Engineering mathematics II

This page intentionally left blank Copyright © 2009, New Age International (P) Ltd., Publishers Published by New Age International (P) Ltd., Publishers All 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 or mechanical, without the written permission of the publisher All inquiries should be emailed to rights@newagepublishers.com ISBN (13) : 978-81-224-2882-7 PUBLISHING FOR ONE WORLD NEW AGE INTERNATIONAL (P) LIMITED, PUBLISHERS 4835/24, Ansari Road, Daryaganj, New Delhi - 110002 Visit us at www.newagepublishers.com Preface This Textbook has been prepared as per the syllabus for the Engineering Mathematics Second semester B.E classes of Visveswaraiah Technological University The book contains eight chapters, and each chapter corresponds to one unit of the syllabus The topics covered are: Unit I and II— Differential Calculus, Unit III and IV—Integral Calculus and Vector Integration, Unit V and VI— Differential Equations and Unit VII and VIII—Laplace Transforms It gives us a great pleasure in presenting this book In this edition, the modifications have been dictated by the changes in the VTU syllabus The main consideration in writing the book was to present the considerable requirements of the syllabus in as simple manner as possible This will help students gain confidence in problem-solving Each unit treated in a systematic and logical presentation of solved examples is followed by an exercise section and includes latest model question papers with answers from an integral part of the text in which students will get enough questions for practice The book is designed as self-contained, comprehensive and friendly from students’ point of view Both theory and problems have been explained by using elegant diagrams wherever necessary We are grateful to New Age International (P) Limited, Publishers and the editorial department for their commitment and encouragement in bringing out this book within a short span of period AUTHORS Acknowledgement It gives us a great pleasure to present this book ENGINEERING MATHEMATICS-II as per the latest syllabus and question pattern of VTU effective from 2008-2009 Let us take this opportunity to thank one and all who have actually given me all kinds of support directly and indirectly for bringing up my textbook We whole heartedly thank our Chairman Mr S Narasaraju Garu, Executive Director, Mr S Ramesh Raju Garu, Director Prof Basavaraju, Principal Dr T Krishnan, HOD Dr K Mallikarjun, Dr P.V.K Perumal, Dr M Surekha, The Oxford College of Engineering, Bangalore We would like to thank the other members of our Department, Prof K Bharathi, Prof G Padhmasudha, Mr Ravikumar, Mr Sivashankar and other staffs of The Oxford College of Engineering, Bangalore for the assistance they provided at all levels for bringing out this textbook successfully We must acknowledge Prof M Govindaiah, Principal, Prof K.V Narayana, Reader, Department of Mathematics, Vivekananda First Grade Degree College, Bangalore are the ones who truly made a difference in our life and inspired us a lot We must acknowledge HOD Prof K Rangasamy, Mr C Rangaraju, Dr S Murthy, Department of Mathematics, Govt Arts College (Men), Krishnagiri We are also grateful to Dr A.V Satyanarayana, Vice-Principal of R.L Jalappa Institute of Technology, Doddaballapur, Prof A.S Hariprasad, Sai Vidya Institute of Technology, Prof V.K Ravi, Mr T Saravanan, Bangalore college of Engineering and Technology, Prof L Satish, Raja Rajeshwari College of Engineering, Prof M.R Ramesh, S.S.E.T., Bangalore A very special thanks goes out to Mr K.R Venkataraj and Bros., our well wisher friend Mr N Aswathanarayana Setty, Mr D Srinivas Murthy without whose motivation and encouragement this could not have been completed We express our sincere gratitude to Managing Director, New Age International (P) Limited, and Bangalore Division Marketing Manager Mr Sudharshan for their suggestions and provisions of the font materials evaluated in this study We would also like to thank our friends and students for exchanges of knowledge, skills during our course of time writing this book AUTHORS Dedicated to my dear parents, Shiridi Sai Baba, my dear loving son Monish Sri Sai G and my wife and best friend S Mamatha — A Ganesh & my dear parents, and my wife S Geetha — G Balasubramanian This page intentionally left blank QUESTION PAPER LAYOUT Engineering Mathematics-II O6MAT21 Units-1, 2, 3, PART-A Qns Qn from each unit PART-B Qns Units-5, 6, 7, Qn from each unit To answer fivefull questions choosing at leasttwo questions from each part Time: Hrs Unit/Qn No Max Marks: 100 Topics DIFFERENTIAL CALCULUS-I Unit/Qn No Topics DIFFERENTIAL EQUATIONS-I Linear differential equation with constant coefficients, Solution of homogeneous and non homogeneous linear D.E., Inverse differential operator and the Particular Integral (P.I.) Radius of Curvature: Cartesian curve Parametric curve, Pedal curve, Polar curve and some fundamental theorems Method of undetermined coefficients DIFFERENTIAL CALCULUS-II Taylor’s, Maclaurin’s Maxima and Minima for a function of two variables INTEGRAL CALCULUS-II Method of variation of parameters, Solutions of Cauchy's homogeneous linear equation and Legendre’s linear equation, Solution of initial and Boundary value problems VECTOR INTEGRATION AND ORTHOGONAL CURVILINEAR COORDINATES LAPLACE TRANSFORMS Periodic function, Unit step function (Heaviside function), Unit impulses function Double and triple integral, Beta and Gamma functions DIFFERENTIAL EQUATIONS-II INVERSE LAPLACE TRANSFORMS Applications of Laplace transforms 433 MODEL QUESTION PAPER—I (b) Use the line integral, compute work done by a force F = (2y + 3) i + xz j + ( yz – x) k when it moves a particle from the point (0, 0, 0) to the point (2, 1, 1) along the curve x = 2t2, y = t, z = t3 [VTU, Jan 2009] (04 marks) Solution Refer Unit IV (c) Verify Green’s theorem for zd c i xy + y dx + x dy where c is bounded by y = x and y = x2 [VTU, Jan 2009] (06 marks) Solution Refer Unit IV (d ) Prove that the cylindrical coordinate system is orthogonal [VTU, Jan 2009] (06 marks) Solution Refer Unit IV PART B dy = y is: dx (a) Linear (b) Non-linear (c) Quasilinear (d ) None of these (a) (i) The differential equation (ii) The particular integral of d2y + y = cos x is: dx 1 1 x sin x sin x (b) cos x (c) x cos x (d ) 2 2 (iii) The general solution of the D.E (D2 + 1)2 y = is: (a) c1 cos x + c2 sin x (b) (c1 + c2 x) cos x + (c3 + c4 x) sin x (c) c1 cos x + c2 sin x + c3 cos x + c4 sin x (d ) (c1 cos x + c2 sin x) (c3 cos x + c4 sin x) (a) (iv) The set of linearly independent solution of the D.E (a) {1, x, ex, e–x} (c) {1, x, ex, xex} (b) {1, x, e–x, xe–x} (d ) {1, x, ex, xe–x} [Ans d] [Ans b] d4y d2y − = is: dx dx [Ans a] (04 marks) d y d y dy = e–x + sin 2x +2 + dx dx dx Solution Here, the A.E is m3 + 2m2 + m = ⇒ m (m2 + 2m + 1) = ⇒ m (m + 1)2 = whose roots are 0, –1, –1 ∴ C.F = C1 e0x + (C2 + C3 x) e–x ∴ C.F = C1 + (C2 + C3 x) e–x (b) Solve [Ans b] [VTU, July, 2008] (04 marks) 434 ENGINEERING MATHEMATICS—II Also P.I = (e–x + sin 2x) D + D2 + D sin x e−x + 3 D + D + D D + D2 + D P.I = P.I.1 + P.I.2 = ∴ P.I.1 = = e− x D3 + D2 + D (D → –1) e− x b−1g + 2b−1g + b−1g (Dr = 0) Diefferentiate the denominator and multiply ‘x’, we get ∴ P.I.1 = = xe − x 3D + D + (D → –1) xe − x b g b g (Dr = 0) −1 + −1 + Again differentiate and multiply x ∴ ⇒ P.I.1 = x 2e − x 6D + P.I.1 = x2e− x −2 P.I.2 = sin x D + D2 + D (D → –1) (D2 → – 22 = – 4) = sin x − 4D − + D = sin x −3 D − = − = − sin x 3D − × 3D + 3D − b6 cos x − sin xg (D2 → – 22) D − 64 (6 cos 2x – sin 2x) 100 ∴ The general solution is y = C.F + P.I ∴ P.I.2 = y = C1 + (C2 + C3 x) e–x – 1 –x x e + (3 cos 2x – sin 2x) 50 435 MODEL QUESTION PAPER—I d2y − y = cos h (2x – 1) + 3x (c) Solve dx Solution We have (D2 – 4) y = cos h (2x – 1) + 3x A.E is m2 – = or (m – 2) (m + 2) = ⇒ m = 2, – ∴ C.F = C1 e2x + C2 e–2x P.I = b [VTU, Jan 2009] (06 marks) g cos h x − + x LM MN D −4 OP PQ e x −1 e b 1g 3x + + = 2 D −4 D −4 D −4 − x− = P.I.1 + P.I.2 + P.I.3 P.I.1 = = e x −1 e x −1 = D2 − 22 − (Dr = 0) x e x −1 x e x −1 = = x e x −1 2D e b g D2 − − x −1 P.I.2 = (D → – 2) e −b −1g −2 − x = b g (Dr = 0) − −1 xe b g 2D x = P.I.2 = − x – (2x e (D → – 2) – 1) 3x e b log 3g = P.I.3 = D2 − D2 − x = = e b log 3g blog 3g (D → log 3) x −4 3x blog 3g −4 Complete solution: y = C.F + P.I = C1 e2x + C2 e–2x + x 2x – x – (2x – 1) e – e + 8 3x blog 3g −4 436 ENGINEERING MATHEMATICS—II (d) Solve by the method of undetermined coefficients; (D2 + 1) y = sin x [VTU, July 2008] (06 marks) Solution Refer Unit V (a) (i) The homogeneous linear differential equation whose auxillary equation has roots 1, and – is: (a) (D3 + D2 + 2D + 2) y = (b) (D3 + 3D – 2) y = (c) (D3 – 3D + 2) y = (d ) (D + 1)2 (D – 2) y = (ii) The general solution of (x2 D2 [Ans c] – xD) y = is : (a) y = C1 + C2 ex (b) y = C1 + C2 x (c) y = C1 + C2 x2 (d ) y = C1 x + C2 x2 [Ans c] (iii) Every solution of y′′ + ay′ + by = where a and b are constants approaches to zero as x → ∞ provided (a) a > 0, b > (b) a > 0, b < (c) a < 0, b < (d ) a < 0, b > [Ans a] (iv) By the method of variation of parameters the W is called (a) Work done (b) Wronskian (c) Euler’s (d ) None of these [Ans b] (04 marks) (b) Solve (1 + x)2 y″ + (1 + x) y′ + y = sin [log (1 + x)] [VTU, Jan 2009] (04 marks) Solution t = log (1 + x) or et = + x Put Then we have (1 + x) dy = 1·Dy dx d2y (1 + = 12·D (D – 1) y dx Hence the given D.E becomes x)2 [D (D – 1) y + D + 1] y = sin t i.e, A.E is ∴ (D2 + 1) y = sin t m2 + = ⇒ m = ± i C.F = C1 cos t + C2 sin t P.I = sin t D2 + (D2 → –12 = –1) = sin t −1 + (Dr = 0) = sin t D × 2D 2D 437 MODEL QUESTION PAPER—I = cos t D2 (D2 → –12 = –1) cos t −4 P.I = – cos t = ∴ ∴ The complete solution is y = C.F + P.I y = C1 cos t + C2 sin t – cos t where t = log (1 + x) y = C1 cos [log (1 + x)] + C2 sin [log (1 + x)] – cos [log (1 + x)] (c) Solve x FG H IJ K d 3y d y x + + y = 10 x + · x dx dx [VTU, Jan 2009] (06 marks) Solution Refer Unit VI d2y + y = tan x dx (d) Solve, by the method of variation of parameters Solution Refer Unit VI (a) (i) The Laplace transform of t2 et is : (a) (c) b s − 2g (b) (d ) bs − 2g b s − 2g bs − 1g [Ans b] (ii) L [e–t sin ht] is: (a) (c) bs + 1g +1 s s+2 b g (b) (d ) bs − 1g +1 s −1 bs − 1g +1 [Ans c] (iii) L [e–3t cos 3t] = (a) s−3 s − 6s − 18 (b) s+3 s + 6s + 18 (c) s+3 s − 6s + 18 (d ) s−3 s + 6s − 18 (iv) L 2 LM sin t OP = Nt Q s +1 (c) cot–1 (s – 1) (a) 2 [Ans b] (b) cot–1 s (d ) tan–1 s [Ans b] (04 marks) 438 ENGINEERING MATHEMATICS—II (b) Find the Laplace transform of cos 2t − cos 3t ⋅ t Solution Refer Unit VII 2t + Put [VTU, Jan 2009] (04 marks) a = 2, Ans b= s + 32 1 + log s − log 2 s + 22 (c) Find the Laplace transform of the periodic function with period R|sin wt , < t < π w f (t) = S 2π π ||0, 2π Solution Let ⇒ and i.e., ∴ But f (t) = cos t + (cos 2t – cos t) u (t – π) + (cos 3t – cos 2t) u (t – 2π) L [f (t)] = L (cos t) + L [(cos 2t – cos t) u (t – π)] + L [(cos 3t – cos 2t) u (t – 2π)] (1) F (t – π) = cos 2t – cos t G (t – 2π) = cos 3t – cos 2t F (t) = cos (t + π) – cos (t + π) G (t) = cos (t + 2π) – cos (t + 2π) F (t) = cos 2t + cos t G (t) = cos 3t – cos 2t bg G b sg F s = s s + s + s +1 = s s − s +9 s +4 2 bg G b sg LM s + s OP N s + s + 1Q LM s − s OP Ns + s + 4Q L [F (t – π) u (t – π)] = e–πs F s and L [G (t – 2π) u (t – 2π)] = e–2πs i.e., − πs L [(cos 2t – cos t) u (t – π)] = e and [VTU, Jan 2009] (06 marks) L [(cos 3t – cos 2t) u (t – 2π)] = e −2 πs 2 2 439 MODEL QUESTION PAPER—I Hence (1) becomes Thus LM N OP Q s s s s s + e −2 π s − + e − πs + s +1 s + s +1 s +9 s +4 L [f (t)] = 5s e −2 πs 1 s − πs s e − + + s2 + s2 + s + s2 + s2 + (a) (i) Given L−1 LM Ns 2 + a2 OP = Q FG sin at IJ H a K LM e OP is: Ns Q LM N sin at then L−1 a (a) cos at −1 (ii) L bt − 1g cos at a (d ) sin at a (b) u (t – 1) t2 LM s e OP is: Ns + 9Q [Ans a] bt − 1g (d ) None of these [Ans a] (b) – cos 3t u (t – π) (d ) e–as L [f (t – a)] [Ans b] − sπ (a) cos 3t u (t – π) (c) cos 3t u (t – π)/3 (iv) The Laplace inverse of (c) i (c) u (t) (a) OP Q −s (a) u (t – 1) (iii) L−1 OP Q d id LM s OP is: Ns + a Q (b) (c) LM N L [f (t)] = t is: π s π 2s 32 LM FG s + a IJ OP N H s + bKQ F s + a IJ Solution Let F b sg = log G H s + bK (b) Find L−1 log (b) π s (d ) Does not exist [Ans c] (04 marks) [VTU, Jan 2009] (04 marks) = log (s + a) – log (s + b) 440 ENGINEERING MATHEMATICS—II bg −F ′ s Now i.e., Thus, LM N b gOQP L−1 − F ′ s RS − UV Ts + a s +bW L OP − L LM OP L M Ns + bQ Ns + aQ = − = −1 −1 t f (t) = e–bt – e–at f (t) = e −bt − e −at · t (c) Apply Convolution theorem to evaluate L−1 LM MM d s N OP P⋅ +a i P Q s 2 [VTU, Jan 2009] (06 marks) Solution Refer Unit VIII (d) Solve the differential equation by Laplace transform onethod, y″ + 4y′ + 3y = e–t and the initial conditions y (0) = y′ (0) = [VTU, Jan 2009] (06 marks) Solution Refer Unit VIII GGG MODEL QUESTION PAPER–II 06 MAT 21 Second Semester B.E Degree Examination Engineering Mathematics–II Time : hrs Max Marks : 100 Note: Answer any five full questions selecting at least two questions from each part Answer all objective type questions only in first and second writing pages Answer for objective type questions shall not be repeated PART A (a) (i) The radius of curvature of y = e–x at (0, 1) is (a) (b) 1 (d ) (ii) The radius of curvature of the circle of curvature is (a) (b) ρ (c) (c) ρ [Ans c] (d ) ρ2 [Ans b] (iii) The first three non-zero terms in the expansion of ex tan x is (a) x + x2 + x (b) x + x3 + x 5 x3 x + x (d ) x + 6 (iv) The radius of curvaluve r = a sin θ at (r, θ) is (a) (b) (c) (d ) None of these (c) x + x + (b) Find the radius of curvature at x = πa on y = a sec [Ans c] FG x IJ H aK [Ans d ] (04 marks) [VTU, July 2008] (04 marks) Solution Refer Unit I (c) Verify Rolle’s theorem for the function f (x) = (x – a)m (x – b)n in [a, b] where m > and n > [VTU, Jan 2008] (06 marks) Solution Refer Unit I (d ) Expand esin x up to the term containing x4 by Maclaurin’s theorem [VTU, Jan 2008] (06 marks) Solution Refer Unit I 441 442 ENGINEERING MATHEMATICS—II (a) (i) The value of lim x log x is x→0 (a) (b) (c) (d ) ∞ [Ans b] LM FG IJ OP cot (x – a) is N H KQ x log − (ii) The value of xlim →a a −1 (b) a (c) (d) None of these [Ans a] (iii) If f (x, y) has derivatives upto any order with in a neighbourhood of a point (a, b) then f (x, y) can be extended to the (a) Finite series (b) Infinite series (c) Some extension limits (d) None of these [Ans b] (iv) If AC – B < then f has neither a maximum nor a minimum at (a, b) the point (a, b) is called : (a) Saddle point (b) Maximum at (a, b) (c) Minimum at (a, b) (d ) Both (a) and (b) [Ans a] (04 marks) (a) (b) Evaluate lim x →0 FG − cot xIJ Hx K [VTU, Jan 2005] (04 marks) Solution Refer Unit II (c) Expand ex log (1 + y) by Maclaurin’s theorem up to the third degree term [VTU, Jan 2008] (06 marks) Solution Refer Unit II (d ) Determine the maxima/minima of the function sin x + sin y + sin (x + y) [VTU, Jan 2005] (06 makrs) Solution Refer Unit II (a) (i) For (a) (c) zz zz zz b g f b x , y g dx dy f b x , y g dx dy ∞ ∞ x ∞ ∞ x ∞ y 0 f x , y dx dy , the change of order is (ii) The value of the integral (b) z (d ) −2 dx is x2 zz zz ∞ ∞ x ∞ x 0 b g f b x , y g dx dy f x , y dx dy [Ans c] (a) (b) 0.25 (c) (d ) ∞ [Ans d ] (iii) The volume of the tetrahedron bounded by the coordinate planes and the plane x y z + + = is a b c 443 MODEL QUESTION PAPER—II (iv) (a) abc (b) abc (c) abc (d ) 24 abc bg Γ b5g Γ7 [Ans c] is (a) 30 (c) 48 (b) Find the value of x2 zz (b) 42 (d ) 17 b [Ans a] (04 marks) g xy x + y dx dy taken over the region enclosed by the curve y = x [VTU, Jan 2008, July 2008] (04 marks) and y = Solution Refer Unit III x2 y2 z2 + + = a b2 c2 [VTU, Jan 2008] (06 marks) (c) Using the multiple integrals find the volume of the ellipsoid Solution Refer Unit III (d ) With usual notation show that β (m, n) = b g bg Γ b m + ng Γ m Γ n [VTU, Jan, 2008] (06 marks) (a) (i) If all the surfaces are enclosed in a region containing volume V then the following theorem is applicable (a) Stoke’s theorem (b) Green’s theorem (c) Gauss divergence theorem (d ) Only (a) and (b) [Ans c] (ii) The component of ∇φ in the direction of a unit vector a is ∇φ · a and is called (a) The directional derivative of φ in the direction a (b) The magnitude of φ in the direction a (c) The normal of φ in the direction a (d ) None of these [Ans a] (iii) For a vector function F , there exists a scalar potential only when (a) div F = (b) gred (div F ) = (c) cur F = (iv) Which of the following is true: (a) curl FH A ⋅ BIK (d ) F curl F = = curl A + curl B (b) div curl A = ∇ A [Ans c] 444 ENGINEERING MATHEMATICS—II FH A ⋅ BIK (c) div = div A div B (d ) div cur A = (b) Evaluate z [Ans d ] (04 marks) F ⋅ dr where F = x2 i + y2 j + z2 k and c is given by x = cos t, y = sin t, z = t, ≤ t ≤ π Solution Refer Unit IV (c) Evaluate zd [VTU, Jan 2006] (04 marks) i xy − x dx + x y dy where c is the closed formed by y = 0, x = and y = x (a) directly as a line integral (b) by employing Green’s theorem [VTU, Jan 2007] (06 marks) Solution Refer Unit IV (d ) If f and g are continuously differentiable show that ∇f × ∇g is a solenoidal Solution Refer Unit IV PART B (a) (i) The general solution of the differential equation (D4 – 6D3 + 12D2 – 8D) y = is (a) (b) (c) (d) y y y y = = = = c1 + [c2 + c3 x + c4 x2) e2x (c1 + c2 x + c3 x2) e2x c1 + c2 x + c3 x2 + c4 x4 c1 + c2 x + c3 x2 + c4 e2x [Ans a] (ii) The particular integral of (a) x2 + 4x d y dy + = x2 + 2x + is dx dx (b) x3 +4 x3 x3 + 4x + 4x2 (d) 3 (iii) The particular integral of (D2 + a2) y = sin ax is (c) (a) −x cos ax 2a (b) [Ans c] x cos ax 2a −ax ax cos ax (d) cos ax 2 (iv) The solution of the differential equation (D2 – 2D + 5)2 y = is (c) (a) (b) (c) (d) y y y y = = = = e2x {(c1 + c2 x) cos x + (c3 + c4 x) sin x} ex {(c1 + c2 x) cos 2x + (c3 + c4 x) sin 2x} (c1 ex + c2 e2x) cos x + (c3 ex + c4 e2x) sin x ex {4 cos x + c2 cos 2x + c3 sin x + c4 sin 2x} [Ans a] [Ans b] (04 marks) 445 MODEL QUESTION PAPER—II (b) Solve: d 3y − y = (ex + 1)2 dx Solution Refer Unit V (c) Solve : [VTU, July 2007] (04 marks) d2y dy − + y = 3x2 e2x sin 2x dx dx Solution Refer Unit V (d ) Solve by the method of undetermined coefficients y″ – 3y′ + 2y = x2 + ex Solution Refer Unit V [VTU, July 2007] (06 marks) [VTU, Jan 2008] (06 marks) (a) (i) The general solution of (x2 D2 – xD) y = is (a) y = c1 + c2 ex (b) y = c1 + c2 x2 (c) y = c1 + c2 x2 (d ) y = c1 x + c2 x2 [Ans c] (ii) For the variation of parameters the value of W is (a) y1 y2′ − y y ′ (b) y2 y2′ − y1 y2′ (c) y2 y1′ − y2 y2′ (d ) y2 y1′ − y y1 [Ans a] (iii) The DE in which the conditions are specified at a single value of the independent variable say x = x0 is called (a) Initial value problem (b) Boundary value problem (c) Final value (d ) Both (a) and (b) [Ans a] (iv) The DE in which the conditions are specified for a given set of n values of the independent variable is called a (a) Intial value problem (b) Boundary value problem (c) Final value (d ) Both (a) and (b) [Ans b] (04 marks) (b) Solve by the method of variation parameters y″ + y = tan x [VTU, Jan 2008] (04 marks) Solution Refer Unit VI (c) Solve: d2y dy +x + y = x cos (log x) dx dx Solution Refer Unit VI (d) Solve: x2 [VTU, July 2008] (06 marks) d2y dy +4 + y = e–x subject to the conditions is y (0) = y′ (0) dx dx [VTU, Jan 2008] (06 marks) Solution Refer Unit VI 446 ENGINEERING MATHEMATICS—II (a) (i) The Laplace transform of sin2 3t is: (a) (c) s + 36 (b) 18 d i s s + 36 d i s s + 36 (d ) 18 s + 36 [Ans c] (ii) L [e–3t cos 3t] is: (a) s−3 s − s − 18 (b) s+3 s + 65 + 18 (c) s+3 s + 6s + 18 (d ) s−3 s + 6s − 18 2 [Ans b] (iii) L [(t2 + 1) u (t – 1)] d1 + x + s i d1 + s + s i 2 (a) e – s (b) e – s s3 d1 + s + s i s3 (c) 2e s (d ) None of these s3 (iv) The Laplace transform of a function f (t) exists of [Ans a] (a) It is uniformly continuous (b) It is piecewise continuous (c) It is uniformly continuous and of exponential order (d ) It is piecewise continuous of exponential order [Ans d ] (04 marks) (b) Prove that z ∞ · 50 Solution Refer Unit VII (c) A periodic function f (t) of period 2a is defined by e –3t t sin t dt = f (t) = RS a T–a for for [VTU, Jan 2006] (04 marks) 0≤t [...]... 69 74 74 74 77 78 (xii) Worked Out Examples Exercise 2.4 2.3 Maxima and Minima of Functions of Two Variables 2.3.1 Necessary and Sufficient Conditions for Maxima and Minima Worked Out Examples Exercise 2.5 2.4 Lagrange’s Method of Undetermined Multipliers Working Rules Worked Out Examples Exercise 2.6 Additional Problems (from Previous Years VTU Exams.) Objective Questions UNIT III Integral Calculus... then But the radius of a circle is non-negative So to take ρ = i.e., k = dψ ds 1 X Fig 1.1 dψ is negative, i.e., k is negative ds 1 ds = some authors regard k also as non-negative dψ k 2 ENGINEERING MATHEMATICS II dψ indicates the convexity and concavity of the curve in the neighbourhood of ds ds the point Many authors take ρ = and discard negative sign if computed value is negative dψ The sign of... 2 R|FG dx IJ + FG dy IJ U| S|H dt K H dt K V| T W 2 ρ= U| V| W |UV |W RS dx ⋅ d y – dy ⋅ d x UV / FG dx IJ T dt dt dt dt W H dt K 2 ∴ 2 2 3 2 dx d 2 y dy d 2 x ⋅ − ⋅ dt dt 2 dt dt 2 .(6) 4 ENGINEERING MATHEMATICS II 2 2 dx dy , y′ = , x″ = d x , y″ = d y 2 dt dt dt dt 2 where x′ = ρ= o x′ 2 + y′ 2 t 3 2 x ′ y ″ – y′ x ″ This is the cartesian form of the radius of curvature in parametric form WORKED... 3x + 1 y 1 = 2x – 3, y2 = 2 –1) = – 1 –1) = 2 –1) b1 + 1g = 3 2 2 = = 2 2 2 2 4 Find the radius of curvature at (a, 0) on y = x3 (x – a) Solution We have ρ = d 1 + y12 y2 i 3 2 at (a, 0) 6 ENGINEERING MATHEMATICS II y = x3(x – a) = x4 – x3a y 1 = 4x3 – 3ax2 y 2 = 12x2 – 6ax ( y1)(a, 0) = 4a3 – 3a3 = a3 ( y2)(a, 0) = 12a2 – 6a2 = 6a2 Here, and Now RS1 + da i UV T W = 2 3 ∴ ρ(a, 0) 6a 2 o 1 + a6 = 6a... 3 2 at on x3 + y3 = 3axy FG 3a , 3a IJ H2 2K y3 + = 3axy Here, Differentiating with respect to x 3x2 + 3y2 y1 = 3a (xy1 + y) 3 ( y2 – ax) y1 = 3 (ay – x2) ⇒ y1 = ay – x 2 y 2 − ax .(1) 8 ENGINEERING MATHEMATICS II Again differentiating w.r.t x ⇒ y2 = Now, from (1), at dy 2 ib g d − ax ⋅ ay1 − 2 x − ay − x 2 FG 3a , 3a IJ H 2 2K F 3a I F 3a I aG J −G J H 2K H 2K y = FG 3a IJ − a FG 3a IJ H 2K H 2K... at (0, b) is = ∴ ρ(0, b1 + 0g FG – b IJ Ha K 3 2 b) = = – a2 b 2 ∴ Radius of curvature is a2 b Next consider (0, – b), y1 = – b2 0 × =0 2 –b a y2 = – b2 a2 FG – b – 0IJ = a H b K b 2 2 10 ENGINEERING MATHEMATICS II ρ(0, – b) = b1 + 0g FG b IJ Ha K 3 2 = a2 b 2 ∴ Radius of curvature of (0, – b) is a2 b 9 Show that at any point P on the rectangular hyperbola xy = c2, ρ = r3 where r is the 2c 2 distance... y0) to the ellipse x x0 y y0 + 2 = 1 a2 b 2 + b4 x 2 2 a6 y3 2 3 2 4 2 = 2 da y − OP Q y2 =1 b2 i 3 2 i 3 2 + b4 x 2 × 3 a2 y3 b4 a 4b 4 + b4 x 2 a 4b 4 i 3 2 x2 y2 + = 1 is a 2 b2 .(1) 12 ENGINEERING MATHEMATICS II Length of perpendicular from (0, 0) upon this tangent 1 = FG x IJ + FG y IJ Ha K Hb K 2 0 2 2 0 2 a 2b2 = a 4 y02 + b 4 x02 So, the length of perpendicular from the origin upon the tangent... + xK W ba + x g + a2 ∴ L.H.S = R.H.S using (2) and (3) (2) 2 2 = 2 3 2 2 2 R.H.S U| |V || W 4 2 = 3 2 3 3 4 = .(1) 3 2 2 2 R| F ax I U| S GH a + x JK V| + | T x W a2 ba + x g 2 2 .(3) 14 ENGINEERING MATHEMATICS II 12 Find ρ at any point on x = a (θ + sinθ) and y = a (1 – cosθ) Solution Here x = a (θ + sinθ), y = a (1 – cosθ) Differentiating w.r.t θ dx dy = a (1 + cos θ), = a sin θ dθ dθ dy dy a sin... tan θ y 1 = tan θ y2 = = b g d2y d tanθ = dx 2 dx b g d dθ tan θ ⋅ dθ dx 2 = sec θ × = Now, ρ = sec 2 θ a tan θ o1 + y t 3 2 2 1 y2 d1 + tan θi F sec θ I GH tan θ JK 2 = 1 a tan θ 2 3 2 16 ENGINEERING MATHEMATICS II sec 3 θ × a tan θ sec 2 θ ρ = a sec θ tan θ = 14 For the curve x = a (cos θ + θ sin θ), y = a (sin θ – θ cos θ), show that the radius of curvature at ′θ′ varies as θ Solution x = a (cos... 2 3 = × 1 2 1 1 × b2ag d1 + t i 2 3 2 1 t12 b2ag d1 + t i 1 R 1 ST1 + t + 1 +t t UVW b2 a g b2ag × d11 ++ tt i d i 2 3 2 1 2 1 2 3 –2 3 b2 a g –2 3 2 1 2 1 2 1 2 1 · Hence proved 1 · t1 18 ENGINEERING MATHEMATICS II EXERCISE 1.1 1 Find ρ at any point on y = log sin x 2 Find ρ at x = 1 on y = Ans cosec x log x x FG x IJ · H 2K π 3 Find the radius of curvature at x = on y = log tan 4 x2 y2 + 4 Find the ... LMwhere cos nθ = r OP a Q N n n 24 ENGINEERING MATHEMATICS II Find the radii of curvature of the following curves: (ii) r (1 + cos θ) = a (i) r = aeθ cot α (iii) θ = r2 − a2 – cos –1 FG a IJ ·... covered are: Unit I and II Differential Calculus, Unit III and IV—Integral Calculus and Vector Integration, Unit V and VI— Differential Equations and Unit VII and VIII—Laplace Transforms It... (iii) r2 = a2 cos 2θ (v) r2 cos 2θ = a2 (vii) r = a sec 2θ LMAns a OP N 2Q (ii) r = a cos θ LM Ans a OP (iv) = sin nθ MN bn + 1g r PQ LMAns ar OP a (vi) r = b1 − cos θg PQ MN LM Ans na OP (viii)