceS and SerieS   n 1.25 =

1 ⋅ 3 ⋅ 5… n(4 − 7n ) . 2 ⋅ 4 ⋅ 6 … n 2 (4 − 6n )(4 − n4 )

1

lim(un ) n = lim

n →∞

Take vn = 2⋅4⋅6…n . Then uvn → A (finite and n nonzero). Hence, the series converges for x2 = 1. 1⋅3⋅5…n

Therefore, the series converges when x2 < 1 and diverges when x2 > 1. 1.11  CAUCHY’S ROOT TEST If Sun is a series of positive terms, then

n →∞

un +1 , un

provided the latter limit exists. It follows from this result that if D’Alembert. Ratio Test is applicable to any series, then Cauchy’s Root Test is also applicable. But the converse is not true. To show that Cauchy’s Root Test is more powerful than D’Alembert’s Ratio Test, consider the series with nth term un defined b n

1 n

(i) the series Sun converges if lim(un ) < 1. n →∞

un = 2− n − ( −1) . Then 1

1 n

un ) > 1. (ii) the series Sun diverges if nlim( →∞

lim(u n ) n =

n →∞

1

(When lim(un ) n = 1, the root test fails to give n →∞ any information regarding convergence of the series Sun ). 1 Proof: Suppose that lim(un ) n = r. Then, by n →∞ definition, to each e > 0, there exists a positive integer m such that 1 | (un ) n − r |< ε for all n > m or r − ε < (un ) n < r + ε for all n > m 1

or

1 2

Hence, by Cauchy’s Root Test, the series Sun converges. On the other hand n n +1 un +1 2 if n is even = 2−1+ ( −1) − ( −1) =  1 un  23 if n is odd.

Thus,

lim

un+1 un

= 2 and lim

un+1 un

= 18 .

Therefore,

D’Alembet’s Ratio Test yields no definite result. (ii) The two tests are, of course, completely un+1 equivalent if, lim un exists. n →∞

(r − ε ) n < u n < (r + ε ) n . Case (i). If r < 1, choose e > 0 such that r + e < 1. Taking r + e = t < 1, we have un < tn for all n > m. But the geometric series, S tn,t < 1 is convergent. Hence, by Comparison Test, Sun converges.

EXAMPLE 1.39

Examine the convergence of the series 2 32 2 43 3 x + x + 4 x +… 12 23 2



Solution. The nth term of the given series is

Case (ii). If r > 1, choose e > 0 such that, r – e > 1. Taking r – e = T > 1, we have un < Tn for all n > m. But the geometric series STn, T > 1 is divergent. Hence, by Comparison Test, the series Sun diverges. Remark 1.8. (i) If un is positive for all values of n, then, by Cauchy’s Limit Theorem,

M01_Baburam_ISBN _C01.indd 25

un = Therefore, 1

(n + 1) n n n + 1 n xn x =( ) ⋅ . n +1 n n n

(1 + 1n ) = x. n +1 x . 1 = lim 1 n →∞ n n n n →∞ n n

lim unn = lim

n →∞

Hence, by Cauchy’s Root Test, the given series converges if x < 1 and diverges if x > 1. When x = 1, we have

1/2/2012 11:33:52 AM

1.26  n  Chapter One EXAMPLE 1.41

n

 n +1 1 un =   ⋅ .  n  n 1 Take vn = n . Then,

Examine the convergence of the series

n

un  1 = lim  1 +  = e (finite and non-zero). n →∞ v n →∞  n n lim

But Σvn = Σ 1n is divergent. So Sun diverges for x = 1. Hence, the given series converges if x < 1 and diverges if x > 1. EXAMPLE 1.40

Examine the convergence of the following series (i)

2 (n + 1)n ∑ 2 n n .3n

(ii) 1 + 2 x + ( 3 )2 x2 + ( 4 )2 x3 2



3

4

(i) (ii)

2 3 1 + x + x + x + … ( x > 0) 2 32 43

∑ (1 + ) 1 n

3

−n2

Solution. (i). Neglecting the first term, the nth term of the given series is

u = n

Therefore,

Solution. (i) The nth term for the series is

(n + .1) n . un = 2 n n 3n

n →∞

n

(n + 1) n 1 1  1 = lim 1 +  = e < 1. n →∞ n →∞ 3 n 3 3n n  

lim unn = lim

Hence, by Cauchy’s Root Test, the given series is convergent. (ii) Ignoring the first term, the nth term of the series is n  n +1  n un =   x . n+2 Therefore, 1 1 + 1n  n +1  lim unn = lim  x = lim = x = x.  n →∞ n →∞ n + 2 n →∞ 1 + n2   Hence, the series converges if x < 1 and diverges if x > 1. If x = 1, then

n (1 + 1n ) = (1 + 1n )  n +1  un =  n  = n n+2 (1 + n2 ) [(1 + n2 ) 2 ]2 n

and so lim un = n →∞

e e2

n

un ≠ 0, the = 1e ≠ 0. Since nlim →∞

Hence, the given series is convergent by Cauchy’s Root Test. (ii) For this series,   u = 1 + 1  n  n

M01_Baburam_ISBN _C01.indd 26

3 −n 2

1

=

 1  1 +  n

n .n

.

Therefore, 1

lim unn = lim

n →∞

n →∞

1

(1 + )

n

1 n

=

1 < 1. e

Hence, the given series converges by Cauchy’s Root Test. EXAMPLE 1.42 Test the convergence of the series  n  ∑  n + 1  Solution. For the given series,

series diverges for x = 1. Thus the given series converges for x < 1 and diverges for x > 1.

n   

x = 0 < 1. n (1 + 1n )

= lim

2

n →∞

1

 xn = lim  n →∞  n 1 n  n (1 + n )

5

()

1

.

(n + 1)n

1 1  xn  n lim (u ) n = lim   n→∞ n n → ∞  (n + 1)n 

4 + 5 x 4 + … for x > 0. 6

Therefore,

xn

n2

n2

Therefore,

 n  un =   .  n +1

1/2/2012 11:33:53 AM

SequenceS and SerieS   n 1.27 n

and so

p ( p − 1) 1 p ( p − 1)( p − 2 ) 1   lim  p + . + . 2 + … 2! n 2! n  

n

1 1  n   1  unn =   =  = n  n + 1   1 + 1n  + 1 ( 1n )

1

1

lim unn = lim

n →∞

n →∞

(1 + 1n )

n

=

n →∞

or if

1 < 1. e

Hence, by Cauchy’s Root Test, the given series converges. 1.12  RAABE’S TEST Let Sun be a series of positive terms, then

( diverges if lim n (

n (i) Sun converges if nlim →∞

un un+1

(ii) Sun

un un+1

(This lim n n →∞

(

n →∞

test gives un − 1 = 1 ). un+1

)

(

un un+1

no

) − 1) < 1.

n →∞

)

p

un  n +1  1 >  = 1 +  un +1  n   n



= 1+ +

or if

p

p p ( p − 1) 1 + ⋅ 2 n 2! n

p ( p − 1)( p − 2 ) 1 . 3 +… 3! n

p ( p − 1) 1  u  ⋅ n  n − 1 > p + 2! n  un +1  or if

p ( p − 1)( p − 2 ) 1 + . 2 +… 2! n

 u  lim n  n − 1 > n →∞  un +1 

M01_Baburam_ISBN _C01.indd 27

)

To prove (ii), we use the fact that ∑ vn = ∑ n1p diverges if p < 1. Then, Sun will also diverge if un v < n un +1 vn +1 or if

information

un v > n un +1 vn +1

or if

(

−1 > 1

un p p ( p − 1) 1 < 1+ + . 2 2! un +1 n n

if

− 1 = k . If k > 1, choose a number p so that k > p > 1. Now compare the series Sun with the auxiliary series Svn, where vn = n1p . Then Svn is convergent if p > 1. Therefore, Sun is convergent if after some particular term Proof: Let lim n

 u  lim n  n − 1 > p > 1.  un +1  u n unn+1 − 1 > p > 1. Thus, Sun converges if nlim →∞ n →∞



+

or if



p ( p − 1)( p − 2 ) 1 . 3 + 3! n

p ( p − 1) 1  u  . n  n − 1 < p + 2! n  un +1  +

or if

p ( p − 1)( p − 2 ) 1 . 2 + 3! n

 u  lim n  n − 1 < p ≤ 1.  un +1 

n →∞

 un  − 1 < 1.  un +1 

n Thus, the series Su n diverges if nlim →∞

EXAMPLE 1.43

Examine the convergence of the series 1 1⋅ 3 1⋅ 3 ⋅ 5 1 ⋅ 3 ⋅ 5… (2n − 1) + + + + + 4 4 ⋅ 6 4 ⋅ 6 ⋅8 4 ⋅ 6 ⋅ 8… (2n + 2) Solution. The nth term of the given series is

un =

1 ⋅ 3 ⋅ 5… (2n − 1) . 4 ⋅ 6 ⋅ 8… (2n + 2)

Therefore,

un +1 =

1 ⋅ 3 ⋅ 5… (2n + 1) 4 ⋅ 6 ⋅ 8… (2n + 4)

1/2/2012 11:33:54 AM

1.28  n  Chapter One and so

1 1 1  log n + n − 2 n2 + …   =  x  log n + n2 − 1n ( n2 )2 + …   

u (2n + 4) lim n = lim = 1. n →∞ u n →∞ (2n + 1) n +1

Hence, D’Alembert’s Ratio Test is not applicable. Further,

( 2n + 4 ) un = un +1 ( 2n + 1)  u   2n + 4  n  n − 1 = n  − 1 u  2n + 1   n +1   2n + 4 − (2n + 1)  = n  2n + 1  



=

Therefore,

q

1 1 1 1+ n log n − 2 n2 log n + …   . =  2 +…  x  1+ n log2 n − n2 log n  

Hence lim uun = 1x and, by D’Alembert’s Ratio n →∞ n+1 1 Test, the series Sun converges if x > 1 , that is, if 1 x < 1 and diverges if x < 1 , that is, if x > 1. For x = 1, the Ratio Test gives no information. But for x = 1, we have

implies



1+ n log1 n − 2 n2 1log n +…  un  = 2 +…  un +1  1+ n log2 n − n2 log n  

3n 3n . = 2n + 1 n ( 2 + 1n )

1 = 1+ n log − 2 n2 1log n + … n  

 u  3n 3 lim n  n − 1 = lim = > 1. 1 →∞ n →∞ n n (2 + n ) 2  un +1  Hence, by Raabe’s Test, the given series converges.



EXAMPLE 1.44



Examine the following series for convergence (i) x (log 2) + x (log 3) +x (log 4) +… 2

(ii)

q

3

q

q

4

3 5 7 x + 1 ⋅ x + 1⋅ 3 x + 1⋅ 3 ⋅ 5 ⋅ x +  ⋅ 2 3 2⋅ 4 5 2⋅ 4⋅6 7

Solution. (i). The nth term of the given series is

un = x Therefore,

n +1

[log (n + 1)] . q

2 × 1+ n log2 n − n2 log +… n  

= 1+q 



(

1 n log n

× 1+q  = 1+q



(

(

− 2 n2 1log n + … + … 

2 n log n

1 n log n

)

−q

)

− 2 n2 2log n + … + … 

)

− n log2 n + … + …,

which yields un un+1

q − 1 = − n log + term containing higher powers of n

n and log n in the denominator or n

(

un un+1

)

−1 =

−q log n

+

Therefore,  u  lim n  n − 1 = 0 < 1.  un +1 

q

n →∞

Hence, by Raabe’s Test, the given series diverges for x = 1. It follows that the given series converges for x < 1 and diverges for x > 1.

q

1  log n + log (1 + 1n )  =   x  log n + log (1 + n2 ) 

M01_Baburam_ISBN _C01.indd 28

q

denominator.

and so

1  log n (1 + 1n )  =   x  log n (1 + n2 ) 

q

term containing powers of n and log n in the

un +1 = x n + 2 [ log (n + 2)]q un 1  log (n + 1)  =   un +1 x  log (n + 2) 

q

q

(ii) Neglecting the first term, the nth term of the given series is un =

1.3.5… (2n − 1) x 2 n +1 . . 2.4.6 … (2n) (2n + 1)

1/2/2012 11:33:55 AM

SequenceS and SerieS   n 1.29 Thus

Therefore, un +1 =

2n+3

1.3.5… (2n − 1)(2n + 1) x . 2.4.6 … (2n)(2n + 2) 2n + 3

and so un (2n + 2)(2n + 3) 1 = ⋅ un +1 (2n + 1)(2n + 1) x 2 =

2n (1 +

2n (1 +

1 n

) 2n (1 + ) 1 . ⋅ ) 2n (1 + 21n ) x 2 3 2n

1 2n

Therefore, lim

n →∞

(1 + )(1 + ) . 1 = 1 . un = lim →∞ n un +1 (1 + 21n )(1 + 21n ) x 2 x 2 1 n

3 2n

Hence, by D’Alembert’s Raito Test, the series Sun converges if x12 > 1, that is, if x2 < 1, and diverges 1 if x2 < 1, , that is, if x2 > 1. If x2 = 1 the Ratio Test gives no information. But for x2 = 1, we have ( 2n + 2 )( 2n + 3) 4n 2 + 10n + 6 un = = . un +1 ( 2n + 1)( 2n + 1) 4n 2 + 4n + 1 Thus  u   4n 2 + 10n + 6  n  n − 1 = n  − 1 2  4n + 4n + 1   un +1  6 n 2 + 5n = 2 4n + 4n + 1

and so

 u  6n 2 + 5n lim n  n − 1 = lim 2 n →∞  un +1  n →∞ 4n + 4n + 1 n 2 ( 6 + 5n ) 3 = lim 2 = > 1. n →∞ n 4 + 4 + 1 2 n n2

Therefore, by Raabe’s Test, the given series converges for x2 = 1. Hence, Sun converges for x2 < 1 and diverges for x2 > 1. EXAMPLE 1.45

Examine the convergence of the series 1 1⋅ 3 2 1⋅ 3 ⋅ 5 3 1+ x + x + x + 2 2⋅ 4 2⋅ 4⋅6 Solution. The nth term of the series is un =

M01_Baburam_ISBN _C01.indd 29

1 ⋅ 3 ⋅ 5… (2n − 1) n x . 2 ⋅ 4 ⋅ 6 … (2n)

and so

un +1 =

lim

n →∞

1 ⋅ 3 ⋅ 5… (2n − 1)(2n + 1) n +1 x 2 ⋅ 4 ⋅ 6 … (2n)(2n + 2)

un 2n + 2 1 1 = lim ⋅ = . →∞ n 2n + 1 x x un +1

By D’Alembert’s Ratio Test, the series converges if 1x > 1, that is if x < 1 and diverges if 1x < 1, that is if x > 1. The test fails to give any information if x = 1. However, for x = 1, we have un 2n + 2 = un +1 2n + 1

or

 u  n 1 n  n − 1 = . = 1 u n 2 1 2 + +  n +1  n Therefore,  u  1 1 lim n  n − 1 = lim = 1. EXAMPLE 1.46

Use Raabe’s Test to examine the convergence of series of Example 1.36 (iii) for the case x2 = 1. Solution. The series in Example 1.36 (iii) was 1 x2 1⋅ 3 ⋅ 5 4 1⋅ 3 ⋅ 5 ⋅ 7 ⋅ 9 x 6 1+ ⋅ + x + ⋅ + 2 4 2⋅ 4 ⋅6 ⋅8 2 ⋅ 4 ⋅ 6 ⋅ 8 ⋅10 12

The nth term is un =

1⋅ 3 ⋅ 5 ⋅⋅⋅ (4 n − 7) x 2 n − 2 ⋅ . 2 ⋅ 4 ⋅ 6 ⋅⋅⋅ (4 n − 6) 4 n − 4

Then un 16 n 2 − 8n 1 = ⋅ . u n +1 16 n 2 − 32n + 15 x 2

By D’Alembert’s Ratio Test, the series converges for x2 < 1 and diverges for x2 > 1. For x2 = 1, the Ratio Test gives no information. In this case, we have u n +1 =

16 n 2 − 8n 16 n 2 − 32n + 15

1/2/2012 11:33:56 AM

1.30  n  Chapter One and so

Solution. The nth term of the series is

 u  24n − 15   n  n − 1 = n  . 2 u n n 16 32 15 − +    n +1  Therefore,

and so the given series converges for x2 = 1. Hence, the series converges if x2 < 1 and diverges if x2 > 1. EXAMPLE 1.47

Examine the convergence of the series 22 x 2 33 x 3 4 4 x 4 + + + . 2! 3! 4! Solution. The nth term of the series is x+

and so lim

n →∞

Therefore, by D’Alembert’s Ratio Test, the 1 series converges if x > 1, that is, if x < 1 and diverges if x1 < 1, that is, if x > 1. If x = 1, then Ratio Test gives no information. However, for x = 1, we have

and so

un n+2 = u n +1 n  u  n+2  2 n  n − 1 = n  − 1 = n   = 2. u n   n  n +1 

Therefore, u n +1 =

( n + 1) n +1 x n +1 ( n + 1)!

Therefore,  u  lim n  n − 1 = 2  un +1 

n →∞

n

un nn x n ( n + 1).!  n  1 = . =  ⋅ . u n +1 n ! ( n + 1) n +1 x n +1  n + 1 x

Hence, n

 1  1 1 1 u ⋅ = ⋅ ⋅ lim n = lim  n →∞ u n →∞  1 + 1  x e x n +1 n

By D’Alembert’s Ratio Test, the series converges if

1 e

. x1 > 1, that is, if x < e1 , and

1 1 1 diverges if e ⋅ x < 1, that is, if x > e . Since, this question involves e, we do not apply Raabe’s Test. We shall solve it by another test, known as Logarithmic Test (see Example 1.53).

EXAMPLE 1.48

Test the convergence of the series 2

3

M01_Baburam_ISBN _C01.indd 30

and so the series converges by Raabe’s Test for x = 1. Hence, the given series converges if x ≤ 1 and diverges if x > 1. EXAMPLE 1.49

Test the convergence of the series 1+

α α (α + 1) 2 α (α + 1)(α + 2) 3 x+ x + x + β β (β + 1) β (β + 1)(β + 2)

Solution. Neglecting the first term, the nth term of the given series is

un =

α (α + 1)(α + 2) ⋅ . ⋅ . ⋅ .(α + ( n − 1)) n x . β (β + 1)(β + 2)(β + ( n − 1))

Therefore,

4

x x x x + + + + . 1⋅ 2 2 ⋅ 3 3 ⋅ 4 4.5

x n +1 (n + 1)(n + 2)

un n+2 1 1 = lim . = . →∞ n u n +1 n x x

or

nn x n . n!

xn . n ( n + 1)

Therefore, u n +1 =

 u  3 lim n  n − 1 = > 1 n →∞  un +1  2

un =

un =

u n +1 =

α (α + 1)(α + 2) ⋅⋅⋅ (α + ( n − 1))(α + n ) n +1 x β (β + 1)(β + 2) ⋅⋅⋅ (β + ( n − 1))(β + n )

1/2/2012 11:33:57 AM

SequenceS and SerieS   n 1.31 and

un =

un β+n 1 = lim . n →∞ u n →∞ α + n x n +1 lim

= lim

n →∞

(

n 1+

β n

)⋅1 = 1.

n (1 + αn ) x

Therefore, lim

n →∞

x

Thus, by D’Alembert’s Ratio Test, the series converges if x1 > 1, that is, if x < 1 and diverges 1 if x < 1 that is, if x > 1. For x = 1, the Ratio Test gives no infomiation. But for x = 1, we have un β+n = . u n +1 α + n Therefore,

and so  u  lim n  n − 1 = β − α . n →∞  un +1  Thus, by Raabe’s Test, the series converges for b – a > 1 and diverges for b – a < 1, when x = 1. If b – a = 1 then the series becomes 1+

α α (α + 1) α (α + 2) + + + 1 + α (1 + α )(2 + α ) (1 + α )(2 + α )(3 + α )

or 1+

α

α

α

+ + + , 1+ α 2 + α 3 + α which is divergent. Hence, the given series converges for x < 1 and diverges for x > 1. If x = 1, then the series in question converges for b – a > 1 and diverges for b – a < 1. EXAMPLE 1.50

Test for convergence of the series 3 3.6 2 3.6.9 3 x+ x + x + . 7 7.10 7.10.13 Solution. Ignoring the first term, the nth term of the given series is 1+

M01_Baburam_ISBN _C01.indd 31

un 3n + 7 1 1 = lim ⋅ = →∞ n u n +1 3n + 4 x x

and so, by D’Alembert’s Ratio Test, the series converges if x1 > 1, that is, if x < 1 and diverges if 1 < 1, that is, if x > 1. If x = 1, the test fails. But x for x = 1, we get un 3n + 7 = un +1 3n + 3 and thus  u   3n + 7  lim n  n − 1 = lim n  − 1 n →∞ n →∞ u  3n + 3   n +1 

 u   β +1  n  n − 1 = n  −1  α + n   u n +1 

 β −α   β −α  = n   = n  α  α +n   n (1 + n ) 

3.6.9 …3n xn. 7.10.13…(3n + 4)

 4  4 = lim n   = > 1. n →∞  3n + 3  3 Therefore, the series converges for x = 1. Hence, the given series converges for x ≤ 1 and diverges for x > 1. 1.13  LOGARITHMIC TEST A series Σun of positive terms converges or diverges according as u lim n log n > 1 or < 1. n →∞ u n +1 un

n log u (This Test gives no information if lim n →∞

n +1

= 1).

Proof: Consider the auxiliary series Σν n = Σ n1p ,

which converges if p > 1 and diverges if p < 1. Then, p p vn  n + 1  1 =  = 1 +  . v n +1  n  n If p > 1, then Σνn is convergent. The series Σun will be convergent if un v > n un +1 vn +1 or if un  1 > 1 +  u n +1  n 

p

1/2/2012 11:33:58 AM

1.32  n  Chapter One or if

Therefore, p

u 1   1 log n > log  (1 +  = p log 1 +    n u n +1 n

u n +1 =

so that un n! ( n + 2) n +1 xn ⋅ = n u n +1 ( n + 1) ( n + 1)! x n +1

or if log

un 1 1 1  > p  − 2 + 3 −  u n +1 3n  n 2n 

or if n log

un 1 1   > p 1 − + 2 −  u n +1 n 2 n 3  

n →∞

=

( n + 2) n +1 1 ⋅ . ( n + 1) n +1 x

=

( n + 2) n ( n + 2)1 (1 + n2 ) (1 + n2 ) 1 = ⋅ . ( n + 1) n ( n + 1)x (1 + n1 )n (1 + n1 ) x n

or if lim n log

(n + 1)! n +1 x (n + 2) n +1

un > p > 1. u n +1

Thus

(1 + n2 ) ⋅ (1 + n2 ) ⋅ 1 = e 2 ⋅ 1 = e un = lim n →∞ u n →∞ n +1 (1 + n1 )n (1 + n1 ) x e x x n

If p < 1, then Σvn diverges. The series Σun will diverge if un ν < n u n +1 ν n +1 or if

lim

and so the series converges if ex > 1, that is, if x < e and diverges if ex < 1, that is, if x > e. If x = e, u = 1 and so the D’Alembert’s Ratio then lim n →∞ u Test gives no information about convergence. But when x = e, then n

n +1

p

log

un  1  1 < log 1 +  = p log 1 +   n  n u n +1

(1 + n2 ) ⋅ 1 . un = u n +1 (1 + n1 )n +1 e n +1

or if log

un 1 1 1  < p  − 2 + 3 −…. u n +1 3n  n 2n 

un

n +1

or if lim n log n →∞

un 1 1   < lim p 1 − + 2 −  < p < 1. →∞ n u n +1  2n 3n  un

n log u Thus, the series Σun converges if lim n →∞

log and diverges if lim n →∞

un u n +1

< 1.

n +1

>1

Remark 1.9. The Logarithmic Test should be applied after the failure of D’Alembert’s Ratio Test and generally when the ratio uu involves ‘e.’ n

n +1

EXAMPLE 1.51

Test for convergence of the series 3! 4! x 2! 1 + + 2 x 2 + 3 x 3 + 4 x 4 + . 2 3 4 5 Solution. Ignoring the first term, the nth term of the given series is n! un = xn. ( n + 1) n

M01_Baburam_ISBN _C01.indd 32

Since, the expression for u involves e, we do no apply Raabe’s Test and so Logarithmic Test is preferred. Thus, u  2 log n = ( n + 1) log 1 +   n u n +1

 1 −( n + 1) log 1 +  − log e  n

  2  1  = ( n + 1) log 1 +  − log 1 +   − 1    n  n  2 1 4 1 8 = ( n + 1)  − ⋅ 2 + ⋅ 3 −  3 n n 2 n 1 1 1  −  − 2 + 3 +   − 1  n 2n  3n

3 1  = ( n + 1)  − 2 +  − 1 n 2 n   3 1 3 = 1− + − + −1 2n n 2n 2

1/2/2012 11:33:59 AM

SequenceS and SerieS   n 1.33

and so

=−

Hence, by Log Test, the given series diverges for all values of p.

1 3 − + 2n 2n 2

3  1  u lim n log unn+1 = lim n  − − 2 + … n →∞ n →∞ 2 n 2 n   1 = − < 1. 2 Hence, by Logarithmic Test, the series diverges. It follows, therefore, that the given series converges if x < e and diverges if x > e .

EXAMPLE 1.53

Examine the convergence of the series 22 x 2 33 x 2 4 4 x 4 + + + . 2! 3! 4! Solution. As shown is Example 1.47 the given series converges for x < e1 and diverges for x > e1 . If x = e1 , then x+

un e = u n +1 (1 + n1 )n

EXAMPLE 1.52

Test for convergence of the series

and so

 1 u log unn+1 = loge − n log 1 +   n

1 1 1 + ++ + p p ( log 2) ( log 3) ( log n ) p

1 1 1  = 1 − n  − 2 + 3 +   n 2n  3n

Solution. The nth term of the series is

1 . un = ( log n ) p

=

Therefore, un  log ( n + 1)  = u n +1  log n 

Thus

p

 log n + log (1 + 1n )  =  log n  

lim n log n →∞

p

 log n + n1 − 21n2 + 31n2 − ⋅⋅⋅  =  log n  

un

1.14  DE MORGAN–BERIRAND TEST p

= 1. Thus the D’Alembert’s Ratio and so lim n →∞ u Test fails. We have   u 1 1 log n = p log 1 + − 2 +  u n +1 n n log n n 2 log   n +1

 1  = p +   n log n 

u  1  lim n log n = lim p  +  →∞ nn →∞ n u n +1  log n  = 0, for all values of p.

M01_Baburam_ISBN _C01.indd 33

un 1 1  1 = lim  − +  = < 1,  2 u n +1 n →∞  2 3n

and so, by Log Test, the given series diverges, for x = e1 . Hence, the given series converges if x < e1 and diverges if x > e1 .

p

  1 1 = 1 + − 2 +   n log n 2n log n 

and so

1 1 − + . 2n 3n 2

The series Σun of positive terms converges or diverges according to    u   lim log n n  n − 1 − 1 > 1 or < 1. n →∞   un +1    Proof: Consider the Auxiliary Series Σνn=

∑

1 n ( log n ) p

, which is convergent when p > 1. Then

vn ( n + 1)[log ( n + 1)]p = v n +1 n (log n ) p

 1   log n + log (1 + = 1 +    n   log n

1 n

)

p

 

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1.34  n  Chapter One 1 1 1  1   = 1 +  1 + − 2 +     n   log n  n 2n

p

1 p 1  1   = 1 +  1 + +    −  n   log n  n 2n 2 = 1+

1 p + + . n n log n

Solution. The nth term of the series is

un = and so

12 ⋅ 32 ⋅ 52  (2n − 1)2 n −1 x 22 ⋅ 4 2  (2n )2 2

un  2n + 2  1 1 = lim   ⋅ = . n →∞ u n →∞  2n + 1  x x n +1

lim

But, by Comparison Test, the series Σun will converge if

Thus, the series converges for x1 < 1, that is, if x < 1 and diverges when x > 1. When x = 1, the D’Alembert’s Ratio Test fails. For x = 1, we have

un v > n , u n +1 v n +1

un  2n + 2  =  u n +1  2 n + 1 

that is, if un p 1 > 1+ + + u n +1 n n log n or if un p 1 −1 > + + u n +1 n n log n or if  u  p n  n − 1 > 1 + + log n  u n +1 

or or

un 4n 2 + 8n + 4 4n + 3 −1 = −1 = 2 2 u n +1 4n + 4n + 1 4n + 4n + 1

 u  4n 2 + 3n lim n  n − 1 = lim 2 = 1. n →∞  un +1  n →∞ 4n + 4n + 1 Thus Raabe’s Test also fails. Now  u  4n 2 + 3n n  n − 1 − 1 = 2 −1 4n + 4n + 1  un +1  =

or if   u   log n  n  n − 1 − 1 > p +    u n +1  

or if   u   lim log n  n  n − 1 − 1 > p > 1. n →∞   u n +1  

Similarly, it could be proved that the series diverges if   u   lim log n  n  n − 1 − 1 < 1. n →∞   u n +1   EXAMPLE 1.54

Examine the convergence of the series 12 12 ⋅ 32 12 ⋅ 32 ⋅ 52 2 + x + x + 22 22 ⋅ 4 2 22 ⋅ 4 2 ⋅ 6 2

M01_Baburam_ISBN _C01.indd 34

2

−n − 1 . 4n + 4n + 1 2

Therefore,   u   lim log n  n  n − 1 − 1 n →∞   un +1   −n − 1 n log n . n 4n 2 + 4n + 1 2 −n − n log n = lim 2 . n →∞ 4 n + 4 n + 1 n 1 = − (0) = 0 < 1, n = lim

n →∞

and so, by de Morgan–Bertrand Test, the series diverges for x = 1. Hence, the series converges for x < 1 and diverges for x > 1. e 1.15 GAUSS’S TEST Let Σun be a series of positive terms. If

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SequenceS and SerieS   n 1.35 un λ  1  = 1 + + O  p +1  , p > 1, n  u n +1 n then the series Σun converges if l > 1, and diverges if l < 1.

un (γ + n)(n + 1) 1 = ⋅ un +1 (α + n)( β + n) x

and so

(

Proof: We have

un λ  1  and so = 1 + + O  p +1  , p > 1 n  u n +1 n  u   1 n  n − 1 = λ + O  p n  un +1 

 , p > 1 

Thus

diverges if observe that

1 x

(

+

α ⋅β α (α . + 1) β ( B + 1) 3 x+ x 1⋅ γ 1 ⋅ 2 ⋅ γ (γ + 1)

α (α + .1)(α + 2) β ( β + 1)( β + 2) 3 ⋅ x + . 1 ⋅ 2 ⋅ 3 ⋅ γ (γ + 1)(γ + 2)

Solution. Since a + n is positive after a certain stage,

we can assume that a, b, γ are positive. Then ignoring the firstterm, the nth term of the series is un =

α (α + 1) ⋅⋅⋅ (α + n − 1) β ( β + 1) ( β + n − 1) n x . 1 ⋅ 2 ⋅ 3 ⋅ n ⋅ γ (γ + 1) (γ + n − 1)

Therefore,

M01_Baburam_ISBN _C01.indd 35

)

(

) −1

 γ  1  α   β  = 1 + 1 + 1 +  1 +  n  n  n  n 

−1

 1 γ γ  = 1 + + + 2   n n n   α β αβ  × 1 − − + 2 +   n n n  

 log n  = lim  p  → 0 as n → ∞, n →∞  n 

1+

1 x

1 + nγ (1 + 1n ) un = un +1 (1 + αn ) 1 + nβ

  u   lim  n  n − 1 − 1 log n n →∞   un +1  

Discuss the convergence of the Hypergeometric Series

)

> 1 , that is, if x < 1 and < 1 , that is, if x > 1. For x = 1, we

Sun converges if

and by Raabe’s Test, the series Σun converges if l > 1 and diverges if l < 1. When l = 1, then

EXAMPLE 1.55

(

Hence, by D’Alembert’s Ratio Test, the series

 u  lim n  n − 1 = λ n →∞  un +1 

by L’Hospital Rule. Hence, by de Morgan– Bertrand Test, the given series diverges. Thus, the given series converges for l > 1 and diverges for le > x1.

)

1 + nγ (1 + 1n ) 1 1 un = lim ⋅ = n →∞ u n →∞ (1 + αn ) 1 + nβ x x n +1 lim

1  1 = 1 + (1 − γ − α − β ) + O  2 n n

 .  Therefore, by Gauss’s Test, Sun converges if l – γ – a – b > 1, that is, if γ > a + b and diverges if γ < a + b. It follows, therefore, that the Hypergeometric Series converges if x < 1 and diverges if x > 1. For x = 1, the series converges if γ > a + b and diverges if γ < a + b. EXAMPLE 1.56

Discuss the convergence of the Harmonic Series Σ n1 using Gauss’s Test. p

Solution. The nth term of the given series is

un =

1 np

. Therefore, p

p

un p  n + 1  1  1 =  = 1 +  = 1 + + O  2  u n +1  n  n n n

Hence, by Gauss’s Test, the Harmonic Series converges if p > 1 and diverges if p < 1.

1/2/2012 11:34:01 AM

1.36  n  Chapter One and so the expression (1) reduces to

EXAMPLE 1.57

Discuss the convergence of (i) 1 +

f (n) > I n +1 − I n > f (n + 1) .

3 3⋅ 6 3⋅ 6 ⋅ 9 + + + . 7 7 ⋅10 7 ⋅10 ⋅13

12 12 ⋅ 32 12 ⋅ 32 ⋅ 52 (ii) 2 + 2 2 + 2 2 2 +  . 2 2 ⋅4 2 ⋅4 ⋅6 Solution. (i) The nth term is

un =

Hence, by Gauss’s Test, the series Σun converges. (ii) For this series 12.32...(2n. − 1) 2 un = 2 2 2 . 2 .4 6 ...(2n) 2 Therefore,

and so the series diverges by Gauss’s Test. 1.16   CAUCHY’S INTEGRAL TEST Let f be a non-negative monotonic decreasing function of x > 1 and let f(n) = un for all positive integral values of n. Then, the series Sun and the ∞

converge or diverge together.

∫ f ( x)dx 1

Proof: Since f is monotonic in any interval

1 < x < ∞ , it is integrable in that interval. If x lies between n and (n + 1), we have f (n) > f ( x) > f (n + 1) . Integrating, we have

∫ n

(2)

Therefore, un −1 > I n − I n −1 > un , un − 2 > I n −1 − I n − 2 > un −1 , u1 > I 2 − I1 > u2

u1 + u2 +  + un −1 > I n − I1 > u2 + u3 +  + un

or

S n − un > I n − I1 > S n − u1 ,

where S n = u1 + u2 + …+ un is the partial sum of the series Sun. But, 1

un (2n + 2) 2 4n 2 + 8n + 4 1  1 = = = 1+ + O  2  n  un +1 (2n + 1) 2 n 4n 2 + 4n + 1



u n > I n +1 − I n > u n +1 .

Adding, we have

un 3n + 7 4  1 = = 1+ +O 2  . n  un +1 3n + 3 3n

n +1



3 ⋅ 6 ⋅ 9 (3n) 7 ⋅10 ⋅13 (3n + 4)

Therefore,

integral

Since f (n) = un , f (n + 1) = un +1 , we get

f (n)dx >

n +1

∫

f ( x)dx >

n

n

n +1

∫ n

f (n + 1)dx. (1)

Let I n = ∫ f ( x)dx. Then 1

I n +1 − I n =

1

Therefore, or

S n − un > I n > S n − u1 u1 > S n − I n > un > 0 .

Thus, the sequence (Sn – In) is bounded. Also, (Sn – In) – (Sn+1 – In+1) = In+1– In – un+1> 0 by (2) Hence, the sequence (Sn – In) is monotonically decreasing and bounded and so converges to a definite limit between 0 and u1. Therefore Sn and In converge or diverge together. Consequently, the series Sun and the integral converge or diverge together.

∞

∫ f ( x)dx 1

Remark. 1.10 (a) It is important to observe that whether Sn is convergent or divergent, the sequence Sn – In is always convergent and the limit lies between 0 and u1. 1.10 (b) Setting f (n) = 1n , we have

n +1

∫ n

M01_Baburam_ISBN _C01.indd 36

I1 = ∫ f ( x)dx = 0.

f ( x)dx

S n = u1 + u2 +  + un = 1 +

1 1 1 + ++ , 2 3 n

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SequenceS and SerieS   n 1.37 ∞

n

1 I n = ∫ dx = [log x]1n = log n. x 1

Thus

Hence, S n − I n = 1 + 12 + …+ 1n − log n tends to a finite limit g, called Euler’s Constant, whose value 0.55721 (less than u1 = 1) is correct to five places of decimals.

∫ f ( x)dx 1

diverges and so, by Cauchy’s

Integral Test, Sun diverges for p < 1. If p = 1, then ∞ 1 ∫1 f ( x)dx = ∫1 x dx

EXAMPLE 1.58

= [ log x]1∞

Apply Cauchy’s Integral Test to show that the series ∑ n1p converges if p > 1, and diverges if 0 < p < 1.

= ∞ − log1 = ∞−0 =∞

Solution. The nth term of the series is

∞

1 un = p . n As per Cauchy’s Integral Test, we have

Thus, ∫ f ( x)dx diverges in this case and so, by 1 Cauchy’s Integral Test, Sun diverges for p = 1 Hence, Σun = Σ n1 converges for p > 1 and diverges for p < 1. p

1 f ( n) = un = p . n

EXAMPLE 1.59

Therefore,

Apply Cauchy’s Integral Test to show that

1 f ( x) = p . x We observe that f is positive and monotonic decreasing for x > 1. Therefore, by Cauchy’s ∞

∫ f ( x)dx

Integral Test, Sun and diverge together. If p ≠ 1 , then ∞

∫ 1

converge or

1

∞

∞

∞  x1− p  1 f ( x)dx = ∫ p dx = ∫ x − p dx =   . x 1 − p 1 1 1

When p > 1, then ∞

∫

∞

f ( x)dx = −

1

=−

1  1  p − 1  x p −1 1 1 [o − 1] p −1

= ∞

1 (finite). p −1

Thus ∫ f ( x)dx converges and so, by Cauchy’s

∞

n= 2

diverges if 0 < p < 1. Solution. For the given series, the nth term is

1 . n( log n) p Taking un = f(n), one gets un =

1 . x( log x) p For x > 2 and positive p, f is positive and monotonically decreasing. Hence, Cauchy’s Integral Test is applicable. f ( x) =

When p ≠ 1 ,

1

1 1 [ x1− p ]1∞ = [∞ − 1] = ∞ 1− p 1− p

M01_Baburam_ISBN _C01.indd 37

∞

Now if p > 1, then p – 1is positive and so ∞

∞

∫ f ( x) dx = −

Integral Test, Sun also converges for p > 1. When 0 < p < 1, then

∫ f ( x)dx =

∞

 ( log x) − p +1  1 −p = ( log x ) f ( x )dx = dx   . ∫2 ∫2 x  − p + 1 2 ∞

1

∞

1

the series ∑ converges if p > 1 and n( log n) p

2

=−

 1  1 p − 1  ( log x) p −1  2

 1  1 0− p − 1  ( log 2) p −1 

1/2/2012 11:34:03 AM

1.38  n  Chapter One =



∞

1 (finite) ( p − 1)( log 2) p −1

Thus, the integral ∫ f ( x) dx is divergent. Hence, 1

∞

∫ f ( x)

Thus, dx converges and hence, by 2 Cauchy’s Integral Test, the given series Sun converges for p > 1. If p < 1, then 1– p is positive and so ∞

∫ f ( x) dx = 2

1 [( log x)1− p ]∞2 1− p

by Cauchy’s Integral Test, the series

∑

1 n

is divergent. 1.17  CAUCHY’S CONDENSATION TEST If f is positive and monotonic decreasing to the limit 0 as x → ∞ and a is any constant greater than 1, then Sf(x) and Sanf(an) converge or diverge together.

1 [∞ − ( log 2)1− p ] = ∞ p Hence, by Cauchy’s Integral Test, Sun diverges for p < 1.

of n. Also f is monotonic decreasing. Therefore,

1 . Therefore, If p = 1, then f ( x) = x log x

and so

=

∞

∞

2

log x

2

∫

dx If



= [ log log x]



= ∞ − log log2 = ∞ .

∞ 2

∞

an

f (a

n −1

) dx >

∫

a

a n −1

an

f ( x) dx >

n −1

∫

a

f (a n )dx .

n −1

t

Thus, ∫ f ( x) dx diverges and so Sun also diverges 2

I t = ∫ f ( x) dx , 1

then an

for p = 1.

∞

Hence, the series

1

∑ n( log n) n= 2

p

converges if p > 1

and diverges if p < 1.

∫

a n −1

Examine the convergence of the series Solution. We have

un =

1 n

∑

1 n

.

= f ( n) .

Therefore, f ( x) = 1x and it is positive and monotonic decreasing. Therefore, Cauchy’s Integral Test is applicable. We have ∞

∞

∫ f ( x) dx = ∫ 1

1

∞

1 x

dx

1 1  =  x2  = ∞ .  2 1

f ( x) dx = I an − I an−1

and the above expression reduces to (a n − a n −1 ) f (a n −1 ) > I an − I an−1



EXAMPLE 1.60

M01_Baburam_ISBN _C01.indd 38

f (a n −1 ) > f ( x) > f (a n ) for a n −1 < x < a n an

1 x

∫ f ( x) dx = ∫

Proof: Since a > 1, an is an increasing function

> (a n − a n −1 ) f (a n )

or

(a − 1)a n −1 f (a n −1 ) > I an − I an−1



 1 > 1 −  a n f (a n ).  a

Summing up for n = 2,3,…, n–1, n, we have n −1

(a − 1)∑a r f (a r ) > I an − I a ∞ −1



 1 n > 1 −  ∑a r f (a r ) .  a  r=2

If follows that San f (an) and I an converge or diverge together. But

1/2/2012 11:34:03 AM

SequenceS and SerieS   n 1.39 ∞

Condensation Test, the series (ii) Here

lim I an = ∫ f ( x) dx.

n →∞

1

By the Integral Test, this integral and Sf(x) converge or diverge together. Hence, Sf(n) and San f (an) converge or diverge together.

and

EXAMPLE 1.61

Applying Cauchy’s Condensation Test, examine the convergence of the series ∞ 1 ∑ p n = 2 n( log n) Solution. We have

1 , x( log x) p

which satisfies the conditions of the Cauchy’s n Condensation Test. So Sun and ∑ a an ( log1 an ) p converge or diverge together. But n

1 1 =∑ p a ( log a n ) p n ( log a ) p n

Since (log a)p is constant, it follows that Sun and Σ n1 converge or diverge together. But, Σ n1 converges if p > 1 and diverges if p < 1. Hence, the given series converges if p > 1 and diverges if p < 1. p

p

EXAMPLE 1.62 ∞

∑ 2

1 log n

1 (ii) ∑ p 2 ( n log n)

=

1 . a n ( p −1) ⋅ n p ( log a ) p

If p > 1, p – 1 then is positive. Since ∑ n converges when p > 1, the series San f (an) converges and so, by Cauchy’s Condensation Test, the series ∑ ( n log1 n ) p converges for p > 1. 1

For p = 1, we have

a n f (a n ) =

1 n log a

Since log a is constant and Σ 1n is divergent San f (an) diverges for p = 1. Hence, the given series diverges for p = 1. If p < 1, then an(p–1) < 1 since a > 1. Therefore, a n f (a n ) >

1 . n p ( log a ) p

But the series ∑ n diverges when p < 1, therefore, San f (an) diverges for p < 1. Hence, the given series diverges for p < 1. It follows,

f ( n) =

un =

(i) Here, f (n) = log1 n and

Solution.

n

1 log n

.

Therefore,

n

a a 1 since a > 1. = > log a n n log a n log a

But, the series

∑

p

∑

1 ( n log n ) p

converges

(iii) We have

1 ∑2 ( log n) log n .

1 n log a

is divergent due to

the divergence of Σ . Hence, by Cauchy’s 1 n

M01_Baburam_ISBN _C01.indd 39

n

for p > 1 and diverges for p < 1 .

∞

a n f (a n ) =

1 (n log n) p

an 1 = (a log a n ) p a n ( p −1) (n log a ) p

a n f (a n ) =

therefore, that the series

∞

(iii)

is divergent.

1

Examine the convergence of the series (i)

1 log n

p

f ( x) =

∑a

un = f ( n) =

∑

1 . ( log n) log n

Therefore,

∑a n f (a n ) = ∑ =∑

an ( log a n ) log a

n

an . (n log a ) n log a

Calling this series as Svn, we have

1/2/2012 11:34:04 AM

1.40  n  Chapter One 1

vnn =

a → 0 as n → ∞ . (n log a ) log a

Therefore, by Cauchy’s Root Test, San f (an) converges. Hence by Cauchy’s. Condensation Test, the series ∑ (log n1)log n converges.

(b) Let Sun be a series positive terms and let Σ D1 be divergent. If n

  u lim  Dn n − Dn +1  < 0.  un +1  then the series Sun is divergent.

u   n −D  = p and let Proof: (a) Let lim  Dn u n + 1  

EXAMPLE 1.63

n +1

Discuss the convergence of

0 < h < p. Thus h is an inferior number.

1 ∑ n log n( log log n) p .

Dm um − Dm +1um +1 > h um +1 ,

1 f ( n) = . n log n( log log n) p Therefore, an a log a ( log log a n ) p

=

1 n log a ( log (n log a )) p

= =

n

Dm +1um +1 − Dm + 2 um + 2 > h um + 2 , Dn −1un −1 − Dn un > h un . Hence, h(um +1 + um + 2 + …+ un ) < Dm um − Dn un < Dm um , or

n

u m +1 + u m + 2 +  + u n
1 and

diverges for p < 1. (see Example 1.62). Thus Sa n f (a n) converges for p > 1 and diverges for p < 1. Hence, by Cauchy’s Condensation Test, the given series converges for p > 1 and diverges for p < 1.

(a) Let Sun be a series of positive terms and let Dn be a positive sequence. If

then the series Sun is convergent.

M01_Baburam_ISBN _C01.indd 40

n

∑u

n

m +1

remains bounded as n → ∞ and so Sun is convergent. u

n (b) Let lim( Dn un+1 − Dn +1 ) < 0 . Then, 0 is a superior number of this sequence. Therefore,

Dn un − Dn +1un +1 < 0 for some n > m and so Dm um < Dn +1un +1 <  < Dn un , which yields

1.18  KUMMER’S TEST

  u lim  Dn n − Dn +1  > 0,  un +1 

Dm um , h

which is independent of n. Therefore,

1 n log a ( log n) p 1 +

after a certain

value of n, say n = m. Thus,

Solution. Here

a n f (a n ) =

u n −D D >h nu n +1 n +1

Therefore,

un >

Dm um . Dn

Since Σ D diverges, therefore, by Comparison Test, Sun also diverges. 1

n

Deductions: (i) Let Dn = 1 for all n. Then Kummer’s Test reduces to “The series Sun converges if

1/2/2012 11:34:05 AM

SequenceS and SerieS   n 1.41 u

u lim( unn+1 − 1) > 0, that is if lim n+n1 > 1 and

diverges if

(

un un+1

)

u

− 1 < 0 or if lim unn+1 < 1."



lim



Thus D’Alembert’s Ratio Test is a particular case of Kummer’s Test.

(ii) Let Dn = n. Then Kummer’s Test reduces to “Sun converges if u

lim(n unn+1 − (n + 1)) > 0, u

n that is, if lim n( un+1 − 1) > 1

is, if lim n(

− 1) < 1. Thus Raabe’s Test

is a particular case of Kummer’s Test”. (iii) Let Dn = n log n. The series ∑ = ∑ is divergent. Then u n log n n − (n + 1) log (n + 1) u n +1 1 Dn



1 n log n

u = n log n n − (n + 1) log n u n +1



 1 −(n + 1) log 1 +   n

  u   = log n  n  n − 1 − 1   u n +1   n

 1  1 − log 1 +  − log 1 +  .  n  n



+ 1n ) n = e and log e = 1. Hence, Now nlim(1 →∞



Kummar’s Test reduces to “ Sun converges



if ±±±± n 

(

un un+1

)

− −  > and diverges if 

lim log n  n 

(

un un+1

)

− 1 − 1 < 1 

Thus de Morgan–Bertrand Test is also a particular case of Kummer’s Test.

M01_Baburam_ISBN _C01.indd 41

A series in which positive and negative terms occur alternately is called an Alternating Series. Regarding convergence behavior of an alternating series, we have the following theorem, known as Leibnitz’s Rule. Theorem 1.19. (Leibnitz’s Rule). If un is positive and monotonically decreases to the limit zero, then the alternating series u1 − u2 + u3 − u4 + … is convergent. Proof: Consider un − un +1 + un + 2 −  + (−1) p un + p .

and diverges if lim(n uunn+1 − (n + 1)) < 0, that un un+1

1.19  ALTERNATING SERIES

Writing this expression as (un − un +1 ) + (un + 2 − un + 3 ) + , we see that when p is odd, –u n+p occurs in the last bracket; and when p is even, u n+p is the last term. Since u n > u n+1,each bracket is non-negative and so the expression discussed earlier is non-negative. Now we write the same expression as un − (un +1 − un + 2 ) − (un + 3 − un + 4 ) +…, then we observe that when p is even, the term un+p is included in the last bracket and when p is odd, – un+p is the last term. Since all the brackets are non-negative, the expression is certainly less then un. It follows, therefore, that 0 ≤ un − un +1 + un + 2 −  + (−1) p un + p < un . Since, un → 0 as n → ∞ for each e > 0, there exists an integer n0 such that un < e for all n > n0. Thus, |u n – u n+1 + u n+2 –…+(–1) p u n+p| < e for all n > n 0. Hence, by Cauchy’s Principle of Convergence, the alternating series u1 – u2 + u3 – … is convergent. Corollary: If un is positive, monotonically

decreasing and tends to the finite limit a, then the series S(–1)n–1un oscillates finitely.

Proof: Let un = a +vn. Then vn → 0. Since a is the lower bound of {un}, vn is positive for all n.

1/2/2012 11:34:06 AM

1.42  n  Chapter One Also, since un is monotonically decreasing, so is vn. Therefore,

un − un +1 =

S 2 n = (α + v1 ) − (α + v2 ) + …− (α + v2 n )

=

= v1 − v2 + …− v2n , S 2 n +1 = α + (v1 − v2 +…+ v2 n +1 ).

Since vn is positive and monotonically decreasing to zero, the series S(–1)nun converges by Leibnitz’s Rule to the sum V. Then S(–1)n–1un oscillates between V and V + a. Remark 1.11. If the term of a series are alternately positive and negative and the term continuously decreases, we cannot say that the series is un = 0 . For example, convergent unless nlim →∞ consider the series 3 4 5 2 − + − + …. 2 3 4 We note that lim un = lim {(−1) n −1 nn+1} = 1 or n →∞ n →∞ –1 according as n → ∞ through odd or even values. This series is oscillatory. EXAMPLE 1.64 Show that the series 1 − 21k + 31k − 41k +… is convergent for all positive values of k.

1 1 − (2n − 1)(2n) (2n + 1)(2n + 2) 8n + 2 >0 (2n − 1)(2n)(2n + 1)(2n + 1)

for all n. Thus {un} is monotonically decreasing to zero. Therefore, by Leibnitz’s Rule, the given series is convergent. EXAMPLE 1.66 Examine for convergence of the series − x +1 2 + x 1+ 3 − , where x is not a negative integer. (ii) 1− 13 + 15 − 17 + (i)

1 x +1

(iii)

log 2

(iv)

1 6

22

− log32 3 + log42 4 −

− 112 + 163 − 214 + 265 −.

Solution. (i) If x > –1, the terms are altemating

from the beginning. If x < –1 (except negative integers), the terms are ultimately altering their signs. Since, the removal of finite number of terms does not affect the convergence of the series, we may assume the series to be alternating in this case also. The nth term of the series is un = x +1 n and so lim un = 0. Further, n →∞

Solution. For the given series, the nth term is un =

1 nk

.

lim n1k = 0. Therefore, by Then un ≥ un +1 and n→∞ Leibnitze’s Rule, the given series converges. In particular, the series 1− 12 + 13 − 14 + … converges. EXAMPLE 1.65 Examine the convergence of the series 1 1 1 1 − + − + …. 1.2 3.4 5.6 7.8 Solution. For the given series, we have

un =

1 1 = . (2n − 1)(2n) 2n 2 (2 − 1n )

un = 0. Further, Therefore, nlim →∞

M01_Baburam_ISBN _C01.indd 42

u n − u n +1 =

=

1 1 − x + n x + n +1

1 >0 ( x + n)( x + n + 1)

and so {u n } is monotonically decreasing sequence. Hence, by Leibnitz’s Test, the given series converges. (ii) Clearly, {un} is monotonic decreasing and 1 = 0. 2n − 1 Hence, the given alternating series is convergent. (iii) The nth term of the given alternating series is log (n + 1) un = . (n + 1) 2 lim un = lim

n →∞

n →∞

1/2/2012 11:34:07 AM

SequenceS and SerieS   n 1.43 Then,

log (n + 1)  ∞  lim un = lim  form  n →∞ (n + 1) 2  ∞  = lim

n →∞



= lim

n →∞

1 n +1

, 2(n + 1) (L’Hospital Rule) 1 = 0. 2(n + 2) 2

To examine the monotonicity, we make use of a corollary to Mean Value Theorem, according to which “a function is monotonic decreasing if its derivative is negative.” So, let us take log n f ( n) = , n > 0. n2 Then n 2 ( 1n ) − 2n log n f ′(n) = n4 1 − 2 log n < 0 if 1—2 log n < 0, n3 1 that is, if log n > 12 , that is, if n > e 2 = 1.65. But, n > 1 and so the condition is satisfied. Thus, =

f (n + 1) > f (n + 2) for all n > 1,

Then, lim un = lim

n →∞

EXAMPLE 1.67 Examine the convergence of the series 1 1 1 1 − + − +… log 2 log 3 log 4 log 5 Solution. For this series, the nth term is

1 . un = log (n + 1)

M01_Baburam_ISBN _C01.indd 43

1 =0 log (n + 1)

and un − un +1 =

1 1 − > 0. log (n + 1) log (n + 2)

Hence, by Libnitz’s Test, the given series converges. EXAMPLE 1.68 Examine the convergence of the series 1 2 +1

−

1 3 −1

+

1 4 +1

−

1 5 −1

+ .

Solution. The given series is an alternating series. We observe that (−1) n (n > 2) un = n + (−1) n

and that un → 0 as n → ∞. But the series is not convergent. In fact, the Leibnitz’s Test is not applicable because the terms do not decrease monotonically. Further,

which shows that un > un+1. Hence, by Leibnitz’s Test, the given series is convergent. (iv) The nth term of the given series is n 1 1 un = = → as n → ∞. 1 5n + 1 5 + n 5 Thus the terms are monotonically decreasing and un tends to a finite limit. Hence, by corollary to Leibnitz’s Test, the given series oscillates finitely

n →∞

un =

( −1) n n + ( −1) n

=

( −1) n [ n − ( −1) n ] n −1

=

(−1) n n 1 − . n −1 n −1

Therefore,

∑u

n

=∑

(−1) n n 1 −∑ . n −1 n −1

The first series on the right converges by Leibnitz’s Test while the second series diverges. Therefore, Sun, whose terms are the difference of the terms of these two series, diverges to − ∞. 1.20  ABSOLUTE CONVERGENCE OF A SERIES A series Sun containing both positive and negative terms is said to be absolutely convergent, if S|un| is convergent.

1/2/2012 11:34:07 AM

1.44  n  Chapter One Thus, the series which becomes convergent when all its negative terms are made positive is called absolutely convergent series.

Proof: Suppose S|u n| converges. Therefore,

for a given e > 0, there exists a positive integer n 0 such that | un +1 | + | un + 2 | +  + | um |< ε for m, n > n0 .

For example, the series 1−

Therefore,

1 1 1 + − + 2 2 2 23

| un +1 + un + 2 +  + | ≤ | un +1 | + | un + 2 |

is absolutely convergent because the series 1+ + 1 2

+ is convergent. If the series Sun converges and the series S|un| diverges, then the series Sun is said to be non- absolutely or semi-convergent series. For example, the series 1 1 1 1 1 1− + − + − + 2 3 4 5 6 converges by Leibnitz’s Test but the series 1 22

+

1 23

1 1 1 1 1 + + + + + 2 3 4 5 6 is divergent. Hence, the series 1− 12 + 13 − 14 + 15 − 16 + is non-absolutely convergent. 1+

A convergent series which is unaffected by any rearrangement of its terms is called an unconditional convergent series, whereas the convergent series which is affected by any rearrangement of its terms is called conditionally convergent series. For example, the series

+ | un + 3 | +  + | um | < e for m, n > n0. Hence, by Cauchy’s Principle of Convergence, the series Sun is convergent. Remark 1.13. The converse of Theorem 1.20 is not true. For example, the series 1− 12 + 13 − 14 + 15 − 1 6

+ is convergent by Leibnitz’s Test, but the series

∑| u

n

|= 1 + 12 + 13 + 14 + 15 + 16 +  is divergent.

Theorem 1.21. In an absolutely convergent series, the series formed by positive terms only is convergent and the series formed by negative terms only is also convergent. Proof: Let Sun be the series and let S n = u1 + u2 +  + un ,

σ n =| u1 | + | u2 | +  + | un | .

If P(n) and –Q(n) denotes the sum of positive and negative terms, respectively, in S n, then S n = P(n) − Q(n),

1 1 1 1− + 2 − 3 + 2 2 2

is unconditionally convergent since its value is not affected by rearrangement. On the other hand, the series 1−

1 1 1 1 + − + − 2 3 4 5

is conditionally convergent since its value can be affected by rearrangement. Remark 1.12. To test for absolute convergence, we have to apply only the test for series with positive terms, as discussed earlier. Theorem 1.20. An absolutely convergent series is itself convergent.

M01_Baburam_ISBN _C01.indd 44

σ n = P(n) + Q(n). Therefore, P ( n) =

1 (σ n + S n ) 2

and 1 (σ n − S n ). 2 Since the series S|un| converges, both S n and s n tend to a finite limit as n → ∞ . Let these limits be s and t, respectively. Then P(n) → 12 ( s + t ) and Q(n) → 12 ( s − t ), and so the series formed by both the positive and by negative terms are convergent. Q ( n) =

1/2/2012 11:34:08 AM

SequenceS and SerieS   n 1.45 EXAMPLE 1.69 Show that the exponential series

lim

n →∞

x 2 x3 x4 + ++ + 2! 3! n! is absolutely convergent for all values of x. Solution. For the given series, we have 1+ x +

the series diverges by D’Alembert’s Ratio Test. When x > 1,n the nth term does not tend to zero. For if xn = y, then log y = n log x − log n

un x n (n + 1)! n + 1 = = un +1 n ! x n +1 |x|

log n   = n  log x −  → ∞, n  

and so un = ∞ > 1. un +1

lim

n →∞

un 1 = < 1, un +1 | x |

since

log n → 0 as n → ∞. n

Hence, the series is absolutely convergent by extended D’Alembert’s Test.

Therefore, xn → ∞ as n → ∞. The series, therefore, oscillates infinitely

EXAMPLE 1.70 Examine the logarithmic series

EXAMPLE 1.71

2

3

n

x x x x− + − …. + ( −1) n +1 +… n 2 3

n

Show that the series ∑ cosn pnθ and absolutely convergent for p > 1. Solution. Since,

Therefore, lim

n →∞

un 1 = un +1 | x |

and so, by extended D’Alembert’s Ratio Test, the given series converges absolutely 1 if | x| > 1 , that is, if |x| < 1. When x = 1, the series becomes 1− 12 + 13 − + , which converges by Leibnitz’s Test. But the series 1+ 12 + 13 − 14 + diverges.

and since that

∑

When x < –1, the terms are all negative. Removing a common negative sign, the term becomes positive. Since

M01_Baburam_ISBN _C01.indd 45

are

∑

cos nθ np

1 np

converges for p > 1, it follows

and

∑

sin nθ np

converge for p >1.

EXAMPLE 1.72 Examine absolute convergence of the hypergeometric series 1+

α .β α (α + 1) β ( β + 1) 2 x+ x + . 1⋅ γ 1 ⋅ 2(γ + 1)

Solution. We have

1 4

When x = –1, the series becomes −(1 + 12 + 13 + 14 +) and is hence divergent.

sin nθ np

cos nθ 1 sin nθ 1 ≤ p and ≤ p p p n n n n

for absolute convergence. Solution. It is an alternating series for which un x n n + 1 n(1 + 1n ) 1 = = . . |x| un +1 n x n +1 n

∑

lim

n →∞

un (n + 1)(n + γ ) 1 1 = lim ⋅ = . →∞ n (α + n)( β + n) x | x | un +1

Therefore, by D’Alembert’s Ratio Test, the hyper geometric series is absolutely convergent if |1x| > 1, that is, if | x |< 1. When, x = 1, we have

1/2/2012 11:34:09 AM

1.46  n  Chapter One



( )(

(1 + 1n ) 1 + γn un = un +1 1 + γn 1 + βn

(

) )

1− m + ×

1+ γ − α − β = 1+ n



 1  +O  2  by binomial expansion. n  Therefore, by Gauss’s Test, the series is convergent if 1 + g – a – b > 1 or if g > a + b and divergent if 1 + g – a – b < 1 or if g < a + b . When

u

x > 1, lim unn+1 = n →∞

1 | x|

< 1.

Therefore,

the series diverges whatever a , b , and g may be. EXAMPLE 1.73 Discuss convergence of the Binomial Series m(m − 1) 2 x + 2! m(m − 1) … (m − n + 1) n + x + n! Solution. We have 1 + mx +

lim

n →∞

un n 1 = lim un +1 n →∞ m − n + 1 x = lim

n →∞ m n

1 1 1 . = 1 , −1 + n x | x |

Hence, the series is absolutely convergent if |x| < 1. Further, u lim | un |1/ n = lim n +1 =| x | . n →∞ n →∞ u n Therefore, lim | un |= lim | x |n = ∞ if x > 1.

n →∞

M01_Baburam_ISBN _C01.indd 46

m (m − 1) … (m − n − 1) +… n!

Whatever m may be, the terms are of the same sign after a certain value of n. We have un m +1  1 = 1+ + O 2 un +1 n n

 . 

Hence, by Gauss’s Test, the series is convergent if m + 1 > 1, that is, if m > 0 and divergent if m < 0. If m = 0, the series reduce to single term 1. If x = 1, the series is m(n − 1) 1+ m + + 2! +

m(m − 1) … (m − n − 1) +. n!

The terms are alternately positive and negative after a certain value of n. From above, we have un m +1  1 = 1+ + O 2 un +1 n n

 . 

Hence, by Gauss’s Test, the series converges absolutely if m + 1 > 1, that is, if m > 0. EXAMPLE 1.74 Examine the following series for convergence/ absolute convergence. (i) ∑ (ii) (iii)

( −1)n − 1 sin nx n3

2 3 4 5 x − x + x − x + x − 2 3 4 5 2 3 4 x − x + x − x + 2 3 4

Solution. (i) The nth term of the series is

n →∞

Hence, the series cannot converge when | x | > 1. When x < 0, the terms are ultimately of the same sign and hence the series is divergent if x < – 1. If x > 1, the terms are alternately positive and negative after a certain stage and the series oscillates infinitely. When x = –1, the series is

m (m − 1) −…+ (−1) n 2!

un = and so | un |=

(−1) n +1 sin nx n3 | sin nx | 1 ≤ 3. n3 n

Taking vn as n13 , we have ∑ vn = ∑ n13 , which converges. Therefore, by comparison test, S| un|

1/2/2012 11:34:09 AM

SequenceS and SerieS   n 1.47 converges. Hence, the given series is absolutely convergent. n (ii) We have un = xn and so un xn n + 1  n + 1  1 = ⋅ = ⋅ . un +1 n x n +1  n  x

n →∞

∞

2

and absolute convergence

given series converges absolutely if > 1 , that is, if | x | < 1. If x = 1, the series becomes 1 x

1 1 1 1 1 + − + − + , 2 3 4 5 6 which is convergent by Leibnitz’s Test. If x = – 1, then the series becomes −1 − 12 − 13 − 14 − 15 −… = −(1 + 12 + 13 +…), which is divergent. Hence, the given series converges for –1 < x < 1 and converges absolutely for – 1 < x < 1. 1−

xn n

.

n +1 1 ⋅ n x

Solution. The nth term of the given alternating

series is

( −1)n n ( log n )2

1

Hence, by extended D’Alembert Ratio Test, the series converges absolutely if 1x > 1, that is, if |x| < 1, that is if –1 < x < 1. If x = 1, then the series becomes 1−

2

+

3

−

1 4

We observe that lim u n = lim n →∞

1−

M01_Baburam_ISBN _C01.indd 47

2

−

1 3

−

n →∞

1

To examine for absolute convergence, we have 1 un = = f ( n ), say. n ( log n )2 Therefore,

4

1 , x ( log x )2

which is positive and monotonic decreasing. Hence Cauchy’s Integral Test is applicable. We have ∞

∞

∞

dx 1 = ∫ (log x ) −2 ⋅ dx 2 x 2 x ( log x ) 2

∫ f (x ) dx = ∫ 2

∞

 ( log x) −1  =  −1  2 

= Thus,

∞

∫ f (x ) dx converges 2

−

1 =0 n ( log n )2

and un > un+1, n > 2. Therefore, by Leibnitz’s Test, the given series is convergent.

+ ...,

whose nth term is un = 1n → 0 as → ∞. Also un > un+1. Hence, by Leibnitz’s Test, Sun converges. If x = –1, then the series becomes 1

1 . n ( log n )2

f (x ) =

u 1  1 2 1 lim n = lim 1 +  ⋅ = . n →∞ u n →∞  n x | x| n +1

1

. So

un =

and so

1

( −1)n

n=2

Hence, by extended D’Alembert’s Ratio Test, the

un xn n + 1 = = n +1 u n +1 n x

)

Examine the series ∑ n ( log n ) for convergence

un 1  1 1 = 1 +  = . un +1  n  x x

(iii) The nth term of the series is un = Therefore,

(

− 1 + 12 + 13 + 14 + , which is divergent: Hence, the given series converges for –1 < x < 1 and converges absolutely for –1 < x < 1. EXAMPLE 1.75

Thus lim

or

1 (finite). log 2 and so by Cauchy’s ∞

Integral Test, the series ∑ un converges. n=2 Hence, the given series converges absolutely.

1/2/2012 11:34:10 AM

1.48  n  Chapter One 1.21  CONVERGENCE OF THE SERIES OF THE TYPE ∞

∑u ν n =1

n

n

To study the convergence behavior of the series of the type Σu nv n, we have two theorems of great importance known as Abel’s Test and Dirichlet’s Test. The following lemma would be required to establish these tests. Theorem 1.22. (Abel’s Lemma). If v1, v2 …, vn are positive and decreasing, then Bv 1 < u1v 1 + u2v 2 +  + u n v n < A v 1 , where A and B are, respectively, the greatest and the least of the quantities u1 , u1 + u2 , u1 + u2 + u3 , , u1 + u2 +  + u n . Proof: Let

S n = u1 + u2 +  + u n . Then

u1 = S1, u2 = S2 – S1, …, un = Sn – Sn–1.

Since A is greatest of S1, S2,…, Sn, we obtain n

∑ um v m = u1ν1 + u2ν 2 + u3ν3 +  + unν n m =1



= S 1ν1 + (S 2 − S 1 )ν 2 + (S 3 − S 2 )ν 3 +  + (S n − S n −1 )ν n = S 1 (ν1 − ν 2 ) + S 2 (ν 2 − ν 3 ) +  + S n −1 (ν n −1 − ν n ) + S n ν n < A [ν1 − ν 2 + ν 2 − ν 3 +  + ν n −1 − ν n + ν n ] = A ν1 .

Similarly, if A is least of S1, S2, …, Sn, then we get n

∑ um ν m > Bν1 . m =1

Hence Bν1 < u1ν1 + u2ν 2 +  + u n ν n < A ν1 . Theorem 1.23. (Abel’s Test). If {vn} is positive and monotonically decreasing and if Σun converges, then Σun vn is convergent.

M01_Baburam_ISBN _C01.indd 48

Proof: Let {v n} be positive and monotonically decreasing, whereas u n may be positive or negative. Then for the convergence of the series Σu n v n, we must have

u n ν n + u n +1ν n +1 +  + u n + p ν n + p < ε .

By Abel’s Lemma, Bν n < u n ν n + u n +1ν n +1 +  + u n + p ν n + p < A ν n ,

where A and B are respectively the greatest and the least of un, un + un+1, … un + un+1 + … + un+p. Therefore, | u n ν n + u n +1ν n +1 +  + u n + p ν n + p |< α ν n ,

where a is the maximum of |A| and | B |. Since {vn} is positive and decreasing, it is bounded, that is, vn < m for all n. Since Σun converges, α < mε for n > some n0. Thus | un vn + un +1vn +1 +  + un + p vn + p |< ε for n > n . 0 Hence, by Cauchy’s Principle of Convergence, the series Σunvn is convergent.

Theorem 1.24. (Dirichlet’s Test). If {v n} is positive and monotonically decreasing to the limit 0 and if Σu n is either convergent or oscillates finitely, then Σu nv n is convergent. Proof: By Abel’s Lemma, we have Bν n < u n ν n + u n +1ν n +1 +  + u n + p ν n + p < A ν n ,

where A and B are, respectively, the greatest and the least of u n , u n + u n +1 ,  , u n + u n +1 +  + u n + p .

Therefore, if a = max (|A|, |B|), then u n ν n + u n +1ν n +1 +  + u n + p ν n + p < αν n .

Since Σun is convergent or oscillates, a is finite, say less then k for all n. Since vn tends monotonically to 0, we have ν n < kε for some n > n0. Hence,

un vn + un +1vn +1 +…+ un + p vn + p < ε for n > n0 . Hence, the series Σunvn is convergent by Cauchy’s Principle of Convergence.

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SequenceS and SerieS   n 1.49 Remark 1.14. If Σun = 1 – 1 + 1 + …, then it follows that Leibnitz’s Test is a particular case of Dirichlet’s Test.

sequence. Hence, by Abel’s Test, the given series Σunvn is convergent.

EXAMPLE 1.76

and ν n = , then the given series can be written as Σu n v n . But, Σu n is convergent by Leibnitz’s Test and v n decreases monotonically to zero. Hence, by Dirichlet’s Test, the given series converges.

1 1 1 2 1 3 + − + − + − . 2 22 3 32 4 4 2 Solution. Consider the series 0−

1 1 1 1 1 1 − + − + − + . 2 2 3 3 4 4 S 2 n = 0, This series is convergent since lim n →∞ n

= 1−1+

lim S 2 n +1 = lim n →∞

n →∞

1 = 0. Now consider the sequence n +1

{vn} with elements 0,

1 1 1 1 1 + − + − 2 22 3 32 4 3 + 2 − … (given series) 4

and so

( n + 1) − n log n

Solution. (i) If un = (n + 1) − n and ν n = then

Σun vn = Σ

3

1 2 sin θ2

1 3

log n

−n

1 log n

,

(given series). 1 3

 θ 3θ   cos 2 − cos 2   3θ cos 5θ   +  cos −   2 2 

2n − 1 2n + 1    θ − cos θ  +  +  cos   2 2 =

1 2 sin θ2

m =1

The series ∑ u n = ( n 3 + 1) − n is convergent and {vn} is a positive monotonically decreasing

M01_Baburam_ISBN _C01.indd 49

=

θ 2n + 1   cos 2 − cos 2 θ   

∑ sin mθ

1 1 1 + − 2 + 2 2 3.2 5.3 74

(n +1)

θ +  + 2 sin nθ sin  2

n

1 3

θ

2



1 3

3

θ



1



Test the following series for convergence:

(ii) 1 −

− 212 + 312 − 412 + 

Solution. We have

EXAMPLE 1.77

∑

1 12

Examine the convergence of the series ∑ n1 sin nθ and ∑ n1 cos nθ .

m =1

Hence, by Abel’s Test, the given series Σunvn is convergent.

3

=

EXAMPLE 1.78

n

∑ unv n = 0 −

n

∑ sin mθ = 2 sin θ 2 sin θ sin 2 + 2 sin 2θ sin 2

1 1 2 2 3 3 , , , , ,… 2 2 3 3 4 4

This sequence is positive and monotonic decreasing. Then multiplying the terms of the above two series, we get

(i)

∑u

If

1 2 n −1

Test the following series for convergence:

∑u

(ii)

≤

cosec

θ 2

.

Hence the series Σsin mq oscillates finitely. Also, the sequence 1, 12 13 , …, n1 is a decreasing sequence of positive terms which tends to zero as n → ∞. Hence, by Dirichlet’s Test, 1 the series ∑ n sin nθ converges if sin θ2 ≠ 0, that is, if θ ≠ 0 or not an even multiple of p. If q = 0 or 2mp where m is an integer, the series ∑ n1 sin nθ is merely a series of zeros and is, therefore, convergent.

1/2/2012 11:34:12 AM

1.50  n  Chapter One Similarly, n

1 cos mθ = ∑ 2 sin θ2 m =1

θ  2n + 1 sin 2 θ − sin 2   

and so n

∑ cos mθ m =1

≤

cosec

θ 2

.

1.22  DERANGEMENT OF SERIES If a series consists of finite number of terms, then its terms can be arranged in any manner what so ever. On the other hand, if terms of an infinite series are deranged, its sum or even the nature of convergence may alter. However, we will notice that derangement does not affect the sum for convergence of the series, if the infinite series is absolutely convergent. To study the effect of derangement, consider the series 1 1 1 1 1 − + − + − . 2 3 4 5 This series is convergent by Leibnitz’s Test. Further, by the definition of Euler’s Constant γ, we know that 1 + 1 + 1 +  + 1 → log 2n + γ (1) 2 3 2n and 1 1 1 1 + + +  + → log n + γ 2 3 n or 1 1 1 1 1 2  + + + +  +  → log n + γ . (2) 2 4 6 8 2n 

But 1 1 1 1 1 1 + − + − + − 2 3 4 5 6 2n

1  1 1 = 1 + + +  +   2 3 2n 

M01_Baburam_ISBN _C01.indd 50

Therefore, by (1) and (2), we get S 2 n → log (2n ) + γ − log n − γ = log

Therefore, as in the case of series ∑ n1 sin nθ the given series ∑ n1 cos nθ converges. If q = 0 or 2mp where m is integer, then we get the series 1 + 12 + …+ n1 + …+, which is divergent.

1−

1 1 1 1 −2  + + +  +  . 2 4 6 2n  2n = log 2. n

Thus, the sum of the series in question is log 2. Now consider the following derangement of the above series: 1−

1 1 1 1 1 1 1 1 − + − − + − − + 2 4 3 6 8 5 10 12

Then  1 1  1 1 1  1 1  S 3n = 1 −  − +  −  − +  −   2  4  3 6  8  5 10  −

and so

1 1  1  1 ++  − −  2n − 1 4 n − 2  4 n 12

=

1 1 1 1 1 1 − + − ++ − 2 4 6 8 4n − 2 4n

=

1 1 1 1 1 1 −  1 − + − +  + 2 2 3 4 2n − 1 2n 

=

1 S2 n 2

1 S 1 lim S 3n = lim S 2 n = = log 2. n →∞ n →∞ 2 2 2

If follows, therefore, that derangement of the series has changed its sum. We now show that derangement of terms does not affect the sum of an absolutely convergent series. Theorem 1.25. The sum of an absolutely convergent series is not altered by any derangement. Proof: Let Σun = u1 + u2 + u3 + … + un + … be an absolutely convergent series. Let Σvn = v1 + v2 + … + vn + … be the deranged series of Σun. Thus, the terms in Σvn are the same as the terms in Σun but in other order. Let Sn and Tn denote the sum of n terms of the two series Σun and Σvn, respectively. We can find p such that Tp contains

1/2/2012 11:34:12 AM

SequenceS and SerieS   n 1.51 all the terms of Sn. Thus taking r an integer greater then p, Tr – Sn contains a number of terms of the first series beyond un. Since Σun is absolutely convergent, there exists n0 such that un +1 + un + 2 + un + 2 + 
n0 . 2 Since, | Tr – Sn | < the sum of a certain number of terms on the left side of the above expression, it follows that Tr − S n
n0 , r > p. 2 If S denotes the sum of Σun, then S − S n ≤ un +1 + un + 2 + 
n0 .

S − Tr = S − S n + S n − Tr ≤ S − S n + S n − Tr
p.

1.23  NATURE OF NON-ABSOLUTELY CONVERGENT SERIES

r

k

j

+

j =1

r

r + t −1

∑ν

k = r +1

−  + ∑ νk < λ

r + t −1

r

< ∑ νk − ∑ w j + k =1

n −1

k

∑ν

j =1

k = r +1

n

k

−  + ∑ νk

(6)

or r

r

s + t −1

k =1

p =1

k = r +1

∑ νk − ∑ w p + r



< ∑ νk + k =1

∑

r + t −1

∑ν

k = t +1

m

ν k −  − ∑w j < λ m −1

k

− + ∑w j

(7)

1/2/2012 11:34:12 AM

1.52  n  Chapter One If the resulting series be called S and the total number of terms on the left had sides of (6) and (7) be N and M, respectively, then S N < λ < S N + ν n and S M < λ < S M − w m . Since the series Σun is convergent, ν n → 0 as n → ∞ and wm → 0 as m → ∞ . Therefore, S n → λ as n → ∞ . Now to get a divergent or oscillatory series, we replace l in (3), (4), and (5) by the successive terms of a sequence l 1, l 2,…, which may by divergent or oscillatory. Thus, SN < lp < SN + vn and SM < lq < SM + wm. Therefore, lim S N or lim S M = l p or l p . Thus, if l p diverges as p → ∞ , then S N diverges and if l p oscillates, then S N oscillates. 1.24  EFFECT OF DERANGEMENT OF NON-ABSOLUTELY CONVERGENT SERIES The following theorem of Pringsheim tells us about the change in sum due to derangement of terms of a non-absolutely convergent series. Theorem 1.27. (Pringsheim). Let f be a positive function monotonically decreasing to the limit zero and let the series

∞

∑(−1) n =1

n −1

f (n) be

deranged so that in the first p + n terms, there are p positive terms and n negative = k, lim nf ( n ) = g and lim n →∞ n n →∞ then the sum of the series is increased by

terms. 1 2

If

p

g log k .

Proof: In the deranged series, there are p

positive terms and n negative terms in the first p + n terms. If p > n, then the sum of p + n terms of the deranged series is equal to [ f (1) – f (2) + f (3) – … – f (2n)] + [f (2n + 1) + f (2n + 3) +  + f (2 p − 1) ] .

Thus, the excess over the sum of 2n terms of the

M01_Baburam_ISBN _C01.indd 52

original series is E = f (2n + 1) + f (2n + 3) +  + f (2 p − 1).

nf ( n ) = g , for ε > 0, Since lim n →∞ positive integer n0 such that

there exists

nf ( n ) − g < ε for n > n0 or

g − ε < nf ( n ) < g + ε for n > n0

or

g −ε g +ε < f (n) < for n > n0 . n n Substituting n = 2n+1, 2n+2,…, 2p – 1 and adding we get

 1 1 1  + ++ (g − ε )  2 p − 1  2n + 1 2n + 3 < f (2n + 1) + f (2n + 3) +  + f (2 p − 1)

 1 1 1  (1) < (g + ε )  + ++ 2 p − 1  2n + 1 2n + 3

But 1 1 1 + ++ 2n + 1 2n + 3 2p − 1

 1 1 1   1 1  = 1 + + +  +  − 1 + +  +  2 3 2 p 2 2 n    1 1 1 −  + +  2  n +1 p = log 2 p + γ 2 p − ( log 2n + γ 2 n )

1 − (γ p + log p − γ n − log n ). 2

Therefore,  1 1 1  lim  + ++  2n + 1 2n + 3 2 p − 1

p , n →∞

= lim log 2 p + γ 2 p − ( log 2n + γ 2 n ) p , n →∞ 1  − (γ p + log p − γ n − log n )  2 

1/2/2012 11:34:13 AM

SequenceS and SerieS   n 1.53 = lim

p , n →∞

1 p 1 log = log k , by hypothesis. 2 n 2

1 1 log 4 + log k 2 2 1 1 = ( log 4 + log k ) = log 4 k . 2 2 =

Hence, by (1), we have E = f (2n + 1) + f (2n + 3) 1 +  + f (2 p − 1) → g log k . 2 Thus the sum of the series is increased by 1 g log k . If p < n, the sum of p + n terms 2 of the deranged series is [ f (1) – f (2) + f (3)– …+f (2p–1)] – [ f (2p) + f (2p + 2) + … + f (2n)] and the excess E is now [ f (2p) + … + f (2n)]. Proceeding as above, the increase in the sum is found to be 12 g log k . EXAMPLE 1.79

Find the sum of the series 1 1 1 1 1 1+ − + + − + 3 2 5 7 4 Solution. The given series is the derangement of the

series

1−

1 1 1 1 1 + − + − + 2 3 4 5 6

The sum of the latter series is log 2. We observe that in the deranged series, two positive terms are followed by one negative term. Thus p = 2, n = 1 so that p k = n = 2 and g = lim nf ( n ) = lim n. n1 = 1. n →∞ n →∞ Therefore, the increase in the sum due 1 to derangement is g log k = 12 log 2. 2 Hence, the sum of the deranged series is log 2 + 12 log 2 = 23 log 2. Remark 1.15. If we consider the derangement of the series 1 − 12 + 13 − 14 + 15 − 61 + , then the total sum of the deranged series is log 2 + 12 g log k . g = lim nf ( n ) = lim n. n1 = 1. Therefore, But n →∞ n →∞ the total sum of the deranged series is log 2 +

1 1 1 log k = log 4 2 + log k 2 2

M01_Baburam_ISBN _C01.indd 53

EXAMPLE 1.80

Show that 1 1 1 1 1 1 1 − − + + − − +  = log 2. 3 2 4 5 7 6 8 Solution. The given series is derangement of the series 1+

1 1 1 1 1 + − + − + 2 3 4 5 6 Here two positive terms are being followed by two negative terms. Thus, p 2 k = = = 1. n 2 Hence, the sum of the deranged (given) series is 1−

1 1 1 log 4 k = log 4 = log 4 2 = log 2. 2 2

EXAMPLE 1.81

Find the sum of the series 1 1 1 1 1 1 − + − − + − 2 4 3 6 8 5 Solution. The given series is derangement of the series 1−

1 1 1 1 1 + − + − + 2 3 4 5 6 Here one positive term is followed by two p negative terms. Thus k = n = 12 . Hence, the sum of the deranged series is 1−

1 1 log 4k = log 2. 2 2 EXAMPLE 1.82

Show that 1−

1 1 1 1 1 1 1 1 1 1 − − − + − − − − + −  = 0. 2 4 6 8 3 10 12 14 16 5

Solution. The sum of the given series is

where k = . Hence, the sum is 1 4

1 2

1 2

log 4k ,

log1 = 0.

1/2/2012 11:34:14 AM

Soluti

ample S = 0,

the de

1.54  n  Chapter One EXAMPLE 1.83

Investigate what derangement of the series 1− 12 + 13 − 14 + 15 − 16 + will reduce its sum to S. Solution. We know that the sum of the deranged series is 12 log 4k . Thus we have 1 2S 1 log 4k = S or k = e . 4 2 Thus to obtain the deranged series with sum S, the ratio of number of positive terms followed by 2S the number of negative terms should be 14 e . EXAMPLE 1.84

Investigate, what derangement of the series 1−

1 1 1 1 1 + − + − + 2 3 4 5 6

will reduce its sum to zero. Solution. It follows from Example (1.83) that

k = 14 e 2 S . Here S = 0, so that k = 14 e0 = 14 . Thus, the deranged series is 1−

1 1 1 1 1 1 1 1 1 1 − − − + − − − − + − 2 4 6 8 3 10 12 14 16 5

EXAMPLE 1.85

Show that 1 1 1 1 1 1 1 1 1 1 1 + + − − + + + − − +  = log 6. 3 5 2 4 7 9 11 6 8 2

Solution. The given series is the derangement of the series

1−

1 1 1 1 1 + − + − +. 2 3 4 5 6

Here three positive terms are being followed by two negative terms. So k = np = 32 . Hence, the sum of the given series is 12 log 4k = 12 log 6. EXAMPLE 1.86

Criticize the following paradox: (i) 1 − 1 + 1 + 1 + 1 −  = (1 + 1 + 1 + 1 + ) 2 3 4 5 2 3 4 −2( 12 + 14 + 16 +) = (1 + 12 + 13 + 14 + ) − (1 + 12 + 13 + ) = 0.

M01_Baburam_ISBN _C01.indd 54

(ii) 2 ( 12 + 14 + 16 + ) = 1 + 12 + 13 + 14 +  = (1 + 13 + 15 + ) + ( 12 + 14 + 16 + ). Therefore, 1 1 1 1 1 + + + = 1+ + + 2 4 6 3 5 which is absurd since each term of the series on the left is less than the corresponding term in the series on the right. Solution. (i) The series is semi-convergent (conditionally convergent) and so we cannot rearrange its term. The rearrangement of the term changes the sum of the series to zero, which is otherwise log 2. (ii) The series 2( 12 + 14 + 16 +) is a divergent series of positive terms. Therefore, we can derange it in any manner. Thus, the first two steps in the example are correct. However, the third step is wrong. In this step we have taken 1 1 1  1 1 1  2  + + +  −  + + +  2 4 6  2 4 6  1 1 1 = + + + 2 4 6 Both the series in the bracket are divergent and tend to ∞ so that we get the indeterminate form ∞ − ∞. On account of this fallacy, we get the absurd result. 1.25  UNIFORM CONVERGENCE We now consider sequences or infinite series whose terms are functions of a variable, say x, in some interval [a,b]. The convergence of such sequences (or series) in that interval will be called uniform convergence. We define uniform convergence of such sequence {s n(x)} as follows: A sequence {sn(x)}is said to converge uniformly to the limit s(x) in the interval [a,b], if given e > 0, there exists a positive integer m, independent of x such that | sn ( x) − s ( x) |< ε for all n > m and all x ∈ [a, b]. EXAMPLE 1.87

Show that the sequence {sn(x)}, where sn ( x ) =

nx , 0≤ x≤a 1 + n2 x2

1/2/2012 11:34:15 AM

SequenceS and SerieS   n 1.55 is non-uniformly convergent in any interval which includes the origin. Solution. Let e > 0. Then | sn(x) – 0| < e implies

nx 0

or nx >

1 1 1 + −4 2ε 2 ε 2

or n>

 1 1 1 1 − 4.  + 2 x  2ε 2 ε 

Thus, we can find an upper bound for n in any interval 0 < a < x < b, but the upper bound is infinite if the interval includes 0. Hence, the given sequence is non-uniformly convergent in any interval which includes the origin. EXAMPLE 1.88

Show that 1 is the point of non-uniform convergence for the sequence {sn(x)}, where sn ( x) = xn , x ∈ [0,1].

Solution. We note that

0 if 0 ≤ x < 1 s ( x) = lim S n ( x) =  n →∞ 1 if x = 1.

To see whether the convergence is uniform, let e > 0. Then, | sn(x) – 0| < e implies xn < ε or n

1 1   > ε x log ε1 1 1 . > log or n > log 1x x ε Thus, for uniform convergence we have to take

n log

log 1ε log 1x

. If we take 0

o < x < a < 1, then the upper bound of m will be

M01_Baburam_ISBN _C01.indd 55

. But this upper bound becomes infinite if

a = 1. Hence for a = 1, m cannot be selected Therefore, the sequence converges uniformly in [0, a], where a < 1 but it is not uniformly convergent if the interval includes 1. This means that 1 is the point of non-uniform convergence of the sequence {s n(x)}. The following theorem provides characterization for uniform convergence of a sequence {sn(x)}. Theorem 1.28. (Cauchy’s General Principle of Uniform Convergence): The necessary and sufficient condition for the sequence {sn(x)} to converge uniformly in the interval [a,b] is that, given e > 0, there exists positive integer p independent of x, such that

| sn ( x) − sm ( x) |< ε for all n > m ≥ p and all x ∈ [a, b].

The following theorem serves as a test for uniform convergence. Theorem 1.29. (Mn-Test). Let {sn(x)} be a sequence on [a,b] and let Mn = sup {| sn(x) – s(x) |: x ∈ [a, b]}. Then {sn(x)} converges uniformly to s(x) if and only if M n → 0 as n → ∞. Proof: Necessity: Suppose that {sn(x)} converges

uniformly to s(x) in [a,b]. Then for a given e > 0, there exists a positive integer m, independent of x, such that | sn(x) – s(x) | < e for all n > m and x ∈ [a, b]. Since Mn is the supremum of | sn(x) – s(x) | over [a,b], it follows that Mn < e for all n > m. Hence M n → 0 as n → ∞. Sufficiency: Suppose that M n = sup{| sn ( x) − s ( x) |} → 0 as n → ∞. Hence, M n < ε for all

or

the integer m next higher to

log 1ε log 1a

n ≥ m and all x ∈ [a, b]. This implies that

sup{| sn ( x) − s ( x) |} < ε for all n ≥ m and

all x ∈ [a, b]. Hence

{| sn(x) – s(x) |} < e for all n > m and all

x ∈ [a, b].

Hence, {sn(x)} converges uniformly to s(x).

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1.56  n  Chapter One 1.26  UNIFORM CONVERGENCE OF A SERIES OF FUNCTIONS The series Σun ( x) is said to converge uniformly on [a, b] if the sequence {sn ( x)} of its partial sums converges uniformly on [a, b]. Theorem 1.30. (Weierstrass’s M-Test). A series Σun ( x) converges uniformly and absolutely on [a, b] if | un ( x) |≤ M n for all n and all xε [a, b], where Mn is independent of x and ΣM n is convergent. Proof: Since |un (x)| < Mn and SMn converges, the

series S|un (x)| converges by comparison test. To prove uniform convergence, we note that



| sn ( x) − sm ( x) |=| un +1 ( x) + un + 2 ( x) +  + um ( x ) |

≤ M n +1 + M n + 2 +  + M m . Since SMn converges, M n +1 + M n + 2 +  + M m < ε for n > m > p, where p is independent of x. Hence | sn ( x) − sm ( x) |< ε for n > m ≥ p

and all xε [a, b]. Hence, by Cauchy’s General Principle of Uniform Convergence, Sun(x) converges uniformly on [a, b]. EXAMPLE 1.89

Show that the series

converges uniformly on the real line for p > 1. Solution. We have

cos nx 1 ≤ p for all x ∈ R. np n

But the series ∑ n p converges for p > 1. Hence, by Weierstrass’s M-Test, the given series Sun (x) converges uniformly on R for p > 1. 1

EXAMPLE 1.90

and applying Weierstrass’s M-Test. EXAMPLE 1.91

Test for uniform convergence of the series 1 + a cos x + a 2 cos 2 x +  + a n cos nx +  Solution. We have ∞

∑u ( x) = ∑ a n

n

cos nx

n=0

n n n and | un ( x) | = | a cos x |≤| a | = a if a > 0.

∑M

Then,

n

= ∑a n = 1 + a + a 2 +  + a n +

 = 1−1a , 0 < a < 1. Therefore, the series Sa (x) n converges uniformly, by Weierstrass’s M-Test, if 0 < a < 1. EXAMPLE 1.92

If San converges absolutely, show that the series

∑

an xn 1+ x 2 n

converges uniformly for all real x.

Solution. We have

un ( x ) =

an xn . 1 + x2n

Substituting the first differential of un(x) to zero, we get d 2 un

Also

dx 2

Prove that the series ∑ uniformly for all real values of x.

converges

< 0 when, x n = 1. Therefore,

M n = max | un ( x) |= max

an x n |a | = n 2n 2 1+ x

Thus, | un ( x) |≤ M n and ∑ M n = 12 ∑ an < ∞ (by hypothesis). Hence, by Weierstrass’s M-Test, the given series converges uniformly for all real values of x. EXAMPLE 1.93

Examine the series ∞

sin ( x + nx ) n ( n +1)

M01_Baburam_ISBN _C01.indd 56

1 n ( n +1)

n 1 + x 2 n − 2 x 2 n = 0, which yields x = 1.

cos x cos 2 x cos nx + ++ + 1p 2p np

| un |=

Solution. The result follows taking M n =

∑u n =1

n

∞  nx (n − 1) x  − ( x) = ∑   2 2 1 + (n − 1) 2 x 2  n =1 1 + n x

for uniform convergence.

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SequenceS and SerieS   n 1.57 Solution. We have

u1 ( x) =

x − 0, 1 + x2

2x x u2 ( x ) = − , 2 2 1 + 2 x 1 + x2

un ( x ) =

nx (n − 1) x − . 1 + n 2 x 2 1 + (n − 1) 2 x 2

Therefore, the partial sum S n(x) is given by nx S n ( x) = u1 ( x) + ( x) +…+ un ( x) = . 1 + n2 x2

{

}

The sequence 1+ nx has 0 as the point of nonn x uniform convergence (see Example 1.87). 2 2

1.27  PROPERTIES OF UNIFORMLY CONVERGENT SERIES (A) We know that the sum of two continuous functions is also continuous and that this result can be extended to the sum of a finite number of functions. The question arises, “Can we extend this result for infinite number of functions?” The following theorem provides a sufficient condition for the sum function of an infinite series of continuous functions to be continuous. Theorem 1.31. If u1(x), u2(x),…, un(x) are all continuous functions of x ∈[a, b] and if S(x) = Sun(x) is uniformly convergent in [a,b], then S(x) is continuous in [a,b] . Remark 1.16. The condition of uniform convergence of Sun(x) is sufficient but not necessary for the sum S(x) to be continuous. For example, consider the series,  n2 x (n − 1) 2 x  ∑ 1 + n3 x 2 − 1 + (n − 1)3 x 2  .  

For this series, the partial sum sn(x) is given by sn ( x ) =

n2 x . 1 + n3 x 2

We have n2 x  ∞   form  n →∞ 1 + n 3 x 2 ∞  

S ( x) = lim sn ( x) = lim n →∞

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x = 0, x ∈[0,1]. + nx 2

= lim



n →∞ 1 n2

Hence, S(x) is continuous at 0 on [0, 1] But it can be seen that 0 is a point of non-uniform convergence for the given series. (B) We know that the sum of two integrable functions is integrable and that the result can be extended to the sum of a finite number of functions. The following theorem provides a sufficient condition to extend this result to an infinite number of functions Theorem 1.32. (Term-by-Term Integration): If u1(x), u2(x),…, un(x) are all continuous (hence, integrable) functions of x in [a,b] and if S(x) = Sun(x) is uniformly convergent in [a,b],then b

b

b

a

a

a

∫S ( x)dx = ∫[∑u

n

( x)]dx

b

b

a

a

= ∫u1 ( x)dx + ∫u2 ( x)dx b

+  + ∫un ( x)dx + .



a

Remark 1.17. The condition of uniform convergence in the above theorem is sufficient but not necessary. For example, we have 1 − x + x 2 − x3 +  = The result 1

dx

1

1

1 ,0 ≤ x < 1 1+ x 1

∫ 1 + x = ∫dx − ∫xdx + ∫x dx − . 0

2

0

0

0

is correct because 1 1 1 + − +. 2 3 4 Thus, for this series termby-term integration is valid but the series is not convergent at x =1. (C) We now extend the result log 2 = 1 −

d [ f1 ( x) + f 2 ( x) + f n ( x)] dx = f1′ ( x) + f 2′( x) +  + f n′( x) for the sum of an infinite serie

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1.58  n  Chapter One Theorem 1.33. (Term-by-Term Differentiation): If (i) Sun(x) converges to the sum S(x) in [a,b], (ii) Σun′ ( x) converges uniformly to the sum F(x) and (iii) un′ ( x) is a continuous function of x in [a, b] for all n, then F ( x) = S ′( x). Remark 1.18. Again the condition of uniform convergence of Σun′ ( x) is sufficient but not necessary. For example, consider the series for which n 1 log (1 + n 2 x 2 ). S n ( x) = ∑uk ( x) = 2n k =1 Then log (1 + n 2 x 2 ) ∞ S ( x) = lim S n ( x) = lim ( form) n →∞ n →∞ 2n ∞

series is convergent, is called the interval of convergence of the power series and r is called the radius of convergence of the power series. 1 Further, if lim an n = 0, then r = ∞ and, thus, in that case, the series converges for all finite values of x. The function represented by the series in such a case is an entire function. For example, the series x 2 x3 x 4 xn + + +… = ∑ n! 2! 3! 4! has infinite radius of convergence and as such the function ex, representing this series is an entire function. Differentiating the power series (1) term by term, we obtain the series 1+ x +

nx 2 = 0 for 0 ≤ x ≤ 1 n →∞ 1 + n 2 x 2 and so S ′( x) = 0. Also

Then the nth term of the series is

nx = 0 for 0 ≤ x ≤ 1. lim S n′ ( x) = lim n →∞ n →∞ 1 + n 2 x 2

We have

un = nan x n −1 .

Therefore,

lim nan

∞

S ′( x) = lim S n′ ( x) = ∑un′ ( x). n →∞

n =1

Thus term-by-term differentiation is valid, but the series Σun′ ( x) has 0 as the point of nonuniform convergence. 1.28 POWER SERIES A series of the form

∞

n a0 + a1 x + a2 x 2 + …+ an x n + … = ∑ an x

(1)

n=0

is called a power series. The numbers an are called coefficients of the power series. If we apply Cauchy’s root test for the convergence of the power series, we observe that the series is absolutely convergent if x < 1l , where l = lim an

1 n

and divergent if x < 1l . Thus, if r = l , it follows that the power series converges absolutely if | x | < r and diverges if | x | > r. If a1, a2,… are all real and x is real, then the interval –r < x < r, inside which the power 1

M01_Baburam_ISBN _C01.indd 58

(2)

a1 + 2a2 x + 3a3 x 2 + …+ nan x n −1 + … a

= lim

1 n

1

= lim n n .lim an

1 n

= lim an

1 n

and so the radius of convergence of the differentiated series (2) is the same as that of the series (1). Thus, the differentiated series has same interval of convergence as the original series. In general, “A real power series can be differentiated any number of times inside the interval of convergence”. Furthermore, a power series Sanxn with radius of convergence r converges uniformly on [– r + e, r + e], e > 0. Thus, if f is the sum of the series Sanxn, then x

∞

x

∫ f ( x ) dx =∑ ∫ a x dx, a

n =1 a

n

n

x < r.

Hence, “A real power series can be integrated term by term any number of times inside its interval of convergence”. EXAMPLE 1.94

Examine for convergence the series convergence.

∞

∑ n=0

xn nn

for

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SequenceS and SerieS   n 1.59 Solution. The given series is a power series with

an =

1 nn

.

Therefore, its radius of convergence is 1 1 r= = = ∞. 1 1 n lim an lim 1n n Hence, the series converges absolutely for all values of x. EXAMPLE 1.95

Examine for convergence the series x 2 x3 x 4 x − + − +… 2 3 4 Solution. The radius of convergence for this series is

r=

1 lim

1 n

1 n

= 1.

5. Show that the sequence {an}, where 1 1 1 an = 1 + + +  + 4 7 3n − 2 cannot converge 6. Using Cauchy’s General Principal of Convergence, show that the sequence { n+n 1} is convergent n + 1 − n ] = 0. 7. Show that nlim[ →∞ II Comparison Test Examine the convergence of the following series: 8.

∞

1

∑(n2 + 1) 2 − n

9.

∞

∑ n= 2

10.

EXERCISES

1 log n

∞

∑n n =1

p

1. Show that the sequence {rn} converges if -l < r < 1. 2. Show that lim 3+ 2n n = 2.

p> 11.

Ans. (i) {1 − 1n } , (ii) {n}. 4. If an = (−1) (1 +

1 n

)

for all natural number

n, show that lim an = 1 and lim an = −1



Hint: The sequence is {−2, 32 , −34 , 54 , −56 , …}. The set inferior numbers and superior numbers are and {−2, − 34 , − 65 , …} 3 5 { 2 , 4 , …}. Therefore, 4 6  lim an = lub −2, − , −  = −1, 3 5 

M01_Baburam_ISBN _C01.indd 59

∞

∑(n + 1)

1 3

Ans. Converges for 1 2

and diverges for p ≤

1

1 2

Ans. Divergent.

− n3

n =1

n →∞

3. Give an example of a monotonically increasing sequence that is (i) convergent (ii) not convergent

Ans. Divergent.

1 (n + 1) p

Sequences

n

Ans. Divergent.

n =1

Therefore, it converges absolutely on the interval –1 < x < 1. It converges at x = 1 by Leibnitz test but diverges at x = –1.

I

 3 5  lim an = glb − , − , … = 1,  2 4 

12. 13. 14.



1 1+ 2 1+ 2 + 3 + 2 + 2 +… 2 2 1 1 + 2 1 + 22 + 32 Ans. Divergent. 2 −1 3 −1 4 −1 + 3 + 3 +… 3 −1 4 −1 5 −1 Ans. Convergent. 2

1 2 3 n −+ + + + +… 4 6 8 2(n + 1)

Ans. Divergent. 15.

1 1+ 2

+

2 1+ 2 3

+

3 1+ 3 4

+… Ans. Divergent.

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1.60  n  Chapter One 16.

1 1 1 + + +… 12 2.3 34

17.

∑ (n

18.

∑ sin



Hint:

Ans. Convergent.

(n + 1)(n + 2) 2 + 1)(n 2 + 2)

Ans. Converges for x < 1 and diverges if x > 1. 28.

un = sin 1n , vn = 1n ,

take

u

∑

1 n

Ans. Converges for x < 1

diverges.

and diverges if x > 1.

Examine the following series for convergence 2! 3! 4! + + +… 22 33 44

Ans. Convergent.

∑3

IV

Cauchy’s Root Test

21.

∑(

n

Ans. Convergent.



Examine the following series for convergence 30.

(3 x + 5) n (n + 1)!

∑

n

2

29.

D’AIemberl’s Ratio Test

20.

X x 2 x3 + + + 13 35 57

then

Ans. Divergent.

19. 1 +

2 n +1 − 2 n x n +1 +1

∑2

Ans. Convergent.

1 n

lim vnn = 1, but

III

27.

(n + 1) n n ∑ n n +1 x

Ans. Converges for x < 1 and diverges if x > 1.

Ans. Convergent for all values of x.

22.

2

n 2n

)

+ n12

∞

Ans. Convergent.

nx n

∑ (n + 1)(n + 2)

Ans. Convergent for x < 1, divergent for x ≤ 1.



24.

n3 − n + 1 ∑ n! x 2 3

+

x2 3 4

+

25.

32. 33.

Ans. Convergent. x3

4 5

369 ⋅⋅3n 5n ∑ 4.7.10 (3n + 1) . 3n + 2 n

n



Ans. Convergent.



n

− n − ( −1)n

Ans. Convergent.



3 2 32 (n + 1) n x + 3 x 2 + 4¯ x 3 +  + + 2 3 1 2 n n +1 Ans. Converges for x < 1 and

∑ ( )

35.

∑

Ans. Convergent.

n +1 n +1 n

− nn+1  

−n

Ans. Convergent.

x > 1 and divergent for x > 1.

∑ 1+ 2

∑5

34.

(1+ nx )n nn

Ans. converges for x < 1, diverges for x > 1.

Ans. Convergent. 26.

1

diverges if x > 1. +

Ans. Convergent for

∞

∑ ( log n) 2

n =1

23.

31.

V

Raabe’s Test, Logarithmic Test, deMorgonBertrand Test, and Gauss Test

Examine for the convergence of the series

M01_Baburam_ISBN _C01.indd 60

1/2/2012 11:34:20 AM

SequenceS and SerieS   n 1.61 36.

∑

4.7.10 …( 3n + 1) 1.2.3… n

xn

Ans. Converges for x < 13 , 1 diverges for x ≥ 3 . 1.3.5…( 2n − 1)

37.

∑

38.

∑ 7.10.13…(3n + 4) x

2.4.6 … 2n

Ans. Divergent.

3.6.9 … (3n)

9  2 3  3  = 1 + + 2 1 − + 2 −… n n 4 n n   



1 3 − 2 + n 4n By Gauss’s Test, the series diverges. = 1+

44.

x 1 x 3 1.3 x 5 1.3.5 x 7 + . + . + . + 1 2 3 2.4 5 2.4.6 7

Ans. Converges for x < 1, and diverges for x > 1. 2 3 n x x x x 45. + + + + 1.2 3.4 5.6 7.8

n

Ans. Converges for x < 1, diverges for x > 1.

Ans. Converges for x < 1, diverges for x > 1.

12 ⋅ 32 … (2.n − 1) 2 n x 39. ∑ 2 2 2 2 ⋅ 4 ⋅ 6 … (2n) 2 Ans. Converges for x < 1, diverges for x > 1.

Hint: Use Ratio Test and then Gauss’s Test as Raabe’s Test fails

40.

∑

(a + nx) n n!

Ans. Converges for x < 1e ,

n +1

x 41. ∑ (n + 1) log (n + 1) Ans. Converges for x < 1, diverges for x > 1.

Ans. Converges for x < 14 , diverges for x ≥ 14 .

43.

22 22.42 22.42.62 + + + 32 32.52 32.52.7 2



22.42 … (2n) 2 , Hint: un = 2 2 2 3 .5 .7 …(2n + 1) 2 2

un 3   1  = 1 +  1 +  un +1  2n   n 

M01_Baburam_ISBN _C01.indd 61

Cauchy’s Integral Test, Cauchy’s Condensation Test

47. Apply Cauchy’s Integral Test to examine the convergence of the series.

(a)

∞

∑n n =1



(b)

1 +1

Ans. Convergent.

1 +n

Ans. Convergent.

2

∞

∑n n =1

2

48. Prove that

∞



(1!) 2 (2!) 2 4 (3!) 2 6 x + x + 1 x + 2! 4! 6

Ans. Converges for x2 < 4, diverges for x2 > 4. VI

diverges for x ≥ 1e .

(2n)! n x 42. ∑ 2 n =1 ( n !)

46. 1 +

−2

∑

1 ( n log nlog 2 n(log k n ) p

, where log2n

denotes log log n, etc, converges if p > 1, diverges if p < 1. VII Alternating Series 49. Examine the convergence of x x2 x3 x4 − + − + (a) 2 3 1 + x 1 + x 1 + x 1 + x4 Ans. Convergent. Hint: Terms decrease monotonically, un = xn = 1+1 1 → 0 as n → ∞ . 1+ x n xn

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1.62  n  Chapter One

Hence, by Leibnitz’s Test, the given alternating series converges.

∑(−1)

n −1

n n2 + 1

Ans. Convergent.



(b)



(c)  1 − 1  −  1 − 1  +  1 − 1  − 2



log 2   2

log 3   2

1 1   −  +  2 log 5 

Hint: un → 12 (finite) and so the series oscillates finitely 1 1 1 (d) 1 − + − + 2 2 3 3 4 4 Ans. Convergent.

VIII Absolute Convergence / Conditional Convergence 50. Show that the series 1 1 1 1 1 1 1 (a) − . 2 + . 3 − . 4 + … 2 2 2 3 2 4 2 n



 1  3.6 … (3n) (b) ∑  −  .  2  2.5… (3n − 1)



(c) (−1) n +1 2n



(d)

3

n

xn

∑ n +1,0 < x < 1 ( −1) n



(a)

∑



(b)

∑ n − log n



n ( −1) n

and

( −1) n +1 (c) ∑ log (n + 1) are conditionally convergent.

52. Show that the series

cos x cos 2 x cos 3 x − + − 13 23 33 converges absolutely.

M01_Baburam_ISBN _C01.indd 62

IX

cos nx 1 ≤ 3 and 3 n n

1

∑n

3

converges.

Derangement of Series

53. Show that 1 1 1 1 1 1 1 1 − − + − − + −  = log 2 2 4 3 6 8 5 2 54. Find the sum of the series 1 1 1 1 1 1 1 1 − − + − − + + − 2 4 6 3 8 10 5 Ans. X

1 2

log 34 .

Uniform Convergence

55. Show that the sequence {sn ( x)} = { x +1 n } converges uniformly in [a, b], a > 0. 56. The partial sum sn(x) of a series Sun(x) is

sn ( x) = 1+nn4xx2 . Show that its convergence is 2

non-uniform on [0,1].

57. Examine for uniform convergence the ∞

series ∑x.e − nx in the interval [0,1]

and

are absolutely convergent. 51. Show that the series



log 4 

Hint:



n=0

Ans. 0 is point of non-uniform convergence.

58. Show that the series 2x 4 x3 8x7 + + + 2 4 1 + x 1 + x 1 + x6 is uniformly convergent in−l < x < 1 . 59. Test for uniform convergence of the series x ∑ (n + x 2 ) 2 Ans. Converges uniformly for all values of x.

{ }

60. Show that the sequence {sn ( x)} = 1+ ,xnx2 converges uniformly on the real line.

61. Apply Weierstrass’s M-Test to show that the series [1, ∞] .

∞

∑ n −1

1 1+ n 2 x

converges uniformly on

Hint: | un ( x) | ≤ 1+ 1n2 x
0, such that f(x) < f(c) = m for x ∈ (c,c + d). This contradicts the fact that m is the infimum. If f ′(c) > 0, then there exists d > 0, such that f(x) < f(c) = m for x ∈ (c – d,c), which again contradicts the fact that m is the infimum. Hence, f ′(c) = 0. A.  Geometric Interpretation of Rolle’s Theorem If we draw the curve y = f (x), which is continuous on [a, b] and derivable on (a, b), then Rolle’s Theorem states that between two points with equal ordinates on the graph of f, there exists at least one point where the tangent is parallel to the x-axis.

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2.10  n  chapter two Y

EXAMPLE 2.18

Y

Verify Rolle’s Theorem in case of the function f (x) = 2x3 + x2 – 4x–2.

c

c

A

A f(a)

0

Solution. The function f is continuous and its derivative exists for any real value of x. If f(x) = 0, then 2x3 + x2 – 4x –2 = 0 or (x2 – 2)(2x + 1) = 0, which yields x = ± 2, − 12 . Thus,

f(a)

a0

a

B

B

f(b)

f(b)

b

f

We choose the interval  − 2, 2  so that in this   interval all the conditions of Rolle’s Theorem are Xsatisfied. Now

X

b

f ′(c) = 6x2 + 2x – 4 = (3x – 2)(x + 1), which vanishes for x = –1 and x = 32 . Both the points lie in the interval − 2, 2 Hence, the Rolle’s Theorem is verified

Y

Y

( 2 ) = f ( − 2 ) = f  − 12  = 0.

(

)

EXAMPLE 2.19

A

A

0

a

f(a)

0 Y

B

f(b)

f(b)

c

c f(a)

B

Verify Rolle’s Theorem when f (x) = (x–a)m (x–b)n, where m and n are positive integers. Solution. f (x) = 0 yields x = a and x = b. So, we consider the interval [a, b]. The given function is clearly continuous and differentiable in [a, b] and f (a) = f (b) = 0. Also

a

b

X

X

b

f ′(x) = (x – a)m.n(x–b)n–1 + m(x–a)m–1(x–b)n. Substituting f ′(x) = 0, we get

Y

x = a, b, and

na + mb . m+n

na + mb

c1 c1

Out of these values m + n lies in the open interval (a, b) and hence, the Rolle’s Theorem is verified

c3 c3

A A

B B

EXAMPLE 2.20

cc22 0

0

X

B.  Algebraic Interpretation of Rolle’s Theorem In Rolle’s Theorem, f (a) = f (b). So if a and b are zeros of f, then Rolle’s Theorem says that between two zeros a and b of f, there exists at least one zero of f ′(x).

M02_Baburam_ISBN _C02.indd 10

X

Discuss the application of Rolle’s Theorem to the function 2 f ( x) = 2 + ( x − 1) 3 in the interval [ 0, 2]. Solution. We note that 2

f (0) = 2 + (−1) 3 and f (2) = 2 + 1 = 3 and so, f(0) ≠ f(2). Hence, the Rolle’s Theorem is not applicable to this function.

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SucceSSive differentiation, Mean value theoreMS and expanSion of functionS   n 2.11 EXAMPLE 2.21

If f, f, and y are continuous on [a, b] and derivable on (a, b), show that there exists a point c ∈ (a, b) such that f (a) φ (a ) ψ (a ) f (b) φ (b) ψ (b) = 0. f ′(c) φ ′(c) ψ ′(c) Solution. Consider the function F defined b

φ (a ) φ (b) φ ( x)

f (a) F ( x) = f (b) f ( x)

ψ (a ) ψ (b) . ψ ( x)

In F(a) and F(b), two rows in the determinant become identical in each case. Therefore, F(a) = F(b) = 0. Since f, f, and y are continuous on [a, b] and differentiable on (a, b), F also has these properties. Hence, all the conditions of Rolle’s Theorem are satisfied for F in this interval. Therefore, there exists c ∈ (a, b) such that F′(c) = 0. But f (a) φ (a) ψ (a ) F ′( x) = f (b) φ (b) ψ (b) . ′ ′ f ( x) φ ( x) ψ ′( x) Hence, F′(c) = 0 implies f (a) φ (a ) f (b) φ (b) f ′(c) φ ′(c)

ψ (a ) ψ (b) = 0. ψ ′(c)

EXAMPLE 2.22 C

C

C

If C0 + 21 +…+ nn−1 + n +n1 = 0, where C0, C1,…,Cn are real constants, show that the equation C0 + C1x + … + Cn-1 xn–1 + Cnxn = 0 has at least one root between 0 and 1. Solution. Let f(x) = C0 + C1x + … + Cn-1 xn–1 + Cnxn. Consider x2 x3 x n +1 φ ( x) = C0 x + C1 + C2 + …+ Cn . n +1 2 3 Cn −1 C1 Then, f (0) = 0 and φ (1) = C0 + + …+ n 2 Cn + = 0 (given). n +1

M02_Baburam_ISBN _C02.indd 11

Therefore, f(0)=f(1)=0. Also, f(x) is continuous in [0,1] and derivable in (0, 1). Thus, f(x) satisfies all the conditions of Rolle’s Theorem. Hence, there exists x ∈ (0,1) such that f ′(x) = 0. But,

f ′(x) = f(x)

and so, 0 = f ′(x) = f(x). This implies that x ∈ (0,1) is a root of f(x) = 0, which is the thing we wished to highlight. EXAMPLE 2.23

If f is continuous in [a, c] and derivable in (a, c), deduce from Rolle’s Theorem by considering

φ ( x) = f ( x) − f (a) −

x−a [ f (c ) − f ( a ) ] , c−a

that f(c) – f(a) = (c–a)f ′(x), a < x < c. Solution. We note that

(i) being the sum of continuous functions, f (x) is continuous, (ii) f (x) is differentiable for all real values of x, and (iii) f (a) = 0 = f (c). Thus, all conditions of Rolle’s Theorem are satisfied. Therefore, there exists a point x ∈ (a,c) such that f′(x) = 0. But, 1 [ f (c) − f (a )] φ ′( x) = f ′( x) − c a − and so f′(x) = 0 implies f(c) – f(a) = (c–a)f ′(x), x ∈ (a,c).

Theorem 2.3. (Lagrange’s Mean Value Theorem). If a function f defined on a, b] is (i) continuous on [a, b] and (ii) derivable in (a, b), then there exists at least one real number c∈ (a,c) such that f (b) − f (a ) = f ′(c). b−a Proof: Consider the function

f (x) = f(x) + Ax, where A is a constant to be determined such that f (a) = f (b). Since f (a) = f(a) + Aa, and f (b) = f(b) + Ab, therefore, f (a) = f (b) yields

1/2/2012 11:53:00 AM

Y

2.12  n  chapter two A=

Y

f (b) − f (a ) . b−a

B

The function f is (i) continuous on [a, b], being the sum of two continuous functions, (ii) derivable on (a, b), and (iii) f (a) = f (b). Thus, f satisfies the conditions of Rolle’s Theorem. Hence, there exists a real number c ∈ (a,b) such that f′(c) = 0. But, f′(x) = f ′(x) + A. Therefore, f ′(c) = 0 implies f (b) − f (a ) 0 = f ′(c) + A = f ′(c) − . b−a Hence,

Remark 2.1 (i) Lagrange’s Mean Value Theorem is also known as First Mean Value Theorem of Differential Calculus. (ii) If we put b = a + h, then the number between a and b can be written as a + qh, 0 < q < 1. Thus, Lagrange’s Mean Value Theorem takes the form f(a + h) – f (a) = h f ′(a + qh) or f(a + h) – f (a) = h f ′(a + qh), 0 < q 0 for all x ∈ [a,b], then f is strictly increasing on [a, b]. In fact, if x1, x2 ∈ [a,b], then by Lagrange’s Mean Value Theorem,

X



f ( x2 ) − f ( x1 ) = f ′(c) > 0, x1 < c < x2 . x2 − x1

1/2/2012 11:53:01 AM

SucceSSive differentiation, Mean value theoreMS and expanSion of functionS   n 2.13 Hence, f(x2) – f(x1) > 0 or f(x2) > f(x1) for x2 > x1 and so, f is strictly increasing on [a, b]. Similarly, if f ′(x) < 0 for x ∈ [a,b], then f is strictly decreasing in [a, b]. Theorem 2.4. (Cauchy’s Mean Value Theorem). If two functions f and F defined on a, b] are (i) continuous on [a, b], (ii) derivable on (a, b), and (iii) F′(x) ≠ 0 for any x ∈ (a,b), then there exists a point c ∈ (a,b) such that f (b) − f (a ) f ′(c) = , a < c < b. F (b) − F (a ) F ′(c)

[Obviously F(b) ≠ F(a). Otherwise, F satisfies all the conditions of Rolle’s Theorem and as such, F′(c) = 0, which contradicts condition (iii) of the statement]. Proof: Consider the function f defined b f (x) = f(x) + AF(x), where A is a constant to be determined such that f (a) = f (b). Since f (a) = f(a) + AF(a) and f (b) = f(b) + AF(b), f (a) = f(b) implies f(a) + AF(a) = f(b) + AF(b). This relation yields A=−

f (b) − f (a ) . F (b) − F (a )

The function f, being the sum of two continuous and derivable functions is itself (i) continuous on [a,b], (ii) derivable on (a,b), and (iii) f (a) = f (b). Hence, by Rolle’s Theorem, there exists a point c ∈ (a,b) such that f′(c) = 0. But f′(x) = f ′(x) + AF′(x). Therefore, f′(c) = 0 implies f ′(c) + AF′(c) = 0

or

f ′(c) −

M02_Baburam_ISBN _C02.indd 13

f (b) − f (a ) F ′(c) = 0 F (b) − F (a )

or

f (b) − f (a ) f ′(c) = . F (b) − F (a ) F ′(c)

Another Form of the Statement: If two functions f and F are continuous in [a, a + h], derivable in [a, a + h], and F′(x) ≠ 0 for any xe (a ,a + h), then there exists at least one number q between 0 and 1 such that f ( a + h) − f ( a ) f ′(a + θ h) = . F (a + h) − F (a ) F ′(a + θ h)

[f ′(x) and F′(x) should not vanish for the same value of x]. Deduction. If we define F as F(x) = x, then Cauchy’s Mean Value Theorem reduces to the Lagrange’s Mean Value Theorem. In fact, in that case F′(x) = 1, F(b) = b, F(a) = a and so, f (b) − f (a ) = f ′(c), a < c < b. b−a EXAMPLE 2.24

A twice-differentiable function f is such that f(a) = f(b) = 0 and f(c) > 0 for a < c < b. Show that there is atleast one value x between a and b for which f ″(x) < 0. Solution. Since f ″ exists, f and f ′ both exists and are continuous on [a,b]. As a < c < b, then by applying Lagrange’s Mean Value Theorem to f on the intervals [a, c] and [c, b], we have f (c ) − f ( a ) = f ′(ξ1 ), a < ξ1 < c c−a

and f (b) − f (c) = f ′(ξ 2 ), c < ξ 2 < b. b−c

Since f(a) = f(b) = 0, these equations reduce to f ′(ξ1 ) =

f (c ) , a < ξ1 < c c−a

and f (c ) , c < ξ 2 < b. b−c Since, f ′ is continuous and derivable on [x1, x2], the use of Lagrange’s Mean Value Theorem yields f ′(ξ 2 ) = −

1/2/2012 11:53:02 AM

2.14  n  chapter two Solution. Let f(x) = cos x, a < x < b. This function

f ′(ξ 2 ) − f ′(ξ1 ) = f ′′(ξ ), ξ1 < ξ < ξ 2 ξ 2 − ξ1

or f ′′(ξ ) =

− f (c) b−c

− cf −( ca) (b − a ) f (c) =−

x sin x 2 Solution. We want to show that

Show that

Show that

Since x sin x > 0 for 0 < x < π2 , it is sufficient to show that tan x sin x − x 2 > 0, 0 < x
0, 0 < x < . x sin x 2

π

satisfies the conditions of Lagrange’s Mean Value Theorem. Therefore, using the theorem, we get f (b) − f (a ) = f ′(c) = − sin c, b−a

.

v−u v−u < tan −1v − tan −1u < , 0 < u < v. 2 1+ v 1+ u2 Solution. Consider the function f defined by f(x) =

tan–1x. We have

f ′( x) =

Let f(x) = tan x sin x–x2. Then,

1 . 1 + x2

Lagrange’s Mean Value Theorem is applicable to f and we have = sin x + sin x sec 2 x − 2 x. tan −1v − tan −1u 1 = , u < ξ < v. The function f ′ is continuous and derivable on v−u 1+ ξ 2 π . Therefore, ( 0, 2 ) Since x < ν, we have f ′′( x) = cos x + cos x sec 2 x + 2 sin x sec 2 x tan x − 2 1 1 . > 2 2 2 1 1 v2 ξ + + ′′ f ( x) = cos x + cos x sec x + 2 sin x sec x tan x − 2 Similarly, x > u, yields = sec x + cos x − 2 + 2 sin x tan xsec 2 x 1 1 2 . < 2 = sec x − cos x 1+ ξ 1+ u2 f ′( x) = cos x tan x + sin xsec 2 x − 2 x

(

)

+2 sin x tan x sec 2 x > 0 for 0 < x
0, it follows that f ′ is strictly increasing in ( 0, π2 ) . Also f ′(0) = 0. Therefore, f ′(x) > 0 for 0 < x < π2 . This implies that f is strictly increasing in 0, π2  . Also f(0)=0 and so, f(x) > 0 for 0 < x < π2 . Hence, tanx sin x– x2 > 0 for 0 < x < π2 , which in turn implies that x tan x π ,0 < x < . > x sin x 2 EXAMPLE 2.26

Show that |cos b – cos a| < |b – a|.

M02_Baburam_ISBN _C02.indd 14

Therefore, 1 tan −1v − tan −1u 1 < < 2 v−u 1+ v 1+ u2 or

v−u v−u < tan −1v − tan −1u < , 0 < u < v. 2 1+ v 1+ u2

EXAMPLE 2.28

Show that sin α − sin β π = cot θ , 0 < α < θ < β < . cos β − cos α 2 Solution. Let f(x) = sinx and F(x)= cosx, x ∈

(a, b). Since f and F are continuous in [a, b],

1/2/2012 11:53:03 AM

SucceSSive differentiation, Mean value theoreMS and expanSion of functionS   n 2.15 the conditions of the Cauchy’s Mean Value Theorem are satisfied. Therefore, there exists q ∈ (a, b) such that f ( β ) − f (α ) f ′(θ ) = , F ( β ) − F (α ) F ′(θ )

that is, sin β − sin α cos θ =− , cos β − cos α sin θ

Theorem 2.5. (Taylor’s Theorem). If a function f defined on a,a + h] is such that (i) the (n – 1)th derivative f(n – 1) is continuous in [a, a + h] and (ii) the n th derivative f (n) exists in (a, a + h), then there exists at least one real number q between 0 and 1 such that f (a + h) = f (a ) + hf ′(a ) +

that is, sin α − sin β π = cot θ , 0 < α < θ < β < . cos β − cos α 2

+

EXAMPLE 2.29

If a function f is such that its derivatives f ′ is continuous on [a, b] and derivable on (a, b), show that there exists a number c ∈ (a,b) such that 1 f (b) = f (a ) + (b − a ) f ′(a ) + (b − a ) 2 f ′′(c). 2 Solution. Consider the function f(x) = f(b) –f (x) –(b–x) f ′(x)–(b–x)2A, where A is the constant such that f(a)= f(b). The function f satisfies all the conditions of Rolle’s Theorem. Since f(a)= f(b), we have f (b) − f (a ) − (b − a ) f ′(a ) − (b − a ) 2 A = 0. (1)

Further,

φ ′( x) = f ′( x) − f ′( x) − (b − x) f ′′( x) + 2(b − x) A = −(b − x) f ′′( x) + 2(b − x) A. By Rolle’s Theorem, there exists c ∈ (a,b) such that f′(x) = 0. Therefore, −(b − c) f ′′(c) − 2(b − c) A = 0 or

+…+

(b − c)[2 A − f ′′(c)] = 0

or A=

1 f ′′(c), b ≠ c. 2

Then (1) yields 1 f (b) = f (a ) + (b − a ) f ′(a ) + (b − a ) 2 f ′′(c). 2

M02_Baburam_ISBN _C02.indd 15

h2 f ′′(a ) 2!

h n −1 f ( n −1) (a ) (n − 1)!

h n (1 − θ ) n − p ( n ) f (a + θ h), p (n − 1)!

where p is a positive integer.

Proof: The condition (i) of the statement implies

that all the derivatives f′, f ″,…, f(n–1) exist and are continuous on [a, a + h]. Let a + h = b. Consider the function F ( x) = f ( x) + (b − x) f ′( x) +…+

(b − x) n −1 ( n −1) f ( x) + A(b − x) p , (n − 1)!

where A is a constant to be determined, such that F(a) = F(b). But, F (a ) = f (a ) + (b − a ) f ′(a )

+…+

(b − a ) n −1 ( n −1) f (a ) + A(b − a ) p , (n − 1)!

F(b) = f(b).

Therefore, F(a) = F(b) implies f (b) = f (a ) + (b − a ) f ′(a ) + …+

(b − a ) n −1 (n − 1)!

(1) × f ( n −1) (a ) + A(b − a ) p The function F is continuous in [a, a + h], derivable in (a, a + h), and F(a) = F(a + h). Hence, by Rolle’s Theorem, there exists at least one real number q between 0 and 1 such that F′(a + qh) = 0. But, F ′( x) =

(b − x) n −1 ( n ) f ( x) − pA(b − x) p −1 . (n − 1)!

1/2/2012 11:53:03 AM

2.16  n  chapter two Therefore, F′(a + qh) = 0 implies n −1

h f ( n ) (a + θ h) − pA(h − θ h) p −1 = 0 (n − 1)!

or h n −1 f ( n ) (a + θ h) − pAh p −1 (1 − θ ) p −1 = 0 (n − 1)!

and so, A=

h

(1 − θ ) p (n − 1)!

n− p

n− p

f ( n ) (a + θ h), h ≠ 0, θ ≠ 1.

Substituting this value of A in (1), we get f (b) = f (a ) + (b − a ) f ′(a ) +

(b − a ) 2 f ′′(a ) 2!

(b − a ) n −1 ( n −1) +…+ f (a) (n − 1)!

+

or

h

(1 − θ ) p (n − 1)!

n −1

n− p

f

f (a + h) = f (a ) + hf ′(a ) + +…+



+

The term Rn =

(n)

(a + θ h) and

h2 f ′′(a ) 2!

h n −1 f ( n −1) (a ) (n − 1)!

h (1 − θ ) f ( n ) (a + θ h). p (n − 1)! n

n− p

h n (1 − θ ) n − p ( n ) f ( a + θ h) p (n − 1)!

is called Taylor’s remainder after n terms and the theorem with this form of remainder is known as Taylor’s Theorem with Schlomilch and Roche form of remainder. Substituting p = 1, we get Rn =

h n (1 − θ ) n −1 ( n ) f (a + θ h), (n − 1)!

which is called Cauchy’s form of the remainder. Substituting p = n, we get Rn =

M02_Baburam_ISBN _C02.indd 16

hn (n) f (a + θ h), n!

which is called Lagrange’s form of the remainder. If we put a + h = x, then the Taylor’s Theorem reads ( x − a)2 f ( x) = f (a ) + ( x − a ) f ′(a ) + f ′′(a ) 2! +…+

+

( x − a ) n −1 ( n −1) f (a) (n − 1)!

( x − a)n (1 − θ ) n − p f ( n ) (a + θ ( x − a )). p (n − 1)!

Substituting a = 0 in this last expression, we have x2 f ( x) = f (0) + xf ′(0) + f ′′(0) 2!

+…+ +

x n −1 f ( n −1) (0) (n − 1)!

xn (1 − θ ) n − p f ( n ) (θ x), p (n − 1)!

which is known as Maclaurin’s Theorem with Schlomilch and Roche form of remainder. Substituting p = 1, we get xn Rn = (1 − θ ) n −1 f ( n ) (θ x), (n − 1)!

which is Cauchy’s form of remainder for Maclaurin’s Theorem. Substituting p = n, we get xn (n) Rn = f (θ x), n! which is Lagrange’s form of remainder for Maclaurin’s Theorem. 2.4  TAYLOR’S INFINITE SERIES AND POWER SERIES EXPANSION The Taylor’s Theorem asserts that h2 f ( a + h) = f ( a ) + h f ′ ( a ) + f ′′ (a ) 2! +…+

h n −1 f ( n −1) (a ) + Rn , (n − 1)!

Where Rn is the remainder after n terms. Thus, f (a + h) = S n + Rn ,

1/2/2012 11:53:03 AM

SucceSSive differentiation, Mean value theoreMS and expanSion of functionS   n 2.17 Since f (x) = log sin x, we have

where h2 Sn = f (a) + h f ′(a) + f ′′ (a ) 2!



h n −1 f +…+ ( n − 1)!

( n −1)

( a)

. Thus, if Rn → 0 as n → ∞ , then lim S n = lim [ f (a + h) − Rn ] = f (a + h), n →∞

n →∞

and so the series

cos x = cot x, sin x



f ′′ ( x) = −cosec 2 x,



f ′′′ ( x) = −2 cosec x ( − cosec x cot x)

= 2cosec 2 x cot x ,



and so on. Hence,

h2 f (a) + h f ′(a) + f ′′ (a ) 2!

log sin x = f (3 + h) = log sin 3 + ( x − 3) cot 3

h n −1 f ( n −1) (a ) + … +…+ (n − 1)! ,

called Taylor’s Series, converges to f (a + h). On the other hand, if we put a + h = x, then f ( x) = f (a) + ( x − a ) f ′ (a ) +



f ′ ( x) =

( x − a)

2

f ′′ (a ) + …, 2! which is the expansion of f (x) in powers of (x – a).

−



( x − 3)2 cosec2 3 2

3 x − 3) ( + cosec 2 3 cot 3 + …

3



EXAMPLE 2.31

If 0 < θ < 1 , show that log (1 + x) = x −

2.5  MACLAURIN’S INFINITE SERIES If the remainder Rn in the Maclaurin’s expansion of a function tends to zero, then we get 2

and hence, log (1 + x) < x −

3

x x f ′′ (0) + f ′′ (0) + …, 2! 3! which is called Maclaurin’s Infinite Expansion of f in power of x. The series 2 is called f (0) + x f ′ (0) + x2! f ′ (0) + … f ( x) = f (0) + xf ′ (0) +

Maclaurin’s Infinite Series.

x2 x3 + 2 3 (1 + θ x )3

x 2 x3 + , x > 0. 2 3

Solution. By Maclaurim’s Theorem with remainder R3, we have



f ( x) = f (0) + xf ′ (0) + +

x2 f ′′ (0) 2

x3 f ′′′ (θ x), 0 < θ < 1. 3!

2.6  EXPANSION OF FUNCTIONS



EXAMPLE 2.30

It is given that f ( x) = log (1 + x), which gives f (0) = 0. Therefore,

Expand log sin x in powers of (x – 3). Solution. If f ( x) = log sin x , then

f ( x) = f (3 + ( x − 3)) = f (3 + h) , h = x − 3 . Therefore, by Taylor’s Series Expansion, f (3 + h) = f (3) + hf ′ (3) +

M02_Baburam_ISBN _C02.indd 17

h2 h3 f ′′ (3) + f ′′ (3) + … 2! 3!

1 ⇒ f ′ (0) = 1, 1+ x 1 f ′′ ( x) = − ⇒ f ′′ (0) = −1, and (1 + x )2

f ′ ( x) =

f ′′′ ( x) =

2

(1 + x )

3

⇒ f ′′′ (0) =

2

(1 + θ x )3

.

1/2/2012 11:53:04 AM

2.18  n  chapter two Hence, x2 x3 f ( x) = log (1 + x) = x − + . 2 3 (1 + θ x )3 Further, since x > 0 and θ > 0 , we have θ x > 0 and so,

(1 + θ x )

3

> 1 or

Therefore,

1

(1 + θ x )3

< 1.

x 2 x3 log (1 + x) < x − + , for x > 0 . 2 3

= 0.84857 − 0.01745



−0.0006 (0.86602) + …

= 0.8481, correct to four decimal places. EXAMPLE 2.33

Expand loge x in powers of ( x − 1) and hence evaluate loge 1.1, correct to four decimal places. We have given that f ( x) = log e x . Therefore, f (1) = 0. Further Solution.

f ′ ( x) =

EXAMPLE 2.32

1 and so f ′ (1) = 1, x

Use Taylor’s Series Expansion to compute the value of cos 32º, correct to four decimal places.

f ′′ ( x) = −

1 and so f ′′ (1) = −1, x2

Solution. Let f (x) = cos x. Then, f (x + h) = cos (x

f ′′′ ( x) =

2 and so f ′′′ (1) = 2, x3

+ h). By Taylor’s Expansion,

h2 cos ( x + h) = f ( x + h) = f ( x) + hf ′ ( x) + f ′′ ( x) 2 +



f iv ( x) = −

Substituting these values in the Taylor’s Series

3

h f ′′′ ( x) + … 3!

f ( x) = f (a) + ( x − a ) f ′ (a ) +

But,

f ′′′ ( x) = sin x, …

we get

Substituting x = 30 and h = 2 = we have o

o

cos ( x + h) = cos x − h sin x −

π

90

= 0:03490,

h2 h3 cos x + sin x + … 2 6

or

cos 32 = cos 30 − 0.03490 sin 30 o

o

(0.03490)2 − 2





=

3  1 − 0.03490    2 2 −

M02_Baburam_ISBN _C02.indd 18



o

(0.03490) sin 30o + … + 6

0.0006 3 +… 2

( x − a )3 3!

( x − a)2 f ′′ (a ) 2!

f ′′′ (a ) + …,

log e x = f ( x) = f (1) + ( x − 1) f ′ (1)



cos 30o

3



+

f ′ ( x) = − sin x, f ′′ ( x) = − cos x, and



6 and so iv f (1) = −6 , and so on. x4



+ +

( x − 1)2 2

( x − 1)4 24

= ( x − 1) − −

( x − 1)

f ′′ (1) +

( x − 1)3 6

f ′′′ (1)

f iv (1) + …

( x − 1)2 + ( x − 1)3 2

3

4

+ …, 4 which is the required expansion. Substituting x = 1.1in the expansion, we get log e 1.1 = 0.1 − 0.005 + 0.0003 − 0.00002 + …

= 0.0953.

1/2/2012 11:53:05 AM

SucceSSive differentiation, Mean value theoreMS and expanSion of functionS   n 2.19 EXAMPLE 2.34

Expand (i) ex in power of (x – 1) up to four terms. (ii) tan −1 x in power of ( x − π4 ) .

(iii) sin x in power of ( x − π2 ) and determine sin 91º, correct to four decimal places. Solution. (i) Let f ( x) = e x . Then, expanding in

f ′′ ( x) =

−2 x

(1 + x )

2 2

e x = f ( x) = f [1 + ( x − 1) ]

= f (1) + ( x − 1) f ′ (1) +

But,

+

( x − 1)3 3!

( x − 1)

f ′′′ ( x) = e x and so f ′′ (1) = e,

and so on. Hence, e x = e + ( x − 1)e +



+

( x − 1)

( x − 1) 2

2

( x − 1)

3

e+

3!

e

4

4!

e +…

4   ( x − 1) 2 ( x − 1)3 ( x − 1) = e 1 + ( x − 1) + + + + … . 2 3! 4!  

(ii) Let f ( x) = tan −1 x . Then, by Taylor’s Series Expansion, we have

π 2

But,

 π f ′′   + …  4



 π −1 π , f ( x) = tan −1 x and so f   = tan  4 4



f ′ ( x) =

1  π 1 and so f ′   = 2 , 2   4 1 + π16 1+ x

M02_Baburam_ISBN _C02.indd 19

(

2

2 x − π2 ) ( +

2

2

)

 π  (x − 2 ) π f ′′   + f ′′′   + …  2  2 3! π 3

But,

π π f ( x) = sin x and so f   = sin = 1, 2 2   π π   f ′( x) = cos x and so f ′   = cos = 0, 2 2 π π   f ′′( x) = − sin x and so f ′′   = − sin = −1, 2 2 π   f ′′′( x) = − cos x and so f ′′′   = 0, 2 and so on. Hence,

( x − 2 ) (−1) π  sin x = 1 +  x −  (0) + 2 2  π 2

π  π  tan −1 x = f ( x) = f  +  x −    4 4  

π  π (x − 4 )  π  = f   + x −  f ′  +  4  4  4 2!

( x − π4 ) + … π  π 1 +x− . π −π 2 π 4  4  1 + 16 4 1 + 16

π  π  π   = f   +  x −  f ′  2 2 2 



f (x) = ex and so f (1) = e,



,

π  π  sin x = f ( x) = f  +  x −   2  2 

f ′′′ (1) + …

f ′ ( x) = e x and so f ′ (1) = e,

2

2

tan −1 x = tan −1

f ′′ (1)



)

(iii) Let f (x) = sin x. Then, by Taylor’s Series Expansion, we get

2

2

(

and so on. Substituting these values in the above expansion, we get

Taylor’s Series, we have

−π  π and so f ′′   = 2 4 2 1 + π16

+



( x − π2 )

= 1−



3

3!

( x − π2 ) 2

(0) + 2

+

( x − π2 ) 4!

( x − π2 ) 4!

4

(1) +…

4

−…

Substituting x = 91o , we get x−

π 2

= 91o − 900 = 1o =

π 180

= 0:01746:

1/2/2012 11:53:05 AM

2.20  n  chapter two Hence, sin 91o = 1 −

( 0.01746 )

2

+

2

( 0.01746 )

 x2   x   f  = f  x +−   1+ x   1+ x 

4

24

−…

= 0.9998.

x   = f ( x) +  −  f ′( x)  1+ x 



2

EXAMPLE 2.35

Expand 2 x 3 + 7 x 2 + x − 1 in power of (x – 2).



Solution. Let f ( x) = 2 x + 7 x + x − 1 . Then,

= f ( x) −

3

2

the expansion of f (x), by Taylor’s Theorem, in powers of (x – 2) is f (x) = f (2+(x – 2)) = f (2) + ( x − 2) f ′(2) + +

But,

( x − 3)

( x − 2)

f ′′(2)

x2 x f ′′( x) +…. f ′( x) + 2 1+ x 2 (1 + x )

(ii) Let f ( x) = tan −1 x . Then

= f ( x) + hf ′( x) +



2

f ′′′(2) +…

3!

+…+

But, f ( n ) ( x) = D n tan −1 x

f ( x) = 2 x 3 + 7 x 2 + x − 1 giving f (2) = 45, f ′( x) = 6 x 2 + 14 x + 1 giving f ′(2) = 53,

f ′′( x) = −sin 2θ sin 2θ ,

and f ( n ) (2) = 0 for n ≥ 4 . Substituting these values in the above expansion, we get f ( x) = 45 + 53( x − 2) + 19 ( x − 2 ) + 2 ( x − 3) . 2

3

EXAMPLE 2.36

Use Taylor’s Theorem to show that (i) f 1x+ x = f ( x) − 1+x x f ′( x) + 1+x x . 2!1 f ′( x) −…. )2

(ii) tan −1 ( x + h) = tan −1 x + h sin θ ⋅ sin1 θ −



−…+ ( −1)

where θ = cot −1 x .

( h sin θ )

2

sin 2θ

2

n −1

+

( h sin θ )

( h sin θ )

n

3



and so on. Substituting these values in the above expression, we get tan −1 ( x + h) = tan −1 x + h sin θ sin θ −

sin nθ n

( m −1)2 2



+ …,



x 2 f ′′( x) +….

Solution. (i) From the statement it is clear that we wish to expand f 1x+ x in power of ( − 1+x x ) . We have, by Taylor’s Theorem,

( )



sin 3θ

3

(iii) f (mx) = f ( x) + (m − 1) xf ′( x) +

f ′′′( x) = 2!( sin 3θ sin 3θ ) ,

2

(

h2 h3 sin 2θ sin 2θ + 2!sin 3θ sin 3θ 2 3!

−…+

hn n −1 ( −1) n!

× (n − 1)! sin nθ sin nθ +…

= tan −1 x + h sin θ

sin θ sin 2θ 2 ( h sin θ ) . 1 2

sin 2θ 3 sin 3θ + ( h sin θ ) +… 2 3 .

2

M02_Baburam_ISBN _C02.indd 20

hn (n) f ( x) + … n!

where θ = cot −1 x . So substituting n = 1, 2, 3,…, we get f ′( x) = sin θ sin θ ,

f iv ( x) = 0 giving f iv (2) = 0 ,

2

h2 f ′′( x) 2

= ( −1) n −1 (n − 1)! sin nθ sin nθ ,

f ′′( x) = 12 x + 14 giving f ′′(2) = 38, f ′′′( x) = 12 giving f ′′′(2) = 12,

( )

x  1 −  f ′′( x) + … 2!  1 + x 

tan −1 ( x + h) = f ( x + h)

2

2!

+



+(−1) n −1 ( h sin θ )

n

sin nθ +… n

1/2/2012 11:53:07 AM

SucceSSive differentiation, Mean value theoreMS and expanSion of functionS   n 2.21 But,

(iii) We have f (mx) = f ( x + (m − 1) x) = f ( x) + (m − 1) xf ′( x) +



( m − 1) 2!

2

x 2 f ( x) +….

If f (x) = x3 + 8x2 + 15x – 24, calculate the value 11 of f ( 10 ) by using Taylor’s Series. Solution. We have f ( x) = x 3 + 8 x 2 + 15 x − 24 . Then, by Taylor’s Series, we have f ( x + h) = f ( x) + h f ′( x) + +



h2 f ′′( x) 2!

h3 h 4 iv f ′′′( x) + f ( x) + − 3! 4!

We take x = 1 and h = 101 = 0.1. Also f ′( x) = 3 x 2 + 16 x + 15,

1 , h

1 2 , and f ′′′(h) = 3 , …. h2 h

log (h + x) = log h +

f ′′′( x) = 6, and

EXAMPLE 2.39

Expand the following by Maclaurin’s Theorem: (i) tan x. (ii) ax. (iii) log sec x. (iv) log (1 + x), log (1 – x), and log 11+− xx . Solution. Let f (x) = tan x. Then, by Maclaurin’s Theorem,

f ( x) = f (0) + xf ′(0) +

x2 x3 f ′′(0) + f ′′′(0) + …. 2! 3!

y1 = f ′( x) = sec 2 x = 1 + tan 2 x

f ( x) = 0 . iv

= 1 + y2 giving y1(0) = 1,



Therefore, 0.01  11  f   = f (1) + 0.1 f ′(1) + f ′′(1) 2  10 

0.001 f ′′′(1) 6 = 0 + 3.4 + 0.11 + 0.001 = 3.511 . +

y2 = f ′′( x) = 2 yy1 giving

y2 (0) = 2 ( y )0 ( y1 )0 = 0 , y3 = f ′′′( x) = 2 y + 2 yy2 giving y3 (0) = 2 , 2 1

y4 = f iv ( x) = 4 y1 y2 + 2 y1 y2 + 2 yy3

= 6 y1 y2 + 2 yy3 giving y4 (0) = 0 , y5 = f

EXAMPLE 2.38

Show that x x2 x3 − 2 + 3 −… h 2h 3h Solution. We wish to expand log (x + h) in the ascending power of x. So, let f (h) = log(h). Then log (h + x) = f (h + x) log ( x + h) = log h +

x2 = f (h) + xf (h) + f ′′(h) 2! +

x x2 x3 − 2 + 3 −…. h 2h 3h

But, y = f (x) = tan x giving y(0) = tan 0 = 0,

f ′′( x) = 6 x + 16,

M02_Baburam_ISBN _C02.indd 21

f ′′(h) = −

f ′ ( h) =

Hence,

EXAMPLE 2.37



f (h) = log h,

x3 f ′′′(h) + …. 3!

(5)

( x) = 6 y22 + 8 y1 y3 + 2 yy4 giving

y5 (0) = 16, and so on. Therefore, f (x) = tan x = 0+ x+



x2 x3 x4 (0) + (2) + (0) 2! 3! 4! +

x5 (16) + …. 5!

x3 2 5 + x +…. 3 15 (ii) Let f (x) = ax. Then, by Maclaurin’s Theorem, we have = x+

1/2/2012 11:53:08 AM

2.22  n  chapter two f ( x) = f (0) + xf ′(0) + +

But,

(iv) Let

x2 f ′′(0) 2!

3

f (x) = log(1 + x) so that f (0) = log 1 = 0,

4

x x (n) f ′′′(0) + …+ f (0) + …. 3! n!

f ′ ( x) =

1 so that f ′(0) = 1, 1+ x

f ′′( x) = −

f ( x) = a so that f (0) = 1, f ′( x) = a x log a so that f ′(0) = log a, x

f ′′′( x) =

f ′′( x) = a (log a ) so that f ′′(0) = ( log a ) , 3 f ′′′( x) = a x ( log a ) so that f ′′′ (0) = (log a )3 , and f ( n ) ( x) = a x (log a ) n so that f n (0) = ( log a ) n . 2

x

2

Hence,

f ( x) = a x = 1 + x log a + +

x2 (log a ) 2 2!

x3 xn 3 n ( log a ) + …+ ( log a ) + …. n! 3!

(iii) Let y = log sec x so that y(0) = log sec 0 = log 1 = 0. Then, 1 y1 = . sec x tan x = tan x sec x so that y1(0) = tan 0 = 0, y2 = sec 2 x = 1 + tan 2 x

1

(1 + x ) 2

(1 + x )

f iv ( x) = −

3

2

so that f ′′(0) = −1,

so that f ′′′(0) = 2,

−6

iv so that f (0) = −6 ,

(1 + x ) and so on. Therefore, by Maclaurin’s Theorem, we have log (1 + x) = f (x)

4

= f (0) + xf ′(0) +

= x−

+

x2 x3 f ′′(0) + f ′′′(0) 2! 3!

x 4 iv f (0) +… 4!

x 2 x3 x 4 + − + …. 2 3 4

Now changing x to –x, we get x 2 x3 x 4 log (1 − x) = − x − − − …. 2 3 4

y3 = 2 y1 y2 so that y3 (0) = 2 y1 (0) y2 (0) = 0 ,

Further, 1+ x 1 1+ x 1 = log = [ log (1 + x) − log(1 − x)] log 1− x 2 1− x 2

y4 = 2 y22 + 2 y1 y3 so that y4 (0) = 2 + 0 = 2 ,

=

= 1 + y so that y2(0) = 1, 2 1

y5 = 4 y2 y3 + 2 y2 y3 + 2 y1 y4

  1 x 2 x3 x 4 x 2 x3 x 4  x − + − +…−  − x − − − −…  2 2 3 4 2 3 4  

  1 x 2 x3 x 4 x 2 x3 x 4 = 6 y2 y3 + 2 y1 y4 so that y5 = (0) = x0−, + − +…−  − x − − − −…  2 2 3 4 2 3 4   y6 = 6 y32 + 8 y2 y4 + 2 y1 y5 so that y6 (0) = 16 , 3 5  1 2x 2x = 2 x + + +… and so on. Using Maclaurin’s Theorem, we have 2 3 5  2 3 x x 3 5 x x y = y (0) + xy1 (0) + y2 (0) + y3 (0) = x + + +…. 2! 3! 3 5 x4 x5 x6 + y4 (0) + y5 (0) + y6 (0) + … EXAMPLE 2.40 4! 5! 6! Show that 3 5 7 x2 x4 x6 (i) sin −1 x = x + 12 . x3 + 1.3 . x + 1.3.5 . x +… = + + +… 2.4 5 2.4.6 7 2 12 45 and hence, find the value of π

M02_Baburam_ISBN _C02.indd 22

1/2/2012 11:53:08 AM

SucceSSive differentiation, Mean value theoreMS and expanSion of functionS   n 2.23 or

(ii) tan −1 x = x − x3 + x5 −…. 3

5

(iii) log (1 + sin x) = x − x2 + 2

x3 6

x x − 12 + 24 +…. . 4

5

(iv) e x sec x = 1 + x + 22!x + 43!x +…. 2

3

Solution. (i) Let y = sin–1 x.

1

Then, y1 =

(1)



1 − x2

(2)

2 2 so that y 1 (1 − x ) − 1 = 0.

Differentiating again, we get

(1 − x ) 2 y y 2

1 2

π = 0.5 + 0.0208 + 0.0023 + 0.0003 +… ≈ 0 : 5234. 6

− 2 xy = 0 2 1

or

(3) (1 − x ) y2 − xy1 = 0. Differentiating n times, by Leibnitz’s Theorem, we have

Hence, π ≈ 3.1404 . (ii) Let y = tan −1 x . 1. Then, 1 y1 = or (1 + x 2 ) y1 − 1 = 0. 1 + x2 2. Differentiating once again, we get

(2)

(1 + x 2 ) y2 + 2 xy1 = 0. (3) Differentiating (2) n times by Leibnitz’s Theorem, we get

2

(1 − x 2 ) yn + 2 − (2n + 1) xyn +1 − n2 yn = 0. (4) Substituting x = 0 in (1), (2), and (3), we get

(1)

yn + 2 (1 + x 2 ) + nyn +1 (2 x) +

or

(1 + x ) y 2

n+2

n(n − 1) yn (2) 2!

+2 xyn +1 + 2nyn = 0

+ 2(n + 1) xyn +1 + n(n + 1) yn = 0. (4)

Substituting x = 0 in (1), (2), (3), and (4), we get

y (0) = 0, y1 (0) = 1, y2 (0) = 0. Substituting n = 1, 2, and 3, in (4), we get

y (O) = 0 , y (0) = 0, y1 (0) = 1, y2 (0) = 0, and

y3 (0) = y1 (0) = 1 , y4 (0) = 22 y2 (0) = 0 ,

yn + 2 (0) = − n(n + 1) yn (0) . Taking n = 1,2,3,…, we get

y5 (0) = 32 y3 (0) = 9 , y6 (0) = 42 y4 (0) = 0 , and y7 (0) = 5 y5 (0) = 5 .3 .1 . 2

2

2

2

Therefore, by Maclaurin’s Theorem, we have

Therefore, by Maclaurin’s Theorem, we get sin −1 x = y = y (0) + xy1 (0) +



4

x2 x3 y2 (0) + y3 (0) 2! 3!

5

6

+

x x x y4 (0) + y5 (0) + y6 (0) 4! 5! 6!

+

x7 y7 (0) +  7!

Substituting x = 12 , we get 3

5

1 1 11 3 1 5 1 = +   +   +   2 2 6  2  40  2  112  2 

M02_Baburam_ISBN _C02.indd 23

tan −1 x = y = y (0) + xy1 (0) +

x2 x3 y2 (0) + y3 (0) + 2! 3!

x4 x5 y4 (0) + y5 (0) + … 4! 5! x3 x5 = x− + −… . 3 5 (iii) Let y = log (1 + sin x) so that y (0) = log 1 = 0.



+

cos x Then, y1 = 1 + sin x so that y1 (0) = 1 , (1 + sin x)(− sin x) − cos 2 x y2 = 2 (1 + sin x )

1 x 3 1.3 x 5 1.3.5 x 7 = x+ . + . + . + 2 3 2.4 5 2.4.6 7

sin −1

y3 (0) = −2, y4 (0) = 0, y5 (0) = 4! , and so on.

7

=

−(1 + sin x) 1 =− so that y2 (0) = −1 , 2 1 + sin x (1 + sin x)

1/2/2012 11:53:08 AM

2.24  n  chapter two y3 =

cos x

(1 +

sin x )

2

=

y2 = y1 cos x − y sin x so that y2 (0) = 1 − 0 = 1,

cos x 1 . 1 + sin x 1 + sin x

= − y1 y2 so that y3 (0) = 1 ,

y3 = y2 cos x − y1 sin x − y1 sin x − y cos x = y2 cos x − 2 y1 sin x − y cos x so that y3 (0) = 1 − 0 − 1 = 0 ,

y4 = − y1 y3 − y 22 so that y4 (0) = −1 − 1 = −2 , y5 = − y1 y4 − 3 y2 y3 so that y5 (0) = 5 , and so on. Therefore, Maclaurin’s Theorem implies x2 y2 (0) log (1 + sin x) = y = y (0) + xy1 (0) + 2! x3 x4 x5 + y3 (0) + y4 (0) + y5 (0) + … 3! 4! 5! = x−

x 2 x3 x 4 x5 + − + +…. 2 6 12 24

x (iv) Let y = e sec x so that y (0) = 1. Then,

y4 = y3 cos x − 3 y2 sin x − 3 y1 cos x + y sin x so that y4 (0) = −3 , and so on. Therefore, Maclaurin’s Theorem yields x2 e sin x = y = y (0) + xy1 (0) + y2 (0) 2!

y1 = e sec x + e sec x tan x x

x

= y + y tan x so that y1 (0) + 0 = 1 , y2 = y1 + y1 tan x + ysec 2 x so that y2 (0) = 1 + 0 + 1 = 2 ,

Then,

so that y3 (0) = 4 ,

y1 =

and so on. Therefore, application of Maclaurin’s Theorem yields ex sec x = y x2 x3 = y (0) + xy1 (0) + y2 (0) + y3 (0) + … 2! 3! 2 x 2 4 x3 . = 1+ x + + +… 2! 3! EXAMPLE 2.41

Expand the following functions by Maclaurin’s Theorem: (i) esin x. (ii) loge (1+ex). (iii) ex cos x. ex 1+ e x

Solution. (i) Let y = e so that y (0) = e = 1. Then, y1 = esin x cos x = y cos x so that y1 (0) = y (0) cos 0 = 1,

M02_Baburam_ISBN _C02.indd 24

= 1+ x +

x2 x4 + ( −3) + … 2! 4!

= 1+ x +

x2 x4 − + …. 2 8

eX 1 1 1 = 1− so that y1 (0) = 1 − = , x 1+ e 1+ eX 2 2 ex ex 1 y2 = = . 2 1 + ex 1 + ex 1 + ex

(

0

)

1 , 4 1 1 y3 = y2 − 2 y1 y2 so that y3 (0) = − = 0 , 4 4 y4 = y3 − 2  y 22 + 2 y1 y3  so that y4 (0) = − 1 , 8 and so on. Therefore, Maclaurin’s Theorem yields x2 y2 (0) log e (1 + e x ) = y = y (0) + xy1 (0) + 2! = y1 (1 − y1 ) = y1 − y 12 so that y2 (0) =



. sin x

x3 x4 y3 (0) + y4 (0) + … 3! 4!

(ii) We have y = log e (1 + e x ) so that y(0)=loge 2.

y3 = y2 + y2 tan x + 2 y1sec 2 x + 2 ysec 2 x tan x

(iv)

+

+

x3 x4 y3 (0) + y4 (0) + … 3! 4!

x x2 x4 + − +… . 2 8 192 (iii) Let y = ex cos x so that y (0) = e0 = 1. Then, = log e 2 +

y1 = e x cos x − e x sin x

1/2/2012 11:53:09 AM

SucceSSive differentiation, Mean value theoreMS and expanSion of functionS  n 2.25 = e x ( cos x − sin x) so that y1 (0) = 1 ,

and so on. Hence Maclaurin’s Theorem yields

y2 = e ( cos x − sin x) + e (− sin x − cos x)

eX x2 x3 = y = y (0) + xy1 (0) + y2 (0) + y3 (0) + … x 2! 3! 1+ e

x

x

= −2e x sin x so that y2 (0) = 0 , y3 = −2e x sin x − 2e x cos x = −2e x ( sin x + cos x) so that y3 (0) = −2 , y4 = −2e x ( sin x + cos x) − 2e x ( cos x − sin x) = −4e cos x = −2 y so that y4 (0) = −22 , 2

x

y5 = −22 y1 so that y5 (0) = −22 ,

EXAMPLE 2.42

Use Maclaurin’s Theorem to expand (i) sin (m sin–1 x) and hence, expand sin mθ in powers of sin θ . (ii) e asin

y6 = −22 y2 so that y6 (0) = 0 , n nπ   yn = 2 2 cos  x +  4  so that  n nπ yn (0) = 2 2 cos . 4 Therefore, by Maclaurin’s Theorem, we get x2 x3 e x cos x = y = y (0) + xy1 (0) + y2 (0) + y3 (0) 2! 3!

x4 x5 x5 x7 + y4 (0) + y5 (0) + y6 (0) + y7 (0) + … 4! 5! 6! 7!

= 1+ x −

x yn (0) + … n! 2

4

1 2 sin 2θ + sin 3θ + …. 2! 3!

Solution. (i) Let

y = sin (m sin −1 x) so that y (0) = 0.

Then, y1 = cos (m sin −1 x).

or

(1 − x ) y 2

or

(1 − x ) y

2 1

2 1

5

3

nπ x n +2 cos + …. . 4n ! ex 1 = 1− (iv) Let y = so that y (0) = 12 . x 1+ e 1 + ex Then, ex ex 1 y1 = = . = y (1 − y ) x 2 x 1 + e 1 + ex 1+ e

)

2

= y − y 2 so that y (0) = 1 −  1  = 1 , 1   2 2 4

= m 2 cos 2 ( m sin −1 x ) = m 2 (1 − y 2 )

+ m 2 y 2 = 0 so that y1 (0) = m .

or

2 (1 − x 2 ) y2 − xy1 + m 2 y = 0 so that y2 (0) = 0

Now differentiating n times by Leibnitz’s Theorem, we get

(1 − x ) y 2

n+2

+ nC1 yn +1 (−2 x) + nC2 yn (−2) − xyn +1 − nC1 yn + m 2 yn = 0

or

(1 − x ) y 2

n+2

− (2n + 1) xyn +1 − ( n 2 − m 2 ) yn = 0.

y2 = y1 − 2 yy1 so that

Substituting x = 0, we get

1  1  1  y2 (0) = − 2    = 0, 4  2  4 

Taking n = 1,2,3,…, we get

1 y3 = y2 − 2 y − 2 yy2 so that y3 (0) = − , 8

M02_Baburam_ISBN _C02.indd 25

1 − x2

2 (1 − x 2 ) y1 y2 − 2 xy 12 + 2m 2 yy1 = 0

7

2x 2 x 2 x 2 x − − + +… 3! 4! 5! 7!

2 1

m

Differentiating again, we get 2

n 2

(

θ and hence, show that e = 1

x

2

n

2

−1

+ sin θ +

y7 = −22 y3 so that y7 (0) = 23 , and

+

1 1 1 3 + x− x + …. 2 4 48

=

yn + 2 (0) = ( n 2 − m 2 ) yn (0). y3 ±±±± = ( − m 2 ) y1

= m ( − m2 )

1/2/2012 11:53:10 AM

2.26  n  chapter two y4 (0) = ( 22 − m 2 ) y2 (0) = 0, y5 (0) = ( 32 − m 2 ) y3 (0) = ( 32 − m 2 )(12 − m 2 ) m,

and so on. Hence, the application of Leibnitz’s Theorem yields x2 x3 y2 (0) + y3 (0) sin (m sin −1 x) = y (0) + xy1 (0) + 2! 3! +



3! m (1 − m 2

+

2

x2

)( 3

−m

2

)x

5

+ …,

5! which is the required expansion. Substituting x = sin θ , we get



+



asin (ii) Let y = e Then,

x

(1 − x ) y

sin 2θ

sin 5θ + ….

so that y (0) = 1.

+

1− x

a ( 22 + a 2 ) 4!

2 2 a 2 x 2 a (1 + a ) 3 + x 2! 3!

x4 +

a (12 + a 2 )( 32 + a 2 ) 5!

x5 + …

Show that (i) log[ x + 1 + x 2 ] = x − x3! 12 + x5! ( 32.12 ) − x7! 2 2 2 3

2

5

7

(5 .3 .1 ) + …

(ii) log (1 + tan x) = x − 12 x 2 + 32 x 3 − 127 x 4 + … Solution. (i) Let

y = log  x + 1 + x 2  so that y (0) = 0.   Then,  1 2  1 y1 = 1 + 2 = 2 x + 1+ x  x +1  x2 + 1

or

a

−1

2

3!

5!

y1 = e asin x .

or

m (12 − m 2 )

m (12 − m 2 )( 32 − m 2 ) −1

= 1 + ax +

x4 x5 y4 (0) + y5 (0) + … 4! 5!

EXAMPLE 2.43 2

sin mθ = m sin θ +

+



x4 x5 y4 (0) + y5 (0) + … 4! 5!

m (12 − m 2 )

= mx +

Hence, using Maclaurin’s Theorem, we get −1 x2 x3 e asin x = y (0) + xy1 (0) + y2 (0) + y3 (0) 2! 3!

(1 + x )

2

y12 − 1 = 0 so that y1(0) = 0.

Differentiating again, we get 2 1

− a y = 0 so that y1 (0) = a. 2

2

Differentiating once more, we get (1 − x 2 ) y2 − xy1 − a 2 y = 0 so that y2 (0) = a 2 . Now differentiating n times by Leibnitz’s Theorem, we have

(1 − x 2 ) yn + 2 − (2n + 1) xyn +1 − ( n2 + a 2 ) yn = 0

Substituting x = 0, we get yn + 2 (0) = ( n 2 + a 2 ) yn (0).

Now taking n = 1,2,3… we get y3 (0) = a (1 + a

), y (0) = a ( 2 + a ) , and y (0) = a (1 + a )( 3 + a ) . 2

2

2

2

2

4

2

5

M02_Baburam_ISBN _C02.indd 26

2

2

2

(1 + x ) y 2

2

+ xy1 = 0 so that y2(0) = 0.

Differentiating n times by Leibnitz’s Theorem, we get

(1 + x ) y 2

n+2

+ ( 2n +1 ) xyn +1 + n 2 yn = 0.

Substituting x = 0, we get yn + 2 (0) = − n 2 yn (0) .

Taking n = 1,2,3,…, we get y3 (0) = −12 , y4 (0) = 0 , 6 y5 (0) = 32.12 , y (0) = 0 ,

y7 (0) = −52.32.12 , and y8 (0) = 0 .

Thus we note that yn (0) = 0, when n is even and yn (0) = ( −1) 2 ( n − 2 ) ( n − 4 ) ….52.32.12 , n −1

2

2

1/2/2012 11:53:11 AM

SucceSSive differentiation, Mean value theoreMS and expanSion of functionS  n 2.27 when n is odd.



Hence, by Maclaurin’s Theorem, we have x2 y2 (0) log  x + 1 + x 2  = y (0) + xy1 (0) +   2! 3

4

x x + y3 (0) + y4 (0) 3! 4!



+



= x− −

(ii) Let

x5 y5 (0) + … 5! 3

5

x 2 x 1 + ( 32.12 ) 3! 5!

x7 2 2 2 ( 5 .3 .1 ) + … 7!

y = log (1 + tan x) so that y(0) = log 1 = 0. We have e y = 1 + tan x . Differentiating with respect to x, we get e y y1 = sec 2 x so that y1(0) = 1. Differentiating once more, we have e y ( y 12 + y2 ) = 2sec 2 x tan x and so y (0) = –1. 2

found by simply substituting that value of the variable in the function. Let f and g be real-valued functions. We know that lim x→a

But if both lim f ( x) and lim g ( x) are zero, x→a x→a f ( x) then (1) gives no information about the lim g ( x )

Special methods are required to evaluate these indeterminate forms. These methods are called L’Hospital Rule after the name of the French mathematician L’Hospital. Theorem 2.6. Let f and g be functions such that (i) lim f ( x) = lim g ( x) = 0. x→a



= 16sec 4 x tan x + 8sec 2 xtan 2 x = 16 sec4 x tan x + 8 sec2 x tan2 x

and so, y4(0)= –14. Hence, Maclaurin’s Theorem yields 1 2 7 log (1 + tan x) = x − x 2 + x 3 − x 4 + … 2 3 12 .

Proof: Since, by hypothesis of the theorem, f and g are derivable at x = a, they are also continuous at x = a and so,

lim f ( x) = f (a ) and lim g ( x) = g (a ) . x→a

M02_Baburam_ISBN _C02.indd 27

x→a

Therefore, by condition (i), f (a ) = g (a ) = 0 . On the other hand, f ( x)− f (a) x−a

= lim

g ( x)− g (a) x−a x→a

= lim

f ′(a ) = lim x→a

f ( x) x→a x −a

and

(2) g ′(a ) = lim

2.7  INDETERMINATE FORMS A function is said to assume an indeterminate form if it involves the independent variable in such a way that for a certain assigned value of the independent variable its value cannot be

x→a

(ii) f ′(a ) and g ′(a ) exist and g ′(a ) ≠ 0. Then, f ( x) f ′(a ) lim . = x → a g ( x) g ′(a )

which yields y3(0) = 4.

x→a

lim f ( x )

→a since the quotient xlim in that case takes g ( x) x →a 0 the form 0 , which is meaningless. Thus, 00 is one of the indeterminate forms. Similarly, if f ( x) lim f ( x) = lim g ( x) = ∞ , then the quotient g ( x ) x→a x→a assumes the indeterminate value ∞∞ . The other indeterminate forms are ∞ − ∞ , 0 × ∞ , 00 ,1∞ , and ∞ 0 .

e y ( y 31 + 3 y1 y2 + y3 ) = 2sec 4 x + 4sec 2 x tan 2 x,

e y ( y 14 + 6 y 12 y2 + 4 y1 y3 + 3 y 22 + y 4 )

(1)

x→a

Differentiating again, we get

If we differentiate once again, we have

f ( x) f ( x) lim = x→a , lim g ( x) ≠ 0 . g ( x) lim g ( x) x → a

g ( x) x→a x −a

Consequently, f ( x) x→a g ( x )

lim



(2)

f ( x)  fx −( xa)  lim x−a x→a = lim  g ( x )  = g ( x) x→a x−a  x − a  lim x→a

1/2/2012 11:53:12 AM

2.28  n  chapter two

If f ′(a ) = g ′(a ) = 0, then by Taylor’s Theorem (assuming second-order derivatives of f and g exist), we have

f ′(a ) , using (2). g ′(a )

=

Theorem 2.7. (L’Hospital Rule for Let f and g be functions such that f ( x) = lim g ( x) = 0, (i) lim x→a x→a

0 0

form).

' (ii) f ′( x) and g ( x) exist and g ′( x) ≠ 0 for x ∈ (a − δ , a + δ ) , δ > 0 except possibly

at x = a, and f ′( x)

(iii) lim g ′ ( x ) exists. x→ a Then,

lim x→a

f ( x) f ′ ( x) , = lim g ( x) X → a g ′ ( x) g ′(a) ≠ 0.

Proof: Under the given hypothesis, f and g are

f (a + h) = f (a ) + hf ′(a ) +

x→a

By Taylor’s Theorem truncated at the first derivative term, we have

=

and

h2 f ′′ ( a + θ 3 h ) 2!

and

g (a + h) = g (a ) + hg ′ ( a + θ 2 h ) = hg ′ ( a + θ 2 h ) ,

0 < θ2 < 1 .

Therefore, lim x→a

f ′ ( a + θ1h ) f ( x) f ( a + h) = lim = lim g ( x ) h → 0 g ( a + h) h → 0 g ′ ( a + θ 2 h )

=

f ′(a ) f ′( x) , g ′(a ) ≠ 0 = lim . x → 0 g ′( x ) g ′(a )

Hence, lim x→a

Therefore, lim x→a

f ′ ( a + θ3 h ) f ( x) f ( a + h) = lim = lim g ( x ) h → 0 g ( a + h) h → 0 g ′ ( a + θ 4 h )

f ′′(a ) f ′′( x) = lim , g ′′(a ) ≠ 0. x → a ′′ g (a) g ′′( x)

=

Hence, in general, if

f (a ) = f ′(a ) = f ′(a ) = … = f ( n −1) (a ) = 0, g (a ) = g ′(a ) = g ′′(a ) = … = g ( n −1) (a ) = 0,

and g ( n ) (a ) ≠ 0 , then using Taylor’s Theorem with a remainder after n terms, we have f ( x) f ( n ) ( x) , g ( n ) (a) ≠ 0 , = lim ( n ) x → a g ( x) x→ a g ( x) which is called the generalized L’Hospital Rule. lim

Remark 2.2. L’Hospital Rule is also applicable if lim f ( x) = lim g ( x) = 0 . The proof follows the x →∞ x →∞ same lines as the proof given in Remark after Theorem 2.8. EXAMPLE 2.44

f ( x) f ′ ( x) , g ′ (a ) ≠ 0. = lim g ( x) x → a g ′ ( x)

Obviously, this relation fails if g ′(a ) = 0. If g ′(a ) = 0 and if f ′(a ) ≠ 0, then

h2 g ′′ ( a + θ 4 h ) , 2!. 0 < θ4 < 1

g (a + h) = g (a ) + hg ′(a ) +

f (a + h) = f (a ) + hf ′ ( a + θ1h ) = hf ′ ( a + θ1h ) ,

0 < θ1 < 1

0 < θ3 < 1



continuous at x = a and so,

g ( x) = 0 . f (a ) = lim f ( x) = 0 and g (a ) = Xlim →a

h2 f ′′ ( a + θ3 h ) , 2!

lim x→a

f ′( x) = +∞ or − ∞. g ′( x)

Evaluate lim e t+anlogx (− xe ) . x →0 Solution. Since 1e = log1 − log e = 0 − 1 = −1, x

lim

e x + log ( 1−e x )

is of

0 form: 0

tan x − x Therefore, using L’Hospital Rule, we have x→a

lim x→a

M02_Baburam_ISBN _C02.indd 28

1− x

e x + log ( 1−e x ) tan x − x

e x + 1−ex ( − 1e ) 1

= lim x →0

sec 2 x − 1

1/2/2012 11:53:13 AM

SucceSSive differentiation, Mean value theoreMS and expanSion of functionS  n 2.29 e x − 1−1 x  0   form  x → 0 tan 2 x 0  

= lim

e − x

= lim x →0

x

2

= lim



Thus, (1) reduces to

1 1− x

( )

tan x 2 x

e −( x

1 1− x

lim

)

e x − (1 − x )

2 = lim e x − x→0 (1 − x )2

x

− xa − xχ

lim x→a

x→a

a −1

a

2 x + 3x 2

x →0

= −e lim

log (1 + x)  0   form  2 x + 3x 2  0 

= −e lim

1 e =− (1 + x)(2 + 6 x) 2

x →0

x→0

−x

Evaluate lim e x−−e sin−x2 x . x

Solution. We have, by L’Hospital Rule,

e x − e− x − 2 x  0   form  x →0 x − sin x  0 

lim

(1 + x ) x − e . x

Solution. Since lim (1 + x ) = e, the given limit is 1 x

x →0

of form. Therefore, using L’Hospital Rule, we have 0 0

1 x

(1 + x) − e = lim x →0 x →0 x

lim

1 − 11++ xx + log (1 + x) 

EXAMPLE 2.47

1

x→a

x − (1 + x) log (1 + x) x 2 + x3

x →0

EXAMPLE 2.46 lim

x →0

a

log ae log e − log a . = = log e + log a log ea

d dx

(1 + x )

1 x

.

1

(1 + x ) , let y = (1 + x ) . Then, 1 log y = log (1 + x) . x Differentiating with respect to x yields 1 dy 1 1 = − 2 log (1 + x) y dx x(1 + x) x To find

= e lim



a log a − ax a log a − a 1 − log a = = − x x (1 + log x ) − a a (1 + log a ) 1 + log a

Evaluate

x 2 (1 + x)

x →0

0   form  0 



a −x 0  form  a X  a −x 0 

= lim



a

x

= lim

x

x →0

.

Solution. We have x

(1 + x ) [ x − (1 + x) log (1 + x)] 1 x

−e

= lim (1 + x ) x lim

a

1 x



EXAMPLE 2.45 a Evaluate lim x →0 a

(1 + x )

1

−2

0   form  x →0 2x 0  e0 − 2 1 − 2 1 = = =− 2 2 2 = lim

x →0

tan x since lim = 1. x →0 x

x2

x →0

1  x − (1 + x ) log (1 + x )  = (1 + x) x  . x 2 (1 + x)  

d dx

1 x

1 x

or  1  dy 1 = y − 2 log (1 + x)  dx  x(1 + x) x 

M02_Baburam_ISBN _C02.indd 29

(1)



e x + e− x − 2  0   form  x → 0 1 − cos x 0 

= lim

e x − e− x x →0 sin x

= lim

0   form  0 

e x + e− x 1 + 1 = =2 x→0 cos x 1

= lim

EXAMPLE 2.48

Evaluate lim xx x→a

a x

−ax − aa

.

Solution. Using L’Hospital Rule, we have

lim x→a



xa − a x xx − aa

0   form  0 

= lim x→a

ax a −1 − a x log a a a − a a log a = x x (1 + log x) − 0 a a (1 + log a )

1/2/2012 11:53:14 AM

2.30  n  chapter two =



a a (1 − log a ) 1 − log a = a a (1 + log a ) 1 + log a .

lim

X →a

1 g ( x)

f ( x) = lim g ( x) x → a

1 f ( x)

− g ′( x)

EXAMPLE 2.49

= lim

xx − x x →1 −1+ x − log x

Evaluate lim

.



xx − x 0   form  x →1 −1 + x − log x 0  

x (1 + log x)(1 + log x) + x . 1 x2

=

1+1 = 2. 1

l = Xlim →a

that is, x→a

l + 1 = lim

e x − e − x − 2 log (1 + x)  0   form x→a x sin x 0 e x + e − x − 1+2x  0  = lim  form x → a x cos x + sin x  0 lim

e x + e− x −

2

=

− x sin x + cos x + cos x



2 = 1. 2

). Let lim f ( x) = ∞, lim g ( x) = ∞, x→a

f ′( x) and Xlim exist (finite or infinite). Then, g ′( x) →a

lim x→a

f ( x) f ′ ( x) . = lim g ( x) x → a g ′ ( x)

Proof: Since lim f ( x) = lim g ( x) = ∞ , we have x→a

f ( x)+ g ( x)

l + 1 = lim

Therefore, using L’Hospital Rule for obtain

M02_Baburam_ISBN _C02.indd 30

f ′ ( x) + g ′ ( x) f ′ ( x) = lim +1 x → a g ′ ( x) g ′ ( x)

or l = lim

X →a

f ′ ( x) f ( x) f ′ ( x) , . = lim g ′ ( x) that is, lim x → a g ( x) X → a g ′ ( x)

Case 3. If l = ∞ , then lim gf (( xx)) = 0. Therefore, x→a by case 2, we have or

x→a

1 1 lim = lim = 0. x → a f ( x) x → a g ( x)

f ( x) f ( x) + g ( x) + 1 = lim . x → a g ( x) g ( x)

Thus, lim is neither zero nor infinity g ( x) x→a Therefore, by case 1,

Theorem 2.8. (L’Hospital Rule for the indetermix→a

x→a

x→a

(1+ x )2

0 = lim x→a

∞ = lim x→a

f ′ ( x) g ′ ( x)

that is, 0 0

f ( x) f ′ ( x) = lim x → a g ( x) g ′ ( x)

Case 2. If l = 0, then,

e x − e − x − 2 log (1 + x) . x→a x sin x Solution. Use of L’Hospital Rule yields

Evaluate lim

∞ ∞

g ′ ( x) 2 1 f ′ ( x) g ′ ( x) l or = lim , or l = lim x → a X → a f ′ ( x) g ′ ( x) l f ′ ( x)

lim

EXAMPLE 2.50

nate form



x→a

x 1 x

x→a

(1)

Case 1. If l is neither zero nor infinite, then (1) reduces to

x (1 + log x) − 1  0   form  1 − 1x 0 

= lim

x→ a

x→ a

x

x →1

f ′( x)

[ f ( x )]2

g ′ ( x)  f ( x)  . f ′ ( x)  g ( x) 

Let lim gf (( xx)) = l. Then three cases arise.

x

= lim

= lim

Since lim exists by the given hypothesis, it follows from (1) that lim gf (( xx)) exists.

lim

x →1

2

[ g ( x )]2

f ′( x ) ′ x→a g ( x)

Solution. We have

= lim

x→ a

0   form 0

form, we

lim x→a

g ( x) g ′( x) = lim x → a f ( x) f ′( x) or l = lim f ′ ( x) , x→a

g ′ ( x)

f ( x) f ′( x) = lim . x → a g ( x) g ′( x)

1/2/2012 11:53:14 AM

SucceSSive differentiation, Mean value theoreMS and expanSion of functionS  n 2.31 Remarks 2.3. The L’Hospital Rule, proved above, can be extended to the case where x → ∞ . Thus, if lim f ( x) = ∞, lim g ( x) = ∞ , and x →∞ x →∞ f ′( x) lim exists, then x →∞ g ′( x ) f ( x) f ′( x) lim = lim . x →∞ g ( x ) x →∞ g ′( x )

= lim

x →1−



EXAMPLE 2.52

Evaluate

f (1) f ( x) lim = lim 1z x →∞ g ( x ) z →0 g ( ) z



= lim z →0



= lim z →0

2. If we have lim x→a change it to

f ( x) g ( x)

( ) g′( ) ( − ) 1 z2

f ′ ( 1z ) f ′( x) = lim . g ′ ( 1z ) x→∞ g ′( x) in

∞ ∞

form, we should

form to avoid differentiation of

0 0

numerators and denominators again and again.

x →1−

log(1− x ) cot(π x )

.

x →1



log (1 − x)  ∞  form  , since log 0 = −∞ cot (π x)  ∞ 

= lim

x →1−



= lim

x →1−

−1 1− x

−π cos ec (π x) 2

= lim x →1

M02_Baburam_ISBN _C02.indd 31

.

log ( x − π2 )  ∞  form   tan x  ∞ 

limπ x→ 2

1 x − π2

∞  = limπ form  = limπ  2 x → 2 sec x ∞   x→ 2

1 x − π2 1 cos 2 x

cos 2 x  0  form   π x→ 2 x −  2 0 −2 cos x sin x = limπ = 0. x→ 2 1

EXAMPLE 2.53

Evaluate lim x →0

log x cot x

.

Solution. We note that

lim x →0

log x  ∞   form  cot x  ∞  ∞  form   x →0 −cosec x ∞   1 x

2

sin 2 x  0  form   x →0 x 0  2 sin x cos x = lim = 0. x→0 1

= lim −

Solution. We have

lim

tan x

x→ 2

= lim

EXAMPLE 2.51

Evaluate lim

log ( x − π2 )

= limπ

f ′ ( 1z ) − z12 1 z

limπ

Solution. We have

In fact, substituting x = 1z , we have z → 0 as x → ∞ and therefore, 1 1 lim f   = ∞ and lim g   = ∞. z → 0 z →0 z z f ( 1z ) Applying L’Hospital Rule to , we have g ( 1z )

2π sin π x cos π x = 0. −π

−1 1− x

−π cos ec 2 (π x)

sin 2 (π x)EXAMPLE  0 2.54  form  = x log sin x, x ≠ 0, f(0) = 0 is x →1 −π (1 − xShow )  0 that f(x) 

= lim

sin (π x)  0   form  −π (1 − x)  0  2

continuous at x = 0.

Solution. To show that f is continuous at x = 0, we have to prove that lim f ( x) = f (0) = 0 . We note x →0 that

1/2/2012 11:53:14 AM

2.32  n  chapter two lim f ( x) = lim x log sin x x →0

x →0

log sin x  ∞   form  1 ∞   x

= lim x →0



cot x x = lim − x. x →0 tan x − x12

= lim x →0



x = − lim x.lim = 0 = f (0). x →0 x → 0 tan x Hence f is continuous at x = 0. EXAMPLE 2.55

Evaluate lim x →0

x →0

= lim

1 x→0 log cos x

= lim x →0

1 log(1− x 2 ) 1 log cos x

⋅ (1−−2xx2)

.

− sin x cos x

= lim x →0

= lim

2 x cos x log cos x  ∞   form  (1 − x 2 ) log (1 − x 2 )  ∞ 

−2 sin x log cos x + 2 cos x (− sin x) x →0 (1 − x 2 ). (1−−2xx2) + log (1 − x )(−2 x) 1 cos x 2

= lim

−2 sin x log cos x − 2 sin x  0  = lim  form  2 x →0 (−2 x) − 2 x log (1 − x )  0 

2 sin x − 2 cos x log cos x − 2 cos x = lim cos x x→0 −2 log (1 − x 2 ) − 2 x. (1−−2xx2) 2

s x log cos x − 2 cos x −2 = = 1. −2 1 − x 2 ) − 2 x. (1−−2xx2)

The indeterminate form ∞ – ∞ arises when we want to evaluate lim[ f ( x) − g ( x)] and when x→a

1 f ( x)

1 f ( x) g ( x)

0   form  , 0 

which can be evaluated using L’Hospital Rule for 00 form. EXAMPLE 2.56

Evaluate lim  x −1 2 − log (1x −1)  . x→2

( ∞ − ∞ form )

  −1 1 = lim  =− . x → 2 1 + 1 + log ( x − 1 2   EXAMPLE 2.57

Evaluate lim( 1x − cot x). x →0

Solution. We have

1  lim  − cot x  x →0 x  

(∞ − ∞

form )

 1 cos x  = lim  −  x →0 x sin x   = lim

sin x − x cos x  0   form  x sin x 0 

= lim

x sin x − x cos x .lim sin x x → 0 x2

= lim

sin x − x cos x  0   form  x2 0 

x →0

Indeterminate form ∞ – ∞.

M02_Baburam_ISBN _C02.indd 32

x→a

−

  0 2− x  = lim    0 form  x → 2 x − 2 + ( x − 1) log ( x − 1)   

2 x cos x log cos x (1 − x 2 ) sin x log (1 − x 2 ) x 2 x cos x log cos x lim sin x x →0 (1 − x 2 ) log (1 − x 2 )

x →0

x→a

1 g ( x)

1   −1 = lim  x − 2 x −1  x→2  x −1 + log ( x − 1) 

− sin x cos x

= lim x →0

lim[ f ( x) − g ( x)] = lim

 log ( x − 1) − ( x − 2)   0  = lim   form   x→2   ( x − 2) log ( x − 1)   0

⋅ (1−−2xx2)

.

x→a

 1  1 − lim   x→2 x − 2 log ( x − 1)  

log log (1 − x 2 )  ∞   form  log log cos x  ∞  1 log(1− x 2 )

x→a

write

Solution. We have

log log (1 − x 2 ) . log log cos x

Solution. We have

lim

lim f ( x) = lim g ( x) = ∞. In such a case, we may

x→0

x →0

1/2/2012 11:53:15 AM

SucceSSive differentiation, Mean value theoreMS and expanSion of functionS   n 2.33 = lim x→0

cos x + x sin x − cos x sin x = lim =0 x→0 2x 2

cos x + x sin x − cos x sin x = lim =0. x→0 2x 2 EXAMPLE 2.58

Evaluate lim x→ 0

(

1 x2

−

1 sin 2 x

).

x sin x − x lim sin 2 x x → 0 x4 2

2

x →0

= lim x→0

2

−2 sin 2 x − sin 2 x = lim x→0 12 x 6x

x →0

1 x2

)

− cot 2 x .

Solution. We have

 1  lim  2 − cot 2 x  ( ∞ − ∞ form ) x →0 x   1  tan 2 x − x 2  0  1  = lim  2 − = lim   form  x →0 x tan 2 x  x → 0 x 2 tan 2 x  0  

M02_Baburam_ISBN _C02.indd 33

EXAMPLE 2.60

= lim

x − sin x  0   form  x sin x  0 

= lim

x − sin x  x  x − sin x  x    = lim   lim x2 x2  sin x  x →0  sin x  X →0

= lim

x − sin x  0   form  x2 0 

= lim

1 − cos x  0  sin x = 0.  form  = lim x →0 2x 0  2

x →0

x →0

x →0

x →0

The Indeterminate Form 0 × ∞ . If

−1 sin x 1 lim lim cos x = − . x → 0 x → 0 3 x 3

(

x4

x →0

x →0

2 cos 2 x − 2  0   form  0 12 x 2  

Evaluate lim

2

2  2 = lim  + terms having power of x  = . x →0 3   3

2

2 sin x cos x − 2 x sin 2 x − 2 x  0  lim  form  x →0 0 4 x3 4 x3  

EXAMPLE 2.59

)

+ 152 x 5 + … − x 2

lim ( cos ec x − 1x ) ( ∞ − ∞ form )

1 2 sin x cos x = − lim 6 x→0 x =

x3 3

x →0

sin 2 x − x 2  0   form  x →0 0 x4  

= lim

x →0

tan 2 x − x 2 x4

Solution. We have

= lim

x →0

( ) (x + = lim

lim

tan x 2 x → 0 x

Evaluate lim ( cos ecx − 1x ) .

2

sin x − x  x  = lim   lim x →0 x →0 sin x x4  

= lim

( )

1

= lim

4

sin 2 x − x 2  0   form  x → 0 x 2 sin 2 x 0   x→0

x4

tan x 2 x

 x 2 + 23x + terms having higher power of x  − x 2  = lim  x →0 x4

= lim

= lim

x →0

tan 2 x 2

tan 2 x − x 2 = lim x →0 x4

Solution. To determine the required limit, we have 1   1 lim  2 − 2  ( ∞ − ∞ form ) x →0 x sin x  

2

= lim

and

lim f ( x) = 0 x →0

lim g ( x) = ∞ , x →0

then

lim f ( x) g ( x) is indeterminate form of 0 × ∞ . To x →0 find this limit, the product f(x) g(x) is written in one of the following form: (i) f ( x) g ( x) =

f ( x)

(ii) f ( x) g ( x) =

g ( x)

0   form  0  

1 g ( x)

1 f ( x)

∞   form  . ∞ 

EXAMPLE 2.61

x tan 1x . Evaluate Xlim →∞

1/2/2012 11:53:16 AM

2.34  n  chapter two Solution. For the given limit, we have

lim x tan x →∞



1 x

Solution. We have

lim ( x − a )

( 0 × ∞ form )

tan 1  0  = lim 1 x  form  = lim tanh = 1, h = 1 . x →∞ 0   h →0 h x x

0

form ) .

Put y = (x– a)x– a . Then log y = ( x − a ) log ( x − a ) =

x log x . Evaluate xlim →0+

lim log y = lim x→a

Solution. We have

lim x log x ( 0 × ∞ form )

log (x − a ) 1 x−a

.

x→a

log x  ∞   form  = lim 1 ∞  x

1 x x → 0 + −1 x2

Thus,

The Indeterminate Forms 0 ,1 , and ∞ . These indeterminate forms occur when evaluation of limits of function of the type g ( x) [ f ( x)] is required when x → a . Thus,

lim [ f ( x) ]

g ( x)

x→a

x→a

x→ 0

g ( x)

and so, lim log y = lim g ( x) log f ( x) x→a

tan x x

x−a

.

. 1

 tan x  x2 y=  ,  x  and we have 1 tan x . log 2 x x

Thus, lim log y = lim x →0

( 0 × ∞ form )

= lim x →0

= lim x →0

Evaluate lim ( x − a) x →0

1 x2

log tanx x x2

tan x 0  = 1  form since lix →m0 x 0 

Since lim (log y) = log (lim y), lim y can be found. EXAMPLE 2.63

x→a

∞ Solution. The given limit is of 1 form. So, let

x →0

is of ∞ 0 form.

log y = g(x) log f(x)

M02_Baburam_ISBN _C02.indd 34

( )

log y =

For f(x) > 0, let y = [f(x)]g(x) Then,

x→a

Evaluate lim

∞ is of 1 form.

f ( x) = ∞ and lim g ( x) = 0 , then (iii) if lim x→a x→a lim [ f ( x) ]

)

EXAMPLE 2.64

is of 00 form.

f ( x) = 1 and lim g ( x) = ∞ , then (ii) if lim x→a x→a lim [ f ( x) ]

x→a

x →0

x→a

x→a

= lim − ( x − a ) = 0.

lim ( x − a )( x − a ) = 1.

0

∞

(i) if lim f ( x) = lim g ( x) = 0 , then g ( x)

−

1 x−a 1 ( x − a )2

(

that is,

x→0+

x→a

1 x−a

lim log y = log lim y = 0 or lim y = e0 = 1 , x→a x→a

= lim ( − x) = 0 . 0

log (x − a )

x→a

= lim

x →0 +

x →0 +

(0

Therefore,

EXAMPLE 2.62

= lim

x−a

x→a

sec2 x tan x

− 1x

2x

= lim x →0

x sec 2 x − tan x  0   form  2 x 2 tan x 0 

sec 2 tan x tan x  0  = lim  form  2 tan x + x sec 2 x x → 0 sin 2 x + x  0 

sec 2 x 1 = . x → 0 2 cos 2 x + 1 3

= lim

1/2/2012 11:53:17 AM

SucceSSive differentiation, Mean value theoreMS and expanSion of functionS  n 2.35 and so,

Hence,

(

)

1 lim log y = log lim y = , x →0 x →0 3

x →0

which implies lim y = e 3 or lim  x →0 x →0  1

= lim

1 x2

1 tan x  3  =e . x 

= lim

EXAMPLE 2.65

= lim

1 x

x →0

= lim − ( x→0

x→0

(

x →0

1 6

1

1 − sin x  x2 6 lim y = e or lim  e . =  x→0 x →0  x 

EXAMPLE 2.67 1

cos x) x2 . Evaluate lim( x →0

)

x→0

1 e

Solution. The given limit is of 1∞ form. Let 1

y = ( cos x) x2 . Therefore,

1 1 lim(cot x) log x = . x →0 e

log y =

log cos x x2

and so,

EXAMPLE 2.66 x→ 0

)

1 6

or

Evaluate lim

x →0

or

−x = lim x → 0 sin x cos x

x →0

(

lim log y = log lim y = −

∞   form  ∞ 

lim log y = log lim y = −1 or lim y = e −1 = x →0

x cos x − sin x  0   form  3 2x 0 

Thus,

x 1 ) lim = −1 . sin x x → 0 cos x

Therefore,

 x cos x2−sin x   x  2x

= lim

1

let y = (cot x) log x and we have 1 log y = log cot x . log x log cot x lim log y = lim x →0 x →0 log x

x sin x

x( − sin x) + cos x − cos x 6x2 − sin x − cos x 1 = lim = lim =− . x→0 x → 0 6x 6 6

1 log x

x) . Evaluate lim(cot x →0 Solution. The given limit is of the form ∞ 0 . So,

Thus,

x →0

x →0

x →0

− cos ec 2 x cot x

log sinx2 x  0   form  x 0 

lim log y = lim

( ) sin x x

1 x2

Solution. Since lim x →0

∞

lim log y = lim x →0

sin x x

= 1 , the given limit is of

the form 1 . So, let

x→0

1

 sin x  x2 y=  ,  x  which yields log y =

M02_Baburam_ISBN _C02.indd 35

= lim −

1 sin x log x x2

x →0

log cos x  0   form  0 x2  

tan x 1 =− . 2x 2

Thus,

(

)

1 lim log y = log lim y = − , x →0 x →0 2 which implies lim y = e x →0

− 12

1

or lim( cos x) x2 = e − 2 . 1

x →0

1/2/2012 11:53:18 AM

2.36  n  chapter two EXERCISES

10. If y = A sin mx + B cos mx , show that y2 + m 2 y = 0 . 2 2 2 2 2 11. If p = a cos θ + b sin θ , show that

Successive Derivatives 1. If x 3 + y 3 = 3axy, show that



d2y 2a 2 xy =− . 2 3 dx ( y 2 − ax )



Leibnitz’s Theorem

ax 2. If y = e sin bx , show that



12. Find the nth differential coefficient of e χ log x .

d2y dy − 2a + ( a 2 + b 2 ) y = 0. dx dx 2

Ans: e x log x + nc1 x −1 − nc2 x −2 + nc3 .2! x −3 n −1 −…+ ( −1) .(n − 1)! x − n  . 

−1 3. If y = tan ( sinh x) , show that 2 . y2 − y1 tan y = 0

4. Find the nth differential coefficient o

d2 p a 2b2 +p= 3 . 2 dθ p

1 x2 + a2

Ans: yn = ( −1) n n !a − ( n + 2) sin (n + 1)φsin n +1φ , a where φ = tan −1   .  x 2x 2 e cos x sin x. 5. Find the nth derivative of n 1 −1  3  Ans: yn = [(13) 2 sin (3 x + n tan   4 2 n 1   − 1 + ( 5 ) 2 sin ( x + n tan   . 2  

2 13. If y = ( x − 1) , show that n



(x

2

− 1) yn + 2 + 2 xyn +1 − n(n + 1) yn = 0.

2 x 14. If y = x e , show that 1 yn = n(n − 1) y2 − n(n − 2) y1 2 1 + (n − 1)(n − 2) y . 2

Hint: By Leibnitz’s Theorem yn = x 2 e x + 2nxe x + n ( n − 1) yn = x 2 e x + 2nxe x + n ( n − 1) e x . Find y1 and y2. 2 6. Find the nth derivative of ( x + 2)x(2 −3 x ) Substitute the values of x2ex, 2xex and ex from y, y1 and y2 respectively in yn and get n −1 ( −3)  (−1) n  1 the required result. Ans: −  . 2  ( x + 2 )n +1 ( 2 − 3 x )n +1  2 −1 15. If y = ( sin x ) , show that 7. Find the nth derivative of 2 x2 +x3 x +1 d2y dy (i) (1 − x 2 ) 2 − x − 2 = 0, n   1 2 n dx dx − . Ans: ( −1) n!  n +1 n +1 2  ( x + 1) ( 2 x + 1)  (ii) (1 − x ) yn + 2 − (2n + 1) xyn +1 − n 2 yn = 0. 8. Find the nth derivative of sin 2 xcos3 x Ans:

1  nπ  2 cos  x + 16  2 

nπ   n   − 3 cos  3 x +  2    nπ    −5n cos  5 x +  . 2  

9. Find the nth derivative of tan −1 11+− xx Ans: ( −1)

n −1

M02_Baburam_ISBN _C02.indd 36

(n − 1)! sin nφ sin n φ , φ = cot −1 x.



16. If y = sin −1 x , show that

( yn ) 0 = ( n − 2 ) ( n − 4 )



Deduce that

2

2

( yn − 4 ) 0 .

(i) ( yn )0 = ( n − 2 ) ( n − 4 ) …52.32.1 for odd n, 2



2

(ii) ( yn )0 = 0 for even n.

1/2/2012 11:53:19 AM

SucceSSive differentiation, Mean value theoreMS and expanSion of functionS  n 2.37 a sin 17. If y = e



−1

x

find ( yn )0 .

Hint: From Example 2.11, we have ( yn + 2 ) 0 = ( n 2 + a 2 ) ( yn ) 0 . 2 2 2 2 Ans: ( yn )0 = ( n − 2 ) + a  …( 3 + a )

× (12 + a 2 ) .a if n is odd

( yn )0 = ( n − 2 )

2

+ a 2  …( 4 2 + a 2 ) 

× ( 22 + a 2 ) − a 2 if n in even

1

23. Using Lagrange’s Mean Value Theorem, show that π 1 3 π 1 − > cos −1 > − 3 5 3 5 3 8. 24. If f (x) = sin–1 x, 0 < a < b < 1, use Mean Value Theorem to prove that b−a b−a < sin −1 b − sin −1a < . 2 1− a 1 − b2 Hint: Consider f (x) = sin–1 x. Then, the Mean Value Theorem implies

1

18. If y m + y m = 2 x, show that

(x



2

− 1) yn + 2 + ( 2n + 1) xyn +1 + ( n 2 − m 2 ) yn = 0.





1−ξ 2

Ans: ( yn )0 = ( −1) 2 ( n − 2 ) ( n − 4 ) … 2

( yn )0 = 0 if n is even

21. In the Mean Value Theorem: f ( x + h) = f ( x) + hf ′( x + θ h),



for f ( x) = ax 2 + bx + c in

show that θ = [0,1]. Hint: Since f ′( x) = 2ax + b, the given theorem f ( x + h) − f ( x) = hf ′( x + θ h) implies a ( x + h ) + b ( x + h ) + c − ( ax 2 + bx + c ) 2



= h [ 2a ( x + θ h) + b ]



⇒ ah 2 + 2axh + bh = 2axh + 2aθ h 2 + bh

x

is

M02_Baburam_ISBN _C02.indd 37

1 2.

positive,

x > log (1 + x) > x −

1 1− a 2

and as ξ > b ,

. Hence,

which gives the required result.

=

1 (c − a )(c − b) f ′′(ξ ), 2

Where c and x both lie in [a, b]. 26. Using Lagrange’s Mean Value Theorem, show that x < log (1 + x) < x, x > 0 . 1+ x

⇒ ah 2 = 2aθ h 2 ⇒ θ =

22. If

1 1− b2

>

25. Assuming f ′′ to be continuous in [a, b], show that b−c c−a f (c ) − f (a) − f (b) b − a b −a

20. Verify Rolle’s Theorem for −x f ( x) = x( x + 3)e 2 in [–3, 0]



>

1 1−ξ 2

1 sin −1b − sin −1a 1 , < < 2 b−a 1− a 1 − b2

2

× 32.12 if n is odd

1 2

since ξ > a , 1

2 19. If y = log  x + 1 + x  , find ( yn )0 . n −1

sin −1b − sin −1a 1 = a 1+x x . Thus, log (1 + x) > 1+x x . Similarly, prove the second half of the inequality.

27. If in the Cauchy’s Mean Value Theorem, −χ f ( x) = e χ and F ( x) = e , show that c is the arithmetic mean between a and b. Hint: eb − e a f ′(c) ec = = = −e 2 c ⇒ −e a + b e − b − e − a F ′(c) e − c = −e 2 c ⇒ c =

a +b 2

.

28. Expand log sin (x + h) in power of h by Taylor’s Theorem. 29. Show that sin (x + h) = sin x + h + cos x − h2! sin x −….

36. Apply Maclaurin’s Theorem to obtain terms up to x4 in the expansion of log (1+sin2 x). 2 3 4 Ans. x − 12 x + 32 x − 127 x + …. 37. Show that e ax sin x = x + x 2 −



n

m

2 38. Expand  x + 1 + x  in ascending power   of x. 2 2 m2 2 m ( m − 1 ) 3 Ans. 1 + mx + x + x 2! 3!

+

31. Calculate the approximate value of 10 to four decimal places. 32. Using Maclaurin’s Theorem, find the expansion of log (sec x + tan Ix). x x Ans. x + 6 + 24 + ….



5



33. Expand y = sin (e –1) by Maclaurin’s Theorem. Ans. x − x2 − 245 x 4 −…. 2

sin −1 x 1− x 2

in power of x up to three terms. 2.4 5 x + …. Ans. x + 32 x 3 + 3.5

35. Apply Maclaurin’s Theorem to prove that

e ax cos bx = 1 + ax +



+



M02_Baburam_ISBN _C02.indd 38

a 2 − b2 2 x 2

a ( a 2 − 3b 2 ) 3!

x

3

(a +…+



2

+ b2 ) 2 n

n!

  b  x n cos  ntan −1    + ….  a  

4!

x4 +

m ( m 2 − 12 )( m 2 − 32 ) 5!

x5 + …

39. Evaluate the following limits:

x

34. Expand

m 2 ( m 2 − 22 )

+… Indeterminate form 00 .

 h 2 h3 π  1  cos  + h  = 1 − h − + +… . 2! 3! 2 4  

3

 nπ  2 2 n + sin   x +….  4  n!



2

30. Show that

2 3 22 5 x − x −… 3! 5!



x − sin x . 2 (i) lim x→0 tan x

Ans. 16 . π lim log (1 − x ) cot x . (ii) x →0 2 Ans. − π2 . 1+ sin x − cos x + log(1− x ) . 2 (iii) lxim x tan x →0 Ans. 1. x cos x − log(1+ x ) . (iv) lix →m0 x2 Ans. 12 . e x sin x − x − x 2 lim 2 . (v) x →0 x + x log(1− x ) Ans. − 32 . e x sin x − x − x 2 li m . (vi) x →0 x3 Ans. 13 . cosh x − cos x . (vii) lim x sin x x →0 Ans. 1. 1 1 2 x −3 x x (viii) lim log x →∞ x −1 Ans. log 32 .

x (1+ a cos x ) − b sin x = 1, find the value of a 40. If lim x3 x →0

and b.

1/2/2012 11:53:22 AM

SucceSSive differentiation, Mean value theoreMS and expanSion of functionS  n 2.39

Hint: It is of 00 form. After differentiation, the denominator becomes zero for x = 0. So, to find the limit the numerator should also be zero. This will yield 1 + a – b = 0. Again differentiating twice we get the denominator 6 at x = 0 with the numerator b – 3a. But the limit is 1. Thus, b −63a = 1 or b – 3a = 6. Solving for a and b, we get a = − 52 and b = − 32 .

Indeterminate form ∞ − ∞ . 43. Evaluate (i) lim x →1

(



41. Point out the fallacy in the following use of L’Hospital Rule:



x − 7x + 6 3x − 7 6x lim 2 = lim = lim = 6. x→2 x − 5x + 6 x→2 2 x − 5 x→2 2 3



2

3 x2 − 7 x → 2 2 x −5

Ans. lim

is not of

0 0

form.

So,we cannot use L’ Hospital



log sin x . (i) lim x→ 0 cot x

(ii) lim x → π2

tan x tan 3 x

.



cos ec x x→ 0 + log x



(iii) lim



x (iv) lim x , where n is a positive integer. x→∞ e Ans. 0.



lim logcotsinx x . (v) x→ 0+ Ans. −∞ .



. Ans. −∞ .

im (v) lx→ 0

( ( (

1 x

1 x2

1 x2

(

(vii) lxi→m0



(viii) lix →m1

)

Ans. 1.

2

log x (vii) lim log cot 2 x . x→ 0 Ans. –1. log( x − a ) . x a (viii) lim x → a log ( e − e )

M02_Baburam_ISBN _C02.indd 39

Ans. 1.

Ans. 0. Ans. 0.

)

Ans. 12 .

− x tan1 x .

)

Ans. 13 .

− cosxec x . Ans. − 16 .

1 e x −1

(

x x −1

)

− 1x .

)

− log1 x .

Ans. − 12 . Ans. 12 .

Indeterminate form 0 × ∞ . 44. Evaluate

x m log x , m > 0. (i) xlim →0+



− sin x) tan x . (ii) lim(1 π x→ 2

Ans. 0.

n

log x sin x. (vi) xlim →0 +



Ans. 12 .

− x12 log (1 + x) .



Ans. 0. Ans. 3.

(iv) lim x →0



42. Evaluate the following:

(cosec x − cot x). (iii) lim x →0

(vi) lim x→0

Indeterminate form ∞∞ .

)

− x42−1 .

(sec x − tan x). π (ii) lim x→ 2



Rule further.The answer is −5.

1 x 2 −1

Ans. 0. x



2 sin (iii) lim x →∞



x sin . (iv) lim x →∞



(v) ±±±± x

a 2x

, a ≠ 0. Ans. a.

1 x

x →0

( 2 − x)

Ans. 1.

π

Ans. 1.

1/2/2012 11:53:23 AM

2.40  n  chapter two

(vi) lim x cot x . x→0

Ans. 1. Indeterminate form 00 ,1∞ , and ∞ 0 .

(i) lim x →0

( )

lim x→ 0

log

1 x

sin x x

x



( )

1 x

, then sin x log y = log x . Therefore, lim x→0 sin x x

and so on.

(ii) lim ( π2 − tan −1 x ) x . x →∞ 2a



M02_Baburam_ISBN _C02.indd 40

(vi) lim(cos x) . x →0

Ans. 1. 2

1 log x

Ans. e .

(vii) lim (1 + 3 ) x .



(viii) lim(1 + x) x .



(ix) lim  12 ( a x + b x )  x .   x →0

x

x→∞

Ans. e3.

1

x →∞

1

(iii) lim ( 2 − ax ) tan . x→a Ans. e π . (iv) lim (cosec x) . x →0 + πx



Ans. e .

1 x2



1



Ans. 1e .

.

− 12

Hint: It is of 1 form. Let y = log y =

1 x2

1 x

∞



( ) sinh x x

1 6

45. Evaluate sin x x



(v) lim x→0

(x) lim (1+ x ) x − e + 12 ex − 1124 ex2 . x2 1





Ans. 1.

Ans.

ab .

x →0

Ans.

−7 e 16

.

1/2/2012 11:53:23 AM

3

Curvature

Let P be any point on a given curve and Q a neighboring point of P such that the arc PQ is concave towards its chord. Let the normals at P and Q intersect at N.

tangents at P and Q make angles y and y + δy with the x-axis so that the angle between the tangents at P and Q is δy. Let R be the point of intersection of the tangents at P and Q and let N be the intersection of normals at P and Q. We have ∠PNQ = ∠SRT = δy.

y y

C C N

A

When Q → P, N tends to a definite position C, called the center of curvature of the curve at P. The distance CP is called the radius of curvature of the curve at the point P and is denoted by r. The circle with center at C and the radius r, equal to CP, is called the circle of curvature of the given curve at the point P. Any chord of the circle of curvature drawn through the point P is called the chord of curvature. The reciprocal of the radius of curvature is called the curvature of the curve at the point P and is denoted by K. 3.1 RADIUS OF CURVATURE OF INTRINSIC CURVES Let A be a fixed point on a given curve and P and Q be two neighboring points on the curve. Let arcs AP and PQ be s and δs, respectively. Let the

M03_Baburam_ISBN _C03.indd 1

0

S

δψ s

A

Q

δψ

N

s

Q

s s P δψ

ψ

P δψ

ψ

T

0 S From the triangle PNQ,

R R ψ δψ x

ψ δψ

x

T

chord PQ chord PQ = _________ = ________     sin ∠NQP sin ∠PNQ sin δY and so, chord PQ PN = ________    · sin ∠NQP sin δY _________ PN

chord PQ δs ______ = ________ ____   δY   sin ∠NQP. δs δY  sin δY  If Q→P, δy→0, and δs→0, and the chord PQ tends to the tangent at P and QN tends to the p . normal at P. Consequently, ∠NQP→ __ 2

12/30/2011 6:15:54 PM

3.2 n chapter three Suppose N approaches C as Q approaches P. Then the radius of curvature at P is given by r = lim PN Q→P

( 

chord PQ = lim ________ arc PQ Q→P

) (  lim δδsy  ) (  lim sinδyδ y )  ____

( 

δy→0

______

δy→0

lim sin ∠NQP × Q→P

)

ds ds . p = ____   = 1· ____ ·1·sin__ 2 dy   dy  Consequently, y. 1 d___ K = __ r = ds   Since the relation between arc length s and y is called the intrinsic equation of a curve, the formula ds r = ___   is known as the intrinsic formula for the dy radius of curvature of the curve. Since the unit of δy is radian, the unit of curvature is radian per unit length. 3.2

RADIUS OF CURVATURE FOR CARTESIAN CURVES Let the equation of the curve be y = f(x). Let y be the angle which the tangent at any point (x, y) makes with the x-axis. Then dy tan y = __ . dx Differentiating with respect to s, we have dy __ dy __ dy d __ d __ dx sec2 y  ___ = __ = · ds ds dx dx dx ds 2 or d y d 2y __ __ ___ dx 1 sec2 y = ___ · = cos y, r dx2 ds dx2 since __ dx = cos y ds 3 __ or 3 sec y (1 + tan2 y  ) 2 ______ ___________ r= 2   = d y d 2y ___ ___ 2 dx dx 2

(  ) (  )

[ (  ) ]

3 __

dy 2 3 __ 1 + __ ( 1 + y21 )2. dx __________ ________ = = 2 y2 d___ y 2 dx 2

Remark 3.1 (i) If the tangent at the point is parallel to the dy y-axis, then __ is infinite and so the above dx

M03_Baburam_ISBN _C03.indd 2

formula for r does not hold good in that case. (ii) The value r is positive or negative in accordance with y2 being positive or negative. However, we will ignore the negative sign as we shall be interested only in finding the length of the radius of curvature. (iii) The value of r depends on the curve and not on the coordinate axes chosen. Therefore, interchanging x and y in the above formula, we get 3

[ (  ) ] dx __

1+

2

__

2

dy , r = __________ 2 d x ___ dy 2 dy the formula which is used when __ is infinite, dx that is, when the tangent is parallel to the y-axis. (iv) The point (x, y) on a curve is called a point d 2y of inflexion if ___2 = 0 at that point. dx (v) If the equation of the curve is given in an implicit form f(x, y) = 0, then since f dy __ = - __x, fy ≠ 0, fy dx we have (f )2 - 2fx fy fxy + fyy (fx)2 d 2y f____________________ xx y ____ . = (fy)3 dx 2 Hence, the formula for radius of curvature reduces to

[ (  ) ] 2

r=

[ (  ) ]

3 __

dy 2 1 + __ dx __________ d y ___ 2

1+

=-

dx 2

( fy)3

fxx( fy )2 - 2fx fy fxy + fyy( fx)2 3 __ 2 2

[( f ) + ( f ) ] ____________________ x

f

3 __ 2 2

y _____________________

2

=

-f_x

y

fxx( fy)2 - 2fx fy fxy + fyy( fx)2

(in magnitude).

EXAMPLE 3.1

Find the radius of curvature at the point (s, y ) on the following curves: (i) s = c log sec y. __ y  + __  . (ii) s = a log tan  p 4 2

(

)

12/30/2011 6:15:55 PM

3.3 curvature n 

(

) 

sin y . __ y __ ______ (iii) s = a log cot  p 4  - 2  + a cos2 y     2 (iv) s = 8ay.

( 

Solution.

(i) We have s = c log sec y. Therefore, 1 ds = c _____         r = ___ · sec y  sec y tan y = c tan y. dy (ii) We are given that __ p + y s = a log tan __  . 4 2 Therefore, ds   r = ___ dy p y 1 1 · sec2  __ + __  · __ = a ___________ p __ y 4 2 2 tan + __  4 2 p y cos __  + __  4 2 1 __________ ___________ __ 1 =a p __ y · 2 p y ·2 __ __ __ sin   +   cos   +    4 2 4 2 ______________________ a =     __ __ p  + y p + y 2 sin __   cos __   4 2 4 2 a = ___________ p y sin 2 __  + __   4 2 a = __________ p + y     sin __ 2 a   = _____ cos y

( 

)

(

(  (  ( 

) ) )

( 

) ( 

( 

( 

)

( 

)

)

)

)

= a sec y.

( 

)

sin y , __ p - y (iii) Since s = a log cot __   + a ______     2 4 2 cos y  we have

[ 

ds  = ___ d   a log cot r = ___ dy dy

[ 

( 

) + a sin y sec y ] 

__ __ p y  -   4 2

( 

2

) ](  )

__ p  - y 1 1 = a · _________ -cosec2 __   - __ y 4 2 2 __ __ p cot  -   4 2 + a sin y (2 sec y sec y tan y ) + a cos y sec2 y 2asin2 y _____ a a   3      = _____________________ +  _______ + cos y  y y __ __ __ __ p p cos y 2sin  -   cos 4  - 2   4 2 2a sin2 y _____ a _______ +  = ___________     + cosa y  __ y __ cos3y sin 2 p  -   4 2

( 

)

( 

) ( 

( 

)

M03_Baburam_ISBN _C03.indd 3

)

2a sin2 y _____ a + _______  + cosa y  = __________ 3    __ p cos y sin 2  - y    sin2y _______ 2a  + 2a    = _____      cos y cos3y

)

2a  (cos2y + sin2y) + = 2a sec3y. = _____ cos3y (iv) We are given that s2 = 8 a y. Differentiating with respect to s, we get dy dy 2s = 8a __ = 8a sin y since sin y = __ ds ds or s = 4a sin y. Therefore, ________ ds  = 4a cos y = 4a 1- sin2 y  r = ___ √ dy _______ ________ ______ 2 8ay y ____ ____ s = 4a 1 = 4a 1 = 4a 1 - ___ . 2a 16a2 16a2

√

√

√

EXAMPLE 3.2

Show that for any curve the following relation holds: dy __ d __ 1 __ r = dx ds . Solution. We know that dy __ = sin y. ds Therefore,

(  )

(  ) = dxd [sin y] = cos y ddxy = cos y ddsy · dxds

dy __ d __ dx ds

__

___

___ __

__ ds 1 = cos y · __ r sec y, sec y = dx 1 = __ r , which proves the result.

EXAMPLE 3.3

Find the radius of curvature of the curve, x __ s = ae a.

x __

Solution. The equation of the curve is s = ae a.

Differentiating with respect to x, we get x s __ ds = a e __a · __ 1 = e __a = __s a a dx or ds = a sec y, since sec y = __ ds . s = a __ dx dx

12/30/2011 6:15:58 PM

3.4 n chapter three Therefore, ds   r = ___   dy = a sec y tan y _________ = s tan y = s √sec2 y - 1 _____ ______ s2 - 1 = __s √s2 - a2 . = s __ a a2

[ (  ) ] (  )

√

EXAMPLE 3.4

Find the radius of curvature of the following curves: 2 __

2 __

2 __

=

(i) x 3 + y 3 = a 3 at (x, y). (ii) xy = c2 at (x, y). p . (iii) y = 4 sin x - sin 2x at x = __ 2

( 

)

3a 3a (iv) x3 + y3 = 3axy at the point ___, ___ . 2 2 Solution.

(i) The equation of the given curve is 2 __

2 __

2 __

x 3 + y 3 = a 3. Any point on this curve may be taken as (a cos3 t, a sin3 t) where t is a parameter. Then, dy ___ dx = - 3a cos2 t sin t and __ = 3a sin2 t cos t. dt dt Therefore, dy/dt dy _____ 3a sin2 t cos t = - tan t ___ = = - ____________ dx dx/dt 3a cos2 t sin t and 2 d___y __ d (-tan t) __ dt = - sec2 t __ dt = d (-tan t) = ___ dt dx dx dx2 dx 1 1 sec4 t cosec t. = sec2 t ______________ =  ___ 3a - 3a cos2 t sin t Hence, r at the point (a cos3 t, a sin3 t) is given 3 by __ dy 2 2 3 __ __ 1+ 3a (1 + tan2 t) 2 dx __________ ____________ r= = d 2y sec4 t cosec t ___ 2 dx 3a sec3 t = 3a sin t cos t. = ___________ sec4 t cosec t

[ (  ) ]

1 __

But x = a cos3 t and y = a sin3 t imply cos t = ( _ax )3 1 __

and sin t = ( _a )3. y

x __13 __y Hence, r = 3a( __ a) a

M03_Baburam_ISBN _C03.indd 4

(  )

1 __ 3

1 __

(ii) The given curve is c2 . xy = c2 or y = __ x Therefore, 2 dy __ c2 and d___y = ___ 2c2 . = - __ 2 2 dx x x3 dx Hence, 3 __ 3 __ dy 2 2 c4 2 1 + __ 1 + __ dx x4   r = __________ = ________ 2 ___ d y 2c2 ___ x3 dx 2

1 __

1 __

= 3a 3x 3y 3.

3 __

(x + c ) 2 ________ 4

4

2c2x3

=

3 __

(x + x y ) 2 _________ 4

2 2

2c2x3

, since xy = c2

3 __ r 3 , where r 2 = x2 + y2. = (x2 + y 2) 2/2c2 = ___ 2c2 (iii) The equation of the given curve is y = 4 sin x sin 2x Therefore, dy __ = 4 cos x - 2 cos 2x and dx 2 d___y = - 4 sin x + 4 sin 2x dx 2 and so, r - 2 cos r = 2 and dy __  = 4 cos __ 2 p __ dx x = 2

(  ) ( ddxy ) 2

___ 2

__ r    =  - 4 sin 2  + 4 sin r = - 4.

p __ x=2

Hence,

[ (  ) ] 2

r=

3 __

dy 2 1 + __ dx __________ d y ___ 2

3 __

=

(1 + 4) 2 _______ -4

dx __ 5√5 (ignoring sign). = ____ 4 (iv) The given curve is x3 + y3 = 3axy. Differentiating both sides with respect to x, we get dy dy 3x2 + 3y2 __ = 3a x __ + y dx dx or dy (y2- ax) __ = ay - x2 (1) dx or dy ay - x2 __ = ______ 2 dx y - ax 2

( 

)

12/30/2011 6:16:00 PM

3.5 curvature n 

( dx )( 

so that,

dy __

3a ___ 3a ___ 2 2

)

(2)

= -1.

Further, differentiating (1) again, we get

( 

) or d y dy dy (ax - y ) = 2x + 2y (  ) - 2a . dx dx dx 3a 3a Therefore at the point (  , ), we have 2 2 ( ddxy )(  ) = -   3a32, [using (2)].

dy d 2y dy dy (y2 -ax) ___2 + 2y __ - a __ = a __ - 2x dx dx dx dx 2

2

___

__

2

2

__

___ ___

2

___

2

2 2

( 

__ 3a __ 3a

(  )

Thus, r at (0, c) is

)

3a__ (in magnitude). = ____ 8√2

which yields

r=

d y ___ 2

dx 2

2

M03_Baburam_ISBN _C03.indd 5

(  ) ] cosh ( c )

3 __

( 

)

 

(  )

 

 1  1 - __ - __ 2

 __ 1

 __ 1

 __ 1

3 __

(-y 2 x- 2) + __21 y 2 x- 2

 __ 1

(  )  __ 1

( 

[ (  ) ]

)

3 __

dy 2 1 + __ dx __________ 2

r=

 

y2 1 1 + __ y2 = + ______ = __  __ 3  __ 2x 2x 1 2x 2 x2 __ __ __ √x + √y √a . __ _______ ___ 1 __ = =  __3 √x 2x 2x 2 __ 1

Hence,

d y ___ 2

dx 2

]

[ 

 __ 3

 __ 3 y 2 1 + __ 2 2(x + y) x ________ ________ __ __ . = =  √a √a ___  __ 3

2x 2

EXAMPLE 3.6

3 __

_ x 2 1 + sinh c = _____________ __ _ x 1

c cosh2 _xc 2 ____________ = _ x

[

(  ) ] c cosh ( c )

= 0,

1 - __

dy __

1 y 2x = - __ 2

(i) The equation of the curve is y = c cosh ( _xc ). Therefore, dy __ 1 = sinh  _x = c sinh ( _cx ) __ ( c) c dx and 2 d___ y __ = 1 cosh ( _cx ). dx 2 c Hence, the radius of curvature is 2

dx

1 __ dy - __1 1x- __32 1 y- __12 __ = -  __ x 2 - y 2 - __ 2 dx 2 dx 2

d y ____

Solution.

[ (  ) ] [ 

dy __

1 - __ 2

Differentiating again with respect to x, we have

(  )

3 __

1 - __

1 1 __ - __ x 2 = _____1 = -y 2 x 2. dx y 2

Find the radius of curvature of the following curves: x _ (i) y __= c cosh __ c__ at (0, c). (ii) √x + √y = √a at (x, y).

2

2

(ii) The equation of the curve is __ __ __ √x + √y = √a . Differentiating with respect to x, we get

EXAMPLE 3.5

dy 2 1 + __ dx __________

c = c. r = __ c

1y x 2 + __ 2 2

Hence, the radius of curvature r at 2 , is 2 3 __ dy 2 2 3 __ 1 + __ (1 + 1) 2 dx __________ _______ r= = 32 d 2y ___ - ___ 3a 2 dx

[ (  ) ]

y 2 y2 . = c _c = __ c

__ 1

___

3a ___ 3a ___

= c cosh2 ( _xc )

y2 x2 + __ Prove that for the ellipse ___ = 1, the radius 2 a b2 2 2 ____ b a of curvature r = 3 , where p is the perpendicup lar from the center to the tangent at (x, y). y2 x2 + __ Solution. The equation of the ellipse is ___ = 1,  2 a b2 2 2 2 2 2 2 which implies b x + a y = a b . Differentiating with respect to x, we get 2y __ dy __ 2x + __ =0 a2 b2 dx

12/30/2011 6:16:04 PM

3.6 n chapter three or

(  )

dy __

b2 __x . = - ___ dx a2 y Differentiating again with respect to x, we have

(  )

[  (  ) ]

dy __ 2 2 y-x d___ y __ _______ dx b b2 y - x -   ____  b2x = - ____ 2 2 2 = - 2 2 a ay dx y a2y

[ 

[  ]

]

2 2 2 2 y + b2x2 b = - ____ b2 a_________ b2 a____ b4 . = - ____ = - ____ 2 2 2 2 2 2 ay ay ay ay a2y3

and

(  ) (  ) [  (  ) ]

2 dy y′ y′ __ y __ d___ d __ d __    = __   dt = d __ = __ dx x′ dt x′ dx dx 2 dx dx x′y″ - y′x″ . x′y″- y′x″ __ = _________   · 1 = _________ 2 x′ (x′) (x′ )3 Therefore, 3 __ dy 2 2 1 + __ dx   r = __________ 2 d y ___ dx 2

[ (  ) ]

[ (  ) ]

 

3 __ 2 2

r=

3 __ 4 2 2

[a y + b x ] ___________ 4 2

(in magnitude). a4b4 The equation of the tangent to the given ellipse Yy Xx + ___ at (x, y) is ___ = 1. Therefore, p = length of a2 b2 the perpendicular from the center (0, 0) to the tangent a2b2 1 __________ _________ . = ___________ = __________ 4 2 2 2 y __ __ x + b x + a4y2 √ a2 b2 a6b6 Thus, p3 = _____________3 (b4x2 + a4y2) 2 so that, 3 __ 6 6 b. (b4x2 + a4y2) 2 = a____ p3 Substituting this value in the expression for r, we get 6 6 b a____ =

=

bx 1 + -   ___ a2y _____________ b4 - ____ a2y3 2

√(  ) (  )

2 2 p3 b . r = ____ = a____ a4b4 p3

3.3

RADIUS OF CURVATURE FOR PARAMETRIC CURVES Let the parametric equation of the curve be x = f (t) and y = f (t). dy Then if x′ = and y′ = __, we have dt dt dy _____ dy/dt __ y′ __ = =      dx dx/dt x′ __ dx

M03_Baburam_ISBN _C03.indd 6

[  (  ) ] 2

Therefore r at the point (x, y) is

3 __

y′ 2 1 +   __  x′ __________

x′y″ - y′x″ ___________

-x′ 3 3 __ (x′ 2 + y′ 2) 2 , _________ = x′y″ - y′x″

x′y″ - y′x″ ≠ 0.

EXAMPLE 3.7

Solve Example 3.6 using parametric equation of the ellipse. y2 x2 + __ Solution. Any point on the ellipse __ = 1 is 2 b2 a (a cos q, b sin q ) and the parametric equation of the ellipse is x = a cos q and y = b sin q. Therefore, x′ = -a sin q and y′ = b cos q, x″ = -a cos q and y″ = -b sin q. Hence, r at (a cos q, b sin q ) is 3 __

3 __

(a2 sin2 q + b2 cos2 q ) 2 r=   = __________________ x′y″ -  y′x″ ab sin2 q + ab cos2 q  (x′ + y′ ) 2 _________ 2

2

3 __

(a2 sin2 q + b2 cos2 q ) 2 = __________________. (1) ab Now the equation of tangent to the ellipse at (a cos q, b sin q ) is yb sin q _______ xa cos q  x cos q + __y sin q = 1. + _______   = 1 or __ 2 2 a b a b Therefore, p = length of the perpendicular from the center (0, 0) on the tangent ______________ ab 1   =   _________________   ________________ =      _____________ 2 2 2 2 2 2 _____ _____ q  q a sin q + b cos q   cos sin √ +   a2 b2

√

12/30/2011 6:16:08 PM

3.7 curvature n  Thus,

a3b3 p3 = ____________________3 . (a2 sin2 q + b2 cos2 q) 2

Putting this value in (1), we get b. r = a____ p3 2 2

EXAMPLE 3.8

Therefore, x′ = a (1 + cos t) and y′ = a sin t, x″ = -a sin t and y″ = a cos t. Hence, 3 __ (x′ 2 + y′ 2) 2 _________   r= x′y″ - y′x″ 2

2

Find the radius of curvature at any point q of the curve x = a (q - sin q ) and y = a (1 - cos q ).

= 

Solution. The equation of the curve is

1 + cos t __ = (2√2 )a 2 cos2__t 2

x = a (q - sin q ) and y = a (1 - cos q ). Therefore, __ x′ = a(1 - cos q) = 2a sin2 q  and 2 __ __ .  cos q   y′ = a sin q = 2a sin q 2 2 Thus, __ q cos q sin__   y′ 2a dy __ ____________ __ . __ 2 2 = cot q = =   __ 2 dx x′ 2 q 2a sin   2 Further, 2 dy d___y __ __ d cotq = d __ = __   2 dx dx 2 dx dx

(  ) ( 

 

)

__ d q  d   cot q = ___   ___ 2 dx dq __ ________ 1 1 cosec2 q  · = -__ 2 2 2a sin2 q_  2 __ q ___ 4 = -   1 cosec  . 2 4a

  Hence,

(  )

[ (  ) ]

3 __

3 __ dy 2 2 __ 2 1 + __   1 + cot2 q dx 2 r = __________ = ___________ 2 __ 1 __ d y ___ -  cosec4 q   2 4 2 dx __ 4a = 4a sin q  (in magnitude). = - _______ __ 2 q cosec   2

( 

)

EXAMPLE 3.9

Show that the radius of curvature at any point of the cycloid x = a (t + sin t) and y = a (1 - cos t) is 4acos __t . 2 Solution. The equation of the curve is x = a (t + sin t) and y = a (1 - cos t).

M03_Baburam_ISBN _C03.indd 7

2

2

a(1 + cos t) a cos t - a sin t (-a sin t) 3 __

a [2 (1 + cos t)] 2 ______________

=

3 __

[a (1 + cos t) + a sin t] 2 ______________________________

( 

)

__ 1 __ = (2 √2 )a (1 + cos t) 2 1 __ 2

= 4a cos __t . 2

EXAMPLE 3.10

If r1 and r2 are the radii of curvatures at the extremities of a focal chord of the parabola y2 = 4ax, prove that  2 2 2 - __ - __ - __ r1 3 + r2 3 = (2a) 3. Solution. The parametric equation of the parabola y2 = 4ax is x = at2 and y = 2at. Therefore, x′ = 2at and y′ = 2a, x″ = 2a and y″ = 0.

Therefore, r at the point (at2, 2at) is   r =  

3 __

[x′ + y′ ] 2 __________ 2

2

x′ y″ - y′ x″

=

3 __

[4a t + 4a ] 2 ___________ 2 2

2

- 4a2

3 __

3 __ 8a3(t 2  + 1) 2 = 2a(1 + t2) 2 (in magnitude). = __________ 2 - 4a Let the extremities of the focal chord be

P (at21, 2at1) and Q (at22, 2at2). Then, by the property of focal chord, t1t2 = -1. Further, by the above-obtained expression for r, we have        r1 = radius of curvature at P (at21, 2at1) 3 __

 

= 2a(1 + t21) 2 and r2 = radius of curvature at Q (at22, 2at2) 3 __

= 2a (1 + t22) 2.

12/30/2011 6:16:11 PM

3.8 n chapter three Therefore, since t21 t22 = 1, we have  2 - __ 3

 2 - __ 3

 r1 + r2 = (2a)

 

 2 - __ 3

 2 - __ 3

= (2a)

[ 

_____ 1

1- t21

[ 

1 + _____ 1 - t22

]

]

( 

))

(

 2 t21 + t21 + 2 - __ ________ = (2a) 3. 2 2 t1 + t1 + 2

If r1 and r2 are the radii of curvatures at the extremities of two conjugate semi-diameters of an y2 x2 + __ = 1, show that ellipse __ 2 a b2

(   r

2 __

1

3

2 __

)

2 __

+ r23 (ab)3 = a2 + b2.

Solution. The parametric equations of the ellipse are x = acos q and y = bsin q. As in Example 3.7, 3 __ we have 2 sin2 q + b2 cos2 q ) 2 (a __________________ r= ab and so, 3 __ rab = (a2 sin2 q + b2 cos2 q ) 2 or 2 2 __ __ (ab) 3 r 3 = a2 sin2 q + b2 cos2 q.

If C is the center of the ellipse and CP and CQ be a pair of conjugate semi-diameters of the ellipse, __ then the eccentric angles of P and Q are q and q + p  , 2 respectively. Thus, if r is the radius of curvature at P (a cos q, b sin q ), then the curvature at Q is __ obtained by replacing q by q + p  . Thus, if r1 and 2 r2 are the radii of curvatures at P and Q, then from the above expressions we get

3 __

(a2 sin2 q + b 2 cos2 q ) 2 (CQ)3 r = __________________ = _____ . ab ab EXAMPLE 3.13

Show that the radius of curvature at the end of y2 x2 + __ the major axis of the ellipse ____ = 1 is equal 2 a b2 to the semi-latus rectum of the ellipse. Solution. As in the above examples, we have 3 __

(a2 sin2 q + b2 cos2 q ) 2 r = __________________. ab Now at the ends of the major axis, q = 0 and p. Therefore, putting q = 0, we get

r=

3 __

(b ) 2 __ ____ b2 2

ab

= a = Semi-latus rectum of the ellipse.

3.4 RADIUS OF CURVATURE FOR PEDAL CURVES Let P(r, q ) be a given point on the curve such that the tangent to the curve at P makes an angle y with ythe x-axis. Let f be the angle between the tangent and the radius vector. Then y = q + f. y

2  __

(ab) 3 r 13  = a2 sin2 q + b2 cos2 q 2 __ __  2 p  + b2 cos2 q + __ p  (ab) 3 r 23  = a2 sin2 q + __ 2 2 = a2 cos2 q + b2 sin2 q. Adding these two relations, we get

( 

(

2  __

)

(

Hence, from Example 3.7, we have

EXAMPLE 3.11

2 __

Solution. If P (a cos q, b sin q ) is the point on the __ ellipse, then the point Q shall be Q a cos q + p  , 2 __ p b sin q +   = Q (-a sin q b cos q ). Therefore, 2 ________________ CQ = √a2 sin2 q + b2 cos2 q  or 3 __ (CQ)3 = (a2 sin2 q + b 2 cos2 q ) 2.

2  __

2 __

)

( 

φ

)

) 

r 13  + r 23  (ab) 3 = a2(sin2 q + cos2 q ) + b2(sin2 q + cos2 q ) = a2 + b2.

EXAMPLE 3.12

Show that the radius of curvature at a point P y2 (CQ)3 , _____ x2 + __ on an ellipse ___ 2 2 = 1 is given by r = ab a b where CQ is the semi-diameter conjugate to CP.

M03_Baburam_ISBN _C03.indd 8

r θ

0

r θ

φ

Ρ (γ,θ )

φ

Ρ (γ,θ )

φ

ψ

p

90¡

Nψ

0 Differentiating with respect 90¡ to s, we get p df . y d___ d___ N   = q +  ___ ds ds ds

x x

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curvature n  3.9

q , we have Moreover, since sin f = r d___ ds sin f __ 1 r ________ q = _____ d___ 1 _______ ______ ______ . r  = r · r 2 + r2 = r 2 + r2 ds √ √ 1 1 ds

d

Hence from (1), we have

[ 

2 r_______ - rr2 1 1 _______ ______ =   1 + 2 r 2 2 r1 + r 2 √r + r

__ 1

1

r 2 - rr2 + 2r21 = ___________ 3 __ (r 2 + r21) 2



dr

Also p = rsin f. Differentiating with respect to r, we get df dp df ___ q + r ___ dr · ___ = sin f + r cos f   ___    = r d___   ds dr ds dr dr df dy r q +  ___ ds   .   = r ___ = __  , r = ___ = r d___ ds ds ds r dy

[ 

]

Hence,

dr . r = r ___ dp

3.5 RADIUS OF CURVATURE FOR POLAR CURVES From the figure of article 3.4, we have y = q + f and so differentiation yields df y d___ d___  = q + ___  ds ds ds or df d___ __ q ___ q. 1 d___ (1) r = ds  + dq   · ds   But, __ r ___ ___ q d dr   . dr r , where r = ___ tan f = r  =  = __ 1 dr dq r1 dq Differentiating with respect to q, we get 2 df r1r1 - rr2 d___r . sec2 f   ___     = ________ = , where r 2 dq 2 dq r21

Therefore,

r - rr r - rr   =  = dq r sec f (1 + tan f) 2 1 _______ 2 2 2 1

df ___

r - rr

2 1 ________ 2 2 __ 2 2 1 1

2 1 _________ 2 2

___ q =   __ , since tan f = rd =  r . dr r1 r r 1+ r

(  )

= 

M03_Baburam_ISBN _C03.indd 9

2

r1 - rr2 _______ . r21

+r

2

]

which yields

3 __ 2 2

(r + r1) ___________ . 2

r=

r 2 - rr2 + 2r21

3.5.1 Second Method The relation between p, r, and q is dr   . 1 + __ 1 r2, r = ___ = __ p2 r 2 r4 1 1 dq Differentiating with respect to r, we get __ 1

dp d r 2 - __ 4 r2 + __ 2 r __ 2 ___ = - __ - __ p3 dr r 3 r 5 1 r 4 1 dr 1

[  (  ) ]

d   ___ 4 r2 + __ 2 r ___ 2 - __ q  dr      d___              = - __ r 3 r 5 1 r 4 1 dq dq dr 4 r2 + __ 2 r · r · __ 2 - __ 1              = - __ 5 1 3 r r 4 1 2 r1 r 2 - __ 4 r2 + __ 2r.              = - __ r3 r5 1 r4 2 Therefore, dp __ 1 ___ p3 dr

2 r2 - __ 1 r2 1 + __ = __ 5 1 3 r r r4 r 2 + 2r12 - rr2 . = ___________ r5

Since, __ 1

p

3

=

we have,

[ ( r

__ 1

[ ( 

2

+

__ 1 2

r

4

r1

)] 2

3 __ 2

(1)

dp r , and ___ = __    dr r

3 __ 2 2

)]

3 __ 1 r2 1 + __ 2 2 2 r6 __ 4 1 2 + r ) (r r r 1 . r = ______________ = ___________ r 2 + 2r21 - rr2 r 2 + 2r21 - rr2

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3.10 n chapter three Corollary. Let the equation of the curve be __ 1 or r = __ 1 = f(q ). Put u = __ 1. Then, r r u

Hence,

du  1 · ___  = - __ dq u2 dq

     r =

___ dr

and

(  )

2 d 2u + __ d___ du  2  . r = - __ 1 · ___ 2 ___ 2 3 2 2 u dq  u dq dq Therefore,

[  (  ) ]

=

( dq )  -r dq 

r +2

(  ) ]

3 __

(  ) [  (  ) ] __ du    1 [u +  ___ du        u + ( ___ ( ) [ [ ] d q u dq ) ] ____________ ______________ = = __ u u 1 + __ 1 d____ u ( u + d___ u u dq  dq  ) 3 __

3 __

2 2

2

2

2

= where

3

2

2 2

2

2

2

2

3 __ 2 2

[ u + u1 ] , _________ 2

u3(u + u2)

du and u = d___ u. u1 = ___ 2 dq dq 2 2

EXAMPLE 3.14

Find the radius of curvature at the point (r, q ) of the following curves: (i) r 2 cos 2q = a2. (ii) r n = an sin nq. (iii) r = a (1 - cos q ). Solution.

(i) The equation of the curve is r 2 cos 2q = a2. (1) Taking logarithm, we get 2 log r + log cos 2q = log a2. (2) Differentiating (2) with respect to q, we get ________ -2 sin 2q  = 0 r dq + cos 2q   

__ dr 2 ___

dr  = r tan 2q. or r1 = ___ dq

Differentiating again with respect to q, we get 2 dr   r2 = d___r2 = 2r sec2 2q + tan 2q ___ dq dq 

M03_Baburam_ISBN _C03.indd 10

3 __

(r + r tan 2q) 2 _________________________________ 2

2

2

r2 + 2r2 tan2 2q - (2r2 sec2 q + r2 tan2 2q) 3

(  )

du  2  2 1 ___    __  12 + __ 4 dq u u _________________________________ =    2 __ du  2  - __ du  2  u + __ 1 + 2 · __ 1 ___ 1 -  __ 2 ___  12 d___ 2 4 u u u dq u dq 2 u3 dq 6

r 2 + 2r12 - rr2

3

2

[ 

3 __

(r 2 + r12) 2 ___________

sec 2q = - r________    = - r sec2q r 2 sec2 2q

3 __

dr   2  2 r + ___ d q ________________ r= 2 2 ___ dr d___ r 2 2

 

= 2r sec2 2q + r tan2 2q.

2 = - r r__2 , using(1) a __3 r = 2 (in magnitude). a

(ii) The given curve is r n = an sin nq Taking log, we get nlog r = nlog a + log sin nq. Differentiating with respect to q, we get __ dr n ___ r dq = n cot nq. Therefore, dr = r cot nq. r1 = ___ dq Differentiating once more, we get

(1)

2 dr   r2 = d___r2 = nr cosec2 nq + cot nq ___ dq dq  = -nr cosec2nq + r cot2 nq. Hence, 3 __ 2 2 2 + r ) (r 1 r =  ___________  2 r + 2r 21 - rr2 3 __     ( r 2 + r 2 cot2 nq ) 2 _____________________________________  =      2 r + 2r 2 cot2 nq - r (- nr cosec2 nq + r cot2 nq) r 3 cosec3 nq       = ________________________ 2 2 r + r cot2 nq + nr 2 cosec2 nq  r 3cosec3nq = _________________________       2 r (1 + cot2 nq) + nr 2 cosec2 nq  3 r 3 cosec = _______________   nq  r 2cosec2nq (1 + n)

cosec nq  = r_________ n+1 _________ an , using (1). = (n + 1)r n -1

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curvature n 3.11 (iii) The equation of the given curve is (1) r = a (1 - cos q). , Differentiating with respect to q we get d 2r dr  = a sin q and r = ___ r1 = ___ = a cos q. 2 dq dq 2 Therefore, 3 __

(r + r 1) ___________ 2

  r=

2 2

r 2 + 2r 21 - rr2

3 __

[a (1 - cos q  ) + a sin q  ]2 _______________________________________ 2

2

2

2

  =   2 a (1 - cos q  )2 + 2a2,sin2 q - a2cos q (1 - cos q  ) using (1) 3 __

a (1 + cos q + sin q - 2cos q  ) 2 __________________________________ 3

2

2

=     2 a [1 + cos2 q - 2cos q + 2sin2 q - cos q + cos2q  ] 3 __

3 __

_ ___ =   =     =   __  2 √2ar . 3 (1 - cos q) 3 3√a It also follows that the radius of curvature at any _ point of the cardioid r = a(1 - cos q ) varies as √r . 2 2 a (1 - cos q) 2 ______________

 

3 __

2 2a√ _____ __r

EXAMPLE 3.15

Show that the radius of curvature ___ at any point of2 r  __ 2 the cardioid r = a(1 + cos q ) is √2ar and that __ r 3 is constant. Solution. The equation of the curve is r = a (1 + cos q ). Therefore, 2 r = - a cos q. dr  = - a sin q and r = d___ r1 = ___ 2 dq 2 dq Therefore, 3 __ 2 2

(r + r1) __________ 2

r=

3 __

[a (1 + cos q  ) + a sin q  ] 2   =__________________________________ 2 a (1 + cos q  )2 + 2a2 sin2 q + a 2 cos q (1 + cos q  ) 2

2

2

2

3 __

a (1 + cos q + sin q + 2cos q  )2 ______________________________________ 2

2

=     2 a [1 + cos2 q + 2cos q + 2sin2 q + cos q + cos2 q  ] _ 2 2 a (1 + cos q) 2 =    ______________ =    2 a __r 3√a 3 (1 + cos q) ___ __ 4a cos q 2 √2ar = ___  (in terms of q ). = __ 3 2 3 3 __

M03_Baburam_ISBN _C03.indd 11

EXAMPLE 3.16

Find the radius of curvature of the curve _l r = 1 + e cos q. Solution. The equation of the given curve is _l r = 1 + e cos q. __ 1 . Let r = u Then the equation of the curve becomes 1 (1 + e cos q ). u = __ l Therefore, ___ 1 (- e sin q ), u1 = du = __ dq l 2 u = __ 1 (- e cos q ). u2 = d___ dq2 l Hence, the radius of curvature is (u2 + u21)     r = _________ u3 (u + u2)

( l1 )[(1 + e cos q ) + e sin q ]

3 __

__ 2 2 2 2 3 _________________________________      = 1 __ 3

( l ) (1 + e cos q ) [(1 + e cos q ) - e cos q ] 4

3 __

l[1 + e2 + 2e cos q ] 2 .      = _________________ (1 + e cos q )3 EXAMPLE 3.17

q , If r is the radius of curvature and tan f = r d___ dr show that df ___ __ r r = sin f 1 + dq    for any curve.

( 

)

Solution. We know that

r2 + 2r21 - rr2

3

_ It follows that r varies as √r . Further, 2 __ ___ 8ar or r  8a r2 = ___ r = 9 (constant). 9

3 __

3 __ 2 √ _____

y = q + f. Differentiating with respect to s, we get df d___ df y d___ d___ q + ___  = q + ___ =    d___   · q  ds ds ds ds dq ds  

( 

)

df , q  1 + ___ d___    = ds dq

( 

)

sin f df , ___ 1 _____ that is, __ r = r   1 + dq  , ds   and since r = ___ dy

q . sin f = r d___ ds

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3.12 n chapter three Hence, df r cosec f . ___ ________ __ r      r = sin f 1 + dq    so that r = df 1 + ___   dq

( 

)

EXAMPLE 3.18

If r1 and r2 are the radii of curvatures at the extremities of any chord of the cardioid r = a (1 + cos q ) which passes through the pole, show that 16a2 . r 12 + r 22 = ____ 9 Solution. As in Example 3.15, the radius of curvature at any point of the cardioid r = a (1 + cos q) is ____ r =   __  2√2ar so that r 2 =   __  8 ar. 3 9 If PQ is any chord of the cardioid passing through the pole and if P and Q are the points (r1, q1) and (r2, q2), respectively, then q2 = p + q1 ,Hence, if r1 and r2 are the radii of curvatures at P and Q, respectively, then 8 ar and r 2  = __ 8 r 2 1 = __ 2 9 ar2. 9 1 Therefore, 8a [r + r ]  r 2 1 + r 22  = ___ 9 1 2 ___ 8a   =     [a (1 + cos q1) + a (1 + cos q2)] 9 ___ 8a2 [2 + cos q + cos q ]   = 1 2 9 2 ___ 8a [2 + cos q + cos (p + q )]   = 1 1 9 2 2 ___ 8a ____ 16a .   = [2 + cos q1 - cos q1] =     9 9 EXAMPLE 3.19

Find the radius of curvature at any point ( p, r) of the following curves. 1 = __ 1 + __ 1 - ____ r2 . (i) ellipse __ p2 a2 b2 a2b2 (ii) pa2 = r 3. (iii) pan = r m + 1 Solution.

(i) The equation of the ellipse is __ 1 = __ 1 + __ 1 - ____ r2 . p2 a2 b2 a2b2 Differentiating with respect to p, we get 2 2 dr = a____ dr or r ___ b . 2r · ___ 2 = - ____ - __ dp p3 p3 a2b2 dp

M03_Baburam_ISBN _C03.indd 12

Hence,

dr = a____ b . r = r ___ dp p3 (ii) The equation of the curve is pa2 = r 3. Differentiating with respect to p, we have dr = __ a2 . dr or r ___ a2 = 3r 2 ___ dp 3r dp Hence, dr = __ a2 r = r ___ dp 3r (iii) The given equation is pan = r m + 1. Therefore, differentiation with respect to p yields dr = __________ an dr or r ___ . an = (m + 1)r m ___ dp dp (m + 1)r m - 1 Hence, an dr = __________ . r = r ___ dp (m + 1) rm - 1 Thus, radius of curvature varies inversely as the (m - 1)th power of the radius vector. 2 2

EXAMPLE 3.20

Show that the radius of curvature of the equiangular spiral r = aeq cotα at any point (r, q ) is r = r cosec α Solution. The given curve is r = aeq cotα. Differentiating with respect to q, we have dr  = aeq cotα cot α = r cot a and r1 = ___ · dq 2 r2 =    d___r2 = r cot 2α. dq  Therefore, 3 3 __ __ 2) 2 (r + r (r 2 + r 2 cot2 α) 2 ____________________ 1 ___________ r =     2 =  r + 2r21 - rr2 r 2 + 2r 2 cot2 α - r 2 cot2α 3 __

r (1 + cot α)       = ____________    = r (cosec2α) 2 = r cosec α. 3

2

2

1 __

r2 (1 + cot2α)

3.6 RADIUS OF CURVATURE AT THE ORIGIN The radius of curvature at the origin is determined by the following methods: 1. Newton’s Method: Suppose that a curve passes through the origin and the axis of x is tangent at the origin. Then,

12/30/2011 6:16:33 PM

curvature n 3.13

(  )

dy slope of the tangent at the origin = __ (0, 0) dx = y1 (0, 0). Since the tangent at the origin is x-axis, the slope of the tangent should be zero. Therefore y1 (0, 0) = 0. The radius of curvature at the origin is  

r at (0, 0) = 

3 __

2

[1 + y1 (0, 0)] 2 ____________ y2 (0, 0)

On the other hand,

lim __ 2x x2 = lim ___ 2y x→0 2y y→0 1

x→0 y→0

1 . (1) = _______ y2 (0, 0)

( __00 form ) (2)

y→0

__ x2 .

2y

Similarly, if the curve passes through the origin and y-axis is tangent to it at the origin, then y2 . lim __ r at (0, 0) = x→0 2x y→0

If the initial line is tangent at the pole (origin), then 2 x2 lim r_______ __ cos2 q     r = lim x→0 2y = r→0 2r sin q  y→0 θ→0

( 

lim ___ q   · cos2 q = lim ___ 0 form r  · ____ r   __ = r→0 sin q 2 q r→0 2q 0 q→0  

___ dr   lim d___ q =   __ dr   . =  1 · ___ r→0 q→0

2

q→0

)

2 dq

2. Method of Expansion: This method is used when neither x-nor y-axis is tangent to the curve at the origin. Suppose that the equation of the curve is y = f (x). Since the curve passes through the origin, f (0) = 0. By Maclaurin’s expansion, we have 2 x3 f ″′ (0) + . . . y = f (0) + x f ′(0) + x__ f ″(0) + __ 2! 3! 2 3 __ __   = xf ′(0) + x f ″(0) + x f ″′ (0) + . . . 2! 3! __ __ (1)   = px + 1 qx2 + 1 rx3 + . . . 2! 3!

M03_Baburam_ISBN _C03.indd 13

Differentiating (1) with respect to x, we get y1 = p + qx +  __  1 r x2 + . . . and 2 y2 = q + rx + . . .. Thus, y1 (0, 0) = p, y2 (0, 0) = q. 2

3 __

[1 + y1 (0, 0)] 2 ____________

r at the origin = 

y2(0)

=

3 __

(1 + p ) 2. _______ 2

q

EXAMPLE 3.21

From (1) and (2), it follows that

lim r at (0, 0) = x→0

p = f ′ (0) = y1 (0), q = f ″ (0) = y2 (0), r = f ″′ (0) = y3 (0), and so on.

Hence,

lim __ 1 ______ 1 . = x→0 y = y (0, 0) 2 y→0 2

 

where,

Find the radius of curvature at the origin for the curve 5x3 + 7y2x2 + 4x2y+xy2 + 2x2 + 3xy + y2 + 4x = 0. Solution. The given curve passes through the origin. The lowest-degree term in the given equation is 4x. Therefore, the tangent at the origin is 4x = 0 or x = 0, that is, the y-axis. Therefore, by Newton’s Method, y2 . ___ r at the origin = lim x→0 2x y→0 Dividing the equation of the curve throughout by 2x, we get 7y2x2 y2 y2 3 y + __ x2 + _____ + 2xy + __ + x + __ + 2 = 0. 2 2x 2 2 2x

__ 5

Taking limit as x → 0 and y → 0, we get y2 lim __ x→0 2x + 2 = 0 y→0 or r = - 2 = 2 (numerically). EXAMPLE 3.22

Show that the radii of curvatures at the origin 3a. on the curve x3 + y3 = 3axy is each equal to ___ 2 Solution. The equation of the given curve is x3 + y3 = 3axy. The curve passes through the origin. Further, the tangent at the origin is given by 3axy = 0. Thus, x = 0 and y = 0 are the tangents at the origin.

12/30/2011 6:16:35 PM

3.14 n chapter three Therefore, the radii of curvatures are given by __ x2 . r1 at the origin = lim x→0 2y y→0

and

y . __ r2 at the origin = lim x→0 2x y→0 2

Dividing the given equation throughout by 2xy, we get y2 ___ __ x2 + __ = 3a . 2y 2x 2 Taking limit as x → 0 and y → 0, we have y2 ___ lim __ 3a x2 lim __ x→0 2y + x→0 2x = 2 y→0 y→0 or 2y ___ lim __ 3a x2 lim __ 1 __ x→0 2y + x→0 4 xy x2 = 2 y→0 y→0

or

___ 3a 3a . 1 ___ r1 + 0 __ r1 = 2 or r1 = 2 3a . Similarly, we can show that r2 = ___ 2

EXAMPLE 3.23

Show that the radius of curvature of the curve a+x y 2 = x 2 _____ a-x __ __ is ± a√2 (or a√2 in magnitude). Solution. The curve is

y 2 (a - x) = x 2(a + x). Obviously, the curve passes through the origin. But equating to zero the lowest-degree terms, we get a ( y 2 - x 2) = 0. Therefore y = ± x is the tangent at the origin. Hence, Newton’s method cannot be applied to this curve. So let 1 qx 2 + __ 1 rx3+ ... y = px + __ 2! 3! Substituting this value of y in the given equation of the curve, we get

( px + __21 qx + __31! r x + . . . ) (a - x) 2

3

2

  = x 2(a + x). Equating coefficients of x 2 and x 3 on both sides, we get ap2 = a and - p2 + apq = 1,

M03_Baburam_ISBN _C03.indd 14

which yield 2 p = 1, q = __ a Therefore,

r1 =

___ 2. p = - 1, q = a

and 3 __

(1 + p ) 2 _______ 2

q

3 __

___

2 2 = 2a = __ √ __ 2

a ___ ___ (1 + 1) 2 r2 = _______ = - √2a = √2a (in magnitude). 2 - __ a

and

3 __

EXAMPLE 3.24

Find the radius of curvature at the origin of the cycloid x = a (q + sin q ) and y = a (1 - cos q ). Solution. The given curve is

x = a (q + sin q ) and y = a (1 - cos q ). Therefore, differentiation with respect to q yields dy ___ dx  = a (1 + cos q ) and ___  = a sin q. dq dq Then, __ __  cos q   2 sin q q __________ dy dy/d _____ __________ q a sin 2 2 ___    =  = = __  dx dx/dq a (1 + cos q ) 2 cos2 q 2 __ q   = tan  . 2 dy Since q = 0 at (0, 0), we have __ (0,0)= 0. dx Therefore, x-axis is tangent to the curve at the origin and so, the Newton’s method is applicable. Therefore, r at the origin is given by

(  )

__ x2 r (0,0) = lim x→0 2y y→0

[ 

2 (q + sinq )2 a___________   = lim q →0 2a (1 - cos q )

] ( 

__ 0

0

form

)

] (  ) [  a (q + sin q ) (- sin q ) + (1 + cos q )   = lim [  ]  cos q 

q + sin q ) (1 + cos q )} __ _____________________ __ a {2(   = lim       0 form q →0 2 0 sin q 2

____ _____________________________ q →0

  = 4a. 3.7 CENTER OF CURVATURE Let P(x, y) be any point on the curve and let PT be the tangent at P making an angle y with the

12/30/2011 6:16:37 PM

curvature n 3.15 positive direction of x-axis. Let C(X, Y ) be the center of curvature corresponding to P(x, y). y

ρ

N

P (X, Y)

ψ

T

0

M

L

  and

x

(1)

= y + r cosy

Therefore,

or

√1 + y21

y2

3.9 EQUATION OF THE CIRCLE OF CURVATURE Let (X, Y ) be the coordinates of the center of curvature and r the radius of curvature at any point (x, y) of the curve. Then the equation of the circle of curvature is (x - X)2 + (y - Y)2 = r 2. EXAMPLE 3.25

Y = CM = NM + CN = NM + r cos y   But,

y2

3.8 EVOLUTES AND INVOLUTES The locus of the center of curvature for a curve is called the evolute of the curve and the curve itself is called the involute of its evolute.

Draw PL and CM perpendicular to the x-axis from P and C , respectively. Draw PN ⊥ CM. Then X = OM = OL - ML = OL - NP = x - r sin y

3 __

1 + y2 ( 1 + y21 ) 2 · _______ 1 _____ Y = y +  ________ = y + _____1 .

Also (1) and (2) can be expressed as dy dy ds   · __ = x - ___   X = x - r siny = x - ___ dy ds dy and dx   . ds   · __ dx = y + ___ Y = y + r cosy = y + ___ dy ds dy

C (X, Y ) ψ

and

(2)

dy tan y = __ = y1. dx ________ sec y = √1 + tan2 y  _____ = √1 + y21

1 _____ cos y = ______ √1 + y21 _________ and so, ________ 1 2 sin y = √1 - cos y  = 1 - ______ 1 + y21 y1 _______ _____ .  = √1 + y21 Also, 3 __ 1 + y21 ) 2 (  ________ . r= y2 Hence (1) yields

√

3 __

 1 + y21 ) y1 y________ ( 1 + y21 ) 2 · _______ 1( _____ X = x - ________ = x y2 y2 1 + y2

√

M03_Baburam_ISBN _C03.indd 15

1

Find the center of curvature of the following curves. (i) y = x3 - 6x2 + 3x + 1 at (1, - 1). (ii) the parabola y2 = 4ax at (x, y). Also, find the equation of the evolute of the given parabola. y2 x2 + __ (iii) the ellipse __ = 1 at (x, y). Also, find a2 b2 the evolute. Solution.

(i) The given curve is y = x 3 - 6x2 + 3x + 1. Therefore, dy y1 = __ = 3x 2 - 12x + 3 and so y1 (1, - 1) = - 6, dx d 2y y2 = ___2 = 6x - 12 and so y2 (1, - 1) = - 6, dx y1(1 + y21) X = x - ________ y2 (- 6)[1 + (- 6)2]   =1 - _____________ = - 36, and -6 + (- 6)2 1________ 43 . Y=-1+ = - ___ -6 6

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3.16 n chapter three Hence, the center of curvature is 43 (X, Y ) = -36, - ___ . 6 (ii) The equation of the given parabola is y2 = 4ax. Therefore, __  __ 3 __ 2a = __ 2a = _____ a 1 __ - 2 ____ y1 = ___ y √4ax √ x and y2 = - 2√a x . __ Therefore, a a x____________ - __x 1 + __x X=  3 - __ 1 __ - __ √a x 2 2   = x + 2 (x + a) = 3x + 2a (1) and a y + 1 + __x Y = __________3 1 __ - __ √a x 2 2 __ __ 2(x + a)   = 2 √a √x - _______ __ - __1 √a x 2 __ __ x + a)   = 2 √a √x ( 1 - _____ a

( 

)

√ ( 

)

3 __

x__2 . = - 2 ___ (2) √a Hence, the center of curvature of the given parabola is 3 __ 2x__2 . (X, Y ) = 3x + 2a, - ___ √a ______ X 2a . Putting this value From (1), we have x = 3 in (2), we get  

( 

)

3 __

or

2(X -__2a) 2 Y = -  _________ √a

or

aY 2 = 4

( 

______ X - 2a

3

)

3

3

27aY 2 = 4 (X - 2a) .

Therefore, the locus of the center of curvature (X,Y ) is 27ay2 = 4 (x - 2a)3, which is the required evolute. (iii) The equation of the ellipse is y2 __ x2 + __ 2 = 1. 2 a b Therefore, b2x and y = - ____ b4 . y1 = - ___ 2 2 a2y3 ay

M03_Baburam_ISBN _C03.indd 16

Therefore, and

y1(1 + y21) b4x2 + a4y2 X = x - ________ = x - _________ y2 a4b2

4 2 1 + y21 x + a4y2 . b_________ Y = y + ________ y2 = y + a2b4

Second Method

Equations of the ellipse are x = a cos q and y = b sin q. Therefore, dy ___ dx  = - a sin q and ___  = b cos q dq dq and so, dy dy/dq b  = - __ y1 = __ = _____  a cot q and dx dx/dq

(  ) [ 

)]

( 

d 2y d __ dy d   - __ b  d___ q y2 = ___2 = __ = ___ a cot q      dx  dx dx dq dx

( 

)

_______ __ __ b b 1 2 3 = a cosec q - a sin q   = - a2 cosec q. Therefore, y1(1 + y21) X = x - ________ y2 b2 cot2 q    __ b - a cot q 1 + ___ a2     = x - ___________________       __ b - 2 cosec3 q a

 

( 

   

)

(a2sin2q + b2cos2q ) = a cos q - _______________ cos q a 2 2 a - b cos3q =    ______ (1) a

and

b2 cot2q 2 1 + __ + y 1 __________ a2 1     Y = y + ________ y2 = y + __ b - 2 cosec3 q a ____ q sin   = b sin q  (a2sin2 q + b2cos2 q ) b 1 sin q [b2 - a2sin2 q - b2cos2 q ] = __   b 2 2 - a2 sin3 q = - a______ - b2 sin3 q (2) = b______   b b Hence the center of curvature is

( 

)

2 2 - b2 cos3 q, - a______ - b2 sin3 t . (X, Y ) = a______ a b

From (1) and (2), we have 2 2 __ __ aX = (a2 - b2) cos3q or (aX ) 3 = (a2 - b2) 3 cos2 q

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curvature n 3.17 and

The locus of (X, Y) is x = a (q + sinq ) and y = - a (1 - cos q ), which is another equal cycloid.

bY = - (a2 - b2) sin3 q or 2 __

2 __

(bY ) 3 = - (a2 - b2) 3 sin3 q. Adding both we get

EXAMPLE 3.27

(aX ) 3 + (bY ) 3 = (a2 - b2) 3. Therefore, the evolute of the ellipse is

Show that the evolute of the rectangular hyperbola xy = c2 is the curve

2 __

2 __

2 __

2 __

2 __

2 __

2 __

(ax) 3 + (by) 3 = (a2 - b2) 3. EXAMPLE 3.26

Show that the evolute of the cycloid x = a(q - sin q ) and y = a(1 - cos q ) lies on an equal cycloid. Solution. For the given cycloid, we have dy __ dx  = a (1 - cos q ) and ___  = a sin q. dq dq Therefore, dy   dy/dq __________ __ y1 = __ =   _____  = a sin q  = cot q   2 dx dx/dq a (1 - cos q ) and q  d (y ) = ___ d (y ) d___ y2 = __ dx 1 dq 1 dx 1 1 cosec2 q __ · __________   = - __ 2 2 a (1 - cos q ) -1 . = _______ __  4a sin4 q 2 Therefore, X=x-

y1(1 + y21) ________ y2

[ 

]

__ __    1 + cot2 q cot q 2 ______________ 2   = a (q - sin q ) __ 1 4q __ - cosec   4 2 __ __      = a (q - sin q ) + 4a cos q  sin q   2 2

   

(2)

Hence, the center of curvature of the given cycloid is (X, Y ) = (a (q + sinq ), - a (1 - cos q )).

M03_Baburam_ISBN _C03.indd 17

    and

)

4 + x2y2 3x ________ = , since xy = c2 2x3 y2 __ 3x + ___ = 2 2x

c 1 + ___ 4 x4 + c4 x_____ Y = y + y = y + _____ 2 = y + ___ 2c2x 2c 2 3 x x4 + x2y2 , since xy = c2 = y + _______ 2x2y 4

2

1 + y1 _____

x2 + y2 3y2 + x2 3 x2 . = y + _____ = ______ = __ y + __ 2y 2y 2 2y Hence, the center of curvature is

 

2

__  = a (1 - cos q ) - 4a sin2 q 2 = a - a cos q - 2a + 2a cos q = - a (1 - cos q ).

2 __

( 

 

  = aq - a sin q + 2a sin q = a (q + sinq ) (1) and 1 + y21 Y = y + _____ y  

2 __

(x + y) 3 - (x - y) 3 = (4c) 3 . Solution. We have c2. xy = c2 or y = __ x Therefore, 2 dy d 2y ___ c2 and y =  ____ = 2c . y1 = __ = - __ 2 2 dx x dx2 x3 Let (X, Y ) be the coordinate of the center of curvature. Then, ______ c2 1 +  ___ c4 2 2 (1 + y ) y________ x 1 x4 X=x- 1 y = x - __________ 2 ___ 2c 2 x3 4 4 4 4 4 __________ + c = 2x +x +c   = x + x_____ 2x3 2x3

( 

)

2 3y __ y , __ 3x + __ +x . (X, Y ) = __ 2 2x 2 2y Further, we note that 1 [x3 + y3 + 3x2y + 3xy2] X + Y = ___ 2xy ___ 1 (x + y)3   = 2c2 and so, 2 __ 1 (x + y)2. (X + Y ) 3 = _____ 2

2 __

(2c2) 3

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3.18 n chapter three Similarly, 2 __ 1 (x - y)2. (X - Y ) 3 = - ______ 2 __ (2c2)) 3 Therefore, 2 2 __ __ (X + Y ) 3 - (X - Y ) 3 1 [(x + y)2 + (x - y)2] = _____   2 __

 

(2c2) 3 1 (4xy) = _____ 1 (4c2) = _____ 2 __

2 __

(2c2) 3

(2c2) 3

2 __

  = (4c) 3 . The locus of (X,Y) is therefore, 2 __

2 __

Therefore, the equation of the circle of curvature is (x - X )2 + ( y - Y )2 = r 2 or am(1 + m2) 2 a(1 + m2) 2 x + _________ + y - ________ 2 2 a2(1 + m2)3 = _________ 4 or 2 2 4x + 4y = a(1 + m2) + [a(1 + m2)2 - am2(1 + m2)     - a (1 + m2) + 4 ( y - mx)]

[ 

] [ 

= 4a (1 + m2) ( y - mx), 2 __

(x + y) 3 - (x - y) 3 = (4c) 3s. Remark 3.2 The question can also be solved by taking a parametric equation of the curve as x = ct and c y = _t , and then eliminating t from the expression for X and Y. EXAMPLE 3.28

Show that the circle of curvature at the origin of x2 is the parabola y = mx + __ a

which yields x2 + y2 = a (1 + m2) ( y - mx). 3.10

CHORDS OF CURVATURE PARALLEL TO THE COORDINATE AXES Let P (X, Y ) be a point on the curve and let the tangent at P make an angle y with the positive direction of x-axis. Then CP = r and PD = 2r. Let PA and PB be the chords of curvature parallel to the x- and y-axis, respectively. y

x + y = a (1 + m ) (y - mx). 2

2

]

2

B

D

Solution. The equation of the given parabola is

x . y = mx + __ a 2

Therefore,

am + 2x 2x = _______ y1 = m + __ a a

so that y1 (0, 0) = m and __ 2 2. y2 = __ a so that y2 (0, 0) = a Let (X, Y ) be the center of curvature at the origin. Then, y1(1 + y21) m(1 + m2) -am (1 + m2) = 0- ________ = ___________ X = x-________ y2 __ 2 2 a and 1 + y21 a(1 + m2) . ______ 1 + m2 = ________ Y = y + _____ y2 = 0 + __ 2 2 a Also, 3 __ 3 __ ( 1 + y21 )) 2 ________ (1 + m2) 2 ________ r at (0, 0) =  = y2 __ 2 a 3 __ a(1 + m2) 2 _________   = 2

M03_Baburam_ISBN _C03.indd 18

C ψ 90¡− ψ ψ

A

P

ψ

x

0

p - y with CP The chord PA makes an angle __ 2 and the chord PB makes an angle y with CP. Let Cx denote the length of PA and Cy denote the length of the chord PB. Then p - y   = 2r sin y Cx = PD cos __ 2

( 

=

3 __ 2 2

)

2(1 + y1) ______ y1 ________ _____ y2

·

√1 + y21

2y1(1 + y21) = _________ y 2

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curvature n 3.19 and

 

Cy = PD cos y = 2 rcos y =

3.11

3 __ 2 2

2 (1 + y1) . ______ _________ 1 _____ y2

√1 + y21

EXAMPLE 3.29

2(1 + y21) . = _______ y 2

CHORD OF CURVATURE IN POLAR COORDINATES

1. Chord of Curvature Through the Pole (Origin)

Let P (X,Y ) be a point on the curve and let the tangent at P make an angle f with the radius vector OP. Then PA, the chord of curvature p  - f with through the pole O makes an angle __ 2 CP, the radius of curvature. y B D C

φ

φ 90˚Ð φ

 

r

0

P

T

x

)

3 __

(r2 + r21) 2 r , since tan f = __ r _____        = 2 ___________ · _______ r1 r2 + 2r21 - rr2 √r2 + r21 2r ( r + r1 ) __________ . 2

       =

2

r2 + 2r21 - rr2

2. Chord of Curvature Perpendicular to the Radius Vector

The chord PB of curvature perpendicular to the radius vector makes an angle f with CP. Therefore, the length Cp of this chord is given by Cp = PB = PD cos f = 2r sin f 3 __

2(r 2 + r21) 2 _______ r1 = ___________ · _____ 2 2 2 2 r + 2r1 - rr √r + r21

M03_Baburam_ISBN _C03.indd 19

x Show that in the curve y = a log sec __ a, the chord of curvature parallel to the axis of y is of constant length. Solution. The equation of the curve is x. y = a log sec __ a Therefore, dy __ x, 1 sec __x tan __x · __ 1 y1 = __ = a _____ a a a = tan a x dx sec __ a 2 d___ y __ 1 x y2 = 2 = a sec2 __ a. dx Therefore, the length of chord of curvature parallel to the y-axis is x 2(1 + tan2 __ 2(1 + y21) __________ a) = Cy = _______ y2 __ 1 sec2 __x a a   = 2a (constant). EXAMPLE 3.30

If Co denotes the length of the chord of curvature through the pole, then r  - f   = 2r sin f Co = PA = PD cos __ 2

(

(r 2 + r21) 2r _________ 1 . = 2 r + r21 - rr2

Find the length of the chord of curvature through the pole of the cardioid r = a (1 + cos q ). Solution. The cardioid is r = a (1 + cos q ). Therefore, 2 dr  = - a sin q and r = d___ r = - a cos q. r1 = ___ 2 dq dq2 Hence, 2r(r2 + r21) Co = ___________ r2 + 2r21 - rr2 2r[a2 (1 + cosq )2 + a2 sin2q ]  = _____________________________________   2 a (1 + cosq)2 + 2a2 sin2q + a2 (1 + cos q ) cosq 2r (a2 + a2 cos2q + 2a2 cos q + a2 sin2q )  = _______________________________     3a2 + 3a2 cos q 2 2r[2a2 + 2a2 cos q ] ____ 4r.  = ________________ = 4r a = __ 2 3ar 3 3a (1 + cos q ) EXAMPLE 3.31

If Cx and Cy are the chords of curvature paralx __ lel to the axes at any point of the curve y = ae a, show that ___ 1 + ___ 1 = ____ 1 . C2x C2y 2aCx

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3.20 n chapter three Solution. We know that

2y1(1 + y21) 2(1 + y21) . Cx = _________ and Cy = _______ y y 2

2

x __

The given curve is y = ae . Therefore, a

d 2y __x 1 . and y2 = ___2 = e a · __ a dx

__ dy 1 = e __a y1 = __ = ae a · __ a dx

x

Hence,

and

x

x __ a

( 

2x __ a

)

( 

2x __ 2e 1 + e = 2a 1 + e a Cx = ___________ x __ __ 1 ea a

( 

2x __

)

( 

)

)

2x __ 2 1+ e a 2a Cy = ________ = _____x 1 + e a . x __ 1 ea __ ea a

Then, 1 1 ___ 1 + ___ 1 = __________ + ___________ 2x 2 __ 2x 2 2 __ C2x C2y 4a ___ a 4a2 1 + e a 1 + e 2x __ a e __ 2x 1 ___________ 1+ e a .   = ___ 2a 2 4a2 1 + e x 1 ___________ 1 .   = = ____ 2x ___ 2aCx 2 4a 1 + e a

( 

)

( 

)

( 

( 

( 

)

)

)

EXAMPLE 3.32

Find the length of the chord of curvature through the pole of the curve r n = ancos nq.. Solution. The equation of the given curve is r n =

an cos nq. Differentiating with respect to q, we have dr  = - nan sin nq nr n - 1 ___ dq or n n sin nq = - r a_______ sin nq  dr  = - a________ r1 = ___ n-1 rn dq r an sin nq     = - r ________ = - r tan nq and an cos nq r2 = - r1 tan nq - rn sec nq 2

  = r tan2 nq - rn sec2 nq. Therefore, 2r(r 2 + r21) Co = __________ r 2 - rr2 + 2r21  

2r(r2 + r2 tan nq) ________________________________   = 2 2 r - r (r tan nq - r n sec2 nq) + 2r2 tan2 q

M03_Baburam_ISBN _C03.indd 20

     

_____________________________ 2r sec nq  =   r2[1 - tan2 nq + n sec2 nq + 2 tan2 nq] __________________ 2r  sec2 nq    = 1 + tan2 nq + n sec2 nq _______________ 2r   sec2 nq   = _____ 2r . = 2 sec nq + n sec2 nq n + 1 3

2

EXAMPLE 3.33

If Co and Cp denote the length of the chords of curvature of the cardioid r = a (1 + cos q), along and perpendicular to the radius vector through any point, show that 3 (C o2 + C 2p ) = 8aCo. Solution. From Example 3.30, we note that

dr  = - a sin q, r1 = ___ dq 2 4r . r2 = d___r2 = - a cos q, and Co = __ 3 dq  Further, 2r1(r 2 + r12) Cp = ___________ r 2 - rr2 + 2r12

2a sin q (2a2 + 2a2 cos q ) = - ____________________ 3a2 (1 + cos q) 4a2 sin q (1 + cos q ) _______ sin q  .     = - ________________ = - 4a 3 3a2 (1 + cos q ) Therefore,

 

[ 

]

16a2 sin2 q    16r2 + ____ 3(C2o + C2p) = 3 ____ 9 9 ___ 16 [r 2 + a2 sin2 q ]   = 3 ___ 16[a2(1 +cos2q + 2cosq ) + a2sin2 q ]   = 3 ___ 16 [2a2 (1 + cos q )] = ____ 32ra = 8aC .   = o 3 3 EXAMPLE 3.34

Prove that the points on the curve r = f(q ), is the circle of curvature at which it passes through the origin (pole) are given by the equation f(q ) + f ″ (q ) = 0. Solution. Since the circle of curvature passes through the pole, the chord of curvature through the pole is r. But the chord of curvature Co is given by 2r (r2 + r12) . Co = ___________ r 2 - rr2 + 2r21

12/30/2011 6:16:51 PM

curvature n 3.21 Therefore, or

EXAMPLE 3.37

2r(r2 + r21) r = __________ 2 r - rr2 + 2r21 r 2 - rr2 + 2r21 = 2 (r 2 + r21 )

or

r 2 + rr2 = 0 or r (r + r2) = 0

and so, r + r2 = 0. Hence, f(q ) + f ″(q ) = 0. EXAMPLE 3.35

Show that in any curve the chord of curvature perpendicular to the radius vector is 2r _____ ___ 2 2 r  √r - p . Solution. The chord of curvature perpendicular to the radius vector is ________ Cp = 2r cos f = 2r  √1 - sin2 f  _______ p 2 = 2r  1 - __r , since p = r sin f 2r _____ ___ 2 2   = r  √r - p .

√ (  )

3.12

Find the curvature of the curve 2x + 2y + 5x 2y + 1 = 0. Solution. Let f (x, y) = 2x2 + 2y2 + 5x - 2y + 1. Then, fx = 4x + 5, fy = 4y - 2, fxx = 4, 2

fxy = 0.

and

_____ a

__

2 y (a + x)a - ax ______ = a 2 = _x = ___________ 2 dx (a + x) (a + x)

(  )

(  )

y   [ ] ( ) 1 + ( x ) ] [  r= =     3 __

dy 2 1 + __ dx __________ 2

1 __

1 [ (4x + 5)2 + (4y - 2)2] 2. = __ 4 The curvature is given by 4 1 ___________________ K = __  . 1 r = __ 2 2 2 [(4x + 5) + (4y - 2) ]

2r ( ___ a  )

(  )

3 __ 4 2

[ 1 + ( _yx ) ] 2r __________ ___  =     a

or

3 __

_ 4 2 __________ (in magnitude) __ 2 _y 3

a x

dx 2

fxx (fy)2 - 2fx fy fxy + fyy (fx)2

[(4x + 5)2 + (4y - 2)2] 2 =    _______________________ 4 (4y - 2)2 + 0 + 4 (4x + 5)2

2

2 y d___ ______ 2a2 = - __ 2 _y 3. 2 = 3 a x dx (a + x) Therefore, the radius of curvature r is given by

or

[ (f )2 + (f )2] 2

x y _____________________

M03_Baburam_ISBN _C03.indd 21

2

dy __

2

3 __

 

2

3

d y ___

3 __

 

2 __

(  a  ) = ( _xy ) + ( _xy ) . y ax 2r ___

Then,

2

r=

ax , For the curve y = _____ a + x show that _____

EXAMPLE 3.36

Therefore,

EXAMPLE 3.38

Solution. We have y = a + x or x = a + x .

MISCELLANEOUS EXAMPLES

fyy = 4,

Find the points on the parabola __ y2 = 4x at which the radius of curvature is 4√2 . Solution. The parametric equations of the parabola y2 = 4x are x = t 2 and y = 2t. Then, (see Example 3.10), r at (t2, 2t) is given by 1 __ 2 2 r = 2 (1 + t ) . __ But r = 4 √2 (given). Therefore, __ 1 __ 2 (1 + t 2) 2 = 4√2 or 4 (1 + t 2) = 32 or __ t = √7 . __ Hence, the required point is (t 2, 2t) = (7, 2√7 ).

2 __ 3

( _yx )

3

y 4 1_______ + _x y 2. _ _ x 2 = y 2 = y + x _ x

(  )   (  ) (  ) ( )

EXAMPLE 3.39

With the usual notation, prove that 1 + __ __ dr   2  . 1 = __ 1 ___ p2 r 2 r4 dq

(  )

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3.22 n chapter three which yield

Solution. From figure of article 3.4, we have

dy __

Therefore, __ 1 ·cosec2 f = __ 1 (1 + cot2 f) 1 = __ p2 r 2 r2 __ dr   2  , since tan f =  ____ 1 1 + __ 1 ___ rdq      = 2 2 r r dq dr 2 dr     . 1 + __ 1 ___     =__ r 2 r 4 dq

__ .   = cot q 2 dx Hence, from (2), we get ________ __ __ . ds = 1 + cot2 q __  = cosec q   2 dx 2 Similarly, from (3), we have ________ __ __ . ds = 1 + tan2q __  = sec q   2 dy 2

EXAMPLE 3.40

EXAMPLE 3.41

p = r sin f.

[

√

(  ) ] (  )

√

For the cycloid x = a (q - sin q), y = a (1 - cos ds and __ ds . q), find __ dx dy Solution. For the points P(x, y) and Q (x + δx, y + δy) in the figure of article 3.1, we have (chord PQ)2 = (δx)2 + (δy)2

or or

( chordδx PQ ) = 1 + ( δxδy ) PQ arc PQ δy = [ 1 + (  ) ] · ( chord δx arc PQ δx ) ________

__

________ ______

2

( 

chord PQ ________ arc PQ

2

__

2

or

) (  ) [ (  ) ] 2

(1)

2

δy = 1 + __ δx δx

__ δs

2

2

2

.

Letting Q → P, that is, δx → 0 , we get

( 

)

[  (  ) ]

(  )

2 chord PQ 2 lim __ δy lim _________ δs lim 1 + __ Q→P arc PQ · δx→0 δx = δx→0 δx

or or

2

(  ) [  (  ) ]

dy ds = 1 + __ 1 __ dx dx 2

_ __ __ Find the radius of curvature of √a = √r cosq   at 2 (r, q). Solution. We have  1  1 - __ - __ __ . r 2 = a 2 cos q 2 Taking log, we get __ . 1 log r = - __ 1 log a + log cos q   - __ 2 2 2 Differentiating, we get __ __ . dr  = - __ dr  = r tan q 1 tan q 1 ___   or ___   - __ 2r dq 2 2 2 dq Differentiating once more, we get d 2r __ ___ __ __ ___  = 1 rsec2 q  + tan q   dr   2 2 2 2 dq dq  __ __ __ . 1rsec2 q   =  + r tan2 q   2 2 2 Therefore,  

2

[  (  ) ] .

dy = 1 + __ dx dx

2 2

2

(2)

We shall use (3) and (4) to solve our problem. We have x = a (q - sin q) and y = a (1 - cos q ). Therefore, as in Example 3.8, we have dy ___ __ __ , q cos q dx  = 2 asin2 q   and ___ = 2a sin __   2 2 2 dq dq

M03_Baburam_ISBN _C03.indd 22

r =     2 r + 2 r21 - rr2

2

Similarly, if we divide (1) throughout by (δy)2, 1 we get __ dx 2 2 __ ds = 1 + __ . (3) dx dy

[ (  ) ]

2

 r + r tan q__2 ) ( ________________________________         = __ - r  __ r + 2r tan q ( 12 rsec q__2 + r tan q__2 ) 2

1 __

__ ds

3 __ 2  

(r + r1) 2 ___________

 

     

2

2

3 __

2

2

2

2

2

__  r 3 sec3q ____________________ 2 = __ __ __ 1 2 2 2 2 q 2 q

r + r tan

 - r sec   2 2 2

__   r 3 sec3 q _________________ 2 = __ 1 r 2 sec2 q __ - __ r 2 sec2 q   2 2 2 _ __ r √__  = 2r ___ = 2r secq √a 2  __ 3

 1 - __

= 2r 2a 2.

12/30/2011 6:16:57 PM

curvature n 3.23 EXAMPLE 3.42

__ x

y __

Find the envelope of the straight lines a + = 1, b where a2 + b2 = 4. _____ Solution. We have (using b = √ 4 - a2 ) y x ______ _____ f (x, y, a) = __ a + √4 - a2 - 1 = 0. Therefore, df a ___ x + y _______ =0 = - __ 3 __ da a2 (4 - a2) 2 or 3 __ - a2) 2 __ a3 =    (4 _______ x y or __________ 1 __ a2 +______ (4 - a2) _______ - a2) 2 √__________ a (4 2 . __ _______ = = = ______ 1 1 __ __ 2 2 2 2 __ __ __ __ x3 y3 x3 + y3 x3 + y3 Therefore 1 __ 2x 3 , _______ ______ a =  2 2 __ __ x3 + y3 x __y Putting these values of a and b in __ a + b = 1, we get 3 __

[

]

√

√

√

( x

and so,

2 __ 3

) =2

1 Further, curvature K = __ r , that is, (3 - y) (5 - y + x) . K = ______________ (3 - y)2 + (x + 2)2 EXAMPLE 3.44

Find the envelope of y cot2 α + x - α cosec2α = 0 where α is the parameter. Solution. We have y cot 2α + x - a cosec2α = 0, where α is a parameter. Differentiating w.r.t. α, we get 2y cot α (- cosec2 α) - 2a cosec α (-cosecα · cot α) = 0       ⇒ (2a - 2y) (cosec2α · cot α) = 0   ⇒ cosec2α · cot α = 0   ⇒ (1 + cot 2α). cot α = 0   ⇒ cotα + cot 3α = 0   ⇒ cot 2α = - 1. Substituting in given equations, we get y(- 1) + x - a(1 + cot 2α) = 0 ⇒ y(- 1) + x - a(1 - 1) = 0 ⇒ x - y = 0, which is the required envelope.

2 2 __

+ y3 2 __ 3

2 __ 3

EXAMPLE 3.45

Find the radius of curvature of the curve xy 2 = a 3x 3 at (a, 0).

2 __ 3

x +y =2 .

Solution. Let f (x, y) = xy 2 + x 3 - a 3.

EXAMPLE 3.43

Find the curvature of x2 + y2 + 4x - 6y - 1 = 0. Solution. We have x2 + y2 + 4x - 6y - 1 = 0 and so y2 = 1 + 6y - 4x - x2. Differentiating this equation, we get dy _____ __ (1) =x+2. dx 3 - y Further, Differentiating (1), we get - y + x. d___y 5_______ 2 = dx (3 - y)2

Then, fx = y 2 + 3x 2, fy = 2xy, fxx = 6x,

fx(a, 0) = 3a 2, fy (a, 0) = 0,

[ 1 + ( 3 - y ) ] _____ x+2

3 __ 2 2

2

Hence,

3 __

  [ ( fx)2 + ( fy)2 ] 2

  r (a, 0) =   ______________________   fxx ( fy)2 - 2 fx fy fxy + fyy ( fx)2 at (a, 0) in magnitude

3 __

[(3 - y) + (x + 2) 2] =   ________________ . (3 - y) (5 - y + x)

M03_Baburam_ISBN _C03.indd 23

fxx (a, 0) = 6a,

fyy (a, 0) = 2a, fxy (a, 0) = 0.

(1 + y21) _____________ r = ______ = (5 - y + x) y2 _________ (3 - y)2  

fxy = 2y.

Therefore,

2

Then,

fyy = 2x,

 

2 2 2

[(3a ) ] ______________ ____ 27a6 = ___ 3a . = 2 2 = 0 - 0 + 2a (3a ) 18a5 2

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3.24 n chapter three Solution. Differentiating the given equation with

EXAMPLE 3.46

Obtain the equation of the evolute of the curve x = a(cos q + q sin q ), y = a (sin q - q cos q ). Solution. Evolute is the locus of the center of curvature. Therefore, we find the center of curvature of the given curve. We have x = a (cos q + q sin q ), y = a (sin q - q cos q ). Therefore, ___ dx  = a (-sin q + sin q + q cos q ) = aq cos q, dq dy ___  = a (cos q + sin q - cos q ) = aq sin q, dq and

dy dy/dq _______  = aq sin q  y1 = __ = _____   = tan q , dx dx/dq aq cos q

(  ) [ 

]

d 2y d __ dy q  d  (ta n q) d___ y2 = ___2 = __ = ___ dx dq dx dx dx

( 

)

1    = ___ 1  sec3q. = sec2 q _______ aq cos q aq Therefore, if (X, Y ) is the center of curvature, we have y1(1 + y21) X = x - ________ = a(cosq + q sinq )  y2

 

[tanq (1 + tan2q)] aq - _________________        sec3q (sin q)aq            = a (cos q + q sin q) - __________     cos3q sec3q            = a cos q (1) and (1 + y21) Y = y + ______ y2 = a (sin q - q cos q) (1 + tan2 q) aq   + ____________        sec3 q    = a (sinq - q cosq) + aq cosq = a sinq. (2) From (1) and (2), it follows that X 2 + Y 2 = a2. Hence, the evolute of the curve, being the locus of (X, Y ), is given by x 2 + y 2 = a2. EXAMPLE 3.47

Find the circle of curvature at (0, 0) for x + y = x 2 + y 2 + x 3.

M03_Baburam_ISBN _C03.indd 24

respect to x, we have 2 dy __________ + 2x - 1 = - 1 at (0, 0) y1 = __ = 3x 1 - 2y dx

d 2y y2 = ___2 dx (1 - 2y) (6x + 2) - (3x2 + 3x - 1) (- 2y1) ________________________________   = (1 - 2y)2   = 4 at (0, 0). Therefore, the radius of curvature is 3 __ 3 __ __  1 + y21 ) 2 _______ (________ (1 + 1) 2 ____ 2 √ 2 ___ r= = = = 1__ . y2 4 4 √2 The center of curvature at (0, 0) is given by y1(1 + y21) (- 1) [1 + 1] 1 = 0 - __________ = __ X = x - ________ 4 2 y2 1 + y21 __ 2 __ 1. Y = y + _____ y2 = 0 + 4 = 2 Therefore, the required circle of curvature at (0, 0) of the given curve is (x - X)2 + ( y - Y )2 = r2 or 1 2 + y - __ 1 2 = ___ 1__ 2 x - __ 2 2 √2 or __ 1 1 2 2 (x + y - x - y) + = __ 2 2 or x 2 + y 2 - x - y = 0.

( 

) ( 

) (  )

EXAMPLE 3.48

Find the equation of the circle of curvature of the curve x = a (cos q + q sin q ), y = a (sin q - q cos q ). Solution. Differentiating the given equation with

respect to q, we have ___ dx

 = a (- sin q + q cos q + sin q) = aq cos q dq

and dy ___  = a (cos q + q sin q - cos q) = aq sin q. dq Therefore, dy dy/dq aq sin q  y1 = __ = _____  = _______  = tan q dx dx/dq   aq cos q

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curvature n 3.25 and

( 

d 2y dy q = sec2q    _______   1    y2 = ___2 = ___1   · d___ dq dx dx aq cos q

)

1  sec3 q.  = ___ aq The center of curvature (X, Y ) is given by  

y1(1 + y21) X = x - ________ y 2

__________ q sec  q   = a (cos q + q sin q) - tan  (aq) sec3 q          = a (cos q + q sin q ) - aq sin q = a cos q, 2

 

y1(1 + y21) Y = y + ________ y    = a (sin q - q cos q ) + aq cos q    = a sin q. Hence, the equation of the circle of curvature is

or

(x - X )2 + (y - Y )2 = r2 (x - a cos q )2 + (y - a sin q )2 = a2q 2 . x 2 + y 2 + a 2 - 2a (x cos q + y sin q ) = a2q 2.

EXAMPLE 3.49

__ __ For the curve √x + √y = 1, find the equation of 1 , __ 1 . the circle of curvature at __ 4 4 __ __ Solution. The given curve is √ x + √ y = 1. From Example 3.5 (ii), we have 1 1 __ dy - __ y1 = __ = - y 2 x 2, dx

(  )

1 , and y2 = ___ 3 __ 2x 2 3 __

r(x, y) = 2(x + y)2. Therefore,

__ 1 2 __ __ 1 __ 1 y1 , = - = - 1, 1 __ 4 4

(  ) 2 __ _____ 1 __ 1 y (  , ) = 1 = 4 , and 4 4 __ 1

( 4 ) 1, __ 1__ . 1 = 2  __ r ( __ ( 41 + __41 ) = ___ 4 4) √2 2

 

M03_Baburam_ISBN _C03.indd 25

Therefore, the equation of the circle of curvature is (x - X )2 + (y - Y )2 = r2 or 3 2 + y - __ 3 2 = __ 1. x - __ 4 4 2

( 

) ( 

)

EXAMPLE 3.50

2

or

If(X, Y ) denotes the center of curvature, then y1 (1 + y21) __ -1 (1 + 1) 3 , = 1 -  ________ = __ X = x - _________ y2 4 4 4 2 1 + y 1 __ 3 1 __ 2 __ Y = y + _____ y2 = 4 + 4 = 4.

2

3 __ 2

3 __ 2

x __y Find the equation of the envelope of __ a + b = 1, where the parameters a and b are connected by the relation a2 + b2 = c2 and c is a constant. Solution. Assume that a and b are functions of x + __y = 1 and a2 + parameter t. Then differentiating __ a b b2 = c2 with respect to t, taking x and y as constants, we have, respectively, y ___ __ x ___ da + __ db = 0 and so, a2 dt b2 dt y __ da ___ dt b2 ___ = - __ (1) x __ db ___ 2 a dt da + b ___ db = 0 and so, a ___ dt dt da ___ dt ___ b = - __ (2) a. db ___ dt From (1) and (2), we get y x __ __ b2 __ a2 __ __ __ a. b y =b x = a or __ __ a2 b2 Therefore, x __y __x + __y __ a __ b a b __ __ 1, = 2 = _____ 2 = 2 2 a b a2 + b c

which yields a3 = c2x

and

b3 = c2y.

Putting these values in a2 + b2 = c2, we get the 2 __

2 __

2 __

envelope x 3 + y 3 = c 3 .

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3.26 n chapter three EXAMPLE 3.51

y Find the envelope of _x + __ = 1 where l and m are l m __l __ m connected by a + = 1 and a, b are constants. b Solution. We assume that l and m are functions of y l a parameter t. Differentiating _x + __ = 1 and __ a m l m = 1 with respect to t taking x and y as con+ __ b stants, we have y ___ __ dl + ___ dm = 0 x __ 2 l dt m2 dt and __ dm dl __ 1 __ 1 ___ a dt + b dt = 0 y or ___ dl __ dt m2 ___ ___ (1) = - __ x dm ___ 2 l dt and dl __ dt ___ a = - __ (2) b dm ___ dt From (1) and (2), we have y x ___ __ m2 __ l2 __ ___ ___ b. a y =a x = b or ___ __ l2 m2 Therefore, y x __ _ y _x + __ m _____ l m __ __l __ = 1 = 1. =m= m 1 __l __ __l __ a b a+ b Hence, __ ax = 1 , which yields l = √___ ax l2 and ___ by ___ 2 = 1, which yields m = √ by . m l __ m Putting these values of l and m in __ a + b = 1, __ __ we get __ x + __y = 1. √a b EXAMPLE 3.52

√

x __y m n Find the envelope of __ a + b = 1, where a b = m+n c . Solution. Assume that a and b are functions of a x + __y = 1, and parameter t. Then differentiating __ a b ambn = cm+n with respect to t, treating x and y as constants, we have respectively

M03_Baburam_ISBN _C03.indd 26

__ x ___ da

a2 dt

+

y ___ __ db b2 dt

=0

or

dt

and or

y __ da ___ dt b2 . __ ___ = - __ x db ___

da + nambn - 1 ___ db = 0 mbnan - 1 ___ dt dt

(1)

a

2

or

da ___ dt ___ nambn - 1 . = - _______ mbnam - 1 db ___

(2)

dt From (1) and (2), we get y x __ __ m n-1 n m-1 b2 _______ a2 mb __ __ _______ b a . (3) na x = mbnam - 1 or __ x = nambn - 1 __ a2 b2 Relation (3) yields x __y __x + __y __ a b _____ a __ b _____ __ 1 m = n = m+n = m + n. These relations yield m + n y. m + n x and b = _____ a = _____ m n Putting the values of a and b in ambn = cm + n, we get or

m + n m _____ m + n n m+n xmyn ( _____ m ) (  n ) = c , c m + n. xmyn = mnnn( _____ m + n)

EXERCISES 1. Find the radius of curvature at any point (x, y) of the following curves: 3 __ __ √ x (4a + 9x) 2 ___________ 2 3 . (i) ay = x Ans. (i) 6a x2 + y2. Ans. (ii) ______ (ii) xy = c 2 2c2 2. Find the radius of curvature at a given point of the following curves: (i) y = e x at the point where it cross the y-axis. Hint: The point is x = 0 and y = e0 = 1.

(

x _

x -_

)

(ii) y = __c  e c + e c at (x, y). 2 Hint: The curve is y = c cosh _cx.

12/30/2011 6:17:06 PM

curvature n 3.27 __ __ a a + √y = √a at __, __ ). 4 4 a 2 (a - x) (iv) y 2 = ________ at (a, 0). x __ y2 ___ __ a__ a. Ans. (i) 2√2 (ii) __ c (iii) √2 (iv) 2 (iii)

__

√x

(

9. Find the radius of curvature of the point (p, r) of the following curves: (i) r3 = 2ap2. r . (ii) p2 =  _________ r2 + a2 4

2 __

2 __

3 __

___ (r2 + a2) 2 . Ans. (i) √2ar (ii) ________ 3 r2 + 2a2

3. Show that the radius of curvature at any

__ 2

2 __

point of the astroid x 3 + y 3 = a 3 is three times the length of the perpendicular from the origin to the tangent at that point.

10. Find the radius of curvature at any point of the curve r n = an cos nq.

Hint: If parametric equations are x = a cos2 t and y = a sin3 t, then r = 3 a sin t cos t, tangent is dy y - a sin3 t = __ (x - a cos3t), and length of dx perpendicular from (0, 0) on tangent is sin t cos t.

Hint: r1 = - r tan nq and r2 = r tan2q - nr (r2 + r21) , we get sec2 nq. Putting in r = ___________ r 2 + 2r21 - rr2 an r = _________ . (n + 1)rn - 1

4. If r is the radius of curvature of the parabola y 2 = 4 ax at a point P, show that r 2 varies as (SP)3, where S is the focus.

11. Find the radius of curvature at (r, q ) of the __  . curve r = 6 1 - sin2 q 2 __ .   Ans. r = 4 cos q 2

3 __

Hint: r at (at2, 2at) is 2a(1 + t2) 2, S(a, 0) is the focus and so, _________________ SP = √(at2 - a)2 + (2at - 0)2 = a (1 + t2). 4 3 Then r 2 = __ a (SP) . 5. Show that for the curve x = a cos q (1 + sin q ) and y = a sin q (1 + cos q ), the radius of __ curvature at q = - r  is a. 4 6. Prove that for the curve r = a (sec3 y - 1) the radius of curvature is r = 3 a tan y sec3 d 2y dy ___ y and hence, show that 3a __ · 2 = 1. dx dx 7. Prove that for the curve 4 x = q + sin q and 4y = 1 - cos q, the radius of curvature is r q = cos __  . 2 8. Find the radius of curvature at the point q of the curve x = a (cos q + log tan q__ ) and 2 y = a sin q. dy dy a cos2  q  dx  = _______ Hint: ___  , ___ = a cos q, __ = tan q, dx dq sin q dq d 2y sin q   . and ___2 = _______ dx a cos4 q Ans. a cos q.

M03_Baburam_ISBN _C03.indd 27

(

)

 

12. Show that the radius of curvature at the point (r, q) on the curve 1 1 a is r = (r2 - a2) __2. 1(r2 - a2) __2 - cos - 1  __ q = __ ( ) a r 1 __

2 - a2) 2 q =    (r ar . dr  = _______ ________ ______ Hint: d___ and so, ___ ar dr dq √r2 - a2

(  )

1 and so pedal 1 = __ 1 + __ 1 ___ dr   2  = ______ Also, __ 2 p2 r2 r4 dq r - a2 equation is p2 = r2 - a2. Differentiating, we ______ dr = p = √r2 - a2 . dr . Then r = r ___ get 2p = 2r ___ dp dp 13. Show that the curvatures of the curves r = aq and rq = a at their common point are in the ratio 3:1. Hint: Eliminating r between the two equations, we get q 2 = 1. Thus, at the common point q 2 = 1, for the first curve r__1 = a and 2a√2 . For the r2 = 0 and so, r1 (at q 2= 1) = _____ 3 a and r = ___ 2a and so, second curve r1 = - ___ 2 2 q 3 __q  2 r2 (at q  = 1) = 2a√2 .

12/30/2011 6:17:08 PM

3.28 n chapter three 2 sin2 f _____ sin f   14. Prove that for any curve d___r2 = _____ - r    r   ds dr = cos f, r = ___ ds  , f = y - q, and Hint: __ ds dy q . sin f = r d___ ds

15. Find the radius of curvature for the curve ______ x = c log {s + √s2 + c2 } and ______ y = √s2 + c2 . dy _____ d2y __ dx = - _______ c s ___ ______ and = Hint: __ = ds ds s2 + c2 ds2 √ s2 + c 2 dx __ 2 dy ________ ds y . c2 __ ___ d __ and so, r = 2 = __ = 3 c __ ds ds d y ___ (s2 + c2) 2 ds2 16. Find the radius of curvature at the origin of the following curves:

(  )

(i) x - y - 2x + 6y = 0. 3

3

2

(ii) x - 2x y + 3xy - 4y + 5x - 6xy + 7y2 - 8y = 0. 3 4. Ans. (i) r (0, 0) = __ (ii) r (0, 0) = __ 5 2 3

2

2

3

2

17. Find the radius of curvature at the pole for the curve r = a sin nq. na. dr  = ___ 1 ___ Ans. r = __ 2 dq 2 18. Find the radius of curvature at the origin for the curve a (y 2 - x 2) = x 3. ___ Ans. 2 √2a .

M03_Baburam_ISBN _C03.indd 28

19. Find the coordinates of the center of curvature for the curve a 2y = x 3.

(  ( 

) ( 

))

9x4 , ___ 5x3 + __ a2 . Ans. __x 1 - ___ 4 3 6x 2 a 2a 20. Show that the evolute of the tractrix x = c cos t + c log tan __t and y = c sin t is the catenary 2 y = c cosh _cx. 21. Find the circle of curvature for the curve 3, __ 3 . x3 + y3 = 3xy at the point __ 2 2 Ans. 8 (x2 + y2) - 21 (x + y) + 27 = 0.

(  )

22. Find the length of the chord of curvature through the pole for the curves r = aemq. Ans. 2r. 23. Find the coordinates of the center of curvature at (at 2, 2at) on the parabola y 2 = 4ax. Ans. (2a + 3at 2, - 2at 3). x 24. Show that in the curve y = a cosh __ a , the chord of curvature parallel to the axis of x is 2x . of length a sinh __ a 1 cosh __x . x and y = __ Hint: y1 = sinh __ a a a 2

(  )

(  )

2y (1 + y21 ) 2x.   1 y = a sinh __ Therefore, Cx =   ___________ a 2

25. If Cx and Cy are the chords of curvature parallel to the axis of x and y, respectively, at x any point of the curve y = a cosh __ a, show that 4a2(C2x + C2y ) = C 4y .

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4

Asymptotes and Curve Tracing

The aim of this chapter is to study the shape of a plane curve y=f(x). For this purpose, we must investigate the variation of the function f, in the case of unlimited increase and absolute value and of x or y, or both, of a variable point (x, y) on the curve. The study of such variation of the function requires the concept of an asymptote. Before defining an asymptote to a curve, let us define finite and infinite branches of a plane curve as follows: Consider the equation of the ellipse

x2 a2

+

y2 b2

= 1.

Solving this equation, we get y = b 1−

x2 x2 or y = − b 1 − . a2 a2

curve is said to tend to infinity along the curve if either x or y, or both, tend to infinity as P(x, y) moves along the branch of the curve. Now we are in a position to define an asymptote to a curve. A straight line, at a finite distance from the origin, is said to be a rectilinear asymptote (or simply asymptote) of an infinite branch of a curve if the perpendicular distance of a point P on that branch from this straight line tends to zero as P tends to infinity along the branch of the curve. For example, the line AB will be asymptote of the curve in the following figure if the perpendicular distance PM from the point P to the line AB tends to zero as P tends to infinity along the curve. Y

The first equation represents the upper half of the ellipse while the second equation represents the lower half of the ellipse. Thus, the earlier equation represents two branches of the ellipse. Further, both these branches lie within the finite part of the xy-plane bounded by x = + a and y = + b. Hence, both these branches of the ellipse are finite. Consider now the equation of the hyperbola y2 x2 2 − 2 = 1 . Its solution is a b y=

b b x 2 − a 2 or y = − x2 − a2 . a a

Therefore, y tends to ± ∞ as x → ± ∞ . Hence, both branches this hyperbola extend to infinity and are therefore called the infinite branches of the rectangular hyperbola. A variable point P(x, y) moves along a curve to infinity if the distance of the point from the origin increases without bound. In other words, a point P(x, y) on an infinite branch of a

M04_Baburam_ISBN _C04.indd 1

B

M

0

4.1

A

P

X

DETERMINATION OF ASYMPTOTES WHEN THE EQUATION OF THE CURVE IN CARTESIAN FORM IS GIVEN

Let

y = mx + c (1) be the equation of a straight line. Let P (x, y) be an arbitrary point on the infinite branch of the

12/30/2011 4:15:24 PM

4.2 n chapter four curve f(x, y) = 0. We wish to find the values of m and c so that (1) is an asymptote to the curve. Let PM = p be the perpendicular distance of the point P (x, y) from (1). Then p=

y − mx − c 1+ m

2

.

The abscissa x must tend to infinity as the point P (x, y) recedes to infinity along this line. Thus, p → 0 as x → ∞. Therefore, lim ( y − mx − c) = 0

X →∞

or

lim ( y − mx) = c.

 +…+ xφ1  

() y

or lim

y = m. x

Solving the equation (2), we get the slope m of the asymptote y = mx + c. But lim( y − mx) = c x →∞ . Let y – mx = p so that x → ∞, p → c . But y p y–mx = p implies x = m + x . Substituting this in equation (1), we have p p   x φ n  m +  + x n −1φ n −1  m +    x x

Thus, to find asymptotes which are not parallel y to the y-axis, we fin lim x and lim( y − mx). If x →∞ x →∞ these limits are, respectively, m and c, then y = mx + c is an asymptote. THE ASYMPTOTES OF THE GENERAL RATIONAL ALGEBRAIC CURVE

Let f (x, y) = 0 be the equation of any rational algebraic curve of the nth degree. Arranging this equation in groups of homogeneous terms in x and y, we get (a0 x n + a1 x n −1 y + a2 x n − 2 y 2 + …+ an y n ) +…+ (b1 x

+ b2 x

+ c3 x

n −3

n−2

y + …+ bn x

y + …+ cn y

+…+ (k0 x + k1 y ) + K = 0.

M04_Baburam_ISBN _C04.indd 2

y x

n

y and c = lim ( y − mx). X →∞ x

n −1

(2)

φn (m) = 0.

value of

Hence,

+ (c2 x

x →∞

which yields

1 y  lim  − m = lim ( y − mx) lim = c(0) = 0 X →∞  x X →∞ x  X →∞

n−2

y

  y 1 1   y  y lim φ n   + φ n −1   + 2 φ n − 2   + … = 0, x →∞  x x  x   x x 

Therefore,

X →∞

(1)

where φr x is a polynomial in x of degree r. Suppose y = mx + c as an asymptote of the curve, where m and c are finite. We have to find m and c. Dividing both sides of equation (1) by xn, we get  y 1  y 1  y φn   + φn −1   + 2 φn − 2   + … = 0.  x x  x x  x we have

y 1 − m = ( y − mx) . x x

X →∞

y  = 0, x

y

On the other hand,

m = lim

y   + φ0  x

Proceeding to limits as x → ∞ so that lim x = m,

X →∞

4.2

This equation can be written as  y  y  y x nφ n   + x n −1φ n −1   + x n − 2φn − 2    x  x  x

n−2

)

n −1

)

p  + x n − 2φn − 2  m +  + … = 0.  x

(3)

Taylor’s Theorem expansion of equation (3) yields   p p2 x n φn (m) + φn′ (m) + 2 φn′′(m) + … x 2x   p   + x n −1 φn −1 (m) + φn′ −1 (m) + … x   p   + x n − 2 φ n − 2 (m) + φ n′ − 2 (m) + … + … = 0. x   Using equation (2), the said equation reduces to  p2 x n −1 [ pφn′ (m) + φn −1 (m) ] + x n − 2  φn′′(m)  2!  + pφn′′−1 (m) + φn − 2 (m)  + … = 0. 

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4.3 aSyMptoteS and curve tracing n  (ii) Asymptotes parallel to y-axis cannot be found by the said method as the equation of a straight line parallel to y-axis cannot be put in the form y = mx + c.

or 1  p2 pφn′ (m) + φn −1 (m) +  φn' (m) + pφn′−1 (m) x  2!  + φn − 2 (m)  +… = 0. 

(4)

Since x → ∞, p → c, we have cφ n′ (m) + φ n −1 (m) = 0.

(5)

Case (i): If fn (m) has no repeated root, then f′n (m) ≠ 0. Hence, in that case, equation (5) implies c=−

φn −1 (m) . φn′ (m)

ASYMPTOTES PARALLEL TO COORDINATE AXES

(i) Asymptotes parallel to y-axis of a rational algebraic curve: Let f(x, y) = 0 be the equation of any algebraic curve of the mth degree. Arranging the equation in descending powers of y, we get y m φ0 ( x) + y m −1φ1 ( x) + y m − 2φ 2 ( x)

(6)

If m1, m2, m3,… are the distinct roots of fn (m) = 0 and c1, c2, c3,… are the corresponding values of c determined by equation (6), then the asymptotes are y = m1x + c1, y = m2x + c2, y = m3x + c3,… Case (ii): If f′n (m) = 0, that is, f′n (m) has a repeated root and if f′n–1 (m) ≠ 0, then equation (6) implies that c is undefined. Hence, there exists no asymptote to the curve in this case. Case (iii): If f′n (m) = fn-1 (m) = 0. Then equation (5) reduces to an identity and equation (4) reduces to 2

p 1 φn′′(m) + pφn′ −1 (m) + φn − 2 (m) + […] + … = 0. 2! x As x → ∞, p → c we have c2 φn′′(m) + cφn′ −1 (m) + φn − 2 (m) = 0. 2 If f″n (m) ≠ 0, then this last quadratic in c gives two values of c. Therefore, there are two asymptotes y = mx + c1 and y = mx + c2, corresponding to the slope m. Thus, in this case, we have two parallel asymptotes. Remark 4.1 (i) Since the degree of fn(m) = 0 is n at the most, the number of asymptotes, real or imaginary, which are not parallel to y-axis, cannot exceed n. In case the curve has asymptotes parallel to y-axis, then the degree of fn(m) is smaller than n by at least the number of asymptotes parallel to y-axis. Thus, the total number of asymptotes cannot exceed the degree n of the curve.

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4.3

+…+ φm ( x) = 0,

(1)

where f0(x), f1(x), f2(x),… are polynomials in x. Dividing the equation (1) by ym, we get 1 y

φ0 ( x) + φ1 ( x) +

1 φ2 ( x) y2

+…+

1 φm ( x) = 0. ym

(2)

If x = c be an asymptote of the curve parallel to y-axis then lim x = c, where (x, y) lies on the y →∞

curve (1). Therefore,   1 1 lim φ0 ( x) + φ1 ( x) + 2 φ2 ( x) + … = 0 y →∞ y y   or f0(c) = 0 so that c is a root of the equation f0(x) = 0. If c1,c2,… are the roots of f0(x) = 0, then (x – c1), (x – c2),… are the factors of f0(x). Also f0(x) is the coefficient of the highest power of y, that is, of ym in equation (1). Thus, we have the following simple rule to determine the asymptotes parallel to y-axis. The asymptotes parallel to the y-axis are obtained by equating to zero the coefficient of the highest power of y in the given equation of the curve. In case the coefficient of the highest power of y is a constant or if its linear factors are imaginary, then there will be no asymptotes parallel to the y-axis. (ii) Asymploles parallel to the x-axis of a rational algebraic curve: Proceeding exactly as in case (i) mentioned earlier, we arrive at

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4.4 n chapter four the following rule to determine the asymptotes parallel to the x-axis: The asymptotes parallel to the x-axis are obtained by equating to zero the coefficient of the highest power of x in the given equation of the curve. In case the coefficient of the highest power of x is a constant or if its linear factors are imaginary, then there will be no asymptotes parallel to the x-axis. 4.4

WORKING RULE FOR FINDING ASYMPTOTES OF RATIONAL ALGEBRAIC CURVE In view of the mentioned discussion, we arrive at the following working rule for finding the asymptotes of rational algebraic curves: 1. A curve of degree n may have atmost n asymptotes. 2. The asymptotes parallel to the y-axis are obtained by equating to zero the coefficient of the highest power of y in the given equation of the curve. In case the coefficient of the highest power of y is a constant or if its linear factors are imaginary, then there will be no asymptotes parallel to the y-axis. The asymptotes parallel to the x-axis are obtained by equating to zero the coefficient of the highest power of x in the given equation of the curve. In case the coefficient of the highest power of x is a constant or if its linear factors are imaginary, then there will be no asymptotes parallel to the x-axis. If y = mx + c is an asymptote not parallel to the y-axis, then the values of m and c are found as follows: (i) Find fn(m) by putting x = 1, y = m in the highest-degree terms of the given equation of the curve. Solve the equation f0(m) = 0 for slope (m). If some values are imaginary, reject them. (ii) Find fn-1(m) by putting x = 1, y = m in the next lower-degree terms of the equation of the curve. Similarly fn-2(m) may be found taking x = 1, y = m in the next lower-degree terms in the curve and so on. (iii) If m1, m2,… are the real roots of fn(m), then the corresponding values of c, that is, c1,c2,… are given by φ ( m) c = − n −1 , m = m1 , m2 , … φn′ (m)

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Then the required asymptotes are y = m1x + c1, y = m2x+c2, … (iv) If f′n(m) = 0 for some m but fn-1 (m) ≠ 0, then there will be no asymptote corresponding to that value of m. (v) If f′n (m) = 0 and fn-1 (m) = 0 for some value of m, then the value of c is determined from c2 c φn′ (m) + φn′ −1 (m) + φn − 2 (m) = 0. 2! 1! This equation will yield two values of c and thus, we will get atmost two parallel asymptotes corresponding to this value of m, provided f″n (m) ≠ 0. (vi) Similarly, if f″n (m) = f′n-1 (m) = fn-2 (m)=0, then the value of c is determined from c3 c2 c φn′′(m) + φn′′−1 (m) + φn′ − 2 (m) 3! 2! 1! +φn − 3 (m) = 0. In this case, we get atmost three parallel asymptotes corresponding to this value of m. EXAMPLE 4.1

Find the asymptotes of the curve y2 (x2 – a2) = x2(x2 – 4a2). Solution. The equation of the curve is y2 (x2 – a2) = x2(x2 – 4a2) or y2 x2 – x4 – a2y2 + 4a2x2 = 0. Since the degree of the curve is 4, it cannot have more than four asymptotes. Equating to zero, the coefficient of the highest power of y, the asymptote parallel to the y-axis is given by x2 – a2 = 0. Thus, the asymptotes parallel to the y-axis are x = + a. Since the coefficient of the highest power of x in the given equation is constant, there is no asymptote parallel to the x-axis. To find the oblique asymptotes, we put x = 1 and y = m in the highest-degree term, that is fourth- degree term y2x2 – x4 in the given equation and get f4(m) = m2–1. Therefore, slopes of the asymptotes are given by f4(m) = m2 – 1 = 0. Hence, m = +1. Again putting y = m and x = 1 in the next highest-degree term, that is,

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aSyMptoteS and curve tracing n  4.5 third-degree term, we have f3(m) = 0 (since there is no term of degree 3). Now c is given by c=−

φ3 ( m) 0 = = 0. φ4′ (m) 2m

Therefore, the oblique asymptotes are y = x + 0 and y = –x + 0. Hence, all the four asymptotes of the given curve are x = +a and y = + x. EXAMPLE 4.2

Find all the asymptotes of the curve f(x, y) = y3 – xy2 – x2y + x3 + x2 – y2 –1 = 0. Solution. The given curve is of degree 3 and so, it may have atmost three asymptotes. Since the coefficients of the highest power of x and y are constants, the curve has no asymptote parallel to the coordinate axes. To find the oblique asymptotes, we put x = 1 and y = m in the expression containing thirddegree terms of f (x, y). Thereby we get f3(m) = m3 – m2– m+1 = 0. This equation yields m = 1, 1, –1. Further, putting x = 1, y = m in the next highest-degree term, we get f2(m) = 1–m2. Therefore for m = –1, the expression 1 − m2 c=− 2 3m − 2m − 1 yields c = 0 and the corresponding asymptote is y = –x + 0 or y + x = 0. For m = 1, the denominator is zero and so, c cannot be determined by the preceding formula. Putting x = 1, y = m in the first-degree terms, we have f1(m) = 0 (since there is no first-degree term). Now for m = 1, the constant c is given by c2 φ2′′(m) + cφ2′ (m) + φ 1 (m) = 0 2 or (3m–1)c2 – 2mc = 0 or 2c2 – 2c = 0 for m = 1 or c(c – 1) = 0. Hence, c = 0 and c = 1. So the two parallel asymptotes corresponding to m = 1 are y = x and y = x + 1. Therefore, the asymptotes to the curve are y + x = 0, y = x and y = x + 1.

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EXAMPLE 4.3

Find the asymptotes of the curve y2 (x – 2a) = x3 – a3. Solution. The degree of the curve is 3. So, there cannot be more than three asymptotes. There is no asymptote parallel to the x-axis. The asymptote parallel to the y-axis is given by x–2a = 0, that is, x = 2a. To find the oblique asymptotes, we put x = 1, y = m in the third-degree term and get f3(m) = m2–1 and so, the slope m is given by f3(m) = m2–1 = 0. Thus, m = +1. Further, putting x = 1, y = m in the second-degree terms, we get f2(m) = –2am2. Therefore for m = 1 and m = –1, the expression φ (m) 2am 2 = = am c=− 2 φ3 ( m) 2m yields c = a, and –a respectively. Hence, the oblique asymptotes are y = x + a and y = –x–a. Hence, the three asymptotes of the curve are x = 2a, x–y + a = 0, and x + y + a = 0. EXAMPLE 4.4

Find the asymptotes of the curve x3 + 3x2y – 4y3 – x + y + 3 = 0. Solution. There is no asymptote parallel to the coordinate axes. To find the oblique asymptotes, we have f3(m) = 1 + 3m –4m3 and so, the slope m is given by f3(m) = 1 + 3m –4m3 = 0. Therefore, m = 1, − 12 , and − 12 . For m = 1, the value of c is given by φ ( m) 0 c=− 2 =− = 0. φ3′′(m) −12m 2 + 3 Thus, the asymptote corresponding to m = 1 is y = x or x – y = 0. For m = − 12 , φ3′ (m) = 0. So we find f1(m), which is equal to f1(m) = –1 + m. Hence, c is given by, c 2φ3′′(m) + cφ2′ (m) + φ 1 (m) = 0 2 or 6c 2 −

3 1 1 = 0 or c 2 = or c = ± . 2 4 2

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4.6 n chapter four Thus, the asymptotes corresponding to m = − 12 are 1 1 1 1 y = − x + and y = − x − 2 2 2 2 or x + 2y – 1 = 0 and x + 2y + 1 = 0. Hence, the three asymptotes of the curve are x – y = 0, x + 2y –1 = 0, and x + 2y + 1 = 0. EXAMPLE 4.5

Find the asymptotes of the curve (x – y)2 (x2 + y2) – 10 (x – y) x2 + 12y2 + 2x + y = 0. Solution. The equation of the given curve is (x – y)2 (x2 + y2) – 10 (x – y) x2 + 12y2 + 2x + y = 0. The coefficient of x4 and y4 are constant. Therefore, the curve has no asymptotes parallel to the axes. Putting x = 1 and y = m in the fourth-, third- and second-degree terms, we have f4(m) = (1 – m)2(1 + m2) = m4 –2m3 + 2m2 –2m+1 f3(m) = 10 (m–1), and f2(m) = 12m2. The slopes of the asymptotes are given by f4(m) = (1 – m)2(1 + m2) = 0. Therefore, m = 1, 1 are the real roots. Further we have f ′n (m) = 4m3 –6m2 + 4m – 2, so that f′4 (m) = 0 for m = 1. Therefore, values of c are given by c2 φ4′′(m) + cφ3′ (m) + φ2 (m) = 0, 2 that is,

To find the oblique asymptotes, we put x = 1 and y = m in third-, second- and first-degree terms and get φ3 (m) = (1 + m) 2 (1 + 2m) = 2m3 + 5m 2 + 4m + 1 φ2 (m) = 2(1 + m) 2 = 2m 2 + 4m + 1

φ1 (m) = −(1 + 9m) = −9m − 1. Thus,

φ3′′(3) = 12m + 10, and φ2′ (m) = 4m + 4.

The slopes of the asymptotes are given by φ3 (m) = (1 + m) 2 (1 + 2m) = 0, which yields m = –1,–1, and − 12 . The value of c is given by c=−

EXAMPLE 4.6

Find the asymptotes of the curve (x + y)2 (x + 2y) + 2(x + y)2 – (x + 9y) – 2 = 0. Solution. Since the coefficients of the highestdegree term of x and y are constant, the given curve does not have asymptotes parallel to the axes.

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φ 2 ( m) 2m 2 + 4m + 1 =− . φ3′ (m) 6m 2 + 10m + 4

For m = –1, f′3(m) = 0 and so, c cannot be found from this equation. For m = − 12 we have c = –1. Thus, the asymptotes corresponding to m = − 12 is 1 x − 1 or x + 2 y + 2 = 0. 2 For m = –1, the value of c is calculated from the relation y=−

c2 φ3′′(m) + cφ2′ (m) + φ1 (m) = 0 2 or

2

c (12m 2 − 12m + 4) + 10c + 12m 2 = 0. 2 For m = 1, this equation yields 2c2 + 10c +12 = 0 or c2 + 5c + 6 = 0 This equation gives c = –2, –3. Putting the values of m and c in y = mx + c, the asymptotes are given by y = x – 2 and y = x – 3.

φ3′ (m) = 6m 2 + 10m + 4,

c2 (12m + 10) + c (4m + 4) − 9m − 1 = 0 2 or or or

c2 (6m + 5) + c (4m + 4) –9m –1 = 0 c2 (–1) + 9 – 1 = 0

c2 = 8, which yields c = ±2 2. Thus, the two parallel asymptotes corresponding to the slope m = –1 are y = − x + 2 2 and y = − x − 2 2. Hence, the asymptotes of the curve are x + 2y + 2 = 0, y + x = 2 2, and y + x = −2 2.

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4.7 aSyMptoteS and curve tracing n  EXAMPLE 4.7

Find the asymptotes of the curve 6x2 + xy – 2y2 + x + 2y + 1 = 0 Solution. Since the coefficients of the highest powers of x and y are constants, there is no asymptotes parallel to the axes. To find the oblique asymptotes, we put x = 1 and y = m in second- and first-degree terms and ge f2(m) = 6 – 2m2 + m, f1(m) = 2m +1 and f ′2(m) = – 4m + 1. The slopes of the asymptotes are given by f2(m) = 6 – 2m2 + m = 0 and so, m = 2, − . The value of c is given by φ ( m) 2m + 1 =− c=− 1 . −4m + 1 φ2′ (m) 3 2

For m = 2 and m = − 32 , the value of c are 75 and respectively. Therefore, the asymptotes are 5 3 2 y = 2 x + and y = − x + 7 2 7 or 14x – 7y + 5 = 0 and 21x + 14y – 4 = 0

2 7

EXAMPLE 4.8 a2 x2

b2 y2

Find the asymptotes of the curve − = 1. Solution. The equation of the given curve is a 2 b2 − =1 x2 y 2 or

x2y2 – a2y2 + b2x2 = 0. Since the curve is of degree 4, it cannot have more than four asymptotes. Equating the coefficient of the highest power of x to zero, we get y2 + b2 = 0, which yields imaginary asymptotes. Equating the coefficient of the highest power of y to zero, we get x2 – a2 = 0 or (x – a)(x + b) = 0. Hence, the asymptotes parallel to the y-axis are x = a and x = –a. Thus, the only real asymptotes are x – a = 0 and x + a = 0. 4.5

INTERSECTION OF A CURVE AND ITS ASYMPTOTES We have seen that the equation of a curve of degree n can be expressed in the form

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 y  y x n φn   + x n −1φn −1    x  x  y + x n − 2φ n − 2   + … = 0.  x

(1)

Let

y = mx + c (2) be an asymptote to the curve (1). Eliminating y from (1) and (2), we get c c   x n φn  m +  + x n −1φn −1  m +     x x  + x n − 2φ n − 2  m + 

c  + … = 0. x

Expanding by Taylor’s Theorem, we get   c c2 x n φn (m) + φn′ (m) + φ ′ (m) + … 2 n x 2! x   c  + x n −1 φn −1 (m) + φ n′ −1 (m) x 

+

 c2 φ ′′ (m) + … 2 n−2 2! x 

c  + x n − 2 φ n − 2 (m) + φ n′ − 2 (m) x 

+

 c2 φn′′− 2 (m) + … = 0, 2! x 2 

(3)

that is, x n φ n (m) + x n −1 [φ n′ (m) + φ n −1 (m)]  c2  + x n − 2  φn′′(m) + cφn′ −1 (m) + φn − 2 (m)  + … = 0. 2!   But equation (2) being an asymptote of equation (1), the values of m and c are given by φn (m) = 0 and cφn′ (m) + φn −1 (m) = 0. Hence, equation (3) reduces to  c2  x n − 2  φn′′(m) + cφn′ −1 (m) + φn − 2 (m) + … = 0,  2!  which is of degree n – 2 and so, yields (n – 2) values of x. Hence, the asymptote (2) cuts the curve (1) in (n – 2) points. If the curve has n asymptotes, then they all will intersect the curve in n(n – 2) points.

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4.8 n chapter four Further, if the equation of the curve of the nth degree can be put in the form Fn + Fn–2 = 0, where Fn–2 is of degree n – 2 at the most and Fn consists of n distinct linear factors, then the n(n – 2) points of intersection of the curve Fn + Fn–2 = 0 and its n asymptotes (given by Fn = 0) lie on the curve Fn–2 = 0. EXAMPLE 4.9

Find the asymptotes of the curve x2y – xy2 + xy + y2 + x – y = 0 and show that they cut the curve in three points that lie on the straight line x + y = 0. Solution. Equating to zero the coefficient of highest power of x, we get y = 0. Thus, x-axis is an asymptote to the given curve. Similarly, equating to zero the coefficien of the highest power of y, we get – x+ 1 = 0 or x = 1. Thus, x = 1 is the asymptote parallel to y-axis. To find the oblique asymptotes, we put x = 1 and y = m in the third- and second- degree terms and get

f3(m) = m – m , f2(m) = m + m , and f′3(m) = 1 – 2m. Then the slopes of the asymptotes are given by f3(m) = m – m2 = 0, which implies m = 0 and m = 1. The values of c are given by φ ( m) m + m2 =− c=− 2 . φ3′ (m) 1 − 2m 2

2

Thus, the values of c corresponding to m = 0 and m = 1 are c = 0 and c = 2, respectively. Therefore, the oblique asymptotes are y = 0 and y = x + 2. Hence, the asymptotes of the curve are y = 0, x = 1, and x – y + 2 = 0. The joint equation of the asymptotes is (x – 1) y (x – y + 2) = 0 or x2y – xy2 + xy + y2 – 2y = 0. On the other hand, the equation of the curve can be written as (x2y – xy2 + xy + y2 – 2y) + y + x = 0, which is of the form Fn + Fn–2 = 0. Hence, the points of intersection which are n(n – 2) = 3(1) = 3 in number lie on the curve Fn–2 = x + y = 0, which is a straight line.

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EXAMPLE 4.10

Show that the four asymptotes of the curve (x2 – y2) (y2 – 4x2) + 6x3 – 5x2y – 3xy2 + 2y3 – x2 + 3xy – 1 = 0 cut the curve in eight points which lie on the circle x2 + y2 = 1. Solution. Substituting x = 1and y = m in the fourthand third-degree terms, respectively, we get f4(m) = (1 – m2) (m2 – 4) and Thus,

f3(m) = 6 – 5m – 3m2 + 2m3.

f′4(m) = 10m – 4m3. The slopes of the asymptotes are given by f4(m) = (1 – m2) (m2 – 4) = 0 and so, m = +1, and +2. The value of c is given by the expression φ (m) 6 − 5m − 3m 2 + 2m3 c=− 3 . = φ4′ (m) 4m3 − 10m The value of c corresponding to m = 1, –1, 2, and –2 are respectively 0,1,0, and 1. Hence, the asymptotes are y = x, y = –x + 1, y = 2x, and y = –2x + 1. Since the degree of the given curve is 4, the number point of intersection is equal to (n – 2) = 4(4 – 2) = 8. The joint equation of the asymptotes is (y – x)(y + x – 1)(y – 2x)(y + 2x – 1) = 0 or (y2 – x2) (y2 – 4x2) – 6x3 + 5x2y + 3xy2 – 2y3 + y2 – 3xy + 2x2 = 0 or (x2 – y2) (y2 – 4x2) – 6x3 + 5x2y – 3xy2 + 2y3 – y2 + 3xy – 2x2 = 0 The given equation of the curve can be written as (x2 – y2) (y2 – 4x2) + 6x3 – 5x2y – 3xy2 + 2y3 – y2 + 3xy – 2x2 + (x2 + y2 – 1) = 0, which is of the form Fn + Fn–2 = 0. Hence, the points of intersection lie on Fn–2 = 0, that is, on the circle x2 + y2 – 1 = 0. EXAMPLE 4.11

Find the equation of the cubic which has the same asymptotes as the curve x3 – 6x2y + 11xy2 –6y2 + x + y + 1 = 0, and which touches the axis of y at the origin and passes through the point (3, 2). Solution. The equation of the curve is x3 – 6x2y + 11xy2 –6y2 + x + y +1 = 0.

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4.9 aSyMptoteS and curve tracing n  The curve has no asymptote parallel to the axes. To find the oblique asymptotes, we hav f3(m) = 1 – 6m + 11m2 – 6m2 = (1 – m) (1 – 2m)(1 – 3m), f2(m) = 0, f′3(m) = 10m – 6. The slopes of the asymptotes are given by f3(m) = (1–m)(1 – 2m)(1–3m) = 0 and so, m = 1, 12 , and 13 . Further,

φ 2 ( m) = 0. φ3′ (m) Therefore, the asymptotes are x x y = x, y = , and y = . 2 3 The joint equation of the asymptotes is (x – y) (x – 2y) (x – 3y) = 0. The most general equation of any curve having these asymptotes is Fn + Fn–2 = 0, that is, F3 + F1 = 0 or (x – y) (x – 2y) (x – 3y) + ax + by + k = 0, since F1 is of degree 1. Since the curve passes through the origin, putting x = 0, y = 0, in the preceding equation, we get k = 0. Thus, the equation of the curve becomes (x – y) (x – 2y) (x – 3y) + ax + by = 0. (1) Equating to zero, the lowest-degree term in (1), we get ax + by = 0 as the equation of the tangent at the origin. But y-axis, that is, x = 0 is tangent at the origin. Therefore, b = 0 and the equation of the curve reduces to (x – y) (x – 2y) (x – 3y) + ax = 0. Since the curve passes through (3, 2), we have (3 – 2) (3 – 4) (3 – 6) + 3a = 0 and so, a = –1. Hence, the required curve is (x – y) (x – 2y) (x – 3y) –x = 0. or x3 – 6x2y + 11xy2 – 6y3 – x = 0. c=−

EXAMPLE 4.12

Show that the eight points of the curve x4 + 5x2y2 + 4y4 + x2 – y2 + x + y + 1 = 0 and its asymptotes lie on a rectangular hyperbola. Solution. The equation of the curve is of degree 4. Therefore, the number of points of intersection with the asymptotes is n(n – 2) = 4(4 – 2) = 8. Further, the equation of the given curve can be written as

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or

(x2–y2) (x2–4y2) + x2–y2 + x + y + 1 = 0

Fn + Fn–2 = 0, where Fn = (x2–y2) (x2–4y2) is of degree 4 and Fn–2 = x2 – y2 + x + y + 1 = 0 is of degree 2. The asymptotes are given by Fn = 0, that is, by (x2–y2) (x2–4y2) = 0. Thus, the asymptotes are x = +y and x = +2y. The equation Fn–2 = 0, that is, x2 –y2 + x + y + 1 = 0 is the equation of the curve on which the points of intersection of the asymptotes and the given curve lie. The conic x2 – y2 + x + y + 1 = 0 is a hyperbola since the sum of the coefficients of x2 and y2 is zero. Hence, the eight points of intersection of the given curve with its asymptotes lie on a rectangular hyperbola. EXAMPLE 4.13

Find the asymptotes of the curve a (1 − t 2 ) at (1 − t 2 ) , . x= y = 1+ t2 1+ t2 Solution. The equation of the curve is given in parametric form. We eliminate t by dividing and get x 1 y = so that t = . y t x Substituting this value of t in x = x=

a (1− t 2 ) 1+ t 2

, we obtain

a( x − y ) x2 + y 2 2

2

or

y2 (a + x) = x2 (a – x). (1) Equating to zero the highest power of y in the equation (1) of the curve, we have x + a = 0. Hence, x + a = 0 is the asymptote parallel to the y-axis. To find the oblique asymptotes, we put x = 1 and y = m in the highest-degree term of f(x, y) to get f3(m) = m2 + 1 = 0. But the roots of the equation m2 + 1 = 0 are imaginary. Therefore, there is no oblique asymptote. Hence, the only asymptote is x + a = 0. 4.6

ASYMPTOTES BY EXPANSION

Let the equation of the given curve be of the form y = mx + c +

A B C + + +… x x 2 x3

(1)

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4.10 n chapter four Dividing both sides by x, we get y c A B C = m + + 2 + 3 + 4 +… x x x x x Taking limit as x → ∞, we have  y lim   = m. x →∞  x 

Solution. The equation of the given curve can be written as 1 sin θ = = f (θ ). r 2 cos θ (2)

Therefore, f(θ) = 0 yields sinθ = 0 and so, θ = nπ, where n is an integer. Since

The equation (1) can also be written as A B C + + +… x x 2 x3 Taking limit as x → ∞, we get lim( y − mx) = c. y − mx = c +

(3) It follows (see Article 4.1) from (2) and (3) that y = mx + c is an asymptote of the curve (1). Hence, y = mx + c is an asymptote of a curve, whose equation can be expressed in the form (1) given earlier. For example, consider the curve f(x, y) = 2x3 + x2 (2 – y) + x + 1 = 0. The given equation can be written as x2y = 2x3 + 2x2 + x +1 or 1 1 y = 2x + 2 + + 2 . x x Hence, y = 2x + 2 is an asymptote of the given curve. x →∞

4.7

ASYMPTOTES OF THE POLAR CURVES

If α is a root of the equation f(θ) = 0, then r sin(θ − α ) = f ′1(α ) is an asymptote of the polar curve 1r = f (θ ). Thus, to find the asymptotes of a polar curve, first write down the equation of the curve in the form 1r = f (θ ). Then find the roots of the equation f(θ) = 0. If the roots are θ1, θ2, θ3,…, find f′ (θ) at θ = θ1, θ2, θ3,…. Then the asymptotes of the curve shall be 1 r sin(θ − θ1 ) = , f ′ (θ1 ) r sin(θ − θ 2 ) =

1 , and so on. f ′ (θ 2 )

EXAMPLE 4.14

Find the asymptotes of the curve r sin θ = 2 cos2θ.

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f ′ (θ ) =

1  cos 2θ cos θ − sin θ ( −2sin 2θ )   , 2  cos 2 2θ

we have 1 2 cos 2 (2nπ ) = f ′ (nπ ) cos(2nπ ) cos 2nπ + 2sin nπ sin 2nπ =

2 2 . = cos nπ ( −1) n

Hence, the required asymptotes are r sin(θ − nπ ) =

2 ( −1) n 2 ( −1) n

or

− r sin(nπ − θ ) =

or

− r ( −1) n −1 sin θ  =

or

r sin θ = 2.

2 ( −1) n

EXAMPLE 4.15 a Show that the curve r = 1− cos θ has no asymptotes.

Solution. The equation of the given curve can be written in the form 1 1 − cos θ = = f (θ ). r a Then f(θ) = 0 implies cos θ =1 and so, θ = 2nπ, where n is an integer. Further, 1 f ′ (θ ) = sin θ a and so, 1 f ′ (2nπ ) = sin (2nπ ) = 0. a We know that if α is a root of the equation f(θ) = 0, then asymptote corresponding to this asymptotic direction α is given by

f′ (α).r sin(θ – α) = 1.

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4.11 aSyMptoteS and curve tracing n  So for α = 2nπ, the equation of the asymptote is f′ (2nπ).r sin(θ – 2nπ) = 1. But, we have shown that f′(2nπ) = 0. Thus, 0 = 1, which is impossible. Hence, there is no asymptote to the given curve. EXAMPLE 4.16

Find the asymptotes of the curve r = a (sec θ+ tan θ). Solution. We are given that sin θ  a (1 + sin θ )  1 + = r = a .  cos θ cos θ  cos θ Thus,

1 cos θ = = f (θ ). r a (1 + sin θ )

But f(θ ) = 0 yields a (1+cossinθ θ ) = 0 or cosθ = 0 or θ = (2n + 1) π2 .

or r cos θ =

Putting n = 0, 1, 2,…, the asymptotes of the curve are given by r cos θ = 2a and r cos θ = 0. Thus, we note that there are only two asymptotes of the given curve. EXAMPLE 4.17

Find the asymptotes of the curve r = a tan θ. Solution. The equation of the given curve may be written as 1 1 cos θ = = f (θ ). r a sin θ Therefore, f(θ)= 0 implies cos θ = 0 and so,

θ = (2n + 1) π2 . Also 1 f ′ (θ ) = − cosec 2θ. a

Also, f ′ (θ ) =

1  (1 + sin θ )( − sin θ ) − cos θ cos θ   a  (1 + sin θ ) 2 

=−

( sin θ + 1) . a (1 + sin θ ) 2

Therefore, 1 sin (2n + 1) π2 + 1 π  f ′ (2n + 1)  = − a [1 + sin (2n + 1) π2 ] 2 2  1 ( −1) n + 1 =− a 1 + ( −1) n  2   and so, the asymptotes are

π 1  r sin θ − (2n + 1)  = 2  f ′ [ (2n + 1) π2 ]  or

or

π a[1 + ( −1) n ]2   − r sin  (nπ + − θ  = −   2 ( −1) n + 1 n 2 π  a [1 + ( −1) ] ( −1) n r sin  − θ  = 2  ( −1) n + 1

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a[1 + ( −1) n ]2 = a [1 + ( −1) n ] 1 + ( −1) n

Therefore, π 1 −1  = f ′ (2n + 1)  = − 2 π 2 − ( 1) 2 n a   a [ sin (2n + 1) 2 ] Thus, 1 = a ( −1) 2 n −1 = ± a. f ′ [ (2n + 1) π2 ] The asymptotes are now given by

π  r sin  θ − (2n + 1)  = ± a.  2 Proceeding as in the earlier example, we get the asymptotes as r cosθ = a and r cosθ = –a. EXAMPLE 4.18

Find the asymptotes of the following curves: (i) rθ = a a (ii) r = 1− 22cos θ (iii) r = sin nθ = a. Solution. (i) From the given equation, we get 1 θ = = f (θ ). r a

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4.12 n chapter four Therefore, f(θ) = 0 yields f ′ (θ ) =

θ

a

= 0 or θ = 0. Also

1 1 and so, = a. a f ′ (θ )

Thus, the asymptotes are given by 1 r sin (θ − 0) = = a or r sin θ = a. f ′ (0) (ii) From the given equation, we get 1 1 − 2 cos θ = = f (θ ). r 2a Therefore, f(θ)= 0 gives 1–2cosθ = 0 or cos θ = 12 and so, θ = 2nπ ± π3 , where n is an integer. Further, f ′ (θ ) =

1 sin θ (2 sin θ ) = . 2a a

This gives

π 1 π 1 π   f ′  2nπ ±  = sin  2nπ ±  = ± sin   3 a 3 a 3 3 . 2a Hence, the asymptotes are given by   π 1 2a =± r sin θ −  2nπ ±   = π 3   f ′ ( 2nπ ± 3 ) 3   =±

or on simplification π  2a π 2a   r sin  θ −  = and r sin  θ +  = − .   3 3 3 3 (iii) The equation of the curve may be written as 1 sin nθ = = f (θ ). r a Therefore, f(θ) = 0 implies that sin nθ = 0 and so, nθ = mπ, where m is an integer. Thus, θ = mnπ . Also, f ′ (θ ) =

and so,

n cos nθ a

 mπ  n cos mπ f ′ . =  n  a

Hence, the asymptotes are given by mπ  1 a  r sin  θ − , = =  n  f ′ ( mnπ ) n cos mπ where m is an integer.

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4.8

CIRCULAR ASYMPTOTES

Let the equation of a curve be r = f(θ). If lim f (θ ) = a, then the circle r = a is called the θ →∞ circular asymptote of the curve r = f(θ). EXAMPLE 4.19

Find the circular asymptotes of the curves (i) r(eθ – 1) = a (eθ + 1). (ii) r (θ + sin θ) = 2θ + cosθ. aθ (iii) r = θ −1 . Solution. (i) The given equation is r(eθ – 1) = a (eθ + 1) or a (eθ − 1) r= θ = f (θ ). e −1 Now a (eθ − 1) 1 + e −θ lim θ = a lim = a. θ →∞ e − 1 θ →∞ 1 − e −θ Hence, r = a is the circular asymptote. (ii) The equation of the given curve is 2θ + cos θ r= = f (θ ). θ + sin θ Further, 2 + θ1 θ 2θ + cos θ lim f (θ ) = lim = lim θ →∞ θ →∞ θ + sin θ θ →∞ 1 + sin θ θ 2 = 2. 1+ 0 Hence, r = 2 is the required circular asymptote. (iii) The given equation is aθ r= θ −1 and aθ 1 lim = a lim = a. θ →∞ θ − 1 θ →∞ 1 − 1 θ =

Hence, r = a is the circular asymptote of the given curve. 4.9

CONCAVITY, CONVEXITY AND SINGULAR POINTS

Consider the curve y = f(x), which is the graph of a single-valued differentiable function in a

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4.13 aSyMptoteS and curve tracing n  plane. The curve is said to be convex upward or concave downward on the interval (a, b) if all points of the curve lie below any tangent to it on this interval. We say that the curve is convex downward or concave upward on the interval (c, d) if all points of the curve lie above any tangent to it on this interval. Generally, a convex upward curve is called a convex curve and a curve convex down is called a concave curve. For example, the curves in figures (a) and (b) are respectively convex and concave curves.

For example, the point P, in the figure shown below is a point of inflexion y

P

x

0

P

Tangent at P P

a

b (a) Convex Curve

c

Tangent at P

d (b) Concave Curve

The following theorems tell us whether the given curve is convex or concave in some given interval. Theorem 4.1. If at all points of an interval (a, b) the second derivative of the function f (x) is negative, that is, f ″(x) < 0, then the curve y = f(x) is convex on that interval. Theorem 4.2. If at all points of an interval (c, d) the second derivative of the function f(x) is positive, that is, f ″(x) > 0, then the curve y = f(x) is concave on that interval. A point P on a continuous curve y = f(x) is said to be a point of inflexion if the curve is convex on one side and concave on the other side of P with respect to any line, not passing through the point P. In other words, the point that separates the convex part of a continuous curve from the concave part is called the point of inflexion The following theorem gives the sufficient conditions for a given point of a curve to be a Y point of inflexion Theorem 4.3. Let y = f(x) be a continuous curve. If f ″(p) = 0 or f ″(p) does not exist and if the derivative f ″(x) changes sign when passing through x = p, then the point of the curve with abscissa x = p is the point of inflexion 0 Thus at a point of inflexion P, f ′(x) is positive on one side of P and negative on the other side. The above theorem implies that at a (Origin as a Node) point of inflexion f ″(x) = 0 and f ″′ (x) ≠ 0.

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A point through which more than one branches of a curve pass is called a multiple point on the curve. If two branches of curve pass through a point, then that point is called a double point. If r branches of a curve pass through a point, then that point is called a multiple point of order r. If two branches of a curve through a double point are real and have different tangents, then the double point is called a node. For example, the curve in the figure below has a node at the origin. Y

0

Y

X

0

(Origin as a Node)

(Origin as a Cusp)

If two branches through a double point P are real and have coincident tangents, then P is called a cusp. For example, the curve in the figure below has a cusp at the origin. Y

X

0

X

(Origin as a Cusp)

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4.14 n chapter four Let P(x,y) be any point on the curve f(x, y) dy = 0. The slope dx of the tangent at P is given by ∂f dy ∂f ∂f dy = − ∂x or + ⋅ = 0, ∂ f ∂x ∂y dx dx ∂y which is a first degree equation in dx . Since at a multiple point, the curve must have at least dy two tangents, therefore dx must have at least two values at a double point. It is possible if and only if dy

∂f = 0 and ∂x

∂f = 0. ∂y

Hence the necessary and sufficient conditions for the existence of multiple points are ∂f = 0 and ∂x

∂f = 0. ∂y

If x = 0, then

d3 y dx3

If x = 3a , then

d3 y dx3

If x = − 3a , then

6 a6 a8

=

=

6 a2

≠ 0.

= − 34 a 2 ≠ 0.

d3 y dx3

= − 34 a 2 ≠ 0.

Thus all the three values of x corresponds to the points of inflexion When x = 0, the given equation yields y = 0. When x = 3a , the given equation yields y = 3 43a When x = − 3a, the given equation yields y = − 3 43 a. Hence the point of inflexionof the given curve are   3 3  3 3  (0, 0),  3a , a  and  − 3a , a . 4  4    EXAMPLE 4.21

EXAMPLE 4.20

Does the curve y = x4 have points of inflexion

Find the points of inflexion of the curv y (a2 + x2) = x3. Solution. The equation of the given curve is

Solution. The equation of the given curve is y = x4. Differentiating with respect to x, we have

y=

3

x . a2 + x2

Therefore dy (a + x )3 x − 2 x x + 3a x = = 2 . dx (a 2 + x 2 ) 2 (a + x 2 )2 2

2

2

4

4

2

2

Differentiating once more with respect to x, we get d 2 y x(6a 4 + 10a 2 x 2 + 4 x 4 − 12a 2 x 2 − 4 x 4 ) = dx 2 ( a 2 + x 2 )3 =

2 xa 2 (3a 2 − x 2 ) . ( a 2 + x 2 )3 d2 y

At the point of inflextion, we must have dx2 = 0 and so 2 xa (3a − x ) = 0 and 2 xa 2 (3a 2 − x 2 ) = 0, ( a 2 + x 2 )3 2

2

2

which yields x = 0 , ± 3a . Further, d 3 y 6a 2 ( x 4 − 6a 2 x 2 + a 4 ) = . dx 3 (a 2 + x 2 ) 4

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dy d2y d3y = 4 x 3 , 2 = 12 x 2 , 3 = 24 x. dx dx dx Then for the points of inflexion, we must hav d2y = 0, that is, 12 x 2 = 0. dx 2 which yields x = 0. But d2 y for x < 0, dx2 > 0 and therefore the curve is concave, d2 y

for x > 0, dx2 > 0 and therefore the curve is concave. Since the second derivative does not change sign passing through x = 0, the curve has no points of inflexion EXAMPLE 4.22

Find the points of inflexion on the curve y2 = x (x +1)2. Solution. The given curve is symmetrical about x axis and gives 1 y = ± x 2 ( x + 1). So, we can proceed with 1

y = x 2 ( x + 1).

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4.15 aSyMptoteS and curve tracing n  Then

for x > −

1 dy 1 −1 3x + 1 = x 2 + x 2 ( x + 1) = , 1 dx 2 2x 2 1

d2 y

To determine the point of inflexion, we put dx2 equal to 0 . Therefore 3x – 1 = 0 or x = 13 . Further d2y 3 1 = 5 (1 − x) ≠ 0 at x = . 3 3 dx 8x 2 Therefore the curve has point of inflexion corresponding to x = 13 . Putting x = 13 in the equation of the curve, we have y = ± 3 43 . Hence the points of inflexion on the curve ar 4  1 4  1  3 ,  and  3 , − . 3 3 3 3 Find the points of inflexion and the intervals of convexity and concavity of the Gaussian curve

(

the second point of inflexion is −

1 2

,e

− 12

).

EXAMPLE 4.24

Determine whether the curve y = e x is concave or convex. Solution. The given exponential curve is y = e x. dy x x Then dy , for the all values of x. dx = e , dx = e > 0 Hence the curve is everywhere concave. 2

2

EXAMPLE 4.25

Determine the existence and nature of the double points on the curve Solution. We have

f ( x, y ) = y 2 − ( x − 2) 2 ( x − 1) = 0,

∂f = −( x − 2)(3 x − 4), ∂x ∂f = 2 y. ∂y

2

y = e− x . Solution. The equation of the Gaussian curve is 2

y = e − x . Therefore 2 d2y = 2e − x [2 x 2 − 1]. 2 dx

2 dy = −2 xe − x , dx

For the existence of points of inflexion, we must d2 y have dx2 = 0, which yields x = ± 12 . Now, since 1 d2y for x < , we have

, we have > 0, dx 2 2 therefore the point of inflexion exists for x = 12 . −1 Putting x = 12 in the given equation, y = e 2 . 1 − Therefore 12 , e 2 is a point of inflexion on the curve. Also 1 d2y for x < − , we have >0 dx 2 2

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d2y < 0. dx 2

f ( x, y ) = y 2 − ( x − 2) 2 ( x − 1) = 0.

EXAMPLE 4.23

(

2

, we have

Thus another point of inflexion exists for the value x = − 12 . Putting x = − 12 in the equation −1 of the Gaussian curve, we get y = e 2 . Hence

d 2 y 1  3 x 2 − (3 x + 1) 12 x 2  3 x − 1 =  . = 3 x dx 2 2  4x 2  1

1

)

Now for the existence of double points, we must have ∂f ∂f = = 0. ∂x ∂y Hence

( x − 2)(3 x − 4) = 0 and 2 y = 0,

which yield 4 and y = 0. 3 Thus the possible double points are (2, 0) and ( 34 , 0) . But, only (2, 0) satisfies the equation of the curve. To find the nature of the double point (2,0), we shift the origin to (2, 0). The equation reduces to y2 = (x + 2 – 2)2 (x + 2 – 1) = x2 (x + 1) = x3 + x2. x = 2,

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4.16 n chapter four Equating to zero the lowest degree term, we get y2 – x2 = 0, which gives y = + x as the tangent at (2,0). Therefore, at the double point (2,0), there are two real and district tangents. Hence the double point (2, 0) is a node on the given curve. EXAMPLE 4.26

Does the curve x 4 − ax 2 y + axy 2 + a 2 y 2 = 0 have a node on the origin? Solution. Equating to zero the lowest degree term in the equation of the given curve, we have a2 y2 = 0, which yields y = 0, 0. Therefore there are two real and coincident tangents at the origin. Hence the given curve has a cusp or conjugate point at the origin and not a node. 4.10

CURVE TRACING (CARTESIAN EQUATIONS)

x3 + y3 = 3axy is symmetrical about the line y = x. 2. Origin: (i) If the equation of a curve does not contain a constant term, then the curve passes through the origin. In other words, a curve passes through the origin if (0, 0) satisfies the equation of the curve. (ii) If the curve passes through the origin, find the equation of the tangents at the origin by equating to zero the lowest-degree terms in the equation of the curve. In case there is only one tangent, determine whether the curve lies below or above the tangent in the neighbourhood of the origin. If there are two tangents at the origin, then the origin is a double point; if the two tangents are real and distinct, then the origin is a node; if the two tangents are real and coincident, then the origin is cusp; if the two tangents are imaginary, then the origin is a conjugate point or an isolated point.

The aim of this section is to find the appropr ate shape of a curve whose equation is given. We shall examine the following properties of the curves to trace it. 1. Symmetry: (i) If the equation of a curve remains unaltered when y is changed to –y, then the curve is symmetrical about the x-axis. In other words, if the equation of a curve consists of even powers of y, then the curve is symmetrical about the x-axis. For example, the parabola y2 = 4ax is symmetrical about the x-axis. (ii) If the equation of a curve remains unaltered when x is changed to –x, then the curve is symmetrical about the y-axis. Thus, a curve is symmetrical about the y-axis, if its equation consists of even powers of x. For example, the curve x2 + y2 = a2 is symmetrical about the y-axis. (iii) If the equation of a curve remains unchanged when x is replaced by –x and y is replaced by –y, then the curve is symmetrical in the opposite quadrants. For example, the curve xy = c2 is symmetrical in the opposite quadrants. (iv) If the equation of a curve remains unaltered when x and y are interchanged, then the curve is symmetrical about the line y = x. For example, the folium of Descarte’s

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3. Intersection with the Coordinate Axes: To find the points where the curve cuts the coordinate axes, we put y = 0 in the equation of the curve to find where the curve cuts the x-axis. Similarly, we put x = 0 in the equation to find where the curve cuts the y-axis 4. Asymptotes: Determine the asymptotes of the curve parallel to the axes and the oblique asymptotes. 5. Sign of the Derivative: Determine the dy points where the derivative dx vanishes or becomes infinite. This step will yield the points where the tangent is parallel or perpendicular to the x-axis. 6. Points of Inflexion A point P on a curve is said to be a point of inflexion if the curve is concave on one side and convex on the other side of P with respect to any line AB, not passing through the point P.

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4.17 aSyMptoteS and curve tracing n  or

a 2 y 2 = ( x + a ) ( −2ax − x 2 ) . 2

Equating to zero the lowest-degree term, the tangent at the new origin is given by 4a2x2 = 0 or x = 0. Thus, the tangent at (a, 0) is parallel to the y-axis. (v) The given equation can be written as y = 2

There will be a point of inflexion at a point on the curve if ddx y = 0 but ddx y ≠ 0. 2

3

2

3

7. Region, Where the Curve Does Not Exist: Find out if there is any region of the plane such that no part of the curve lies in it. This is done by solving the given equation for one variable in terms of the other. The curve will not exist for those values of one variable which make the other variable imaginary.

x2 ( a2 − x2 )

. a2 When x = 0, y = 0 and when x = a, y = a. When 0 < x < a, y is real and so, the curve exists in this region. When x > a, y2 is negative and so, y is imaginary. Hence, the curve does not exist in the region x > a.

(vi) The given curve has no asymptote. Hence, the shape of the curve is as shown in the following figure

EXAMPLE 4.27

Trace the curve a 2 y 2 = x 2 (a 2 − x 2 ).

Solution. The equation of the curve is a 2 y 2 = x 2 (a 2 − x 2 ).

We observe the following: (i) Since powers of both x and y are even, it follows that the curve is symmetrical about both the axes. (ii) Since the equation does not contain constant terms, the curve passes through the origin. To find the tangent at the origin, we equate to zero the lowest-degree terms in the given equation. Thus, the tangents at the origin are given by a 2 y 2 − a 2 x 2 = 0 or y = ± x.

Since tangents are distinct, the origin is a node. (iii) Putting y = 0 in the given equation, we get x = 0 and x = ± a. Therefore, the curve crosses the x-axis at (0,0), (a,0), and (–a,0). (iv) Shifting the origin to (a, 0), the given equation reduces to 2 2 a2 y 2 = ( x + a ) a2 − ( x + a )   

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EXAMPLE 4.28

Trace the curve xy 2 = 4a 2 (2a − x) (Witch of Agnesi).

Solution. We note that (i) The curve is symmetrical about the x-axis because the equation contains even powers of y. (ii) Since the equation consists of a constant term, 8a3, the curve does not pass through the origin. (iii) Putting y = 0 in the equation, we get x = 2a. Therefore, the curve crosses the x-axis at (2a, 0). When x = 0, we do not get any value of y. Therefore, the curve does not meet the y-axis. Shifting the origin to (2a, 0), the equation of the curve reduces to ( x + 2a ) y 2 = 4a 2 (2a − x − 2a )

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4.18 n chapter four or

y 2 x + 2ay 2 + 4a 2 x = 0 .

Equating to zero, the lowest-degree terms of this equation, the equation of the tangent at this new origin is given by 4a2 x = 0 or x = 0. Hence, the tangent at the point (2a, 0) to the curve is parallel to y-axis. (iv) Equating to zero the coefficient of highest power of y, the asymptote parallel to the y-axis is x = 0, that is, the y-axis. Further, the curve has no other real asymptote. (v) The equation of the given curve can be written as 4a 2 (2a − x) y2 = . x Therefore, when x → 0 , y approaches ∞ and so, the line x = 0 is an asymptote. When x = 2a, y = 0. When 0 < x < 2a, the value of y is real and so, the curve exists in the region 0 < x < 2a. When x > 2a, y is imaginary and so, the curve does not exists for x > 2a. Similarly, when x is negative, again y is imaginary. Therefore, the curve does not exist for negative x. In view of the mentioned points, the shape of the curve is as shown in the following figure

EXAMPLE 4.29

through the origin. Equating to zero the lowest-degree term in the equation, the tangent at the origin is given by 2ay2 = 0. Thus, y = 0, y = 0 and so, there are two coincident tangents at the origin. Hence, the origin is a cusp. (iii) Putting x = 0 in the equation, we get y = 0 and similarly, putting y = 0, we get x = 0. Therefore, the curve meets the coordinate axes only at the origin. (iv) Equating to zero the highest power of y in the equation of the curve, the asymptote parallel to the y-axis is x = 2a. The curve does not have an asymptote parallel to the x-axis or any other oblique asymptote. (v) The given equation can be written as y2 =

x3 . 2a − x

When x → 2a , y 2 → ∞ and so, x = 2a is an asymptote. If x > 2a, y is imaginary. Therefore, the curve does not exist beyond x = 2a. When 0 < x < 2a, y2 is positive and so, y is real. Therefore, the curve exists in the region 0 < x < 2a. When x < 0, again y is imaginary. Therefore, the curve does not exist for a negative x. In view of the said observations, the shape of the curve is as shown in the following figure

EXAMPLE 4.30

Trace the curve

Trace the curve

Solution. We note that (i) Since the powers of y in the given equation of the curve are even, the curve is symmetrical about the x-axis. (ii) Since the equation of the curve does not contain a constant term, the curve passes

Solution: We observe that (i) The curve is not symmetrical about the axes. However, the equation of the curve remains unaltered if x and y are interchanged. Hence, the curve is symmetrical about the line y = x. It meets this line at ( 32a , 32a ) .

y 2 (2a − x) = x 3 (Cissoid).

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x 3 + y 3 = 3axy (Folium of Descartes).

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4.19 aSyMptoteS and curve tracing n  (ii) Since the equation does not contain a constant term, the curve passes through the origin. Equating to zero the lowest-degree term, we get 3axy = 0. Hence, x = 0, y = 0 are the tangents at the origin. Thus, both yand x-axis are tangents to the curve at the origin. Since there are two real and distinct tangents at the origin, the origin is a node of the curve. (iii) The curve intersects the coordinate axes only at the origin. (iv) If, in the equation of the curve, we take both x and y as negative, then the right-hand side becomes positive while the left-hand side is negative. Therefore, we cannot take both x and y as negative. Thus, the curve does not lie in the third quadrant. (v) There is no asymptote parallel to the axes. Further, putting x = 1, y = m in the highestdegree term, we have φ3 ( m) = m 3 + 1 The slope of the asymptotes are given by m3 + 1 = 0. The real root of this equation is m = –1. Also, putting x = 1, y = m in the second-degree term, we have φ2 (m) = −3am and further,

φ ′3 (m) = 3m 2 .

Therefore, c=−

φ2 (m) 3am a = = . φ3′ (m) 3m 2 m

For m = –1, we have c = –a. Hence, the oblique asymptote is y = –x –a or x + y + a = 0. In view of the earlier facts, the shape of the curve is as shown in the following figure

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EXAMPLE 4.31

Trace the curve y2 (a + x) = x2 (a – x). Solution. We note that (i) The equation of the curve does not alter if y is changed to –y. Therefore, the curve is symmetrical about the x-axis. (ii) Since the equation does not contain a constant term, the curve passes through the origin. The tangents at the origin are given by ay2 – ax2 = 0 or y = +x. Thus, there are two real and distinct tangents at the origin. Therefore, the origin is a node. (iii) Putting y = 0, we have x2 (a – x) = 0 and so, the curve intersects the x-axis at x = 0 and x = a, that is, at the points (0, 0) and (a, 0). Putting x = 0, we get y = 0. Thus, the curve intersects the y-axis only at (0, 0). Shifting the origin to (a, 0), the equation of the curve reduces to y 2 (2a + x) = − x ( x 2 + 2ax + a 2 ) .

Equating to zero the lowest-degree term, we get a2 x = 0. Hence, at the new origin, x = 0 is the tangent. Thus, the tangent at (a, 0) is parallel to the y-axis. (iv) The equation of the curve can be written as x 2 (a − x) . a+x When x lies in 0 < x < a, y2 is positive and so, the curve exists in this region. But when x > a, y2 is negative and so, y is imaginary. Thus, the curve does not exist in the region x > a. Further, if x → − a , then y 2 → ∞ and so, x = –a is an asymptote of the curve. If –a < x < 0, y2 is positive and therefore, the curve exists in –a < x < 0. When x < –a, y2 is negative and so, the curve does not lie in the region x < –a. (v) Equating to zero the coefficient of the highest power of y in the equation of the curve, we have x + a = 0. Thus, x + a = 0 is the asymptote parallel to the y-axis. To see whether oblique asymptotes are there or y2 =

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4.20 n chapter four not, we have φ3 (m) = m 2 + 1 . But the roots of m2 + 1 = 0 are imaginary. Hence, there is no oblique asymptote. Thus, the shape of the curve is as shown in the following figure

EXAMPLE 4.32

Trace the curve x = (y – 1) (y – 2) (y –3). Solution. We note that (i) The equation of the curve has odd powers of x and y. Therefore, the curve is not symmetrical about the axes. It is also not symmetrical about y = x or in the opposite quadrants. (ii) The curve does not pass through the origin. (iii) Putting x = 0 in the given equation, we get y = 1, 2 and 3. Thus, the curve cuts the y-axis at y = 1, 2 and 3. Similarly, putting y = 0, we see that the curve cuts the x-axis at x = –6. (iv) The curve has no linear asymptotes since y → ±∞ , x → ±∞ . (v) When 0 < y < 1, then all the factors are negative and so, x is negative. When 1 < y < 2, x is positive. Similarly, when 2 < y < 3, then x is negative. At y = 3, x = 0. When y > 3, x is positive. When y < 0, x is negative. Hence, the shape of the curve is as shown in the following figure

EXAMPLE 4.33

Trace the curve x3 + y3 = a2x. Solution. We note the following characteristics of the given curve: (i) Since the equation of the curve contains odd powers of x and y, the curve is not symmetrical about the axes. But if we change the sign of both x and y, then the equation remains unaltered. Therefore, the curve is symmetrical in the opposite quadrants. (ii) Since the equation of the curve does not have a constant term, the curve passes through the origin. The tangent at the origin is given by a2 x = 0. Thus, x = 0, that is, y-axis is tangent to the curve at the origin. (iii) Putting y = 0 in the equation, we get x (x2 – a2) = 0 or x (x – a) (x + a) = 0. Hence, the curve cuts the x-axis at x = 0, x = a, and x = –a, that is, at the points (0, 0), (a, 0), and (–a, 0). On the other hand putting x = 0 in the equation, we get y = 0. Therefore, the curve cuts the y-axis only at the origin (0, 0). (iv) The curve does not have any asymptote parallel to the axes. But φ3 ( m) = m 3 + 1 , φ 2 ( m) = 0 . Thus, the slope of the oblique asymptotes is given by m3 + 1 = 0. Thus, the real root is m = –1. Also φ ( m) c=− 2 = 0. φ3 (m) Therefore, the curve has an oblique asymptote y = –x. (v) From the equation of the curve, we have y3 = a2 x – x3. Differentiating with respect to x, we get dy dy a 2 − 3x 2 = a 2 − 3 x 2 or = . dx dx 3y2 Thus,  dy  = −∞    dx ( a ,0) 3y

M04_Baburam_ISBN _C04.indd 20

2

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4.21 aSyMptoteS and curve tracing n  and so, the tangent at (a, 0) is perpendicular dy = −∞ to the x-axis. Similarly, dx ( − a ,0) and so, the tangent at (–a, 0) is also perpendicular to the x-axis. dy Also we note that dx = 0 implies x = ± a3 . Therefore, the tangents at these points are parallel to the x-axis. (vi) Also y3 = a2 x – x3 = x (a2 – x2) implies that y3 is positive in the region 0 < x < a. But y3 is negative in the region x > a. The earlier facts imply that the shape of the given curve is as shown in the following figure

4. Special Points on the Curve: Solve the equation of the curve for r and find how r varies as θ increases from 0 to ∞ and also as θ decreases from 0 to −∞ . Form a table with the corresponding values of r and θ. The points so obtained will help in tracing the curve. 5. Region: Find the region, where the curve does not exist. If r is imaginary in α < θ < β, then the curve does not exist in the region bounded by the lines θ = α and θ = β. 6. Value of tanf : Find tan f, that is, r ddrθ , which will indicate the direction of the tangent at any point. If for θ = α, f = 0 then θ = α will be tangent to the curve at the point θ = α. On the other hand if for θ = α, φ = π2 , then at the point θ = α, the tangent will be perpendicular to the radius vector θ = α. 7. Cartesian Form of the Equation of the Curve: It is useful sometimes to convert the given equation from polar form to cartesian form using the relations x = r cos θ and y = r sin θ.

4.11

EXAMPLE 4.34

( )

CURVE TRACING (POLAR EQUATIONS)

To trace a curve with a polar form of equation, we adopt the following procedure: 1. Symmetry: If the equation of the curve does not change when θ is changed into –θ the curve is symmetrical about the initial line. If the equation of the curve remains unchanged by changing r into –r, then the curve is symmetrical about the pole and the pole is the center of the curve. If the equation of the curve remains unchanged when θ is changed to –θ and r is changed in to –r, then the curve is symmetrical about the line θ = π2 . 2. Pole: By putting r = 0, if we find some real value of θ, then the curve passes through the pole which otherwise not. Further, putting r = 0, the real value of θ , if exists, gives the tangent to the curve at the pole. 3. Asymptotes: Find the asymptotes using the method to determine asymptotes of a polar curve.

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Trace the curve r = a sin 3θ. Solution. We note that (i) The curve is not symmetrical about the initial line. But if we change θ to –θ and r to –r, then the equation of the curve remains unchanged. Therefore, the curve is symmetrical about the line θ = π2 . (ii) Putting r = 0, we get sin 3θ = 0. Thus, 3θ = 0, π or θ = 0, π3 . Thus, the curve passes through the pole, and the lines θ = 0 and θ = π3 are tangents to the curve at the pole. (iii) r is maximum when sin 3θ = 1 or 3θ = π2 or θ = π6 . The maximum value of r is a. dr (iv) We have and so, = 3a cos 3θ dθ dθ 1 tan φ = r dr = 3 tan 3θ . Thus, φ = π2 when 3θ = π2 or θ = π6 , and the tangent is perpendicular to the radius vector θ = π6 . (v) Some points on the curve are given below:

θ : 0

π 6

π 3

r

a

0 –a

:

0

π 2

2π 3

5π 6

π

0

a

0

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4.22 n chapter four One loop of the curve lies in the region 0 < θ < π3 . The second loop lies in the region π3 < θ < 23π in the opposite direction because r is negative there. The third loop lies in the region 23π < θ < π as r is positive (equal to a) there. When θ increases from π to 2π, we get again the same branches of the curve. Hence, the shape of the curve is shown in the following figure

greater than 2a. Hence, no portion of the curve lies to the left of the tangent at (2a, 0). Since | r | < 2a, the curve lies entirely within the circle r = 2a. (vi) There is no asymptote to the curve because for any finite value of θ, r does not tend to infinity. Hence, the shape of the curve is an shown in the following figure

EXAMPLE 4.36 EXAMPLE 4.35

Trace the curve r = a(1– cosθ) (Cardioid). Solution. The equation of the given curve is r = a (1– cosθ). We note the following characteristics of the curve: (i) The equation of the curve remains unchanged when θ is changed to –θ. Therefore, the curve is symmetrical about the initial line. (ii) When r = 0, we have 1–cosθ = 0 or θ = 0. Hence, the curve passes through the pole and the line θ = 0 is tangent to the curve at the pole. (iii) The curve cuts the line θ = π at (2a, π). (iv)

dr

= a sin θ and so, tan φ = r ddrθ = a sinr θ = = tan θ2 . If θ2 = π2 , then f = 90º.

dθ a (1− cos θ ) a sin θ

Thus, at the point θ = π, the tangent to the curve is perpendicular to the radius vector. (v) The values of θ and r are: θ : 0 π3 π2 23π π r : 0 a3 a 33a 2a We observe that as θ increases from 0 to π, r increases from 0 to 2a. Further, r is never

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Trace the curve r = a + b cosθ, a < b (Limacon). Solution. The given curve has the following characteristics: (i) Since the equation of the curve remains unaltered when θ is changed to –θ, it follows that the curve is symmetrical about the initial line. (ii) r = 0 when a + b cosθ = 0 or θ = cos −1 ( − ba ) . Since ba < 1, cos −1 ( − ba ) is real. Therefore, the curve passes through the pole and the radius vector θ = cos −1 ( − ba ) is tangent to the curve at the pole. (iii) We note that r is maximum when cosθ = 1, that is when θ = 0. Thus, the maximum value of r is a + b. Thus, the entire curve lies within the circle r = a + b. Similarly, r is minimum when cosθ = –1, that is when θ = π. Thus, the mimimum value of r is a – b, which is negative. (iv) ddrθ = −b sin θ and so, tan φ = r ddrθ = − b sinr θ = − a +bbsincosθ θ . Thus, f = 90º when θ = 0, π. Hence, at the points θ = 0 and θ = π, the tangent is perpendicular to the radius vector. (v) The following table gives the value of r corresponding to the value of θ :

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4.23 aSyMptoteS and curve tracing n  −1 −1 a a π θ : 0 2 cos ( − b ) cos ( − b ) < θ < π θ r: a+b a 0 negative a–b (vi) Since r is not infinite for any value of θ, the given curve has no asymptote: Hence, the shape of the curve is as shown in the following figure

> a. Further, when x → ∞, y 2 → ∞. In view of the mentioned facts, the shape of the curve is as shown in the following figure

y

EXAMPLE 4.37

Trace the curve r2 cos 2θ = a2. Solution. The equation of the given curve can be written as r2 (cos2θ – sin2θ) = a2 or x2 – y2 = a2 since x = r cosθ, y = r sin θ. Therefore, the given curve is a rectangular hyperbola. We note that (i) The curve is symmetrical about both the axes. (ii) It does not pass through the origin. (iii) It cuts the x-axis at (a, 0) and (–a, 0). But it does not meet y-axis. (iv) Shifting the origin to (a, 0), we get (x + a)2 – y2 = a2 or x2 – y2 + 2ax = 0. Therefore, the tangent at (a, 0) is given by 2ax = 0 and so, the tangent at (a, 0) is x = 0, the line parallel to the y-axis. (v) The curve has no asymptote parallel to coordinate axes. The oblique asymptote (verify) are y = x and y = –x. (vi) The equation of the curve can be written as y2 = x2 –a2. When 0 < x < a, the y2 is negative and so, y is imaginary. Therefore, the curve does not lie in the region 0 < x < a. But when x > a, y2 is positive and so, y is real. Thus, the curve exists in the region x

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4.12 CURVE TRACING (PARAMETRIC EQUATIONS) If the equation of the curve is given in a parametric form, x = f(t) and y = f(t), then eliminate the parameter and obtain a cartesian equation of the curve. Then, trace the curve as dealt with in case of cartesian equations. In case the parameter is not eliminated easily, a series of values are given to t and the dy corresponding values of x, y, and dx are found. Then we plot the different points and find the slope of the tangents at these points by the values of dy at the points. dx EXAMPLE 4.38

Trace the curve x = a(t + sin t) y = a(1 + cos t). Solution. We note that (i) Since y = a(1 + cos t) is an even function of t, the curve is symmetrical about the y-axis. (ii) We have y = 0 when cos t = –1, that is when t = –π, π. When t = π, we have x = aπ. When t = –π, x = – aπ. Thus, the curve meets the x-axis at (aπ, 0) and (–aπ, 0). (iii) Differentiating the given equation, we get dx dy = a (1 + cos t ), = − a sin t. dt dt Therefore,

dy = dx

dy dt dx dt

=−

a sin t a (1 + cos t )

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4.24 n chapter four =−

This implies that the curve lies within the square bounded by the lines x = +a, y = + a. (ii) The equation of the curve can be written as

2a sin 2t cos 2t t = − tan . 2 2acos 2 2t

Now

π  dy    = − tan = −∞ 2 dx  t = x Thus, at the point (aπ, 0), the tangent to the curve is perpendicular to the x-axis. Similarly, at the point (–aπ, 0), dy = ∞ and dt hence, at the point (–aπ, 0), the tangent to the curve is perpendicular to the x-axis. (iv) y is maximum when cos t = 1, that is, t = 0. When t = 0 x = 0 and y = 2a. Thus, the curve cuts the y-axis at (0, 2a). Further,  dy    =0  dx ι = 0

 x2 3  y 2 3  2  +  2  = 1. a  a  This equation shows that the curve is symmetrical about both the axes. Also it is symmetrical about the line y = x since interchanging of x and y do not change the equation of the curve. (iii) The given curve has no asymptote. (iv) The curve cuts the x-axis at (a, 0) and (–a, 0). It meets the y-axis at (0, a) and (0, –a). For x = a, we have cos3t = 1 or t = 0. Therefore,

and so, at the point (0, 2a), the tangent to the curve is parallel to the x-axis. (v) It is clear from the equation that y cannot be negative. Further, no portion of the curve lies in the region y > 2a. (vi) There is no asymptote parallel to the axes. (vii) The values of x, y corresponding to the values of t are as follows: t –π 0 π2 π − π2 π π a + 1 ( ) x –aπ −a ( 2 + 1) 0 aπ 2 y 0 a 2a a 0 Hence, the shape of the curve is as shown in the following figure

 dy   dy  dt =  dx  = (− tan t )t = 0 = 0.    dx t = 0  dt t = 0 Hence, at the point (a, 0), the x-axis is the tangent to the curve. Similarly, at the point (0,a), the y-axis is the tangent to the curve. Hence, the shape of the curve is as shown in the following figure

1

1

EXERCISES EXAMPLE 4.39 2 3

2 3

2 3

Trace the curve x + y = a . Solution. (i) The parametric equation of the curve are x = a cos3t, y = a sin3t. Therefore, | x | < a and y < a.

M04_Baburam_ISBN _C04.indd 24

Find the asymptotes of the following curves: 1. Test the curve y = x3 for concavity/ convexity. Ans. Concave for x > 0 convex for x < 0. 2.

Find the points of inflexion on the curve y(a2 + x2) = a2x. Ans. (0, 0) , (0, 0),

(

3a,

3a 4

),(−

3a,

3a 4

).

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4.25 aSyMptoteS and curve tracing n  3.

Show that the points of inflexion on the curve y2 = (x – a)2 (x – b) lie on the line 3x + a = 4b. 2 Hint: d 2y = 0 yields 3x +a = 4b dx

4.

Find the points of inflexion on the curve x = a(2θ – sinθ), y = a(2–cosθ). Ans.  a 4nπ ± 23π  23 , 32a    Show that origin is a node on the curve 2 a2 − by 2 = 1. x2

(

5.

6.

)

Find the asymptotes of the following curves: y3 + x2y + 2xy2 – y+1=0 Ans. y = 0, y + x–1 = 0, y + x + 1 = 0.

7.

x3 + y3 – 3axy = 0

8.

x – 2y + 2x y – xy + xy – y2 + 1 = 0 3

3

Ans. x + y + a = 0

2

2

Ans. x–y = 0, x + y + 1 = 0, x + 2y –1 = 0 9.

3x3 + y –7xy2 + 2y3 – 14xy+7y2+ 4x + 5y = 0 Ans. 6x–6y–7 = 0, y = 3x –1, 3x + 6y + 5 = 0

10. y3 = x3 + ax2

Ans. y = x + a3

11. y3–3x2y + xy2 – 3x3 + 2y2 + 2xy + 4x + 5y + 6=0 Ans. y = x − 2, y = x 3 − 1, y = − x 3 − 1 12. x2y2 = a2(x2 + y2)

Ans. x = + a, y = + a

13. x y + x y = x + y 2 3

3 2

3

3

Ans. x = + 1, y = + 1, y = –x 14. x3 + 2x2y + xy2 – x2 – xy + 2 = 0 Ans. x = 0, x+ y = 0, x+ y–1= 0 15. x3 + x2y –xy2– y3– 3x – y – 1 = 0 Ans. y = x, y = –x+1, x + y + 1 = 0 16. (x2 – y2)(x +2y + 1) + x + y + 1 = 0 Ans. x + y = 0, x – y = 0, x + 2y + 1 = 0 17. (i) y2 (x – 2) = x2(y – 1) Ans. x = 2, y = 1, y = x + 1

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(ii) x =

a (t + t3 ) 1+ t 4

,y=

a (t − t3 ) 1+ t 4

.

Hint: Eliminating t, we get (x2 + y2)2 = a2(x2–y2) Ans. No asymptote. 18. Show that the asymptotes of the curve x2y2 – a2(x2+y2)–a3(x + y) + a4 = 0 form a square and that the curve passes through two angular points of that square. Hint: The four asymptotes are x = + a, y = + a. They form a square of length 2a. The angular points are (a, a), (a, –a), (–a, a), and (–a, –a). The curve passes through two angular points (–a, a), and (–a, –a). 19. Show that the points of intersection of the curve 2y3 – 2xy – 4xy2 + 4x3 – 14xy + 6y2 + 4x2 + 6y + 1 = 0 and its asymptotes lie on the straight line 8x + 2y + 1 = 0. 20. Show that the asymptotes of the cubic x3 – 2y3 + xy(2x – y) + y(x – y) + 1 = 0 cut the curve again in three points which lie on the straight line x – y + 1 = 0. Hint: The asymptotes are y = x, y = –x–1, and y = − 12 x + 12 . Their joint equation is x2 – 2y3 + 2x2y –x y2 + xy – y2– x + y = 0. Subtracting this equation from the equation of the curve, we get x – y + 1 = 0. 21. Show that the point of intersection of the curve 4(x4 + y4) – 17x2y2–4x(4y2 – x2) + 2(x2 – 2) = 0 and its asymptotes lie on the ellipse x2 + 4y2 = 4. 22. Find the equation of the hyperbola passing through the origin and having asymptotes x + y –1 = 0 and x – y + 2 = 0. Hint: joint equation of asymptotes is F2 = (x – y –1)( x – y –2) = 0. Equation of the curve is Fn + Fn-2 = 0, that is, F2 + F0 = 0. Thus, F0 is of a zero degree and so, is a constant. Thus, the equation of the curve is (x + y – 1) (x – y + 2) + k = 0. It passes through the origin. So k = 2. Hence, the curve is (x + y –1) (x – y + 2) + 2 = 0 or x2 – y2 + x + 3y = 0.

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4.26 n chapter four 23. Find the asymptotes of the curve xy (x2 – y2) + x2 + y2 + a2 and show that the eight points of intersection of the curve and its asymptotes lie on a circle with the origin at the center. 24. Find the equation of the cubic which has the same asymptotes as the curve x3 – 6x2y + 11xy3 + 4x + 5y + 7 = 0 and which passes through the points (0,0,)(0,2), and (2,0).

28. Trace the curve x2y2 = a2(x2 + y2)

Ans.

Ans. The joint equation of the asymptote is (x – y)( x – 2y)( x – 3y) = 0. The cubic is x3 – 6x2y + 11xy2 + –6y3 – 4x + 24y = 0. 25. Find the equation of the straight line on which lie the three points of intersection of the curve x3 + 2x2y –xy2 – 4y2 + 2xy +y – 1= 0 and its asymptotes. Ans. x + 3y = 1.

29. Trace the curve y2(x2 + y2) + a2(x2 – y2) =0

Ans.

26. Find the asymptotes of the following polar curves: (i) rθ cosθ = a cos2θ

Ans. r cosθ =

2a (2 k +1)π

30. Trace the curve ay2 = x(x – a)2

(ii) r = a cosecθ + b Ans. r sinθ = a (iii) r =

a log θ

Ans. r sin (θ –1) = a

Ans.

(iv) r = a sec θ + b tan θ Ans. r = cosθ = a + b, r cosθ = a–b (v) r (1–e ) = a θ

31. Trace the curve xy = a2(a – x)

Ans. r sinθ = –a θ (vi) r = 2a θ 2 −H

Ans. r sinθ = –a

27. Find the circular asymptote of the curves: (i) r =

Ans. r = (ii) r =

Ans:

3θ 2 + 2θ +1 2θ 2 +θ +1 3 2

2

6θ + 5θ −1 2θ 2 − 3θ + 7

Ans. r = 3

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4.27 aSyMptoteS and curve tracing n  35. Trace the curve r = a cos2θ.

32. Trace the curve y2(a – x) = x2(a + x)

Ans:

Ans:

33. Trace the curve r = a (cosθ + secθ) Hint: r2 = ar (cosθ + secθ). Therefore, cartesian form is y2(x – a) = x2(2a – x)

Ans:

Ans:

34. Trace the curve r = 2 sin θ Hint: r = arr cos θ = 2 y (a – x) = x 2

3

2

36. Trace the curve x = a(θ –sinθ), y = a (1 + cosθ).

a sin 2θ cos θ ay x

2

.

or x 2 + y 2 =

ay 2 x

or

Ans:

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5

Functions of Several Variables

Let n be a positive integer and ℜ be the set of n real numbers. Then, ℜ is the set of all n-tuples (x1, x2,…, xn), xn ∈ ℜ. Thus, ℜ is the set of all real numbers called, the real line, ℜ2 = ℜ × ℜ = {(x, y): x, y ℜ} is a twodimensional Cartesian plane, ℜ3 = ℜ × ℜ × ℜ = {(x, y, z): x, y, z ℜ} is a three-dimensional Euclidean space. n Let A be a nonempty subset of ℜ . Then, a function f : A → ℜ is called a real-valued function of n variables defined on the set A. Thus, f maps (x1, x2, … , xn ), xi ∈ ℜ into a unique real number f (x1, x2, … , xn). A function f of n variables x1, x2,…, xn is said to tend to a limit l as (x1, x2,… , xn) → (an, a2,…, an ) if given e > 0, however small, there exists a real number d  > 0 such that | f (x1, x2, … , xn) - l| < e whenever |(x1, x2, … , xn) - (a1, a2, … , an)| < d 

or

| f (x1, x2, … , xn) - l| < e,

whenever ______________________________ 0 < ​√(x1 - a1)2 + (x2 -     a2)2 + … + (xn - an)2 ​< d.

In this case, we write

If we put y = mx, then xy mx2 ​  lim  ​______      ​  2  2 ​  =     ​lim  ​ ​ ________    ​  2 x →0 x + m2x2 (x, y) → (0, 0) x + y m . =     ​lim  ​______ ​     ​  x →0 1 + m2 Thus the limit depends upon the value of m and so is not unique for various path followed from (x, y) to (0, 0). Hence the limit does not exist. EXAMPLE 5.2

Show that

x2 - y2 ​  lim  ​xy ​ ______         ​= 0. (x.y) →(0, 0) x2 + y2 Solution.  Putting x = r cos q, y = r sin q, we have x2 - y2 - 0 ​= ​ r 2 sin q cos q cos 2q ​ ​xy​  ______   ​  2 2 x +y 2 2 x2 + y2. __ __ ______ = ​ ​ r    ​ sin 4q ​0 x0 2 fy (x, y) =  __ 1 2 - __ ​ 1 ​ ​ ​_​ xy ​ ​  ​​   ​ , y 0         __ fx (x, x) = fy (x, x) = 2     - __ ​ 1 ​ , x < 0. 2 If Taylor’s expansion about (x, x) for n = 1 were possible, then we should have f  (x + h, x + h) = f  (x, x) + h [ fx  (x + q h, x + q h) + fy (x + q h, x + q h)] or   |x| + h if x + q h > 0 (1) |x + h| = |x| - h if x - q h < 0 . |x| if x - q h = 0

( |  | ) ( |  | )

{

{

Now if the domain (x, x; x + h, x + h) includes the origin, then x and x + h are of opposite signs. Thus, either | x + h| = x + h, | x| = - x or |x + h| = - (x + h), | x| = x. But under these conditions, none of the inequalities (1) holds. Hence, the expansion is not valid.

f  (x, y) = f  (a + h, b + k) = f  (a, b) + df   (a, b) 2 + __ ​ 1  ​ d  f  (a, b) + __ ​ 1  ​ d 3f  (a, b) 2! 3! __ 1 4 + ​    ​ d  f  (a, b) + … 4! In the present example, a = 1 and b = - 2. Thus, f  (x,y) = x2y +  3y - 2   which yields f (1, - 2) = -10, fx  (x, y) = 2xy which yields fx  (1, - 2) = - 4, 2 fy  (x, y) = x + 3 which yields fy  (1, - 2) = 4, which yields fxx  (1, - 2) = - 4, fxx  (x, y) = 2y fxy  (x, y) = 2x which yields fxy  (1, - 2) = 2, fyy (x, y) = 0 which yields fyy  (1, - 2) = 0, fxxx  (x, y) = 0 which yields fxxx  (1, - 2) = 0, fyyy  (x, y) = 0 which    yields fyyy  (1, - 2) = 0, and

fyxx  (1, - 2) = fxxy  (1, - 2) = 2. All other higher derivatives are zero. Hence, ∂f f  (x, y) = f  (a, b) + ​ (x - 1) ​ __ ​ (1, - 2) ∂x

[ 

]

∂f     + ( y + 2) ​ __  ​  (1, - 2)  ​ ∂y ∂ ∂ 2      + __ ​ 1  ​ ​ (x - 1) ​ __  ​+ ( y + 2) ​ __  ​ ​ f  (1, - 2) ∂x ∂y 2!

[  [ 

] ]

∂ ∂ 3      + __ ​ 1  ​ ​ (x - 1) ​ __  ​+ (y + 2) ​ __  ​ ​ f  (1, - 2) ∂x ∂y 3!     = - 10  - 4  (x - 1) + 4  ( y + 2)      + __ ​ 1  ​ [ - 4 (x - 1)2 + 4  (x - 1)  ( y +2)] 2!     + __ ​ 1  ​ [ (x - 1)3  (0) + 3  (x - 1)2  ( y + 2)  (2) 3!       + 3  (x - 1)  ( y + 2)2  (0) + ( y + 2)2  (0)]     = -10  - 4  (x - 1) + 4  (y + 2) - 2  (x - 1)2       + 2  (x - 1) (y + 2) + (x - 1)2  ( y + 2). EXAMPLE 5.50

( 

)

EXAMPLE 5.49

Expand sin (xy) in power of (x - 1) and ​ y - __ ​ p ​   ​ 2 up to and including second-degree terms.

Expand x2 y + 3y - 2 in powers of (x - 1) and (y + 2) using Taylor’s theorem for several variables. Solution.  For all points in the domain a ≤ x < a + h and b ≤ y ≤ b + k, the Taylor’s theorem asserts that

Solution.  We want to expand sin (xy) about the p point ​ 1, __ ​   ​   ​. By Taylor’s theorem, we have 2 f (x, y) = f (a + h, b + k) ∂ ∂ = f (a, b) + ​ h ​__    ​  + k​ __ ​  ​f (a, b) ∂x ∂y

M05_Baburam_ISBN _C05.indd 22

(  )

[ 

]

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functionS of Several variableS  n 5.23

[ 

]

∂ ∂ 2 + __1​   ​ ​ h​ __  ​+   k​ __  ​  ​ f (a, b) +... ∂y 2 ∂x But f (x, y) = sin xy        implies f ​ 1, __ ​ p ​   ​= 1, 2 fx  (x, y) = y cos xy      implies fx  ​ 1, __ ​ p ​   ​= 0, 2 __ fy  (x, y) = x cos xy     implies fy  ​ 1, ​ p ​   ​= 0, 2 __ p  __2 ​ , 2 fxx  (x, y) = - y sin xy   implies fxx  ​ 1, ​   ​   ​= - ​ p  2 4 p  p  fxy  (x, y) = - xy sin xy   implies fxy  ​ 1, __ ​   ​  ​= ​ y - __ ​   ​  ​, 2 2 and __ fyy  (x, y) = - x2  sin xy  implies fyy  ​ 1, ​ p   ​   ​= - 1. 2 Hence, __2 ​ (x - 1)2 - ​ p  __ ​ (x - 1) ​ y - ​ p  __ ​   ​ f (x, y) = 1 - ​ p  8 2 2 __ ​   2​ . - __ ​ 1 ​ ​ y - ​ p  2 2

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( 

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)

)

EXAMPLE 5.51

Expand e xy at (1, 1) in powers of (x - 1) and (y - 1). Solution.  We have f (x, y) = e . By Taylor’s theorem, we have xy

f (x, y) = f (a + h, b + k) ∂ ∂ = f (a, b) + ​ h ​__     ​+ k ​__     ​ ​ f (a, b) ∂x ∂y ∂ ∂ 2 + __ ​ 1  ​ ​ h​ __   ​+ k​ __  ​ ​    f (a, b) + . . . . ∂y 2! ∂x But, xy f (x, y) = e       implies f (1, 1) = e, fx  (x, y) = y e xy  implies fx  (1, 1) = e, implies fy  (1, 1) = e, fy  (x, y) = x e xy fxx  (x, y) = y2 e xy implies fxx  (1, 1) = e,

( 

( 

)

)

fxy  (x, y) = xy e xy + e xy

implies fxy  (1, 1) = e + e = 2e,

and fyy  (x, y) = x2 e xy

implies fyy  (1, 1) = e.

We have h = x - a = x - 1 and k = y - b = y - 1. Hence, f (x, y) = f (1, 1) + (x - a) fx(1, 1) + ( y - k)  ×  fy  (1, 1) + __ ​ 1  ​ (x - 1)2 fxx  (1, 1) 2!        + ( y - 1)2   fyy  (1, 1)       



+ 2  (x - 1) ( y - 1) fxy  (1, 1)

M05_Baburam_ISBN _C05.indd 23

      = e + (x - 1) e + ( y - 1) e + __ ​ 1  ​ [(x - 1)2 e 2!        + 4  (x - 1) ( y - 1) e + ( y - 1)2 e] + …

{ 

1  ​ (x - 1)2        = e ​ 1 + (x - 1) + ( y - 1) + ​ __ 2!         + 4 (x - 1) ( y - 1) + ( y - 1)2  ​+ …

}

EXAMPLE 5.52

Expand eax sin by in power of x and y as far as terms of third degree. Solution.  We have f (x, y) = e ax sin by. By Taylor’s theorem for function of two variables,

f (x, y) = f (a + h, b + k) ∂ ∂ = f (a, b) + ​ h​ __   ​+ k​ __   ​ ​f (a, b) ∂x ∂y

( 

)

( 

)



∂ ∂ 2 + __ ​ 1  ​ ​ h​ __   ​+ k​ __   ​ ​ f (a, b) ∂y 2! ∂x



∂ ∂ 3 + __ ​ 1  ​ ​ h​ __  ​+   k​ __  ​  ​ f (a, b) + … . 3! ∂x ∂y

( 

)

We wish to expand the function about (0, 0). So h = x - 0 = x and k = y - 0 = y. f (x, y) = f (0 + h, 0 + k)

(

)



∂ ∂ = f (0, 0) + ​  x __ ​    ​+ y​ __  ​ ​  f (0, 0) ∂x ∂y



∂ ∂ + __ ​ 1  ​ ​  x​ __  ​+ y​ __  ​ ​ f (0, 0) ∂y 2! ∂x



1 ∂ ∂ 3 + ​ __  ​ ​ x __ ​     ​+ y​ __   ​ ​ f (0, 0) + … ∂y 3! ∂x

( ( 

) )

2

But, f (x, y) = eax sin by implies f (0, 0) = 0, fx (x, y) = aeax sin by implies fx (0, 0) = 0, fy (x, y) = beax cos by implies fy (0, 0) = b, fxx (x, y) = a2eax sin by implies fxx (0, 0) = 0, fxy (x, y) = abeax cos by implies fxy (0, 0) = ab, fyy (x, y) = - b2eax sin by implies fyy (x, y) = 0, fxxx (x, y) = a3eax sin by implies fxxx (0, 0) = 0, fxxy (x, y) = a2beax cos by implies fxxy (0, 0) = a2b, fxyy (x, y) = - b2aeax sin by implies fxyy (0, 0) = 0, fyyy (x, y) = - b3eax cos by implies fyyy (0, 0) = - b3,

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5.24  n  chapter five and so on. Hence,

f (x, y) = by + ab xy + __ ​ 1  ​ (3a2 bx2 y - b3 y3) + … 3!

EXAMPLE 5.53

y Expand f (x, y) = tan-1 _​ x ​in the neighborhood of (1, 1) up to third-degree terms. Hence, compute f (1.1, 0.9) approximately. Solution.  We note that y __ ,  ​   f (x, y) = tan- 1 ​ _x ​implies f (1, 1) = ​ p  4 -y fx (x, y) = ​ ______    ​  implies fx (1, 1) = - __ ​ 1 ​  , 2 x2 + y2 fy (x, y) = ______ ​  2 x  2   ​implies fy (1, 1) = __ ​ 1 ​  , 2 x +y 2xy ________ fxx (x, y) = ​  2      ​implies fx (1, 1) = __ ​ 1 ​  , 2 (x + y2)2 y2 - x2 ________ fxy (x, y) = ​  2 2 2   ​implies fxy (1, 1) = 0, (x + y ) 2xy fyy (x, y) = - ________ ​  2      ​implies fyy (1, 1) = - __ ​ 1 ​  , 2 (x + y2)2 2 2 2y(y - 3x ) fxxx (x, y) = ​ __________  ​    implies fxxx (1, 1) = - __ ​ 1 ​  , 2 (x2 + y2)3 2 2x(x - 3y2) __________ __ fxxy (x, y) = ​  2 2 3 ​    implies fxxy (1, 1) = - ​ 1 ​  , 2 (x + y ) 2y(3x2 - y2) __________ __ 1 fxyy (x, y) = ​  2 2 3 ​    implies fxyy (1, 1) = ​   ​  , and 2 (x + y ) 2x (3y 2 - x 2) ___________ fyyy (x, y) = ​      ​  implies fyyy (1, 1) = __ ​ 1 ​ . 2 (x2 + y2)3 Therefore, by Taylor’s theorem, we have y f (x, y) = tan-1 _x​  ​  ∂ ∂ = f (1, 1) + ​ (x - 1)​ __  ​+   ( y - 1)​ __ ​  ​ f (1, 1) ∂x ∂y ∂ ∂ 2 + __ ​ 1  ​  (x - 1)​ __  ​+ ( y - 1) __ ​    ​ f (1, 1) ∂x ∂y 2! 1 ∂3 ∂ __ __ __ + ​    ​  (x - 1) ​   ​  + ( y - 1) ​   ​   f (1, 1) + … ∂y ∂x 3! __ = ​ p   ​ + (x - 1) ​ - __ ​ 1 ​   ​+ (y - 1) ​ __ ​ 1 ​   ​ 4 2 2 + __ ​ 1 ​  (x - 1)2 ​ __ ​ 1 ​   ​+ 2  (x - 1) (y - 1) (0) 2 2 __ 1 2 + ( y - 1) ​ - ​   ​   ​ 2 __ 1 3  + ​    ​  (x - 1) ​ - ​ __1 ​   ​+ 3(x - 1)2 ( y - 1) ​ - ​ __1 ​   ​ 2 2 3! + 3  (x - 1) ( y - 1)2 ​ __1​   ​   ​+ ( y - 1)3 ​ __1​   ​   ​ + … 2 2

[

[

[

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[ 

(  ) (  ) (  )] (  ) (  )

]

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]

(  )]

(  )

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]

__ __ = ​ p   ​ - ​ 1 ​  (x - 1) - ( y - 1) 4 2 + __ ​ 1 ​  (x - 1)2 + ( y - 1)2 4 ___ - ​ 1  ​  (x - 1)3 + 3 (x - 1)2 ( y - 1) 12



[

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]

- 3 (x - 1) ( y - 1)3 - ( y - 1)3 + . . . . Putting x = 1.1 and y = 0.9, we get f (1.1, 0.9) = 0.6857. 5.11  APPROXIMATION OF ERRORS In numerical computation, the quantity [True value - Approximate Value] is called the error. We come across the following types of errors in numerical computation. 1. Inherent Error (initial error). Inherent error is the quantity which is already present in the statement (data) of the problem before its solution This type of error arises due to the use of approximate value in the given data because there are limitations of the mathematical tables and calculators. This type of error can also be there due to mistakes by human. For example, some one can write, by mistake, 67 instead of 76. The error in this case is called transposing error. 2. Round - off Error. This error arises due to rounding off the numbers during computation and occur due to the limitation of computing aids. However, this type of error can be minimized by (i) Avoiding the subtraction of nearly equal numbers or division by a small number. (ii) Retaining at least one more significant figure at each step of calculation. 3. Truncation Error. If is the error caused by using approximate formulas during computation such as the one that arise when a function f (x) is evaluated from an infinite series for x after truncating it at certain stage. For example, we will see that in NewtonRaphson Method for finding the roots of an equation, if x is the true value of the root of f (x) = 0 and x0 and h are approximate value and correction respectively, then by Taylor’s Theorem, f (x0 + h) = f (x0) + hf ′  (x0) h2  ​f  ''  (x ) + … + = 0. + ​ __ 0 2!

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functionS of Several variableS  n 5.25 To find the correction h, we truncate the series just after first derivative. Therefore some error occurs due to this truncation. 4. Absolute Error. If x is the true value of a quantity and x0 is the approximate value, then | x - x0 | is called the absolute error. 5. Relative Error. If x is the true value of a quantity x -and x0 x0 is the approximate value, then ​ ​ _____   ​  ​is called the relative error.   x  

( 

)

6. Percentage Error. If x is the true value of quanx-x0 tity and x0 is the approximate value, then ​ ____ ​  x  ​   ​ × 100 is called the percentage error. Thus, percentage error is 100 times the relative error.

(  )

5.12  GENERAL FORMULA FOR ERRORS Let u = f (u1, u2, … , un) (1) be a function of u1, u2, . . . , un which are subject to the errors D u1, D u2, . . . , D un , respectively. Let D u be the error in u caused by the errors D u1, D u2, . . . , D un in u1, u2, . . . , un , respectively. Then u + D u = f  (u1 + D u1, u2 + D u2, . . . , un + D un) (2) Expanding the right hand side of (2) by Taylor’s Theorem for a function of several variables, we have u + D u = f (u1, u2, . . . , un )

( 

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∂   ​ + … + D u  ​ ___ ∂    ​ ​f + ​ D u1​ ___ n



∂   ​  + . . . + D u ​ ___ ∂    ​ 2​ f + … + __ ​ 1 ​ ​ D u1​ ___ n ∂u 2 ∂u

( 

∂u1

∂un

 

1

n

)

Since the errors are relatively small, we neglect the squares, product and higher powers and have u + D u = f (u1, u2, … , un)

( 

)

∂   ​  + … + D u ​ ___ ∂    ​ ​ f + ​ D u1​ ___ n ∂u ∂u1 n

(3)

Subtracting (1) from (3), we have

or

∂ f ∂ f ∂ f D u = ​ ___  ​ D u1 +​  ___  ​ D u2 + … +​  ___   ​D un ∂u ∂u   ∂u1  2 n

∂u   ​D u +​ ___ ∂u   ​D u + … + ___ D u =​ ___ ​ ∂u  ​D un, ∂u1 1 ∂u2 2 ∂un which is known as general formula for error. We note that the right hand side is simply the total

M05_Baburam_ISBN _C05.indd 25

derivative of the function u. For a relative error Er of the function u, we have D u2 un . D u ∂u D u1 ___ ∂u   ​​ D  ______ Er = ​ ___   ​= ​ ___   ​​ ______     ​+​  ∂u   ​​ ______     ​+ … +​  ___     ​ u ∂u1 u ∂u2 u ∂un u EXAMPLE 5.54

5xy2 If u = ____ ​  3 ​   and error in x, y, z are 0.001, compute z the relative maximum error (Er)max in u when x = y = z = 1. 5xy2 Solution.  We have u = ____ ​  3 ​   . Therefore z 2 10xy ___ 15xy2 ∂u 5y ___ ∂u ____ ∂u ___ ​    ​= ​ ___ ​= ​  3 ​   , ​    ​= - ​ _____  ​    3 ​ , ​     ∂x z ∂y z ∂z z4 and so 5y2 10xy 15xy2 ____ D u = ​ ___  D y - ​ _____  ​   D z 3 ​ D x + ​  3 ​  z z z4 But it is given that D  x = D  y = D  z = 0.001 and x = y = z = 1. Therefore

 

  

  

5y2 10xy 15xy2 ____ _____ (D u)max ≈ ​​ ___  ​   D  x  ​ + ​   ​  ​     D  y  ​ + ​ ​   ​    D z ​ z3 z3 z4 = 5 (0.001) + 10 (0.001) + 15 (0.001) = 0.03. Thus the relative maximum error (Er)max is given by (D u)max ____ 0.03 ____ 0.03 (Er )max = ​ ______       ​= ​       ​= ​   ​   = 0.006. u u 5



EXAMPLE 5.55

The diameter and altitude of a can in the shape of a right circular cylinder are measured as 4 cm and 6 cm respectively. The possible error in each measurement is 0.1 cm. Find approximately the maximum possible error in the values computed for the volume and the lateral surface. Solution.  If x and y denote the diameter and the height of the can, then volume of the can is given by p  V = ​ __ ​ x2y and so 4 __ __ 2 p  ∂V  ​= ​ p  ​ ___  ​ xy and ___ ​ ∂V ∂y  ​= ​ 4 ​ x . ∂x 2 Therefore the error formula yields __ __ 2 ∂V  ​D x +​ ___ ∂V  ​D y = ​ p  D V =​ ___  ​ (xy D x) + ​ p   ​ (x D y). 2 4 ∂x ∂y Putting x = 4, y = 6, D x = D y = 0.1, we get D V = (1.2) p  + (0.4) p  = 1.6 p  cm3. Further, the lateral surface is given by S = p xy and so

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5.26  n  chapter five

Therefore

∂S ∂S  ​= p y  and ​ ___ ​ __   ​= p x. ∂y ∂x

D S = p  ( yD x + xD y). Putting the values of x, y, D x and D y, we get D S = p (0.6 + 0.4) = p  cm2. EXAMPLE 5.56

The height h and the semi-vertical angle a of a cone are measured and from them the total area A of the cone (including the base) is calculated. If h and a are in error by small quantities d h and d a respectively, find corresponding error in p  the area. Show further that a = ​ __ ​ , an error of 1 6 percent in h will be approximately compensated by an error of -0.33 degree in a. Solution.  Radius of the base = r = h tan a. Further, slant height = l = h sec a. Therefore, Total area = p r2 + p r l = p  r (r + l ) = p h tan a (h tan a + h sec a) = p h2 (tan2 a + sec a tan a). Then the error in A is given by ∂A ∂A ​   d  h+​ ___ d A =​ ___   ​d a ∂a ∂h  = 2p h (tan2 a + sec a tan a ) d h + p h2 (2 tan a sec2 a + sec3 a + sec a tan2 a ) d a For the second part of the question, __ a = ​ p   ​ , d h = ____ ​  h   ​  . 6 100 Therefore d A = 2p h ​ __ ​ 1 ​ + __ ​ 2 ​   ​____ ​  h   ​  3 3 100 + p h2 ​ ___ ​  2__   ​ ​ __ ​ 4 ​   ​+ ____ ​  8__    ​ + ____ ​  2__    ​  ​d a ​ 3 ​  3 3​√3 ​  3​√3 ​  √ __ ___ h2 ​ + 2​√3 ​ p h2 d a = ​ p  (1) 50 But after compensation d A = 0. Therefore (1) implies d a = - ______ ​  1 __ ​  radians = - _____ ​ 57.3°   ​= - 0.33°. 173.2 100​√3 ​ 

[ 

] (  (  )

)

EXAMPLE 5.57

The time T of a complete oscillation of a simple pendulum of length L is governed by the equation ___ T = 2p  ​√L/g ​ , g is constant. Find the approximate

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error in the calculated value of T corresponding to the error of 2% in the value of L. Solution.  We have

__ . T = 2p  ​ L__ g ​

√

Taking logarithm, we get log T = log 2p  + __ ​ 1 ​ log l - __ ​ 1 ​ log g (1) 2 2 Differentiating (1), we get d g __ ​ 1 ​d T = __ ​ 1 ​ ​ d __L  ​  - __ ​ 1 ​ ​ ___ ​  T 2 L 2 g or d g ___ __ T ​ × 100 = __ ​ d  ​ 1 ​ ​ ​ dL  ​  × 100 - __ ​ 1 ​ ​ g__ ​ ×  100  ​ T 2 L 2

[ 

]

= __ ​ 1 ​ [2 - 0] = 1. 2 Hence the approximate error is 1%.

EXAMPLE 5.58

The radius of circle is found to be 100 cm. Find the approximate error in the area of the circle due to an error of 1 mm in the measured value of the radius. Also, find relative error. Solution. Area of circle is A = p r2 (1) On differentiation d A = 2p r d r where r = 100 cm, and d r = 1 mm = 0.1 cm

∴ d A = 2p  × 100 × 0.1   = 20p  sq. cm.

Now taking log both sides of Eq. (1), we have log A = log p  + 2 log r On differentitation, we get 1 2 __ __ ​    ​d A = 0 + ​ r ​  d r A Putting r = 100 cm and d r = 0.1 cm, we have 2 2 _____ d A ____ Relative Error = ​ ___ ​ = ​     ​ × 0.1 = ​     ​  = 0.002 cm 1000 A 100 EXAMPLE 5.59

The radius of a sphere is found to be 10 cm with a possible error of 0.02 cm. What is relative error in computing the volume?

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functionS of Several variableS   n 5.27 Solution. We know that the volume of sphere is ​ 4 ​p r3 V = __ 3 4 __ i.e., log V = log ​ ​   ​  p  ​+ 3 log r 3 Given r = 10 cm. and d r = 0.02 cm On differentitation of Eq. (1), we have 3 1 __ __ ​ v ​d V = 0 + ​ r ​  d r 0.02 ____ __r Relative Error ​ d   ​ = 3 × ​   ​  10 V = 0.006 cm.

= 1% + 1% = 2% Percentage Error in A = 2% (1)

EXAMPLE 5.60

Find the percentage error in the area of an ellipse when an error of 1% is made in measuring the major and minor axis. Solution.  The area of an ellipse is given by A = p xy where x and y are the semi-major and semi-minor axis. Taking both side log log A = logp  + log x + log y and so 1 1 1 __ __ __ ​   ​ d A = 0 + ​ x ​d x + ​ y ​d y A Now, multiplying throughout by 100, we have d  d  1 __ ___x ___y ​   ​ d A × 100 = ​  x ​ × 100 + ​  y ​ × 100 A Hence Percentage Error = 1% + 1% = 2%. EXAMPLE 5.61

Find the percentage error in the area of rectangle when an error of 1% is made in measuring its length and breadth. Solution.  Let x and y be the length of sides the area of rectangle A = xy. Taking log of both sides, we get log A = log x + log y Therefore 1 1 1 __ __ __ ​   ​  d A = ​ x ​ d x + ​ y ​ d y A On multiplying both side by 100, we get y d  A ___ d x ​ 100 + ​ d  ___ ​   ​ × 100 = ​ ___ x y ​ × 100 A Hence Percentage Error in A = Percentage Error in x + Percentage Error in y

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EXAMPLE 5.62

Deflection of a beam varies as the cube of the depth. What is the percentage error in the calculated deflection due to on error of 1.75% in the measurement of depth? Solution.  Let y be the deflection and D be the depth of beam then y = KD3, where K is a constant. Now taking log of both sides, we have log  y = log k + 3 log D On differention, we have 3 1 __ __ ​ y ​d y = 0 + ​    ​d D. D Therefore d  ___y D ​ 100 ___ ​  y ​ × 100 = 3 × ​ d  D = 3 × (% Error in depth) = 3 × 1.75 = 5.25%, which is required percentage error. EXAMPLE 5.63

The indicated horse-power I of an engine is calPLAN culated from the formula I = ______ ​   ​  . Where A 33000 p  __ 2 = ​   ​  d  . Assuming that errors of r percent may 4 have been made in measuring P, L, N and d. Find the greatest possible error in I. Solution.  Here,

PLAN ______ I = ​   ​  33000 Putting p  __ A = ​   ​  d 2 we get 4 p  __ PLN I = ​   ​  ​ ______  ​  4 33000 or p PLd 2N d 2 = ​ _______ ​  132000 Taking log on both sides, we have log I = log p  + log P + log L + 2log d + log N - log 132000 On differentitation, we have d d ​ d N ​ - 0 __ ​ 1 ​ d I = 0 + __ ​ 1 ​d P + __ ​ 1 ​d L + 2 ___ ​   ​ + ___ I P L N d Multiplying both sides by 100, we get

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5.28  n  chapter five

d  d L d  d __I ___ ___ d P ​   ​ × 100 = ​ ___ ​ × 100 + ​   ​ × 100 + 2 ​   ​ × 100 I L d P d  N ___ + ​   ​ × 100 N Now putting, d L ___ N ​  d ​ = 100 × ​ d  ___ ___ 100 × ___ ​ d P ​ = 100 × ​   ​ = 100 × ​ d  L N d P = r. Therefore, percentage error is d  __I ∴ ​   ​ × 100 = r + r + 2r + r I = 5r% Thus the greatest possible error in I = 5r%. EXAMPLE 5.64

The side a and the opposite angle A of a triangle ABC remains constant, show that when the others sides and angles are slightly varied, d b d c _____ _____ ​    ​ + ​ cos  c ​ = 0 cos b Solution.  We know that by the sine formula, a b c _____ _____ _____ ​     ​  = ​     ​  = ​     ​  sin A sin B sin C ∴ a sin B = b sin A (1) and a sin C = c sin A (2) Since a and A are constants, we have from Eq. (1) and (2) on taking differentials   a cos B d  B = sin  A d b and

a cos C d  C = sin A d c

f (x + d x) - f (x) = f ′ (x) d x

or

f (x + d x) - f (x) = ___ ​  1  __2 ​ · d x

3x ​ 3 ​  We may write 127 = 125 + 2 Taking x = 125 and d x = 2, then we have f (127) - f (125) = _______ ​  1   ​  ×2  __ 2

3(125) ​ 3 ​ 

Therefore

2 ______ f (127) = f (125) + ​     ​  3 × 25  __ 1 2 ___   = (125) ​ 3 ​ + ​    ​  75 377 2 ____ ___        = 5 + ​    ​ = ​   ​  75 75 377 ____         = ​   ​  75

5.13  TANGENT PLANE AND NORMAL TO A SURFACE Let P(x,y,z) and Q(x+ d x, y+d y, z+d z) be two neighboring points on given surface F(x,y,z) = 0. Let d s and d c be the length of the are PQ and δs that of the chord PQ. Then lim δc = 1. Q−P The direction cosines of the chord PQ are

δx δ y δz δ x δs δ y δs δ z δs , , or . , , . δc δc δc δs δc δs δc δs δc Z

ad B d c ad c _____ _____ _____ d b  ​  \ ​ _____ = ​    ​   and ​     ​  = ​    ​  cos C sin A cos B sin A

Q

d A + d B + d C = 0 ⇒ d  B = - d  C Hence (3) yields d b d c _____ _____ ​    ​  + ​     ​  =0 cos B cos C EXAMPLE 5.65

Find the approximate value of cube root of 127.  __ 1

Solution.  Let f (x) = x ​ 3 ​  Now

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N

P

o

F(x,y,z)=o

s

c

d c a _____ _____ d b  ​     \ ​ _____ + ​     ​  = ​     ​  (d B + d C ) (3) cos B cos C sin A Since A + B + C = 180°  and  A is a constant, we note that

T

y

x

If d sà0, then QàP and PQ tends to tangent line PT. Since co-ordinates of points on the are PQ are functions of s, the direction cosines of PT dy dz are dx Differentiating F(x,y,z) = 0 with ds , ds , ds . respect to s, we have

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functionS of Several variableS  n 5.29 ∂F dx ∂F dy ∂F dz . + . + . = 0. ∂x ds ∂y ds ∂z ds

(1)

Expression (1) shows that the tangent line PT dy dz with direction cosines dx is perpendicular ds , ds , ds . to the line having direction ratios ∂∂Fx , ∂∂Fy , ∂∂Fz . Corresponding to different curves joining P and Q, there are distinct tangent lines at P to which the line with direction ratios ∂∂Fx , ∂∂Fy , ∂∂Fz is perpendicular. Thus all tangent lines at P lie in a plane through P perpendicular to the line with direction ratios ∂∂Fx , ∂∂Fy , ∂∂Fz . Hence the equation of the tangent plane to the surface F(x,y,z) = 0 is given by ∂F ∂F ∂F ( X − x) + (Y − y ) + ( Z − z ) = 0. ∂x ∂x ∂z

Then the equation of the normal to the surface at P is given by X −x ∂F ∂x

=

Y−y ∂F ∂y

=

Z − z. ∂F ∂z

EXAMPLE 5.67

Find the equations of the tangent plane and the normal to the surface z2 = 4(x2 + y2 + 1) at the point (2, 2, 6). Solution. The given surface is F(x, y, z) = 4x2 + 4y2 – z2 + 4. Therefore, ∂F = 8 x = 16 at (2, 2, 6) ∂x ∂F −= 8 y − 16 at (2, 2, 6) ∂y ∂F = −2 z = −12 at (2, 2, 6). ∂z Therefore the equation of the tangent plane is 16(x – 2) + 16(y – 2) – 12(z – 6) = 0 or

4x + 4y – 3z + 2 = 0.

The equation of the normal to the surface is x−2 y−2 z−6 = = . 18 16 −12

EXAMPLE 5.66

Find the equations of the tangent plane and the normal to the surface 2xz2 – 3xy – 4x = 7 at the point (1, –1, 2).

or

x−2 y−2 z−6. = = . 4 4 −3

Solution. The given surface is F(x, y, z) = 2xz2 – 3xy – 4x – 7. Therefore, ∂F = 2 z 2 − 3 y − 4 = 7 at (1, −1, 2) ∂x ∂F = −3 x = −3 at (1, −1, 2) ∂y ∂F = 4 xz = 8 at (1, −1, 2). ∂z

Therefore equation of the tangent plane at (1, –1, 2) is or

7(x – 1) –3(y + 1) + 8(z – 2) = 0 7x – 3y + 8z – 26 = 0.

The equation of the normal to the surface at (1, –1, 2) is x7−1 = y−+31 = z −8 2 .

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EXAMPLE 5.68

Find the equation of the tangent plane to the surface 2x2 + y2 – 3 + 2z = 0 at (2, 1, –3). Solution. We have

F(x, y, z) = 2x2 + y2 + 2z – 3.

Therefore, ∂F ∂F ∂F = 4 x, = 2 y, = 2. ∂x ∂y ∂z At (2, 1, –3), we have ∂F ∂F ∂F = 8, = 2, = 2. ∂x ∂y ∂z

Therefore, the equation of the tangent plane to the given surface at the point (2, 1, –3) is 8(x – 2) + 2(y – 1) + 2(z + 3) = 0

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5.30  n  chapter five or

4x + 2y + 3z – 6 = 0

EXAMPLE 5.69

Show that the plane 3x + 12y – 6x – 17 = 0 touches the conicoid 3x2 – 6y2 + 9z2 + 17 = 0. Find also the point of contact. Solution. Comparing the equation of the plane with lx + my + nz = p, we get l = 3, m = 12, n = –6, p = 17.



The equation of the given conicoid in standard form is −

3 2 6 2 9 2 x + y − z =1 17 17 17 (ax2 + by2 + cz2 = 1 form).

The plane will touch the conicoid if l 2 m2 n2 + + = p 2 (condition of tangency) a b c or if

9 −3 17

or if

+

144 6 17

+

36 −9 17

= (17) 2

–51 + 408 – 68 = 289, which is true.

Now let P(x1, y1, z1) be the point of contact. The equation of the tangent plane to the conicoid at (x1, y1, z1) is 3xx1 – 6yy1 + 9zz1 = –17. Comparing it with 3x + 12y – 62 = 17, we get x1 6 y 9z −17 =− 1 = 1 = . 1 12 −6 17 Therefore x1 = –1, y1 = 2, z1 = 32 . Hence, the point of contact is (–1, 2, 32 ). . 5.14  JACOBIANS If u1, u2,… ,un are n functions of n variables x1, x2, … , xn, then the determinant

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∂u ∂u ∂u … . ​ ___1  ​ ​ ___1 ​  ​ ___1 ​  … . ∂x1 ∂x ∂x ∂u22 ∂un2 ∂u2 ___ ___ ___ … . ​    ​ ​   ​  ​   ​  … . ∂x2 ∂xn ∂x1 … . … . … . … . … . … . … . … . … . … . ∂u ∂u ∂u … . ​ ___n  ​ ​ ___n ​  ​ ___n ​  … . ∂x1 ∂x2 ∂xn is called the Jacobian of u1, u2, … , un with regard to x1, x2, … , xn. This determinant is often denoted by​ ∂ (u1,  u2, … ,  un) _____________    ​or J (u1, u2, … , un). ∂ (x1,  x2, … ,  xn) 5.15  PROPERTIES OF JACOBIAN Theorem 5.7. If U, V are functions of u and v, where u and v are themselves functions of x and y, then ∂ (u, v) . ∂ (U, V) _______ ∂ (U, V) ______ ​ _______ ​  = ​   ​  · ​   ​   ∂ (x, y) ∂ (u, v) ∂ (x, y) Proof:  Let U = f (u, v), V = F  (u, v), u = f  (x, y), and v = ψ  (x, y). Then, ___ ∂u  ​+ ___ ∂U ​ · ​ ___ ​ ∂U ​ =​ ___ ​ ∂U ​ · __ ​ ∂v ​ , ∂x ∂u ∂x ∂v ∂x ∂U ∂v , ___ ∂U  ​· ___ ​ ∂U ​ =​ ___ ​ ∂u  ​+ ___ ​   ​ · __ ​   ​  ∂y ∂u ∂y ∂v ∂y ∂V ___  ​· ___ ​ ___  ​=​  ∂V ​ ∂u  ​+ ___ ​ ∂V  ​· __ ​ ∂v ​ , ∂x  ∂u ∂x ∂v ∂x ∂V  ​= ___ ∂V  ​· __ ​ ___ ​ ∂V  ​· ___ ​ ∂u  ​+​ ___ ​ ∂v ​ . ∂y ∂v ∂y ∂y ∂u and so, ∂ (U, V ) ______ ∂ (u, v) ​ _______ ​  · ​   ​  ∂ (u, v) ∂ (x, y) ∂U ​  ​ ___ ∂U ​  ___ ​ ___ ​ ∂u  ​ ___ ​ ∂u  ​ ∂u ∂v ∂x ∂y  = ∂V  ​ ​ __ ∂v ​ ​ __ ∂v ​ ∂V  ​  ​ ___ ​ ___ ∂u ∂v ∂x ∂y ∂u  ​+ ​ ___ ∂U ​ · ​ __ ∂U ​  · ​ ___ ∂v ​ ∂u  ​+ ___ ∂U ​ · ​ ___ ​ ∂u  ​· __ ​ ∂v ​ ​ ___ ​ ___ ∂u ∂y ∂v ∂y ∂u ∂x ∂v ∂x ∂u  ​+ ​ ___ ∂v ​ ∂V  ​· ​ ___ ∂V  ​· ​ ___ ∂u  ​+​  ∂V ___  ​· __ ∂V ​ · ​ __ = ​ ___ ​ ∂v ​ ​ ___ ∂u ∂x ∂v ∂x ∂u ∂y ∂v ∂y ___ ∂U ​  ​ ∂U ​  ​ ___

= 

∂x ∂y

∂V  ​ ​ ___ ∂V  ​ ​ ___ ∂x ∂y

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functionS of Several variableS  n 5.31 ∂  (U, V) ∂  (x, y)

. = ​ _______ ​  Theorem 5.8. If J is the Jacobian of the system u, v with regard to x, y, and J’ is the Jacobian of x, y with regard to u, v, then JJ ′  = 1. Proof:  Let u = f (x, y) and v = F  (x, y). Suppose that these are solved for x and y giving x = f   (u,v) and y = ψ (u, v). Differentiating u = f (x, y) with respect to u and v, we have ∂y ∂y ∂u ∂u ∂u ∂u ∂x​  + ___ ∂x 1 = ___ ​   ​ ·​ ___ ​   ​ · ___ ​   ​   and  0 = ___ ​   ​ · __ ​  ​  + ___ ​   ​ · ​ __ ​  . ∂x ∂u ∂y ∂u ∂x ∂v ∂y ∂v Similarly, differentiating v = F (x, y) with respect to u and v, we get y ∂​  y​   . ∂x __​  · ​ ___ __​  · __ __ ​ · __ ∂x ​+ ∂v 0 =​  ∂v   __ ​   ​ . ___ ​ ∂ ​    and  1 =​  ∂v ​    ​ +​  ∂v ∂x ∂u ∂y ∂u ∂x ∂v ∂y ∂v Therefore, ∂x ​ ​ __ ∂u  ​ ___ ∂x ​ ​ __ ​ ∂u  ​ ​ ___ ∂x ∂y ∂u ∂v JJ ′  = __ ∂y __ ∂v ​ ​ __  ​ ​ ∂v ​ ​ ∂v ​ ​ __ ∂x ∂y ∂u ∂v ∂y __ __ ∂u  ​+ ___ ∂u ∂u  ​· ___ ​ ∂u  ​· ___ ​   ​ ​ __ ​ ∂u  ​· ​ ∂v ​    ​· ​ ___ ​ ∂x  ​+ ___   ​ ∂x ∂v ∂x ∂u ∂x ∂v ∂y ∂v = y ∂y ​ ​ __ ∂y ​· ___ ∂v ​· ___ __ ​· __ ​ ∂v ​· ​ ___ ​ ∂x ​+ __ ​ ∂x  ​+​  ∂v ​ ∂  ​ ​ __ ∂x ∂u ∂y ∂u ∂x ∂v ∂y ∂v

1 = 0

0 1 = 1.

EXAMPLE 5.70

If x = r cos q and y = r sin q, show that ∂ (x, y) ∂ (r, q ) __ (i) ​ ______   ​= r  and  (ii) ​ ______ ​  = ​ 1 ​ . ∂ (r, q ) ∂ (x, y) r Solution.  (i) We have ∂ x  ​ ​ ___ ∂ x   ​ cos q -r sin q  ​ ___ ∂ (x, y) ∂ r ∂ q  ______ ​     ​= = ∂ (r, q ) ∂ y ∂ y ​ ___  ​ ​ ___  ​ sin q r cos q ∂ r ∂ q = r cos2 q + r sin2 q = r. (ii) We have y r 2 = x 2 + y 2 and tan q = _​ x .​ Differentiating partially with respect to x and y, we get ∂r ∂r ​= _​ x ​ , 2r ​ __ ​ = 2x and so, ​ __ ∂x r ∂x ∂r ​= 2y and so, ​ __ ∂r ​= _​ y ​ , 2r ​ __ ∂y ∂y r

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y y ∂___ q __ q _______ sec2 q ​ ∂___  ​ ∂x  ​= - ​ x2  ​ and so,​  ∂x  ​= - ​ x2 sec 2 q  y y = - ____________ ​  2      ​= - ​ __2  ​  , r cos2 q sec2 q r  2 _____ q  ​= _______ q​  q  ​= __ sec2 q ​ ∂___ ​ 1 ​, and so, ​ ∂___ ​  1   ​   = ​ cos x    ∂y x sec2 q ∂y x 2 1 ​= __ = __ ​ x2 ​ ​ __ ​ x  ​  , r x r2 Therefore,

∂r ​ ​ __ ∂   (r, q ) ∂x ______ ​   ​  = ∂  (x, y) q  ​ ​ ∂___ ∂r

_ ∂r ​ ​ xr ​ ​ __ ∂y = y q  ​ - ​ __   ​  ​ ∂___ r2 ∂y

y  ​_r ​ __ x

 ​ 2  ​  r

2 y2 x______ + y2 __ x2 ​ + ​ __ r2 __ 1 = ​ __  ​   = ​   = ​  3  ​= ​   ​ . 3 3 3 ​  r r r r r

EXAMPLE 5.71

If x = r sin q cos f , y = r sin q sin f , and z = r cos q, show that ∂ (x, y, z) ​ ________     ​= r 2 sin q. ∂ (r, q, f ) Solution.  We have ∂x ​ ​ __ ∂r ∂y ∂ (x, y, z) __ ​ ________     ​= ​   ​ ∂r ∂ (r, q, f ) ∂z ​ ​ __ ∂r sin q cos f  = sin q sin f  cos q

∂x ​  ​ __ ∂q ∂y ​ ___ ​  ∂q ∂z ​  ​ ___ ∂q

∂x ​   ​ ___ ∂f ∂y ​ ___ ​  ∂f ∂z ​  ​ ___ ∂f

r cos q cos f  - r sin q sin f  r cos q sin f 

r sin q cos f 

- r sin q

0

= cos q (r2  sin q cos q cos2f  + r2  sin q cos q sin2f  ) + r sinq (r sin2q cos2 f  + rsin2 q sin2f) = r2  sin q cos2 q + r2  sin3 q = r2  sin q (cos2  q + sin2 q ) = r2  sin q. EXAMPLE 5.72

If u = x + y + z, uv = y + z, and uvw = z, show that ∂ (x, y, z) ​ _________   ​  = u2 v. ∂ (u, v, w)

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5.32  n  chapter five

z = uvw, y = uv - z = uv - uvw and x = u - y - z = u - uv + uvw - uvw = u - uv. Therefore, ∂x ​ ​ __ ∂x ​ ​ ___ ∂x   ​ ​ ___ 1 - v -u 0 ∂u ∂v ∂w ∂  (x, y, z) ___ ∂y ___ ∂y = ​ ________   ​= ∂y __ ∂  (u, v,w) ​ ∂u ​ ​ ∂v ​ ​ ∂w   ​ v - vw u - uw -uv ∂z ​ ​ __ ∂z   ​ ∂z ​ ​ ___ vw uw uv ​ ___ ∂u ∂v ∂w

1 - v -u 0 1 - v -u v u 0 = uv v u vw uw uv = uv (u - uv + uv) = u2v. =

x2x3 x3x1 x x2 If u1 = ____ ​   ​  , u2 = ____ ​   ​  , and u3 = ____ ​  1  ​   , show that x1

x2 x3 ∂   ( u , u , u ) 1 2 3 ​ __________  ​  = 4. ∂  (x1, x2, x3)

Solution.  We have ∂u ∂u ∂u ​ ___1 ​  ​ ___1 ​  ​ ___1 ​  ∂x1 ∂x2 ∂x3 ∂  (u1, u2, u3) ∂u ∂u ∂u __________ ​   ​  = ​ ___2 ​  ​ ___2 ​  ​ ___2 ​  ∂  (x1, x2, x3) ∂x1 ∂x2 ∂x3 ∂u3 ___ ∂u ∂u ___ ​   ​  ​  3 ​  ​ ___3 ​  ∂x1 ∂x2 ∂x3 x x3 x3 x2 - ____ ​  2 2  ​     ​  ​ __ ​ __ x x1 ​  x​1 ​  1 x x x x = __ ____ 3 1 __ ​ x3 ​  - ​  x​2  ​     ​ x1 ​   ​  2 2 2 x2 x1 x2 x____ __ __ ​ x  ​  ​ x  ​  - ​  1 2  ​    x​3 ​  3 3 -x2x3 = ______ ​   2 1 2   2 ​ x x x​1 ​ x​2 ​ x​3 ​  2 3 x2x3 0



2x1x2 x2x3 -x3x1 = ​ ______     ​ x​ 21 ​x​ 22 ​x​ 23 ​  x x x3x1 2 3 2x x = ______ ​  2  12  22   ​  (2x1x2x​ 23​)  = 4. x​1 ​ x​2 ​ x​3 ​ 

EXAMPLE 5.74

∂  (u,  v) y2 x2 + y2, . If u = ​ __  ​and v = ​ ______     ​ find ______ ​   ​  2x 2x ∂  (x,  y) Solution.  We have y y2 ___ _ ___ ∂x ​ ​ ___ ∂u  ​ -​  ​ x ​ ​  2  ​  ∂  (u, v) ______ 2x ​   ​  = ∂u ∂y = ∂  (x, y) y2 _y 1  ​- ​ ___ ∂v ​ ​ __ ∂v ​ ​ __   ​  ​   ​ ​ __ 2 2x2 x ∂x ∂y 3 3 y y y y = - ​ ___3  ​ - ​ __  ​+ ___ ​  3  ​ = - ​ __  ​ . 2x 2x 2x 2x EXAMPLE 5.75

EXAMPLE 5.73



using R1 → R1 + R2



Solution.  We have

= ______ ​   2 1 2   2 ​ x2x3 x​1 ​ x​2 ​ x​3 ​  x2x3

M05_Baburam_ISBN _C05.indd 32

x3x1

x1x2

x3x1

-x1x2

-x3x1 x1x2 0

2x1x2

-x3x1 x1x2

x3x1 -x1x2

∂ (u, v). x+y If u = _____ ​   ​  and v = tan- 1x + tan-1y, find ______ ​   ​  ∂ (x, y) 1 - xy Solution.  We have (1 - xy) - (x + y)( - y) 1_____________ - xy + xy + y2 ∂u  ​= ___________________ ​ ___ ​   ​    = ​      ​   2   ∂x (1 - xy) (1 - xy)2 1 + y2 , = _______ ​    ​  (1 - xy)2 - xy + x2 + xy - xy) - (x + y) (-x) 1_____________ ∂u  ​= ​ (1 __________________ ​ ___        ​= ​      ​  2 (1 - xy)2    ∂y (1 - xy)

1+x  = ​ _______    ​, (1 - xy)2 ∂v ​= _____ ​ __ ​  1   ​, and __ ​ ∂v ​= _____ ​  1   ​.  ∂x 1 + x2 ∂y 1 + y2 2

Therefore, 1+y _______ 2

∂ (u, v) ​ ______ ​  = ∂ (x, y)

2 ​      ​ _______ ​  1 + x     ​ (1 - xy)2 (1 - xy)2 

_____ ​  1  2 ​ 

1+x



1   ​  ​ _____ 1 + y2

1    = _______ ​  1  2 ​  - ​ _______  ​= 0. (1 - xy) (1 - xy)2

EXAMPLE 5.76

If u = 2xy, v = x2 - y2, x = r cosq, and y = r sin q, ∂  (u, v) . find ​ ______   ​  ∂  (r, q )

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functionS of Several variableS  n 5.33 a relation between u, v, and w.

Solution.  We have

Solution.  We have

∂  (x, y) ∂  (u, v) ______ ∂  (u, v) . ______ ​ ______   ​= ​   ​ ​     ​ ∂  (r, q ) ∂  (x, y) ∂  (r, q )

∂u  ​ ___ ∂u  ​ ​ ___ ​ ∂u  ​ ​ ___ ∂x ∂y ∂z ∂ (u, v, w) __ ∂v ∂v __ __ ​ ________ ​  = ​   ​ ​ ∂v ​ ​   ​ ∂ (x, y, z) ∂x ∂y ∂z ∂w ​ ​ ___ ∂w ​ ​ ___ ∂w ​  ​ ___ ∂x ∂y ∂z

∂x ​ ​ __ ∂x  ​ ∂u  ​ ​ __ ​ ___ ∂y . ∂r ∂q ∂v ​ ∂v ​  ∂v ​ ​ __ ​ __ ​ __ ∂y ∂r ∂q



∂u  ​ ​ ___ = ∂x ∂v ​ ​ __ ∂x



=



= - 4  ( y2 + x2) . r = - 4r3.

2y 2x

2x  . -2y

cos q sin q

-r sin q r cos q

5.16 NECESSARY AND SUFFICIENT CONDITIONS FOR JACOBIAN TO VANISH The following two theorems, stated without proof, provide necessary and sufficient condition for the Jacobian to vanish. Theorem 5.9. If u1, u2, …, un are n differentiable functions of the n independent variables x1, x2, … , xn and there exists an identical, differentiable functional relation  f (u1, u2 , …, un)  =  0,  which  does  not involve ∂ (u1, u2, … , un) xi explicitly, then the Jacobian ​ _____________    ​ ∂ (x1, x2, … , xn) vanishes identically provided that f  as a function of the ui has no stationary values in the domain considered. Theorem 5.10  If u1, u2, … , un are n functions of the n variables x1, x2,. . ., xn, say, um = fm (x1, x2,… , xn), m = 1, 2, … , n, and ∂ (u1, u2, … , un) if ​ _____________    ​ = 0, then if all the differen∂ (x1,  x2, … , xn) tial coefficients are continuous, there exists a functional relation connecting some or all of the variables and which is independent of x1, x2,…, xn. EXAMPLE 5.77

If u = x + 2y + z, v = x - 2y + 3z, and w = 2xy ∂ (u, v, w) xz + 4yz - 2z2, show that ________ ​   ​  = 0 and find ∂ (x, y, z)

M05_Baburam_ISBN _C05.indd 33

=

1 2 1 -2 2y - z 2x + 4z

1 3 - x + 4y - 4z

=

1 0 0 1 -4 2 = 0. 2y - z 2x + 6z - 4y -x - 2y - 3z

Hence, a relation between u, v, and w exists. Now, u + v = 2x + 4z = 2 (x + 2z) u - v = 4y - 2z = 2 (2y - z) w = x (2y - z) + 2z (2y - z) = (x + 2z) (2y - z) = __ ​ 1 ​ (u + v) (u - v). 4 Therefore, 4w = (u + v) (u - v) is the required relation connecting u, v, and w. EXAMPLE 5.78

, show that If f  (0) = 0 and f ′ (x) = _____ ​  1  2 ​   1+x x+y f (x) + f ( y) = f ​ _____ ​      ​  ​. 1 - xy

( 

)

Solution.  Suppose that

x+y . u = f (x) + f ( y) and v = ______ ​      ​ 1 - xy

Then, _____ 1   ​  ​  1  2 ​  ​ _____ ___ ∂u  ​ ​ ___ ∂u  ​ 1 + x 1 + y2 ​  ∂ (u, v) ∂x ∂y ______ ​   ​  = = = 0. ∂ (x, y) __ 1 + y2 _______ ∂v ​ _______ 1 + x2 ​ ∂v ​ ​ __      ​ ​       ​ ​  ∂x ∂y (1 - xy)2 (1 - xy)2

Therefore, u and v are connected by a functional relation. Let u = f  (v), that is, x+y f  (x) + f  ( y) = f  ​ _____ ​      ​  ​ . 1 - xy

( 

)

1/2/2012 11:58:46 AM

5.34  n  chapter five Putting y = 0, we get f (x) + f (0) = f  (x) or f  (x) = f  (x), since f  (0) = 0. Hence, x+y . f (x) + f ( y) = f ​ _____ ​      ​  ​  1 - xy

Remark 5.3  If the limits of integration a and b are not independent of a, then

5.17  DIFFERENTIATION UNDER THE INTEGRAL SIGN In the following theorem of Leibnitz, we shall show that under suitable conditions, the derivative of the integral and the integral of the derivative are equal. The result is useful to determine the value of a definite integral by differentiating the integrand with respect to a quantity of which the limits of integration are independent.



( 

[ ∫ 

]

b

b

___ ∂   ​[ f (x, a )] dx, ​  d    ​​ ​  ​ f (x, a) dx  ​= ​  ​​ ​ ___

da

a

∫  ∂a a

where the limits a and b are independent of a. Proof:

a

b

b

F (a + d a) - F (a) = ​∫  ​​  [ f (x, a + d a) - f (x, a)] dx. (1)

But by Lagrange’s mean value theorem, we have f (a + d a) - f (a) = d a ___ ​  ∂  ​   f (x, a + d a), 0 < q < 1. ∂a

__ tan-1 ax  ​∫  ​ ​ ​  ________ dx = ​ p   ​ log (1 + a), a > 0. 2  ​  2 0 x (1 + x )  

Solution.  Let

Therefore, F  (a  +  d  a)  -  F (a) ________________

​ lim ​​      d a→0

b

∂    ​  f (x, a ) dx  ​     = ​∫  ​  ​___ d a a ∂a

or

[ ∫  b

Then by Leibnitz’s Rule,

da

M05_Baburam_ISBN _C05.indd 34

a

[ ∫ 

]

∞

___ tan-1 ax  ​ d  ​ ​ ​  ​ ​ ​ ________ dx  ​ 2  ​ 

da

 

0

x (1 + x ) ∞



[ 

]

∂  ​​ ​ ________ tan-1 ax  = ​∫  ​ ​​ ​ ___ ​  dx 2  ​   0 ∂a x (1 + x )



= ​∫  ​ ​​_______ ​  1  2  ​  · _______ ​  1 2 2   ​ · xdx x(1 + x ) 1+ax 0



dx  = ​∫  ​  ______________ ​     ​ 2 (1 + x ) (1 + a2 x2) 0

∞

 

∞



] ∫    

a

∂a

[ 

]

a2    = _____ ​  1  2   ​​∫  ​ ​ ​ _____ ​  1   ​  - ​ _______  ​  ​  dx 1 - a 0 1 + x2 1 + a2x2 (by partial fractions)  

∞

_______ dx    a2        = _____ ​  1  2   ​[tan-1x​]​∞0​ ​- ​ _____ 2 2  ​ 2 ​​  ​ ​ ​ 

1-a

∫  1 + a x  

0

∞



b

___ ∂   ​  f (x, a ) dx. d    ​​ ​  ​​ f (x, a ) dx  ​= ​  ​​  ​ ___ ​  d    ​[F (a)] = ​ ___

da

-1

 

b

∂   ​  f (x, a + qd a) dx.  ​     = ​∫  ​​ ​ ___ d a a ∂a

​ 

∞

tan ax  F  (a) = ​∫  ​ ​ ​  ________ dx. 2  ​  0 x (1 + x )

1-a

Hence, (1) reduces to F  (a  +  d a)  -  F  (a) ________________

db + ___ ​   ​ f (b, a ) - ___ ​ da  ​ f (a, a ). da da

∞

 

a

∂a 

Show that

 



∫ 

EXAMPLE 5.79

F (a + d a) = ​∫  ​​   f (x, a + d  a) dx a

 

a

∞

 

and so,

da

b

Let F (a) = ​∫  ​​   f  (x, a ) dx. Then

]

b

)

Theorem 5.11. (Leibnitz’s Rule): Let f  (x, a) and fx  (x, a) be continuous functions of x and a. Then,

[ ∫ 

___ ∂    ​[ f (x, a )]dx ​  d    ​​ ​  ​  f (x, a) dx  ​= ​ ​ ​​​ ___



dx p     = ​ ________  ​- _____ ​  1     ​ ______ ​     ​  2 (1 - a2) 1 - a2 ​∫0  ​ ​ 2 __ x + ​ 12  ​  a x ∞ 1 __ ________ p  _____ 1 = ​   2   ​- ​    2   ​· ​   ​​ tan-1 __ ​   ​  ​  ​  2 (1 - a ) 1 - a __ __ ​ 1a ​ ​ 1a ​ 0  

[  ]

__ a    p     = ​ ________  ​- ​ _____  ​​ p   ​  2 (1 - a2) 1 - a2 2 p     p    ​   . = ​ ________  ​[1 - a] = ​ _______ 2 (1 + a) 2 (1 - a2)

1/2/2012 11:58:46 AM

functionS of Several variableS  n 5.35 Integrating both sides with respect to a, we get ∞

p  tan-1 ax  F (a) = ​∫ ​  ​​ _______  ​  dx = __ ​   ​ log (1 + a) + c. (1) 2 x(1 + x2) 0 p  Also F (0) = 0. Therefore, (1) yields 0 = ​ __ ​ log 1 2 + c and so, c = 0. Hence,

 

∞

__ tan ax  ​∫  ​ ​ ​ ________ dx = ​ p   ​ log (1 + a). 2  ​  2 0 x (1 + x )

Then by Leibnitz’s Rule, we have da

da

[ 

]

∞

 

∞

 

1

]

 

0

_____  ∂    ​​ e-x ​ sin a x   ​  dx = ​∫  ​ ​​ ___ x  ​   ∂  a 0



-1

EXAMPLE 5.80

[ ∫  ∞

d _____ ___ a​ dx  ​ ​  d    ​F (a) = ___ ​      ​​ ​  ​ ​e-x ​ sin x   

= ​∫  ​  e-x __ ​ 1x ​cos a x . x dx



 

0

∞

______ xa - 1

= ​∫  ​  e-x cos a x dx = _____ ​  1  2   ​ . 1 + a  0

Evaluate ​∫  ​​​  ​   ​  dx,   a > 0 using differentiation 0  log x



under the integral sign. 1

Integrating, we get

a - 1  Solution.  Let F (a) = ​∫  ​ ​​​  x_____  ​ dx. Then by log x 0 Leibnitz’s Rule,  

[ ∫  1

]

∫  ∂a [  log x ]

 

F (a) = tan-1 a + c.

But,

∞

1

a ___ ∂  ​​ ​ x_____ xa - 1 ​  - 1 ​  ​  d    ​​ ​  ​​  ​  ​  _____ dx  ​= ​  ​​  ​ ___   ​dx

da

0

log x

 

1

 

[ 

1

]

x    1    = ​∫  ​​  xa  dx = ​ ​ _____  ​  ​​​= ​ _____  ​. a + 1 1 + a 0



 

a+1

 

0

= ​∫  ​​  _____ ​  1   ​ · xa  log x dx log x 0



∞

F (0) = ​∫  ​  e-x _____ ​ sinx 0x  ​ dx = ​∫  ​  0 dx = 0.

 

0

(1)

1     0

Therefore, (1) yields 0 = tan-1 0 + c and so, c = 0. Hence,

∞

sin a  x​ dx = tan-1 a. F (a ) = ​∫  ​  e-x ​ ______ x   

Integrating with respect to a, we get

 

0

1

a F (a) = ​∫  ​​ _____ ​ x - 1 ​  dx 0 log x

0

EXAMPLE 5.82

 

a

= log  (1 + a ) + c (a constant of integration) 1 But when a = 0, F (0) = ​∫  ​​ ​ 0 dx = 0. Therefore, 0 = log1 + c = c. Hence, 0

log(1 +  ax) Evaluate ​∫  ​  _________ dx and hence, show that 1+ x2 ​  0 ​   

 

 1

xa - 1  F  (a ) = ​  ​​ ​ ​ ______  ​  dx = log (1 + a ).

∫   log x

    0

EXAMPLE 5.81

Evaluate the integral ∞

_____ ax ​dx. ​∫  ​ ​ e-x ​ sin x   

1

log  (1 + x) ​ _________ __ ​ log 2. ∫  ​​  dx = ​ p  2 1 + x 0 ​   ​    8 Solution.  We note that the limits of integration are not independent of the parameter a. Therefore, the formula mentioned in Remark after Theorem 5.11 is applicable. Let  

a

 

0

Solution.  We cannot compute this integral directly because the anti-derivative of the function sin a  e -x   ​ ______    x  ​ is not expressible in terms of elex mentary functions. So we use Leibnitz’s Rule to evaluate it. Let ∞

sin a  x​ dx. F  (a ) = ​∫  ​ ​ e-x · ​ ______ x   

log (1 + ax) F (a) = ​∫  ​  __________ ​   ​    dx. 1 + x2 0  

Then,

[ 

] log (1 + ax) = ​∫ ​​​ ​ ∂  ​​[ ​   ​   ​ d  ​(a) ]​dx + ​ log1(1+ a+ a ​)  da 1+x ∂a a

log (1 + ax) ___ ​ d  ​ ​  ​  ​ __________ ​    dx  ​ 2 ​  da ∫   

0 a

 

0

1+x

___ __________     2

2

__________ ___    2

 

0

M05_Baburam_ISBN _C05.indd 35

1/2/2012 11:58:47 AM

5.36  n  chapter five log (1 + a.0) ___ - ___________ ​   ​    · ​ d  ​ (0) 1+0 da a log (1 + a2) x    ​dx + ​ __________ = ​∫  ​  _____________  ​    ​​ (1 + x2)(1 + ax) 1 + a2 0  

 

a

[ 

]



2

+ ​ 

 ​    (by partial fractions)

1 + a2

[ 

= _____ ​  1  2 ​  ​ -log (1 + ax) + __ ​ 1 ​ log (1 + x2) 1+a

2

+ a tan

       

-1

[ 

]

log  (1 + a ) x  ​​  + __________ ​    2 ​ 

EXAMPLE 5.83

__ ​∫  ​​sin ​  q cos-1 (cosa cosecq )dq = ​ p   ​ (1- cos a ). 2

  __ ​ p    ​- a 2

Solution.  Here the limits involve the parameter a. Let p  __ ​    ​ 2

F(a ) = ​∫  ​​​ sin q cos-1 (cos a cosec q ) dq.

2

a  

1+a

0

]

= _____ ​  1  2 ​  ​ - __ ​ 1 ​ log (1 + a2) + a tan-1a  ​ 2 1+a log (1 + a2) + ​ __________  ​    1 + a2 = _____ ​  1  2 ​  ​ __ ​ 1 ​ log (1 + a2) + a tan-1 a  ​. 1+a 2 Integrating with respect to a, we get   

[ 

]

a



 

​ p__  ​ 2

log  (1 + a ) __________

 

1

log (1 + x) __ p  ​∫  ​​  _________ ​      = ​   ​ log 2. 2 ​  dx 8 1 + x 0 Prove that

-a   ​  = _____ ​  1  2 ​  ​ ​ ______ + _____ ​ x + a   ​  ​dx 1 + a ​∫0  ​​​  1 + ax 1 + x2  

Substituting a = 1, we get

log (1 + ax) F(a) = ​∫  ​ ​__________ ​   ​    dx 1 + x2 0

Then, ___ ​  d    ​[F (a )] da __ ​ p    ​ 2

∂   ​[sin q cos-1 (cos a cosec q )]dq = ​∫  ​​  ​ ​ ___ ∂ a p  __ ​    ​- a 2

(  ) [ 

da 2

2

[ 

 

]

      - ​∫  ​​​ _____ ​  2a 2 ​  tan-1 a da  ​   1 + a



a

M05_Baburam_ISBN _C05.indd 36

( 

( 

)]

)

))]

( 

)

__ sin a dq   ______________ = ​∫  ​  ​ ​​  _______________  2  ​+ sin ​ ​ p   ​ - a  ​ 2 2 __    q ​ ​ p    ​- a ​√ 1 - cos a cosec 2 __ × cos-1 ​ cos a cosec ​ ​ p   ​ - a  ​  ​ 2 p  __   ​​ 

[ 

( 

)]

2

a tan a ​        + ​∫  ​​​ ​ ________ da + c 1 + a2

F (a) = ​∫  ​​​ __________ ​ log (1 + ax)  ​    dx = __ ​ 1 ​ log (1 + a2) tan-1a. 2 2 1+x 0

) [  ( 

p  __   ​ ​  2

-1

      = __ ​ 1 ​ log (1 + a2) tan-1 a + c. 2 Substituting a = 0, we have F (0) = 0. Therefore, 0 = __ ​ 1 ​ log 1 tan-1 0 + c and so, c = 0. 2 Hence,

( 

2

__ × cos-1 ​ cos a cosec ​ ​ p   ​ - a  ​  ​  ​ 2

-1

__ 1       = ​   ​ ​ log (1 + a2) tan-1 a

( 

__ __ - ___ ​  d    ​​ ​ p   ​ - a  ​​ sin ​ ​ p   ​ - a  ​ 2 da 2



+ ​∫ ​​  ​ ________ ​ a tan a ​    da + c 1 + a2



2

__ __ __ + ___ ​  d    ​​ ​ p   ​   ​​ sin ​ p   ​ cos-1 ​ cos a cosec ​ p   ​   ​  ​

 

1   ​      = __ ​ 1 ​ ​∫   ​​​ log  (1 + a2) ​ ______ da 2 1 + a2

p  ​ __  ​- a 2

sin a sin   q  = ​∫   ​​  ​  ____________ ​  ___________  ​dq + cos a cos-1 (1) 2 2 p  __ ​ sin q - cos a   ​ ​    ​- a √ 2

sin a



sin a    ________ = ​∫  ​  ​​​ _________  ​dt taking cos q = t 2 2   ​ 0 ​√ sin a - t  

[ 

( 

)]

sin a __ = ​ sin a sin-1 ​ _____ ​  t     ​  ​  ​​       ​= ​ p    ​sin a. sin a 0 2 Integrating with respect to a, we get __ ​ cos a + c. F (a) = - ​ p  2 But F (0) = 0, therefore,



1/2/2012 11:58:48 AM

functionS of Several variableS  n 5.37 __. __ ​ + c  or  c = ​ p   ​  0 = - ​ p  2 2

Hence,

  dm ( m + 1 )

__ p  __ F (a) = - ​ p    ​cos a + ​ p   ​ = ​ __ ​ [1 - cos a]. 2 2 2

 

x

y = ​∫  ​​  f (t) sin [k (x - t)] dt.

( 

x

∂ ​   ​= ​  ​  ___ ​    ​ [ f (t) sin  [k (x - t)] dt dx

( 

dm (m + 1)



d  ​(x) + f (x) sin [k (x - x)] ​ __ dx d  ​(0) - f (0) sin [k (x - 0)] ​ __ dx x

x

∫  ∂x 0



d  ​(x) + k f (x) cos [k (x - x)] ​ __ dx d  ​(0). - k f (0) cos [k (x - 0)] ​ __ dx

x

= - k 2 ​∫  ​​  f (t) sin [k (x - t)] dt + kf (x)  

0

= - k  y + kf (x).

Hence,

1

EXAMPLE 5.85

1

or

n

(m + 1)

If z = f (x , y) and u, v are two variables such that u = lx + my, v = ly - mx. Prove that



Solution.  We have

  m + 1 

1 xm-1  ​   I = ​  ​​  x dx = ​​  _____ ​​​  = _____    ​  1   ​. 

m+1 Therefore, using Leibnitz’s Rule, we get  

0

M05_Baburam_ISBN _C05.indd 37

0

]

Solution.  We have

 

0

 

0

EXAMPLE 5.86

 

0

evaluate ​∫  ​​ xm (log x)n dx.

∫ 

5.18  MISCELLANEOUS EXAMPLES

By successive use of Leibnitz’s Rule to ​∫  ​​ xm dx, 1

1

(-1)   n! _________ ​   ​  = ​  ​​  xm  (log x)n  dx. n+1 

2 ∂2z  ​ + ​ ___ ∂2z  ​= (l 2 + m2) ​ ​ ___ ∂2z  ​  ​ ​ ∂___z2  ​+ ​ ___ 2 2 ∂u ∂x ∂y ∂v2

1

∫ 

 

0

(-1) (-2) (-3) . . . (-n) ​ __________________     ​    = ​∫  ​​  xm (log x)n dx (m + 1)n+1 0

2

d  y ​ ___2  ​+ k 2y = k f (x). dx

m

∫ 

[ 

2

1

∂m

 

d  y ___ ∂  ​[kf (t) cos [k (x - t)] dt ​  2 ​ = ​  ​​ ​ __

 

0

 

0

Using once more the Leibnitz’s Rule, we get dx

∫ 

1

 

2

)

(-1) (-2) (-3) ​ ___________     ​  = ​∫  ​​  xm  (log x)3  dx (m + 1)4 0 …………………………………… ……………………………………

= ​∫  ​​  kf (t) cos [k (x - t)] dt.



1

Repeated use of Leibnitz’s Rule yields

 



 

0

(-1) (-2) ________ ​  = ​  ​​  xm  (log x)2  dx. 3 ​  (m + 1)

∫  ∂x 0

1

2

∂ m ___ ___ -1   ​   ​(x log x)  dx ​  d   ​ ​ ​ _______ 2 ​= ​  ​​  ​      

0

The upper limit in this integral involves the parameter x. So, using Leibnitz’s Rule, we have

)

Applying again the Leibnitz’s Rule, we get

or

 

0

- ​ ______ ​  1   ​  ​ = ​∫  ​​  xm log xdx. m+1

x

If y = ∫​   ​​ f (t) sin [k (x - t)] dt, show that it satisfies 0 d 2y the differential equation ___ ​  2 ​ + k 2 y = kf  (x). dx Solution.  We have

∫  ∂m

or

EXAMPLE 5.84

dy __

1

∂ ___ ​  d   ​ ​ _____ ​  1   ​  ​= ​  ​​​ ​ ___    ​(xm)  dx

Therefore

u = lx + my,  v = ly - mx, ___ ​ ∂u  ​= l, ___ ​ ∂v ​ = - m ∂x ∂x ___ ​ ∂u  ​= m, ___ ​ ∂v ​ = l. ∂y ∂y

∂z ∂z ∂z ∂z ∂u ∂z ∂v ​ ___ ​= ​ ___  ​· ​ ___  ​+ ​ ___ ​· ​ ___ ​= l ​ ___  ​- m ​ ___ ​ ∂u ∂v ∂x ∂u ∂x ∂v ∂x

(1)

1/2/2012 11:58:48 AM

5.38  n  chapter five ∂z ∂z ∂z ∂z ∂u ___ ∂z ∂v ​ ___ ​= ​ ___  ​· ​ ___   ​+ ​   ​· ​ ___ ​= m ​ ___  ​+ l  ​ ___ ​ (2) ∂y ∂y ∂u ∂v ∂y ∂u ∂v From (1) and (2), we have ∂ ∂ ∂ ​ ___   ​= l ​ ___   ​– m ​ ___   ​ ∂x ∂u ∂v and ∂ ∂ ∂ ​ ___   ​= m ​ ___   ​+ l ​ ___   ​ . ∂y ∂u ∂v Therefore ∂ ∂ ∂z ∂z ∂ ∂z ∂2z ​ ___2  ​= ___ ​    ​ ​ ___ ​   ​ ​= ​  l ​___     ​  - m ​___     ​  ​​ l ​___     ​  - m ​___    ​ ​ ∂u ∂v ∂u ∂v ∂x ∂x ∂x

 5 ∂2f - __ ​ ___2  ​= - __ ​ 3 ​ y (1 - 2xy + y2) ​ 2 ​ (-2y) 2 ∂x  5 - __ = 3y2  (1 - 2xy + y2) ​ 2 ​ . Therefore







(  ) ( ) ) (  ∂ ∂z ∂z ∂ ∂z ∂z = l ​___     ​ ​( l ​ ___  ​  - m ​___    ​)​- m ___ ​    ​ ​( l ​___     ​  - m ​___    ​)​ ∂u ∂u ∂v ∂v ∂u ∂v ∂2z 2 ____ ∂2z ∂2z ∂2z = l    ​  2   ​ - lm ​ ____    ​- lm ​ ____    ​+ m2 ​ ___  ​ ∂u∂v ∂v∂u ∂u ∂v2 2 ∂ z    ∂2z  ​+ m2  ​ ___ ∂2z  ​- 2lm ​ ____ = l 2  ​ ___ ​ (3) ∂u∂v ∂u2 ∂v2

and ∂ ∂  ∂z ∂z ∂2z ∂ ∂z  ​ ___2  ​=​ ___  ​ ​ __ ​   ​ ​= ​  m ​___     ​ + l ​___     ​  ​​ m​ ___   ​+ l ​___    ​ ​ ∂y ∂u ∂u  ∂v ∂v ∂y ∂y ∂ ∂z ∂z ∂ ∂z ∂z = m ___ ​    ​ ​ m​ ___   ​+ l ​___    ​ ​+ l ___ ​    ​ ​ m ​___    ​ + l ​___    ​ ​ ∂u ∂u ∂v ∂v ∂u ∂v

(  ) ( ( 



) ( 

)

( 

)

)

2 ∂2z ∂2z 2∂ z ∂z = m2 ​ ____  ​ + lm ​ _____    ​+ lm ​ _____    ​+ l  ​ ___  ​ 2 ∂u∂v ∂v∂u ∂u ∂v2 2

∂2z ​. ∂2z ∂2z  ​ = l 2  ​ ___   ​+ m2 ​ ___  + 2lm ​ _____     ∂u∂v ∂v2 ∂u2 Adding (3) and (4), we get ∂2z ∂2z ∂2z ∂2z ​ ___2  ​+ ​ ___  ​= (l 2 + m2) ​ ​ ___2  ​ + ​ ___2  ​  ​. ∂v ∂u ∂x ∂y2

( 

EXAMPLE 5.87

(4)

)

 1 - __ 2

(a)  If f  (x, y) = (1 - 2xy + y  ) ​   ​ , show that 2

[ 

] [  ]

∂f ∂f ∂ ∂ ​ ___  ​ ​ (1 - x2) ​ ___ ​  ​+ ___ ​    ​ ​ y2​ ___  ​ ​= 0. ∂x ∂x ∂y ∂y (b) If V = f (2x - 3y, 3y - 4z, 4z - 2x), compute the value of 6Vx + 4Vy + 3Vz. 1 - __ 2

Solution.  (a) We have, f = (1 - 2xy + y2) ​   ​ . Then  3



- __ ∂f 2 ​ ___ ​ = - __ ​ 1 ​ (1 - 2xy + y2) ​   ​ (-2y) ∂x 2   __ 3 2 = y (1 - 2xy + y2) ​   ​ ,

M05_Baburam_ISBN _C05.indd 38

[ 

]

∂ ∂f ​ ___  ​ ​ (1 - x2)​ ___ ​   ​ ∂x ∂x ∂f ∂2f ∂ ___ 2 ___ 2 ___ =​     ​ (1 - x )​   ​ + (1 - x ) ​  2  ​ ∂x ∂x ∂x ∂f ∂2f ___ 2 ___ = (-2x) ​     ​+ (1 - x )​  2  ​ ∂x  ∂x  3 - __

= -2xy (1 - 2xy + y2) ​ 2 ​ 

 5 - __

+ 3 (1 - x2) y2 (1 - 2xy + y2) ​ 2 ​ . (1) Similarly, differentiating partially with respect to y, we get  3 - __ ∂f ​ __   ​= (x - y) (1 - 2xy + y2) ​ 2 ​ , ∂y  3 - __ ∂2f ​ ___2  ​= - (1 - 2xy + y2) ​ 2 ​  ∂y 5 - __ + 3(x - y)2  (1 - 2xy + y2) ​ 2 ​ . Therefore,

[  ]

∂f ∂2f ∂ ∂f ​  ___     ​​ y2​  __   ​ ​= __ ​ ∂  ​( y2) __ ​   ​ + y2 ​ ___   ​ ∂y ∂y ∂y ∂y ∂y2

 3 - __



= 2y (x - y) (1 - 2xy + y2) ​ 2 ​ 



+ y2 - (1 - 2xy + y2) ​ 2 ​ 



+ 3(x - y)2  (1 - 3xy + y2) ​ 2 ​  



= y (1 - 2xy + y ) ​   ​ [3y (x - y)2

[

 3 - __

 3 - __ 2 2

5 - __

]

× (1 - 2xy + y2)-1 + (2x - 3y)] (2) Adding (1) and (2), we get the required result. (b)  We have V = f (2x - 3y, 3y - 4z, 4z - 2x). Let r = 2x - 3y, s = 3y - 4z and t = 4z - 2x. Then V = f (r, s, t). Further, ∂V ∂V ∂r ∂V ∂s ∂V ∂t ​ ___ ​ = ​ ___ ​ · ​ ___ ​+ ​ ___ ​ ​ ___ ​+ ​ ___ ​ · ​ ___  ​ ∂x ∂r ∂x ∂s ∂x ∂t ∂x

1/2/2012 11:58:49 AM

functionS of Several variableS  n 5.39 ∂V ∂V ∂V ∂V = 2 ​ ___ ​ + 0 - 2 ​ ___ ​ = 2 ​ ___ ​ - 2 ​ ___ ​  (1) ∂t ∂r ∂t ∂r ∂V ∂t ∂V  ​= ___ ​ ___ ​ ∂V ​ · ___ ​ ∂r ​+ ___ ​ ∂V ​ ___ ​ ∂s ​+ ___ ​   ​ · ___ ​   ​  ∂y ∂r ∂y ∂s ∂y ∂t ∂y ∂V ∂V ∂V ∂V = - 3 ​ ___ ​ + 3 ​ ___ ​ + 0 = 3 ​ ___ ​ + 3 ​ ___ ​  (2) ∂s ∂r ∂r ∂s and   ∂s  ∂V ∂t ∂V ∂V ∂r  ∂V ​ ___ ​ =​ ___ ​ · __ ​   ​ +​ ___ ​​   __ ​ +​ ___ ​ · __ ​   ​  ∂z ∂r ∂z ∂s ∂z ∂t ∂z ∂V ∂V ∂V ∂V = 0 - 4 ___ ​   ​ + 4 ___ ​   ​ = - 4 ___ ​   ​ + ∂s ∂s ___. ∂t 4 ​   ​  ∂t The relations (1), (2) and (3) yields ∂V ∂V 6Vx + 4Vy + 3Vz =6 2​ ___  ​- 2 ​ ___ ​   ​ ∂r ∂t ∂V ∂V + 4 ​ - 3 ​ ___ ​ + 3 ​ ___ ​  ​ ∂r ∂s ∂V ∂V + 3 ​ - 4 ​ ___ ​ + 4 ​ ___ ​  ​ ∂s ∂t = 0.

( 

( 

)

( 

EXAMPLE 5.88

)

)

( 

)

3x2 + 4y2 (a) If u = sin-1 ​ ​ ________ ​    ​, prove that x ___ ​ ∂u  ​ + ∂x 3x + 4y ___ ∂u y ​    ​= tan u. ∂y (b)  If u = x3 + y3 + z3 + 3xyz, show that x ___ ​ ∂u  ​+ y ___ ​ ∂u  ​+ z ___ ​ ∂u  ​= 3u. ∂x ∂y ∂z

( 

)

x2 + y2 (c)  If u = log ​ ​ ______   ​, prove that x ___ ​ ∂u  ​ + y ___ ​ ∂u ​  x + y ​  ∂x ∂y = 1.

 y 2 −1  x   − y tan   , show x  y

−1 2 (d)  If u =​ xx tan 

that x 2

2 ∂2u ∂2u 2 ∂ u + 2 xy + y = 2u. ∂x ∂y ∂x 2 ∂y 2

Solution.  (a) We have 3x2 + 4y2 sin u = ________ ​     ​  = z, say. 3x + 4y Thus 2 y 2 4 ​ ​ _​ y ​ ​ __ 1 + ​  1 + __ ​ 4 ​ ​ _​ x ​ ​ 2 x 3 3 3x  ​ ​ ​ ________ z = ​ ___    ​  ​= x ​ ​ ________   ​  ,​ y y 3x 4 4 __ _ __ _ 1 + ​   ​ ​ ​ x ​ ​ 1 + ​   ​ ​ ​ x ​ ​ 3 3

(  ) (  ) (  ) (  )

M05_Baburam_ISBN _C05.indd 39

(  ) (  )

and so z is a homogeneous function of degree 1 in x and y. Hence, by Euler’s Theorem, we have ∂z ∂z x ___ ​   ​+ y ___ ​   ​= z. (1) ∂x ∂y But ∂u ∂z ∂z ​ ___ ​= cos u ___ ​ ∂u  ​  and ​ ___ ​= cos u ___ ​    .​ ∂y ∂x ∂y ∂x Hence (1) reduces to or

​ ∂u  ​+ y cos u ___ ​ ∂u  ​= sin u x cos u ___ ∂x ∂y x ___ ​ ∂u  ​+ y ___ ​ ∂u  ​= tan u. ∂x ∂y

(b) We have u = x3 + y3 + z3 + 3xyz. Replacing x by tx, y by ty and z by tz, we get  u (tx, ty, tz) = t 3 x3 + t 3 y3 + t 3 z3 + 3tx ty tz = t 3(x3 + y3 + z3 + 3 xyz) = t 3u  (x, y, z). Hence, u (x, y, z) is a homogeneous function of degree 3 in u  (x, y, z). Therefore, by Euler’s Theorem, x ___ ​ ∂u  ​+ y ___ ​ ∂u  ​+ z ___ ​ ∂u  ​= 3u. ∂z ∂y ∂z (c)  We have x2 + y2 . u = log ​ ​ _____  ​ x + y ​  Therefore x2 + y2 , eu = _____ ​ x + y ​    which is homogeneous function of degree 1 in x and y. Therefore, by Euler’s Theorem, we have ∂ ∂ ​    ​ (eu) + y ___ ​    ​ (eu) = eu x ___ ∂x ∂y or xeu ___ ​ ∂u  ​+ yeu ___ ​ ∂u  ​= eu ∂x ∂y or x ___ ​ ∂u  ​+ y ___ ​ ∂u  ​= 1. ∂x ∂y

( 

)

(d) We have u = u1 + u2 , where

 x  y u1 = x 2 tan −1   , u2 = y 2 tan −1   .  x  y we note that u1 and u2 are both homogeneous functions of degree 2. Therefore (see Example 5.26), we have

1/2/2012 11:58:50 AM

5.40  n  chapter five

x2

2 ∂ 2 u1 ∂ 2 u1 2 ∂ u1 + 2 xy + y = 2(2 − 1)u1 ∂x ∂y ∂x 2 ∂y 2 = 2u1 (1)

and x2

∂ 2 u2 ∂ 2 u2 ∂ 2 u2 + 2 xy + y2 = 2(2 − 1)u2 ∂x2 ∂x ∂y ∂y 2

= 2u2 Adding (1) and (2) we get



(2)

∂u  ​ ___ ∂u ​  ​ ___ ​ ∂u  ​ ​ ___ ∂x ∂y ∂z 2x 2y 2z ∂  (u, v,w) __ ________ ∂v ∂v ∂v __ __ ​     = ​ = 1 1 1 ∂  (x, y, z) ​ ∂x ​ ​ ∂y ​ ​ ∂z ​ z+y z+x x+y ∂w ​  ___ ​ ∂w ​  ___ ​ ∂w ​  ​ ___ ∂x ∂y ∂z    = 2x [ (x + y) - (z + x)] - 2y [(x + y)      - (z + y)] + 2z [(z + x) - (z + y)] = 0.

Since, Jacobian J (u, v, w) = 0, there exists a 2  ∂ 2 u1 ∂ 2 u2   ∂ 2 u1 ∂ 2 u2  ∂ 2 u2  relation connecting some or all of the 2  ∂ ufunctional 1 x  + + 2 xy + + y + = 2 (u1 + u2 )  ∂x ∂y ∂x ∂y   ∂y variables ∂x2  ∂y2  x, y and z. Hence u, v, w are not inde ∂x2 2 pendent. 2 or  ∂ 2u 2 ∂ 2 u 2   ∂ 2 u1 ∂ 2 u2  ∂ 2 u2  ∂ 2 2 ∂ 2  ∂ u1 1 u1 2 u2 x+ y 2 + + 2 + 2=xy2(u12 + u2+) + y +  ∂y EXAMPLE  = 2 (u1 + u2 ) ∂ ∂ux 2 ∂x ∂ 2u  ∂xu∂y ∂x ∂y  ∂y2 5.90 2 x 2   2+∂y2 xy 2∂y  + y 2 2 = 2u.  (a) If x = eu cosv and y = eu sinv, show that J·J ' = 1. ∂x ∂y ∂y or ∂x2 (b) Verify the chain rule for Jacobians if x = u, 2 2 2 u u u ∂ ∂ ∂ y = u tan v, z = w. x2 + 2 xy + y 2 2 = 2u. ∂x2 ∂x ∂y ∂y Solution.  (a) We have 2





EXAMPLE 5.89

(a) If u = x + y and y = uv, find the Jacobi∂  (x,  y) . an ______ ​     ​ ∂  (u,  v) (b) Show that the functions u = x2 + y2 + z2 , v = x + y + z , w = yz + zx + xy are not independent of one another. Solution.  (a) We are given that u = x + y, y = uv. Therefore x = u - y = u - uv  and  y = uv. Therefore ∂x ∂x ​ ​ ___ ​ ​ __ ∂u ∂v ∂  (x, y) 1 - v -u ​ ______   ​= ∂y = ∂y ___ __ ∂  (u, v) v u ​   ​ ​   ​ ∂u ∂v = u  (1 - v) + uv = u. (b) We have u = x2 + y2 + z2,  v = x + y + z,  w = yz + zx + xy. Then

x = eu cos v,  y = eu sin v. Then

∂x ​ ​ __ ∂x ​ ​ ___ eu cos v -eu sin v ∂   ( x, y) ∂u ∂v J = ​ ______   ​= = u e sinv eu cos u ∂y ∂y ∂  (u, v) ___ ​   ​ ​ __ ​ ∂u ∂v 2u 2 = e cos u + e2u sin2 v = e2u.

One the other hand the given equations yield x2  + y2  = e2u Therefore

and

y. v = tan-1 ​ __ x ​ 

2e2u ___ ​ ∂u  ​= 2x which yields ___ ​ ∂u  ​= ___ ​  x    ​, ∂x ∂x e2u y 2e2u ___ ​ ∂u  ​= 2y which yields ___ ​ ∂u  ​= ___ ​      ,​ ∂y ∂y e2u -y y ∂v ​= _____ ∂v ​= ​ _____ ​ __    ​  = - ___ ​  2u    ​, ​ __ ​  x     ​= ___ ​  x    .​ ∂x x2 + y2 ∂y x2 + y2 e2u e Therefore ∂  (u, v) J ′= ​ ______ ​  = ∂  (x, y)

x y ___ ​  2u    ​ ​ ___     ​ 2u

2u 2y2 e___ e ___ 2x2  ​  + ​ ___ = ​    = ​  4u ​  = ___ ​ 12u   .​ 4u 4u ​  x ___ y e e e e ___ ​     ​  ​  2u    ​ e2u e

e

Hence JJ ′ = 1. (b) We are given thet

M05_Baburam_ISBN _C05.indd 40

1/2/2012 11:58:51 AM

functionS of Several variableS   n 5.41 Then

x = u, y = u tan v, z = w

∂x ___ ∂x ​ ​ __ ​ ___  ​ ​ ∂x   ​ ∂u ∂v ∂w ∂  (x, y, z) ∂y __ ∂y ∂y J = ​ _________   ​  = ​ ___  ​ ​   ​ ​ ___   ​ ∂  (u, v, w) ∂u ∂v ∂w ∂z ​ ​ __ ∂z ​ ​ ___ ∂z   ​ ​ ___ ∂u ∂v ∂w 1 0 0 = tan v u sec2 v 0 = u sec2 v. 0 0 1 Also, from (1), we have y u = x , v = tan-1 _​ x ​and w = z. Therefore ∂  (u, v, w) J ′ = ​ _________ ​  ∂  (x, y, z) ∂u ∂u ​    ​ ​ ___  ​ ___ ​    ​ ∂u ___

1 0 0 ∂y ∂y ∂z -y x ______ ______ ∂v __ ∂v ∂v = ​     ​  ​     ​  0 = ​ __  ​ ​   ​ __ ​   ​ x2 + y2 x2 + y2 ∂x ∂y ∂z 0 0 1 ∂w ​  ___ ​ ___ ​ ∂w ​  ___ ​ ∂w ​  ∂x ∂y ∂z 1    = _____ ​  2 x  2   ​= __________ ​   ​ x +y y 2 _ x  ​ 1 + ​ ​ x ​ ​   ​ y 1   ​,  = ___________ ​  since __ ​ x ​= tan v 2 u  (1 + tan v) ______ = ​  1 2    ​. u sec v Hence J J′ = 1, which proves the chain rule.

[  (  ) ]

EXAMPLE 5.91

__ ​___ p  ​  √ Assuming ​∫  ​ ​ e dx = ​   ​  , prove that 2 0 __ ∞ ​___ p  ​ ​   e-a2. √ -x e cos 2ax dx = ​  ​∫  ​ ​ 2 ∞

2

-x

 

2

 

0

Solution.  Let

∞

F (a) = ​∫  ​ ​ e -x cos 2ax dx. 2

 

Then

0

∞

2 F′  (a) = ​∫  ​ ​ e -x ___ ​ ∂  ​ (cos 2ax) dx  ∂a 0  

M05_Baburam_ISBN _C05.indd 41

(1)

∞



= ​∫  ​ ​ -2xe-x sin 2ax dx 2

 

0

2

= [e-x sin 2ax​]​∞0​ ​- 2a ​∫  ​ e-x cos 2ax dx 2



= - 2aF  (a).

Therefore

F′  (a) _____

 ​  ​ = - 2a. F  (a) Integrating, we get, a2 log F  (a) = - ​ 2__ ​  = -a2 + log c 2 or F  (a) 2 log _____ ​  c     ​= -a or 2

F (a) = c e -a .

∞ __ ​___ p  ​ ​   . Putting a = 0, we get c = F (0) = ​∫  ​ ​ e-x dx = ​ √ 2 0 Hence __ ∞ 2 ​___ p  ​ ​   e -a2. F (a) = ​∫  ​ ​ e- x cos 2ax dx = ​ √ 2 2

 

 

0

5.19  EXTREME VALUES A function f (x, y) of two independent variables x and y is said to have an extreme value at the point (a, b) if the increment Δ f = f (a + h, b + k) - f (a, b) preserves the same sign for all values of h and k whose moduli do not exceed a sufficiently small positive number η. If Δf is negative, then the extreme value is a maximum and if Δf is positive, then the extreme value is a minimum. Necessary and Sufficient Conditions for Extreme Values By Taylor’s theorem, we have Δ f = f (a + b, b + k) - f (a, b) ∂f ∂f  = h ___ ​   ​ (a, b) + k ___ ​   ​ (a, b) ∂x ∂y + terms of second and higher order. Now by taking h and k sufficiently small, the first-order terms can be made to govern the sign of the right-hand side and therefore, of the lefthand side of the previous expansion. Hence, the

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5.42  n  chapter five change in the sign of h and k would change the sign of the left-hand side, that is, of Δf. But if the sign of Δf changes, f (x, y) cannot have an extreme point at (a, b). Hence, as a first condition for the extreme value, we must have ∂f ∂f h ___ ​    ​(a, b) + k ___ ​    ​(a, b) = 0. ∂x ∂y Since the arbitrary increments h and k are independent of each other, we must have ∂f ___

∂f  ​  ​ (a, b) = 0 and ___ ​   ​ (a, b) = 0, ∂x ∂y which are necessary conditions for the existence of extreme points. However, these are not sufficient conditions for the existence of extreme points. Further, a point (a, b) is called a stationary point if fx (a, b) = fy (a, b) = 0. The value f (a, b) is called a stationary value. To find sufficient conditions, let (a, b) be an interior point of the domain of f such that f admits the second-order continuous partial derivatives in the neighborhood of (a, b). Suppose that fx (a, b) fy (a, b) = 0. We further, suppose that Thus,

∂2f ∂2f , r = ___ ​  2  ​  , s = _____ ​      ​ and ∂x∂y ∂x 2 ∂f t = ___ ​  2   ​ , when x = a and y = b. ∂y

fxx (a, b) = r,  fxy (a, b) = s, and fyy (a, b) = t. If (a + b, b + k) is any point in the neighborhood of (a, b), then by Taylor’s theorem, we have

Δ f = f (a + b, b + k) - f (a, b)    = hfx (a, b) + kfy (a, b) + __ ​ 1 ​[h2  fxx (a, b) 2



+ 2hkfxy (a, b) + k2 fyy (a, b)] + R3

= __ ​ 1 ​[rh2 + 2hks + tk 2] + R3, 2 where R3 consists of terms of third and higher orders of small quantities. Thus, by taking h and k sufficiently small, now the second order terms can be made to govern the sign of the right-hand side and therefore, of the left-hand side of the previous expansion. But 1 1 2 2 ___ 2 2 2 ​ __ 2 ​[rh + 2hks + tk ] = ​ 2r  ​ [r h + 2hkrs + rtk ]

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​ 1  ​ [r2h2 + 2hkrs = ___ 2r + rtk2 + k2s2 - k2s2] = ___ ​ 1  ​ [(rh + sk)2 + k2 (rt - s2)]. 2r Since (rh + sk)2 is always positive, it follows that Δf is positive if rt - s2 is positive. Now rt - s2 > 0 if both r and t have the same sign. Thus, the sign of Δf shall be that of r. Therefore, if rt - s2 is positive, we have a maximum or a minimum accordingly, as both r and t are either negative or positive. This condition was first pointed out by Lagrange and is known as Lagrange’s condition. However, if rt = s2, then rh2 + 2hks + tk 2 be1 2 comes ​ __ r  ​(hr + ks) and is, therefore, of the same sign as r or t unless 2 h s __ ​   ​= - _​ r ​, say, for which (hr + ks) vanishes. k In such a case, we must consider terms of higher order in the expansion of f (a + h, b + k). Thus, we may state that 1. The value f (a, b) is an extreme value of f (x, y) if fx  (a, b) = fy  (a, b) = 0 and if rt - s2 > 0. The value is maximum or minimum accordingly as fxx (a, b) or fyy  (a, b) is negative or positive. 2. If rt - s2 < 0, then f (x, y) has no extreme value at (a, b). The point (a, b) is a saddle point in this case. 3. If rt - s2 = 0, the case is doubtful and requires terms of higher order in the expansion of the function. EXAMPLE 5.92

Show that the function f (x, y) = y2 + x2 y + x4 has a minimum value at the origin. Solution.  We have f (x, y) = y2 + x2y + x4. Therefore, fx  = 2xy + 4x3 which yields fx (0, 0) = 0, fy  = 2y + x2 which yields fy (0, 0) = 0,   fxx  = 2y + 12x2 which yields fxx (0, 0) = 0,  fyy  = 2 which yields fyy (0, 0) = 2, and  fxy  = 2x which yields fxy (0, 0) = 0. Hence, at the origin, we have rt - s2 = 0. Thus, further investigation is needed in the case. We write

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functionS of Several variableS  n 5.43

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2 3x4 ​ 1 ​x2  ​ + _____ ​     ​ . f (x, y) = y2 + x2 y + x4 = ​ y + __ 4 2 Then, 2 2 h 3h2 Δf = f (h, k) - f (0, 0) = ​ k + __ ​    ​  ​ + ___ ​     ​ , 2 4 which is always greater than zero for all values of h and k. Hence, f (x, y) has a minimum value at the origin.

EXAMPLE 5.93

Show that the function 3 a a3 ​ x  ​+ __ ​ y  ​ u = xy + __ has a minimum value at (a, a). Solution.  We have Therefore,

a3 a3 . u = xy + __ ​ x  ​+ __ ​ y  ​ 

a  ​   ​= y - __ ​  2 ​yields fx (a, a) = 0, ∂x x



∂u ___



∂u ___



3

a  ​   ​= x - __ ​  2 ​yields fy (a, a) = 0, ∂y y 3

∂2u ___

2a3  ​ 2  ​ = ___ ​  3  ​yields fxx (a, a) = 2, ∂x x

∂f ___ ​  2  ​   = 2, which implies fyy (0, 0) = 2. 2

∂y Thus, rt - s2 = 0 and so, further investigation is required. We have f (x, y) = (x2 - y) (2x2 - y),  f (0, 0) = 0. Therefore, Δf = f (x, y) - f (0, 0) = (x2 - y) (2x2 - y). Thus, Δf is positive, for y < 0 or x2 > y >0 and Δf y is negative, for y > x2 > __ ​   ​> 0. Thus, Δf does not 2 keep the same sign in the neighborhood of (0, 0). Hence, the function does not have a maximum or a minimum at (0, 0). EXAMPLE 5.95

Examine the function sin x + sin y + sin (x + y) for extreme points. Solution.  The given function is f (x, y) = sin x + sin y + sin (x + y). Therefore, fx  = cos x + cos (x + y),  fy  = cos y + cos (x + y),   fxx  = - sin x - sin (x + y),

∂2u _____

 ​   ​ = 1 yields fxy (a, a) = 1, and ∂x∂y ∂u ___ 2

2a  ​ 2  ​ = ___ ​  3 ​ and so, fyy (a, a) = 2. y ∂y We observe that rt - s2 = 4 - 1 = 3 (positive) and r and t too positive. Therefore, u has the minimum at (a, a). Thus, the minimum value of u is u (a, a) = a2 + a2 + a2 = 3a2.

EXAMPLE 5.94

Show that the function f (x, y) = 2x4 - 3x2 y + y2 does not have a maximum or a minimum at (0, 0). Solution.  The given function is f (x, y) = 2x4 - 3x2y + y2. Therefore, ∂f ___ ​    ​= 8x3 - 6xy, which implies fx(0, 0) = 0, ∂x ∂f ___ ​    ​= -3x2 + 2y, which implies fy(0, 0) = 0, ∂y ∂2f ___ ​   2   ​ = 24x2 -6y, which implies fxx(0, 0) = 0, ∂x ∂2f _____ ​      ​= - 6x, which implies fxy (0, 0) = 0, and ∂x∂y

M05_Baburam_ISBN _C05.indd 43

fxy  = - sin (x + y), and

3

  fyy  = - sin y - sin (x + y). For extreme points, we must have fx = fy = 0 and so, cos x + cos (x + y) = 0  (1) and

cos y + cos (x + y) = 0 

(2)

Subtracting (2) from (1), we get cos x = cos y and so, x = y. Also then, cos x + cos 2x = 0 which yields cos 2x = - cos x = cos (π - x) and so, 2x = π - x or π π   ​ is a stationary point. Now x = __ ​   .​ Thus, ​ __ ​ π ​, __ __ 3 3 ​ 3 ​ __ __ ​√3 ​  ___ ​√3 ​  π π ___ __ __ r = fxx   ​   ​, ​   ​  ​= - ​  __   ​- ​     ​= - √ ​ 3 ​ (negative), 2 2 3 3 ​√3 ​  π π ___ __ __ s = fxy ​ ​   ​, ​   ​ ​= ​     ​ 2__ 3 3 __ __ ​√3 ​  ___ ​√3 ​  π π ___ __ __ t = fyy ​ ​   ​, ​   ​ ​= ​     ​- ​     ​= - √ ​ 3 ​ (negative). 2 2 3 3 Thus, 3 ​ 9 ​(positive) rt - s2 = 3 - __ ​   ​= __ 4 4 and r is negative. Hence, the given function has a maximum value at ​ __ ​ π ​, __ ​π ​ ​given by 3  3 

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5.44  n  chapter five

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π ​ π ​  ​= sin __ π π 2π f ​ __ ​   ​, __ ​   ​+ sin __ ​   ​+ sin ___ ​    ​ 3 3 3 __ __ 3 __ 3 __ ​√3 ​  ​√3 ​  ___ ​√3 ​  ____ 3​√3 ​ . = ___ ​     ​+ ___ ​   2 ​  2   ​+ ​  2   ​= ​  2   



EXAMPLE 5.96

Examine the following surface for high- and low points: z = x2 + xy + 3x + 2y + 5. Solution.  We have

∂z ___

∂z  ​  ​= 2x + y + 3, ___ ​   ​= x + 2, ∂x ∂y ∂2z  ​ = 1, and ​ ___ ∂2z  ​= 0. ∂2z  ​ = 2, ​ _____ ​ ___ 2 ∂x∂y ∂y2 ∂x ∂z ∂z For an extreme point, we must have ​ ___  ​= ​ ___  ​= 0 ∂x ∂y and so, 2x + y + 3 = 0 and x + 2 = 0. Solving these equations, we get x = - 2, y = 1. Thus, z can have a maximum or a minimum only at (- 2, 1). Further, ∂2z r = ​ ___2  ​  (- 2, 1) = 2, ∂x 2 _____ s = ​  ∂ z    ​(- 2, 1) = 1, and ∂x∂y 2 ∂z t = ​ ___2  ​  (- 2, 1) = 0. ∂y Therefore, rt - s2 = - 1 (negative) and so, the stationary value of z at (- 2, 1) is neither a maximum nor a minimum. Hence, the surface has no high- or low point.

EXAMPLE 5.97

Locate the stationary points of x 4 + y 4 - 2x 2 + 4xy - 2y2 and determine their nature. Solution.  We have f (x, y) = x4 + y4 - 2x2 + 4xy - 2y2. Therefore,

fxx  = 12x2 - 4, fxy  = 4, and fyy  = 12y2 - 4. At (0,0), we have

r = fxx (0, 0) = - 4,  s = fxy (0, 0) = 4, and



t = fyy (0, 0) = - 4,

and so, rt - s = 0. Thus, at (0, 0), the case is doubtful. The given equation can be written as 2

2

f (x, y) = x4 + y4 - 2 (x - y) . So, 2

f (0, 0) = 0 and f (h, k) = h4 + k4 - 2 (h - k) . We observe that for small quantities of h and k, 2

Δ f = f (h, k) - f (0, 0) = h4 + k 4  - 2 (h - k)

is greater than 0, if h = k and less than 0, if h ≠ k. Since Δf does not preserve the sign, the function has no extreme __ __ value at the origin. At (​√2 ​,  - ​√2 ​)  , we have r = 20, s = 4, and t = 0 so that rt - s2 = 384 (positive). Since r is positive, f (x, y) has__a minimum at this point. __ At (-​√2 ​,  ​√2 ​)  , we have r = 20, s = 4, and t = 20. Thus, rt - s2 is positive. Since __ __ r is positive, f (x, y) has a minimum at (-​√2 ​,  √ ​ 2 ​)  also. Find the minimum value of x2 + y2 + z2 when ax + by + cz = p.

fx  = 4x - 4x + 4y and fy  = 4y   + 4x - 4y. 3

The stationary points are given by fx  = 4x3 - 4x + 4y = 0 fy  = 4y3 + 4x - 4y = 0

(1) (2)

Adding (1) and (2), we get 3

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Solution.  Let f (x, y, z) = x2 + y2 + z2. From the p - ax - by

.​ relation ax + by + cz = p, we get z = ​ ________     c  Putting this value of z in f (x, y, z), we get

x + y = 0 or (x + y) (x - xy + y ) = 0. 3

Also,

EXAMPLE 5.98 3



Therefore, either y = - x or x 2 - xy + y2 = 0. Putting y = - x in __ (1), we__get x (x2 - 2) = 0, which yields x = 0, √ ​ 2 ​,  or - ​√2 ​.  The value __ of y corresponding __ to these values __are 0,__- √ ​ 2 ​,  and ​√__2 ​.  Thus, the __ points (0, 0), (​√2 ​,  - √ ​ 2 ​)  , and (- √ ​ 2 ​,  √ ​ 2 ​)  satisfy (1) and (2). On the other hand, from equation (1) and x2 - xy + y2 = 0 we get (0, 0) as the only real root. Thus, the stationary points are __ __ __ __ (0, 0), (​√2 ​,  -​√2 ​)  , and (-​√2 ​,  √ ​ 2 ​)  .

2

2

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p - ax - by 2 f (x, y, z) = x2  + y2 + ​ ​ __________     ​  ​ c 

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functionS of Several variableS  n 5.45 as a function of two variables x and y. Then, 2a ​  2  ​( p - ax - by) fx  = 2x - ___ c and 2b fy  = 2y - ___ ​  2  ​( p - ax - by). c For extreme points, we must have fx = fy = 0. Thus, 2a 2x - ___ ​  2  ​  ( p - ax - by) = 0 c and 2b 2y - ___ ​  2  ​( p - ax - by) = 0. c Solving these equations, we get ap bp . x = _________ ​  2 2   2 ​    and y = _________ ​  2 2   2 ​   a +b +c a +b +c Now, 2a2 2ab 2b2 fxx  = 2 + ___ ​  2  ​ , fxy = ____ ​  2   ​ , and fyy = 2 + ___ ​  2 ​  , c c c so that

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a2 b2 4a2b2 rt - s2 = 4 ​ 1 + __ ​  2 ​  ​​ 1 + __ ​  2 ​  ​- _____ ​  4    ​  c c c

( 

a2 __

b2 __

)

= 4 ​ 1 + ​  2 ​+ ​  2 ​  ​(positive). c c



2x - 2y + 3y2 + 5x4 = 0 and - 2x + 2y - 3y2 = 0. The origin (0, 0) satisfies these equations. Further, r = fxx (0,0) = 2, s = fxy (0,0) = - 2, t = fyy (0,0) = 2, and so, rt - s2 = 0. Hence, further investigations are required. We rewrite the equation as f (x, y) = (x - y)2 + (x - y) (x2 + xy + y2) + x5. We note that f (0, 0) = 0. But, Δf = f (h, k) - f (0, 0) = f (h, k)



= (h - k)2 + (h - k) (h2 + hk + k2) + k5. In the neighborhood of (0, 0), if h = k, then Δf = k5; which is positive, when k > 0 and negative, when k < 0. Thus, Δf does not keep the same sign in the neighborhood of (0, 0). Hence, f (x, y) cannot have a maximum or a minimum at the point (0, 0). EXAMPLE 5.100

Find the dimensions of the rectangular box, open at the top, of maximum capacity whose surface is 432 sq. cm.

Also, r = fxx is positive. Therefore, f (x, y) has a ap bp minimum at ​ __________ ​  2      ​ , ​ __________      ​  ​and the a + b2 + c2 a2 + b2 + c2

Solution.  Let x, y, and z cm be the dimensions of the box and S be its surface. Then

minimum value is

and V = xyz. (2) We have to maximize V. From (1), we have

EXAMPLE 5.99



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)

p2 . Min.  f (x, y, z) = _________ ​  2 2  2 ​   a +b +c

Show that the function f (x, y) = x2 - 2xy + y2 + x3 - y3 + x5



S = xy + 2yz + 2zx = 432 (given)

432 - xy . z = ​ _______   2y + 2x ​  

( 

Now, ∂V (2y + 2x) (432y - 2xy ) - 2 (432xy - x y ) ___ ​    ​= ​ __________________________________         ​ 2  2

4 

fxx  = 2 + 6x + 20x and  fxy  = - 2, fyy = 2 - 6y.

∂x

3 

For a stationary value of f (x, y), we must have fx  = fy = 0. Thus,

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)

432 - xy 432xy - x2y2 . V = xy ​ ​ _______   ​  ​= ​ __________ ​   2y + 2x    2y + 2x

fx  = 2x - 2y + 3x + 5x and  fy  = - 2x + 2y 3y 2. 2

(3)

Therefore, (2) reduces to

has neither a maximum nor a minimum at (0, 0). Solution.  For the given function,

(1)



2 2

(2y + 2x)

864y2 - 4xy3 - 2x2y2 , = ​ ________________         ​  (2x + 2y)2

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5.46  n  chapter five (2x + 2y) (432x - 2x2y) - 2 (432xy - x2y2)  ​   ​= ​ __________________________________           ​ ∂y (2y + 2x)2 ∂V ___

864x2 - 4x3y - 2x2y2 . = ​ ________________         ​  (2x + 2y)2.

∂V ∂V For stationary points, we must have ___ ​    ​= ___ ​    ​= 0. ∂x ∂y So, 864 - 4xy - 2x2 = 0 (4) 2 864 - 4xy - 2y = 0. (5) Subtracting (5) from (4), we get y = ± x. Substituting x = y in (5), we get 864 864 - 4y2 - 2y2 = 0 or y2 = ____ ​      ​= 124. 6 Thus, x = y = 12 and (3) implies z = 6. It can be verified that rt - s2 > 0 and that r is positive for these values. Hence, the dimensions of the box are x = y = 12cm and z = 6cm. EXAMPLE 5.101

Examine x 3 y 2 (1 - x - y) for extreme points. Solution.  We have f (x, y) = x 3 y 2 (1 - x - y). Therefore,



∂f  ​___  ​= 3x2y2 (1 - x - y) + x 2 y 2  (-1) ∂x = 3x2y2 - 4x3y2 - 3x2y3 and ∂f ___ ​    ​= 2x3y (1 - x - y) + x2y2  (- 1) ∂y = 2x3y - 2x4y - 3x3y2.

For a maximum or a minimum of f, we must ∂f ∂f have ___ ​     ​= ___ ​    ​= 0. Therefore, ∂x ∂y 2 2 x  y (3 - 4x - 3y) = 0 and x 3 y (2 - 2x - 3y) = 0. Solving these equations, we get the stationary 1 points (0, 0) and ​ __ ​ 1  ,​ __ 2 ​ 3 ​  ​ . Further, ∂2f   r =​ ___2   ​= 6xy2-12x2y2 - 6xy3 = 6xy2(1- 2x- y), ∂x ∂2f   s = ​ _____    ​= x2y (6 - 8x - 9y), and ∂x∂y

(  )

∂2f    t = ​ ___2   ​= 2x3 (1 - x - 3y). ∂y Therefore,

M05_Baburam_ISBN _C05.indd 46

(i)  at (0,0), r = 0, t = 0, and s = 0, and so, rt - s2 = 0.  But, Δf = f (h, k) - f (0, 0) = h3k 2 (1 - h - k). Sign is governed by h3k2 which is positive, if h > 0 and negative, if h < 0. Since Δf does not keep the same sign in the neighborhood of (0, 0), the given function does not have a maximum or a minimum value at (0, 0). 1 (ii) at ​ __ ​ 1 ​, __ 2 ​ 3 ​  ​, we have ∂2f 1  , __       r = ​ ___2   ​  ​ __ ​   ​ ​ 1 ​  ​= - __ ​ 1 ​, 9 ∂x 2 3 ∂2f __ 1 1 __ _____       s = ​      ​​ ​    ,​ ​   ​  ​= - ___ ​  1  ​,  and 12 ∂x∂y 2 3 2 ∂ f        t = ​ ___2   ​  ​ __ ​ 1 ​ , __ ​ 1 ​  ​= - __ ​ 1 ​ . 8 ∂y 2 3 Therefore, 2 1   ​  rt - s2 = ​ - __ ​ 1 ​  ​​ - __ ​ 1 ​  ​- ​ - ___ ​  1  ​   ​ = ___ ​  1  ​ - ____ 9 8 12 72 ​ 144 = ____ ​  1     144 ​(positive). But r is negative. Hence, f (x, y) has a maximum 1 at ​ __ ​ 1 ​, __ 2 ​ 3 ​  ​. The maximum value is 1 1 . __ 1 1 __ 1 1   ​  . __ __ ____ f ​ __ ​ 1 ​, __ 2 ​ 3 ​  ​= ​ 8 ​  ​ 9 ​​ 1 - ​ 2 ​- ​ 3 ​  ​= ​ 432

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EXAMPLE 5.102

Find the points where the function x3 + y3 - 3axy has a maximum or a minimum. Solution.  We have

fx = 3x2 - 3ay, fy = 3y2 - 3ax, fxx = 6x, fyy = 6y, and fxy = - 3a.

For extreme points, we have fx = fy = 0 and so, 3x2 - 3ay = 0 and 3y2 - 3ax = 0. Solving the earlier equations, we get two stationary points (0, 0) and (a, a). Further, rt - s2 = 36xy - 9a2. At (0, 0), rt - s2 = - 9a2 (negative). Therefore, there is no extreme point at the origin. At (a, a), we have rt - s2 = 36a2 - 9a2 = 27a2 > 0. Also r at (a, a) is equal to 6a. If a is positive, then r is positive and f (x, y) will have a minimum at (a, a). If a is negative, then r is negative

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functionS of Several variableS  n 5.47 and so, f (x, y) will have a maximum at (a, a) for a < 0. EXAMPLE 5.103

Prove that the rectangular solid of maximum volume which can be inscribed in a sphere is a cube. Solution.  Let x, y, and z be the length, breadth, and height of a rectangular solid. Then, the volume of the solid is

V = xyz.

(1)

Now each diagonal of the rectangular solid passes through the center of the sphere. Therefore, each diagonal is the diameter of the sphere, that is, _________ 2 2 2 =d √​ x + y + z  ​  or

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48d 8 _____ 16d 8 ____8 ____ _____  8  ​- ​ 16d    = ​      ​  = ​      ​   0. Therefore, rt-s2 =​ 64d  81    81 81 27 ​> Since r is negative, it follows that f (x, y) or V 2 d d__ has a maximum value at ​ ___ ​  __    ​, ​ ___     ​  ​. Hence, V ​√3 ​  ​√3 ​  is maximum when x = y = z. Consequently, the

( 

)

solid is a cube. EXAMPLE 5.104

x2 + y2 + z2 = d 2 or

_________ z = ​√d 2 - x2 - y2   ​.

(2)

Therefore, (1) reduces to _________ V = xy ​√d 2 - x2 - y2   ​ or

Solving the preceding equations, we get y = x. d Substituting y = x in (3), we get x = ___ ​  __   ​ .Thus, ​√3 ​  d x = y = ___ ​  __    .​ Hence, from (2), we have z = ___ ​  d__    .​ Thus, ​√3 ​  ​√3 ​  d__ ___ d__ d d__ ___ the stationary point is ​ ​      ​ , ​      ​  ​. At ​ ___ ​  __    ​ , ​ ___    ​  ​, ​√3 ​  ​√3 ​  ​√3 ​  ​√3 ​  4 4 8d  4d  8d 4 r = - ​ ___   ​(negative), s = - ____ ​     ​  , and t = - ​ ___   ​ . 9 9 9

V 2 = x2y2 (d 2 - x2 - y2) = x2 y2d 2 - x4 y2 - x2 y4 = f (x, y).

Then, ∂f    ​ ___  ​= 2xy2d 2-4x3y2-2xy4 = 2xy2 (d 2-2x2 - y2), ∂x ∂f    ​ ___  ​= 2x2yd 2- 2x4y-4x2y3 = 2x2y  (d 2-x2-2y2), ∂y ∂ 2f    ​ ___2  ​= 2d 2y2 - 12x2y2 - 2y4, ∂x ∂ 2f    ​ ___2  ​= 2d 2x2 - 12x2y2 - 2x4, and ∂y ∂ 2f  ​ _____    ​= 4xyd 2 - 8x3y - 8xy3. ∂x∂y

∂f ∂f For stationary points, we have ___ ​    ​= ​ ___  ​= 0. ∂x ∂y Therefore, 2 2 2 d - 2x - y = 0 and d 2 - x2 - 2y2 = 0. (3)

M05_Baburam_ISBN _C05.indd 47

A rectangular box, open at the top, is to have a volume of 32 cubic feet. Determine the dimensions of the box requiring least material for its construction. Solution.  Let S be the surface, and x, y, and z in feet be the edges of the box. Then, S = xy + 2yz + 2zx  (1) and V = xyz = 32 cubic feet (given). (2) 32 ___ From (2), we have z = ​ xy  ​and so,

( 

)

32 S = xy + 2 ( y + x) ___ ​ xy  ​= xy + 64 ​ __ ​ 1x ​+ __ ​ 1y ​ ​. Then, ∂S 64 ∂S 64      ​ ___  ​= y - ___ ​  2  ,​ ​ ___  ​= x - ___ ​  2  ​ , x ∂y y ∂x 2 2 ∂2S 128 . 128 ∂ S ​= 1, and ___ ∂ S  ​= ____ ​ ___ ​  3   ,​ ​ _____    ​  2  ​= ____ ​  3   ​  2 y x ∂x∂y ∂y ∂x

The stationary values are given by ∂S ∂S 64 64  ​ ___  ​= y - ___ ​  2  ​= 0 and ___ ​    ​= x - ___ ​  2   ​= 0. x y ∂x ∂y Solving these equations, we get x = y = 4. Putting these values in (1), we get z = 2. Further, at (4, 4), we have rt - s2 = 3 (positive) and r at (4, 4) is 2 (positive). Therefore, S is minimum for (4, 4). The dimensions of the box are x = 4, y = 4, z = 2.

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5.48  n  chapter five EXAMPLE 5.105

Find the points on the surface z = xy + 1 nearest to the origin. 2

= 0 ∂f ∂f ∂f ∂f df2  = ​ ___2  ​  dx1 + ​ ___2  ​  dx2 + ​ ___2   ​  dx3 + . . . + ​ ___2   ​dxn ∂x1 ∂x2 ∂x3 ∂xn =0

Solution.  If r is the distance from (0, 0, 0) of any point (x, y, z) on the given surface, then

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

  r2  = (x - 0) 2  + ( y - 0) 2  + (z - 0)2 = x2  + y2  + z2

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

= x + y + xy + 1, using the equation of the given surface. Thus, we have a function of two variables given by r2  = x2  + y2  + xy + 1 = f (x, y), say. Then, ∂f ∂f  ​ ___  ​= 2x + y, ​ ___  ​= 2y + x, ∂x ∂y

∂fm ∂f ∂f ∂f dfm  = ​ ___   ​  dx + ​ ___m  ​  dx   + ​ ___m  ​  dx + . . . + ​ ___m  ​  dxn ∂x1 1 ∂x2 2 ∂x3 3 ∂xn



2

2

∂2f ∂2f ∂2f  ​___2   ​ = 2, ​ ___2   ​= 2, and _____ ​      ​= 1. ∂y ∂x∂y ∂x ∂f ∂f The stationary points are given by ​ ___  ​= 0 and ___ ​    ​ ∂x ∂y = 0 and therefore, 2x + y = 0 and 2y + x = 0. Solving the preceding equations, we get x = y = 0 and then, z2 = xy + 1 yields z = ±1. Thus, the stationary points are (0, 0, ±1). Further, at these points, r = 2, s = 1, and t = 2 and so, rt - s2 = 3 (positive). Since r is positive, the value is minimum at (0, 0, ±1). 5.20 LAGRANGE’S METHOD OF UNDETERMINED MULTIPLIERS Let u = f (x1, x2, . . . , xn) be a function of n variables x1, x2,..., xn, which are connected by m equations f1 (x1, x2, . . . , xn) = 0,  f2 (x1, x2, . . . , xn) = 0,..., fm (x1, x2, . . . , xn) = 0, so that only n - m of the variables are independent. For a maximum or a minimum value of u, we must have ∂u ∂u ∂u ∂u du = ​ ___   ​  dx1  + ​ ___   ​  dx2 + ​ ___   ​  dx3 +  . . . + ​ ___   ​  dxn ∂x1 ∂x2 ∂x3 ∂xn = 0. Also, differentiating the given m equations connecting the variables, we have ∂f ∂f ∂f ∂f df1  = ​ ___1  ​  dx1 + ​ ___12   ​dx2 + ​ ___1  ​dx3 + . . . + ​ ___1   ​dxn ∂x1 ∂x3 ∂x ∂xn

M05_Baburam_ISBN _C05.indd 48

= 0. Multiplying the earlier (m + 1) equations, obtained on differentiation, by 1, λ1, λ2, . . . , λm, respectively, and then adding all, we get an equation which may be written as P1dx1  + P2dx2  + P3dx3  +. . . + Pndxn  = 0,   (1) where,

∂f ∂f ∂f ∂f ∂u Pr  = ___ ​     ​+ λ1 ​ ___1  ​  + λ2  ​ ___2 ​  + λ3  ​ ___3  ​  + . . . + λm  ​ ___m  ​ . ∂xr ∂xr ∂xr ∂xr ∂xr The m quantities λ1, λ2,..., λm are at our choice. Let us choose them so as to satisfy the m linear equations. P1 = P2 = P3 = . . . = Pm = 0. Then, the equation (1) reduces to Pm + 1 dxm + 1  + Pm + 2 dxm + 2  +... + Pndxn  = 0. It is indifferent which of the n - m of the n variables are regarded as independent. So, suppose that the variables xm + 1, xm + 2,..., xn are independent, then as the n - m quantities dxm + 1, dm + 2,..., dxn are all independent, their coefficients must be separately zero. Thus, we obtain the additional n - m equations as follows: Pm + 1  = 0, Pm + 2  = 0,..., Pn  = 0. In this way, we get (m + n) equations

f1  = 0, f2  = 0,..., fm  = 0 and P1  = 0,  P2  = 0,..., Pn  = 0,

which together with relation u = f  (x1, x2, ... , xn) determine the m multipliers λ1, λ2, ... ,

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functionS of Several variableS  n 5.49 λm  and the values of n variables x1, x2, ... , xn for which the maximum and minimum values of u are possible. The drawback of the Lagrange’s method of undetermined multipliers is that it does not determine the nature of the stationary point. EXAMPLE 5.105

Find the point of the circle x2 + y2 + z2 = k2 and lx + my + nz = 0 at which the function u = ax2  + by2  + cz2 + 2 fyz + 2 gzx + 2 hxy attains its greatest and the least value. Solution.  We have    u = ax2  + by2  + cz2  + 2fyz + 2gzx + 2hxy, (1)    f1  = lx + my + nz, and

(2)

   f2  = x   + y   + z - k .

(3)

2

2

2

2

For extreme points, we must have du = 0. So, (ax + gz + hy) dx + (hx + by + fz) dy + (gx + fy + cz) dz = 0.

(4)

Also differentiating (2) and (3), we get

ldx + mdy + ndz = 0, and

(5)



xdx + ydy + zdz = 0.

(6)

Multiplying (4), (5), and (6) by 1, λ1, and λ2, respectively, and then by adding all and equating to zero the coefficients of dx, dy, and dz, we get

ax + hy + gz + λ1l + λ2 x = 0,

(7)



hx + by + fz + λ1m + λ2 y = 0, and

(8)



gx + fy + cz + λ1n + λ2z = 0. 

(9)

Multiplying (7), (8), and (9) by x, y, and z, respectively, and then adding all, we get u + λ2  = 0  or  λ2  = - u. Putting λ2 = - u in (7), (8), and (9), we obtain

(a - u) x + hy + gz + λ1l = 0,



hx + (b - u) y + fz + λ1m = 0, and (11)



gx + fy + (c - u) z + λ1n = 0.

Also,

M05_Baburam_ISBN _C05.indd 49

lx + my + nz + λ1 . 0 = 0.

(10) (12) (13)

Eliminating x, y, z, and λ1 from (10), (11), (12), and (13), we get a–u h g h b – u f g f c–u l m n

l m = 0, n 0

which gives the maximum or minimum value of u. EXAMPLE 5.106

Prove that the volume of the greatest rectangular parallelopiped that can be inscribed in the ellipsoid 2 8abc ​ . x2 y z2 ​= 1 is _____ __ ​  2 ​+ ​ __2 ​+ ​ __ ​  __   a b c2 3​√3 ​  Solution.  Let (x, y, z) denote the coordinates of the vertex of the rectangular parallelopiped which lies in the positive octant and let V denote its volume. Volume V is given by V = 8xyz. Its maximum value is to be determined under the x2 condition that it is inscribed in the ellipsoid​ ___2 ​  a y2 z2 + ​ __2 ​+ ​ __ 2 ​= 1. Thus, we have b c  V  = 8xyz, and (1)

2 x2 ​ y  ​+ ​ __ z2 ​- 1. f1  = ​ __2 ​+ __ 2 a b c2

(2)

For an extreme value, we must have

dV = yzdx + zxdy + xydz.

(3)

Also differentiating (2), we get y x df1 = __ ​  2  ​dx + __ ​  2  ​  dy + __ ​ z2  ​  dz. (4) a c b Multiplying (3) and (4) by 1 and λ, respectively, and then adding both and equating the coefficients of dx, dy, and dz to zero, we get λx yz + __ ​  2  ​= 0,  (5) a λy zx + __ ​  2  ​  = 0, (6) b and λz xy + __ ​  2  ​  = 0.  (7) c From (5), (6), and (7), we get

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5.50  n  chapter five

a2yz c2xy b2zx  ​= - ​ ____ ____ λ = - ​ ____  ​= - ​  y    ​  z    x    and so,

dS = ( y + 2z) dx + (x + 2z) dy + 2 (x + y) dz = 0.  (3)

a2yz ____ c2xy . b2zx ____  ​____       ​ = ​        ​ = ​  ​   z    y x

Also, from (2), since V is constant, we have

Dividing throughout by xyz, we get b2 ​ = ​ ___ c2  ​ . a2 ​= ​ __ ​ __ 2 2 y z2 x Then, equation (2) yields a x2 3 ​ __2 ​  = 1  or  x = ___ ​  __    ​ , a 2 ​√3 ​  y b 3 ​ __2 ​  = 1  or  y = ___ ​  __   ​  , and b ​√3 ​  c z2 ​  = 1  or  z = ___ 3 ​ __ ​  __   ​  . c2 ​√3 ​  Thus, the stationary value is at the point a b__ , ___ c ​ ___ ​  __   ​  , ___ ​     ​  ​  __   ​ ​ . ​√3 ​  ​√3 ​  ​√3 ​  Differentiating partially the equation (2) with respect to x, taking y as constant, we get c2x ___ ∂z ∂z ___ ​ 2x2  ​+ __ ​ 2z2  ​___ ​   ​= 0 and so, ​ ___ ​= - ​  2   ​ . a z ∂x ∂x c a Now,

( 

)

(  a z )

∂V ∂z c2x ___ ​    ​= 8yz + 8xy ​ ___ ​ = 8yz + 8xy ​ - ___ ​  2  ​  ​ ∂x

∂x



8c2x2y = 8yz - ______ ​  2     ​  az

and so, 16c2xy ______ 8c2x2y ___ ∂2V c2x  ​ ​- ​ ______ c2x ​ ___2  ​= 8y ​ - ​ ___   ​- ​  2       ​  ·  ​  2   ,​ 2 2     a z a z a z az ∂x which is negative. Hence, V is maximum at a b__ ___ c ​ ___ ​  __    ​, ___ ​      ​, ​  __   ​ ​and ​√3 ​  ​√3 ​  ​√3 ​ 

(  )

( 

)

8abc ​ . Max V = _____ ​  __   3 ​√3 ​ 



yz dx + zx dy + xy dz = 0.

(4)

Multiplying (3) by 1 and (4) by λ and then adding both and equating to zero the coefficients of dx, dy, and dz, we get ( y + 2z) + λ yz = 0

(5)

 (x + 2z) + λ xz = 0

(6)

   2x + 2y + λ xy = 0.

(7)



Multiplying (5) by x and (6) by y and subtracting, we get 2zx - 2zy = 0 or x = y, since z = 0 is not admissible due to the fact that depth cannot be zero. Similarly, from the equations (6) and (7), we get y = 2z. Thus, for a stationary value, the dimensions of the box are x = y = 2z = 4, [using (2)]. Proceeding, as in Example 5.102, we note that ∂ 2f ​ ___2  ​= 2 (positive) and rt - s2 > 0. Thus, at (4, 4, ∂x 2), S has a minimum. Hence, the required dimensions are x = 4, y = 4, and z = 2. EXAMPLE 5.108

Investigate the maximum- and minimum radii vector of the sector of ‘‘surface of elasticity’’ (x2 + y2 + z2) = a2x2 + b2y2 + z2c2, made by the plane lx + my + nz = 0. Solution.  On differentiating, we get

EXAMPLE 5.107

Solve Example 5.100 using Lagrange’s method of undetermined multipliers. Solution.  We have

S = xy + 2yz + 2zx  and

(1)



V = xyz = 32.

(2)

For S to be minimum, we must have

M05_Baburam_ISBN _C05.indd 50



  xdx + ydy + ndz = 0

(1)



a2 xdx + b2 ydy + c2 zdz = 0

(2)



 ldx + mdy + ndz = 0.

(3)

Multiplying (1), (2), and (3) by 1, λ1, and λ2, respectively, and adding and equating to zero the coefficients of dx, dy, and dz, we get      x + a2xλ1 + lλ2 = 0

(4)

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functionS of Several variableS  n 5.51      y + b2yλ1 + mλ2 = 0 

(5)

    

(6)

 z + c2zλ1 + nλ2 = 0. 

Multiplying (4), (5), and (6) by x, y, and z, respectively, and adding we get (x2 + y2 + z2) + (a2 x2 + b2 y2 + c2 z2) λ1 + (lx + my + nz) λ2 = 0 or

​ 12  .​ r2  + λ1 r4  = 0 or λ1 = - __ r Putting this value of λ1 in (4), (5), and (6), we get

λ lr2 λ mr2 λ2nr2 x = ______ ​  22    ,​ y = ______ ​  22 2     ,​ and z =​ _____      ​ . 2  a -r b -r c2 - r2 Substituting these values of x, y, and z in lx + my nz = 0, we get λmr λnr λ l  r ______ ​  2      ​+ ______ ​  2      ​  + _____ ​  2      ​  = 0, 2 2

a - r2 2

or

2 2

2 2

b2 - r2

c2 - r2

n  ​  = 0, m l     ​______  ​+ ______ ​  2  2   ​  + ​ _____ 2  2  a2 - r2 b - r  c - r which is an equation in r giving the required values. 2

2

2

EXAMPLE 5.109

Find the length of the axes of the section of the ­ellipsoid 2 y2 z2 __ ​ x 2 ​ + __ ​  2 ​+ ​ __ 2 ​= 1 by the plane lx + my + nz = 0. a b c

y y + λ1 __ ​  2  ​+ λ2m = 0, and  (5) a z + λ1 __ ​ z2  ​+ λ2n = 0.  (6) a Multiplying (4), (5), and (6) by x, y, and z and adding, we obtain x y2 z2 (x2 + y2 + z2) + λ1 ​ ​ ____22  ​+​ ___2  ​+ ​ __  ​  ​ a b c2

( 

)

+ λ2  (lx + my + nz) = 0

or

r2 + λ1 = 0, which gives λ1 = - r2. Hence, from (4), (5), and (6), we have

λ 2l λ2n λm , x = ​ _____     ,​ y = _____ ​  2 2     ​ and z = ​ _____     .​ 2 r​  2  ​- 1 r __ __ r __ ​  2 ​- 1 ​  2 ​ - 1 2

b

a

c

Putting these values of x, y, and z in lx + my + nz = 0, we get

(

 

)

n2c2   ______ l 2a2   m2b2   ______ λ2 ​ ​ ______ 2 ​  ​= 0. 2 2 ​+ ​  2 2 ​+ ​ r2 - c  r -a r -b Since λ2 ≠ 0, the equation giving the values of r2, the squares of the length of the semi-axes, is

m2b2   n2c2   ______ _____ l 2a2    ​______  ​= 0. 2 ​+ ​  2 2 2 ​+ ​ r2 - b  r - c2 r -a EXAMPLE 5.110

If a, b, and, c are positive and a2 x2  + b2 y2  + c2 0z2 , u = _______________ ​         ​ x2 y2 z2

Solution.  We have to find the extreme values of the function r 2 = x 2 + y 2 + z 2 subject to the conditions 2 x2 y z2 __ ​  2 ​+ ​ __2 ​+ ​ __  ​= 1 and lx + my + nz = 0. a b c2 Differentiation yields

show that a stationary value of u is given by



where μ is the positive root of the cubic



x dx + y dy + z dz = 0 y ​  2  ​dx + __ ​  2  ​dy + __ ​ z2  ​  dz = 0 a b c ldx + mdy + ndz = 0. x __

(1) (2) (3)

Multiplying (1), (2), and (3) by 1, λ1, and λ2, respectively, adding and then equating to zero the coefficients of dx, dy, and dz, we get x x + λ1 __ ​  2  ​+ λ2l = 0,  (4) a

M05_Baburam_ISBN _C05.indd 51

ax2 + by2 + cz2 = 1,

μ μ μ x2=  _________    ​,  y2=​ _________    ​,  and z2=  _________    ​,  2a (μ + a) 2b (μ + b) 2c (μ + c) μ3 - (bc + ca + ab) μ - 2abc = 0. Solution.  We have

a2x2  + b2y2  + c2z 2 , u = ______________ ​       ​   x2 y2 z2

(1)



ax2  + by2  + cz2  = 1

(2)

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5.52  n  chapter five Differentiating (1), we get

Solution.  We have f (x, y) = x2  - xy + y2   - 2x + y,

1  b2 c2  ∑ x3  z 2 + y 2  dx = 0,  

fx  = 2x - y - 2, fy  = - x + 2y + 1. Therefore the stationary points are given by

which on multiplication by x2 y2 z2 yields

fx  = fy  = 0

1 ∑ x ( b2 y 2 + c 2 z 2 ) dx = 0.

(3)

and so 2x - y - 2 = 0  and  - x + y + 1 = 0.

(4)

Solving these equations, we get the stationary 3 ___ 4 point as ​ __ ​ 5 ​, ​ -   5   ​  ​.

Differentiating (2), we get

∑ axdx = 0.

Using Lagrange’s multipliers 1 and μ, we get __ ​ 1x ​(b2 y2 + c2 z2) = μ ax or b2y2 + c2z2 = μ ax2,  (5) __ ​ 1y ​(c2 z2 + a2 x2) = μ by or c2z2 + a2x2 = μ by2,



(6)

and __ ​ 1z ​(a2 x2 + b2 y2) = μcz or a2 x2 + b2 y2 = μcz2. (7) Then, (6) + (7) - (5) yields 2a2 x2  = μ (by2 + cz2 - ax2)  = μ (1 - 2ax2), using (2).

  Thus,

μ 2a (a + μ) x = μ  or  x = _________ ​     .​  2

2

2a (μ + a)

Similarly, we obtain μ μ y2 = _________ ​       ​  and  z2 = _________ ​       .​ 2b ( μ + b) 2c ( μ + c) Substituting these values of x , y , and z in (2), we have μ μ μ ________ ​       ​+ ________ ​       ​+ ________ ​       ​= 1 2 (a + μ) 2 (b + μ) 2 (c + μ) 2

or

2

2

μ3 - (bc + ca + ab) μ - 2abc = 0.

(8)

Since a, b, and c are positive, any one of (5), (6), or (7) shows that μ must be positive. Hence, μ is a positive root of (8). EXAMPLE 5.111

Find the stationary points of x2  - xy + y2   - 2x + y.

M05_Baburam_ISBN _C05.indd 52

( 

)

EXAMPLE 5.112

The temperature u (x, y, z) at any point in space is u = 400 xyz2. Find the highest temperature on surface of the sphere x2  + y2  + z2  = 1. Solution.  We have     u (x, y, z) = 400 xyz2  = 400 xy (1 - y2 - x2 ) = 400 xy - 400 xy3 - 400 x3y, which is a function of two variables x and y. Then ux  = 400y - 400y3  - 1200x2y, uy  = 400x - 1200xy2  - 400x3. For extreme points, we must have ux = uy = 0. Thus, 1 - y2 - 3x2 = 0 and 1 - 3y2 - x2 = 0. Solving these equations, we get x = ± __ ​ 1 ,​ y = ± __ ​ 1 .​ 2 2 1 ​  ​,​ Thus, we have four stationary points ​ __ ​ 1 ​, ​ __ 2 2 1 ,​ __ - __ ​ 1 ,​ - __ ​ 1 ​  ,​ ​ __ ​ 1 ,​ - __ ​ 1 ​  ​and ​ - ​ __ ​ 1 ​  .​ Also, 2 2 2 2 2 2    uxx  = - 2400xy, uxy  = 400 - 1200 y2  - 1200 x2

( 

) ( 

) ( 

uyy  = - 2400xy.

(  )

( 

)

(  )

)

1 1 , __ 1 , __ 2 At ​ __ ​ 1 ,​ ​ __ 2 2 ​  ​and ​ - ​ 2 ​ - ​ 2 ​  ​ rt - s is positive and r is negative. Therefore, maximum exist at these _________ 1 1 __ __ , , points. Further, x = ​   ​ y = ​   ​ give z = ​√1 - x2 - y2 ​   2 2 = ___ ​  1__   . ​ Therefore, ​√2 ​  1 __ 1 max u (x, y, z) = 400 ​ __ ​ 1 ​  ​​ __ 2 ​ 2 ​  ​​ ​ 2 ​  ​= 50.

(  )(  ) (  )

EXAMPLE 5.113

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functionS of Several variableS  n 5.53 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. Solution.  We have u (x, y) = x2 + 2y2 - x, so that ux = 2x - 1, uy = 4y. 1 Then ux = uy = 0 imply x = __ 2​   ​, y = 0. Also uxx = 2, uyy = 4 and uxy = 0. Then rt - s2 = 8 (+ ve) and r = 2. ​ 1 ​, 0  ​. Therefore the Therefore u is minimum at ​ __ 2 1 __ coldest point is ​ ​   ​, 0  ​. 2

(  )

(  )

EXAMPLE 5.114

Find the minimum value of x2 + y2 + z2 when x + y + z = 3a. Solution.  Special case of Example 5.94 (putting a = b = c = 1 and p = 3a in that example). EXERCISES 1. Let f : ℜ2 → ℜ be a continuous function. Define f  : ℜ2 → ℜ by

{

    f (x, y) if (x, y) ≠ (0, 0) f (x, y) =  f (x, y) + 1 if (x, y) = (0, 0). Show that f  is not continuous at (0, 0). Hint: f  (x, y) = f (x, y) + 1 for (0, 0) and so, f  (0, 0) = f (0, 0) + 1. Since f is continuous, it is continuous at (0, 0) also. So ​      lim  ​f (x, y) = (x, y)→(0,0)

f  (0, 0). Then, ​      lim ​f  (x, y) =      ​  lim ​f (x, y) = (x, y)→(0,0)

(x, y)→(0,0)

lim  ​f  (x, y) ≠ f (0,0). Hence, f (0, 0). Thus, ​      (x,y)→(0,0)

f  is not continuous at (0, 0). 2. Show that f (x, y) = x2 + y - 1 is continuous at (1, - 2).  __ 1

3. Prove that the function f (x, y) = (| xy |) ​ 2 ​ is not ∂f differentiable at the point (0, 0) but that __ ​   ​  ∂x ∂f and ​ __  ​exist at the origin and have value zero. ∂y Deduce that these two partial derivatives are continuous except at the origin. Hint: If f (x, y) is differentiable at (0, 0),

M05_Baburam_ISBN _C05.indd 53

then

______ ∂f ∂f D f = f (h, k) = h ​ __  ​+ k ​ __  ​+ e ​√h2 + k2 ​,  ∂x ∂y ______ 2 2 where e→0 as ​√h + k  ​  → 0. Since fx = fy = 0 ______ at (0, 0), f__1 (h, k) = 0 + 0 + e  ​√h2 + k2 ​  and so,  2 [|hk|] ​   ​  . ______  e = ​ _______  ​  Put h = r cos q and k = r sin q, so 2 ​√h______   +  k2 ​  __________ that ​√h2 + k2 ​  = r. Then, e = √ ​ |sin q cos q | ​  and __________ so,     ​lim  ​e = ​√|sin q cos q | ​=  0, which is absurd. r→0

Hence, f is not differentiable at (0, 0). y y 4. If u = xf  ​ _​ x ​ ​+ ψ ​ _​ x ​ ​, prove that

(  ) (  )

∂ u    ∂2u ​ + 2xy ​ ____ ∂2u ​ = 0. x2  ​ ___ ​+ y2  ​ ___ 2 ∂x∂y ∂x ∂y2 2

∂3u ∂3u . 5. If u = x y, show that _____ ​  2     ​= ​ ______      ​ ∂x ∂y ∂x∂y∂x

(  )

y ∂z ∂z 6. If z = f ​ _​ x ​ ​, show that x ​ __ ​+ y ​ __ ​= 0. ∂x ∂y x2 + y2 7. If z = ______ ​     ​, show that x+y

( 

( 

)

)

∂z ∂z 2 ∂z ∂z ​ ___ ​   ​- ​ ___ ​ ​ = 4 ​ 1 - ​ ___ ​- ​ ___ ​ ​. ∂x ∂y ∂x ∂y  1 - __

8. If u = (1 - 2xy + y2) ​ 2 ​ , show that

{ 

}

{  }

∂   ​ (1 - x2) ___ ∂  ​​ y2 ___ ​ __ ​ ∂u  ​ ​+​ __ ​ ∂u  ​ ​= 0. ∂x ∂y ∂x ∂y ∂2z  ​+ __ ∂2z  ​, where a2x2 9. Find the value of __ ​ 12  ​ ​ ___ ​ 12  ​ ​ ___ 2 2 2 2 2 2 + b y - c z = 0. a ∂x b ∂y Ans. ___ ​  12    ​.  cz y _z _x ∂u ∂u _ ___ ___ 10. If u = ​ z ​ + ​ x ​ + ​ y ​, show that x ​    ​ + y ​    ​ + ∂x ∂y ∂u z ​ ___  ​= 0. ∂z 11. If u = ex  (x cos y - y sin y), show that uxx + uyy = 0. 12. Find the envelope of a system of concentric and coaxial ellipses of constant area.

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5.54  n  chapter five 2 x2 y Hint: Area of the ellipse __ ​  2  ​+ __ ​  2  ​= 1 is p ab, a b given to be constant say equal to p c2 and so, y2 c2 c2 x2 __ ab = c2 or b = __ ​ a ​  . Putting b = __ ​ a ​ in __ ​  2  ​+ ​  2  ​= 1, we 2 a b x2 y get ​ __2  ​+ ​ ____   ​ = 0. Differentiating partially with 4 2 a ca c2x respect to a gives a = ​ ___ y    ​. Putting this value y2 x2 ____ __ of a in ​  2  ​+ ​  4  2 ​ = 0, we get 2xy = c2, which a ca is the required equation of the envelope. x __y 13. Find the envelope of the straight line ​ __  ​+ ​   ​= a b 1, where a2 + b2 = c2. __ __ __ Ans. √ ​ x ​ + √ ​ y ​ = √ ​ c ​.  14. Show that the evolute of the rectangular hyperbola xy = c 22 is  2  2 __

__

x4 + y4 ∂u ∂u 15. If u = log ​ ______ , show that x ​ ___  ​+ y ​ ___  ​= 3. x + y ​  ∂x ∂y x4 + y4 u ______ Hint: e = ​   ​(A   homogeneous function of x+y degree 3 in x and y). x+y __ ​  16. If u = sin-1 ​ _______ ​  __   ​, show that ​√x ​ + ​√y ​ 

)

x ___ ​ ∂u  ​+ y ___ ​ ∂u  ​= __ ​ 1 ​ tan u. ∂x ∂y 2 x+y __   __  Hint: sin u = ​ _______  ​ is a homogeneous ​ x ​  + √ ​ y ​  √ function of degree __ ​ 1 ​ . 2 x+y -1 _______ __ ​  17. If u = sin ​ ​  __   ​, show that ​√x ​ + ​√y ​  ∂2u    ∂2u ​ + 2xy ​ ____ ∂2u ​ = ___________ .​ x2 ​ ____ ​+ y2  ​ ___ ​ - sin u cos2u        2 ∂x∂y 4  cos3 u ∂x ∂y2 y2 18. If u = tan-1 ​ __ ​ x ​  ​, show that ∂2u    ∂2u ​ + 2xy ​ ____ ∂2u ​ = - sin 2u sin2 u. x2  ​ ___ ​+ y2 ​ ___ 2 ∂x∂y ∂x ∂y2

( 

)

(  )

( 

)

x +y +z 19. If u = sin ​ __________ ​     ​  ​, show that -1

3

3

3

ax + by + cz

∂u  ​+ y ​ ___ ∂u  ​+ z ​ ___ ∂u  ​= 2 tan u. x ​ ___ ∂x ∂y ∂z x+y -1 _______ __ ​,  20. If u = cos ​  __   show that ​ x ​ + ​√y ​  √

M05_Baburam_ISBN _C05.indd 54

(  )

√

d  y 2a2 x2 ___ ​  2 ​ + _____ ​  4 ​   = 0. 2

dx

__

(x + y) ​ 3 ​ - (x - y) ​ 3 ​ = (4c) ​ 3 ​ .

( 

​ ∂u  ​+ y ___ ​ ∂u  ​= - __ ​ 1 ​ cot u. x ___ ∂x ∂y 2 y _ 21. If u = tan-1 ​ ​ x ​ ​, where x = e t - e -t and y = e t du + e -t, find ___ ​   ​ . dt -2    Ans. ​ _______  ​. e2x + e-2x ax + by 22. If z = e f (ax - by), show that ∂z ​+ a ​ __ ∂z ​= 2ab z. b ​ __ ∂x ∂y _____ _____ dy 23. If ​√1 - x2   ​+ ​√1 - y2   ​= a(x - y), find __ ​   ​. dx______ 1 - y2 . Ans. ​ ​ _____2   ​ ​  1-x 24. If y3 - 3ax2 + x3 = 0, show that y

25. If x = r cosq and y = r sin q, show that 2

2

[ (  ) (  ) ]

∂r 2 ∂r 2 ∂ r ∂ r 1 ___ ​ ___2  ​+ ​ ___2  ​= __ ​ r ​​ ​ ​   ​ ​ + ​ ___ ​   ​ ​   ​. ∂y ∂x ∂x ∂y ___ 26. If x + y = 2eq cosf  and x - y = 2 ​√-1 ​ eq sin f , show that 2 ∂2V  ​  ∂2V  ​= 4xy ​ ____ V  ​+ ​ ___ . ​ ∂___ 2 2 ∂x∂y ∂q  ∂f  27. Expand ex log (1 + y) in a Taylor’s series in the neighborhood of the point (0, 0). y2 x2y ___ xy2 1 2 Ans. y + xy - ​ __ ​ + ___ ​   ​ - ​   ​ + __ ​   ​ y - … . 2 2 2 3 28. Expand e x cos y in powers of x and y up to third-degree terms. Ans. 1 + x + __ ​ 1 ​ (x2 - y2) + __ ​ 1 ​ (x3 - 3xy2) + … . 2 6 1 ​  29. show that for 0 < q < 1, sin x sin y = xy - ​ __ 6 [(x3 + 3xy2) cos q x sin q y + (y3 + 3x2y) sin q x cos q y]. Hint: Use Maclaurin’s theorem. 30. If x = c cos u cosh v and y = c sin u sinh v, show that ∂(x, y) __ ______ ​     ​= ​ 1 ​ c2 (cos 2u - cosh 2v). ∂(u, v) 2 y , x , z    31. If u = ​ ____     ​ v = ​ ____     ​ and w = ​ ____  ​, show y - z z - x z - y that ∂  (u, v, w) ​ _________ ​  = 0. ∂  (x, y, z)

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functionS of Several variableS  n 5.55 32. If x = r cos θ, y = r sin θ, and z = z, e­ valuate ∂  (x, y, z) . ​ ________   ​ ∂  (r, θ, z) Ans. r.

log  (1 + sin a cos x) 38. Show that ∫​   ​ _________________ ​  dx = p a. cos x    ​ 

∂  (u, v) . ______ u  +  v 33. If x = uv and y = _____ ​     ​, determine ​   ​ 

e-x -ax 39. Show that ∫​   ​ ​ __ x ​ (1 - e ) dx = log (1 + a),

u  -  v

∂  (x, y) (u  -  v)2 . Ans. ​ ______       ​ 4uv

34. The roots of the equation in λ (λ - x )3 + (λ - y )3 + (λ - z)3 = 0 are u, v, and w. Show that

Hint: The equation simplifies to λ3 - (x + y + z)  λ2 + (x2 + y2 + z2) λ - __ ​ 1 ​ (x3 + y3 + z3) = 0. 3 Let x + y + z = ξ, x2 + y2 + z2 = h, and x3 + y3 + z3 =

ξ. Then u + v + w = ξ, uv + vw + uw =η, and uvw ∂  (ξ,  η,  ζ ) ∂  (ξ,  η,  ζ ) . = ξ. Find ​ _________ ​  and ​ _________ ​   Then ∂  (x,  y,  z) ∂  (u,  v,  w) ∂  (u, v, w) _________ ∂  (ξ, η, ζ ) . ∂  (u, v, w) . _________ _________ ​   ​  = ​     ​ ​   ​  ∂  (x, y, z) ∂  (ξ, η, ζ ) ∂  (x, y, z) 35. If U = x + y - z, V = x - y + z, and W = x2 + y2 + z2 - 2yz, show that U, V and W are connected by a functional relation, and find that functional relation. ∂ (U, V, W) __________ Hint: Show that ​   ​    = 0. Further, U ∂  (x, y, z) + V = 2x, U - V = 2  ( y - z). Then (U + V)2 + (U - V )2 = 4 (x2 + y2 + z2 - 2yz) = 4W. 36. If u = x2 - 2y, v = x + y + z, and w = x - 2y = ∂  (u, v, w). _________ x - 2y + 3z find ​   ​  ∂  (x, y, z) Ans. 10x - 2. p

37. Evaluate ∫​   ​  ​log (1 + a cos x) dx using Leibnitz 's Role 0 _____ Ans. p log ​ __ ​ 1 ​ + __ ​ 1 ​ √ ​ 1 - a2 ​  ​. 2 2

M05_Baburam_ISBN _C05.indd 55

0

∞ 0

(a > -1).

x

dx x 40. Differentiating ∫​   ​ ______ ​  2   2 ​  = __ ​ 1a ​tan-1 __ ​ a ​under the 0 x + a x . ​  dx   ​  integral sign , find the value of ∫​   ​ ________ 2 0 (x2  +  a2) __  ​p   ​ 2

(y  -  z) (z  -  x) (x  -  y) . ∂ (u,  v,  w) ​ _________ ​  = -2 ​ __________________       ​ (v  -  w) (w  -  u) (u  -  v) ∂ (x,  y,  z)

( 

∞

)

dx 41. Evaluate ∫​   ​  ​________________ ​  2 2       ​. 2 2 2 0 (a  cos x + b  sin x) Hint: Putting tan x = t, we get  ​p__  ​ 2

dx p    ​ . ​∫  ​  ​________________ ​  2       ​= ​ ____ 2 2 2 2ab 0 a cos x + b sin x  se Leibnitz's Rule, first differentiating with U respect to a and then with respect to b. We shall get  ​p__  ​ 2

cos2 x dx p   ​ and __________________        ​= ​ ____ ​∫  ​  ​​  2 2 2 2 2 4a3b 0 (a  cos  x + b  sin  x)  ​p__  ​ 2

sin2 x dx p   ​  . __________________        ​= ​ ____ ​∫  ​  ​​  2 2 2 2 2 3 (a  cos  x + b  sin  x) 4ab 0 Adding these two results, we get the value of p (a2 + b2). the given integral as ​ ________  ​    4a3b3 42. If the perimeter of a triangle is constant, show that its area is maximum when the triangle is equilateral. Hint: 2s = a + b + c c = 2s - a - b and __________________ Δ=√ ​ s_____________________ (s - a) (s - b)   (s - c) ​ = ​√s (s - a) (s - b) (a   + b - s) ​ Take f (a, b) = Δ2 = s (s - a) (s - b) (a + b - s) and find fa and fb etc, and proceed. 43. Find the points (x, y), where the function xy (1 - x - y) is either maximum or minimum.

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5.56  n  chapter five

(  )

1 . Ans. ​ __ ​ 1  ​, __ 3 ​ 3 ​  ​  44. Find a point within a triangle such that the sum of the squares of its distances from the three vertices is the minimum. x + x + x _________ y +y +y Ans. ​ _________ ​  1 2  3  ​, ​  1 2  3    ​  ​   3 3 (centroid of the triangle).

( 

)

45. Find the point on the plane 2x + 3y - z = 12 that is nearest to the origin. Hint: Distance form the origin is ____________________ l = ​√x2 + y2  + (2x + 3y   - 12)2 ​. Put f (x, y) = x2 + y2 + (2x + 3y - 12)2 and proceed. 18 , -6 ___ . , ___ Ans. ​ ___ ​ 12 7  ​ ​  7  ​ ​  7  ​  ​

( 

)

46. Discuss the maxima and minima of u = 2 sin __ ​ 1 ​(x + y) cos __ ​ 1 ​(x - y) + cos (x + y). 2 2 47. Find the extreme values of 2 (x - y ) - x + y4. Ans. Max at (± 1,  0) and Min at (0, ± 1). 2

M05_Baburam_ISBN _C05.indd 56

2

4

48. Find the maximum value of xm  yn  zp  subject to the condition x + y + z = a. Hint: Use Lagrange’s method of undetermined multipliers. ma     na    _________ , Ans. x = _________ ​ m + n + p ​, y = ​ m + n   + p​ pa .​ and z = _________ ​ m + n   + p   mmnnpm + n + p . Max. value = _______________ ​        ​  (m + n + p)m + n + p 49. Divide 24 into three parts such that the continued product of the first, square of the second, and the cube of the third part may be maximum. Hint: Find the Max of xy2z3 subject to the condition x + y + z = 24. Also, can be obtained from Exercise 7 by putting a = 24, m = 1, n = 2, and p = 3. 50. The temperature T at any point (x, y, z) in space is T = 400 xyz2. Find the highest temperature on the surface of the unit sphere x2 + y2 + z2 = 1. Ans. 50.

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6 6.1

Tangents and Normals Hence, the equation of the tangent at P is

INTRODUCTION

We know that a tangent line to a curve C at a point P is the line, in the limiting position, if it exists, of the secant through the point P and a variable point Q on C as Q moves along C closer to P. Also we know that the slope of tangent to the curve y = f ( x) at ( x , y ) is f '( x).

y − y1 =

dy ( x − x1 ) . dx

(1)

On the x – axis, we have y = 0 and so from (1) x = x1 −

y y dx , = x1 − dy dy dx

which is intercept cut off by the tangent on the x-axis. Similarly, on the y-axis, we have x = 0 and so y = y1 = x

dy dx

is the intercept cut off by the tangent on the y-axis. 6.3

In the figure above , the line P T is tangent to the curve at P, T R is called sub-tangent and RM is called sub – normal. 6.2

EQUATION OF THE TANGENT AT A POINT OF A CURVE

Let P(x1, y1) be a point in a curve C. Then equation of a line passing through P(x1, y1) and with slope m is given by y − y1 = m( x − x1 ), where m = tan φ . If the line is tangent to the curve at P, then m = tan φ =

M06_Baburam_ISBN _C06.indd 1

dy dx

.

EQUATION OF THE NORMAL AT A POINT P ( x1 , y1 ) OF A CURVE We know that the normal to the curve C at a point P is the line that passes through P and is perpendicular to the tangent line to the curve at the point P. Thus if m is the slope of the tangent line, then slope m′ of the normal satisfie 1 1 . m' = − = − dy m dx Hence the equation of the normal at P(x1, y1) is y − y1 = −

dx ( x − x1 ) . dy

EXAMPLE 6.1 −x

Prove that ax + b = 1 touches the curve y = be a at the point where the curve crosses the axis of y. y

12/7/2011 5:56:44 PM

6.2 n chapter Six For intercept on the x – axis, y = 0 and so (1) yields

Solution. The given curve is y = be

x − a

.

1

Differentiating with respect to x, we have dy b −x =− e a. (1) dx a The given curve cross the axis of y at the point (0, b). The slope at this point is

1 2  1  = x3  x3 + y 3   

2

1 2  2  = y3  x3 + y3  .  

2

X 2 +Y2 = =

Hence the line ax + b = 1 touches the curve at the point (0, b), which is the point at which the curve crosses the y – axis. y

2

4

Solution. The give curve is 2 3

2

= a 3 .a 3

= a 2 = a (constant). EXAMPLE 6.3

n

Differentiating with respect to x, we get 2 − 13 2 − 13 dy x + y =0 3 3 dx

()

n

n

n

 x  y   +   = 2 . a b

or

Differentiating we get

1

dy  y 3 = −   (slope of the tangent).  x dx

n n −1 n n −1 dy x + n y =0 dx an b

Hence the equation of the tangent at (x, y) is  y Y − y = −   ( X − x) .  x

4

Show that the curve ( ax ) + by = 2 touches the y straight line ax + b = 2 at the point (a, b), whatever be the value of n. Solution. The equation of the curve is

2 3

x +y =a .

1 3

2

x3a3 + y3a3

4

Find the equation of 2 the 2tangent at any point 2 (x, y) to the curve x 3 + y 3 = a 3 . Show that the portion of the tangent intercepted between the axes is of constant length.

M06_Baburam_ISBN _C06.indd 2

2 2 2 2  2   2  x3  x3 + y3  + y3  x3 + y3     

4 2  2  = a3  x3 + y3   

EXAMPLE 6.2

2 3

(3)

Hence the portion of the tangent intercepted between the axes is equal to

x y + = 1. a b

or

1

Y = y + x3 y3

b y − b = − ( x − 0) a y x −1 = − b a

or

(2)

For intercept of the y – axis, x = 0 and so (1) yields

dy b b = − e0 = − . dx a a Thus the equation of the tangent at (0, b) is or

2

X = x + x3 y 3

or (1)

dy b n x n −1 = − n . n −1 . dx a y

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6.3 tangentS and norMalS n  or

b  dy  =− .   dx ( a ,b ) a

Therefore

Hence the equation of the tangent to the curve at the point (a, b) is b y − b = − ( x − a) a or x y + = 2. a b Hence the line ax + b = 2 touches the given curve at (a, b) for any arbitrary value of n. y

EXAMPLE 6.4

The straight line + = 2 touches the curve n ( ax )n + by = 2 for all values of n. Find the point of contact. y b

x a

()

Solution. Let (x1, y1) be the point of contact. Since the point of contact lies on both the curve and the tangent, we have n

n

 x1   y1    +   = 2 for all n. a b

y2 +

Solution. The given curve cuts the x – axis at x = 7, y = 0. Thus we wish to find the equation of the tangent and normal to the given curve at the point (7, 0). Differentiating the equation of the given curve, we have 2 xy + x 2

dy 2 [ x − 5 x + 6] = 1 − 2 xy + 5 y dx

(1)

The equation of the tangent at (x, y) is Y − y = cos x( X − x) . Since tangent passes through the origin, we have 0 − y = cos x(0 − x)

M06_Baburam_ISBN _C06.indd 3

dy 1 − 2 xy + 5 y . = 2 dx x − 5x + 6

Therefore 1 1 .  dy  = =   dx (7,0) 49 − 35 + 6 20

EXAMPLE 6.5

Solution. The equation of the curve is y = sin x and so dy = cos x . dx

dy dy  dy  − 5  y + x  + 6 − −1 = 0 dx dx  dx 

or

or

Tangents are drawn from the origin to the curve y = sin x. Show that the point of contact lies on x2y2 = x2 – y2.

x2 y 2 = x2 − y 2 .

Find the equation of the tangent and normal to the curve y (x – 2) (x – 3) – x + 7 = 0 at the point where it cuts the x axis.

(2)

The only solution of (1) and (2) is x1= a, y1 = b. Hence (a, b) is the required point of contact.

y2 = 1 or x2

EXAMPLE 6.6

(1)

and x1 y1 + = 2. a b

y = cos x . (2) x The point of contact lie on the locus given by (1) and (2). Squaring and adding (1) and (2), we get

Hence the equation of the tangent at (7, 0) is y−0 = or

1 ( x − 7) 20

x − 20 y = 7 .

For normal, the slope is – 20. Therefore the equation of the normal is y − 0 = −20( x − 7) or 20 x + y = 140 .

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6.4 n chapter Six EXAMPLE 6.7

EXAMPLE 6.9

At what point is the tangent to the curve y = log x parallel to the chord joining the points (0, 0) and (0, 1).

If the normal to the curve x 3 + y 3 = a 3 makes an angle φ with the axis of x, show that its equation is

Solution. The chord joining (0, 0) and (0, 1) is y – axis. The given curve is y = log x. Differentiating, we get dy 1 . = dx x Therefore slope of the tangent is 1x . But it is parallel to the y – axis. But slope of y – axis is

y cos φ − x sin φ = a cos 2φ . Solution. The equation of the curve is

= ∞ or x = 0. Putting x = 0 in the equation y = log x, we get y = −∞ . Hence the point of contact is (0, − ∞).

or

tan π2 = ∞. Therefore

1 x

EXAMPLE 6.8

Find the equation of the tangent at the point (x, y) to the curve m

m

2

m

m

 x  y   +   = 1 . a b Differentiating both sides with respect to x, we get m m −1 m m −1 dy x + m y =0 dx am b or

2 − 13 2 − 13 dy x + y =0 3 3 dx 1

dy y3 =− 1. dx x3 Thus the slope of the normal is

dy b x = − m . m −1 . dx a y Therefore, the equation of the tangent is Y−y=−

m

or X x m −1 Y y m −1 x m y m + = m + m =1, am bm a b since (x, y) lies on the curve. Hence the equation of the tangent is Xx m −1 Y y m −1 + = 1. am bm

M06_Baburam_ISBN _C06.indd 4

.But it is given

1

that slope of the normal is tan φ . Therefore 1

x3 1

y3

= tan φ.

(2)

Squaring (2), we get 2

2

y 3 tan 2 φ = x 3 . 2 3

Putting this value of x in (1), we get 2

2

2

y 3 tan 2 φ + y 3 = a 3 or 2

y 3 = a 3 cos 2 φ or

y = a cos3 φ.

Putting this value of y in (2), we get

m −1

b x . ( X − x) a m y m −1

1

x3 y3

2

m −1

m

2

2 2 2 (1) x3 + y3 = a3 . Differentiating both sides with respects to x, we have

 x  y   +   = 1 . a b Solution. The given curve is

2

1

1

x 3 = a 3 sin φ or

x = a sin 3 φ.

Thus tan φ is the slope of the normal at (a sin 3 φ , a cos3 φ ) .Hence the equation of the normal at this point is y − a cos3 φ = tan φ ( x − a sin 3 φ ) =

sin φ ( x − a sin 3 φ ) cos φ

12/7/2011 5:56:46 PM

6.5 tangentS and norMalS n  or or

y cos φ − a cos 4 φ = x sin φ − a sin 4 φ x sin φ − y cos φ + a (cos 2 φ − sin 2 φ ) (cos 2 φ + sin 2 φ ) = 0

or 6.4

Solution. The equation of the curve is xm y m = am+ n . Taking log of both sides, we have m log x + n log y = ( m + n ) log a . Differentiating both sides with respect to x, we get m n dy + =0 x y dx

x sin φ − y cos φ + a cos 2φ = 0 . LENGTHS OF TANGENT, NORMAL, SUBTANGENT AND SUBNORMAL AT ANY POINT OF A CURVE

(i) Length of the tangent. The length of the normal is given by PT = T R cosec φ = y 1 + cot 2 φ 2

 dx  dy = y 1 +   , since tan φ =  dy  dx

(ii) Length of the normal We have PM = PR sec φ

dy my . =− dx nx Therefore the length of the sub-tangent is equal is y

2

 dy  = y 1+   .  dx  (iii) Length of the sub-tangent We have T R = P R cot φ

Thus, we note that length of the sub – tangent n = − (constant).. x m Hence the result.

dx . dy

Find the lengths of the tangent, sub-tangent and sub-normal at the point t = π2 on the cycloid x = a (t + sin t ) , y = a (1 − cos t ). Solution. The equation of the curve is x = a (t + sin t ) , y = a (1 − cos t ) . Therefore dx dy = a (1 + cos t ), and = a (sin t ).. dt dt Thus dy dy dt a sin t = = dx dx a (1 + cos t ) dt 2 a sin 2t cos 2t = = tan 2t 2 a cos 2 2t

(iv) Length of the sub-normal We have R M = P R tan φ =y

dy . dx

EXAMPLE 6.10

Show that the sub-tangent at any point of the curve xmyn = am+n varies as the abscissa of the point of contact.

M06_Baburam_ISBN _C06.indd 5

dx n = − x. dy m

EXAMPLE 6.11

= y 1 + tan 2 φ

=y

or

and so  dy    = 1.. dx t = π 2

12/7/2011 5:56:47 PM

6.6 n chapter Six Also taking t = π2 , we get y2 = a from the equation of the curve. Hence (i)

 dx  Length of tangent = y 1 +    dy 

2

Ans. y − x = 2a − π2a

dx Length of sub − tangent = y dy = a.1 = a.

 dy  (iii) Length of the normal = y 1 +    dx 

2

= a 1 + 1 = a 2. (iv)

Length of the sub − normal = y dx dy

= a.1 = a. EXAMPLE 6.12

In the catenary y = c cosh cx , show that the length of the sub-tangent is c coth cx and that of c the sub-normal is 2 sinh 2 x. Solution. The equation of the curve is x y = c cosh . c Therefore dy x = sinh . dx c Hence dx c cosh cx = Length of the sub − tangent = y dy sinh cx = c coth

x c

dy Length of the sub-normal = y dx x x = c cosh sinh c c c = sinh 2 x . 2

M06_Baburam_ISBN _C06.indd 6

1. Find the equation of the tangent at t = π2 to the cycloid x = a (t − sin t ) , y = a (1 − cos t ) .

= a 1 + 1 = a 2.

(ii)

EXERCISES

2. Find the condition for the line x cos θ + y sin θ = p to touches the curve ax + by = 1. Hint. Find equation of the tangent and equate it to x cos θ + y sin θ = p .Find x and y from that equation.Then (x, y) lies on the curve. m

m

m

m

m

m

m

Ans. (a cos θ ) m −1 + (b − sin θ ) m −1 = p m −1 . 3. Show that the sum of intercepts on the axes of any tangent to the curve x + y = a is constant. 4. Find the equation of the normal to the parabola y2 = 4ax in the form Ans. y = mx – 2am – am2, where m denotes the slope of the normal. 5. If α and β be the intercepts on the axes of x and y respectively cut off by the tangent n n  x  y to the curve   +   = 1 , show that  a  b  n

n

 a  n −1  b  n −1 =1   +   α  β 6. For the catenary y = c cosh cx , show that the 2 length of the normal is yc .. 7. Find the lengths of tangent, normal, subtangent and sub-normal at the point θ on the cycloid x = a (θ − sin θ ) , y = a (1 − cos θ ) . Ans. 2a sin θ2 tan θ2 normal : 2a sin θ2 sub − tangent : 2a sin 3 θ2 sec θ2 sub − normal : a sin θ. 8. Show that the sub-normal to the curve xy = c2 varies as the cube of the ordinate.

12/7/2011 5:56:48 PM

7

Beta and Gamma Functions

The beta and gamma functions, also called Euler’s Integrals, are the improper integrals, which are extremely useful in the evaluation of integrals. 7.1  BETA FUNCTION 1

The integral ∫​   ​​ ​ xm-1(1 - x)n-1 dx, which converges

1



= ​∫  ​ ​ xn-1 (1 - x)m-1 dx

Thus,

= b (n, m), m > 0, n > 0.

 

0

b (m, n) = b (n, m), m > 0, n > 0.

 

0

for m > 0 and n > 0 is called the beta function and is denoted by b (m, n). Thus,  

0

Beta function is also known as Eulerian Integral of First Kind. As an illustration, consider the integral ​∫ ​​ x​   ​(1 - x) dx. We can write this integral as 4

1

3 __ -1

 2 5-1 ​∫  ​ ​​ x​   ​ (1 - x) dx, 0

(  ) 3 __

which is a beta function, denoted by b ​ ​   ​  , 5  ​. 2 1 1 __

But, on the other hand, the integral ​∫  ​​ x​   ​(1 - x) dx 2

-3

 

Putting x = sin2q, we get dx = 2 sinq cosq dq and therefore, (1) becomes p ​  __  ​  2

2 b (m, n) = ​∫  ​  (sin q)m-1 (1 - sin2 q )n-1 ​​

is not a beta function as n - 1 = -3 implies n = -2 (negative). 7.2  PROPERTIES OF BETA FUNCTION 1

m-1 (1 - x)n-1 dx b (m, n) = ​∫  ​ ​x​    

0



= 2 ​∫  ​  ​​sin2m-2 q cos2n-2 q sin q cos q dq



= 2 ​∫  ​  ​​sin2m-1 q cos2n-1 q dq.

 

a

​∫  ​  f (x) dx = ​∫  ​​​ f (a - x) dx,  

0

M07_Baburam_ISBN _C07.indd 1

p ​ 0__  ​  2  

Thus,

0

1

b (m, n) = ​∫  ​​  xm-1 (1 - x)n-1 dx  

0 p__ ​    ​  2

= 2 ​∫ ​   ​​sin2m-1 q cos2n-1 q dq .



 

0

3. Let m and n be positive integers. By definition, 1

b (m, n) = ​∫  ​​  xm-1(1 - x)n-1 dx, m > 0, n > 0.  

= ​∫  ​​  (1 - x)m-1 (1 - (1 - x))n-1dx, since 0 a

· 2 sin q cos q dq

p ​  __  ​  2

0

1



0



0

1. We have

(1)

 

1

  

1

b (m, n) = ​∫  ​​  xm-1 (1 - x)n-1 dx. 0

b (m, n) = ​∫  ​​ xm-1 (1 - x)n-1 dx, m > 0, n > 0.

1 __ 1 2     0

2. We have

0

Integration by parts yields

[ 

]

1

(1 - x)n   b (m, n) = ​ xm -1 ​ ______ ​   ​​​  n (- 1)  0

12/9/2011 3:38:13 PM

7.2  n  chapter Seven

[ 

1

]

Hence,

(1 - x)n - ​∫  ​​ (m - n) xm-2 ​ ​ ______ ​   ​dx n (-1) 0



 

∞

xm-1 ​     ​  dx. b (m, n) = ​∫  ​ ​​ _________ (1 + x)m + n 0

1



- 1 xm-2(1 - x)n dx _____ = ​ m ​ ​∫  ​​  n   



- 1 b (m - 1, n + 1). _____ = ​ m ​  n   

Since b (m, n) = b (n, m), we have

 

0

∞

Similarly, m-2 b (m - 2, n + 2) b (m - 1, n + 1) = ​ _____ ​  n+1 1 b (2, m + n - 2) = ________ ​     ​  b (1, m + n - 1). m+n-2 Multiplying the preceding equations, we get

× b (1, m + n - 1) (m - 1)! [1.2.3.... (n - 1)]

=_________________________________ ​            ​ 1.2....(n - 1) n (n + 1) (n + 2)...(m + n - 2)



1



∞

xm - 1 b (m, n) = ​∫  ​ ​​ ​ _________ dx m  + n ​  0 (1 + x) xm - 1 xm - 1   ​  = ​∫  ​​ ​ ​ ________    ​  dx + ​∫  ​ ​​ ​ ________ m+n m+n dx (1 + x) 0 1 (1 + x)



= I1 + I2, say. 

1

 

(m - 1)! (n - 1)! = ​ ______________        ​∫​   ​ ​ (1 - x)m + n -2 dx (m + n - 2)! 0  

1

]

(m - 1)! (n - 1)! . = ​ ______________        ​ (m + n - 1)!

 

1

t 1 4. Put x = ____ ​     ​  so that dx = ______ ​    2 ​  dt. Then the 1+t (1 + t) expression for beta function reduces to

)

t m-1 t n-1 ______ 1 ____ ____ dt b (m, n) = ​∫  ​ ​​ ​ ​ 1 +  t ​ ​ ​ 1- ​ 1 +  t ​ ​ · ​ (1 +  t)2 ​  0 ∞



 

1

x ________

1

xn - 1 b (m, n) = ​∫  ​​  ​ ​ (1 + x)m+n    ​  dx + ​∫  ​​  ​ ________ ​ (1 + x)m+n    ​  dx m - 1

0

0

1 xm - 1 + xn - 1 _________ = ​∫  ​​​  ​ (1 + x)m + n ​  dx, m > 0, n > 0.



0

EXAMPLE 7.1

Show that ∞

x + x  ​  ∫​   ​ _________  ​​ ​  m + n dx = 2b (m, n). 0 (1 + x) m -1

n - 1

t m-1 1 1 = ​∫  ​ ​​_________ ​     ​  · ________ ​   n-1 ​   · ______ ​    2 ​  dt m - 1 (1 + t) (1 + t) (1 + t) 0

Solution.  We know that

t  x = ​∫  ​ ​​________ ​     ​   dt = ​∫  ​ ​​ _________ ​     ​  dx. m + n m + n 0 (1 + t) 0 (1 + x)

xm - 1 b (m, n) = ​∫  ​ ​​ ​ _________   ​  dx.  (1 + x)m + n 0

∞



n-1

Hence, (2) reduces to

(m - 1)! (n - 1)! . b (m, n) = ​ ______________        ​ (m + n - 1)!

(  ) ( 

1

t ________

xn - 1   ​  = ​∫  ​​ ​  ​     ​dt  = ​∫  ​​  ​  ​ _________ m + n m + n dx. 0 (1 + t) 0 (1 + x)

Hence, if m and n are positive integers, then

 

(  )   (  ) ( )

tm + n 1 __ = ​∫  ​​ ​  ​ ___________      ​· ​    ​dt m - 1 (t + 1)m+n t2 0 t

1

(m - 1)! (n - 1)! ____________ (1 - x)m + n -1   = ​ ______________        ​​ ​        ​  ​​  ​ (m + n - 2)! (m + n -1)(-1)   0 (m - 1)! (n - 1)! ________ 1 = ​ ______________        ​· ​     ​  m+n-1 (m + n - 2)!

∞

(2)

1 1 In I2, put x = __ ​ t ​  so that dx = - __ ​  2  ​dt and so, t 0

 

[ 

∞



1 m-1 __ ​ ​  t ​   ​ 1 I2 = ​∫  ​​ ​ ​ _________     ​​ - ​ __2  ​  ​dt t 1 m+n __ 1 ​ 1 + ​ t ​  ​

× ​∫  ​​  x1-1(1 - x)m+n-2 dx 0

2. From the property (4), we have

 1

(m - 1) (m - 2) ..... (2) (1) b (m, n) =​  __________________________          ​ n (n + 1) (n + 2) ..... (m + n - 2)

∞

xm-1 xn-1 b (m, n) = ​∫  ​ ​​ _________ ​     ​  dx = ​∫  ​ ​​ _________ ​     ​  dx. m + n m + n 0 (1 + x) 0 (1 + x)

M07_Baburam_ISBN _C07.indd 2

m-1

∞

m-1

∞

(3)

12/9/2011 3:38:17 PM

beta and gaMMa functionS  n  7.3 Since b (m, n) = b (n, m), we have xn - 1 b (m, n) = ​∫  ​ ​​  ​ _________ dx. m  + n ​  0 (1 + x)

(4)

Adding (3) and (4), we get ∞

2b (m, n) = ​∫  ​ ​​  _________ ​ x + xm + n ​  dx. 0 (1 + x) m - 1

[ 

1

∞

n - 1

a

 

 

0

0

m = __ ​ n ​  [b (n, m) - b(n, m + 1)] m m = __ ​ n ​  b  (m, n) - __ ​ n ​  b(m + 1, n). Thus,

m m ​( 1 + __ ​ n ​ ) ​b (m + 1, n) = __ ​ n ​  b (m, n)

EXAMPLE 7.2

Show that

or

∫​   ​​ x

(a - x)n-1 dx = am+n-1b (m, n).

m-1

  

(n + m) b (m + 1, n) = m b (m, n)

or

b(m + 1, n) _____ m . ​  __________  ​    = ​ m + n ​  b (m, n)

0

Solution. Putting x = ay, we get a

∫​   ​​ x



(a - x)n-1 dx

m-1

 

0

1

= ​∫  ​ ​(ay)m-1 (a - ay)n-1· a dy



EXAMPLE 7.4

Prove that p ​  __  ​  2

 

0

1



= ​∫  ​ ​​(ay)m-1an-1 (1 - y)n-1· a dy



= ​∫  ​ ​am-1 + n -1 + 1 ym-1(1 - y)n-1dy

 

0 1

 

0

= am + n -1​∫  ​​  ym -1(1 - y)n-1dy



=a

 

0

b (m, n).

m + n-1

EXAMPLE 7.3

Show that

b (m + 1, n) _____ m . ​  __________  ​     = ​ m + n ​  b (m, n)

Solution.  We have

1

b (m + 1, n) = ​∫  ​ ​xm(1 - x)n-1dx  

0

1

= ​∫  ​​ (1 - x)mxn-1 dx, since b (m + 1, n) 0

= b (n, m + 1)



( 

Solution. We have p ​  __  ​  2

m n ​∫  ​  ​​sin pq__ cos q dq ​    ​  2

0



m-1 = ​∫  ​  sin q cosn-1 q  · sin q cos q dq ​​



= ​∫ ​   ​​sinm-1 q (1 - sin2 q)​  2   ​sin q cos q dq.

0 p ​  __  ​  2

n____ −1

0

Putting sin2q = x so that 2sin q cos q dq = dx, we get

p ​  __  ​  2

​∫  ​  ​​sinm q cosn q dq  

0

1

m-1 ____

1

n-1 ____

[ 

]

1

xn = ​ (1 - x) ​ xn ​  ​  ​- ​∫  ​ ​m(1 - x)m-1(-1) · ​ __ n ​  dx 0  

  0

1

m-1 m n -1  = __ ​ n ​  ​∫  ​​ x · x(1 - x) dx 0 1

m = __ ​ n ​  ​∫  ​​ ​ xn - 1[1 - (1 - x)](1 - x)m -1 dx  

0

M07_Baburam_ISBN _C07.indd 3

m+1 ____

n+1 ____

- 1 -1 1 1 = __ ​   ​  ​∫  ​​ ​ x​  2   ​(1 - x)​  2 ​ dx = __ ​   ​  ​∫  ​​ ​ x​  2   ​  (1 - x)​  2   ​  dx 20 20  

n 1 m __

)

n+1 1 1  ​∫ ​   ​​sinm q cosn q dq = __ ​   ​  b ​ _____ ​  m + ​ , _____ ​   ​    ​, 2 2 2 0 m > - 1 and n > - 1.

1



]

1

m = __ ​ n ​  ​ ​∫  ​​ ​ xn -1(1 - x)m -1 dx - ​∫  ​​ ​ xn -1(1 - x)m dx  ​

 

( 

)

n+1 1 1  = __ ​   ​  b ​ _____ ​  m + ​ , _____ ​   ​    ​, m > -1 and n > -1. 2 2 2 EXAMPLE 7.5

Show that b (m, n) = b (m + 1, n) + b (m, n + 1).

12/9/2011 3:38:20 PM

7.4  n  chapter Seven p ​  __  ​  2

Solution. By definition, 1

 

1

= ​∫  ​​  xm (1 - x)n-1 dx + ​∫  ​​ ​ xm-1 (1 - x)n dx 0

1

m-1 n-1 = ​∫  ​​ x (1 - x) [x + (1 - x)] dx

1 = _____ ​  2m - 1    ​ · 2 ​∫  ​  ​​sin2m-1 2q dq 2 0



1 = _____ ​  2m  - 1   ​​  ​ ​sin2m-1 f df, f = 2q

1

m-1 n-1 = ​∫  ​​ x ​  (1 - x) dx = b (m, n). 0

EXAMPLE 7.6

x Express ​  ​ ​​  ​ __________   dx, m, n, a, b > 0 in terms m + n ​  m - 1

∫ (a + bx) 0

of beta function.

(  )

∞

y 1 1 = ____ ​  n  m    ​ ​  ​ ​​ ​ ________   dy = ____ ​  n  m   ​b (m, n), m+n ​  m-1

∫ (1 + y)

ab

ab

0

using property (4) of beta function. EXAMPLE 7.7

p  ​ __  ​  2



1 = _____ ​  2m  - 1   ​· 2 ​∫  ​  ​​sin2m-1 f df 2 0



1 1 = _____ ​  2m  - 1   ​b ​ m, __ ​   ​   ​, using (6), 2 2

 

(  )

and so,

(  )

1 b ​ m, __ ​   ​   ​= 22m-1b (m, m). 2

7.3  GAMMA FUNCTION The gamma function is defined as the definite integral ∞ Γ(n) = ​∫  ​ ​​e- xxn-1 dx, n > 0. 0

The gamma function is also known as Euler’s Integral of Second Kind. 7.4  PROPERTIES OF GAMMA FUNCTION

(  )

∞

Show that b ​ m, ​   ​   ​= 2 2

b (m, m).

p ​  __  ​  2

b (m, n) = 2 ​∫  ​  ​​sin2m-1 q cos2n-1 q dq.  

0

(5)

0

1 Putting n = __ ​   ​ , we get 2

(  ) 1 __

p ​  __  ​  2  

Now, putting n = m in (5), we have p ​  __  ​  2

b (m, m) = 2 ​∫  ​  ​sin2m-1 q cos2m-1 q dq  



M07_Baburam_ISBN _C07.indd 4

∞

Hence,

∞

  0

= n ​∫  ​ ​​ e x

-x n-1

0

dx = nΓ(n).

0

Γ(n + 1) = nΓ(n),

which is called the recurrence formula for Γ(n).

b ​ m, ​   ​   ​= 2 ​∫  ​  ​​sin2m-1 q dq. 2 0

0

∞

Γ(n + 1) = ​∫  ​ ​​ e -xxn dx = [ - xne -x ] ​   ​+ n ​∫  ​ ​​ xn-1e-x dx

2m-1

Solution. We know [see property (2)] that



 

0

1. We have 1 __



∫ 

2

a __

Solution. Put bx = ay so that dx = ​   ​ dy in the b given integral. This gives ay m - 1 ∞ ∞ ​ ___ ​   ​  ​ m - 1 x b a ∫​   ​ ​​   ​ __________   dx = ​∫  ​ ​​  ​ __________    ​    ·  ​__    ​dy m + n ​  m + n b 0 (a + bx) 0 (a + ay)

 

p

 

0

∞

)

p ​  __  ​  2



 

 

0

( 

2m-1 = 2 ​∫ ​   ​​​ __ ​ 1 ​  sin 2q  ​ dq 0 2



b (m + 1, n) + b (m, n + 1)

p ​  __  ​  2

= 2 ​∫  ​  ​​(sin q cos q)2m-1 dq 0

(6)

2. Let n be a positive integer. By property (1), we have Γ(n + 1) = nΓ(n) = n (n - 1) Γ(n - 1)  = n (n - 1) (n - 2) Γ(n - 2)  = n (n - 1) (n - 2). . .3.2.1 Γ(1)  = n!Γ(1). But, by definition, ∞

∞

Γ(1) = ​∫  ​  ​​ e -x dx = [ -e -x ] ​   ​= 1.  

0

  0

12/9/2011 3:38:27 PM

beta and gaMMa functionS   n  7.5 Hence, Γ(n + 1) = n!, when n is a positive integer. If we take n = 0, then 0! = Γ(1) = 1, and so, gamma function defines 0! Further, from the relation Γ(n + 1) = nΓ(n), we deduce that Γ(2) = 1·Γ(1) = 1!, Γ(3) = 2·Γ(2) = 2.1 = 2!, Γ(4) = 3·Γ(3) = 3.2.1 = 3!, and so on.

Therefore,

Moreover, Γ(0) = ∞ and Γ(-n) = - ∞ if n > 0. Also,



Γ(n + 1) Γ(n) = ​ _______ ​, n ≠ 0 n   



= 2 ​∫  ​  ​​​ 2 ​∫  ​ ​​ e-r r2 (m + n)-1dr  ​



× cos2m-1 q sin2n-1 q dq

(n + 1) Γ(n + 1) = ​ _____________     ​  n(n + 1) Γ(n + 2) = ​ _______   ​,  n ≠ 0 and n ≠ -1 n(n + 1)



(n + 2) Γ(n + 2) = ______________ ​        ​



Γ(n + 3) =​ _____________      ​ , n ≠ 0, n ≠ -1,



and n ≠ -2

Γ(n + k + 1) = ​ _____________________       ​, n ≠ 0, n ≠ -1, n(n + 1) (n + 2) ... (n + k)

n ≠ -2, and n ≠ -k. Thus, Γ(n) for n < 0 is defined, where k is a least-positive integer such that n + k + 1 > 0. 7.5 RELATION BETWEEN BETA AND GAMMA FUNCTIONS We know that

∞

Γ(m) = ​∫  ​ ​​ e t  dt. - t m-1

0

Putting t = x2 so that dt = 2x dx, we get ∞



2

Γ(m) = 2 ​∫  ​ ​​ e-x x2m-1 dx. 

0 ∞ ∞

0

M07_Baburam_ISBN _C07.indd 5

2

0

= 4 ​∫  ​ ​​​∫  ​ ​​ e



2

2

x

y

dx dy.

0 0

Taking x = r cos q and y = r sin q, we have dx dy = rdq dr. Therefore, p ​  __  ​  2 ∞

Γ(m)Γ(n) = 4 ​∫  ​  ​​​∫  ​ ​​ e -r r2(m + n)-1 cos2m-1 q 2

 

0 0

× sin2n-1 q drdq

[ 

p ​  __  ​  2  

0

[ 

]

∞

2

0

]

p ​  __  ​  2

(7)

 

0

using (7)

= Γ (m + n) b   (m, n) using property (2) of beta function. Hence, Γ(m)Γ(n) . b (m, n) = ​ _________ ​  Γ(m + n) EXAMPLE 7.8

(  )

1 ​   ​= ​ __ Show that Γ​ ​ __ √ p ​ . 2 Solution.  We know that  Γ(m)Γ(n) . b (m, n) = ​ _________ ​  Γ(m + n) 1 __ Putting m = n = ​   ​ , we get 2 1 ​    ​Γ​ ​ __ 1 ​   ​ Γ​ ​ __ 1 2 2 __ __________ 1 __ b ​ ​   ​ , ​   ​   ​= ​   ​    Γ(1) 2 2

     ( ) ( ) ( ) = ​[ Γ​( ​ 1 ​  )​ ]​ . 2 __

Thus,

[  (  ) ] (  )

1

1 __

 

2

1 __

-1 -1 1 1 ​   ​  2​ = b ​ ​  __ 1 ​ , __ ​ Γ​ ​ __ ​   ​   ​= ​∫  ​​ ​ x ​ 2  ​  (1 - x) ​ 2 ​   dx 2 2 2 0

Similarly, we can have ∞

2

-(x + y ) 2m-1 2n-1

0

Γ(n) = 2 ​∫  ​ ​​ e-y y2n-1dy.

∞

2

= Γ (m + n) ​ 2 ​∫  ​  ​​cos2m-1 q sin2n-1 q dq  ,​

n (n + 1) (n + 2)

n(n + 1) (n + 2) . . . . . . .

∞

Γ(m) Γ(n) = 4 ​∫  ​  ​​e-x x2m-1 dx .​∫  ​ ​​ e-y y2n-1dy

1



 

1

dx _____ _____ __ dx = ​∫  ​​ ​ ​ _________      ​= ​∫  ​​ ​ ​ ______     ​ ​ x ​ ​√1 - x   ​ 0 ​√x - x2 ​  0 √  

 

12/9/2011 3:38:30 PM

7.6  n  chapter Seven

[)   (  )]

1​ 1 __ x ​     ​ 1 ____________ dx 2  ​  __________     2  ​= ​ sin -1​ ​ _____  ​  ​​​ ​   ​  = ∫ __ 1 __ 1 1 __ ​     ​ ​0   ​​  ​ ​   ​  - ​ x - ​   ​   ​   ​ 2 0 4 2  

√ ( 

Similarly, we can show that p ​  __  ​  2

(  )

p p = sin-1(1) - sin-1 (-1) = ​ __  ​- ​ -​ __  ​  ​= p. 2 2 Hence, __ 1 ​   ​= ​ p ​ . Γ​ ​ __ √ 2

(  )

Second Method

Example 7.10

p Show that Γ(n) Γ(1 - n) = ​ ______    ​  , 0 < n < 1. sin np (Euler’s Reflection Formula) Solution.  We know that ∞

xn - 1 b (m, n) = ​∫  ​ ​​​ _________ dx. m  + n ​  0 (1 + x)

We know that (see Example 7.4) p ​  __  ​  2

(  ) Γ ​  ​ m+1    ​  ​Γ ​  ​ n+1    ​  ​ 1 ( 2 ) ( 2 ). = ​   ​  ​      ​ 2 Γ ​( ​ m + n  + 2  ​ )​ 2

n+1 1 1  ​∫  ​  ​​sinm q cosn q dq = __ ​   ​  b ​ _____ ​  m + ​ , _____ ​   ​    ​ 2 2 2 0  

__



____ ____     _______________      _________  

Putting m = n = 0, we get

[  (  ) ] [  (  ) ] 2

p ​  __  ​  2

1 ​   ​  ​ ​ Γ​ ​ __ 2 _______

​∫  ​  ​​dq = ​   

0

2Γ(1)

2

1 ​   ​  ​  ​ Γ​ ​ __ 2 . _______  ​    = ​   ​   

2

Hence,

∞

Γ(m) Γ(n) xn-1  ​_________ ​  = ​∫  ​ ​​ _________ ​     ​  dx. Γ(m + n) (1 + x)m + n 0 Putting m = 1 - n so that m > 0 implies n < 1, we get ∞

or

2

∞

p xn - 1 , 0 < n < 1. Γ(n) Γ(1 - n) = ​∫  ​ ​​​ _____  ​ dx = ​ ______    ​  1 + x sin np 0

2

1 ​   ​= ​ __ Γ​ ​ __ √ p ​ . 2

EXAMPLE 7.11

Show that

p ​  __  ​  2

EXAMPLE 7.9

Γ(m)Γ(n) b (m, n) = ​ ________ ​  , m > 0 and n > 0. Γ(m + n) Therefore,

2

[  (  ) ]

(  )

Also,

Γ(n) Γ(1 - n) xn - 1  ​____________  ​      = ​∫  ​ ​​​ ______   ​  dx Γ(1) 0 (1 + x)

1  ​  ​  ​ Γ​ ​ __ p  ​_______ 2 . __ ​    ​= ​   ​   

Thus,

p ​  __  ​  2

p ​  __  ​  2

Γ(m) Γ(n) . ​∫ ​   ​sin2m-1 q cos2n-1 q dq = ​ _________   ​ 2Γ(m + n) 0

_____ _____ Express the integrals ​∫  ​  ​​​√tan q ​ dq and ​∫  ​  ​​​√cot q ​   

dq in terms of gamma function.

0

p ​  __  ​  2

Hence, evaluate ​∫  ​  ​​sinp q dq and ​∫  ​  ​cosp q dq. 0

 1

p ​  __  ​

__ 2 2  __ 1 1 _____ - __ sin​ 2 ​  q ​∫  ​  ​​​√tan q ​ dq = ∫​   ​  ​​​ ______       ​ dq = ∫ ​   ​  ​sin ​ 2 ​  q cos ​ 2 ​  q dq  __ 1 0 0 0 cos ​ 2  ​  q   1 ​  + 1 1 __ ​ __ - ​   ​  + 1 3 2 1 _____ ______ __ __ 2  ​    Γ​ ​   ​    ​Γ​ ​   ​ Γ​ ​    ​  ​Γ​ ​   ​   ​ 2 2 4 4 ________________ __________  = ​         ​= ​   ​    2Γ(1) 1 __ 1 __ ​   ​  - ​   ​  + 2 2 2 ​  2Γ​ ​ ________    ​ 2 1 3  ​  ​Γ​ ​ __ 1 ​   ​. = ​ __ ​   ​   ​Γ​ ​ __ 2 4 4

 

(  ) (  ) (  )

(  ) (  ) (  )

(  ) (  )

0

Solution.  We know that p ​  __  ​  2

2 ​∫  ​  ​​sin2m-1 q cos2n-1 q dq = b (m, n)

 

.

M07_Baburam_ISBN _C07.indd 6

p ​  __  ​  2

 

Solution.  We have p ​  __  ​ 

 

 

0

p ​  __  ​  2

(  ) (  )

_____ 1 3 1 ​   ​. ​∫  ​  ​​​√cot q ​ dq = __ ​   ​  Γ​ ​ __  ​   ​Γ​ ​ __ 2 4 4 0

 

or

0

p ​  __  ​  2

1 ​∫  ​  ​​sin2m-1 q cos2n-1 q dq = __ ​   ​  ·b (m, n). 2 0  

Γ(m)Γ(n) But, b (m, n) = ​ ________ ​  . Therefore, Γ(m + n)

12/9/2011 3:38:34 PM

beta and gaMMa functionS  n  7.7 p ​  __  ​  2

∞

Γ(m)Γ(n) . ​∫  ​  ​sin2m-1 q cos2n-1 q dq =​  ________     ​ 2Γ(m + n) 0

Γ(n) (i) ∫​   ​ ​​ e-axxn-1 cos bx dx = ​ ____ rn    ​cos nq 0

 

If we put 2m - 1 = p and 2n - 1 = q, then this result reduces to

(  ) (  ) (  )

p + 1 q +1 Γ ​ ​  ____    ​   ​Γ ​ ​ ____    ​   ​ 2 2 _______________ ​∫  ​  ​​sin  q cos q dq = ​         ​. p+q+2 0  2Γ ​ ​ ________     ​  ​ (8)   2 Putting q = 0 in (8), we get p ​  __  ​  2

p

q

 

(  ) (  ) (  ) (  ) (  )

p + 1 __ p+1 1 ​  ​ Γ​ ​ _____     ​  ​ ​ p ​  Γ ​ ​ _____     ​  ​Γ ​ ​ __ √ . 2   _________ 2 2 ______________ ​∫  ​  ​​sin q dq = ​          ​= ​   ​​ ___  ​    2 p + 2 p + 2 _____ 0 _____ Γ​ ​      ​  ​ 2Γ ​ ​      ​  ​   2 2 Similarly, taking p = 0, we get p ​  __  ​  2

p

 

(  ) (  )

(q+1) __ Γ​ ​ _____  ​    ​ ​ p ​  √ 2 ________ ___ q ​∫  ​  ​​cos q dq = ​   ​  ​   ​  . 2 q+2 0 Γ​ ​ ____     ​  ​ 2

p ​  __  ​  2  

EXAMPLE 7.12

__ Show that ​√p ​  1 __ ____ Γ(m) Γ ​ m + ​    ​  ​= ​  2m-1    ​Γ(2m) 2 2 (Duplication Formula).

( 

)

Solution.  In Example 7.7, we have shown that

(  ) 1 __

b ​ m, ​   ​   ​= 2 b (m, m). 2 Converting into gamma function, we get 2m-1

(  ) (  ) (  ) (  ) (  )

1 ​  ​ Γ(m)Γ​ ​ __ Γ(m) Γ(m) . 2    ​ _________  ​= 22m-1 ​ _________  ​    Γ(2m) 1 __ Γ​ m + ​   ​  ​ 2 __ 1 __ Since Γ​ ​   ​   ​= √ ​ p ​ , we get 2 __ ​√p ​  Γ(m) ________ ​       ​= 22m-1 ​ _____    ​ 1 __ Γ(2m) Γ​ m + ​   ​  ​ 2 or __ ​√p ​  _____ 1 __ Γ(m) Γ​ m +​   ​   ​= ​  2m - 1   ​Γ(2m). 2 2 EXAMPLE 7.13

Show that

∞

Γ(n) ​∫  ​ ​​e-axxn-1 dx = ​ ____ an   ​, 0 where a and n are positive. Deduce that

M07_Baburam_ISBN _C07.indd 7

∞

Γ(n) (ii) ∫​   ​ ​​ e-axxn-1 sin bx dx = ​ ____ rn    ​sin nq, 0 where r2 = a2 + b2 and q = tan-1 __ ​ b a ​. Also evaluate ∞ ∞ -ax -ax ​∫  ​ ​​ e cos bx dx and ​∫  ​ ​​ e sin bx dx. 0

0

Solution.  Put ax = z, so that adx = dz, to get ∞

∞

n-1 dz ​∫  ​ ​​e-axxn-1 dx = ​∫  ​ ​​e-z ​( __ ​ az  ​)​ · __ ​ a ​  0 0 ∞



Γ(n) . 1 = __ ​  n  ​ ​  ​ ​​e -zzn-1dz = ​ ____ ​ n   

(9)



Γ(n) . ​∫  ​ ​​e - (a + ib)xx n-1 dx = ​ _______   ​  (a + ib) n 0

(10)

a ∫0  a Replacing a by a + ib in (9), we get ∞

But as, e-(a + ib)x = e-ax· e-ibx = e-ax (cos bx - i sin bx) and taking a = r cos q and b = r sin q, DeMoivre’s Theorem implies (a + ib)n = (r cos q + ir sin q)n = rn (cos q + i sin q)n = rn (cos nq + i sin q ). Therefore, (10) reduces to ∞

​∫  ​ ​​ [e-ax (cos bx - isin bx)]xn-1 dx 0

Γ(n) Γ(n) ____ = ​ ________________       ​ = ​  n    n r  ​(cos nq + isin r-1 (cos nq + isin nq) nq ) Γ(n) = ​ ____ rn    ​(cos nq - isin nq). Equating real-and imaginary parts on both sides, we get ∞ Γ(n) ∫​   ​ ​​ e -axxn-1 cos bx dx = ​ ____  ​cos nq rn    0 and ∞ Γ(n) ∫​   ​ ​​ e -axxn-1 sin bx dx = ​ ____ rn    ​sin nq. 0 If we put n = 1, then p ​  __  ​  2

Γ(1) r______ cos q _____ a ​∫  ​  ​​e-ax cos bx dx = ​ ____   = ​  2   2   ​ r   ​ cos q = ​  r2 ​  a +b 0

12/9/2011 3:38:38 PM

7.8  n  chapter Seven Hence,

and p/2

​∫  ​  ​​e

-ax

0

Γ(1) r______ sin q _____ b . sin bx dx = ​ ____  = ​  2 +   2 ​  r    ​sin q = ​  r2 ​  a b

EXAMPLE 7.14

Show that

( 

)

Example 7.15

Show that p ​  __  ​  2

Solution.  In Example 7.9, we have proved that p ​  __  ​  2

p ​  __  ​  2  

1 __

But,

)

p ​  __  ​  2

( 

(  ) (  )

__ 3 ​   ​= p ​√2 ​ (Example 7.14.) 1 ​   ​ Γ ​ ​ __ Γ ​ ​ __ 4 4 Hence, p__ p__

Therefore, b ​ n + ​   ​ , n +​   ​   ​ 2 2 p ​  __  ​  2

​    ​  2

 

)

 

p

1 1 = _____ ​  2n  - 1   ​​  ​  ​​sin2n 2q dq = ___ ​  2n   ​·​  ​ ​​ sin2n f df, f = 2q 2

∫ 

2

0

p ​  __  ​  2

∫   

0

[ 

(  ( 

]

) )

__ 2n + ​  1  Γ​ ​ ______  ​​ p ​  √___ 1 1 2 ___ _____ _________ 2n = ​  2n   ​· 2 ​  ​  ​​sin f df = ​  2n  - 1   ​​ ​     ​​  2 ​   ​ 2 2 2n +2 0 Γ​ ​ _____  ​    ​

∫   

(see Example 7.11) __ 1 ​√p ​ Γ ​ n + __ ​   ​   ​ 2 . ____________ = ​  2n     ​   2 Γ(n +1)

( 

Also,

2

)

(11)

( 

) ( 

)

1 1 Γ​ n + __ ​   ​   ​Γ​ n + __ ​   ​   ​ 2 2 ________________ b ​ n + ​   ​ , n + ​   ​   ​= ​      ​   

( 

1 __

1 __

2

2

)

Γ(2n + 1)

[  (  ) ]

1 ​  ​ Γ ​ n + __ ​   ​   ​  2 . = ​ ___________  ​      (12)



​    ​ 

2 _____ _____ p __  ​ . ​∫  ​  ​​​√tan q ​ dq = ​∫  ​  ​​​√cot q ​ dq = ​ ___ ​ 2 ​  √ 0 0

2n _____ = 2 ​∫  ​  ​​ sin2n q cos2n q dq = 2 ​∫  ​  ​​​ ​ sin2q  ​    ​ dq 2 0 0 p__ ​    ​  2

2

 

EXAMPLE 7.16

Show that

( 

1

From (11) and (12), we have __ ​___ p ​  Γ(2n + 1) √ 1 __ Γ ​ n + ​   ​    ​= ​  2n  ​ ·​ ________ ​  . 2 Γ(n + 1) 2 1 Further, putting n = __ ​   ​ , we have 4 1 __ __ 3  ​  ​ __ __ Γ​ ​ __  ​  Γ​ ​ 1 ​   ​ ​___ p ​__  _____ p ​ _______ 2 2 . √ 2 __ 3 __ Γ ​ ​   ​   ​= ​    ​ ·  ​     ​= ​  ​   ​ ​​     ​  2 __ 1 1 4 5 ​   ​ ​ 2 ​  Γ​ ​ __ √ ​   ​  Γ​ ​ __  ​   ​ 4 4 4

(  )

M07_Baburam_ISBN _C07.indd 8

)

(  ) (  ) √

(  ) (  )

)

p-1 Γ(p) 1 ​∫  ​​ ​ y q - 1 ​ log __ ​ y ​ ​ dy = ​ ____ q p ​ , p > 0, q > 0. 0 1 1 x __ Solution.  Putting log ​ __ y  ​= x, we have ​ y ​= e or y = -x -x e and so, dy = -e dx. Therefore,  

1

( 

)

p-1

1 ​∫  ​​  yq-1​ log __ ​ y ​ ​ dy



 

0

∞



= ​∫  ​ ​​e -(q-1)xx p-1 (-e -x) dx

= ​∫  ​ ​​ e -qxx p-1 dx

0 ∞

0 ∞

(  )

p-1 dt = ​∫  ​ ​​ e-t ​ __ ​ qt  ​ ​ .  ​ __ q ​ , putting qx = t

Γ(2n + 1)

( 

(  ) (  )

 

b (m, n) = 2 ​∫  ​  ​​sin2m-1 q cos2n-1 q dq.

( 

p ​  __  ​ 

2 _____ _____ 1 3 ​   ​. 1 ​   ​Γ​ ​ __ ​∫  ​  ​​​√tan q ​ dq = ​∫  ​  ​​​√cot q ​  dq = __ ​   ​  Γ ​ ​ __ 2 4 4 0 0

Solution.  We know that

0

(  ) (  )

 

(  ) (  )

1 __

p ​  __  ​ 

2 ____ ____ p__ . 1 3 ​   ​= ​ ___ 1 ​   ​Γ ​ ​ __ ​∫  ​  ​​​√tanq ​ dq = ​∫  ​  ​​​√cotq ​ dq = __ ​   ​  Γ ​ ​ __   ​  2 4 4 ​√2 ​  0 0

__ ​___________ p ​ Γ(2n + 1) √ . Γ ​ n + ​   ​    ​= ​  2n  ​  2 2 Γ(n + 1) __ 3 ​   ​= p​√2 ​.  1 ​   ​Γ​ ​ __ Hence, deduce that Γ ​ ​ __ 4 4 1 __

(  ) (  )

__ 3 ​   ​Γ​ ​ __ 1 ​   ​= p​√2 ​.  Γ​ ​ __ 4 4

0



∞

Γ(p) . 1 = __ ​ q p  ​​∫  ​ ​​e -tt p-1dt = ​ ____ q p  ​ 0

EXAMPLE 7.17

Show that p ​  __  ​  2

p ​  __  ​ 

2 _____ dq _____ ​∫  ​  ​​​√sin q ​ dq . ​∫  ​  ​​​ ______    ​  = p. ​ sin q ​  0 0 √  

 

12/9/2011 3:38:40 PM

7.9 beta and gaMMa functionS  n  ∞

Solution.  We have p ​  __  ​  2

p ​  __  ​ 

2 ____ dq ____ ​∫  ​  ​​​√sinq ​ dq . ​∫  ​  ​​​ _____    ​  ​ sinq ​  0 0 √  

p ​  __  ​  2

 

p ​  __  ​  2

1 __

1 - __ 2

= ​∫  ​  ​​sin ​   ​  q dq · ​∫  ​  ​​sin ​    ​q dq. 2

 

0

6 1 1 = ____ ​     ​ ​   ​ ​​ e-t·t3dt = ____ ​     ​ Γ(3) = ____ ​     ​ . 625

2

(  ) (  ) (see Example 7.11) ( 4 ) ( 4 ) 1 ​  ​ 1 __ Γ​( ​ __ ) p Γ​( ​ 4 ​ )​ p 4 = ​    ​. ​    ​= ​    ​·​    ​= p. 5 ​  ​ 4 __ Γ​( ​ __ )4 ​ 41 ​Γ​( ​ __14 ​ )​ 4 3 ​ ​ Γ​ ​ __ 1 ​ ​ Γ​ ​ __ p 4 4 . __ ___________ = ​     ​     ​    ​ 4 5 3 __ __ Γ​ ​   ​ ​Γ​ ​   ​ ​

EXAMPLE 7.20

Show that (-1)nn! . ​∫  ​​ ​ xm(log x)n dx = ​ _________    ​  (m + n)n + 1 0  

Solution.  Putting log x = -y, we have x = e-y and so, dx = -e-y. When x = 1, y = 0 and when x → 0, y → ∞. Therefore, 1

​∫  ​​ ​ xm(log x)n dx  

0

1

 

( 

)

1 Solution.  Putting log ​ __ y ​= x, that is, e-x, we have dy = -e-x dx. Hence, 1

( 

1 x ​ __ y ​= e or y =

0

)

n-1 -x 1 n-1 ​∫  ​​  log __ ​ y ​ ​ dy = -​∫  ​​​ x e dx ∞  

0

∞



= ​∫  ​ ​​e-xxn-1 dx



= Γ(n), n > 0.

0

0

0

putting (m + 1)y = t

( 

1 __

)

(-1) (-1)n _________ -t n = ​ _________    ​   ∫ ​     ​  ​​ e t dt = ​    ​  Γ(n + 1) (m + 1)n + 1 0 (m + 1)n + 1 (-1)n n! . = ​ _________    ​  (m + 1)n + 1 EXAMPLE 7.21

Prove that

( 

1 __

 

1

)

∞

-4y 3 -y -5y 3 ​∫ ​​  x ​ log ​ x ​ ​ dx = -​∫  ​ ​e ·y ·e dy = ​∫  ​ ​​ e ·y dy  

  



M07_Baburam_ISBN _C07.indd 9

∞

t __ dt = ​∫  ​ ​​ e - t·​ ____    ​·​  5 ​ , putting 5y = t 125 0

2

4



0

 ∞

 

p ​  __  ​  4

q ____ 1 1 dq _____ 1 sec = __    ​· ​ sec  q  ​dq = __ ​   ​  ​∫  ​  ​​​ _________ ​   ​ ​∫  ​  ​​​  ​ ____  ​ 1   __ 1   __ 2  ​ 2 0 √tan q  2 2 0 p__ p__ sin​    ​q cos​   ​ q ​    ​  ​    ​ 

 

0

​    ​  4

sec2 q 1 ______ ​∫  ​​ ​ ​   4  ​  = ​∫  ​  ​​​ ________   ​  · __________ ​  ________  2  ​  dq 0 ​√ 1 + x  ​  0 ​√ 2 tan q ​  ​√ 1 + tan q ​  dx _______ _____ p ​  __  ​  4

 0

3

(  )

1

1 1 dx   ​  1  ​, __ _______ = ____ ​  __    ​ b ​ ​  __ ​   ​   ​. ​∫  ​​ ​  _____ 4 2 4 4​ 2 ​   √ ​ 1 + x  ​   √ 0 2 Solution.  Putting x = tan q, we have 2xdx = sec2 q dq. Therefore, p__

3

e-y, we have dx = -ey dy. Therefore,

∞

n

1 ​= y, that is, ​ __ 1 ​ = ey or x = Solution.  Putting log ​ __ x x 4

∫ 

-t

0

Evaluate ∫​   ​​​ x ​ log ​ x ​ ​ dx. 4

( m + 1 ) m + 1

0

t n dt = (-1) ​  ​ ​​e ·​ _____ ​     ​  ​ _____ ​     ​, 

 

1

1

∞

n

EXAMPLE 7.19

1

∞

______ __   

1 n-1 ​∫  ​​  log __ ​ y ​ ​ dy = Γ(n), n > 0. 0

∞

= ​∫  ​ ​​ e-my (-y)n(-e- y)dy = (-1)n ​∫  ​ ​​ e-(m + 1)y.yndy

EXAMPLE 7.18

Prove that

625

1

(  ) (  ) (  ) (  )

_____ __    

625

0

0

1 ​ + 1 1 ​ + 1 - ​ __ ​ __ _ 2     ______ 2 _____ ​  ​ __ Γ ​   ​   ​  ​ √ Γ​ ​      p    ​ ___________ ​ __ p ​  2 2 ________ . .​  ​√___ .   ​   ​    ​    ​       ​ 2 ​  = ​  1 __ 2 1 __ - ​   ​ + 2 ​   ​ + 2 2 ______ 2      Γ​ ​      ​  ​ Γ​ ​ _____   ​  ​ 2

0

∫ 

3



4

1 dθq dq 1 ________ __ = __ ​   ​ ​∫___________       __1  ​  1  ​ = ​   ​​∫  ​  ​​​  2   ​  ​​​  __ 2 0  0 (sinqθ cosqθ)​ 2 ​  sin 2q 2 _____ ​        ​ ​   ​ 2 p p ​  __  ​  ​  __  ​  4 2 df dq 1 1 ______ _______ ___ ___      ​, f = 2q    ​  = ​  __   ​ ∫​   ​  ​ ​​ ______ = ​  __   ​ ∫​   ​  ​ ​​ ____ ​ ​ 2 ​  0 ​√sin 2q ​  √ ​ 2 ​  0 2​√sinf   √

(

)

12/9/2011 3:38:42 PM

7.10  n  chapter Seven 1__ ____

p ​  __  ​  2

1 - __ 2 

7.6  DIRICHLET’S AND LIOUVILLE’S THEOREMS

= ​     ​ ​∫  ​  ​​sin ​   ​  f cos f df 2​√2 ​  0



0

The following theorems of Dirichlet and Liouville are useful in evaluating multiple integrals.

p ​  __  ​  2



1 2 - __ 0 2 f cos f df = ____ ​   __ ​ ​∫  ​  sin ​​ ​   ​  4√ ​ 2 ​  0



1 1 1 ​ , __ = ____ ​   __ ​ b ​ ​  __ ​   ​   ​, 4 2 4√ ​ 2 ​ 

Theorem 7.1  (Dirichlet). If V is the region, where x ≥ 0, y ≥ 0, z ≥ 0, and x + y + z ≤ 1, then

(  )

Γ(p)Γ(q)Γ(r) ​∫ ​​​∫  ​   ​​​∫  ​   ​​​ x p-1y q-1z r-1 dx dy dz = ​ _____________       ​. Γ(p + q + r + 1)   V  

p ​  __  ​  2

  

Show that

(The Dirichlet’s Theorem can be extended to a finite number of variables).

0

Proof:  Since x + y + z ≤ 1, we have y + z ≤ 1 - x = a.

p ​  __  ​  2

_____ _____ ​∫  ​  ​​(​√tan q ​ + ​√secq  ​  )dq

[ 

 

]

Therefore,

_ ​√p   ​ 1 __ 3 1 __ __ _____ = ​   ​  Γ ​ ​   ​   ​​ Γ​ ​   ​   ​+ ​      ​  ​. 2 4 4 3 ​   ​ Γ​ ​ __ 4 Solution.  We have 0

p ​  __  ​  2

____

(  ) (  )   ( )

_____

p ​ __ ​  2

 

p-1 q-1 r-1 ​∫ ​ ​​​∫ ​​ ​​∫  ​ x y z dx dy dz  

 

1

 

0

1 __ 2

p ​ __ ​  2

 

0

1 - __ 2 

 

(  ) (  ) 1 1 Γ​  ​ 3 ​  ​Γ​  ​ 1 ​  ​ Γ​( ​   ​  )​Γ​( ​   ​ )​ 2 4 1 ( 4) ( 4) = ​   ​ ​ ​   ​  + ​   ​   ​ 2 Γ(1) 3 Γ​( ​   ​ )​ 4 Γ​( ​ 1 ​  )​ 2 1 1 3 = ​   ​  Γ​( ​   ​  )​​ Γ​( ​   ​  )​+ ​     ​  ​ 2 4 4 Γ​( ​ 3 ​  )​ 4 1 1 1 1 __ 1 3 ​ , __ = __ ​   ​  b ​ ​  __ ​    ​  ​+ __ ​   ​ b ​ ​  __  ​ , ​   ​   ​ 2 2 2 4 4 4

[ 

]

__ __ __ __         __ _________ _________       __   __

 

__  

[ 

__  

[ 

(  ) (  )

]

__   _____   __  

__

]

​√p ​  3 ​   ​+ ​  _____ 1 ​   ​​ Γ​ ​ __ = ​   ​  Γ​ ​ __     ​  ​. 2 4 4 3 ​   ​ Γ​ ​ __ 1 __

M07_Baburam_ISBN _C07.indd 10

]

 

0

0

(13)

a a-y

( 4 )

Putting y = aY and z = aZ, this integral reduces to   I = ​∫  ​​ (aY )q-1(aZ )r-1. a2dZ dY,  

1 - __   2 = ​   ​  ​∫  ​  ​​sin​    ​q cos ​   ​ q dq + __ ​   ​  ​∫  ​  ​​sin0 q cos ​ 2  ​q dq 20 20

2 __

[ 

= ​∫  ​​  x p-1 ​ ​∫  ​   ​​∫  ​ ​yq-1zr-1dz dy  ​dx.

 

 

p ​ __ ​  2

0

a a-y

0 0

1 __ - __ - __ = ​∫  ​  ​​sin​ 2 ​ q cos ​ 2 ​q dq + ​∫  ​  ​​sin0 q cos ​ 2 ​ q dq 0

x

I = ​∫  ​ ​  ​∫  ​ ​ yq-1zr-1dz dy.

p ​ __ ​  2

1

V

1-x 1-x-y

0

 

1

 

 

 

Let

____ ____ = ​∫  ​  ​​​√tanq ​ dq + ​∫  ​  ​​​√secq ​ dq 0

 

p-1 q-1 r-1 = ​∫  ​ ​  ​​∫  ​ ​  ​​∫  ​  ​x y z dx dy dz



p ​  __  ​  2

0 p ​ __ ​  2

 

 

1

)​dq ​ sec q  ​   ​∫  ​  (​​​  ​√tanq ​ + √ 0

 

 

since b (m, n) = 2 ​∫  ​  ​​sin2m-1 f cos2n-1 f df. EXAMPLE 7.22

 

 

D

where D is the domain where X ≥ 0, Y ≥ 0, and Y + Z ≥ 1. Thus, 1 1-Y

I = a

q+r

​∫  ​ ​  ​​∫  ​  ​Y   Z  dZ dY q-1

r-1

 

0

0

1

[  ] r

1-Y



q-1 __ = aq+r ​∫  ​ ​ Y  ​ ​ Z  r ​  ​​ ​  ​dY 0



q-1 a = ___ ​  r   ​ ​∫  ​​  Y  (1 - Y )rdY

 

q+r

0

1

 

0

 Γ(q)Γ(r + 1) aq+r aq + r ___________ = ___ ​  r   ​ b (q, r + 1) = ​ ____ ​ ​     ​  r    Γ(q + r + 1)

Γ(q) Γ(r) = a q+r ​ __________    ​.  Γ(q + r + 1)

12/9/2011 3:38:44 PM

beta and gaMMa functionS  n 7.11 Therefore, for the first quadrant,

Hence, (13) yields  

p-1 q-1 r-1 ​∫  ​ ∫  ​ ∫  ​ x y z dx dy dz  

 

 

 

 

 

 

​

 

    V

1

Γ(q) Γ(r) = ​∫  ​​ ​​ ___________    ​  x p-1a q+r dx 0 Γ(q + r + 1)

 

 

∫  ​ ∫  ​ ∫  ​ xyz dx dy dz  

 

 

 

 

     

 



= ​∫    ​ ∫    ​ ∫  ​ (x dx) (ydy) (zdz)



a2 b2 c2 = ​∫    ​ ∫    ​ ∫  ​  ​ __ ​ dX  ​​ ​ __ ​ dY  ​​ ​ __ ​  dZ  ​ 2 2 2  

Γ(q)Γ(r)    ​  b (p, q + r + 1) = ​ __________ Γ(q + r + 1)



a2b2c2 = ​ _____  ​    X 1-1Y 1-1Z1-1dX dY dZ 8 ​∫    ​ ∫    ​ ∫    ​ 

Γ(q)Γ(r) ______________ Γ(p)Γ(q + r + 1) _____________ Γ(p)Γ(q)Γ(r) . = ​ __________    ​.  ​     ​=​        ​ Γ(q + r + 1) Γ(p + q + r + 1) Γ(p + q + r + 1)



Γ(1) Γ(1) Γ(1) a2b2c2 ______________ = ​ _____  ​   ​        ​, 8 Γ(1 + 1 + 1 + 1)

 

Γ(q)Γ(r) __________

 

= ​     ​  ∫​   ​​x​  p-1 (1 - x)q + r dx, since a = 1 - x Γ(q + r + 1)  

Remark 7.1. x If x + y + z ≤ h, then by putting ​ __ ​ = X, h y h z __ ​   ​= Y, and ​ __  ​= Z, we have X + Y + Z ≤ ​ __ ​= 1 and h h h so, the Dirichlet’s Theorem takes the form  

 

 

​∫  ​ ∫  ​ ∫  ​  x p-1y q-1z r-1 dx dy dz  

 

 

 

 

    V

Theorem 7.2  (Liouville). If x, y, and z are all positive such that h1 < x + y + z < h2, then p-1 q-1 r-1 ​∫    ​ ∫    ​ ∫    ​  f (x + y + z) x y z dx dy dz  

  

 

Γ(p)Γ(q) Γ(r) ____________

h2

= ​     ​  ​∫  ​f (h)h Γ(p + q + r) h ​  

 

   

 

 

  

 

  

( 

) ( 

) (  )   

  



 

  

 

 

by Dirichlet’s Theorem a2b2c2 ____ a2b2c2 _____ a2b2c2 . 1 _____ = ​ _____  ​   ·​      ​= ​   ​ = ​   ​    8 48 8.3! Γ(4)



Therefore, value of the integral for the whole of a2b2c2 a2b2c2 . _________ the ellipsoid is 8 ​ _____ ​   ​    ​= ​   ​    6 48

( 

)

y __z x __ The plane ​ __ a ​ + ​ b ​ + ​ c ​ = 1 meets the axes in A, B, and C. Find the volume of the tetrahedron OABC. Solution.  We wish to evaluate  

dh.

p+q + r-1

1

(Proof, not provided here, is a slight modification of the proof of Dirichlet’s Theorem). EXAMPLE 7.23

Evaluate ∫∫∫x y z dx dy dz taken throughout the ellipsoid 2 x2 y z2 ​ __2  ​+ ​ __2  ​+ ​ __2  ​ ≤ 1. a b c y2 x2 z2 __ Solution.  Put ​  2  ​= X, ​ __2  ​= Y, and ​ __2  ​= Z to get a __1 b __1 c 1 __ 2 x = aX ​ 2 ​, y = bY ​   ​, and z = cZ ​ 2 ​ and a2 b2 c2 x dx = ​ __ ​ dX, ydy = ​ __ ​ dY, and zdz = ​ __ ​  dZ. 2 2 2 The condition, under this substitution, becomes X + Y + Z ≤ 1.

M07_Baburam_ISBN _C07.indd 11

 

  

EXAMPLE 7.24

Γ(p)Γ(q)Γ(r) p + q + r = ​ ___________ ​   ·h . Γ(q + r + 1)

  

 

  

 

 

​∫  ​ ∫  ​ ∫  ​ dx dy dz       x y __z x under the condition __ ​ a ​+ __ ​   ​+ ​ c ​= 1. Putting ​ __ a ​= X, b y z __ ​    ​= Y, and __ ​ c ​= Z, we get X + Y + Z = 1. Also dx b = adX, dy = bdY, and dz = cdZ. Therefore, using Dirichlet’s Theorem, the required volume of the ­tetrahedron is  

 



 

 

 

 

 

V = ​∫  ​ ∫  ​ ∫  ​ dx dy dz  

 

 

 

 

        



 

 

= ​∫  ​ ∫  ​ ∫  ​ abc dX dY dZ  

 

 

 

 

       

 

 

= abc ​∫  ​ ∫  ​ ∫  ​ X 1-1Y 1-1Z 1-1dX dY dZ  

 

 

 

 

     



Γ(1)Γ(1)Γ(1) = abc ​ ______________       ​ Γ(1 + 1 + 1 + 1) abc abc ____ abc = ____ ​     ​= ____ ​   ​ = ​   ​  . 6 Γ(4) 3!

12/9/2011 3:38:46 PM

7.12  n  chapter Seven EXAMPLE 7.25

dx dy dz, where x > 0, ∫∫∫ x p​ y q y > 0, and z > 0 under the condition ​( __ ​ a ​ + ​ __ ​  ​  ​ + b z r __ ​( ​ c ​)​ ≤ 1. y q x p z r Solution.  Put (​  __ ​ a ​ ​ = X, ​ __ ​   ​ ​ = Y, and ​( __ ​ c ​)​ = Z b so that 1 a __p -1 dx = __ ​ p ​X ​   ​  dX, 1 b __q -1 dy = __ ​ q ​Y ​   ​  dY , and x y l-1

Evaluate

z

m-1 n-1

)

(  )

(  )

)

1

c __r -1 dz = __ ​ r ​Z ​   ​  dz .  

​∫  ​ ∫  ​ ∫  ​ x l-1y m-1z n-1 dx dy dz  

 

 

 

 

     

1__ m-1 q

= ​∫    ​ ∫    ​ ∫    ​  aX  p ​   ​   

 

  

 

 

( 

1 __ p -1

1 __ q -1

∞

x dx   ​ using Beta and Gamma Evaluate ​∫​​  ​  ​  ______ 6 0 1 + x functions.  

1 __

∞

 

 

l __ -1

∫  1 + x

r

 

 

 

1 __

 

( 

y

n-1

)

1 - __

Y   ​ 2 ​dY. Therefore,  

 

 

   

 

m-1 ____ 2

n-1 ____ 2

 

 

 

 

m __ -1 2

a __

1 - __ __ 2   

1

b - __ = ​∫  ​ ∫  ​  am-1X ​     ​bn-1Y​   ​ ​   ​  X  ​   ​​   ​  Y ​ 2 ​dXdY 2 2      

 

 

 

 

(  ) (  ) (  )

M07_Baburam_ISBN _C07.indd 12

1 __

∞



1 2 1 ​ , __ = __ ​   ​  b ​ ​  __ ​   ​   ​, 6 3 3

(  )

1 __ -1

t ​ 3 ​   _________

   ​dt 1 __ 2  __ +  (1 + t) ​ 3 ​    ​ 3 ​ ∞

(  ) (  ) ( 3 3 ) 1 1 ​   ​Γ​  ​ __ 2  ​  ​ = __ ​   ​  Γ ​( ​ __ 6 3) ( 3)

(  ) (  )

1 ​ ​Γ ​ ​ __ 2 ​ ​ 1 ​   ​Γ​ ​ __ 2 ​   ​ Γ ​ ​ __ Γ​ ​ __ 3 3 3 3 1 __________ 1 ___________ __ __ = ​   ​  ​   ​    = ​   ​  ​   ​     6 6 Γ(1) 2 __ 1 __ Γ​ ​   ​ + ​   ​   ​



 (

)

1 p = __ ​   ​ ​ ​ _____   p   ​  ​, 6 sin ​ __  ​ 3



p using Γ (n) Γ (1 – n)​= ​ ______    ​  (0 < n). sin np

0

n __ -1 2

ambn = ​ ____  ​   ​∫  ​ ∫  ​ X ​   ​   Y ​   ​  dX dY 4    n n ​   ​ m ​   ​Γ​ ​ __ m ​   ​Γ​ ​ __ n  ​  ​__ Γ​ ​ __ Γ​ ​ __ ​   ​  ambn ___________ ambn ____________ 2 2 2 2 2 ____ ____ ·​    ​= ​   ​ ​  _____     ​ = ​   ​  m    n  m + n     4 Γ​ 1 + __ 4n ​   ​  + __ ​   ​   ​ ___ + 1  ​ ​   ​  Γ​ ​  2 ​  2 2 2  

2 - __

t ​ 3  ​ _____

a ______ Evaluate ​∫​​  ​ x4 ​√a2 - x2 ​  dx using Gamma function.

  

 

0

∞

EXAMPLE 7.28

m-1 n-1 ​∫  ​ ∫  ​ x y dxdy  

5 __

xm - 1 since b (m, n) = ​∫  ​ ​​​ _________ dx m  + n ​  0 (1 + x)

dx dy over the posi∫∫ 2 ambn x2 y tive octant of the ellipse ​ ___2  ​ + __ ​  2  ​ = 1 is ____ ​      ​ b​ 2n a b n m ​ , ​ __ ​ __  ​  + 1  ​. 2 2 y2 x2 Solution.  Putting __ ​  2  ​= X and __ ​  2  ​= Y, we get x = a​ b 1 __ __ a a - __2 b __ __ √ X  ​ and y = b​√ Y ​ and dx = ​    ​X  ​   ​dX and dy = ​   ​  2 2 x

Show that

∫  1 + t

= ​   ​  ​∫  ​ ​​​    ​ dt = ​   ​  ​∫  ​ ​​​  6 0 1+t 6 0

EXAMPLE 7.26 m-1

1 __



n __ -1

(  ) (  ) (  )

 

6

0

1 __ r -1

m __ -1

∞

x dx __ t​ 6 ​  -  1 ​  ​ ​​_____ ​    = ​   ​  ​  ​ ​​_____ ​    ​ t  ​ 6 ​dt 6 ​ 

ab c = ​ _____ ​ ​∫  ​ ∫  ​ ∫  ​ X ​ p  ​  Y ​ q ​   Z ​ r ​  dX dY dZ pqr          l m n __ __ __ l m n Γ​ ​ p  ​ ​Γ​ ​ q ​  ​Γ​( ​ r ​ )​ a b c _______________ . _____ = ​  pqr    ​· ​         ​ l m __ n Γ​ 1 + __ ​ p  ​+ __ ​ q ​  + ​ r ​  ​  

)

EXAMPLE 7.27

1 n-1 __

abc × ____ ​ pqr  ​X ​   ​  Y ​   ​  Z ​   ​  dX dY dZ l m n

( 

)

7.7  MISCELLANEOUS EXAMPLES

( ) (bY ​   )​ (cZ ​   ​ ) 1 l-1 __

)

Solution.  Putting x6 = t, we have x = t ​ 6 ​  and so 5 1 - __ dx = ​ __  ​t   ​ 6 ​ . When x = 0, t = 0 and when x = ∞, t = 6 ∞. Therefore,

Therefore,    

(  ) ( 

m ​   ​Γ​ ​ __ n ​  + 1  ​ Γ​ ​ __ ambn m n ambn 2 2 ____ = ​ _______   ​· ​  _____________       ​= ​      ​b ​ __ ​   ​ , ​ __  ​+ 1  ​. m + n _____ 2n 2n 2 2 Γ​ ​   ​   + 1  ​ 2

(  ) (  )

( 

)

Solution.  Putting x = a sin q, we get dx = a cos q ______ __________ dq and √ ​ a2 - x2 ​ = ​√a2 - a2 sin2 q ​  = a cos q. When p x = 0, q = 0 and when x = a, q = ​ __  ​. Therefore 2 a ______ 4 2 2 I = ∫​   ​​ x ​√a - x    ​dx  

0

12/9/2011 3:38:49 PM

beta and gaMMa functionS   n 7.13 p ​ __  ​ 2



= ​∫  ​ ​​a sin q (a cos q) (a cos q) dq 4

4

p ​  __  ​  2

0

= a6 ​∫  ​  ​​ sin4 q cos2 q dq.



 

0

(  ) (  ) (  2 )

m+1 n+1 Γ ​ ​ ____    ​  ​Γ ​ ​ ____    ​   ​ 2 2 1 _______________ __ m n Since ​∫  ​  ​​sin q cos q dq = ​   ​  ​        ,​ 2 + n   + 2  0 Γ ​ ​ m_________ ​  ​ p ​  __  ​  2  

we have

[  (  ) (  ) ]

5 ​   ​Γ​ ​ __ 3 ​   ​ Γ​ ​ __ 2 2 ​   _________ I = a ​​    ​ 2Γ(4) 6

[ 

]

3 1 __ __ 1 __ __ ​   ​  · __ ​   ​  ​√p ​ · ​   ​ ​√p ​  pa6 6 a 2_____________ 2 2 = ​ __ ​  ​ ​     ​   ​= ___ ​   ​ . 2

3!

32

(n - 1) (n - 3) (n - 5). . . 6.4.2 = ​ _________________________         ​, (n odd) n (n - 2). . . 5.3.1

which was to be established. EXAMPLE 7.30

( 

∞

(  )

1 _____

2.4. ... (n - 1) xn _____ Show that ∫​   ​​ ​ ______   2   ​dx = ​ ____________      ​,  where 1.3. 5. ... n   0 ​√ 1 - x  ​ n is odd integer.

p ​  __  ​  2

1

sinn q xn _____ ∫​   ​​​ ​ _______  2    ​dx = ​∫  ​  ​​​ _____   cosq dq cosq ​    0 ​√ 1 - x  ​ 0  

p ​  __  ​  2

n = ​∫  ​  ​​sin q dq



 

0

(  ) (  2 )

n + ​  1  Γ​ ​ _____  ​ __ 2 _________ = ​       ​​√p ​ (see Example 7.11). n + ​  2  2Γ​ ​ _____  ​ Since n is odd, we take n = 2m + 1. Therefore 1 xn _______ _____  ​dx ∫​   ​​ ​​ ​√1 - x2 ​   0 __ ​___ p ​  Γ(m + 1) √ ___________ = ​       ​​   ​   2 3 __ 2 Γ ​ m + ​   ​   ​ 2 __ ​√ p ​  m(m - 1)(m - 2). . . 3.2.1 ________________________________ ___ = ​            ​ ​   ​   2 3 3 1 1 1 1 ​   ​ ​ m + __ ​   ​   ​ ​ m - __ ​   ​   ​ ​ m - __ ​   ​   ​. . . __ ​   ​ · __ ​   ​ Γ​ ​ __ 2 2 2 2 2 2

( 



)

) ( 

) ( 

)

(  )

2m (2m - 2) (2m - 4). . . 6.4.2 = ​ ____________________________         ​ (2m + 1) (2m - 1) (2m - 3). . . 3.1

M07_Baburam_ISBN _C07.indd 13

∞

m + 1 _____



-1 1 = _____ ​  m + 1 ​  ​∫  ​ ​​e- zz​  2 ​      dz 2a 0



1 1  = _____ ​  m + 1 ​  Γ ​ _____ ​ m + ​   ​. 2a 2

( 

)

EXERCISES 1. Show that p (i)  b (2.5, 1.5) = ​ ___  ​ . 16 9 7 _____ 5p . (ii)  b __ ​   ​  , __ ​   ​  = ​    ​  2 2 2048 __ 3 ​  Γ  ​ __ 1 ​  = p √ (iii)  Γ ​ __ ​ 2 ​.  4 4

( ) () ()

2. Show that

__ __ -x3 ​√ p ​  ___ ​∫  ​ ​​​√x ​e  dx = ​   ​   . 3 0 ∞

3. Show that ∞

Γ(n) 2 ​∫  ​ ​​x2n - 1e-ax dx = ​ ____n ​ . 2a 0

 

( 

m - 1 _____

= ​  m + 1 ​  ​∫  ​ ​​e-zz​  2 ​ dz 2a 0

  

Solution.  Putting x = sin q, we have dx = cosq dq. Therefore

∞



EXAMPLE 7.29 1

)

2 2 1 m + ​  1  Show that ​∫  ​ ​​xme-a x dx = _____ ​  m + 1 ​  Γ ​ ​ _____  ​. 2a 2 0 _ 1 - __1 Solution.  Putting ax = ​√z ​,  we have a dx = ​ __ ​  z ​ 2 ​  2 dz Therefore ∞ ∞ _ m __1 2 2 1 ​ z  ​​   ​ z- ​ 2 ​  dz ​∫  ​ ​​xme-a x dx = __ ​   ​  a ​∫  ​ ​​e-z ​ ___ ​ √ a 2 0 0

4. Show that ∞

Γ(a + 1) xa ​∫  ​ ​​__ ​  x ​dx = ​ _________    ​  , if a > 1. a (log a)a + 1 0 5. Show that p ​ __  ​ 2

8 ​∫  ​ ​  ​ sin3 x cos 5/2 x dx = ___ ​ 77  ​ .  

0

12/9/2011 3:38:52 PM

7.14  n  chapter Seven 14. Show that yb (x + 1, y) = xb (x, y + 1).

6. Show that b

m n m+n+1 b (m+1, n+1). ​∫  ​ (x - a) (b - x) dx = (b - a)  

a

7. Prove that

( 

∞

)

2 2 1 1  ​∫  ​ ​​xne-a x dx = _____ ​  n  + 1 ​  Γ ​ _____ ​ n + ​   ​, n > - 1. 2a 2 0

8. Show that 1

1

x2dx p__ . _______ _____ dx    ​· ​∫  ​​ _______ ​ _____  4  ​  = ​ ____    ​  ​∫  ​​   ​​√1 - x  4  ​   4​ 2 ​  √ ​ 1 + x  ​   √ 0 0  

∞

y x z __ __ 16. The plane ​ __ a ​ + ​ b ​ + ​ c ​ = 1 meets the axes in A, B, and C, respectively. Find the mass of the t­etrahedron OABC if the density at any point is ρ = μ xyz. y __ x __ z ρ dx dy dz, 0 0  0

(

)

1 n-1 (ii) ​∫ ​​ ​  log ​ __ x ​ dx  0

∞

∞

(iii) ​∫  ​ ​​e-yyn-1dy = Γ(n). 0

(  )

∞

1 2 -    __ 1 1 1  ​  ​. (iv) ​∫  ​ ​​e-x dx   (v) ​ __ ​  ​∫  ​ ​​e-tt ​ 2 ​ dt = __ ​   ​  Γ ​ ​ __ 2 2 2 0 0 ∞

(  )

1 1 ​   ​. dx = __ ​   ​  Γ ​ ​ __ 9 3 Γ(n) . 2 13. Show that ∫​   ​ ​​x2n-1e-ax dx = ​ ____   ​ 2an 0 12. Show that ∫​   ​ ​​x3e

-x2

0 ∞

M07_Baburam_ISBN _C07.indd 14

 

  

= ​∫    ​​ ​​∫  mxyz dx dy dz    ​​ ​​∫     ​​    

 

  

 

  

 

17. Show that the volume of the solid bounded __ x by the coordinate planes and the surface ​ __ √__​ a ​  __ y abc z + ​ __ ​  ​ + ​√__c ​  ​ = 1 is ____ ​   ​ . 90 b 2 x2 y 18. Find the volume of the ellipsoid __ ​  2  ​ + ​ __2  ​ + a b z2 __ ​  2  ​= 1. c

√

 

1

 

  

y x z Put __ ​ a ​= X, __ ​   ​= Y, and __ ​ c ​= Z and proceed. b μa2b2c2 Ans. ​ ______  ​  .  720

xdx 1 1 _______ 2  ​, __  5    ​= __ ​ 5 ​  b ​ ​  __ ​   ​   ​. ​∫ ​​​   ​  _____ 2 5 ​ 1 x  ​   √ 0

10. Show that

 

  



9. Prove that

∞

2 __ 2 p__ . e- __x 15. Show that ∫​   ​ ​​​√x ​  e-x dx · ​∫  ​ ​​​ ___  ​  ​ dx = ​ ____    ​  ​ x  √ 2​√2 ​  0 0

 

 

Hint: V = 8 ​∫  ​​ ​​∫  ​​ ​​∫  ​​ dx dy dz.  

 

 

 

 

 

      2

y x2 z2 Put ​ __2  ​= X, ​ __2  ​= Y, and ​ __2  ​= Z, a b c and use Dirichlet’s Theorem to get 4p V = ​ ___ ​ abc. 3 19. Show that the entire volume of the solid

(  ) (  ) 2

2

2

__ x __3 y __3 z ​3 4 ​ __ ​   ​ ​    ​+ ​ __ ​   ​ ​   ​  + ​( __ ​ c ​)​   ​  = 1 is ___ ​    ​ p abc. 35 a b  

12/9/2011 3:38:54 PM

8 8

Reduction Formulas

A reduction formula for a given integral is an integral which is of the same type as the given integral but of a lower degree (or order). The reduction formula is used when the given integral cannot be evaluated otherwise. The repeated application of the reduction formula helps us to evaluate the given integral. In what follows, we shall observe that the reduction formulas are obtained by repeated application of integration by parts. 8.1  REDUCTION FORMULAS FOR ∫ sin xdx AND n ∫ cos x dx n

∫ sin nxdx =∫ sin n −1 x .sin xdx = sin n −1 x( − cos x)

− ∫ (n − 1) sin n − 2 x cos x( − cos x)dx = − sin

x cos x

+ (n − 1)∫ sin n − 2 x cos 2 xdx = − sin n −1 x cos x + (n − 1)∫ sin



n−2

x(1 − sin x)dx 2

= − sin n −1 x cos x + (n − 1)

× ∫ sin

n−2

xdx − (n – 1)∫ sin n xdx.

Therefore, transposing the last term to the lefthand side, we get n∫ sin n xdx = − sin n −1 x cos x + (n − 1)∫ sin n − 2 xdx.

M08_Baburam_ISBN _C08.indd 1

− sin n −1 x cos x n − 1 + sin n − 2 xdx n n ∫ (ii) Following the steps of part (i), we get sin x cos n −1 x n − 1 n + cos n − 2 xdx ∫ cos xdx = n n ∫ Deduction: We have n ∫ sin xdx =

π

2

∫ sin

n

xdx

0

π

2

= ∫ cos n xdx 0

(i) Integration by parts yields

n −1

Hence,

 (n − 1)(n − 3)(n − 5) … 2 , if n is odd.  n(n − 2)(n − 4)3  =  (n − 1)(n − 3)(n − 5) … 2 π , if n is even.  n(n − 2)(n − 4)3 2

∫

Proof: Using reduction formula for sin n xdx we have π

2

I n = ∫ sin nxdx. 0

=−

sin

n −1

x cos x n

π

2

0

π

n − 1 2 n−2 + sin xdx 3 n ∫0

n −1 In−2 . n When n is odd, we have =

In−2 =

n−3 n−5 In−4 , In−4 = In−6 , n−2 n−4

1/2/2012 12:01:58 PM

8.2  n  chapter eight Therefore,

π

4 2 22 I 5 = I 3 , and I 3 = I1 = ∫ sin xdx 5 3 30 =

π

6

x dx =

0

π 2 2 [ − cos x]02 = . 3 3

=

Hence, In =

(n − 1)(n − 3)(n − 5) … 2 . n(n − 2)(n − 4) …3

When n is even, we note that n−3 n−5 In−2 = In−4 , In−4 = In−6 , n−2 n−4 3 1 I 4 = I 2 , and I 2 = I 0 4 4

∞

Evaluate

x = ∞, θ = π2 . Therefore,

12 0 1 π sin xdx = . . 2 ∫0 2 2

xdx

Solution. Putting sin x = t , we get cos xdx = dt. So,

∫ cos

5

xdx = ∫ cos 4 x cos x dx

= ∫ (1 − sin 2 x) 2 cos xdx

π

∞

=

3

t t =t + −2 5 3 2 3 1 5 = sin x − sin x + sin x . 3 5

EXAMPLE 8.2 Evaluate ∫ sin 6 x dx. 0

Solution. For even n, we have 2

∫ sin xdx = n

0

M08_Baburam_ISBN _C08.indd 2

a

2

∫ cos

2 n −1

2

θ dθ ( even power of cosine )

0

1 (2n − 3)(2n − 5) …3.1 π . . . a 2 n −1 (2n − 2)(2n − 4) … 4.2 2

π 2

Evaluate ∫ cos 7 x dx 0

Solution. We have, for odd power of cosine, π

2

∫ cos

n

x dx =

0

(n − 1)(n − 3)(n − 5) … 2 . n(n − 2)(n − 1) …3

π

2

Therefore, ∫ cos 7 x dx = 0

6.4.2 16 = . 7.5.3 35

EXAMPLE 8.5 Evaluate

2a

∫x

7 2

−1

(2a − x) 2 dx.

0

π 2

π

π

1

EXAMPLE 8.4

= ∫ (1 − t ) dt 5

π

2 2 dx a sec 2θ dθ a sec 2θ dθ ∫0 (a 2 +x 2 )2 = ∫0 [a 2 (1 + tan 2 θ )]n = ∫0 a 2nsec2n θ

2 2

= ∫ (1 + t 4 − 2t 2 )dt

dx . + x 2 )n

θ d θ. When x x= =0,0, θ = 0 and when x = ∞, θ = π2 .

= 5

2

Solution. Putting x = a tan θ , we have dx=a sec2

EXAMPLE 8.1

∫ cos

∫ (a 0

Hence, (n − 1)(n − 3)(n − 5) 3.1 π In = . n(n − 2)(n − 4) … 4.2 2

Evaluate

5.3.1 π 5π = . 6.4.2 2 32

EXAMPLE 8.3

π

=

(6 − 1)(6 − 3)(6 − 5) π . 6(6 − 2)(6 − 4) 2

2

∫ sin

(n − 1)(n − 3)(n − 5) …3.1 π . . 2 n(n − 2)(n − 4) … 2

Solution. Putting x = 2a sin 2 θ , we get dx = 4a sin θ cos θ dθ . Therefore, the given integral reduces to π

2

∫ (2a sin 0

2

7 2

θ) .

4a sin θ cos θ 2a 1 − sin 2 θ

dθ

1/2/2012 12:01:59 PM

reduction forMulaS  n 8.3 π

π

2

= ∫ 2 a sin θ dθ = 32 a 5

4

8

0

= 32a 4

2

4

8 ∫ sin θ dθ 0

7.5.3.1 π 35π a 4 . = . 8.6.4.2 2 8

Proof: Using reduction formula for

∫ sin

∫ sin

m

x cos n x dx

= ∫ sin m −1 x.cos n x sin xdx  cos x  = sin m −1 x  −   n +1  n +1

π

m − 1 2 m−2 m −1 + sin x cos n x dx = I m − 2, n . ∫ m+n 0 m+n When m is odd, we get m−3 I m − 2, n = I m − 4, n , m+n−2 I m − 4, n = I 5, n =

sin m −1 x cos n +1 x m − 1 + sin m − 2 x cos n xdx n +1 n +1 ∫ m −1 − sin m x cos n x dx. n +1 ∫

Hence, sin m −1 x cos n +1 x sin x cos xdx = − ∫ m+n m

n

+

m −1 sin m − 2 x cos n xdx, m+n∫

which is the required reduction formula. Deduction: We have

I 3, n

2 2 2 sin x cos n x dx = I1, n = n+3 n + 3 ∫0 π

Hence, Im, n =

(m − 1)(m − 3)(m − 5) … 4.2 . (m + n)(m + n − 2)(m + n − 4) …(n + 3)(n + 1)

When m is even, we have m−3 I m − 2, n = I m − 4, n , m+n−2 I m − 4, n = I 4, n =

m−5 I m − 6, n , m+n−4 3 I 2, n , and n+4 π

2

m

x cos n x dx.

0

 ( m − 1)( m − 3) …( n − 1)( n − 3) …  ( m + n )( m + n − 2)( m + n − 4) …  =  ( m − 1)( m − 3) …( n − 1)( n − 3) … . π ,   ( m + n )( m + n − 2)( m + n − 4) … 2 if both m and n are even.

M08_Baburam_ISBN _C08.indd 3

4 I 3, n , and n+5

2  cos n +1 x  2 2 . = −  = n + 3  n + 1  0 (n + 3)(n + 1)

π

∫ sin

m−5 I m − 6, n , m+n−4

π

sin m −1 x cos n +1 x m − 1 + sin m − 2 x cos n + 2 xdx n +1 n +1 ∫

=−

π

π

 cos n −1 x  − ∫ ( m − 1) sin m − 2 x cos x  −  dx  n +1  =−

x cos n x dx, we have

2  sin m −1 x cos n +1 x  2 I m , n = ∫ sin m x cos n x dx =  −  m+n  0 0

m n 8.2  REDUCTION FORMULA FOR ∫ sin x cos x dx

Integration by parts yields

m

I 2, n

1 1 2 cos n x dx. = I 0, n = n+2 n + 2 ∫0

Hence, I m,n =

(m − 1)(m − 3)(m − 5) …1 (m + m)(m + n − 2)(m + n − 4) … (n + 2) π



× ∫ cos n x dx 0

1/2/2012 12:02:00 PM

8.4  n  chapter eight (m − 1)(m − 3)(m − 5) …1   (m + n)(m + n − 2)(m + n − 4) … (n + 2)  (n − 1)(n − 3) …  . if n is odd.  n(n − 2) …  = (m − 1)(m − 3)(m − 5) …1   (m + n)(m + n − 2)(m + n − 4) … (n + 2)  (n − 1)(n − 3) … π  . if n is even.  n(n − 2) … 2 Hence,

I m,n

π

π

82 3 cos 3 sin 6 d sin x cos 7 x dx θ θ θ = ∫0 3 ∫0 6

4

π 2

∫ sin

3

2a

Therefore, 2a

∫x

5 2

2a − xdx

0

π

2

= ∫ 2a (2a sin 2 θ ) 2 (4a sin θ cos θ ) cos θ dθ 5

0

π

2

= 32a

4

= 32a4 EXAMPLE 8.8

0

Evaluate

Solution. (i) Since m = 3 is odd, we have π

2

3 4 ∫ sin x cos x dx = 0

(3 − 1)(4 − 1)(4 − 3) 2 = . 7.5.3 35

(ii) We have π

4

5

∫ x 2 2a − xdx .

6

4 3 (ii) ∫ cos 3θ sin 6θ dθ.

∫ cos

1 . 15

Solution. Substituting x = 2a sin 2 θ , we have dx = 4a sin θ cos θdθ .

x cos 4 x dx and

π

6

=

∫ sin

6

θ cos 2 θ dθ

0

0



8  2.6.4.2  3 10.8.6.4.2 

0

EXAMPLE 8.6 Evaluate (i)

=

EXAMPLE 8.7 Evaluate

(m − 1)(m − 3)(m − 5) …1   (m + n)(m + n − 2)(m + n − 4) … (n + 2)  (n − 1)(n − 3) …  . if at least one  n(n − 2) …  of m and n is odd.  = (m − 1)(m − 3)(m − 5) …1   (m + n)(m + n − 2)(m + n − 4) … (n + 2)  (n − 1)(n − 3) … π  . if both m  n(n − 2) … 2  and n are even. 

3

3θ sin 3 6θ dθ

5.3.1.1 π 5π a4 = . 8.6.4.2 2 8

1

∫x

6

1 − x 2 dx .

0

Solution. Putting x = sin θ , we get dx = cos θ dθ . Therefore, the given integral reduces to π

π

2

∫ sin 0

2

6

θ.cos θ.cos θ dθ = ∫ sin 6 θ.cos 2 θ dθ 0

0

π

6

= ∫ cos 4 3θ (2sin 3θ cos 3θ )3 dθ 0

π

6

= 8∫ sin 3 3θ cos 7 3θ dθ. 0

Putting 3θ = x, we get 3 dθ = dx , and so,

M08_Baburam_ISBN _C08.indd 4

=



5.3.1.1 π 5π . = . 8.6.4.2 2 256

EXAMPLE 8. 9

Evaluate π

(i) ∫ sin 0

4

θ 2

cos

3

θ 2

π

dθ and (ii)

2

∫ sin

15

x cos3 x dx.

0

1/2/2012 12:02:00 PM

reduction forMulaS  n 8.5 Solution.

EXAMPLE 8.10

(i) Putting

θ

2

= φ , we get dθ = 2d φ . Hence, the

integral becomes π

3.1.2 4 2∫ sin 4 φ cos3 φ d φ = 2 = . 7.5.3.1 35 0 2

(ii) We have π

2

15 3 ∫ sin x cos x dx = 0

=

14.12.10.8.6.4.2.2 18.16.14.12.10.8.6.4.2 1 . 144

n 8.3  REDUCTION FORMULAS FOR ∫ tan x dx n AND ∫ sec x dx

I n = ∫ tan n x dx = ∫ tan n − 2 x tan 2 x dx = ∫ tan n − 2 x(sec 2 x − 1)dx = ∫ tan n − 2 x sec 2 xdx − ∫ tan n − 2 x dx n −1

tan x − In−2 . n −1

– ∫ (n − 2) cosec n − 3 x cosec x 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 − 2 x cot x − (n − 2) I n + (n − 2) I n − 2 .

Hence,

(n − 1) I n = − cosec n − 2 cot x + (n − 2) I n − 2

n−2 or I n = − cosec cot x + n − 2 I n − 2 . n −1 n −1

4 Evaluate ∫ sec x dx.

I n = ∫ sec n xdx

n Solution. The reduction formula for ∫ sec x dx is

n−2 = ∫ sec x sec 2 x dx = ∫ sec n − 2 x tan x − ∫ (n − 2) sec n − 3 x

. sec x tan x .tan xdx

= 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 − sec n − 2 x) dx

= sec n − 2 x tan x − (n − 2) I n + (n − 2) I n − 2 . Hence, (n − 1) I n = sec n − 2 x tan x + (n − 2) I n − 2 n−2 or I n = sec x tan x + n − 2 I n − 2 . n −1 n −1

M08_Baburam_ISBN _C08.indd 5

= – cosec n − 2 x cot x

EXAMPLE 8.11

(ii) Let



I n = ∫ cosec n xdx = ∫ cosec n − 2 x cosec2 xdx

× ∫ cosec n x dx + (n − 2)∫ cosec n – 2 x dx

(i) Let

=

Obtain a reduction formula for cosecn xdx. Solution. We have

I n = ∫ sec n x dx =

sec n − 2 x tan x n − 2 + In−2 . n −1 n −1

Therefore, I4 = =

sec 2 x tan x 2 + ∫ sec 2 x dx. 3 3 sec 2 x tan x 2 + tan x 3 3

1 = [tan x sec 2 x + 2 tan x]. 3 EXAMPLE 8.12 π 4

n If I n = ∫ tan x dx show that (n–1)(In +In–2) =1. 0

Hence, evaluate I5.

1/2/2012 12:02:01 PM

8.6  n  chapter eight Solution. We have π

 − cos mx  n −1  − cos mx  In = xn   − nx   dx  m  ∫ m 

π

4

4

I n = ∫ tan n x dx = ∫ tan n − 2 x tan 2 x dx 0

0

π

=−

x n cos mx n + ∫ x n −1 cos mx dx m m

=−

x n cos mx n  n −1 sin mx  +  x  m m  m 

4

= ∫ tan n − 2 x(sec 2 x − 1) dx 0 π

4

= ∫ tan 0

π

n−2

4

x sec x dx − ∫ tan n − 2 x dx 2

− ∫ ( n − 1)x n − 2

0

π

 tan n −1 x  4 =  − In−2 .  n −1  0

=−

Hence,

−

π

(n − 1) I n + (n − 1) I n − 2 = [tan n −1 x] 40 = 1 or

=−

(n − 1) ( I n + I n − 2 ) = 1.

(1) −

Putting n = 5 in (1) we get 4( I 5 + I 3 ) = 1 or I 5 = 14 − I 3 . But, putting n = 3 in (1), we have 2 ( I 3 + I1 ) = 1 or I 3 = 12 − I1 .

(2)

Now,

x n cos mx n n −1 + 2 x sin mx m m

n(n − 1) n − 2 x sin mx dx m2 ∫ x n cos mx n n −1 + 2 x sin mx m m

n ( n − 1) I n −2 . m2

(ii) Let I n = ∫ x n cos mxdx. Then proceeding exactly as in part (i), we get the following reduction formulas: In =

π

4

π

I1 = ∫ tan xdx = [ − log cos x] 40 = − log 0

1 2

= −[log1 − log 2] = log 2 = 12 log 2. Therefore, (2) yields 1 1 I 3 = − log 2. 2 2 Hence, 1 1 1 1 1 I 5 = − + log 2 = log 2 −  . 4 2 2 2 2 n 8.4  REDUCTION FORMULAS IFOR n = ∫ x sin mx dx

AND I n = ∫ x n cos sin mx dx

(i) Put I n = ∫ x n sin mx dx. Then integration by parts yields the following reduction formulas:

M08_Baburam_ISBN _C08.indd 6

sin mx  dx  x 

x n sin mx n n −1 + 2 x cos mx m m

−



n ( n − 1) I n −2 . m2

EXAMPLE 8.13 π

2

If I n = ∫ x n sin xdx, n > 1, show that 0

π I n + n(n − 1) I n − 2 = n    2

n −1

.

Hence, evaluate I5. Solution. The reduction formula for dx is

∫x

n

sinmx

x n cos mx n n −1 + 2 x sin mx m m n(n − 1) In−2 . − m2

In = −

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reduction forMulaS  n 8.7 Here m = 1and the limits are from 0 to π2 . Therefore, π

I n = [ − x n cos x + nx n −1 sin x] 20 − n(n − 1) I n − 2

=

n −1

.

Putting n = 5 in (1), we get

(1)



x m +1 1 x m +1 dx − ∫ n(log x) n −1 . m +1 x m +1

I m , n = (log x) n

or π I n + n (n − 1) I n − 2 = n    2

Then integrating by parts, we get

x m +1 n x m (log x) n −1 dx (log x) n − m +1 m +1∫

x m +1 n I m , n −1 , (log x) n − m +1 m +1 which is the required reduction formula. =

4

π I 5 + 20 I 3 = 5   .  2

(2)

Putting n = 3 in (1), we have

EXAMPLE 8.14 ∞

2

π I 3 + 6 I1 = 3   .  2

(3)

Evaluate ∫ e − x x n dx , where n is a positive integer. 0

Solution. We have

But, π

π

2

π

2

I1 = ∫ x sin x dx = [ − x cos x] 0 − ∫ cos x dx 2

0

0

∞

∫e

−x

x n dx = Γ (n + 1)

0

= n! since n is a positive integer.

π

= [sin x]02 = 1. EXAMPLE 8.15

Therefore, (3) yields I3 = 3

π2 4

a

− 6.

Now putting this value of I3 in (2), we get  3π 2  5π 4 5π 4 − 20 I 3 = − 20  − 6 16 16  4 

I5 =

5π 4 − 15π 2 + 120. 16

=

n ax 8.5  REDUCTION FORMULAS FOR ∫ x e dx AND

∫x

m

0

I n − (n + a ) I n −1 + a (n − 1) I n − 2 = 0. Solution. We have a

a −x  e− x  n −1 e In =  xn nx dx −  ∫ −1  −1  0 0 a

= − a n e − a + n∫ x n −1e − x dx 0

n

(log x ) dx

I n = ∫ x e dx. n ax

= −a e n

Then integration by parts yields the following reduction formulas: e ax e ax x n e ax n dx = − ∫ nx n −1 − I n −1 . a a a a

(ii) Let

a

−a

0

0

+ (n + a ) I n −1

a n −1 − x a    x e   ( n − 1)x n − 2e − x dx  −a   +  ∫ 1 − 0 0    

= − a n e − a + (n + a ) I n −1 − a (n − 1) I n − 2 + a(an −1e − a )

I m , n = ∫ x (log x) dx. m

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a

= − a n e − a + (n + a )∫ x n −1e − x dx − a ∫ x n −1e − x dx

(i) Let

In = xn −

If I n = ∫ e − x x n , show that

n

= (n + a ) I n −1 − a (n − 1) I n − 2 .

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8.8  n  chapter eight Hence, I n − ( n + a)I n −1 + a( n − 1)I n − 2 = 0. = ∫ cos m x sin nx dx. 8.6  REDUCTION FORMULAI mn FOR Let I mn = ∫ cos m x sin nx dx. Integrating by parts, we get I mn = − cos m x

cos nx n

2na 2 ∫

which in tum implies

∫ (x

2

∫ (x

2

=

x 1 + 2n ∫ 2 dx 2 n (x + a ) ( x + a 2 )n 2

dx . ( x + a 2 ) n +1 2

Integrating (1) by parts, we get dx x ∫ ( x 2 + a 2 )n = ( x 2 + a 2 )n − ∫ x( − n)( x 2 + a 2 ) − n −1 2 x dx

dx x = + (2n − 1) ( x 2 + a 2 ) n +1 ( x 2 + a 2 ) n ×∫

or

∫ (x

2

dx ( x + a2 ) n 2

dx x = 2 n +1 2 +a ) 2na ( x 2 + a 2 ) n +

2n − 1 dx (2) 2 ∫ 2 2na ( x + a2 ) n

Changing n to n–1, the expression (2) reduces to the required Reduction Formula:

∫ (x

2

dx x = 2 n 2 +a ) 2(n − 1)a ( x 2 + a 2 ) n −1 +

2n − 3 dx . (3) 2 ∫ 2 2( n − 1)a ( x + a2 ) n −1

Evaluate the integral

∫ (x

dx . + 1)3

2

Solution. By the reduction formula, we have

dx + a 2 )n

dx 1 = ∫ 1. 2 dx 2 n +a ) ( x + a2 ) n

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x x2 + a2 − a2 + 2n ∫ 2 dx 2 n (x + a ) ( x + a 2 ) n +1 2

EXAMPLE 8.16

m m cos x cos nx I m −1, n −1 . + m+n m+n

We write

=

Thus dx

1 = − cos m x cos nx n m − ∫ cos m −1 x cos nx sin x dx n 1 = cos m x cos nx n m − ∫ cos m −1 x [sin nx cos x− sin(n − 1) x] dx n 1 m = − cos m x cos nx − [ I m , n − I m −1, n −1 ]. n n Thus, 1 m  m m 1 +  I m , n = − cos x cos nx + I m −1, n −1 , n n n

8.7  REDUCTION FORMULA FOR

x x2 dx + 2n ∫ 2 2 n (x + a ) ( x + a 2 ) n +1 2

−2na 2 ∫

 cos nx  − ∫m cos n −1 x ( − sin x )  −   n 

I m,n = −

=

dx x = 3 + 1) 2(3 − 1)1( x 2 + 1) 2 3 dx + ∫ 2 2.2.1 ( x − 1)2

∫ (x

(1)

2

=

x 3 dx + . 4( x 2 + 1) 2 4 ∫ ( x 2 + 1) 2 (1)

Further, the reduction formula applied to the integral on the right hand side of (1) yields

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reduction forMulaS  n 8.9 π 4

dx x 1 dx ∫ ( x 2 + 1)2 dx = 2( x 2 + 1) + 2 ∫ x 2 + 1

7. If I n = ∫ tan nθ dθ , 0

1 = + tan −1 x. 2 2( x + 1) 2 x



π

Hence (1) reduces to

∫ (x

2

show that n[I n −1 + I n +1 ] = 1. 2

8. ShowI mthat cosm x cosnx dx , n = ∫ cos x cos nxdx , m

 dx x x 3 1 0 = +  + π tan −1x  π 3 2 2 2 + 1) 4( x + 1) 4  2( x + 1) 2  m 2 m m −1 ∫ cos x cos nxdx = m + n ∫0 cos x cos( n − 1)xdx . x 3x 3 0 −1 = + + tan x. 4( x 2 + 1) 2 8( x 2 + 1) 8 π

2

9. If I n = ∫ x sin n xdx ( n > 1), show that

EXERCISES

0

π 6



1. Evaluate (i)∫ sin 53θ dθ and

I5.

0

1

3 2 2

(ii)∫ x 4 (1 − x ) dx . 0

2. Evaluate

Ans. (i)

1

∫x

5

−1 sin xdx .

0

3. Evaluate

∫ cos

3 4

1

∫x

3 2

11π . Ans. 192 1 x 2 − x +1 log 2 . 2 x + x +1

Ans.

Ans.

2

10. Prove that ∫ cos

n −2

x sin nxdx =

0

149 . 225

1 , n > 1. n +1

π

2

11. If I m , n = ∫ cos cosm xxcosnx dx, ,show that cos nxdx m



m ( m − 1) = 2 I m − 2, n . m − n2

I m ,n

3π . 128

1 3 3 sec x tan x + sec x tan x 4 8 3 + log (sec x + tan x ). 8 5

6. Evaluate ∫ (a2 + x 2 ) 2 dx . 0

 67 2 5  + log(1 + 2)  a6 . Ans.  16  48 

M08_Baburam_ISBN _C08.indd 9

π

12. Evaluate

(1 − x ) dx .

5. Evaluate ∫ sec 5xdx .

Ans.

0

3 2

0

a

8 3π (ii) . 45 256

x sin 5 xdx . Ans.

4. Evaluate

n I n = n ( n − 1)I n − 2 + 1. Hence, evaluate 2

∫ (x

dx + 1) 4

2

Ans.

x 5x + + 3 6( x + 1) 24( x 2 + 1)2 5x 5 + tan −1X 16( x 2 + 1) 16 2

13. Show that ∞

∞

dx 2n − 3 dx ∫0 (x 2 + 1)n = 2n − 2 ∫0 (x 2 + a2 )n −1 and deduce that ∞

∫ (x 0

π 1.3.5.7 dx = . . 3 2 2.4.6.8 + 1)

2

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9

Quadrature and Rectification

In this chapter, we discuss the processes to find area of a given bounded region of a curve and to find length of an arc of a curve between two given points. Of course, these concepts are nothing but applications of definite integral 9.1  QUADRATURE The process of finding the area of a bounded region of a curve is called quadrature. 9.1.1  Area of a Curve Given by the Cartesian Equation Let AB be the arc of the curve y = f (x), between two ordinates x = a and x = b represented by AC and BD respectively.

that is, δ A lies between yδ x and ( y + δ y )δ x. Thus, ( y + δ y )δ x > δ A > yδ x or y+δy >

δA >y δx

Taking limit as Q → P, that is, as δ x → 0 and δ y → 0, we get dA = y = f ( x) dx

or dA = ydx. (1) Integrating both sides of (1) between the limits x = a and x = b, we get b

b

b

a

a

a

b

b

a

a

∫ dA = ∫ ydx = ∫ f ( x)dx or | A |ba = ∫ ydx = ∫ f ( x)dx

or Let P(x, y) and Q( x + δ x, y + δ y ) be two neighboring points on the given curve. Draw PR and QS perpendiculars to the x-axis. Then,

b

b

a

a

Area ACDB = ∫ ydx = ∫ f ( x)dx

(2)

PR = x, QS = y + δ y and RS = δ x.

Remark 9.1. 1. The area bounded by a curve, the x-axis, and the two ordinates and given by (2) is called the area under the curve.

Let A be the area ACRP and A + δ A be the area ACSQ. Then the area PRSQ is given by A + δ A − A = δ A. We observe that δ A lies between areas of the rectangles PRSM and NRSQ,

2. It can be shown similarly that the area bounded by the curve x = f ( y ) , the axis of y and the abscissae y = c and y = d is given by

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9.2   n  chapter nine

d

d

| A |cd = ∫ xdy = ∫ f ( y )dy. c

(3)

c

3. The area is considered to be positive if its boundary is described in the anticlockwise direction and it will be considered negative if its boundary is described in the clockwise direction. EXAMPLE 9.1

(ii) whole area of the ellipse

x2 y 2 + =1 a 2 b2

Solution. (i) The equation of the ellipse is x2 y 2 + = 1. a 2 b2 which yields b 2 a − x 2 (for the first quadrant ). a

Therefore, Area of quadrant a

= ∫ ydx 0

a

b a 2 − x 2 dx a ∫0

b x 2 a2 x a − x 2 + sin −1 = a 2 a 2 =

M09_Baburam_ISBN _C09.indd 2

(ii) The whole area of the ellipse is 4 times the area of one quadrant. Hence,  π ab  Area of the ellipse = 4  = π ab.  4 

Trace the curve ay 2 = x 2 (a − x) and show that 8 2 a . the area of its loop is 15 Solution. We note that (i) The equation of the curve does not alter if y is changed to –y. Therefore, the curve is symmetrical about the x-axis. Further, since the curve does not have a constant term, it passes through the origin. (ii) Equating to zero, the lowest degree term in the equation of the curve, the tangents at the origin are given by y2 – x2 = 0, that is, y = + x. Thus there are two real and district tangents at the origin. Therefore the origin is a node. (iii) Putting y = 0 in the equation, we get x2 (a – x) = 0. Thus x = 0 and x = a. Therefore, the curve intersects the x-axis at (0, 0) and (a, 0). Shifting the origin to (a, 0), the equation of the curve reduces to ay2 = – x (x2 + 2ax + a2). Equating to zero the lowest degree term, we get a2x = 0. Hence, at the new origin (a, 0), the tangent is x = 0. Thus at (a, 0), the tangent is parallel to the y-axis.

x2 y 2 (i) quadrant of the ellipse 2 + 2 = 1 a b

=

b  a 2   π  π ab ⋅  = a  2   2  4

EXAMPLE 9.2

Find the area of

y=

=

b  a2 −1   sin 1 a 2 

a

0

(iv) The equation of the curve can be written x 2 (a − x) . When 0 < x < a, y2 is as y 2 = a positive and so the curve exists in the region 0 < x < a. But when x > a, y2 is negative and so y becomes imaginary. Thus, the curve does not exist in the region x > a. (v) Equating to zero, the coefficient of highest power of y in the given equation, we see that above that there is no asymptote to the

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quadrature and rectification  n  9.3 curve. Thus, the shape of the curve is as shown below:

Further, from the given equation, we have y=

2a 2a − x x

.

Also, x = 0, that is, y-axis is the asymptote of the curve. Hence, the required area is 2a

A = 2 ∫ ydx 0

2a

2a 2a − x

= 2∫ From the given equation, we have y=

x a−x a

x

0

dx

π

2

= 4a ∫ 4a cos 2θ dθ , putting x = 2a sin 2θ ,

.

0

π

2

Therefore,

= 16a a

x a−x a

0

dx

π

2

= 4a 2 ∫ sin 3θ cos 2θ dθ , 0

EXAMPLE 9.4

Trace the curve a 2 y 2 = x 2 (a 2 − x 2 ) and find its whole area. Solution. From the given equation, we have

x = a sin 2θ = 4a 2

2

1 π  = 16a 2  .  = 4π a 2 . 2 2

0

a

∫ cos θ dθ 0

Area of the loop = 2∫ ydx

= 2∫

2

2.1 8 = a2 . 5.3.1 15

EXAMPLE 9.3

y=

x a2 − x2 . a

Also, Whole area = 4 times the area of half loop. a

Trace the curve xy 2 = 4a 2 (2a − x) and find the area bounded by this curve and its asymptote. Solution. The shape of the curve is

= 4∫ 0

x a2 − x2 dx a π

2

= 4a ∫ cos 2θ sin θ dθ , 0

Putting x = a sin θ  1  4 = 4a 2   = a 2 .  3.1 3 EXAMPLE 9.5

Find the area enclosed by the curve a2x2 = y3 (2a–y). Solution. We note that (i) The curve is symmetrical about y-axis and passes through the origin.

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9.4 n  chapter nine   (ii) Tangents at the origin are given by a2x2 = 0, that is, x = 0. Thus, the origin is a cusp. (iii) The curve cuts the y-axis at (0, 0) and (0, 2a). (iv) The curve exists in the region 0 < y < 2a but not for y > 2a and y < 0. (v) Shifting the origin to (0, 2a), we see that y = 2a is tangent at (0, 2a). Hence, the shape of the curve is as shown below:

(ii) The tangent at the origin are given by a ( y 2 − a 2 ) = 0. Thus y = x and y = –x are tangents at the origin (iii) Equating to zero, the coefficient of the highest power of y, the asymptotes parallel to the y-axis are given by x2 – a2 = 0. Thus, asymptotes parallel to the y-axis are x = a and x = –a. There is no real asymptote parallel to the x-axis. Thus, the shape of the curve is as shown below:

From the given equation, we have 3

y 2 2a − y x= . a

Therefore, Required Area = twice the area of half loop 2a

= 2 ∫ xdy

From the given equation, we have

0

π

2

= 32a

y=

3

y 2 2a − y = 2∫ dy a 0 2a

2

∫ sin θ dθ , 4

a2 x2 . a2 − x2

Then the required area is given by A = 4 times the area enclosed in first quadran

0

putting x = 2asin 2θ  3.1.1 π  . , = 32a 2   6.4.2 2  using reduction formula 2 = πa .

a

= 4∫ 0

M09_Baburam_ISBN _C09.indd 4

a − x2 a

= −2a ∫ 0

EXAMPLE 9.6

Find the area enclosed between the curve x2y2 = a2 (y2 – x2) and its asymptotes. Solution. To trace the curve, we note that (i) The given curve is symmetrical about both axes and passes through the origin.

ax 2

dx

−2 xdx a2 − x2 a

= −2a

(a − x ) 1 2 2

2

1 2

= −4a (0 − a ) = 4a 2 . 0

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quadrature and rectification  n  9.5 EXAMPLE 9.7

Find the common area between the curves y2 = 4ax and x2 = 4ay. Solution. Solving the given equations of the parabolas, we note that (4a, 4a) and (0, 0) are the points of their intersections.

Therefore, Common area = 2 [ Area OPQ + Area PQA] 4a 3a  = 2  ∫ axdx + ∫ 4ax − x 2 dx  3a 0  3a

= 2 a∫ 0

4a

xdx + 2 ∫ [4a 2 − ( x − 2a ) 2 ] 2 dx 1

3a

3a

2 3  1 = 2 a  x 2  + 2  ( x − 2a ) 4a 2 − ( x − 2a ) 2 3  0 2 3a

Then the required area is given by 4a

A=

∫ 0

4a

=

∫ 0

4a

y1dx − ∫ y2 dx

4a

+

0

4a

4axdx − ∫ 0

3

=2 a =

2 3  1 = 2 a  x 2  + 2  ( x − 2a ) 4a 2 − ( x − 2a ) 2 3  0 2

x2 3 2

4a

0

x2 dx 4a

1 x3 − 4a 3

4a

0

32 2 16 2 16 2 a − a = a . 3 3 3

EXAMPLE 9.8

Find the area common to the curves y2 = ax and x2 + y2 = 4ax. Solution. The equation y2 = ax represents a parabola with vertex at the origin, axis along x-axis and latus rectum a. On the other hand, x2 + y2= 4ax represents a circle with center (2a, 0) and radius 2a. Both curves are symmetrical about xaxis. Their points of intersection are x = 0 and x = 3a as shown in the figure below

M09_Baburam_ISBN _C09.indd 5

4a 2 x − 2a  sin 2 2a  3a

4 = 4a 3a 2 − 3a 2 + π a 2 3

4   = a2  3 3 + π  .  3  EXAMPLE 9.9 2 2 Find the area common to the circle x + y = 4 2 2 and the ellipse x + 4 y = 9.

Solution. The point of intersection of the two curves is x = 73 (for the first quadrant). Fur1 9 − x 2 from the equation of ellipse ther, y = 2 and y = 4 − x 2 from the equation of the circle.

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9.6 n  chapter nine   Therefore, Area common to the curves = 4 × [Area O APB]

π

= 2∫ a (1 − cos θ ).a (1 − cos θ ) dθ 0

= 4 [Area O M PB + AreaM AP]  73 1 = 4 ∫ 9 − x 2 dx + 0 2 

2

∫

7 3

 4 − x 2 dx   

9 x x = 2  9 − x 2 + sin −1  2 3 0 2

π

= 2a 2 ∫ (1 − cos θ ) 2 dθ 0

π

2

θ  = 2a 2 ∫  2sin 2  dθ  2 0 π

= 8a 2 ∫ sin 4

7 3

0

θ 2

dθ

π

θ

2

4 x x 4 − x 2 + sin −1  +4 2 3 2 =

= 8a

2

−8sin

7 7 . − 8sin −1 27 12

EXAMPLE 9.10

Find the area included between the cycloid x = a (θ − sin θ ) , y = a (1 − cos θ ) and its base. Solution. We note that y = 0 if cos θ = 1 , that is, θ = 0 . When θ = 0 . we have x = 0 . Thus, the curve passes through the origin. Further, y-axis is tangent at (0, 0). Also y is maximum when cos θ = −1, that is θ = π . When θ = π , we get x = aπ and y = 2a . The shape of the curve is as shown below:

π

π

0

0

M09_Baburam_ISBN _C09.indd 6

π

2

= 16 a

2

∫ sin t dt 4

0

 3.1 π  = 16a 2  ⋅  4.2 2 

= 3π a 2 . 9.1.2  Area of a Curve Given by Polar Equation Let AB be the curve r = f (θ ) and let OA and OB be the radii vectors θ = α and θ = β Let P ( r , θ ) be any point on the curve and Q(r + δ r ,θ + δθ ) be its neighboring point. Let the areas of the sectors AOP and AOQ be A and A + δ A respectively. Then, Curvilinear area OPQO = A + δ A − A = δ A.

Thus, Area of circular sector OPP ′ < δ A < Area of circular sector OQQ′ or

The required area is = 2∫ ydx = 2∫ y ⋅

4

0

7 12

= 4π + 9sin −1

∫ sin t (2dt ), t = 2

7 3

7 20 5 7 8π ⋅ −2 + 9sin −1 + 3 3 3 57 2 −1

2

dx dθ dθ

1 1 r (rδθ ) < δ A < (r + δ r )(r + δ r )δθ 2 2

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quadrature and rectification   n  9.7 or

π

1 2 1 r δθ < δ A < (r + δ r ) 2 δθ 2 2 or 1 2 δA 1 r < < (r + δ r ) 2 . 2 δθ 2

Taking limit as δθ → 0 . we have δ r → 0 and so dA 1 2 = r . dθ 2 Integrating both sides, between the limits α and β , we have β

dA

β

1

∫α dθ dθ = ∫α 2 r or [ A]αβ =

2

dθ

β

1 2 r dθ. 2 ∫α

= ∫ a 2 (1 + cos θ ) 2 dθ 0

π

2

θ  = a 2 ∫  2cos 2  dθ  2 0 π

= 4a 2 ∫ cos 4 0

θ 2

dθ

π

2

= 8a 2 ∫ cos 4φ d φ , φ = 0

θ 2

 3.1 π  3 2 . = 8a 2  = πa .  4.2 2  2 EXAMPLE 9.12

Trace the curve r = a (1 − cos θ ) and find its area. Solution. The shape of the curve is

EXAMPLE 9.11

Find the area of the cardioid r = a (1 + cos θ ). Solution. The given curve is symmetrical about the initial line. We have r = 0 when θ = π . Thus, θ = π is tangent at the pole. Further, r is maximum when cos θ = 1 , that is, θ = 0 and then r = 2a . When θ increases from 0 to p, r decreases from 2a from 0. The shape of the curve is shown below:

Then π

1 Required area = 2∫ r 2 dθ 2 0 π

= ∫ a 2 (1 − cos θ ) 2 dθ 0

π

Therefore, the required area = 2 × area of the upper half of the cardioid π 1 = 2∫ r 2 d θ 2 0

M09_Baburam_ISBN _C09.indd 7

2

θ  = a 2 ∫  2sin 2  dθ  2 0 π

2

= 4a

2

θ

∫ sin φ d φ, φ = 2 4

0

12/9/2011 12:40:29 PM

9.8   n  chapter nine  3.1 π  = 8a 2  ⋅  4.2 2 

=

 π a2  = 3  12 

3 2. πa . 2

=

π a2 . 4

.

EXAMPLE 9.14

EXAMPLE 9.13

Find the area of one loop of the curver r==aa sin 3θ. Solution. The shape of the curve is

Find the area enclosed by the loop of the curve x 3 + y 3 = 3axy (Folium of Descartes). Solution. The shape of the curve is

Putting x = cos θ , y = r sin θ . we see that 3a sin θ cos θ . r= . cos3θ + sin 3θ We note that r = 0, when 3a sin θ cos θ = 0 , that is, θ = 0 , π2 . Therefore,

Then π

13 Area of one loop = ∫ r 2 dθ 20 π

=

13 2 2 a sin 3θ dθ 2 ∫0

π

π

a2 3 = (1 − cos 6θ ) dθ 4 ∫0

a2 = 4 = =

a2 4

π

1  3 sin 6 θ − θ   6 0 π   3

π a2.

.

12 Also, Total area enclosed by the curve = 3 × (Area of one loop)

M09_Baburam_ISBN _C09.indd 8

12 Area of the loop = ∫ r 2 dθ 20 π

1 2 9a 2 sin 2θ cos 2θ dθ = ∫ 2 0 (cos3θ + sin 3θ ) 2 = =

π

tan 2θ sec 2θ

9a 2 2

∫ (1 + tan θ )

3a 2 2

∫t

2

3

2

dθ

0

∞

dt 2

,

1

where t = 1 + tan 3θ ∞ 3a 2  −1 = 2  t  1 3 = a 2 .. 2

12/9/2011 12:40:30 PM

quadrature and rectification   n  9.9 EXAMPLE 9.15

Find the area outside the circle r = 2a cos θ and inside the cardioid r = a (1 + cos θ ). Solution. From the given equations, we have 2a cos q = a (1 + cos q). which yields cos θ = 1 , that is, θ = 0 . Since 2a cos θ ≤ a (1 + cos θ ), therefore the circle lies entirely within the cardioid. Thus, Required area = Area of the cardioide – Area of the circle π 1 = 2∫ r 2 dθ − π a 2 .. 2 0 3 = π a2 − π a2 2 1 = π a 2 .. 2

2



(ii) Parametric form: If the equation of the curve is given in the parametric form x = f(t), y = φ (t ) . Then the arc length s is a function of t and 2

ds  dx   dy  =   +     dt  st dt 2

t

2

ds  dy  = 1 +   dx.  dx  dx

a

a

ds  ds  = r2 +    dθ  dθ 2

 dr  ds= r 2 +   dθ .  dθ 

θ2



s=

∫

2

 dr  r 2 +   dθ .  dθ 

r2

2

 dy  [ s ] ba = ∫ 1 +   dx  dx  a

(4)

In case the curve is given in the form θ = f (r ). then (1)

Hence, the length s of the arc between x = a and x = b is given by the value of the integral on the right hand side of the formula (1). Similarly, the length of the arc of the curve x = f(y) between y = a and y = b is equal to

M09_Baburam_ISBN _C09.indd 9

2

or

θ1

 dy  1 +   dx  dx  b

2

(iii) Polar form: If the equation of the curve is given in the polar form r = f (θ ) and we measure the arc length s in the direction of θ increasing, then

2

or

2

2  dx   dy  s = ∫   +   dt.. (3)  dt   dt  t1

Integrating between the limits θ1 and θ 2 , we have

Integrating between the limits a and b, we have ds

2

 dx   dy  ds =   +   dt..  dt   dt 



(i) Cartesian form: Let s denote the length of the arc of a curve y = f ( x) between any two points. Then s is a function of x and from a result from Differential calculus, we have

∫ dx dx = ∫

2

or

9.2.1  Length of a Curve

b

(2)

Integrating between the limits t1 and t2 m we have

9.2  RECTIFICATION The process of finding the length of an arc of a curve between two given points is called rectification.

b

b  dx  [ s ] ba = ∫ 1 +   dy..  dy  a



2

 dθ  s = ∫ 1 +  r  dr..  dr  r

(5)

1

(iv) Pedal form: If p = f (r ) is equation of the curve and r1 and r2 be the values of r at two given points of the curve, then ds = dr

r r − p2 2

12/9/2011 12:40:31 PM

9.10   n  chapter nine or

2a

1 1 y y y 2 + 4a 2 + 2a 2 sin −1  r 2a  2 2a  0 ds = sr.. r 2 − p2 2a 1 1 y  Integrating between the limits r1 and r2, we get = y y 2 + 4a 2 + 2a 2 sin −1  2a  2 2a  0    r2 r 1  s=∫ dr.. 2 2 a 8a − 2a sinh −1 1 − 0 =  r 2 − p2 2a  r1 =

EXAMPLE 9.16

Find the length of arc of the parabola y 2 = 4ax: (i) from the vertex to an extremity of the latus rectum (ii) cut off by the latus rectum. Solution. Let O be the vertex and L be an extremity of the latus rectum (line through the focus and perpendicular to the x-axis). The co-ordinates of the extremities of latus rectum of the parabola y 2 = 4ax are (a, 2a ) and (a, −2a ) .



= a  2 + sin h −11



= a  2 + log (1 + 1 + 1) 

= a  2 + log (1 + 2)  . (ii) Length of the arc cut off by the latus rectum is equal to 2 × arc OL = 2a [ 2 + log (1 + 2).



EXAMPLE 9.17

Find the perimeter of the circle x1 + y1 = a1. Solution. The center of the given circle is (0,0) and its radius is a. It is symmetrical about both axes. Therefore, Perimeter = 4 × Arc in the first quadrant from the point (0, 0) to the point (a, 0). Differentiating the given equation, we get 2x + 2 y

dy =0 dx

or dy x x =− =− . 2 dx y a − x2

Hence, Then, differentiating the given equation, we dy dx y have 2 y = 4a, that is, = .Therefore, dx dy 2a 2a

(i) Length OL =

∫ 0

2

a

 dy  Perimeter = 4∫ 1 +   dx  dx  0 a

2

 dx  1 +   dy  dy 

= 4∫ 1 +

y2 1 + 2 dy 4a

= 4a ∫

0

a

2a



M09_Baburam_ISBN _C09.indd 10

=

∫ 0

1 = 2a

0

dx a − x2 2

a

2a

∫

0

x2 dx a − x2 2

y + 4a dy 2

2

  x  = 4a sin −1     a  0 

12/9/2011 12:40:32 PM

quadrature and rectification   n  9.11 = 4a [sin −11 − sin −1 0]

π  = 4a  − 0 = 2π a. 2  EXAMPLE 9.18

Find the length of the loop of the curve 3ay2 = x(x – a)2. Solution. The given curve is symmetrical about the x-axis. It cuts the x-axis at (0, 0) and (a, 0). Thus, the loop varies from x = 0 to x = a. The given equation is 3ay2 = x(x – a)2. Taking logarithm, we get log 3a + 2 log y = log x + 2 log ( x − a ).

1  32  4a . 4a = .  3a  3

= EXAMPLE 9.19

2

2

2

Show that the length of the asteroid x 3 + y 3 = a 3 is 6a. Solution. The shape of the curve is

Differentiating with respect to x, we get 2 dy 1 2 3x − a = + = y dx x x − a x( x − a )

or

dy 3x − a y . = ⋅ . dx x( x − a ) 2

Thus,

Differentiating the given equation with respect to x, we get

2

(3 x − a ) 2 y 2  dy  .   = 2 dx x ( x − a)2 4 (3 x − a ) 2 x( x − a ) 2 . 12a x 2 ( x − a)2

=

=

(3 x − a ) 2 . . 12ax

Therefore, Total length of the loop = 2 × length of the loop in first quadrant  dy  = 2∫ 1 +   dx  dx  0 a

= 2∫ 0

=

1

3x + a 12ax a

∫ (3 3a

dy  y 3 = −   ..  x dx

Then, Required length = 4 × length of curve in first quadrant a

0

2

 dy  1 +   dx  dx  1

2  y3  = 4 ∫ 1 + 2  dx x3  0 2

a

dx

a

1 2

x + ax ) dx

0

= 4∫ 0

a

1  1  x 32 2 = 3 3 + ax  3a  2 0

M09_Baburam_ISBN _C09.indd 11

1

or

= 4∫

2

a

2 − 13 2 13 dy x + y =0 3 3 dx

(x

2 3

2

+ y3 x

1 3

) dx

1

(a) 3  2 = 4∫  1  dx, using the given equation x3 0 2

a

12/9/2011 12:40:33 PM

9.12   n  chapter nine a

a

= 4a 3 ∫ x 3 dx 1

= 2a ∫

−1

0

2(a 2 − x 2 ) + a 2 a2 − x2

0

a

dx

1 3 2  = 4a 3  x 3  2  0

=

a  2  a2 2 2 2 a − x + ∫   dx 2 2 a 0 a −x 

1 3 2  = 4a 3  a 3  = 6a.. 2  

=

2  2 a2 x −1 x  2 + a 2 sin −1   2  xa − x 2 + sin  a   2 a a 0

a

a

2  2 a2 x x 2 = − + 2 xa x 2 sin −1  + a 2 sin −1    a   2 a a0 Show that the whole length of the curve x 2 (a 2 − x 2 ) = 8a 2 y 2 is π a 2 . a 2 2 2 2 −1 x  x ( a x ) 2 a sin = − + Solution. The shape of the curve is a  a  0 EXAMPLE 9.20

=

2 [2a 2 sin −11] a

=

2 a

 2 π  2a . 2 

= π a 2. EXAMPLE 9.21

The whole curve lies between the lines x = – a and x = a. The given equation yields 1

y=

8a

Rectify the cycloid x = a (t + sin t ) , y = a (1 + cos t ) . Solution. The shape of the curve is

x a 2 − x 2 ..

Therefore, dy 1  a2 − 2x2  =   dx 8a  a 2 − x 2  Then, Whole length of the curve = 4 × length in the first quadran 2

a

 dy  = 4∫ 1 +   dx  dx  0 a

= 4∫ 1 + 0

=

42

a

∫ 2a 0

M09_Baburam_ISBN _C09.indd 12

(a 2 − 2 x 2 ) 2 dx 8a 2 ( a 2 − x 2 )

3a 2 − 2 x 2 a2 − x2

Further, dx dy = a (1 + cos t ) and = − a sin t.. dt dt Therefore, 2

dx

2

 dx   dy  2 2 2 2   +   = a (1 + cos t ) + a sin t dt dt

12/9/2011 12:40:34 PM

quadrature and rectification   n  9.13

Note: If we find the length of the upper half from 0 to π3 . then

= 2a 2 (1 + cos t )

t = 4a 2 cos 2 − . 2

π

Therefore, π

2

 π  4a = 4a sin  = 6 2 

2

 dx   dy  Length of cycloid = 2∫   +   dt  dt   dt  0 π

t 4a cos dt 2

= 2∫

2

0

2

π

t = 4a ∫ cos dt 2 0 π

 sin t  = 4a  1 2   2 0

 π = 8a sin  = 8a..  2 EXAMPLE 9.22

Find the whole length of the cardioid r = a (1 + cos θ ) Solution. For the shape of the curve, please see Example 9.11. Also, dr = − a sin θ . dθ Therefore, Length of whole cardioid = 2 × length of the upper half π

2

 dr  r +   dθ  dθ 

= 2∫

= 2a = half the length of upper point. Thus, θ = π3 bisects the arc of upper half of the cardioid. EXAMPLE 9.23

Find the perimeter of the cardioid r = a (1– cos q) and show that the arc of the upper half is bisected by θ = 23π .. Solution. For the shape of the curve, please see Example 9.12. Due to symmetry, the perimeter of the given cardioid is twice the length of the upper half of the cardioid. We have dr = a sin θ dθ

and so 2

 dr  r 2 +   = a 2 (1 − cos θ ) 2 + a 2 sin 2θ  dθ 

θ = 4a 2 sin 2 .. 2 Therefore, π

Required perimeter = 2∫ 4a 2 sin 2 0

2

0

π

= 2∫ a 2 (1 + cos θ ) 2 + a 2 sin 2θ dθ π

0

0

2

dθ

π

 sin θ  π = 4a  1 2  = 8a sin = 8a.. 2  2 0

M09_Baburam_ISBN _C09.indd 13

dθ

θ

π

= 2a ∫ 2(1 + cos θ ) dθ = 2a ∫ 2 cos

2

 − cos θ  = 4a  1 2   2 0

0

θ

θ

= 4a ∫ sin dθ 2 0

π

π

π

 sin θ  3 Length = ∫ 2a cos dθ = 2a  1 2  2  2 0 0

θ

3

Thus,

= 8a.

Length of the upper half = 4a. On the other hand, the arc length of the cardioid from the point θ = 0 to the point θ = 23π is equal to

12/9/2011 12:40:34 PM

9.14   n  chapter nine 2π 3

2. Show that the area of a loop of the curve π a2 a 4 y 2 = x 4 (a 2 − x 2 ) is . 8 3. Find the area bounded by the curve xy 2 = 4a 2 (2a − x) and its asymptote.

2π

 − cos θ  3 2a ∫ sin dθ = 2a  1 2  2  2 0 0

θ

π   = −4a cos − cos θ  3  

Ans. 4π a 2 .

1  = −4a  − 1 2  = 2a. Hence, θ = 23π bisects the arc length of the upper half of the cardioid. EXAMPLE 9.24

Find the length of the arc of the equiangular spiral r = aeθ cot α between the points for which radii vectors are r1 and r2. Solution. Differentiating the given equation, we have dr = aeθ cot α . cot α = r cot α dθ or dθ 1 dθ or r = = tan α .. dr r cot α dr Therefore,

4. Find the area included between y 2 = 4ax and y = mx. 8a 2 . Ans. . 3m3 5. Show that the larger of the two areas 2 2 2 into which the circle x + y = 64a is 2 divided by the parabola y = 12ax is 16 2 a (8π − 3). 3 6. Show that the area bounded by the Cissoid a sin 3t and its asymptote is x = a sin 2 t , y = cos t 2 3π a . . 4 7. Find the area enclosed by the spiral 1

r2

rθ 2 = a between θ = α to θ = β .

2

 dθ  Required length = ∫ 1 +  r  dr  dr  r

Ans.

1

r2

1 2  β a log   .. α 2

8. Find the area of the loop of the curve r =

= ∫ 1 + tan 2α dr

aθ cos θ between θ = 0 and θ =

r1

r2

= sec α ∫ dr = sec α [r ] rr21 r1

= sec α [r2 − r1 ]..

π a2

(π 2 − 6). 96 9. Find the whole area of the curve r = sin 2q. Ans.

EXERCISES

1 2 πa . 2

10. Find the area of the curve r 2 = a 2 cos 2θ .

Quadrature 1. Find the area of a loop of the curve x 4 + y 4 = 4a 2 xy . Ans.

M09_Baburam_ISBN _C09.indd 14

Ans.

π . 2

πa . . 2

2

Ans. a2. 11. Find the area common to the circles r = a 2 and r = 2a cos θ. Ans. a 2 (π − 1).

12/9/2011 12:40:35 PM

12. Find the area of the portion included between the cardioids r = a (1 + cos θ ) and r = a (1 − cos θ ).  3π  Ans. 2a 2  − 2 ..  4  Rectification 13. Find the perimeter of the loop of the curve 3ay 2 = x 2 (a − x).

4a . . 3 14. Show that the length of the arc of the 2 parabola y = 4ax intercepted between the points of intersection of the parabola and 15   the straight line 3 y = 8 x is a  log 2 +  ..  16  Ans.

15. Find the whole length of the curve x = acos3t , y = bsin 3t. b 2 + ab + a 2 . Ans. 4 . b+a

M09_Baburam_ISBN _C09.indd 15

quadrature and rectification  n  9.15

16. Find the length of the arc of the curve x = t et sin t , y = e cos t between t = 0 and π t= . 2 π Ans. 2 e 2 − 1 ..   17. Find the length of the cardioid r = a (1 − cos θ ) lying outside the circle r = a cos θ . Ans. 4a 3. 18. Show that the perimeter of the circle r = a  b2  cos θ is approximately 2π a 1 + 2 ..  4a  19. Find the length of a loop of the curve r 3 = sin 3θ . 1

Ans. 2

∫ 0

dr 1 − r6

..

12/9/2011 12:40:36 PM

This page is intentionally left blank.

M09_Baburam_ISBN _C09.indd 16

12/9/2011 12:40:36 PM

10

Centre of Gravity and Moment of Inertia and

10.1  CENTRE OF GRAVITY A point fixed in the body through which the weight of the body passes in whatever position the body may be placed is called the centre of gravity of that body. (1) Centre of Gravity of a System of Particles: Consider a system of particles of masses m1, m2, .., mn situated at the points ( x1 , y1 ), ( x2 , y2 ) , …, ( xn yn ) respectively. Then the centre of gravity (C.G) of this system is the point ( x , y ), where nn

∑ ∑mm xx

xx ==

ii==11 nn

ii ii

nn

,,, yy ==

∑ ∑mm yy ii==11 nn

ii ii

(1)

∑ ∑mmii ∑ ∑mmii ii==11 ii==11 Due to relation (1), the centre of gravity is also known as centre of mass or Centre of Inertia

(2) Centre of Gravity of a Curve: Let δ s be a small portion of the curve between the points (x, y) and ( x + δ x, y + δ y ). Let ρ be the mass per unit length of the matter of the curve y = f (x). Then mass of the small portion is δ m = ρδ s . Therefore ds yy..ρρds ds ∫ xx..ρρds ,,, yy == ∫ ds ds ∫ ρρds ∫ ρρds

xx == where

(2)

in case of Cartesian 2  dy  ds = 1 +   in caseco-ordinates. of Cartesian  dx  in case of parametric  dx   dy  co-ordinates. ds =   +    dt   dt  2

M10_Baburam_ISBN _C10.indd 1

2

 dr  ds = r +    dθ 

2

2

in case of polar co-ordinates.

When ρ is constant, then (2) reduces to xdA ydA ydA ∫∫xdA ,, ,yy==∫∫ . .. dA dA dA dA ∫∫ ∫∫

xx==

(3) Centre of Gravity of an Area: If δ A is an elementary area an ρ is the mass per unit area, then mass of the elementary area is δ m = ρδ A. Therefore, the centre of gravity is given by x=

∫ xρdA ∫ ρdA

and y ==

∫ yρdA .. ∫ ρdA

For uniform area, ρ is constant and so in that case, ∫ xdA , y = ∫ ydA .. x= ∫ dA ∫ dA

Special Cases (i) When the area is bounded by the curve y = f ( x) , x-axis and the two ordinates x = a and x = b, then the area of the elementary strip of width δ x is δ A = yδ x. The centre y of gravity of the elementary strip is x, 2 . Therefore centre of gravity of the area is given by

( )

b

x=

∫xydx a b

∫ y dx a

b

and y =

y

∫2 a

b

..

∫ y dx a

12/8/2011 1:09:22 PM

10.2  n  chapter ten (ii) When the area is bounded by the curve x = f (y), y-axis, and the ordinates y = a and y = b, then the area of the elementary strip of width δ y is δ A = xδ y. Therefore centre of gravity of the area is given by b

x=

b

∫

x 2

xdy

1 2

=

a

b

∫ xdy

2

a b

∫ xdy

a

y=

∫ x dy

∫r

2 x= α 3

cos θ dθ

β

y=

3

,,

∫α r dθ 3

sin θ dθ

β

..

∫α r dθ 2

∫ xρ dV = ∫ x dV ∫ ρ dV ∫ dV

∫ yρdV = ∫ ydV .. y= ∫ ρdV ∫ dV In case of the volume of revolution about x-axis, the volume of the circular disc of thickness

M10_Baburam_ISBN _C10.indd 2

2

2

(5) Centre of Gravity of a Surface of Revolution: For surface of revolution about the x-axis, the centre of gravity is given by x=

∫ x ⋅ 2π yds = ∫ xyds ,, ∫ 2π yds ∫ yds

y = 0.

For surface of revolution about y-axis, we have

∫ xyds .. ∫ xds

EXAMPLE 10.1

Find the centre of gravity of the arc of the astroid 2 2 2 x 3 + y 3 = a 3 in the first quadrant

(4) Centre of Gravity of Volume of Revolution: Let δV be the element of volume of body with uniform density ρ . Then mass of this element isδ m = ρdV . Therefore the centre of gravity ( x , y ) is given by x=

∫ yπ x dy ∫ π x dy

x = 0 and y =

2

β

∫r 2α

In case of the volume of revolution about y-axis, the centre of gravity is x = 0, y =

(iii) When the area is sectorial area bounded by a curve r = f (θ ) and two radii vectors θ = α and θ = β . Then the area of the elementary sartorial area is δ A = 12 r 2δθ. The centre of gravity of this elementary sectorial area is ( 32 (r32 cos θ , θ32 ,r32 sin θ ).θ ). r cos r sin Therefore centre of gravity of the sartorial area is given by 3

gravity lies on the x-axis. Therefore the centre of gravity ( x , y ) 2 ∫ x ⋅ π y dx , y = 0.. x= 2 ∫ π y dx

a

∫ yxdy .. ∫ xdy

β

δ x and radii y is δV = π y 2δ x. The centre of

Solution. The parametric equations of the curve are x = a cos3θ , y = a sin 3θ. Therefore dx = −3a cos 2θ sin θ , dθ dy = 3a sin 2θ cos θ , dθ 2

2

ds  dx   dy  =   +  .  dθ   dθ  sθ Therefore the centre of gravity is π

π

∫ xds

∫x

2

x=

0

π

2

∫ ds 0

2

=

ds dθ

dθ

0

π

2

∫

ds dθ

⋅ dθ

0

12/8/2011 1:09:23 PM

centre of gravity and MoMent of inertia  n 10.3 π

= a 2 (1 + cos θ ) 2 + a 2 sin 2θ

2

=a

4 ∫ cos θ sin θ dθ 0

π

2

=

∫ cos θ sin θ dθ

   and similarly,

2a , , 5

θ = 2a cos .. 2 Therefore

0

π

2a , y= , 5

x=

∫ xds ∫ ds

EXAMPLE 10.2

y2

2

the ellipse ax 2 + b2 = 1 in the first quadrant Solution. From the equation of the ellipse, we have b y= a 2 − x 2 .. a Therefore the centre of gravity of the ellipse in the first quadrant is given b a

x=

∫ xyds 0 a

=

∫ ydx 0

a 2 − x 2 dx

b a

=

0

π

2

∫

b a

⋅ a 2 − x 2 dx

y=

∫ y. ydx 0 a

=

∫ ydx

∫

b a

a 2 − x 2 dx =

0

a

b a

0

∫

b a

⋅ a 2 − x 2 dx

4b . . 3π

0

Hence

 4a 4b  ( x , y ) =  ,  ..  3π 3π  EXAMPLE 10.3

Find the centre of gravity of the arc of the cardioid r = a (1 + cos θ ). Solution. We have r = a + a cos θ , dr = − a sin θ , dθ

ds  dr  = r2 +    dθ  dθ

M10_Baburam_ISBN _C10.indd 3

∫ 2a cos θ dθ 2

π

=

a ∫ (1 + cos θ ) cos θ cos θ2 dθ 0

2 π

=

θ  a  2 θ  2 θ  ∫ 2cos  cos − 1 cos dθ  2 0 2 2 2 

=

4a (using the substitution θ2 = φ ). 5

Solution. Hemi-sphere is obtained by revolving the circle x 2 + y 2 = a 2 about the x-axis. Thus

a

b2 2 a2

π

Find the centre of gravity of a uniform hemispherical shell.

4a . . 3π

0

a

1 2

0

EXAMPLE 10.4

a

∫x

2

=

0

Find the centre of gravity of a uniform lamina bounded by the co-ordinate axes and the arc of

∫ r cos θ (2a cos θ ) dθ

dy x =− dx y 22

ds ds dy xx22 ++ yy22  dy == 11++   ,,,==  dx dx dx dx yy22

=

a. . y

Hence the centre of gravity ( x , y ) is given by a

x=

∫ x ⋅ 2π yds 0

∫ 2π yds a

2

=

∫ xy

ds dx dx

0 a

∫y

ds dx

dx

0

12/8/2011 1:09:24 PM

10.4  n  chapter ten a

=

∫ ax dx 0 a

a

=

∫ a dx

∫ x dx

Hence, Moment of inertia about OY = lim

0

a

∫ dx

=

0

2

∫ ρ x dx 2

−a

a

=

∑ ρx δ x

a

 x2  2  0

=ρ

a = .. 2

[ x] 0a

∫ x dx 2

−a

=

10.2  MOMENT OF INERTIA Let a particle of mass m be at a distance r from a given line. Then mr2 is called the moment of inertia (M.I) of that particle about the given line.

2 3 ρa 3

The moment of inertia of a system of particles of masses m1, m2,… mn placed at distances r1, r2, …, rn is given by

n

∑m r

2 i 1

Figure 10.1

.

i =1

Further, if m is the mass of a body and IG is the moment of inertia of the body about an axis through the centre of gravity of the body, then moment of inertia I of that body about an axis parallel to the axis through centre of gravity and at a distance d from centre of gravity is given by I = IG + md 2.

Similarly, if OX and OY are perpendicular axes in a plane and Ix and Iy are moment of inertia of a plane area about OX and OY, then moment of inertia of that plane about an axis perpendicular to both OX and OY is given by I=I +I . x

y

EXAMPLE 10.5

Find the moment of inertia of a thin uniform rod of length 2a about a perpendicular axis through its centre of gravity and about an perpendicular axis on the extremities of the rod.

Further, Moment of inertia about OY’ through the extremity A is given by 1 4 I = I G + Md 2 = Ma 2 + Ma 2 = Ma 2 .. 3 3 EXAMPLE 10.6

Find the moment of inertia of a circular disc about an axis perpendicular to the disc at its centre and about a tangent. Solution. Let ρ be the mass per unit area of the circular disc of radius a. Thus mass of the disc 2 is M = π a ρ. Let OY be the axis perpendicular to the plane of the disc through the centre O (see Fig. 10.2). Then mass of the elementary ring of radius x and thickness δ x is 2ρπ xδ x. Therefore, moment of inertia of this elementary ring about OY is 2π xδ xρ ⋅ x 2..

Solution. Let M be the mass of the rod AB of length 2a and let O be the centre of AB. We first find moment of inertia about the line OY perpendicular to AB (see Fig. 10.1). If ρ is the mass per unit length of the rod then the mass of an element length δ x is ρδ x. Therefore 2 the M.I of the element about OY is ρδ x.x .

M10_Baburam_ISBN _C10.indd 4

Figure 10.2

12/8/2011 1:09:24 PM

centre of gravity and MoMent of inertia  n 10.5 Hence moment of inertia of the disc about OY is a

I = ∫ 2π x 3 ρdx a

1 (π ρ a 4 ) 2 1 = (π a 2 ρ). a 2 2 1 = Ma 2 .. 2 Then moment of inertia about a diameter is 12 (moment of inertia about the perpendicular to the 2 disc) = Ma4 .. Then using the concept of parallel axis, the moment of inertia of the circular disc 2 2 . about a tangent to the disc is Ma4 + Ma 2 = 5 Ma 4 . =

EXAMPLE 10.8

Find the moment of inertia of a solid right circular cone about it axis. Solution. Consider the right circular cone with vertex O, height h, radius r as shown in Figure 10.4. Its mass is 13 π r 2 hρ.

EXAMPLE 10.7

Find moment of inertia of a uniform rectangular lamina about axis through the centre.

Figure 10.4

Let y be the radius of an elementary disc perpendicular to the x-axis at a distance x from O and of width δ x. Then y x rx = or y = .. r h h

Figure 10.3

2 Mass of this strip is π y 6 xρ2 . Its moment of yy22 inertia about the x-axis is π y 22δ xρ.. 22 .. Therefore moment of inertia of the solid right circular cone is

Solution. Consider the rectangular lamina ABCD with AB = 2a and CD = 2b (see Fig. 10.3). If ρ is mass per unit area, then mass of the lamina is 4abρ. The mass of the elementary strip at a distance x from the centre of the lamina and perpendicular to OX is 2bρδ x. Therefore its 2 moment of inertia is 2bρδ x. b3 . Hence the moment of inertia of the lamina about OX is a

2 4 Mb 2 . b3 ρdx = ab3 ρ = . ∫ 3 −a 3 3

Similarly, taking strip perpendicular to OY, the moment of inertia of the lamina about OY is 2 also Mb3 ..

M10_Baburam_ISBN _C10.indd 5

hh

22 ∫ ππ yy 66xxρρ... 00

44

yy22 ρπ ρπ  rx rx dx dx == dx   dx 22 22 ∫00  hh  bb

=

πρhr 4 b 4 x dx 2π ∫0

=

πρhr 4 3Mr 2 . . = 10 10

10.3  MEAN VALUES OF A FUNCTION Let f (x) be a continuous function defined on (a, b). Then the mean value off in (a, b) is defined b

by

∫ f ( x ) dx a

b−a

..

12/8/2011 1:09:25 PM

10.6  n  chapter ten Thus mean value of f is equal to

EXERCISES 1. Find the centre of gravity of a uniform solid cone of height h.

area enclosed by f from a to b . . length of the interval (a, b )

h

Further, Mean Square Value of f is (a, b) is b

defined by

1 b−a

∫ [ f ( x)]

2

dx and root mean

a

square value of f in (a, b) is defined by b

1 b−a

∫ [ f ( x)]

2

dx..

a

EXAMPLE 10.9

Find the mean value of sin2wt from t = 0 to t = 2wπ .. Solution. The required mean value is 22ππ

1 ww 2 w sin wt dt = ∫ 2π 0 2 π − w 0

22ππ w w

1

∫ 2 (1 − cos 2wt )dt 0

w  w w  ∫ dt − ∫ cos 2 wt  dt = 4π  0  0  2π

2π

2π

w  2π 1 w ( sin 2 wt )  = −  4π  w 2 w 0 =

M10_Baburam_ISBN _C10.indd 6

w  2π  1 . = . 4π  w  2



Hint: x =

∫ π xy

2

0 h

∫π y

dx

Ans.

2

3 h 4

0

2. Find the C.G. of the area of a loop of the curve r = sin 2θ in the first quadrant  128 , 128  , Ans. ( x , y ) =   105π 105π  3. Find the C.G. of a uniform solid hemisphere 3a ,y =0 Ans. x = 8 4. Find the moment of inertia of a solid right circular cone about the diameter of its base M Ans. (3r 2 + 2h 2 ) 20 5. Show that the moment of inertia of a hollow right circular cone about its axis is 12 Mr 2,, where r is the base radius and M is the mass of the cone. 6. Find the mean value of kx(l – x) in the interval (0, l). k Ans. (3l − 2) 6 7. Find the root mean square value of a function e–kt sin wt in 0, πw  . Ans.

(

w2 1 − e

−2 k πws

).

4k π ( w + k ) 2

2

12/8/2011 1:09:26 PM

11

Volumes and Surfaces of Solids of Revolution

The body generated by the revolution of a plane area, about a fixed line lying in its own plane, is called a solid of revolution. On the other hand, the surface generated by the boundary of the plane area is called the surface of revolution. The fixed line, lying in the plane of plane area, about which the plane area revolves is called the axis of revolution. The section of a solid of revolution by a plane, perpendicular to the axis of revolution, is a circle having its center on the axis of revolution.

We further observe that the volume of the solid generated by the revolution of the area PMNQP lies between the volumes of the right circular cylinder generated by the revolution of the areas PMNR and PMNQ. Thus, δ V lies between π y 2 dx and π ( y + δ y ) 2 δ x . y B

Q

S

11.1  VOLUME OF THE SOLID OF REVOLUTION (CARTESIAN EQUATIONS) (A) Revolution About x-axis: Let AB be the arc of the curve y = f (x) lying between the ordinates x = a and x = b. We assume that f is continuous in (a, b) and that it does not meet the x-axis. Let P(xy) and Q(x + δx, y + δy) be any two neighboring points on the curve y = f (x). Draw PM ⊥ OX and QN ⊥ OX Draw PR ⊥ QN and SQ ⊥ PM. Let V be the volume of the solid generated by the revolution about the x-axis of the area ACMP. As x increases, MP moves toward the right and the volume increases. Let the volume of revolution obtained by revolving the area ACNQ about x-axis be V + δ V so that the volume of the solid generated by the revolution of the strip PMNQ about the x-axis is δ V . We now have PM = y and QN = y + δ y and MN = x + δ x − x = δ x . Then, the volume of the solid generated by revolving the area PMNR is π y 2δ x and the volume of the solid generated by revolving the area PMNQ is π ( y + δ y ) 2 δ x .

M11_Baburam_ISBN _C11.indd 1

P

A

0

Therefore, and so,

C δV ∂x

R

M

N

D

x

lies between π y and π ( y + δ y ) 2

π y2
0) about the axis of y. If the depth of the basin is 8 cm, how many cu. cm of water will it hold? Solution. The given curve is x3 = 64y. The curve passes through the origin and is symmetrical in the opposite quadrants. Since the depth is 8 cm, that is, y = 8, we have x = 8. Thus, the point A (8, 8) is at a height of 8 cm. The basin is formed by the revolution of the arc OA about the y-axis, where A is (8, 8). y A(8, 8)

A P(x, y

0

α

M

8cm

)

a

h

C

x

0

x

Therefore, the required volume is given by 8

B

 In DOMP , we have tan α =

M11_Baburam_ISBN _C11.indd 4

PM OM

=

y x

and so,

8

∫π x dy = ∫π (64 y) 2

0

0

2 3

8

dy = 16π ∫ y 3 dy 2

0

12/14/2011 3:48:10 PM

voluMeS and SurfaceS of SolidS of revolution  n 11.5 π

 y3  48π 53 = 16π  5  = (8) 5  3  0 48π 1536π (32) = = cu cm. 5 5 5

π

2 2 6 a6 2a Required = 2πvolume θ dθ4 ⋅ a sec 2θ dθ ∫0 a 4sec4=θ 2⋅πa ∫seca 4sec θ 0

8

π

= 2π a

EXAMPLE 11.7

Show that the volume of the solid generated by the revolution of the curve (a − x) y 2 = a 2 x 1 about its asymptote is π 2 a 3 . 2 Solution. The given curve (a − x) y 2 = a 2 x is symmetrical about the x-axis, passes through the origin, and its asymptotes parallel to the y-axis is a – x = 0 or x = a.

π

2 1 π 1 1 π 1 cos θ d θ π a 32θ⋅ dθ⋅ = =2π aπ3 2⋅a 3 ⋅. = π 2 a 3 . = 2 π a = ∫0 ∫02cos 2 2 2 2 2 2 2

3

2

3

EXAMPLE 11.8

Find the volume of the reel-shaped solid formed by the revolution about the y-axis, of the part of the parabola y2 = 4ax, cut off by the latus rectum. Solution. Let O be the vertex and L one extremity of the latus rectum of the given parabola y2 = 4ax. For the arc OL, y varies from 0 to 2a. y

x=a

y

L(a, 2a)

P

X

0

M a–x

N

a

A

Let P (x, y) be any point on the curve and let PM be the perpendicular on the asymptote. Then, PM = AN = a – x. Also AM = y. The point A is (a, 0). Then, the required volume is given by ∞

2∫ π ( PM ) 2 d ( AM ) = 2∫ π (a − x) 2 dy 0

2

∞

 ay 2  = 2π ∫  a − 2 dy, a + y 2  0 using equation of the curve ∞

2

∞

 a  a = 2π ∫  2 dy = 2π ∫ 2 dy. 2 a +b  (a + y 2 ) 2 0 0 3

6

Put y = a tan θ so that dy = asec2 q dq. Therefore,

M11_Baburam_ISBN _C11.indd 5

0

x

x

Therefore, Required volume = 2 times the volume generated by revolution about the y-axis of the area OLC: 2

2a 2a  y2  2 = 22∫a π x 22 dy = 2∫a π  y 2  dy = 2 ∫0 π x dy = ∫0 π  4a  dy  4a  0 0 π 22 aa 4 π 32a 55 4π a 33 = π 2 ∫ y 4 dy = π 2 . 32a = 4π a . = 8a 2 ∫0 y dy = 8a 2 . 5 = 5 . 5 5 8a 0 8a

EXAMPLE 11.9

Find the volume of the solid generated by the revolution of the plane area bounded by y2 = 9x and y = 3x about the x-axis. Solution. The parabola y2 = 9x is symmetrical about the x-axis. The line y = 3x cuts the

12/14/2011 3:48:11 PM

11.6  n  chapter eleven parabola at P(1, 3). Therefore, the required volume is given by y

2

y

x

The given curve is symmetrical about both the axes and the asymptote to the curve is y = 0, that is, x-axis. For the portion of the curve lying in the second quadrant, y varies from a to 0, t varies from π2 to 0, and x varies from 0 to −∞ . Thus,

A(1,3)

=9

y=

3x x

0

1 x = a cos t + a log tan 2 2t and y = a sin t. 2

0

Required volume = 2 ∫ π y 2 dx −∞ π

2

= 2 ∫ π y2 −∞

1

y

1

∫π ( y

− y22 )dx = π ∫ (9 x − 9 x 2 )dx

2 1

0

(0, a) t

0

= π[

9x 9x 1 9 3π . ]0 = π ( − 3) = − 2 3 2 2 2

dx ⋅ dt. dt

=

π 2

3

0

x

11.2  VOLUME OF THE SOLID OF REVOLUTION (PARAMETRIC EQUATIONS) Let the curve be given by the parametric equations x = φ (t) and y = ψ (t ) . Then, the volume of the solid generated by revolution about x-axis of the area bounded by the curve, the axis of x, and the ordinates at the points where t = a and t = b is given by b

2 ∫π y a

b

dx dt = π ∫[ψ (t )]2 .φ ' (t )dt . dt a

Similarly, the volume of the solid generated by the revolution about the y-axis of the area between the curve x = φ (t ) and y = ψ (t ) , the y-axis, and the abscissae at the points t = a and t = b is given by b

∫π x

b

2

a

dy dt = π ∫[φ (t )]2 .ψ ' (t )dt . dt a

EXAMPLE 11.10

Find the volume of the solid generated by the 1 revolution of the tractrix x = a cos t + a 2 t log tan 2 and y = a sin t about its asymptote. 2 Solution. The equations of the tractrix are

M11_Baburam_ISBN _C11.indd 6

But,

dx a t t 1 1 = −a sin t + . .2 tan sec 2 . dt 2 tan 2 t 2 2 2 2 a = −a sin t + t t 2 sin cos 2 2 a = − a sin t + t sin 2 a (1 − sin t ) = sin t

= a

cos 2 t . sin t

Therefore, Required volume π

2

= 2π ∫ a 2 sin 2 t. 0

a cos 2 t dt sin t

π

2

= 2π a 3 ∫ cos 2 t sin tdt = 2π a 3 . 0

1 2 = π a3 . 3.1 3

12/14/2011 3:48:12 PM

voluMeS and SurfaceS of SolidS of revolution  n 11.7 EXAMPLE 11.11

Find the volume of the solid formed by revolving the cycloid x = a (θ − sin θ ) and y = a (1 − cos θ ) about its base.

The curve is symmetrical about both the axes. It cuts the axes at t = 0 and t = π2 . For the portion of the curve in the first quadrant, t varies from 0 to π2 . Therefore, y

Solution. The given equations of the cycloid are x = a (θ − sin θ ) and y = a (1 − cos θ ) . The cycloid is symmetrical about the line through the point, where θ = π , perpendicular to x-axis. For the first half, OBH of the cycloid, θ varies from 0 to π .

t= π 2

y

0

x

t =0

θ=π

B

π

π

2 dx dx Required volume = 2∫ π y dt = 2π ∫ y 2 dt dt dt 0 0 π 2

2

θ = 0

0

H

θ = 2π

x

2

= 2π ∫ a 2 sin 6 t.( −3a cos 2 t sin t )dt 0

π

The required volume is given by π

π

= 2π ∫ a 2 (1 − cos θ ) 2 .a (1 − cos θ )dθ 0

2

θ  θ  = 2π a 2 ∫  2sin 2  .a  2sin 2  dθ    2 2 0 π

2

= 2π a 3 ∫ (2sin 2 t )3 2dt , 0

θ 2

=t

π

2

= 32π a 3 ∫ sin 6 tdt = 32π a 3 . 0

5.3.1 π . = 5π 2 a 3 . 6.4.2 2

EXAMPLE 11.12

Find the volume of the spindle-shaped solid 2 2 2 generated by revolving the asteroid x 3 + y 3 = a 3 about the x-axis. Solution. The parametric equations of the asteroid are x = acos3t and y = asin 3t.

M11_Baburam_ISBN _C11.indd 7

2

6.4.2.1

∫ sin t.cos tdt = −6π a . 9.7.5.3.1 7

2

3

0

−32 3 32 = πa = π a 3 (in magnitude). 105 105

dx 2∫ π y . dθ dθ 0 2

π

= −6π a

3

EXAMPLE 11.13

Find the volume of the solid formed by the revolution of an arch of the cycloid x = a (θ + sin θ ) and y = a (1 − cos θ ) about the tangent at the vertex. Solution. The cycloid x = a (θ + sin θ ) and y = a (1 − cos θ ) is symmetrical about the y-axis. Further, tangent at the vertex is x-axis. For the portion of the curve iny the first quadrant, θ varies from 0 to π . Therefore, y

0 = −π

0= π

0 = −π

0= π

0

θ=0

0

θ=0

x x

12/14/2011 3:48:13 PM

11.8  n  chapter eleven y

π

dx Required volume = 2∫π y dθ d θ 0 2

θ  π /2

π

= 2π ∫ a 2 (1 − cos θ ) 2 .a (1 + cos θ )dθ

θπ

π

θ0

0

0

x

2

θ  θ  = 2π a 3 ∫  2sin 2   2cos 2  dθ    2 2 0 = 4π a

π

3

∫ (2sin φ ) 2

2

(2cos φ ).d φ , φ = 2

0

θ 2

π

π

2 3 2π 3 3 ∫0 3 π r sin θ dθ = 3 ∫0 a (1 + cos θ ) sin θ dθ =−

π

2

= 32π a ∫ 3 sin 4φ cos 2φ d φ

π

0

3.1 π = 32π a 3 . . = π 2 a3 . 6.4.2 2 11.3  VOLUME OF THE SOLID OF REVOLUTION (POLAR CURVES) The volume of the solid generated by the revolution of the area bounded by the curve r = f (θ ) and the radii vectors θ = α and θ = β about the initial line (θ = 0) is given by β

2

∫α 3 π r

3

sin θ dθ.

Similarly, the volume of the solid generated by the revolution of the area bounded by the curve r = f (θ ) and the radii vectors θ = α and θ = β about the line θ = π is given by 2 β 2 3 r c os π θ dθ. ∫α 3

−

Solution. The cardioid r = a (1 + cos θ ) is symmetrical about the initial line. We note that r = 0 when θ = π and r = 2a when θ = 0 . When θ increases from 0 to π , r decreases from 2a to 0. Thus, the required volume is given by

M11_Baburam_ISBN _C11.indd 8

2π a 3  (1 + cos θ ) 4    3  4 0

1 8 = − π a 3 (0 − 24 ) = π a 3 . 6 3 EXAMPLE 11.15

Find the volume of the solid generated by revolving one loop of the lemniscate π r 2 = a 2 cos 2θ about the line θ = . 2 Solution. The given curve r 2 = a 2 cos 2θ is symmetrical about the initial line. Putting r = 0 in the equation of the curve, we get cos 2θ = 0 π π or θ = ± . The revolution is about θ = . 2 4 y

 π /4  0

(–a, 0)

EXAMPLE 11.14

The cardioid r = a (1 + cos θ ) revolves about the initial line. Find the volume of the solid thus formed.

π

2π a 3 (1 + cos θ )3 ( − sin θ )dθ 3 ∫0

0

x

(a, 0)

Therefore, the required volume is given by π 4

2 2∫ π r 3 cos θ dθ 3 0 π

3 4π 4 3 a ( cos 2θ ) 2 cos θ dθ = ∫ 3 0

12/14/2011 3:48:14 PM

voluMeS and SurfaceS of SolidS of revolution   n 11.9 π

π

3 4π a 3 4 (1 − 2sin 2θ ) 2 cos θ dθ = ∫ 3 0

2π 2 = ∫ {2a(1 + cos θ )}3  3 0

π

3 4π a 3 4 1 (1 − sin 2φ ) 2 cos φ d φ , = ∫ 3 0 2

2 sin θ = sin φ =

4π a 3

π

2

∫ cos φ d φ = 2

3

4

0

π

3 2π 2  2a    3 = {2a (1 + cos θ )} −   sin θ dθ  1 + cos θ   3 ∫0 



4π a 3 3.1 π π 2 a 3 . . = . 3 2 4.2 2 4 2

=

EXAMPLE 11.16

Show that if the area, lying within the cardioid r = 2a (1 + cos θ ) and not included in the parabola r (1 + cos θ ) = 2a, revolves about the initial line, the volume generated is 18π a 3. Solution. Both the curves are symmetrical about the initial line. Eliminating r from the given equations, we get 2a 2a (1 + cos θ ) = or 1 + cos θ (1 + cos θ ) 2 = 1 or cos θ (2 + cos θ ) = 0 . Since cos θ cannot be −2, we have cos θ = 0 and so, θ = ±

π

. So, for the upper half of the 2 area, θ varies from 0 to π . 2 r (1

+c

os

u)



16π a 3 = 3

y uπ 2

2a

r

2a

(1+

co

s )

π

 (1 + cos θ ) 4 (1 + cos θ ) −2  2 + −  4 −2  0

16π a 3 15 3  − = 18π a 3.. 3  4 8 

11.4  SURFACE OF THE SOLID OF REVOLUTION (CARTESIAN EQUATIONS) Let AB be the arc of the curve y = f ( x) included between the ordinates x = a and x = b. Assume that the curve does not cut the x-axis. Let P (x, y) and Q( x + δ x, y + δ y ) be any two neighboring points on the curve. Let length of the arc AP be s and that of the arc AQ be s + δ s so that the arc PQ is of length δ s . Draw the ordinates PM and QN. Let S denote the curved surface of the solid generated by the revolution of the area CMPA about the x-axis. The curved surface of the solid generated by the revolution of the area MNQP is δ S . Without any loss of generality, we assume that the curved surface of the solid generated by the revolution of the area MNQP about the x-axis lies between the curved surfaces of the right circular cylinders whose radii are PM and NQ and which are of the same thickness (height) δ s . y

(a, 0) 0

u 0

x B

A

π

2

∫

– (r of the inner curve )3 ] sin θ dθ

M11_Baburam_ISBN _C11.indd 9

Q

x a

2π [(r of outer curve)3 3 0

s P δ

x  b

Therefore, the required volume is given by

s

0

C

M

N

D

x

12/14/2011 3:48:15 PM

11.10  n  chapter eleven Therefore, δ S lies between 2π ( y + δ y )δ s, that is, or

2π yδ s

and

2π yδ s < δ S < 2π ( y + δ y )δ s δS 2π y < < 2π ( y + δ y ) . δs

1

2   dy   2 y h 2 π 1 + ∫0   dx   dx   h

h

= 2π ∫ x tan α [1 + tan 2α ] 2 dx 0

As Q → P , that is, as δ s → 0, δ y will tend to 0. Therefore,

h

x=b

∫

x=b

∫

2π yds =

x=a

0

where

a

a

EXAMPLE 11.18

2

ds  dy  = 1+   .  dx  dx

Similarly, the curved surface of the solid generated by the revolution about the y-axis, of the area bounded by the curve x = f ( y ) , the lines y = a and y = b , and the y-axis, is b

2π ∫ xds. 0

Find the surface generated by the revolution of xX an arc of the catenary yy == cc cosh from x = 0. cC x = 0 to any point ( x, y ) about the axis of x. Solution. The equation of the catenary is x y = c cosh . c Therefore, dy  x 1  x = c sinh   ⋅ = sinh      c dx c c and so,

EXAMPLE 11.17

Find the area of the surface of a cone with a semi-vertical angle α and a base as a circle of radius a. Solution. The cone is obtained by the revolution of the right-angled triangle OAC about its base OC (x-axis). We have y = x tan α . Therefore, dy = tan α . Hence, the required surface area is A dx

2

ds x x  dy  = 1 +   = 1 + sinh 2 = cosh .  dx  dx c c Hence, the required surface area is x

∫ 2π y 0

x

ds x x dx = ∫ 2π c cosh . cosh dx dx c c 0 x

) P(x, y

0

α

M

= ∫ 2π c cosh 2

a

h

C

B

M11_Baburam_ISBN _C11.indd 10

h 2

= π a 2 cosecα .

ds

∫ 2π yds = ∫ 2π y dx dx,

2

 a2  a = π tan α sec α  , tan α = 2  h  tan α 

dS = [ S ]ba .

x=a

b

0

= 2π tan α sec α .

Hence, the required curved surface is given by b

h

= 2π ∫ x tan α sec α dx = 2π tan α sec α ∫ xdx

dS = 2π y or dS = 2π yds. ds

Hence,

1

0

x

x dx c

x

2x   = π c ∫ 1 + cosh  dx c  0  c 2x   = π c  x + sinh  . 2 c  

12/14/2011 3:48:16 PM

voluMeS and SurfaceS of SolidS of revolution   n 11.11

Solution. The equation of the parabola is x2 = 4ax. Differentiating with respect to x, we get dy 2a dy = . 2y = 4a or dx y dx

11.5  SURFACE OF THE SOLID OF REVOLUTION (PARAMETRIC EQUATIONS) The area of the surface of the solid generated by the revolution about the x-axis, of the area bounded by the curve x = f (t ) , y = φ (t ), the x-axis, and the ordinates at the points, where t = a and t = b is b ds ∫a 2π y dt dt ,

Therefore,

where

EXAMPLE 11.19

Find the area of the surface formed by the revolution of y 2 = 4ax about the x-axis, by an arc from the vertex to one end of the latus rectum.

2

ds  dx   dy  =   +   dt   dt  dt

2

ds 4a 2 4a 2  dy  = 1+   = 1+ 2 = 1+  dx  dx 4ax y x+a . x

=

EXAMPLE 11.20

For the arc from the vertex O to L, the end of the latus rectum, x varies from 0 to a. Therefore, a

Required surface = ∫ 2π y 0

ds dx dx

Solution. The equations of the tractrix are a t x = a cos t + log tan 2 and y = a sin t. 2 2 This curve is symmetrical about both the axes and its asymptote is y = 0 , that is, x-axis. We have (see Example 11.11)

L

0

Show that the surface generated by revolution a t of the tractrix x = a cos t + log tan 2 and 2 2 y = a sin t about its asymptote is equal to the surface of a sphere of radius a.

y

2

= M

= 2π ∫ 4ax . 0

= 4π a ∫ 0

x+a dx x

a 2 cos 2 t (cos 2 t + sin 2 t ) a cos t = . sin t sin 2 t

π

2

3    ( x + a) 2  x + adx = 4π a    3   2 0

2∫ 2π y 0

ds dt dt π 2

= 4π ∫ a sin t. 0

a cos t dt sin t

π

2

3 3 8π a  8π a (2a ) 2 − a 2  = (2 2 − 1). =   3 3

2

M11_Baburam_ISBN _C11.indd 11

a 2 cos 4 t + a 2 cos 2 t sin 2 t

For the curve in the second quadrant, t varies from 0 to π . Hence, the required surface is given by 2

a

a

2

ds  dx   dy  =   +  =  dt   dt  dt

x

S

a

2

π

= 4π a 2 ∫ cos tdt = 4π a 2 [sin t ] 20 = 4π a 2 0

= the surface of a sphere of radius a.

12/14/2011 3:48:17 PM

11.12  n  chapter eleven EXAMPLE 11.21

Find the surface of the solid generated by revolving the loop of the curve 9 y 2 = x( x − 3) 2 about the x-axis. Solution. The parametric equations of the curve are 1 3 x = t 2 and y = t − t . 3 We have

Solution. The curve (see Example 11.14) is symmetrical about the y-axis and tangent at the vertex is x-axis. We have dy dx = a sin θ . = a (1 + cos θ ) and dθ dθ Therefore, 2

ds  dx   dy  =   +   dθ   dθ  dθ

dy 1 dx = 1 − 3t 2 = 1 − t 2 . = 2t and dt 3 dt

= a 2 (1 + cos θ ) 2 + a 2 sin 2θ

= a 1 + cos 2θ + 2 cos θ + sin 2θ

Therefore, 2

2

= a 2(1 + cos θ )

2

ds  dx   dy  =   +   = 4t 2 + (1 − t 2 ) 2 = 1 + t 2 .  dt   dt  dt

= a 2.2cos 2

Y

= 2a cos

θ.

θ 2

2

Therefore, Required surface π

= 2∫ 2π y 0

0

t 3

ds dθ dθ

π

X

θ

= 4π ∫ a (1 − cos θ )2a cos dθ 2 0

θ θ  = 4π ∫ a  2sin 2  .2a cos dθ   2 2 0 a

The curve cuts the x-axis at 0 and 3. So, as x varies from 0 to 3, t varies from 0 to 3 . Hence, the required surface is 3

∫ 0

3

ds 2π y dt = 2π ∫ dt 0 3

= 2π ∫ 0

 t3  2  t − 3  (1 + t )dt

 2t 3 t 5   t 2 2t 4 t 6  t dt 2 + − = −  π    + 3 3   2 12 18  0

0

= 16a 2

sin 3 θ2 3. 12

θ 2

θ

cos dθ 2

π

= 0

32 2 πa . 3

3

 3 18 27  = 2π  + −  = 3π .  2 12 18  EXAMPLE 11.22

Find the surface area of the solid generated by revolving the cycloid x = a (θ + sin θ ) and y = a (1 − cos θ ) about the tangent at the vertex.

M11_Baburam_ISBN _C11.indd 12

π

= 16π a 2 ∫ sin 2

EXAMPLE 11.23

Find the surface of the solid generated by the 2 2 2 revolution of the asteroid x 3 + y 3 = a 3 or 3 3 x = acos t and y = asin t about the x-axis. Solution. As in the Example 11.13, we have dx dy = −3acos 2 t sin t and = 3asin 2 t cos t dt dt Therefore,

12/14/2011 3:48:17 PM

voluMeS and SurfaceS of SolidS of revolution  n 11.13 Y

ds = 9a 2 cos 4 tsin 2 t + 9a 2 sin 4 tcos 2 t dt = 9a 2 cos 2 tcos 2 t (cos 2 t + sin 2 t ) = 3a sin t cos t .

Hence, the required surface is given by π

X

0

π

2 ds 2∫ 2π y .dt = 4π ∫ a sin 3t.3a sin t cos tdt dt 0 0 2

π

= 12π a

2

2

∫ sin t cos tdt 4

0

π

 sin t  = 12π a 2    5 0 5

=

2

Therefore, the required surface area is equal to π

1

π

2  2  dr  2  2 ds 2 π y . d θ 2 π r sin θ =  r +    dθ ∫0 dθ ∫0  dθ    2

12 2 12 π a (1 − 0) = π a 2. 5 5

π

2

= ∫ 4aπ cos θ sin θ 4a 2 cos 2θ + 4a 2 sin 2θ dθ 0

11.6  SURFACE OF THE SOLID OF REVOLUTION (POLAR CURVES) The curved surface of the solid generated by the revolution, about the initial line, of the area bounded by the curve r = f (θ ) and the radii vector θ = α and θ = β , is β

∫ 2π y α

where

ds .dθ , dθ 2

ds  dr  = r 2 +   , y = r sin θ.  dθ  dθ

EXAMPLE 11.24

Find the area of the surface of revolution formed by revolving the curve r = 2a cos θ about the initial line. the curve is dr r = 2a cos θ . Therefore, = −2a sin θ . dθ The curve is symmetrical about the initial line. Solution.

The

equation

of

For the upper half of the circle, θ varies from 0 to π . 2

M11_Baburam_ISBN _C11.indd 13

π

π

 sin 2θ  2 = 8π a ∫ cos θ sin θ dθ = 8π a    2 0 0 = 4π a 2. 2

2

2

EXAMPLE 11.25

Find the surface of the solid formed by the revolution of the cardioid r = a (1 + cos q) about the initial line. Solution. We have r = a (1 + cos θ ). Therefore, dr = −a sin θ . For the upper half of the curve dθ (see Example 11.15), q varies from 0 to π . Hence, the required surface area is 1

π  2  dr  2  2 ds 2 π y . d θ 2 π r sin θ =  r +    dθ ∫0 dθ ∫0  dθ    π

π

= 2π ∫ a (1 + cos θ ) sin θ [a 2 (1 + cos θ ) 2 0

1

+ a 2 sin 2θ ] 2 dθ π

= 16π a 2 ∫ cos3 0

θ 2

sin

θ 2

θ

cos dθ 2

12/14/2011 3:48:17 PM

11.14  n  chapter eleven π

= 16π a 2 ∫ cos 4 0

θ 2

θ

sin dθ



2

−a

π

θ 32 2  2 πa . = 16π a 2  − cos5  = 2 0 5  5 EXAMPLE 11.26

Find the surface area generated by the revolution of the loops of the lemniscate r 2 = a 2 cos 2θ about the initial line. Solution. The given curve is symmetrical about the initial line and the line θ = π2 (see Example 11.16). It consists oftwo loops and the tangents at the pole are θ = − π4 and θ = π4 . For half of the loop in the first quadrant, θ varies from 0 to θ = π4 . Further, 2 dr a sin 2θ ds  dr  2 = = r +  and dθ  dθ  cos 2θ dθ

cos 2 2θ + sin 2 2θ = a = cos 2θ

a

2

cos 2θ

4

2∫ 2π y 0

π

4 ds dθ = 4π ∫ r sin θ dθ 0

a cos 2θ

π

4

= 4π ∫ a cos 2θ sin θ.

.

0

cos 2θ

= 4π a

dθ

4

∫ sinθ dθ 0

π

= 4π a 2 [ − cos θ ] 04 = 2π a 2 (2 − 2). EXERCISES 1. Show that the volume of the solid formed by the revolution of the loop of the curve x 2 (a − x) about the . x-axis is 2 2π (1 og 2 − )a 3 3 2. Find the volume of a sphere of radius a. Hint: The sphere is the solid of revolution generated by the revolution of a semicircular

M11_Baburam_ISBN _C11.indd 14

4. Find the volume when the loop of the curve y 2 = x(2 x − 1) 2 revolves about the x-axis. Ans. 48π . 5. Find the volume generated by revolving the portion of the parabola y 2 = 4ax cut off by its latus rectum about the axis of the parabola a Hint: V = ∫π y 2 dx. Ans. 2π a 3. 0

7. The area bounded by the parabola y = 4 x and the straight line 4 x − 3 y + 2 = 0 is rotated about the y-axis. Find the volume of the solid so generated. Ans. 20π . 2

π

2

3. Find the volume of the solid generated by revolving the curve xy 2 = 4(2 − x) about the y-axis. Ans. 4π 2.

Ans. 2ab 2π .

dθ a

Ans. 43 π a3.

6. Find the volume of the paraboloid generated by the revolution of the parabola y 2 = 4ax about the x-axis from x = 0 to x = b.

Therefore, the required surface is given by π

area about its bounding diameter. Thus, a 2 V = ∫ π y dx , x 2 + y 2 = a 2 .

8. Find the volume of the reel formed by the revolution of the cycloid x = a (t + sin t ) and y = a (1 − cos t ) about the tangent at the vertex. Ans. π 2 a 3. 9. Prove that the volume of the solid generated by the revolution about the x-axis of the loop of the curve 9 y 2 = x( x − 3) 2 or x = t 2 1 3 and y = t − t . 3 Ans. 3π . 4

10. Find the volume of the solid generated by the revolution of the cycloid x = a (θ − sin θ ) and y = a (1 − cos θ ) about y-axis. Ans. 6π 3 a 3.

12/14/2011 3:48:18 PM

voluMeS and SurfaceS of SolidS of revolution   n 11.15 11. Find the volume of the solid generated by revolving the lemniscate r 2 = a 2 cos 2θ about the line θ = π2 . 2 3 Ans. 2π8 a 12. Show that the volume of the solid formed by the revolution of the curve r = a + b cos θ (a > b) about the initial line is 34 π a(a 2 + b 2 ). 13. Find the surface of the solid generated by the revolution of the ellipse x 2 + 4 y 2 = 16 about its major axis Ans. 8π [1 + 34π3 ]. 14. Find the surface area of the solid generated by revolving the cycloid x = a (θ − sin θ ) and y = a (1 − cos θ ) about the x-axis. Ans. 643 π a2.

M11_Baburam_ISBN _C11.indd 15

15. Show that the surface area of the solid generated by revolving one complete arc of the cycloid x = a (θ − sin θ ) and y = a (1 − cos θ ) about the line y = 2a is 32 π a2. 3 16. Find the area of the surface of the solid formed by the revolution of the cardioid r = a (1 − cos θ ) about the initial line. Ans.

32π a2 5

.

17. The lemniscate r 2 = a 2 cos 2θ revolves about a tangent at the pole. Show that the surface of the solid so generated is 4 π a 2.

12/14/2011 3:48:19 PM

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M11_Baburam_ISBN _C11.indd 16

12/14/2011 3:48:19 PM

12

Multiple Integrals

The aim of this chapter is to study double- and triple integrals along with their applications. Thus, we shall consider here the integrals of the functions of two- and three variables. 12.1  DOUBLE INTEGRALS The notion of a double integral is an extension of the concept of a definite integral on the real line to the case of two-dimensional space. Let f (x, y) be a continuous function of two independent variables x and y inside and on the boundary of a region R. Divide the region R into subdomains R1, R2,. . ., Rn of areas d R1, d R2,. . ., d Rn, respectively. Let (xi, yi) be an arbitrary point inside the ith elementary area, d Ri. Consider the sum Sn = f (xi, yi)d R1 + f (x2, y2) d R2+. . .

+ f (xi, yi) d Ri +. . . + f (xn, yn)d Rn. n



= ​    ​ f (xi, yi)d Ri.

   

i=1

dRi→0

the double integral of the function f (x, y) over the region (domain) R and is denoted by  

∫​∫  ​ f (x, y) dR. R

If the region R is divided into rectangular meshes by a network of lines parallel to the coordinate axes and if dx and dy be the length and breadth of a rectangular mess, then dxdy is an element of area in Cartesian coordinates. In such a case, we have

 

 

 

R

R

We now state, without proof, two theorems that provide sufficient conditions for the existence of a double integral over a closed region R. Theorem 12.1. Let φ and y be two continuous functions defined on a closed interval [a, b] such that φ (x) ≤ y (x) for all x ∈ [a, b]. Let f be a continuous function defined over   R = {(x, y): a ≤ x ≤ b, φ (x) ≤ y ≤ y  (x)}. Then, ∫​∫  ​  f(x, y)dxdy and

[ 

]

ψ (x)

a

 

R

​∫  ​  ​ ​∫  ​ ​ f (x, y) dy  ​dx exist and are equal.  

b

  

φ (x)

Theorem 12.2. Let φ and y be two continuous functions defined on a closed interval [c, d ] such that φ (y) ≤ y (y) for y ∈ [c, d]. Let f be a con  = {(x, y) : c ≤ y tinuous function defined over R ≤ d, φ (y) ≤ x ≤ y (y)}. Then, ∫∫​   ​ f (x, y) dxdy and d

When n → ∞, the number of subregions increases indefinitely such that the largest of the areas d Ri approaches zero. The     lim ​​Sn, if exists, is called ​ ​ n→∞     

M12_Baburam_ISBN_C12.indd 1

 

∫​∫  ​  f (x, y) dR = ∫​∫  ​  f (x, y) dx dy.

[ 

]

ψ (y)

R

​∫  ​  ​ ​∫  ​ ​ f (x, y) dx  ​dy exist and are equal. c

 

φ (y)

EXAMPLE 12.1 Show that 1

[ 

1

​∫  ​​  ​ ​∫  ​​ ​   

0

 

0

(x + y)

Solution.  We have 1

] [  1

1

]

x - y  x - y  ______ dx  ​dy ≠ ​  ​​  ​ ​  ​​  ​  ______ dy  ​dx. 3 ​  3 ​ 

[ 

1

∫  ∫   

0

0

 

(x + y)

]

x-y ​∫  ​ ​ ​ ​∫  ​​ ​​  ______  3 ​  dx  ​dy 0 0 (x + y)  

 

12/9/2011 4:47:30 PM

12.2  n  chapter twelve

[ 

]

1

1

x + y - 2y = ​∫  ​​  ​ ​∫ ​​  ​ ​ ________  ​    dx  ​dy (x + y)3 0 0



 

 

[  { 

1

1

12.3 EVALUATION OF DOUBLE INTEGRALS (CARTESIAN COORDINATES)

} ]

2y - ______ = ​  ​​  ​ ​  ​ ​ ​ ______ ​  1  2 ​  ​    3 ​  ​dx  ​dy

∫  ∫   

(x + y)

 

0

0

[ 

1

(x + y)

]

The double integrals can be evaluated using Theorems 12.1 and 12.2. In fact, (i) If the limits in the inner integral are functions ψ (x)

2y -1 _______ = ​∫  ​​​  ​ ​  ____  ​   ​​​ ​​ dy x + y ​ + ​ 2(x +   y)2 0 0



 

1

[ 

1

] [  ]

of x, then we evaluate ∫​   ​ ​f (x, y) dy, first φ (x)

taking x as a constant and then evaluate the integrand (function of x), obtained in the first step, integrating it with respect to x between the limits a and b. Thus,

-1   ​   = ​dy = ​ _____ ​  1   ​ ​ ​ = - __ ​ 1 ​. ∫​   ​​​  ​ ​  ______ 2 2 1 + y (1 + y) 0 0



 

[ 

1

Similarly, we can show that 1

]

1

b ψ (x)

x - y   ​    ​ ​∫  ​ ​ ​ ​∫  ​​ ​ ______ ​ 1 ​ . dy dx = __ 3 2 0 0 (x + y)  

Hence,

[ 

1

 

] [ 

1

1

1

 

0

 

∫  ∫  (x + y)  

   

0

0

 

0

 

]

ψ (y) )

of y, then we evaluate ∫​  ​ f (x, y) dx, first tak 

φ (y)

ing y as a constant and then evaluate the integrand (function of y) , obtained as a result of the first step, integrating it with respect to y between the limits c and d. Thus, d ψ (y)

∫​∫  ​  K  f (x, y) dx dy = K ∫​∫  ​  f (x, y) dx dy.

∫​∫  ​ [  f1(x, y) + f2 (x, y) + . . . + fn(x, y)] dx dy R



= ∫​∫  ​f1 (x, y) dx dy + ∫​∫ ​f2 (x, y) dx dy R

R

 

+. . . + ∫​∫  ​ fn (x, y) dx dy. R

3. If the region R is partitioned into two regions R1 and R2, then

EXAMPLE 12.2 Show that 2

[ 

R



R1

+ ∫​∫ ​​f (x, y) dx dy. R2

M12_Baburam_ISBN_C12.indd 2

]

 

] [ 

4

4

]

2

​∫  ​​  ​ ​∫  ​​  (xy + e ) dx  ​dy = ​∫  ​​  ​ ​∫  ​​  ​(xy + e y ) dy  ​dx.  

1

y 

 

 

3

3

1

Solution.  The function f (x, y) = xy + e y is a continuous function over the rectangle R = {(x, y) : 1 ≤ x ≤ 2, 3 ≤ y ≤ 4}. Therefore, the values of these integrals are equal. In fact, we note that 4

​∫  ​​  (xy + e y ) dx = __ ​ 7 ​ y + e y 2 3

 

∫​∫ ​f (x, y) dx dy = ∫​∫  ​ f (x, y) dx dy

[

ψ (y)

c φ (y)

R

 



 

c φ (y)

 

R

d

​∫  ​ ​∫  ​ ​ f (x, y) dx dy = ​∫  ​ ​ ​∫  ​ ​​ f (x, y) dx  ​dy.

 

2. The double integral of the algebraic sum of a finite number of functions fi is equal to the sum of the double integrals taken for each function. Thus,

 

a φ (x)

(ii) If the limits in the inner integral are functions

12.2  PROPERTIES OF A DOUBLE INTEGRAL 1. Let K ≠ 0 be any real number. Then,  

 

 

a φ (x)

x-y The reason is that the function f (x, y) = ______ ​     ​  is (x + y)3 not continuous in R = {(x, y): 0 ≤ x ≤ 1, 0 ≤ y ≤1}.

 

]

​∫  ​ ​​∫  ​ ​ f(x, y) dx dy = ​∫  ​ ​ ​∫  ​ ​ f (x, y) dy  ​dx.

x - y  x - y  dx  ​dy ≠ ​  ​​  ​ ​  ​​ ​ ______ dy  ​dx. ​  ​​  ​ ​  ​​ ​ ______ 3 ​  3 ​ 

∫  ∫  (x + y)

[ 

b ψ (x)

 

and so 2 4

2

[ 

]

​∫  ​​  ​∫  ​​  (xy + e ) dx dy = ​∫  ​​  ​ __ ​ 7 ​ y + e y  ​dy 2 1 3 1 y 

 

 

 

12/9/2011 4:47:34 PM

Multiple integralS   n 12.3 __ of integration for x are x = -y, x = √ ​ y ​,  y = 0, and y = 2. Therefore,

___ = ​ 21 ​ + e 2 - e. 4 One the other hand,



2

2

 

R

 

4

[ 

]

2

4

[ 

__ 3

]

​∫  ​​  ​ ​∫  ​​  ​(xy + e  )dy  ​dx = ​∫  ​​  ​ ​   ​ x + e  - e  ​dx 3 1 3 2  

y

 

2

= ___ ​ 21 ​ + e 2 - e EXAMPLE 12.3 Evaluate ∫∫ x 2y 2 dx dy over the circle x 2 + y 2 ≤ 1.

Solution.  Since x 2 + y 2 ≤ 1, it follows that 2 x 2 ≤ 1  and y  ≤ 1 - x 2 or ______ |x| ≤ 1  and  |y| ≤ ​√1 - x 2    ​ or _____ ______ - 1 ≤ x ≤ 1   and   - √ ​ 1 - x 2   ​≤ y ≤ √ ​ 1 - x 2    ​. The integrand f (x, y) = x 2y 2 is continuous over the region ______ _____ R = {(x, y): -1 ≤ x ≤ 1, -​√1 - x 2  ​  ≤ y ≤ ​√1 - x 2    ​}. Therefore, _____  

1

[ 

​√1-x 2 ​ 

]

2 2 ∫​∫  ​  x 2y 2 dx dy = ​∫   ​​​ ​   ​ ∫_____ ​   ​​ x    y  dy  ​dx  

R

-1

{  }

1

_____

-​√1-x 2     ​

1

 __ 3

y 3 ​√1  -  x   ​ ​ 2  ​ x 2 (1 - x 2) ​ 2 ​ dx = ​∫   ​​ ​ x 2​​ ​ __ ​   ​​ _____ ​​ dx = ​∫   ​​​ __ 3 3 - ​√1  -  x  ​  -1 -1  

1



2

 

2

[ 

]

__ ​√y ​ 

0

2

 

[ (  ∫  (  [  2

) ( 



)]



__ y __ y 2 = ​∫  ​​  ​ ​ ​√y ​ + __ ​    ​+ y ​√y ​  ​- ​ - y + ​ ___ ​ - y 2  ​  ​dy 2 2 0



__ __  y 2 3y = ​  ​​  ​ ___ ​   ​ + __ ​   ​ + ​√y ​ + y ​√y ​  ​dy 2 4 0

 

 

) ]

 _ 5 2 y 3 3y 2 2  _32 __ = ​ ___ ​   ​ + ​ ___ ​  + __ ​   ​ y ​   ​+ ​ 2 ​ y ​ 2 ​  ​ ​​ ​​  5 2 4 3 0 __ 13 44 ___ = ​   ​ + ___ ​   ​ √ ​ 2 ​.  3 15



EXAMPLE 12.5 Evaluate ∫​∫   ​  ydx dy, where R is the region bounded R

by the parabolas y 2 = 4x and x 2 = 4y. Solution.  The given parabolas are y 2 = 4x and x 2 = 4y. Solving these equations, we get x = 0 and y = 4. The corresponding values of y are y = 0 and y = 4. Both the curves pass through the origin and the points of intersection are (0, 0) and (4, 4). Thus, the limits of integration are x = 0 to x = 4 __ x 2 and y = ___ ​   ​ to y = 2​√x ​.  Thus, 4 __  x 2 R = ​ (x, y) : 0 ≤ x ≤ 4; ___ ​   ​ ≤ y ≤ 2​√x ​  ​. 4

{ 

}

y

 __ 3

__ = ​ 4 ​ ∫​   ​​  x 2(1 - x 2) ​ 2 ​ dx, since integrand is even 30

x 2  4y

 

__ ​ p   ​ 2



 

-y

 x 2 = ​∫  ​​  ​​ x + ___ ​   ​ + xy  ​ ​  dy -y 2 0



1

4

Hence, the result.

]

∫​∫   ​ (1 + x + y) dx dy = ​∫  ​​  ​ ​∫   ​ ​ (1 + x + y) dx  ​dy

​∫  ​​  (xy + e y ) dy = __ ​ 3 ​ x + e 2 - e 2 1

and so,

[ 

__ ​√y ​ 

A(4, 4)

__ = ​ 4 ​ ​∫  ​  ​​ sin2q cos4q dq, substituting x = sin q 30 __ 1 3.1 . p p  ​ . __ = ​ 4 ​  . ​ _____ ​   ​    ​= ​ ___ 3 6.4.2 2 24  

EXAMPLE 12.4 Evaluate the double integral of the function f (x, y)__= 1 + x + y over a region bounded by y = -x, x = ​√y ​,  and y = 2. Solution.  The region R is bounded by y = -x, the parabola y 2 = x, and the line y = 2. Thus, the limits

M12_Baburam_ISBN_C12.indd 3

0

x

y 2  4x

Therefore,

[  ] __ 2​√x ​ 

4

∫ ​∫   ​  ydx dy = ​∫  ​​  ​ ​∫ ​ ​ y dy  ​dx  

R

0

 

x2 __

​    ​ 4

12/9/2011 4:47:39 PM

12.4  n  chapter twelve 4

[  ]

__ 2 2​√x ​ 

4

(  2

their center at the origin and with sides parallel to the axes of coordinates, if each side of the inner square is equal to 2 and that of the outer square is 4.

)

y 4x x 4 = ​  ​​​​​  ​ __ ​   ​​__ ​  ​dx = ​  ​​ ​ __ ​   ​ - ​ ___   ​  ​dx x2

∫  2 0

4

( 

​   ​  4

∫  0

32

) [ 

]

x 4 x 5 4 48. = ​∫  ​​​ ​ 2x - ​ ___  ​   ​dx = ​ x 2 - ​ ____  ​  ​​​ ​​ = ___ ​   ​  32 160 0 5 0

Solution.  The region R is irregular. However, the straight lines x = -1 and x = 1 divide this region into four regular subregions R1, R2, R3, and R4.

EXAMPLE 12.6 Calculate the double integral

y

1 x 2

​∫  ​​ ​∫  ​ ​(x2 + y2) dx dy 0 0

R2

and determine the region of integration. Solution.  The region of integration is bounded by the lines x = 0, x = 1, y = 0, and the parabola x 2 = y. Thus, the region is R = {(x, y) : 0 ≤ x ≤ 1; 0 ≤ y ≤ x 2}, and is shown in the following figure:

Ð2

Ð1

0

R1

2

x

R4

R3

Y x2 = y

1

Therefore,

∫​∫   ​e   x + y dR = ∫​∫ ​e x + y dx dy + ∫​∫  ​ e x + y dx dy R

x=1

-1

[ 

]

x 2

​∫  ​​ ​ ​∫  ​ ​​(x 2 + y 2) dy  ​dx. 0

0

So, we evaluate the inner integral first. We have

[ 

x 2

 y  ___

]

3 x2

x6 ​∫  ​ ​​(x  + y  ) dy = ​ x  y + ​   ​  ​​​ ​ ​= x4 + ​ __ ​ . 3 3 0 0 2

2

Therefore, 1

(  [ 

2

) ]

x6 ​∫    ​∫ ​ (x2 + y2) dx dy = ​∫  ​​ ​ x4 + ​ __ ​   ​dx 3 0 x5 ___ x7 0 __ 26 __ = ​​ ​   ​ + ​    ​  ​​​ ​= ​ 1 ​ + ___ ​ 1  ​ = ____ ​    ​.  5 21 1 5 21 105

      R  

EXAMPLE 12.7 Evaluate the double integral ∫∫​ ​   e x + y dR over the  

R

region R, which lies between two squares with

M12_Baburam_ISBN_C12.indd 4

+ ∫​∫  ​ e 

X

The given integral can be expressed as

R2

x+y



0

1

R1

[ 

2

= ​∫   ​  ​​​ ​∫   ​e​    -2

x+y

[ 

-2

1

R3

]

1

[ 

dx dy + ∫​∫  ​e   x + y dx dy

]

2

R4

dy  ​dx + ​∫   ​​​ ​ ​∫  ​​ e x + y dy  ​dx

-1

] [  -1

1

2

2

]

+ ​∫   ​ ​ ​ ​∫   ​  ​ e x + y dy  ​dx + ​∫ ​​   ​ ​∫   ​ ​ e x + y dy  ​dx  

-1

 

 

-2

1

 

-1

= (e 2 - e -2) (e -1 - e -2) + (e 2 - e) (e - e -1) + (e -1 - e -2) (e - e -1) + (e 2 - e -2) (e 2 - e ) = (e 3 - e -3) (e - e -1) = 4 sinh 3 sinh 1. EXAMPLE 12.8 Evaluate ∫∫ xy dx dy over the positive quadrant of the circle x 2 + y 2 = a 2. Solution.  The region of integration is ______ R = {​  (x, y): 0 ≤ x ≤ a; 0 ≤ y ≤ √ ​ a 2 - x 2    ​ }​. The integrand f (x, y) = xy is continuous over R. ______ Therefore,

[ 

 ​ a ​√ a 2 - x 2  

∫​∫   ​  xy dx dy = ​∫  ​​     ​∫  ​   

R

 

0

0

]

​xy dy  ​dx  

12/9/2011 4:47:42 PM

Multiple integralS   n 12.5

[  ]

_____ 2 ​√a2-x2   ​

a

y __

 

1 __

[ 

Solution.  The region of integration is bounded by x = 0, x = 1, y = x, and the parabola y2 = x. Thus, __ R = {(x, y) : 0 ≤ x ≤ 1; x ≤ y ≤ √ ​ x ​}  . Therefore,

a

​= __ ​ 1 ​ ​∫  ​​  x[a 2 - x 2] dx 20

= ​∫  ​​  ​​ x ​   ​   ​​ ​  2 0 0



 

a

a

]

= ​   ​ ​ ​∫  ​​ a  x dx - ​∫  ​​  x 3dx  ​ 2 0 0



2

 

 

[ 

] [  ]

2 a

4 a

__ 1 ​√x ​ 

4

4

0 x

__ 3 ​√x ​ 

y = ​  ​ ​​ ​ x y + __ ​   ​   ​​​​  ​  ​dx

∫ 

EXAMPLE 12.9 Evaluate ∫​∫  ​ ​xy dx dy, where R is the domain

1

R

}

x2 R = ​ (x, y) : 0 ≤ x ≤ 2a; 0 ≤ y ≤ ​ ___  ​  ​. 4a The region is bounded by y = 0, x = 2a, and the parabola x 2 = 4ay. y

x 2 = 4ay

x 5 2a x

{  [  ] }

0 0

x 2 2 ___ ​    ​

2a

y 4a x5 dx = ​  ​ ​​ x​ __ ​   ​   ​​ ​  ​  d​ x = ​  ​ ​​ ____  2 ​ 

∫0 

2

0

∫0  32a

[  ]

[  ]

6 a4 x6 ​    2a __ 64a ​    = ____ ​  1 2 ​ ​ ​ __ ​​​ ​  ​= ____ ​  1 2 ​ ​ ​ ____  ​= ​   ​ . 3 6 32a 6 0 32a

EXAMPLE 12.10 __ 1 ​√x ​ 

Evaluate ∫​   ​​ ​  ​∫ ​ ​(x 2 + y 2) dy dx. 0

x

M12_Baburam_ISBN_C12.indd 5

 5 __

7 __

1

9 x ​ 2 ​  x​ 2 ​  ___ 4x4 3 = ___ ​​   ​ + ___ ​    ​ - ​   ​ ​   ​​ = __ ​ 2 ​ + ___ ​ 2  ​ - __ ​ 1 ​ = ____ ​     ​ = ___ ​    ​.  7 ___ 7 15 3 105 35 15 12 __ ​   ​  ​   ​  2

2

0

EXAMPLE 12.11 Evaluate ∫∫ (x + y)2dx dy over the area bounded 2 x2 y by the ellipse __ ​  2  ​+ __ ​  2  ​= 1. 2 a b x2 y Solution.  The equation of the ellipse is __ ​  2 ​ + __ ​  2 ​ = 1. a b Therefore, _____ 2 y2 y x2 __ __ x __ ​  2  ​= 1 - ​ a ​   or  ​   ​= ± ​ 1 - __ ​  2  ​ ​   or b b a ______ b y = ± __ ​ a ​​√a2 - x2 ​  .

√

{ 

∫​∫   ​ xy dx dy = ​∫  ​ ​​∫  ​ ​xy dy dx 2a

 __3  __52 ___ x ​ 2 ​  ___ 4x3 = ​∫  ​​  x ​   ​ + ​   ​ - ​   ​  dx. 3 3 0

The ellipse cuts the x-axis at x = ± a. Therefore, the region of integration is ______ R = ​ (x, y):  - a ≤ x ≤ a; - __ ​ ba ​√ ​ a2 - x2   ​≤ y ______ b 2 2 . ≤ __ ​ a ​√ ​ a -x   ​  ​

x2 ___ ​    ​ 

2a 4a

R

 5

x

 __ 3 __2 

1

Solution.  The region of integration is

Therefore,

3

__ x ​   ​  x3 = ​  ​​ ​ ​ x ​ 2 ​ + ​   ​   ​- ​ x3 + ​ __ ​   ​  ​dx 3 3 0

bounded by the x-axis, ordinate x = 2a, and the curve x2 = 4ay.

2

0

 

0

]

 

x

{ [  ] } ∫ [ (  ) (  ) ] 0

1

{ 

[ 

__ 1 ​√x ​ 

​∫  ​​ ​∫ ​ ​(x 2 + y 2) dy dx = ​∫  ​​ ​ ​∫ ​ ​(x 2 + y 2) dy  ​dx

 x  x a a a __ = ​ 1 ​ ​ a 2 ​ ___ ​  ​​​ ​​ - __ ​ 1 ​ ​ ​ __ ​   ​​​ ​​ = ​ __ ​ - ​ __ ​ = ​ __ ​ . 2 2 0 2 4 0 4 8 8



4

}

Since x2 + y2 is an even function of y and xy is an odd function of y, we have

∫​∫   ​(  x + y)2 dx dy R

______ ​ a ​​√a2 - x2 ​  b __

a

= ​∫   ​         ​∫   

-a

​  ​​(x2 + y2 + 2xy) dy dx

_____    - __ ​ ba ​​√a2-x2 ​ 

12/9/2011 4:47:46 PM

12.6  n  chapter twelve ______   ​ a ​​√a2 - x2 ​ 

= ​∫   ​         ​∫   

-a

​ ​​(x + y ) dy dx 2

  _____   - ​ a ​​√a2-x2 ​  ______ b 2    2 __ ​ a ​√ ​ a -  x  ​  a

2

1

​  1  [tan-1 1 - tan-1 0]dx _______ = ​∫  ​​  _____2  ​  0 ​√ 1 + x  ​  _____ 1 1 x + ​√1 + x2 ​  dx p p __ ______ __ _________ _____ = ​    ​∫​   ​​  ​     ​  = ​    ​​ log ​ ​   ​    ​  ​​​ ​​  4 0 ​√1 + x2 ​  4 1 0 __ __ p p __ __ = ​    ​[log (1 + √ ​ 2 ​)  - log 1] = ​    ​log (1 + √ ​ 2 ​)  . 4 4

b __

   + ​∫   ​       ​∫   

-a

​  ​2xy dy dx  

_____

_____ ​ a ​​√a2-x2   ​ 

= ​∫   ​  2 ​∫ ​   ​​(x2 + y2) dy dx + 0  

-a

a

[ 

0

]

____ b 2 2 3 __ __ ​ a √​​ a -x    ​

y = 2 ​∫   ​  ​ x2y + ​   ​   ​ ​ ​  ​dx 3 0 -a a  __ 3 ______ x2b 2 2 ___ b3 = 2 ∫​    ​  ___ ​  a    ​√ ​ a -x   ​+ ​  3  ​ (a2 - x2) ​ 2 ​  dx -a 3a a 3 ______ __ b b3 = 4 ∫​    ​  __ ​ a ​x2 √ ​ a2 - x2 ​  + ___ ​  3  ​ (a2 - x2)​ 2 ​  dx. 3a -a Substituting x = a sin q, we have dx = a cos q dq and so,  

 

∫​∫   ​  (x + y)2 dx dy  

p ​ __  ​ 2

[ 

]

ab3 = 4 ​∫  ​  ​​​ a3b sin2 q cos2 q + ​ ___ ​  cos4 q  ​dq 3 0 3 3.1 ab 1 p ___ __ ___ ___ 3 = 4 ​ a b . ​     ​  .​    ​ + ​   ​   . ​   ​  . ​ p__  ​   ​ 4.2 2 3 4.2 2 p ab 2 p __ 3 = ​    ​ (a b + ab3) = ​ ____  ​   (a + b2). 4 4

[ 

]

 

[  ( 

 

- __ ​ ba ​​√a2-x2 ​ 

b __

a

R

dx ______ x = __ ​ 1 ​tan-1 __ ​ a ​ since ∫ ​  2  2 ​  a +x a

b __

a

12.4 EVALUATION OF DOUBLE INTEGRALS (POLAR COORDINATES) q2 r2

We wish to evaluate ​∫  ​  ​ ​∫ ​ ​ f (r, q ) dr dq over the  

q1

 

r1

region bounded by the straight lines q = q1 and

q = q2 and the curves r = r1 and r = r2. To do so, we first integrate f (r, q ) with respect to r between the limits r = r1 and r = r2. The resulting integrand is then integrated with respect to q between the limits q = q1 and q = q2. EXAMPLE 12.13 Evaluate ∫∫ r sin q dr dq over the area of the cardioid r = a (1 + cos q ) above the initial line. Solution.  To evaluate the given integral above the initial line, we note that the limits of integration for r are r = 0 and r = a (1 + cos q ) , whereas the limits of q are q = 0 to q = p. y

θ  π 2

EXAMPLE 12.12

_____ 1 ​√ 1 + x2 ​ 

r=

1   ​  Evaluate ​  ​ ​  ​ ​  ​  ​_________ ​​  dy dx. 2 2

∫  ∫  0

0

 

_____ 1 ​√ 1 + x2 ​ 

θ=π

1   ​  ​∫  ​​     ​  ​∫ ​   ​_________ ​  dy dx. 1 + x 2 + y2 0 0

]



x

p a (1 + cos q)

= ​∫  ​  ​  ​ ​∫  ​ 

2

0



​ r sin q dr dq  

a (1 + cos q )

r2 ​  ​​​ ​  = ​∫  ​  ​ sin q ​ ​ __ 20 0

  

 

p

____

y ​√1+x  ​  1   ​  = ​∫  ​​  ______ ​  _____ ​​ tan-1 ______ ​  _____    ​   ​​ ​  ​dx, 2 0 √ ​√1 + x2 ​ 0 ​ 1 + x  ​ 

M12_Baburam_ISBN_C12.indd 6

θ=0

0

p 0

1   ​  = ​∫  ​​​    ​∫ ​   ​​__________ ​  dy dx. (1 + x 2) + y2 0 0

[ 

θ)

 

R

 

_____ ​√1 + x2 ​ 

 

os

∫​∫  ​  rsin q dr dq

1   ​  = ​∫  ​​      ​∫ ​   ​ _________ ​  dy dx. 1 + x 2 + y2 0 0

1

c

Therefore,  

_____ 1  ​√ 1 + x2 ​ 

1

a (1

1 + x  + y

Solution.  We have

 

)]

​dx

__ = ​ 1 ​ ∫​   ​  ​ a2 (1 + cos q ) 2 sin q dq 20  

12/9/2011 4:47:51 PM

Multiple integralS   n 12.7 a2 __

p

( 

)

EXAMPLE 12.15

2

__ __ __   ​ 2 sin ​ q   ​cos ​ q   ​dq = ​   ​ ∫​   ​  ​ ​ 2 cos2 ​ q 2 0 2 2 2  

dr dq  Evaluate ∫∫ _______ ​ r______  ​  over the loop of the lemnis ​ a2 + r 2 ​  √ cates r2 = a2 cos 2q.

p

__ __ = 4a2 ​∫  ​  ​cos5 ​ q   ​sin ​ q   ​dq 2 2 0



 

__ ​ p   ​ 2

Solution.  From the figure of the curve, we note that______ in the region of integration, r varies from 0 to __ __ a√ ​ cos 2q ​  and q varies from - ​ p   ​to ​ p   ​ 4 4

= 4a2 ​∫  ​  ​​cos5 φ sin φ . 2dφ, q = 2φ



0 __ ​ p   ​ 2

= 8a2 ​∫  ​  ​cos5 φ sin φ d φ



0

y

__ ​ p   ​ 2

  π 4

= - 8a ​∫  ​  ​​cos φ (-sin φ) dφ



2

r 2  a2 cos 2

5

0

[  ]

0

__ ​ p   ​

cos6 φ 2 ___ 4a2 = - 8a2 ​​ ​ _____  ​    ​ ​   ​= ​     ​. 3 6 0



(a, 0)

Therefore,

Solution.  The region of integration is between the circles as shown in the following figure:

rdr dq ______   ​  = ​∫    ​ ​​     ∫​  ​   ∫∫ ​  _______ ​√ a2 + r2 ​   p

p ______ ​ __  ​ ​ 4 a​√cos 2q  

__

- ​    ​ 4

__ ​ p   ​ 4

p ​ __  ​ 4

r = 2sin θ

∫  [ 4 ] 2sin q

4sin q

p

r4 ​   ​​​4sin q​  ​dq r  dr dq = ​  ​  ​  ​​ ​ ​​r dr dq = ​  ​  ​ ​ __  

3

 

0 2sin q

0

p



__ = ​ 1 ​ ​∫  ​  ​(256 sin4 q -p__ 16 sin4 q ) dq ​    ​ p 4 0 2

= 60 ​∫  ​  ​ sin4 q dq = 120 ​∫ ​   ​​sin4 φ d φ, q = 2φ 0

3.1 __ __ . = 120 · ​ ___ ​  · ​ p   ​= 45 ​ p   ​ 4.2 2

M12_Baburam_ISBN_C12.indd 7

2

______ 1 a​√cos 2q  ​ __   2 2  



​dq

1 __

    = ​∫  ​ ​​a [(1 + cos 2q )​ 2 ​ - 1] dq θ =0

∫  ∫ 

 

p ​ __ ​  4

θ=π 0

∫∫

   __ 1

(a + r  )​   ​  __     = ​ 1 ​ ∫​    ​ ​​ ________ ​   ​     ​​ ​  2 - ​ p__  ​ __ ​ 1 ​  4 2 0  

3

 

_____ a​√cos2q ​ 

4

p

r   ​  ​ _______ ​  ______ dr dq 2 ​ a + r2 ​  √

0

    = ​∫   ​ ​​ __ ​ 1 ​      ​∫  ​  ​2r (a2 + r2) ​ 2 ​ dr dq __  ​ 2 0 - ​ p

r = 4sin θ

Therefore,

x

  π 4

EXAMPLE 12.14 Calculate ∫∫ r 3dr dq over the area included between the circles r = 2sin q and r = 4 sin q.

y

(a, 0)

0

2

0

x

p - ​ __ ​  4

p ​ __ ​  4

1 __

    = a ​∫   ​ ​​a [(2 cos2 q )​ 2 ​ - 1] dq p - ​ __ ​  4

__ ​ p  ​  4

__     = a ​∫ p__ ​ ​​ (​√2 ​ cos q - 1) dq - ​   ​  4 p __ ​  ​  4

__     = 2a ​∫p__ ​  ​​(​√2 ​ cos q - 1) dq - ​   ​  4

__ ​ p_ ​ __     = 2a ​​ ​√2 ​ sin q - q ​ 04 ​ ​= 2a ​  1 - ​ p   ​  ​. 4

(

)

12/9/2011 4:47:55 PM

12.8  n  chapter twelve EXAMPLE 12.16 Evaluate ∫∫ r 3dr dq over the area included between the circles r = 2a cos q, r = 2b cos q, where b < a. Solution.  The region of integration between the given circles is shown in the following figure: y π  2

Solution.  The region of integration is shown in the following figure. In this region r varies from __ 0 to 2acos q, whereas q varies from 0 to ​ p   ​. 2 y

r  2acos 

r  2a cos  r  2b cos θ

0

 0

Therefore,

  π 2

In the region of integration, r varies from 2b cos __ __ q to 2a cos q, whereas q varies from - ​ p   ​to ​ p   ​. 2 2 Therefore, p__ ​   ​ 2

[ 

2a cos q

]

∫∫r  dr dq = ​∫   ​ ​​​     ​∫  ​ ​ r3dr  ​dq 3

 

p __ - ​    ​ 2

 

2b cos q

__ ​ p   ​ 2



∫  [ 4 ]2b cos q

r4 ​   ​2a cos q​ dq = ​ ​  ​​​​ ​ __ p __ 2

- ​    ​



__ ​ p   ​ 2

__ = ​ 1  ​ ∫​   ​  ​​ [16a4 cos4 q - 16b4 cos4 q] dq 4 - ​ p__  ​  

2

__ ​ p   ​ 2



x

0

x

= 4 ​∫  ​  ​​ (a4 - b4) cos4 q dq  

p __ 2 p __ ​    ​ 2

- ​    ​



= 8 ​∫ ​   ​​(a4 - b4) cos4 q dq



3.1 __ 3p = 8 (a4 - b4)· ___ ​   ​  . ​ p   ​= ​ ___ ​ (a4 - b4). 4.2 2 2

0

EXAMPLE 12.17 2a3 Show that ∫∫r 2 sin q dr dq = ​ ___ ​ ,  where R is the 3 region bounded by the semi-circle r = 2acos q, above the initial line.

M12_Baburam_ISBN_C12.indd 8

[ 

]

__ ​ p   ​ 2      2acos q

∫​∫   ​  r2 sin q dr dq = ​∫  ​  ​​​ ​∫  ​   

 

R

0

0 __ ​ p   ​ 2

​r 2 sin q dr  ​dq  

  

__3 2acos q = ​∫  ​  ​​sin q ​​ r   ​   ​​​ ​  ​dq 30 0



__ ​ p   ​ 2

8a3 = ​ ___ ​  ∫​   ​  ​​sin q cos3 q dq 3 0 3 8a 2a3 = ​ ___ ​  . ___ ​  2   ​ = ​ ___ ​  . 3 4.2 3



 



12.5  CHANGE OF VARIABLES IN A DOUBLE INTEGRAL Regarding the change of variables in a double integral, we have the following theorem stated here without proof. Theorem 12.3  Let D be a domain in ℜ2 and f : D→ ℜ be integrable over D. Suppose that D be mapped on to a set A of the uv-plane by the transformation x = φ (u, v) and y = y (u, v) , where φ and y have continuous partial derivatives in A and the Jacobian ∂ (x, y) ______ ​     ​≠ 0 for all (u, v) ∈ A. Then, ∂ (u, v)  

∫​∫ ​f (x, y) dx dy = ∫​∫  ​  f (φ (u, v), y (u, v)) ​ J ​dudv. D

A

Deduction. If there is a change in variable from Cartesian- to polar coordinates, then x = rcos q, y = rsin q, and

12/9/2011 4:47:59 PM

Multiple integralS  n 12.9 ∂x __

∂x  ​__  ​ cos q r sin q ∂q = ∂y ​ ___   ​ sin q r cos q ∂q

 ​  ​  ∂r J= ∂y ​ __ ​  ∂r

__ ​ p   ​ 2 ∞

∞ ∞

​∫  ​ ​ ​∫  ​ ​e 

-(x2

 

 

+ y2)

2 ∂ (x, y) dy dx = ​∫ ​   ​​ ​∫  ​ ​e -r ​ ______   ​dr dq ∂ (r, q ) 0 0  

0 0

__ ​ p   ​ 2 ∞

= r (cos2 q + sin2 q ) = r. Therefore,



= ​∫ ​  ​​​∫  ​ ​​re -r dr dq

∫​∫  ​  f (x, y) dx dy = ∫​∫  ​ f (r cos q , r sin q )r dq dr.



dt = ​∫ ​  ​​​∫  ​ ​e -t· __ ​   ​ dq, r 2 = t 2 0 0

 

 

 

D

 

A

EXAMPLE 12.18 Calculate the integral (x - y)2  ​  dx dy ∫∫​  ______ x 2 + y 2 over the circle x 2 + y 2 ≤ 1. Solution.  The value of the given integral is four times the value of the integral taken over the positive quadrant of the circle x 2 + y 2 = 1. Substituting x = rcos q and y = rsin q, the given integral is equal to __ ​ p   ​ 1 2

∂ (x, y) 4​∫  ​​ ​​∫  ​  ​​__ ​ 1   ​(rcos q - rsin q )2 ​ ______   ​dr dq 2 ∂ (r, q ) 0 0 r 

__ ​ p   ​ 2

__ ​ p   ​ 2

___ __ . = ​ -1  ​ ∫​   ​ ​​(0 - 1) dq = ​ p   ​ 2 0 4 On the other hand, ∞∞

​∫  ​ ​∫  ​ ​e -(x

2

 

∞

(1)

∞

dy dx = ​∫  ​ ​e -x dx ​∫  ​ ​e -y dy

+ y2)

2

 

2

 

00

[ 

0

 

0

∞

]

2

= ​ ​∫  ​ ​e  dx  ​ .

Hence, (1) implies

0

2

-x 

 

__ ​___ p  ​ √ ​∫  ​ ​ e  dx = ​   ​ . 2 0 ∞

-x2

__ ​ p   ​ 2

 

= 4 ​∫  ​​  r ​∫  ​  ​​(cos2 q + sin2 q - 2sin q cos q ) dr dq  

0 __ ​ p   ​ 2

  

e -t ∞ __ = ​ 1 ​ ​∫ ​  ​​​​ ___  ​ ​​​ ​ ​dq 2 0 -1 0



 

1

2

 

0 0 __ ​ p   ​ 2 ∞

0

[  ] 1

√

= 4 ​∫  ​ ​​(1 - sin 2q ) ​ ​∫  ​​  rdr  ​dq  

0

0 __ ​ p   ​ 2

= 2 ​∫  ​ ​​(1 - sin 2q ) dq 0

[ 

]

__ ​ p   ​

cos 2q 2 = 2​​ q + ​ _____  ​    ​​ ​   ​= p - 2. 2 0 EXAMPLE 12.19 ∞∞

Evaluate ∫​   ​​ ​∫  ​​ e -(x

2

+ y2)

dy dx by changing topolar __ ∞  0  0 ​√___ p ​  ​ -x2 coordinates. Hence, deduce that ∫​   ​​ e dx = ​   ​ . 2  0 Solution. In the given integral, both x and y vary from 0 to ∞. Hence, the region of integration is xy-plane. Changing to polar coordinates by substituting x = r cos q and y = r sin q, we get x2 + y2 = r2; and in the region of integration, r varies from 0 to ∞ and q __ varies from 0 to ​ p   ​. Thus, 2  

EXAMPLE 12.20 _______________ a2b2 - b2x2 - a2y2 Evaluate ∫∫ ​ ​ _______________          ​ ​dx dy over the a2b2 + b2x2 + a2y2 2 x2 y positive quadrant of the ellipse ​ __2  ​+ ​ __2  ​= 1. a b y x Solution.  Substituting __ ​ a ​ = X and __ ​   ​ = Y, the b problem reduces to the evaluation of ∫∫ ab​ ___________ 1 - X 2 -Y 2  ​ __________  ​ ​    dx dy over the positive quadrant of 1 + X 2 + Y 2 the circle X 2 + Y 2 = 1. Substituting X = r cos q and

 

 

√

Y = r sin q, we have dXdY = rdr dq. In the region of integration r varies from 0 to1 and q varies from __ 0 to ​ p   ​Hence, the given integral reduces to 2 __ ​ p   ​ 2 1 _______ - r 2  ​  ​ ab ​∫ ​  ​​​∫  ​ ​​ ​  1_____     rdr dq 1 + r 2 0 0  

√

__ ​ p   ​ ______ 2 2 1 r  _____ = ab ​∫  ​​ ​ ​    ​ ​    rdr ​∫ ​  ​​dq 1 + r 2 0 0

1

 

M12_Baburam_ISBN_C12.indd 9

√

12/9/2011 4:48:04 PM

12.10  n  chapter twelve 1 _____ abp - r 2  ____ = ​   ​  ∫​   ​ ​​ ​ 1_____  ​ ​rdr 2 0 1 + r 2 __ ​ p   ​ 2 _______ 1 - sin t __ abp ____ = ​   ​  ​∫ ​  ​​​ ​ _______   ​ ​ 1 ​ cos t dt, r 2 = sin t 2 0 1 + sin t 2 __ ​ p   ​ _______ 2 _______ ​√1_______ - sin t   ​ 1_______ - sin t _________ p ab ____ = ​   ​   ​∫  ​ ​​​ ​     ​ ​        ​cos t dt 4 0 1 + sin t ​√1 - sin t ​   

√

√

__ ​ p   ​ 2

√

polar coordinates.

______ Solution.  The______ limits of y are from 0 to √ ​ 2x - x 2   ​ 2 2 2 . But y = √ ​ 2x - x    ​ implies y  = 2x - x  or x 2 + y 2 = 2x. Thus, the region of integration is bounded by x = 0, x = 2, y = 0, and x 2 + y 2 = 2x. Changing to polar coordinates x 2 + y 2 = 2x transforms to r 2 = 2rcos q or r = 2cos q or r = 2cos q. y

- sin t p ab 1_______  ​   ​∫ ​  ​​​      ​  cos t dt = ​ ____ 4 0 cos t

r  2cos 

__ ​ p   ​ 2

 π 2

p ab = ​ ____  ​   ​∫ ​  ​​(1 - sin t) dt 4 0 __ ​ p   ​ p ab p ab ____ p __ = ​ ____  ​   [t + cos t​]​02​   ​= ​   ​   ​ ​    ​- 1 ]​ 4 4 [2 p ab = ​ ____  ​   (p - 2). 8 Remark 12.1  Substituting a = b ___________ = 1, the problem   1​  - x 2 - y ​ ​ 2 dx reduces to the evaluation of ∫∫ ​__________       dy 2 2 1 + x  + y  over the positive quadrant of the circle x2 + y2 = 1. __2 p The value of the integral in that case is ​ p   ​  - ​ __  ​. 8 4 EXAMPLE 12.21

√

Evaluate ∫  

∫ sin p (x2 + y2 ) dx dy.

x2 + y2 ≤ 1

Solution.  Changing to polar form yields

∫  ∫sin p (x2 + y2) dx dy x2 + y2 ≤ 1

2p 1

= ​∫  ​  ​ ​∫  ​​ sin (p r ) rdr dq



2

 

 

0 0 2p 1



dt = ​∫  ​  ​ ​∫  ​​  sin (p t) ​ __ ​ dq 2 0 0



cos p t 1 = ​   ​ ∫​   ​  ​ ​ - ​ _____   ​ ​​​ ​ ​dq p    20 0

 

1 __

 

2p

 

2p

1 ___



= - ​    ​ ​∫  ​  ​[cos p - cos 0] dq 2p 0



___ = ​  2  ​ ​∫  ​  ​ dq = 2. 2p 0

x2

 0 x

0

For the region of integration, r varies from 0 to __ . 2 cos q, whereas q varies from 0 to ​ p   ​ There2 fore, _____ 2  ​ 2 ​√2x - x 

x ​∫  ​ ​  ​​∫  ​  ​​_______ ​  ______      ​dy dx 0 0 ​√x2 + y2 ​  __ ​ p   ​ 2 2 cos q

r cos q = ​∫  ​ ​​  ​∫  ​  ​ ______ ​ r dr dq r    0 0 __ ​ p   ​ 2

2 cos q

0

0

= ​∫  ​ ​​cos q ​∫  ​  ​ r dr dq  

__ ​ p   ​ 2

  

2cos q

r2 ​ ​ ​  = ​∫  ​ ​​cos q ​​​ __ 20 0

​dq

__ ​ p   ​ 2

= 2 ​∫  ​ ​​cos3 q dq = 2 __ ​ 2 ​ = __ ​ 4 ​ . 3 3 0

 

2p

 

p a

EXAMPLE 12.22

_____ 2 ​√2x - x 2 ​

Evaluate ∫​     ​​ ​ ∫​   ​       

0

0

M12_Baburam_ISBN_C12.indd 10

EXAMPLE 12.23 Transform to the Cartesian form and hence,

x ​_______ ​​  ______      ​ dy dx by changing to ​√x2 + y2   ​

evaluate the integral ∫​   ​  ​​∫  ​​ r3 sin q cos q dr dq. 0

0

 

12/9/2011 4:48:06 PM

Multiple integralS   n 12.11 Solution.  We are given that p a

I = ​∫  ​  ​​∫  ​​  r3 sin q cos q dr dq. 0

0

Put x = r cos q and y = r sin q so that dx dy = r dr dq. The region of integration is shown in the following figure: y

one. In order to exclude (0, 0) , we note that the given integral exists as the limit, when h→0 of the integral over the region is bounded by x + y = 1, x = h, and y = 0 where h > 0. The transformed region is then bounded by the lines u = 1, v = 0, and u (1 - v) = h. When h → 0, the new region of the uv-plane tends, as its limit, to the square bounded by the lines u = 1, v = 1, u = 0, and v = 0. Thus, the region of integration in the xy- and uv-planes are as shown in the following figures: y

x 2y 2  a 2

v

(0, 1) v 1 x

a

 y  1

u 0

π

0

a

0

x

0

x

(1, 0)

0

u 1

u

v0

Therefore, 1 1-x

Therefore, the Cartesian form of the given integral is a a

p a

p

  

r​  4 ​  a​​​​  dq I = ​∫ ​   ​​​∫ ​​  r3 sin q cos q dr dq. = ​∫ ​   ​sinqcos q ​__ 40 0 0 0  

 

 

p

  4

2 p q a4 a4 sin __ = ​   ​ ​  ​  ​ sin q cos q dq = __ ​   ​ ​​  ____  ​   ​​​ ​ ​= 0.

∫  4



 

0

2



0

EXAMPLE 12.24 Use the transformation x + y = u and y = uv to 1

1-x

y  ____ x  +  y

e - 1. show that ∫​   ​ ​  ​​∫  ​ ​  ​e ​     ​ dy dx = ​ ____  ​    2 0 0 Solution.  We have x = u - y = u - uv = u (1 - v) and y = uv. Therefore, ∂x ∂x __ ​   ​  ​ __ ​ 1 v u ∂(x, y) ∂r ∂v ______ ​     ​  = = ∂(u, v) ∂y ∂y ___ v u ​   ​ ___ ​   ​  ∂u ∂v = u (1 - v) - (-uv) = u. The Jacobian vanishes when u = 0, that is, when x = y = 0, but not otherwise. Also the origin (0, 0) corresponds to the whole line u = 0 of the uv-plane so that the correspondence ceases to be one-to-

M12_Baburam_ISBN_C12.indd 11

 

 

 

 

00 1

1

1

[  ]

u2 ​   1​​​ ​​ dv = ​∫  ​​ e ​∫  ​​ u du dv = ​∫  ​​ ev​ ​ __ 2 0 0 0 0 1 __ = ​ 1 ​ ∫​   ​​ ev dv = __ ​ 1 ​ [ev]​ 10​   = __ ​ 1 ​ (e - 1). 20 2 2 v



-a 0

 __ uv

​∫  ​ ​ ​∫  ​ ​e ​     ​ dy dx = ​∫  ​​ ​​∫  ​​ e ​ u ​ .u du dv 0 0

I = ​∫   ​ ​∫  ​​ xy dx dy.

Further,

11

y  ____ x  +  y

 



 

 

 

EXAMPLE 12.25 a

a

x dx dy Evaluate ∫​   ​​  ​∫  ​  ​ ______  ​by changing into polar 2 2  0 y x + y coordinates.  

 

Solution.  The region of integration is shown in the following figure: y

yx ya

xa

0

 0

x

12/9/2011 4:48:07 PM

12.12  n  chapter twelve The region is bounded by x = y, x = a, y = 0, and y = a. Changing topolar coordinates, we have x = r cos q, y = r sin q, and dx dy = r dr dq. Further, in the region of integration q varies __ from 0 to ​ p   ​. Also, x = a implies r cos q = a or a _____ a 4 _____  ​. r = ​ cos  q    ​. Therefore, r varies from 0 to ​ cos q  Hence,

EXAMPLE 12.27 Change the order of integration in the integral ______ 2  2  ​ a ​√ a - y  

__ _____ ​ p   ​ a   ​  4 ​ cos q

a a

x dx dy r cos q ​∫  ​​  ​∫  ​  ______ ​  2 2 ​  = ​∫ ​  ​​  ​∫ ​   ​​​ ______  ​   r dr dq x + y r 2 0 y 0 0  

are variable, a change in the order of integration requires a change in the limits of integration. Some integrals are easily evaluated by changing the order of integration in them.

 

I = ​∫    ​  ​∫  ​  ​ f (x, y) dx dy.  

-a

 

__ ​ p   ​ 4

__ ​ p   ​ 4

a ____ ap = ​∫ ​  ​​cos q [r​]​​ 0cos q ​      ​​dq = a ​∫ ​  ​​dq = ​ ___ ​ . 4 0 0



Solution.  The region of integration is bounded by y = - a, y = a, x = 0, and x 2 + y 2 = a 2. We have

 __n

]

I = ​∫   ​ ​​ ​∫  ​  ​f (x, y) dx  ​dy.  

Evaluate ∫∫ xy (x 2 + y 2)2​  ​ dx dy over the positive octant of the circle x 2 + y 2 = 4, supposing n + 3 > 0.

[ 

______ 2 2  ​ a     ​√ a - y  

-a

EXAMPLE 12.26

0

 

0

Thus, in the given form, we first integrate with respect to x and then with respect to y. y

Solution.  The region of integration is bounded by x = 0, x = 2, y = 0, and y = 2. Changing to polar coordinates, we have x = r cos q and y = r sin q and so, rdq dr = dxdy. The limits of integration in the first quadrant of the given circle are now r = 0 to __ r = 2 and q = 0 to q = ​ p   ​. Hence, 2

S

 __ n

∫∫ xy (x 2 + y 2) ​ 2 ​ dx dy

0

__ ​ p   ​ 2 2

(a, 0)

x

 __ n

= ​∫  ​ ​​​∫  ​​  r cos q.r sin q (r2) ​ 2 ​ . r dr dq  

0 0

[ 

__ ​ p   ​ 2

]

2

= ​∫  ​ ​​sin q cos q ​ ​∫  ​ ​ r n + 3 dr  ​dq  

0

0

__ ​ p   ​ 2

[ 

]

r n + 4  ​  2​​​ ​​ dq = ​∫  ​ ​​sin q cos q ​ ​ _____ n +4 0 0 __ ​ p   ​ 2

1.1 2n + 3 . 2n + 4  ​ = ​∫  ​ ​​sin q cos q dq = ​ _____ 2n + 4  ​   . ___ = ​ _____ ​   ​  = ​ _____   ​  n+4 0 n+4 2 n+4

R

On changing the order of integration, we first integrate with respect to y, along______ a vertical ship 2 RS, which extends from y = ​ a - x2   ​ to y =​ √ ______ 2 2  ​. To cover the whole region of integra√a - x   tion, we then integrate with respect to x from x = 0 to x = a. Thus, ______ ​√a2 - x2   ​

a

I = ​∫   ​  dx ​∫______   ​ ​f (x, y) dy  

12.6  CHANGE OF ORDER OF INTEGRATION We have seen that, in a double integration, if the limits of both variables are constant, then we can change the order of integration without affecting the result. But if the limits of integration

M12_Baburam_ISBN_C12.indd 12

-a



 

[ 

- ​√a2 - x2     ​ ______ ​ a2 - x2 ​  √ a

]

= ​∫   ​   ​​   ​∫______   ​ ​f(x, y) dy  ​dx. -a

 

- ​√a2 - x2   ​

12/9/2011 4:48:10 PM

Multiple integralS  n 12.13 EXAMPLE 12.28 1 2-x Change the order of integration in I = ​∫  ​ ​​  ​∫ ​ ​ ​xydx 0 x dy and hence, evaluate the same.



2

Solution.  For a given integral, the region of integration is bounded by x = 0, x = 1, y = x2, (parabola), and the line y = 2 - x. Thus, the region of integration OABO is as shown in the following figure: y

x2  y

∫  [  1



2]

__ ​ y ​ 

2

[  2 ]

yx x2 √ = ​  ​ ​ ​​ y ​ __ ​   ​ ​  ​dy + ​  ​​  ​ ​ ___ ​  ​​​ ​  ​  dy  

0

1

∫   

0

1

2

2 2-y

__ = ​ 1 ​ ​∫  ​​​ y2dy + __ ​ 1 ​ ​∫  ​​  y (2 - y)2dy 20 21 3 1 y y2 y4 y3 2 __ = ​ 1 ​ ​​ ​ __ ​   ​​ ​​ + __ ​ 1 ​ ​ ​ __ ​ + 4 ​ __ ​ - 4 ​ __ ​   ​​ ​​  2 3 0 2 4 3 1 2 5 3. __ = ​ 1 ​ + ___ ​    ​ = __ ​   ​  6 24 8  

[  ] [ 

]

EXAMPLE 12.29 Changing the order of integration, find the value

B

∞ ∞

e -y of the integral ∫​   ​ ​ ​∫ ​  ​ ___ y ​ dy dx.  

 

0 x

A(1, 1)

C

x  y  2

0

x

Solution.  The region of integration is bounded by x = 0 and y = x. The limits of x are from 0 to ∞ and those of y are from x to ∞. The region of integration is shown in the following figure: y

In the given form of the integral, we have to integrate first with respect to y and then with respect to x. Therefore, on changing the order of integration, we first integrate the integrand, with respect to x and then, with respect to y. The integration with respect to x requires the splitting-up of the region OABO into two parts OACO and the triangle ABC. For the subregion OACO, __ the limit of integration are from x = 0 to x=√ ​ y ​ and y = 0 to y = 1. Thus, the contribution to the integral I from the region OACO is __

[  ] ​√y ​ 

1

I1 = ​∫  ​ ​ ​ ​∫  ​ ​ xy dx  ​dy.  

0

 

0

For the subregion ABC, the limits of integration are from x = 0 to x = 2 - y and y = 1 to y = 2. Thus, the contribution to I from the subregion ABC is 2 2-y

[  ]

I2 = ​∫  ​​  ​ ​∫  ​  ​xy dx  ​dy.  

1

 

0

Hence, on changing the order of integration, we get __

[  ] [  ] ​√y ​ 

1

2-y

2

I = ​∫  ​​  ​ ​∫  ​ ​ xy dx  ​dy + ​∫  ​ ​ ​ ​∫  ​  ​xy dx  ​dy  

0

 

0

M12_Baburam_ISBN_C12.indd 13

0

 

1

 

0

yx

R

S

x

0

On changing the order of integration, we first integrate the integrand, with respect to x, along a horizontal strip RS, which extends from x = 0 to x = y. To cover the region of integration, we then integrate, with respect to y, from y = 0 to y = ∞. Thus, ∞

[  ] y

∞

e -y e-y y I = ​∫  ​ ​ ​ ​∫  ​​  ​  __  ​ dx  ​dy = ​∫  ​ ​ ​ ___ y y ​ [x​]​0 ​​ dy 0 0  

 

0

∞

 

[  ]

∞  = ​∫  ​ ​e-y dy = [-e-y]​​∞0​ ​= - ​ __ ​ 1y  ​ ​ ​ ​ e  0 0  = - (0 -1) = 1.  

EXAMPLE 12.30 Change the order of integration in the integral

12/9/2011 4:48:13 PM

12.14  n  chapter twelve ___ 4a 2​√ ax ​ 

_____ 1 ​√2 -x2 ​ 

​∫  ​ ​​  ​∫ ​ ​ dy dx and evaluate.

x dy dx ​∫  ​ ​  ​​∫  ​  ​_______ ​  ______     ​ x 0 ​√x2 + y2 ​ 

 

 

x2 ___

0

___ 4a2​√ ax ​ 

​      4a ​

Solution.  The given integral is ∫​   ​ ​​   ​∫ ​ ​ dy dx. The inx2 0 ___ ​      4a ​

tegration is first carried out with respect to y and then with respect to x. The region of integration is bounded by x = 0, x = 4a, and the parabolas x2 = 4ay and y2 = 4ax. Thus, the region of integration is as shown in the following figure: Y

by changing the order of integration. Solution.  For the given integral, the region of integration is bounded by x = 0, x = 1, y = x, and the circle x2 + y2 = 2. Thus, the region of integration is as shown in the following figure: y

yx

x 2  4ay

xa B  (0, √2)

A(4a, 4a) R

A(1, 1)

C

S

X

0

x

0 y

2

4ax

The coordinates at the point of intersection of the parabolas are A (4a, 4a). On changing the order of integration, we first integrate the integrand with respect to x along the y2 ___ horizontal strip RS, which extends from x = ​     ​to ____ 4a ___ x=√ ​ 4ay ​ = 2​√ay ​. To cover the region of integration, we then integrate with respect to y from y = 0 to y = 4a. Thus, ___ 4a 2​√ ax ​ 

4a

[  ] ___ 2​√ay ​ 

4a

___ ​ 2​√ay ​     ​∫  ​ ​​  ​∫  ​dy dx = ​∫  ​ ​​   ​∫ ​ ​dx  ​dy = ​∫  ​ ​[x​]​y ​  ​dy  

 

0

x2 ___

​    ​  4a

0

 

 

0

y ___ 2

[ 

​      4a ​

[ 

2

​ ___  ​  4a

]



4a __ __3 ___ ___ y2 2​√a ​y  ​ 2 ​  ____ y3 ______ = ​∫  ​ ​ 2​√ay ​ - ​    ​ ​d y = ​ ​   ​   - ​    ​ ​ ​  ​ 4a 3 12a __ 0 ​   ​  2 0



32a2 ____ 16a2 ____ 16a2 . = ​ ____  ​   - ​   ​   = ​   ​    3 3 3

4a

 

EXAMPLE 12.31 Evaluate the integral

M12_Baburam_ISBN_C12.indd 14

]

 

  

The point of intersection of the circle x2 + y2 = 2 and x = y is A (1, 1). Draw AC || OX. Thus, the region of integration is divided intotwosubregions ABCA and ACO. On changing the order of integration, we first integrate with respect to x, along the strips parallel to the x-axis. In the subregion ABCA, the strip extends from _____ x = 0 to x = ​√ 2 - y2   ​ To cover the subregion, we then __ integrate with respect to y from y = 1 to y = ​√ 2 ​ . Thus, the contribution tothe integral due to this subregion is __ ​√2 ​ 

[ 

]

_____ ​√2 - y2 ​ 

x I1 = ​∫  ​ ​ ​   ​∫  ​  ​_______ ​​  ______      ​ dx  ​dy. 2 1 0 √ ​ x + y2 ​   

 

On the other hand, in the subregion ACO, the strip extends from x = 0 to x = y. To cover this subregion, we then integrate with respect to y from y = 0 to y = 1. Thus, the contribution to the integral by this subregion is 1

[ 

y

]

y I2 = ​∫  ​​  ​ ​∫  ​ _______ ​​  ______      ​dx  ​dy. 2 0 0 √ ​ x + y2 ​   

12/9/2011 4:48:16 PM

Multiple integralS  n 12.15 Hence, the given integral is equal to

[  [ 

x _______ ______

 

]

y

1

 

2

__ ​√2 ​ 

[ 

_____ 2  __ 1 ​√2 -y  ​  2 2  0

]

 

1

1

[ 

 __ 1

]

y

​dy + ​∫  ​​​ ​​ (x2 + y2) ​ 2 ​   ​​0 ​  ​ dy

= ​∫  ​ ​​​ (x2 + y ) ​   ​   ​​ ​ 



__ ​√2 ​ 

0

__ ___ = ∫​   ​ ​ ( ​​ √​ 2 ​ - y )​dy + ​∫  ​ ​ ​( ​√2y ​ - y )​dy



1

 

1

[ 

0

]

__ 2 ​√2 ​ 

[  ]

___ y __ y2 1 = ​ ​√2y ​ - ​ __ ​   ​​​ ​  ​ + (​√2 ​ - 1) ​ ​ __ ​   ​​​ ​​  2 0 __ 2 1 2______ -√ ​ 2 ​  1 ___ = ​   ​   = 1 - ​  __   ​ . 2 ​ 2 ​  √



EXAMPLE 12.32

_____  ​ 3 ​√4 - y 

Evaluate the integral ​∫  ​ ​  ​ ​∫  ​  ​(x + y) dx dy by  

0

1

changing the order of integration.

[ 

[  ( 

]

)

( 

)

2a ​√2ax ​ 

I = ​∫  ​  _______ ​ ​∫    ​ ​φ (x, y)dy dx.  

0 ​√2ax - x2 ​ 

Solution.  The region of integration is bounded by x = 0 and x = 2a, the circle x 2 + y 2 = 2ax, and the parabola y 2 = 2ax. The equation of the circle can be written as (x - a)2 + y 2 = a2 and so, has the center at (a, 0). The region of integration is as shown in the following figure: y

2

y

E

y3

]

2

EXAMPLE 12.33 Change the order of____integration in

y (1, 3)

]

x2 x4 x5 4x3 = ​​ 4 ​ __ ​ - ​ __ ​ + 8x + ​ ___  ​ - ​ ___ ​    ​​ ​​  2 4 10 3 1 32 ___ 32 ___ = ​ 8 - 4 + 16 + ​   ​ - ​   ​   ​ 10 3 - ​ 2 - __ ​ 1 ​ + 8 + ___ ​ 1  ​ - __ ​ 4 ​   ​= ____ ​ 241 ​ . 4 10 3 60

Solution.  The region of integration is bounded by x = 1, x 2 = 4 - y, y = 0, and y = 3, as shown in the following figure:

(0, 4)

[ 

2 2 4-x

y  __

x4 = ​∫  ​​​ ​ 4x - x3 + 8 + ​ __ ​ - 4x2  ​dx 2 1

x    + ​∫  ​​​ ​ ​∫  ​​​   _______ ​  ______  ​dx  ​dy 0 0 ​ x2 + y2 ​  √



]

2

= ​∫  ​ ​ ​  ​∫  ​  ​​(x + y) dy  ​dx = ​∫  ​​​ ​​ xy + ​   ​   ​​ ​  ​dx 2 0 1 1 0

 ​dx  ​dy I = I1 + I2 = ​∫  ​ ​​​ ​∫  ​  ​​​  2   2   1 0 √ ​ x + y  ​   

[ 

4 - x 2

2

]

__ ____2 ​√2 ​  ​√2-y  ​ 

ax 2 D

A(2a, 2a)

F

2

x

x  2a

4

y

0

(1, 0)

(2, 0)

0

x

On changing the order of integration, we first integrate the integrand, with respect to y, by taking the strip parallel to the axis of y. In the region of integration, y varies from 0 to 4 - x2 and x varies from 1 to 2. Therefore, _____ ​ 3 ​√4 - y  

​∫  ​ ​  ​​∫  ​  ​(x + y) dx dy  

0

0

M12_Baburam_ISBN_C12.indd 15

C

B(2a, 0)

x

We divide the region of integration intothree parts by drawing the line EDF through D parallel to the x-axis. On changing the order of integration, we first integrate the integrand, with respect to x and then integrate the resultant integrand, with respect to y. So, we draw horizontal strips parallel to the xaxis. y2 In the subregion OEDO, x varies from ​ ___   ​ to ______ 2a a - ​√a2 - y2   ​and y varies from 0 to a. Thus, the

12/9/2011 4:48:20 PM

12.16  n  chapter twelve

[ 

contribution to the integral due to this subregion ______ is a - ​ a - y  ​  a

√

2

]

2

I1 = ​∫  ​​  ​  ​∫ ​  ​​φ (x, y) dx  ​dy.  

0

 

y ___ 2

​      2a ​

∞ x

[ 

]

2a

I2 = ​∫  ​​ ​    ​∫   ​ φ (x, y) dx  ​dy, 0

 

______ a + ​√a2 - y2 ​ 

[ 

2a

 

Hence,

 

y ___ 2

​    ​  2a

______ 2 2 a a- ​√ a - y  ​ 

I = ​∫  ​​​  0

a

]

y

0

2

 x - __ y 

(1)

[  ] t -_

∞

t -_ y   

∞

e ​ y ​ ​  ​ ​ xe ​   ​dx = ​   ​ ​ ​ e ​  ​dt = __ ​ 1 ​ ​ ​ ____   ​ ​ ​  ​​ = __ ​ 1 ​ y e - y. 2 y2 2 2 - __ 1 y ​ y ​

∫   

1 __

∫   

Therefore, (1) reduces to x

 

y2

∞

2

 

 

[  ]

​∫ ​  ​​φ (x, y) dx dy



y ​ ___  ​  2a 2a 2

[  ]

 

e-y __ = ​ 1 ​ ​ ​ ____   ​  ​​​ ​ ​= __ ​ 1 ​ . 2 -1 0 2 ∞

12.7  AREA ENCLOSED BY PLANE CURVES (CARTESIAN AND POLAR COORDINATES) (A) Cartesian Coordinates: The area A of the region R = {(x, y): a ≤ x ≤ b; f1 (x) ≤ y ≤ f2(x)}

 

______ a + ​√a2 - y2 ​  2a 2a

+ ​∫  ​ ​ ​∫  ​ φ (x, y) dx dy. y2 0 ___ ​    ​  2a

is given by the double integral

EXAMPLE 12.34 ∞ x

]

2

x - __ y

We first evaluate the inner integral. Substituting x2 = t, we have 2x dx = dt. When x = y, t = y2 and when x = ∞, t = ∞, Therefore,

∞

+ ​∫  ​ ​  ​ ​∫   ​ ​φ (x, y) dx dy 0

∞

x - __ ​∫  ​ ​  ​​∫  ​ ​x e  ​y ​  dy dx = __ ​ 1 ​ ​∫  ​ ​​y e - y dy 20 0 0 ∞ y____ e-y ∞ __ 1 1 __ = ​   ​ ​ ​   ​  ​​​ ​ ​+ ​   ​ ∫​   ​ ​e -y dy 2 -1 0 2 0

I3 = ​∫  ​ ​ ​ ​∫  ​  φ (x, y) dx  ​dy. 0

0 0

∞

and the contribution to the integral due to the subregion AEFA is a

[ 

∞

 x2 - __ y 

​∫  ​ ​​​∫  ​ ​​xe ​   ​dy dx = ​∫  ​ ​ ​​ ​∫  ​ ​​x e ​   ​  dx  ​dy. 

Similarly, the contribution to the integral due to the subregion DBFD is a

On changing the order of integration, we first integrate, with respect to x and then, with respect to y. Thus,

f2(x)

b

Evaluate ​∫  ​ ​​ ​∫  ​​ e ​   ​  dy dx by changing the order of

A = ​∫  ​    ​∫  ​ dy dx.

integration. Solution.  The region of integration is bounded by the lines x = 0, x = ∞, y = 0,  and  y = x. Therefore, the region of integration is as shown in the following figure:

Similarly, the area A of the region R = {(x, y): c ≤ y ≤ d; f1 (y) ≤ x ≤ f2 (y)}is given by the double integral d f (y)

 x2 - __ y

 

0 0

y

yx

 

a

2

A = ​∫  ​​    ​∫   ​ ​​dx dy. c

f1(y)

(B) Polar Coordinates: The area A of the region R = {(r, q ); α ≤ q ≤ β; f1 (q ) ≤r ≤ f2 (q )} is given by f (q ) b

R

 

f1(x)

2

A = ​∫a  ​  ​ ​∫  ​   ​​r dr dq.

S

f1(q )

Similarly, the area A of the region R = {(r, q ); r1 ≤ r ≤ r2; f1(r) ≤ q ≤ f2(r)}is given by 0

M12_Baburam_ISBN_C12.indd 16

x

12/9/2011 4:48:24 PM

Multiple integralS  n 12.17 r2

f2(r)

r1

f1(r)

4x < x2 is greater than x. Therefore, the region of integration is as shown in the following figure:

A = ​∫ ​ ​​   ​∫  ​ ​​dq r dr.

y

EXAMPLE 12.35 Find the area of a plate in the form of a quadrant 2 x2 y of the ellipse ​ __2  ​ + __ ​  2  ​ = 1. Hence, find the area a b enclosed by the given ellipse. Solution.  From the figure, we note that the required area is ______ bounded by yx = 0, x = a, y = 0, and b 2 2 __ y = ​ a ​√ ​ a - x  ​  . y

(a, 0)

Thus,

______ 2 2 a ​ a ​√ a - x  ​  b __

0 a

0 a

0

b __

 

 

0

(a, 0)

x

(a, 0)

x

​dx

______ b __ = ​ a ​​∫  ​​ ​​√a2 - x2 ​    dx 0 ______ a √a2 - x2 ​  __ b x​ a2 -1 __x ________ __ = ​ a ​​ ​   ​    + ​   ​ sin ​ a ​ ​ ​  2 2 0 2 b a __ = ​ a ​​ ​ 0 + ​ __ ​ sin-1 1  ​- 0  ​ 2 ba -1 ba __ ___ = ​   ​ sin 1 = ___ ​   ​  ·​  p   ​ 2 2 2 p ab = ​ ____  ​   sq units. 4 Hence, the total area enclosed by the given ellipse is four times the area enclosed by the plate in the form of one quadrant = p ab sq units.

 

[  [ ( 

) ]

Thus, the required area lies between y = x, y = 4x - x2, x = 0, and x = 3. Therefore,

[  ] 4x - x2

3

[ 

x

 

0

​

3

]

2

 

0

27 9 3x2 x3 3 27 ___ = ​ ​ ___ ​  - ​ __ ​   ​​​ ​​ = ___ ​   ​ - ​   ​ = __ ​   ​ . 2 3 0 2 3 2

EXAMPLE 12.37 Find the area lying between the parabola y2 = 4ax and x2 = 4ay. Solution.  Solving the equation of the given parabola, we have O(0, 0) and A(4a, 4a) as the points of intersection. The region of integration is shown in the following figure: y

]

EXAMPLE 12.36 Find the area lying between the parabola y = 4x - x2 and the line y = x. Solution.  The parabola passes through the origin. Solving y = 4x - x2 and y = x for x, we get x = 0 and x = 3. Thus, the curves y = 4x - x2 and y = x intersect at x = 0 and x = 3. When 0 < x < 3,

M12_Baburam_ISBN_C12.indd 17

A(3, 3)

x

 

______ 2 - x2 ​ 

A = ​∫  ​​    ​∫ ​   ​dy dx = ​∫  ​​ ​​ ​[ y ]​ ​0a​  ​√a

x

0

0

x2 a2  y2

0

4x

4x - x A = ​∫   ​ ​ ​​ ​∫  ​  ​dy  ​dx = ​∫  ​​  ​​[ y ]​x​  ​dx = ​∫  ​​  (3x - x2) dx

b2 1

(a, 0)

y

3

x2 a2  y2

b2 1

yx 2

x 2  4ay

A(4a, 4a)

x

0

[  ]

y 2  4ax

Therefore, the required area is 4a

___ 2​√ax ​ 

4a

x2 ___

0

​

___ 2​√ax ​ 

A = ​∫  ​ ​​ ​∫ ​ ​ dy  ​dx = ​∫  ​ ​​​[ y ]​___ ​  ​ x  

0



    ​  4a 4a ​

[ 

2

​ 4a  ​ 

]

___ x2 = ​∫  ​  ​ 2​√ax ​ - ​ __   ​  ​dx 4a 0

12/9/2011 4:48:28 PM

12.18  n  chapter twelve EXAMPLE 12.39 Find the area lying inside the circle r = a sin q and outside the cardioid r = a (1 - cos q ).

4a

[  ]

 

 

 __ 3 4a

[  ]

4a __ x ​ 2 ​  x3 = 2​√a ​ ​ __ ​   ​   ​​​ ​  ​- ___ ​ 1  ​ ​ ​ __ ​   ​ 3 4a 3 0 __ ​   ​  2 0 3 __ __ = __ ​ 4 ​ √ ​ a ​ ​ 8a ​ 2 ​   ​- ____ ​  1   ​ (64a3) 3 12a 32a2 ____ 16a2 ___ 16 ____ = ​   ​   - ​   ​   = ​   ​ a2. 3 3 3



Solution.  We have r = a sin q and r = a(1 - cos q ). Therefore, a sin q = a (1 - cos q ), which yields sin q + cos q = 1 or sin2q + cos2q + 2 sin q cos q = 1 or sin 2q = 0. Hence, 2q = 0 and p and so, q = __ 0 or ​ p   ​. Further, from the region of integration, it 2 is clear that r varies from a (1 - cos q ) to a sin q.

(  )





y

EXAMPLE 12.38 Find the area of the cardioid r = a(1 + cosq). Solution.  The curve passes through the origin and cuts the x-axis at x = 2a. Clearly, q varies from 0 to p and r varies from 0 to a(1 + cos q) in the upper-half part of the integration region. y

  π/2

a(1

π 2

)



x a(

1

0

co

s

)



 π

r

r

 cos

r  a sin 

4a

__ __ = 2​√a ​ ​∫  ​ ​​√x ​ dx - ___ ​ 1  ​ ​∫  ​ ​​x2dx 4a 0 0



 0

0

(2a, 0)

Therefore,

x

__ ​ p   ​ 2

[ 

a sin q

]

__ ​ p   ​ 2

 0

[  ]

asin q

__2 ​  dr  ​r dq = ​∫ ​  ​​​ ​ r   ​   ​​​   ​​  dq 2 a (1-cos q ) 0    a(1 - cos q) 0

A = ​∫ ​  ​​​ ​∫  __ ​ p   ​ 2

The required area is given by

[ 

 a (1+cos q)

p

0

]

)

A = 2 ​∫  ​  ​  ​ ​∫  ​ 

 

0

p



p

( 

__ 2 = 4a ​∫  ​  ​ cos ​ q   ​  ​ dq 2 0 2

2

)

= 4a2 ​∫  ​  ​cos4 ​ q__  ​ dq 2 0



a = ​ __ ​ ∫​  ​  ​​[sin2 q - cos2 q + 2 cos q - 1] dq 2 0



a = ​ __ ​ ∫​  ​  ​​(-2 cos2 q + 2 cos q ) dq 2 0



1 ​  · ​ p __ __ = a2 ​ - ​ __   ​+ 1  ​= a2 ​  1 - ​ p   ​  ​ 2 2 4

a (1 + cos q )

​dq

2

2

__ ​ p   ​ 2

__ ​ p   ​ 2

[ 

] (

)

 

= 8a2 ​∫  ​ ​​cos4 φ dq, q = 2φ 0



a2 = ​ __ ​ ​∫ ​  ​​[sin2 q - (1 - cos q ) 2]dq 2 0

 

0

__ ​ p   ​ 2



 



= ​∫  ​  ​a2 (1 + cos q )2 dq

p



[  ]

p

r2 ​   ​​​ ​  ​r dr  ​dq = 2 ​∫  ​  ​​​ ​ __ 2 0 0

 

2 3 __ 3pa = 8a2 · ​ ___   ​  · ​ p   ​= ​ ____  ​   . 4.2 2 2

M12_Baburam_ISBN_C12.indd 18

EXAMPLE 12.40 Find the area of one loop of the lemniscates r 2 = a2 cos 2q. Solution.  The region of integration is shown in the following figure:

12/9/2011 4:48:29 PM

Multiple integralS  n 12.19 __ ​ p   ​ 2

y

 π 4

=a2 ​∫ ​  ​​(sec2 q + cos2 q + 2 - sec2 q ) dq  0

(a, 0)

0 __ ​ p   ​ 2

r 2  a 2 cos 2

0

(a, 0)

x

  π/4

The required area is given by

[ 

]

__ ______ ​ p   ​ 4 a​√cos 2q ​ 

__ ​ p   ​ 4

[  ]​ r2 __

______ a​√cos2q  ​ 

A = 2 ​∫ ​  ​​​ ​∫   ​  ​​r dr  ​dq = 2 ​∫ ​  ​​ ​ ​ 2 ​   ​​ ​  0

  0

0

0

​ dq

__ ​ p   ​ 4   0

a a = a2 ​∫ ​  ​​cos 2q dq = __ ​   ​ [sin 2q​]​ ​  ​= ​ __ ​ . 2 2 0



2

2

]

EXAMPLE 12.42 Find the area bounded by the parabolas y2 = 4 - x and y2 = 4 - 4x.

(  )

Solution.  The required area is given by 4 - y2

2

__ ​ p   ​ 4

[ 

5p a2 __ 2p =a2 ​∫ ​  ​​(cos2 q + 2) dq = a2 ​ __ ​ 1 ​  · ​ p   ​+ ​ ___ ​   ​= ​ _____  ​     2 2 2 4 . 0

2

[ 

]

y2 4 - y2 A = 2 ​∫  ​​​ ​ ​∫   ​  ​​dx  ​dy = 2 ​∫  ​​  ​ 4 - ​ __ ​ - ​ _____  ​    ​dy 4 4 0 4   0 -  y _____  

2

 ​   

​ 

2

4

( 

)

[ 

]

EXAMPLE 12.41 Find the area included between the curve r = a (sec q + cos q ) and its asymptote.

y3 2 3 =2 ​∫  ​​ ​ 3 - __ ​   ​ y2  ​dy = 2 ​ 3y - ​ __ ​   ​​​ ​​  4 4 0 0 =2 [6 - 2] = 8.

Solution.  The curve r = a (sec q + cos q) is symmetrical about the initial line. The equation of the asymptote is r = a sec q.

12.8 VOLUME AND SURFACE AREA AS DOUBLE INTEGRALS

y

r  a (sec  cos  )

0

a

x

(A) Volume as a Double Integral: Consider a surface z = f (x, y). Let the region S be the orthogonal projection of the portion S ′ of z = f (x, y) on the xy-plane. Divide S into elementary rectangles of area d x d y by drawing lines parallel tothe x- and y-axis. On each of these rectangles, erect a prism which has a length parallel to Oz. Then, the volume of the prism between S’ and S is z d x d y. z

r  a sec 

2a

 

z = f (x, y) S′

The required area is

[  ∫ 

]

__ ​ p   ​ 2 a (sec q + cos q)

A = 2 ​∫ ​  ​​​ 0 __ ​ p   ​ 2

​

​  ​ r dr  ​dq

[  ]

0

y

a(sec q + cos q )

r2 ​   ​​​ ​  =2 ​∫ ​  ​​​ ​ __ 0 2 a sec q __ ​ p   ​ 2

 

a sec q

​dq

=​∫ ​  ​​[a2(sec q + cos q )2 - a2 sec2 q ] dq 0

M12_Baburam_ISBN_C12.indd 19

x

S

Therefore, the volume of the solid cylinder with S as base, is composed of these prisms and so,

12/9/2011 4:48:32 PM

12.20  n  chapter twelve  

​  ​z d x d y = ∫∫ z dx dy

V =     ​lim  ​​   ​  ​dx→0 dy→0  

 

 

 

= ∫​∫  ​​ f (x, y) dx dy.



 

S

In the polar coordinates, the region S is divided intoelements of area r d r dq and so, the volume in that case is given by

(C) Surface Area as a Double Integral: Let z = ψ (x, y) be a surface bounded by a curve C. Let the projection of C on the xy-plane be bounded by Γ and let D be the domain on the xy-plane bounded by Γ. z

 

C

V = ∫​∫  ​ f (r cos q, r sin q ) rdr dq.  

S

S

(B) Volumes of Solids of Revolution: Let  P(x, y) be a point in a plane area R. Suppose that the elementary area d x d y at P(x, y) revolves about the x-axis. This will generate a ring of radius y. The elementary volume of this ring is dV = 2p y d y d x. Hence, the total volume of the solid formed by the revolution of the area R about the x-axis is given by

0

D x

 

V = 2p ∫​∫  ​​  y dy dx.

y

Then, the area of the surface S is given by ______________   2 ∂z 2 S = ∫∫​   ​​  ​ 1 + ​ __ ​ ∂z ​ ​ +   ​ __ ​   ​ ​  ​dx dy. ∂x ∂y D

 

R

Changing to polar coordinates, we get  

 

V = 2p ∫​∫  ​​ r sin q r dr dq  

R

√ (  ) (  )

EXAMPLE 12.43 Find the volume of the sphere x 2 + y 2 + z 2 = a 2 using polar coordinates.

 

= 2p ∫​∫  ​​  r2 sin q dr dq.  

R

y

Solution.  The solid under consideration is bounded above by z 2 = a 2 - (x2 + y2) = a2 - r 2. The sphere cuts the xy-plane in the circle x2 + y2 = a2 or r 2 = a2. Because of symmetry, the required volume is given by 2p a ______ V = 2 ∫​   ​  ​∫​   ​ ​​√a2 - r2 ​  r dr dq

R

P(x, y) y x

 

 

0 0 2p a

x

0

Similarly, the volume V of the area R revolved about the y-axis is given  by V = 2p ∫​∫  ​​​ x dx dy. R

Changing topolar coordinates, we have   V = 2p ∫​∫  ​​  r cos q r dr dq  

R  



= 2p ∫​∫  ​​  r2 cos q dr dq.

M12_Baburam_ISBN_C12.indd 20

 

R

______ = ​∫  ​  ​​∫  ​​ ​​√a2 - r2 ​   · 2r dr dq

[  ]  

 

0

2p

0

 __ 3 a 2 2

2p

(a   -  r ) ​   ​  2a3 4pa ___ ____3 . = ​∫  ​  ​​ ​ ________  ​    ​​ ​​ dq = ​   ​  ∫​   ​  ​ dq = ​   ​    3 3 0 3 __ 0 ​   ​  2 0 2

 

 

EXAMPLE 12.44 2 2 x2 y ___ Find the volume of the ellipsoid __ ​  2  ​+ __ ​  2  ​+ ​ z2  ​= 1. a b c Solution.  Due to symmetry, the volume of the given ellipsoid is eight times the volume of the portion

12/9/2011 4:48:35 PM

Multiple integralS   n 12.21 of the ellipsoid in the first octant. For the positive octant, the given equation yields _________ 2 x2 y . z = c ​ 1 - ​ __2  ​- ​ __2    ​ ​ a b The region in this octant is bounded by _____ x2 . x = 0, x = a, y = 0,  and  y = b ​ 1 - ​ __2  ​ ​  a Hence, the required volume is given by _____

√

√

a 

x b ​ 1 - __ ​  2  ​ ​   a

√

V = 8 ​∫  ​  ​​ ​∫ ​   ​z dy dx  

0

0_____ x b ​ 1 - __ ​  2  ​ ​   a a 



_____ x b ​ 1 - __ ​  2  ​ ​   a a 

√

√

[  (  ) ]

 __ 1 2

[ (  ) (  )

]

x x V = 8 ​∫  ​​ ​∫ ​  ​​c​ ​ 1 - ​ __2  ​  ​- ​ 1 - ​ __2  ​  ​sin2 q  ​​   ​  a a 0 0 ______ x2   × b ​ 1 - ​ __2  ​ ​  cos q dq a 2

2

√

__ ​ p   ​ 2





 __ 1

x2 = 8bc ​∫  ​​  ​ 1 - ​ ___2  ​  ​∫​  ​  ​​[1 - sin2 q] ​ 2 ​ cos q dq a 0 0  

a



2

(  ) x = 8bc ​∫ ​​  ​( 1 - ​    ​ )​​∫​  ​​cos q dq a x = 8bc ​∫ ​​  ​( 1 - ​    ​ )​ · ​ 1 ​  · ​ p  ​ a 2 2 x = 2p bc ​[ x - ​    ​ ]​​= ​ 4 ​ p abc. 3 3a a

__ ​ p   ​ 2

2

   

0 a

   

0

___

2 

__2

2 

3 ___

 

2

0

__ __    

a

2 ​ ​  0

__  

EXAMPLE 12.45 Find the volume contained between the ellipsoid 2 y2 __x x2 y z2 x2 __ __ . ​ __2  ​+ ​ __2  ​+ ​ __ 2  ​= 1 and the cylinder ​  2  ​+ ​  2  ​= ​ a  ​ a b c a b Solution.  The equation of the given elliptical cylinder is

M12_Baburam_ISBN_C12.indd 21

√

0 __ ​ p   ​ 2

[  ]

 __ 3 cos q

(1  -  r2) ​ 2 ​  = - ​   ​   ​∫  ​ ​​ ​​ ​ _______  ​    ​​ ​  ​dq 2 0 3 __   ​    ​ __ 2 0 ​ p   ​ 2 4abc = - ____ ​   ​   ∫​   ​ ​​(sin3 q - 1) dq 2 0 4  abc _____

√

a 2

The required volume is given by _________ 2 x2 y  ​dy dx V = 4 ∫∫c ​ 1 - ​ __2  ​- ​ __2  ​  a b p __ ​    ​ 2 cos q _____ = 4abc ∫​   ​ ​​ ∫​   ​  ​​​√1 - r2 ​ r dr dq 0

y2 x2 = ​∫  ​  ​  ​​∫ ​   ​ c ​ ​ 1 - ​ __2 ​   ​- ​ __2 ​   ​​   ​ dy dx. b a 0 0 ______ y x2 Substituting ​ __ ​ = ​ 1 - __ ​  2     ​ ​ sin q, we get dy = b​ ______ b a 2 x 1 - ​ __2  ​   ​cos q dq (as x is a constant). Therefore, a __  __ ​ p   ​ 1

√

r2 = r cos q or r = cos q.

 

_________ √ 2 x2 y = 8 ​∫  ​  ​  ​​∫ ​   ​c ​ 1 - ​ ___2  ​- ___ ​  2    ​ ​dy dx a b 0 0  

2 x2 y x , ​ __2  ​+ ​ __2  ​= __ ​    ​ a b a y x __ Substituting ​ a ​ = r cos q and __ ​   ​ = r sin q, this b equation yields

[ 

]

4abc __ 2 p 2 __ __ = - ____ ​   ​   ​ ​   ​ - ​    ​  ​= ​   ​ abc [3p - 4]. 3 3 2 9 EXAMPLE 12.46 Find the volume common toa sphere x2 + y2 + z2 = a2 and a circular cylinder x2 + y2 = ax. (Particular case of Example 12.45, taking a = b = c). Solution.  The required volume is the part of the sphere lying within the cylinder and is given by     _________ V = 4 ∫∫​   ​ zdy dx = 4 ∫∫​   ​ √ ​ a2 - x2 - y2   ​dy dx, R

R

where R is the half of the circle lying in the first quadrant. Substituting x = r cos q and y = r sin q,the equation x2 + y2 = ax yields r 2 = a r cos q  or  r = a cos q. Thus, the region of integration is bounded by __ r = 0, r = acos q, q = 0,  and  q = ​ p   ​ 2 Hence, __ ​ p   ​ 2 a cos q

______ V = 4 ​∫  ​ ​​ ​∫  ​  ​√a2 - r2 ​  r dr dq 0

0 __ ​ p   ​ 2 a cos q

______ =___ ​ 4   ​​∫  ​ ​​ ​∫  ​  ​​​√a2 - r2 ​  (-2r) dr dq -2 0 0

12/9/2011 4:48:39 PM

12.22  n  chapter twelve __ ​ p   ​ 2

[  ]  __ 3

​    ​ 2

[ 

​dq

∫ 

]

_____ =16 ​∫  ​ ​​​√4 - y2 ​ dy, because of even integrand 0 _____ 2 y ​ 4 - y2   ​ __ y √ 4 ________ -1 __ =16​ ​   ​    + ​   ​ sin ​    ​  ​​​ ​​  2 2 20 32p =16 [2sin-1 1] = ​ ____  ​  = 16p. 2

Find the volume bounded by the cylinder x2 + y2 = 4 and the planes y + z = 4 and z = 0. Solution.  Tofind the required volume, z = 4 - y is to be integrated over the circle x2 + y2 = 4 in the xy-plane. z

x

To cover the area (half of the circle) in the xy-plane, _____ x varies from 0 to ​√4 - y2 ​  and y varies from - 2 to 2. Thus,

[  ] ∫  [  ∫  ] _____ ​√4 - y2 ​ 

V = 2 ​∫  ​​  ​  ​∫  ​  ​ z dx  ​dy ____ ​√4-y2 ​ 

=2 ​  ​​ ​ ​ ​  ​  ​​(4 - y) dx  ​dy  

-2 2

0

​ _____2 ​  √4 - y

[]

=2 ​∫  ​​  (4 - y) x ​​ ​   

-2 2

0

​dy

______ =2 ​∫  ​​  (4 - y) ​√4 - y2    ​dy  

-2

M12_Baburam_ISBN_C12.indd 22

z = φ1 (x) = 4 - y 2 and

z = φ2 (x) = x2 + 3y 2. Therefore, φ1(x) - φ2(x) = 4 - y2 - x2 - 3y2 = 4 - 4y2 - x2 = 4(1 - r 2). Also, ∂ (x, y) = 2. J = ​ ______ ​  ∂ (r, q) Since the solid is symmetrical about x- and yaxis, we have __ ​ p   ​ 2 1

V = 4 ​∫ ​  ​​​∫  ​​  4 (1 - r 2) 2r dr dq  

0 0 __ ​ p   ​ 2 1

 

2

]

Solution.  The two surfaces intersect in a space curve, whose projection on the xy-plane is the ellipse 2 x 2 + 4y 2 = 4 or ​ x__ ​  + y2 = 1. Substituting x = 2r 4 cos q and y = r sin q, the ellipse becomes r 2 = 1.

y

0

 

EXAMPLE 12.48 Find the volume of the solid bounded above by the parabolic cylinder z = 4 - y2 and bounded below by the elliptic paraboloid z = x2 + 3y2.

Further,

-2

2

[ 

EXAMPLE 12.47

 

 

-2

 

-2

30 4a3 2 p 2a3 ___ =- ​   ​  ​ __ ​   ​ - ​ __  ​  ​= ​ ___ ​  (3p - 4). 3 3 2 9

2

 

-2

_____ =8 ​∫  ​​  ​√4 - y2 ​ dy, second integrand being odd 2

=- __ ​ 4 ​ ​  ​ ​​(a3 sin3 q - a3) dq

[ 

]

2 2 _____ _____ =2 ​ 4 ​∫  ​ ​ ​√4 - y2 ​ dy - ​∫  ​ ​ y​√4 - y2 ​ dy  ​

a cos q

(a2  -  r2) ​ 2 ​  =- 2 ​∫  ​ ​​ ​​ ​ ________  ​    ​​ ​  3 __ 0 ​   ​  2 0 p __



= 32 ​∫ ​  ​​​∫  ​​  (r - r3) dr dq  

__ ​ p   ​ 2

0 0

[ 

]

r4 ​   1​​​ ​​ dq = ​ 32p ____ r2 ​ - ​ __ = 32 ​∫ ​  ​​​ ​ __  ​  = 4p. 8 2 4 0 0 EXAMPLE 12.49 Find the volume bounded by xy-plane, the cylinder x2 + y2 = 1, and the plane x + y + z = 3.

12/9/2011 4:48:43 PM

Multiple integralS  n 12.23 Solution.  We have to integrate z = 3 - x - y over the circle x 2 + y 2 = 1. Substituting x = r cos q and y = r sin q, so that x2 + y2 = r 2, the integrand reduces to 3 - r cos q - r sin q = 3 - r (cos q + sin q ) and the circle x 2 + y 2 = 1 reduces to r 2 = 1. Thus, to cover half of the region, r varies __ from 0 to 1 and q varies from 0 to ​ p   ​. Hence, 2 p __

[ 

]

​    ​ 2 1

V = 4 ​∫ ​  ​​​ ​∫  ​​ {3 - r (cos q + sin q )}r dr  ​dq 0



__ ​ p   ​ 2



]

[ 

]

__ ​ p   ​ 2

Solution.  The required volume is given by

[ 

]

______ V = 8 ​∫  ​​  ​ ​∫  ​  ​ ​√a2 - x2 ​  dy  ​dx  



 

0

______ ______ = 8 ​∫  ​​ ​√a2 - x2    ​​​[ y ]​​√0​ a - x  ​ ​dx 2

__ ​ p   ​



___ 2 2a cos q _ _____  = 4 ​√2a ​ ∫​   ​ ​​ ​∫  ​  ​ ​√r ​ ​√cos q    ​r dr dq  

__ ​ p   ​



0



0

[  ]

 

 __ 5 2a cos q

___ 2 ____ 2  = 4 ​√2a ​ ∫​   ​ ​​​√cos q ​   ​ __ ​ r ​  ​  ​   ​​ ​  ​ 5 __ 0 ​   ​  2 0 p __ ___ ​ 2  ​  __  __ 5 5 _____ 8​√2a ​   = _____ ​   ​   ​∫  ​ ​​(2a) ​ 2 ​ √ ​ cos q ​ (cos q ) ​ 2 ​ dq 5 0

EXAMPLE 12.52 Find, by double integration, the volume generated by revolving the cardioid r = a (1 + cos q ) about the initial line. Solution.  We observe that the upper and lower halves of the cardioid r = a (1 + cos q ) generate the same volume. Therefore, it is sufficient to consider the revolution of the upper-half cardioid only, for which r varies from 0 to a (1 + cos q ) and q varies from 0 to p. y

2

r

]

3 a

x __

a(1



16a . = 8 ​  ​​  (a - x ) dx = ​ 8a x - ​   ​   ​​​ ​​ = ​ ____  ​   

∫ 

2

 

0

2

2

3

0

3

co

s

3

EXAMPLE 12.51 Prove that the volume, enclosed between x2 + y2 = 3 128a ​ . 2ax and z 2 = 2ax is ​ _____     15 ____ Solution.  To find the required volume, z = √ ​ 2ax ​  is to be integrated over the curve x2 + y2 = 2ax in the xy plane. Changing to polar coordinates by substituting x = r cos q and y = r sin q, the required volume is given by

M12_Baburam_ISBN_C12.indd 23

]

 

0

)



[ 

[ 

___ 2 _____ 2a cos q  __3  = 4 ​√2a ​ ∫​   ​ ​​​√cos q  ​ ​   ​∫  ​  ​r ​ 2 ​ dr  ​dq

 

a

0

0

8 128a3 . 64a3 . __ 2 _____ __    = ​   ​ (2a)3 ​∫  ​ ​​cos3 q dq = ​ _______  ​     ​   ​ = ​   ​    5 5 3 15 0

EXAMPLE 12.50 Find the volume common to the cylinders x 2 + y 2 = a2 and x2 + z2 = a2.

0 a

0

__ ​ p   ​ 2

]

______ a ​√a2 - r2 ​ 

 

0



= ​∫ ​  ​​6dq - __ ​ 4 ​ ​∫ ​  ​​(cos q + sin q )dq 30 0 __ ​ p   ​ 6p 4 ___ __ = ​   ​ - ​   ​  sin q - cos q​ ​2​   ​ 0 2 3 6p 8 4 ___ __ = ​   ​ - ​   ​ [1 + 1] = 3p - __ ​   ​ . 2 3 3

[

________ V = 4 ​∫  ​ ​​    ​∫  ​  ​​√2ar cos q  ​   r dr dq

__ ​ p   ​

1

= 4 ​∫ ​  ​​​ __ ​ 3 ​ - __ ​ 1 ​ (cos q + sin q )  ​dq 3 0 2 __ ​ p   ​ 2



[ 

r __

r __

3

2

= 4 ​∫ ​  ​​​ 3 ​   ​ - ​   ​ (cos q + sin q )  ​​​ ​​ dq 2 3 0 0 __ ​ p   ​ 2



 

0

__ ​ p   ​ 2 2a cos q

 0  π

Hence,

0

(2a, 0)

x

p a (1 + cosq )

Volume of Revolution = 2p ​∫ ​   ​ ​∫ ​   ​​r2 sinq dr dq 0

0

12/9/2011 4:48:47 PM

12.24  n  chapter twelve

[  ]

p

3 a(1 + cos q )

r  ​   ​​​ ​  = 2p ​∫  ​  ​sin q ​ ​ __ 3 0 0  

​

p

2pa3 = ​ ____  ​   ​∫  ​  ​ sin q (1 + cos q )3 dq 3 0  

[ 

EXAMPLE 12.54 Find the area of the surface of the paraboloid x2 + y2 = z, which lies between the planes z = 0 and z = 1. Solution.  The required surface area is given by ______________ ∂z 2 __ ∂z 2 S = ∫∫ ​ 1 + ​ ​ __ ​ ​ +    ​ ​   ​ ​  ​dx dy. ∂x ∂y But, __ ∂z ​= 2y. ​ ∂z ​= 2x  and ​ __ ∂x ∂y Therefore, ___________ S = ∫∫ ​√1 + 4(x2 +   y2) ​dx dy ______ = ∫∫ ​√1 + 4r2 ​ r dr dq (changing to polar coordinates).

√ (  ) (  )

]

+ cos q )4 p 2pa3 (1 __________ = - ​ ____  ​   ​ ​   ​     ​​​ ​   ​ 3 4 0 2pa3 8 __ = - ​ ____  ​   (-24) = ​   ​ pa3. 3 3 EXAMPLE 12.53 Find the volume of the solid generated by rey2 x2 __ __ volving the ellipse ​  2  ​+ ​  2  ​= 1 about the x-axis. a b Solution.  Due tosymmetry, it is sufficient to calculate the volume obtained on revolving the upper half of the ellipse. For this, x_____ varies from x2 . - a to a and y varies from 0 to b ​ 1 - ​ __2  ​ ​  a

√

y

To find the limits, we see that the projection on the plane z = 1 is the circle x2 + y2 = 1 or r2 = 1 and this circle lies between q = 0 and q = 2p. Hence, __ ​ p   ​ 2 1

______ S = ​∫ ​  ​​​∫  ​​  ​√1 + 4r2 ​  r dq dr  

0 0 __ ​ p   ​ 2 1

x2 a 2 y 2 b 2 1

(a, 0)

0



_____ 2 b ​ 1 - __ ​ x2  ​ ​  a a

√ volume of revolution = 2p ​∫   ​  ​∫ ​   a

[  ]

-a

_____ x2 2 b​ 1 - ​ __   ​ ​  __ a2

y  = 2p ​∫  ​ ​ ​   ​   ​ √​  -a 2 0

0

a

​y dy dx

( 

)

x2 ​dy = p ​∫  ​b  2​ 1 - ​ __2  ​   ​dx -a a

[ 

a

]

a



2p b2 2p b x3 _____2 2 __ = ​ _____  ​   ∫​   ​ ​(a2 - x2) dx = ​  2 ​   ​ a x - ​   ​   ​​​ ​​  2 3 0 a 0 a



2p b2 a3 __ 4 __ = ​ _____  ​   ​∫  ​​ ​ a3 - ​   ​   ​= ​   ​ p ab2. 2 3 3 a 0

a

M12_Baburam_ISBN_C12.indd 24

[ 

]

______ __ = ​ 1 ​ ​∫ ​  ​​​∫  ​​​ ​√1 + 4r2 ​  8r dq dr 80 0 __ ​ p   ​ 2

(a, 0) x

Therefore,





[  ]

 __ 3 1

2p __ (1  +  4r2) ​ 2 ​  1 1 __ ___ ________ = ​   ​ ​∫ ​  ​​​​ ​   ​    ​​ ​​ dq = ​    ​ ​∫  ​  ​​(5​√ 5 ​ - 1) dq __ 80 12 0 ​ 3 ​  0 __ 2 __ 5​√5 ​ - 1 2p p = ​ _______  ​    [ q]​​0​   ​= ​ __  ​[5​√5 ​ - 1]. 12 6  

EXAMPLE 12.55 Compute the surface area of the sphere x 2 + y 2 + z 2 = a 2. Solution.  The surface area of the sphere is twice the surface area of the upper-half sphere _________ z = ​√a2 - x2 - y2   ​. We have ∂z x __ ​   ​= - ___________ ​  _________      ​and 2 2 2 ∂x  ​ √​ a - x - y   y ∂z __ ​   ​= - __________ ​  _________      ​. 2 2 2 ∂y  ​ √​ a - x - y  

12/9/2011 4:48:53 PM

Multiple integralS   n 12.25 Therefore,



______________ ∂z 2 __ ∂z 2 ​   ​ ​ +    ​ ​   ​ ​  ​dx dy. S = ∫∫ ​ 1 + ​ __ ∂x ∂y  ___________ a    ​ _________  ​ = ∫∫  2 2 2 dx dy.  ​ √​ a - x - y  

√ (  ) (  )

The domain of integration is the circle x2 + y2 = a2 on the xy-plane. Therefore, a

[ 

]

______ ​√a2 - x2 ​ 

a ___________ ​∫   ​  ​​​  _________      ​dy  ​dx. S = 2 ​∫  ​  ​   ______ 2 -a ​ a - x2 - y2   ​ -​√a - x  ​ √  

2

2

Changing topolar coordinates, we have

[ 

__ ​ p   ​ 2 a

] ]

a S = 2 ​∫  ​ ​​​ ​∫  ​ ​ _______ ​  ______   2   ​r dr  ​dq 2   0 0 ​√ a - r  ​

[ 

a

2p

a _______ ______

= ​∫  ​  ​ ​ ​∫  ​​ ​  2   2    ​2r dr  ​dq = 4pa2. 0 0 ​√ a - r  ​      

EXAMPLE 12.56 Find the area of the spherical surface x2 + y2 + z2 = a2 inside the cylinder x2 + y2 = ax.

______________ ∂y 2 __ ∂y 2 a . ​ 1 + ​ __ ​   ​ ​ +    ​ ​    ​ ​  ​= _______ ​  ______      ​ ∂x ∂z ​√a2 - x2 ​ 

√ (  ) (  )

The domain of integration is a quarter circle x2 + z2 = a2 where x ≥ 0 and z ≥ 0 on the xz plane. Therefore,

[ 

]

______ a ​√a2 - x2 ​ 

a    S = 8 ​∫  ​​  ​ ​∫  ​  ​​_______ ​  ______  ​dz  ​dx = 8a2. 0 0 ​√a2 - x2 ​   

EXAMPLE 12.58 2 y2 ​ xa ​  + ​ __ ​  inFind the area of the paraboloid 2z = __ 2 b x2 y side the cylinder ​ __2  ​+ ​ __2  ​= 1 a b Solution.  The required area is ______________ ∂z 2 __ ∂z 2 S = 4 ∫∫ ​ 1 + ​ __ ​   ​ ​ +   ​ ​   ​ ​  ​dx dy, ∂x ∂y where the integration extends over the positive 2 x2 y octant of the ellipse __ ​  2  ​+ __ ​  2  ​= 1. a b

√ (  ) (  ) z

Solution.  We have ______________ ∂z 2 __ ∂z 2 S = 4 ∫∫ ​ 1 + ​ __ ​   ​ ​ +   ​ ​   ​ ​  ​dx dy ∂x ∂y

√ (  ) (  )

 ​over x2 + y2 = ax adx dy   _________ = 4 ∫∫ ​ ___________ 2 2 2  ​ √​ a - x - y   __ ​ p   ​ 2 a cos q

0

r ______ dr dq ,     ​ x = r cos q, y = r sin q = 4a ​∫  ​ ​​  ​∫  ​  ​  _______ 0 ​√a2 - r2 ​  0

y

 



= 2a2(p - 2).

EXAMPLE 12.57 Find the area of that part of the cylinder x2 + y2 = a2, which is cut off by the cylinder x2 + z2 = a2. Solution.  The equation of the surface has the form ______ y = ​√a2 - x2 ​  so that ∂y ∂y x __ ​   ​= - _______ ​  ______  2   ,​  __ ​    ​= 0 2 ∂x ​√a - x  ​ ∂z and

x

∂z x ∂z y We have __ ​   ​= __ ​ a ​and __ ​   ​= __ ​   ​. Therefore, ∂x ∂y b S = 4 ∫∫

( 

 __ 1

)

2 2 x y ​ 1 + __ ​  2  ​+ __ ​  2  ​  ​​   ​ dx dy a b 2

=4ab ∫∫ (1 + ξ 2 + η 2) dξ dη, where x = aξ, y = bξ, so that ξ 2 + η 2 = 1

__ ​ p   ​ 2 1

=4ab ​∫  ​ ​​​∫  ​​  (1 + r2) r dr dq,  

0 0

M12_Baburam_ISBN_C12.indd 25

12/9/2011 4:49:09 PM

12.26  n  chapter twelve ξ = r cos q, η = r sin q

=​   ​ p ab (​  2 ​   ​ - 1 )​. 3  __ 3 2

2 __

log2 x x + log y

12.9  TRIPLE INTEGRALS AND THEIR EVALUATION Let f (x, y, z) be a continuous function in a finite region V of ℜ3. Divide the region V into n subregions of respective volumes d V1, d V2,. . ., d Vn. If (xi, yi, zi) be an arbitrary point of the ith subregion, then n ​lim  ​​   ​   f (xi, yi, zi) dVi,    

 

n→∞    ​ →0  ​ i=1 dV i

 

 

V

​∫ ​​​∫  ​   ​​​∫​    ​  f (x, y, z) dx dy dz. V

Evaluation of Triple Integrals (a) If the region V is specified by the inequalities, a ≤ x ≤ b, c ≤ y ≤ d,  and  e ≤ z ≤ f, and if a, b, c, d, e, and f are constants, then bd f

 

 

​∫  ​​ ​​∫  ​ ​  ​​∫  ​  f (x, y, z) dx dy dz = ​∫  ​ ​∫  ​ ​​∫  ​​   f (x, y, z) dx dy dz  

I = ​∫  ​  ​∫  ​   ​​∫  ​  ​e x + y + z dz dy dx.  

 

 

 

 

a c e

    V

b

f

d

 

a

 

c

log 2 x

I = ​∫  ​  ​ ​∫  ​  [​ e x + y + z  ]​​0​ 

x + log y

 

0

0

log 2 x

=​∫  ​  ​​​∫  ​ [e x + y + x + log y - e x + y] dy dx  

 

0

log 2 x

=​∫  ​  ​  ​​∫  ​ [e 2x  · e y  · e log y - e x·e y] dy dx  

0

0

log 2 x

=​∫  ​  ​​​∫  ​ [e 2xye y - e x·e y] dy dx  

0

 

0

[  ∫  [ 

log 2

x

 

 

 

    V b

[  {  y2(x)

z2(x,y)

 

 

}]

= ​∫  ​  ​   ​∫  ​ ​ ​∫  ​ f (x, y, z) dz  ​ dy  ​dx.  

a

y1(x)

z1(x,y)

Thus, the integration with respect toz is performed first regarding x and y as constants, then the integration with respect to y is performed regarding x as constant and in the last, the integration with respect to x is performed.

M12_Baburam_ISBN_C12.indd 26

]

x

=​∫  ​  ​ ​ ​∫  ​ ​ e  ye  dy - ​∫  ​ e xe ydy  ​dx 2x

 

0

y

 

 

0

0

log 2

x

x

]

=​  ​  ​​​ e  { ye  }​​ ​​ -e  ​∫  ​​ ​ e  dy - e  ​∫  ​ ​ e ydy  ​dx y x   0

2x

0

y

2x

x

 

0

0

log 2

=​∫  ​  ​[e 2x  · xe x - e 2x (e x - 1) - e x(e x - 1)] dx  

0

log 2

=​∫  ​  ​[xe 3x - e 3x + e x] dx  

0

log 2

log 2

log 2

=​∫  ​  ​xe  dx - ​∫  ​  ​e  dx + ​∫  ​  ​e xdx 3x

3x

 

 

0

0

 

log 2

 

 

​dy dx

 

 

e

 ince a, b, c, d, e, and f are constant, the orS der of integration is immaterial, and the integration with respect to any of x, y, and z can be performed first. (b) If the limits of z are given as functions of x and y, and the limits of y as functions of x while x takes the constant values say from a to b, then     ​∫  ​​ ​​∫  ​​ ​​∫  ​  f (x, y, z) dx dy dz

0

Solution.  We have

 

= ​∫  ​  dx ​∫  ​​  dy ​∫  ​​  f (x, y, z) dz.



 

0 0

0

if exists, is called the triple integral  of f (x, y, z) over the  region V, and is denoted by ∫​ ​​​∫  ​   ​​​∫​    ​ f (x, y, z)dV or

 

EXAMPLE 12.59 Evaluate

0

log 2

log 2

=__ ​ 1 ​ ​[ xe 3x ]log ​​  ​ 2​- __ ​ 1 ​  ​  ​  ​e 3xdx - ​  ​  ​e 3xdx + ​  ​  ​e x dx 3

0

3

∫  0

 

∫  0

 

[  ] [ ]

∫ 

 

0

log 2 e 3x log2 =__ ​ 1 ​ log 2 e 3 log2 - __ ​ 4 ​ ​ ​ ___ ​   ​​​ ​  ​+ e x ​​0​  ​ 3 3 3 0 =__ ​ 1 ​ log 2 e log 8 - __ ​ 4 ​ (e log 8 - 1) + (e log 2 - 1) 3 9 8 4 __ __ =​   ​ log 2 - ​   ​ (8 - 1) + (2 - 1) 3 9

12/9/2011 4:49:11 PM

Multiple integralS   n 12.27 19 8 28 8 =__ ​   ​ log 2 - ___ ​   ​ + 1 = __ ​   ​ log 2 - ___ ​   ​ . 3 9 3 9 EXAMPLE 12.60 Evaluate

y =​   ​ + e - ​∫  ​​  ​ __ ​    ​+ 1  ​dy - (e 2 - 1) + (e - 1)2 2 2 1 e y2 e2 __ __ = ​   ​ + e - ​ ​   ​ + y  ​ ​ - 2e + 2 2 4 1  

(  ) ( 

I = ​∫  ​​ ​ ​∫  ​  ​ ​∫  ​ ​ log z dz dx dy.    

 

1

 

1

Solution.  We have e log y 

EXAMPLE 12.61 Evaluate the _____ integral _________

e x

I = ​∫  ​​ ​ ​∫  ​  ​ ​∫  ​ ​ log z ·1 dz dx dy  

 

  

1

1

e

log y

1

∫   ∫ 

{ 

​∫  ​   ​​∫  ​  ​ ​​∫  ​  ​ xyz dz dy dx.

=​  ​ ​ ​​   ​ ​​​ [z log z]​​ ​ ​- ​  ​ ​ z · ​__  1z ​dz  ​dx dy 1

1

 

∫   

1

0

 

∫  ∫ 

x

0

0 _____ 1 ​√1 - x2 ​ 

e log y

        

 

1 1

log y

log y

 

 

1

1

e

log y

[ 

=​∫  ​​  __ ​ 1 ​ x​ ​∫  ​  ​(y - x 2y - y 3) dy  ​dx 0 2 0  

1

e

2

 

{ 

}

=​∫  ​​  ​ [(x - 1)e x]​​log ​  y​- ​∫  ​  ​​e x dx + log y - 1  ​dy 1  

1

 

1

e

=​∫  ​​  [(log y - 1)e log y - (e log y - e) + log y - 1] dy

 

0

[ 

2

]

_____ 4 ​√1 - x2 ​ 

2

2

4

0

​dx

=__ ​ 1 ​ ​  ​​  x (1 - x 2)2 dx 8

=​∫  ​​  [ y (log y - 1) - (y - e) + log y - 1]dy

∫   

0 __ ​ p   ​ 2

=__ ​ 1 ​ ∫​   ​ ​​sin q (1 - sin2 q )2 cos q dq, x = sin q 80 __ ​ p   ​ 2

 

1

e

∫ 

2

1

1

log y

 

y y y =__ ​ 1 ​ ​  ​​  x ​ ​ __ ​ - x2 __ ​   ​ - ​ __ ​   ​ ​ 

=​∫  ​​     ​∫  ​  ​ (x - 1)e x dx + log y - 1 dy 1

]

_____ ​√1 - x2 ​ 

1

 

0

 

1

=​∫  ​​ ​ ​∫  ​  ​(x - 1)e x dx + ​∫  ​  ​ dx dy  

_________

=​∫  ​​  ​​∫  ​  ​​__ ​ 1 ​ xy(1 - x 2 - y 2) dy dx 2 0 0

=​∫  ​​ ​​∫  ​  ​​[(x - 1)e x + 1] dx dy e

[ 2 ]

 

 

 

1

 

 

0

​ 1 - x2 - y2   ​ z2 ​   √​​​ ​  I = ​  ​​  xy ​   ​ ​​​ ​ __ ​dy dx

=​∫  ​​ ​ ​∫  ​  ​ [e  log e  - 0 - e  + 1] dx dy 1

_____ ​√1 - x2 ​ 

1

x

0

Solution.  The given triple integral is

e log y

x

 

2 2  ​ 1 ​√1 - x2 ​ ​√1 - x - y  

}

e x

e x 0

) (  )

e2 e2 = ​ __ ​ + e - ​ ​ __ ​ + e  ​- ​ __ ​ 1 ​ + 1  ​- 2e + 2 2 4 4 2 e 13 1 2 = ​ __ ​ - 2e + ___ ​   ​ = __ ​   ​ (e - 8e + 13). 4 4 4

e log y e x

1

(  )

e

e2 __

1.4.2 ___ =__ ​ 1 ​ ​∫  ​ ​​sin q cos5q dq = __ ​ 1 ​ · _____ ​   ​ = ​ 1  ​ . 80 8 6.4.2 48

 

1 e

=​∫  ​​  [( y + 1) log y - 2y + e - 1] dy  

1

[  (  ) ] [  ]

e

e

(  )

y2 y2 =​ log y ​ ​ __ ​ + y  ​  ​ ​ - ​∫  ​​  __ ​ 1y ​​ ​ __ ​ + y  ​dy 2 2 1 1 2 e y - ​ 2 ​ __ ​   ​​​ ​​ + (e - 1) [y]​​e1 ​​  2 1  

EXAMPLE 12.62 a a-x a-x-y Evaluate I = ​∫  ​​    ​∫  ​  ​  ​​∫  ​  ​ x2 dx dy dz.      

 

0

Solution.  We have a a-x

[ 

a-x-y

0

]

I = ​∫  ​ ​  ​∫  ​  ​x 2​   ​∫  ​  ​ ​dz dx  ​dy. 0

M12_Baburam_ISBN_C12.indd 27

    

0

0

0

12/9/2011 4:49:14 PM

12.28  n  chapter twelve a a-x

1 1 - x  1 - x - y



= ​∫  ​ ​  ​∫  ​  ​x2 [z​]a​0 ​- x - y​dx dy



= ​∫  ​​ x 2 ​ ​∫  ​  ​(a - x - y) dy  ​dx

[ 

0

0

a

a-x

 

0

[  [ 

a

0

]

]

 

a

]



(a - x) = ​∫  ​​ x2 ​ a2 - ax - ax + x2 - ​ ______  ​    ​dx 2 0



__ = ​ 1 ​ ​∫  ​​  (x2a2 - 2ax3 + x4) dx 20

 

a

2

[ 

]

1-x

0

0

1-x-y

[

11-x

0

]

1-x-y + y + z  ​  +   1)- 2  _____________ = ​∫  ​ ​ ​​∫  ​  ​ ​ ​ (x ​​​ ​  ​dy dx 0 0 -2 0  

11-x

[ (x + y + 1) 4 ]

1   ​  1 __ ​ 1 ​ ​  ​ ​ ​​  ​  ​  ​​ _________ ​  = __ 2 - ​   ​   ​dy dx 2

 

a

x3 x4 x5 a5 __ = ​ 1 ​ ​ a2 ​ __ ​ - 2a ​ __ ​ + ​ __ ​   ​​​ ​​ = ​ ___  ​ . 2 3 4 5 0 60



1

 

= ​∫  ​ ​  ​​∫  ​  ​  ​    ​∫  ​  ​(x + y + z + 1)- 3 dz dy dx

2 a-x

y = ​∫  ​​ x 2 ​ ay - xy - ​ __ ​   ​​ ​  ​dx 2 0 0



 

1    ​dz dy dx = ​∫  ​ ​    ​​∫  ​  ​     ​∫  ​  ​____________ ​  (x + y + z + 1)3 0 0 0

∫  ∫  0

0

[ 

1

]

y 1-x - 1   ​  = __ ​ 1 ​ ∫​   ​​ ​​ ​ _________ - __ ​    ​  ​​​ ​  ​dx 2 0 (x + y + 1) 4 0  

EXAMPLE 12.63 dx dy dz ______________ Evaluate ​∫  ​ ​​​∫  ​ ​​​∫         ​ over a tetrahedron ​  ​​​  (x + y + z + 1)3 bounded by coordinate planes and the plane x + y + z = 1.

∫  (  1

)

1 - x _____ 1 = __ ​ 1 ​ ​  ​​  ​ -__ ​ 1 ​ - ​ ____  ​   + ​     ​  ​dx 2

 

0

2

x+1

4

[ 

1

]

3 x _____ = __ ​ 1 ​ ∫​   ​​  ​ - __ ​   ​ + __ ​    ​+ ​  1   ​   ​dx 20 4 4 x+1  

Solution.  The region of integration is bounded by the coordinate planes x = 0, y = 0, and z = 0 and the plane x + y + z = 1. Thus,

1 3x x2 = __ ​ 1 ​ ​ - __ ​   ​ + ​ __ ​ + log (x + 1)  ​​​ ​​  2 4 8 0

R = {(x, y, z); x > 0, y > 0, z > 0, x + y + z < 1} ={(x, y, z); 0 < x < 1, 0 < y < 1 - x, 0 < z < 1 - x - y}.

3 1 5 = __ ​ 1 ​ ​ - __ ​   ​ + __ ​   ​ + log 2  ​= __ ​ 1 ​ log 2 - ___ ​    ​.  2 4 8 2 16

[ 

]

[ 

]

z

EXAMPLE 12.64

C(0, 0, 1)

a x x+y

Evaluate ∫​   ​ ​ ​∫  ​  ​​∫  ​  ​  ​e x + y + z dz dy dx. 0 0

0

Solution.  We have a x x+y

​∫  ​ ​ ​∫  ​  ​​∫  ​  ​  ​e x + y + z dz dy dx.

B (0, 1 ,0) 0

x

A(1, 0, 0)

Therefore,  

dx dy dz ​∫ ​​​∫  ​   ​​​∫  ​   ​  ____________ ​       3 ​ R (x + y + z + 1)  

M12_Baburam_ISBN_C12.indd 28

y

0 0 a x

0



= ​∫  ​​ ​∫  ​​ ​[e x + y + z​]x​0​ + y​dy dx



= ​∫  ​ ​ ​∫  ​ ​ ​[​ e 2(x + y) - e x + y ]​dy dx

 

 

0 0 a x

a

[ 

0 0

x

x

]

= ​∫  ​​ ​ ​∫  ​​  e 2 (x + y) dy - ​∫  ​​  e x + y dy  ​dx 0

0

 

 

0

12/9/2011 4:49:18 PM

Multiple integralS   n 12.29 a

{ [  ]

}

[ 

]

dx dy dz _____________ “Evaluate ∫∫∫ ​ ______________     2  ​ over the posi2 ​√a2 - x2 - y   - z  ​

2(x + y) x x+y x = ​∫  ​​  ​ ​ ​ e_____  ​    ​​​ ​​ - [e  ]​​0 ​ ​  ​dx 0 2 0  

a

tive octant of the sphere x 2 + y 2 + z 2 = a2”.

e e = ​∫  ​​  ​ ​ ___ ​ - ​ ___ ​ - e 2x + e x  ​dx 2 2 0  

[ 

4x

2x

EXAMPLE 12.66 Evaluate

]

a e4x e2x e2x = ​ ​ ___ ​ - ​ ___ ​ - ​ ___ ​ + e x  ​ ​  8 4 2 0 4a 2a e___ 3e ____ a = ​   ​ - ​   ​  + e  . 8 4

c b a

I = ​∫  ​ ​∫  ​ ​∫  ​  (x2 + y2 + z2) dz dy dx  

 

 

-c -b -a

Solution.  We observe that the integrand x2 + y2 + z2 is symmetrical in x, y, and z. Therefore, the limits of integration can be assigned as per our preference. We have

EXAMPLE 12.65 Evaluate the triple integral _________ _________ 2 2 2  ​ a ​√a2 - x   2 ​ ​√a - x - y  

c b

a

1     ​dz dy dx. I = ​∫  ​​      ​∫  ​  ​      ​​∫  ​  ​______________ ​  _____________ 2 2 2 ​ a x - y   - z2 ​ 0 0 0 √

I = ​∫  ​ ​∫  ​  ​∫  ​  (x2 + y2 + z2) dx dy dz

Solution.  We have

=2 ​∫  ​   ​∫  ​ ​ ​∫  ​ ​ (x2 + y2 + z2) dx  ​dy dz,

 

1     ​ dz dy I = ​∫  ​ ​  ​   ​∫  ​  ​   ​∫  ​  ________________ ​​​  ______________ 2 2 2 0 0 0 ) - z2 ​ √​ (a - x - y     

dx

[ 

]

_________ ​√a2 - x2 - y2   ​

z    =​∫  ​ ​  ​  ​∫  ​  ​ ​ sin-1 __________ ​  _________  ​  ​​​ ​  2 2 2 0 0  ​ 0 √​ a - x - y    

​dy dx

______ ​√a2 - x2  ​ 

=​∫  ​ ​  ​  ​∫  ​  ​​[sin- 1 1] dy dx 0

______ ______ __ __ =​ p   ​​∫  ​​ [ y​]​√​0​ a - x   ​​dx = ​ p   ​​∫  ​​ √a2 - x2   ​dx 20 20 ______ a √a2 - x2 ​  __ a2 x p __ x​ ________ =​    ​​ ​   ​    + ​   ​ sin-1 __ ​ a ​ ​ ​  2 2 2 0 a

2

2

 

 

[ 

]

[ 

]

a2 pa2 __ =​ p   ​​ 0 + ​ __ ​ sin-1 1 - 0  ​= ​ ___ ​  sin- 1 1 2 2 4 pa __ p a . =​ ___ ​   · ​ p   ​= ​ ____  ​    4 2 8 2

2 2

Note: The earlier example may be restated as

M12_Baburam_ISBN_C12.indd 29

-c -b

[ 

]

a 0

since x2 + y2 + z2 is even in x

[  ] a =2 ​∫ ​ ​​∫  ​ ​[ ​   ​ + ay + az  ]​dy dz 3 c b

a x3 =2 ​∫  ​ ​∫  ​ ​​ ​ __ ​ + y2x + z2x  ​ ​ dy dz 3 0 -c -b  

 

b

     -c -b

__3  

2

2

[  ( 

c

b

)

]

a3 =4 ​∫  ​  ​​ ​∫  ​​​ ​ ​ __ ​ + ay2 + az2  ​dy  ​dz, 3 -c 0 since integrand is even in y

[ 

c

0

a

 

c

______ a ​√a2 - x2 ​ 

a

        

-c -b -a c b



_________ ______ 2 2 2  ​ a ​√a2 - x2 ​  ​√a - x - y  

 

       

]

b

a3y ay =4 ​∫  ​  ​ ​ ___ ​  + ​ ___ ​ + az2y  ​ ​​ dz 3 -c 3 0 3

[  ] ba ab =8 ​∫ ​ ​  ​​[ ​   ​ + ​   ​ + abz  ]​dz, 3 3 c

ba3 ab3 =4 ​∫  ​ ​​ ​ ___ ​  + ​ ___ ​  + abz2  ​dz 3 3 -c  

c

___3

  0

 

___3  

2

since integrand is even in z

[  ] ba c ab c abc =8 ​[ ​   ​ + ​   ​ + ​   ​  ]​ 3 3 3

c

ba ab z ____ abz3 =8 ​ ​ ___ ​  z + ​ ____  ​   + ​   ​    ​​​ ​​  3 3 3 0 3

3 ____

3

 

3 ____

 

____3

 

8abc 2 =____ ​   ​   [a + b2 + c2]. 3

12/9/2011 4:49:22 PM

12.30  n  chapter twelve z

EXAMPLE 12.67 Evaluate  ​∫   ​ ​​​∫   ​ ​​​∫   ​ ​​ xyz dx dy dz over the ellipsoid 2 x2 y z2 ​ __2  ​+ ​ __2  ​+ ​ __   ​= 1. a b c2

P(x, y, z)

Solution.  The region of integration is bounded by _________ _________ 2 y y2 . x2 __ __ x2 ​ - __ z = - c ​ 1 - ​ ___ ​       ​ ​   and z = c ​ 1 ​     ​ ​      ​ ​ a2 b2 a2 b2 The projection on the xy-plane is the ellipse 2 x2 y ​ __2  ​+ ​ __2  ​= 1. Hence, the _____ limits of integration _____for y a b x2 x2 __ and x are from y = - b ​ 1 - ​  2  ​ ​  to y = b ​ 1 - __ ​  2  ​  and a a x = - a to x = a. Thus,

√

√

√

}

______ _____ x2 x2 __ , y, z); - a ≤ x ≤ a, - b ​ 1 - ​  2  ​ ​  ≤ y ≤ b​ 1 - ​ __2  ​ ​,  a a ​ ​ _________     _________                    2 2 2 2 y y x x - c ​ 1 - ​ __2  ​- ​ __2  ​ ​≤   z ≤ c ​ 1 - ​ __2  ​- ​ __2  ​ ​  a b a b Hence,  

√

√

√

√

∫∫​∫  ​ ​xyz dx dy dz  

R

_____ x2 b ​ 1 - __ ​  2  ​ ​  a  

2

 

M

x

2

 

r sin  sin 



90¡

y

90¡ r sin 

N

x

Then, x = r sin q cos φ, y = r sin q sin φ, z = r cos q and so, x2 + y2 + z2 = r 2. Under these transformations, the region R = (x, y, z); x2 + y2 + z2 ≤ a2} is mapped intothe region Also,

dy dx

2

__2

z

R′ = {(r, q, φ); 0 ≤ r ≤ a, 0 ≤ q ≤ p, 0 ≤ φ ≤ 2p}.

__________ 2 x2 y c ​ 1 - __ ​  2  ​- __ ​    ​   ​ a b2

√ √ ​xy     __________ = ​∫  ​      ​​∫  ​  ​∫  ​  ​​ z dz _____ -a x x y -b ​ 1 - ​    ​ ​  -c ​ 1 - ​    ​- ​    ​   ​ √ a √ a b __2



0

√

R = (x,

a

r

__   2

= 0, since the integrand z is an odd function. 12.10 CHANGE TO SPHERICAL POLAR ­COORDINATES FROM CARTESIAN ­COORDINATES IN A TRIPLE INTEGRAL Let P (x, y, z) be any point in ℜ3. Then, the position of this point is determined by the following three numbers: _________ (i) The distance r = ​√x2 + y2 + z2   ​of P (x, y, z) from the origin (0, 0, 0). (ii) The polar distance q, where q is the angle between the radius vector OP and the positive direction of z-axis. (iii) The angle φ, which the projection of the radius vector OPon the xy-plane makes with the x-axis.

∂x __ ​   ​ 

∂r ∂y ∂ (x, y, z) __ ________ ​     ​= ​   ​  ∂r ∂ (r, q, φ) ∂z __ ​   ​ ∂r

=

∂x  ​  ​___ ∂q ∂y ​ ___  ​ ∂q ∂z  ​ ​ ___ ∂q

∂x ___ ​    ​ ∂φ

∂y ___ ​    ​ ∂φ

∂z ___

 ​   ​ ∂φ

sin q cos φ

r cos q cos φ – r sin q sin φ

sin q sin φ

r cos q sin φ

r sin q cos φ

cos q

– r sin q

0

= r 2 sin q. Hence,  

I = ∫∫​∫  ​  ​f (x, y, z) dx dy dz R  

= ∫∫​∫  ​  ​f (r sin q cos φ, r sin q sin φ, r cos q) R′



× r 2 sin q dr dq dφ.

The polar spherical coordinates are useful when the region of integration is a sphere or a part of

M12_Baburam_ISBN_C12.indd 30

12/9/2011 4:49:24 PM

Multiple integralS   n 12.31 it. If the region of integration is a whole sphere, then 0 ≤ r ≤ a, 0 ≤ q ≤ p, and 0 ≤ φ ≤ 2p. But if the region of integration is the positive octant of the sphere, then 0 ≤ r ≤ a, 0 ≤ q ≤ p, and 0 ≤ __ . φ ≤ ​ p   ​ 2 Remark 12.2. If the region of integration is a right circular cylinder, then the Cartesian coordinates are changed to cylindrical polar coordinates (r, q, z) because the position of P (x, y, z) is determined by r, q, and z as shown in the following figure:

Solution.  The region of integration is V = {(x, y, z); x2 + y2 < 1, 2 < z < 3}. Using the transformation x = r cos q, y = r sin q, and z = z (cylindrical polar coordinates), we have

[ 

]

3

1 2p

1

2p

[  ]

z2 ​   3​​​ ​​ dq dr I = ​∫  ​  ​∫  ​  ​ ​ ​∫  ​​  zr2·r dz  ​dq dr = ​∫  ​  r3 ​∫  ​  ​ ​ ​ __ 2 2 2 0 0 0 0  

 

 

1

2p

 

 

[ 2 2 ]

1

9 4 5 = ​  ​   ​r  ​  ​  ​​​ __ ​   ​ - __ ​   ​   ​dq dr = __ ​   ​ ​  ​ ​r3[q ]​​2p ​   ​dr 0

∫ 

3

0

∫  0

z

 

2

[  ]

1

P(x, y, z) z y

0 x

 

0

r4 ​   1​​​ ​​ = ​ 5p ___ . = 5p ​∫  ​ ​r 3dr = 5p ​ ​ __  ​  4 0 4 0  

M

∫ 

90¡



y

r

90¡

r sin 

N

EXAMPLE 12.69 Evaluate I = I = over the region

∫∫∫

_____________ 2 2 x2 y __ ​ 1 - __ ​  2  ​ - __ ​  2     ​ - ​ z2  ​  ​dx dy dz a b c

√

2 2 x2 y __ V = {(x, y, z); x > 0, y > 0, z > 0, __ ​  2 ​ + __ ​  2 ​ + ​ z2 ​ < 1} a b c y __ __ x Solution.  Substituting ​ a ​= X, ​   ​= Y, and _​ cz ​= Z b so that dx = adX, dy = bdY, dz = cdZ and hence, dx dy dz = abc dX dY dZ. Therefore,   1 __

x

I = abc ∫∫∫ (1 - X 2 - Y 2 - Z 2) ​ 2 ​ dx dy dz,

Then, x = r cos q, y = r sin q, and z = z,

and

∂x __

 ​  ​  ∂r ∂y ∂ (x, y, z) ________ ​ __ ​  ​   ​=     ∂ (r, q, z) ∂r ∂z ​ __ ​ ∂r

∂x  ​__  ​ ∂q ∂y ​ __  ​ ∂q ∂z ​ __  ​ ∂q

∂x ___ ​    ​

cos φ -r sin q 0 ∂φ ∂y ___ ​    ​ = sin q r cos q 0 ∂φ 0 0 1 ∂z ___ ​    ​ ∂φ

= r cos2 q + r sin2 q = r. Hence,  

I = ∫∫​∫  ​  ​f (x, y, z) dx dy dz R  

= ∫∫​∫  ​  ​f (r cos q, r sin q, z) r dr dq dz. R

EXAMPLE 12.68

over the region V ′ = {(X, Y, Z ); X ≥ 0, Y ≥ 0, Z ≥ 0, X 2 + Y 2 + Z 2 ≤ 1}. Using spherical polar coordinates, X = r sin q cos φ, Y = r sin q sin φ, and Z = r cos q, the region of integration becomes __ __ V ″ = {​  (r, q, φ ); 0 ≤ r ≤ 1, 0 < 0, ≤ ​ p   ​, 0 < φ ≤ ​ p   ​  ​ 2 2} . Hence, p__ p__ ​    ​ ​    ​ 1 2 2

 __ 1

I = abc ​∫  ​ ​​​∫ ​  ​​​∫ ​  ​​(1 - r 2) ​ 2 ​ r 2 sin q dr dq dφ  

0 0 0 1

__ ​ p   ​  __ 1 2 2 2     0

[  ] __ ​ p   ​ 2

=abc ​∫  ​​ ​r 2 (1 - r ) ​   ​ ​∫​  ​​sin q ​  ​∫ ​  ​​dφ  ​dq dr  

0

0

Evaluate I = ∫∫∫ z (x2 + y2 ) dx dy dz over x2 + y2 ≤ 1 and 2 ≤ z ≤ 3.

M12_Baburam_ISBN_C12.indd 31

12/9/2011 4:49:26 PM

12.32  n  chapter twelve __ ​ p   ​  __ 1 2 2 2     0

1

__ ​ p   ​ 2   0

 

1

 __ 1 2 2

[ 

__ ​ p   ​ 2

 __ 1

∫ 

]

0

 

(1)

__0 ​ p   ​ 2

__ ___ = ​∫ ​  ​​sin2 t cos2 t dt = ___ ​  1   ​  · ​ p   ​= ​ p  ​.  4.2 2 16 0

Hence (1) reduces to 2 p abc ___ p  abc p _____ I = ​ _____  ​   ​  ​    ​   ​= ​   ​    2 16 32

( )

EXAMPLE 12.70 Evaluate _____

_________ 2 2  ​  ​√1 - x - y   1 ​√1 - x2 ​  

dz dy dx I = ​∫  ​ ​  ​​∫  ​  ​  ​​∫  ​  ​_____________ ​  ____________     2  ​ 0 0 0 ​√1 - x2 - y2   - z  ​

_____ , V = {(x, y, z); 0 ≤ x ≤ 1, 0 ≤ y ≤ ​√1 - x2 ​  _________ 2 2 0 ≤ z ≤ ​√1 - x - y    ​}. Now, we transform the region by using spherical polar coordinates, by substituting x = r sin q cos φ, y = r sin q sin φ, and z = r cos q. The transformed region is __ __ V ′= ​{ (r, q, φ); 0 ≤ r ≤ 1, 0 < q ≤ ​ p   ​, 0 ≤ φ ≤ ​ p   ​  ​. 2 2} Therefore, __ p ​ p   ​ ​ __  ​ 1 2 2

sin q  ______ _____ I = ​∫  ​ ​ ​​∫   ​  ​​​∫  ​ ​​ ​ r   ​dr dq dφ 0 0 0 ​√ 1 - r 2   ​ 1

[ ∫  ] __ ​ p   ​ 2

r 2    _____ =​  ​ ​ ​​ ______  ​​  ​ ​​sin q ​ ​  ​ ​​dφ  ​dq dr 2

∫  0

∫ 

​√1 - r     ​ 0

M12_Baburam_ISBN_C12.indd 32

0

1

[ 

]

1 1 _____ p __ ​ _______ dr _____ =​    ​​ ​∫   ​ ​​       ​- ​∫  ​ ​ ​​√1 - r 2   ​  ​dr 2 0 ​√1 - r 2   ​ 0 ____ 1 r√ ​ 1 - r   ​ __ 1 __ =​ p   ​​ ​ sin-1 r - ​ ​ _______  ​   + ​   ​ sin-1 r  ​  ​ ​ ​  ​ 2 2 2 0 2 p p p  __ p __ p __ __ p __ ___ =​    ​​  ​    ​- ​    ​  ​= ​    ​ · ​    ​= ​   ​ . 2 2 4 2 4 8

{ [  {

( 

}

)] }

Note: This example is a particular case of Example 12.65 for a = 1. EXAMPLE 12.71   Evaluate I = ​∫ ​​​∫  ​   ​​​∫  ​   ​   ​(x2 + y2 + z2)m dx dy dz, m > 0 V

over the region V = {(x, y, z); x2 + y2 + z2 ≤ 1}. Solution.  The given region of integration is V = {(x, y, z); x2 + y2 + z2 ≤ 1}. Changing tospherical polar coordinates by substituting x = r sin q cos φ, y = r sin q sin φ, and z = r cos q, ∂ (x, y, z) we get x2 + y2 + z2 = r2 and ________ ​     ​= r 2 sin q. ∂ (r, q, φ) Therefore, the region of integration reduces to V ′= {(r, q, φ); 0 ≤ r ≤ 1, 0 ≤ q ≤ p, 0 ≤ φ ≤ 2p}. Hence,

1

p 2p

I = ​∫  ​ ​ ​​∫  ​  ​  ​∫  ​  ​  ​r 2m + 2 sin q dr dq dφ 0

2

__ ​ p   ​ 2

p __

- (1 - r 2) __ ​ _______ __ ​ 1_________ r 2   ​  =​ p   ​​∫   ​ ​​  _____ dr = ​ p   ​​∫  ​ ​​  ______2  ​    dr 2 0 ​√1 - r 2  2 0 ​√1 - r     ​  ​

by changing tospherical polar coordinates. Solution.  The region of integration is

]

1

 



[ 

__ ​ p   ​ 2

1

But, substituting r = sin t so that dr = cos t dt, we have __ ​ p   ​ 1 2  __ 1 ________ ​ 2 2 2 ∫​   ​ ​r  (1 - r ) ​   ​ dr = ∫​   ​ ​​sin2 t ​√1 - sin2 t   ​+ cost dt 0

2

​    ​ __ ​ ______ r 2    =​ p   ​ ​  _____  ​[- cos q ​]​02​   ​dr 2 ​∫0   ​ ​ ​√1 - r 2   ​

p __

 __ 1

​√1 - r     ​ 0

__ ​ _______ r      =​ p   ​​∫   ​ ​​  ______  ​​  ​∫ ​  ​​sin q dq  ​dr 2 0 ​√1 - r 2    ​ 0

 

1

∫ 

1

​    ​ abcp ​ 2 =​ ____  ​   ∫​   ​ ​r  (1 - r 2) ​ 2 ​ [-cos q ​]​02​   ​dr 2 0

abcp ​ 2 =​ ____  ​   ∫​   ​ ​r  (1 - r2) ​ 2 ​ dr.  2 0

p __

​    ​ r 2    _____ =​  ​ ​ ​​ ______  ​​  ​ ​​sin q [φ​]​02​   ​dq dr 2

abcp ​ 2 =​ ____  ​   ∫​   ​ ​r  (1 - r  ) ​   ​ ​  ​∫ ​  ​​sin q dq  ​dr 2 0 0 1

__ ​ p   ​ 2

1

=abc ​∫ ​ ​r  (1 - r  ) ​   ​ ​∫​  ​​sin q [φ​]​ ​  ​dq dr ​ 2   0

0

0

1

p

[  ] 2p



= ​∫  ​ ​ ​r  ​∫  ​  ​  ​sin q ​ ​∫  ​  ​  ​dφ   ​dq dr



= ​∫  ​ ​ ​r 2m + 2 ​∫  ​  ​  ​sin q [φ]​​2p ​   ​dq dr 0

2m + 2

0 1

0 p

0

0

0

12/9/2011 4:49:29 PM

Multiple integralS   n 12.33

[ 

1



0

0

1



]

p

0

[ 

1



]

1

]

0 p ​ __  ​ 2

= 4p ​∫  ​ ​ ​r 2 ​ ​∫ ​  ​​sin q dq  ​dr



r  = 4p ​∫  ​ ​  ​r 2m + 2 dr = 4p ​ ​ ______   ​   ​ 2m + 3 0 0 4p   ​  . = ​ ______ 2m + 3 2m + 3

[ 

0

a

p   0

__ ​ p   ​

= 8 ​∫  ​ ​ ​r 2 ​∫ ​  ​​sin q [φ​]​02​   ​dq dr



= 2p ​∫  ​ ​  ​r  [- cos q ]​​ ​  ​dr 2m + 2

__ ​ p   ​ 2

a

= 2p ​∫  ​ ​  ​r 2m + 2 ​ ​∫  ​  ​  ​sin q dq  ​dr

0 a

0

__ ​ p   ​

a



= 4p ​∫  ​ ​ ​r2 [-cos φ​]​02​   ​dr = 4p ​∫  ​ ​ ​r2 dr



a3 . _____ r  ​   ​ = ​ 4p = 4p ​ ​ __  ​    3 3 0

[  ] 0

3

0

a

12.11  VOLUME AS A TRIPLE INTEGRAL In Cartesian coordinates, the volume of a region V is given by the triple integral ∫∫∫ dx dy dz,

EXAMPLE 12.73 2 x2 y z2 Find the volume of the ellipsoid ​ __2  ​+ ​ __2  ​+ ​ __   ​= 1. a b c2 y Solution.  Substituting __ ​ ax ​ = X, ​ __ ​ = Y, and ​ _cz ​ = Z, b we have dx = adX, dy = adY, and z = adZ. Therefore, the volume is given by

∫∫∫ r 2 sin q dr dq dφ,

taken over X  + Y 2 + Z 2 = 1. Changing tospherical polar coordinates by substituting X = r sin q cos φ , y = r sin q sin φ, and z = r cos q, we have

where the limits of integration are chosen to cover the entire region V. In spherical polar coordinates, the volume of a region V is given by the triple integral

where the limits of integration are chosen to cover the entire region V. In cylindrical coordinates, the volume of a region V is given by the triple integral

∫∫∫ r dr dq dz,

where the limits of integration are chosen tocover the entire region V.

V = abc ∫∫∫dX dY dZ, 2

__ p ​ p   ​ ​ __  ​ 1 2 2

V = 8abc ​∫  ​ ​ ​​∫  ​ ​​​∫ ​  ​​r 2 sin q dr dq dφ 0 0 0



Solution.  The required volume is given by



V = 8 ∫∫∫ dx dy dz, taken over the positive octant of the given sphere. Changing to spherical polar coordinates, we put x = r sin q cos φ, y = r sin q sin φ, and z = r cos q. So, x2 + y2 + z2 = r2. In the positive octant, we have __ __ . 0 ≤ r ≤ a, 0 ≤ q ≤ ​ p   ​, and 0 ≤ φ ≤ ​ p   ​ 2 2 __ p Therefore, ​ p   ​ ​ __  ​ a 2 2 V = 8 ​∫  ​​ ​​∫ ​  ​​​∫ ​  ​​r 2 sin q dr dq dφ 0 0 0



a

__ ​ p   ​ 2

0

0

[  ] __ ​ p   ​ 2

= 8 ​∫  ​ ​  ​r 2 ​∫ ​  ​​sin q ​ ​∫ ​  ​​dφ  ​dq dr

M12_Baburam_ISBN_C12.indd 33

0

__ ​ p   ​ 2

= 8abc ​∫  ​ ​ ​r 2 ​∫ ​  ​​sin q ​ ​∫ ​  ​​dφ  ​dq dr 0

0 __ ​ p   ​ 2

1

EXAMPLE 12.72 Find the volume of the sphere x2 + y2 + z2 = r2.

[  ]

__ ​ p   ​ 2

1

0

__ ​ p   ​

= 8abc ​∫  ​ ​ ​r 2 ​∫ ​  ​​sin q [φ​]​02​   ​dq dr 0

[ 

0

1

]

__ ​ p   ​ 2

= 4p abc ​∫  ​ ​ ​r 2 ​ ​∫ ​  ​​sin q dq  ​dr 0

1

0

1

__ ​ p   ​ 2   0

= 4p abc ​∫  ​ ​ ​r  [- cos q ​]​ ​  ​dr = 4p abc ​∫  ​ ​ ​r2 dr 2

0



[  ] r 3 __

1

0

4 __

= 4p abc ​ ​   ​   ​​ ​​ = ​   ​ p abc. 3 0 3

EXAMPLE 12.74 Find the volume of the solid bounded by the surface  __ 2

(  )

 __ 2

 __ 2

3 x 3 y 3 ​ __ ​ a ​ ​​   ​ + ​ __ ​   ​ ​​   ​ + ​ _​ cz ​  ​​   ​  = 1. b  __2  __ 2 y 3 x 3 Solution.  Substituting ​ __ ​ a ​ ​​   ​  = X, ​ __ ​   ​  ​​   ​  = Y, and b

(  )

(  )

(  )

(  )

12/9/2011 4:49:32 PM

12.34  n  chapter twelve  __ 2

(  )

3 ​ _​ cz ​ ​​   ​  = Z, that is, x = aX 3, y = bY 3, and z = cZ 3,

we get dx = 3aX 2 dX, dy = 3bY 2dY, and dz = 3cZ 2 dZ. Then, the required volume is given by  

V = ∫∫​∫  ​   ​dx dy dz V

= ∫∫∫ 27abc X 2Y 2Z 2 dX dY dZ, taken throughout the sphere X 2 + Y 2 + Z 2 = 1. Changing to spherical polar coordinates (r, q, φ), we have   V = 8 ∫∫​∫  ​​  ​ 27abc r 2 sin2 q cos2 φ V ′

where

× r 2 sin2q sin2φ r 2 cos2q · r 2 sinq dr dq dφ,

{ 

p p V ′ = ​ (r, q, φ); 0 ≤ r ≤ 1, 0 ≤ q ≤ __ ​   ​ , 0 ≤ φ ≤ __ ​   ​   ​.

Thus,

2}

2

1

__ ​ p   ​ 2

0

0

V = 216abc ​∫  ​ ​ ​r 8 ​∫ ​  ​​sin5 q cos2 q

[ 

1

[ 

]

__ ​ p   ​ 2

0

]

__ =216abc ​∫  ​ ​ ​r 8 ​  ​∫ ​  ​​sin5 q cos2 q · ___ ​  1   ​  · ​ p   ​dq  ​dr 4.2 2 0 0

216 ____

1

[ 

__ ​ p   ​ 2

1

5

[ 

2

]

 __ 1

3 __ - 1

3 __ - 1

3 __ - 1

2 2 2 27 = ___ ​   ​ abc ∫∫∫ X ​   ​  Y ​   ​  Z ​   ​  dX dY dZ 8 3 3 3 Γ​ __ ​   ​   ​Γ​ __ ​   ​   ​Γ​ __ ​   ​   ​ 27 2 2 2 ___ ______________ =​   ​ abc   ​        8 3 __ 3 __ 3 __ Γ​ 1 + ​   ​ + ​   ​ + ​   ​   ​ 2 2 2

(  ) (  ) (  ) (  ) 3 ​  Γ​( __ ​   ​  )​ )​ 27abc (_______ 2   _____  ​ =​   ​   ​     ​ 8 9 __ Γ​( ​   ​ + 1 )​ 2 3

(  )

__ 2​ ​ __ ​ 1  ​·​√p  ​  2  27 =___ ​   ​ abc  _____________ ​       ​ 8 9 __ 7 __ 5 3 1 __ __ ​   ​  · ​   ​  · ​   ​  · ​ __ ​  · ​ __  ​  · ​√p  ​ 2 2 2 2 2

4p  abc. p abc ___ Hence, the total volume is = 8 _____ ​   ​  ·  ​ 4  ​ = ______ ​   ​    8 35 35 EXAMPLE 12.75 Find the volume of the portion cut off from the sphere x 2 + y 2 + z 2 = a 2 by the cylinder x 2 + y 2 = ax. Solution.  The required volume is

=​   ​ p abc ​∫  ​ ​ ​r  ​  ​∫ ​  ​​sin q cos q dq  ​dr 16 0 0 8

 __ 1

4  ​ . __ =​ p   ​abc · ​ ___ 8 35

× · ​  ​∫ ​  ​​sin2 φ cos2 φ dφ  ​dq dr __ ​ p   ​ 2

 __ 1

V = ∫∫∫ ___ ​ 27 ​ abc X ​ 2 ​Y ​ 2 ​Z ​ 2 ​dX dY dZ 8

z

]

216 =____ ​   ​ p abc ​∫  ​ ​ ​r 8​ ______ ​  4.2.1  ​  ​dr 16 7.5.3.1 0

[  ]

1 216 __9 =____ ​   ​ p abc ​ ​ r   ​   ​​​ ​​  210 9 0 216p abc ________ =​   ​    1890 =___ ​ 4  ​ p abc. 35

0

y

Second Method: For the positive octant, we have V = ∫∫∫dx dy dz.  

(  )

 __2 3

x  __2

2 __ y x Substituting ​ a__​  ​  3​​  ​  = X, ​ __​  ​   ​​  ​  = Y, and ​ c_​ z​  3​​  ​  = Z, we b   __1 1 1 __ __ 3 3 3 have dx = __​   ​ aX 2​  ​ dX, dy = __​   ​ bY 2​  ​ dY, dz = __​   ​ cZ 2​ dZ,  ​  2 2 2

(  )

(  )

______ _______ 2 2 2  ​ ​√a -x -y  ​  a ​√ ax -x2  

V = 4 ​∫  ​ ​  ​    ​∫ ​   ​​      ​∫  ​  ​  ​dx dy dz 0

0

0

and X + Y + Z ≤ 1. Therefore, by Dirichlet’s Theorem,

M12_Baburam_ISBN_C12.indd 34

12/9/2011 4:49:34 PM

Multiple integralS  n 12.35 ______ a ​√ax - x2 ​ 

 __ 1 2 2



= 4 ​∫  ​       ​  ​∫  ​  ​  ​(a2 - x2 - y ) ​   ​ dy dx



______ = 4 ​∫ ​   ​​ ​∫  ​  ​  ​​√a2 - r2 ​  r dr dq,

0 0 __ ​ p   ​ 2 a cos q

changing to polar coordinates __ ​ p   ​ 2 a cos q

______ __ = ​ 4 ​ ​∫ ​  ​​   ​∫  ​  ​  ​2r ​√a2 - r2 ​  dr dq 20 0



__ = ​ 4 ​ ​∫ ​  ​​[- a3 sin3 q + a3]dq 30 __ __ = ​ 4 ​ a3 ​ - __ ​ 2 ​ + ​ p   ​  ​ 3 3 2

x2 + y2 = a2 and x2 + z2 = a2.

[  ] = ​ 2 ​ a ​( p - ​ 4 ​  )​. 3 3 __

__

3

 

 

EXAMPLE 12.76 Prove that the volume of the wedge intercepted between the cylinder x 2 + y 2 = 2ax and the planes z = mx and z = nx is p (m - n)a3.

Therefore, the region of integration is defined by ______ ______ -​√a2 - x2 ​  ≤z≤√ ​ a2 - x2 ​  , ______ ______ 2 2 2 2 -​√a - x  ​  ≤y≤√ ​ a - x  ​  , and -a ≤ x ≤ a. Hence, the required volume is

______ 2a ​√2ax-x2 ​ mx

V = ​∫  ​   ​______ ​∫  ​    ______ ​​∫  ​    ​dx dy dz -a

0

0

nx



= 2 ​∫  ​ ​  ​​∫  ​  ​​[z]​​mx ​ ​dy dx nx 0



______ 2a ​√2ax-x2 ​ 



= 2 ​∫  ​ ​  ​​∫  ​  ​​(m - n) x dy dx 0

0

2a

______ ​√2ax-x2 ​  0



= 2 ​∫  ​ ​​(m - n) x[ y]​​ ​ 



2a _______ = 2 (m - n) ​∫  ​ ​​x​√2ax - x2 ​  dx



2a  __ 3 ______ = 2 (m - n) ​∫  ​ ​  ​x ​ 2 ​ ​√2a - x ​  dx.

0

​dx

0

M12_Baburam_ISBN_C12.indd 35

= 8 ​∫  ​ ​  ​∫  ​  ​  ​​∫  ​  ​​dx dy dz 0

0

0

_____ a ​√a2 -x2 ​ 

______ 2a ​√2ax-x2 ​  0

-​√a2 - x2 ​  -​√a2 - x2 ​ 

______ _____ ​√a2 -x2 ​  a ​√a2 -x2  ​ 



V = 2 ​∫  ​ ​  ​​∫  ​  ​​ ​∫ ​  dx dy dz

_____ _____ ​√a2 -x2 ​  ​√a2 -x2 ​ 

a

Solution.  The required volume is given by



3.1 p = 32 (m - n) · ​ _____  ​   · ​ __  ​= p (m - n) a3. 6.4.2 2

[  ]

 __ 3 a cosq

- (a2  -  r2) ​ 2 ​  = 2 ∫​   ​ ​​​​ _________ ​   ​     ​​ ​  ​dq 3 __ 0 ​   ​  2 0 __ ​ p   ​ 2



0



EXAMPLE 12.77 The axes of two right circular cylinders of the same radius a, intersect at right angles. Prove that 16a3. ​   ​   the ­volume inside both the cylinders is ____ 3 Solution.  Let the equations of the cylinders be

__ ​ p   ​ 2



V = 2 (m - n) ​∫ ​  ​​16a3 sin4 q cos2 q dq

0

0



Substituting x = 2a sin2 q, we get dx = 4asin q cos q dq. The limits of integration are q = 0 to __ q = ​ p   ​. Therefore, p__ ​    ​ 2 2



______ = 8 ​∫  ​ ​  ​∫  ​   ​​​√a2 - x2  ​  dy dx 0 a

0

_____ ______ = 8 ​∫  ​ ​  ​√a2 - x2 ​  [ y]​​​√0​ a -x  ​ ​dx 2

2

0 a

= 8 ​∫  ​ ​ ​(a2 - x2) dx

[ 

0

] [  a

]

x3 a3 16a3 . = 8 ​ a2x - ​ __ ​   ​​​ ​​ = 8 ​ a3 - ​ __ ​   ​= ​ ____  ​    3 0 3 3 EXAMPLE 12.78 Find the volume in the positive octant bounded by the coordinate planes and the plane. x + 2y +3z = 4.

0

12/9/2011 4:49:36 PM

12.36  n  chapter twelve z

Solution.  The region of integration is

{ 

4-x V = ​ (x, y, z); 0 ≤ x ≤ 4, 0 ≤ y ≤ ​ ____  ​   , 2 4 - x - 2y 0 ≤ z ≤ _________ ​   ​    .​ 3 Therefore, the required volume is

C

}

[  ]

-  x  -  2y 4  -  x ​ 4   ________    ​  ​ ____  ​    3 2

4

V = ​∫  ​ ​  ​​∫ ​   ​​​ ​∫ ​   ​​ dz  ​dy dx

0

 

0

0

0

B

y

4  -  x ​ ____  ​    4 2

= __ ​ 1 ​ ​∫  ​ ​ ​ ​∫ ​   ​​(4 - x - 2y) dy dx 30 0

( 

4

A

)

x

-  x ____  ​    y2 ​ 4   = __ ​ 1 ​ ​∫  ​ ​ ​​​ 4y - xy - 2 ​ __ ​   ​​ ​ 2 ​dx 30 2 0

Hence, the volume of the tetrahedron OABC is

(  ) x x = ​ 1 ​ ​∫ ​​ ​​( 4 - 2x + ​   ​  )​dx = ​ 1 ​ ​( 4x - x + ​    ​ )​​​ ​​  3 4 3 12 4

x2 x2 = __ ​ 1 ​ ​∫  ​ ​ ​​ 8 - 2x - 2x + ​ __ ​ - 4 - ​ __ ​ + 2x  ​dx 30 2 4 __

4

 

__2

  

0

[ 

__

 

]

 

2

3

12

3 4 ___  

( 

9

[ 

0

x V = ​ (x, y, z); 0 ≤ x ≤ a, 0 ≤ y ≤ b ​ 1 - ​ __ a ​ ​, x x 0 ≤ Z ≤ c ​ 1 - __ ​ a ​- __ ​   ​ ​.  ​ b

)}

)

)

 

( 

0

0

​​dz dy dx

)​dy dx     

y c ​ 1 - __ ​ ax ​- __ ​   ​ ​ b

b ​ 1- __ ​ ax ​ ​

a

)

( 

)

   = ​∫  ​​    ​∫ ​   ​​[z​]​0​ ( 0

( 

0

x __ a b ​ 1- ​ a ​ ​

0

Solution.  The region of integration is bounded x __y _z by four planes x = 0, y = 0, z = 0, and ​ __ a ​+ ​ b ​+ ​ c ​ = 1. It is bounded below and above by z = 0 and x y z = c ​ 1 - __ ​ a ​- __ ​   ​ ​. Its projection on the xy-plane b x y is a triangle bounded by x = 0, y = 0, and __ ​ a ​+ __ ​   ​= b 1. Therefore, the region is

{ 

)

 

V

EXAMPLE 12.79 Find the volume of the tetrahedron bounded by x y the planes x = 0, y = 0, z = 0, and ​ __a  ​+ ​ __  ​+ ​ _zc  ​= 1; b a, b, c ≥ 0.

( 

( 

V = ​∫ ​​​∫  ​   ​​​∫  ​   ​  dx dy dz = ​∫  ​   ​  ​∫ ​   ​​       ​∫ ​  

64 16 . = ​   ​ ​ 16 - 16 + ___ ​   ​  ​= ___ ​   ​  1 __

( 

y x c ​ 1 - __ __ ​ ax ​- __ ​   ​ ​ b a b ​ 1- ​ a ​ ​

 

)

( 

)

]

x y    = c ​∫  ​​ ​   ​∫ ​   ​​​ 1- __ ​ a ​- __ ​   ​ ​dy  ​dx b 0 0

 [(  (  ) ) ]

x __ x y 2 b ​( 1  - a​  ​ )​ a ​ 1  -  ​ a__​   -  ​ __​  ​ x2 b bc    = c ​∫ ​​  __________ ​ ​    ​     ​ ​  ​dx = __ ​   ​  ∫​ ​ ​  ​ 1  - a__​  ​  ​ dx 2 0 0 2  ​ -  ​ __1​  ​ b 0 x 3a 1  -  ​ __ a ​ ​ bc ​_______ abc __    = ​   ​ ​ ​   ​   ​​​ ​​ = ___ ​   ​  [0-1] 2 6 1 __ 3  ​ -  ​ a  ​  ​ 0 a

( 

)

[  (  ) ] ( 

)

abc    = ___ ​   ​  cu units. 6 y x Second Method: Substituting a__​  ​  = u, __​  ​  = v, and b _z c​  ​ = w, we have u ≥ o, v ≥ 0, w ≥ 0, and u + v + w ≤ 1. Therefore  

V = ∫∫​∫  ​   ​dx dy dz V  

   = ∫∫​∫  ​ ​  ​(a d u) . (b d u) . (c d u), u + v + w ≤ 1 V′

 

    = abc ∫∫​∫  ​ ​ ​0 u1-1 v1-1 w1-1 du dv dw V′

M12_Baburam_ISBN_C12.indd 36

12/9/2011 4:49:38 PM

Multiple integralS   n 12.37 Γ(1) Γ(1) Γ(1) ,   = abc _____________ ​        ​  by Dirichlet’s Theorem Γ(1 + 1 + 1 + 1) abc abc.    = ___ ​   ​ = ___ ​   ​   6 3!

Solution.

EXAMPLE 12.80 Find the volume of the portion cut off from a sphere x2 + y2 + z2 = a2 by a cone x2 + y2 = z2. Solution.  The origin is the center of the sphere and the vertex of the cone x2 + y2 = z2. Therefore, the volume is symmetrical about the plane z = 0. Hence, V = 2 ∫∫∫ dx dy dz. Changing the coordinates to spherical polar, x2 + y2 + z2 = a2 reduces to r2 = a2 or r = a. Further, x2 + y2 = z2 reduces to



r  sin q cos φ + r  sin q sin φ = r  cos q 2

or

2

2

2

2

2

2

2

r 2 sin2 q (cos2 φ + sin2 φ) = r 2 cos2 q

or

__. sin2 q = cos2 q, which yields q = ​ p   ​  4 p __ Thus, q varies from 0 to ​    ​ and φ varies from 0 4 top. Therefore, __ ​ p   ​ a 4 2p

V = 2 ​∫  ​​ ​∫ ​   ​​​∫  ​  ​  ​r2sin q dr dq dφ 0 0

a

__ ​ p   ​ 4

0

3

[ 

]

2

​∫  ​​​ ​ ​∫  ​​​ (x + y)-2 dx  ​dy 4

1

3

[ 

]

(x + y)-1 2 = ​∫  ​​​ ​ ​ _______  ​    ​​​ ​​ dy -1 0 4 3

[ 

]



= -​∫  ​​​ ​ _____ ​  1   ​ - _____ ​  1   ​ ​dy 1+y 4 2 + y



= -[log (2 + y) - log (1 + y)​]​34​​ 



= log 4 + log 6 - 2 log 5



= log ___ ​ 24 ​ . 25

EXAMPLE 12.82

a y

x Evaluate the integral ​∫  ​​ ​∫  ​  ______ ​  2   2   ​dxdy by chang0 a x + y ing the order of integration. Solution.  The given integral is a y

x I = ​∫  ​ ​​∫  ​ ______ ​  2   2   ​dxdy. 0 a x + y The region of integration is bounded by the lines x = y, x = a, y = 0 and y = a. Thus the region of integration is shown in the figure below: Y

= 2 ​∫  ​ ​  ​r 2 ​∫ ​  ​​sin q [φ]​​2p ​   ​dq dr 0 0

a

0 __ ​ p   ​ 4

0

0

x=y (a,a)

= 4p ​∫  ​ ​ ​r 2 ​∫ ​  ​​sin q dq a

x=a a

__ ​ p   ​

( 

)

1__   ​   ​dr = 4p ​∫  ​ ​ ​r 2[-cos q ​]​04​   ​dr - 4p ​∫  ​​ r2 ​ 1 - ​ ___ ​ 2 ​  √ 0 0 3 a 4pa 1 r 1 ___ __ ____3 ___   __ __ = 4p ​ 1 + ​     ​   ​​ ​   ​   ​​ ​ = ​   ​   ​ 1 - ​     ​   ​ 3 ​ 2 ​  3 0 ​ 2 ​  √ √ __ 2p a3 ( _____ = ​   ​   ​  2 - √ ​ 2 ​  )​. 3

( 

) [  ]

( 

12.12  MISCELLANEOUS EXAMPLES 2

Evaluate ∫​   ​ ​ ​​∫  ​ ​ ​(x + y) dx dy. -2

4

)

O

X

y=0

On changing the order of integration, we first integrate with respect to y along the strip parallel to y axis. The strip extends from y = 0 to y = x. To cover the whole region, we then integrate with respect to x from x = 0 to x = a. Hence a x

a

[ 

]

y x I = ​  ​ ​ ​​  ​  ​______ ​  2   2   ​dx dy = ​  ​ ​  ​x ​ __ ​ 1x ​tan-1 _​ x ​ ​ ​ dx

∫  ∫  x 0 0 a

EXAMPLE 12.81 3

y=a

+y

∫  0

x

0

__ __ = ​∫  ​​ ​ p   ​dx = ​ p   ​a. 4 0 4

1

M12_Baburam_ISBN_C12.indd 37

12/9/2011 4:49:40 PM

12.38  n  chapter twelve EXAMPLE 12.83 By changing the order of integrations, evaluate ______ a ​√ a2 - x2     ​

Solution. We have

∞ ∞

I = ​∫  ​ ​  ​​∫  ​ ​  ​e -xy sin nx dx dy

​∫  ​ ​  ​​∫  ​  ​​(a2 - x2 - y2) dy dx. 0

0

[ 

0

∞

0

∞

]

Solution.  The region of integration is bounded by x = 0, x = a, y = 0 and the circle x2 + y2 = a2.



= ​∫  ​ ​  ​sin nx ​ ​∫  ​ ​  ​e -xy dy  ​dx

y (0,a)



 e -xy ∞ = ​∫  ​ ​  ​sin nx ​ ​ ____   ​  ​ ​  dx -x   0 0

0 ∞

0

[ 

]

∞

sin nx = ​∫  ​ ​  ​_____ ​  x    ​ dx. 0 On the other hand,

O

x

(a,0)

∞ ∞

I = ​∫  ​ ​​​∫  ​ ​​e -xy sin nx dx dy 0 0

After changing the order of integration, we have to integrate the integrand first with respect to x and then with respect to y. We take a strip ______ parallel to x-axis. The limit of x varies from 0 to √ ​ a2 - y2 ​. To cover the whole region, the limits of y will vary from 0 to a. Hence the given integral is ______ 2 2   ​ a ​√ a - y  

I = ​∫  ​ ​  ​​∫  ​  ​​[(a2 - y2) - x2] dxdy 0



a



[ 

[ 

]

______ 2

2

0

[ 

]

]

EXAMPLE 12.84

Changing the order of integration of ​∫  ​ ​  ​​∫  ​ ​  ​e  sin -xy

0

(

)

sin nx __ . nx dx dy, show that ∫​   ​ ​  ​ _____ ​  x    ​  dx = ​ p   ​ 2 0

M12_Baburam_ISBN_C12.indd 38

[ 

]

∞ e -xy = ​∫  ​ ​​​ ​  ______  2   ​(n cos nx + y sin nx)  ​​​ ​ ​dy 2 n +y 0 0

∞

[ 

]

y ∞ __ n = ​∫  ​ ​​______ ​  2   2   ​dy = ​ tan-1 __ ​ n ​ ​0​  = ​ p   ​ 2 0 n + y



(2)

0

EXAMPLE 12.85 By transforming into polar co-ordinates, evalux2y2 ate ∫∫ ​ ______     ​ dxdy over the annular region be2 x + y2 tween the circles x2 + y2 = a2 and x2 + y2 = b2, Solution.  Putting x = r cos q, y = r sin q, we have dx dy = rdrdq. Therefore

∞ ∞

∞

∞

0

where b > a.

y3 a __ __ 2 = ​ p   ​​∫  ​​ (a2 - y2) dy = ​ p   ​​ a y - ​ __ ​   ​​​ ​​  40 4 3 0 a3 p p __ 3 __ __ 3 = ​    ​​ a - ​   ​   ​= ​    ​a . 4 3 6

[ 

0

sin nx __ . = ​∫  ​ ​  ​_____ ​  x    ​ dx = ​ p   ​ 2 0

p (a2 - y2) = ​∫  ​​ ​ 0 + ​ ________  ​    - 0  ​dy 4 0 a



]

]

∞

∞

​​√a  - y   ​​ a2 - y2 x _______ + ​ ______  ​   sin -1 ​  ______      ​  ​ dy 2 ​ a2 - y2 ​ 

√

[ 

= ​∫  ​ ​  ​​ ​∫  ​ ​​e -xy sin nx dx  ​dy



From (1) and (2), it follows that

0

__________ = ∫​   ​​ ​ __​ 2x  ​√​ (a2 - y2) - x2   ​ 0 a

∞

b

2p

xy r4 sin2 q cos2 q ​ ______  2   ​dxdy = ​  ​   ​​  ​  ​  ​​ ____________  ​     rdr dq 2 2

∫∫ x

2 2

+y

∫a  ∫  0

r

∫  (  2p



)

1 - cos 4q 1  ​ (b4 - a4) ​  ​  ​​​ ​ _________ = ​ ___  ​    ​dq 16

0

2

12/9/2011 4:49:43 PM

Multiple integralS  n 12.39

[ 

]



4 4 1 1 __ __ - a ​ = ​  b______    ​ ​   ​ q - ​   ​ sin 4q  ​​​ ​  ​ 8 16 2 0



p  ​ (b4 - a4). = ​ ___ 16

EXAMPLE 12.86

____ 1 ​√1 - x 2 ​ 0

Evaluate the integral ∫​   ​ ​  ​​∫  ​  ​​​∫   0

0

2p

0

__ ​ p   ​ 2

dz dy dx . ​​​ __________ ​  _________    ​ 



​√x2 + y2 + z2 ​ 

______ ​√x2 + y2 ​ 



I = ​∫  ​ ​  ​​∫ ​   ​​​∫   ​​​ ​  2    ​  dz dy dx 2 2 ______ 0 0 ​ x + y  ​ ​√ x + y + z  ​  √ 2

1 __________ _________

2

_____  ​ 0 1 ​√ 1 - x2 

1    ​dz dy dx ____________ ______ = ​∫  ​ ​  ​​∫ ​   ​​​∫   ​​​ ​ ______________ ______ 2 0 0 ​ x + y  ​ ​ (​ x2 + y2 ​)      + z2 ​ √ √√ 2



2

_____  ​ 1 ​√ 1 - x2 

_________ 0 = ∫​   ​ ​  ​∫​  ​   ​​​[ log ​ z + ​√x2 + y2 + z2 ​   ​ dxdy ​ ]​​​​  ______ x + y  ​  0

√

0

2

2

_____  ​ 1 ​√ 1 - x2 

______ = ​∫  ​ ​ ​  ​∫ ​   ​​​ log ​ ​√x2 + y2 ​   ​ 0

0

[ 

2

_____ __ = - (log (​√2 ​ + 1) ​∫  ​​​ ​√1 - x2 ​ dx

[  [ 

_____ __ x1 ​ 1 ​ x ​√1 - x2  ​+ __ ​ 1 ​ sin-1 __ ​    ​  ​​​ ​​  = - (log (​√2 ​ + 1) ​ __ 2 2 10 __ __ __ = - (log (​√2 ​ + 1) ​ __ ​ 1 ​ sin-1 1  ​= - ​ p   ​log (​√2 ​ + 1). 2 4 EXAMPLE 12.87 Find the area in the first quadrant enclosed by y β x α the curve ​ __ ​ a ​ ​ + ​ __ ​   ​ ​ = 1, where α > 0, β > 0. b Solution.  The equation of the curve is y β α ​ __ ​ ax ​ ​ + ​ __ ​   ​ ​ = 1, α, β > 0. b The parametric form of the curve is

(  )

(  )

 __ 2

]

(  ) (  )

 __ 2

x = a cos ​ a ​t, y = b sin ​ b ​t. Therefore, the required area is

M12_Baburam_ISBN_C12.indd 39

) ( 

__ ​ p   ​ 2

)

  _____ 2

(  )

( ) 2ab ____ = ​  α   ​ ​∫ ​   ​​sin​ ​ β  +   1 ​  ​t cos​  ​ α  -   1  ​  ​t dt 0 2 ​  -  1  +  1  ​ __ ​ __ ​ 2 ​  +  2  ​ ​_________ ​ α β Γ  ​ ​ ______  ​    ​  Γ  ​ ​   ​    ​ 2ab 2 2 ____ = ​  α   ​  ​  ______________________         ​ 2 __ ​ 2 ​  +  1  +  ​ __ α ​  -  1  +  2 β______________ 2Γ  ​ ​   ​      ​ 2 1 ​ ​Γ ​ __ Γ​ __ ​ α ​ 1 ​ ​ β 2ab ____ = ​    ​ ​ ___________     ​ 2αβ 1 ​+ __ Γ​ __ ​ α ​ 1 ​+ 1  ​ β 1 __ Γ​ ​ α ​ ​Γ ​ __ ​ 1 ​ ​ β ab _________ _____ = ​    ​ ​   ​  α+β 1 ​+ __ F ​ __ ​ α ​ 1 ​ ​ β 2 _____

2 _____

( (  ) ) ( (  ) ) (  ) (  )

( 

(  )

(  ) (  )

(  )

)

D

0

0

2

Evaluation of a Double Integral 1. Evaluate ∫​ ∫  ​(4 - x2 - y2 ) dx dy if the region

1 ​√1 - x  ​ __ = - (log (​√2 ​ + 1) ​∫  ​​​    ​∫  ​  ​  ​dy dx 0 1



 __ 2 β

EXERCISES

______ _______     - log ​ ​√x2 + y2 ​  + ​√2(x2 + y2 ​   ​ ]​dxdy ____

( 

2

( α  -  1 ) 2a = ​∫  ​​​ ​ b sin ​   ​t  ​​ - ​ ___ α ​ cos ​  ​      ​  ​t  ​sin t dt 0

Solution.  We have _____  ​ 0 1 ​√ 1 - x2 

0

dx A = ​∫   ​​​ y dx = ​∫  ​​​ y __ ​    ​dt dt p p __ __ ​    ​ ​    ​

]

D is bounded by the lines x = 0, x = 1, y = 3 0, and y = __ ​   ​ . 2 35 Ans. ___ ​   ​ . 8 2. Evaluate ∫∫ e 2x+3y dx over the triangle bounded by x = 0, y = 0, and x + y = 1. Ans. __ ​ 1 ​ (2e + 1) (e - 1)2. _____ 6 a ​√a - y  ​ 2

2

 __ 1

3. Evaluate ∫​   ​ ​  ​∫ ​  ​(a2 - x2 - y2) ​ 2 ​ dx dy. 0

0

p a3 Ans. ​ ___ ​ .  6

a b

dx dy . 4. Evaluate ∫​   ​​ ​∫  ​​ _____ ​  xy     ​ 1 1

Ans. log b log a.

____ 1 ​√1+x2 ​ 

dx dy . 5. Evaluate ∫​   ​​​    ​∫ ​  _________ ​​​   ​ 2   2  0 0 1 + x + y

12/9/2011 4:49:46 PM

12.40  n  chapter twelve

[ 

1

]

p (1+cos q )

_____

y ​√1 + x  ​  1   ​  Hint: I = ​∫  ​​​ ______ ​  _____ ​​ tan-1 _______ ​  _____   2  ​   ​​ ​  ​dx 2 0 ​√ 1 + x  ​  ​√1 + x  ​  ]0

10. Evaluate ∫​   ​  ​  ​​∫  ​  ​​r 2 cos q dq dr.

2

a

0

5pa3. Ans. ​ ____  ​    8

1

1   ​  = ​∫  ​​​ ______ ​  _____ [tan-1 1 - tan-1 0 ]dx 2 0 ​√ 1 + x  ​ 



11. Evaluate ∫∫  r sin q dr dq over the cardioid r = a (1 - cos q) above the initial line.

1

1   ​  __ ______ = ​ p   ​∫​   ​​​ ​  _____ dx 4 0 ​√1 + x2 ​  _____ __ = ​ p   ​[log {x + ​√1 + x2 ​  }]​​10​​  4 __ __ = ​ p   ​log[1 + √ ​ 2 ​]  . 4



p a (1 - cos q )

Hint: ∫​ ∫  ​r sinq drdq = ​∫  ​  ​  ​​∫  ​ 

6. Evaluate ∫∫ x y dx dy over the region bounded by x = 0, y = 0, and x2 + y2 = 1. _____ 2  ​ 1 ​√ 1 - y  

∫   ∫ 

p  ​ . Ans. ​  ​ ​ ​​  ​ ​​x y dx dy = ​ ___ 0

2

0

{ 

}

√

D

common to the circle x2 + y2 = x and x2 + y2 = y.

​​r sin q dr dq a (1 - cos q )



r2 ​   ​ ​​ ​  = ​∫  ​  ​sin q ​ ​ __ ​dq 2 0 0



a2 = ​ __ ​ ​  ​  ​sin q (1 - cos q)2 dq



3 p a2 (1 - cos q ) 4a2 = ​ __ ​ ​ ​ __________  ​     ​​​ ​   ​= ​ ___ ​  . 2 3 3 0

p

2

96

7. Evaluate ∫​  ∫ ​ y dx dy, where R is the region in R x2 the first quadrant bounded by the ellipse ​ __2  ​ a y2 + ​ __2  ​= 1. b _____ x2 Hint: R = ​ (x,y);0 ≤ x ≤; a, 0 ≤ y ≤ b ​ 1 - __ ​  2  ​ ​  ​ a ab2 . ___ Ans. ​   ​   3 8. Evaluate ∫∫ xy (x + y) dx dy over the area between y = x2 and y = x. 3 ​    ​ . Ans. ___ 56 9. Evaluate ∫​  ∫ ​ xy dx dy, where A is the region

0

[  ]

p

2 2

2

0

R

∫  0

[ 

]

12. Evaluate ∫∫ r 2dq dr over the area of the circle r = acos q. 4a3 Ans. ​ ___ ​  . 9 __ ​ p   ​ 2 a cos q ______ 13. Evaluate ∫​   ​  ​​  ∫​   ​  ​​r​√a2 - r2   ​dr dq. 0 0 a3 Ans. ​ ___  ​ (3p - 4). 18 r dr dq ​  14. Evaluate ∫∫________ ​ _______ over   one loop of r2 = a2 2 2 cos2 2q. ​√a + r  ​   a Ans. __ ​   ​ (4 - p). 2

tion of two circles are (0, 0) and​

Change of Variable in a Double Integral 15. Transform the following double integral to polar coordinates and hence, evaluate the same. ______

1 ​   .​ The limits are x = 0 to x = ​ __ 1 ​  and __ ​ 1 ​ ,  ​ __

I = ​∫  ​ ​  ​​∫ ​   ​​(a2 - x2 - y2) dx dy.

Hint:

The

( 2 2 )

points

of

______ _____ 1__________ - ​√1 - 4x2   ​ y = ​   ​    to y = ​√x - x2    ​. 2

M12_Baburam_ISBN_C12.indd 40

intersec-

2

Ans. ___ ​ 1  ​ . 96

2 2  ​ a ​√ a - y  

0

0 __ ​ p   ​ 2 1

Hint: I = ​∫ ​   ​​​∫  ​ ​  ​(a2 - r2) r dr dq.

pa4 Ans. ​ ___ ​ .  8 16. Changing to polar coordinates, evaluate 0

0

12/9/2011 4:49:47 PM

Multiple integralS   n 12.41 _________ 1_________ - x2 - y2 ​ ​   ​ ​dx dy over the positive quadrant ∫∫  1 + x2 + y2 

√

of the circle x + y = 1. __ __ ​ p   ​ ​ p   ​ 2 1 ______ 2 1 2 1 r 1_____ - r2   ______ Hint: I = ​∫ ​   ​​​∫  ​​​  ​ ​   ​ ​    r dr dq = ​∫ ​   ​​​∫  ​​​  ​ ______  ​ × r 2 1+r 0 0 0 0 √ ​ 1 - r4   ​ dr dq. __ p__ Ans. ​ p   ​​  ​    ​ - 1 )​. 4( 2 2

and hence, evaluate the same. cos  q _______ __ ​ 4a        ​ ​ p   ​ 2 sin2  q

√

17. Using the transformation x + y = u and y = uv, show that

4

21. Changing to polar coordinates, evaluate ∫∫ xy 3 __

(x2 + y2) ​ 2 ​ dx dy over the positive quadrant of the circle x2 + y2 = 1.

∫∫

[xy (1 - x - y )] ​   ​  dx dy,

1 1

 __ 1 2

Change of Order of Integration 22. Change the order of integration in 1 x (2 - x)

​∫  ​ ​  ​​∫ ​  ​​f (x, y) dx dy. 0

x

 

0

 

0 0

1

 __ 1

 __ 1

 __ 1

 

0

0

1

1

3  _____

3  _____

a

3  _____

5a4 Ans. ​ ___ ​  . 6 24. Changing the order of integration, evaluate 0

 

0

0

(  ) (  )

3 3 3 = β ​ 3, __ ​   ​   ​β ​ __ ​   ​ , __ ​   ​   ​, convert to gamma function. 2 2 2 _________ 18. Evaluate ∫∫ √ ​ a2 - x2 - y2   ​dx dy over the semicircle x2 + y2 = ax in the positive quadrant. __ ​ p   ​ 2 a cos q

______ Hint: I = ​∫ ​   ​​ ​∫  ​  ​​​√a2 - r2 ​  r dr dq. 0 0 a3 p 2 . __ __ __ __________ Ans. ​ 3 ​ ​  ​ 2  ​- ​ 3 ​   ​ 2 x2 y 1- ​ __2  ​ - ​ __2  ​ a b 19. Evaluate ∫∫ ​ ​  __________       ​ ​ dx dy over the 2 y2 x __ __ 1 + ​  2  ​ + ​  2  ​ a b 2 x2 y positive quadrant of the ellipse ​ __2  ​+ ​ __2  ​= 1. a b Hint: Another from of Example 12.20.

( 

√

4a y

)

x -y 20. Change ​∫  ​   ​​∫  ​ ______ ​  2 2    ​  dx dy in polar coordinates y x +y 0 ___     ​  2

 

2

2

2a  -  x

the integral ∫​   ​ ​  ​​∫ ​  ​xy dy dx.

2  -  1 2  -  1 2  -  1 = ​∫  ​​  u3-1(1 - u) ​      ​du. ​∫  ​​  v  ​     ​(1 - v) ​      ​dv  

1 - ​√1 - y   ​

23. Changing the order of integration, evaluate

= ​∫  ​​ u2 (1 - u) ​ 2 ​ du. ​∫  ​​ v ​ 2 ​ (1 - v) ​ 2 ​ dv  

y

1

Ans. ​∫  ​ ​  ​​∫_____    ​  ​ f (x, y) dy dx.

 __ 1 2

 __ 1 2

I = ​∫  ​​  ​∫  ​​  u(1 - u) ​   ​ v ​   ​ (1 - v) ​   ​ u du dv 1

Ans. ___ ​ 1  ​.  14

 __ 1 2

taken over the area of triangle bounded by 2p  ​.  the lines x = 0, y = 0, and x + y = 1, is ​ ____ 105 Hint: x = u - y = u - uv and y = uv. Therefore, Jacobian J = u. So, dx dy = u du dv. Further, 0 ≤ u ≤ 1 and 0 ≤ v ≤ 1.Thus,  

(  )

5 __ __ Ans. ​∫  ​  ​​ ​∫ ​   ​​r cos 2 q dq dr = 8a2 ​ ​ p   ​- ​   ​   ​. 2 3 p __ 0 ​    ​

2

a

x2 ​ __ a ​ 

a

x dx dy . ​∫  ​ ​  ​​∫  ​   ​______ ​  2 2 ​  0 y x + y

pa Ans. ​ ___ ​ . 4

25. Change the order of integration a2 __ a  ​x ​ 

​∫  ​ ​  ​​∫ ​   (x + y) dx dy. and hence evaluate it. 0 x Ans. ∞. ______ a __  

​   ​​ b2 - y2  ​  b b√

26. Evaluate the integral ∫​   ​ ​  ​ ​∫ ​   ​​xy dy dx. 0

0

a2b2 . Ans. ​ ____  ​    8 27. Changing the order of integration, evaluate a__ _____ ​ ___    ​  2 2

​√2 ​  ​√a - y  ​

the integral ∫​  ​  ​​     ∫​   ​  ​​log (x2 + y2) dx dy, a > 0. 0

y

( 

)

p a 1 __ Ans. ​ ____  ​   ​ log a - ​   ​   .​ 2 4 4

4a ​

M12_Baburam_ISBN_C12.indd 41

12/9/2011 4:49:50 PM

12.42  n  chapter twelve Area Enclosed by Curves 28. Find the area bounded by the parabola y = x2 and the line y = 2x + 3. 32 Ans. ___ ​   ​ . 3 29. Find the area of the region bounded by the lines, x = -2 and x = 2 and the circle, x2 + y2 = 9. __ Ans. 4​√5 ​ + 18 sin-1 __ ​ 2 ​ sq. units. 3 30. Find the area of the cardioid r = a (1 - cos q). p a (1 - cos q )

Ans. Area = 2 ​∫  ​  ​  ​​∫  ​  0

0

3p a2 ​  ​r dr dq = ​ ____  ​   sq. 2 units.

31. Find the area outside the circle r = a and inside the cardioid. r = a (1 + cosq ). p a2 . Ans. ​ ____  ​   2 32. Find by double integration, the area lying inside the cardioid r = a (1 + cos q ) and outside the parabola r (1 + cos q ). = 1. Hint: Eliminating r between the two equations, we get cos2  q + 2 cosq = 0, which implies __. q = ± ​ p   ​ __ ​ p   ​ 2 2 1 + cos q 9p +16 . Then Area = ​∫  ​  ​​  ​∫  ​  ​​r dr dq = ​ ______  ​    12 1    __ ________ - ​ p   ​ ​  2 1  +  cos  q  ​

a

[ 

( 

x b ​ 1 - __ ​ a ​ ​

)

( 

)

]

x y abc Ans. ​∫  ​​ ​    ​∫ ​   ​​c ​ 1 - __ ​ a ​- __ ​   ​ ​dy  ​dx = ___ ​   ​ .  6 b 0 0 36. Find the volume of the region bounded by the surfaces y = x2 and x = y2 and the planes z = 0 and z = 3. __ 1 ​√x ​ 

Hint: V = ​∫  ​ ​ ​​∫ ​ ​  ​3 dy dx = 1. 0 x2

37. Calculate the volume of the solid bounded by the surfaces x = 0, y = 0, z = 0, and x + y + z = 1. Ans. __ ​ 1 ​ . 6 38. Find the volume of the cylinder x2 + y2 - 2ax = 0 intercepted between the paraboloid x2 + y2 = 2az and the xy-plane. Ans. 3p a3. 39. Find the volume bounded by the xy-plane, the paraboloid 2z = x2 + y2 = 2ax, and the cylinder x2 + y2 = 4. Ans. 4p. 40. Find the volume common to the surface y2 + z2 = 4ax and x2 + y2 = 2ax, the axis being rectangular. Hint: _______  ​ 2a ​√ 2ax - x   _______ 2 V = ∫​   ​  ​ _______ ​∫​   ​   ​​√ 4ax - y    ​ dy dx 2

0



-​√2ax - x2   ​ _______ ​ 2ax - x2   ​ 2a √

________ = 2​∫  ​ ​  ​​∫ ​   ​  ​​√4ax - y2    ​dy dx.  

0

0

33. Find, using double integration, the smaller of the areas bounded by the circle x2 + y2 = 9 and the line x + y = 3. 9 Ans. __ ​   ​ (p - 2). 4

Ans. __ ​ 2 ​ (3p + 8)a3. 3 2 41. Find the volume of the sphere x + y2 + z2 = 9. Ans. 36p.

Volume and Surface Areas as Double Integrals 34. Find the volume of the solid region under the surface z = 3 - x2 - 2y2 for x2 + y2 ≤ 1. 9p Ans. ​ ___ ​ . 4 35. Using double integration, find the volume of the tetrahedron bounded by the coordinate x y _z planes and the plane __ ​ a ​+ __ ​   ​+ ​ c ​= 1. b

42. 42. Find the area of the surface z2 = 2xy included between x = 0, x = a, y = 0, and y = b. __ ___ 2​√2 ​  Ans. ____ ​   ​   ​√ ab ​ (a + b). 3

M12_Baburam_ISBN_C12.indd 42

43. Find the area of the portion of the sphere x 2 + y 2 + z 2 = 9 lying inside the cylinder x 2 + y 2 = 3y. Hint: z 2 = 9 - x2 - y2. Then

12/9/2011 4:49:51 PM

Multiple integralS   n 12.43

(  ) (  )

∂z 2 ∂z 2 9    ​   ​ ​ + ​ __ ​   ​ ​ = _________ ​   ​. Change to 1 + ​ __ ∂x ∂y 9 - x2 - y2

_____ __ ​ p   ​ ​ a2 - r2   ​ 2 a cos q √

50. Evaluate ∫​  ​   ​​  ​∫  ​  ​​ ​∫ ​   ​​r dz dr dq.

polar coordinates. Surface area 3 = 4 ​∫ ​   ​​   ​∫  ​  ​ ______ ​  _____      ​dr dq = 18p - 36. 0 0 ​ 9 - r2 ​  √

51. Evaluate

44. Find the area of the portion of the cylinder x2 + y2 = 4y lying inside the sphere x2 + y2 + z2 = 16. Ans. 64 sq. units. 45. Using double integration, find the volume generated by the revolution of the cardioid r = a (1 - cos q ) about its axis. Hint: Volume of revolution p a (1 - cos q)

0

8 ​​r2 sin q dr dq = __ ​   ​ pa3. 3

0

46. Find the volume generated by revolving the 2 x2 y ellipse ​ __2  ​+ ​ __2  ​= 1 about the y-axis. a b Ans. __ ​ 4 ​ pa2b. 3 Evaluation of Triple Integral

x2 + y2 + z2 ≤ a2, 0 ≤ z ≤ h}, using cylindrical polar coordinates. __ ​ p   ​ 2 2p h

Ans. ​∫ ​   ​​​∫  ​  ​​​∫  ​ ​​r dz dq dr = p a2h. 0 0 0

Volume as a Triple Integral 52. Find the volume bounded by the surface y2 x2 __ x2 + y2 = a2 and ​ __  ​   + ​  p q ​ = 2z, p > 0, q > 0. p a4 1 __ Ans. ​ ___ ​  ​ __ ​   ​+ ​ 1 ​ ​. 8 p q

(  )

53. Find the volume of the paraboloid of revolution x2 + y2 = 4z cut off by the plane z = 4. ______  ​ 4 4 ​√ 16 - x2 

Hint: V = 4 ​∫  ​ ​  ​​∫ ​   ​  ​​∫   ​​​  dx dy dz = 32p.

0

49. Evaluate ∫​   ​ ​ ​​∫  ​​​  ​∫ ​  ​  ​xyz dz dy dx.

( 

0

x   +  y ______ ​   ​    2

2

4

Ans. ____ ​  1   ​ . 720 48. Evaluate ∫∫∫z 2dx dy dz over the sphere. x2 + y2 + z2 = 1. ___ Ans. ​ 4p   ​.  __ 15 3 1 ​√ xy ​ 

M12_Baburam_ISBN_C12.indd 43

 

V

 

47. Evaluate ∫​   ​​​  ​∫  ​  ​  ​∫  ​  ​​xyz dx dy dz.

1 __ ​ 1x ​ 0

)

(

__ p Ans. a3 ​  ​ p   ​- ​ __  ​  .​ 6 9

∫∫​∫  ​ ​dx dy dz, where V = {(x, y, z);

0

1 1-x1-x-y 0 0

0

 

 

= 2p ​∫  ​  ​  ​​∫  ​ 

0

0

__ ​ p   ​ 2 3 sin q

)

13 1 Ans. __ ​ 1 ​ ​ ___ ​   ​ - __ ​   ​ log 3  .​ 3 3 2

54. Find the volume bounded above by the sphere x2 + y2 + z2 = 2a2 and below by the paraboloid az = x2 + y2. __ 4√ ​ 2 ​  __ 7 3 ____  - ​   ​   ​. Ans. p a ​ ​   ​  3 6

( 

)

55. Show that the volume enclosed by the cylin128a3 . der x2 + y2 = 2ax and z2 = 2ax is ​ _____  ​    15 56. Show that the volume of the wedge intercepted between the cylinder x2 + y2 = 2ax and the planes z = x and z = 2x is p a3. Hint: See Example 12.76.

12/9/2011 4:49:53 PM

This page is intentionally left blank.

M11_Baburam_ISBN _C11.indd 16

12/14/2011 3:48:19 PM

13

Vector Calculus

We know that scalar is a quantity that is characterized solely by magnitude whereas vector is a quantity which is characterized by both magnitude and direction. For example, time, mass, and temperature are scalar quantities whereas displacement, velocity, and force are vector quantities. We represent a vector by an arrow over it. Geometrically, we represent   a a vector  by a directed line segment PQ , where a has direction from P to Q. The point P is called the initial point and the point Q is  called the terminal point of a . The length PQ  of this line segment  is the magnitude of a . Two  vectors a and b are said to be equal if they have the same magnitude and direction. The  a and a scalar m is a vector product of a vector  m a with magnitude |m| times the magnitude of  a with direction, the same or opposite to that of  a , according as m > 0 or m < 0. In particular,  if m = 0, then m a is a null vector 0 . A vector with unit magnitude is called a unit vector. If   a is non-zero vector, then aa = aa is a unit vector  having the same direction as that of a and is denoted by aˆ .   If a , b and c are vectors and m and n are scalars (real or complex), then addition and scalar multiplication of vectors satisfy the following properties:   (i) a + b = b + a (Commutative law for addition).       (ii) a + b + c = a + b + c (Associative law for addition).      (iii) m a + b = ma + na (Distributive law for addition).    (iv) (m + n)a = ma + na (Distributive law for scalars).

(

(

) (

     a + 0 = a = 0 + a (Existence of identity for addition).

(v)









(Existence of inverse for addition).   ma = m a .

(vii)

  (viii) m(na ) = (mn)a.   (ix) n(ma ) = m(na ). The unit vectors in the directions of positive x-, y-, and z- axes of a three-dimensional, rectangular coordinate system are called the rectangular unit vectors and are denoted, respectively, by iˆ, ˆj , and kˆ. Let a1, a2, and a3 be the rectangular coordinates of the terminal point of vector  a with the initial point at the origin O of a rectangular coordinate system in three dimensions. Then, the vectors a1iˆ, a2 ˆj , and a3 kˆ are called rectangular component vectors or  simply component vectors of a in the x, y, and z directions, respectively. Z

→

a

)

a2iˆ

)

M13_Baburam_ISBN _C13 Part I.indd 1





(vi) a + (−a ) = 0 = (−a ) + a

O

a2kˆ Y

a2 jˆ Z

1/2/2012 12:16:02 PM

13.2  n  chapter thirteen The resultant (sum) of a1iˆ, a2 ˆj , and a3 kˆ is the  vector a and so,  a = a1iˆ + a2 ˆj + a3 kˆ.  Further, the magnitude of a is  a = a 12 + a22 + a32 . In particular, the radius vector or position  vector r from O to the point (x, y, z) in a threedimension space is expressed as  r = xiˆ + yjˆ + zkˆ

and

M

θ

Projection of a Vector on another Vector.   Let OM be the projection of the vector a on the vector b . Then    a.b = a b cos θ  a.b   = a cos θ b = OM, Length of the projection. a

M

Thus    a.b . Length of projection of a on b = b Further, the projection as a vector is

M13_Baburam_ISBN _C13 Part I.indd 2

a

Thus, “The  scalar (dot) product of two vectors  a and b is the product of the length of the vector a and the length of the projection  of the vector b in the direction of the vector a ” Properties of Scalar product of two vectors following properties.   (i) a.b = b .a. In fact,    a.b = a b cos θ =

B

Product. The scalar   a and b satisfies the

   b a cos θ = b .a.

(ii) Two vectors are perpendicular if and only if their scalar product is zero.   In fact, assume first that a and b are perpendicular. Then angle between them is q = 90º and so      a.b = a b cos θ = a b cos 90º = 0 

A

θ

L

O

The scalar product or dot  or inner  product product of two vector a and b is a scalar defined b    a.b = a b cos θ , where q is the angle between the vectors a and  b and 0 < – q< – π.

b

 a.b   2 b. b

Geometrical Interpretation of Scalar Product     Let a = OL, b = OM and let q be the angle  between a and b . Then    a.b = a b cos θ    = a (Projection of b in the direction of a )

 r = r = x2 + y 2 + z 2 .

O

 =  

b

Scalar Product or Dot Product or Inner Product of Vectors

or

    a.b  b   OM = OM eb = b b 



Conversely, if a.b = b = 0, then   a . b cos θ = 0 ⇒ cos θ = 0 ⇒ θ = 90º   and so a and b are perpendicular to each other. In particular, if iˆ, ˆj and kˆ are mutually perpendicular unit vectors, then iˆ. ˆj = ˆj. kˆ = kˆ.iˆ = (1)(1) cos 90º = 0

And

iˆ.iˆ = ˆj. ˆj = kˆ.kˆ = (1)(1) cos 0º = 1.

1/2/2012 12:16:04 PM

vector calculuS  n 13.3   (iii) a. b + c = a.b + a.c.

(

since all other terms are zero by property (ii). As a deduction, we have    2 a.a = a a cos 0º = a

)

b

= a12 + a22 + a32 .

c

(v) For any scalar t      t a.b = ta.b a. t b = a.b t.

( ) (

b+c

Proof: follows directly from the definition of dot a



In fact, let ea be a unit vector in the direction of     a . Then Projection of b + c on a = projection     of b on a a + Projection of c on a . or or

(

       b + c .ea = b .ea + a. ea

)





( b + c ) . a e

a

or

(

      = b . a ea + c . a . ea

)

 (iv) If a = a1iˆ + a2 ˆj + a3 kˆ and bˆ = b1iˆ + b2 ˆj + b3 kˆ,  a.b = a1b1 + a2 b2 + a3b3 .

(

(

)(

= a1iˆ ⋅ b1iˆ + b2 ˆj + b3 kˆ



)

)

( ) + a kˆ. ( b iˆ + b ˆj + b kˆ ) 1

2

3

= a1b1iˆ ⋅ iˆ + a1b2 iˆ. ˆj + a1b3iˆ.kˆ



+ a2 b1 ˆj.iˆ + a2 b2 ˆj. ˆj + a2 b3 ˆj.kˆ

ˆˆ ˆˆ ˆˆ + a3b1k .i + a3b2 k . j + a2 b3 k .k = a1b1 + a2 b2 + a3b3 ,

M13_Baburam_ISBN _C13 Part I.indd 3

  a = 3iˆ + 4 ˆj and b = 6iˆ − 3 ˆj + 2kˆ .

Solution. By dot product, we have

   a.b = a b cos θ º

 a.b 3(6) − 4(3) + 4(2) 2 cos θ =   = = 2 2 a b 3 + 4 36 + 9 + 4 5(7) =

2 35

Find the value of a for which the vectors 3iˆ + 2 ˆj + 9kˆ and iˆ + a ˆj + 3 kˆ are perpendicular. Solution. The given vectors will be perpendicular if their dot product is zero, that is, if 3iˆ + 2 ˆj + 9kˆ . iˆ + ajˆ + 3kˆ = 0 or 3 (1) + 2(a) + 9(3) = 0 which yields a = –15.

(

+ a2 ˆj. b1iˆ + b2 ˆj + b3 kˆ 3

Find the angle between the vectors

EXAMPLE 13.2

)

In fact, using (iii), we have  a.b = a1iˆ + a2 ˆj + a3 kˆ ⋅ b1iˆ + b2 ˆj + b3 kˆ

EXAMPLE 13.1

and so θ = cos −1 ( 352 ) .

or by property (i),      a. b + c = a.b + a.c . then

product.

or

     b + c .a = b .a + c .a

(

) ( ) ( )

)(

)

EXAMPLE 13.3

(Cauchy-Schwarz Inequality). Show that  2 2 2 a.b ≤ a b , that is 2 ( a1b1 + a2b2 + a3b3 ) ≤ ( a12 + a22 + a32 )

( )

2 2 2 × ( b1 + b2 + b3 ) . Solution. Since cos θ ≤ 1 , we have 2  2   2 2 a.b ≤  a b cos θ  ≤ a b   2

( )

1/2/2012 12:16:05 PM

13.4  n  chapter thirteen or

(a1b1 + a2 b2 + a3b3 )

  OA = iˆ − 2 ˆj + 2kˆ, OB = 2iˆ + ˆj − kˆ and  OC = 3iˆ − ˆj + 2kˆ.

2

= (a + a22 + a32 )(b12 + b22 + b32 ). 2 1

A

EXAMPLE 13.4

Determine the projection of the vector iˆ − 2 ˆj + kˆ on the vector 6iˆ − 3 ˆj + 2kˆ. Solution. Let aˆ = iˆ − 2 ˆj + kˆ and b = 6iˆ − 3 ˆj + 2kˆ.  Then, the unit vector in the direction of b is   b 6iˆ − 3 ˆj + 2kˆ 6 3 2 eb =  = = iˆ − ˆj + kˆ. 7 7 36 + 9 + 4 7 b  Therefore, the length of projection of a on the  vector b is 3 2    6 a.eb = iˆ − 2 ˆj + kˆ .  iˆ − ˆj + kˆ  7 7 7  

(

=

)

6 6 2 + + = 2. 7 7 7

The projection (as a vector) is given  a.b  6 + 6 + 2 ˆ 6i − 3 ˆj + 2kˆ ⋅ 2b = 72 b

)

(

=

(

)

2 ˆ 12 6 4 6i − 3 ˆj + 2kˆ = iˆ − ˆj + kˆ. 7 7 7 7

C

O

Then    (i) OA + AB →= OB or   

AB = OB − OA = iˆ + 3 ˆj − 3kˆ.

Therefore AB = 1 + 9 + 9 = 19.    (ii) OC + CA = OA or   

CA = OA − OC = −2iˆ − ˆj.

Therefore CA = 4 + 1 = 5.    (iii) OB + BC = OC or   

BC = OC − OB = iˆ − 2 ˆj + 3kˆ.

EXAMPLE 13.5

  If a and b are two non-zero vectors, show that 2  2  2 2 a +b + a −b = 2 a + 2 b . Solution. We have

 2  2 a +b + a −b         a +b . a +b + a −b . a −b         = a.a + a.b + b .a + b .b + a.a − a.b − b .a + b .b  2  2 = 2 ( a.a ) + 2 b .b = 2 a + 2 b .

(

B

)(

) (

)(

)

( )

EXAMPLE 13.6

Find the sides and angles of the triangle whose vertices are iˆ − 2 ˆj + 2kˆ, 2 iˆ + ˆj − kˆ 3 iˆ − ˆj + 2kˆ. Solution. The position vector of the vertices A, B and C are

M13_Baburam_ISBN _C13 Part I.indd 4

Therefore BC = 1 + 4 + 9 = 14. Further, angle between AB and BC is given by   AB.BC cos θ =   AB BC =

=

( iˆ + 3 ˆj − 3kˆ ) ⋅ ( iˆ − 2 ˆj + 3kˆ ) iˆ + 3 ˆj − 3kˆ iˆ − 2 ˆj + 3kˆ

1− 6 − 9

=

−14

19 14 19 14 Hence θ = cos −1 14 . 19

=−

14 . 19

 Similarly, the angles between the vectors BC    and CA and between CA and AB are given respectively by cos −1

5 19

and cos–1 (0).

1/2/2012 12:16:07 PM

vector calculuS  n 13.5 or

EXAMPLE 13.7

Show that the triangle formed by the vectors   a = 3iˆ − 2 ˆj + kˆ, b = iˆ − 3 ˆj + 5kˆ and  a = 2iˆ + ˆj − 4kˆ is a right triangle. Solution. We note that

 a.b = 3 + 6 + 5 = 14  b .c = 2 − 3 − 20 = −21

    a −b ≤ a − b

(2)

Combining (1) and (2), we get       a −b ≤ a − b ≤ a −b or

    a − b ≤ a −b .

 c .a = 6 − 2 − 4 = 0.   Since dot product of a and c is zero, it follows

Vector Product or Cross Product of Two Vectors. The

EXAMPLE 13.8   If a and b are two vectors, show that

Where q in the angle between a and b such that 0< –q< – π and eˆin the unit vector perpendicular   to both a and b . The direction of a × b is  b a perpendicular    to the plane of and such that a , b and a × b form a right-handed triad of vectors.

that they are perpendiculars to each other. Thus the angle between two sides formed by the   vector a and c is 90º. Hence the triangle formed by the given vectors is a right triangle.

    (i) a + b ≤ a + b   (ii) a − b ≤ a − b .

  Solution. (i) Let q be the angle between a and b . Then

vector product or cross product of two vectors   a and b is a vector defined by     a × b = a b sin θ eˆ, 

ê

 2         a + b = a + b . a + b = a.a + a.b + b .a + b .b 2 2    = a + b + 2a.b ,since a.b = b .a

(

)(

)

B b

2 2   = a + b + 2 a . b cos θ 2 2   ≤ a + b + 2 a b ,since cos θ ≤ 1



θ O

a

A

The equality in this relation holds if and only if cos q = 1, that is, if and only if q = 0, that is, if and only if the vectors are parallel and in the same direction. In other words, equality holds if  and only if the vectors a and b are like vectors. (ii) To prove (ii) We note that        a = a −b +b ≤ a −b + b

    In particular, if a = b or if a is parallel to b ,   then a × b = 0.   Geometrical Interpretation of a × b . Let eˆ be a unit vector normal to the plane of the triangle   OAB having a and b as its adjacent sides. Then     1    a × b = a b sin θ eˆ = 2  a b sin θ  eˆ 2   = 2∆OAB.   Thus a × b represents twice the vector area of  the triangle having a and b as adjacent sides.

using (i) proved above or (1)   Interchanging a and b , we get       b − a ≤ b −a = a −b

Properties of Vector Product. The vector product satisfies the following properties: (i) The vector product is anti-commutative. In fact, we have     b × a = b a sin θ ( −eˆ )

Hence

  = a+b

(

). 2

    a +b ≤ a + b .

    a − b ≤ a −b .

M13_Baburam_ISBN _C13 Part I.indd 5

1/2/2012 12:16:09 PM

13.6  n  chapter thirteen     = − a b sin θ eˆ = − a × b

(

Hence

)

    a × b = − b × a (anti-commutative law)

(

)

(ii) The vector product is distributive over vector addition.    In fact, if a , b and c are three vectors, then        a × b × c = a × b + a × c (distributive law)

(

)









  Show that for vector a and b ,  2  2 2 2 a × b + a.b = a b





 2  2 a × b + a.b



2     = a b sin θ eˆ + a b cos θ

(iv) If iˆ, ˆj and kˆ are three mutually perpendicular unit vectors, then  iˆ × iˆ = ˆj × ˆj = kˆ × kˆ = 0

ˆj × iˆ = − kˆ, kˆ × ˆj = −iˆ, iˆ × kˆ = − ˆj.   (v) If a = a1iˆ + a2 ˆj + a3 kˆ and b = b1iˆ + b2 ˆj + b3 kˆ,

then

kˆ a3 . b3

In fact, using distributive law, we have   a × b = a1iˆ × a2 ˆj + a3 kˆ × b1iˆ × b2 ˆj + b3 kˆ

(

) (

(

)

)

2 2 2 2 a b (sin 2 θ + cos 2 θ ) = a b . EXAMPLE 13.10

Find a unit vector perpendicular to both   a = 2iˆ + ˆj − kˆ and b = iˆ + ˆj + 2kˆ.   Solution. The vector a × b is perpendicular to both   a and b . But   a × b = 2iˆ + ˆj − kˆ × iˆ + ˆj + 2kˆ

(

= a1iˆ × b1iˆ × b2 ˆj + b3 kˆ + a2 ˆj

(

× b1iˆ × b2 ˆj + b3 kˆ



)



) = a b ( iˆ × iˆ ) + a b ( iˆ × ˆj ) + a b ( iˆ × kˆ ) + a b ( ˆj × iˆ ) + a b ( ˆj × ˆj ) + a b ( ˆj × kˆ ) + a b ( kˆ × iˆ ) + a b ( kˆ × ˆj ) + a b ( kˆ × kˆ ) (

+ a3 kˆ × b1iˆ × b2 ˆj + b3 kˆ

1 1



1 2

1 3

2 1

2 2

2 3

3 1

3 2

3 3

= a1b2 kˆ − a1b3 ˆj − a2 b1kˆ + a2 b3iˆ ˆ

ˆ

+ a3b1 j − a3b2 i = ( a2 b3 − a3b2 ) iˆ + ( a3b1 − a1b3 ) ˆj + ( a1b2 − a2 b1 ) kˆ

M13_Baburam_ISBN _C13 Part I.indd 6

2

2 2 2 2 = a b sin 2 θ + a b cos 2 θ

iˆ × ˆj = kˆ, ˆj × kˆ = iˆ, kˆ × iˆ = ˆj

ˆj a2 b2

kˆ a3 . b3

EXAMPLE 13.9

t (a× b) = (t a ) × b = a× (t b) = (a× b)t

iˆ   a × b = a1 b1

ˆj a2 b2

Solution. We have

(iii) For a scalar t, we have 



iˆ = a1 b1



) (

)

iˆ ˆj kˆ = 2 1 −1 1 −1 2 = iˆ(2 − 1) − ˆj (4 + 1) + kˆ(−2 − 1) ˆ

ˆ

ˆ

= i − 5 j − 3k .  Therefore unit vector in the direction of a × b is iˆ − 5 ˆj − 3kˆ iˆ − 5 ˆj − 3kˆ iˆ − 5 ˆj − 3kˆ = = eˆ = . 1 + 25 + 9 35 iˆ − 5 ˆj − 3kˆ EXAMPLE 13.11

Show that the area of a parallelogram with sides     a and b is a × b .   Solution. Let q be the angle between a and b .  Then Area of the parallelogram = (height h) b

1/2/2012 12:16:10 PM

vector calculuS  n 13.7     = a sin θ b = a × b .

or

EXAMPLE 13.12

   In DABC with sides a , b and c , prove the law of sines: a b c = = . sin A sin B sin C    Solution. Let a , b and c be the sides of the triangle as shown below:

      a b sin C = b c sin A = c a sin B

or    Dividing throughout by a b c , , we get sin C sin A sin B  =  =  c a b or a b c = = . sin A sin B sin C

A

c

EXAMPLE 13.13  If a = 3iˆ − ˆj + 2 kˆ , b = 2iˆ + ˆj − kˆ and  ˆ c = i − 2 ˆj + 2 kˆ , determine

b



B

C

or

      a× a +b + c = a×0

(

)

     a ×b + a ×c = 0 Similarly,       b × a +b + c = b ×0

(

or and or

iˆ ˆj kˆ   b × c = 2 1 −1 = −5 ˆj − 5kˆ . 1 −2 2

       a ×a + a ×b + a ×c = 0

or

or

iˆ ˆj kˆ   a × b = 3 −1 2 = −iˆ + 7 ˆj + 5kˆ , 2 1 −1

    a + b + c = 0.

Therefore

(1)

       b ×a + b ×b + b ×c = 0

ˆj kˆ iˆ    a × b × c = −1 7 5 = −iˆ + 7 ˆj + 5kˆ 1 −3 2

)

= iˆ(14 + 10) − ˆj ( −2 − 5) + kˆ (2 − 7) (2)

= 24iˆ + 7 ˆj − 5kˆ , iˆ ˆj kˆ    a × b × c = 3 −1 2 0 −5 −5

(

      c × a +b + c = c ×0

(

Therefore

(

)

     −a × b + b × c = 0.

)

)

       c ×a + c ×b + c ×c = 0

or      c × a − b × c = 0. From (1), (2) and (3), we have       a ×b = b ×c = c ×a

M13_Baburam_ISBN _C13 Part I.indd 7



Solution. We have

a

Then



( a × b ) × c, a × ( b × c ) and a.( b × c ) .

= iˆ(5 + 10) − ˆj ( −15) + kˆ ( −15) (3)

= 15iˆ + 15 ˆj − 15kˆ . Thus, we observe that

      a × b × c ≠ a × b × c.

(

) (

)

1/2/2012 12:16:11 PM

13.8  n  chapter thirteen Furthermore,    a. b × c = 3iˆ − ˆj + 2 kˆ ⋅ −5 ˆj − 5kˆ .

) (

(

)(

= 5 – 10 = – 5.

+ (b1c 2 − b2c1 ) kˆ  = a1 (b2c3 − b3c 2 ) + a2 (b3c1 − b1c3 )

)

 



Triple Product of Vectors. Let a, b and c be three vectors. Then the products of the form

        a.b c , a. b × c and a × b × c are called triple    products of the vectors   a, b and c . Note that a.b .c is not defined

( )

(

)

(

( )



 

( ) 

a1 a2 = b1 b2 c1 c 2

)

These products follow the following laws:

( )

(i) a.b c ≠ a b .c .  In fact, if a = a1iˆ + a2 ˆj + a3 kˆ  b = b1iˆ + b2 ˆj + b3 kˆ and



( a.b ) c

(

= (a1b1 + a2 b2 + a3b3 ) c1iˆ + c2 ˆj + c3 kˆ

)

= (a1b1c1 + a2b2c1 + a3b3c1 )iˆ



+(a1b1c 2 + a2b2c 2 + a3b3c3 ) ˆj

and

  a b .c = (a1iˆ + a2 ˆj + a3 kˆ )(b1c1 + b2c 2 + b3c3 )

( )

and also,



+(a2b1c1 + a2b2c 2 + a2b3c3 ) ˆj



+(a3b1c1 + a3b2c 2 + a3b3c3 ) kˆ .

We note that

In fact,

)

(



c1 c 2 c3    = a1 a2 a3 = c . a × b . b1 b2 b3

)

(

= a1iˆ + a2 ˆj + a3 kˆ

(

 

(

)



(

)



) ( ) 

(iii) a. b × c = a × b .c In fact

)

(

)

   = a × b .c (commutation of dot product).

(

kˆ b3 c3

)

× (b2c3 − b3c 2 )iˆ + (b3c1 − b1c3 ) ˆj

M13_Baburam_ISBN _C13 Part I.indd 8

)

      a ⋅ b × c = b ⋅ (c × a ) = c ⋅ a × b .

(

( ) ( )

iˆ ˆj    ˆ ˆ ˆ a. b × c = (a1i + a2 j + a3 k ) ⋅ b1 b2 c1 c 2

c3 b3 a3

      a. b × c = c . a × b (proved above)

    a.b c ≠ a b .c .    (ii) a. b × c = b . ( c × a ) = c. a × b .

( )

a3 c1 c 2 b3 = − b1 b2 c3 a1 a2

(

Hence

b3 a3 c3

b3    c 3 = b . (c × a ) , a3

a1 a2 = b1 b2 c1 c 2

= (a1b1c1 + a1b2c 2 + a1b3c3 )iˆ

(

b1 b2 = c1 c 2 a1 a2

+(a1b1c3 + a2b2c 2 + a3b3c3 ) kˆ ,



a3 b3 = ∆, say. c3

But, by property of determinant, a1 a2 a3 b1 b2 = b1 b2 b3 = − a1 a2 c1 c 2 c3 c1 c 2

 c = c1iˆ + c2 ˆj + c3 kˆ,

then

+ a3 (b1c2 − b2 c1 )



)



(





) (





)



(iv) a × b × c ≠ b × c × c . In fact,

   a × b ×c

(

)

iˆ ˆj  = a × b1 b2 c1 c 2

kˆ b3 c3

1/2/2012 12:16:11 PM

vector calculuS  n 13.9  = a × (b2c3 − c 2b3 )iˆ − (b1c3 − c1b3 ) ˆj

= (a2b1c 2 − a2b2c1 − a3b3c1 + a3b1c3 ) iˆ 

 +(b1c 2 − b2c1 ) kˆ ] 

iˆ = a1 b 2 c 3 − b 3c 2

+(a3b2c3 − a3b3c 2 − a1b1c 2 − a1b2c1 ) ˆj

ˆj kˆ a2 a3 , −b1c3 + b3c1 b1c 2 − b2c1

   a ×b ×c

)

iˆ = a1 b1

(

(

iˆ = a2b3 − a3b2 c1

+(a1b3c1 − a1b1c3 − a2b2c3 + a2b3c 2 ) kˆ . Hence

)

ˆj kˆ −a1b3 + a3b1 a1b2 − a2b1 c2 c3

      a × b × c ≠ a × b × c.        (v) a × b × c = ( a.c ) b − a.b c

)

(

(

) (

) ( )

       a × b × c = ( a.c ) b − b .c a.

)

( )

In fact, we note that, as in part iv, the left hand side of first relation i    a × b ×c

(

)

iˆ ˆj kˆ = ( a1i + a2 j + a 3 k ) × b1 b2 b3 c1 c2 c3

ˆj iˆ kˆ = a1 a2 a3 (b2c3 − b3c 2 ) (b3c1 − b1c3 ) (b1c 2 − b2c1 )

M13_Baburam_ISBN _C13 Part I.indd 9

)

+(a3b2c3 − a3b3c 2 − a1b1c 2 − a1b2c1 ) ˆj

Expanding these determinants, we find tha

and

)

= (a2b1c 2 − a2b2c1 − a3b3c1 + a3b1c3 )iˆ 

)

 +(a1b2 − a2b1 ) kˆ  × c1iˆ + c 2 ˆj + c3 kˆ 

(

(

(

 = (a2b3 − a3b2 )iˆ − (a1b3 − a3b1 ) ˆj 





( ) 

−(a1b1 + c 2b2 + a3b3 ) c1iˆ + c 2 ˆj + c3 kˆ

kˆ a3 × c1iˆ + c2 ˆj + c3 kˆ b3

ˆj a2 b2

 

Also b (a.c ) − a.b c

= b1iˆ + b2 ˆj + b3 kˆ (a1c1 + a2c 2 + a3c3 )

where as

(

+(a1b3c1 − a1b1c3 − a2b2c3 + a2b3c 2 ) kˆ .

       a × b × c = (a.c ) b − a.b c .

(

)

( )

The second relation can be established similarly.    The product a. b × c is called the scalar triple product or box product and is denoted by [abc].

(



)

(





The product a × b × c triple product.

)

is called the vector

EXAMPLE 13.14

Show that           a × b × c + b × (c × a ) + c × a × b = 0.

(

)

(

)

Solution. By property (v), we have

       a × b × c = b ( a.c ) − c a.b ,

(1)

       b × (c × a ) = c b .a − a b .c ,

(2)

(

)

( )

( ) ( ) ) ( )

       c × a × b = a c .b − b (c .a ) .

(



(3) Adding(l), (2), (2) and (3), we have          a × b × c + b × (c × a ) + c × a.b = 0.

(

)

( )

Geometrical Interpretation of Scalar Triple Product    The scalar triple product a. b × a equals the volume of the parallelepiped having the vectors    a, b and c as its three coterminus edges.

(

)

1/2/2012 12:16:12 PM

13.10  n  chapter thirteen a1 a2    [ab c ] = a. b × c = b1 b2 c1 c 2

b×c

(

A a

C

O

1( λ − 1) − 1(2λ + λ ) + 1( −2 − λ ) = 0 or 3λ − 3 = 0 or λ =1.

   = a cos θ b × c    = a. b × c .

(

EXAMPLE 13.15

)





Show that three vectors a, b and c are coplanar if and only if [abc] = 0.    Solution. Suppose that the vectors a, b and c are   coplanar. Since  a × b is perpendicular to the   plane of a and b and since a, b ,c are coplanar,    it follows the a × b is perpendicular to c .    Hence a × b .c = 0 or [abc] = 0. Conversely, suppose that [abc] = 0, that is      a × b .c = 0. Therefore is perpendicular a × b    to c . But  a × b isperpendicular to the plane of  c lies in the plane of a and b . a and b . Hence    Hence a, b and c are coplanar.

)

)

EXAMPLE 13.16

Find the value of l which makes the vectors iˆ − ˆj + kˆ , 2iˆ + ˆj − kˆ and λ iˆ − ˆj + λ kˆ coplanar Solution. The necessary and sufficient condition   for three vectors a, b and c to be coplanar is that the box product [abc] = 0. But

M13_Baburam_ISBN _C13 Part I.indd 10

−1 1 1 −1 = 0 −1 λ

or

 h = a cos θ

Therefore Volume of the parallelepiped = h × area of parallelogram OBCD

(

λ

B

b

then

1 2

D

  The vector b × c is perpendicular to the vectors     b and c . Further, b × c gives the area of the  parallelogram OBCD having adjacent sides b  and c . If h is the height of the parallelepiped,

(

)

Thus for coplanarity, we must have

h c

θ

a3 b3 = 0 c3

EXAMPLE 13.17

Find the volume of a parallelepiped whose three coterminus edges are represented by   a = 2iˆ − 3 ˆj + 4 kˆ , b = iˆ − 2 ˆj − kˆ and  c = 3iˆ − ˆj + 2 kˆ . Solution. The volume of the parallelepiped with the given edges is

      V =  a b c  = a. b × c

(

a1 a2 = b1 b2



c1

c2

)

a3 2 −3 4 b3 = 1 2 −1 c3

3 −1

2

= 2(4 − 1) + 3(2 + 3) + 4(−1 − 6) = 6 + 15 − 28 = −7 cubic units.    The negative sign shows that a, b , c , constitute

a left hand triad of vectors. Thus V = 7 cubic units EXAMPLE 13.18

Find the volume of a tetrahedron whose coterminus edges OA,  OB and OC are   represented by vectors a, b and c Solution. Let OABC be the tetrahedron in which the  edges OA, OB and OC are represented by a, b and c respectively.

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vector calculuS  n  13.11 function of time t. In rectangular coordinate  system, the vector function f can be expressed in a component form as   r = f (t ) = f 1iˆ + f 2 ˆj + f 3 kˆ ,

A

a b

c

O

B

C

Then

1   b × c and 2 Volume of the tetrahedron OABC 1 = height × (area of the base) 3 1    1  = a. b × c =  a b c  . 6 6 Area of ∆OBC =

(

(

)

)

EXAMPLE 13.19

Find volume of the tetrahedron formed by the points (1,1,1), (2,1,3),(3,2,2) and (3,3,4). Solution. Let A(1, 1, 1), B(2, 1, 3), C(3, 2, 2) and D(3, 3, 4) be the given points. Then  A B = 2iˆ + ˆj + 3kˆ − iˆ + ˆj + kˆ = iˆ + 2 kˆ

( ( (

) ( ) ( ) (

)

 AC = 3iˆ + 2 ˆj + 2kˆ − iˆ + ˆj + kˆ = 2iˆ + ˆj + kˆ  A D = 3iˆ + 3 ˆj + 4 kˆ − iˆ + ˆj + kˆ = 2iˆ + 2 ˆj + 3kˆ .

) )

Hence Volume of tetrahedron 1    =  A B A C A D  6 1 0 2 1 1 = 2 1 1 = [1 + 0 + 4] 6 6 2 2 3 =

5 cubic units. 6

13.1  DIFFERENTIATION OF A VECTOR  A vector r is said to be a vector function of a scalar variable t if to each value of t there  corresponds a value of r .   A vector function is denoted by r = r (t ) or    r = f (t ). For example, the position vector r of a particle moving along a curved path is a vector

M13_Baburam_ISBN _C13 Part I.indd 11

where f1, f2, and f3 are scalar functions of t and  are called components of f .   Let r = f (t ) be a vector function of the scalar variable t. If Dt denotes a small increment in t   and ∆r the correspondingincrement in r , then   dr ∆r f (t + ∆t ) − f (t ) = lim = lim , dt ∆t → 0 ∆t t → 0 ∆t  if exists, is called the ordinary derivative of r with respect to the scalar t.  Since dr is itself a vector depending on t, we can dt further consider its derivative with respect to t. 2 If this derivative exists, it is denoted by d 2r . dt  Similarly, higher derivatives of r can be defined.    Geometric Significance of ddrt : Let r = f (t ) be the vector equation of a curve C in space. Let P and Q be two neighboring points on C    with position vectors r and r + δ r . Then,      and so, OP = r , OQ  = r + δr     PQ = OQ − OP = r + δ r − r = δ r . 

Therefore, δδrt is directed along the chord PQ. As δ t → 0 , that is, as Q → P , the chord PQ tends to the tangent to the curve C at P. Hence,   dr dr dt = lim δt is a vector along the tangent to the δt → 0 curve at P. dr

r r

O

r

Q

r P

Unit Tangent Vector to a Curve: Suppose that we take

an arc length s from any point, say A, on the curve C, up to the point P as the parameter, instead of t. Then, AP=s, A Q = s + δ s , and so,

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13.12  n  chapter thirteen 

PQ = δ s . In this case, dr ds will be vector along the tangent at P. Further,   dr δr chord PQ = lim = lim = 1. ds δ s → 0 δ s Q → P Arc PQ  dr ds

Hence,

is the unit vector tˆ along the tangent at P.

Theorem 13.1.    If a, b , and c are differentiable vector functions of a scalar t and f is a differentiable scalar function of t, then   d   da db (i) a ±b = + . dt dt dt   d    db da  (ii) a ⋅b = a ⋅ + ⋅b. dt dt dt   d    db da  a ×b = a × + × b. (iii) dt dt dt  d  da d φ  (φa ) = φ + a. (iv) dt dt dt   d         dc    db   a ⋅ b × c  = a ⋅ b ×  + a ⋅  × c (v)  dt  dt   dt    da  + ⋅ (b × c ). dt  d         dc  a × b × c = a × b × (vi)     dt  dt      db   da   +a ×  × c + × b ×c .  dt  dt Proof: We prove (i), (iii), and (v). The other parts may similarly be proved by the readers themselves. d   a +b (i) dt    (a + ∆a ) + b + ∆b  − a + b   = lim ∆t → 0 ∆t     ∆a + ∆b ∆a ∆b = lim = lim + lim ∆t → 0 ∆t → 0 ∆t ∆t → 0 ∆t ∆t   da db = + . dt dt d   a ×b (iii) dt

(

)

( ) (

)

(

)

(

)

(

(

)

(

(

)

M13_Baburam_ISBN _C13 Part I.indd 12

) (

)

)

   ( a + ∆a ) × b + ∆b  − a × b  = lim  ∆t → 0 ∆t           a × b + a × ∆b + ∆a × b + ∆a × ∆b − a × b = lim ∆t → 0 ∆t       a × ∆b + ∆a × b + ∆a × ∆b = lim ∆t → 0 ∆t       ∆b ∆a  ∆a = lim a × + ×b + × ∆b  ∆t → 0 ∆t ∆t ∆t        da  db da =a× + ×b + ×0 dt dt dt      db da    db da  =a× + ×b + 0 = a × + × b. dt dt dt dt (v) Using (ii) and (iii), we have d     a⋅ b ×c  dt    d   da   = a⋅ b ×c + ⋅ b ×c dt dt       dc db   da   = a ⋅ b × + ×c + ⋅ b ×c dt dt   dt       dc    db   da   = a ⋅ b ×  + a ⋅  × c  + ⋅ b × c . dt    dt  dt

(

(

)

)

(

)

(

)

(

)

(

)

Theorem 13.2. The derivative of a constant vector is the zero vector. Proof: We know that a vector is called constant if its magnitude and direction do not change.    Let c be a constant vector and let r = c . Then,        r + δ r = c and so, δ r = 0. Thus, δδ rt = δ0t = 0     δr = lim 0 = 0. and hence, drdt = δlim t →0 δ t δ t →0 Theorem 13.3.  A vector function f of a scalar variable t is  constant if and only if dfdt = 0. Proof: If

 f is a constant vector, then, by Theorem

 13.2, = 0.   Conversely, suppose that dfdt = 0. If f1, f2 and f3 are the components of f along x-, y-, and  df dt

1/2/2012 12:16:16 PM

13.13 vector calculuS  n   z-axes, then f = f1iˆ + f 2 ˆj + f 3 kˆ. Hence,   df df1 df df 0= = iˆ + 2 ˆj + 3 kˆ. dt dt dt dt Therefore, equality of two vectors implies df1 df 2 df 3 = = =0 dt dt dt Therefore, f1, f2 and f3 are constant scalars, independent of t. Hence, f is a constant vector function. Theorem 13.4.

   f has a constant direction. Since f = fF , it   follows that f and F have the same direction.  Thus, F has a constant magnitude, equal to unity and a constant direction too and so, is a constant  = 0. Differentiating vector. Therefore, dF dt   f = fF with respect to t, we have   df df  dF = F + f .    df dt dt  df dt dF  f× = ( fF ) ×  F + f  dt dt   dt

Now,



A vector function f of a scalar variable t has a constant magnitude if and only if

  f . dfdt = 0. 

Proof: First, suppose that the vector function

has a constant magnitude c. Then,    f ⋅ f =| f |2 = c 2

f

and so, d   d ( f ⋅ f ) = (c 2 ) = 0 dt dt

But,

    df df   df d   (f ⋅ f)= f ⋅ + ⋅ f =2f ⋅ . dt dt dt dt

Hence,

   df  df 2f ⋅ = 0 or f ⋅ = 0. dt dt   Conversely, suppose that f . dfdt = 0. Therefore,         d 2 f ⋅ dfdt = 0 or f ⋅ dfdt + dfdt ⋅ f = 0 or dt f , f = 0,   which implies f ⋅ f is constant = c2, say.    Hence, | f |2 = c 2 or | f |= c, , that is, f has a constant magnitude.

(

)

Theorem 13.5. The necessary and sufficient condition for a  vector function f of a scalar variable t to have a constant direction is that

  df  f× = 0. dt

 Proof: Let F be a vector function of modulus    unity for all t. Let f = f . Then, f = f F.

The Condition is Necessary: Suppose that

M13_Baburam_ISBN _C13 Part I.indd 13



  dF df   2 = f F×F + f F× dt dt    dF = 0 + f 2F × , dt    since F × F = 0   dF 2 = f F× dt    2 = f F × 0 = 0,

since dF = 0 (as shown earlier). df

The  Condition is Sufficient Suppose that  df  f× = 0. Therefore, as shown previously, dt     dF  dF  = 0. Also, since f 2F × = 0 and so F × dt dt  

F is of constant magnitude, 

two facts imply that

dF  = 0. dt

 dF  F. = 0. dt

These

 Therefore, F is a

constant vector. But magnitude of F is constant  (unity). Therefore, F has a constant direction.    But f = fF . Therefore, direction of f is also constant. Corollary 13.1: The derivative of a vector function of a scalar variable t having a constant direction is collinear with it.   Proof: Since f has a constant direction, f × df = 0  dt  and so, f and df are collinear. This completes dt the proof of the corollary. FromTheorems 13.3−13.5, we conclude that  df  (i) = 0 if and only if f is a constant vector dt

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13.14  n  chapter thirteen functionin both magnitude and direction   df (ii) f . = 0 if and only if f has a constant dt magnitude.   df = 0 if and only if f has a constant (iii) f × dt direction. Theorem 13.6.  If f = f1iˆ + f 2 ˆj + f 3 kˆ is a vector function of the scalar variable t, then  df = f1′ (t )iˆ + f 2′ ( t ) ˆj + f 3′ (t )kˆ. dt

Proof: We have

 f = f1iˆ + f 2 ˆj + f 3 kˆ,

where f1, f2, and f3 are scalar functions of t. Therefore,  df d d d = ( f1iˆ) + ( f 2 ˆj ) + ( f 3 kˆ) dt dt dt dt diˆ df djˆ df dkˆ df 3 ˆ = f1 + 1 iˆ + f 2 + 2 ˆj + f 3 + k dt dt dt dt dt dt  df  df  df = 0 + 1 iˆ + 0 + 2 ˆj + 0 + 3 kˆ dt dt dt df df df = 1 iˆ + 2 ˆj + 3 kˆ. dt dt dt

Thus, to differentiate a vector, it is sufficient to differentiate its components.  Velocity and Acceleration. Let r be the position  vector of a moving particle P, and let δ r be the displacement of the particle in time dt, where  t denotes time. Then, the vector δδ rt denotes the average velocity of the particle during the interval dt of time. Therefore, the velocity  vector v of the particle at P is given by   δ r dr  v = lim = , δ t →0 δ t dt and its direction is along the tangent  at P.  Further, if δ v is the change in velocity v during the time interval dt, then the rate of change of velocity, that is, δδvt is the average acceleration of the particle during the interval dt. Thus, the acceleration of the particle at P is

M13_Baburam_ISBN _C13 Part I.indd 14

    δ v dv d  dr  d 2 r  a = lim = =  = 2 . δ t →0 δ t dt dt  dt  dt Tangential and Normal Acceleration. Let

 r be the

position vector of a point P moving in a plane curve at any time t. Then the velocity of the moving point is given by    dr dr ds v= = ⋅ . dt ds dt  But dr = Tˆ is a unit vector along the tangent at ds



P. Therefore, v = dsdt Tˆ. Thus, the acceleration is  dv d  ds ˆ  d 2 s ˆ ds dTˆ (1) =  T= T+ ⋅ . dt dt  dt  dt 2 dt dt But,

y

^ T ^ N P A r

S x

O

dTˆ dTˆ dψ dψ ˆ N, = ⋅ = dt dψ dt dt where Nˆ is a unit vector along the normal at P. Therefore, dTˆ dψ ds ˆ 1 ds ˆ = ⋅ N= N, ρ dt dt ds dt where r is the radius of curvature at P. Hence, (1) reduces to  2 dv d 2 s ˆ 1  ds  ˆ dv ˆ v 2 ˆ = 2 T +   N = T + N. dt dt dt ρ  dt  ρ Therefore, Tangential acceleration = Normal acceleration =

v2

ρ

dv d 2 s = and dt dt 2 .

Radial and Transverse Acceleration of a Moving Particle.

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13.15 vector calculuS   n   Let r be the position vector of a moving particle P(r, q). Suppose that Rˆ and Tˆ are the unit vector in radial- and transverse directions, respectively. Then, rˆ = rRˆ and  dr ˆ dRˆ  dr d ˆ = (rR ) = R+r velocity v = dt dt dt dt dr ˆ dRˆ dθ dr ˆ dθ ˆ R+r . R+r T. = = dt dθ dt dt dt Therefore, the components of the velocity in the radial- and transverse directions are dr dθ vR = and vT = r . dt dt

 X

dTˆ dR ˆ = − Rˆ , we have Further, since = T and dθ  dθ acceleration a  dv d  dr ˆ dθ ˆ  T = =  R+r dt dt  dt dt  =

d 2 r ˆ dr dRˆ dr dθ ˆ d 2θ dθ dTˆ R+ . T + r 2 Tˆ + r . + . 2 dt dt dt dt dt dt dt dt

=

d 2 r ˆ dr dRˆ dθ  dr dθ d 2θ  + + r 2  Tˆ R+ . 2 dt dθ dt  dt dt dt dt 



dθ dTˆ dθ . dt dθ dt

 dr dθ d 2r dr dθ d 2θ  = 2 Rˆ + . Tˆ +  . + r 2  Tˆ dt dt dt dt   dt dt 2



 dθ  ˆ −r   R  dt 

M13_Baburam_ISBN _C13 Part I.indd 15

))

Solution. We are given that

r

+r

EXAMPLE 13.20

( (

P(r,  )

O

Transverse acceleration dr dθ d 2θ = aT = 2 . +r 2 . dt dt dt

If a = sin θ iˆ + cos θ ˆj + θ kˆ,     b = cos θ iˆ − sin θ ˆj − 3kˆ, and c = 2i + 3 j − k ,    fin ddθ a × b × c at θ = 0.

^ R

^ T

2  d 2r d 2θ  ˆ  dθ   ˆ  dr dθ =  2 −r + + R 2 r   T.  dt 2   dt    dt  dt dt Hence, 2 d 2r  dθ  Radial acceleration = aR = 2 − r   dt  dt  and

a = sin θ iˆ + cos θ ˆj + θ kˆ,      b = cos θ iˆ − sin θ ˆj − 3kˆ, and c = 2i + 3 j − k . Therefore, ˆj iˆ kˆ   b × c = cos θ − sin θ −3 −1 2 3 = ( sin θ + 9 ) iˆ − ( − cos θ + 6 ) ˆj + ( 3 cos θ + 2 sin θ ) kˆ.    Then, a × (b × c ) ˆj iˆ kˆ =

sin θ 9 + sin θ

θ cos θ − cos θ + 6 3cos θ + 2sin θ

= (3cos 2θ + sin 2θ − θ cos θ + 6θ )iˆ 3  −  sin 2θ + 2sin 2θ − 9θ − θ sin θ  ˆj 2  ˆ + ( −6 sin θ − 9 cos θ ) k .



Therefore, d     a× b ×c  dθ 

(

)

= ( −6cosθ sinθ + 2cos 2θ − cosθ + θ sinθ + 6 ) iˆ

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13.16  n  chapter thirteen − ( 3cos 2θ + 4sinθ cosθ − 9 − θ cos θ − sin θ ) ˆj +(−6cosθ + 9sinθ )kˆ. Putting θ = 0, we get d     a × b × c = (2 − 1 + 6)iˆ − (3 − 9) ˆj − 6kˆ  dθ  = 7iˆ + 6 ˆj − 6kˆ.

(

)

EXAMPLE 13.21

 If r = (cos nt )iˆ + (sin nt ) ˆj , where n is a constant   dr and t varies, show that r × = nkˆ. dt Solution. We have  r = ( cos nt )iˆ + ( sin nt ) ˆj. Therefore,  dr = (−n sin nt )iˆ + (n cos nt ) ˆj. dr Therefore, iˆ

  dr = cos nt r× dt −n sin nt

ˆj

kˆ

sin nt n cos nt

0 0

= kˆ(ncos 2 nt + nsin 2 nt ) = nkˆ. EXAMPLE 13.22   If a and b are constant vectors, ω is a constant    scalar, and r = a sin ωt +b cos ωt , show that

 d 2r   = +ω 2 r = 0 and 2 dt  dr   r× = −ω a × b . dt   Solution. (i) Since a and b are constant vectors, we have   db  da  = 0. (1) = 0 and dt dt    Now it is given that r = a sin ωt +b cos ωt. Therefore,  dr da d  = sin ωt + ( sin ωt )a dt dt dt

M13_Baburam_ISBN _C13 Part I.indd 16

  db d + cos ωt + ( cos ωt )b dt dt  d d  = 0 + ( sin ωt )a + 0 + ( cos ωt )b , using(1) dt dt   = (ω cos ωt )a − (ω sin ωt )b and

  d 2r  = (−ω 2 sin ωt )a − (ω 2 cos ωt )b 2 dt    = −ω 2 a sin ωt + b cos ωt = −ω 2 r .

(

Hence,

)

 d 2r   + ω 2 r = 0. 2 dt      a × a = 0, (ii) Since b × b = 0,     a × b = −b × a , we have    dr  = ( cos ωt ) a + ( sin ωt ) b  r× dt

and

  × ( −ω sin ωt ) a + (ω cos ωt ) b      = (ω cos 2ωt ) a × b − (ωsin 2ωt ) b × a     = (ω cos 2ωt ) a × b + (ωsin 2ωt ) a × b   = ω ( cos 2ωt + sin 2ωt )  a × b   = ω a ×b .

( (

) )

( (

(

(

) )

)

)

EXAMPLE 13.23

Show that     d   db da    d 2 b d 2 a  − × b  = a × 2 − 2 × b. a× dt  dt dt dt dt  Solution. We have   d   db da   a × − ×b   dt  dt dt       2 2 da db  d b d a  da db = × + a × 2 − 2 ×b − × dt dt dt dt dt dt  2 2   d b d a = a × 2 − 2 × b, dt dt which proves our assertion.

1/2/2012 12:16:20 PM

vector calculuS  n  13.17 EXAMPLE 13.24

 2 Let r = t iˆ − 3tˆj + (2t + 1)kˆ, find at t = 0, the  d 2r . value of dt 2  2 Solution. Let r = t iˆ − 3tˆj + (2t + 1)kˆ. Then,  dr d 2 ˆ d d = ( t ) i − ( 3t ) ˆj + ( 2t + 1) kˆ dt dt dt dt = 2tiˆ − 3 ˆj + 2kˆ and 2 d r d d d = ( 2t ) iˆ − (3) ˆj + (2)kˆ = 2iˆ. dt dt dt 2 dt  d 2r When t = 0, we have 2 = 2iˆ. . Further, dt  d 2r = 22 + 02 + 02 = 2. dt 2 EXAMPLE 13.25

  2 2 If a = 5t iˆ + tˆj − t kˆ. and b = sin tiˆ − cos tjˆ, d  d   a.b and a ×b . find dt dt  a = 5t 2 iˆ + tˆj − t 2 kˆ Solution. Let and  b = sin tiˆ − cos tjˆ.

( )

Then,

(

)

 a.b = 5t 2 sin t − t cos t.

Therefore, d  d a.b = = (5t 2 sin t − t cos t ) dt dt

( )

= 5t 2 cos t + 10t sin t + t sin t − cos t

= (5t 2 − 1) cos t + 11t sin t Also, iˆ   a × b = 5t 2 sint



ˆj t −cost

kˆ −t 0

= (−t 2 cos t )iˆ − (t 3 sin t ) ˆj

Therefore,

M13_Baburam_ISBN _C13 Part I.indd 17

+(−5t 2 cos t − t sin t )kˆ.

d   a ×b dt

(

=

)

d d (−t 3 cos t )iˆ − (t 3 sin t ) ˆj dt dt +

d (−5t 2 cos t − t sin t )kˆ dt

= (t 3sin t − 3t 2 cos t )iˆ − (t 3 cos t + 3t 2 sin t ) ˆj

+ (5t 2 sin t − 11tcos t − sin t )kˆ .

EXAMPLE 13.26

Find a unit tangent vector to any point on the curve x = a cos ωt , y = a sin ωt , and z = bt , where a, b, and ω are constants.  Solution. Let r be the position vector of any point (x, y, z) on the given curve. Then,

 r = x iˆ + y ˆj + z kˆ

= (a cos ωt )iˆ + (a sin ωt ) ˆj + (bt )kˆ. Therefore,  dr = (−aω sin ωt )iˆ + (aω cos ωt ) ˆj + bkˆ. dt  dr

The vector dt is along the tangent at the point (x, y, z) to the given curve. Hence, unit tangent vector is given by   dr (−aω sin ωt )iˆ + (aω cos ωt ) ˆj + bkˆ T = drdt = . a 2ω 2 + b 2 dt EXAMPLE 13.27

A particle moves along the curve x = 3t 2 , y = t 2 − 2t and z = t 3 . Find its velocity and acceleration at t = 1 in the direction of iˆ + ˆj − kˆ.  Solution. Let r be the position vector of any point (x, y, z) on the given curve. Then,  r = xiˆ + yˆj + zkˆ = 3t 2 iˆ + (t 2 − 2t ) ˆj + t 3 kˆ and so, the velocity and acceleration of the particle are, respectively,   dr v= = 6tiˆ + (2t − 2) ˆj + 3t 2 kˆ = 6iˆ + 3kˆ at t = 1 dt and

1/2/2012 12:16:21 PM

13.18  n  chapter thirteen   d 2r a = 2 = 6iˆ + 2 ˆj + 6tkˆ = 6iˆ + 2 ˆj + 6kˆ at t = 1. dt The unit vector in the direction of iˆ + ˆj − kˆ is iˆ + ˆj − kˆ iˆ + ˆj − kˆ = nˆ = . 3 iˆ + ˆj − kˆ Therefore, the components of velocity and acceleration in the direction of iˆ + ˆj − kˆ are (iˆ + ˆj − kˆ) 3  v .nˆ = (6iˆ + 3kˆ) ⋅ = = 3 and 3 3 (iˆ + ˆj − kˆ) 2 2 3.  a.nˆ = (6iˆ + 2 ˆj + 3kˆ) ⋅ = = 3 3 3 EXAMPLE 13.28

Find the angle between the tangents to the curve  r = t 2 iˆ + 2tˆj − t 3 kˆ at the points t = ±1 .

 r = (t 3 − 4t )iˆ + (t 2 + 4t ) ˆj + (8t 2 − 3t 3 )kˆ. Therefore,   dr v= = (3t 2 − 4)iˆ + (2t + 4) ˆj + (16t − 9t 2 )kˆ dt = 8iˆ + 8 ˆj − 4kˆ at t = 2 and   d 2r acceleration a = 2 = 6tiˆ + 2 ˆj + (16 − 18t ) kˆ dt = 12iˆ + 2 ˆj − 20kˆ at t = 2. The velocity is along the tangent to the curve. Therefore,  Component of a along the tangent  8iˆ + 8 ˆj − 4kˆ  v = a.  = (12iˆ + 2 ˆj − 20kˆ) |v | 8iˆ + 8 ˆj − 4kˆ 8iˆ + 8 ˆj − 4kˆ = (12iˆ + 2 ˆj − 20kˆ) 64 + 64 + 16

Solution. We have

 r = t 2 iˆ + 2tˆj − t 3 kˆ.

Therefore, the vector along the tangent at any point is  dr = 2tiˆ + 2 ˆj − 3t 2 kˆ. dt Thus, the vectors along the tangents at t = ±1 are   T1 = 2iˆ + 2 ˆj − 3kˆ and T2 = −2iˆ + 2 ˆj − 3kˆ. The angle q between the tangents is given by   T1 .T2 2(−2) + 2(2) − 3(−3) 9 = . cos θ =   = 4 + 4 − 9. 4 + 4 + 9 17 T1 T Hence,  9  θ = cos   .  17  −1

EXAMPLE 13.29

A particle moves along the curve  r = (t 3 − 4t )iˆ + (t 2 + 4t ) ˆj + (8t 2 − 3t 3 )kˆ., where t denotes time. Find the magnitude of acceleration along the tangent and normal at time t =2. Solution. The curve is

M13_Baburam_ISBN _C13 Part I.indd 18

=

96 + 16 + 80 = 16 12

and  Component of a along the normal 



=| a – resolved part of a along the tangent | 8iˆ + 8 ˆj − 4kˆ =|12iˆ + 2 ˆj − 20kˆ − 16 64 + 64 + 16 =

1 ˆ | 4i − 26 ˆj − 44kˆ | 3

1 = [ 16 + 676 + 1936] = 2 73. 3

13.2  PARTIAL DERIVATIVES OF A VECTOR FUNCTION   Let f be a vector function of x, y, and z. Let δ f be  the change in f corresponding to a small change    δ x in x. Then, δ f = f ( x + δ x, y, z ) − f ( x, y, z ) The limit    δf f ( x + δ x, y , z ) − f ( x, y , z ) lim = lim , δ x →0 δ x δx δx if it exists, is called the partial derivative of the vector function   f with respect to x and is ∂f denoted by ∂x or f x .

1/2/2012 12:16:22 PM

vector calculuS   n  13.19  Similarly, the partial derivatives of f with respect to y and z are defined b   f ( x , y + δ y , z ) − f ( x, y , z ) f y = lim δ y →0 δy and

   f ( x , y , z + δ z ) − f ( x, y , z ) f z = lim , δ z →0 δz

( ) ( )

(iii)

      ( f .g ) = f . ∂∂gx + ∂∂fx .g .

(iv)

∂ ∂x

( f × g ) = f ×





 ∂g ∂x

  + ∂∂fx × g .

Similar expressions for partial derivatives with respect to y and z are valid. Higher partial derivatives of f may also be defined in the same way. For example,    2 f xx = ∂∂x2f = ∂∂x ∂∂fx .

( )

EXAMPLE 13.30

 2 3 If f = xyziˆ + xz ˆj − y kˆ, find

∂f ∂x∂y

at the origin.

Solution. We have

 f = xyziˆ + xz 2 ˆj − y 3 kˆ.

Therefore,   ∂f = xziˆ + 0 − 3 y 2 kˆ and ∂y   ∂2 f = ziˆ = 0 at (0, 0, 0). ∂x∂y EXAMPLE 13.31

 a = x 2 yziˆ − 2 xz 3 ˆj + xz 2 kˆ If and 2  ∂   b = 2 ziˆ + yˆj − x 2 kˆ, find the value of 2 a × b ∂x at the point (1, 0, 1). Solution. We have   2 a = x 2 yziˆ − 2 xz 3 ˆj + xz 2 kˆ and b = 2 ziˆ + yˆj − x kˆ.

(

M13_Baburam_ISBN _C13 Part I.indd 19

iˆ   2 a × b = x yz 2z

)

kˆ

ˆj −2 xz y

3

xz 2 − x2

= (2 x 3 z 3 − xyz 2 )iˆ − (− x 4 yz − 2 xz 3 ) ˆj



provided these limits exist.   If f and g are differentiable vector functions of the independent variables x, y, and z and f is a differentiable scalar function of x, y, and z, then     ∂f ∂g (i) ∂∂x f + g = ∂x + ∂x .    ∂f ∂φ ∂ (ii) ∂x φ f = φ ∂x + ∂x . f ∂ ∂x

Therefore,

+ ( x 2 y 2 z + 4 xz 4 )kˆ.

Hence,

∂   (a × b ) = (6 x 2 z 3 − yz 2 )iˆ − (−4 x 3 yz − 2 z 3 ) ˆj ∂x +(2 xy 2 z + 4 z 4 )kˆ and ∂2   a × b = 12 xz 3 iˆ ∂x 2 − ( −12 x 2 yz ) ˆj + ( 2 y 2 z ) kˆ = 12iˆ at (1, 0, 1).

(

) (

)

13.3  GRADIENT OF A SCALAR FIELD A variable quantity whose value at any point in a region of space depends upon the position of the point is called a point function. If for each point P(x, y, z) of a region R, there corresponds a scalar φ ( x, y, z ) , then f is called a scalarpoint function for the region R. The region R is then called a scalar field. For example, the temperature at any point within or on the surface of the earth is a scalar-point function. Similarly, atmospheric pressure in the space is a scalar-point function. On the other hand, if for each pointP(x, y, z), of a region R, there exists  a vector f ( x, y, z ) , then f is called a vectorpoint function and the region R is then called a vector field. For example, the gravitational force is a vector-point function. Let f ( x, y, z ) , be a scalar-point function. Then, the points satisfying an equation of the type f ( x, y, z ) = c (constant) constitute a family of surface in a three-dimensional space. The surfaces of this family are called level surfaces. Since the value of the function f at any point of the surface is the same, these surfaces are also called iso-f-surfaces. The operator ∇, defined b

1/2/2012 12:16:23 PM

13.20  n  chapter thirteen ∇ = iˆ

∂ ˆ ∂ ˆ ∂ + j +k , ∂x ∂y ∂z

is called the vector differential operator and is read as del or nabla. Let f be a scalar function defined and differentiable at each point (x, y, z) in a certain region of space. Then, the vector defined b  ∂ ∂ ∂  ∇φ =  iˆ + ˆj + kˆ  φ , ∂y ∂z   ∂x ∂φ ˆ ∂φ ˆ ∂φ = iˆ +j +k ∂x ∂y ∂z

is called the gradient of the scalar function f and is denoted by grad f or ∇ f. ∂φ ∂φ Thus, grad f is a vector with components ∂x , ∂y , ∂φ

and ∂z . We note that f is a scalar-point function, whereas ∇ f is a vector-point function. 13.4  GEOMETRICAL INTERPRETATION OF A GRADIENT  Let r = xiˆ + yjˆ + zkˆ be the position vector of a

point P through which a level surface f (x, y, z) = c (constant) passes. Then, differentiating f (x, y, z) = c with respect to t, we get ∂φ dx ∂φ dy ∂φ dz dφ . + . + . =0 = 0 or ∂x dt ∂y dt ∂z dt dt or  ∂φ ∂φ ˆ ∂φ ˆ   dx ˆ dy ˆ dz ˆ  j+ k  . i + j+ k =0  iˆ + ∂ ∂y ∂z   dt x dt dt   or

(i) ∇(φ ± ψ ) = ∇φ ± ∇ψ . (ii) ∇(φψ ) = φ∇ψ + ψ∇φ .  θ  ψ∇φ − φ∇ψ (iii) ∇   = , ψ2 ψ  provided that ψ ≠ 0 . (iv) ∇(cφ ) = c∇φ . (v) ∇ f is a constant if and only if f is a constant.

Proof: (i). By the definition of a gradient, we have ∇(φ ± ψ )

 ∂ ∂ ∂  =  iˆ + ˆj + kˆ  (φ ± ψ ) ∂y ∂z   ∂x ∂ ∂ ∂ = iˆ (φ ± ψ ) + ˆj (φ ± ψ ) + kˆ (φ ± ψ ) ∂x ∂y ∂z

 ∂  ∂ ∂ ∂  ∂ ∂  =  iˆ + ˆj + kˆ  φ ±  iˆ + ˆj + kˆ ψ ∂y ∂z  ∂y ∂z   ∂x  ∂x = ∇φ ± ∇ψ  ∂ ∂ ∂  (ii) ∇(φψ ) =  iˆ + ˆj + kˆ  (φψ ) ∂y ∂z   ∂x ∂ ∂ ∂ = iˆ (φψ ) + ˆj (φψ ) + kˆ (φψ ) ∂x ∂y ∂z

∂φ   ∂ψ = iˆ φ +ψ + ∂ ∂x  x 

 dr = 0. dt  Since drdt is the vector tangent to the curve at P and since P is an arbitrary point on f (x, y, z) = c, it follows that ∇ f is perpendicular to f (x, y, z) = c at every point. Hence, ∇ f is normal to the surface f (x, y, z) = c.



13.5  PROPERTIES OF A GRADIENT The following theorem illustrates the properties satisfied by a gradient Theorem 13.7. If f and ψ are two scalar-point functions, and c is a constant, then,

φ (iii) ∇  ψ

∇φ .

M13_Baburam_ISBN _C13 Part I.indd 20



ˆj φ ∂ψ + ψ ∂φ   ∂ ∂y   y

∂φ   ∂ψ + kˆ φ +ψ ∂z   ∂z

 ∂ψ ˆ ∂ψ ˆ ∂ψ  = φ iˆ +j +k ∂y ∂z   ∂x  ∂φ ˆ ∂φ ˆ ∂φ  +ψ iˆ +j +k  ∂y ∂z   ∂x



= φ∇ψ + ψ∇φ

  ˆ ∂ ˆ ∂ ˆ ∂  φ  + j + k    = i ∂y ∂z   ψ    ∂x

= iˆ

∂ φ  ∂x  ψ

 ˆ ∂ φ + j  ∂y  ψ 

 ˆ ∂ φ  +k   ∂z  ψ  

1/2/2012 12:16:24 PM

13.21 vector calculuS  n  ∂ψ  ∂φ ψ ∂x − φ ∂x = iˆ  ψ2  

  +  

∂ψ  ∂φ ψ ∂y − φ ∂y ˆj  ψ2  

∂ψ  ∂φ ψ ∂z − φ ∂z ˆ +k  ψ2  



    

    

 ˆ ∂φ ˆ ∂φ ˆ ∂φ  i ∂x + j ∂y + k ∂z   =ψ  2

ψ



 ˆ ∂ψ ˆ ∂ψ ˆ ∂ψ  i ∂x + j ∂y + k ∂z   −φ  2

ψ



=

ψ ∇φ − φ ∇ψ , ψ ≠ 0. ψ2

(iv) We have  ∂ ∂ ∂  ∇(cφ ) =  iˆ + ˆj + kˆ  (cφ ) ∂y ∂z   ∂x ∂ ∂ ∂ = iˆ (cφ ) + ˆj (cφ ) + kˆ (cφ ) ∂x ∂y ∂z ∂ ∂ ∂ = ciˆ φ + cjˆ φ + ckˆ φ ∂x ∂y ∂z

 ∂φ ˆ ∂φ ˆ ∂φ  = c  iˆ +j +k  = c∇φ . ∂y ∂z   ∂x (v) We note that  ∂φ ˆ ∂φ ˆ ∂φ  ∇φ = 0 ⇔ iˆ +j +k =0 ∂x ∂y ∂z



⇔

∂φ ∂φ = 0, ∂φ = 0. = 0, ∂y ∂x ∂z

⇔ φ is a constant.

13.6  DIRECTIONAL DERIVATIVES Let A be any given point in the region of definition of a scalar-point function f. Let P be a point on any line drawn on one side of A. Then

M13_Baburam_ISBN _C13 Part I.indd 21

lim

P→ A

φ ( P) − φ ( A) if exists, is called the directional , AP

derivative of the scalar-point function f at A in the direction of AP. The length of AP is regarded as positive. The direction derivative in the direction of AP¢, where P¢ is a point on the other side of A, is negative of that in the direction of AP. P A P

The directional derivative of the vector function  f at A in the direction of AP is defined as f ( P) − f ( A) , provided the limit exists. lim P→ A

AP

13.6.1  Directional Derivatives along Coordinate Axes Let A(x, y, z) be a point and let P(x+ dx, y, z) be a point on a line drawn through A and parallel to the positive direction of x-axis. Then, AP = δ x > 0. Therefore, directional derivative of a scalar-point function at A along AP is defined a lim

P→ A

φ ( P) − φ ( A) AP

= lim

δ x →0

φ ( x + δ x , y , z ) − φ ( x, y , z ) δx

∂φ . ∂x Thus, the directional derivative of a scalarpoint function f along the x-axis is the partial derivative of f with respect to x. Similarly, directional derivatives of f along y=

and z-axis are, respectively,

∂φ ∂y

and

∂φ . ∂z

The directional derivatives of a vector-point  function f along the coordinate axes are 





similarly ∂∂fx , ∂∂fy , and ∂∂fz , respectively. Further, if l, m and n are direction cosines of AP = r, then the coordinates of P are x + lr, y + mr, and z + nr and so, the directional derivative of the scalar-point function f along AP becomes

1/2/2012 12:16:25 PM

13.22  n  chapter thirteen lim

φ ( P) − φ ( A)

AP φ ( x + lr , y + mr , z + nr ) − φ ( x, y, z ) = lim r →0 r φ( x , y , z ) + ( lr ∂∂φx + mr ∂∂φy + nr ∂∂φz ) + … − φ( x , y , z ) P→ A

= lim

r

r →0

=l

∂φ ∂φ ∂φ , +m +n ∂x ∂y ∂z

by the application of Taylor’s Theorem for function of several variables, under the assumption that f has a continuous first-order partial derivatives. Similarly, the directional derivative of a  vector-point function f along any line with   direction cosines l, m and n is l ∂∂fx + m ∂∂fy + n ∂∂fz . Theorem 13.8 The directional derivative of a scalar-point  function f along the direction of unit vector a is ∇φ .aˆ. Proof: The unit vector aˆ along a line whose direction cosines are l, m and n is aˆ = liˆ + mˆj + nkˆ.

Therefore,  ∂φ ˆ ∂φ ˆ ∂φ  ˆ ˆ ∇φ .aˆ =  iˆ +j +k  li + mjˆ + nk ∂y ∂z   ∂x

(

)

∂φ ∂φ ∂φ =l +m +n , ∂x ∂y ∂z which is nothing but directional derivative of f in the direction of the unit vector aˆ.

f (x, y, z) is zero along directions perpendicular to grad f (since cos π2 = 0 ) and is maximum along the direction parallel to grad f. Since grad f acts along the normal direction to the level surface of f (x, y, z), the directional derivative is maximum along the normal to the surface. The maximum value is gradφ = ∇φ . EXAMPLE 13.32   If r = xiˆ + yˆj + zkˆ and r = r , , show that (i)   ∇f (r ) = f ′(r )∇r and (ii) ∇f (r ) × r = 0. Solution. (i) By the definition of gradient

∂ ∂ ∂ ∇f (r ) = iˆ f (r ) + ˆj f (r ) + kˆ f (r ) ∂x ∂y ∂z = ifˆ ′(r )

∂r ˆ ∂r ˆ ∂r ′(r ) + jf ′(r ) + kf ∂x ∂y ∂z

 ∂r = f ′(r )  iˆ +  ∂x

ˆj ∂r + kˆ ∂r  = f ′(r )∇r. ∂y ∂z 

(ii) As in part (i), we have  ∂r ∂r ∂r  ∇f (r ) = f ′(r )  iˆ + ˆj + kˆ  . ∂ ∂ ∂z  x y   Since r = r = x 2 + y 2 + z 2 , we have ∂r 1 = (2 x) = 1 2 ∂x 2( x + y 2 + z 2 ) 2

x +y +z 2

∂r ∂x

y r

and

=

x , r

∂r ∂x

 f ′(r )    Hence, ∇f (r ) × r = (r × r ) = 0. r

change of f (x, y, z) in the direction of the unit vector aˆ, it follows that the rate of change of

Solution.

M13_Baburam_ISBN _C13 Part I.indd 22

2

= rz . Therefore,  r y z  x ∇f (r ) = f ′(r )  iˆ + ˆj + kˆ  = f ′(r ) . r r r  r

and similarly,

=

x 2

Theorem 13.9 Grad f is a vector in the direction of which the maximum value of the directional derivative of f occurs. Hence, the directional derivative is maximum along the normal to the surface and the maximum value is gradφ = ∇φ .  Proof: Recall that a.b = a b cosθ , where  q is  the angle between the vectors a and b . Since ( gradφ ) .aˆ gives the directional derivative in the direction of unit vector aˆ , that is, the rate of

.

EXAMPLE 13.33  If f (r ) = x 2 yz 2 , find ∇f ( at the point (1,2,3).

Hence calculate  (i) the directional derivative of f (r ) at (1, 2, 3) in the direction of the vector (–2, 3, 6). (ii) the maximum rate of change of the function at (1,2,3) and its direction. ∂f ∂z

f Since ∂∂x = 2 xyz 2 , = 2 x 2 yz , we have

∂f ∂y

= x2 z 2 ,

and

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vector calculuS  n  13.23 ∇f = 2 xyz 2 iˆ + x 2 z 2 ˆj + 2 x 2 yzkˆ.

Therefore, at the point (1, 2, 3),

 1 ∂r  = iˆ  − 2 +  r ∂x 

ˆj  − 1 ∂r  + kˆ  − 1 ∂r  2 2  r ∂z   r ∂y 

(i) The unit vector aˆ in the direction of the

 1 x = iˆ  − 2 .  +  r r

ˆj  − 1 . y  + kˆ  − 1 . z  2 2  r r  r r

grad f = ∇f = 36iˆ + 9 ˆj + 12kˆ.

vector (–2, 3, –6) is −2iˆ + 3 ˆj − 6kˆ −2 ˆ 3 ˆ 6 ˆ = i + j − k. 7 7 7 4 + 9 + 36

=−

Therefore, the directional derivative at (1, 2, 3) in the direction of the vector (–2, 3, –6) is 2 3 6 ∇f .aˆ = (36iˆ + 9 ˆj + 12kˆ).(− iˆ + ˆj − kˆ) 7 7 7 =−

36iˆ + 9 ˆj + 12kˆ

36iˆ + 9 ˆj + 12kˆ 12iˆ + 3 ˆj + 4kˆ = = . 39 13 1296 + 144 + 81  The maximum rate of change of f ( r ) is grad f = 1296 + 144 + 81 = 39. EXAMPLE 13.34   If r is the usual position vector r = xiˆ + yˆj + zkˆ  with r = r , evaluate

(i) ∇r, (ii) ∇ ( 1r ) , (iii) ∇rn, and (iv) ∇  Solution. Since r = r = ∂r x ∂r y = , = , ∂x r ∂y r

and

( ) 1 r2

x + y + z , we have 2

2

∂r z = . ∂z r

 ∂ ∂ ∂  ∂r ∂r ∂r ∇(r ) =  iˆ + ˆj + kˆ  (r ) = iˆ + ˆj + kˆ ∂y ∂z  ∂x ∂y ∂z  ∂x

x y z xiˆ + yjˆ + zkˆ rˆ = iˆ + ˆj + kˆ = = . r r r r r

 ∂  ∂  ∂ 1 1 (ii) ∇( ) = (i + j + k )( ) ∂x ∂y ∂z r r

= iˆ

= iˆ

∂ 1 ˆ ∂ 1 ˆ ∂ 1  + j  +k   ∂x  r  ∂y  r  ∂z  r 

M13_Baburam_ISBN _C13 Part I.indd 23

)

∂ n ∂ ∂ r ) + ˆj ( r n ) + kˆ ( r n ) ( ∂x ∂y ∂z

∂r   = iˆ  nr n −1  + ∂x  

ˆj  nr n −1 ∂r  ∂y  

∂r   + kˆ  nr n −1  ∂z  



x  = iˆ  nr n −1 .  + r  

ˆj  nr n −1 . y  + kˆ  nr n −1 . z  r r  

 = nr n − 2 ( xiˆ + yˆj + zkˆ) = nr n − 2 r .

1 (iv) ∇  2 r

  ˆ ∂ ˆ ∂ ˆ ∂  1  + j + k  2   = i ∂y ∂z   r    ∂x

= iˆ

2

Therefore, (i)

.

(

 ∂ ∂ ∂  (iii) ∇r n =  iˆ + ˆj + kˆ  (r n ) ∂y ∂z   ∂x

72 27 72 117 + − =− . 7 7 7 7

(ii) The maximum rate of change of the function at (1, 2, 3) occurs along the direction parallel to ∇f at (1, 2, 3), that is, parallel to 36iˆ + 9 ˆj + 12kˆ. The unit vector in that direction is

 1 ˆ ˆ ˆ =− r . xi + yj + zk r3 r3



∂ 1  ∂x  r 2

 ˆ ∂ 1 + j  2 ∂y  r 

 2 ∂r  = iˆ  − 3 +  r ∂x 



 2 x = iˆ  − 3 .  +  r r

 ˆ ∂1 +k  2  ∂z  r  

ˆj  − 2 ∂r  3  r ∂y 

 2 ∂r  + kˆ  − 3   r ∂z  ˆj  − 2 . y  3  r r

 2 z + kˆ  − 3 .   r r  2 2r = − 4 ( xiˆ + yjˆ + zkˆ) = − 4 . r r

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13.24  n  chapter thirteen EXAMPLE 13.35

Find the directional derivative of f (x,y,z) = xy2 + yz3 at the point (2, –1, 1) in the direction of the vector iˆ + 2 ˆj + 2kˆ. Solution. We have  ∂ ∂ ∂  ∇f =  iˆ + ˆj + kˆ  ( xy 2 + yz 3 ) ∂ ∂ ∂ x y z  = y 2 iˆ + (2 xy + z 3 ) ˆj + (3 yz 2 )kˆ = iˆ − 3 ˆj − 3kˆ at the point (2, –1, 1). The unit vector in the direction of the vector iˆ + 2 ˆj + 2kˆ is iˆ + 2 ˆj + 2kˆ 1 ˆ aˆ = = (i + 2 ˆj + 2k ). 1+ 4 + 4 3 Therefore, the directional derivative of f at (2, –1, 1) in the direction of iˆ + 2 ˆj + 2kˆ is 1 ∇φ .aˆ = iˆ − 3 ˆj − 3kˆ . iˆ + 2 ˆj + 2kˆ 3

(

=

) (

)

1 11 (1 − 6 − 6 ) = − . 3 3

EXAMPLE 13.36

Find the directional derivative of f (x, y, z) = xy 2 + yz 3 at the point (2, –1, 1) in the direction of the normal to the surface x log z –y2 + 4 = 0 at (2, –1, 1). ∂φ 2 3 ∂φ = = y2 , Solution. We have φ = xy + yz , ∂y ∂x ∂ φ = 3 yz 2 . Therefore, as in 2xy + z 3, and ∂z Example 13.16, ∇φ = iˆ − 3 ˆj − 3kˆ at (2, –1, 1). On the other hand, ∇( x log z − y 2 + 4)  ∂ ∂ ∂  =  iˆ + ˆj + kˆ  ( x log z − y 2 + 4) ∂y ∂z   ∂x = iˆ

∂ ∂ ( x log z − y 2 + 4) + ˆj ( x log z − y 2 + 4) ∂x ∂y

+ kˆ

∂ ( x log z − y 2 + 4) ∂z

x = log z iˆ − 2 yˆj + kˆ = −4 ˆj − kˆ at (–1, 2, 1). Z

M13_Baburam_ISBN _C13 Part II.indd 24

But ∇ ( x log z = y 2 + 4 ) is normal to the surface x log z = y 2 + 4. Unit vector along ∇ ( x log z = y 2 + 4 ) is aˆ =

−4 ˆj − kˆ 16 + 1

=

−4 ˆj − kˆ 17

.

Therefore, the directional derivative of f at (2, –1, 1) in the direction of the normal to the 2 surface x log z = y + 4 at (–1, 2, 1) is  −4 ˆj − kˆ  12 + 3 15 ∇φ .aˆ = (iˆ − 3 ˆj − 3kˆ)  = = .  17  17 17   EXAMPLE 13.37

Find the angle between the surfaces 2 2 x 2 + y 2 + z 2 = 9 and z = x + y − 3 at the point (2, –1, 2). Solution. Let φ ( x, y, z ) = x 2 + y 2 + z 2 − 9 and ψ ( x, y, z ) = x 2 + y 2 − 3 − z . Then, the angle between the surfaces at the given point (2,−1,2) is the angle between the normal to the surfaces at that point. Also ∇f and ∇ψ are along the normal to f and ψ, respectively. But,  ∂ ∂ ∂  ∇φ =  iˆ + ˆj + kˆ  ( x 2 + y 2 + z 2 − 9). ∂y ∂z   ∂x = 2 xiˆ + 2 yˆj + 2 zkˆ and

= 4iˆ − 2 ˆj + 4kˆ at the point (2,−1,2)

 ∂ ∂ ∂  ∇ψ =  iˆ + ˆj + kˆ  ( x 2 + y 2 − 3 − z ) ∂ ∂ ∂ x y z  = 2 xiˆ + 2 yˆj − kˆ = 4iˆ − 2 ˆj − kˆ at the point (2, −1, 2) . If q is the angle between ∇f and ∇ψ , then ∇φ .∇ψ (4iˆ − 2 ˆj + 4kˆ).(4iˆ − 2 ˆj − kˆ) = cos θ = ∇φ ∇ψ 16 + 4 + 16 16 + 4 + 1 =

16 + 4 − 4 6 21

=

8 3 21

.

 8  Hence, the required angle is θ = cos −1  .  3 21 

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13.25 vector calculuS  n  Hence,

EXAMPLE 13.38

In what direction from (3, 1,−2) is the directional derivative of φ = x 2 y 2 z 4 maximum and what is its magnitude? Solution. The directional derivative at a given point of a given surface f is maximum along the normal to the surface and grad f acts along the normal. Therefore, the directional derivative is maximum along ∇f. We have  ∂ ∂ ∂  ∇φ =  iˆ + ˆj + kˆ  ( x 2 y 2 z 4 ) ∂ ∂ ∂ x y z  = (2 xy 2 z 4 )iˆ + (2 x 2 yz 4 ) ˆj + (4 x 2 y 2 z 3 )kˆ = 96iˆ + 288 ˆj − 288kˆ at the point (3, 1, –1). Thus, the directional derivative is maximum in the direction of 96iˆ + 288 ˆj − 288kˆ. The magnitude of the maximum directional derivative is | ∇φ |= 96 1 + 9 + 9 = 96 19. Find the angles between the normal to the 2 surface xy = z at the points (4, 1, 2) and (3, 3, −3). 2 Solution. Let φ ( x, y, z ) = xy − z . Since ∇f is along the normal, it is sufficient to find angle between ∇f at (4, 1, 2) and ∇f at (3, 3, −3). Now,  ∂ ∂ ∂  ∇φ =  iˆ + ˆj + kˆ  ( xy − z 2 ) ∂ ∂ ∂ x y z  = yiˆ + xjˆ − 2 zkˆ.

−1 22

EXAMPLE 13.40

Find the constants a and b so that the surface ax 2 − byz = (a + 2) x is orthogonal to the surface 4x 2 y + z 3 = 4 at the point (1, −1, 2). Solution. The two given surfaces will be orthogonal if the angle between the normal to the surfaces π at the point (1, −1, 2) is 2 . Since ∇f acts along the normal, it is sufficient to find ∇f and ∇ψ at 2 (4, 1,2) where φ = ax − byz − (a + 2) x and ψ = 4 x 2 y + z 3 − 4 . We have  ∂ ∂ ∂  ∇φ =  iˆ + ˆj + kˆ  (ax 2 − byz − (a + 2) x) ∂ ∂ ∂ x y z  = iˆ(2ax − a − 2) + ˆj (−bz ) + kˆ(−by )

.

M13_Baburam_ISBN _C13 Part II.indd 25

= (a − 2)iˆ − 2bˆj + bkˆ at (1, −1, 2),

 ∂ ∂ ∂  ∇ψ =  iˆ + ˆj + kˆ  (4 x 2 y + z 3 − 4) ∂y ∂z   ∂x 2 = iˆ ( 8 xy ) + ˆj ( 4 x ) + kˆ ( 3z 2 ) = −8iˆ + 4 ˆj + 12kˆ at (1, −1, 2). Since θ =

π

, we have 2 π ((a − 2)iˆ − 2bjˆ + bkˆ).(−8iˆ + 4 ˆj + 12kˆ) cos = . 2 | (a − 2)iˆ − 2bjˆ + bkˆ || −8iˆ + 4 ˆj + 12kˆ |

(a − 2)(−8) − 8b + 12b = 0

Hence, the required angle q is the angle between iˆ + 4 ˆj − 4kˆ and 3iˆ + 3 ˆj + 6kˆ. Therefore, (iˆ + 4 ˆj − 4kˆ).(3iˆ + 3 ˆj + 6kˆ) = | iˆ + 4 ˆj − 4kˆ || 3iˆ + 3 ˆj + 6kˆ |

1  . 22 

Hence, ((a − 2)iˆ − 2bjˆ + bkˆ).(−8iˆ + 4 ˆj + 12kˆ) = 0 or

Therefore, ∇f at (4, 1, 2) is iˆ + 4 ˆj − 4kˆ and ∇f at (3, 3, −3) is 3iˆ + 3 ˆj + 6kˆ.

=



and

EXAMPLE 13.39

cos θ =



θ = cos −1  −

−9 33 54

or −8a + 4b = −16. (1) Since the points (1, −1, 2) lie on both surfaces f and ψ, we have from the surface f, (2) a + 2b = a + 2 or b = 1 . Putting the value of b from (2) in (1), we get 20 5 = . −8a = −16 − 4 or a = 8 2

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13.26  n  chapter thirteen Hence, a =

5 and b = 1 . 2

EXAMPLE 13.41

A paraboloid of revolution has the equation 2z = x 2 + y 2 . Find the equation of the normal and the tangent plane to the surface at the point (1, 3, 5). 2 2 Solution. Let φ = x + y − 2 z . Then, ∇f gives the vector normal to the surface. Thus, the normal vector to the surface is  ∂ ∂ ∂  ∇φ =  iˆ + ˆj + kˆ  ( x 2 + y 2 − 2 z ) ∂ ∂ ∂ x y z 

 ∂ ∂ ∂  ∇ψ =  iˆ + ˆj + kˆ  ( x 2 y − 2 + z ) ∂ ∂ ∂ x y z  = 2 xyiˆ + x 2 ˆj + kˆ = 2iˆ + ˆj + kˆ at the point (1, 1, 1). Therefore, the required angle is given by cos θ =

(−2 ˆj + kˆ).(2iˆ + ˆj + kˆ) 5. 6

x + 3y − z = 5 . EXAMPLE 13.42

Find the angle between the tangent planes to the 2 2 surfaces x log z = y − 1 and x y = 2 − z at the point (1, 1, 1). Solution. The required angle will be the angle between the vectors normal to the given surfaces at the given point. The normal vectors to the surfaces f = x log z − y 2 + 1 and ψ = x 2 y − 2 + z are given by  ∂ ∂ ∂  ∇φ =  iˆ + ˆj + kˆ  ( x log z − y 2 + 1) ∂y ∂z   ∂x

and

x = log z iˆ − 2 yˆj + kˆ z ˆ = −2 ˆj + k at the point (1, 1, 1)

M13_Baburam_ISBN _C13 Part II.indd 26

−2 + 1 30

=

−1 30

.

Hence,  −1  θ = cos −1  .  30 

= 2 xiˆ + 2 yˆj − 2kˆ = 2iˆ + 6 ˆj − 2kˆ , at the point (1, 3, 5). Therefore, the unit normal vector at the point (1,3,5) is 2iˆ + 6 ˆj − 2kˆ iˆ + 3 ˆj − kˆ aˆ = = . 4 + 36 + 4 11 The equation of the line through the point (1, 3, 5) in the direction of this normal vector is x −1 y − 3 z − 5 = = . −1 1 3 Therefore, the equation of the tangent plane to the surface at the point (1,3,5) is 1 ( x − 1) + 3( y − 3) + (−1)( z − 5) = 0 or

=

13.7  DIVERGENCE OF A VECTOR-POINT FUNCTION If we want to consider the rate of change of a vector point function f , there are two ways  of operating the vector operator ∇ to the vector f . Thus, we have   two cases to consider, namely, ∇ ⋅ f and ∇ × f . These two cases lead us to the two concepts called Divergence of a Vector Function and curl of a Vector Function. If we consider a vector field as a fluid flow, then at every point in the flow, we need to measure the rate of flow of the fluid from that point and the amount of spin possessed by the particles of the fluid at that point. The above two concepts provide respectively, the two measures called divergence   of f and curl of f .  Let f = f iˆ + f ˆj + f kˆ be a vector function, 1

2

3

where f1 , f 2 , and f 3 are scalar-point functions, which is defined and differentiable at each point of the region of space. Then,  the divergence of f, denoted by ∇. f or div f , is a scalar given by   ∂ ∂ ∂  ∇ ⋅ f =  iˆ + ˆj + kˆ  .( f1iˆ + f 2 ˆj + f 3 kˆ) ∂y ∂z   ∂x ∂f1 ∂f 2 ∂f 3 + + . ∂x ∂y ∂z   The vector f is called Solenoidal if ∇ ⋅ f = 0. =

13.8  PHYSICAL INTERPRETATION OF DIVERGENCE Consider the steady motion of the fluid having  velocity v = v X iˆ + v y ˆj + vz kˆ at a point P ( x, y, z ) Consider a small parallelopiped with edges dx,

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13.27 vector calculuS  n  dy, and δ z parallel to the axes, with one of its comer at P ( x, y, z ) . The mass of the fluid entering through the face PQRS per unit time is v yδ xδ z and the mass of the fluid that flows out through the opposite face ABCD is v y +δ y δ xδ z . Therefore, the change in the mass of fluid flowing across these two faces is equal ∂v y   v y + δ y δ xδ z − vy δ xδ z =  vy + .δ y  δ x δ z ∂y   −v yδ xδ z =



∂v y ∂y

δ yδ xδ z .

Z R

C

y

S vy

z

D vy  y

P Q O

 Find div v , where v = 3 x 2 yiˆ + zjˆ + x 2 kˆ. Solution. We know that  ∂v ∂v ∂v divv = 1 + 2 + 3 . ∂x ∂y ∂z Here, v1 = 3 x 2 y , v2 = z , Therefore,  div v = 6 xy:.

and

v3 = x 2 .

EXAMPLE 13.44

Similarly, the changes in the mass of the fluid for the other two pairs of faces are ∂v X δ xδ yδ z and ∂vZ δ xδ yδ z . ∂x ∂z

x

EXAMPLE 13.43

A

B Y

Find the value of the constant l such that the vector field defined  f = (2 x 2 y 2 + z 2 )iˆ + (3 xy 3 − x 2 z ) ˆj

+(λ xy 2 z + xy )kˆ is solenoidal. Solution. We have

f1 = 2 x 2 y 2 + z 2 , f 2 = 3xy 3 − x 2 z , and f3 = lxy2z + xy. Therefore,  ∂f ∂f ∂f div f = 1 + 2 + 3 = 4 xy 2 + 9 xy 2 + λ xy 2 . ∂x ∂y ∂z The vector field shall be a solenoidal if  div f = 0. So, we must have 4 xy 2 + 9 xy 2 + λ xy 2 = 0 , which yields λ = −13 . EXAMPLE 13.45 

X

Therefore, the total change in the mass of the fluid inside the parallelopiped per unit time is equal to  ∂vx ∂v y ∂vZ  + +   δ xδ yδ z. ∂z   ∂x ∂y Hence, the rate of change of the mass of the fluid per unit time per unit volume i ∂vx ∂v y ∂vZ  + + = ∇ ⋅ v, ∂x ∂y ∂z by the definition of divergence. Hence, div v gives the rate at which the fluid (the vector field) is flowing away at a point of the flui

M13_Baburam_ISBN _C13 Part II.indd 27



Find div f , where f = grad ( x3 + y 3 + z 3 − 3 xyz ) . Solution. We have  f = ∇( x 3 + y 3 + z 3 − 3 xyz )  ∂ ∂ ∂  =  iˆ + ˆj + kˆ  ( x 3 + y 3 + z 3 − 3 xyz ) ∂y ∂z   ∂x 2 = (3 x − 3 yz )iˆ + (3 y 2 − 3 xz ) ˆj + (3 z 2 − 3 xy )kˆ = f1iˆ + f 2 ˆj + f3 kˆ, say. Then,  ∂f ∂f ∂f divf = 1 + 2 + 3 ∂x ∂y ∂z

= 6 x + 6 y + 6 z = 6( x + y + z ).

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13.28  n  chapter thirteen EXAMPLE 13.46

Find div (3 x iˆ + 5 xy 2 ˆj + xyz 3 kˆ) at the point. (1, 2, 3). Solution. Let  f = 3x 2 iˆ + 5 xy 2 ˆj + xyz 3 kˆ 2

= f1iˆ + f 2 ˆj + f 3 kˆ, say. Then,  ∂f ∂f ∂f div f = 1 + 2 + 3 = 6 x + 10 xy + 3 xyz 2 ∂x ∂y ∂z

ˆj iˆ    v = ω × r = ω1 ω 2 x

 ∂ ∂ ∂  =  iˆ + ˆj + kˆ  × ( f1iˆ + f 2 ˆj + f 3 kˆ) ∂y ∂z   ∂x ˆj kˆ iˆ =

∂ ∂x

∂ ∂y

∂ ∂z

f1

f2

f3

 ∂f ∂f   ∂f ∂f  =  3 − 2  iˆ +  1 − 3  ˆj  ∂y ∂z   ∂z ∂x   ∂f ∂f  +  2 − 1  kˆ.  ∂x ∂y   Obviously, curl f is a vector-point function. 13.10  PHYSICAL INTERPRETATION OF CURL Consider a rigid body rotating about a fixed axis through the origin with angular velocity   ω = ω1iˆ + ω 2 ˆj + ω 3 kˆ. Let r = xiˆ + yˆj + zkˆ be the position vector of any point P(x, y, z) on the body. Then, the velocity v of P is given by

M13_Baburam_ISBN _C13 Part II.indd 28

ω3 z

= (ω2 z − ω3 y )iˆ + (ω3 x − ω1 z ) ˆj + (ω1 y − ω2 x)kˆ. Therefore,   Curl v = ∇ × v iˆ

ˆj

kˆ

=

∂ ∂y

∂ ∂z

∂ ∂x

ω2 z − ω3 y ω3 x − ω1 z ω1 y − ω2 x

= 6 + 20 + 54 = 80 at (1, 2, 3). 13.9  CURL OF A VECTOR-POINT FUNCTION  Let f = f1iˆ + f 2 ˆj + f 3 kˆ be a vector-point function, where f1, f2, and f3 are scalar-point functions. If f is defined and differentiable at each point (x, y, z) of the region of space, then   the curl (or rotation) of f , denoted, by curl f     ∇ × f , or rot f is defined by Cur f = ∇ × f

y

kˆ

= 2(ω1iˆ + ω 2 ˆj + ω 3 kˆ), since ω1 , ω2 , and

ω3 are constants  = 2ω.   Hence, ω = 12 curl v . It follows, therefore, that the angular velocity at any point is equal to half the curl of the linear velocity at that point of the body.  Thus, curl is a measure of rotation. If curl v = 0,  then the vector v is called an irrotational vector. 13.11  THE LAPLACIAN OPERATOR ∇ 2 If f is a scalar-point function, then ∂φ ˆ ∂φ ˆ ∂φ +j +k grad φ = ∇φ = iˆ ∂x ∂y ∂z (vector-point function) and then, div[grad φ ] = ∇ ⋅∇φ =

∂  ∂φ  ∂  ∂φ  ∂  ∂φ   +  +   ∂x  ∂x  ∂y  ∂y  ∂z  ∂z 

=

∂ 2φ ∂ 2φ ∂ 2φ + + ∂x 2 ∂y 2 ∂z 2



 ∂2 ∂2 ∂2  =  2 + 2 + 2  φ = ∇ 2φ , ∂y ∂z   ∂x

where ∇ 2 = ∂22 + ∂ 22 + ∂ 22 is called Laplacian ∂x ∂y ∂z operator. A scalar-point function possessing second-order continuous partial derivatives and satisfying the Laplacian equation ∇ 2φ = 0 is called a harmonic function.

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vector calculuS  n  13.29 EXAMPLE 13.47

  Find curl F , whereF = grad ( x3 + y 3 + z 3 − 3 xyz ). Solution. We have

 F = grad ( x 3 + y 3 + z 3 − 3 xyz )  ∂ ∂ ∂  =  iˆ + ˆj + kˆ  ( x 3 + y 3 + z 3 − 3 xyz ) ∂ ∂ ∂ x y z  = (3 x 2 − 3 yz )iˆ + (3 y 2 − 3 xz ) ˆj + (3 z 2 − 3 xy )kˆ.

Therefore,    ∂ ∂ ∂   Curl F = ∇ × F =  iˆ + ˆj + ˆj  × F ∂y ∂z   ∂x ˆj iˆ kˆ =

∂ ∂x

∂ ∂y

∂ ∂z

3 x 2 − 3 yz 3 y 2 − 3 xz 3 z 2 − 3 xy = iˆ(−3 x + 3 x) + ˆj (−3 y + 3 y ) + kˆ(−3 z + 3 z )  = 0. EXAMPLE 13.48

 Show that the vector v = ( yz )iˆ + ( zx) ˆj + ( xy )kˆ is irrotational. Solution. It is sufficient to show that th  curl v = 0. We have ∂ ∂     ∂ Curl v = ∇ × v =  iˆ + ˆj + kˆ  ∂y ∂z   ∂x × ( yz )iˆ + ( zx) ˆj + ( xy )kˆ  ˆj kˆ iˆ =

∂ ∂x

∂ ∂y

∂ ∂z

yz

zx

xy

 = iˆ ( x − x ) + ˆj ( y − y ) + kˆ ( z − z ) = 0.

ˆj  ∂ ( x) − ∂ ( z )   ∂z ∂x  

∂  ∂ + kˆ  ( y ) − ( x)  ∂ x ∂ y    = iˆ(0) + ˆj (0) + kˆ(0) = 0. EXAMPLE 13.50

  f = 0, Show that curl curl  f = ziˆ + xˆj + ykˆ.  Solution. Let f = ziˆ + xˆj + ykˆ. Then iˆ ˆj kˆ  curl f = ∂∂x ∂∂y ∂∂z z

x

where

y

 ∂y ∂x   ∂z ∂y   ∂x ∂z  =  −  iˆ +  −  ˆj +  −  kˆ  ∂y ∂z   ∂z ∂x   ∂x ∂y  = (1 − 0)iˆ + (1 − 0) ˆj + (1 − 0)kˆ = iˆ + ˆj + kˆ. Hence,  curl curl f =

iˆ

ˆj

kˆ

∂ ∂x

∂ ∂y

∂ ∂z

1

1

1

∂  ∂ =  (1) − (1)  iˆ ∂z   ∂y

∂ ∂  +  (1) − (1)  ∂ z ∂ x   ˆj +  ∂ (1) − ∂ (1)  kˆ  ∂y   ∂x  = 0iˆ + 0 ˆj + 0kˆ = 0. EXAMPLE 13.51

EXAMPLE 13.49

  If r = xiˆ + yˆj + zkˆ, show that curl r = 0. Solution. We have

  curl r = ∇ × r =

 ∂ ∂ = iˆ  ( z ) − ( y )  + ∂ y ∂ z  

iˆ

ˆj

kˆ

∂ ∂x

∂ ∂y

∂ ∂z

x

y

z

M13_Baburam_ISBN _C13 Part II.indd 29

 If all second-order derivatives of f and v are continuous, show that  (i) curl ( grad φ ) = 0,    (ii) curl (curl v ) = grad div v − ∇ 2 v ,  (iii) div (curl v )=0, and  (iv) grad (div v )

1/2/2012 12:20:20 PM

13.30  n  chapter thirteen  ∂ ˆ ∂ ˆ ∂   ∂v1 ∂v2 ∂v3  +j + k  + + . ∂y ∂z   ∂x ∂y ∂z   ∂x



( div v ) =  iˆ

Solution. (i) We have

 ∂  ∂v ∂v   ∂ 2 v ∂ 2 v   = ∑iˆ   2 + 3  −  21 + 21   ∂z    ∂x  ∂y ∂z   ∂y

Curl (grad f) = ∇ × ∇φ  ∂ ∂ ∂  =  iˆ + ˆj + kˆ  ∂y ∂z   ∂x



iˆ ∂ = ∂x ∂φ ∂x



ˆj ∂ ∂y ∂φ ∂y



2 2 ˆj  ∂ φ − ∂ φ   ∂z∂x ∂x∂z 

 (ii) If v = v1iˆ + v2 ˆj + v3 kˆ, then ˆj kˆ iˆ ∂ ∂ ∂   curl v = ∇ × v = ∂x ∂y ∂z v1 v2 v3  ∂v ∂v  = ∑i 3 − 2   ∂y ∂z  and so,   curl (cur1 v ) = ∇ × (∇ × v )

=

=

+ =

∂v ∂v 3− 2 ∂y ∂z

∂v ∂v 1− 3 ∂z ∂x

∂v ∂v 2− 1 ∂x ∂y

∂  ∂v2 ∂v1  −   ∂z  ∂x ∂y 

∂ 2 v3 ∂ 2 v2 ∂ 2 v1 ∂ 2 v3 − + − ∂x∂y ∂x∂z ∂y∂z ∂y∂x



+

∂ 2 v2 ∂ 2 v1 − ∂z∂x ∂z∂y

∂ 2 v3 ∂ 2 v2 ∂ 2 v1 ∂ 2 v3 − + − ∂x∂y ∂x∂z ∂y∂z ∂x∂y +

kˆ ∂ ∂z

∂v3   ∂ 2 v1 ∂ 2 v2 ∂ 2 v3   ∂v2 + + +  − ∂y ∂z   ∂x 2 ∂y 2 ∂z 2  

∂  ∂v3 ∂v2  ∂  ∂v1 ∂v3   − −   + ∂x  ∂y ∂z  ∂y  ∂z ∂x  

=

ˆj ∂ ∂y

+

 ∂v ∂v  + kˆ  2 − 1  .  ∂x ∂y  Therefore,   div (curl v ) = ∇ ⋅ (∇ × v )



iˆ ∂ ∂x

M13_Baburam_ISBN _C13 Part II.indd 30





 ∂ ∂ ∂   =  iˆ + ˆj + kˆ  (div v ) ∂y ∂z   ∂x 2 ˆ ˆ ) = grad div v − ∇ 2 v. −∇ (iv1 + ˆjv2 + kv 3 (iii) As in (ii),  ∂v ∂v  ∂v ∂v  curl v = iˆ  3 − 2  + ˆj  1 − 3  ∂ ∂ y z  ∂z ∂x   

 ∂ 2φ ∂ 2φ  + kˆ  −   ∂x∂y ∂y∂x   = iˆ(0) + ˆj (0) + kˆ(0) = 0.



 ∂  ∂v1

∑iˆ  ∂x  ∂x

∂  = ∑iˆ (divv ) − ∑∇ 2 iˆv1 ∂x

kˆ ∂ ∂z ∂φ ∂z

 ∂ 2φ ∂ 2φ  = iˆ  − +  ∂y∂z ∂z∂y 



=

 ∂φ ˆ ∂φ ˆ ∂φ  ×  iˆ +j +k  ∂y ∂z   ∂x



 ∂  ∂v ∂v  ∂  ∂v ∂v   = ∑ iˆ   2 − 1  −  1 − 3    ∂y  ∂x ∂y  ∂z  ∂z ∂x  

∂ 2 v2 ∂ 2 v1  − since v is continuous ∂x∂z ∂y∂z

= 0.  (iv) If v = v1iˆ + v2 ˆj + v3 kˆ, then  ∂v ∂v ∂v div v = 1 + 2 + 3 . ∂x ∂y ∂z Therefore,

1/2/2012 12:20:21 PM

vector calculuS   n  13.31 ∂v   ∂ ∂ ∂   ∂v1 ∂v2  + ˆj + kˆ + + 3 grad(div v ) =  iˆ  ∂y ∂z   ∂x ∂y ∂z   ∂x

 curl curl v =

EXAMPLE 13.52

  If a is a constant vector and r = x iˆ + y ˆj + zkˆ , show that      (i) div(a × r ) = 0 and (ii) curl (a × r ) = 2a.  Solution. (i) We have a = a1iˆ + a2 ˆj + a3 kˆ . Therefore, iˆ ˆj kˆ   a × r = a1 a2 a3 x y z = iˆ(a2 z − a3 y ) − ˆj (a1z − a3 x ) + kˆ (a1 y − a2 x ). Hence, ∂ ∂   div (v × r ) = (a2 z − a3 y ) − (a1z − a3 x ) ∂x ∂y ∂ (a1 y − a2 x) = 0 − 0 + 0 = 0 ∂z     (ii) curl (a × r ) = ∇ × (a × r ) +

iˆ

ˆj

kˆ

∂ ∂x

∂ ∂y

∂ ∂z

a2 z − a3 y

a3 x − a1z

a1 y − a2 x

=



= iˆ (a1 + a1 ) + ˆj (a2 + a2 ) + kˆ (a3 + a3 )



 = 2(a1iˆ + a2 ˆj + a3 kˆ ) = 2a.

EXAMPLE 13.53

  Determine curl curl v if v = x 2 y iˆ + y 2 z ˆj + z 2 ykˆ .  2 2 2 Solution. Let v = x y iˆ + y z ˆj + z y kˆ . Then

  curl v = ∇ × v =

iˆ

ˆj

∂ ∂x

∂ ∂y

kˆ

x 2y y 2z z 2y

( = iˆ ( z

) + ˆj (0) + kˆ ( −x ) − y ) + 0 ˆj − x kˆ

= iˆ z − y Hence,

M13_Baburam_ISBN _C13 Part II.indd 31

2

2

2

2

kˆ

∂ ∂y

∂ ∂z

= (2z + 2x ) ˆj + 2 ykˆ

EXAMPLE 13.54

 Show that r n r is irrotational.



Solution. It is sufficient to show that curl r n r = 0 . We

have curl r n r = ∇ × r n r = ∇ ×  r n xiˆ + yjˆ + zkˆ    ˆj iˆ kˆ ∂ ∂ ∂ = ∂x ∂y ∂z

)

(



rnx

rny

rnz

∂ n ˆ ∂ n r z − r y i = ∂z  ∂x 

( )

( )



∂ n ˆ ∂ n r x − r z j + ∂x  ∂z 



∂ n ∂ n ˆ + r y − r x k ∂y  ∂x 



 ∂r ∂r  =  znr n −1 − ynr n −1  iˆ ∂y ∂z  

( )

( )

( )

( )



∂r ∂r   +  xnr n −1 − znr n −1  ˆj ∂z ∂x  



 ∂r ∂r  +  ynr n −1 − xnr n −1  kˆ ∂ ∂ x y 



2

2

ˆj

∂ ∂x

z 2 − y 2 0 − x2



∂ ∂z

iˆ



y z  =  znr n −1 − ynr n −1  iˆ r r  z x  +  xnr n −1 − znr n −1  ˆj r r  x y  +  ynr n −1 − xnr n −1  kˆ r r   = 0iˆ + 0 ˆj + 0 kˆ = 0.

Hence, r n r is irrotational.

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13.32  n  chapter thirteen 13.12  PROPERTIES OF DIVERGENCE AND CURL   (A) Properties of Divergence. Let f and g be two Vector-point functions and f a scalarpoint function. Then, the divergence has the following properties.       (i) div f + g = ∇ ⋅ f + g = ∇ ⋅ f + ∇ ⋅ g .     (ii) div φf = ∇ ⋅ φf = (∇φ ).f + φ ∇ ⋅ f   = ( grad φ ) ⋅ f + φ divf .   (iii) div f = 0 if f is a constant vector.       (iv) div f × g = g . curl f − f . curl g .  Proof: Let f = f 1iˆ + f 2 ˆj + f 3 kˆ and  g = g1iˆ + g 2 ˆj + g 3 kˆ .

( ) ( ) ( ) ( ) (

(

)

)

(i) We have

  f + g = (f 1 + g1 ) iˆ + (f 2 + g 2 ) ˆj + (f 3 + g 3 ) kˆ .

∂f   ∂φ + f3 +φ 3  ∂z ∂z 



 ∂φ ∂φ ∂φ  =  f1 + f2 + f3  ∂x ∂y ∂z 



∂f  ∂f  ∂f + φ 1 + φ 2 + φ 3   ∂x ∂y ∂z 







 ∂φ ∂φ ˆ ∂φ ˆ  =  iˆ + j+ k ⋅ f 1iˆ + f 2 ˆj + f 3 kˆ  ∂x ∂y ∂z 

(

)

∂f  ∂f  ∂f +φ  1 + 2 + 3   ∂x ∂y ∂z    = ( ∇φ ) ⋅ f + φ ∇ ⋅ f   = (grad φ ).f + φ divf .



(



)

 (iii) Let f = f 1iˆ + f 2 ˆj + f 3 kˆ . . Then

 ∂f  ∂f ∂f ∇ ⋅ f = 1 + 2 + 3 = 0, since f is constant. ∂x ∂y ∂z Therefore,    ∂      ∂ ∂ ∇ ⋅ f + g =  iˆ + ˆj + kˆ  . (iv) div f × g = ∇ ⋅ f × g  ∂x ∂y ∂z   ∂ ˆ ∂ ∂   =  iˆ +j + kˆ  . f × g ( f1 + g1 ) iˆ + ( f 2 + g 2 ) ˆj + ( f 3 + g 3 ) kˆ   ∂x ∂y ∂z     ∂ ∂ ∂ ∂  = = ∑ iˆ. f ×g ( f 1 + g 1 ) + ∂y ( f 2 + g 2 ) + ∂z ( f 3 + g 3 ) ∂x ∂x    ∂f   ∂g   ∂f 1 ∂f 2 ∂f 3  ˆ = i . × g + f × = + +  ∑  ∂x ∂x   ∂x ∂y ∂z     ∂g 1 ∂g 2 ∂g 3   ∂f   + + + = ∑ iˆ.  × g   ∂x ∂y ∂z   ∂x      = ∇⋅f + ∇⋅ g ∂g   + ∑ iˆ.  f ×   ∂x    ∂ ∂ ∂ (ii) ∇ ⋅ φf =  iˆ  + ˆj + kˆ  . φf 1iˆ + φf 2 ˆj + φf 3 kˆ  ∂f    ∂x ∂y ∂z  = ∑  iˆ ×  .g ∂x     ∂ ∂ ∂ ∇ ⋅ φf =  iˆ + ˆj + kˆ  . φf 1iˆ + φf 2 ˆj + φf 3 kˆ   ∂x ∂y ∂z   ∂g   − ∑ iˆ.  ×f   ∂x  ∂ ∂ ∂ = φ f 1 ) + (φ f 2 ) + (φ f 3 ) (  ∂x ∂y ∂z  ∂f   = ∑  iˆ ×  .g ∂ f ∂ f   ∂ ∂ φ φ    ∂x  =  f1 + φ 1  +  f 2 + φ 2   ∂x ∂x   ∂y ∂y 

(

)

(

)

(

)

(

(

(

( )

( )

M13_Baburam_ISBN _C13 Part II.indd 32

(

)

)

)

)

1/2/2012 12:20:24 PM

13.33 vector calculuS  n     ∂g   − ∑  i ×  .f  ∂x      = ∇ × f .g − ( ∇ × g ) .f     = g . ∇ × f − f (∇ × g )



(



)

(

)

(commutativity of dot product)     = g curl f − f .curl g .   (B) Properties of Curl. Let f and g be two vector-point functions and f a scalar-point function, all having continuous second-order partial derivatives. Then,     (i) curl f + g = curl f + curl g .    (ii) curl φ f = ( grad φ ) × f + φ curl f .    (iii) curl f = 0 , if f is a constant vector.     (iv) ∇ × f × g = ( g .∇) f − f .∇ g     +f (∇ ⋅ g ) − ∇ ⋅ f g .  Proof: If f = f 1iˆ + f 2 ˆj + f 3 kˆ and  g = g iˆ + g ˆj + g kˆ , then

( ) ( ) (

1

)

2

(

( )

)

3

  (i) f + g = (f 1 + g1 ) iˆ + (f 2 + g 2 ) ˆj + (f 3 + g 3 ) kˆ and so,     curl f + g = ∇ × f + g

(



)

(

iˆ ∂ = ∂x f1 + g1

kˆ ∂ ∂z f3 + g3

 ∂  ∂ =  (f 3 + g 3 ) − (f 2 + g 2 ) iˆ ∂ y ∂ z   ∂ ∂  +  (f 1 + g1 ) − (f 3 + g 3 ) ˆj ∂x  ∂z  ∂  ∂ +  (f 2 + g 2 ) − f 1 + g1 ) kˆ ( ∂y  ∂x    ∂f ∂f  ∂f   ∂f =  3 − 2  iˆ +  1 − 3  ˆj  ∂z ∂x    ∂y ∂z 

M13_Baburam_ISBN _C13 Part II.indd 33

  ∂g ∂g   ∂g ∂g  +  3 − 2  iˆ +  1 − 3  ˆj  ∂z   ∂x  ∂ y ∂ z 

∂g    ∂g +  2 − 1  kˆ   ∂x ∂y       = ∇ × f + ∇ × g = curl f + curl g .   (ii) curl φf = ∇ × φf

( )

( ) = ∇ × (φf iˆ + φf 1

=

iˆ

ˆj

kˆ

∂ ∂x

∂ ∂y

∂ ∂z

2

ˆj + φf kˆ 3

)

φf 1 φf 2 φf 3  ∂  ∂ =  (φf 3 ) − (φf 2 ) iˆ ∂z  ∂y  ∂ ∂  +  (φf 1 ) − (φf 3 ) ˆj ∂x  ∂z  ∂  ∂ +  (φ f 2 ) − φf 1 ) kˆ ( ∂y  ∂x  ∂f  ∂f ∂φ ∂φ  ˆ f3 −φ 2 − f2 i = φ 3 + ∂ ∂ ∂ ∂z  y y z 

) ˆj ∂ ∂y f2 + g2

∂f    ∂f +  2 − 1  kˆ   ∂x ∂y  

∂f  ∂f ∂φ ∂φ  ˆ f1 − φ 3 − f3 j + φ 1 + ∂ ∂ ∂ ∂x  y z x 

∂f  ∂f ∂φ ∂φ  ˆ + φ 2 + f2 −φ 1 − f1 k ∂ ∂ ∂ ∂ z x y y     ∂f ∂f  ∂f  ∂f    ∂f  ∂f = φ  3 − 2  iˆ +  1 − 3  ˆj +  2 − 1  kˆ   ∂z ∂x   ∂x ∂y     ∂y ∂z 

  ∂φ ∂φ  ˆ  ∂φ ∂φ  ˆ  ∂φ ∂φ  ˆ  +  f 3 − f 2 i +  f1 − f 3 j +  f 2 − f1 k     ∂ ∂  ∂ ∂  ∂ ∂ y z z x x y   

 = φ ∇×f +

(

)

iˆ

ˆj kˆ

∂φ ∂x

∂φ ∂y

∂φ ∂z

f1 f 2 f 3   = φ ∇ × f + ( ∇φ ) × f   = φ curl = φ curl f + ( grad φ ) × f .

(

)

1/2/2012 12:20:25 PM

13.34  n  chapter thirteen  (iii) Let f = f 1iˆ + f 2 ˆj + f 3 kˆ be a constant vector. Then,   curl f = ∇ × f  ∂ ˆ ∂ ∂ =  iˆ +j + kˆ  × f 1iˆ + f 2 ˆj + f 3 kˆ  ∂x ∂y ∂z 

(

=

iˆ

ˆj

kˆ

∂ ∂x

∂ ∂y

∂ ∂z

f1

f2

f3

∂f   ∂f = iˆ  3 − 2  +  ∂y ∂z 

    ∂f  ∂f    ˆ ˆ = ∑  i .g − i. g ∂x  ∂x      ∂g    ∂g   + ∑  iˆ.  f − iˆ.f ∂x    ∂x 

( )

( )

)

(by property of vector triple product)  ∂f  ∂f 1   = ∑ g1 − ∑ g ∂x  ∂x   ∂g  ∂g   +  ∑ 1  f − ∑ f1  ∂x  ∂x

ˆj  ∂f 1 − ∂f 3   ∂z ∂x 

   ∂ ∂ ∂  =  g1 + g2 + g3  f − ∇ ⋅ f g  ∂x ∂y ∂z 

(

∂f   ∂f + kˆ  2 − 1   ∂x ∂y 

∂ ∂    ∂ + ( ∇⋅ g ) f −  f 1 + f 2 +f3  g  ∂x ∂y ∂z 

 = iˆ(0) + ˆj (0) + kˆ (0) = 0. 

(iv) If f = f 1iˆ + f 2 ˆj + f 3 kˆ and g1iˆ + g 2 ˆj + g 3 kˆ , then    ∂    ∂ ∇ × f × g =  iˆ + ˆj + kˆ  × f × g  ∂x ∂y 

(





)

(

= iˆ ×

)

∂   ˆ ∂   f ×g + j× f ×g ∂x ∂y

(

)

(

∂   f ×g ∂z    ∂f   ∂g  = iˆ ×  × g + f ×  ∂x   ∂x    ∂f   ∂g  + ˆj ×  × g + f ×  ∂y   ∂y    ∂f   ∂g  + kˆ ×  × g + f ×  ∂z   ∂z   ∂f   = ∑ iˆ ×  × g  ∂x     ∂g  + ∑ iˆ ×  f ×   ∂x  + kˆ ×

M13_Baburam_ISBN _C13 Part II.indd 34

(

)

)

)

        = ( g.∇) f − ∇⋅ f g + (∇⋅ g ) f − f .∇ g.

( )

( )

EXAMPLE 13.55

Show that div (grad rn) = ∇ ⋅ (∇r n ) = ∇ 2 (r n ) = n(n + 1)r n − 2 . Deduce that ∇ 2 ( 1r ) = 0. Solution. From the definition of the Laplacian operator ∇ 2 , we have

(

n div (grad rn) = ∇ ⋅ ∇r

)

 ∂ ∂2 ∂2  = ∇2 r n . =  2 + 2 + 2  r n . ∂y ∂z   ∂x 2

( )

( )

(1)

But, ∂2 n ∂  ∂ n  ∂  n −1 ∂r  r =  r =  nr  ∂x  ∂x  ∂x  ∂x  ∂x 2

( )

=

∂  n −1 x  ∂r x =  nr .  , since ∂x  r ∂x r

=

∂ ∂r   nr n − 2 x = n  (n − 2)r n − 3 .x + r n − 2  ∂x ∂x  

(

)

1/2/2012 12:20:26 PM

vector calculuS  n  13.35 Therefore,

x = n ( n − 2) r n − 3 . .x + nr n − 2 r = n ( n − 2) r n − 4 x 2 + nr n − 2 . Similarly, ∂2 n r = n(n − 2)r n − 4 y 2 + nr n − 2 and ∂y 2

( )

∂2 n r = n(n − 2)r n − 4 z 2 + n r n − 2 . ∂z 2

( )

(

)

∇ 2 r n = n ( n − 2) r n − 4 x 2 + y 2 + z 2 + 3r n − 2  = n ( n − 2) r n − 2 + 3r n − 2 θ  = n(n + 1)r n − 2 .

Putting n = −1, we get ∇ 2 ( 1r ) = 0. Second Method: From Example 13.34, we have ∇r n = nr n − 2 r. Therefore,

( )

(

∇ 2 r n = ∇ ⋅ ∇r n

)

 = ∇ ⋅ nr n − 2 r

(

)

 r = 0. 



(2) 



But, div(φ r ) = ∇ ⋅ (φ r ) = (∇φ ). r + φ (∇ ⋅ r ) . Therefore, (2) becomes 



∇ 2 (r n ) = n[(∇r n − 2 ). r + r n − 2 (∇ ⋅ r )] 



= n[(n − 2)r n − 4 r . r + 3r n − 2 ] since ∇ ⋅ r = 3 = n[(n − 2)r

r + 3r

n−2





] , since r . r = r

2

= n(n + 1)r n − 2 .   Show that curl ( grad r ) = ∇ × ∇r n = 0.

y z  = ∑  nz (n − 2)r n − 3 − ny (n − 2)r n − 3  iˆ r r 

= ∑  n ( n − 2)r n − 4 yz − n ( n − 2)r n − 4 yz ] iˆ  = 0iˆ + 0 ˆj + 0 kˆ = 0. EXAMPLE 13.57

    If f and g are irrotational, show that f × g is solenoidal.   Solution. Since f and g are irrotational, we have     ∇ × f = 0 and ∇ × g = 0. Now,       div f × g = g . ∇ × f − f . ( ∇ × g )    = g .0 − f .0 = 0.   Hence, f × g is solenoidal.

(

)

(

)

Show that

2 ∇2 f ( r ) = f ′′( r ) + f ′( r ), r  where r = xiˆ + yjˆ + zkˆ .

∂r x ∂r y ∂r z = , = , and = . Then, ∂x r ∂y r ∂z r

Solution. We have seen that

)

 ∇r n = nr n − 2 r = nr n − 2 xiˆ + yjˆ + zkˆ .

M13_Baburam_ISBN _C13 Part III.indd 35

 ∂  ∂ = ∑  ( nr n − 2 z ) − ( nr n − 2 y )  iˆ ∂z  ∂y 

Solution. We have r 2 = x 2 + y 2 + z 2 and so,

EXAMPLE 13.56

(

kˆ ∂ ∂z nr n − 2 z

EXAMPLE 13.58



n−4 2

ˆj ∂ ∂y nr n − 2 y

 ∂r ∂r  = ∑  nz ( n − 2)r n − 3 − ny ( n − 2)r n − 3  iˆ ∂y ∂z  

Hence, (1) reduces to

( )

iˆ ∂ ∇ × ∇r n = ∂x nr n − 2 x

 ∂ ∂ ∂ ∂r grad r =  i + j + k  r = iˆ +  ∂x ∂y ∂z  ∂x

ˆj ∂r + kˆ ∂r ∂y ∂z

1/2/2012 2:35:54 PM

13.36  n  chapter thirteen  x y z 1 r = iˆ + ˆj + kˆ = xiˆ + yjˆ + zkˆ = . r r r r r

)

(

(1)

grad f ( r ) = f ′( r )∇r = f ′( r )grad r .

(2)

Therefore, ∇2 f ( r ) = ∇ ⋅ (∇f ( r )) = div(gradf ( r )) = div(f ′( r )∇r )

 3   d 1  f ′( r ) + r .   f ′( r ) grad r  ,   r  dr r 

 using (2) and div r = 3.

=

using (l)

1  1   − r 2 f ′( r ) + r f ′′( r )   

3 1 2 = f ′( r ) + f ′′( r ) − f ′( r ) = f ′( r ) + f ′′( r ). r r r EXAMPLE 13.59

 Show that div r n r = ( n + 3) r n .

( )

Solution. We know that

   div φ f = ( grad φ ) .f + φ div f .

( )

Therefore,

   div r n r = grad r n .r + r n div r .

( ) (

)

  But, grad r n = nr n − 2 r and div r = 3. Therefore,   div r n r = nr n − 2 r .r + 3r n = nr n − 2 r 2 + 3r n

( )

= (n + 3) rn.

M13_Baburam_ISBN _C13 Part III.indd 36

and so, ∂φ ˆ ∂φ ˆ ∂φ ˆ i +φ j +φ k. ∂x ∂y ∂z

iˆ curl (φ grad φ) = ∂ ∂x ∂φ φ ∂x

ˆj ∂ ∂y ∂φ φ ∂y

kˆ ∂ ∂z ∂φ φ ∂z

 ∂  ∂φ  ∂   ∂φ   ==∑ ∑ 0 iˆ = 0φiˆ + 0 ˆj +−0 kˆ = φ0.   iˆ

 3 1  1 r f ′( r ) + r  − 2 f ′( r ) + f ′′( r )  , r r  r r

3 r2 f ′( r ) + r r

∂φ ˆ ∂φ ˆ ∂φ ˆ i+ j+ k ∂x ∂y ∂z

Hence,

by divergent property

 3 r .r  1 1  = f ′( r ) + − 2 f ′( r ) + f ′′( r )   r r  r r 

grad φ = ∇φ =

φ grad φ = φ

 1 = div  f ′( r )r  , using (1) r  1   1  = f ′( r )div r + r .grad  f ′( r ) , r  r

=

 Show that curl (φ grad φ ) = 0. Solution. We have

Also, by Example 13.32,

=

EXAMPLE 13.60

 ∂y  ∂z  ∂z  ∂y   2  2   ∂φ ∂φ ˆj +∂0 kˆφ =−0.∂φ ⋅ ∂φ − φ ∂ φ  iˆ ∑ 0iˆ = ⋅0iˆ + +0 φ ==∑ ∂y ∂z ∂z ∂y ∂z ∂y   ∂y ∂z   = ∑ 0iˆ = 0iˆ + 0 ˆj + 0 kˆ = 0.

Also, it follows that f grad f is irrotational. EXAMPLE 13.61

 Show that the vector f ( r )r is irrotational.    Solution. A vector f is irrotational if curl f = 0. Also, we know that    r × r = 0, curl r = 0 and  r grad f ( r ) = f ′( r ) grad r = f ′( r ) . r Therefore,    curl [f ( r )r ] = [grad f ( r )] × r + f ( r ) curl r   = [f ′( r ) grad r ]] × r + 0  r   = f ′( r ) × r + 0 r 1    = f ′( r )( r × r ) = 0. r  Hence, f ( r ) r is irrotational.

1/2/2012 2:35:55 PM

vector calculuS   n  13.37 13.13  INTEGRATION OF VECTOR FUNCTIONS   If f and F are two vector functions of the  scalar variable t such that dF then = f , F is dt  called the indefinite integral of f with respect to t. Thus,   ∫ f dt = F.

 If f = f 1iˆ + f 2 ˆj + f 3 kˆ , where f1 f2, and f3 are scalar functions of the scalar t, then 

∫ f dt = iˆ∫ f dt + ˆj ∫ f 1

2

dt + kˆ ∫ f 3dt .

Hence, to integrate a vector function, we integrate its components. EXAMPLE 13.62

The acceleration of a particle at any time t > 0 is given by   dv a= = 12 cos 2t iˆ − 8 sin 2t ˆj + 16t kˆ . dt   If the velocity  ν and  displacement r are zero at t = 0, find ν and r at any time t. Solution. We have

  dν = 12 cos 2t iˆ − 8 sin 2t ˆj + 16t kˆ . a= dt Integration with respect to t yields   ν = 6 sin 2t iˆ + 4 cos 2t ˆj + 8t 2 kˆ + c ,    w here c is a constant of integration. But ν = 0 when t = 0. Therefore, 

0 = 0iˆ + 4 ˆj + 0 kˆ + cˆ  and so, c = −4 ˆj . Therefore,  ν = 6 sin 2t iˆ + (4 cos 2t − 4) ˆj + 8t 2 kˆ or

 dr = 6 sin 2t iˆ + (4 cos 2t − 4) ˆj + 8t 2 kˆ . dt Integrating again with respect to t, we get 8   r = −3 cos 2t iˆ + (2 sin 2t − 4t ) ˆj + t 3 kˆ + p , 3   where p is a constant of integration. But r = 0 when t = 0. Therefore,   0 = −3iˆ + 0 ˆj + 0 kˆ + p

M13_Baburam_ISBN _C13 Part III.indd 37

 and so, p = 3iˆ. Hence, 8  r = (3 − 3 cos 2t )iˆ + (2 sin 2t − 4t ) ˆj + t 3 kˆ . 3 EXAMPLE 13.63   If r = 0 when t = 0 and

 = u when t = 0, find 2   the value of r satisfying the equation ddt 2r = a,  where a is a constant vector. 2  Solution. Integrating ddt 2r = a with respect to t, we get    dr = at + c , dt  dr dt



where c is a  constant vector of integration.  dr When t = 0, dt = u . Therefore,      u = a(0) + c and so c = u . Therefore,

 dr   = at + u dt Integrating again with respect to t, we get  1   r = at 2 + ut + p , 2  where p is the constant vector  of integration.  When t = 0, r = 0. Therefore, 0 = p. Hence,   1 r = ut + at 2 . 2 13.14  LINE INTEGRAL An integral which is evaluated along a curve is called a line integral. Note, however, that a line integral is not represented by the area under the curve. Consider any arc of the curve C enclosed between two points A and B. Let a and b be the values of the parameter t for A and B, respectively. Partition the arc between A and B into n parts as given in the following equation: A = P0, P1, Pn = B. Let r0 , r1 ,…, rn be the position vectors of the points P0, P1, Pn respectively. Let xi be any point on the     subarc P1–1Pi and let δ ri = ri − ri −1 . Let f ( r ) be a continuous vector-point function. Then, n   lim ∑ f (ξi ) . δ ri , (1) n →∞  δ ri → 0 i =1

1/2/2012 2:35:57 PM

13.38  n  chapter thirteen  if exists, is called a line integral of f along C    d r and is denoted by ∫ f .d r or ∫ f . dt dt . Thus, the C

(

C

line integral is a scalar and  is also called the tangential line integral of f along the curve C.

0

1

 9t 3 28t 7 60t10  = − +  = 5. 7 10  0  3

  If f = f 1iˆ + f 2 ˆj + f 3 kˆ and r = xiˆ + y ˆj + zkˆ , then

 d r = iˆ dx + ˆj dy + kˆ dz and so,

EXAMPLE 13.65  If f = ( sin y )iˆ + x (1 + cos y ) ˆj , evaluate the line

  ∫ f .d r = ∫ (f 1dx + f 2dy + f 3dz )

C

)

. iˆ + 2t ˆj + 3t 2 ˆj + 3t 2 kˆ  dt  1 = ∫ (9t 2 − 28t 6 + 60t 9 )dt

 

C

integral ∫ f .d r along the circular path C given by x2 + y2 = a2 and z = 0. Solution. The parametric equation of the circular path C are x = a cos t, y = a sin t, and z = 0, where t varies from 0 to 2p. Now,  r = x iˆ + y ˆj + z kˆ = (a cos t ) iˆ + (a sin t ) ˆj . C

b



∫  f

1

a

dx dy dz  + f2 + f 3  dt , dt dt dt 

where a and b are, respectively, values of the parameter t at the points A and B. If we replace the dot product in (1) by a vector product, then the vector line integral is   defined as ∫ f × d r , which is a vector. If C is a simple closed curve, then the  tangent line integral of the vector function f  around C is called the circulation of f around C  and denoted by ∫ f .d r .  The vector function f is said to be  irrotational in a region R if the circulation of f around any closed curve in R is zero. C

EXAMPLE 13.64    If f = (3x 2 + 6 y )iˆ − 14 yz ˆj + 20xz 2 kˆ , evaluate ∫ f .d r ,

Therefore,  dr = ( −a sin t ) iˆ + (a cos t ) ˆj . dt  Also, f , in terms of parameter t, is given by

 f = sin (a sin t )iˆ + (a cos t )(1 + cos (a sin t )) ˆj .

Therefore,

   dr 

 

∫ f .d r = ∫  f . dt  dt

C

C

=

= (3t + 6t )iˆ − 14t 2 .t 3 ˆj + 20t .t 6 kˆ 2

2

=

2π

∫ [ − a sin t sin(a sin t ) 0

+ a 2 cos 2 t (1 + cos (a sin t ))]dt =

2π

∫  −a sin t sin (a sin t ) + a cos t 2

2

0

+a2 cos 2 t cos (a sin t )  dt =

2π

∫ {d [a cos t sin (a sin t )] + a cos t } dt 2

2

0

2π

= 9t 2 iˆ − 14t 5 ˆj + 20t 7 kˆ .

= [a cos t sin (a sin t )]20π + ∫ a2 cos 2t dt 0

Therefore,

     dr 2 5 7 ∫C f .d r = C∫  f . dt  dt =  9t iˆ − 14t ˆj + 20t kˆ

(

M13_Baburam_ISBN _C13 Part III.indd 38

cos (a sin t )) ˆj )]

. ( −a sin t )iˆ + (a cos t ) ˆj  dt



Now, r = xiˆ + yjˆ + zkˆ = tiˆ + t 2 ˆj + t 3 kˆ . Therefore,  dr ˆ = i + 2tjˆ + 3t 2 kˆ . dt Further,  f = (3x 2 + 6 y )iˆ − 14 yz ˆj + 20 xz 2 kˆ

∫ [sin (a sin t )iˆ + (a cos t (1 + 0

C

where C is given by x = t, y = t2, and z = t3, and t varies from 0 to 1. Solution. The parametric equation of C is x = t, y = t2, and z = t3, where t varies from 0 to 1.

2π

)

2π

= a2 ∫ 0

2π

1 + cos 2t a  sin 2t  dt = t + 2 2  2  0 2

1/2/2012 2:35:58 PM

vector calculuS   n  13.39 =

Therefore,

a2 (2π ) = π a2 . 2

EXAMPLE 13.66

 

Calculate ∫ f .d r , where C is the part of the  C spiral r = (a cos θ , a sin θ , aθ ) corresponding to  0 ≤ θ ≤ π2 and f = r 2 iˆ. Solution. We have

ˆj kˆ −t 3 t 4 2 3t 2

= ( −3t 5 − 2t 4 )iˆ + ( −4t 5 ) ˆj + (4t 3 + 2t 4 ) kˆ . Hence,

 1     dr f × d r = f × ∫ ∫0  dt  dt C

 r = a cos θ iˆ + a sin θ ˆj + aθ kˆ so that  dr = −a sin θ iˆ + a cos θ ˆj + a kˆ . dθ

Also,

iˆ  d r f × = 2t 2 dt 2t

1

= ∫ ( −3t 5 − 2t 4 )iˆ + ( −4t 5 ) ˆj + (4t 3 + 2t 4 ) kˆ  dt 0

 f = r 2 iˆ = (a2 cos 2θ + a2 sin 2θ + a2θ 2 ) iˆ

1

1

 3t 6 2t 5  ˆ  t 6   ˆ = − −  i +  −4   j 5 0 6  0  6 

= [a2 (1 + θ 2 )iˆ. Therefore, π 2     dr  f d r f d θ . . {[a2 (1 + θ 2 )] iˆ = =   ∫C ∫ dθ 0

1

 t 4 2t 5  ˆ + 4 +  k 5 0  4

=−

.[( −a sin θ )iˆ + (a cos θ ) ˆj + akˆ ]}dθ π

9 ˆ 2ˆ 7ˆ i − j + k. 10 3 5

EXAMPLE 13.68    If f = (5xy − 6x 2 )iˆ + (2y − 4x ) ˆj , evaluate ∫ f .d r along

2

= − ∫ a3 (1 + θ 2 ) sinθ dθ 0

C

the curve C in the xy plane y = x3 from the point (1, 1) to (2, 8).

π

2

= −a3 ∫ (sinθ + θ 2 sinθ ) dθ 0

π

= − a [ cos θ + 2θ sin θ − θ 2 cos ] 02 3

= − a (π − 1). 3

Solution. Substituting x = t, we get y = t3. When x

= 1, t =l and when x = 2, t = 2. Then,   3 ˆ dr ˆ ˆ ˆ ˆ r = xi + yj + zk = ti + t j , = iˆ + 3t 2 ˆj and dt  2 ˆ f = (5xy − 6x )i + (2y − 4x ) ˆj = (5t 4 − 6t 2 )iˆ + (2t 3 − 4t ) ˆj .

EXAMPLE 13.67  2 If f = xyiˆ − z ˆj + x 2 kˆ and C is the curve 2x =ˆ t , f = (5 xy − 6 x ) i + (2y − 4x ) ˆj = (5t 4 − 6t 2 )iˆ + (2t 3 − 4t ) ˆj . 3 t = 0 t = 1 , and from to , find the y = 2t z =t 

vector line integral ∫ f

 × d r.

C

Solution. We have

 r = x iˆ + y ˆj + z kˆ = t 2 iˆ + 2t ˆj + t 3 kˆ,  dr = 2t iˆ + 2 ˆj + 3t 2 kˆ , dt

and

 f = xy iˆ − z ˆj + x 2 kˆ = 2t 3 iˆ − t 3 ˆj + t 4 kˆ .

M13_Baburam_ISBN _C13 Part III.indd 39

Hence,

2   2   d r  4 2 f d r f dt . . = = ∫C ∫1  dt  ∫1 {[(5t − 6t )iˆ

(

+ (2t 3 − 4t ) ˆj  ⋅ iˆ + 3t 2 ˆj

) } dt

2

= ∫ [(5t 4 − 6t 2 ) + 3t 2 (2t 3 − 4t )]dt 1

1/2/2012 2:35:58 PM

13.40  n  chapter thirteen Along AB, we have x = 1 and so, dx = 0. Also along AB, y varies from −l to 1. Thus,

2

= ∫ [6t 5 + 5t 4 − 12t 3 − 6t 2 ]dt 1

∫ [(x

2

 t6 t5 t 4 6t 3  = 6 + 5 − 12 −  5 4 3 1  6 5

4

+ xy )dx + ( x 2 + y 2 )dy ] 1

1  y3 8 = ∫ (1 + y 2 )dy = [ y +  = . 3  −1 3  −1

= [t + t − 3t − 2t ] = 35. 6

2

AB

3 2 1

y

EXAMPLE 13.69 B

Evaluate ∫ [2xydx + (x − y ) dy ] along the arc of the circle x2 + y2 = 1 in the first quadrant from A (1. 0) to B (0, 1). 2

2

C(–1, 1)

B(1, 1)

A

2 Solution. On the circle, y = 1 − x so that

(

)

− x 1 − x2 Therefore,

− 12

or

dy dx

= −x

dy = −2x (1 − x ) dx . 1 2 −2

x

0

B

∫ [2xydx + (x

2

− y 2 )dy ]

A

0

−1

1

1

0

0

3  1  2  −1 =  − (1 − x 2 ) 2  − ∫  x 2 −  2x (1 − x 2 ) 2 dx   3 2  1 1

0

1  2   1 = − + 2  x 2 −  (1 − x 2 ) 2  3   2 1

0

Along BC, we have y = 1 so that dy = 0. Also along BC, x varies from 1 to −l. Thus,

∫ [(x

1

=

1 2

 (1 − x 2 ) 2  2 = − −1+ 2   3 3 2   3

0

∫ [(x

1

)

+ xy )dx + ( x 2 + y 2 )dy ]

2

CD

−1

= ∫ (1 + y 2 )dy = − 1

EXAMPLE 13.70

Evaluate ∫ [( x + xy )dx + ( x + y )dy ], where C is the C square formed by the lines y = + 1 and x = + 1. Solution. The curve C is as shown in the following figure We note that 2

2

2 2 2 ∫ [(x + xy )dx + (x + y )dy ] = ∫ + ∫ + ∫ + ∫ .

M13_Baburam_ISBN _C13 Part III.indd 40

−1

x3 x2  2 x 2 + x dx =  +  = − . 3 2 3  1

Similarly,

2 4 1 = − −1+ = − . 3 3 3

C

∫(

−1

1

2

+ xy )dx + ( x 2 + y 2 )dy ]

2

BC

−2∫ 2x (1 − x ) dx 2

A(1, –1)

D(–1, –1)

= ∫ [2x (1 − x 2 ) 2 dx − (2x 2 − 1)x (1 − x 2 ) 2 dx ]

AB

BC

CD

∫ [(x

2

8 and 3

+ xy )dx + ( x 2 + y 2 )dy ]

DA

1

=

∫ (x

−1

2

2 + x )dx = . 3

Hence,

∫ [(x c

2

+ xy )dx + ( x 2 + y 2 )dy ] =

8 2 8 2 − − + = 0. 3 3 3 3

DA

1/2/2012 2:35:59 PM

vector calculuS   n  13.41 13.15  WORK DONE BY A FORCE

EXAMPLE 13.71

The work  done as the point of application of a force f moves along a given path C can be expressed as a line integral. In fact, the work done,  when the point  of applicationmoves from P( r ) to Q ( r + δ r ), where PQ = δ r , is     δW =| δ r | f cos θ = f .δ r .

Find the total work done in moving a particle in  a force field, given by f = 3xy iˆ − 5z ˆj + 10x kˆ , 2 along the curve x = t 2 + 1 , y = 2t , and z = t 3 , from t = 1 to t = 2 . Solution. The parametric equation of the curve is x = t 2 + 1 , y = 2t 2 , and z = t 3 ,1 ≤ t ≤ 2 . We have  f = 3xy iˆ − 5z ˆj + 10x kˆ .

f (r ) P

tangent

A B

r

C

Therefore, the total work done as P moves from A to B is B   W = ∫ f .d r . A

 Now, suppose that force f is conservative. Then, there exists a scalar func tion f such that f = − grad φ, that is,   ∂φ ˆ ∂φ ˆ ∂φ  +j +k  f = −∇φ = − iˆ ∂y ∂z   ∂x Therefore, the work done in this case is given by

 ∂φ + W = ∫ − iˆ  ∂x A B

ˆ ) ˆj ∂φ + kˆ ∂φ  (idx ˆ + ˆj dy + kdz ∂y ∂z 

 ∂φ ∂φ ∂φ  = ∫  dx + dy + dz ∂x ∂y ∂z  B  A

A

= ∫ d φ = [φ ]AB = φA − φB . B

Hence, in a conservative field, the work done depends on A and B and is the same for all paths joining Aand B. Thus, in the case of conservative force f ( r ).d r is an exact differential – df. In such a case, f is called the potential energy. The forces which do not have this property are said to be dissipative or nonconservative.

M13_Baburam_ISBN _C13 Part III.indd 41

= 3(t 2 + 1)(2t 2 )iˆ − 5t 3 ˆj + 10(t 2 + 1) kˆ = 6(t 4 + t 2 )iˆ − 5t 3 ˆj + 10(t 2 + 1) kˆ and

 r = xiˆ + y ˆj + z kˆ = (t 2 + 1)iˆ + 2t 2 ˆj + t 3 kˆ . 

Therefore, ddtr = 2tiˆ + 4t ˆj + 3t 2 kˆ and so, the total work done is given by   2   d r  W = ∫ f .d r = ∫ f .   dt  dt  1 C 2

= ∫ 6(t 4 + t 2 )iˆ − 5t 3 ˆj + 10(t 2 + 1) kˆ  1

⋅ 2tiˆ + 4t ˆj + 3t 2 kˆ  dt 2

= ∫ [12(t 5 + t 3 ) − 20t 4 + 30(t 4 + t 2 )] dt 1

2

= ∫ (12t 2 + 10t 4 + 12t 3 + 30t 2 ) dt 1

2

 t6 t5 t4 t3  = 12 + 10 − 12 + 30  5 4 3 1  6 = 320 – 17 = 303. EXAMPLE 13.72

Find the work done by the force  f = (2 y + 3)iˆ + xz ˆj + ( yz − x ) kˆ , when it moves a particle from the point (0,0,0) to the point (2, 1, 1) along the curve x =2t2, y = t, and z = t3.

1/2/2012 2:36:00 PM

13.42  n  chapter thirteen Solution. The parametric equations of the curve are x =2t2, y = t, and z = t3. Further,  f = (2 y + 3)iˆ + xz ˆj + ( y z − x ) kˆ

and

= (2t + 3)iˆ + 2t 5 ˆj + (t 4 − 2t 2 ) kˆ  r = xiˆ + yjˆ + zkˆ = 2t 2 iˆ + t ˆj + t 3 kˆ .

Therefore,

 dr = 4t iˆ + ˆj + 3t 2 kˆ . dt The given points (0,0,0) and (2, 1, 1) correspond to t = 0 and t = 1. Therefore, the work done by the force is given by   W = ∫ f .d r c

1

= ∫ (2t + 3) iˆ + 2t 5 ˆj + (t 4 − 2t 2 ) kˆ 

Solution. (i) The curve C is the line joining

(0,0,0) to (2, 1, 3) whose equation is

x −0 y −0 z −0 x y Z = = or = = = t , say. 2 − 0 1− 0 3 − 0 2 1 3 Thus, x = 2t, y = t, and z = 3t are the parametric equations of the line. The point (0,0,0) corresponds to t = 0 and the point (2, 1, 3) corresponds to t = 1. Also,  f = 3x 2 iˆ + (2xz − y ) ˆj + z kˆ = 12t 2 iˆ + (12t 2 − t ) ˆj + 3tkˆ and



Therefore, ddtr = 2iˆ + ˆj + 3kˆ . Hence,   1   d r  W = ∫ f .d r = ∫  f .  dt  dr  0 C 

⋅  4t iˆ + ˆj + 3t 2 kˆ  dt = ∫ [(2t + 3)4t + 2t + 3(t − 2t )t ] dt 5

4

2

2

}

1

= ∫ [24t 2 + (12t 2 − t ) + 9t ]dt

1

0

= ∫ [8t 2 + 12t + 2t 5 + 3t 6 − 6t 4 ] dt

1

= ∫ (36t 2 + 8t )dt = [12t 3 + 4t 2 ]10 = 16.

0

1

0

= ∫ [3t 6 + 2t 5 − 6t 4 + 8t 2 + 12t ] dt 0

1

 t7 t6 t5 t3 t2  = 3 + 2 − 6 + 8 + 12  6 5 3 2 0  7 3 1 6 8 288 + − + +6 = . 7 3 5 3 35

EXAMPLE 13.73

Find the work  done in moving a particle in the force field f = 3x 2 iˆ + (2x z − y ) ˆj + zkˆ along (i) the straight line from (0,0,0) to (2, 1, 3) and (ii) the curve defined by x2 = 4y and 3x2 = 8z from x = 0 to x = 2.

M13_Baburam_ISBN _C13 Part III.indd 42



⋅ 2iˆ + ˆj + 3kˆ  dt



0

=



= ∫{[12t 2 i + (12t 2 − t ) j + 3t k ]

0

1

 r = xiˆ + yjˆ + zkˆ = 2t iˆ + t ˆj + 3t kˆ .

(ii) Putting x = t in the given curve, we get 3 2 y = t4 and z = 33t , where 0 ≤ t ≤ 2 . Then,  f = 3x 2 iˆ + (2xz − y ) ˆj + z kˆ 3 t2  3 = 3t 2 iˆ +  t 4 −  ˆj + t 3 kˆ 4 8 4

and t2 3  r = xiˆ + y ˆj + z kˆ = t iˆ + ˆj + t 3 kˆ , 0 ≤ t ≤ 2. 4 8

Therefore,

 dr ˆ 1 ˆ 9 2ˆ = i + t j + t k. dt 2 8

Hence,

  2   d r  W = ∫ f .d r = ∫  f .  dt  dt  0 C

1/2/2012 2:36:01 PM

vector calculuS  n  13.43 2

3 1 27 = ∫ (3t 2 + t 5 − t 3 + t 5 ) dt 8 8 64 0 2

 t6 t4 9 6 = t 3 + − + t  = 16. 16 32 128  0 13.16 SURFACE INTEGRAL An integral evaluated over a surface is called a surface integral. Two types of surface integral exist: (i) ∫ ∫ f ( x , y , z )dS S

and      (ii) ∫ ∫ f ( r ).nˆ dS = ∫ ∫ f ( r ).dS . S

expressing them as double integrals, taken over an orthogonal projection of the surface S on any of the coordinate planes. But, the condition for this is that any line perpendicular to the coordinate plane chosen meets the surface S in not more than one point. However, if S does not satisfy this condition, then S can be subdivided into surfaces satisfying this condition. Let S be the surface such that any line perpendicular to the xy-plane meets S in not more than one point. Then, the equation of the surface S can be written as z = h(xy). Let R1 be the projection of S on the xy-plane. Then, the projection of dS on the xy-plane is dS cos g, where g is the acute angle which the normal nˆ at P to the surface S makes with z-axis. Therefore, dS cos g = dxdy.

S

In case (i), we have a scalar field f, whereas  in case (ii), we have a vector field f ( r ), vector  element of area dS = nˆ dS , and nˆ the outwarddrawn unit normal vector to the element dS. (i) Let f(x, y, z) be a scalar-point function defined over a surface S of finite area. Partition the area S into n subareas d S1, d S2,…,d Sn. In each area d Si, choose an arbitrary point Pi (xi, yi, zi). Defin f(Pi) = fi (xi, yi, zi) and form the sum ∑f (x , y , z )δ S . Then, the limit of this sum as n → ∞ in such a way that the largest of the subarea δSi approaches zero is called the surface integral of f (x,y,z) over S and is denoted by  dS . y, z) dS. ∫∫ ff. nˆ(x,  (ii) Now, let f be a vector-point function defined and continuous over a surface S. Let P be any point on the surface S and let nˆ be the unit vector at P in the direction of the outward drawn normal to the surfaceS at P. Then, f . nˆ is the normal component of f at P. The integral of f . nˆ over S is called the normal surface  integral of f over S and is denoted by ∫∫ f . nˆ dS . S This integral is also known as flux of f over S. If we associate with the differential of surface area dS, a vector dS , with magnitude dS, and  whose direction is that of nˆ , then dS = ndS ˆ and hence,    ∫∫ f .nˆ dS = ∫∫ f .d S .

z s

y

0 R1

n

1

i

i

i

x

i =1

S

S

The surface integrals are easily evaluated by

M13_Baburam_ISBN _C13 Part III.indd 43

dxdy

 But cos γ = nˆ . kˆ ,where k is, as usual, a unit vector along the z-axis. Thus, dS = Hence,

dxdy . nˆ . kˆ



 dxdy . nˆ . kˆ

∫∫ f .nˆ dS = ∫∫ f .nˆ S

R1

Similarly, if R2 and R3 are projections of S on the yz, and zx-plane, respectively, then   dxdy ∫∫S f .nˆ dS = ∫∫R f . nˆ nˆ . iˆ , 2

and



 dxdy . nˆ . ˆj

∫∫ f .nˆ dS = ∫∫ f .nˆ S

R3

1/2/2012 2:36:02 PM

13.44  n  chapter thirteen EXAMPLE 13.74

1

 Evaluate ∫∫ f = yziˆ + zxjˆ + xykˆ and S is that part of the surface of the sphere x 2 + y 2

=

3 x(1 − x 2 ) dx ∫ 20

+z2 = 1 ,

=

3  x2 x4  3  −  = . 2 2 4 0 8

S

 f .nˆ dS , where

Solution. A vector normal to the surface of the given sphere is

∇( x + y + z − 1) 2

2

2

 ∂ ∂ ∂ =  iˆ + ˆj + kˆ  x 2 + y 2 + z 2 − 1  ∂x ∂y ∂z 

(

)

Therefore, the unit normal to any point (x, y, z) of the surface is 2 xiˆ + 2 yjˆ + 2 zkˆ 2 xiˆ + 2 yjˆ + 2 zkˆ = nˆ = 2 xiˆ + 2 yjˆ + 2 zkˆ 4x2 + 4 y 2 + 4z 2 2 xiˆ + 2 y ˆj + 2 z kˆ 2 x2 + y 2 + z 2

Solution. A vector normal to the surface S is

Therefore, the unit normal vector to the surface S is nˆ =

2 iˆ + 3 ˆj + 6 kˆ 4 + 9 + 36

=

2 iˆ + 3 ˆj + 6 kˆ 7

and so,  2 iˆ + 3 ˆj + 6 kˆ  ˆ 6 nˆ . kˆ =   . k = .  77 7  

Also,

 f .nˆ = yz iˆ + zx ˆj + xy kˆ xiˆ + y ˆj + z kˆ

(

)(

)

= xyz + xyz + xyz = 3xyz

(

)

 n . kˆ = . xiˆ + y ˆj + z kˆ .kˆ = z ,

which gives





Evaluate ∫∫ f . nˆ dS , where f = 18 z iˆ − 12 ˆj + 3 y kˆ and S S is the surface 2 x + 3 y + 6 z = 12 in the first octant.

= xiˆ + y ˆj + z kˆ,

since x 2 + y 2 + z 2 = 1 on S. Now,

and

EXAMPLE 13.75

∇(2 x + 3 y + 6 z = 12) = 2 iˆ + 3 ˆj + 6 kˆ.

= 2 xiˆ + 2 y ˆj + 2 z kˆ.

=

1

which lies in the first octant

nˆ . kˆ = z. Hence, in the first quadrant,





∫∫ f . nˆ dS = ∫∫ f . nˆ S

S

∫∫ R

  2 iˆ + 3 ˆj + 6 kˆ  f . nˆ = (18 z iˆ − 12 ˆj + 3 y kˆ) ⋅    7  

=

36 z 36 18 − + y 7 7 7

=

36 12 − 2 x − 3 y  36 18 − 7 + 7 y 7  6 

=

36 12 x − . 7 7

dxdy nˆ . kˆ

z

3 xyz dxdy z

= 3∫∫ xy dx dy R

1  = 3∫  0  

M13_Baburam_ISBN _C13 Part III.indd 44

0 1− x 2

∫ 0

 xy dy  dx 

x

(0, 4, 0)

Y

(6, 0, 0)

1/2/2012 2:36:04 PM

vector calculuS   n  13.45 Hence,

∫∫ S

A vector normal to the surface S is given by ∇( x 2 + y 2 − 16) = 2 x iˆ + 2 y ˆj ,

  dxdy f . nˆ dS = ∫∫ f . nˆ nˆ . kˆ R =

so that the unit normal vector nˆ at any point of S is 2 xiˆ + 2 yjˆ 2 x iˆ + 2 y ˆj = nˆ = 2 2 4x + 4 y 2 x2 + y 2

7  36 12 x   −  dx dy 6 ∫∫ 7 7  R 

= ∫∫ (6 − 2 x)dx dy. R

But, R is the region of projection of S (triangle) on the xy plane. Thus, the projection is a triangle bounded by x-axis, y-axis, and the line 2 x + 3 y = 12 and z = 0 . Hence, the limits of x are from 0 to 6 and that of y are from 0 to 12 −32 x . Therefore,

∫∫ S

6    f . n dS = ∫   0 

12 − 2 x 3

∫ 0

6

 (6 − 2 x ) dy  dx  

2 xiˆ + 2 yjˆ x iˆ + y ˆj = . 8 4

Also,

  xiˆ + yj  1 f . nˆ = ( z iˆ + x ˆj − 3 y 2 z kˆ).   = ( xz + xy ).  4  4

Let R be the projection of the surface S on xzplane. Then 1 y nˆ = ( xiˆ + yjˆ). ˆj = . 4 4 Hence   xz + xy dxdz = f dS f . ∫∫S ∫∫S .nˆ dS = ∫∫R 4 ⋅ nˆ. ˆj

()

= ∫ [ 6 y − 2 xy ]0 3 dx 12− 2 x

0

6

=

=

1 (72 − 36 x + 4 x 2 ) dx 3 ∫0 6

x 2 4 x3  1 = 72 x − 36 +  3 2 3 0

R

= 144 − 216 + 96 = 24. EXAMPLE 14.76 





xz + xy dxdz xz + xy . y = ∫∫ .dxdz , y 4 R 4

= ∫∫

Evaluate f . d S , where f = z iˆ + x ˆj − 3 y 2 zkˆ and S is the surfaces of the cylinder x 2 + y 2 = 16 in the first octant between z = 0 and z = 5. Solution. The surface S is shown in the following figure z

where R is the rectangular region in the xzplane bounded by 0 ≤ x ≤ 4 , 0 ≤ z ≤ 5 . Since the integrand is still evaluated on the surface, we have y = 16 − x and so, 2

  4 5  XZ   f .d S = ∫  ∫  x +  dz  dx 16 − x 2   0  0  5

  xz 2 = ∫  xz +  dx 2 2 16 − x  0 0  4

(0, 0, 5)

4  25 x = ∫  5 x + 2 16 − x 2 0

dS

  dx  4

(0, 4, 0)

(4, 0, 0) x

M13_Baburam_ISBN _C13 Part III.indd 45

 x 2 25  = 5 − 16 − x 2  = 90. 2 2  0 y

EXAMPLE 13.77

  Evaluate ∫∫ f .nˆ dS , where f = 4 xiˆ − 2 y 2 ˆj + z 2 kˆ and S is S the surface bounding the region

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13.46  n  chapter thirteen x 2 + y 2 = 4, z = 0 , and z = 3 . Solution. The region is bounded by the cylinder x 2 + y 2 = 4, z = 0 , and z = 3 . Therefore, the

surface S consists of three parts: (i) S1, the circular base of the cylinder in the plane z = 0, (ii) S2, the circular top in the plane z =3, and (iii) S3, the curved surface of the cylinder given by x 2 + y 2 = 4 . Now, for the subsurface S1, we  z = 0, nˆ = −kˆ, and f = 4 xiˆ − 2 y 2 ˆj. Therefore,  f .nˆ = (4 xiˆ − 2 y 2 ˆj ). −kˆ = 0.  Hence, ∫∫ f . nˆ dS = 0.

dS = 2dθ dz . For this surface, z varies from 0 to 3 and θ varies from 0 to 2p. Therefore, 

2π 3

∫ ∫ f .nˆ dS = ∫ ∫ 2 ( 2 cos θ ) − ( 2 sin θ )  2dθ dz 2π

2π

= 48 ∫ ( cos 2θ − sin 3θ ) dθ 0

2π

Hence,

ˆ = ∫ ∫ 9 dxdy ∫ ∫ fˆ.nds

S2

0

Therefore, the unit normal vector to the surface S3 is given by 2 x iˆ + 2 y ˆj 2 x iˆ + 2 y ˆj = nˆ = 4x2 + 4 y 2 2 x2 + y 2

0

π

2

= (48)(4) ∫ cos 2θ dθ − 0 0

1 π = 192. . 2 2 = 48π .

Hence,     ∫ ∫ f .nˆ dS = ∫ ∫ f . n dS + ∫ ∫ f .nˆ dS s

s1

s2



∫ ∫ f .nˆ dS s3

= 0 + 36 p + 48 p = 84 p.

S2

= 9(4π ) = 36π For the surface S3, which is the curved surface of the cylinder and is given by x 2 + y 2 = 4 , the vector normal to the surface is ∇( x 2 + y 2 − 4) = 2 x iˆ + 2 y ˆj.

2π

= 48 ∫ cos 2θ dθ − 48 ∫ sin 3θ dθ

S1

)( )

3

0

( )

(

3

= ∫ 16 ( cos 2θ − sin 3θ ) [ z ]0 dθ

have

On S2, we have z = 3, nˆ = kˆ, and  2ˆ ˆ ˆ f = 4 x i − 2 y j + 9k . Therefore,  f . nˆ = 4 xiˆ − 2 y 2 ˆj + 9kˆ . kˆ = 9.

2

0 0

s3

EXAMPLE 13.78

  2 If f = 4 xz iˆ − y ˆj + yz kˆ, evaluate ∫ f .nˆ dS , where S is the surface of the cube bounded by x = 0, x = 1, y = 0, y = 1, z = 0, and z = 1. Solution. The surface of the cube is bounded by x = 0, x = 1, y = 0, y = 1, z = 0, and z = 1 and so, the surface can be subdivided into six parts in the following manner: z F

C

2 x iˆ + 2 y ˆj 2.2 x iˆ + y ˆj = . 2 =

Therefore,   xiˆ + yjˆ  f . nˆ = (4 x iˆ − 2 y 2 ˆj + z 2 kˆ).  = 2 x2 − y3 .  2  Now, on S3, x = 2 cos θ , y = 2 sin θ , and

M13_Baburam_ISBN _C13 Part III.indd 46

G E

a A x

B y

O

D

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vector calculuS   n  13.47 (i) S1 is the surface formed by the face OADB, where  z = 0, nˆ = −kˆ, and f = − y 2 ˆj , so that  f .nˆ = − y 2 ˆj . −kˆ = 0

)( )

(

and

∫∫

S1

1 1  f .nˆ dS = ∫ ∫ 0dxdy = 0.

 f .nˆ = 4 xiˆ − y 2 ˆj + ykˆ . kˆ = y.

(

)( )

Hence,

∫∫

S2

1 1 1 1   y2  f . nˆ dS = ∫ ∫ ydxdy = ∫   dx 2 0 0 0 0  1

1 1 = ∫ dx = . 20 2

(iii) S3 is the surface formed by the face ADEG, where nˆ = iˆ, x = 1, and dx = 0. On this face,  f = 4 xz iˆ − y 2 ˆj + yz kˆ = 4 z iˆ − y 2 ˆj + yz kˆ and  f .nˆ = 4 z iˆ − y 2 ˆj + yz kˆ .iˆ = 4 z.

(

)

Hence, 1 1  ˆ f . n ds = ∫∫ ∫ ∫ 4 z dy dx 0 0

S3

1

1

1

= 4∫  z2  dy = 2∫ dy = 2. 0 2

0 0 (iv) S4 is the surface formed by OBFC, where nˆ = −iˆ x = 0, and dx = 0. On this face,  f = 4 xz iˆ − y 2 ˆj + yz kˆ = − y 2 ˆj + yz kˆ and  f .nˆ = − y 2 ˆj + yz kˆ ⋅ −iˆ = 0.

(

)( )

Hence,

∫∫

S4

where nˆ = − ˆj , y = 0, and dy = 0. On this face,  f = 4 xz iˆ − y 2 ˆj + yz kˆ = 4 xz iˆ and so,  f .nˆ = 4 xz iˆ . − ˆj = 0.

(

)( )

Hence,



∫ ∫ f .nˆ dS = 0.

0 0

(ii) S2 is the surface formed by the face GEFC, where z = 1 nˆ = kˆ, and dz = 0. On this face, we have  f = 4 xz iˆ − y 2 ˆj + yz kˆ = 4 x iˆ − y 2 ˆj + ykˆ and so,

(v) S5 is the surface formed by the face OCGA,

1 1  f .nˆ dS = ∫ ∫ 0 dx dy = 0.

M13_Baburam_ISBN _C13 Part III.indd 47

S5

(vi) S6 is the surface formed by the face DBFE, where nˆ = ˆj , y = 1, and dy = 0. On this face,  f = 4 xz iˆ − y 2 ˆj + yz kˆ = 4 xz iˆ − ˆj + z kˆ and  f .nˆ = 4xz iˆ − ˆj + z kˆ ⋅ ˆj = −1.

(

)( )

Therefore, 1 1  ˆ f . n ±±±± = ∫∫ ∫ ∫ (−1) S6

1

= −∫ [

0 0

]0 1

0

1

= −1∫ dx = −1. 0 Hence,  ∫ ∫ f .nˆ dS = ∫ ∫ + ∫ ∫ + ∫ ∫ + ∫ ∫ S



S1

= 0+

S2

S3

+∫ ∫ +∫ ∫

S4

S5

S6

1 3 + 2 + 0 + 0 + −1 = . 2 2

13.17  VOLUME INTEGRAL Let f be a scalar-point function defined throughout a given region of volume V. Partition the given region into n subregions of volumes dV1, dV2, …, dVn. Let P(xi yi zi) be any point inside or on the boundary of the subregion of volume dVi. Then the limit n

lim

n →∞

∑ φ ( P )δV , i

i

δ Vi → 0 i =1

if exists, for all mode of subdivision (partition), is called the volume integral of f over the volume V, and this integral is denoted by

∫ ∫ ∫ φ dV = ∫ ∫ ∫ φ dx dy dz. V

V

0 0

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13.48  n  chapter thirteen  Similarly, if f = f1 iˆ + f 2 ˆj + f 3 kˆ is a vectorpoint function, then  ∫ ∫ ∫ f dV = iˆ ∫ ∫ ∫ f1 ( x, y, z ) dx dy dz V

V

+ ˆj



∫∫∫ f

2

( x, y, z )dx dy dz

2 2− x

=∫

∫ ( 4 − 2 x − 2 y ) ˆj − (8 y − 4 xy + 4 y ) kˆ  dx dy 2

x 0

2  y2 = ∫  4 y − 2 xy − 2 2 0 

2− x

 y2 y2 y3   −  8 − 4 x + 4  kˆ  2 3  0  2

V

+ kˆ



∫ ∫ ∫ f ( x, y, z )dx dy dz. 3

V

 f = (2 x 2 − 3 z ) iˆ − 2 xy ˆj − 4 x kˆ, evaluate ∫ ∫ V∫ ∇ × f dV , where V is the region bounded by the coordinate planes and the plane 2x + 2y + z = 4. Solution. We have If

iˆ

ˆj

kˆ

∂ ∂x

∂ ∂y

∂ ∂z

2 x 2 − 3 z −2 xy

−4 x

∂  ∂ = iˆ  ( −4 x ) − ( −2 xy )  ∂ ∂ y z   ∂ ∂  + ˆj  ( 2 x 2 − 3 z ) − ( −4 x )  ∂x  ∂z 

∂  ∂ + kˆ  ( −2 xy ) − ( 2 x 2 − 3z )  ∂y  ∂x  = ˆj − 2 y kˆ.

The region V is bounded by the planes x = 0, y = 0, z = 0 and the plane 2x + 2y + z = 4. Therefore, the limits of integration are: z varies from 0 to 4 – 2x – 2y, y varies from 0 to 2 – x, and x varies from 0 to 2. Hence,  ∫ ∫ ∫ ∇ × f dV V

2 2− x 4−2 x −2 y

=∫

∫ ∫

0 0

2 2− x

=∫

∫(

0 0

0

2

EXAMPLE 13.79

 ∇× f =

ˆ j 

( ˆj − 2 y kˆ ) dx dy dz )

2 2 3   = ∫ ( 2 − x ) ˆj − ( 2 − x ) kˆ  dx 3  0  2

4  ( 2 − x )3 2 − x ) ˆ ˆj − 2 ( = k 3 −4  0  −3

=

(

)

8ˆ 8ˆ 8 ˆ ˆ j− k = j−k . 3 3 3

EXAMPLE 13.80

 2 If f = ( 2 x − 3z ) iˆ − 2 xy ˆj − 4 x kˆ, then evaluate  ∫ ∫ ∫ ∇ ⋅ f dV , where V is bounded by the V

coordinate planes and the plane 2x + 2y + z = 4. Solution. We have

  ∂ ∂ ˆ ∂ ˆ ∇ ⋅ f =  iˆ + j+ k ∂ ∂ ∂z  x y  . ( 2 x 2 − 3 z ) iˆ − 2 xy ˆj − 4 x kˆ  = 4x – 2x = 2x. The limits of integration are as mentioned in Example 13.79. Therefore, 2 2− x  ∫∫∫ ∇. f dv = ∫ ∫ v

0

4−2 x −2 y

∫

0

2x dx dy dz

0

2 2− x



= 2∫ 0

∫ x ( 4 − 2 x − 2 y ) dx dy 0

2 2− x



= 4∫

∫ x ( 2 − x − 2) dx dy

0 0 2

=

4 2 x ( 2 − x ) dx ∫ 20

ˆj − 2 ykˆ [ z ]4 − 2 x − 2 y dx dy 0

M13_Baburam_ISBN _C13 Part III.indd 48

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vector calculuS   n  13.49 2

 x 2 4 x3 4 x 2  8 = 2 − +  = . 4 3 2  0 3

EXAMPLE 13.81

  Evaluate ∫ ∫ ∫ ∇ ⋅ f dV if f = 4xy iˆ + yz ˆj − xy kˆ V

and V is bounded by x = 0, x = 2, y = 0, y = 2, z = 0 and z = 2. Solution. We have  ∫ ∫ ∫∇⋅ f V

 ∂  ∂ ∂ = ∫ ∫ ∫  (4 xy ) + ( yz ) + (− xy )  dV ∂ ∂ ∂ x y z  V 2 2 2

= ∫ ∫ ∫ (4 y + z )dV = ∫ ∫ ∫ (4 y + z )dz dy dx 0 0 0

V

2

2 2  z  = ∫ ∫  4 yz +  dy dx = 2 ∫ ∫ ( 4 y + 1) dy dx 2 0 0 0  0 0 2 2

2

2

2 2  4 y2  2 = 2∫  + y  dx = 4 ∫ 5dx = 20 [ x ]0 = 40. 2 0 0  0

13.18  GAUSS’S DIVERGENCE THEOREM The following theorem of Gauss is useful in evaluating the surface integral over a closed surface by reducing it to a volume integral (triple integral) and vice versa. Theorem 13.10. (Gauss’s Divergence Theorem).  Let f be a vector-point function possessing continuous first-order partial derivatives at each point of a three-dimensional region V enclosed in a closed surface S. Then,    ∫ ∫ f .nˆ dS = ∫ ∫ ∫ div f dV = ∫ ∫ ∫ ∇ ⋅ f dV , S

V

V

where nˆ is the outward-drawn unit normal vector to the surface S. The divergence theorem can be expressed in the form of Cartesian coordinates as follows:  Let f = f1 iˆ + f 2 ˆj + f 3 kˆ. Then   ∂f ∂f ∂f div f = ∇ ⋅ f = 1 + 2 + 3 . ∂x ∂y ∂z

M13_Baburam_ISBN _C13 Part III.indd 49

Let the outward-drawn unit normal vector nˆ makes angles a, b, and g, respectively, with positive directions of x-, y-, and z-axis. Thus, cos a, cos b, and cos g are the direction cosines of nˆ and so, nˆ = cos α iˆ + cos β ˆj + cos γ kˆ and then,

 f .nˆ = f1iˆ + f 2 ˆj + f 3 kˆ . cos α iˆ + cos β ˆj + cos γ kˆ

(

)(

)

= f1 cos α + f 2 cos β + f 3 cos γ .

Hence, the Gauss’s Divergence Theorem takes the form  ∂f1 ∂f 2 ∂f 3  ∫ ∫ V∫  ∂x + ∂y + ∂z  dx dy dz = ∫ ∫ ( f1 dy dz + f 2 dz dx + f 3 dx dy ) , S since cos a dS = dy dz, cos b dS = dz dx, and cos g dS = dxdy. This form of Gauss’s Divergence Theorem is also known as Green’s Theorem in Space. Proof: Consider a closed surface S, which is such that it is possible to introduce a rectangular coordinate system, such that any line parallel to any coordinate axis cuts S in, at the most, two points. Let R be the projection of the surface S on the xy-plane. Then, in accordance to our assumption, a line through a point (x, y, 0) of R meets the boundary of S in two points. Suppose that the z coordinates of these points are z = f1 (x, y) and z = f2(x, y), where φ2 ( x, y ) ≥ φ1 ( x, y ) Then, ∂f 3 ∂f 3 ∫ ∫ V∫ ∂z dV = ∫ ∫ V∫ ∂z dz dy dx



φ2 ( x , y ) ∂f 3  = ∫∫ ∫ dz  dy dx ∂ z  R φ ( x , y ) 1 = ∫ ∫ [ f 3 ( x, y, z ) ]φ2 ( xy ) dy dx φ ( xy ) 1

R = ∫ ∫  f 3 ( x, y, φ2 ) − f 3 ( x, y, φ )1  ]dy dx.

(1)

R

Let S1 and S2 be the portion of the surface S corresponding to z = φ1 ( x, y ) and z = φ2 ( x, y ), respectively. Let nˆ2 be the outward-drawn unit normal vector to S2, making an acute angle g2 with the positive direction kˆ of z-axis. If dS2

()

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13.50  n  chapter thirteen is projected on the xy-plane, then this projection dy dx of dS2 is dy dx = cos γ 2 dS 2 = kˆ.kˆ2 dS 2 . In the same fashion, let nˆ1 be the outward-drawn unit normal vector to S1, making an obtuse angle g1 with kˆ. Then, dy dx = cos (π − γ 1 ) dS1 = − cos γ 1dS1 = −kˆ.nˆ dS1 . Therefore,

∫∫ R

and

S2

∫ ∫ f ( x, y,φ ) dy dx = − ∫ ∫ f kˆ.nˆ dS . 1

3

R

1

(2)

S

Adding (2), (3), and (4), we obtain  ∂f1 ∂f 2 ∂f 3  ∫ ∫ V∫  ∂x + ∂y + ∂z  dV

(

)

= ∫ ∫ f1iˆ + f 2 ˆj + f 3 .kˆ .nˆ dS S

 = ∫ ∫ f .nˆ dS .

S This proves the theorem. The Gauss’s Divergence Theorem can be extended to the surfaces which are such that lines parallel to the coordinate axes meet them in more than two points. For this, the region enclosed by S is partitioned into subregions whose surfaces satisfy the condition assumed in

M13_Baburam_ISBN _C13 Part III.indd 50

  = ∫ ∫ ∫ φ div f + f .∇φ dV V

(

)

 = ∫ ∫ ∫ φ div f dV



S1

( )

V



Similarly, projecting S on the remaining two coordinate planes, we have ∂f ∫ ∫ ∫ 1 dV = ∫ ∫ f1iˆ. nˆ dS and (3) ∂x V S ∂f 2 ∫ ∫ V∫ ∂y dV = ∫ ∫S f 2 ˆj.nˆ dS . (4)



V

S

+ ∫ ∫ f 3 kˆ . nˆ1 dS 1



 − ∫ ∫ ∫ φ div f dV .

S

Hence, (1) reduces to ∂f 3 ∫ ∫ V∫ ∂z dV = ∫ S∫ f3 kˆ.nˆ2 dS2 2

= ∫ ∫ f 3 kˆ. nˆ dS .

S

  Putting F = φ f , we have   ∫ ∫ φ f .nˆ dS = ∫ ∫ ∫ div φ f dV

1

S1



V

V Proof: By Gauss’s Divergence Theorem,   ∫ ∫ F .nˆ dS = ∫ ∫ ∫ div F dV .

f 3 ( x, y,φ2 ) dy dx = ∫ ∫ f 3 kˆ.nˆ2 dS 2

3

the above proof. Applying the theorem to each subregion and adding will yield the required result. Deductions: (i) If nˆ is the outward-drawn unit normal vector to S, then   ∫ ∫ ∫ f .∇φ dV = ∫ ∫ φ f .nˆ dS

V

 + ∫ ∫ ∫ f .∇φ dV

V and so,    ∫ ∫ ∫ f .∇φ dV = ∫ ∫ φ f .nˆ dS − ∫ ∫ ∫ φ div f dV .

V

(ii)

S

∫∫ S

V

  f × nˆ dS = − ∫ ∫ ∫ curl f dV . V

Proof: By Gauss’s Divergence Theorem,





∫ ∫ F .nˆ dS = ∫ ∫ ∫ div F dV . S

V

    Putting F = a × f , whose a is an arbitrary constant vector, we have     ∫ ∫ a × f .nˆ dS = ∫ ∫ ∫ ∇ ⋅ a × f dV S

or or

(

(

V

 

)





∫ ∫ a ( f × nˆ ) dS = −∫ ∫ ∫ a. ( ∇ × f ) dV S

S

    a.∫ ∫ f × nˆ dS = −a.∫ ∫ ∫ ∇ × f dV S

or

)

(

)

V

(

)

    a.  ∫ ∫ f × nˆ dS + ∫ ∫ ∫ ∇ × f dV  = 0 V  S 

(

)

(

)

1/2/2012 2:36:09 PM

vector calculuS  n  13.51 or

∫ ∫( S

that is,

  f × nˆ ds + ∫ ∫ ∫ ∇ × f dV = 0,

)

(

V

)



∫ ∫ ( f × nˆ ) dS = −∫ ∫ ∫ curl

 f dV .

V

∫ ∫ φ nˆ dS = ∫ ∫ ∫ grad φ dV . S

V

Proof: By Gauss’s Divergence Theorem



∫ ∫ f .nˆ dS = ∫ ∫ ∫ div S

 f dV .



S

V

 = ∫ ∫ ∫ ∇ ⋅ ( a φ ) dV . V

  a.∫ ∫ (φ nˆ ) dS = a.∫ ∫ ∫ ( ∇φ ) dV S



= 3∫ ∫ ∫ dV = 3V . V (ii) By the divergence theorem,

∫ ∫ ∫ div nˆ dV = ∫ ∫ nˆ. nˆ dS V

S

= ∫ ∫ dS = S . S (iii) If aˆ is any constant vector, then   a.∫ ∫ nˆ dS = ∫ ∫ a.nˆ dS S



V

by divergence theorem    = 0, because div a = 0.

S

or

V

∫ ∫ φ nˆ dS = ∫ ∫ ∫ ∇φ dV . S

V

EXAMPLE 13.82

If S is a closed surface, nˆ is the outward-drawn normal to S and V is the volume enclosed by S, show that  (i) ∫ ∫ r .nˆ dS = 3V , S

= S, ∫ ∫ ∫ div nˆ dV  (iii) ∫ ∫ nˆ dS = 0, and (ii)

V

S

(iv)

(iv) By the Gauss’s Divergence Theorem   ∫ ∫ f .nˆ dS = ∫ ∫ ∫ ∇ ⋅ f dV S

∫ ∫ (φ nˆ ) dS − ∫ ∫ ∫ (∇φ )dV = 0



∫ ∫ f .nˆ dS = 6V , S

 where f = x iˆ + 2 y ˆj + 3 z kˆ.

M13_Baburam_ISBN _C13 Part III.indd 51

S

 = ∫ ∫ ∫ div a dV ,

or

or

V

 ∂ ∂ ∂  = ∫ ∫ ∫  iˆ + ˆj + kˆ  ∂x ∂y ∂z  V  . ( x iˆ + y ˆj + z kˆ)dV

V

  a.  ∫ ∫ (φ nˆ ) dS − ∫ ∫ ∫ ( ∇φ ) dV  = 0 V  S 

 dV = ∫ ∫ ∫ ∇ ⋅ r dV

V

V

   Putting f = a φ , , where a is on arbitrary constant vector, we get   ∫ ∫ ( a φ ) .nˆ dS = ∫ ∫ ∫ div ( a φ ) dV Thus,



∫ ∫ r .nˆ ds = ∫ ∫ ∫ div r S

S

(iii)

Solution. (i) By the divergence theorem,

V

 ∂ ∂ ∂  = ∫ ∫ ∫  iˆ + ˆj + kˆ  ∂ ∂ ∂ x y z V  . x iˆ + 2 y ˆj + 3 z kˆ dV

(

)

= ∫ ∫ ∫ (1 + 2 + 3)dV V

= 6 ∫ ∫ ∫ dV = 6V . V

EXAMPLE 13.83

Verify the divergence theorem for  2 f = 4 x iˆ − 2 y 2 ˆj + z kˆ, taken over the region bounded by the cylinder x2 + y2 = 4, z = 0 and z = 3. Solution. In Example 13.77, we have shown that  ∫ ∫ f .nˆ dS = 84π . S

On the other hand,

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13.52  n  chapter thirteen   f dV = ∫ ∫ ∫ ∇ ⋅ f dV

∫ ∫ ∫ div V

verified for the given function

V

EXAMPLE 13.84

∂ ∂ = ∫ ∫ ∫  ( 4 x ) + ( −2 y 2 ) ∂x ∂y V  +

Verify Gauss’s Divergence Theorem for the  2 function f = 4xz iˆ − y ˆj + yz kˆ over the surface S of the cube bounded by x = 0, x = 1, y = 0, y = 1, z = 0, and z = 1.

∂ 2  ( z ) dV ∂z

Solution. In Example 13.78, we have shown that

= ∫ ∫ ∫ (4 − 4 y + 2 z )dx dy dz.



V

Since z varies from 0 to 3, y varies from − 4 − x to 4 − x 2 , and x varies from –2 to 2, we have  ∫ ∫ ∫ div f dV V

=

∫

V

 = ∫ ∫ ∫ ∇ ⋅ f dV V

∂  ∂ ∂ = ∫ ∫ ∫  (4 xz ) + ( − y 2 ) + ( yz )  dV ∂x ∂y ∂z  V 

3

4 − x2

 z2   ∫ 2  4 z − 4 yz + 2 2   dy dx 0 4− x 

2

∫

−2 −

= ∫ ∫ ∫ (4 z − 2 y + y )dV = ∫ ∫ ∫ (4 z − y )dV

4 − x2

2

=

On the other hand,  ∫ ∫ ∫ div f dv

3   ∫ 2  ∫0 ( 4 − 4 y + 2 z ) dz  dy dx 4− x

−2 −

=

S

4 − x2

2

V

∫ ∫ [12 − 12 y + 9] dy dx

 4 − x2  = 21∫  ∫ dy  dx, −2   − 4 − x2  2

4− x

∫ 0

 dy  dx = 42 

1

2

∫

4 − x 2 dx

−2

Therefore,

S

V

EXAMPLE 13.85



  f .nˆ dS = .∫ ∫ ∫ div f dV V

and thus, Gauss’s Divergence Theorem is

M13_Baburam_ISBN _C13 Part III.indd 52

S

 f dV

and thus, the Gauss’s Divergence Theorem is verified

2

 x 4 − x 2 4 −1 x  = 84  + sin  2 2 2   0

 2π  = 84 0 + 2sin −11 = 84   = 84π.  2 



∫ ∫ f .nˆ dS = ∫ ∫ ∫ div

= 84 ∫ 4 − x dx, since integrand is even 0

1

1 3 3  = ∫  2 −  dx = ∫ dx = . 2 2 2   0 0

2

∫∫

1

1 1   y2  = ∫  ∫ (2 − y )dy  dx = ∫  2 y −  dx 2 0 0 0 0  

2

Therefore,

1

11

1

1

since 12 y is an odd function of y

2  = 42 ∫  −2  

V

   4z2  − yz  dy dx = ∫∫  ∫ (4 z − y )dz  dy dx = ∫∫  2 0 0 0 0 00   11

−2 − 4 − x 2

2

3

∫ ∫ f .nˆ dS = 2.

2

  Evaluate ∫ ∫ f .nˆ dS , where f = ax iˆ + by ˆj + cz kˆ S and S is the surface of the sphere x2 + y2 + z2 = 1.

Solution. Since S is closed, by the divergence

theorem, we have   ∫ ∫ f .nˆ dS = ∫ ∫ ∫ div f dV S

V

1/2/2012 2:36:11 PM

vector calculuS   n  13.53  = ∫ ∫ ∫ ∇ ⋅ f dV

=∫

∂ ∂ = ∫ ∫ ∫  (ax) + (by ) ∂x ∂y V 

= ∫∫ (−2)dxdy = −2 ∫ [ y ]0 dx

V

+

3

OADB

( )

− yz ) iˆ − 2 x 2 y ˆj + 2 kˆ  . −kˆ dS

aa

a

00

a

0

a

∂  (cz )  dV ∂z 

= −2a ∫ dx = −2a [ x ]0

a

0

= −2a . 2

= ∫ ∫ ∫ (a + b + c)dV

(1) (ii) For the face CGEF, we have nˆ = kˆ and z = a. Therefore, aa  ˆ f n dS = . ∫ ∫ ∫∫ 2dx dy

V

= (a + b + c) ∫ ∫ ∫ dV V

4  = (a + b + c)  π  , 3 

00

CGEF

a

since ∫ ∫ ∫ dV = volume of the sphere x2 + y2 + z2 = 3 4 1, which is 3 π .1 .

= 2a ∫ dx 0

V

EXAMPLE 13.86

Verify Gauss’s Divergence Theorem for  f = ( x 3 − yz ) iˆ −2 x 2 y ˆj + 2kˆ, taken over the cube bounded by the planes x = 0, x = a, y = 0, y = a, z = 0, and z = a. Solution. The surface F is a cube with six faces as shown in the following figure

= 2a 2 .

(2) ˆ (iii) For the face ADEG, we have nˆ = i , x = a, and dx = 0. Therefore, aa  3 ˆ f . n dS = ∫ ∫ ∫ ∫ ( a − yz ) dydz 00

ADEG

a

 z2  = ∫  a 3 z − y  dy 2 0 0  a

z

1  a2 y  = ∫  a4 −  dy 2  0

F

C

∫ ( x

a

G

 a2 y2  = a4 y −  4 0 

E

B y

O A

D

x  f .nˆ dS ,

 f .nˆ dS

To calculate ∫ ∫S we evaluate ∫ ∫S over the six faces and then add those values. (i) For the face OADB, we have nˆ = −kˆ and z = 0. Therefore,  ∫ ∫ f .nˆ dS OADB

M13_Baburam_ISBN _C13 Part III.indd 53

a4 (3) . 4 (iv) For the face OBFC, we have nˆ = −iˆ, x = 0, and dx = 0. Therefore, = a5 −

∫ ∫

OBFC

aa  f .nˆ dS = ∫∫ ( yz )dydz 00

a

a  z2  = ∫y  2 0 0 a

=

1 a 2 ydy 2 ∫0

1/2/2012 2:36:12 PM

13.54  n  chapter thirteen and thereby, Gauss’s Divergence Theorem is verified

a

=

a2  y2    2  2 0

a2 = . (4) 4 (v) For the face OAGC, we have nˆ = − ˆj, y = 0, and dy = 0. Therefore, aa  ˆ f . n dS = (5) ∫ ∫ ∫∫ 0 dxdz = 0. 00

OAGC

(vi) For the face DBFE, we have dy = 0. Therefore,  ∫ ∫ f .nˆ dS

nˆ = ˆj ,

y = a, and

EXAMPLE 13.87

Evaluate ∫ ∫ ( x3 − yz ) dy dz − 2 x 2 ydzdx + zdxdy over the surface Sbounded by the coordinate planes and the planes x = y = z = a. Solution. By giving divergence theorem in Cartesian form, we have

∫ ∫ ( f dydz + f dzdx + f dxdy ) 1

q

∂  ∂ ∂ = ∫ ∫ ∫ [  ( x 3 − yz ) + ( −2 x 2 y ) + ( z ) dV ∂ ∂ ∂ x y z   V

a

= ∫∫ − 2 x 2 a dxdz = −2a ∫  x 2 z  dx 00

0

0

a

aaa

a  x3  = −2a 2 ∫ x 2 dx = −2a 2    3 0 0

= ∫ ∫ ∫ ( x 2 + 1) dV = ∫∫∫ ( x 2 + 1) dx dy dz

2a 3 Adding (1) – (6), we get  2 2 5 ∫ ∫ f .nˆ dS = −2a + 2a + a

a a   a = ∫∫  x 2 z + z  dydz = ∫  ∫ a ( x 2 + 1) dy  dx 0 00 0 0  aa

(6)

a

EXAMPLE 13.88

∂  ∂ ∂ = ∫ ∫ ∫  ( x 3 − yz ) + ( −2 x 2 y ) + (2)  dV ∂x ∂y ∂z  V  aaa

= ∫ ∫ ∫ ( 3 x 2 − 2 x 2 ) dV = ∫∫∫ x 2 dx dy dz 000

a

a

= a ∫  x y  dx = a 0 2

0

Thus,

a

2

∫ 0

a



∫ ∫ f .nˆ dS = ∫ ∫ ∫ div S

M13_Baburam_ISBN _C13 Part III.indd 54

V

2

 f dV

∫ ∫ ( 4 xzdy dz − y dz dx + yzdx dy ), where S is the 2

S

surface of a cube bounded by the planes x = 0, y = 0, z = 0, x = 1, y = 1, and z = 1. f1 = 4 xz , f 2 = − y 2 , and f 3 = yz.

 x3  a5 x dx = a   = .  3 0 3 2

Using Green’s Theorem of in space, evaluate

Solution. Let

a  = ∫∫  x z  dy dx = ∫  ∫ x 2 a dy  dx 0 00 a 0  a

a

0

 x3  a5 = a2  + x = + a3 . 3 0 3

V

2

0

a

On the other hand,   ∫ ∫ ∫ div f dV = ∫ ∫ ∫ ∇ ⋅ f dV

aa

a

0

a4 a2 2a 5 a 5 − + +0− = . 4 4 3 3

V

a

= a ∫  x 2 y + y  dx = a 2 ∫ ( x 2 + 1) dx

S

V

000

V

5

=−

3

∂f   ∂f ∂f = ∫ ∫ ∫  1 + 2 + 3  dV ∂ ∂ ∂z  x y V 

DBFE

aa

2

S

Then, by Green’s Theorem in space, we have

∫ ∫ ( f dydz + f dzdx + f dxdy ) 1

2

3

S

∂f   ∂f ∂f = ∫ ∫ ∫  1 + 2 + 3  dx dy dz ∂ x ∂ y ∂z  V 

1/2/2012 2:36:13 PM

vector calculuS   n  13.55  ∂  ∂ ∂ 2 =∫ ∫ ∫  (4 xz ) + ( − y ) + ( yz )  dx dy dz ∂ ∂ ∂ x y z  V  = ∫ ∫ ∫ (4 z − y ) dx dy dz

that f (x) > ψ (x) for all x ∈ [a, b] . Let f be a realvalued continuous function defined in R, and let ∂f ∂y exists and is continuous in R. Then, y

V

y   (x)

1

11 1 1    z2  = ∫∫  4 − yz  dy dx = ∫  ∫ (2 − y )dy  dx 2 0 00  0 0 

y   (x) xa

1

1  y2  3 3 = ∫  2 y −  = ∫ dx = . 2 0 2 0 2 0  1

13.19 GREEN’S THEOREM IN A PLANE A domain D is said to be a quadratic with respect to y-axis, if it is bounded by the curves of the form

y = f (x), y = ψ (x) : x = a, x = b,

where f and ψ are continuous functions and f (x) > ψ (x) for all x ∈ [a, b] . Thus, a domain which is quadratic with respect to y-axis is such that a line parallel to y-axis and lying between x = a and x = b meets the boundary of D in just two points. Similarly, we can define domains which are quadratic with respect to x-axis. The Green’s Theorem is useful in changing a line integral around a closed curve C into a double integral over the region R enclosed by C. ∂f

∂f

 ∂g ∂f  C∫ [ f ( x, y)dx + g ( x, y)dy ] = ∫ ∫R  ∂x − ∂y  dx dy,

where the integral on the left is a line integral around the boundary C of the region, taken in such a way that the interior of the region remains on the left as the boundary is described. Proof: Consider the region R bounded by the

curves x = a, x = b, y = f(x), and y = ψ(x), such

M13_Baburam_ISBN _C13 Part III.indd 55

x

0

∂f ∫ ∫R ∂y dxdy = ∫a b

 φ ( x ) ∂f  dy  dx  ∫ ψ ( x ) ∂y 

b

b

= − ∫ f ( x, .φ ( x))dx − ∫ f ( x,ψ ( x) ) dx a

a

a

b

= − ∫ f ( x, φ( x) ) dx − ∫ f ( x, ψ ( x) ) dx b

a

a   = −  ∫ f ( x,ψ ( x) ) dx + ∫ f ( x,φ ( x))dx  b a  b

= − ∫ f ( x, y)dx. C

Therefore,

∫ f ( x, y)dx = − ∫ ∫

Theorem 13.11. (Green’s Theorem). Let f , g , ∂y , and ∂y are continuous in a region R, which can be split up in finite number of regions quadratic with respect to either axis. Then,

xb

C

R

∂f dx dy. ∂y

(1)

Similarly, it can be shown that ∂g

∫ g ( x, y)dy = ∫ ∫ ∂x dx dy. C

(2)

R

Adding (1) and (2), we obtain

 ∂g

∂f 

∫ [ f ( x, y)dx + g ( x, y)dy ] =∫ ∫  ∂x − ∂y  dx dy. C

R

Deductions:

(i) If f (x, y) = –y and g (x, y) = x, then by Green’s Theorem, we have ∫ ( xdy − ydx ) = ∫ ∫ (1 + 1) dxdy C

R

1/2/2012 2:36:14 PM

13.56  n  chapter thirteen For the line integral along C1, we have y = x2 and so, dy = 2xdx and x varies from 0 to 1. Thus,

= 2 ∫ ∫ dxdy = 2 A, S

where A denotes the area of the region R. Thus, 1 A=  [ xdy − ydx ]. 2 C∫

∫ ( xy + y ) dx + x dy  2

1

= ∫ ( x 3 + x 4 ) dx + x 2 (2 x)dx 

(ii) Putting f (x, y) = –y and g (x, y) = 0, the Green’s Theorem implies − ∫ ydx =

0

1

∫ ∫ dxdy = Area of the region R.

1  x5 3x 4  19 = ∫ ( x 4 + 3 x 3 ) dx =  +  = . 5 4   0 20 0

R

(iii) Putting g (x, y) = x and f (x, y) = 0, we get

∫ xdy = ∫ ∫ dx dy = Area of the region R. C

R

Hence, the area of a closed region R is given by any of the three formulae 1 C∫ xdy, − C∫ ydx, or 2 C∫ ( xdy − ydx ) ,

2

C1

For the line integral along (2), we have y = x and so, dy = dx, and x varies from 1 to 0. Therefore, 0

2 2 2 2 2 ∫ ( xy + y ) dx + x dy  ∫ ( x + x ) dx + x dx  1

C2

0

0  3x3  = ∫ 3 x 2 dx =   = −1.  3 1 1

where C denotes the boundary of the closed region R described in the positive sense.

Hence, (1) yields

EXAMPLE 13.89

Verify Green’s theorem in the plane for 2 2 C∫ ( xy + y ) dx + x dy , where C is the closed curve of the region bounded by y = x and y = x2. Solution. The region is bounded by the straight line y = x and the parabola y = x2. The point of intersection of y = x and y = x2 are (0, 0) and (1, 1). We note that y

∫ ( xy + y ) dx + x 2

C

19 1 −1 = − . 20 20

On the other hand,  ∂f 2

∂f1 

∫ ∫  ∂x − ∂y  dx dy S

∂  ∂ = ∫ ∫  ( x 2 ) − ( xy + y 2 )  dx dy ∂ ∂ x y  S  =∫

∫ [ 2 x − ( x + 2 y)] dy dx

0 y = x2

 x

y

dy  =

1 y=x

(1, 1)

1 x 1   x = ∫  ∫ ( x − 2 y ) dy  dx = ∫  xy − y 2  2 dx x 0  0  x2 

x2  y

C2

2

C1 x

0

1

1  x5 x 4  1 1 1 = ∫ ( x 4 − x 3 ) dx =  −  = − = − . 20  5 4 0 5 4 0

2 2 = ∫ ( xy + y ) dx + x dy  C

Hence,

= ∫ ( xy + y 2 ) dx + x 2 dy  C1

+ ∫ ( xy + y 2 ) dx + x 2 dy  C2

M13_Baburam_ISBN _C13 Part III.indd 56

∫ [ f (1)

C

1

 ∂f ∂f  ( x, y )dx + f 2 ( x, y )dy ] = ∫ ∫  2 − 1  dx dy, ∂x ∂y  S 

and thus, Green’s theorem is verified

1/2/2012 2:36:15 PM

13.57 vector calculuS   n  EXAMPLE 13.90

Apply Green’s Theorem to show that the area bounded by a simple closed curve C is given by

1 2

∫ ( xdy − ydx). Hence, find the area of the C

x2 a2

ellipse

+

y2 b

2

= 1.

Solution. In Deduction (i) of Green’s Theorem,

∫ [ f dx + f dy] = ∫ + ∫ + ∫ 2

.

1

C

OA

AB

BO

Along OA, we have y = 0 so that dy = 0 and x varies from 0 to 1. Hence, 1

1  x3  2 ∫OA = ∫O3x dx = 3  3  = 1. 0

we have shown that the area A bounded by a simple closed curve C is equal to 12 ( xdy − ydx). For the second parts, we know that the 2 2 parametric equations of the ellipse ax 2 + by2 = 1 are x = a cos θ and y = b sin θ . Thus, dx = −a sin θ dθ and dy = b cos θ dθ . Therefore, the area A of the ellipse is given by 1 A = ∫ ( xdy − ydx) 2C

y B (0, 1)

x  y

x0

 1

2π

1 [(a cos θ )(b cos θ )dθ 2 ∫0

=

O

−(b sin θ )(−a sin θ )dθ ]

1 [ab cos 2θ + absin 2θ ]dθ 2 ∫0

=

ab ab dθ = [θ ]θ2π = π ab . ∫ 2 0 2

∫

2π

AB

0

= ∫[3 x 2 − 8(1 − x)]2 dx 1

+[4(1 − x) − 6 x(1 − x)](−1)dx

EXAMPLE 13.91

0

Verify Green’s theorem in the plane for 2 2 ∫ [(3x −8 y )dx + (4 y − xy)dy], where C is

= ∫ (−11x 2 + 26 x − 12)dx 1

0

the boundary of the region bounded by x = 0 , y = 0 , and x + y = 1 .

  x3 x2 =  −11 + 26 − 12 x  3 2  1

Solution. We have

=

C

f1 ( x, y ) = 3 x − 8 y and f 2 ( x, y ) = 4 y − 6 xy:. 2

2

Therefore,

 ∂f 2

x

Along AB, we have y = 1 − x and so dy = −dx and x varies from 1 to 0 . Therefore,

2π

=

A(1, 0)

y0

∂f1 

∫∫  ∂x − ∂y  dx dy S

Along BO, we have x = 0 so that dx = 0 and y varies from 1 to 0 . Therefore, 0

 y2  ∫BO = ∫1 4 ydy = 4  2  = −2. 1 Hence, 0

1− x

1 1− x 1    y2  = ∫  ∫ (−6 y + 16 y )dy  dx = 10 ∫   dx 2 0 0  0 0   1 5 5 = 5∫ (1 − x) 2 dx = − [(1 − x)3 ]10 = . (1) 3 3 0     Further, the line integral splits into three parts:

M13_Baburam_ISBN _C13 Part IV.indd 57

11 8 − 13 + 12 = . 3 3

8 5 [ f 2 dx + f1dy ] = 1 + − 2 = . (2) 3 3

From (1) and (2), it follows that

1/2/2012 2:37:29 PM

13.58  n  chapter thirteen

π 2  2  2 = − [ x( x − cos x)]0 − ∫ ( x − cos x)dx  π 0   π

EXAMPLE 13.92

Using Green’s theorem in a plane, evaluate C

π

π

2

∫( x − cos x)dx 0

π

2 2  x2 = − +  − sin x  2 π2 0

Solution. We have

Y

2

+

2

π

triangle with vertices (0,0), ( π2 , 0) , and ( π2 ,1) . f1 ( x, y ) = y − sin x and f 2 ( x, y ) = cos x . The closed curve C is the triangle with vertices (0,0), ( π2 , 0) , and ( π2 ,1) as shown in the following figure. The equation of the line OB is y−0 = 0 , that is, y = 2x . x − 0 π2 − 0 π

π

=−

cos xdy ], where C is the

π

=−

2

=−

π 2

+ +

2 π 2  − 1 π  8 

π 4

−

2

π

=−

π 4

2 − .

π

EXAMPLE 13.93

Evaluate, by Green’s Theorem,

B ( π , 1)

∫ [(3x − y)dx + C

(2 x + y )dy ], where C is the curve x + y 2 = a 2

2

y

x

2

sin x)dx

0

and thus, Green’s theorem is verified

∫ [( y − sin x)dx +

2

x(1 + π∫

=−

 ∂f 2 ∂f1  ∫C [ f 2 dx + f1dy] = ∫∫S  ∂x − ∂y  dx dy,

π

2

π

Solution. We have

x  π2

A( π 2 , 0)

O (0, 0)

f 2 ( x, y ) = 2 x + y . and f1 ( x, y ) = 3 x − y By Green’s Theorem, we have X

By Green’s Theorem, we have

 ∂f 2

∂f1 

∫ [ f dx + f dy] = ∫∫  ∂x − ∂y  dx dy 1

2

C

∫( f dx + f dy) 1

2

C

 ∂f ∂f  = ∫∫  2 − 1  dx dy = ∫∫ (2 − 1)dx dy ∂x ∂y  R R

S

∂  ∂ = ∫∫  (cos x) − ( y − sin x)  dx dy ∂x ∂y  S  = ∫∫ −( sin x + 1)dx dy π  = − ∫  ∫ (1 + sin x)dy  dx  0  0 π

2

2x

π

2

M13_Baburam_ISBN _C13 Part IV.indd 58

∫ ∫

− a − a2 − x2

a

a2 − x2

0

0

dx dy = 4∫

∫

dx dy

a

= 4 ∫ a 2 − x 2 dx 0

π 2

= 4a 2 ∫cos 2 θd θ, x = sin θ, 0

2x

= − ∫[ y + y sin x]0 dx 0

a2 − x2

a

=

π

1 π = 4a 2 . . = π a 2 . 2 2

1/2/2012 2:37:30 PM

vector calculuS   n  13.59 EXAMPLE 13.94

 ∂f ∂f  = iˆ  3 − 2  +  ∂y ∂z 

Compute the area of the loop of Descartes’s Folium, x 3 + y 3 = 3axy .

 ∂f ∂f  + kˆ  2 − 1   ∂x ∂y 

Solution. Putting y = tx , we get the parametric

equations of the contour of the folium as x=

and so,

3at 2 3at y = . and 1+ t3 1+ t3

  ∂f ∂f  curl f .nˆ  3 − 2  cos α  ∂y ∂z 

The loop is described as t varies from 0 to ∞ , since

from 0 to

t= π

2

y x

= tan θ ,

. Thus,

4 dy = 3a 2t − t dt . (1 + ι3 )2

where θ 3 dx = 3a 1 − 2t dt 3 (1 + ι )2

 ∂f ∂f  +  1 − 3  cos β  ∂z ∂x 

varies

 ∂f ∂f  +  2 − 1  cos γ.  ∂x ∂y 

and

Hence, by Green’s Theorem, ∞ 9a 2 t 2 dt 3 Area = 1 (xd y– ydx) = = a2 . 3 2 ∫ 2 0 (1 + t ) 2 2

ˆj  ∂f1 − ∂f 3   ∂z ∂x   

On the other hand,   ˆ ) ˆ + ˆjdy + kdz f . d r = ( f1iˆ + f 2 ˆj + f 3 kˆ) . (idx

= f1dx + f 2 dy + f 3 dz. 13.20  SIOKE’S THEOREM Therefore, Stoke’s Theorem takes the form The Stoke’s Theorem provides a relation between a surface integral taken over a surface C∫ ( f1dx + f 2 dy + f3 dz ) to a line integral along the boundary curve of the surface.   ∂f ∂f   ∂f ∂f   = ∫∫  3 − 2  cos α +  1 − 3  cos β Theorem 13.12. (Stoke’s Theorem). Let f be  ∂z ∂x   ∂ ∂  y z S  vector-point function possessing continuous  ∂f and∂fS be a surface   ∂f ∂f   ∂f ∂f  first-order partial derivatives = ∫∫  3 − 2  cos α +  1 − 3  cos β +  2 − 1  cos γ  dS . bounded by a closed curve  ∂z ∂x   ∂C. ∂z  y Then,  ∂x ∂y   S     We now prove the theorem in this form. Suppose ∫C f .dr = ∫ ∫ curl f .nˆdS z = f(x, y) be the equation of the surface S and S R be the projection of S on the xy-plane. Then, where nˆ is a unit normal vector at any point of the projection of the curve C on the xy-plane S, drawn in the sense in which a right-handed shall be the curve C1, which enclose the region screw would move when rotated in the sense of R. Therefore, description of the curve C. C∫ f1 ( x, y, z )dx = C∫ f1 ( x, y,φ ( x, y))dx Proof: Let the unit normal vector nˆ makes 1 angles a , b , and g with the positive directions of coordinate axes x, y, and z, respectively. = ∫ f1 ( x, y,φ )dx − 0dy ] Then, nˆ = cos α iˆ + cos β ˆj + cos γ kˆ. Since C1   ˆ ˆ . r = xiˆ + yˆj + zkˆ, we have dr = idx + ˆjdy + kdz  ∂  ∂ Let f = f1iˆ + f 2 ˆj + f 3 kˆ. Then, =  (0) − f1 ( x, y,φ )  dx dy   Curl f = ∇ × f =

M13_Baburam_ISBN _C13 Part IV.indd 59

iˆ

ˆj

kˆ

∂ ∂x

∂ ∂y

∂ ∂z

f1

f2

f3

∫∫  ∂x R

∂y



(by Green’s theorem in plane)

∂ f1 ( x, y,φ )dx dy ∂y R

= − ∫∫

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13.60  n  chapter thirteen

 ∂f ∂f ∂φ  = − ∫∫  1 + 1  dx dy. ∂y ∂z ∂y  R 

(1)

Since the direction ratios of the normal nˆ to the ∂φ ∂φ surface S are ∂x , ∂y , and –l, we have cos α ∂φ θr

cos β

=

∂φ ∂y

=

cos γ ∂φ cos β and so =− . −1 ∂y cos γ

Moreover, dxdy being the projection of dS on the xy-plane, we have dxdy = cos γ dS .

EXAMPLE 13.95

Verify Stoke’s Theorem for the ffinctio  f = x 2 iˆ + xyˆj , integrated around the square in the plane z = 0 , whose sides are along the lines x = 0 , x = a , y = 0 , and y = a .  Solution. Since f = x 2 iˆ + xyˆj , we have   ˆ + ˆjdy ) = x 2 dx xydy. f .dr = ( x 2 iˆ + xyjˆ) . (idx Therefore,

 

∫ f .dr = ( x dx + xydy),

Hence, (1) reduces to

∫ f ( x, y, z )dx

C

1



where C is the square shown in the figure

C

Y

 ∂f ∂f cos β  ) cos γ dS = − ∫∫  1 + 1 (− cos γ  ∂y ∂z S 



 ∂f  ∂f = ∫∫  1 cos β − 1 cos γ  dS . ∂ ∂ z y  S 

 ∂f 3

∫ f ( x, y, z )dz = −  ∂y 3

cos α −

C

B(a, a)

(2)

∂f 3  cos β  dS . (4) ∂x 

Adding (2), (3), and (4), we get ∫ ( f1dx + f 2 dy + f3 dz ) C

 ∂f ∂f  = ∫∫  3 − 2  cos α ∂y ∂z  S    ∂f ∂f   ∂f ∂f  +  1 − 3  cos β +  2 − 1  cos γ  dS .  ∂z ∂x   ∂x ∂y   This completes the proof of the theorem. Remark 13.1. The equivalent statement of Stoke’s Theorem is that the line integral of the tangential component of a vector-point function  f taken around a simple closed curve C is equal to the surface integral of the normal component  of the curl of f taken over any surface S having C as its boundary.

M13_Baburam_ISBN _C13 Part IV.indd 60

y5a

C(0, a)

Similarly, it can be established that ∂f  ∂f  f 2 ( x, y, z )dx = ∫∫  2 cos γ − 2 cos α dS (3) ∂ ∂ x z   and

2

x50

x5a

O (0, 0)

Thus,   ∫ f .dr = C

y50

∫ + ∫ + ∫ + ∫.

OA

AB

BC

CA

X A(a, 0)



(1)

Along OA, we have y = 0 and so, dy = 0. Thus, a   a 2  x3  a3 f d r = x dx = = . .   ∫ ∫0 3  3 0 AB

Along AB, x = a and so, dx= 0. Thus,

  a 2 f ∫AB .dr = ∫0 aydy = a2 . Along BC, we have y = a and so, dy = 0. Thus,

  0 2 a3 f . d r = x d x = − . ∫ ∫a 3 BC

1/2/2012 2:37:32 PM

vector calculuS   n  13.61 Along CO, we have x = 0 and so, dx = 0. Thus, 



∫ f .d r = ∫ 0dy = 0. Hence, (1) yields

∫

CO

xy-plane. Thus, the parametric equations of C are x = cos t, y = sin t, z = 0, and 0 < t < 2p. Therefore,   ∫ f dr = ∫ ( f1dx + f 2 dy + f3 dz ) C

C

  a3 a 2 a3 a 2 f .dr = + − = . 3 2 3 2

= ∫ [ ydx + zdy + xdz ] C

2p

On the other hand,

 curl f =

iˆ

ˆj

kˆ

∂ ∂x

∂ ∂y

∂ ∂z

x2

xy

0

= ∫ [ sin t (− sin t )]dt 0

2π

= ykˆ

Since the square (surface) lies in the xy-plane, nˆ = kˆ . Therefore,  curl f .nˆ = ykˆ.kˆ = y

= − ∫ sin 2 tdt 0

π

2 1 π = −4∫sin 2 t dt = −4. . = −π. 2 2 0

On the other hand,

 curl f =

and so, a aa a   y2  ˆ ydxdy = ∫   dx ∫∫S curl f .ndS = ∫∫ 2 0 00 0  a

= Hence,

∫ C

a2 a3 dx = . ∫ 2 0 2

   ˆ . f .dr = ∫∫ curlf .nds S

EXAMPLE 13.96

 Verifies Stoke’s Theorem for f = yiˆ + zˆj + xkˆ , where S is the upper-half surface of the surface 2 2 x 2 + y + z = 1 and C is its boundary. Solution. Here, C is the boundary of the upper-

half surface of

x 2 + y 2 + z 2 = 1 , that is, C is

the boundary of the circle x 2 + y 2 = 1 in the

M13_Baburam_ISBN _C13 Part IV.indd 61

iˆ

ˆj

kˆ

∂ ∂x

∂ ∂y

∂ ∂z

y

z

x

 ∂x ∂z   ∂z ∂y   ∂y ∂x  = iˆ =  −  = + ˆj  −  + kˆ  −   ∂z ∂x   ∂y ∂z   ∂x ∂y  ˆ = −(iˆ + ˆj + k ) and so,

 curl f .nˆ = −(iˆ + ˆj + kˆ).kˆ = −1. Therefore,  ˆ = − ∫∫ dxdy = −π (1) 2 = −π . ∫∫ curl f .ndS S

Hence,

 



∫ f .dr = ∫∫ curl f .nˆ ds C

S

and thus, Stoke’s theorem is verified EXAMPLE 13.97

Verify Stoke’s theorem for the vector field  f = ( x 2 − y 2 )iˆ + 2 xyˆj , integrated around the rectangle z = 0, and bounded by the lines x = 0, y = 0, x = a, and y = b.

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13.62  n  chapter thirteen



Solution. Since f = ( x 2 − y 2 )iˆ + 2 xyˆj , we have

  ˆ + ˆjdy ) f .dr = [( x 2 − y 2 )iˆ + 2 xyjˆ] ⋅ (idx

∫ C

= ( x − y )dx + 2 xydy. 2

Hence,

2

  a3 a3 f .dr = + ab 2 − + ab 2 = 2ab 2 . 3 3

On the other hand,

y

iˆ

ˆj

kˆ

∂ ∂x

∂ ∂y

∂ ∂z

x2 − y 2

2 xy

0

  curl f = ∇ × f = y5b

C(0, b)

B(a, b)

Therefore,

x5a x50

0

And so, (0, 0)

A(a, 0)

y50

x

 curl f .nˆ = (4 y kˆ).(kˆ) = 4 y

b a b b     y2  ∫∫S curl f .nˆ ds = 4∫0  ∫0 ydy dx = 4∫0  2  dx 0

a

Therefore,   2 2 ∫ f dr = ∫ [( x − y )dx + 2 xydy] c

c

=

AB

BC

Along AB, we have x = a and dx = 0. Therefore,   b 2 f ∫ dr = ∫ 2aydy = ab 0

Along BC, we have y = b and dy = 0. Therefore,   0 2 2 ∫ f .dr = ∫ x − b dx BC

a

(

)

0

 x3  =  − b2 x 3  a − a3 + ab 2 3 Along CO, we have x = 0 and dx = 0. Therefore =

∫

CO

M13_Baburam_ISBN _C13 Part IV.indd 62

0

 

  f .dr = 0.



∫ f .dr = ∫∫ curl f .nˆ ds

CO

Along OA, we have y = 0 and dy = 0 . Therefore, a   a 2  x3  a3 f dr = x dx = .   = ∫OA ∫0 3  3 0

AB

= 2d 2 ∫ dx = 2ab 2 Hence

∫+∫+∫+∫

OA

k = 4 ykˆ.

C

S

And thus, Stoke’s theorem is verified EXAMPLE 13.98

Apply Stoke’s Theorem to evaluate

∫ [( x + y)dx + (2 x − z )dy + ( y + z )dz ] ,

C

where C is the boundary of the triangle with vertices (2,0,0), (0,3,0), and (0,0,6). Solution. Taking projection on three planes, we note that the surface S consists of three triangles, OAB in xy-plane, OBC in yz-plane, and OAC in xz-plane. Using two-point formula (or intercept form), the equations of the lines AB, BC, and CA are respectively 3x + 2y = 6, 2y + z = 6 and 3x + z = 6. We have  f = ( x + y )iˆ + (2 x − z ) ˆj + ( y + z )kˆ. Therefore,  curl f =

iˆ

ˆj

kˆ

∂ ∂x

∂ ∂y

∂ ∂z

x+ y

2x − z

y+z

1/2/2012 2:37:33 PM

13.63 vector calculuS  n  = 3 + 18 = 21.

∂  ∂ = iˆ  ( y + z ) − (2 x − z )  ∂ ∂ y z  

EXAMPLE 13.99 

∂ ∂  + ˆj  ( x + y ) − ( y + z )  ∂x  ∂z 

∂  ∂ + kˆ  (2 x − z ) − ( x + y )  = 2iˆ + kˆ. ∂ ∂ x y  

Evaluate ∫ f .d r by Stoke’s theorem, where  C f = y 2 iˆ + x 2 ˆj − ( x + z ) kˆ and C is the boundary of the triangle with vertices at (0, 0, 0), (1, 0, 0), and (1, 1, 0). Solution. We have

z

 curl f =

z

z

3x 

6

2y 

6

y2

x2 − ( x + z)

= ˆj + 2( x − y )kˆ.

∫ +∫ ∫ +∫ ∫

∫

( 2iˆ + kˆ ).( kˆ ) ds

∫

( 2iˆ + kˆ ).( ˆj ) ds

∫

(2iˆ + kˆ). iˆ ds

OAB

OBC

∫ 0

x1

()

OAC 6− 2 y 3

B(1, 1)

OAC

x

OBC

y

OAB

 = ∫ 0   3

∂ ∂z

y

S

+∫

∂ ∂y

We note that the z-coordinate of each vertex of the triangle is zero. Therefore, the triangle lies in the xy-plane. Hence, nˆ = kˆ. Thus,  curl f . nˆ =  ˆj + 2( x − y )kˆ  . kˆ = 2( x − y ).

Now, by Stoke’s Theorem,    ∫ f .d r = ∫ ∫ curl f .nˆ ds

+∫

∂ ∂x

 curl f . nˆ =  ˆj + 2( x − y )kˆ  .nˆ.  

B(0, 3, 0)

x

=∫

kˆ

Therefore,



3x  A(2, 0, 0)

=∫

ˆj



0

C

iˆ

2y

6

C(0, 0, 6)

6− z 6  2    dx dy + 0 + ∫  ∫ dy  dz   0  0 

3

y0

O

C

1x    f . d r = ∫ ∫ curl f . nˆ dS = ∫∫ 2( x − y )dy dx

3

6

y  2 z  1 = 6 y − 2  + 6 z −  3 2 0 2  2 0

M13_Baburam_ISBN _C13 Part IV.indd 63

2



00

S

x

1   y2  x2  = 2 ∫  xy −  dx = 2 ∫  x 2 −  dx 2 0 2  0  0 1

2

x

The equation of OB is y = x. Therefore, by Stoke’s theorem, we have

∫

6

6 − 2y 6− z ∫0 3 dy + 2 ∫0 2 dz

A(1, 0)

1/2/2012 2:37:34 PM

13.64  n  chapter thirteen 1

1

1  x3  1 = ∫ x 2 dx =   = . 3  0 3 0



= ∫ t 2 − t 4 + t − 4t 4 − 2t 3  dt 0

1

= ∫ ( −5t 4 − 2t 3 + t 2 + t ) dt

13.21  MISCELLANEOUS EXAMPLES

0

EXAMPLE 13.100

Evaluate

∫(x

2

C

1

+ yz ) dz ,

where C is the curve

defined by x = t, y = t , z = 3t for t lying in the interval 1 ≤ t ≤ 2 . 2

Solution. The parametric equation of the curve C are x = t, y = t2 and z = 3t. Therefore

∫(x

2

2

+ yz ) dz = 3∫ ( t + 3t ) dt 2

1 1 1 2 + + =− . 2 3 2 3

EXAMPLE 13.102

If f = ( x + 3 y )iˆ + ( y − 2 z ) ˆj + ( x + λ z ) kˆ is Solenoidal, find l

2

t3 t4  = 3 + 3  4 1 3 =

= −1 −



3

1

C

 −5t 5 2t 4 t 3 t 2  = − + +  4 3 2 0  5

Solution. As in Example 13.44, we have

f1 = x + 3 y ,

163 4

f 2 = y − 2 z,

EXAMPLE 13.101

 Find the work done when a force F = ( x 2 − y 2 + x ) i − (2 xy + y) j moves a particle from the origin to the point (1, 1) along y = x2. Solution. We put x = t and get y = t2. Then x = 0 ⇒

t = 0 and x = 1 ⇒ t = 1 . Thus

 dr ˆ  r = xiˆ + yjˆ + zkˆ = tiˆ + t 2 ˆj implies = i + 2t ˆj dt  and F = t 2 − t 4 + t iˆ − 2t 3 + t 2 ˆj .

(

) (

)

Therefore

{

1



div f =

∂f1 ∂f 2 ∂f 3 + + = 1 + 1 + λ = 2 + λ. ∂x ∂y ∂z 

The vector will be solenoidal if div f = 0, that is, if 2 + λ = 0 or if λ = −2. EXAMPLE 13.103

 (a) Show that the vector field F = 2 x y 2 + z 3 iˆ + 2 x 2 y ˆj + 3 x 2 z 2 kˆ is conservative.

(

(

)

)

2 xy ) ˆj + (3 xy − 2 xz + 2 z ) kˆ is both solenoidal

(

)}

= ∫ (t 2 − t 4 + t )iˆ − ( 2t 3 + t 2 ) ˆj  ⋅ iˆ + 2t ˆj dt 0

Then

2 2 (b) Prove that y − z + 3 yz − 2 x iˆ + (3xz +

  1   d r  W = ∫ F ⋅d r = ∫ F ⋅  dt dt  c 0 1

f3 = x + λ z .

= ∫ ( t 2 − t 4 + t ) − 2t ( 2t 3 + t 2 )  dt

and irrotational. Solution. (a) Since every irrotational field is

conservative, it is sufficient to show that  curl F = 0. We note that

0

M13_Baburam_ISBN _C13 Part IV.indd 64

1/2/2012 2:37:35 PM

vector calculuS   n  13.65

 curl F =

(

iˆ

ˆj

kˆ

∂ ∂x

∂ ∂y

∂ ∂z

2x y2 + z3

)

= iˆ

+ kˆ

2 x 2 y 3x 2 z 2

∂  ∂ = iˆ  ( 3 x 2 z 2 ) − ( 2 x 3 y )  ∂z  ∂y 

)

∂  ∂ + kˆ  ( 2 x 2 y ) − 2x ( y2 + z3 )  ∂y  ∂x  2 2 ˆ = 0 − 6 xz − 6 xz j + kˆ(4 xy − 4 xy ) = 0.

(

(

)

)

 Hence the force F is conservative.  (b) f = y 2 − z 2 + 3 yz − 2 x iˆ

(

)

iˆ ∂ ∂x

∂ ∂z

EXAMPLE 13.104

Find the gradient of the sealer field f (x, y) = y2 – 4xy at (1, 2).

M13_Baburam_ISBN _C13 Part V.indd 65

nˆ =

.

= iˆ [3 x − 3 x ] − ˆj [3 y − 2 z + 2 z − 3 y ]  + kˆ [3 z + 2 y − 2 y − 3 z ] = 0  Hence f is irrotational.

 ∂ ∂ ∂  ∇f =  iˆ + ˆj + kˆ  ( y 2 − 4 xy ) ∂y ∂z   ∂x

)

The unit vector in the direction of iˆ − 3 ˆj + 2kˆ is

y 2 − z 2 + 3 yz − 2 x 3xy + 2 xy 3xy − 2 xz + 2 z

Solution.

A particle moves on the curve x = 2t2, y = t2 – 4t, z = 3t – 5, where t is time. Find the components of velocity and acceleration at time t = 1 in the direction of I – 3j + 2k. Solution. Proceeding as in Example 13.27, we have  r = x iˆ + y ˆj + z kˆ = 2t 2 iˆ + t 2 − 4t ˆj + (3t − 5)kˆ.

  d 2r a = 2 = 4iˆ + 2 ˆj dt

= –2 + 2x + 2 – 2x = 0.  f is solenoidal. Further, = ˆj kˆ ∂ ∂y

EXAMPLE 13.105

Therefore   dr v= = 4t iˆ + (2t − 4) ˆj + 3kˆ dt = 4 iˆ − 2 ˆj + 3 kˆ at t = 1,

  ∂ ∂ ∂   ∇ ⋅ f =  iˆ + ˆj + kˆ  . f ∂y ∂z   ∂x

Hence  curl f

= −8iˆ + 0 ˆj = −8iˆ at (1, 2).

(

+(3 xz + 2 xy ) ˆj + (3 xy − 2 xz + 2 z )kˆ Then

∂ 2 ( y − 4 xy ) ∂z

= −4 yiˆ + (2 y − 4 x) ˆj + 0

∂ ∂  − ˆj  ( 3 x 2 z 2 ) − 2x ( y2 + z3 )  ∂z  ∂x 

(

∂ 2 ( y − 4 xy ) + ˆj ∂∂y ( y 2 − 4 xy ) ∂x

iˆ − 3 ˆj + 2kˆ iˆ − 3 ˆj + kˆ = . 14 iˆ − 3 ˆj + 2kˆ

Therefore the components of velocity and acceleration in the direction of iˆ − 3 ˆj + 2kˆ are iˆ − 3 ˆj + 2kˆ 4 + 6 + 6  v .nˆ = 4iˆ − 2 ˆj + 3kˆ ⋅ = 14 14 16 = 14

(

or

)

iˆ − 3 ˆj + 2kˆ 2  a.nˆ = 4iˆ + 2 ˆj ⋅ =− . 14 14

(

)

EXAMPLE 13.106

Find the values of a and b so that the surfaces 2 2 ax 3 − by z = (a + 3) x and 4 x 2 y − z 3 = 11 may

1/2/2012 12:48:22 PM

13.66  n  chapter thirteen cut orthogonally at (2, –1, –3).

= 8iˆ − ˆj − 10kˆ at (1, −2, −1).

Solution. Following Example 13.40, we have

The unit vector in the direction of the given vector 2iˆ − ˆj − 2kˆ is

φ = ax3 − by 2 z − (a + 3) x 2 Then

ψ = 4 x 2 y − z 3 − 11 .

 ∂ ∂ ∂  ∇φ =  iˆ + ˆj + kˆ  ( ax 3 − by 2 z − a + z ) x 2 ∂ ∂ ∂ x y z  = iˆ 3ax 2 − 2(a + 3) x  + ˆj [ −2byz ] + kˆ  −by 2  = iˆ(8a − 12) − 6b ˆj − b kˆ at (2, −1. − 3)

 ∂ ∂ ∂  ∇ψ =  iˆ + ˆj + kˆ  ( 4 x 3 y − z 3 − 11) ∂y ∂z   ∂x 2 = iˆ(8 xy ) + ˆj ( 4 x ) + kˆ ( −3 z 2 ) Then ∇φ ⋅∇ψ = 0 implies (8a − 12)iˆ − 6b ˆj − b kˆ].[−16iˆ + 16 ˆj − 27 kˆ  = 0  

8a + 3b – 4a – 12 = 0 or 4a + 3b = 12 Solving (1) and (2), we get a = − 73 , b =

2iˆ − ˆj − 2kˆ 4 + 4 +1

=

2iˆ − ˆj − 2kˆ . 3

Therefore the directional derivative of f at (1, –2, –1) in the direction of 2iˆ − ˆj − 2kˆ is

 2iˆ − ˆj − 2kˆ  37 ∇f .nˆ = 8iˆ − ˆj − 10kˆ .   = .  3   3

(

)

(b) Similar to Exercise 24 of Chapter 13.

= 16iˆ + 16 ˆj − 27kˆ at (2, −1, −3).

⇒ 128a + 69b = 192: Also (2, –1, –3) lies on f. Therefore

nˆ =

(1)

Let φ = z 2 − 4 x 2 − 4 y 2 . Then ∇φ is along the normal vector. But  ∂ ∂ ∂  ∇φ =  iˆ + ˆj + kˆ  ( z 2 − 4 x 2 − 4 y 2 ) ∂ ∂ ∂ x y z 

= −8 x iˆ − 8 y ˆj + 2 z kˆ = −8 iˆ + 4 kˆ at the point (1, 0, 2).

(2) 64 9

.

EXAMPLE 13.107

(a) Find the directional derivative of ϕ = x 2 yz + 4 xz 2 at the point (1, –2, 1) in the direction of the vector 2iˆ − ˆj − 2kˆ. (b) Find a unit normal vector nˆ of the cone of revolution z 2 = 4 ( x 2 + y 2 ) at the point P(1, 0, 2). (c) Find the directional derivative of f (x, y, z) = 2x2 + 3y2 + z2 at the point P (2, 1, 3) in the direction of the vector aˆ = iˆ − 2kˆ.

Therefore unit normal vector nˆ to the given cone at (1,0,2) is ˆ = nn∧

−8 iˆ + 4 kˆ 64 + 16

=

−8 iˆ + 4 kˆ 80

=

−2 iˆ + kˆ 5

(c) Similar to Example 13.35. We have

 ∂ ∂ ∂  ∇f =  iˆ + ˆj + kˆ  ( 2 x 2 + 3 y 2 + z 2 ) ∂y ∂z   ∂x

= 4 x iˆ + 6 y ˆj + 2 z kˆ

Solution. (a) We have

 ∂ ∂ ∂  ∇f =  iˆ + ˆj + kˆ  ( x 2 yz + 4 xz 2 ) ∂ ∂ ∂ x y z 

= iˆ ( 2 yz +4 z 2 ) + ˆj ( x 2 z ) + kˆ ( x 2 y + 8 xz )

M13_Baburam_ISBN _C13 Part V.indd 66

= 8 iˆ + 6 ˆj + 6 kˆ at the point (2, 13). Now unit vector in the direction of iˆ − 2kˆ is

1/2/2012 12:48:22 PM

13.67 vector calculuS  n  uˆ =

iˆ − 2 kˆ 1+ 4

=

(

)

1 ˆ i − 2kˆ . 5

Therefore, the directional derivative at (2, 1, 3) in the direction of iˆ − 2 kˆ is ∇f .uˆ =

=

5

5

Therefore     aa××r r  −−nn  nn      22aa ∇∇××  n n  == == n naa++ n +n2+ 2( a( a.r.r) r) r++ n n rr rr  rr  rr

(8 iˆ + 6 ˆj + 6 kˆ )(iˆ − 2 kˆ )

1

1

Also by Example 13.52, (ii)      ∇ × ( a × r ) = curl ( a × r ) = 2a.

[8 + 0 − 12] = −

4 5

EXAMPLE 13.108

  If r = r and a is a constant vector, prove that   n    a×r  2 − n  ∇ ×  n  = n a + n+2 ( a ⋅ r ) r . r r  r 

  Solution. If r = r and a is a constant vector, then using 13.12 (B) (ii), we have

=

which is the required result. EXAMPLE 13.109

Show that        grad f .g = f × curl g + g × curl f + f .∇ g   + ( g .∇ ) f . Solution. We have

( )

    nr    =  − n + 2  × ( a × r ) + r − n ∇ × ( a × r )   r  =−

n r n+2

n r n+2 n = − n+2 r =−

      r × ( a × r )  + r − n ∇ × ( a × r )      . ) − ( . )  + ( ±±±±

−n

  ∇ × ( × ) 

    2 −n  r a − ( a. r ) r  + r ∇ × ( a × r ) 

n  n     = − n a + n + 2 ( a. r ) r ] + r − n ∇ × ( a × r )  r r

M13_Baburam_ISBN _C13 Part V.indd 67

( )

  ∂   grad f ⋅ g = ∑ iˆ f ⋅g ∂x

(

)

(

)

  ∂   ∂g ∂f   ˆ = ∑ i  f ⋅ + .g  ∂x  ∂x ∂x    ∂f     ∂g  ˆ ˆ = ∑ i  f ⋅  + ∑ i  .g  . (1)  ∂x   ∂x 

    a×r  ∇ ×  n  = ∇ ×  r − n ( a × r )   r 

    = ( ∇r − n ) × ( a × r ) + r − n [∇ × ( a × r )

2−n  n   a + n + 2 ( a. r ) r , rn r

On the other hand   ∂g    ∂g   ∂g f ×  iˆ ×  =  f .  iˆ − f .iˆ ∂x  ∂x   ∂x 

( )

or

     ∂g  ˆ   ˆ ∂g   ˆ ∂g  f ⋅ i = f ×i ×  + f ⋅i ∂x  ∂x   ∂x 

( )

Therefore      ∂g ˆ   ˆ ∂g   ˆ ∂g ∑  f . ∂x i = f ×  i × ∂x  + f .i ∂x .

( )

    = f × curl g + f ⋅∇ g .

(

)

(2)

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13.68  n  chapter thirteen

  Interchanging f and g in (2), we get       ∂f   ∑  g. ∂x  iˆ = g × curl f + ( g.∇ ) f  

Z

(3) G

From (1), (2) and (3), it follows that       grad f ⋅ g = f × curl g + g × curl f

(

E

)

    + f ⋅∇ g + ( g .∇ ) f .

(

)

A

Verify divergence theorem for  F = x 2 − yz iˆ + ( f − zx) ˆj + z 2 − xy kˆ

)

(

)

taken over the rectangular parallelopiped 0≤ x≤a, 0≤ y≤b, 0≤ z≤c.

D

x

(i) For the face OADB, we have nˆ = −kˆ , z = 0. Therefore  F .nˆ. ds =ds = ∫

∫ ∫

OADB

Firstly,

S



∫ ∫ ∫ div F dv V

∂ ∂ = ∫∫∫  ( x 2 − yz ) + ( y 2 − zx ) ∂ ∂ x y 000  cba

+

∂ 2  z − xy )  dxdy ( ∂z 

000

= a 2 bc + ab 2 c + abc 2 = abc(a + b + c)  Now to calculate F ⋅ nˆ ds, we divide the surface s of the parallelepiped 0 ≤ x ≤ a , 0 ≤ y ≤ b , 0 ≤ z ≤ c into six parts.

M13_Baburam_ISBN _C13 Part V.indd 68

2

00

2

)( )

ˆj − xykˆ . − kˆ ds

a 2b2 . 4

(ii) For the face CGEF, we have z = c, nˆ = kˆ . Therefore  ∫ ∫ F .nˆ. ds CGEF

=∫

∫ ( x

CGEF

2

− cy ) iˆ + ( y 2 − cx ) ˆj

+ ( c 2 − xy ) kˆ)  ⋅ kˆ ds ba

= ∫∫ ( c 2 − xy ) dxdy = abc 2 − 00

cba

= ∫∫∫ 2( x + y + z )dxdydz

( x iˆ + y

= ∫∫xydxdy =

have to show that

  ∫ ∫ ∫ div F dv = ∫ ∫ F .nˆ ds.

∫

OADB

ba

Solution. To verify Gauss divergence theorem, we

V

B y

O

a

EXAMPLE 13.110

(

F

C

a 2b2 . 4

(iii) For the face ADEG, we have nˆ = iˆ, x = a and dx = 0 . Therefore

∫ ∫

ADEG

cb  F . nˆ. ds = ∫∫ ( a 2 − yz ) dy dz . 00

= a 2 bc −

b2c2 4

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vector calculuS  n  13.69 (iv)

For

the

face

OBFC,

we

have

nˆ = −iˆ, x = 0, dx = 0 . Therefore

∫ ∫

OBFC

Using Divergence Theorem, we have

dy = 0 Therefore





∫∫ (v ⋅ nˆ )dA = ∫ ∫ ∫ div v dv v

ac  a2c2 F . nˆ . ds = ∫∫zx dz dx = . 4 00

 = ∫∫∫ dv, since div v = 1. v

4 y − y2

4

(vi) For the face DBFE, we have nˆ = ˆj , y = b,

=∫

ac  a2c2 2 2 ˆ F n ds = b − zx dzdx = ab c − . . . ( ) ∫ DBF∫ E ∫∫ 4 00

Hence adding the values of the above integrals, we get  ∫ ∫ F . nˆ. ds = abc(a + b + c).

∫

4y

∫

dz dx dy

0 − 4 y − y 2 x2 + y 2

dy = 0 Therefore

4 y − y2

4

=∫

∫

[4 y − x 2 − y 2 ]dx dy

0 − 4 y − y2

4

= 2∫ 0

4 y − y2

∫

[4 y − x 2 − y 2 ]dx dy

0

(even integrand in x )

S

 x3  = 2 ∫ (4 y − y 2 ) x −  3 0 0  4

Hence   ∫ ∫ ∫ div F dv = ∫ ∫ F . nˆ ds, V

4

S

=

which verifies the Gauss’s divergence theorem EXAMPLE 13.111

∫∫



(v ⋅ nˆ )dA, Evaluate using divergence theorem S  2 2 where v = x ziˆ + yˆj − xz kˆ and S is the boundary 2 2 of the region bounded by the paraboloid z = x + y and the plane z = 4 y .

Solution. We have

 v = x 2 ziˆ + yˆj − xz 2 kˆ .

M13_Baburam_ISBN _C13 Part V.indd 69

)

= 2 xz + 1 − 2 xz = 1.

(v) For the face OAGC, we have nˆ = − ˆj , y = 0,

OAGC

)(

(

= iˆ ∂∂x + ˆj ∂∂y + kˆ ∂∂z . x 2 ziˆ + yjˆ − x 2 zkˆ

ab  b2c2 F .nˆ. ds = ∫∫ yzdydz = . 4 00

∫ ∫

Therefore   div v = ∇, v

4 y−J 2

dy

4

3 3 4 4 (4 y − y 2 ) 2 dy = ∫[4 − ( y − 2) 2 ] 2 dy 3 ∫0 30

Substituting y – 2 = 2 sin t, we have dy = 2 cos tdt and so π

8 2  (4 − sin 2 t ) cos t dt divvdv = ∫ ∫ ∫v 3 −∫π 2

π

64 2 cos3t cos tdt = 3 −∫π 2

π

128 2 4 128 3 π cos tdt = = ⋅ ⋅ = 8π . ∫ 3 0 3 8 2

1/2/2012 12:48:23 PM

13.70  n  chapter thirteen EXAMPLE 13.112

(a) Using Green’s theorem in the plane evaluate

∫ [(2 x

− y 2 )dx + ( x 2 + y 2 )dy ] where C is the

2

C

boundary of the region bounded by x = 0, y = 0, x + y = 1. (b) Using Green’s Theorem find the area of the region in the first quadrant bounded by the curves 1 x y = x, y = , y = . x 4 Solution. (a) We have  ∂f 2

2

1

1   −1 dx = dx  x x  2

∫ ( xdy − ydx) = ∫  c2

∂f1 

1 1− x

1− x

 y  = ∫ ∫ 2( x − y )dy dx = 2 ∫  xy −  dx 2 0 0 0 0  1

= ∫ [2 xy − y ]

2 1− x 0

2

1

= ∫[2 x(1 − x) − (1 − x) ]dx 2

0

1

= ∫[2 x − 2 x 2 − 1 − x 2 + 2 x]dx 0

1

= ∫ (4 x − 3 x 2 − 1)dx 0

1

 4 x 2 3x3  = − − x = 2 −1 −1 = 0 . 3  2 0 (b) Using Green’s Theorem, 1 ( x dy − ydx), A=  2 ∫c 1  ∫ + ∫ + ∫ 2   C1 C2 C3

 , 

1  ( xdy − ydx) + ∫ ( xyd − ydx) 2  C∫1 C2  + ∫ ( xdy − ydx  ,  c3

M13_Baburam_ISBN _C13 Part VI.indd 70

Along c3, we have y = x so that dy = dx and x varies from 1 to 0. Therefore

∫ ( xdy − ydx) = ∫ ( xdx − xdx) = 0. c3

c3

Hence

dx

0

=

Along c2 we have y = 1x so that dy = − x1 dx and x varies from 2 to 1. Therefore

1

s

=

2

x  x c∫ ( xdy − ydx) = ∫0  4 dx − 4 dx  = 0. 1

1 = −2 ∫ dx = 2 log 2. x 0

∫∫  ∂x − ∂y  dx dy

1

where c1 is y = 4x , c2is y = 1x , and c3 is y = x. Along c1 , we have y = 4x so that dy = 14 dx and x varies from 0 to 2. Therefore

∫

1 (xdy-ydx) = [0 + 2 log 2 + 0] = log 2 . 2

c3

EXAMPLE 13.113

Verify Gauss divergence theorem for the function F = yi + xj + z 2 k over the cylindrical region bounded by x2 + y2 = 9, z = 0 and z = 2. Solution. We have 

∫∫∫div f dv 3

=

∫

9 − x2

2  ∫ 2  ∫0 2 zdz  dydx 9− x

−3 −

3  9 − x2 3   4 ∫  ∫ dy  dx = 8 ∫  −3  −3   − 9 − x2   3

= 16 ∫ 0

9 − x2

∫ 0

 dy  dx  3

 x 9 − x 2 9 −1 x  9 − x dx = 16  + sin  2 2 3   0 2

9 3  16 × 9 × π  = 16 0 + sin −1  = = 36π 2 3 2× 2  

.

1/2/2012 12:51:14 PM

13.71 vector calculuS  n  Similarly (Proceeding as in Example 13.77), we have  ˆ = 36π . ∫∫F ⋅ nds

Therefore   2 2 ∫ f .dr = ∫ ( xy dx + ydy + z xdz ) C

C

S

EXAMPLE 13.114

Evaluate and

 

∫∫ f .dS

∫+∫+∫+∫

=

Hence the theorem is verified

OA

 if f = yziˆ + wy 2 ˆj + xz 2 kˆ S is

AB

BC

CO

y=2 C (0,2)

B(1,2)

S

the surface of the cylinder x2 + y2 = 9 contained in the first octant between the planes z = 0 and z = 2. x= 1

Solution. By Gauss’s Divergence Theorem

= ∫∫ S

  ˆ = ∫∫∫ divfdv f .nds

O (0,0)

V

∂  ∂ ∂ = ∫∫∫  ( yz ) + (2 y 2 ) + ( xz 2 )  dv x y z ∂ ∂ ∂  V 

= ∫∫∫ [4 y + 2 xz ]dz dy dx 3

=∫ 0

3

=∫ 0

Along OA, we have y = 0 and dy = 0 . Therefore

∫

OA

∫ 0

9− x2

∫

0

(8 y + 4 x) dy dx

1

  f .dr = ∫ 0dx = 0. 0

Along AB, we have x = 1 and dx = 0 . Therefore

9− x2 2

∫ (4 y + 2 xz )dz dy dx

∫

AB

= ∫ [4(9 − x) + 4 x(9 − x ) ]dx = 108. 2

2

0

0   x2   f ⋅ d r = ∫ 4 xdx = 4   = −2 .  2 1 1

∫

BC

Along CO, we have x = 0 and dx = 0. Therefore 0

EXAMPLE 13.115

 Verify Stoke’s theorem for F = xy 2 iˆ + yjˆ + z 2 xkˆ for the surface of a rectangular lamina bounded by x = 0, y = 0 , x = 1, y = 2, z = 0. Solution. Similar to Example 13.97. We have 

2



Along BC, we have y = 2 and dy = 0. Therefore

1 2

0

2

 y2  f ⋅ d r = ∫ ydy =   = 2.  2 0 0 

0

3

A (1,0)

f = xy i + yjˆ + z xkˆ. 2

2

∫

CO

0    y2  f ⋅ d r = ∫ ydy =   = −2 .  2 2 2

Hence 



∫ f ⋅ d r = 0 + 2 − 2 − 2 = −2 . C

On the other hand,

Therefore 



ˆ ) ˆ + ˆjdy + kdz f ⋅ d r = ( xy 2 iˆ + yjˆ + z 2 xkˆ).(idx = xy 2 dx + ydy + z 2 xdz .

M13_Baburam_ISBN _C13 Part VI.indd 71

  curl f = ∇ × f =

iˆ

ˆj

kˆ

∂ ∂x

∂ ∂y

∂ ∂z

xy 2

y

z2 x

1/2/2012 12:51:14 PM

13.72  n  chapter thirteen 5. Find the area of a triangle having vertices A(2,3, −1) , B (1, −2,3) and C (−2,1, 4) . Ans. 12 734 sq.units

= iˆ[0] + ˆj[0 − z 2 ] + kˆ[−2 xy ]

= − z 2 ˆj + 2 xykˆ. Therefore

     6. If | a |= 3,| b |= 4, a.b = 9, find a × b .



curl f ⋅ nˆ ⋅ ds = ( − z 2 ˆj + 2 xykˆ)kˆ = −2 xy

Ans. 63

and so

7. Show that the vectors 2iˆ + kˆ, 3 ˆj + 4kˆ and 8iˆ − 3 ˆj are coplanar. Also express the last vector as a linear combination of first two Ans. 8iˆ − 3 ˆj = 4(2iˆ + kˆ) − (3 ˆj + 4kˆ)

1 2    ˆ ⋅ ⋅ = − 2 c url f n ds ∫ ∫S ∫0  ∫0 xydy  dx 2

1  y2  = −2 ∫  x  dx 2 0 2

8. Find the volume of the parallelepiped whose coterminus edges are iˆ + ˆj + 2kˆ, iˆ + 2 ˆj + 4kˆ and 2iˆ + ˆj + 3kˆ . Ans. 1 cubic unit  9. If rˆ is a unit vector in the direction of r ,    dr 1  dr show that rˆ × = 2 r × , where | r |= r .

1

 x2  = −2 ∫ 2 xdx = −4   = −2 .  2 0 0 1

Hence





∫ f ⋅ dr = ∫∫ C

S

 ˆ curlf ⋅ nds

dt

and Stoke’s Theorem is verified EXERCISES Vector Product and Differentiation of Vectors 1. Find the length of the projection of  ˆ the vector a = i − 2 ˆj + kˆ on the vector 

b = 4iˆ − 4 ˆj + 7 kˆ.

Ans. 2. Find the value of λ so that 2iˆ + λ ˆj + kˆ and 4iˆ − 2 ˆj − 2kˆ are perpendicular. Ans. λ = 3 19 9

3. Using dot product, find the angle between the lines AB and AC, where A, B, C are the points (1,2–1), (2,0,3) and(3,–1,2) respectively. Ans. θ = cos −1 2120 22  4. Dot products of a vector a with the vectors iˆ − ˆj + kˆ , 2iˆ + ˆj − 3kˆ and iˆ + ˆj + kˆ are  respectively 4, 0 and 2.Find the vector a . Ans. 2iˆ − ˆj + kˆ

M13_Baburam_ISBN _C13 Part VI.indd 72

r

dt

  10. If a = t 2 iˆ − tˆj + (2t + 1)kˆ and b = (2t − 3)iˆ +     ˆj − tkˆ , find (i ) d (a . b) and (ii ) d (a× b) , dt dt when t = 1. Ans. (i ) − 6 , (ii) 7 ˆj + 3kˆ .  11. If the vector a has a constant magnitude,   show that a and da are perpendicular, dt  provided | da . | ≠ 0 dt









a . a =| a |2 = Hint: constant implies  da  da d     (a . a) = 0 or a. dt + a dt = 0 or 2a. da =0 dt dt    and so, a is orthogonal to da , if | da |≠ 0 . dt dt

  12. If a , b , and c are constant vectors, show    that the vector r = at 2 + b t + c is the position vector of a point moving with a constant acceleration.   Hint: ddtr = 2a (constant). 13. A particle moves along the curve x = t3+1, y = t2, and z = 2t + 5, where t is the time. Find the component of its velocity and acceleration at time t = 1 in the direction i + ˆj + 3kˆ . Ans. 11 , 811 . 2

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13.73 vector calculuS  n  14. A particle moves so that its position vector is given by r = cos ωtiˆ + sin ωtjˆ , where  ω is constant. Show that (i) the velocity v  of the particle is perpendicular to r , (ii) the acceleration is directed toward the origin and has a magnitude proportional to the   distance from the origin, and (iii) r × v is a constant vector.     Hint: r .v = 0 and a = dvdt = −ω 2 cos tiˆ −  ω 2 sin ωtjˆ = −ω 2 .r . 15. Find the unit tangent vector at any point on the curve x = t 2 + 2 , y = 4t − 5 , and z = 2t 2 − 6t , where t is any variable. 



 Hint: Tˆ = d r , where r = xiˆ + yˆj + zkˆ . dra ˆ ˆ ˆ | dt | Ans. 2i + 23 j + k .

16. Find the angle between the tangents to the curve x = t , y = t 2 , and z = t 3 at t = ±1 .

19. The position vector of a particle at time t is  r = cos (t − 1)iˆ + sinh (t − 1) ˆj + Kt 3 kˆ . Find the value of K such that at time t = 1 , the  acceleration is normal to the position vector r . 2 Hint: ddt 2r at t = 1 is −iˆ + 6 Kkˆ and r at t = 1 is iˆ + Kkˆ . Therefore, cos θ =

 2 3 2 17. If a = x yziˆ − 2 xz ˆj + xz kˆ , find the value  2  of ∂∂x∂y (a × b) at the point (1, 0, −2) .

(3iˆ + 2 ˆj + 2kˆ)





M13_Baburam_ISBN _C13 Part VI.indd 73

iˆ + ˆj + 3 kˆ 11

)=

11 .

Similarly,



8 11

.



and  g = x 3iˆ −  ∂2f ∂2 g xyzˆj + x 2 zkˆ , then determine at × ∂y 2 ∂y 2 the point (1, 1, 0).

21. If

F = xyziˆ + xz 2 ˆj − y 3 kˆ

Ans. −36 ˆj . Gradient and Fractional Derivatives  22. If r is the position vector of a point and a     is any vector, show that grad [rab ] = a × b

 dr

Hint: v = = et ( cos t − sin t ) ˆj + et ( sin t + cos t ) ˆj, dt  a = sinh (t − 1)iˆ + 2et cos tˆj .    Clearly a = 2(v − r ) .

(

proceed for acceleration, which will be

Ans. −4iˆ − 8 ˆj . 18. The position vector of a point at any time  t is given by r = et ( cos tjˆ + sin tˆj ) . Show      that (i) a = 2(v − r ) , where a and v are respectively acceleration and velocity of the particle and (ii) the angle between the radius vector and the acceleration is constant.

1

7(1 + K 2 ) 2

The unit vector in the direction of iˆ + ˆj + 3kˆ is iˆ + ˆj + 3k . Therefore, the  11 component of v along iˆ + ˆj + 3kˆ is

. Ans. cos −1 73 .

6K 2 − 1

1 . 6 3 20. A particle moves along the curve x = t + 1, y = t2 , and z = 2t + 5, where t represents the time. Find the component of its velocity and acceleration at time t = 1 in the direction of iˆ + ˆj + kˆ .  Hint: r = (t 3 + 1)iˆ + t 2 ˆj + (2t + 5)kˆ and   dr = 3t 2 iˆ + 2tˆj + 2kˆ .At t = 1, v = 3iˆ + 2 ˆj + 2kˆ dt



T 1T 2 cos θ =   T 1T 2

=

normality, 6K 2–1 = 0 and so, K = ±

and put t = 1 and t = −1 to get T 1 and    T2 . Then, the angle between T and T is   given by

1 2 2

7(1 + k )



 Hint: r = xiˆ + yˆj + zkˆ = tiˆ + t 2 ˆj + t 3 kˆ. Find  dr dt

(−iˆ + 6 Kkˆ)(i + Kk )





Hint: r = xiˆ + yˆj + zkˆ and  a = a1iˆ + a2 ˆj + a3 kˆ . Then,  a.r = a1 x + a2 y + a3 z and ∇(a.r ) = a. Therefore,

          ∇( rab ) = ∇[r .(a × b )] = ∇[(a × b ).r ] = a × b .

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13.74  n  chapter thirteen 23. If φ ( x, y, z ) = 3 xy 2 − y 3 z 2 , find ∇φ at the point (−1, 2, −1) . Ans. 12 iˆ − 24 ˆj + 16 kˆ .

28. If φ ( x, y, z ) = 2 xy + z 2 , find the directional derivative of φ in the direction of iˆ + 2 ˆj + 2kˆ .

24. Find a unit normal to the surface x2y + 2xy = 4 at the point (2, –2,3). Hence, find the equation of the normal to the surface at (2, –2,3).





Hint: Let f = x2y + 2xz – 4. Then, Df = ∇φ = 2( xy + z )iˆ + x 2 ˆj + 2 xkˆ and ∇φ at (2, –2, 3) = −2iˆ + 4 ˆj + 4kˆ . Thus, the unit normal vector to the surface at (2, –2, 3) is −2i + 4 j + 4k 1 ˆ = 3 (−i + 2 ˆj + 2kˆ) . Thus, the 36 equation of normal is x−−12 = y +2 2 = z −2 3 .

25. Find a unit normal vector to the surface x3 + y3 + 3xyz = 3 at the point (1, 2 –1), Ans. − i + 314j + 2 k .

3

required directional derivative is ∇φ .aˆ = ˆ ˆ ˆ (−2iˆ + 2 ˆj + 6kˆ). i + 2 j + 2 k = 14 .

(









O Q − O P = (5iˆ + 4 ˆj ) − (iˆ + 2 ˆj + 3kˆ) = 4iˆ − 2 ˆj + kˆ .



Unit vector aˆ in the direction of PQ is 4 iˆ − 2 ˆj + kˆ .Then, the required directional 21 derivative = ∇φ (at (1, 2,3)).aˆ = 34 21 . It will be maximum in the direction of the normal to f, that is, in the direction of ∇φ , which is equal to 2iˆ − 4 ˆj + 12kˆ . Its maximum value is | ∇φ |= 4 + 16 + 144 = 164 = 2 41 .

M13_Baburam_ISBN _C13 Part VI.indd 74

)

3

3

29. Find the greatest rate of increase of 2 u = x 2 + yz at the point (1, −1,3) . Ans. | ∇u |= 121 . 30. Find the equation of the tangent plane to the 2 2 surface z = x + y at the point (2, −1,5) . Hint: ∇φ at (2, –1, 5) is 4 iˆ − 2 ˆj − kˆ. The unit normal vector at (2, –1, 5) is aˆ = 4i − 221j − k . The equation of the line through (2, –1, 5) in the direction of normal vector aˆ is x−2 = y−+21 = z−−15 . Therefore, the equation of 4 tangent plane to the surface at (2, –1, 5) is 4(x–2)– 2(y+1)– (z–5) = 0 or 4x – 2y – z = 5. We may also find a tangent plane using   (r − a ).∇φ = 0. Therefore, in the present case, we have [( xiˆ + yjˆ + zkˆ) − (2iˆ − ˆj + 5kˆ)]. (4iˆ − 2 ˆj − kˆ) = 0 or 4(x – 2) –2(y + 1) –2(y + 1) – (z – 5) = 0 or 4 x − 2 y − z = 5 .

26. Find the directional derivative of f = (x,y,z) = x2yz + 4xz2 at the point (1, 2 –1) in the direction of 2 iˆ − ˆj − 2kˆ . Hint: Proceed as in Example 13.16. Ans. 373 . 27. Find the directional derivative of the function f (x, y, z) = x2 – y2 + 2z2 at P (1,2,3), in the direction of the line PQ, where Q is the point (5, 0, 4). In what direction the directional derivative will be maximum? Hint: ∇φ = 2 xiˆ − 2 yˆj + 4 zkˆ . Therefore,  ∇φ at (1, 2, 3) is 2 iˆ − 4 ˆj + 12kˆ . Also, PQ =

Hint: ∇φ = 2 yiˆ + 2 xˆj + 2 zkˆ = −2iˆ + 2 ˆj + 6kˆ at (1, –1, 3). Unit vector aˆ in the direction of iˆ + 2 ˆj + 2kˆ is iˆ + 2 ˆj + 2 kˆ . Therefore, the

Divergence and Curl of Vector-point function 31. Show that the vector (− x 2 + yz )iˆ + (4 y − z 2 x) ˆj + (2 xz − 4 z )kˆ is solenoidal.  Hint: Show that ∇ ⋅ f = 0 . 32. If f = ( x 2 + y 2 + z 2 ) − n , find div grand f and also n, so that div grand f = 0. Ans. ( x22+ny(22 +n −z 21))n+1 and n = 12 . 33. Show that div  r = xiˆ + yˆj + zkˆ .

( )=0,  r r3





where 

Hint: Use div (φ f ) = φ div f + gradφ . f . We get

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13.75 vector calculuS   n 

div

( ) = div(r  r r3

−3

   r ) = r −3 divr + r .gradr −3 =



  3r −3 + r .(−3r −4 gradr ) = 3r −3 + r .



−  3r  =  3r −3 − 3r −5 (r .r ) = 3r −3 − 3r −5 (r 2 ) = 0 .



Thus, it also follows that

−4 r r

 r r3

is solenoidal.  34. Show that the function , where r =| r |= x 2 + y 2 + z 2 , is a harmonic function, if r ≠0.

=

3 r2 3 1 2 − = − = r r3 r r r  4 2 39. Show that ∇ (rr ) =   r . r Hint: =

1 r



35. If f = ∇v , where u and v are scalar  fields and f is a vector field, show that   f curl f = a .  Hint: curl f = ∇ × ( u1 ∇u ).∇ u1 × ∇v + u1 ∇ × ∇v = ∇ u1 × ∇v + 0 . Hence,  f . Curl f = u1 ∇v.(∇ u1 × ∇v) = 0. 36. Show that the vector ∇φ × ∇ψ is solenoidal. 37. Find the value of a so that  f = (ax 2 y + yz )iˆ + ( xy 2 − xz 2 ) ˆj +(2 xyz − 2 x 2 y 2 )kˆ is solenoidal. Also find



=∑

  2 2 2 38. If r = xiˆ + yˆj + zkˆ andr =| r |= x + y + z , 2 show that div(rˆ) = . r  ˆ ˆ xi + yj + zkˆ r Hint: rˆ = = and x2 + y 2 + z 2 r  1  1  div( rr ) = divr + grad .r r r 3  = + [(−1)r −2 grad r ].r r

=∑

∂ x   r + ri   ∂x  r 

  r x x  x   x  = ∑   + (i ) − 2   r  + i  r r  r r r 4 =  r . r

40. Show that the vector fiel

 v = ( sin y + z )iˆ + ( x cos y − z ) ˆj + ( x − y ) kˆ

is irrotational.   r  41. If r = xi + yj + zkˆ , determine ∇ ⋅   . r (See Exercise 33). Vector Integration and Line Integrals  42. If r (t ) = 2iˆ − ˆj + 2kˆ for t = 2 and  r (t ) = 4iˆ − 2 ˆj + 3kˆ for t = 3 , show that 3 

∫[r .

 dr dt

]dt = 10.

2



     1d  Hint: dtd [(r ) 2 ] = 2r drdt implies r drdt = [(r )2 ] 2dt Therefore, 3



M13_Baburam_ISBN _C13 Part VI.indd 75

∂  ∂    (rr )  ∂x  ∂x 

 ∂  ∂r  ∂r  =∑  r +r  ∂x  ∂x ∂x 

1 4

the curl of this solenoidal vector.   Hint: divf = 2(a + 2) xy . Now, f will  be solenoidal if div f = 0 , which yields  a = −2 . curl f can be found.

.

∂2   ∇ 2 (rr ) = ∑ 2 (rr ) ∂x

Hint: Show that ∇ 2 ( 1r ) = 0 (see Example 13.36). 

 3 1 r   3 1  − 2   . r = − 3 (r .r ) r r r r r

 dr

∫[r 2

dt

1  1 ]dt = [(r ) 2 ]32 = [29 − 9] = 10 , using 2 2

 r (t ) = 4iˆ − 2 ˆj + 3kˆ for t = 3 and 2iˆ − ˆj + 2kˆ for t = 2 .

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13.76  n  chapter thirteen

43. Evaluate ∫

and x varies from 0 to 2. The equation of AB is y = 2x – 4 so that dy = 2dx and on this line, x varies from 2 to 3. Therefore,   f . dr ∫ = ∫ + ∫ , which will come out to be

  f dr , where f = ( x 2 + y 2 )iˆ − 2 xyjˆ



C

and the curve C is the rectangle bounded by y = 0, x = a, y = b, and x = 0. Ans. – 2ab2.  44. If f = 2 yiˆ − zjˆ + xkˆ , find the vector line   integral ∫ f × dr along the curve x = cos t, C

y = sin t, andc z = 2 cos t and from t = 0 to t = π2 . 1 Ans. (2 − π4 )iˆ +  π −  ˆj . 2     45. Evaluate f .dr , where f = yziˆ + zxˆj + xykˆ

∫

C

 and C is the portion of the curve r = a cos tiˆ + b sin tˆj + ctkˆ from t = 0 to t = π . 2

Hint: Parametric equations of the curve are x = a cos t, y = b sin t, and z = ct. Also  dr = − a sin tiˆ + b cos tjˆ + ckˆ . Putting the dt  values of x, y, z (in terms of t) in f , we see     that ∫ f .dr = ∫ f . ( drdt )dt = abc(0) = 0. C

C

2 2 46. Evaluate ∫[ y dx − x dy ] along the triangle C

whose vertices are (1, 0), (0,1) and (–1,0). Hint: Find the equations of three sides by a two-point formula and evaluate the integral over those sides. 2 Ans. − . 3  47. If f = (2 x + y )iˆ + (3 y − x) ˆj , evaluate   ∫ f .dr , where C is the curve in the xy-plane

C



Hint: If O (0,0), A(2,0), and B(3,2) are the points, then C consists of two lines OA and AB. On OA, we have y = 0 so that dy = 0

M13_Baburam_ISBN _C13 Part VI.indd 76

AB

 48. Find the circulation of f around the curve  C, where f = yiˆ + zˆj + xkˆ and C is the circle x 2 + y 2 = 1 and z = 0 . Hint: Parametric equations of C are x = cos t , y = sin t , and z = 0, where t varies from 0 to 2π . Then, 2π   1 − cos 2t C∫ f d . r = − ∫0 2 dt = −π .  49. Find the work done when a force f = ( x 2 − y 2 + x)iˆ − (2 xy + y ) ˆj moves a particle in xy-plane from (0, 0) to (1, 1) along the parabola y2 = x. Hint: Proceed as in Example 13.49. Ans. − 32 .

 50. Compute the work done by a force f = xiˆ − zˆj + 2 ykˆ to displace a particle along a closed path C consisting of the segments C1, C2, and C3, such that 0 < x < 1, y = x, z = 0 on C1, 0 < z < 1, x = 1, z = 1 on C2, and 0 < x < 1, y = z = x on C3. Ans. 32 . 51. Find the work done in moving a particle once around a circle C in the xyplane, if the circle has its center at the origin with a radius 3, and if the force  field is given by f = (2 x − y + z )iˆ + ( x + y − z 2 ) ˆj + (3 x − 2 y + 4 z )kˆ .

C

consisting of straight lines from (0,0) to (2,0) and (2, 0) to (3, 2).

OA

4 + 7 =11.



Hint: Parametric equations of C are x = 3 cos t , y = 3 sin t , and 0 ≤ t ≤ 2π . Ans. 18π .

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vector calculuS  n  13.77 Surface Integrals

  52. Evaluate ∫∫ f .nˆ ds, where f = 12 x 2 yjˆ − 3 yzjˆ + 2 zkˆ and S is the portion of the plane x + y + z = 1 , included in the first octant 

=

3 xz ( xiˆ + yˆj )dxdz 8 ∫∫R

=

3 ( x 2 ziˆ + xz 16 − x 2 ˆj )dxdz = 100(iˆ + ˆj ) . 8 ∫∫ 00

S



Hint: =

1

3 ˆ ˆ. nk =

nˆ =

iˆ + ˆj + kˆ 3

and

 1 f .nˆ = (12 x 2 y − 3 yz + 2 z ) 3

1 3

. Evaluate

∫∫



∫∫ f .nˆdS , where

 f = yiˆ + 2 xˆj − zkˆ

S

And S is the surface of the plane 2 x + y = 6

[12 x y − 3 y (1 − x − y ) + 2(1 − x − y ) .

∫∫

fˆ . nˆ ds. Ans.

49 . 120

  f . nˆ dS , where f = ( x + y 2 )iˆ −

S



56. Evaluate

2

53. Evaluate

54

2 xˆj + 2 yzkˆ and S is the surface of the plane 2x + y + 2z = 6, in the first octant Hint: Proceed as in Example 13.56.

in the first octant cut off by the plane z = 4 . Ans. 108. Volume Integral 57. Evaluate

∫∫ φ dV ,

2 where φ = 45x y and V

S

is the region bounded by the planes 4x + 2y + z = 8, x = 0, y = 0, and z = 0. Ans. 128.

Ans. 81. 

54. Evaluate ∫∫ f . nˆ dS , where fˆ = 4 xyiˆ + yzjˆ − S xykˆ and S is the surface bounded by the planes x = 0, x = 2, y = 0, y = 2, z = 0, and z =2.

Hint: Proceed as in Example 13.59.

58. Evaluate

∫∫ φ nˆ dS , where φ = S

3 8

xyz and S

is the surface of the cylinder x + y2 = 16 2



3 ( xiˆ + yˆj ) dxdz = ∫∫ xyz . y 8 4 R 4



∫∫∫ div fdV , where

 f = ( x 2 − yz )iˆ −

V

and z = 5. Hint: ∇( x 2 + y 2 − 16) = 2 xiˆ + 2 yjˆ, nˆ = 2 xi + 2 yj , 4x2 + 4 y 2 ˆ + yjˆ) ( xi y and nˆ. ˆj = . ˆj = Therefore; 4 4 dxdz ˆ ˆ φ n ds = φ n ∫∫S ∫∫S | nˆ. ˆj |

closed Region bounded by the cylinder z = 4 – x2 and the planes x = 0, y = 0, y = 2, and z = 0. Hint: The limits of integration are x = 0 to x = 2, y = 0 to y = 2, and z = 0 to z = 4 – x2. 80 Ans. . 3

59. Evaluate

included in the first octant between z = 0



where V is the

V

Ans. 40. 55. Evaluate

∫∫∫ (2 x + y)dV ,



2 x 2 yˆj + 2kˆ and the region V is enclosed by the planes x = 0, x = a, y = 0, y = a. and z = 0, and z = a. Hint: See Example 13.87. a5 Ans. . 3

 2 60. If f = 2 xziˆ − xˆj + y kˆ, evaluate



∫∫∫ fdV , V

where V is the region bounded by the surface x = 0, x = 2, y = 0, y = 6, z = x2 and z = 4. Ans. 128iˆ − 24 ˆj + 384kˆ .

M13_Baburam_ISBN _C13 Part VI.indd 77

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13.78  n  chapter thirteen Gouss’s Divergence Theorem  61. If f = xiˆ + 2 yˆj + 7 zkˆ , evaluate

∫∫

= ∫∫∫3dV

 ˆ , fndS

V

4  = 3V = 3  π (3)3  = 108π 3 

S



Where S is the surface enclosing volumes V. Hint: By Divergence Theorem,   ∫∫ f .nˆdS = ∫∫∫(∇. f )dV S

V

∂  ∂ ∂ = ∫∫∫  ( x) + (2 y ) + (7 z )  dV ∂x ∂y ∂z  V 

= ∫∫∫ (1 + 2 + 7)dV = 10V

.

V

62. Verify divergence theorem for  f = ( x 2 − yz )iˆ + ( y 2 − zx) ˆj + ( z 2 − xy )kˆ,





S



S

V

= ∫∫∫ (3 x 2 + 3 y 2 + 3 z 2 )dV V

= abc(a + b + c) .

.Thus, the theorem is verified   ˆ , where f = (2 x + 3 z )iˆ − 63. Evaluate ∫∫ f .ndS ( xz + y ) ˆj + ( y 2 s + 2 z )kˆ and S is the surface of the sphere having a radius 3. Hint: By divergence theorem,   ˆ f ndS = f . div ∫∫ ∫∫∫ dV S

V

∂ ∂ = ∫∫∫  (2 x + 3 z ) + (− xz + y ) ∂ x ∂ y V  +

∂ 2  ( y + 2 z )  dV ∂z 

= 3a 2 ∫∫∫dV = 3a 2V V

4 = 3a 2 . π a 3 = 4π a 5 . 3   ˆ 67. Evaluate ∫∫ f ⋅ nds for f = xiˆ − yˆj + 2 zkˆ 2 2 S over the sphere x + y + ( z − 1) = 1 . 8 Ans. π . 3 Green’s Theorem 68. Verify Green’s theorem in the xy-plane for

∫ [( xy

2

− 2 xy )dx + ( x 2 y + 3)dy ] around the

C

boundary C of the region enclosed by y2 = 8x and x = 2 Ans.

 ∂f 2

∂f1 

∫ ( f dx + f dy) = ∫∫  ∂x − ∂y dxdy = 1

C

M13_Baburam_ISBN _C13 Part VI.indd 78

 f = x 3iˆ + y 3 ˆj + z 3 kˆ

and S is the surface of the sphere x2 + y2+z2 = a2. Hint: By divergence theorem,   ∫∫ f .nˆdS = ∫∫∫div f dV

000

 ˆ on all the six faces and Evaluate ∫∫ f ⋅ nds  ˆ = abc(a + b + c) add. We shall get ∫∫ f .ndS



ˆ , where ∫∫ f .nds S

 Hint: divf = 2( x + y + z ). Therefore, abc  div f dV = ∫∫∫ ∫∫∫ 2( x + y + z )dxdydz

S



66. Evaluate



64. Verify Divergence Theorem for the function  f = yiˆ + xˆj + z 2 kˆ over the cylindrical region bounded by x2 + y2 = 9, z = 0, and z = 2. Hint: Proceed as in Example 13.83.  65. Verify Divergence Theorem for f = yiˆ + xˆj + z 2 kˆ over the cylindrical region bounded by x2 + y2 = a2, z = 0, and z = h. (Similar to Exercise 64.)

taken over the rectangular parallel-opiped 0 ≤ x ≤ a , 0 ≤ y ≤ b , and 0 ≤ z ≤ c .

V

.

2

S

128 . 5

1/2/2012 12:51:23 PM

vector calculuS  n  13.79 69. Evaluate

∫ e

−x

by

Green’s

Theorem

( sin ydx + cos ydy ) , where C is

C

the rectangle with vertices (0,0), (π , 0) ,  π  π  π ,  , and  0,  . 2    2 Ans. 2e − χ − 2 . 70. Verify Green’s theorem in the plane for

∫ [(2 xy − x

2

0







Along the lower portion C1, we have x2 = y, so that 2xdx = dy and x varies from 0 to 1.



l



3

2



 4x 2x x  = + −  5 6 3 0 





1 1 = 1+ − = 1. 3 3 Along the upper portion, we have y2 = x so that 2 ydy = dx and y varies from 1 to 0. Thus,

∫ ( f dx + f dy) 1

2

C2

0



= ∫[(2 y 3 − y 4 )2 y + ( y 4 + y 2 )]dy 1

0



= ∫[4 y 4 − 2 y 5 + y 4 + y 2 ]dy 1

M13_Baburam_ISBN _C13 Part VI.indd 79

1

∂f 2 x

2

S

−

∂f1 ∂y

) dxdy =

3 2

.

4 2( x + y )dydx = . 3

−1 0 72. Verify Green’s theorem in the plane for

∫(3x

2

− 8 y 2 )dx + (4 y − 6 xy )dy, where C is

C

the boundary of the region bounded by the parabolas y = x and y = x 2 .  ∂f 2 ∂f1  3 − dxdy = . ∂ 2 y θ r  

Ans. ∫ ( f1dx + f 2 dy ) = ∫∫ 

0

1

− y 2 )dx + ( x 2 + y 2 )]dy, where C

∫ ( f dx + f dy) = ∫∫ ( θ

1− x 2

∫ ∫

1

3

Hint: 1

= ∫ (4 x 3 + 2 x 5 − x 2 )dx 6

2

C

− x 2 + ( x 2 + x 4 )2 xdx

4

2

C

0

C1

1

C

is the boundary in the xy-plane of the area enclosed by the x-axis and the semi-circle x 2 + y 2 = 1 in the upper half of the xy-plane.

1

∫ ( f dx + f dy) = ∫[2 x

∫ ( f dx + f dy) = 1 − 1 = 0.

∫ [(2 x

is the boundary of the region enclosed by y = x 2 and y 2 = x .

1 x  ∂f 2 ∂f1  ∫∫S  θ r − ∂y dxdy = ∫0 ∫ 2 (2 x − 2 x)dxdy = 0. y=x

So,

71. Using Green’s theorem in a plane, evaluate

)dx + ( x 2 + y 2 )dy ] , where C

Hint: The two parabolas intersect at (0,0) and (1,1).

1 1 = −1 − + = −1 . 3 3



C



0  5 y5 2 y6 y3  = ∫ (5 y 4 − 2 y 5 + y 2 )dy =  − +  6 3 1  2 1

C

S

Sloke’s Theorem 73. Verify Stoke’s Theorem for the function  f = ( x 2 + y 2 )iˆ − 2 xyjˆ ( x 2 + y 2 )iˆ − 2 xyˆj

taken around the rectangle bounded by x = + a, y = 0, and y = b. ˆj iˆ kˆ  ∂ ∂ ∂ ˆ Hint: curl f = ∂x ∂y ∂z = −4 yk .



x 2 + y −2 xy 0 For the given surface, nˆ = kˆ . Therefore, b a  2 ˆ curl f ndS = . ∫∫ ∫ ∫ − 4 yxdy = −4ab 0 −a

S

It can beseen that the line integral

 

∫ f .dr = −4ab . 2

C

1/2/2012 12:51:24 PM

13.80  n  chapter thirteen 74. Evaluate by Stoke’s Theorem, the integral (e x dx + 2 ydy − dz ) , where C is the curve x 2 + y 2 = 4 and z = 2 .   Hint: curl f = 0 and so, curlf .nˆ = 0 . Hence,  curl f ∫∫S .nˆds = 0. 75. Verify Stoke’s Theorem for the function  f = (2 x − y )iˆ − yz 2 ˆj − y 2 zkˆ, where S is the upper-half surface of the sphere x2 + y2 + z2 = 1, bounded by its projection on the xy-plane. Hint: Parametric equations of C are x = cos t , y = sin t , z = 0 , and 0 ≤ t ≤ 2π . Therefore,   ∫ f .dr = ∫( f1dx + f 2 dy + f3 dz ) 2π



= ∫ (−2 sin t cos t + sin 2 t )dt = π .



  Further, curl f = kˆ . Therefore, curlf .ndS ˆ =



∫∫ nˆ.kˆ | n.k | ,

dxdy

R

where

is

the

R

projection of S on xy-plane. Then, dxdy ∫∫R nˆ.kˆ | n.k | = dxdy = ∫∫R dxdy = area of

R = π (1) 2 = π .

76. Transform the integral



ˆ ∫∫ curlf .ndS

into

S



a line integral, if S is a part the surface of the paraboloid z = 1– x2–y2 for which, z > 0  and f = yiˆ + zˆj + xkˆ . Hint: Surface S is x2 + y2 = 1 and z = 0 with parametric equations x = cos q, y = sin θ , z = 0, and 0 < q 0. Thus, PQ = s nˆ = s is the shortest distance between the lines. Let  θ be the angle between AC and PQ. Since PQ is a projection of AC along the line of shortest distance, we have   PQ = AC cos θ = c − a cos θ     PQ. AC = c −a   PQ AC     snˆ. ( c − a )   = c −a   = nˆ. ( c − a ) s c −a     b × d .(c − a ) = .   b ×d

(

)

1/2/2012 4:00:36 PM

14.12  n  chapter fourteen We have taken an absolute value because distance is always nonnegative. Thus, the shortest distance is given by     b × d .(c − a ) s=   b ×d    cbd  −  abd      (Vector form).   = b ×d To obtain the shortest distance between the skew lines in Cartesian form, let A(x1, y1, z1) and C(x2, y2, z2) be the points with position   vectors a and c , respectively on the lines l1 and l2, respectively. Let the direction ratios of the lines be b1, b2, b3 and d1, d2, d3 respectively. The equations of the lines in Cartesian form are x − x1 y − y1 z − z1 and = = b1 b2 b3 x − x2 y − y2 z − z2 = = . d1 d2 d3    In vector form, the lines are r = a + λb

(

)

 = x1i + y1 ˆj + z1kˆ + λ b1iˆ + b2 ˆj + b3 k

   and r = c + µd

(

)

 = x2 i + y2 ˆj + z2 kˆ + λ d1iˆ + d 2 ˆj + d3 k

(

)

Therefore, iˆ ˆj kˆ   b × d = b1 b2 b3 d1

d2

d3

= iˆ (b2 d3 − d 2 b3 ) + ˆj ( d1b3 − b1d3 )



and so,   b ×d =

+ kˆ (b1d 2 − d1b2 )

(b2 d3 − d 2b3 )2 + (d1b3 − b1d3 )2 + (b1d 2 − d1b2 )2 .

Further, x2    cbd  −  abd  = b1     d1



M14_Baburam_ISBN _C14.indd 12

y2

z2

b2 d2

b3 d3 x1

y1

z1

− b1 d1

b2 d2

b3 d3

=

x2 − x1

y2 − y1

z3 − z1

b1 d1

b2 d2

b3 d3

Hence, the shortest distance is given by    cbd  −  abd        s= b ×d

=

=

(

x −x y −y z −z 2 1 2 1 3 1 b b b 1 2 3 d d d 1 2 3 2 2 2 b d −b d + b d −b d + b d −b d 2 3 3 3 3 1 1 3 1 2 2 1

) (

) (

)

( x2 − x1) (b2d3 − b3d2 ) + ( y2 − y1)(b3d1 − b1d3 ) + ( z2 − z1)(b1d2 − b2d1) 2 2 (b2d3 − b3d2 ) + (b3d1 − b1d3 ) + (b1d2 − b2d1)2

= l ( x2 − x1 ) + m ( y2 − y1 ) + n ( z2 − z1 ) ,

where l, m and n are direction cosines of the line perpendicular   to both lines l1 and l2, that is, of the line b × d . Deductions: Intersection of Lines The shortest distance between two lines       r = a + λb and r = c + µd is    cbd  −  abd     s=  b ×d The lines will intersect if and only if the shortest distance between them is zero. Hence, the two lines l1 and l2 will intersect if        cbd  −  abd  = 0 or c − a b d  = 0     or, in Cartesian form, if x2 − x1 b1

y2 − y1 b2

z2 − z1 b3 = 0

d1 d2 d3 The above condition can be obtained in the following   also:    manner Let r = a + λb and r = c + µd be two lines. The first line passes through a point with a  position vector a and is parallel to the vector   line passes through c b , whereas the second  and is parallel to d . Let these lines intersect at some point. Then, thelines  will lie in a plane. Therefore, the vector c − a , b and d are planar.       Hence, ( c − a ) . b × d = 0 or c − a , b d  = 0 .

(

)

1/2/2012 4:00:37 PM

three-diMenSional geoMetry  n 14.13 Remark 14.2.     c − a , b d  = 0 even if the lines are parallel     because in that case b × d = 0 and so,     (c − a ) . b × d = 0 . So, we must check whether In case they are not b and d are parallel.     parallel c − a , b d  = 0 , then the given line       r = a + λb and r = c + µd intersect. The distance between the parallel lines is given 

(

by s =

)

  b × (c − a ) .  b

To find the equation of the line of shortest distance between the lines x − x1 y − y1 z − z1 = = (1) b1 b2 b3 and x − x2 y − y2 z − z2 = = , (2) d1 d2 d3 we observe that the line of shortest distance is coplanar with both the said lines. The equation of the plane containing the line (1) and the line of shortest distance, x−α y − β z −γ (3) = = , l m n where b2 d3 − b3 d 2 b3 d1 − b1d3 l= , m= , 2 (b2 d3 − b3 d 2 ) (b3 d1 − b1d3 )2 n= is

x − x1 b1

y − y1 b2

b1d 2 − b2 d1

(b1d 2 − b2 d1 )2 z − z1 b3 = 0

Find the magnitude and equation of the line of shortest distance between the lines x − 8 y + 9 z − 10 and = = −16 3 7 x − 15 y − 29 z − 5 = = . 3 8 −5 Solution. Let l1 , m1 , and n1 be the direction ratios of the line of shortest distance. Since it is perpendicular to both the given lines, we have 3l1 − 16m1 + 7n1 = 0 and 3l1 + 8m1 − 5n1 = 0 . Therefore, l1 m1 n1 l m n = = or 1 = 1 = 1 . 80 − 56 21 + 15 24 + 48 2 3 6 Therefore, the direction cosines of the line of shortest distance between the given lines are 2 3 l= , m= , and 2 2 2 2 2 +3 +6 2 + 32 + 62 6 n= 2 2 + 32 + 62 or 6 2 3 , m = , and n = 7 7 7 Hence, the shortest distance is l=

s = l ( x2 − x1 ) + m( y2 − y1 ) + n( z2 − z1 ) 2 3 6 (15 − 8) + (29 + 9) + (5 − 10) = 14 . 7 7 7 Further, the equation of the line of shortest distance is x − 8 y + 9 z − 10 =

3 −16 7 = 0 and 2 3 6 7 7 7 x − 15 y − 29 z − 5 3 8 −5 = 0 2 3 6 7 7 7

(4)

l m n and that of the plane containing line (2) and the line (3) is x − x2 y − y2 z − z2 (5) d1 d2 d3 = 0 l m n

Hence (4) and (5) are the two equations of the line of shortest distance with l , m, n given in (3).

M14_Baburam_ISBN _C14.indd 13

EXAMPLE 14.20

or

117 x + 4 y − 41z − 49 = 0 and 9 x − 4 y − z − 14 = 0

1/2/2012 4:00:38 PM

14.14  n  chapter fourteen Second Method: Let P and Q be points on the given lines, respectively. Then, the general coordinates of P and Q are P(3λ + 8, −16λ − 9, 7λ + 10) and Q(3µ + 15,8µ + 29, −5µ + 5) Therefore, the direction cosines of the line PQ are proportional to 3λ − 3µ − 7, −16λ − 8µ − 38 , and 7λ + 5µ + 5 Now, PQ will be line of shortest distance if it is perpendicular to both the given lines. Therefore, by the condition of perpendicularity, we have and

3(3λ − 3µ − 7) − 16(−16λ − 8µ − 38) + 7(7λ + 5µ + 5) = 0

and

3(3λ − 3µ − 7) + 8(−16λ − 8µ − 38) − 5(7λ + 5µ + 5) = 0

or 157λ + 77 µ + 311 = 0 and 11λ + 7 µ + 25 = 0 Solving these equations, we get λ = −1 and µ = −2 . Hence, the points are P(5, 7,3) and Q(9,13,15) . The shortest distance PQ is PQ = (9 − 5) 2 + (13 − 7) 2 + (15 − 3) 2 = 14 . The equation of PQ, by the two-point formula, is x−5 y −7 z −3 or x − 5 = y − 7 = z − 3 . = = 9 − 5 13 − 7 15 − 3 2 3 6 EXAMPLE 14.21 x − 4 y + 3 z +1 and = = −4 7 x − 1 y + 1 z + 10 1 = = intersect and find 2 −3 8

Show that the lines

the coordinates of their point of intersection. Solution. Any point on the first line is (λ + 4, − 4λ − 3, 7 λ − 1) whereas any point on the second line is (2 µ + 1, − 3 µ − 1, 8 µ − 10) At the point of intersection, we must have λ + 4 = 2 µ − 1, − 4λ − 3 = −3 µ − 1 , and 7λ − 1 = 8µ − 10 . The first two members yield λ = 1 and µ = 2 . These values satisfy 7λ − 1 = 8µ − 10 Hence, the given lines intersect at the point (λ + 4, − 4λ − 3, 7 λ − 1) = (5, −7, 6). EXAMPLE 14.22

Find the equation of the perpendicular to both the lines

M14_Baburam_ISBN _C14.indd 14

straight

line

x −1 y −1 z + 2 x+ 2 y −5 z +3 = = = = and 1 2 3 −1 2 2 and passing through their point of intersection. Solution. Any point on the first lines is (λ + 1, 2λ + 1, 3λ − 2) and any point on the second line is (2 µ − 2, − µ + 5, 2 µ − 3). At the point of intersection, we must have λ + 1 = 2 µ − 2, 2λ + 1 = − µ + 5 and 3λ − 2 = 2 µ − 3. From the first two members, we get λ = 1 and µ = 3 . These values also satisfy 3λ − 2 = 2 µ − 3 . Hence, the given lines intersect at (λ + 1, 2λ + 1, 3λ − 2) = (2,3,1) . Therefore, the required line is x − 2 y − 3 z −1 = = l m n This line will be perpendicular to the given lines if l + 2m + 3n = 0 and 2l − m + 2n = 0 Therefore, l m n = = 7 4 −5

Hence, the required line is x − 2 y − 3 z −1 = = −5 7 4 EXAMPLE 14.23

Find the magnitude and equation of the shortest distance between the lines x +1 y +1 z −1 . x−3 y −5 z −7 = = = = and −6 7 1 −2 1 1 Also, find the points where the line of shortest distance intersects the given lines. Solution. Any point P on the first line is P(λ + 3, − 2λ + 5, λ + 7) and any point Q on the second line is Q(7 µ − 1, − 6 µ − 1, µ − 1) Therefore, direction cosines of PQ are proportional to λ + 3 − 7 µ + 1, − 2λ + 5 + 6 µ + 1 and λ + 7 − µ + 1 or λ − 7 µ + 4, − 2λ + 6 µ + 6 , and λ − µ + 8 . The line PQ will be of shortest distance if it is perpendicular to both the given lines. Therefore, the condition for perpendicularity yields λ − 7 µ + 4 − 2(−2λ + 6 µ + 6) + λ − µ + 8 = 0 and 7(λ − 7 µ + 4) − 6(−2λ + 6 µ + 6) + λ − µ + 8 = 0

1/2/2012 4:00:39 PM

three-diMenSional geoMetry  n 14.15 or

6λ − 20 µ = 0 and 20λ − 86 µ = 0 Thus, we get trivial solution λ = 0 and µ = 0 Thus, the points P and Q are (3, 5, 7) and (−1, −1, −1) , respectively. The magnitude of PQ is

2

EXAMPLE 14.24

Find the shortest distance between the lines x −1 y − 2 z − 3 x−2 y −4 z −5 = = and = = 2 3 4 3 4 5 Solution. The general points on the given lines are respectively P(2λ + 1, 3λ + 2, 4λ + 3) and Q(3 µ + 2, 4 µ + 4, 5 µ + 5) Therefore, the direction cosines of PQ are proportional to 2λ + 1 − 3 µ − 2, 3λ + 2 − 4 µ − 4 , and 4λ + 3 − 5 µ − 5 or

2λ − 3µ − 1,3λ − 4 µ − 2 , and 4λ − 5µ − 2 . The line PQ will be perpendicular to both the lines if 2(2λ − 3 µ − 1) + 3(3λ − 4 µ − 2) +4(4λ − 5 µ − 2) = 0 3(2λ − 3 µ − 1) + 4(3λ − 4 µ − 2) +5(4λ − 5 µ − 2) = 0 or 29λ − 38µ − 16 = 0 and 38λ − 50 µ − 21 = 0 . Solving these equations for λ and µ , we get 1 1 λ = and µ = − . 6 3 Thus, the points and are  3 10 25  13  5 P  , 3,  and Q  , ,  2 3 6 3 3

M14_Baburam_ISBN _C14.indd 15

2

 3 5   10   25 13  | PQ | =  −  +  − 3  +  −  3  2 3  3   6

=

s = (3 + 1) 2 + (5 + 1) 2 + (7 + 1) 2

= 16 + 36 + 64 = 10.7703 . The equation of PQ is x−3 y −5 z −7 x−3 y −5 z −7 = = = = or 3 +1 5 +1 7 +1 2 3 4

and

Hence, the shortest distance is 2

1 1 1 1 + + = 36 9 36 6

EXAMPLE 14.25

Show

that

the

x−6 y +8 z +5 = = −4 2 3

lines

x y z +3 = = 1 −1 1

and

do not intersect. Find the

equation of a straight line parallel to the second line which does meet the first li e. Solution. The coordinates of the general points P and Q on the given lines, respectively, are P(λ, − λ, λ − 3) and Q(2 µ + 6, − 4 µ − 8, 3 µ − 5) .

The given lines will intersect only if there exist λ and µ such that λ = 2 µ + 6, − λ = −4 µ − 8, and λ − 3 = 3µ − 5 . Solving the first and the last equations, we get λ = 22 and µ = 8. But the equation −λ = −4 µ − 8 is not satisfied for these values. Hence, the given lines do not intersect. Further, the equation of the line parallel to the second line shall be of the type x − α y − β z − γ . = = (1) −4 2 3 Taking, α = β = 0 , the parametric equation of the line is x = 2λ , y = −4λ , and z = 3λ + γ. The parametric equation of the first line is x = µ, y = − µ and z = µ − 3. Therefore, (1) with α = β = 0 will intersect the first given line if 2λ = µ , −4λ = − µ , and 3λ + γ = µ − 3 The first two equations implies λ = 0 and µ = 0 and then the third equation implies γ = −3 . Hence, one of the required equation is x y z +3 = = 2 −4 3 EXAMPLE 14.26

Find the shortest distance between the lines  r = (1 − t ) iˆ + (t − 2) ˆj + (3 − 2t ) kˆ and  r = ( s + 1) iˆ + ( 2s − 1) ˆj − ( 2s + 1) kˆ.

1/2/2012 4:00:41 PM

14.16  n  chapter fourteen Solution. The given lines can be expressed as

   r = iˆ − 2 ˆj + 3kˆ + t −iˆ + ˆj − 2kˆ = a + tb , say and    r = iˆ − ˆj − kˆ + s iˆ + 2 ˆj − 2kˆ = c + sd , say Thus,   a = iˆ − 2 ˆj + 3kˆ, b = −iˆ + ˆj − 2kˆ,   c = iˆ − ˆj − kˆ, d = iˆ + 2 ˆj − 2kˆ.

) ( ) (

( (

)

Therefore, the shortest distance (S.D) between the given lines is     (c − a ). b × d (1) S .D = .   b ×d ˆi ˆj kˆ   Now, b × d = −1 1 −2 = 2iˆ − 4 ˆj − 3kˆ, 1 2 2   c − a = iˆ − ˆj − kˆ − iˆ − 2 ˆj + 3kˆ = ˆj − 4kˆ ,

(

)

) (

(

) (

)

)(

) (

Hence, (1) implies 8 S .D = = 2iˆ − 4 ˆj − 3kˆ

)

8 4 + 16 + 9

EXAMPLE 14.27

=

(

8 29

.

)

 Show that the line r = 7iˆ − ˆj − kˆ + λ 2iˆ + 3kˆ and r = iˆ + ˆj − kˆ + µ 3iˆ − kˆ intersect and find their point of intersection. Solution. The given lines are    r = 7iˆ − ˆj − kˆ + λ 2iˆ + 3kˆ = a + λb , say and    r = iˆ + ˆj − kˆ + µ 3iˆ − kˆ = c + µd , say Thus,    a = 7iˆ − ˆj − kˆ, b = 2iˆ + 3kˆ, c = iˆ + ˆj − kˆ, a n d  d = 3iˆ − ˆj. The lines will intersect if    (1) cbd  −  abd  = 0.     We have 1 1 −1   and   cbd  = 2 0 3 = 3 + 9 + 2 = 14, 3 −1 0

)

(

)

(

(

M14_Baburam_ISBN _C14.indd 16

)

Thus, (1) is satisfied and so, the given lines intersect. Now, at the point of intersection, we     have r = a + λb = c + µd , which gives 7iˆ − ˆj − kˆ + λ 2iˆ + 3kˆ = iˆ + ˆj − kˆ + µ(3iˆ − ˆj ). Comparing the coefficients of iˆ , ˆj , and kˆ on both sides, we have 7 + 2λ = 1 + 3 µ, − 1 = 1 − µ, and −1 + 3λ = −1. These equations yield µ = 2, λ = 0. Hence, the  vector of point of intersection is  position  r = a + 0b = 7iˆ − ˆj − kˆ. Therefore, the point of intersection is (7,–1, –1).

(

)

EXAMPLE 14.28

Find the shortest distance between the lines x+3 y+7 z−6 x −3 y −8 z −3 = = . = = and 2 4 −3 3 −1 1

and     (c − a ) . b × d = ˆj − 4kˆ . 2iˆ − 4 ˆj − 3kˆ = 8.

(

7 −1 −1   abd  = 2 0 3 = 21 − 9 + 2 = 14.   3 −1 0

)

Also, find the equation of the line of shortest distance. Solution. Let PQ be the shortest distance between the given lines with point lying on the first line and lying on the second line. Then, we have P (3λ + 3, − λ + 8, λ + 3) and Q ( −3 µ − 3, 2 µ − 7, 4 µ + 6) for some scalars λ and µ . The direction cosines of PQ are proportional to −3 µ − 3 − 3λ − 3, 2 µ − 7 + λ − 8, and 4µ + 6 − λ − 3 or −3λ − 3 µ − 6, λ + 2 µ − 15, and − λ + 4 µ + 3. The line PQ will be of shortest distance if it is perpendicular to both the given lines. So, we must have 3 ( −3λ − 3 µ − 6) − 1( λ + 2 µ − 15) +1( − λ + 4 µ + 3) = 0 and

−3 ( −3λ − 3 µ − 6) + 2 ( λ + 2 µ − 15) +4 ( − λ + 4 µ + 3) = 0

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three-diMenSional geoMetry  n 14.17 or −11λ − 7 µ = 0 and 7 λ + 29 µ = 0. These equations have a trivial solution λ = µ = 0. Therefore, the points P and Q are P (3, 8, 3) and Q (–3, –7, 6). Hence, the shortest distance is PQ =

(3 + 3)2 + (8 + 7)2 + (3 − 6)2

= 270 = 3 30. Also, using the two-point formula, the equation of PQ is x−3 y −8 z −3 or = = −3 − 3 −7 − 8 6 − 3 x −3 y −8 z −3 . = = −1 2 5 EXAMPLE 14.29

Find the direction cosines of the line bisecting the angles between the lines whose direction ratios are l1, m1, n1 and l2, m2, n2, respectively. Solution. If L = L (l1 , m1 , n1 ) and L ′ = L ′ (l2 , m2 , n2 )

then OL and OL′ are the lines from the origin O with the given direction ratios. Also, OL = 1 and OL ′ = 1 . The midpoint M of LL′ has the n +n l + l m + m2 coordinates 1 2 , 1 and 1 2 . 2 2 2 L (l1, m1, n1)

O

 2  2

M

(l

l

1 1

2

m m n n , 1 1 , 1 1 2 2

)

l1 − l2 m1 − m2 n −n , , and 1 θ2 . 2 cos θ2 2 cos θ2 2 cos 2 Remark 14.3. It follows from the previous example that the direction cosines of the bisectors are proportional to l1 + l2 , m1 + m2 , n1 + n2 and l1 − l2 , m1 − m2 , n1 − n2 . EXAMPLE 14.30

Find the equations of the lines bisecting the angle between the lines x −1 y + 2 z − 3 x −1 y + 2 z − 3 = = . and = = 12 2 −3 −2 2 1 Solution. The direction ratios of the bisectors are proportional to l1 + l2 , m1 + m2 , n1 + n2 and l1 − l2 , m1 − m2 , n1 − n2 . In the present case, the direction cosines of the given lines are respectively l1 = 32 , m1 = − 32 , and −3 2 , m2 = 157 , and n2 = 157 . n1 = 13 and l2 = 12 157 Therefore, the direction cosines of the bisectors are proportional to l1 + l2 , m1 + m2 , n1 + n2 and l1 − l2 , m1 − m2 , n1 − n2 . or proportional to 61, − 19, 3.5 and −11, − 31, 21.5. Since the bisectors pass through (1, − 2, 3) their equations are x − 1 y + 2 z − 3 and x − 1 y + 2 z − 3 = = . = = −11 −31 21.5 −19 61 3.5 EXAMPLE 14.31

L‘ (l2, m2, n2)

Find the equations of two straight lines that pass through the origin and intersect the line x − 3 y − 3 z at an angle of 60º. = =

then OM = cos θ , that is OL 2 θ OM = OL cos . Hence, the direction cosines of 2 OM (the bisector of the angle θ ) are n1 + n2 l1 + l2 m1 + m2 . , , and θ θ 2 cos θ2 2 cos 2 2 cos 2

x y z = = . a b c Any point on this line is given by

Similarly, the direction cosines of the other bisector are

x = aλ, y = bλ, and z = cλ. The given line is

If

∠LOL ′ = θ

M14_Baburam_ISBN _C14.indd 17

2

1

1

Solution. Any line that passes through the origin

(0, 0, 0) is

(1)

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14.18  n  chapter fourteen x−3 y −3 z = = 2 1 1 Any point on this line is

(2)

x = 2 µ + 3, y = µ + 3, and z = µ. For intersection, we have aλ = 2 µ + 3, bλ = µ + 3, and cλ = µ. We note that aλ − bλ = µ = cλ and so, a = b + c or a − b − c = 0. (3) The angle between (1) and (2) is 60° . Therefore, 2a + b + c cos 60° = 6 a 2 + b2 + c2 or 1 3b + 3c = , using a = b + c or 2 6 2b 2 + 2c 2 + 2b 2

6b + 6c = 6 2b 2 + 2c 2 + 2b 2 Squaring both sides, we get 2b 2 + 2c 2 + 5bc = 0 or (2b + c)(b + 2c) = 0 Thus, 2b + c = 0 or b + 2c = 0 that is, 0a + 2b + c = 0 or 0a + b + 2c = 0 (4) If (using (3) and (4)) a − b − c = 0 and 0a + 2b + c = 0 , then a b c. a b c = or = = = 1 −1 2 −1 + 2 0 − 1 2 − 0 If a − b − c = 0 and 0a + b + 2c = 0 , then a b c a b c = = . or = = −1 −2 1 −2 + 1 0 − 2 1 − 0 Hence, (1) gives the required equations as x y = = 1 −1 that is, x y = = 1 −1

x y z z = = , and −1 −2 1 2 x y z z and = = . 1 2 −1 2

M14_Baburam_ISBN _C14.indd 18

14.8  EQUATION A PLANE Let O be the origin of the three-dimensional coordinate system and P(x, y, z) any point on the plane. Let ON = d be a perpendicular drawn from O to the plane. Let n denote the unit normal  vector to the plane. Then ON = ON ( nˆ ) = dnˆ. P

N n^



O

Since P is an arbitrary point on the plane, the  vector PN is in the plane and is perpendicular     to ON . Therefore, PN .ON = 0 or PN .dnˆ = 0 or  PN .nˆ = 0.     But PN = ON − OP = dnˆ − r . Therefore, ( dnˆ − rˆ) .nˆ = 0 or

rˆ.nˆ = dnˆ.nˆ = d . Hence, the required equation of the plane is (1) rˆ.nˆ = d (Vector form), where nˆ is the unit normal to the plane and d is the perpendicular distance from the origin to the plane. To obtain the Cartesian form, we have  r = xiˆ + yjˆ + zkˆ.   Let the normal ON be ON = aiˆ + bjˆ + ckˆ. Then, the unit vector nˆ is aiˆ bjˆ nˆ = + 2 2 2 2 a +b +c a + b2 + c2 ckˆ + = liˆ + mjˆ + nkˆ, 2 2 2 a + b + c where l, m, and n are direction cosines of the normal to the plane. Hence, the equation (1) of the plane reduces to

( xiˆ + yjˆ + zkˆ).(liˆ + mjˆ + nkˆ) = d

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three-diMenSional geoMetry   n 14.19 or lx + my + nz = d (Normal form). (2) Multiplying (2) throughout by a 2 + b 2 + c 2 , we get ax + by + cz = d a 2 + b 2 + c 2 = D say, (3) as the equation of the plane. Hence, in a threedimensional system, any linear equation of the form ax + by + cz = D represents a plane. A plane parallel to the plane ax + by + cz = D has the same normal. Hence, the direction cosines of the normal to the two planes are same. However, the distance from the origin to the two planes is different. Hence the equation of the plane parallel to the plane ax + by + cz = D is ax + by + cz = p , where a2 +Pb2 + c2 is the perpendicular distance from the origin to the parallel plane. If D and p are of opposite signs, then the two planes lie on the opposite sides with respect to the origin. If the plane ax + by + cz = D intersects the axes at, (A, 0, 0), (0, B, 0) and (0, 0, C) , then, since these points lie on the plane, we have Aa = D, Bb = D , and Cc = D , which yields a=

D A

,b =

D B

, and c =

D C

. Hence, the equation

of the plane reduces to x y z Dx Dy Dz + + = D or + + = 1 , (4) A B C A B C which is called the intercept form of the equation of the plane. 14.9  EQUATION OF A PLANE PASSING THROUGH A GIVEN POINT AND PERPENDICULAR TO A GIVEN DIRECTION Let the plane pass through a fixed  point A(a1 , a2 , a3 ) , whose position vector is a . Let  N be the given direction and let r be the position vector of a point P( x, y, z ) on the plane. Then, PA lies in the plane. Since the plane is perpendicular to N , we have      PA.N = 0 or (r − a ).N = 0 (Vector form), (1) which is the required equation of the plane.

M14_Baburam_ISBN _C14.indd 19

P(x, y, z)

A(a1, a2, a3)

a



N

O

Substituting

  r = xiˆ + yjˆ + zkˆ, a = a1iˆ + a2 ˆj + a3 kˆ, and  N = n1iˆ + n2 ˆj + n3 kˆ in (1), we have

( x − a1 )n1 + ( y − a2 )n2 + ( z − a3 )n3 = 0 , (2) which is the equation of the plane, in Cartesian form, passing through (a1 , a2 , a3 ) and perpendicular to the direction N . 14.10  EQUATION OF A PLANE PASSING THROUGH THREE POINTS Let P ( x1 , y1 , z1 ), Q( x1 , y2 , z2 ), and R( x3 , y2 , z3 ) be the points and n1 , n2 , n3 be the direction ratios of the normal to the plane. The equation of the plane passing through P( x1 , y1 , z1 ) and perpendicular to the line with direction ratios n1 , n2 , n3 is n1 ( x − x1 ) + n2 ( y − y1 ) + n3 ( z − z1 ) = 0 (1)

The plane also passes through Q( x2 , y2 , z2 ) and R( x3 , y2 , z3 ) Therefore, n1 ( x2 − x1 ) + n2 ( y2 − y1 ) + n3 ( z2 − z1 ) = 0 (2) and n1 ( x3 − x1 ) + n2 ( y3 − y1 ) + n3 ( z3 − z1 ) = 0 .

(3)

Eliminating n1 , n2 , n3 from (1), (2), and (3), we get x − x1 y − y1 z − z1 (4) x2 − x1 y2 − y1 z2 − z1 = 0 , x3 − x1 y3 − y1 z3 − z1 which gives the required equation of the plane. Second Method: We want to find the equation of the plane passing through three points P( x1 , y1 , z1 ), Q(x2, y2, z2) and R(x3, y3, z3).

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14.20  n  chapter fourteen )

Q(x2, y2, z2)

,z1 ,y1

x1

P(

R

b

z 3) y 3, , (x 3

   The vector AB = b − a lies in the given plane  and hence is parallel to c . A

c

a

a

O



b



The vector PQ and PR lie in a given  plane. Therefore, the cross product PQ × PR is in the of the normal to the plane. Let  direction  a , b , c be the position vectors of P, Q, and R respectively. The equation of the plane passing through P(x1, y1, z1) and normal to the vector       PQ × PR is ( r − a ) . ( PQ × PR ) = 0       or ( r − a ) .  b − a × ( c − a )  = 0 (5)         or ( r − a ) .  a × b + b × c + c × a  = 0. (6)

(

)

14.11  EQUATION OF A PLANE PASSING THROUGH A POINT AND PARALLEL TO TWO GIVEN VECTORS Let the plane pass through the point A with a  position vector a and let the plane  be parallel  to the vectors b and c . Now b × c is in the direction of the normal to the plane. Therefore, the equation of the plane passing through A and  parallel to b and c is     (r − a ). b × c = 0 (Scalar Product form). (1)

(

B

)

c P

O

Therefore, the normal to the required plane is    given by (b − a ) × c . Therefore, the equation of the plane passing through A and perpendicular    to (b − a ) × c is      (r − a ). (b − a ) × c  = 0    r − a

or

  b −a

 c  = 0

EXAMPLE 14.32

Find the direction cosines of the normal to the plane. What is the distance of the plane from the origin? Solution. The equation of the plane is

3x + 4y + 5z = 5

Dividing throughout by 32 + 42 + 52 = 50, we get 3 4 5 5 x+ y+ z= 50 50 50 50

Second Method: Any parallel to the  vector     vectors b and c is λ b + µ c . Since r − a is  the vector in the plane and hence, is parallel to b and  c . Therefore,     (2) r − a = λ b + µc is the required equation of the plane.

Comparing it with the normal form lx + my + nz = d, we get 3 4 5 l= ,m = , and n = 50 50 50

14.12  EQUATION OF A PLANE PASSING THROUGH TWO POINT AND PARALLEL TO A LINE Let A and B be two points  on the plane with   position vectors a and b , respectively. If c is the direction of the given line and P is any point on the line with the position vector p . Then the    given line is r = p + λ c .

d=

M14_Baburam_ISBN _C14.indd 20

and

(direction cosines of the normal) 5 50

=

1 2

(the distance from the origin).

EXAMPLE 14.33

Find the equation of a plane which is at a distance of 7 units from the origin and is perpendicular to 4iˆ + 2 ˆj − 3kˆ, directed away from the origin.

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three-diMenSional geoMetry   n 14.21 Solution. Let nˆ be the unit vector in the direction of the given vector 4iˆ + 2 ˆj − 3kˆ, directed away

from the origin. Then 4iˆ + 2 ˆj − 3kˆ 4iˆ 2iˆ 3kˆ nˆ = = + − . 29 29 29 42 + 22 + 32  If r = xiˆ + yjˆ + zkˆ is the position vector of a general point (x, y, z) on the plane, then the equation of the plane is 3 ˆ  4 ˆ 2 ˆ xiˆ + yjˆ − zkˆ .  i+ j− k=7 29 29   29

(

4 x + 2 y − 3 z = 7 29. EXAMPLE 14.34

Find the equation of the plane passing through the  point (1, 2, 3) and having the vector N = 2iˆ − ˆj + 3kˆ normal to it. Solution. We know that if n1, n2, n3 are direction

ratios of the normal to the plane, then the equation of the plane passing through (a1, a2, a3) is (x – a1) n1 + (y – a2)n2 + (z – a3)n3 = 0. In the present case, we have (a1, a2, a3) = (1, 2, 3) and n1 = 2, n2 = –1, n3 = 3. Therefore, the equation of the plane is 2(x – 1) – (y –2) + 3(z –3) = 0 or 2x – y + 3z = 9. EXAMPLE 14.35

Find the equation of the plane passing through the point (3, –3,1) and perpendicular (normal) to the line joining the points (3,4,–1) and (2, –1,5). Solution. The direction of the line is  N = 2iˆ − ˆj + 5kˆ − 3iˆ + 4 ˆj − kˆ

(

) (

ˆ

ˆ

)

ˆ

= − i − 5 j + 6k . We know that the equation of the plane passing  through the point with the position vector a and perpendicular to the direction N is    ( r − a ) .N = 0.

Therefore, the required equation is  xiˆ + yjˆ + zkˆ − 3iˆ − 3 ˆj + kˆ  . −iˆ − 5 ˆj + 6kˆ = 0  

) (

M14_Baburam_ISBN _C14.indd 21

) (

)

(

)

 ( x − 3)iˆ + ( y + 3) ˆj + ( z − 1)kˆ  . −iˆ − 5 ˆj + 6kˆ = 0  

or (x – 3) (–1) + (y + 3) (–5) + (z – 1) (6) = 0 or x + 5y – 6z + 18 = 0. Second Method: Equation of the line joining (3,4,–1) and (2, –1, 5) is

)

or

(

or

or

x − 3 y − 2 z +1 = = 2 − 3 −1 − 4 5 + 1

x − 3 y − 2 z +1 = = . −1 −5 6 The direction ratios of the normal to the above line are −l, –5, and 6. Therefore, the equation of the plane passing through (3, –3, 1) is n1(x – 3) + n2 (y + 3) + n3(z – 1) = 0, that is, –1 (x – 3) – 5 (y + 3) +6 (z – 1) = 0 or x + 5y –6z + 18 = 0. EXAMPLE 14.36

Find the equation of the plane which passes through the point (3, –3, 1) and is normal to the line joining the points (3, 2, –1) and (2, –1, 5). Solution. As in Example 14.35, the required

equation is  xiˆ + yjˆ + zkˆ − 3iˆ − 3 ˆj + kˆ  . −iˆ − 3 ˆj + 6kˆ = 0  

(

) (

) (

)

or  ( x − 3)iˆ + ( y + 3) ˆj + ( z − 1)kˆ  . −iˆ − 3 ˆj + 6kˆ = 0  

(

)

or (x – 3) (–1) + (y + 3) (–3) + (z – 1) (6) = 0 or x + 3y – 6z + 12 = 0. EXAMPLE 14.37

Find the equation of the plane passing through the point (1, 2, 3) and parallel to the plane 4x + 5y – 3z = 7. Solution. The equation of the plane parallel to the plane 4x + 5y – 3z = 7 is 4x + 5y – 3z + k = 0. Since it passes through (1,2,3), we have

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14.22  n  chapter fourteen 4 + 10 – 9 + k = 0  or  k = – 5 Hence, the equation of the required plane is 4x + 5y – 3z = 5 EXAMPLE 14.38

Find the equation of the plane passing through the point (3, 4, –1), which is parallel to the plane  r . 2iˆ − 3 ˆj + 5kˆ + 7 = 0.

(

)

Solution. The equation of the given plane is

or

( xiˆ + yjˆ + zhˆ).(2iˆ − 3 ˆj + 5kˆ) + 7 = 0.

2x – 3y + 5z = 0. The equation of the plane parallel to this plane is 2x – 3y + 5z + k = 0. (1) Since the plane (1) passes through (3,4,–1), we have 6 – 12 – 5 + k = 0  or  k = 11. Hence, the equation of the required plane is 2x – 3y + 5z + 11 = 0.

(

)

 ( x − 4)iˆ + ( y + 2) ˆj + ( z + 5)kˆ  . 4iˆ − 2 ˆj − 5kˆ = 0  

or

4 (x – 4) –2 (y + 2)–5(z +5) = 0

or

4x – 2y – 5z = 45.

Second Method: The distance OA = d = (4 − 0) 2 + (−2 − 0) 2 + (5 − 0) 2 = 16 + 4 + 25 = 45.

 The unit normal vector along OA is nˆ =

4iˆ − 2 ˆj − 5kˆ 16 + 4 + 25

=

4iˆ − 2 ˆj − 5kˆ 45

.

Therefore, the equation of the plane, in the  normal form, is r .nˆ = d or  ˆ

ˆ

ˆ

( xiˆ + yjˆ + zkˆ ) . 4i − 245j − 5k  = 



45

or 4 x − 2 y − 5 z = 45. 45 = 45.

EXAMPLE 14.39

The foot of the perpendicular from the origin to a plane is (4,–2, –5). Find the equation of the plane. Solution. The position vector of the foot of the perpendicular A drawn from the origin to the plane is  a = 4iˆ − 2 ˆj − 5kˆ. P(x, y, z)

Third Method: Equation of the plane passing through (4,–2,–5) is n1 (x – 4) + n2 (y + 2)+ n3(z +5) = 0. The direction ratios of the normal are 4,–2, and –5. Therefore, the equation of the required plane is 4 (x – 4) –2 (y + 2)–5(z +5) = 0 or

A(4, 2,5)

4x – 2y – 5z = 45.

EXAMPLE 14.40 

The position vectors of two points A and B are 3iˆ + ˆj + 2 kˆ and iˆ − 2 ˆj − 4 kˆ , respectively. Find the equation of the plane passing through B and perpendicular to AB.

a

O(0, 0, 0)

Also

Solution.

  OA = N = 4iˆ − 2 ˆj − 5kˆ.

P(x, y, z)

Thus, the plane passes through A and is  perpendicular to the direction OA. Therefore, its   equation is ( r − a ) .N = 0 or

(

) (

) (

)

 xiˆ + yjˆ + zkˆ − 4iˆ − 2 ˆj − 5kˆ  . 4iˆ − 2 ˆj − 5kˆ = 0  

or

M14_Baburam_ISBN _C14.indd 22

B(1, 2, 4)



A(3, 1, 2)

 Let r = xiˆ + yjˆ + zkˆ . Then

1/2/2012 4:00:49 PM

three-diMenSional geoMetry   n 14.23   A B = N = iˆ − 2 ˆj − 4 kˆ − 3iˆ + ˆj + 2 kˆ

(



) (

)

= −2iˆ − 3 ˆj − 6 kˆ .

The plane is to  pass through B, whose position vector is a = iˆ − 2 ˆj − 4 kˆ . Therefore, the    required equation is ( r − a ) .N = 0 or

(

) (

) (

)

 xiˆ + yjˆ + zkˆ − iˆ − 2 ˆj − 4 kˆ  . −2iˆ − 3 ˆj − 6 kˆ = 0  

or

(

)

( x − 1) iˆ + ( y + 2) ˆj + ( z + 4 ) kˆ  . −2i − 3 ˆj − 6 kˆ = 0  

or –2 (x – 1) – 3 (y + 2) – 6 (z + 4) = 0 or 2x + 3y + 6z + 28 = 0. In Vector form, we can write  r . 2iˆ + 3 ˆj + 6 kˆ = −28.

(

)

EXAMPLE 14.41

Find the equation of the plane passing through the points (–2, 6, –6), (–3, 10, –9), and (–5, 0, –6). Solution. The equation of the plane passing through (–2, 6, –6) is a (x + 2) + b (y – 6) + c (z + 6) = 0. (1) Since (–3, 10, –9) and (–5, 0, –6) lie on it, we have a(–3 + 2) + b (10 – 6) + c (–9 + 6) = 0 and a(–5 + 2) + b(0 – 6) + c(–6 + 6) = 0, that is –a + 4b – 3c = 0. (2) and – 3a – 6b + 0c = 0. (3) From (2) and (3), we have a b c a b c = = or = = . −18 9 18 −2 1 2 Hence, the equation of the plane (1) becomes –2 (x + 2) + (y – 6) + 2 (z + 6) = 0 or

M14_Baburam_ISBN _C14.indd 23

EXAMPLE 14.42

Show that the four points, (0, –1, 0), (2, 1, –1), (1, 1, 1), and (3, 3, 0) are coplanar. Find the equation of the plane passing through them. Solution. The equation of the plane passing through the first three points i x − 0 y +1 z − 0 x y +1 z 2 − 0 1 + 1 − 1 − 0 = 0 or 2 2 − 1 = 0 1− 0 1+1 1− 0 1 2 1 or

x(2 + 2) + (y + 1) (–1 –2) + z(4 – 2) = 0 or 4x – 3y + 2z = 3. (1) We note that (1) is satisfied by the fourth point (3, 3, 0) also. Hence, all the four points lie on the plane 4x – 3y + 2z = 3 and so, are coplanar. EXAMPLE 14.43

Find the equation of the plane which meets the coordinate axes at P, Q and R respectively and if the centroid of the triangle PQR is the point ( a , b, c ) . Solution. Suppose that the points P, Q and R are respectively P(A, 0, 0), Q(0, B, 0) and R(0, 0, C). Since the centroid of the triangle PQR is (a, b, c) we have

A+ 0+ 0 A 0+ B+0 B = , b= = , and 3 3 3 3 0+0+C C c= = . 3 3 Hence A = 3a, B = 3b, and C = 3c. Also, the equation of the plane, in the intercept form, is x y z + + =1 A B C a=

Putting the values of A, B and C, we get x y z x y z + + = 1 or + + = 3 . a b c 3a 3b 3c EXAMPLE 14.44

2x – y – 2z – 2 = 0.

Find the image of the point P(1,3, 4) in the plane 2x − y + z + 3 = 0 .

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14.24  n  chapter fourteen Solution. Let Q ( x1 , y1 , z1 ) be the image of the point

P(1,3, 4) in the plane 2 x − y + z + 3 = 0 . Then the direction ratios of PQ are x1 − 1, y1 − 3, and z1 − 4. Since PQ must be perpendicular to the given plane, we have x1 − 1 y1 − 3 z1 − 4 = = = λ , say −1 2 1 P(1, 3, 4)

M

2x  y  z  3 = 0

Q(x1, y1, z1)

Thus, Q is (2λ + 1, − λ + 3, λ + 4). Now the midpoint M of PQ is 2λ + 1 + 1 − λ + 3 + 3 λ + 4 + 4 , , . 2 2 2 λ λ   But M  λ + 1, − + 3, + 4, lies on the given  2 2  plane. Therefore,  λ  λ  2(λ + 1) −  − + 3 +  + 4 + 3 = 0.  2  2  This yields λ = −2 . Hence, the point Q is (2λ + 1, − λ + 3, λ + 4) = ( −3, 5, 2). EXAMPLE 14.45

x−2

y −1

z −3

Show that the line −1 = 6 = −6 lies in the plane that passes through the points A(1, − 2, 3), B(1, 1, 1) , and C (0,1, −1) . Solution. The equation of the plane passing through the three given points is x −1 y + 2 z − 3 0 3 −2 = 0 −1 3 −4 or Any

−6 x + 2 y + 3z + 1 = 0 point on the given

M14_Baburam_ISBN _C14.indd 24

line

(1) is

(−λ + 2, 6λ + 1, −6λ + 3) . This point clearly satisfies the equation (1). Hence, the given line lies in the plane − 6 x + 2 y + 3 z + 1 = 0 . 14.13  ANGLE BETWEEN TWO PLANES Let the equation of the two given planes be a1 x + b1 y + c1 z = D1 and a2 x + b2 y + c2 z = D2 Therefore, the direction ratios of the normal to these planes are respectively  a1 , b1 , c1 , and a2 , b2 , c2 .The normals are N1 = a1iˆ + b1 ˆj + c1kˆ and N 2 = a2 iˆ + b2 ˆj + c2 kˆ . Therefore, the angle between the planes is the angle θ between these normals and  so  N1 .N 2 a1a2 + b1b2 + c1c2 cos θ =   = 2 | N1 .N 2 | a1 + b12 + c12 a22 + b22 + c22 The two planes will be perpendicular to each other if a1a2 + b1b2 + c1 + c2 =0 2 a1 + b12 + c12 a22 + b22 + c22 or if a1a2 + b1b2 + c1c2 = 0 (1) The two planes will be parallel to each other if the vectors a1iˆ + b1 ˆj + c1kˆ and a2 iˆ + b2 ˆj + c2 kˆ are parallel. Therefore, for some scalar λ , we have a2 = λ a1 , b2 = λ b1 , and c2 = λ c1 or a2 b2 c2 = = =λ a1 b1 c1 Therefore, the planes a1 x + b1 y + c1 z = D1 and a2 x + b2 y + c2 z = D2 are parallel if a2 b2 c2 = = . a1 b1 c1 If we put a2 = λ a1 , b2 = λ b1 , and c2 = λ c1 in a2 x + b2 y + c2 z = D2 , we get

λ (a1 x + b1 y + c1 z ) = D2 or D2 a1 x + b1 y + c1 z = = k , say. λ Thus, the equation of any plane parallel to a given plane x + by + cz = D is ax + by + cz = k .

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three-diMenSional geoMetry   n 14.25 EXAMPLE 14.46

Find the equation of the plane through the points (2, 2, 1) and (9, 3, 6) and perpendicular to the plane 2 x + 6 y + 6 z = 9 . Solution. The equation of the plane passing through (2, 1, 0) is (1) a ( x − 2) + b( y − 2) + c( z − 1) = 0 It passes through the point (9, 3, 6) if a (9 − 2) + b(3 − 2) + c(6 − 1) = 0

Eliminating a,b and c from (1), (2), and (3), we get

EXAMPLE 14.48

Find the equation of the plane passing through the point (2, 3, 4) and parallel to the plane

If the plane passes through the point (2, 3, 4), we have Hence, the required plane is 5 x − 6 y + 7 z = 20

3x + 4 y − 5 z = 9 ,

EXAMPLE 14.49

which is the equation of the required plane. EXAMPLE 14.47

Find the equation of the plane through the point (2,1, 0) and perpendicular to the planes 2 x − y − z = 5 and x + 2 y − z = 5. Solution. The equation of the plane passing

through (2, 1, 0) is

a ( x − 2) + b( y − 1) + c( z − 0) = 0. (1) The plane (1) will be perpendicular to the given planes if (2) 2a − b − c = 0

M14_Baburam_ISBN _C14.indd 25

5 x + 5 y + 5 z = 15 or x + y + z = 3

5(2) − 6(3) + 7(4) = k or k = 20

−24( x − 2) − 32( y − 2) + 40( z − 1) = 0

a + 2b − 3c = 0.

5( x − 2) + 5( y − 1) + 5 z = 0

Solution. The equation of the plane parallel to the given plane is 5 x − 6 y + 7 z = k .

or

and

or

( x − 2)(3 + 2) + ( y − 1)(−1 + 6) + z (4 + 1) = 0

5x − 6 y + 7 z = 3 .

y − 2 z −1 1 5 =0 6 6

or ( x − 2)(6 − 30) + ( y − 2)(10 − 42) + ( z − 1)(42 − 2) = 0 or

or or

or if (2) 7 a + b + 5c = 0 Also (1) is perpendicular to the plane 2 x + 6 y + 6 z = 9 if (3) 2a + 6b + 6c = 0

x−2 7 2

Eliminating a, b and c from (1), (2), and (3), we get x − 2 y −1 z 2 −1 −1 = 0 2 2 −3

(3)

Find the equation of the plane passing through the point (3, 4, –1) which is parallel to the plane  r . 2iˆ − 3 ˆj + 5kˆ + 7 = 0.

(

)

Solution. The given plane is

( xiˆ + yjˆ + zkˆ).(2iˆ − 3 ˆj + 5kˆ) + 7 = 0

or 2x - 3y + 5z + 7 = 0. The plane parallel to the above plane is 2x – 3y + 5z = k. It passes through (3, 4, –1), if 6 – 12 – 5 = k or k = –11. Hence, the equation of the required plane is 2x – 3y + 5z + 11 = 0

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14.26  n  chapter fourteen or or

ˆ + ˆ + ˆ ) . ( 2 ˆ − 3 ˆ + 5 ˆ ) + 11 = 0 (±±±±

(

)

 r . 2iˆ − 3 ˆj + 5kˆ + 11 = 0.

EXAMPLE 14.50

Find the angle between the planes   r . iˆ + ˆj = 1 and r . iˆ + kˆ = 3.

(

)

(

)

Solution. Let   N 1 = iˆ + ˆj and N 2 = iˆ + kˆ .

Then, the angle q between the given planes is given by   iˆ + ˆj . iˆ + kˆ N 1 .N 2 1 = . cos θ =   = 2 N 1 .N 2 1+1 1+1 Hence, q = 60º.

(

)(

)

14.14  ANGLE BETWEEN A LINE AND A PLANE      Let r = a + λb be the givenline and r .N = d be the given plane, where N is normal to the plane. Let the line meet the plane at P. Line



or

  b .N

θ = sin −1   (Vector form).

Cartesian Form.

N

x − x1 y − y 1 z − z 2 = = = λ. b1 b2 b3



On the other hand, the equation of the plane is n1x + n2 y + n3 z = d .



Thus, the direction ratios of the line are b1, b2, b3 and the direction ratios of the normal to the plane are n1, n2, n3. Thus, the angle f between the line and the normal to the plane is given by cos φ = cos(90 − θ ) = sin θ or

θ = sin −1

 Let f be the angle between the normal N and the line. Then   b .N cos φ =   . (1) b .N As we know, the angle between a line and a plane is defined as the angle between the line and the projection of the line on the plane. Let this angle be q. Then, we note that φ = π2 − θ . Therefore, (1) yields

M14_Baburam_ISBN _C14.indd 26

b .N

  Putting r = xiˆ + yjˆ + zkˆ , a = x 1iˆ + x 2 ˆj + x 3 kˆ ,  and b = b1iˆ + b2 ˆj + b3 kˆ , the line becomes

=



P

  b .N π  cos φ = cos  − θ  = sin θ =   2  b .N

n1b1 + n2b2 + n3b3 n + n 22 + n 23 b 21 + b 22 + b 23 2 1

n1b1 + n2b2 + n3b3 n + n 22 + n 23 2 1

b 21 + b 22 + b 23

,

where q is the angle between the given line and the plane. The line will be parallel to the plane if and only if q = 0, that is, if and only if n1b1+n2b2+n3b3 = 0. 14.15  PERPENDICULAR DISTANCE OF A POINT FROM A PLANE 

Let r .nˆ = d

be the given plane and let

A ( x 1 , y 1 , z 1 ) be the point with the position  vector a. We have OB = d.

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three-diMenSional geoMetry   n 14.27 C

A Plane P2 Passing Through A and Parallel to P1

B

O

Let P2 be a plane parallel to the given plane P1 and passing through the point A. Extend the normal OB to meet the new plane P2 at C. Then OC is the perpendicular distance from the origin to the plane P2. Hence, Distance BC between the planes = OC – OB. (1) The equation of the plane  P2 passing through A and perpendicular to N is     (2) r − a ) .nˆ = 0 or r .nˆ = .a.nˆ , (  where nˆ is the unit vector along N . Thus,  OC = a.nˆ . Hence, Perpendicular distance of A from Plane P1 = OC − OB = a.nˆ − d . Since distance is always non-negative, the  required perpendicular distance is a.nˆ − d . If the equation of the given plane is of the form,  where N is normal to the plane, then the Perpendicular distance of A from   a.N − d P1 =  (Vector form). N To obtain the Cartesian form, let ax + by + cz = D be the given plane and A ( x 1 , y 1 , z 1 ) be the  given point. Then N = aiˆ + bjˆ + ckˆ and so Perpendicular distance of A form P1 1

=

1

)(

a2 + b 2 + c 2

ax 1 + by 1 + cz 1 − D

M14_Baburam_ISBN _C14.indd 27

)

ˆj + z kˆ . aiˆ + bjˆ + zkˆ − D 1

a2 + b 2 + c 2

a1 x + b1 y + c1 z = D1

.

(1)

a2 x + b2 y + c2 z = D2 (2) be the given planes. Let P(x, y, z) be any point on either of the two planes bisecting the angle between the given planes (1) and (2). Then Perpendicular distance of P from (1) = perpendicular distance of P from (2) and so,

Plane P1 (Given)

(x iˆ + y =

Let and

a

N

14.16  PLANES BISECTING THE ANGLES BETWEEN TWO PLANES

a1x + b1 y + c1z − D1 a +b +c 2 1

2 1

2 1

=±

a2 x + b2 y + c 2 z − D 2 a22 + b22 + c 22

,

which are the equations of the two bisector planes. EXAMPLE 14.51

Find the angle between the line x − 2 y +1 z − 3 = = −1 3 2 and the plane 3x + 4y + z + 5 = 0. Solution. The direction ratios of the given line are 3, –1, and 2, whereas the direction ratios of the given plane are 3, 4, and 1. Hence, the angle q between the given line and the given plane is given by 3(3) + ( −1)(4) + 2(1) sin θ = 2 32 + ( −1) + 22 32 + 42 + 12 =



7 14 26

=

7 2 13

.

 7  Hence, θ = sin −1  .  2 13  EXAMPLE 14.52

Find the distance between the parallel planes   r .(iˆ + 2 ˆj − 3kˆ) = 6 and r .(3iˆ + 6 ˆj − 9kˆ) = −15 . Solution. The standard forms of the given planes

are

 r .(iˆ + 2 ˆj − 3kˆ) 1 + 2 + ( −3) 2

2

2

=

6 and 12 + 22 + ( −3)2

1/2/2012 4:00:53 PM

14.28  n  chapter fourteen  r .(3iˆ + 6 ˆj − 9kˆ) 32 + 62 + ( −9) 2

=

−15 2 3 + 62 + ( −9)2

or

6  −15 −5  r .nˆ = = and r .nˆ = 14 3 14 14 The distances of the planes from the origin are 6 and −145 , which are the opposite sides of the 14 origin. Hence, the distance between the planes is | d1 | + | d 2 | = 614 + 514 = 1114 . EXAMPLE 14.53

Find the distance between the parallel planes 2 x − 2 y + z + 3 = 0 and 4 x − 4 y + 2 z + 5 = 0 . Solution. The distance between the parallel planes

is equal to the distance of any point lying on one plane from the second plane. We note that the point (0, 0, −3) satisfies the equation of the first plane. Thus, the required distance is the distance of (0, 0, −3) from the plane 4 x − 4 y + 2 z + 5 = 0 Thus, 4 x − 4 y1 + 2 z1 + 5 Required distance = 1 16 + 16 + 4 4(0) − 4(0) + 2(−3) + 5 = 6

=

−1 1 = 6 6

This equation yields 21λ + 63 = 0 or λ = −3. Therefore, the foot of the perpendicular is (2λ + 7, 4λ + 14, − λ + 5) = (1, 2,8). Now, the length of the perpendicular is the distance between (7, 14, 5) and (1, 2, 8), which is p = (7 − 1) 2 + (14 − 2) 2 + (5 − 8) 2 = 189 = 3 21. EXAMPLE 14.55

Find the equation of the plane passing through the points (2, 2, 1) and (1, −2,3) , and parallel to the line joining the points (2,1, −3) and (−1,5, −8) . Solution. The equation of the plane passing through the point (2, 2, 1) is (1) a ( x − 2) + b( y − 2) + c( z − 1) = 0

It passes through (1, −2,3) and so, a (1 − 2) + b(−2 − 2) + c(3 − 1) = 0 or (2) a + 4b − 2c = 0 The direction ratios of the line joining (2,1, −3) and (−1,5, −8) are 3, −4, 5 . The plane (1) will be parallel to the line joining (2,1, −3) and (−1,5, −8) if (3) 3a − 4b + 5c = 0

From (2) and (3), we have

a b c = = . 20 − 8 −6 − 5 −16

EXAMPLE 14.54

Find the length and foot of the perpendicular from the point (7, 14, 5) to the plane 2 x + 4 y − z = 2 . Solution. Since the line passing through (7, 14, 5) is perpendicular to the plane 2 x + 4 y − z = 2 , its direction ratios are the direction ratios of the normal to the plane. Thus, the direction ratios of the perpendicular are 2, 4, and −1 . So, the equation of the perpendicular is x − 7 y − 14 z − 5 = = −1 2 4 Any point on this line is (2λ + 7, 4λ + 14, − λ + 5). Since the foot of the perpendicular lie on the plane, we have 2(2λ + 7) + 4(4λ + 14) − (−λ + 5) = 2

M14_Baburam_ISBN _C14.indd 28

Hence, the equation of the required plane is 12( x − 2) − 11( y − 2) − 16( z − 1) = 0 or

12 x − 11y − 16 z + 14 = 0.

EXAMPLE 14.56

Find the equations of the planes bisecting the angle between the planes 6 x + 3 y + 2 z = 5 and 4 x + 3 y + 12 z + 3 = 0 . Specify the plane which bisects the acute angle. Solution. The equations of the bisector planes are 6x + 3y + 2z − 5 36 + 9 + 4

=±

4 x + 3 y + 12 z + 3 16 + 9 + 144

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three-diMenSional geoMetry   n 14.29 or

13 (6 x + 3 y + 2 z − 5) = ± 7(4 x + 3 y + 12 z + 3) which yields the equation of the bisectors as 25 x + 9 y − 29 z − 43 = 0 and

x1 =

53 x + 30 y + 55 z − 22 = 0. Further, the angle between the plane 6 x + 3 y + 2 z = 5 and 25 x + 9 y − 29 z − 43 = 0 is given by cos θ =

6(25) + 3(9) + 2(−29) 36 + 9 + 4 625 + 81 + 841

≈ 0.432
45° . Therefore, the second bisector plane 53 x + 30 y + 55 z − 22 = 0 bisects the acute angle. EXAMPLE 14.57

 Show that the line r = (iˆ + ˆj ) + λ (3iˆ − ˆj + 2kˆ)  is parallel to the plane r .(2 ˆj + kˆ) = 3 . Also fmd the distance between the line and the plane. Solution. The equation of the plane is   r .(2 ˆj + kˆ) = 3 . Therefore, 2 ˆj + kˆ = N is normal to the plane. Further, the line is parallel to the vector 3iˆ − ˆj + 2kˆ and we note that (3iˆ − ˆj + 2kˆ).(2 ˆj + kˆ) = −2 + 2 = 0 Therefore, the angle between the line and the normal to the plane is 90°. Hence, the line is parallel to the given plane. The position vector of the point through which the line passes is iˆ + ˆj . Therefore, the distance between the line and the plane is (iˆ + ˆj ).(2 ˆj + kˆ) − 3 2−3 1 = = . ˆ ˆ 4 +1 5 |2j+k|

14.17  INTERSECTION OF PLANES Two planes intersect in a line. Therefore, a1 x + b1 y + c1 z = D1 the equations and a2 x1 + b2 y1 + c2 z = D2 , taken together, represent a line. To find the equation of this line in a standard form, we take any one of the coordinates of a point on the line as 0. Let us

M14_Baburam_ISBN _C14.indd 29

take z1 = 0 . Then we have a1 x1 + b1 y1 = D1 and a2 x1 + b2 y1 = D2 . Solving by Cramer’s Rule, we get b2 d1 − b1d 2 a d − a2 d1 , y1 = 1 2 , and z1 = 0 . a1b2 − a2 b1 a1b2 − a2 b1

Further, if A, B, and C are the direction ratios of the line, then since the line is perpendicular to the normals of the planes, we have a1 A + b1 B + c1C = 0 and a2 A + b2 B + c2 C = 0 Solving these equations, we get A B C = = . b1c2 − b2 c1 c1a2 − a1c2 a1b2 − a2 b1

Therefore, the equation of the line in a standard form is x − x1 y − y1 z − 0 = = . A B C 14.18  PLANES PASSING THROUGH THE INTERSECTION OF TWO GIVEN PLANES   We know that the two planes r .N1 = d1 and   r .N 2 = d 2 intersect along a line. The points on this line are common to both the planes. Further, the points that satisfy the equations of both the planes, also satisfy the equation     ( r . N − d ) + λ ( r .N 2 − d 2 ) = 0 (1) 1 1 Hence, the equation of a plane passing through the  line of intersection   of the two given planes r .N1 = d1 and r .N 2 = d 2 is given by (1). Thus, if P1 = 0 and P2 = 0 represent the two given planes, then P1 + λ P2 = 0 represent the equation of the plane passing through the line of intersection of the given planes. The value of the parameter can be found by additional conditions like “plane passes through a point” or “the required plane is perpendicular to another plane,” and so on. EXAMPLE 14.58

Find the equation of the plane passing through the intersection of the planes x + y + z = 6 and 2 x + 3 y + 4 z + 5 = 0 and passing through the point (1, 1, 1).

1/2/2012 4:00:55 PM

14.30  n  chapter fourteen Solution. The equation of the plane passing through the intersection of two given planes is x + y + z − 6 + λ (2 x + 3 y + 4 z + 5) = 0

Since this plane passes through (1, 1, 1), we have 1 + 1 + 1 − 6 + λ (2 + 3 + 4 + 5) = 0 and so, λ = 14 . Hence, the required plane is 3

x+ y+ z −6+

or

3 (2 x + 3 y + 4 z + 5) = 0 14

Solution. The equation of the plane passing through the intersection of first two planes i

2 x + 5 y + 3z + λ ( x − y + 4 z − 2) = 0

(1)

The three given planes will have a common line of intersection if for some value of λ , the equation (1) yields the equation of the third plane. Comparing the coefficie ts of x, y and z and the constant terms in (1) and the equation of the third plane, we get 2 + λ = 1, 5 − λ = 6, 3 + 4λ = −1, and −2λ = 2.

20 x + 23 y + 26 z − 69 = 0.

EXAMPLE 14.59

Find the equation of the plane passing through the intersection of the planes x + 2 y + 3 z − 4 = 0 and 2 x + y − z + 5 = 0 perpendicular to the plane 5 x + 3 y + 6 x + 1 = 0 .

We note that all these equations yields λ = −1 Hence, for λ = −1 , the equation (1) reduces to x + 6 y − z + 2 = 0 . Therefore, the three given planes have a common line of intersection. EXAMPLE 14.61

Solution. The equation of the plane passing through the intersection of the given planes is

Find the equation of the plane passing through the line of intersection of the planes x + y + z = 1 and 2 x + 3 y − z + 4 = 0 and perpendicular to the plane 2 y − 3 z = 4.

x + 2 y + 3z − 4 + λ (2 x + y − z + 5) = 0

Solution. The equation of the plane passing

or (1 + 2λ) x + (2 + λ) y + (3 − λ) z − 4 + 5λ = 0.

through the line of intersection of the given planes is

(1)

Since the plane (1) is perpendicular to the plane 5 x + 3 y + 6 x + 1 = 0 , we have

5(1 + 2λ ) + 3(2 + λ ) + 6(3 − λ ) = 0

This equation yields λ = − 297 . Hence, the required plane is 29 x + 2 y + 3z − 4 − (2 x + y − z + 5) = 0 7 or

x + y + z − 1 + λ (2 x + 3 y − z + 4) = 0 or (1 + 2λ) x + (1 + 3λ) y + (1 − λ) z − 1 + 4λ = 0. (1) The plane (1) will be perpendicular to the plane 2 y − 3 z − 4 = 0 if 0(1 + 2λ) + 2(1 + 3λ) − 3(1 − λ) = 0, 1 that is, if, 9λ = 1 or λ = 9 . Hence, the required plane is

1 x + y + z − 1 + (2 x + 3 y − z + 4) = 0 9

51x + 15 y − 50 z + 173 = 0 EXAMPLE 14.60

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

M14_Baburam_ISBN _C14.indd 30

or 11x + 12 y + 8 z − 5 = 0

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three-diMenSional geoMetry   n 14.31 14.19  SPHERE A sphere is the locus of a point in a threedimensional space, which remains at a constant distance from a fixed point. The fixed point is called the Center of the sphere and the constant distance is called the radius of the sphere.  Let r be the position vector of any point P(x, y, z) on the surface of a sphere with respect to the origin O. Let the  position vector of the center C (c1 , c 2 , c3 ) be c . Thus,      CP = OP − OC = r − c .

C

C

P(X, Y, Z )

which is called the Cartesian equation of the sphere with center (c1, c2, c3) and radius a. The equation (3) can be expressed as x 2 + c12 − 2c1x + y 2 + c 22 − 2c 2 y

(

) (

(

)

)

+ z 2 + c32 − 2c3 z = a2 or x 2 + y 2 + z 2 − 2c1 x − 2c2 y − 2c3 z +c12 + c 22 + c32 − a2 = 0. (4) If we put g = −c1 , f = −c2 , h = −c3 , and c = c12 + c22 + c32 − a 2 , then (4) reduces to x2 + y2 + z2 + 2gx + 2fy + 2hz + c = 0. (5) Equation (5) is called the general form of the equation of the sphere with center (c1 ,c2 ,c3 ) , = ( − g , − f , − h ) and radius a = c 21 + c 22 + c32 − c = g 2 + f 2 + h 2 − c . Hence, the equation (5) represents a sphere with center (− g , − f , −h) and radius



C

g 2 + f 2 + h2 − c .

O

 If CP = a be the radius of the sphere, then or



  r −c = a







( r − c ) . ( r − c ) = a2



(1)

2   or ( r ) − 2r .c + c 2 − a2 = 0. (2)

(

)

Equations (1) and (2) are called the vector equations of the sphere with center C and radius a. To obtain the Cartesian form of the equation of the sphere, we have   r = xiˆ + yjˆ + zkˆ and c = c1iˆ + c 2 ˆj + c3 kˆ .

We note that the equation of the sphere is a second-degree equation in x, y, and z, where the coefficients of x2, y2, and z2 are equal, and there is no term involving xy, yz, and zx. Remarks 14.4. (i) If the center of the sphere is its origin, then (1) transforms to   (6) r = a or r .r = a2 , which, in the Cartesian form, is x2 + y2 + z2 = a2. (ii) Equation (6) can be rewritten as     (7) ( r + a ) . ( r − a ) = 0.

Therefore, (1) reduces to xiˆ + yjˆ + zkˆ − c1iˆ + c 2 ˆj + c3 kˆ = a

(

or or

( x − c1 ) iˆ + ( y − c2 ) ˆj + ( z − c3 ) kˆ

( x − c1 )2 + ( y − c2 )2 + ( z − c3 )2 or

=a

 A

=a

( x − c1 )2 + ( y − c2 )2 + ( z − c3 )2 = a2 ,

M14_Baburam_ISBN _C14.indd 31

P

)

 

a

O

a

B

(3)

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14.32  n  chapter fourteen 











We have A P = r + a and BP = r − a. 



Therefore, (7) implies A P.BP = 0, which shows that A P is perpendicular to BP. Hence, the diameter of the sphere subtends a right angle at any point P on the surface of the sphere. Further, this result is true even if the center is not the origin.

O 5 A

(iii) If the origin lies on the sphere, then

B

   r −c = c

OA = Length of the perpendicular from the origin O to the plane

or       r + c 2 − 2r .c = c 2 or r 2 − 2r .c = 0. EXAMPLE 14.62

Find the center and radius of the sphere x2 + y 2 + z 2 + 4x − 8 y + 6z + 4 = 0 . Solution. Comparing the given equation with the general equation of the sphere x2 + y2 + z2 + 2gx + 2fy + 2hz + c = 0, we have g = 2, f = –4, and h = 3. Hence, the center of the given sphere is (–g, –f, –h) = (–2, 4, –3) and the radius of the sphere is r = g 2 + f 2 + h 2 − c = 4 + 16 + 9 − 4 = 5. EXAMPLE 14.63

Find the radius of the circular   of the   section sphere r = 5 by the plane r . i + j + k = 3 3.

(

)

Solution. The equation of the given sphere is  r = 5 or x 2 + y 2 + z 2 = 5 or

x 2 + y 2 + z 2 = 25. Therefore, the center C = (0, 0, 0) and radius r = 5. The equation of the plane is  r . iˆ + ˆj + kˆ = xiˆ + yjˆ + zkˆ . iˆ + ˆj + kˆ = 3 3

(

) (

or

)(

)

1(0) + 1(0) + 1(0) − 3 3 1+1+1

=3

and so, the radius of the circle is AB =

(OB )2 − (OA )2

= 52 − 32 = 4.

EXAMPLE 14.64

Find the equation of the sphere for which the circle x 2 + y 2 + z 2 + 10 y − 4z − 8 = 0, x + y + z = 3, is a great circle. Solution. Section of a sphere by a plane through

its center is called a great circle. Further, if S = 0 is a sphere and U = 0 is a plane, then the equation S+lU = 0 represent a sphere; and the points of intersection of the sphere S = 0 and the plane U = 0 satisfy it. In view of this argument, the equation of any sphere through the given circle is x 2 + y 2 + z 2 + 10 y − 4 z − 8 + λ ( x + y + z − 3) = 0 (1)

or x+ y+z =3 3 .

Let A be the center of the circle and AB be the radius. Then

M14_Baburam_ISBN _C14.indd 32

=

x 2 + y 2 + z 2 + λ x + (10 + λ ) y − (4 − λ )z − (8 + 3λ ) = 0.

The center of this sphere is ( − λ2 , − 102+ λ , 4 −2λ ) . According to the definition of the great circle, this center must lie on the plane x + y + z = 3. Therefore,

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three-diMenSional geoMetry  n 14.33

λ 10 + λ

P

4−λ = 3 and so, l = –4. 2 2 2 Hence, the equation (1) of the sphere reduces to x2 + y 2 + z 2 − 4 x + 6 y − 8z + 4 = 0. −

−

+

A

EXAMPLE 14.65

B

Find the equation of the sphere whose center has the position vector 3iˆ + 6 ˆj − 4 kˆ and which  touches the plane r . 2iˆ − 2 ˆj − kˆ ) = 10.

(

)



b

Solution. The center of the sphere is (3, 6,

–4). Since the sphere touches the plane  r . 2iˆ − 2 ˆj − kˆ = 10, the radius of the sphere is equal to the length of the perpendicular from the center of the sphere to the plane. The Cartesian form of the plane is

(

)

a

O



2x – 2y – z – 10 = 0.

(1)

Therefore, the length of the perpendicular from (3, 6, –4) to the plane (1) is its radius =



=

2(3) + 6( −2) − 4( −1) − 10 22 + 22 + 12

6 − 12 + 4 − 10 = 4. 3

Hence, the equation of the sphere with radius 4 and center (3, 6, –4) is

( x − 3)2 + ( y − 6)2 + ( z + 4)2 = 42

or x 2 + y 2 + z 2 + −6 x − 12 y + 8 z + 45 = 0 . 14.20  EQUATION OF A SPHERE WHOSE DIAMETER IS THE LINE JOINING TWO GIVEN POINTS Let AB be the line joining A and B   two points a and with position vectors respectively. b     Thus, OA = a and OB = b . Let AB be the diameter of the sphere. Further, let r be the position vector of an arbitrary point P on the sphere.

M14_Baburam_ISBN _C14.indd 33













Then we have A P = r − a and BP = r − b .   Since, by property of a sphere, A P and BP are perpendicular to each other,     (1) ( r − a ) . r − b = 0, which is the required equation of the sphere in Vector form. Substituting

(

)

  r = xiˆ + yjˆ + zkˆ , a = a1iˆ + a2 ˆj + a3 kˆ , and  b = b1iˆ + b2 ˆj + b3 kˆ

in (1), we get

( x − a1 )( x − b1 ) + ( y − a2 )( y − b2 ) + ( z − a3 )( z − b3 ) = 0,

(4) which is the Cartesian form of the equation of a sphere with (a1 , a2 , a3 ) and (b1 ,b2 ,b3 ) as extremities of the diameter. 14.21  EQUATION OF A SPHERE PASSING THROUGH FOUR POINTS Let the equation of the sphere be x 2 + y 2 + z 2 + 2 gx + 2 fy + 2hz + c = 0 . (1) Let this sphere pass through four points ( x 1 , y 1 , z 1 ) , ( x 2 , y 2 , z 2 ) , ( x 3 , y 3 , z 3 ) , and ( x 4 , y 4 , z 4 ) . Thus,

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14.34  n  chapter fourteen x 12 + y 12 + z 12 + 2 gx 1 + 2fy 1 + 2hz 1 + c = 0, (2) x 22 + y 22 + z 22 + 2 gx 2 + 2fy 2 + 2hz 2 + c = 0, (3) x 32 + y 32 + z 32 + 2 gx 3 + 2fy 3 + 2hz 3 + c = 0,

(4)

and x + y + z + 2gx 4 + 2fy 4 + 2hz 4 + c = 0.   (5) Eliminating g, f, h, and c from (1), (2), (3), (4)and (5), we get 2 4

2 4

2 4

x 2 + y 2 + z 2 2x 2 y 2z

1

x 21 + y 21 + z 21 2x 1 2 y 1 2z

1

x +y +z

2 2

2x 2 2 y 2 2z 2 1 = 0.

x +y +z

2 3

2x 3 2 y 3 2z 3 1

2 2

2 2

2 3

2 3

x +y +z 2 4

2 4

2 4

2x 4 2 y 4 2z 4 1

EXAMPLE 14.66

Find the center and radius of the circle x 2 + y 2 + z 2 − 2 y − 4 z = 11 , x + 2y + 2z = 15. Solution. The center of the sphere is (0,1,2). The perpendicular distance from (0, 1, 2) to the plane x + 2y + 2z = 15 is

1(0) + 2(1) + 2(2) − 15 12 + 22 + 22

=

9 = 3. 3

Radius of the sphere is r=

g 2 + f 2 + h 2 − c = 1 + 4 + 11 = 4

O(0, 1, 2) 3

4

A

B

Therefore, the radius of the circle ( A B ) = (OB )2 − (OA )2 = 16 − 9 = 7

Since OA is normal to the plane, its direction ratios are 1, 2, and 2. Therefore, the equation of OA is x − 0 y −1 z − 2 = = . 1 2 2

M14_Baburam_ISBN _C14.indd 34

Any point on this line is (l, 2l + 1, 2l + 2). Since this point lies on the plane, we have l + 2(2l + 1) + 2(2l + 2) = 15 and so, l = 1. Therefore, the point A is (1, 3, 4). Hence, the center of the circle is (1, 3, 4) and its radius is 7 . EXAMPLE 14.67

Find the equation of the sphere whose diameter is join of the points A and B, with position vectors 2iˆ + 6 ˆj − 7 kˆ and −2iˆ + 4 ˆj − 3kˆ , respectively. Solution. The equation of the sphere in a Vector form is  r − 2iˆ + 6 ˆj − 7 kˆ  ⋅  r − −2iˆ + 4 ˆj − 3kˆ  = 0,    

(

)

(

)

whereas the Cartesian form of the equation of the sphere is (x – 2) (x + 2)+ (y – 6) (y – 4) + (z + 7) (z + 3) = 0 or x2 + y2 + z2 – 10y + 10z + 41 = 0, whose center is (–g, –f –h) = (0, 5, –5) and radius is g 2 + f 2 + h 2 − c = 3 . EXAMPLE 14.68

Find the equation of the sphere passing through the points (0, 0, 0), (–1, 2, 0), (0, 1, –1), and (1, 2, 5). Solution. Let the equation of the sphere be x 2 + y 2 + z 2 + 2 gx + 2 fy + 2hz + c = 0 Since it passes through (0, 0, 0), (–1, 2, 0), (0, 1, –1), and (1, 2, 5), we have c = 0, –2g + 4f = –5, 2f – 2h = –2, 2g + 4f + 10h = –30. Solving these equations, we get 3 5 5 f = − , g = − , and h = − . 2 2 2 Hence, the equation of the required sphere is x 2 + y 2 + z 2 − 5x − 5y − 3z = 0. EXAMPLE 14.69

A plane passes through a fixed point (a, b, c) and cuts axes in A, B, and C. Show that the locus of the center of the sphere OABC is a b c + + =2. x y z

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three-diMenSional geoMetry   n 14.35 Solution. Let the equation of the plane be

px + qy + rz = 1. (1) Since it passes through (a, b, c), we have ap + bq + cr = 1. (2) Putting y = z = 0 in (1), we get 1 px = 1 or x = . p 1 1 Similarly, y = q and z = r . Hence, the points A, B and C are 1   1  1  A  , 0, 0 , B  0, , 0 , and C  0, 0,  .  p   q  r Let the equation of the sphere OABC be x 2 + y 2 + z 2 + 2 gx + 2 fy + 2hz + c = 0 . (3) Since the origin O lies on the sphere, we have c = 0. Since A 1p , 0, 0 , B 0, q1 , 0 , and C ( 0, 0, 1r ) lie on the sphere, we have 1 1 2g + = 0 or 1 + 2 gp = 0 or g = − , 2 2p p p

(

) (

)

1 1 2f , and + = 0 or f = − 2q q q2 1 2h + = 0 or h = − 1 . r2 r 2r Putting these values of c, f , g , and h in (3), we get 1 1 1 x2 + y 2 + z 2 − x − y − z = 0 p q r The center of the sphere is  1 1 1   2 p , 2q , 2r  = (α , β , γ ) say. Then   1 1 1 p= ,q = , and r = . 2α 2β 2γ Thus, (2) reduces to a b c + + =2

α

β

γ

Therefore, the locus of the center is a b c + + =2 x y z EXAMPLE 14.70

x 2 + y 2 + z 2 − 3 x + 4 y + 5 z − 6 = 0, x + 2 y − 7 z = 0

lie on the same sphere and find the equation for the same. Solution. The equation of any sphere through the first circle i x 2 + y 2 + z 2 − 2 x + 3 y + 4 z − 5 + λ (5 y + 6 z + 1) = 0 (1) and the equation of any sphere through the second circle is x 2 + y 2 + z 2 − 3 x − 4 y + 5 z − 6 + µ ( x + 2 y − 7 z ) = 0 (2) The equations (1) and (2) can be written respectively as x 2 + y 2 + z 2 + −2 x + (5λ + 3) y + (6λ + 4) z + (λ − 5) = 0 (3) and

(4) The equations (3) and (4) will represent the same sphere if −2 = µ − 3,5λ + 3 = 2 µ − 4, 6λ + 4 = −7 µ + 5 , and λ − 5 = −6 From the last equation, we get λ = −1 and from the first equation, we get µ = 1 . These values of λ and µ satisfy the second- and third equation also. Hence, the given circles lie on the sphere x 2 + y 2 + z 2 + ( µ − 3) x + (2 µ − 4) y + (−7 µ + 5) z − 6 = 0

x 2 + y 2 + z 2 − 2 x + 3 y + 4 z − 5 − 1(5 y + 6 z + 1) = 0

or

x2 + y 2 + z 2 − 2x − 2 y − 2z − 6 = 0

EXAMPLE 14.71

Find the equation of the spheres passing through 2 2 2 the circle x + y + z − 6 x − 2 z + 5 = 0, y = 0 , and touching the plane 3 y + 4 z + 5 = 0 . Solution. The equation of any sphere through the given circle is x 2 + y 2 + z 2 − 6 x − 2 z + 5 + λ y = 0 (1)  −λ  ,1 . The Its center is (− g , − f , −h) =  3, sphere (1) will touch the plane 3 y2+ 4 z + 5 = 0

if the perpendicular distance from the center to the plane is equal to the radius of the sphere. Thus, we must have  −λ  3  + 4(1) + 5  2  = g 2 + f 2 + h2 − d 2 2 3 +4

Show that the two circles

x + y + z − 2 x + 3 y + 4 z − 5 = 0,5 y + 6 z + 1 = 0 and 2

2

2

M14_Baburam_ISBN _C14.indd 35



= 9+

λ2

λ2

+1− 5 = 5 + 4 4

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14.36  n  chapter fourteen Squaring yields 11 . 4 Substituting these values of λ in (1), we get the required equations of the spheres as x 2 + y 2 + z 2 − 6 x − 4 y − 2 z + 5 = 0 and 11 x2 + y 2 + z 2 − 6x − y − 2z + 5 = 0 4 4λ + 27λ + 44 = 0 , so that λ = −4, − 2

EXAMPLE 14.72

Find the equation of the sphere through the 2 2 2 circle x + y + z = 1, x + 2 y + 3z = 4 and the point (1, −1, 2) . Solution. The equation of the sphere through the given circle is x 2 + y 2 + z 2 − 1 + λ ( x + 2 y + 3z − 4) = 0 Since it passes through (1, −1, 2) , we have 1 + 1 + 4 − 1 + λ (1 − 2 + 6 − 4) = 0, which yields l = –5. . Hence, the equation of the sphere is x 2 + y 2 + z 2 − 1 − 5( x + 2 y + 3 z − 4) = 0 or

x 2 + y 2 + z 2 − 5 x − 10 y − 15 z + 19 = 0

14.22  EQUATION OF THE TANGENT PLANE TO A SPHERE A line which meets a sphere in two coincident points is called the tangent line to the sphere. The locus of the tangent lines at a point P of a sphere is a plane known as tangent plane to the sphere at P. Let A (x, y, z) be any point on the tangent plane at P (x1, y1, z1) to the sphere x2 + y2 + z2 + 2gx + 2fy + 2hz + c = 0. Then, the direction ratios of PA are x – x1, y – y1, and z – z1. Since the center of the sphere is (–g, –f, –h), the direction ratios of the radius OP are x1 + g, y1 + f, and z1 + h., Since OP is normal to the tangent plane at P1, we have OP ⊥ PA and so

( x1 + g )( x − x1 ) + ( y 1 + f )( y − y 1 ) + ( z 1 + h )( z − z 1 ) = 0

or or

xx1 + yy1 + zz1 + gx + fy + hz = x 21 + y 21 + z 21 + gx 1 + fy 1 + hz 1

M14_Baburam_ISBN _C14.indd 36

xx 1 + yy 1 + zz 1 + g ( x + x 1 ) + f ( y + y 1 ) + h ( z + z 1 )

= x 21 + y 21 + z 21 + 2 gx 1 + 2fy 1 + 2hz 1 = –c, since ( x1 , y 1 , z 1 ) lies on the sphere (1) or xx 1 + yy 1 + zz 1 + g ( x + x 1 ) + f ( y + y 1 ) + h ( z + z 1 ) + c = 0,

(2) which is the desired equation of the tangent plane at the point ( x 1 , y 1 , z 1 ) of the sphere. If the equation of the sphere is x 2 + y 2 + z 2 = a 2 , then the center is O (0, 0, 0) that is, g = f = h = 0. Therefore, (1) reduces to xx1 + yy1 + zz1 = x12 + y12 + z12 = a 2 , that is, (3) xx1 + yy1 + zz1 = a 2 ,

which is the required equation of the tangent plane. 14.23  CONDITION OF TANGENCY The plane ax + by + cz = D will be a tangent plane to the given sphere x 2 + y 2 + z 2 + 2 gx + 2fy + 2hz + c = 0, if the perpendicular distance of the plane from the center of the sphere is equal to the radius of the sphere, that is, if a ( − g ) + b( − f ) + c ( − h) − D a 2 + b2 + c2 or

=

(ag + bf

g 2 + f 2 + h2 − c ,

(

+ ch + D ) = a2 + b 2 + c 2 2

(

)

)

× g 2 + f 2 + h2 − c ,

which is the required condition of tangency.

14.24  ANGLE OF INTERSECTION OF TWO SPHERES The angle of intersection of two spheres is the angle between their tangent planes at a common point of intersection of the spheres. Also we know that the angle between two planes is the angle between their normals. It follows, therefore, that the angle between two spheres at a point of intersection P is equal to the angle between their radii at P.

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three-diMenSional geoMetry   n 14.37 EXAMPLE 14.73

P

1



c1

2 d

c2

Thus, if c1 and c2 are the centers and r1 and r2 are the radii, respectively, of the given spheres, then the ∠c1 Pc2 is the angle between the spheres at P. Let this angle be q. Then, the cosine formula, when applied to the triangle Pc1c2 , yields cos θ =

r 21 + r22 − d 2 2r1r2

,

where d is the distance between their centers. 14.25  CONDITION OF ORTHOGONALITY OF TWO SPHERES Let the spheres x 2 + y 2 + z 2 + 2 g1 x + 2 f1 y + 2h1 z + c1 = 0

(1)

and

x 2 + y 2 + z 2 + 2 g 2 x + 2 f 2 y + 2h2 z + c2 = 0 (2)

cut orthogonally as shown in the following figure P

1 c1

90°

2 d

c2

Then the angle c1 Pc2 is 90º. The centers of the spheres are c1 ( − g1 , −f 1 , − h1 ) and c 2 ( − g 2 , −f 2 , − h2 ) . The triangle Pc1c2 is a right2 angled triangle. Therefore, (c1c 2 ) = r12 + r22 or

( g1 − g 2 )2 + (f 1 − f 2 )2 + ( h1 − h2 )2

= g12 + f12 + h12 − c1 + g 22 + f 22 + h22 − c2

or

2 g1 g 2 + 2 f1 f 2 + 2h1h2 = c1 + c2

, which is the required condition of orthogonality.

M14_Baburam_ISBN _C14.indd 37

Find the equation of the sphere which is tangential to the plane x – 2y – 2z = 7 at (3, –1, –1) and passes through (1, 1, –3). Solution. Let the equation of the sphere be x2 + y2 + z2 + 2gx + 2fy + 2hz + c = 0. (1) Since it passes through (1, 1, –3), we have 2g + 2f – 6h + 11 + c = 0. (2) The equation of the tangent plane at (3, –1, –1) is 3x – y – z + g(x+3) + f(y – 1) + h(z – 1) + c = 0 or (g+3) x + (f–1) y + (h–1) z + 3g–f– h+c = 0. (3) But the given tangent plane is x – 2y – 2z – 7 = 0. (4) The equations (3) and (4) will be the equations of the same plane if g + 3 f − 1 h − 1 3g − f − h + c = = = = k , say. 1 −2 −2 −7 Then g = k – 3, f = –2k + 1, h = –2k + 1 (5) and 3g – f – h + c = –7k. (6) Substituting the values of g, f, and h from (5) in (2) and (6), we get 10k + 1 + c = 0 (7) and 14k – 11 + c = 0. (8) Subtracting (7) from (8), we get 4k – 12 = 0 and so, k = 3. Putting this value in (7) and (5), we get g = 0, f = –5, h = –5, and c = –31. Substituting these values in (1), the required equation of the sphere is x2 + y2 + z2 – 10y – 10z – 31 = 0. EXAMPLE 14.74

Show that the spheres x2 + y2 + z2 +6y + 2z + 8 = 0 and x2 + y2 + z2 + 6x + 8y + 4z + 20 = 0 intersect at right angles. Find their plane of intersection. Solution. Comparing with the standard form, we have g1 = 0, f1 = 3, h1 = 1 and c1 = 8 and g2 = 3, f2 = 4, h2 = 2 and c2 = 20. Therefore, 2g1 g2 +2f1 f2+2h1h2 = 0 + 24 + 4 = 28

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14.38  n  chapter fourteen and c1 + c2 = 8 + 20 = 28. Thus, the condition for orthogonality 2g1 g2 +2f1 f2+2h1h2 = c1 + c2 is satisfied. The plane of intersection of these spheres is x2 + y2 + z2 + 6y + 2z + 8 –( x2 + y2 + z2 + 6x + 8y + 4z + 20) = 0 or 3x + y + z + 6 = 0. EXAMPLE 14.75

If any tangent plane to the sphere x2 + y2 + z2 = r2 makes intercepts a, b, and c on the coordinate axes, show that 1 1 1 1 + + = . a 2 b2 c2 r 2 Solution. Let (x1, y1, z1) be any point on the given sphere x2 + y2 + z2 = r2. The equation of the tangent plane to the sphere at the point (x1, y1, z1) is x y z (1) xx1 + yy1 + zz1 = r 2 or r 2 + r 2 + r 2 = 1. x1 y1 z1 But the tangent plane to the given sphere makes intercepts a, b, and c, respectively, on the coordinate axes. Therefore, its equation is x y z (2) + + = 1. a b c Comparing (1) and (2), we get r2 r2 r2 = a, = b, and =c x1 y1 z1 or r2 r2 r2 x1 = , y1 = , and z1 = . a b c Since (x1,y1, z1) lies on the given sphere, we have x12 + y12 + z12 = r 2

or

r4 r4 r4 1 1 1 1 + + = r 2 or 2 + 2 + 2 = 2 . a 2 b2 c2 a b c r

EXAMPLE 14.76

Two spheres of radii r1 and r2 cut orthogonally. Show that the radius of the common circle is r1r2 . r12 + r22

M14_Baburam_ISBN _C14.indd 38

Solution. Let the common circle in the xy-plane be x2 + y2 + z2 = a2, z = 0, where a is the radius of the circle. The equations of any two spheres through this circle are x2 + y2 + z2 – a2 + k1z = 0 (1) and x2 + y2 + z2 – a2 + k2z = 0. (2) The spheres (1) and (2) will cut orthogonally if 2g1 g2 +2f1 f2+2h1h2 = c1 + c2, that is, if k k 2 1 . 2 = −a 2 − a 2 or if k1k2 = −4a 2 . (3) 2 2 On the other hand, the radius r1 and r2 of the spheres (1) and (2), respectively, are

r1 = g12 + f12 + h12 − c1 =

k12 + a2 . 4

r2 = g 22 + f 22 + h22 − c2 =

k22 + a2 . 4

and

Thus, k12 k2 + a 2 , r22 = 2 + a 2 , and k12 k22 = 16a 4 . 4 4 We have from the firs - and second equation, r12 =

k12 = 4(r12 − a 2 ) and k22 = 4(r22 − a 2 ).

Substituting these values in the third equation, we get 16 (r12 − a 2 ) (r22 − a 2 ) = 16a 4

or or

r12 r22 − a 2 (r12 + r22 ) + a 4 = a 4

a2 =

2 2 1 2 2 2 1 2

r r or a = r +r

r1r1 r12 + r22

.

EXAMPLE 14.77

Find the equation of the sphere which touches the plane 3x + 2y – z + 2 = 0 at the point (1, –2, 1) and cuts orthogonally the sphere x2 + y2 + z2 – 4x + 6y + 4 = 0. Solution. Let the equation of the required sphere be x2 + y2 + z2 + 2gx + 2fy + 2hz + c = 0. (1)

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three-diMenSional geoMetry   n 14.39 The equation of the tangent plane at (1, –2,1) to the sphere (1) is x − 2 y + z + g ( x + 1) + f ( y − 2) + h( z + 1) + c = 0

or

( g + 1) x + ( f −2) y + (h + 1) z



+ ( g − 2 f + h + c) = 0. But the tangent plane is given to be 3 x + 2 y − z + 2 = 0.

orthogonally if  3λ − 2   −4λ + 3   5λ − 4  2  (1) + 2   (2) + 2   (−3) 2 2      2  = 6 − 15λ + 11

(2)

or if λ = − 15 . Putting this value of l in (1), the equation of the required sphere is

(3)

EXAMPLE 14.79

Since (2) and (3) represent the same planes, we have g +1 f − 2 h +1 g − 2 f + h + c = = = = k , say. −1 3 2 2 This yields g = 3k – 1, f = 2k + 2, and h = – k – 1 (4) and g – 2f + h + c = 2k. (5) Since the sphere (1) cut orthogonally the sphere x2 + y2 + z2 – 4x + 6y + 4 = 0, we have 2g(–2) + 2f(3) +2h(0) = c + 4 or (6) –4f + 6f = c + 4. Substituting the values of g, f, and h from (4) in equations (5) and (6), we obtain –4k + c = 6 and c = 12, respectively. Therefore, k = 32 . 7 Thus, (4) yields g = 2 , f = 5, and h = − 52 . Substituting these values of g, f, h, and c in (1), the required equation of the sphere is

5( x 2 + y 2 + z 2 ) − 13 x + 19 y − 25 z + 45 = 0.

Show that the spheres x2 + y2 + z2 = 25 and x2 + y2 + z2 – 24x – 40y –18z + 225 = 0 touch externally. Find their point of contact and the equation of the common tangent plane. Solution. The centers of the given spheres are (0,0,0) and (12,20,9), respectively. Their radii are r1 = 5 and r2 = 144 + 400 + 81 − 225 = 20. If c1 and c2 are the centers of the spheres, then c1c2 = (12) 2 + (20) 2 + 92 = 25. Also,

rl + r2 = 5 + 20 = 25. Thus, c1c2 = r1 + r2 and so, the given spheres touch externally. P 5 c1(0, 0, 0)

20 c2(12, 20 ,9)

x 2 + y 2 + z 2 + 7 x + 10 y − 5 z + 12 = 0. EXAMPLE 14.78

Find the equation of a sphere which passes through the circle x2 + y2 + z2 – 2x + 3y – 4z + 6 = 0, 3x – 4y + 5z – 15 = 0 and cuts orthogonally the sphere x2 + y2 + z2 +2x +4y – 6z + 11 = 0. Solution. The equation of the sphere through the given circle is x2 + y 2 + z 2 − 2x + 3 y − 4z + 6



+ λ (3 x − 4 y + 5 z − 15) = 0

or

(1)

x 2 + y 2 + z 2 + (3λ − 2) x + (−4λ + 3) y + (5λ − 4) z + 6 − 15λ = 0.

It cuts the sphere

x + y + z + 2 x + 4 y − 6 z + 11 = 0 2

2

2

M14_Baburam_ISBN _C14.indd 39

If P is the point of contact, then P divides c1c2 internally in the ratio 5:20. Therefore, the coordinates of P are  60 − 0 100 − 0 45   12 9  , ,  or  , 4,  .  25 25  5  25  5 The equation of the tangent plane common to the given sphere is  12  9 x   + y (4) + z   = 25 5   5 or 12x + 20y + 9z = 125. EXAMPLE 14.80

Find the center and the radius of the circle x 2 + y 2 + z 2 − 8 x + 4 y + 8 z − 45 = 0,

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14.40  n  chapter fourteen = 0, x − 2 y + 2 z − 3 = 0. Solution. The center of the sphere is (–g, –f, –h) = (4,–2,–4). The radius of the sphere is

r=

g 2 + f 2 + z 2 − c = 16 + 4 + 16 + 45

=9 The distance of the plane from the center of the sphere is 4+ 4−8−3 d= = 1. 1+ 4 + 4 Therefore, the radius of the circle is a = r 2 − d 2 = 81 − 1 = 4 5. Further, the center of the circle is the foot of the perpendicular drawn from the center of the sphere to the plane. The equations of the perpendicular from the center are x−4 y+2 z+4 = = = λ , say. 1 −2 2 Any point on this perpendicular is (l + 4, –2l –2, 2l – 4). This point will be the foot of the perpendicular, that is, the center of the circle if it lies on the plane. Therefore, (λ + 4) − 2(−2λ − 2) + 2(2λ − 4) − 3 = 0. The above equation yields λ = 13 . Hence, the center of the circle is 2 2  1 (λ + 4, −2λ − 2, 2λ − 4) =  + 4, − − 2, − 4 3 3 3   13 8 10  =  , − , . 3 3 3 EXAMPLE 14.81

Find the equation of the tangent line to the circle x2 + y2 + z2 = 3, 3x – 2y + 4z + 3 = 0 at the point (1,1,–1). Solution. The equation of the sphere is x2 + y2 +

z2 = 3 and the equation of the plane is 3x – 2y + 4z + 3 = 0. The tangent plane to the sphere at (1,1,–1) is x(1) + y(1) + z(1) = 3 or x + y + z = 3. The tangent line is the intersection of the given plane 3x – 2y + 4z + 3 = 0 and the tangent plane x + y + z = 3. Taking z = 0, the point of

M14_Baburam_ISBN _C14.indd 40

intersection of these planes is given by x + y = 3 and 3x – 2y = –3. Solving the equations, we get x = 0.6 and y = 2.4. The direction ratios of the tangent line are given by x + y + z = 0 and 3x – 2y + z = 0. Thus, x y Z x y z = = or = = . 4 − 2 −3 − 4 −2 − 3 2 −7 −5 Hence, the tangent line through (0.6, 2.4, 0) with direction ratios 2, –7, and −5 is x − 0.6 y − 2.4 Z = = . −7 −5 2 14.26  CYLINDER The surface generated by a variable line which remains parallel to a fixed line and intersects a given curve (or touches a given surface) is called a cylinder. The variable line is called the generator, the fixed line is called the axis, and the given curve is called the guiding curve. 14.27  EQUATION OF A CYLINDER WITH GIVEN AXIS AND GUIDING CURVES Let the equation of the axis be x y z = = (1) l m n and let the equation of the guiding curve be ax 2 + 2 xy + by 2 + 2 gx + 2 fy + c = 0, z = 0. (2)

Let (x1,y1,z1) be any point on the cylinder. Then, the equation of the generator (being parallel to (1)) is x − x1 y − y1 z − z1 (3) = = . l m n The generator (3) meets the plane z = 0 at the point given by x1 − lzn1 , y1 − mzn1 , 0 . Since the generator intersect the conic (2), the point lz mz x1 − n1 , y1 − n1 , 0 lies on (2). Therefore,

(

(

)

)

2

lz  lz   y − mz1    a  x1 − 1  + 2h  x1 − 1   1  n n  n     2

mz  lz    +b  y1 − 1  + 2 g  x1 − 1  n n    

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three-diMenSional geoMetry   n 14.41 ( x − x1 ) 2 + ( y − y1 ) 2 + ( z − z1 ) 2 − ( AM ) 2 = r 2 . (1)

mz   +2 f  y1 − 1  + c = 0 n  

or

a (nx1 − lz1 ) 2 + 2h(nx1 − lz1 )(ny1 − mz1 ) +b(ny1 − mz1 ) 2 + 2 gn(nx1 − lz1 )



+2 fn (ny1 − mz1 ) + cn = 0. Thus, the locus of (x1,y1,z1) is a(nx – lz)2 + 2h(nx – lz) (ny – mz) + b(ny – mz)2 + 2gn(nx – lz) + 2fn(ny – mz) + cn2 = 0 which is the required equation of the cylinder. 2

14.28  RIGHT CIRCULAR CYLINDER A surface generated by a line which intersects a fixed circle (called the guiding curve) and is perpendicular to the plane of the circle is called the right circular cylinder. The normal to the plane of the circle through its center is called the axis of the cylinder and the radius of the circle is called the radius of the cylinder. To find the equation of a right circular cylinder with radius r, let A(x1,y1,z1) be a fixed point on the axis of the cylinder. Then, the equations of the axis are x − x1 y − y1 z − z1 = = . l m n The direction cosines of the axis are l l +m +n 2

2

2

m

,

l +m +n 2

P

2

r

2

, and

n l + m2 + n2 2

.

( AP ) − ( AM ) = ( PM )

or

M14_Baburam_ISBN _C14.indd 41

2

+( y − y1 )

=

m l + m2 + n2 2

+( z − z1 )

n l 2 + m2 + n2

1 l + m2 + n2 2

×[l ( x − x1 ) + m( y − y1 ) + n( z − z1 )] . Therefore, (1) reduce to ( x − x1 ) 2 + ( y − y1 ) 2 + ( z − z1 ) 2

−

[l ( x − x1 ) + m( y − y1 ) + n( z − z1 )]2 = r2 l 2 + m2 + n2

or

(l 2 + m 2 + n 2 )[( x − x1 ) 2 + ( y − y1 ) 2 + ( z − z1 ) 2 ]



−[l ( x − x1 ) + m( y − y1 ) + n( z − z1 )]2 = r 2 (l 2 + m 2 + n 2 ),

which is the required equation of the right circular cylinder. If the axis of the cylinder is z-axis, then A is (0, 0, z), l= 0, m = 0, and n = 1and so, the equation of the right circular cylinder reduces to x2 + y2 = r2. Similarly, the equation of a right circular cylinder whose axis is xl = my = nz is

A

(l 2 + m 2 + n 2 )( x 2 + y 2 + z 2 − r 2 ) = (lx + my + nz ) 2 .

M

EXAMPLE 14.82

Let P(x,y,z) be any point on the cylinder and let M be the foot of the perpendicular from P on the axis. Then, PM = r and 2

But AM = length of projection of AP on the axis of the cylinder l = ( x − x1 ) 2 l + m2 + n2

2

Find the equation of the right circular cylinder whose guiding curve is the circle passing through the points (1,0,0), (0,1,0), and (0,0,1). Solution. The equation of the plane passing through A(1,0,0), B(0,1,0), and C(0,0,1) is

x −1 y z −1 1 0 = 0 −1 0 1

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14.42  n  chapter fourteen

or x y z x + y + z = 1 or + + = 1. 1 1 1 Let O be the origin. Then OA = OB = OC = 1. Therefore, a sphere with center O (0, 0, 0) and radius unity passes through A, B, and C. Thus, the guiding curve is the circle x2 + y2 + z2 = 1, x + y + z = 1. The equation of the axis of the cylinder (which passes through (0, 0, 0) and is perpendicular to the plane x + y + z = 1) is x−0 y−0 z −0 x y z = = or = = . 1 1 1 1 1 1 The perpendicular distance from the center of the sphere x2+ y2 + z2 = 1 to the plane x + y + z = 1 is 1 1 p= = . 1+1+1 3

The equation of the generators through this point is x − x1 y − y1 z − z1 = = = λ , say. −1 1 1 Any point on the generator is (l + x1, –l+y1, l+z1). This point lies on the guiding curve if (λ + x1 ) 2 + (−λ + y1 ) 2 + (λ + z1 ) 2 = 9, (λ + x1 ) − (−λ + y1 ) + (λ + z1 ) = 3

or

x12 + y12 + z12 + 2λ ( x1 − y1 + z1 ) + 3λ 2 = 9, x1 − y1 + z1 + 3λ = 3.

x − y + z −3 . From second member, we have λ = −3 Substituting this value of l in the first member, we get 2( x1 − y1 + z1 − 3)( x1 − y1 + z1 ) x12 + y12 + z12 − 3 1

+

Therefore, the radius of the circle is or

Therefore, the equation of the right circular cylinder is (l2 + m2 + n2) (x2 + y2 + z2 – r2) = (lx + my + nz)2 or 2  (1 + 1 + 1)  x 2 + y 2 + z 2 −  = ( x + y + z ) 2 3  or 3x2 + 3y2 + 3z2 – 2 = x2+y2+z2+ 2xy + 2yz + 2xy or x2 + y2 + z2 – xy – yz – xy – 1 = 0.

Hence, the locus of (x1, y1, z1) is

Find the equation of the right circular cylinder whose guiding curve is x2 + y2 + z2 = 9, x – y + z = 3. Solution. Proceeding as in the above example, the required equation of the right circular cylinder is x2 + y2 + z2 + xy + yz + zx + 9 = 0. Second Method: The direction ratios of the plane x – y + z = 3 are 1, –1, and 1. Since the axis of the cylinder is perpendicular to the plane, its direction cosines are proportional to 1, –1, and 1. Let (x1, y1, z1) be any point on the cylinder.

M14_Baburam_ISBN _C14.indd 42

1

3( x1 − y1 + z1 − 3) 2 =9 9

1 2 r = (radius of the sphere) 2 − p 2 = 1 − = . 3 3

EXAMPLE 14.83

1

x12 + y12 + z12 + y1 z1 − z1 x1 + x1 y1 − 9 = 0. x 2 + y 2 + z 2 + yz − zx + xy − 9 = 0,

which is the required equation of the right circular cylinder. EXAMPLE 14.84

Find the equation of the right circular cylinder of radius 2 whose axis passes through (1, 2, 3) and has direction ratios (2,–3,6). Solution. The equation of the axis of the cylinder is x −1 y − 2 z − 3 = = . −3 2 6 Thus, the point on the axis is A(1, 2, 3) and l = 2, m = –3, n = 6, and r = 2. The standard equation of the right circular cylinder is (l 2 + m 2 + n 2 )[( x − x1 ) 2 + ( y − y1 ) 2 + ( z − z1 ) 2 ] −[l ( x − x1 ) + m( y − y1 ) + n( z − z1 )]2 = r 2 (l 2 + m 2 + n 2 ).

Putting the value of l, m, n, r, x1, y1, and z1, we get 49[( x − 1) 2 + ( y − 2) 2 + ( z − 3) 2 ]

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three-diMenSional geoMetry   n 14.43 −[2( x − 1) − 3( y − 2) + 6( z − 3)]2 = 4(49)

or

45 x 2 + 40 y 2 + 13 z 2 + 12 xy + 36 yz

−24 zx − 42 x − 112 y − 126 z − 392 = 0, which is the required equation of the right circular cylinder. EXAMPLE 14.85

Find the equation of the cylindery whose generators are parallel to the line 1x = −2 = 3z and whose guiding curve is the ellipse x2 + 2y2 = 1, z = 3. Solution. Let (x1, y1, z1) be any point on the cylinder. The generator is parallel to the line x = y = z . Therefore, the direction ratios of the 1 −2 3 generator are 1, –2, and 3. Thus, the equation of the generator is x − x1 y − y1 z − z1 (1) = = . −2 1 3 The generator (1) meets the plane z = 3 in the 3 − z1 2 , y = y1 − (3 − z1 ), z = 3. point x = x1 + 3 3 This point will lie on the guiding curve if 2 2 3 − z1  2     x1 + 3  + 2  y1 − 3 (3 − z1 )  = 1     or 3 ( x12 + 2 y12 + z12 ) + 8 y1 z1 − 2 z1 x1 +6 x1 − 24 y1 − 18 z1 + 24 = 0. Hence, the locus of (x1, y1, z1) is

3( x 2 + 2 y 2 + z 2 ) + 8 yz − 2 zx + 6 x − 24 y − 18 z + 24 = 0.

14.29 CONE A cone is a surface generated by a variable line which passes through a fixed point and intersects a given curve (or touches a given surface). The fixed point is called the vertex, the variable line is called the generator, and the given curve is called the guiding curve. 14.30  EQUATION OF A CONE WITH ITS VERTEX AT THE ORIGIN Let f(x,y,z) = 0 be the equation of a cone with its vertex at the origin and let P(x1, y1, z1) be any point on the cone. Therefore, f(x1, y1, z1) = 0. Let

M14_Baburam_ISBN _C14.indd 43

the vertex of the cone be O (0,0,0). Then, the equation of the generator OP is x−0 y−0 z−0 x y z (1) = = or = = x1 − 0 y1 − 0 z1 − 0 x1 y1 z1 Any point on the generator (1) is (lx1, ly1, lz1). Since the generator lies on the cone, the point (lx1, ly1, lz1) lies on the cone and so, f(lx1, ly1, lz1) = 0 for all values of l. But this is possible only if f(x, y, z) = 0 is homogeneous. It follows, therefore, that the equation of the cone f(x, y, z) = 0, with vertex at the origin, is a homogeneous equation in x, y, and z. Conversely, suppose that f(x, y, z) = 0 is homogeneous in x, y and z, and let P(x1, y1, z1) be a point on this surface. Then, f(x1, y1, z1) = 0 and due to homogeneity f(tx1, ty1, tz1) = 0 for any real t. In particular, for t = 0, we have f(0,0,0) = 0. Therefore, O(0,0,0) is a point on the locus of f(x, y, z) = 0. The equation of OP is x−0 y−0 z−0 x y z = = or = = . (1) x1 − 0 y1 − 0 z1 − 0 x1 y1 z1 Any point on this line is (tx1, tx2, tx3). Since f(tx1, tx2, tx3) = 0, the point (tx1, tx2, tx3) lies on the surface of f(x1, x2, x3) = 0. Therefore, the line (1) lies wholly on f(x, y, z) = 0. Thus, the surface f(x, y, z) = 0 is generated by a straight line through the fixed point O (0,0,0). Hence the homogeneous equation f(x, y, z) = 0 represents a cone with vertex at the origin. 14.31  EQUATION OF A CONE WITH GIVEN VERTEX AND GUIDING CURVE Let A(a, b, g) be the vertex and let the equation of the basic conic be ax 2 + 2hxy + by 2 + 2 gx + 2 fy + c = 0, z = 0. (1) The equation of the generator through the vertex (a, b, g) is x −α y − β z − γ (2) = = . l m n

This line meets the plane z = 0 at the point P α − lnγ , β − mnγ , 0 .

(

)

Since this point lies on the conic (1), we have 2

lγ  lγ   mγ   a  α −  + 2h  α −   β − n n  n  

mγ     + b β −  n   

2

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14.44  n  chapter fourteen lγ  mγ    (3) +2 g  α −  + 2 f  β −  + c = 0. n n    To eliminate l, m, and n from (2) and (3), we note that (2) gives l x −α m y−β = = and . n z −γ n z −γ

5l2 + 3m2 + n2 – 6mn – 4nl – 2lm = 0. (4) Eliminating l, m, and n from (2) and (4), we obtain 5(x–1)2 + 3(y – 2)2 + (z–3)2–6(y–2)(z – 3) – 4 (z – 3)(x – 1)–2(x – 1) (y – 2) = 0 or

Substituting these values of and in (3), we get 2   x −α  x − α  y−β  a α − γ  + 2h  α − γ  β − γ  z −γ  z − γ  z −γ   



l n

m n

2

  y−β  x −α  +b  β − γ  + 2g α − γ  − γ z z −γ      y−β  +2 f  β − γ +c = 0 z −γ   or

a (α z − γ x) 2 + 2h(α z − γ x)( β z − γ y ) + b( β z − γ y ) 2

+2 g (α z − γ x)( z − γ ) + 2 f ( β z − γ y )( z − γ ) + c( z − γ ) 2 = 0,

5 x 2 + 3 y 2 + z 2 − 6 yz − 4 zx − 2 xy

+6 x + 8 y + 10 z − 26 = 0.

EXAMPLE 14.87

The plane ax + b + cz = 1 meets the coordinate axes in A, B, and C. Show that the equation of the cone generated by the lines drawn from O to meet the circle ABC is b c c a a b yz  +  + zx  +  + xy  +  = 0. c b a c     b a Solution. The given plane meets the axes in A(a, 0, 0), B(0, b, 0), and C(0, 0, c). The equation of the sphere OABC is x 2 + y 2 + z 2 − ax − by − cz = 0. (1) Therefore, the circle ABC is y

x y z + + = 1. (2) a b c

which is the required equation of the cone.

x 2 + y 2 + z 2 − ax − by − cz = 0,

EXAMPLE 14.86

Since the vertex is the origin (0,0,0), the equation of the required cone is obtained by making equation (1) homogeneous with the y help of ax + b + cz = 1. Towards this task, we have x y Z x 2 + y 2 + z 2 − (ax + by + cz )  + +  = 0 a b c  or b c c a a b yz  +  + zx  +  + xy  +  = 0, c b a c     b a as the required equation of the cone.

Find the equation of the cone whose vertex is the point (1,2,3) and guiding curve is the circle x2 + y2 + z2 = 4, x+ y+ z =1. Solution. The equation of the guiding curve is x2 + y2 + z2 = 4, x+ y+ z =1. (1) The vertex is (1,2,3). The equation of the generators is x −1 y − 2 z − 3 = = = t , say. (2) l m n

Any point on the line (2) is (lt + 1, mt + 2, nt + 3). The line (2) will be a generator if it intersects the circle (1). Thus, the point (lt + 1, mt + 2, nt + 3) should satisfy (1). Therefore, (lt + 1)2 + (mt + 2)2 + (nt + 3)2 = 4 and lt + 1 + mt + 2 + nt + 3 = 1. (3) From the second member of (3), we have −5 t= . l +m+n Putting this value of t in the first member of (3), we have

M14_Baburam_ISBN _C14.indd 44

EXAMPLE 14.88

Find the equation of the cone whose vertex is the point (–1,1,2) and whose guiding curve is 3x2 – y2 = 1, z = 0. Solution. The equation of the guiding curve is 3x2 – y2 = 1, z = 0. (1) The equation of the generator through (–1,1,2) is x +1 y −1 z − 2 (2) = = . l m n

This line meets the plane z = 0 in the point

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three-diMenSional geoMetry   n 14.45 l m   n + 2l n − 2m   , , 0  or  −1 − 2 ,1 − 2 , 0  . − n n n n    

This point lies on (1), and so, 2 2 l  m  3  −1 − 2  − 1 − 2  = 1. n  n  But from (2), we have l x +1 m y −1 = = and . n z−2 n z−2

(3)

The section of a cone whose2 vertex is P and y2 guiding curve is the ellipse ax 2 + b2 = 1, z = 0 by the plane x = 0 is a rectangular hyperbola. Show that the locus of the vertex P is ax + y b+ z = 1. Solution. Let P(a, b, g) be the vertex of the cone. The equation of the generator through (a, b, g) is x −α y − β z − γ (1) = = . l m n This line meets the plane z = 0 in the point (α − lnγ , β − mnγ , 0 ) . This point will lie on the ellipse if 2

2

2

2

2

lγ  1  mγ   α −  + 2  β −  = 1. n n  b   

Putting these values of 2

and

m n

14.32  RIGHT CIRCULAR CONE A right circular cone is a surface generated by a straight line which passes through a fixed point and makes a constant angle with a fixed line through the fixed point. The fixed point is called the vertex of the cone, the fixed line is called the axis of the cone, and the constant angle is called the semivertical angle of the cone. Let P(x, y, z) be any point on the cone so that the semivertical angle ZOP = q. Z

P



O(0, 0, 0)

in (2), we get

Y

2

1  x −α  1  y−β  α −γ  + 2 β −γ  =1 2  z −γ  b  z −γ  a  or 1 1 (α z −γ x) 2 + 2 ( β z − γ y ) 2 = ( z − γ ) 2 , 2 a b

which is the equation of the cone with vertex (a, b, g). The section cone (1) by the plane

M14_Baburam_ISBN _C14.indd 45

α2 β2  2 βγ yz +  2 + 2 − 1 z 2 − + 2γ z − γ 2 = 0. (3) b b2 a 

(2)

From (1), we have l x −α m y−β = = and . n z −γ n z −γ l n

2

Hence, the locus of (a, b, g) is x2 y 2 z 2 x2 y 2 + z 2 + + = 1 or + = 1. a 2 b2 b2 a2 b2

EXAMPLE 14.89

2

γ 2 y2

The equation (3) will represent a rectangular hyperbola if coeff. of y2 + coeff. of z2 = 0, that is, if γ 2 α2 β2  + + − 1 = 0. b2  a 2 b2 

12 x 2 − 4 y 2 + z 2 + 4 yz + 12 zx + 4 z − 4 = 0.

1 a2

or b

Hence, (3) reduces to 2 2 x −1   y −1   − − 3  −1 − 2 1 2    =1 z−2  z−2  or

2

x = 0 is given by α 2 z2 1 + 2 (β z − γ y)2 = ( z − γ )2 b2 b

X

The direction ratios of OP are x – 0, y – 0, and z – 0 and the direction cosines of the axis (z–axis) of the cone are 0,0, and 1. Therefore, cos θ =

x(0) + y (0) + z (1) 0 + 02 + 12 x 2 + y 2 + z 2 2

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14.46  n  chapter fourteen = Thus, cos 2θ =

z x2 + y 2 + z 2

( x 2 + y 2 + z 2 )cos 2θ = (lx + my + nz ) 2 ,

.

z x +y +z or sec 2θ = . x2 + y 2 + z 2 z2 2

2

2

2

This yields (sec 2θ − 1) z 2 = x 2 + y 2 or x 2 + y 2 = z 2 tan 2θ , which is called the standard form of the right circular cone. 14.33  RIGHT CIRCULAR CONE WITH VERTEX (a, b, g), SEMI-VERTICAL ANGLE q, AND l , m, n THE DIRECTION COSINES OF THE AXIS. Let P(x, y) be any point on the right circular cone whose vertex is A (a, b, g) and the direction cosines of its axis AL are . Let q be the semi vertical angle of the cone. The direction ratios of AP are . Therefore, L

P(x, y, z)

 A(,  , )

cos θ =

( x − α )l + ( y − β )m + ( z − γ )n ( x − α )2 + ( y − β )2 + ( z − γ )2 . l 2 + m2 + n2

or cos 2θ =

[( x − α )l + ( y − β )m + ( z − γ )n]2 , ( x − α )2 + ( y − β )2 + ( z − γ )2

since l2 + m2 + n2 = 1, or [( x − α ) 2 + ( y − β ) 2 + ( z − γ ) 2 ]cos 2θ 2 (1) = [( x − α )l + ( y − β )m + ( z − γ )n] , which is the required equation of the cone.

Corollary 14.1: If the vertex is the origin A(0,0,0), then taking a = b = g = 0 in the equation (1), we have

M14_Baburam_ISBN _C14.indd 46

which the equation a right circular cone with vertex (0,0,0). EXAMPLE 14.90

Find the equation of the right circular cone which passes through the point (1,1,2) and has y vertex at the origin and axis x = = 3z . −4 2 Solution. The direction cosines of the axis are < l , m, n > =
. 29 29 29 We know that the equation of the cone through the origin, with semivertical angle q and direction cosines of axis l, m, and n, is (1) ( x 2 + y 2 + z 2 ) cos 2θ = (lx + my + nz ) 2 . ,

,

Since the cone passes through (1,1,2), we have 1 16 6 cos 2 θ = (l + m + 2n) 2 = (2 − 4 + 6) 2 = . 29 29 Thus, cos 2θ = 878 . Putting the values of l, m, n, and cos2q in (1), we get 8 x 2 + 8 y 2 + 8 z 2 − 3(2 x − 4 y + 3 z ) 2 = 0

or

8 x 2 + 8 y 2 + 8 z 2 − 3[4 x 2 + 16 y 2 + 9 z 2

−16 xy − 24 yz + 12 xz ] = 0 or

4 x 2 + 40 y 2 + 19 z 2 − 48 xy − 72 yz + 36 xz = 0,

which is the required equation of the cone. EXAMPLE 14.91

If a right circular cone has three mutually perpendicular generators, Show that the −1 semivertical angle is tan 2 . Solution. The equation, in a standard form, of a right circular cone with vertex at the origin, semivertical angle q, and axis along z-axis, is (1) x 2 + y 2 − z 2 tan 2θ = 0. The cone (1) will have three mutually perpendicular generators if coefficient of x2 + coefficient of y2+ coefficient of z2 = 0, that is, if 1 + 1 − tan 2θ = 0 or tan θ = 2 or θ = tan −1 2.

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three-diMenSional geoMetry  n 14.47 or

EXAMPLE 14.92

Find the equation of the right circular cone, whose vertex is the origin, axis is the line x = 2y = 3z , and which makes semivertical angle 1 of 30º. Solution. We know that the equation of the right circular cone with vertex (0,0,0), is (1) ( x 2 + y 2 + z 2 )cos 2θ = (lx + my + nz ) 2 , where q is the semivertical angle and < l, m, n > are direction cosines of the axis. The direction 3 2 , 14 > and q = cosines of the axis are < 114 , 14 30º. Therefore, (1) reduces to 3 1 ( x 2 + y 2 + z 2 ) = ( x + 2 y + 3z )2 4 14 or 42 ( x 2 + y 2 + z 2 ) = 4( x + 2 y + 3 z ) 2

or

−1

(9 − 4 3)cos 2θ = 1 or cos θ = (9 − 4 3) 2 . EXAMPLE 14.94

Find the equation of the right circular cone generated by rotating the line x = y = z , about 1 2 3 the line x = y = z . −1 1 2 y Solution. The equation of the axis is x = = z , −1

19 x 2 + 13 y 2 + 3 z 2 − 24 yz − 12 zx − 8 xy = 0,

If q is the semivertical angle of the right circular cone passing through the lines OX, OY, and x = y = z, show that −1

2 cos θ = 9 − 4 3  . Solution. Since the circular cone passes through OX, OY, and 1x = 1y = 1z , and all the above lines pass through (0,0,0), the origin is the vertex of the cone. Let be the direction cosines of the axis of the cone. The direction cosines of OX, OY, and the given line are respectively, , , and < 13 , 13 , 13 > . Since the axis makes the same angle q with each of OX, y OY, and 1x = 1 = 1z , we have

= l (0) + m(1) + n(0) =

l (1) + m(1) + n(1) 3

.

Therefore, l = cosq, m = cosq, and The last member n = 3 cos θ − l − n = 3 − 2 cos θ . yields Since l, m, and n are direction cosines, l2 +m2 + n2 = 1, that is cos 2θ + cos 2θ + ( 3 − 2) 2 cos 2θ = 1

(

M14_Baburam_ISBN _C14.indd 47

)

6 14

and so,

EXAMPLE 14.93

cos q = l(1) + m(0) + n(0)

2

7

=

which is the required equation of the cone.

1

whose direction cosines are < −61 , 16 , 26 > . The point of intersection of the rotating line x = y = z with the axis is clearly (0,0,0). 1 2 3 Therefore, the vertex of the cone is the origin (0,0,0). The angle q (semivertical angle of the cone) between the rotating line and the axis is given by 1(−1) + 2(1) + 3(2) cos θ = 2 1 + 22 + 32 (−1) 2 + 12 + 22

49 . 84 The equation of the right circular cone with vertex as the origin, semivertical angle q, and direction cosines of the axis as < l, m, n > is 2 (1) x 2 + y 2 + z 2 cos 2θ = (lx + my + nz ) . cos 2θ =

(

)

Putting the values of cos2 q and < l, m, n > = < −61 , 16 , 26 >, in (1), we get 49 1 2 x2 + y 2 + z2 = ( − x + y + 2z ) 84 6 or

(

)

(

)

49 x 2 + y 2 + z 2 = 14 ( − x + y + 2z )

or

2

5x2 + 5y2 – z2 + 4xy – 8yz + 8zx = 0.

14.34  CONICOIDS The locus of the general equation of the second degree ax2 + by2 + cz2 + 2fyz + 2gzx+ 2hxy + 2ux + 2vy + 2wz + d = 0 (1) in x, y, and z is called a quadric or conicoid. By the transformation of coordinate axes, the equation (1) can be transformed to any of

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14.48  n  chapter fourteen following important surfaces: y2

+ b 2 + = 1, surface is called the ellipsoid. y2 x2 z2 2. a2 + b 2 − c 2 = 1, surface is called the hyperboloid of one sheet. 1.

x2 a2

≠ d

2

3.

y − b 2 − zc 2 = 1, surface is hyperboloid of two sheets.

4.

y + b 2 = 2cz , surface is called the elliptic paraboloid.

5.

y − b 2 = 2cz , surface hyperbolic paraboloid.

6.

y + b 2 = 1 , surface is called the elliptic cylinder.

7.

− b 2 = 1, surface is called the hyperbolic cylinder.

x2 a2 x2 a2 x2 a2

x2 a2

x2 a2

2

called

the

2

2

is

called

the

2

y2

y2

+ b 2 = −1, surface is called the imaginary cylinder. A conicoid whose center is the origin, that is, whose equation remains unchanged on changing x to –x, y to –y, and z to –z, is called a central conicoid. For example, the ellipsoid, the hyperboloid of one sheet, and the hyperboloid of two sheets are central conicoids. 8.

x2 a2

lengths a, b, and c. These lengths are called the semiaxes of the ellipsoid. The largest of lengths, 2a, 2b, 2c is called the major axis, the next to largest is called the mean axis, and the smallest is called the minor axis of the ellipsoid. (iii) Its sections by the coordinate planes xy-, yz-, and zx-plane, respectively, are the ellipses x2 z 2 x2 y 2 y2 z2 + 2 = 1, 2 + 2 = 1 and 2 + 2 = 1 . 2 a c a b b c (iv) The plane section by the plane z = k (arbitrary fixed constant) are the ellipse x2 y 2 a2 − k 2 , z = k , and k < c. + = a 2 b2 c2 The size of the ellipses decreases as |k| increases. Similarly, the plane sections by the planes x = k and y = k are ellipses. In view of the above facts, the sketch of the surface is as shown below: z

O

14.35  SHAPE OF AN ELLIPSOID The equation of the ellipsoid is x2 y 2 z 2 (1) + + = 1. a2 b 2 c 2 We note that (i) The equation (1) is symmetrical with respect to all the three coordinate axes since the equation remains unaltered by changing x to –x, y to –y, and z to –z. (ii) It meets the x-axis at (a, 0, 0), (–a, 0, 0), the y-axis at (b, 0, 0), (–b, 0, 0), and the z-axis at (0, 0, c), (0, 0, –c). These six points are called the vertices of the ellipsoid. The three-line segments joining the pair of vertices on each coordinate axis are called the axes of the ellipsoid. The three axes intersect at the point (0,0,0), called the center of the ellipsoid. The line segments joining the center to the vertices have

M14_Baburam_ISBN _C14.indd 48

y

x

14.36  SHAPE OF THE HYPERBOLOID OF ONE SHEET The equation of the hyperboloid of one sheet is x2 y 2 z 2 (1) + − = 1. a 2 b2 c2

We observe that (i) The equation (1) is symmetrical with respect to all the three coordinate axes. (ii) The intersection of the surface with x- and y-axis are real, (± a, O, O) and (0, ±b, O) respectively; but the intersection with z-axis is imaginary. (iii) The plane section by the yz-plane is the

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three-diMenSional geoMetry  n 14.49 hyperbola

y2 z2 − = 1. b2 c2

Similarly,

the

plane sections by the zx- and xy-plane are respectively

2

2

x z − = 1 (hyperbola) and a2 c2

x2 y 2 (ellipse). + =1 a 2 b2 (iv) The plane sections by the plane z = k (plane parallel to xy-plane) are the ellipses x2 y 2 k2 , z = k + = 1 + a 2 b2 c2 The sizes of these ellipses increase as |k| increases. The size becomes infinite as | k |→ ∞ For k = 0, we get the smallest ellipse. The shape of the surface is as shown below:



x2 z 2 k2 − 2 = 1+ 2 , y = k , 2 a c b respectively. The plane x = k does not meet the surface if –a < k < a. Hence, the surface does not exist between the planes x = –a and x = a. But when k2 > a2, that is, when k ≥ a or k ≥ −a , the plane x = k cuts the surface in the ellipse y2 z2 k2 + = − 1, x = k . b 2 c 2 a2

The sizes of these ellipses increase as k2 increases. Hence, the sketch of the surface is as shown below. z

z

O

x

14.37  SHAPE OF THE HYPERBOLOID OF TWO SHEETS The equation of the hyperboloid of two sheets is x2 y 2 z 2 (1) − − = 1. a2 b 2 c 2 We note that (i) The surface is symmetrical about all the three coordinate axes. (ii) The surface meets the x-axis at the point (a, 0, 0) and (–a, 0, 0), whereas the y- and z-axis do not meet the surface at all. (iii) The section by the planes z = k and y = k are the hyperbolas

x2 y 2 k2 − 2 = 1 + 2 , z = k and 2 a b c

M14_Baburam_ISBN _C14.indd 49

x

O

y

y

14.38  SHAPE OF THE ELLIPTIC CONE The equation of the elliptic cone is x2 y 2 z 2 + − =0, a 2 b2 c2

(1) where a, b, and c are all positive. We observe that (i) The surface passes through the origin. (ii) The surface is symmetrical about all the three coordinate axes. (iii) The surface meets each of the three coordinate axes only at the origin. (iv) The

surface

(1)

meets

yz-plane

in

− zc 2 = 0 or = ± zc , which is a pair of straight lines through the origin. Similarly, y2

b2

2

y b

it meets zx- plane in a pair of lines 2

2

2

2

x a

= ± zc

y and the xy- plane in xa + b = 0, which is the

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14.50  n  chapter fourteen point ellipse x = 0 and y = 0 in the xy-plane. Thus, it meets the xy-plane in (0,0,0). (v) The surface (1) is generated by a variable curve. The plane z = k meets the surface (1) in the curve

x2 y 2 k2 + = , z = k. a2 b 2 c 2 Thus, the2 surface is generated by a variable 2 2 ellipse xa2 + by 2 = kc 2 .

(vi) The surface (1) extends to infinity both above and below the xy-plane. Hence, the shape of the surface is as shown below:

The line (3) is a quadratic in l and so, yields two values l1 and l2 of l. Hence, a line intersects a conicoid in two points which may be real and distinct, imaginary or coincident. If λ1 = λ2 , then these two points are coincident and so, the line becomes a tangent line to the conicoid. Thus, the condition that the line (1) is a tangent line to conicoid (2) is that the discriminant of (3) is equal to zero, that is 4(al α +bm β + cnγ )2 − 4(al 2 + bm 2 + cn 2 ) ×(aα 2 + bβ 2 + cγ 2 − 1) = 0

or

(alα +bmβ + cnγ ) 2 = (al 2 + bm 2 + cn 2 )(aα 2 + bβ 2 + cγ 2 − 1) .

z

14.40  TANGENT PLANE AT A POINT OF CENTRAL CONICOID

O

The locus of the tangent lines to a conicoid at a point on it is called the tangent plane at that point of the conicoid. Let (1) ax 2 + by 2 + cz 2 = 1

y

x

14.39  INTERSECTION OF A CONICOID AND A LINE Let the equation of given line and conicoid be respectively x −α y − β z − γ − = = λ , say m n l and ax 2 + by 2 + cz 2 = 1.

(1) (2)

Any point on the line (1) is (α + l λ , β + mλ , γ + nλ ) . This point will be on the conicoid (2) if a (α + l λ ) 2 + b( β + mλ ) 2 + c(γ + nλ ) 2 = 1

or

λ 2 (al 2 + bm 2 + cn 2 ) + 2λ (aα l + bβ m + cγ n) + aα 2 + b β 2 + cγ 2 − 1 = 0.

M14_Baburam_ISBN _C14.indd 50

(3)

be the equation of the central conicoid and let (a, b, g)be a point on it. Equation of any line through the point (a, b, g) is x −α y − β z − γ = = = λ , say. l m n Any point on this line is (α + l λ , β +, mλ ,γ + nλ ) . This point will lie on the conicoid (1) if 2 2 2 a ( α + l λ ) + b ( β + m λ ) + c (γ + n λ ) − 1 = 0 or λ 2 (al 2 + bm 2 + cn 2 ) + 2λ (aα l + b β m + cγ n ) + (3) aα 2 + b β 2 + cγ 2 − 1 = 0. Since (a, b, g) is a point on the conicoid, 2 aα 2 + bβ 2 + cγ = 1 and so, (3) reduces to λ 2 (al 2 + bm 2 + cn 2 ) + 2λ (aα l + b β m + cγ n ) = 0.

Thus, λ = 0 is one of the root, and the reduced equation is λ (al 2 + bm 2 + cn 2 ) + 2(aα l + b β m + cγ n ) = 0 and so,

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three-diMenSional geoMetry  n 14.51 2(aα l + b β m + cγ n ) . al 2 + bm 2 + cn 2 The line (2) will be the tangent line if both roots are equal. But as one of the root is zero, the second root should also be zero only. For this, we must have aa + bbm + cgn = 0. (4) Eliminating l, m, and n from (2) and (4), we get the locus of the tangent line as aα ( x − α ) + bβ ( y − β ) + cγ ( z − γ ) = 0

λ=−

or or

aα x + bβ y + cγ z = aα 2 + bβ 2 + cγ 2

aα x + bβ y + cγ z = 1 . (5) The equation (5) represents a plane, called the tangent plane to the conicoid ax 2 + by 2 + cz 2 = 1.

14.41  CONDITION OF TANGENCY Let (1) lx + my + nz = p be a plane. The equation of the tangent plane to the conicoid (2) ax 2 + by 2 + cz 2 = 1

at any point (a, b, g) is aα x + b β y + cγ z = 1. (3) The plane (1) will be a tangent plane to (2) if lx + my + nz − p ≡ aα x + b β y + c λ z − 1, that is, if

aα bβ cγ −1 = = = l m n −p

or if n l , β = m , and γ = . cp ap bp But (a, b, g) is a point on the conicoid (2). Therefore,

α=

2

or

2

2

 l  m  n a  + b   + c   = 1  ap   bp   cp  l 2 m2 n2 + + = p 2 , a b c

M14_Baburam_ISBN _C14.indd 51

which is the required condition of tangency. From equation (4), we have l 2 m 2 n2 + + . a b c 2 l2 m2 Thus, lx + my + nz = a + b + nc is always a 2 tangent plane to the conicoid ax + by 2 + cz 2 = 1 whatever be the values of l, m, and n. This shows that it is a family of tangent planes to the central conicoid ax 2 + by 2 + cz 2 = 1. p=±

14.42  EQUATION OF NORMAL TO THE CENTRAL CONICOID AT ANY POINT (a, b, g) ON IT A line passing through a point P on a surface, which is perpendicular to the tangent plane, to the surface at P is called the normal to the surface 2at P. 2 2 Let ax + by + cz = 1 be a central conicoid. Then, the equation of the tangent plane at (a, b, g) to this conicoid is aax + bby + cgz = 1. The direction cosines of the normal to this plane are proportional to aa, bb, and cg. Hence, the equation of the normal to the conicoid at (a, b, g) is x −α y −β z −γ = = . aα bβ cγ EXAMPLE 14.95

Show that from any point (f, g, h), six normals can be drawn to a conicoid. Solution. The equation of the normal at any point 2 2 2 (a, b, g) on the conicoid ax + by + cz = 1 is x −α y −β z −γ = = . aα bβ cγ It passes through (f, g, h) if f −α g −β h −γ = = = λ , say. aα bβ cγ This yields h f . , β = g , and γ = 1 + cλ 1 + aλ 1 + bλ Since (a, b, g) lies on ax 2 + by 2 + cz 2 = 1 , we get af 2 bg 2 ch 2 + + = 1, (1 + aλ )2 (1 + b λ )2 (1 + c λ )2

α=

(4)

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14.52  n  chapter fourteen which is of sixth degree in l and so, yields six values of l. Hence, there are six normals that can be drawn from a point (f, g, h) to the conicoid. EXAMPLE 14.96

What surfaces are represented by the following equations? x2 y 2 2 2 2 − = z. (i) z = 4 (1 + x + y ) and (ii) 2 3 Solution. (i) The given equation can be written as

4 x 2 + 4 y 2 − z 2 = −4

x2 y 2 z 2 x2 y 2 z 2 + − = −1 or + + = 1, 1 1 4 −1 −1 4 which is the hyperboloid of two sheets. (ii) Taking a = 2 , b = 3 , and c = 2 , the given equation can be written as x2 y 2 2z , − = c a 2 b2 which is the hyperbolic paraboloid.

or

EXAMPLE 14.97

Show that the equation of the tangent plane to x2 y 2 z 2 the ellipsoid 2 + 2 + 2 = 1 at the point (a, b, a b c αx βy γ z g) on it is 2 + 2 + 2 = 1. a b c Solution. We know that the equation of the tangent 2 2 2 plane to the conicoid Ax + By + Cz = 1 at the point (a, b, g) is A α x + B β y + C γ z = 1. (1) 2 y x2 z2 But for the ellipsoid a2 + b 2 + c 2 = 1, we have 1 1 1 A = 2 , B = 2 , and C = 2 . b c a Substituting the values of A, B, and C in (1), we get the equation of the tangent plane as αx β y γ z + + =1. a 2 b2 c2

EXAMPLE 14.98

Find the equation of two tangent plane passing through the line 7x + 10y – 30 = 0, 5y – 3z = 0 and which touch the conicoid 7x 2 + 5y 2 + 3z 2 − 60 = 0.

M14_Baburam_ISBN _C14.indd 52

Solution. The equation of the conicoid is

x 2 + 605 y 2 + 603 z 2 = 1. Comparing it with the standard equation of the central conicoid ax 2 + by 2 + cz 2 = 1 , we have a = 607 , b = 121 , and c = 201 . Now, the equation of any plane through the given line is 7x + 10y – 30 + l(5y – 3z) = 0 or 7 60

7x + (10 + 5l) y – 3lz – 30 = 0. (1) Comparing this with lx + my + nz = p, we get l = 7, m = 10 + 5l, n = –3l, and p = 30. The condition for tangency is l 2 m2 n2 + + = p2 a b c and so,

49 7 60

2 2 10 + 5λ ) ( −3λ ) ( + + 1 12

1 20

= (30 )

2

or 2l 2 + 5l + 3 = 0. Solving this quadratic, we get l = –1 and − 23 . Substituting these values of l in (1), we get 7x + 5y + 3z – 30 = 0 and 14x + 5y + 9z – 60 = 0 as the equations of the two tangent planes. EXAMPLE 14.99

Show that the plane lx + my + nz = p x2 y 2 z 2 touches the ellipsoid + 2 + 2 = 1 if 2 2 2 2 2 2 2 2. a b c p = a l +b m +c n Solution. The plane lx + my + nz = p touches the conicoid

Ax 2 + By 2 + Cz 2 = 1 if

Here,

l 2 m 2 n2 + + = p 2 . (1) A B C

1 1 1 , B = 2 , and C = 2 . 2 a b c Therefore, the condition of tangency (1) becomes a2 l2 + b2 m2 + c2n2 = p2. A=

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three-diMenSional geoMetry   n 14.53 EXAMPLE 14.100

Tangent planes are drawn to the ellipsoid y2 z2 x2 = 1 through the point (a, b, 2 + 2 + c2 a b g). Show that the perpendiculars to these planes through the origin generate the cone (ax + β y + γ z )2 = a2 x 2 + b 2 y 2 + c 2 Solution. The equation of the plane through the

Hence, the given plane touches the given conicoid. Now let P (x1, y1, z1) be the point of contact. The equation of tangent plane to the conicoid at (x1, y1, z1) is xx1 – 2yy1 + 3zz1 = 2.

point (a, b, g) is

Comparing it with x + 2y + 3z = 2, we have

l (x – a) + m (y – b) + n (z – g) = 0 or lx + my + nz = al + bm + g n. (1) By the condition of tangency, this plane will touch ellipsoid if (see Example 14.99)

x1 −2 y1 3 z1 2 = = = or x 1 = y 1 = z 1 = 1. 1 2 3 2 1 −1 1 Therefore, x1 = 1, y1 = –1, and z1 = 1. Hence, the point of contact is (1, –1, 1).

(α l + β m + γ n ) = a 2 l 2 + b 2 m 2 + c 2 n 2 . (2) On the other hand, the equation of the perpendicular to the plane (1) passing through the origin is

EXAMPLE 14.102

2

x y z = = . (3) l m n Eliminating l, m, and n from (2) and (3), we get (α x + β y + γ z )2 = a2 x 2 + b 2 y 2 + c 2 z 2 , which is the required equation of the cone. EXAMPLE 14.101

Show that the plane x + 2y + 3z = 2 touches the conicoid x2 – 2y2 + 3z2 = 2. Find also the point of contact. Solution. Equation of the given plane is x + 2y + 3z = 2. Comparing it with lx + my + nz = p, we have l = 1, m = 2, n = 3, and p = 2. The equation of the given conicoid is x2 – 2y2 + 3z2 = 2 or x2 − y 2 + 23 z 2 = 1. Comparing with the standard 2 equation of the central conicoid ax2 + by2 + cz2 = 1, we have a = 12 , b = −1, and c = 23 . The given plane will touch the given conicoid if l 2 m2 n2 + + = p 2 (Condition of tangency) a b c or if 1 4 9 + + 3 = (2)2 1 − 1 2 2 or if 2 – 4 + 6 = 4, which is true.

M14_Baburam_ISBN _C14.indd 53

Show that the locus of the foot of the perpendicular drawn from the center of the 2 2 2 ellipsoid xa2 + by 2 + zc 2 = 1 to any of its tangent planes is

(

)

a2 x 2 + b 2 y 2 + c 2 z 2 = x 2 + y 2 + z 2 .2 Solution. Let P ( x 1 , y 1 , z 1 ) be the foot of the perpendicular from the center (0,0,0) of the given ellipsoid to any of its tangent plane. The equation of the plane to the ellipsoid at ( x1 , y 1 , z 1 ) is

l x − x1 ) + m ( y − y1 ) + n ( z − z1 ) = 0. (1) ( The direction ratios of the perpendicular line OP are < x 1 − 0, y 1 − 0, z 1 − 0 > . Since OP is perpendicular to the plane (1), we have l = x1 , m = y1 , and n = z1 . Therefore, (1) transforms to x1 ( x − x1 ) + y 1 ( y − y 1 ) + z 1 (z − z 1 ) = 0 or

xx 1 + yy 1 + zz 1 = x 21 + y 21 + z 21 .

(2)

Since the plane (2) touches the ellipsoid, the tangency condition a 2 l 2 + b 2 m 2 + c 2 n 2 = p 2 implies

(

a2 x 12 + b 2 y 12 + c 2 z 12 = x 12 + y 2 + z 2

)

2

.

Therefore, the locus of the foot of the perpendicular is

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14.54  n  chapter fourteen

(

a2 x 2 + b 2 y 2 + c 2 z 2 = x 2 + y 2 + z 2

)

2

.

14.43  MISCELLANEOUS EXAMPLES EXAMPLE 14.103

Find the equation of the plane parallel to 2x – 3y + z = 6 and which passes through the point (1,–2,1). Solution. The equation of the plane parallel to 2x – 3y + z = 6 is 2x – 3y + z + k = 0. Since it passes through (1, –2,1), we have 2 +6 + 1 + k = 0 or k = –7. Hence the equation of the required plane is 2x – 3y + z – 7 = 0. EXAMPLE 14.104

Find the equation of the plane which passes through (2, 1, 3) and which contains the line x = y2−1 = z−+21 . 1 Solution. The equation of the plane containing the given line is a(x – 0) + b (y – 1) + c (z + 1) = 0. Since the plane passes through (2, 1, 3), we have 2a + 0b + 4c = 0. (1) Also, direction cosines of the line are proportional to 1, 2 and −2 and so a + 2b – 2c = 0. (2) From (1) and (2), we have a b c = = 8 −8 −4 or a b c = = . 2 −2 −1 Hence the equation of the required plane is 2 (x – 0) + 2 (y – 1) – 1 (z + 1) = 0 or 2x – 2y – z + 1 = 0. EXAMPLE 14.105

Find the image of the point (4,5,–2) in the plane x – y + 3z – 4 = 0. Solution. Let (x1, y1, z1) be the image of the point P(4,5,–2) in the plane x – y + 3z – 4 = 0. Then the direction ratios of PQ are x1 – 4, y1 – 5, z1 + 2. Since PQ must be perpendicular to the given plane, we have

M14_Baburam_ISBN _C14.indd 54

x1 − 4 y1 − 5 z1 + 2 = = = λ , say. 1 −1 3 \ Therefore Q = (l + 4, – l + 5, 3l – 2). Then midpoint M of PQ is

( λ + 24 + 4 , −λ +25 + 5 , 3λ −22 − 2 ) = ( λ 2+ 8 , λ +210 , 3λ2− 4 ). But M lies on the given plane. Therefore, λ + 8  −λ + 10   3λ − 4  −  + 3 −4 = 0 2  2   2  or or

l + 8 + l – 10 + 9l – 12 – 8 = 0

11l – 22 = 0 and so l = 2. Hence the image Q of P is Q = (6,3,4) EXAMPLE 14.106

Find the equation to the plane parallel to x + y – z + 1 = 0 and is 3 units from the origin. Solution. The equation of the plane parallel to the plane x + y – z + 1 = 0 is p x + y – z = p, where 1+1+1 = perpendicular distance from the origin to the parallel plane = 3 (given). Thus p = 3 3 = 3. Hence the equation of the parallel plane is x + y – z = 3.

( )

EXAMPLE 14.107

y+7 Find the image of the line x + 5 = 6 = z in the plane 2x – y + z + 3 = 0. Solution. Let P(p,q,r) be any point on the given y+7 line x 1+ 5 = 6 = 1z . The equation of the line through P and perpendicular to the given plane 2x – y + z + 3 = 0 is x− p y−q z−r = = = λ , say. −1 2 1 Any point on this line is Q (2l + p, – l + q, l + r). If this point is image of P, then the midpoint M of PQ is M 2 λ +2 2 p , − λ 2+ 2 q , λ +22 r . Then M lies on the plane 2x – y + z + 3 = 0 and so  −λ + 2q   λ + 2r  2λ + 2 p −  + +3= 0 2   2   or 3l + 2p – q + r + 3 = 0, which yields

(

)

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three-diMenSional geoMetry  n 14.55

λ=

or

−2 p + q − r − 3 . 3

EXAMPLE 14.109

Putting this value of l in the coordinates of Q, we get  − p + 2q − 2r − 6 2 p + 2q + r + 3 Q , , 3 3  −2 p + q + 2r − 3   . 3

Two points on the given line are P1 (–5,–7,0) and P2(–4, –1,1). Their images are Q1 (–5,–7,0) and Q2(–2, –2,2) Hence equation of the image line (using two point formula) is x+5 y+7 z = = −2 + 5 −2 + 7 2 or

x+5 y+7 z = = . 3 5 2

Find the length and equations of the shortest distance between x +1 y +1 z +1 x +1 y z = = = = . and 2 3 4 3 4 5 Solution. Let PQ be the shortest distance with l, m, n as its direction cosine. Then PQ is perpendicular to both lines. Therefore 2l + 3m + 4n = 0 and 3l + 4m + 5n = 0. Solving these equations, use get l 2 + m2 + n2

=

1

. 1+ 4 +1 6 The given lines pass through (–1,–1,–1) and (–1,0,0). Therefore the shortest distance between them is PQ = (−1 + 1)

1 6

+ (0 + 1)

1 6

+ (0 + 1)

1 6

=

2 6

.

The equations of the line of shortest distance are x y z x y Z 2 3 4 = 0 and 3 4 5 = 0 −1 2 −1 −1 2 −1

M14_Baburam_ISBN _C14.indd 55

Find the image of the point (1, –1, 2) in the plane 2x + 2y + z = 1. Solution. Let Q(x1, y1, z1) be the image of the point

P(1,–1, 2) in the plane 2x + 2y + z –1 = 0. Then the direction rations of PQ are x1 – 1, y1 + 1, z1 – 2. Since PQ is perpendicular to the given plane, we have x1 − 1 y1 + 1 z1 − 2 = = = λ , say. 2 2 1 Thus Q is (2l + 1, 2l – 1, l + 2). Now midpoint M of PQ is ( 2 λ +21+1 , 2 λ −21−1 , λ +22 + 2 ) . But M (l + 1, l – 1, λ2 +2) lies on the given plane. Therefore λ  2(λ + 1) + 2(λ − 1) +  + 2  − 1 = 0, 2   which yields λ = − 92 . Hence the point Q is ( 95 , −913 , 149 ) .

EXAMPLE 14.108

l m n = = = −1 2 −1

11x + 2y + 7z = 0 and 7x + y – 5z = 0

EXAMPLE 14.110

Find the magnitude and equations of the shortest y −2 distance between the lines x2−1 = 3 = z −4 3 and y − 4 x−2 = 4 = z −5 5 . 3 Solution. The shortest distance is

. Also, using two point formula, the equation of the line is 1 6

x − 53 y − 3 z − 133 = 10 = 25 13 3 5 2 − 3 3 −3 6 − 3 or

6 x − 10 3 y − 9 6 z − 26 = = . −1 −1 1

EXAMPLE 14.111

Find the equation of the sphere whose centre is the same as that of the sphere x2 + y2 + z2 – 2x –4y – 6z + 7 = 0 and which passes through (1,–1,1). Solution. The center of the given sphere is (–g, –f, –h) = (1,2,3). The equation of the sphere with centre (1, 2, 3) is x2 +y2 + z2 –2x –4y –6z + k = 0. Since it passes through (1,–1,1), we have

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14.56  n   chapter fourteen 1 + 1 + 1 – 2 + 4 – 6 + k = 0 or k = 1. Hence the required equation of the sphere is x2 +y2 + z2 –2x –4y –6z + 1 = 0. EXAMPLE 14.112

Find the equation of the tangent planes to the sphere x2 +y2 + z2 –2x –4y –6z – 2 = 0 which passes through the line 9x – 3y + 25 = 0 = 3x + 4z + 9. Solution. Taking z = 0, we have 9x – 3y + 25 = 0, 3x + 9 = 0. These equations give x = –3, y = − 32 , z = 0. The equation of the sphere is x2 + y2 + z2 – 2x – 4y – 6z – 2 = 0. Comparing it with x2 + y2 + z2 – 2gx + 2fy + 2hz + c = 0, we get g = – 1, f = –2, h = –3, c = –2. The equation of the tangent plane at (x1, y1, z1) is xx1 + yy1 +zz1 + g(x + x1) + f(y + y1) + h (z + z1) + c = 0. Therefore the equation of the required plane is 2  2  x(−3) + y  −  + z (0) − ( x − 3) − 2  y −  –3z–2 = 0 3  3 

or –9x – 2y – 3(x – 3) – 2(3y – 2) – 9z – 6 = 0 or 12x + 8y + 9z – 7 = 0. EXAMPLE 14.113

Find the centre, radius and area of the circle which is the intersection of the sphere x2 +y2 + z2 – 8x + 4y + 8z = 45 and the plane x – 2y + 2z = 3. Solution. The given sphere is x2 +y2 + z2 – 8x + 4y + 8z – 45 = 0. Its centre is (–g, –f, –h) = (4, –2,–4). Further, the length of the perpendicular from (4, –2,–4) to the plane x – 2y + 2z = 3 is p=

4(1) − 2(−2) − 4(2) − 3 1+ 4 + 4

=1

The radius of the sphere is a=

g 2 + f 2 + h 2 − c = 16 + 4 + 16 + 45 = 9.

Therefore the radius r of the circle is r = a 2 − p 2 = 81 − 1 = 80 = 4 5.

The area of the circle is A = pr2 = 80p. Further, the perpendicular from (4, –2,–4), being perpendicular to the plane, has direction ratios 1, –2, 2. Therefore the equation of the

M14_Baburam_ISBN _C14.indd 56

perpendicular is

x−4 y+2 z+4 = = = λ , say. 1 2 −2

Any point on this line is (l + 4, –2l –2, 2l – 4). Since this point lies on the plane, we have l + 4 – 2(–2l –2) + 2(2l – 4) = 3, and so λ = 13 . Hence centre of the circle is (l + 4, –2l –2, 2l – 4) = ( 133 , −38 , −310 ) . EXAMPLE 14.114

The point (2, 3, 4) is one end of the diameter of a sphere x2 +y2 + z2 – 2x – 2y + 4z – 1 = 0, find the other end. Solution. The equation of the sphere is x2 +y2 + z2 – 2x – 2y + 4z – 1 = 0 (1) and (2, 3, 4) is one of the ends of the diameter of (1). Let the other end of the diameter be (x, y, z). The centre of the sphere (1) is (–g, –f, –h) = (1,1,–2). Since centre is the midpoint of the y +3 diameter, we have x +2 2 = 1, 2 = 1 and z +2 4 = −2. Solving these equations, we get the other end of the diameter as (x,y,z) = (0,–1,–8). EXERCISES Direction Ratios and Direction Cosines 1. The vertices of a triangle are A (1,3, 2), B (2, −1,3) and C (−1, 4, 2). Find the direction cosines of the vectors AB and  AC and find the angle A Ans.
,
, A = cos −1 −

2 5

)

whether

the points and R (−8,5,5) are collinear. Find the ratio in which R divides PQ. Ans. Collinear, R divides PQ externally in the ratio 2:1. P (2, −1, 3), Q (−3, 2, 4)

3. A line makes angles α, β, γ and δ with four diagonals of a cube. Show that 4 3 4. Find the ratio in which the line joining (2, 4,16) and (3,5, −4) is divided by the plane 2 x − 3 y + z + 6 = 0 Ans. cos 2 α + cos 2 β + cos 2 γ + cos 2 δ =

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three-diMenSional geoMetry  n 14.57

Hint: See Example 14.9. Ans. 2:1. ˆ ˆ ˆ ˆ 5. Show that the points i − 2 j + 3k , 2i + 3 ˆj − 4kˆ and −7 ˆj + 10kˆ are collinear. 6. Show that the vectors   a = 3iˆ + ˆj − 2kˆ, b = iˆ + 3 ˆj + 4kˆ, and  c = 4iˆ − 2 ˆj − 6kˆ form the sides of a triangle. 7. Find the coordinates of the foot of the perpendicular drawn from the point A (1,8, 4) to the line joining B (0, −1,3) and C (2, −3, −1) . Ans. ( − 53 , 32 , 193 ) 8. Find the angle between the line of intersection of the planes 3 x + 2 y + z = 5 and x + y − 2 z = 3 and the line of intersection of the planes 2x = y + z and 7 x + 10 y = 8 z .

Hint: Planes intersect in a line (see Example 14.19). Ans. 90º 9. Find the image of the point ( p, q, r ) in the plane 2 x + y + 2 = 6 . Ans.

(

1 3

[12 − p − 2q − 2r ], 13 [6 − 2 p + 2q − r ], 13 [6 − 2 p − q + 2r ])

10. Find the distance of the point (3, −4,5) from the plane 2 x + 5 y − 6 z = 16 , measured y parallel to the line 2x = 1 = −z2 .

Hint: Line through (3, −4,5) having direction y+4 ratios 2, 1, − 2 is x 2− 3 = 1 = z−−25 = λ. . Any point on this line is (2λ + 3, λ − 4, −2λ + 5) This point lies on the given plane if 2 (2λ + 3) + 5 (λ − 4) − 6 (−2λ + 5) = 16 that is, if λ = 207 . Thus, the point is ( 617 , −78 , −75 ) The distance between (3, −4,5) and this point is

( 7 − 3) + ( 7 + 4) + ( 7 − 5) = 7 = 8.571. 11. Find the angle between the lines, whose direction cosines are given by 3l + m + 5n = 0 and 6mn − 2nl + 5lm = 0 61

2

M14_Baburam_ISBN _C14.indd 57

−8

2

−5

2

60



Hint: Solving the given equations, we get two sets of the direction cosines < 1, 2, −1 > and < −2,1,1 > . Therefore, 1( −2) + 2(1) + ( −1)(1) cos θ = 1+ 4 +1 4 +1+1 = 16 . or cos −1  1  . 6 12. Find the perpendicular distance of the point (2, 4, −1) from the line x + 5 = y + 3 = z − 6 −9 1 4 Hint: See Example 14.17. Ans. 7. 13. Find the foot of the perpendicular from the point (0, 2,3) to the line x +5 3 = y2−1 = z +3 4 . Also find the length of the perpendicular Ans. (2,3, −1) , Length 21. 14. Find the y +1 x−2 3 = −2 =

z 2

angle between the lines y +3 and x1−1 = 3 = z +2 5 . 1 Ans. θ = cos −1 ( 238 ).

A(4,5,10), B(2,3, 4) 15. The points and C(1, 2, −1) are three vertices of a parallelogram ABCD. Find vector equation for the side AB and the coordinates of D.        Ans. r = (4i + 5 j + 10k ) + λ( −2i + 2 j − 6k ), D(3, 4,5) 16. Determine the value of α for which the y− β z −γ lines x −1 α = −2 = 0 and 1x = y−−13 = z +2 2 intersect. Ans. 54 . 17. Find the shortest distance between the        lines r = 3i + 8 j + 3k + λ (3i − j + k ) and        r = −3i − 7 j + 6k + µ (−3i + 2 j + 4k ) Ans. 3 30 . 18. Find the shortest distance between the lines

x −3 y −8 z −3 = = −1 3 1

x+3

y+7

z−6

= = and −3 2 4 Also find the equation of the line of shortest distance. y −8 Ans. 3 30 , Equation: x 2− 3 = 5 = z−−13 .

19. Find the shortest distance between the lines  r = (3 − t )iˆ + (4 + 2t ) ˆj + (t − 2)kˆ and  r = (1 + s )iˆ + (3s − 7) ˆj + (2 s − 2)kˆ, where t and s are scalars. Ans. 35 .

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14.58  n  chapter fourteen 20. Find the shortest distance between the lines x−5 y −4 z −4 = = 1 1 −2

and

are coplanar. Further,

x −1 y + 2 z + 4 = = −6 7 1

Ans. 2 29 .

21. Determine whether or not the following pair of lines intersect:  r = iˆ − 2 ˆj + 3kˆ + λ (−iˆ + ˆj − 2kˆ) and  r = iˆ − ˆj − kˆ + µ (iˆ + 2 ˆj − 2kˆ) Ans. Do not intersect. 22. Show that the lines, x+5 y+4 z −7 = = ,3 x + 2 y + z = 2, a n d −2 3 1 x − 3 y + 2 z = 13 are coplanar. Hint: Show that these lines intersect, that is, the shortest distance between them is zero. 23. Find the condition that the y− β z −γ x−α lines and and l = m = n ax + by + cz + d = 0 = a ′x + b′y + c′z + d ′ are coplanar. aα + bβ + cγ + d a ′α + b ′β + c ′γ + d ′ Ans. al + bm + cn = a ′l + b ′m + c ′n . 24. Find the magnitude and the equation of the shortest distance between the lines x y z x − 2 y −1 z + 2 and . = = = = −5 2 −3 1 3 2 Hint: Proceed as in Example 14.23. Ans. 13 , 4 x + y − 5 z = 0 = 7 x + y − 8 z − 31







2 −1 3 − 2 4 − 3 1 1 1 = 2 3 4 = 2 3 4 =0 3 4 5 3 4 5 Therefore, the given lines intersect and so,

M14_Baburam_ISBN _C14.indd 58

l2

m2

z − z1 x −1 y − 2 z − 3 n1 = 2 3 4 n2

3

4

5

=0 yields ( x − 1)(−1) + ( y − 2)(2) + ( z − 3)(−1) = 0 or x − 2 y + z = 0 , which is the equation of the plane containing the given lines.

26. Find the equation of the plane passing through (1, 2, 2) and perpendicular to the planes 22 x + 3 y − z = 3 y − z = 2 and 5x + 3 y − 4 z = 5 . Ans. 3 x + y + 3 z = 7 . 27. Show that the points, (0, –1, –1), (4, 5, 1), (3, 9, 4) and (−4, 4, 4) lie on a plane. Hint: See Example 14.42. 28. Find the vector equation of the plane  r = iˆ + ˆj + λ (iˆ + ˆj + kˆ) + µ (iˆ − 22 ˆj + 3kˆ) in the scalar‐product form:   Hint: Setting a = iˆ − ˆj , b = iˆ + ˆj + kˆ , and      c = iˆ − 2 ˆj + 3kˆ , we have r = a + λ b + µc

2 3 4 x−2 y −3 z + 4 = = are coplanar, and also 3 4 5



y − y1 m1

Plane

25. Show that the lines x − 1 = y − 2 = z − 3 and

find the equation of the plane containing them. Hint: We note that x2 − x1 y2 − y1 z2 − z1 l1 m1 n1 l2 m2 n2

x − x1 l1



Therefore, the given plane represents a plane which passes through A(1,–1, 0) and   is parallel to the vectors b and c . Thus,     the vectors r − a , b and c are coplanar.   Therefore  r − a b c  = 0 , that is, x −1 y − 2 z − 3 1 1 1 = 0 or 1 −2 3 5 x − 2 y − 3z = 7

( (

)( )

)

or xiˆ + yjˆ + zkˆ . 5iˆ − 2 ˆj − 3kˆ = 7 or  r . 5iˆ − 2 ˆj − 3kˆ = 0, which is the required form.

29. Find the equation of the plane passing through the points, (−3, −2, −4), (2, 4,1) and (4, 2, 4). Hint: The equation of the plane is

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three-diMenSional geoMetry  n 14.59





x − x1 x2 − x1 x3 − x1

y − y1 y2 − y1 y3 − y1

32. Find the equation of the plane passing through (3, 4, 2) and (7, 0, 6) perpendicular to the plane 2 x − 5 y = 15 .

z − z1 z2 − z1 = 0 or z3 − z1



x+3 y+2 z +4 5 6 5 =0 7 4 8 or 28 x − 5 y − 22 z − 14 = 0.

30. A variable plane which remains at a constant distance 3p from the origin cuts the coordinate axes at A, B, and C. Show that the locus of the centroid of the triangle ABC is x12 + y12 + z12 = p12 .

Hint: If < l. m, n >are direction cosines of the normal to the plane, then since distance remains 3p, the equation of the plane is lx + my + nz = 3 p. Since it cuts the axes at A, B, and C, respectively, we have A

(

3p l

) (

, 0, 0 , B 0,

3p m

)

(

, 0 , and C = 0, 0,

3p n

If ( x1 , y1 , z1 ) is centroid of ∆ABC , then 3p 3l

, y1 =

3p 3m

x1 =



or l =



Since, l 2 + m 2 + n 2 = 1, we have

,m=

2

p y1

, z1 =

3p 3n



p x1

).

, n = z1 .

2

p

2

 p  p  p  x  +  y  +  z  = 1 or 1 1 1 1 1 1 1 + + = 2. x12 y12 z12 p



Therefore, the locus of the centroid ( x1 , y1 , z1 ) is 12 + 12 + 12 = 12 . x y z p

31. Show that the plane through the points (1,1,1), (1, −1,1) , and (−7,3, −5) is perpendicular to the xz‐plane.

Hint: Equation of the plane through the given points is 3 x − 4 z + 1 = 0 . The given plane is y = 0 . Use orthogonality condition a1a2 + b1b2 + c1c2 = 0 to get 3(0) + 0(1) + (−4)(0) = 0 .

M14_Baburam_ISBN _C14.indd 59

Hint: Plane passing through (3, 4, 2) A( x − 3) + B( y − 4) + C ( z − 2) = 0 is As it passes through (7,0,6), we have A − B + C = 0 . Since it is perpendicular to the given plane, we have 2 A − 5 B = 0 . Eliminating A, B, and C from these three equations, we get 5 x + 2 y − 3 z = 17.

33. Find the angle between the lines 3 x + 2 y + z − 5 = 0, x + y − 2 z − 3 = 0 , and 8 x − 4 y − 2 z − 2 = 0, 7 x + 10 y − 8 z = 0

Hint: The lines are perpendicular to the normal of the planes. 99 Ans. θ = cos −1 5 12651 .

(

)

34. Find the angle between the planes   r .(2iˆ − 3 ˆj + 4kˆ) = 1 and r .(−iˆ + ˆj ) = 4

Hint: See Example 14.50. Ans. θ = cos −1

( ). −5 58

35. Find the equation of the plane through the point (1, 1, 1) and perpendicular to x + 2 y + 3z − 7 = 0 and 2 x − 3 y + 4 z = 0 . Hint: The plane passing through (1,1,1) is A( x − 1) + B( y − 1) + C ( z − 1) = 0 . It is perpendicular to the given Planes. Therefore, and A + 2 B + 3C = 0 . Eliminating A, B, 2 A + 3B + 4C = 0 and C between these equations we get 17 x + 2 y − 7 z − 12 = 0. 36. Find the angle between the line x −1 y + 2 z − 5 and the plane = = 2 3 4. 3x + y − z + 3 = 0

Hint: See Example 14.51. Ans. θ = sin −1

( ). 5 319

37. Show that the lines x +2 4 = y 5+ 6 = z−−21 and 3x − 2 y + z + 5 = 0 = 2 x + 3 y + 4 z − 4 intersect (are coplanar). Find their point of

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14.60  n  chapter fourteen



intersection and the plane in which they lie. Hint: Any point on the first line is (3λ − 4, 5λ − 6, − 2λ + 1) . This point will lie on the planes 3 x − 2 y + z + 5 = 0 2 x + 3 y + 4 z − 4 = 0, and if 3 (3λ − 4) − 2 (5λ − 6) + ( −2λ + 1) + 5 = 0 ,

2 (3λ − 4) − 2(5λ − 6) + 4( −2λ + 1) − 4 = 0



Both these equations yield λ = 2. Therefore, this given lines intersect, that is, they are coplanar. The point of intersection is (3λ − 4, 5λ − 6, − 2λ + 1) = ( 2, 4, − 3) . The equation of plane through the second line is 3x − 2 y + z + k ( 2 x + 3 y + 4 z − 4) = 0. (1)



This plane will contain the first line if the point (2,4,–3) lies on it. Putting x = 2, y = 4, and z = −3 in the above equation of the plane (1), we get k = 133 . Substituting this value in (1), we get the plane as 45 x − 17 y + 25 z + 53 = 0. 38. Show that the line  ˆ) r = 2iˆ − 3 ˆj + 5kˆ + λ(iˆ − ˆj + 2ks  lies in the plane r . 3iˆ + ˆj − kˆ = −2.

Hint: Comparing the equation of the line     with r = a + λb , we have b = iˆ − ˆj + 2kˆ. Comparing the equation of the plane with    r .N = d , we have N = 3iˆ + ˆj − kˆ. The angle θ between the line and the plane is given by   iˆ − ˆj + 2kˆ . 3iˆ + ˆj − kˆ b .N sin θ =   = 1+1+ 4 9 +1+1 b N

)(

=

)

3 −1− 2

= 0. 6 11 Hence, the line is parallel to the plane. The perpendicular distance from (2,–3,5) to the plane is

(



)(

)

  2iˆ − 3 ˆj + 5kˆ . 3iˆ + ˆj − kˆ + 2 a. N − D = = 0.  9 +1+1 N

M14_Baburam_ISBN _C14.indd 60

)

(

)

(

)(

(



)

Hence, they are parallel. Distance between them   2iˆ − 2 ˆj + 3kˆ . iˆ + 5 ˆj + kˆ + 2 a. N − D =  N 12 + 52 + 12

)(

(



2 − 10 + 3 − 5

=

)

10

. 27 3 3 40. Find the length and the coordinates of the foot of the perpendicular from the point (1, 1, 2) to the plane 2 x − 2 y + 4 z + 5 = 0 Hint: Proceed as in Example 14.54.

)

(

(



Therefore, the line lies in the plane. 39. Show that the line  r = 2iˆ − 2 ˆj + 3kˆ + λ iˆ − ˆj + 4kˆ  is parallel to the plane r . iˆ + 5 ˆj + kˆ = 5. . Find the distance between them. Hint: The angle between the line and the plane should be zero degree. Therefore,   we should have b .N = 0. We note that   b .N = iˆ − ˆj + 4kˆ . iˆ + 5 ˆj + kˆ = 0.

Ans. Length

13 24

, foot

(

1 12

25 −1 , 12 , 6 ).

41. Find the equation of the plane passing through the intersection of the planes x + 2 y + 3x + 5 = 0 and 2 x − 4 y + z − 3 = 0 and the point (0,1, 0). Hint: Proceed as in Example 14.58. Ans. 3 x − 2 y + 4 z + 2 = 0 . 42. Find the equation of the plane passing through the intersection of the planes 2 x + 3 y + z = 5 and 3 x − y + 4 z + 3 = 0 and perpendicular to the plane x + y − 3 z = 6 .

Hint: Proceed as in Example 14.59. Ans. 13 x + 14 y + 9 z − 22 = 0 .

43. Find the equation of the planes bisecting the angle between the planes 3 x − 4 y + 5 z = 3 and 5 x + 3 y − 4 z = 9. Specify which of the planes bisect the acute angle. Hint: Proceed as in Example 14.56.

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three-diMenSional geoMetry  n 14.61 Ans. 2 x + 7 y − 9 z − 6 = 0 and

this plane. The triangles ABC, BCD, CAD, and ABD are the four faces and the lines AB, CD, BC, AD, CA, and BD are the six edges. The edges AB and CD are called opposite edges. Similarly, BC, AD and CA, BD are opposite edges. Thus, the graph of the tetrahedron is as shown in the following figure

8 x − y + z − 12 = 0, the plane 2 x + 7 y − 9 z − 6 = 0 bisects the acute angle. 44. Find the symmetrical form of the equation of the line represented by the planes 3 x − 4 y + 5 z = 3 and 5 x + 3 y − 4 z = 9.

Hint: Putting z = 0 in these equations, we get 3 x − 4 y − 3 = 0 and 5 x + 3 y − 9 = 0. y Solving, we get 45x = 12 = 291 and so, a 45 12 point on the line is ( 29 , 29 , 0) . The line lies on both the planes. Therefore, it is perpendicular to their normal. Hence, are direction cosines of the line, then 3l − 4m + 5n = 0 and 5l + 3m − 4n = 0. Therefore, 1l = 27m = 29n , Thus, the line is 45 x − 29 y − 12 z 29 . = = 1 27 29

A

C P





Hint: The line passing through the point y −3 (–2, 3, 4) is x +l 2 = m = z −n 4 . Since the above line is parallel to the planes, it is perpendicular to the normals to the given planes. Therefore, 2l + 3m + 4n = 0 and 4l + 3m + 5n = 0. Solving these, we have l m n 3 = 6 = −6 . Hence, the equation of the line is x+2 y −3 z −4 or = = 3 6 −6 x+2 y −3 z −4 = = . 1 2 −2

46. Find the volume of a tetrahedron in terms of the coordinates of its vertices. Solution. Let A( x1 , y1 , z1 ), B( x2 , y2 , z2 ), C ( x3 , y3 , z3 ), and D( x4 , y4 , z4 ), be the vertices of the tetrahedron. Then, D does not lie on the plane determined by the points A, B, and C. Thus, the three points may be assumed as lying on the plane of the paper and the point D as lying above

M14_Baburam_ISBN _C14.indd 61

D

or A

P

45. Find the equation of the line passing through the point (–2, 3, 4) and parallel to the planes 2 x + 3 y + 4 z = 5 and 4 x + 3 y + 5 z = 6.

L

B

D L B







Let V be the volume of the tetrahedron ABCD. If p is the length of the perpendicular AL from vertex A on the opposite face BCD and ∆ is the area of the triangle BCD, the volume (V) = 13 p∆. Now, the equation of the plane BCD passing through three points B( x2 , y2 , z2 ), C ( x3 , y3 , z3 ), and D( x4 , y4 , z4 ) is x y z 1 x2 y2 z2 1 =0 x3 y3 z3 1 x4 y4 z4 1 or y2



C

z2 1

x2

z2 1

x2

y2 1

x y3 z3 1 − y x3 z3 1 + z x3 y3 1 y4 z 4 1 x4 z4 1 x4 y4 1 x2 y2 z2 − x3 y3 z3 = 0. (1) x4 y4 z4

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14.62  n  chapter fourteen

Now the length of the perpendicular from A to the plane (1) is equal to y2

x1 y3 y4

z2 1





x2

y2 1

x2

y2

z3 1 − y1 x3

z3 1 + z1 x3

y3 1 − x3

y3

z3

z4 1

y4 1

y4

z4

x4 2

x4

2 y2 1   y3 1  y4 1  

2

z2 1 x2 z3 1 + x3 z4 1 x4

x4

z2 1 x2 z3 1 + x3 z4 1 x4

1 2

We note that numerator of p is x1 x2 x3 x4

y1 y2 y3 y4

z1 z2 z3 z4

1 1 . 1 1

z2 1 x2 1 z3 1 , ∆ y = x3 2 z4 1 x4

z2 1 z3 1 , z4 1





and ∆z =

x2 1 x3 2 x4

y2 1 y3 1 .

4 Therefore, the denominator of p is equal to

p=

x1

y1

z1 1

x2 x3 x4

y2 y3 y4

z2 1 z3 1 z4 1

From the first two members of equation (2), we have 19λ = 38 or λ = 2. This value of λ satisfies all equations in (2). Hence, the three given planes intersect in a line.

2∆

49. Find the equation of the sphere concentric with the sphere x 2 + y 2 + z 2 − 6 x + 2 y − 4 z = 2 and double its radius.

.

Hence, the volume V of the tetrahedron is

M14_Baburam_ISBN _C14.indd 62

The given three planes will intersect in a common line if (1) and the third plane represent the same plane. Therefore, 2 + 3λ 3 + 14λ 7 + 13λ (2) = = . 8 31 33

Ans. (−2, 4, −3),5



or ( 2 + 3λ) x − (3 + 14λ) y − (7 + 13λ) z = 0.

48. Find the center and the radius of the sphere x2 + y 2 + z 2 + 4x − 8 y + 6z + 4 = 0

1

 4 ∆ 2x + 4 ∆ 2y + 4∆ 2z  2 = 2 ∆ 2x + ∆ 2y + ∆ 2z = 2 ∆.

Hence,

z1 1 z2 1 . z3 1 z4 1

Sphere

1

y

1 6

y1 y2 y3

x4 y4 47. Show that the three planes 2 x − 3 y − 7 z = 0, 3 x − 14 y − 13 z = 0 and 8 x − 31 y − 33 z = 0 pass through one line. Hint: Any plane that passes through the line of intersection of first two planes is 2 x − 3 y − 7 z = 0 + λ(3x − 14 y − 13z ) = 0



Further, if ∆x, ∆y, and ∆z are projections of the ∆BCD on the YZ, ZX, and XY planes, then y2 1 ∆ x = y3 2 y4

V = 13 p ∆ =

z2

z4 1  y2   y3 y  4



z2 1

x2

x1 x2 x3



Hint: The radius of the given sphere = g 2 + f 2 + h2 − c = 9 + 1 + 4 + 2 = 4 a n d its center is ( − g , − f , − h ) = (3, −1, 2) . Therefore, the center of the new sphere shall be (3, −1, 2) and its radius be 8. Thus, the equation of the required sphere is

( x − 3)2 + ( y + 1)2 + ( z − 2)2 = 82

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three-diMenSional geoMetry  n 14.63

or

(given).

x 2 + y 2 + z 2 − 6 x − 2 y − 4 z = 50.

50. Find the equation of the sphere for which the circle x 2 + y 2 + z 2 − 3 x + 4 y − 2 z − 5 = 0, 5 x + 2 y + 4 z + 7 = 0 is a great circle.

Hint: Proceed as in Example 14.64 Ans. x 2 + y 2 + z 2 + 2 x + 2 y + 2 z + 2 = 0.

51. Find the equation of the sphere having its center on the plane 4 x − 5 y − z = 3 and passing through the circle



x + y + z − 2 x − 3 y + 4 z + 8 = 0, 2

2

2

If

( x1 , y1 , z1 ) is the centroid of the triangle ABC, then α +0+0 α x1 = = , y1 = 0 + β + 0 = β , 3 3 3 3 0+0+γ γ = . and z1 = 3 3 α = 3x1 , β = 3 y1 , and γ = 3z1 . Thus, 2 2 2 2 Putting these values in α + β + γ = 4k , we get 9 x12 + 9 y12 + 9 z12 = 4k 2 and hence, the locus of ( x1 , y1 , z1 ) is

(

)

x − 2 y + z = 8.



Hint: Here the circle becomes great circle Ans. 2 2 2

55. Find the equation of the spheres which 2 2 passes through the circle x + y = 4, z = 0 and is cut by the plane x + 2 y + 2 z = 0 in a circle of radius 3. Ans. x 2 + y 2 + z 2 ± 6 z − 4 = 0

13 x + 13 y + 13 z + −35 x − 21 y + 43 z + 176 = 0.

52. Show that the plane 2 x + 2 y − z + 10 = 0 touches the sphere x 2 + y 2 + z 2 − 4 x + 2 y − 6 z + 5 = 0 . Also find its point of contact Ans. (0, −3, 4) 53. Show that the two circles x 2 + y 2 + z 2 − y + 2 z = 0, x − y + z − 2 = 0

and



x 2 + y 2 + z 2 + x − 3 y + z = 5, 2 x − y + 4 z − 1 = 0

lie on the same sphere and find its equation. 2 2 2 Ans. x + y + z + 3x − 4 y + 5 z − 6 = 0

54. A sphere of constant radius k passes through the origin and meets the axes in A, B and C. Show that the centroid of the triangle ABC 2 2 2 2 lies on the sphere 9 x + y + z = 4k . Hint: Let the coordinates of the points O, A, B and C be (0,0,0), A ( α, 0, 0) , B ( 0, β, 0) , and C ( 0, 0, γ ) . Using the four-point formula, the equation of the sphere through these points is x 2 + y 2 + z 2 − α x − β y − γ z = 0.

(





α 2 + β 2 + γ 2 = 4k 2 .

Thus

The radius of this sphere is

M14_Baburam_ISBN _C14.indd 63

)

α2 4

2

2

+ β4 + γ4 = k

9 x 2 , y 2 , z 2 = 4k 2 .

56. A sphere of constant radius k passes through the origin ) and meets the axes in A, B, and C. Prove that the locus of the foot of the perpendicular from O to the plane ABC is given by ( x 2 + y 2 + z 2 )( x −2 + y −2 + z −2 ) = 4k 2 . (1) Hint: As in Exercise 54, the equation of the sphere through O, A, B, and x 2 + y 2 + z 2 −α x − β y − γ z = 0. C is Since the radius is k, we also have α 2 + β 2 + γ 2 = 4k 2 . The intercept form of the plane ABC is

x α

+ βy + γz = 1. The

direction cosines of the normal to this plane are proportional to

1

α

, β1 , γ1 . Therefore, the

equation of the perpendicular from O to this plane is

x 1

α

+

y 1

β

+

z 1

γ

= λ , say. Any point

(

)

+ βλ + γλ . This point lies on the plane if αλ2 + βλ2 + γλ2 = 1 and so λ = α −2 + β1−2 +γ −2 . Therefore, the foot on this perpendicular is

λ α

of the perpendicular is

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14.64  n  chapter fourteen



(

Then x12 + y12 + z12 = −2 1

−2 1

x +y +z



)

2x + y + z = 4 3 x + 4 y = 14.

1

α −2 + β −2 + γ −2

and −2 1



59. Find the equations of the spheres which pass through the circle x 2 + y 2 + z 2 − 2 x + 2 y + 4 z − 3 = 0,

λ λ λ , , = α β γ  β −1 γ −1 α −1 , , . α −2 + β −2 + γ −2 α −2 + β −2 + γ −2 α −2 + β −2 + γ −2

( x1 , y1 , z1 ) = 

=

(α

−2

+ β −2 + γ −2 )

−2

(x



2

+ y 2 + z 2 )( x −2 + y −2 + z −2 ) = 4k 2 .

Hence, the locus of ( x1 , y1 , z1 ) is

Hint: Proceed as in Exercise 54. The only difference is that the controid of the tetrahedron is

58. The plane ABC, whose equation is y x z meets the axes in A, B, and C. a + b + c =1 Find the equation of the circumcircle of the triangle ABC.



2 2 2 and x + y + z − 2 x + 4 y + 6 z − 11 = 0 .

60. Find the equation of the sphere that passes through the circle

 x1 + x2 + x3 + x4 y1 + y2 + y3 + y4 z3 + z2 + z3 + z4  , ,    4 4 4 



plane

Putting these values in (1) and using α 2 + β 2 + γ 2 = 4k 2 , we get

( x 2 + y 2 + z 2 )( x −2 + y −2 + z −2 ) = 4k 2 . 57. A sphere of constant radius k passes through the origin and meets the axes in A, B, and C. Show that the controid of the tetrahedron OABC lies on the sphere k2 . x2 + y 2 + z 2 = 4

the

2 2 2 Ans. x + y + z − 2 x + 2 y + 4 z − 3 = 0



.

touch

Hint: Proceed as in Example 14.71.



α2 + β2 +γ 2

and

Hint: The circumcircle of the triangle ABC is the intersection of the plane ABC and any sphere through A, B, and C. We consider the sphere OABC, where O is the origin. As in Exercise 54, the equation of the sphere is x 2 + y 2 + z 2 − ax − by − cz = 0. Hence, the equation of the circumcircle of ∆ABC is y x 2 + y 2 + z 2 − ax − by − cz = 0, x + + z = 1 a b c

M14_Baburam_ISBN _C14.indd 64

x 2 + y 2 + z 2 − 2 x + 3 y − 4 z + 6 = 0, 3 x − 4 y + 5 z = 15 and cuts orthogonally the 2 2 2 sphere x + y + z + 2 x + 4 y − 6 z + 11 = 0.





Hint: First find the sphere through the circle and then use the condition of orthogonality to get the value of λ. Ans. 5 x 2 + y 2 + z 2 − 13 x + 19 y − 25 z + 45 = 0.

(

)

61. Find the center and radius of the circle x 2 + y 2 + z 2 = 49, 2 x + 3 y + 6 z = 14.

Hint: See Example 14.66.  4 6 12  Ans.  , ,  , 45 7 7 7 



62. Prove that the spheres x 2 + y 2 + z 2 + 2ax + c = 0 x 2 + y 2 + z 2 + 2by + c = 0

and (a , b > c > 0)

touch if and only if and only if

2

2

1 1 1 + 2 = 2 c a b

Cylinder 63. Find the equation of the right circular cylinder of radius 2 and whose axis is the x −1 y − 2 z − 3 line = = 2 1 2 Ans. 5 x 2 + 8 y 2 + 5 z 2 − 4 xy − 4 yz − 8 zx + 22 x − 16 y − 14 z − 10 = 0.

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three-diMenSional geoMetry  n 14.65 Ans. 10 x 2 + 5 y 2 + 13 z 2 − 12 xy − 6 yz − 4 zx − 8 x +30 y − 74 z + 59 = 0.

64. Obtain the equation of the right circular cylinder whose generating circle is x 2 + y 2 + z 2 = 4, x + y + z − 3 = 0.

Hint: Proceed as in Example 14.83.

65. Find the equation of the cylinder whose y generators are parallel to the line 1x = 2 = 3z and which passes through the curve x 2 + y 2 = 16, z = 0.

Hint: Direction ratios of the generator are . If ( x1 , y1 , z1 ) is any point on the cylinder, then the equation of generator through ( x1 , y1 , z1 ) is x − x1 y − y1 z − z1 = = . Equation of the 1 2 2 3 2 guiding curve is x + y = 16, z = 0.



The line must intersect the guiding curve. Taking z = 0, the line becomes x − x1 y − y1 z − z1 The result = = . 1 2 z 3 2z 1 1 yields x = x1 − 3 and y = y1 − 3 . Putting 2 2 these values in x + y = 16, we get 2 2 2 9 x1 + z1 − 6 z1 x1 + 9 y1 +4 z12 − 12 y1 z1 = 144 Therefore, the locus of ( x1 , y1 , z1 ) is 9 x 2 + z 2 − 6 zx + 9 y 2 + 4 z 2 − 12 yz = 144.

66. Find the equation of the cylinder whose generators are parallel to the line y−4 x z +1 and whose guiding curve is 3 = 5 = −4 the hyperbola 3 x 2 − 4 y 2 = 5, z = 2. Ans. 48 x 2 − 64 y 2 − 73 z 2 − 160 yz + 72 zx − 144 x +320 y + 292 z − 372 = 0. 67. Find the equation of the right circular cylinder whose radius is 2 and axis is x −1 y z − 3 . = = 2

3

M14_Baburam_ISBN _C14.indd 65

1

Cone 68. Find the equation of the cone with vertex at the origin and which passes through the curve ax 2 + by 2 + cz 2 = 1, lx + my + nz = p. 2 2 2 2 2 Ans. p (ax + by + cz ) = (lx + my + nz ) . 69. Find the equation of the cone with vertex (1, 1, 1) and passing through the curve 2 2 2 of intersection of x + y + z = 1 and x + y + z = 1. Ans. x 2 + y 2 + z 2 − 2 xy − 2 yz − 2 zx + 2 x +2 y + 2 z − 3 = 0.

70. Find the equation of the cone with vertex at (2, 3, 1) and passing through the curve of intersection of x2 + y 2 + z 2 − 2x + 4 y − 6x + 7 = 0

and x + 2 y + 2 z = 5 . Ans. 38 x 2 + 17 y 2 + 157 z 2 + 22 xy + 124 yz +92 zx − 310 z − 270 y − 870 z + 1150 = 0.

71. Find the equation of the right circular cone whose vertex is at the origin, axis is x y z , and vertical angle is 60° . = = 1 2 3 Ans. 2 2 2 19 x + 13 y + 3z − 24 yz − 12 zx − 8 xy = 0 . (see Example 14.92) 72. Find the equation of the right circular cone whose vertex is (2, −3,5) , axis makes equal angles with the axes, and the cone passes through (1, −2,3) . Ans. x 2 + y 2 + z 2 + 6( yz + zx + xy ) − 16 x − 36 y − 4 z − 28 = 0. 73. Find the equation of the right circular cone generated when the straight line

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14.66  n  chapter fourteen 2 y + 3 z = 6, x = 0 revolves about z‐axis.

Hint: Vertex is the point of intersection of the given line with z‐axis. Thus, putting x=0 and y = 0 in the line, we get vertex as (0, 0, 2). One of the generator is x−0 y−0 z−2 = = . 0 3 −2

(

)

Ans. 4 x 2 + y 2 − 9 z 2 + 36 z − 36 = 0. Conicoids 74. Name the conicoid represented by (i) x 2 + y 2 = 9 z 2 and (ii) x 2 + y 2 = 5 − 2 y . Ans. (i) Cone, (ii) Right circular cylinder 75. Show that the equation

2 y 2 − 8 yz − 4 zx − 8 xy

+6 x − 4 y − 2 z + 5 = 0 represents a cone. Find the coordinates of its vertex. Hint: Make the given equation homogeneous by introducing another variable t. Thus, we have F ( x, y, z , t ) = 2 y 2 − 8 yz − 4 zx − 8 xy +6 xt − 4 yt − 2 zt + 5t 2 .





This will represent a cone if and only if Fx = Fy = Fz = Ft = 0 has a unique solution (means these equations are consistent). The unique solution to these equations will yield the vertex. These equations will come out to be Fx = −4 z − 8 y + 6t = 0, Fy = 4 y − 8 z − 8 x − 4t = 0,

Fz = −8 y − 4 x − 2t = 0, and Ft = 6 x − 4 y − 2 z + 10t = 0.

Taking t = 1 and solving firstthree equations, we get ( x, y, z ) = ( −67 , 13 , 56 ) , which satisfies the last equation 6 x − 4 y − 2 z = 0. 76. Show that the equation x 2 − 2 y 2 + 5 z 2 − 4 xy − 5 yz − 6 zx

+8 x − 19 y − 2 z − 20 = 0 represents a cone. Hint: Proceed as in Exercise 75.

M14_Baburam_ISBN _C14.indd 66

77. Find the points of intersection of the line joining (4, 3, –1) and (1, 0, 2) with the 2 2 2 hyperboloid x − y + 2 z = 5. Hint: Equation of the line joining the given points is x −3 4 = y 3− 3 = z−+31 or y −3 x−4 z +1 1 = 1 = −1 = λ, say. Any point on this line is ( λ + 4, λ + 3, − λ − 1) . That point lies on the hyperboloid and so, ( λ + 4)2 − ( λ + 3)2 + 2 ( − λ − 1)2 = 5, w h i c h yields λ = −1, − 20. Hence, the points of intersection are (3,2,0) and (2,1,1). 78. Find the points of intersection of the line x + 5 y − 4 z − 11 with the central = = −3 1 2 72 2 conicoid 12 x − 17 y − +8 z − 7 = 0 Hint: Proceed as in Exercise 77. Ans. (−2,3, 4), (1, 2, −3) . 79. Find the equation to the tangent planes to the surface 4 x 2 − 5 y 2 + 7 z 2 + 13 = 0 parallel to the plane 4 x + 20 y − 21z = 0.

Hint: We know that the tangent planes to 2 2 2 the conicoid ax + by + cz = 1 are given

x + my + nz = ±

as

l2 a

+

m2 b

+

n2 c

. Here,

l

= 4, m = 20, n = –21 and the conicoid is a = − 134 ,

− 134 x 2 + 135 y 2 − 137 z 2 = 1. Thus, b=

5 13

, and c =

−7 13

tangent planes are

4 x + 20 y − 21z = ±

. Hence, the required 16 4 − 13

441 + 400 5 + −7 13

13



= ± 169 = ±13.

80. Show that the plane 3 x + 12 y − 6 z − 17 = 0 touches the conicoid 3 x 2 − 6 y 2 + 9 z 2 + 17 = 0 . 81. Prove that the lines drawn from the origin parallel to the normals to

x2 y 2 z 2 + + =1 a 2 b2 c2

at the point of intersection with the plane lx + my + nz = p ,

generate

the

cone

P 2 (a 2 x 2 + b 2 y 2 + c 2 z 2 ) = (a 2 lx + b 2 my + c 2 nz ) 2 .

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15

Logic

Logic is the study of reasoning and is specifically concerned with whether a particular reasoning is valid. It is a science of the necessary laws of thought, without which no employment of the understanding and the reason takes place. 15.1  PROPOSITIONS Definition 15.1  A declarative sentence that is either true or false, but not both is called a Proposition or a Statement. EXAMPLE 15.1

Which of the following are propositions? (i) London is the capital of France (ii) Open the door (iii) Take two tablets of medicine (iv) x 1 y > 0, x, y ∈ Z (v) The only positive integers that divide a prime number are 1 and the number itself. (vi) The sun is hot Solution. (i) The sentence is declarative and false. Hence, it is a proposition. (ii) The sentence “Open the door” is not declarative. It is rather a command. Therefore, it is not a proposition. (iii) The given sentence is not declarative and so it is not a proposition. (iv) The sentence is not a statement because it is true for some values of x and y whereas for other values of x and y it is false. For example, if x 5 2, y 5 1, then it is true but if x 5 −2, y 5 1, then it is false.

M015_Baburam_ISBN _C15.indd 1

(v) The given sentence is declarative as well as true. Hence it is a statement. (vi) The sentence “The sun is hot” is both a declarative sentence and true. Hence it is a ­proposition. Notations.  The propositions are represented by lower case letters such as p, q and r. Thus, the notation p: 3 1 7 5 10 means that p is a proposition 3 1 7 5 10. Many propositions are composite, that is, composed of sub-propositions and various connectives. Thus, we have Definition 15.2 Composite propositions are called compound propositions. A proposition which is not compound is said to be primitive. Thus, a primitive proposition cannot be broken into simpler propositions. EXAMPLE 15.2

Consider the sentence: The sun is shining and it is cold. This is a compound proposition composed of two propositions: The sun is shining and It is cold connected by the connective “and”. On the other hand, the proposition London is in Denmark is a primitive statement. Definition 15.3 The truth values of a compound statement in terms of its component parts is called a truth table.

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15.2  n  chapter fifteen 15.2  BASIC LOGICAL OPERATIONS The three basic logical operations are 1. Conjunction 2. Disjunction 3. Negation which correspond, respectively, to “and”, “or” and “not”.

Definition 15.5 The disjunction of two proposition p and q is the proposition p or q and is denoted by p ∨ q. The compound statement p ∨ q is true if at least one of p or q is true. It is false when both p and q are false. The truth value of the compound proposition p ∨ q is defined by the following truth table: p T T F F

Definition 15.4 The conjunction of two propositions p and q is the proposition p and q. It is denoted by p ∧ q. EXAMPLE 15.3

Let Then

p: This child is a boy q: This child is intelligent.

p ∧ q: This child is a boy and intelligent. Thus, p ∧ q is true, if the child is a boy and intelligent both. Even if one of the component is false, p ∧ q is false. Thus, “the proposition p ∧ q is true if and only if the propositions p and q are both true”. The truth value of the compound proposition p ∧ q is defined by the truth table given below: p

q

p∧q

T T F F

T F T F

T F F F

EXAMPLE 15.4

If

p: London is capital of India q: A decade is 10 years

then p is false, q is true and the conjunction p ∧ q: London is capital of India and a decade is 10 years is false.

M015_Baburam_ISBN _C15.indd 2

q T F T F

p∨q T T T F

For example, if p: London is capital of India q: A decade is 10 years, then p is false, q is true. The disjunction p ∨ q: London is capital of India or a decade is 10 years is true. EXAMPLE 15.5

Form the disjunction of p and q for each of the following: (a) p: 2 is a positive integer, q: √2 is a rational number (b) p:   2 1 3 ≠ 5,   q:   London is the capital of France. Solution. (a) p ∨ q: 2 is a positive integer or √2 is a rational number. Since p is true, the disjunction p ∨ q is true, even though q is false. (b) p ∨ q: 2 1 3 ≠ 5 or London is the capital of France. Since both p and q are false, p ∨ q is false. Remark 15.1. It is clear from the above example that in logic, unlike in ordinary English, we may join totally unrelated statements by the connective “or”.

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logic   n 15.3 EXAMPLE 15.6

Remark 15.2.

Consider the following four statements: (i) Taj is in Agra and 7 is a prime number (ii) Paris is in France and √2 is a rational number (iii) Paris is in England and 2 1 2 5 4 (iv) Paris is in England and 2 1 2 5 5 Solution. Here the propositions are combined by connective “and”. Therefore, the compound proposition shall be conjunction. We observe that in (i) both sub-propositions are true. In the rest three statements, at least one subproposition is false. Hence only (i) is true. The truth table is p T T F F

q T F T F

p∧q T F F F

Definition 15.6  If p is a statement, the negation of p is the statement “not p”, denoted by ~ p. Thus, ~ p is the statement “it is not the case that p”. Hence, if p is true, then ~ p is false and if p is false, then ~ p is true. The truth table for negation is p T F

~ p F T

EXAMPLE 15.7

Give the negation of the following statements:

(a) p: 2 1 3 > 1 (b)  q: It is cold.

Solution. (a) ~ p:  2 1 3 is not greater than 1, that is, ~ p: 2 1 3 ≤ 1. Since p is true in this case, ~ p is false. (b) ~ q: It is not the case that it is cold. More simply, ~ q: It is not cold.

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(i) In expressions that include the symbol ~ as well as ∧ or ∨, the order of operation is that ~ is performed first. For example, ~ p ∧ q 5 (~ p) ∧ q. (ii) An expression such as p ∧ q ∨ r is considered ambiguous. This expression must be written as either (p ∧ q) ∨ r or p ∧ (q ∨ r) to have meaning. 15.2.1  Translating from English to Symbols We consider EXAMPLE 15.8

Write each of the following sentences symbolically, letting p: “It is hot” and q: “It is sunny”: (a)  It is not hot but it is sunny (b)  It is neither hot nor sunny Solution. (a) In logic, the words “but” and “and” mean the same thing. Generally, “but” is used in place of “and” when the part of the sentence that follows is in some way unexpected. The given sentence is equivalent to “ It is not hot and it is sunny” which can be written symbolically as ~ p ∧ q. (b)  The phrase neither A nor B means the same as not A and not B. Thus, to say “It is neither hot nor sunny” means that it is not hot and it is not sunny. Therefore, the given sentence can be written symbolically as ~ p ∧ ~ q. Remark 15.3. The notation for inequalities involves “and” and “or” statements. For example, if x, a and b are particular real numbers, then x ≤ a means x < a or a ≤ x ≤ b means a ≤ x and

x 5 a, x ≤ b.

We note that 2 ≤ x ≤ 1 means 2 ≤ x and x ≤ 1, which is false, no matter what x happens to be. We have taken x, a and b as particular real numbers to ensure that sentences such as x  0. Ans. The argument is valid. The logical expression of the argument is

p∨q



p→r



q→r



∴ r.



2

p: Julia live in Italy q: Julia speaks Italian r: Julia drives a car s: Julia travels by a train

2

    Inverse:  If ABC is not a right triangle, then |AB|2 + |BC|2 ≠ |AC|2     

Contrapositive:  If |AB|2 + |BC|2 ≠ |AC|2, then triangle ABC ia not a right triangle

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(p → r) → r is a tautology

8. Verify the validity of the following argument by using rules of inference: If Julia does not live in Italy, then she does not speak Italian. Julia does not drive a car If Julia lives in Italy, then she travels by train. Either Julia speaks Italian or she drives a car. Therefore, Julia travels by train. Ans. Let

Ans.  Converse:  If |AB| + |BC| = |AC| , then triangle ABC is a right triangle 2

This is because ( p ∨ q) ∧ (q → r) ∧





Then the logical form of the given statement is: (i)  ~ p → ~ q (ii)  ~ r (iii)  p → s (iv)  ~q → r



The following deduction can be made:

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15.20  n  chapter fifteen (i)  ~ q → r by (iv)  ~ r by (ii)

(ii)  ~ p → ~ q by (i) q by conclusion of (i) ∴p

q (iii)  p → s p

by (iii) by conclusion (ii)

∴s "Hence Julia travels by a train." 9. Verify the validity of the following ­argument: ~  r p→q q→r ∴ ~ p. Ans. We note that

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(a)  q → r   and   (b)  p → q     ~  r ~ p by (a)    ∴ ~ q ∴ ~ p Hence the argument is valid 10. Fill in the blanks: (i) If the graphs are isomorphic, then their degree spectrum will be the same    Their degree spectrums are different    Hence _______________________ (ii) If it rains, Bill will be happy    Bill is not happy    Therefore ____________________ Ans. (i)  The graphs are not isomorphic

 (ii)  It did not rain

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16

Elements of Fuzzy Logic

According to Georg Cantor, a set may be viewed as a well-defined collection of objects, called the element or member of the set. The term ‘well defined’ means that it is possible to decide if a given object belongs to the collection or not. Thus, if S is a set, then an element s is either a member of S or it is not. Let χS be a characteristic (indicator) function of S. Then, by definition, χS is a function from universe of discourse to the set {0, 1} such that 1 if  s∈S χS (s) = 0 if  s∉S

Let S and T be two fuzzy subsets of a universal set X. If mS (x) ≤ mT (x) for every x ∈ X, then S is called a subset of T and we write S ⊆ T. For example, if X = {x, y, z} and

Hence, a set S can be represented in term of ordered pairs as

Hence S is a subset of T. Obviously, every fuzzy set is a subset of itself. The fuzzy set S will be called proper subset of a fuzzy set T if mS (x) ≤ mT (x) for every x ∈ X and mS (x) < mT (x), for at least one x ∈ X. In this case, we write S ⊂ T. For example, if X = {x, y, z} is universal set and S = {(x, 0.4), ( y, 0.7), (z, 0.9)} T = {(x, 0.4), ( y, 0.8), (z, 0.9)} then we note that



{(s, χS ): χS (s) = 1}.

If we call χS as a membership function, then, in standard set theory, the membership has only two values 1 and 0. The membership of an elements s is 1 if s ∈ S and 0 if s ∉ S. 16.1  FUZZY SET The fuzzy set is a generalization of the standard set in which membership has values in the closed interval [0, 1]. Let X be universe of discourse. Then a fuzzy subset S of X is a set of ordered pairs (s, mS(s)), s ∈ X, where membership function mS assumes the values in [0, 1] called the degree of membership or level of membership in S. For example, let X = {x, y, z}. Then S = {(x, 0.2}, ( y, 0.5), (z, 0.7)} is a fuzzy subset of X in which 0.2 is the degree of membership of x in S, 0.5 is the degree of membership of y in S, 0.7 is the degree of membership of z in S.

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S = {x, 0.2}, (y, 0.5), (z, 0.7)}, T = {x, 04}, (y, 0.7), (z, 0.8)} then, we note that

mS (x) < mT (x), mS ( y) < mT ( y), and mS (z) < mT (z)



mS (x) 5 mT (x), mS ( y) < mT ( y), mS (z) 5 mT (z).

Hence mS (u) ≤ mT (u), for every u ∈ X and mS ( y) < mT ( y), y ∈ X. Hence S is a proper subset of T. Two fuzzy subset S and T of a universal set X are called equal if S  T and T  S, that is, if mS (x) = mT (x) for every x ∈ X. Let S be a fuzzy subset of a finite universal set X. Then

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16.2  n  chapter Sixteen

{ ( 

 

 ​  ​m  (x)

| S | = ​

   x∈X

​ 

is called the scalar cardinality of S. Thus the scalar cardinality of a fuzzy set S is the sum of degree of membership of its members. For example, if X = (x, y, z) is universal set and S = {(x, 0.2), (y, 0.5), (z, 0.7)} is a fuzzy subset of X, then scalar cardinality of S is | S | = mS (x) + mS ( y) + mS (z) = 0.2 + 0.5 + 0.7 = 1.4. Let S be a fuzzy subset of universal set X. Then the support of S in X is the set of all elements of X that have a non-zero degree of membership. Thus supp (s) = {x ∈ X : mS (x) > 0}. For example, let X = {x, y, z, t} be universal set and let S = {(x, 0), ( y, 0.7), (z, 0), (t, 0.9)} be a fuzzy subset of X. Then supp (S ) = {u ∈ X: mS (u) > 0} = { y, t}.



The height of a fuzzy set S is the largest degree of membership attained by any element in S. For example, the height of the fuzzy set S = {(x, 0.2), (y, 0.4), (z, 0.9)} is 0.9. A fuzzy set S is called normal if at least one of its member attains the membership value1. Thus a fuzzy set of height 1 is normal. A given fuzzy set S can be normalized by dividing each degree of membership by the height of that fuzzy set. The normalized version of a given fuzzy set S is denoted by SN. Obviously, SN is also a fuzzy set. EXAMPLE 16.1 Normalize the fuzzy set S = {(x, 0.2), ( y, 0.5), (z, 0.7)}. Solution.  The height of the fuzzy set S is 0.7. Dividing each degree of membership of the members of S by 0.7, we get the normalized set

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) ( 

) ( 

)}

0.7 0.2 0.5 SN = ​ ​ x, ___ ​    ​  ​, ​ y, ___ ​    ​  ​, ​ z, ___ ​    ​  ​  ​ 0.7 0.7 0.7

S



= {(x, 0.287), (y, 0.714), (z, 1)}.

Let S be fuzzy subset of a universal set X and a ∈ [0, 1]. Then the set Sa = {x ∈ X : mS (x)) ≥ a} is called the a-cut of the fuzzy set S. EXAMPLE 16.2 Show that a fuzzy set A is normal if and only if A1.0 ≠ f. Solution. We know that a-cut of a fuzzy set A is defined by Aa = {x: mA(x)) ≥ a}. Thus A1.0 = {x: mA(x)) ≥ 1}. If A is normal, then by definition, there is at least one element x in the universal set such that mA(x)) = 1. Hence A1.0 is not an empty set, that is, A1.0 ≠ f. Conversely, if, A1.0 ≠ f, then there is at least one element x in A1.0 such that mA(x) ≥ 1. But, by the definition of membership function mA(x) cannot be greater than 1. Hence mA(x) = 1. Thus A1.0 contains at least one element x such that mA(x) = 1. Hence A is normal. EXAMPLE 16.3 If X = {x, y, z} is universal set and S = {(x, 0. 2), (y, 0.5), (z, 0.7)} be fuzzy subset of X, determine a-cut of S for a = 0.5 and a = 0.1. Solution. We have S = {(x, 0, 2), (y, 0.5), (z, 0.7)} and a = 0.5. Therefore Similarly,

S0.5 = {u ∈ X: mS (u) ≥ 0.5} ={ y, z}.

S0.1 = {u ∈ X: mS (u) ≥ 0.1} ={x, y, z}. If S is a fuzzy subset of a universal set X and a ∈ [0, 1]. Then the set

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eleMentS of fuzzy logic   n 16.3 Sa+ = {x ∈ X: mS (x) > a} is called the strong a-cut of the fuzzy set S. In Example 16.3, we have

S0.5+ = {x ∈ X: mS (x) > 0.5} = {z}.

The set of all levels a ∈[0, 1] that represent acuts of a given fuzzy set S is called a level set of the fuzzy set S and is denoted by ∧S. Thus ∧S = {a: mS (x) = a for some x ∈ X }. An ordinary (Crisp) set X is said to be convex if lx + (1 - l)y ∈ X for all x, y ∈ X and l ∈ [0, 1]. Let S be a fuzzy subset of a universal set X. Then S is called convex if each of its a-cuts is a convex set. Theorem 16.1 A fuzzy set S on a universal set X is convex if and only if

mS [ lx1 + (1- l) x2] ≥ min [ mS (x1), mS (x2)] for x1, x2 ∈ X and 0 ≤ l ≤1.

Remark 16.1  Since l ∈ [0, 1], x1, x2 ∈ X implies that lx1 + (1 - l)x2 ∈ [x1, x2]. Thus the condition (1) implies that degree of membership of any element in [x1, x2] is never less than the degree of membership of x1 or x2. For example, let R be universal set of real

{ 

}

__ 1 ​, 1  ​. If m  (x) numbers and S = ​ -1, -​  1  ​, 0, ​ __ S 2 2 1 ______ = ​    2   ​is the membership function, then 1 + 5x  4 1 __ 4 1 __ __ S = ​ ​ -1, __ ​ 1 ​  ​, ​ - __ ​ 1 ​, __ 6 2 ​ 9 ​  ​, (0.1),​ ​ 2 ​, ​ 9 ​  ​, ​ 1, ​ 6 ​  ​  ​ 1 ​, 0.44  ​, (0, 1), ​ (-1, 0.16), ​ - ​ __ 2 1 __ ​   ​, 0.44  ​(1,0.16)  ​ 2

{ ( 

) ( 

( 

)

)

(  ) (  ) } (  )

}

is a fuzzy subset of R. We note that for any x, y ∈ R, the membership value of any element in [x, y] is greater than or equal to mS (x) and mS ( y). Hence S is convex in this case. Graph of the function mS is shown below:

Proof: Suppose first that S is convex. Therefore each a-cut Sa is convex. Let a = mS (x1) < mS (x2). Then x1, x2 ∈ Sa. Since Sa is convex, we have

l x1 + (1 - l) x2∈ Sa Therefore, by definition of a-cut mS [lx1 + (1- l) x2] ≥ a = mS (x1) = min[ mS (x1), mS (x2)]. Conversely, suppose that

mS ( lx1 + (1- l) x2] ≥ min [mS (x1), mS (x2)] (1) for x1, x2 ∈ X and 0 ≤ l ≤ 1. We wish to show that each a-cut of S is convex. Let x1, x2 ∈ Sa. Therefore mS (x1) ≥ a and mS (x2) ≥ a. Now Sa will be convex if l x1 + (1- l) x2 ∈ Sa, that is, if

mS [lx1 + (1- l) x2] ≥ a.

(2)

Since (1) holds, we have

mS ( lx1 + (1- l) x2] ≥ min [ mS (x1), mS (x2)] ≥ min [a, a] = a. Hence (2) holds and so Sa is convex.

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_

A normalized convex fuzzy set S on the set ℜ of real numbers is called a fuzzy number if its membership function mS is continuous. Let S = {(u, mS (u), u ∈ X )} be a fuzzy subset of the universal set X. Then the set {x ∈ X, mS (x) = 1} set is called the Core of the fuzzy set S and is denoted by core (S ). For example, core of the fuzzy set S = {(x, 1), ( y, 0.5), (z, 1)}is core (S ) = {x, z}. 16.2  STANDARD OPERATIONS ON A FUZZY SET (a) Complement of a Fuzzy set: Let S = {(S, ( mS (S )), s ∈ X }be a fuzzy set on the universal set X. Then the complement of S, denoted by S′, is the set S ′ = {(s, 1 - mS (s), s ∈ X }.

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16.4  n  chapter Sixteen EXAMPLE 16.4 Let X = {x, y, z} be universal set and

Solution. The given fuzzy sets are A = {(x, 0.4), (y, 0.6), (z, 0.2), (t, 0)}

S = {(x, 0.2), (y, 0.6), (z, 0.7)}

and

be a fuzzy set on X. Find the complement of S.

B = {(x, 0.7), (y, 0), (z, 0.4), (t, 0.3)}

Solution. We have S = {(x, 0.2), (y, 0.5), (z, 0.7)}. Therefore complement of S is the set S′ = {(x, 1 - 0.2), ( y, 1 - 0.5), (z, 1 - 0.7)} = {(x, 0.8), ( y, 0.5), (z, 0.3)} (b) Union of Fuzzy Sets: Let S = {(s, mS (s)), s ∈ X } and T = {t, mT (t), t ∈ X} be two fuzzy sets defined on a universal set X. Then union of S and T, denoted by S ∪ T , is the set defined by S ∪ T = {(u, max (mS (u), mT (u)), u ∈ X }. EXAMPLE 16.5 Let X = {x, y, z} be universal set and S = {(x, 0.2), (y, 0.5), (z, 0.7)} and T = {(x, 0.4), (y, 0.3), (z, 0.6)} be two fuzzy subsets of X. Find S ∪ T. Solution. We have S = {(x, 0.2), (y, 0.5), (z, 0.7)} T = {(x, 0.4), (y, 0.3), (z, 0.6)}.

Therefore A ∪ B = {u, max (XA(u), XB(u)), u ∈ X} = {(x, 0.7), (y, 0.6), (z, 0.4), (t, 0.3)}. (c) Intersection of Fuzzy sets: Let X be universal set and let S = {(s, mS (s), s ∈ X} and T = {(t, mt (t), t ∈ X}



be two fuzzy subsets of X. Then the intersection of S and T, denoted by S ∩ T, is the set defined by S ∩ T = {(u, min (mS (u), mT (u)), u ∈ X}. EXAMPLE 16.7 Let X = {x, y, z} be universal set and let A = {(x, 0.2), (y, 0.5), (z, 0.7)} and B = {(x, 0.4), (y, 0.3), (z, 0.6)} be fuzzy subsets of X, Find A ∩ B. Solution. We have A = {(x, 0.2), (y, 0.5), (z, 0.7)}

Since max (0.2, 0.4) = 0.4, max (0.5, 0.3) = 0.5, max (0.7, 0.6) = 0.7, it follows that S ∪ T = {(x, 0.4), (y, 0.5), (z, 0.7)}. EXAMPLE 16.6 If X = {x, y, z, t} be universal set and A = {(x, 0.4), (y, 0.6), (z, 0.2), (t, 0)} and B = {(x, 0.7), (y, 0), (z, 0.4), (t, 0.3)} are two fuzzy subsets of X, find A ∪ B.

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B = {(x, 0.4), (y, 0.3), (z, 0.6)}. Since min (0.2, 0.4) = 0.2, min (0.5, 0.3) = 0.3, min (0.7, 0.6) = 0.6, the intersection of A and B is

A ∩ B = {(x, 0.2), ( y, 0.3), (z, 0.6)}.

EXAMPLE 16.8 Let X be a universal set and let A and B be two fuzzy subset of X ? Show that (i) supp (X ) = X

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eleMentS of fuzzy logic   n 16.5 (ii) supp (A ∪ B = supp (A) ∪ supp (B) (iii) supp (A ∩ B = supp (A) ∩ supp (B). Solution. (i) Let X be a universal set, then X is a fuzzy set with all its membership values equal to 1. Hence Supp ( X ) = {x: xS (x) > 0} = X, since χX (x) = 1 for all x ∈ X. (ii) If A and B are two fuzzy subsets of a universal set X, then A ∪ B = {(x, max( mA(x), mB(x)), x ∈ X } Thus supp (A ∪ B) = {(x: max( mA(x), mB(x)) > 0} (1) On the other hand,

and so

supp (A) = {x: mA(x) > 0} supp (B) = {x: mB(x) > 0},

supp (A) ∪ supp (B) = {x, max(mA(x), mB(x)) > 0)} Now (1) and (2) imply that

(2)

supp (A ∪ B) = supp (A) ∪ supp (B). (iii) We have A ∩ B = {x, min ( mA(x), mB(x)), x ∈ X}. Therefore supp (A ∩ B) = {x: min(mA(x), mB(x)) > 0}  (3) On the other hand,

and so

_

​p​ = 1 - p p ∧ q = min ( p, q) p ∨ q = max ( p, q) p ⇒ q = min (1, 1 + q - p) p ⇔ q = 1 - |1 - q|. Thus, for this particular 'three value logic', the truth tables for p ∧ q, p ∨ q, p ⇒ q and p ⇔ q turn out to be as shown in the below tables: p 0

supp (A) ∩ supp (B) = {x: min ( mA(x), (4)

supp (A ∩ B) = supp (A) ∩ supp (B). 16.3  MANY VALUED LOGIC The classical logic deals with the propositions which are either true or false. Thus classical logic is two valued logic. The concept of

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(  )

0

supp (A) = {x: mA(x) > 0} supp (B) = {x: mB(x) > 0}

mB(x)) > 0} Form (3) and (4), it follows that

classical logic has been extended to three value logic by allowing a third truth value called indeterminate. Thus in three value logic a proposition may be true, false or indeterminate. Five best known three value logics are: 1. Lukasiewicz three value logic 2. Bochvar three value logic 3. Kleene three value logic 4. Heyting three value logic 5. Reichenback three value logic. In all these “three value logic”, we denote truth, falsity and indeterminacy by 1, 0 and __ ​ 1 ​ 2 respectively. If p is a_ proposition, then negation​ _ p​ of p is defined by p​ ​  = 1 - p._ Thus _ _ ​   ​  ​= __ ​  = 0, 0​ 1​ ​  = 1 and ​ __ ​ 1​ ​ 1 ​. 2 2 However operations of conjunctivity (∧) and disjunctivity (∨) differ from one logic to another in the above mentioned “three value logics”. In Lukasiewiez logic, if p and q are two propositions, then the logic operations are defined by

0 __ ​ 1 ​ 2 __ ​ 1 ​ 2 __ ​ 1 ​ 2 1 1 1

q

p q

p

0

0

0

​   ​ 2 1

0

0

0

0

1 __

1 ​ ​ __ 2 __ ​ 1 ​ 2 0 __ ​ 1 ​ 2 1

1 __

​   ​ 2 1 0

1 __

​   ​ 2 1

0

0

1 __

​   ​ 2 __ ​ 1 ​ 2 __ ​ 1 ​ 2 1 1 1

q 0

1 __

​   ​ 2 1

p∨q 0

1 __

​   ​ 2

​   ​ 2 1 1 ​ ​ __ 2 __ ​ 1 ​ 2

1

1

0

1

1 __

1

0

1 __

​   ​ 2 1

1

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16.6  n  chapter Sixteen p

q

p q

p

q

p⇔q

0

0 __ ​ 1 ​ 2 1

1

0

1

0

1 __ ​ 1 ​ 2

0 __ ​ 1 ​ 2 __ ​ 1 ​ 2 __ ​ 1 ​ 2 1

0 __ ​ 1 ​ 2 1

1 1 ​ ​ __ 2 0 1 ​ ​ __ 2

0 0 __ ​ 1 ​ 2 __ ​ 1 ​ 2 __ ​ 1 ​ 2 1 1 1

0

1 ​ ​ __ 2 1 0

1 ​ ​ __ 2 1

1 1 0

1 __

​   ​ 2 1

1 1

0

1 __

​   ​ 2 1 0

1 __

​   ​ 2 1

1

1 ​ ​ __ 2 0 1 ​ ​ __ 2 1

The concept “three-valued logic” has further been generalized into “ n-valued logic” for which the truth values are: 0 _____ n - 2  ​, ​  n_____ - 1  ​ 1  ​, _____ _____ . . . _____ ​      ​  2    n - 1 ​, ​ n - 1  n - 1 ​,  , ​  n - 1  n - 1  or n - 2  2  ​, . . .,​  _____ _____ 0, _____ ​  1     ​, 1. n - 1 ​, ​ n - 1  n - 1  These values are called degree of truth. 16.4  FUZZY LOGIC Logic based on fuzzy set theory is called fuzzy logic. It is an extension of n-valued logics. In fact fuzzy logic is any logic having the closed interval [0, 1] as its truth value set. If p and q are fuzzy propositions, then logical connectives are defined by T ( p ∧ q) = min [T ( p), T(q)] T ( p ∨ q) = max [T ( p), T(q)] _ T ( ​p​ ) = 1 - T ( p),



where T ( p), T (q) ∈ [0, 1]. However, the conditional proposition is defined in different ways. For example, (i) T ( p → q) = min[1 - T ( p) + T (q)] and (ii) T ( p → q) = max[min{T ( p), T (q)}, 1 - T ( p)]. The definition (ii) is due to L. A. Zadeh.

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16.5  FUZZY PROPOSITIONS Linguistic variables. Linguistic variables are those variables, where the meaning remains constant but form (values) varies. For example, 1. Pressure is a linguistic variable with linguistic values low pressure and high pressure etc. 2. Temperature is a linguistic variable with linguistic values cold, hot, cool, warm etc. 3. Age is a linguistic variable with linguistic values infant, young, middle-aged, old etc. Fuzzy Predicate. The linguistic values of a linguistic variable are called fuzzy predicates. For example, consider the propositions (1) Today is hot. (2) Ram is young. (3) Sohan is tall. In (1), hot is a linguistic value of the linguistic variable temperature and so it is a fuzzy predicate. In (2), young is a linguistic value of the linguistic variable age and so it is a fuzzy predicate. In (3), tall is a linguistic value of the linguistic variable height and so it is a fuzzy predicate. Fuzzy proposition. If P is a fuzzy predicate, then a proposition of the form ‘x is P’ is called a fuzzy proposition. Using the logical connectives ∧, ∨ and-, compound fuzzy preposition can be obtained. EXERCISES 1. Let X = {x, y, z, t} be universal set and S = {(x, 0), (y, 0.4), (z, 0.8), (t, 1)} be a fuzzy subset of X. Find (i) S ′, (ii) height (S), (iii) supp (S) and (iv) core (S). Ans. (i) S ′ = {(x, 1), (y, 0.6), (z, 0.2), (t, 0)} (ii) height (S) = 1 (iii) supp (S) = { y, z, t} (iv) core (S) = {t}. 2. Let X = {a, b, c} be universe of discourse and A = {(a, 0.5), (b, 1), (c, 0.4)} and B = {(a, 0.6), (b, 0.4), (c, 0.7)}

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eleMentS of fuzzy logic   n 16.7  int: Height (A) = 0.8. So dividing the H memberships by 0.8, we get

be fuzzy subsets of X. Find A ∪ B, A ∩ B, Aa for a = 0.4 and BN (normalized). Ans. A ∪ B = {(a, 0.6), (b,1), (c, 0.7)} A ∩ B = {(a, 0.5), (b, 0.4), (c, 0.4)} B0.4 = {a, b, c} 0.6 0.4 BN = ​ ​ a, ​ ___  ​  ​, ​ b, ​ ___  ​  ​, (c, 1)  ​ 0.7 0.7

{ ( 

) ( 

)

}

AN = {(x, 0), (y, 1), (3, 0.5)} as the normalized fuzzy set. 5. Let X = {x, y, z}be universal set and let and

= {(a, 0.86), (b, 0.56), (c,1)}



3. Let X = {x, y, z}be universal set and let and

be two fuzzy subsets of X. Show that supp (A ∪ B) = supp (A) ∪ supp (B). Hint: A ∪ B = {(x, 0.4), (y, 0.5), (z, 0.6)} Therefore supp (A ∪ B) = {x, y, z} Also supp (A) = {x, y, z} supp (B) = {x, y, z} Therefore supp (A) ∪ supp (B) = {x, y, z}. Hence supp (A ∪ B) = supp (A) ∪ supp (B). 4. Normalize the fuzzy set A = {(x, 0), (y, 0.8), (3, 0.4)}.

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B = {(x, 0.3), (y, 0.2), (z, 0.6)}

be two fuzzy subsets of X. Show that

A = {(x, 0.4), (y, 0.2), (z, 0.6)} B = {(x, 0.3), (y, 0.5), (z, 0.2)}

A = {(x, 0.2), (y, 0.4), (z, 0.7)}

(A ∪ B) ′ = A′ ∩ B ′ (De-Morgan Law)

Hint:

(A ∪ B) = {(x, 0.3), (y, 0.4), (z, 0.7)} (A ∪ B) ′ = {(x, 0.7), (y, 0.6), (z, 0.3)} A ′ = {(x, 0.8), (y, 0.6), (z, 0.3)} B ′ = {(x, 0.7), (y, 0.8), (z, 0.4)} A ′ ∩ B ′ = {(x, 0.7), (y, 0.6), (z, 0.3)}

Hence the result. 6. For the fuzzy sets in exercise 5, verify A ∪ (A ∩ B) = A (absorption law) Hint: A ∩ B = {(x, 0.2), (y, 0.2), (z, 0.6)} and so A ∪ (A ∩ B) = {(x, 0.2), (y, 0.4), (z, 0.7)} = A.

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17

Graphs

The interest in the study of graph theory has ­increased due to its applicability in so many fields like artificial intelligence, electrical ­engineering, transportation system, scheduling problems, ­economics, chemistry and operations research. 17.1  DEFINITIONS AND BASIC CONCEPTS

Definition 17.7  Two vertices that are connected by an edge are called adjacent. Definition 17.8  An endpoint of a loop is said to be adjacent to itself. Definition 17.9  An edge is said to be incident on each of its endpoints.

Definition 17.1  A graph G 5 (V, E) is a ­mathematical structure consisting of two finite sets V and E. The elements of V are called vertices (or nodes) and the elements of E are called edges. Each edge is associated with a set consisting of either one or two vertices called its endpoints.

Definition 17.10  Two edges incident on the same endpoint are called adjacent edges.

The correspondence from edges to endpoints is called edge-endpoint function. This function is generally denoted by g. Due to this function, some authors denote graph by G 5 (V, E, g).

Definition 17.12  A vertex of degree zero is called an isolated vertex.

Definition 17.2  A graph consisting of one vertex and no edges is called a trivial graph. Definition 17.3  A graph whose vertex and edge sets are empty is called a null graph. Definition 17.4  An edge with just one endpoint is called a loop or a self-loop. Thus, a loop is an edge that joins a single endpoint to itself. Definition 17.5  An edge that is not a self-loop is called a proper edge. Definition 17.6  If two or more edges of a graph G have the same vertices, then these edges are said to be parallel or multi-edges.

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Definition 17.11  The number of edges in a graph G which are incident on a vertex is called the degree of that vertex.

Thus, a vertex on which no edges are incident is called isolated. Definition 17.13  A graph without multiple edges (parallel edges) and loops is called simple graph. Notations: In pictorial representations of a graph, the vertices will be denoted by dots and edges by line segments. EXAMPLE 17.1 (a) Let V 5 {1, 2, 3, 4}  and  E 5 {e1, e2, e3, e4, e5}. Let g be defined by g (e1) 5 g (e5) 5 {1, 2},  g (e2) 5 {4, 3}, g (e3) 5 {1, 3},  g (e4) 5 {2, 4}.

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17.2  n  chapter Seventeen We note that both edges e1 and e5 have same endpoints {1, 2}. The endpoints of e2 are {4, 3}, the endpoints of e3 are {1, 3} and endpoints of e4 are {2, 4}. Thus the graph is as shown in the Figure 17.1. e5

e5 1

1 e 1

e4 4

4

e4 e2

2

e1 e3 e2

2 or

e3 3

or

3

1

1

e3

e3

3

3

e5

e5

e1

e1

e2

e2

(iv) It is a multi-graph because it has multiedges and a loop. (v) A and B are adjacent vertices because they are connected by the edge e2 or e1, whereas A and E are not adjacent.

2

2

e4

e4

4

4

The edges e2 and e3 are adjacent edges because they are incident on the same vertex B. (c) Consider the graph with the vertices A, B, C, D and E pictured in Figure 17.3.

A

B

Figure 17.1 (b) For the graph pictured in the Figure 17.2, (i) Write down the degree of the vertices A, B, D. (ii) Which of the edges are parallel? (iii) Which vertex is isolated? (iv) Point out whether it is a simple graph or a multi-graph. (v) Point out one pair of adjacent vertices and one pair of adjacent edges. e1 e2

A

B

e3

e4

E

C e5

F

E

D

Figure 17.3 In this graph, we note that Number of edges 5 5 Degree of vertex A 5 4 Degree of vertex B 5 2 Degree of vertex C 5 3 Degree of vertex D 5 1 Degree of vertex E 5 0 Sum of the degree of vertices    5 4 1 2 1 3 1 1 1 0 5 10. Thus, we observe that 5

D

​ ​​   deg (v ) 5 2e, ​

e6

Figure 17.2 In this graph, (i)  We notice that Degree of the vertex A 5 2 Degree of the vertex B 5 4 Degree of the vertex D 5 3   (because loop e6 has degree 2). The vertex D has degree 3 (because e6 is a loop and so has degree 2) (ii)  The edges e1 and e2 are parallel (iii) The vertex F is isolated because no edge is ­incident on F

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C

  

i51

i 

where deg(vi) denotes the degree of vertex vi and e denotes the number of edges. Euler’s Theorem 17.1 (The First Theorem of Graph Theory) The sum of the degrees of the vertices of a graph G is equal to twice the number of edges in G. Thus, total degree of a graph is even. Proof:  Each edge in a graph contributes a count of 1 to the degree of two vertices (endpoints of the edge), that is, each edge contributes 2 to the degree sum. Therefore, the sum of degrees

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graphS   n 17.3 of the vertices is equal to twice the number of edges.

1 5 1 7 1 8 1 4 5 37, which is odd. This contradicts the first theorem of graph theory, according to which, total degree of a graph is always even.

Corollary 17.1  There can be only an even number of vertices of odd degree in a given graph G.

17.2  SPECIAL GRAPHS

Proof:  We know, by the fundamental theorem, that

Definition 17.14  A graph G is said to simple if it has no parallel edges or loops. In a simple graph, an edge with endpoints v and w is denoted by {v, w}.

n

​ ​​   deg (v ) 5 2 3 number of edges. ​

  

i51

i 

Thus the right-hand side is an even number. Hence to make the left-hand side an even number there can be only even number of vertices of odd degree.

Definition 17.15  For each integer n ≥ 1, let Dn denote the graph with n vertices and no edges. Then Dn is called the discrete graph on n vertices. For example, we have   •   •   •   and   •   •   •   •   • D3 D5

Remarks 17.1 (i) A vertex of degree d is also called a -valent vertex. (ii) The degree (or valence) of a vertex v in a graph G is the number of proper edges incident on v plus twice the number of selfloops.

Definition 17.16  Let n ≥ 1 be an integer. Then a simple graph with n vertices in which there is an edge between each pair of distinct vertices is called the complete graph on n vertices. It is denoted by Kn. For example, the complete graphs K2, K3 and K4 are shown in Figure 17.4.

Theorem 17.2  A nontrivial simple graph G must have at least one pair of vertices whose degrees are equal. Proof:  Let the graph G has n vertices. Then there appear to be n possible degree values, namely 0, 1, . . ., n − 1. But there cannot be both a vertex of degree 0 and a vertex of degree n − 1 because if there is a vertex of degree 0 then each of the remaining n − 1 ­vertices is adjacent to at most n − 2 other ­vertices. Hence the n vertices of G can realize at most n − 1 possible values for their degrees. Hence, the pigeonhole principle implies that at least two of the vertices have equal degree. EXAMPLE 17.2  Is there a graph with eight vertices of degree 2, 2, 3, 6, 5, 7, 8, 4? Solution.  The answer is No. In such a graph, there will be three vertices of odd degree which is impossible. We can also argue as follows: Total degree of the graph (if possible) is equal to 2 1 2 1 3 1 6 

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v1

v1

K2

v2 K2

v2

v1

v3

v3

v1 K3

Kv32

v4

v2 v1

v4

v1 K 4

v3

v3

K4 v2

v2

Figure 17.4 Definition 17.17  If each vertex of a graph G has the same degree as every other vertex, then G is called a regular graph. A k-regular graph is a regular graph whose common degree is k. For example, consider K3. The degree of each vertex in K3 is 2. Hence K3 is regular. Similarly, K4 is regular. Also the graph shown in Figure 17.5 is regular because degree of each vertex here is 2. But this graph is not complete because v2 and v4 have not been connected through an edge. Similarly, v1 and v3 are not connected by any edge.

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17.4  n  chapter Seventeen v4

v3

v1

v2

2-regular graph

,

Figure 17.8 EXAMPLE 17.5  The smallest possible simple graph that is not bipartite is the complete graph K3 shown in ­Figure 17.9.

Figure 17.5 Thus, a complete graph is always regular but a regular graph need not be complete. EXAMPLE 17.3  The oxygen molecule O2, made up of two oxygen atoms linked by a double bond can be represented by the regular graph shown in Figure 17.6.

K3

Figure 17.9 Definition 17.20  A complete bipartite graph G is a simple graph whose vertex set V can be partitioned into two subsets U 5 {v1, v2, . . ., vm ) and W 5 {w1, w2, . . ., wn} such that for all i, k in {1, 2, . . ., m} and j, l in {1, 2, . . ., n}

Figure 17.6 Definition 17.18  Let n ≥ 1 be an integer. Then a graph Ln with n vertices {v1, v2, . . ., vn} and with edges {vi,vi + 1} for 1 ≤ i  n − 1, then there will be more than n vertices in the sequence. But the graph is with n vertices. Therefore some vertex,

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v6

(c) v1

v2

v3

v4

v5

v6 v1

(d)

v7 v3 v6

v5 v2

v4

Figure 17.51

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graphS   n 17.13 Solution.  In Figure 17.51, we observe that (a) Since every two vertices in the given graph are connected by a path, therefore the given graph is connected. (b) This graph is also connected. (c) The graph in (c) is disconnected. It has two connected components. (d) In this graph, the edge (v5, v6) cross the edges (v1, v2) and (v3, v4) at points which are not vertices. Therefore the graph can be redrawn as shown in Figure 17.52. v1

v3

v1

EXAMPLE 17.17  Find the connected components of the graphs given in the following (Figure 17.54). (a) v1 v1 v3 v3 v1 v1

v4

v3 v5 v4

v2 v2

e1 e1 e2 e2

v6

v4v5



v4 v4

e6 e6

v5 v5 e5 e5 v6 v6

v2

v1

v1

v3

v3

v2

v2

v3v4

v4

e1

v3

e2



H1: with vertex set {v1, v2, v3} and edge set {e1, e2} H2: with vertex set {v4} and edge set f H3: with vertex set {v5} and edge set f H4: with vertex set {v6, v7, v8,v9} and edge set {e3, e4, e5} e3

v8 e5

v5

(a)          (b)

v1

Figure 17.55

v6 v4

e4 e4

v2

v2

v3

v2 v2

v8 ve8 5 ev5 7 v7

Solution. (a) The graph in (a) has four connected components (Figures 17.55 and 17.56)

EXAMPLE 17.16  Which of the graphs shown in Figure 17.53 are ­connected? v1

v9 v9

v5 v5

e3 e3

e3 e3 e4 e4

Figure 17.54

Figure 17.52 There is no path from v2 to v4, etc. Hence the given graph is disconnected and has three connected components.

v1

v6 v6

v4 v4

v3 v3

v6

v2

e1 e1 e2 e2

(b) v5

v2

Definition 17.47  If a graph G is disconnected, then the various connected pieces of G are called the connected ­components of the graph.

v9

e4

v7

Figure 17.53

Figure 17.56

Solution.  Graph (a) is not connected as there is no walk from any of v1, v2, v3, v4 to the vertex v5. The graph (b) is clearly connected.

(b) The graph in (b) is disconnected and have two connected components (Figures 17.57(a), 17.57(b))

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17.14  n  chapter Seventeen H1:

e1

v1 e2

v2 e3

v3

Figure 17.57(a)

with vertex set {v1, v2, v3} and edge set {e1, e2, e3} H2: v5

e4 e5

v4 e6

v6

Figure 17.57(b)

with vertex set {v4, v5, v6} and edge set {e4, e5, e6}.

Remark 17.3  If a connected component has n vertices, then degree of any vertex cannot exceed n − 1. 17.6  EULERIAN PATHS AND CIRCUITS Definition 17.48  A path in a graph G is called an ­Euler path if it includes every edge exactly once. Definition 17.49  A circuit in a graph G is called an Euler circuit if it includes every edge exactly once. Thus, an Euler circuit (Eulerian trail) for a graph G is a sequence of adjacent vertices and edges in G that starts and ends at the same vertex, uses every vertex of G at least once, and uses every edge of G exactly once. Definition 17.50  A graph is called Eulerian graph if there exists an Euler circuit for that graph. EXAMPLE 17.19  Find which of the following are Euler paths and ­Euler circuits in the graph given below (Figure 17.60) v1

EXAMPLE 17.18  Find the number of connected components in the graph shown below (Figure 17.58).

e1 v2 e2

e5 v5

v3 e3

e4 v4

Figure 17.60 Figure 17.58 Solution.  The connected components are shown in Figure 17.59.

and and

Figure 17.59

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(a) v1 e1 v2 e2 v3 e3 v4 e4 v5 e5 v2 (b) v1 e1 v2 e2 v3 e3 v4 e4 v5 e5 v2 e1 v1 (c) v2 e2 v3 e3 v4 e4 v5 e5 v2 Solution.

(a) The walk in (a) is an Euler path but it is not Euler circuit because it is not closed. (b) It is not an Euler circuit because the edge e1 is covered twice. (c) It is neither an Euler path nor an Euler circuit because the vertex v1 of G has not been used.

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graphS  n 17.15 Theorem 17.4  If a graph has an Euler circuit, then every vertex of the graph has even degree. Proof:  Let G be a graph which has an Euler circuit. Let v be a vertex of G. We shall show that degree of v is even. By definition, Euler circuit contains every edge of graph G. Therefore the Euler circuit contains all edges incident on v. We start a journey ­beginning in the middle of one of the edges ­adjacent to the start of Euler circuit and continue around the Euler circuit to end in the middle of the starting edge. Since Euler circuit uses every edge exactly once, the edges incident on v occur in entry/exist pair and hence the degree of v is a multiple of 2. Therefore, the degree of v is even. This completes the proof of the theorem. Starting point

In graph (b), we have deg(v2) 5 3  and  deg(v4) 5 3. Since there are vertices of odd degree in the given graph, therefore it does not have an Euler circuit. Remark 17.4  The converse of Theorem 17.4 is not true. There exist graphs in which every vertex has even degree but the Euler circuits do not exist. For example,

and

v

Figure 17.61 We know that contrapositive of a conditional statement is logically equivalent to the statement. Thus, Theorem 17.4 is equivalent to the following: Theorem 17.5  If a vertex of a graph is not of even ­degree, then it does not have an Euler ­circuit. Thus, If some vertex of a graph has odd degree, then that graph does not have an Euler ­circuit. EXAMPLE 17.20  Show that the graphs shown in Figure 17.62 do not have Euler circuits. v1

v2

v1

v2

v3

v4

v4

v3

      (a)          (b) Figure 17.62 Solution.  In graph (a), degree of each vertex is 3. Hence this does not have an Euler circuit.

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Figure 17.63 are graphs (Figure 17.63) in which each vertex has degree 2 but these graphs do not have Euler circuits since there is no path which uses each vertex at least once. Theorem 17.6  If G is a connected graph and every vertex of G has even degree, then G has an Euler circuit. Proof:  Let every vertex of a connected graph G has even degree. If G consists of a single vertex v, the trivial walk from v to v is an Euler circuit. So suppose that G consists of more than one vertices. We start from any vertex v of G. Since the degree of each vertex of G is even, if we reach each vertex other than v by travelling on one edge, the same vertex can be reached by travelling on another previously unused edge. Thus a sequence of distinct ­adjacent edges can be produced indefinitely as long as v is not reached. Since the number of edges of the graph is finite (by definition of graph), the sequence of distinct edges will terminate. Thus the sequence

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17.16  n  chapter Seventeen must return to the starting vertex. We thus obtain a sequence of adjacent vertices and edges starting and ending at v without repeating any edge. Thus we get a circuit C. If C contains every edge and vertex of G, then C is an Euler circuit. If C does not contain every edge and vertex of G, remove all edges of C from G and also any ­vertices that become isolated when the edges of C are ­removed. Let the resulting subgraph be G′. We note that when we removed edges of C, an even number of edges from each vertex have been removed. Thus degree of each remaining vertex remains even. Further, since G is connected, there must be at least one vertex common to both C and G′. Let it be w (in fact there are two such vertices). Pick any ­sequence of adjacent vertices and edges of G′ starting and ending at w without repeating an edge. Let the resulting circuit be C′. Join C and C′ together to create a new circuit C″. Now, we observe that if we start from v and follow C all the way to reach w and then follow C′ all the way to reach back to w. Then continuing travelling along the untravelled edges of C, we reach v. If C ″ contains every edge and vertex of C, then C ″ is an Euler circuit. If not, then we again repeat our process. Since the graph is finite, the process must terminate. The process followed has been described in the graph G shown below (Figure 17.64). G′ C′

v w

Theorem 17.7 (Euler’s Theorem)  A finite ­connected graph G has an Euler circuit if and only if every vertex of of G has even degree. Thus, finite connected graph is Eulerian if and only if each vertex has even degree. Theorem 17.8  If a graph G has more than two vertices of odd degree, then there can be no Euler path in G. Proof:  Let v1, v2 and v3 be vertices of odd degree. Since each of these vertices had odd degree, any possible Euler path must leave (arrive at) each of v1, v2, v3 with no way to return (or leave). One vertex of these three vertices may be the beginning of Euler path and another the end, but this leaves the third vertex at one end of an untravelled edge. Thus, there is no Euler path. v1 v2

v1 v2

or

v3

v3

(Graphs having more than two vertices of odd degree).

Figure 17.65 Theorem 17.9  If G is a connected graph and has ­exactly two vertices of odd degree, then there is an Euler path in G. Further, any Euler path in G must begin at one vertex of odd degree and end at the other. Proof:  Let u and v be two vertices of odd degree in the given connected graph G (see Figure 17.66) u

C v

(Graph G)

u e

v

v

w

C′′

Figure 17.64 Theorems 17.4 and 17.6 taken together imply:

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G

G′

Figure 17.66

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graphS   n 17.17 If we add the edge e to G, we get a connected graph G′ all of whose vertices have even degree. Hence there will be an Euler circuit in G′. If we omit e from Euler circuit, we get an Euler path beginning at u (or v) and ending at v (or u). EXAMPLE 17.21  Has the graph given in Figure 17.67 an Eulerian path? B

A

C

D

Solution.  The given graph is connected. Further deg(v1) 5 3,  deg(v2) 5 4, deg(v3) 5 3,  deg(v4) 5 4. Since this connected graph has vertices with odd ­degree, it cannot have Euler circuit. But this graph has Euler path, since it has exactly two vertices of odd degree. For example, v3 e2 v2 e7 v4 e6 v2 e1 v1 e4 v4 e3 v3 e5 v1 is an Euler path. EXAMPLE 17.23  Consider the graph given below (Figure 17.70):

Figure 17.67 Solution.  In the given graph, deg(A) 5 1,  deg(B) 5 2, deg(C) 5 2,  deg(D) 5 3. Thus the given connected graph has exactly two vertices of odd degree. Hence, it has an Eulerian path. If it starts from A (vertex of odd degree), then it ends at D (vertex of odd degree). If it starts from D (vertex of odd degree), then it ends at A (vertex of odd degree). But, on the other hand, let the graph be as shown below (Figure 17.68) B e2

e1

A

e4

C

Then deg(A) 5 1, deg(B) 5 3, deg(C) 5 1, degree of D 5 3 and so we have four vertices of odd degree. Hence it does not have an Euler path. EXAMPLE 17.22  Does the graph given in Figure 17.69 possess an ­Euler circuit?

e4

e5

v1

e6 e1

Figure 17.69

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e2 v2

v4

EXAMPLE 17.24  Is it possible to trace the graph in Figure 17.71 without lifting the pencil?

v4

v2

v3

v1

v1

Figure 17.68

e3

v3

Figure 17.70 Here, deg(v1) 5 4, deg(v2) 5 4, deg(v3) 5 2, deg(v4) 5  2. Thus degree of each vertex is even. But the graph is not Eulerian since it is not connected.

D e3

v4

v2

e7

v9

v6 v5

v3

v8 v7

Figure 17.71 Solution.  The problem is equivalent to say that “Is it possible for this graph to have an Eulerian circuit”? We observe that deg(v1) 5 3, deg(v2) 5 2, deg(v3) 5 2, deg(v4) 5 3, deg(v5) 5 4, deg(v6) 5 2, deg(v7) 5 2, deg(v8) 5 4, deg(v9) 5 2. Since the graph contains vertex of odd degree, therefore it cannot have Euler’s circuit and ­therefore can not be traced without lifting the pencil.

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17.18  n  chapter Seventeen EXAMPLE 17.25  Find the Euler’s circuit for the graph given below (Figure 17.72): v2

e1 v1

e13

e14

e16

e12 e15

v6

e3

e4

v3

v12

v11

v4

e2

e11

e10

e9 v9

v10

v5

e5

e6 v7 e7

e8 v8

Figure 17.72 Solution.  The given connected graph has 12 vertices and degree of each vertex is even. Hence, by Euler’s theorem, this has an Euler’s circuit. For example, we observe that v1 e1 v2 e13 v11 e12 v10 e15 v12 e14 v2 e2 v3 e3 v4 e4 v5 e5 v6 e6 v7 e7 v8 e8 e9 v9 e10 v3 e11 v10 e16 v1 is an Euler’s circuit. In short we can represent it by e1, e13, e12, e15, e14, e2, e3, e4, e5, e6, e7, e8, e9, e10, e11, e16. EXAMPLE 17.26 (The bridges of Konigsberg)  The graph theory began in 1736 when Leonhard ­Euler solved the problem of seven bridges on Pregel River in the town of Konigsberg in Prussia (now Kaliningrad in Russia). The two islands and seven bridges are shown in Figure 17.73. Island A

Bridge

Bridge C

D

Bridge

Bridge Island

“Beginning anywhere and ending any where, can a person walk through the town of Konigsberg crossing all the seven bridges exactly once?” Euler showed that such a walk is impossible. He replaced the islands A, B and the two sides (banks) C and D of the river by vertices and the bridges as edges of a graph. We note then that deg(A) 5 3,  deg(B) 5 5, deg(C) 5 3,  deg(D) 5 3. Thus the graph of the problem is as shown in ­Figure 17.74. A (Island)

B (Island) (Euler’s graphical representation of seven bridges problem)

Figure 17.74 The problem then re1duces to “Is there any Euler’s path in the above ­diagram?”. To find the answer, we note that there are more than two vertices having odd degree. Hence there exists no Euler path for this graph. EXAMPLE 17.27  The floor plan shown in Figure 17.75 is for a house that is open for public viewing. Each room is connected to every room with which it has a common wall and to the outside along each wall. Is it possible to begin in a room or outside and take a walk which goes through each door exactly once?

Bridge

e1

B Bridge

C (Side of the river)

(Side of the river) D

D

e10

Room A e2

Bridge

D

e3 Room B e6

e5

River

e7

Figure 17.73 The people of Konigsberg posed the following question to famous Swiss Mathematician Leonhard Euler:

M17_Baburam_ISBN _C17.indd 18

Room C

e4 D e9

Doorway

e8 D (Outside)

Figure 17.75

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graphS  n 17.19 If each room and out side constitute a vertex and each door corresponds to an edge, then the floor plan converts into the graph. There are two edges e1 and e2 from A to D, two edges e3 and e4 from B to D, three edges e7, e8, e9 from C to D, one edge e10 from A to B, one edge e5 from A to C, one edge e6 from B to C. Hence the graph is as shown below (Figure 17.76):

In this graph, if we remove the edge e3, then the graph breaks into two connected components given below (Figure 17.78): e1 e5

v1

e2

e1

e3

e2 e5

D e9 e e8 7

B e4 e6

Figure 17.76 We note that

Figure 17.78 Hence the edge e3 is a bridge in the given graph. Method 1:  We know that if every vertex of a non-empty connected graph has even degree, then the graph has an Euler circuit. We shall make use of this result to find an Euler path in a given graph. Consider the graph shown in Figure 17.79. v2

deg(A) 5 deg(B) 5 4,  deg(C) 5 5,  deg(D)  5 7. Since the graph is connected and degree of exactly two vertices C and D is odd, there exists an Euler’s path and that path should start from one vertex of odd degree and end to the other vertex of odd degree. Therefore, either the path will begin from C and end at D or it will begin from D and end at C. An Euler’s path is shown (which begins from C). Definition 17.51  An edge in a connected graph is called a bridge or a cut edge if deleting that edge creates a disconnected graph. For example, consider the graph shown below (Figure 17.77): e1 e5 e3

v4

v2 e2 v3

e4 v5

Figure 17.77

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v5

17.6.1  METHODS FOR FINDING EULER CIRCUIT

C

v1

e4

v3

e10 A

v4

v2

e1 v1

v6 e2

e3 e4

e9 v3

e6

e5

v5 e10

e8 e7

v7

e11 v8

v4 e12

Figure 17.79 We note that deg(v2) 5 deg(v4) 5 deg(v6) 5 deg(v8) 5 2, deg(v1) 5 deg(v3) 5 deg(v5) 5 deg(v7) 5 4. Hence all vertices have even degree. Also the given graph is connected. Hence the given graph has an Euler circuit. We start from the vertex v1 and let C be C: v1 v2 v3 v1.

Then C is not an Euler circuit for the given graph but C intersects the rest of the graph at v1 and v3. Let C ′ be C ′: v1v4 v3 v5 v7 v6 v5 v8 v7 v1.

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17.20  n  chapter Seventeen (In case we start from v3, then C′ will be v3 v4 v1 v7 v6 v5 v7 v8 v5). Path C′ into C and obtain

v1 e1

C ″: v1v2 v3 v1 v4 v3 v5 v7 v6 v5 v8 v7 v1,

e7

e2

that is, C ″: e1e2 e3 e4 e5 e6 e7 e8 e9 e10 e11 e12.

e9 v3 e6

v5

(If we had started from v2, then C ″: v1v2 v3 v4 v1 v7 v6 v5 v7 v8 v5 v3 v1 or e1e2 e5 e4 e12 e8 e9 e7 e11 e10 e6 e3). In C ″ all edges are covered exactly once. Also every vertex has been covered at least once. Hence C″ is an Euler circuit. Method 2:  Fleury’s Algorithm: This algorithm is used to find Euler’s circuit for a connected graph with no vertices of odd degree. We shall illustrate the method with the help of the following example: Let G be the connected graph as shown in the Figure 17.80.

Current Path

e8

v2

e3

e5 e4 v6

v4

Figure 17.80 Step 1.  We begin from any vertex, say v1 Step 2.  We then construct the table given below. Hence one possible Euler circuit is C: v1 v2 v4 v5 v6 v3 v5 v2 v3 v1. In term of edges, this Euler circuit can be expressed as e1 e2 e3 e4 e5 e6 e7 e8 e9.

Next Edge

Reason

v1

{v1, v2}

No Edge from v1 is a bridge, so choose any edge, say {v1, v2}

v1 v2

{v2, v4}

(v2, v5) a bridge, so choose {v2, v4}

v1 v2 v4

{v4, v5}

only one edge {v4, v5} from v4 remains

v1 v2 v4 v5

{v5, v6}

Since {v2,v5} and {v5, v3} are bridges, choose {v5, v6}

v1 v2 v4 v5 v6

{v6, v3}

only one edge {v6, v3} remains

v1 v2 v4 v5 v6 v3

{v3, v5}

Since {v1, v3} is a bridge, choose either {v3, v2} or {v3, v5}

v1 v2 v4 v5 v6 v3 v5

{v5, v2}

only one edge {v5, v3} remains

v1 v2 v4 v5 v6 v3 v5 v2

{v2, v3}

only one edge {v2, v3} remains

v1 v2 v4 v5 v6 v3 v5 v2 v3

{v3, v1}

only one edge {v3, v1} remains

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graphS   n 17.21 17.7  HAMILTONIAN CIRCUITS Definition 17.52  A Hamiltonian path for a graph G is a sequence of adjacent vertices and distinct edges in which every vertex of G appears exactly once. Definition 17.53  A Hamiltonian circuit for a graph G is a sequence of adjacent vertices and distinct edges in which every vertex of G appears exactly once, ­except for the first and the last which are the same. Definition 17.54  A graph is called Hamiltonian if it admits a Hamiltonian circuit.

EXAMPLE 17.30  The graph shown below does not have a Hamiltonian circuit. v4

v1

v3 v2

v5

Figure 17.83 EXAMPLE 17.31  The graph shown in Figure 17.84 does not have a Hamiltonian circuit.

EXAMPLE 17.28  The wooden graph shown in Fig. 17.81 and constructed by William Hamilton in the shape of a regular dodecahedron is a Hamiltonian c­ ircuit. v1 v20 v13

v2

v3

v19

v14 v18

v12 v11

Figure 17.84

v15 v10

v16

v6

v17

v9

v5

v7 v8 v4

Figure 17.81 The Hamilton circuit is v1 v2 v3 . . . v18 v19 v20 v1. EXAMPLE 17.29  A complete graph Kn has a Hamiltonian circuit. In particular, the graphs shown in Figure 17.82 are Hamiltonian. and K3

K4

Figure 17.82

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Remark 17.5  It is clear that only connected graphs can have Hamiltonian circuit. However, there is no simple criterion to tell us whether or not a given graph has Hamiltonian circuit. The following results give us some sufficient conditions for the existence of Hamiltonian circuit/path. Theorem 17.10  Let G be a linear graph of n vertices. If the sum of the degrees for each pair of vertices in G is greater than or equal to n − 1, then there exists a Hamiltonian path in G. Theorem 17.11  Let G be a connected graph with n vertices. If n ≥ 3 and deg(v) ≥ n for each vertex v in G, then G has a Hamiltonian circuit. Theorem 17.12  Let G be a connected graph with n vertices and let u and v be two vertices of G that are not adjacent. If deg(u) 1 deg(v) ≥ n, then G has a Hamiltonian circuit.

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17.22  n  chapter Seventeen Corollary 17.2  Let G be a connected graph with n vertices. If each vertex has degree greater than or equal to n/2, then G has a Hamiltonian circuit. Proof:  It is given that degree of each vertex is greater than or equal to n/2. Hence the sum of the degree of any two vertices is greater than or equal to n/2 1 n/2 5 n. So, by the above theorem, the graph G has a Hamiltonian circuit. Theorem 17.13  Let n be the number of vertices and m be the number of edges in a connected 1 ​ (n2 – 3n 1 6), then G has a graph G. If m ≥ ​ __ 2 Hamiltonian c­ ircuit. The following example shows that the above conditions are not necessary for the existence of Hamiltonian path. EXAMPLE 17.32  Let G be the connected graph shown in the ­Figure 17.85. v8

v7

v6

v1

v5

v2

We note that

v3

v4

Figure 17.85

n 5 Number of vertices in G 5 8, m 5 Number of edges in G 5 8, Degree of each vertex 5 2. Thus, if u and v are non-adjacent vertices, then deg(u) 1 deg(v) 5 2 1 2 5 4  8. Also, __ ​ 1 ​(n2 – 3n 1 6) 5 __ ​ 1  2 ​ (64 –  24 1 6) 5 23. 2 1 ​ (n2 – 3n 1 6). Therefore the Clearly m  ​ __ 2 above two theorems fail. But the given graph has ­Hamiltonian circuit. For example, v1, v2, v3, v4, v5, v6, v7, v8, v1 is an Hamiltonian circuit for the graph.

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Proposition 17.1  Let G be a graph with at least two vertices. If G has a Hamiltonian circuit, then G has a subgraph H with the following properties: 1. H contains every vertex of G 2.  H is connected 3.  H has the same number of edges as v­ ertices 4.  Every vertex of H has degree 2 The contrapositive of this proposition is “If a graph G with at least two vertices does not have a subgraph H satisfying (1) – (4), then G does not have a Hamiltonian circuit”. Also we know that contrapositive of a statement is logically equivalent to the statement. Therefore the above result can be used to show non-existence of a Hamiltonian circuit. EXAMPLE 17.33  Consider the graph G shown below (Figure 17.86): v3

v1

v2 v5

v4

Figure 17.86 Let us verify whether this graph has a Hamiltonian circuit. Suppose that G has a Hamiltonian circuit. Then it has a subgraph H such that 1.  H contains every vertex of G, i.e., H has 5 vertices v1, v2, v3, v4, v5 2. H is connected 3. H has 5 edges 4. Every vertex of H has degree 2 Since the degree of v2 in G is 4 and every vertex of H has degree 2, two edges incident on v2 must be removed from G to create H. But we note that the edge {v1, v2} cannot be removed, because the removal of {v1, v2} will decrease the degree of v1 to 1, which is less than 2. Similarly the edge {v2, v3}, {v2, v4} and {v2, v5} cannot be removed. Hence the degree of v2 in H must

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graphS   n 17.23 stand at 4 which contradicts the condition that every vertex in H has degree 2 in H. Therefore, no such subgraph H exists. Hence G does not have a Hamiltonian circuit. EXAMPLE 17.34  Does the graph G given below (Figure 17.87) have Hamiltonian circuit? a

b

e c

d

Figure 17.87 Solution.  The given graph has   n 5 Number of vertices 5 5   m 5 Number of edges 5 8   deg(a) 5 deg(b) 5 deg(c) 5 deg(d) 5 3   deg(e) 5 4 We observe that (i) Degree of each vertex is greater than n/2 (ii) The sum of degrees of any non-adjacent pair of vertices is greater than n 1 ​(n2 – 3n 1 6) 5 ​ __ 1 ​(25 – 15 1 6) 5 8. (iii) ​ __ 2 2 1 ​ (n2 − 3n 1 6) is satisThus the condition m ≥ ​ __ 2 fied.

(iv) The sum of degrees of each pair of ­vertices in the given graph is greater than n − 1 5 5 − 1 5 4. Thus four sufficiency conditions are satisfied (whereas one condition out of these four conditions is sufficient for the existence of ­Hamiltonian path/circuit). Hence the graph has a Hamiltonian circuit. For example, the following circuits in G are Hamiltonian (Figure 17.88). a a

a a

b b andand

e e c c

d d

e e c c

Figure 17.88

M17_Baburam_ISBN _C17.indd 23

b b

d d

EXAMPLE 17.35  Does the graph shown below (Figure 17.89) has Hamiltonian circuit? a

b

e c

d

Figure 17.89 Solution. Here   Number of vertices (n) 5 5   Number of edges (m) 5 4   deg(a) 5 deg(b) 5 deg(c) 5 deg(d) 5 1   deg(e) 5 4 We note that (i) deg(a) 5 deg(b) 5 deg(c) 5 deg(d)  __ ​ 5 ​ 2 (ii) deg(a) 1 deg(b) 5 2  5, that is sum of any non-adjacent pair of vertices is not greater than 5 1 ​(n2–3n 1 6) 5  __ (iii) ​ __ ​ 1 ​(25 –15 1 6) 5 8 2 2 1 ​ (n2 – 3n 1 6) is Therefore the condition m ≥  ​ __ 2 not satisfied, (iv) deg(a) 1 deg(b) 5 2  4, i.e., the condition that sum of degrees of each pair of vertices in the graph is not greater than or equal to n − 1. Hence no sufficiency condition is satisfied. So we try the Proposition 17.1. Suppose that G has a Hamiltonian circuit. Then G should have a subgraph which contains every vertex of G, and ­number of vertices and number of edges in H should be same. Thus, H should have five vertices a, b, c, d, e and five edges. Since G has only four edges, H cannot have more than four edges. Hence no such subgraph is possible. Hence, the given graph does not have Hamiltonian circuit. EXAMPLE 17.36  Does the graph shown below (Figure 17.90) ­possess a Hamiltonian circuit? A

D

B

C

Figure 17.90

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17.24  n  chapter Seventeen Solution.  In the given graph   Number of vertices (n) 5 4   Number of edges (m) 5 8   Degree of each vertex 5 4 Thus we see that (1) Degree of each vertex is greater than n/2 (ii) Sum of degree of each pair of vertices is greater than n − 1,

1 ​ (n2 – 3n 1 6) 5​ __ 1 ​ (16 – 12 1 6) 5 5 and (iii) ​ __ 2 2 1 __ so m ≥ ​   ​ (n2 – 3n 1 6). 2 Hence the given graph has a Hamiltonian path. For example, ABCDA is a Hamiltonian path in the given graph (see Figure 17.91).

A

D

B

C

So suppose that G has a Hamiltonian circuit. Then it should have a connected subgraph H containing six vertices, six edges and degree of each vertex should be 2. To have degree of each vertex equal to 2, we should remove two edges from C and two edges from B. For example, now, if we remove CE from C, then deg(E) ≠ 2, If we remove CF, then deg(F) ≠ 2. So, we cannot remove CE and CF. If we remove CD, then deg(D) ≠ 2. If we remove CA, then deg(A) ≠ 2. Hence no such subgraph H exists. So, G cannot have a Hamiltonian circuit. Remark 17.6  Since the degree of each vertex in the above graph is even, it has Eulerian circuit. Thus the graph in EXAMPLE 17.37 is Eulerian but not ­Hamiltonian.

Figure 17.91 EXAMPLE 17.37  Is the graph shown in Figure 17.92 a ­Hamiltonian? B

A

(iii) deg(A) 1 deg(E) 5 2 1 2 5 4 and so the condition that “the sum of degrees of non-adjacent vertices is greater than or equal to n” is not satisfied (iv)  “The sum of degrees of any pair of vertices is greater than or equal to n − 1” is not satisfied.

EXAMPLE 17.38  Is the graph given in Figure 17.93 Hamiltonian? A

B

C

E

F

E F D

C

Figure 17.92 Solution.  We note that deg(A) 5 deg(D) 5 deg(E) 5 deg(F) 5 2, deg(B) 5 deg(C) 5 4. Further,   Number of vertices (n) 5 6   Number of edges (m) 5 8. So,

1 ​ (36 – 18 1 6) 5 12. __ ​ 1 ​ (n2 – 3n 1 6) 5 ​ __

2 2 Thus the conditions (i) Degree of each vertex is greater than or equal to n/2 is not satisfied __ 2 (ii)  m  ≥ ​  1 2 ​ (n  – 3n 1 6) is not satisfied

M17_Baburam_ISBN _C17.indd 24

D

Figure 17.93 Solution.  In this graph,   Number of vertices (n) 5 6   Number of edges (m) 5 9. So, 1 ​(36 – 18 1 6) 5 12, __ ​ 1 ​(n2 – 3n 1 6)  5 ​ __ 2 6 deg(A) 5 deg(C) 5 2, deg(D) 5 deg(F) 5 3 5 deg(E), deg(B) 5 5. Thus, no sufficient condition is satisfied in this case. But, the graph has Hamiltonian circuit

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graphS   n 17.25 ADEFCBA shown in the Figure 17.94.

weight. For example, if vertices in a graph denote recreational sites of a town and weights of edges denote the distances in kilometers between the sites, then the graph shown in ­Figure 17.96 is a weighted graph.

A

B

C

D

E

F

Figure 17.94 EXAMPLE 17.39  Show that the graph G shown in Figure 17.95 is not Hamiltonian. a e

d

g

c

b

i

Figure 17.95 Solution.  If G is Hamiltonian, then it has a subgraph H that 1.  Contains every vertex of G 2.  Is connected 3.  Has the same number of edges as vertices 4.  Is such that every vertex has degree 2 Thus, if such a graph exists, then it will have vertices (a, b, c, d, e, f, g, h, i), will be connected, will have six edges and the degree of each vertex shall be 2. We note that deg(d ) 5 d eg(e) 5 deg(f   ) 5 2. The degree of the vertex h in G is 3, one edge incident on h must be deleted from G to create H. The edge {h, e} is required since otherwise deg(e) ≠ 2. So we have to delete either {g, h} or {h, i}. If we delete {h, i} and retain {g, h}, then {g, i} has to be deleted to keep g of degree 2. In such a case, there is only one edge {i, f } on i and so there is a contradiction to (4). So we cannot remove {h, i}. Similarly, we see that {g, h} cannot be deleted. Hence no such subgraph H exists and so G does not have a Hamiltonian circuit. Definition 17.55  A weighted graph is a graph for which each edge or each vertex or both is (are) ­labelled with a numerical value, called its

M17_Baburam_ISBN _C17.indd 25

7 A

C 14

11 9

5

B

Figure 17.96 Definition 17.56  The weight of an edge (vi, vj) is called distance between the vertices vi and vj.

f

h

7

D

Definition 17.57  A vertex u is a nearest neighbour of vertex v in a graph if u and v are adjacent and no other vertex is joined to v by an edge of lesser weight than (u, v). For example, in the above example, B is the nearest neighbour of C, whereas A and C are both nearest neighbours of the vertex D. Thus, nearest neighbour of a set of vertices is not unique. Definition 17.58  A vertex u is a nearest neighbour of a set of vertices {v1, v2, . . ., vn} in a graph if u is adjacent to some member vi of the set and no other vertex adjacent to member of the set is joined by an edge of lesser weight than (u, vi). In the above example, if we have set of vertices as {B, D}, C is the neatest neighbour of {B, D} ­because the edge (C, B) has weight 5 and no other vertex adjacent to {B, D} is linked by an edge of lesser weight than (C, B). Definition 17.59  The length of a path in a graph is the sum of lengths of edges in the path. Definition 17.60  Let G 5 (V, E) be a graph and let lij denote the length of edge (vi, vj) in G. Then a shortest path from vi to vk is a path such that the sum l12 1 l23 1  1 lk − 1, k of lengths of its edges is minimum, that is, total edge weight is minimum.

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17.26  n  chapter Seventeen 17.7.1  Travelling Salesperson Problem This problem requires the determination of a shortest Hamiltonian circuit in a given graph of cities and lines of transportation to minimize the total fare for a travelling person who wants to make a tour of n cities visiting each city exactly once before returning home. The weighted graph model for this problem consists of vertices representing cities and edges with weight as distances (fares) between the cities. The salesman starts and ends his journey at the same city and visits each of n − 1 cities once and only once. We want to find minimum total distance. We discuss the case of five cities and so ­consider the weighted graph shown in the ­Figure 17.97.

a 7

c

b

(ii) From d, the nearest vertex is c, so we have a path shown below: a 7

6

12 9

d

c

10

6

5

13

(iii) From c, the nearest vertex is e. So we have the path as shown below: a

8 11

e

Figure 17.97 We shall use nearest neighbour algorithm to solve the problem: Algorithm: Nearest neighbour (closest insertion) Input: A weighted complete graph G. Output: A sequence of labelled vertices that forms a Hamiltonian cycle. Start at any vertex v. Initialize l(v) 5 0. Initialize i 5 0. While there are unlabelled vertices i: 5 i 1 1 Traverse the cheapest edge that join v to an unlabelled vertex, say w Set l(w) 5 i. v: 5 w. For the present example, (i) Let us choose a as the starting vertex. Then d is the nearest vertex and then (a, d) is the corresponding edge. Thus we have the figure

M17_Baburam_ISBN _C17.indd 26

e

d

14 b

c

b

a 7

e

d

7

c

b 6

8 e

d

(iv) From e, the nearest vertex is b and so we have the path a 7

b 5

d

c

6 8 e

(v) Now, from b, the only vertex to be covered is a to form Hamiltonian circuit. Thus we have a Hamiltonian circuit as given below. The length of this Hamiltonian circuit is 7 1 6 1 8 1 5 1 14 5 40.

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graphS   n 17.27 a 14

7

10

c

b

6

5

10

b

8

9

15

a

8

c

d

12

Figure 17.100 e

d

However, this is not Hamiltonian circuit of minimal length. The total distance of a minimum Hamiltonian circuit (Figure 17.98) is 37.

Definition 17.61  A k-factor of a graph is a spanning subgraph of the graph with the degree of its vertices being k. Consider the graph shown in Figure 17.101.

a

7

9

b

10

c 6

5

Solution.  Starting from the point a and using nearest neighbour method, we have the required Hamiltonian circuit as a b c d a with total length as 10 1 10 1 8 1 12 5 40.

a

b

c

d

e

e

d

Figure 17.98 Total length 5 7 1 6 1 9 1 5 1 10 5 37.

f

Remark 17.7  Unless otherwise stated, try to start from a vertex of largest weight. EXAMPLE 17.40

Figure 17.101  Then the graph shown in Figure 17.102 shows a 1-factor of the given graph.

Find a Hamiltonian circuit of minimal weight for the graph given below (Figure 17.99) b

2

d 5

3

3

4

2 a

2

f

6

c

5

4 e 6

g

5

4 h

Figure 17.102  Also then, (see Figure 17.103)

Figure 17.99 Solution.  Starting from c and applying ­nearest neighbour method, we have the required ­Hamiltonian circuit as c a b d e f g h c with total length as 2 1 3 1 2 1 5 1 4 1 5 1 4 1 6 5 31. EXAMPLE 17.41 Find a Hamiltonian circuit of minimal weight for the graph shown below (Figure 17.100)

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Figure 17.103  is a 2-factor of the given graph. 17.8  MATRIX REPRESENTATION OF GRAPHS A graph can be represented inside a computer by using the adjacency matrix or the incidence matrix of the graph.

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17.28  n  chapter Seventeen Definition 17.62  Let G be a graph with n ordered vertices v1, v2, . . ., vn. Then the adjacency matrix of G is the n × n matrix A(G) 5 (aij) over the set of non-negative integers such that

EXAMPLE 17.43  Find the adjacency matrix of the graph given ­below:

aij5 the number of edges connecting vi and vj for all i, j 5 1, 2, . . ., n. We note that if G has no loop, then there is no edge joining vi to vi, i 5 1, 2, . . ., n. Therefore, in this case, all the entries on the main diagonal will be 0. Further, if G has no parallel edge, then the entries of A(G) are either 0 or 1. It may be noted that adjacent matrix of a graph is symmetric. Conversely, given a n × n symmetric matrix A(G) 5 (aij) over the set of non-negative integers, we can associate with it a graph G, whose adjacency matrix is A(G), by letting G have n vertices and joining vi to vertex vj by aij edges. EXAMPLE 17.42  Find the adjacency matrix of the graph shown below (Figure 17.104): v1

v4

v5

aij 5 Number of edges connecting vi and vj.

]

So, for the given graph, the adjacency matrix is

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1 0 1 1 1

1 1 0 0 0

1 1 0 0 0

v3

Figure 17.105  Solution.  We note that there is a loop at v1 and parallel edges between v1, v3, and v3, v4. So the adjacency matrix A(G) is given by the following 4 × 4 matrix:

[

1 0 2 0

0 0 1 0

2 1 0 2

0 0 2 0

]

EXAMPLE 17.44  Find the graph that has the following adjacency

Solution.  The adjacency matrix A(G) 5 (aij) is the matrix such that

[

v4

A(G) =

Figure 17.104 

A(G) =

v1

v2

v3

0 1 1 1 1

v2

1 1 0 0 0

­matrix:

[

1 2 1 2

2 0 2 1

1 2 1 0

2 1 0 0

]

Solution.  We note that there is a loop at v1 and a loop at v3. There are parallel edges between v1, v2; v1, v4; v2, v1; v2, v3; v3, v2; v4, v1. Thus the graph is as shown in Figure 17.106.

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v1

v2

A3 =

v3

v4

Figure 17.106

A4 =

The following theorem is stated without proof. Theorem 17.14  Let G be a graph with n vertices v1, v2, . . ., vn and let A(G) denote the matrix of G. If B 5 (bij) is the matrix B 5 A 1 A2 1 . . . 1 An − 1, then the graph G is a connected graph if and only if bij ≠ 0 for i ≠ j, that is, B has no zero entry off the main diagonal. EXAMPLE 17.45  A graph G has the following adjacency matrix. ­Verify whether it is connected.

A=

[ ] 0 0 1 0 0

[

0 0 0 1 0

1 0 0 0 1

0 1 0 0 1

0 0 1 1 0

Solution.  We have

A2 =

M17_Baburam_ISBN _C17.indd 29

1 0 0 0 1

0 1 0 0 1

0 0 2 1 0

0 0 1 2 0

1 1 0 0 2

]

.

[ [

graphS   n 17.29 0 0 2 1 0

0 0 1 2 0

2 1 0 0 3

1 2 0 0 3

0 0 3 3 0

2 1 0 0 3

1 2 0 0 3

0 0 5 4 0

0 0 4 5 0

3 3 0 0 6

Therefore,

[

B = A + A2+ A3+ A4 = 3 1 3 1 4 1 3 1 3 4 3 1 7 5 4 1 3 5 7 4 4 4 4 4 8

]

] ]

,

,

,

Since B has no zero entry off the main diagonal, the graph is connected. Definition 17.63  Suppose a graph G has n vertices v1, v2, . . ., vn and t edges e1, e2, . . ., et. The incidence ­matrix B(G) of G is the n × t matrix B(G) 5 (bij), where bij 5 the number of times that the vertex vi is incident with the edge ej, that is, 0  if vi is not end of ej bij =  1  if vi is an end of the non-loop ej 2  if vi is an end of the loop ej



17.9  PLANAR GRAPHS ,

Definition 17.64  A graph which can be drawn in the plane so that its edges do not cross is said to be planar. For example, the graph shown in Figure 17.107 is planar:

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17.30  n  chapter Seventeen But the map K5 is not planar because in this case, the edges cross each others (see Figure 17.110).

A

B

C

D

E

B

A E

Figure 17.107 

Figure 17.110 

C

Also the complete graph K4 shown below (­Figure 17.108) is planar. A

B

C

D

Figure 17.108  In fact, it can be redrawn as shown below in Figure 17.109 so that no edges cross. A

D

Definition 17.65  An area of the plane that is bounded by edges of the planar graph and is not further subdivided into subareas is called a region or face of the planar graph. A face is characterized by the cycle that forms its boundary. Definition 17.66  A region is said to be finite if its area is finite and infinite if its area is infinite. ­Clearly a planar graph has exactly one infinite region. For example, consider the graphs shown in ­Figure 17.111.

B

D

C

Figure 17.109  1 2

G1 1

3

2

4 6

5

5

6

7 9

G2

4

8

3

10

Figure 17.111

M17_Baburam_ISBN _C17.indd 30

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graphS   n 17.31 We note that in the graph G1, there are five regions A, B, C, D, E as shown below: 1

4

3

5

B C

6

A

7 3

4

Finite region

Finite region

8

Finite region

1 E 3 3

4

D

9

8 9

Infinite region

10

Infinite region

8

Infinite region

In graph G2, there are four region A, B, C, D (Figure 20.112)

1

2 1

2 5

B

C

6

A 6

3

4

4 Finite region

Finite region 1

E

Finite region 2

D

5

3

4 Infinite region

Figure 17.112

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17.32  n  chapter Seventeen Definition 17.67  Let f be a face (region) in a planar graph. The length of the cycle (or closed walk) which borders f is called the degree of the region f. It is denoted by deg(f ). In a planar graph we note that each edge either borders two regions or is contained in a region and will occur twice in any walk along the border of the region. Thus we have the following: Theorem 17.15  The sum of the degrees of the regions of a map is equal to twice the number of edges. For example, in the graph G2, discussed above, we have deg(A) 5 4, deg(B) 5 3, deg(C) 5 4, deg(D) 5 5. The sum of degrees of all regions 5 4 1 3 1 4 1 5 5 16. Number of edges in G2 5 8. Hence the sum of degrees of regions is twice the number of edges. Theorem 17.16  (Euler’s Formula for Connected Planar Graphs) If G is a connected planar graph with e edges, v vertices and r regions, then v – e1 r 5 2. Proof:  We shall use induction on the number of edges. Suppose that e 5 0. Then the graph G consists of a single vertex, say P. Thus, G is as shown below: •P and we have e 5 0, v 5 1, r 5 1. Thus 1 – 0 1 1 5 2 and the formula holds in this case. Suppose that e 5 1. Then the graph G is one of the two graphs shown below: , e = 1, v = 2, r = 1

a x

G

Figure 17.113  We delete “a” and the edge x incident on “a” from the graph G. The resulting graph G′ (­Figure  17.114) has n edges and so by induction hypothesis, the formula holds for G′. Since G has one more edge than G′, one more vertex than G′ and the same number of faces as G′, it follows that the formula v – e 1 r 5 2 holds also for G.

G′

Figure 17.114 Now suppose that G contains a cycle. Let x be an edge in a cycle as shown in Figure 17.115. x

G

Figure 17.115  Now the edge x is part of a boundary for two faces. We delete the edge x but no vertices to obtain the graph G′ as shown in Figure 17.116.

e = 1, v = 1, r = 2

We see that, in either case, the formula holds. Suppose that the formula holds for connected planar graph with n edges. We shall prove that this holds for graph with n 1 1 edges. So, let G be the graph with n 1 1 edges. Suppose first that G contains no cycles. Choose a vertex v1 and

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trace a path starting at v1. Ultimately, we will reach a vertex a with degree 1, that we cannot leave (see Figure 17.113).

G′

Figure 17.116 

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graphS   n 17.33 Thus G′ has n edges and so by induction hypothesis the formula holds. Since G has one more face (region) than G′, one more edge than G′ and the same number of vertices as G′, it follows that the formula v – e 1 r 5 2 also holds for G. Hence, by mathematical induction, the theorem is true.

class consists of a set of mutually homeomorphic graphs. EXAMPLE 17.46  Show that the graph K3,3, given in Figure 17.117.­below, is not planar.

Remark 17.8  Planarity of a graph is not affected if (i) An edge is divided into two edges by the insertion of new vertex of degree 2 (Figure 17.117).

c1

c2

c3

c4

c5

c6

K3,3

Figure 17.117  (ii) Two edges that are incident with a vertex of degree 2 are combined as a single edge by the removal of that vertex (Figure 17.118).

Figure 17.118  Definition 17.68  Two graphs G1 and G2 are said to be isomorphic to within vertices of degree 2 (or ­homeomorphic) if they are isomorphic or if they can be transformed into isomorphic graphs by ­repeated insertion and/or removal of vertices of ­degree 2. Definition 17.69  The repeated insertion/removal of vertices of degree 2 is called sequence of series ­reduction. For example, the graphs shown in ­Figure 17.119 are isomorphic to within vertices of degree 2.

Figure 17.117. A problem based on this example can be stated as “Three cities c1, c2 and c3 are to be directly connected by express ways to each of three cities c4, c5 and c6. Can this road system be designed so that the express ways do not cross?” This example shows that it cannot be done. Solution.  Suppose that K3,3 is planar. Since ­every ­cycle in K3,3 has at least four edges, each face (­region) is bounded by at least four edges. Thus the number of edges that bound regions is at least 4r. Also, in a planar graph each edge belongs to at most two bounding cycles. Therefore, 2e ≥ 4r (sums of degrees of region is equal to twice the number of edges) But, by Euler’s formula for planar graph, r 5 e – v 1 2. Hence,

2e ≥ 4(e – v 1 2)

(1)

In case of K3,3 we have e 5 9, v 5 6 and so (1) yields and Figure 17.119  If we define a relation R on the set of graphs by G1 R G2 if G1 and G2 are homeomorphic, then R is an equivalence relation. Each equivalence

M17_Baburam_ISBN _C17.indd 33

18 ≥ 4 (9 2 6 1 2) 5 20 which is a contradiction. Therefore K3,3 is not planar. Remark 17.9  By a argument similar to the above example, we can show that the graph K5 (Figure 17.121) is not planar.

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17.34  n  chapter Seventeen a

f

b

g

(Non-planar graph K5)

h

Figure 17.121  We observe that if a graph contains K3,3 or K5 as a subgraph, then it cannot be planar. The following theorem, which we state without proof, gives necessary and sufficient condition for a graph to be planar. Kuratowski’s Theorem 17.17  A graph G is planar if and only if G does not contain a subgraph homeomorphic to K3,3 or K5. The complete graph K5 and the complete bipartite graph K3,3 are called the Kuratowski graphs. EXAMPLE 17.47  Using Kuratowski’s theorem show that the graph G, shown in Figure 17.122, is not planar. a

f

e c d

(Graph obtained by deleting edges (a, b) and (f, e)). a

f

h e c d

(Graph obtained by eliminating the edge (g, h). Performing series reduction now, we obtain an isomorphic copy of K3,3 (Figure 17.123).

b

g

b

g

a

h e c

f

b

d G

Figure 17.122 

e c

Solution.  Let us try to find K3,3 in the graph G. d We know that in K3,3, each vertex has degree 3. (Isomorphic copy of K3,3, obtained by series reduction) (Isomrophic copy of K3,3, obtained by series reducBut we note that in G, the degree of a, b, f and e tion) each is 4. So we eliminate the edges (a, b) and (f, e) so that all vertices have degree 3. If we Figure 17.123 eliminate one more edge, we will obtain two Hence, by Kurtowski’s theorem, the given graph vertices of degree 2 and we can then carry out G is not planar. series reduction. The resulting graph will have nine edges. EXAMPLE 17.48  Also we know that K3,3 has nine edges. So If the graph given in Figure 17.124 is planar, this approach seems promising. Using trial and redraw it so that no edges cross, otherwise find error, we find that the edge (g, h) should be rea subgraph homeomorphic to either K5 or K3,3. moved. Then g and h have degree 2.

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graphS   n 17.35 a

b

f

c

e

d

Figure 17.124 Solution.  The given graph can be redrawn as shown in the Figure 17.125.

Definition 17.70  Any graph homeomorphic either to K5 or to K3,3 is called Kuratowski subgraph. EXAMPLE 17.49  Find Kuratowski subgraph of the following graph (Figure 17.126): a

b

b

a

e

f

g

h

d f

c

Figure 17.126 

c

e

d

Figure 17.125  We then observe that no edges cross. Hence the given graph is planar. Theorem 17.18  Let G be a connected planar graph with v vertices, e edges, where v ≥ 3. Then e ≤ 3v – 6 or 6 ≤ 3v – e. (Note that the theorem is not true for K1, where v 5 1 and e 5 0 and is not true for K2, where v 5 2 and e 5 1). Proof:  Let r be the number of regions in a planar representation of the graph G. The sum of the degrees of the regions is equal to 2e. But each region has degree greater than or equal to 3. 2e Hence 2 e ≥ 3 r, that is, r  ≤ ​ ___  ​ . But, by Euler’s 3 formula, v  2  e 1 r 5 2. So, 2e e 2 5 v – e 1 r ≤ v – e 1 ​ ___  ​ 5 v − ​ __ ​. 3 3 Hence,

Solution.  In the given graph,   Number of vertices 5 8   Number of edges 5 12. Firstly, we remove the edge (g, h). Then deg(g) 5 2, deg(h) 5 2 and so the vertices g and h can be ­removed. Then, we have   Number of vertices 5 6   Number of edges 5 9 and the graph reduces to a

b e

f

d

, c

which is homeomorphic to K3,3 (Kuratowski graph) shown in Figure 17.127. a

c

f

6 ≤ 3 v – e  or  e ≤ 3v – 6

Remark 17.10  Using above theorem, let us consider planarity of K5. Here v 5 5, e 5 10. Suppose K5 is planar, then by the above theorem, 6 ≤ 3(5) – 10 5 5, which is impossible. Hence K5 is non-planar.

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d

e K3, 3

b

Figure 17.127  Thus the given graph is non-planar by Kuratowski theorem.

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17.36  n  chapter Seventeen EXAMPLE 17.50  Find a subgraph homeomorphic to either K5 or K3,3 in the graph given below (Figure 17.128): a e2 e12

f

e7

e1 e9

b e13

e10 e6

e8

e3

e4

e

c

e11

e5

Solution. Here,    Number of vertices 5 6    Number of edges 5 13 deg(a) 5 5,  deg( f ) 5 4,  deg(c) 5 4,  deg(d ) 5 5 But, in K3,3, the degree of each vertex is 3. We delete the edges (a, c), (f, d), (a, e), (b, d), then degree of each vertex a, f, c and d becomes 3. As such, the subgraph becomes as given below (Figure 17.129), which has six vertices each with degree 3 and has 9 edges. b c

d

Figure 17.129 This subgraph is homeomorphic to K3,3 as shown in Figure 17.130. a

f

e

b K3,3

c

d

Figure 17.130 This also shows, by Kuratowski theorem, that the given graph is not planar. 17.10  COLOURING OF GRAPH Definition 17.71  Let G be a graph. The assignment of colours to the vertices of G, one colour

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EXAMPLE 17.51  Find the chromatic number for the graph shown in the Figure 17.131. b a

c d e G

Figure 17.131 

f e

Definition 17.72  A graph G is n-colourable if there exists a colouring of G which uses n colours. Definition 17.73  The minimum number of colours required to paint (colour) a graph G is called the chromatic ­number of G and is denoted by x (G).

d

Figure 17.128 

a

to each vertex, so that the adjacent vertices are assigned different colours is called vertex colouring or colouring of the graph G.

Solution.  The triangle a b c needs three colours. Suppose that we assign colours c1, c2, c3 to a, b and c respectively. Since d is adjacent to a and c, d will have different colour than c1 and c3. So we paint d by c2. Then e must be painted with a colour different from those of a, d and c, that is, we cannot colour e with c1, c2 or c3. Hence, we have to give e a fourth colour c4. Hence x (G) 5 4. Welsh–Powell algorithm to determine upper bound to the chromatic number of a given graph. The input is a given graph G. 1. Order the vertices of G according to decreasing degree. 2. Assign the first colour, say c1, to the first vertex and then, in sequential order, assign c1 to each vertex, which is not adjacent to a previous vertex assigned c1. 3. Repeat Step 2 with a second colour c2 and the subsequence of the remaining non-painted vertices.

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graphS   n 17.37 4. Repeat Step 3 with a third colour c3, then a fourth colour c4 and so on until all vertices are coloured. 5.  Exit. EXAMPLE 17.52  Use Welsh-Powell algorithm to determine an ­upper bound to the chromatic number of the “Wheel” graph shown in Figure 17.132. a

e

b

f

d

c

Figure 17.132 Solution.  We note that deg(f  ) 5 5, deg(a) 5 deg(b) 5 deg(c) 5 deg(d) 5 deg(e) 5 3. Step 1. Ordering the vertices according to decreasing degree yields f, a, b, c, d, e Step 2.  Paint f with colour c1. Step 3.  Paint a, d with colour c2. Step 4.  Paint b, e with colour c3. Step 5.  Paint c with colour c4. Hence x(G) ≤ 4. Also since there is a triangle in the given graph so we have x(G) ≥ 3. Thus 3 ≤ x(G) ≤ 4. But we do not yet know exactly what the chromatic number is. We try to build a 3-colouring of G. Let us start colouring the triangle abf with the colours c1, c2, c3, respectively. Since c is adjacent to the vertices b and f of colour c2 and c3, respectively, c is forced to be coloured c1 and then d is forced to be c2. However, now the adjacent vertices a and e cannot both have colour c1. Thus the graph cannot be 3-coloured. But using a fourth colour c4 for e gives us 4-colouring of G. Hence, x(G) 5 4. EXAMPLE 17.53  Use Welsh-Powell algorithm to colour the graph shown in Figure 17.133.

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b

a

c

e

d g

f h

17.133 Solution.  Ordering the vertices according to decreasing degrees, we get the sequence e, c, g, a, b, d, f, h.   Use the colour c1 to colour (paint) e and a.   Use the colour c2 to paint c, d and h.   Use third colour c3 to paint vertices g, b and f. Thus all the vertices are painted such that no adjacent vertices get the same colour. Hence x(G) 5 3. Some rules for colouring a graph 1. x(G) ≤ | V |, where | V | is the number of vertices of G. 2. A triangle always requires three colours, that is, x(K3) 5 3. Similarly x(Kn) 5 n, where Kn is the complete graph of n ­vertices. 3. If some subgraph of G requires k-colours, then x(G) ≥ k. 4. If deg(v) 5 n, then at most n colours are required to colour the vertices adjacent to v. 5.  x(G) 5 max {x(C): C is a connected component of G}. 6. Every n-colourable graph has at least n vertices v such that deg(v) ≥ n − 1. 7.  x(G) ≤ 1 1 Δ(G), where Δ(G) is the largest ­degree of any vertex of G. 8. The following statements are equivalent: (i)  A graph G is 2-colourable (ii)  G is bipartite (iii)  Every cycle of G has even length 9. If d(G) is the smallest degree of any vertex of G, then |v| x(G)  _________ ​       ​ |v| 2 d(G) EXAMPLE 17.54 Find chromatic number of the graphs shown in ­Figure 17.134.

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17.38  n  chapter Seventeen c2

It is equivalent to the following bipartite graph shown in Figure 17.137.

c1

c1 c2

c2

c3

a

c

b

d

and

c1 c2

c1

c2

c1 (i)

Figure 17.137 

(ii)

Therefore, it is 2-colourable.

Figure 17.134. Solution.  The graph (i) is 2-colourable whereas the graph (ii) is 3-colourable. Theorem 17.19  Bipartite graph is 2-colourable unless G is edgeless. Proof:  A two colouring is obtained by assigning one colour to every vertex in one of the bipartition parts and another colour to every vertex in the other partition part as shown in Fig. 17.135. c1

c2

c1

c2

c2

Figure 17.135  Corollary 17.3  Even cycle graphs C2n have x(C2n ) 5 2.

Proposition 17.2  Odd cycle graph C2n 1 1 has x(C2n 1 1) 5 3. Proof:  Let v1, v2, . . ., v2n, v2n 1 1 be vertices of a cycle graph c2n 1 1. If two colours were sufficient, then they would have to alternate around the cycle. Thus, the odd subscripted would have to be one colour and the even subscripted vertices have to be second colour but vertex v2n 1 1 is adjacent to v1, and so according to this scheme two adjacent vertices v1 and v2n 1 1 have the same colour. This is a contradiction and so c2n 1 1 is not 2-colourable but 3-colourable. It follows, therefore, that “The chromatic number of a cycle is either two or three, depending on whether its length is even or odd.” Definition 17.74  The join G 1 H of the graphs G and H is obtained from the graph G  H by adding an edge between each vertex of H. For example, if we have a graph as shown below (Figure 17.138):

Proof:  An even cycle graph is bipartite and therefore, by Theorem 17.19, its chromatic number is 2. For example consider the graph shown in ­Figure 17.136. a

(G)

b

Figure 17.138  and a graph

K1, c

d

Figure 17.136 

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then the join G 1 K1 of the graphs G and K1 is the graph as shown below (Figure 17.139):

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graphS   n 17.39

K1

K1

Figure 17.139

Figure 17.141  (W6)

Regarding the chromatic number of the join G 1 H of the graphs G and H we have the ­following:

EXAMPLE 17.55 Find chromatic number of the graph shown in Figure 17.142:

Proposition 17.3  The join of a graph G and H has chromatic number

a

d

x(G 1 H) 5x(G) 1 x(H). Definition 17.75  The n vertex wheel graph Wn is called an odd-order wheel if n is odd and an even-order wheel if n is even. Proposition 17.4  Odd-order wheel graph has x(W2m 1 1) 5 3,  for all m ≥ 1. Proof:  Using the fact that the wheel graph W2m 1 1 is the join C2m 1 K1, it follows that x(W2m 1 1) 5 x(C2m) 1 x(K1) 5 2 1 1 5 3.

K1

b

c

Figure 17.142  Solution.  Here degree of each vertex is 3. Therefore, x(G) ≤ 1 1 3 5 4. Since it has a triangle, x(G) ≥ 3. Thus, 3 ≤ x (G) ≤ 4. We first colour the triangle abd with colours c1, c2, c3, respectively. Since c is adjacent to each of a, b and d, therefore a different colour c4 have to be given to it. Hence the graph is 4-colourable and so x(G) 5 4. 4   ​  Note:  If we apply rule 9, then x  ​ _______ = 4. (4 2 3) Hence x(G) = 4. Theorem 17.20.(Four-Colour Theorem of Appel and Haken) Any planar graph is 4-colourable. (Proof of the theorem is out of the scope of this book).

Figure 17.140  (W5) Proposition 17.5  Even-order wheel graph has x(W2m) 5 4. Proof:  Using the fact that W2m 5 c2m −1 1 K1, it follows that x(W2m) 5 x(c2m − 1) 1 x(K1) 5 3 1 1 5 4.

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17.11  DIRECTED GRAPHS Definition 17.76  A directed graph or digraph consists of two finite sets:   (i)  A set V of vertices (or nodes or points).   (ii) A set E of directed edges (or arcs), where each edge is associated with an ordered pair (v, ) of vertices called its endpoints. If edge e is associated with the ordered

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17.40  n  chapter Seventeen pair (v, w), then e is said to be directed edge from v to w. The directed edges are indicated by arrows. We say that edge e 5 (v, w) is incident from v and is incident into w. The vertex v is called initial vertex and the vertex w is called the terminal vertex of the directed edge (v, w). Definition 17.77  Let G be a directed graph. The outdegree of a vertex v of G is the number of edges beginning at v. It is denoted by outdeg(v). Definition 17.78  Let G be a directed graph. The ­indegree of a vertex v of G is the number of edges ­ending at v. It is denoted by indeg(v). EXAMPLE 17.56 Consider the directed graph shown below (Figure 17.143):

aij 5 the number of arrows from vi to vj, i, j 5 1, 2,. . ., n. EXAMPLE 17.57 Find the adjacency matrices for the graphs given below (Figure 17.144):

c

a

v2 e2 v4

e3 e4

(i)

(ii)

Figure 17.144  Solution.  The edges in the directed graph are (a, a), (b, b), (c, c), (d, d), (c, a), (c, b) and (d, b). Therefore the adjacency matrix A 5 (aij) is

[

e5 e6

Figure 17.143 Here edge e1 is (v2, v1) whereas e6 is denoted by (v5, v5) and is called a loop. The indegree of v2 is 1, outdegree of v2 is 3.

Definition 17.80  If the edges and/or vertices of a directed graph G are labelled with some type of data, then G is called a labelled directed graph. Definition 17.81  Let G be a directed graph with ordered vertices v1, v2, . . ., vn. The adjacency matrix of G is the matrix A 5 (aij) over the set of non-negative integers such that

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1 0 1 0

0 1 1 1

0 0 1 0

0 0 0 1

]

(iii) The edges in the graph in (ii) are (v2, v3), (v1, v1), (v1, v3), (v3, v1), (v3, v4), (v4, v3). Hence the adjacency matrix is

[

Definition 17.79  A vertex with 0 indegree is called a source, whereas a vertex with 0 out­ degree is called a sink. For instance, in the above example, v1 is a sink.

v3

v4

v3

v5

v2

,

d

v1 e1

v1

b

1 0 1 0

0 0 0 0

1 1 0 1

0 0 1 0

]

EXAMPLE 17.58 Find the directed graph represented by the adjacency matrix:

[

0 0 0 1 0

1 0 0 1 0

0 1 0 0 0

0 0 1 0 0

0 0 1 0 0

]

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graphS   n 17.41 Solution.  We observe that a12 5 1, a23 5 1, a34 5 1, a35 5 1, a41 5 1, a42 5 1. Hence the digraph is as shown in Figure 17.145. v2

v1

v4

v3

v5

Figure 17.145  Definition 17.82  In a directed graph, if there is no more than one directed edge in a particular direction between a pair of vertices, then it is called simple directed graph. For example, the graph shown in Figure 17.146 is a simple directed graph.

pij =



1 if there is a path from vi to vj 0 otherwise

is called path matrix or reachability matrix of G. The path matrix can be given in term of adjacency matrix: P 5 A ∨ A(2) ∨ A(3) ∨ . . . ∨ A(n) and A(2) 5 A  A,   A(r) 5 A(r 2 1)  A, where ∨ represents Boolean addition and  denotes the Boolean matrix multiplication. Warshall’s algorithm for finding path matrix from adjacency matrix. We discuss this algorithm with the help of the following example. EXAMPLE 17.59 Find the adjacency matrix and the path matrix for the digraph shown in Figure 17.148. b

Figure 17.146  A directed graph which is not simple is called ­directed multi-graph. Definition 17.83  A simple digraph is said to be ­strongly connected if for each pair of vertices v, w, there are path, from v to w and w to v. For example, the graphs shown in Figure 17.147 are strongly connected.

v

w

,

Figure 17.147 Definition 17.84  Let G be a simple digraph with n vertices. Then a n 3 n matrix P 5 (pij) such that

M17_Baburam_ISBN _C17.indd 41

c a d

Figure 17.148  Solution.  The adjacency matrix of this digraph is:

A=

[

0 1 0 0

Then the path matrix lows: We take 0 1 1 0 P0 = 0 0 0 0

[

1 0 0 0

0 1 0 0

0 0 1 0

]

.

P 5 (pij) is found as fol0 1 0 0

0 0 1 0

]

= A itself.

Now we find P1 by consulting column 1 and row 1. We note that P0 has 1 in location 2 of

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17.42  n  chapter Seventeen column 1 and location 2 of row 1. Thus the element p22 is replaced by 1 in P0. Thus we have

P1 =

[

0 1 0 0

1 1 0 0

0 1 0 0

0 0 1 0

]

To find P2 we consult column 2 and row 2 of P1. We note that P1 has 1 in locations 1 and 2 of column 2 and locations 1, 2 and 3 of row 2. Thus, to obtain P2 we should put 1 in positions p11, p12, p13, p21, p22 and p23 of matrix P1. We thus have

P2 =

[

1 1 0 0

1 1 0 0

1 1 0 0

0 0 1 0

]

P3 =

[

1 1 0 0

1 1 0 0

1 1 1 0

]

P = (pij) =

[

1 1 0 0

1 1 0 0

.

1 1 1 0

]

Remark 17.11  Given an adjacency matrix of a graph G, we first draw digraph of G and then apply definition of path matrix. We will obtain path matrix. Now, we state a result without proof.

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Theorem 17.22  Let A be adjacency matrix of a graph G with n vertices and let Bn 5 A 1 A2 1 . . . 1 An. Then G is strongly connected if and only if Bn has no zero entries.

A(G) =

[

0 1 1

1 0 1

1 1 0

]

.

Solution.  To find the path matrix, we first consider column 1 and row 1 of

Finally, we consult column 4 and row 4 of P3. We have 1 in location 1, 2, 3 of column 4 and there is no 1 in row 4. Hence P4 5 P3. Thus the path matrix is 1 1 0 0

Let G be a strongly connected directed graph. Then for any pair of vertices v and w in G, there is a path from v to w and from w to v. Accordingly, G is strongly connected if and only if the path matrix P of G has no zero entries. Thus, in view of the above theorem, we have:

EXAMPLE 17.60  A directed graph has the following adjacency ­matrix. Check whether it is strongly connected.

.

We now consult column 3 and row 3 of P2. The ­matrix P2 has 1 in locations 1 and 2 of column 3 and 1 in location 4 of row 3. Thus we shall obtain P3 by putting 1 in positions p14, p24. Thus 1 1 0 0

Theorem 17.21  Let A be the adjacency matrix of a graph G with n vertices and let Bn 5 A 1 A2 1 A3 1 . . . 1 An. Then the path matrix P and Bn have the same non-zero entries.

P0 =

[ ] 0 1 1

1 0 1

1 1 0

= A(G).

In column 1, the entry 1 is located at 2 and 3 positions and 1 is at locations 2 and 3 in row 1. Thus put 1 at p22, p23, p32, p33 and get

P1 =

[ ] 0 1 1

1 1 1

1 1 . 1

We note that 1 is located in column 2 of P1 at position 1, 2, 3, and 1 is located at position 1, 2, 3 of row 2. Thus put 1 at p11, p12, p13, p21, p22, p23, p31, p32, p33, and get

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P2 =

[

1 1 1

1 1 1

1 1 1

]

graphS  n 17.43

. (i) Trivial tree

(ii) Tree of 3 vertices

(iii) Tree of 4 vertices

The next step will not change the position. Hence, P=

[ ] 1 1 1

1 1 1

1 1 = (all non-zero entries). 1

Therefore, G is strongly connected. The graph of G is shown in Figure 17.149.

(iv) Tree of 13 vertices

Figure 17.150  But the graphs shown in Figure 17.151 are not trees:

(i) Has a cycle and so is not a tree

(i) Has a cycle and

is not a tree Figureso17.149 

(ii) Has a cycle and so is not a tree

Remark 17.12  In the above example, we also note that B3 5 A + A2 + A3 1 1 1 2 1 1 2 3 3 5 1 0 1 + 1 2 1 + 3 2 3 1 1 0 1 1 2 3 3 2





 

 



4 5 5 5 5 4 5 (all non-zero entries) 5 5 4





showing that G is strongly connected. 17.12  TREES Definition 17.85  A graph is said to be a tree if it is a connected acyclic graph. A trivial tree is a graph that consists of a single vertex. An empty tree is a tree that does not have any vertices or edges. For example, the graphs shown in Figure 17.150 are all trees.

M17_Baburam_ISBN _C17.indd 43

(v) Tree of 8 vertices

(ii) Has a cycle and so is not a tree

(iii) Disconnected g and so is not a t

(iii) Disconnected graph and so is not a tree

Figure 17.151  Definition 17.86  A collection of disjoint trees is called a forest. Thus a graph is a forest if and only if it is circuit free. Definition 17.87  A vertex of degree 1 in a tree is called a leaf or a terminal node or a terminal ­vertex. Definition 17.88  A vertex of degree greater than 1 in a tree is called a branch node or internal node or internal vertex. Consider the tree shown in Figure 17.152. b

c

f a

e

d

g

h

i

Figure 17.152 

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17.44  n  chapter Seventeen In this tree, the vertices b, c, d, f, g, i are leaves whereas the vertices a, e, h are branch nodes. 17.12.1  Characterization of Trees We have the following interesting characterization of trees: Lemma 17.1  A tree that has more than one vertex has at least one vertex of degree 1. Proof:  Let T be a particular but arbitrary ­chosen tree having more than one vertex (Figure 17.153).

e

v

e′

v′ e′′

T

v′′

Figure 17.153  1. Choose a vertex v of T. Since T is connected and has at least two vertices, v is not isolated and there is an edge e incident on v. 2. If deg(v) > 1, there is an edge e  e because there are, in such a case, at least two edges incident on v. Let v be the vertex at the other end of e. This is possible because e is not a loop by the definition of a tree. 3. If deg(v) > 1, then there are at least two edges incident on v. Let e be the other edge ­different from eand v be the vertex at other end of e. This is again possible because T is acyclic. 4. If deg(v) > 1, repeat the above process. Since the number of vertices of a tree is finite and T is circuit free, the process must terminate and we shall arrive at a vertex of degree 1. Remark 17.13  In the proof of the Lemma 17.1, after finding a vertex of degree 1, if we return to v and move along a path outward from v starting with e, we shall reach to a vertex of degree 1 again. Thus it follows that “Any tree that has more than one vertex has at least two vertices of degree 1”.

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Lemma 17.2  There is a unique path between every two vertices in a tree. Proof:  Suppose on the contrary that there are more than one path between any two vertices in a given tree T. Then T has a cycle which contradicts the definition of a tree because T is acyclic. Hence the lemma is proved. Lemma 17.3  The number of vertices is one more than the number of edges in a tree. Equivalently, we may state this lemma as “For any positive integer n, a tree with n vertices has n − 1 edges”. Proof:  We shall prove the lemma by mathematical induction. Let T be a tree with one vertex. Then T has no edges, that is, T has 0 edge. But 0 5 1 – 1. Hence the lemma is true for n 5 1. Suppose that the lemma is true for k  > 1. We shall show that it is then true also for k 1 1. Since the lemma is true for k, the tree has k vertices and k − 1 edges. Let T be a tree withk 1 1 vertices. Since k is 1ve, k 1 1 ≥ 2 and so T has more than one vertex. Hence, by Lemma 17.1, T has a vertex v of degree 1. Also there is another vertex w and so there is an edge e connecting v and w. Define a subgraph T of T so that V(T ) 5V(T) – {v}, E(T ) 5 E(T) – {e}. Then number of vertices in T5 (k 1 1) 2 1 5k and since T is circuit free and T has been obtained on removing one edge and one vertex, it follows that T is acyclic. Also T is connected. Hence T is a tree having k vertices and therefore by induction hypothesis, the number of edges in T is k 2 1. But then Number of edges in T 5 number of edges in T1 1  5 k 21 1 1 5 k. Thus the Lemma is true for tree having k 1 1 vertices. Hence the lemma is true by mathematical induction. Corollary 17.4  Let C(G) denote the number of components of a graph. Then a forest G on n vertices has n 2 C(G) edges.

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graphS   n 17.45 Proof:  Apply Lemma 17.3 to each component of the forest G. Corollary 17.5  Any graph G on n vertices has at least n 2 C(G) edges. Proof:  If G has cycle-edges, remove them one at a time until the resulting graph G* is acyclic. Then G* has n 2 C(G*) edges by Corollary 17.4. Since we have removed only circuit, C(G*) 5 C(G). Thus G* has n 2 C(G) edges. Hence G has at least n 2 C(G) edges. Lemma 17.4  A graph in which there is a unique path between every pair of vertices is a tree (This lemma is converse of Lemma 17.2). Proof:  Since there is a path between every pair of points, therefore the graph is connected. Since a path between every pair of points is unique, there does not exist any circuit because existence of circuit implies existence of distinct paths between pair of vertices. Thus the graph is connected and acyclic and so is a tree. Lemma 17.5  (Converse of Lemma 17.3) A connected graph G with e 5 v 2 1 is a tree First Proof:  The given graph is connected and e 5 v 2 1. To prove that G is a tree, it is sufficient to show that G is acyclic. Suppose on the contrary that G has a cycle. Let m be the number of vertices in this cycle. Also, we know that number of edges in a cycle is equal to number of vertices in that cycle. Therefore number of edges in the present case is m. Since the graph is connected, every vertex of the graph which is not in cycle must be connected to the vertices in the cycle (see Figure 17.154).

Figure 17.154 

M17_Baburam_ISBN _C17.indd 45

Now each edge of the graph that is not in the cycle can connect only one vertex to the vertices in the cycle. There are v 2 m vertices that are not in the cycle. So the graph must contain at least v 2 m edges that are not in the cycle. Thus we have e ≥ v 2 m 1 m 5 v, which is a contradiction to our hypothesis. Hence there is no cycle and so the graph is a tree. Second Proof:  We shall show that a connected graph with v vertices and v 2 1 edges is a tree. It is sufficient to show that G is acyclic. Suppose on the ­contrary that G is not circuit free and has a ­nontrivial circuit C. If we remove one edge of C from the graph G, we obtain a graph G which is connected (see Figure 17.155).

C

G

G′

Figure 17.155  If G still has a nontrivial circuit, we repeat the above process and remove one edge of that circuit obtaining a new connected graph. Continuing this process, we obtain a connected graph G* which is circuit free. Hence G* is a tree. Since no vertex has been removed, the tree G* has v vertices. Therefore, by Lemma 17.3, G* has v 2 1 edges. But at least one edge of G has been removed to form G*. This means that G* has not more than v 2 1 – 1 5 v 2 2 edges. Thus, we arrive at a contradiction. Hence our supposition is wrong and G has no cycle. Therefore G is connected and cycle free and so is a tree. Lemma 17.6  A graph G with e 5 v 2 1, that has no circuit is a tree. Proof:  It is sufficient to show that G is connected. Suppose G is not connected and let G, G, . . . be the connected components of G. Since each of G, G, . . . is connected and has no cycle, they all are tree. Therefore, by Lemma 17.3,

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17.46  n  chapter Seventeen e 5 v 2 1, e 5 v 2 1, ———— ————, where e, e, . . . are the number of edges and v, v, . . . are the number of vertices in G, G, . . ., ­respectively. We have, on adding e1e 1 . . . 5 (v 2 1) 1 (v 2 1) 1 . . . Since e 5 e 1 e″ 1 . . ., v 5 v 1 v 1 . . ., we have e  0 with p leaves. We consider the ­following two cases: (a) Suppose that the root of T has only one child (Figure 17.180).

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graphS   n 17.53 Root

Root

Figure 17.180  Eliminating the root and the edge incident on the root, we get a tree T of height h 2 1 in which number of leaves is same as in T. Using induction ­hypothesis on T, we get p ≤ 2h2 1. Since 2h 21