Transcription
SECOND EDITIONA Textbook ofENGINEERINGMATHEMATICS-IH.S. Gangwar l Prabhakar Gupta
A Textbook ofENGINEERINGMATHEMATICS-I
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A Textbook ofENGINEERINGMATHEMATICS-I(SECOND EDITION)Prabhakar GuptaH.S. GangwarM.Sc. (Math.), M.Tech., Ph.D.Dean AcademicsSRMS College of Engineeringand Technology, Bareilly (U.P.)M.Sc., Ph.D.LecturerDeptt. of MathematicsSRMS College of Engineeringand Technology, Bareilly (U.P.)UPTUAn Imprint of
Copyright 2010, 2009 New Age International (P) Ltd., PublishersPublished by New Age International (P) Ltd., PublishersAll rights reserved.No part of this ebook may be reproduced in any form, by photostat, microfilm, xerography,or any other means, or incorporated into any information retrieval system, electronic ormechanical, without the written permission of the publisher. All inquiries should beemailed to rights@newagepublishers.comISBN (13) : 978-81-224-2847-6PUBLISHING FOR ONE WORLDNEW AGE INTERNATIONAL (P) LIMITED, PUBLISHERS4835/24, Ansari Road, Daryaganj, New Delhi - 110002Visit us at www.newagepublishers.com
Preface to theSecond Revised EditionThis book has been revised exhaustively according to the global demands of the students.Attention has been taken to add minor steps between two unmanageable lines where essential sothat the students can understand the subject matter without mental tire.A number of questions have been added in this edition besides theoretical portion wherevernecessary in the book. Latest question papers are fully solved and added in their respective units.Literal errors have also been rectified which have been accounted and have come to ourobservation. Ultimately the book is a gift to the students which is now error free and user- friendly.Constructive suggestions, criticisms from the students and the teachers are always welcomefor the improvement of this book.AUTHORS
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Some Useful Formulae1.2.sin ix i sin hx3.sin x e ix e ix2i4.cos x e ix e ix25.Sin h2x 1(cosh 2x – 1)26.cos h2x 1(cosh 2x 1)27.za x dx axa 1, a 0log azzzsin haxdx 1cos haxacos haxdx 1sin haxatan haxdx 1log cos haxa8.9.10.11.12.13.14.15.16.17.cos ix cos hxzzzzzzz1a2 x212x a12x a22dx sin 1xx arc sinaadx log x x 2 a 2dx 1xxtan 1 arc tanaaaa 2 x 2 dx x2a2 x2 a2xsin 1a2e ax sin bx dx e ax(a sin bx – b cos bx)a2 b2e ax cos bx dx e ax(a cos bx b sin bx)a 2 b2sec ax dx 1log sec ax tan ax a
1log cosec ax – cot ax a18.z19.sin x x x 3 x5 .3520.cos x 1 x2 x 4 .2421.tan x x x32 5 17 7 x x .3 1531522.log (1 x) x 23.log (1 – x) x 24.sin hx x 25.d xa ax loge adx26.1dcos 1 x dx1 x227.1dcot 1 x dx1 x228.29.30.cosec ax dx ddxddxx2 x3 .23x 3 x5 .35cosec 1 x log a x x2 x3 x 4 .2341x1x x2 1log a edxatan 1. 2dxaa x2
ContentsPREFACE TO THE SECOND REVISED EDITIONSOME USEFUL FORMULAEU NIT I. Differential Calculus-I1.0 Introduction1.1 nth Derivative of Some Elementary FunctionsExercise 1.11.2 Leibnitz’s TheoremExercise 1.2Exercise 1.3Partial Differentiation1.3 Function of Two Variables1.4 Partial Differential CoefficientsExercise 1.41.5 Homogeneous Function1.6 Euler’s Theorem on Homogeneous FunctionsExercise 1.51.7 Total Differential CoefficientExercise 1.6Curve Tracing1.8 Procedure for Tracing Curves in Cartesian FormExercise 1.71.9 Polar CurvesExercise 1.81.10 Parametric CurvesExercise 1.9Expansion of Function of Several Variables1.11 Taylor’s Theorem for Functions of Two VariablesExercise 1.10Objective Type QuestionsAnswers to Objective Type QuestionsU NIT II. Differential Calculus-II2.1 JacobianExercise 2.11 9495 15095109
2.2 Approximation of ErrorsExercise 2.22.3 Extrema of Function of Several VariablesExercise 2.32.4 Lagrange’s Method of Undetermined MultipliersExercise 2.4Objective Type QuestionsAnswers to Objective Type QuestionsU NIT III. tion of MatrixTypes of MatricesOperations on MatricesTrace of MatrixProperties of TransposeProperties of Conjugate MatricesSingular and Non-Singular MatricesAdjoint of a Square MatrixInverse of a Matrix (Reciprocal)Exercise 3.1Elementary Row and Column TransformationsMethod of Finding Inverse of a Non-Singular Matrix by ElementaryTransformationsExercise 3.2Rank of a MatrixExercise 3.3System of Linear Equations (Non-Homogeneous)System of Homogeneous EquationsGaussian Elimination MethodExercise 3.4Linear Dependence of VectorsExercise 3.5Eigen Values and Eigen VectorsExercise 3.6Cayley-Hamilton TheoremExercise 3.7Diagonalization of a MatrixApplication of Matrices to Engineering ProblemsExercise 3.8Objective Type QuestionsAnswers to Objective Type Questions111119121134135145147150151 6188197200206210214214230232238239249253255257
U NIT IV. Multiple 134.144.154.164.17Multiple IntegralsDouble IntegralsWorking RuleDouble Integration for Polar CurvesExercise 4.1Change of the Order of IntegrationChange of Variables in a Multiple IntegralExercise 4.2Beta and Gamma FunctionsTransformations of Gamma FunctionTransformations of Beta FunctionRelation between Beta and Gamma FunctionsSome Important DeductionsDuplication FormulaEvaluate the IntegralsExercise 4.3Application to Area (Double Integrals)Exercise 4.4Triple IntegralsExercise 4.5Application to Volume (Triple Integrals)Exercise 4.6Dritchlet’s TheoremExercise 4.7Objective Type QuestionsAnswers to Objective Type QuestionsU NIT V. Vector CalculusVector Differential Calculus5.1 Vector Function5.2 Vector Differentiation5.3 Some Results on DifferentiationExercise 5.15.4 Scalar Point Function5.5 Vector Point Function5.6 Gradient or Slope of Scalar Point Function5.7 Geometrical Meaning of Gradient, Normal5.8 Directional Derivative5.9 Properties of GradientExercise 5.25.10 Divergence of a Vector Point Function5.11 Physical Interpretation of Divergence5.12 Curl of a Vector258 9300311312314315322323329330332333 418333333333334336337337337338338339350351352353
5.13 Physical Meaning of Curl5.14 Vector IdentitiesExercise 5.35.15 Vector Integration5.16 Line Integral5.17 Surface Integral5.18 Volume IntegralExercise 5.45.19 Green’s TheoremExercise 5.55.20 Stoke’s Theorem5.21 Cartesian Representation of Stoke’s TheoremExercise 5.65.22 Gauss’s Divergence Theorem5.23 Cartesian Representation of Gauss’s TheoremExercise 5.7Objective Type QuestionsAnswers to Objective Type 0401413414418Unsolved Question Papers (2004 2009)419 431Index433 434
UNIT1Differential Calculus-I1.0INTRODUCTIONCalculus is one of the most beautiful intellectual achievements of human being. The mathematicalstudy of change motion, growth or decay is calculus. One of the most important idea of differentialcalculus is derivative which measures the rate of change of a given function. Concept of derivativeis very useful in engineering, science, economics, medicine and computer science.dyd2 y, second order derivative, denoted bydxdx 2d3 ythird order derivative byand so on. Thus by differentiating a function y f(x), n times,dx 3dnyor Dny or yn(x). Thus, the processsuccessively, we get the nth order derivative of y denoted bydx nof finding the differential co-efficient of a function again and again is called SuccessiveDifferentiation.The first order derivative of y denoted by1.1 nth DERIVATIVE OF SOME ELEMENTARY FUNCTIONS1. Power Function (ax b)mLety (ax b)my1 ma (ax b)m–1y2 m (m–1)a2 (ax b)m–2. . . . .yn m(m–1) (m–2) . (m – n – 1) an (ax b)m–nCase I. When m is positive integer, thenm (m – 1).(m – n 1)(m – n).3 2 1 nyn a (ax b)m–n(m – n).3 2 1yn mdnm( ) axba n ( ax b) m nm ndx n1
2A TEXTBOOK OF ENGINEERING MATHEMATICS—ICase II. When m n ve integernyn 0a n (ax b)0 n an dn( ax b)n n a nndxCase III. When m –1, theny (ax b)–1 1( ax b)yn (–1) (–2) (–3) . (–n) an (ax b)–1–n RSTdn1ndx ax bUVW( 1)n n an (ax b)n 1Case IV. Logarithm case: When y log (ax b), thenaax bDifferentiating (n–1) times, we gety1 yn and n 1( ax b) 1dx n 1Using case III, we obtain lqdnlog( ax b)dx n ( 1) n 1 ( n 1) a n( ax b) n2. Exponential Function(i) Considery amxy1 mamx. loge ay2 m2amx (loge a)2.yn mn amx (loge a)n(ii) ConsiderPuttingy emxa e in aboveyn mnemx3. Trigonometric Functions cos (ax b) or sin (ax b)Lety cos (ax b), thenF ax b π IH2K2π IFcos ax b H2 K3π IFcos ax b H2 Ky1 – a sin (ax b) a cosy2 – a2 cos (ax b) a2y3 a3 sin (ax b) a3.
3DIFFERENTIAL CALCULUS-Iyn Similarly,yn FIHKnπ IFsin ax b H2Knπdncos ( ax b) a n cos ax b n2dxdndxnsin( ax b) a n4. Product Functions eax sin (bx c ) or e ax cos (bx c )Consider the function y eax sin (bx c)y1 eax·b cos (bx c) aeax sin (bx c) eax [b cos (bx c) a sin (bx c)]To rewrite this in the form of sin, puta r cos φ, b r sin φ, we gety1 eax [r sin φ cos (bx c) r cos φ sin (bx c)]y1 reax sin (bx c φ)Here,r –1a 2 b 2 and φ tan (b/a)Differentiating again w.r.t. x, we gety2 raeax sin (bx c φ) rbeax cos (bx c φ)Substituting for a and b, we gety2 reax. r cos φ sin (bx c φ) reax r sin φ cos (bx c φ)y2 r2eax [cos φ sin (bx c φ) sin φ cos (bx c φ)] r2 eax sin (bx c φ φ)y2 r2 eax sin (bx c 2φ)Similarly,y3 r3eax sin (bx c 3φ).yn dndxne ax sin( bx c ) r n e ax sin (bx c nφ)In similar way, we obtaind n axe cos(bx c) r ne ax cos (bx c nφ)yn dx n11 5x 6x 21 ( 2 x 1)( 3 x 1)Example 1. Find the nth derivative ofSol. Letor 11 5x 6x 223 y 2x 1 3x 1y yn 2(By Partial fraction)dndn–1(2x–1)–3(3x – 1)–1dx ndx n
4A TEXTBOOK OF ENGINEERING MATHEMATICS—IL( 1) n 2 OP LM( 1) n 3 OP2 MMN (2x 1) PQ – 3 MN (3x 1) PQL 2 3 OP .( 1) n MN (2x 1) ( 3x 1) Qn oryn nnn 1n 1n 1Example 2. Find the nth derivative of eax cos2 x sin x.(1 cos 2 x)sin xSol. Lety eax cos2 x sin x eax21 ax1 axe sin x e ( 2 cos 2x sin x) 22 21 ax1e sin x e ax sin( 3x) sin x 241 ax1 axe sin x e sin 3xy 441 n ax1 yn r e sin (x nφ) r1 n e ax sin ( 3x nθ) .44whereand( 1) n n a ndn 1(ax b) dx n( ax b) n 1n 1lorAsn 1n 1nnr qa 2 1 ; tan φ 1/aa 2 9 ; tan θ 3/a.2x, find yn.Example 3. If y tan–11 x22xSol. We havey tan 11 x2Differentiating y w.r.t. x, we getr1 y1 11 F 2x IH1 x K2 (U.P.T.U., 2002)FGHd2xdx 1 x 222(1 x 2 ) 2 IJK e1 x(1 x 2 ) 24 2x 2 4x 2(1 x 2 ) 2y1 1 11 , (by Partial fractions)i x i x iLMNOPQDifferentiating both sides (n–1) times w.r. to ‘x’, we getyn LMMNn 1n 11 ( 1) ( n 1) ( 1) ( n 1) i( x i) n( x i) n( 1)n 1 ( n 1)i 2(1 x 2 ) 4x 2(1 x 2 ) 22( x i)( x i )y1 (1 x 2 )j ( x i ) n ( x i ) n( 1)n 1 ( n 1)iOPPQr n (cos θ i sin θ) n r n (cos θ i sin θ) n(where x r cos θ, 1 r sin θ)
5DIFFERENTIAL CALCULUS-I ( 1) n 1 ( n 1) r nin 1 r–n sin nθ, where r yn 2(–1)n–1x 1. Show thatx 1LM x nN (x 1)n 2yn (–1)n–2Sol. We havex2 1F 1I.H xKθ tan–1Example 4. If y x logcos nθ i sin nθ cos nθ i sin nθn x n( x 1)nOPQ(U.P.T.U., 2002)x 1 x [log(x – 1) – log (x 1)]x 1y x logDifferentiating w.r. to ‘x’, we getLM 1 1 OPN x 1 x 1QF 1 I F 1 1 I log (x – 1) – log (x 1) 1 H x 1K H x 1 Ky1 log (x – 1) – log (x 1) xory1 log (x –1) – log (x 1) 11 x 1 x 1Differentiating (n–1) times with respect to N, we getyn d n 1dx n 1log (x 1) d n 2d n 1log (x 1) dx n 1d n 1d n 1dxdx n 1(x 1) 1 n 1RS d log(x 1)UV d RS d log(x 1)UV ( 1) n 1 ( 1) n 1TdxW dx TdxW (x 1)(x 1)n 2dxn 2(x 1) 1n 1n 2FHIKFHn 1nnIK( 1)n 1 n 1 ( 1) n 1 n 1d n 2d n 211 (x 1) n( x 1) ndx n 2 x 1dx n 2 x 1( 1)n 2 n 2(x 1) n 1 ( 1) n 2 n 2(x 1)n 1LM x 1N (x 1)L x nn 2MN (x 1)n 2n 2 ( 1)n n 2 ( 1)n Example 5. Find yn (0) if y Sol. We havey x 1( x 1)nx n( x 1)x3x 12n( 1)n 1 (n 1) n 2 OP .Q(x 1)n( n 1)( x 1)n ( n 1)( x 1)n ( 1)n 1 (n 1) n 2OPQ.x3x 3 1 1 ( x 1)( x 2 x 1)1 2 22( x 1)( x 1)x 1x 1x 1(x 1)n
6A TEXTBOOK OF ENGINEERING MATHEMATICS—Iory x2 x 11 ( x 1)( x 1)( x 1)y x2 1 11 1 x 1x 1 x 1y ory a fa f11L 11 O M x x 1 2 N x 1 x 1 PQ1L 11 O x M2 N x 1 x 1 PQ( 1) n O1 L ( 1) nP 0 M2 NM ( x 1)( x 1) QP( 1) n L11MN (x 1) (x 1) OPQ2( 1) n L 11 O M2N ( 1) (1) PQn yn nn 1n 1noryn n 1n 1nAtx 0, yn (0) When n is odd, yn(0) When n is even, yn(0) n 1( 1) n n2( 1) n n2n 11 1 n 1 1 0 .EXERCISE 1.11 . If y x2, find nth derivative of y.( x 1)2 ( x 2)(U.P.T.U., 2002)LMAns. yMN2 . Find the nth derivative ofx2.( x a)( x b)3 . Find the nth derivative of tan–1LM 1 x OP .N1 x Qn ( 1) n n 13( x 1) n 2LMAns.MN ( 1) n n( a b)5( 1) n n9( x 1) n 1LM aN ( x a)2n 1 OPPQOPOPQPQ4( 1) n n9( x 2) n 1b2(x b)n 1[Ans. ( 1)n 1 n 1 sin n θ sin nθ where θ cot–1x ]4 . If y sin3 x, find yn.5 . Find nth derivative of tan–1F xI .H aKLMAns.NF IH KAns. a 1fn 1 aFHππ31 . 3n . sin 3x nsin x n4242n 1 nI OPKQsin n θ sin nθ
7DIFFERENTIAL CALCULUS-Ia f2( 1) n OPP( x 1)QAns. e x x n6 . Find yn, where y ex.x.LMAns.MNAns. a 1fn 1 2xn1 x7 . Find yn, when y .1 xn 1n 18 . Find nth derivative of log x2.LMAns.N9 . Find yn, y ex sin2 x.ex1 5n/2 cos( 2x n tan 1 2 )21 0 . If y cos x · cos 2x · cos 3x find yn.1(cos 6 x cos 4x cos 2x 1)[Hint: cos x · cos 2x · cos 3x 4LMAns. 1 L6N 4 MNnFHcos 6x n nIKFHIKFHπππ 4n cos 4x n 2n cos 2x n222OPQI OPOPK QQ1.2 LEIBNITZ'S* THEOREMStatement. If u and v be any two functions of x, thenDn (u.v) nc0 Dn (u).v nc1Dn–1(u). D(v) nc2 Dn–2(u).D2 (v) . ncr Dn–r (u).Dr(v) . ncn u. Dn v .(i)(U.P.T.U., 2007)Proof. This theorem will be proved by Mathematical induction.D (u.v) D (u).v u.D(v) 1c0 D (u).v 1c1 u.D(v)Now,.(ii)This shows that the theorem is true for n 1.Next, let us suppose that the theorem is true for, n m from (i), we haveDm (u.v) mDifferentiating w.r. to x, we haveDm 1 (uv) mc0 Dm(u).v mc1 Dm–1 (u) D (v) mc2 Dm–2(u) D2 (v) . mcrDm–r(u) Dr (v) . mcm u Dm(v)rmrc0 D m 1 ( u) v D m ( u) D( v) mc1 D m ( u) D( v) Dm 1 ( u) D 2 ( v)m mc2mDm 1rmm . mcm D( u) D ( v ) uDBut from Algebra we know that mcr mcr 1 Dm 1(uv) m 1c 0 D m 1 ( u) v . cr( v )r( u) D 2 ( v ) D m 2 ( u). D 3 ( v) . m c r Dm r 1 ( u) Dr v D m r ( u) Dr 1 ( v)mcmhhm 1cr 1 and mc0 c 0 m c 1 D m ( u ) D( v ) cmmm 1m 1c0 1hc1 m c 2 D m 1 u D 2 vc r m c r 1 D m r ( u) D r 1 ( v ) . m 1 c m 1 u D m 1 ( v )c Asmcm m 1 cm 1 1h* Gottfried William Leibnitz (1646 1716) was born Leipzig (Germany). He was Newton’s rival inthe invention of calculus. He spent his life in diplomatic service. He exhibited his calculating machine in1673 to the Royal society. He was linguist and won fame as Sanskrit scholar. The theory of determinantsis said to have originated with him in 1683. The generalization of Binomial theorem into multinomialtheorem is also due to him. His works mostly appeared in the journal ‘Acta eruditorum’ of which hewas editor-in-chief.
8A TEXTBOOK OF ENGINEERING MATHEMATICS—I Dm 1(uv) m 1c0 D m 1 (u) v m 1 c1D m (u) D(v) m 1 c2 D m 1 (u) D 2 ( v) . m 1 c r 1 D m r (u) D r 1 (v) . m 1 cm 1u D m 1 (v).(iii)Therefore, the equation (iii) shows that the theorem is true for n m 1 also. But from (2)that the theorem is true for n 1, therefore, the theorem is true for (n 1 1) i.e., n 2, and sofor n 2 1 3, and so on. Hence, the theorem is true for all positive integral value of n.Example 1. If y1/m y–1/m 2x, prove that(x2 – 1) yn 2 (2n 1) xyn 1 (n2 – m2) yn 0.Sol. Given y1/m 1y 1/ m(U.P.T.U., 2007) 2xy2/m – 2xy1/m 1 0(y ) – 2x(y1/m) 1 0(y1/m z)z2 – 2xz 11/m 2or z 2x 4x 2 4 x 2x2 1x x2 1 y x Differentiating equation (i) w.r.t. x, we get y1 orx2 1y1/m m x x 1y1 2mym 1 y1LM1 N 22xx2mOP 1 Q.(i)m x x2 1x2 1x 2 1 myx 1y (x – 1) m yDifferentiating both sides equation (ii) w.r.t. x, we obtain2y1y2(x2 – 1) 2xy21 2m2 yy121 22m2 2.(ii)y2 (x2 – 1) xy1 – m2y 0Differentiating n times by Leibnitz's theorem w.r.t. x, we getDn (y2) · (x2 – 1) nc1 Dn–1y2·D2(x2 – 1) nc2 Dn 2 y2 D2 (x2 1) Dn (y1)x nc1 Dn–1 (y1) Dx–m2yn 0 yn 2 (x2 – 1) nyn 1· 2x n(n 1)yn · 2 yn 1 · x nyn – m2yn 02 (x2 – 1)yn 2 (2n 1) xyn 1 (n2 – n n – m2)yn 0 (x2 – 1) yn 2 (2n 1) xyn 1 (n2 – m2) yn 0. Hence proved.Example 2. Find the nth derivative of ex log x.Sol. Let u ex and v log xnxnThen D (u) e and D (v) ( 1) n 1 n 1xnD n ( ax b) 1 ( 1) n n( ax b)n 1
9DIFFERENTIAL CALCULUS-IBy Leibnitz’s theorem, we haveDn (ex log x) Dnex log x nc1 Dn–1 (ex) D(log x) nc2 Dn–2 (ex)D2 (log x) . ex Dn (log x)FGH1 n(n 1) x1e 2 e · log x ne · x2xx xLMMNDn (ex log x) ex log x IJKx . eOPPQxn( 1) n 1 n 1n n( n 1) .x2x 2xnExample 3. Find the nth derivative of x2 sin 3x.Sol. Let u sin 3x and v x2 ( 1)n 1 n 1F 3x n π IH 2KDn(u) Dn (sin 3x) 3n sinD(u) 2x, D2 (v) 2, D3 (v) 0By Leibnitz’s theorem, we haveDn (x2 sin 3x) Dn (sin 3x)x2 nc1 Dn–1 (sin 3x) · D (x2) nc2 Dn–2(sin 3x) · D2(x2)FH 3n sin 3x nπ2IK FH 3nx2 sin 3x FG 3x n 1π IJ · 2xH2 Kn(n 1)F n 2π IJ 2· 3 sin G 3x H22 KF n 1π IJsin G 3x H2 KF n 2π IJ . 3 n(n 1) · sin G 3x H2 K· x2 n3n–1 sinIKnπ 2nx · 3n-12n–2n 2Example 4. If y x log (1 x), prove thatyn ( 1) n 2 n 2 (x n)(x 1) n.Sol. Let u log (1 x), v xDn (u) dnd n 1log(1 )x dx ndx n 1d n 1dx n 1 (U.P.T.U., 2006)F d log (1 x)IH dxKd n 11( x 1) 1 x 1dx n 1( 1) n 1 n 1nD (u) ( x 1) nand D(v) 1, D2(v) 0By Leibnitz’s theorem, we haveyn Dn (x log (1 x) Dn (log (1 x)) x nc1 Dn–1 (log (1 x)) Dx x( 1) n 1 n 1( x 1) n n( 1) n 2 n 2( x 1) n 1
10A TEXTBOOK OF ENGINEERING MATHEMATICS—ILM x(n 1) n(x 1) OPN (x 1) (x 1) QL xn x xn n OPn 2 MN (x 1) QL x n OP . Hence proved.n 2 MN (x 1) Qyn (–1)n–2 n 2 (–1)n–2 (–1)n–2nnnnExample 5. If y a cos (log x) b sin (log x). Show thatx2y2 xy1 y 0x2yn 2 (2n 1) xyn 1 (n2 1) yn 0.y a cos (log x) b sin (log x)andSol.Given ory1 – a sin (log x)(U.P.T.U., 2003)F 1 I b cos (log x) F 1 IH xKH xKxy1 – a sin (log x) b cos (log x)Again differentiating w.r.t. x, we getxy2 y1 – a cos (log x) F 1IH xK– b sin (log x)F 1IH xKx2y2 xy1 – {a cos (log x) b sin (log x)} – yx2y2 xy1 y 0. Hence proved.(i)Differentiating (i) n times, by Leibnitz’s theorem, we haveyn 2 · x2 n yn 1 · 2x n(n 1)yn · 2 yn 1 · x nyn yn 02 x2yn 2 (2n 1) xyn 1 (n2 – n n 1) yn 0 x2yn 2 (2n 1) xyn 1 (n2 1) yn 0.Hence proved.Example 6. If y (1 – x)–α e–αx, show that(1 – x)yn 1 – (n αx) yn – nαyn–1 0.Sol. Given y (1 – x)–α. e–αxDifferentiating w.r.t. x, we gety1 α (1 – x)–α–1 e–αx – (1 – x)–α e–αx·αy1 (1 – x)–α e–αx ·α y1 (1 – x) αxyLM 1 1OPN1 x Q yαLM x OPN1 x QDifferentiating n times w.r.t. x, by Leibnitz’s theorem, we getyn 1 (1 – x) – nyn αyn· x nαyn–1 (1 – x)yn 1 – (n αx) yn – nαyn–1 0.FG y IJ log F x IH bKH mKFG y IJ log F x IH bKH mKHence proved.mExample 7. If cos–1Sol. We have cos–1, prove that x2yn 2 (2n 1)xyn 1 (n2 m2) yn 0.m m logxm
11DIFFERENTIAL CALCULUS-I y b cosF m log x IH mKOn differentiating, we haveF m log x I · mH mK xxIF– mb sin H m log Km2y1 – b sin xy1 Again differentiating w.r.t. x, we getFHxy2 y1 – mb cos m logx (xy2 y1) – m2b cos 1mIK11xm··xmmmF m log x IH mK –m2yx2y2 xy1 m2y 0orDifferentiating n times with respect to x, by Leibnitz’s theorem, we getyn 2 · x2 nyn 1 · 2x n(n 1) · 2yn xyn 1 nyn m2yn 02 x2yn 2 (2n 1) xyn 1 (n2 – n n m2)yn 0 x2yn 2 (2n 1) xyn 1 (n2 m2) yn 0.Hence proved.Example 8. If y (x2 – 1)n, prove that(x2 – 1)yn 2 2xyn 1 – n (n 1)yn 0(U.P.T.U., 2000, 2002)UVWRSTdPddn(1 – x 2 ) n n(n 1) Pn 0.(x2 – 1)n, show thatndxdxdx2nSol. Giveny (x – 1)Differentiating w.r. to x, we getHence, if Pn y1 n(x2 – 1)n–1. 2x 2nx (x 2 1)n(x 2 1)(x2 – 1) y1 2nxyAgain differentiating, w.r.t. x, we obtain(x2 – 1) y2 2xy1 2nxy1 2nyNow, differentiating n times, w.r.t. x by Leibnitz's theorem(x2 – 1)yn 2 2nxyn 1 oror2n(n 1)yn 2xyn 1 2nyn 2nxyn 1 2n2yn 2nyn2(x2 – 1)yn 2 2xyn 1 (n 1 – n) (n2 – n 2n – 2n2 – 2n)yn 0(x2 – 1) yn 2 2xyn 1 – (n2 n)yn 0 (x2 – 1) yn 2 2xyn 1 n(n 1) yn 0. Hence proved.(i)
12A TEXTBOOK OF ENGINEERING MATHEMATICS—ISecond part: Let y (x2 – 1)n NowPn RSTddyn(1 x 2 )dxdxUV Wdny yndx nnd(1 x 2 )yn 1dxs (1 – x2)yn 2 – 2xyn 1 – (x 2 1) y n 2 2xyn 1 RSUV – n(n 1)yTWd RS(1 x ) dPdx UVW n (n 1)y 0.dx Tdd(1 x 2 )Pndxdx2or[Using equation (i)]nnnHence proved.Example 9. Find the nth derivative of y xn–1 log x at x 1.2Sol. Differentiatingy1 (n – 1) xn–1–1 log x xn–1or1x(n 1)x n 1 log x x n 1 xy1 (n – 1)y xn–1xxDifferentiating (n–1) times by Leibnitz's theorem, we gety1 c1 yn-1 (n – 1)yn–1 n 1 xyn (n – 1)yn–1 (n – 1)yn–1 n 1 xyn xyn n–1Atx ynFG 1 IJH 2Kn 1 i.e. y nd n 1 n 1 ( n 1)( n 2).2.1 n 1xdx n 1n 1x12 2n 1 .Example 10. If y (1 – x2)–1/2 sin–1x, when –1 x 1 and ππ sin–1x , then show22that (1 – x2)yn 1 – (2n 1) xyn – n2 yn–1 0.Sol. Given y (1 – x2)–1/2 sin–1xDifferentiatingy1 –12 2x(1 x )xy 1sin 1 x1 22(1 x 2 )(1 x )(1 x )y1 (1–x2) xy 1y1 1(1 – x2)–3/2 (–2x) sin–1x (1 – x2)–1/2·211 x2
13DIFFERENTIAL CALCULUS-IDifferentiating n times w.r.t. x, by Leibnitz's theorem, we getn(n 1)yn 1 (1 – x2) nyn (–2x) yn–1 · (–2) xyn nyn–12(1 – x2)yn 1 – (2n 1)xyn – (n2 – n n)yn–1 0 Hence proved.(1 – x2)yn 1 – (2n 1)xyn – n2yn–1 0.Example 11. If y xn log x, then prove thata fn(ii) yn nyn–1 n 1 .xSol. (i) We have y xn log xDifferentiating w.r. to x, we get(i) yn 1 y1 nxn–1 · log x xnxxy1 nxn . log x xnxy1 ny xn.(i)Differentiating equation (i) n times, we getnxyn 1 nyn nyn yn 1 (ii)yn nProved.xFG d x . log xIJH dxKF x nx . log xIGH xJKex .log xj dxd . xdnd n 1x n .log x ndxdx n 1ed n 1dx n 1 njnd n 1dx n 1 nyn–1 nn 1n 1n 1an 1f .n 1n 1Proved. yn 1ejdnx n log xndxd n 1 n 1 x n 1 log xdxAs yn eEXERCISE 1.2Find the nth derivative of the following:1 . ex log x.2 . x2 ex.LMAns. e Llog x MNNxnc1 1 n11 c 2 2 2 n c 3 3 . ( 1) n 1 n 1 n c n x nxxxOP OPQQAns. e x x 2 2nx n(n 1)j
14A TEXTBOOK OF ENGINEERING MATHEMATICS—ILMAns. 6( 1) n 4 OPxMNPQLMAns. 2 ( 1) n OPMN (1 x) PQn3 . x log x.34.n 31 x.1 xnn 15 . x2 sin 3x.LMAns. 3 xNn2FHsin 3x IKLMNUVOPWQ 12(n 1)( 2x 3) 8n(n 1)(n 2)tOPQRSTπ1nπ 2nx 3 n 1 sin{ 3x ( n 1) π } 3 n 2 n( n 1) sin 3x ( n 2)222oAns. e x ( 2x 3) 2 6n( 2x 3) 26 . ex (2x 3)3.7 . If x tan y, prove that(1 x2) yn 1 2nxyn n(n–1)yn–1 0.(U.P.T.U., 2006)8 . If y ex sin x, prove that y′′–2y′ 2y 0.9 . If y sin (m sin–1x), prove that(1 – x2)yn 2 – (2n 1)x · yn 1 (m2 – n2)yn 0.1 0 . If x cos hLMF 1 I log yOP , prove that (x – 1)y xy – m y 0 and (x – 1)yNH m K Q2FG y IJH bK22 (n2 – m2)yn 0.1 1 . If cos–1(U.P.T.U., 2004, 2002) logF xIH nK21n 2 (2n 1)xyn 1n, prove that x2yn 2 (2n 1)xyn 1 2n2yn 0.1 2 . If y etan–1x, prove that (1 x2)yn 2 {2(n 1) x –1}yn 1 n(n 1)yn 0.1 3 . If sin–1y 2 log (x 1), show that(x 1)2yn 2 (2n 1)(x 1)yn 1 (n2 4)yn 0.d1 4 . If y C1 x x 2 1ind C2 x x 2 1i , prove that (xn2– 1)yn 2 (2n 1)xyn 1 0.1 5 . If x cos [log (y1/a)], then show that (1 – x2)yn 2 – (2n 1) xyn 1 – (n2 a2) yn 0.1.2.1 To Find (yn)0 i.e., nth Differential Coefficient of y, When x 0Sometimes we may not be able to find out the nth derivative of a given function in a compact formfor general value of x but we can find the nth derivative for some special value of x generallyx 0. The method of procedure will be clear from the following examples:–1Example 1. Determine yn(0) where y em.cos x.Sol. We havey em.cos–1xDifferentiating w.r.t. x, we gety 1 emorcos–1xmFGH 11 x2IJK.(i) 222 21 x 2 y1 – my (1 – x )y1 m ym1 x 2 . y1 – mecos–1x
15DIFFERENTIAL CALCULUS-IDifferentiating again(1 – x2) 2y1 y2 – 2xy21 2m2yy1 (1 – x2)y2 – xy1 m2y.(ii)Using Leibnitz's rule differentiating n times w.r.t. x2n(n 1)yn – xyn 1 – nyn m2yn(1 – x2)yn 2 – 2nxyn 1 –2(1 – x2)yn 2 – (2n 1)xyn 1 – (n2 m2)yn 0orPutting x 0yn 2 (0) – (n2 m2)yn (0) 0 replace n by (n – 2)yn 2 (0) (n2 m2)yn(0).(iii)yn (0) {(n – 2)2 m2} yn–2 (0)replace n by (n – 4) in equation (iii), we getyn–2 (0) {(n – 4 )2 m2} yn–4 (0) yn(0) {(n – 2)2 m2} {(n – 4)2 m2} yn–4 (0)Case I. When n is odd:yn(0) {(n – 2)2 m2} {(n – 4)2 m2} . (12 m2)y1 (0)[The last term obtain putting n 1 in eqn. (iii)]1m cos 1 xm Now we havey1 e1 x2At x 0, y1(0) memπ2.(iv).(v) As cos 1 0 Using (v) in (iv), we getyn(0) {an 2f2 m2} {an 4f2} ej m 2 . 1 2 m 2 memπ2π2.Case II. When n is even:yn(0) {(n – 2)2 m2} {(n – 4)2 m2} . (22 m2)y2 (0)[The last term obtain by putting n 2 in (iii)]From (ii),y2(0) m2(y)0 y2(0) m2 emπ/2.(vi)As y e m cos.(vii) y(0) e m cos 10 1x emπ/2From eqns. (vi) and (vii), we getyn(0) {(n – 2)2 m2} {(n – 4)2 m2} . (22 m2) m2emπ/2.Example 2. If y (sin–1x)2. Prove that yn(0) 0 for n odd and yn(0) 2.22.42.(n – 2)2,n 2 for n even.(U.P.T.U., 2005, 2008)–12Sol. We havey (sin x).(i)1 1On differentiatingy 1 2 sin–1 x · y1 1 x 2 2 y , As y sin x21 xSquaring on both sides,y21 (1 – x2) 4yej
16orA TEXTBOOK OF ENGINEERING MATHEMATICS—IAgain differentiating2(1 – x2) y1 y2 – 2xy12 4y1(1 – x2) y2 – xy1 2.(ii)Differentiating n times by Leibnitz’s theorem2n n 1(1 – x2) yn 2 – 2nxyn 1 –yn – xyn 1 – nyn 02a f(1 – x2) yn 2 – (2n 1) xyn 1 – n2yn 0Putting x 0 in above equationyn 2 (0) – n2 yn (0) 0 yn 2 (0) n2 yn(0)replace n by (n – 2)yn(0) (n – 2)2 yn –2 (0)or.(iii)Again replace n by (n – 4) in (iii) and putting the value of yn–2 (0) in above equationyn (0) (n – 2)2 (n – 4)2 yn–4 (0)Case I. If n is odd, thenyn(0) (n – 2)2 (n – 4)2 (n – 6)2 . 12·y1 (0)1Buty1(0) 2 sin–1 0· 01–0 yn(0) 0. Hence proved.Case II. If n is even, thenyn(0) From (ii)y2 (0) Using this value in eqn. (iv), weyn(0) .(iv)yn(0) 2.22.42 . (n – 2)2, n 2 otherwise 0. Proved.orExample 3. Ify x 1 x2Sol. Giveny x 1 x2 y1 1 x 2orSquaringmm x 1 x21 x, find yn 0).m2y1 m x 1 x m 1mLM1 N 2x1 x2OPQ.(i)my1 x2 myy21(1 x2) m2y221or(n – 2)2 (n – 4)2 . 22· y2 (0)2get(n – 2)2 (n – 4)2 . 22·2.(ii)22Again differentiating, y (2x) (1 x )2y1y2 m ·2yy1y2 (1 x2) xy1 – m2y 0Differentiating n times by Leibnitz’s theorem2n(n 1)(1 x2)yn 2 2nxyn 1 yn xyn 1 nyn – m2yn 02.(iii)
17DIFFERENTIAL CALCULUS-I(1 x2)yn 2 (2n 1)xyn 1 (n2 – m2)yn 0Putting x 0, we getyn 2 (0) (n2 – m2) yn(0) 0 yn 2 (0) – (n2 – m2) yn (0)replace n by n – 2yn (0) – { (n – 2)2 – m2} yn–2 (0)Again replace n by (n – 4) in (iv) and putting yn–2 (0) in above equationor.(iv)yn(0) (–1)2 {(n – 2)2 – m2} {(n – 4}2 – m2} yn–4 (0)Case I. If n is oddButor Case II. If n is evenyn(0)y1(0)y1(0)yn(0) – {(n – 2)2 – m2} {(n – 4)2 – m2} . {12 – m2} y1 (0)my(0)m (As y (0) 1){m2 – (n – 2)2}{m2 – (n – 4)2} . (m2 – 12)·m.yn (0) {m2 – (n – 2)2} {m2 – (n – 4)2}. (m2 – 22) y2 (0) yn (0) {m2 – (n – 2)2} {m2 – (n – 4)2} . (m2 – 22)·m2.(As y2 (0) m2).–1Example 4. Find the nth differential coefficient of the function on cos (2 cos x) at the pointx 0.orSol. Lety cos (2 cos–1 x)On diffentiating,y1 – sin (2 cos–1 x)LMMN 21 x2OPPQ.(i)–11 x 2 2 sin (2 cos x)Squaring on both sides, we gety1e 1 4 sin2 2 cos xy21 (1 – x2){ejj}2 1 4 1 cos 2 cos xor2122y (1 – x ) 4 (1 – y )Again differentiating w.r.t. x, we get2y1y2 (1 – x2) – 2xy21 – 8yy1ory2 (1 – x2) – xy1 4y 0.(ii)Differentiating n times by Liebnitz’s theorem2n n 1(1 – x2) yn 2 – 2nxyn 1 –yn – xyn 1 – nyn 4yn 02a f(1 – x2) yn 2 – (2n 1) xyn 1 – (n2– 4) yn 0Putting x 0 in above equation, we getyn 2 (0) – (n2 – 4) yn (0) 0oryn 2 (0) (n2 – 4) yn(0)Replace n by n – 2, we getyn (0) {(n – 2)2 – 4} yn–2 (0).(iii)
18A TEXTBOOK OF ENGINEERING MATHEMATICS—IAgain replace n by (n – 4) in (iii) and putting yn –2 (0) in above then, we getyn (0) {(n – 2)2 – 4} {(n – 4)2 – 4} yn – 4 (0)Case I. If n is oddyn (0) {(n – 2)2 – 4} {(n – 4)2 – 4} . (12 – 4) y1 (0)Buty1 (0) 2 sin (2 cos–1 0) 2 sin (π) 0 yn (0) 0.Case II. If n is evenyn (0) {(n – 2)2 – 4}{(n – 4)2 – 4} . {22 – 4} y2 (0)yn (0) 0Hence for all values of n, even or odd,yn (0) 0.Example 5. Find the nth derivative of y x2 sin x at x 0.Sol. We have(U.P.T.U., 2008)y x2 sin x sin x. x2.(i)Dif
2.2 Approximation of Errors 111 Exercise 2.2 119 2.3 Extrema of Function of Several Variables 121 Exercise 2.3 134 2.4 Lagrange’s Method of Undetermined Multipliers 135 Exercise 2.4 145 Objective Type Questions 147 Answers to Objective Type Questions 150 ˇ ˇ 3.0 Introduction 151 3.1 Definition
