ADVANCED ENGINEERING MATHEMATICS Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ ADVANCED ENGINEERING MATHEMATICS [For the Students of M.E., B.E. and other Engineering Examinations] H.K. DASS M.Sc. Diploma in Specialist Studies (Maths.) University of Hull (England) Secular India Award - 98 for National Integration and Communal Harmony given by Prime Minister Shri Atal Behari Vajpayee on 12th June 1999. S. CHAND & COMPANY LTD. (AN ISO 9001 : 2008 COMPANY) RAM NAGAR, NEW DELHI-110055 Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ S. CHAND & COMPANY LTD. (An ISO 9001 : 2008 Company) Head Office: 7361, RAM NAGAR, NEW DELHI - 110 055 Phone: 23672080-81-82, 9899107446, 9911310888; Fax: 91-11-23677446 Shop at: schandgroup.com; e-mail: info@schandgroup.com Branches : AHMEDABAD : 1st Floor, Heritage, Near Gujarat Vidhyapeeth, Ashram Road, Ahmedabad - 380 014, Ph: 27541965, 27542369, ahmedabad@schandgroup.com BENGALURU : No. 6, Ahuja Chambers, 1st Cross, Kumara Krupa Road, Bengaluru - 560 001, Ph: 22268048, 22354008, bangalore@schandgroup.com BHOPAL : Bajaj Tower, Plot No. 243, Lala Lajpat Rai Colony, Raisen Road, Bhopal - 462 011, Ph: 4274723. bhopal@schandgroup.com CHANDIGARH : S.C.O. 2419-20, First Floor, Sector - 22-C (Near Aroma Hotel), Chandigarh -160 022, Ph: 2725443, 2725446, chandigarh@schandgroup.com CHENNAI : 152, Anna Salai, Chennai - 600 002, Ph: 28460026, 28460027, chennai@schandgroup.com COIMBATORE : 1790, Trichy Road, LGB Colony, Ramanathapuram, Coimbatore -6410045, Ph: 0422-2323620, 4217136 coimbatore@schandgroup.com (Marketing Office) CUTTACK : 1st Floor, Bhartia Tower, Badambadi, Cuttack - 753 009, Ph: 2332580; 2332581, cuttack@schandgroup.com DEHRADUN : 1st Floor, 20, New Road, Near Dwarka Store, Dehradun - 248 001, Ph: 2711101, 2710861, dehradun@schandgroup.com GUWAHATI : Pan Bazar, Guwahati - 781 001, Ph: 2738811, 2735640 guwahati@schandgroup.com HYDERABAD : Padma Plaza, H.No. 3-4-630, Opp. Ratna College, Narayanaguda, Hyderabad - 500 029, Ph: 24651135, 24744815, hyderabad@schandgroup.com JAIPUR : 1st Floor, Nand Plaza, Hawa Sadak, Ajmer Road, Jaipur - 302 006, Ph: 2219175, 2219176, jaipur@schandgroup.com JALANDHAR : Mai Hiran Gate, Jalandhar - 144 008, Ph: 2401630, 5000630, jalandhar@schandgroup.com JAMMU : 67/B, B-Block, Gandhi Nagar, Jammu - 180 004, (M) 09878651464 (Marketing Office) KOCHI : Kachapilly Square, Mullassery Canal Road, Ernakulam, Kochi - 682 011, Ph: 2378207, cochin@schandgroup.com KOLKATA : 285/J, Bipin Bihari Ganguli Street, Kolkata - 700 012, Ph: 22367459, 22373914, kolkata@schandgroup.com LUCKNOW : Mahabeer Market, 25 Gwynne Road, Aminabad, Lucknow - 226 018, Ph: 2626801, 2284815, lucknow@schandgroup.com MUMBAI : Blackie House, 103/5, Walchand Hirachand Marg, Opp. G.P.O., Mumbai - 400 001, Ph: 22690881, 22610885, mumbai@schandgroup.com NAGPUR : Karnal Bag, Model Mill Chowk, Umrer Road, Nagpur - 440 032, Ph: 2723901, 2777666 nagpur@schandgroup.com PATNA : 104, Citicentre Ashok, Govind Mitra Road, Patna - 800 004, Ph: 2300489, 2302100, patna@schandgroup.com PUNE : 291/1, Ganesh Gayatri Complex, 1st Floor, Somwarpeth, Near Jain Mandir, Pune - 411 011, Ph: 64017298, pune@schandgroup.com (Marketing Office) RAIPUR : Kailash Residency, Plot No. 4B, Bottle House Road, Shankar Nagar, Raipur - 492 007, Ph: 09981200834, raipur@schandgroup.com (Marketing Office) RANCHI : Flat No. 104, Sri Draupadi Smriti Apartments, East of Jaipal Singh Stadium, Neel Ratan Street, Upper Bazar, Ranchi - 834 001, Ph: 2208761, ranchi@schandgroup.com (Marketing Office) SILIGURI : 122, Raja Ram Mohan Roy Road, East Vivekanandapally, P.O., Siliguri-734001, Dist., Jalpaiguri, (W.B.) Ph. 0353-2520750 (Marketing Office) VISAKHAPATNAM: Plot No. 7, 1st Floor, Allipuram Extension, Opp. Radhakrishna Towers, Seethammadhara North Extn., Visakhapatnam - 530 013, (M) 09347580841, visakhapatnam@schandgroup.com (Marketing Office) © 1988, H.K. Dass All rights reserved. No part of this publication may be reproduced or copied in any material form (including photo copying or storing it in any medium in form of graphics, electronic or mechanical means and whether or not transient or incidental to some other use of this publication) without written permission of the copyright owner. Any breach of this will entail legal action and prosecution without further notice. Jurisdiction : All desputes with respect to this publication shall be subject to the jurisdiction of the Courts, tribunals and forums of New Delhi, India only. First Edition 1988 Subsequent Editions and Reprints 1990, 92, 93, 94, 96, 97, 98, 99, 2000 (Twice), 2001 (Twice), 2002, 2003 (Twice), 2004, 2005, 2006, 2007, 2008, 2009, 2010, 2011, 2012 Twentyfirst Revised Edition 2013 ISBN : 81-219-0345-9 Code : 10A 110 PRINTED IN INDIA By Rajendra Ravindra Printers Pvt. Ltd., 7361, Ram Nagar, New Delhi -110 055 and published by S. Chand & Company Ltd., 7361, Ram Nagar, New Delhi -110 055. Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ PREFACE TO THE TWENTYFIRST REVISED EDITION I am happy to be able to bring out this revised edition. Misprints and errors which came to my notice have been corrected. Suggestions and healthy criticism from students and teachers to improve the book shall be personally acknowledged and deeply appreciated to help me to make it an ideal book for all. We are thankful to the Management Team and the Editorial Department of S. Chand & Company Ltd. for all help and support in the publication of this book. D-1/87, Janakpuri New Delhi-110 058 Tel. 28525078, 32985078, 28521776 Mob. 9350055078 hk_dass@yahoo.com H.K. DASS Disclaimer : While the author of this book have made every effort to avoid any mistake or omission and have used their skill, expertise and knowledge to the best of their capacity to provide accurate and updated information. The author and S. Chand do not give any representation or warranty with respect to the accuracy or completeness of the contents of this publication and are selling this publication on the condition and understanding that they shall not be made liable in any manner whatsoever. S.Chand and the author expressly disclaim all and any liability/responsibility to any person, whether a purchaser or reader of this publication or not, in respect of anything and everything forming part of the contents of this publication. S. Chand shall not be responsible for any errors, omissions or damages arising out of the use of the information contained in this publication. Further, the appearance of the personal name, location, place and incidence, if any; in the illustrations used herein is purely coincidental and work of imagination. Thus the same should in no manner be termed as defamatory to any individual. (v) Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ PREFACE TO THE FIRST EDITION It gives me great pleasure to present this textbook of Mathematics to the students pursuing I.E.T.E and various engineering courses. This book has been written according to the new revised syllabus of Mathematics of I.E.T.E. and includes topics from the syllabi of the other engineering courses. There is not a single textbook which entirely covers the syllabus of I.E.T.E. and the students have all along been facing great difficulties. Endeavour has been made to cover the syllabus exhaustively and present the subject matter in a systematic and lucid style. More than 550 solved examples on various topics have been incorporated in the textbook for the better understanding of the students. Most of the examples have been taken from previous question papers of I.E.T.E. which should make the students familiar with the standard and trend of questions set in the examinations. Care has been taken to systematically grade these examples. The author possesses very long and rich experience of teaching Mathematics to the students preparing for I.E.T.E. and other examinations of engineering and has first hand experience of the problems and difficulties that they generally face. This book should satisfy both average and brilliant students. It would help the students to get through their examination and at the same time would arouse greater intellectual curiosity in them. I am really thankful to my Publishers, Padamshree Lala Shyam Lal Gupta, Shri Ravindra Kumar Gupta for showing personal interest and his General Manager, Shri P.S. Bhatti and Km. Shashi Kanta for their co-operations. I am also thankful to the Production Manager, Shri Ravi Gupta for bringing out the book in a short period. Suggestions for the improvement of the book will be gratefully acknowledged. D-1/87, Janakpuri New Delhi-110 058 H.K. DASS (vi) Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ FOREWORD On my recent visit to India, I happened to meet Prof. H.K. Dass, who has written quite a number of successful books on Mathematics for students at various levels. During my meeting, Prof. H.K. Dass presented me with the book entitled ‘‘Advanced Engineering Mathematics” I am delighted to write this Foreword, as I am highly impressed on seeing the wide variety of its contents. The contents includes many key topics, for examples, advanced calculus, vector analysis, tensor analysis, fuzzy sets, various transforms and special functions, probability (curiously some tests of significance are given under that chapter), numerical methods; matrix algebra and transforms. In spite of this breadth , the development of the material is very lucid, simple and in plain English. I know of quite a number of other textbooks on Engineering Mathematics but the material that has been included in this textbook is so comprehensive that the students of all the engineering streams will find this textbook useful. It contains problems, questions and their solutions which are useful both to the teachers and students, and I am not surprised that it has gone through various editions.The style reminds me of the popular books of Schaum’s Series. I believe that this book will be also helpful to non-engineering students as a quick reference guide. This book is a work of dedicated scholarship and vast learning of Mr. Dass, and I have no hesitation in recommending this book to the students for any Engineering degree world-wide. Prof. K.V. Mardia M.Sc. (Bombay), M.Sc.(Pune) Ph.D. (Raj.), Ph.D. (N’cle),D.Sc.(N’cle) Senior Research Professor University of Leeds, LEEDS (England) (vii) Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ CONTENTS Chapter Pages 1. Partial Differentiation 1–90 1.1 Introduction (1); 1.2 Limit (1); 1.3 Working Rule to Find the Limit (1); 1.4 Continuity (3); 1.5 Working Rule for Continuity at a Point (a, b) (4) 1.6 Types of Discontinuity (4); 1.7 Partial Derivatives (6); 1.8 Partial Derivatives of Higher Orders (8); 1.9 Which Variable is to be Treated as Constant (13); 1.10 Homogeneous Function (16); 1.11 Euler’s Theorem on Homogeneous Function (16); 1.12 Total Differential (26); 1.13 Total Differential Co-efficient (26); 1.14 Change of two Independent Variables x and y by any other Variable t. (26); 1.15 Change in the Independent Variables x and y by other two Variables u and v. (27); 1.16 Change in both the Independent and Dependent Variables, (Polar Coordinates) (31); 1.17 Important Deductions (37); 1.18 Typical z z Cases (41); 1.19 Geometrical Interpretation of x and y (44); 1.20 Tangent Plane to a Surface (44); 1.21 Error Determination (46); 1.22 Jacobians (53); 1.23 PRoperties of Jacobians (56); 1.24 Jacobian of Implicit Functions (60); 1.25 Partial Derivatives of Implicit Functions By Jacobian (64) 1.26 Taylor’s series of two Variables (67); 1.27 Maximum Value (74); 1.28 Conditions for Extremum Values (75); 1.29 Working rule to find Extremum Values (76); 1.30 Lagrange Method of Undetermined Multipliers (81). 2 . Multiple Integral 91–137 2.1 Double Integration (91); 2.2 Evaluation Of Double Integral (91); 2.3 Evaluation of double Integrals in Polar Co-ordinates (96); 2.4 Change of order of Integration (99); 2.5 Change of Cariables (103); 2.6 Area in Cartesian Co-ordinates (105); 2.7 Area in polar Co-ordinates (106); 2.8 Volume of solid by rotation of an area (double integral) (109); 2.9 Centre of Gravity (110); 2.10 Centre of Gravity of an arc (112); 1.11 Triple Integration (114); 2.12 Integration by change of Cartesian Coordinates into Spherical Coordinates (117) 2.13 Volume = dx dy dz. (120); 2.14 Volume of Solid bounded by Sphere or by Cylinder (121); 2.15 Volume of Solid bounded by Cylinder or Cone (123); 2.16 Surface Area (128); 2.17 Calculation of Mass (131) 2.18 Centre of Gravity (132); 2.19 Moment of inertia of a Solid (133); 2.20 Centre of Pressure (135). 3. Differential Equations 138 – 222 3.1 Definition (138); 3.2 Order and Degree of a Differential Equation (138); 3.3 Formation of Differential Equations (138); 3.4 Solution of a Differential Equation (140); 3.5 Differential Equations of the First Order and First Degree (140); 3.6 Variables Separable (140); 3.7 Homogeneous Differential Equations (142); 3.8 Equations Reducible to Homogeneous Form (144); 3.9 Linear Differential Equations (147); 3.10 Equations Reducible To The Linear Form (Bernoulli Equation) (150); 3.11 Exact Differential Equation (154); 3.12 Equations Reducible to the Exact Equations (157); 3.13 Equations of First order and Higher Degree (161); 3.14 Orthogonal Trajectories (163); 3.15 Polar Equation of the Family of Curves (165); 3.16 Electrical Circuit Kirchhoff’s Laws (166); 3.17 Vertical Motion (168); 3.18 Linear Differential Equations of Second order with Constant Coefficients (174); 3.19 Complete Solution = Complementary Function + Particular Integral (174); 3.20 Method for finding the Complementary Function (175); 3.21 Rules to find Particular Integral (177); 1 1 1 ax x n [ f ( D)]1 x n . e ax e 3.22 (178); 3.23 (180); f (D) f ( D) f (a) (viii) Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ 3.24 3.25 1 2 f (D ) sin ax 1 sin ax 2 2 f (D ) f (a ) cos ax cos ax f (–a2 ) (181); 1 ax 1 1 .e (x) eax . .(x) (184); 3.26 To find the Value of x n sin ax. f (D) f (D) f (D a) (187); 3.27 General Method of Finding the Particular Integral of any Function f (x) (188); 3.28 Cauchy Euler Homogeneous Linear Equations (189); 3.29 Legendre's Homogeneous Differential Equations (190); 3.30 Method of Variation of Parameters (193); dn y 3.31 Simultaneous Differential Equations (195); 3.32 Equation of the Type f ( x) n dx n d y (202); 3.33 Equation of the Type f ( y) (203); 3.34 EQUATION WHICH DO dxn NOT CONTAIN ‘y’ DIRECTLY (205); 3.35 EQUATION WHICH DO NOT CONTAIN ‘x’ DIRECTLY (207); 3.36 EQUATION WHOSE ONE SOLUTION IS KNOWN (208); 3.37 NORMAL FORM (REMOVAL OF FIRST DERIVATIVE) (213); 3.38 Method of solving linear differential equations by changing the independent variable (216); 3.39 Application of Differential Equations of Second Order (220); 4. Determinants and Matrices 223–371 4.1. Introduction (223); 4.2. Determinant (223); 4.3. Determinant as Eliminant (224); 4.4. Minor (225); 4.5. Cofactor (225); 4.6 Rules of Sarrus (230); 4.7. Properties of Determinants (231); 4.8. Factor Theorem (248); 4.9 Pivotal Condensation Method (250); 4.10 Conjugate Elements (253); 4.11. Special Types of Determinants (254); 4.12 Laplace Method For The Expansion of A Determinant In Terms of First Two Rows (255); 4.13.Application of Determinants (256); 4.14. Solution of Simultaneous Linear Equations By Determinants (Cramer’s Rule) (257); 4.15 Rule for multiplication of two Determinants (262); 4.16. Condition for Consistency of a System of Simultaneous Homogeneous equations (263); 4.17. For A System of Three Simultaneous Linear Equations with Three Unknowns (264); 4.18 Matrices (269); 4.19 Various types of matrices (269); 4.20 Addition of Matrices (272); 4.21 Properties of matrix Addition (274); 4.22 Subtraction of matrices (274); 4.23 Scalar Multiple of a matrix (274); 4.24 Multiplication (275); 4.25 (AB)´ = B´A´ (275); 4.26 Properties of Matrix Multiplication (275); 4.27 Mathematical Induction (282) 4.28. Adjoint of a square matrix (283); 4.29 Property of Adjoint matrix (283); 4.30 Inverse of a matrix (284); 4.31 Elementary Transformations (287) 4.32 Elementry matrices (288); 4.33 Theorem (288); 4.34 To compute the inverse of a matrix from elementary matrices (Gauss Jordan method) (289); 4.35 The Inverse of a Symmetric Matrix (289); 4.36 Rank of a matrix (292); 4.37 Normal Form (Canomical Form) (292); 4.38 Rank of Matri by triangular form (297); 4.39 Solution of simultaneous equations (301); 4.40 Gauss-Jordan Method (302); 4.41 Types of Linear Equations (304); 4.42 Consistency of a system of Linear equations (304); 4.43 Homogeneous equations (309); 4.44 Cramer’s Rule (311); 4.45 Linear Dependence and independence of vectors (313); 4.46 Linearly Dependence and Independence of Vectors by Rank Method (315); 4.47 Another Method (Adjoining Method) to solve Linear Equation (317); 4.48 Partitioning of matrices (320); 4.49 Multiplication by Sub-Matrices (321); 4.50 Inverse by Partitioning (321); 4.51 Eigen Values (325); 4.52 Cayley Hamilton Theorem (329); 4.53 Power of matrix (Cayley Hamilton Theorem) (333); 4.54 Characteristic Vectors or Eigen Vectors (335); 4.55 Properties of Eigen Vectors (336); 4.56 Non Symmetric matrices with nonrepeated eigen values (336); 4.57 Non Symmetric matrices with repeated eigen values (338); 4.58 Symmetric matrices with non-repeated eigen values (340); 4.59 Symmetric matrices with repeated eigen values (342); 4.60 Diagonalisation of a matrix (344); 4.61. Theorem on diagonalisation of a matrix (344); 4.62 Powers of a matrix (by diagonalisation) (348); 4.63 Sylvester’s Theorem (350); (ix) Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ 4.64 Quadratic forms (351); 4.65 Quadratic form expressed in matrices (351); 4.66 Linear transformation of Quadratic form (353); 4.67 Conical Form of the Sum of the Squares form using Linear ransformation (353); 4.68 Canonical Form of Sum of the Square for m using orthogonal Transformation (353); 4.69 Classification of definiteness of a Quadratic form A (354); 4.70 Differentiation and integration of matrices (357); 4.71 Complex Matrices (362); 4.72 Theorem (362); 4.73 Transpose of Conjugate of a Matrix (363); 4.74 Hermitian Matrix (363); 4.75 SkewHermitian Matrix (365); 2.76 Periodic Matrix (367); 2.77 Idempotent Matrix (367); 4.78 Unitary Matrix (368) 4.79 The Modules of each Characteristic Roots of a Unitary Matrix is Unity 370 5. Vectors 372–466 5.1 Vectors (372); 5.2 Addition of Vectors (372); 5.3 Rectangular resolution of a vector (372); 5.4 Unit Vector (372); 5.5 Position vector of a point (373); 5.6 Ratio formula (373); 5.7 Product of two vectors (374); 5.8 Scalar, or dot product (374); 5.9 Useful Results (374); 5.10 Work Done as a scalar product (374); 5.11 Vector Product or cross product (375); 5.12 Vector product expressed as a determinant (375); 5.13 Area of parallelogram (375); 5.14 Moment of a force (376); 5.15 Angular velocity (376); 5.16 Scalar triple product (376); 5.17 Geometrical interpretation (377); 5.18 Coplanarity questions (378); 5.19 Vector product of three vectors (379); 5.20 Scalar product of four vectors (381); 5.21 Vector product of four vectors (381); 5.22 Vector Function (383); 5.23 Differentiation of vectors (383); 5.24 FormulaE of differentiation (383); 5.25 Scalar and Vector point functions (385); 5.26 Gradient of a Scalar Function (386); 5.27 Geometrical meaning of gradient, Normal (386); 5.28 Normal and directional derivative (387); 5.29 Divergence of a vector function (398); 5.30 Physical interpretation of Divergence (398); 5.31 Curl (403); 5.32 Physical meaning of curl (403); 5.33 Line integral (421); 5.34 Surface integral (428); 5.35 Volume integral (430); 5.36 Green’s Theorem (for a plane) (431); 5.37 Area of the plane region by Green’s Theorem (434); 5.38 Stoke’s theorem (Relation between Line Integral and Surface Integral) (436); 5.39 Another method of proving stoke’s theorem (437); 5.40 Gauss’s theorem of divergence (452). 6. Complex Numbers 467–505 6.1 Introduction (467); 6.2 Complex Numbers (467); 6.3 Geometrical Representation of Imaginary Numbers (467); 6.4 Argand Diagram (467); 6.5 Equal Complex Numbers (467); 6.6 Addition of complex numbers (468); 6.7 Addition of Complex Numbers by Geometry (468); 6.8 Subtraction (468); 6.9 Powers of i (468); 6.10 Multiplication (469); 6.11 i (Iota) as an operator (470); 6.12 Conjugate of a complex number (470); 6.13 Division (470); 6.14 Division of Complex numbers by Geometry (471); 6.15 Modulus and argument (474); 6.16 Polar form (479); 6.17 Types of Complex Numbers (479); 6.18 Square root of a complex number (480); 6.19 Exponential and circular functions of complex variables (481); 6.20 De moivre’s theorem (By Exponential Function) (482); 6.21 De moivre’s theorem (by induction) (482); 6.22 Roots of a complex number (486); 6.23 Circular functions of complex Numbers (489); 6.24 Hyperbolic Functions (489); 6.25 Relation between circular and Hyperbolic Functions (490); 6.26 Formulae of hyperbolic functions (490); 6.27 Separation of Real and Imaginary parts of circular functions (493); 6.28 Separation of Real and Imaginary Parts of Hyperbolic Functions (494); 6.29 logarithmic function of a complex variable (498); 6.30 Inverse functions (500); 6.31 Inverse Hyperbolic Functions (500); 6.32 Some other inverse functions (502). 7. Functions of a Complex Variable 506–617 7.1 Introduction (506); 7.2 Complex variable (506); 7.3 Functions of a complex variable (506); 7.4 Neighbourhood of Z0 (506); 7.5Limit of a function of a complex variable (507); 7.6 Continuity (508); Continuity in terms of Real and imaginary parts (508); (x) Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ 7.8 Differentiability (509); 7.9 Analytic function (512); 7.10 The necessary condition for f (z) to be analytic (512); 7.11 Sufficient condition for f (z) to be analytic (513); 7.12 C–R Equations in Polar Form (520); 7.13 Derivative of w or f (z) in polar form (521); 7.14 Orthogonal Curves (522); 7.15 Harmonic function (523); 7.16 Application to flow problems (525); 7.17 Velocity Potential Function (526); 7.18 Method to find the conjugate function (526); 7.19 Milne thomson method (To construct an Analytic function) 533; 7.20 Working Rule: to construct an analytic function by Milne Thomson Method (533); 7.21 Partial differentiation of function of complex variable (539); 7.22Introduction (line integral) (544); 7.23 Important Definitions (547); 7.24 Cauchy’s integral theorem (548); 7.25 Extension of cauchy’s theorem to multiple connected region (550); 7.26 Cauchy integral formula (550); 7.27 Cauchy integral formula for the derivative of an analytic function (551); 7.28 Geometrical representation (558); 7.29 Transformation (558); 7.30 Conformal transformation (559); 7.31 Theorem. If f (z) is analytic, mapping is conformal 560; 7.32 Theorem (561); 7.33 Translation w = z + C, (562); 7.34 Rotation w = zeiq (563); 7.35 magnification (563); 7.36 Magnification and rotation (564); 7.37 Inversion and reflection (566); 7.38 Bilinear transformation (Mobius Transformation) (569); 7.39 Invariant points of bilinear transformation (569); 7.40 Cross-ratio (570); 7.41 Theorem (570); 7.42 Properties of bilinear transformation (570); 7.43 Methods to find bilinear transformation (570); 7.44 Inverse point with respect to a circle (575); 7.45 Transformation: w = z2 (580); 7.46 Transformation: w = zn (581); 7.49 Transformation: (584); 7.50 Zero of analytic Function (585); 7.51Principal Part (585); 7.52 Singular point (585); 7.53 Removable Singularity (586); 7.54Working Rule to find singularity (586); 7.55Theorem (589); 7.56 Definition of the residue at a pole (589); 7.57 Residue at infinity (590); 7.58 Method of finding residues (590); 7.59 Residue by definition (591); 7.60 Formula: Residue (592); 7.61 Formula: Residue of (593); 7.62 Formula: Res. (at z = a) (594); 7.63 Formula: Residue = Coefficient of (594); 7.64 Cauchy’s Residue theorem (596); 7.65 Evaluation of real definite integrals by contour integration (600); 7.66 Integration round unit circle of the type (600); 7.67 Evaluation of where are polynomials in x. (609) 8. Special Functions 618–670 8.1 Special functions (618); 8.2 Power series solution of Differential equations (618); 8.3 Ordinary point (618); 8.4 Solution about singular point (622); 8.5 Frobenius Method (623); 8.6 Bessel’s Equation (632); 8.7 Solution of Bessel’s Equation (632); 8.8 Bessel’s functions, Jn (x) (633); 8.9 Recurrence Formulae (635); 8.10 Equations Reducible to Bessel’s Equation (640); 8.11 Orthogonality of Bessel Functions (641); 8.12 A Generating Function of Jn (x) (642); 8.13 Trigonometric Expansion involving Bessel functions (643); 8.14 Bessel Integral (645); 8.15 Fourier-Bessel Expansion (647); 8.16 Ber and Bei Functions (649); 8.17 Legendres Equation (651); 8.18 Legendre’s polynomial Pn (x) (653); 8.19 Legendre’s function of the second kind (653); 8.20 General solution of Legendre’s Equation (654); 8.21 Rodrigue’s Formula (654); 8.22 Legendre Polynomials (656); 8.23 A generating function of Legendre’s polynomial (657); 8.24 Orthogonlity of Legendre polynomials (659); 8.25 Recurrence Formulae for Pn (x) (662); 8.26 Fourier-Legendre Expansion (666); 8.27 Laguerres Differential Equation (668); 8.28 Strum Liouville Equation (668); 8.29 Orthogonality (669); 8.30 Orthogonality of Eigen Functions (669). 9. Partial Differential Equations 671–734 9.1 Partial Differential Equations (671); 9.2 Order (671); 9.3 Method of forming Partial Differential Equations (671); 9.4 Solution of Equation by Direct Integration (672); 9.5 Lagrange’s Linear equation (674); 9.6 Working Rule (675); 9.7 Method of Multipliers (677); 9.8 Partial Differential Equations non-Linear in p, q (683); 9.9 Charpits Method (xi) Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ (688); 9.10 Linear Homogeneous Partial Diff. Eqn. (691); 9.11 Rules for finding the complementary function (691); 9.12 Rules for finding the particular integral (692); 9.13 Non-Homogeneous Linear Equations (700); 9.14 Monge’s Method (704); 9.15 Introduction (707); 9.16 Method of Separation of Variables (707); 9.17 Equation of vibrating string (710); 9.18 Solution of Wave equation by D’Almbert’s method (718); 9.19 One dimensional Heat flow (720); 9.20 Two dimensional Heat Flow (725); 9.21 Laplace Equation in polar co-ordinates (729); 9.22 Transmission line Equations (732). 10. Statistics 735–762 10.1 Statistics (735); 10.2 Frequency distribution (735); 10.3 Graphical Representation (735); 10.4 Average or Measures of Central Tendency (736); 10.5 Arithmetic Mean (736); 10.6 Median (737); 10.7 Mode (738); 10.8 Geometric Mean (739); 10.9 Harmonic Mean (739); 10.10 Average Deviation or Mean Deviation (740); 10.11 Standard Deviation (740); 10.12 Shortest method for calculating Standard Deviation (740); 10.13 Moments (742); 10.14 Moment generating function (743); 10.15 Skewness (743); 10.16 Correlation (745); 10.17 Scatter diagram or Dot-diagram (746); 10.18 Karl Pearson’s Coefficient of Correlation (746); 10.19 Short cut Method (748); 10.20 Spearman’s Rank Correlation (750); 10.21 Spearman’s Rank Correlation Coefficient (750); 10.22 Regression (752); 10.23 Line of Regression (752); 10.24 Equations to the lines of Regression (753); 10.25 Error of Prediction (759). 11. Probability 763–849 11.1 Probability (763); 11.2 Definitions (763); 11.3 Addition law of Probability (765); 11.4 Multiplication law of Probability (767); 11.4 (b) Baye’s Theorem (779); 11.5 Binomial Distribution (781); 11.6 Mean of Binomial Distribution (787); 11.7 Standard Deviation of Binomial Distribution (787); 11.8 Central Moments (790); 11.9 Moment Generating Functions (791); 11.10 Recurrence Relation for Binomial Distribution (792); 11.11 Poisson Distribution (794); 11.12 Mean of Poisson Distribution (794); 11.13 Standard deviation of Poisson Distribution (795); 11.14 Mean Deviation (796); 11.15 Moment Generating Function (797); 11.16 Cumulants (797); 11.17 Recurrence Formulae (798); 11.18 Continuous Distribution (806); 11.19 Moment Generating Function (808); 11.20 Normal Distribution (809); 11.21 Normal Curve (809); 11.22 Mean for Normal Distribution (810); 11.23 Standard Deviation for Normal Distribution (810); 11.24 Median of the Normal Distribution (811); 11.25 Mean Deviation (811); 11.26 Mode of the Normal Distribution (811); 11.27 Moment of Normal Distribution (812); 11.28 Area under the normal curve (815); 11.29 Other Distributions (823); 11.30 Population (824); 11.31 Sampling (824); 11.32 Parameters and statistics (824); 11.33 Aims of a sample (825); 11.34 Types of sampling (825); 11.35 Sampling Distribution (825); 11.36 Standard error (825); 11.37 Sampling Distribution of Means (825); 11.38 Sampling Distribution of Variance (827); 11.39 Testing a Hypothesis (827); 11.40 Null Hypothesis (827); 11.41 Errors (827); 11.42 Level of significance (827); 11.43 Test of significance (828); 11.44 Confidence limits (828); 11.45 Test of significance of Large samples (828); 11.46 Sampling Distribution of the proportion (829); 11.47 Estimation of the parameters of the population (829); 11.48 Comparison of Large Samples (830); 11.49 The t Distribution (small sample) (831); 11.50 Working Rule (832); 11.51 Testing for Difference between two t samples (836); 11.52 The Chi-square Distribution (839); 11.53 Degree of freedom (839); 11.54 x2 curve (840); 11.55 Goodness of fit (840); 11.56 Steps for testing (840); 11.57 F-Distribution (846); 11.58 Fisher z Distribution (847). (xii) Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ 12. Fourier Series 850–884 12.1 Periodic Functions (850); 12.2 Fourier series (850); 12.3 Dirichlet’s Conditions (851); 12.4 Advantages of Fourier Series (851); 12.5 Useful Integrals (851); 12.6 Determination of Fourier constants (Euler’s Formulae) (851); 12.7 Functions defined in two or more sub spaces (855); 12.8 Even Functions (861); 12.9 Half Range’s series (864); 12.10 Change of Interval (866); 12.11 Parseval’s Formula (874); 12.12 Fourier series in Complex Form (879); 12.13 Practical Harmonic Analysis (880). 13. Laplace Transformation 885–932 13.1 Introduction (885); 13.2 Laplace Transform (885); 13.3 Important Formulae (885); 13.4 Properties of Laplace Transforms (888); 13.5 Laplace Transform of the Derivative of f (t) (889); 13.6 Laplace Transform of Derivative of order n (890); 13.7 Laplace Transform of Integral of f (t) (890); 13.8 Laplace Transform of t · f (t) (Multiplication by t) (891); 13.9 Laplace Transform of 1 f (t) (Division by t) (893); 13.10 Unit step function (895); t 13.11 Second shifting theorem (896); 13.12 Theorem (896); 13.13 Impulse Function (898); 13.14 Periodic Functions (899); 13.15 Convolution Theorem (903); 13.16 Laplace Transform of Bessel function (903); 13.17 Evaluation of Integral (904); 13.18 Formulae of Laplace Transform (905); 13.19 Properties of Laplace transform (906); 13.20 Inverse of Laplace Transforms (906); 13.21 Important Formulae (907); 13.22 Multiplication by s (908); 13.23 Division by s (Multiplication by 1 (909); 13.24 First shifting property (910); s 13.25 Second shifting property (911); 13.26 Inverse Laplace transformation of Derivatives (913); 13.27 Inverse Laplace Transform of Integrals (913); 13.28 Partial Fraction Method (914); 13.29 Inverse Laplace transformation (915); 13.30 Solution of Differential Equations (916); 13.31 Solution of simultaneous equations (924); 13.32 Inversion Formula for the Laplace transform (927). 14. Integral Transforms 933–981 14.1 Introduction (933); 14.2 Integral Transforms (933); 14.3 Fourier Integral Theorem (934); 14.4 Fourier sine and cosine Integrals (935); 14.5 Fourier’s Complex Integral (936); 14.6 Fourier Transforms (938); 14.7 Fourier sine and cosine Transforms (939); 14.8 Properties of Fourier Transform (947); 14.9 Convolution (951); 14.10 Parseval’s Identity for Fourier Transform (951); 14.11 Parseval’s identity cosine Transform (952); 14.12 Parseval’s identity for sine Transform (952); 14.13 Fourier Transforms of Derivative of a function (958); 14.14 Relationship Between Fourier and Laplace Transforms (959); 14.15 Solution of Boundary value problems by using integral transform (959); 17.16 Fourier Transforms of Partial Derivative of a Function (965); 14.17 Finite Fourier Transforms (969); 14.18 Finite Fourier sine and Cosine transforms of Derivatives (976). 15. Numerical Techniques 982–1025 15.1 Introduction (982); 15.2 Solution of the equations graphically (982); 15.3 NewtonRaphson Method or Successive substitution method (984); 15.4 Rule of False position (Regula False) (989); 15.5 Iteration Method (993); 15.6 Solution of Linear systems (994); 15.7 Crout’s Method (996); 15.8 Iterative Methods or Indirect Methods (1000); 15.9 Jacobi’s Method (1000); 15.10 Gauss-Seidel Method (1002); 15.11 Solution of Ordinary Differential Equations (1006); 15.12 Taylor’s Series Method (1006); 15.13 Picard’s (xiii) Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ method of successive approximations (1010); 15.14 Euler’s method (1014); 15.15 Euler’s Modified formula (1016); 15.16 Runge’s Formula (1017); 15.17 Runge’s Formula (Third order) (1018); 15.18 Runge’s Kutta Formula (Fourth order) (1019); 15.19 Higher order Differential Equations (1023). 16. Numerical Method for Solution of Partial Differential Equation 1026–1041 16.1 General Linear partial differential equations (1026); 16.2 Finite-Difference Approximation to Derivatives (1026); 16.3 Solution of Partial Differential equation (Laplace’s method) (1027); 16.4 Jacobi’s Iteration Formula (1029); 16.5 Gauss-Seidel method (1029); 16.6 Successive over-Relaxation or S.O.R. Method (1029); 16.7 Poisson Equation (1034); 16.8 Heat equation (Parabolic Equations) (1036); 16.9 Wave equation (Hyperbolic Equation) (1039). 17. Calculus of Variations 1042–1054 17.1 Introduction (1042); 17.2 Functionals (1042); 17.3 Definition (1042); 17.4 Euler’s Equation (1043); 17.5 Extremal (1045); 17.6 Isoperimetric Problems (1049); 17.7 Functionals of second order derivatives (1053). 18. Tensor Analysis 1055–1084 18.1 Introduction (1055); 18.2 Co-ordinate Transformation (1055); 18.3 Summation Convention (1056); 18.4 Summation of co-ordinates (1056); 18.5 Relation between the direction cosines of three mutually perpendicular straight lines (1057); 18.6 Transformation of velocity components on change from one system of rectangular axes to another (1057); 18.7 Rank of a tensor (1058); 18.8 First order tensors (1058); 18.9 Second order tensors (1058); 18.10 Tensors of any order (1059); 18.11 Tensor of zero order (1059); 18.12 Algebraic operations on tensors (1059); 18.13 Product of two tensors (1059); 18.14 Quotient law of tensors (1060); 18.15 Contraction theorem (1060); 18.16 Symmetric and antisymmetric tensors (1061); 18.17 Symmetric and skew symmetric tensors (1061); 18.18 Theorem (1062); 18.19 A fundamental property of tensors (1062); 18.20 Zero tensor (1062); 18.21 Two special tensors (1063); 18.22 Kronecker tensor (1063); 18.23 Isotropic Tensor (1064); 18.24 Relation between alternate and kronecker tensor (1064); 18.25 Matrices and tensors of first and second order (1065); 18.26 Scalar and vector products of two vectors (1065); 18.27 The three scalar Invariants of a second order tensor (1065); 18.28 Singular and non-singular tensors of second order (1066); 18.29 Reciprocal of a Non-singular tensor (1066); 18.30 Eigen values and Eigen Vectors of a tensor of second order (1067); 18.31 Theorem (1067); 18.32 Reality of the eigen values (1068); 18.33 Association of a skew symmetric tensors of order two and vectors (1068); 18.34 Tensor fields (1069); 18.35 Gradient of tensor fields:gradient of a scalar function (1069); 18.36 Gradient of vector (1069); 18.37 Divergence of vector point function (1069); 18.38 U curl of a vector point fuion (1069); 18.39 Second order differential operators (1071); 18.40 Tensorial form of Gauss’s and Stoke’s theorem (1071); 18.41 Stoke’s theorem (1071); 18.42 Relation between alternate and kronecker tensor (1072); 18.43 The three scalar invariants of a second order tensor (1073); 18.44 Tensor analysis (1073); 18.45 Conjugate or reciprocal tensors (1078); 18.46 Christoffel symbols (1078); 18.47 Transformation law for second kind (1079); 18.48 Contravariant, covariant and mixed tensor (1082). 19. Z-transforms 1085–1118 19.1 Introduction (1085); 19.2 Sequence (1085); 19.3 Representation of Sequence (1085); 19.4 Basic Operations on Sequences (1086); 19.5 Z-Transforms (1086); 19.6 Properties (xiv) Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ of Z-Transforms (1087); 19.7 Theorem (1094); 19.8 Change of Scale (1095); 19.9 Shifting Property (1096); 19.10 Inverse Z-Transform (1096); 19.11 Solution of Difference Equations (1096); 19.12 Multiplication by K (1097); 19.13 Division by K (1097); 19.14 Initial Value (1098); 19.15 Final Value (1098); 19.16 Partial Sum (1098); 19.17 Convolution (1099); 19.18 Convolution Property of Casual Sequence (1099); 19.19 Transform of Important Sequences (1100); 19.20 Inverse of Z-Transform by division (1102); 19.21 By Binomial Expansion and Partial Fraction (1104); 19.22 Partial Fractions (1105); 19.23 Inversion by Residue Method (1111); 19.24 Solution of Difference Equations (1114). 20. Infinite Series 1119–1157 20.1 Sequence (1119); 20.2 Limit (1119); 20.3 Convergent Sequence (1119); 20.4 Bounded Sequence (1119); 20.5 Monotonic Sequence (1119); 20.6 Remember the following limits (1120); 20.7 Series (1120); 20.8 Convergent, Divergent and Oscillatory Series (1120); 20.9 Properties of Infinite Series (1120); 20.10 Properties of Geometric Series (1121); 20.11 Positive Term Series (1122); 20.12 Necessary Conditions for Convergent Series (1123); 20.13 Cauchy’s Fundamental Test for Divergence (1123); 20.14 p-Series (1124); 20.15 Comparison Test (1125); 20.16 D’Alembert’s Ratio Test (1129); 20.17 Raabe’s Test (1132); 20.18 Gauss’s Test (1139); 20.19 Cauchy’s Integral Test (1140); 20.20 Cauchy’s Root Test (1142); 20.21 Logarithmic Test (1144); 20.22 DeMorgan’s and Bertrand’s Test (1147); 20.23 Cauchy’s Condensation Test (1148); 20.24 Alternating Series (1148); 20.25 Leibnitz’s Rule for Convergence of an Alternating Series (1148); 20.26 Alternating Convergent Series (1149); 20.27 Power Series in X (1151); 20.28 Exponential Series (1152); 20.29 Logarithmic Series (1152); 20.30 Binomial Series (1152); 20.31 Uniform Convergence (1153); 20.32 Abel’s Test (1153); 20.33 Brief Procedure for Testing a Series for Convergence (1154); 20.34 List of the Tests for Convergence (1155). 21. Gamma, Beta Functions, Differentiation under the Integral Sign 1158–1188 21.1 Gamma Function (1158); 21.2 Transformation of Gamma Function (1160); 21.3 Beta Function (1161); 21.4 Evaluation of Beta Function (1162); 21.5 A property of Beta Function (1162); 21.6 Transformation of Beta Function (1163); 21.7 Relation between Beta and Gamma Functions (1164); 21.8 Liouville’s Extension of Dirichlet Theorem (1174); 21.9 Elliptic Integrals (1176); 21.10 Definition and property (1176); 21.11 Error Function (1179); 21.12 Differentiation under the integral sign (1181); 21.13 Leibnitz’s Rule (1181); 21.14 Rule of differentiation under the integral sign when the limits are functions of parameter (1185). 22. Chebyshev Polynomials 1189 — 1202 22.1 Introduction (1189); 22.2 Chebyshev Polynomials (Tchebcheff Or Tschebyscheff Polynomials) (1189); 22.3 Orthogonal Properties of Chebyshev Polynomials. (1190); 22.4 Recurrence Relation of Chebyshev Polynomials (1191); 22.5 Powers of X in Terms of T2 (X) (1192); 22.6 Recurrence formulae for Un (x) (1194); 22.7 Generating Function for Tn (x) (1199). 23. Fuzzy sets 1203 – 1207 23.1 Introduction (1203); 23.2 Fuzzy set (1203); 23.3 Equality of two fuzzy sets (1204); 23.4 Complement of a fuzzy set (1204); 23.5 Union of two fuzzy sets (1204); 23.6 Intersection of two fuzzy sets (1204); 23.7 Truth value (1205); 23.8 Application (1206). 24. Hankel Transform 1208 – 1229 24.1 Hankel Transform (1208); 24.2. The Formulae used in Finding the Hankel Transforms (1208); 24.3 Some More Integrals Involving Exponential Functions and Bessel’s Function (1209 ); 24.4 Inversion formula for Hankel Transform (1215 ); (xv) Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ 24.5 Parseval’s Theorem for Hankel Transform (1215); 24.6. Hankel Transformation of the Derivative of a function (1215); 24.7. Finite Hankel Transmission formation (1221); 24.8 Another form of Hankel transform (1222). 25. Hilbert Tranform 1230 – 1231 25.1 Introduction (1230); 25.2. Elementary Function and their Hilbert Transform (1230); 25.3. Properties (1231) 25.4. Applications (1231). 26. Empirical Laws and Curve Fitting (Method of Least Squares) 1232 – 1242 26.1 Empirical Law (1232); 26.2. Curve Fitting (1232); 26.3. Graphical Method (1232); 26.4 Determination of other Empirical Laws Reducible to Linear form (1232); 26.5 Principle of Least Squares (1234); 26.6 Method of Least Squares (1235); 26.7 Change of Scale (1238). 27. Linear Programming 1243 – 1303 27.1 Introduction (1243); 27.2 Some definitions (1244); 27.3 Graphical method (1251); 27.4 Corner point Method (1251); 27.5 Iso-profit or Iso-cost method (Maximum Z) (1256); 27.6 Iso-profit or iso-cost method (Minimum Z) (1256); 27.7 Solution of linear programming problems (1263); 27.8 Simplex method (1277); 27.9 Degeneracy (1287); 27.10 Duality (1292); 27.11 Dual of L.P.P. (1292); 27.12 North West Corner Method, (1297); 12.13 Vogel’s approximation method (VAM) (1299). Useful Formulae 1305–1312 Solved Question Paper 2007 1313–1335 Question Papers, 2006, 2005 and 2004 1336–1353 Index 1355–1358 (xvi) Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ 1 Partial Differentiation 1.1 INTRODUCTION Area of a rectangle depends upon its length and breadth, hence we can say that area is the function of two variables i.e. its length and breadth. z is called a function of two variables x and y if z has one definite value for every pair of values of x and y. Symbolically, it is written as z = f (x, y) The variable x and y are called independent variables while z is called the dependent variable. Similarly, we can define z as a function of more than two variables. Geometrically: Let z = f (x, y) where x, y belong to an area A of the xy-plane. For each point (x, y) corresponds a value of z. These values of (x, y, z) form a surface in space. Hence, the function z = f (x, y) represents a surface. 1.2 LIMIT The function f (x, y) is said to tend to the limit l as x a and y b if and only if the limit l is independent of the path followed by the point (x, y) as x a and y b. Then lim f ( x, y ) = l x a y b The function f (x, y) in region R is said to tend to the limit l as x a and y b if and only if corresponding to a positive number (a, b), there exists another positive number such that f (x, y) – l < for 0 < (x – a)2 + (y – b)2 < 2 for every point (x, y) in R. 1.3 WORKING RULE TO FIND THE LIMIT Step 1. Find the value of f (x, y) along x a and y b. Step 2. Find the value of f (x, y) along y b and x a. If the values of f (x, y) in step 1 and step 2 remain the same, the limit exists otherwise not. Step 3. If a 0 and b 0, find the limit along y = mx or y = mxn. If the value of the limit does not contain m then limit exists. If it contains m, the limit does not exist. Note. (i) Put x = 0 and then y = 0 in f. Find its value f1. (ii) Put y = 0 and then x = 0 in f. Find the value f2. If f1 f2, limit does not exist. If f1 = f2, then 1 Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ Partial Differentiation 2 (iii) Put y = mx and find the limit f3. If f1 = f2 f3, then limit does not exist. If f1 = f2 = f3, then 2 (iv) Put y = mx and find the limit f4. If f1 = f2 = f3 f4, then limit does not exist. If f1 = f2 = f3 = f4, then limit exists. 2 Example 1. Evaluate lim x0 y0 Solution. (i) lim x y (ii) lim x y x y 4 x y2 x2 y 4 x y 2 x2 y 4 x y 2 0 lim = 0 = f1 (say) 0 y2 y 0 0 lim = 0 = f2 (say) 4 x 0 x 0 Here, f1 = f2, therefore (iii) Put y = mx lim x y x 2 mx mx = 0 = f3 lim 2 4 2 2 x 0 x m 2 x m x (say) Here, f1 = f2 = f3, therefore (vi) Put y = mx2 lim x y x 2 mx 4 2 4 x m x m 1 m2 = f4 Here, f1 = f2 = f3 f4 Thus, limit does not exist. 3 Ans. 3 Example 2. Evaluate lim ( x y ). x 0 y0 Solution. (i) lim ( x 3 y 3 ) lim (0 y 3 ) = 0 = f1 (say) (ii) lim ( x 3 y 3 ) lim ( x 3 0) = 0 = f2 (say) x y y0 x y x 0 Here, f1 = f2, therefore (iii) Put y = mx lim ( x 3 y 3 ) lim lim ( x 3 y 3 ) lim ( x 3 m 3 x 3 ) = 0 = f3 (say) x 0 x 0 y mx x y Here, f1 = f2 = f3, therefore (iv) Put y = mx2 lim ( x3 y 3 ) lim lim ( x 3 y 3 ) lim ( x3 m3 x 6 ) x x 0 x 0 y mx2 y = lim x 3 (1 m 3 x 3 ) = 0 = f4 x0 Here, f1 = f2 = f3 = f4 Thus, limit exists with value 0. Example 3. Evaluate lim x 1 y 2 3x 2 y x2 y 2 5 (say) Ans. . Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ Partial Differentiation lim Solution. x 1 y 2 3 3x 2 y 3x 2 y 3 x 2 (2) lim lim 2 lim 2 2 x y 5 x 1 y 2 x y 5 x1 x (2)2 5 2 2 = lim x 1 Example 4. Evaluate lim x y 3 Solution. 2x 3 3 x 4 y3 6x2 2 x 9 6 3 19 5 Ans. . 2x 3 lim lim 3 y 3 x x 4y x 4 y 3 2 3 3 2 00 x = lim lim x lim 0 f1 3 y 3 x y y 3 1 4(0) 1 4 x 2x 3 2x 3 lim lim 3 (ii) lim 3 y 3 x 4 y 3 x y 3 x 4 y 3 x 2 3 3 2 2x 3 00 x x lim 0 f2 = lim 3 108 1 0 x x 108 x 1 3 x Here, f1 = f2. (i) lim x y 3 2x 3 3 3 Hence, the limit exists with value 0. (say) (say) Ans. EXERCISE 1.1 Evaluate the following limits: 1. 3. lim x 1 y2 lim x y3 5. lim x 0 y0 6. lim x 1 y 1 8. lim x 0 y 0 10. lim x 1 y 1 1.4 2 x2 y 2 2 xy 2 xy 3 3 x 4y xy x2 y 2 3 3 4 2. lim Ans. 0 4. lim Ans. , x2 y3 x2 y2 x 0, y 0 3x ( y 2) 2 y ( x 2) x 0 y 0 x2 y xy y x2 , Ans. 17 ; x 0, y 0 Ans. Limit does not exist Ans. Limit does not exist ; x 0, y 0 xy 2 x xy 2 y x 2 y 3 x3 y 2 Ans. 1 7. lim Ans. 0 9. lim x 0 y 0 x 0 y0 Ans. x3 2 y3 x 0, y 0 Ans. 0 , x 0, y 0 Ans. 0 x2 4 y2 xy 2 x2 y 2 1 2 CONTINUITY A function f (x, y) is said to be continuous at a point (a, b) if lim ( x, y ) ( a , b ) f ( x, y ) = f (a, b) A function is said to be continuous in a domain if it is continuous at every point of the domain. Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ Partial Differentiation 4 1.5 WORKING RULE FOR CONTINUITY AT A POINT (a, b) Step 1. f (a, b) should be well defined Step 2. lim f ( x, y ) should exist. ( x, y ) ( a , b ) Step 3. f ( x, y ) = f (a, b) lim ( x, y ) ( a , b ) x3 y 3 when x 0, 2 2 Example 5. Test the function f (x, y) = x y when x 0, 0 for continuity. Solution. Step 1. The function is well defined at (0, 0). y0 y0 x3 y 3 lim lim lim f ( x, y ) = ( x, y ) (0, 0) x 2 y 2 x 0 y mx x 2 y 2 Step 2. ( x , y ) (0, 0) 3 3 3 3 x m x x (1 m ) lim = xlim 0 x 2 m2 x 2 = x 0 1 m2 = 0 Thus, limit exists at (0, 0). Step 3. limit of f (x) at origin = value of the function at origin. lim lim x3 y 3 ( x , y ) (0, 0) x2 y 2 x3 y3 = f (0, 0) = 0 Hence, the function f is continuous at the origin. Ans. x , x 0, y 0 2 Example 6. Discuss the continuity of f (x, y) = x y 2 x 0, y 0 2, at the origin. x , x 0, y 0 2 2 Solution. Here, we f (x, y) = x y x 0, y 0 2, Step 1. The function f (x, y) at (0, 0) is well defined. 1 x x x lim lim lim lim Step 2. ( x, y )lim = = x0 (0, 0) x 0 y mx 1 m2 x2 y 2 x 2 y 2 x 0 x 2 m 2 x 2 For different values of m the limit f is not unique. so the ( x, y )lim (0, 0) x 2 x y2 does not exist. Hence f (x, y) is not continuous at origin. 1.6 TYPES OF DISCONTINUITY (Gujarat Univ. I sem. Jan. 2009) Ans. Y 1. First Kind. f (x) is said to have discontinuity of first kind at 0) f (x 1– + x1 f( the point x = x1 if Right limit f (x1 + 0) and left limit f (x1 – 0) exist but are not equal. O 0) x1 (First kind) X Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ Partial Differentiation 5 Y 2. Second Kind. f (x) is said to have discontinuity of the second kind at x = x1 if neither right limit f (x1 + 0) exists nor left limit f (x1 – 0) exists. f (x 1– +0 x1 f( O 0) ) x1 (Second kind) X 3. Third Kind (Mixed discontinuity). f (x) is said to have mixed discontinuity at the point x = x1 if only one of the two limits right limit f (x1 + 0) and left limit f (x1 – 0) exists and not the other. Y Y + x1 f( 0) 0) f (x 1– OR X x1 Third kind (i) O f( + x1 0) f (x 1– 0) X x1 Third kind (ii) O 4. Fourth Kind (Infinite discontinuity). f (x) is said to have infinite discontinuity at the point x = x1 if either one or both limits right limit and left limit f (x1 – 0) is infinite. If both limits do not exist and if f (x1 ± h) oscillates between limits one of which is infinite as ± h 0. It is also a point of infinite discontinuity. X x (Fourth kind) ) +0 1 f (x f (x1 + 0) O OR X O 5. Fifth Kind (Removable discontinuity). If right limit f (x1 + 0) is equal to left limit f (x1 – 0) is not equal to f (x1), then f (x) is said to have removable discontinuity. EXERCISE 1.2 ) –0 OR x1 f( 1 x1 Y ) –0 f (x O ) f (x 1– 0 Y f (x1 + 0) Y X x Y + x1 f( O 0) ) –0 f (x 1 x1 (Fifth kind) X Test for continuity: xy ( x 2 y 2 ) , when x 0, y 0 1. f (x, y) = x 2 y 2 when x 0, y 0 0, at origin. x2 y2 , when x 0, y 0 2. f (x, y) = x 2 y 2 when x 0, y 0 0, at origin. x3 y3 , when x 0, y 0 3. f (x, y) = x3 y 3 at origin. when x 0, y 0 0, Ans. Continuous at origin. Ans. Not continuous at origin. Ans. Not continuous at origin. Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ Partial Differentiation 6 4. 5. 6. 7. 8. 1.7 xy , when x 0, y 0 2 2 f (x, y) = x y 0, when x 0, y 0 at origin. Ans. Not continuous at origin. x3 y 3 , when x 0, y 0 f (x, y) = when x 0, y 0 0, at origin. Ans. Continuous at origin. x2 2 y , f (x, y) = x y 2 when x 1, y 2 1 at the point (1, 2). Ans. Continuous at (1, 2). 2 x 2 y , ( x, y ) (1, 2) Show that the function f (x, y) = ( x, y ) (1, 2) 0, is discontinuous at (1, 2). 1 ( x y ) sin , x y 0 Show that the function f (x, y) = x y x y0 0, is continuous at (0, 0) but its partial derivatives of first order do not exist at (0, 0). (A.M.I.E.T.E., Dec. 2007) PARTIAL DERIVATIVES Let z = f (x, y) be function of two independent variables x and y. If we keep y constant and x varies then z becomes a function of x only. The derivative of z with respect to x, keeping y as constant is called partial derivative of ‘z’, w.r.t. ‘x’ and is denoted by symbols. z f , , f (x, y) etc. x x x f ( x x, y ) f ( x, y ) z = lim x 0 x x The process of finding the partial differential coefficient of z w.r.t. ‘x’ is that of ordinary differentiation, but with the only difference that we treat y as constant. Similarly, the partial derivative of ‘z’ w.r.t. ‘y’ keeping x as constant is denoted by Then z f , , fy (x, y) etc. y y z = y Notation. z = p, x lim y 0 f ( x , y y ) f ( x, y ) y z = q, y 2z x 2 = r, 2 z = s, x y 2 z y 2 =t u u 1 x 1 y Example 7. If u = sin tan , then find the value of x y . y x x y 1 x 1 y Solution. u = sin tan y x u 1 1 1 1 y y . . 2 2 = 2 2 y 2 2 x x y2 y x y x x 1 1 x y Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ Partial Differentiation x u = x u = y 7 x 2 y x 2 xy ...(1) 2 x y2 1 1 1 x x x 2 2 2 2 2 x x y2 y y x x y 1 y 1 x y u x xy y. 2 = 2 2 y x y2 y x 2 On adding (1) and (2), we have x. u u y. =0 x y ...(2) Ans. u u and if u = er cos . cos (r sin ) r u = er cos . cos (r sin ) u = er cos . [– sin (r sin ).sin ] + [cos .er cos ].cos (r sin ) r (keeping as constant) = er cos .[– sin (r sin ).sin + cos (r sin ).cos ] = er cos .cos (r sin + ) Ans. u = er cos .[– sin (r sin ).r cos ] + [–r sin .er cos ].cos (r sin ) (keeping r as constant) = – r er cos . [sin (r sin ).cos + sin cos (r sin )] = – r er cos . sin (r sin + ) Ans. Example 8. Find Solution. Example 9. If u = (1 – 2xy + y2)–1/2 prove that, x u u y = y2 u3. x y Solution. u = (1 – 2xy + y2)–1/2 Differentiating (1) partially w.r.t. ‘x’, we get ...(1) u 1 2 3/2 ( 2 y ) = (1 2 xy y ) x 2 u x = xy (1 – 2xy + y2)–3/2 x Differentiating (1) partially w.r.t. ‘y’, we get u 1 2 3/2 ( 2 x 2 y ) = (1 2 xy y ) y 2 u y = (xy – y2) (1 – 2xy + y2)–3/2 y Subtracting (3) from (2), we get u u x y = xy (1 – 2xy + y2)–3/2 – (xy – y2) (1 – 2xy + y2)–3/2 x y = y2 (1 – 2xy + y2)–3/2 = y2u3. Example 10. If z = eax + by.f (ax – by), prove that b ...(2) ...(3) Proved. z z a = 2abz. x y (A.M.I.E.T.E., Summer 2004) Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ Partial Differentiation 8 ax + by Solution. z = e .f (ax – by) Differentiating (1) w.r.t. ‘x’, we get z = a eax + by.f (ax – by) + eax + by.a f (ax – by) x z b = a b eax + by.f (ax – by) + a b eax + by.f (ax – by) x Differentiating (1) w.r.t. ‘y’, we get z = b eax + by.f (ax – by) + eax + by.(– b) f (ax – by) y z = a b eax + by.f (ax – by) – a b eax + by.f (ax – by) y On adding (2) and (3), we get z z b a = 2ab eax + by f (ax – by) x y a 1.8 b z z a = 2a b z x y ...(1) ...(2) ...(3) Proved. PARTIAL DERIVATIVES OF HIGHER ORDERS z z and being the functions of x and y can further be differentiated y x partially with respect to x and y. Symbolically Let z = f (x, y) then z 2z = x x x 2 z 2 z = y x y x z 2 z = x y x y or or or 2 f x 2 2 f y x 2 f x y or fxx or fyx or fxy 2 z 2 z = yx xy Example 11. Prove that y = f (x + at) + g(x – at) satisfies Note. 2 2 y a = x 2 t 2 where f and g are assumed to be at least twice differentiable and a is any constant. (U.P., I Sem; Jan 2011, A.M.I.E., Summer 2000) Solution. y = f (x + at) + g(x – at) ...(1) Differentiating (1) w.r.t. ‘x’ partially, we get y = f (x + at) + g(x – at) x 2 y = f (x + at) + g(x – at) x 2 Differentiating (1) w.r.t. ‘t’ partially, we get y = f (x + at).a + g(x – at) (– a) t 2 y Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ Partial Differentiation 9 2 y t 2 = a2f (x + at) + g(x – at) a2 2 = a2 [f (x + at) + g(x – at)] = a 2 y Hence t 2 2 = a 2 y x 2 2 y Proved. dx 2 2 z x2 y2 2 1 y 2 1 x 2 . Example 12. If z = x tan y tan , prove that y x x y 2 x y y x Solution. z = x 2 tan 1 y 2 tan 1 (U.P., I Semester Comp. 2002) x y z 1 y 1 1 y 2 y2 = 2 x tan 1 x 2 2 x x y x x2 y 1 2 1 2 x y x2 y y3 1 y = 2 x tan 2 x x y 2 x2 y 2 ( x2 y 2 ) y 1 y = 2 x tan y 2 = 2 x tan 1 y 2 x x y x 2 z = 2 x. y x x2 x2 y2 1 . 1 2 1 y2 x x2 y 2 x2 y 2 1 2 x 3u xyz Example 13. If u = e , find the value of . x y z Solution. u = exyz 1 Proved. (A.M.I.E. Winter 2000) u = exyz (x y) z 2u = exyz (x) + exyz (x z) (x y) = exyz (x + x2y z) yz 3u = exyz (1 + 2x y z) + exyz (y z).(x + x2y z) xyz = exyz [1 + 2 x y z + x y z + x2y2z2] = exyz [1 + 3 x y z + x2y2z2] Ans. m Example 14. If v = ( x 2 y 2 z 2 ) 2 , then find the value of m (m 0) which will make 2v x 2 2v y 2 2v z 2 = 0. m Solution. We have, 2 2 2 v = (x y z ) 2 m m 1 1 v m 2 ( x y 2 z 2 ) 2 (2 x) = mx ( x2 y 2 z 2 ) 2 = x 2 2v x 2 m m 2 1 m 2 2 2 2 2 2 = m 1 x( x y z ) 2 (2 x ) m( x y z ) 2 2 Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ Partial Differentiation 10 m 2 2 2 2 = m(m 2) x ( x y z ) 2 m = m( x 2 y 2 z 2 ) 2 m 2v Similarly, y 2 2 2 = m( x y z ) 2 2 m 2v 2 2 2 = m( x y z ) 2 2 2 2 2 z On adding (1), (2) and (3), we get 2v x 2 2v y 2 2v z 2 m 2 2 2 = m( x y z ) 2 2 m 0 = m (x2 y 2 z2 ) 2 2 ...(1) .[(m – 2) y2 + x2 + y2 + z2] ...(2) .[(m – 2)z2 + x2 + y2 + z2] ...(3) [(m – 2) (x2 + y2 + z2) + 3(x2 + y2 + z2)] 1 1 m = 0, – 1 1 2 Example 15. If u = (1 2 xy y ) 2 , prove that 2 Solution. We have, u = (1 2 xy y ) u 1 = (1 2 xy x 2 (1 x 2 ) 2v 2v 2v 2 2 2 0 y z x [m – 2 + 3] m 0 = m (m + 1) m = –1 1 [(m 2) x x2 y 2 z 2 ] 0 = m(m + 1) ( x 2 y 2 z 2 ) 2 Hence, m m( x y 2 z 2 ) 2 (m 0) Ans. x 2 u 2 u y = 0. (1 x ) x y y 1 2 ...(1) 3 2 2 y ) y ( 2 y ) 3 (1 2 xy y 2 ) 2 (1 x 2 ) y u = x (1 2 xy 3 2 2 y ) (1 2 xy 3 2 2 y ) 2 u (1 x ) = x x 1 3 2 2 (2 xy ) (1 x ) y (1 2 xy y ) (2 y ) 2 (1 2 xy y 2 )3 2 1 Cancelling (1 2 xy y 2 ) 2 from numerator and denominator, we have = (1 2 xy y 2 ) (2 xy ) 3(1 x 2 ) y 2 5 = 2 xy 4 x 2 y 2 2 xy 3 3 y 2 3 x 2 y 2 5 (1 2 xy y 2 ) 2 = (1 2 xy y 2 ) 2 x 2 y 2 2 xy 3 2 xy 3 y 2 ...(2) 5 2 2 y ) (1 2 xy Differentiating (1) partially w.r.t. y, we get 3 1 u 2 = (1 2 xy y ) 2 (2 x 2 y ) 2 y xy (1 2 xy 3 2 2 y ) Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ Partial Differentiation y2 u = y 11 xy 2 y 3 3 (1 2 xy y 2 ) 2 3 1 3 (1 2 xy y 2 ) 2 (2 xy 3 y 2 ) ( xy 2 y 3 ) (1 2 xy y 2 ) 2 (2 x 2 y ) 2 2 u y = 2 3 y y (1 2 xy y ) 1 Dividing numerator and denominator by (1 2 xy y 2 ) 2 , we get 2 u (1 2 xy y 2 )(2 xy 3 y 2 ) ( xy 2 y 3 ) 3 ( x y ) y = 5 y y (1 2 xy y 2 ) 2 (2 xy 4 x 2 y 2 2 xy 3 3 y 2 6 xy 3 3 y 4 ) 3x 2 y 2 3xy 3 3xy 3 3 y 4 = 5 (1 2 xy y 2 ) 2 = x 2 y 2 2 xy 3 2 xy 3 y 2 ...(3) 5 y2 )2 (1 2 xy On adding (2) and (3), we get 2 2 3 2 u 2 u x 2 y 2 2 xy 3 2 xy 3 y 2 x y 2 xy 2 xy 3 y (1 x 2 ) y = 5 =0 5 x x y y 2 2 2 2 (1 2 xy y ) (1 2 xy y ) Proved. 1 Example 16. Prove that if f (x, y) = y fxy (x, y) = fyx (x, y). 1 e ( x a )2 4y then ( x a )2 4y e y Differentiating f (x, y) partially w.r.t. x, we get Solution. f (x, y) = ...(1) ( x a)2 1 [2 ( x a)] 4 y ( x a ) . e e fx (x, y) = 4y y 2 y 3/ 2 Differentiating again partially w.r.t. ‘y’ by product rule, we have fyx (x, y) = 3( x a) 4 y 5/ 2 ( x a) .e ( x a)2 4y ( x a)2 4y .e 8y7 / 2 Differentiating (1) partially w.r.t. ‘y’, we have = fy (x, y) = 1 2 y 3/ 2 .e ( x a)2 4y ( x a)3 8y7 / 2 e ( x a )2 4y .[6 y ( x a)2 ] ( x a )2 4 y5 / 2 e ( x a )2 4y ...(2) ( x a )2 4y Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ Partial Differentiation 12 Differentiating again partially w.r.t. ‘x’, we have fxy(x, y) = = ( x a) 4 y 5/ 2 ( x a) 8y7 / 2 ( x a) e e ( x a)2 4y ( x a )2 4y ( x a )2 4y ( x a) 2 y 5/ 2 e ( x a)2 4y ( x a )3 8 y7 / 2 e [2 y 4 y ( x a)2 ] e [6 y ( x a )2 ] 8y7 / 2 From (2) and (3), we have fxy (x, y) = fyx (x, y) = ( x a )2 4y ...(3) Proved. 3u 3u y . Example 17. If u = x , show that x 2 y xyx Solution. u = xy log u = log xy = y log x Differentiating partially, we have y 1 u , . = x u x y u = u , x x and 1 u . = log x u y u = u log x y 2u 1 u u y u u uy.log x = . u y . . = yx x y x x y x x u u x. .log x u log u 1 u u x x = 2 . y. x x xyx x x2 u 1 u y log x u uy uy log x = 2 x x x x x 2 x x2 2 u uy uy log x uy uy log x 2 = 2 2 x x x2 x x2 u 2uy uy 2 log x uy log x = 2 2 x x x2 x2 u = u log x y 3 2u u u log x. = x x xy = u u log x As y x u uy log x. x x ...(1) u uy x x u u x. log x. u log x (1) 1 u x x = 2 . y. 2 x x x x2 x y u uy uy y log x u uy log x = 2 2 2 x x x x x x2 u 2uy y log x uy uy log x = 2 2 x x x x x2 3u u Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ Partial Differentiation 13 = u x2 From (1) and (2), we get 2uy x2 3u x 2 y uy 2 log x x2 = uy log x ...(2) x2 3u xyx Proved. Example 18. If u = log (x3 + y3 + z3 – 3xyz), show that 2 9 u = ( x y z)2 x y z 3 3 Solution. u = log (x + y + z3 – 3xyz) Differentiating (1) partially w.r.t. ‘x’, we get Similarly, (U.P. I Semester, winter 2003) ...(1) u 3x 2 3 yz = 3 x x y 3 z 3 3xyz ...(2) u 3 y 2 3zx = y x3 y 3 z 3 3xyz ...(3) 3z 2 3xy u = 3 z x y 3 z 3 3xyz On adding (2), (3) and (4), we get u u u 3( x 2 y 2 z 2 xy yz zx ) = x y z x 3 y 3 z 3 3 xyz = 3( x 2 y 2 z 2 xy yz zx) 2 2 2 ( x y z ) ( x y z xy yz zx ) ...(4) = 3 x yz 3 u = x yz x y z 2 3 u = x y z x y z x y z 3 3 3 = x x y z y x y z z x y z = – 3(x + y + z)– 2 – 3(x + y + z)– 2 – 3 (x + y + z)– 2 = 9 Proved. ( x y z )2 1.9 WHICH VARIABLE IS TO BE TREATED AS CONSTANT Let x = r cos , r To find , we need a relation between r and x. x r = x sec Differentiating (1) w.r.t. ‘x’ keeping as constant r = sec x 2 Here, we have r = x2 + y2 Differentiating (3) w.r.t. ‘x’ keeping y as constant. y = r sin ...(1) ...(2) ...(3) Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ Partial Differentiation 14 2r r = 2x x r x = cos x r or ...(4) r r r = sec and from (4), = cos . These two values of make confusion. x x x To avoid the confusion we use the following notations. r Notation. (i) x means the partial derivative of r with respect to x, keeping as constant. From (2), r From (3), = sec x r (ii) means the partial derivative of r with respect to x keeping y as constant. x y r From (4), = cos x y (iii) When no indication is given regarding the variables to be treated as constant means , means . x x y y y x means , means . r r r Example 19. If x = r cos , y = r sin , find x r y (i) (ii) (iii) r x y r Solution. constant. (iv) y x x (i) means the partial derivative of x with respect to r, keeping as r x = cos r x = r cos y (ii) means the partial derivative of y with respect to , treating r as constant. r y y = r sin = r cos r r (iii) means the partial derivative of r with respect to x, treating y as constant. We have x y to express r as a function of x and y. r = 1 r = 2 x y x2 y 2 (From the given equations) 1 2 x y 2 .2 x x 2 x y2 (iv) Before finding we have to express in terms of x and y. y x 1 y = tan (From the given equations) x 1 1 x . 2 Ans. = 2 y x x y2 y x 1 2 x Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ Partial Differentiation 15 EXERCISE 1.3 1. If z3 – 3yz – 3x = 0, show that z 2. If z(z2 + 3x) + 3y = 0, prove that 2 z z 2 z z 2 z ;z 2 x y xy x y 2z 2 2 z x y 2 x y 2 3. If z = log (e + e ), show that rt – s = 0. 2 z ( x 1) ( z 2 x)3 . 1 1 4. If f (x, y) = x3y – xy3, find f f x y x 1 Ans. 13 22 y 2 5. If = t n e r2 4t , find what value of n will make 1 r 2 r 2 r . r t 6. Show that the function u = arc tan (y/x) satisfies the Laplace equation Ans. n = 2u x 2 2u y 2 3 2 0. z z xz 7. If z = y f (x2 – y2) show that y x x y y . 8. Show that 2z x 2 2 2 z 2 z = 0, where z = x . f (x + y) + y . g(x + y). xy y 2 y 2 z 2u = 0. . Show that x x 2 y 2 1 9. If u = log (x2 + y2) + tan 10. If u (x, y, z) = 1 2 2 2 , find the value of x y z 11. If x = er cos cos (r sin ) and y = er x 1 y y 1 x , Prove that = r r r r 2 2u x y sin (r sin ) 2 2u . z 2 Ans. 2 ( x 2 y 2 z 2 )2 1 x 1 2 x = 0 r r r 2 r 12. If x = r cos , y = r sin , prove that Hence deduce that (i) r x 1 x , r. . x r x r 2x cos 2u 2 (ii) 2 x 2 13. If v = (x2 – y2). f (xy), prove that 2v x 2 2 y 2 = 0 2v y 2 (c) 2r x 2 2r y 2 2 2 r 1 r r x y = (x4 – y4) f (xy) u v u v x 2 y 2 14. If ux + vy = 0 and x y = 1, show that x y y 2 x 2 15. If z = xy + yx, verify that 2 z 2z x y y x 16. If u = f (ax2 + 2h x y + by2) and v = (ax2 + 2hx y + by2) show that 17. If u = rm, where r2 = x2 + y2 + z2, find the value of 18. If x = r (e e ), 2 y r (e e ) 2 2u x 2 2u 2u y2 z 2 x r prove that r x . v v u u . y x x y Ans. m(m + 1)r m – 2 Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ Partial Differentiation 16 u 2u a2 2 , show that ag = t x u u sin 2 y = log (tan x + tan y), prove that, sin 2 x = 2. x y 2 x2u 2 u 1 x = [(x – y) + (x + y)], then show that x x y 2 x xz = ex y z f , prove that y u u u u z x y = 2 x y z u, (ii) y = 2xyzu y z x y 19. If u (x, t) = a e– 20. If u 21. If u 22. If u (i) gx sin (nt – gx), satisfies the equation 2u 2u y zx zy 23. If u = f (x, y), x = r cos , y = r sin , then show that Also deduce that x 2 u u x y 2 2 1 u = 2 r r u n . 2 (A.M.I.E., Summer 2001) 2 1.10 HOMOGENEOUS FUNCTION A function f (x, y) is said to be homogeneous function in which the power of each term is the same. A function f (x, y) is a homogeneous function of order n, if the degree of each of its terms in x and y is equal to n. Thus a0 xn + a1xn – 1y + a2xn – 2 y2 + ... + an – 1 xyn – 1 + an yn ...(1) is a homogeneous function of order n. The polynomial function (1) which can be written as 2 n 1 n y y y y y n x n a0 a1 a2 ... an 1 an = x x x x x x ...(2) 2 3 y y y 3 x 1 3 5 (i) The function is a homogeneous function of order 3. x x x y y x 1 1 x x y x 3/ 2 x . (ii) is a homogeneous function of order – 3/2. 2 2 x2 y 2 y y 2 1 x 1 x x x y 1 (iii) sin is not a homogeneous function as it cannot be written in the form of x2 y 2 y x n f so that its degree may be pronounced. It is a function of homogeneous x expression. 1.11 EULER’S THEOREM ON HOMOGENEOUS FUNCTION (U.P. I Semester, Dec. 2006) Statement. If z is a homogeneous function of x, y of order n, then x. z z y. = nz x y Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ Partial Differentiation 17 Proof. Since z is a homogeneous function of x, y of order n. z can be written in the form y n z = x .f x Differentiating (1) partially w.r.t. ‘x’, we have y y y n x . f 2 x x x z y n 1 y n2 y. f = nx . f x x x x Multiplying both sides by x, we have z y y n n 1 x = n x . f x y. f x x x Differentiating (1) partially w.r.t. ‘y’, we have z y 1 n = x f . y x x Multiplying both sides by y, we get z y n 1 y. = x y. f y x Adding (2) and (3), we have z z n y x. y. = n.x f x y x z z x. y. = nz x y ...(1) z n 1 = nx . f x ...(2) ...(3) Proved. Note. If u is a homogeneous function of x, y, z of degree n, then u u u y z = nu x y z I. Deduction from Euler’s theorem If z is a homogeneous function of x, y of degree n and z = f (u), then u u f (u ) x y n = (Nagpur University, Winter 2003) x y f (u ) Proof. Since z is a homogeneous function of x, y of degree n, we have, by Euler’s theorem, z z x y = nz ...(1) x y Now z = f (u), given z u z u = f (u ) and y = f (u ) y x x Substituting in (1), we get u u x f (u ) y f (u ) = nf (u) x y f (u ) u u x y = n f (u ) x y x Note. If v = f (u) where v is a homogeneous function in x, y, z of degree n, then x u u u y z x y z = nf (u) f (u ) Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ Partial Differentiation 18 x1/3 y1/3 Example 20. Verify Euler’s theorem for z = Solution. Here, we have x1/3 1 = x1/ 2 1 x1/ 2 y1/ 2 . (U.P. Ist Semester, Dec. 2009) 1/ 3 y x 1 6 y x z = 1/ 2 1/ 2 1/ 2 x x y y x Thus z is homogeneous function of degree 1 . 6 z z 1 By Euler’s theorem x y = z. x y 6 Differentiating (1) w.r.t. ‘x’, we get x1/ 3 y1/ 3 z = x 1 2 1 1 1 1 1 1 x2 y2 x 3 x3 y3 x 2 3 2 1 1 x2 y2 5 x z = x z = y 1 1 1 6 1 3 2 1 x x y x 3 3 2 5 6 2 1 ...(1) ...(2) 1 2 1 1 1 1 1 2 3 x y 2 ...(3) 2 1 1 x2 y2 1 2 1 1 1 1 1 1 x2 y2 y 3 x3 y3 y 2 3 2 1 1 1 1 x2 y2 2 5 1 1 1 1 6 1 3 2 1 6 1 2 3 x x y x x y 3 2 2 = 3 1 2 1 x2 y2 1 = 1 2 x y 3 2 3 1 1 1 1 6 1 3 2 1 y x y y 3 2 2 1 1 x2 y2 1 6 2 5 1 2 3 1 6 1 3 2 1 6 x y y x y y z 3 2 2 y = 3 2 y 1 1 x2 y2 Adding (3) and (4), we get 5 1 1 5 1 1 1 1 5 1 1 5 1 6 1 3 2 1 6 1 2 3 1 2 3 1 6 1 3 2 1 6 x x y x x y x y y x y y z z 3 3 2 2 3 3 2 2 x y = 2 x y 1 1 x2 y2 1 1 1 1 1 1 5 5 1 1 1 1 1 1 x 2 x3 y 3 y 2 x 3 y 3 x 6 y 6 x3 y 2 x 2 y 3 6 6 = = 2 2 1 1 1 1 x2 y2 x2 y2 ...(4) Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ Partial Differentiation = 19 1 1 1 1 1 x2 y2 x3 y3 6 2 1 1 1 x3 y3 = 1 6 1 2 x y2 1 1 x2 y2 1 z z x y = z 6 x y From (2) and (5), Euler’s theorem is verified. x y 1 , show that Example 21. If u = cos x y u u 1 x y cot u 0. x y 2 xy 1 Solution. Here, we have, u = cos x y u is not a homogeneous function but if z = cos u, then y y x 1 1 1 x y x x x2 u = cos–1 z = = x y y y x 1 1 x x 1 z is a homogeneous function in x, y of degree . 2 z z 1 By Euler’s theorem, we have x x y y = z 2 z u z u 1 x y = z u x u y 2 u u x ( sin u ) y ( sin u ) = 1 cos u x y 2 u u u 1 x y = cot u. x x y x y 2 1 x 2 y 3 z , show that Example 22. If u = sin 8 8 8 x y z u u u x y z 3 tan u = 0. x y z 1 x 2 y 3 z Solution. We have, u = sin 8 8 8 x y z ...(5) Verified. (U.P. Ist Semester, Dec. 2009) 1 y x 2 . x u 1 cot u = 0 y 2 Proved. (U.P. I Sem., Winter 2003) Here, u is not a homogeneous function but if v = sin u = x 2 y 3z x8 y 8 z 8 then v is a homogeneous function in x, y, z of degree – 3. v v v By Euler’s Theorem x y z = nv x y z x v u v u v u y z = –3 v u x u y u z ...(1) Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ Partial Differentiation 20 Putting the value of v in (1), we get u u u u = –3 sin u x cos u y cos u z cos u x y z u u u sin u y z = 3 = –3 tan u x y z cos u u u u x y z 3 tan u = 0 Proved. x y z x u u x4 y4 Example 23. If u = loge , show that x x y y = 3. x y (Nagpur University, Summer 2008, Uttarakhand, I Semester 2008) x4 y 4 u = loge x y Here, u is not a homogeneous function but if Solution. We have, y 4 x 4 1 x4 y 4 x x 3 y z = eu = x y y x x 1 x Then z is a homogeneous function of degree 3. By Euler’s Deduction formula I u u f (u ) eu x y n 3 3 = x y f (u) eu Example 24. If f (x, y) = x Solution. 1 x 2 Proved. 1 log x log y , prove that xy x2 y 2 f f y 2 f = 0. x y 1 1 log x log y f (x, y) = 2 xy x x2 y2 (A.M.I.E. Summer 2004) y 0 log 1 y 1 1 1 x = 2 2 x x x y x 2 y 2 1 x x f (x, y) is a homogeneous function of degree – 2. By Euler’s Theorem f f f f x y y 2f = 0 = –2.f x Proved. x y x y Example 25. If z be a homogeneous function of degree n, show that 2 z 2 z z 2 z 2 z z y. 2 (n 1) (i) x. 2 y. (n 1) (ii) x. xy y y xy x x (iii) x 2 . 2 z x 2 2 xy. 2 z 2 z y 2 . 2 n(n 1) z. xy y (Uttarakhand Ist Semester, Dec. 2006) Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ Partial Differentiation 21 Solution. By Euler’s Theorem x. z z y =nz x y ...(1) Differentiating (1), partially w.r.t. ‘x’, we get z z 2 z 2 z x. 2 y = n x x x y x 2 z z = (n 1) 2 xy x x Differentiating (1), partially w.r.t. ‘y’, we have x. x. 2 z y Proved (i) ...(2) Proved (ii) ...(3) z 2 z z 2 z y 2 = n y yx y y x 2 z y 2 z z = (n 1) 2 xy y y Multiplying (2) by x, we have 2 z z = ( n 1) x 2 xy x x Multiplying (3) by y, we have z 2 z 2 z xy. y 2 . 2 = (n 1) y y yx y Adding (4) and (5), we get x2 . 2 z ...(4) ...(5) 2 z z 2 z 2 z y . = (n 1) x y 2 2 x y xy x y = (n – 1) n z [From (1)] = n(n – 1) z Proved (iii) Example 26. If f (x, y) and (x, y) are homogeneous functions of x, y of degree p and q respectively and u = f (x, y) + (x, y), show that x2 . 2 z xy. 2 xy. 2 2 2u 1 2u q 1 u u 2 u x 2 xy y x y f (x, y) = P ( P q) 2 2 xy y y P ( P q) x x (A.M.I.E.T.E. Winter 2000) Solution. Since f and are homogeneous functions of degree p and q respectively, we have f f x y = P.f ...(1) x y x y = q. ...(2) x y On adding (1) and (2), we get u f ( x, y ) ( x, y ) f u f f x y = P f + q x x y y x x x u f u u x y i.e., = P f + q ...(3) y y y x y Also x2 2 f x 2 2 xy 2 f 2 f y2 2 xy y = P (P – 1) f ...(4) Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ Partial Differentiation 22 2 2 2 y And = q (q – 1) xy x 2 y 2 On adding (4) and (5), we obtain x2 2 2 xy 2 f 2 2 f 2 f 2 2 x 2 2 2 2 xy y2 2 2 x xy xy y x y = P(P – 1) f + q (q – 1) u f 2 u x 2 2u 2 f x 2 2 f 2 x 2 2 2 2u 2 u 2 y = P(p – 1) f + q (q – 1) y 2 y 2 y xy x 2 y 2 Dividing by P (P – q), we get 2 2 2u 1 2u 1 2 u 2 xy y [ P ( P 1) f q(q 1)] P ( P q) x 2 2 = x y P ( P q) y x q 1 u u y from both sides, we get x Subtracting P ( P q) x y x2 2u ...(5) 2 xy 2 2 2u 1 2u (q 1) 2 u 2 xy y x 2 P ( P q) x 2 xy P ( P q) y u u x x y y 1 (q 1) u u [ P ( P 1) f q ( q 1) ] y = x P ( P q) P ( P q ) x y 1 [ P ( P 1) f q(q 1) (q 1) [ Pf q]] = P ( P q) 1 [ P 2 P Pq P ) f (q 2 q q 2 q)] = P ( P q) 1 P( P q ) 2 = P ( P q) [( P Pq ) f ] P ( P q) f = f (x, y) [From (3)] Proved. y n n x Example 27. If z = x f y then prove that x y 2 2 2 z z z z z 2 x 2 2 2 xy y2 2 x y (Nagpur University, Summer 2003) xy x y = n z. x y y n n x z = x f y x y z = u+v y n x n where, u = x f and v = y x y Since u is a homogeneous function of x, y of degree n. u u x y = nu x y Solution. 2 2u 2 u y and = n(n – 1)u xy x 2 y 2 As v is a homogeneous function of x, y of degree –n. v v x y = –nv x y x2 2u 2 xy ...(1) ...(2) ...(3) ...(4) Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ Partial Differentiation x2 and 2v x 2 23 2 xy 2v 2v y 2 2 = –n (– n – 1)v = n (n + 1)v xy y On adding (2) and (4), we get u v u v x y = nu – nv x x y y z z x y = nu – nv x y On adding (3) and (5), we get ...(5) zuv z u v x x x z u v y y y ...(6) [From (1)] 2u 2 v 2u 2u 2v 2v x 2 2 2 2 xy y 2 2 2 = n(n – 1) u + n (n + 1)v x xy xy y x y 2 z 2u 2v x 2 x 2 x 2 2 2 2 z z z x 2 2 2 xy y 2 2 = n(n – 1)u + n (n + 1)v ...(7) [From (1)] xy x y On adding (6) and (7), we have 2 2 z z z 2 z y x y 2 2 x y x y = n(n – 1)u + n (n + 1) v + nu – nv x y = nu (n – 1 + 1) + nv (n + 1 – 1) = n2u + n2v = n2 (u + v) = n2z II. Deduction: Prove that x2 2 z 2 xy x2 2u x 2 2 xy Proved. 2u 2u y 2 2 = g(u) [g(u) – 1] (Nagpur University, Winter 2003) xy y f (u ) g(u) = n f (u ) Proof. By Euler’s deduction formula I u u f (u ) x y n. = x y f (u ) where, u u y = g(u) x y Differentiating (1) partially w.r.t. ‘x’, we have 2 u u 2u u x 2 .1 y g (u ) = x xy x x x 2u u = [ g (u ) 1] xy x x 2 Similarly, on differentiating (1) partially w.r.t. ‘y’, we have x 2u 2u u [ g (u ) 1] = 2 y x y y Multiplying (2) by x, (3) by y and adding, we get y x2 2u x 2 2 xy 2u y. x. 2u 2u u u y 2 2 = [ g (u ) 1] x y xy x y y = [g (u) – 1]g(u) = g(u) [g(u) – 1] f (u ) g (u ) Given n f (u ) ...(1) ...(2) ...(3) [From (1)] Proved. Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ Partial Differentiation 24 3 3 1 x y , prove that Example 28. If u = tan x y u u (i) x. y. = sin 2u (A.M.I.E., Winter 2001) x y 2 2u 2u 2 u y 2 2 = 2 cos 3u sin u. (M.U. 2009; Nagpur University, 2002) (ii) x . 2 2 xy x y x y Solution. Here u is not a homogeneous function. We however write 3 y 3 3 y x 1 1 x3 y 3 x x y x 2 . x 2 z = tan u = x y y y x 1 x 1 x x so that z is a homogeneous function of x, y of order 2. (i) By Euler’s Theorem [Here f (u) = tan u] x u u n f (u ) y. = f (u) x y = 2 tan u 2 sec u ...(1) 2 sin u cos 2 u = 2 sin u cos u = sin 2u cos u (ii) By deduction II x2 . 2u x 2 2 xy Here x2 . 2u x 2 2 xy 2u 2u y 2 2 = g(u)[g(u) – 1] xy y sin 2u = g(u) 2u 2u y 2 2 = sin 2u (2 cos 2u – 1) = 2 sin 2u cos 2u – sin 2u xy y = sin 4u – sin 2u = 2 cos 3u sin u Proved. xy 1 Example 29. If u = sin x y 2 2u sin u cos 2u 2 u y . 2 2 xy x y 4 cos3 u x y 1 Solution. We have, u = sin x y y x 1 x y x x1/ 2 ( x) Let z = sin u = x y y x 1 x z = f (u) = sin u 1 z is a homogeneous function of degree . 2 By Euler’s deduction I u u f (u ) u u 1 sin u x y y = n x = x y f (u ) x y 2 cos u u u 1 x y tan u = x y 2 Prove that x 2 2u 2 xy Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ Partial Differentiation 25 Let 1 tan u 2 g(u) = By Euler’s deduction II x2 2u 2 xy x 2 2u 2u 1 1 y 2 2 = g(u) [g(u) – 1] = tan u sec 2 u 1 xy y 2 2 1 sin u 1 sin u cos 2u 1 sin u 2 (1 2 cos 2 u ) = 4 cos u cos 2 u 4 cos3 u 4 cos3 u Proved. EXERCISE 1.4 1. Verify Euler’s theorem in case (i) f (x, y) = ax2 + 2hxy + by2 2. If v = x3 y3 3 x y 3. If u = log 3 , show that x. x3 y3 2 x y 2 (ii) u = ( x y ) ( xn y n ) v v y. 3v . x y , prove that x. u u y. 1 . x y z z 3 4. If z = ( x 2 y 2 ) / ( x y), prove that x y. z. x x 2 y f f 5. If f (x, y) = x4y2 sin–1 , then find the value of x y . x x y (A.M.I.E.T.E., Winter 2001) Ans. 6 f (x, y) u y u 1 x y . 6. If u = sin , show that x x y x y 3 3 u u 1 x y 7. If u = sec , show that x. y. = 2 cot u, then evaluate x y x y 2u 2u y2 2 (A.M.I.E.T.E., Winter 2001) Ans. –2 cot u (2 cosec2 u + 1). x y x y 8. If x = eu tan v, y = eu sec v, find the value of x2 2u 2 2 xy u u v v x. y. . x. y. . (A.M.I.E., Summer 2001) Ans. 0 y x y x [Hint: Eliminate u and apply formula I. Again eliminate v and apply the formula] x1/ 4 y1/ 4 9. If u = sin 1 , prove that 1./6 y1/ 6 x x2 2u x 2 2 xy 2u 2u 1 y2 2 = tan u [tan 2 u 11]. x y 144 y 2 10. Find the value of x 2u x 2 2 xy 2u 2u y 2 2 if u = sin–1 (x3 + y3)2/5. x y y Ans. 2 1 y , find 11. If u = tan x u u (i) x. x y. y , and 1 12. If u = tan x3 y 3 x y 2 (ii) x . 2u x 2 2 xy. , find the value of x 2 2u 2u y2. 2 xy y 2u x 2 2 xy Ans. (i) 2u 2u y2 2 x y y 5 6 tan u sec2 u 1 6 5 xy 2 x2 y4 (ii) 2 xy 6 ( x2 y 4 )2 Ans. –2 sin3 u cos u Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ Partial Differentiation 26 y 13. If u = f x 2 2 2 x 2 y 2 , find the value of x 2 u 2 xy u y 2 u . x y x 2 y 2 2 2z 2 z y . Ans. 0 x y x 2 y 2 2 2 4 15. Verify Euler’s theorem on homogeneous function when f (x, y, z) = 3x yz + 5xy z + 4z 2 14. If z = xy/(x + y), find the value of x Y 16. If u = x X X2 2u x 2 Y X 2 XY 2 z Ans. 0 2 xy , prove by Euler’s theorem on homogeneous function that 2u 2u Y 2 2 = 0. x y y 17. Given F (u) = V(x, y, z) where V is a homogeneous function of x, y, z of degree n, prove that x u u u F (u ) y z =n x y z F (u ) x y z 18. State and prove Euler’s theorem, and verify for u = y z x 19. If u = x2 y 2 z 2 x2 y 2 z 2 cos xy yz x2 y2 z , x 2 , show that (A.M.I.E., Summer 2000) u u u 4 x2 y 2 z 2 y z 2 x y z x y 2 z 2 1.12 TOTAL DIFFERENTIAL In partial differentiation of a function of two or more variables, only one variable varies. But in total differentiation, increments are given in all the variables. 1.13 TOTAL DIFFERENTIAL CO-EFFICIENT Let z = f (x, y) ...(1) If x, y be the increments in x and y respectively, let z be the corresponding increment in z. Then z + z = f (x + x, y + y) ...(2) Subtracting (1) from (2), we have z = f (x + x, y + y) – f (x, y) ...(3) Adding and subtracting f (x, y + y) on R.H.S. of (3), we have z = f (x + x, y + y) – f (x, y + y) + f (x, y + y) – f (x, y) f ( x x, y y ) f ( x, y y ) f ( x, y y ) f ( x, y ) x y z = x y On taking limit when x 0 and y 0 f f d x dy dz = ...(4) [Remember] x y dz is called as the total differential of z. 1.14 CHANGE OF TWO INDEPENDENT VARIABLES x AND y BY ANY OTHER VARIABLE t. Differentiation of composite function If z = f(x, y) Where x = (t) y = (t) Here z is composite function of t. Dividing (4) by dt , we have Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ Partial Differentiation 27 z d x z dy dz = x d t y dt dt Then ...(5) [Remember] dz is called the total differential co-efficient of z. dt 1.15 CHANGE IN THE INDEPENDENT VARIABLES x AND y BY OTHER TWO VARIABLES u AND v. Let z = f (x, y) where x = (u, v) y = (u, v) Then from (5), we obtain f z = x u f z = x v and . x f y . u y u ...(6) . x f y . v y v ...(7) Example 30. If u = x3 + y3 where, x = a cos t, y = b sin t, find du and verify the result. dt Solution. We have, u = x3 + y3 x = a cos t y = b sin t dz u d x u d y = x dt y dt dt = = Verification. u = = du = dt Results (1) and (2) (3 x2) (– a sin t) + (3 y2) (b cos t) –3 a3 cos2 t sin t + 3 b3 sin2 t cos t x3 + y3 a3 cos3 t + b3 sin3 t ...(1) – 3a3 cos2t sin t + 3b3 sin2 t cos t ...(2) are the same. Verified. u u Example 31. If z = f (x, y) where x = e cos v and y = e sin v, show that (i) y z z 2u z . x = e y u v (M.U. 2009; Nagpur Univesity 2002) 2 2 2 2 z z z 2u z e (ii) = (M.U. 2009) u v x y Solution. (i) We have, x = eu cos v, y = eu sin v z z x z y z u z z z u e cos v e sin v x y = = u x u y u x y x y y And z = xy u z z = v x z z y2 x y ...(1) x z y v y v Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ Partial Differentiation 28 = z u z z z (eu sin v) (e cos v) y x x y x y z z z = x y x2 v x y On adding (1) and (2), we get z z z 2 2 z y x (e2u cos 2 v e 2u sin 2 v ) = (x y ) u v y y x 2u 2 2 = e (cos v sin v ) z z e 2u y y = z u z u (e cos v) e sin v x y y eu sin v z z z cos v sin v = u x y On squaring, we get e u 2 Proved. x eu cos v x x eu cos v and eu sin v u v z z x z y = u x u y u (ii) 2 y y eu sin v and eu cos v u v 2 z z z z z 2 2 e 2u = cos v sin v 2 sin v cos v u x y x y z x z y z Again = x v y v v z z u = ( eu sin v) (e cos v) x y z z z e u sin v cos v = v x y On squaring, we get 2 2 e ...(3) 2 z z z z z 2 2 e 2u = sin v cos v 2 sin v cos v v x y x y On adding (3) and (4), we get 2u ...(2) ...(4) 2 2 z 2 z 2 z z 2 2 2 2 = (sin v cos v ) (sin v cos v ) u v x y 2 z z = x y Example 32. If u = u (y – z, z – x, 2 x – y), prove that Proved. u u u =0 x y z (Nagpur University, Winter 2002, U.P., I Sem., Winter 2002, A.M.I.E winter 2001) Solution. Let r = y – z, s = z – x, t=x–y so that u = u (r, s, t) u u r u s u t = x r x s x t x Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ Partial Differentiation 29 = u u u u u (0) (1) (1) r s t s t u r u s u t u = r y s y t y y u u u u u (1) (0) (1) = = r s t r t u r u s u t u = r z s z t z z u u u u u (1) (1) (0) = r s t r s ...(1) ...(2) ...(3) u u u Adding (1), (2) and (3), we get x y z = 0 Proved. y x z x u u 2 u y2 z2 , Example 33. If u = u =0 , show that x x y z x y x z (U. P. I. Sem., Dec. 2004, Nagpur University, Summer 2000) y x z x , Solution. Here, we have u = u = u (r, s) zx xy where r= yx , xy and s = zx zx r= 1 1 x y and s = 1 1 x z r 1 = 2 x x 1 r = 2 y y r =0 z and and and s 1 = 2 x x s 1 = 2 z z s =0 y We know that, u u r u s = x r x s x = u r 1 u 1 u 1 u 1 2 2 2 s x x x r x2 s x2 u u u = x r s u r u s u 1 u 1 u u 0 2 = r y s y r y2 s y y r y2 u u = y r u u r u s u u 1 1 u 0 2 2 = z r z s z r s z z s ...(1) ...(2) Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ Partial Differentiation 30 u u = ...(3) z s On adding (1), (2) and (3), we get u u u x2 y2 z2 =0 Proved. x y z Example 34. If (cx – az, cy – bz) = 0 show that ap + bq = c : z z and q where p = x y Solution. Here, we have [x and y are independent but z is dependent on x and y] (cx – az, cy – bz) = 0 (r, s) = 0 where r = cx – az, s = cy – bz r z z r a , = ca y y x x z2 s z = b , x x We know that, r s = r x s x r 0 = s z cb y y z z c a b r x s x 0 = c z b a r x r s z b c = a x r s r Again z a = x ac a r b r s ...(1) r s = y r y s y z z a c b r y s y z z b 0 = c = a c s y r s y s 0 = bc s z b = y a b r s Adding (1) and (2), we get ac bc r s z z a b = x y a b r s z z a b =c a p b q c x y b a s r ...(2) Proved. Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ Partial Differentiation 31 1.16 CHANGE IN BOTH THE INDEPENDENT AND DEPENDENT VARIABLES, (POLAR COORDINATES) Example 35. If w = f (x,y), x = r cos , y = r sin , show that 2 2 2 f f w 1 w 2 = r x y r Solution. Here, x = r cos , x = cos r 2 x = – r sin y = r sin y = sin r y = r cos f x f y w . . = x r y r r w f f . (cos ) . (sin ) = r x y Now, ...(1) w f f f x f y . ( r sin ) . (r cos ) = . . = x y x y 1 w f f = sin cos r x y Squaring (1) and (2) and adding, we obtain 2 w 1 w 2 r r 2 2 f f = x y Example 36. Transform the equation Solution. We r2 u x 2 u x2 2 u ...(2) 2 Proved. 2 u = 0 into polar co-ordinates. x2 y 2 have, x = r cos , y = r sin y = x2 + y2, = tan–1 x u r u u x u y u u sin cos = 2 2 = r x x r r x y r r = = = = = u x x sin u sin u cos cos r r r r u sin u sin u sin u cos cos cos r r r r r r 2 u sin u sin 2 u cos cos 2 2 r r r r sin u 2 u cos u sin 2 u cos sin r r r r r 2 2 u sin cos u sin cos 2u cos 2 2 r r r r2 sin 2 u sin cos 2 u sin cos u sin 2 2 u r r r r r2 r 2 2 Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ Partial Differentiation 32 2 = cos 2u 2 r u = y 2 u = y2 = = 2 sin cos u 2 sin cos 2 u sin 2 u sin 2 2 u 2 ...(1) 2 r r r r r r 2 u y u x u u cos u r u sin = r r x2 y2 r r r y y u cos u cos u = sin sin y y r r r r u cos u cos u cos u sin sin sin + r r r r r r 2 u cos u cos 2 u sin sin 2 2 r r r r cos u 2 u sin u cos 2 u sin cos r r r r r 2 2 u sin cos u sin cos 2 u 2 = sin 2 r r r r2 2 = sin 2 u r2 cos 2 u sin cos 2 u sin cos u cos 2 2 u r r r r r2 r 2 2 2sin cos u 2 sin cos 2 u r r r2 2 cos u cos 2 2 u r r r 2 2 ...(2) Adding (1) and (2), we get 2 u x 2 2 u y (sin 2 cos 2 ) = 2 r2 (sin 2 cos 2 ) 1 u 1 2 u (sin 2 cos 2 ) 2 r r r 2 1 u 1 2 u r 2 r r r 2 2 Example 37. If u = f (r) and x = r cos , y = r sin , prove that = 2 u x 2 2 u 2 u 2 u y 2 f " (r ) 1 f ' (r ). r Ans. (Nagpur University, Winter 2004) (A.M.I.E.T.E., Winter 2003, U.P. I Semester, Winter 2005, 2000) Solution. Here, we have x = r cos y = r sin r2 = x2 + y2 so that r x x r u = f (r) d f r u . = dr x x Differentiating again w.r.t. x, we get u d f x . = x dr r r r.1 x 2 d f r x d f u x d f x x d f . . . . = 2 = d r 2 x r d r d r 2 r r d r x r2 2 2 r.1 x . . r2 x r Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ Partial Differentiation d 2 f x2 = d r2 r2 33 d f r 2 x2 d 2 f x2 d f y 2 . dr d r r3 r3 d r2 r2 ...(1) Similarly, 2 u d f x2 d r r3 y d r2 r2 On adding (1) and (2), we get 2 d 2 f y2 = 2 u x2 2 u y2 = = d 2 f x2 y2 d r2 d2 f dr 2 r2 ...(2) d f x2 y 2 dr r3 d f 1 1 f (r ) f '(r ) dr r r Proved. Example 38. A function f (x, y) is rewritten in terms of new variables u = ex cos y, v = ex sin y f f f f f f Show that (i) u v and (ii) v u x u v y u v and hence deduce that 2 2 f 2 f 2 f 2 2 f ( u v ) (iii) u2 x2 y 2 v 2 u Solution. u = excos y, u e x cos y u, = – ex sin y = – v y x v v e x sin y v , e x cos y u x y f f u f v (i) We know that = x u x v x v = ex sin y, (ii) (iii) f f f = u v x u v f u f v f f f f f = u y v y u .(v ) v u – v . u u . v y 2 f x 2 = x ... (1) Proved. ... (2) Proved. f x f f v. v . = u . u . v u v u f f f f v v = u u v u u u v v u v f f f f = u u u v v u v v u u u v v u v v 2 f f 2 f 2 f 2 f f = u u 2 uv v u vv u u v v u v 2 v u 2 = u 2 f u2 u f 2 f 2 f 2 f f u v uv v2 2 v u vu uv v v [From (1)] ...(3) Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ Partial Differentiation 34 2 f y f f f u u v v y y u v u v = 2 [From (2)] f f f f u u v u v u u v v u v f f f f = v v v u u v u u u u u v v u v v 2 f 2 f f 2 f f 2 f = v v 2 v u u v uu u u v v v u u v 2 = v 2 2 f f 2 f f 2 f v u v u u uv v u v u u2 v2 On adding (3) and (4), we obtain 2 = v 2 f x2 2 f y2 2 = u 2 f 2 f u2 u v v2 2 f v2 2 f u2 u2 v2 2 2 V 2 f Proved. 2 V (A.M.I.E.T.E., Winter 2007) x y (Nagpur University, Summer 2001, Winter 2000, U.P., I Semester, Winter 2001) Solution. We have, x + y = 2 e cos ...(1) x – y = 2 i e sin ...(2) By adding and subtracting equations (1) and (2), we have 2 x = 2e (cos + i sin ) x = e+i and 2 y = 2e (cos – i sin ) y = e–i ...(3) It is clear that V = f (x, y) and x, y are functions of and . Hence V is a composite function of and . We want to convert V, , in V, x, y respectively. From equation (3), we have x x i e i i x e i x, = Show that : 2 V (u 2 v 2 ) ...(4) 2 f 2 2 f 2 2 f ( u v ) u2 v2 u2 v2 Example 39. If x + y = 2 e cos and x – y = 2 ie sin 2 2 = (u v ) 2 f v2 2 4xy y e – i y , Now, and and y i iy = ie V V x V y V V . . x y = x y x y 2 V 2 = V V V y y x x x y x y = x x V V y x y x y y V V y x x y 2 V V 2 V 2 V 2 V V x x y y x y = 2 x x y y2 y x x y Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ Partial Differentiation 2 V Again 2 35 2 V 2 = x x 2 y2 2 V y 2 V 2 V V x y x y x y 2xy ...(4) V V V x V y V i x y = x y y x y = i x y x 2V 2 = = ix V V V = i x x y y i x x y y V V V V .i x y .i x y iy x x y y x y = x x V V y x y y y x V V y x y x V 2 V 2 V 2 V 2 V V = x x y y x y x y x2 y2 y x xy 2 V 2 2 V V V 2 V 2 V = x y x2 y2 2x y 2 y xy x y2 x Adding (4) and (5), we get 2 V 2 2 V 2 = 4x y 2 V xy ...(5) Proved. EXERCISE 1.5 1. If z = u2 + v2 and u = at2, v = 2 at, find dz . dt Ans. 4 a2t (t2 + 2) dz 3 . dt 1 t2 dw dw dw 3. If w = f (u, v), where u = x + y and v = x – y, show that 2 dx dy du 2. If z = sin–1 (x – y), x = 3 t, y = 4 t3; show that 4. If u = xey z, where y = du 3 a 2 x 2 , z = sin x. Find d x 5. If u = x 2 + y2 + z2 – 2 xyz = 1, show that [Hint. du = x2 Ans. eyz 1 3x cot y dx dy dz 0 2 2 1 x 1 y 1 z2 x u u u dx dy dz 0 x y z = 2 (x – yz) dx + 2 (y – zx) dy + 2 (z – xy) dz = 0 But x2 + y2 + z2 – 2 xyz = 1, y2 – 2xyz = 1 – x2 – z2 2 2 2 2 2 2 2 y – 2xyz + x z = 1 + x z – x – z (y – xz)2 = (1 – x2) (1 – z2)] 2 2 6. If z = z (u, v), u = x – 2xy – y and v = y . Show that z z z ( x y) 0 is equivalent to = 0 x y v 7. If u = f (x2 + 2 y z, y2 + 2 z x), prove that u du du ( y 2 z x) ( x 2 y z) ( z 2 xy ) 0 x dy dz (x + y) Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ Partial Differentiation 36 8. By changing the independent variables x and t to u and v by means of the relationships u = x – at, v = x + at 2 y a2 Show that x2 2 y t2 4 a2 2 y u v du dx 1 v y 9. If x2 = au + bv, y = au – bv, prove that . . d x d u y v 2 y x v u uv z z z , show that 2 x u v . u v dx du dv y u u 2 3 2 2 find , . 11. If u = x cos , x 3r 2 s, y 4 r 2 s , z 2r 3s z r s 10. If z = f (x, y) where x = uv, y = u y 4x y 4x y r y u y 6 x s2 y 6 xys y 6 r cos sin 2 sin , 2 cos sin 2 sin r z z z z s z z z z z z (x, y) where x = eu cos v, y = eu sin v. Prove that 2 2 2 f f 2u f e u y v Ans. 12. If z = f f x 2 cos sin ,y and z f ( x, y ) , then show that u u 13. If x = 2 z x2 2 z y2 1 14. If x = z sin u4 2 z u2 u3 z 2 z u4 u 2 y u u where x 3 r 2 2 s , y 4 r 2 s 3 , z 2 r 2 3 s 2 , find , x r s u 6r y z 4z 4 r sin 2 2 2 r x x y x y2 15. If z = f (u, v) where u = x cos – y sin , v = x Ans. y , x u 2yz 6 s2z y 6 s sin 2 2 s x x2 y 2 x x y sin + y cos , show that z z x z y u v , being constant. x y u v 16. Given the transformation x = cosh cos , y = sinh sin establish the following equation for the function u (function of x, y and also of ): x 2 u 2 u 2 u 2 2 (sinh sin ) 2 2 2 y 2 x 17. If z = f (x, y), where x = r cos , y = r sin , prove that 2 u 2 2 2 18. If by substituting u = x2 – y2, 2 f 2 1 2 z 1 z x 2 y 2 r 2 r 2 2 r r v = 2 xy, f (x, y) = (u, v). Show that z z z 1 z 2 x y r r and 2 z 2 z 2 z 2 2 4 (u 2 v 2 ) 2 2 u x y v 19. If x = r sin cos , y = r sin sin , z = r cos , v = v(x, y, z), prove that 2 2 2 f 2 2 2 2 2 v v v v 1 v 1 v x y z r r r sin 2 Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ Partial Differentiation 37 2 2 2 2 20. If v be a potential function such that v = v(r) and r = x + y + z , show that 2 v x 2 2 v y 2 2 v z 2 d2v d r2 2 dv r dr 21. Given that w = x + 2y + z2, x = r/s, y = r2 + es, and z = 2r, show that r w w s 12 r 2 2 s e s . r s w when u = 0, v = 0 v If w = (x2 + y – 2)4 + (x – y + 2)3 , x = u – 2 v + 1, and y = 2 u + v – 2 Ans. 99 23. If x = u + v + w , y = v w + w u + u v, z = uvw and F is a function of x, y, z, then show that F F F F F F u v w x 2y 3z u v w x y z 24. If u = x + a y and v = x + b y, transform the equation 22. Find 2 2 z x2 5 2 z 2 z 2 z = 0, find the values of a and b. 3 2 0 into the equation u v x y y 2 2 (A.M.I.E.T.E., Summer 2000) Ans. a 1, b , a , b 1 3 3 1.17 IMPORTANT DEDUCTIONS Let z = f (x, y), then f f dx dy dz = x y If z = 0, dz = 0 f f dx dy 0 = x y f dy x = – dx f y 2 d y We can find by differentiating (1). d x2 f 2 f f q, r, Let = p, y x x2 y p From (1) = . x q On differentiating again, we obtain But dp p p dy = dx x y dx dp f = dx x x y d y d x2 f x [Remember] 2 f s, x y q 2 f dy f . y dx x 2 2 y2 t dp dq p dx dx ...(2) q2 dy . dx f f f dy f f x dp = dx x 2 y x dx x 2 y x f y 2 2 f ...(1) 2 p qr ps r s q q Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ Partial Differentiation 38 dq q q d y f f p = dx x y d x x y y y q 2 f 2 f p t p qst p s 2 xy y q q q Making substitutions in (2), we obtain qr ps qs tp q p 2 q 2 r pqs pqs p 2 t q q d y = = q3 q2 d x2 = d2 y = – d x2 q 2 r – 2 pqs + p 2 t q3 dy . dx 3 2 2 3 f (x, y) = x + 3x y + 6xy + y – 1 = 0 f = 3x2 + 6xy + 6y2 x Example 40. If x3 + 3 x2y + 6 xy2 + y3 = 1, find Solution. Let f = 3x2 + 12xy + 3y2 y f x 3 x 2 6 xy 6 y 2 x 2 2 xy 2 y 2 dy = f 2 dx 3 x 12 xy 3 y 2 x 2 4 xy y 2 y Example 41. If y3 – 3 ax2 + x3 = 0, then prove that d2 y = Ans. 2 a 2 x2 d x2 y5 3 Solution. Let f (x, y) = y – 3ax2 + x3 f f 6 a x 3 x2 , q 3 y2 p= x y r= 2 y 2 f x2 = 6 a 6 x, s 2 f 0, x y ...(1) t 2 f y2 6y q 2 r 2 p q s p 2t x2 q3 Putting the values of p, q, r, s and t in (2), we get 2 y x2 = = 2 y 2 = ...(2) [Art 1.17] (3 y 2 )2 ( 6 a 6 x ) 2( 6 a x 3 x 2 ) (3 y 2 ) (0) ( 6 a x 3 x 2 ) 2 (6 y ) (3 y 2 )3 54 y 4 ( a x) 54 ( 2 a x x 2 ) 2 y 27 y 6 2 y 3 ( a x ) 2 (4 a 2 x 2 x 4 4 a x3 ) x y5 Putting the value of y3 = 3 ax2 – x3 from (1) in (3), we get 2 y 2(3 a x 2 x 3 ) ( a x) 2 (4 a 2 x 2 x 4 4 a x 3 ) = x2 y5 ...(3) Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ Partial Differentiation = 2 y x2 39 6 a 2 x 2 6 a x 3 2 a x3 2 x 4 8 a 2 x 2 2 x 4 8 a x 3 2 = 2a x y5 2 Proved. y5 d2y dy . and dx dx 2 3 3 Solution. Let f (x, y) = x + y – 3axy = 0 f f 3 x 2 3 ay , q 3 y 2 3 ax p = x y Example 42. If x3 + y3 – 3axy = 0, find r= 2 f 6 x, s 2 f 3a, x y x2 f dy x 3 x 2 3 ay a y x 2 = 2 f dx 3 y 3 ax y 2 ax y d2 y 2 = t 2 f y2 6y q 2 r 2 p q s p 2t dx q3 Putting the values of p, q, r, s and t in (1), we get d2 y d x2 = = d2 y dx 2 = ...(1) [Art. 1.17] (3 y 2 3 ax) 2 6 x 2(3x 2 3 ay ) (3 y 2 3 ax) ( 3a) (3 x 2 3 ay )2 (6 y ) (3 y 2 3 ax)3 2 x ( y 2 a x )2 2 a ( x 2 a y ) ( y 2 a x ) 2 y ( x 2 a y )2 ( y 2 a x )3 2 a3 x y ( a x y 2 )3 dy when (cos x) y = (sin y) x dx Solution. Given equation can be written as : (cos x)y – (sin y)x = 0 Here f (x, y) = (cos x)y – (sin y)x = 0 f = y (cos x)y – 1 (– sin x) – (sin y)x log sin y x = – [y sin x (cos x)y – 1 + (sin y)x log sin y] f = (cos x)y log cos x – x (sin y)x –1 cos y y d f dy dx = d f dx dy Ans. Example 43. Find dy y sin x (cos x) y 1 (sin y ) x log sin y = dx (cos x ) y log cos x x(sin y ) x 1 cos y y In (1), put (cos x) for (sin y)x [Art. 1.17] ...(1) y sin x (cos x ) y 1 (cos x) y log sin y dy = x (cos x) y dx (cos x ) y log cos x . cos y sin y Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ Partial Differentiation 40 = (cos x) y [ y tan x log sin y ] (cos x ) y [log cos x x cot y ] y tan x log sin y log cos x x cot y f d z Ans. f Example 44. If f (x, y) = 0 and (y, z) = 0, show that y . z . d x x . y . Solution. f (x, y) = 0 (y, z) = 0 Differentiating (1) w.r.t. x, we get ...(1) ...(2) f f dy . 0 = x y dx Differentiating (2) w.r.t. ‘y’, we get 0 = d z . y z d y f dy x = f dx y dz y = dy z Multiplying (3) and (4), we get f x y dy dz = dx dy f y z f d z f . . = . y z dx x y ...(3) ...(4) f dz x y = f dx y z Proved. du dx (U.P. I Sem., Dec. 2005, Com. 2002) Example 45. If u = x log xy where x3 + y3 + 3 xy = 1. Find Solution. We have, u = x log xy 1 u = x . y 1log xy x xy u = 1 + log xy ...(1) x 1 x u = x .x xy y y x3 + y3 + 3 xy = 1 On differentiating, we get dy dy 3 x2 3 y2 3x 3y = 0 dx dx dy x2 y = dx x y2 We know that ...(2) ...(3) u d x u d y du = x dx y dx dx = (1 log xy ) .1 = 1 log xy x x2 y y x y 2 x x2 y . y x y2 [From (1), (2), (3)] Ans. Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ Partial Differentiation 41 EXERCISE 1.6 Find dy in the following cases : dx 1. x sin (x – y) – (x + y) = 0 2. xy = yx Ans. [y + x2 cos (x – y)] / [x + x2 cos (x – y)] Ans. y (y – x log y)/ x (x – y log x) 3. If ax2 + 2 hxy + by2 = 1, find d2 y Ans. h 2 ab ( hx by )3 d x2 4. If u = x2y + y2z + z2x and if z is defined implicitly as a function of x and y by the equation x2 + yz + z3 = 0 u find , where u is considered as a function of x and y alone. x 2x u 2 2 Ans. x 2 xy z ( y 2 zx) 2 y 3z x dy y3 1 0 , if tan–1 y dx 6. If f (x, y, z) = 0, prove that 5. Find Ans. dx dy dz –1 dy z dz x dx y y 2 2 x 3x y 3 y 4 dx f f Hint . y x dy z 1.18 TYPICAL CASES Example 46. If x = f (u, v), y = (u, v), find u u v v , , , . x y x y Solution. x = f (u, v) y = (u, v) Differentiating (1), (2) w.r.t. x (treating y as constant), we obtain f u f v . . 1 = u x v x 0 = u v . . u x v x v u and , we obtain x x v f f . . u v v u u f f . . u v v u (2) w.r.t. y, we get f u f v . . u y v y ...(1) ...(2) ...(3) ...(4) Solving the equations (3) and (4) for u = x v = x Similarly, differentiating (1) and 0 = 1 = u v . . u y v y Solving the equations (5) and (6) for Ans. Ans. ...(5) ...(6) u v and , we obtain y y Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ Partial Differentiation 42 f v u = f f y . . u v v u f u v = f f y . . u v v u Example 47. If x = u2 – v2 and y = uv, find v u u v and , , y x y x Solution. Here, we have x = u2 – v2 y = uv x y = 2 u; = v u u Ans. Ans. (Nagpur University, Winter 2003) x y = – 2 v, = u v v Differentiating (1) w.r.t. x, we get u v 1 = 2u 2v x x Similarly differentiating (2) w.r.t. x, we get u v u 0 = v x x On solving (3) and (4), we get u u v v , = 2 x 2u 2 2v 2 x 2u 2v 2 On differentiating (1) and (2) w.r.t. y, we get u v 0 = 2u 2v y y u v 1 = v y u y On solving (5) and (6), we get v u u v = 2 2, and = 2 2 y y u v u v Example 48. Find p and q, if x = a (sin u + cos v) ...(1) ...(2) ...(3) ...(4) Ans. ...(5) ...(6) Ans. y = a (cos u sin v) z = 1 + sin (u – v) z z where p and q mean x and y respectively.. Solution. We have, x = a (sin u cos v) y = a (cos u sin v) z = 1 + sin (u – v) Differentiating (3) w.r.t. x, we get u v z = cos (u – v) x x x ...(1) ...(2) ...(3) ...(4) Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ Partial Differentiation 43 Differentiating (1) partially w.r.t. x, we get u v 1 = a cos u sin v x x Differentiating (2) partially w.r.t. x, we get u v cos v 0 = a sin u x x Solving (5) and (6) for ...(6) v u and , we get x x u = x Putting the values of ...(5) 1 cos v v sin u , a cos u v x a cos u v u v and in (4), we get x x p 1 cos v sin u z = cos u v a cos u v x a cos u v 1 z sin u cos v = x a Differentiating (3) w.r.t. y, we get p u v z = cos u v y y y Differentiating (1) and (2) partially w.r.t. y, we get u v a cos u sin v y y u v = a sin u cos v y y v and y , we get sin v v cos u = , and y a cos u v a cos u v Ans. ...(7) 0 = ...(8) 1 ...(9) u Solving (8) and (9) for y u y Putting the values of u , and v in (7), we have y y sin v cos u z q = cos (u – v) y a cos u v a cos u v 1 q = sin v cos u a Ans. EXERCISE 1.7 1. Fill in the blanks (i) (ii) x y z is equal to...... · · y z x dz If z = f (x, y), where x = (t), y = (t), then =...... dt If f (x, y, z) = 0, then Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ Partial Differentiation 44 dy = ...... dx (iii) If f (x, y) = 0, then (iv) If u = x 2 + y 2, x = s + 3t, y = 2s – t, then (v) (vi) du = ds f z · · = If f (x, y) = 0 and (y,z) = 0, then y z x f d2 y z dx z dy (iii ) – x If f (x, y) = 0, then = .......Ans. (i) – 1 (ii) f dx 2 x dt y dt y 2 2 f q r – 2 pqs p t · (iv)2x + 4y (v) (vi) – x y q3 1.19 GEOMETRICAL INTERPRETATION OF z z AND y x (Gujarat, I Semester, Jan. 2009) Let z = f (x, y) be a surface S. Z Let y = k be a plane parallel to XZ – plane, passing through P (x, k, z) cutting the surface z = f (x, y) along the curve APB. B P This section APB is a plane curve whose equations are (x, k, z) z = f (x, y) A y = k X O z The slope of the tangent to this curve is given by . x Y z Similarly, is the slope of the tangent to the curve y of intersection of the surface z = f (x, y) with a plane parallel to YZ-plane. 1.20 TANGENT PLANE TO A SURFACE Let f (x, y, z) = 0 be the equation of a surface S. Now we wish to find out the equation of a tangent plane to S at the point P (x1, y1, z1). Let Q (x1 + x1, y1 + y1, z1 + z1) be a neighbouring point to P. Let the arc PQ be s and the chord PQ be c. The direction cosines of PQ are Z Q T P x y z , , O c c c x s y s z s Y . , . , . s c s c s c As s 0, Q P and PQ tends to a tangent line PT. The direction cosines of PT are dx dy dz , , ...(1) ds ds ds Differentiating F (x, y, z) = 0 w.r.t. ‘s’, we get F dx F dy F dz =0 ...(2) x ds y ds z ds From (1) and (2) it is clear that the tangent whose direction cosines are dx dy dz , , is ds ds ds Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ Partial Differentiation 45 perpendicular to a line having direction ratios F F F , , ...(3) x y z There are a number of tangent lines at P to the curves joining P and Q. All these tangents will be perpendicular to the line having direction ratios as given by (3). Hence all these tangent lines will lie in a plane known as tangent plane. Equation of tangent plane F F F + y – y1 + z – z1 x y z Equation of the normal to the plane. x – x1 y – y1 z – z1 = = F F F x y z x – x1 = 0 Example 49. Find the equation of the tangent plane and normal line to the surface x2 + 2 y2 + 3 z2 = 12 at (1, 2, – 1). Solution. F (x, y, z) = x2 + 2 y2 + 3 z2 – 12 F = 2 x, x F 4 y, y F F = 2, x y Hence the equation of the tangent plane at (1, 2 (x – 1) + 8 (y – 2) – 6 (z + 1) 2 x + 8 y – 6z = 24 x + 4y – 3z At the point (1, 2, – 1) F 6z z F 6 z 2, – 1) is = 0 = 12 8, x 1 y 2 z 1 x 1 y 2 z 1 Ans. 1 4 3 2 8 6 Example 50. Show that the surface x2 – 2 yz + y3 = 4 is perpendicular to any number of the family of surfaces x2 + 1 = (2 – 4a) y2 + a z2 at the point of intersection (1, – 1, 2). Solution. f (x, y, z) = x2– 2 yz + y3 – 4 = 0 ...(1) 2 2 2 F (x, y, z) = x + 1 – (2 – 4 a) y – az = 0 ...(2) f f f = 2x, 2z 3 y2 , 2y x y z Direction ratios to the normal of the tangent plane to (1) are 2 x, – 2 z + 3y2, – 2 y DRs at the point (1, – 1, 2) are 2, – 1, 2. Now differentiating (2), we get F F F = 2 x, 2 2 4 a y, 2 a z. x y z Direction ratios to the normal of the tangent plane to (2) are 2 x, (– 4 + 8 a) y, – 2az. DRs at the point (1, – 1, 2) are 2, 4 – 8a, – 4a Now l1 l2 + m1 m2 + n1 n2 = (2) (2) + (– 1) (4 – 8 a) + 2 (– 4 a) = 4 – 4 + 8 a – 8 a = 0. Hence, the given surfaces are perpendicular at (1, – 1, 2). Ans. Equation of normal is Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ Partial Differentiation 46 EXERCISE 1.8 1. Find the equation of tangent plane and the normal line to the surface x –1 y – 2 z – 3 6 3 2 2 2 Find the equations of the tangent plane and the normal to the surface z = 4 (1 + x + y2) at x2 y2 z6 (2, 2, 6). Ans. 4x + 4y – 3z + 2 = 0, 4 4 3 Find the equations of the tangent plane and the normal to the surface x2 y3 z 6 x2 y 2 z at (2, 3, – 1) Ans. 2x – 2y – z + 1 = 0, 2 2 1 2 3 Show that the plane 3x + 12y – 6z – 17 = 0, touches the conicoid 3x2 – 6y2 + 9z2 + 17 = 0. 2 Find also the point of contact. Ans. 1, 2, 3 Show that the plane ax + by + cz + d = 0 touches the surface px2 + qy2 + 2z = 0, a2 b2 if + 2 c d = 0. p q x y z = 6 at (1, 2, 3). 2. 3. 4. 5. Ans. 6x + 3y + 2z = 18, Applications of differential Calculus (Error, Jacobians, Taylor’s Series, Maxima and Minima) 1.21 ERROR DETERMINATION y lim We know that = x 0 x y = x Definitions: (i) dy dx dy dy approximately y = . x approximately dx dx x is known as absolute error in x. (ii) x is known as relative error in x. x x (iii) 100 is known as percentage error in x. x E2 . Find by using Calculus R the approximate percentage change in P when E is increased by 3% and R is decreased by 2%. (A.M.I.E., Summer 2001) E2 Solution. Here, we have P = log P = 2 log E – log R R On differentiating, we get Example 51. The power dissipated in a resistor is given by P = P 2 R E– P E R 100 P 100 E 100 R 2 – P E R P 100 E 100 R 2 (3) – (– 2) 8 2%, 2% Given, E P R Percentage change in P = 8% Ans. Example 52. The diameter and altitude of a can in the shape of a right circular cylinder are measured as 40 and 64 cm respectively. The possible error in each measurement is ± 5%. Find approximately the maximum possible error in the computed value for the volume and the lateral surface. Find the corresponding percentage error. Solution. Here we have, Diameter of the can (D) = 40 cm 100 Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ Partial Differentiation 47 100 D 100 h = = ± 5% D h V = r2h (D2 ) h 2 D h 4 4 2 log D log h 4 V 2D h = 0 V D h 2 D h V 100 100 2( 5) ( 5) 15 100 = D h V S=2rl= Dh log S = log + log D + log h log V = log Again Ans. D h S = 0 D h S D h S 100 100 ( 5) ( 5) 10 100 = D h S Example 53. The period T of a simple pendulum is Ans. 1 . 8 Find the maximum error in T due to possible errors upto 1% in l and 2% in g. (U.P. I semester winter 2003) l . Solution. We have, T = 2 g 1 1 log T = log 2 log l – log g 2 2 Differentiating, we get T 1 l 1 g – = 0 T 2 l 2 g g 1 l T 100 = 2 l 100 – g 100 T l g 100 = 1, 100 2 But l g 1 3 T 100 = 2 [1 2] 2 T Maximum error in T = 1.5% Ans. Example 54. A balloon is in the form of right circular cylinder of radius 1.5 m and length 4 m and is surmounted by hemispherical ends. If the radius is increased by 0.01 m and the length by 0.05 m, find the percentage change in the volume of the balloon. (U.P. I Sem., Dec., 2005, Comp 2002) Solution. Radius of the cylinder (r) = 1.5 m Length of the cylinder (h) = 4 m Volume of the balloon = Volume of cylinder + Volume of two hemispheres 1.5 m 2 2 4 2 3 3 2 3 Volume (V) = r h r r r h r 4m 3 3 3 4 2 2 V = 2 r r. h r . h 3r . r 3 T = 2 Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ Partial Differentiation 48 r [2 r. h r. h 4 r r ] 2. r. h r. h 4r. r V = 4 4 V r 2h r3 r h r2 3 3 2 0.01 4 1.5 0.05 4 1.5 0.01 = 4 1.5 4 (1.5) 2 3 0.08 0.075 0.06 0.215 = 63 9 V 100 0.215 21.5 100 2.389% = Ans. V 9 9 Example 55. In estimating the number of bricks in a pile which is measured to be (5m × 10m × 5m), count of bricks is taken as 100 bricks per m3. Find the error in the cost when the tape is stretched 2% beyond its standard length. The cost of bricks is ` 2,000 per thousand bricks. (U.P., I Semester, Winter 2000) Solution. Volume V = x y z log V = log x + log y + log z Differentiating, we get x y z V = x y z V 100 x 100 y 100 z V 100 = =2+2+2 x y z V 100 V =6 V 6V 6 (5 10 5) V= = 15 cubicmetre. 100 100 Number of bricks in V = 15 × 100 = 1500 1500 2000 3000 1000 Thus error in cost, a loss to the seller of bricks = ` 3000. Ans. Example 56. The angles of a triangle are calculated from the sides a, b, c. If small changes a, b, c are made in the sides, show that approximately a a – b. cos C – c. cos B A= 2 where is the area of the triangle and A, B, C are the angles opposite to a, b, c respectively. Verify that A + B + C = 0 (U.P., I Sem., Winter 2001, A.M.I.E.T.E., 2001) Solution. We know that Error in cost = b2 c 2 – a 2 cos A = 2bc 2 2 a = b + c2 – 2 b c cos A ...(1) Differentiating both sides of (1), we get 2a a = 2b b + 2 c c – 2b c cos A – 2 b c cos A + 2b c sin A A (approx.) a a = b b + c c – b c cos A – b c cos A + b c sin A A bc sin A A = a a – (b – c cos A) b – (c – b cos A) c 2 A = a a – (a cos C + c cos A – c cos A) b – (a cos B + b cos A – b cos A) c 1 2 bc sin A b cos C c cos B a Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ Partial Differentiation 49 2 A = a a – a b cos C – a c cos B = a (a – b cos C – c cos B) A= a [a – b. cos C – c. cos B] 2 ...(2) Similarly, b [b – c. cos A – a. cos C ] 2 c [c – a. cos B – b. cos A] C= 2 On adding (2), (3) and (4), we get B= ...(3) ...(4) 1 [(a – b cos C – c cos B ) a (b – a cos C – c cos A) b 2 + (c – a cos B – b cos A) c] 1 [(a – a) a (b – b) b (c – c) c] = 2 =0 [ b cos C + c cos B = a] Verified. Example 57. The height h and semi-vertical angle of a cone are measured, and from there A, the total area of the cone, including the base, is calculated. If h and are in error by small quantities h and respectively, find the corresponding error in the area. Show further that, if = , an error of + 1 per cent in h will be approximately compensated by an error of 6 – 19.8 in . (A.M.I.E.T.E., Summer 2003) A Solution. Let l be the slant height of the cone and r its radius l = h sec [ A + B + C] = r = h tan A = r2 + r l l h = h2 tan2 + (h tan ) (h sec ) = h2 [tan2 + tan sec ] A = 2 h h [tan2 + tan sec ] + h2[2 tan sec2 B + A = 2h [tan + sec ] tan . h + h2 sec2 r D C . . sec + tan sec tan ] [2 tan sec + sec2 + tan2]. sec . A = 2h [tan + sec ] tan . h + h2 [tan + sec ]2 sec . h (tan sec ) sec . A = h2 [tan + sec ] 2 tan h h 100 1, we get On putting A = 0, , 6 h 1 tan sec 0 = h 2 tan sec 2 tan 6 6 6 100 6 1 tan sec sec 0 = 2 tan 6 100 6 6 6 2 1 1 2 2 2 1 1 0 = 3 100 3 3 3 3 100 3 sec 6 6 2 2 3 3 Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ Partial Differentiation 50 1 1 2 1 3 1 = 100 100 3 3 100 3 3 180 9 60 degree – minutes = – 19.8 minutes Ans. 100 3 5 3 Example 58. Find the possible percentage error in computing the parallel resistance r of 1 = – three resistances r1, r2, r3 from the formula 1.2%. 1 1 1 1 = + + if r1, r2, r3 are each in error by plus r r1 r2 r3 1 1 1 1 = r r1 r2 r3 Solution. Here, Differentiating, we get 1 1 2 radius remains constant, prove that r12 dr1 – 1 r22 dr2 – 1 dr3 r32 r 1 100 dr 1 100 dr1 1 100 dr2 1 100 dr3 = r r r1 r1 r2 r2 r3 r3 1 1 1 1 1 1 = (1.2) (1.2) (1.2) (1.2) r1 r2 r3 r1 r2 r3 1 = 1.2 [From (1).] r 100 dr 1.2% Ans. r Example 59. If the sides and angles of a plane triangle vary in such a way that its circum – dr = – ...(1) da db dc + + = 0, where da, db, dc are small cos A cos B cos C increments in the sides, a, b, c respectively. Solution. From the sine rule, a b c = sin A sin B sin C a We know that R = 2 sin A , ...(1) Differentiating, we get a cos A R = – 2 sin 2 A A 1 R = 2 sin A a R R dA da By total differentiation dR = A a a cos A 1 . dA . da , 0=– 2 sin A 2 sin 2 A A R B O a C R being constant cos A 1 dA da sin A sin A Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ Partial Differentiation 51 a da . dA 2R. dA = sin A cos A da =2RdA cos A db Similarly, =2RdB cos B dc and =2RdC cos C Adding (1), (2) and (3), we have da db dc = 2R [dA + dB + dC] cos A cos B cos C [Using (1)] ...(1) ...(2) ...(3) ...(4) But in any triangle ABC, A + B + C = Hence, dA + dB + dC = 0 Putting value of dA + dB + dC = 0 in (4), we get da db dc da db dc 0 = 2 R (0) 0 cos A cos B cos C cos A cos B cos C Proved. Example 60. Compute an approximate value of (1.04)3.01. Solution. Let We have Here, let Now f (x, y) = xy f f x y log x y xy –1 , y x x = 1, x = 0.04, y = 3, x = 0.01 ...(1) f f df = x dx y dy ...(2) = y x y – 1 x y log x Substituting the values from (1) in (2), we get d f = (3) (1)3–1 (0.04) + (1)3 log (1) (0.01) = 0.12 (1.04)3.01 = f (1, 3) + d f = 1 + 0.12 = 1.12 Ans. 1 Example 61. Find [(3.82)2 + 2(2.1)3 ] 5 1 2 3 Solution. Let f (x, y) = ( x 2 y ) 5 Taking x = 4, x = 3.82 – 4 = – 0.18 y = 2, x = 2.1 – 2 = 0.1 4 4 – – 1 2 2 8 1 1 f 3 3 = [ x 2 y ] 5 (2 x ) (4) [16 2(2) ] 5 5 5 5 x 16 10 4 4 – – f 1 2 6 24 1 3 3 2 2 3 = [ x 2 y ] 5 (6 y ) (2) [16 2(2) ] 5 y 5 5 5 16 10 By total differentiation, we get f f df x y = 1 (– 0.18) 3 (0.1) – 0.018 0.03 0.012 x y 10 10 Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ Partial Differentiation 52 1 [(3.82)2 2(2.1)3 ]5 = f (4, 2) + df 1 = [(4)2 2(2)3 ]5 0.012 2 0.012 2.012 Ans. EXERCISE 1.9 1. If the density of a body be inferred from its weights W, in air and water respectively, show that – W . . the relative error in due to errors W in W, is W – W W – 2. The period of oscillation of a pendulum is computed by the formula T = 2 l . g Show that the percentage error in T = 1 [% error in l – % error in g] 2 1 , find the error in the determination of T.. 160 (Given g = 981 cm/sec2) Ans. – 0.00153 3. The indicated horse power I of an engine is calculated from the formula. I = PLAN/33000 If l = 6 cm and relative error in g is equal to 2 d . Assuming that errors of r percent may have been made in measuring P. L, N and 4 d. Find the greatest possible error in I. Ans. 5 r % The dimensions of a cone are radius 4 cm, height 6 cm. What is the error in its volume if the scale used in taking the measurement is short by 0.01 cm per cm. Ans. 0.96 cm3. The work that must be done to propel a ship of displacement D for a distance s in time t is proportional to s2 D2/3 t2. Find approximately the percentage increase of work necessary when the displacement is increased 14 by 1%, the time is diminished by 1% and the distance is increased by 3%. Ans. % 3 The power P required to propel a ship of length l moving with a velocity V is given by P = kV3 t2. Find the percentage increase in power if increase in velocity is 3% and increase in length is 4%. Ans. 17% In estimating the cost of a pile of bricks measured as 2m × 15 m × 1.2 m, the tape is stretched 1% beyond the standard length if the count is 450 bricks to 1 m3 and bricks cost ` 1300 per 1000, find the approximate error in the cost. Ans. ` 631.80 In estimating the cost of a pile of bricks measured as 6 × 50 × 4, the tape is stretched 1% beyond the standard length. If the count is 12 bricks to ft 3, and bricks cost ` 100 per 1000, find the approximate error in the cost. (U.P. I Sem., Dec. 2004) Ans. 720 bricks, ` 25.20 The sides of a triangle are measured as 12 cm and 15 cm and the angle included between them as 60°. If the lengths can be measured within 1% accuracy while the angle can be measured within 2% accuracy. Find the percentage error in determining (i) area of the triangle (ii) length of opposite side of the triangle. (A.M.I.E.T.E., Winter 2002) The voltage V across a resistor is measured with error h, and the resistance R is measured with an where A = 4. 5. 6. 7. 8. 9. 10. error k. Show that the error in calculating the power W(V, R) = V2 generated in the resistor is R V (2 Rh – V k ). If V can be measured to an accuracy of 0.5 p.c. and R to an accuracy of 1 p.c., R2 what is the approximate possible percentage error in W ? Ans. Zero percent Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ Partial Differentiation 53 11. Find the possible percentage error in computing the parallel resistance r of two resistance r1 and r2 1 1 1 from the formula , where r1 and r2 are both is error by + 2% each. Ans. 2% r r1 r2 12. In the manufacture of closed cylindrical boxes with specified sides a, b, c (a b c), small changes of A%, B%, C% occurred in a, b, c, respectively from box to box from the specified dimension. However, the volume and surface area of all boxes were according to specification, show that: A B C a (b – c ) b(c – a ) c (a – b ) 13. Find the percentage error in calculating the area of ellipse x2/a2 + y2/b2 = 1, when error of + 1% is made in measuring the major and minor axes. Ans. 2% (U.P., I Sem, Jan 2011) 1 2 2 14. If f = x y z 10 , find the approximate value of f, when x = 1.99, y = 3.01 and z = 0.98. Ans. 107.784 15. A diameter and altitude of a can in the form of 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 value computed for the volume and lateral surface. (A.M.I.E., Summer 2001) Ans. 5.0336 cm3, 3.146 cm2 16. Prove that the relative error of a quotient does not exceed the sum of the relative errors of the dividend and the divisor. (A.M.I.E., Winter 2001) 1.22 JACOBIANS If u and v are functions of the two independent variables x and y, then the determinant u u x y v v x y is called the jacobian of u, v with respect to x, y and is written as u, v (u , v ) or J ( x, y ) x, y Similarly, the jacobian of u, v, w with respect to x, y, z is u x (u, v, w) v = ( x, y , z ) x w x u y v y w y u z v z w z Example 62. If x = r cos , y = r sin ; evaluate Solution. We have, ( r , ) ( x, y ) , and ( x, y ) ( r , ) y = r sin y = sin r y = r cos x = r cos , x = cos , r x = – r sin , x x ( x, y ) r cos r sin = y y = (r , ) sin r cos r Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ Partial Differentiation 54 = r cos2 + r sin2 = r (cos2 + sin2 ) = r y Now, r2 = x2 + y2, = tan 1 x y r x y = , = 2 = x r x y2 x r2 x r x y = , y = 2 = 2 y x y2 r r r r x y x y 1 r r x2 y2 x2 y 2 r2 ( r , ) = = = 3 3 = = 3 = Ans. 3 y x r ( x, y ) r r r r x y r2 r2 1 ( x, y ) (r , ) Note : = r = 1 r (r , ) ( x, y ) Example 63. If x = a cosh cos , y = a sinh sin , show that ( x, y ) a2 ( cosh 2 cos 2 ) = (, ) 2 Solution. Here, we have, x = a cosh cos y = a sinh sin x x a sinh cos a cosh sin ( x, y ) = = a cosh sin a sinh cos (, ) y y 2 = a sinh cos cosh sin = a2 [sinh2 cos2 + cosh2 sin2 ] cosh sin sinh cos = a2 [sinh2 (1 – sin2) + (1 + sinh2 ) sin2 ] = a2 [sinh2 – sinh2 sin2 + sin2 + sinh2 sin 2 ] a2 a2 [ cosh 2 1 1 cos 2 ] = [ cosh 2 cos 2 ] Proved. = a2 [sinh2 + sin2 ] = 2 2 x1 x2 x2 x3 x3 x1 Example 64. If y1 = , y2 = , y3 = . x3 x1 x2 Show that the Jacobian of y1, y2, y3 with respect to x1, x2, x3 is 4. (U.P. I Sem. Jan 2011; 2004, Comp. 2002, A.M.I.E., Summer 2002, 2000, Winter 2001) x2 x3 x3 x1 x1 x2 Solution. Here, we have y1 = x , y2 = x , y3 = x 1 2 3 x2 x3 y1 x1 y1 x2 y1 x3 y2 ( y1 , y2 , y3 ) = x1 ( x1 , x2 , x3 ) y3 x1 y2 x2 y2 = x3 x3 x2 y3 x2 y3 x3 x2 x3 = 1 x12 x22 x32 x12 x3 x1 x2 x1 x3 x1 x1 x2 x22 x1 x3 x2 x3 x3 x1 x1 x2 x2 x3 x3 x1 x1 x2 = x1 x2 x32 x12 x22 x32 x12 x22 x32 1 1 1 1 1 1 1 1 x2 x3 x3 x1 x1 x2 = – 1 (1 – 1) – 1 (– 1 – 1) + 1 (1 + 1) = 0 + 2 + 2 = 4 1 Proved. Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ Partial Differentiation 55 Example 65. If x = r sin cos , y = r sin sin , z = r cos , ( x, y , z ) Show that (r , , ) = r2 sin . Solution. We have, x = r sin cos , x = sin cos , r (U.P., I Semester, Winter 2000) x = r cos cos , x =– x r ( x, y , z ) y = (r , , ) r z r = r2 sin = r2 sin = r2 sin = r2 sin r sin sin , x y z If u = x2, v = y2, find 2. If u = 4. 5. z = r cos z = cos r y = r cos sin , z = – r sin y = r sin cos , z =0 x sin cos r cos cos r sin sin y = sin sin r cos sin r sin cos cos r sin 0 z sin cos cos cos sin 2 = r sin sin sin cos sin cos cos sin 0 [sin cos (0 + sin cos ) – cos cos (0 – cos cos ) – sin (– sin2 sin –cos2 sin )] [sin2 cos2 + cos2 cos2 + sin2 sin2 + cos2 sin2 ] [(sin2 + cos2 ) cos2 + (sin2 + cos2 ) sin2 ] [cos2 + sin2 ] = r2 sin Ans. EXERCISE 1.10 1. 3. y = r sin sin , y = sin sin , r (u , v) ( x, y ) Ans. 4xy yx (u , v) and v = tan–1 y – tan–1 x , find Ans. 0 1 xy ( x, y ) ( u , v , w) If u = xyz, v = xy + yz + zx, w = x + y + z, compute Ans. (x – y) (y – z) (z – x) ( x, y , z ) ( x, y , z ) . If x = r cos , y = r sin , z = z find ( r , θ, z) ( u1, u2 ,......un 1 ) xn 1 x1 x2 If u1 = x , u2 = x , ...... un–1 = and x12 x22 x32 ...... xn2 = 1 find ( x , x ,...... x ) x n n n 1 2 n 1 1 Ans. 6. 7. 8. xnn 1 ( y1 , y2 , y3 ) , y = (1 – x1), y2 = x1 (1 – x2), y3 = x1x2 (1 – x3) Ans. x12 x2 ( x1, x2 , x3 ) 1 x y (u , v, w) z If u = ,v= ,w= then show that =0 yz zx x y ( x, y , z ) Fill in the blanks ( x, y ) (i) If x = r cos , y = r sin , then the value of Jacobian is ........ Ans. r ( r ,θ) Find the value of Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ Partial Differentiation 56 (ii) If u = x (1 – y), v = xy, then the value of the Jacobian (u , v) = ........ ( x, y ) Ans. x 1.23 PROPERTIES OF JACOBIANS (1) First Property If u and v are the functions of x and y, then (u, v ) ( x, y ) =1 ( U. P. I Semester Dec. 2005 ) ( x , y ) (u , v ) Proof. Let u = f (x, y) ...(1) v = (x, y) ...(2) u u x x x y u v (u, v ) ( x, y ) = v v y y ( x , y ) (u , v ) x y u v On interchanging the rows and columns of second determinant u x u y x u y u u x u y x u y u = v v x y v x v y x y v v x u y u On differentiating (1) and (2) w. r. t. u and v, we get = u x u y x v y v v x v y x v y v ...(3) u u x u y 1 u x u y u u u x u y 0 v x v y v ...(4) v v x v y 1 v x v y v v v x v y 0 u x u y u On making substitutions from (4) in (3), we get 1 0 (u, v ) ( x, y ) = =1 Proved. 0 1 ( x, y ) (u, v ) uv (u , v ) Example 66. If x = uv, y = , find . uv ( x, y ) x x y y u u v v Solution. Here it is easy to find , , , . But to find , , , is u v u v x y x y ( x, y ) comparatively difficult. So we first find ( u, v ) x x v u u v 1 1 uv 4u v uv ( x, y ) ( 2 2) = 2v 2u = = = = 2 2 y y ( u v) ( u v )2 ( u , v) ( u v) 2 2 (u v)2 ( u v)2 u v ( u , v ) ( x, y ) ( u, v ) 4u v ( u , v ) ( u v) 2 But = 1 = 1 Ans. ( x, y ) ( u , v ) ( x, y ) ( u v )2 ( x, y ) 4u v Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ Partial Differentiation 57 ( x, y, z ) . ( u, v, w) Solution. Since u, v, w are explicitly given, so first we evaluate (U.P. I Sem., Winter 2002) ( u, v, w) J = ( x, y, z ) Example 67. If u = xyz, v = x2 + y2 + z2, w = x + y + z, find J = u u u x y z yz zx xy v v v J = = 2x 2y 2 z x y z 1 1 1 w w w x y z = yz ( 2y – 2z ) – zx ( 2x – 2z ) + xy (2x – 2y) = 2 [yz ( y – z ) – zx ( x – z) + xy ( x – y)] = 2 [x2y – x2z – xy2 + xz2 + y2z – yz2] = 2 [ x2 ( y – z ) – x ( y2 – z2 ) + yz ( y – z)] = 2 ( y – z ) [x2 – x ( y + z ) + yz] = 2 ( y – z) [ y ( z – x) – x ( z – x )] = 2 ( y – z ) ( z – x ) ( y – x ) = – 2 ( x – y) ( y – z ) ( z – x ) Hence, by JJ = 1, we have ( x, y , z ) 1 J= = Ans. ( u , v, w ) 2( x y ) ( y z ) ( z x ) EXERCISE 1.11 ( x, y ) ( u, v ) 1. Given u = x2 – y2, v = 2xy, calculate 2. If x = uv, y = 3. If x = r sin cos , y = r sin sin , z = r cos , find 4. Verify JJ = 1, if x = uv, y = 5. 6. Ans. uv ( u, v) , find u v ( x, y ) Ans. ( r,θ,φ) ( x, y , z ) Ans. 1 2 4( x y 2 ) ( u v)2 4 uv 1 r 2 sin θ u v Verify JJ = 1, if x = ev sec u, y = ev tan u. Verify JJ = 1, if x = sin cos , y = sin sin (2) Second Property (Chain Rule) If u, v are the functions of r, s where r, s are functions of x, y, then ( u, v ) ( r , s ) ( u , v) = ( r , s ) ( x, y ) (U.P. I Sem. Jan 2011) ( x, y ) u u r r r s x y ( u , v) ( r , s ) Proof. = v v s s ( r, s ) ( x, y) r s x y On interchanging the columns and rows in second determinant u u r s u r u s u r u s r s x x r x s x r y s y = v v r s = v r v s v r v s r s y y r x s x r y s y Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ Partial Differentiation 58 u x = v x Similarly, u y (u , v) = v ( x, y ) y (u, v, w) (u, v, w) (r , s, t ) = ( x, y , z ) ( r , s , t ) ( x, y, z ) Example 68. Find the value of the Jacobian Proved. (u, v) , wheree u = x2 – y2, v = 2x y and (r , ) x = r cos , y = r sin . Solution. u = x2 – y2, v = 2 x y u u 2x 2 y x y (u , v) = = = 4 (x2 + y2) = 4 r2 2 y 2x v v ( x, y ) x y x x cos r sin ( x, y ) r = y y = = r cos2 + r sin2 = r ( r , ) sin r cos r (u, v) (u , v ) ( x , y ) = = 4r2 . r = 4r3 Ans. (r , ) ( x, y ) (r , ) EXERCISE 1.12 1. If u = ex cos y, v = ex sin y, where x = lr + sm, y = mr – sl, verify 2 2. If u = x (1 r ) w = z (1 Show that 1 r2 ) 2 1 2, 2 v y (1 r ) ( u , v ) ( x, y ) (u , v) = ( x, y ) ( r , s) (r, s) 1 2 where r 2 x2 y 2 z 2 5 (u, v, w) = (1 r 2 ) 2 ( x, y , z ) (Q. Bank, U. P. 2001) 3. If u = x + y + z, u2 v = y + z, u3 w = z, show that (u, v, w) = u–5 ( x, y , z ) Hint. Put r = x + y + z, s = y + z, t = z u = r u2 v = s u3w = t (u, v, w) (u , v, w) ( r , s, t ) ( r , s, t ) 1 = = ( x, y , z ) ( r , s , t ) ( x, y , z ) ( r , s, t ) ( x, y , z ) (u , v, w) ( x, y , z ) 4. If u = x + y + z, uv = y + z, uvw = z. Evaluate ( u , v , w) 5. If u3 + v3 = x + y, u2 + v2 = x3 + y3, show that (3) ( U. P. I Sem. Winter 2003 ) Ans. u2v (u , v ) 1 ( y2 x2 ) = ( x, y ) 2 u v (u v) Third Property If functions u, v, w of three independent variables x, y, z are not independent, then (u , v, w) =0 ( x, y , z ) Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ Partial Differentiation 59 Proof. As u, v, w are not indepentent, then f (u, v, w) = 0 Differentiating (1) w.r.t x, y, z, we get f u f v f w =0 u x v x w x ...(1) ...(2) f u f v f w =0 u y v y w y ...(3) f u f v f w =0 u z v z w z ...(4) Eliminating f f f , , from (2), (3) and (4), we have u v w u x u y u z v x v y v z w x w y w z =0 On interchanging rows and columns, we get u x v x w x u y v y w y u z v =0 z w z ( u , v, w ) =0 Proved. ( x, y , z ) Converse (The sufficient condition) ( u , v, w ) If it is given that = 0 and u, v, w are not independent of one another then they are ( x, y , z ) connected by a relation f (u, v, w) = 0. Example 69. If u = x y + y z + z x, v = x2 + y2 + z2 and w = x + y + z, determine whether there is a functional relationship between u, v, w and if so, find it. Solution. We have, u = x y + y z + z x, v = x2 + y2 + z2, w = x + y + z u u u x y z yz zx x y ( u , v, w ) v v v 2y 2z = = 2x ( x, y , z ) x y z 1 1 1 w w w x y z x y z x yz x y z R R + R yz zx x y l = 2 x y z 1 1 1 = 2 x y z 1 1 1 1 2 Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ Partial Differentiation 60 1 1 1 = 2 ( x y z ) x y z = 0 (R1 = R3) 1 1 1 Hence, the functional relationship exists between u, v and w. Now, w2 = (x + y + z)2 = x2 + y2 + z2 + 2(x y + y z + z x) w2 = v + 2u w2 – v – 2u = 0 which is the required relationship. Ans. Example 70. Verify whether the following functions are functionally dependent, and if so, find the relation between them. xy , v tan 1 x tan 1 y u= 1 x y u 1 y2 1 x2 y 1 1 (1 xy )2 (1 xy )2 Solution. = = =0 v 1 1 (1 xy )2 (1 xy )2 y 1 x2 1 y2 Hence u, v are functionally related. 1 x y tan–1 x + tan–1 y = tan 1 xy v = tan– 1 u u = tan v. Ans. EXERCISE 1.13 u x (u , v) = v ( x, y ) x 1. Verify whether u = x y , x y v x y are x functionally dependent, and if so, find the relation between them. 2. 4. 5. xy , x y v xy Ans. 4v = 1– u2 ( x y) 2 Are x + y – z, x – y + z, x2 + y2 + z2 – 2yz functionally dependent ? If so, find a relation between them. Ans. u2 + v2 = 2w 2 2 2 3 3 3 If u = x + y + z, v = x + y + z , w = x + y + z – 3xyz, prove that u, v, w are not independent and find the relation between them. Ans. 2w = u (3v – u2) Are the following two functions of x, y, z functionally dependent ? If so find the relation between them. u 6. 2v v Determine functional dependence and find relation between u 3. Ans. u = x y , xz If u = v xz yz Ans. v = 1 1u y z y ( x y z) x y ,v= ,w= , show that u, v, w are not independent and find the x xz z relation between them. (U.P., Ist Semester, 2009) 1.24 JACOBIAN OF IMPLICIT FUNCTIONS The variables x, y, u, v are connected by implicit functions f1 (x, y, u, v) = 0 f2 (x, y, u, v) = 0 where u, v are implicit functions of x, y. Ans. uv – w = 1 ... (1) ... (2) Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ Partial Differentiation 61 Differentiating (1) and (2) w.r.t. x and y, we get f1 f1 u f1 v =0 x u x v x f1 f1 u f1 v =0 y u y v y f 2 f 2 x u f 2 f 2 y u Now, we have u f 2 x v u f 2 y v ... (3) ... (4) v =0 x v =0 y f1 ( f1 , f 2 ) (u, v ) u = (u , v ) ( x, y ) f 2 u f1 u = f 2 u ... (5) ... (6) u f1 x v f 2 v v x u f1 x v u f 2 x v 2 = ( 1) v x v x u y v y f1 u f 2 u u f1 y v u f 2 y v v y f1 f 1 x y = v f f 2 2 y x y [From (3), (4), (5), (6)] ( f1 , f 2 ) ( x, y ) (u, v) 2 ( f1 , f 2 ) / ( x, y ) = ( 1) ( f1 , f 2 ) / (u , v ) ( x, y ) In general, the variables x1, x2, ... xn are connected with u1, u2, ... un implicitly as f1(x1, x2, ... xn, u1, u2, ... un) = 0, f2(x1, x2, ... xn, u1, u2, ... un) = 0 ... ... ... ... ... fn (x1, x2, ... xn, u1, u2, ... un) = 0 Then we have (u1 , u2 , ... un ) n ( f1 , f 2 , ... f n ) / ( x1 , x2 , ... xn ) = ( 1) ( x1 , x2 , ... xn ) ( f1 , f 2 , ... f n ) / (u1 , u2 , ... un ) Example 71. If x2 + y2 + u2 – v2 = 0 and uv + xy = 0, prove that Solution. Let But (u , v ) x 2 y 2 ( x, y ) u 2 v 2 f1 = x2 + y2 + u2 – v2, f2 = uv + xy f1 f1 x y 2x 2 y ( f1 , f 2 ) = = = 2 (x2 – y2) f 2 f 2 y x ( x, y ) x y f1 f1 2u 2v ( f1 , f 2 ) u v = = = 2 (u2 + v2) f 2 f 2 (u , v ) v u u v ( f1 , f 2 ) x2 y2 (u , v) 2 (x2 y2 ) ( x, y ) = (1)2 = = ( f1 , f 2 ) ( x, y ) u 2 v2 2 (u 2 v 2 ) (u , v ) Proved. Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ Partial Differentiation 62 Example 72. If u3 + v + w = x + y2 + z2, u + v3 + w = x2 + y + z2, u + v + w3 = x2 + y2 + z, prove that 1 4 ( xy yz zx ) 16 xyz (u, v, w) = ( x, y , z ) 2 3 (u 2 v 2 w2 ) 27 u 2 v 2 w2 Solution. Let Now, and f1 = u3 + v + w – f2 = u + v3 + w – f3 = u + v + w3 – 1 2 y ( f1 , f 2 , f 3 ) = 2 x 1 ( x, y , z ) 2x 2 y ( f1 , f 2 , f 3 ) = (u , v, w) 3u 2 1 2 1 3v 1 1 x – y2 – z2 x2 – y – z2 x2 – y2 – z 2 z 2 z = – 1 + 4 (yz + zx + xy) – 16 xyz 1 1 1 = 2 – 3 (u2 + v2 + w2) + 27 u2 v2 w2 3w2 (u, v, w) 1 4 ( yz zx xy ) 16 xyz 3 ( f1 , f 2 , f 3 ) / ( x, y , z ) = (1) = Proved. ( x, y , z ) ( f1 , f 2 , f 3 ) / (u, v, w) 2 3 (u 2 v 2 w2 ) 27 u 2 v 2 w2 ( x, y , z ) Example 73. If x + y + z = u, y + z = uv, z = uvw, show that (u, v, w) = u2 v Solution. Let f1 = x + y + z – u f2 = y + z – uv f3 = z – uvw f1 x f 2 ( f1 , f 2 , f 3 ) = x ( x, y , z ) f3 x f1 y f 2 y f3 y f1 z 1 1 1 f 2 = 0 1 1 =1 z 0 0 1 f 3 z f1 u ( f1 , f 2 , f 3 ) f 2 = (u , v, w) u f3 u f1 v f 2 v f 3 v f1 w 1 0 0 f 2 u 0 = – u2 v = v w vw uw uv f 3 w ( f1 , f 2 , f3 ) u2 v ( x, y , z ) (u, v, w) 3 But = (–1) ( f , f , f ) = = u2 v Proved. 1 (u, v, w) 1 2 3 ( x, y , z ) Example 74. If u, v, w are the roots of the equation ( – x)3 + ( – y)3 + ( – z)3 = 0 in find (u , v, w) . ( x, y , z ) (U.P. I Sem. Jan, 2011; Winter 2001) Solution. ( – x)3 + ( – y)3 + ( – z)3 = 0 Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ Partial Differentiation 63 3 3 – 3(x + y + z) 2 + 3(x2 + y2 + z2) – (x3 + y3 + z3) = 0 Sum of the roots = u + v + w = x + y + z Product of the roots = uv + vw + wu = x2 + y2 + z2 ... (1) ... (2) 1 3 ( x y3 z3 ) 3 Equations (1), (2) and (3) can be rewritten as f1 = u + v + w – x – y – z f2 = uv + vw + wu – x2 – y2 – z2 uvw = f3 = uvw f1 x ( f1 , f 2 , f 3 ) f 2 = x ( x, y , z ) f3 x 1 1 = (1) ( 2) (1) x y 2 2 x = 2 ( x y) ( y z) y 1 3 ( x y 3 z3 ) 3 f1 f1 y z 1 f 2 f 2 = 2x y z x2 f3 f 3 y z 1 0 z = 2 x y z 2 2 x y 0 0 1 1 1 z 1 y2 y z z2 1 yz 2 1 2 y 2z 0 2 ... (3) z 2 z 2 C1 C1 C2 C2 C2 C3 2 ( x y) ( y z) ( y z x y) 2 f1 u f 2 ( f1 , f 2 , f 3 ) = u (u , v, w) f3 u x y y z z 2 ( x y) ( y z) ( z x) f1 f1 v w 1 1 1 f 2 f 2 v w u w u v = v w vw wu uv f3 f 3 v w = 0 0 0 0 vu wv 1 C1 C1 C2 u v C2 C2 C3 w ( v u ) u ( w v) uv 1 = ( v u ) ( w v ) 1 1 u v ( v u ) ( w v ) ( u w) w u uv (u v) (v w) ( w u ) ( f1 , f 2 , f3 ) 2 ( x y ) ( y z ) ( z x) (u, v, w) 2 ( x y) ( y z) ( z x) ( x, y , z ) =– =– = – (u v) (v w) ( w u ) ( x, y , z ) ( f1 , f 2 , f3 ) (u v) (v w) (w u ) (u, v, w) Ans. Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ Partial Differentiation 64 EXERCISE 1.14 1 y 2 x2 (u , v) = 2 2 uv(u v ) ( x, y ) y + z, u2 + v2 + w2 = x3 + y3 + z3, u + v + w = x2 + y2 + z2, ( x y) ( y z) ( z x) = (u v ) (v w) ( w u ) y z , w , show that 2 1 r 1 r2 1. If u3 + v3 = x + y, u2 + v2 = x3 + y3, then prove that 2. If u3 + v3 + w3 = x + ( u , v , w) show that ( x, y , z ) x , v If u = 1 r2 3. 4. 1 ( u , v , w) = where r2 = x2 + y2 + z2 ( x, y , z ) (1 r 2 )5 2 If u1 = x1 + x2 + x3 + x4, u1u2 = x2 + x3 + x4, u1u2u3 = x3 + x4, u1u2u3u4 = x4 ( x1 , x2 , x3 , x4 ) show that = u13 u22 u3 (u1 , u 2 , u3 , u4 ) 5. If u, v, w are the roots of the equation in and x y z ( x, y , z ) = 1, then find . aλ bλ cλ (u , v, w) (u v ) (v w) ( w u ) Ans. (a b) (b c ) (c a ) 1.25 PARTIAL DERIVATIVES OF IMPLICIT FUNCTIONS BY JACOBIAN Given f1 (x, y, u, v) = 0, f2 (x, y, u, v) = 0 f1 u f1 v f1 1 = 0 u x v x x f 2 u f 2 v f 2 1 = 0 u x v x x Solving (1) and (2) , we get u x f1 f 2 f1 f 2 v x x v = v x f1 f 2 f1 f 2 x u u x f1 u v = f1 x u and if, f 2 f1 x x f f 2 1 v v = ... (1) ... (2) 1 f1 f 2 f1 f 2 u v v u ( f1 , f 2 ) f 2 ( x, v ) v =– f 2 ( f1 , f 2 ) u (u , v ) ( f1 , f 2 ) f1 f 2 f1 f 2 ( x, u ) v = x u u x = ( f1 , f 2 ) f1 f 2 f1 f 2 x (v , u ) u v v u f1 (x, y, z, u, v, w) = 0, f2 = (x, y, z, u, v, w) = 0 f3 (x, y, z, u, v, w) = 0 x ( f1 , f 2 , f3 ) / (u, y, z ) = u ( f1 , f 2 , f3 ) / ( x, y, z ) and so on. Note. First we write the Jacobian in the denominator and then we write the Jacobian in the numerator by replacing x by u. Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ Partial Differentiation 65 u Example 75. Use Jacobians to find if : x v 2 2 2 u + xv – xy = 0 and u + xyv + v2 = 0 Solution. Let f1 = u2 + xv2 – xy, f2 = u2 + xyv + v2 f1 f1 u v ( f1 , f 2 ) = f f 2 = 2 (u, v) u v f1 f1 ( f1 , f 2 ) x v = = f 2 f 2 ( x, v) x v = x y v2 + 2v3 – 2u 2 xv xy 2v = 2uxy + 4uv – 4uxv 2u v2 y 2 xv yv xy 2v xy2 – 2yv – 2xyv2 = – xyv2 + 2v3 – xy2 – 2yv u ( f1 , f 2 ) / ( x, v) xyv 2 2v 3 xy 2 2 yv = = x ( f1 , f 2 ) / (u, v) 2uxy 4uv 2 xuv Proved. Example 76. If u = x + y2, v = y + z2, w = z + x2, prove that x 1 = u 1 8 xyz Solution. (i) Here (i) Now (ii) Also find 2x . u 2 f1 u – x – y2, f2 = v – y – z2 f3 w – z – x2. 1 2y 0 ( f1 , f 2 , f 3 ) = 0 1 2 z = 1 ; (u , y , z ) 0 0 1 ( f1 , f 2 , f 3 ) = ( x, y , z ) 1 2 y 0 0 1 2 z = – 1 (1 + 0) + 2y (0 – 4zx) = – 1 – 8 xyz 0 1 2x x ( f1 , f 2 , f 3 ) / (u, y , z ) = u ( f1 , f 2 , f 3 ) / ( x, y , z ) 1 x 1 = = 1 8 xyz u 1 8 xyz (ii) 2x u 2 = u = = We have, ( f1 , f 2 , f 3 ) = ( x, u , z ) Proved. 1 1 x (1 8 xyz ) = u 1 8 xyz = 2 u (1 8 xyz ) u 1 y z x 0 8 u y z u z x u x y (1 8 xyz ) 2 8 y z x u y z u z x u x y (1 8 xyz ) 1 1 0 2 ... (1) 0 0 2 z = 4 zx ; 2 x 0 1 Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ Partial Differentiation 66 ( f1 , f 2 , f 3 ) = ( x, y , u ) y u y u z u z u Substituting in (1), we have Now, 2x u 2 1 2y 1 0 1 2x 0 0 = – 2x. 0 ( f1 , f 2 , f 3 ) / ( x, u, z ) ( f1 , f 2 , f 3 ) / ( x, y , z ) 4 zx 4 zx = = 1 8 xyz 1 8 xyz = ( f1 , f 2 , f 3 ) / ( x, y, u ) ( f1 , f 2 , f3 ) / ( x, y , z ) 2x 2x = = 1 8 xyz 1 8 xyz = = = yz 4 z 2 x2 2 x2 y (1 8 xyz ) 1 8 xyz 1 8 xyz 1 8 xyz 8 2 8 ( yz 4 z 2 x 2 2 x 2 y ) Ans. (1 8 xyz )3 Example 77. Given, x = u + v + w, y = u2 + v2 + w2, z = u3 + v3 + w3 vw u show that = (u v ) (u w ) x Solution. Let f1 = u + v + w – x = 0 f2 = u2 + v2 + w2 – y = 0 f3 = u3 + v3 + w3 – z = 0 ( f1 , f 2 , f 3 ) / ( x, v, w) u = x ( f1 , f 2 , f 3 ) / (u, v, w) 1 1 ( f1 , f 2 , f 3 ) = 0 2v ( x, v, w) 0 3 v2 1 1 ( f1 , f 2 , f 3 ) = 2u 2v (u , v, w) 3 u 2 3 v2 Thus from (1), (2) and (3), we get ... (1) 1 1 1 2 w = 6 vw = 6 vw (v – w) v w 2 3w ... (2) 1 2 w = 6 (v – u) (w – u) (w – v) 3w ... (3) 2 vw 6 vw (v w) u = = (u v ) (u w ) x 6 (v u ) ( w u ) ( w v ) Proved. EXERCISE 1.15 1. If u2 + xv2 – uxy = 0, v2 – xy2 + 2uv + u2 = 0 find 2. If x = u + e–v sin u, y = v + e–v cos u, find 3. u . x Ans. (v 2 uy ) (u v ) xvy 2 (u v ) (2 u xy 2 xv) u v e v sin u u v . Ans. , y x 1 e 3v y x u u v v (u , v) ; ; ; and If x = u2 – v2, y = 2uv, find x y x y ( x, y ) u v v u 1 Ans. ; ; ; ; 2 (u 2 v3 ) 2 (u 2 v 3 ) 2 (u 2 v 2 ) 2 (u 2 v 2 ) 4 (u 2 v 2 ) Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ Partial Differentiation 67 u x 4. If u3 + xv2 – uy = 0, u2 + xyv + v2 = 0 , find 5. If u2 + xv2 = x + y, v2 + yu2 = x – y, find 6. If u = xyz, v = x2 + y2 + z2, w = x + y + z find 7. If u = x2 + y2 + z2, v = xyz, find u v , x y x u x u Ans. xyv2 2 v3 2 x y u 2 x y2 6 u 2 v 2 v y 4 x u v Ans. 1 x v2 1 + y + u 2 , 2 u (1 xy ) 2v (1 xy ) Ans. 1 ( x y) ( x z) Ans. x 2 (2 x 2 y 2 ) 1.26 TAYLOR’S SERIES OF TWO VARIABLES If f (x, y) and all its partial derivatives upto the nth order are finite and continuous for all points (x, y), where a x a + h, b y b + k 2 3 1 1 Then f (a + h, b + k) = f (a , b ) h k f h k f h k f ... x y 2! x y 3! x y Proof. Suppose that f (x + h, y + k) is a function of one variable only, say x where y is assumed as constant. Expanding by Taylor’s Theorem for one variable, we have f (x + x, y + y) = f ( x, y y ) x f ( x, y y ) (x )2 2 f ( x, y y ) ...... x 2! x 2 Now expanding for y, we get f ( x, y ) (y )2 2 f ( x, y ) f ( x, y ) ..... x f ( x, y ) y ...... = f ( x, y ) y y 2 2! x y y y 2 2 ( x ) f ( x, y ) ..... ..... f ( x, y ) y 2! x 2 y 2 2 f ( x, y ) f ( x, y ) ( y ) . ..... = f ( x, y ) y y 2! y2 2 f ( x, y ) f ( x, y ) ( x ) 2 2 f ( x , y ) x y. ... ... 2 xy 2! x x 2 2 f ( x, y) f ( x , y ) 1 f ( x, y ) 2 f ( x, y ) y. 2 x. y. (x ) = f ( x , y ) x 2 x y 2! x y x 2 f ( x, y ) ( y ) 2 . ... y2 f f 1 2 2 f 2 f 2 f f (a + h, b + k) = f (a, b) h k 2h k k2 ..... h 2 y 2! x x y y2 x 2 1 k k f (a + h, b + k) = f (a, b) h f h f ..... y 2! x y x On putting a = 0, b = 0, h = x, k = y, we get 2 f f 1 2 2 f 2 f 2 f f (x, y) = f (0, 0) x y 2 x y y x y 2! x y x 2 y2 x ..... Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ Partial Differentiation 68 Example 78. Expand ex. sin y in powers of x and y, x = 0, y = 0 as far as terms of third degree. Solution. x = 0, y = 0 f (x, y) ex sin y, 0 fx (x, y) ex sin y, 0 fy (x, y) ex cos y, 1 fxx (x, y) ex sin y, 0 fxy (x, y) ex cos y, 1 fyy (x, y) – ex sin y, 0 fxxx (x, y) ex sin y, 0 fxxy (x, y) ex cos y, 1 fxyy (x, y) – ex sin y, 0 fyyy (x, y) – ex cos y, –1 By Taylor’s theorem 2 1 y y f (x, y) = f (0, 0) x f (0, 0) x f (0, 0) y 2! x y x 3 1 x y f (0, 0) ... 3! x y 2 x 2 xy y2 = f (0, 0) x f x (0, 0) y f y (0, 0) f xx (0, 0) f xy (0, 0) f yy (0, 0) 2! 2! 2! 1 3 3x 2 y 3 1 x f xxx (0, 0) f xxy (0, 0) x y 2 f xyy (0, 0) y 3 f yyy (0, 0) ... 3! 3! 3! 3! x2 y2 x3 3x 2 y 3xy 2 y3 (0) x y(1) (0) (0) (1) (0) (1) ... 2 2 6 6 6 6 x2 y y3 ....... = yxy Ans. 2 6 Example 79. Find the expansion for cos x cos y in powers of x, y upto fourth order terms. Solution. By Taylor’s Series 1 2 2 2 f (x, y) = f (0, 0) x f x (0, 0) y f y (0, 0) 2 ! x f x (0, 0) 2 x y f xy (0, 0) y f yy (0, 0) 1 x3 f x3 (0, 0) 3 x 2 y f x2y (0, 0) 3 xy 2 f x 2y (0, 0) y 3 f y3 (0, 0) 3! ex sin y = 0 x (0) y (1) 1 x 4 f x4 (0, 0) 4 x3 y f x3 y (0, 0) 6 x 2 y 2 f 2 2 (0, 0) 4 xy 3 f 3 (0, 0) y 4 f 4 (0, 0) + ... xy xy y 4! 1 4 2 2 4 1 1 cos x cos y = 1 0 0 ( x 2 0 y 2 ) (0 0 0 0) ( x 0 6 x y 0 y ) 24 2 6 x2 y2 x4 x2 y2 y2 = 1 Ans. ... 2 2 24 4 24 Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ Partial Differentiation 69 x = 0, y = 0 f (x, y) fx fy fxx cos x cos y, – sin x cos y, – cos x sin y, – cos x cos y, 1 0 0 –1 fxy fyy fxxx fxx y sin x sin y, – cos x cos y, sin x cos y, cos x sin y, 0 –1 0 0 fx yy fyyy fxxxx fxxx y fxx yy fx yyy sin x cos y, cos x sin y, cos x cos y, – sin x sin y, cos x cos y, – sin x sin y, 0 0 1 0 1 0 fyyyy cos x cos y, 1 Example 80. Find the first six terms of the expansion of the function ex log (I + y) in a Taylor’s series in the neighbourhood of the point (0, 0). Solution. x = 0, y = 0 x Taylor’s series is f (x, y) e log (1 + y) 0 f f f y f (x, y) = f (0, 0) x ex log (1 + y) 0 y x x 2 1 2 2 f 2 f 2 f 2 x y y x 2 ! x2 x y y2 ... ex log (1 + y) = 0 ( x 0 y 1) 1 2 [ x (0) 2 xy 1 y 2 ( 1)] .. 2! 2 ex log (1 + y) = y x y y 2 Ans. f y 2 f x2 2 f y 2 2 f x y ex 1 y 1 ex log (1 + y) 0 ex (1 y )2 ex (1 y ) –1 1 EXERCISE 1.16 1 2 ( x y 2 ) ........ 2 cos 3y in power series of x and y upto quadratic terms. (AMIE Summer 2004) 9 2 2 Ans. 1 2 x 2 x y ... 2 1. Expand ex cos y at (0, 0) upto three terms. 2. Expand z = e2x 3. Show that ey log (1 + x) = x xy – 4. Verify sin (x + y) = x y Ans. 1 x x2 approximately.. 2 ( x y )3 ....... 3 Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ Partial Differentiation 70 Example 81. Expand sin (xy) in powers of (x – 1) and y as far as the terms of second 2 degree. (Nagpur University, Summer 2003) Solution. We have, f (x,y) = sin (xy) x = 1, y = 2 a h x and h x 1 Here a ( x 1) x a 1 f (x, y) sin (x y) 1 b k y and k y 2 b y y b 2 2 By Taylor’s theorem for a function of two variables, we have f (a + h, b + k) = f (a, b) + hfx (a, b) fx (x, y) y cos (xy), 0 fy(x, y) x cos (xy), 0 fxx (x, y) – y2 sin (xy), fxy (x, y) cos (xy) – xy sin (xy), fyy (x, y) – x2 sin (xy), 2 4 2 –1 + kfy (a, b) 1 2 h f xx (a, b ) 2hkf xy (a , b) k 2 f yy (a, b ) 2! f (x, y)= f 1, ( x 1) f x 1, y f y 1, 2 2 2 2 2 1 ( x 1)2 f xx 1, 2 ( x 1) y f xy 1 , y f yy (1, ) 2! 2 2 2 2 2 sin (xy) = 1 ( x 1) . 0 y . 0 2 2 2 1 2 ( x 1) 2 ( x 1) y y ( 1) ... 2! 2 2 2 4 2 2 1 ( x 1)2 ( x 1) y y ... 8 2 2 2 2 x Example 82. Expand e cos y near the point 1, by Taylor’s Theorem. 4 (U.P., I Semester Dec. 2007) k Solution. f (x + h, y + k) = f (x, y) h f x y f (x, y) ex cos y sin (xy) = 1 2 3 1 1 h k k f h f .... 2! x y 3! x y ex cos y = f (x, y)= f 1 ( x 1), y 4 4 where h = x – 1, k = y = f 1 h, k 4 4 Putting these values in Taylor’s Theorem, we get e e e ( x 1) y ex cos y = 4 2 2 2 f x f y 2 f x 2 2 f y 2 2 f x y ex cos y, – ex sin y, ex cos y, – ex cos y, –1 ex sin y, Ans. x = 1, y = e 4 2 e 2 e 2 e 2 e 2 e 2 Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ Partial Differentiation 71 2 1 e e e ( x 1)2 2( x 1) y y .... 2! 4 2 4 2 2 2 2 e ( x 1) ( x 1) y y .... 1 ( x 1) y = Ans. 4 2 4 4 2 Example 83. If f (x, y) = tan–1 (x y), compute an approximate value of f (0.9, – 1.2). Solution. We have, f (x, y) = tan–1 (x y) Let us expand f (x, y) near the point (1, – 1) f (0.9, – 1.2) = f (1 – 0.1, – 1 – 0.2) 2 f f 1 2 f ( 0.2) = f (1, 1) ( 0.1) ( 0.1) x y 2 ! x 2 2 ( 0.1) ( 0.2) f (x, y) tan–1 (xy) f x f y 2 f x 2 2 f y x y 1 x2 y2 x 1 x2 y2 , (1 x 2 y 2 ) 2 (1 x 2 y 2 )2 2 2 2 (1 x y ) Substituting the values of 1 2 1 2 , 1 x 2 y 2 x (2 x y 2 ) x (2 x 2 y ) ...(1) 1 2 (2 x) y 2 y , 2 f 2 f 2 f ( 0.2)2 ... x y y 2 x = 1, y = – 1 4 1 x2 y2 (1 x 2 y 2 )2 0 1 2 , f f , etc. in (1), we get x y f (0.9, – 1.2) = 1 1 1 1 ( 0.1) ( 0.2) ( 0.1) 2 4 2 2 2 2 1 2 ( 0.1) ( 0.2) 0 ( 0.2) 2 ... 2 = 22 1 0.05 0.1 (0.005 0.02) 28 2 = 0.786 0.05 0.1 0.0125 0.8235 Ans. Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ Partial Differentiation 72 Example 84. Obtain Taylor’s expansion of tan–1 degree terms. Hence compute f (1. 1, 0.9). Solution. f (x, y) f x f y 2 f x 2 2 f y 2 2 2 2 2 (x y ) 2 f y x x = 1, y = 1 4 y tan x 1 y y 2 2 , 2 y x x y2 1 2 x 1 1 x 2 , 2 y x x y2 1 2 x y (2 x ) 2 xy 2 , 2 2 2 (x y ) ( x y 2 )2 1 x (2 y ) 2 y about (1, 1) upto and including the second x (U.P., I Sem. Winter 2005, 2002) 2 xy 2 (x y ) ( x y ) ( x ) (2 x ) 2 2 2 2 2 (x y ) 1 2 1 2 , y2 x2 ( x2 y2 )2 1 2 , 1 2 0 By Taylor’s Theorem 2 f f 1 2 f ( y b) ( x a ) f (x, y) = f (a, b) ( x a) x y 2! x2 2 f 2 f 2 ( x a ) ( y b) ( y b) 2 ... x y y2 Here, a = 1, b = 1 1 1 1 1 1 y ( x 1) 2 tan = ( x 1) ( y 1) 4 2 2! x 2 2 1 2 ( x 1) ( y 1) (0) ( y 1) 2 ... 2 1 1 1 1 y tan 1 = ( x 1) ( y 1) ( x 1)2 ( y 1)2 ... ...(1) x 4 2 2 4 4 Putting (x – 1) = 1.1 – 1 = 0.1, (y – 1) = 0.9 – 1 = – 0.1 in (1), we get f (1.1, 0.9) = 1 1 1 1 (0.1) ( 0.1) (0.1)2 ( 0.1)2 4 2 2 4 4 = 0.786 – 0.05 +0.05 + 0.0025 – 0.0025 = 0.786 Ans. ( x h) ( y k ) in powers of h, k upto and inclusive of the second xh yk degree terms. (A.M.I.E., Summer 2001) ( x h) ( y k ) Solution. f (x + h, y + k) = xh yk xy f (x, y) = x y Example 85. Expand Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ Partial Differentiation 73 y2 ( x y) y x y f = = x ( x y )2 ( x y) 2 x2 ( x y) x x y f = = ( x y) 2 y ( x y )2 2 f x2 = 2 y2 ( x y )3 ( x y )2 2 x 2 ( x y ) x 2 ( x y ) 2 x 2 x 2 2x y 2 f = 4 3 y x (x y) (x y) ( x y )3 2 f y2 = 2x2 ( x y )3 2 1 k k f (x + h, y + k) = f ( x , y ) h f ( x, y) h f ( x, y ) ... y 2! x y x xy y2 x2 ( x h) ( y k ) h k = x y xh yk ( x y) 2 ( x y )2 h2 ( 2 y 2 ) 1 2x y 1 2 ( 2 x 2 ) 2 h k k ... 2 ! ( x y )3 2 ! ( x y )3 2 ! ( x y )3 xy h y2 kx 2 h2 y2 2h kxy k 2 x2 ... Ans. 2 2 3 3 x y ( x y) ( x y) ( x y) (x y) ( x y )3 Example 86. Expand x2y + 3y – 2 in powers of x – 1 and y + 2 using Taylor’s Theorem. (A.M.I.E.T.E., Winter 2003, A.M.I.E., Summer 2004, 2003) Solution. f (x, y) = x2y + 3 y – 2 Here a + h = x and h = x – 1, so a = 1 b + k = y and k = y + 2 so b = – 2 x = 1, y = – 2 = f (x, y) x2y + 3y – 2, fx (x, y) 2xy, fy (x, y) x2 + 3, – 10 –4 4 fxx (x, y) 2 y, –4 fxy (x, y) 2 x, 2 fyy (x, y) 0, 0 fxxx (x, y) 0, 0 fxxy (x, y) 2, 2 fxyy (x, y) 0, 0 fyyy (x, y) 0, 0 Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ Partial Differentiation 74 Now Taylor’s Theorem is f f 1 2 2 f 2 f k 2 h k k2 h f (a + h, b + k) = f (a, b) h y ( a, b ) 2 ! x 2 x y x 1 3 f 3 f 2 f 2 2 h3 3 h k 3 h k k3 3! x3 x2 y x y2 Putting the values of f (a, b) etc. in Taylor’s Theorem, we get 2 f y 2 (a, b) 3 f ... y 3 x2y + 3y – 2 = – 10 + [(x – 1) (– 4) + (y + 2) (4)] 1 [(x – 1)2 (– 4) + 2 (x – 1) (y + 2) (2) + (y + 2)2 (0)] 2! 1 [(x – 1)3 (0) + 3 (x – 1)2 (y + 2) (2) + 3 (x – 1) (y + 2)2 (0) + (y + 2)3 (0)] 3! x2y + 3y – 2 = – 10 – 4 (x – 1) + 4 (y + 2) – 2 (x – 1)2 + 2 (x – 1) (y + 2) + (x – 1)2 (y + 2)Ans. EXERCISE 1.17 exy 1. Expand at (1, 1) upto three terms. 2. 3. 1 [( x 1) 2 4 ( x 1) ( y 1) ( y 1)2 ] 2! Expand yx at (1, 1) upto second term Ans. 1 + (y – 1) + (x – 1) (y – 1) + ....... Expand eax sin by in powers of x and y as far as the terms of third degree. (U.P. I sem. Jan 2011) Ans. e [1 ( x 1) ( y 1) Ans. by abxy 4. Expand (x2y + sin y + ex) 1 3a 2 bx 2 y – b3 y 3 .... 3! in powers of (x – 1) and (x – ). 1 ( x 1) 2 (2 e) 2 ( x 1) ( y ). 2 x 1 ( x 1) 2 y 2 ... Ans. 2 1 4 32 4 Ans. e ( x 1) (2 e) 5. Expand (1 + x + y2)1/2 at (1, 0). 6. Obtain the linearised form T(x, y) of the function f (x, y) = x2 – xy + 7. 8. 1 2 y + 3 at the point (3, 2), using 2 the Taylor’s series expansion. Find the maximum error in magnitude in the approximation f (x, y) T (x, y) over the rectangle R: | x – 3 | < 0.1, | y – 2 | < 0.1. Ans. 8 + 4 (x – 3) – (y – 2)., Error 0.04. Expand sin (x + h) (y + k) by Taylor’s Theorem. 1 Ans. sin xy h ( x y ) cos xy hk cos xy h 2 ( x y ) 2 sin xy ... 2 Fill in the blank: f (x, y) = f (2, 3) + ............ Ans. ( x 2) 9. If f (x) = f (0) kf1 (0) given as .......... 2 ( y 3) f + 1 ( x 2) ( y 3) f ... x y 2! x y k2 f 2 (k ), 0 1 then the value of when k = 1 and f(x) = (1 – x)3/2 is 2! (U.P. Ist Semester, Dec 2008) 1.27 MAXIMUM VALUE A function f (x, y) is said to have a maximum value at x = a, y = b, if there exists a small neighbourhood of (a, b) such that, f (a, b) > f (a + h, b + k) Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ Partial Differentiation 75 Minimum Value. A function f (x, y) is said to have a minimum value for x =a, y = b, if there exists a small neighbourhood of (a, b) such that f (a, b) < f (a + h, b + k) The maximum and minimum values of a function are also called extreme or extremum values of the function. f(x, y) f(x, y) Maximum Minimum O O Y X Y X Maximum value of f(x, y) at (a, b) Minimum value of f(x, y) at (a, b) Saddle point or Minimax. It is a point where a function is neither maximum nor minimum. Geometrical Interpretation. Such a surface (looks like the leather seat on the back of a horse) forms a ridge rising in one direction and falling in another direction. 1.28 CONDITIONS FOR EXTREMUM VALUES If f (a + h, b + k) – f (a, b) remains of the same sign for all values (positive or negative) of h, k then f (a, b) is said to be extremum value of f (x, y) at (a, b) (i) If f (a + h, b + k) – f (a, b) < 0, then f (a, b) is maximum. (ii) If f (a + h, b + k) – f (a, b) > 0, then f (a, b) is minimum. By Taylor’s Theorem 2 f f 1 2 f 2 f 2 f k h2 2 hk k ... y ( a , b ) 2! x 2 x y y 2 x f (a + h, b + k) = f (a, b) + h f f 1 2 f 2 f 2 f h2 2hk k2 ...(1) f (a + h, b + k) – f (a, b) = h x k y 2 2! x y x y 2 ( a , b ) f (a + h, b + k) – f (a, b) = h f f k x y ( a , b ) ...(2) For small values of h, k, the second and higher order terms are still smaller and hence may be neglected. The sign of L.H.S. of (2) is governed by h f k f which may be positive or negative x y depending on h, k. Hence, the necessary condition for f (a, b) to be a maximum or minimum is that f f f f k = 0, =0 h =0 y y x x By solving the equations, we get, point x = a, y = b which may be maximum or minimum value. Then from (1) f (a + h, b + k) – f (a, b) = 2 1 2 2 f 2 f 2 f 2 hk k h 2! x 2 x y y 2 Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ Partial Differentiation 76 = 1 2 [h r 2 h k s k 2 t ] 2! ...(3) 2 where r = f x2 2 ,s 2 f f ,t at ( a, b ) x y y2 Now the sign of L.H.S. of (3) is sign of [rh2 + 2hks + k2 t] 1 2 2 [ r h 2hkrs k 2 rt ] = sign of 1 [(r 2 h 2 2 hkrs k 2 s 2 ) ( k 2 s 2 k 2rt )] r r 1 2 2 2 = sign of [(hr ks ) k ( rt s )] r 1 2 2 = sign of [(always + ve) k ( rt s )] [(hr + ks)2 = + ve] r 1 2 2 = sign of [ k (rt s )] = sign of r if rt – s2 > 0 r = sign of Hence, if rt – s2 > 0, then f (x, y) has a maximum or minimum at (a, b), according as r < 0 or r > 0. Note: (i) If rt – s2 < 0, then L.H.S. will change with h and k hence there is no maximum or minimum at (a, b), i.e., it is a saddle point. (ii) If rt – s2 = 0, then rh2 + 2shk + tk2 = 1 [(rh sk ) 2 k 2 ( rt s 2 )] r 1 ( rh sk ) 2 which is zero for values of h, k, such that r h s = k r = This is, therefore, a doubtful case, further investigation is required. 1.29 WORKING RULE TO FIND EXTREMUM VALUES (i) f f 2 f 2 f 2 f , , Differentiate f (x, y) and find out x , y , x2 x y y 2 (ii) Put f f = 0 and y = 0 and solve these equations for x and y. Let (a, b) be the values x of (x, y). (iii) Evaluate r = 2 f ,s 2 f 2 f , t 2 for these values (a, b). x y y x 2 (iv) If rt – > 0 and (a) r < 0, then f (x, y) has a maximum value. (b) r > 0, then f (x, y) has a minimum value. (v) If rt – s2 < 0, then f (x, y) has no extremum value at the point (a, b). (vi) If rt – s2 = 0, then the case is doubtful and needs further investigation. s2 Note: The point (a, b) which are the roots of f f 0, = 0, are called stationary points. x y Example 87. Discuss the maximum and minimum of x2 + y2 + 6x + 12. Solution. We have, f (x, y) = x2 + y2 + 6x + 12 f f 2 f 2 f 2 f 2 y, 2 2, 2 2, 0 = 2 x 6, x y x y x y For maxima and minima, f f = 0 and =0 y x Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ Partial Differentiation 77 At 2x + 6 = 0, and 2y = 0 x = – 3, and y = 0 (– 3, 0) rt – s2 = 2 × 2 – 0 = 4 > 0 r=2>0 Hence f (x, y) is minimum when x = – 3 and y = 0 Minimum value = f (– 3, 0) = 9 + 0 – 18 + 12 = 3 Ans. Example 88. Find the absolute maximum and minimum values of f (x, y) = 2 + 2x + 2y – x2 – y2 on triangular plate in the first quadrant, bounded by the lines x = 0, y = 0 and y = 9 – x. (Gujarat, I semester, Jan. 2009) 2 2 Solution. We have, f (x, y) = 2 + 2x + 2y – x – y f f = 2 – 2x, = 2 – 2y x y 2 f x 2 = – 2, For maxima and minima, 2 f 2 f 0, 2 xy y 2 f = 0 2 – 2x = 0 x = 1 x f = 0 2 – 2y = 0 y = 1 y rt – s2 = (– 2) (– 2) – 0 = 4 > 0 At (1, 1) Hence f (x, y) is maximum at (1, 1). Maximum value of f (x, y) = 2 + 2 + 2 – 1 – 1 = 4 Ans. 3 3 Example 89. Examine f (x, y) = x + y – 3 a x y for maximum and minimum values. (U.P. I Sem., Dec. 2004), (M.U. 2004, 2003) 3 3 Solution. We have, f (x, y) = x + y – 3axy p= r= f = 3x2 – 3ay,, x 2 f x 2 6 x, q= f = 3y2 – 3ax y 2 f 2 f 3 a , t 6y s = xy y 2 For maxima and minima f =0 y 3y2 – 3 ax = 0 f =0 x 3x2 – 3 ay = 0. x2 = ay y = x2 a and ...(1) y2 = ax ...(2) Putting the value of y from (1) in (2), we get x4 = a3x x(x3 – a3) = 0 2 x(x – a)(x + ax + a2) = 0 x = 0, a Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ Partial Differentiation 78 Putting x = 0 in (1), Putting x = a in (1), we get y = 0, we get y = a, Stationary pairs (0, 0) (a, a) r 0 6a s – 3a – 3a t 0 6a rt – s2 –9a2 < 0 27 a2 > 0 At (0, 0) there is no extremum value, since rt – s2 < 0. At (a, a), rt – s2 > 0, r > 0 Therefore (a, a) is a point of minimum value. The minimum value of f (a, a) = a3 + a3 – 3 a3 = – a3 Example 90. Show that the function f (x,y) = x3 + y3 – 63 (x + y) + 12 xy is maximum at (– 7, – 7) and minimum at (3, 3). Solution. We have, f (x,y) = x3 + y3 – 63 (x + y) + 12 xy f = 3x2 – 63 + 12y, x 2 f x 2 2 f 12, = 6x, xy Ans. ...(1) f = 3y2 – 63 + 12x y 2 f y 2 6y For extremum, we have f = 3x2 – 63 + 12y = 0 x f q= = 3y2 – 63 + 12x = 0 y p= x2 + 4y – 21 = 0 ...(2) y2 + 4x – 21 = 0 ...(3) 2 f 2 f 2 f 12 6y 6 x s = t = xy y 2 x 2 We have to solve (2) and (3) for x and y. On subtracting (3) from (2), we have x2 – y2 – 4 (x – y) = 0 (x – y) (x + y – 4) = 0 x = y and x + y = 4 ...(4) 2 If x = y then (2) becomes, x + 4x – 21 = 0, (x + 7)(x – 3) = 0 x = – 7, and x = 3 y = – 7, and y = 3 Two stationary points are (–7, –7) and (3, 3). On solving (2) and (4), we get x2 + 4 (4 – x) – 21 = 0, x2 – 4x – 5 = 0 (x – 5) (x + 1) = 0 x = – 1, x = 5 y = 5, y = – 1 Two more stationary points are (–1, 5) and (5, –1). Hence four possible extremum points of f (x, y) are (– 7, – 7), (3, 3), (–1, 5) and (5, – 1) may be. r= Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ Partial Differentiation 79 Stationary pairs r = 6x – 42 (– 7, – 7) + 18 (3, 3) –6 (– 1, 5) (5, – 1) 30 s = 12 12 12 12 12 t=6y r t – s2 – 42 + 1620 18 + 180 30 – 324 –6 – 324 At (– 7, – 7) r = – ve, and rt – s2 = + ve Hence, f (x, y) is maximum at (– 7, – 7), At (3, 3) r = + ve, and rt – s2 = + ve Hence, f (x, y) is minimum at (3, 3). Proved. 2 2 2 2 Example 91. Find the extreme values of u = x y – 5x – 8xy – 5y . Solution. We have, u u = x2 y2 – 5x2 – 8xy – 5y2 p = = 2xy2 – 10x – 8y x u 2u q = = 2x2 y – 8x – 10y r = = 2y2 – 10 y x 2 2u 2u 4 xy 8 s = t = = 2x2 – 10 y 2 x y u u = 0, =0 y x For extreme values of u, 2xy2 – 10x – 8y = 0 x = 8y 2 2 y 10 2x2 y – 8x – 10y = 0 2 8y 8y 2 2 y 8 2 10 y 0 2 y 10 2 y 10 128 y 2 2 Now, 64 16 y 2 2 2 4y y2 5 If y = 1 then x = – 1; If y = 3 then x = 3; 10 0, y = 0 then x = 0 2 2 16 50 ( y 5) y 5 (2 y 10) 2 y 10 16 y2 – 16 (y2 – 5) – 5 (y2 – 5)2 = 0 16 y2 – 16 y2 + 80 – 5 (y2 – 5)2 = 0 (y2 – 5)2 = 16 y2 – 5 = ± 4 y2 = 9 and 1 and y = ± 3, ± 1 2 x = If y = – l then x = 1 If y = – 3 then x = – 3 Stationary pairs (0, 0) (1, – 1) (– 1, 1) (3, 3) (– 3, – 3) r = 2 y2 – 10 – 10 –8 –8 8 8 s=4xy–8 –8 – 12 – 12 28 28 t = 2 x – 10 – 10 –8 –8 8 8 rt – s2 + 36 – 80 – 80 – 720 – 720 2 Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ Partial Differentiation 80 At (0, 0), r is – ve. Origin (0, 0) is the only point at which r t – s2 > 0. Hence, the function u is maximum at origin. Ans. Example 92. A rectangular box, open at the top, is to have a volume of 32 c.c. Find the dimensions of the box requiring least material for its construction. (M.U. 2009; U.P. I semester Jan. 2011; Dec. 2005, A.M.I.E Summer 2001) Solution. Let l, b and h be the length, breadth, and height of the box respectively and S its surface area and V the volume. V = 32 c.c. 32 lh S = 2 (l + b) h + l b Putting the value of b in (1), we get 32 32 S = 2 l h l lh lh 64 32 S = 2lh+ l h Differentiating (2) partially w.r.t. l, we get S 64 = 2h– 2 l l Differentiating (2) partially w.r.t. h, we get l b h = 32 b = S 32 = 2l– 2 h h For maximum and minimum S, we get S 64 = 0 2h– 2 =0 l l S 32 = 0 2l– 2 =0 h h From (5) and (6), l = 4, h = 2 and b = 4 2 S 128 128 = =2 l2 l3 64 2 S = 2 l h ...(1) ...(2) ...(3) ...(4) h= 32 l2 ...(5) 16 h2 ...(6) l= 2 S 64 64 8 = 3 2 h h 8 2 2 S 2 S 2 S . = (2) (8) – (2)2 = + 12 l 2 h2 l h 2 S = + 2, so S is minimum for l = 4, b = 4, h = 2 l2 Ans. EXERCISE 1.18 Find the stationary points of the following functions 1. f (x, y) = y2 + 4 xy + 3 x2 + x3 2 4 Ans. , , Minimum 3 3 Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ Partial Differentiation 81 1 1 [A.M.I.E., Summer 2004] Ans. , , Maximum 2 3 2. f (x, y) = x3 y2 (1 – x – y) 3. f (x, y) = x3 + 3 xy2 – 15 x2 – 15 y3 + 72x. (M.U. 2007, 2005, 2004) Ans. (6, 0), (4, 0) 4. f (x, y) = x2 + 2xy + 2y2 + 2x + 3y such that x2 – y = 1. 5. xy e– (2x + 3y) 7 155 3 . Ans. , , 4 16 128 (A.M.I.E., Winter 2000) 2 2 6. Find the extreme value of the function f (x, y) = x + y + xy + x – 4y + 5. State whether this value is a relative maximum or a relative minimum. Ans. Minimum value of f (x, y) at (– 2, 3) = – 2. 2 2 7. Find the values of x and y for which x + y + 6 x = 12 has a minimum value and find this minimum value. Ans. (– 3, 0), 3. 8. Find a point within a triangle such that the sum of the square of its distances from the three angular points is a minimum. 9. A tapering log has a square cross-section whose side varies uniformly and is equal to a at the top 3a and b b at the bottom. Show that the volume of the greatest conical frustum that can be 2 b3l obtained from the log is , where l is the length of the log. 27(b a ) 10. A tree trunk of length l metres has the shape of a frustum of a circular cone with radii of its ends a and b metres where a > b. Find the length of a beam of uniform square cross section which can be cut from the tree trunk so that the beam has the greatest volume. Ans. 8a 3 l 27( a b) 1.30 LAGRANGE METHOD OF UNDETERMINED MULTIPLIERS Let f (x, y, z) be a function of three variables x, y, z and the variables be connected by the relation. (x, y, z) = 0 ...(1) f (x, y, z) to have stationary values, f f f = 0, y = 0, = 0 x z f f f dx dy dz = 0 x y z dx dy dz = 0 x y z Multiplying (3) by and adding to (2), we get By total differentiation of (1), we get ...(2) ...(3) f f f dx dy dy dz dz = 0 dx x y y z z x f f dx x x y This equation will hold good if f x x f y y f z z f dy dz = 0 y z z = 0 ...(4) = 0 ...(5) = 0 ...(6) Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ Partial Differentiation 82 On solving (1), (4), (5), (6), we can find the values of x, y, z and for which f (x, y, z) has stationary value. Draw Back in Lagrange method is that the nature of stationary point cannot be determined. Example 93. Find the point upon the plane ax + by + cz = p at which the function f = x2 + y2 + z2 (Nagpur University, Winter 2000) has a minimum value and find this minimum f. Solution. We have, f = x2 + y2 + z2 ...(1) ax + by + cz = p = ax + by + cz – p ...(2) f = 0 x x 2x + a = 0 x = a 2 f = 0 y y 2y + b = 0 y = b 2 z = c 2 f = 0 2z+c=0 z z Substituting the values of x, y, z in (2), we get a b c a b c = p 2 2 2 (a2 + b2 + c2) = – 2p x = The minimum value of f = = ap 2 2 a b c a2 p2 2 2 p = , y (a 2 b2 c 2 )2 p 2 (a 2 b2 c 2 ) 2 a b2 c 2 bp 2 2 a b c b2 p 2 2 , z (a 2 b 2 c 2 )2 cp 2 a b2 c 2 c2 p 2 (a 2 b 2 c 2 )2 p2 Ans. (a 2 b2 c 2 ) 2 a 2 b2 c 2 p q r Example 94. Find the maximum value of u = x y z when the variables x, y, z are subject to the condition ax + by + cz = p + q + r. Solution. Here, we have u = xp yq zr ...(1) If log u = p log x + q log y + r log z ...(2) 1 u p = u x x u pu x x 1 u q = uy y u qu y y 1 u u ru r = u z z z z ax + by + cz = p + q + r (x, y, z) = ax + by + cz – p – q – r = a, = b, =c x y z Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ Partial Differentiation 83 Lagranges equations are u = 0 x x u = 0 y y pu a = 0 x qu b = 0 y u = 0 z z ru c = 0 z x = pu a y = qu b z = ru c Putting in (2), we have pu qu ru = p+q+r u u ( p q r) = p + q + r = 1 pu pu p x = a ua a qu qu q y = b ub b ru ru r z = c uc c Putting in (1), we have p q =–u r p q r u = Ans. a b c Example 95. Show that the rectangular solid of maximum volume that can be inscribed in a sphere is a cube. Solution. Let 2x, 2y, 2z be the length, breadth and height of the rectangular solid. Let R be the radius of the sphere. Volume of solid V = 8 x . y . z ...(1) 2 2 2 2 x +y +z = R ....(2) z (x, y, z) = x2 + y2 + z2 – R2 = 0 Maximum value of V + = 0 x x From (3) From (4) From (5) y 8yz + (2x) = 0 z V = 0 y y 8x z + (2 y) = 0 ...(4) V z z 8xy + (2 z) = 0 ...(5) = 0 2x 2y 2z 2 x2 x2 x Hence, rectangular solid = = = = = = is y ...(3) O – 8y z – 8x z – 8x y 2 y2 = 2 z2 y2 = z2 y= z a cube. y R x 2 2 x = – 8x y z 2 y2 = – 8x y z 2 z2 = – 8 x y z Proved. Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ Partial Differentiation 84 Example 96. A rectangular box, which is open at the top, has a capacity of 256 cubic feet. Determine the dimensions of the box such that the least material is required for the construction of the box. Use Lagrange’s method of multipliers to obtain the solution. Solution. Let x, y, z be the length, breadth and height of the box. Volume = xyz = 256 xyz – 256 = 0 ...(1) (x, y, z) = x y z – 256 Let S be the material surface of the box. x S = xy+2yz+2zx y S = y + 2z and = yz x x z S = x + 2z and = xz y y S z = 2y + 2x and = xy z By Lagrange’s method of multiplier, we have S x S y = 0 x = 0 y y + 2z + yz = 0 ...(2) x + 2z + xz = 0 ...(3) S = 0 2y + 2x + xy = 0 z z Multiplying (2) by x, we get xy + 2 xz + xyz = 0 xy + 2 xz + 256 = 0 (xyz = 256) xy + 2 xz = – 256 Multiplying (3) by y, we get xy + 2 yz + xyz = 0 xy + 2 yz + 256 = 0 xy + 2 yz = – 256 Multiplying (4) by z, we get 2 yz + 2 xz + xyz = 0 2 yz + 2 xz + 256 = 0 2 yz + 2 zx = – 256 From (5) and (6), we have xy + 2 xz = xy + 2 yz 2 xz = 2 yz x = y From (6) and (7), we have xy + 2 yz = 2 yz + 2 xz xy = 2 xz y = 2z From (1) xyz = 256 y (y) (y) = 256 y3 = 512 y = 8 2 x = 8, y = 8, z = 4 Hence, length = breadth = 8, height = 4. ...(4) ...(5) ...(6) ...(7) Ans. Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ Partial Differentiation 85 Example 97. Use the method of the Lagrange’s multipliers to find the volume of the largest x2 y 2 z2 rectangular parallelopiped that can be inscribed in the ellipsoid 2 2 2 1 . a b c (Nagpur Univesity, Summer 2008, Winter 2003) (A.M.I.E.T.E., Summer 2004, U.P., I Semester, Winter 2002, 2000) x2 y 2 z2 Solution. Here, we have = 1 a 2 b2 c 2 x2 y 2 z2 (x, y, z) = 2 2 2 1 0 ...(1) a b c Let 2x, 2y, 2z be the length, breadth and height of the rectangular parallelopiped inscribed in the ellipsoid. V = (2x) (2y) (2z) = 8 xyz V V V = 8 yz; 8xz, 8 xy x y z 2x = a2 , x 2 y , y b 2 2 z z c 2 Lagrange’s equations are V x x V y y V z z Multiplying (1), (2) and (3) 2x = 0 a2 2y = 0 8 xz + 2 = 0 b 2z = 0 8 xy + 2 = 0 c by x, y, z respectively and adding, we get = 0 8 yz + x2 y 2 z 2 x2 y 2 24 xyz 2 2 2 2 = 0 2 2 b c b a a 24 xyz + 2 (1) = 0 = – 12 xyz Putting the value of in (1), we get 2x 3x 2 8 yz + (– 12 xyz) 2 = 0 1– 2 =0 x= a a Similarly from (2) and (3), we have b c , z y = 3 3 Volume of the largest rectangular parallelopiped = 8 xyz ...(1) ...(2) ...(3) z2 1 c2 a 3 a b c 8 abc = 8 Ans. 3 3 3 3 3 Example 98. The shape of a hole pored by a drill is a cone surmounted by cylinder. If the cylinder be of height h and radius r and the semi-vertical angle of the cone be , where h show that for a total height H of the hole, the volume removed is maximum if r h = H ( 7 1) / 6. (R.G.P.V., Bhopal I sem. 2003) tan Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ Partial Differentiation 86 Solution. Let ABCD be the given cylinder of height ‘h’ and radius ‘r’ and DPCO be the cone of course, of radius r. Now, since is the semi-vertical angle of the cone. PC r OP OP h but, given that tan = ...(2) r h r r2 From (1) and (2), we have OP = ...(3) r OP h Total height of the hole = H H = h + OP OP = H – h ...(4) From (3) and (4), we have r2 =H–h ...(5) h 2 r Again, let =H–h– ...(6) h In drilling a hole, the volume of the removed portion 2r r2 = – , = 1 2 r h h h V = Volume of the cylinder + Volume of the cone. 1 2 1 2 r2 2 2 = r h r (OP ) r h r . 3 3 h 4 r V = r2h , 3h V 4r 3 2rh r 3h By Lagrange’s Method tan = V = 0 r r 2r h A B h H D P r r2 1 2 h 2r 2 6 r4 2r 2 2h = 0 3h h h 2 r4 + (– 4r2 + 2 h2) = 0 r4 + (– 2 r2 + h2) = 0 C O [From (3)] 4 r3 2r = 0 3h h V r4 = 0 r2 h h 3 h2 Multiplying (8) by r and (9) by 2h, we get 2r 2 4 r4 2r 2 h = 0 3h h 2 r4 r2 2r 2 h 2 h = 0 3h h On subtracting, we get ...(1) = 0 ...(7) ...(8) ...(9) 4 r2 6 r4 2h = 0 3h h = r4 h2 2 r 2 Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ Partial Differentiation 87 Putting the value of in (8), we get r2 2r 4 r3 r 4 2 r2 r4 H h 2 h = 0 = 0 2 3h 3 h h (h 2 2 r 2 ) h h 2r h 2 2 2 2 2 (H h 2 h H ) 2 h ( H h) h ( H h) = 0 h ( H h) =0 2 3 3 h 2 H 2h h [h 2h ( H h)] 2r h 2 H 2 h 2 2hH ( H h) =0 3 3h 2 H 2 3h2 – 2H h + (H – h) (3 h – 2 H) + H2 + h2 – 2 h H = 0 3 9h2 – 6H h + 6H h – 4H2 – 6h2 + 4Hh + 3H2 + 3h2 – 6h H = 0 6h2 – 2h H – H2 = 0 h 2 H 4 H 2 24 H 2 12 H H 7 [ 7 1] h = =H (–ve is not possible) Proved. 6 6 Example 99. A tent of a given volume has a square base of side 2a, has its four-side vertical of length b and is surmounted by a regular pyramid of height h. Find the values of a and b in terms of h such that the canvas required for its construction is minimum. h = Solution. Let V be the volume and S be the surface of the tent. 1 V = 4 a 2b (4 a 2 ) h 3 1 Area of the base × height] 3 1 S = 8 a b 4 a a2 h2 [Surface Area of pyramid = perimeter × slant height] 2 S V = 0 a a From (2) From (3) [Volume of pyramid = 8b 4 a 2 h 2 4 a2 8a h 8 a b = 0 3 a h ...(1) 8 a + 4 a2 = 0 ...(2) 2 2 S V b b = 0 S V h h = 0 a+2 = 0 12ah 4a 2 a 2 h 2 = 0 3h a a 2 h 2 = 0 4 ah 4 a2 = 0 3 a h a = –2 2 2 ...(3) ...(4) ...(5) Substituting the value of a from (4) in (5), we get 3h 2 a 2 h2 = 0 a = 9 h2 = 4a2 + 4h2 4a2 = 5h2 5 h 2 Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ Partial Differentiation 88 5 h in (1) and simplifying, we get 2 Substituting a = – 2 and a = 8b 4 5h 2 h2 4 5h 2 2 8h 2 8 b = 0 3 5h h2 4 10 h 16 h h 8 b + 6h + – 16 b – =0 – 8b + 4h = 0 b= . 3 3 2 h 5 Thus, when a = Ans. h and b = we get the stationary value of S. 2 2 Example 100. Find the maximum and minimum distances of the point (3, 4, 12) from the sphere x2 + y2 + z2 = 1. Solution. Let the co-ordinates of the given point be (x, y, z), then its distance (D) from (3, 4, 12). D = ( x 3)2 ( y 4)2 ( z 12)2 F (x, y, z) = (x – 3)2 + (y – 4)2 + (z – 12)2 x + y2 + z2 = 1 (x, y, z) = x2 + y2 + z2 – 1 2 F = 2 (x – 3) + 2 x = 0 x x F = 2 (y – 4) + 2 y = 0 y y F = 2 (z – 12) + 2 z = 0 z z Multiplying (1) by x, (2) by y and (3) by z and adding, we get (x2 + y2 + z2) – 3x – 4y – 12z + (x2 + y2 + z2) = 0 1 – 3x – 4y – 12 z + = 0 ...(2) ...(3) ...(4) 3 1 4 From (2) y = 1 12 From (3) z = 1 Putting these values of x, y, z in (4),we have From (1) ...(1) x = ...(5) ...(6) ...(7) 9 16 144 2 0 (1 + ) = 169 1 1 1 Putting the value of 1 + in (5), (6) and (7) we have the points 1 1 + = ± 13 3 4 12 3 4 12 , , , and , . 13 13 13 13 13 13 2 The minimum distance 3 4 12 3 4 12 13 13 13 = 3 4 12 3 4 12 13 13 13 2 The maximum distance 2 = 2 2 = 12 2 = 14 Ans. Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ Partial Differentiation 89 2 2 2 2 2 2 Example 101. If u = ax + by + cz where x + y + z = 1 and lx + my + nz = 0 prove that stationary values of ‘u’ satisfy the equation l2 m2 n2 = 0 a u bu cu Solution. We have, Let u u x x x = ax2 + by2 + cz2 = x2 + y2 + z2 – 1 = lx + my + nz u = 2 a x, y = 2 b y, = 2 x, = 2 y, y = l, = m, y ...(1) ...(2) ...(3) u = 2cz z = 2z z = n z By Lagrange’s method u 1 2 = 0, x x x 2 a x + 2 x l + 2 l = 0 ...(4) u 1 2 = 0, y y y 2 b y + 2 y l + 2 m = 0 ...(5) u 1 2 = 0, z z z 2 c z + 2 z l + 2 n = 0 ...(6) Multiplying (4), (5) and (6) by x, y and z respectively and adding, we get (2 ax2 + 2 by2 + 2 cz2) + (2 x2 + 2 y2 + 2 z2) 1 + (lx + my + nz) 2 = 0 2u + 21 = 0, 1 = – u Putting the value of 1 in (4), (5) and (6), we get 2 a x – 2 x u + 2 l = 0, x= 2l 2 a u 2 b y – 2 y u + 2 m = 0, y= 2 m 2 b u 2 c z – 2 z u + 2 n = 0, z= 2 n 2(c u ) Putting the values of x, y, z in (3), we get 2 m2 2 l 2 2 n2 =0 2 a u 2 b u 2 c u l2 m2 n2 = 0 au bu cu Proved. EXERCISE 1.19 1. Show that the greatest value of xm yn where x and y are positive and x + y = a is where a is constant. mm . n n . a m n ( m n) m n , 2. Using Lagrange’s method (of multipliers), find the critical (stationary values) of the function f (x, y, z) = x2 + y2 + z2, given that z2 = xy + 1. Ans. (0, 0, – 1), (0, 0, 1). Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ Partial Differentiation 90 3. Decompose a positive number ‘a’ into three parts so that their product is maximum. a a a , , 3 3 3 4. The sum, of three numbers is constant. Prove that their product is a maximum when they are equal. 5. Using the method of Lagrange’s multipliers, find the largest product of the numbers x, y and z when Ans. x + y + z2 = 16. Ans. 4096 25 5 6. Using the method of Lagrange’s multipliers, find the largest product of the numbers x, y and z when x2 + y2 + z2 = 9. Ans. 3 3 7. Find a point in the plane x + 2y + 3z = 13 nearest to the point (1, 1, 1) using the method of 3 5 Ans. ,2, 2 2 8. Using the Lagrange’s method (of multipliers), find the shortest distance from the point (1, 2, 2) to the sphere x2 + y2 = 36. Ans. 3 Lagrange’s multipliers. 9. Find the shortest and the longest distances from the point (1, 2, – 1) to the x2 + y2 + z2 = 24. Ans. 6,3 6 10. The sum of the surfaces of a sphere and a cube is given. Show that when the sum of the volumes is least, the diameter of the sphere is equal to the edge of the cube. 11. The electric time constant of a cylindrical coil of wire can be expressed approximately by mxyz K = ax by cz where z is the axial length of the coil, y is the difference between the external and internal radii and x is the mean radius ; a, b, m and c represent positive constants. If the volume of the coil is fixed, find the values of x and y which make the time constant K as large as possible. a3 12. If u = x x 2 b3 y 2 c3 z2 , where x + y + z = 1, prove that the stationary value of u is given by a b c , y , z abc abc a b c n n i 1 i 1 2 13. Find maximum value of the expression ai xi with xi 1, 1 where a1 , a2, a3.......an are positive constants. Ans. ( a12 a22 ....... an2 ) 2 2 2 2 14. If r is the distance of a point on conic ax + by + cz = 1, lx + my + nz = 0 from origin, then the stationary values of r are given by the equation. l2 m2 n2 0 (A.M.I.E.T.E., Winter 2002) 1 ar 2 1 br 2 1 cr 2 2 2 15. If x and y satisfy the relation ax + by = ab, prove that the extreme values of function u = x2 + xy + y2 are given by the roots of the equation 4 (u – a) (u – b) = ab (A.M.I.E.T.E., Winter 2000) 16. Use the Lagranges method of undetermined multipliers to find the minimum value of x2 + y2 + z2 subject to the conditions x + y + z = 1, xyz + 1 = 0. 2 2 17. Test the function f ( x, y ) ( x 2 y 2 ) e ( e y ) for maxima and minima for points not on the circle x2 + y2 = 1. 18. Find the absolute maximum and minimum values of the function f (x, y) = cx2 + y2 – x over the region to 2x2 + y2 1 (AMIETE, Dec. 2008) Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ Multiple Integral 91 2 Multiple Integral 2.1 DOUBLE INTEGRATION We know that b a f ( x) dx = lim [ f ( x1 ) δx1 f ( x2 ) δx2 f ( x3 ) δx3 +... n x 0 + f (xn) xn] Let us consider a function f (x, y) of two variable x and y defined in the finite region A of xy-plane. Divide the region A into elementary areas. A1, A2, A3, ...... An Then A f ( x, y) dA f ( x1 , x1 ) δA1 f ( x2 , y2 ) δA2 ..... f ( xn , yn ) δAn = nlim δ A 0 2.2 EVALUATION OF DOUBLE INTEGRAL Double integral over region A may be evaluated by two successive integrations. If A is described as f1 (x) y f2 (x) [y1 y y2] and a x b, Then b y2 A f ( x, y) dA = a y1 f ( x, y ) dx dy (1) First Method b y2 A f ( x, y) dA = a y1 f ( x, y ) dy dx f (x, y) is first integratred with respect to y treating x as constant between the limits a and b. In the region we take an elementary area xy.Then integration w.r.t y (x keeping constant). converts small rectangle xy into a strip PQ (y x). While the integration of the result w.r.t. x corresponding to the sliding to the strip PQ, from AD to BC covering the while region ABCD. Second method A f ( x, y) dxdy = d x2 c x f ( x, y ) dx dy 1 91 Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ 92 Multiple Integral Here f (x,y) is first integrated w.r.t x keeping y constant between the limits x1 and x2 and then the resulting expression is integrated with respect to y between the limits c and d Take a small area xy. The integration w.r.t. x between the limits x1, x2 keeping y fixed indicates that integration is done, along PQ. Then the integration of result w.r.t y corresponds to sliding the strips PQ from BC to AD covering the whole region ABCD. Note. For constant limits, it does not matter whether we first integrate w.r.t x and then w.r.t y or vice versa. 1 x 0 0 Example 1. Evaluate ( x 2 y 2 ) dA, where dA indicates small area in xy-plane. (Gujarat, I Semester, Jan. 2009) x 1 2 1 x y3 2 2 I = 0 0 ( x y ) dy dx 0 x y dx 3 0 1 1 1 x3 x 2 ( x 0) ( x 3 0) dx x 3 dx 0 0 3 3 1 4 1 4 3 4 x 1 1 x dx = [1 0] sq. units. Ans. 0 3 3 4 0 3 3 Solution. Let 1 1 x 1 0 Example 2. Evaluate x1/3 y 1/ 2 (1 x y)1/ 2 dy dx . (M.U., II Semester 2002) Solution. Here, we have I= Putting 1 1 x 1 0 x1/3 y 1/ 2 (1 x y)1/ 2 dy dx ...(1) (1 – x) = c in (1), we get I= 1 1 Again putting y = ct I = = = = = = 1 1 1 x 3 dx x1/3 dx 1 1 c y 1/ 2 (c y)1/ 2 dy 0 1 c 0 c x1/ 3 dx 1 0 1 2 t 1 2 1 (c c t ) 2 c dt c 1/2 t 1/ 2 c1/ 2 (1 t )1/ 2 c dt c 0 t 1/ 2 (1 t )1/ 2 dt 1 1 1 1/3 1 1 cx c x1/3 1 3 dx 2 2 1 3 2 2 dx = 2 2 1 1 c x1/ 3 dx 1 3 c x 3 dx , 1 2 2 1 ...(2) dy = c dt in (2), we get 1 1 x1/3 dx = 1 1 1 1 1/3 cx 1 0 1 0 t1/ 2 1 (1 t )3 / 2 1 dt xl 1 (1 x)m 1 dx (l , m) 1 1 1 . dx 2 2 2 2 1 1 1/ 3 cx dx 1 2 1 x1/3 . c dx Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ Multiple Integral 93 Putting the value of c, we get 1 1/3 x (1 x) dx I = 2 1 2 1 x 4/3 x 7/3 ( x x ) dx 7 2 4 1 3 1 3 3 3 3 3 9 9 (1) (1) (1) (1) = = 2 4 7 4 7 2 14 28 R ( x y) dy dx, R Example 3. Evaluate 1 1/3 4/3 Ans. is the region bounded by x = 0, x = 2, y = x, y = x + 2. Solution. Let I ( x y ) dy dx (Gujarat, I Semester, Jan. 2009) Y R The limits are x = 0, x = 2, y = x and y = x + 2 I 0 dx x 2 ( x y ) dy 0 x2 y2 xy 2 x dx y= 1 x2 2 2 x ( x 2) ( x 2) x dx 2 2 2 1 2 x2 2 x 2 x ( x 4 x 4) x dx 2 2 2 0 2 0 2 [2x + 2x + 2] dx 0 2 2 0 X´ 2 (2 x 1) dx 2 [ x Example 4. Evaluate x ]02 x+ 2 2 x=a x 2 x= 2 2 –2 –1 y=x 0 1 X 2 = 2 [4 + 2] = 12 Ans. R xy dx dy where R is the quadrant of the circle x2 + y2 = a2 where x 0 and y 0. (A.M.I.E.T.E, Summer 2004, 1999) Solution. Let the region of integration be the first quadrant of the circle OAB. y 2 a 2 , y a 2 x2 ) First we integrate w.r.t. y and then w.r.t. x. R xy dx dy ( x 2 2 The limits for y are 0 and = a 0 x dx a 2 x2 0 a x y dy = a 0 2 Y B and for x, 0 to a. y2 x dx 2 0 2 a x P y 2 = 2 a 2 –x 2 dy dx a 4 1 a 2 x 2 x4 1 a 2 2 = a Ans. x ( a x ) dx = 0 = 2 2 4 0 2 8 Example 5. Evaluate O Q y=0 A X xy y 2 dy dx, s where S is a triangle with vertices (0, 0), (10, 1) and (1, 1). Solution. Let the verties of a triangle OBA be (0, 0) (10, 1) and (1, 1). Equation of OA is x = y. Equation of OB is x = 10 y. The region of OBA, given by the limits y < x < 10 y and 0 < y < 1. s xy y 2 dy dx = 1 10 y 0 dy y ( xy y 2 )½ dx Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ 94 Multiple Integral 10 y 2 1 ( xy y 2 )3/ 2 y y 1 0 dy 3 = 12 1 1 (9 y 2 )3/ 2 dy 18 y 2 dy 0 3 y 0 1 y3 18 6 = 18 3 0 3 Example 6. Evaluate A x 2 Ans. dx dy, where A is the region in the first quadrant bounded by the hyperbola xy = 16 and the lines y = x, y = 0 and x = 8. (A.M.I.E., Summer 2001) Solution. The line OP, y = x and the curve PS, xy = 16 intersect at (4, 4). The line SN, x = 8 intersects the hyperbola at S (8, 2). y = 0 is x-axis. The area A is shown shaded. Divide the area in to two part by PM perpendicular to OX. For the area OMP, y varies from 0 to x, and then x varies from 0 to 4. For the area PMNS, y -series from 0 to 16/x and then x varies from 4 to 8. A x = = 4 2 0 x 4 3 x 0 2 dx dy = 4 x 2 0 0 x x 8 16 / x 0 4 0 dx dy x2 dx dx dy 8 16/ x 4 0 dy = 4 x 2 dx dxy 8 x 16/ x 2 2 0 x y 0 dx 4 x y 0 dx 4 8 x4 x2 dx 16 x dx 16 = 64 + 8 (82 – 42) = 64 + 384 = 448. Ans. 4 4 0 2 4 8 Example 7. Evaluate 2 ( x y) dx dy over the area bounded by the ellipse x2 y2 1 a2 b2 (U.P. Ist Semester Compartment 2004) Solution. For the ellipse x2 a2 y2 b2 1 Y y x2 b y a 2 x2 = 1 2 b a a The region of integration can be expressed as b b a x a and a2 x2 y a2 x2 a a ( x y ) 2 dx dy = ( x 2 y 2 2 xy ) dx dy = b / a a2 x 2 a a (b / a) = a a 2 x2 b / a a 2 x2 2 a ( b / a ) a x = a b / a a 2 x2 a 0 2 X x=a x = –a O ( x 2 y 2 2 xy ) dx dy ( x 2 y 2 ) dx dy a b / a a 2 x2 a ( b / a ) a 2 x2 2 xy dy dx 2 ( x 2 y 2 ) dy dx 0 [Since (x2 + y2) is an even function of y and 2xy is an odd function of y] b 2 y 3 a = a 2 x y 3 0 a a 2 x2 dx Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ Multiple Integral 95 a 2 b 1 b3 2 2 2 2 3/ 2 = 2 a x a a x 3 3 (a x ) dx a 3 a b 2 b 2 2 2 2 3/ 2 = 4 0 a x a x 3 (a x ) dx 3a [On putting x = a sin and dx = a cos d] b b3 = 4 2 . a 2 sin 2 . a cos 3 a3 cos3 0 a 3a 3 3 ab 2 2 cos 4 d = 402 a b sin cos 3 3 3 2 2 = (a b ab ) ab (a b ) 4 4 Example 8. Evaluate ( x 2 a cos d 3 1 1 ab3 3 1 . . . = 4 a b . . . 4 2 2 3 4 2 2 Ans. y 2 ) dx dy throughout the area enclosed by n the curves y = 4x, x + y = 3, y = 0 and y = 2. Solution. Let OC represent y = 4x; BD, x + y = 3; OB, y = 0, and CD, y = 2. The given integral is to be evaluated over the area A of the trapezium OCDB. Area OCDB consists of area OCE, area ECDF and area FDB. The co-ordinates of C, D and B are 1 , 2 (1, 2) and (3, 0) respectively.. 2 2 2 A ( x y ) dy dx 2 2 2 2 2 2 = OCE ( x y ) dy dx ECDE ( x y ) dy dx FDB ( x y ) dy dx = ½ 4x 1 Now, I1 = = I2 = = 2 3 3 x ( x 2 y 2 ) dy dx ( x 2 y 2 ) dy dx ( x 2 y 2 ) dy ½ 0 1 0 I1 I2 I3 4x ½ 4x 2 ½ 2 ½ 76 3 y3 2 0 dx 0 ( x y ) dy 0 x y 3 dx 0 3 x dx 0 ½ 76 ½ 3 76 x 4 76 1 1 19 x dx . 3 0 3 4 0 3 4 16 48 2 1 1 1 2 1 y3 8 2 2 2 dx ( x y ) dy x y dx ½ 2 x dx ½ ½ ½ 3 0 3 1 2 x3 8 2 8 2 1 8 1 23 x . . 3 3 ½ 3 3 3 8 3 2 12 0 dx 0 3 x 2 3 2 y3 (3 x)3 I3 = dx ( x y ) dy = x y dx x (3 x ) dx 1 0 1 0 3 0 3 3 3 x 4 (3 x )4 3 2 (3 x)3 3 = 1 3x x dx x 3 4 3 1 3 3 x 2 2 3 Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ 96 Multiple Integral 81 1 16 22 0 1 = 27 4 4 12 3 19 23 22 463 31 2 2 A ( x y ) dy dx I1 I 2 I3 48 12 3 48 9 48 . Ans. EXERCISE 2.1 Evaluate 2 x2 1. 0 0 3. 0 0 a y e x dy dx 0 0 0 y2 (1 xy Ans. a2 4 4. xy dy dx Ans. 2a 4 3 6. 2 a 2 ax x 5. 0 0 2 2 a a x 7. a a 2 x2 0 9. Ans. 0 0 2 a 4 0 0 2 2 ax x 2 1 (1 y 2 ) 1 2 3 2e a log 2 1 ea 1 y 2a 8. ) dx dy Ans. 41 210 x 2 dy dx Ans. 5a 4 8 Ans. 4 1 x2 y 2 0 aa a4 6 dx dy 0 Ans. xy dx dy x dx dy Ans. a2 log ( 2 1) 2 (A.M.I.E.T.E., June 2009) Ans. 14 3 where A is given by y = 0, x + 2y = 3, x = y2. Ans. dx dy y 2 (1 e ) a x y 0 1 11 a 2 x2 y 2 dy dx ay 2. dx dy a2 y 2 a Ans. e2 – 1 Ans. 2 10. 0y 2 x y 2 2 ( x 2 3x y 2 ) dx dy x 0 y 0 12. (5 2 x y) dx dy, A 13. xy dx dy, where A is given by x2 + y2 – 2x = 0, y2 = 2x, y = x. A 14. 1 3 33 2 4 x 2 y 2 dx dy , where A is the triangle given by y = 0, y = x and x = 1. Ans. x 2 xy (1 x y ) dx dy where A is the area bounded by x = 0, y = 0 and x + y = 1. Ans. A 15. 217 60 7 Ans. 12 dx dy , where R is the two-dimensional region bounded by the curves y = x and y = x2. Ans. R 16. A 1 20 2 105 2.3 EVALUATION OF DOUBLE INTEGRALS IN POLAR CO-ORDINATES We have to evalaute 2 r ( ) 1 r12() f (r , ) dr d over the region bounded by the staight lines = 1 and = 2 and the curves r = r1 (q) and r = r2 (). We first integrate with respect to r between the limits r = r1() and r = r2() and taking as constant. Then the resulting expression is integrated with respect to between the limits = 1 and = 2. The area of integration is ABCD. On integrating first with respect to r, the strip extends from P to Q and the integration with respect to means the rotation ot this strip PQ from AD to BC. Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ Multiple Integral 97 Example 9. Transform the integral to cartesian form and hence evaluate a 0 0 r 3 sin cos dr d. (M.U., II Semester 2000) B Solution. Here, we have a 0 0 r 3 sin cos dr d r=a ...(1) Here the region i.e., semicircle ABC of integration is bounded by r = 0, i.e., x-axis. r = a i.e., circle, = 0 and = i.e., x-axis in the second quadrant. C A O (r sin ) (r cos ) (r d dr) Putting x = r cos , y = r sin , dx dy = r d dr in (1), we get a 2 x2 a a = a a 1 = 2 0 y2 x dx 2 0 a a xy dy dx = a 2 x2 = a x dx a 2 a a x dx a2 x 2 0 Y y dy r = 2 cos = — 2 O 2 (a x ) 2 Since f ( x) is odd function (a x x ) dx = 0 Ans. a f ( x) dx 0 a 2 3 Example 10. Evaluate Solution. 2 0 0 2 x – x2 2 0 0 2 x – x2 rd dr 0 0 2 x – x2 2 X Y ( x 2 y 2 ) dy dx ( x 2 y 2 ) dy dx Limits of y = 2 x – x 2 y2 = 2x – x2 x2 + y2 – 2x = 0 (1) represents a circle whose centre is (1, 0) and radius = 1. Lower limit of y is 0 i.e., x-axis. Region of integration is upper half circle. Let us convert (1) into polar co-ordinate by putting x = r cos , y = r sin r2 – 2 r cos = 0 r = 2 cos Limits of r are 0 to 2 cos Limits of are 0 to 2 2 x=2 = 0 2 ( x y ) dy dx = 2 cos 2 2 r 0 0 (r d dr ) = 4 = 4 02 cos d 4 Example 11. Evaluate 2 0 0 2 x – x2 x dy dx x2 y2 2 cos 3 r 0 dr 2 d 0 2 cos r4 4 0 3 1 3 422 4 Ans. by changing to polar coordinates. Solution. In the given integral, y varies from 0 to y= 2 d 0 ...(1) 2 x – x 2 and x varies from 0 to 2. 2 x – x2 Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ 98 Multiple Integral y2 = 2x – x2 x2 + y2 = 2x In polar co-ordinates, we have r2 = 2r cos r = 2 cos . . 2 In the given integral, replacing x by r cos , y by r sin , dy dx by r dr d, we have For the region of integarion, r varies from 0 to 2 cos and varies from 0 to I= = 2 cos r /2 0 0 / 2 0 / 2 2 cos cos . r dr d r cos dr d 0 0 r 2 cos r2 cos 2 0 d / 2 0 2 cos3 d 2. 2 4 . 3 3 Ans. EXERCISE 2.2 Evaluate the folloing: a (1 – cos ) 1. 0 0 2. 0 0 3. Ans. a (1 cos ) 2 r cos dr d r dr d r 2 a2 A 4. 8 3 a 3 5 3 Ans. a 8 a Ans. 2a – 2 2 r 2 sin d dr r 2 where A is a loop of r2 = a2 cos 2 sin d dr where A is r = 2a cos above initial line. (A.M.I.E. Winter 2001) Ans. A 5. Calculate the integral 6. (x 2 ( x y)2 x2 y 2 dx dy over the circle x2 + y2 1. Ans. – 2 y 2 ) x dx dy over the positive quadrant of the circle x2 + y2 = a2 by changing to polar coordinates. Ans. 7. 8. x 2 y 2 dx dy by changing to polar coordinates, R is the region in the xy-plane bounded by the 38 circles x2 + y2 = 4 (AMIETE, Dec. 2009) Ans. 3 Convert into polar coordinates 0 0 10. a2 5 R 2 a 2 ax – x 2 9. 2a3 3 3 / 2 2a cos dx dy Ans. 0 r d dr 0 3 4 4 r dr d, over the area bounded between the circles r = 2b cos and r = 2b cos . Ans. 2 (a – b ) 5 r sin dr d over the area of the cardiod r = a (1 + cos ) above the initial line. Ans. 8 a3 Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ Multiple Integral 2 99 11. Ax 12. Transform the integral dr d, where A is the area between the circles r = a cos and r = 2a cos . 1 x 0 0 Ans. 28a 3 9 f ( x, y ) dy dx to the integral in polar co-ordinates. /4 sec 0 0 Ans. f ( r , ) r d dr 2.4 CHANGE OF ORDER OF INTEGRATION On changing the order of integration, the limits of integration change. To find the new limits, we draw the rough sketch of the region of integration. Some of the problems connected with double integrals, which seem to be complicated, can be made easy to handle by a change in the order of integration. a a x dx dy by changing the order of integration. Example 12. Evaluate 0 y 2 x y2 (AMIETE, June 2010, Nagpur University, Summer 2008) Solution. Here we have Y a a x y=a dx dy B I = 0 y 2 x y2 Here x = a, x = y, y = 0 and y = a The area of integration is OAB. x On changing the order of integration Lower limit of = y y = 0 and x=a upper limit is y = x. Lower limit of x = 0 and upper limit is x = a. yx 1 a X dy I = xdx 0 O y=0 A 2 0 x y2 yx Y y 1 xdx tan 1 0 x0 x a x x dx tan 1 tan 1 0 0 x x a π a aπ π dx [ x] Ans. 0 4 0 4 4 Example 13. Change the order of integration in a I= 1 2– x 0 x2 xy dx dy and hence evaluate the same. y=a y = B (a, a) x x=a O y=0 A X (A.M.I.E.T.E., June 2010, 2009, U.P. I Sem., Dec., 2004) Solution. I = 1 2– x 0 x2 xy dx dy Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ 100 Multiple Integral The region of integration is shown by shaded portion in the figure bounded by parabola y = x2 and the line y = 2 – x. The point of intersection of the parabola y = x2 and the line y = 2 – x is B (1, 1). In the figure below (left) we have taken a strip parallel to y-axis and the order of integration is 1 2– x 0 x dx x2 y dy In the second figure above we have taken a strip parallel to x-axis in the area OBC and second strip in the area ABC. The limits of x in the area OBC are 0 and area ABC are 0 and 2 – y. = 1 0 1 = 2 y dy 1 0 y 0 2 x dx y dx 1 y 2 dy 1 2 2 1 2– y 0 x2 1 x dx y dy 0 2 0 y (2 – y ) 2 dy 2 1 2 1 1 2 4 3 y4 1 1 = 2 y – y 6 2 3 4 6 2 1 = 1 y3 1 3 0 2 y 2 y 0 1 (4 y – 4 y 2 y and the limits of x in the 2– y x2 y dy 2 0 y 3 ) dy 32 4 1 8 – 3 4 – 2 3 – 4 1 1 96 – 128 48 – 24 16 – 3 1 5 9 3 6 24 24 8 6 2 12 Example 14. Evaluate the integral x 0 dy y – xe x2 y 0 0 x exp – integration Solution. Limits are given y = 0 and y = x x = 0 and x = Here, the elementary strip PQ extends from y = 0 to y = x and this vertical strip slides from x = 0 to x = . The region of integration is shown by shaded portion in the figure bounded by y = 0, y = x, x = 0 and x = . On changing the order of integration, we first integrate with respect to x along a horizontal strip RS which extends from x = y to x = and this horizontal strip slides from y = 0 to y = to cover the given region of integration. New limits : x = y and x = y = 0 and y = We first integrate with respect to x. Thus, Ans. x2 dx dy by changing the order of y (U.P. I Semester Dec., 2005) x2 y 2x – y dx = dy – – e dx 0 y 2 y 2 x – y – y = 0 dy – e y 0 dy 0 e 2 2 y y2 2 y e – y dy 0 2 Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ Multiple Integral 101 y = 2 (– e –y 1 ) – (e – y ) 2 0 1 (Integrating by parts) 1 = (0 – 0) 0 2 2 Ans. Example 15. Change the order of the integration x x y 0 0e y dy dx (B.P.U.T.; I Semester 2008) Solution. Here, we have 0 y dy dx Here the region OAB of integration is bounded by y = 0 (x-axis), y = x (a straight line), x = 0, i.e., y axis. A strip is drawn parallel to y-axis, y varies 0 to x and x varies 0 to . On changing the order of integration, first we integrate w.r.t. x and then w.r.t. y. A strip is drawn parallel to x-axis. On this strip x varies from y to and y varies from 0 to . x 0 0 Hence B Y x x y e 0 e xy y dy dx = 0 y dy 0 = 0 O 2 y dy [0 e y ] y 0 X A A x 1 Ans. 2 O Example 16. Change the order of integration in the double integral = x y=0 Y e xy dx = e xy y dy y y = y y = y x= 2 e y dy 2a 0 2 ax 2 ax – x 2 B X V dx dy Solution. Limits are given as x = 0, x = 2a y = 2 ax and y = 2 ax – x 2 y2 = 2 ax and (x – a)2 + y2 = a2 The area of integration is the shaded portion OAB. On changing the order of integration first we have to integrate w.r.t. x, The area of integration has three portions BCE, ODE and ACD. 2a 0 = dx 2a 0 dy 2 ax 2 ax – x 2 V dy 2a a a a 2 y2 y 2 / 2 aV dx 0 dy y 2 / 2 a a 2a 0 a a 2 – y2 dy V dx V dx Ans. Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ 102 Multiple Integral EXERCISE 2.3 Change the order of integration and hence evaluate the following: a cos y dy x 1. 0 0 2. 0 x2 (a – x) (a – y) 2a 3a – x cos y dx a a 2 ay 0 0 ( x 2 y 2 ) dy dx Ans. (a) dy (a – x ) (a – y ) ( x2 y 2 ) dx 3a a dy (b) 2 sin a. 3a – y 0 ( x 2 y 2 ) dx (b) 314 a 4 . 35 4a 1 x a ya 3. 0 x2 (x 4. 0 2 Ans. 0 dy y f ( x, y ) dx dy Ans. 0 dx a 2 – y2 a – a 0 6. 0 x 1 y 2 ) –1/ 2 dy dx a2 – y 2 5. 1 a Ans. ( a ) 0 dy y dx f ( x , y ) dx dy x dy dx y x dy dx a a Ans. 0 Ans. 0 dy y b a a bx / a 5 2x 2 a a a a2 – y 2 2a ( x 2 y 2 ) – 1 / 2 dx. a f ( x, y) dy a 2 – x2 a 2 – x2 dx a 2 x y – a 2 – x2 y 2 xdx 0 1 2a dx a a x–a f ( x, y) dy f ( x, y ) dy 2 y dy y xdx; log 0 4 e 7. 0 y x2 y2 8. 0 0 9. Ans. 0 dy 2 – y f ( x, y ) dx 2 dy y – 2 f ( x, y ) dx 0 2 – x f ( x, y) dx dy y x 2 2 –y 2 2 –y Ans. – dx – x ( y – x ) e dy (A.M.I.E., Summer 2000) 0 – y ( y – x ) e dx dy 1 2 y (A.M.I.E.T.E., June 2009) y 0 x y yx dy a 2a – x 0 x2 xy dx dy (U.P. I Semester, Dec., 2007) Ans. a ay xy dx dy 2a – y xy dx dy, 3a 2 10. 11. 12. 13. 0 a – (M.P. 2003) b a2 – y2 a Ans. ( a ) 0 dy ay / b x dx x dy dx 5 7 0 xy dx dy Ans. 0 1 2 a b 3 (b ) x dx 5 0 0 a 2 – ( x – a )2 0 y dy, 8 2 4 a 3 [Hint: Put x = a sin2 dx = 2 a sin cos d ] 14. 1 0 2a 1 – y 1/ 3 x –1 0 16. 0 0 ( x 18. 19. 0 1 y 2 y 2 ) dx dy 2 1 2– y 0 1 1 x 3 dx Ans. –1 Ans. 0 dy x2 dx 4a ( x y )3 dy 15. 17. y – 1/ 2 (1 – x – y )1/ 2 dx dy ( x 2 y 2 ) dx dy a 1– x – 0 2a 4a y y 1 2 1 (1 – x – y ) 2 dy , – 3 7 ( x y )3 dx Ans. 1 2– x 0 dx x ( x 2 y 2 ) dy, 5 3 a5 20 (U.P., I Semester, Dec. 2007, A.M.I.E., Summer 2001) 1 2 2 2 1 1 1 0 1 x 2 y 2 dx dy 0 y x2 y 2 dx dy R x2 y 2 dy dx Recognise the region R of integration on the R.H.S. and then evaluate the integral on the right in the order indicated. (AMIETE, Dec. 2004) Ans. Region R is x = 0, x = y, y = 1 and y = 2, log 2. 4 Express as single integral and evaluate : a 0 0 a a 2 – x2 x y2 x 2 y 2 dx dy by changing into polar coordinates. Ans. a 0 2 0 x dx dy a 0 2 a2 – x 2 x dx dy Ans. 0 a 2 dy a2 – y 2 y x dx, 5a 3 6 2 Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ Multiple Integral 20. Express as single integral and evaluate : 1 y 0 0 ( x 21. 103 2 y 2 ) dx dy 2 1 2– y 0 ( x 2 y 2 ) dx dy Ans. 1 2–x 0 dx x ( x 2 y 2 ) dy, If f(x, y) dx dy, where R is the circle x2 + y2 = a2, is R equivalent to the repeated integral. 5 3 2 1 (AMIE winter 2001) [Ans. (r, ) r dr d . ] 0 0 2.5 CHANGE OF VARIABLES Sometimes the problems of double integration can be solved easily by change of independent variables. Let the double integral as be f ( x, y ) dx dy. It is to be changed by the new variables R u, v. The relation of x, y with u, v are given as x = f(u, v), y = (u, v). Then the double integration is converted into. f {(u, v), (u, v)} | J | du dv, where R x x ( x, y ) u v du dv du dv dx dy= | J | du dv = y y (u , v ) u v 2 Example 17. Evaluate ( x y ) dx dy, where R is the parallelogram in the xy-plane with R vertices (1, 0), (3, 1), (2, 2), (0, 1), using the transformation u = x + y and v = x – 2y. (U.P., I Semester, 2003) Solution. The region of integration is a parallelogram ABCD, where A (1, 0), B (3, 1), C (2, 2) and D (0, 1) in xy-plane. The new region of integration is a rectangle ABCD in uv-plane xy-plane A (x, y) A (1, 0) A (u, v) uv-plane A (x + y, x – 2y) A (1 + 0, 1 – 2 × 0) A (1, 1) and u x y v x – 2 y B (x, y) B (3, 1) B (u, v) B (x + y, x – 2y) B (3 + 1, 3 – 2 × 1) C (x, y) C (2, 2) C (u, v) C (u, v) C (2 + 2, 2 – 2 × 2) D (x, y) D (0, 1) D (u, v) B (4, 1) C (4, – 2) D (1, – 2) D (0 + 1, 0 – 2 × 1) 1 (2u v ) 3 1 y (u – v ) 3 1 1 3 – 1 3 – 3 x x ( x, y ) u J = (u , v ) y u and x 2 v 3 y 1 v 3 Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ 104 Multiple Integral dx dy = | J | du dv 1 du dv 3 4 u3 1 1 dv – 2 7 dv 7 [v]2 7 3 21 Ans. 3 R 1 Example 18. Using the transformation x + y = u, y = uv, show that 2 ( x y ) dx dy = 1 4 2 u . –2 1 1 du dv = 3 1 – 2 3 1 2 , integration being taken over 105 the area of the tringle bounded by the lines x = 0, y = 0, x + y = 1. 1/ 2 [ xy (1 – x – y)] Solution. dx dy 1/ 2 [ xy (1 – x – y)] dx dy x + y = u or x = u – y = u – uv, x x ( x, y ) u v du dv dx dy du dv = y y (u , v ) u v 1– v –u du dv u du dv. dx dy = v u x=0 u (1 – v) = 0 u = 0, v = 1 y=0 uv = 0 u = 0, v = 0 x+y=1 u=1 Hence, the limits of u are from 0 to 1 and the limits of v are from 0 to 1. The area of integration is a square OPQR in uv-plane. On putting x = u – uv, y = uv, dx dy = u du dv in (1), we get 1/ 2 (u – uv) (uv)1/ 2 (1 – v )1/ 2 u du dv 3 3 3 2 1/ 2 (1 – v )1/ 2 dv 2 2 2 = 0 u (1 – u ) du 9 5 2 3 1 1 1 1 1 1 2. . 1 2 2 2 2 2 2 2 2 = 7 5 3 2 1 105 7 5 3 3 2 2 2 2 2 2 2 1 1 1/ 2 v 0 3 Ans. EXERCISE 2.4 1. 2. 3. y sin dx dy by means of the transformation u = x + y, v = y from (x, y) to x y 1 (u, v) Ans. 1 1– x y 1 dy dx (e – 1) Using the transformation x + y = u, y = uv, show that 0 0 2 ex y (A.M.I.E. Winter 2001) x– y 1 dx dy sin 1 where R is bounded Using the transformation u = x – y, v = x + y, prove that cos x y 2 R by x = 0, y = 0, x + y = 1 1 1 1 Hint : x 2 (u v ), y 2 (v – u ) so that | J | 2 Evaluate –( x y ) 0 0 e Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ Multiple Integral 105 2.6 AREA IN CARTESIAN CO-ORDINATES Let the curves AB and CD be y1 = f1 (x) and y2 = f2 (x). Let the ordinates AD and BC be x = a and x = b. So the area enclosed by the two curves y1 = f1 (x) and y2 = f2(x) and x = a and x = b is ABCD. Let P(x, y) and Q(x + x, y + y) be two neighbouring points, thent the area of the small rectangle PQ = x. y. Area of the vertical strip = y2 y x y x y12 dy x y 0 lim the width of the strip is constant y1 throughout. If we add all the strips from x = a to x = b, we get b y b y x y12 dy a dx y12 dy The area ABCD = xlim 0 a Area b a y2 y1 dx dy Example 19. Find the area bounded by the parabola y2 = 4ax and its latus rectum. Solution. Required area = 2 (area (ASL) a = 2 0 2 ax 0 dy dx a = 2 0 2 ax dx a x3 / 2 8a 2 = 4 a 3 3/ 2 0 Example 20. Find the area between the parabolas y2 = 4 ax and x2 = 4 ay. Solution. y2 = 4ax x2 = 4ay On solving the equations (1) and (2) we get the point of intersection (4a, 4a). Divide the area into horizontal strips of width y, x varies from P, ...(1) ...(2) y2 to Q, 4 ay and then y varies from O(y = 0) to 4a A (y = 4a). The required area = 4a 0 dy 4ay y 2 / 4a dx 4a 4a 4a y y 3/ 2 y3 4ay = 0 dy x y2 / 4 a 0 dy 4ay – 4a 3 – 4 a 12 a 2 0 3 4 a (4 a ) 32 2 16 2 16 2 3/ 2 = 3 (4a) – 12a 3 a – 3 a 3 a 2 Ans. Example 21. Find by double integration the area enclosed by the pair of curves y = 2 – x and y2 = 2(2 – x) Solution. y= 2 – x y2 = 2 (2 – x) Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ 106 Multiple Integral On solving the equations (1) and (2), we get the points of intersection (2, 0) and (0, 2). A= The required area = = 2 2 (2 – x ) 0 dx y 2 – x dx dy 2 2 (2 – x ) 0 dx 2 – x dy 2 dx [ 4 – 2 x – 2 x] 0 2 2 x2 (4 2 x )3/2 – 2 x = 2 0 3 – 2 2 1 4 8 2 x2 3/ 2 = – (4 – 2 x) – 2 x = – 4 2 3 3 2 0 3 Ans. EXERCISE 2.5 Use double integration in the following questions: 1. Find the area bounded by y = x – 2 and y2 = 2x + 4. x2 + y2 = a2 Ans. 18. 2. Find the area between the circle and the line x + y = a in the first quadrant. Ans. ( – 2)a2/4 3. Find the area of a plate in the form of quadrant of the ellipse 4. Find the area included between the curves y2 = 4 a(x + a) and y2 = 4 b (b – x). x2 a 2 y2 b 2 1. Ans. Ans. (A.M.I.E.T.E., Summer 2001) 5. Find the area bounded by (a) y2 = 4 – x and y2 = x. (b) x – 2y + 4 = 0, x + y – 5 = 0, y = 0 (A.M.I.E., Winter 2001) Find the area enclosed by the leminscate r2 = a2 cos 2 . 7. Find the area common to the circles x2 + y2 = a2 and x2 + y2 = 2ax. 8. Find the area included between the curves y = x2 – 6x + 3 and y = 2x + 9. (A.M.I.E., Summer 2001) 2.7 8 ab 3 16 2 3 27 Ans. 2 Ans. a2 Ans. 6. 9. ab 4 Determine the area of region bounded by the curves xy = 2, 4y = x2, y = 4. 3 2 Ans. – a 4 3 Ans. 88 22 3 28 Ans. – 4 log 2 3 (U.P. I Semester 2003) AREA IN POLAR CO-ORDINATES Area = r d dr Let us consider the area enclosed by the curve r = f (). Let P (r, ), Q(r + r, + ) be two neighbouring points. Draw ares PL and QM, radii r and r + r. PL = r, PM = r Area of rectangle PLQM = PL × PM = (r) (dr) = r r. The whole area A is composed of such small rectangles. Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ Multiple Integral 107 Hence, A= lim r 0 r . r r d dr 0 Example 22. Find by double integration, the area lying inside the cardioid r = a (1 + cos ) and outside the circle r = a. (Nagpur University, Winter 2000) Solution. r = a (1 + cos ) ...(1) r=a ...(2) Solving (1) and (2), by eliminating r, we get a(1 + cos ) = a 1 + cos = 1 cos = 0 – or 2 2 limits of r are a and a(1 + cos ) to 2 2 Required area = Area ABCDA limits of are – = /2 a (1 cos ) = – / 2 a = a2 2 2 = a / 2 – / 2 / 2 0 /2 for cardioid – / 2 r for circle r d dr r2 = – / 2 2 /2 r d dr [(1 cos )2 – 1] d (cos 2 2 cos ) d = a2 2 a (1 cos ) a / 2 d – / 2 (cos 2 2 cos ) d / 2 2 / 2 2 = a 0 cos d 2 0 cos d a2 2 / 2 2 ( 8) = a 2 (sin )0 a 2 = Ans. 4 4 4 Example 23. Find by double integration, the area lying inside the circle r = a sin and outside the cardioid r = a (1 – cos ). Solution. We have, r = a sin ...(1) r = a (1 – cos ) ...(2) Solving (1) and (2) by eliminating r, we have sin = 1 – cos sin + cos = 1 Squaring above, we get sin2 + cos2 + 2 sin cos = 1 1 + sin 2 = 1 sin 2 = 0 2 = 0 or = = 0 or 2 The required area is shaded portion in the fig. Limits of r are a(1 – cos ) and a sin , limits of are 0 and a sin 2 r 0 a (1 – cos ) a sin Required area = = 2 0 r2 1 d 2 2 a (1 – cos ) . 2 dr d / 2 2 0 a [sin 2 – (1 – cos )2 ] d Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ 108 Multiple Integral a2 / 2 (sin 2 – 1 – cos 2 2 cos ] d 0 2 a2 / 2 (– 2 cos 2 2 cos ) d = 2 0 / 2 a2 / 2 – 2 cos 2 d 2 cos d = 0 0 2 2 a – 2. 2 (sin )0 / 2 = 2 4 a2 a2 – 2 sin – sin 0 = – 2 = a 2 1 – = Ans. 2 2 2 2 2 4 Example 24. Find by double integration, the area lying inside a cardioid r = 1 + cos and outside the parabola r (1 + cos ) = 1. Solutio. We have, r = 1 + cos ...(1) r (1 + cos ) = 1 ...(2) Solving (1) and (2), we get (1 + cos ) (1 + cos ) = 1 (1 + cos )2 = 1 1 + cos = 1 cos = 0 2 1 limits of r are 1 + cos and limits of are – to . 1 cos 2 2 Required area = Area ADCBA (Shaded portion) = /2 = – / 2 1 cos r 1 1 cos d dr = r2 2 – 2 2 1 cos 1 1 cos d = 1 2 / 2 – / 2 (1 cos ) 2 – d (1 cos ) 1 2 1 1 2 d (1 cos 2 cos ) – = – /2 2 2 2 2 cos 2 1 / 2 1 2 4 = 2 (1 cos 2 cos ) – sec d 0 2 4 2 /2 1 2 2 2 = 0 (1 cos 2 cos ) – 1 tan sec d 4 2 2 / 2 / 2 = 0 = 0 1 1 cos 2 1 2 cos – 1 tan 2 sec 2 d 2 2 2 4 /2 1 cos 2 1 2 2 2 1 2 2 2 cos – 4 sec 2 tan 2 sec 2 d sin 2 1 2 sin – 2 tan = 2 4 4 2 1 1 3 = 0 2 sin – tan – tan 2 4 2 2 4 6 2 2 tan 3 3 2 0 1 1 3 4 3 = 2– – 4 2 6 4 3 4 Ans. Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ Multiple Integral 109 EXERCISE 2.6 3a 2 2 2 Ans. a Ans. a2 Ans. 11 1. Find the area of cardioid r = a(1 + cos ). 2. 3. 4. 5. Find Find Find Find 6. Ans. ( a 2 b 2 ) 2 Show that the area of the region included between the cardioides r = a(1 + cos ) and r = a (1 – cos ) the the the the Ans. area of the curve r2 = a2 cos 2. area enclosed by the curve r = 2 a cos area enclosed by the curve r = 3 + 2 cos . area enclosed by the curve r3 = a2 cos2 + b2 sin2. 7. a2 (3 – 8). 2 Find the area outside the circle r = 2 and inside the cardioid r = 2(1 + cos ). Ans. ( + 8) 8. Find the area inside the circle r = 2a cos and outside the circle r = a. is 9. 2.8 3 Ans. 2a 2 4 3 Find the area inside the circle r = 4 sin and outside the lemniscate r2 = 8 cos 2 . 8 Ans. 4 3 – 4 3 VOLUME OF SOLID BY ROTATION OF AN AREA (DOUBLE INTEGRAL) When the area enclosed by a curve y = f (x) is revolved about an axis, a solid is generated, we have to find out the volume of solid generated. Volume of the solid generated about x-axis = b y2 ( x ) a y ( x ) 2 PQ dx dy 1 Example 25. Find the volume of the torus generated by revolving the circle x2 + y2 = 4 about the line x = 3. Solution. x2 + y2 = 4 V= (2 PQ ) dx dy 2 (3 – x) dx dy 2 4 – x2 = 2 – 2 dx – = 2 = 2 2 –2 2 –2 4 – x2 (3 – x ) dy dx 3 y – x y 4 – x2 – 4 – x2 dx [3 4 – x 2 – x 4 – x 2 3 4 – x 2 – x 4 – x 2 ] x 2 2 = 4 [3 4 – x – x 4 – x ] dx 4 3 2 2 4 – x2 3 4 x 1 sin –1 (4 – x 2 )3 / 2 2 2 3 – 2 2 = 4 6 6 24 Ans. 2 2 Example 26. Calculate by double integration the volume generated by the revolution of the cardioid r = a (1 – cos ) about its axis. (AMIETE, June 2010) Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ 110 Multiple Integral Solution. r = a (1 – cos ) V = 2 = y dx dy V 2 (r d dr ) y 2 d r dr ( r sin ) = 2 sin d 0 a (1 – cos ) 2 r dr 0 3 a (1 – cos ) r 2 3 3 = 2 0 sin d 0 a (1 – cos ) sin d 3 3 0 3 2 a (1 – cos )4 2 a 3 8 [16] a3 = Ans. 3 4 12 3 0 Example 27. A pyramid is bounded by the three co-ordinate planes and the plane x + 2y + 3z = 6. Compute this volume by double integration. Solution. x + 2y + 3z = 6 ...(1) x = 0, y = 0, z = 0 are co-ordinate planes. The line of intersection of plane (1) and xy plane (z = 0) is x + 2y = 6 ...(2) The base of the pyramid may be taken to be the triangle bounded by x-axis, y-axis and the line (2). An elementary area on the base is dx dy. Consider the elementary rod standing on this area and O having height z, where 6 – x – 2y 3z = 6 – x – 2y or z = 3 6 – x – 2y Volume of the rod = dx dy, z, Limits for z are 0 and . 3 6– x Limits of y are 0 and and limits of x are 0 and 6. 2 6– x 6– x 6 6 6 – x – 2y 2 z dx dy dx dy Required volume = 0 0 0 0 2 3 6–x 2 1 6 1 6 6(6 – x) x(6 – x ) 6 – x 2 – – = 0 dx 6 x – xy – y 2 = 0 dx 0 3 3 2 2 2 1 6 36 – 6 x 6 x – x 2 36 x 2 –12 x – – dx 3 0 2 2 4 1 6 2 2 (72 – 12 x – 12 x 2 x – 36 – x 12 x) dx = 12 0 6 1 6 2 1 x 3 12 x 2 1 ( x – 12 x 36) dx – 36 x [72 – 216 216] 6 Ans. = = 12 0 12 3 2 12 0 EXERCISE 2.7 = 1. Find the volume of the sphere x2 + y2 + z2 = a2 by revolving area of the circle x2 + y2 = a2. Ans. 4 3 a 3 2.9 CENTRE OF GRAVITY x = x dx dy , y y dx dy dx dy dx dy Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ Multiple Integral 111 Example 28. Find the position of the C.G. of a semi-circular lamina of radius a if its density varies as the square of the distance from the diameter. (AMIETE, Dec. 2010) Solution. Let the bounding diameter be as the x-axis and a line perpendicular to the diameter and passing through the centre is y-axis. Equation of the circle is x2 + y2 = a2. By symmetry x 0. y dx dy y dx dy = a2 – x 2 2 a a ( y ) dx dy – a dx 0 y4 dx – a 4 2 a –x a = a – a dx 0 2 ( y ) y dx dy – x2 y 2 dy 2 0 y3 a – a dx 3 0 2 y 3 dy a 2 – x2 3 a –a a 4 –a (a 2 – x 2 )2 dx 2 ( a 2 – a 2 sin 2 ) 2 a cos d – 2 2 ( a 2 – a 2 sin 2 ) 3/ 2 a cos d – 2 3 = 4 Put x = a sin (a 2 – x 2 )3/ 2 dx 3 42 3a 5 3 3a 8 16 32a = 3 1 4 4 15 3 15 42 2 4 2 a5 – 2 4 a4 – 2 cos5 d cos 4 d 32 a Hence C.G. is 0, 15 Example 29. Find C.G. of the area in the positive quadrant of the curve x2/3 + y2/3 = a2/3. x dx dy , y y dx dy Solution. For C.G. of area; x dx dy dx dy a x= 0 a ( a 2 / 3 – x 2 / 3 )3 / 2 0 ( a 2 / 3 – x 2 / 3 )3 / 2 0 dx 0 a = x dx 0 x dx (a a 0 2/3 a dy – x 2 / 3 )3/ 2 dx (a 2 / 3 – x 2 / 3 )3 / 2 ( a 2 / 3 – x 2 / 3 )3 / 2 0 x dx y 0 dy Ans. a ( a 2.3 – x 2 / 3 )3 / 2 dx y 0 0 0 3 2/3 – a cos ( a 2 0 2/3 – a2 / 3 (a 2 [Put x = a cos3] a 2 / 3 cos 2 )3/ 2 (– 3a cos2 sin d ) cos 2 )3/ 2 (– 3a cos 2 sin d ) 5 6 2 2a 4 3 3 3 2 4 5 02 3 a cos sin cos sin d a 02 sin cos d 2 2 = 5 3 2 3 2 4 2 02 3a sin cos sin d 02 sin cos d 2 2 2 4 Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ 112 Multiple Integral = 3 4a 3 11 2 2 (2) (6) a 256 a , Similarly, y 256 a 1 9 7 5 3 1 315 315 2 2 2 2 2 2 256 a 256 a , Hence, C.G. of the area is . 315 315 Example 30. Find by double integration, the centre of gravity of the area of the cardioid r = a (1 + cos ). Solution. Let ( x , y ) be the C.G. the cardioid By Symmetry, y 0. x dx dy x = A = a (1 cos ) – 0 A dx dy A (r cos ) (r d dr ) a (1 cos ) – 0 r = dx dy A x dx dy 3 a ( a cos ) 0 a (1 cos ) a (1 cos ) d 0 r d dr – cos d 3 a (1 cos ) 2 – cos d 0 – cos d . r dr r dr a3 (1 cos )3 3 a2 r2 d (1 cos )2 d – – 2 2 0 3 a3 2 2 2 cos – 1 1 2 cos – 1 d 3 – 2 2 = a2 2 – 1 d 1 2 cos 2 – 2 a3 a2 2 6 2 cos – 1 8 cos d 4 cos 4 d = – 3 – 2 2 2 2 3 8a 8 – cos6 d 2a 2 cos 4 d = 2 cos – 3 – 2 2 2 2 8a 3 8 – cos6 d 4 a 2 cos 4 d = 2 cos 0 0 3 2 2 2 / 2 16a3 / 2 (2 cos8 t – cos6 t ) (2 dt ) 4a 2 cos4 t (2 dt ) = 0 3 0 3 3 1 32 a 2 7 5 3 1 5 3 1 – 8a 2 = 3 8 6 4 2 2 6 4 2 2 42 2 32a3 35 5 8a 3 15 16 5a 2 3 = 3 128 – 32 8a 16 3 128 2 8a 3 24 2.10 CENTRE OF GRAVITY OF AN ARC Example 31. Find the C.G. of the arc of the curve x = a ( + sin ), y = a(1 – cos ) in the positive quadrant. Solution. We know that, x Ans. xds , y yds ds ds Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ Multiple Integral 113 2 2 dx dy d d d Now, ds = {a 2 (1 cos )2 a 2 sin 2 } d a 1 2 cos cos2 sin 2 d = d 2a cos d 2 2 a 2 sin cos d 0 2 2 2 = a 1 2 cos 1 d a 2(1 cos ) d a 4 cos x = = x dx ds a 2 0 a ( sin ) 2a cos 2 d 0 2a cos d 2 a 2 cos 2 2 sin 2 cos 2 d 2 0 2 sin 2 0 2 (2t 0 cos t 2 sin t cos 2 t ) 2 dt cos3 t 2 1 4 = 2a t sin t cos t – 2a – 1 a – 3 2 3 3 0 a (1 – cos ) 2a cos d a 2 sin 2 cos d y ds 0 0 2 2 2 y= ds 0 2a cos 2 d 0 cos 2 d a r sin 3 20 4a 2a 4 2a = Hence, C.G. of the arc is a – , 3 2 3 3 3 3 2 sin 2 Ans. 0 EXERCISE 2.8 1. 2. 3. Find the centre of gravity of the area bounded by the parabola y2 = x and the line x + y = 2. 1 8 Ans. , – 2 5 Find the centroid of the tetrahedron bounded by the coordinate planes and the plane x + y + z = 1, the 1 1 2 density at any point varying as its distance from the face z = 0. Ans. , , 5 5 5 Find the centroid of the area enclosed by the parabola y2 = 4 ax, the axis of x and latus rectum. 3a 3a Ans. , 20 16 4. 5. a 2 , 0 Ans. 8 Find the centroid of solid formed by revolving about the x-axis that part of the area of the ellipse Find the centroid of the loop of curve r2 = a2 cos 2 . x2 6. 7. y2 3a Ans. , 0 1 which lies in the first quadrant. a b2 8 Find the average density of the sphere of radius a whose density at a distance r from the centre of the k r3 0 1 sphere is 0 1 k 3 . 2 a The density at a point on a circular lamina varies as the distance from a point O on the circumference. Show that the C.G. divides the diameter through O in the ratio 3 : 2. 2 Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ 114 Multiple Integral 1.11 TRIPLE INTEGRATION Let a function f(x, y, z) be a continuous at every point of a finite region S of three dimensional space. Consider n sub-spaces S1, S2, S3, ... Sn of the space S. If (xr, yr, zr) be a point in the rth subspace. n f (x , y , z ) S , as n , S 0 is known as the triple integral of The limit of the sum r r r r r r 1 f (x, y, z) over the space S. Symbolically, it is denoted by f ( x, y, z) dS S x2 y2 z2 x1 y1 z1 ( x) dx It can be calculated as f ( x, y, z) dx dy dz. First we integrate with respect to z treating x, y as constant between the limits z1 and z2. The resulting expression (function of x, y) is integrated with respect to y keeping x as constant between the limits y1 and y2. At the end we integrate the resulting expression (function of x only) within the limits x1 and x2. x2 b x1 a y2 2 ( x ) y1 1( x ) z2 f2 ( x, y ) ( x, y) dy f ( x, y, z) dz z1 f1( x, y ) First we integrate from inner most integral w.r.t. z, then we integrate with respect to y and finally the outer most with respect to x. But the above order of integration is immaterial provided the limits change accordingly. Example 32. Evaluate Solution. 1 R ( x y z ) dx dy dz, where R : 0 x 1, 1 y 2, 2 z 3. 3 ( x y z) 2 dy 2 2 1 2 0 1 1 2 1 2 1 1 dx dy [( x y 3) 2 ( x y 2)2 ] dx (2 x 2 y 5) .1. dy = 2 0 2 0 1 1 2 0 dx dy 3 1 ( x y z) dz = dx 2 1 = 2 1 = 8 2 (2x 2 y 5) 2 1 dx 4 1 8 0 1 1 0 (4x 16) . 2 dx Example 33. Evaluate the integral : Solution. log 2 0 = = = = = x x log y 1 0 1 0 dx [(2x 4 5)2 (2x 2 5)2 ] 1 x2 1 9 (x 4) dx 4 x 4 2 0 2 2 log 2 0 x 0 x log y 0 Ans. e x y z dz dy dx. xyz e dz dy dx. e dz e dx e dy(e ) e dx e dy 1) e dx e dy (e . e 1) e dx e dy (e e dx e ( y e 1) dy = e dx ( ye 1) e e . e dy e dx ( ye 1) e e = e dx [(xe 1) e e 1 e ] e dx [ xe e e 1 e ] ( xe e e ) dx 0 0 log 2 0 log 2 0 log 2 0 log 2 0 log 2 0 x x x x 0 x 0 x y x log y 0 y y x log y 0 log 2 x x x x 0 log 2 y z x log y 0 0 x x y log y x 0 x x y x y 0 0 x log 2 z x 2x y x 2x 0 log 2 xy x 0 x x x x 2x x 0 x log 2 3x 3x x 0 Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ Multiple Integral 115 log 2 e3 x e3 x e3x 1. dx ex 3 3 3 0 = x log 2 x e 3x e3x e 3x ex 3 9 3 0 log 2 3 log 2 e3 log 2 e3 log 2 1 1 e e log 2 1 3 9 3 9 3 3 3 log 2 log 23 elog 2 elog 2 1 1 = e elog 2 1 3 9 3 9 3 8 8 8 1 1 8 19 = log 2 2 1 log 2 3 9 3 9 3 3 9 = Example 34. Evaluate Solution. I = = log 2 0 0 = log 2 x 0 x 0 log 2 0 e x 0 xy x y 0 zx y e 0 e x y z dx dy dz. (M.U. II Semester, 2005, 2003, 2002) dx dy e x y (e x y 1) dx dy = log 2 0 4x 2x log 2 2x 0 x 2 x e2 y x y e . 2 e . e dx = 0 log 2 Ans. x e 2( x y ) e ( x y ) dx dy 0 e4 x e2x e2x e x dx 2 2 log 2 0 e e 1 1 1 e e e 2 log 2 e 2 log 2 x elog 2 1 = 8 2 4 e 8 2 4 8 2 4 0 elog 16 elog 4 elog 4 1 1 1 elog 2 1 = 8 2 4 8 2 4 16 4 4 1 1 1 5 = 2 1 Ans. 8 2 4 8 2 4 8 Example 35. Evaluate R 4 log 2 ( x 2 y 2 z 2 ) dx dy dz where R denotes the region bounded by x = 0, y = 0, z = 0 and x + y + z = a, (a > 0) Solution. R Y ( x 2 y 2 z 2) dx dy dz x+y+z=a or z=a–x–y Upper limit of z = a– x – y On x-y plane, x + y + z = a becomes x + y = a as shown in the figure. Upper limit of y = a – x Upper limit of x = a = = = = a 0 a 0 a 0 a x0 dx dx dx ax y0 ax 0 ax 0 dy a x y z0 2 2 2 ( x y z ) dz = x=0 x+ O a 0 dx y=0 ax 0 y= a x=a X ax y z3 dy x 2 z y 2 z 3 0 (a x y)3 dy x 2 (a x y) y 2 (a x y) 3 2 (a x y)3 2 2 3 x (a x) x y (a x) y y dy 3 ax x2 y2 y 3 y 4 (a x y ) 4 dx x 2 (a x ) y (a x ) 2 3 4 12 0 Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ 116 Multiple Integral = = x2 (a x)3 (a x)4 (a x)4 dx x 2 (a x)2 (a x) 2 (a x) 2 3 4 12 0 2 4 4 a x a 1 (a x) (a x) 2 2 2 3 4 (a x) dx (a x 2 a x x ) dx 2 6 2 6 0 0 a 5 a 1 x 3 ax 4 x5 (a x) a5 a5 a5 a5 a5 = a2 Ans. 3 4 10 30 6 4 10 30 20 2 0 dx dy dz Example 36. Compute if the region of integration is bounded by the ( x y z 1)3 coordinate planes and the plane x + y + z = 1. (M.U., II Semester 2007, 2006) Solution. Let the given region be R, then R is expressed as 0 z 1 – x – y, 0 y 1 – x, 0 x 1. R = 1 0 dx = 1 2 = 1 2 = 1 2 dx dy dz = ( x y z 1)3 1 0 dx 1 x 0 1 x y dy 1 x y dz ( x y z 1)3 0 1 dy 2 2( x y z 1) 0 1 1 x 1 1 dx dy 2 2 0 0 ( x y 1) ( x y 1 x y 1) 1 x 1 1 x 1 1 1 1 y 1 dx dy dx 4 2 0 4 x y 1 0 0 0 ( x y 1) 2 1 1 1 1 1 1 x 1 1 1 x dx dx x 1 1 x x 1 2 0 4 2 x 1 0 4 1 x 0 1 (1 x) 2 x 1 1 log( x 1) log 2 8 2 2 2 0 1 5 = log 2 2 16 = 1 2 Example 37. Evaluate 1 1 5 log 2 8 2 8 Ans. 2 x yz dx dy dz throughout the volume bounded by the planes x = 0, x y z = 1. a b c y = 0, z = 0, (M.U. II Semester 2003, 2002, 2001) Z Solution. Here, we have I = 2 x yz dx dy dz ...(1) Putting x = au, y = bv, z = cw dx = a du, dy = b dv, dz = c dw in (1), we get I = c a2bc u 2 v w a bc du dv dw Limits are for u = 0, 1 for v = 0, 1 – u and for w = 0, 1 – u – v u+v+w = 1 I= 1 uo 1 u v0 = a3b2c 2 2 = a3b 2c 2 2 1 u v w0 1 1 u 0 0 1 1 u 0 0 3 2 2 2 a b c u vw du dv dw = 1 1u 0 0 a X b 1 u v w2 ab c u v 2 0 3 2 2 Y O 2 du dv u 2v(1 u v)2 du dv u 2v (1 u)2 2(1 u)v v 2 du dv Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ Multiple Integral 117 = a3b 2c 2 2 = a3b2c 2 2 = a3b 2c 2 2 3 2 2 = abc 2 1 1 u 0 0 1 0 1 0 1 0 u 2 [(1 u)2 v 2 (1 u) v 2 v 3 ] du dv 1u v2 v3 v4 u 2 (1 u)2 2(1 u) 2 3 4 0 du (1 u)4 2(1 u) 4 (1 u)4 u2 du 2 3 4 a3b2c 2 u 2 (1 u)4 du = 24 12 1 0 u3 1 (1 u)5 1 du a 3b2c 2 2! 4! a 3 b 2 c 2 abc a3b2c 2 3 5 . = Ans. (3, 5) . 2520 . 24 24 24 7! 8 2.12 INTEGRATION BY CHANGE OF CARTESIAN COORDINATES INTO SPHERICAL COORDINATES Sometime it becomes easy to integrate by changing the cartesian coordinates into spherical coordinates. The relations between the cartesian and spherical polar co-ordinates of a point are given by the relations x = r sin cos y = r sin sin z = r cos dx dy dz = | J | dr d d = r 2 sin dr d d Note. 1. Spherical coordinates are very useful if the expression x2 + y2 + z2 is involved in the problem. 2. In a sphere x2 + y2 + z2 = a2 the limits of r are 0 and a and limits of are 0, and that of are 0 and 2. 3 2 2 = (x Example 38. Evaluate the integral 2 y 2 z 2 ) dx dy dz taken over the volume enclosed by the sphere x2 + y2 + z2 = 1. Solution. Let us convert the given integral into spherical polar co-ordinates. By putting x = r sin cos ; y = r sin sin ; z = r cos 2 = 2 4 2 0 5 5 0 = d 0 sin d 1 0 Example 39. Evaluate x2 + y2 + z2 = a2. Solution. Here, we have r 4 dr 2 1 0 0 0 ( x 2 y 2 z 2 ) dx dy dz = 2 0 d 0 r 2 (r 2 sin d d dr) 1 r5 1 sin d = 5 5 0 2 0 d cos 0 2 5 2 0 d Ans. (x 2 y 2 z 2 ) dx dy dz over the first octant of the spheree (M.U. II Semester 2007) I = (x 2 y 2 z 2 ) dx dy dz ...(1) 2 Putting x = r sin cos , y = r sin sin , z = r cos and dx dy dz = r sin dr d d in (1), we get Limits of r are 0, a for are 0, for are 0, . 2 2 Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ 118 Multiple Integral 2 2 a I= 0 0 0 r 2 . r 2 sin dr d d = 2 0 d 2 0 sin d a 0 r 4 dr x 2 y 2 z 2 r 2 sin 2 cos 2 r 2 sin 2 sin 2 r 2 cos 2 r 2 sin 2 r 2 cos2 r 2 a 5 5 5 a a /2 /2 r . . = 0 cos 0 = . (1) . Ans. 2 5 10 5 0 dx dy dz Example 40. Evaluate throughout the volume of the sphere x2 + y2 + z2 = a2. 2 x y2 z2 (M.U. II Semester 2002, 2001) Solution. Here, we have dx dy dz I = ...(1) 2 x y2 z2 2 Putting x = r sin cos , y = r sin sin , z = r cos and dx dy dz = r sin dr d d in (1), we get The limits of r are 0 and a, for are 0 and for are 0 and in first octant. 2 2 2 I= 8 2 I= 8 0 2 0 d 0 2 0 a 0 r 2 sin dr d d r2 sin d a 0 /2 dr = 8 0 = 8 .1. a = 4a 2 [Sphere x2 + y2 + z2 lies in 8 quadrants] cos 0 /2 r a0 8 2 0 (0 1)(a 0) Ans. EXERCISE 2.9 Evaluate the following : 1 2 3 1 2 3 4 x x y 0 0 1. 2. 3. 2 dx dy dz z dz dy dx (M.U., II Semester 2002) Ans. 48 (R.G.P.V. Bhopal I Sem. 2003) Ans. 70 0 1 1 1 0 1 1 1 1 1 z xz ( x 2 y 2 z 2 ) dx dy dz 2 Ans. 6 y 2 z 2 ) dz dy dx 4. 0 0 0 (x 5. 1 0 x z ( x y z ) dx dy dz 6. (x y z) (AMIETE, June 2006) Ans. 1 (AMIETE, Summer 2004) Ans. 0 dx dy dz, where R : 1 x 2; 2 y 3; 1 z 3 Ans. 2 R 7. 2 2 1 xy 2 z dx dy dz (AMIETE, Dec. 2007) dx 1 1 Ans. 26 8. 0 0 dy 2 2 1 2 (M.U. II Semester, 2003) 1 10. 2 x yz dz Ans. 1 x yz dx dy dz throughout the volume bounded by x = 0, y = 0, z = 0, x + y + z = 1. 0 9. 3 0 1 x 0 1 x2 y2 0 dz dy dx Ans. 1 3 e 11. 1. 1 log y 1 ex 1 Ans. log z dz dx dy Ans. 1 2520 1 2 (e 8e 13) 2 Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ Multiple Integral 12. 119 y dx dy dz, where T is the region bounded by the surfaces x = y2, x = y + 2, 4z = x2 + y2 and T z = y + 3. 2 x 2x 2 y 0 0 0 13. 14. (x y z) ex y z (AMIETE Dec. 2008) 12 6 1 e e 1 1 1 dz dy dx Ans. [e4 1] [e2 1] (M.U. II Sem., 2003) 3 6 3 6 3 2 dx dy dz over the tetrahedron bounded by the planes x = 0, y = 0, z = 0 and x + y + z = 1. 15. 17. 18. a ax 0 0 a x y 0 1 z xz 1 2 0 y xz x y 0 0 xy Ans. 5 x 2 dx dy dz Ans. a 16. 60 2 (4 x 2) /2 2 (4 x 2) /2 8 x2 y 2 x2 3 y 2 ( x y z) dz dx dy (M.U. II Semester, 2000, 02) ( x y z) dx dy dz (M.U. II Semester 2004) 20. a b c dx dy dz over the volume of the ellipsoid a b x y z dx dy dz throughout the volume of the tetrahedron 21. x2 y2 z2 x2 y2 2 2 2 2 2 Ans. 0 Ans. 16 z2 c2 24. 25. dx dy dz taken throughout the volume of the sphere x2 + y2 + z2 = 1, lying in the first 1 x2 y 2 z 2 2 8 a 2 Ans. h 2 0 Ans. 2d a(1 cos ) 0 r dr h 0 /2 a sin (a 2 r 2 )/ a 0 0 0 r 1 a(1 cos ) dz z dx dy dz over the volume common to the sphere x Ans. r d dr dz 2 2 dx dy dz (1 x 2 y 2 z 2 ) 2 dx dy dz V 27. 2 2 2 3/ 2 2a5 15 2 Ans. 8 Ans. where V is the volume in the first octant. over the volume bounded by the spheres x 2 + y 2 + z 2 = 16 and (x y z ) x2 + y2 + z2 = 25. 28. 2 T 5a 3 64 + y2 + z2 = a 2 and the cylinder x2 + y2 + z2 = ax. 26. 4 abc 3 1 l m n Ans. (l m n) . lmn octant. 23. = 1. Ans. l 1 m 1 n 1 x 0, y 0, z 0, x + y + z 1. 22. dz dy dx Ans. 8 2 x2 y 2 z 2 x2 y2 z2 dx dy dz throughout the volume of the ellipsoid 2 2 2 = 1. a b c a 2 b2 c 2 2 Ans. abc 4 19. 1 1 8 (M.U. II Semester, 2001, 03) 2 2 Ans. 4 log (5/4) 2 z dx dy dz over the volume bounded by the cylinder x + y = a and the paraboloid x2 + y2 = z and the plane z = 0. Ans. a 8 12 Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ 120 Multiple Integral 2.13 VOLUME = dx dy dz. The elementary volume v is x . y . z and therefore the volume of the whole solid is obtained by evaluating the triple integral. V = x y z V dx dy dz. ρ dx dy dz if is the density.. (ii) In cylindrical co-ordinates, we have V r dr dφ dz V Note : (i) Mass = volume × density = (iii) In spherical polar co-ordinates, we have V r 2 sin θdr dθ dφ V Example 41. Find the volume of the tetrahedron bounded by the planes x = 0, y = 0, z = 0 and x + y + z = a. (M.U. II Semester, 2005, 2000) Solution. Here, we have a solid which is bounded by x = 0, y = 0, z = 0 and x + y + z = a Z planes. The limits of z are 0 and a – x – y, the limits of y are 0 and 1 – x, the limits of x are 0 and a. z = a–x–y V= = = = = = a x0 a x0 x0 a 0 a 0 y0 ax y dx dy dz = z0 a x0 ax y0 x+y+z=a z a0 x y dx dy ax (a x y) dx dy O y0 2 a 0 a ax Y ax y ay xy 2 0 dx z=0 dxdy 2 (a x ) dx a(a x) x(a x) 2 2 a2 x2 2 a ax ax x 2 ax 2 dx a2 x2 ax dx 2 2 X O y=a–x y=x a 3 a2 ax 2 x 3 1 1 1 a a3 = .x . 2 6 2 2 6 6 2 0 Y (a, 0, 0) Ans. X Example 42. Find the volume of the cylindrical column standing on the area common to the parabolas y2 = x, x2 = y and cut off by the surface z = 12 + y – x2. (U.P., II Sem., Summer 2001) Y Solution. We have, y2 = x x2 = y z = 12 + y – x2 V= = 1 0 dx x x2 dy 2 y 12 y x 2 1 dz = 0 0 x 2 (12 y x 2 ) dy 2 X x =y O x y2 dx 12 y x2 y 2 2 0 x 1 x dx =x Y Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ X Multiple Integral = 1 0 121 x x4 5/ 2 2 x 4 dx 12 x x 12 x 2 2 1 2 x2 2 7 / 2 x5 x 5 3/2 x 4x3 = 12x 4 7 10 5 3 0 1 2 1 1 1 2 1 1 560 35 40 14 28 569 4 4 = Ans. 4 7 10 5 4 7 10 5 140 140 Example 43. A triangular prism is formed by planes whose equations are ay = bx, y = 0 and x = a. Find the volume of the prism between the planes z = 0 and surface z = c + xy. (M.U. II Semester 2000; U.P., Ist Semester, 2009 (C.O) 2003) = 8 Z bx a a Solution. Required volume = 0 a = = = 0 a 0 0 bx a c xy dz dy dx (c xy) dy dx 0 x=a bx a O xy 2 cy dx 2 0 0 z=0 Y ay = bx X cbx b 2 3 bc 2 x dx a 2a a a z = c + xy 0 a x2 b2 2 2 0 2a a x4 4 0 a b c b2 a 2 a b (4c a b) 2 8 8 2.14 VOLUME OF SOLID BOUNDED BY SPHERE OR BY CYLINDER = Ans. We use spherical coordinates (r, , ) and the cylindrical coordinates are (, , z) and the relations are x = cos , y = sin . Example 44. Find the volume of a solid bounded by the spherical surface x2 + y2 + z2 = 4a2 and the cylinder x2 + y2 – 2 a y = 0. x2 + y2 + z2 = 4a2 Solution. 2 ...(1) 2 x +y –2ay = 0 ...(2) Considering the section in the positive quadrant of the xy-plane and taking z to be positive (that is volume above the xy-plane) and changing to polar co-ordinates, (1) becomes r2 + z2 = 4a2 z= 2 4a r z2 = 4 a2 – r2 Y (0, a) X 2 O (2) becomes r2 – 2 a r sin = 0 r = 2a sin Volume = dx dy dz = 4 /2 d 0 Y 2a sin r dr 0 4a 2 r 2 dz 0 (Cylindrical coordinates) Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ X 122 Multiple Integral = 4 = 4 = = 4 3 /2 d 0 /2 0 32 a 3 0 r dr z 0 4a 2 r 2 2a sin = ( 8a 3 cos3 8a 3 ) d / 2 /2 d 0 4 3 8 4a 3 3 /2 0 / 2 2a sin r dr . 4a 2 r 2 0 (4a 2 4a 2 sin 2 )3/2 8a3 d (1 cos3 ) d 0 1 3 1 4 cos 3 4 cos d 0 / 2 3 = = 4 1 d (4a 2 r 2 )3/2 3 0 / 2 0 3 2a sin 32 a 1 3 sin 3 sin 3 12 4 0 32 a 3 3 32 a 3 1 3 2 12 4 = 3 2 2 3 Ans. Example 45. Find the volume enclosed by the solid 2/3 2/3 2/3 x y z =1 a b c Solution. The equation of the solid is 2/3 2/3 2/3 x y z a b c = 1 1/3 x Putting = u x = a u3 a 1/3 y = v y = b v3 b 1/3 z = w z = c w3 c The equation of the solid becomes u2 + v2 + w2 = 1 V = d x = 3 au2 du d y = 3 bv2 dv d z = 3 c w2 dw ...(1) dx dy dz ...(2) On putting the values of dx, dy and dz in (2), we get V = 27abc u v w du dv dw 2 2 2 ...(3) (1) represents a sphere. Let us use spherical coordinates. u = r sin cos , v = r sin sin , w = r cos , du dv dw = r2sin dr d d On substituting spherical coordinates in (3), we have V = 27abc . 8 = 216 a b c 1 r0 1 r0 /2 0 r 8 dr /2 r 2 sin 2 cos 2 . r 2 sin 2 sin 2 0 /2 . r2 cos2 . r2 sin dr d d sin 2 cos2 d 0 3 3 1 r9 2 2 3 = 216 a b c . 9 0 2 3 2 / 2 sin 5 cos 2 d 0 3 3 3 3 3 1 2 2 1 2 2 . . = 24 a b c . . 2 2 9 3 9 2 2 Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ Multiple Integral 123 2 1 1 3 2 2 2! 2 . = 6abc . = 2! 7 5 3 3 2 2 2 2 1 1 4 . abc 4 7 5 3 35 222 Ans. Example 46. Find the volume bounded above by the sphere x2 + y2 + z2 = a2 and below by the cone x2 + y2 = z2. (U.P. II Semester 2002) 6abc . Solution. The equation of the sphere is x2 + y2 + z2 = a2 ...(1) x2 + y2 = z2 and that of the cone is ...(2) In polar coordinates x = r sin cos , y = r sin sin , z = r cos The equation (1) in polar co-ordinates is Z (r sin cos )2 + (r sin sin )2 + (r cos )2 = a2 r2 sin2 cos2 + r2 sin2 sin2 + r2 cos2 = a2 r2 sin2 (cos2 + sin2 ) + r2 cos2 = a2 r2 sin2 + r2 cos2 = a2 O r2 (sin2 + cos2 ) = a2 r2 = a2 r = a The equation (2) in polar co-ordinates is X (r sin cos )2 + (r sin sin )2 = (r cos )2 r2 sin2 (cos2 + sin2 ) = r2 cos2 r2 sin2 = r2 cos2 x2 + y2 + z2 = a2 Y x2 + y2 = z2 4 Thus equations (1) and (2) in polar coordinates are respectively, r = a and = 4 The volume in the first octant is one fourth only. tan2 = 1 tan = 1 = and from 0 to . 4 2 The required volume lies between x2 + y2 + z2 = a2 and x2 + y2 = z2. Limits in the first octant : r varies 0 to a, from 0 to V = 4 2 4 0 = 4 0 2 d 0 4 a 0 r 2 sin dr d d = 4 d 0 3 sin d . 0 2 a 4a 3 3 3 2 0 4 0 a r3 sin d 30 d cos 04 2 3 1 a 1 3 2 2.15 VOLUME OF SOLID BOUNDED BY CYLINDER OR CONE = 4a3 02 3 1 2 1 Ans. We use cylindrical coordinates (r, , z). Example 47. Find the volume of the solid bounded by the parabolic y2 + z2 = 4x and the plane x = 5. Solution. y2 + z2 = 4x, x = 5 5 V= 0 dx 2 x 2 x dy 4x y2 4x y 2 dz 4 5 0 dx 2 x 0 dy 4 x y2 dz 0 Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ 124 Multiple Integral = 4 = 4 5 dx 2 x 0 0 5 y dx 2 0 dy z 0 4 x y2 5 4 dx 0 2 x dy 4 x y 2 0 2 x 4x y 4x y 2 sin 1 2 2 x 0 = 4 5 0 2 x 2 dx 4 0 5 x dx 0 5 x2 = 4 50 2 0 Example 48. Calculate the volume of the solid bounded by the following surfaces : x2 + y2 = 1, z = 0, 2 Solution. x+y+z=3 2 x +y = 1 x+y+z = 3 z = 0 Required Volume = Ans. ...(1) ...(2) ...(3) 3x y 0 dx dy dz = dx dy z (3 x y) dx dy On putting x = r cos , y = r sin , dx dy = r d dr, we get = = 2 0 (3 r cos r sin ) r d dr = 2 d 0 1 3r 2 r 3 r3 d cos sin = 2 3 3 0 2 0 1 (3r r 2 cos r 2 sin ) dr 0 1 3 1 2 3 cos 3 sin d 2 1 1 1 1 1 3 = sin cos = 3 sin 2 cos 2 3 Ans. 3 3 3 2 3 3 0 Example 49. Find the volume bounded by the cylinder x2 + y2 = 4 and the planes y + z = 4 and z = 0. Z Solution. x2 + y2 = 4 y = 4 x 2 y + z = 4 z = 4 – y and z = 0 y x varies from –2 to + 2. V= = = = dx dy dz = 2 dx 2 2 dx 2 2 2 4 x2 2 dx 2 4x 4 x dy 4 y dy (4 y) = z= 4 dz 0 O dy z 0 4 x2 4x 2 4 y 4 x2 2 2 + 2 Y 2 x +y =4 y2 dx 4 y 2 2 2 4 x2 X 4 x2 1 1 dx 4 4 x 2 (4 x 2 ) 4 4 x 2 (4 x 2 ) 2 2 2 2 4 x x 4 x 2 dx 8 4 x 2 sin 1 = 16 Ans. 2 2 2 2 2 Example 50. Find the volume in the first octant bounded by the cylinder x2 + y2 = 2 and the planes z = x + y, y = x, z = 0 and x = 0. (M.U. II Semester 2005) Solution. Here, we have the solid bounded by = 8 Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ Multiple Integral 125 2 Z 2 x + y = 2 (cylinder) (or r2 = 2) z = x + y z = r (cos + sin ) y = x r sin = r cos tan = 1 q= 4 x = 0 r cos = 0 cos = 0 = 2 z varies from 0 to r (cos + sin ) r varies from 0 to 2 (plane) (plane) O X varies from to 4 2 V = = = = / 2 / 4 / 2 / 4 / 2 / 4 Y Y x = y=x y = /2 2 r0 2 r0 2 r (cos sin ) == / / 44 Y x=0 r dr d dz z0 r (cos sin ) 0 r z X O dr d x2 + y2 = 2 r= 2 r 2 (cos sin ) dr d r 0 2 r3 2 2 (cos sin ) d = 3 0 3 /4 /2 / 2 (cos sin ) d / 4 1 2 2 1 Ans. (1 0) 3 2 2 Example 51. Show that the volume of the wedge intercepted between the cylinder x2 + y2 = 2ax and planes z = mx, z = nx is (m – n) a3. (M.U. II Semester, 2000) Solution. The equation of the cylinder is x2 + y2 = 2 a x z = mx we convert the cartesian coordinates into cylindrical coordinates. x = r cos y = r sin = 2 2 2 2 /2 sin cos / 4 = 3 3 x2 + y2 = 2ax r2 = 2ar cos r = 2a cos r varies from 0 to 2a cos z = nx to 2 2 z varies from z = nx (z = nr cos ) to z = m x (z = m r cos ) varies from and V= 2 = 2 = 2 /2 0 / 2 0 /2 0 = 2 (m n) 2a cos r0 2a cos r 0 mr cos r dr d dz Y z nr cos r = 2a cos mr cos r z nr cos dr d 2a cos r .(m n) r cos dr d O X r0 /2 0 2a cos r 2 cos dr d r0 Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ 126 Multiple Integral = 2 (m n) / 2 0 2a cos r3 3 0 cos d = 2 (m n) /2 0 8a3 cos3 cos d 3 16 (m n) 3 / 2 16 (m n) 3 3 1 a cos 4 d = . a . . . (m n) a3 Ans. 3 4 2 2 3 0 Example 52. A cylindrical hole of radius b is bored through a sphere of radius a. Find the volume of the remaining solid. (M.U. II Semester 2004) Z Solution. Let the equation of the sphere be 2 2 2 2 x +y +z = a z = a2 – r 2 Now, we will solve this problem using cylindrical coordinates x = r cos y = r sin z = z = a 2 ( x 2 y 2 ) i.e., Limits of z are 0 and Oz=0 a2 r 2 a Y Limits of r are a and b. 2 and the limits of are 0 and /2 = a 8 (a V = 8 X 2 0 rb /2 a 0 a r 2 r dr d dz = 8 z0 2 /2 0 rb z 0 a 2 r2 r dr d r 2 )1/2 . r dr d r b a (a 2 r 2 )3/2 3/ 2 8 1 . d = 3 0 2 b 3 3 8 2 4 2 / 2 (a b 2 ) 2 0 (a b 2 ) 2 = 3 3 Example 53. Find the volume cut off from the paraboloid = 8 a /2 / 2 3 (a 2 b2 ) 2 d 0 Ans. y2 z = 1 by the plane z = 0. 4 Solution. We have y2 x2 z = 1 (Paraboloid) 4 z = 0 (x-y plane) x2 z varies from 0 to 1 – x2 – (M.U. II Semester 2005) ...(1) ...(2) 2 y 4 Z y varies from 2 1 x 2 to 2 1 x 2 x varies from –1 to 1. V= dx dy dz = 1 = 1 2 1 x 2 2 1 x 1 = 4 2 dx 1 2 1 x2 2 1 x2 dy y2 2 1 x 4 0 2 y 2 1 x dx dy 4 0 1 2 1 x2 0 y2 2 1 x dx dy 4 dz –1 –2 O 2 1 X Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ Y Multiple Integral 127 Y = 4 2 1 x2 y3 2 (1 x ) y 12 0 1 0 y = 2 1 – x2 . dx x2 + 1 8 (1 x 2 ) . 2 1 x 2 (1 x 2 )3/ 2 d x 12 0 X 1 2 2 3/ 2 2 3/ 2 = 4 2(1 x ) 3 (1 x ) d x 0 On putting x = sin , we get = 4 y2 =1 4 X O V = 4 1 4 16 (1 x 2 )3/2 dx 3 3 0 /2 y = –2 1 – x2 ( sin 2 )3/2 cos d Y 0 /2 16 16 3 1 cos 4 d . . . Ans. 3 0 3 4 2 2 Example 54. Find the volume enclosed between the cylinders x2 + y2 = a x, and z2 = a x. Solution. Here, we have x2 + y2 = ax ...(1) z2 = ax ...(2) = V= = = = 2 2 dx dy dz a dx 0 a dx 0 a ax x 2 ax x 2 dy ax ax ax x 2 ax x dz = 2 ax dy z 0 = 2 2 a dx 0 dx 0 ax dx 2 ax x 2 4 a 0 a ax x 2 2 a x ax x 2 ax x 2 2 dy ax dz 0 dy ax = 2 a ax dx y ax x 2 0 Z a x a x dx 0 2 Putting x = a sin so that dx = 2a sin cos d, we get V= 4 a = 8a 3 /2 =a 2 z x a sin 2 a a sin 2 . 2a sin cos d 0 /2 O sin 3 cos 2 d 0 2 X (a, 0) 2 2 x + y = ax 3 3 3 3 2 2 16a = 8a 4a 15 7 5 3 3 2 . 2 2 2 2 EXERCISE 2.10 3 ax x 2 Y Ans. x y z abc = 1 Ans. a b c 6 2. Find the volume bounded by the cylinders y2 = x and x2 = y between the planes z = 0 and 1. Find the volume bounded by the coordinate planes and the plane. x + y + z = 2. Ans. 11 30 3. Find the volume bounded by the co-ordinate planes and the plane. 1 6l m n 4 4. Find the volume of the sphere x2 + y2 + z2 = a2 by triple integration. (AMIETE, June 2009) Ans. a 3 3 lx+ my+n z= 1 (A.M.I.E.T.E. Winter 2001) Ans. Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ 128 Multiple Integral 6. 7. 8. 9. 10. 11. 12. x2 y2 z2 4 a b c = 1 Ans. a b c2 3 Find the volume bounded by the cylinder x2 + y2 = a2 and the planes y + z = 2a and z = 0. (M.U. II Semester 2000, 02, 06) Ans. 2a3 2 2 2 Find the volume bounded by the cylinder x + y = a and the planes z = 0 and y + z = b. Ans. a2b Find the volume of the region bounded by z = x2 + y2, z = 0, x = – a, x = a and y = – a, y = a. 8 4 Ans. a 3 Find the volume enclosed by the cylinder x2 + y2 = 9 and the planes x + z = 5 and z = 0. Ans. 45 – 36 Compute the volume of the solid bounded by x2 + y2 = z, z = 2x.(A.M.I.E., Summer 2000) Ans. 2 Find the volume cut from the paraboloid 4 z = x2 + y2 by plane z = 4. (U.P. I Semester, Dec. 2005) Ans. 32 By using triple integration find the volume cut off from the sphere x2 + y2 + z2 = 16 by the plane 5. Find the volume of the ellipsoid 2 2 z = 0 and the cylinder x2 + y2 = 4 x. Ans. 13. The sphere x2 + y2 + z2 = a2 is pierced by the cylinder x2 + y2 = a2 (x2 – y2). 64 (3 4) 9 8 5 4 2 3 Prove that the volume of the sphere that lies inside the cylinder is 3 4 3 3 a . 14. Find the volume of the solid bounded by the surfaces z = 0, 3 z = x2 + y2 and x2 + y2 = 9. (A.M.I.E.T.E., Summer 2005) Ans. 27 2 x y 15. Obtain the volume bounded by the surface z = c 1 a 1 b and a quadrant of the elliptic cylinder x2 y2 = 1, z > 0 and where a, b > 0. (A.M.I.E.T.E., Dec. 2005) a 2 b2 2 2 16. Find the volume of the paraboloid x + y = 4z cut off by the plane z = 4. Ans. 32 17. Find the volume bounded by the cone z2 = x2 + y2 and the paraboloid z = x2 + y2. Ans. 6 128a 3 15 19. Find the volume of the solid bounded by the plane z = 0, the paraboloid z = x2 + y2 + 2 and the cylinder x2 + y2 = 4. Ans. 16 18. Find the volume enclosed by the cylinders x2 + y2 = 2ax and z2 = 2 a x. 20. The triple integral dx dy dz (a) Volume of region (c) Area of region T Ans. gives (b) Surface area of region T (d) Density of region T. (A.M.I.E.T.E., 2002) Ans. (a) 2.16 SURFACE AREA Let z = f(x,y) be the surface S. Let its projection on the x-y plane be the region A. Consider an element 8x. y in the region A. Erect a cylinder on the element x. y having its generator parallel to OZ and meeting the surface S in an element of area s. x y = s cos , Where is the angle between the xy-plane and the tangent plane to S at P, i.e., it is the angle between the Zaxis and the normal to S at P. Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ Multiple Integral 129 The direction cosines of the normal to the surface F (x, y, z) = 0 are proportional to F F F , , x y z The direction of the normal to S [F = f (x, y) – z] are proportional to those of the Z-axis are 0, 0 , 1. z x 2 2 , 1 , 2 2 2 2 z z z z 1 1 x x y y 1 Hence cos = (cos = l1l2 + m1 m2 + n1 n2) z 2 z 2 1 x y z 2 z 2 z 2 z 2 x y S = cos = 1 x y ; S = A 1 dx dy x y x y 2 2 Example 55. Find the surface area of the cylinder x + z = 4 inside the cylinder x2 + y2 = 4. Solution. x2 + y2 = 4 z z x z 2x2z 0 or , 0 x x z y Direction cosines = , z y z z , , 1 and x y z z 1 x y 2 2 x2 z 2 x2 z 4 z 1 1 = = = 2 2 y z z 4 x2 x 2 2 2 4 x 2 z z Hence, the required surface area = 8 0 0 1 dx dy x y 2 2 2 1 2 2 4 x 1 4x 2 [ y ]0 dx = 16 [ 4 x 2 ] dx = 8 dx dy = 16 0 2 0 0 0 2 2 4x 4x 4 x 2 2 = 16 0 dx = 16 ( x)0 = 32 Ans. 2 2 2 Example 56. Find the surface area of the sphere x + y + z = 9 lying inside the cylinder x2 + y2 = 3y. Solution. x2 + y2 + z2 = 9 z z x 2x 2 z = 0, x x z z z y 2x 2 z = 0, y y z 2 2 2 z 2 z 2 x r cos x y x y2 z2 9 9 1 = 2 2 1 = = = z z z2 9 x 2 y 2 9 r 2 y r sin x y x2 + y2 = 3y or r2 = 3 r sin or r = 3 sin . Hence, the required surface area /2 / 2 3sin z 2 z 2 3 sin r dr 3 r d dr = 12 d 0 = 1 dx dy = 4 x 9 r 2 y 9 r 2 0 0 0 /2 = 12 0 d [ 9 r 2 ]30 sin = 12 /2 [ 9 9 sin 2 3] d 0 Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ 130 Multiple Integral /2 / 2 ( cos 1) d = 36 ( sin )0 = 36 1 = 18 ( – 2) 2 0 Example 57. Find the surface area of the section of the cylinder x2 + y2 = a2 made plane x + y + z = a. Solution. x2 + y2 = a2 x+ y+ z=a The projection of the surface area on xy-plane is a circle x2 + y2 = a2 z z 1 1 = 0 or x x z z 1 = 0 or 1 y y = 36 a 2 x2 = 4 ( 1)2 ( 1)2 1 = 2 2 a z z 1 dx dy = 4 x y 0 0 by the ... (1) ... (2) 2 2 z z 1 = x y Hence the required surface area a Ans. 0 a = 4 3 [ y ]0 a2 x 2 a dx = 4 3 0 3 a 2 x2 3 dx dy 0 a 2 x 2 dx 0 a a2 x a2 a x 2 a2 x2 sin 1 = 4 3 0 = 4 3 = 3 a Ans. = 4 3 4 2 2 2 2 a 0 Example 58. Find the area of that part of the surface of the paraboloid of the paraboloid y2 + z2 = 2 ax, which lies between the cylinder, y2 = ax and the plane x = a. Solution. y2 + z2 = 2 ax ... (1) y2 = ax ... (2) x= a ... (3) Differentiating (1), we get z z a 2z = 2 a, x x z z z y 2y 2 z = 0, y y z 2 y 2 z 2 2 ax 2 2 z 2 ax y 2 2 a2 y2 a2 y2 z z 1 1 = 2 2 1 = z2 z z x y a2 y2 a2 2 a x a2 y2 2 a x y2 1 = = = 2 a x y2 2 a x y2 2 a x y2 a S = z z 1 dx dy = x y 0 ax a = a 2 2 ax ax 0 ax ax 0 ax a 2 2 ax 2 ax y 2 a a2x 2 ax y a 2 dx dy = a 0 ax a 2 x dx ax y 2 ax y ax dx dy 1 2 ax y 2 dy Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ Multiple Integral 131 a = a 0 a = a 0 = = 1. 2. 3. 5. 2 2ax y ax ax 1 1 a 2 x dx sin 1 sin 1 = 2 2 a a 2 x dx = 0 a a 0 a 2 x dx 4 4 a 2 [(a 2 x )3 / 2 ]0a 2 2 3 a a2 [(3a)3/ 2 a3/ 2 ] = [3 3 1] 6 6 EXERCISE 2.11 Ans. Find the surface area of sphere x2 + y2 + z2 = 16. Ans. 64 Find the surface area of the portion of the cylinder x 2 + y 2 = 4 y lying inside the sphere x2 + y2 + z2 = 16. Ans. 64. Show that the area of surfaces cz = xy intercepted by the cylinder x2 + y2 = b2 is 4. a a 2 x dx sin 1 c2 x2 y2 dx dy , where A is the area of the circle x2 + y2 = b2, z = 0 c A 1 Ans. 2 π (c 2 b 2 ) 2 c 2 3 c 2 2 2 2 2 2 Find the area of the portion of the sphere x + y + z = a lying inside the cylinder x + y = ax. Ans. 2 ( – 2) a2 Find the area of the surface of the cone z2 = 3 (x2 + y2) cut out by the paraboloid z = x2 + y2 using surface integral. Ans. 6 2.17 CALCULATION OF MASS We have, Volume = V dx dy dz [Density = Mass per unit volume] Mass = Density = = f (x, y, z) Mass = Volume × Density Mass f ( x, y, z ) dx dy dz V dx dy dz V Example 59. Find the mass of a plate which is formed by the co-ordinate planes and the plane x y z 1, the density is given by = k x y z. a b c (U.P., I Semester, Dec., 2003) Solution. The plate is bounded by the planes x = 0, y = 0, z = 0 and Mass = dx dy dz = z y z b 1 a 1 c b c dx 0 0 0 c x y z 1. a b c dy dz (k xyz ) y z a 1 x 2 b c y dy dx = k z dz y dy = k 0 z dz 0 2 0 2 2 z z 2 2 b 1 c b 1 ka c z y a y z c y z dz 1 dy 1 = k 0 z dz 0 c y dy = 0 2 0 c b 2 b c 2 z b 1 k a2 c z y3 2 y 2 z c y 1 z dz = 1 dy 2 0 0 2 c b c b c b 1 0 z c y z a 1 b cx 0 c z b 1 c 0 Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ 132 Multiple Integral k a2 = 2 k a2 = 2 c 0 c 0 2 y2 z dz 2 z y4 2 y3 1 c 3b 4 b2 b2 z dz 2 z b4 1 c 4 b2 4 z b 1 c z 1 c 0 4 4 z 1 c z 2 b3 1 c 3 b 4 4 b2 b 2 2b 2 k a 2 b2 c z z 1 dz z 1 dz = [Put z = c sin2 ] 0 2 4 3 c 2 12 0 c k a 2 b2 c 2 2 2 2 4 = 0 c sin (1 sin ) (2 c sin cos d ) 12 k 2 a2 b 2 c 2 / 2 2 k 2 a2 b 2 c 2 / 2 3 8 9 sin (cos ) sin cos d = = 0 0 sin cos d 12 12 3 1 9 1 k a 2 b2 c2 2 5 k a 2 b 2 c2 k 2 a2 b 2 c 2 2 k a 2 b2 c 2 (1) ( 5) 2 = = = = Ans. 720 12 12 12 392 2 7 265 5 2 2 2.18 CENTRE OF GRAVITY = k a2 2 c x x dx dy dz , y y dx dy dz , z z dx dy dz dx dy dz dx dy dz dx dy dz Example 60. Find the co-ordinates of the centre of gravity of the positive octant of the sphere x2 + y2 + z2 = a2, density being given = k xyz. V x dx dy dz V dx dy dz Solution. x = = z dx dy dz dx dy dz 2 = V x yz dx dy dz V xyz dx dy dz Converting into polar co-ordinates, x = r sin cos , y = r sin sin , z = r cos , dx dy dz = r2 sin dr d d / 2 / 2 a x = = = = Similarly, 2 2 0 0 0 (r sin cos ) ( r sin sin ) (r cos ) (r sin dr d d ) /2 /2 a 2 0 0 0 (r sin cos ) (r sin sin ) (r cos ) (r sin dr d d ) / 2 / 2 a 6 4 2 0 0 0 r sin cos sin cos dr d d / 2 / 2 a 5 3 0 0 0 r sin cos sin cos dr d d / 2 /2 a 6 2 4 0 sin cos d 0 sin cos d 0 r dr / 2 / 2 a 5 3 0 sin cos d 0 sin cos d 0 r dr / 2 sin 5 5 0 /2 sin 4 4 0 cos3 3 0 cos2 2 0 16 a ; y = z = 35 / 2 r7 7 0 a / 2 r6 6 0 a 7 1 1 a 3 5 7 16 a = = 6 35 11 a 24 6 16 a 16 a 16 a , , Hence, C.G. is 35 35 35 Ans. Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ Multiple Integral 133 2.19 MOMENT OF INERTIA OF A SOLID Let the mass of an element of a solid of volume V be x y z. Perpendicular distance of this element from the x-axis = M.I. of this element about the x-axis = x y z y2 z2 y2 z2 M.I. of the solid about x-axis = V ( y 2 z 2 ) dx dy dz M.I. of the solid about y-axis = V ( x 2 z 2 ) dx dy dz M.I. of the solid about z-axis = V ( x 2 y 2 ) dx dy dz The Perpendicular Axes Theorem If Iox and Ioy be the moments of inertia of a lamina about x-axis and y-axis respectively and Ioz be the moment of inertia of the lamina about an axis perpendicular to the lamina and passing through the point of intersection of the axes OX and OY. IOZ = IOX + IOY The Parallel Axes Theorem M.I. of a lamina about an axis in the plane of the lamina equals the sum of the moment of inertia about a parallel centroidal axis in the plane of lamina together with the product of the mass of the lamin a and square of the distance between the two axes. IAB = IXX + My–2 Example 61. Find M.I. of a sphere about diameter. Solution. Let a circular disc of x thickness be perpendicular to the given diameter XX at a distance x from it. The radius of the disc = a 2 x2 Mass of the disc = (a2 – x2) Moment of inertia of the disc about a diameter perpendicular on it 1 1 1 2 2 2 2 2 2 2 2 = MR = [ (a x )] (a x ) = (a x ) 2 2 2 a 1 1 a 4 2 2 4 2 2 2 M.I. of the sphere = a (a x ) dx = 2 0 [a 2 a x x ] dx 2 2 a 5 2 a5 a5 4 2 a 2 x3 x5 = a = a x 3 5 3 5 0 2 4 3 2 2 8 a a = M a2 a5 = = Ans. 5 3 5 15 Example 62. The mass of a solid right circular cylinder of radius a and height h is M. Find the moment of inertia of the cylinder about (i) its axis (ii) a line through its centre of gravity perpendicular to its axis (iii) any diameter through its base. Solution. To find M.I. about OX. Consider a disc at a distance x from O at the base. a4 dx ( a 2 dx ) a 2 = 2 2 M.I. of the cylinder about OX M.I. of the about OX, = (i) h 0 4 2 2 a4 a 4 dx x 0h = a h = ( a 2 h) a = M a = 2 2 2 2 2 Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ 134 Multiple Integral (ii) M.I. of the disc about a line through C.G. and perpendicular to OX. IOX + IOY = IOZ IOX + IOX = IOZ 1 IOZ 2 M.I. of the disc about a line through 1 M a2 M a2 C.G. = 2 2 = 4 IOX = a 2 dx 2 M.I. of the disc about the diameter = a 4 2 a 2 dx h 2 ( a dx ) x M.I. of the disc about line GD = 4 2 2 2 ha h h dx ( a 2 dx) x Hence, M.I. of cylinder about GD = 0 0 4 2 3 3 3 h 2 2 2 2 a h a h a2 h a a h h x x 0 = = 4 3 2 4 2 4 3 2 0 M a2 M h2 a 2 h a 2 h3 = 4 12 4 12 (iii) M.I. of cylinder about line OB (through) base = 2 M a 2 M h2 M a 2 M h2 M h2 h IOB = IG M = = Ans. 4 3 4 12 4 2 Example 63. Find the moment of inertia and radius of gyration about z-axis of the region in x y z the first octant bounded by 1 . a b c Solution. Let r be the density. M.I. of tetrahedron about z-axis = ( dx dy dz ) ( x 2 y 2 ) x b 1 2 a (x 0 a = dx 0 y 2 ) dy x y c 1 a b dz 0 x x y c 1 a b = a dx b 1 a ( x 2 y 2 ) dy ( z ) 0 0 0 x b 1 2 a (x 0 x y y 2 ) dy c c 1 a b x a b 1 2 x x2 y x y3 y 2 1 dy = c 0 dx 0 a x 1 a b a b a = 0 dx x b 1 a 2 x x 2 y 2 y3 x y4 c dx x 1 y 1 = 0 a 2b 3 a 4 b 0 2 2 2 a b3 x x x 2 x b 1 = c 0 dx x 1 b 1 3 a a 2b a 2 2 4 a 2 x x2 x b2 x b2 1 1 = bc 0 x 1 a 2 a 3 a 4 a 3 4 x x b4 x 1 1 1 a a 4b a 4 x 1 dx a Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ Multiple Integral 135 2 ax = bc 0 2 2 x b2 1 a 12 4 x 1 dx a a1 2 2 x3 x 4 b 2 bc x 2 = 0 2 a a 12 a 1 = bc 0 x3 x 4 x5 2 2 3 2 a 5 a 4 x 6 x 2 4 x3 x4 1 2 3 4 dx a a a a b2 12 a 2 x2 6 x2 4 x3 x4 2 3 4 dx x a a a a 0 3 a 1 a a3 a3 b 2 a = bc 0 a 2a 2a a 2 5 12 5 2 3 a3 ab2 abc 2 (a b2 ) = bc 60 60 = 60 abc 2 (a b2 ) 1 2 M .I . 60 (a b2 ) Radius of gyration = = = Ans. 10 abc Mass 6 2.20 CENTRE OF PRESSURE The centre of pressure of a plane area immersed in a fluid is the point at which the resultant force acts on the area. Consider a plane area A immersed vertically in a homogeneous liquid. Let x-axis be the line of intersection of the plane with the free surface. Any line in this plane and perpendicular to x-axis is the y-axis. Let P be the pressure at the point (x, y). Then the pressure on elementary area x y is P x y. Let ( x , y) be the centre of pressure. Taking moment about y-axis. x P dx dy = A A Px dx dy x = A Px dx dy A P dx dy y = A Py dx dy A P dx dy Similarly, Example 64. A uniform semi-circular lamina is immersed in a fluid with its plane vertical and its bounding diameter on the free surface. If the density at any point of the fluid varies as the depth of the point below the free surface, find the position of the centre of pressure of the lamina. Solution. Let the semi-circular lamina be x2 + y2 = a2 By symmetry its centre of pressure lies on OY. Let ky be the density of the fluid. Py dx dy = A (y) y dx dy y = A ( = ky) P dx dy (y) dx dy A A A A (ky . y ) dx dy = a y 3 dx dy (ky . y ) y dx dy = A 2 A y dx dy = a dx 0 a a dx 0 a 2 x2 a 2 x2 y 3 dy y 2 dy Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ 136 Multiple Integral y4 a dx 4 0 a = y3 dx a 3 0 a a 2 x2 a a 2 x2 / 2 3 = 4 / 2 (a cos d ) (a / 2 / 2 2. 3. 2 a dx (a a a dx (a 2 2 x 2 )2 x 2 )3 / 2 a 2 sin 2 )2 (Put x = a sin ) (a cos d ) (a 2 a 2 sin 2 )3/ 2 42 /2 5 32 a 3 a 2 0 cos d 3a 53 = = = /2 / 2 3 1 4 4 15 4 / 2 cos d 4 2 0 cos d 42 2 EXERCISE 2.12 / 2 3a = 4 1. 3 = 4 / 2 cos 5 d Find the mass of the solid bounded by the ellipsoid x2 a2 y2 b2 z2 c2 Ans. 1 and the co-ordinate planes, where the density at any point P (x, y, z) is k xyz. Ans. P If the density at a point varies as the square of the distance of the point from XOY plane, find the mass of the volume common to the sphere x2 + y2 + z2 = a2 and cylinder x2 + y2 = ax. 4 k 5 8 a Ans. 15 2 15 2 2 Find the mass of the plate in the form of one loop of leminscate r = a sin 2 , where = k r2. k π a4 16 Find the mass of the plate which is inside the circle r = 2a cos and outside the circle r = a, if the density varies as the distance from the pole. Find the mass of a lamina in the form of the cardioid r = a (1 + cos ) whose density at any point varies 21 π k a 4 as the square of its distance from the initial line. Ans. 32 x y z a b c Find the centroid of the region in the first octant bounded by + + = 1 . Ans. , , a b c 4 4 4 4 Find the centroid of the region bounded by z = 4 – x2 – y2 and xy-plane. Ans. 0, 0, 3 Find the position of C.G.. of the volume intercepted between the parallelepiped x2+y2 = a(a – z) and the a plane z = 0. Ans. 0, 0, 3 A solid is cut off the cylinder x2 + y2 = a2 by the plane z = 0 and that part of the olane z = mx for which z is positive. The density of the solid cut off at any point varies as the height of the point above plane Ans. 4. 5. 6. 7. 8. 9. 64 ma 45 π If an area is bounded by two concentric semi-circles with their common bounding diameter in a free Ans. z = z = 0. Find C.G. of the solid. 10. surface, prove that the depth of the centre of pressure is 11. An ellipse x2 a2 y2 b2 3 π ( a b) ( a 2 b 2 ) 16 a 2 ab b 2 1 is immersed vertically in a fluid with its major axis horizontal. If its centre be at depth h, find the depth of its centre of pressure. Ans. h b2 4h Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ Multiple Integral 12. 13. 137 A horizontal boiler has a flat bottom and its ends are plane and semi-circular. If it is just full of water, show that the depth of centre of pressure of either end is 0.7 × total depth approximately. A quadrant of a circle of radius a is just immersed vertically in a homogeneous liquid with one edge in 3a 3π a , Ans. 8 16 Find the product of inertia of an equilateral triangle about two perpendicular axes in its plane at a vertex, one of the axes being along a side. Find the M.I. of a right circular cylinder of radius a and height h about axis if density varies as distance the surface. Determine the co-ordinates of the centre of pressure. 14. 15. 2 k π a5 h 5 Compute the moment of inertia of a right circular cone whose altitude is h and base radius r, about (i) from the axis. 16. Ans. π h r4 π h r2 (ii) (2 h 2 3 r 2 ) 10 60 Find the moment of inertia for the area of the cardioid r = a (1 – cos ) relative to the pole. 35 π a 4 Ans. 16 π Find the M.I. about the line = of the area enclosed by r = a (1 + cos ). 2 Find the moment of inertia of the uniform solid in the form of octant of the ellipsoid the axis of symmetry (ii) the diameter of the base. 17. 18. 19. x2 z2 M 2 (b c 2 ) 1 about OX Ans. 5 a b c2 Prove that the moment of inertia of the area included between the curves y2 = 4 ax and x2 = 4 ay about 2 20. y2 Ans. (i) 2 144 M a 2, , where M is the mass of area included between the curves. 35 A solid body of density p is the shape of solid formed by revolution of the cardioid r = a (1 + cos ) about the initial line. Show that its moment of inertia about a straight line through the pole perpendicular the x-axis is 21. 352 5 l a . 105 to the initial line is 22. (U. P. II Semester, Summer 2001) Find the product of inertia of a disc in the form of a quadrant of a circle of radius ‘a’ about bounding (U. P. II Semester, Summer 2002) Ans. radii. 23. Show that the principal axes at the origin of the triangle enclosed by x = 0, y = 0, 1 ab π 1 to the x-axis, where a = tan 2 2 2 2 a b Choose the correct answer: at angles and α 24. The triple integral a4 4 x y 1 are inclined a b (U.P. II Semester Summer 2001) dx dy dz gives T (i) Volume of region T (ii) Area of region T (ii) Surface area of region T (iv) Density of region T. (A.M.I.E.T.E. 2002) Ans. (i) 25. The volume of the solid under the surface az = x2 + y2 and whose base R is the circle x2 + y2 = a2 is given as a 3 (i) (ii) Ans. (ii) 2a 2 4 3 (iii) a (iv) None of the above. [U.P., I. Sem. Dec. 2008] 3 Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ 138 Differential Equations 3 Differential Equations 3.1 DEFINITION An equation which involves differential co-efficient is called a differential equation. For example, dy 2 3. 1 dx d2y dy 1 x 2 1. dx 1 y 2 dy 2 8y 0 2. 2 dx dx 3 2 k d2y dx 2 2 z z u u y nu, 5. x y y x y There are two types of differential equations : 4. x (1) Ordinary Differential Equation A differential equation involving derivatives with respect to a single independent variable is called an ordinary differential equation. (2) Partial Differential Equation A differential equation involving partial derivatives with respect to more than one independent variable is called a partial differential equation. 3.2 ORDER AND DEGREE OF A DIFFERENTIAL EQUATION The order of a differential equation is the order of the highest differential co-efficient present in the equation. Consider 1. L d 2q dt 2 R dq q E sin wt. dt c 3 dy 2 d 2 y 3. 1 2 dx dx 2. cos x d2y 2 dy sin x 8 y tan x 2 dx dx 2 The order of the above equations is 2. The degree of a differential equation is the degree of the highest derivative after removing the radical sign and fraction. The degree of the equation (1) and (2) is 1. The degree of the equation (3) is 2. 3.3 FORMATION OF DIFFERENTIAL EQUATIONS The differential equations can be formed by differentiating the ordinary equation and eliminating the arbitrary constants. Example 1. Form the differential equation by eliminating arbitrary constants, in the following cases and also write down the order of the differential equations obtained. (a) y = A x + A2 (b) y = A cos x + B sin x (c) y2 = Ax2 + Bx + C. (R.G.P.V. Bhopal, June 2008) 138 Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ Differential Equations 139 2 Solution. (a) y = Ax + A dy A On differentiation dx ... (1) 2 dy dy dx dx On eliminating one constant A we get the differential equation of order 1. (b) y = A cos x + B sin x dy A sin x B cos x On differentiation dx Again differentiating Putting the value of A in (1), we get y x d2y dx 2 d2 y A cos x – B sin x d2y dx 2 Ans. ( A cos x B sin x ) d2y y0 Ans. dx 2 dx 2 This is differential equation of order 2 obtained by eliminating two constants A and B. (c) y2 = Ax2 + Bx + C On differentiation 2 y y dy 2 Ax B dx Again differentiating 2 y d2y 2 dy 2 2A 2 dx dx 3 2 2 d3y dy d 2 y On differentiating again y d y dy d y 2 dy d y 0 y 3 3 0 Ans. dx dx 2 dx dx 2 dx dx 3 dx dx 2 This is the differential equation of order 3, obtained by eliminating three constants A, B, C. EXERCISE 3.1 1. Write the order and the degree of the following differential equations. (i) d2y dx 2 3 2 a x 0; 2 dy 2 d 2 y (ii) 1 ; dx 2 dx 3 4 d2y dy 4 (iii) x 2 y y 0. dx dx 2 Ans. (i) 2,1 (ii) 2,2 (iii) 2,3 2. Give an example of each of the following type of differential equations. (i) A linear-differential equation of second order and first degree Ans. Q, 1 (i) (ii) A non-linear differential equation of second order and second degree (iii) Second order and third degree. Ans. Q, 1 (ii) Ans. Q 1 (iii) 2 3. Obtain the differential equation of which y = 4a(x + a) is a solution. 2 dy dy Ans. y 2 2 xy y 2 0 dx dx 4. Obtain the differential equation associated with the primitive Ax² + By² = 1. 2 d2y dy dy 0 Ans. xy 2 x y dx dx dx 5. Find the differential equation corresponding to d2 y dy 4 3y 0 y = a e3x + bex. Ans. 2 dx dx Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ 140 Differential Equations 6. By the elimination of constants A and B, find the differential equation of which d2 y dy 2y 0 dx dx 7. Find the differential equation whose solution is y = a cos (x = 3). (A.M.I.E., Summer 2000) dy tan ( x 3) Ans. dx 1 8. Show that set of function x, forms a basis of the differential equation x2y + xy – y = 0. x 3x 1 Obtain a particular solution when y (1) = 1, y (1) = 2. Ans. y 2 2x y = ex (A cos x + B sin x) is a solution. Ans. 2 2 3.4 SOLUTION OF A DIFFERENTIAL EQUATION In the example 1(b), y = A cos x + B sin x, on eliminating A and B we get the differential equation d2y y0 dx 2 d2y y0. y = A cos x + B sin x is called the solution of the differential equation dx 2 d2y y 0 is two and the solution The order of the differential equation dx 2 y = A cos x + B sin x contains two arbitrary constants. The number of arbitrary constants in the solution is equal to the order of the differential equation. An equation containing dependent variable (y) and independent variable (x) and free from derivative, which satisfies the differential equation, is called the solution (primative) of the differential equation. 3.5 DIFFERENTIAL EQUATIONS OF THE FIRST ORDER AND FIRST DEGREE We will discuss the standard methods of solving the differential equations of the following types: (i) Equations solvable by separation of the variables. (iii) Linear equations of the first order. (ii) Homogeneous equations. (iv) Exact differential equations. 3.6 VARIABLES SEPARABLE If a differential equation can be written in the form f ( y ) dy ( x ) dx We say that variables are separable, y on left hand side and x on right hand side. We get the solution by integrating both sides. Working Rule: Step 1. Separate the variables as Step 2. Integrate both sides as f ( y ) dy ( x ) dx f ( y) dy ( x) dx Step 3. Add an arbitrary constant C on R.H.S. Example 2. Solve : Solution. We have, dy x(2 log x 1) dx sin y y cos y dy x (2log x 1) dx sin y y cos y (UP, II 2008, U.P.B. Pharm (C.O.) 2005) Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ Differential Equations 141 Separating the variables, we get (sin y + y cos y) dy = {x (2 log x + 1)} dx x(2log x 1) dx C cos y y sin y (1) sin y dy 2 log x x dx x dx C Integrating both the sides, we get (sin y y cos y )dy x2 1 x2 x2 cos y y sin y cos y 2 log x dx C 2 x 2 2 x2 x2 y sin y 2log x x dx C 2 2 2 2 2 x x x y sin y 2 log x C 2 2 2 y sin y x 2 log x C Ans. Example 3. Solve the differential equation. dy x4 x3 y sec( x y ). (A.M.I.E.T.E., Winter 2003) dx 3 dy 4 dy x 3 y sec( x y ) Solution. x x x y sec xy dx dx dv dy 3 dv x y sec v Put v = xy, x dx dx dx dv dx dx 3 cos v dv 3 c sec v x x 1 1 c c Ans. sin v = sin xy = 2x2 2x2 Example 4. Solve : cos (x + y)dy = dx Solution. cos (x + y) dy = dx On putting So that Separating the variables, we get x+y=z dy dz 1 dx dx dz 1 sec z dx dy sec ( x y ) dx dy dz 1 dx dx dz 1 sec z dx dz dx 1 sec z On integrating, 1 cos z 1 dz dx dz x C cos z 1 cos z 1 1 1 dz x C 2cos 2 z 1 1 2 z 1 z 2 z tan x C 1 sec dz x C 2 2 2 Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ 142 Differential Equations x y xC 2 x y y tan C 2 x y tan Ans. Example 5. Solve the equation. (2x2 + 3y2 – 7) x dx – (3x2 + 2y2 – 8) y dy = 0 (U.P. II Semester, Summer 2005) Solution. We have (2x2 + 3y2 – 7) x dx – (3x2 + 2y2 – 8) y dy = 0 x dx 3x 2 2 y 2 8 Re-arranging (1), we get y dy 2 2x 3y 2 7 Applying componendo and dividendo rule, we get x dx y dy x dx y dy 5 x 2 5 y 2 –15 x dx y dy 5 2 2 2 2 2 2 x dx – y dy x – y –1 x y 3 x y 1 Multiplying by 2 both the sides, we get 2 x dx 2 y dy 2 x dx 2 y dy 2 5 2 2 2 x y 3 x y 1 Integrating both sides, we get log (x² + y² – 3) = 5 log (x² – y² – 1) + log C x² + y² – 3 = C (x² – y² – 1)5 Ans. where C is arbitrary constant of integration. EXERCISE 3.2 Solve the following differential equations : dx tan y dy 1. x 1 y2 dy 2. dx 1 x2 Ans. x cos y = C 3. y (1 x 2 )1/ 2 dy x 1 y 2 dx 0 4. sec² x tan y dx + sec² y tan x dy = 0 5. (1 + x²) dy – x y dx = 0 6. (ey + 1) cos x dx + ey sin x dy = 0 7. 3 ex tan y dx + (1 – ex) sec² y dy = 0 8. (ey + 2) sin x dx – ey cos x dy = 0 9. Ans. sin–1 y = sin–1 x + C 1 y2 1 x2 C Ans. tan x tan y = C Ans. y² = C (1 + x²) Ans. (ey + 1) sin x = C Ans. (1 – ex)³ = C tan y Ans. (ey + 2) cos x = C Ans. dy e x y x 2 e y dx y x Ans. e e x3 C 3 dy 10. d x 1 tan ( y x ) [Put y – x = z] Ans. sin (y – x) = ex+c 2 d x 1 4 x y 1 2x C 11. (4 x y ) Ans. tan dy 2 dy 2 1 4 x y 1 2x C 12. d x (4 x y 1) [Hint. Put 4x + y +1 = z] Ans. tan 2 3.7 HOMOGENEOUS DIFFERENTIAL EQUATIONS A differential equation of the form dy f ( x, y ) dx ( x, y ) Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ Differential Equations 143 is called a homogeneous equation if each term of f (x, y) and (x, y) is of the same degree i.e., dy 3 x y y 2 dx 3 x 2 x y dv dy v x In such case we put y = vx. and dx dx The reduced equation involves v and x only. This new differential equation can be solved by variables separable method. Working Rule dv dy vx Step 1. Put y = vx so that Step 2. Separate the variables. dx dx Step 4. Put v Step 3. Integrate both the sides. y and simplify.. x Example 6. Solve the following differential equation (2xy + x2) y = 3y2 + 2xy (A.M.I.E.T.E. Dec. 2006) 2 dy dy 3 y 2 xy 3 y 2 2 xy Solution. We have, (2xy + x2) dx dx 2 xy x 2 dy dv v x Put y = vx so that dx dx dv 3v 2 x 2 2vx 2 3v 2 2v On substituting, the given equation becomes v x dx 2v 1 2vx 2 x 2 dv 3v 2 2v 2v 2 v dx 2v 1 dx 2v 1 v 2 v dv x x v2 + v = cx x log (v2 + v) log x + log c y2 + xy = cx3 Example 7. Solve the equation : dy y x sin dx x dy y x sin dx x Solution. Put dv v 2 v dx 2v 1 2 dv dx 2v 1 x v v y2 x 2 y cx x y x y x ... (1) y = vx in (1) so that dv v x sin v dx dv x x sin v dx Separating the variable, we get dv dx sin v v log tan x C 2 dy dv v x dx dx v x dv sin v dx cosec v dv dx C log tan y xC 2x Ans. Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ 144 Differential Equations EXERCISE 3.3 Solve the following differential equations: 2. (x² – y²) dx+2xy dy = 0 (AMIETE, June 2009) x Ans. y log x C Ans. x² + y² = ax dy y ( y x). dx 4. x (x – y) dy + y² dx = 0 Ans. y log xy a x Ans. y = x log C y 1. (y² – xy) dx + x²dy = 0 3. x ( y x) 5. (U.P. B. Pharm (C.O.) 2005) dy x 2 y 0 Ans. y – x = C (x + y)³ dx 2 x y dy 3 x y y 2 7. dx 3 x2 6. dy y y tan dx x x Ans. sin Ans. 3x + y log x + Cy = 0 8. dy x 2 2 y 2 dx 2x y Ans. 4y² – x² = Ans. 9. (x² + y²) dy = xy dx x2 2 y2 y Cx x C x2 log y C x3 log y C 3 y3 11. (y² + 2xy) dx + (2 x² + 3xy) dy = 0 (AMIETE, Summer 2004) Ans. xy² (x + y) = C 12. (2xy² – x³) dy + (y³ – 2yx²) dx = 0 Ans. y² (y² – x²) = Cx–2 10. x²y dx – (x³ + y³) dy = 0 Ans. 13. (x³ – 3 xy²) dx + (y³ – 3 x²y) dy = 0, y (0) = 1 14. 2 xy² dy – (x³ + 2y³) dx = 0 Ans. x4 – 6x² y² + y4 = 1 Ans. 2y³ = 3x³ log x + 3x³ + C 15. x sin y y dy y sin x dx x x Ans. cos y y y y dy 0 16. x cos y sin y y sin x cos x x x x x dx y x2 y 2 17. dy dx x dy y (log y log x 1) (AMIETE, Summer 2004) 18. x dx 2 x x 2 19. xy log dx y x log dy 0 given that y (1) = 0 y y Ans. x y log x C x Ans. x y cos y a x Ans. y x 2 y 2 C x 2 Ans. log x2 2 y2 log x y Cx x x x2 3 2 log y 1 2 y 4y 4e x x x y y (1 e y ) dx e y 1 dy 0 (AMIETE, June 2009) Ans. e e C y y 3.8 EQUATIONS REDUCIBLE TO HOMOGENEOUS FORM 20. a b A B The equations of the form dy ax + by + c = dx Ax + By + C Case I. can be reduced to the homogeneous form by the substitution if a b A B Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ Differential Equations 145 x = X + h, y = Y + k (h,k being constants) d y dY dx dX The given differential equation reduces to dY a ( X h) b (Y k ) c a X bY a h b k c d X A ( X h) B (Y k ) C A X BY A h B k C Choose h, k so that ah+bk+c=0 Ah+Kk+C=0 dY a X bY Then the given equation becomes homogeneous d X A X BY a b Case II. If then the value of h, k will not be finite. A B a b 1 (say) A B m A = a m, B = b m The given equation becomes dy a x b y c d x m (a x b y ) c Now put ax + by = z and apply the method of variables separable. dy x 2 y 3 dx 2 x y 3 Solution. Put x = X + h, y = Y + k. The given equation reduces to d Y ( X h) 2(Y k ) 3 d X 2( X h) (Y k ) 3 Example 8. Solve : 1 2 2 1 X 2 Y (h 2 k 3) 2 X Y (2 h k 3) Now choose h and k so that h + 2k – 3 = 0, 2h + k – 3 = 0 Solving these equations we get h = k = 1 dY X 2Y d X 2 X Y .... (1) ... (2) dY dv Put Y = v X, so that d X v X d X The equation (2) is transformed as d v X 2 v X 1 2 v v X d X 2X v X 2v X d v 1 2v 1 v2 v d X 2v 2v 1 1 3 1 dX dv dv 2 (1 v) 2 1 v X On integrating, we have 1 3 log(1 v ) log(1 v ) log X log C 2 2 dX 2v dv 2 X 1 v (Partial fractions) Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ 146 Differential Equations Put log 1 v 3 log C 2 X 2 (1 v ) Y 1 X C2 X 2 3 Y 1 X X = x – 1 and Y = y – 1 1 v (1 v) 3 C2X 2 X Y ( X Y )3 C 2 or X + Y = C 2 (X – Y)³ x + y – 2 = a (x – y)³ Ans. Example 9. Solve : (x + 2y) (dx – dy) = dx + dy Solution. (x + 2y) (dx – dy) = dx + dy (x + 2y – 1) dx – (x + 2y + 1) dy = 0 dy x 2 y 1 dx x 2 y 1 a b Hence i.e., 1 2 A B 1 2 dy dz Now put x + 2y = z so that 1 2 dx dx Equation (1) becomes 1 d z 1 z 1 dz ( z 1) 3z 1 2 1 2 d x 2 z 1 dx z 1 z 1 z 1 dz dx 3z 1 On integrating, ...(1) (Case of failure) 1 4 1 dz dx 3 3 3z 1 z 4 log (3 z 1) x C 3 9 3z + 4 log (3z – 1) = 9x + 9C 3 (x + 2y) + 4 log (3x + 6y – 1) = 9x + 9C 3x – 3y + a = 2 log (3x + 6y – 1) Ans. EXERCISE 3.4 Solve the following differential equations : dy 2 x 9 y 20 1. Ans. (2x – y)² = C (x + 2y – 5) dx 6 x 2 y 10 dy y x 1 1 y 3 a 2. dx y x 5 Ans. log[( y 3)² ( x 2)²] 2 tan x2 dy x y 2 3. dx x y 6 Ans. (y + 4)² + 2 (x + 2) (y + 4) – (x + 2)² = a² dy y x 2 4. dx y x 4 (AMIETE, Dec. 2009) Ans. – (y – 3)² + 2(x + 1) (y – 3) + (x + 1)² = a dy 2x 5y 3 5. dx 2x 4y 6 Ans. (x – 4y + 3) (2x + y – 3) = a 6. (2x + y + 1) dx + (4x + 2y – 1) dy = 0 Ans. 2 (2x + y) + log (2x + y – 1) = 3x + C 7. (x – y – 2) dx – (2x – 2y – 3) dy = 0 Ans. log (x – y – 1) = x – 2y + C (U.P. B. Pharm (C.O.) 2005) 8. (6x – 4y + 1) dy – (3x – 2y + 1) dx = 0 (A.M.I.E.T. E., Dec. 2006) Ans. 4x – 8y – log (12x – xy + 1) = c Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ Differential Equations 147 dy 3y 2x 7 9. dx 7 y 3x 3 (A.M.I.E.T.E., Summer 2004) dy 2 y x 4 10. (AMIETE, Dec. 2010) dx y 3 x 3 Ans. (x + y – 1)5 (x – y – 1)2 = 1 2Y (5 21) X 1 X x2 2 2 , Ans. X 5 XY Y c 2Y (5 21) X 21 Y y 3 3.9 LINEAR DIFFERENTIAL EQUATIONS A differential equation of the form dy PyQ dx ... (1) is called a linear differential equation, where P and Q, are functions of x (but not of y) or constants. In such case, multiply both sides of (1) by e Pdx Pdx dy Pdx e Py Q e dx The left hand side of (2) is d Pdx y.e dx d P dx P dx y.e (2) becomes Q.e dx Integrating both sides, we get y.e P dx Q. e This is the required solution. Note. e P dx P dx ... (2) dx C is called the integrating factor.. y × [I.F.] = Q [I.F.] dx + C Solution is Working Rule Step 1. Convert the given equation to the standard form of linear differential equation dy Py Q i.e. dx Step 2. Find the integrating factor i.e. I.F. = e Pdx Step 3. Then the solution is y ( I .F .) Q ( I .F .)dx C Example 10. Solve: Solution. dy y e x ( x 1)2 dx dy y e x ( x 1) dx x 1 ( x 1) Integrating factor = e The solution is (A.M.I.E.T.E., Summer 2002) dx x 1 e log( x 1) elog( x 1)1 1 x 1 1 1 y. e x .( x 1). dx e x dx x 1 x 1 y ex C x 1 Ans. Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ 148 Differential Equations Example 11. Solve a differential equation dy ( x 3 x) (3 x 2 1) y x 5 2 x3 x. (Nagpur University, Summer 2008) dx dy Solution. We have ( x3 x ) (3 x 2 1) y x 5 2 x 3 x dx 2 dy 3x 1 x 5 2 x3 x dy 3x 2 1 3 y = y = x² – 1 3 dx x x dx x 3 x x x 3 x 2 1 I.F. = e x3 x dx e log( x 3 x) e log( x 3 x ) –1 Its solution is y 3 x x y x3 x = x2 1 x( x 2 1) dx C = log x + C Example 12. Solve sin x y 3 x x = x x x2 1 1 = y 3 x x y I.F. Q( I .F .) dx C 1 3 x 3 x dx C 1 x dx C y = (x³ – x) log x + (x³ – x) C dy x 2 y tan 3 dx 2 Ans. (Nagpur University, Summer 2004) x tan 3 dy dy 2 3 x 2 Solution. Given equation : sin x 2 y tan y dx 2 dx sin x sin x dy Py Q This is linear form of dx x tan 3 2 2 P and Q sin x sin x I.F. e Pdx 2 dx e 2 cosec x dx e e sin x y.(I.F.) I.F.(Q dx) C Solution is 2log tan x 2 tan 2 x tan 4 1 2 dx C C = x x x 2 cos 2 2 sin cos 2 2 2 1 x x tan 4 .sec 2 dx C = 2 2 2 x 1 2x Putting tan t so that sec dx dt on R.H.S. (1), we get 2 2 2 x 1 t5 2 x y.tan 2 t 4 (2dt ) C C y tan 2 2 2 5 5 x tan x 2 C y tan 2 2 5 EXERCISE 3.5 Solve the following differential equations: x y tan = 2 2 x tan 2 2 tan3 x 2 x 2 ... (1) 1. dy 1 y x3 3 dx x Ans. xy Ans. x5 3x 2 C 5 2 Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ Differential Equations 149 2. (2y – 3x) dx + x dy = 0 dy y cot x cos x 3. dx dy y sec x tan x 4. dx 2 dy 5. cos x y tan x dx dy 5 6. ( x a ) 3 y ( x a ) dx dy 7. x cos x y ( x sin x cos x ) 1 dx dy 8. x log x y 2 log x dx dy 2 9. x 2 y x log x dx Ans. y x² = x³ + C sin 2 x C Ans. y sin x 2 Cx 1 Ans. y sec x tan x Ans. y tan x 1 Ce tan x Ans. 2y = (x + a)5 + 2C (x + a)³ Ans. x y = sin x + C cos x Ans. y log x = (log x)² + C x4 x4 log x C 4 16 4 sin C Ans. r sin 2 2 2 Ans. y x 10. dr (2r cot sin 2 ) d 0 dy 1 y cos x sin 2 x Ans. y sin x 1 Ce sin x dx 2 2 dy 2 1/ 2 12. (1 x ) 2 xy x(1 x ) Ans. y 1 x 2 C (1 x 2 ) dx dy y sin x (A.M.I.E.T.E., Dec 2005) 13. sec x Ans. y = –sin x – 1 + cesinx dx 14. y y tan x cos x, y (0) 0 (A.M.I.E.T.E., June 2006) Ans. y = x cos x 11. 1 15. Solve (1 + y2) dx = (tan–1 y – x) dy (AMIETE, Dec. 2009) Ans. x = – tan–1 y – 1 + ce tan y 16. Find the value of so that e2 is an integrating factor of differential equation x (1 – y) 1 dx – dy = 0. (A.M.I.E.T.E., Summer 2005) Ans. 2 dy 17. Slove the differential equation cot 3 x –3y = cos 3x + sin 3x, 0 < x < . dx 2 1 (AMIETE, Dec. 2009) Ans. y cos 3 x 6 x sin 6 x cos 6 x 12 y2 18. The value of so that e is an integrating factor of the differential equation y2 (e 2 (A.M.I.E.T.E. Dec., 2005) xy ) dy dx 0 is 1 1 (d) Ans. (c) 2 2 dy a 2 is given by 19. The solution of the differential equation (y + x)2 dx yc y c (a) y x a tan (b) y x tan a a c (c) y x a tan ( y c) (d) a ( y x ) tan y Ans. (a) a (AMIETE, June 2010) (a) –1 (b) 1 (c) Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ 150 Differential Equations 3.10 EQUATIONS REDUCIBLE TO THE LINEAR FORM (BERNOULLI EQUATION) The equation of the form dy + Py = Qy n ...(1) dx where P and Q are constants or functions of x can be reduced to the linear form on dividing 1 by yn and substituting n 1 z y On dividing bothsides of (1) by yn, we get 1 dy 1 n 1 P Q ...(2) n y dx y 1 1 dy dz (1 n ) dy dz Put n 1 z , so that n n y dx dx y dx 1 n y 1 dz dz Pz Q or P(1 n ) z Q(1 n) 1 n dx dx which is a linear equation and can be solved easily by the previous method discussed in article 3.8 on page 144. (2) becomes Example 13. Solve x²dy + y(x + y) dx = 0 (U.P. II Semester Summer 2006) Solution. We have, x² dy + y (x + y) dx = 0 1 dy 1 1 dy y y2 2 = 2 2 dx xy y x dx x x 1 dy dz 1 Put z so that 2 y dx dx y The given equation reduces to a linear differential equation in z. dz z 1 2 dx x x I.F. e 1 x dx e log x elog1/ x 1 . x Hence the solution is 1 1 1 z. 2 . dx C x x x 2 1 x C xy 2 dy Example 14. Solve: x y log y xy e x dx dy x y log y xy e x Solution. dx Dividing by xy, we get 1 dy 1 log y e x y dx x 1 dy dz Put log y = z, so that y dx dx dz z Equation (1) becomes, ex dx x z x 3 dx C x 1 1 2 C xy 2x Ans. (A.M.I.E., Summer 2000) ...(1) Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ Differential Equations 151 1 dx I.F. e x elog x x z x x ex d x C Solution is z x x ex ex C x log y x e x e x C Ans. dy tan y (1 x )e x sec y. dx 1 x dy tan y (1 x )e x sec y Solution. dx 1 x dy sin y cos y (1 x ) e x dx 1 x dy dz Put sin y = z, so that cos y dx dx dz z (1 x)e x (1) becomes dx 1 x Example 15. Solve: (Nagpur University, Summer 2000) ...(1) 1 1 1 1 dx I .F . e 1 x e log(1 x) elog x 1 x 1 1 x x z. (1 x) e . dx C e dx C Solution is 1 x 1 x sin y ex C Ans. 1 x dy Example 16. Solve: tan y tan x cos y cos 2 x (Nagpur University, Summer 2000) dx dy 2 Solution. tan y tan x cos y cos x dx dy sec y tan y sec y tan x cos 2 x dx dz dy sec y tan y Writing z = sec y, so that dx dx dz The equation becomes z tan x cos 2 x dx I.F. e elog sec x sec x The solution of the equation is tan x dx z sec x cos 2 x sec x dx C sec y sec x cos x dx C sin x C sec y (sin x C ) cos x Example 17. Solution. Ans. dx x y 1 y dy dy x y (1 y ) dx dy 1 y y dx x x (Nagpur University, Summer 2004) dy 1 1 1 y dx x x Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ 152 Differential Equations which is in linear form of Its solution is dy Py Q. dx 1 P 1 , x I.F. e Pdx Q 1 e 1 x 1 x dx e x log x 1 y ( x . e ) ( x . e ) dx C x = e x .elog x e x . x x e x y (I.F.) I.F.(Q dx) C x x x y ( x . e ) e dx C x x x y (x . e ) = e + C 1 C y e x Ans. x x Example 18. Solve the differential equation. y log y dx + (x – log y) dy = 0 (Uttarakhand II Semester, June 2007) Solution. We have, y log y dx + (x – log y) dy = 0 dx x log y dx x log y dy y log y dy y log y y log y dx x 1 dy y log y y 1 I.F. e Its solution is x.log y y log y dy e log (log y ) log y 1 y (log y ) dy (log y ) C Ans. 2 –1 Example 19. Solve: (1 + y²) dx = (tan y – x) dy. (AMIETE, June 2010, 2004, R.G.P.V., Bhopal, April 2010, June 2008, U.P. (B. Pharm) 2005) Solution. (1 + y²) dx = (tan–1 y – x) dy x.log y dx tan 1 y x dy 1 y2 This is a linear differential equation. dx x tan 1 y dy 1 y 2 1 y 2 1 I.F. e Its solution is Put x . e tan 1 y 1 y2 dy e tan 1 y 1 e tan tan 1 y 1 y2 y dy C tan–1 y = t on R.H.S., so that x . e tan 1 y 1 1 y2 dy dt et .t dt C t .et et C e tan x tan 1 y 1 Ce tan 1 y 1 y tan 1 y 1 C Ans. Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ Differential Equations 153 Example 20. Solve : r sin dr cos r 2 d (Nagpur University, Summer 2005) Solution. The given equation can be written as dr cos r sin r 2 d 2 dr r 1 tan sec Dividing (1) by r 2 cos , we get r d dv 2 dr Putting r 1 v so that r d d in (2), we get dv v tan sec d I.F. e Solution is tan d 2. 3. 4. 5. 6. 7. v sec sec , sec C 1 dy 1 2 x e x y 2 dx y dy y 3 3 2 x4 y 4 dx x dy y tan x y 2 sec x dx dy 2 y tan x y 2 tan 2 x, if y 1 at x 0 dx dy tan x tan y cos x sec y dx dy + y tan x . dx = y² sec x . dx (x² y² + xy) y dx + (x² y² – 1) x dy = 0 8. (x² + y² + x) dx + xy dy = 0 9. dy y 3e x y 3 dx 10. (x – y²) dx + 2 x y dy = 0 y dy x 11. e 1 e dx 2 3 dy y 4 cos x 12. x y x dx dy 2 x2 13. 3 dx x 1 . y 2 y dy 14. cos x 4 y sin x 4 y sec x dx ... (2) elog sec sec v sec sec2 d C sec tan C r 1 r sin C cos EXERCISE 3.6 Solve the following differential equations: 1. ... (1) r 1 (sin C cos ) Ans. Ans. ex + x²y + Cy = 0 Ans. 1 y3 x5 Cx3 Ans. sec x = (tan x + C) y 1 2 tan 3 x sec x 1 Ans. y 3 Ans. sin y sec x = x + C Ans. y (x + C) + cos x = 0 Ans. x y = log C y 2 2 Ans. x y Ans. 1 y2 x 4 2 x3 C 2 3 6 e x C e2 x y2 log x C x e2 x C Ans. e x y 2 Ans. Ans. x3 = y3(3 sin x – C) x5 x 4 x 3 C 5 2 3 3 tan x y sec 2 x 2 tan x C 3 3 2 Ans. y ( x 1) Ans. Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ 154 15. 16. 17. 18. 19. Differential Equations dy x sin 2 y x3 cos 2 y dx 1 dy 2 x tan 1 y x 3 1 y 2 dx e–y sec² y dy = dx + x dy dy ( x y 1) 1 dx dy y3 2x dx e y 2 2 1 2 ( x 1) C e x 2 1 2 1 x2 Ans. tan y ( x 1) C e 2 Ans. x ey = tan y + C Ans. tan y Ans. x + y + 2 = C ey Ans. e–2x y² + 2 log y + C = 0 20. dx – xy (1 + xy2) dy = 0 21. dy y y log y 2 (log y )2 dx x x dy xy xy 2 dx dy y x3 y 6 27. x dx 22. 3 Ans. 2 1 y 2 2 Ce y / 2 x (A.M.I.E.T.E., Summer 2004, 2003, Winter 2003, 2001) 1 1 Ans. x log y 2 C 2x (A.M.I.E.T.E., June 2009) Ans. y 3 1 Ce x (AMIETE, June 2010) Ans. 1 5 5 y x 5 2 x2 2 /2 C dx 23. General solution of linear differential equation of first order dy Px Q (where P and Q are constants or functions of y) is P . dx P. dx P . dy P. dy (a) ye dx + c (b) xe dy + c Q e Q e (c) y Q e P. dx P. dy dx + c (d) x Q e dy + c (AMIETE, June, 2010) Ans. (b) 3.11 EXACT DIFFERENTIAL EQUATION An exact differential equation is formed by directly differentiating its primitive (solution) without any other process Mdx + Ndy = 0 is said to be an exact differential equation if it satisfies the following condition M N = y x where M denotes the differential co-efficient of M with respect to y keeping x constant and N , y x the differential co-efficient of N with respect to x, keeping y constant. Method for Solving Exact Differential Equations Step I. Integrate M w.r.t. x keeping y constant Step II. Integrate w.r.t. y, only those terms of N which do not contain x. Step III. Result of I + Result of II = Constant. Example 21. Solve : (5x4 + 3x2y2 – 2xy3) dx + (2x3y – 3x2y2 – 5y4) dy = 0 Solution. Here, M = 5x4 + 3x2y2 – 2xy3, N = 2x3y – 3x2y2 – 5y4 Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ Differential Equations 155 M = 6x2y – 6xy2, y N M = , the given equation is exact. x y Since, Now N = 6x2y – 6xy2 x M dx (terms of N is not containing x) dy C 5x 3x y 2 xy dx 5 y dy C 4 2 2 3 (y constant) 4 x5 + x3y2 – x2y3 – y5 = C 2 2 Ans. 2 Example 22. Solve: 2 xy cos x 2 xy 1 dx sin x x 3 dy 0 (Nagpur University, Summer 2000) Solution. Here we have 2 xy cos x 2 xy 1 dx sin x 2 x 2 3 dy 0 M dx + N dy = 0 Comparing (1) and (2), we get M M = 2xy cos x² – 2xy + 1 = 2x cos x² – 2x y N N = sin x² – x² + 3 = 2x cos x² – 2x x M N Here, = y x So the given differential equation is exact differential equation. Hence solution is 2 M dx ... (1) ... (2) (terms of N not containing x ) dy C y as const Put (2 xy cos x 2 2 xy 1) dx 3 dy C 2 [ y(2 x cos x ) y(2x) 1] dx 3 dy C y 2 x cos x dx y 2 x dx 1 dx 3 y dy C 2 x2 = t so that 2x dx = dt x2 x 3y C 2 y sin t – x² y + x + 3y = C y sin x² – yx² + x + 3y = C y cos t dt 2 y Example 23. Solve : Solution. We have, x 1 e y Ans. x dy (1 e x / y ) e x / y 1 0 y dx (Nagpur University, Summer 2008, A.M.I.E.T.E. June, 2009) x x dy e y 1 0 y dx x 1 e y x x dx e y e y x dy 0 y x M 1 e x y M x 2 ey y y Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ 156 Differential Equations x y x x y x x x N 1 y 1 y x y x x e e 2 e 2 ey N e e x y y y y y M N = y x Given equation is exact. x Its solution is 1 e y dx (terms of N not containing x) dy C x x 1 e y dx 0 dy C Ans. x ye y C x Example 24. Solve: [1 log ( x y )] dx 1 dy 0 (Nagpur University, Winter 2003) y x Solution. [1 log x y ] dx 1 dy 0 y x [1 log x log y ] dx 1 dy 0 y which is in the form M dx + N dy = 0 x M = [1 + log x + log y] and N 1 y M N N 1 M 1 and = y x x y y y Hence the given differential equation is exact. Solution is M dx N (terms not containing x) dy C y constant y constant Now, (1 log x log y) dx dy C x log x dx log y dx y C 1 d log x dx log x .(1) dx (log x) x dx (log x) x dx x log x x .x dx x log x dx x log x x x[log x 1] Equation (1) becomes ... (1) x + x log x – x + x log y + y = C x [log x + log y] + y = C x log xy + y = C Ans. EXERCISE 3.7 Solve the following differential equations (1 – 12). 1. (x + y – 10) dx + (x – y – 2) dy = 0 2. (y2 – x2) dx + 2x y dy = 0 3. 1 3e dx 3e x/ y x/ y x2 y2 xy 10 x 2y C 2 2 x3 x y2 C Ans. 3 Ans. x x/y 1 dy 0 (R.G.P.V. Bhopal, Winter 2010) Ans. x + 3y e = C y Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ Differential Equations 157 y2 C 2 Ans. y tan x + sec x + y2 = C Ans. xy x 2 4. (2x – y) dx = (x – y) dy 5. (y sec2 x + sec x tan x) dx + (tan x + 2y) dy = 0 6. (ax + hy + g) dx + (hx + by + f) dy = 0 Ans. ax² + 2h xy + by2 + 2gx + 2fy + C = 0 8. (2xy + ey) dx + (x2 + xey) dy = 0 x5 x 2 y 2 xy 4 cos y C 5 Ans. x2y + xey = C 9. (x2 + 2ye2x) dy + (2xy + 2y2e2x) dx = 0 Ans. x2y + y2 e2x = C 7. (x4 – 2xy2 + y4) dx – (2x2y – 4xy3 + sin y) dy = 0 Ans. 1 10. y 1 cos y dx ( x log x x sin y ) dy 0 (M.D.U., 2010) x Ans. y (x + log x) + x cos y = C 11. (x3 – 3xy2) dx + (y3 – 3x3y) dy = 0, y(0) = 1 Ans. x4 – 6x2y2 + y4 = 1 12. The differential equation M (x, y) dx + N (x, y) dy = 0 is an exact differential equation if M N M N M N (a) y x = 0 (b) y x = 0 (c) y x 1 (d) None of the above (A.M.I.E.T.E. Dec. 2010, Dec 2006) Ans. (b) 3.12 EQUATIONS REDUCIBLE TO THE EXACT EQUATIONS Sometimes a differential equation which is not exact may become so, on multiplication by a suitable function known as the integrating factor. M N y x f ( x ) dx Rule 1. If is a function of x alone, say f (x), then I.F. e N Example 25. Solve (2x log x – xy) dy + 2y dx = 0 Solution. M = 2y, N = 2x log x – xy M N 2, 2(1 log x ) y y x M N 1 y x 2 2 2 log x y (2 log x y ) Here, f ( x) N 2 x log x xy x (2 log x y ) x I.F. e f ( x ) dx ... (1) 1 e x dx e log x elog x1 x –1 1 x 1 On multiplying the given differential equation (1) by , we get x 2y 2y dx (2 log x y )dy 0 dx y dy c x x 1 2 y log x y 2 c 2 EXERCISE 3.8 Solve the following differential equations: 1. (y log y) dx + (x – log y) dy = 0 1 3 1 2 1 2 2. y y x dx 1 y x dy 0 3 2 4 Ans. Ans. 2x log y = c + (log y)² Ans. yx 4 y 3 x 4 x 6 c 4 12 12 Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ 158 Differential Equations y y2 x2 c x 2 1 3 Ans. tan y x sin y c x Ans. y2 = cx – x log x 3. (y – 2x3) dx – x (1 – xy) dy = 0 Ans. 4. (x sec2y – x2 cos y) dy = (tan y – 3x4) dx 5. (x – y2) dx + 2xy dy = 0 N M x y Rule II. If is a function of y alone, say f (y), then M I.F. e 4 Example 26. Solve (y + 2y) dx + (xy3 + 2y4 – 4x) dy = 0 f ( y ) dy Solution. Here M = y4 + 2y; N = xy3 + 2y4 – 4x ...(1) M N 4 y 3 2; y3 4 y x N M ( y 3 4) (4 y 3 2) 3( y 3 2) 3 x y f ( y) 4 3 M y y 2y y ( y 2) I.F. e f ( y ) dy 3 e y dy e 3log y elog y On multiplying the given equation (1) by 1 y3 3 y 3 y3 we get the exact differential equation. 2 4x y 2 dx x 2 y 3 dy 0 y y 2 y 2 dx 2 y dy c y EXERCISE 3.9 Solve the following differential equations: 1 2 x y 2 y 2 c y 1. (3x2y4 + 2xy) dx + (2x3y3 – x2) dy = 0 Ans. x3 y 2 2. (xy3 + y) dx + 2(x2y2 + x + y4) dy = 0 Ans. 3. y(x2y + ex)dx – exdy = 0 Ans. x2 c y x2 y 4 y6 xy 2 c 2 3 x3 e x Ans. c 3 y x2 x 3 c y y Rule III. If M is of the form M = y f1 (xy) and N is of the form N = x f2(xy) 4. (2x4y4ey + 2xy3 + y) dx + (x2y4ey – x2y2 – 3x) dy = 0 Then I.F. = Ans. x 2 e y 1 M .x N . y Example 27. Solve y (xy + 2x2y2) dx + x (xy – x2y2) dy = 0 Solution. y (xy + 2x2y2) dx + x (xy – x2y2) dy = 0 ... (1) Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ Differential Equations 159 Dividing (1) by xy, we get y (1 + 2xy) dx + x (1 – xy) dy = 0 M = y f1 (xy), N = x f2 (xy) I.F. On multiplying (2) by ... (2) 1 1 1 Mx Ny xy (1 2 xy ) xy (1 xy ) 3x 2 y 2 1 3x 2 y 2 , we have an exact differential equation 1 1 2 1 dy 0 2 dx 2 3y 3x y 3x 3xy 1 2 1 log x log y c 3 xy 3 3 1 2 1 3x y 3x dx 3 y dy c 2 1 2log x log y b xy Ans. EXERCISE 3.10 Solve the following differential equations 1. (y – xy2) dx – (x + x²y) dy = 0 2. y (1 + xy) dx + x(1 – xy) dy = 0 2. y (1 + xy) dx + x (1 + xy + x2y2) dy = 0 4. (xy sin xy + cos xy) y dx + (xy sin xy – cos xy) x dy = 0 x Ans. log xy A y y Ans. xy log c xy 1 x 1 1 log y c Ans. 2 x 2 y 2 xy Ans. y cos xy = cx Rule IV. For of this type of xm y n (ay dx bx dy) xm y n (a y dx b x dy) 0, the integrating factor is xh yk. Example 28. Solve m h 1 n k 1 m h 1 n k 1 , and a b a b (y3 – 2x2y) dx + (2xy2 – x3) dy = 0 Solution. (y3 – 2x2y) dx + (2xy2 – x3) dy = 0 where y2 (ydx + 2xdy) + x2 (–2ydx – xdy) = 0 Here m = 0, h = 2, a = 1, b = 2, m 2, n 0, a 2, b 1 0 h 1 2 k 1 2 h 1 0 k 1 and 1 2 2 1 2h + 2 = 2 + k + 1 and h + 3 = 2k + 2 2h – k = 1 and h – 2k = –1 On solving h = k = 1. Integrating Factor = x y Multiplying the given equation by x y, we get (xy4 – 2x3y2) dx + (2x2y3 – x4y) dy = 0 which is an exact differential equation. ( xy 4 2 x 3 y 2 )dx C x2 y 4 2 x4 y 2 C 2 4 Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ 160 Differential Equations x2y4 – x4y2 = C 3 x2 y2 (y2 – x2) = C 2 2 Example 29. Solve (3y – 2xy ) dx + (4x – 3x y ) dy = 0. 3 Solution. Ans. (U.P., II Semester, June 2007) 2 2 (3y – 2xy ) dx + (4x – 3x y ) dy = 0 (3y dx + 4x dy) + xy2(–2y dx – 3x dy) = 0 Comparing the coefficients of (1) with ...(1) xm y n (a y dx b x dy) x m y n (a y dx b x dy ) 0, we get m = 0, n = 0, a = 3, b = 4 m 1, n 2, a 2, b 3 To find the integrating factor xh yk m h 1 n k 1 m h 1 n k 1 and a b a b 0 h 1 0 k 1 1 h 1 2 k 1 and 3 4 2 3 h 1 k 1 h 2 k 3 and 4h – 3k + 1 = 0 3 4 2 3 and 3h – 2k = 0 h ... (2) 2k 3 ... (3) Putting the value of h from (3) in (2), we get 8k k – 3k 1 0 1 0 3 3 2k 2 3 Putting k = 3 in (2), we get h 2 3 3 I.F. = xhyk = x2y3 k=3 On multiplying the given differential equation by x2y3, we get x2y3 (3y – 2xy3)dx + x2y3(4x–3x2y2) dy = 0 (3x2y4 – 2x3y6) dx + (4x3y3 – 3x4y5) dy = 0 This is the exact differential equation. Its solution is (3x 2 y 4 2 x 3 y6 )dx 0 x3 y 4 x4 6 y C 2 Ans. EXERCISE 3.11 Solve the following differential equations. 1. (2y dx + 3x dy) + 2xy (3y dx + 4x dy) = 0 Ans. x2y3 (1 + 2xy) = c 3/2 2. (y2 + 2yx2) dx + (2x3 – xy) dy = 0 2 Ans. 4( xy )1/ 2 3 y x 3. (3x + 2y2)y dx + 2x (2x + 3y2) dy = 0 Ans. x2y4 (x + y2) = c 4. (2x2y2 + y) dx – (x3y – 3x) dy = 0 Ans. 5. x (3y dx + 2x dy) + 8y4 (y dx + 3x dy) = 0 Ans. x3y2 + 4x2y6 = c c 7 10 / 7 5 / 7 7 4 / 7 –12 / 7 x y x y c 5 4 Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ Differential Equations 161 Rule V. If the given equation M dx + N dy = 0 is homogeneous equation and Mx + Ny 0, then 1 is an integrating factor.. Mx Ny dy x 3 y 3 dx xy 2 (x3 + y3) dx – (xy2) dy = 0 M = x3 + y3, N = –xy2 Example 30. Solve Solution. Here I.F. Multiplying (1) by 1 x 4 ... (1) 1 1 1 Mx Ny x ( x3 y 3 ) xy 2 ( y ) x 4 1 4 ( x 3 y 3 ) dx 1 ( xy 2 ) dy 0 x x4 1 y3 y2 4 dx 3 dy 0, which is an exact differential equation. x x x 1 y3 y3 log x 3 c Ans. 4 dx c 3x x x we get EXERCISE 3.12 Solve the following differential equations: 1. x2y dx – (x3 + y3) dy = 0 Ans. 2. (y3 – 3xy2) dx + (2x2y – xy2) dy = 0 Ans. x3 3 y3 log y c y 3log x 2 log y c x x Ans. 2log x 3log y c y Ans. x2y4 – x4y2 = c 3. (x2y – 2xy2) dx – (x3 – 3x2y) dy = 0 4. (y3 – 2yx2) dx + (2xy2 – x3) dy = 0 3.13 EQUATIONS OF FIRST ORDER AND HIGHER DEGREE dy dy The differential equations will involve in higher degree and will be denoted by p. dx dx The differential equation will be of the form f (x, y, p) = 0. Case 1. Equations solvable for p. Example 31. Solve : x2 = 1 + p2 Solution. x2 = 1 + p2 p x2 1 p2 = x2 – 1 dy x2 1 dx dy x 2 1 dx x 2 1 x 1 log x x 2 1 c 2 2 Case II. Equations solvable for y. (i) Differentiate the given equation w.r.t. “x”. (ii) Eliminate p from the given equation and the equation obtained as above. (iii) The eliminant is the required solution. Example 32. Solve: y = (x – a) p – p2. Solution. y = (x – a) p – p2 which gives on integration y Ans. ... (1) Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ 162 Differential Equations Differentiating (1) w.r.t. “x” we obtain dy dp dp p ( x a) 2p dx dx dx dp dp p p ( x a) 2 p dx dx dp dp 0 ( x a) 2p dx dx dp 0 [ x a 2 p] dx On integration, we get p = c. Putting the value of p in (1), we get y = (x – a) c – c2 Case III. Equations solvable for x dp 0 dx Ans. (i) Differentiate the given equation w.r.t. “y”. (ii) Solve the equation obtained as in (1) for p. (iii) Eliminate p, by putting the value of p in the given equation. (iv) The eliminant is the required solution. Example 33. Solve: y = 2px + yp2 Solution. y = 2px + yp2 2 2px = y – yp Differentiating (2) w.r.t. “y” we get dx 1 y dp dp 2 p y dy p p 2 dy dy 2 1 y dp dp 2 p y p p p dy dy 1 dp 1 p y 2 1 p p dy dy dp y dp 1 y p p dy log p y = log c p y = c Putting the value of p in (1), we get c2 c y 2 x y 2 y y 2 y = c(2x + c) Class IV. Clairaut’s Equation. y 2 x yp p ... (2) 1 y dp dp p 2 y p dy p dy 1 p2 1 p 2 dp y p p 2 dy log y log p log c c p y y2 = 2 cx + c2 ... (1) The equation y = px + f (p) is known as Clairaut’s equation. Ans. ... (1) Differentiating (1) w.r.t. “x”, we get dy dp dp px f ( p ) dx dx dx dp dp dp dp p px f ( p ) 0x f ( p ) dx dx dx dx dp dp [ x f ( p )] 0 0 p = a (constant) dx dx Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ Differential Equations 163 Putting the value of p in (1), we have y = ax + f (a) which is the required solution. Method. In the Clairaut’s equation, on replacing p by a (constant), we get the solution of the equation. Example 34. Solve : p = log (p x – y) Solution. p = log (p x – y) or ep = p x – y or y = p x – ep Which is Clairaut’s equation. Hence its solution is y = a x –ea Ans. EXERCISE 3.13 Solve the following differential equations. 1. xp2 + x = 2yp Ans. 2cy = c2x2 + 1 2. x(1 + p2) = 1 2 1 Ans. y c x x tan c 2 Ans. y 3 , y c1 x x 3. x2p2 + xyp – 6y2 = 0 dy dx x y dx dy y x 5. y = px + p3 6. x2 (y – px) = yp2 1 x x Ans. xy = c, x2 – y2 = c 4. Ans. y = ax + a3 Ans. y2 = cx2 + c2 Y 3.14 ORTHOGONAL TRAJECTORIES Two families of curves are such that every curve of either family cuts each curve of the other family at right angles. They are called orthogonal trajectories of each other. Orthogonal trajectories are very useful in engineering problems. For example: (i) The path of an electric field is perpendicular to equipotential curves. (ii) In fluid flow, the stream lines and equipotential lines are orthogonal trajectories. O (iii) The lines of heat flow is perpendicular to isothermal curves. Working rule to find orthogonal trajectories of curves 90° X Step 1. By differenciating the equaton of curves find the differential equations in the form dy f x, y, 0 dx dy dx Step 2. Replace by (M1. M2 = –1) dx dy dx Step 3. Solve the differential equation of the orthogonal trajectories i.e., f x, y – 0 dy Self-orthogonal. A given family of curves is said to be ‘self-orthogonal’ if the family of orthogonal trajectory is the same as the given family of curves. Example 35. Find the orthogonal trajectories of the family of curves xy = c. Solution. Here, we have xy = c ... (1) Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ 164 Differential Equations Differentiating (1), w.r.t., “x”, we get dy y x 0 dx dx dy On replacing by , we get dy dx dx y dy x y dy = x dx 2 dy y dx x dy x dx y ... (2) 2 y x c 2 2 y2 – x2 = 2c Integrating (2), we get Ans. Example 36. Show that the family of porabolas y2 = 2cx + c2 is “self-orthogonal.” Solution. Here we have y2 = 2cx + c2 ... (1) dy dx 2 dy dy 2 Putting the value of c in (1), we have y 2 y x y dx dx dy Putting p in (2), we get dx y2 = 2ypx + y2p2 Differentiating (1), we get 2 y dy 2c dx c y ... (2) ... (3) This is differential equation of give n family of parabolas. For orthogonal trajectories we put 1 for p in (3) p 1 1 y2 2 y x y2 p p y2p2 = – 2pyx + y2 2 y2 2 yx y 2 2 p p Rewriting, we get y2 = 2ypx + y2p2 Which is same as equation (3). Thus (2) is D.E. for the given family and its orthogonal trajectories. Hence, the given family is self-orthogonal. Proved. EXERCISE 3.14 Find the orthogonal trajectories of the following family of curves: 1. y2 = cx3 Ans. (x + 1)2 + y2 = a2 3. x2 – y2 = c 2 2. x2 – y2 = cx Ans. y (y2 + 3x2) = c Ans. xy = c 2 4. (a + x) y = x (3a – x) Ans. (x2 + y2)5 = cy3 (5x2 + y2) 5. y = ce–2x + 3x, passing through the point (0, 3) Ans. 9x – 3y + 5 = –4e6(3 – y) 6. 16x2 + y2 = c Ans. y16 = kx 7. y = tan x + c Ans. 2x + 4y + sin 2x = k 8. y = ax2 Ans. x2 + 2y2 = c Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ Differential Equations 2 2 9. x + (y – c) = c 165 2 2 Ans. x + y = cx 10. x2 + y2 + 2gx + 2fy + c = 0 3.15 2 Ans. x2 + y2 + 2fy – c = 0 POLAR EQUATION OF THE FAMILY OF CURVES Let the polar equation of the family of curves be f (r , , c ) 0 ... (1) Working Rule Step 1. On differentiating and eliminating the arbitrary constant c between (1) and f (r , , c ) 0 we get the differential equation of (1) i.e., dr F r , , ... (2) 0 d dr 2 d Step 2. Replace by r in (2). Here we will get the differential equation of orthogonal d dr trajectory i.e., d F r, r 2 ... (3) 0 dr Step 3. Integrating (3) to get the equation of the orthogonal trajectory. Example 37. Find the orthogonal trajectory of the cardioids r = a (1 – cos ). Solution. We have, r = a(1 – cos ) ... (1) dr a sin Differentiating (1) w.r.t. , we get ... (2) d Dividing (2) by (1) to eliminate a, we get 2sin cos 1 dr sin 2 2 cot ... (3) r d 1 cos 2 1 1 2 sin 2 2 which is the differential equation of (1). 1 2 d dr 2 d Replacing by r in (3), we get – r cot r dr 2 d dr d cot dr 2 dr Separating the variables we get tan d r 2 2 Integrating (4), we get log r 2 log cos log c log c cos 2 2 r c cos 2 2 Which is the required trajectory. Example 38. Find the orthogonal trajectory the family of curves r2 = c sin2 Solution. We have r2 = c sin2 dr 2c cos 2 Differentiating (1), we get 2r d 2 dr 2 cot 2 Dividing (2) by (1), to eliminate ‘c’ we get r d r ... (4) r c (1 cos ) 2 Ans. ... (1) ... (2) ... (3) Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ 166 Differential Equations 2 2 d dr 2 d by r in (3), we have r 2 cot 2 r dr d dr d 2 r 2 cot 2 dr dr tan 2 d Separating the variables of (4), we obtain r 1 Integrating (5), we get log r log cos 2 log c 2 2 log r = log c cos r2 = c cos2 which is the required trajectory Replacing ... (4) ... (5) Ans. EXERCISE 3.15 Find the orthogonal trajectory of the following families of the curves: 1. r ce 3. r a (1 cos ) Ans. r ke Ans. r c (1 cos ) 2. r c 4. r n sin n a n 2 2 4 Ans. r ke Ans. r n cos n c n 5. r a cos 2 Ans. r 2 c sin 6. r 2a (sin cos ) Ans. r 2c (sin cos 7. r c (1 sin 2 ) Ans. r 2 k cos .cot a 8. r Ans. r 2 sin 3 (1 cos ) 1 2 cos 3.16 ELECTRICAL CIRCUIT We will consider circuits made up of (i) Voltage source which may be a battery or a generator. (ii) Resistance, inductance and capacitance. The formation of differential equation for an electric circuit depends upon the following laws. (i) i dq , dt (ii) Voltage drop across resistance R = Ri di dt q (iv) Voltage drop across capacitance C = C Kirchhoff’s laws (iii) Voltage drop across inductance L = L. I. Voltage law. The algebraic sum of the voltage drop around any closed circuit is equal to the resultant electromotive force in the circuit. II. Current law. At a junction or node, current coming is equal to current going. (i) L - R series circuit. Let i be the current flowing in the circuit containing resistance R and inductance L in series, with voltage source E, at any time t. di di R E Ri L By voltage law =E ...(1) (M.U. II Semester, 2009) i dt dt L L This is the linear differential equation R I.F. = e L dt R eL t Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ Differential Equations 167 Its solution is R t i. e L i. e L = R R R E t = e L dt A L L R t E L Lt e A L R E R i At t = 0, i= 0 i Rt – Ae L – E ...(2) A– + E R i E R O t Rt E L Thus, (2) becomes i = 1 e R (ii) C-R series circuit. Let i be current in the circuit containing resistance R, L, and capacitance C in series with voltage source E, at any time t. By voltage law q =E C dq q R =E dt C dq i dt Ri di Ri E0 sin wt dt Example 39. Solve the equation L where L, R and E0 are constants and discuss the case when t increases indefinitely. Solution. L di Ri E0 sin wt dt di R E i = 0 sin wt dt L L R I.F. = e Solution is R t i.e L R t i. e L = L dt E0 L R eL R E0 sin wt t e L sin wt dt A eL E0 = L 2 t b sin b x – tan –1 a a2 b2 ea x Lw sin wt – tan –1 A R R w2 L2 R – t E0 Lw L sin wt – tan –1 Ae 2 2 2 R R L w As t increases indefinitely, then Ae i= R t e a x sin bx dx R i= so L – Rt L tends to zero. E0 Lw sin wt – tan –1 2 2 2 R R L w Ans. Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ 168 Differential Equations EXERCISE 3.16 1. A coil having a resistance of 15 ohms and an inductance of 10 henries is connected to 90 volts supply. Determine the value of current after 2 seconds. (e–3 = 0.05) Ans. 5.985 amp. 2. A resistance of 70 ohms, an inductance of 0.80 henry are connected in series with a battery of 10 volts. 175 – t 1 2 1 – e 7 3. A circuit consists of resistance R ohms and a condenser of C farads connected to a constant e.m.f. E; if q q is the voltage of the condenser at time t after closing the circuit Show that E – Ri and hence C C t – show that the voltage at time t is E 1 – e CR . Determine the expression for current as a function of time after t = 0. Ans. i t 4. Show that the current i q – R.C. e during the discharge of a condenser of charge Q coulomb through CR a resistance R ohms. 5. A condenser of capacity C farads with voltage v0 is discharged through a resistance R ohms. Show that if q coulomb is the charge on the condenser, i ampere the current and v the voltage at time t. 1 – dq , hence show that v = v e Rc . 0 dt q = Cv, v = Ri and i = – 6. Solve L di Ri E cos wt dt Ans. i E L2 w2 R2 ( R cos wt Lw sin wt – Re – Rt L ) 7. A circuit consists of a resistance R ohms and an inductance of L henry connected to a generator of E cos (wt + ) voltage. Find the current in the circuit. (i = 0, when t = 0). Ans. i E 2 2 2 R Lw cos [ wt – tan –1 Lw ]– R E 2 2 2 R L w .e – R t L Lw cos – tan –1 R 3.17 VERTICAL MOTION Example 40. A body falling vertically under gravity encounters resistance of the atmosphere. If the resistance varies as the velocity, show that the equation of motion is given by du = g – ku dt where u is the velocity, k is a constant and g is the acceleration due to gravity. Show that as dx t increases, u approaches the value g/k. Also, if u = where x is the distance fallen by the dt body from rest in time t, show that gt g – (1 – e – kt ) x= k k2 Solution. Let the mass of the falling body be unity. du Acceleration = dt du du Force acting downward = 1. dt dt Force of resistance = ku du g – ku Net force acting downward = g – ku ...(1)Proved. dt du = dt g – ku Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ Differential Equations 169 du g – ku Integrating, we get = dt 1 log ( g – ku ) log A log ( g – ku ) – 1/ k A k A(g – ku)–1/k = et (g – ku) = Ak e–kt t= – u= If t increases very large then g Ak – kt – e k k Ak – kt e 0 k g u= k Given when t Proved. u = dx dt du d 2 x dt dt 2 du and u in (1), we get dt Putting the values of d2x dx =g (D2 + kD) x = g dt dt A.E. is m(m + k) = 0 m = 0, m = – k C.F. = A1 + A2e–kt 1 1 g t g P.I. = 2 2D k D kD 2 k 1 t 2D t 1 t 2D g 1 = g g = 1 2 D k k k k k 1 k gt x = A1 A2 e – kt k Putting the values of t = 0 and x = 0 in (2), we get 0 = A1 + A2 A2 = – A1 gt (2) becomes x = A1 – A1 e – kt k dx g On differentiating (3), we get A1 ke – kt dt k dx g g On putting 0, when t = 0 in (4), we get 0 A1k A1 – 2 dt k k Putting the value of A1 in (3), we get tg k ...(2) ...(3) ...(4) gt g gt – 2 (1 – e – kt ) x Proved. k k k k k Example 41. The acceleration and velocity of a body falling in the air approximately satisfy the equation : Acceleration = g – kv2, where v is the velocity of the body at any time t, and g, k are constants. Find the distance traversed as a function of the time, if the body falls from rest. x– g 2 g 2 e – kt Show that value of v will never exceed g . k Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ 170 Differential Equations Solution Acceleration = g – k v2 dv g – k v2 dt 1 2 g g k . v 1 dv g – k v2 dt . dv dt g – k .v 1 On integrating, we get 1 1 1 log ( g k . v) – log ( g – k . v) t A 2 g k 2 gk 1 2 gk g k. v log =t+A g – k. v On putting t = 0, v = 0 in (1), we get Equation (1) becomes 1 log 1 0 A A 0 2 gk g – k.v g k .v 1 g k .v log 2 gk ...(1) 2 = e t log g k .v g – k .v 2 gk t gk t g – k .v By componendo and dividendo, we have k .v g = v = e 2 gkt e2 gkt –1 1 e gk t –e – gk t e gk t e– gk t tan h gk t g tan h gk t k Whatever the value of t may be tanh gk t 1. Hence the value of v will never exceed dx = dt g . k Proved. g tanh gk t k g k tanh Integrating again, we get x = when t = 0, B =0 1 x = log cosh gk t k x = 0 then gk t dt = 1 log cosh gk t B k Ans. EXERCISE 3.17 1. A moving body is opposed by a force proportional to the displacement and by a resistance proportional to the square of velocity. Prove that the velocity is given by cx c VdV V 2 ae – – K1 x – K 2 V 2 Hint. Equation of motion is m b ab2 dx 2. A particle of mass m is projected vertically upward with an initial velocity v0. The resisting force at any time is K times the velocity. Formulate the differential equation of motion and show that the distance s covered by the particle at any time t is given by v0 g g – Kt )– t s = 2 (1 – e K K K Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ Differential Equations 171 3. A particle falls in a vertical line under gravity (supposed constant) and the force of air resistance to its motion is proportional to its velocity. Show that its velocity cannot exceed a particular limit. g Ans. V K n2 4. A body falling from rest is subjected to a force of gravity and an air resistance of times the square of g g velocity. Show that the distance travelled by the body in t seconds in log cosh nt. n2 5. A body of mass m, falling from rest is subject to the force of gravity and an air resistance proportional to the square of the velocity Kv2. If it falls through a distance x and possesses a velocity v, at the instant, prove that a2 2kx log 2 where a – v2 m mg a2 k (A.M.I.E.T.E., June 2009) HEAT CONDUCTION Example 42. The rate at which a body cools is proportional to the difference between the temperature of the body and that of the surrounding air. If a body in air at 25°C will cool from 100° to 75° in one minute, find its temperature at the end of three minutes. Solution. Let temperature of the body be T°C. dT k (T – 25) dt or log (T – 25) = kt + log A or dT k dt T – 25 T – 25 kt log A T – 25 = A e kt ...(1) When t = 0, then T = 100, from (1) A = 75 When t = 1, then T = 75 and A =75, From (1) (1) becomes When t = 3, 2 k =e 3 T = 25 + 75 ekt then T = 25 + 75 e3k = 25 + 75 × 8 / 27 = 47.22 Ans. Example 43. The rate at which the ice melts is proportional to the amount of ice at the instant. Find the amount of ice left after 2 hours if half the quantity melts in 30 minutes. Solution. Let m be the amount of ice at any time t. dm dm km k dt dt m dm ...(1) m k dt C log m = kt + C At t = 0, m = M log M = 0 + C C = log M On putting the value of C, (1) becomes, log m = kt + log M ...(2) m =M/2 when t =1/2 hour M k log M log 2 2 M k 2M 2 1 k 1 log or k 2 log 2 2 2 log Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ 172 Differential Equations 1 On putting the value of k in (2), we have log m= 2 log t+ log M ...(3) 2 1 On putting t= 2 hours in (3), we have log m = 4 log + log M 2 4 m m 1 M 1 log or or m log M M 16 16 2 1 After 2 hours, amount of ice left = of the amount of ice at the beginning. Ans. 16 CHEMICAL ACTION: Example 44. Under certain conditions, cane sugar is converted into dextrose at a rate, which is proportional to the amount unconverted at any time. If out of 75 grams of sugar at t = 0, 8 grams are 1 converted during the first 3 minutes, find the amount converted in 1 hours. 2 Solution. Let M be the amount of cane sugar initially, m be the amount of cane sugar converted into dextrose. Then according to problem, dm dm K ( M – m ) or Km KM dt dt which is a linear differential equation. I.F. = e Kdt = e k.t Solution is m e k.t = KMekt dt = Mekt + C m = M + Ce–k.t (i) At t = 0, m= 0, M = 75 (1) becomes m = 75 –75 e –k.t 0 = 75 + C C = –75 (ii) At t = 30, m = 8 8 = 75 –75 e–30k 67 = 75e–30k e–30k = (iii) At t = 90, (2) becomes 67 75 ...(1) ...(2) ...(3) 3 67 m = 75 –75 e– 90 k = 75 –75 from(3) ...(4) 75 3 67 300763 = 75 – 2 = 75 – = 75 – 53.45 = 21.55 Ans. 5625 75 Example 45. Uranium disintegrates at a rate proportional to the amount present at any instant. If m1 and m2 grams of uranium are present at time t 1 and t 2 respectively, show that half life of uranium is (t1 – t2 )log 2 m log 1 m2 Solution. Let m be the amount of uranium at any time t. dm – km dt m2 dm t2 m1 m1 m – k t1 dt log m k (t 2 – t1 ) ...(1) 2 Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ Differential Equations 173 m 2 m t2 dm – k dt 0 m m log 2 log – log m = – kt kt = log 2 = log 2 k 2 t Substituting the value of k in (1), we get Let the mass m reduce to log 1. 2. 3. m in time t. 2 Also m1 log 2 (t2 – t1 ), t (t 2 – t1 ) log 2 m2 t m log 1 m2 Proved. EXERCISE 3.18 Radium decomposes at a rate proportional to the amount present. If 5% of the original amount disappears in 50 years, how much will remain after 100 years? Ans. 90.25% If a thermometer is taken outdoors where the temperature is 0°C from a room in which the temperature is 21°C and the reading drops to 10°C is 1 minute, how long after its removal will the reading be 5°C ? Ans. 2 minutes, 13 seconds. In one dimensional steady state heat conduction for a hollow cylinder with constant thermal conductivity k in the region a r b , the temperature T r at a distance r (a r b) is given by d dTr r 0, dr dr with Tr = T1 where r = a and Tr = T2 where r = b. Use this to determine steady state temperature T1 – T2 T2 log r1 – T1 log r2 distribution Tr in the cylinder in terms of r. Ans. Tr log(r / r ) log r log(r1 / r2 ) 1 2 MISCELLANEOUS QUESTION Example 46. If the population of a country doubles in 50 years, in how many years will it treble, assuming that the rate of increase is proportional to the number of inhabitants? Solution. Let t = time in years, y = population after t years P= original population (when t= 0). The rate of increase of population is proportional to the population, so that dy = ky, where k is a constant. or k dt = dy/y dt Integrating, kt = c + log y ... (1) When t = 0, y = P When t = 50, y = 2P Substituting in (1), Solving, 0 = c + log P, and 50k = c + log 2P c = – log P 50k = – log P + log 2P = log 2 or k= 1 log 2 50 The value of t when population has trebled is obtained by putting y = 3P in (1). We get kt = c + log 3P = – log P + log 3P = log 3 t= 50 1 log 3 = · log 3 years. log 2 k Ans. Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ 174 Differential Equations 3.18 LINEAR DIFFERENTIAL EQUATIONS OF SECOND ORDER WITH CONSTANT COEFFICIENTS The general form of the linear differential equation of second order is d2y dy Qy R dx dx where P and Q are constants and R is a function of x or constant. 2 P Differential operator. Symbol D stands for the operation of differential i.e., dy d2y Dy , D2 y dx dx 2 1 stands for the operation of integration. D 1 stands for the operation of integration twice. D2 d2y dx 2 P dy Qy R can be written in the operator form. dx D2y + P Dy + Q y = R (D2 + PD + Q) y = R 3.19 COMPLETE SOLUTION = COMPLEMENTARY FUNCTION + PARTICULAR INTEGRAL Let us consider a linear differential equation of the first order dy Py Q dx Its solution is ye Pdx ...(1) (Q e Pdx ) dx C Pdx Pdx Pdx y Ce e (Qe ) dx y = cu + v (say) Pdx where u e and v e Pdx Q e Pdx ...(2) dx Pdx du Pe Pu (i) Now differentiating u e Pdx w.r.t. x, we get dx du d (cu ) Pu 0 P (cu ) 0 dx dx dy Py 0 which shows that y = c.u is the solution of dx (ii) Differentiating v e Pdx (Qe Pdx dx with respect to x, we get Pdx Pdx Pdx Pdx dv Pe (Qe ) dx e Qe dx dv Pv Q dx dy dv Pv Q which shows that y = v is the solution of dx Py Q dx Solution of the differential equation (1) is (2) consisting of two parts i.e. cu and v. cu is the solution of the differential equation whose R.H.S. is zero. cu is known as complementary function. Second part of (2) is v free from any arbitrary constant and is known as particular integral. Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ Differential Equations 175 Complete Solution = Complementary Function + Particular Integral. y = C.F.+ P.I. 3.20 METHOD FOR FINDING THE COMPLEMENTARY FUNCTION (1) In finding the complementary function, R.H.S. of the given equation is replaced by zero. (2) Let y = C1 emx be the C.F. of d2y dx 2 P 2 dy Qy 0 dx ...(1) d y dy mx and (m2 + Pm + Q) = 0 2 in (1) then C1e dx dx m2 + Pm + Q = 0. It is called Auxiliary equation. (3) Solve the auxiliary equation : Case I : Roots, Real and Different. If m1 and m2 are the roots, then the C.F. is Putting the values of y, y C1em1x C2 em2 x Case II : Roots, Real and Equal. If both the roots are m1, m1 then the C.F. is y (C1 C2 x) em1 x Equation (1) can be written as (D – m1)(D – m1)y = 0 Replacing (D – m1)y = v in (2), we get (D – m1)v = 0 dv – m1v 0 dx dv m1dx v ... (2) ... (3) log v m1 x log c2 v c2em1 x v c2em1x ( D –1) y c2 em1 x This is the linear differential equation. From (3) Solution is I.F. e – m1 dx e – m1 x y.e – m1x (c2 em1 x ) (e – m1 x ) dx c1 c2 dx c1 c2 x c1 y (c2 x c1 )em1 x C.F. (c1 c2 x) em1 x d2 y dy 8 15 y 0. Example 47. Solve: 2 dx dx Solution. Given equation can be written as (D2 – 8D + 15) y = 0 2 Here auxiliary equation is m – 8m + 15 = 0 (m – 3) (m – 5) = 0 Hence, the required solution is y = C1 e3x + C2e5x 2 d y dy 6 9y 0 Example 48. Solve: dx dx 2 Solution. Given equation can be written as m = 3, 5 Ans. (D2 – 6D + 9) y = 0 Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ 176 Differential Equations A.E. is m2 – 6m + 9 = 0 (m – 3)2 = 0 m = 3, 3 Hence, the required solution is y = (C1 + C2x) e3x 2 d y dy 4 5 y 0, Example 49. Solve: 2 dx dx dy d 2 y y = 2 and when x = 0. dx dx 2 Solution. Here the auxiliary equation is m2 + 4m + 5 = 0 Its root are 2 i The complementary function is y = e–2x (A cos x + B sin x) On putting y = 2 and x = 0 in (1), we get 2=A On putting A = 2 in (1), we have y = e–2x [2 cos x + B sin x] On differentiating (2), we get dy e 2 x [2sin x B cos x ] 2e2 x [2 cos x B sin x ] dx = e–2x [(– 2B – 2) sin x + (B – 4) cos x] d2y dx 2 Ans. ...(1) ...(2) e 2 x [(2 B 2) cos x ( B 4)sin x] = e–2x – 2e–2x [(– 2B – 2) sin x + (B – 4) cos x] [( – 4B + 6) cos x + (3B + 8) sin x] dy d 2 y dx dx 2 e–2x [(–2B –2) sin x + (B – 4) cos x] = e–2x [(– 4B + 6) cos x + (3B + 8) sin x] On putting x = 0, we get B – 4 = – 4B + 6 B=2 (2) becomes, y = e–2x [2 cos x + 2 sin x] y = 2e–2x [sin x + cos x] But Ans. Exercise 3.19 Solve the following equations : 1. 3. d2 y dx 2 d2y 2 dx d2y 3 d 2 y dy dy 30 y 0 Ans. y = C1e5x + C2e–6x 2 y 0 Ans. y = C1 ex + C2 e2x 2. dx dx 2 dx 8 dy 16 y 0 dx Ans. y = (C1 + C2x) e4x Ans. y C1 cos x C2 sin x 2 y 0 dx 2 5. ( D 2 2 D 2) y 0, y(0) 0, y(0) 1 (A.M.I.E.T.E., June 2006) Ans. y = e–x sin x 4. 6. 7. d3 y dx 3 d4y dx 4 2 d2y dx 32 2 4 d2y dx 2 dy 8y 0 dx Ans. y = C1e2x + C2cos 2x + C3sin 2x 256 0 (A.M.I.E.T.E., Dec. 2004) Ans. y= (C + x) cos 4x + (C3 + C4x) sin 4x 1 Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ Differential Equations 4 8. 9. d y 4 dx d4y dx 4 3 –4 d y dx d2y 3 dx 2 177 2 8 d y dx 2 8 dy 4y 0 dx Ans. y = ex [(C1 + C2 x) cos x + (C3 + C4 x) sin x] 0, y (0) y (0) y (0) 0, y (0) 1 10. The equation for the bending of a strut is EI d2y dx 2 Ans. y = x – sin x Py 0 P x EI y Ans. P 1 sin EI 2 a sin 1 If y = 0 when x = 0, and y = a when x , find y.. 2 11. 12. d3 y dx 3 d3 y dx 3 6 d2y dx d2y dx 2 2 12 dy 8 y 0, y(0) = 0, and y(0) 0 and y(0) 2 dx (A.M.I.E.T.E. Dec. 2008) Ans. y = x2e–2x 4dy 1 4 y 0, y(0) = 0, y(0) 0, y(0) 5, Ans. y = e x cos 2 x sin 2 x dx 2 13. ( D8 6 D6 – 32 D 2 ) y 0 (A.M.I.E.T.E., Summer 2005) Ans. y = C1 + C2x + C3 e 2x C4 e 2x (iv ) 14. Show that non-trivial solutions of the boundary value problem y C5 cos 2 x C6 sin 2 x – w4 y 0, y(0) 0 y(0), nx where Dn are constants. (AMIETE, Dec. 2005) L y(L) = 0, y ( L) 0 are y ( x ) Dn sin n 1 15. Solve the initial value problem y 6 y 11y 6 y 0, y(0) = 0, y(0) 1, y(0) –1. (A.M.I.E.T.E., Dec. 2006) Ans. y = 2e–x – 3e–2x+ e–3x. 16. Let y1, y2 be two linearly independent solutions of the differential equation yy – ( y ) 2 0. Then, c1 y1 + c2 y2, where c1, c2 are constants is a solution of this differential equation for (a) c1 = c2 = 0 only. (b) c1 = 0 or c2 = 0 (c) no value of c1, c2. (d) all real c1, c2 (A.M.I.E.T.E., Dec. 2004) 3.21 RULES TO FIND PARTICULAR INTEGRAL 1 1 ax eax e If f (a) = 0 then 1 e ax x 1 e ax (i) f (D) f (a) f (D) f (a) 1 1 If f (a ) 0 then eax x 2 eax f ( D) f (a) 1 (ii) x n [ f ( D )]1 x n Expand [f (D)]–1 and then operate. f (D) 1 1 1 1 cos ax cos ax sin ax sin ax and (iii) 2 2 2 f (D ) f (a 2 ) f (D ) f (a ) 1 1 sin ax x sin ax If f (– a2) = 0 then f (D 2 ) f ( a 2 ) 1 1 ax ax (iv) f ( D) e ( x ) e f ( D a) ( x ) 1 ( x ) e ax e ax ( x ) dx (v) Da Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ 178 3.22 Differential Equations 1 1 ax e ax e f ( D) f (a ) We know that, D.eax = a.eax, D2eax = a2.eax,…............., Dn eax = an eax ax n n–1 Let f (D) e = (D + K1D + … + Kn) eax = (an + K1an–1 +…+ Kn)eax = f (a) eax. 1 Operating both sides by f ( D) 1 1 f ( D) e ax f (a) e ax f (D) f (D) 1 e ax f ( D) If f (a) = 0, then the above rule fails. e ax f (a) Then 1 1 1 e ax x e ax x e ax f ( D) f ( D) f (a) 1 1 ax e ax e f ( D) f (a) 1 1 e ax x . e ax f ( D) f (a ) 1 1 e ax = x 2 e ax f ( D) f (a ) Example 50. Solve the differential equation If f (a ) 0 then d2x g g x L l dt 2 t where g, l, L are constants subject to the conditions, dx 0 at t = 0. dt d2x g g x L l dt 2 t x = a, Solution. We have, m2 A.E. is g 0 l C.F. = C1 cos g t C2 sin l g 2 g D x L l l m i g l g t l g g g 1 L L e0 t L L g l g g l l 2 D D 0 l l l General solution is = C.F. + P.I. P.I. = 1 2 g g x C1 cos t C2 sin tL l l [D = 0] ...(1) g g dx g g C1 sin cos t t C2 l dt l l l Put t = 0 and dx 0 dt 0 C2 g l C2 = 0 Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ Differential Equations 179 g tL l x C1 cos (1) becomes Put x = a and t = 0 in (2), we get a = C1 + L or ...(2) C1 = a – L g On putting the value of C1 in (2), we get x (a L) cos t L l d2 y dy 9 y 5e3 x dx dx Solution. (D2 + 6D + 9)y = 5e3x Auxiliary equation is m2 + 6m + 9 = 0 C.F. = (C1 + C2x) e–3x Example 51. Solve : 2 6 1 P.I. = D 2 6D 9 (m + 3)2 = 0 .5.e3 x 5 e3 x (3)2 6(3) 9 y (C1 C2 x )e 3 x The complete solution is d2y 2 m = – 3, – 3, 5e3 x 36 5e3 x 36 dy 9 y 6e3 x 7e 2 x log 2 dx dx Solution. (D2 – 6D + 9)y = 6e3x + 7e–2x – log 2 2 A.E. is (m – 6m + 9) = 0 (m – 3)2 = 0, Example 52. Solve : Ans. Ans. 6 m = 3, 3 3x C.F. (C1 C2 x) e 1 1 1 6e3 x 2 7 e 2 x 2 ( log 2) P.I. = 2 D 6D 9 D 6D 9 D 6D 9 1 1 1 6e3 x 7e 2 x log 2 2 e0 x = x 2D 6 4 12 9 D 6D 9 2 = x 1 7 7 1 1 6 e3 x e2 x log 2 3x 2 e3 x e 2 x log 2 2 25 9 25 9 Complete solution is y (C1 C2 x ) e3 x 3 x 2 e3 x 7 2 x 1 e log 2 25 9 Ans. EXERCISE 3.20 Solve the following differential equations: ex 12 1. [D2 + 5D + 6] [y] = ex Ans. C2 e 2 x C2 e 3 x 2 2. d y 3 dy 2 y e3 x dx dx 2 Ans. C 1e x C2 e 2 x 3. (D3 + 2D2 – D – 2) y = ex Ans. C 1e x C 2 e – x C3 e – 2 x d2y dy 2 y sinh x dx dx d2y dy 4 5 y 2 cosh x 5. 2 dx dx 4. 2 2 e3 x 2 (A.M.I.E.T.E. June 2010, 2007) x x e 6 x x Ans. e x [C1 cos x C2 sin x ] e e 10 2 1 e x 2 x x Ans. e (C1 cos x C2 sin x ) e 10 2 Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ 180 Differential Equations 6. (D3 – 2D2 – 5D + 6) y = e3x 7. d3 y dx 3 d2y d2y 4 Ans. C1e x C2 e – 2 x C3 e3 x dy 4 y ex dx x Ans. C1e C2 cos 2 x C3 sin 2 x dx 2 dy 6 9 y e3 x 8. 2 dx dx 3 d y d2y dy 3 3 y e x 9. 2 dx dx dx 2 d y dy 6 y e x cosh 2 x 10. 2 dx dx Ans. (C1 C2 x)e3 x x ex 5 x2 3x e 2 x3 x e 6 1 1 xe3 x e x 10 8 2 x Ans. (C1 C2 x C3 x )e Ans. C1e3 x C2 e –2 x x 2 x ex 2x x Ans. C1e (C2 C3 x)e e 9 4 Ans. (C1 + C2x + C3 x2) ex + 2e3x 11. (D – 2) (D + 1)2 y = e2x + ex 12. (D – 1)3 y = 16 e3x 3.23 x.e3 x 10 1 xn [ f (D)]1 xn . f ( D) Expand [f (D)]–1 by the Binomial theorem in ascending powers of D as far as the result of operation on xn is zero. d2y a2 R (l x) Example 53. Solve the differential equation p dx 2 dy where a, R, p and l are constants subject to the conditions y = 0, 0 at x = 0. dx d2y a2 a2 2 2 2 Solution. a y R ( l x ) ( D a ) y R (l x ) p p dx 2 A.E. is m2 + a2 = 0 m ia C.F. = C1 cos ax + C2 sin ax 1 1 a2 a2 R 1 1 R D2 R (l x) = (l x) 1 2 (l x) P.I. = 2 p a2 D2 p a D a2 p 1 2 a R D2 R = 1 2 (l x) (l x) p a p y = C1 cos ax C2 sin ax a2 y R (l x) p On putting y = 0, and x = 0 in (1), we get 0 = C1 ...(1) R l p C1 Rl p dy R On differentiating (1), we get dx a C1 sin ax a C2 cos ax p dy On putting 0 and x = 0 in (2), we have dx R R 0 = a C2 C2 p a. p On putting the values of C1 and C2 in (1), we get R sin ax R R R l cos a.x l x sin a.x (l x) y = y = l cos a.x p a p a. p p ...(2) Ans. Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ Differential Equations 181 EXERCISE 3.21 Solve the following equations : Ans. C1e x C2 e 4 x 1 (11 4 x) 8 Ans. (C1 + C2 x) e–x + x – 2 1. (D2 + 5D + 4) y = 3 – 2x 2. d2y dx 2 2 dy yx dx 3 x 4 [ A cos 23 23 1 x B sin x ] [8x 2 28 x 13] 4 4 32 1 4. (D2 – 4D + 3) y = x3 Ans. C1e x C2 e3 x (9 x 3 36 x2 78 x 80). 27 d3 y d 2 y dy 1 3 25 2 2 x 3x 2 1 x . 5. 5 3 2 6 Ans. A Be C e 2 x x x dx 36 3 dx dx 4 1 4 d y x x 4 y x4 6. Ans. e (C1 cos x C2 sin x ) e (C3 cos x C4 sin x ) ( x 6) 4 4 dx 3. (2D2 + 3D + 4) y = x2 – 2x 7. dy dx 2 2p Ans. e dy ( p 2 q 2 ) y ecx p.q x 2 dx eCx 2 4p x 6 p 2 2q 2 x ( p C )2 q 2 p 2 q 2 p 2 q 2 ( p 2 q 2 )2 Ans. C1 + C2x + C3 cos 2x + C4 sin 2x + 2x2 (x2 – 3) Ans. e px [C1 cos qx C2 sin qx] 8. D2 (D2 + 4) y = 96 x2 1 1 sin ax sin ax 2 f (D ) f (a2 ) 3.24 pq 2 f (D ) cos ax cos ax f ( –a 2 ) D (sin ax) = a.cos ax, D2 (sin ax) = D (a cos ax) = – a2. sin ax D4 (sin ax) = D2.D2 (sin ax) = D2 (– a2 sin ax) = (– a2)2 sin ax (D2)n sin ax = (– a2)n sin ax Hence, f (D2) sin ax = f ( – a2) sin ax 1 2 f (D ) f ( D 2 ) sin ax sin ax f (– a 2 ) Similarly, If If f (a 2 ) 0 then, 1 2 cos ax 1 f (D2 ) 1 2 f (D ) . f ( a 2 ).sin ax sin ax 1 f ( D2 ) sin ax sin ax f (a 2 ) cos ax f (D ) f ( a2 ) f (– a2) = 0 then above rule fails. 1 sin ax sin ax x 2 f (D ) f ( a 2 ) 1 2 f (D ) sin ax x 2 sin ax f ( a 2 ) Example 54. Solve : (D2 + 4) y = cos 2x (R.G.P.V., Bhopal June, 2008, A.M.I.E.T.E. Dec 2008) Solution. (D2 + 4) y = cos 2x Auxiliary equation is m2 + 4 = 0 m 2i , C.F. = A cos 2x + B sin 2x Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ 182 Differential Equations 1 x1 x cos 2 x sin 2 x sin 2 x 2D 22 D 4 4 x Complete solution is y A cos 2 x B sin 2 x sin 2 x 4 P.I. = Example 55. Solve : 3 d3 y 3 dx 3 1 cos 2 x = x. 2 d2 y 3 dx 4 Ans. dy 2 y ex cos x (U.P., II Semester, Summer 2006, 2001) dx 2 Solution. Given (D – 3D + 4D – 2) y = ex + cos x A.E. is m3 – 3m2 + 4m – 2 = 0 (m – 1) (m2 – 2m + 2) = 0, i.e., m = 1, 1 i C.F. = C1ex + ex (C2 cos x + C3 sin x) 1 1 ex 3 cos x P.I. 2 2 ( D 1) ( D 2 D 2) D 3D 4 D 2 1 1 ex cos x = ( D 1) (1 2 2) (1) D 3(1) 4 D 2 1 1 1 3D 1 cos x ex cos x = x e x = 1 ( D 1) 3D 1 9D 2 1 x = e .x ( 3sin x cos x) 1 x = e .x (3sin x cos x) 9 1 10 Hence, complete solution is y = C1e x e x (C2 cos x C3 sin x ) x e x x Example 56. Solve : ( D 3 1) y cos 2 e x 2 Solution. A.E. is m3 + 1 = 0 Ans. (Nagpur University, Summer 2004) x ( D 3 1) y cos 2 e x 2 (m + 1) (m2 – m + 1) = 0 m or 1 (3sin x cos x ) 10 (1) 1 4 1 i 3 2 2 m=–1 1 3 m i 2 2 x 3 3 x C3 sin x C.F. = C1e x e 2 C2 cos 2 2 1 2 x x 1 1 x cos 2 3 e x [Put D = – 1] cos 2 e = 3 D3 1 2 D 1 D 1 1 1 cos x 1 e x = 3 2 2 D 1 3D 1 1 1 1 1 1 e0 x cos x e x = 1 1 1 cos x 1 e x = 3 3 2 D 1 2 D 1 3(1) 2 1 2 2 D 1 4 P.I. = = 1 1 ( sin x cos x) 1 x 1 1 ( D 1) cos x 1 x e e = 2 2 4 2 2 ( D 1) ( D 1) 4 ( D 2 1) Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ Differential Equations 183 1 1 sin x 1 1 1 cos x e x 2 2 2 2 ( D 1) 2 ( D 1) 4 1 sin x cos x 1 x 1 1 sin x 1 1 1 e cos x e x = Put D2 = – 1 = 2 4 4 4 2 2 (1 1) 2 (1 1) 4 1 1 x P.I. = (cos x sin x e ) 2 4 Hence, the complete solution is = x 3 3 1 1 y = C1e x e 2 C2 cos x C3 sin x (cos x sin x e x ) 2 2 2 4 Ans. EXERCISE 3.22 Solve the following differential equations : 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. d2y 1 sin 4 x dx 10 2 d x dx 1 2 3x sin t Ans. e t [ A cos 2t B sin 2t ] (cos t sin t ) 2 dt 4 dt d2x dx dx 2 5 x sin 2t , given that when t = 0, x = 3 and 0 dt dt dt 2 53 55 1 Ans. e t cos 2t sin 2t (4cos 2t sin 2t ) 34 17 17 d2y dy dy 7 6 y 2sin 3x, given that y = 1, 0 when x = 0. dx dx dx 2 13 27 1 Ans. e6 x e x (7 cos 3 x sin 3 x ) 75 25 75 (D3 + 1) y = 2cos2 x 1 x 3 3 1 Ans. C1e x e 2 C2 cos x C sin x 1 (8sin 2 x cos 2 x) 3 2 2 65 x 2 2 (D + a ) y = sin ax (A.M.I.E.T.E., June 2009) (Ans. C1 cos ax C2 sin ax cos ax 2a x2 4 2 2 4 (D + 2a D + a ) y = 8 cos ax Ans. (C1 C2 x C3 cos ax C4 sin ax) 2 cos ax a d2y dy 3 2 y sin 2 x (A.M.I.E.T.E., Summer 2002) dx dx 2 1 Ans. C1e x C2 e 2 x (3cos 2 x sin 2 x ) 20 d2y 1 y sin 3 x cos 2 x Ans. C1 cos x C2 sin x [ sin 5 x 12 x cos x ] dx 2 48 d2y dy 2 3 y 2e 2 x 10sin 3x given that y (0) = 2 and y(0) 4 2 dx dx 29 3 x 1 x 2 2 x 1 Ans. e e e [cos 3 x 2sin 3 x ] 12 12 3 3 2 d y dy 2 3 2 y 4cos x (R.G.P.V., Bhopal, I Semester, June 2007) dx dx 2 Ans. C1e x C2 e 2 x e 2 x 1 (3sin 2 x cos 2 x ) 1 10 d2y dy 2 2 3 y cos x x dx 1 1 4 2 dx 2 Ans. e x [C1 cos 2 x C2 sin 2 x ] (cos x sin x ) ( x 2 x ) 4 3 3 9 2 6 y sin 4 x Ans. C1 cos 6 x C2 sin 6 x Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ 184 Differential Equations 1 13. ( D 3 3D 2 4 D 2) y e x cos x Ans. (C1 C2 cos x C3 sin x ) e x (3sin x cos x) 3 2 10 14. (D – 4D + 13 D) y = 1 + cos 2x 1 x Ans. C1 e 2 x (C2 cos 3 x C3 sin 3 x ) (9 sin 2 x 8 cos 2 x ) 290 13 15. (D2 – 4D + 4) y = e2x + x3 + cos 2x 1 1 1 Ans. (C1 C2 x ) e2 x x 2 e 2 x (2 x 3 6 x2 9 x 6) sin 2 x 2 8 8 d2y 2 n y h sin px 16. ( P n) dx 2 where h, p and n are constants satisfying the conditions b dy ph h sin px b for x = 0 y = a, Ans. a cos nx sin nx 2 2 2 dx (n p 2 ) n n( n p ) 17. y y 2 y 6sin 2 x 18cos 2 x, y (0) = 2, y(0) 2 3.25 Ans. – e–2x + 3 cos 2x 1 1 .e ax ( x ) e ax . .( x ) f ( D) f ( D a) D[eax ( x)] eax D ( x) aeax ( x) eax ( D a ) ( x) ax 2 ax D 2 [eax ( x)] D[eax ( D a) ( x)] = e ( D aD) ( x) ae ( D a) ( x) = eax ( D 2 2a D a 2 ) ( x) eax ( D a)2 ( x) D n [eax ( x)] eax ( D a )n ( x) Similarly, f ( D)[eax ( x)] eax f ( D a) ( x) e ax ( x) 1 .[eax f ( D a) ( x)] f ( D) Put f ( D a) ( x) X , so that ( x) Substituting these values in (1), we get e ax ...(1) 1 .X f ( D a) 1 1 X [e ax . X ] f ( D a) f ( D) 1 1 [eax . ( x )] e ax ( x) f ( D) f ( D a) Example 57. Solve : (D2 – 4D + 4) y = x3 e2x Solution. (D2 – 4D + 4) y = x3 e2x A.E. is m2 – 4m + 4 = 0 (m – 2)2 = 0 m = 2, 2 C.F. = (C1 + C2 x) e2x 1 1 x 3 e2 x e 2 x x3 P.I. = 2 2 D 4D 4 ( D 2) 4( D 2) 4 4 5 1 1 x 2x 3 2x 2x x x e . e = e D 4 20 D2 2x 2x The complete solution is y (C1 C 2 x) e e . Example 58. Solve the differential equation : d3 y dx 3 7 d2y dx 2 10 x5 20 Ans. dy e2 x sin x (AMIETE, June 2010, Nagpur University, Summer 2005) dx Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ Differential Equations d3 y Solution. A.E. is 185 dx 3 7 d2y dx 2 10 dy e2 x sin x dx m3 – 7m2 + 10 m = 0 m (m – 2) (m – 5) = 0 D3y – 7D2y + 10 Dy = e2x sinx (m – 2) (m2 – 5m) = 0 m = 0, 2, 5 C.F = C1e0 x C2e2 x C3e5 x P.I. = 1 3 2 D 7 D 10 D e2 x e2 x e2x 2x e 2 x sin x e 1 3 ( D 2) 7 ( D 2)2 10 ( D 2) . sin x 1 .sin x D 6 D 12 D 8 7 D 2 28 D 28 10 D 20 1 1 2x sin x sin x e 2 3 2 ( 1 ) D (12 ) 6 D D D 6D 3 2 1 1 sin x e2 x sin x e2 x 1 7 D sin x e2 x 1 7 D sin x D 1 6D 1 7D 1 49 D2 1 49 ( 12 ) 1 7D e2 x sin x (sin x 7 cos x) 50 50 Complete solution is y = C.F. + P.I. e2 x 2x 5x y C1 C2 e C3 e 2 Example 59. Solve ( D 6 D 9) y e2 x (sin x 7 cos x ) 50 Ans. e 3 x . x3 (Nagpur University, Summer 2002, A.M.I.E.T.E., June 2009) Solution A.E. is m2 + 6m + 9 = 0 (m + 3)2 = 0 m = – 3, – 3 C.F. = (C1 + C2x) e– 3x P.I. = 1 e 3 x D2 6 D 9 x3 3 x = e 3 x = e e 3 x 1 1 ( D 3) 2 6( D 3) 9 x3 1 2 1 3 x = e 1 D 6 D 9 6 D 18 9 x D2 2 1 3 x 1 3 x 1 x 3 x x e x e e D 2 (2) (1) 2 2x ( x 3 ) e 3 x Ans. 2x Example 60. Solve (D2 + 5D + 6) y = e–2x sec2x (1 + 2 tan x) (A.M.I.E.T.E., Summer 2003) Solution. (D2 + 5D + 6) y = e–2x sec2x (1 + 2 tan x) Auxiliary Equation is m2 + 5m + 6 = 0 (m + 2) (m + 3) = 0 m = –2, and m = –3 Hence, complementry function (C.F.) = C1e–2x + C2e–3x Hence, the solution is y = (C1 C2 x )e 3 x P.I. = 1 2 D 5D 6 e 2 x sec 2 x (1 2 tan x ) e 2 x 1 2 ( D 2) 5( D 2) 6 sec 2 x (1 2 tan x) Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ 186 Differential Equations 2 x = e 1 sec 2 x(1 2 tan x ) 2 D 4 D 4 5 D 10 6 2 1 2 tan x sec 2 x 2 2 x sec x e sec x (1 2 tan x ) e 2 D2 D D2 D D D 1 1 e 2 x sec 2 2 tan x sec2 x D ( D 1) D ( D 1) 1 1 2 1 1 2 e 2 x sec x 2 tan x sec x D D 1 D D 1 1 1 1 2 x 1 2 2 e sec x sec x 2 tan x sec2 x 2 tan x sec 2 x D D 1 D D 1 e 2 x tan x e x ex .sec 2 x dx tan 2 x e x 2e x tan x sec 2 x dx Now, e 2 x e x sec 2 x dx e x sec2 x e x .2 sec x sec x tan x . dx 2 x e x sec 2 x 2 ex sec 2 x .tan x dx 2 x e x 2 x x 2 x x 2 P.I. = e tan x x .e sec x 2e e sec x tan x dx tan x 2e e sec x tan x dx e2 x [tan x sec2 x tan 2 x] e2 x [tan x (sec2 x tan 2 x] e2 x (tan x 1) Complete solution is y C.F . P.I . C1e2 x C2 e3x e2 x (tan x 1) Example 61. Solve the differential equation (D2 – 4D + 4) y = 8x2 e2x sin 2x (U.P. II Semester, Summer 2008, Uttrakhand 2007, 2005, 2004; Nagpur University June 2008) Solution. (D2 – 4D + 4) y = 8x2 e2x sin 2x A.E. is (m2 – 4m + 4) = 0 (m – 2)2 = 0 m = 2, 2 2x C.F. = (C1 + C2x) e 1 P.I. = = 8e 2 x 2 D 4D 4 1 ( D 2 2)2 8 x 2 e 2 x sin 2 x = 8 x 2 sin 2 x = 8e 2 x 1 D2 1 ( D 2) 2 x 2 e 2 x sin 2 x x 2 sin 2 x 2 1 2 ( cos 2 x) cos 2 x x sin 2 x cos 2 x sin 2 x 2x 1 x x 2 x 2 8 e cos 2 x = D 2 4 8 D 2 2 4 x2 sin 2 x 2 x cos 2 x sin 2 x x cos2 x 1 sin 2 x sin 2 x = 8e2 x (1) 4 8 2 2 2 4 8 2 2 2 2x = 8e = e2x [ – 2x2 sin 2x – 2x cos 2x + sin 2x – 2x cos 2x + sin 2x + sin 2x] = e2x [– 2x2 sin 2x – 4x cos 2x + 3 sin 2x] = – e2x [4x cos 2x + (2x2 – 3) sin 2x] Complete solution is, y = C.F. + P.I. y = (C1 + C2x) e2x – e2x [4x cos 2x + (2x2 – 3) sin 2x] Ans. EXERCISE 3.23 Solve the following equations : 2x 3x 1. (D2 – 5D + 6) y = ex sin x Ans. y C1e C2 e ex (3cos x sin x ) 10 Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ Differential Equations 2. 3. 4. 5. 6. 7. 8. 9. 187 e2 x dy 2x y C1e2 x C2 e5 x (3cos x sin x) 10 y e sin x Ans. 10 dx dx 2 d3 y dy xe x x 2 x x 2 4 y e cos x y C e e ( C cos x C sin x ) (3sin x cos x) Ans. 1 2 3 dx 20 dx 3 (D2 – 4D + 3) y = 2xe3x + 3e3x cos 2x 1 3x 2 3 3x x 3x Ans. y C1e C2 e e ( x x ) e (sin 2 x cos 2 x ) 2 8 d2 y dy e x 2 y 2 Ans. y = (C1 + C2 x) e–x – e–x log x dx dx 2 x e3 x 2 12 x 62 2x 2 x x (D2 – 4) y = x2 e3x Ans. y C1e C2 e 5 5 25 4x e 5 [18 x 2 30 x 19] e3 x (D2 – 3D + 2) y = 2x2 e4x + 5e3x Ans. y C1e x C2 e 2 x 54 2 d2y x 2 4 y x sinh x Ans. y C1e 2 x C2 e 2 x sinh x cosh x 3 9 dx 2 k – ht d2y dy 2h (h2 p 2 ) y ke ht cos pt Ans. y e ht [ A cos pt B sin pt ] te sin pt 2 dt 2p dt d2y 7 1 x n sin ax . f ( D) 1 1 1 x n (cos ax i sin ax) x n eiax eiax xn f ( D) f ( D) f ( D ia) 1 1 x n sin ax Imaginary part of e iax xn f ( D) f ( D ia ) 3.26 TO FIND THE VALUE OF Now 1 1 x n cos ax Real part of e iax xn f ( D) f ( D ia ) d2y dy 2 y x sin x dx dx 2 Solution. Auxiliary equation is m2 – 2m + 1 = 0 or m = 1, 1 C.F. = (C1 + C2 x) ex 1 x sin x (eix = cos x + i sin x) P.I. = 2 D 2D 1 1 1 x eix x(cos x i sin x) = Imaginary part of 2 = Imaginary part of 2 D 2 D 1 D 2D 1 1 1 ix x = Imaginary part of eix x = Imaginary part of e 2 2 D 2(1 i ) D 2i ( D i ) 2( D i ) 1 1 1 1 2 1 (1 i ) D D = Imaginary part of eix x – 2i 2i 1 i = Imaginary part of (cos x i sin x) 1 (1 i) D x = Imaginary part of (i cos x sin x) [ x 1 i] 2 2 1 1 1 P.I. = x cos x cos x sin x 2 2 2 1 x Complete solution is y = (C1 C2 x)e ( x cos x cos x sin x) Ans. 2 Example 62. Solve Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ 188 Differential Equations EXERCISE 3.24 Solve the following differential equations : 1. (D2 + 4)y = 3x sin x 2. 3. d2y dx 2 d2y 2 Ans.C1 cos 2x + C2 sin 2x + x sin x – 1 3 cos 3x x sin 3x 5cos x 10 5 1 x x 2 [ x sin x cos x ] e (2 x 3x 9) 2 12 y x sin 3 x cos x Ans. C1e x C2 e x y x sin x e x x 2 e x Ans. C1e x C2 e x dx 4. (D4 + 2D2 + 1) y = x2 cos x 2 cos x 3 1 3 1 x sin x ( x 4 9 x 2 ) cos x 12 48 3.27 GENERAL METHOD OF FINDING THE PARTICULAR INTEGRAL OF ANY FUNCTION (x) 1 ( x ) y P.I. = ...(1) Da 1 ( D a) ( x) ( D a ) y or Da ( x) ( D – a ) y or ( x) Dy – ay dy ay ( x ) which is the linear differential equation. dx Ans. (C1 C2 x ) cos x (C3 C4 x ) sin x a dx a dx ye e ( x ) dx Its solution is or yeax e ax ( x) dx ax ax y = e e ( x) dx 1 ( x ) = e ax e –ax ( x ) dx D–a 2 d y 9 y sec 3x. dx 2 Solution. Auxiliary equation is m2 9 0 or m 3i , C.F. = C1 cos 3x + C2 sin 3x 1 1 1 1 1 sec 3x sec 3x sec 3x = P.I. = 2 6i D 3i D 3i ( D 3i ) ( D 3i ) D 9 1 1 1 1 sec 3 x sec 3 x = 6i D 3i 6i D 3i Example 63. Solve Now, ...(1) 1 ax ax 1 sec 3 x e3ix e 3ix sec3 x dx D a ( x) e e ( x) dx D 3i i 3ix cos 3 x i sin 3 x dx e3ix (1 i tan 3 x) dx e3ix ( x log cos 3 x ) = e cos 3 x 3 1 i sec 3 x e 3ix ( x log cos 3 x) D 3i 3 Putting these values in (1), we get Changing i to – i, we have P.I. = = 1 3ix i i e x log cos 3 x e 3ix x log cos3 x 6i 3 3 x 3ix e3ix log cos 3x xe3ix e 3ix e log cos3 x 6i 18 6i 18 x 1 x e3ix e 3ix 1 e3ix e 3ix . log cos 3x = sin 3 x cos 3 x log cos 3 x 3 9 3 2i 9 2 x 1 Hence, complete solution is y C1 cos 3 x C2 sin 3 x sin 3 x cos3 x log cos3 x 3 9 = Ans. Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ Differential Equations 189 EXERCISE 3.25 Solve the following differential equations : d2y a y sec ax 1. (R.G.P.V., Bhopal April, 2010) dx 2 x 1 Ans. C1 cos ax C2 sin ax sin ax 2 cos ax log cos ax a a d2y y cosec x 2. Ans. C1 cos x + C2 sin x – x cos x + sin x log sin x dx 2 1 3. (D2 + 4) y = tan 2x Ans. C1 cos 2 x C2 sin 2 x cos 2 x log (sec 2 x tan 2 x) 4 d2y y ( x cot x ) 4. (A.M.I.E. Winter 2002) dx 2 Ans. C1 cos x + C2 sin x – x cos 2x – sin x log (cosec x – cot x) 3.28 CAUCHY EULER HOMOGENEOUS LINEAR EQUATIONS an x n dny an 1 x n 1 d n 1 y ... (1) .... a0 y ( x ) dx n d xn 1 where a0, a1, a2, ... are constants, is called a homogeneous equation. d x ez , z log e x, D Put dz dy dy dz 1 dy dy dy dy . x x Dy dx dz dx x dz dx dz dx Again, d2y dx 2 d dy d 1 dy 1 dy 1 d 2 y dz 2 dx dx dx x dz x dz x dz 2 dx = 2 1 dy 1 d 2 y 1 1 d 2 y dy 1 2 d y 2 x (D2 D) y ; ( D D ) y dx2 x2 dz x dz 2 x x2 dz 2 dz x2 x2 d2 y x3 d3 y D ( D 1) ( D 2) y Similarly. dx 2 dx 3 The substitution of these values in (1) reduces the given homogeneous equation to a differential equation with constant coefficients. or D ( D 1) y Example 64. Solve: x2 Solution. We have, x2 d2y 2 dx d2y dx 2 2x dy 4 y x4 dx 2x dy 4 y x4 dx (A.M.I.E. Summer 2000) ... (1) d dy d2y , x Dy , x 2 D ( D 1) y in (1), we get dz dx dx 2 or D ( D 1) y 2 Dy 4 y e4 z ( D 2 3D 4) y e4 z Putting x e z , D A.E. is m2 3m 4 0 (m 4) (m 1) 0 C.F. = C1 e z C2 e4 z P.I. = = z 1 2 D 3D 4 e4 z m = –1, 4 [Rule Fails] 1 1 z e4 z e4 z z e4 z 2D 3 2 (4) 3 5 Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ 190 Differential Equations Thus, the complete solution is given by C1 1 z e4 z C2 x 4 x 4 log x y Ans. x 5 5 2 dy 2 d y x y sin ( log x 2 ) (Nagpur University, Summer 2005) Example 65. Solve x 2 dx dx 2 dy 2 d y x y sin ( log x 2 ) Solution. We have, x ... (1) 2 dx dx y C1 e z C2 e 4 z Let x = ez, so that z = log x, D d dz (1) becomes D (D – 1) y + Dy + y = sin (2z) A.E. is m2 + 1 = 0 or (D2 + 1) y = sin 2z m=±i C.F. = C1 cos z + C2 sin z 1 P.I = 2 D 1 1 1 sin 2 z sin 2 z 41 3 sin 2 z 1 sin 2 z 3 1 2 = C1 cos ( log x ) C 2 sin ( log x ) sin ( log x ) 3 y = C.F. + P.I. = C1 cos z C2 sin z 2 Example 66. Solve: x Solution. We have, x3 d3 y dx 3 d y 3x 2 d2y dx 2 d2y 2 dy x 2 log x (Nagpur University, Summer 2003) dx x dy x3 log x dx dx d Let x = ez so that z = log x, D dz The equation becomes after substitution [D (D – 1) (D – 2) + 3D (D – 1) + D ] y = z e3z Auxiliary equation is m3 = 0 m = 0, 0, 0. C.F. = C1 + C2 z + C3 z2 = C1 + C2 log x + C3 (log x)2 P.I. = 1 D 3 . z e3 z e3 z . dx 3 3 3x 1 ( D 3)3 Ans. D3y = ze3z .z 3 1 D e3z e3z x3 (1 D) z ( z 1) (log x 1) 1 z 27 3 27 27 27 x3 y C1 C2 log x C3 (log x)2 (log x 1) Complete solution is 27 3z = e Ans. 3.29 LEGENDRE'S HOMOGENEOUS DIFFERENTIAL EQUATIONS A linear differential equation of the form (a bx)n dn y a1 (a bx)n 1 d n 1 y ... an y X ... (1) dx n dx n 1 where a, b, a1, a2, .... an are constants and X is a function of x, is called Legendre's linear equation. Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ Differential Equations 191 Equation (1) can be reduced to linear differential equation with constant coefficients by the substitution. a + bx = ez z = log (a + bx) b dy dy dy dz . so that = a bx . dz dx dz dx where dy dy b b Dy , dx dz ( a bx ) d2 y Again dx 2 d2y (a + bx) dy = b Dy dx dy b d 2 y dz . . (a bx )2 dz (a bx ) dz 2 dx b2 dy b d2y b . 2 . (a bx ) dz (a bx ) dz (a bx ) 2 b2 dx 2 b2 = (a bx)2 d dz d dy d b dy . dx dx dx a bx dz = D dy d2y b2 2 dz dz 2 dy 2 d y 2 2 = b 2 b ( D y D y ) b 2 D ( D 1) y dz dz (a bx) 2 d2y dx 2 Similarly, (a bx)3 b 2 D ( D 1) d3y b3 D ( D 1) ( D 2) y dx3 ...................................................................... (a bx)n dn y n b n D ( D 1) ( D 2) ..... ( D n 1) y dx Substituting these values in equation (1), we get a linear differential equation with constant coefficients, which can be solved by the method given in the previous section. 2 Example 67. Solve (1 x ) 2 Solution. We have, (1 x) Put 1 + x = ez or d2y 2 dx d2y 2 (1 x) dy y sin 2 {log (1 x )} dx (1 x) dy y sin 2 { log (1 x ) } dx dx log (1 + x) = z dy d d2y Dy and (1 x)2 D ( D 1) y, where D 2 dx dz dx Putting these values in the given differential equation, we get D (D – 1) y + D y + y = sin 2z or (D2 – D + D + 1)y = sin 2z 2 (D + 1)y = sin 2z A.E. is m2 + 1 = 0 m=±i C.F. = A cos z + B sin z 1 1 1 sin 2 z sin 2 z sin 2 z P.I. = 2 4 1 3 D 1 (1 x) Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ 192 Differential Equations Now, complete solution is y = C.F. + P.I. 1 sin 2 z 3 1 y A cos {log (1 x)} B sin {log (1 x )} sin 2 {log (1 x)} 3 y A cos z B sin z Ans. EXERCISE 3.26 Solve the following differential equations: 2 1. x d2y dx 4x 2 dy 42 6y 4 dx x 2 3 Ans. C1 x C2 x 2. ( x 2 D 2 3x D 4) y 2 x 2 2 3. x 4. d2 y dx d2y dx 2 2 x d2y dx 2 dy y log x (AMIETE, June 2010) dx 1 dy 12 log x x dx x2 2x Ans. (C1 + C2 log x) x + log x + 2 Ans. C1 C2 log x 2 (log x)3 C1 x2 2 C2 x 3 log x x 3 3 (A.M.I.E. Winter 2001, Summer 2001) Ans. dy 2 y x 2 sin (5 log x) dx 2 2 Ans. c1 x c2 x x log x d2y x4 Ans. (C1 C2 log x) x2 x2 (log x) 2 5. ( x 2 D 2 x D 3) y x2 log x 2 6. x 1 1 [ 15 cos (5 log x ) 23 sin (5 log x) ] 754 dy sin (logx) 1 y log x (AMIETE, Dec. 2009) dx x dx 1 382 54 2 3 C2 x 2 3 cos log x Ans. y C1 x sin (log x) + 6 log x cos (log x) + x 61 61 1 5 log x sin (log x)] + 6x 2 dy 2 d y (1 x) y 2 sin log (1 x) 8. (1 x) dx dx 2 Ans. y C1 cos log (1 x) C2 sin log (1 x) log (1 x ) cos log (1 x) 2 7. x 2 3x 2 9. Which of the basis of solutions are for the differential equation x 1 1 (c) , 2 , x x (b) I n x, e x (a) x, x I n x, 2 10. The general solution of x d2y 2 d2y dx (d) 2 x 1 2 dy y0 dx ex , x In x x (A.M.I.E., Winter 2001) Ans. (a) dy 9 y 0 is dx x) x3 (c) (C1 + C2x) x3 (d) (C1 + C2 ln x) e x3 (AMIETE, Dec. 2009) Ans. (b) 5x dx (b) (C1 + C2n (a) (C1 + C2x) e3x d2y dy 1 into a linear differential equation with constant coefficients, dx x dx the required substitution is (a) x = sin t (b) x = t2 + 1 (c) x = log t (d) x = et (AMIETE, June 2010) Ans. (d) 11. To transform x 2 Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ Differential Equations 193 3.30 METHOD OF VARIATION OF PARAMETERS d2 y dy b cy X dx dx 2 Let complementary function = Ay1 + By2, so that y1 and y2 satisfy To find particular integral of a d2y dy cy0 dx dx Let us assume particular integral y = uy1 + vy2, a b 2 ... (1) ... (2) ... (3) where u and v are unknown function of x. Differentiation (3) w.r.t. x, we have y' = uy1' + vy2' + u'y1 + v'y2 assuming that u,v satisfy the equation u'y1 + v'y2 = 0 then ... (4) y' = uy1' + vy2' Differentiating (5) w.r.t.x, we have ... (5) y'' = uy1'' + u'y1' + vy2'' + v'y2' Substituting the values of y, y' and y'' in (1), we get a(uy1'' + u'y1' + vy2'' + v' y2') + b (uy1' + vy2') + c (uy1 + vy2) = X u(ay1'' + by1' + cy1) + v(ay2'' + by2' + cy2) + a (u' y1' + v' y2') = X ... (6) y1 and y2 will satisfy equation (1) and ay1'' + by1' + cy1 = 0 ... (7) ay2'' + by2' + cy2 = 0 ... (8) Putting the values of expressions from (7) and (8) in (6), we get u' y1' + v'y2' = X ... (9) Solving (4) and (9), we get u' 0 y2 X y2 ' y v' 1 y1 u y 1 0 X y1 y2 y1 ' y2 ' y1 y2 y1 ' y2 ' y2 X y1 y2 ' y1 ' y2 y1 X y1 y2 ' y1 ' y2 y2 X dx y2 ' y1 ' y2 v y 1 y1 X dx y2 ' y1 ' y2 General solution = complementary function + particular integral. Working Rule Step 1. Find out the C.F. i.e., A y1 + B y2 Step 3. Find u and v by the formulae y2 X u dx, y1 y2 ' y1 ' y2 A.E. is v d2y y cosec x. dx 2 (D2 + 1) y = cosec x Example 68. Solve Solution. Step 2. Particular integral = u y1 + v y2 m2 + 1 = 0 y1 X dx y1 y2 ' y1 ' y2 (Nagpur University, Summer 2005) m=±i C.F. = A cos x + B sin x Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ 194 Differential Equations Here y1 = cos x, y2 = sin x P.I. = y1 u + y2 v y2 . cosec x dx sin x . cosec x dx y ' . y cos x (cos x ) ( sin x ) (sin x ) 1 2 1 2 y . y' u where sin x . v cos 2 1 dx sin x x sin 2 x y1 . X dx cos x . cosec x dx cos x (cos x ) ( sin x ) (sin x ) 1 2 y '1 y2 1 cos x . sin x cot x dx dx log sin x 2 2 1 cos x sin x y . y' dx x P.I. = uy1 + vy2 = – x cos x + sin x (log sin x) General solution = C.F. + P.I. y = A cos x + B sin x – x cos x + sin x .(log sin x) Ans. Example 69. Apply the mehtod of variation of parameters to solve d2y dx 2 y tan x (A. M. I. E. T. E., Dec. 2010, Winter 2001, Summer 2000) d2y y tan x dx 2 (D2 + 1)y = tan x Solution. We have, m2 = – 1 A.E. is C. F. or m=±i y = A cos x + B sin x Here, y1 = cos x, y2 = sin x y1 . y'2 – y'1 . y2 = cos x (cos x) – (– sin x) sin x = cos2 x + sin2 x = 1 P. I. = u. y1 + v. y2 where y2 tan x dx y1 . y '2 y '1 . y2 sin x tan x dx 1 sin 2 x dx cos x 1 cos 2 x dx cos x ( cos x sec x ) dx sin x log ( sec x tan x ) y tan x v dx cos x tan x dx sin x dx cos x 1 y . y' y' . y u 1 1 2 1 2 P. I. = u. y1 + v. y2 = [ sin x – log (sec x + tan x) ] cos x – cos x sin x = – cos x log ( sec x + tan x) Complete solution is y = A cos x + B sin x – cos x log (sec x + tan x) Ans. Example 70. Solve by method of variation of parameters: d2y dx 2 y 2 1 ex (Uttarakhand, II Semester, June 2007, A.M.I.E.T.E., Summer 2001) (Nagpur University, Summer 2001) Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ Differential Equations 195 2 d y Solution. 2 y 2 dx 1 ex 2 (m – 1) = 0 m2 = 1, m=±1 A. E. is C. F. = C1 ex + C2 e–x P.I. = uy1 + vy2 y 1 = e x, Here, y2 = e–x y1 . y'2 – y'1 . y2 = – ex . e–x – ex.e–x = – 2 and u y2 X e x 2 dx dx y1. y '2 y '1 . y2 2 1 e x e x = 1 e e v x x dx = dx e e dx x e x x 1 x (1 e ) 1 e x dx 1 e 1 x dx e x log (e x 1) y1 X ex 2 ex dx dx dx log (1 ex ) = y1 . y '2 y '1 . y2 2 1 e x 1 ex P.I. = u. y1 v. y2 [ e x log (e x 1) ] e x e x log (1 e x ) = 1 e x log (e x 1) e x log (e x 1) Complete solution = y C1ex C2e x 1 ex log (e x 1) e x log (e x 1) Ans. EXERCISE 3.27 Solve the following equations by variation of parameters method. 1. 2. 3. 4. d2 y 2 dx d2y dx 2 d2y 2 dx d2y dx 2 4 y e2 x y sin x 3 dy 2 y sin x dx y sec x tan x 5. y´´ 6 y´ 9 y x 2 x e2 x e 4 16 x 1 Ans. y C1 cos x C2 sin x cos x sin x 2 4 1 x 2x (3 cos x sin x ) Ans. y C1e C2 e 10 2x 2 x Ans. y C1e C2 e e3 x x2 Ans. y C1 cos x C2 sin x x cos x sin x log sec x sin x (AMIETE, June 2010, 2009) Ans. y = (C1 + x C2) e3x – e3x log x 3.31 SIMULTANEOUS DIFFERENTIAL EQUATIONS If two or more dependent variables are functions of a single independent variable, the equations involving their deriavatives are called simultaneous equations, e.g. dx 4y = t dt dy 2 x = et dt The method of solving these equations is based on the process of elimination, as we solve algebraic simultaneous equations. Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ 196 Differential Equations Example 71. The equations of motions of a particle are given by dx dy y 0 x = 0 dt dt Find the path of the particle and show that it is a circle. (R.G.P.V. Bhopal, Feb. 2006, U.P. II Semester summer 2009) d D in the equations, we have dt Dx + y = 0 –x + Dy = 0 On multiplying (1) by w and (2) by D, we get Dx + 2y = 0 –Dx + D2y = 0 On adding (3) and (4), we obtain 2 y + D2 y = 0 (D2 + 2) y = 0 Now, we have to solve (5) to get the value of y. A.E. is m2 + 2 = 0 m2 = – 2 m = ± i y = A cos wt + B sin wt Dy = – A sin t + B cos wt On putting the value of Dy in (2), we get – x – A sin t + B cos t = 0 x = – A sin t + B cos t x = – A sin t + B cos t On squaring (6) and (7) and adding , we get x2 + y2 = A2(cos2t + sin2t) + B2 (cos2t + sin2t) x2 + y2 = A2 + B2 This is the equation of circle. Example 72. Solve the following differential equation Solution. On putting dx dy = y + 1, =x+1 dt dt Solution. Here, we have Dx – y = 1 – x + Dy = 1 Multiplying (1) by D, we get D2x – Dy = D.1 Adding (2) and (3), we get (D2 –1) x = 1 + D.1 (D2 –1) x = 1 or (D2 – 1)x = e0 A.E. is m2 – 1 = 0 m = ± 1 C.F. = c1et + c2 e– t 1 1 0 . e0 e 1 P.I. = 2 0 1 D 1 x = C.F. + P.I. = c1et + c2 e– t – 1 d dx (c1et c2 e t 1) 1 1 From (1), y = y= dt dt y = c1et – c2 e– t – 1 and x = c1et + c2e–t – 1 ...(1) ...(2) ...(3) ...(4) ...(5) ...(6) ...(7) Proved. (U.P. II Semester, 2009) ... (1) ...(2) ...(3) [D. (1) = 0] Ans. Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ Differential Equations 197 Example 73. where y (0) = 0, x (0) = 2 t t ...(5) Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ 198 Differential Equations Example 74. Solve: dx 4x 3y = t dt dy 2 x 5 y = et dt Solution. Here, we have (D + 4)x + 3y = t [U.P. II Semester, 2006] ...(1) d ...(2) D dt To eliminate y, operating (1) by (D + 5) and multiplying (2) by 3 then subtracting, we get t (D + 5) (D + 4) x + 3 (D + 5) y – 3 (2x) – 3 (D + 5) y = (D + 5)t – 3e t [(D + 4) (D + 5) – 6]x = (D + 5)t – 3e t (D2 + 9D + 14)x = 1 + 5t – 3e Auxiliary equation is m2 + 9m + 14 = 0 m = – 2, – 7 C.F. = c1e–2t + c2e–7t 2x + (D + 5)y = e t 1 (1 5t 3et ) D 9 D 14 1 1 1 e0 t 5 2 t 3 2 et = 2 D 9 D 14 D 9 D 14 D 9 D 14 1 1 1 e0t 5 . t 3 2 et = 2 9D D2 0 9(0) 14 1 9(1) 14 14 1 14 14 P.I. = 2 1 1 5 9D D2 1 t 1 5 9D D 1 t = 1 1 t e t e 14 14 14 14 8 14 14 14 14 8 1 5 9 1 t 5 31 1 t t e = t e 14 14 14 8 14 196 8 5 31 1 t 2 t 7 t e x = c1e c2 e t 14 196 8 dx 4x 3y = t [From (1)] dt d 2 t 5 31 1 t 5 31 et 7 t t– e 4 c1e2t c2 e 7 t t = t c1e c2 e dt 14 196 8 14 196 8 3y = t 2c1e2t 7c2 e 7t y= Hence, 1 2c1e 2t 3c2 e 7t 3 5 1 t 10 31 1 t e 4c1e 2t 4c2 e 7t t e 14 8 7 49 2 3 27 5 t – t e 7 98 8 2 t 7 t x = c1e c2 e y= 5 31 1 t t e 14 196 8 2 2 t 1 9 5 t c1e c2 e 7 t – t e 3 7 98 24 Ans. Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ ... (5) Differential Equations 199 dx 2y , dt Solution. Here, we have dx dt dy dt dz dt dx From (1), we have dt Example 75. Solve d2x dt 2 d3x dt 3 dy dz 2z , 2x dt dt (Uttarakhand, II Semester, June 2007) = 2y Dx = 2y ...(1) = 2z Dy = 2z ...(2) = 2x Dz = 2x ...(3) = 2y = 2 dy = 2 (2z) = 4z dt dy Using (2), dt 2 z dz = 4 (2x) = 8x dt dz Using (3), dt 2 x = 4 3 d x 8x = 0 dt 3 m3 – 8 = 0 m– 2=0 (D3 – 8) x = 0 (m – 2) (m2 + 2m + 4) = 0 m=2 m2 + 2m + 4 = 0 m= A.E. is or So the C.F. of x is 2 4 16 2 i 12 = = 1 i 3 2 2 x = C1 e2t et ( A cos 3 t B sin 3 t ) ...(4) B tan A tan 1 B A [A = C2 cos , B = C2 sin ] x = C1 e2t et [C2 cos cos 3 t C2 sin sin 3 t ] x = C1 e2t et C2 cos ( 3 t ) = C1 e2t C2 et cos ( 3 t ) From (3), we have dz = 2x dt dz 2t t = 2C1 e 2C2 e cos ( 3 t ) dt 2t z = C1 e 2C2 e t 1 3 e t cos ( 3 t ) 2 cos 3 t 3 1 3 4 2t t z = C1 e C2 e cos 3 t 3 2t z = C1 e 2C2 [On putting the value of x] e ax cos bx dx cos (bx ) a2 b2 2 1 3 tan 1 3 2 4 ...(5) 3 3 e ax Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ 200 Differential Equations dy = 2z dx From (2), we have dy 4 2t t = 2C1 e 2 C2 e cos 3 t dt 3 4 y = 2C1 e2 t dt 2C2 e t cos 3 t dt 3 e x 4 y = C1 e 2t 2 C2 cos ( 3 t ) 3 1 3 e t 4 2 cos 3 t y = C1 e 2t 2 C2 3 3 13 2 2t t y = C1 e C2 e cos 3 t 3 Relations (4), (5) and (6) are the required solutions. Example 76. Solve the following simultaneous equations : d2x dt 2 Solution. We have, 3 x 4 y = 0, d2x dt 2 d2y dt 2 x y =0 [On putting the value of z] 3 2 tan 1 1 3 ...(6) Ans. (U.P. II Semester, Summer 2005) 3x 4 y = 0 d2y x y =0 dt 2 (D2 – 3)x – 4y = 0 x+ (D2 ...(1) + 1) y = 0 ...(2) Operating equation (2) by (D2 – 3), we get (D2 – 3)x + (D2 – 3) (D2 + 1) y = 0 ...(3) Subtracting (3) from (1), we get – 4y – (D2 – 3) (D2 + 1) y = 0 – 4y – (D4 – 2D2 – 3) y = 0 (D4 – 2D2 – 3 + 4) y = 0 (D2 – 1)2 y = 0 A.E. is (D4 – 2D2 + 1) y = 0 (m2 – 1)2 = 0 (m2 – 1) = 0 y = (c1 + c2t)et + (c3 + c4t)e–t m=±1 ...(4) From (2), we have x = – (D2 + 1)y = – D2y – y = – D2 [(c1 + c2t) et + (c3 + c4t)e–t] – [(c1 + c2t)et + (c3 + c4t)e–t] = – D [{(c1 + c2t )et + c2et} + {(c3 + c4t) (–e–t) + c4e–t}] – [(c1 + c2t)et + (c3 + c4t)e–t] = – [(c1 + c2t) et + c2et + c2et + (c3 + c4t) (e–t) – c4e–t – c4e–t] – [(c1 + c2t) et + (c3 + c4t)e–t] = – [(c1 + c2t + 2c2 + c1 + c2t) et + (c3 + c4t – 2c4 + c3 + c4 t) e–t] = – [(2c1 + 2c2 + 2c2t)et + (2c3 – 2c4 + 2c4t)e–t] Relations (4) and (5) are the required solutions. ...(5) Ans. Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ Differential Equations 201 EXERCISE 3.28 Solve the following simultaneous equations: 1. 2. 3. dx dy 2 x 3y 0 , 3x 2 y 0 dt dt d2y dt 2 x, d 2x Ans. x = c1 et – c2 e–5t , y = c1 et + c2 e–5t y dt 2 Ans. x = c1 et + c2 e– t + (c3 cos t + c4 sin t) y = c1 et + c2 e– t – (c3 cos t + c4 sin t) dx dy 5x 2 y t , 2x y 0 dt dt So that x = y = 0 when t = 0 1 1 (1 6t ) e 3t (1 3t ) 27 27 (AMIETE, June 2009, U.P., II Semester, June 2008) Ans. x = Ans. y = 4. 5. dy 2 dx t x yt, dt dt dx 2 y sin t 0 dt Ans. x = c1 cos t + c2 sin t + t2 –1 ; y = –c1 sin t + c2 cos t + t dy 2 x cos t 0 dt 4 7. dy x and dx 8. dx y t, dt 9. 10. Ans. x = c1 cos 2t + c2 sin 2t – cos t ; y = c1 sin 2t – c2 cos 2t – sin t dx dy 3 x sin t ; dt dt 6. 2 2 (2 3t ) e 3t (2 3t ) 27 27 dx y cos t dt dx y e2t dt Ans. x = c1 e–t + c2 e–3t, y = c1 e–t + 3 c2 e–3t + cos t t t Ans. x = C1 e C2 e dy 2x 3 y 1 dx dx dy y 0, t x0 dt dx given x(1) = 1 and y(–1) = 0 t t Ans. x = c1 e 2 2t 1 e , y C1 et C2 e t e 2t 3 3 1 3 5 3 c2 e 2t t , y c1 et c2 e 2t t 2 2 4 2 Ans. x = c1 t + c2 t –1, y = c2 t –1 – c1 t dx dy y sin t , x cos t , given that x = 2, y = 0 when t = 0 (U.P., II Semester, 2004) dt dt Ans. x = et + e–t, y = sin t – et + e–t 11. (D – 1) x + Dy = 2t + 1; (2D + 1) x + 2 Dy = t 2 t2 4 Ans. x t , y t C 3 2 3 12. dx 2 ( x y ) 1, dt t (U.P., II Semester, Summer (C.O.) 2005) dy 1 ( x 5 y) t dt t 13. (D2 – 1)x + 8Dy = 16et and Dx + 3(D2 +1)y = 0 Ans. y = c1 cos x = 14. 4 3 Ans. x = At Bt t 3 3 c1 sin c2 sin t 3 t 3 c3 cosh 3 t c4 sin h 3 t 2et – 3 c2 cos t 3 3 3 c3 sinh 3 t 3 3 c4 cosh 3 t 6et 3t. dx 2 ( x y ) 1, dt t dy 1 ( x 5 y ) t. dt t 2t 2 t t 2 3y 4 1 3 , y = At Bt 2 15 20 15 10 (Q. Bank U.P.T.U. 2001) (U.P. II Semester, 2005) Ans. x = At 4 Bt 3 t 2 3t 1 2t 2 t , y = At 4 Bt 3 15 10 2 15 20 Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ 202 Differential Equations 3.32 EQUATION OF THE TYPE dn y dx n = f ( x) This type of exact differential equations are solved by successive integration. d2 y Example 77. Solve = x 2 sin x. dx 2 d2y Solution. We have ...(1) x 2 sin x 2 dx Integrating the differential equation (1), we get dy = x2(– cos x) – (2x) (– sin x) + (2) (cos x) + c1 dx dy = x2 cos x + 2x sin x + 2 cos x + c1 dx Integrating again, we have y = [(– x2) (sin x) – (– 2x) (– cos x) + (– 2) (– sin x)] + [(2x) (– cos x) – 2(– sin x)] + 2 sin x + c1x + c2 = – x2 sin x – 4x cos x + 6 sin x + c1x + c2 Ans. d3 y Example 78. Solve x log x. dx 3 d3 y Solution. We have, 3 x log x ...(1) dx d2 y x2 1 (log x ) ( x ) – x dx c1 Integrating the differential equation (1), we get = 2 2 x dx d2 y x2 x log x – x c1 = ...(2) 2 dx 2 Again integrating (2), we have, x2 x3 1 x2 x2 dy (log x) – dx – c1 x c2 = 2 6 x 2 2 dx dy x3 x 2 x 2 x2 log x – – c1 x c2 = dx 6 2 4 2 Again integrating (3), we obtain ...(3) x4 x3 1 x3 x3 x 3 x2 (log x ) – dx – – c1 c2 x c3 24 6 x 6 12 6 2 x 4 x3 x3 x3 x 3 c1 x 2 log x – – – c2 x c3 y= 24 6 18 12 6 2 y= y= x 4 x3 11x3 x2 log x – c1 c2 x c3 24 6 36 2 Ans. EXERCISE 3.29 Solve the following differential equations: 1. d5 y 2. dx5 d2 y 3. d4y dx 2 dx 4 x6 c x 4 c2 x3 c3 x 2 1 c4 x c5 720 24 6 2 x Ans. y x ex Ans. y = (x – 2) ex + c1 x + c2 x e – x – cos x Ans. y x5 x3 e – x – cos x c1 120 6 Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ Differential Equations d y 4. x2 5. d3 y dx 6. 203 2 dx 3 d3 y dx 3 2 Ans. y – log x 1 (log x) 2 log x – c1x c2 2 log x Ans. y 1 [6 x3 log x – 11x3 c1x 2 c2 x c3 ] 36 sin 2 x Ans. y x3 sin 2 x c1 x 2 c2 x c3 12 16 2 3.33 EQUATION OF THE TYPE Multiplying by 2 dn y dx n = f ( y) dy d 2 y dy dy 2 f ( y) , we get 2 dx dx 2 dx dx ...(1) 2 dy Integrating (1), we have = 2 f ( y ) dy c ( y ) (say) dx dy dy dx = ( y ) dx ( y) Example 79. Solve Solution. We have d2y 2 dx d2 y y , under the condition y 1. dy ( y) =x+c dy 2 at x = 0 dx 3 y dx 2 dy dy d 2 y dy Multiplying (1) by 2 , we get 2 = 2 y 2 dx dx dx dx 2 4 3/ 2 dy Integrating (2), we get y c1 3 dx dy 2 On putting y = 1 and , we have c1 = 0 dx 3 ...(1) ...(2) ...(3) 2 2 dy 2 3/ 4 dy 4 – 3/ 4 dy dx Equation (3) becomes y 3/ 2 or y or y dx 3 3 3 dx 1 1/ 4 2 Again integrating y 2 x c2 ...(4) 4y4 x c2 1 3 3 4 On putting x = 0, y = 1, we get c2 = 4 2 x4 (4) becomes 4y1/4 = Ans. 3 2 d y dy sec2 y tan y under the condition y = 0 and 1 when x = 0. Example 80. Solve dx dx 2 2 d2y dy 2y tan y 2 dy d y = 2 sec 2 y tan y Solution. = sec 2 2 dx dx dx dx dy dy d 2 y 2 2 dx dx2 = 2 sec y tan y dx 2 dy dy 2 tan 2 y c1 = tan y c1 or dx dx dy 1, we get c1 = 1 On putting y = 0, and dx Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ 204 Differential Equations dy = tan 2 y 1 sec y dx cos y dy = dx On integrating we get sin y = x + c On putting y = 0, x = 0, we have c = 0 sin y = x y = sin–1 x Now, 2 Example 81. Solve d y dx 2 dy 1 when x = 0. dx (U.P., II Semester, Summer 2003) 2( y 3 y ), under the condition y 0, d2y Solution. dx Integrating, we get On putting y = 0 and Ans. 2 = 2(y3 + y) or 2 dy d 2 y dy 4( y 3 y ) dx dx 2 dx 2 y4 y2 dy 4 2 = 4 4 2 c1 y 2 y c1 dx dy = 1 in (1), we get 1 = c1 dx ...(1) 2 dy Equation (1) becomes = y4 + 2y2 + 1 = (y2 + 1)2 dx dy dy = y2 + 1 or dx dx 1 y2 Again integraiting, we get tan–1 y = x + c2 On putting y = 0 and x = 0 in (2), we have 0 = c2 Equation (2) is reduced to tan–1 y = x y = tan x ...(2) Ans. 2 dx Example 82. A motion is governed by d x 36 x –2 , given that at t = 0, x = 8 and 0, find 2 dt dt the displacement at any time t. d2x Solution. We have dt 2 = 36x–2 2 d 2 x dx –2 dx = 2 36 x 2 dt dt dt ...(1) 2 dx –1 Integrating (1), we hae – 72 x c1 dt 72 dx c1 or c1 9 Putting x = 8 and 0 in (2), we get 0 = – 8 dt 2 2 72 – 72 9 x dx dx – 9 or (2) becomes = x x dt dt x dx x dx 3 dt c = = 3t + c2 x –8 2 x2 – 8 x 1 2x – 8 8 dx = 3t + c2 2 x 2 – 8x 1 2 2x – 8 2 x – 8x dx + 4 1 ( x – 4) 2 – (4) 2 x 2 – 8 x 4 cos h –1 x–4 3t c2 4 ...(2) dx ( x – 8) = 3 dt x dx 3t c2 ...(3) Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ Differential Equations 205 On putting x = 8 and t = 0 in (3), we get c2 = 0 x–4 (3) becomes x 2 – 8 x 4 cos h –1 3t 4 EXERCISE 3.30 d2y 1. y3 3. sin3 y 4. 5. dx 2 a d2y dx 2 2. e 2 y Ans. c1 y2 = (c1 x + c2)2 Ans. d2y dx 2 1 Ans. c1 ey = cosh (c1 x + c2) 1 c1 Ans. sin [( x c2 ) (1 c1 )] cos y 0 c1 cos y a4 A particle is acted upon by a force x 3 per unit mass towards the origin where x is the distance x from the origin at time t. If it starts that it will arrive at the origin in time . 4 In the case of a stretched elastic string which has one end fixed and a particle of mass m attached to the d 2s mg (s – l ) e dt where l is the natural length of the string and e its elongwith due to a weight mg. Find s and v determining the constants, so that s = s0 at the time t = 0 and v = 0 when t = 0. g g 2 2 1/ 2 Ans. v – e [( s0 – l ) – ( s – l ) ] , s – l ( s0 – l ) cos e . t other end, the equation of motion is 3.34 2 – EQUATIONS WHICH DO NOT CONTAIN 'y' DIRECTLY The equation which do not contain y directly, can be written dn y dn –1y dy f n , n – 1 , ..... , dx dx dx x = 0 ...(1) d n –1 P dy d 2 y dP d 3 y d 2 P P i.e., 2 , 3 2 etc. in (1), we get f n – 1 , ....... P, x = 0 dx dx dx dx dx dx 3 2 d y dy dy Example 83. Solve 0 dx 2 dx dx On substituting d 2 y dP dy , equation (1) becomes P and dx dx 2 dx dP dP P P3 = 0 or P(1 P 2 ) 0 dx dx dP dP 2 – dx 1 – P dP = – dx = – P (1 P ) or 2 2 dx P (1 P ) P 1 P 1 P On integrating, we have log P – log (1 P 2 ) = – x + c1 or log – x c1 2 1 P2 Solution. On putting P 1 P 2 = e– x + c1 or P2(1 – a2 e–2x) = a2 e–2x P 1 P2 a 2 e –2 x P 2 (1 P 2 ) a 2 e – 2 x a e– x 1 – a2 e– 2x dy a e– x dx 1 – a 2 e– 2 x a e– x dx 1 – a 2 e– 2 x y = – sin–1 (a e– x) + b dy = On integration, we get P2 Ans. Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ 206 Differential Equations 1/ 2 dy 2 Example 84. Solve 1 – dx 2 dx d2y (U.P. Second Sem., 2002) 1/ 2 dy 2 1 – Solution. We have, = dx 2 dx 2 dP dy dP d y Putting P 2 in (1), we get = dx dx dx dx d2y ... (1) 1 – P2 dP 1 – P2 dx On integrating, we have sin–1 P = x + c P = sin (x + c) dy = sin (x + c) dx On integrating, we have y = – cos (x + c) + c1 2 Example 85. Solve x d y dx 2 Ans. 2 dy dy x – 0 dx dx (U.P. II Semester, 2010) d 2 y dP dy in the given equation, we get P and dx dx 2 dx 1 dP 1 1 dP – –1 x x P2 – P = 0 2 dx P dx P x 1 1 dP dz Again putting z so that – 2 P P dx dx – dz z dz z Equation (1) becomes – –1 1 dx x dx x Solution. On putting ...(1) 1 I.F. = e x dx elog x x x2 1 x2 c1 or x c1 2 P 2 2x dy 2x 2x x x 2 2c1 2 dy 2 dx P = 2 dx x 2c1 P 2 x 2c1 x 2c1 On integrating, we have y = log (x2 + 2c1) + c2 Hence, solution is zx= x dx c1 or z x Ans. EXERCISE 3.31 Solving the following differential equations: 1. 2. 3. (1 x 2 ) d2y dx 2 x dy ax 0 dx 2 d2y dy 1 0 2 dx dx 4 3 d y d y – cot x 3 0 4 dx dx (1 x 2 ) Ans. y c2 – ax c1 log [ x (1 x 2 ) ] Ans. y – x 1 k2 log (1 kx ) a k k2 Ans. y = c1 cos x + c2x2 + c3x + c4 2 4. 5. d3y d2 y d2 y 2 –a dx3 dx 2 dx 2 d 2 y dy 2 e x / 2 x 2 – x3 dx dx 2x 2 2 5/ 2 c2 x c3 Ans. 15 c1 y 4(c1 x a ) –x Ans. y e 2 /2 c1 x2 c2 2 Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ Differential Equations 3.35 207 EQUATIONS THAT DO NOT CONTAIN 'x' DIRECTLY The equations that do not contain x directly are of the form dn y d n –1y dy f n , n – 1 , ...... , y 0 dx dx dx dy d 2 y dP dP dy dP On substituting P, 2 P in the equation (1), we get dx dx dx dx dy dx ...(1) dP n – 1 n – 1 ,..... P, y = 0 dy Equation (2) is solved for P. Let ...(2) P = f1 ( y ) Example 86. Solve y dy dy f1 ( y ) or dx dx f1 ( y ) dy f1 ( y) =x+c 2 d2y dy dy 2 dx dx dx ...(1) dy d 2 y dP dP dy dP P, 2 P in equation (1) dx dx dy dx dy dx dP yP P 2 = P y dP 1 – P dy dy dp dy – log (1 – P ) log y log c1 = 1– P y 1 dy c1 y – 1 1 or = c1 y P 1 – c1 y dx c1 y 1– P c1 y 1 dy = dx 1 dy dx c1 y – 1 c1 y – 1 1 y log (c1 y – 1) = x + c1 c1 Solution. Put Example 87. Solve y Solution. Put Ans. 2 d2y dy y2 dx 2 dx ...(1) dy d 2 y dP dP dy dP P, 2 P in (1) dx dx dy dx dy dx dP dP P 2 P 2 = y2 or P y dy dy y dP dz 1 dz z dz 2 z Put P2 = z or 2P in (2), y or 2y dy dy 2 dy y dy y yP ...(2) 2 y dy Hence, the solution is 2 e2 log y elog y y 2 I.F. = e z y2 = 2 y. ( y 2 ) dy c P2 y2 = y4 c 2 2 P2 y2 = y4 + k or 2yP y4 k [Put 2c = k] Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ 208 Differential Equations 2y 1 dt 2 2 t k sin h –1 dy = dx y2 k y 4 k or 2 y dy y4 k dx = dx [Put y2 = t, 2y dy = dt] 1 2 sin h –1 t k =x+c 2 x c or y 2 k sin h ( 2 x c) = Ans. EXERCISE 3.32 Solve the following differential equations: 1. 2 d2y dy 1 Ans. y2 = x2 + ax + b 2 dx dx 3. 2 y d2y 2 dy dy – 2 0 Ans. cy + 2 = d ecxa 2 dx dx dx 2 d2y 3 d 2 y dy dy dy 2 2 (x + b) 4. y – 3 – 4 y 0 0 Ans. y = a – sin–1 (b e–x) Ans. y = a sec dx dx 2 dx 2 dx dx 5. y 6. y 3.36 2. y d2y 2 d2y 2 dy 2 dx dx dy dy 1 – dx cos y y dx sin y Ans. x = c1 + c2 log y + sin y dy – y 2 log y dx 2 dx Ans. log y = b. ex + a e– x EQUATION WHOSE ONE SOLUTION IS KNOWN If y = u is given solution belonging to the complementary function of the differential equation. Let the other solution be y = v. Then y = u. v is complete solution of the differential equation. d2y dy Qy R ....(1), be the differential equation and u is the solution included in dx dx the complementary function of (1) Let 2 P d 2u dx 2 P du Qu = 0 dx y = u. v dy du dv u = v dx dx dx d2y dx 2 = v d 2u 2 ...(2) dv du d 2v u 2 dx dx dx dx 2 dy d 2 y Substituting the values of y, , 2 in (1), we get dx dx v d 2u dx 2 2 dv du d 2v dv du u 2 Pv u Q u. v R dx dx dx dx dx On arranging d 2u d 2v du dv du dv Qu + u P 2 R v 2 P 2 dx dx dx dx dx dx The first bracket is zero by virtue of relation (2), and the remaing is divided by u. d 2v 2 dx d 2v dx 2 P R dv 2 du dv = u dx u dx dx 2 du dv R P dx = u u dx ...(3) Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ Differential Equations 209 dv d 2 v dz = z, so that dx dx 2 dx dz 2 du R Equation (3) becomes dx P u dx z u This is the linear differential equation of first order and can be solved (z can be found), which will contain one constant. dv On integration z , we can get v. dx Having found v, the solution is y = uv. Note: Rule to find out the integral belonging to the complementary function Let Rule Condition u 1 1+P+Q =0 ex 2 1–P+Q =0 e–x 4 P Q =0 a a2 P + Qx = 0 5 2 + 2Px + Qx2 = 0 x2 6 n (n – 1) + Pnx + Qx2 = 0 xn 1 3 eax x Example 88. Solve y – 4xy + (4x2 – 2)y = 0 given that y= ex2 is an integral included in the complementary function. (U.P., II Semester, 2004) Solution. y – 4xy + (4x2 – 2)y = 0 ...(1) On putting y = v.ex2 in (1), the reduced equation as in the article 3.36 d 2v 2 du dv P =0 u dx dx dx [P = – 4x, Q = 4x2 – 2, R = 0] 2 d 2v 2 dv 2 – 4 x 2 (2 x e x ) =0 x dx e dx 2 d 2v dx 2 [– 4 x 4 x] dv =0 dx y = uv d 2v dx 2 0 dv c, v c1 x c2 dx 2 [u e x ] 2 x y = e (c1 x c2 ) Ans. 2 dy ( x 1) y 0 dx dx given that y = ex is an integral included in the complementary function. Example 89. Solve x Solution. x d2y 2 d y 2 – (2 x – 1) – (2 x – 1) dy ( x – 1) y 0 dx dx 2 x – 1 dy x – 1 – y =0 [1 + P + Q = 0] 2 x dx x dx By putting y = vex in (1),we get the reduced equation as in the article 3.36. d2y d 2v 2 du dv P =0 u dx dx dx 2 ...(1) ...(2) Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ 210 Differential Equations dv z in (2), we get dz – 2 x – 1 2 e x z = 0 dx dx x e x dz – 2 x 1 2 x dz z z =0 0 dx x dx x dx dz log z – log x log c1 = – x z c1 dv c1 dx or or dv c1 v c1 log x c2 z= x dx x x x y = u. v = e (c1 log x + c2) Putting u = ex and Example 90. Solve x 2 2 Solution. x d2y dx 2 2 d y – 2 x [1 x ] dx 2 – 2 x (1 x) Ans. dy 2(1 x ) y x 3 dx dy 2(1 x ) y x3 dx 2 2 x (1 x ) dy 2(1 x ) y =x dx dx x2 x2 2 x (1 x) 2(1 x) x0 Here P + Qx = – x2 x2 Hence y = x is a solution of the C.F. and the other solution is v. Putting y = vx in (1), we get the reduced equation as in article 3.36 d y 2 – ...(1) d 2v x 2 du du P = u u dx dx dx 2 d 2v x – 2 x (1 x) 2 dv (1) = 2 x dx x dx x dz d 2v dv – 2z 1 –2 =1 2 dx dx dx 2 dv dx z which is a linear differential equation of first order and I.F. = e Its solution is z e–2x = e –2 x – 2 dx e – 2 x dx c1 e –2 x –1 c1 or z c1 e 2 x –2 2 dv – x c1 2 x 1 1 2x 2x e c2 = – c1 e or dv – c1 e dx v= dx 2 2 2 2 – x c1 2 x e c2 y = uv x Ans. 2 2 Example 91. Verify that y = e2 x is a solution of (2x + 1) y´´ – 4 (x + 1) y´ + 4y = 0. Hence find the general solution. z e–2x = Solution. We have (2 x 1) d2 y 2 4 ( x 1) dy 4y 0 dx dx y = e2x, y´ = 2e2x, y´´ = 4e2x Substiting the values of y, y´ and y´´ in (1), we get (2x + 1) 4e2x – 4 (x + 1) 2e2x + 4e2x = 0 or [8x + 4 – 8x – 8 + 4] e2x = 0 ... (1) 0=0 Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ Differential Equations 211 y1 = e2x is a solution 4 ( x 1) 4 y´ y =0 Equation (1) in the standard form is y´´ (2 x 1) (2 x 1) 4 ( x 1) 1 . So P (x) Then (x) 2 e P dx (2 x 1) y Thus 1 2 4 ( x 1) 4x 2 dx P dx dx (2 x 1) 2x 1 2x 1 = 2x + 1n (2x + 1) Now Then (x) Integrating by parts 1 (e 2 x ) 2 2x 1 e 2x e2x + 1n (2x + 1) Now e2 x = v (x) = ( e2 x ) 2 . (2 x 1) (x) dx = 2x 1 e2 x dx e 2 x e 2 x 2. 2 4 The required second solution 1 1 1 2x 2 x 1 . 2x y2 (x) = y1 (x) v (x) = e = – x – 1 = – (x + 1) 2 2 e 2 x e Then the general solution is y (x) = c1 y1 (x) + c2y2 (x) = c1e2x – c2 (x + 1) Ans. 2 2 3 x Example 92. Solve x y – (x + 2x)y + (x + 2)y = x e given that y = x is a solution. v (x) = (2x + 1) Solution. x2 y – (x2 + 2x)y + (x + 2)y = x3 ex y – x2 2x 2 y x2 2 y x ex x x On putting y = vx in (1), the reduced equation as in the article 3.36. 2 du dv R d 2 v x 2 2 x 2 dv xe x P – (1) = = u dx dx u x dx x dx 2 dx 2 x2 2 dz d v dv x – z ex – =e 2 dx dx dx which is a linear differential equation – dx –x I.F. = e e z e–x = e x . e – x dx c ...(1) d 2v dv z dx dv = e x. x + c e x dx v = x.ex – ex + c ex + c1 v = (x – 1) ex + c ex + c1 y = vx = (x2 – x + cx) ex + c1x z e–x = x + c or z = ex. x + c ex Ans. 2 dy 2 y ( x 1)e x dx d y 2 x 5 dy 2y ( x 1) e x – Solution. x 2 dx x 2 x2 dx 2 P Q 1 2 = 0, Choosing a = 2 Here a a Example 93. Solve ( x 2) d y dx 2 – (2 x 5) ...(1) Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ 212 Differential Equations 2x 5 2 =0 2x 4 4 x 8 Hence y = e2x is a part of C.F. Putting y = e2xv in (1), the reduced equation as in the article 3.36. 1– d 2v d 2v 2 du P 2 u dx dx 2x 5 2 2 x 2e 2 x x2 e – 2 dx d 2v ( x 1) e x dv = u ( x 2) dx ( x 1) e x dv = 2x dx e ( x 2) x 1 –x 2x 5 dv – 4 = x 2e dx x2 dx x 1 –x dz 2 x 3 e z = x 2 dx x 2 which is a linear differential equation, 2 I.F. = e 2x 3 dx x2 e dv z dx 1 2 – x 2 dx e 2 x – log( x 2) e2 x x 1 – x x 2 x 2 e dx c e x ( x 1) 1 x 1 dx c e – dx c = = 2 ( x 2)2 x 2 ( x 2) Its solution is z. e2 x x2 e2 x = x2 e x dx e x dx x 2 – ( x 2)2 c ex e x dx e x dx ex – c c 2 2 x2 x2 ( x 2) ( x 2) dv z = e– x + c (x + 2)e–2x = e–x + c(x + 2)e–2x dx ( x 2) e – 2 x e – 2 x –x –x –2 x – e c – c1 v = e dx c ( x 2)e dx c1 = –2 4 ce –2 x [2 x 5] c1 = – e– x 4 y = u.v c e –2 x 2x –x e – e (2 x 5) c1 y = – e x c (2 x 5) c1 e 2 x y= 4 4 EXERCISE 3.33 = Ans. Solve the following differential equations: d2y dy 1. (3 – x) 2 – (9 – 4 x ) (6 – 3x ) y 0, given y = ex is a solution. dx dx c1 3 x 3 2 x Ans. y e (4 x – 42 x 150 x – 183) c2e 8 d 2 y dy 2. x 2 – (1 – x ) y x 2 e – x given y = ex is an integral included in C.F.. dx dx 1 x –x 2 –x Ans. y c2 e c1 (2 x 1) e – (2 x 2 x 1)e 4 d2y dy 3. (1 – x 2 ) 2 x – y x (1 – x 2 )3 / 2 , given y = x is part of C.F.. dx dx x 2 3/ 2 – c1 [ (1 – x 2 ) x sin –1 x] c2 x. Ans. y – (1 – x ) 9 Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ Differential Equations 4. 5. 6. 7. 8. sin 2 x d2y 213 2 d y dx 2 2 y , given that y = cot x is a solution Ans. cy = 1 + (c1 – x) cot x x3 dy 1 –x xy x, given y = x is a part of C.F.. Ans. y 1 c1x 2 e 3 dx c2 x 2 dx dx x d2y dy ( x sin x cos x ) 2 – x cos x y cos x 0 given y = x is solution. dx dx Ans. y = c2x – c1 cos x 2 1 c1 1 dy 2d y x x – y 0, given that y x is one integral. Ans. y c2 x 2 x x x dx dx 2 d y dy x 2 2 2 x – 12 y x3 log x (U.P., II Semester 2004) dx dx 2 dv z] dx 3 3 c2 x log x (7 log x – 2) Ans. y c1x 4 x 98 NORMAL FORM (REMOVAL OF FIRST DERIVATIVE) [Hint. (n (n – 1) + pnx + Qx2 = 0), n = 3, satisfies this equation. Put y v x 3 , 3.37 d2y dy Qy R dx dx Put y = uv where v is not an integral solution of C.F. du du dy u = v dx dx dx Consider the differential equation d2 y = u 2 P d2 v 2 du dv d 2u v 2 dx dx dx d x2 d x2 dy d 2 y On putting the values of y, , 2 in (1) we get dx dx d 2v dv du d 2u du du v 2 Pu v Q.uv = R u 2 2 dx dx dx dx dx dx 2 2 d v d u du dv dv Q.v = R v 2 Pv 2 u 2 P dx dx dx dx dx R d 2 u du 2 dv u d 2 v dv P Q.v = P u dx v dx 2 dx dx 2 dx v Here in the last bracket on L.H.S. is not zero y = v is not a part of C.F. Here we shall remove the first derivative. P ...(1) ...(2) 2 dv dv 1 –1 = 0 or – P dx or log v P dx v dx v 2 2 1 – P dx v= e 2 2 In (2) we have to find out the value of the last bracket i.e., d v P dv Qv dx dx 2 1 dv v e – 1/ 2 pdx P – 2 P dx 1 = – e – Pv dx 2 2 2 1 dP P dv 1 dP P 1 1 dP 1 d v – v– – v – – Pv – v P2 v 2 = 2 dx 2 dx 2 dx 2 2 2 dx 4 dx Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ 214 Differential Equations d 2v 1 dP 1 2 1 dP 1 dv 1 – P v P 2 v P – Pv Qv = v Q – Qv = – 2 dx 4 2 dx 4 dx dx 2 Equation (1) is transformed as d 2 u u 1 dP P 2 R v Q – – 2 dx 4 = v dx 2 v 1 d 2u 1 dP P 2 P dx u Q – – = Re2 2 2 dx 4 dx 1 2 P dx 1 dP P 2 d u R 2 Q Q – – Q u R R e or = R where , 1 1 1 2 1 2 dx 4 v dx 2 P y = uv and ve – 1 P dx 2 Ans. d 2 dy cos x cos 2 x. y 0 dx dx d 2 dy 2 Solution. We have, cos x cos x. y 0 dx dx Example 94. Solve Here, d2y dx 2 dy d2 y dy (cos 2 x) y = 0 – 2 tan x. y = 0 2 dx dx dx P = – 2 tan x, Q = 1, R = 0 cos 2 x – 2 cos x sin x 1 dP P 2 1 4 tan 2 x – = 1 – (– 2 sec 2 x) – 2 dx 4 2 4 2 = 1 + sec x – tan2x = 1 + 1 = 2 Q1 = Q – 1 R1 = R e 2 P dx 0 v= e – 1 P dx 2 e – 1 (– 2 tan x ) dx 2 e tan x dx elog sec x sec x Normal equation is d 2u dx 2 d 2u dx 2 Q1u = R1 2u = 0 or (D2 + 2) u = 0 D i 2 u = c1 cos 2 x c 2 sin 2 x y = u.v = [c1 cos 2 x c2 sin 2 x] sec x dy – 2( x 2 x) ( x 2 2 x 2) y 0 dx dx 2 d 2 y 2( x 2 x ) dy x 2 2 x 2 Solution. We have, – y 0 dx dx 2 x2 x2 2 Example 95. Solve x Ans. 2 d y ...(1) 1 x2 2 x 2 , R0 Here p = – 2 1 , Q x x2 In order to remove the first derivative, we put y = u.v in (1) to get the normal equation d 2v Q1v = R1 ...(2) dx 2 Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ Differential Equations 215 1 where v e – 2 pdx Q1 = Q – 1 1 – – 21 dx 2 x e = 1 1 x dx e e x . elog x x e x 2 2 2 1 1 2 1 dp p 2 x 2 2 x 2 1 2 4 1 = 1 2 – 2 –1– 2 – – – – 1 x x x x x 2 dx 4 2 x2 4 x x2 1 R1 = R e 2 p dx 0 On putting the values of Q1 and R1 in (2), we get d 2u dx 2 0(u ) = 0 du = c1 u = c1x + c2 dx y = u.v = (c1x + c2) x ex 2 Example 96. Solve d y dx 2 d2y – 4x d 2u dx 2 0 Ans. 2 dy (4 x 2 – 1) y – 3e x sin 2 x (U.P. II Semester, (C.O.) 2004) dx 2 dy (4 x 2 – 1) y – 3 e x sin 2 x dx dx 2 2 Here p = – 4x, Q = 4x – 1, R = – 3ex sin 2x Solution. We have, 2 4x 1 ...(1) 1 – pdx – – 4 x dx 2 2 x dx e 2 e ex In order to remove the first derivative, v e 2 2 d u Q1u = R1 On putting y = uv, the normal equation is dx 2 1 dp p 2 1 16 x 2 – (4 x 2 – 1) – (– 4) – where Q1 = Q – = 4x2 – 1 + 2 – 4x2 = 1 2 dx 4 2 4 ...(2) 2 R – 3e x sin 2 x – 3 sin 2 x R1 = 2 v ex d 2u u – 3 sin 2 x dx 2 (D2 + 1)u = – 3 sin 2x 2 A.E. is D + 1 = 0 D = i C.F. = c1 cos x + c2 sin x 1 – 3 sin 2 x (– 3 sin 2 x) sin 2 x P.I. = D2 1 – 4 1 u = c1 cos x + c2 sin x + sin 2x y = u. v = (c1 cos x + c2 sin x + sin 2x)ex2 Equation (2) becomes Ans. EXERCISE 3.34 Solve the following differential equations: d2 y 1. – 2 tan x. y – 5 y 0 dx 2 2 d2 y dy 2. – 4 x (4 x 2 – 3) y e x 2 dx dx 3. 4. 5. 2 d y 1 2 ( x 2 x) e2 dy ( x 2 2) y dx dx d2y dy 2 2 – 2x n 2 y 0 dx dx 2 x d 2 y 2 dy – n2 y 0 dx 2 x dx 2 – 2x Ans. y = (a e2x + e–3x)sec x Ans. y = (c1ex + c2 e–x – 1) x2 Ans. y = (c1 cos 3 x c2 sin 3 x2 1 x 2 e .e x) e 2 4 Ans. y = (c1 cos nx + c2 sin nx)x nx – nx ) Ans. y (c1 e c2 e 1 x Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ 216 Differential Equations 2 6. 7. d2y 1 dy 1 1 6 1 2/3 – 4/3 – 2 y 0 2 dx 4 x dx x x x3 2 d y 1 dy y – 2 (– 8 x x ) 0 2 dx x dx 4x Ans. y (c1 x3 c2 3 – x3 x –2 ) e 4 2 –1 Ans. y (c1x c2 x ) e x 3.38 METHOD OF SOLVING LINEAR DIFFERENTIAL EQUATIONS BY CHANGING THE INDEPENDENT VARIABLE d2y dy P Qy = R Consider, ...(1) 2 dx dx Let us change the independent variable x to z and z = f (x) 2 d 2 y dz dy d 2 z 2 2 = dz dx 2 dz dx dx dy d2y Putting the values of and in (1), we get dx dx 2 d 2 y dz 2 dy d 2 z dy d 2 z 2 P Qy = R 2 2 dz dx dz dx dz dx dy dy dz = dx dz dx d2y 2 dz d 2 z dy d 2 y dz 2 Qy = R P dz 2 dx dx dx dz dz d 2 z R P 2 2 2 dx dx d y dy Qy = dz 2 d y P dy Q y = R ...(2) 1 1 1 2 dz d z2 dz dz 2 dz 2 dz dx dx dx dz d 2 z P 2 dx dx Q R , where P1 = Q1 = R1 = 2 , 2 2 dz dz dz dx dx dx Equation (2) is solved either by taking P1 = 0 or Q1 = a constant. Equation (2) can be solved by by two methods, by taking First Method, P1 = 0 Second Method, Q = constant Working Rule 2 Step 1. Coefficient of d y should be made as 1 if it is not so. dx 2 Step 2. To get P, Q and R, compare the given differential equation with the standard form y + P y + Qy = R. Step 3. Find P1, Q1 and R1 by the following formulae. d2y dz p R dx dx 2 , R = P1 = 2 1 2 dz dz dx dx Step 4. Find out the value of z by taking First Method, P1 = 0 Second Method. Q1 = constant Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ Differential Equations 217 d2y dy P1 Q1 y = R1 dz dz 2 On solving this equation we can find out the value of y in terms of z. Then write down the solution in terms of x by replacing the value of z. Step 5. We get a reduced equation Example 97. Solve d2y 2 cot x dy 4 y cosec 2 x = 0 dx dx 2 d y dy Solution. We have, cot x 4 y cosec 2 x = 0 2 dx dx Here, P = cot x, Q = 4 cosec2 x and R = 0 Changing the independent variable from x to z, the equation becomes d2y dx 2 P1 dy Q1 y = 0 dz P where P1 = Case I. Let us take P1 = 0 dz d 2 z p dx dx 2 dz dx Put (3) becomes 2 ...(1) ...(2) dz d 2 z dx dx 2 Q = Q , 1 2 2 dz dz dx dx = 0 or P dz d 2 z d2z dz 2 =0 cot x =0 2 dx dx dx dx ...(3) d2z dv dz = v, 2 = dx dx dx dv dv (cot x ) v = 0 = – cot x. dx dx v log v = – log sin x + log c = log c log c cosec x v = c cosec x dz x = c cosec x dz = (c cosec x) dx z = c log tan dx 2 Case II. Now let us take Q1 = Constant. Q1 = Q dz dx 2 = 4 cosec2 x 2 2 c cosec x = 4 c2 which is constant Hence the equation (2) reduces to 4 d2y dy 4 d2y 4 0 2 y = 0 or 2 y =0 P1 0, Q1 c 2 2 2 dz c dz dz c 2 4 4 2 m2 2 = 0 m = i A.E. is D 2 y = 0, c c c x 2z 2z C.F. = c1 cos c2 sin z c log tan 2 c c x x y = c1 cos 2 log tan c2 sin 2 log tan Ans. 2 2 Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ 218 Differential Equations 2 1 Example 98. Solve x 6 d y 3x 5 dy a 2 y = 2 2 x dx dx 1 d 2 y 3 dy y Solution. We have, a2 6 = 8 x dx 2 x dx x 3 a2 Here P = and Q 6 x x On changing the independent variable x to z, the equation (1) is reduced to d2 y P1 2 dz Using Second Method dy Q1 y = R1 dz Let Q1 = a2 (constant) Q1 = ...(2) Q dz dx 2 = a2 dz x6 dx 2 = constant = a2 (say) 2 dz x 6 = 1 x3 dz = 1 dz = 1 dx dx x3 dx 2 3 d z On differentiating twice, we have 2 = x4 dx 3 1 3 dz d 2 z P x x3 x 4 dx dx 2 P1 = = = 0 R1 = 2 2 1 dz 3 x dx On putting the values of P1, Q1 and R1 in (2), we get d2y 2 a 2 y = – 2z dz A.E. is m2 + a2 = 0, m = ± i a, P.I. = ...(1) dz = dx x3 z= x 2 c 2 1 1 x8 = = 2 = – 2z 2 1 x dz 6 x dx R (D2 + a2) y = – 2 z C.F. = c1 cos az + c2 sin az 2 1 1 D2 ( 2 z ) = 1 1 D ( 2 z ) = 1 ( 2 z ) = 2z = 1 2 2 2 2 D a a a a a a2 a 2 x2 y = C.F. + P.I. a a 1 y = c1 cos 2 c2 sin 2 2 2 Ans. 2x 2x a x 1 2 Example 99. Solve 2 d2 y 2 dx d2 y 1 dy 4 x 2 y = x4 x dx (U.P., I Semester Summer 2003, 2002) 1 dy 4 x 2 y = x4 x dx dx 1 Hence P = , Q = 4x2, R = x4 x On changing the independent variable x to z, the equation (1) is transformed as Solution. We have, d2y 2 dy P1 Q1 y = R1 dz dz 2 Using Second Method Let Q1 = 1 (constant) ...(1) ...(2) Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ Differential Equations 219 but Q1 = 2 Q 2 = 4x 2 2 dz dx dz = 4x2 = 2x dx dz dx = constant = 1 (say) dz dx dz = 2x dx z = x2 + c x2 = z – c ...(3) 2 dz d z 1 P 2 (2 x ) 2 dx dx x P1 = = =0 2 (2 x)2 dz dx R1 = R = x4 = zc x2 = 4 4 [Using (3)] 2 4x2 dz dx On putting the value of P1, Q1 and R1 in (2), we get d2 y zc zc dy zc y = (1) y = (D2 + 1) y = 2 2 dz 4 4 dz 4 dz A.E. is m2 + 1 = 0 m = ± i C.F. = c1 cos z + c2 sin z 1 z c zc zc 2 –1 z c = P.I. = 2 = (1 D 2 ) = (1 + D ) 4 D 1 4 4 4 Now complete solution = C.F. + P.I. d2y (0) z c x2 y = c1 cos x 2 c2 sin x 2 4 4 EXERCISE 3.35 y = c1 cos z c2 sin z Ans. Solve the following differential equations: d2y 1. x4 2. cos x 3. 4. dx 2 d2 y dx 2 x d y 6. dx 2 d2y 8. dx 2 tan x 2 – dy a2 y 0 dx dy sin x – 2 y cos3 x 2 cos5 x dx dy y cos 2 x 0 dx dy – 4 x 3 y 8x 2 sin x 2 dx 2 5. 7. d2y d2y dx 2 x3 dx 2 d2y dx 2 d2 y dx 2 (3sin x – cot x ) (tan x – 1)2 – cot x dy 2 y sin 2 x e – cos x sin 2 x dx dy – n ( n – 1) y sec 4 x 0 dx dy – y sin 2 x cos x – cos3 x dx (3 sin x – cot x ) dy 2 y sin 2 x e – cos x sin 2 x dx Ans. y c1 cos Ans. y c1 e 2 sin x c2e – a a c2 sin x x 2 sin x sin 2 x Ans. y = c1 cos (sin x) + c2 sin (sin x) x cos x Ans. y c1 e c2e 1 – cos x e 6 2cos x c2 ecos x Ans. y c1 e 1 – cos x e 6 2 Ans. y = C1 e–n tan x + C2 e(n – 1) tan x Ans. y = C1 e– cos x + C2 ecos x – cos x cos x C2 e 2 cos x Ans. y C1e 1 – cos x e 6 Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ 220 Differential Equations 3.39 Applications of Differential Equations of Second Order Example 100. The differential equation satisfied by a beam uniformly loaded (W kg/metre), with one end fixed and the second end subjected to tensile force P, is given by d2y 1 = Py – Wx 2 2 dx 2 Show that the elastic curve for the beam with conditions dy at x 0, is given by y= 0 dx W Wx 2 P (1 – cosh nx ) where n 2 y= 2P EI Pn2 E.I . Solution. We have, E.I . d2y 2 = Py – 1 W . x2 2 dx 2 d y P W – y = – x2 2 E.I . dx 2 E.I . P P 0 m2 n2 E.I . E.I . C.F. = c1 enx + c2 e– nx A.E. is m2 – P.I. = P W 2 x2 D – y– E.I . 2 E.I . 2 P mn n EI W 2 W 1 . x2 – x – 2 2 P 2 E . I . 2 E . I . 2 D – n D – E .I . 1 –1 D2 W D2 2 W 2 2 2 1 – . x 1 . x = = x 2 2 2n 2 . E.I . n 2 2n2 E.I . n 2 2 n E.I . n W 2 2 y = c1 enx c2 e – nx 2 x 2 2n E.I . n W ...(1) Differentiating (2) w.r.t. x, we get dy W (2 x ) = n c1 enx – n c2 e – nx 2 dx 2n E.I . dy 0 in (3), we get Putting x = 0, dx 0 = n c1 – n c2 c1 = c2 Putting x = 0, y = 0 in (2), we get W 2 W 0 c1 c2 4 0 = c1 c2 2 2n E .I . n 2 n E.I . W W Putting c1 = c2 in (4), we get 0 2 c1 4 c1 – , n E.I . 2 n 4 E.I . P n 2 E.I . P Now, n2 = E.I . W c1 = c2 – 2 n2 P Putting the values of c1 and c2 in (2), we get –W W 2 2 (e nx e – nx ) y= x 2 2 2 P 2n P n ...(2) ...(3) ...(4) Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ Differential Equations y = –W n2 P 221 cosh nx W 2 W x 2P P n2 y= W P. n 2 (1 – cosh nx ) W x2 2P Ans. EXERCISE 3.36 1. A beam of length l and of uniform cross-section has the differential equation of its elastic curve as w l2 – x2 2 4 dx where E is the modulus of elasticity, I is the moment of inertia of the cross-section, w is weight per unit length and x is measured from the centre of span. E.I . d2y 2 dy 0 . Prove that the equation of the elastic curve is dx 1 2 l 3. x2 x 4 5w. l 4 y – 2 E.I . 8 12 384 E.I . A horizontal tie rod of length l is freely pinned at each end. It carries a uniform load w kg per unit length and has a horizontal pull P. Find the central deflection and the maximum bending moment, taking the If at x 0, 2. Ans. w sec h al – 1 where a 2 P a 2 EI A light horizontal strut AB is freely pinned at A and B. It is under the action of equal and opposite l compressive forces P at its ends and it carries a load W at its centre. Then for 0 x , 2 origin at one of its ends. 3. EI d2y dx 2 Py 1 Wx 0 2 dy W sin ax l P 0 at x . Prove that y – x , where a 2 dx 2 P a cos al 2 EI 2 A horizontal tie-rod of length 2l with concentrated load W at its centre and ends freely hinged satisfies dy d2 y W 0. Prove that Py – x. With conditions x = 0, y = 0 and x l , the differential equation EI dx 2 dx 2 the deflection and bending moment M at the centre (x = l) are given by W ( nl – tan nl ) and 2 Pn W M – tan h nl , where n2 EI = P. 2n Also y = 0 at x = 0 and 4. Example 101. The voltage V and the current i at a distance x from the sending end of the transmission line satisfy the equations. dV di Ri , GV dx dx where R and G are constants. If V = V0 at the sending end (x = 0) and V = 0 at receiving end (x = 1). Show that sinh n(l x ) V V0 sinh nl When n2 = RG Solution. dV Ri dx . . . (1) di GV dx . . . (2) Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ 222 Differential Equations When x = 0, V = V0 When x = l V =0 Putting the value of i from (1) into (2), we get d dV 1 GV dx dx R d 2V RGV dx 2 d 2V ( RG )V 0 (D2 – RG)V = 0 (RG = n2) dx 2 A. E. is m2 – n2 = 0, m=n V = Ae nx + Be– nx Now, we have to find out the value of A and B with the help of given conditions. On putting x = 0 and V = V0 in (3), we get V0 = A + B On putting x = 1 and V = 0 in (3), we get 0 = Ae nl + Be– nl . . . (3) ... (4) V0 V0 e 2 nl B , 1 e 2nl 1 e 2nl Substituting the values of A and B in (3), we have On solving (4) and (5), we have A V V0 e nx V0 e 2 nl e nx V0 [e nx e2 nl nx ] V0 [e nl nx ) e ( nl nx ) ] sinh n(l x) V0 nl nl 2 nl 2 nl 2 nl e e 1 e 1 e 1 e sinh nl Proved. EXERCISE 3.37 1. An e.m.f. E sin pt is applied at t = 0 to a circuit containing a condenser C and inductance L in series. The current x satisfies the equation dx 1 L xdt E sin pt dt C 1 2 If p and initially the current x and the charge q are zero, show that the current in the LC dq circuit at time t is given by E t sin pt, where x . dt 2l Fill in the blanks: dy 2. The integrating factor of cos 2 x y tan x is ............... dx dy 3. The integrating factor of x( x 1) ( x 2) y x 2 (2 x 1) is ............. dx dy 4. Solution of ( x y 1) 1 is ............ dx 5. Mdx + Ndy = 0 is an exact differential equation if ....... 6. P. I. of (D2 + 4)y = sin 3x is ........... 7. P. I. of (D2 – 2D + 1)y = ex is .......... 8. 9. d 2 y xdy y x is ............. dx 2 dx If the C.F. of ay+ by + cy + X is Ay1 + By2, then P. I. = uy1 + vy2 where u = ....... and v= ....... 2 On putting x = eZ, the transformed differential equation of x Ans. 2. etan x 3. M N x 1 4. x + y + 2 = cey 5. dy x 2 x Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ 4 4 Determinants and Matrices DETERMINANTS AND MATRICES 4.1 INTRODUCTION In Engineering Mathematics, solution of simultaneous equations is very important. In this chapter we shall study the system of linear equations with emphasis on their solution by means of determinants. 4.2 DETERMINANT The notation of determinants arises from the process of elimination of the unknowns of simultaneous linear equations. Consider the two linear equations in x, a1 x + b1 = 0 ... (1) a2 x + b2 = 0 ... (2) b1 a1 Substituting the value of x in (2); we get the eliminant x From (1) b a2 1 b2 0 a1 or a1b2 – a2b1 = 0 From (1) and (2) by suppressing x, the eliminant is written as a1 b1 a2 b2 ... (3) ... (4) 0 when the two rows of a1, b1 and a2, b2 are enclosed by two vertical bars then it is called a determinant of second order. a1 b1 and a2 b2 Column 1 Column 2 a1 b1 Row 1 .... a2 b2 Row 2 .... Each quantity a1, b1, a2, b2 is called an element or a constituent of the determinant. From (3) and (4), we know that both expressions are eliminant, so we equate them. a1 b1 a1 b1 a1 b2 a2 b1 = a1b2 – a2b1 or a2 b2 a2 b2 223 Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ 224 Determinants and Matrices a1b2 – a2b1 is called the expansion of the determinant of Example 1. Expand the determinant + 3 3 2 6 7 a1 b1 a2 b2 . . – 2 Solution. Ans. = (3) × (7) – (2) × (6) = 21 – 12 = 9. 6 7 EXERCISE Expand the following determinants : 4 6 1. 2 5 Ans. 8 2. 8 5 3. 3 1 Ans. – 7 4. 4.1 3 7 2 4 5 2 4 3 Ans. – 26 Ans. 23 4.3. DETERMINANT AS ELIMINANT Consider the following three equations having three unknowns, x, y and z. a1 x + b1 y + c1 z = 0 a2 x + b2 y + c2 z = 0 a3 x + b3 y + c3 z = 0 From (2) and (3) by cross-multiplication, we get ...(1) ...(2) ...(3) y x z k (say) b2 c3 b3 c2 a3 c2 a2 c3 a2 b3 a3 b2 x = (b2 c3 – b3 c2) k y = (a3 c2 – a2 c3) k and z = (a2 b3 – a3 b2) k Substituting the values of x, y and z in (1), we get the eliminant a1 (b2c3 – b3c2) k + b1 (a3c2 – a2c3) k + c1 (a2b3 – a3b2) k = 0 or a1 (b2c3 – b3c2) – b1 (a2c3 – a3c2) + c1 (a2b3 – a3b2) = 0 ...(4) From (1), (2) and (3) by suppressing x, y, z the remaining can be written in the determinant as a1 b1 c1 a2 b2 c2 0 a3 b3 c3 ...(5) This is determinant of third order. As (4) and (5) both are the eliminant of the same equations. a1 b1 c1 a2 b2 c2 a1 (b2c3 b3c2 ) b1 (a2c3 a3c2 ) c1 (a2b3 a3b2 ) 0 a3 b3 c3 Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ Determinants and Matrices a1 b1 c1 a2 b2 c2 a1 or a3 b3 4.4. 225 c3 b2 c2 b3 c3 b1 a2 c2 a3 c1 c3 a2 b2 a3 b3 MINOR The minor of an element is defined as a determinant obtained by deleting the row and column containing the element. Thus the minors a1, b1 and c1 are respectively. b2 c2 b3 c3 b1 c1 a2 c2 , a3 c3 and a2 b2 a3 b3 Thus a1 a2 b2 c2 a3 b3 c3 = a1 (minor of a1) – b1 (minor of b1) + c1 (minor of c1). 4.5. COFACTOR Cofactor = (– 1)r+c Minor where r is the number of rows of the element and c is the number of columns of the element. The cofactor of any element of jth row and ith column is (– 1)i+j minor 1+1 Thus the cofactor of a1 = (– 1) (b2c3 – b3c2) = + (b2c3 – b3c2) The cofactor of b1 = (– 1)1+2 (a2c3 – a3c2) = – (a2c3 – a3c2) The cofactor of c1 = (– 1)1+3 (a2b3 – a3b2) = + (a2b3 – a3b2) The determinant = a1 (cofactor of a1) + a2 (cofactor of a2) + a3 (cofactor of a3). Example 2. Write down the minors and cofactors of each element and also evaluate the determinant. 1 3 –2 4 –5 6 3 Solution. M11 = Minor of element (1) 5 2 3 2 1 4 5 6 3 5 2 5 6 ( 5) 2 6 5 10 30 40 5 2 Cofactor of element (1) = A11 = (– 1)1 + 1 M11 = (– 1)2 (– 40) = – 40 M12 = Minor of element (3) 1 3 2 4 6 4 5 6 4 2 3 6 8 18 10 3 2 3 5 2 Cofactor of element (–2) = A12 = (– 1)1 + 2 (– 10) = 10 M13 = Minor of element (– 2) Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ 226 Determinants and Matrices 1 3 2 = 4 5 3 4 5 6 3 2 5 5 4 5 (5) 3 20 15 35 Cofactor of element (– 2) = A13 = (– 1)1+3 M13 = (–1)4 35 = 35 M21 = Minor of element (4) 1 3 2 3 2 3 2 (2) 5 6 10 16 = 4 5 6 5 2 3 5 2 Cofactor of element (4) = A21 = (– 1)2+1 M21 = (– 1)2+1 (16) = – 16 M22 = Minor of element (– 5) 1 3 2 1 2 1 2 (2) 3 2 6 8 = 45 6 3 2 3 5 2 Cofactor of element (– 5) = A22 = (– 1)2+2 M22 = (– 1)2+2 (8) = 8 M23 = Minor of element (6) 1 3 2 1 3 4 5 6 1 5 3 3 5 9 4 = 3 5 3 5 2 Cofactor of element (6) = A23 = (–1)2+3 M23 = (– 1)2+3 (– 4) = 4 M31 = Minor of element (3) 1 3 = 2 3 2 4 5 6 3 6 (2) (5) 18 10 8 5 6 3 5 2 Cofactor of element (3) = A31 = (– 1)3+1 M31 = (– 1)3+1 8 = 8 M32 = Minor of element (5) 1 3 2 1 2 4 5 6 1 6 (2) 4 6 8 14 4 6 35 2 Cofactor of element (5) = A32 = (– 1)3+2 M32 = (– 1)3+2 14 = – 14 M33 = Minor of element (2) = 1 = 3 2 1 3 4 5 6 1 (5) 4 3 5 12 17 4 5 3 5 2 Cofactor of element (2) = A33 = (– 1)3+3 M33 = (– 1)3+3 (– 17) = – 17. 1 3 2 4 5 6 = 1 × (cofactor of 1) + 3 × (cofactor of 3) + (– 2) × [cofactor of (– 2)]. 3 2 5 = 1 × (– 40) + 3 × (10) + (– 2) × (35) = – 40 + 30 – 70 = – 80 Ans. Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ Determinants and Matrices 227 Example 3. Find : (i) Minors (ii) Cofactors of the elements of the first row of the determinant 2 3 5 4 1 0 6 2 7 Solution. (i) The minor of the element (2) is 235 1 0 2 7 4 1 0 6 2 7 (1) (7) (0) (2) 7 0 7 The minor of the element (3) is 23 5 4 0 6 7 4 1 0 6 2 7 (4) (7) (0) (6) 28 0 28 The minor of the element (5) is 23 5 4 1 6 2 4 1 0 6 2 7 (4) (2) (1) (6) 8 6 2 The cofactor of (2) = (– 1)1+1 (7) = + 7 The cofactor of (3) = (– 1)1+2 (28) = – 28 The cofactor of (5) = (– 1)1+3 (2) = + 2. 6 2 3 Example 4. Expand the determinant 2 3 5 (ii) 4 2 1 6 2 3 Solution. Ans. 2 3 5 = 6 (cofactor of 6) + 2 (cofactor of 2) + 3 (cofactor of 3). 4 2 1 = 6 (3 × 1 – 5 × 2) – 2 (2 × 1 – 4 × 5) + 3 (2 × 2 – 3 × 4) = = = = 6 (3 – 10) – 2 (2 – 20) + 3 (4 – 12) 6 (– 7) – 2 (– 18) + 3 (– 8) – 42 + 36 – 24 – 30. 1 0 4 3 5 –1 . Example 5. Evaluate the determinant 0 1 2 Ans. (i) With the help of second row, (ii) with the help of third column. Solution. 1 0 4 (i) 3 5 1 = 3 × (cofactor of 3) + 5 × (cofactor of 5) + (– 1) (cofactor of – 1). 0 1 2 Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ 228 Determinants and Matrices 0 4 1 4 1 0 = 3 × (– 1)2+1 1 2 + 5 × (– 1)2+2 0 2 + (– 1) × (– 1)2+3 0 1 = – 3 × (0 – 4) + 5 (2 – 0) + (1 – 0) = 12 + 10 + 1 = 23 1 0 (ii) Ans. 4 3 5 1 = 4 × (cofactor of 4) + (– 1) (cofactor of (– 1)) + 2 × (cofactor of 2) 0 1 2 3 5 1 0 1 0 = 4 × (– 1)1+3 + (– 1) (– 1)2+3 + 2 × (– 1)3+3 0 1 0 1 3 5 = 4 × (3 – 0) + (1 – 0) + 2 (5 – 0) = 12 + 1 + 10 = 23 Ans. 0 1 2 3 1 0 2 0 Example 6. Expand the fourth order determinant 2 0 1 3 1 2 1 0 0 2 0 Solution. Given determinant = (0) (–1)1 + 1 1 2 0 1 2 0 1 3 1 (–1) 2 1 3 2 1 0 1 1 0 1 0 0 + 2 (–1)1+ 3 Now 1 = 0 2 1 1 2 0 2 1 3 1 1 0 1 0 0 2 0 3 1 2 0 1 0 2 2 0 1 1 2 1 1 0 2 2 0 3 3 (–1)1 4 2 0 1 1 2 0 2 0 1 3 2 1 0 0 1 0 2 2 0 3 3 2 0 1 1 0 1 2 0 1 2 1 1 2 1 1(1 0 3 1) 2 (2 0 3 1) 0 (2 1 1 1) 3 6 03 1 (0 0 3 2) 0 (2 0 3 1) 0(2 2 0 1) 6 1 ( 0 1 1 2) 0(2 1 1 1) 2 (2 2 0 1) 2– 0 8 6 0 1 2 3 1 0 2 0 3 2( 6 ) 3(6) Now 2 0 1 3 3 – 12 18 33 1 2 1 0 Ans. Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ Determinants and Matrices 229 EXERCISE 4.2 Write the minors and co factors of each element of the following determinants and also evalutate the determinant in each case : 1. 2. 2 M11 = 3 A11 = – 9, A12 = – 4, A21 = – 3, A22 = – 2 4 9 cos sin sin cos 28 7 4 14 3 2 4. 1 a 1 b bc ca 1 c ab |A|=6 Ans. M11 = cos , M12 = sin M21 = – sin , M22 = cos A11 = cos , A12 = – sin , A21 = sin , A22 = cos | A | = 1 M11 M23 A11 A23 42 1 6 3. – 9, M12 = 4, M21 = 3, M22 = – 2 Ans. = 2, M12 = 0, M13 = – 14, M21 = – 16, M22 = 0 = 112, M31 = – 38, M32 = 0, M33 = 266 = 2, A12 = 0, A13 = – 14, A21 = 16, A22 = 0 = – 112, A31 = – 38, A32 = 0, A33 = 266, | A | = 0 Ans. M11 = (ab2 – ac2), M12 = (ab – ac), M13 = (c – b), M21 = a2b – bc2 M22 = (ab – bc), M23 = (c – a), M31 = (ca2 – cb2), M32 = ca – bc, M33 = (b – a), A11 = (ab2 – ac2), A12 = (ac – ab), A13 = (c – b), A21 = bc2 – a2b A22 = (ab – bc), A23 = (a – c), A31 = (ca2 – cb2), A32 = (bc – ca), A33 = (b – a) | A | = (a – b) (b – c) (c – a). Ans. Expand the following determinants : 2 3 4 5 1 6 5. 7 8 Ans. | A | = 5 5 6. 9 0 7 8 6 4 2 3 9 Ans. | A | = 42 a h g 7. h b f g f c Ans. | A | = abc + 2fgh – af 2 – bg2 – ch2 Expand the following determinants by two methods : (i) along the-third row. (ii) along the-third column. 8. 1 3 2 4 1 2 3 5 2 Ans. | A | = 40 11. 12. log 3 512 log 4 3 log 3 8 log 4 9 9. 3 2 4 1 2 1 0 1 1 2 3 2 10. 1 2 2 Ans. | A | = – 7 Ans. | A | = 3 1 3 Ans. | A | = – 37 15 2 If a, b, c are all positive and are the pth, qth, rth terms of a G.P. respectively; then prove that log a p 1 log b q 1 0 13. log c r 1 3 2 5 1 4 3 6 2 4 1 7 0 2 1 0 3 Ans. 96 Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ 230 4.6 Determinants and Matrices RULESOF SARRUS (For third order determinants only). After writing the determinant, repeat the first two columns as below = (a1b2c3 + b1c2a3 + c1a2b3) + (– c1b2a3 – a1c2b3 – b1a2c3) Example 7. Expand the determinant 2 3 4 1 5 3 by Rule of Sarrus. 3 0 5 Solution. =(2) × (5) × (5) + (3) × (3) × (3) + (4) × (1) × (0) – (4) × (5) × (3) – (2) × (3) × (0) – (3) × (1) × (5) = 50 + 27 + 0 – 60 – 0 – 15 = 2 Ans. EXERCISE 4.3 Expand the following determinants by Rule of Sarrus. 3 2 4 1. 5 1 1 2 6 7 Ans. – 155 1 4 2 2. 2 5 3 3 6 4 Ans. 0 6 3. 3 7 32 13 37 10 4 11 Ans. 10 4. 9 25 6 7 13 5 9 23 6 Ans. 6 ax c b bx a 0 5. If a + b + c = 0, solve the equation c b a cx Ans. x 2 2 2 (a b c ab bc ca) , x = 0 Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ Determinants and Matrices 231 4.7 PROPERTIES OF DETERMINANTS Property (i) The value of a determinant remains unaltered, if the rows are interchanged into columns (or the columns into rows). Consider the determinant. a1 b1 c1 a2 b2 c2 a3 b3 c3 = a1 (b2c3 – b3c2) – b1 (a2c3 – a3c2) + c1 (a2b3 – a3b2) = a1b2c3 – a1b3c2 – a2b1c3 + a3b1c2 + a2b3c1 – a3b2c1 = (a1b2c3 – a1b3c2) – (a2b1c3 – a2b3c1) + (a3b1c2 – a3b2c1) = a1 (b2c3 – b3c2) – a2 (b1c3 – b3c1) + a3 (b1c2 – b2c1) a1 a2 a3 b1 b2 b3 Proved. c1 c2 c3 Property (ii) If two rows (or two columns) of a determinant are interchanged, the sign of the value of the determinant changes. Interchanging the first two rows of , we get a2 b2 ' a1 c2 b1 c1 a3 b3 c3 = a2 (b1c3 – b3c1) – b2 (a1c3 – a3c1) + c2 (a1b3 – a3b1) = a2b1c3 – a2b3c1 – a1b2c3 + a3b2c1 + a1b3c2 – a3b1c2 = – [(a1b2c3 – a1b3c2) – (a2b1c3 – a3b1c2) + (a2b3c1 – a3b2c1)] = – [(a1 (b2c3 – b3c2) – b1 (a2c3 – a3c2) + c1 (a2b3 – a3b2)] a1 b1 c1 = a2 b2 c2 a3 b3 c3 Proved. Property (iii) If two rows (or columns) of a determinant are identical, the value of the determinant is zero. a1 b1 Let a1 b1 c1 c1 , so that the first two rows are identical. a3 b3 c3 By interchanging the first two rows, we get the same determinant . By property (ii), on interchanging the rows, the sign of the determinant changes. or =– or 2=0 or = 0 Proved. Property (iv) If the elements of any row (or column) of a determinant be each multiplied by the same number, the determinant is multiplied by that number. ka1 kb1 kc1 a2 b2 c2 a3 b3 c3 Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ 232 Determinants and Matrices = ka1 (b2c3 – b3c2) – kb1 (a2c3 – a3c2) + kc1 (a2b3 – a3b2) = k [a1 (b2c3 – b3c2) – b1 (a2c3 – a3c2) + c1 (a2b3 – a3b2)] a1 b1 c1 k a2 b2 c2 k . a3 b3 c3 Example 8. Prove that a2 b a 2 b c2 a2 1 1 2 2 ca a b a3 c ab b3 1 c 2 c3 bc 2 b ca c2 c ab b Solution. a bc By multiplying R1, R2, R3 by a, b and c respectively we get a3 a2 abc a3 3 2 1 b b abc abc c3 c2 abc 3 abc b abc c3 a3 a2 1 b 2 1 c 2 1 a2 1 1 a2 a3 2 1 1 b2 b3 c3 c2 1 1 c2 c3 1 1 1 2 2 c2 b3 c3 b 3 a b a3 b By changing rows into columns Proved Example 9. Without expanding and or evaluating, show that a 2 b 2 c 2 d 2 a 1 bcd b 1 cda c 1 dab d 1 abc a 3 a 2 a 1 b 3 b 2 b 1 c 3 c 2 c 1 d 3 d 2 d 1 Solution a 2 a 1 bcd a3 a2 a abcd b 2 b 1 cda 3 2 b abcd c 2 c 1 dab c3 c2 c abcd d2 d 1 abc d3 d2 d abcd 1 abcd b b R1 aR1 R2 bR2 R3 cR3 R4 dR4 Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ Determinants and Matrices abcd abcd a3 a2 3 2 b b c3 c2 3 2 d d 233 a 1 a3 a2 a 1 b 1 3 2 b 1 c3 c2 c 1 3 2 d 1 1 C4 c 1 C4 abcd d 1 1 a a2 Example 10. Prove that 1 b b 2 1 c 2 c 1 a bc 1 b ca 1 c b d b d Proved (Try yourself) ab Property (v) The value of the determinant remains unaltered if to the elements of one row (or column) be added any constant multiple of the corresponding elements of any other row (or column) respectively. Let a1 b1 c1 a2 b2 c2 a3 b3 c3 On multiplying the second column by l and the third column by m and adding to the first column we get a1 lb1 mc1 b1 c1 ' a2 lb2 mc2 b2 c2 a3 lb3 mc3 b3 c3 a1 b1 b1 c1 b1 c1 c1 b1 c1 a2 b2 c2 l b2 b2 c2 m c2 b2 c2 a3 b3 c3 b3 b3 c3 c3 b3 c3 = + 0 + 0 = (Since columns are identical) Proved 9 9 12 Example 11. Evaluate, using the properties of determinant 1 3 – 4 1 9 12 9 9 12 Solution. Let 1 3 4 1 9 12 Applying : R1 R1 3R 2 and R 3 R 3 3R 2 , we get 12 18 1 2 3 0 3 4 6 2 1 3 4 4 18 Expand by C3 0 = 6×2×4 0 2 9 0 2 3 2 9 = 48 (2 × 9 – 2 × 3) = 48 × 12 = 576. Ans. Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ 234 Determinants and Matrices 265 240 219 Example 12. Without expanding evaluate the determinant 240 225 198 219 198 181 Solution. Applying C1 C1 C3 and C 2 C 2 C3 , we get 46 21 219 42 27 198 38 17 181 Applying C1 C1 2C2 and C3 C3 10 C2 , we get 4 21 9 12 27 72 4 17 11 Applying R1 R1 R3 and R2 R2 3R3 0 4 2 0 0 78 39 2(39) 0 4 17 11 2 1 2 1 [Taking 2 common from R1 and 39 common from R2] 4 17 11 = 78 × 0 = 0 (Since R1 and R2 are identical) Ans. bc ca ab c a a b bc =0 Example 13. Show that ab bc ca bc ca ab Solution. Let ca ab ab bc bc ca Applying C1 C1 C2 C3 , we get 0 ca ab 0 ab bc 0 0 bc ca [C1 consists of all zeros.] sin cos sin( + ) Example 14. Without expanding, evaluate the determinant sin cos sin( + ) . sin cos sin( + ) sin cos sin ( ) sin cos sin ( ) Solution. Let sin cos sin ( ) sin cos sin cos cos sin sin cos sin cos cos sin sin cos sin cos cos sin [ sin (A + B) = sin A cos B + cos A sin B] Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ Determinants and Matrices 235 sin cos 0 sin cos 0 sin cos 0 [Applying C3 C3 cos . C1 sin . C2 ] =0 [ C3 consists of all zeros] Ans. 2x – 1 x + 7 x 6 2 x–1 x+1 3 Example 15. Solve the determinantal equation 2x 1 x 7 0 x4 x 6 2 x 1 x 1 3 Solution. Given equation x+4 By applying R1 R1 – (R2 + R3), we get 0 0 0 x 1 x 6 2 x 1 x 1 0 3 On expanding by first row, we get (x – 1) (x2 + x – 6x + 6) = 0 (x – 1) (x – 2) (x – 3) = 0 x = 1, 2, 3 Example 16. Using the properties of determinants, show that x+ y x x 3 5x + 4y 4x 2x = x . 10x + 8y 8x 3x x y 5x 4 y Solution. Let Ans. x x 4x 2x 10 x 8 y 8x 3x Operate : R2 R2 2 R1 ; R3 R3 3R1 x y x x 3x 2 y 2x 0 a 2b bc 2 Expand by C3 x 3x 2 y 2x 7 x 5 y 5x 7 x 5 y 5x 0 = x [5x (3x + 2y) – 2x (7x + 5y)] = x [15x2 + 10 xy – (14x2 + 10 xy)] = x3. Proved. Example 17. Using the properties of determinants, evaluate the following : 2 2 0 ab ac 0 2 a c cb 0 Solution. Let a 2b ab 2 0 a 2c cb2 ac 2 bc 2 2 0 0 Take a2, b2 and c2 common from C1, C2 and C3 respectively, 0 a a 2 2 a b c 2 b 0 b c c 0 Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ 236 Determinants and Matrices 0 Operate : C2 C2 C3 , 2 2 0 a 2 a b c b b b c 2 2 2 a b c .a c 0 b b 3 2 2 3 3 3 a b c (bc bc) 2a b c . Ans. c c Example 18. Using properties of determinants, prove that Expand by R1, Solution. Let x y x 2 y 2 z x 3 y 3 z 2 z 3 = x y z (x – y)(y – z) (z – x). x y z 1 1 1 x 2 y 2 z 2 xyz x y z x 3 y 3 z 3 2 2 x y z 2 0 xyz Operate : C1 C1 C2 ; C2 C2 C3 , x y 2 x y On expanding by R1, xyz x y 2 x y yz 2 2 y z 0 yz 2 2 y z xyz ( x y) ( y z) 2 = xyz (x – y) (y – z) (z – x). Example 19. Using the properties of determinants, show that Solution. Let a+x y z x a+ y z x y a+z a Operate : R1 R1 R2 , a Operate : C2 C2 C1 , On expanding by R1 a 2 1 1 x y yz Proved. 2 az a 0 y z az 0 0 x a y x x z z z y x a y x z 2 = a (a + x + y + z). ax y x a y x 1 y x a yx z a z z = a [(a + y + x) (a + z) – (y + x) z] yx az = a [a2 + az + (y + x) a + (y + x) z – (y + x) z] = a2 (a + x + y + z). Proved. Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ Determinants and Matrices 237 Example 20. If is the one of the imaginary cube roots of unity, find the value of the determinant 2 1 Solution. The given determinant 1 2 1 2 By R1 R1 R2 R3, we get 1 2 1 2 1 2 2 2 1 0 0 0 1 2 1 2 1 =0 [1 + + 2 = 0] (Since each entry in R1 is zero) Ans. Example 21. Without expanding the determinant, show that (a + b + c) is a factor of b c a . c a b a b c b c a Solution. Let a b c c a b abc b 1 b c c c a (a b c) 1 c a 1 a b abc a b abc Operate : C1 C1 C2 C3 , (a + b + c) is a factor of . Example 22. Using properties of determinants, prove that : Solution. Let x+4 x x x x+4 x x x x+4 x4 x x x x4 x x x x4 3x 4 Operate : C1 C1 C2 C3 , 1 x (3x 4) 1 x 4 1 16(3x 4) x = 16 (3x + 4) x 3x 4 x 4 3x 4 x x x x4 x 1 x x (3x 4) 0 4 0 R2 R1 x4 Proved. x 0 0 4 R3 R1 (3x 4) 4 0 0 4 Proved. Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ 238 Determinants and Matrices 1 a b+c Example 23. Without expanding the determinant, prove that 1 b c + a = 0. 1 c a +b 1 a bc Solution. Let 1 b ca 1 c ab 1 a abc Operate : C3 C3 C2 , 1 a 1 1 b a b c (a b c) 1 b 1 1 c abc 1 c 1 =0 ( C1 and C3 are identical). Proved. Example 24. Without expanding the determinant, prove that 1 a2 a 2 Solution. Let 1 b b 1 c2 c Multiply R1 by a, R2 by b and R3 by c. bc 1 a 1 b 1 c a2 b 2 c2 bc ca = 0 ab ca ab 1 a3 1 1 a 3 abc 3 1 1 b abc abc 1 c 3 abc 3 1 . abc 1 b 1 1 0 0. abc 1 c3 1 (Since C1 and C3 are identical) Proved. 1 a a 2 Example 25. Evaluate 1 b b2 1 c c 2 Solution. Let be the given determinant. Applying R2 R2 R1 and R3 R3 R1, we get, 1 a2 a 2 0 ba b a 1 a 2 0 c a c2 a 2 1 a a 2 (b a) (c a) 0 1 b a 0 1 ca a [Taking out (b – a) common from R2 and (c – a) from R3] 2 = (b – a) (c – a) 0 1 b a 0 0 cb [Applying R3 R3 R2 ] Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ Determinants and Matrices = (b – a) (c – a) 239 1 ba [Expanding along C1] 0 cb = (b – a) (c – a) (c – b). Example 26. Using properties of determinants, prove that : Ans. 1 a a3 1 b b 3 = (a – b)(b – c)(c – a)(a + b + c) 1 c c3 1 a a3 Solution. Let = 1 b b 1 c 3 c3 0 a b a3 b3 3 Operate : R1 R1 R2 ; R2 R2 R3 , 0 b c b c 1 (a b) (b c) 3 1. c3 c 2 2 2 2 1 a ab b 1 b bc c 2 3 3 3 3 ab a b bc b c (Expanding by C1) 2 0 (a c ) (ab bc) Operate : R1 R1 R2 , (a b) (b c) 2 2 1 b bc c = (a – b) . (b – c) . (– 1) [(a2 – c2) + b (a – c)] = (a – b) . (b – c) (c – a) (a + b + c). Example 27. Evaluate a–b–c 2a 2a 2b b–c–a 2b 2c 2c c–a–b Proved. abc abc abc Solution. By R1 R1 + R2 + R3, we get 2b bca 2b 2c 2c cab 1 1 1 (a b c) 2b b c a 2c 2b 2c cab 0 0 1 (a b c) 2b (a b c) 2c 0 On expanding by first row = (a + b + c) (a + b + Example 28. Show, without expanding x c)2 0 C2 C1 (a b c) C3 C1 = (a + b + c)3. 1 1 1 x y z = (x – y)(y – z)(z – x) 2 y 2 z Ans. 2 Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ 240 Determinants and Matrices Solution. By C1 – C2, C2 – C3, we get 0 0 1 x y yz z 2 x y ( x y) ( y z ) Proved. 1 1 x y yz Solution. Let 2 2 2 x y 2 x y yz 2 2 y z 2 y z z On expanding by first row, we get = (x – y) (y – z) (y + z – x – y) = Example 29. Prove that 2 (x – y) (y – z) (z – x). 2 2 2 . 2 2 2 Applying R3 R1 R3 = ( + + ) 2 2 2 [Taking out (+ + ) common from R3] 1 1 1 2 ( ) 1 2 0 2 2 0 2 ( ) ( ) ( ) 1 ( ) ( ) ( ).1 2 Applying C2 C2 C1 C3 C3 C1 1 1 0 0 1 1 [Expanding along R3] = (+ + ) (– ) (– ) (+ – – ) = (+ + ) (– ) (– ) (– ) –a Example 30. Prove that 2 ba – b ac a2 Solution. Let ab ac 2 2 2 2 bc = 4 a b c bc – c 2 ab ac 2 bc ba b ac Proved. bc c 2 Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ Determinants and Matrices 241 a Taking a, b, c common from R1, R2 and R3 respectively, we get 1 2 2 2 abc 1 1 1 1 1 2 2 2 a a b b b c 1 0 2 [Applying C2 C2 C1, C3 C3 C1 ] 1 2 0 0 2 = a2b2c2 (– 1) 2 0 [Expanding along R1] = a2b2c2 (– 1) (0 – 4) = 4a2 b2 c2 Example 31. Show that Solution. Let c c [Taking a, b, c common from C1, C2 and C3 respectively] 1 1 1 1 0 0 abc abc Proved. 3a –a + b –b + a 3b –c + a –c + b –a + c – b + c = 3(a + b + c) (ab + bc + ca) 3c 3a a b a c b a 3b b c c a c b 3c a b c a b a c abc Applying C1 C1 C2 C3, we get abc 3b b c c b 3c 1 a b a c (a b c) 1 3b b c 1 c b [Taking (a + b + c) common from C1] 3c 1 a b a c = (a b c) 0 2b a b a [Applying R2 R2 R1 , R3 R3 R1 ] 0 c a 2c a (a b c) 2b a b a [Expanding along C1] c a 2 c a = (a + b + c) [(2b + a) (2c + a) – (– b + a) (– c + a)] = (a + b + c) {(4bc + 2ab + 2ca + a2 – (bc – ab – ac + a2)} = (a + b + c) (3bc + 3ab + 3ca) = 3 (a + b + c) (ab + bc + ca) Proved. Property (vi) If each element of a row (or column) of a determinant consists of the algebraic sum of n terms, the determinant can be expressed as the sum of n determinants, Let a1 p1 q1 b1 c1 a 2 p2 q 2 b2 c2 . a3 p3 q3 b3 c3 Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ 242 Determinants and Matrices = (a1 + p1 + q1) (b2c3 – b3c2) – (a2 + p2 + q2) (b1c3 – b3c1) + (a3 + p3 + q3) (b1c2 – b2c1) = a1 (b2c3 – b3c2) – a2 (b1c3 – b3c1) + a3 (b1c2 – b2c1) + p1 (b2c3 – b3c2) – p2 (b1c3 – b3c1) + p3 (b1c2 – b2c1) + q1 (b2c3 – b3c2) – q2 (b1c3 – b3c1) + q3 (b1c2 – b2c1) a1 b1 c1 b1 c1 a2 b2 c2 p2 b2 c2 q2 b2 c2 a3 b3 a a Example 32. If a a2 b b 2 c 2 c c3 c1 p3 b3 q1 c3 q3 b3 Proved. c3 3 b3 – 1 = 0, prove that abc = 1. 2 3 c –1 a3 1 3 b 1 0 3 c 1 1 a a2 b1 a –1 b b2 c c Solution. 2 p1 abc 1 b b 2 1 c c 2 a a2 a3 a a2 b b 2 b 3 b b 2 1 0 c 2 c 3 c 2 1 c c 1 a a2 1 b b 2 1 0 c c 2 1 (Taking out common a, b, c from R1, R2 and R3 from 1st determinant) 1 a a2 abc 1 b b 1 c c 2 a 1 a2 b 1 b 2 c 1 c 2 2 1 a a2 abc 1 b b 1 c c ( a b c 1) 2 (Interchanging C2 and C3) 1 a a2 1 b b 2 1 c c 2 1 a a 2 1 b b 2 c 2 2 1 0 c 0 (Interchanging C1 and C2 ) 0 abc 1 0 abc 1 Example 33. Show that Proved. b+c c+a a+b a b c q+r r+ p p+q =2 p q r y+z z+x x+ y y z x Solution. The above determinant can be expressed as the sum of 8 determinants as given below: bc ca ab b c a b a a b c b b a b qr r p pq q r p q p p q r q q p q y z z x x y y z x y x x y z y y x y Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ Determinants and Matrices 243 c c b c a c a a c c b r r p r p p r z z x x x z a r q z z y c a b q r p 000000 r p q y z x x y = (1) 2 z a b c p q r (1) x y z 2 a b c p q a + c a b r p q z x y b c r 2 p q r x y z x y z 0 Example 34. Prove that Solution. Given determinant The above determinant can be expressed as the sum of 8 determinants. Proved. Each of the 8 determinants has either two identical columns or identical rows. Each of the resulting determinant is zero. Hence the result. Example 35. Prove that Solution. x l m 1 x n 1 x 1 1 Proved. (x ) (x ) (x ) x l m 1 x l m 1 x n 1 0 x n 1 (C1 C1 – C4) x 1 0 x 1 [On expanding by first column we get] 1 0 1 x n 1 x n 1 (x ) x 1 (x ) 0 x 1 1 0 1 ( x ) ( x ) ( x ) (C1 C1 – C3) [On expanding by first column] Proved. Example 36. Show that x = – (a + b + c) is one root of the equation: x+a b c b x+c a c a x+b = 0 and solve the equation completely.. xabc Solution. By C1 C1 + C2 + C3, we get b xabc xc xabc a c a 0 xb Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ 244 Determinants and Matrices 1 c ( x a b c) 1 x c 1 1 b a b 0 xb c ( x a b c) 0 x b c 0 a ab ac 0, R2 R2 R1; R3 R3 R1 xbc On expanding by first column, we get (x + a + b + c) [(x – b + c) (x + b – c) – (a – b) (a – c)] = 0 (x + a + b + c) [x2 – (b – c)2 – (a2 – ac – ab + bc)] = 0 (x + a + b + c) (x2 – b2 – c2 + 2bc – a2 + ac + ab – bc] = 0 (x + a + b + c) (x2 – a2 – b2 – c2 + ab + bc + ca) = 0 Either x + a + b + c = 0 or x2 x = – (a + b + c) – a2 – b2 – c2 + ab + bc + ca = 0 x= 2 2 2 a b c ab bc ca Hence, x = – (a + b + c) is one root of the given equation. (b + c)2 Example 37. Find the value of b 2 a2 Proved. a2 2 b c2 (a + b)2 c2 (b c)2 a 2 2 b b Solution. By C1 – C3, C2 – C3, we get 2 (c + a) 2 a2 a2 2 (c a) b c 2 (a b) 2 a2 2 b c 2 (a b) 2 2 (a b)2 (a b c) (b c a) 0 a2 0 (a b c) (c a b) b 2 (a b c) (c a b) (a b c) (c a b) (a b)2 On taking out (a + b + c) as common from 1st and 2nd column, we get (a b c ) bca 0 a2 0 cab b 2 cab a b c (a b c) 2 0 2b 2 c a b (a b ) 2 0 a2 2 a b c b R3 R3 (R1 R2 ) 2a 2ab Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ Determinants and Matrices 245 a b c 2 (a b c) 2 a2 0 2 0 b abc b a ab On expanding by first row, we get = – 2 (a + b + c)2 [(– a + b + c) {– ab (a – b + c) – ab2} + a2 {0 – b (a – b + c)}] = – 2 (a + b + c)2 [(– a + b + c) (– a2b – abc) – a2b (a – b + c)] = – 2ab (a + b + c)2 [(– a + b + c) (– a – c) – a (a – b + c)] = – 2ab (a + b + c)2 (a2 + ac – ab – bc – ac – c2 – a2 + ab – ac] = – 2ab (a + b + c)2 (– bc – ac – c2) = 2abc (a + b + c)2 (b + a + c) = 2abc (a + b + c)3. Ans. a+x a – x a – x Example 38. Using properties of determinants, solve for x : a – x a+x a – x =0 a – x a– x a+x ax ax ax Solution. Given that ax ax ax 0 ax ax ax 3a x a x a x 3a x a x a x 0 Applying C1 C1 C2 C3 3a x a x a x 1 ax ax (3a x) 1 a x a x 0 1 ax ax Now, R2 R2 R1 and R3 R3 R1, 1 ax ax (3a x) 0 2x 0 0 0 2x 0 2 Expanding by C1, we get (3a x) (4x 0) 0 2 2 4x (3a x) 0 If 4 x 0, then x 0 If 3a x 0, then x 3a Hence, x=0 or 3a Ans. Example 39. Using properties of determinants, prove the following Solution. Let 1+ a 1 1 1 1+ b 1 1 1 1+c 1 a 1 1 1 b 1 1 = abc 1 + 1 + 1 + 1 a b c 1 1 1c Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ 246 Determinants and Matrices 1 a a abc 1 b 1 c 1 a 1 b 1c c 1 1 1 1 a b c 1 Operate : R1 R1 R2 R3 , abc b 1 c 1 1 1 Taking 1 common from R1, we get a b c 1 a 1b b 1 c 1 1 1 1 a a a 1 1 1 1 abc b b b 1 1 1 1 c c c 1 1 1 a b c 1 1 b 1 c 1 abc 1 1 1 1 a b c Operate : C2 C2 C1 ; C3 C3 C1 , abc 1 1 1 1 (On expanding by R ) 1 a b c Example 40. Prove that : Solution. Let 1 b 1 1 c 1 0 0 1 1 0 b 1 0 1 c Proved. a a 2 bc b b 2 ac = (a – b)(b – c)(c – a)(ab + bc + ac). c 2 ab c a a2 bc 2 ac b b c c2 ab a2 a3 2 1 b abc c2 b 3 c 3 a 2 b2 Operate : R1 R1 R2 ; R2 R2 R3, 2 b c c a2 abc 2 abc 1 . abc b abc abc c2 2 3 b c 2 c 3 2 0 2 2 0 (a b) (b c). 1 c 3 1 c 3 1 1 2 2 2 2 a b a ab b bc 3 1 2 c b 0 3 a b a ab b 2 a3 1 a 3 b3 0 (a b) (b c) b c b bc c Expand by C3 1 1 1 a b c 1 1 1 1 1 1 1 b b b 1 1 1 1 c c c abc 1 1 1 1 a b c 1 b bc c Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ Determinants and Matrices 247 2 ab Operate : R2 R2 R1 (a b) (b c) a ab b 2 2 2 c a b (c a) (c a ) (a b) (b c) (c a) a b a 2 ab b2 1 b c a = (a – b) (b – c) (c – a) [(a + b) (a + b + c) – 1 . (a2 + ab + b2)] = (a – b) (b – c) (c – a) (ab + bc + ac). Proved. EXERCISE 4.4 Expand the following determinants, using properties of the determinants : 1 3 7 x a a 2 4 9 1 1. Ans. 51. 2. Prove that a x a ( x 2a) ( x a) . 2 7 6 a a x x3 a3 x2 x a3 3 3 2 , x b, x c 3. Solve the equation b a b b 0, b c, bc 0 Ans. x bc 3 3 2 c a c c 0 xa xb xa 0 xc 0 4. Show that zero is one of the roots of the equation x b x c 0 1 a 1 b 1 c x 5. Without expanding the determinant, prove that 6. Without expanding the determinant, prove that : 8. c 9. a 1 x y x2 y2 1 yz y z 1 zx 2 2 2 2 z x b b b ca 0 c ab y yz zx z x y 1 1 1 x4 2x 2x 2x x4 2x 2x 2x x4 7. Using properties of determinant prove that : a b 2c a c b c 2a a bc 0. (5x 4) (4 x) 2 2(a b c)3 c a 2b ( x y) ( y z) ( z x). Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ 248 Determinants and Matrices 10. 2 2 bc bc c2a 2 ca ca 0 2 2 ab a 12. 14. 2 11. ab a b ab abc 2a 3a 2b 3 4a 3b 2c a . 3a 6a 3b 10a 6b 3c 1 a a 2 1 b b 2 1 c 16. 1 a a 2 bc b c c2 1 a bc 1 b ca 1 c ab 13. 15. abc c b c abc a b a abc 1 b b ca 0. 1 c c 2 ab 1 1 1 ( ) ( ) ( ). a2 bc ac c 2 a 2 ab b2 ac ab b 2 bc c2 4a 2 b 2 c 2 2 (a b) (b c) (c a). 4.8 FACTOR THEOREM If the elements of a determinant are polynomials in a variable x and if the substitution x = a makes two rows (or columns) identical, then (x – a) is a factor of the determinant. When two rows are identical, the value of the determinant is zero. The expansion of a determinant being polynomial in x vanishes on putting x = a, then x – a is its factor by the Remainder theorem. Example 41. Show that 1 1 1 x y z = (x – y)(y – z)(z – x) 2 2 2 x y z Solution. If we put x = y, y = z, z = x then in each case two columns become identical and the determinant vanishes. (x – y), (y – z), (z – x) are the factors. Since the determinant is of third degree, the other factor can be numerical only k (say). 1 x 1 y 2 2 1 z k ( x y ) ( y z ) ( z x) ... (1) 2 x y z This leading term (product of the elements of the diagonal elements) in the given determinant is yz2 and in the expansion k (x – y) (y – z) (z – x) we get kyz2 Equating the coefficient of yz2 on both sides of (1), we have k=1 Hence the expansion = (x – y) (y – z) (z – x). Proved. Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ Determinants and Matrices 249 1 1 2 2 Example 42. Factorize = a b 3 1 c 3 2 3 a b c Solution. Putting a = b, C1 = C2 and hence = 0. a – b is a factor of . Similarly b – c, c – a are also factors of . (a – b) (b – c) (c – a) is a third degree factor of which itself is of the fifth degree as is judged from the leading term b2c3. The remaining factor must be of the second degree. As is symmetrical in a, b, c the remaining factor must, therefore, be of the form k (a2 + b2 + c2) + l (ab + bc + ca) = (a – b) (b – c) (c – a) {k (a2 + b2 + c2) + l (ab + bc + ca)} If k 0, we shall get terms like a4b, b4c etc. which do not occur in . Hence, k must be zero. = (a – b) (b – c) (c – a) {0 + l (ab + bc + ca)} or = l (a – b) (b – c) (c – a) (ab + bc + ca) The leading term in = b2c3. The corresponding term on R.H.S = l b2c3 l =1 Hence, = (a – b) (b – c) (c – a) (ab + bc + ca). Example 43. Show that Solution. x x2 x3 y y2 y 3 xyz ( x y ) ( y z ) ( z x ). z z2 z3 x x2 x3 y y2 y 3 xyz 1 y z z2 z3 Example 44. Show that 1 x 1 z x2 y 2 = xyz (x – y ) (y – z) (z – x) (see example 42). z2 Proved. x3 x2 x 1 3 2 1 2 1 3 Ans. ( x ) ( x ) ( x ) ( ) ( ) ( ) 3 2 1 Solution. If we put x = ; x = ; x = ; = , = ; then two rows become identical and the determinant vanishes. (x – ) ; (x – ; (x – ; – ; (– ; – are the factors. Since the determinant is of six degree the other factor can be numerical only say k. x3 x2 x 1 3 2 1 3 2 1 3 2 1 k ( x ) ( x )( x ) ( ) ( ) ( ) The leading term is x3 . And in the expansion it is kx3 (– 2 ). Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ 250 Determinants and Matrices k = –1 Hence the expansion = – (x – ) (x – (x – – (– – Proved. EXERCISE 4.5 2 3 a a 1 a 2 3 1. Evaluate, without expanding b b 1 b 2 3 c c 1c 2. Without expanding, show that (a x)2 (a y)2 (b x) 2 (c x)2 (b y) 2 (c y) 2 Ans. (a – b) (b – c) (c – a) (1 + abc) (a z ) 2 (b z) 2 (c z)2 = 2 (a – b) (b – c) (c – a) (x – y) (y – z) (z – x). 3. Show (without expanding) that bc a 2 b 2 c2 a2 bc ab ca ab ca bc ca bc ab ab ca b c2 2 1 2 2 2 (ab bc ca)[(ab bc) (bc ca) (ca ab) ] 2 4.9 PIVOTAL CONDENSATION METHOD The condensation process of reducing nth order determinant to (n – 1)th order determinant is shown below : a1 b1 c1 d1 a2 b2 c2 d 2 a3 b3 c3 d3 th Consider n order determinant D a4 b4 c4 d 4 an bn cn dn Add such a multiple of first column in the other columns so that at the places of b1, c1, d1 ......., we get zero. Hence subtracting columns respectively, we get a1 0 b a2 b2 1 . a2 a1 D b1 c1 d1 , , , ..., times the first column from the 2nd, 3rd, 4th... a1 a1 a1 0 0 c2 c1 a2 a1 d2 d1 . a2 a1 a3 b3 b1 . a3 a1 c3 c1 a3 a1 d3 d1 . a3 a1 a4 b4 b1 . a4 a1 c4 c1 a4 a1 d4 d1 . a4 a1 an b bn 1 . an a1 c cn 1 . an a1 d d n 1 . an a1 Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ Determinants and Matrices 251 b2 b1 . a2 a1 c2 c1 a2 a1 d2 d1 . a2 a1 b3 b1 . a3 a1 c3 c1 a3 a1 d3 d1 . a3 a1 D a1 b4 b1 . a4 a1 c4 c1 a4 a1 d4 d1 . a4 a1 b1 . an a1 cn bn c1 . an a1 dn d1 . an a1 Which is a determinant of (n – 1)th order. Now, a1b2 b1a2 a1c2 c1a2 a1d2 d1a2 a1 a1 a1 a1b3 b1a3 a1 D a1 a1c3 c1a3 a1 a1b4 b1a4 a1 a1c4 c1a4 a1 1 (a1 )n 1 a1d 4 d1a4 a1 a1bn b1an a1 D a1 . a1d3 d1a3 a1 On expanding along the first row a1dn d1an a1 a1cn c1an a1 a1b2 b1a2 a1b3 b1a3 a1b4 b1a4 a1c2 c1a2 a1c3 c1a3 a1c4 c1a4 a1bn b1an a1cn c1an a1d2 d1a2 a1d3 d1a3 a1d 4 d1a4 a1dn d1an 1 as the determinant is of (n – 1)th order and a is common in every row (or column) 1 a1 b1 a1 c1 a1 d1 a2 b2 a2 c2 a2 d 2 1 (a1 ) n2 a1 b1 a1 c1 a1 d1 a3 b3 a3 c3 a3 d3 a1 b1 a1 c1 a1 d1 a4 b4 a4 c4 a4 d4 a1 b1 a1 c1 a1 d1 an bn an cn an dn Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ 252 Determinants and Matrices Thus, the nth order determinant is condensed to (n – 1)th order determinant. Repeated application of this method ultimately results in a determinant of 2nd order which can be evaluated. It is obvious that the leading element a1 behaves like a pivot in the condensation process (i.e., reduction from n to (n – 1) and hence the method is pivotal condensation. If the leading element is zero, it can be made non-zero by interchanging the columns. Example 45. Condense the following determinants to second order and hence evaluate them: 2 1 3 5 10 2 3 4 2 7 6 D 5 12 15 (i) (ii) 8 3 1 0 7 6 4 5 7 2 6 Solution. (i) Using the leading element as pivot, we get D 32 (10) 1 10 D= (ii) 1 1 (2)4 2 120 10 150 15 60 14 40 21 110 165 74 61 order = 3 55 2 3 11 11 344 11172 1892. Ans. = 122 222 = 10 74 61 2 2 4 4 14 12 12 20 68 2 24 0 40 14 5 4 15 12 25 as the order is 4. 8 2 8 4 1 4 52 7 80 28 1 2 2 1 = 4 14 26 40 4 7 13 20 3 2 44 9 148 36 (4) 9 11 37 9 11 37 = 1 59 52 4 35 184 = 4 59 52 59 46 13 35 2259. 4 35 46 Ans. 0 4 1 2 5 3 7 8 Example 46. Condense and hence evaluate the determinant , 4 1 2 3 1 2 5 5 Solution. As the leading element is zero, hence interchanging the 1st and second columns, we get 0 4 1 2 4 0 1 2 20 0 28 3 32 6 20 25 26 5 3 7 8 3 5 7 8 1 16 0 8 1 12 2 1 16 7 10 4 1 2 3 = 16 42 1 4 2 3 4 0 20 2 20 4 4 18 16 1 2 5 5 2 1 5 5 5 25 13 4 2 4 7 5 1 . 1 35 100 25 52 1 65 27 1 65 27 0. 16 2 5 90 25 40 13 10 65 27 10 65 27 1 18 8 Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ Determinants and Matrices 253 x 1 1 7 9 3 0 2 5 Example 47. By condensing the given determinant evaluate x, 2 x 2 6 8 3 2 1 1 0 Solution. D x 1 1 7 9 3 0 2 5 2x 2 6 8 3 2 1 1 0 7 14 7 0 x 1 1 0. 9 3 2 5 6 2x 2 8 3 1 2 1 0 35 1 2 5 1 1 2 14 x 14 6 x 6 56 54 21 18 2 7 8 x 20 2 3 7 7 14 x 1 7 9 03 x 13 2 3 1 1 2 8 x 20 1 7 x 13 1 5 3 3 2 8 x 19 40 x 97 5 x 62 7 x 12 2 1 8 x 20 3 40 x 100 3 5 x 65 7 1 x 13 2 8 x 19 (5 x 62) (40 x 97) ( x 12) 7 2 40 x 2 95 x 496 x 1178 40 x 2 97 x 480 x 1164 7 Thus 4x + 4 = 0 x+1=0 2 [14 x 14] 4 x 4 7 x = –1 Ans. EXERCISE 4.6 Using the leading element as pivots, condense the following determinants to second order and hence evaluate them. 1 2 1 3 1 3 7 2 0 2 5 2 7 3 4 2 5 4 9 1 3 7 4 9 1 10 1. 2. 3. 4. 6 1 7 1 2 7 6 2 5 1 2 3 4 4 3 9 2 Ans. 51 Ans. 52 Ans. – 39 Ans. 75 1 2 3 4 5 4 2 3 0 3 7 2 1 1 1 0 2 7 9 4 1 2 3 5. 5 1 6 1 Ans. –1334 6. 8 1 3 7 2 Ans. –2276 2 3 5 4 4 2 0 3 1 7. Condense the following determinant and hence evaluate x, 3 2 1 5 4 7 6 2 0. 2 1 x 1 4 Ans. x = 4 5 3 4 1 4.10 CONJUGATE ELEMENTS Two equidistant elements lying on a line perpendicular to the leading diagonal are said to be conjugate. Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ 254 Determinants and Matrices a1 In the determinant b1 c1 a2 b2 c2 , a3 c3 b3 a2 , b1 ; a3 , c1 ; b3 , c2 ; are pairs of conjugate elements. 4.11 SPECIAL TYPES OF DETERMINANTS (i) Orthosymmetric Determinant. If every element of the leading diagonal is the same and the conjugate elements are equal, then the determinant is said to be orthosymmetric determinant. a h g h a f g f a (ii) Skew-Symmetric Determinant. If the elements of the leading diagonal are all zero and every other element is equal to its conjugate with sign changed, the determinant is said to be Skewsymmetric. 0 a b a 0 c b c 0 Property 1. A Skew-symmetric determinant of odd order vanishes. 0 –a –b Example 48. Prove that Δ = a 0 – c = 0 b c 0 Solution. Taking out (– 1) common from each of the three columns 0 a b 3 (1) a 0 c b c 0 0 a b Changing rows into columns (1)3 a 0 c (1)3 b c 0 or 2 = 0 or = 0 Property 2. A skew-symmetric determinant of even order is a perfect square. 0 x Example 49. Prove that x 0 y c z b y c 0 a z b a 0 Proved. (ax by cz )2 0 ax y z 0 c b 1 x Solution. Multiplying column 2 by a the given determinant is y ac 0 a a On expanding by column 2, we get z ab a 0 x c b (ax by cz ) y 0 a a z a 0 (ax by cz ) a x c b y 0 a z a 0 Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ Determinants and Matrices ax ac ab (ax by cz ) aa 255 (ax by cz ) y 0 a z a 0 a2 ax by cz 0 0 y 0 a a2 z ax by cz ac ac ab ab (ax by cz ) y 0 a z a 0 R1 R1 bR2 cR3 ( ax by cz ) a2 a 0 ( ax by cz ) ( a 2 ) = (ax – by + cz)2 Proved. 4.12 LAPLACE METHOD FOR THE EXPANSION OF A DETERMINANT IN TERMS OF FIRST TWO ROWS (i) Make all possible determinants from first two rows by taking any two columns. (ii) Multiply each of them by corresponding determinant which is left by suppressing the rows and columns intersecting at them. (iii) Add them with proper signs. Here we count the number of movements of columns of the determinant by shifting to the place of the first determinant. If the number of movement is odd then negative sign, if even then positive sign. a1 b1 c1 d1 a2 b2 c2 d 2 Example 50. Expand the determinant a b c d by Laplace method. 3 3 3 3 a4 b4 c4 d 4 Solution . a1 a2 b1 c3 b2 b1 d1 a3 c3 b2 d2 a4 c4 d3 c4 d4 a1 a2 c1 b3 c2 b4 c1 d1 a3 b3 c2 d 2 a4 b4 d3 d4 a1 d1 b3 c3 a2 d2 b4 c4 a1 Explanation : a 2 b1 c1 a3 d3 b2 c 2 a4 d4 c1 c2 Now the c column being 3rd can be made 2nd by one movement of column; “a” column is in the position of first column so that the total number of movements is one i.e. odd; hence the sign will be –ve. Ans. a1 b1 0 0 a2 b2 0 0 Example 51. Expand the following determinant by Laplace method : a3 b3 c3 d3 a4 b4 c4 d 4 a1 b1 c3 d3 a1 0 b3 d 3 Solution . a b c d a 0 b d 2 2 4 4 2 4 4 a1 0 b3 c3 a2 0 b4 c4 a1 b1 c3 d3 a2 b2 c4 d4 b1 0 a3 d3 b2 0 a4 d4 b1 0 a3 c3 b2 0 a4 c4 ( a1b2 2 b1 ) (c3 d 4 c4 d 3 ) 0 0 a3 b3 0 0 a4 b4 Ans. Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ 256 Determinants and Matrices 4.13 APPLICATION OF DETERMINANTS Area of triangle. We know that the area of a triangle, whose vertices are (x1, y1), (x2, y2) and (x3, y3) is given by 1 x ( y y3 ) x2 ( y1 y3 ) x3 ( y1 y2 ) 2 1 2 x1 y2 1 y1 1 y1 1 1 1 x1 x2 x3 2 x2 2 y3 1 y3 1 y2 1 x3 y1 1 y2 1 y3 1 Note. Since area is always a positive quantity, therefore we always take the absolute value of the determinant for the area. Condition of collinearity of three points. Let A (x1, y1), B (x2, y2) and C (x3, y3) be three points. Then, A, B, C are collinear area of triangle ABC 0 x1 y1 1 x1 y1 1 1 x2 y2 1 0 x2 y2 1 0 Proved. 2 x3 y3 1 x3 y3 1 Example 52. Using determinants, find the area of the triangle with vertices (– 3, 5), (3, – 6) and (7, 2). 1 Solution. The area of the given triangle 2 3 5 1 3 6 1 7 2 1 6 11 0 1 4 8 0 Operate : R1 R1 R2 ; R2 R2 R3 2 7 2 1 6 11 1 1 Expand by C3 2 .1. 4 8 2 (48 44) 46 sq. units Ans. Example 53. Using determinants, show that the points (11, 7), (5, 5) and (– 1, 3) are collinear. Solution. The area of the triangle formed by the given points 1 2 11 7 1 5 5 1 1 3 1 Operate : R1 R1 R2 ; R2 R2 R3 1 2 6 2 0 6 2 0 1 .0 0. 2 1 3 1 ( R1 and R2 are identical) The three given points are collinear. Proved. Example 54. Using determinants, find the area of the triangle whose vertices are (1, 4) (2, 3) and (– 5, – 3). Are the given points collinear? Solution. Area of the required triangle 1 2 1 4 1 2 3 1 5 3 1 Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ Determinants and Matrices 257 1 1 13 [1(3 3) 4 (2 5) 1( 6 15)] (6 28 9) 0 2 2 2 Hence, the given points are not collinear. Ans. EXERCISE 4.7 Using determinants, find the area of the triangle with vertices: 1. (2,– 7), (1, 3), (10, 8). Ans. Area 95 2 2. (– 2, 4), (2, – 6) and (5, 4). Ans. Area = 35 3. (– 1, – 3), (2, 4) and (3, – 1). Ans. Area = 11 4. (1, – 1), (2, 4) and (– 3, 5). Ans. Area = 13 5. Using determinants, show that the points (3, 8), (– 4, 2) and (10, 14) are collinear. 6. Find the value of , so that the points (1, – 5), (– 4, 5) and (, 7) are collinear. Ans. = –5 7. Find the value of x, if the area of is 35 square cms with vertices (x, 4), (2, – 6), (5, 4). Ans. x = – 2, 12 8. Using determinants find the value of k, so that the points (k, 2 – 2k), (– k + 1, 2k) and (– 4 – k, 6 – 2 k) may be collinear. 1 2 Ans.x = 3 Ans. x = 1 Ans. k = –1, 9. If the points (x, – 2), (5, 2) and (8, 8) are collinear, find x using determinants. 10. If the points (3, – 2), (x, 2) and (8, 8) are collinear, find x using determinants. 4.14. SOLUTION OF SIMULTANEOUS LINEAR EQUATIONS BY DETERMINANTS (CRAMER’S RULE) Let us solve the following equations. a1 x + b1 y + c1 z = d1 a2 x + b2 y + c2 z = d2 Let a3 x + b3 y + c3 z = d3 a1 b1 c1 D a2 b2 c2 b1 c1 or x D a2 x b2 a1 x c2 a3 b3 c3 a3 x b3 c3 Multiplying the 2nd column by y and 3rd column by z and adding to the 1st column, we get b1 c1 x D a2 x b2 y c2 z b2 a1x b1 y c1 z c2 a3 x b3 y c3 z b3 c3 d1 d1 b1 c1 d 2 b2 c2 x d3 a1 b1 b1 c3 D 1 D c1 a2 b2 c2 a3 b3 c3 b1 c1 x D d2 b2 c2 d3 b3 c3 Similarly, y D2 D a1 d1 c1 a2 a3 d 2 c2 d3 c3 a1 b1 c1 a2 b2 c2 a3 b3 c3 Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ 258 Determinants and Matrices a1 b1 d1 a2 b2 d2 a b d3 D3 3 3 D a1 b1 c1 z a2 b2 c2 a3 b3 c3 D1 D D , y 2, z 3 D D D Example 55. Solve the following system of equations using Cramer’s rule : 5x – 7y + z = 11 6x – 8y – z = 15 x Ans. 3x + 2y – 6z = 7 Solution. The given equations are 5x 7 y z 11 6x 8 y z 15 3x 2 y 6 z 7 5 7 Here, D 6 8 1 1 = 5 (48 + 2) + 7 (– 36 + 3) + 1 (12 + 24) = 55 ( 0) 3 2 6 11 7 1 D1 15 8 1 7 2 6 5 11 1 D2 6 15 1 = 5 (– 90 + 7) – 11 (– 36 + 3) + 1 (42 – 45) = – 55 3 7 6 5 7 11 D3 6 8 15 3 2 = 11 (48 + 2) + 7 (– 90 + 7) + 1 (30 + 56) = 55 = 5 (– 56 – 30) + 7 (42 – 45) + 11 (12 + 24) = – 55 7 D1 55 D 55 1, y 2 1 , D 55 D 55 y = – 1, z=–1 By Cramer’s Rule x Hence, x = 1, z D3 55 1 D 55 Ans. Example 56. Solve, by determinants, the following set of simultaneous equations : 5x – 6y + 4z = 15 7x + 4y – 3z = 19 2x + y + 6z = 46 5 6 Solution. D 7 2 4 4 3 419 1 6 Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ Determinants and Matrices 15 6 D1 19 46 259 5 15 4 Hence, x = 3, 2 46 6 By Cramer’s Rule: x D1 1257 3, D 419 y = 4, 5 6 15 D2 7 19 3 1676 4 3 1257 , 1 4 y 6 , D3 7 4 19 2514 2 D2 1676 4, D 419 z 1 46 D3 2514 6. D 419 z=6 Ans. Example 57. Solve the following system of equations using Cramer’s Rule : 2x – 3y + 4z = – 9 – 3x + 4y + 2z = – 12 4x – 2y – 3z = – 3 Solution. The given equations are 2x 3y 4z 9 3x 4 y 2z 12 4 x 2 y 3z 3 Here 2 3 D 3 4 4 2 =2 (– 12 + 4) + 3 (9 – 8) + 4 (6 – 16) = – 53 4 2 3 9 3 4 D1 12 4 2 3 2 3 2 9 4 D2 3 12 2 4 2 D3 3 3 3 3 =2 (36 + 6) + 9 (9 – 8) + 4 (9 + 48) = – 321 9 4 12 4 2 =– 9 (– 12 + 4) + 3 (36 + 6) + 4 (24 + 12) = – 342 3 By Cramer’s Rule, D 342 342 x 1 , D 53 53 =2 (– 12 – 24) + 3 (9 + 48) – 9 (6 – 16) = – 189 y D2 321 321 , D 53 53 z D3 189 189 D 53 53 342 y 321 , z 189 Hence, x 53 , 53 53 Example 58. Solve the following system of equations by using determinants : Ans. x+ y+z=1 ax + by + cz = k 2 2 2 a x+b y+c z=k Solution. We have D 1 1 1 a b c 2 2 a b c 2 2 Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ 260 Determinants and Matrices 1 a a 2 0 ba 2 b a 2 0 c a [Applying C2 C2 C1 and C3 C3 C1] 2 c a (b a) (c a) 2 1 0 0 a 1 1 a 2 (b a) (c a).1. ba ca 1 1 ba ca = (b – a) (c – a) (c + a – b – a) = (b – c) (c – a) (a – b) D1 k D2 and 1 1 k b c (b c) (c k ) (k b) 2 2 2 b c 1 1 1 a k c (k c) (c a) (a k ) 2 2 2 a D3 1 k c 1 1 1 a b k (a b) (b k ) (k a) 2 2 2 a b k [Expanding along R1] ...(1) [Replacing a by k in (1)] [Replacing b by k in (1)] [Replacing c by k in (1)] D1 (b c) (c k ) (k b) D (k c) (c a) (a k ) , y 2 D (b c) (c a) (a b) D (b c) (c a) (a b) D (a b) (b k ) (k a) z 3 and D (a b) (b c) (c a) (c k ) (k b) (k c) (a k ) (b k ) (k a) x , y and z Hence, (c a) (a b) (b c) (a b) (b c) (c a) Ans. Example 59. The sum of three numbers is 6. If we multiply the third number by 2 and add the first number to the result, we get 7. By adding second and third numbers to three times the first number we get 12. Use determinants to find the numbers. Solution. Let the three numbers be x, y and z. Then, from the given conditions, we have x y z 6 x yz6 x 2z 7 or x 0. y 2 z 7 3x y z 12 3x y z 12 x Here, 1 1 1 D 1 0 2 1(0 2) 1(1 6) 1(1 0) 2 5 1 4 3 1 1 6 1 1 D1 7 0 2 6 (0 2) 1(7 24) (7 0) = – 12 + 17 + 7 = 12 1 1 12 Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ Determinants and Matrices 1 6 261 1 D2 1 7 2 1(7 24) 6 (1 6) 1(12 21) = – 17 + 30 – 9 = 4 3 12 1 1 1 6 7 1(0 7) 1(12 21) 6(1 0) = – 7 + 9 + 6 = 8 3 1 12 D D D x 1 12 3, y 2 4 1, and z 3 8 2 D 4 D 4 D 4 Thus, the three numbers are 3, 1 and 2. and D3 1 0 Ans. EXERCISE 4.8 Using Cramer’s Rule, solve the following system of equations : 1. 2x – 3y = 7 7x – 3y = 10 4. 7. 10. 13. 2. 2x + y = 1 x – 2y = 8 3 29 Ans. x 5 , y 15 5x + 2y = 3 5. 3x + 2y = 5. Ans. x = – 1, y = 4 x – 4y – z = 11 8. 2x – 5y + 2z = 39 – 3x + 2y + z = 1. Ans. x = – 1, y = – 5, z = 8 x+y+z=1 11. 3x + 5y + 6z = 4 9x + 2y – 36x = 17 1 1 Ans. x 3 , y 1, z 3 x+y+z=1 x + 2y + 3z = k 3. 2x + 3y = 10 x + 6y = 4. 16 2 Ans. x 3 , y 9 7x – 2y = – 7 6. x – 2y = 4 2x – y = 1. – 3x + 5y = – 7 Ans. x = – 3, y = – 7 Ans. x = – 6, y = – 5 x + 3y – 2z = 5 9. x + 2y + 5z = 23 2x + y + 4z = 8 3x + y + 4z = 26 6x + y – 3z = 5. 6x + y + 7z = 47 Ans. x = 1, y = 2, z = 1 Ans. x = 4, y = 2, z = 3 2y – z = 0 12. x + y + z = – 1 x + 3y = – 4 x + 2y + 3z = – 4 3x + 4y = 3 x + 3y + 4z = – 6 Ans. x 2, y 3 Ans. x = 5, y = – 3, z = – 6 Ans. x = 1, y = – 1, z = – 1 (2 k ) (3 k ) (1 k ) (3 k ) (1 k ) (2 k ) , y , z 2 1 2 14. Show that there are three real values of for which the equations: (a – ) x + by + cz = 0 bx + (c – ) y + az = 0 cx + ay + (b – ) z = 0 a b c 12x + 22y + 32z = k2 Ans. x are simultaneously true, and that the product of these values of is b c a c a b 15. Solve the following system of equations by using the Cramer’s Rule x1 + x2 = 1; x2 + x3 = 0; x3 + x4 = 0; x4 + x5 = 0; x5 + x1 = 0 (A.M.I.E.T.E., Summer 2005) 1 1 1 1 1 Ans. x1 , x2 , x3 , x4 , x5 2 2 2 2 2 Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ 262 Determinants and Matrices 4.15 RULE FOR MULTIPLICATION OF TWO DETERMINANTS Multiply the elements of the first row of 1 with the corresponding elements of the first, the second and the third row of 2 respectively. Their respective sums form the elements of the first row of 12. Similarly multiply the elements of the second row of 1 with the corresponding elements of first, second and third row of 2 to form the second row of 12 and so on. a Example 60. Find the product c c b a b a c b Solution. Product of the given determinants a11 b11 c11 a1 2 b12 c1 2 a1 3 b13 c1 3 a21 b21 c2 1 a2 2 b22 c2 2 a2 3 b23 c2 3 a31 b31 c31 a3 2 b32 c3 2 a b c Example 61. Show that –a Ans. a33 b33 c3 3 c b b c a × –b a c c a b –c b a 2bc a c 2 2 c2 b2 2ca b b2 2 a2 a 2 3 3 3 (a b c 3abc) 2 2ab c 2 Solution. Product of the given determinants a b c a c b a 2 bc bc ab ab c 2 2 ac b2 ac 2 b c a b a c ab c ab b ac ac c a b c b a ca ca b2 c 2 b 2 Now, b a c (1) 2 a a b c 2 c b a c a b b2 2 a 2 2ab c ... (1) 2 b c a c = a (bc – a b c b c a c2 2ca b a c b 2 bc a 2 bc c 2 ab ab 2bc a 2 bc bc a a2) a b – b (b2 – ac) + c (ab – c2) = – (a3 + b3 + c3 – 3abc) a c b b a c = (a3 + b3 + c3 – 3abc)2 c b a From (1) and (2), we get the required result. ... (2) Proved. Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ Determinants and Matrices 263 Example 62. Prove that the determinant 2b1 + c1 c1 + 3a1 2a1 + 3b1 2b2 + c2 c2 + 3a2 2a2 + 3b2 2b3 + c3 is a multiple of the determinant c3 + 3a3 2a3 + 3b3 a1 b1 c1 a2 b2 c2 a3 b3 2b1 c1 Solution. c1 3a1 2a1 3b1 2b2 c2 c2 3a2 2a2 3b2 2b3 c3 2a3 3b3 a1 c3 3a3 b1 c1 a2 b2 0 2 1 c2 3 0 1 a3 b3 Example 63. Prove that and find the other factor.. c3 c3 cos cos cos cos cos cos cos sin 0 Solution. cos sin 0 sin 0 cos sin 0 0 cos sin sin 0 0 cos cos cos sin sin cos cos sin sin 2 2 cos cos sin sin cos sin cos cos sin sin cos cos sin sin 1 or cos cos 2 sin 2 or Ans. 2 3 0 cos cos sin sin 0 cos2 sin 2 cos ( ) cos ( ) cos ( ) 1 cos ( ) 0 Proved. cos ( ) cos ( ) 1 Example 64. If A1, A2, A3, B1, B2, B3, C1, C2, C3 are cofactors of the elements a1, a2, a3, b1, b2, a1 b1 c1 b3, c1, c2, c3 respectively of the determinant a2 b2 c2 a3 b3 c3 , show that A1 A2 B1 C1 a1 b1 B2 C2 = a2 b2 c1 c2 A3 B3 C3 c3 a3 b3 2 (Try Yourself) 4.16 CONDITION FOR CONSISTENCY OF A SYSTEM OF SIMULTANEOUS HOMOGENEOUS EQUATIONS Case I For a system of homogeneous equations. Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ 264 Determinants and Matrices a11 x a12 y a13 z 0 a11 a21 x a22 y a23 z 0 a31 x a32 y a33 z 0 , a12 a13 D a21 a22 a23 a31 a32 a33 1. If D 0, then the system of equations are consistent with infinite solutions. 2. If D 0, then the system of equations is consistent with trivial solution. Example 65. Find values of , for which the following system of equations is consistent and has nontrivial solutions: (– 1) x + (3 + 1) y + 2z = 0 (– 1) x + (4 – 2) y + (+ 3) z = 0 2x + (3 + 1) y + 3 (– 1) z = 0 Solution. (– 1) x + (3+ 1) y + (– 1) x + (4 – 2) y + 2x + (3 + 1) y + This is a system of homogeneous equations. 2z=0 (+ 3) z = 0 3 (– 1) z = 0 For infinite solutions, 1 3 1 D 1 4 2 2 0 0 3 3 3 0 3 1 3 3 0 1 5 1 2 3 0 3 1 3 3 1 4 2 2 2 [ R1 R1 R2 ] 3 3 0 6 2 3 3 [C2 C2 + C3) (– 3) [(– 1) (6 – 2) – 2 (5 + 1)] = 0 [6 2 – 8 + 2 – 10 – 2] = 0 6 2 – 18 = 0 6 (– 3) = 0 = 3, 0 Ans. 4.17 FOR A SYSTEM OF THREE SIMULTANEOUS LINEAR EQUATIONS WITH THREE UNKNOWNS (i) If D 0, then the given system of equations is consistent and has a unique solution given D D D by x 1 , y 2 , and z 3 . D D D (ii) If D = 0 and D1 = D2 = D3 = 0, then the given system of equations is consistent, and it has infinitely many solutions. (iii) If D = 0 and at least one of the determinants D1, D2, D3 is non zero, then the given system of equations is inconsistent. Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ Determinants and Matrices 265 Equations with three unknowns D0 Consistent with unique solution D=0 D1 = D2 = D3 = 0 Consistent with infinitely many solutions D1 or D2 or D3 0 Inconsistent Example 66. Test the consistency of the following equations and solve them if possible: 3x + 3y + 2z = 1, x + 2y = 4, 10y + 3z = – 2 Solution. The system of equations is 3x + 3y + 2z = 1 x + 2y + 0z = 4 0x + 10y + 3z = – 2 Therefore 3 3 2 D 1 2 0 0 10 3 = 3 (6 – 0) – 3 (3 – 0) + 2 (10 – 0) = 18 – 9 + 20 = 29 0 Since D 0, so the system of simultaneous equations is consistent with unique solution. Now let us solve the system of equation. D1 1 3 2 4 2 0 1(6 0) 3(12 0) 2(40 4) 2 10 3 6 36 88 58 3 D2 1 1 2 4 0 3(12 0) 1(3 0) 2 ( 2 0) 0 2 3 36 3 4 29 3 D3 1 3 2 1 4 3( 4 40) 3( 2 0) 1 (10 0) 0 10 2 132 6 10 116 By Cramer’s Rule D1 58 D D 116 2, y 2 29 1, z 3 4 D 29 D 29 D 29 Hence x = 2, y = 1, z = – 4. x Ans. Example 67. Show that the system of equations 2x + 6y = – 11, 6x + 20y – 6z = – 3, 6y – 18z = –1 is not consistent. Solution. 2x + 6y + 0z = – 11 6x + 20y – 6z = – 3 0x + 6y – 18z = – 1 Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ 266 Determinants and Matrices 2 6 0 D 6 20 6 0 11 D1 =2 (– 360 + 36) – 6 (– 108) = – 648 + 648 = 0 6 18 6 0 3 20 6 1 6 18 = – 11 (– 360 + 36) – 6 (54 – 6) = 3564 – 288 = 3276 Here D = 0 and D1 0 Hence the system of equations is not consistent. Proved EXERCISE 4.9 Find, whether the following system of equations is consistent or inconsistent. If consistent solve them. 1. Find the value of k, for which the following system of equations 3x1 – 2x2 + 2x3 = 3, x1 + kx2 – 3x3 = 0, 4x1 + x2 + 2x3 = 7 is consistent. 1 Ans. k , 4 2. Find the value of , for which the system of equations 7 x + y + 4z = 1, x + 2y – 2z = 1, x + y + z = 1 will have a unique solution. Ans. 10 3. For what values of and , the following system of equations 2x + 3y + 5z = 9, 7x + 3y – 2z = 8, 2x + 3y + z = will have (i) unique solution; (ii) no solution. Ans. (i) 5 (ii) 5, 9 3 2 1 x b 5 8 9 y 3 4. Determine the values of a and b for which the system 2 1 a z 1 (i) has a unique solution, (ii) has no solution and (iii) has infinitely many solutions. 1 1 Ans. (i) a 3 (ii) a 3, b 3 (iii) a 3, b 3 5. Find the condition on for which the system of equations 3x – y + 4z = 3, x + 2y – 3z = – 2, 6x + 5y + z = – 3 has a unique solution. Find the solution for = – 5. 5k 4 13k 9 Ans. 5, x 7 7 , y 7 7 , z k EXERCISE OF OBJECTIVE QUESTIONS Choose the Correct Answers : 1. (a) 52 52 53 54 The value of 53 54 55 54 56 57 (b) 0 is (c) 513 (d) 59 Ans. (b) Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ Determinants and Matrices a1 b1 c1 2. If a2 b2 a3 b3 c2 b2 c3 b3c2 = 5 then the value of b3c1 b1c3 b1c2 b2 c1 c3 (a) 5 1 ax 267 (b) 25 1 bx a3c2 a2 c3 a1c3 a3c1 a2 c1 a1c2 (c) 125 a2 b3 a3b2 a1b1 a1b3 is a1b2 a2 b1 (d) 0 Ans. (b) 1 cx 3. If 1 a1x 1 b1x 1 c1x = A0 A1x A2 x2 A3 x3 , then A1 is equal to 1 a2 x 1 b2 x 1 c2 x (a) abc (b) 0 (c) 1 (d) none of these 1 n 2n 4. If 1, , 2 are the cube roots of unity, then n 2 n 1 2n (a) 0 (c) (b) 1 xp y 5. The determinant yp z 0 (a) x, y, z are in A.P. (c) x, y, z are in H.P. 6. If the determinant 1 x y y z Ans. (b) is equal to n (d) 2 Ans. (a) = 0 if xp y yp z (b) x, y, z are in G.P. (d) xy, yz, zx are in A.P. a b 2a 3b b c 2b 3c 0, then 2a 3b 2b 3c Ans. (b) 0 (b) is root of 4ax 2 12bx 9c 0 or a, b, c are in G.P. (d) a, b, c are in A.P. Ans. (a) (a) a, b, c are in H.P. (c) a, b, c are in G.P. only log l p 1 7. If l, m, n are the p , q and r term of a G.P. all positive, then log m q 1 equals th th th log n (a) 3 (b) 2 8. If the system of linear equations x 2ay az 0 r 1 (c) 1 (d) zero Ans. (d) (c) are in H.P. (d) satisfy a 2b 3c 0 Ans. (c) x 3by bz 0 x 4cy cz 0 has a non-zero solution, then a, b, c (a) are in A.P. (b) are in G.P. 9. If , and are the roots of the equation x 3 px q 0 , then the value of the Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ 268 Determinants and Matrices determinant is (a) q (b) 0 (d) p2 2q . (c) p 10. The number of values of k for which the system of equations (k 1) x 8y 4k , kx (k 3) y 3k 1 has infinitely many solutions is (a) 0 (b) 1 (c) 2 Ans. (a) (d) infinite Ans. (b) [Hint : Here 0 for k = 3, 1, x 0 for k = 2, 1, y 0 for k = 1. Hence k = 1. Alternatively, for infinitely many solutions the two equations become identical k 1 8 4k k k 3 3k 1 k = 1] 11. The system x y 0 , y z 0 , z x 0 has infinitely many solutions when (a) 1 (b) 1 (c) 0 (d) no real value of 1 Ans. (b) [Hint : 0 0 1 0 . Solve for .] 0 1 12. If the system of equations x ky z 0 , kx y z 0 , x y z 0 has a non-zero solution, then the possible values of k are (a) – 1, 2 (b) 1, 2 (c) 0, 1 (d) – 1, 1 Ans. (d) 13. The value of for which the system of equations 2 x y 2z 2 , x 2 y z 4 , x y z 4 has no solution is (a) 3 (b) – 3 (c) 2 (d) – 2 2 1 2 1 2 1 Ans. (b) [Hint : The Coefficient determinant = 3 9 . 1 1 For no solution the necessary condition is –3 – 9 = 0 = –3 For = – 3, there is no solution for the given system of equations]. 14. If the system of equations x 2 y 3z 1 , ( 3) z 3 , (2 1) x z 0 is inconsistent, then the value of is equal to 1 (a) (b) – 3 (c) 2 (d) 0 Ans. (b) 2 a 0 1 a 1 1 15. A 1 c b , B 0 1 d b f f a2 c d , U g , V 0 . If there is a vector matrix X such that h g h 0 AX = U has infinitely many solutions, then prove that BX = V cannot have a unique solution. If a f d 0 , prove that BX = V has no solution. Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ Determinants and Matrices 269 4.18 MATRICES Let us consider a set of simultaneous equations, x+2y+3z+5t=0 4x+ 2y+ 5z+ 7 t= 0 3 x + 4 y + 2 z + 6 t = 0. Now we write down the coefficients of x, y, z, t of the above equations and enclose them within brackets and then we get 1 2 3 5 A = 4 2 5 7 3 4 2 6 The above system of numbers, arranged in a rectangular array in rows and columns and bounded by the brackets, is called a matrix. It has got 3 rows and 4 columns and in all 3 × 4 = 12 elements. It is termed as 3 × 4 matrix, to be read as [3 by 4 matrix]. In the double subscripts of an element, the first subscript determines the row and the second subscript determines the column in which the element lies, aij lies in the ith row and jth column. 4.19 VARIOUS TYPES OF MATRICES (a) Row Matrix. If a matrix has only one row and any number of columns, it is called a Row matrix, e.g., [2 7 3 9] (b) Column Matrix. A matrix, having one column and any number of rows, is called a Column 1 matrix, e.g., 2 3 (c) Null Matrix or Zero Matrix. Any matrix, in which all the elements are zeros, is called a Zero matrix or Null matrix e.g., 0 0 0 0 0 0 0 0 (d) Square Matrix. A matrix, in which the number of rows is equal to the number of columns, is called a square matrix e.g., 2 5 1 4 (e) Diagonal Matrix. A square matrix is called a diagonal matrix, if all its non-diagonal elements are zero e.g., 1 0 0 0 3 0 0 0 4 (f ) Scalar matrix. A diagonal matrix in which all the diagonal elements are equal to a scalar, say (k) is called a scalar matrix. For example; 0 0 0 6 2 0 0 0 0 0 2 0 , 0 6 0 0 6 0 0 0 2 0 0 6 0 Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ 270 Determinants and Matrices 0, when i j i.e., A = [aij]n × n is a scalar matrix if aij = k , when i j (g) Unit or Identity Matrix. A square matrix is called a unit matrix if all the diagonal elements are unity and non-diagonal elements are zero e.g., 1 0 0 0 1 0 , 1 0 0 1 0 0 1 (h) Symmetric Matrix. A square matrix will be called symmetric, if for all values of i and j, aij = aji i.e., A = A a h g h b f e.g., g f c (i) Skew Symmetric Matrix. A square matrix is called skew symmetric matrix, if (1) aij = – aji for all values of i and j, or A = –A (2) All diagonal elements are zero, e.g., 0 h g h 0 f g f 0 (j) Triangular Matrix. (Echelon form) A square matrix, all of whose elements below the leading diagonal are zero, is called an upper triangular matrix. A square matrix, all of whose elements above the leading diagonal are zero, is called a lower triangular matrix e.g., 1 3 2 2 0 0 0 4 1 4 1 0 0 0 6 5 6 7 Upper triangular matrix Lower triangular matrix (k) Transpose of a Matrix. If in a given matrix A, we interchange the rows and the corresponding columns, the new matrix obtained is called the transpose of the matrix A and is denoted by A or AT e.g., 2 3 4 2 1 6 A = 1 0 5 , A 3 0 7 6 7 8 4 5 8 (l) Orthogonal Matrix. A square matrix A is called an orthogonal matrix if the product of the matrix A and the transpose matrix A’ is an identity matrix e.g., A. A = I if | A | = 1, matrix A is proper. (m) Conjugate of a Matrix Let 4 1 i 2 3i A= i 3 2 i 7 2 i Conjugate of matrix A is A 4 1 i 2 3i A = i 3 2 i 7 2 i (n) Matrix A. Transpose of the conjugate of a matrix A is denoted by A. Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ Determinants and Matrices 271 4 1 i 2 3i A= i 3 2 i 7 2 i 4 1 i 2 3i A = 7 2 i i 3 2 i 1 i 7 2 i 2 3 i i ( A ) = 4 3 2 i 1 i 7 2 i 2 3 i i A = 4 3 2 i (o) Unitary Matrix. A square matrix A is said to be unitary if A A = I Let 1 i 1 i 1 i 1 i 2 2 2 2 A e.g. A= , , A A I 1 i 1 i 1 i 1 i 2 2 2 2 (p) Hermitian Matrix. A square matrix A = (aij) is called Hermitian matrix, if every i-jth element of A is equal to conjugate complex j-ith element of A. In other words, aij = a ji 2 3i 3 i 1 2 3 i 2 1 2 i 3 i 1 2 i 5 e.g. Necessary and sufficient condition for a matrix A to be Hermitian is that A = A i.e. conjugate transpose of A A = ( A) . (q) Skew Hermitian Matrix. A square matrix A = (aij) will be called a Skew Hermitian matrix if every i-jth element of A is equal to negative conjugate complex of j-ith element of A. In other words, aij = a j i All the elements in the principal diagonal will be of the form aii = aii or aii aii 0 If aii = a + ib then aii a ib (a + ib) + (a – ib) = 0 2a=0a=0 So, aii is pure imaginary aii = 0. Hence, all the diagonal elements of a Skew Hermitian Matrix are either zeros or pure imaginary. i 2 3 i 4 5 i (2 3 i ) 0 2i e.g. 2i 3 i (4 5 i ) The necessary and sufficient condition for a matrix A to be Skew Hermitian is that A = – A ( A ) = – A Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ 272 Determinants and Matrices A2 (r) Idempotent Matrix. A matrix, such that = A is called Idempotent Matrix. 2 2 4 2 2 4 2 2 4 2 2 4 1 3 4 , A2 1 3 4 1 3 4 1 3 4 A e.g. A = 1 2 3 1 2 3 1 2 3 1 2 3 (s) Periodic Matrix. A matrix A will be called a Periodic Matrix, if Ak+1 = A where k is a +ve integer. If k is the least + ve integer, for which Ak+1 = A, then k is said to be the period of A. If we choose k = 1, we get A2 = A and we call it to be idempotent matrix. (t) Nilpotent Matrix. A matrix will be called a Nilpotent matrix, if Ak = 0 (null matrix) where k is a +ve integer ; if however k is the least +ve integer for which Ak = 0, then k is the index of the nilpotent matrix. ab b 2 2 ab b 2 ab b 2 0 0 , A e.g., A = 0 2 2 2 a ab a ab a ab 0 0 A is nilpotent matrix whose index is 2. (u) Involuntary Matrix. A matrix A will be called an Involuntary matrix, if A2 = I (unit matrix). Since I2 = I always Unit matrix is involuntary. (v) Equal Matrices. Two matrices are said to be equal if (i) They are of the same order. (ii) The elements in the corresponding positions are equal. 2 3 2 3 A = 1 4 , B 1 4 Here A=B (w) Singular Matrix. If the determinant of the matrix is zero, then the matrix is known as Thus if 1 2 singular matrix e.g. A = is singular matrix, because |A| = 6 – 6 = 0. 3 6 Example 1. Find the values of x, y, z and ‘a’ which satisfy the matrix equation. x 3 2 y x 0 7 z 1 4 a 6 3 2a Solution. As the given matrices are equal, so their corresponding elements are equal. x+3=0 x=–3 2y + x = – 7 z– 1=3 z=4 4 a – 6= 2 a a=3 Putting the value of x = – 3 from (1) into (2), we have 2y – 3 = – 7 y=–2 Hence, x = – 3, y = – 2, z = 4, a = 3 ...(1) ...(2) ...(3) ...(4) Ans. 4.20 ADDITION OF MATRICES If A and B be two matrices of the same order, then their sum, A + B is defined as the matrix, each element of which is the sum of the corresponding elements of A and B. 5 4 2 1 0 2 Thus if A = 1 3 6 , B 3 1 4 Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ Determinants and Matrices 273 5 2 5 2 7 4 1 2 0 A + B = 1 3 3 1 6 4 4 4 2 If A = [aij], B = [bij] then A + B = [aij + bij] Example 2. Show that any square matrix can be expressed as the sum of two matrices, one symmetric and the other anti-symmetric. Solution. Let A be a given square matrix. 1 1 Then A = ( A A ) ( A A ) 2 2 Now, (A + A) = A + A = A + A. A + A is a symmetric matrix. Also, (A – A) = A – A = –(A – A) 1 A – A or (A – A ) is an anti-symmetric matrix. 2 1 1 A = ( A + A) + ( A – A) 2 2 Square matrix = Symmetric matrix + Anti-symmetric matrix Proved. Example 3. Write matrix A given below as the sum of a symmetric and a skew symmetric matrix. 1 2 4 A = 2 5 3 1 6 3 then 1 2 4 Solution. A = 2 5 3 On transposing, we get A = 1 6 3 On adding A and A, we have 1 2 4 1 2 5 A + A = 2 5 3 2 1 6 3 4 3 1 2 1 2 5 6 4 3 3 1 2 0 3 6 0 10 9 3 3 9 6 ...(1) On subtracting A from A, we get 5 1 2 4 1 2 1 0 4 2 5 3 2 4 0 3 5 6 A – A = 1 6 3 4 3 3 5 3 0 On adding (1) and (2), we have ...(2) 5 2 0 3 0 4 0 10 9 4 0 3 2A= 0 3 9 6 5 3 3 5 1 0 2 0 2 2 9 3 A= 0 5 2 0 2 2 3 9 3 5 3 0 2 2 2 2 A = [Symmetric matrix] + [Skew symmetric matrix.] Ans. Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ 274 Determinants and Matrices 4.21 PROPERTIES OF MATRIX ADDITION Only matrices of the same order can be added or subtracted. (i) Commutative Law. A + B = B + A. (ii) Associative law. A + (B + C) = (A + B) + C. 4.22 SUBTRACTION OF MATRICES The difference of two matrices is a matrix, each element of which is obtained by subtracting the elements of the second matrix from the corresponding element of the first. A – B = [aij – bij] Thus 4.23 1 3 8 6 4 3 5 1 8 3 6 5 4 1 5 1 2 0 7 6 2 = 1 7 2 6 0 2 6 4 2 Ans. SCALAR MULTIPLE OF A MATRIX If a matrix is multiplied by a scalar quantity k, then each element is multiplied by k, i.e. 2 3 4 A = 4 5 6 6 7 9 2 3 4 3 2 3 3 3 4 6 9 12 3 A = 3 4 5 6 3 4 3 5 3 6 12 15 18 6 7 9 3 6 3 7 3 9 18 21 27 EXERCISE 4.10 1. (i) 1 7 1 If A = 2 3 4 , represent it as A = B + C where B is a symmetric 5 0 5 and C is a skew-symmetric matrix. 1 2 0 (b) Express 3 7 1 as a sum of symmetric and skew-symmetric matrix. 5 9 3 9 5 1 2 3 0 2 2 1 9 5 5 3 2 0 2 ( b) A Ans. (i) A 2 2 2 3 2 5 2 2 0 5 2 2. Matrices A and B are such that 5 2 7 5 5 1 5 0 2 2 2 1 5 0 4 2 5 3 4 0 2 1 2 1 2 3A–2B= and – 4 A + B = 4 3 2 1 0 1 1 2 Ans. A , B 4 1 2 1 Find A and B. x y x 3. Given 3 z w 1 Find x, y, z and w. 0 2 0 4. If A 1 0 3 , B 1 1 2 6 4 x y 2w z w 3 1 2 0 Find (i) 2 A + 3 B (ii) 3 A Ans. x = 2, y = 4, z = 1, w = 3 2 1 1 0 0 3 3 10 3 4 2 4 Ans. (i) 8 3 6 , (ii ) 5 4 9 2 2 13 3 3 6 – 4 B. Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ Determinants and Matrices 4.24 275 MULTIPLICATION The product of two matrices A and B is only possible if the number of columns in A is equal to the number of rows in B. Let A = [aij] be an m × n matrix and B = [bij] be an n × p matrix. Then the product AB of these matrices is an m × p matrix C = [cij] where cij = ai1 b1j + ai2 b2j + ai3 b3j + .... + ain bnj 4.25 (AB) = BA If A and B are two matrices conformal for product AB, then show that (AB) = BA, where dash represents transpose of a matrix. Solution. Let A = (aij) be an m × n matrix and B = (bij) be n × p matrix. Since AB is m × p matrix, (AB) is a p × m matrix. Further B is p × n matrix and A an n × m matrix and therefore B A is a p × m matrix. Then (AB) and B A are matrices of the same order. n Now the (j, i)th element of (AB) = (i, j)th element of (AB) = aik bkj ...(1) k 1 Also the jth row of B is b1j, b2j .... bnj, and ith column of A is ai1, ai2, ai3.... ain. n (j, i)th element of BA = bkj aik ...(2) k 1 From (1) and (2), we have (j, i)th element of (AB) = (j, i) th element of BA. As the matrices (AB) and BA are of the same order and their corresponding elements are equal, we have (AB) = BA. Proved. 4.26 PROPERTIES OF MATRIX MULTIPLICATION 1. Multiplication of matrices is not commutative. AB BA 2. Matrix multiplication is associative, if conformability is assured. A (BC) = (AB) C 3. Matrix multiplication is distributive with respect to addition. A (B + C) = AB + AC 4. Multiplication of matrix A by unit matrix. AI = IA = A 5. Multiplicative inverse of a matrix exists if |A| 0. A . A–1 = A–1 . A = I 6. If A is a square then A × A = A2, A × A × A = A3. 7. A0 = I 8. In = I, where n is positive integer. 0 1 2 1 –2 1 2 3 and B –1 0 Example 4. If A = 2 3 4 2 1 obtain the product AB and explain why BA is not defined. Solution. The number of columns in A is 3 and the number of rows in B is also 3, therefore the product AB is defined. C1 C2 0 1 2 R1 1 2 R1 C1 1 2 3 R 1 0 R C 2 2 1 AB = 2 3 4 R3 2 1 R3 C1 R1 C2 R2 C2 R3 C2 Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ 276 Determinants and Matrices R1, R2, R3 are rows of A and C1, C2 are columns of B. 1 2 0 0 1 2 1 0 1 2 2 1 1 2 0 = 1 2 3 1 1 2 3 2 1 1 2 2 3 4 1 2 3 4 0 2 1 For convenience of multiplication, we write the columns in horizontal rectangles. = 0 1 2 1 1 2 1 1 3 1 1 2 2 0 1 2 0 1 2 2 0 3 4 2 3 1 1 2 2 0 2 1 0 1 1 (1) 2 2 0 (2) 1 0 2 (1) 3 = 1 1 2 (1) 3 2 1 (2) 2 0 3 (1) 1 2 1 3 (1) 4 2 2 (2) 3 0 4 (1) 4 1 0 1 4 0 0 2 3 2 1 2 6 2 0 3 5 5 = Ans. 2 3 8 4 0 4 7 8 Since, the number of columns of B is (2) the number of rows of A is 3, BA is not defined. 1 2 3 1 0 2 2 3 1 and B = 0 1 2 Example 5. If A = 3 1 2 0 1 2 from the products AB and BA, and show that AB BA. Solution. Here, 1 0 3 0 2 6 1 2 3 1 0 2 2 0 1 0 3 2 2 3 1 0 1 2 AB = = 3 0 2 0 1 4 3 1 2 1 2 0 1 0 2 1 2 3 1 0 6 2 0 2 0 1 2 2 3 1 BA = = 0 2 6 0 3 2 1 2 0 3 1 4 0 2 6 0 1 2 AB BA 2 4 0 4 4 2 4 6 0 1 1 10 6 2 0 1 5 4 3 0 4 5 0 7 0 1 4 4 5 3 3 2 0 5 4 1 Proved. 1 2 2 1 –3 1 Example 6. If A = , B and C –2 3 2 3 2 0 Verify that (AB) C = A (BC) and A (B + C) = AB + AC. Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ Determinants and Matrices 277 Solution. We have, (1) (1) (2) (3) 6 7 1 (1) (2) (2) (2) = (2) (2) (3) (2) (2) (1) (3) (3) 2 7 3 1 6 2 2 0 4 2 0 6 6 2 0 0 2 1 0 1 1 1 2 3 1 3 4 AC = 2 3 2 0 6 6 2 0 12 2 2 ( 3) 1 1 1 2 B + C = 2 2 3 0 4 3 6 7 3 1 18 14 6 0 4 6 (i) (AB) C = 2 7 2 0 6 14 2 0 8 2 2 4 4 6 1 2 4 2 4 0 and A (BC) = 2 3 0 2 8 0 4 6 8 2 Thus from (1) and (2), we get (AB) C = A (BC) 1 2 2 AB = 2 3 2 2 1 3 BC = 2 3 2 (ii) ...(1) ...(2) 1 2 1 2 A (B + C) = 2 3 4 3 2 6 7 8 1 8 = 2 12 4 9 14 5 6 1 7 1 7 8 AB + AC = 2 12 7 2 14 5 Thus from (3) and (4), we get A (B + C) = AB + AC ...(3) ...(4) Verified. 1 2 2 Example 7. If A = 2 1 2 show that A2 – 4 A – 5 I = 0 where I, 0 are the unit matrix and 2 2 1 the null matrix of order 3 respectively. Use this result to find A–1. (A.M.I.E., Summer 2004) 1 2 2 2 1 2 Solution. Here, we have A= 2 2 1 1 2 2 1 2 2 9 8 8 2 1 2 2 1 2 8 9 8 A2 = 2 2 1 2 2 1 8 8 9 9 8 8 1 2 2 1 0 0 8 9 8 4 2 1 2 5 0 1 0 A2 – 4 A – 5 I = 8 8 9 2 2 1 0 0 1 9 4 5 8 8 0 8 8 0 0 0 0 8 8 0 9 4 5 8 8 0 0 0 0 A2 – 4 A – 5 I = 8 8 0 8 8 0 9 4 5 0 0 0 A2 – 4 A – 5 I = 0 5 I = A2 – 4 A Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ 278 Determinants and Matrices Multiplying by A–1, we get 5 A–1 = A – 4 I 1 2 2 1 0 0 3 2 2 2 1 2 4 0 1 0 2 3 2 = 2 2 1 0 0 1 2 2 3 A–1 = 2 2 3 1 2 3 2 5 2 2 3 Ans. Example 8. Show by means of an example that in matrices AB = 0 does not necessarily mean that either A = 0 or B = 0, where 0 stands for the null matrix. 1 1 1 1 2 3 Solution. Let A = 3 2 1 , B 2 4 6 2 1 2 3 1 0 242 3 6 3 1 2 1 3 4 1 6 8 2 9 12 3 AB = 2 2 0 4 4 0 6 6 0 0 0 0 0 0 0 0 0 0 AB = 0. But here neither A = 0 nor B = 0. Proved. Example 9. If AB = AC, it is not necessarily true that B = C i.e. like ordinary algebra, the equal matrices in the identity cannot be cancelled. 2 1 4 1 0 3 3 0 1 1 3 2 1 3 2 1 1 1 1 15 0 5 Solution. Let AB = 4 3 1 1 2 1 2 3 15 0 5 2 2 1 1 2 3 3 0 1 1 3 2 1 3 3 2 1 1 1 15 0 5 AC = Proved. 4 3 1 2 5 1 0 3 15 0 5 Here, AB = AC. But B C. Example 10. Represent each of the transformations x1 = 3y1 + 2y2, y1= z1 + 2z2 and x2 = – y1 + 4y2 , y2 = 3z1 by the use of matrices and find the composite transformation which expresses x1, x2 in terms of z1, z2. Solution. The equations in the matrix form are x1 3 2 y1 x = 1 4 y 2 2 y1 1 2 z1 y = 3 0 z 2 2 Substituting the values of y1, y2 in (1), we get ...(1) ...(2) 6 z1 9 z1 6 z2 x1 3 2 1 2 z1 9 x = 1 4 3 0 z 11 2 z 11z 2 z 2 2 2 1 2 x1 = 9z1 + 6z2, x2 = 11z1 – 2z2 Ans. Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ Determinants and Matrices 279 Example 11. Prove that the product of two matrices cos 2 cos sin cos 2 cos sin cos sin sin 2 is zero when and differ by an odd multiple of . 2 2 2 cos cos sin cos cos sin × Solution. = 2 2 cos sin sin cos sin sin cos2 cos 2 cos sin cos sin cos 2 cos sin cos sin sin 2 = 2 2 2 2 cos sin cos sin cos sin cos sin cos sin sin sin cos sin and sin2 cos cos (cos cos sin sin ) cos sin (cos cos sin sin ) = sin cos (cos cos sin sin ) sin sin (cos cos sin sin ) cos cos cos ( ) cos sin cos ( ) = sin cos cos ( ) sin sin cos ( ) Given – = (2 n + 1) 2 cos ( – ) = cos (2n + 1) 0 The product = 0 Example 12. Verify that A = =0 2 0 0. 0 Proved. 2 1 2 1 2 1 – 2 is orthogonal . 3 – 2 2 – 1 2 2 2 1 2 1 1 1 1 2 Solution. A = 2 1 2 A 2 3 3 2 2 1 2 2 1 2 1 2 2 9 0 0 1 0 0 1 2 1 1 AA = 2 1 2 2 1 2 = 0 9 0 0 1 0 I 9 9 0 0 9 0 0 1 2 2 1 2 2 1 Hence, A is an orthogonal matrix. Verified. Example 13. Determine the values of , , when 0 2 is orthogonal. 0 2 Solution. Let A = On transposing A, we have Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ 280 Determinants and Matrices 0 A = 2 If A is orthogonal, then AA = I 0 1 0 0 0 2 2 0 1 0 0 0 1 4 2 2 2 2 2 2 2 2 1 0 0 2 2 2 2 2 2 2 2 2 0 1 0 2 2 2 2 2 2 2 2 0 0 1 2 Equating the corresponding elements, we have 1 1 4 2 2 1 , 6 3 2 2 2 0 1 1 1 , , But 2 + 2 + 2 = 1 as Ans. 6 3 2 Example 14. Prove that (AB)n = An . Bn, if A . B = B . A Solution. (AB)1 = AB = (A) . (B) (AB)2 = (AB) . (AB) = (ABA) . B = { A (AB) } . B = (A2B) . B = A2 (B . B) = A2.B2 Suppose that (AB)n = An . Bn (AB)n+1 = (AB)n . (AB) = (An . Bn) . (AB) = An . (BnA) . B = An . (Bn–1 . BA) . B = An . (Bn–1 . AB) . B = An . (Bn–2 . B . AB) . B = An . (Bn–2 . AB . B) . B = An . (Bn–2 . AB2) . B, continuing the process n times. = An . (A . Bn) . B = An . (A . Bn+1) = An+1 . Bn+1 Hence, taking the above to be true for n = n, we have shown that it is true for n = n + 1 and also it was true for n = 1, 2, .... so it is universally true. Proved. EXERCISE 4.11 1. Compute AB, if 2 5 3 3 6 4 20 38 26 Ans. 47 92 62 4 7 5 0 2 3 4 1 2 3 1 . From the product AB and BA. Show that AB BA. ,B= 1 1 2 2 0 0 0 0 1 0 0 1 ,B= 0 0 1 0 1 2 3 A= and B = 4 5 6 1 3 1 2 2. If A = 0 0 0 1 0 0 3. If A = 0 0 (i) Calculate AB and BA. Hence evaluate A2 B + B2 A (ii) Show that for any number k, (A + kB2)3 = KI, where I is the unit matrix. Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ Determinants and Matrices 281 0 1 2 4. If A = choose and so that ( I + A) = A 1 0 1 Ans. = = 2 3 0 2 . B 1 1 0 5. If A = and C 2 1 0 0 1 T T T T verify that ABC C B A , where T denotes the transpose. 6. Write the following transformation in matrix form : 3 1 1 3 y1 y2 ; x2 = y1 y2 2 2 2 2 Hence, find the transformation in matrix form which expresses y1, y2 in terms of x1, x2. x1 = Ans. y1 = 7. y1 0 0 x1 The linear transformation y 0 1 x ,represents 2 2 3 1 1 3 x1 x2 , y2 = x1 x2 2 2 2 2 (a) reflection about x1 -axis (b) reflection about x2 -axis (c)clockwise rotation through angle (d) orthogonal projection on to x2 axis. 9. 10. 11. 12. (A.M.IE.T.E., Summer 2005) Ans. (d) cos sin 2 and I is a unit matrix, show that I + A = ( I A) sin cos 0 1 3 1 3 3 If f (x) = x3 – 20 x + 8, find f (A) where A = 1 2 4 4 1 1 tan 1 tan 2 2 cos sin Show that = tan sin cos 1 tan 1 2 2 3 3 4 If A = 2 3 4 then show that A3 = A–1. 0 1 1 2 1 2 1 2 1 2 is orthogonal. Verify whether the matrix A = 3 2 2 1 0 8. If A = tan 2 2 tan Ans. 0 1 2 2 1 2 1 2 13. Verify that 3 is an orthogonal matrix. 2 2 1 cos sin sin 14. Show that cos sin cos 0 15. Show that A = sin 0 sin cos sin cos is an orthogonal matrix. sin cos cos 0 sin 1 0 is an orthogonal matrix. 0 cos (A.M.I.E., Summer 2004) 16. If A and B are square matrices of the same order, explain in general (i) (A + B)2 A2 + 2 AB + B2 (ii) (A – B)2 A2 – 2 AB + B2 (iii) (A + B) (A – B) A2 – B2 Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ 282 Determinants and Matrices 17. Let A and B be any two matrices such that AB = 0 and A is non- singular. then (a) B = 0; (b) B is also non-singular; (c) B = A; (d) B is singular. Ans. (d) 18. If A2 = A then matrix A is called (a) Idempotent Matrix (b) Null Matrix (c) Transpose Matrix (d) Identity Matrix (A.M,I.E.T.E.,Dec.,2006) Ans. (a) 4.27 MATHEMATICAL INDUCTION By mathematical induction we can prove results for all positive integers. If the result to be proved for the positive integer n then we apply the following method. Working Rule: Step 1. Verify the result for n = 1 Step 2. Assume the result to be true for n = k and then prove that it is true for n = k + 1. Explanation. By step 1, the result is true for n = k = 1 By step 2, the result is true for n = k + 1 = 1 + 1 = 2 (k = 1) Again, the result is also true for n = k + 1 = 2 + 1 = 3 (k = 2) Similarly, the result is also true for n = k + 1 = 3 + 1 = 4 (k = 3) Hence, in this way the result is true for all positive integer n. Example 15. By mathematical induction, cos n sin n cos sin , show that An = if A = – sin n cos n – sin cos Where n is a positive integer. Solution. We prove the result by mathematical induction : cos n sin n An = – sin n cos n Let us verify the result for n = 1. cos1 sin1 cos sin A1 = A – sin1 cos1 – sin cos The result is true when n = 1. Let us assume that the result is true for any positive integer k. [Given] cos k sin k Ak = – sin k cos k Now, cos k sin k cos sin Ak+ 1 = Ak. A = – sin k cos k – sin cos cos k sin sin k cos cos k cos – sin k sin = – sin k cos – cos k sin – sin k sin cos k cos cos(k 1) sin(k 1) cos(k ) sin( k ) = = – sin(k 1) cos(k 1) – sin(k ) cos( k ) The result is true for n = k + 1. Hence, by mathematical induction the result is true for all positive integer n. Proved. Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ Determinants and Matrices 283 4.28 ADJOINT OF A SQUARE MATRIX Let the determinant of the square matrix A be | A |. a1 a2 a3 a1 a2 a3 b b b , b b b . 2 3 2 3 If A = 1 Than | A | = 1 c1 c2 c3 c1 c2 c3 The matrix formed by the co-factors of the elements in A1 A2 A3 B B B . 2 3 | A | is 1 C1 C2 C3 where A1 A3 b2 c2 b3 b2 c3 b3c2 , c3 b1 b2 A2 b1 b3 b1c3 b3c1 c1 c3 a2 c2 a3 a2 c3 a3 c2 c3 a1 a2 c1 c2 a1 b1 a3 a1b3 a3b1 b3 c1 c2 b1c2 b2 c1 , B1 B2 a1 c1 a3 a1c3 a3 c1 , c3 B3 C1 a2 b2 a3 a2 b3 a3b2 , b3 C2 a1c2 a2 c1 a2 a1b2 a2 b1 b1 b2 Then the transpose of the matrix of co-factors A1 B1 C1 A B C 2 2 2 A3 B3 C3 is called the adjoint of the matrix A and is written as adj A. C3 = a1 4.29 PROPERTY OF ADJOINT MATRIX The product of a matrix A and its adjoint is equal to unit matrix multiplied by the determinant A. Proof. If A be a square matrix, then (Adjoint A) . A = A . (Adjoint A) = |A| . I A1 B1 C1 a1 a2 a3 A B C b b b Let A= 1 2 2 2 3 and adj . A = 2 A B C 3 3 3 c1 c2 c3 a1 a2 a3 A1 B1 C1 A . (adj. A) = b1 b2 b3 A2 B2 C2 c1 c2 c3 A3 B3 C3 a1 A1 a2 A2 a3 A3 a1 B1 a2 B2 a3 B3 a1 C1 a2 C2 a3 C3 = b1 A1 b2 A2 b3 A3 b1 B1 b2 B2 b3 B3 b1 C1 b2 C2 b3 C3 c A c A c A c1 B1 c2 B2 c3 B3 c1 C1 c2 C2 c3 C3 2 2 3 3 1 1 0 | A | 0 1 0 0 0 | A| 0 = (A.M.I.E., Summer 2004) = | A | 0 1 0 = | A | I 0 0 | A | 0 0 1 Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ 284 Determinants and Matrices 4.30 INVERSE OF A MATRIX If A and B are two square matrices of the same order, such that AB = BA = I (I = unit matrix) then B is called the inverse of A i.e. B = A–1 and A is the inverse of B. Condition for a square matrix A to possess an inverse is that matrix A is non-singular, i.e., | A | 0 If A is a square matrix and B be its inverse, then AB = I Taking determinant of both sides, we get | AB | = | I | or | A | | B | = I From this relation it is clear that | A | 0 i.e. the matrix A is non-singular. To find the inverse matrix with the help of adjoint matrix We know that A . (Adj. A) = | A | I 1 A ( A dj. A) = I [Provided | A | 0] ...(1) | A| and A . A–1 = I ...(2) From (1) and (2), we have 1 A –1 = ( Adj. A) | A| 3 2 Example 16. If A = 0 3 – 3 2 – 3 Solution. A = 0 – 1 – 3 4 – 3 4 , find A1 . (A.M.I.E. Summer 2004) – 1 1 4 4 1 | A | = 3 (– 3 + 4) + 3 (2 – 0) + 4 (– 2 – 0) = 3 + 6 – 8 = 1 The co-factors of elements of various rows of | A | are ( 2 0) ( 2 0) ( 3 4) (3 4) (3 0) (3 0) (12 12) ( 12 8) ( 9 6) Therefore, the matrix formed by the co-factors of | A | is 0 1 1 1 2 2 2 3 4 1 3 3 , Adj. A = 2 3 3 0 4 3 0 1 1 0 1 1 1 2 3 4 1 1 2 3 4 A Adj. A = 1 Ans. | A| 2 3 3 2 3 3 1 4 – 8 1 4 7 , prove that A–1 = A, A being the transpose of A. Example 17. If A = 4 9 1 – 8 4 (A.M.I.E., Winter 2000) Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ Determinants and Matrices 285 1 4 8 4 4 7 , A= 9 1 8 4 1 4 – 8 1 1 4 4 7 AA = 9 9 1 – 8 4 1 8 4 1 1 4 8 Solution. We have, 9 4 7 4 1 8 4 1 4 8 4 7 4 64 1 16 32 4 28 8 8 16 1 32 4 28 16 16 49 4 32 28 = 81 8 8 16 4 32 28 1 64 16 81 0 0 1 0 0 1 0 81 0 0 1 0 or AA I = 81 0 0 81 0 0 1 A = A–1 Proved. 2 Example 18. If a matrix A satisfies a relation A A I 0 proved that A1 exists and that A –1= I + A, I being an identity matrix. (A M I E Winter 2003) A A I I Solution. Here A2 A I 0 or A2 AI I or A A I I 1 Again 2 A AI 0 A A 0 and so A–1exists. A2 A I or ...(1) Multiplying (1) by A ,we get A1 A2 A A1 I or A I A1 A–1 = I + A Proved. Example 19. If A and B are non-singular matrices of the same order then, (AB)–1 = B–1 . A–1 Hence prove that (A–1)m = (Am)–1 for any positive integer m. Solution. We know that, (AB) . (B–1 A–1) = [(AB) B–1] . A–1 = [A (BB–1] . A–1 = [AI] A–1 = A . A–1 = I 1 Also, B–1 A–1 . (AB) = B–1[A–1 . (AB)] = B–1 [(A–1 A) . B] = B–1 [I . B] = B–1 . B = I By definition of the inverse of a matrix, B–1 A–1 is inverse of AB. B–1 A–1 = (AB)–1 (Am)–1 = [A . Am–1]–1 = (Am–1)–1 A–1 Proved. = (A . Am–2)–1 . A–1 = [(Am–2)–1 . A–1] . A–1 = (Am–2)–1 (A–1)2 = (A . Am–3)–1 . (A–1)2 = [(Am–3)–1 . A–1] (A–1)2 = (Am–3)–1 (A–1)3 = A–1 (A–1)m–1 = (A–1)m Proved. Example 20. Find A satisfying the Matrix equation. 2 4 2 1 – 3 – 2 3 2 A 5 – 3 = 3 – 1 2 2 1 3 2 4 Solution. 3 2 A 5 3 = 3 1 Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ 286 Determinants and Matrices 2 1 2 1 i.e., Both sides of the equation are pre-multiplied by the inverse of 3 2 3 2 2 2 1 2 4 2 1 2 1 3 3 2 3 2 A 5 3 = 3 2 3 1 2 7 9 1 0 3 0 1 A 5 3 = 12 14 2 9 3 7 A = 5 3 12 14 2 3 Again both sides are post-multiplied by the inverse of i.e. 5 3 2 3 2 9 3 2 3 7 A = 5 3 5 3 12 14 5 3 3 2 5 3 13 13 1 0 24 24 A = A = 0 1 34 18 34 18 EXERCISE 4.12 Ans. Find the adjoint and inverse of the following matrices: (1 - 3) 1. 3. 4. 2 5 3 3 1 2 1 2 1 0 1 3 4 0 6 3 If A 1 1 5 7 1 7 3 1 5 Ans. 1 1 4 1 13 5 6 4 2 1 21 7 8 Ans. 20 18 6 4 2. 1 1 2 1 9 3 1 4 2 6 15 6 1 0 1 Ans. 1 3 8 5 3 4 1 2n 4n , then show that An 1 1 2n n 1 0 0 1 1 2 1 1 3 1 2 1 , P 0 3 2 , show that P–1 AP = 0 2 0 5. If A = 0 0 1 0 1 1 1 1 1 1 1 1 2 5 3 6. If A = 1 2 3 , B 3 1 2 , show that (AB)–1 = B–1 A–1. 1 4 9 1 2 1 3 2 2 3 4 2 7. Given the matrix A = 1 3 1 compute det (A), A–1 and the matrix B such that AB = 1 6 1 5 3 4 5 6 4 Also compute BA. Is AB = BA ? 9 2 4 1 0 0 1 2 1 . B 0 2 0 , AB BA Ans. 5, 1 5 12 0 0 1 1 7 8. Find the condition of k such that the matrix 29 17 14 1 3 4 3 1 1 –1 5 6 A = 3 k 6 has an inverse. Obtain A for k = 1. Ans. k , A 9 5 8 16 8 8 1 5 1 Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ Determinants and Matrices 287 9. Prove that (A–1)T = (AT)–1. 0 1 2 1 a b 10. If A 1 0 where A , then A is 2 1 c d 2 1 (a) 0 0 0 1 (b) 2 1 2 1 (c) 1 0 1 2 (AMIETE, June 2010) Ans. (d) (d) 1 1 2 2 2 3 x 2 2 4 x 1 11. For what values of x, the matrix is singular ? (A.M.I.E.T .E . Summer 2004) Ans. 0,3 2 4 1 x 12. Prove that (A–1)T = (AT)–1 13. Let I be the unit matrix of order n and adj. (2I) = 2k I. Then k equals (a) 1 (b) 2 (c) n – 1 (d) n. Ans (c) 14. Let T be a linear transformation defined by 1 1 1 1 1 1 0 0 0 1 0 0 4 5 2 , T 2 , T 2 , T 2 , T 1 1 1 1 1 1 0 1 Find T 3 8 . 3 3 3 3 (AMIETE Dec. 2005) 4.31 ELEMENTARY TRANSFORMATIONS Any one of the following operations on a matrix is called an elementary transformation. 1. Interchanging any two rows (or columns). This transformation is indicated by Rij, if the ith and jth rows are interchanged. 2. Multiplication of the elements of any row Ri (or column) by a non-zero scalar quantity k is denoted by (k.Ri). 3. Addition of constant multiplication of the elements of any row Rj to the corresponding elements of any other row Rj is denoted by (Ri + kRj). If a matrix B is obtained from a matrix A by one or more E-operations, then B is said to be equivalent to A. The symbol ~ is used for equivalence. i.e., A ~ B. Example 21. Reduce the following matrix to upper triangular form (Echelon form) : 1 2 3 2 5 7 3 1 2 Solution. Upper triangular matrix. If in a square matrix, all the elements below the principal diagonal are zero, the matrix is called an upper triangular matrix. 2 3 3 1 2 3 1 1 2 2 5 7 ~ 0 R2 R2 2 R1 ~ 0 1 1 1 1 R R3 5 R2 3 1 2 0 5 7 R3 R3 3 R1 0 0 2 3 1 3 3 Example 22. Transform 2 4 10 into a unit matrix. 3 8 4 Ans. (Q. Bank U.P., 2001) Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ 288 Determinants and Matrices Solution. 3 3 1 3 3 1 2 4 10 ~ 0 2 4 R2 R2 2 R1 3 8 4 0 1 5 R3 R3 3 R1 3 9 R1 R1 3 R2 1 3 1 0 1 ~ 0 1 2 R2 R2 ~ 0 1 2 2 0 1 5 0 0 7 R3 R3 R2 9 1 0 1 0 0 R1 R1 9 R3 ~ 0 1 2 ~ 0 1 0 R2 R2 2 R3 1 0 0 1 R3 R3 0 0 1 7 4.32 ELEMENTARY MATRICES A matrix obtained from a unit matrix by a single elementary transformation is called elementary matrix. 1 0 0 I = 0 1 0 0 0 1 Consider the matrix obtained by R2 + 3 R1 1 0 0 3 1 0 is called the elementary matrix. 0 0 1 4.33 THEOREM Every elementary row transformation of a matrix can be affected by pre-multiplication with the corresponding elementary matrix. Consider the matrix 2 3 4 A = 5 6 7 3 5 9 Let us apply row transformation R3 + 4 R1 and we get a matrix B. 2 3 4 5 6 7 B= 11 17 25 Now we shall show that pre-multiplication of A by corresponding elementary matrix R3 + 4 R1 will give us B. 1 0 0 1 0 0 0 1 0 then, Elementary matrix = 0 1 0 Now, if I = 0 0 1 4 0 1 (R 4 R ) 3 1 1 0 0 2 3 4 2 3 4 Elementary matrix × A = 0 1 0 5 6 7 = 5 6 7 = B 4 0 1 3 5 9 11 17 25 Similarly, we can show that every elementary column transformation of a matrix can be affected by post-multiplication with the corresponding elementary matrix. Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ Determinants and Matrices 289 4.34 TO COMPUTE THE INVERSE OF A MATRIX FROM ELEMENTARY MATRICES (Gauss-jordan Method) If A is reduced to I by elementary transformation then PA = I where P = PnPn–1 ... P2 P1 –1 P=A = Elementary matrix. Working rule. Write A = IA. Perform elementary row transformation on A of the left side and on I of the right hand side so that A is reduced to I and I of right hand side is reduced to P getting I = PA. Then P is the inverse of A. 4.35 THE INVERSE OF A SYMMETRIC MATRIX The elementary transformations are to be transformed so that the property of being symmetric is preserved. This requires that the transformations occur in pairs, a row transformation must be followed immediately by the same column transformation. Example 24. Find the inverse of the following matrix employing elementary transformations: 3 3 4 2 3 4 (U.P., I Semester, Compartment 2002) 0 1 1 3 3 4 Solution. The given matrix is A = 2 3 4 0 1 1 R1 1 4 3 0 0 R1 3 1 1 3 A 2 3 4 = 0 1 0 0 1 1 0 0 1 1 4 4 1 0 0 1 1 3 0 0 3 1 1 3 3 2 4 2 1 0 4 0 = 0 1 1 = 3 1 0 A A 3 3 3 R2 R2 2R1 0 0 1 R2 R2 0 1 1 0 0 1 0 1 1 4 1 4 1 0 0 0 0 1 1 1 1 3 3 3 3 4 2 2 4 0 A 1 1 0 1 0 A 1 = 0 = 3 3 3 3 1 0 0 1 2 3 3 R3 3 R3 2 1 1 R3 R3 R2 0 0 3 3 3 3 4 1 0 0 2 3 4 = 0 1 0 A 0 0 1 0 1 1 4 4 3 4 R1 R1 R3 1 1 0 3 0 3 4 A 4 1 0 = 2 R2 R2 R3 2 3 3 3 0 1 0 Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ 290 Determinants and Matrices 0 R1 R1 R2 1 1 0 1 1 1 0 0 0 1 0 = 2 –1 3 4 2 3 4 Hence, A = Ans. A 2 3 3 2 3 3 0 0 1 Example 23. Find the inverse of the matrix M by applying elementary transformations 0 1 1 –1 2 1 2 1 Solution. Here, we Let 1 3 –1 –2 . 0 1 2 6 0 1 have A = 1 –1 1 [U.P.T.U.(C.O.) 2003] 2 1 3 1 –1 –2 2 0 1 1 2 6 0 1 1 –1 2 3 1 1 –1 –2 0 ~ 2 0 1 0 1 2 6 0 0 0 0 1 0 0 A 0 1 0 0 0 1 1 0 1 –1 1 –1 –2 2 1 3 ~ 2 0 1 1 2 6 0 1 0 0 1 0 0 R1 R2 0 0 0 0 1 0 A 0 0 1 1 0 0 0 1 –1 –2 2 1 3 ~ 1 1 3 2 1 4 0 1 0 1 0 0 0 –1 1 0 1 0 1 0 0 0 1 –1 –2 0 1 0 0 0 –1 1 0 A 1 1 3 = 1 0 0 0 R3 R2 2 1 3 2 1 4 0 1 0 1 1 0 0 0 1 –1 –2 0 1 0 0 1 1 3 0 –1 1 0 A = 0 –1 –3 1 2 –2 0 R R –2R 3 3 2 0 –1 –2 0 3 –2 1 R R – 2R 4 2 4 1 0 0 0 1 –1 –2 0 1 0 0 1 1 3 0 –1 1 0 A = 0 –1 –3 1 2 –2 0 0 0 1 –1 1 0 1 R4 R4 – R3 0 0 A 0 R3 R3 – R1 1 R4 R4 + R1 Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ Determinants and Matrices 1 0 0 0 291 1 –1 –2 0 1 0 0 1 1 3 = 0 –1 1 0 A 0 1 3 –1 –2 2 0 R3 – R3 0 0 1 –1 1 0 1 1 –1 0 –2 3 0 2 R1 R1 2 R4 3 –4 1 –3 R R – 3 R 1 1 0 2 4 2 = 2 –5 2 –3 R3 R3 – 3R4 0 1 0 1 A 0 0 1 –1 1 0 1 0 0 0 1 0 0 0 1 0 0 1 0 0 = 0 1 0 0 0 1 1 0 0 0 0 0 0 1 0 0 = 0 1 0 0 0 1 0 –2 2 –1 R1 R1 R3 1 1 –1 0 R R R 2 3 2 2 –5 2 –3 A 1 –1 1 0 –1 1 2 –1 I = A–1 A –3 3 –1 R1 R1 – R2 1 –1 0 –5 2 –3 A 1 0 1 –1 –3 3 –1 1 1 –1 0 A–1 = 2 –5 2 –3 –1 1 0 1 Hence, Ans. EXERCISE 4.13 Reduce the matrices to triangular form: 1 2 3 1. A = 2 5 7 3 1 2 1 2 3 Ans. 0 1 1 0 0 –2 3 1 4 2. 1 2 5 0 1 5 0 1 4 Ans. 0 5 –19 0 0 22 Find the inverse of the following matrices: 3. 5. 1 3 3 1 4 3 1 3 4 1 – 1 1 1 0 4. 4 8 1 1 1 1 3 Use elementary row operations to find inverse of A 1 3 3 2 4 4 7 – 3 – 3 Ans. – 1 1 0 – 1 0 1 2 1 Ans. – 4 – 7 – 4 – 9 12 4 1 Ans. 5 1 4 1 1 – 1 4 5 6 3 1 (AMIETE, June 2010) 6. 1 –1 2 2 1 3 2 – 3 – 1 2 1 – 1 4 2 – 3 – 1 5 –7 1 2 5 –1 5 2 1 Ans. 5 11 10 18 – 7 1 – 2 10 5 Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ 292 Determinants and Matrices 7. 1 1 2 1 8. 1 –1 2 2 1 2 5 –7 1 3 2 –3 5 –1 5 –2 Ans. 1 –1 2 1 –1 18 –7 5 11 10 2 –3 –1 4 1 –2 10 5 10. 3 3 1 1 4 3 1 3 4 1 1 1 1 – 2 – 1 11. 2 3 4 1 3 3 3 1 1 2 (Q. Bank U.P. II Semester 2001) 3 1 1 0 1 –2 1 –2 2 –3 Ans. 0 1 –1 1 –2 3 –2 3 2 –6 –2 –3 –2 5 –13 –4 –7 Ans. 1 9. –1 –4 4 1 2 1 0 1 0 –1 1 0 1 0 2 –1 1 –3 1 0 –2 2 30 – 20 – 15 25 – 5 30 – 11 – 18 7 – 8 1 – 30 Ans. 12 21 – 9 6 15 – 15 12 6 – 9 6 15 – 7 – 6 – 1 – 1 2 1 3 – 1 1 1 1 – 1 2 2 X If X, Y are non-singular matrices and B = O matrix. X –1 O O where O is a null –1 = , show that B Y O Y –1 4.36 RANK OF A MATRIX The rank of a matrix is said to be r if (a) It has at least one non-zero minor of order r. (b) Every minor of A of order higher than r is zero. Note: (i) Non-zero row is that row in which all the elements are not zero. (ii) The rank of the product matrix AB of two matrices A and B is less than the rank of either of the matrices A and B. (iii) Corresponding to every matrix A of rank r, there exist non-singular matrices P and Q such I r 0 that PAQ = 0 0 4.37 NORMAL FORM (CANONICAL FORM) By performing elementary transformation, any non-zero matrix A can be reduced to one of the following four forms, called the Normal form of A : (i) Ir (ii) [Ir 0] Ir (iii) 0 I (iv) r 0 0 0 I r 0 The number r so obtained is called the rank of A and we write (A) = r. The form is 0 0 called first canonical form of A. Since both row and column transformations may be used here, the element 1 of the first row obtained can be moved in the first column. Then both the first row and first column can be cleared of other non-zero elements. Similarly, the element 1 of the second row can be brought into the second column, and so on. Example 24. Reduce to normal form the following matrix 1 2 3 4 A 2 1 4 3 3 0 5 10 R 3 – 2R 2, R 3 – 3R 1 Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ Determinants and Matrices Solution. 293 4 1 2 3 4 1 2 3 A 2 1 4 3 0 3 2 5 3 0 5 10 0 6 4 22 1 1 C2 2C1 , C3 3C1 , C4 4C1 , R2 , R3 R 3 R 2 3 6 1 0 0 0 1 0 0 0 0 1 0 0 2 5 2 5 0 3 2 5 0 1 0 1 3 3 3 3 0 6 4 22 0 0 0 2 11 2 0 1 3 3 2 5 1 C C3 C2 , C4 C2 C3 C4 2 3 3 3 1 0 0 0 1 0 0 0 1 0 0 0 0 1 0 0 0 1 0 0 0 1 0 0 0 0 0 2 0 0 2 0 0 0 1 0 = [I3 0] is the normal form of A. Ans. Example 25. Find the rank of the following matrix by reducing it to normal form – 2 – 1 3 1 4 1 2 1 A= (U.P. I Sem., Com. 2002, Winter 2001) 3 – 1 1 2 2 0 1 1 2 –1 3 1 2 – 1 3 1 4 0 –7 6 – 11 R2 R2 – 4 R1 1 2 1 Solution. ~ 0 – 7 4 – 7 R3 R3 – 3 R1 3 – 1 1 2 0 1 – 2 R4 R4 – R1 2 0 1 0 1 0 0 0 0 0 0 1 1 0 – 7 6 – 11 0 – 7 6 – 11 ~ 0 –2 4 R3 R3 – R2 0 – 7 4 – 7 0 0 0 1 – 2 0 1 – 2 0 C2 C2 – 2 C1, C3 C3 + C1, C4 C4 – 3C1 0 0 0 0 1 1 0 – 7 0 – 7 0 0 ~ 0 0 0 –2 4 0 0 0 1 – 2 0 0 6 C3 C3 + C , C C4 – 7 2 4 0 0 0 1 0 – 7 0 0 C4 C4 + 2C3 0 0 – 2 0 0 0 0 0 Rank of A = 3 0 0 0 0 – 2 4 1 0 0 R4 R4 R3 2 11 C, 7 2 1 0 0 0 0 1 0 0 R – 1/ 7 R 2 2 ~ 0 0 1 0 R3 – 1/ 2 R3 0 0 0 0 Ans. Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ 294 Determinants and Matrices Example 26. Reduce the matrix A to its normal form, when Hence, find the rank of A. 2 –1 4 1 2 4 3 4 A= 1 2 3 4 – 1 – 2 6 – 7 (U.P., I Semester, Dec. 2004, Winter 2001) 2 –1 4 1 2 4 3 4 Solution. The given matrix is A = 1 2 3 4 6 – 7 – 1 – 2 1 0 0 0 2 –1 4 1 0 R R – 2 R 0 5 – 4 2 2 1 0 0 4 0 R3 R3 – R1 R R R 0 5 – 3 4 0 4 1 1 0 0 0 1 0 0 0 0 5 0 – 4 C3 C2 4 0 0 0 5 0 – 3 0 1 0 0 0 1 0 0 0 0 0 0 C2 C2 – 2 C1 0 5 – 4 C3 C3 C1 0 4 0 C4 C4 – 4 C1 0 5 – 3 0 0 0 5 0 – 4 16 4 0 0 R3 R3 – R2 5 5 0 0 1 R4 R4 – R2 0 0 0 0 1 0 0 5 0 5 – 4 0 16 C4 C3 16 0 0 0 0 5 5 0 0 0 0 1 0 0 0 0 1 0 0 R2 1/ 5 R2 I3 0 1 0 R3 5 /16 R3 0 0 0 0 0 5 0 R2 R2 R3 4 0 5 R3 0 R4 R4 – 16 0 0 Which is the required normal form. And since, the non-zero rows are 3 hence, the rank of the given matrix is 3. Example 27. Find non-singular matrices P, Q so that PAQ is a normal form where Ans. 1 – 3 – 6 2 1 2 A = 3 – 3 (R.G.P.V., Bhopal, April, 2010, U.P., I Sem. Winter 2002) 1 1 1 2 and hence find its rank. Solution. Order of A is 3 × 4 Total number of rows in A = 3; Consider unit matrix I3. Total number of columns in A = 4 Hence, consider unit matrix I4, A 3 × 4 = I3 A I4 Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ Determinants and Matrices 295 1 1 – 3 – 6 2 1 0 0 0 3 – 3 1 2 = 0 1 0 A 0 1 0 0 1 1 1 2 0 1 1 1 2 1 0 0 1 0 3 – 3 1 2 = 0 1 0 A 0 1 0 0 2 1 – 3 – 6 0 0 0 0 1 0 0 0 1 0 0 0 1 0 0 0 1 0 0 R1 R3 0 1 0 0 0 1 1 1 1 2 0 0 1 1 0 – 6 – 2 – 4 0 = 0 1 – 3 A 0 0 – 1 – 5 – 10 1 0 – 2 0 0 0 0 1 0 0 R2 R2 – 3 R1 0 1 0 R3 R3 – 2 R1 0 0 1 C2 C2 – C1, C3 C3 – C1, C4 C4 – 2 C1 1 1 1 2 1 0 0 0 0 0 1 0 1 0 0 0 – 6 – 2 – 4 = 0 1 – 3 A 0 0 1 0 0 – 1 – 5 – 10 1 0 – 2 0 0 1 0 1 – 1 – 1 – 2 0 0 1 0 1 0 0 1 0 0 R2 (1) R2 A 0 0 1 3 0 6 2 0 4 = 0 1 0 R3 (1) R3 1 0 2 0 1 5 10 0 0 1 0 1 – 1 – 1 – 2 1 0 0 0 0 1 0 0 1 5 10 0 1 0 0 R2 R3 = – 1 0 2 A 0 0 1 0 0 6 2 4 0 – 1 3 0 0 1 0 1 – 1 – 1 – 2 0 0 0 1 1 0 0 1 0 0 0 1 – 1 0 A 0 5 10 2 R3 R3 – 6 R2 = 0 0 1 0 6 – 1 – 9 0 0 – 28 – 56 0 0 1 0 C3 C3 – 5 C2, C4 C4 – 10 C2 4 8 1 – 1 0 1 0 0 0 1 0 1 – 5 – 10 – 1 0 A 0 0 1 2 = 0 0 0 0 1 0 6 – 1 – 9 0 0 – 28 – 56 0 0 1 0 1 0 0 0 0 1 0 0 = 0 0 1 2 – 0 0 –1 0 6 28 1 28 1 – 1 4 8 1 0 1 – 5 – 10 1 2 A R3 – R3 0 0 1 0 28 9 0 0 0 1 28 Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ 296 Determinants and Matrices 0 0 1 0 0 0 0 1 0 0 – 1 0 = 0 0 1 0 6 1 – 28 28 N = PAQ 1 – 1 4 0 1 0 1 –5 0 2 A C4 C4 – 2 C3 0 0 1 – 2 9 0 0 0 1 28 4 0 1 – 1 0 0 1 0 1 –5 0 – 1 0 2 , Q P= Ans. 0 0 1 – 2 3 1 9 – 0 0 1 0 14 28 28 Note. P and Q are not unique. 1 0 0 0 Normal form of the given matrix is 0 1 0 0 0 0 1 0 The number of non zero rows in the normal matrix = 3 Hence Rank = 3 Ans. 3 – 3 4 Example 28. If A = 2 – 3 4 , Find two non singular matrices P and Q such that 0 – 1 1 PAQ = I. Hence find A–1. Solution. A 3 × 3 = I3 A I3 3 – 3 4 1 0 0 1 0 0 2 – 3 4 0 1 0 A 0 1 0 = 0 – 1 1 0 0 1 0 0 1 0 0 1 – 1 0 1 0 0 1 2 – 3 4 1 0 A 0 1 0 R1 R1 – R2 = 0 0 – 1 1 0 0 1 0 0 1 0 0 1 1 – 1 0 1 0 0 0 – 3 4 3 0 A 0 1 0 R3 R2 – 2R1 = – 2 0 0 1 0 0 1 0 – 1 1 1 0 0 1 – 1 0 1 0 0 0 3 4 – 2 3 0 A 0 – 1 0 C2 – C2 = 0 1 1 0 0 1 0 0 1 1 0 0 1 – 1 0 1 0 0 0 1 1 0 0 1 A 0 – 1 0 R2 R3 = 0 3 4 – 2 3 0 0 0 1 0 1 0 0 1 0 0 1 – 1 0 1 1 0 0 1 A 0 – 1 0 R3 R3 – 3 R2 = 3 3 0 0 1 0 0 1 – 2 Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ Determinants and Matrices 297 0 1 0 0 1 0 0 1 – 1 0 1 0 = 0 0 1 A 0 – 1 1 C3 C3 – C2 3 – 3 0 0 1 0 0 1 – 2 I3 = PAQ A–1 = QP, A–1 = 0 1 0 0 1 – 1 0 – 1 1 0 0 1 0 0 1 – 2 3 – 3 A–1 = 0 1 –1 – 2 3 – 4 – 2 3 – 3 I P A Q –1 P A Q –1 –1 P Q A ( P –1 Q –1 ) –1 A –1 QP A–1 Ans. Exercise 4.14 Find non singular matrices P and Q such that PAQ is normal form 1. 1 2 3 3 1 2 2. 2 1 1 1 2 3 0 1 1 3. 1 2 3 2 2 2 1 3 3 0 4 1 4.38 2 1 1 5 5 1 0 1 7 p , Q 0 Ans. 3 1 5 5 0 0 1 1 0 0 1 1 1 Ans. p 1 1 0 , Q 0 1 1 1 1 1 0 0 1 1 4 1 1 3 15 21 0 1 1 1 1 0 0 6 6 6 Ans. P 2 1 0 , Q 0 0 1 0 1 1 1 5 1 0 0 0 7 RANK OF MATRIX BY TRIANGULAR FORM Rank = Number of non-zero row in upper triangular matrix. Note. Non-zero row is that row which does not contain all the elements as zero. Example 29. Find the rank of the matrix 1 2 3 2 2 3 5 1 1 3 4 5 (U.P., I Semester, Winter 2003, 2000) Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ 298 Determinants and Matrices 3 2 1 2 3 2 1 2 2 3 5 1 0 – 1 1 – 3 R2 R2 – 2 R1 1 3 4 5 0 1 1 3 R3 R3 – R1 Solution. 3 2 1 2 0 – 1 – 1 – 3 0 0 0 0 R3 R3 R2 Rank = Number of non zero rows = 2. Ans. 2 3 – 2 – 1 2 –5 1 2 Example 30. Find the rank of the matrix 3 –8 5 2 6 5 – 12 – 1 2 3 – 2 2 3 – 2 – 1 – 1 0 – 1 7 – 2 R R 2 R 2 –5 2 1 1 2 2 Solution. ~ R R 3 R 3 –8 3 1 5 2 0 – 2 14 – 4 3 0 – 2 14 – 4 R R 5 R 4 1 6 5 – 12 – 1 4 – 1 2 3 – 2 0 – 1 7 – 2 R R – 2 R 3 2 3 ~ 0 0 0 0 R4 R4 2 R2 0 0 0 0 Here the 4th order and 3rd order minors are zero. But a minor of second order 3 –2 7 –2 – 6 14 = 8 0 Rank = Number of non-zero rows = 2. Ans. Example 31. Find the rank of matrix 3 – 2 4 2 3 –2 1 2 3 2 3 4 4 0 5 – 2 Solution. Multiplying R1 by (U.P., I Semester, Dec., 2006) 1 , we get 1 as pivotal element 2 3 – 1 2 1 2 1 2 3 –2 3 2 3 4 4 0 5 – 2 Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ Determinants and Matrices 299 3 0 0 0 2 –1 2 1 1 2 R2 – 13 R2 8 8 0 1 – 0 – 13 4 – 4 R2 R2 – 3R1 13 13 C C – 3 C 2 2 1 2 2 5 5 0 – 6 – 2 0 – 6 – 2 R3 R3 – 3R1 2 C3 C3 C1 2 0 7 –2 9 C4 C4 – 2C1 0 7 –2 9 R4 R4 2 R1 1 0 0 0 0 0 0 1 8 8 0 1 – 13 13 5 58 6 0 0 – R3 R3 R2 2 13 13 30 61 0 0 R R –7R 13 13 4 4 2 1 0 0 0 0 1 0 0 0 0 0 1 0 0 0 1 0 1 0 0 3 R 13 R 0 1 – 3 3 0 58 29 30 61 0 0 13 13 0 1 0 0 58 13 30 0 13 0 0 0 6 8 – C3 C3 C2 13 13 61 8 C C4 – C2 13 4 13 0 0 0 1 0 0 3 0 1 – 29 30 143 R4 R4 – R3 0 0 13 29 0 1 0 0 0 1 0 0 143 0 0 0 3 C3 29 C4 C 4 29 0 0 0 1 0 0 0 1 0 29 R R4 0 0 1 4 143 I4 Hence, the rank of the given matrix = 4 Ans. Example 32. Use elementary transformation to reduce the following matrix A to triangular from and hence find the rank of A. 3 –1 2 1 – 1 – 2 A= 3 1 3 6 3 0 1 – 4 – 2 – 7 (R.G.P.V., Bhopal, June 2007, Winter 2003, U.P., I Semester, Dec. 2005) Solution. We have, 3 –1 2 1 – 1 – 2 A= 3 1 3 3 0 6 1 1 – 1 – 2 – 4 – 4 2 3 – 1 – 1 R1 R2 – 2 3 1 3 – 2 – 7 6 3 0 – 7 Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ 300 Determinants and Matrices –2 – 4 1 – 1 – 2 – 4 1 – 1 0 0 5 3 7 5 3 7 R2 R2 – 2 R1 0 4 9 10 R3 R3 3R1 0 0 33 / 5 22 / 5 R3 R3 – 4 / 5 R2 9 12 17 R4 R4 – 6 R1 0 0 33 / 5 22 / 5 R4 R4 – 9 / 5 R2 0 –2 – 4 1 – 1 0 5 3 7 0 0 33 / 5 22 / 5 0 0 0 R4 R4 – R3 0 R(A) = Number of non-zero rows. R(A) = 3 Ans. EXERCISE 4.15 Find the rank of the following matrices: 1. 1 2 3 2 4 7 3 6 10 3. 0 1 2 – 2 4 0 2 6 2 1 3 1 Ans. 2 1 1 2 2. – 1 0 2 2 1 – 3 Ans. 3 Ans. 2 4 3 – 2 2 – 3 – 2 – 1 4 4. 6 – 1 7 2 Ans. 3 4 3 – 2 1 1 4 1 1 3 – 2 – 3 – 1 2 4 3 4 3 6 5. Ans. 4 6. –1 – 1 – 2 6 6 7 2 9 4 3 6 6 12 1 – 1 2 – 3 – 3 Reduce the following matrices to Echelon form and find out the rank: 1 0 0 Ans. 0 1 0 , Rank = 3 0 0 1 7. 1 1 2 1 2 2 2 2 3 9. 3 2 5 7 12 I2 1 1 2 3 5 Ans. 0 3 3 6 9 15 0 , Rank = 2 0 1 2 8. 3 6 2 3 0 4 3 2 2 1 3 8 7 5 Ans. 2 I Ans. 3 0 0 , Rank = 3 0 3 1 0 2 – 4 1 – 2 I 0 1 – 4 2 10. Ans. 3 , Rank = 3 0 1 –1 3 1 0 0 4 – 4 5 4 – 7 Using elementary transformations, reduce the following matrices to the canonical form (or row-reduced Echelon form): 0 0 0 0 0 1 0 0 0 0 1 2 3 4 Ans. 0 1 0 0 11. A 0 2 3 4 1 0 0 1 0 0 3 4 1 2 4 – 12 8 0 0 2 –6 2 12. A 0 1 –3 6 24 3 0 – 8 9 1 0 0 0 5 Ans. 0 1 0 0 4 0 0 1 0 1 Using elementary transformations, reduce the following matrices to the normal form: 1 0 0 0 1 2 0 – 1 13. A 2 Ans. 0 1 0 0 3 4 1 0 0 1 0 – 2 3 2 5 1 2 3 4 14. A 3 4 1 2 4 3 1 2 1 0 0 0 Ans. 0 1 0 0 0 0 1 0 Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ Determinants and Matrices 301 Obtain a matrix N in the normal form equivalent to 0 0 0 0 0 0 4 5 0 0 15. A 0 9 1 – 1 2 1 11 0 10 0 Hence find non-singular matrices P and Q such that PAQ = N. 1 – 3 0 1 3 4 Find the rank 16. 1 2 17. A 1 0 1 2 2 3 1 – 2 of the following matrix by reducing it into normal form: 1 2 3 1 3 2 5 1 1 3 3 2 2 –1 6 3 Ans. 3 18. A 2 4 3 3 1 2 3 – 1 2 5 2 – 3 1 1 1 1 1 2 3 19. Rank of matrix A 1 4 2 is 2 6 5 (a) 0 (b) 1 (d) 2 1 5 4 20. For which value of ‘b’ the rank of the matrix A 0 3 2 is b 13 10 (a) 1 (b) 2 (c) 3 (c) 3 (d) 0 Ans. 4 (AMIETE, June 2009) Ans. (d) (AMIETE, Dec. 2009) Ans. (b) 4.39 SOLUTION OF SIMULTANEOUS EQUATIONS The matrix of the coefficients of x, y, z is reduced into Echelon form by elementary row transformations. At the end of the row transformation the value of z is calculated from the last equation and value of y and the value of x are calculated by the backward substitution. Example 33. Solve the following equations x – y + 2z = 3, x + 2y + 3z = 5, 3x – 4y – 5z = – 13 Solution. In the matrix form, the equations are written in the following form. 2 x 1 1 3 2 x 1 1 3 1 2 3 y = 5 or 0 3 1 y 2 R2 R2 R1 3 4 5 z R R3 3 R1 13 0 1 11 z 22 3 1 1 2 x 3 1 1 y = 2 R3 R3 R2 0 3 3 32 z 64 0 0 3 3 x– y+ 2 z=3 3 y + z= 2 ...(1) ...(2) 32 64 z = z=2 3 3 Putting the value of z in (2), we get 3y + 2 = 2 y = 0 Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ 302 Determinants and Matrices Putting the value of y, z in (1), we get x – 0 + 4 = 3 x = – 1 x = – 1, y = 0, z = 2 Ans. Example 34. Find all the solutions of the system of equations x1 + 2x2 – x3 = 1, 3x1 – 2x2 + 2x3 = 2, 7x1 – 2x2 + 3x3 = 5 Solution. 1 1 2 1 x1 2 3 2 2 x 2 = 5 7 2 3 x3 R2 R2 – 3R1, R3 R3 – 7R1 2 1 x1 1 1 1 1 2 1 x1 1 R R 2 R 0 8 5 x 1 , 0 8 5 x 3 2 2 = 2 = 3 0 2 0 0 0 x3 0 16 10 x3 x1 + 2 x2 – x3 = 1 – 8x2 + 5x3 = – 1 Let ...(1) ...(2) x3 = k Putting x3 = k in (2), we get – 8x2 + 5k = – 1 x2 = 1 (5k 1) 8 1 (5k 1) k = 1 4 5k 1 k 3 = x1 = 1 k 4 4 4 4 k 3 5k 1 , x3 k x1 = , x2 = 4 4 8 8 The equations have infinite solution. Substituting the values of x3, x1 in (1), we get x1 Ans. 4.40 Gauss - Jordan Method (R.G.P.V., Bhopal, III Semester, Dec. 2007) This is modification of the Gaussian elimination method. By this method we eliminate unknowns not only from the equations below but also from the equations above. In this way the system is reduced to a diagonal matrix. Finally each equation consists of only one unknown and thus, we get the solution. Here, the labour of backward substitution for finding the unknowns is saved Gauss-Jordan method is modification of Gaussian elimination method. Example 35. Express the following system of equations in matrix form and solve them by the elimination method ( Gauss Jordan Method) 2x1 + x2 + 2x3 + x4 = 6 6x1 – 6x2 + 6x3 + 12x4 = 36 4x1 + 3x2 + 3x3 – 3x4 = – 1 2 x1 + 2 x2 – x3 + x4 = 10 Solution. The equations are expressed in matrix form as 6 1 2 1 x1 2 36 6 6 6 12 x 2 = 1 4 3 3 3 x3 10 2 1 1 x4 2 Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ Determinants and Matrices 303 1 2 1 x1 2 0 9 0 9 x2 = 0 1 1 5 x3 1 3 0 x4 0 2 0 0 0 1 2 1 1 0 1 1 1 5 1 3 0 x1 x 2 x3 x4 6 18 R2 R2 3 R1 R R 2 R 3 1 13 3 R4 R4 R1 4 6 2 R R2 2 13 = 9 4 6 2 1 x1 2 1 2 R R R 0 1 0 1 x2 3 2 3 = 11 0 0 1 4 x3 1 x4 6 R4 R4 R2 0 0 3 1 x1 6 2 1 2 0 1 0 1 x 2 2 = 0 0 1 4 x3 11 R4 R4 3 R3 0 0 0 13 x4 39 2 x1 + x2 + 2 x3 + x4 = 6 x2 – x4 = – 2 – x3 – 4 x4 = – 11 13 x4 = 39 x4 = 3 Putting the value of x4 in (3), we get – x3 – 12 = – 11 x3 = – 1 Putting the value of x4 in (2), we get x2 – 3 = – 2 x2 = 1 Substituting the values of x4, x3 and x2 in (1), we get 2 x1 + 1 – 2 + 3 = 6 or 2 x1 = 4 x1 = 2 x1 = 2, x2 = 1, x3 = – 1, x4 = 3 Example 36. Find the general solution of the system of equations: 3x1 + 2x3 + 2x4 = 0 – x1 + 7x2 + 4x3 + 9x4 = 0 7 x1 – 7x2 – 5x4 = 0 Solution. The system of equations in the matrix form is expressed as ...(1) ...(2) ...(3) Ans. x1 0 2 2 0 3 x 2 1 7 4 9 = 0 x3 0 7 7 0 5 x4 x1 7 4 9 0 1 x2 3 0 2 2 = 0 R1 R2 x3 0 7 7 0 5 x4 Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ 304 Determinants and Matrices x1 1 7 4 9 0 0 21 14 29 x2 R2 R2 3 R1 x = 0 R R 7 R 3 3 1 0 3 0 42 28 58 x4 x1 1 7 4 9 0 0 21 14 29 x2 0 R R 2 R 3 2 x = 3 0 0 0 0 3 0 x4 – x1 + 7 x2 + 4 x3 + 9 x4 = 0 21 x2 + 14 x3 + 29 x4 = 0 Let x4 = a, x3 = b 2 b 29 a From (2), 21 x2 + 14 b + 29 a = 0 or x2 = 3 21 2 b 29 a From (1), x1 7 4b 9a = 0 3 21 ...(1) ...(2) 2a 2b 3 3 2 1 (29 a 14 b) x 1 = ( a b) , x 2 = 3 21 x3 = b, x4 = a Ans. 4.41 TYPES OF LINEAR EQUATIONS (1) Consistent. A system of equations is said to be consistent, if they have one or more solution i.e. x + 2y = 4 x + 2y = 4 3x + 2y = 2 3x + 6y = 12 Unique solution Infinite solution (2) Inconsistent. If a system of equation has no solution, it is said to be inconsistent i.e. x +2 y = 4 3x + 6y = 5 4.42 CONSISTENCY OF A SYSTEM OF LINEAR EQUATIONS a11 x1 + a12 x2 + . . . a1n xn = b1 a21 x1 + a22 x2 + . . . a2n xn = b2 x1 = .......................................................................... am1 x1 + am2 x2 + ... amn xn= bm a11 a12 .......... a1n a 21 a22 .......... a2 n .............................. am1 am 2 .......... amn x1 b1 x 2 = b2 ... .... xm bm a11 a12 .......... a1n b1 a 21 a22 .......... a2 n b2 and C= [A, B] = ..................................... am1 am2 .......... amn bm AX = B is called the augmented matrix. [ A : B] C Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ Determinants and Matrices 305 (a) Consistent equations. If Rank A = Rank C (i) Unique solution: Rank A = Rank C = n (ii) Infinite solution: Rank A = Rank C = r, r < n (b) Inconsistent equations. If Rank A Rank C. In Brief : where n = number of unknown. A system of non-homogeneous linear equations AX = B Find R (A) and R (C) R (A) = R (C) R (A) R (C) Solution exists, system is consistent R (A) = R (C) = n (no. of unknowns) Unique solution No solution, system is inconsistent R (A) = R (C) < n (no. of unknowns) Infinite no. of solutions Example 37. Show that the equations 2x + 6y = – 11, 6x + 20y – 6z = – 3, 6y – 18z = – 1 are not consistent. Solution. Augmented matrix C = [A, B] 0 : 11 2 6 0 : 11 2 6 6 20 6 : 3 ~ 0 2 6 : 30 R R 3 R 2 1 2 = 0 6 18 : 1 0 6 18 : 1 0 : 11 2 6 0 2 6 : 30 R R – 3 R 3 3 2 0 : 91 0 0 The rank of C is 3 and the rank of A is 2. Rank of A Rank of C. The equations are not consistent. Example 38. Test the consistency and hence solve the following set of equation. x1 + 2x2 + x3 = 2 Ans. 3x1 + x2 – 2x3 = 1 4x1 – 3x2 – x3 = 3 2x1 + 4x2 + 2x3 = 4 (U.P., I Semester, Compartment 2002) Solution. The given set of equations is written in the matrix form: 2 1 2 1 x1 1 3 1 2 = x 2 3 4 3 1 x3 4 2 4 2 AX = B 2 1 2 1 3 1 2 1 [ A , B ] ~ Here, we have augmented matrix C = 4 3 1 3 4 2 4 2 Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ 306 Determinants and Matrices 2 1 2 1 0 5 5 5 ~ 0 11 5 5 0 0 0 0 2 1 2 1 1 R2 R2 3 R1 0 1 1 1 R2 R2 ~ 5 R3 R3 4 R1 0 11 5 5 R4 R4 2 R1 0 0 0 0 1 0 ~ 0 0 2 1 2 1 2 1 2 0 1 1 1 1 1 1 ~ 1 0 6 6 R3 R3 11 R2 0 0 1 1 R3 R3 6 0 0 0 0 0 0 0 Number of non-zero rows = Rank of matrix. R(C) = R(A) = 3 Hence, the given system is consistent and possesses a unique solution. In matrix form the system reduces to 2 1 2 x1 1 1 1 x2 = 0 1 1 x3 0 0 0 x1 + 2x2 + x3 = 2 ...(1) x2 + x3 = 1 ...(2) x3 = 1 From (2), x2 + 1 = 1 x2 = 0 From (1), x1 + 0 + 1 = 2 x1 = 1 Hence, x1 = 1, x2 = 0 and x3 = 1 Ans. Example 39. Test for consistency and solve : 5x + 3y + 7z = 4, 3x + 26y + 2z = 9, 7x + 2y + 10 z = 5 Solution. The augmented matrix C = [A, B] (R.G. P.V. Bhopal I. Sem. April 2009-08-03) 1 0 0 0 3 7 4 1 : 5 3 7 : 4 5 5 5 3 26 2 : 9 3 26 2 : 9 R 1 R 1 1 5 7 2 10 : 5 7 2 10 : 5 3 7 4 3 7 4 : 1 : 1 5 5 5 5 5 5 121 11 33 121 11 33 R2 R2 3 R1 0 0 : : 5 5 5 5 5 5 1 R2 1 3 0 0 0 : 0 R3 R3 0 11 : R3 R3 7 R1 11 5 5 5 Rank of A = 2 = Rank of C Hence, the equations are consistent. But the rank is less than 3 i.e. number of unknows. So its solutions are infinite. Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ Determinants and Matrices 307 3 7 4 1 5 5 5 x 0 121 11 y 33 = 5 5 5 z 0 0 0 0 3 7 4 x y z = 5 5 5 33 121 11z y = or 111y – z = 3 5 5 5 3 k Let z = k then 11y – k = 3 or y = 11 11 33 k 7 4 16 7 x k = k or x = Ans. 5 5 11 11 5 11 11 Example 40. Discuss the consistency of the following system of equations 2x + 3y + 4z = 11, x + 5y + 7z = 15, 3x + 11y + 13z = 25. If found consistent, solve it. (A.M.I.E.T.E., Winter 2001) Solution. The augmented matrix C = [A, B] 2 3 4 11 1 5 7 15 1 5 7 15 ~ 2 3 4 11 R R 2 1 3 11 13 25 3 11 13 25 5 7 15 1 0 7 10 19 0 4 8 20 1 1 R 2 , R 3 R 3 , R3 R3 – R2 7 4 1 5 7 15 1 5 7 15 0 1 10 19 ~ 0 1 10 19 7 7 7 7 0 1 2 5 0 0 4 16 7 7 R2 R2 R2 – 2R1, R3 R3 – 3R1, ~ Rank of C = 3 = Rank of A Hence, the system of equations is consistent with unique solution. 15 1 5 7 x 10 19 0 1 y Now, = 7 7 16 z 0 0 4 7 7 x + 5y + 7z = 15 y 10 z 19 = 7 7 ...(1) ...(2) 4z 16 = z=4 7 7 Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ 308 Determinants and Matrices 19 10 y=–3 4 = 7 7 From (1), x + 5 (– 3) + 7 (4) = 15 x = 2 x = 2, y = – 3, z = 4 Ans. Example 41. Find for what values of and the system of linear equations: x+y+z=6 x + 2y + 5z = 10 2x + 3y + z = has (i) a unique solution (ii) no solution (iii) infinite solutions. Also find the solution for = 2 and = 8. (Uttarakhand, 1st semester, Dec. 2006) 1 1 1 x 6 Solution. 1 2 5 y = 10 2 3 z AX = B 1 1 1 : 6 1 : 6 1 1 1 2 5 : 10 0 1 4 : 4 R2 R2 R1 C = (A, B) = ~ 2 3 : 0 1 2 : 12 R3 R3 2 R1 From (2), y 1 : 6 1 1 0 1 4 : 4 ~ 0 0 6 : 16 R3 R3 R2 ...(1) (i) A unique solution If R (A) = R (C) = 3 then – 6 0 6 and – 16 0 16 (ii) No solutions If R (A) R (C), then R (A) = 2 and R (C) = 3 – 6 = 0 = 6 and – 16 0 16 (iii) Infinite solutions If R (A) = R (C) = 2 then – 6 = 0 and – 16 = 0 = 6 and = 16 (iv) Putting = 2 and = 8 in (1), we get 1 : 6 1 x 1 1 1 1 6 0 1 4 : 4 0 1 4 y 4 0 0 4 : 8 0 0 4 z 8 x+ y+ z=6 y + 4z = 4 – 4z = – 8 z=2 Putting z = 2 in (3), we get y+8=4 y=–4 Putting y = – 4, z = 2 in (2), we get x–4+2=6 x=8 Hence, x = 8, y = – 4, z = 2 Ans. Example 42. Find for what values of k the set of equations 2x – 3y + 6z – 5t = 3, y – 4z + t = 1, 4x – 5y + 8z – 9t = k has (i) no solution (ii) infinite number of solutions. (A.M.I.E.T.E., Summer 2004) Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ Determinants and Matrices 309 Solution. The augmented matrix C = [A, B] R3 R3 – 2 R1 . 6 5 3 6 5 . 3 2 3 2 3 0 . 1 ~ 0 . 1 1 4 1 1 4 1 4 5 0 8 9 . k 1 4 1 . k 6 6 5 . 3 2 3 ~ 0 1 4 1 . 1 0 0 0 0 . k 7 R3 R3 R2 (i) There is no solution if R (A) R (C) k – 7 0 or k 7, R (A) = 2 and R (C) = 3. (ii) There are infinite solutions if R (A) = R (C) = 2 k – 7= 0 k = 7 x 3 6 5 y 2 3 = 1 0 1 4 1 z t 2x – 3y + 6z – 5t = 3 y – 4z + t = 1 Let t = k1 and z = k2. From (2), y – 4k2 + k1 = 1 or y = 1 + 4k2 – k1 From (l), 2x – 3 – 12k2 + 3k1 + 6k2 – 5k1 = 3 2x = 6 + 6k2 + 2k1 x = 3 + 3k2 + k1 y = 1 + 4k2 – k1 z = k2, t = k1 4.43. HOMOGENEOUS EQUATIONS Ans. ...(1) ...(2) Ans. For a system of homogeneous linear equations AX = O (i) X = O is always a solution. This solution in which each unknown has the value zero is called the Null Solution or the Trivial solution. Thus a homogeneous system is always consistent. A system of homogeneous linear equations has either the trivial solution or an infinite number of solutions. (ii) If R (A) = number of unknowns, the system has only the trivial solution. (iii) If R (A) < number of unknowns, the system has an infinite number of non-trivial solutions. A system of homogeneous linear equations AX = O Always has a solution Find R (A) R (A) = n (no. of unknowns) Unique or trivial solution (each unknown equal to zero) R (A) < n (no. of unknowns) Infinite no. of non-trivial solutions Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ 310 Determinants and Matrices Example 43. Determine ‘b’ such that the system of homogeneous equations 2x + y +2z = 0 ; x + y +3z = 0 ; 4x +3y + bz = 0 has (i) Trivial solution (ii) Non-Trivial solution . Find the Non-Trivial solution using matrix method. (U.P., I Sem Dec 2008) Solution. Here, we have 2x + y + 2z = 0 x + y + 3z = 0 4x + 3y + bz = 0 (i) For trivial solution: We know that x = 0, y = 0 and z = 0. So, b can have any value. (ii) For non-trivial solution: The given equations are written in the matrix form as : 2 1 2 x 1 1 3 y = 4 3 b z AX=B R1 R2, 2 1 2 : C = 1 1 3 : 4 3 b : 0 1 1 3 : 0 ~ 2 1 2 : 0 4 3 b : 0 0 0 R2 R2 – 2R1, R3 R3 – 4R1, R3 R3 – R2 0 1 1 3 : 0 1 1 3 : 0 0 ~ 0 1 4 : 0 ~ 0 1 4 : 0 0 0 1 b 12 : 0 0 0 b 8 : 0 For non trivial solution or infinite solutions R (C) = R (A) = 2 < Number of unknowns b – 8 = 0, b = 8 Ans. Example 44. Find the values of k such that the system of equations x + ky + 3z = 0, 4x + 3y + kz = 0, 2x + y + 2z = 0 has non-trivial solution. Solution. The set of equations is written in the form of matrices 0 1 k 3 x 4 3 k y = 0 , AX = B, C = [A : B] = 0 2 1 2 z On interchanging first and third rows, we have 1 k 3 : 0 4 3 k : 0 2 1 2 : 0 2 1 2 : 0 4 3 k : 0 1 k 3 : 0 R2 R2 – 2 R1, R3 R3 2 1 ~ 0 1 1 0 k 2 2 k 4 : : 2 : 1 R1 2 1 R3 R3 k R2 2 0 2 1 2 0 ~ 0 1 k 4 1 0 0 0 2 k (k 4) 2 : 0 : 0 : 0 Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ Determinants and Matrices 311 For a non-trivial solution or for infinite solution, R (A) = R (C) = 2 1 k 2 k (k 4) = 0 2 k 2 4k 2 = 0 2 2 so 9 9 9 k = 0 k k = 0 k = , k = 0 Ans. 2 2 2 4.44 CRAMER’S RULE Example 45. Find values of for which the following system of equations is consistent and has non-trivial solutions. Solve equations for all such values of . k2 ( – 1) x + (3 + 1) y + 2z = 0 ( – 1) x + (4 – 2) y + ( + 3) z = 0 2x + (3 + 1) y + 3 ( – 1) z = 0 2 x ( 1) (3 1) 0 ( 1) (4 2) ( 3) y = 0 (3 1) (3 3) z 0 2 AX = 0 Solution. For infinite solutions, 1 3 1 0 2 3 = 0, 3 1 3 3 0 0 3 1 4 2 2 3 1 3 3 = 0, 3 3 3 1 5 1 2 ...(1) |A|=0 1 4 2 2 (A.M.I.E.T.E., Summer 2010, 2001) 3 = 0, 6 2 3 3 ( – 3) [( – 1) (6 – 2) – 2 (5 + 1)] = 0 [62 – 8 + 2 – 10 – 2] = 0 or 62 – 18 = 0 or 6 ( – 3) = 0, = 3 On putting = 3 in (1), we get 0 2 10 6 x 0 2 10 6 x 0 0 0 0 y 0 2 10 6 y = 0 0 0 0 z 0 2 10 6 z 2x + 10y + 6z = 0 x + 5y + 3z = 0 Let x = k1, y = k2, 3z = – k1 – 5 k2 z = k1 5k 2 3 3 Ans. EXERCISE 4.16 Test the consistency of the following equations and solve them if possible. 1. 3x + 3y + 2z = 1, x + 2y = 4, 10y + 3z = – 2, 2x – 3y – z = 5 Ans. Consistent, x = 2, y = 1, z = – 4 2. (R.G.P.V. Bhopal 1st Sem 2001) x1 – x2 + x3 – x4 + x5 = 1, 2x1 – x2 + 3x3 + 4x5 = 2, 3x1 – 2x2 + 2x3 + x4 + x5 = 1, x1 + x3 + 2x4 + x5 = 0 (A.M.I.E.T.E., Winter 2003) Ans. x1 = – 3k1 + k2 – 1, x2 = –3k1 – 1, x3 = k1 – 2k2 + 1, x4 = k1, x4 = k1, x5 = k21 Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ 312 Determinants and Matrices 3. Find the value of k for which the following system of equations is consistent. 3x1 – 2x2 + 2x3 = 3, x1 + kx2 – 3x3 = 0, 4x1 + x2 + 2x3 = 7 Ans. k = 1 4 4. Find the value of for which the system of equations x + y + 4z = 1, x + 2y – 2z = 1, x + y + z = 1 will have a unique solution. (A.M.I.E., Winter 2000) Ans. 7 10 3 2 1 x b 5. Determine the values of a and b for which the system 5 8 9 y 3 2 1 a z 1 (i) has a unique solution, (ii) has no solution and, (iii) has infinitely many solutions. 1 1 , (iii) a = –3, b = 3 3 6. Choose that makes the following system of linear equations consistent and find the general solution of the system for that . x + y – z + t = 2, 2y + 4z + 2t = 3, x + 2y + z + 2t = Ans. (i) a –3, (ii) a = –3, b Ans. = 7 1 3 , x = 3 k 2 , y = 2 k2 k1 , z = k2, t = k1 2 2 2 7. Show that the equations 3x + 4y + 5z = a, 4x + 5y + 6z = b, 5x + 6y + 7z = c don’t have a solution unless a + c = 2b. Solve the equations when a = b = c = – 1 Ans. x = k + 1, y = – 2 k – 1,z = k 8. Find the values of k, such that the system of equations 4 x1 + 9x2 + x3 = 0 , kx1 + 3x2 + kx3 = 0, x1 + 4x2 + 2x3 = 0 has non-trivial solution. Hence, find the solution of the system. Ans. k = 1, x1 = 2 , x2 = –, x3 = 9. Find values of for which the following system of equations has a non-trivial solution. 3x1 + x2 – x3 = 0, 2x1 + 4x2 + x3 = 0, 8x1 – 4x2 – 6x3 = 0 Ans. = 1 10. Find value of so that the following system of homogeneous equations have exactly two linearly independent solutions x1 – x2 – x3 = 0, – x1 + x2 – x3 = 0, – x1 – x2 + x3 = 0, Ans. = – l 11. Find the values of k for which the following system of equations has a non-trivial solution. (3k – 8) x + 3y + 3z = 0, 3x + (3k – 8) y + 3z = 0, 3x + 3y + (3k – 8) z = 0 (AMIETE, June 2010) Ans. k = 12. Solve the homogeneous system of equations : 4x + 3y – z = 0, 3x + 4y + z = 0, x – y – 2z = 0, 5x + y – 4z = 0 2 11 , 3 3 Ans. x = k, y = – k, z = k 1 2 1 13. If A = 3 1 2 Ans. (i) 1, (ii) – 1 1 0 find the values of for which equation AX = 0 has (i) a unique solution, (ii) more than one solution. 14. Show that the following system of equations: x + 2y – 2u = 0, 2x – y – u = 0, x + 2z – u = 0, 4x – y + 3z – u = 0 do not have a non-trivial solution. 15. Determine the values of and such that the following system has (i) no solution (ii) a unique solution (iii) infinite number of solutions: 2x – 5y + 2z = 8, 2x + 4y + 6z = 5, x + 2y + z = Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ Determinants and Matrices 313 5 5 (ii) 3, (iii) = 3, = 2 2 16. Test the following system of equations for consistency. If possible, solve for non-trivial solutions. Ans. (i) = 3, 3x + 4y – z – 6t = 0, 2x + 3y + 2z – 3t = 0, 2x + y – 14z – 9t = 0, x + 3y + 13z + 3t = 0 (A.M.I.E.T.E., Winter 2000) Ans. x = 11k1 + 6k2, y = –8k1 – 3k2, z = k1 t = k2 17. Given the following system of equations 2x – 2y + 5z + 3z = 0, 4x – y + z + w = 0, 3x – 2y + 3z + 4w = 0, x – 3y + 7z + 6w = 0 Reduce the coefficient matrix A into Echelon form and find the rank utilising the property of rank, test the given system of equation for consistency and if possible find the solution of the given system. (A.M.I.E.T.E., Summer 2001) Ans. x = 5k, y = 36k, z = 7k, w = 9k 18. Find the values of for which the equations (2 – ) x + 2y + 3 = 0, 2x + (4 – ) y + 7 = 0, 2x + 5y + (6 – ) = 0 are consistent and find the values of x and y corresponding to each of these values of . (R.G.P.V, Bhopal I sem. 2003, 2001) Ans. = 1, – 1, 12. 4.45 LINEAR DEPENDENCE AND INDEPENDENCE OF VECTORS Vectors (matrices) X1, X2, .... Xn are said to be dependent if (1) all the vectors (row or column matrices) are of the same order. (2) n scalars 1, 2, ... n (not all zero) exist such that 1 X1 + 2 X2 + 3 X3 + ..... + n Xn = 0 Otherwise they are linearly independent. Remember: If in a set of vectors, any vector of the set is the combination of the remaining vectors, then the vectors are called dependent vectors. Example 46. Examine the following vectors for linear dependence and find the relation if it exists. X1 = (1, 2, 4), X2 = (2, –1, 3), X3 = (0, 1, 2), X4 = (–3, 7, 2) (U.P., I Sem. Winter 2002) Solution. Consider the matrix equation 1 X1 + 2 X2 + 3 X3 + 4 X4 = 0 1 (1, 2, 4) + 2 (2, –1, 3) + 3 (0, 1, 2) + 4 (– 3, 7, 2) = 0 1 + 22 + 03 – 34 = 0 21 – 2 + 3 + 74 = 0 41 + 32 + 23 + 24 = 0 This is the homogeneous system 1 1 2 0 3 0 2 1 1 7 2 = 0 or A = 0 3 2 0 4 3 2 4 1 1 2 0 3 0 0 5 1 13 2 = 0 3 0 5 2 14 0 4 1 1 2 0 3 0 0 5 1 13 2 = 0 3 0 0 1 1 0 4 R2 R2 2 R1 R3 R3 4 R1 R3 R3 R2 Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ 314 Determinants and Matrices 1 + 2 2 – 3 4 = 0 –5 2 + 3 + 13 4 = 0 3 + 4 = 0 4 = t, 3 + t = 0, 3 = – t Let 12 t 5 9 t 24 t 1 3 t = 0 or = 1 5 5 Hence, the given vectors are linearly dependent. Substituting the values of in (1), we get – 52 – t + 3 t = 0, 2 = 9 X 1 12 X 2 9 t X 1 12 t X3 X4 = 0 X2 t X3 t X4 = 0 5 5 5 5 9 X1 – 12 X2 + 5 X3 – 5 X4 = 0 – Ans. Example 47. Define linear dependence and independence of vectors. Examine for linear dependence [1, 0, 2, 1], [3, 1, 2, 1], [4, 6, 2, –4], [–6, 0, –3, –4] and find the relation between them, if possible. Solution. Consider the matrix equation 1 X1 + 2 X2 + 3 X3 + 4 X4 = 0 ...(1) 1 (1, 0, 2, 1) + 2 (3, 1, 2, 1) + 3 (4, 6, 2, – 4) + 4 (– 6, 0, – 3, – 4) = 0 1 + 3 2 + 4 3 – 6 4 = 0 0 1 + 2 + 6 3 + 0 4 = 0 2 1 + 2 2 + 2 3 – 3 4 = 0 1 + 2 – 4 3 – 4 4 = 0 1 0 2 1 3 4 6 1 0 0 1 6 0 2 = 2 2 3 3 0 1 4 4 4 0 0 1 3 4 6 1 0 0 1 6 0 2 = 0 R3 R3 2 R1 0 4 6 9 3 2 4 0 R4 R4 R1 0 2 8 1 3 4 6 1 0 0 1 6 0 0 2 = 0 0 18 9 3 0 R3 R3 4 R2 2 4 0 0 4 0 R4 R4 2 R2 1 0 0 0 3 4 6 1 0 0 1 6 0 2 = 0 18 9 3 0 2 0 0 0 4 0 R4 R4 R3 9 Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ Determinants and Matrices 315 1 + 3 2 + 4 3 – 6 4 = 0 2 + 6 3 = 0 18 3 + 9 4 = 0 t 2 2 – 3 t = 0 or 2 = 3 t 1 + 9 t – 2t – 6t = 0 1 = – t Substituting the values of 1, 2, 3 and 4 in (1), we get Let 4 = t, 18 3 + 9 t = 0 or 3 = t X + t X4 = 0 or 2 X1 – 6 X2 + X3 – 2 X4 = 0 Ans. 2 3 4.46 LINEARLY DEPENDENCE AND INDEPENDENCE OF VECTORS BY RANK METHOD 1. If the rank of the matrix of the given vectors is equal to number of vectors, then the vectors are linearly independent. 2. If the rank of the matrix of the given vectors is less than the number of vectors, then the vectors are linearly dependent. Example 48. Is the system of vector – t X1 + 3 t X2 T T T X 1 2, 2,1 , X 2 1, 3,1 , X 3 1, 2, 2 linearly dependent . 2 1 1 2 , X 3 X 2 X Solution . Here 1 2 3 1 1 2 Consider the matrix equation (T stands for transposition ) 1 X 1 2 X 2 3 X 3 0 ...(1) 2 1 1 0 1 2 2 3 3 2 0 1 1 2 0 21 2 3 0 21 3 2 23 0 1 2 23 0 which is the homogeneous equation. R1 , R3 2 1 1 1 0 2 3 2 0 2 or 1 1 2 3 0 R 2 – 2R 1, R 3 – 2R 1 1 1 2 1 0 0 1 2 0 2 0 1 3 3 0 1 1 2 1 0 2 3 2 0 2 2 1 1 3 0 R3 + R2 1 1 2 1 0 or 0 1 2 2 0 0 0 5 3 0 Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ 316 Determinants and Matrices 1 + 2 + 23 = 0 2 – 23 = 0 –53 = 0 3 = 0 2 = 0 and 1 = 0 Thus non-zero values of 1, 2, 3 do not exist which can satisfy (1). Hence by definition, the given system of vectors is not linearly dependent. Ans. Example 49. Show using a matrix that the set of vectors X = [1, 2,– 3, 4], Y = [3, – 1, 2, 1] , Z = [1, –5, 8, –7] is linearly dependent. Solution. Here, we have X = [1, 2, –3, 4], Y = [3, –1, 2, 1], Z = [1, –5, 8, –7] Let us form a matrix of the above vectors 4 1 2 3 4 1 2 3 1 3 1 2 1 0 7 11 11 R2 R2 3R1 0 1 5 8 7 0 7 11 11 R3 R3 R1 0 Here the rank of the matrix = 2 < Number of vectors Hence, vectors are linearly dependent. Example 50. Show using a matrix that the set of vectors [4, 5, 14, 14], [5, 10, 8, 4] is linearly independent. Solution. Here, the given vectors are [2, 5, 2, –3], [3, 6, 5, 2], [4, 5, 14, 14], [5, 10, 8, 4] Let us form a matrix of the above vectors : 2 3 4 7 11 11 0 0 0 R3 R3 R2 Proved. : [2, 5, 2, –3], [3, 6, 5, 2], 2 5 2 3 2 5 2 3 3 6 5 2 1 1 3 5 R2 R2 R1 4 5 14 14 1 1 9 12 R3 R3 R2 5 10 8 4 1 5 6 10 R4 R4 R3 5 R1 R2 1 3 5 1 1 3 1 0 2 5 2 3 R R 3 4 13 R2 R2 2 R1 1 2 0 2 1 1 9 12 6 7 R3 R3 R1 4 9 15 R4 R4 R1 0 1 5 6 10 3 5 1 1 1 1 0 3 4 13 0 3 10 5 2 0 0 R3 R3 R2 0 0 3 3 3 7 4 0 0 11 0 0 R4 R4 R2 3 3 3 Here, the rank of the matrix = 4 = Number of vectors Hence, the vectors are linearly independent. 3 5 4 13 10 5 3 3 11 1 R4 R4 R3 0 10 2 Proved. EXERCISE 4.17 Examine the following system of vectors for linear dependence. If dependent, find the relation between them. 1. X1 = (1, –1, 1), X2 = (2, 1, 1), X3 = (3, 0, 2). 2. X1 = (1, 2, 3), X2 = (2, – 2, 6). Ans. Dependent, X1 + X2 – X3 = 0 Ans. Independent Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ Determinants and Matrices 317 3. X1 = (3, 1, – 4), X2 = (2, 2, – 3), X3 = (0, – 4, 1). Ans. Dependent, 2 X1 – 3 X2 – X3 = 0 4. X1 = (1, 1, 1, 3), X2 = (1, 2, 3, 4), X3 = (2, 3, 4, 7). Ans. Dependent, X1 + X2 – X3 = 0 5. X1 = (1, 1, –1, 1), X2 = (1, –1, 2, –1), X3 = (3, 1, 0, 1). Ans. Dependent, 2 X1 + X2 – X3 = 0 6. X1 = (1, –1, 2, 0), X2 = (2, 1, 1, 1), X3 = (3, –l, 2, –l), X4 = (3, 0, 3, 1). Ans. Dependent, X1 + X2 – X4 = 0 7. Show that the column vectors of following matrix A are linearly independent: 1 0 0 A = 6 2 1 4 3 2 8. Show that the vectors x1 = (2, 3, 1, –1), x2 = (2, 3, 1, –2), x3 = (4, 6, 2, 1) are linearly dependent. Express one of the vectors as linear combination of the others. 9. Find whether or not the following set of vectors are linearly dependent or independent: (i) (1, –2), (2, 1), (3, 2) (ii) (1, 1, 1, 1), (0, 1, 1, 1), (0, 0, 1, 1), (0, 0, 0, 1). Ans. (i) Dependent (ii) Independent 10. Show that the vectors x1 = (a1, b1), x2 = (a2, b2) are linearly dependent if a1 b2 – a2 b1 = 0. 4.47 ANOTHER METHOD (ADJOINT METHOD) TO SOLVE LINEAR EQUATIONS Let the equations be a1 x + a2 y + a3 z = d1 b1 x + b2 y + b3 z = d2 c1 x + c2 y + c3 z = d3 We write the above equations in the matrix form a1 a2 a1 x a2 y a3 z d1 b b b x b y b z d 2 2 3 1 = 2 or 1 c1 c2 c1 x c2 y c3 z d3 AX = B a3 x d1 b3 y = d2 d3 c3 z ...(1) d1 x a1 a2 a3 where A = b1 b2 b3 , X = y and B = d2 d3 z c1 c2 c3 – 1 Multiplying (1) by A . A– 1 AX = A– 1 B or IX = A–1 B or X = A–1 B. Example 51. Solve, with the help of matrices, the simultaneous equations x + y + z = 3, x + 2y + 3z = 4, x + 4y + 9z = 6 (A.M.I.E., Summer 2004, 2003) Solution. The given equations in the matrix form are written as below: 1 1 1 x 3 1 2 3 y = 4 1 4 9 z 6 AX = B 1 1 1 x 1 2 3 , X where A= = y , B = 1 4 9 z – 1 Now we have to find out the A . 3 4 6 Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ 318 Determinants and Matrices | A | = l × 6 + l × (– 6) + l × 2 = 6 – 6 + 2 = 2 1 2 6 5 6 6 6 5 8 2 8 3 , Adjoint A = Matrix of co-factors = 1 1 2 3 1 2 1 6 5 1 1 – 1 8 2 A = Adjoint A = 6 2 | A| 1 2 3 1 3 6 5 1 8 2 4 X = A–1 B = 6 2 1 6 2 3 x 18 20 6 4 2 1 1 y = 18 32 12 2 1 2 2 z 0 0 6 12 6 x = 2, y = 1, z = 0 Example 52. Given the matrices 1 2 3 x 1 3 –1 1 , X y and C 2 A 4 2 1 z 3 Ans. Write down the linear equations given by AX = C and solve for x, y, z by the matrix method. Solution. AX = C 1 2 3 x 1 3 –1 1 y = 2 4 2 1 z 3 X = A–1 . C x 1 2 3 y = 3 –1 1 z 4 2 1 1 1 2 3 1 10 3 4 11 6 Matrix of co-factors of A = 5 8 7 | A | = 1 (–3) + 2 (1) + 3 (10) = –3 + 2 + 30 = 29 5 3 4 1 11 8 Adj. A = 10 6 7 5 3 4 1 1 Adj. A = 1 11 8 |A| 29 10 6 7 X = A–1 C A–1 = Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ Determinants and Matrices 319 5 1 3 4 x 1 y 1 11 8 2 = 29 10 6 7 3 z 20 29 3 8 15 20 x 1 1 3 y 1 22 24 = 3 = = 29 29 29 10 12 21 1 z 1 29 20 3 1 Hence, x = ,y= ,z= 29 29 29 Example 53. Let y1 5 x1 3x2 3x3 , y2 3x1 2 x2 2 x3 , y3 2 x1 x2 2 x3 Ans. be a linear transformation from x1 , x2 , x3 to y1 , y2 , y3 and z1 4 x1 2 x3 , z2 x2 4 x3 , z3 5x3 be a linear transformation from x1 , x2 , x3 to z1 , z2 , z3 .Find the linear transformation from z1 , z2 , z3 to y1 , y2 , y3 by inverting appropriate matrix and matrix multiplication. (A.M.I.E.T.E.,Dec. 2004) y1 5 3 3 x1 y 3 2 2 x2 Solution: Hence 2 ...(1) y3 2 1 2 x3 z1 4 0 2 x1 z 0 1 4 x2 and 2 z3 0 0 5 x3 1 x1 4 0 2 z1 5 0 2 z1 1 x2 0 1 4 z 2 0 20 16 z2 20 x3 0 0 5 z3 0 0 4 z3 ...(2) x1 x Putting the value of 2 from (2) in (1) , we get x3 y1 5 3 3 5 0 2 z1 25 60 46 z1 1 1 y2 3 2 2 20 0 20 16 z2 20 15 40 46 z2 y3 2 1 2 0 0 10 20 20 z3 4 z3 Ans. EXERCISE 4.18 Solve the following equations 1. 3x + y + 2z = 3, 2x – 3y – z = – 3, x + 2y + z = 4 2. x + 2y + 3z = 1, 2x + 3y + 8z = 2, x + y + z = 3 (A.M.I.E. Winter 2001) Ans. x = 1, y = 2, z = – 1 9 1 Ans. x , y 1, z 2 2 Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ 320 Determinants and Matrices 3. 4. 5. 4x + 2y – z = 9, x – y + 3z = – 4, 2x + z = 1 5x + 3y + 3z = 48, 2x + 6y – 3z = 18, 8x – 3y + 2z = 21 x + y + z = 6, x – y + 2z = 5, 3x + y + z = 8 Ans. x =1, y = 2, z = – 1 Ans. x = 3, y = 5, z = 6 Ans. x = 1, y = 2, z = 3 6. x + 2y + 3z = 1, 3x – 2y + z = 2, 4x + 2y + z = 3 Ans. x 7. 9x + 4y + 3z = – 1, 5x + y + 2z = 1, 7x + 3y + 4z = 1 8. x + y + z = 8, x – y + 2z = 6, 9x + 5y – 7z = 14 9. 3x + 2y + 4z = 7, 2x + y + z = 4, x + 3y + 5z = 2 10. Represent each of the transformations x1 = 3 y1 + 2 y2, x2 = – y1 + 4 y2 and 7 3 1 ,y ,z 10 40 20 Ans. x = 0, y = – 1, z = 1 5 4 Ans. x = 5, y = , z = 3 3 Ans. x 9 9 5 ,y ,z 4 8 8 y1 = z1 + 2z2, y2 = – 3 z1 by the use of matrices, find the composite transformation which expresses x 1, x 2 in terms of z1, z2. Ans. x1 = – 3z1 + 6 z2, x2 = – 13z1 – 2z2 4.48 PARTITIONING OF MATRICES Sub matrix. A matrix obtained by deleting some of the rows and columns of a matrix A is said to be sub matrix. 4 1 0 4 1 5 2 1 0 For example, A = 5 2 1 , then , , are the sub matrices. 5 2 6 3 2 1 6 3 4 Partitioning: A matrix may be subdivided into sub matrices by drawing lines parallel to its rows and columns. These sub matrices may be considered as the elements of the original matrix. 2 1 : 0 4 1 1 0 : 2 3 4 For example, A= .... .... : .... .... .... 4 5 : 1 6 5 2 1 0 4 1 A11 = 1 0 , A12 = 2 3 4 A21 = [4 5], A22 = [1 6 5] A11 A12 A = A 21 A22 So, the matrix is partitioned. The dotted lines divide the matrix into sub-matrices. A11, A12, A21, A22 are the sub-matrices but behave like elements of the original matrix A. The matrix A can be partitioned in several ways. Addition by submatrices: Let A and B be two matrices of the same order and are partitioned identically. For example; 3 1 4 6 2 3 4 5 2 1 0 4 0 1 2 3 B A= 4 5 1 2 3 4 5 6 4 5 0 1 1 3 4 5 Then we may write Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ Determinants and Matrices 321 A11 A12 B11 B12 A , B B A B22 A = 21 22 21 A31 A32 B31 B32 A11 B11 A12 B12 A B A22 B22 21 A + B = 21 A31 B31 A32 B32 4.49 MULTIPLICATION BY SUB-MATRICES Two matrices A and B, which are conformable to the product AB are partitioned in such a way that the columns of A partitioned in the same way as the rows of B are partitioned. But the rows of A and columns of B can be partitioned in any way. For example, Here A is a 3 × 4 matrix and B is 4 × 3 matrix. 4 5 6 3 2 1 1 2 3 4 0 1 2 3 and B 1 0 4 A = 1 4 1 2 2 5 3 The partitioning of the columns of A is the same as the partitioning of the rows of B. Here, A is partitioned after third column, B has been partitioned after third row. Example 54. If C and D are two non-singular matrices, show that if C A = 0 Solution. Let Then C 1 0 1 , then A D 0 E F A–1 = G H C 0 E AA–1 = 0 D G CE 0G CF 0 H 0 E DG 0 F DH CE + 0G CF + 0H 0E + DG OF + DH Since, C is non singular and CF CE Similarly, D is non singular and DG Putting these values in (1), we get So that 0 D 1 ...(1) F CE 0G CF 0 H H 0 E DG 0 F DH I 0 = 0 I = I CE = I = 0 CF = 0 = 0 DG = 0 = I DH = I = 0, F=0 = I E = C–1 = 0 G = 0 and DH = I H = D–1 C 1 0 = Proved. 1 0 D 4.50 Inverse By Partitioning: Let the matrix B be the inverse of the matrix A. Matrices A and B are partitioned as A11 A12 B11 B12 A = A , B B A 21 22 21 B22 Since, AB = BA = I A–1 Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ 322 Determinants and Matrices A11 A 21 A11 B11 A12 B21 A B A B 21 11 22 21 A12 B11 A22 B21 B12 B11 = B B22 21 B12 A11 B22 A21 A11 B12 A12 B22 B11 A11 B12 A21 = B A B A A21 B12 A22 B22 21 11 22 21 A12 I 0 A22 0 I B11 A12 B12 A22 I 0 B21 A12 B22 A22 0 I Let us solve the equations for B11, B12, B21 and B22. Let, B22 = M –1 From (2), 1 1 B12 = A11 ( A22 B22 ) ( A11 A22 ) M 1 From (3), 1 1 1 B21 = ( B22 A21 ) A11 M ( A21 A11 ) From (1), 1 1 1 1 B11 = A11 A11 ( A12 B21 ) A11 ( A11 A12 ) B21 1 1 1 = A11 ( A11 A12 ) M 1 ( A21 A11 ) 1 M = A22 – A21 ( A11 A22 ) Here Note: A is usually taken of order n – 1. Example 55. Find the inverse of the following matrix by partitioning 1 3 3 1 4 3 1 3 4 Solution. Let the matrix be partitioned into four submatrices as follows: Let 1 3 3 A = 1 4 3 1 3 4 1 3 3 ; A12 A11 = 1 4 3 A21 = [1 B11 We have to find A–1 = B21 3]; A22 = [4] B12 where B22 1 1 1 B11 = A11 ( A11 A12 ) ( M 1 ) ( A21 A11 ) 1 B21 = M 1 ( A21 A11 ) B12 = A111 A12 M 1 ; B22 = M and Now –1 M = A22 A21 ( A111 A12 ) 4 3 4 3 3 3 A111 = 1 1 ; A111 A12 = 1 1 3 = 0 4 3 A21 A111 = 1 3 1 1 = 1 0 Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ Determinants and Matrices 323 3 M = [4] [1 3] 0 = [4] – [3] = [1] M–1 = [3] 4 3 3 4 3 3 0 7 3 B11 = 1 1 0 [1 0] = 1 1 0 0 B111 = 1 1 B21 = [1] [1 0] = [1 0] 3 B12 = 0 B22 = [1] B11 A–1 = B 21 7 3 3 B12 = 1 1 0 B22 1 0 1 Ans. 1 2 3 1 1 3 3 2 by partitioning. Example 56. Find the inverse of A = 2 4 3 3 1 1 1 1 1 2 3 1 3 3 and partition so that Solution. (a) Take G3 = 2 4 3 1 2 3 A11 = 1 3 , A12 = 3 , A21 = 2 4 , and A22 = [3] Now, 3 2 3 3 3 2 1 , A11 A111 = A12 = 1 1 3 = 0 , 1 1 3 2 1 = 2 4 1 1 = 2 0 A21 A11 3 1 M = A22 A21 ( A11 A12 ) = [3] [2 4] 0 = [–3], And M–1 = [1 3] Then 3 2 3 1 3 2 2 0 1 1 1 B11 = A11 ( A11 A12 ) M 1( A21 A11 ) = 1 1 0 3 [2 0] = 1 1 0 0 = 1 3 6 3 3 3 1 3 1 B12 = ( A11 A12 ) M 1 = 3 0 1 1 B21 = M 1 ( A21 A11 ) = [2 0] 3 Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ 324 Determinants and Matrices 1 B22 = M–1 = 3 1 3 and G 3 6 3 B12 1 = 3 3 0 B22 3 2 0 1 B11 = B 21 1 2 3 1 3 3 , A12 = (b) Partition A so that A11 = 2 4 3 1 Now, A11 1 2 , A21 = 1 1 1 , and A22 = [1]. 3 3 6 3 0 1 1 1 1 1 3 3 0 = , A11 A12 = 3 3 , A21 A11 = 3 2 3 2 3 2 0 1 1 0 1 1 M = [1] [1 1 1] 3 = . and M–1 = [3] 3 3 1 Then B11 3 1 = 3 3 2 3 1 = 3 3 2 6 3 0 1 1 3 0 3 [3] [2 3 2] 3 3 1 0 1 6 3 3 0 0 1 0 1 0 6 9 6 1 3 2 0 1 3 2 0 2 2 1 2 1 1 0 3 , B12 = B21 = [– 2 3 – 2], B22 = [3] 1 1 0 1 2 2 3 B11 B12 1 2 A–1 = 1 1 1 B21 B22 0 3 2 3 2 Ans. EXERCISE 4.19 1. Compute A + B using partitioning 4 6 A= 0 1 1 0 5 3 1 7 8 1 , B 2 2 1 1 2 0 1 0 2 1 1 0 1 1 1 2 1 1 2 3 2. Compute AB using partitioning 2 1 2 0 1 0 A = 4 1 3 2 , B 4 2 1 3 0 2 3 1 4 6 11 1 4 24 18 18 Ans. 1 2 16 10 12 1 2 A B 3. Find the inverse of C 0 where B, C are non-singular. 0 Ans. 1 B C 1 B 1 AC 1 Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ Determinants and Matrices 325 Find the inverse of the following metrices by partitioning: 4. 2 1 1 1 3 2 1 2 1 6. 2 3 4 4 3 1 1 2 4 8. 3 2 52 2 Ans. 1 3 5 1 3 1 5 10 5 5 5 4 9 10 1 Ans. 15 4 14 5 5 1 6 5. 7. 1 2 1 3 1 5 1 1 2 1 Ans. 5 3 1 14 2 1 1 1 5 3 1 2 3 2 4 5 3 5 6 4 2 7 3 3 2 7 3 9 3 2 3 1 3 2 3 3 1 Ans. 2 1 0 1 11 7 26 16 1 1 7 3 Ans. 1 1 0 2 1 2 1 1 1 Choose the correct answer: 9. If 3x + 2y + z = 0, x + 4y + z = 0, 2x + y + 4z = 0, be a system of equations then (i) System is inconsistent (ii) it has only trivial solution (iii) it can be reduced to a single equation thus solution does not exist (iv) Determinant of the coefficient matrix is zero. (AMIETE, June 2010) Ans. (ii) 4.51 EIGEN VALUES y1 a1n x1 y a2 n x2 2 a3 n x3 = y3 yn ann xn AX = Y ...(1) Where A is the matrix , X is the column vector and Y is also column vector. Here column vector X is transformed into the column vector Y by means of the square matrix A. Let X be a such vector which transforms intoX by means of the transformation (1). Suppose the linear transformation Y = AX transforms X into a scalar multiple of itself i.e. X. AX = Y = X AX – IX = 0 (A – I) X = 0 ...(2) Thus the unknown scalar is known as an eigen value of the matrix A and the corresponding non zero vector X as eigen vector. The eigen values are also called characteristic values or proper values or latent values. a11 a12 a13 a 21 a22 a23 Let a31 a32 a33 an1 an 2 an 3 Let 2 2 1 A 1 3 1 1 2 2 1 2 2 1 1 0 0 2 2 A I 1 3 1 0 1 0 1 3 1 1 2 2 0 0 1 1 2 2 characteristic matrix Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ 326 Determinants and Matrices (b) Characteristic Polynomial: The determinant | A – I | when expanded will give a polynomial, which we call as characteristic polynomial of matrix A. 2 For example; 1 1 2 1 3 1 2 2 = ( 2 – ) (6 – 5 + 2 – 2) – 2 (2 – – 1) + 1( 2 – 3 + ) = – 3 + 7 2 – 11 + 5 (c) Characteristic Equation: The equation | A – I | = 0 is called the characteristic equation of the matrix A e.g. 3 – 72 + 11 – 5 = 0 (d) Characteristic Roots or Eigen Values: The roots of characteristic equation | A – I | = 0 are called characteristic roots of matrix A. e.g. 3 – 7 2 + 11 – 5 = 0 ( – 1) ( – 1) ( – 5) = 0 = 1, 1, 5 Characteristic roots are 1, 1, 5. Some Important Properties of Eigen Values (AMIETE, Dec. 2009) (1) Any square matrix A and its transpose A have the same eigen values. Note. The sum of the elements on the principal diagonal of a matrix is called the trace of the matrix. (2) The sum of the eigen values of a matrix is equal to the trace of the matrix. (3) The product of the eigen values of a matrix A is equal to the determinant of A. (4) If 1, 2, ... n are the eigen values of A, then the eigen values of m (i) k A are k 1, k 2, ....., k n (ii) Am are 1m , m 2 ,......., n (iii) A–1 are 1 1 , , ...., . 1 2 n 6 2 2 Example 57. Find the characteristic roots of the matrix 2 3 1 2 1 3 Solution. The characteristic equation of the given matrix is 6 2 2 2 3 1 0 2 1 3 (6 – ) (9 – 6 + 2 – 1) + 2 (–6 + 2 + 2) + 2(2 – 6 + 2) = 0 –3 + l2 2 – 36 + 32 = 0 By trial, = 2 is a root of this equation. ( – 2) (2 – 10 + 16) = 0 – 2) ( – 2) ( – 8) = 0 = 2, 2, 8 are the characteristic roots or Eigen values. 1 2 3 Example 58. The matrix A is defined as A 0 3 2 0 0 2 Find the eigen values of 3 A3 + 5 A2 – 6A + 2I. Solution. | A – I | = 0 Ans. Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ Determinants and Matrices 327 1 2 0 3 3 2 0 0 0 2 (1 – ) (3 – ) (–2 – ) = 0 or = 1, 3, – 2 Eigen values of A3 = 1, 27, –8; Eigen values of A2 = 1, 9, 4 Eigen values of A = 1, 3, –2; Eigen values of I = 1, 1, 1 3 2 Eigen values of 3 A + 5A – 6A + 2I First eigen value = 3 (1)3 + 5 (1)2 – 6 (1) + 2(1) = 4 Second eigen value = 3 (27) + 5 (9) – 6 (3) + 2(1) = 110 Third eigen value = 3 (–8) + 5 (4) – 6 (–2) + 2 (1) = 10 Required eigen values are 4, 110, 10 m Ans. Example 59. If 1, 2, .... n are the eigen values of A, find the eigen values of the a r t r i x 2 (A – I) . Solution. (A – I)2 = A2 – 2 AI + 2 I2 = A2 – 2 A + 2 I Eigen values of A2 are 12 , 22 , 33 ... 2n Eigen values of 2 A are 2 1, 2 2, Eigen values of 2 I are 2. Eigen values of A2 – 2 A + 2 I 12 21 2 , 2 2 3 ... 2 n. 22 2 2 2 , 32 23 2 ... ..... 2 2 Ans. 1 , 2 , 3 , ... ( n )2 Example 60. Prove that a matrix A and its transpose A have the same characteristic roots. Solution. Characteristic equation of matrix A is | A – I | = 0 ... (1) Characteristic equation of matrix A is | A – I | = 0 ...(2) Clearly both (1) and (2) are same, as we know that | A | = | A | i.e., a determinant remains unchanged when rows be changed into columns and columns into rows. Proved. Example 61. If A and P be square matrices of the same type and if P be invertible, show that the matrices A and P–1 AP have the same characteristic roots. Solution. Let us put B = P–1 AP and we will show that characteristic equations for both A and B are the same and hence they have the same characteristic roots. B –I = P– 1 AP – I = P–1 AP – P–1 lP = P–1 (A – I) P | B – I | = |P–1 (A – I) P | = | P –1 | |A – I| | P | = |A – I | | P–1 | | P | = |A – I| | P–1P| = |A –I | | I | = | A – I| as | I | = 1 Thus the matrices A and B have the same characteristic equations and hence the same characteristic roots. Proved. Example 62. If A and B be two square invertible matrices, then prove that AB and BA have the same characteristic roots. Solution. Now AB = IAB = B–1 B (AB) = B –1 (BA) B ...(1) But by Ex. 8, matrices BA and B–1 (BA) B have same characteristic roots or matrices BA and Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ 328 Determinants and Matrices AB by (1) have same characteristic roots. Proved. Example 63. If A and B be n rowed square matrices and if A be invertible, show that the matrices A–1 B and BA–1 have the same characteristics roots. Solution. A–1 B = A–1 BI = A–1 B (A–1A) = A–1 (BA–1) A. ...(1) –1 –1 –1 But by Ex. 8, matrices BA and A (BA )A have same characteristic roots or matrices BA–1 and A–1 B by (1) have same characteristic roots. Proved. Example 64. Show that 0 is a characteristic root of a matrix, if and only if, the matrix is singular. Solution. Characteristic equation of matrix A is given by | A – I | = 0 If = 0, then from above it follows that | A | = 0 i.e. Matrix A is singular. Again if Matrix A is singular i.e., | A | = 0 then | A – I | = 0 | A | – | I | = 0, 0 – · 1 = 0 = 0. Proved. Example 65. Show that characteristic roots of a triangular matrix are just the diagonal elements of the matrix. Solution. Let us consider the triangular matrix. a11 0 a 21 a22 A a31 a32 a 41 a42 Characteristic equation is |A – I| = 0 a11 a21 a31 a41 or 0 a22 a32 a42 0 0 a33 a43 0 0 0 a44 0 0 a33 a43 0 0 0 0 a44 On expansion it gives (a11 – ) (a22 – ) (a33 – ) (a44 – ) = 0 = a11, a22, a33, a44 which are diagonal elements of matrix A. Proved. 1 Example 66. If is an eigen value of an orthogonal matrix, then is also eigen value. [Hint: AA = I if is the eigen value of A, then 1, 1 ] Example 67. Find the eigen values of the orthogonal matrix. 1 2 2 1 B= 2 1 –2 3 1 2 – 2 Solution. The characteristic equation of 2 2 1 A 2 1 2 2 2 1 is 1 2 2 1 2 2 2 2 0 1 1 1 1 4 2 2 1 4 2 4 2 1 0 (1–) (1 – 2 + 2 – 4) – 2 (2 – 2 + 4) + 2 (– 4 – 2 + 2) = 0 Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ Determinants and Matrices 329 3 3 2 9 27 0 3 2 3 0 The eigen values of A are 3, 3, –3, so the eigen values of B Note. If = 1 is an eigen value of B then its reciprocal 1 A are 1, 1, –1. 3 1 1 1 is also an eigen value of B. Ans. 1 EXERCISE 4.20 Show that, for any square matrix A. 1. If be an eigen value of a non singular matrix A, show that | A| is an eigen value of the matrix adj A. 2. There are infinitely many eigen vectors corresponding to a single eigen value. 3 3 3 3. Find the product of the eigen values of the matrix 2 1 1 1 5 6 Ans. 18 3 2 1 4. Find the sum of the eigen values of the matrix 1 3 2 4 1 5 Ans. 11 4 5. Find the eigen value of the inverse of the matrix 1 1 6 3 4 6 2 3 1 0 1 6. Find the eigen values of the square of the matrix 1 2 1 2 2 3 3 3 1 4 7. Find the eigen values of the matrix 0 2 6 0 0 5 Ans. –1, 1, 1 4 Ans. 1, 4, 9 Ans. 8, 27, 125 2 2 1 8. The sum and product of the eigen values of the matrix A 1 3 1 are respectively 1 2 2 (a) 7 and 7 (b) 7 and 5 (c) 7 and 6 (d) 7 and 8 (AMIETE, June 2010) Ans. (b) 4.52 CAYLEY-HAMILTON THEOREM Satement. Every square matrix satisfies its own characteristic equation. n If | A I | 1 n a1 n 1 a2 n 2 an be the characteristic polynomial of n n matrix A = (aij), then the matrix equation X n a1 X n 1 a2 X n 2 an I 0 is satisfied by X = A i.e., An a1 An1 a2 An 2 an I 0 Proof. Since the elements of the matrix A – I are at most of the first degree in , the elements of adj. (A – I) are at most degree (n –1) in . Thus, adj. (A – I) may be written as a matrix polynomial in , given by Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ 330 Determinants and Matrices Adj A I B0 n 1 B1 n 2 Bn 1 where B0 , B1 , , Bn 1 are n n matrices, their elements being polynomial in . We know that A I Adj A I | A I | I A I B0 n1 B1 n – 2 .... Bn 1 1n n a1 n 1 ... an I Equating coefficient of like power of on both sides, we get n IB0 1 I n AB0 IB1 1 a1 I n AB1 IB2 1 a2 I ................................. n ABn 1 1 an I On multiplying the equation by An , An 1 ,..., I respectively and adding, we obtain n 0 1 An a1 An 1 ... an I Thus An a1 An1 ... an I 0 for example, Let A be square matrix and if ...(1) 3 2 2 3 4 0 be its characteristic equation, then according to Cayley Hamilton Theorem (1) is satisfied by A. ...(2) A3 2 A2 3 A 4 I 0 We can find out A1 from (2). On premultiplying (2) by A1 , we get A2 – 2 A 3I 4 A1 0 1 A1 A2 2 A 3I 4 Example 68. Find the characteristic equation of the symmetric matrix 2 1 1 A 1 2 1 1 1 2 and verify that it is satisfied by A and hence obtain A–1. Express A6 – 6A5 + 9A4 – 2A3 – 12A2 + 23A – 9I in linear polynomial in A. (A.M.I.E.T.E., Summer 2000) Solution. Characteristic equation is |A – I| = 0 2 1 1 1 2 1 1 1 0 2 (2 – ) [(2 – )2 – 1] + 1 [–2 + + 1] + 1 [1 – 2 + ] = 0 or (2 – )3 – (2 – ) + – 1 + – 1 = 0 or (2 – )3 – 2 + + – 1 + – 1 = 0 or (2 – )3 + 3 – 4 = 0 Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ Determinants and Matrices 331 or 8 – 3 – 12 + 2 + 3– 4 = 0 or – 3 + 2 – 9+ 4 = 0 or 3 – 2 + 9– 4 = 0 By Cayley-Hamilton Theorem A3 – 6A2 + 9A – 4I = 0 ... (1) Verification: 2 1 1 2 1 1 A 1 2 1 1 2 1 1 1 2 1 1 2 2 2 2 1 2 1 2 6 5 4 1 1 2 2 1 1 4 1 1 2 2 5 6 1 1 4 5 5 2 1 2 1 2 2 6 5 A 5 6 5 5 3 5 2 1 5 1 2 6 1 1 5 5 6 1 1 2 6 5 10 22 21 12 5 5 6 10 5 10 6 5 5 12 5 5 6 10 21 22 = 5 5 12 21 21 10 5 6 5 10 6 21 21 22 A3 – 6A2 + 9A – 4I 22 21 21 22 = 21 21 21 6 5 21 6 5 6 22 5 5 22 36 18 4 21 30 9 0 = 21 30 9 0 22 36 18 4 21 30 9 0 21 30 9 0 5 2 1 5 9 1 2 6 1 1 1 1 0 1 4 0 1 2 0 0 0 0 1 21 30 9 0 0 0 0 21 30 9 0 0 0 0 0 22 36 18 4 0 0 0 So it is verified that the characteristic equation (1) is satisfied by A. Inverse of Matrix A, A3 – 6A2 + 9A – 4I = 0 On multiplying by A–1, we get A2 – 6A + 9I – 4A–1 = 0 6 5 4 A 5 6 5 5 1 or 5 2 1 5 6 1 2 6 1 1 6 12 9 5 6 0 5 6 0 6 12 9 5 6 0 5 6 0 or 4A–1 = A2 – 6A + 9I 1 1 0 0 1 9 0 1 0 2 0 0 1 560 3 1 1 1 5 6 0 , A1 1 3 1 4 6 12 9 3 1 1 A6 – 6A5 + 9A4 – 2A3 – 12A2 + 23A – 9I = A3 (A3 – 6A2 + 9A – 4I) + 2(A3 – 6A2 + 9A – 4I) + 5A – I = 5A – I Ans. Ans. Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ 332 Determinants and Matrices 2 1 1 Example 69. Find the characteristic equation of the matrix A 0 1 0 1 1 2 Verify Cayley Hamilton Theorem and hence prove that : 8 5 5 A8 5 A7 7 A6 3 A5 A4 5 A3 8 A2 2 A I 0 3 0 5 5 8 (Gujarat, II Semester, June 2009) Solution. Characteristic equation of the matrix A is 2 1 1 0 1 0 1 1 0 2 2 [1 2 ] 1 0 1 0 1 0 3 5 2 7 3 0 According to Cayley-Hamilton Theorem ...(1) A3 5 A2 7 A 3I 0 We have to verify the equation (1). 2 1 1 2 1 1 5 4 4 A 0 1 0 0 1 0 0 1 0 1 1 2 1 1 2 4 4 5 2 2 1 1 5 4 4 14 13 13 A3 A2 . A 0 1 0 0 1 0 0 1 0 1 1 2 4 4 5 13 13 14 14 13 13 5 4 4 2 1 1 1 0 0 A 5 A 7 A 3I 0 1 0 5 0 1 0 7 0 1 0 3 0 1 0 13 13 14 4 4 5 1 1 2 0 0 1 3 2 14 25 14 3 13 20 7 0 13 20 7 0 0 0 0 0 0 0 0 1 5 7 3 0 0 0 0 0 0 0 0 13 20 7 0 13 20 7 0 14 25 14 3 0 0 0 Hence Cayley Hamilton Theorem is verified. Now, A8 5 A7 7 A6 3 A5 A4 5 A3 8 A2 2 A I 5 =A A 3 5 A2 7 A 3I A A3 5 A2 7 A 3 I A2 A I 5 2 2 A O A O A A I A A I 5 4 4 0 1 0 4 4 5 5 2 1 0 0 0 4 1 0 2 1 1 1 0 0 0 1 0 0 1 0 1 1 2 0 0 1 4 1 0 1 1 1 4 1 0 4 1 0 8 5 5 0 0 0 0 3 0 5 5 8 5 2 1 Proved. Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ Determinants and Matrices 333 4.53 POWER OF MATRIX (by Cayley Hamilton Theorem) Any positive integral power Am of matrix A is linearly expressible in terms of those of lower degree, where m is a positive integer and n is the degree of characteristic equation such that m n. Example 70. Find A4 with the help of Cayley Hamilton Theorem, if 1 0 1 A 1 2 1 2 2 3 Solution. Here, we have 1 0 1 A 1 2 1 2 2 3 Characteristic equation of the matrix A is 1 0 1 3 6 2 11 6 0 1 2 1 0 1 2 3 0 2 2 3 Eigen values of A are 1, 2, 3. Let 4 3 6 2 11 6 Q a 2 b c 0 Put = 1 in (1), (1)4 = a + b + c 4 Put = 2 in (1), (2) = 4a + 2b + c Put = 3 in (1), (3)4 = 9a + 3b + c ...(1) a+b+c=1 4a + 2b + c = 16 9a + 3b + c = 81 (where Q () is quotient) ...(2) ... (3) ... (4) Solving (2), (3) and (4), we get a = 25, b = –60, c = 36 Replacing by matrix A in (1), we get A4 A3 6 A2 11A 6 Q A aA2 bA c 2 = O + aA + bA + cI 1 0 1 1 0 1 1 0 –1 1 0 0 25 1 2 1 1 2 1 60 1 2 1 36 0 1 0 2 2 3 2 2 3 2 2 3 0 0 1 0 60 36 0 0 25 50 100 60 125 150 100 60 120 60 0 36 0 250 250 225 120 120 180 0 0 36 50 0 0 100 60 0 49 50 40 25 60 36 125 60 0 150 120 36 100 60 0 65 66 40 250 120 0 250 120 0 225 180 36 130 130 81 (It is also solved by diagonalization method on page 496 Example 38.) Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ 334 Determinants and Matrices EXERCISE 4.21 1. Find the characteristic polynomial of the matrix 3 1 1 1 5 1 A = 1 1 3 Ans. A1 Verify Cayley-Hamilton Theorem for this matrix. Hence find A–1. 7 2 3 1 1 4 1 20 2 2 8 2. Use Cayley-Hamilton Theorem to find the inverse of the matrix cos sin sin cos 3. Using Cayley-Hamilton Theorem, find A–1, given that cos sin Ans. sin cos 2 1 3 1 0 2 A = 4 2 1 4 5 2 1 7 10 1 Ans. 5 0 1 2 4. Using Cayley-Hamilton Theorem, find the inverse of the matrix 5 5 1 0 2 0 5 3 15 3 0 1 Ans. 1 0 5 0 10 1 1 1 5. Find the characteristic equation of the matrix 1 3 7 A 4 2 3 1 2 1 (R.G.P.V., Bhopal, Summer 2004) Ans. 3 – 4 2 – 20 – 35 0 and show that the equation is also satisfied by A. 6. Find the eigenvalues of the matrix 2 3 1 3 1 3 5 2 4 Ans. Eigenvalues are 0, +1, –2 7. Using, Cayley-Hamilton Theorem obtain the inverse of the matrix 1 1 3 1 3 3 (R.G.P.V. Bhopal, I Sem., 2003) 2 4 4 1 2 2 8. Show that the matrix A 1 2 3 0 1 2 satisfies its characteristic equation. Hence find A–1. 24 1 10 8 2 7 1 Ans. 2 9 1 Ans. 8 12 2 6 2 2 2 10 2 1 1 4 9. Use Cayley Hamilton Theorem to find the inverse of 1 2 4 A 1 0 3 3 1 2 8 6 3 1 Ans. A 7 14 7 7 1 5 2 1 Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ Determinants and Matrices 335 10. Verify Cayley-Hamilton Theorem for the matrix 5 1 2 1 1 2 1 1 3 5 A= 3 1 1 Hence evaluate A–1. Ans. 11 7 1 2 2 3 1 1 4 5 4 3 11. If A 1A2 – A –10I in terms of A. , then express A – 4A – 7A + 11 2 3 (A.M.I.E.T.E., Winter 2001) 12. If 1, 2 and 3 are the eigenvalues of the matrix Ans. A + 5 I 5 2 9 5 10 7 then 1 + 2 + 3 is equal to 9 21 14 (i) –16 (ii) 2 (iii) –6 (iv) –14 Ans. (ii) 1 0 –1 13. The matrix A is given. The eivenvalues of 4A + 3A + 2l are 2 4 (A) 6, 15; (B) 9, 12 (C) 9, 15; (D) 7, 15 Ans. (C) 14. A(3 × 3) real matrix has an eigenvalue i, then its other two eigenvalues can be (A) 0, 1 (B) –1, i (C) 2i, –2i (D) 0, –i (A.M.I.E.T.E, Dec. 2004) 15. Verify Cayley-Hamilton theorem for the matrix 1 2 3 2 4 2 A = 1 1 2 16. Find adj. A by using Cayley-Hamilton thmeorem where A is given by 1 2 1 0 0 1 1 A = (R.G.P.V., Bhopal, April 2010) Ans. 3 3 1 1 3 3 3 2 1 7 1 1 0 0 1 0 0 32 17. If a matrix A 0 1 0 , find the matrix A , using Cayley Hamilton Theorem. Ans. 0 1 0 32 0 1 1 0 1 4.54 CHARACTERISTIC VECTORS OR EIGEN VECTORS As we have discussed in Art 21.2, A column vector X is transformed into column vector Y by means of a square matrix A. Now we want to multiply the column vector X by a scalar quantity so that we can find the same transformed column vector Y. i.e., AX = X X is known as eigen vector. Example 71. Show that the vector (1, 1, 2) is an eigen vector of the matrix 3 1 1 A 2 2 1 corresponding to the eigen value 2. 2 2 0 Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ 336 Determinants and Matrices Solution. Let X = (1, 1, 2). Now, 3 1 1 1 3 1 2 2 1 AX 2 2 1 1 2 2 2 2 2 1 2 X 2 2 0 2 2 2 0 4 2 Corresponding to each characteristic root , we have a corresponding non-zero vector X which satisfies the equation [ A I ] X 0. The non-zero vector X is called characteristic vector or Eigen vector. 4.55 PROPERTIES OF EIGEN VECTORS 1. The eigen vector X of a matrix A is not unique. 2. If 1 , 2 , .... , n be distinct eigen values of an n × n matrix then corresponding eigen vectors X1, X2, ......., Xn form a linearly independent set. 3. If two or more eigen values are equal it may or may not be possible to get linearly independent eigen vectors corresponding to the equal roots. 4. Two eigen vectors X1 and X2 are called orthogonal vectors if X 1 X 2 0. 5. Eigen vectors of a symmetric matrix corresponding to different eigen values are orthogonal. a Normalised form of vectors. To find normalised form of b , we divide each element by c a 2 b2 c 2 . 1 1/ 3 For example, normalised form of 2 is 2 / 3 2 2 / 3 12 22 2 2 3 4.56 NON-SYMMETRIC MATRICES WITH NON-REPEATED EIGEN VALUES 3 1 4 Example 72. Find the eigen values and eigen vectors of matrix A 0 2 6 0 0 5 3 1 4 Solution. | A I | 0 0 2 6 (3 ) (2 ) (5 ) 0 5 Hence the characteristic equation of matrix A is given by | A I | 0 (3 ) (2 ) (5 ) 0 2, 3, 5. Thus the eigen values of matrix A are 2, 3, 5. The eigen vectors of the matrix A corresponding to the eigen value is given by the nonzero solution of the equation ( A I ) X 0 or 4 x1 0 3 1 0 x 0 2 6 2 0 0 5 x3 0 ... (1) When 2, the corresponding eigen vector is given by Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ Determinants and Matrices 3 2 0 0 337 x1 0 2 2 6 x2 0 0 5 2 x3 0 1 4 1 1 4 x1 0 0 0 6 x 0 2 0 0 3 x3 0 x1 x2 4 x3 0 0 x1 0 x2 6 x3 0 x x1 x 2 3 k 60 06 00 k Hence X1 k k 0 x1 x2 x3 k 1 1 0 x1 k , x2 k , x3 0 1 1 can be taken as an eigen vector of A corresponding to the eigen 0 value 2 When 3, substituting in (1), the corresponding eigen vector is given by 4 x1 0 33 1 0 2 3 6 x2 0 0 0 5 3 x3 0 0x1 + x2 + 4x3 = 0 0 1 4 x1 0 0 1 6 x 0 2 0 0 2 x3 0 0x1 – x2 + 6x3 = 0 x x1 x 2 3 6 4 00 00 x1 = k, x2 = 0, x3 = 0 x1 x2 x3 k 10 0 0 10 k 1 Hence, X 2 0 k 0 can be taken as an eigen vector of A corresponding to the 0 0 eigen value = 3. When 5. Again, when 5, substituting in (1), the corresponding eigen vector is given by 4 x1 0 1 4 x1 3–5 1 –2 0 0 2 – 5 6 x2 0 0 –3 6 x2 = 0 0 5 – 5 x3 0 0 0 0 0 x3 0 –2 x1 x2 4 x3 0 –3x2 6 x3 0 By cross-multiplication method, we have x x1 x 2 3 6 12 0 12 6 – 0 x1 = 3k, x2 = 2k, x3 = k x1 x2 x3 18 12 6 x1 x2 x3 =k 3 2 1 Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ 338 Determinants and Matrices 3k 3 Hence, X 3 2k k 2 can be taken as an eigen vector of A corresponding to the eigen k 1 value 5. Ans. EXERCISE 4.22 Non-symmetric matrix with different eigen values: Find the eigen values and the corresponding eigen vectors for the following matrices: 1 1 2 1. 1 2 1 0 1 1 (A.M.I.E.T., June 2006) 2 2 3 1 1 2. 1 1 3 1 4 2 2 2 1 0 2. 5 3 2 Ans. 1, 2, 5; 1 , 1 , 1 2 4 1 4 2 0 1 Ans. 1, 1, 2, 0 , 1 3 2 , 1 1 3 1 11 1 1 Ans. 2, 1, 3; 1 , 1 , 1 14 1 1 9 2 6 4 6 6 6 0 3 2 1 2 1 3 2 5 0 3 3. Ans. 1, 1, 4; 2 , 1 , 1 Ans. 1, 1, 2; 1 , 1 , 1 4. 1 4 3 16 4 11 7 1 1 3 2 4 2 1 1 1 1 1 0 1 4 0 1 2 11 4 5 1 2 1 0,1,5; 1 , 0 , 5 Ans. –1, 1, 2; 1 , 2 , 3 4. Ans. 5. 1 1 0 3 2 3 1 1 11 1 1 1 T 8. Show that the matrices A and A have the same eigenvalues. Further if l, m are two distinct eigenvalues, then show that the eigenvector corresponding to l for A is orthogonal to eigenvector corresponding to m for AT. 4.57 NON-SYMMETRIC MATRIX WITH REPEATED EIGEN VALUES Example 73. Find all the Eigen values and Eigen vectors of the matrix 2 2 3 A 2 1 6 1 2 0 Solution. Characteristic equation of A is 2 2 3 2 1 1 6 2 0 (AMIETE, Dec. 2009) 0 2 (2 ) [ 12] 2(2 6) 3 (4 1 ) 0 3 2 – 21 – 45 0 By trial: If 3, then 27 9 63 45 0, so ( 3) is one factor of (1). The remaining factors are obtained on dividing (1) by 3. –3 1 1 –21 –45 –3 6 45 1 –2 –15 0 .... (1) 2 2 15 0 ( 5) ( 3) 0 ( 3) ( 3) ( 5) 0 5, 3, 3 Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ Determinants and Matrices 339 To find the eigen vectors for corresponding eigen values, we will consider the matrix equation (A I)X 0 i.e., 2 2 1 2 1 2 3 6 x 0 y 0 0 z 0 ... (2) 7 2 3 x 0 On putting 5 in eq. (2), it becomes 2 4 6 y 0 1 2 5 z 0 We have – 7x + 2y – 3z = 0, 2x – 4y – 6z = 0 x y z or 12 12 6 42 28 4 x = k, y = 2k, z = – k x y z 24 48 24 or x y z k 1 2 1 k 1 Hence, the eigen vector X 1 2k = k 2 k 1 1 2 3 x 0 Put 3 in eq. (2), it becomes 2 4 6 y 0 1 2 3 z 0 We have x + 2y – 3z = 0, 2x + 4y – 6z = 0, – x – 2y + 3z = 0 Here first, second and third equations are the same. 1 Let x = k1, y = k2 then z (k1 2k 2 ) 3 k1 Hence, the eigen vector is k2 1 (k1 2k 2 ) 3 0 Let k1 0, k 2 3, Hence X 2 3 2 Since the matrix is non-symmetric, the corresponding eigen vectors X2 and X3 must be linearly independent. This can be done by choosing 3 k1 = 3, k2 = 0, and Hence X 3 0 1 1 Hence, X 1 2 , 1 0 X 2 3 , 2 3 X 3 0 . 1 Ans. Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ 340 Determinants and Matrices EXERCISE 4.23 Non-symmetric matrices with repeated eigen values Find the eigen values and eigen vectors of the following matrices: 4 0 2 2 2 1. 1 1 1 Ans. –2, 2, 2; 1 , 1 7 1 1 3 1 2 1 1 3. 2 3 2 3 3 4 1 1 0 0 1 0 5. 0 0 1 2 2 1 2. 1 3 1 1 2 2 1 1 2 , 1 Ans. 1, 1, 5; 5 1 0 1 1 9 4 4 1 , 1 , 1 8 3 4 1, 1, 3; 4. Ans. 16 8 7 1 1 2 0 1 1 1 , 0 , 2 1, 1, 7; Ans. 1 1 3 1 Ans. 1, 1, 1, 0 1 (AMIETE, Dec. 2010) 4.58 SYMMETRIC MATRICES WITH NON REPEATED EIGEN VALUES Example 74. Find the eigen values and the corresponding eigen vectors of the matrix 2 5 4 5 7 5 4 5 2 Solution. | A I | 0 2 5 5 7 4 By trial: Take 5 4 5 3 32 90 216 0 0 2 3, then – 27 – 27 + 270 – 216 = 0 By synthetic division –3 1 1 –3 –90 –216 –3 18 216 –6 –72 0 2 6 72 0 ( 12) ( 6) 0 3, 6, 12 Matrix equation for eigen vectors [ A I ] X 0 5 4 x 0 2 5 7 5 y 0 4 5 2 z 0 Eigen Vector On putting 3 in (1), it will become 1 5 4 x 0 5 10 5 y 0 4 5 1 z 0 ...(1) x 5 y 4z 0 5 x 10 y 5z 0 Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ Determinants and Matrices 341 x y z 25 40 20 5 10 25 1 Eigen vector X 1 1 . 1 or x y z 1 1 1 Eigen vector corresponding to eigen value – 6. Equation (1) becomes 4 5 4 x 0 5 13 5 y 0 4 5 4 z 0 or x y z 25 52 20 20 52 25 4x 5 y 4 z 0 5 x 13 y 5z 0 or x y z 1 0 1 1 eigen vector X 2 0 1 Eigen vector corresponding to eigen value = 12. Equation (1) becomes 4 x 0 14 5 5 5 5 y 0 4 5 14 z 0 or 14 x 5 y 4 z 0 5x 5 y 5z 0 x y z 25 20 20 70 70 25 or x y z 1 2 1 1 Eigen vector X 3 2 1 Ans. EXERCISE 4.24 Symmetric matrices with non-repeated eigen values Find the eigen values and eigen vectors of the following matrices: 5 0 1 0 2 0 1. 1 0 5 8 6 2 3. 6 7 4 2 4 3 0 1 1 Ans. 2, 4, 6; 1 , 0 , 0 0 1 1 (U.P., I Semester, Jan 20111) 1 2 1 2 4 6 4. 4 2 6 Ans. –2, 9, –18; 1 , 2 , 1 0 1 4 6 6 15 1 1 1 3 1 1 2. 1 5 1 Ans. 2, 3, 6; 0 , 1 , 2 1 1 1 1 1 3 1 2 2 Ans. 0, 3, 15; 2 , 1 , 2 2 2 1 1 1 1 1 1 3 5. 1 5 1 Ans. 2, 3, 6; 0 , 1 , 2 1 1 1 3 1 1 Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ 342 Determinants and Matrices 4.59 SYMMETRIC MATRICES WITH REPEATED EIGEN VALUES Example 75. Find all the eigen values and eigen vectors of the matrix 2 1 1 1 2 1 1 1 2 Solution. The characteristic equation is 2 1 1 1 2 1 1 1 2 0 (2 )[(2 )2 1] 1[2 1] 1[1 2 ] 0 (2 ) (4 4 2 1) ( 1) 1 0 8 8 2 2 2 4 4 2 3 2 2 0 3 6 2 9 4 0 ... (1) 3 62 9 4 0 On putting 1 in (1), the equation (1) is satisfied. So 1 is one factor of the equation (1). The other factor ( 2 5 4) is got on dividing (1) by 1. ( 1) ( 2 5 4) 0 or ( 1) ( 1) ( 4) 0 The eigen values are 1, 1, 4. 1 1 24 When 4 1 24 1 1 1 2 4 2 x1 x2 x3 0 = 1, 1, 4 2 1 1 x1 0 x1 0 1 2 1 x2 0 x2 0 x 0 1 1 2 x3 0 3 x1 x2 2 x3 0 x1 x x 2 3 2 1 1 4 2 1 x1 k , x2 k , x3 k k X 1 k k k 1 1 1 1 X 1 1 1 or 2 1 1 1 When 1 x1 x2 x3 k 1 1 1 1 1 1 x1 1 1 1 x2 0 1 1 1 x 3 1 2 1 1 1 1 2 1 x1 x2 0 x 3 1 1 1 x1 0 0 0 x2 0, R2 R2 R1 0 0 0 x R3 R3 R1 3 x1 x2 x3 0 Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ Determinants and Matrices 343 Let x1 = k1 and x2 = k2 k1 – k2 + x3 = 0 or k1 X 2 k2 k2 k1 x3 = k2 – k1 1 X 2 1 0 k1 1 k 1 2 l Let X 3 m n As X3 is orthogonal to X1 since the given matrix is symmetric l [1, 1, 1] m 0 n or l–m+n=0 ... (2) As X3 is orthogonal to X2 since the given matrix is symmetric l [1, 1, 0] m 0 n Solving (2) and (3), we get or l+m+0=0 l m n 0 1 1 0 11 ... (3) l m n 1 1 2 1 X 3 1 2 Ans. EXERCISE 4.25 Symmetric matrices with repeated eigen values Find the eigen values and the corresponding eigen vectors of the following matrices: 1. 1 2 3 2 3 1 2 4 6 1 , 6 , 2 Ans. 0, 0, 14; 3 6 9 0 5 3 3. 6 2 2 2 3 1 2 1 3 4. 2 0 1 2. 0 3 0 1 0 2 2 1 1 Ans. 8, 2, 2; 1 , 0 , 2 1 2 0 1 1 1 Ans. 1, 3, 3; 0 , 1 , 2 1 1 1 6 3 3 4. 3 6 3 3 3 6 Ans. 3, 3, 12 Choose the correct or the best of the answers given in the following Parts; (i) Two of the eigenvalues of a 3 × 3 matrix, whose determinant equals, 4, are –1 and +2 the third eigen value of the matrix is equal to (a) –2 (b) –1 (c) 1 (d) 2 (ii) If a square matrix A has an eigenvalue , then an eigenvalue of the matrix (kA)T where, k 0,is a scalar is (a) k (b) k / (c) k (d) None of these (iii) An eigenvalue of a square matrix A is Then (a) | A | (b) A is symmetric (c) A is singular; Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ 344 Determinants and Matrices (d) A is skew-symmetric; (e) A is an even order matrix; 1 (iv) The matrix A is defined as A = 2 1 (a) –1, –9, –4, (b)1, 9, 4 1 (v) If the matrix is A= 0 0 0 3 4 0 0 .The eigenvalues of A2 are 2 (c) –1, –3, 2, 2 3 0 (f) A is an odd order matrix. (d) 1, 3, –2. 3 5 then the eigenvalues of A3 + 5A + 8 I, are 2 (a) –1, 27, –8; (b) –1, 3, –2; (c) 2, 50, –10, (d) 2, 50, 10. (vi) The matrix A has eigen values i 0 .Then A–1 – 2I + A has eigenvalues 1 (a) 1 + 2 i +i 2 (b) 2 i 2 1 i i2 i (viii) The eigen values of a matrix A are 1,–2, 3. The eigen of 3I–2A + A2 are (a) 2, 11, 6 (b) 3, 11, 18 (c) 2, 3, 6 (d) 6, 3, 11 (c) 1–2i +i 2 Ans. (i)(b), (ii)(c), (iii)(c), (iv)(b), (v)(c), (d) 1 (vi)(b), (vii)(a) 4.60 DIAGONALISATION OF A MATRIX Diagonalisation of a matrix A is the process of reduction of A to a diagonal form ‘D’. If A is related to D by a similarity transformation such that D = P–1 AP then A is reduced to the diagonal matrix D through modal matrix P. D is also called spectral matrix of A. 4.61 THEOREM ON DIAGONALIZATION OF A MATRIX Theorem. If a square matrix A of order n has n linearly independent eigen vectors, then a matrix P can be found such that P–1 AP is a diagonal matrix. Proof. We shall prove the theorem for a matrix of order 3. The proof can be easily extended to matrices of higher order. a1 b1 c1 A a2 b2 c2 Let a3 b3 c3 and let 1 , 2 , 3 be its eigen values and X1, X2, X3 the corresponding eigen vectors, where x1 x2 X 1 y1 , X 2 y2 , z1 z2 For the eigen value 1 , the eigen vector is given by We have (a1 1 ) x1 b1 y1 c1 z1 0 a2 x1 (b2 1 ) y1 c2 z1 0 a3 x1 b3 y1 (c3 1 ) z1 0 a1 x1 b1 y1 c1 z1 1 x1 a2 x1 b2 y1 c2 z1 1 y1 a3 x1 b3 y1 c3 z1 1 z1 x3 X 3 y3 z3 ...(1) ...(2) Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ Determinants and Matrices 345 Similarly for 2 and 3 we have a1 x2 b1 y2 c1 z2 2 x2 a2 x2 b2 y2 c2 z2 2 y2 a3 x2 b3 y2 c3 z2 2 z2 a1 x3 b1 y3 c1 z3 λ3 x3 a and 2 x3 b2 y3 c2 z3 λ 3 y3 a3 x3 b3 y3 c3 z3 λ 3 z3 x1 x2 x3 y y y 2 3 We consider the matrix P= 1 z1 z 2 z3 Whose columns are the eigenvectors of A. a1 b1 c1 x1 a b c y Then A P = 2 2 2 1 a3 b3 c3 z1 ...(3) ...(4) x2 y2 z2 x3 y3 z3 a1 x1 b1 y1 c1 z1 a1 x2 b1 y2 c1 z2 a1 x3 b1 y3 c1 z3 a2 x1 b2 y1 c2 z1 a2 x2 b2 y2 c2 z2 a2 x3 b2 y3 c2 z3 a x b y c z a x b y c z a3 x3 b3 y3 c3 z3 3 2 3 2 3 2 3 1 3 1 3 1 1 x1 2 x2 3 x3 1 y1 2 y2 3 y3 [Using results (2), (3) and (4)] z z z 2 2 3 3 11 x1 x2 x3 1 0 0 y1 y 2 y3 0 2 0 PD z z z3 0 0 3 2 1 1 0 0 where D is the Diagonal matrix 0 2 0 0 0 3 AP = PD P–1 AP = P–1 PD = D Notes 1. The square matrix P, which diagonalises A, is found by grouping the eigen vectors of A into square-matrix and the resulting diagonal matrix has the eigen values of A as its diagonal elements. 2. The transformation of a matrix A to P–1 AP is known as a similarity transformation. 3. The reduction of A to a diagonal matrix is, obviously, a particular case of similarity transformation. 4. The matrix P which diagonalises A is called the modal matrix of A and the resulting diagonal matrix D is known as the spectra matrix of A. 6 2 2 Example 76. Let A 2 3 1 Find matrix P such that P–1 AP is diagonal matrix. 2 1 3 Solution. The characteristic equation of the matrix A is Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ 346 Determinants and Matrices 6 2 2 3 2 1 2 1 0 3 2 (6 )[9 6 1] 2 [6 2 2] 2 [2 6 2] 0 (6 )( 2 6 8) 8 4 8 4 0 6 2 36 48 3 62 8 16 8 0 3 12 2 36 32 0 ( 2)2 ( 8) 0 Eigen vector for = 2 3 12 2 36 32 0 = 2, 2, 8 2 1 1 x1 0 4 2 2 x1 0 2 1 1 x 0 or 2 1 1 x 0 R2 R1 R2 2 R R R 2 2 3 2 1 1 x3 0 3 2 1 1 x3 0 2 1 1 x1 0 0 0 0 x 0 or 2x – x + x = 0 1 2 3 2 0 0 0 x3 0 This equation is satisfied by x1 = 0, x2 = 1, x3 = 1 0 X 1 1 1 and again x1 = 1, x2 = 3, x3 = 1. 1 X 2 3 1 Eigen vector for = 8 2 2 2 x1 0 2 5 1 x 0 2 2 1 5 x3 0 2 x1 2 x2 2 x3 0 2 x1 5x2 x3 0 x1 x2 x x x x x 3 x1 x2 3 1 2 3 2 10 4 2 10 4 2 1 1 12 6 6 2 X 3 1 1 0 1 2 P 1 3 1 , 1 1 1 P 1 1 7 4 1 2 2 2 6 2 1 1 Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ Determinants and Matrices 347 1 7 6 2 2 0 1 2 2 0 0 4 1 Ans. P AP 2 2 2 2 3 1 1 3 1 0 2 0 6 2 1 1 2 1 3 1 1 1 0 0 8 a h –1 Example 77. The matrix A T, where is transformed to the diagonal form D = T AT h b Now 1 cos sin T Find the value of which gives this diagonal transformation. sin cos cos sin cos sin 1 T Solution. T sin cos sin cos cos sin a h cos sin Now T 1 AT sin cos h b sin cos a cos h sin h cos b sin cos sin a sin h cos h sin b cos sin cos a cos 2 2h sin cos b sin 2 (a b)sin cos h sin 2 h cos 2 2 2 a sin 2 2h sin cos b cos2 (a b)sin cos h cos h sin a cos 2 h sin 2 b sin 2 (a b) sin cos h cos 2 (a b)sin cos h cos 2 a sin 2 h sin 2 b cos 2 0 d 1 being diagonal matrix 0 d2 (a b)sin cos h cos 2 0 a b a b sin 2 h cos 2 sin 2 h cos 2 0 2 2 2h 1 2h tan 2 tan 1 ba 2 ba Ans. EXERCISE 4.26 1. Find the 8 A 4 3 matrix B which transforms the matrix 8 2 3 2 to a diagonal matrix. 4 1 4 3 2 Ans. B 3 2 1 2 1 1 4 1 0 2. For the matrix A 1 4 1 , determine a matrix P such that P–1AP is diagonal matrix. 0 1 4 1 1 1 2 Ans. P 0 2 1 1 1 5 7 5 3. Determine the eigen values and the corresponding eigen vectors of the matrix A 0 4 1 2 8 3 2 1 1 Hence find the matrix P such that P–1AP is diagonal matrix. Ans. P 1 1 1 3 2 1 Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ 348 Determinants and Matrices 4. Reduce the following matrix A into a diagonal matrix 8 6 2 A 6 7 4 2 4 3 0 0 0 Ans. 0 3 0 0 0 15 5. Prove that similar matrices have the same eigenvalues. Also give the relationship between the eigenvectors of two similar matrices. (A.M.I.E.T.E, June 2005) 6. Let a 4 × 4 matrix A have eigenvalues 1, –1, 2, –2 and matrix B = 2A + A–1 – I Find (i) determinant of matrix B. (ii) trace of matrix B. (A.M.I.E.T.E, June 2005) 4.62 POWERS OF A MATRIX (By diagonalisation) We can obtain powers of a matrix by using diagonalisation. We know that D = P–1 AP Where A is the square matrix and P is a non-singular matrix. D2 = (P–1 AP) (P–1 AP) = P–1 A (P P–1) AP = P–1 A2 P Similarly D3 = P–1 A3 P In general Dn = P–1 An P Pre-multiply (1) by P and post-multiply by P–1 P Dn P–1 = P (P–1 An P) P–1 = (P P–1) An (P P–1) = An Procedure:(1) Find eigen values for a square matrix A. (2) Find eigen vectors to get the modal matrix P. (3) Find the diagonal matrix D, by the formula D = P–1 AP (4) Obtain An by the formula An = P Dn P–1. ...(1) 1 0 1 Example 78. Find a matrix P which transform the matrix A 1 2 1 to diagonal 2 2 3 form. Hence A4. Solution. Characteristic equation of the matrix A is 1 0 1 or 3 62 11 6 0 1 2 1 0 or ( 1) ( 2) ( 3) 0 2 2 3 1, 2, 3 For 1, eigen vector is given by 1 x1 0 1 1 0 0 0 1 x1 0 1 1 1 1 x 0 2 1 1 x2 0 2 2 2 2 2 x3 0 2 3 1 x3 0 0 x1 0 x2 x 3 0 x x1 x2 3 or x1 = 1, x2 = –1, x3 = 0 x1 x2 x3 0 0 1 1 0 0 Eigen vector is [1, –1, 0]. For = 2, eigen vector is given by 1 2 1 2 0 22 2 1 x1 0 x 0 1 2 3 2 x3 0 1 1 2 0 1 x1 0 0 1 x2 0 2 1 x3 0 Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ Determinants and Matrices 349 0 0 0 x1 0 1 0 1 x 0 R R + R 1 2 2 1 2 2 1 x3 0 x1 0 x2 x3 0 2 x1 2 x2 x3 0 x x1 x 2 3 0 2 2 1 2 0 Eigen vector is [–2, 1, 2]. For = 3, eigen vector is given by 1 1 3 0 1 23 1 2 2 33 x1 = – 2, x1 0 x 0 2 x3 0 x2 = 1, x3 = 2 2 0 1 x1 0 1 1 1 x 0 2 2 2 0 x3 0 2 x1 0 x2 x3 0 x1 x2 x3 0 x x1 x2 3 0 1 1 2 2 0 Eigen vector is [–1, 1, 2]. x1 = – 1, 0 1 2 1 1 1 Modal matrix P 1 1 1 and P 2 2 2 0 2 2 1 0 1 2 1 0 1 1 2 Now P 1 AP 1 1 0 1 2 1 1 1 1 2 2 3 0 2 1 1 2 A4 PD 4 P 1 x2 = 1, x3 = 2 2 1 2 0 2 1 1 1 0 0 1 0 2 0 D 2 0 0 3 0 1 1 2 1 1 0 0 1 1 1 0 16 0 1 1 0 2 2 0 0 81 1 1 EXERCISE 4.27 1 2 49 50 40 0 65 66 40 1 130 130 81 2 Ans. Find a matrix P which transforms the following matrices to diagonal form. Hence calculate the power matrix. 1 1 3 1. If A = 1 5 1 , calculate A4. 3 1 1 2. If 3 1 1 A 1 5 1 , calculate A4. 1 1 3 2 1 1 3. If A 1 2 1 , calculate A6. 1 1 2 251 405 235 Ans. 405 891 405 235 405 251 251 405 235 Ans. 405 891 405 235 405 251 1366 1365 1365 Ans. 1365 1366 1365 1365 1365 1366 Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ 350 Determinants and Matrices 1 1 1 0 2 1 , 8 A 4. If calculate A . 4 4 3 12099 12355 6305 Ans. 12100 12356 6305 13120 13120 6561 3 1 1 5. Show that the matrix A is diagonalisable A 2 1 2 . If so obtain the matrix P such that 0 1 2 –1 P AP is a diagonal matrix. (AMIETE, June 2010) 4.63 SYLVESTER THEOREM Let and P(A) = C0 An + C1 An–1 + C2 An–2 + … + Cn–1 A + Cn I | I A | f () and Adjoint matrix of [ I A] [ f ()] z ( ) [ f ()] Adjoint matrix of [ I A] f () f () Then according to Sylvester’s theorem P ( A) P (1 ). Z (1 ) P ( 2 ). Z ( 2 ) P(3 ). Z (3 ) n P ( ). Z ( ) r r r 1 2 0 100 Example 79. If A , find A . 0 1 Solution. f ( ) | I A | 0 0 2 0 0 1 2 0 0 1 0 f () ( 2) ( 1) 0 or 1 1, 2 2 f ( ) 2 3 2, f () 2 3 f (2) = 4 – 3 = 1, f (1) = 2 – 3 = – 1 0 1 [ f ()] = Adjoint matrix of the matrix [ I A] 0 2 Z ( 1 ) Z (1) [ f (1)] 1 0 0 0 0 f (1) 1 0 1 0 1 Z ( 2 ) Z (2) [ f (2)] 1 1 0 f (2) 1 0 0 By Sylvester theorem P ( A) P(1 ). Z (1 ) P( 2 ). Z ( 2 ) A100 P( 1 ) Z (1 ) P( 2 ) Z ( 2 ) 0 0 100 1 0 100 0 0 100 1 0 100 1 2 0 0 1 0 1 2 0 0 0 1 0 0 2100 0 1 0 0 2100 0 0 0 1 Ans. Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ Determinants and Matrices 351 EXERCISE 4.28 1 4 1. Verify Sylvesters theorem for A3, where A 3 2 Use sylvesters theorem in solving the following: 1 0 256 2. Given A , find A . 0 3 e 1 1 0 0 A , 3. Given A show that e . 0 2 0 e 2 1 3 4. Given A , show that 2 sin A = | sin 2 | A. 1 1 1 4 5. Prove that 3 tan A = A tan (3) where A 2 1 1 2 6. Prove that sin2A + cos2 A = 1, where A 1 4 1 2 3 7. Given A 0 2 0 , find A–1. 0 0 3 0 1 Ans. 256 0 3 1 1 1 1 0 Ans. 0 2 1 0 0 3 1 20 0 1 , find tan A. 8. Given A 1 7 3 0 2 18 60 20 20 80 20 20 100 20 tan 3 Ans. tan1 0 0 0 tan 2 1 4 1 2 10 2 1 2 2 18 60 20 15 60 15 12 60 12 4.64 QUADRATIC FORMS The quadratic forms are defined as a homogeneous polynomial of second degree in any number of variables. For example 1. Two variables ax2 + 2hxy + by2 = Q (x, y) 2. Three variables ax2 + 2hxy + by2 + cz2 + 2hxy + 2gyz + 2fzx = Q (x, y, z) 3. Four variables ax2 + by2 + cz2 + dw2 + 2hxy + 2gyz + 2fzx + 2lxw + 2myw + 2nzw = Q (x, y, z, w) 4. n variables = Q (x1, x2, ........xn.) 4.65 QUADRATIC FORM EXPRESSED IN MATRICES Quadratic form can be expressed as a product of matrices. Quadratic form = Q (x) = X AX x1 a11 a12 a13 a X x A where 2 and 21 a22 a23 x3 a31 a32 a33 X is the transpose of X. a11 a12 a13 x1 X AX x1 x2 x3 a21 a22 a23 x2 a31 a32 a33 x3 Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ 352 Determinants and Matrices x1 a11 x1 a21 x2 a31 x3 a12 x1 a22 x2 a32 x3 a13 x1 a23 x2 a33 x3 x2 x3 2 2 a11 x1 a21 x1 x2 a31 x1x3 a12 x1 x2 a22 x2 a32 x2 x3 a13 x1 x3 a23 x2 x3 a33 x32 a12 a11 x12 a22 x22 a33 x32 (a12 a21 ) x1 x2 (a23 a32 ) x2 x3 (a31 a13 ) x1 x3 and a21 are the coefficients of x1x2 it means (a12 + a21) are the coefficient of x1x2. In general aij and aji are the both coefficients of xi,xj (i j ). So (a ij + a ji ) are the coefficient of xi xj Let us have new coefficients of xi xj. cij = cji = 1 (a + aji) 2 ij 1 ( A A) = symmetric matrix C 2 Thus, the coefficient matrix in quadratic form is always symmetric matrix without loss of generality. We know Then X AX c11 x12 c22 x22 c33 x2 2c12 x1 x2 2c23 x2 x3 2c31 x1 x3 Matrix A is known as the coefficient matrix or matrix of quadratic form and (R) is the discriminant of the quadratic form. Example 80. Write down the quadratic form corresponding to the matrix 1 2 5 A 2 0 3 5 3 4 Solution. Quadratic form = X AX 1 2 5 x1 x1 [ x1 x2 x3 ] 2 0 3 x2 [ x1 2 x2 5 x3 , 2 x1 3 x3 , 5 x1 3 x2 4 x3 ] x2 = 5 3 4 x3 x3 2 2 x1 2 x1 x2 5 x3 x1 2x1x2 3x2 x3 5 x1 x3 3x2 x3 4 x3 x12 4 x32 4 x1 x2 10 x1 x3 6 x2 x3 Ans. Example 81. Find a real symmetric matrix C of the quadratic form: Q ( x1, x2 , x3 ) x12 4 x22 6 x32 2 x1 x2 x2 x3 3x1x3 Solution. On Comparing the coefficients in the given quadratic form, with the standard quadratic form, we get Here a11 1, a22 4, a 33 6, a12 2, a21 0, a23 1, a32 0, a13 3, a31 0 Q ( x1 , x2 , x3 ) [ x1 x2 1 2 3 x1 x3 ] 0 4 1 x 2 x1 0 0 6 x 3 2x1 4 x2 x1 3x1 x2 6 x3 x 2 x 3 x12 2 x1 x2 4 x22 3x1 x3 x2 x3 6 x32 1 2 3 1 0 0 2 2 3 1 1 1 C ( A A) 0 4 1 2 4 0 2 8 1 2 2 2 0 0 6 3 1 6 3 1 12 1 1 3 2 1 4 1 2 3 2 1 2 6 Ans. Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ Determinants and Matrices 353 4.66 LINEAR TRANSFORMATION OF QUADRATIC FORM (Diagonalisation of the matrix) Let the given quadratic form be XAX where A is a symmetric matrix. Consider the linear transformation X = PY Then X = (PY) = Y P XAX = (Y P) A (PY) = Y (P AP) Y = Y BY where B = PAP (Transformed quadratic form) Now B = (PAP) = PAP = B Rank (B) = Rank (A) Therefore, A and B are congruent matrices and the transformation X = PY is known as congruent transformation. 4.67 CANONICAL FORM OF SUM OF THE SQUARES FORM USING LINEAR TRANSFORMATION When a quadratic form is linearly transformed then the transformed quadratic of new variable is called canonical form of the given quadratic form. When XAX is linearly transformed then the transformed quadratic YBY is called the canonical form of the given quadratic X AX. n If B = PAP = Diag (1 , 2 , 3 ....... n ) than XAX = YBY = Y i i 2 i 1 Remarks. (1) i (eigen values) can be positive or negative or zero. (2) If Rank (A) = r, then the quadratic form XAX will contain only r terms. 4.68 CANONICAL FORM OF SUM OF THE SQUARES FORM USING ORTHOGONAL TRANSFORMATION Real symmetric matrix A can be reduced to a diagonal form MAM = D ...(1) where M is the normalised orthogonal modal matrix of A and D is its spectoral matrix. Let the orthogonal transformation be X = MY Q = XAX = (MY) A (MY) = (YM) A (MY) = Y (MAM) Y [ MAM = D] = YDY =Y Diag. (1 2 n ) Y 1 0.....0 y1 0 .....0 y 2 2 y1 y2 ..... yn 0 0 ..... n yn 1 y1 y1 y 2 2 y2 ..... n yn yn 1 y12 2 y22 ..... n yn2 , which is called canonical form. Now, we have seen that quadratic form XAX can be reduced to the sum of the squares by the transformation X = PI where P is the normalised modal matrix of A. Canonical form. B is a diagonal matrix, then the transformed quadratic is a sum of square terms, known as canonical form. Index. The number of positive terms in canonical form of a quadratic form is known as index (s) of the form. Rank of form. Rank (r) of matrix B (or A) is called the rank of the form. Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ 354 Determinants and Matrices Signature of quadratic form. The difference of positive terms (s) and negative terms (r–s) is known as the signature of quadratic form. Signature = s – (r – s) = s – r + s = 2s – r 4.69 CLASSIFICATION OF DEFINITENESS OF A QUADRATIC FORM A Let Q be XAX and variables (x1, x2, x3 … xn), Rank (A) = r, Index = s 1. Positive definite If rank and index are equal i.e., r = n, s = n or if all the eigen values of A are positive. 2. Negative definite If index = 0, i.e., r = n, s = 0 or if all the eigen values of A are negative. 3. Positive semi-definite If rank and index are equal but less than n, i.e., s=r<n [ | A | = 0] or all eigen values of A are positive at least one eigen value is zero. 4. Negative semi-definite If index is zero, i.e., s = 0, r<n [ | A | = 0] or all eigen values of A are negative and at least one eigen value is zero. 5. Indefinite If some eigen values are positive and some eigen values are negative. 6. Notes : (1) If Q is negative definite (semi-definite) then – Q is positive definite (semi-definite). (2) The classification of the definiteness of a quadratic form depends upon the location of eigen values of A. Example 82. Reduce to diagonal form the following symmetric matrix by congruent transformation and interpret the result in terms of quadratic form 3 2 1 A 2 2 3 1 3 1 2 1 3 Solution. A 2 2 3 1 3 1 Let us reduce A into diagonal matrix. IAI = A 1 0 0 1 0 0 3 2 1 0 1 0 A 0 1 0 2 2 3 | A| 0 0 0 1 0 0 1 1 3 1 R S Row transformation carried out on R.H.S. will be applied on R prefactor matrix. Column transformation applied on R.H.S. will be applied on S post factor matrix. 1 0 0 3 2 1 1 0 0 2 1 0 A 0 1 0 0 2 11 3 3 3 1 0 0 1 11 2 0 0 1 3 3 3 2 R2 R1 , 3 R R3 1 3 Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ Determinants and Matrices 355 1 0 0 1 2 1 0 A 3 0 1 0 1 0 3 1 0 0 1 2 1 0 A 0 3 11 4 1 0 2 1 0 0 1 – 2 1 0 A 0 3 11 0 4 – 1 2 2 3 1 0 2 3 1 0 – 2 3 1 0 1 3 0 3 2 0 0 3 11 1 0 3 1 3 0 3 2 0 0 3 1 0 0 4 3 11 0 – 2 1 0 0 11 3 2 3 2 1 C2 C1 , C3 C1 3 3 0 11 3 39 2 0 2 3 0 R3 11 R2 2 0 0 39 – 2 C3 – 11 C2 2 Thus the matrix A is reduced to the diagonal form B. 3 P AP 0 0 0 2 3 0 2 1 3 0 0 where P 0 1 39 0 0 2 The canonical form (sum of the squares) is 3 0 2 Q Y BY [ y1 y2 y3 ] 0 3 0 0 1 x1 X = PY i.e. x2 0 x3 0 x1 y1 2 y2 4 y3 , 3 4 11 2 1 0 y1 0 y2 3 y 2 2 y 2 39 y 2 1 2 3 3 2 y3 39 2 2 3 1 0 x2 y2 4 y1 11 y2 2 y3 1 11 y3 , 2 x3 = y3 The rank of A (r) = 3 The index of quadratic form (s) = 2 The signature of quadratic form [r – (r – P)] = 2 – (3 – 2) = 1 Ans. Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ 356 Determinants and Matrices Example 83. Reduce the quadratic form 6 x12 3 x22 3x32 4 x1 x2 2 x2 x3 4 x3 x1 to the sum of square, by Lagrange Reduction Method Solution. Q 6 x12 3 x22 4 x1 x2 4 x1 x3 2 x2 x3 2 6 x12 x1 ( x2 x3 ) 3x22 3 x32 2 x2 x3 3 2 1 2 6 x1 ( x2 x3 ) 3x22 3x32 2 x2 x3 ( x2 x3 )2 3 3 2 1 1 7 2 7 6 x1 x2 x3 x22 x2 x3 x32 3 3 3 3 3 2 1 1 7 2 7 6 x1 x2 x3 x22 x2 x3 x32 3 3 3 7 3 2 2 2 2 1 1 7 1 7 7 1 6 x1 x2 x3 x2 x3 x32 x32 3 3 3 7 3 3 49 1 1 7 1 16 7 16 6 x1 x2 x3 x2 x3 x32 6 y12 y22 y32 3 3 3 7 7 3 7 where 1 1 y1 x1 x2 x3 3 3 1 y2 x2 x3 7 y3 x3 x3 y3 1 y3 7 1 2 x1 y1 y2 y3 3 7 x2 y2 EXERCISE 4.29 1. Express the quadratic form x12 2 x22 2 x32 2 x1x2 2 x2 x3 as product of matrices. Ans. x1 x2 1 1 0 x1 x3 1 2 1 x2 0 1 2 x3 2. Write down the matrix of the quadratic form 1 2 4 0 2 2 0 0 x12 2 x22 7 x32 x42 4 x1 x2 8 x1 x3 6 x3 x4 Ans. 4 0 7 3 0 3 1 0 3. Find the transformation that will transform 10x2 + 2 y2 + 5 z2 + 6 yz – 10 z x – 4 x y into a sum of square and find its reduced form. 1 8 Ans. Q 10 y12 y22 , P 0 5 0 1 5 1 4 5 1 4 0 1 Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ Determinants and Matrices 357 4. Find the transformation which will transform the following form into a sum of squares and find the reduced form : 1 1 3 2 Q = 4x2 + 3y2 + z2 – 8 xy – 6 yz + 4 xz Ans. Q 4 y12 y22 y32 , P 0 1 1 0 0 1 5. Reduce to sum to squares 1 2 4 0 1 4 2 2 2 2 2 2 Ans. P = Q x1 2 x2 7 x3 4 x1 x2 8 x1 x3 Q y1 2 y2 9 y3 , 0 0 1 6. Express the following quadratic form as “sum of squares” by congruent transformation and write down the corresponding linear transformation Q 10 x12 x22 x32 6 x1 x2 2 x2 x3 x3 x1 1 2 3 y2 , x1 y1 y2 , x2 = y2 + y3, x3 = y3 10 10 7. Reduce to the diagonal matrix by rational congruent transformation and interpret the result in terms of quadratic form. Ans. 10 y12 1 2 1 A 2 1 3 Ans. Q x12 x22 x32 4 x1 x2 6 x2 x3 2 x3 x1 ; Q y12 4 y22 25 y32 4 1 3 1 Determine the definiteness of the quadratic forms. 8. Q x12 2 x22 3x32 2 x2 x3 2 x3 x1 2 x1 x2 Ans. Indefinite. 9. Q 4 x12 x22 15 x32 4 x1 x2 . Ans. Positive semi-definite 10. Q 5 x12 26 x22 10 x32 4 x2 x3 14 x3 x1 6 x1 x2 Ans. Positive semi-definite 11. Q x12 5 x22 x32 2 x1 x2 2 x2 x3 6 x3 x1. Ans. Indefinite 12. Q 8 x12 7 x22 3x32 12 x1 x2 8 x2 x3 4 x3 x2 Ans. Positive semi-definite 13. Q – 4 x12 – 2 x22 –13x32 4 x1 x2 8 x2 x3 – 4 x3 x1. Ans. Negative definite 3 2 4 2 2 6 A 14. Find the eigenvalues and corresponding eigen vector of 4 6 1 Verify that the eigen vectors are orthogonal and write down an orthogonal matrix M such that MAM = D, where D is diagonal matrix. Ans. – 9, 6, 3, [12 – 2], [212], [–2 2 1], M P , where P is modal matrix 3 4.70 DIFFERENTIATION AND INTEGRATION OF MATRICES If the elements of a matrix A are function of scalar variable t, the matrix is called a matrix function of t. A = A(t) = [aij (t)] Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ 358 Determinants and Matrices The differential coefficient of A w.r.t. “t” is defined as d d A aij dt dt dA Hence the elements of the differentiated matrix are the derivatives of the corresponding dt elements of A. da1n da11 da12 ... dt dt dt d da da22 da2 n A t 21 ... dt dt dt dt dan1 dan 2 ... dann dt dt dt It is easy to prove that d dB da A B A. B dt dt dt The integral of the matrix A is defined as Adt a dt ij Thus the integral of A is obtained by integrating each element of A. Power series. Let A be a square matrix with all eigenvalues less than 1 in absolute value, then a0I + aa A + a2A 2 + ... is convergent. The following series are also convergent A 1 e A I A2 .... 1 2 cos A 1 1 2 1 4 A A .... 2 4 sin A A 1 3 1 5 A A .... 3 5 1 A 1 1 A A2 .... d tA tA 1 1 3 tA tA 2 Example 84. Prove that dt e Ae if e 1 1 2 (t A) 3 t A .... 1 d tA d tA 1 e 1 t 2 A2 t 3 A3 .... 1 2 3 dt dt Solution. d d d1 1 1 t A t 2 A2 t 3 A3 ..... dt dt dt 2 3 1 1 0 A t A2 t 2 A3 .... 1 2 1 1 A 1 t A t 2 A2 .... AetA 2 1 d2x dx 4 12 x 0. dt 2 dt x 0 0; x 0 8 by Matrix Method . Ans. Example 85. Solve ...(1) Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ Determinants and Matrices 359 x x1 and Solution. Let dx1 x2 dt ...(2) (1) becomes or dx1 d dx1 12 x1 4 dt dt dt ...(3) dx2 12 x1 4 x2 dt (2) and (3) are written in matrix form. d x1 0 1 x1 dt x2 12 4 x2 From R.H.S. we have to find eigenvector. 1 0 Characteristic equation is 12 4 0 or λ 4 12 0 or 2 4 12 0 or 2 6 0 2, 6 1 1 2 and 6 For = 2 and = –6, eigenvectors are 1 1 1 1 1 6 Matrix of eigen vectors = P = ,P 8 2 1 2 6 1 1 e 2t Pe λ t P 1 2 6 0 Now 0 1 6 1 e 6t 8 2 1 e 6t 6 1 1 6e 2t 2e 6t 1 e 2t = 8 2t 6e 6t 2 1 8 12e 2t 12e 6t 2e By initial conditions x (0) = 0 , x(0) = 8 x1 1 6e 2t 2e 6t x 2t 6 t 2 8 12e 12e e 2t e 6t 2e 2t 6e 6t e 2t e 6t 0 e 2t e 6t 2e 2t 6e 6t 8 2e 2t 6e6t dx 2e 2t 6e 6t dt Example 86. Solve by matrix method. x1 x e 2t e 6t , x2 Ans. d2x dx 5 6 x 0, x(0) 1. x '(0) 2. 2 dt dt Solution. d2x dx 5 6x 0 2 dt dt dx1 x2 dt On substitution (1) becomes Let x = x1 and dx2 dx2 5 x2 6 x1 or 6 x1 5 x2 dt dt Equations (2) and (3) are written in a single matrix equation ...(1) ...(2) ...(3) Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ 360 Determinants and Matrices d x1 0 1 x1 dt x2 6 5 x2 From R.H.S we have to find eigen vector. 0 λ 1 Characteristic equation is 0 5 λ 6 –( 5 – ) + 6 = 0 or – 5+ 6 = 0 = 2, 3 1 1 Eigen vectors for and = 3 are and 2 3 1 1 Matrix of eigen vectors = P = , 2 3 3 1 P 1 2 1 1 1 e 2t Pe λ t P 1 2 3 0 0 3 1 e3t 2 1 e 2t e3t 3 1 3e 2t 2e3t = 2t 3e3t 2 1 6e 2t 6e3t 2e By initial conditions x (0) = 1 and x(0) = 2 x1 3e 2t 2e3t x 2t 3t 2 6e 6e e 2t e3t 2e 2t 3e3t e 2 t e 3 t 1 2e 2t 3e 3t 2 3e 2t 2e3t 2e 2t 2e3t e 2t = 2t 3t 4e 2t 6e3t 2e 2t 6e 6e x1 = x = e2t dx x2 2e 2 t dt Example 87. Use matrices to solve the differential equation d2 y 4 y 0, y (0) 1, y' (0) 0 dx 2 dy1 Solution. Let y = y1, y2 dx dy2 d2 y d dy1 4 y 0, or 4 y1 4 y1 0 or 2 dx dx dx dx Differential equations (1) and (2) are written in matrix form Ans. ...(1) ...(2) d y1 0 1 y1 dx y2 4 0 y2 0 1 The characteristic equation of is 4 0 0 λ 4 1 0 λ 2 4 0 λ = ±i 2 0 λ Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ Determinants and Matrices 361 Eigenvector for= – i 2 0 i2 4 1 x1 0 i 2 or 0 i 2 x2 0 0 1 x1 0 0 x2 0 R2 R2 + 2iR1 x1 1 –i 2 x1 + x2 = 0 or x2 i 2 Eigenvector for = –i 2 or 0 i2 4 1 x1 0 i 2 or 0 i 2 x2 0 4 i 2 0 1 x1 0 x 1 or i 2 x1 x2 0 or 1 0 x2 0 x2 i 2 1 2 1 1 , P 1 Let P 1 2i 2i 2 λx Pe P 1 1 ei 2 x 1 2i 2i 0 1 i 2x 1 i2 x 2 e 2 e = i2x ie ie i 2 x 1 x1 0 i 2 x2 0 1 4i 1 4i 1 0 2 ei 2 x 1 2 1 i2x 4i e i2 x 1 2ie 4i 1 e i 2 x 2 2ie i 2 x 1 2 1 i 2 x 1 i 2 x e e cos 2 x 4i 4i 1 i2 x 1 i2 x 2 sin 2 x e e 2 2 1 4i 1 4i 1 sin sx 2 cos 2 x Applying the initial conditions, we get 1 sin 2 x 1 cos 2 x or y1 cos 2 x and y2 2 sin 2 x 2 0 2 sin 2 x cos 2 x EXERCISE 4.30 Solve the following differential equations by matrix method: y1 cos 2 x y 2 2 sin 2 x 1. 2. 3. 4. 5. d 2 y 4dy 3 y 0, y (0) 2, y' (0) 1 dx 2 dx d 2 y 3dy 2 y 0, y (0) 5, y' (0) 8 dx 2 dx d 2 y 5dy 14 y 0, y (0) 2, y' (0) 5 dx 2 dx d2 y μ 2 y 0, y (0) 1, y' (0) μ 2 dx d2 y 9 y 0, y (0) 1, y' (0) 3 dx 2 Ans. y Ans. 5 x e3 x e 2 2 Ans. y = 2e x + 3e 2x Ans. y = e 2x + e –7x Ans. y = cosx + sinx Ans.y = cosx + sinx Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ 362 Determinants and Matrices 4.71 COMPLEX MATRICES Conjugate of a Complex Number z = x + i y is called a complex number where 1 = i, x, y are real numbers. z x iy is called the conjugate of the complex number z, e.g., Complex number 2 + 3i – 4 – 5i 6i 2 Conjugate number 2 – 3i – 4 + 5i – 6i 2 Conjugate of a matrix. The matrix formed by replacing the elements of a matrix by their respective conjugate numbers is called the conjugate of A and is denoted by A . A (aij ) m n , then A (aij )m n Example 3 4i 2 i 4 3 4i 2 i 4 If A then A i 2 3 i 2 3 i i 4.72 THEOREM If A and B be two matrices and their conjugate matrices are A and B respectively, then (i) ( A) A Proof. Let (ii) ( A B) A B A = [aij]m × n, then (iii) (k A) k A (iv) ( AB) A B A [a ij ]m n where aij is the conjugate complex of a . ij The (i, j) th element of ( A) = the conjugate complex of the (i, j)th element of A = the conjugate complex of aij = aij = the (i, j)th element of A. Hence (ii) Let ( A) A. A = [aij]m × n and B = [bij]m × n Proved. A [ a ij ]m n and B [bij ]m n (i, j) th element of ( A B) = conjugate complex of (i, j) th element of (A + B) = conjugate complex of (aij + bij) ( aij bij ) a ij b ij = (i, j)th element of A + (i, j)th element of B = (i, j)th element of ( A B) Hence, ( A B) A B (iii) Let A = [aij]m × n, let k be any complex number. Proved. The (i, j)th element of (kA) = conjugate complex of the (i, j)th element of kA = conjugate complex of kaij = kaij k aij k (i, j )th element of A (i, j)th element of k . A Hence, (iv) Let kA k A A = [aij]m × n, B = [bij]n × p Proved. Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ Determinants and Matrices Then 363 A [ aij ]m n , B [bij ]n p The (i, j)th element of ( AB) = conjugate complex of (i, j)th element of AB aij b jk = conjugate complex of j 1 n aij b jk j 1 n n a ij b jk j 1 = (i, j)th element of A B Hence, Proved. ( AB) A B 4.73 TRANSPOSE OF CONJUGATE OF A MATRIX The transpose of a conjugate of a matrix A is denoted by A or A* . ( A) A The (i, j)th element of A = (j, i)th element of A = conjugate complex of (j, i)th element of A. 2 3i 1 2i 2 4i Example 88. If A 3 4i 4 3i 2 6i , find A 5 5 6i 3 2 3i 1 2i Solution. We have, A 3 4i 4 3i 5 5 6i 2 3i A ( A) 1 2i 2 4i 2 4i 2 3i 1 2i 2 4i 2 6i A 3 4i 4 3i 2 6i 5 3 5 6i 3 3 4i 5 4 3i 5 6i 2 6i 3 Ans. EXERCISE 4.31 1 i 3 5i 1. If the matrix A , find (i) A (ii) ( A) (iii) A (iv) ( A ) 2 i 5 1 i 2i 1 i 3 5i Ans. (i) A (ii) ( A) 3 5i 5 2 i 5 1 i 3 5 i 1 i 2 i (iv) ( A ) (iii) A 2i 3 5i 5 5 4.74 HERMITIAN MATRIX Definition. A square matrix A = [aij] is said to be Hermitian if the (i, j)th element of A, i.e., aij a ji for all i and j. b id 3 4i a 2 For example, , c 1 b id 3 4i Hence all the elements of the principal diagonal are real. A necessary and sufficient condition for a matrix A to be Hermitian is that A A . Example 89. The characteristic roots of a Hermitian matrix are all real. (A.M.I.E.T.E., June 2006) Solution. We know that matrix A is Hermitian if A A i.e., where A ( A ') or ( A) ' Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ 364 Determinants and Matrices Also ( A) A and ( AB ) B A If is a characteristic root of matrix A then AX = X. (AX) = (X) or XA X. But A is Hermitian A = A. ... (1) X A X X AX X X Again from (1) I XAX = XX = XX. Hence from (2) and (3) we conclude that showing that is real. ... (2) ... (3) Deduction 1.From above we conclude that characteristic roots of real symmetric matrix are all real, as in this case, real symmetric matrix will be Hermitian. For symmetric, we know that A = A. ( A ') A . or A A A A as A is real. Rest as above. Example 90. Prove that the following (i) ( A ) A (ii) ( A B) A B (iii) (kA) k A (iv) ( AB) B A where A and B be the transposed conjugates of A and B respectively, A and B being conformable to multiplication. Solution. (ii) as ( A) A ( A ) [{( A )}] [ A] A (i) ( A B) ( A B) ( A B) ( A) ( B) A B (kA) (kA) (k A) k ( A) k A (iii) (iv) ( AB) ( AB) ( A B) ( B ) ( A) B A 1 Example 91. Prove that matrix A 1 i 2 Solution. 1 1 i 2 A 1 i 3 i 2 i 0 A A 1 i 3 i Proved. 2 i is Hermitian. 0 1 1 i 2 ( A) 1 i 3 i 2 i 0 A is Hermitian matrix. Proved. 3 2i 2 i i 0 3 4i is Skew-Hermitian matrix. Example 92. Show that A 3 2i 2 i 3 4i 2i Solution. 3 2i 2 i i A 3 2i 0 3 4i 2 i 3 4i 2i Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ Determinants and Matrices 365 3 2i 2 i i ( A) 3 2i 0 3 4i 2i 2 i 3 4i 3 2i 2 i i A 3 2i 0 3 4i 2 i 3 4i 2i [ A ( A) ] 3 2i 2 i i 3 2i 0 3 4i A 2 i 3 4i 2i A = – A A is Skew-Hermitian matrix. Proved. Example 93. Show that the matrix B AB is Hermitian or Skew-Hermitian according as A is Hermitian or Skew-Hermitian. Solution. (i) Let A be Hermitian A A Now ( B A B) ( AB) ( B ) B A B B A B ( A A) Hence, A AB is Hermitian. (ii) Let A be Skew-Hermitian A A Now, ( B AB) ( AB) ( B ) B A B B A B ( A A) Hence, B AB is Skew-Hermitian. Proved. 4.75 SKEW-HERMITIAN MATRIX Definition. A square matrix A = (aij) is said to be Skew-Hermitian matrix if the (i, j)th element of A is equal to the negative of the conjugate complex of the (j, i)th element of A, i.e., aij a ji for all i and j. If A is a Skew-Hermitian matrix, then aii a ii aii aii 0 Obviously, aii is either a pure imaginary number or must be zero. a ib 3 4i 0 0 For example, and are Skew-Hermitian matrixes. 0 0 a ib 3 4i A necessary and sufficient condition for a matrix A to be Skew-Hermitian is that A A. Deduction 2. Characteristic roots of a skew Hermitian matrix is either zero or a pure imaginary numbers. If A is skew Hermitian, then iA is Hermitian. Also be a characteristic root of A then AX = X. (i .A) X = (i) X. Above shows that i is characteristic root of matrix iA, which is Hermitian and hence i should be real, which will be possible if is either pure imaginary or zero. Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ 366 Determinants and Matrices Example 94. Show that every square matrix can be expressed as R + iS uniquely where R and S are Hermitian matrices. Solution. Let A be any square matrix. It can be rewritten as 1 1 A ( A A ) i ( A A ) R iS 2 2i 1 1 where R ( A A ), S ( A A ) 2i 2 Now we have to show that R and S are Hermitian matrices. 1 1 1 1 R ( A A ) [ A ( A ) ] ( A A) ( A A ) R 2 2 2 2 Thus R is Hermitian matrix. 1 1 S ( A A ) ( A A ) 2i 2i 1 1 1 [ A ( A ) ] ( A A) ( A A ) S 2i 2i 2i Thus S is a Hermitian matrix. Hence A = R + iS, where R and S are Hermitian matrices. Now, we have to show its uniqueness. Let A = P + iQ be another expression, where P and Q are Hermitian matrices, i.e., Now, P P, Q Q Then A ( P iQ) P (iQ ) P iQ P iQ A = P + iQ and A P iQ 1 1 ( A A ) R and Q ( A A ) S 2i 2 Hence A = R + iS is the unique expression, where R and S are Hermitian matrices. Proved. P 2 5 5i 1 i 2 i 4 2i as the sum of Hermitian matrix Example 95. Express the matrix A 2i 7 1 i 4 and Skew-Hermitian matrix. 2 5 5i 2 5 5i 1 i 1i A 2i 2 i 4 2i A – 2i 2 i 4 2i Solution. ...(1) 1 i 4 1 i 4 7 7 2i 1 i 2i 1 i 1 i 1 i ...(2) ( A) 2 2i 4 A 2 2i 4 5 5i 4 2i 5 5i 4 2i 7 7 On adding (1) and (2), we get Let 2 2i 4 6i 2 A A 2 2i 4 2i 4 6i 2i 14 1 i 2 3i 1 1 R (A A ) 1 i 2 i 2 2 3i i 7 ...(3) Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ Determinants and Matrices 367 On subtracting (2) from (1), we get 2 2i 6 4i 2i A A 2 2i 2i 8 2i 6 4i 8 2i 0 1 i 3 2i i 1 1 i ( A A ) i 4 i S= 2 3 2i 4 i 0 Let ...(4) From (3) and (4), we have 1 i 2 3i i 1 i 3 2i 1 A 1 i 2 i 1 i i 4 i 7 3 2i 4 i 0 2 3i i Hermitian matrix Skew-Hermitian matrix Ans. Example 96. For any square matrix, if AA I show that A A I . Solution. AA I So A is invertible. Let B be another matrix such that (given) AB = BA = I ...(1) Now B = BI = B ( AA ) ( AA I ) = (BA) A = IA [Using (1)] We know that BA = I [From (1)] Putting the value of B from (2) in (1), we get A A = I Proved. 4.76 PERIODIC MATRIX A square matrix is said to be periodic, if Ak+1 = A, where k is a positive integer. If k is the least positive integer for which Ak+1 = A, then A is said to be of period k. 4.77 IDEMPOTENT MATRIX A square matrix is said to be idempotent provided A2 = A. Example 97. Determine all the idempotent diagonal matrices of order n. Solution. Let A = diag. [d1, d2, d3, ... dn] be an idempotent matrix of order n. Here, for the matrix ‘A’ to be idempotent A2 = A d1 0 0 d 2 0 0 0 0 0........0 d1 0 0........0 0 d 2 d3 .......0 0 0 0........dn 0 0 d12 0 0 0 0 d 22 0 0 0........0 d1 0........0 0 d3 .......0 0 0........dn 0 0........0 d1 0........0 0 d32 .......0 0 0........d n2 0 0 d2 0 0 0 d2 0 0 0........0 0........0 d3 .......0 0........d n 0........0 0........0 d3 .......0 0........d n Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ 368 Determinants and Matrices d12 d1; d22 d 2 .........dn2 dn i.e., d1 = 0, 1; d2 = 0, 1; d3 = 0, 1 ............ dn = 0, 1. Hence diag. [d1, d2, d3 … dn], is the required idempotent matrix where d1 = d2 = d3 = ... dn = 0 or 1. Ans. EXERCISE 4.32 1. Which of the following matrices are Hermitian: 2 i 3 i 2i 1 (a) 2 i (b) 4 2 4i 3 3 i 4 i 3 2. Which of the following matrices are 2 i 5 2i 4 1 0 i 3 2i 1 2 5 i 6 (c) (d) 7 0 5i Ans. (c) 2 7 2i 3i 1 0 5 2i 2 5i Skew-Hermitian: 3 1 2 3i 1 3 7 i 1 1 i 2 3i 0 1 2i 6 3i i 6 (b) (d) 0 6i (c) 1 i 4 6 3i 0 2 3i 6i 4i 7 i 8 Ans. (a), (c) 3. Give an example of a matrix which is Skew-symmetric but not Skew-Hermitian. 2i 3 4 (a) 3 3i 5 4 5 4i 2 3i 0 Ans. 2 3 i 0 4. If A be a Hermitian matrix, show that iA is Skew-Hermitian. Also show that if B be a SkewHermitian matrix, then iB must be Hermitian. 5. If A and B are Hermitian matrices, then show that AB + BA is Hermitian and AB – BA is SkewHermitian. 6. If A is any square matrix, show that A A is Hermitian. 5 2i 3 3 7. If H 5 2i 7 4i , show that H is a Hermitian matrix. 3 4i 5 Verify that iH is a Skew-Hermitian matrix. 8. Show that for any complex square matrix A, T * (i) ( A A* ) is a Hermitian matrix, where A A (ii) ( A A* ) is Skew-Hermitian matrix. (iii) AA* and A* A are Hermitian matrices. 9. Show that any complex square matrix can be uniquely expressed as the sum of a Hermitian matrix and a Skew-Hermitian matrix. 2 3i 4 5i i 10. Express A 6 i 0 4 5i as the sum of Hermitian and Skew-Hermitian matrices. 2 i 2 i i 11. Prove that the latent roots of a Hermitian matrix are all real. 2 i 3 1 3i 12. If A = show that AA* is a Hermitian matrix; where A* is the conjugate 5 i 4 2i transpose of A. (AMIETE, June 2010) 4.78 UNITARY MATRIX A square matrix A is said to be unitary matrix if A A A A I Example 98. If A is a unitary matrix, show that AT is also unitary. Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ Determinants and Matrices 369 Solution. A A A A I , since A is a unitary matrix. ( AA ) ( A A) I (I I ) ( AA ) ( A A) I ( A ) A A ( A ) I AA A A I [since ( A ) A ] ( AA )T ( A A)T ( I )T ( A )T AT AT ( A )T I ( AT ) AT AT ( AT ) I Hence, AT is a unitary matrix. Example 99. If A is a unitary matrix, show that A–1 is also unitary. Proved. Solution. AA A A I , since A is a unitary matrix. ( AA )1 ( A A)1 ( I )1 taking inverse ( A )1 A1 A1 ( A )1 I ( A1 ) A1 A1 ( A1 ) I Hence, A–1 is a unitary matrix. Proved. Example 100. If A and B are two unitary matrices, show that AB is a unitary matrix. Solution. A A A A I since A is a unitary matrix. ...(1) B B B B I Similarly, Now, ...(2) ( AB)( AB) ( AB)( B A ) A( BB ) A A I A [From (2)] [From (1)] AA I Again, ( AB) ( AB) ( B A ) ( AB) B ( A A) B [From (1)] B IB B B =I [From (2)] Hence, AB is a unitary matrix. Proved. 1 1 1 i Example 101. Prove that the matrix is unitaryy. 3 1 i 1 Solution. Let A 1 1 1 i 3 1 i 1 1 1 1 i 3 1 i 1 1 1 1 i 1 1 1 i A A 3 1 i 1 3 1 i 1 (1 i ) (1 i ) 1 3 0 1 0 1 1 (1 1) I (1 1) 1 3 0 3 0 1 3 (1 i ) 1(1 i ) Hence, A is a unitary matrix. A Proved. Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ 370 Determinants and Matrices 1 2i 0 Example 102. Define a unitary matrix. If N is a matrix, then show that 0 1 2i (I – N) (I + N)-1 is a unitary matrix, where I is an identity matrix. (U.P., I Semester, Winter 2000) Solution. Unitary matrix: A square matrix ‘A’ is said to be unitary if A A I , where A ( A)T and I is an identity matrix. we have 1 2i 0 N 1 2 i 0 1 2i 1 1 2 i 1 0 0 I N 0 1 – 2i 1 0 1 1 2i Now we have to find (I + N)–1 ...(1) 1 2i 1 1 2i 1 0 0 IN 0 1 2i 1 0 1 1 2i | I + N | = 1 – (– 1 – 4) = 6 1 2i 1 Adj. ( I N ) 1 1 2i ( I N ) 1 1 2i Adj ( I N ) 1 1 1 |IN| 6 1 2i ...(2) For unitary matrix, A A I From (1) and (2), we get ( I – N ) ( I N ) 1 Now ( B )T ( B )T B 1 6 1 2i 1 1 1 2i 1 1 2i 1 2i 1 4 2 4i B (say) 1 6 2 4i 4 2 4i 1 4 2 4i 4 6 1 36 2 4i 4 2 4i 1 36 0 4 I. 2 4i 2 4i 4 4 36 0 36 Hence the result. Proved. 4.79 THE MODULUS OF EACH CHARACTERISTIC ROOT OF A UNITARY MATRIX IS UNITY. (U.P., I Semester, Compartment 2002) Solution. Suppose A is a unitary matrix. Then A A I . Let be a characteristic root of A. Then AX X Taking conjugate transpose of both sides of (1), we get ( AX ) X X A X ...(1) ...(2) Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ Determinants and Matrices 371 From (1) and (2), we have ( X A ) ( AX ) X X X ( A A) X X X X IX X X X X X X ( A . A I ) X X ( 1) 0 ...(3) Since, X X 0 therefore (3) gives 1 0. or 1 or | |2 1 | | 1 Proved. EXERCISE 4.33 1. Show that the matrix A 1 1 2 i i is unitary.. 1 2. Prove that a real matrix is unitary if it is orthogonal. 3. Prove that the following matrix is unitary: 1 2 (1 i ) 1 (1 i ) 2 1 1 1 4. Show that U 1 3 2 1 1 ( 1 i ) 2 1 (1 i ) 2 1 2 is a unitary matrix, where is the complex cube root of unity.. 5. Prove that the latent roots of a unitary matrix have unit modulus. 6. Verify that the matrix A 1 1 i 1 i 2 1 i 1 i has eigen values with unit modulus. Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ 372 Vectors 5 Vectors 5.1 VECTORS A vector is a quantity having both magnitude and direction such as force, velocity acceleration, displacement etc. 5.2 ADDITION OF VECTORS Let a and b be two given vectors B OA = a and AB = b then vector OB is called the — b — — O OA AB = OB A a — b a+ sum of a and b . Symbolically a b = OB 5.3 RECTANGULAR RESOLUTION OF A VECTOR ^ ^ ^ Let OX, OY, OZ be the three rectangular axes. Let i , j , k be three unit vectors and parallel to three axes. If OP = n and the co-ordinates of P be (x, y, z) ^ OA = x i , ^ OB = y j and Z C ^ OC = z k k OP = OF FP OP = OA OB OC r ^ xi ^ ^ 2 OP = OF + FP zk O yj j A = xi + y j + zk 2 – r OP = (OA AF ) FP (x, y, z) P X 2 B F (x, y) Y xi = (OA2 + AF2) + FP2 = OA2 + OB2 + OC2 = x2 + y2 + z2 OP = x 2 y2 z 2 x 2 y2 z 2 |r| = 5.4 UNIT VECTOR ^ ^ Let a vector be x i + y ^j + z k . ^ Unit vector = ^ ^ xi y j zk x 2 y 2 z2 372 Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ 373 Vectors Example 1. If a and b be two unit vectors and be the angle between them, then find the value of such that a + b is a unit vector. (Nagpur, University, Winter 2001) Solution.Let OA = a be a unit vector and AB = b is another unit vector and A be the angle between a and b . If OB = c = a + b is also a unit vector then, we have |OA| = 1 a b | OB| = 1 c | OB| = 1 OAB is an equilateral triangle. O B a+b=c Hence each angle of OAB is 3 5.5 Ans. POSITION VECTOR OF A POINT The position vector of a point A with respect to origin O is the vector OA which is used to specify the position of A w.r.t. O. — To find AB if the position vectors of the point A and point B are given. If the position vectors of A and B are a and b . Let the origin be O. OA = a , Then B (b) OB b OA AB = OB AB = OB OA A (a) O AB = b a AB = Position vector of B – Position vector of A Example 2. If A and B are (3, 4, 5) and (6, 8, 9), find AB . Solution. AB = Position vector of B – Position vector of A = (6 iˆ 8 ˆj 9 kˆ ) (3 ˆi 4 ˆj 5 kˆ ) = 3 iˆ 4 ˆj 4 kˆ 5.6 Ans. RATIO FORMULA To find the position vector of the point which divides the line joining two given points. Let A and B be two points and a point C divides AB in the ratio of m : n. Let O be the origin, then OA = a , and OB b , OC ? A (a) m C n B (b) OC = OA AC m m AB AC m n AB mn m ( AB b a ) = a m n . (b a) = OA O Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ 374 Vectors mb na OC = mn Cor. If m = n = 1, then C will be the mid-point, and a b OC = 2 5.7 PRODUCT OF TWO VECTORS The product of two vectors results in two different ways, the one is a number and the other is vector. So, there are two types of product of two vectors, namely scalar product and vector product. They are written as a . b and a b . 5.8 SCALAR, OR DOT PRODUCT The scalar, or dot product of two vectors a and b is defined to be a b cos i.e., scalar where is the angle between a and b . Symbolically, a . b = a b cos B Due to a dot between a and b this product is also called dot product. The scalar product is commutative Proof. a . b = b . a To Prove. b . a = = b b a cos ( ) a b cos 0 A a = a .b Proved. Geometrical interpretation. The scalar product of two vectors is the product of one vector and the length of the projection of the other in the direction of the first. B OA = a and OB b Let then b a . b = (OA) . (OB) cos ON = OA . OB . OB = OA . ON 5.9 0 a = (Length of a ) (projection of b along a ) A N USEFUL RESULTS ^ ^ ^ ^ i . i = (1) (1) cos 0° = 1 i . j = (1) (1) cos 90° = 0 ^ ^ ^ ^ Similarly, j . j = 1, ^ ^ ^ ^ k.k =1 Similarly, j . k = 0, k.i =0 Note. If the dot product of two vectors is zero then vectors are prependicular to each other. 5.10 WORK DONE AS A SCALAR PRODUCT F If a constant force F acting on a particle displaces it from A to B then, Work done = (component of F along AB). Displacement = F cos . AB = F . AB Work done = Force . Displacement A Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ B 375 Vectors 5.11 VECTOR PRODUCT OR CROSS PRODUCT 1. The vector, or cross product of two vectors a and b is defined to be a vector such that b (i) Its magnitude is a b sin , where is the angle between a and b . (ii) Its direction is perpendicular to both vectors a a and b . (iii) It forms with a right handed system. ^ Let be a unit vector perpendicular to both the vectors a and b . 2. a b sin . a b = Useful results ^ ^ ^ Since i , j , k are three mutually perpendicular unit vectors, then ^ ^ ^ ^ ^ ^ ^ ^ i i = j j kk0 ^ ^ ^ ^ ^ ^ ^ ^ ^ ^ ^ ^ ^ ^ ^ i j = j i k ^ ^ j i i j ^ ĵ kˆ = k j i and k j jk ^ ^ ^ kˆ iˆ = i k j 5.12 i k k i VECTOR PRODUCT EXPRESSED AS A DETERMINANT If ^ ^ ^ ^ ^ ^ a = a1 i a 2 j a3 k b = b1 i b2 j b3 k ^ ^ ^ ^ ^ ^ a b = ( a1 i a 2 j a3 k ) ( b1 i b2 j b3 k ) ^ ^ ^ ^ ^ ^ ^ ^ ^ ^ = a1 b1 ( i i ) a1 b2 ( i j ) a1b3 ( i k ) a2 b1 ( j i ) a2 b2 ( j j ) ^ ^ ^ ^ ^ ^ ^ ^ a2 b3 ( j k ) a3 b1 ( k i ) a3 b2 ( k j ) a3 b3 ( k k ) ^ ^ ^ ^ ^ ^ = a1 b2 k a1b3 j a2 b1 k a2 b3 i a3 b1 j a3 b2 i ^ ^ ^ = ( a2 b3 a3 b2 ) i ( a1 b3 a3 b1 ) j ( a1 b2 a2 b1 ) k ^ ^ j k a1 b1 a2 b2 a3 b3 i = 5.13 ^ AREA OF PARALLELOGRAM Example 3. Find the area of a parallelogram whose adjacent sides are i – 2j + 3 k and 2i + j – 4k. ^ Solution. Vector area of gm = ^ ^ i j k 1 2 3 2 1 4 Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ 376 Vectors ^ ^ ^ ^ ^ ^ = (8 3) i ( 4 6) j (1 4) k = 5 i 10 j 5 k (5)2 (10)2 (5)2 = 5 6 Area of parallelogram = 5.14 Ans. O MOMENT OF A FORCE Let a force F ( PQ ) act at a point P. r Moment of F about O = Product of force F and perpendicular ^ distance (ON. ) N P F Let a rigid body be rotating about the axis OA with the angular velocity which is a vector and its magnitude is radians per second and its direction is parallel to the axis of rotation OA. ^ ^ Q = (PQ) (ON)( ) = (PQ) (OP) sin ( ) = OP PQ 5.15 M r F ANGULAR VELOCITY V B P A Axis Let P be any point on the body such that OP = r and AOP = and AP OA. Let the velocity of P be V. be a unit vector perpendicular to and r . Let ^ ^ r = ( r sin ) = ( AP) = (Speed of P) r = Velocity of P to and r 5.16 Hence V = r O SCALAR TRIPLE PRODUCT Let a , b , c be three vectors then their dot product is written as a . ( b c ) or [ a b c ] . ^ ^ ^ ^ ^ ^ ^ ^ ^ a = a1 i a2 j a3 k , b b1 i b2 j b3 k , and c c1 i c 2 j c3 k If ^ ^ ^ ^ ^ ^ ^ ^ ^ ^ ^ ^ a . ( b c ) = ( a1 i a 2 j a3 k ) . [( b1 i b2 j b3 k ) ( c1 i c 2 j c 3 k )] ^ ^ ^ = ( a1 i a2 j a3 k ) . [( b2 c3 b3 c2 ) i ( b3 c1 b1 c3 ) j ( b1c2 b2 c1 ) k ] = a1 (b2c3 – b3c2) + a2 (b3c1 – b1c3) + a3 (b1c2 – b2c1) = a1 a2 a3 b1 b2 b3 c1 c2 c3 Similarly, b . ( c a ) and c . ( a b ) have the same value. a . (b c ) = b . (c a) = c . (a b) The value of the product depends upon the cyclic order of the vector, but is independent of the position of the dot and cross. These may be interchanged. The value of the product changes if the order is non-cyclic. Note. a ( b . c ) and ( a . b ) c are meaningless. Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ 377 Vectors 5.17 GEOMETRICAL INTERPRETATION The scalar triple product a . ( b c ) represents the volume of the parallelopiped having a , b , c as its co-terminous edges. F G ^ a . ( b c ) = a . Area of gm OBDC = Area of gm OBDC × perpendicular distance A – between the parallel faces OBDC and AEFG. a = Volume of the parallelopiped ^ n E B D – b Note. (1) If a . ( b c ) = 0, then a , b , c are O coplanar. 1 (a b c). (2) Volume of tetrahedron 6 c– C Example 4. Find the volume of parallelopiped if ^ ^ ^ ^ ^ ^ ^ ^ ^ a 3 i 7 j 5 k, b 3 i 7 j 3 k , and c 7 i 5 j 3 k are the three co-terminous edges of the parallelopiped. Solution. Volume = a . ( b c ) 3 7 5 = 3 7 3 = – 3 (–21 – 15) – 7 (9 + 21) + 5 (15 – 49) 7 5 3 = 108 – 210 – 170 = – 272 Volume = 272 cube units. Ans. Example 5. Show that the volume of the tetrahedron having A B , B C , C A as concurrent edges is twice the volume of the tetrahendron having A, B , C as concurrent edges. 1 Solution. Volume of tetrahendron = ( A B ) . [( B C ) ( C A )] 6 1 ( A B ) . [ B C B A C C C A] = [ C C 0] 6 1 (A B) . ( B C B A C A ) = 6 1 [ A . ( B C ) A . ( B A ) A . ( C A) B . ( B C ) B . ( B A ) B . ( C A )] 6 1 1 = [ A . ( B C ) B . (C A)] A . ( B C ) 6 3 1 = 2 [A B C] 6 = = 2 Volume of tetrahedron having A, B , C , as concurrent edges. Proved. EXERCISE 5.1 1. Find the volume of the parallelopiped with adjacent sides. OA = 3 i j , OB j 2 k , and OC i 5 j 4 k extending from the origin of co-ordinates O. Ans. 20 2. Find the volume of the tetrahedron whose vertices are the points A (2, –1, –3), B (4, 1, 3) C (3, 2, –1) and D (1, 4, 2). Ans. 7 1 3 Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ 378 Vectors ^ ^ ^ ^ ^ 3. Choose y in order that the vectors a 7 i y j kˆ , b 3 i 2 j k , ^ ^ ^ c 5 i 3 j k are linearly dependent. 4. Prove that Ans. y = 4 [a b , b c , c a] 2 [a b c] 5.18 COPLANARITY QUESTIONS Example 6. Find the volume of tetrahedron having vertices ^ ^ ^ ^ ^ ^ ^ ^ ^ ^ ^ ( j k ), ( 4 i 5 j q k ), ( 3 i 9 j 4 k ) and 4 ( i j k ) . Also find the value of q for which these four points are coplanar. (Nagpur University, Summer 2004, 2003, 2002) ^ ^ ^ ^ ^ ^ ^ ^ ^ ^ ^ Solution. Let A = j k , B 4 i 5 j q k , C 3 i 9 j 4 k , D 4( i j k ) ^ ^ ^ ^ ^ ^ ^ AB = B A 4 i 5 j q k ( j kˆ ) 4 i 6 j (q 1) k ^ ^ ^ ^ ^ ^ ^ AC = C A (3 i 9 ˆj 4 k ) ( j k ) 3 i 10 j 5 k ^ ^ ^ ^ ^ ^ ^ ^ AD = D A 4 ( i j k ) ( j k ) 4 i 5 j 5 k 1 Volume of the tetrahedron = [ AB AC AD ] 6 4 6 q1 1 1 3 10 5 {4 (50 25) 6 (15 20) ( q 1) (15 40)} = = 6 6 4 5 5 1 1 ( 110 55 55 q) {100 210 55 ( q 1)} = = 6 6 1 55 ( 55 55 q) ( q 1) = 6 6 If four points A, B, C and D are coplanar, then ( AB AC AD ) = 0 i.e., Volume of the tetrahedron = 0 55 ( q 1) = 0 q=1 Ans. 6 Example 7. If four points whose position vectors are a , b , c , d are coplanar, show that [ a b c ] [ a d b ] [ a d c ] [ d b c ] (Nagpur University, Summer 2005) Solution. Let A, B, C, D be four points whose position vectors are a , b , c , d . AD = d a , BD d b and CD d c If AD , BD , CD are coplanar, then AD . (BD CD ) = 0 ( d a ) . [( d b ) ( d c )] = 0 (d a) . [d d d c b d b c ]= 0 ( d a ) . [ d c b d b c ] = 0 d . ( d c ) d . ( b d ) d . ( b c ) a . ( d c ) a . ( b a ) a .( b c ) = 0 0 0 [d b c ] [ d d c ] [ d b d ] [ a b c ] = 0 [ a b c ] [ a b d ] [ a d c ] [ d b c ] Proved. Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ 379 Vectors EXERCISE 5.2 1. Determine such that ^ ^ ^ ^ ^ ^ ^ ^ a i j k , b 2 i 4 k , and c i j 3 k are coplanar. 2. Show that the four points ^ ^ ^ ^ ^ ^ ^ ^ ^ ^ Ans. = 5/3 ^ ^ 6 i 3 j 2 k , 3 i 2 j 4 k , 5 i 7 j 3 k and 13 i 17 j k are coplanar. 3. Find the constant a such that the vectors ^ ^ ^ ^ ^ ^ ^ ^ ^ 2 i j k , i 2 j 3 k , and 3 i a j 5 k are coplanar. 4. Prove that four points ^ ^ ^ ^ ^ ^ ^ ^ ^ ^ Ans. – 4 ^ 4 i 5 j k , ( j k ), 3 i 9 j 4 k , 4 ( i j k ) are coplanar. 5. If the vectors a , b and c are coplanar, show that a a . a c a . b b . a 5.19 b a . c b . b =0 b . c VECTOR PRODUCT OF THREE VECTORS (A.M.I.E.T.E., Summer, 2004, 2000) Let a , b and c be three vectors then their vector product is written as ^ ^ ^ ^ ^ ^ a (b × c ). a = a1 i a2 j a3 k , Let b = b1 i b2 j b3 k , ^ ^ ^ c = c1 i c2 j c3 k ^ ^ ^ ^ ^ ^ ^ ^ ^ ^ ^ ^ a ( b c ) = ( a1 i a2 j a3 k ) (b1 i b2 j b3 k ) ( c1 i c2 j c3 k ) ^ ^ ^ = ( a1 i a2 j a3 k ) [( b2 c 3 b3 c 2 ) i ( b3 c1 b1c 3 ) j ( b1c 2 b2 c1 ) k ] ^ ^ = [ a2 ( b1c2 b2 c1 ) a3 ( b3 c1 b1 c3 )] i [ a3 ( b2 c3 b3 c2 ) a1 ( b1 c2 b2 c1 )] j ^ [ a1 ( b3 c1 b1c 3 ) a2 ( b2 c3 b3 c 2 ) k ] ^ ^ ^ ^ ^ ^ = ( a1 c1 a2 c 2 a3 c 3 ) ( b1 i b2 j b3 k ) ( a1 b1 a2 b2 a3 b3 ) ( c1 i c 2 j c 3 k ) = (a . c) b (a . b) c . Ans. Example 8. Prove that : a ( b c ) b ( c a ) c ( a b ) 0 (Nagpur University, Winter 2008) Solution. Here, we have a (b c ) b (c a) c ( a b) = [( a . c ) b ( a . b ) c ] [( b . a ) c ( b . c ) a ] [( c . b ) b ( c . a ) b ] = [( b . a ) c ( a . b ) c ] [( c . b ) a ( b . c ) a ] [( a . c ) b ( c . a ) b ] = [( a . b ) c ( a . b ) c ] [( b . c ) a ( b . c ) a ] [( c . a ) b ( c . a ) b ] =0+0+0=0 Proved. Example 9. Prove that : ^ ^ ^ ^ ^ ^ i ( a i ) j ( a j) k ( a k) 2 a ^ ^ (Nagpur University, Winter 2003) ^ Solution. Let a = a1 i a2 j a3 k Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ 380 Vectors Now, ^ ^ ^ ^ ^ ^ L.H.S. = i ( a i ) j ( a j ) k ( a k ) ^ ^ ^ ^ ^ ^ ^ ^ ^ ^ = i ( a1 i a2 j a3 k ) i j ( a1 i a2 j a3 k ) j ^ ^ ^ ^ ^ k ( a1 i a2 j a3 k ) k ^ ^ ^ ^ ^ ^ ^ ^ ^ ^ ^ ^ ^ ^ = i a1 ( i i ) a2 ( j i ) a3 ( k i ) j a1 ( i j ) a2 ( j j ) a3 ( k j ) ^ ^ ^ ^ ^ ^ ^ k a1 ( i k ) a2 ( j k ) a3 ( k k ) ^ ^ ^ ^ ^ ^ ^ ^ ^ = i 0 a2 k a3 j j a1 k 0 a3 i k a1 j a2 i 0 ^ ^ ^ ^ ^ ^ ^ ^ ^ ^ ^ ^ = a2 ( i k ) a3 ( i j ) a1 ( j k ) a3 ( j i ) a1 ( k j ) a2 ( k i ) ^ ^ ^ ^ ^ ^ ^ ^ ^ = a2 j a3 k a1 i a3 k a1 i a2 j = 2 a1 i 2 a2 j 2 a3 k ^ ^ ^ = 2 ( a1 i a2 j a3 k ) 2 a Proved. Example 10. Show that for any scalar , the vectors x , y given by q (a b) x a a2 , y (1 p ) p ( a b ) a satisfy the equations q a2 p x q y a and x y b . Solution. The given equations are (Nagpur University, Winter 2004) px qy = a ...(1) x y = b ...(2) Multiplying equation (1) vectorially by x , we get x ( px qy ) = x a p (x x) q (x y) = x a as x x 0 q (x y) = x a, x a = qb , [From (2) x y b ] ...(3) Multiplying (3) vectorially by a , we have a (x a) = a q b ( a . a) x ( a . x) a = q ( a b) x = ( a . x) a a2 a2 x = ( a . x ) a q ( a b ) a2 x ( a . x ) a = q ( a b ) q (a b) a2 Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ 381 Vectors q (a b) x = a a2 a . x where = a2 q ( a b ) p a Substituting the value of x in (1), we get q y = a 2 a q (a b) q y = a p a a2 (1 p ) a p (a b) q a2 EXERCISE 5.3 y = Ans. 1. Show that a ( b a ) ( a b ) a 2. Write the correct answer (a) ( A B ) C lies in the plane of (i) A and B (iii) C and A (ii) B and C Ans. (ii) (b) The value of a . ( b c ) ( a + b c ) is 5.20 (ii) [ a , b , c ] [ b , c , a ] (iii) [ a , b , c ] (iv) None of these Ans. (ii) SCALAR PRODUCT OF FOUR VECTORS (i) Zero Prove the identity ( a b) . ( c d) = ( a . c ) (b . d) ( a . d) (b . c ) Proof. ( a b ) . ( c d ) = ( a b ) . r = a . ( b r ) dot and cross can be interchanged. Put c d r = a . [ b ( c d )] = a . [( b . d ) c ( b . c ) d ] = (a . c ) (b . d) (a . d) (b . c ) a . c = a .d b . c Proved. b .d EXERCISE 5.4 1. If a 2 i 3 j k , b i 2 j 4 k , c i j k , find ( a b ) . ( a c ). 2 Ans. –74 2. Prove that ( a b ) . ( a c ) a ( b . c ) ( a . b ) ( a . c ) . 5.21 VECTOR PRODUCT OF FOUR VECTORS Let a , b , c and d be four vectors then their vector product is written as ( a b) ( c d) Now, ( a b ) ( c d ) = r ( c d ) [Put a b r ] = (r . d) c (r . c) d Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ 382 Vectors = [( a b ) . d ] c [( a b ) . c ] d = [ a b d] c [a b c ] d ( a b ) ( c d ) lies in the plane of c and d . ...(1) Again, ( a b ) ( c d ) = ( a b ) s [Put = c d s ] = s (a b) = ( s . b) a ( s . a) b = [( c d ) . b ] a [( c d ) . a ] b = [( b c d ] a [ a c d ] b ( a b ) ( c d ) lies in the plane of a and b . ...(2) Geometrical interpretation : From (1) and (2) we conclude that ( a b ) ( c d ) is a vector parallel to the line of intersection of the plane containing a , b and plane containing c , d . Example 11. Show that ( B C ) ( A D ) (C A) ( B D ) ( A B ) (C D ) 2 ( A B C ) D Solution. L.H.S. = ( B C ) ( A D ) ( C A ) ( B D ) ( A B ) ( C D ) = [( B C D ) A ( B C A) D ] [( C A D ) B (C A B ) D] [( B C D ) A ( A C D ) B ] = ( B C D ) A ( B C D ) A (C A D ) B ( A C D ) B ( B C A) D (C A B ) D = ( A C D) B ( A C D) B ( A B C ) D ( A B C) D = 2 ( A B C ) D = R.H.S. Proved. EXERCISE 5.5 Show that: 1. ( b c ) ( c a ) c ( a b c ) when ( a b c ) stands for scalar triple product. 2. [ b c , a b )] [ a v c ]2 c a, 3. d [ a { b ( c d )}] [( b . d )[ a . ( c d )] 4. a [ a [ a ( a b )] a 2 ( b a ) 5. [( a b ) ( a c )] . d ( a . d ) [ a b c ] 2 6. 2a 7. 8. ^ 2 a i ^ ^ 2 a j ^ ^ ^ 2 a k ^ ^ ^ a b [( i a ) . b] i [( j a ) . b ] j [( k a ) . b ] k p [( a q ) ( b r )] q [( a r ) ( b p )] r [( a p ) ( b q )] 0 Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ Vectors 383 5.22 VECTOR FUNCTION If vector r is a function of a scalar variable t, then we write r = r (t ) If a particle is moving along a curved path then the position vector r of the particle is a function of t. If the component of f (t) along x-axis, y-axis, z-axis are f1(t), f2(t), f3(t) respectively. Then, — f1 (t ) i f 2 (t ) j f3 (t ) k f (t ) = 5.23 DIFFERENTIATION OF VECTORS Let O be the origin and P be the position of a moving particle at time t. — Let OP = r Let Q be the position of the particle at the time t + t and Q t en ng Ta — the position vector of Q is OQ = r r — — — PQ = OQ OP r P (r) = (r r) r r r+r r is a vector. As t 0, Q tends to P and the chord t becomes the tangent at P. r dr lim We define = t 0 t , then dt r O dr is a vector in the direction of the tangent at P. dt dr is also called the differential coefficient of r with respect to ‘t’. dt d2 r Similarly, is the second order derivative of r . 2 dt dr d2 r gives the velocity of the particle at P, which is along the tangent to its path. Also dt dt 2 gives the acceleration of the particle at P. 5.24 FORMULAE OF DIFFERENTIATION d d F dG (F G) (i) dt dt dt d d dF (ii) ( F ) F dt dt dt (U.P. I semester, Dec. 2005) d dG dF d dG dF (iii) (F . G) F . .G (iv) (F G) F G dt dt dt dt dt dt d d a d b d c (v) [a b c ] b c a c a b dt dt dt dt d da d b d c (vi) [ a ( b c )] (b c ) a c a b dt dt dt dt The order of the functions F , G is not to be changed. Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ 384 Vectors Example 12. A particle moves along the curve r (t 3 4 t ) i (t 2 4 t ) j (8 t 2 3 t 3 ) k , where t is the time. Find the magnitude of the tangential components of its acceleration at t = 2. (Nagpur University, Summer 2005) Solution. We have, r = (t 3 4t ) i (t 2 4t ) j (8t 2 3t 3 ) k dr (3t 2 4) i (2t 4) j (16t 9t 2 ) k Velocity = dt t = 2, Velocity = 8 i 8 j 4 k At Acceleration = a = At d2 r dt 2 t = 2 6t i 2 j (16 18t ) k a 12 i 2 j 20 k The direction of velocity is along tangent. So the tangent vector is velocity. Unit tangent vector, v 8 i 8 j 4k 8 i 8 j 4k 2 i 2 j k T =|v| 12 3 64 64 16 Tangential component of acceleration, at = a.T 24 4 20 48 2i 2 j k = (12 i 2 j 20 k ). = = 16 Ans. 3 3 3 d da db [a b] u (a b) u a and u b then prove that dt dt dt (M.U. 2009) Solution. We have, Example 13. If d b da d b [a b] = a dt dt dt = a (u b) (u a) b = a ( u b ) b (u a ) = ( a . b ) u ( a . u ) b [( b . a ) u (b . u ) a ] (Vector triple product) = (a . b ) u (u . a ) b (a . b ) u (u . b ) a = (u . b ) a (u . a ) b = Proved. u (a b) 2 2 2 2 Example 14. Find the angle between the surface x + y + z = 9 and z = x + y2 – 3 at (2, –1, 2). (M.D.U. Dec. 2009) Solution. Here, we have x2 + y2 + z2 = 9 ...(1) z = x2 + y2 – 3 ...(2) 2 2 2 Normal to (1) 1= (x + y + z – 9) 2 2 2 ˆj kˆ = iˆ ( x y z – 9) = 2 x iˆ + 2 y ĵ + 2 z k̂ x y z Normal to (1) at (2, – 1, 2), 1 = 4 iˆ – 2 ĵ + 4 k̂ ...(3) Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ Vectors 385 2 = (z – x2 – y2 + 3) Normal to (2), 2 2 ˆj kˆ ( z – x – y 3) = – 2 x iˆ – 2 y ĵ + k̂ = iˆ y z x Normal to (2) at (2, – 1, 2), 2 = – 4 iˆ + 2 ĵ + k̂ ...(4) 1.2 = | 1 | | 2 | cos 1.2 (4iˆ – 2 ˆj 4kˆ).(– 4iˆ 2 ˆj kˆ) – 16 – 4 4 = = ˆ ˆ ˆ ˆ ˆ ˆ | 1 | | 2 | 16 4 16 16 4 1 |4i – 2 j 4k | | – 4i 2 j k | – 16 –8 = = 6 21 3 21 –1 – 8 = cos 3 21 –8 Hence the angle between (1) and (2) cos –1 Ans 3 21 cos = EXERCISE 5.6 1. The coordinates of a moving particle are given by x = 4t t2 t3 and y = 3 6t . Find the 2 6 Ans. 4.47, 2.24 velocity and acceleration of the particle when t = 2 secs. 2. A particle moves along the curve x = 2t2, y = t2 – 4t and z = 3t – 5 where t is the time. Find the components of its velocity and acceleration at time t = 1, in the 8 14 14 , 7 7 3. Find the unit tangent and unit normal vector at t = 2 on the curve x = t2 – 1, y = 4t – 3, 1 1 z = 2t2 – 6t where t is any variable. Ans. (2 i 2 j k ), (2 i 2 k ) 3 3 5 direction i 3 j 2 k . (Nagpur, Summer 2001) Ans. 4. Prove that d dG d F ( F G) F G dt dt dt 5. Find the angle between the tangents to the curve r t 2 i 2t j t 3 k , at the points t = ± 1. 9 Ans. cos 1 17 6. If the surface 5x2 – 2byz = 9x be orthogonal to the surface 4x2y + z3 = 4 at the point (1, –1, 2) then b is equal to (a) 0 (b) 1 (c) 2 (d) 3 (AMIETE, Dec. 2009) Ans. (b) 5.25 SCALAR AND VECTOR POINT FUNCTIONS ^ Point function. A variable quantity whose value at any point N R in a region of space depends upon the position of the point, is Q + called a point function. There are two types of point functions. n r d = (i) Scalar point function. If to each point P (x, y, z) of a c = P region R in space there corresponds a unique scalar f (P), then f is c called a scalar point function. For example, the temperature distribution in a heated body, density of a body and potential due to gravity are the examples of a scalar point function. (ii) Vector point function. If to each point P (x, y, z) of a region R in space there corresponds a unique vector f (P), then f is called a vector point function. The velocity of a moving fluid, gravitational force are the examples of vector point function. (U.P., I Semester, Winter 2000) Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ 386 Vectors Vector Differential Operator Del i.e. The vector differential operator Del is denoted by . It is defined as j k = i x y z 5.26 GRADIENT OF A SCALAR FUNCTION If (x, y, z) be a scalar function then i is called the gradient of the scalar j k x y z function . And is denoted by grad . Thus, grad = i j k x y z j k ( x, y , z ) gard = i y z x gard = ( is read del or nebla) 5.27 GEOMETRICAL MEANING OF GRADIENT, NORMAL (U.P. Ist Semester, Dec 2006) If a surface (x, y, z) = c passes through a point P. The value of the function at each point on the surface is the same as at P. Then such a surface is called a level surface through P. For example, If (x, y, z) represents potential at the point P, then equipotential surface (x, y, z) = c is a level surface. Two level surfaces can not intersect. Let the level surface pass through the point P at which the value of the function is . Consider another level surface passing through Q, where the value of the function is + d. — Let r and r r be the position vector of P and Q then PQ r j k .(i dx j dy k dz ) .dr = i y z x = dx dy dz d x y z If Q lies on the level surface of P, then d = 0 Equation (1) becomes . dr = 0. Then is to dr (tangent). ...(1) Hence, is normal to the surface (x, y, z) = c Let = || N is a unit normal vector. Let n be the perpendicular distance N , where between two level surfaces through P and R. Then the rate of change of in the direction of the normal to the surface through P is d = dn . n .d r lim n 0 n n 0 n lim | | N .d r = lim n 0 n = lim n 0 N . r | N | | r | cos | r | cos n | | n | | n Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ Vectors 387 n Hence, gradient is a vector normal to the surface = c and has a magnitude equal to the rate of change of along this normal. || = 5.28 NORMAL AND DIRECTIONAL DERIVATIVE (i) Normal. If (x, y, z) = c represents a family of surfaces for different values of the constant c. On differentiating , we get d = 0 But d = . d r so . d r = 0 The scalar product of two vectors and d r being zero, and d r are perpendicular to each other. d r is in the direction of tangent to the given surface. Thus is a vector normal to the surface (x, y, z) = c. (ii) Directional derivative. The component of in the direction of a vector d is equal to .d and is called the directional derivative of in the direction of d . = lim where, r = PQ r 0 r r is called the directional derivative of at P in the direction of PQ. r Let a unit vector along PQ be N . n = cos r = r Now r = lim r 0 n .N N = N . N | | n cos n ...(1) N . N N . N n n From (1), r N.N = N . ( N | | ) , directional derivative is the component of in the direction N . r = N . | | cos | | r Hence, is the maximum rate of change of . Hence, Example 15. For the vector field (i) A miˆ and (ii) A m r . Find . A and A . Draw the sketch in each case. (Gujarat, I Semester, Jan. 2009) Solution. (i) Vector A m i is represented in the figure (i). (ii) (iii) A = m r is represented in the figure (ii). . A = i x j y k z .( x i y j z k ) 1 1 1 3 (iv) . A = 3 is represented on the number line at 3. j k A = i (x i y j z k) y z x Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ 388 Vectors i j = x y x y are represented in the adjoining figure. k z z =0 k k k^ O O . i^ ^i r m r j j 0 1 2 3 Number line m i (i) (ii) 2 j O i (iii) (iv) 3 2 Example 16. If = 3x y – y z ; find grad at the point (1, –2, –1). (AMIETE, June 2009, U.P., I Semester, Dec. 2006) Solution. grad = j k (3 x 2 y y 3 z 2 ) = i y z x 2 3 2 2 3 2 2 3 2 = i x (3x y y z ) j y (3x y y z ) k z (3x y y z ) = i (6 xy ) j (3 x 2 3 y 2 z 2 ) k ( 2 y 3 z ) grad at (1, –2, –1) = i (6) (1) (2) j [(3) (1) 3(4) (1)] k ( 2)( 8) (1) = 12 i 9 j 16 k Ans. Example 17. If u = x + y + z, v = x2 + y2 + z2, w = yz + zx + xy prove that grad u, grad v and grad w are coplanar vectors. [U.P., I Semester, 2001] Solution. We have, grad u = i grad v = i grad w = i j k ( x y z) i j k x y z j k ( x2 y 2 z 2 ) 2 x i 2 y j 2 z k x y z j k ( yz zx xy ) i ( z y ) j ( z x ) k ( y x) x y z [For vectors to be coplanar, their scalar triple product is 0] Now, grad u.(grad v × grad w) = 1 = 2 x yz zy = 2( x y z ) 1 1 1 2x 2y 2z zy zx yx 1 1 x yz x y z z x 1 1 1 1 yz zx yx 1 1 2 1 1 1 x y z zy zx yx [Applying R2 R2 + R3] 0 x y Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ Vectors 389 Since the scalar product of grad u, grad v and grad w are zero, hence these vectors are coplanar vectors. Proved. Example 18. Find the directional derivative of x2y2z2 at the point (1, 1, –1) in the direction of the tangent to the curve x = et, y = sin 2t + 1, z = 1 – cos t at t = 0. (Nagpur University, Summer 2005) Solution. Let = x2 y2 z2 Directional Derivative of 2 2 2 j k = = i (x y z ) y z x = 2 xy 2 z 2 i 2 yx 2 z 2 j 2 zx 2 y 2 k Directional Derivative of at (1, 1, –1) = 2(1)(1) 2 (1)2 i 2(1)(1) 2 (1) 2 j 2(1)(1)2 (1)2 k = 2i 2 j2k t = x i y j z k e i (sin 2t 1) j (1 cos t ) k = r Tangent vector, ...(1) T d r e t i 2 cos 2t j sin t k dt Tangent(at t = 0) = e0 i 2 (cos 0) j (sin 0) k i 2 j Required directional derivative along tangent = (2 i 2 j 2 k ) ...(2) (i 2 j) 1 4 [From (1), (2)] 240 6 5 5 Example 19. Find the unit normal to the surface xy3z2 = 4 at (–1, –1, 2). Solution. Let (x, y, z) = xy3z2 = 4 We know that is the vector normal to the surface (x, y, z) = c. Normal vector = = i x j y k z = Now = i (M.U. 2008) ( xy 3 z 2 ) j ( xy 3 z 2 ) k ( xy 3 z 2 ) x y z Ans. 3 2 2 2 3 Normal vector = y z i 3 xy z j 2 xy z k Normal vector at (–1, –1, 2) = 4 i 12 j 4 k Unit vector normal to the surface at (–1, –1, 2). 4 i 12 j 4 k 1 (i 3 j k ) = Ans. | | 16 144 16 11 Example 20. Find the rate of change of = xyz in the direction normal to the surface x2y + y2x + yz2 = 3 at the point (1, 1, 1). (Nagpur University, Summer 2001) Solution. Rate of change of = j k ( x y z ) i yz j xz k xy = i y z x Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ 390 Vectors Rate of change of at (1, 1, 1) = ( i j k ) Normal to the surface = x2y + y2x + yz2 – 3 is given as 2 j k ( x y y 2 x yz 2 3) = i x y z 2 2 2 = i (2 xy y ) j ( x 2 xy z ) k 2 yz ()(1, 1, 1) = 3 i 4 j 2 k Unit normal = 3i 4 j 2k 9 16 4 Required rate of change of = ( i j k ). (3 i 4 j 2 k ) = 3 42 9 Ans. 9 16 4 29 29 Example 21. Find the constants m and n such that the surface m x2 – 2nyz = (m + 4)x will be orthogonal to the surface 4x2y + z3 = 4 at the point (1, –1, 2). (M.D.U. Dec. 2009, Nagpur University, Summer 2002) Solution. The point P (1, –1, 2) lies on both surfaces. As this point lies in mx2 – 2nyz = (m + 4)x, so we have m – 2n (–2) = (m + 4) m + 4n = m + 4 n=1 Let 1 = mx2 – 2yz – (m + 4)x and 2 = 4x2y + z3 – 4 Normal to 1 = 1 j k [ mx 2 2 yz ( m 4) x ] = i x y z = i (2mx m 4) 2 z j 2 y k Normal to 1 at (1, –1, 2) = i (2m m 4) 4 j 2 k = (m 4) i 4 j 2 k Normal to 2 = 2 j k (4 x 2 y z 3 4) = i 8 xy 4 x 2 j 3 z 2 k = i y z x Normal to 2 at (1, –1, 2) = – 8 i 4 j 12 k Sinec 1 and 2 are orthogonal, then normals are perpendicular to each other. 1 . 2 = 0 [( m 4) i 4 j 2 k ].[ 8 i 4 j 12 k ] = 0 – 8 (m – 4) – 16 + 24 = 0 m – 4 = –2 + 3 m=5 Hence m = 5, n = 1 Ans. Example 22. Find the values of constants and so that the surfaces x2 – yz = ( + 2) x, 4x2y + z3 = 4 intersect orthogonally at the point (1, – 1, 2). (AMIETE, II Sem., Dec. 2010, June 2009) Solution. Here, we have x2 – yz = ( + 2) x ...(1) 4x2 y + z3 = 4 ...(2) Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ Vectors 391 Normal to the surface (1), λx yz (λ 2) x ˆ 2 iˆ ˆj k x yz ( 2) x y z x iˆ (2x 2) ˆj (z ) kˆ (y ) Normal at (1, –1, 2) = iˆ (2 – – 2) – ĵ (–2) + k̂ ...(3) = iˆ ( – 2) + ĵ z (2) + k̂ Normal at the surface (2) iˆ ˆj kˆ (4 x 2 y z 3 4) y z x = iˆ (8 × y) + ĵ (4x2) + k̂ (3z2) Normal at the point (1, –1, 2) = – 8 iˆ + 4 ĵ + 12 k̂ Since (3) and (4) are orthogonal so ...(4) iˆ ( 2) ˆj (2 ) kˆ . 8 iˆ 4 ˆj 12 kˆ 0 8 ( 2) 4(2) 12 0 8 16 8 12 0 8 20 16 0 2 5 4 0 Point (1, – 1, 2) will satisfy (1) 4(2 5 4) 0 2 5 4 ...(5) (1)2 (1) (2) ( 2) (1) + 2 = + 2 = 1 Putting = 1 in (5), we get 9 2 5 4 2 9 and =1 Hence Ans. 2 2 2 2 2 2 Example 23. Find the angle between the surfaces x + y + z = 9 and z = x + y – 3 at the point (2, –1, 2). (Nagpur University, Summer 2002) Solution. Normal on the surface (x2 + y2 + z2 – 9 = 0) 2 j k ( x y 2 z 2 9) (2 x i 2 y j 2 z k ) = i y z x Normal at the point (2, –1, 2) = 4 i 2 j 4 k 2 j k ( x y 2 z 3) Normal on the surface (z = x2 + y2 – 3) = i y z x ...(1) = 2x i 2 y j k Normal at the point (2, –1, 2) = 4 i 2 j k Let be the angle between normals (1) and (2). (4 i 2 j 4 k ).(4 i 2 j k ) = ...(2) 16 4 16 16 4 1 cos 16 + 4 – 4 = 6 21 cos 16 = 6 21 cos Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ 392 Vectors 8 cos = = cos 1 8 Ans. 3 21 3 21 1 Example 24. Find the directional derivative of in the direction r wheree r x i y j z k . r (Nagpur University, Summer 2004, U.P., I Semester, Winter 2005, 2002) Solution. Here, (x, y, z) = 1 r 1 x2 y 2 z 2 (x2 y2 z2 ) 1 2 1 2 j k ( x y2 z2 ) 2 Now 1 = i y z x r 1 1 1 2 ( x y 2 z2 ) 2 i ( x2 y 2 z2 ) 2 j (x2 y 2 z2 ) 2 k x y z 3 3 3 1 1 1 = ( x 2 y 2 z 2 ) 2 2 x i ( x 2 y 2 z 2 ) 2 2 y j ( x 2 y 2 z 2 ) 2 2 z k 2 2 2 = = (x i y j z k ) ...(1) ( x 2 y 2 z 2 )3/ 2 r = unit vector in the direction of x i y j z k and = xi y j zk ...(2) x2 y 2 z2 So, the required directional derivative = . r xi y j zk ( x2 y 2 z 2 ) 1 1 2 = 2 2 2 x y z r . 3/ 2 xi y j zk ( x 2 y 2 z 2 )1/ 2 x2 y 2 z 2 ( x 2 y 2 z 2 )2 [From (1), (2)] Ans. Example 25. Find the direction in which the directional derivative of (x, y) = x2 y 2 at xy (1, 1) is zero and hence find out component of velocity of the vector r (t 3 1) i t 2 j in the same direction at t = 1. (Nagpur University, Winter 2000) x2 y2 i j k Solution. Directional derivative = = y z xy x 2 2 2 2 = i x y.2 x ( x y ) y j xy.2 y x ( y x ) 2 2 2 2 x y x y x2 y y3 xy 2 x3 = i j 2 2 2 2 x y x y Directional Derivative at (1, 1) = i 0 j 0 0 Since ()(1, 1) = 0, the directional derivative of at (1, 1) is zero in any direction. Again r = (t 3 1) i t 2 j Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ Vectors Velocity, 393 v = dr 3t 2 i 2t j dt Velocity at t = 1 is = 3 i 2 j The component of velocity in the same direction of velocity 3 i 2 j 9 4 13 (3 i 2 j ). = Ans. 9 4 13 Example 26. Find the directional derivative of (x, y, z) = x2 y z + 4 x z2 at (1, –2, 1) in the direction of 2 i j 2 k . Find the greatest rate of increase of . (Uttarakhand, I Semester, Dec. 2006) Solution. Here, (x, y, z) = x2y z + 4xz2 2 j k ( x yz 4 xz 2 ) Now, = i y z x = (2 xyz 4 z 2 ) i ( x 2 z ) j ( x 2 y 8 xz ) k at (1, – 2, 1) = {2(1) (2)(1) 4(1)2 } i (1 1) j {1( 2) 8(1)(1)} k = (4 4) i j (2 8) k = j 6 k 1 (2 i j 2 k ) = unit vector = a 3 4 1 4 So, the required directional derivative at (1, –2, 1) 1 1 13 = . a ( j 6 k ). (2 i j 2 k ) = (1 12) 3 3 3 Let 2 i j 2k Greatest rate of increase of = j 6 k = 1 36 = 37 Ans. Example 27. Find the directional derivative of the function = x2 – y2 + 2z2 at the point P (1, 2, 3) in the direction of the line PQ where Q is the point (5, 0, 4). (AMIETE, Dec. 20010, Nagpur University, Summer 2008, U.P., I Sem., Winter 2000) Solution. Directional derivative = 2 j k ( x y2 2z2 ) 2x i 2 y j 4z k = i y z x Directional Derivative at the point P (1, 2, 3) = 2 i 4 j 12 k ...(1) PQ = Q P = (5, 0, 4) – (1, 2, 3) = (4, –2, 1) Directional Derivative along PQ = (2 i 4 j 12 k ). = 8 8 12 ...(2) (4 i 2 j k ) 16 4 1 [From (1) and (2)] 28 Ans. 21 x Example 28. For the function (x, y) = 2 , find the magnitude of the directional x y2 derivative along a line making an angle 30° with the positive x-axis at (0, 2). (A.M.I.E.T.E., Winter 2002) 21 Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ 394 Vectors Solution. Directional derivative = x = i x j y k z 2 x y2 = i y2 x2 1 x (2 x ) x (2 y ) 2 j 2 = i 2 2 ( x y 2 )2 ( x y2 )2 x y j 2 xy j A ( x 2 y 2 )2 ( x 2 y 2 )2 Directional derivative at the point (0, 2) = i 4 0 (0 4) 2 j 2(0) (2) (0 4) 2 1 (0, 2) i 4 1j — 2 30° C i 3 — i 2 — Directional derivative at the point (0, 2) in the direction CA i.e. 3 i 1 j 2 2 CA OB BA i cos 30 j sin 30 i 3 1 3 1 . i j = i j 4 2 2 2 2 3 = Ans. 8 2 Example 29. Find the directional derivative of V , where V xy 2 i zy 2 j xz 2 k , at the point (2, 0, 3) in the direction of the outward normal to the sphere x2 + y2 + z2 = 14 at the point (3, 2, 1). (A.M.I.E.T.E., Dec. 2007) Solution. V2 = V .V 2 2 2 2 2 2 = ( xy i zy j xz k ).( x y i z y j xz k ) = x2y4 + z2y4 + x2z4 Directional derivative = V 2 2 4 j k (x y z2 y4 x2 z4 ) = i x y z = (2 xy 4 2 xz 4 ) i (4 x 2 y 3 4 y 3 z 2 ) j (2 y 4 z 4 x 2 z 3 ) k Directional derivative at (2, 0, 3) = (0 2 2 81) i (0 0) j (0 4 4 27) k = 324 i 432 k 108 (3 i 4 k ) ...(1) Normal to x2 + y2 + z2 – 14 = 2 j k ( x y 2 z 2 14) = i y z x = (2 x i 2 y j 2 z k ) Normal vector at (3, 2, 1) Unit normal vector = = 6 i 4 j 2k 6 i 4 j 2k 36 16 4 ...(2) 2(3 i 2 j k ) 2 14 Directional derivative along the normal = 108(3 i 4 k ). = 108 (9 4) 14 1404 14 3i 2 j k 14 3i 2 j k 14 [From (1), (2)] . Ans. Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ Vectors 395 Example 30. Find the directional derivative of ( f) at the point (1, – 2, 1) in the direction of the normal to the surface xy2z = 3x + z2, where f = 2x3y2z4. (U.P., I Semester, Dec 2008) Solution. Here, we have f = 2x3 y2 z4 3 2 4 f = i x j y k z (2 x y z ) = 6 x 2 y 2 z 4 i 4 x3 yz 4 j 8 x3 y 2 z 3 k 2 2 4 3 4 3 2 3 (f) = i x j y k z (6 x y z i 4 x yz j 8 x y z k ) = 12xy2z4 + 4x3z4 + 24x3y2z2 Directional derivative of ( f ) 2 4 3 4 3 2 2 = i x j y k z (12 xy z 4 x z 24 x y z ) = (12 y 2 z 4 12 x 2 z 4 72 x 2 y 2 z 2 )i (24 xyz 4 48 x3 yz 2) j + (48xy2z3 + 16x3z3 + 48x3y2z) k Directional derivative at (1, – 2, 1) = (48 + 12 + 288) i + (– 48 – 96) j + (192 + 16 + 192) k = 348i – 144 j 400k Normal to(xy2z – 3x – z2) = (xy2z – 3x – z2) 2 2 = i x j y k z ( xy z – 3x – z ) = ( y 2 z – 3)i (2 xyz ) j ( xy 2 – 2 z )k Normal at(1, – 2, 1) = i – 4 j 2k i – 4 j 2k 1 Unit Normal Vector = = (i – 4 j 2k ) 1 16 4 21 Directional derivative in the direction of normal 1 (i – 4 j 2k ) = (348i – 144 j 400k ) 21 1 1724 (348 576 800) = = Ans. 21 21 2 2 2 Example 31. If the directional derivative of = a x y + b y z + c z x at the point x 1 y 3 z , (1, 1, 1) has maximum magnitude 15 in the direction parallel to the line 2 2 1 find the values of a, b and c. (U.P. I semester, Winter 2001) 2 2 2 Solution. Given = ax y+by z+cz x j k (a x2 y + b y2 z + c z2 x) = i x y z 2 2 2 = i (2a x y c z ) j (a x 2 b y z ) k (b y 2 c z x) at the point (1, 1, 1) = i (2a c) j (a 2b) k (b 2c) ...(1) We know that the maximum value of the directional derivative is in the direction of . i.e. || = 15 (2a + c)2 + (2b + a)2 + (2c + b)2 = (15)2 But, the directional derivative is given to be maximum parallel to the line Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ 396 Vectors x 1 y 3 z i.e., parallel to the vector 2 i 2 j k . = 2 2 1 On comparing the coefficients of (1) and (2) 2a c 2b a 2c b = 2 2 1 2a + c = – 2b – a 3a + 2b + c = 0 and 2b + a = – 2(2c + b) 2b + a = – 4c – 2b a + 4b + 4c = 0 Rewriting (3) and (4), we have 3a 2b c 0 a b c = k (say) 4 11 10 a 4b 4 c 0 a = 4k, b = –11k and c = 10k. Now, we have (2a + c)2 + (2b + a)2 + (2c + b)2 = (15)2 (8k + 10k)2 + (–22k + 4k)2 + (20k – 11k)2 = (15)2 k = 5 9 20 55 50 a = , b= and c = 9 9 9 ...(2) ...(3) ...(4) Ans. Example 32. If r x i y j z k , show that : r 1 (ii) grad 3 . r r (i) grad r = r r (Nagpur University, Summer 2002) Solution. (i) r = x i y j z k r = x 2 y 2 z 2 r2 = x2 + y2 + z2 r r x 2r = 2x x x r y r r z Similarly, = and r z r y r r r j k ri j k grad r = r = i y z x y z x x y z xi y j zk r j k Proved. r r r r r 1 1 1 1 1 1 (ii) grad = i j k = i j k x r y r z r y z r r r x 1 r 1 r 1 r = i 2 k 2 j 2 x y r r z r = i 1 x 1 y 1 z xi y j zk r = i Proved. 3 j 2 k 2 = 2 3 r r r r r r r r 2 2 Example 33. Prove that f (r ) f ´´ (r ) f ´ (r ) . (K. University, Dec. 2008) r Solution. Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ Vectors 397 j k f (r ) f (r ) i y z x 2 r r x r y 2 2 2 2x , r x y z 2r x x r y r r r r y x i f ´ (r ) j f ´(r ) k f ´(r ) f ´ (r ) i j k x y z r r xi y j zk f ´ (r ) r xi yj zk j k f ´(r ) 2 f (r ) [ f (r )] i y z r x x y z f ´(r ) f ´ (r ) f ´(r ) x r y r z r r r r.1 y r1 x y y r x x f ´´(r ) r f ´´(r ) f ´(r ) r f ´(r ) 2 x r y r r2 and r z z r z r z r.1 z r z r f ´´( r ) f ´( r ) z r r2 x2 y2 z2 r r r r f ´´(r ) y y f ´ (r ) r f ´´ (r ) z z f ´(r ) r = f ´´(r ) x x f ´(r ) 2 2 r r r r r r r r r2 zz r 2 z2 x x r 2 x2 y r 2 y2 f ´´ ( r ) f ´ ( r ) f ´´( r ) f ´( r ) = f ´´(r ) f ´( r ) r r r rr r3 r3 r3 2 2 2 2 2 2 2 2 2 x y z y x z z x y f ´´(r ) 2 f ´(r ) f ´´(r ) 2 f ´(r ) f ´´(r ) 2 f ´(r ) 3 3 r r r r r r3 2 2 2 2 2 2 2 2 2 x y z y z z x x y = f ´´(r ) 2 2 2 f ´ (r ) 3 3 r r r r3 r r x2 y2 x2 2( x 2 y 2 z 2 ) r2 2r 2 f ´´ (r ) f ´( r ) f ´´( r ) f ´ ( r ) = r2 r3 r2 r3 2 = f ´´(r ) f ´(r ) Ans. r EXERCISE 5.7 1. Evaluate grad if = log (x2 + y2 + z2) Ans. 2( x i y j z k ) x2 y 2 z 2 1 ˆ 2. Find a unit normal vector to the surface x2 + y2 + z2 = 5 at the point (0, 1, 2). Ans. ( j 2kˆ ) 5 (AMIETE, June 2010) 3. Calculate the directional derivative of the function (x, y, z) = xy 2 + yz 3 at the point 5 (1, –1, 1) in the direction of (3, 1, –1) (A.M.I.E.T.E. Winter 2009, 2000) Ans. 11 4. Find the direction in which the directional derivative of f (x, y) = (x2 – y2)/xy at (1, 1) is zero. (Nagpur Winter 2000) Ans. i j 2 Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ 398 Vectors 5. Find the directional derivative of the scalar function of (x, y, z) = xyz in the direction of the outer 27 normal to the surface z = xy at the point (3, 1, 3). Ans. 11 6. The temperature of the points in space is given by T(x, y, z) = x2 + y2 – z. A mosquito located at (1, 1, 2) desires to fly in such a direction that it will get warm as soon as possible. In what direction 1 should it move? Ans. 3 (2 i 2 j k ) 7. If (x, y, z) = 3xz2y – y3z2, find grad at the point (1, –2, –1) Ans. (16 i 9 j 4 k ) 8. Find a unit vector normal to the surface x2y + 2xz = 4 at the point (2, –2, 3). 1 ( i 2 j 2 k ) 3 9. What is the greatest rate of increase of the function u = xyz2 at the point (1, 0, 3)? Ans. 9 10. If is the acute angle between the surfaces xyz2 = 3x + z2 and 3x2 – y2 + 2z = 1 at the point Ans. (1, –2, 1) show that cos = 3/7 6 . 11. Find the values of constants a, b, c so that the maximum value of the directional directive of = axy2 + byz + cz2x3 at (1, 2, –1) has a maximum magnitude 64 in the direction parallel to the axis of z. Ans. a = b, b = 24, c = –8 12. Find the values of and µ so that surfaces x2 – µ y z = ( + 2)x and 4 x2 y + z3 = 4 intersect 9 , 1 2 13. The position vector of a particle at time t is R = cos (t – 1) i + sinh (t – 1) j + at2k. If at t = 1, the acceleration of the particle be perpendicular to its position vector, then a is equal to orthogonally at the point (1, –1, 2). (a) 0 (b) 1 (c) Ans. = 1 2 (d) 1 2 (AMIETE, Dec. 2009) Ans. (d) 5.29 DIVERGENCE OF A VECTOR FUNCTION The divergence of a vector point function F is denoted by div F and is defined as below.. Let F = F1 i F2 j F3 k F1 F2 F3 div F = . F i x j y k z ( i F1 j F2 k F3 ) = x y z It is evident that div F is scalar function. 5.30 PHYSICAL INTERPRETATION OF DIVERGENCE Let us consider the case of a fluid flow. Consider a small rectangular parallelopiped of dimensions dx, dy, dz parallel to x,y and z axes respectively. Let V Vx i V y j Vz k be the velocity of the fluid at P(x, y, z). Mass of fluid flowing in through the face ABCD in unit time = Velocity × Area of the face = Vx (dy dz ) Mass of fluid flowing out across the face PQRS per unit time = Vx (x + dx) (dy dz) O Vx dx ( dy dz ) = Vx Y x Net decrease in mass of fluid in the parallelopiped corresponding to the flow along x-axis per unit time Z C S D Vx A R dz B Q P X Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ Vectors 399 Vx dx dy dz = Vx dy dz Vx x Vx = dx dy dz x (Minus sign shows decrease) Similarly, the decrease in mass of fluid to the flow along y-axis = and the decrease in mass of fluid to the flow along z-axis = Vx Total decrease of the amount of fluid per unit time = x Vx V y Thus the rate of loss of fluid per unit volume = x y Vy y dx dy dz Vz dx dy dz z V y Vz dx dy dz y z V z z j k .( i Vx jVy k Vz ) = .V div V = i y z x If the fluid is compressible, there can be no gain or loss in the volume element. Hence div V = 0 and V is called a Solenoidal vector function. Equation (1) is also called the equation of continuity or conservation of mass. Example 34. If v xi y j zk , find the value of div v . x2 y 2 z 2 Solution. We have, v = ...(1) (U.P., I Semester, Winter 2000) xi y j zk x2 y 2 z2 x i y j zk i j k . div v = . v = y z ( x 2 y 2 z 2 )1/ 2 x x y z = 2 2 2 1/ 2 2 2 2 1/ 2 2 2 x ( x y z ) y ( x y z ) z ( x y z 2 )1/ 2 1 1 ( x 2 y 2 z 2 )1/ 2 x. ( x2 y 2 z 2 ) 2 .2 x 2 = ( x2 y 2 z 2 ) 1 1 1 ( x 2 y 2 z 2 ) 2 y. ( x 2 y 2 z 2 ) 2 2 y 2 (x2 y 2 z2 ) = = ( x 2 y 2 z 2 ) x2 ( x 2 y 2 z 2 )3/ 2 ( x2 y 2 z 2 ) y 2 ( x 2 y 2 z 2 )3/ 2 y 2 z 2 x2 z 2 x2 y 2 2 2 1 2 2 2 2 1/ 2 2 2 1/ 2 .2 z ( x y z ) z. 2 ( x y z ) ( x2 y 2 z 2 ) = 2 3/ 2 (x y z ) (x2 y 2 z2 ) z2 ( x 2 y 2 z 2 )3/ 2 2( x 2 y 2 z 2 ) 2 2 2 3/ 2 (x y z ) 2 (x2 y2 z2 ) Ans. Example 35. If u = x2 + y2 + z2, and r x i y j z k , then find div (ur ) in terms of u. (A.M.I.E.T.E., Summer 2004) Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ 400 Vectors j k .[( x 2 y 2 z 2 ) ( x i y j z k )] div (u r ) = i y z x j k .[( x 2 y 2 z 2 ) x i ( x 2 y 2 z 2 ) y j ( x 2 y 2 z 2 ) z k ] = i y z x Solution. 3 ( x xy 2 xz 2 ) ( x 2 y y 3 yz 2 ) ( x 2 z y 2 z z 3 ) x y z = (3x2 + y2 + z2) + (x2 + 3y2 + z2) + (x2 + y2 + 3z2) = 5 (x2 + y2 + z2) = 5 u = Ans. Example 36. Find the value of n for which the vector r n r is solenoidal, wheree r xi y j zk. n 2 2 2 n/2 F = . F .r r .( x y z ) ( x i y j z k ) Solution. Divergence 2 2 2 n/2 2 2 2 n/ 2 2 2 2 n/2 = i j k .[( x y z ) x i ( x y z ) y j ( x y z ) z k ] y z x n n 2 = (x2 + y2 + z2)n/2 – 1 (2x2) + (x2 + y2 + z2)n/2 + (x + y2 + z2)n/2 – 1 (2y2) 2 2 n 2 + (x2 + y2 + z2)n/2 + (x + y2 + z2)n/2 – 1 (2z2) + (x2 + y2 + z2)n/2 2 = n(x2 + y2 + z2)n/2 – 1 (x2 + y2 + z2) + 3 (x2 + y2 + z2)n/2 = n(x2 + y2 + z2)n/2 + 3(x2 + y2 + z2)n/2 = (n + 3) (x2 + y2 + z2)n/2 If r n r is solenoidal, then (n + 3) (x2 + y2 + z2)n/2 = 0 or n + 3 = 0 or n = –3. Ans. Example 37. Show that ( a . r ) a n ( a . r ) r . (M.U. 2005) r n r n rn 2 Solution. We have, a. r rn = (a1 i a2 j a3 k ).( x i y j z k ) r n = a1 x a2 y a3 z rn Let = = x a . r a1 x a2 y a3 z rn rn n r .a1 (a1 x a2 y a3 z ) n r n 1(r / x) But r2 = x2 + y2 + z2 r 2n r 2r = 2x x r x x r a1r n (a1 x a2 y a3 z ).n r n 2 .x n (a1 x a2 y a3 z ) x a = = 1n x r 2n r rn 2 i j k = x y z 1 n = n (a1 i a2 j a3 k ) n 2 [(a1 x a2 y a3 z ) ( x i y j z k )] r r a n = n n 2 ( a .r ) r r r Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ Vectors 401 Example 38. Let r x i y j z k , r | r | and a is a constant vector. Find the value of ar div n r a = a1 i a2 j a3 k Solution. Let ar = a1 x a2 y i ar | r |n = (a1 i a2 j a3 k ) ( x i y j z k ) j k a3 = (a2 z a3 y ) i (a1 z a3 x ) j (a1 y a2 x) k z (a2 z a3 y ) i (a1 z a3 x ) j (a1 y a2 x) k = ( x2 y2 z 2 )n / 2 a r ar = . div n | r |n |r| (a2 z a3 y ) i (a1 z a3 x ) j (a1 y a2 x) k j k . = i y z ( x2 y 2 z 2 ) n / 2 x a2 z a3 y a1 z a3 x ( a1 y a2 x) = 2 2 2 n/ 2 2 2 2 n/2 2 x ( x y z ) y ( x y z ) z ( x y 2 z 2 )n / 2 n (a2 z a3 y ) 2 x n (a1 z a3 x) 2 y n (a1 y a2 x ) 2 z = n2 n2 n2 2 2 2 2 2 2 (x y2 z2 ) 2 (x y2 z2 ) 2 (x y2 z2 ) 2 n = [(a2 z a3 y ) x (a1 z a3 x) y (a1 y a2 x ) z ] n2 = ( x2 y 2 z 2 ) n 2 2 2 2 n2 ) 2 [a2 zx a3 xy a1 yz a3 xy a1 yz a2 zx] = 0 Ans. (x y z Example 39. Find the directional derivative of div ( u ) at the point (1, 2, 2) in the direction of the outer normal of the sphere x2 + y2 + z2 = 9 for u x 4 i y 4 j z 4 k . Solution. div ( u ) = . u 4 j k .( x i y 4 j z 4 k ) 4 x 3 4 y 3 4 z 3 = i y z x Outer normal of the sphere = (x2 + y2 + z2 – 9) 2 j k ( x y 2 z 2 9) 2 x i 2 y j 2 z k = i y z x Outer normal of the sphere at (1, 2, 2) = 2 i 4 j 4 k ...(1) Directional derivative = (4 x3 4 y 3 4 z 3 ) j k (4 x3 4 y 3 4 z 3 ) 12 x 2 i 12 y 2 j 12 z 2 k = i y z x Directional derivative at (1, 2, 2) = 12 i 48 j 48 k ...(2) Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ 402 Vectors Directional derivative along the outer normal = (12 i 48 j 48 k ). 24 192 192 = 68 6 n n–2 Example 40. Show that div (grad r ) = n (n + 1)r , where 2 i 4 j 4k 4 16 16 [From (1), (2)] = x2 y 2 z 2 r = 1 Hence, show that 2 = 0. r Solution. (U.P. I Semester, Dec. 2004, Winter 2002) n n n r j r k r by definition x y z r r r r n 1 r n 1 r j n r n 1 k .n r n 1 i j k = i nr = nr x y z y z x x y z = n r n 1 i j k nr n 2 ( x i y j z k ) nr n 2 r . r r r r r x 2 2 2 2 r x y z 2r x 2 x x r etc. grad (rn) = i Thus, grad (rn) = n r n 2 x i n r n 2 y j n r n 2 z k Ans. ...(1) div grad rn = div [ n r n 2 x i n r n 2 y j n r n 2 z k ] j k .( nr n 2 x i nr n 2 y j nr n 2 z k ) = i [From (1)] y z x ( n r n 2 x) (n r n 2 y ) (n r n 2 z ) = (By definition) x y z r n 2 r ny (n 2) r n 3 = n r n 2 nx (n 2) r n 3 n r x y = 3n r n 2 n (n 2)r n 3 r n r n 2 nz (n 2) r n 3 z r r r z x y y z x x y z n2 n ( n 2)r n 3 x y z = 3n r r r r r r x 2 2 2 2 r x y z 2r x 2 x x r etc. = 3nrn – 2 + n (n – 2)rn – 4 [x2 + y2 + z2] = 3nrn – 2 + n (n – 2) rn – 4.r2 ( r2 = x2 + y2 + z2) n–2 2 n–2 2 = r [3n + n – 2n] = r (n + n) = n(n + 1) rn – 2 If we put n = –1 div grad (r– 1) = –1 (–1 + 1) r– 1 – 2 1 2 = 0 r r Ques. If r x i y j z k , and r = |r| find div 2 . (U.P. I Sem., Dec. 2006) r Ans. 1 r2 Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ Vectors 403 EXERCISE 5.8 r 1. If r = x i y j z k and r = | r | , show that (i) div 3 = 0, | r | (ii) div (grad rn) = n (n + 1) rn – 2 (AMIETE, June 2010) (iii) div (r ) = 3 + r grad . 2. Show that the vector V = ( x 3 y ) i ( y 3 z ) j ( x 2 z ) k is solenoidal. (R.G.P.V., Bhopal, Dec. 2003) 3. Show that .( A) = .A + (.A) 4. If , , z are cylindrical coordinates, show that grad (log ) and grad are solenoidal vectors. 5. Obtain the expression for 2f in spherical coordinates from their corresponding expression in orthogonal curvilinear coordinates. Prove the following: 6. .( F ) ( ). F ( . F ) (b) ( A R ) (2 n ) A n ( A . R ) R , r | R | n n n2 r r r 8. div ( f g) – div (g f) = f 2g – g 2 f 7. (a) .() = 2 5.31 CURL (U.P., I semester, Dec. 2006) The curl of a vector point function F is defined as below curl F = F ( F F1 i F2 j F3 k ) j k ( F1 i F2 j F3 k ) = i y z x i j k F F F F F F i 3 2 j 3 1 k 2 1 = x y z z z y x y x F1 F2 F3 Curl F is a vector quantity.. 5.32 PHYSICAL MEANING OF CURL (M.D.U., Dec. 2009, U.P. I Semester, Winter 2009, 2000) We know that V r , where is the angular velocity, V is the linear velocity and r is the position vector of a point on the rotating body. 1 i 2 j 3 k r x i y j z k Curl V = V = ( r ) = [( 1 i 2 j 3 k ) ( x i y j z k )] i j k = 1 2 3 x y z = [(2 z 3 y ) i (1 z 3 x ) j (1 y 2 x) k ] j k [( 2 z 3 y ) i ( 1 z 3 x ) j (1 y 2 x ) k ] = i y z x Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ 404 Vectors j k x y z i = 2 z 3 y 3 x 1 z 1 y 2 x = (1 2 ) i (2 2 ) j (3 3 ) k = 2(1 i 2 j 3 k ) 2 Curl V = 2 which shows that curl of a vector field is connected with rotational properties of the vector field and justifies the name rotation used for curl. If Curl F = 0, the field F is termed as irrotational. Example 41. Find the divergence and curl of v ( x y z ) i (3 x 2 y ) j ( xz 2 y 2 z ) k at (2, –1, 1) (Nagpur University, Summer 2003) Solution. Here, we have 2 2 2 v = ( x y z ) i (3 x y ) j ( xz y z ) k Div. v = ( x y z) (3x 2 y ) ( xz 2 y 2 z ) x y z = yz + 3x2 + 2x z – y2 = –1 + 12 + 4 – 1 = 14 at (2, –1, 1) Div v = Curl v = i x j y xyz 3 x 2 y k z xz 2 y 2 z Curl at (2, –1, 1) = 2 yz i ( z 2 xy ) j (6 xy xz ) k = 2 yz i ( xy z 2 ) j (6 xy xz ) k = 2(1)(1) i {(2) (1) 1} j {6(2)(1) 2(1)} k = 2 i 3 j 14 k Ans. Example 42. If V x i y j z k , find the value of curl V . x2 y 2 z 2 (U.P., I Semester, Winter 2000) Solution. Curl V = V x i y j zk j k = i y z ( x 2 y 2 z 2 )1/ 2 x = i x x j y y k z z ( x 2 y 2 z 2 )1/ 2 ( x 2 y 2 z 2 )1/ 2 ( x 2 y 2 z 2 )1/ 2 Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ Vectors 405 z y z = i 2 2 j 2 2 2 1/ 2 2 2 1/ 2 2 2 1/ 2 y z x ( x y z ) ( x y z ) ( x y z ) x y x k 2 2 2 2 2 1/ 2 2 2 1/ 2 2 2 1/ 2 z ( x y z ) x ( x y z ) y ( x y z ) yz y.z zx zx 2 j 2 2 = i 2 2 2 3/ 2 2 2 3/ 2 2 2 3/ 2 2 2 3/ 2 (x y z ) (x y z ) (x y z ) (x y z ) xy xy k 2 2 0 2 2 3/ 2 2 2 3/ 2 (x y z ) (x y z ) Ans. Example 43. Prove that ( y 2 – z 2 3 yz – 2 x ) i (3xz 2 xy ) j (3xy – 2 xz 2 z ) k is both solenoidal and irrotational. Solution. Let (U.P., I Sem, Dec. 2008) 2 2 F = ( y – z 3 yz – 2 x ) i (3xz 2 xy ) j (3xy – 2 xz 2 z ) k For solenoidal, we have to prove . F = 0. 2 k ( y – z 2 3 yz – 2 x) i (3xz 2 xy ) j (3xy – 2 xz 2 z ) k Now, . F = i j y z x = – 2 + 2x – 2x + 2 = 0 Thus, F is solenoidal. For irrotational, we have to prove Curl F = 0. j k x y z i Now, Curl F = y 2 – z 2 3 yz – 2 x 3xz 2 xy 3xy – 2 xz 2 z = (3z 2 y – 2 y 3z ) i – (– 2 z 3 y – 3 y 2 z ) j (3 z 2 y – 2 y – 3z ) k = 0 i 0 j 0k = 0 Thus, F is irrotational. Hence, F is both solenoidal and irrotational. Proved. Example 44. Determine the constants a and b such that the curl of vector A = (2 xy 3 yz ) i ( x 2 axz – 4 z 2 ) j – (3 xy byz ) k is zero. o. (U.P. I Semester, Dec 2008) k [(2 xy 3 yz ) i ( x 2 axz – 4 z 2 ) j Solution. Curl A = i j y z x = (3xy byz ) k ] i x j y k z 2 xy 3 yz x 2 axz – 4 z 2 – 3 xy – byz Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ 406 Vectors = [– 3 x – bz – ax 8 z ] i – [– 3 y – 3 y ] j [2 x az – 2 x – 3 z ] k = [– x(3 a ) z ( 8 – b )] i 6 y j z (– 3 a ) k =0 i.e., 3 + a = 0 and 8 – b = 0, –3+a=0 a = – 3, 3 b =8 a=3 Example 45. If a vector field is given by (given) Ans. F ( x 2 – y 2 x) i – (2 xy y ) j . Is this field irrotational ? If so, find its scalar potential. (U.P. I Semester, Dec 2009) Solution. Here, we have F = ( x 2 – y 2 x ) i – (2 xy y ) j Curl F = F j k ( x 2 – y 2 x ) i – (2 xy y ) j = i y z x j k x y = i (0 – 0) – j (0 – 0) k (– 2 y 2 y ) = 0 z i = x 2 – y 2 x – 2 xy – y 0 Hence, vector field F is irrotational. To find the scalar potential function F = d = j k ( i dx j dy k dz ) dx dy dz = i x y z x y z = i x j y k z ( d r ) = d r = F d r = [( x 2 – y 2 x )i – (2 xy y ) j ] ( i dx j dy k dz ) = (x2 – y2 + x)dx – (2xy + y)dy. = [( x 2 – y 2 x ) dx – (2 xy y ) dy ] c 3 2 2 x 2 d x x dx y dy y 2 dx 2 xy dy c = x x – y – xy 2 c 3 2 2 3 2 2 x x y Hence, the scalar potential is Ans. – – xy 2 c 3 2 2 = Example 46. Find the scalar potential function f for A y 2 i 2 xy j z 2 k . (Gujarat, I Semester, Jan. 2009) Solution. We have, 2 2 A = y i 2 xy j z k j k ( y 2 i 2 xy j z 2 k ) Curl A = A = i y z x Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ Vectors 407 i = x j y y2 k z = i (0) j (0) k (2 y 2 y ) = 0 2 xy z 2 Hence, A is irrotational. To find the scalar potential function f. A =f f f f f f f dx dy dz j k .( i dx j dy k dz ) df = = i x y z y z x j k f .dr = f .d r = i y z x = A .dr (A = f) 2 2 = ( y i 2 xy j z k ).( i dx j dy k dz ) 2 = y dx + 2xy dy – z2 dz = d (xy2) – z2 dz f= = xy 2 2 2 d ( xy ) z dz z3 C 3 Ans. Example 47. A vector field is given by A = (x2 + xy2) i + (y2 + x2y) j . Show that the field is irrotational and find the scalar potential.(Nagpur Univeristy, Summer 2003, Winter 2002) Solution. A is irrotational if curl A = 0 i j x y x 2 xy 2 y2 x2 y 0 k z Curl A = A = i (0 0) j (0 0) k (2 xy 2 xy ) 0 Hence, A is irrotational. If is the scalar potential, then A = grad dx dy dz [Total differential coefficient] x y z j k .( i dx j dy k dz ) = grad . dr = i y z x d = 2 2 2 2 = A . dr = [( x xy ) i ( y x y ) j ].( i dx j dy k dz ) 2 2 2 2 = (x + xy ) dx + (y + x y) dy = x2 dx + y2 dy + (x dx)y2 + (x2) (y dy) = 2 2 2 2 x dx y dy [( x dx) y ( x ) ( y dy)] = x3 y 3 x 2 y 2 c Ans. 3 3 2 Example 48. Show that V ( x, y, z ) 2 x y z i ( x 2 z 2 y ) j x 2 y k is irrotational and find a scalar function u(x, y, z) such that V = grad (u). Solution. 2 2 V (x, y, z) = 2 x y z i ( x z 2 y ) j x y k Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ 408 Vectors Curl V j k [2 x y z i ( x 2 z 2 y ) j x 2 y k ] = i y z x i x j y k z 2x y z x2 z 2 y x2 y = 2 2 = ( x x ) i (2 xy 2 xy ) j (2 xz 2 xz ) k 0 Hence, V (x, y, z) is irrotational. To find corresponding scalar function u, consider the following relations given V = grad (u) V = (u ) u u u du = dx dy dz x y z or ...(1) (Total differential coefficient) u u u j k = i .( i dx j dy k dz ) y z x = u .d r V . d r = = = = u = Integrating, we get [From (1)] [2 x y z i ( x 2 z 2 y ) j x 2 y k ].( i dx j dy k dz ) 2 x y z dx + (x2z + 2y) dy + x2y dz y(2x z dx + x2 dz) + (x2z) dy + 2y dy [yd (x2z) + (x2z) dy] + 2y dy = d(x2yz) + 2y dy x2yz + y2 Ans. Example 49. A fluid motion is given by v ( y z ) i ( z x ) j ( x y ) k . Show that the motion is irrotational and hence find the velocity potential. (Uttarakhand, I Semester 2006; U.P., I Semester, Winter 2003) Solution. Curl v = v j k [( y z ) i ( z x) j ( x y ) k ] = i y z x = i x yz j y zx k = (1 1) i (1 1) j (1 1) k 0 z x y Hence, v is irrotational. To find the corresponding velocity potential , consider the following relation. v = d = dx dy dz [Total Differential coefficient] x y z Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ Vectors 409 = i j k .( i dx j dy k dz ) = i j k . d r . d r = v . d r y z y z x x = [( y z ) i ( z x ) j ( x y ) k ].( i dx j dy k dz ) = (y + z) dx + (z + x) dy + (x + y) dz = y dx + z dx + z dy + x dy + x dz + y dz ( y dx x dy) ( z dy y dz ) ( z dx x dz ) = = xy + yz + zx + c Velocity potential = xy + yz + zx + c Example 50. A fluid motion is given by Ans. 2 v = (y sin z – sin x) i + (x sin z + 2yz) j + (xy cos z + y ) k is the motion irrotational? If so, find the velocity potential. Curl v = v j k (y sin z sin x) i + (x sin z + 2yz ) j + (xy cos z + y 2 ) k = i y z x Solution. j k x y z i = xy cos z y 2 y sin z sin x x sin z 2 yz = (x cos z + 2y – x cos z – 2y) i – [y cos z – y cos z] j + (sin z – sin z) k = 0 Hence, the motion is irrotational. So, v = where is called velocity potential. dx dy dz [Total differential coefficient] x y z j k .( i dx j dy k dz ) = .d r = v . d r = i y z x = [(y sin z – sin x) i + (x sin z + 2yz) j + (xy cos z + y2) k ]. [ i dx j dy k dz ] = (y sin z – sin x) dx + (x sin z + 2 y z) dy + (x y cos z + y2) dz = (y sin z dx + x dy sin z + x y cos z dz) – sin x dx + (2 y z dy + y2 dz) = d (x y sin z) + d (cos x) + d (y2 z) d = = d ( xy sin z ) d (cos x) d ( y 2 z ) = xy sin z + cos x + y2z + c Hence, Velocity potential = xy sin z + cos x + y2z + c. Ans. Example 51. Prove that F r 2 r is conservative and find the scalar potential such that F = . j k x y z i (Nagpur University, Summer 2004) 2 2 2 2 F = r r = r (x i y j z k) = r x i r y j r z k Solution. Given Consider F = 2 r 2 x r 2 y r2 z Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ 410 Vectors 2 2 2 2 2 2 = i r z r y j r z r x k r y r x z x z y y x r r r r r r 2ry j 2rz 2rx k 2ry 2rx = i 2rz y z x z x y x r y r z 2 2 2 2 r But r x y z , x r , y r , z r y z x z x y = i 2 rz 2ry j 2 rz 2rx k 2 ry 2 rx r r r r r r = i (2 yz 2 yz ) j (2 zx 2 zx) k (2 xy 2 xy ) = 0 i 0 j 0 k 0 F = 0 F is irrotational F is conservative. Consider scalar potential such that F = . d = dx dy dz x y z [Total differential coefficient] j k .( i dx j dy k dz ) = i y z x j k .( i dx j dy k dz ) = .( i dx j dy k dz ) = i y z x 2 = r r .( i dx j dy k dz ) = F .( i dx j dy k dz ) ( = F ) = ( x 2 y 2 z 2 ) ( i x j y k z ).( i dx j dy k dz ) = (x2 + y2 + z2) (x dx + y dy + z dz) = x3 dx + y3 dy + z3 dz + (x dx) y2 + (x2) (y dy) + (x dx)z2 + z2 (y dy) + x2 (z dz) + y2 (z dz) = x3 dx y 3 dy z 3 dz [( x dx ) y 2 ( y dy ) x 2 ] [( x dx) z 2 ( z dz ) x 2 ] [( y dy ) z 2 ( z dz ) y 2 ] x4 y 4 z 4 1 2 2 1 2 2 1 2 2 x y x z y z c 4 4 4 2 2 2 1 4 = (x + y4 + z4 + 2x2y2 + 2x2z2 + 2y2 z2) + c 4 = Ans. r Example 52. Show that the vector field F 3 is irrotational as well as solenoidal. Find |r| the scalar potential. (Nagpur University, Summer 2008, 2001, U.P. I Semester Dec. 2005, 2001) Solution. F = r | r |3 xi y j zk ( x 2 y 2 z 2 )3/2 x i y j zk j k Curl F = F = i y z ( x 2 y 2 z 2 )3/ 2 x Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ Vectors 411 i x x j y y k z z ( x 2 y 2 z 2 )3/ 2 ( x 2 y 2 z 2 )3/ 2 ( x 2 y 2 z 2 )3/ 2 = 3 2 yz 3 2 yz = i 2 2 2 5/2 2 2 2 5/2 2 2 (x y z ) (x y z ) 3 2 xz 2 xz 3 j 2 2 2 2 5/2 2 2 5/2 2 2 (x y z ) (x y z ) 3 2 xy 2 xy 3 k 2 2 2 2 5/2 2 2 5/2 2 (x y z ) 2 (x y z ) = 0 Hence, F is irrotational. F = , where is called scalar potential dx dy dz d = [Total differential coefficient] x y z j k .( i dx j dy k dz ) = .d r F .d r = i y z x = xi y j zk 2 2 2 .( i dx j dy k dz ) 3/ 2 (x y z ) 1 2 x dx 2 y dy 2 z dz = 2 ( x 2 y 2 z 2 )3/ 2 1 1 2 2 2 2 = (x y z ) 2 2 1 Now, = x dx y dy z dz ( x 2 y 2 z 2 )3/ 2 1 (x2 y2 1 z2 ) 2 1 |r | Ans. Div F = . F x i y j z k = i j k . 2 y z ( x y 2 z 2 )3/ 2 x x y z = x ( x 2 y 2 z 2 )3/ 2 y ( x 2 y 2 z 2 )3/ 2 z ( x 2 y 2 z 2 )3/ 2 3 ( x 2 y 2 z 2 )3/ 2 (1) x ( x 2 y 2 z 2 )1/ 2 (2 x ) 2 = ( x 2 y 2 z 2 )3 3 ( x 2 y 2 z 2 )3/ 2 (1) y ( x 2 y 2 z 2 )1/ 2 (2 y ) 2 2 ( x y 2 z 2 )3 3 ( x 2 y 2 z 2 )3/ 2 (1) z ( x 2 y 2 z 2 )1/ 2 (2 z ) 2 2 ( x y 2 z 2 )3 Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ 412 Vectors = ( x 2 y 2 z 2 )1/ 2 [x2 + y2 + z2 – 3x2 + x2 + y2 + z2 – 3y2 + x2 + y2 + z2 – 3z2] ( x 2 y 2 z 2 )3 = 0 Hence, F is solenoidal. Proved. Example 53. Given the vector field V ( x 2 y 2 2 xz ) i ( xz xy yz ) j ( z 2 x 2 ) k find curl V. Show that the vectors given by curl V at P0 (1, 2, –3) and P1 (2, 3, 12) are orthogonal. Curl V = V j k [( x 2 y 2 2 xz ) i ( xz xy yz ) j ( z 2 x 2 ) k ] = i y z x Solution. j k x y z xz xy yz z 2 x2 i curl V = x 2 y 2 2 xz = ( x y ) i (2 x 2 x) j ( z y 2 y ) k = ( x y ) i ( y z ) k curl V at P0 (1, 2, –3) = (1 2) i (2 3) k 3 i k curl V at P1 (2, 3, 12) = (2 3) i (3 12) k 5 i 15 k The curl V at (1, 2, –3) and (2, 3, 12) are perpendicular since ( 3 i k ).( 5 i 15 k ) = +15 – 15 = 0 Example 54. Find the constants a, b, c, so that Proved. F = ( x 2 y az ) i (bx 3 y z ) j (4 x cy 2 z ) k ...(1) is irrotational and hence find function such that F = . (Nagpur University, Summer 2005, Winter 2000; R.G.P.V., Bhopal 2009) Solution. We have, i j k F = x y z ( x 2 y az ) (bx 3 y z ) (4 x cy 2 z ) = (c 1) i (4 a ) j (b 2) k As F is irrotational, F 0 i.e., (c 1) i (4 a ) j (b 2) k 0 i 0 j 0 k c + 1 = 0, 4–a=0 and i.e., a = 4, b = 2, c = –1 Putting the values of a, b, c in (1), we get b–2=0 F = ( x 2 y 4 z ) i (2 x 3 y z ) j (4 x y 2 z ) k Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ Vectors 413 Now we have to find such that F = We know that dx dy dz d = [Total differential coefficient] x y z j k .( i dx j dy k dz ) = i y z x j k .( i dx j dy k dz ) = .( i dx j dy k dz ) = i y z x = F .( i dx j dy k dz ) = [( x 2 y 4 z ) i (2 x 3 y z ) j (4 x y 2 z ) k )].( i dx j dy k dz ) = (x + 2y + 4z) dx + (2x – 3y – z) dy + (4x – y + 2z) dz = x dx – 3y dy + 2z dz + (2y dx + 2x dy) + (4z dx + 4x dz) + (–z dy – y dz) = x dx 3 y dy 2 z dz (2 y dx 2 x dy ) (4 z dx 4 x dz ) ( z dy y dz ) = x2 3 y 2 + z2 + 2xy + 4zx – yz + c 2 2 Ans. Example 55. Let V (x, y, z) be a differentiable vector function and (x, y, z) be a scalar function. Derive an expression for div ( V ) in terms of . V , div V and . (U.P. I Semester, Winter 2003) Solution. Let V = V1 i V2 j V3 k div ( V ) = .( F ) j k .[V1 i V2 j V3 k ] = (V1 ) (V2 ) ( V3 ) = i y z x y z x V1 V2 V3 V1 V2 V3 = y z z x x y V1 V2 V3 = x V1 y V2 z V3 x y z = i x j y k z .(V1 i V2 j V3 k ) i x j y k z .(V1 i V2 j V3 k ) = (.V ) ( ).V (div V ) (grad ). V Ans. Example 56. If A is a constant vector and R = x iˆ + y ĵ + z k̂ , then prove that Curl A . R A A R (K. University, Dec. 2009) Solution. Let A = A1 iˆ + A2 ĵ + A3 k̂ , R = x iˆ + y ĵ + z k̂ A . R (A1 iˆ + A2 ĵ + A3 k̂ ) . (x iˆ + y ĵ + z k̂ ) = A1 x + A2 y + A3 z [ A . R ] R = (A1 x + A2y + A3 z) (x iˆ + y ĵ + z k̂ ) = (A1 x2 + A2 xy + A3 zx) iˆ + (A1 xy + A2 y2 + A3 yz) ĵ + (A1 xz + A2 yz + A3z2) k̂ Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ 414 Vectors iˆ Curl ( A . R ) R = x A1 x 2 A2 xy A3 zx ˆj y kˆ z A2 xy A2 y 2 A3 yz A1 xz A2 yz A3 z 2 = (A2 z – A3 y) iˆ – [A1 z – A3 x) ĵ [A1 y – A2 x] k̂ ... (1) L.H.S. = A R = (A1 iˆ + A2 ˆj iˆ = A1 A 2 x y ĵ + A3 k) ×(x iˆ + y ĵ + z k̂ ) k A3 z = (A2 z – A3 y) iˆ – (A1 z – A3 x) ĵ + (A1 y – A2 x) k̂ = R.H.S. [From (1)] Example 57. Suppose that U , V and f are continuously differentiable fields then Prove that, div (U V ) V .curl U U .curl V . (M.U. 2003, 2005) U = u1 i u2 j u3 k , V v1 i v2 j v3 k Solution. Let i U V = j k u1 u2 v1 v2 u3 v3 = (u2v3 u3v2 ) i (u1v3 u3v1 ) j + (u1v2 u2v1 ) k j k .[(u2 v3 u3v2 ) i (u1v3 u3v1 ) j + (u1v2 u2 v1 ) k ] div (U V ) = i y z x = x (u2 v3 u3 v2 ) y (u1v3 u3 v1 ) z (u1v2 u2 v1 ) u v u u v u v v3 v3 2 u3 2 v2 3 u1 3 v3 1 u3 1 v1 3 = u2 x x x y y y y x u1 v1 u v2 u1 v2 u2 v1 2 z z z z u u u u u u 3 2 v2 3 1 v3 2 1 = v1 y z x z x y v3 v2 v3 v1 v1 v2 u1 u 2 x z u3 y x y z u u u u u u = (v1 i v2 j v3 k ) . i 3 2 j 1 3 k 2 1 y z z x x y v v v v v v (u1 i u2 j u3 k ). i 2 3 j 3 1 k 1 2 y x z x y z = V .( U ) U .( V ) V .curl U U .curl V Proved. Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ Vectors 415 Example 58. Prove that Solution. ( F G ) = F ( . G ) G ( . F ) (G . ) F ( F . ) G (M.U. 2004, 2005) ( F G) = i ( F G ) x F G F G = i G F G i F i x x x x F F G G = ( i . G ) i G i F (i . F ) x x x x F F G G = (G . i ) G i . F i . ( F . i ) x x x x G F F G ( F . i ) = F i G i. (G . i ) x x x x = F ( G ) G ( . F ) (G . ) F ( F . ) G Questions for practice: Prove that Proved. ( F . G ) = (G . ) F ( F . ) G G ( F ) F ( G ) Example 59. Prove that, for every field V ; div curl V = 0. (Nagpur University, Summer 2004; AMIETE, Sem II, June 2010) Solution. Let V = V1 i V2 j V3 k div (curl V ) = .( V ) i = . x V1 j k y V2 z V3 V3 V2 V3 V1 V2 V1 j k . i = i j k y z y z z y x x x V3 V2 V3 V1 V2 V1 = x y z y x z z x y = 2 V3 2 V2 2 V3 2 V1 2 V2 2 V1 x y x z y x y z z x z y 2 V1 2 V1 2 V2 2 V2 2 V3 2 V3 = y z z y z x x z x y y x =0 Ans. Example 60. If a is a constant vector, show that a ( r ) = ( a . r ) ( a . ) r . Solution. (U.P., Ist Semester, Dec. 2007) a = a1 i a2 j a3 k , r r1 i r2 j r3 k Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ 416 Vectors j k x y z r2 r3 i r = r1 j k a1 a2 a3 i a ( r ) = r3 r2 r3 r1 r2 r1 i = j x y k z x z y r3 r2 y z r3 r1 x z r2 r1 x y r r r r r r r r = a2 2 a2 1 a3 3 a3 1 i a1 2 a1 1 a3 3 a3 2 j y x z x y y z x r r r r a1 3 a1 1 a2 3 a2 2 k x z y z r1 r2 r3 r1 r2 r3 = a1 i x a2 i x a3 i x a1 j y a2 j y a3 j y r r r r r r a1 k 1 a2 k 2 a3 k 3 a1 i 1 a1 j 2 a1 k 3 z z z x x x r r r r r r a2 i 1 a2 j 2 a2 k 3 a3 i 1 a3 j 2 a3 k 3 y y y z z z = i j k (a1r1 a2 r2 a3 r3 ) a1 a2 a3 ( r1 i r2 j r3 k ) y z y z x x = ( a . r ) ( a .) r Proved. Example 61. If r is the distance of a point (x, y, z) from the origin, prove that 1 1 Curl k grad grad k .grad = 0, where k is the unit vector in the direction OZ. r r (U.P., I Semester, Winter 2000) 2 2 2 Solution. r = (x – 0) + (y – 0) + (z – 0)2 = x2 + y2 + z2 1 = (x2 + y2 + z2)– 1/2 r 1 1 2 j k ( x y 2 z 2 )1/ 2 grad = = i r r x y z 1 = ( x 2 y 2 z 2 )3/ 2 (2 x i 2 y j 2 z k ) 2 = – ( x 2 y 2 z 2 )3/ 2 ( x i y j z k ) k × grad 1 = k [( x 2 y 2 z 2 ) 3/ 2 ( x i y j z k )] r = ( x 2 y 2 z 2 )3/ 2 ( x j y i ) 1 1 curl k grad = k grad r r Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ Vectors 417 j k × [–(x2 + y2 + z2)–3/2 ( x j y i ) ] = i y z x i x y j y x ( x 2 y 2 z 2 )3/ 2 ( x 2 y 2 z 2 )3/ 2 = k z 0 3 ( x) (2 z ) 3 y (2 z ) ( x)(2 x ) 3 = 2 i j 2 2 5/2 2 2 2 5/2 2 2 2 5/ 2 2 (x y z ) 2 (x y z ) 2 (x y z ) 1 (3 / 2) ( y ) (2 y ) 1 2 2 2 k 2 2 3/ 2 2 2 5/ 2 2 2 3/ 2 (x y z ) (x y z ) (x y z ) = 3xz 3 yz (3x 2 x 2 y 2 z 2 3 y 2 x 2 y 2 z 2 ) i j k ( x 2 y 2 z 2 )5 / 2 ( x 2 y 2 z 2 )5 / 2 ( x 2 y 2 z 2 )5 / 2 = 3xz i 3 yz j ( x 2 y 2 2 z 2 ) k ( x 2 y 2 z 2 )5 / 2 1 z k . grad = k .[ ( x 2 y 2 z 2 )3/ 2 ( x i y j z k )] 2 r ( x y 2 z 2 )3/ 2 z 1 j k 2 grad k .grad = i y z ( x y 2 z 2 )3/ 2 r x ...(1) 3 i ( z )(2 x) 3 j ( z )(2 y ) = 2 2 2 5 / 2 2 2 (x y z ) 2 ( x y 2 z 2 )5 / 2 ( z )(2 z ) 1 3 2 k 2 2 2 5/2 2 2 3/ 2 (x y z ) 2 (x y z ) 2 2 2 2 2 2 2 = 3xz i 3 yz j (3z x y z ) k 3xz i 3 yz j ( x y 2 z ) k ...(2) 2 2 2 5/ 2 2 2 2 5/2 (x y z ) (x y z ) Adding (1) and (2), we get 1 1 Curl k grad grad k .grad = 0 Proved. r r a r (2 n ) a n ( a . r ) r Example 62. Prove that . r n rn rn 2 (M.U. 2009, 2005, 2003, 2002; AMIETE, II Sem. June 2010) Solution. We have, ar rn = = 1 rn 1 r n j k a1 x a2 y a3 z i (a2 z a3 y ) i 1 r n (a3 x a1 z ) j 1 r n (a1 y a2 x) k Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ 418 Vectors i (a r) = x n r a2 z a3 y j y a3 x a1 z k z a1 y a2 x rn rn rn a y a x a3 x a1 z a1 y a2 x a2 z a3 y = i 1 n 2 j x r rn rn rn z z y a xa z a z a y k 3 n 1 2 n 3 r r y x r r x Now, r2 = x2 + y2 + z2 2r = 2x x x r y r r z , Similarly, = r y z r 1 ar n 1 y . (a1 y a2 x) n a1 n = i nr r r r 1 z nr n 1 (a3 x a1 z ) n (a1 ) + two similar terms r r a a n n 2 2 = i n 2 (a1 y a2 xy ) n1 n 2 (a3 xz a1 z ) n1 r r r r + two similar terms 2a n n 2 2 1 = i n n 2 a1 ( y z ) n 2 (a2 xy a3 xz ) + two similar terms r r r n 2 Adding and subtracting n 2 a1 x to third and from second term, we get r a r 2a na1 n n = i 1 ( x 2 y 2 z 2 ) n 2 (a1 x 2 a2 xy a3 xz ) n n2 r r r r + two similar terms 2a na1 2 n 1 = i n n 2 r n 2 x (a1 x a2 y a3 z ) + two similar terms r r r 2a 2a na n na n j n2 n2 n 2 y (a2 y a3 z a1 x) = i n1 n1 n 2 x(a1 x a2 y a3 z ) r r r r r r 2a na n k n3 n3 n 2 z ( a3 z a1 x a2 y ) r r r n 2 n = n (a1 i a2 j a3 k ) n (a1 i a2 j a3 k ) n 2 (a1 x a2 y a3 z ) ( x i y j z k ) r r r 2n n (a1 i a2 j a3 k ) n 2 (a1 x a2 y a3 z ) ( x i y j z k ) = n r r 2n n a n 2 (a . r ) r = Proved. rn r Example 63. If f and g are two scalar point functions, prove that div (f g) = f 2g + f g. (U.P., I Semester, compartment, Winter 2001) Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ Vectors 419 g g g i j k x y z g g g f g = f i f j f k x y z g g g (f g) = f f f x x y y z z 2 g 2 g 2 g f g f g f g f 2 2 2 y z x x y y z z x 2 f f f g g g 2 2 f 2 2 2 g i j k . i j k y z x y z y z x 2x f g + f.g Proved. Solution. We have, div = = = g = Example 64. For a solenoidol vector F , show that curl curl curl curl F = 4 F . (M.D.U., Dec. 2009) Solution. Since vector F is solenoidal, so div F = 0 ... (1) We know that curl curl F = grad div ( F – 2 F ) ... (2) Using (1) in (2), grad div F = grad (0) = 0 On putting the value of grad div F in (2), we get ... (3) curl curl F = – 2 F ... (4) Now, curl curl curl curl F = curl curl (– 2 F ) [Using (4)] = – curl curl ( 2 F ) = – [grad div ( 2 F ) – 2 ( 2 F ) ] [Using (2)] [ . F = 0] = – grad ( . 2 F ) + 2 ( 2 F ) = – grad ( 2 . F ) + 4 F = 0 + 4 F = 4 F [Using (1)] EXERCISE 5.9 Proved. 1. Find the divergence and curl of the vector field V = (x2 – y2) i + 2xy j + (y2 – xy) k . Ans. Divergence = 4x, Curl = (2y – x) i + y j + 4y k 2. If a is constant vector and r is the radius vector, prove that (ii) div ( r a ) 0 (i) ( a . r ) a (iii) curl ( r a ) 2 a where r = x i y j z k and a a1 i a2 j a3 k . 3. Prove that: (i) .(A) = .A + (.A) (ii) (A.B) = (A.)B + (B.)A + A × ( × B) + B × ( × A) (iii) × (A × B) = (B.)A – B(.A) – (A.)B + A(.B) (R.G.P.V. Bhopal, June 2004) 4. If F = (x + y + 1) i + j – (x + y) k , show that F.curl F = 0. (R.G.P.V. Bhopal, Feb. 2006, June 2004) Prove that 5. ( F ) ( ) F ( F ) 7. Evaluate div ( A r ) if curl A = 0. 6. .( F G ) G .( F ) F .( G ) 8. Prove that curl ( a r ) = 2a Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ 420 Vectors 9. Find div F and curl F where F = grad (x3 + y3 + z3 – 3xyz). (R.G.P.V. Bhopal Dec. 2003) Ans. div F = 6(x + y + z), curl F = 0 = (x + y + az) i + (bx + 3y – z) j + (3x + cy + z) k 10. Find out values of a, b, c for which v is irrotational. Ans. a = 3, b = 1, c = –1 11. Determine the constants a, b, c, so that F = (x + 2y + az) i + (bx – 3y – z) j + (4x + cy + 2z) k is irrotational. Hence find the scalar potential such that F = grad . (R.G.P.V. Bhopal, Feb. 2005) Ans. a = 4, b = 2, c = 1 x2 3 y 2 z 2 2 xy yz 4 zx Potential = 2 2 Choose the correct alternative: 12. The magnitude of the vector drawn in a direction perpendicular to the surface x2 + 2y2 + z2 = 7 at the point (1, –1, 2) is 2 3 (i) (ii) (iii) 3 (iv) 6 (A.M.I.E.T.E., Summer 2000) Ans. (iv) 3 2 13.If u = x2 – y2 + z2 and V xi y j zk then (uV ) is equal to (i) 5u (ii) 5 | V | (iii) 5(u | V |) (iv) 5(u | V |) 14.A unit normal to x2 + y2 + z2 = 5 at (0, 1, 2) is equal to 1 1 1 ( i j k ) (ii) ( j 2 k ) (iv) (i) ( i j k ) (iii) 5 5 5 (A.M.I.E.T.E., June 2007) 1 ( i j k) 5 (A.M.I.E.T.E., Dec. 2008) 15.The directional derivative of = x y z at the point (1, 1, 1) in the direction i is: 1 1 (i) –1 (ii) (iii) 1 (iv) Ans. (iii) 3 3 (R.G.P.V. Bhopal, II Sem., June 2007) 16.If r x i y j z k and r = | r | then (r) is: (i) (r) r ( r ) r r (ii) (iii) ( r ) r r (iv) None of these (R.G.P.V. Bhopal, II Semester, Feb. 2006) 17. If r = x i y j z k is position vector, then value of (log r) is r (i) r r 2 (iii) – r (iv) none of the above. r3 (iv) –2 Ans. (ii) (R.G.P.V. Bhopal, II Semester, Feb. 2006) 19. If V xy 2 i 2 yx 2 z j 3 yz 2 k then curl V at point (1, –1, 1) is (i) ( j 2 k ) 20. If A is such that A = 0 then A is called (i) Irrotational (ii) Solenoidal (iii) Rotational (iii) ( i 2 k ) (iv) ( i 2 j k ) (R.G.P.V. Bhopal, II Semester, Feb 2006) Ans. (iii) (ii) ( i 3 k ) Ans. (ii) 18. If r x i y j z k and | r | = r, then div r is: (i) 2 (ii) 3 (iii) –3 (U.P., I Sem, Dec 2008) r (ii) Ans. (iii) (iv) None of these (A.M.I.E.T.E., Dec. 2008) 21. If F is a conservative force field, then the value of curl F is (i) 0 (ii) 1 (iii) F (iv) –1 (A.M.I.E.T.E., June 2007) Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ Vectors 421 22.If 2 [(1 – x) (1 – 2x)] is equal to (i) 2 (ii) 3 (iii) 4 (iv) 6 (A.M.I.E.T.E., Dec. 2009) Ans. (iii) 23.If R = xi + yj + zk and A is a constant vector, curl ( A R ) is equal to (ii) 2 R (i) R (iii) A (iv) 2 A (A.M.I.E.T.E., Dec. 2009) Ans. (iv) 1 24. If r is the distance of a point (x, y, z) from the origin, the value of the expression ˆj grad 2 equals (i) ( x 2 y 2 z 2 ) 3 2 ( ˆj z kˆ x ) (ii) ( x 2 y 2 z 2 ) (iii) zero (iv) (x2 y2 3 2 3 z2 ) 2 ( ˆj z iˆ z ) ( ˆj y kˆ x ) (AMIETE, Dec. 2010) Ans. (ii) 5.33 LINE INTEGRAL Let F ( x, y , z) be a vector function and a curve AB. Line integral of a vector function F along the curve AB is defined as integral of the component of F along the tangent to the curve AB. Component of F along a tangent PT at P = Dot product of F and unit vector along PT = F dr dr is a unit vector along tangent PT ds ds dr Line integral = F from A to B along the curve ds F dr ds Line integral = c = c F d r ds Note (1) Work. If F represents the variable force acting on a particle along arc AB, then the total work done = B A F dr (2) Circulation. If V represents the velocity of a liquid then c V dr is called the circulation of V round the closed curve c. If the circulation of V round every closed curve is zero then V is said to be irrotational there. (3) When the path of integration is a closed curve then notation of integration is in place of . Example 65. If a force F 2 x 2 yiˆ 3 xyjˆ displaces a particle in the xy-plane from (0, 0) to (1, 4) along a curve y = 4 x2. Find the work done. Solution. Work done = = = c F . dr c (2 x y iˆ 3 xy ˆj) . (dx iˆ dy ˆj) 2 c (2 x y dx 3 xy dy) 2 ˆ ˆ r xi yj dr dxiˆ dy ˆj Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ 422 Vectors y 4 x2 dy 8 x dx Putting the values of y and dy, we get 1 0 [2 x = 2 (4 x 2 ) dx 3 x (4 x 2 ) 8 x dx] 1 x5 1 4 104 = 104 0 x dx 104 5 5 0 Ans. Example 66. Evaluate F . dr where F x 2iˆ xyjˆ and C is the boundary of the square in the C plane z = 0 and bounded by the lines x = 0, y = 0, x = a and y = a. (Nagpur University, Summer 2001) Solution. C F . d r OA F . dr AB F . dr BC F . dr CO F . dr Here r xiˆ yjˆ, d r dxiˆ dyjˆ , F x 2iˆ xy ˆj Y F . dr = x2dx + xydy On OA, y = 0 ...(1) F . dr x dx OA F . dr On AB, x = a (1) becomes = a 2 x dx 0 a x3 a3 3 0 3 ...(2) dx = 0 O F . dr = aydy A On BC, y = a X a y2 a3 aydy a F . dr = 0 Ab 2 2 0 dy = 0 a B C 2 ...(3) (1) becomes F . dr x 2 dx BC F . dr = 0 2 x dx a 0 x3 – a3 3 3 a ...(4) On CO, x = 0, (1) becomes F . dr 0 CO F . dr =0 ...(5) On adding (2), (3), (4) and (5), we get C F . dr Example 67. A vector field is given by F = (2 y 3) iˆ xzjˆ ( yz – x) kˆ. Evaluate y = t, z = t3 from t = 0 to t = 1. Solution. C F . dr = = a3 a3 a3 a3 – 0 3 2 3 2 Ans. C F . dr along the path c is x = 2t, (Nagpur University, Winter 2003) C (2 y 3) dx ( xz) dy ( yz – x) dz Since x 2t dx 2 dt yt z t3 dy dz 1 3t 2 dt dt Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ Vectors 423 = 1 0 (2t 3) (2 dt ) (2t ) (t 3 ) dt (t 4 – 2t ) (3t 2 dt ) = 1 0 (4t 6 2t 4 3t 6 – 6t 3 ) dt 1 1 t2 2 5 3 7 6 4 2 5 3 7 3 4 2 = 4 2 6t 5 t 7 t – 4 t = 2t 6t t t – t 5 7 2 0 0 2 3 3 – = 7.32857. 5 7 2 Example 68. The acceleration of a particle at time t is given by = 26 Ans. a = 18 cos 3t iˆ 8 sin 2t ˆj 6t kˆ. If the velocity v and displacement r be zero at t = 0, find v and r at any point t. d2 r Solution. Here, a = = 18 cos 3t iˆ 8 sin 2t ˆj 6t kˆ. dt 2 On integrating, we have = dr iˆ 18 cos 3t dt ˆj 8 sin 2t dt kˆ 6t dt v dt v = 6 sin 3t iˆ 4 cos 2t ˆj 3t 2 kˆ c At ...(1) t = 0, v = 0 Putting t = 0 and v = 0 in (1), we get 0 = 4 ˆj c c 4 ˆj dr 6 sin 3t iˆ 4(cos 2t 1) ˆj 3t 2 kˆ v = dt Again integrating, we have = iˆ 6 sin 3t dt ˆj 4 (cos 2t 1) dt kˆ 3t 2 dt r r = 2 cos 3t iˆ (2 sin 2t 4t ) ˆj t 3 kˆ c1 ...(2) t = 0, r = 0 At, Putting t = 0 and r = 0 in (2), we get 0 = 2iˆ C1 C1 2iˆ 3 r = 2 (1 cos 3t ) iˆ 2 (sin 2t 2t ) ˆj t kˆ Hence, Ans. Example 69. If A (3x 2 6 y ) iˆ – 14 yzjˆ 20 xz 2 kˆ, evaluate the line integral om A. dr from (0, 0, 0) to (1, 1, 1) along the curve C. x = t, y = t2, z = t3. (Uttarakhand, I Semester, Dec. 2006) Solution. We have, C A . dr = C [ (3x 2 6 y ) iˆ – 14 yzjˆ 20 xz 2 kˆ ] . [iˆ dx ˆj dy kˆ dz ] = C [ (3x 2 6 y ) dx – 14 yzdy 20 xz 2 dz ] If x = t, y = t2, z = t3, then points (0, 0, 0) and (1, 1, 1) correspond to t = 0 and t = 1 respectively. Now, C A . dr = = t 1 t 0 [ (3 t t 1 t 0 [ 9t 2 2 6 t 2 ) d (t ) – 14 t 2 t 3 d (t 2 ) 20 t (t 3 ) 2 d (t 3 )] dt – 14 t 5 . 2 t d t 20 t 7 . 3 t 2 dt ] = 1 0 (9 t 2 – 28 t 6 60 t 9 ) dt Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ 424 Vectors t3 t7 = 9 – 28 3 7 Example 70. Evaluate 1 t10 60 10 0 =3–4+6=5 S A . nˆ ds where A ( x y 2 Ans. ) iˆ – 2 xjˆ 2 yzkˆ and S is the surface of the plane 2x + y + 2z = 6 in the first octant. (Nagpur University, Summer 2000) Solution. A vector normal to the surface “S” is given by ˆ ˆj kˆ (2 x y 2 z ) = iˆ (2 x y 2 z ) 2iˆ ˆj 2k x y z And n̂ = a unit vector normal to surface S Z N 2 iˆ ˆj 2 kˆ 2 ˆ 1 ˆ 2 ˆ i j k = 3 3 3 4 1 4 1 2 2 kˆ . nˆ = kˆ . iˆ ˆj kˆ 3 3 3 dx dy S A . nˆ ds = R A . nˆ kˆ . n Where R is the projection of S. 1 2 2 2 Now, A . nˆ = [ ( x y ) iˆ – 2 xjˆ 2 yzkˆ] . iˆ ˆj 3 3 3 2 2 4 2 2 4 2 = ( x y ) – x yz y yz 3 3 3 3 3 Putting the value of z in (1), we get – n K 2 3 M O 3 R Y L X kˆ ...(1) on the plane 2 x y 2 z 6, 2 4 6 2 x y 2 A . nˆ = y y z (6 2 x y ) 3 3 2 2 2 4 ...(2) A . nˆ = y ( y 6 – 2 x – y ) y (3 – x) 3 3 M dx dy ...(3) S A . nˆ ds = R A . n | kˆ . n | Hence, + 2x = 3y 6 – 2x y2 2 (3 – x ) = 0 2 0 3 = 6 Putting the value of A . nˆ from (2) in (3), we get 3 6 2 x 4 3 S A . nˆ ds = R 3 y (3 – x) . 2 dx dy 0 0 2 y (3 x) dydx 3 0 (3 – x) (6 – 2 x) 2 O dx X L 3 dx 4 (3 – x)3 dx 0 3 4 = 4. (3 – x ) – (0 – 81) 81 4 (– 1) 0 Example 71. Compute c F . dr , where F ˆ ˆjx iy x2 y 2 Ans. and c is the circle x2 + y2 = 1 traversed counter clockwise. Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ Vectors 425 r = iˆ x ˆj y kˆ z, d r iˆ dx ˆj dy kˆ dz ˆ ˆjx iy ˆ ) ˆ ˆjdy kdz c F . d r = c x 2 y 2 (idx ydx xdy ( ydx xdy ) = c 2 ...(1) [ x2 + y2 = 1] c x y2 Parametric equation of the circle are x = cos , y = sin . Putting x = cos , y = sin , dx = – sin d , dy = cos d in (1), we get Solution. C F d r = 2 0 sin ( sin d ) cos (cos d ) 2 2 2 2 2 = 0 (sin cos ) d 0 d = 0 2 Ans. 2 3 2 2 2 Example 72. Show that the vector field F 2 x( y z )iˆ 2 x yjˆ 3 x z kˆ is conservative. Find its scalar potential and the work done in moving a particle from (–1, 2, 1) to (2, 3, 4). (A.M.I.E.T.E. June 2010, 2009) Solution. Here, we have F 2 x( y 2 z 3 ) iˆ 2 x 2 y ˆj 3x 2 z 2 kˆ Curl F F iˆ x ˆj y kˆ z 2 x( y 2 z3 ) 2 x2 y 3x2 z 2 (0 0)i (6 xz 2 6 xz 2 ) ˆj (4 xy 4 xy )kˆ = 0 Hence, vector field F is irrotational. To find the scalar potential function F ˆ ˆ d dx dy dz iˆ ˆj kˆ . idx ˆjdy kdz x y z y z x iˆ ˆj kˆ . d r .d r F . d r y z x 2 3 ˆ ) ˆ ˆjdy kdz 2 x ( y z )iˆ 2 x 2 yjˆ 3x2 z 2 kˆ (idx 2 x ( y 2 z 3 ) dx 2 x 2 y dy 3 x 2 z 2 dz = 2 x( y (2 xy 2 2 z 3 )dx 2 x 2 ydy 3x 2 z 2 dz C dx 2 x 2 ydy ) (2 xz 3dx 3 x 2 z 2 dz ) + C = x2y2 + x2z3 + C Hence, the scalar potential is x2y2 + x2z3 + C Now, for conservative field Work done = (2, 3, 4) F. d r ( 1, 2,1) (2, 3, 4) d (2,3,4) (1,2,1) (2,3,4) x2 y 2 x2 z3 c ( 1,2,1) ( 1, 2,1) = (36 + 256) – (2 – 1) = 291 Ans. Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ 426 Vectors Example 73. A vector field is given by F (sin y ) iˆ x (1 cos y ) ˆj. Evaluate the line integral over a circular path x2 + y2 = a2, z = 0. Solution. We have, `(Nagpur University, Winter 2001) Work done = = C F . d r C F . dr C [ (sin y ) iˆ x (1 cos y ) ˆj ] . [dxiˆ dyjˆ] ( z = 0 hence dz = 0) = C sin y dx x (1 cos y) dy C (sin y dx x cos y dy x dy) = C d ( x sin y) C x dy (where d is differential operator). The parametric equations of given path x2 + y2 = a2 are x = a cos , y = a sin , Where varies form 0 to 2 C F . d r 2 = 0 = 0 2 d [ a cos sin ( a sin ) ] d [ a cos sin ( a sin ) ] 2 1 0 2 0 2 2 a 0 2 2 = [a cos sin (a sin ) ] 0 = 0 a2 2 0 a cos . a cos d a 2 cos 2 . d cos 2 d 2 cos 2 a sin 2 d 2 2 2 0 2 a . 2 a 2 = Ans. 2 Example 74. Determine whether the line integral 2 2 2 2 (2 x y z ) dx ( x z z cos y z ) dy (2 x yz y cos yz) dz is independent of the path of integration ? If so, then evaluate it from (1, 0, 1) to 0, , 1 . 2 2 2 2 2 (2 xy z ) dx ( x z z cos y z ) dy (2 x yz y cos yz ) dz Solution. c 2 2 2 2 ˆ ˆ ˆjdy kdz ˆ ) = c [(2 xy z iˆ) ( x z z cos y z ) ˆj (2 x yz y cos yz ) k ].(idx = c F dr This integral is independent of path of integration if F = F 0 iˆ F = x 2 x yz 2 ˆj y kˆ z x 2 z 2 z cos y z 2 x 2 y z y cos y z = (2x2z + cos yz – yz sin yz – 2x2z – cos yz + yz sin yz) = iˆ – (4 xyz – 4 x yz ) ˆj (2 xz 2 – 2 xz 2 ) kˆ =0 Hence, the line integral is independent of path. dx dy dz d = (Total differentiation) x y z Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ Vectors 427 ˆ ˆ ˆ ˆ j k (idx ˆjdy kdz ) = dr F d r = iˆ y z x 2 ˆ 2 2 ˆ ) ˆ ˆjdy kdz = [(2 xyz ) i ( x z z cos y z ) ˆj (2 x 2 yz y cos yz ) kˆ]. (idx = 2xyz2 dx + (x2z2 + z cos y z) dy + (2x2yz + y cos yz) dz = [(2x dx) yz2 + x2 (dy) z2 + x2y (2z dz)] + [(cos yz dy) z + (cos yz dz) y] = d (x2yz2) + d (sin yz) = BA d (x 2 yz 2 ) d (sin yz ) x 2 yz 2 sin yz = (B) – (A) [ x 2 yz 2 sin yz ](1, 0,1) = 0 sin ( 1) [0 0] 2 =1 Ans. Example 75. Evaluate A . nˆ d S , where A 18 ziˆ – 12 ˆj 3 y kˆ and S is the part of the S plane 2x + 3y + 6z = 12 included in the first octant. (Uttarakhand, I semester, Dec. 2006) Solution. Here, A = 18 ziˆ – 12 ˆj 3 ykˆ Given surface f (x, y, z) = 2x + 3y + 6z – 12 ˆj kˆ (2 x 3 y 6 z – 12) = 2 iˆ 3 ˆj 6 kˆ Normal vector = f = iˆ x y z n̂ = unit normal vector at any point (x, y, z) of 2x + 3y + 6z = 12 2 iˆ 3 ˆj 6 kˆ 1 ˆ (2 i 3 ˆj 6 kˆ) = 7 4 9 36 2 2 = [ x yz sin yz ] (0, ,1) 2 dx dy dx dy dxdy 7 dx dy 6 6 nˆ . kˆ 1 (2 iˆ 3 ˆj 6 kˆ) . kˆ 7 7 1 7 Now, A . nˆ dS = (18 z iˆ – 12 ˆj 3 y kˆ) . (2 iˆ 3 ˆj 6 kˆ) dx dy 7 6 dx dy = (36 z – 36 18 y ) = (6 z – 6 3 y ) dx dy 6 Putting the value of 6z = 12 – 2x – 3y, we get Y dS = 6 = 0 = 6 = 0 6 0 6 1 (12 – 2 x ) 3 (12 0 1 (12 – 2 x ) 3 (6 0 – 2 x – 3 y – 6 3 y ) dx dy 2x + – 2 x ) dx dy 1 (12 – 2 x ) 3 dy 0 1 (12 – 2 x ) ( y)3 0 3y = 12 (6 – 2 x) dx = 0 = 0 (6 – 2 x) 3 (12 – 2 x) dx 6 B (6 – 2 x ) dx 1 = O A X 1 6 (4 x 2 – 36 x 72) dx 3 0 6 1 4 x3 – 18 x 2 72 x = 1 [4 36 2 – 18 36 72 6] = 72 [4 – 9 6] 24 Ans. 3 3 3 0 3 Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ 428 Vectors EXERCISE 5.10 1. 2. 3. Find the work done by a force yiˆ xjˆ which displaces a particle from origin to a point (iˆ ˆj ). Ans. 1 Find the work done when a force F ( x2 – y 2 x) iˆ (2 xy y) ˆj moves a particle from origin to 2 (1, 1) along a parabola y2 = x. Ans. 3 Show that V (2 xy z 3 ) iˆ x 2 ˆj 3 xz 2 kˆ is a conservative field. Find its scalar potential such that V = grad . Find the work done by the force V in moving a particle from (1, – 2, 1) to (3, 1, 4). 2 3 Ans. x y + xz , 202 4. Show that the line integral c (2 xy 3) dx ( x 2 4 z ) dy 4 y dz where c is any path joining (0, 0, 0) to (1, – 1, 3) does not depend on the path c and evaluate the line integral. Ans. 14 5. 6. 7. x2 y 2 1 , z = 0, under the field of 25 16 force given by F = (2x – y + z) iˆ + (x + y – z2) ĵ + (3x – 2y + 4z) kˆ. Is the field of force conservative? (A.M.I.E.T.E., Winter 2000) Ans. 40 z4 x y 2 x2 y z3 If = (y2 – 2xyz3) iˆ + (3 + 2xy – x2z3) ĵ + (z3 – 3x2yz2) kˆ, find . Ans. 3 y 4 Find the work done in moving a particle once round the ellipse R . dR C is independent of the path joining any two point if it is. (i) irrotational field (ii) solenoidal field (iii) rotational field (A.M.I.E.T.E., June 2010) Ans. (i) (iv) vector field. 5.34 SURFACE INTEGRAL A surface r = f(u, v) is called smooth if f (u, v) posses continous first order partial derivative. Let F be a vector function and S be the given surface. Surface integral of a vector function F over the surface S is defined as the integral of the components of F along the normal to the surface. Component of F along the normal = F . n̂ , where n is the unit normal vector to an element ds and grad f dx dy ds = n̂ = | grad f | ( nˆ kˆ ) Surface integral of F over S = F nˆ Note. (1) Flux = S ( F nˆ ) ds S ( F nˆ ) ds S ( F nˆ ) d s where, F represents the velocity of a liquid. If = = 0, then F is said to be a solenoidal vector point function. Example 76. Evaluate S ( yziˆ zxjˆ xykˆ) ds where S is the surface of the spheree x2 + y2 + z2 = a2 in the first octant. Solution. Here, = x2 + y2 + z2 – a2 (U.P., I Semester, Dec. 2004) Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ Vectors 429 ˆ ˆ j k Vector normal to the surface = = iˆ x y z ˆj kˆ ( x 2 y 2 z 2 a 2 ) 2 xiˆ 2 yjˆ 2 zkˆ = iˆ x y z ˆ ˆ ˆ ˆ ˆ ˆ n̂ = 2 x i 2 y j 2 z k xi yj zk | | 4 x2 4 y 2 4 z 2 x2 y 2 z 2 xiˆ yjˆ zkˆ = [ x2 + y2 + z2 = a2] a ˆ ˆ ˆ Here, F = yz i zx j xy k ˆ ˆ ˆ ˆ xi yj zk 3xyz ˆ ˆ F nˆ = ( yz i zx j xy k ) a a 2 2 a a x 3 xyz dx dy dx dy Now, S F nˆ ds = S ( F nˆ ) ˆ 0 0 z | k . nˆ | a a a 2 x2 y2 xy dy dx 3 x dx = 3 0 0 0 2 0 a 3 a 3 a 2 x2 x4 3 a 4 a 4 3a 4 2 2 x ( a x ) dx . = 0 2 2 2 4 0 2 2 4 8 a a 2 x2 3 Example 77. Show that a S F nˆ ds 2 , where F = 4 xz iˆ – y2 ĵ + yz k̂ and S is the surface of the cube bounded by the planes, x= 0, x = 1, y = 0, y = 1, z = 0, z = 1. S.No. Surface Outward normal Solution. S F nˆ ds = OABC F nˆ ds 1 OABC –k 2 DEFG k F nˆ ds F nˆ ds DEFG OAGF 3 OAGF –j 4 BCED j F nˆ ds F nˆ ds BCED ABDG 5 ABDG i 6 OCEF – i F nˆ ds ...(1) OCEF Now, Ans. ds dx dy dx dy dx dz dx dz dy dz dy dz z= z= y= y= x= x= 0 1 0 1 1 0 1 1 OABC F n ds = OABC (4 xziˆ y 2 ˆj yz kˆ) ( k ) dx dy = 0 0 yz dx dy 0 (as z = 0) DEFG (4 xziˆ y 2 ˆj yz kˆ ) kˆ dx dy = 1 1 DEFG yz dx dy 0 0 y (1) dx dy OAGF 1 y2 1 1 1 = 0 dx [ x ]0 2 2 2 0 2 (4 xz iˆ y 2 ˆj yz kˆ) ( j ) dx dz = OAGF y dx dz 0 1 (as y = 0) Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ 430 Vectors BCED (4 xz iˆ y ˆj yz kˆ) ˆj dx dz = 2 1 BCED ( y 2 ) dx dz 1 1 1 = 0 dx 0 dz ( x )0 ( z ) 0 1 ABDG (4 xziˆ y 2 ˆj yzkˆ) iˆ dy dz = = 4 OCEF (4 xz iˆ y 2 ( y )10 z2 2 1 (as y = 1) 1 1 4 xz dy dz 0 0 4 (1) z dy dz 1 4 (1) 2 2 0 ˆj yz kˆ ) ( iˆ ) dy dz = 1 1 0 0 4 xz dy dz 0 (as x = 0) On putting these values in (1), we get 1 3 S F nˆ ds = 0 2 0 1 2 0 = 2 EXERCISE 5.11 1. 2. Proved. where A = ( x y 2 ) iˆ 2 xjˆ 2 yzkˆ and S is the surface of the plane 2x + y + 2z = 6 in the first octant. Ans. 81 Evaluate Evaluate S A . nˆ ds, S A . nˆ ds, where A = ziˆ xjˆ 3 y 2 zkˆ and S is the surface of the cylinder x2 + y2 = 16 included in the first octant between z = 0 and z = 5. Ans. 90 2 3. 2 If r = tiˆ t ˆj (t 1) kˆ and S = 2t 2iˆ 6tkˆ, evaluate 4. Evaluate S F nˆ dS , 0 r S dt. Ans. 12 where, F = 18 z iˆ 12 ˆj 3 ykˆ and S is the surface of the plane 2x + 3y + 6z = 12 in the first octant. 5. Evaluate Ans. 24 S F nˆ ds, where, F = 2 yx iˆ yzjˆ x 2 kˆ over the surface S of the cube bounded by the coordinate planes and planes x = a, y = a and z = a. 6. Ans. 1 4 a 2 If F 2 yiˆ 3 ˆj x 2 kˆ and S is the surface of the parabolic cylinder y2 = 8x in the first octant bounded by the planes y = 4, and z = 6, then evaluate S F nˆ dS. Ans. 132 5.35 VOLUME INTEGRAL Let F be a vector point function and volume V enclosed by a closed surface. The volume integral = V F dv Example 78. If F = 2 z iˆ – x ĵ + y k̂ , evaluate V F dv where, v is the region bounded by the surfaces x = 0, y = 0, x = 2, y = 4, z = x2, z = 2. (2 z iˆ xjˆ ykˆ ) dx dy dz Solution. F dv = V = = 2 4 2 4 2 0 dx 0 dy x2 (2 ziˆ xjˆ ykˆ) dz 0 dx 0 dy [4 iˆ 2 xjˆ 2 ykˆ x = 2 4 0 dx 0 dy [ z 2 2 iˆ xzjˆ yzkˆ ] 2 x i x3 ˆj x 2 ykˆ ] 4ˆ Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ Vectors 431 4 x2 y 2 ˆ 2ˆ 4 3 k = 0 dx 4 yiˆ 2 xyjˆ y k x yiˆ x yjˆ 2 0 2 = 2 0 (16 iˆ 8 xjˆ 16 kˆ 4 x 4 iˆ 4 x3 ˆj 8 x 2 kˆ ) dx 2 4 x5 ˆ 8 x3 i x 4 ˆj = 16 xiˆ 4 x 2 ˆj 16 xkˆ 5 3 kˆ 0 128 ˆ 64 ˆ 32 ˆ 32 iˆ 32 kˆ i 16 ˆj k = (3 i 5 kˆ) = 32 iˆ 16 ˆj 32 kˆ = 5 3 15 5 3 EXERCISE 5.12 1. If F = (2 x2 3z ) iˆ 2 xy ˆj 4 x kˆ, then evaluate V F dV , Ans. where V is bounded by the plane Ans. 8 3 x = 0, y = 0, z = 0 and 2x + 2y + z = 4. 2. 3. Evaluate V dV , where = 45 x2y and V is the closed region bounded by the planes 4x + 2y + z = 8, x = 0, y = 0, z = 0 If F = (2x2 – 3z) iˆ 2 xy ˆj 4 xkˆ, then evaluate V F dV , where V is the closed region bounded by the planes x = 0, y = 0, z = 0 and 2x + 2y + z = 4. 4. Evaluate V (2 x y) dV , Ans. 8 ˆ ˆ ( j k) 3 where V is closed region bounded by the cylinder z = 4 – x2 and the planes x = 0, y = 0, y = 2 and z = 0. 5. Ans. 128 Ans. 80 3 2 If F = 2 xz iˆ xjˆ y kˆ, evaluate F d V over the region bounded by the surfaces x = 0, y = 0, y = 6 and z = x2, z = 4. Ans. (16iˆ 3 ˆj 48 kˆ ) 5.36 GREEN’S THEOREM (For a plane) Statement. If (x, y), (x, y), and be continuous functions over a region R bounded y x by simple closed curve C in x – y plane, then C ( dx dy) = R x y dx dy (AMIETE, June 2010, U.P., I Semester, Dec. 2007) Proof. Let the curve C be divided into two curves C1 (ABC) and C1 (CDA). Let the equation of the curve C1 (ABC) be y = y1 (x) and equation of the curve C2 (CDA) be y = y2 (x). Let us see the value of x c y y2 ( x ) c y y ( x) R y dx dy = x a y y1 ( x) y dy dx = a ( x, y ) y y12 ( x ) dx = c a ( x, y2 ) ( x, y1 ) dx a a c = c ( x, y2 ) dx a ( x, y1 ) dx c = ( x, y2 ) dx ( x, y1 ) dx a c = ( x, y) dx ( x, y ) dx = – c1 c2 c ( x, y) dx Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ 432 Vectors c dx Thus, = dx dy y R ...(1) Similarly, it can be shown that c dy = x dx dy ...(2) On adding (1) and (2), we get ( dx dy) = R x y dx dy Proved. Note. Green’s Theorem in vector form c F d r = R ( F ) kˆ d R where, F iˆ ˆj , r xiˆ yjˆ, kˆ is a unit vector along z-axis and dR = dx dy. Example 79. A vector field F is given by F sin yiˆ x (1 cos y ) ˆj. Evaluate the line integral C F dr where C is the circular path given by x2 + y2 = a2. Solution. F sin yiˆ x (1 cos y ) ˆj ˆ ˆjdy ) = = C [sin yiˆ x (1 cos y ) ˆj ] (idx On applying Green’s Theorem, we have C F dr c ( dx dy) dx dy y = s x = s [(1 cos y) cos y] dx dy C sin y dx x (1 cos y) dy where s is the circular plane surface of radius a. = s dx dy = Area of circle = a2. Ans. Example 80. Using Green’s Theorem, evaluate c (x 2 ydx x 2 dy ), where c is the boundaryy described counter clockwise of the triangle with vertices (0, 0), (1, 0), (1, 1). (U.P., I Semester, Winter 2003) Solution. By Green’s Theorem, we have Y A (1, 1) ( dx dy ) dx dy = R c x y x = c (x = 1 2 y dx x 2 dy ) = 0 (2 x x 2 ) dx x 0 dy = R (2 x x 1 0 (2 x x 2 2 y ) dx dy ) dx [ y ]0x 1 2 = 0 (2 x x ) ( x ) dx = O (0, 0) 2 x3 x 4 2 3 0 (2 x x ) dx = 3 4 1 5 2 1 = = 12 3 4 Example 81. State and verify Green’s Theorem in the plane for X (1, 0) 1 0 Ans. (3x 2 – 8 y 2 ) dx (4 y – 6 xy ) dy where C is the boundary of the region bounded by x 0, y 0 and 2x – 3y = 6. (Uttarakhand, I Semester, Dec. 2006) Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ Vectors 433 Solution. Statement: See Article 24.4 on page 576. Here the closed curve C consists of straight lines OB, BA and AO, where coordinates of A and B are (3, 0) and (0, – 2) respectively. Let R be the region bounded by C. Then by Green’s Theorem in plane, we have [ (3x 2 – 8 y 2 ) dx (4 y – 6 xy ) dy ] – 8 y 2 ) dx dy 10 y dx dy = R x (4 y – 6 xy) – y (3x = R (– 6 y 16 y) dx dy R 3 0 1 (2 x – 6) 3 = 10 dx 0 2 0 y2 5 y dy 10 dx = – 0 9 2 1 (2 x – 6) 3 ...(1) 3 0 dx (2 x – 6) 2 3 3 5 5 (2 x – 6)3 5 (216) – 20 ...(2) (0 6)3 = – – 54 9 3 2 0 54 Now we evaluate L.H.S. of (1) along OB, BA and AO. Along OB, x = 0, dx = 0 and y varies form 0 to –2. 1 3 Along BA, x = (6 3 y ), dx dy and y varies from –2 to 0. 2 2 and along AO, y = 0, dy = 0 and x varies from 3 to 0. = – L.H.S. of (1) = = OB [ (3x [ (3x 2 2 – 8 y 2 ) dx (4 y – 6 xy ) dy ] – 8 y 2 ) dx (4 y – 6 xy ) dy ] 2 2 BA [ (3x – 8 y ) dx (4 x – 6 xy) dy ] [ (3 x 2 – 8 y 2 ) dx (4 y – 6 xy ) dy ] AO 0 3 (6 3 y )2 – 8 y 2 dy [4 y – 3 (6 3 y ) y ] dy 3 x 2 dx 3 2 0 9 = [ 2 y 2 ]0– 2 (6 3 y )2 – 12 y 2 4 y – 18 y – 9 y 2 dy ( x 3 ) 03 –2 8 0 9 2 2 = 2 [4] (6 3 y ) – 21y – 14 y dy (0 – 27) –2 8 0 9 (6 3 y )3 216 3 2 3 2 = 8 8 3 3 – 7 y – 7 y – 27 – 19 8 7 (– 2) 7 (– 2) –2 = –2 0 4 y dy 0 3 –2 4 = – 19 + 27 – 56 + 28 = – 20 ...(3) With the help of (2) and (3), we find that (1) is true and so Green’s Theorem is verified. Example 82. Apply Green’s Theorem to evaluate C [(2 x 2 y 2 ) dx ( x 2 y 2 ) dy ], where C is the boundary of the area enclosed by the x-axis and the upper half of circle x2 + y2 = a2. (M.D.U. Dec. 2009, U.P., I Sem., Dec. 2004) 2 2 2 2 [(2 x y ) dx ( x y ) dy ] Solution. C By Green’s Theorem, we’ve ( dx dy) C = x y dx dy S Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ 434 Vectors = = a a 2 x2 a a 2 x2 a 0 a 0 (x2 y2 ) (2 x 2 y 2 ) dx dy y x (2 x 2 y ) dx dy = 2 y2 = 2 a dx xy 2 0 a a 2 a 2 x2 a dx a 2 x2 0 ( x y ) dy a a 2 x2 2 2 dx = 2 a x a x 2 a 2 a 2 a f ( x ) dx 2 a f ( x) dx, f is even 0 a 0, f is odd 2 = 2 a x a x dx a ( a x ) dx a 2 3 a3 a x3 4a3 2 2 2 a x 2 0 2 ( a x ) dx a = = = Ans. 0 3 0 3 3 y x Example 83. Evaluate 2 dx 2 dy, where C C1 U C2 with C1 : x2 + y2 = 1 C x y2 x y2 and C2 : x = 2, y = 2. (Gujarat, I Semester, Jan 2009) y x Solution. 2 dx 2 dy 2 Y C x y x y2 y=2 = x x 2 x y 2 y dx dy 2 y x y 2 X ( x y ) 1 – 2 x ( x) ( x 2 y 2 ) 1 – 2 y ( y ) dx dy = x=–2 ( x 2 y 2 )2 ( x 2 y 2 )2 x2 y 2 – 2 x 2 x 2 y 2 – 2 y 2 dx dy = 2 2 2 ( x 2 y 2 )2 (x y ) y 2 – x2 0 x2 – y2 dx dy 0 2 dx dy = 2 = 2 2 2 2 2 ( x y 2 )2 (x y ) (x y ) 5.37 AREA OF THE PLANE REGION BY GREEN’S THEOREM Proof. We know that N M Mdx Ndy = A x – y dx dy C 2 On putting 2 x 2+ y 2=1 X x=2 O y=–2 Y Ans. ...(1) N M N=x 1 and M = – y 1 in (1), we get x y – y dx x dy = A [1 – (–1) ] dx dy = 2 dx dy = 2 A C Area = 1 ( x dy – y dx) 2 C Example 84. Using Green’s theorem, find the area of the region in the first quadrant bounded by the curves 1 x y = x, y = , y = (U.P. I, Semester, Dec. 2008) x 4 Solution. By Green’s Theorem Area A of the region bounded by a closed curve C is given by Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ Vectors 435 A = 1 2 Y C ( xdy – ydx) ) y= x 1 1, x 1 B( Here, C consists of the curves C1 : y = , C2 : y = 4 x 1— and C3 : y = x So y= x 1 1 1 1 ) C2 A (2, — ( I1 I 2 I3 ) 2 A 2 C C C C 1 2 3 2 2 C3 C1 x 1 x Along C1 : y = , dy dx, x : 0 to 2 y=— 4 4 4 x 1 I1 = C ( xdy – ydx) C x dx – dx 0 X 1 1 4 4 O (0, 0) 1 1 Along C2 : y = , dy – 2 dx, x : 2 to 1 x x 1 1 1 1 I2 = C2 ( xdy – ydx) 2 x – 2 dx – dx = [– 2log x ]2 2 log 2 2 x Along C3 : y = x, dy = dx ; x : 1 to 0 ; I3 = ( xdy – ydx ) ( xdx – xdx ) 0 C3 A = 1. Evaluate c [(3x 2 1 1 ( I1 I 2 I 3 ) (0 2 log 2 + 0) log 2 2 2 EXERCISE 5.13 Ans. 6 yz ) dx (2 y 3 xz ) dy (1 4 xyz 2 ) dz ] from (0, 0, 0) to (1, 1, 1) along the path c given by the straight line from (0, 0, 0) to (0, 0, 1) then to (0, 1, 1) and then to (1, 1, 1). 2. C (x Verify Green’s Theorem in plane for 2 2 xy ) dx ( y 2 x 3 y ) dy , where c is a square with the Ans. vertices P (0, 0), Q (1, 0), R (1, 1) and S (0, 1). 3. Verify Green’s Theorem for y2 = 8 x and x = 2. c ( x 2 1 2 2 xy ) dx ( x 2 y 3) dy around the boundary c of the region c [(2 x 2 y 2 ) dx ( x 2 y 2 ) dy ] , 4. Use Green’s Theorem in a plane to evaluate the integral 5. where c is the boundary in the xy-plane of the area enclosed by the x-axis and the semi-circle x2 + y2 =1 4 in the upper half xy-plane. Ans. 3 [( y sin x ) dy cos x dy ], Apply Green’s Theoem to evaluate where c is the plane triangle enclosed 6. 2x and y . by the lines y = 0, x = 2 Either directly or by Green’s Therorem, evaluate the line integral c Ans. c e x where c is the rectangle with vertices (0, 0), (, 0,), , and 0, . 2 2 7. Verify the Green’s Theorem to evaluate the line integral of the closed region bounded by y = x and y = x2. c (2 y 2 2 8 4 (cos y dx sin y dy ), Ans. 2 (1 – e– ) (AMIETE II Sem June 2010) dx 3 x dy ), where c is the boundary (U.P., I Semester, Dec. 20005, AMIETE Summer 2004, Winter 2001) Ans. 27 4 Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ 436 8. Vectors Evaluate s F . nˆ.ds, where F x y iˆ x 2 ˆj ( x z ) kˆ and s is the region of the plane 2x + 2y + z = 6 in the first octant. 9. (A.M.I.E.T.E., Summer 2004, Winter 2001) Ans. Verify Green’s Theorem for C 27 4 ( xy y 2 ) dx x 2 dy where C is the boundary by y = x and y = x2. (AMIETE, June 2010) 5.38 STOKE’S THEOREM (Relation between Line Integral and Surface Integral) (Uttarakhand, I Sem. 2008, U.P., Ist Semester, Dec. 2006) Statement. Surface integral of the component of curl F along the normal to the surface S, taken over the surface S bounded by curve C is equal to the line integral of the vector point function F taken along the closed curve C. Mathematically F .d r = S curl F nˆ ds where n̂ = cos iˆ + cos ĵ + cos k̂ is a unit external normal to any surface ds, Proof. Let r = xiˆ yjˆ zkˆ d r = iˆ dx ˆj dy kˆ dz ˆ F = F1 iˆ F2 ˆj F3 k On putting the values of F , d r in the statement of the theorem c ( F1 iˆ F2 ˆj F3 kˆ) (iˆ dx ˆj dy kˆ dz) k ˆ ˆ ˆ ˆ ˆ ˆ y z ( F1 i F2 j F 3 k ). (cos i cos j cos k ) ds F3 F2 F2 F1 F1 F3 ( F1 dx F2 dy F3 dz ) = S y z iˆ z x ˆj x y kˆ . (iˆ cos ˆj cos kˆ cos ) ds = S i x j F3 F2 F2 F1 F1 F3 cos cos x y cos ds ...(1) y z z x Let us first prove F1 F1 ...(2) c F1 dx = S z cos y cos ds Let the equation of the surface S be z = g (x, y). The projection of the surface on x – y plane is region R. F1 ( x, y, z ) dx = F1 [ x, y, g ( x, y)] dx = S c c F1 ( x, y , g ) dx dy [By Green’s Theorem] y F1 F1 g = R ...(3) dx dy z y y The direction consines of the normal to the surface z = g(x, y) are given by cos cos cos = g g 1 x y = R Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ Vectors 437 And dx dy = projection of ds on the xy-plane = ds cos Putting the values of ds in R.H.S. of (2) F1 dx dy F1 cos y cos F1 g F1 F1 cos F1 dx dy dx dy = R = R z cos y z y y F1 F1 g dx dy = R z y y From (3) and (4), we get S z cos F1 cos ds = y F1 R z ...(4) F1 cos ds y F F Similarly, c F2 dy = S x2 cos z2 cos ds F F and c F3 dz = S y3 cos x3 cos ds On adding (5), (6) and (7), we get c F1 dx c ( F1 dx F2 dy F3 dz ) = = F1 cos S z F1 S z ...(5) cos cos ...(6) ...(7) F1 F F cos 2 cos 2 cos y x z F3 F cos 3 cos ds Proved. y x 5.39 ANOTHER METHOD OF PROVING STOKE’S THEOREM The circulation of vector F around a closed curve C is equal to the flux of the curve of the vector through the surface S bounded by the curve C. = S curl F n d s S curl F d S Proof : The projection of any curved surface over xy-plane can be treated as kernal of the surface integral over actual surface c F d r S ( F ) kˆ d S Now, = S ( F ) ( i [ kˆ iˆ ˆj ] j ) dx dy = S [( iˆ) ( F ˆj ) ( ˆj ) ( F iˆ)] dx dy = S x ( Fy ) y ( Fx ) dx dy S [ Fx dx Fy dy] [By Green’s theorem] = S [iˆ Fx = ˆj Fy ] (iˆ dx ˆj dy ) = S curl F nˆ dS = c F . dr c F . d r. where, F = Fx iˆ Fy ˆj Fz kˆ and dr dx iˆ dy ˆj dz kˆ Example 85. Evaluate by Strokes theorem x2 + y2 = 1, z = y2. ( yz dx zx dy xy dz ) where C is the curve C (M.D.U., Dec 2009) Solution. Here we have yz dx zx dy xy dz ˆ ˆjdy kdz ) = ( yziˆ zxjˆ xykˆ ). (idx Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ 438 Vectors = iˆ Curl F = x yz F .dx kˆ z xy ˆj y zx = (x – x) iˆ + (y – y) ĵ + (z – z) k̂ =0 =0 Example 86. Using Stoke’s theorem or otherwise, evaluate 2 2 [(2 x y ) dx yz dy y z dz ] = curl F . ds Ans. c where c is the circle x2 + y2 = 1, corresponding to the surface of sphere of unit radius. (U.P., I Semester, Winter 2001) 2 2 [(2 x y ) dx yz dy y z dz ] Solution. c = ˆj y 2 z kˆ] (iˆ dx ˆj dy kˆ dz ) S Curl F n ds 2 F d r = By Stoke’s theorem c [(2 x y) iˆ yz Curl F = F = iˆ x ˆj y 2 x y yz 2 ...(1) kˆ z 1 O y2 z = (– 2 yz 2 yz ) iˆ – (0 – 0) ˆj (0 1) kˆ kˆ Putting the value of curl F in (1), we get = kˆ nˆ ds = kˆ nˆ dx dy = nˆ kˆ Example 87. Evaluate dx dy dx dy ds (nˆ kˆ) = Area of the circle = C F . d r , where F ( x, y, z ) – y iˆ xjˆ z 2 2 kˆ and C is the curve of intersection of the plane y + z = 2 and the cylinder x2 + y2 = 1. (Gujarat, I sem. Jan. 2009) Solution. C F . dr S curl F . nˆ ds S curl (– y 2 iˆ x ˆj z 2 kˆ) nˆ ds 2 2 F (x, y, z) = y iˆ x ˆj z kˆ ˆj iˆ kˆ Curl F = x y z – y2 x ...(1) (By Stoke’s Theorem) z2 = iˆ (0 – 0) – ˆj (0 – 0) kˆ (1 2 y) (1 2 y) kˆ Normal vector = F ˆ ˆj kˆ = iˆ ( y z – 2) ˆj k x y z ˆj kˆ Unit normal vector n̂ = 2 dx dy ds = ˆ . kˆ 3y + z= 2 x2 + y2 = 1 Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ Vectors 439 On putting the values of curl F , nˆ and ds in (1), we get ˆj kˆ dx dy C F . dr = S (1 2 y) kˆ . 2 ˆj kˆ . kˆ 2 1 2 y dx dy 2 1 (1 2 y ) dx dy 1 = = (1 2 r sin ) r d d r 2 0 0 Y 2 = = 2 1 0 0 (r 2r 2 0 2 sin ) d d r 1 r 2 2r 3 2 d sin 0 3 2 0 r ddr 1 2 2 3 sin d 2 2 2 2 = – cos – – 0 = 3 3 2 3 0 Example 88. Apply Stoke’s Theorem to find the value of X O Ans. c ( y dx z dy x dz ) where c is the curve of intersection of x2 + y2 + z2 = a2 and x + z = a. (Nagpur, Summer 2001) Solution. c ( y dx z dy x dz ) ˆ ˆ = c ( yiˆ zjˆ xk ) (iˆ dx ˆj dy k dz ) = S curl ( yiˆ zjˆ xkˆ) nˆ ds = S iˆ x ˆj y kˆ z ( yiˆ zjˆ xkˆ) nˆ ds C ( yiˆ zjˆ xkˆ) dr = (By Stoke’s Theorem) = S (iˆ ˆj kˆ) nˆ ds ...(1) where S is the circle formed by the intersection of x2 + y2 + z2 = a2 and x + z = a. n̂ = | | iˆ n̂ = 2 Putting the vlaue of n̂ ˆ ˆj kˆ ( x z a ) i iˆ kˆ x y z = = 11 | | kˆ 2 in (1), we have iˆ kˆ ds = S (iˆ ˆj kˆ) 2 2 1 1 = S ds 2 2 2 2 2 a a2 ds = 2 S 2 2 2 Example 89. Directly or by Stoke’s Theorem, evaluate a2 a2 2 2 2 2 Use r R p a 2 2 Ans. s curl v nˆ ds, v iyˆ ˆjz kxˆ , s is the surface of the paraboloid z = 1 – x2 – y2, z3 > 0 and n̂ is the unit vector normal to s. Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ 440 Vectors Solution. iˆ ˆj v = x y y z kˆ iˆ ˆj kˆ z x n̂ = kˆ. Obviously ( v ) nˆ = (– iˆ – ˆj – kˆ). kˆ – 1 Therefore ( v ) nˆ ds Hence S = ( 1) dx dy S = dx dy S = – (1)2 = – . Example 90. Use Stoke’s Theorem to evaluate c v dr , (Area of circle = r2) Ans. where v y 2 iˆ xyjˆ x z kˆ, and c is the bounding curve of the hemisphere x2 + y2 + z2 = 9, z > 0, oriented in the positive direction. Solution. By Stoke’s theorem c v dr = S (curl v ) nˆ ds S ( v ) nˆ ds iˆ v = x ˆj y kˆ z y2 xy xz (0 0) iˆ ( z 0) ˆj ( y 2 y ) kˆ zjˆ ykˆ 2 2 2 i j k ( x y z 9) x y z n̂ = = | | | | ˆ 2 xiˆ 2 yjˆ 2 zk xiˆ yjˆ zkˆ xiˆ yjˆ zkˆ = 3 4 x2 4 y 2 4 z2 x2 y 2 z 2 xiˆ yjˆ zkˆ yz yz 2 yz ( v ) nˆ = ( zjˆ ykˆ) 3 3 3 ˆ yjˆ zkˆ z xi . kˆ dx = dx dy ds = dx dy n̂ kˆ ds = dx dy 3 3 3 ds = dx dy z 2 yz 3 S ( v ) nˆ ds = 3 z dx dy = 2 y dx dy 2 = 2 r sin r d dr = 2 3 0 3 sin d r 2 dr 0 r3 = 2 ( cos )20 = – 2 (– 1 + 1) 9 = 0 Ans. 3 0 Example 91. Evaluate the surface integral S curl F . nˆ d S by transforming it into a line integral, S being that part of the surface of the paraboloid z = 1 – x2 – y2 for which z 0 and F y iˆ z ˆj x kˆ . (K. University, Dec. 2008) Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ Vectors 441 Solution. iˆ ˆj F = x y y z kˆ iˆ ˆj kˆ z x n̂ = kˆ. Obviously ( F ) nˆ = (– iˆ – ˆj – kˆ). kˆ – 1 Therefore ( F ) nˆ ds Hence S = ( 1) dx dy S = dx dy = – (1)2 = – . Example 92. Evaluate C F dr S (Area of circle = r2) Ans. by Stoke’s Theorem, where F y 2 iˆ x 2 ˆj ( x z ) kˆ and C is the boundary of triangle with vertices at (0, 0, 0), (1, 0, 0) and (1, 1, 0). (U.P., I Semester, Winter 2000) Solution. We have, curl F = F iˆ = x kˆ z ˆj y 0. iˆ ˆj 2 ( x y ) kˆ. y 2 x2 ( x z) We observe that z co-ordinate of each vertex of the triangle is zero. Therefore, the triangle lies in the xy-plane. n̂ = k̂ curl F nˆ [ ˆj 2 ( x y ) kˆ] . kˆ 2 ( x y ). In the figure, only xy-plane is considered. The equation of the line OB is y = x By Stoke’s theorem, we have F dr = C (curl F nˆ) ds S x 1 y2 x 0 y 0 2 ( x y) dx dy = 2 0 xy 2 dx 0 1 2 1 2 x2 1x 1 2 3 dx = x dx = x = 1 . = 2 0 x dx = 2 0 2 0 2 3 3 0 = 1 x Ans. by Stoke’s Theorem, where F ( x 2 y 2 ) iˆ 2 xy ˆj and C is the boundary of the rectangle x = a, y = 0 and y = b. (U.P., I Semester, Winter 2002) Solution. Since the z co-ordinate of each vertex of the given rectangle is zero, hence the given rectangle must lie in the xy-plane. Here, the co-ordinates of A, B, C and D are (a, 0), (a, b), (– a, b) and (– a, 0) respectively. Example 93. Evaluate C F dr iˆ x ˆj y x2 y 2 2x y Curl F = kˆ =–4yk z 0 Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ 442 Vectors Here, nˆ kˆ, so by Stoke’s theorem, we’ve C F dr = = S curl F nˆ d s S ( 4 y kˆ) (kˆ) d x d y = 4 a =4 b y2 2 2 dx = 2 b 0 a a b y dx dy x a y 0 a dx 4 a b 2 Ans. a Example 94. Apply Stoke’s Theorem to calculate c 4 y dx 2 z dy 6 y dz where c is the curve of intersection of x2 + y2 + z2 = 6 z and z = x + 3. c F dr Solution. = c 4 y dx 2 z dy 6 y dz = ˆ ) ˆ ˆjdy kdz c (4 yiˆ 2 zjˆ 6 ykˆ) (idx F = 4 yiˆ 2 zjˆ 6 ykˆ ˆj iˆ kˆ (6 2) iˆ (0 0) ˆj (0 4) kˆ F = x y z 4 iˆ 4 kˆ 4y 2z 6y S is the surface of the circle x2 + y2 + z2 = 6z, z = x + 3, n̂ is normal to the plane x – z + 3 = 0 ˆ ˆj kˆ ( x z 3) i iˆ kˆ iˆ kˆ y z x = = n̂ = | | 11 2 | | iˆ kˆ 44 ( F ) nˆ = (4 iˆ 4 kˆ) = = 4 2 2 2 c F dr = S (curl F ) nˆ ds = S 4 2 ( dx dz ) = 4 2 (area of circle) Centre of the sphere x2 + y2 + (z – 3)2 = 9, (0, 0, 3) lies on the plane z = x + 3. It means that the given circle is a great circle of sphere, where radius of the circle is equal to the radius of the sphere. Radius of circle = 3, Area = (3)2 = 9 S ( F ) nˆ ds = 4 2(9 ) 36 2 Ans. Example 95. Verify Stoke’s Theorem for the function F z iˆ x ˆj y kˆ, where C is the unit circle in xy-plane bounding the hemisphere z = Solution. Here Also, (1 x 2 y 2 ). (U.P., I Semester Comp. 2002) F = z iˆ x ˆj y kˆ. r = xiˆ y ˆj z kˆ F dr = z dx + x dy + y dz. ...(1) dr = d xiˆ dy ˆj dz kˆ. C F dr = C ( z dx x dy y dz ). ...(2) Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ Vectors 443 On the circle C, x2 + y2 = 1, z = 0 on the xy-plane. Hence on C, we have z = 0 so that dz = 0. Hence (2) reduces to C F dr = C x dy. x2 Now the parametric equations of C, i.e., x = cos , y = sin . Using (4), (3) reduces to C F . dr = ...(3) + y2 = 1 are ...(4) 2 0 cos cos d = 21 0 cos 2 d 2 2 1 sin 2 = = ...(5) 2 2 0 2 2 2 Let P(x, y, z) be any point on the surface of the hemisphere x + y + z = 1, O origin is the centre of the sphere. Radius = OP = xiˆ y ˆj z kˆ Normal = xiˆ y ˆj z kˆ ˆ ˆ ˆ n̂ = x i y j z k x iˆ y ˆj z kˆ x2 y2 z2 (Radius is to tangent i.e. Radius is normal) x = sin cos , y = sin sin , z = cos ...(6) n̂ = sin cos iˆ + sin sin ĵ + cos k̂ ˆj iˆ kˆ Curl F = / x / y / z iˆ ˆj kˆ z x y Also, s ...(7) Curl F nˆ = (iˆ ˆj kˆ) . (sin cos iˆ sin sin ˆj sin kˆ) = sin cos + sin sin + cos / 2 2 (iˆ ˆj kˆ) Curl F nˆ dS = 0 0 . (sin cos iˆ + sin sin ĵ + cos k̂ ) sin d d = /2 2 0 sin d 0 (sin cos sin sin cos ) d [ dS = Elementary area on hemisphere = sin d d] = = /2 0 /2 0 sin d [sin sin sin ( cos ) cos ]20 = / 2 (0 0 2 sin cos ) d = 0 = – (/2) [– 1 – 1] = . From (5) and (8), C F dr = S curl F nˆ d S , /2 0 sin d / 2 sin 2 d = cos 2 2 0 which verifies Stokes’s theorem. Example 96. Verify Stoke’s theorem for the vector field F (2 x – y ) iˆ – yz 2 ˆj – y 2 z kˆ over the upper half of the surface x2 + y2 + z2 = 1 bounded by its projection on xy- plane. (Nagpur University, Summer 2001) 2 Solution. Let S be the upper half surface of the sphere x + y2 + z2 = 1. The boundary C or S is a circle in the xy plane of radius unity and centre O. The equation of C are x2 + y2 = 1, z = 0 whose parametric form is x = cos t, y = sin t, z = 0, 0 < t < 2 C F . dr = C [ (2 x – y) iˆ – yz 2 ˆj – y 2 z kˆ ].[iˆ dx ˆj dy kˆ dz ] Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ 444 Vectors = C [ (2 x – y ) dx – yz 2 dy – y 2 z dz ] = C (2 x – y) dx, since on C, z = 0 and 2z = 0 2 dx dt (2 cos t – sin t ) (– sin t ) dt = 0 (2 cos t – sin t ) 0 dt 2 2 1 – cos 2t 2 = 0 (– sin 2t sin t ) dt 0 – sin 2t dt 2 2 1 1 cos 2t t sin 2t – = 2 – 2 2 2 4 0 ˆj iˆ kˆ 2 x y z 2x – y – yz 2 – y2 z Curl F = ...(1) = (– 2 yz 2 yz ) iˆ (0 – 0) ˆj (0 1) kˆ kˆ Z Curl F . nˆ = kˆ . nˆ nˆ . kˆ nˆ . kˆ ds Curl F . nˆ ds = dx dy . nˆ kˆ Where R is the projection of S on xy-plane. S = S 1 1– x –1 – 2 1 – x2 dx dy = 1 –1 2 R nˆ . kˆ . 2 1 – x dx 4 1 0 1 O 1 – x dx 1 x 1 2 –1 = 4 2 1 – x 2 sin x 4 2 . 2 0 From (1) and (2), we have C F . dr = Y 2 X C ...(2) Curl F . nˆ ds which is the Stoke's theorem. Ans. Example 97. Verify Stoke’s Theorem for F ( x 2 y 4) iˆ 3 xyjˆ (2 xz z 2 ) kˆ over the surface of hemisphere x2 + y2 + z2 = 16 above the xy-plane. Solution. c F dr, where c is the boundary of the circle x2 + y2 + z2 = 16 = Putting (bounding the hemispherical surface) ˆ ˆjdy ) [( x 2 y 4) iˆ 3 xyjˆ (2 xz z 2 ) kˆ ] (idx c 2 c [( x y 4) dx 3 xy dy)] = x = 4 cos , y = 4 sin , dx = – 4 sin d , dy = 4 cos d = 2 0 [(16 cos 2 4 sin 4) ( 4 sin d ) (192 sin cos 2 d )] 2 2 2 2 = 16 0 [ 4 cos sin sin sin 12 sin cos ] d 2 2 2 = 16 0 (8 sin cos sin sin ) d 2 2 = 16 0 sin d 1 2 = 16 4 02 sin d = 64 = – 16 . 2 2 ˆj iˆ kˆ To evaluate surface integral F = x y z 2 sin n cos d 0 2 n cos sin d 0 0 0 x 2 y 4 3 xy 2 xz z 2 Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ Vectors 445 = (0 – 0) iˆ – (2 z – 0) ˆj (3 y – 1) kˆ – 2 zjˆ (3 y – 1) kˆ ˆ ˆj kˆ ( x 2 y 2 z 2 16) i x y z n̂ = = | | | | ˆ ˆ ˆ ˆ xi yjˆ zkˆ 2 xi 2 yj 2 zk xiˆ yjˆ zkˆ = = = 4 x2 y 2 z2 4 x2 4 y 2 4 z2 xiˆ yjˆ zkˆ 2 yz (3 y 1) z = ( F ) nˆ = [– 2 zjˆ (3 y – 1) kˆ] 4 4 z xiˆ yjˆ zkˆ k̂ n ds = dx dy . k ds = dx dy ds = dx dy 4 4 4 ds = dx dy z 2 yz (3 y 1) z 4 dx dy = [ 2 y (3 y 1)] dx dy = ( F ) nˆ ds = 4 z On putting x = r cos , y = r sin , dx dy = r d dr, we get = (r sin 1) r d dr4 2 2 d (r 2 sin r ) dr 2 64 = d sin 8 3 0 0 0 2 64 64 64 16 cos 8 = = = – 16 3 3 3 0 The line integral is equal to the surface integral, hence Stoke’s Theorem is verified. Proved. = r 3 = ( y 1) dx dy d 3 sin r 2 Example 98. Verify Stoke’s theorem for a vector field defined by F ( x 2 – y 2 ) iˆ 2 xy ˆj in the rectangular in xy-plane bounded by lines x = 0, x = a, y = 0, y = b. (Nagpur University, Summer 2000) Solution. Here we have to verify Stoke’s theorem C F . dr = S ( F ) . nˆ ds Where ‘C’ be the boundary of rectangle (ABCD) and S be the surface enclosed by curve C. F = ( x2 – y 2 ) iˆ (2 xy ) ˆj 2 2 F . dr = [ ( x – y ) iˆ 2 xy ˆj] . [iˆ dx ˆj dy] F . dr = (x2 + y2) dx + 2xy dy ...(1) C F . dr = OA F . d r AB F . d r BC F . d r CO F . d r Now, Along AB, put x = a so that dx = 0 in (1), we get Where y is from 0 to b. b AB F . d r 0 2ay dy = [ay 2 ] b0 ab 2 ...(2) Along OA, put y = 0 so that k dy = 0 in (1) and F . d r = x2 dx, Where x is from 0 to a. a x3 a3 a 2 OA F . dr 0 x dx = 3 0 3 F .dr ...(3) = 2ay dy ...(4) Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ 446 Vectors Along BC, put y = b and dy = 0 in (1) we get F . dr = (x2 – b2) dx, where x is from a to 0. 0 x3 – a3 ( x 2 – b2 ) dx – b2 x b2a F . d r = a BC 3 3 a 0 Along CO, put x = 0 and dx = 0 in (1), we get F . d r 0 ...(5) Y C ...(6) CO F . d r = 0 Putting the values of integrals (3), (4), (5) and (6) in (2), x=0 we get a3 a3 ab2 – ab2 0 2ab2 ...(7) = F . d r O C 3 3 Now we have to evaluate R.H.S. of Stoke’s Theorem i.e. S y=b B (a, b) x=a A X y=0 ( F ) . nˆ ds We have, iˆ x ˆj y x2 – y 2 2 xy F = kˆ (2 y 2 y ) kˆ 4 y kˆ z 0 Also the unit vector normal to the surface S in outward direction is n̂ k ( z-axis is normal to surface S) Also in xy-plane ds = dx dy S ( F ) . nˆ. ds = R 4 y kˆ . kˆ dx dy R 4 y dx dy. Where R be the region of the surface S. Consider a strip parallel to y-axis. This strip starts on line y = 0 (i.e. x-axis) and end on the line y = b, We move this strip from x = 0 (y-axis) to x = a to cover complete region R. S ( F ) . nˆ. ds = = C a b a 0 0 4 y dy dx 0 [2 y a 0 2b 2 2 b ]0 dx dx 2b 2 [ x] a0 2 ab 2 ...(8) From (7) and (8), we get F . dr = S ( F ) . nˆ ds and hence the Stoke’s theorem is verified. Example 99. Verify Stoke’s Theorem for the function 2 F = x iˆ – xyjˆ integrated round the square in the plane z = 0 and bounded by the lines x = 0, y = 0, x = a, y = a. 2 Solution. We have, F = x iˆ – xyjˆ ˆj iˆ kˆ F = x y z x2 xy 0 = (0 – 0) iˆ – (0 – 0) ˆj (– y – 0) kˆ – ykˆ ( n̂ to xy plane i.e. k̂ ) Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ Vectors 447 S ( F ) nˆ ds = S ( yk ) k dx dy a a y2 a2 a3 ( x)a0 = = = dx ydy = dx 2 2 2 0 0 0 0 To obtain line integral a a F d r = C (x 2 iˆ xyjˆ) (iˆ dx ˆj dy ) = C (x 2 ...(1) dx xy dy ) C where c is the path OABCO as shown in the figure. Also, Along F d r = C OABCO OA, y = 0, dy = 0 OA F d r = F dr = OA ( x 2 OA AB BC a Along AB F dr = 2 AB ( x 2 dx x y d y ) dy = 0 dx = 0 dy= 0 dx = 0 Lower Upper limit limit x=0 x=a y=0 y=a x=a x=0 y=a y=0 a y2 a3 = 0 a y d y = a = 2 2 0 y = a, dy = 0 a Along BC, BC ...(2) CO line Eq. of line OA y = 0 AB x = a BC y = a CO x = 0 dx xydy ) x3 a3 = 0 x dx = = 3 3 0 AB, x = a, dx = 0 a F dr F dr F dr F dr 2 F dr = BC ( x dx xy dy ) = Along CO, 0 2 x dx a x = 0, dx = 0 0 x3 a3 = = 3 3 a 2 = CO ( x dx xy dy ) = 0 Putting the values of these integrals in (2), we have CO F dr C F dr = a3 a3 a3 a3 0 = 2 3 2 3 From (1) and (3), ( F ) nˆ ds = S ...(3) F dr C Hence, Stoke’s Theorem is verified. Ans. Example 100. Verify Stoke’s Theorem for F = (x + y) iˆ + (2x – z) ĵ + (y + z) k̂ for the surface of a triangular lamina with vertices (2, 0, 0), (0, 3, 0) and (0, 0, 6). (Nagpur University 2004, K. U. Dec. 2009, 2008, A.M.I.E.T.E., Summer 2000) Solution. Here the path of integration c consists of the straight lines AB, BC, CA where the co-ordinates of A, B, C and (2, 0, 0), (0, 3, 0) and (0, 0, 6) respectively. Let S be the plane surface of triangle ABC bounded by C. Let n̂ be unit normal vector to surface S. Then by Stoke’s Theorem, we must have c F dr = s curl F nˆ ds ...(1) Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ 448 Vectors L.H.S. of (1)= c ABC F dr = AB F dr BC F dr CA F dr Along line AB, z = 0, equation of AB is x y =1 2 3 3 3 (2 x), dy = dx 2 2 At A, x = 2, At B, x = 0, r = xiˆ yjˆ y = AB F dr = = = ˆ ˆjdy ) AB [( x y) iˆ 2 xjˆ ykˆ] (idx AB ( x y) dx 2 xdy AB x 3 3x 3 dx 2 x dx 2 2 0 7 x2 7x 3 dx 3x = 2 4 2 2 = (7 – 6) = + 1 0 line AB BC CA Eq. of Lower line limit At A 3 x y 1 dy – dx 2 x2 2 3 z=0 At B y z 1 dz = – 2dy y3 3 6 x=0 At C x z 1 dz = – 3dx x0 2 6 y=0 Along line BC, x = 0, Equation of BC is Upper limit At B x0 At C y0 At A x2 y z = 1 or z = 6 – 2y,, dz = – 2dy 3 6 At B, y = 3, At C, y = 0, r = yjˆ zkˆ BC F dr = BC [ yi zj ( y z )k ] ( jdy kdz) = BC zdy ( y z) dz 0 = 3 ( 6 2 y ) dy ( y 6 2 y ) ( 2dy ) 0 2 0 = 3 (4 y 18) dy (2 y 18 y ) 3 = 36 x z Along line CA, y = 0, Eq. of CA, = 1 or z = 6 – 3x, dz = – 3dx 2 6 At C, x = 0, at A, x = 2, r = xiˆ zkˆ CA F dr = CA [ xiˆ (2 x z) ˆj zkˆ] [dxiˆ dzkˆ] = CA ( xdx zdz ) = 0 xdx (6 3x) ( 3dx) 2 = 2 0 (10 x 18) dx 2 2 = [5 x 18 x]0 = – 16 Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ Vectors 449 L.H.S. of (1) = ABC F dr = AB F dr BC F dr CA F dr = 1 + 36 – 16 = 21 ...(2) ˆj kˆ [( x y ) iˆ (2 x z ) ˆj ( y z ) k ] Curl F = F = iˆ y z x ˆj iˆ kˆ (1 1) iˆ (0 0) ˆj (2 1) kˆ 2iˆ kˆ y z = x x y 2x z y z Equation of the plane of ABC is x y z =1 2 3 6 Normal to the plane ABC is ˆ ˆ ˆ x y z ˆj kˆ 1 = i j k = iˆ x y z 2 3 6 2 3 6 ˆi ˆj kˆ Unit Normal Vector = 2 3 6 1 1 1 4 9 36 1 n̂ = (3iˆ 2 ˆj kˆ) 14 R.H.S. of (1) = s curl F n ds = s (2iˆ kˆ) 1 4 (3iˆ 2 ˆj kˆ) dx dy 1 (3iˆ 2 ˆj kˆ).kˆ 14 (6 1) dx dy = 7 dx dy = 7 Area of OAB 1 14 14 1 = 7 2 3 = 21 2 with the help of (2) and (3) we find (1) is true and so Stoke’s Theorem is verified. Example 101. Verify Stoke’s Theorem for = s ...(3) F = (y – z + 2) iˆ + (yz + 4) ĵ – (xz) k̂ over the surface of a cube x = 0, y = 0, z = 0, x = 2, y = 2, z = 2 above the XOY plane (open the bottom). Solution. Consider the surface of the cube as shown in the figure. Bounding path is OABCO shown by arrows. ˆ ) ˆ ˆjdy kdz F d r = [( y z 2) iˆ ( yz 4) ˆj ( xz ) kˆ] (idx c = ( y z 2) dx ( yz 4) dy xzdz c F dr c = OA F dr AB F dr BC F dr (1) Along OA, y = 0, dy = 0, z = 0, dz = 0 F dr ...(1) CO Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ 450 Vectors Line of line y 0 dy 0 1 OA 2 AB 3 BC 4 CO Equ. z0 x2 z 0 y2 z0 x0 z 0 dz 0 dx 0 dz 0 dy 0 dz 0 dx 0 dz 0 2 F dr = Lower Upper F . dr limit limit x=0 x=2 2 dx y=0 y=2 4 dy x=2 x=0 4 dx y=2 y=0 4 dy 2 2 dx [2 x]0 =4 0 OA (2) Along AB, x = 2, dx = 0, z = 0, dz = 0 2 F dr = 2 4 dy 4 ( y)0 =8 0 AB (3) Along BC, y = 2, dy = 0, z = 0, dz = 0 2 F dr = 0 (2 0 2) dx (4 x) 2 =–8 0 BC (4) Along CO, x = 0, dx = 0, z = 0, dz = 0 F dr = CO ( y 0 2) 0 (0 4) dy 0 = 4 dy 4 ( y ) 02 = – 8 On putting the values of these integrals in (1), we get F dr =4+8–8–8=–4 c To obtain surface integral iˆ kˆ ˆj x y z = F y z 2 yz 4 xz = (0 – y) iˆ – (– z + 1) ĵ + (0 – 1) k̂ = – y iˆ + (z – 1) ĵ – k̂ Here we have to integrate over the five surfaces, ABDE, OCGF, BCGD, OAEF, DEFG. Over the surface ABDE (x = 2), n̂ = i, ds = dy dz ( F ) nˆ ds = = [ yi ( z 1) j k ] i dx dz z f ( x, y ) R F3 ( x, y, z) z f ( x, y ) dx dy y dy dz 2 1 Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ Vectors 451 Outward normal ABDE i OCGF –i BCGD j OAEF –j DEFG k Surface 1 2 3 4 5 ds dy dz dy dz dx dz dx dz dx dy x=2 x=0 y=2 y=0 z=2 2 2 2 y2 2 = y dy dz [ z ]0 4 2 0 0 0 Over the surface OCGF (x = 0), n̂ = – i, ds = dy dz ( F ) nˆ ds = [ yiˆ ( z 1) ˆj kˆ] (iˆ) dy dz 2 2 2 y2 = y dy dz y dy dz 2 4 2 0 0 0 (3) Over the surface BCGD, (y = 2), n̂ = j, ds = dx dz ( F ) nˆ ds = [ yiˆ ( z 1) ˆj kˆ] ˆj dx dz 2 = ( z 1) dx dz = 2 2 z2 dx ( z 1) dz = ( x)20 2 z = 0 0 0 0 (4) Over the surface OAEF, (y = 0), n̂ = – ĵ , ds = dx dz ( F ) nˆ d s = [ yiˆ ( z 1) ˆj kˆ] ( ˆj) dx dz 2 2 2 z2 = ( z 1) dx dz = – dx ( z 1) dz = ( x)20 z = 0 0 0 2 0 (5) Over the surface DEFG, (z = 2), n̂ = k, ds = dx dy ( F ) nˆ ds = [ yiˆ ( z 1) ˆj kˆ] kˆ dx dy 2 =– dx dy 2 = – dx dy = [ x]20 [ y]02 = – 4 0 0 Total surface integral = – 4 + 4 + 0 + 0 – 4 = – 4 Thus S curl F nˆ ds = F dr = – 4 c which verifies Stoke’s Theorem. Ans. Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ 452 Vectors EXERCISE 5.14 1. y Use the Stoke’s Theorem to evaluate 2 dx xy dy xz dz , C where C is the bounding curve of the hemisphere x2 + y2 + z2 = 1, z 0, oriented in the positive direction. Ans. 0 2. Evaluate s (curl F ) nˆ dA, using the Stoke’s Theorem, where F yiˆ zjˆ xkˆ and s is the paraboloid z = f (x, y) = 1 – x2 – y2 , z 0. 3. Evaluate the integral for C y 2 Ans. 2 2 dx z dy x dz , where C is the triangular closed path joining the points (0, 0, 0), (0, a, 0) and (0, 0, a) by transforming the integral to surface integral using Stoke’s Theorem. Ans. a3 . 3 4. Verify Stoke’s Theorem for A 3 yiˆ xzjˆ yz 2 kˆ, where S is the surface of the paraboloid 2z = x2 + y2 bounded by z = 2 and c is its boundary traversed in the clockwise direction. Ans. – 20 5. Evaluate 6. 7. If S is the surface of the sphere x2 + y2 + z2 = 9. Prove that Verify Stoke’s Theorem for the vector field F (2 y z ) iˆ ( x – z ) ˆj ( y – x) kˆ C F d R where F yiˆ xz 3 ˆj zy 3kˆ, C is the circl x2 + y2 = 4, z = 1.5 S curl F ds Ans. 19 2 = 0. over the portion of the plane x + y + z = 1 cut off by the co-ordinate planes. 8. Evaluate c F dr by Stoke’s Theorem for F yz iˆ zx ˆj xy k and C is the curve of intersection of x2 + y2 = 1 and y = z2. 9. If F ( x – z ) iˆ ( x yz ) ˆj 3 xy kˆ and S is the surface of the cone z = a – xy-plane, show that 10. Ans. 0 3 2 s curl F dS 2 2 ( x y ) above the = 3 a4 / 4. If F 3 yiˆ xyjˆ yz 2kˆ and S is the surface of the paraboloid 2z = x2 + y2 bounded by z = 2, show by using Stoke’s Theorem that 11. s ( F ) dS = 20 . If F ( y 2 z 2 – x 2 ) iˆ ( z 2 x 2 – y 2 ) ˆj ( x 2 y 2 – z 2 ) kˆ, evaluate x2 curl F nˆ ds integrated over y2 the portion of the surface + – 2ax + az = 0 above the plane z = 0 and verify Stoke’s Theorem; where (A.M.I.E.T.E., Winter 20002) Ans. 2 a3 n̂ is unit vector normal to the surface. 12. Evaluate by using Stoke’s Theorem C sin z dx cos x dy sin y dz rectangle 0 x , 0 y 1, z 3 . where C is the boundary of (AMIETE, June 2010) 5.40 GAUSS’S THEOREM OF DIVERGENCE (Relation between surface integral and volume integral) (U.P., Ist Semester, Jan., 2011, Dec, 2006) Statement. The surface integral of the normal component of a vector function F taken around a closed surface S is equal to the integral of the divergence of F taken over the volume V enclosed by the surface S. Mathematically S F . nˆ ds V div Fdw Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ Vectors 453 Proof. Let F F1iˆ F2 ˆj F3 kˆ. Putting the values of F , nˆ in the statement of the divergence theorem, we have S F1 iˆ F2 ˆj F3 kˆ nˆ ds = = We require to prove (1). Let us first evaluate V V V iˆ x ˆj y kˆ z ( F1 iˆ F2 F1 V x ˆj F3 kˆ) dx dy dz. F2 F3 dx dy dz y z ...(1) F3 dx dy dz . z F3 dx dy dz = z z f 2 (x, y ) R z f1 ( x, y ) F3 dz dx dy z = R [ F3 ( x, y , f 2 ) F3 ( x, y, f1 )] dx dy For the upper part of the surface i.e. S2, we have dx dy = ds2 cos r2 = n̂ 2. k̂ ds2 Again for the lower part of the surface i.e. S1, we have, ...(2) dx dy = – cos r1, ds1 = n̂ 1. k̂ ds1 R F3 ( x, y, f2 ) dx dy = S2 F3 nˆ2 kˆ ds2 and R F3 ( x, y, f1 ) dx dy = S1 F3 nˆ1 kˆ ds1 Putting these values in (2), we have F3 V z dv = S2 F3 nˆ2 kˆ ds2 S1 F3 nˆ1 kˆ ds1 = Similarly, it can be shown that F2 ...(4) V y dv = S F2 nˆ ˆj ds S F3 nˆ kˆ ds ...(3) F1 dv = S F1 nˆ iˆ ds ...(5) x Adding (3), (4) & (5), we have F1 F2 F3 V x y z dv ˆ = ( F1 iˆ F2 ˆj F3 k ) nˆ ds V S V ( F ) dv = S F nˆ ds Proved. Example 102. State Gauss’s Divergence theorem S F . nˆ ds Div F dv where S is the surface of the sphere x2 + y2 + z2 = 16 and F 3 x iˆ 4 y ˆj 5 z kˆ. (Nagpur University, Winter 2004) Solution. Statement of Gauss’s Divergence theorem is given in Art 24.8 on page 597. Thus by Gauss’s divergence theorem, S F . nˆ ds = v . F dv Here F 3 x iˆ 4 y ˆj 5 z kˆ Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ 454 Vectors ˆ ˆj kˆ . F = iˆ . (3xiˆ 4 yjˆ 5zk ) x y z . F = 3 + 4 + 5 = 14 Putting the value of . F, we get S F . nˆ ds = v 14 . dv where v is volume of a sphere = 14 v 4 3584 = 14 (4)3 3 3 S F . nˆ ds where F 4 xz iˆ – y Example 103. Evaluate Ans. 2 ˆj yz kˆ and S is the surface of the cube bounded by x = 0, x = 1, y = 0, y = 1, z = 0, z = 1. (U.P., Ist Semester, 2009, Nagpur University, Winter 2003) Solution. By Divergence theorem, S F . nˆ ds = v ( . F ) dv = v iˆ x ˆj y kˆ z . (4 xz iˆ – y = v x (4 xz ) y (– y = v (4 z – 2 y y) dx dy dz 2 1 = v (4 z – y) dx dy dz 0 = 0 0 (2 z 1 1 2 – yz ) 10 dx dy ) 2 ˆj yz kˆ) dv ( yz ) dx dy dz z 1 4z 2 0 2 – yz dx dy 0 1 1 0 1 0 (2 – y) dx dy 1 3 3 3 y2 3 1 1 = 0 2 y – dx 0 dx = [ x ] 0 (1) Ans. 2 2 2 2 2 0 Note: This question is directly solved as on example 14 on Page 574. 1 2 Example 104. Find F nˆ ds, where F (2 x 3 z ) iˆ – ( xz y ) ˆj ( y 2 z ) kˆ and S is the surface of the sphere having centre (3, – 1, 2) and radius 3. (AMIETE, Dec. 2010, U.P., I Semester, Winter 2005, 2000) Solution. Let V be the volume enclosed by the surface S. By Divergence theorem, we’ve S F nˆ ds = V div F dv. 2 Now, div F = x (2 x 3z ) y [ ( xz y )] z ( y 2 z ) = 2 – 1 + 2 = 3 S F nˆ ds = V 3 dv = 3 V dv = 3V. Again V is the volume of a sphere of radius 3. Therefore 4 4 V = r 3 = (3)3 = 36 . 3 3 S F nˆ ds = 3V = 3 × 36 = 108 Ans. Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ Vectors 455 Example 105. Use Divergence Theorem to evaluate S A ds , where A x i y j z kˆ and S is the surface of the sphere x2 + y2 + z2 = a2. (AMIETE, Dec. 2009) Solution. 3ˆ S 3ˆ A ds = 3 v div A dV ˆj kˆ ( x 3 iˆ y 3 ˆj z 3 kˆ) dV = v iˆ y z x 2 2 2 = (3 x 2 3 y 2 3 z 2 ) dV = 3 ( x y z ) dV v v On putting x = r sin cos , y = r sin sin , z = r cos , we get 2 = 3 r 2 (r 2 sin dr d d ) = 3 × 8 v 2 a 4 d sin d 0 r 0 dr 0 a a5 12 a5 r5 = 24 ()02 ( cos )02 = 24 2 ( 0 1) 5 5 5 0 Example 106. Use divergence Theorem to show that Ans. y2 z2 ) d s = 6 V where S is any closed surface enclosing volume V. (U.P., I Semester, Winter 2002) ˆ 2 2 2 ˆj kˆ (x y z ) Solution. Here (x2 + y2 + z2) = i y z x = 2 x iˆ 2 y ˆj 2 z kˆ 2 ( x iˆ y ˆj z kˆ) S ( x 2 S (x 2 y 2 z 2 ) ds = S ( x 2 y 2 z 2 ) nˆ ds n̂ being outward drawn unit normal vector to S = 2 ( x iˆ y ˆj z kˆ) nˆ ds S = 2 div ( x iˆ y ˆj z kˆ) d v V ...(1) (By Divergence Theorem) (V being volume enclosed by S) ˆ ˆj kˆ ( x iˆ y ˆj z k ) div. ( x iˆ y ˆj zkˆ) = iˆ x y z x y z = =3 x y z From (1) & (2), we have Now, ( x 2 y 2 z 2 ) dS = 2 3 dv = 6 V dv = 6 V V Example 107. Evaluate S ( y ...(2) Proved. z i z 2 x 2 ˆj z 2 y 2 kˆ) nˆ dS , where S is the part of the spheree 2 2ˆ x2 + y2 + z2 = 1 above the xy-plane and bounded by this plane. Solution. Let V be the volume enclosed by the surface S. Then by divergence Theorem, we have 2 2 2 2 2 2 div ( y 2 z 2 iˆ z 2 x 2 ˆj z 2 y 2 kˆ) dV ( y z iˆ z x ˆj z y kˆ) nˆ dS = V S = V x ( y 2 2 z ) 2 2 2 2 2 2 (z x ) ( z y ) dV V 2 z y dV 2 V zy dV y z Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ 456 Vectors Changing to spherical polar coordinates by putting x = r sin cos , y= r sin sin , z = r cos , dV = r2 sin dr d d To cover V, the limits of r will be 0 to 1, those of will be 0 to and those of will be 0 to 2 2. 2 2 zy 2 dV = 2 V 0 2 = 2 0 /2 1 /2 1 5 0 0 (r cos ) (r 0 0 r 2 sin 2 sin 2 ) r 2 sin dr d d sin 3 cos sin 2 dr d d 1 2 sin 3 0 r6 = 2 cos sin d d 0 6 0 2 2 2 2 1 2 2 d = sin d = = 0 sin 0 6 4.2 12 12 2 2 Example 108. Use Divergence Theorem to evaluate and S is the surface bounding the region x2 + y2 Solution. By Divergence Theorem, S F dS = V S Ans. 2 2ˆ F d S where F 4 xiˆ – 2 y ˆj z k = 4, z = 0 and z = 3. (A.M.I.E.T.E., Summer 2003, 2001) div F dV = V iˆ x ˆj y kˆ z (4 xiˆ 2 y = V (4 4 y 2 z ) dx dy dz 2 ˆj z 2 kˆ ) dV 3 = dx dy (4 4 y 2 z )dz = 0 = (12 12 y 9) dx dy = Let us put x = r cos , y = r sin 2 = 2 3 ]0 (21 12 y) dx dy 2 d 0 (21 r 12 r (21 12 r sin ) r d dr = 2 sin ) dr 0 2 2 dx dy [4 z 4 yz z 2 21 r 2 4 r 3 sin = = d 2 0 0 = 84 + 32 – 32 = 84 2 d (42 32 sin ) (42 32 cos )0 0 Ans. Example 109. Apply the Divergence Theorem to compute ^ u n ds, where s is the surface of ˆ. ˆ – ˆjy kz the cylinder x2 + y2 = a2 bounded by the planes z = 0, z = b and where u ix Solution. By Gauss’s Divergence Theorem ˆ u nds = = = = V dv = V ( u ) dv V iˆ x ˆj y kˆ z (ixˆ ˆjy kzˆ )dv x y z V x y z dv V dx dy dz = V 1 1 1 dv = Volume of the cylinder = a2b Ans. Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ Vectors 457 Example 110. Apply Divergence Theorem to evaluate V F . nˆ ds, wheree F = 4x3iˆ x 2 y ˆj x 2 zkˆ and S is the surface of the cylinder x2 + y2 = a2 bounded by the planes z = 0 and z = b. (U.P. Ist Semester, Dec. 2006) Solution. We have, 3 2 2 F = 4x iˆ x y ˆj x zkˆ ˆ 3ˆ 2 ˆ 2 ˆ ˆ ˆ div F = i x j y k z (4 x i x yj x zk ) 2 3 2 = x (4 x ) y ( x y) z ( x z ) = 12x2 – x2 + x2 = 12 x2 Now, V div F dV = 12 a 2 x2 a xay a 2 x2 a = 12 x a b a 2 x2 z 0 x 2 dz dy dx a y a 2 x2 x 2 ( z )b0 dy dx = 12 b x 2 ( y ) a a a 2 2 2 = 12 b a x .2 a x dx a 2 = 48 b 0 x a 2 x 2 dx 2 = 24 b a x a 2 x2 a 2 x2 dx a 2 x 2 dx [Put x = a sin , dx = a cos d] / 2 2 a sin 2 a cos a cos d = 48 b 0 = 48 ba 4 / 2 0 2 2 sin cos d = 48 ba 1 1 4 2 = 48 ba 2 = 3 b a4 2 2 4 3 3 2 2 23 Ans. where F = (x2 + y2 + z2) (iˆ ˆj kˆ), S is the surface of the tetrahedron x = 0, y = 0, z = 0, x + y + z = 2 and n is the unit normal in the outward direction to the closed surface S. Solution. By Divergence theorem Example 111. Evaluate surface integral F nˆ ds, S F nˆ ds = V div F dv where S is the surface of tetrahedron x = 0, y = 0, z = 0, x + y + z = 2 ˆj kˆ ( x 2 y 2 z 2 ) (iˆ ˆj kˆ) dv = V iˆ y z x = = V (2 x 2 y 2 z ) dv 2 ( x y z ) dx dy dz V 2 2x = 2 0 dx 0 2 2 x = 2 0 dx 0 dy 2x y 0 ( x y z ) dz 2 x y z2 dy x z y z 2 0 Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ 458 Vectors 2 2x = 2 0 dx 0 (2 x y )2 dy 2 x x 2 xy 2 y xy y 2 2 2 x 2 y 3 (2 x y )3 = 2 dx 2 xy x 2 y x y 2 y 2 0 3 6 0 2 (2 x)3 (2 x )3 = 2 dx 2 x (2 x ) x 2 (2 x ) x (2 x) 2 (2 x)2 0 3 6 3 3 2 (2 x) (2 x ) = 2 4 x 2 x 2 2 x 2 x3 4 x 4x 2 x 3 (2 x )2 0 3 6 2 4 x3 x 4 4 x3 x 4 (2 x)3 (2 x)4 (2 x) 4 = 2 2x2 2x2 3 4 3 4 3 12 24 0 2 (2 x)3 (2 x) 4 (2 x )4 8 16 16 = 2 = 2 3 12 24 = 4 3 12 24 0 Example 112. Use the Divergence Theorem to evaluate Ans. S ( x dy dz y dz dx z dx dy) where S is the portion of the plane x + 2 y + 3 z = 6 which lies in the first Octant. (U.P., I Semester, Winter 2003) Solution. S ( f1 dy dz f 2 f 3 dxd ydz y z where S is a closed surface bounding a volume V. = f1 f 2 dx dz f3 dx dy ) V x S ( x dy dz y dz dx z dx dy) = x y z V x y z dx dy dz = V (1 1 1) dx dy dz = 3 V dx dy dz = 3 (Volume of tetrahedron OABC) 1 = 3 [( Area of the base OAB) height OC ] 3 1 1 = 3 6 3 2 = 18 3 2 Ans. Example 113. Use Divergence Theorem to evaluate : ( x dy dz y dz dx z dx dy ) over the surface of a sphere radius a. (K. University, Dec. 2009) Solution. Here, we have x dy dz y dx dz z dx dy S f f x y z f 1 2 3 dx dy dz dx dy dz V x V y z x y z (1 + 1 + 1) dx dy dz = 3 (volume of the sphere) V 4 3 = 3 a = 4 a3 3 Ans. Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ Vectors 459 Example 114. Using the divergence theorem, evaluate the surface integral ( yz dy dz zx dz dx xy dy dx) where S : x2 + y2 + z2 = 4. S (AMIETE, Dec. 2010, UP, I Sem., Dec 2008) Solution. S ( f1 dy dz f 2 dx dz f3 dx dy ) f1 f 2 f3 dx dy dz y z where S is closed surface bounding a volume V. v x = S ( yz dy dz zx dx dz xy dx dy ) = ( yz ) ( zx ) ( xy ) dx dy dz = x y z v v (0 0 0) dx dy dz = 0 Example 115. Evaluate xz 2 dy dz ( x 2 y z 3 ) dz dx (2 xy y 2 z ) dx dy S where S is the surface of hemispherical region bounded by z = Solution. a 2 x2 y 2 and z = 0. S ( f1 dy dz f 2 dz dx f3 dx dy ) = f1 V x where S is a closed surface bounding a volume V. S xz = 2 f 2 f3 dx dy dz y z dy dz ( x 2 y z 3 ) dz dx (2 xy y 2 z ) dx dy V x ( xz 2 ) 2 (x y z3) (2 xy y 2 z ) dx dy dz y z (Here V is the volume of hemisphere) 2 2 2 = V ( z x y ) dx dy dz Let x = r sin cos , y = r sin sin , z = r cos = Ans. 2 2 r (r sin dr d d ) = 5 2 / 2 r ( ) ( cos ) 0 0 = 5 Example 116. Evaluate S F . nˆ ds 2 0 a d 2 sin d r 4 dr 0 0 a a5 2 a5 = 2 ( 0 1) = 5 5 0 Ans. over the entire surface of the region above the xy-plane bounded by the cone z2 = x2 + y2 and the plane z = 4, if F = 4 xz iˆ xyz 2 ˆj 3z kˆ. Solution. If V is the volume enclosed by S, then V is bounded by the surfaces z = 0, z = 4, z2 = x2 + y2. By divergence theorem, we have S F . nˆ ds = V div F dx dy dz = V x (4 xz ) y ( xyz = V (4 z xz Limits of z are 2 2 ) (3 z ) dx dy dz z 3) dx dy dz x 2 y 2 and 4. Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ 460 Vectors = 4 2 xz 3 3z (4 z xz 3) dz dy dx = 2 z 2 2 3 x y 4 dy dx 2 32 x2 y 2 64 x 12 {2( x 2 y 2 ) x ( x 2 y 2 )3 / 2 3 x 2 y 2 } dy dx 3 64 x 2( x 2 y 2 ) x( x 2 y 2 )3/ 2 3 x 2 y 2 dy dx 3 Putting x = r cos and y = r sin , we have 64 r cos 2r 2 r cos r 3 3r r d dr = 44 3 Limits of r are 0 to 4. and limits of are 0 to 2 2 4 64 r 2 cos 44 r 2r 3 r 5 cos 3r 2 d dr = 0 0 3 = = 44 0 4 64 r 3 cos r 4 r 6 cos r 3 d 22r 9 2 6 0 2 2 64 (4)3 cos (4)4 (4)6 2 cos (4)3 d 22(4) 9 2 6 6 2 64 64 (4) cos 128 cos 64 d = 0 352 9 6 6 64 64 (4) 2 cos d = 0 160 9 6 2 = 0 64 64 (4)6 = 160 9 6 = 320 2 64 64 (4)6 sin = 160 (2) sin 2 6 0 9 Ans. Example 117. The vector field F x 2iˆ zjˆ yzkˆ is defined over the volume of the cuboid given by 0 x a, 0 y b, 0 z c, enclosing the surface S. Evaluate the surface integral S F . d s (U.P., I Semester, Winter 2001) Solution. By Divergence Theorem, we have 2 2 ( x iˆ z ˆj yz kˆ) . ds div ( x iˆ z ˆj yz kˆ) dv, S v where V is the volume of the cuboid enclosing the surface S. 2 ˆ ˆj kˆ . ( x iˆ z ˆj yz k ) dv = v iˆ y z x 2 ( z) ( y z ) dx dy dz y z = v x ( x = x 0 y 0 z 0 (2 x y ) dx dy dz = a b ) c a b a b 0 0 0 0 = a b c 0 dx 0 dy 0 (2 x y) dz c dx [2 xz yz] 0 dy dx (2 xc yc) dy Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ Vectors 461 b a b a a y2 b2 = c dx (2 x y ) dy c 2 xy dx dx c 2bx 2 2 0 0 0 0 0 a 2bx 2 b 2 x 2 ab2 b = c c a b abc a 2 0 2 2 2 Ans. Example 118. Verify the divergence Theorem for the function F = 2 x2yi – y2j + 4 x z2k taken over the region in the first octant bounded by y2 + z2 = 9 and x = 2. Solution. V F dV = iˆ x ˆj y kˆ z (2 x = (4 xy 2 y 8 xz ) dx dy dz = = 0 dx 0 dy (4 xyz 2 yz 4 xz = 0 dx 0 2 3 2 3 2 2 0 )0 2 yiˆ y 2 ˆj 4 xz 2 k ) dV 9 y2 3 dx dy 0 0 (4 xy 2 y 8 xz ) dz 9 y2 [4 xy 9 y 2 2 y 9 y 2 4 x (9 y 2 )] dy 3 4x 2 2 4 xy 3 2 3/ 2 2 3/ 2 dx (9 y ) (9 y ) 36 xy = 0 3 3 0 2 3 2 2 2 x2 108 18 x (0 0 108 x 36 x 36 x 18) dx (108 x 18) dx = 0 = 0 = 2 0 = 216 – 36 = 180 ...(1) 2 Here S F nˆ ds = F nˆ ds = BDEC Normal vector F nˆ ds OABC 2 F nˆ ds OCE F nˆ ds OADE (2 x yiˆ y 2 ˆj 4 xz 2 kˆ) . nˆ ds F nˆ ds ABD F nˆ ds BDEC BDEC ˆj kˆ (y2 + z2 – 9) = = iˆ y z x = 2 yjˆ 2 z kˆ 2 yjˆ 2 zkˆ yjˆ zkˆ Unit normal vector = n̂ = = 2 2 4y 4z y2 z 2 yjˆ zkˆ yjˆ zkˆ = = 3 9 ˆ zk yj 1 2 2 2 (2 x yiˆ y ˆj 4 xz k ) 3 ds = 3 ( y3 4 xz3 ) ds BDEC BDEC yjˆ zkˆ z dx dy kˆ ds or ds dx dy ds (nˆ k ) ds z 3 3 3 2 3 y3 1 3 3 dxdy dx 4 xz 2 dy ( y 4 xz ) = = 0 0 z 3 z BDEC 2 = 0 dx y 3 sin , z 3 cos 3 2 0 3 27 sin 4 x (9 cos 2 ) 3 cos Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ 462 Vectors = 2 2 2 0 dx 27 3 108 x 3 = 2 0 ( 18 72 x) dx 2 2 = 18 x 36 x = 108 0 2 ˆ 2 2ˆ ˆ F nˆ ds = (2 x yi y ˆj 4 xz k ) ( k ) ds ...(2) OABC OABC = 4 x z 2 ds = 0 (2 x 2 yiˆ y 2 ˆj 4 xz 2 kˆ) ( ˆj ) ds = ...(3) because in OABC xy-plane, z = 0 OABC F nˆ ds = OADE (2 x 2 yiˆ y 2 ˆj 4 xz 2 kˆ) ( iˆ) ds = OCE F nˆ ds = (2 x 2 yiˆ y 2 ˆj 4 xz 2 kˆ) (iˆ) ds = ABD = 2 x 2 2 x 2 y ds = 0 OCE ABD ...(4) because in OADE xz-plane, y = 0 F nˆ ds = y 2 ds = 0 OADE OADE OCE 9 z2 3 y dy dz = 0 dz 0 ...(5) because in OCE yz-plane, x = 0 2 x2 y ds ABD 2 (2)2 y dy because in ABD plane, x = 2 9 z2 3 y2 3 z3 = 8 dz = 4 dz (9 z 2 ) = 4 9 z = 4 [27 – 9] = 72 0 0 3 2 0 0 On adding (2), (3), (4), (5) and (6), we get 3 S F nˆ ds = 108 + 0 + 0 + 0 + 72 = 180 V F dV From (1) and (7), we have = ...(6) ...(7) S F nˆ ds Hence the theorem is verified. Example 119. Verify the Gauss divergence Theorem for 2 2 2 F = (x – yz) iˆ + (y – zx) ĵ + (z – xy) k̂ taken over the rectangular parallelopiped 0 x a, 0 y b, 0 z c. (U.P., I Semester, Compartment 2002) Solution. We have 2 2 2 ˆ ˆj kˆ [( x yz ) iˆ ( y zx) ˆj ( z xy ) k ] div F = F = iˆ x y z 2 2 2 ( x yz ) ( y zx) ( z xy ) = 2x + 2y + 2z = x y z Volume integral = V F dV a b = V 2 ( x y z ) dV c a b c = 2 x 0 y 0 z 0 ( x y z ) dx dy dz = 2 0 dx 0 dy 0 ( x y z ) dz a = 2 0 dx b 0 c z2 a dy xz yz = 2 dx 2 0 0 b b 0 dy cx cy c2 2 a y 2 c2 y a b2 c b c 2 = 2 dx bcx = 2 0 dx c x y c 0 2 2 0 2 2 Created with Print2PDF. To remove this line, buy a license at: http://www.software602.com/ Vectors 463 a bc x 2 b 2 cx bc 2 x 2 = [a2bc + ab2c + abc2] = 2 2 0 2 = abc (a + b + c) ...(A) To evaluate S F nˆ ds, where S consists of six plane surfaces. S F nˆ ds = OABC F nˆ ds = OABC F nˆ ds DEFG F nˆ ds OAFG F nˆ ds BCDE F nˆ ds OABC {( x ABEF F nˆ ds OCDG F nˆ ds yz ) iˆ ( y 2 xz ) ˆj ( z 2 xy ) kˆ} ( kˆ ) dx dy 2 a b 2 = ( z xy ) dx dy 0 0 ab a 2 b2 4 = (0 xy ) dx dy = 00 ...(1) DEFG F nˆ ds = DEFG {( x 2 yz ) iˆ ( y 2 xz ) ˆj ( z 2 xy ) kˆ} (kˆ) dx dy ab = 2 ( z xy) dx dy = 00 a = c S.No. Surface Outward normal 1 OABC –k 2 DEFG k 3 OAFG –j 4 BCDE j 5 ABEF i 6 OCDG –i ab 2 (c xy ) dx dy 00 2 y 0 2 b xy dx = 2 0 a 2 c b 0 xb 2 2 dx a 2 x2 b2 a2 b2 c bx = abc 2 = ...(2) 4 0 4 {( x 2 yz ) iˆ ( y 2 zx) ˆj ( z 2 xy ) kˆ} ( ˆj ) dx dz F nˆ ds = OAFG ds dx dy dx dy dx dz dx dz dy dz dy dz z=0 z=c y=0 y=b x=a x=0 OAFG 2 = OAFG ( y zx ) dx dz a c a = dx (0 zx ) dz = 0 0 x z2 dx 2 0 c = 0 a a x2 c 2 x c2 a 2 c2 2 dx = 4 = 4 0 0 BCDE F nˆ ds = {( x2 y z ) iˆ ( y 2 zx) ˆj ( z 2 xy) kˆ} ˆj dx dz = BCDE ( y a c c a 2 x z2 = dx (b x z ) dz = b z dx = 2 0 0 0 0 2 ...(3) 2 xz ) dx dz a 2 x c2 b c dx 2 0 a ABEF a 2 c2 x2 c2 2 = b2 c x = ab c 4 4 0 2 2 2 ˆ F