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If uu K A STROUD mm mm PROGRAMMES AND PROBLEMS PROGRAMMES AND PROBuEMS PROGRAMMES AND PROBLEMS PROGRAMMES AND PROBLEMS PROGRAMMES AND PROBLEMS hi ~ ENGINEERING MATHEMATICS Programmes and Problems K. A. Stroud MACMILLAN © K. A. Stroud 1970 All rights reserved. No part of this publication may be reproduced or transmitted, in any form or by any means, without permission. First published 1970 Published by MACMILLAN AND CO LTD London and Basingstoke Associated companies in New York, Toronto, Melbourne, Dublin, Johannesburg and Madras Printed by photo-lithography and made in Great Britain at the Pitman Press, Bath PREFACE The purpose of this book is to provide a complete year's course in mathematics for those studying in the engineering, technical and scientific fields. The material has been specially written for courses lead- ing to (i) Part I of B.Sc. Engineering Degrees, (ii) Higher National Diploma and Higher National Certificate in techno- logical subjects, and for other courses of a comparable level. While formal proofs are included where necessary to promote understanding, the emphasis throughout is on providing the student with sound mathematical skills and with a working knowledge and appreciation of the basic con- cepts involved. The programmed structure ensures that the book is highly suited for general class use and for individual self-study, and also provides a ready means for remedial work or subsequent revision. The book is the outcome of some eight years' work undertaken in the development of programmed learning techniques in the Department of Mathematics at the Lanchester College of Technology, Coventry. For the past four years, the whole of the mathematics of the first year of various Engineering Degree courses has been presented in programmed form, in conjunction with seminar and tutorial periods. The results obtained have proved to be highly satisfactory, and further extension and development of these learning techniques are being pursued. Each programme has been extensively validated before being produced in its final form and has consistently reached a success level above 80/80, i.e. at least 80% of the students have obtained at least 80% of the possible marks in carefully structured criterion tests. In a research programme, carried out against control groups receiving the normal lectures, students working from programmes have attained significantly higher mean scores than those in the control groups and the spread of marks has been con- siderably reduced. The general pattern has also been reflected in the results of the sessional examinations. The advantages of working at one's own rate, the intensity of the student involvement, and the immediate assessment of responses, are well known to those already acquainted with programmed learning activities. Programmed learning in the first year of a student's course at a college or university provides the additional advantage of bridging the gap between the rather highly organised aspect of school life and the freer environment and which puts greater emphasis on personal responsibility for his own pro- gress which faces every student on entry to the realms of higher education. Acknowledgement and thanks are due to all those who have assisted in any way in the development of the work, including those who have been actively engaged in validation processes. I especially wish to record my sincere lhanks for the continued encouragement and support which I received from my present Head of Department at the College, Mr. J. E. Sellars, M.Sc, A.F.R.Ae.S., F.I.M.A., and also from Mr. R. Wooldridge, M.C., B.Sc, F.I.M.A., formerly Head of Department, now Principal of Derby College of Technology. Acknowledgement is also made of the many sources, too numerous to list, from which the selected examples quoted in the programmes have been gleaned over the years. Their inclusion contributes in no small way to the success of the work. K. A. Stroud CONTENTS Preface v Hints on using the book xii Useful background information xiii Programme 1 : Complex Numbers, Part 1 Introduction: The symbol j; powers ofj; complex numbers 1 Multiplication of complex numbers Equal complex numbers Graphical representation of a complex number Graphical addition of complex numbers Polar form of a complex number Exponential form of a complex number Test exercise I Further problems I Programme 2: Complex Numbers, Part 2 Introduction 37 Loci problems Test exercise II Further problems II Programme 3: Hyperbolic Functions Introduction 73 Graphs of hyperbolic functions Evaluation of hyperbolic functions Inverse hyperbolic functions Log form of the inverse hyperbolic functions Hyperbolic identities Trig, identities and hyperbolic identities Relationship between trigonometric & hyperbolic functions Test exercise III Further problems HI Programme 4: Determinants Determinants \q\ Determinants of the third order Evaluation of a third order determinant Simultaneous equations in three unknowns Consistency of a set of equations Properties of determinants vii Test exercise IV Further problems IV Programme 5: Vectors Introduction: Scalar and vector quantities 141 Vector representation Two equal vectors Types of vectors Addition of vectors Components of a given vector Components of a vector in terms of unit vectors Vectors in space Direction cosines Scalar product of two vectors Vector product of two vectors Angle between two vectors Direction ratios Summary Test exercise V Further problems V ^/Programme 6: Differentiation Standard differential coefficients 1 7 1 Functions of a function Logarithmic differentiation Implicit functions Parametric equations Test exercise VI Further problems VI Programme 7: Differentiation Applications, Part 1 Equation of a straight line 195 Centre of curvature Test exercise VII Further problems VII Programme 8: Differentiation Applications, Part 2 ^-Inverse trigonometrical functions 223 Differentiation of inverse trig, functions ^Differentiation coefficients of inverse hyperbolic functions — • Maximum and minimum values (turning points J Test exercise VIII Further problems VIII Programme 9: Partial Differentiation, Part 1 Partial differentiation 25 1 Small increments Test exercise IX Further problems IX Programme 10: Partial Differentiation, Part 2 Partial differentiation 277 Rates of change problems Change of variables Test exercise X Further problems X Programme 1 1 : Series, Part 1 Series 297 Arithmetic and geometric means Series of powers of natural numbers Infinite series: limiting values Convergent and divergent series Tests for convergence; absolute convergence Test exercise XI Further problems XI Programme 1 2: Series, Part 2 — Power series, Maclaurin 's series 327 Standard series The binomial series Approximate values Limiting values Test exercise XII Further problems XII ^Programme 13: Integration, Part 1 Introduction 357 Standard integrals Functions of a linear function Integrals of the form Integration of products - integration by parts Integration by partial fractions Integration of trigonometrical functions . Test exercise XIII Further problems XIII Programme 14: Integration, Part 2 Test exercise XIV 389 Further problems XIV Programme 15: Reduction Formulae Test exercise XV 419 Further problems XV 1/^Programme 16: Integration Applications, Part 1 x^Parametric equations 435 \^Mean values *-^k.m.s. values Summary sheet Test exercise XVI Further problems XVI Programme 17: Integration Applications, Part 2 Introduction 457 Volumes of solids of revolution Centroid of a plane figure Centre of gravity of a solid of revolution Lengths of curves Lengths of curves - parametric equations Surfaces of revolution Surfaces of revolution - parametric equations Rules of Pappus Revision summary Test exercise XVII Further problems XVII Programme 18: Integration Applications, Part 3 Moments of inertia 483 Radius of gyration Parallel axes theorem Perpendicular axes theorem Useful standard results Second moment of area Composite figures Centres of pressure Depth of centre of pressure Test exercise XVIII Further problems XVIII ^-''Programme 19: Approximate Integration t- Introduction 517 j. Approximate integration 1 Method 1 — by series s/ftethod 2 - Simpson 's rule \ftoof of Simpson 's rule Test exercise XIX Further problems XIX Programme 20: Polar Co-ordinates Systems Introduction to polar co-ordinates 539 Polar curves Standard polar curves Test exercise XX Further problems XX Programme 21: Multiple Integrals Summation in two directions 565 Double integrals: triple integrals Applications Alternative notation Determination of volumes by multiple integrals Test exercise XXI Further problems XXI Programme 22: First Order Differential Equations Introduction 593 Formation of differential equations Solution of differential equations Method 1 - by direct integration Method 2 - by separating the variables Method 3 — homogeneous equations: by substituting y = vx Method 4 - linear equations: use of integrating factor Test exercise XXII Further problems XXII Programme 23: Second Order Differential Equations with Constant Coefficients Test exercise XXIII 637 Further problems XXIII Programme 24: Operator D Methods The operator D 70 1 Inverse operator 7/D Solution of differential equations by operator D methods Special cases Test exercise XXIV Further problems XXIV Answers 707 Index 744 xi HINTS ON USING THE BOOK This book contains twenty-four lessons, each of which has been written in such a way as to make learning more effective and more interesting. It is almost like having a personal tutor, for you proceed at your own rate of learning and any difficulties you may have are cleared before you have the chance to practise incorrect ideas or techniques. You will find that each programme is divided into sections called frames, each of which normally occupies half a page. When you start a programme, begin at frame 1. Read each frame carefully and carry out any instructions or exercise which you are asked to do. In almost every frame, you are required to make a response of some kind, testing your understanding of the information in the frame, and you can immediately compare your answer with the correct answer given in the next frame. To obtain the greatest benefit, you are strongly advised to cover up the following frame until you have made your response. When a series of dots occurs, you are expected to supply the missing word, phrase, or number. At every stage, you will be guided along the right path. There is no need to hurry: read the frames carefully and follow the directions exactly. In this way, you must learn. At the end of each programme, you will find a short Test Exercise. This is set directly on what you have learned in the lesson: the questions are straightforward and contain no tricks. To provide you with the necessary practice, a set of Further Problems is also included: do as many of these problems as you can. Remember that in mathematics, as in many other situations, practice makes perfect — or more nearly so. Even if you feel you have done some of the topics before, work steadily through each programme: it will serve as useful revision and fill in any gaps in your knowledge that you may have. USEFUL BACKGROUND INFORMATION I. Algebraic Identities (a + bf = a 2 + 2ab+b 2 (a + bf = a 3 + 3a 2 b + 3ab 2 + b 3 (a - bf =a 2 - Ixib + b 2 (a- b) 3 =a 3 ~ 3a 2 b + 3ab 2 - b 3 (a + bf = a 4 + 4a 3 b + 6a 2 b 2 + 4ab 3 + 6 4 (a - Z>) 4 = a 4 - 4a 3 * + 6a 2 Z> 2 - 4ab 3 + b 4 a 2 -b 2 = (a- b) (a + b). a 3 -b 3 = (a- b) (a 2 ±ab + b 2 ) a 3 + b 3 =(a + b)(a 2 -ab + b 2 ) II. Trigonometrical Identities (1) sin 2 + cos 2 = 1 ; sec 2 = 1 + tan 2 0; cosec 2 = 1 + cot 2 (2) sin (A + B) = sin A cos B + cos A sin B sin (A — B) = sin A cos B - cos A sin B cos (A + B) = cos A cos B - sin A sin B cos (A - B) = cos A cos B + sin A sin B * a iD ^ tan A + tan B tan (A + B) = -= — — — - 1 - tan A tan B ,. _. tan A - tan B tan (A - B) _ 1 + tan A tan B (3) Let A = B = 0. .'. sin 20 = 2 sin cos cos 20 =cos 2 0-sin 2 = 1-2 sin 2 = 2 cos 2 - 1 , na _ 2tan0 tan 20 - j-^^ xin (4) Letfl=| ;. sin = 2sin|cos| ra 2 ■ 2^ cos = cos^ — - sin 2 — = l-2sin'f .2 = 2cos 2 ^- 1 2 tan § tan0= 2 1-tan^- (5) sin C + sin D = 2 sin — - — cos . n . „ . C+D . C-D sin C - sin D = 2 cos — - — sin n,. r. -, C + D C-D cos C + cos D = 2 cos — — cos 2 2 p. „ . . C + D . C-D cos D - cos C = 2 sin — - — sin — - — (6) 2 sin A cos B = sin (A + B) + sin (A - B) 2 cos A sin B = sin (A + B) - sin (A - B) 2 cos A cos B = cos (A + B) + cos (A - B) 2 sin A sin B = cos (A - B) - cos (A + B) (7) Negative angles: sin (-6) = -sin 9 cos (-8) = cos 9 tan (-6) = -tan 6 (8) Angles having the same trig, ratios: (i) Same sine: 6 and (180° -6) (ii) Same cosine: 6 and (360°- 9), i.e. (-0) (hi) Same tangent: 6 and (180° + 9) xiv (9) a sin + b cos = A sin (0 + a) a sin - b cos = A sin (0 - a) a cos + b sin = A cos (0 - a) fl cos - 6 sin = A cos (0 + a) [A = vV + 6 2 ) where : • (a=tan- 1 |(0°<a<90°) III. Standard Curves (1) Straight line: m 2 -mi Slope, m = *! = ZiZZi dx x 2 -x l Angle between two lines, tan , 1 + m l m 2 For parallel lines, m 2 = m 1 For perpendicular lines, m l m 2 = -1 Equation of a straight line (slope = m) (i) Intercept c on real j-axis: y = mx + c (ii) Passing through (x j, .yj): ^-^i=w(x-Xi) (iii) Joining (x 1 ,y l ) and (x 2 ,.y 2 ): J-^i (2) 0>c/e: Centre at origin, radius r: x 2 + y 2 = r 2 Centre (h,k), radius /•: (x-h) 2 + (y-k) 2 =r 2 General equation: x 2 +y 2 + 2gx + 2fy + c = with centre (-g, -/); radius = \/fe 2 + / 2 - c) Parametric equations : x = r cos , y = r sin (3) Parabola: Vertex at origin, focus (a, 0): y 2 = Aax Parametric equations: x = at 2 , y = 2at xv (4) Ellipse: 2 2 Centre at origin, foci (±\J[a 2 - b 2 ] , 0): ^ +^7= 1 where a = semi major axis, b = semi minor axis Parametric equations: x = a cos 6, y = b sin 8 (5) Hyperbola: x 2 y 2 Centre at origin, foci (± \/a 2 + b 2 , 0): — -p" = 1 Parametric equations: x = a sec 6, y = b tan 6 Rectangular hyperbola: 2 Centre at origin, vertex±/ y-, y-") : xy = — = c 2 where c = -r- i.e. xy = c 2 Parametric equations: x-ct, y = c/t xvi Programme 1 COMPLEX NUMBERS PART1 Programme 1 1 Introduction: the symbol j The solution of a quadratic equation ax 2 + bx + c = can, of course, be u* • au *u f i _ -6±V(6 2 -4gc) obtained by the formula, x = = - For example, if 2x 2 + 9x + 7 = 0, then we have -9±V(81-56) _ -9±V25 -9 + 5 X 4 4 " 4 • v -_4 14 " *" 4 0r ~T :. x =-1 or -3-5 That was straight-forward enough, but if we solve the equation 5x 2 - 6x + 5 = in the same way, we get _ 6 ± \/(36 - 100) ^ 6 ± V(-64) * TO 10 and the next stage is now to determine the square root of (-64). Is it (i)8, (ii)-8, (iii) neither? neither It is, of course, neither, since + 8 and - 8 are the square roots of 64 and not of (—64). In fact, y/(— 64) cannot be represented by an ordinary number, for there is no real number whose square is a negative quantity. However, -64 = "1 X 64 and therefore we can write V(-64) = V(-l X 64) = V(-lK/64 - 8 V(-l)" i.e. V(-64) = 8V(-1) Of course, we are still faced with V( - l), which cannot be evaluated as a real number, for the same reason as before, but, if we write the letter j to stand for VHX then V(~64) = >/(-l) . 8 = j8. So although we cannot evaluate V(-l). we can denote it by j and this makes our working a lot neater. V(-64) = V(-DV64=j8 Similarly, V(~36) = s/(-l )V36 = j6 V(- 7) = V(-1)V 7=j2-646 So V( — 25) can be written Complex numbers 1 J5 We now have a way of finishing off the quadratic equation we started in frame 1. 5x 2 - 6x + 5 = ■ x = 6±V(36-100) _ 6 ± V(~64) 5* tx + i u - x To To : ' x= ^JcT " x = °' 6 * j0 ' 8 .'. x = 0-6+j0-8 or x = 0-6-j0-8 We will talk about results like these later. For now, on to frame 4. Powers of j Since j stands for V( _ l), let us consider some powers of j. j =V(-D 1 ;2v = r=u 2 )i=-i.j = -j j 4 =0 2 ) 2 =(-i) 2 = i j =V(-i) j 2 =-i f=-j j 4 i Note especially the last result: j 4 = 1 . Every time a factor j 4 occurs, it can be replaced by the factor 1 , so that the power of j is reduced to one of the four results above. e.g. j 9 =(j 4 ) 2 j = (l) 2 j = l-j=j j2o =(j 4 )5 =(1)S =1 j 30 =G 4 )'j a =(l) 7 (-l)=l(-l)' and j 15 =(j 4 ) 3 J 3 = lH) = -j So, in the same way, j 5 Programme 1 ) since j s =(j 4 )j = 1 j=j Every one is done in the same way. j 6 =(j 4 )J 2 = Kj 2 )=l(-l) = -l j 7 =G 4 )J 3 = iH) = -J j 8 = (j 4 ) 2 =0) 2 = i So (i) j 42 = 00 j 12 = (iii) J U = and (iv) If x 2 - 6x + 34 = 0, x = (i) -1, (ii) 1, (iii) -j, (iv)x = 3±j5 The working in (iv) is as follows : x 2 - 6x + 34 = .'. x ■■ 6 ± V(36 - 136) _ 6 + V(-100) ,x-l^-3 ±j 5 i.e. * = 3+j5 or x = 3-j5 So remember, to simplify powers of j, we take out the highest power of' j 4 that we can, and the result must then simplify to one of the four results: j, — 1, -j, 1. Turn on now to frame 7. Complex numbers J Complex numbers The result x = 3 + j5 that we obtained, consists of two separate terms, 3 and j5. These terms cannot be combined any further, since the second is not a real number (due to its having the factor j). In such an expression as x = 3 + j5, 3 is called the real part of x 5 is called the imaginary part of x and the two together form what is called a complex number. So, a Complex number = (Real part) + j(Imaginary part) In the complex number 2+j7, the real part = and the imaginary part = real part = = 2; imaginary part = = 7 (NOTJ7!) Complex numbers have many applications in engineering. To use them, we must know how to carry out the usual arithmetical operations. 1 . Addition and Subtraction of Complex Numbers. This is easy, as one or two examples will show. Example 1 (4 +j5) + (3-j2). Although the real and imaginary parts cannot be combined, we can remove the brackets and total up terms of the same kind. (4 + j5) + (3 - j2) = 4 + j5 + 3 - j2 = (4 + 3) + j(5 - 2) = 7+j3 Example 2 (4+j7)-(2-j5) = 4+j7-2+j5 = (4-2)+j(7 + 5) = 2+jl2 So, in general, (a + ]b) + (c + )d) = (a + c) + j(b + d) Now you do this one: (5+j7) + (3-j4)-(6-j3)= 7 8 2+j6 Programme 1 since (5+j7) + (3-j4)-(6-j3) = 5+j7 + 3-j4-6+j3 = (5+3-6)+j(7-4 + 3) = 2+j6 Now you do these in just the same way: (i) (6+j5)-(4-j3) + (2-j7) = and (ii) (3+j5)-(5-j4)-(-2-j3) = 10 (i)4+j (ii)jl2 Here is the working: (i) (6+j5)-(4-j3) + (2-j7) = 6+J5-4+J3 + 2-J7 = (6-4 + 2)+j(5+3-7) = 4+j (ii) £+j5)-(5-j4)-(-2-j3) = 3+J5-5+J4 + 2+J3 = (3- 5 + 2) +j(5 +4 + 3) = 0+J12 = j!2 (Take care with signs!) This is very easy then, so long as you remember that the real and the imaginary parts must be treated quite separately — just like x's andy's in an algebraic expression. On to frame 11. Complex numbers 1 2. Multiplication of Complex Numbers Example: (3 + j4) (2 + j5) These are multiplied together in just the same way as you would deter- mine the product (3x + Ay) (2x + 5y). Form the product terms of (i) the two left-hand terms (ii) the two inner terms (iii) the two outer terms 14 (iv) the two right-hand terms 11 (3 + J4) (2 + j5) ■ .2. "3" 6+j8 +jl5+j 2 20 6+J23-20 (since j 2 =-l) -14+J23 Ukewise, (4-j5)(3 +]2) 12 22 -j7 for: (4-j5)(3+j2)=12-jl5+j8-j 2 10 = 12-J7 + 10 (j 2 =-l) = 22-j7 If the expression contains more than two factors, we multiply the factors together in stages: (3+j4)(2-j5)(l-j2) = (6+j8-jl5-j 2 20)(l-j2) = (6-j7 + 20)(l-j2) = (26-j7)(l-j2) Finish it off. Programme 1 13 12-J59 for: (26-j7)(l-j2) = 26-j7-j52+j 2 14 = 26- j59- 14 = 12-J59 Note that when we are dealing with complex numbers, the result of our calculations is also, in general, a complex number. Now you do this one on your own. (5+j8)(5-j8)= 14 89 Here it is: (5 + j8) (5 - j8) = 25 +j40-j40-j 2 64 = 25+64 = 89 In spite of what we said above, here we have a result containing no j term. The result is therefore entirely real. This is rather an exceptional case. Look at the two complex numbers we have just multiplied together. Can you find anything special about them? If so, what is it? When you have decided, turn on to the next frame. Complex numbers 1 They are identical except for the middle sign in the brackets, i.e. (5+j8) and (5-j8) A pair of complex numbers like these are called conjugate complex numbers and the product of two conjugate complex numbers is always entirely real. Look at it this way — (a + b) (a-b) = a 2 - b 2 Difference of two squares Similarly (5 + j8) (5 -j8) = 5 2 -(j8) 2 =5 2 -j 2 8 2 = 5 2 +8 2 (j 2 =-l) = 25 + 64 = 89 Without actually working it out, will the product of (7 - j6) and (4+j3)be (i) a real number (ii) an imaginary number (iii) a complex number a complex number 15 16 since (7 -j6) (4 + j3) is a product of two complex numbers which are not conjugate complex numbers. Remember: Conjugate complex numbers are identical except for the signs in the middle of the brackets. (4 + j5) and (4 - j5) are conjugate complex numbers (a +]b) and (a -\b) are conjugate complex numbers but (6 +j2)and(2 +j6) are not conjugate complex numbers (5 - j3) and (-5 + j3) are not conjugate complex numbers So what must we multiply (3 - j2) by, to produce a result that is entirely real? Programme 1 17 3+j2 because the conjugate of (3 - j2) is identical to it, except for the middle sign, i.e. (3 + j2), and we know that the product of two conjugate com- plex numbers is always real. Here are some examples: Example 1 • (3 -j2)(3 +j2) = 3 2 -(j2) 2 = 9 - j 2 4 = 9 + 4=13 Example 2 (2 + j7) (2 - j7) = 2 2 - G7) 2 = 4 - j 2 49 = 4+49 = 53 . . . and so on. Complex numbers of the form (a + }b) and (a -]b) are called complex numbers. 18 conjugate Now you should have no trouble with these— (a) Write down the following products (i) (4-j3)(4+j3) (ii) (4+j7)(4-j7) (hi) (fl+j6)(a-j6) (iv) (x-iy)(x+jy) (b) Multiply (3 - j5) by a suitable factor to give a product that is entirely real. When you have finished, move on to frame 1 9. Complex numbers 1 Here are the results in detail. (a) (i) (4-J3) (4+j3) = 4 2 -j 2 3 2 = 16 + 9 = (ii) (4+j7)(4-j7) = 4 2 -j 2 7 2 =16+49 = (iii) (a + ]b) (a - ]b) = a 2 -) 2 b 2 = a 2 + b 2 25 65 (iv) (x - ]y) (x + ]y) = x 2 - ) 2 y 2 = x 1 + y 2 19 (b) To obtain a real product, we must multiply (3 - j5) by its conjugate, i.e. (3 +j5), giving (3-j5)(3 +j5) = 3 2 -j 2 5 2 =9 + 25 = 34 Now move on to the next frame for a short revision exercise. Revision exercise. 1. Simplify (i) j 12 (ii) j'° (iii) j« 2. Simplify: (i) (5-j9)-(2-j6) + (3-j4) (ii) (6-j3)(2+j5)(6-j2) (iii) (4-J3) 2 (iv) (5-j4)(5+j4) 3. Multiply (4 - j3) by an appropriate factor to give a product that is entirely real. What is the result? 20 When you have completed the exercise, turn on to frame 21. ^ 10 £\ Here are the results. Check yours = 12 _ /;4\3 _ 1 3 1. 2. -1 GO j 10 = (j 4 ) 2 j 2 = i 2 (-D= (m) j" = o 4 ) 5 j 3 =j 3 =[T] (i) (5-j9)-(2-j6) + (3-j4) = 5-j9-2+j6 + 3-j4 = (5-2 + 3)+j(6-9-4) = (ii) (6-j3)(2+j5)(6-j2) = (12-j6+j30-j 2 15)(6~j2) = (27+j24)(6-j2) ,_ = 162 +J144-J54 + 48 (iii) (4-J3) 2 = 16-J24-9 6-J7 210+J90 7-J24 (iv) (5-j4)(5+j4) = 25 -j 2 16 = 25 + 16 = 41 Programme 1 Required factor is the conjugate of the given complex number. (4-j3)(4+j3)=16 + 9: 25 All correct? Right. Now turn on to the next frame to continue the programme. 11 Complex numbers 1 Now let us deal with division. ££ Division of a complex number by a real number is easy enough. 5 -J 4 = 5 -4=1.67-jl.33 But how do we manage with 3 3 J 3 7-j4 r 4+j3 ' If we could, somehow, convert the denominator into a real number, we could divide out as in the example above. So our problem is really, how can we convert (4 + j3) into a completely real denominator — and this is where our last piece of work comes in. We know that we can convert (4 + j3) into a completely real number by multiplying it by its c Conjugate i.e. the same complex number but with the opposite sign in the middle, in the case (4 — j3) nDDDanDnnnannnnDDnnnanDDDDnnnDnnanaDan But if we multiply the denominator by (4 - j3), we must also multiply the numerator by the same factor. 7-j4 = (7-j4)(4-j3) = 28-j37-12 _ 16-J37 4+J3 (4+j3)(4-j3) 16 + 9 25 if-j|=0.64-jl48 and the job is done. To divide one complex number by another, therefore, we multiply numerator and denominator by the conjugate of the denominator. This will convert the denominator into a real number and the final step can then be completed. 4-i5 Thus, to simplify . we shall multiply top and bottom by 23 12 Programme 1 24 the conjugate of the denominator, i.e. (1 - j2) DnnannnnDannnnDDnnDDDnnaQDDDDDnnDnnnnD If we do that, we get: 4-j5 _ (4-j5)(l-j2) = 4-J13-10 1+J2 (l+j2)(l-j2) 1+4 — 6 — j 1 3 -6 .13 = -l-2-j2-6 Now here is one for you to do: Simplify 3 +j2 1-J3 When you have done it, move on to the next frame. 25 Result -0-3+jl-l 3+j2 _ (3+,j2)(l+j3) _ 3+jll-6 1-J3 (l-j3)(l+j3) 1+9 = -3+jll = _ Q . 3+jl , 1 10 naDnannDDDnDnnnnnDDDannnnDDDnnnnDDDnnD Now do these in the same way: (i) 4-J5 (ii) 3+j5 2-j v " _/ 5-J3 (2+j3)(l-j2) (iii) 3+j4 When you have worked these, turn on to frame 26 to check your results. 13 jl Complex numbers 1 m Results: Here are the solutions in detail. 26 I m 4-p_(4-j5)(2+j)_8-j6 + 5 1 K> 2-j (2-j)(2+j) 4 + 1 13- j6 _ 2-6-J1-2 (ii) (iii) 3+j5 _ (3+j5)(5+j3) = 15+J34-15 5-j3 (5-j3)(5+j3) 25+9 (2+j3)(l-j2)_2-j+6_8-j (3+J4) 3+J4 3+J4 _ (8-j)(3-j4) (3+j4)(3-j4) . • _24-j35-4_20-j35 9+16 25 = 0-8-J1-4 And now you know how to apply the four rules to complex numbers. Equal Complex Numbers O T Now let us see what we can find out about two complex numbers which we are told are equal. Let the numbers be a + \b and c +]d Then we have Re-arranging terms, we get a +}b = c + \d a-c=](d-b) In this last statement, the quantity on the left-hand side is entirely real, while that on the right-hand side is entirely imaginary, i.e. a real quantity equals an imaginary quantity! This seems contradictory and in general it just cannot be true. But there is one special case for which the statement can be true. That is when 14 Programme 1 28 each side is zero can be true only if a- c=\{d~b) a-c = 0, i.e. a = c and if d~b = 0, i.e. b = d So we get this important result: If two complex numbers are equal (i) the two real parts are equal (ii) the two imaginary parts are equal For example, if x + \y = 5 + j4, then we know* = 5 and.y = 4 and ifa + }b = 6-j3, thena = and b = . 29 a = 6 and b=~3 Be careful to include the sign! aDannnanDnnnnDDnnDnnnDannnDnnnaDnnnDDn Now what about this one? If (a + b)+)(a-b)= 1 +j 2, find the values of a and&. Well now, following our rule about two equal complex numbers, what can we say about (a + b) and (a - b)1 15 Complex numbers 1 « + 6 = 7 and a-b = 2 30 since the two real parts are equal and the two imaginary parts are equal. onaannnnannanaaaaaaaaaaaaaaanaaaaaauaa This gives you two simultaneous equations, from which you can deter- mine the values of a and 6. So what are they? a = 4-5; 6 = 2-5 31 For a + 6=7 a-b = 2 2a = 9 .'. a = 4-5 26 = 5 :. 6 = 2-5 DDDDDDDaDDDDDnDDDDDDDDDDDDDnnDDDDnnnDD We see then that an equation involving complex numbers leads to a pair of simultaneous equations by putting (i) the two real parts equal (ii) the two imaginary parts equal This is quite an important point to remember. 16 Programme 1 32 -3 +3 Graphical Representation of a Complex Number Although we cannot evaluate a complex number as a real number, we can represent it diagrammatically, as we shall now see. In the usual system of plotting numbers, the number 3 could be repre- sented by a line from the origin to the point 3 on the scale. Likewise, ^ a line to represent (-3) would be — j~ drawn from the origin to the point (-3). These two lines are equal in length but are drawn in opposite directions. Therefore, we put an arrow head on each to distinguish between them. A line which represents a magnitude (by its length) and direction (by the arrow head) is called a vector. We shall be using this word quite a lot. Any vector therefore must include both magnitude (or size) and 33 direction DnnDDDnaDnnnnnDnDnnnnnnnDnnDDDDDDDDnnn If we multiply (+3) by the factor (-1), we get (-3), i.e. the factor (--1) has the effect of turning the vector through 1 80° 180° +3 -3 -1 Multiplying by (-1) is equivalent to multiplying by j 2 , i.e. by the factor j twice. Therefore multiplying by a single factor j will have half the effect and rotate the vector through only o j3 \ x l -3 -2 -1 17 Complex numbers 1 90° 34 DDDnDnnDDDDDDnanDaanDnDDnDDDDnna a a a a a a The factor j always turns a vector through 90° in the positive direction of measuring angles, i.e. anticlockwise. xj y -2 -1 If we now multiply j3 by a further factor j, we get j 2 3, i.e. (-3) and the diagram agrees with this result. >«j If we multiply (-3) by a further factor j, sketch the new position of the vector on a similar diagram. Result: 35 J3 *, ~- 1 +3 Let us denote the two reference lines by XXi and YYj as usual. You will see that v ; Y <l (i) The scale on the X-axis represents real numbers. XX! is therefore called the real axis. (ii) The scale on the Y-axis represents imaginary numbers. YY, is therefore called the imaginary axis. On a similar diagram, sketch vectors to represent (i) 5, (ii) -4, (iii) j2, (iv) -j Programme 1 36 Results: Check that each of your vectors carries an arrow head to show direction. aaauaaoaauDaanannnuauanaaonannnnunnaaa If we now wish to represent 3 + 2 as the sum of two vectors, we must draw them as a chain, the second vector starting where the first one finishes. , ,,, , 2 j (3) H- -*■!. I 3+2=5 The two vectors, 3 and 2, are together equivalent to a single vector drawn from the origin to the end of the final vector (giving naturally that 3 + 2 = 5). Continue 37 If we wish to represent the complex number (3 + j2), then we add together the vectors which repre- sent 3 andj2. Notice that the 2 is now multi- plied by a factor j which turns that vector through 90°. The equivalent single vector to represent (3 + j2) is therefore the vector from the beginning of the first vector (origin) to the end of the last one. This graphical representation constitutes an Argand diagram. Draw an Argand diagram to represent the vectors (i) z, =2+j3 (h) z 2 =-3+j2 (iii) z 3 =4-j3 (iv) z 4 =^4-j5 Label each one clearly. 19 Complex numbers 1 Here they are. Check yours. z, = 2+j3 38 Note once again that the end of each vector is plotted very much like plotting x and y co-ordinates. The real part corresponds to the x-value. The imaginary part corresponds to the Rvalue. Move on to frame 39. Graphical Addition of Complex Numbers Let us find the sum of z, = 5 + j2 and z 2 = 2 + j3 by Argand diagram. If we are adding vectors, they must be drawn as a chain. We therefore draw at the end of z l , a vector AP repre- senting z 2 in magnitude and direction, i.e. AP = OB and is parallel to it. Therefore OAPB is a parallelogram. Thus the sum of z x and z 2 is given by the vector join- ing the starting point to the end of the last vector, i.e. OP. The complex numbers z% and z 2 can thus be added together by drawing the diagonal of the parallelogram formed by z x and z 2 . If OP represents the complex number a + jb, what are the values of a and b in this case? 39 20 Programme 1 40 :=5+2=7 Z> = 2 + 3 = 5 :. OP = z = 7+j5 You can check this result by adding (5 + j2) and (2 + j3) algebraically. DnDDDOODDDnnDDDDODDODDODDDDDDDDDDDDDDD So the sum of two vectors on an Argand diagram is given by the of the parallelogram of vectors. 41 diagonal aaaDaannDDnnDDDnnnDnDnnaDaDaaannnnnnaa How do we do subtraction by similar means? We do this rather craftily without learning any new methods. The trick is simply this: Z\ -2 2 ~Z\ +(-Z2) That is we draw the vector representing z, and the negative vector of z 2 and add them as before. The negative vector of z 2 is simply a vector with the same magnitude (or length) as z 2 but pointing in the opposite direction. e.g. Ifzi =5 +j2andz 2 =2+j3 vector OA = z t = 5 +j2 OP =-z 2 =-(2+j3) Then OQ = z 1 + (-z 2 ) = Zi -z 2 Determine on an Argand diagram (4 + j2) + (-2 + j3) ( 1 + j6) 21 Complex numbers 1 P (zi+r 2 ) (* 2 )B 42 1 X OA = z 1 =4+j2 OB =z 2 =-2+j3 OC=-z 3 = l-j6 Then OP = 2! + 2 2 OQ=Z! + z 2 -z 3 3-j Polar Form of a Complex Number It is convenient sometimes to express a complex number a + ]b in a differ- ent form. On an Argand diagram, let OP be a vector a + jb . Let r = length of the vector and 6 the \b angle made with OX. r = ^(a 2 +b 2 ) 6 = tan" 1 ^ 43 a = r cos and 6 = r sin Since z=a +jb, this can be written 2 = r cos + jr sin 6 i.e. z = r(cos 9 + j sin 6) This is called the polar form of the complex number a + ji, where Let us take a numerical example. r = \V + b 2 ) and = tan" 1 - n 22 Programme 1 44 Example: To express z = 4 + j3 in polar form. First draw a sketch diagram (that always helps) We can see that — (i) r 2 = 4 2 + 3 2 =16 + 9 = 25 (ii) tan0=^-=O-75 6 = 36°52 z = a + ]b = K cos + j sin 0) z = 5(cos36°52'+jsin36°52') So in this case Now here is one for you to do— Find the polar form of the complex number (2 + j3) When you have finished it, consult the next frame. 45 z = 3-606 (cos 56°19' + j sin 56°19') Here is the working z = 2 + j3 = /-(cos + j sin 0) r i =4 + 9=13 r = 3-606 tan0 =-= 1-5 = 56°19' z = 3-606 (cos 56°19'+ j sin 56°19') DDDnDDnDODnnnDnnnDDDnnDDDDDnDDDDnnnDDn We have special names for the values of r and . z = a + )b = r(cos 6 + j sin 6) (i) r is called the modulus of the complex number z and is often abbreviated to 'mod z' or indicated by \z\. Thus if z = 2 + j5, then|z| =V(2 2 + 5 2 ) =V(4 + 25) = V29 (ii) d is called the argument of the complex number and can be abbreviated to 'arg z'. So if z = 2 + j5, then argz = 23 Complex numbers 1 argz = 68°12' 46 z = 2 + j5. Then argz = 6 = tan" 1 1 = 68°12' □□DDDDDnnnnnanDnnannDnnnDnnnnnnnnaaaDD Warning. In finding 6, there are of course two angles between 0° and 360°, the tangent of which has the value | We must be careful to use the angle in the correct quadrant. Always draw a sketch of the vector to ensure you have the right one. e.g. Find argz when z =-3 -j4. is measured from OX to OP. We first find E the equivalent acute angle from the triangle shown. tan£ = |=1.333 .\ E = S3°i' Then in this case, = 18O o +£- = 233°7 argz = 233°7' Now you find arg (-5 + j2) Move on when finished. 21°48' x In this particular case, = 1 80° ~E :. = 158°12' □QnnannaDnoaDananonDDnoDDoaDDDoaDDQDOo Complex numbers in polar form are always of the same shape and differ only in the actual values of/- and 6. We often use the shorthand version r\d_\o denote the polar form. e.g. If Z = -5 + j2, r = V(25 + 4) = ^29 = 5-385 and from above 6 = 158°12' -•- The full polar form is z = 5-385 (cos 158°12' + j sin 158°12') and this can be shortened to z = 5-385 |158°12' Express in shortened form, the polar form of (4 - j3) Do not forget to draw a sketch diagram first. 24 Programme 1 48 r = v /(4 2 +3 2 ) r = 5 tan E = 0-75 /. £ = 36°52' = 360°-£ = 323°8' :. z = 5(cos 323°8' + j sin 323°8') = 5 |323°8 ' DDDnanDanDDOnDnaDDDDnDDDDDDDDnDDDDDDDD Of course, given a complex number in polar form, you can convert it into the basic form a + )b simply by evaluating che cosine and the sine and multiplying by the value of r. e.g. z = 5(cos 35° + j sin 35°) = 5(0-8192 + jO-5736) z = 4-0960 +J3-8680 Now you do this one- o Express in the forma + )b, 4(cos 65° + j sin 65 ) 49 z = 1-6904 +J3-6252 for z = 4(cos 65°+j sin 65> 4(0-4226 + J0-9063) = 1-6904 + J3-6252 nDDDDanDDDDnDDDClQ-DDQnDDDDDaaanDDDDDDDD If the argument is greater than 90°, care must be taken in evaluating the cosine and sine to include the appropriate signs, e.g. If z = 2(cos 210° + j sin 210°) the vector lies in the third quadrant. cos 210°= -cos 30° sin210° = -sin30° Then z = 2(-cos 30° -j sin 30 ) = 2(-0-8660-j0-5) = -l-732-j Here you are. What about this one? Express z = 5(cos 140° + j sin 140°) in the form a + }b What do you make it? 25 Complex numbers 1 z =-3-8300 +J3-2140 50 Here are the details - cos 140° = -cos 40° sin 140° = sin 40° z = 5(cos 140° + j sin 140°) = 5(-cos 40° + j sin 40°) = 5(-0-7660 + jO-6428) = -3-8300 +J3-2140 □DaaannnDnnDannnnDnDDnDnnnnnDnnnDnDDDn Fine. Now by way of revision, work out the following, (i) Express -5 + j4 in polar form (ii) Express 3 |300° in the form a + ]b When you have finished both of them, check your results with those on frame 51. r 2 =4 2 +5 2 = 16 + 25 = 41 -•- r = 6-403 tan E = 0-8 :. E = 38°40' -•- e = l41°20' -5 + j4 = 6-403(cos 141°20' + j sin 141°20') = 6-403 |141°20 ' Oi) 3 1300° = 3(cos 300° + j sin 300°) a cos 300° = cos 60° sin 300° = -sin 60° c 3 [300° = 3(cos 60° - j sin 60°) = 3(0-500 -J0-866) Turn to frame 52. 1-500-J2-598 51 26 Programme 1 52 We see then that there are two ways of expressing a c ° m P iex number ■■ (i) in standard form : z=a +)b (ii) in polar form: z = r(cos 6 + j sin 6) where r = \/(a 2 +b 2 ) and = tan" ■i b If we remember the simple diagram, we can easily convert from one system to the other. So on now to frame 53. 53 Exponential Form of a complex number. There is still another way of expressing a complex number which we must deal with, for it too has its uses. We shall arrive at it this way: Many functions can be expressed as series. For example, v 2 Y 3 x 4 x s e* = i+*+%+§7- + fT- + !r + -- sin* -x _ "3f + 3] 7! 9! '" — i X ,x X , cosx- 1 ~2! 4l 6! You no doubt have hazy recollections of these series You had better make a note of them since they have turned up. 27 Complex numbers 1 If we now take the series for e* and write j0 in place of x, we get Jl*| J 2! 3! 4! " •'♦*-!?-!?♦£♦■• • -(-£♦&- ) 2! 4! 4. ua _ 3! 5! +j( *-i; + g-.. ..) = cos + j sin Therefore, /-(cos 9 + j sin 0) can now be written as re > e . This is called the exponential form of the complex number. It can be obtained from the polar form quite easily since the r value is the same and the angle 6 is the same in both. It is important to note, however, that in the exponential form, the angle must be in radians. Move on to the next frame. 55 The three ways of expressing a complex number are therefore (i) z=a+]b (ii) z = r(cos + j sin 0) . . . . Polar form (iii) z = r.e) e Exponential form Remember that the exponential form is obtained from the polar form, (i) the r value is the same in each case. (ii) the angle is also the same in each case, but in the exponential form the angle must be in radians. So, knowing that, change the polar form 5(cos 60° + j sin 60°) into the exponential form. Then turn to frame 56. 28 Programme 1 56 Exponential form for we have 5c 5(cos60° +jsin60°) ■ n J 3 r=5 6 = 60° = | radians .'. Exponential form is 5 e DnDnnnDnnDnnnnaDDnnDDDDnDDnnDanaDnanDn And now a word about negative angles We know eJ e = cos 6 + j sin 6 If we replace 9 by ~6 in this result, we get e"J fl =cos(-0)+j sin(-0) = cos0 — j sinfl So we have t'i e = cos 6 + j sin 6 e~J e = cos 6 - j sin 6 Make a note of these. 57 There is one operation that we have been unable to carry out with complex numbers before this. That is to find the logarithm of a com- plex number. The exponential form now makes this possible, since the exponential form consists only of products and powers. For, if we have .. Then we can say e.g. If then In z = In r + j0 z = 6-42eJ 1 - 57 lnz = ln642+jl-57 = 1-8594 +J1-57 and the result is once again a complex number. And if z = 3-8e-J 0236 , then In z = 29 Complex numbers 1 lnz = ln3-8-jO-236 : 1-3350-J0-236 58 DnDDDnDnnDnDDDDDDnDDnDnDanDDnnDDDDDnDn Finally, here is an example of a rather different kind. Once you have seen it done, you will be able to deal with others of this kind. Here it is. Express e^ 4 in the form a + ]b Well now, we can write e'-J^ 4 as e^ 4 = e(cos tt/4 - j sin n/4) =e U _j >i} V2 (1 - j) This brings us to the end of this programme, except for the test exercise. Before you do that, read down the Revision Sheet that follows in the next frame and revise any points on which you are not completely sure Then turn on and work through the test exercise: you will find the questions quite straightforward and easy. But first, turn to frame 60. 59 30 Programme 1 60 Revision Summary 1 . Powers off i = V(-D, J 2 =-i> 3 3 =n, i 4 = i- A factor j turns a vector through 90° in the positive direction. 2. Complex numbers z = a + \b a = real part b= imaginary part 3. Conjugate complex numbers (a+jb) and (a-)b) The product of two conjugate complex numbers is always real. (a+)b)(a-]b)=a 2 + b 2 4. Equal complex numbers If a + }b = c + }d, then a = c and 6 = d. 5. Polar form of a complex number Y z = a + j& = r(cos0 +j sin0) = r\± r = V(a 2 +& 2 ); = tairl {l) a=rcos6; b=r sin 9 r = the modulus of z, written 'modz' or |z| d = the argument of z, written 'argz' 6. Exponential form of a complex number z=r(cosd +isin6) = rei e and r(cos0-jsin0) = re-J e 7. Logarithm of a complex number z = rei 6 :. lnz = lnr + j0 or if z = re~i e .'. Inz = lnr-j0 i /•IT >"' lb ^e It « a — *1 -X also in radians 31 Complex numbers 1 Test Exercise - I Kl 1. Simplify (i)j 3 , (ii)f, (iii)j 12 , (iv) j 14 . 2. Express in the form a + jb 0) (4-j7)(2+j3) (ii)(-l+j) 2 (iii) (5 + j2) (4 - j5) (2 + j3) (iv) ±1£ ■^ J 3. Express in polar form (i) 3+J5 (ii) -6+J3 (iii) -4-J5 4. Express in the form a + $ (i) 5(cos 225° +j sin 225°) (ii) 4 J330° 5. Find the values of x and y that satisfy the equation (•* + >0+j(*-.>') = 14-8+j6-2 6. Express in exponential form 0) z, = lo|37°15' and (ii) z 2 = lp | 322°45' Hence find In z\ and In z 2 . 7. Express z = e 1+J7r/2 in the forma +j&. Now you are ready to start Part 2 of the work on complex numbers. 32 Programme 1 Further Problems - 1 1 . Simplify (i) (5 + j4) (3 + j7) (2 - j3) .... (2-j3)(3+j2) ,..., cos3x + jsin3x K} (4-j3) KJ cos* +j sin* 2. Express .,- — tt. + — in the form a + }b. 3. If z = x — :- + — r~ , express z in the form a + ]b. 2+j3 1 ]i 4. if z = -^4, find the real and imaginary parts of the complex number z + l z 5. Simplify (2 + j5) 2 + 5 ^jp -j(4-j6), expressing the result in the forma + ]b. 6 If z, = 2 + i, z-, = -2 + i4 and — = — + — , evaluate z 3 in the form 1 J 2 3 Z t Z 2 a +)b. If Zi , z 2 , Z3 are represented on an Argand diagram by the points P, Q, R, respectively, prove that R is the foot of the perpen- dicular from the origin on to the line PQ. 7. Points A, B, C, D, on an Argand diagram, represent the complex numbers 9 + j, 4 + jl3, -8 + j8, -3 - j4 respectively. Prove that ABCD is a square. 8. If (2+j3)(3-j4) = x +jy, evaluate* andj>. 9. If (a + b) + )(a -b) = (2+ j5) 2 + j(2 - j3), find the values of a and b. 10. If x andj> are real, solve the equation )x _ 3x + j4 1 + )y x + 3y 11 if z = a+ $ w here a,b,c,d, are real quantities, show that (i) if z is c+]d 33 Complex numbers 1 a _c real then--— and (ii) if z is entirely imaginary then- = -^. 12. Given that (a + b) + ](a ~b) = (l + j) 2 + j(2 + j), obtain the values of a and b. 13. Express (-1 + j) in the form r e je , where r is positive and ~n < 6 < n. 14. Find the modulus ofz = (2 -j) (5 +jl2)/(l +j2) 3 . 15. If* is real, show that (2 + j)e< 1+ J 3 >* + (2 - j) e* 1 "* 3 * is also real. 16. Given that z x =/?, +R+juL;z 2 =R 2 ;z 3 = ,—^and 24 =R * + j^Q; andalsot hatz 1 z 3 =z 2 z 4 , express/? andZ in terms ofthe real constants .Rl^^Cs andC 4 . 17. Uz-x+iy, where* and y are real, and if the real part of (z +l)/(z + j) is equal to 1 , show that the point z lies on a straight line in the Argand diagram. 18. Whenz 1 =2+j3, z 2 = 3 -j4, z 3 = -5 +J12, thenz = z, + -^J|-. If£" = /z,find£'when/=5+j6. ^ ** 19. tf R i + )° jL ^ ^2 ■]f 3 """; — . where ^i,^2,-R3,^4,co,Z.andCarereal, show that * 4 " j <3c £ = CR2R3 a 2 C 2 Rl + 1 20. If z and z are conjugate complex numbers, find two complex numbers, z = z, and z = z 2 , that satisfy the equation 3zz + 2(z-z) = 39+jl2 On an Argand diagram, these two numbers are represented by the points P and Q. If R represents the number j 1 , show that the angle PRQ is a right angle. 34 Programme 2 COMPLEX NUMBERS PART 2 Programme 2 1 Introduction In> Part 1 of this programme on Complex Numbers, we discovered how to manipulate them in adding, subtracting, multiplying and dividing. We also finished Part 1 by seeing that a complex numbers + \b can also be expressed in Polar Form, which is always of the form r(cos 6 + j sin d). You will remember that values of r and d can easily be found from the diagram of the given vector. r 2 =a 2 + b 2 .'. r=y/(a 2 + b 2 ) and tan 6 = - = tan^ To be sure that you have taken the correct value of 6 , always DRAW A SKETCH DIAGRAM to see which quadrant the vector is in. Remember that 6 is always measured from OX i.e. the positive axis OX. aDDnnnDannnnDnannaDnDDannQnDDDnDDnDnan Right. Just by way of revision and as a warming up exercise, do the following: Express z = 12 — j5 in polar form. Do not forget the sketch diagram. It ensures that you get the correct value for0. When you have finished, and not before, turn on to frame 3 to check your result. 37 Complex numbers 2 Result: 13(cos337 D 23'+jsin337°23') B r ^ 12 -^^£ 1 -j r^V^ 15 ^1 Y z Here it is, worked out in full. Y r 2 = 12 2 + 5 2 = 144+25 = 169 :. r=13 UnE=Y2 =04167 •'• E = 22°3T In this case, 8 = 360° ~E = 360° - 22°37' /. 8 = 337°23' z = r(cos + j sin 8) = 1 3(cos 337°23' + j sin 337°23') aDDDDDDDDDDnDOnDDDPDDDOnDaDDQDOnaDDDDn Did you get that right? Here is one more, done in just the same way. Express -5 - j4 in polar form. Diagram first of all! Then you cannot go wrong. When you have the result, on to frame 4. Result: 2 = 6-403(cos 218°40' + j sin 218°40') Here is the working: check yours. Y r 2 =5 2 +4 2 =25 + 16 = 41 -•- r = V41 =6-403 tan E = j = ■. E = 38°40' In this case, 8 = 180° +£ = 218°40' So z = -5 -j4 = 6-403(cos 218°40' + j sin 218°40') DnDanDDGDnDDDDDDDnnannDnDnnnaDDDaDannD Since every complex number in polar form is of the same shape, i.e. r(cos 8 + j sin 8) and differs from another complex number simply by the values of r and 8, we have a shorthand method of quoting the result in polar form. Do you remember what it is? The shorthand way of writing the result above, i.e. 6-403(cos 218°40' + j sin 218°40') is 4 38 Programme 2 6-403 l218°40' nnnuuDDononnnnaoaaaaaaaaanaoaaauaaaaaa 322°15' 105 c Correct. Likewise: 5 -7 2(cos 322° 1 5 ' + j sin 322° 1 5 ') is written 5-72 5(cosl05°+jsinl05°) " " 5 3-4(cos| + jsing) " " 3-4 They are all complex numbers in polar form. They are all the same shape and differ one from another simply by the values of and r and e DDDDDnnDnDannnDDnnaDnnnnnnDDDDDnDDDDnD Now let us consider the following example. Express z = 4 - j3 in polar form. From this, r=5 tan£ = J = 0-75 /. E = 36°52' 6 = 360° - 36°52' = 323°8' First the diagram. Y 4-j3 = 5(cos323°8'+jsin323°8') or in shortened form, z 39 Complex numbers 2 z = 5 323°8 □DnnanDDnnDnDDnnnnnDannnnnDnDnnDDnDnan In this last example, we have z = 5(cos323°8'+jsin323°8') But the direction of the vector, measured from OX, could be given as -36°52', the minus sign show- ing that we are measuring the angle in the opposite sense from the usual positive direction. We could write z = 5(cos [-36°52'] + j sin [-36°52']). But you already know that cos[-0] = cos d and sin[-0] = -sin 6. z = 5(cos36°52'-jsin36°52') i.e. very much like the polar form but with a minus sign in the middle. This comes about whenever we use negative angles. In the same way , z = 4(cos 250° + j sin 250°) = 4(cos [-110°] + j sin [-1 10°]) = 4( ) z = 4(cos 110°-j sin 110°) since cos(-l 10°) = cos 110° and sin(-110°)=-sin 110° DnnDaDDDDDDDDDnDDnDDDDDnnDDDDDDnDnDDDD It is sometimes convenient to use this form when the value of 6 is greater than 180°, i.e. in the 3rd and 4th quadrants. Ex. 1 z = 3(cos230°+jsin230°) = 3(cos 130° -j sin 130°). Similarly, Ex. 2 z = 3(cos 300° + j sin 300°) = 3(cos 60° - j sin 60°) Ex.3 z = 4(cos 290° + j sin 290°) = 4(cos 70° -j sin 70°) Ex.4 z = 2(cos 215° + j sin 215°) = 2(cos 145° - j sin 145°) and Ex.5 z = 6(cos 310° + j sin 310°) = 8 40 Programme 2 z = 6(cos50°-jsin50°) since cos 310° = cos 50° and sin 310°= -sin 50° DDDDnDDanaDDonDDDnnnDnDaDDnDanDDDDDnna One moment ago, we agreed that the minus sign comes about by the use of negative angles. To convert a complex number given in this way back into proper polar form, i.e. with a '+' in the middle, we simply work back the way we came. A complex number with a negative sign in the middle is equivalent to the same complex number with a positive sign, but with the angles made negative. e.g. z = 4(cos 30° - j sin 30°) = 4(cos[-30°] +jsin[-30°]) = 4(cos 330° + j sin 330°) and we are back in the proper polar form. You do this one. Convert z = 5(cos 40° - j sin 40°) into proper polar form. Then on to frame 10. 10 z = 5(cos320°+jsin320°) since z = 5(cos 40° - j sin 40°) = 5(cos [-40°] + j sin [-40° ] ) = 5(cos 320° + j sin 320°) DDnDDDDDDaDnnDDanDDDnananDDDDaannanDan Here is another for you to do. Express z = 4(cos 100° -j sin 100°) in proper polar form. Do not forget, it all depends on the use of negative angles. 41 Complex numbers 2 z = 4(cos260°+jsin260°) 11 for z = 4(cos 100° -j sin 100°) = 4(cos [-100°] + j sin [-100°]) = 4(cos 260° + j sin 260°) nnDDODDDDDDDDODDDDDDDDDDDDDDDDDDDDDDDD We ought to see how this modified polar form affects our shorthand notation. Remember, 5(cos 60° + j sin 60°) is written 5 1 60° How then shall we write 5(cos 60° - j sin 60°)? 5 160° i 5/ A 60 ° V/-60° 5 \ We know that this really stands for 5(cos [-60°] + j sin [-60°]) so we could write 5 |-60° . But instead of using the negative angle we use a different symbol i.e. 5 [-60° becomes 5 ["60° 5 1-60" Similarly, 3(cos 45° - j sin 45°) = 3 I -45° = 3 [45° 12 DnnDnnnnnDaDDDDnnnDDnnnanDnDDDDDnDDnnn This is easy to remember, for the sign ... Q resembles the first quadrant and indicates measuring angles, \ i.e. in the positive direction, while the sign \j resembles the fourth quadrant and indicates measuring angles J i.e. in the negative direction. e.g. (cos 15° + j sin 15°) is written | 15° but (cos 15° -j sin 15°), which is really (cos [-15°] +j sin [-15°]) is written I 15° So how do we write (i) (cos 120° + j sin 120°) and (ii) (cos 135° -j sin 135°) in the shorthand way? 42 Programme 2 13 (0 120° DDDnnDDDnnaaDnananaDDnDDnDannanannnnDn The polar form at first sight seems to be a complicated way of representing a complex number. However it is very useful as we shall see. Suppose we multiply together two complex numbers in this form. LetZ! =r 1 (cos6 1 +j sin0!)andz 2 = r 2 (cos0 2 +j sin0 2 ) ThenzjZiz =r 1 (cosd 1 +j sin 6 Y ) r 2 (cos 6 2 + j sin0 2 ) = r l r 2 (co%d l cos0 2 +j sin^ cos0 2 +j co%B x sin0 2 + j 2 sin 6 1 sin0 2 ) Re-arranging the terms and remembering that j 2 = -1 , we get Ziz 2 = r x r 2 [(cos0j cos0 2 -sinSj sin 2 ) + j(sin 0! cos 2 + cos^i sin0 2 )] Now the brackets (cos 0, cos0 2 -sin0 x sin B 2 ) and (sin 0j cos0 2 + cos0! sin0 2 ) ought to ring a bell. What are they? 14 cos 6*i cos0 2 - sin0! sin0 2 = cos^j + 2 ) sin^j cos0 2 + cos 0j sin0 2 = sin(0! +6 2 ) DnDaannDnaaDDaDDnnnnaaaanonDDnDnnaDDDD In that case, ZiZ 2 -r x r 2 [cos(0i + 2 )+jsin(0! + 2 )] Note this important result. We have just shown that ^(cosfl! +j sin0 1 ).r 2 (cos0 2 +j sin0 2 ) = r l r 2 [cos(0! +0 2 )+jsin(6» 1 + 2 )] i.e. To multiply together two complex numbers in polar form, (i) multiply the r's together, (ii) add the angles, 6 , together. It is just as easy as that! e.g. 2(cos 30° + j sin 30°) X 3(cos 40° + j sin 40°) = 2 X 3(cos [30° + 40°] + j sin [30° + 40°]) = 6(cos 70° + j sin 70°) So if we multiply together 5(cos 50° +j sin 50°) and 2(cos 65° + j sin 65°) we get 43 Complex numbers 2 10(cos 115° +j sin 115°) □nnnnDnnDannnannDnPDDannDDDnDnaannDDan Remember, multiply the r's; add the 0's. Here you are then; all done the same way: (i) 2(cos 1 20° + j sin 1 20°) X 4(cos 20° + j sin 20°) = 8(cosl40°+jsin 140°) (ii) a(cos 6 + j sin d) X b(cos + j sin ©) = ab(cos[6 + 0] +jsin[0 +0]) (iii) 6(cos 210° +j sin 210°) X 3(cos 80° + j sin 80°) = 18(cos290°+jsin290°) (iv) 5(cos 50° + j sin 50°) X 3(cos [-20°] + j sin [-20°] ) = 15(cos30°+jsin30°) Have you got it? No matter what the angles are, all we do is (i) multiply the moduli, (ii) add the arguments. So therefore, 4(cos 35° + j sin 35°) X 3(cos 20° + j sin 20°) 15 12(cos55°+jsin55°) oaaaanaoannaanaannnanannanaanaaanaaana Now let us see if we can discover a similar set of rules for Division. We already know that to simplify |±i| we first obtain a denominator that is entirely real by multiplying top and bottom by 16 44 Programme 2 17 the conjugate of the denominator i.e. 3 - j4 anDDDaaoooDQcmDananciQDQnnnnoDnnnonDDDD Right. Then let us do the same thing with /•iCcosf?! +j sin^i) r 2 (cos0 2 +j sin0 2 ) r^cosgi +j sin 0i) _ r t (cos t +j sin0i)(cos0 2 -j sin0 2 ) r 2 (cos0 2 +j sin 2 ) r 2 (cos 2 + j sin 2 ) (cos 2 -j sin 2 ) _r 1 (cos0 1 cos0 2 + jsin0 1 cos0 2 -jcos0i sin0 2 +sin0i sin 2 ) ~T 2 (cos' ! 2 + sin'0 2 ) " _ r x [(cos 0! cos 2 + sin x sin 2 ) +j(sin0 t cos 2 -cos0 t sin0 2 )] ~r 2 r So, for division, the rule is =_i[cos(0! -0 2 ) + jsin(0i-0 2 )] 18 divide the r's and subtract the angle DDDDDDDnDDDDDDDDDDDDDDDDnDDaDnnDDDDaDD That is correct. e-g. 6(cos72 +i sin 72) .,, 01 o,. . „o^ -) ..,, . . . l0 ' = 3(cos31 +jsin31) 2(cos 41 + j sin 41 ) So we now have two important rules If Zi =r 1 (cos0 1 +j sin0i) andz 2 =r 2 (cos0 2 +j sin0 2 ) then (i)z!Z 2 =r 1 r 2 [cos(0 t + 2 )+jsin(0! +0 2 )] and (if)£i = ^.[co<fli -0 2 )+j sin(0! -0 2 )] The results are still, of course, in proper polar form. Now here is one for you to think about. If Z! = 8(cos 65° + j sin 65°) andz 2 =4(cos 23° + j sin 23°) then(i)z!Z 2 = and(ii)~ = z 2 45 Complex numbers 2 z x z 2 =32(cos88° + jsin88°) -^=2(cos42°+jsin42°) 19 DDDDDPODDDOaODDDDDDDDDDDDDDDDDDODDDDDD Of course, we can combine the rules in a single example. eg 5(cos 60° + j sin 60°) X 4(cos 30° + j sin 30°) 2(cos50°+jsin50°) = 20(cos90 o +jsin90 o ) 2(cos50 u +jsin50°) = 10(cos40 o +jsin40°) What does the following product become? 4(cos 20° + j sin 20°) X 3(cos 30° + j sin 30°) X 2(cos 40° + j sin 40°) 20 Result: 24(cos90 o +jsin90°) i.e. (4X3X2) [cos(20° + 30° + 40°) + j sin(20° + 30° + 40°)] = 24(cos90 D +jsin90°) DDOOODODDDOOOaDDDDDDDDDDDDDODDDDDDDDDD Now what about a few revision examples on the work we have done so far? Turn to the next frame. 46 Programme 2 21 Revision Exercise Work all these questions and then turn on to frame 22 and check your results. 1 . Express in polar form, z = -4 + j2. 2. Express in true polar form, z = 5(cos 55° - j sin 55°) 3. Simplify the following, giving the results in polar form (i) 3(cos 143° + j sin 143°) X 4(cos 57° + j sin 57°) 10(cosl26°+jsinl26°) Cll) 2(cos72° +jsin72°) 4. Express in the forma + \b, (i) 2(cos30°+jsin30°) (ii) 5(cos 57° -j sin 57°) Solutions are on frame 22. Turn on and see how you have fared. 47 Complex numbers 2 Solutions 1. r 2 =2 2 +4 2 =4+ 16 = 20 .-. r = 4-472 8 tan E = 0-5 :. £ = 26°34' x, —4 ■ o~ ^< ■'• 6 = 153°26' z =-4 + j2 = 4-472(cos 153°26' + j sin 153°26') 2. z = 5(cos 55° -j sin 55°) = 5 [cos(-55°) + j sin(-55°)] = 5(cos305°+j sin 305°) 3. (i) 3(cos 143° + j sin 143°) X 4(cos 57° + j sin 57°) = 3 X 4[cos(143° + 57°) + j sin(143° + 57°)] = 12(cos200°+jsin200 D ) (ii) 10(cosl26°+jsinl26 o ) 2(cos72°+jsin72°) = ~ [cos(126° - 72°) + j sin(126° - 72°)] = 5(cos54°+jsin54°) 4. (i) 2(cos30°+jsin30°) = 2(0-866 +j0-5) = 1-732 +j (ii) 5(cos57°-jsin57°) = 5(0-5446 -J0-8387) = 2-7230 -J4-1935 Now continue the programme on frame 23. 22 48 Programme 2 23 Now we are ready to go on to a very important section which follows from our work on multiplication of complex numbers in polar form. We have already established that — if Z\ =ri(cos Q\ + j sin Oi) andz 2 - r 2 (cos 2 + j sin 2 ) then z t z 2 =r 1 r 2 [cos(6i + 2 )+jsin(0i + 2 )] So if z 3 = r 3 (cos 3 + j sin 3 ) then we have z y z 2 z 3 =r 1 r 2 [cos(0i + 6 2 ) + j sin(0 t + d 2 )] r 3 (cosd 3 + j sin 3 ) 24 ZiZ 2 z 3 =r 1 r 2 r 3 [cos(e 1 + 2 +0 3 ) +j sin^! +0 2 + 3 )] for in multiplication, we multiply the moduli and add the arguments. DDDDDODDnDDDDDDOOODDDDDDDnDDO^nDDnDDQD Now suppose that z x , z 2) z 3 are all alike and that each is equal to z = /-(cos 6 + j sin 6). Then the result above becomes ZiZ 2 z 3 =z 3 =r.r.r[cos(d +6 +0)+j sin(0 + + 0)] = r 3 (cos30 +j sin 30). or z 3 = [r(cos + j sin 0)] 3 = r 3 (cos + j sin 0) 3 = r 3 (cos30 +j sin 30). That is: If we wish to cube a complex number in polar form, we just cube the modulus (r value) and multiply the argument (0) by 3. Similarly, to square a complex number in polar form, we square the modulus (r value) and multiply the argument (0) by 49 Complex numbers 2 i.e. [/-(cos 6 + j sin 6)] 2 = r 2 (cos 26 + j sin 26) □aDaanDnnaDDDDaaaDnDDnnnnnanDDanaPDQan Let us take another look at these results. [r (cos 6 + j sin 6)] 2 = r 2 (cos 2d + j sin 26) 25 Similarly, [/■(cos 6 +j sin 0)] 3 =/- 3 (cos 3(9 + j sin 3(9) [/• (cos 5 + j sin 6)] 4 = /-" (cos 46 + j sin 46) [r (cos + j sin 6)} s = r s (cos 50 + j sin 50) In general, then, we can say [r(cos0+j sine)]" and so on. [/■(cos 5 +j sin^)]" = r"(cosn6+j sinnd) nnnDDDDDDDanDDDDDDDDDnDnnnDnnaDD □ □ a □ □ a This general result is very important and is called DeMoivre 's Theorem it says that to raise a complex number in polar form to any power n we raise the r to the power n and multiply the angle by n . eg. [4(cos 50° + j sin 50°] 2 = 4 2 [cos(2 X 50°) + j sin(2 X 50°)] = 16(cosl00°+jsinl00°) and [3 (cos 1 1 0° + j sin 1 1 0°)] 3 --- 27 (cos 330° + j sin 330°) and in the same way, [2(cos37°+jsin37°)] 4 = 26 50 Programme 2 27 16(cosl48°+jsinl48°) DnnnnanDDDDDDDDnDDDDnnnnDaDDDnnDnnDDnn This is where the polar form really comes into its own! For DeMoivre's theorem also applies when we are raising the complex number to a fractional power, i.e. when we are finding the roots of a complex number, e.g. To find the square root of z = 4(cos 70° + j sin 70°). We have \lz = z^ = [4 (cos 70° + j sin 70°)] i i.e. n = j A, 70° . . 70°, = 45 (cos -~- +j sin ~y ) = 2 (cos 35° + j sin 35°) It works every time, no matter whether the power is positive, negative, whole number or fraction. In fact, DeMoivre's theorem is so important, let us write it down again. Here goes — If z = r(cos d + j sin 6), then z n = 28 z = r(cos 6 + j sin0), then z" = /-"(cos nd + j sin nQ) for any value of n. DnDQaDDDaDQnDnDnnDDDDannDnnDnanDDnDann Look again at finding a root of a complex number,. Let us find the cube Y rootofz = 8(cos 120° +j sin 120°). Here is the given complex number shown on an Argand diagram. z = 8 1 120° Of course, we could say that 9 was '1 revolution + 120°': the vector would still be in the same position, or, for that matter, (2 revs. + 120°), (3 revs. + 120°), etc. i.e.z = 8 1 120° , or 8 |480° ,or8 1 840° , or 8 [ 1200° , etc. and if we now apply DeMoivre's theorem to each of these, we get 480° i i Z3 =8^ if „, 8 ^ or etc. 51 Complex numbers 2 rU 8 4 120° c i -j- or 8^ 480° c i -T- or 83 840° _i —5- or 83 1200 ° 3 29 DDDDDDDnDnnnnnnDDDanDDDnDDnDDnDnnnanna If we simplify these, we get z^ = 2 [40_° or 2 [_160_° or 2 J280^ or 2 |^00_° etc. If we put each of these on an Argand diagram, as follows, 160° we see we have three quite different results for the cube roots of z and also that the fourth diagram is a repetition of the first. Any subsequent calculations merely repeat these three positions. Make a sketch of the first three vectors on a single Argand diagram. Here they are: The cube roots of z = 8(cos 120° + j sin 120°). z, = 2 I 40° z 2 = 2 |160° ?3 = 2 1280° "s aDDDannDDDDDDnDnaDnnnDaDDnQDnnDnnDDDDn We see, therefore, that there are 3 cube roots of a complex number. Also, if you consider the angles, you see that the 3 roots are equally spaced round the diagram, any two adjacent vectors being separated by degrees. 30 52 Programme 2 31 32 120° DnDnDnnnaDDnnnDDDDDDDDnnnnDnnDnnnnaDao That is right. Therefore all we need to do in practice is to find the first of the roots and simply add 120° on to get the next - and so on. Notice that the three cube roots of a complex number are equal in 360° modulus (or size) and equally spaced at intervals of — — - i.e. 1 20°. Now let us take another example. On to the next frame. Example. To find the three cube roots of z = 5(cos 225° + j sin 225°) x i 225° 225° The first root is given by z x = z 3 = 5^(cos^5- + j sin -^- ) = l-71(cos75°+jsin75°) zj = 1-71 | 75° We know that the other cube roots are the same size (modulus), i.e. 1 -71 , and separated at intervals of —5- , i.e. 120°. So the three cube roots are: zi = 1-71 I 75° z 2 = 1-71 [195° z 3 = l-71 |315° It helps to see them on an Argand diagram, so sketch them on a combined diagram. 53 Complex numbers 2 We find any roots of a complex number in the same way. (i) Apply DeMoivre's theorem to find the first of the n roots, (ii) The other roots will then be distributed round the diagram at regular intervals of n A complex number, therefore, has 33 ■\{LC\ Q 2 square roots, separated by — i.e. 180° 3 cube roots, 4 fourth roots, 5 fifth roots, ~~- i.e. 120 360° — i.e. 90° etc. 360° 5 i.e. 72 c There would be 5 fifth roots separated by DDDnnnDnnDDDnDDDDDnanannnnnDaDDannnDDD And now: To find the 5 fifth roots of 1 2 j 300° z=12[300_° --. 2, = 12*" ^°=12Tl60° 34 We now have to find the value of 12^. Do it by logs. Let A = 12s . Then log A ~ log 1 2 = j(l -0792) = 0-2158 Taking antilogs, A = 1 -644 The first of the 5 fifth roots is therefore, z t = 1 .644|60° The others will be of the same magnitude, i.e. 1 -644, and equally separated at intervals of — r^ i.e. 72° So the required 5 fifth roots of 1 2 | 300° are z, = 1 -644 |_60_°, z 2 = 1 .644 |132°, z 3 = 1 -644 [ 204° z 4 = 1 -644 1 276° , z 5 = 1 -644 [348° Sketch them on an Argand diagram, as before. 54 Programme 2 35 "> Jz 1 z 2 X^ i *1 = 1-644 | 60° X, -^ *2 = ■ 644 | 136° 1-644|204° *3 *5 *4 = 1-644 I 2 76° *3 L '5 = 1 644 1 348° > Principal root. Although there are 5 fifth roots of a complex number, we are sometimes asked to find the principal root. This is always the root whose vector is nearest to the positive OX axis. In some cases, it may be the first root. In others, it may be the last root. The only test is to see which root is nearest to the positive OX axis. In the example above, the principal root is therefore 36 Principal root z s = 1 -644 348 DDDDnDnnannDDDDnDnnaDanDnnnDDDDnDDnnaD Good. Now here is another example worked in detail. Follow it. We have to find the 4 fourth roots of z = 7(cos 80° + j sin 80°) The first root, z x = T$ 80 u 4 : lA 20° Now find 7^ by logs. Let A = 7^ Then log A =^log 7 = |(0-845 1) = 0-2113 and A = 1-627 z x = 1 -627 | 20° The other roots will be separated by intervals of -— = 90° Therefore the four fourth roots are — zj = 1-627 20°_ z 3 = 1-627 | 200° And once again, draw an Argand diagram to illustrate these roots. z 2 =1-627 1 110° z 4 = l-627[ 290° 55 Complex numbers 2 Y \ J *i 0\ X Zi = 1627 J 20 2z = 1627 J110 *3 = 1 627 |200 ° z 4 = 1-627 [290° 37 aDDDDnnDDDnnDnnannDDDDDaDnnnnnanDDDaDa And in this example, the principal fourth root is 38 Principal root: z x = 1-627 |20° since it is the root nearest to the positive OX axis. DDDnnDnnnnDnanannnDnnnDnnnnnDnnnnDnnDa Now you can do one entirely on your own. Here it is. Find the three cube roots of 6(cos 240° + j sin 240°). Represent them on an Argand diagram and indicate which is the principal cube root. When you have finished it, turn on to frame 39 and check your results. 56 Programme 2 39 Result: > j 4** z, = 1 817 I 80° K 8o ° z z = 1-817 1 200° z 3 = 1-817 |320° N^O" *2 *3 Principal root : Zj = ■817 |320° <l u □□□□□□□□□□□□□□DDnnnnnnnQnnannnoanaonQO Here is the working. z = 6 I 240° Zi =63 240" 1-817 80° •5zrr\° Interval between roots = — — = 120° Therefore the roots are: 1-817 180 z 2 =1-817 1200 z 3 = 1-817 1 320° The principal root is the root nearest to the positive OX axis. In this case, then, the principal root is z 3 = 1-817 ] 320° On to the next frame. 40 Expansion o/sin nd and cos nd, where n is a positive integer. By DeMoivre's theorem, we know that cos nd + } sin nd = (cos 6 + j sin d) n The method is simply to expand the right-hand side as a binomial series, after which we can equate real and imaginary parts. An example will soon show you how it is done: Ex. 1. To find expansions for cos 38 and sin 3d. We have cos 3d + j sin 30 = (cos 8 + j sin dy = (c + js) 3 where c = cos 8 s = sin d Now expand this by the binomial series — like (a + b) 3 so that cos 30 + j sin 30 = 57 Complex numbers 2 C 3 +j3c 2 s-3cs 2 -js 3 for: j 3 =i cos 30 + j sin 30 = c 3 + 3c 2 Qs) + 3c(js) 2 + (js) 3 = c 3 +j3c 2 s-3cs 2 -js 3 sincej 2 =-l = (c 3 -3cs 2 )+j(3c 2 s~s 3 ) Now, equating real parts and imaginary parts, we get cos 30 - and sin 30 = 41 cos 3d = cos 3 - 3 cos sin 2 sin 30 = 3 cos 2 sin - sin 3 If we wish, we can replace sin 2 by (1 - cos 2 0) and cos 2 by (1 -sin 2 0) so that we could write the results above as cos 30 = (all in terms of cos 0) sin 30 = (all in terms of sin 0) 42 since and cos 30 =4cos 3 0-3 cos0 sin 30 =3 sin0-4sin 3 cos 30 = cos 3 - 3 cos (1 - cos 2 0) = cos 3 -3 cos0 + 3 cos 3 = 4 cos 3 - 3 cos sin 30 = 3(1 - sin 2 0) sin - sin 3 = 3 sin0 -3sin 3 -sin 3 = 3 sin0 -4sin 3 While these results are useful, it is really the method that counts. So now do this one in just the same way: Ex. 2. Obtain an expansion for cos 40 in terms of cos 0. When you have finished, check your remit with the next frame. 43 58 Programme 2 44 cos 40 = 8cos 4 0-8cos 2 + 1 Working: Equating real parts: cos 40 + j sin 40 = (cos + j sin 0) 4 = (c + js) 4 = c 4 + 4c 3 (js) + 6c 2 Cs) 2 + 4c(js) 3 + (js) 4 = c 4 + j4c 3 s - 6c 2 s 2 - j4cs 3 + s 4 = (c 4 - 6c 2 s 2 + s 4 ) + j(4c 3 s - 4cs 3 ) cos 46 = c 4 - 6c 2 s 2 + s 4 „2\2 c 4 -6c 2 (l-c 2 ) + (l-c 2 ) 2 c 4 - 6c 2 + 6c 4 + 1 - 2c 2 + c 4 8c 4 -8c 2 + 1 = 8 cos 4 0-8cos 2 + 1 Now for a different problem. On to the next frame. _ _ Expansions for cos"6 and sin"0 in terms of sines and cosines of iX J) miltiples of 6 . Let z = cos 6 + j sin 6 then -=z x = cos 6 -j sin i .'. z +— = 2 cos 6 and z — = j 2 sin 6 z z Also, by DeMoivre's theorem, z" = cos n# + j sin nd and — n = z~" = cos nd - j sin «0 .'. z" +— = = 2 cos n# and z" --=- = j 2 sin nd z" z" Let us collect these four results together: z = cos 6 + j sin 6 z + — = 2 cos z z - - = i 2 sin z J z" + -t. = 2 COS H0 z" z" — ^ = j 2 sin«0 Mzfe a note of these results in your record book. Then turn on and we will see how we use them. 59 Complex numbers 2 Ex. 1. T5 expand cos 3 From our results, z +— = 2 cos i z 46 L /. (2cos0) 3 = (z+-) 3 ^ -1 ^ 1 = z 3 + 3 z + 3I+ I 3 z z* Now here is the trick: we re-write this, collecting the terms up in pairs from the two extreme ends, thus - ^3_,„3 . K . ~ .U (2cos0) 3 =(z 3 +p) + 3(z+-) And, from the four results that we noted, 1 and z + — = z z" +- z+— =2 cos0;z 3 +-, =2 cos 30 .'. (2 cos 0) 3 = 2 cos 30 + 3.2 cos 6 8 cos 3 = 2 cos 30 + 6 cos 4cos 3 = cos 30 + 3 cos0 , 1 cos 3 0=-(cos30 + 3cos0) Now one for you: Ex.2. Find an expansion for sin 4 Work in the same way, but, this time, remember that z-- = j 2 sin0 andz" -— , =j 2 sin nd z z n When you have obtained a result, check it with the next frame. 47 60 Programme 2 48 for, we have: sin 4 6 = - [cos 46-4 cos 26 + 3] z -— = j 2 sin 6 ; z n - —^ = i 2 sin nd Now Q 2 sin 0) 4 = (z --) 4 1 \ 1 "liM?)-^!?)* ■(2 4 tj«)-* ! +p) + 6 z M + -— = 2 cos nd 1 6 sin 4 6 = 2 cos 40 - 4.2 cos 20 + 6 .'. sin 4 =- [cos 46 -4 cos 26 +3] They are all done the same way: once you know the trick, the rest is easy. Now let us move on to something new. 49 Loci Problems We are sometimes required to find the locus of a point which moves in the Argand diagram according to some stated condition. Before we work through one or two examples of this kind, let us just revise a couple of useful points. You will remember that when we were representing a complex number in polar form, i.e., z =a + )b =r(cos 6 + j sin d), we said that (i) r is called the modulus of z and is written 'mod z' or | z\ and (ii) 6" " " argument of z " " " 'arg z' Also, r = \/(a 2 + b 2 ) and 6 = tan" 1 j-j so that | z | = \J(a 2 + b 2 ) and arg z = tan" 1 — Similarly, if z = x + 'yy, then \z\ = and argz= 61 Complex numbers 2 \z\ = V(* 2 + y 2 ) and arg z = tan" 1 fL\ If 2 =x + jy, Keep those in mind and we are now ready to tackle some examples. Ex. 1. \iz=x + jy, find the locus defined as \z \ = 5. Now we know that in this case, \z\ = yj(x 2 + y 2 ) The locus is defined as \J(x 2 + y 2 ) = 5 50 .'. X 4 + y 2 =25 Locus | z ] = 5 i.e. x 2 +y 2 = 25 This is a circle, with centre at the origin and with radius 5 . That was easy enough. Turn on for Example 2. Ex. 2. If z =x+jy, find the locus defined as arg z =- In this case, arg z = tan ' |Z| ;. tan 1 |Z] = L ••• Z= tanlj_= tan 45° = 1 ;. L = 1 .-. j, = jc So the locus arg z =^ is therefore the straight line y=x All locus problems at this stage are fundamentally of one of these kinds. Of course, the given condition may look a trifle more involved but the approach is always the same. Let us look at a more complicated one. Next frame. 51 62 Programme 2 Jj £ Ex. 3. Uz =x +jy, find the equation of the locus Since z =x + jy, z+l=x+]y+l=(x+l)+]y =r l \9 1 z-l=x+]y-l =(x-l)+]y =r 2 \6 2 z+ 1 : z 2 z+ 1 z - 1 r 2 \ e 2 r 2 z+ 1 J ±= Ifil = y/[(x + l) 2 +y 2 ] r 2 \z 2 \ ^/[{x-lf+y 2 ] = 2 ■Wl(x-\) 2 +y 2 ] ■ (^ + i) 2 +y 2 = " (x-l) 2 +y 2 All that now remains is to multiply across by the denominator and tidy up the result. So finish it off in its simplest form. 53 We had So therefore (x+ l) 2 +y 2 -2x+ 1 +y 2 ) 8x + 4 + 4y 2 ; (x-iy^ry (x+l) 2 +y 2 =4J(x-l) 2 +. x 2 + 2xy + 1 + y 2 = 4(x 2 = 4x 2 - :. 3x 2 - \0x + 3 + 3y 2 This is the equation of the given locus. Although this takes longer to write out than either of the first two examples, the basic principle is the same. The given condition must be a function of either the modulus or the argument. Move on now to frame 54 for Example 4. 63 Complex numbers 2 Ex.4. Ifz=x+j>>,findtheequationofthelocusarg(z 2 )=--. Rtt z~x+iy=r\d :. arg z = d = tan -1 ' y .". tan d =*. {i} .'• By DeMoivre's theorem, z 2 = r 2 [20 ■•• arg(z 2 ) = 20=-?- •". tan 29= tan (-7) =-1 4 • 2 tan 6> 1 -tan 2 = .'. 2tan0 = tan 2 -1 But tan d =L • ^ = z! - 1 X XX ,2 2xy =y 2 -x 2 :. y 2 = x 2 + 2xy In that example, the given condition was a function of the argument. Here is one for you to do: lfz=x+]y, find the equation of the locus arg(z+ 1)=- Do it carefully; then check with the next frame. 55 Here is the solution set out in detail. lfz=x+jy, find the locus arg(z + l)= — z "x+\y :. z+ 1 =x+}y+ 1 =( x + \)+jy arg(z+l) = tan- 1 ( JL.) = 1. lx+11 3 V =tan^ = V3 x+ ] 3 y=\/3(x+l) And that is all there is to that. Now do this one. You will have no trouble with it. lfz=x+jy, find the equation of the locus | z - 1 I =5 When you have finished it, turn on to frame 56. 64 56 Programme 2 Here it is: z = x + \y ; given locus \z — \ [ = 5 z-1 =x+iy-l =(x-l) + iy :. |z-l| =v / [(*-i) 2 +.y 2 ] =5 .-. (^-l) 2 +7 2 =25 :. x 2 -2x+ 1 +j 2 =25 .". x 2 -23c+>> 2 =24 Every one is very much the same. This brings us to the end of this programme, except for the final test exercise. Before you work through it, read down the Revision Sheet (frame 57), just to refresh your memory of what we have covered in this programme. So on now to frame 5 7. 65 Complex numbers 2 Revision Sheet 1 . Polar form of a complex number z=a+]b= /-(cos d + j sin 0) = r | 6 r - mod z = I z | = y/a 2 + b 2 2. Negative angles ' = arg z = tan" z=r(cos [-0] +j sin [-6]) x cos [-0] = COS ,-lf* (f) sin [-0] = -sin 6 .'. z = r(cos d - j sin 0) = r 3. Multiplication and division in polar form If z i =>"! jj^; z 2 = r 2 j^2_ then ^ 2 =/-,/- 2 I 0, + e 2 z 2 r 2 LJ i 4. DeMoivre's theorem If z = r (cos + j sin 0), then z n = /-"(cos n0 + j sin nd) 5. Exponential form of a complex number z -a +]b standard form = r(cos + j sin 0) polar form = r eJtf [0 in radians] .... exponential form Also ei 9 = cos + j sin e"J e = cos0 -j sin0 6. Logarithm of a complex number z = rei e :. In z = In r+jd 7. Loci problems If z= x+]y, \z\ =\/(x 2 +y 2 ) arg z = tan" 1 jZ.1 TTzar 's /Y/ A^o w .yow are razc?y for the Test Exercise on Frame 58. 57 66 Programme 2 Jj Q Test Exercise— II 1 . Express in polar form, z = — 5 — j3. 2. Express in the forma +]b, (i) 2 1 156°, (ii) 5 | 37°. 3. Ifzj = 12(cos 125° + j sin 125°) and z 2 = 3(cos72° +j sin 72°), find (i) z x z 2 and (ii) — giving z 2 the results in polar form. 4. If z = 2(cos 25° + j sin 25°), find z 3 in polar form. 5. Find the three cube roots of 8 (cos 264° + j sin 264°) and state which of them is the principal cube root. Show all three roots on an Argand diagram. 6. Expand sin 40 in powers of sin and cos 8 . 7. Express cos 4 in terms of cosines of multiples of . 8. If z = x + \y, find the equations of the two loci defined by (i)|z-4| = 3 (ii) arg(z + 2)=5 67 Complex numbers 2 Further Problems-II 1. If z =x + iy, where x and y are real, find the values of x and y when 3z + _3z = _4_ J -J J 3-j 2. In the Argand diagram, the origin is the centre of an equilateral triangle and one vertex of the triangle is the point 3 + j-y/3. Find the complex numbers representing the other vertices. 3. Express 2 + j3 and 1 -j2 in polar form and apply DeMoivre's theorem to evaluate A—i-i . Express the result in the forma +J6 and in exponential form. 4. Find the fifth roots of -3 + j3 in polar form and in exponential form. 5. Express 5 + J12 in polar form and hence evaluate the principal value of V(5 + jl2), giving the results in the form a + \b and in form re je . 6. Determine the fourth roots of -16, giving the results in the form a +)b. 7. Find the fifth roots of -1 , giving the results in polar form. Express the principal root in the form r e) e . 8. Determine the roots of the equation x 3 + 64 = in the form a + )b , where a and b are real. 9. Determine the three cube roots of lZJ giving the results in l + j modulus/argument form. Express the principal root in the form a+)b. 10. Show that the equation z 3 = 1 has one real root and two other roots which are not real, and that, if one of the non-real roots is denoted by to, the other is then co 2 . Mark on the Argand diagram the points which represent the three roots and show that they are the vertices of an equilateral triangle. 68 Programme 2 1 1 . Determine the fifth roots of (2 - j5), giving the results in modulus/argument form. Express the principal root in the form a + \b and in the form r ei e . 12. Solve the equation z 2 +2(1 + j)z + 2 = 0, giving each result in the form a + \b, with a and b correct to 2 places of decimals. 13. Express e 1- ^/ 2 in the form a + jb . 14. Obtain the expansion of sin 10 in powers of sin d. 15. Express sin 6 x as a series of terms which are cosines of angles that are multiples of x. 16. If z = x + yy, where x andy are real, show that the locus is a circle and determine its centre and radius. z-2 z +2 17. If z =x + jy, show that the locus are] ^1=— is a circle. Find its \ z -ii 6 centre and radius. 18. If z = x +jy, determine the Cartesian equation of the locus of the point z which moves in the Argand diagram so that |z+j2| 2 +|z-j2J 40 19. If z = x + )y, determine the equations of the two loci: (i) z + 2 = 3 (ii) arg Z + 2) 77 20. If z = x + )y, determine the equations of the loci in the Argand diagram, defined by (0 z + 2 z-1 : 2, and (ii) arg z - 1 1 n z + 2 21. Prove that (i) if | Zj+z 2 ] =|z 1 -z 2 | , the difference of the arguments of Z\ and z 2 is — . 69 Complex numbers 2 (»)if argjiiliij^^henlz,, =,z 2 | 22. If z = x + )y, determine the loci in the Argand diagram, defined by (i) |z+j2l 2 -\z -j2| 2 =24 (ii) \z+]k\ 2 +\z-]k | 2 = 10A: 2 (k>0) 70 Programme 3 HYPERBOLIC FUNCTIONS Programme 3 Introduction When you were first introduced to trigonometry, it is almost certain that you defined the trig, ratios '— sine, cosine and tangent - as ratios between the sides of a right-angled triangle. You were then able, with the help of trig, tables, to apply these new ideas from the start to solve simple right- angled triangle problems and away you went. You could, however, have started in quite a different way. If a circle of unit radius is drawn and various constructions made from an external point, the lengths of the lines so formed can be defined as the sine, cosine and tangent of one of the angles in the figure. In fact, trig, func- tions are sometimes referred to as 'circular functions'. This would be a geometrical approach and would lead in due course to all the results we already know in trigonometry. But, in fact, you did not start that way, for it is more convenient to talk about right-angled triangles and simple practical applications. Now if the same set of constructions is made with a hyperbola instead of a circle, the lengths of the lines now formed can similarly be called the hyperbolic sine, hyperbolic cosine and hyperbolic tangent of a particular angle in the figure, and, as we might expect, all these hyperbolic functions behave very much as trig, functions (or circular functions) do. This parallel quality is an interesting fact and important, as you will see later for we shall certainly refer to it again. But, having made the point, we can say this: that just as the trig, ratios were not in practice defined geometrically from the circle, so the hyperbolic functions are not in practice defined geometrically from the hyperbola. In fact, the defini- tions we are going to use have apparently no connection with the hyper- bola at all. So now the scene is set. Turn on to Frame 1 and start the programme. 73 Hyperbolic Functions You may remember that of the many functions that can be expressed as a series of powers of x, a common one is e x , If we replace x by -x, we get 1 . ,* 2 * 3 x 4 2! 3! 4! 2! 3! 4! and these two functions e x and e x are the foundations of the definitions we are going to use. (i) If we take the value of e x , subtract e' x , and divide by 2, we form what is defined as the hyperbolic sine of x. e — q~ x x = hyperbolic sine of x This is a lot to write every time we wish to refer to it, so we shorten it to sinh x, the h indicating its connection with the hyperbola. We pronounce it 'shine x '. sinh* e y - e -y So, in the same way, -— would be written as sinh y DDnonDOODDOOODDDDDDOODOOQODODODOQDDODQ In much the same way, we have two other definitions: W — 2 = h yP erbo l ic cosine of x = cosh x [pronounced 'cosh x'] ( m ) e x + -x - hyperbolic tangent of x = tanhx [pronounced 'than x'] We must start off by learning these definitions, for all the subsequent developments depend on them. So now then; what was the definition of sinh xl sinh* = 74 1 Programme 3 sinh X = ■ noooanQaoaaannaooDDoaaoonnnQanaoaaaaaQ Here they are together so that you can compare them. sinh x = cosh x = tanh x = Make a copy of these in your record book for future reference when necessary. e* -e x 2 e x + e~ x 2 e x -e x e x + e x sinh x '■ cosh x = - tanhx : DDDnnnanDnDDnDnDnnnnDDDnDDDDDDnDDDDnnn We started the programme by referring to e x and e' x as series of powers of x. It should not be difficult therefore to find series at least for sinh x and for cosh x. Let us try. (i) Series for sinh x *x _ = 1 +X+^7 + X 2 X 3 . X 4 }t 11 e-* = l If we subtract, we get Divide by 2 2 2 3 4 X L X X 2x 3 2x s ■2x +— + — 3! 5! " x x -=sinh x =x + ^T + 5T + (ii) If we add the series for e* and e~ x , we get a similar result. What is it? When you have decided, turn on to Frame 5. 75 Hyperbolic Functions X 2 x 4 X 6 coshx=1+ 2! + 4! + ~6! + DnDn.nnDnDDnnnannnDQnnapnDDnnDDDDnaDDDn For we have : e ' x = l - x+ jr x i +x i'--- x x ■ cosh* = 1 + ^7+-rr+ Move on to Frame 6. So we have: X x^ x^ sinhx=^ +I]+ _ + _ + ... cosh x=l + |! + ^ + 5l + ... Note: All terms positive: sinh x has all the odd powers, coshx has all the even powers. We cannot easily get a series for tanh x by this process, so we will leave that one to some other time. Make a note of these two series in your record book. Then, cover up what you have done so far and see if you can write down the definitions of: (i) sinhx = (ii) C oshx = (iii) tanh x = No looking! 76 Programme 3 sinh X ■ ; cosh x = -; tanh x = e* + e" : All correct? Right. DnnnnDnannnnnDnnnnnnnDDnanoDDanDnnDaDn Graphs of Hyperbolic Functions We shall get to know quite a lot about these hyperbolic functions if we sketch the graphs of these functions. Since they depend on the values of e* and e x , we had better just refresh our memories of what these graphs look like. y = e x and y - e x cross the >>-axis at the pointy = 1 (e° = 1). Each graph then approaches the x-axis as an asymptote, getting nearer and nearer to it as it goes away to infinity in each direction, without actually crossing it. So, for what range of values of x are e* and e' x positive? 8 e x and e x are positive for all values of x Correct, since the graphs are always above the x-axis. DDDnDDDnaDDnDDDDDDDnnnnnDnnnnnDnDannnD At any value of x, e.g. x = x iy e x - v cosh x = ■ , i.e. the value of cosh x is the average of the values of e* and e~ x at that value of x. This is given by P, the mid point of AB. If we can imagine a number of ordinates (or verticals) like AB and we plot their mid-points, we shall obtain the graph of y = cosh x. Can you sketch in what the graph will look like? 77 Hyperbolic Functions Here it is: y = cosh x X 1 ~~ x _ n We see from the graph of y = cosh x that: (i) cosh = 1 (ii) the value of cosh* is never less than 1 (iii) the curve is symmetrical about the >>-axis, i.e. cosh(-x) = coshx (iv) for any given value of coshx, there are two values of x, equally spaced about the origin, i.e. x = ±a. Now let us see about the graph of y = sinh x in the same sort of way. sinha:=- 3*- D' x 10 sinh x = BP e x - e x 2 On the diagram, CA-- --e x CB = ■-e x { = e x -e x _e* -e x The corresponding point on the graph of y = sinh x is thus obtained by standing the ordinate BP on the x-axis at C, i.e. P : . Note that on the left of the origin, BP is negative and is therefore placed below the x-axis. So what can we say about y = sinh x? 78 Programme 3 y = sinh x .* _ .-x From the graph of y = sinh x, we see (i) sinh = (ii) sinhx can have all values from -°° to +°° (iii) the curve is symmetrical about the origin, i.e. sinh(-x) = -sinh x (iv) for a given value of sinh x, there is only one real value of x. If we draw j = sinhx and y = coshx on the same graph, what do we get? 12 y = cosh X y= sinhx Note that y = sinhx is always outsider = coshx, but gets nearer to it as x increases i.e. asx^-°°, sinh x -»■ cosh x And now let us consider the graph of y = tanh x. Turn on. 79 Hyperbolic Functions It is not easy to build y = tanh x directly from the graphs of y = e* and j = e"*. If, however, we take values of e* and e' x and then calculate e x - e' x y = — and plot points, we get a graph as shown. 13 We see (i) tanh = (ii) tanh x always lies between y = -1 and y = 1 (iii) tanh(— x) = —tanh x (iv) as*-* 00 , tanhx^-1 as x -» -°°, tanh x ->• -1 . Finally, let us now sketch all three graphs on one diagram so that we can compare them and distinguish between them. Here they are: y = cosh x sinh x 14 One further point to note: At the origin, y = sinh x and y = tanh x have the same slope. The two graphs therefore slide into each other and out again. They do not cross each other at three distinct points (as some people think). It is worth while to remember this combined diagram: sketch it in your record book for reference. 80 I J) Revision Exercise Fill in the following- Programme 3 e x +e Results on the next frame. Check your answers carefully. Hyperbolic Functions Results: Here they are: check yours. 16 (i) e* + e" : = coshx 00 = tanhx (iii) — — = sinhx (iv) > < 1 ^ y = tcinh x (v) (vi) y = cosh x y = sinh x Now we can continue with the next piece of work. 82 Programme 3 1 / Evaluation of Hyperbolic Functions The values of sinhx, coshx and tanhx for some values of x are given in the tables. But for other values of x it is necessary to calculate the value of the hyperbolic functions. One or two examples will soon show how this is done. Example 1. To evaluate sinh 1 -275 Now sinh* = \(e x - e*) .\ sinh 1-275 = Ke 1 " 275 - e' 1 ' 275 ). We now have to evaluate e 1 ' 275 . Note that when we have done that, e" 1 " 27S is merely its reciprocal and can be found from tables. Here goes then: Let A = e 1 ' 275 :. In A = 1 -275 and from tables of natural logs we now find the number whose log is 1-275. This is 3-579 .". A = 3-579 (as easy as that!) So e 1 " 275 =3-579 and e" 1 " 275 =^ = 0-2794 :. sinh 1-275 =K3-579- 0-279) = 1(3-300)= 1-65 :. sinh 1-275 = 1-65 In the same way, you now find the value of cosh 2-156. When finished, move on to frame 18. 18 cosh 2-156 = 4-377 cosh □ □□DDDDnDDDDDDDDDnDDDDDnnnnDDDDDDDDnnD Here is the working: Example 2. cosh 2-156 =\(e 2 ' 1S6 + e' 2 " 156 ) Let A = e 2 " 156 .'. In A= 2-156 :. A = 8-637 and - = 0-1158 :. cosh 2- 156 = £(8-637 + 0-1 16) = |(8-753) = 4-377 :. cosh 2-156 = 4-377 Right, one more. Find the value of tanh 1 -27. When you have finished, move on to frame 19. 83 Hyperbolic Functions tanh 1-27 = 0-8539 19 DDnaDnnDDDDDnnanDanDnaDDDnnDnDnDQDnnnD Working: e i-27_ e -i-27 Example 3. tanh 1-27 : e - - ■ -r e Let A=e 1-27 ;. In A =1-27 .\ A = 3-561 and \ = 0-2808 A , , „„ 3-561-0-281 _ 3-280 O" 5159 •'• tanh l - 21 ~ 3-561 +0-281 = 3-842 Q'5845 1-9314 tanh 1-27 = 0-8539 iZ£i - i So, evaluating sinh, cosh and tanh is easy enough and depends mainly on being able to evaluate e*, where k is a given number — and that is most easily done by using natural logs as we have seen. And now let us look at the reverse process. So on to frame 20. Inverse Hyperbolic Functions ~ -~ Example 1. To find sinh" 1 1-475, i.e. to find the value of x such that xll sinhx= 1-475. Here it is: sinh x = 1 -475 .'. %(e x - e x ) = 1 -475 .-. e* - -\ = 2-950 Multiplying both sides by e x : (e x f - 1 = 2-95(e*) (e x ) 2 -2-95(e*)-l=0 This is a quadratic equation and can be solved as usual, giving &x = 2-95±V(2-95 2 +4) = 2-95 ± V(8-703 + 4) 2 2 = 2-95 ±y/l 2-703 = 2-95+3-564 2 2 _ 6-514 0-614 _ = — j— or 7y— = 3-257 or -0-307 But e* is always positive for real values of x. Therefore the only real solu- tion is given by e* = 3-257. :. x = In 3-257=1-1809 .-. x= 1-1809 Exercise 2. Now you find cosh" 1 2-364 in the same way. 84 Programme 3 21 cosh -1 2-364 = ±1-507 naaannnnDnannaDDnaannDDnDaDnnnnaaaaDDD For: To evaluate costf 1 2-364, let x = cosh"" 1 2-364 .'. cosh* = 2-364 + e' = 2-364 :. e* + - Y = 4-728 (e x ) 2 -4-728(e*)+l =0 e* = 4-728±V(22-36-4) Vlg . 36 = ^ = i(4-728 ± 4-285) = i(9-013) or ^(0-443) e*= 4-5065 or 0-2215 .'. x = In 4-5065 or In 0-2215 = 1-5056 or 2-4926 i.e. -1-5074 x =±1-507 Before we do the next one, do you remember the exponential defini- tion of tanh x? Well, what is it? 22 tanh x = - DDDDnnDnnaDDDanDDDDannnnanDDnn That being so, we can now evaluate tanh" 0-623. Let x = tanh" 1 0-623 .'- tanh x = 0-623 Dnnnannn • e c =0-6^ e x + e~ x .-. e *- e -*=0-623(e*+e- x ) :. (1- 0-623) e*=(l + 0-623) e -X 0-377 e* = 1-623 e' x 1-623 e* 0-2103 ■ ( s x\2 1-623 "0-377 T-5763 2) 0-6340 :. e x = 2-075 0-3170 :. x = In 2075 = 0-7299 :. tanh" 1 0-623 = 0-730 Now one for you to do on your own. Evaluate sinrf 1 0-5 85 Hyperbolic Functions sinh" 1 0-5 = 0-4810 23 aQnaaonanaanannaaaaQDqnoQQDQDQDOODODOn Check your working. Let x = sinh" 1 0-5 .'. sinh x = 0-5 1 p x _ p -x e e = 0-5 :. e* = 1 /. (e*) 2 - 1 = e* (e*) 2 -(e*)-l=0 a *_ 1±V(1+4)_1±V5 3-2361 or 2 -1-2361 2 "* 2 = 1-6181 or -0-6181 .-. x = ln 1-6181 =0-4810 sinh" 1 0-5 = 0-4810 And just one more! Evaluate tanh" 1 0-75. tanhT 1 0-75 = 0-9731 e* =-0-6181 gives no real value of x. 24 □QnQQnnnnonnaonannnnQnnooonoQannaanaQn Let x = tanh" 1 0-75 .'. tanh x = 0-75 ■ = 0-75 e x - t x = 0-75(e* + e"*) (l-0-75)e*=(l + 0-75)e"* 0-25 e* = l-75e* 1-75 (e x Y 0-25 = 7 e* = ±V7 = ±2-6458 But remember that e x cannot be negative for real values of x. Therefore e* = 2-6458 is the only real solution. .'. x = In 2-6458 = 0-9731 tanh" 1 0-75 = 0-9731 86 Programme 3 25 Log. Form of the Inverse Hyperbolic Functions Let us do the same thing in a general way. To find tanh" 1 * in log. form. As usual, we start off with: Let y = tanh"" 1 * .'. x=tanh^ So that e y - e -y e y + e y ,-y = x(e y + e y ) e y (\-x) = e- y (\ +x)= ^(1 +x) e 2 y 1 +x :. 2j> = ln l-x 1 +x 1 ,-1 1, fl +x y = t<mh l x=$\n\^—^ tanlf 1 0-5=1 In j^) = iln3=i(10986) =0-5493 26 And similarly, tanh ' (-0-6) : tanh" 1 (-0-6) =-0-6932 nnDDDnnDnnannDnDnnnaDnnnnnDnDDnnnnDnnn For, tanh" 1 * = -jln{ t~z — | = \ In 0-25 = i(2-6137) = i(-l-3863) 2-5 0-9163 10 2-3026 2-6137 = -0-6932 Now, in the same way, find an expression for sinh _1 x. Start off by saying: Lety = sinrf 1 * .'. x = sinny y -y & - e^ = x •• e y - € y = 2x (e y ) 2 -2x(e y )-l=0 .e y -- y = 2x Now finish it off. 87 Hyperbolic Functions Result: sinh 1 jc=ln{x+V(^ 2 +l)} 97 nnnnnnnDDnnDDDDDnnDnDDnnnnDnnDnanDDnnD For (e>) 2 -2x(e>)-l=0 y = 2x ± V(4x 2 + 4) = 2x ± 2y/(x 2 + 1) e ^ 2 = x±V(* 2 + 1) e y =x+yj(x 2 + 1) or e y =x-sj(x 2 + 1) At first sight, there appear to be two results, but notice this: In the second result, V(* 2 + 1) >* .'. t y = x - (something > x) i.e. negative. Therefore we can discard the second result as far as we are concerned since powers of e are always positive. (Remember the graph of e x .) The only real solution then is given by t y = x + \J{x 2 + 1) y = sinh" 1 * = \n{x + y/(x 2 +1)} Finally, let us find the general expression for cosh 1 x. , -1 e y + Q y Let y = cosh x .'. x = cosh y =? — :. e y + L = 2x ■•• (e^) 2 - 2x{t y ) +1=0 ■■■e^ W <y- 4) = *±V(*'-l) :. e y = x + V(* 2 - 1) and e y = x-y/(x 2 - 1) Both these results are positive, since \J(x 2 ~ \)<x. u 1 i x-\J(x 2 -\) However, 77-5 — - = 77-5 — - . }L >-% — ~i X+y/(x 2 -l) X+V(* "I) X-\/(x 2 -l) So our results can be written e^W^-Oande ^^^ c y = x + s/(x 2 - 1) or {x + y/(x 2 - l)}" 1 .-. y = \n{x+^(x 2 -])} or -\n{x+^(x 2 -\)} .". cosh" 1 x = ± ln{x + V(* 2 ~ 0} Notice that the plus and minus signs give two results which are symmetri- cal about thejy-axis (agreeing with the graph of y = coshx). 28 88 Programme 3 29 Here are the three general results collected together, sinrf'x = \n{x+\/{x 2 + 1)1 cosh" 1 * = ±ln{* + V(* 2 ~ 1)1 tanh" 1 * = 5 In I -■i-ir^ Add these to your list in your record book. They will be useful. Compare the first two carefully, for they are very nearly alike. Note also that (j) sinh" 1 x has only one value, (ii) cosh" 1 x has two values. So what comes next? We shall see in frame 30. 30 Hyperbolic Identities There is no need to recoil in horror. You will see before long that we have an easy way of doing these. First of all, let us consider one or two relationshi ?s based on the basic definitions. (1) The first set are really definitions themselves. Like the trig, ratios, we have reciprocal hyperbolic functions: (i) coth x (i.e. hyperbolic cotangent) = - — r — (ii) sechx (i.e. hyperbolic secant) = — r — (iii) cosech x (i.e. hyperbolic cosecant) = -r-r — These, by the way, are pronounced (i) coth, (ii) sheck and (iii) co-sheck respectively. These remind us, once again, how like trig, functions these hyperbolic functions are. Make a list of these three definitions: then turn on to frame 31. 89 Hyperbolic Functions (2) Let us consider tanh x '■ sinh* e* - e x e* + e x cosh* 2 ' 2 e* - e~* = — -= tanh * e x + e' x sinh x (Very much like cosh* \ sin 6 . tan 9 = 2 I cos 9) 31 (3) Cosh x = \{e x + e"*); sinh x = ^(e* - e _x ) Add these results: cosh x + sinh x = e* Subtract : cosh x - sinh x = e' x Multiply these two expressions together : (cosh x + sinh x) (cosh x - sinh x) : ■'■ cosh 2 * - sinh 2 * = 1 Jin trig., we have cos 2 + sin 2 = 1 , so there is a difference in | (■ sign here. ) sign here. On to frame 32. (4) We just established that cosh 2 * - sinh 2 * = 1 . Divide by cosh 2 * : 1 r-=- = — r^- cosrr* cosh * .'. 1 - tanh 2 * = sech 2 * .'. sech 2 * = 1 - tanh 2 * {Something like sec 2 = 1 + tan 2 0, isn't it?} (5) If we start again with cosh 2 * - sinh 2 * = 1 and divide this time by sinh 2 *, we get cosh 2 * _ 1 sinh 2 * sinh 2 * .'. coth 2 * - 1 = cosech 2 * .*. cosech 2 * = coth 2 * - 1 I In trig., we have cosec 2 = 1 + cot 2 , so there is a sign difference I I here too. J Turn on to frame 33. 32 90 Programme 3 33 (6) We have already used the fact that cosh x + sinh * = e* and cosh x - sinh x = e~ If we square each of these statements, we obtain (0 (ii) 34 cosh 2 * + 2 sinh x cosh x + sinh 2 * = e 2X cosh 2 * — 2 sinh x cosh x + sinh 2 * = e" 2 * So if we subtract as they stand, we get g 2X _ e -2X 2 sinh x cosh x = = sinh 2x .'. sinh 2* = 2 sinh x cosh x If however we add the two lines together, we get .... 35 2(cosh 2 * + sinh 2 *) = e 2 * + e .'. cosh 2 * + sinh 2 * = cosh 2x .'. cosh 2x = cosh 2 * + sinh 2 * We already know that cosh 2 * - sinh 2 * = 1 .'. cosh 2 * = 1 + sinh 2 * Substituting this in our last result, we have cosh 2* = 1 + sinh 2 * + sinh 2 * .'. cosh 2* = 1 + 2 sinh 2 * Or we could say cosh 2 * - 1 = sinh 2 * .'. cosh 2* = cosh 2 * + (cosh 2 * - 1) .'. cosh 2* = 2 cosh 2 * — 1 Now we will collect all these hyperbolic identities together and com- pare them with the corresponding trig, identities. These are all listed in the next frame, so turn on. 91 Hyperbolic Functions Trig. Identities (1) cot* = 1/tan* sec* = 1/cos* cosec* = 1/sin* (2) cos 2 * + sin 2 * = 1 sec 2 * = 1 + tan 2 * cosec 2 * = 1 + cot 2 * (3) sin 2* = 2 sin * cos * cos 2* = cos 2 * - sin 2 * = 1 — 2 sin 2 * = 2 cos 2 * - 1 Hyperbolic Identities coth* = 1/tanh* sech* = 1/cosh* cosech* = 1/sinh* cosh 2 * - sinh 2 * = 1 sech 2 * = 1 - tanh 2 * cosech 2 * = coth 2 * - 1 sinh 2* = 2 sinh * cosh * cosh 2* = cosh 2 * + sinh 2 * = 1+2 sinh 2 * = 2 cosh 2 * - 1 36 If we look at these results, we find that some of the hyperbolic identities follow exactly the trig, identities: others have a difference in sign. This change of sign occurs whenever sin 2 * in the trig, results is being converted into sinh 2 * to form the corresponding hyperbolic identities. This sign change also occurs when sin 2 * is involved without actually being written as such. For example, tan 2 * involves sin 2 * since tan 2 * could be written as ^~. The change of sign therefore occurs with tan 2 * when it is being converted into tanh 2 * cot 2 * " "" " » » cot^ cosec 2 * " " " " " " cosech 2 * The sign change also occurs when we have a product of two sinh terms, e.g. the trig, identity cos(A + B) = cos A cos B - sin A sin B gives the hyperbolic identity cosh(A + B) = cosh A cosh B + sinh A sinh B. Apart from this one change, the hyperbolic identities can be written down from the trig, identities which you already know. For example: tan 2x = J^£- becomes tanh 2* = 2tanh * 1 - tan 2 * 1 + tanh 2 * So providing you know your trig, identities, you can apply the rule to form the corresponding hyperbolic identities. 92 Programme 3 37 Relationship between Trigonometric and Hyperbolic Functions From our previous work on complex numbers, we know that: e je =cos0 +jsin0 and e~ ,e = cos 6 - j sin 6 Adding these two results together, we have e J e +e-J 9 = 38 2 cos 6 So that. cos 8 e i e + e-J g which is of the form ■= , with jc replaced by G#) 39 cosh j0 Here, then, is our first relationship. cos 9 = cosh j0 Make a note of that for the moment: then on to frame 40. 40 If we return to our two original statements e je = cos# + j sin 6 e~ je = cos0 -j sin0 and this time subtract, we get a similar kind of result eJ e -e-J e = 41 2j sin 6 So that, e> e - e> e j sin 6 = — y 93 Hyperbolic Functions sinh j0 42 So, sinhjfl =j sin Mz&e a note of that also. So far, we have two important results: (i) cosh \8 = cos 8 (ii) sinh j0 = j sin Now if we substitute 8 = jx in the first of these results, we have cosjx = cosh(j 2 x) = cosh(- x) :. cos )x = cosh x [since cosh(-x) = cosh x] Writing this in reverse order, gives cosh x = cos ]x Another result to note. Now do exactly the same with the second result above, i.e. put 8 - jx in the relationship j sin 8 = sinh j0 and simplify the result. What do you get? 43 44 j sinhx = sinjx For we have : j sin 8 = sinh j0 j sin jx = sinh(j 2 x) = sinh(-jc) = -sinhx [since sinh(-x) = -sinhx] Finally, divide both sides by j, and we have sin jx = j sinh x Now on to the next frame. 94 Programme 3 45 Now let us collect together the results we have established. They are so nearly alike, that we must distinguish between them. sin jx = j sinhx sinh jjc = j sin x cosjx = cosh* cosh jx = cos x and, by division, we can also obtain tan jx =j tanhx tanhjx=jtanx Copy the complete table into your record book for future use. 46 Here is one application of these results: Example 1. Find an expansion for sin(x + \y). Now we know that sin(A + B) = sin A cos B + cos A sin B .'. sin(x + ]y) = sin x cos j y + cos x sin ]y so using the results we have listed, we can replace cos \y by and sin \y by 47 cos \y = cosh y sin jy = j sinhj> So that becomes sin(x + ]y) = sin x cos \y + cos x sin \y sin(jc + \y) = sin x cosh y + j cos x sinh y Note: sin(x + \y) is a function of the angle (x + j y), which is, of course, a complex quantity. In this case, (x +jy) is referred to as a Complex Variable and you will most likely deal with this topic at a later stage of your course. Meanwhile, here is just one example for you to work through. Find an expansion for cos(x - ]y). Then check with frame 48. 95 Hyperbolic Functions cos(x - )y) = cos x cosh y + j sin x sinh y 48 Here is the working: cos(A - B) = cos A cos B + sin A sin B .'. cos(x - jy) = cos x cos \y + sin x sin j y But cos jjy = cosh y and sin j.y = j sinh y .'. cos(x - jj>) = cos x cosh ^ + j sin x sinh >> 49 All that now remains is the test exercise, but before working through it, look through your notes, or revise any parts of the programme on which you are not perfectly clear. Then, when you are ready, turn on to the next frame. 96 Programme 3 50 Test Exercise — III 1 . If L = 2C sinh ^, find L when H = 63 and C = 50. 2. If v 2 = 1-8 L tanh-^, find v when d = 40 and L = 315. 3. On the same axes, draw sketch graphs of (i)y = sinh x, (ii)j' = coshx, (iii)^ = tanhx . „. .., 1 + sinh 2A + cosh 2A 4Simpllfy 1 - sinh 2A- cosh 2A 5. Calculate from first principles, the value of (i) sinh -1 1-532 (ii) cosh -1 1-25 6. If tanh x =-r, find e 2x and hence evaluate x. 7. The curve assumed by a heavy chain or cable is y = C cosh-pr If C = 50, calculate (i) the value of y when x = 109, (ii) the value of x whenj> = 75. i 8. Obtain the expansion of sin(x - jj>) in terms of the trigonometric and hyperbolic functions of x and>\ 97 Hyperbolic Functions Further Problems - HI 1 . Prove that cosh 2x = 1 + 2 sinh 2 x. 2. Express cosh 2x and sinh 2x in exponential form and hence solve, for real values of x, the equation 2 cosh 2x - sinh 2x = 2 3. If sinh x = tan.y, show that x = ln(sec^ +tanjO- 4. If a = c cosh x and b = c sinh x, prove that (fl + 6) 2 e- 2x =a 2 -6 2 5. Evaluate (i) tanh" '0-75, (ii) cosh -1 2. 6. Prove that tanrf 1 ! * ~ 1 ) = In x ( * + 1 ) 7. Express (i) cosh —Ll and (ii) sinh ^ in the form a + ]b, giving a and & to 4 significant figures. 8. Prove that (i) sinh (x + y) = sinh x cosh j + cosh x sinh >> (ii) cosh(x +j) = cosh x cosh y + sinh x sinh y Hence prove that tanh(x + y) = tanhx + tanhj _ 1 + tanhx tanhj> 9. Show that the co-ordinates of any point on the hyperbola x 2 v 2 ~2 ~p = 1 can be represented in the formx = a cosh u,y = b sinhw. 10. Solve for real values of x 3 cosh 2x = 3 + sinh 2x 11. Proveth a t- 1+tanh * = e 2 * 1 - tanh x 12. It t = tanhi, prove that sinhx = -~ 2 and coshx = j-^. Hence solve the equation 7 sinh x + 20 cosh x = 24 98 Programme 3 13. If x = ln tan sinh x = tan 6 [54 find e x and e * , and hence show that 14. Given that sinh *x = In [x + \J{x 2 + 1) j, determine sinh * (2 + j) in the forma +]b. 15. If tanj— J = tan A tanh B, prove that sin 2A sinh 2B tan x = ■ 1 + cos 2A cosh 2B 16. Prove that sinh 30 = 3 sinh 0+4 sinh 3 0. cos b 17. If x + )y = tarf 1 (e a+ J*), show that tan 2x = f , and that tanh 2j> : sin ft cosh a ' 18 ifx = a? ' s ' n ^ at + s ' n fl ^ 2 I cosh a/ - cos at sinh a' , calculate \ when a = 0-215 and t = 5. 19. Prove that rantr 1 ! *, a 2 \ = \n- [x +0* ) a 20. Given that sinh 1 x = ln{x + \/(x 2 + 1)}, show that, for small values ofx, 3 5 • t -1 * ■* JX sinh x -x- — + — . 6 40 99 Programme 4 DETERMINANTS Programme 4 1 Determinants You are quite familiar with the method of solving a pair of simultaneous equations by elimination. e.g. To solve 2x + 3y + 2 = ... (i) 3x + Ay + 6 = ... (ii) we could first find the value of x by eliminating y. To do this, of course, we should multiply (i) by 4 and (ii) by 3 to make the coefficient of y the same in each equation. So &c + 12>> + 8 = 9jc + 1 2y + 18 = Then, by subtraction, we get x + 10 = 0, i.e. x = -10. By substituting back in either equation, we then obtain y = 6. So finally, x=-10, y = 6 That was trivial. You have done similar ones many times before. In just the same way, if a x x +b x y +di = ... (i) a 2 x + biy + d 2 = ... (ii) then to eliminate^ we make the coefficients of y in the two equations identical by multiplying (i) by and (ii) by (i) by 6 2 and (ii) by 6 t Correct, of course. So the equations a x x +b^y + di = a 2 x + b 2 y + d 2 = become a x b 2 x + b^b^y + b 2 d\ - a 2 &iX + bib2y + bid 2 = Subtracting, we get (a l b 2 -a 2 bi)x + b 2 di ~bid 2 =0 so that {a l b 2 -a 2 bi)x = bid 2 -b 2 di Then x = 101 Determinants X = b 1 d 2 -b 2 di a l b 2 -a 2 b l In practice, this result can give a finite value for x only if the denominator is not zero. That is, the equations tfi* + b x y + d x = a 2 x + b^y + d 2 = give a finite value for x provided that {a x b 2 ~a 2 bi) f 0. Consider these equations: 3x + 2y - 5 = 4x + 3y - 7 = In this case, a 1 = 3, &i = 2, a 2 = 4, & 2 = 3 0i6 2 ~ a ibi - 3.3 -4.2 = 9-8=1 This is not zero, so there (will | be a finite value of x. will not will The expression 0^2 ~a 2 b\ is therefore an important one in the solu- tion of simultaneous equations. We have a shorthand notation for this. ai bi a \b 2 — a 2 b\ - For a x b x to represent a x b 2 -a 2 b\ then we must multiply the terms diagonally to form the product terms in the expansion: we multiply and then subtract the product bx i.e. + ^ and s e-g- So 3 7 5 2 6 5 1 2 = 3.2- 5.7 = 6- 35 =-29 5 1 102 Programme 4 6 5 1 2 = 12-5 = □DDnaDDnnnnnDnnDnnDDnnnonDDDDnDDnnnDDD is called a determinant of the second order (since it has two a 2 b 2 rows and two columns) and represents a y b 2 -a 2 b\. You can easily remember this as +^^-^-^ r . Just for practice, evaluate the following determinants : 4 ?. 7 4 2 1 (i) , (ii) , Oii) b 3 6 3 4 -3 Finish all three: then turn on to frame 6. 4 ?. (i) 5 3 7 4 (ii) 6 i 2 1 (iii) 4 -3 = 4.3-5.2= 12-10 = = 7.3-6.4 = 21-24 = = 2(-3)-4.1 =-6-4 = -3 -10 anDDnnaDnDDDDnnnanannnDaDnnnDDaDDDnnDQ Now, in solving the equations J a t x + b^y + di =0 { a 2 x + biy + d 2 =0 we found that x = -ij-2 1-1 and the numerator and the denominator aib 2 — a 2 b\ can each be written as a determinant. b 1 d 2 -b 2 di = ; a x b 2 -a 2 b x = 103 Determinants bi di b 2 d 2 i «i b x a 2 b 2 If we eliminate x from the original equations and find an expression for y, we obtain _ Jaid 2 -a 2 di\ ]aib 2 -a 2 b y So, for any pair of simultaneous equations i\X + b\y +di = a 2 x + b^y + d 2 = we have .b y d 2 -b 2 di and _a\d 2 -a 2 d x a\b 2 -a 2 b\ """ ^ fli6 2 -a 2^i Each of these numerators and denominators can be expressed as a determinant. So, x = and y = bi d, ai di X = b 2 d 2 bi and y = - a 2 d 2 «i 61 a 2 b 2 «2 b 2 X 1 and y -1 6, d t ai 61 "i di flj 61 Z> 2 d 2 a 2 b 2 a 2 d 2 a 2 b 2 We can combine these results, thus: X ~y 1 bi d x ai d\ 1 «i 61 b 2 d 2 a 2 d 2 | a 2 6 2 Afafce a note 0/ f«ese results and then turn on to the next frame. 8 104 Programme 4 So if Then a y x +bty +di =0 a 2 x + b^y + d 2 = X -y 1 bi d t «i d x 1\ by b 2 d 2 a 2 d 2 a 2 b 2 Each variable is divided by a determinant. Let us see how we can get them from the original equations. (i) Consider . — . Let us denote the determinant in the denominator b x dx b 2 d 2 bi d, i b 2 d 2 by Aj , i.e. A t To form Ai from the given equations, omit the x-terms and write down the coefficients and constant terms in the order in which they stand. <tix +biy +d t = a 2 x + bjy + d 2 = -y (ii) Similarly for di , let A 2 6, d. gives b 2 d 2 a i di "2 d 2 a 2 d 2 To form A 2 from the given equations, omit the j;-terms and write down the coefficients and constant terms in the order in which they stand. aix + biy +d t = (iii) For the expression a 2 x + b-^y + d 2 = 1 gives A 2 = di a\ by 2 d 2 denote the determinant by Aq. a 2 b 2 To form Aq from the given equations, omit the constant terms and write down the coefficients in the order in which they stand aix + b x y + d x =0 ay b x gives a 2 x + b2y + d 2 = a 2 b 2 .2 =1 A 2 A Note finally that _x A t Now let us do some examples, so on to frame 10. 105 Determinants Example I. To solve the equations j Sx + 2y + 19 = The key to the method is — - ~y - * A, A 2 Aq To find A , omit the constant terms 5 2 •■• Ao 3 4 = 5.4-3.2 = 20-6= 14 .-. Ao = 14 ... (i) Now, to find Ai , omit the x -terms. ■•• A, = for Ai = 2 19 Ai =-42 4 17 Similarly, to find A 2 , omit the j>-terms 5 19 34 -76 = -42 A,= 3 17 85-57 = 28 • (iO (iii) Substituting the values of A t , A 2 , A in the key, we get x _—y _ J_ -42 ~ 28 ~ 14 from which x = and y = -42 28 X = 14 " '■-i; -y- " 14 ,y = ~2 Now for another example. Example 2. Solve by determinants |2x+3>'-14 = \3x-2y+ 5 = First of all, write down the key: x _^y _ J_ Ai _ A 2 "'Aq (Note that the terms are alternately positive and negative.) 2 3 Then A, -4- 9 =-13 3 -2 Now you find Ai and A 2 in the same way. 0) 10 11 12 106 Programme 4 13 Ai =-13; A 2 =-52 For we have 2x+3y-14 = 3x - 2y + 5 = 3 -14 3 -14 • A, = = — -2 5 5 -2 = 15 -28 = -13. :. Aj =-13 ?, -14 ?. -14 A, = = — i 5 t> 3 So that and = 10 -(-42) =52 :. A 2 = 52 x _ -y _ 1 A! " A 2 " A A! =-13; A 2 =52; A =-13 _Ai_-13_, . _, _ A 2 _ 52 . -' = at^3 = " 4 "zn Do not forget the key x _-y _ 1 A^ "A 2 ~A^ with alternate plus and minus signs. Make a note of this in your record book. 14 Here is another one: do it on your own. Example 3. Solve by determinants 4jc - 3y + 20 = 3x + 2y - 2 = First of all, write down the key. Then off you go: find A , Ai and A 2 and hence determine the values of x and.y. When you have finished, turn on to frame 15. 107 Determinants x = -2; y = 4 Here is the working in detail: 4x - 3y + 20 = 3x + 2y- 2 = 4 -3 Ao 15 A t ~ A 2 Ao Ai A 2 = 3 2 -3 20 2 -2 4 20 3 -2 Ai _-34 Ao 17 A 2 -68 . = 8-(-9) = 8 + 9 = 17 = 6 - 40 = -34 = -8 - 60 = -68 -2 :. x = -2 :. j = 4 nDODnnDDDnaDnnnnDDnaDDDnDDDnnannnDnDnn Now, by way of revision, complete the following: (0 00 (iii) (iv) 5 6 7 4 5 -2 ■3 -4 b c P q r s Here are the results. You must have got them correct. (0 20- -42 = -22 (ii) -20 -6 = -26 (iii) ac- bd (iv) ps- rq 16 For the next section of the work, turn on to frame 1 7. 108 Programme 4 Determinants of the third order A determinant of the third order will contain 3 rows and 3 columns, thus: 0i bi Cx a 2 b 2 c 2 17 3 b 3 c 3 Each element in the determinant is associated with its MINOR, which is found by omitting the row and column containing the element concerned. e.g. the minor of a, is the minor of b\ is the minor of c t is b 2 b 3 a 2 a 3 a 2 a 3 ?2 c 3 c 3 b 2 b 3 obtained obtained obtained : «i ! bi Cll \ a 2 \ b 2 c 2 ! «3 ! b 3 c 3 \a l \ bi i C1 i a 2 \b 2 \ c 2 a 3 : b 3 • c 3 01 «2 ~b 2 So, in the same way, the minor of a 2 is a 3 b 3 c 2 \ c 3 i 18 Minor of a 2 is by c t b 3 c 3 since, to find the minor of a 2 , we simply ignore the row and column con- taining a 2 , i.e. Similarly, the minor of b 3 is . ! fli ! bj. Cl 1 a 2 j fc 2 Cl : a 3i ^>3 c 3 19 Minor of b 3 - a x Cy a 2 c 2 i.e. omit the row and column containing b 3 . 01 61 i Cl 02 6 2 ' Ci % b 3 \ c 3 \ J Now on to frame 20. 109 Determinants Evaluation of a third order determinant To expand a determinant of the third order, we can write down each element along the top row, multiply it by its minor and give the terms a plus or minus sign alternately. «i b t Ci ~ ^7 a\ ■b t a 2 c 2 a 3 c 3 + c, a 2 b 2 a 3 b 3 a 3 b 3 c 3 Then, of course, we already know how to expand a determinant of the second order by multiplying diagonally, + Example 1. 1 3 2 1 5 7 -3 4 7 + 2 4 5 4 5 7 = 4 8 2 8 2 4 2 4 8 = 1(5.8 - 4.7) - 3(4.8 - 2.7) + 2(4.4 - 2.5) = l(40-28)-3(32-14) + 2(16-10) = 1(12) -3(18) + 2(6) = 12- 54 + 12 =-30 20 Here is another. Example 2. 3 2 5 3 6 7 -2 4 7 + 5 4 6 4 6 7 — 9 2 ?. 2 ?, 9 2 9 2 = 3(12 -63) -2(8- 14) + 5(36- 12) = 3(-51)-2(-6) + 5(24) = -153 + 12 + 120 = -21 Now here is one for you to do. Example 3. Evaluate 2 7 5 4 6 3 8 9 1 Expand along the top row, multiply each element by its minor, and assign alternate + and — signs to the products. When you are ready, move on to frame 22. 21 no Programme 4 22 Result For 2 4 7 5 6 3 9 1 38 6 3 -7 4 3 + 5 4 6 9 1 8 1 8 9 = 2(6 - 27) - 7(4 - 24) + 5(36 - 48) = 2(-21)-7(-20) + 5(-12) = -42+ 140-60 = 38 We obtained the result above by expanding along the top row of the given determinant. If we expand down the first column in the same way, still assigning alternate + and - signs to the products, we get 2 7 5 2 6 3 -4 7 5 +8 7 5 4 6 3 — 9 1 9 1 6 3 8 9 1 = 2(6 - 27) - 4(7 - 45) + 8(21 - 30) = 2(-21)-4(-38) + 8(-9) = -42 + 152-72 = 38 which is the same result as that which we obtained before. y ^ We can, if we wish, expand along any row or column in the same way, multiplying each element by its minor, so long as we assign to each product the appropriate + or- sign. The appropriate 'place signs' are given by +_ + _ + - + - + - . + - + - + . - + - + - etc., etc The key element (in the top left-hand corner) is always + . The others are then alternately + or - , as you proceed along any row or down any column. So in the determinant 13 7 5 6 9 4 2 8 the "place sign" of the element 9 is Ill Determinants 24 since in a third order determinant, the 'place signs' are + - + - + - + - + Now consider this one Remember that the top left-hand element always has a + place sign. The others follow from it. 3 7 2 6 8 4 1 9 5 If we expand down the middle column, we get 3 7 2 -7 6 4 + 8 3 2 -9 3 2 6 8 4 — 1 5 1 5 6 4 1 9 5 Finish it off. Then move on. Result for -78 6 4 1 5 + 8 3 2 1 5 -9 3 2 6 4 25 = -7(30 -4) + 8(1 5 -2) -9(12- 12) = -7(26) + 8(13)- 9(0) = -182 + 104 = -78 So now you do this one: Evaluate 2 3 4 6 1 3 5 7 2 by expanding along the bottom row. R%e« you have done it, turn to frame 26. 112 Programme 4 26 Answer We have 2 3 4 6 1 3 5 7 2 One more: Evaluate 119 and remember = 5 + - + - + - + - + 3 4 1 3 -7 2 4 6 3 + 2 2 3 6 1 1 2 8 7 3 1 4 6 9 = 5(9 - 4) - 7(6 - 24) + 2(2 - 18) = 5(5)-7(-18) + 2(-16) = 25 + 126-32=119 by expanding along the middle row. 27 143 2 8 6 9 + 3 1 8 4 9 - 1 1 2 4 6 Result For 12 8-7 7 3 1 4 6 9 = -7(1 8 -48) + 3(9 -32) -1(6 -8) = -7(-30) + 3(-23)-l(-2) = 210-69 + 2= 143 DDDDnDDDDnDnDnDnnnDnDnDDDnnDnDDDnDnnDn We have seen how we can use second order determinants to solve simultaneous equations in 2 unknowns. We can now extend the method to solve simultaneous equations in 3 unknowns. So turn on to frame 28. 113 Determinants Simultaneous equations in three unknowns Consider the equations a x x + b x y + c x z + d x = a 2 x + b 2 y + c 2 z + d 2 = a 3 x + b 3 y + c 3 z + d 3 = If we find x, y and z by the elimination method, we obtain results that can be expressed in determinant form thus: x —y _ z _ -1 28 ftl Cl di b 2 c 2 d 2 b 3 c 3 di ci d x c-i d 2 a x bi d t a 2 b 2 d 2 «3 b 3 d 3 a\ bi Cl a 2 b 2 c 2 "3 b 3 c 3 We can remember this more easily in this form:— x Ai ■ZL A 2 z aT ^i_ A where Ai = the det. of the coefficients omitting the x-terms A 2 = " " " " " " " j-terms A 3 = " " " " " " " z-terms Ao = " " " " " " " constant terms. Notice that the signs are alternately plus and minus. Let us work through a numerical example. Example 1. Find the value of x from the equations '2x + 3y- z- 4 = 3x+ y + 2z- 13 = x + 2y - Sz + 1 1 = First the key: 2L=Z2. A, A 2 z : a! ^1 Ao x -l To find the value of x, we use — = — , i.e. we must find A] and A . Ai A (i) to find A , omit the constant terms. A = = -18 + 51-5 = 28 (ii) Now you find Ai , in the same way. 2 3 -1 ?. 1 7. -3 3 7 -1 3 1 3 1 2 = 2 -5 1 -5 1 2 1 2 -5 114 Programme 4 29 for Ai A t =-56 3 -1 -4 = 3(22 -65) + 1(11 +26)-4(-5-4) 1 2 -13 = 3(-43)+l(37)-4(-9) 2-5 11 = -129 + 37 + 36 = -129 + 73 =-56 A i . -1 . x _-l "A " -56 28 But Note that by this method we can evaluate any one of the variables, without necessarily finding the others. Let us do another example. Example 2. Find y, given that f 2x+ ^-5z + ll=0 ! x- y + z- 6 = [4x + 2y-3z+ 8 = First, the key, which is 30 Aj A 2 z a; Ao To find j, we use A 2 A Therefore, we must find A 2 and Ao • 2x+ >>-5z+ 11 =0 x— y + z- 6 = 4.x + 2j> - 3z + 8 = To find A 2 , omit the ^-terms. 2 -5 11 The equations are .". A 2 1 -3 1 -6 -3 8 + 5 + 11 1 -3 2(8 - 18) + 5(8 + 24) + 1 1(-3 - 4) -20 + 160-77 = 63 To find A , omit the constant terms ••• Ao = 115 Determinants for Ao=-21 Ao = 2 1 -5 1 -1 1 4 2-3 = 2 -1 1 2 -3 - 1 1 1 4 -3 -5 1 -1 4 2 = 2(3 - 2) - l(-3 - 4) - 5(2 + 4) = 2 + 7-30 = -21 ~y -i . A 2 63 So we have — = -. — . . y =ir- —z-, A 2 A Aq -21 :.y=-3 The important things to remember are (i) The key X A, ~y _ z A 2 A; "<5 ^0 31 with alternate + and - signs. (ii) To find Ai , which is associated with x in this case, omit the x -terms and form a determinant with the remaining coefficients and con- stant terms. Similarly for A 2 , A 3 , A . Next frame. Here is a short revision exercise on the work so far. Revision Exercise Find the following by the use of determinants. 32 l. x+2y-3z- 3 = 2x- y- i- 11 =0 3x + 2y + z + 5=0 - Ay + 2z + 8 = | + 5y-3z+ 2 = \ J 3x- x + 5y 5x + 3y - z+ 6 Find y. Find* andz. 2x-2y- z- 3 = 4x + 5y - 2z + 3=0 f Find x,y and z. 3x + Ay - 3z + 7 = When you have finished them all, check your answers with those given in the next frame. 116 Programme 4 33 Here are the answers: 1. y=-4 2. x=-2; z = 5 3. x = 2; y = _ 1; z = 3 If you have them a// correct, turn straight on to frame 52. If you have not got them all correct, it is well worth spending a few minutes seeing where you may have gone astray, for one of the main applications of determinants is in the solution of simultaneous equations. If you made any slips, move to frame 34. 34 The answer to question No. 1 in the revision test was y = -4 Did you get that one right? If so, move on straight away to frame 41. If you did not manage to get it right, let us work through it in detail. The equations were ( x + 2y- 3z- 3 = 2x- y- z- 11 =0 3x + 2y+ z + 5 = Copy them down on your paper so that we can refer to them as we go along. The first thing, always, is to write down the key to the solutions. In this case: — - A! To fill in the missing terms, take each variable in turn, divide it by the associated determinant, and include the appropriate sign. So what do we get? On to frame 35. 117 Determinants x _~y_ z _ H_ a7 ~a7~a^"a 35 The signs go alternately + and — . In this question, we have to find y, so we use the second and last terms in the key. • -y -l . A 2 - = — y = ir- 1-6- A A 2 A So we have to find A 2 and A . To find A 2 , we A 36 form a determinant of the coefficients omitting those of the >>-terms. So 1 -3 -3 A 2 = 2-1 -11 3 1 5 Expanding along the top row, this gives A, = -1 -11 1 5 ■(-3) 2 -11 3 5 + (-3) 2 -1 3 1 We now evaluate each of these second order determinants by the usual process of multiplying diagonally, remembering the sign convention that ^and —^^ So we get A 2 = 118 , Programme 4 37 A, = 120 for A 2 = l(-5 + 1 1) + 3(10 + 33) - 3(2 + 3) = 6 + 3(43)- 3(5) = 6 + 129-15= 135- 15 = 120 A A 2 = 120 We also have to find A , i.e. the determinant of the coefficients omit- ting the constant terms. So A = 38 If we expand this along the top row, we get Ao 39 Ao = -1 -1 -2 2 -1 -3 2 -1 2 1 3 1 3 2 Now, evaluating the second order determinants in the usual way gives that Ao= 119 Determinants Ao =-30 40 for So we have A = 1(-1 + 2)- 2(2 + 3)- 3(4 + 3) = 1(1) -2(5) -3(7) = 1-10-21 =-30 So Aq=-30. _A 2 _120_ y A -30 :.y = -A Every one is done in the same way. Did you get No. 2 of the revision questions correct? If so, turn straight on to frame 51. If not, have another go at it, now that we have worked through No. 1 in detail. When you have finished, move to frame 41. The answers to No. 2 in the revision exercise were x = — 2 z = 5 41 Did you get those correct? If so, turn on right away to frame 51 . If not, follow through the working. Here it is: No. 2 The equations were ' 3x - Ay + 2z + 8 = x + 5y-3z + 2 = 5x + 3y- z + 6 = Copy them down on to your paper. The key to the solutions is: x _ _ —-...- ... - ... Fill in the missing terms and then turn on to frame 42. 120 42 A t ~ A 2 A 3 A have to findx andz. .-. We shall use X 1 Ao i.e. x = Ai Ao and z A, -1 Ao i.e. z =- A 3 Ao Programme 4 So we must find Ai , A 3 and A . (i) To find Ai , form the determinant of coefficients omitting those of the .x-terms. /. A x 43 -4 2 8 A t = 5 -3 2 3 -1 6 Now expand along the top row. A, = -3 2 -2 5 2 + 8 5 -3 -1 6 3 6 3 -1 Finish it off: then on to frame 44. 121 Determinants A, =48 44 for A t =-4(-18 + 2)-2(30-6)+8(-5 + 9) = -4(-16)- 2(24) + 8(4) = 64-48 + 32 = 96-48 = 48 :. Aj = 48 (ii) To find A 3 , form the determinant of coefficients omitting the z-terms. ■■• A, A 3 = 3-4 8 1 5 2 5 3 6 Expanding this along the top row gives A 3 = 45 3 A 3 = 5 2 3 6 + 4 1 2 5 6 + 8 1 5 5 3 46 Now evaluate the second order determinants and finish it off. So that A 3 = On to frame 47. 122 47 since Programme 4 -120 A 3 = 3(30 - 6) + 4(6 - 10) + 8(3 - 25) = 3(24) + 4(-4) + 8(-22) = 72-16-176. = 72-192 = -120 :.A 3 =-120 (iii) Now we want to find Ao • Ao = 48 3 -4 2 Ao = 1 5 -3 5 3 -1 Now expand this along the top row as we have done before. Then evaluate the second order determinants which will appear and so find the value of A . Work it right through: so that Ao = 123 Determinants for Ao=24 5 - 3 - -3 -1 + 4 1 - 5 - -3 -1 + 2 1 5 5 3 49 So we have: Also we know that Ao = = 3(-5 + 9) + 4(-l + 1 5) + 2(3 - 25) = 3(4) + 4(14) + 2(-22) = 12 + 56-44 = 68 - 44 = 24 :. A =24 At =48, A 3 =-120, A<> =24 So that x = and z = ... 50 48 x=-2 (-120) _ z — 24 ~ 5 z = 5 Well, there you are. The method is the care not to make a slip with the signs. same ev( :ry time — but take Now what about question No. 3 in the revision exercise. Did you get that right? If so, move on straight away to frame 52. If not, have another go at it. Here are the equations again: copy them down and then find*,.}> andz. 2x-2y- z-3 = Ax + Sy - 2z + 3 = 1 3x + Ay - 3z + 7 = When you have finished this one, turn on to the next frame and check your results. 124 Programme 4 51 Answers to No. 3 2, y=~\, z = 3 Here are the main steps, so that you can check your own working. _^L = IZ = JL =Zi_ Ax A 2 A 3 A -2 -1 -3 A, = 5 -2 3 = 54 4 -3 7 2 -1 -3 A 2 = 4 -2 3 = 27 3 -3 7 2 -2 -3 A 3 = 4 5 3 = 81 3 4 7 2 -2 -1 Ao = 4 5 -2 = -27 3 4 -3 X A, 1 Ao x =- Ao =-ii=2 -27 x=2 -y A 2 1 A y = A 2 A -27 ^=-1 z a; 1 A z =- _A 3 A = li = -3 -27 z =-3 All correct now? On to frame 52 then for the next section of the work. 125 Determinants Consistency of a set of equations Let us consider the following three equations in two unknowns. 52 3x- y-4 = (0 2x + 3y~8 = (ii) x- y-4=Q (iii) If we solve equations (ii) and (iii) in the usual way, we find that x = 1 and y = 2. If we now substitute these values in the left-hand side of (i), we obtain 3x -j-4 = 3-2-4= -3 (and not as the equation states). The solutions of (ii) and (iii) do not satisfy (i) and the three given equations do not have a common solution. They are thus not consistent. There are no values ofx and y which satisfy all three equations. If equations are consistent, they have a 53 common solution Let us now consider the three equations 3x + y - 5 = (i) 2x + 3y - 8 = (ii) x - 2y + 3 = (iii) The solutions of (ii) and (iii) are, as before, x = 1 and y = 2. Substituting these in (i) gives 3x +y -5 = 3 + 2-5 = i.e. all three equations have the common solution x= l,y = 2 and the equations are said to be c 126 Programme 4 54 consistent Now we will take the general case a x x + b±y +di = a 2 x + b 2 y + d 2 - a 3 x + b 3 y +d 3 = If we solve equations (ii) and (Hi), i.e. ( a 2 x + biy + d 2 = a 3 x + 63J + d 3 = we get Ai " A 2 " A (i) 00 (iii) where so that A, = b 2 d 2 , A 2 = a 2 d 2 , A = a 2 b 2 b 3 d 3 a 3 d 3 «3 b 3 A t _ A 2 x - V" and y - ~ -7— Ao Aq If these results also satisfy equation (i), then ay^ + bv^ + d^Q i.e. i.e. i.e. fli-Ai ~b y A 2 +d v A = d 2 d 3 ■61 a 2 <2 2 a 3 d 3 fll 61 C?i a 2 i 2 c? 2 a 3 63 c? 3 + di fl 2 £> 2 fl 3 Z> 3 which is therefore the condition that the three given equations are consistent. So three simultaneous equations in two unknowns are consistent if the determinant of coefficients is 127 Determinants Example 1. Test for consistency For the equations to be consistent 55 must be zero. 1 4 -1 -5 1 -10 = 2 4 1 - 1 1 1 _5 1 1 4 -1 -10 3 -10 13 -1 = 2(-40+l)-l(-10-3)-5(-l-12) = 2(-39)-(-13)-5(-13) = -78 + 13 +65 =-78 + 78 = The given equations therefore consistent. (are/are not) are ;h the equations are consistent. 56 Example 2. Find the value of k or whk (3x+ y + 2 = 3 1 2 J Ax + 2y - k = For consistency, 4 2 -k = [2x- y + 3k = 2 -1 3& :. 3 2 -k -1 4 -k + 2 4 2 = • -1 3k 2 3k 2 -1 3(6& - k) - 1 (1 2k + 2k) + 2(-4 - 4) = /. 15Ar- 14fc- 16 = .\ fc-16 = .\ fc=16 Now one for you, done in just the same way. Example 3. Given ( x + (k + 1 )y + 1 = \ 2kx + Sy -3 = { 3x+ ly + 1 = Find the values of k for which the equations are consistent. 128 Programme 4 57 k = 2 or - 1 The condition for consistency is that 1 k + l 1 2k 5 -3 = 3 7 1 5 -3 -(*+0 2k -3 + 1 2k 5 7 1 3 1 3 7 = (5 + 21)-(fc + 1) (2fc + 9) + (14fc- 15) = 26-2fc 2 -llfc-9+ 14fc-15 = -2k 2 + 3k + 2 = :. 2k 2 -3k~2 = :. (2k+l)(k~2) = Q _1 .". & = 2 or k=-x Finally, one more for you to do. Example 4. Find the values of k for consistency when X + y- k = kx - 3y + 1 1 = 2x + Ay - 8 = 58 k = 1 or -x For 1 1 -k fc -3 11 = 2 4-8 1 -3 11 4 -8 - 1 k 11 2 -8 -k k -3 2 4 = :. (24 - 44) - (-8k - 22) - k(4k + 6) = :. -20 + 8k + 22 - 4/c 2 - 6k = -4k 2 + 2k + 2 = :. 2/fc 2 - A: - 1 = :. (2k + 1) (k - 1) = .'. fc = 1 or k = — y 129 Determinants Properties of determinants Expanding a determinant in which the elements are large numbers can be a very tedious affair. It is possible, however, by knowing something of the properties of determinants, to simplify the working. So here are some of the main properties. Make a note of them in your record book for future reference. 59 1. The value of a determinant remains unchanged if rows are changed to columns and columns to rows. h b. a x b x a 2 b 2 2. If two rows (or two columns) are interchanged, the sign of the determinant is changed. a 2 b 2 at bi a t b\ a 2 b 2 3. If two rows (or two columns) are identical, the value of the deter- minant is zero. = a 2 a 2 4. If the elements of any one row (or column) are all multiplied by a common factor, the determinant is multiplied by that factor. ka x kb-t a 2 b 2 a\ bx a 2 b 2 5. If the elements of any one row (or column) are increased (or decreased) by equal multiples of the corresponding elements of any other row (or column), the value of the determinant is unchanged. a x +kbi b x fii bi a 2 + kb 2 b 2 a 2 b 2 DannnaDnnnnnDnDDnnDDDnDDDnnDnDDDDDDDDD NOTE: The properties stated above are general and apply not only to second order determinants, but to determinants of any order. Turn on now to the next frame for one or two examples. 130 Programme 4 60 427 429 Example 1. Evaluate 369 371 Of course, we could evaluate this by the usual method (427) (371) -(369) (429) which is rather deadly! On the other hand, we could apply our knowledge of the properties of determinants, thus: (Rule 5) 427 429 427 429 - 427 369 371 369 371 -369 427 2 369 2 58 369 2 (Rule 5) = (58)(2)-(0) = 116 Naturally, the more zero elements we can arrange, the better. For another example, move to frame 61. 61 Example 2. Evaluate 1 2 2 4 3 5 4 2 7 1 0. 2 4 -2 5 4 -5 7 1 4 -2 -3 4 -5 -1 -2 - -3 -5 - -1 Next frame. column 2 minus column 3 will give us one zero column 3 minus twice (column 1) will give another zero Now expand along the top row We could take a factor (-1) from the top row and another factor (-1) from the bottom row. (-1X-1) 2 3 5 1 1(2-15) = -13 131 Determinants Example 3. Evaluate You do that one, but by way of practice, apply as many of the listed properties as possible. It is quite fun. When you have finished it, turn on to frame 63. 62 The answer is 32 , but what we are more interested in is the method 63 of applying the properties, so follow it through. This is one way of doing it; not the only way by any means. 4 2 2 2 1 1 2 1 1 -1 -1 1 -3 -1 1 3 1 4 1 4 2 2 4 2 2 4 1 1 2 1 1 2 1 1 1 2 1 1 2 -1 1 -3 -1 -1 1 -4 We can take out a factor 2 from each row, giving a factor 2 3 , i.e. 8 outside the determinant. column 2 minus column 3 will give one zero in the top row. column 1 minus twice (column 3) will give another zero in the same row. Expanding along the top row will now reduce this to a second order determinant. Now row 2 + row 1 = -8 (-4) = 32 132 Programme 4 64 Here is another type of problem. Example 4. Solve the equation x 5 5 x+ 1 -3-4 x - 2 In this type of question, we try to establish common factors wherever possible. For example, if we add row 2 and row 3 to row 1 , we get (x + 2) (x + 2) (x + 2) 5 x+ 1 1 -3 -4 x - 2 Taking out the common factor (pc + 2) gives (*+2) 1 1 5 jc + 1 -3-4 x - 2 Now if we take column 1 from column 2 and also from column 3, what do we get? When you have done it, move on to the next frame. 65 We now have (x + 2) 1 5 jc-4 -4 -3 -1 - x + 1 Expanding along the top row, reduces this to a second order determinant. (x + 2) x - 4 -4 -1 x + l If we now multiply out the determinant, we get (x + 2) [(x-4)(x+ l)-4] =0 :. (x + 2) (x 2 - 3x - 8) = = x + 2 = or jc 2 -3x-8 = which finally gives x = -2 or x .3+V41 Finally, here is one for you to do on your own. Example 5. Solve the equation 5x3 x+2 2 1=0 -3 2 x Check your working with that given in the next frame. 133 Determinants Result: x = -4 or 1 ± \/6 Here is one way of doing the problem 5 x 3 x + 2 2 1=0 -3 2 x x + 4 x +4 x + 4 x + 2 2 1 -3 2 (x+4) (*+4) (x + 4) (x + 4) 1 x + 2 -3 x + 1 -jc — 3 2 - x x + l 1 ■x - 3 2 - x 1 2 2 1 = = = x 1 -5 2-x ,2 :. (x+4)(2x-x 2 + 5) = ;. x + 4 = or x 2 - 2x - 5 = which gives x = -4 or x = 1 ± \J6 Adding row 2 and row 3 to row 1 , gives Take out the common factor (x + 4) Take column 3 from column 1 and from column 2 This now reduces to second order Subtract column 2 from column 1 We now finish it off 66 onaannnnoDDDQaoaaaaonnaDnaDaananonaoaa You have now reached the end of this programme on determinants except for the Test Exercise which follows in frame 67. Before you work through it, brush up any parts of the work about which you are at all uncertain. If you have worked steadily through the programme, you should have no difficulty with the exercise. 134 Programme 4 Q g Test Exercise — IV Answer all the questions. Take your time and work carefully. There is no extra credit for speed. Off you go then. They are all quite straightforward. DnnannnDnnnDnDDaannnnnannnnaDDDnnDnDnn 1. Evaluate (a) 1 1 2 (b) 1 2 3 2 1 1 3 1 2 1 2 1 2 3 1 2. By determinants, find the value of x, given '2x+3y~ z- 13 = x-2y + 2z+ 3=0 3x+ y + z- 10 = 3. Use determinants to solve completely *-3.y+4z-5 = 2x+ y+ z-3 = [4x + 3y + 5z-l=Q 4. Find the values of k for which the following equations are consistent 3x + 5y+k = 2x + y- 5 = (*+ \)x +2y- 10 = 5. Solve the equation x + 1 -5 -6 -\ x 2 -3 2 x + 1 = Now you can continue with the next programme. □ □QnnonaDnaQaQQnnaaoannaciaQLjnannaQanna 135 Determinants Further Problems — IV 1. Evaluate 2. Evaluate 0) (i) 3 5 7 11 9 13 15 17 19 25 3 35 16 10 -18 34 6 38 Oi) 00 1 428 861 2 535 984 3 642 1107 155 226 81 77 112 39 74 HI 37 3. Solve by determinants 4x-5v + 72 =-14 9x + 2v + 3z = 47 x - y - Sz - 11 4. Use determinants to solve the equations 4x - 3y + 2z = -7 2x - Ay - z = -3 5. Solve by determinants 3x + 2y - 2z = 16 4x + 3y + 3z = 2 2x- .y + z=-l 6. Find the values of X for which the following equations are consistent 5x+(\+l)v-5 = (\-1)jc + 7^ + 5 = 3x + 5 v + 1 = 7. Determine the values of k for which the following equations have solutions other than x = y = 4x-(k-2)y- 5 = 2x + y -10 = (fc+l)x - 4j/- 9 = 136 Programme 4 8. (a) Find the values of k which satisfy the equation = k 1 1 k 1 1 k (b) Factorise 1 1 1 a b c a 3 b 3 c 3 9. Solve the equation jc 2 3 2 x + 3 6 = 3 4 x + 6 10. Find the values of x that satisfy the equation x 3 +x 2+ x 3 -3 -1 = 2 -2 -2 11. Express 1 a' b 2 c 2 b 2 c 2 2 <„ j. „\2 („ j. M2 (b+cf {c+af (a+bf as a product of linear factors. 12. A resistive network gives the following equations. 2(i3-/a) + 5(j 3 -/i) =24 (i 2 -«3) + 2i 2 +(i 2 -ii)= 5(i, -i 3 ) + 2(i,-i 2 ) + ;,= 6 Simplify the equations and use determinants to find the value of i 2 correct to two significant figures. 13. Show that (a + b + c) is a factor of the determinant b + c a a 3 c+a b b 3 i a +b c c 3 and express the determinant as a product of five factors. 137 Determinants 14. Find values of fc for which the following equations are consistent. x + (1 + k)y + 1=0 (2 + k)x + 5y - 10 = x + 7y + 9 = 15. Express 1 + x 2 1+y 2 1+z 2 yz 1 zx 1 xy 1 as a product of four linear factors. 16. Solve the equation x+l x+2 3 2 x+3 x+l x+3 1 x+2 = 17. Ifx,.y, z, satisfy the equations (iM, +M 2 >r-M i y = W -M 2 x + 2M 2 >> + (Mi - M 2 )z = ~M 2 y+QU l + M 2 )z = evaluate x in terms of W, M, and M 2 . 18. Three currents, i t , i 2 , i 3 , in a network are related by the following equations. ^ + ^ + ^ = 3Q 6ii - i 2 + 2/ 3 = 4 3r\ - \2i 2 +8i 3 = By the use of determinants, find the value of i t and hence solve com- pletely the three equations. 19. If k(x-a) + 2x-z = k(y-a) + 2y-z = k(z-a)-x~y + 2z = show that *= fq, + 3) k 2 +4k + 2 20. Find the angles between = and = n that satisfy the equation 1 + sin 2 cos 2 4 sin 20 sin 2 sin 2 1 + cos 2 4 sin 20 cos 2 1 + 4 sin 20 = 138 Programme 5 VECTORS Programme 5 | Introduction: scalar and vector quantities Physical quantities can be divided into two main groups, scalar quantities and vector quantities. (a) A scalar quantity is one that is defined completely by a single number with appropriate units, e.g. length, area, volume, mass, time, etc. Once the units are stated, the quantity is denoted entirely by its size or magnitude. (b) A vector quantity is defined completely when we know not only its magnitude (with units) but also the direction in which it operates, e.g. force, velocity, acceleration. A vect<»- quantity necessarily involves direction as well as magnitude. So, (i) a speed of 10 km/h is a scalar quantity, but (ii) a velocity of '10 km/h due North' is a quantity. vector A force F acting at a point P is a vector quantity, since to define it / completely we must give (i) its magnitude, and also (ii) its direction So that: (i) A temperature of 100°C is a quantity. (ii) An acceleration of 9-8 m/s 2 vertically downwards is a quantity. (iii) The weight of a 7 kg mass is a quantity. (iv) The sum of £500 is a quantity. (v) A north-easterly wind of 20 knots is a quantity. 141 Vectors (i) scalar, (ii) vector, (iii) vector, (iv) scalar, (v) vector Since, in (ii), (iii) and (v), the complete description of the quantity includes not only its magnitude, but also its direction Vector representation A vector quantity can be represented graphically by a line, drawn so that: (i) the length of the line denotes the magnitude of the quantity, according to some stated vector scale, (ii) the direction of the line denotes the direction in which the vector quantity acts. The sense of the direction is indicated by an arrow head. e.g. A horizontal force of 35 N acting to the right, would be indicated by a line > and if the chosen vector scale were 1 cm = 10 N, the line would be cm long. 3-5 The vector quantity AB is referred to as AB or a The magnitude of the vector quantity is written |ABl , or|a~|, or simply AB, or a (i.e. without the bar over it). Note that BA would represent a vector quantity of the same magnitude but with opposite sense. B B On to frame 7. AB= • 142 Programme 5 Two equal vectors If two vectors, a and b, are said to be equal, they have the same magnitude and the same direction. If a = b, then (i) a = b (magnitudes equal) (ii) the direction of a = direction of b, i.e. the two vectors are parallel and in the same sense. Similarly, if two vectors a and b are such that b = -a, what can we say about (i) their magnitudes, (ii) their directions? N 8 (i) Magnitudes are equal, (ii) The vectors are parallel but opposite in sense. i.e. if b =—a, then Types of vectors (i) A position vector AB occurs when the point A is fixed. (ii) A line vector is such that it can slide along its line of action, e.g. a mechanical force acting on a body, (iii) A free vector is not restricted in any way. It is completely defined by its magnitude and direction and can be drawn as any one of a set of equal-length parallel lines. Most of the vectors we shall consider will be free vectors. So on now to frame 1 0. M3 Vectors Addition of vectors The sum of two vectors, AB and BC, is defined as the single or equivalent or resultant vector AC i.e. AB + BC = AC 10 or a + b = c A "s B To find the sum of two vectors a and b, then, we draw them as a chain, starting the second where the first ends: the sum c is then given by the single vector joining the start of the first to the end of the second. e.g. If p = a force of 40 N, acting in the direction due East « = a force of 30 N, " " " " due North I then the magnitude of the vector sum r of these two forces will be ttt'then 1 r=50N 11 for *'' ? r 2 =p 2 +q 2 = 1600 + 900 = 2500 r = V2500=50N The sum of a number of v« E ' X jctors a •v D 'c :. a +~b + b +c + J+ (i) Draw the vectors as a chain (ii) Then: a+T= AC AC+c = AD :. a +b +c = AD AD + d = AE + c+d = AE i.e. the sum of all vectors, a, b, c, d, is given by the single vector joining the start of the first to the end of the last - in this case, AE. This follows directly from our previous definition of the sum of two vectors. R Q — ■ — *" A- t Similarly, PQ + QR + RS + ST = 144 Programme 5 12 PT Now suppose that in another case, we draw the vector diagram to find the sum of a", b,c,d, e, and discover that the resulting diagram is, in fact, a closed figure. What is the sum of the vectors a,b,c, d, e~, in this case? Think carefully and when you have decided, move on to frame 13. 13 Sum of the vectors = For we said in the previous case, that the vector sum was given by the single equivalent vector joining the beginning of the first vector to the end of the last. But, if the vector diagram is a closed figure, the end of the last vector coincides with the beginning of the first, so that the resultant sum is a vector with no magnitude. Now for one or two examples. Example 1. Find the vector sum AB + BC + nD + DE+EF. Without drawing a diagram, we can see that the vectors are arranged in a chain, each beginning where the previous one left off. The sum is therefore given by the vector joining the beginning of the first vector to the end of the last. :. Sum = AF In the same way, AK + KL + LP + PQ = . 145 Vectors AQ Right. Now what about this one? Find the sum of AB-CB + CD-ED We must beware of the negative vectors. Remember that -CB = BC, i.e. the same magnitude and direction but in the opposite sense. Also -ED = DE 14 AB-CB + CD-ED AB + BC + CD + DE AE. W Find the vector sum AB + BC - DC - AD r When you have the result, move on to frame 15. >Iow you do this one : Find the vector sum For: 15 AB + BC-DC-AD = AB + BC + CD + DA and the lettering indicates that the end of the last vector coincides with the beginning of the first. The vector diagram is thus a closed figure and therefore the sum of the vectors is 0. Now here are some for you to do: (i) PQ + QR + RS + ST = (ii) AC + CL-ML = (iii) GH + HJ + JK + KL+LG = (iv) AB + BC + CD + DB = When you have finished all four, check with the results in the next frame. 146 Programme 5 16 17 Here are the results: (i) PQ + QR + RS + ST = PT (ii) AC + CL-ML= AC + Cl+LM = AM (iii) GH + HJ + JK + KL+LG = [Since the end of the last vector coincides with the beginning of the first.] (iv) AE + J5C + CB+T5B = SB D The last three vectors form a closed figure and therefore the sum of these three vectors is zero, leaving only AB to be considered. A "^ B Now on to frame 1 7. Components of a given vector Just as AB + BC + CD + T)E can be replaced by AE, so any single vector PT can be replaced by any number of component vectors so long as they form a chain in the vector diagram, beginning at P and ending at T. e.g. i^"''"^\.c \ \ 5 \ >. J_ VY = a + b+c-+d Example 1. ABCD is a quadrilateral, with G and H the mid-points of DA and BC respectively. Show that AB + DC = 2 GH We can replace vector AB by any chain of vectors so long as they start at A and end at B e.g. we could say AB = AC + GH+HB '-=\ Similarly, we could say DC = 147 Vectors DC = DG + GH + HC 18 So we have A AB = AG + GH + HB DC = DG + GH + HC .-. AB + DC = AG + GH + HB + DG + GH + HC = 2GH + (AG + DG) + (HB + HC) Now, G is the mid point of AD. Therefore, vectors AG and DG are equal in length but opposite in sense. .-. DG = -AG Similarly HC=-HB :. AB + DC = 2GH + (AG-AG) + (HB-HB) = 2GH Next frame. 19 r Example 2. Points L, M, N are mid points of the sides AB, BC, CA, of the triangle ABC. Show that (i) AB + BC + CA = (ii) 2AB + 3BC + CA = 2LC (hi) AM + BN + CL = 0. (i) We can dispose of the first part straight away without amy trouble. We can see from the vector diagram that AB + BC + CA = since these three vectors form a 148 Programme 5 20 closed figure Now for part (ii). To show that 2 AB + 3 BC + CA = 2 LC B M From the figure AB = 2AL; BC=BL + LC; CA = CL + LA .'. 2AB + 3BC + CA = 4AL + 3Bl + 3LC + CL+ LA Now BL = -AL; CL = -LC; LA = -AL Substituting these in the previous line, gives 2AB + 3BC + CA = 21 22 2LC For 2 AB + 3 BC + CA = 4 AL + 3 BL + 3 LC + CL + LA = 4AL-3AL + 3LC-LC-AL = 4AL-4AL+3LC-LC = 2LC Now part (iii) To prove that AM + BN + CL = From the figure in frame 20, we can say AM = AB + BM BN=BC +CN Similarly CL= CL = CA + AL So AM + BN + CL = AB + BM + BC + CN + CA + AL = (AB + BT + CA) + (BM + CN"+AL) = (AB + BC + CA) + i(BC + CA + AB) = Finish it off. 149 Vectors 23 AM + BN + CL = Since AM + BN + CL = (AB + BC + CA) + KBC + CA + AB) Now AB + BC + CA is a closed figure .". Vector sum = and BC + CA + AB is a closed figure .'. Vector sum = :. AM + BN + CL = Here is another. Example 3. ABCD is a quadrilateral in which P and Q are the mid points of the diagonals AC and BD respectively. Show that AB + AD + CB + CD = 4PQ First, just draw the figure: then move on to frame 24. 24 A D To prove that AB + AD + CB + CD = 4PQ Taking the vectors on the left-hand side, one at a time, we can write AB = AP + PQ + QB AD = AP + PQ + QD CB = CD= t CB = CP + PQ + QB ; CD = CP + PQ + QD 25 Adding all foi lr lines together, we have AB + AD + CB + CD = 4PQ + 2AP + 2CP+ 2QB + 2QD = 4 PQ + 2 (AP + CP) + 2(QB + QD) Now what can we say about (AP + CP)? ► 150 26 AP+CP = Since P is the mid point of AC .'. AP = PC :. CP = -PC = -AP .'. AP + CP = AP-AP=0. In the same way, (QB + QD) = 27 QB + QD = Since Q is the mid point of BD .'. QD = -QB .'. QB + QD = QB-QB = .'. AB + AD + CB + CD = 4PQ + + = 4PQ 28 Programme 5 Here is one more. Example 4. Prove by vectors that the line joining the mid-points of two sides of a triangle is parallel to the third side and half its length. A Let D and E be the mid-points of AB and AC respectively. DE = DA + AE Now express DA and AE in terms of BA and AC respectively and see if you can get the required results. Then on to frame 29. 151 ) Vectors Here is the working. Check through it. DE = DA + AE = I§A+i-AC = i(BA + AC) .-. DE = ^BC .'. DE is half the magnitude (length) of BC and acts in the same direction. i.e. DE and BC are parallel. Now for the next section of the work: turn on to frame 30. 29 Components of a vector in terms of unit vectors The vector OP is defined by its magnitude (r) and its direction (8). It could also be defined by its two components in the OX and OY directions. i.e. OP is equivalent to a vector a in the OX direction + a vector b in the OY direction. i.e. OP = a (along OX) + b (along OY) If we now define / to be a unit vector in the OX direction, then a=ai Similarly, if we define / to be a unit vector in the OY direction, then b = b j So that the vector OP can be written fc$ r = a i + b j where / and / are unit vectors in the OX and OY directions. Having defined the unit vectors above, we shall in practice omit the bars over the / and /, in the interest of clarity. But remember they are vectors. 30 152 Programme 5 31 Let Zj = 2i + 4/ and z 2 = 5/ + 2/ Y ■• 5 H To find Zi + z 2 , draw the two vectors in a chain. fi + z 2 = OB = (2 + 5)z+(4 + 2)/ = 7/ + 6/ i.e. total up the vector components along OX, and " " " " " " OY Of course, we can do this without a diagram: If Zi = 3/ + 2/ and z 2 = 4/ + 3/ Zi +z 2 =3z+2/+4i + 3/ = li + 5/ And in much the same way, T 2 —T\ = 32 Z 2 -Z! = 1/ + 1/ for z 2 -Zi =(4z + 3/)-(3/ + 2/) = 4/ + 3/-3z-2; = 1/ + 1/ Similarly, if z"i = 5 i - 2/; z 2 = 3/ + 3/; z 3 = 4/ - 1/, then (i)zi +z 2 +Z3 = and (ii)zi _ Z2 - z 3 = When you have the results, turn on to frame 33. 153 Vectors 0)12/ ; (ii)-2/-4/ Here is the working: (i) zj + z 2 + z 3 = 5/ - 2/ + 3/ + 3/+ 4/ - 1/ = (5 +3 + 4)/ + (3 -2-1)/ = 12/ (ii) Z! - z 2 - f 3 = (5/- 2/).- (3/ + 3/) - (4/- 1/) = (5 -3 -4)/ + (-2 -3 + 1)/ = - 2i -4/ Now this one. If OA = 3/ + 5/ and OB = 5/- 2/, find AB. As usual, a diagram will help. Here it is: First of all, from the diagram, write down a relationship between the vectors. Then express them in terms of the unit vectors. AB = AB = 2/-7/ for we have OA + AB = OB (from the diagram) .'. AB = OB-OA = (5/ -2/) -(3/ + 5/) = 2i -7/ On to frame 35. 33 34 Vectors in space z The axes of reference are defined by the 'right-hand' rule. OX, OY, OZ form a right-handed set if rotation from OX to OY takes a "7 right-handed corkscrew action along the positive direction of OZ. 35 Similarly, rotation from OY to OZ gives right-hand corkscrew action along the positive direction of 154 Programme 5 36 ox J—* Vector OP is defined by its components a along OX b " OY c " OZ Let / = unit vector in OX direction, ■ = „ „ „ 0Y k= " " " OZ Then OP = ai + bj + ck Also OL 2 =a 2 +b 2 and OP 2 = OL 2 + c 2 OP 2 =a 2 +ft 2 +c 2 So, if r = ai + 6/ + ck, then r = y/(a 2 +b 2 +c 2 ) This gives us an easy way of finding the magnitude of a vector expressed in terms of the unit vectors. Now you can do this one: If PQ = 4i + 3j+2k, then |pq| = 37 For, if PQ = V29 = = 5-385 PQ =4i + 3j+2k |pq|= V(4 2 +3 2 +2 2 ) = V(16 + 9 + 4) = V29=5-385 Now move on to frame 38. i 155 Vectors Direction cosines J Q The direction of a vector in three dimensions is determined by the angles which the vector makes with the three axes of reference. z Let 0? = r = ai + bj + ck Then a - = cos a .. a = r cos a 7=cos0 : cos 7 b = r cos (3 c-r cos 7 Also If a 1 +b z +<r =r .'. r 2 cos 2 a + r 2 cos 2 (3 + r 2 cos 2 7 =r 2 .". cos 2 a + cos 2 |3 + cos 2 7 = 1 / = cos a w = cosj3 then / 2 +m 2 +« 2 = l n = cos 7 Note: [l,m,n]_ written in square brackets are called the direction cosines of the vector OP and are the values of the cosines of the angles which the vector makes with the three axes of reference. So for the vector r=ai + bj + ck /=-; m=j\ n =j and, of course r = >/(a 2 + b 2 +c 2 ) So, with that in mind, find the direction cosines [l,m,n] of the vector T=3i-2j+6k Then to frame 39. 7=3i-2j + 6k a = 3,b=-2,c = 6 r = V(9 + 4 + 36) /. r = V49 = 7 39 :. /=|; m=-\; n = \ Just as easy as that! On to the next frame. 156 Programme 5 40 Scalar product of two vectors If A and B are two vectors, the scalar product of A and B is definedas A B cos B, where A and B are the magnitudes of the vectors A and B, and 9 is the angle between them. The scalar product is denoted by A.B (sometimes called the 'dot product', for obvious reasons) For example B Al = ABcosfl = A X projection of B on A or BX " " A" B In either case, the result is a scalar quantity. OA.OB = 41 For, we have: OA.OB = ^ Now what about this case: -.90° OA.OB = OA.OB.cos = 5.7.cos45° = « 1 -35V2 V2 2 The scalar product of a and b = a.b = 157 Vectors since, in this case, a.b = a. b. cos 90° = a.b.O = So, the scalar product of any two vectors at right -angles to each other is always zero. And in this case now, with two vectors in the same direction, 6 = 0°, * so a.b = 42 a.b 43 since a.b. = a. b.cos0° = a.b. I = a.b Now suppose our two vectors are expressed in terms of the unit vectors. Let A = a-ii + bij+cik and B = a 2 i + b 2 j + c 2 k Then A.B = (a x i + bij+ c y k).(a 2 i + b2]' + c 2 k) = a l a 2 i.i+ a-ib 2 i.j + aiC 2 i.k + b 1 a 2 /.i + bib 2 j.j + bic 2 j.k + c x a 2 k.i + c 1 b 2 k.j + c±c 2 k.k This will simplify very soon, so do not get worried. For i.i= l.l.cos0° = 1 :.i.i=\; j.j=l; k.k=\ (i) Also /./= 1.1 cos 90° = i.j=0; j.k = 0; k.i = (ii) So, using the results (i) and (ii), we can simplify the expression for A.B above to give AJ = 158 Programme 5 44 A.B = a x a 2 + b x b 2 + c t c 2 since A.B = a x a 2 \ + a x b 2 + a x c 2 Q + bia 2 + b x b 2 \ + b x c 2 + c x a 2 + c x b 2 + c t c 2 \ .'. A.B = a x a 2 +b x b 2 +fiC 2 i.e. we just sum the products of coefficients of the unit vectors alonj; corresponding axes. e.g. If A = 2/ + 3/ + 5k and B = 4/ + lj + 6k then A.B =2.4 + 3.1 +5.6 + 3 +30 =41 A.B =41 Oneforyou: If P= 3z- 2/+ Ik; Q = 2i + 3j~4k, then P.Q= 45 for Now we come to: P.Q = 3.2 + (-2).3 + l(-4) = 6 - 6-4 :. P.Q =-4 Vector product of two vectors The vector product of A and B is written A X B (sometimes called the 'cross product') and is defined as a vector having the magnitude A B sin 8 , where 6 is the angle between the two given vectors. The product_vector acts in a direction perpendicular to A and B in such a sense that A, B, and (AXB) form a right-handed set - in that order (Axe) (BxA) | (AX B)|= ABsin6 Note that BXA reverses the direction of rotation and the product vector would now act downward, i.e. (B X A) = -(A X B) If = 0°,then|(AX B)| = \ and if =90°,then)(AX B)| = 159 Vectors 9= 0°,|(AX B)|=0 0=9O°,|(AX B)|=AB If A and B are given in terms of the unit vectors, then A = a l i + bij+c l k and B = a 2 i + b 2 j + c 2 k Then A X B = (a^ + 6, j + c x k) X (a 2 i + b 2 j+c 2 k) = aia 2 iXi + aib 2 iXj + a 1 c 2 i'>(k + b x a 2 'iXi + bib 2 jXj +J} 1 c 2 j'Xk + Cxa 2 kXi + c x b 2 kXj + CiC 2 kX k But iXi= l.l.sinO° = .-. iXi = jXj = kXk = (i) Also / X / = 1.1. sin 90° = 1 in direction OZ, i.e. iX j = k :. i X / = k }Xk= i \ (ii) k X i = j And remember too that iXj=-(jXi) j X k= -(k X j) kXi = -(i X k) since the sense of rotation is reversed. Now with the results of (i) and (ii), and this last reminder, you can simplify the expression for A X B. Remove the zero terms and tidy up what is left. Then on to frame 47. 46 160 Programme 5 47 A X B = {bic 2 -b 2 c t )i + (a 2 c t - «iC 2 )/ + {a\b 2 ~a 2 bi)k for AX B =a x a 2 Q + dib 2 k + «iC 2 (-/) + bia 2 (-k) + bib 2 + bic 2 i + Cifl 2 ; + Cifc 2 (-z) + CiC 2 = (b l c 2 -b 2 c l )i + (a 2 c l -aic 2 )/ + (fl 1 6 2 -a 2 b x )k Now we could rearrange the middle term slightly and rewrite it thus: AX B = {b\C 2 ~b 2 ci)i-{aic 2 -a 2 c x )i + (flib 2 -a 2 bi)k and you may recognize this pattern as the expansion of a determinant. So we now have that: if A=a t i + bij+cik and B = a 2 i + b 2 j + c 2 k then AX B / / k 0i &1 Cl a 2 b 2 c 2 and that is the easiest way to write out the vector product of two vectors. Note: (i) the top row consists of the unit vectors in order, /, /, k (ii) the second row consists of the coefficients of A (iii) the third row consists of the coefficients of B. Example. If P = 2/ + 4/ + 3 k and Q" = 1 i + 5/ - 2k first write down the determinant that represents the vector product P X Q. 48 PXQ = i j k 2 4 3 1 5 -2 And now, expanding the determinant, we get FXQ= 161 Vectors PX Q=-23/ + 7/ + 6fc 49 PXQ / / k 2 4 3 1 5 -2 4 3 5 -2 + /c 2 4 1 5 2 3 1 -2 = z(-8 - 1 5) - /(-4 - 3) + fc(10 - 4) = -23/ + 1] + 6k So, by way of revision, (i) Scalar product ('dot product') A.B = A B cos 6 a scalar quantity. (ii) Vector Product ('cross product') A X B = vector of magnitude A B sin 9 , acting in a direction to make A, B, (A X B) a right-handed set. Also AX B = z i k ii bx Cx a 2 b 2 c 2 And here is one final example on this point. Example. Find the vector product of P and Q, where P= 3/- 4/ + 2k and ~Q = 2i + 5/- Ik. PX Q = -6z" + 7/ + 23fc 50 for PX Q = / ;' k 3-4 2 2 5-1 -4 2 5 -1 -/ 3 2 2 -1 + k 3 -4 2 5 = z(4 - 10) -/(-3 - 4) + /c(15 + 8) = -6z + 7/ + 23 k On to frame 51. 162 Programme 5 51 Angle between two vectors Let A be one vector with direction cosines [/, m, n] " B be the other vector with direction cosines [/', rri ', ri] We have to find the angle between these two vectors. Let OP and_OP' beunit vectors parallel to A and B respectively. Then P has co-ordinates (/, m, ri) and P' " " (l',m',ri) {m-m') Then (PP') 2 = (/- I'f + (m- m'f + (n - ri) 2 = I 2 - 2.1.1' + l' 2 +m 2 - l.m.rri + m 2 + n 2 - Inn + n' 2 = (l 2 +m 2 +n 2 ) + (I' 2 + m' 2 + ri 2 ) - 2{ll' + mm + nri) But (/ 2 +m 2 +n 2 )= 1 and (/' 2 +m' 2 +n' 2 )= 1 as was proved earlier. :. (PP') 2 = 2-2(//' +mm' + nri) (i) Also, by the cosine rule, (PP') 2 = OP 2 + OP' 2 - 2.0P.0P! cos e = 1 + 1 -2.1.1.cos0 [ OP and OP' are ) = 2-2cos0 (ii) \unitvectors j So, from (i) and (ii), we have: (PP') 2 = 2-2(//' +mm +nri) and (PP') 2 = 2 - 2 cos 6 :. cos 9 = 52 cos 6 = ll' + mm + nri i.e. just sum the products of the corresponding direction cosines of the two given vectors So, if [/, m,n] = [0-5, 0-3, -04] and [l',m',ri] = [0-25, 0-6, 0-2] the angle between the vectors is 6 = 163 Vectors 6= IT for, we have cos 6 = 11' + mm + nri = (0-5) (0-25) + (0-3) (0-6) + (-0-4) (0-2) = 0125 + 0-18 - 0-08 = 0-308 - 0-08 = 0-225 0=77° NOTE: For parallel vectors, d = 0° •'• //' + mm +nri = 1 For perpendicular vectors, 8 = 90°, .'. //' + mm' + nri = Now an example for you to work: Find the angle between the vectors P=2i + 3j + 4k and Q = 4i-3j + 2k First of all, find the direction cosines of P. You do that. 53 for 'vfe' m *<k' n= ^ |P| =V(2 2 + 3 2 + 4 2 ) = V(4 + 9 + 16) = r V29 fe 3 /■ V29 _c_ 4 r \729 ■'■ U,m,n] = "2 3 4 " V29 ' %/29 ' \/29 Now find the direction cosines [/', m, ri] of Q in just the same way. When you have done that, turn on to the next frame. 54 164 Programme 5 55 since r'=|Ql = V(4 2 +3 2 +2 2 ) = V(16 + 9 + 4) = V29 .. [1 ,m,n] = We already know that, for P, [l,m,n] = -3 V29 ' V29 ' V29 _?_ _1 J_ V29 ' V29 ' V29 So, using cos d =11' + mm + nri ', you can finish it off and find the angle 6. Off you go. 56 = 76°2' for a 2 4 ^ 3 (-3) 4 2 COS P = + - - + V29 ' V29 V29 ■ V29 V29 - V29 29 29 29 = 2 1T 0-2414 Now on to frame 5 7. p ^ Direction ratios 3/ If OP = ai + fc/ + cfc, we know that \0P\ = r = ^a 2 +b 2 +c 2 and that the direction cosines of OP are given by I =— , ?n =— , n = — r r r We can see that the components, a,b,c, are proportional to the direction cosines, /, m, n, respectively and they are sometimes referred to as the direction ratios of the vector OP. Note that the direction ratios can be converted into the direction cosines by dividing each of them by r (the magnitude of the vector). Now turn on to frame 58. 165 Vectors Here is a short summary of the work we have covered. Read through it. Summary 1 . A scalar quantity has magnitude only ; a vector quantity has both magnitude and direction. 2. The axes of reference, OX, OY, OZ, are chosen so that they form a right-handed set. The symbols /',/, k denote unit vectors in the direc- tions OX, OY, OZ, respectively. If OP = ai + bj +ck, then | OP| = r = V(tf 2 + b 2 + c 2 ) 3. The direction cosines [I, m, n] are the cosines of the angles between the vector and the axes OX, OY, OZ respectively. i- ^ i i _b _c For any vector l ~ji m-—, n-- and l 2 +m 2 +n 2 =\ 4. Scalar product ('dot product') A.B = A B cos 9 where 9 is angle between A and B. If "K=a 1 i + bij + Cik and B=a 2 i + b 2 j + c 2 k then A.B=a l a 2 +bib 2 +c x c 2 5. Vector product ('cross product') A X B = (A B sin 9) in direction perpendicular to A and B, so that A, B, (A X B) form a right-handed set. Also A X B i i k d\ hi Ci a 2 b 2 c 2 58 6. Angle between two vectors cos =11' + mm' + nn For perpendicular vectors, //' + mm + nn = 0. All that now remains is the Test Exercise. Check through any points that may need brushing up and then turn on to the next frame. 166 Programme 5 59 Now you are ready for the Test Exercise below. Work through all the questions. Take your time over the exercise: the problems are all straight- forward so avoid careless slips. Diagrams often help where appropriate. So off you go. Text Exercise — V 1. If OA = 4/ + 3/, OB = 6/ -2/, OC = 2i~j, find AB, BC and CA, and deduce the lengths of the sides of the triangle ABC. 2. If A = 2/ + 2/ - k and B = 3 i - 6/ + 2k, find (i) A.B and (ii) AXB. 3. Find the direction cosines of the vector joining the two points (4, 2, 2) and (7, 6, 14). 4. If A = 5/ + 4/ + 2k, B = 4z - 5/ + 3k, and C = 2/ -/ - 2k, where ;', /', k, are the unit vectors, determine (i) the value of A.B and the angle between the vectors A and B. (ii) the magnitude and the direction cosines of the product vector (AXB) and also the angle which this product vector makes with the vector C. 167 Vectors Further Problems - V 1. The centroid of the triangle OAB i s den oted by G. If is the origin and OA = Ai + 3/, OB = 6i~j, find OG in terms of the unit vectors, /and/ 2. Find the direction cosines of the vectors whose direction ratios are (3, 4, 5) and (1,2, -3). Hence find the acute angle between the two vectors. 3. Find the modulus and the direction cosines of the vectors 3/ + 7j- 4k, i— 5/— 8k, and 6/ — 2/+ 12k. Find also the modulus and the direction cosines of their sum. 4. If A = 2/ + 4/ -3k, and B = / + 3/ + 2k, determine the scalar and vector products, and the angle between the two given vectors. 5. If OA = 2/ + 3;'- k, OB = i - 2j+3k, determine (i) the value of OA.OB_ (ii) the product OA X OB in terms of the unit vectors (iii) the cosine of the angle between OA and OB 6. Find the cosine of the angle between the vectors 2/ + 3/ - k and 3/-5/ + 2fc. 7. Find the scalar product (A.B) and the vector product (A X B), when (i) A = i + 2/- k, B = 2/ + 3; + k (ii) A=2z + 3/+4£, B = 5i~2j+k 8. Find the unit vector perpendicular to each of the vectors 2/—/+ k and 3/ + 4/ — k, where /',/, k are the mutually perpendicular unit vectors. Calculate the sine of the angle between the two vectors. 9. If A is the point (1,-1, 2) and B is (-LJ2, 2) and C is the point (4, 3, 0), find the direction cosines of BA and BT, and hence show that the angle ABC = 69°1 4'. 10. If A = 3r -/' + 2k, B- i + 3/- 2k, determine the magnitude and direction cosines of the product vector (A X B) and show that it is perpendicular to a vector C = 9/ + 2/ + 2k. 168 Programme 5 11. A, B, C are vectors defined by A = 8/ + 2]- 3k, B = 3/- 6/+ 4/fc, and C = 2/ - 2/ — fc, where z, /', k are mutually perpendicular unit vectors. (i) Calculate A.B and show that A and B are perpendicular to each other (ii) Find the magnitude and the direction cosines of the product vector (A XTT) 12. If the position vectors of P and Q are/ + 3/'— Ik and 5/— 2/ + 4k respectively, find PQ and determine its direction cosines. 13. If position vectors, OA, OB, OC, are defined by OA = 2i- /+ 3k, OB = 3/ + 2/- 4/fc, OC = -/' + 3/ - 2k, determine (i) the vector AB (ii) the vector BC __ (iii) the vector product AB X BC (iv) the unit vector perpendicular to the plane ABC 169 Programme 6 DIFFERENTIATION 1 Programme 6 1 Standard Differential Coefficients Here is a revision list of the standard differential coefficients which you have no doubt use'd many times before. Copy out the list into your note- book and memorize those with which you are less familiar — possibly Nos. 4, 6, 10, 11, 12. Here they are: I 1. 2. 3. 4. 5. 9. 10. 11. 12. 13. 14. y=f(x) dy dx x n nx"- 1 e x t x e kx ke !cx a x a x .\na \nx 1 X \og a x 1 x. In a sinx cosx cosx - sin* tan* sec 2 x cot X — cosec 2 x secx sec x.tanx cosecx - cosec x.cot x sinhx coshx coshx sinhx The last two are proved on frame 2, so turn on. 171 Differentiation The differential coefficients of sinh x and cosh* are easily obtained by remembering the exponential definitions, and also that -r- ie x }=e x and T -{e x } = -e x dx l ' dx (i) y = sinh x y = — ~ ^_e*-(-e-)_e* +e -^ cosh;c "dx 2 2 .'. — (sinh x) = cosh x (ii) y = cosh x e x + e' x y 2 . dy _t x + (~e x ) _ e x - e~ x _ " dx 2 2 .'. — (cosh x) = sinh x sinhx Note that there is no minus sign involved as there is when differen- tiating the trig, function cosx. We will find the differential coefficient of tanhx later on. Move on to frame 3. Let us see if you really do know those basic differential coefficients. First of all cover up the list you have copied and then write down the differential coefficients of the following. All very easy. 1. x s 2. sinx 3. e 3x 4. lnx 5. tan x 6. 2 X 7. secx 8. coshjc 9. logio* 10. e* 11. cosx 12. sinhx 13. cosec x 14. a 3 15. cot X 16. a x 17. x' 4 18. 19. 20. log a x e x/2 When you have finished them all, turn on to the next frame to check your results. 172 Programme 6 Here are the results. Check yours carefully and make a special note of any where you may have slipped up. 1 . 5.x 4 11. — sin x 2. cosx 12. coshjc 3. 3e 3x 13. -cosecx.cotx 4. 1/x 14. 5. sec 2 * 15. - cosec 2 x 6. 2* In 2 16. a x lna 7. secx.tanx 17. -4x' s 8. sinhx 18. l/(xlna) 9. l/(x In 10) 19. W 1 = l/(2\/x) 10. e x 20. \<? h If by chance you have not got them all correct, it is well worth while returning to frame 1 , or to the list you copied, and brushing up where necessary. These are the tools for all that follows. When you are sure you know the basic results, move on. Functions of a function Sin x is a function of x since the value of sin x depends on the value of the angle x. Similarly, sin(2x + 5) is a function of the angle (2x + 5) since the value of the sine depends on the value of this angle. i.e. sin(2x + 5) is a function of (2x + 5) But (2x + 5) is itself a function of x, since its value depends on x. i.e. (2x + 5) is a function of x If we combine these two statements, we have sin(2x + 5) is a function of (2x + 5) " a function of x Sin(2x + 5) is therefore a function of a function of x and such expressions are referred to generally as functions of a function. So e 8 '"^ is a function of a function of 173 Differentiation since e sln y depends on the value of the index sin y and sin y depends on j. Therefore e 81 "-*' is a function of a function of^. DDDnDnDnnnDDnnnnnDnnnnDDnnannannnnDDnn We very often need to find the differential coefficients of such func- tions of a function. We could do them from first principles: Example 1. Differentiate with respect to x, y = cos(5x - 4). Let u = (5x — 4) .'. y = cos u .'. ~- = -sin u - -sin(5x - 4). But this dy dy gives us — , not — . To convert our result into the required coefficient du dx n dy dy du . 1i . . dy , , . , , , . du , . we use — = — . — , i.e. we multiply ~- (which we have) by -=- to obtain — (which we want);-7— is found from the substitution u = (5x~4), du r i.e. ^- = 5. •'• -j- {cos(5x - 4)}= -sin(5x - 4) X 5 = -5 sin(5x - 4) So you now find from first principles the differential coefficient of y = e sm x . (As before, put u = sin x.) ~{e sinx }=cosx.e sinx For: v = e sin x . Put u = sin x .'. y = e" .'. — = e" du _ L dy dy du , du But -r- = -T- — - and -r- = cos x dx du dx dx ■ d {e sinx }=e sinx .cosx dx This is quite general. If J = /(") andw = F(x), then -^- = ^- • —- , i.e. if y = In F, where F r \.- c ^ dx du dx is a function of x, then dy = dy_ dF^ \_dF dx dF dx F dx So,if>' = lnsinx & = _L . C os x = cot x dx sinx dF It is of utmost important not to forget this factor — , so beware! 174 Programme 6 8 Just two more examples: (i) y = tan(5;t ~ 4) Basic standard form is y = tan x, -— = sec 2 x In this case (5x — 4) replaces the single x ;. i^ = sec 2 (5x - 4) X the diff. of the function (5x - 4) dx = sec 2 (5.x - 4) X 5 = 5 sec 2 (5x - 4) Basic standard form isy = X s , — - = 5x A dx (ii) y = (4x- 3) s Here, (4x - 3) replaces the single x :. %= 5(4x - 3) 4 X the diff. of the function (4x - 3) dx = 5(4* - 3) 4 X 4 = 20(4* - 3) 4 So, what about this one? If y = cos(7x + 2), then^ = y = cos(7x + 2) ; -7 sin(7x + 2) DnnnDnaoDDDDnDDDLinnnnDnnnannDDnDnnDana Right, now you differentiate these: 1. y = (4x-5) 6 - 2. y = e 3 ~ x 3. j> = sin 2x 4. y = cos(x 2 ) 5. y = ln(3 -4 cosjc) The results are on frame 10. i Check to see that yours are correct. 175 \ Differentiation Results: 10 - <n 6 ~y.= (s(Av- <n 5 A = 1A(A V - ^\$ 1. j = (4x-5) 6 ^ = 6(4x-5) 5 .4 = 24(4x-5) 2. .y = e 3 - x ^ = e 3 ^(-l)=-e 3 -- >; 3. v = sin 2x -f- = cos 2x.2 = 2 cos 2x ax: 4. ^ = cos(x 2 ) -^- = -sin(x 2 ).2x=-2x sin(x 2 ) 5. jy =ln(3-4cosjc) -7-== — (4smi)- & 3-4cosx 3-4 cos x nnnnnnnnnnnDannnDDnannnnnnDDDDannaDDnD Now do these: 6. _y = e sin2 * 7. y = sin 2 ^ 8. y = In cos 3x 9. j» = cos 3 (3jc). 10. j/=log 10 (2x-l) Take your time to do them. When you are satisfied with your results, check them against the results in frame 11. Results: 6 . ^ = e sin2x ^=e sin2x -2cos2x = 2cos2x.e sin2 * dx •? dy 7. y = sin x -f- = 2 sin x cos x = sin 2x dx 8. y = In cos 3x -/ = — ^- (-3 sin 3x) = -3 tan 3x dx cos 3x v 9. y = cos 3 (3jc) -^- = 3 cos 2 (3x).(-3 sin 3x) = -9 sin 3* cos 2 3x 10. j,=log i0 (2x-l) ^ = (2c _ 1 1 )lnl0 -2= (2x _ 1 2 )lnl0 All correct? Now on with the programme. Next frame please. 11 176 12 Programme 6 Of course, we may need to differentiate functions which are products or quotients of two of the functions. 1. Products Ify = uv, where u and v are functions of x, then you already know that dy _ dv , du e.g. If y = a: 3 , sin 3x then dy dx ax ax ax = x 3 . 3 cos 3x + 3x sin 3x = 3x 2 (x cos 3a: + sin 3x) Every one is done the same way. To differentiate a product (i) put down the first, differentiate the second; plus (ii) put down the second, differentiate the first. So what is the differential coefficient of e 2 * In 5jc? 13 | = «»(i t 21„ 5 ,) for y = e 2x In 5x, i.e. u = t 2x , v = In 5x ^=e 2 *-L.5 + 2e 2 *ln5;t dx 5x = e 2 *(i + 2 In Sx) Now here is a short set for you to do. Find -f- when dx 1. y =x 2 tanx 2. y = e sx (3x + 1) 3. _y = x cos 2x 4. _y = x 3 sin 5x 5. ^ = x 2 In sinh x When you have completed all five move on to frame 14. Ill Differentiation Results: 1. j=x 2 tanx .'. -?- = x 2 sec 2 * + 2x tan x dx = x(x sec 2 x + 2 tanx) 2. >> = e 5 *(3x +1) :.^=e sx .3 + 5e sx (3x + l) = e s *(3 + 15x + 5) = e s *(8 + 15x) 3. .y=xcos2x :. -^- = x(-2 sin 2x) + 1 .cos 2x = cos 2x - 2x sin 2x 4. >> = x 3 sin 5x .'. -f- = x 3 5 cos 5x + 3x 2 sin 5x dx = x 2 (5x cos 5x + 3 sin 5x) 5. j> = x 2 In sinhx .". -f- = x 2 . * cosh x + 2x In sinh x dx sinh x = x(x coth x + 2 In sinh x) So much for the product. What about the quotient? Next frame. 14 2. Quotients In the case of the quotient, if u and v are functions of x, and_y =— v. — — u — then ®= dx 2 dx dx v Example 1 If v = sin 3x <»- (*+ 1)3 cos 3s- sin 3s.l x+1 ' dx (jc + 1> 2 , . e 2 *!-lnx.2e 2x Example 2. Ifj/ = ^ dy = — ? 15 dx e 4x p «(I-21nx) --21nx X If you can differentiate the separate functions, the rest is easy. You do this one. If y = j — > ~T = x 2 ' dx 178 Programme 6 16 d i cos 2* i _ -2(* sin 2* + cos 2*) dx\ * for d ( cos 2x \ _ x 2 (-2 sin 2*) - cos 2x .2x dx\ ■ So: For_y=wv, forv=^ _ — 2x(x sin 2* + cos 2x) _ -2(* sin 2x + cos 2a:) ~ v.3 dy dv , du a* dx dx du dv dy_ _ dx dx dx v ..(0 ■ (ii) Be sure that you know these. You can prove the differential coefficient of tan x by the quotient sin* method, for if y = tan x, y ■ cos* Then by the quotient rule, -J- = (Work it through in detail) 17 y = tan x -f- = sec x dx for y- sin* cos* dy _ cos x. cos * + sin x. sin x dx cos 2 x 1__ 2 — n SGC X cos x In the same way we can obtain the diff. coefft. of tanh x sinh x . dy _ cosh x .cosh x - sinh * .sinh x cosh x " dx ■ cosh 2 * cosh 2 * - sinh 2 * y = tanh * = - cosh 2 * 1 cosh 2 * seen 2 * .-. -j- (tanh *) = sech 2 * Add this last result to your list of differential coefficients in your note-book. So what is the diff. coefft. of tanh(5* + 2)? 179 Differentiation d_ dx (tanh(5x + 2)]= 5 sech 2 (5x + 2) d i i 7 for we have : If — tanh x | = sech x then — [tanh(5x + 2)} = sech 2 (5x + 2) X diff. of (5x + 2) = sech 2 (5x + 2) X 5 = 5 sech 2 (5x + 2) Fine. Now move on to frame 1 9 for the next part of the programme. 18 19 Logarithmic differentiation The rules for differentiating a product or a quotient that we have revised are used when there are just two-factor functions, i.e. uv or—. When there v are more than two functions in any arrangement top or bottom, the diff. coefft. is best found by what is known as 'logarithmic differentiation'. It all depends on the basic fact that ■=— {in jc J = - and that if x is replaced by a function F then — In F = =-■ — . Bearing that in mind, dx \ i r dx let us consider the case where y = — , where u, v and w — and also v — w are functions of x. First take logs to the base e. In y = In u + In v - In w Now differentiate each side with respect to x, remembering that u, v, w and y are all functions of x. What do we get? 180 Programme 6 20 1 dy 1 du 1 dv 1 dw y dx u dx V dx w dx So to get — by itself, we merely have to multiply across by y. Note that when we do this, we put the grand function that y represents. dy _uv t\ du 1 dv 1 dw \ \u dx w \u dv v dx w dx I This is not a formula to memorize, but a method of working, since the actual terms on the right-hand side will depend on the functions you start with. Let us do an example to make it quite clear. jr x 2 sin x _ , dy If y = ~- » find -T- cos 2x dx The first step in the process is 21 To take logs of both sides y = — .". In y = In (x 2 ) + In (sin x) - In (cos 2x) Now diff. both sides w.r.t. x, remembering that j-(ln F) = — •— — I dy 1 . A 1 = — -2x + — cosx- 1 y dx x sinx cos 2x ■ (-2 sin 2x) = —+ cot x + 2 tan 2x x . dy x sin x 1 2 , . , „ . „ .. -r- = — - — - 1- cot a: + 2 tan 2x dx cos 2x \x ) This is a pretty complicated result, but the original function was also somewhat involved! You do this one on your own: If y =x 4 e 3x tanx, then dx 181 Differentiation dy 4 3 x j. \ 4 „ sec 2 x ~x 4 e 3X tanx - + 3 + dx tanx Here is the working. Follow it through. j> = x 4 e 3 *tanx .". \ny = ln(x 4 ) + ln(e 3 *) + In (tanx) 1 dy _ 1 .y dx x _-4-4x j + ir -3e» + 1 3X tanx = 1+3 + sec2jc x tanx . 4y 4 3x j. ( 4 „ sec 2 x ..-f-=x 4 e JJC tanx — + 3 + dx { x tan x 22 There it is. Always use the log. diff. method where there are more than two func- tions involved in a product or quotient (or both). Here is just one more for you to do. Find-^- , given that y = z? x cosh 2x J=^ r (4-^-2tanh2x dx x cosh 2x \ x Working. Check yours. e « y = —5 x cosh 2x In 7 = ln(e 4 *)- ln(x 3 )- In (cosh 2x) 1 • I *1= * -4e«- 1 -3x 2 - " ^ dx e 4 ^ x 3 cosh 2x 2 sinh 2x 23 = 4-— -2tafth2x x • ^_. " I 3 1 ..— -^ r-r- 4---2tanh2x dx x° cosh 2x I x I Well now, before continuing with the rest of the programme, here is a revision exercise for you to deal with. Turn on for details. 182 Programme 6 £i\ Revision Exercise on the work so far. Differentiate with respect to x : 1. (i) ln4x (ii) In (sin 3x) 2. e 3x sin 4x ,, sin 2x 2x + 5 (3x + 1) cos 2x 5. x s sin 2x cos 4x When you have finished them all (and not before) turn on to frame 25 to check your results. 183 Differentiation Solutions r-\ i . . dy \ . \ 1. (!) ,-ln4* - ^%--4=^_ (ii) v = In sin 3x .'. -i- = - — ^— • 3 cos 3x dx sin 3x = 3 cot 3x 2. v = e 3 * sin 4jc .'. -p = e 3x A cos 4x + 3e 3 * sin 4x dx = e 3 *(4 cos Ax + 3 sin 4x) sin 2x . dy _ (2x + 5) 2 cos 2x - 2 sin 2x 2x + 5 " dx (2x + 5) 2 (3x + l)cos2x 4. ^ = i e 2 * \ny = ln(3 x + 1) + ln(cos 2x)-ln(e 2X ) 1 dy 7 dx 1 3.x + 1 3 3jc+ 1 3+ * -(-2 cos 2x - 2 tan 2x - 2 sin 2x) - —■?e 2x dx (3x + 1 ) cos 2x ( 3 e 2 * bx + 1 • 2 tan 2x - 2 I 5. _y = x 5 sin 2x cos Ax .'. In jy = ln(x 5 ) + ln(sin 2x) + In (cos Ax) • 1 d y^\. 5 x^ 2cOS ? C + * (- 4 sin4x) j ox x sin 2x cos 4.x = -+ 2 cot 2x-4tan4x x ~r = X s sin 2x cos 4x — + 2 cot 2x - 4 tan Ax dx I x So far so good. Now on to the next part of the programme on frame 26. 25 184 Programme 6 26 27 Implicit functions If y = x 2 - Ax + 2, y is completely defined in terms of x and >■ is called an explicit function of*. When the relationship between x and y is more involved, it may not be possible (or desirable) to separate y completely on the left-hand side, e.g. x y + sin y = 2. In such a case as this,j is called an implicit function of*, because a relationship of the form^ =f(x) is implied in the given equation. It may still be necessary to determine the differential coefficients of y with respect to x and in fact this is not at all difficult. All we have to remember is that y is a function of x, even if it is difficult to see what it is. In fact, this is really an extension of our 'function of a function' routine. x 2 +y 2 = 25, as it stands, is an example of an function. x 2 +y 2 = 25 is an example of an implicit function. DDDnDaDDnnnDannnDDDDDDnnDDnnnDDDnnnDDn Once again, all we have to remember is that y is a function of x. So, if x 2 +j; 2 = 25,letusfind^. ax If we differentiate as it stands with respect to x, we get 2x + 2y Q = dx Note that we differentiate^ 2 as a function squared, giving 'twice times the function, times the diff. coefft. of the function'. The rest is easy. 2x + 2y^ = dx dy . dy x :. y-f=-x .. -f = dx dx y As you will have noticed, with an implicit function the differential coef- ficient may contain (and usually does) both* and 185 Differen tiation y DnnananoDnDnDDDnnDDDDDDDnnnDnnnDDanDaD Let us look at one or two examples. Example 1. If x 2 + y 2 - 2x~ 6y + 5 = 0, find-^ and-^p at x = 3,j> = 2. Differentiate as it stands with respect to x. dx dx :.(3y-6)g = 2-2* ■ dy _ 2-2x _ 1 -x dx 2y - 6 _>> - 3 ^ v rf,i-r> (y-3)(-l)-(l-x)^ ~, « J' _ d l x | dx Then — 28 dx 2 dx\y-3) iy~3) 2 O-y)-(l-x)^ _ dx (y-3) 2 d 2 y. (3-2)-(l-3)2 _ l-(-4) . dx 2 (2-3) 2 1 ■At (3,2) f = 2, ^ = : Jx dx Now this one. If x 2 + 2xy + 3y 2 = 4, find-^- Away you go, but beware of the product term. When you come to 2xy treat this as (2x)(y). x 2 +2xy + 3y 2 =4 _ 2x + 2xf x + 2y + 6y dx dy __(2x + 2y)__(x+y) dx (2x + 6y) (x + 3y) And now, just one more: If x 3 +y 3 + 3xy 2 = 8, find -^ Tum fQ frame 2Q fof fhe solution 186 Programme 6 30 31 Solution in detail : x z + y i + 3xy 2 3x 2 +3y 2 ^+3x.2y^-+3y 2 = . dy _ (x 2 +y 2 ) " ~dx (y 2 + 2xy) That is really all there is to it. All examples are tackled the same way. The key to it is simply that l y is a function of x' and then apply the 'function of a function' routine. Now on to the last section of this particular programme, which starts on frame 31. Parametric equations In some cases, it is more convenient to represent a function by expressing x and y separately in terms of a third independent variable, e.g. y = cos It, x = sin t. In this case, any value we give to t will produce a pair of values for x and j% which could if necessary be plotted and provide one point of the curve of y = f(x). The third variable, e.g. t, is called a parameter, and the two expressions for x and y parametric equations. We may still need to find the differen- tial coefficients of the function with respect to x, so how do we go about it? Let us take the case already quoted above. The parametric equations of a function are given as y = cos 2t, x = sin t. We are required to find c dy , d 2 y expressions ior—f- and -r-y Turn to the next frame to see how we go about it. 187 Differentiation y = cos 2t, x = sin t. Find % and d -Z dx dx z From>> = cos 2t, we can get -J-= ~2 sin 2t From x = sin t, we can get -^— = cos f We now use the fact that-f =-£■ — ax at dx so that -r-=-2 sin It . dx ' cos t 1 : —4 sin f cos f . cos t dy A ■ -r- = -4 sin f ax That was easy enough. Now how do we find the second diff. coefft.? We d y d X cannot get it by finding — f and -r^- from the parametric equations and joining them together as we did for the first diff. coefft. That method could only give us something called -pf- which has no meaning and is certainly not what we want. So what do we do? On to the next frame and all will be revealed! To find the second differential coefficient, we must go back to the • f d 2 y very meaning of — ^ d 2 y d tdy\ dl , . \ But we cannot differentiate a function of t directly with respect to x. Therefore we say -^(-4 sin t) =—(-4 sin t). — . J dx\ ' dt\ I dx ■ d 2 y A 1 • • —r-n = -4 COS t. = -4 dx cos t ••■g-i Let us work through another one. What about this? The parametric equations of a function are given as y = 3 sin# - sin 3 0, x = cos 3 Find ~ and — - 32 33 dx dx 2 Turn on to frame 34. Programme 6 34 dy .'. -~ = 3 cos 6 - 3 sin 2 cos X = cos c?x d0 ! 3cos 2 0(-sin6l) = -3 cos sin( dy _dy d6 _ . i 3 — ^"T - 3 cos d (1 - sin d) . - dx d6 dx V ^ -3 C os 2 sin 5 3 cos 3 -3 cos 2 fl sin d -j— = —cot l OX Also d^y_d dx (— cosec 2 ) 1 dx 2 -3 cos 2 6 sin 3 cos 2 sir?0 V ' -1 Now here is one for you to do in just the same way 2 - 3t 3 + It If x y- l + ? ' J l + t ! W/?e« >>ow /zflve done it, move on to frame 35. find^ dx 35 dy = }_ dx 5 For . 2-3? 1 +t 3 + 2? dx It dy dt (1+Q (-3) -(2 -3Q (T+Tp (1+0 (2) -(3 +20 1+? dx -3-3/ 2 + 3? * (i +o 2 dy. 2 + 2?- -3-2? dt (H -o 2 dy_ <^ <# _ -1 -5 ^(i+0 2 : (TT77 (1 + ?) 2 _ 1 5 dx dx dt' dx (1+?) 2 ~-5 And now here is one more for you to do to finish up this part of the work. It is done in just the same way as the others. If x = a(cos 6 + 9 sin 0) and y = fl(sin 6 - 6 cos 0) find f£ and ft ax ax 189 Differentiation Here it is, set out like the previous examples. x = a(cos 6 + 6 sin 6) dx .'. -jT- = a(— sin 8 + 8 cos 8 + sin 8) = a 8 cos < do y =a(sin 6-6 cos 6) ■'■ -7q = a(cos 6 + 8 sin 6 - cos 6) =a d sind dy dy dd . . . 1 ~=-^--7- = a8 smd .—. = tan 6 dx dd dx ad cos 6 $ = tan0 dx 36 d y _ d u .. d u .. dd -rr= 7- (tan0) = -^(tan0).— _ sec ! 1 . d*y_ «0 1 COS 8 '• dx 2 00 cos 3 You have now reached the end of this programme on differentiation, much of which has been useful revision of what you have done before. This brings you to the final Test Exercise so turn on to it and work through it carefully. Next frame please. 190 37 Programme 6 Test Exercise — VI Do all the questions. Write out the solutions carefully. They are all quite straightforward. 1 . Differentiate the following with respect to x : (i) tan 2x (ii) (5jc + 3) 6 (iii) cosh 2 x (iv) logioOc 2 —3x— 1) jv) In cos 3x (vi) sin 3 4x (vii) e 2x sin 3x (viii) 4X • e sin x (x + lf j ^ xcos2x dy d 2 y 2./ If x 2 +y 2 - 2x + 2y = 23, find -f- and — -f at the point where / V x =-2, y = 3. dx dx 2 3. Find an expression for J- when x 3 +y 3 +4xy 2 = 5 4. If x = 3(1 - cos 8) and .y = 3(0 - sin 0) find j- and -pr in their simplest forms. 191 Differentiation Further Problems - VI 1 . Differentiate with respect to x _ ^ ^__ ,.. , fcosx + sinx] f ,.. N . , , . >. : ,.... . 4 3 (i)ln{ : — }: (li)ln(secx + tanx) (ni> suvx cos x I cos x sin xj / \ \/ 2. Find & when toy = ^^ (ji)y = ta(j-=4] dx \/~ 1+cosx U +x I e f 3. If v is a function of x, andx = -7 e' + 1 show that -f- = x( l - x) -f- dt dx 4. Find -f- when x 3 + y 3 - 3xv 2 = 8. dx . 5. Differentiate: (i) y = e sln 5* ( u )y = l n |— ^-j-jj (ii)^ = lnjxV(l-x 2 )j 6. Differentiate: (i) y = x 2 cos 2 x (i ..... e 2x lnx 7. If (x - j) 3 = A(x + j>), prove that (2x +y)^= x + 2y. 8y Ifx 2 -xy +y 2 = 7, find ^ and |^ at x = 3,y = 2. <i 2 v --9. If x 2 + 2xy + 3y 2 = l, prove that (x + tyf "di +2 = 0. 10. If x = In tan4- and.y = tan 6 - d, prove that 2 J2 %£ = tan 2 sin 9 (cos 6 + 2 sec 0) (for 192 Programme 6 11. If y = 3 t 2x cos (2x - 3), verify that ^- 4 ^+8v = dx 2 dx * 12. The parametric equations of a curve are* = cos 2B,y=\ + sin 20. c . ,dy d 2 y „ . dx an d?" at = n ' 6 ' Find also the ec l uation of the curve as a relationship between x andj\ 13. n>={* + VU+* 2 )) 3/2 , show that !4. Fi nd^ and-^ifx=a cos 3 , y = a sin 3 <9 . 15. Ifx = 3cos0-cos 3 0,;;=3sin 9 - sin 3 0, express 4^ and %Z in terms offl. dx dx 16. Show that y = e" 2 OTX sin 4mx is a solution of the equation 17. If>> = sec x, prove that )>j^=(—) + y* CMC \ CUC / 18. Prove that x = A e" fcr sin pf , satisfies the equation f^ + 2*f + ( y + *> = 19. If> = e" fcf (A cosh ?f + B sinh qf) where A, B, q and fc are constants, show that 20. If sinh y = {T* X :\ show that£ = , 5 u 4 + 3 sinh x' dx 4 + 3 sinh* 193 Programme 7 DIFFERENTIATION APPLICATIONS PART1 Programme 7 1 Equation of a straight line The basic equation of a straight line is y = mx + c, Y u i fy dy where m = slope =-r- = -f- ox dx c = intercept on real jy-axis Note that if the scales of x and y are identical, -f- = tan 6 ax e.g. To find the equation of the straight line passing through P(3,2) and Q(-2,l), we could argue thus: y = mx + c Line passes through P, i.e. when x = 3,y = 2 .'. 2 = m3 + c Line passes through Q, i.e. when x = ~2,y = 1 .'. 1 = m{—2) + c. So we obtain a pair of simultaneous equations from which the values of m and c can be found. Therefore the equation is We find m = 1/5 and c = 7/5. Therefore the equation of the line is y-^j, i.e. 5y = x + 7 DnanDDnnDaDDnnaDnnnnannnDDnDaDDnDnnDnD Sometimes we are given the slope, m, of a straight line passing through a given point {x x , y\ ) and we are required to find its equation. In that case, it is more convenient to use the form y~y\ =m(x-x l ) For example, the equation of the line passing through the point (5,3) with slope 2 is simply which simplifies to Turn on to the next frame. 195 Differentiation Applications 1 y~3 = 2(x~5) \.e.y-3 = 2x-\0 y = 2x-7 DnnnnDDDnDDnDDnDDDaDaDDDDDnDnnDDDDDnDD Similarly, the equation of the line through the point (-2,-1) and having a slope - is y-(-i)=±{x-(-Z) :. y + l=^(x + 2) 2y + 2 =x + 2 x ■ y= 2 So, in the same way, the line passing through (2,-3) and having slope (-2) is y = i-2x For y-(-S) = -2(x-2) .'• y + 3 = -2x + 4 :. y = 1 - 2x DDDDanaDnDnDaaanDDnanoDDnDDnnnaanDnnnn Right. So in general terms, the equation of the line passing through the point {x l ,y l ) withslope m is Turn on to frame 5. 196 Programme 7 y-yi =m(x-x 1 ) It is well worth remembering. DnDDnnnDnnnDDaDDDDDDDnnDaDnnDnnnDanDnD So for one last time: If a point P has co-ordinates (4,3) and the slope m of a straight line through P is 2, then the equation of the line is thus y - 3 = 2(x - 4) = 2x-8 .'. y=2x-5 The equation of the line through P, perpendicular to the line we have just considered, will have a slope mi , such that m rri\ = -1 i.e. mi =- — . And since m = 2, then m ( =--^-. This line passes through (4,3) and its equation is therefore j,-3 = -I(*-4) = -x/2 + 2 ^ = -^+5 2>> = 10-* If m and rri\ represent the slopes of two lines perpendicular to each other, then mm.\ =—\ or m\ = - — m Consider the two straight lines 2y = Ax - 5 and 6y = 2 — 3x If we convert each of these to the form y = m x + c, we get 5 1 1 (i) y = 2x -— and (ii) j> = - ^ + 3 So in (i) the slope m = 2 and in (ii) the slope mi = - y 1 2 We notice that, in this case, mi =— — or that mm. =-1 Therefore we know that the two given lines are at right -angles to each other. Which of these represents a pair of lines perpendicular to each other: (i) y = 2>x-5 and 3y = x + 2. (ii) 2y = x - 5 and 7 = 6 - x (iii) y-3x-2 = and 3.y+;c + 9 = 0. (iv) 5y-x = 4 and 2j + 10* + 3 = 0. 197 Differentiation Applications 1 Result: (iii) and (iv) DDDnDnDnnDanQanannnnDnapaDDDnDQDDDDDaD For if we convert each to the fornix = mx + c, we get x 2 (i) y = 3x~5 and 7 =-3+ J m = 3;m i =-:.mm 1 f-\ Not perpendicular. (») y = 2~J and y = -x + 6 _ 1 m--;m l --l :. mmif-l Not perpendicular. (iii) j = 3x + 2 and _y = ~ - 3 -•3 1 m- i ;m 1 =-- :. mm 1 =-\ Perpendicular. (iv) y = I + I and ^ = -5a: - 1 m = j ; m, = -5 :. m m l = -1 Perpendicular Do you agree with these? Remember that if y =mx + c and y = m x x + d are perpendicular to each other, then m wj =-1, i.e.m, =-_ Here is one further example: A line AB passes through the point P (3,-2) with slope ~\ . Find its equation and also the equation of the line CD through P perpendicular toAB. When you have finished, check your results with those on frame 9. 8 198 Programme 7 Equation of AB : Equation of CD: So we have: ;y-(-2) = -i(*-3) y + 2 -- 2 + 1 .y = 2 1 "2 2y+x+l-- = slop 2 WIl = 1 _ m 1 = 2 7 -(-2) = ■2(jc -3) ^ + 2 = = 2x- -6 ^ = --2x- -8 D r— /" 2x-8| 10 m rrii =-1 naDnnDanDnnDnnnnnDDDonDDDnnaDDnDDDDnDD And now, just one more to do on your own. The point P(3, 4) is a point on the line y = 5x-\l. Find the equation of the line through P which is perpendicular to the given line. That should not take long. When you have finished it, turn on to the next frame. 199 Differentiation Applications 1 5y + x = 23 For: slope of the given line,;; = 5x - 1 1 is 5. slope of required line = The line passes through P, i.e. when x = 3, y = 4. ^-4=-I(jc-3) Sy - 20 = -x + 3 :. 5y + x = 23 □□□□□nooaaaQOQOonQananaannaanQQonanana Tangents and normals to a curve at a given point. The slope of a curve, y =/(*), at a point P on the curve is given by the slope of the tangent at P. It is also given by the value of ^ at the point P, which we can calculate, knowing the equation of the curve. Thus we can calculate the slope of the tangent to the curve at any point P. What else do we know about the tangent which will help us to determine its equation? *l f 7 y=fio tet.y,) ' X We know that the tangent passes through P, i.e. when x =x i ,y=y 1 . DDnDODDDDnDDDDDDDDDDDDDnDDaDDDDnnnDDDD Correct. This is sufficient information for us to find the equation of the tangent. Let us do an example. e.g. Find the equation of the tangent to the curves = 2x 3 + 3x 2 ~2x -3 at the point P,jc = l,y = 0. h, -2 ~r = 6x 2 + 6x ax Slope of tangent : 6 + 6-2= 10, i.e.m= 10 (^1 \dx( x = 1 Passes through P, i.e. x =1,^ = 0. y~y\ = wOc-Xj) gives j>-0 = 10(x-l) Therefore the tangent is y = 1 Ox - 1 We could also, if required, find the equation of the normal at P which is defined as the line through P perpendicular to the tangent at P. We know for example, that the slope of the normal is 11 12 200 Programme 7 13 Slope of normal = -1 1_ Slope of tangent 10 □DDDDDDaDDDDnDnnnnanDDnDDDDDDnDnDDnaaD The normal also passes through P, i.e. when x = 1 ,y = 0. .'. Equation of normal is: y-0= -— (x - 1) 10y = -x + 1 \0y+x=l That was very easy. Do this one just to get your hand in: Find the equations of the tangent and normal to the curve y = x 3 - 2x 2 + 3x - 1 at the point (2,5). Off you go. Do it in just the same way. When you have got the results, move on to frame 14. 14 Tangent: y = Ix - 9 Normal: ly + x = 37 Here are the details: . dy y = x 3 - 2x 2 + 3x - 1 dy .. ^=3* -4x + 3 /. AtP(2,5),-^= 12-8 + 3 = 7 Tangent passes through (2, 5), i.e. x = 2, y - 5 y — 5 = l{x - 2) Tangent is y = Ix - 9 -1 ~ For normal, slope = - — =-I slope of tangent 7 Normal passes through P (2, 5) :.y-5=-±{x-2) 7y-35=-x + 2 Normal is ly + x = 37 You will perhaps remember doing all this long ago. Anyway, on to frame 15. 201 Differentiation Applications 1 The equation of the curve may, of course, be presented as an implicit function or as a pair of parametric equations. But this will not worry you for you already know how to differentiate functions in these two forms. Let us have an example or two. Find the equations of the tangent and normal to the curve x 2 + y 2 + 3xy - 1 1 = at the point x=l,y = 2. First of all we must find-p at (1 , 2). So differentiate right away. 2x + 2y^- + 3x^+3y = dx dx (2y + 3x)f x = ^2x + 3y) 15 Therefore, at x = l,y = 2, dy _ 2x + 3y dx 2y + 3x dy.. dx dy _ 2 + 6 , dx 4 + 3 ' dy = J&_ dx 1 Now we proceed as for the previous cases. o Tangent passes through (1 , 2) .'. y - 2 = - — (x - 1) 7y-14 = -8x + 8 .'. Tangent is ly + 8x = 22 Now to find the equation of the normal. -1 7 Slope = Normal passes through (1,2) :.y — 2 Slope of tangent 8 7, Now try this one: Jx-l) 8j-16 = 7x-7 Normal is 8y = Ix + 9 16 That's that! Find the equations of the tangent and normal to the curve x 3 +x 2 y+y 3 -7 = at the point x = 2,j = 3. 202 Programme 7 17 Results: Tangent: 31j + 24x = 141 Normal: 24y =31*+ 10 Here is the working: jc 3 +x 2 y+y 3 -7 = 3x 2 +x 2 ^+2xy + 3y 2 ^ = Q dx dx (x 2 +3y 2 )$- = -(3x 2 +2xy) ^L = -(3x 2 +2xv) ■ d£ = _ 3x 2 +2xy dx ^ 3X +lXy) - dx x 2 +3y 2 • A t(2 3) ^i = -i2±i2 = _24 " AU/ '^ dx 4 + 27 31 24 (i) Tangent passes through (2,3) .'. y - 3 = - — (x - 2) 31>-93 = -24x + 48 •'■ 31>> + 24x = 141 31 31 (ii) Normal: slope = — . Passes through (2,3) •'■ ■>> ~ 3 = — (x - 2) 24^-72=31x-62 :. 24y = 3lx + 10 Now on to the next frame for another example. 18 Now what about this one? 3t t 2 The parametric equations of a curve are x = , y = Find the equations of the tangent and normal at the point for which r = 2. d First find the value of — - when t = 3. dx x - 3r . dx_{\ +r)3-3f _3 + 3t - 3t 1 +f " dt (1 + f) 2 (1+0 (1 +0 : r . dy _(1 +t)2t-t 2 _ It + 2f 2 - 1 2 _2t + t y 14-/ " rlt n J- A 2 1 +r '" dr (1 + r) 2 (1 + r) 2 (1 + ff dy ^dy dt _ 2t + t 2 (1 + t) 2 _ 2t + t 2 ;. Att = 2&=- dx dt ' dx (1 + tf ' 3 3 ' <2x 3 To get the equation of the tangent, we must know the x and_y values of a point through which it passes. At P — 3f _ 6 _6_ _ ? 2 _4 1 +r 1+2 3 ' J 1 +r 3 jc : Continued on frame 19. 203 Differentiation Applications 1 So the tangent has a slope of -z and passes through (2, 4) .'. Its equation is y i = 8 3 3 19 (x-2) 3y~4 = 8x-L6 :. 3j; = 8;c-12 (Tangent) -1 3 For the normal, slope = = — slope of tangent 8 Also passes through (2, -|) :. y- 1 = ~\{x - 2) 2Ay - 32 = -9x + 18 ."• 24y + 9x = 50 (Normal) Now you do this one. When you are satisfied with your result, check it with the results on frame 20. Here it is: If>> = cos 2f andx = sin f, find the equations of the tangent and 77 normal to the curve at t = — . 6 Results: Working: Tangent: 2y + Ax = 3 Normal: 4y = 2x + 1 dy y - cos 2f .'. ~=-2 sin 2t = -4 sin t cos t dt dx x = sin t .'. -7T- = cos t dt dy _ dx dy dt -4 sm t cos f „ . = :J7- j- = 1 =-4 sin dt dx cos ? At r=4, 6 |U- 4!l4 ,- 4( . ) = - 2 .'. slope of tangent = -2 Passes through x = sin — = 0-5; y = co 6 0-5 Tangent is y - ^ = -2(jc - ~) .'. 2y - \ = -Ax + 2 .". 2y + Ax = 3 (Tangent) Slope of normal = y • Line passes through (0-5, 0-5) Equation is y \ = \_ 2 2 (*-4) ■'• 4^-2 = 2^-1 •'■ 4y = 2x ± 1 (Normal) 20 204 Programme 7 21 Before we leave this part of the programme, let us revise the fact that we can easily find the angle between two intersecting curves. Since the slope of a curve at (x x ,y± ) is given by the value of -~- at that point, and ~ = tan 6, where is the angle of slope, then we can use these facts to determine the angle between the curves at their point of intersection. One example will be sufficient. e.g. Find the angle between^ 2 = 8x and x 2 + y 2 = 16 at their point of intersection for which y is positive. First find the point of intersection, i.e. solve y 2 = 8x and x 2 +y 2 = 16 Wehavex 2 +8x= 16 :.x 2 + 8x-16 = -8 ± V(64 + 64) _ -8 ± y/128 X 2 2 -8 ±11-314 3-314 -19-314 = ~ = — « — or ■- — x = 1 -657 or [-9-655] Not a real point of intersection. When jc = 1-657, y 2 = 8(1-657) = 13-256, y = 3-641 Co-ordinates of P are x = 1-657, y = 3-641 Now we have to find -f- for each of the two curves. Do that ax 22 (i) y 2 = Sx :. 2y dy. . dy _ 4 _ 1 dx " dx y 3-641 0-910 tan 0! =1099 /. di = 47°42' (ii) Similarly for x 2 + y 2 = 1 6 ^ = _x = _F65_7 dx y 3-641 tan d 2 =-0-4551 /. 8 2 =-24°28' Finally, 6=e l -d 2 = 41° 42' - (-24°28') = 47°42' + 24°28' = 72°10' 1099 2x + 2.y^ = dx = -0-4551 205 Differentiation Applications 1 That just about covers all there is to know about finding tangents and normals to a curve. We now look at another application of differentiation. Curvature The value of— at any point on a curve denotes the slope (or direction) of the curve at that point. Curvature is concerned with how quickly the curve is changing direction in the neighbourhood of that point. Let us see in the next few frames what it is all about. 23 24 Let us first consider the change in direction of a curves =/(*) between the pomts P and Q as shown. The direction of a curve is measured by the slope of the tangent. fix) Slope at P = tan 0! = l&\ Slope at Q = tan 2 =[^\ U*J Q These can be calculated, knowing the equation of the curve. From the values of tan 0, and tan d 2 , the angles 6, and 2 can be found from tables. Then from the diagram, = 2 - 0, . If we are concerned with how fast the curve is bending, we must consider not only the change in direction from P to Q, but also the length of which provides this change in direction. 206 Programme 7 25 26 the arc PQ i.e. we must know the change of direction, but also how far along the curve we must go to obtain this change in direction. Now let us consider the two points, P and Q, near to each other, so that PQ is a small arc (= 6 s). The change in direction will not be great, so that if 8 is the slope at P, Y - ' then the angle of slope at Q can be put as 8 + 88. The change in direction from P to Q is therefore 5 8 . The length of arc from P to Q is 5 s. The average rate of change of direction with arc from P to Q is the change in direction from P to Q _ 5 8 the length of arc from P to Q 8 s This could be called the average curvature from P to Q. If Q now moves down towards P, i.e. 5 s -> 0, we finally get~p which is the curvature at P. It tells us how quickly the curve is bending in the immediate neighbourhood of P. H8 In practice, it is difficult to find — since we should need a relationship between 8 and s, and usually all we have is the equation of the curve, y =/(x) and the co-ordinates of P. So we must find some other way round it. Let the normals at P and Q meet in C. Since P and Q are close, CP - QC (=R say) and the arc PQ can be thought of as a small arc of a circle of radius R. Note that PCQ = 88 (for if the tangent turns through 86 , the radius at right angles to it will also turn through the same angle). You remember that the arc of a circle of radius r which subtends an angle 6 radians at the centre is given by arc = rd . So, in the diagram above, arc PQ = 8s = 207 Differentiation Applications 1 arcPQ = Ss=R50 27 5s = R60 :. |^4 8s R If 5 s -> 0, this becomes— =— which is the curvature at P. That is, we can state the curvature at a point, in terms of the radius R of the circle we have considered. This is called the radius of curvature, and the point C the centre of curvature. So we have now found that we can obtain the curvature — if we have ds some way of finding the radius of curvature R. If R is large, is the curvature large or small? If you think 'large', move on to frame 28. If you think 'small' turn on to frame 29. Your answer was : 'If R is large, the curvature is large.' □nDDDDnnDDQDDDDDDDnanDOODDDDDODDDDnDDD This is not so. For the curvature = — and we have just shown that -j-- g . K is the denominator, so that a large value for R gives a small value for the fraction— and hence a small value for the curvature. You can see it this way. If you walk round a circle with a large radius R then the curve is relatively a gentle one, i.e. small value of curvature, but if R is small, the curve is more abrupt. So once again, if R is large, the curvature is 28 208 Programme 7 29 If R is large, the curvature is small Correct, since the curvature— = — as R DDnnDDDnnDnnnaDnoDDnannnnnnDaDDnnnanDD In practice, we often indicate the curvature in terms of the radius of curvature R, since this is something we can appreciate. Let us consider our two points P and Q again. Since 8 s is very small, there is little difference between the arc PQ and the chord PQ, or between the direction of the chord and that of the tangent. So, when 8s ■ dx ■ tan . Differentiate with respect to s. -0,4^ = tan dx dx — = COS0 ds Then l{f)4H d [dy\ dx d I t A d6 ds d'y dx 2 ' 2o dd = sec 6—r- ds ds dx Now sec 3 = (sec 2 0) 3/2 = (1 + tan 2 0) 3/2 = { 1 + (-^ ) } 3/2 dd ds R" d 2 y 17 ( 1+ €> 2 3/2 R = l 1+ (£) 3/2 £1 dx 2 Now we have got somewhere. For knowing the equation/ = /(x) of the curve, we can calculate the first and second differential coefficients at the point P and substitute these values in the formula for R. This is an important result. Copy it down and learn it. You may never be asked to prove it, but you will certainly be expected to know it and to apply it. So now for one or two examples. Turn on to frame 30. 209 Differentiation Applications 1 Example 1. Find the radius of curvature for the hyperbola xy = A at the «jU R dx 2 So all we need to find are -f and— -=£• at (2,2) dx dx ' a ■ ^ a -i ■ dy . _, -4 xy = 4 .. y = — = Ax 1 ..-*-= -Ax 2 =-s- and S =8 * 3= ^ At (2,2) *: = -! = -!. ^ = 8=i v ' dx 4 ' dx 5 8 , R= {ut!£H'Ui±!r =(2) . ft=2V2 :. R = 2\/2 = 2-828 units. rftere we are. Another example on frame 31. Example 2. If y = x + 3x 2 - x 3 , find R at x = 0. rf>\ 2,3/2 R= — dx 2 *: = i+^_^2 -A^-n^-i • i d y i dx .♦*-*», A„-o£.. ,($-. dx" dx 2 {l + l}3/2 _ 2 3/2 _2x/2_V2 6 6 6 3 ■'. R = 0-471 units Now you do this one: Find the radius of curvature of the curve y 2 =^- at the point (l,4) When you have finished, check with the solution oh frame 32. 31 210 Programme 7 32 R = 5-21 units Here is the solution in full. y =- dy_ = ^_ . dy _ 3x 2 " y dx 4 " dx Sy - At V'V>dx 4 " \dx) 16 dy = 3x^_ . d 2 y = dx Sy " dx 2 8y(6x)-3:c 2 8 dx 64 y 2 R = 1 + At(l,I), 2 > 3/2 1 \ d 2 y _ 24 - 24.f _ 24 - 18 _ 3 dx' 16 16 (dy\ 2 } 312 f. _9J 3/2 {251 3/2 W > _ r + 16) _ \16J _ 8 125 _ 1 dx 2 64 25 _5_ 24 24 •'• R= 5-21 units 33 Of course, the equation of the curve could be an implicit function, as in the last example, or a pair of parametric equations. e.g. \fx = -sin0 andy = 1 - cos 0, find Rwhen0 = 60° =- * = 0-sin0 •-§=l-cosfl > l dy = dy_dB_ d Y I dx dd ' dx y - 1 - cos 8 /. -^ = sin I . dv . „ 1 _ sing dx 1 - cos 1 - cos At = 60°, sin 8 =^j, cos0=i; ^ = ^ dx _d i sin0 \ = d_ ( sing 1 dd_ dx\l-cos0) dd \l-cosd\dx _ (1 - cos 0) cos 8 - sin . sin 6 1 (1 - cos 8) 2 1 - cos 8 _ cos 0- cos 2 8 -sin 2 6 _ cos - 1 = -1 (1 - cos 0) 3 :. At = 60°,-$ = " (i - cos 0) 3 (i - cos ey -l -l dx* (T^l? i . R _0±3p/ 2 _2 3 _8 _ = -4 R = -2 units 211 Differentiation Applications 1 You notice in this last example that the value of R is negative. This merely indicates which way the curve is bending. Since R is a physical length, then for all practical purposes, R is taken as 2 units long. If the value of R is to be used in further calculations however, it is usually necessary to maintain the negative sign. You will see an example of this later. Here is one for you to do in just the same way as before: Find the radius of curvature of the curve x = 2 cos 3 d,y = 2 sin 3 8 , at the point for which 8 = — = 45°. Work through it and then go to frame 35 to check your work. 34 Result: For R = 3 units dx x = 2 cos 3 6 :. — = 6 cos 2 (-sin 6) = -6 sin 8 cos 2 ( do v ' jy = 2sin 3 .'. 4£=6sin 2 0cos0 du dy_dy_ dd _ 6 sin 2 9 cos 6 dx dd ' dx -6 sin d cos 2 6 cos 6 -tan0 At = 45°,^ = -1 /. (^) 2 =1 dx y dx' M» U- $-{■«*.}. U^,} -sec 2 dd _ dx -6 sin 9 cos 2 8 1 6 sin0 cos 4 ■ A tg = 45^= .* =^ = ^- 'dx 2 6(i)(l) 6 R = h&r (■♦■)' ^_Z V2 2V2 dx 2 3 2 3/2 .3.2V2 = 3 2V2 •'• R = 3 units 35 212 Programme 7 36 Centre of curvature. To get a complete picture, we need to know also the position of the centre of the circle of curvature for the point P(x 1 ,j' 1 ). If the centre C is the point (h, k), „- ^ we can see from the diagram that: h = x 1 -L? = x l -Rsin0 k=y l + LC =yi + R cos 6 That is, { h=X! -Rsinfl k =yi + R cos0 where Xi andj'i are the co- ordinates of P, R is the radius of curvature at P, 6 is the angle of slope at P, i.e. tan / / / \ \ 1 \ 1 1 c \ rv \ \ r* \ R 1 / \ // \ \ 'Li— ■-I-- V /\P{x,.y,) 1 l i ■< h — yf "« — j =1 37 Example. Find the radius of curvature and the co-ordinates of the centre of curvature of the curve y = _„ at the point (2,3). 3-x -1 dy (3-jc) M)-(11-4*)(-1) _ -12 + 4jc+U-4x , ±c~ (3^x7 (3-x) 2 (3-x) 2 dx 1 W d 2 y_d -2 dx =^H3-^}=2(3-x)-(-l)=^3 :. Atx = 2, (3/2 ^.=^=-2 K ■V2 dx 2 1 l_ _^y " 2 " 2 dx 2 R = -V2 Now before we find the centre of curvature (h, k) we must find the angle dy of slope from the fact that tan =-j-at P. i.e. tan = -1 .'. 6 = -45° (0 measured between ± 90°) .'. sin = and cos = 213 Differentiation Applications 1 = -45° Sin0= -^ COS6= j2 38 cnnnnnannnnnnDDDnnnnnDnnnDnnnnannnaDDD So we have: Xi = 2, y^ =3 .\ ft = *, - R sin = 2 -(- y/2) (--4) = 2 - 1 = 1 , ft = 1 k =71 + R cos = 3 +(- V2) (-4) = 3- 1 = 2, A: = 2 .'. centre of curvature C is the point (1,2) NOTE: If, by chance, the calculated value of R is negative, the minus sign must be included when we substitute for R in the expressions for ft and k. Next frame for a final example. Example. Find the radius of curvature and the centre of curvature for the curve y = sin 2 , x : 2 cos d , at the point for which d =—. 39 Before we rush off and deal with this one, let us heed an important WARNING. You will remember that the centre of curvature (h, k) is given by h =Xi -Rsin0 \ , . , , , _ .I and in these expressions k=y l + R cos0 J r d is the angle of slope of the curve at the point being considered i.e. tan0 \dx) ] Now, in the problem stated above, 8 is a parameter and not the angle of slope at any particular point. In fact, if we proceed with our usual notation, we shall be using 8 to stand for two completely different things — and that can be troublesome, to say the least. So the safest thing to do is this. Where you have to find the centre of curvature of a curve given in parametric equations involving 8, change the symbol of the parameter to something other than 8. Then you will be safe. The trouble occurs only when we find C, not when we are finding R only. 214 Programme 7 40 So, in this case, we will re-write the problem thus: Find the radius of curvature and the centre of curvature for the curve y = sin 2 t,x = 2 cos f , at the point for which t = -j Start off by finding the radius of curvature only. Then check your result so far with the solution given in the next frame before setting out to find the centre of curvature. 41 R = -2-795, i.e. 2-795 units Here is the working. y = sin 2 f .'. -p = 2 sin t cos t 7 dt dx _ . „ x = 2cosf ..— =-2sinf dt dy _dy dt _2 sin f cos t _ _^ c f ~dx dt' dx -2 sin f A..-"-.f-«M--i.-.f4 A*g-£i-~')-n-~'}-£--r.i,--s . d 2 y_ 1 "" dx 2 2 R= dV " i Wi dx 2 2 _-2-5n/5_-5\/5_ -5 (2-2361) 8 4 4 = -lM805 = _ 27951 4 R =-2-795 All correct so far? Move on to the next frame, then. 215 Differentiation Applications 1 Now to find the centre of curvature (h, k) h=x l - R sin 6 k=y l + Rcos 6 42 where Also tan0 ~=-| .-. 6 = -26°34' (0 between ± 90°) :. sin(-26°34') = -04472; cos(-26°34') = 0-8944 Xi = 2 cos 60° = 2. — = 1 7i = sin' - ^H and you have already proved that R = -2-795. What then are the co-ordinates of the centre of curvature? Calculate them and when you have finished, move on to the next frame. Results: For: and /z=-0-25; fc=-l-75 43 h=\ -(-2-795) (-0-4472) = 1 -1-250 h = -0-25 k = 0-75 + (-2-795) (0-8944) = 0-75-2-50 k = -\-15 0-4464 T-6505 00969 0-4464 1-9515 0-3979 Therefore, the centre of curvature is the point (-0-25 , -1 -75) This brings us to the end of this particular programme. If you have followed it carefully and carried out the exercises set, you must know quite a lot about the topics we have covered. So turn on now and work the Test Exercise. It is all very straightforward. 216 Programme 7 44 Test Exercise— VII Answer all questions 1 . Find the angle between the curves x 2 + y 2 = 4 and 5x 2 + y 2 = 5 at their point of intersection for which x andjy are positive. 2. Find the equations of the tangent and normal to the curve y 2 = 1 1 - j3— at the point (6, 4). 3. The parametric equations of a function are x = 2 cos 3 6,y = 2 sin 3 0. Find the equation of the normal at the point for which 8 = — = 45°. 4. If x = 1 + sin 28, y = 1 + cos 8 + cos 28, find the equation of the tangent at 8 = 60°. 5. Find the radius of curvature and the co-ordinates of the centre of curvature at the point x = 4on the curve whose equation is y = x 2 + 5 In x - 24. i d 2 y 6. Given that x = 1 + sin 8, y = sin 8 -5- cos 28, show that -7-5- = 2. Find the radius of curvature and the centre of curvature for the point on this curve where 8 = 30°. Now you are ready for the next programme. 217 Differentiation Applications 1 Further Problems- VII 1 . Find the equation of the normal to the curve y = 2 2x a t the point (3, 0-6) and the equation of the tangent at the origin. 2. Find the equations of the tangent and normal to the curve 4x 3 + 4xy+y 2 =4 at (0, 2), and find the co-ordinates of a further point of intersection of the tangent and the curve. 3. Obtain the equations of the tangent and normal to the ellipse x 2 y 2 _ 169 + 2T~ at the p0mt ( 13 cos 6 > 5 sin ^ If the tangent and normal meet the x-axis at the points T and N respectively, show that ON.OT is constant, O being the origin of co-ordinates. 4. If x 2 y + xy 2 -x*-y 3 +16 = 0, find-^ in its simplest form. Hence find the equation of the normal to the curve at the point (1,3). 5 . Find the radius of curvature of the catenary y = c cosh (-) at the pointer, ,>>,). 6. If 2x 2 + y 2 - 6y - 9x = 0, determine the equati6n of the normal to the curve at the point (1,7). 7. Show that the equation of the tangent to the curve x = 2a cos 3 /, y = a sin 3 1 , at any point P(0 < r <^) is x sin t + 2y cost -2a sin t cos t = If the tangent at P cuts the j-axis at Q, determine the area of the triangle POQ. 8. Find the equation of the normal at the point x = a cos d,y = b sin 0, of the ellipse ~ +^- = 1 . The normal at P on the ellipse meets the major axis of the ellipse at N. Show that the locus of the mid-point of PN is an ellipse and state the lengths of its principal axes. 218 Programme 7 X ~ X 9. For the point where the curve y =~r— — j passes through the origin, determine: (i) the equations of the tangent and normal to the curve, (ii) the radius of curvature, (iii) the co-ordinates of the centre of curvature. 10. In each of the following cases, find the radius of curvature and the co-ordinates of the centre of curvature for the point stated. (1)^4=1 at (0,4) (ii) y 2 = Ax -x 2 - 3 at x = 2-5 (iii) y = 2 tan 0, x = 3 sec at = 45° 1 1 . Find the radius of curvature at the point (1 , 1) on the curve x 3 -2xy+y 3 =0. 12. If 3ay 2 = x(x -a) 2 with a > 0, prove that the radius of curvature at the point (3a, 2d) is — — . 13. If x = 26- sin 20 and y = 1 - cos 20, show that-/ = cot and that ax -Hr = i—. — z~z ■ If P is the radius of curvature at any point on the dx 2 4sin 4 curve, show that p 2 = 8y. 14. Find the radius of curvature of the curve 2x 2 + y 1 - 6y - 9x = at the point (1,7). 1 5 . Prove that the centre of curvature {h, k) at the point ?(at 2 , 2at) on the parabola y 2 = 4ax has co-ordinates h = 2a + 3at 2 , k = -2at 3 . 16. If p is the radius of curvature at any point P on the parabola x 2 = 4ay, S is the point (0, a), show that p = 2V[(SP) 3 /SO] , where O is the origin of co-ordinates. 17. The parametric equations of a curve are x = cos t + t sin t, y = sin t - 1 cos t. Determine an expression for the radius of curvature (p) and for the co-ordinates (h,k) of the centre of curvature in terms off. 219 Differentiation Applications 1 18. Find the radius of curvature and the co-ordinates of the centre of curvature of the curves = 3 lnx, at the point where it meets the x-axis. 19. Show that the numerical value of the radius of curvature at the point 2(a + x ) 3 / 2 (*i , y i) on the parabola j> 2 = Aax is — — —. \*- — . If C is the centre a i 2 of curvature at the origin and S is the point (a, O), show that OC = 2 (OS). 20. The equation of a curve is 4y 2 =x 2 (2~x 2 ). (i) Determine the equations of the tangents at the origin, (ii) Show that the angle between these tangents is tan" 1 (2\/2). (iii) Find the radius of curvature at the point (1,1/2). 220 Programme 8 DIFFERENTIATION APPLICATIONS PART 2 Programme , 1 Inverse trigonometrical functions You already know that the symbol sin" 1 x (sometimes referred to as 'arcsine x') indicates 'the angle whose sine is the value x\ e.g. sin" 1 0-5 = the angle whose sine is the value 0-5 = 30° There are, of course, many angles whose sine is 0-5, e.g. 30°, 1 50°, 390°, 510°, 750°, 870°, .. .. etc., so would it not be true to write that sin -1 0-5 was any one (or all) of these possible angles? The answer is no, for the simple reason that we have been rather lax in our definition of sin" 1 x above. We should have said that sin" 1 x indicates the principal value of the angle whose sine is the value x; to see what we mean by that, move on to frame 2. The principal value of sin" 1 0-5 is the numerically smallest angle (measured between 0° and 180°, or 0° and -180°) whose sine is 0-5. Note that in this context, we quote the angle as being measured from 0° to 180°, or from 0° to -180°. In this range, there are two angles whose sine is 0-5, i.e. 30° and 150°. The principal value of the angle is the one nearer to the positive OX direction, i.e. 30°. sin" 1 0-5 = 30° and no other angle! ~X to 180° to -180° Similarly, if sin 8 = 0-7071 , what is the principal value of the angle 81 When you have decided, turn on. 223 Differentiation Applications 2 Principal value of 9 = 45 for: sin 9 = 0-7071 .\ In the range 0° to 180°, or 0° to -180°, the possible angles are 45° and 135°. Y The principal value of the angle is the one nearer to the positive OX axis, i.e. 45°. sin" 1 0-7071 =45° DnnnnDDannnDDnDnnDnnnnnDnnnDnanDDnDnnD In the same way, we can find the value of tan" 1 y/3. If tan 6 = V3 = 1 -7321 , then 9 = 60° or 240°. Quoted in the range 0° to 180° or 0° to -180°, these angles are 9 = 60° or -120°. Y p 240°-- ' " / / (\60° x, \ / Y « X The principal value of the angle is the one nearer to the positive OX direction, i.e. in this case, tan" 1 \J3 = tan"V3 = 60° Now let us consider the value of cos 1 0-8 1 92. From the cosine tables, we find one angle whose cosine is 0-8192 to be 35°. The other is therefore 360° - 35°, i.e. 325° (or -35°). Y, " Y, Of course, neither is nearer to OX: they are symmetrically placed. In such a situation as this, it is the accepted convention that the positive angle is taken as the principal value, i.e. 35°, .". cos" 1 0-8192 = 35° So, on your own, find tan"' (-1). Then on to frame 5. 224 Programme 8 tan" 1 (-1) = -45° For, if tan 6 = -1, 6 = 135° or 315° .135° \ In the range 0° to + 180°, these angles are 135° and -45°. The one nearer to the OX axis is -45°. .'. Principal value = -45°. tan -1 (-1) = -45° Now here is just one more: Evaluate cos" 1 (-0-866) Work through it carefully and then check your result with that on frame 6. cos" 1 (-0-866) = 150° For we have cos £' = 0-866 .'. £=30° :. 9 =150° or 210° In the range 0° ±180°, these angles are d = 150° and -150° Neither is nearer to the positive OX axis. So the principal value is taken as 150°. cos" 1 (-0-866)= 150° So to sum up, the inverse trig, functions, sin -1 *, cos" 1 *, tan" 1 * indicate thep v of the angles having the value of the trig, ratio stated. 225 Differentiation Applications 2 principal value Differentiation of inverse trig, functions Sin" 1 *, cos -1 *, tan" 1 * depend, of course, on the values assigned to x. They are therefore functions of x and we may well be required to find their differential coefficients. So let us deal with them in turn. i^2L (i) Let y = sin l x. We have to find dx First of all, write this inverse statement as a direct statement. y = sin" 1 * .". x = siny dx Now we can differentiate this with respect to y and obtain — dx dy -j- = cosy .. -j- = dy dx 1 dy._ dx cosy Now we express cos>" in terms of x, thus: We know that cos 2 y + sin 2 y = 1 .'. cos 2 y = 1 - sin 2 ^ = 1 - x 2 (since x = sin y) .". cos y = \/( 1 ~ x 2 ) "' dx VO -x 2 ) d \ . _, \ 1_ — { sin x }=— z ^ x dx\ j V(l-^ 2 ) 8 Now you can determine — | cos i x J in exactly the same way. Go through the same steps and finally check your result with that on frame 9. 226 Programme 8 d_ { dx [ cos ' lx ] Vo^ Here is the working: Let y = cos x .'. x = cos j dx dy —1 dy dx smy cos 2 y + sin 2 y = 1 .'. sin 2 y = 1 - cos 2 _y = 1 - x 2 siny = V(l -x 2 ) dy -1 . d i _, \ -1 -=- "'cos x]=j^, " dx~sj{\-x 2 )" dx So we have two very similar results 1 ! ) (i) iH sin "*rv(i-* 2 ) (n) — cos x = 7- 7; Different only in sign. Now you find the differential coefficient of tan -1 x. The working is slightly different, but the general method the same. See what you get and then move to frame 10 where the detailed working is set out. 10 d t* -1 ) 1 — { tan x } = 5 dx l+x 2 Working: Let y = tan l x .". x = tan_y. dx -r- = sec 2 y = 1 + tan 2 y = 1 + x 2 ^=i +JC 2 • d y - l dy dx 1 + x 2 — { tan -1 x } = * dx{ j 1 + x 2 Let us collect these three results together. Here they are: d ( . _, 1 1 dx sin 'x (0 Vd-* 2 ) iH cos ~ v } = vfe 2 ) (ii) d U -i 1 1 -v- tan x ) =-. n- dx\ I 1 + x 2 (iii) Copy these results into your record book. You will need to remember them. On to the next frame. 227 Differentiation Applications 2 Of course, these differential coefficients can occur in all the usual 1 1 combinations, e.g. products, quotients, etc. Example 1. Find -^ , given thatj> = (1 ~x 2 ) sin" 1 * Here we have a product •••i = ( 1 -^vF?) +8fa " , *<- 2x) = V(l - x 2 ) - 2xsin _1 x Example 2. Ify = tan" 1 (2x - 1), find ^ This time, it is a function of a function. *- I .2 = 2 ~ dx 1 + (2x - l) 2 1 + 4x 2 - 4x + 1 2 + 4x 2 - Ax 2x 2 - 2x + 1 and so on. 12 Here you are. Here is a short exercise. Do them all: then check your results with those on the next frame. Revision Exercise Differentiate with respect to x: 1 . y = sin" 1 5x 2. y = cos" 1 3* 3. y = tan" 1 2x 4. y = sin' 1 (x 2 ) 5. j=x 2 .sin- 1 (|) JVferc j>om have finished them all, move on to frame 1 3. 228 Programme 8 13 Results: 1 . v = sin" 1 Sx :. -^ = — _ s = _ £ dx V{l-(5^) 2 } V{1 - 25x 2 } 3 v = tan" 1 2v • d V - 1 ? - 2 ' tan 2x -^- 1+(2jc)2 -2- j^^- 4. ^ = sm 1 (x 2 ) •^=- 7r — 1 .2x = 2x ^ V{l-(x 2 ) 2 f~ V(l-x 4 ) 5 '-'■• ta - 1 (f)---S-* a 7p^)-i^^(f) 2V{i-V = vcfe) + ^ sin " 1 (!) Right, now on to the next frame. \J\ Differential coefficients of inverse hyperbolic functions In just the same way that we have inverse trig, functions, so we have inverse hyperbolic functions and we would not be unduly surprised if their differential coefficients bore some resemblance to those of the inverse trig, functions. Anyway, let us see what we get. The method is very much as before. (i) y = sinh" 1 x To find ^ ax First express the inverse statement as a direct statement. y = sinh" 1 x .'. x = sinh y .". — = cosh y ■'■ -f = dy dx cosh j' We now need to express coshj in terms of x We know that cosh 2 y - sinh 2 _y = 1 .'. cosh 2 y = sinh 2 ); + 1 = x 2 + 1 cosh^ =y/(x 2 + 1) dy _ I . d_( ■ ,-i 1 = 1 dx V(* 2 + 1)" dx \ &mh X j V(* 2 + 1) Let us obtain similar results for cosh" 1 * and tanh" 1 * and then we will take a look at them. So on to the next frame. 229 Differentiation Applications 2 We have just established — \ sinh l x ) = ,. 2 , \ dx I J VO + 1) (ii) y = cosh" 1 x .'. x = coshy dx . , dy 1 .. — - = sinhy ••■^- = -r-T— dy dx sinhy Now cosh 2 y - smb. 2 y = 1 .'. smh 2 y = cosh 2 ^ - 1 = x 2 - 1 .'. sinh y =\J{x 2 ~ 1) dy_ " dx VC* 2 " 1 )" d * .'. — { cosh l x \ = V(x 2 - 1) 15 Now you can deal with the remaining one If v = tanh _1 x,-f- = dx Tackle it in much the same way as we did for tan _1 x, remembering this time, however, that sech 2 x = 1 - tanh 2 x. You will find that useful. When you have finished, move to frame 16. y = tanh *x for: dy _ 1 dx 1 -x 2 >' = tanh 1 x .'. x = tanhj> dy >,= l-tanh 2 ,= l-x 2 :.%-^ £ tanh-x U^ Now here are the results, all together, so that we can compare them. 1 (iv) Ijsinh-xj = V(x2 + i} d_ dx 1 (v) { COSh_lx j-V(x 2 -l) sK'+ri? (v ° Make a note of these in your record book. You will need to remember these results. Now on to frame 1 7. 16 230 Programme 8 I f Here are one or two examples, using the last results Example 1. y = cosh -1 j3-2x . dy = 1 (-2) = -2 " dx V((3 - 2x) 2 - 1} 7(9 - 1 2x + Ax 2 - 1) -2 = -2 = -1 V(8 - \2x + Ax 2 ) 2y/(x 2 - 3x + 2) V(* 2 - 3x + 2) Example 2. y = tanrf 1 l^) , dy_ 1 2 = _! 1 " ^ ,_/3*\ 2 4 i _9^ 4 1 U / 16 = 16 1= 12 16 -9x 2 ' 4 16-9jc 2 Example 3. y = sinrT 1 {tan jc } • ty - l 2 _ sec 2 * dx -^(tan 2 * + 1)' VWc 2 * = secx I O Here are a few for you to do. Exercise Differentiate: 1. y = sinh" 1 3x 2. y = cosh ' ( — J 2 3. 7 = tanh" 1 (tanx) A.y = sinh -1 V(* 2 ~ 1) 5. j = cosh" 1 (e 2 *) Finish them all. Then turn on to frame 19 for the results. 231 Differentiation Applications 2 Results: 1. y = sinh" 1 3x . dy _ 1 v 3 = dx V{(3^) 2 + 1}' V(9* 2 +l) 1 5 5 —'(?)■■- t^ (?) ._ ir 2- M ^_ l) 19 2 // 25x 2 -4 \ V(25x 2 -4) 1 j-tanh" (tan*) •• ^= t _ tan ^ ^ = sinh- 1 { v / (x 2 -l)} dy _ 1 J_, , ,ci 2 S6C X 2 . sec x = 1 -tan"* ^V(x 2 -1+1)-2 (X 1} ^)=V(* 2 -1) ^ = cosh 1 (e 2 *) :. -j- = All correct? On then to frame 20. dx V{(e 2 *) 2 -1} ,2e 2 2e 2X V(e 4JC -D- Before we leave these inverse trig, and hyperbolic functions, let us look at them all together. Inverse Trig. Functions Inverse Hyperbolic Functions y dy dx y dy dx sin" 1 x cos" 1 * tan" 1 x 1 sinh" 1 x cosh" 1 * tanh" 1 * 1 V0-* 2 ) -1 V(* 2 + 1) 1 V(l-* 2 ) 1 V(* 2 - 1) 1 l-* 2 1+* 2 It would be a good idea to copy down this combined table, so that you compare and use the results. Do that: it will help you to remember them and to distinguish clearly between them. 20 232 Programme 8 21 Before you do a revision exercise, cover up the table you have just copied and see if you can complete the following correctly. _„ ; „-i„ dy ^ 1. If^sin-*,— 2. If, = co,-',,g = 3. If, = tan-*,g = 4. If y = sinh _1 A:,-r-= J dx 5. If, = cosh" 1 *, -7- = 6. If, = tanh -1 *, -r- = Now check your results with your table and make a special point of brushing up any of which you are not really sure. 22 Revision Exercise Differentiate the following with respect to x: 1. tan -1 (sinh;>c) 2. sinlT 1 (tan x) 3. cosh" 1 (sec x) 4. tanh" 1 (sin x) 5 . sin' If) Take care with these; we have mixed them up to some extent. When you have finished them all - and you are sure you have done what was required — check your results with those on frame 23. 233 Differentiation Applications 2 Solutions 1. j^tan-Csinh*) jLjtan" 1 *} = ^ dy 1 , coshx , ■'• -r = , . , 7 . cosh x = — rj- = seen x dx 1 + sinn x cosh x 2. j, = rinlf l (tan*) fj^ x } = ^Ti) dy 1 i sec 2 * .'. -f- = -. ; .. sec^x = —, — 5 — = sec x dx ^(tan 2 * + 1) yseCjc 3. y = cosh" 1 (sec x) ^( cosh "^J = ^^l) dy 1 se c x , tan x , = ,- — , r; -sec x. tan x = ~r — 5 — dx V(sec 2 x - 1) Vtan z x : secx y = tanlT 1 (sin x) - | tanlT 1 x J = y^j dy 1 cos x :. -r = , rr" ■ cos x = — 5— = sec x dx 1 - sin x cos x 1 x a dT-,\ I 5 - ^ =sin i-i dx" i^" * rvo 7 ^ 5 ) .,}., "dx " 7rTT If you have got those all correct — or nearly all correct — you now know quite a lot about the differential coefficients of Inverse Trig, and Hyperbolic Functions. You are now ready to move on to the next topic of this programme, so off you go to frame 24. 23 234 24 Programme , Maximum and minimum values (turning points) You are already familiar with the basic techniques for finding maximum and minimum values of a function. You have done this kind of operation many times in the past, but just to refresh your memory, let us consider some function,/ = f{x) whose graph is shown below. y = fix) At the point A, i.e. at x = Xi , a maximum value of y occurs since at A, the y value is greater than the y values on either side of it and close to it. Similarly, at B,y is a .'..., since the y value at the point B is less than the y values on either side of it and close to it. 25 y = fix) The point C is worth a second consideration. It looks like 'half a max. and half a min.' The curve flattens out at C, but instead of dipping down, it then goes on with an increasingly positive slope. Such a point is an example of a point of inflexion, i.e. it is essentially a form of S-bend. Points A, B and C, are called turning points on the graph, or stationary values of y, and while you know how to find the positions of A and B, you may know considerably less about points of inflexion. We shall be taking a special look at these. On to frame 26. 235 Differentiation Applications 2 If we consider the slope of the graph as we travel left to right, we can draw a graph to show how this slope varies. We have no actual values for the slope, but we can see whether it is positive or negative, more or less steep. The graph we obtain is the first derived curve of the function and we are really plotting the values of ~- against values of x 26 y = f(x) Y y t + (max) Point of S A ^ inflexion ^/+ ^/S~ \ ^S. c ^^^ r ' \- y^*"" - ] o ! X (min) "^^ i | X^ B ^>f I i ; o ] t dec 1^1 ,JC 2 1^3 X ^iX S^Z ^3 X We see that at x = x t , x 2 , x 3 , (corresponding to our three turning points) the graph of -~ is at the a: -axis — and at no other points. Therefore, we obtain the first rule, which is that for turning points, dx Turn on to frame 27. 236 Programme 27 For turning points, A, B, C, dx If we now trace the slope of the first derived curve and plot this against x, we obtain the second derived curve, which shows values of -—% against jr. dx Y A y = fix) y = f'(oc) y = f"(x) From the first derived curve, we see that for turning points, dx From the second derived curve, we see that d 2 y for maximum y, —r^ is negative f • • *y . .,. for minimum j', -7-5 is positive for P-of-L d 2 y dx 2 is zero Copy the diagram into your record book. It summarizes all the facts on max. and min. values so far. 237 Differentiation Applications 2 From the results we have just established, we can now determine (i) the values of x at which turning points occur, by differentiating the function and then solving the equation -j- = (ii) the corresponding values of y at these points by merely substitut- ing the x values found, in y =f(x) (iii) the type of each turning point (max., min., or P-of-I) by testing dx 2 28 in the expression for -^-^ With this information, we can go a long way towards drawing a sketch of the curve. So let us apply these results to a straightforward example in the next frame. Example. Find the turning points on the graph of the function x 3 x 2 y = -T-— zr~2x + 5 . Distinguish between them and sketch the graph of the function. There are, of course, two stages : (i) Turning points are given by -j- = (ii) The type of each turning point is determined by substituting the dy d y roots of the equation -j- = in the expression for -j-y cue ctx 29 rf d 2 y . If dP ls negative, then y is a maximum, »» •>•> 5» positive, " " " " minimum, 55 »1 55 zero, " " " » point of inflexion. We shall need both the first and second differential coefficients, x x dv them ready. If y = — - — - 2x + 5, then -f = and 3 2 dx so find d 2 y dx 2 238 Programme , 30 dx dx* DaDnDnDDnnDDDnDDnnnDnnDDDnnDnDDnDnaaaD (i) Turning points occur at — = /. x 2 -x-2 = :. (x-2)(x+l) = :. jc = 2andx=-l i.e. turning points occur at x = 2 and x=—l. (ii) To determine the type of each turning point, substitute x = 2 and cfy then x = -1 in the expression for —-j At x - 2, -7^3 =4-1=3, i.e. positive /. x = 2 gives ^ m i n . dx 2 At x = -1 , ^ = -2 -1, i.e. negative /. x = -1 gives y mstx . Substituting in y = /(x) gives x = 2, 7 min = if and x - -1 , j> max - £>\ Also, we can see at a glance from the function, that when x = 0,y = 5. Fow can now sketch the graph of the function. Do it. 31 A 1 2 /3 6V 6 B -— •- We know that (i) at x = -1 , 7 m ax = 65 (ii) atx = 2, ^ m in = l§ 5 x Joining up with a smooth curve gives: Y (iii) at x = 0, y = 5 X, -( \ 2 3 4 5 X There is no point of inflexion like the point C on this graph. Move on. 239 Differentiation Applications 2 All that was just by way of refreshing your memory on work you have done before. Now let us take a wider look at these Points of Inflexion The point C that we considered on our first diagram was rather a special kind of point of inflexion. In general, it is not necessary for the curve at a P-of-I to have zero slope. A point of inflexion is defined simply as a point on a curve at which the direction of bending changes, i.e. from a right-hand bend to a left- hand bend, or from a left-hand bend to a right-hand bend. 32 The point C we considered is, of course, a P-of-I, but it is not essential at a P-of-I for the slope to be zero. Points P and Q are perfectly good points of inflexion and in fact in these cases the slope is ('positive "J J negative | Which? zero At the points of inflexion, P and Q, the slope is in fact positive Correct. The slope can of course be positive, negative or zero in any one case, but there is no restriction on its sign. DnDDnDDDnnDanDDDDnQDnnnnDaDnDDDDDnDDaD A point of inflexion, then, is simply a point on a curve at which there is a change in the d of b 33 240 Programme 8 j(l Point of inflexion: a point at which there is a change in the direction of bending DODDDDQDDDDDnDDDDDDDDDDDDnnnDODDDnDDDD dy_ dx If the slope at a P-of-I is not zero, it will not appear in our usual max. dy and min. routine, for --■ will not be zero. How, then, are we going to find where such points of inflexion occur? Let us sketch the graphs of the slopes as we did before. L % -N\ LH P and Q are points of inflexion. v\_ In curve 1 , the slope is always r.hA - positive, ++ indicating a greater positive slope than +. 5 x Similarly in curve 2, the slope i; always negative. In curve 1 , -r reaches a minimi dx value but not zero. x In curve 2, -r- reaches a maxinr! dx value but not zero. For both points of inflexion, i.J x = X4 and x = x s J d 2 v dx We see that where points of inflexion occur, —-5 = So, is this the clue we have been seeking? If so, it simply means that to find the points of inflexion we differentiate the function of the curve d 2 y twice and solve the equation -72 ~ 0- That sounds easy enough! But turn on to the next frame to see what is involved. 241 Differentiation Applications 2 We have just found that dry where points of inflexion occur, ji = This is perfectly true. Unfortunately, this is not the whole of the story, cPy for it is also possible for -7-^ to be zero at points other than points of inflexion! d 2 y So if we solve -7-5 = 0, we cannot as yet be sure whether the solution x =a gives a point of inflexion or not. How can we decide? Let us consider just one more set of graphs. This should clear the matter up. dx2 The first derived curves could well look like this. 35 Let S be a true point of inflexion and T a point ony =f(x) as shown. Clearly, T is not a point of inflexion. 36 Notice the difference between the two second derived curves. Although Tj is zero for each (at x = x 6 and x = x 7 ), how do they differ? When you have discovered the difference, turn on to frame 37. 242 Programme 8 37 In the case of the real P-of-I d 2 y the graph of ^ crosses the x-axis. In the case of no P-of-I, the d 2 y graph of ^ only touches the x -axis A*y a and —k doe dx* s not change sign. DDDnDDnnDDDnnDDnnnnDnnnnnDDDDnnDDnnnDn This is the clue we have been after, and gives us our final rule. d 2 y d 2 y For a point of inflexion, -pi - and there is a change of sign of -pi as we go through the point. (In the phoney case, there is no change of sign.) So, to find where points of inflexion occur, d 2 y (i) we differentiate y = f(x) twice to get ^ -d 2 v (ii) we solve the equation'-r^ = d 2 y (iii) we test to see whether or not a change of sign occurs in -p\ as we go through this value of x. d 2 v For points of inflexion, then, -H[ - 0, withe of s 38 For a P-of-I, try = dP with change of sign This last phrase is all-important. DGnnnnDnDnnnanDnnDnnnnnDDDnnDDnnnnDDDD Example 1. Find the points of inflexion, if any, on the graph of the function I y-^-j-Tx + 5. (i) Diff. twice. d £=x 2 -x-2, 4$-= 2x - 1 For P-of-I, d 2 y_ dx = 0, with change of sign. .'. 2x - 1 = .'. x = — If there is a P-of-I, it occurs at x = 1 1 (ii) Test for change of sign. We take a point just before x = -z, i.e. x = -r~ a, 11 and a point just after x = x-, i.e. x = — + a, where a is a small positive dry quantity, and investigate the sign of -pi at these two values of x. Turn on. 243 Differentiation Applications 2 dx 2 ^ 1 dP~ 2( 2 39 (i) Atx = --a, ^A = 2(£- a)-l = l-2a-l = -2a (negative) 4 (ii) At x = -+ a,-^£ = 2(^-+ a) - 1 = 1 + 2a - 1 : 2a (positive) r d 2 ^ There is a change in sign of -H, as we go through x = — .'. There is a point of inflexion at x = — If you look at the sketch graph of this function which you have already drawn, you will see the point of inflexion where the right-hand curve changes to the left-hand curve. Example 2. Find the points of inflexion on the graph of the function ZL|] y = 3x s - 5x A + x + 4 d 2 v First, differentiate twice and solve the equation j-^ = 0. This will give the values of x at which there are possibly points of inflexion. We cannot be sure until we have then tested for a change of sign in —^ . We will do that in due course. d 2 v So start off by finding an expression for -Hj and solving the equation d 2 y _ n dx When you have done that, turn on to the next frame. 244 Programme 8 41 We have: y = 3x s - 5x 4 + x + 4 /. ^ =15x 4 -20* 3 + 1 dx %\ = 60x 3 - 60x 2 = 6(k 2 (jc - 1) d 2 y_. For P-of-1, -r-j = 0, with change of sign. :. 60x 2 (x-l) = :. x = 0or;t=l If there is a point of inflexion, it occurs at x = 0, x = 1 , or both. Now comes the test for a change of sign. For each of the two values of x we have found, i.e. x = and x = 1 , take points on either side of it, differing from it by a very small amount. (i) Forx = At jc = -a, %\ = 60(-a) 2 (-a - 1) 'dx 2 At x = +a, dx 2 : (+)(+)(-) = negative 60(+a) 2 (a-l) = (+)(+)(-) = negative J No sign change. No P-of-I. (ii) For x = 1 At x = 1 - a, |^ = 60(1 - a) 2 (l - a - 1) At x = 1 + a, dx 2 = (+)(+)(-) = negative 60(l+a) 2 (l +a-l) = (+)(+)(+) = positive ► Change in sign. :. P-of-I. Therefore, the only point of inflexion occurs when x- 1 , i.e. at the point x=\,y = Z That is just about all there is to it. The functions with which we have to deal differ, of course, from problem to problem, but the method remains the same. Now turn on to the next frame and complete the Test Exercise awaiting you. The questions are all very straightforward and should not cause you any anxiety. 245 Differentiation Applications 2 Test Exercise— VIII Answer all the questions. 1. Evaluate (i) cos -1 (-0-6428), (ii) tan -1 (-0-7536). 2. Differentiate with respect to x: (i) y = sin" 1 (3* + 2) .... cos" 1 * (")/= — (iii) 7=x 2 tan" 1 (|j (iv) y = cosh" 1 (1 - 3x) (v) y = sinlf 1 (cos x) (vi) y = tanh" 1 5x 3. Find the stationary values ofy and the points of inflexion on the graph of each of the following functions, and in each case, draw a sketch graph of the function. (i) y = x 3 - 6x 2 + 9x + 6 (ii) y = x+ — (iii) y =xe' x Well done. You are now ready for the next programme. 42 246 Programme 8 Further Problems— VIII 1 + tan x 1. Differentiate (i) tan -1 {- tanxj (ii) jcV(1 -a: 2 ) -sin" 1 y/(l-x 2 ) sin" 1 * 2. If y =j (l _ x 2y prove that (i) {\-x*) d £=xy+\ 3. Find -f- when (i) y = tan -1 ' dx w J \\-\x 2x ( 2; (ii) ^ = tanh M — 4. Find the co-ordinates of the point of inflexion on the curves (i).y = (x-2) 2 (x-7) (ii) y = 4x 3 +3x 2 -18x-9 5. Find the values of x for which the function^ =f(x), defined by y(3x - 2) = (3x - l) 2 has maximum and minimum values and distinguish between them. Sketch the graph of the function. 6. Find the values of x at which maximum and minimum values of y and points of inflexion occur on the curves = 12 lnx + x 2 - lOx 7. If Ax 2 + &xy + 9y 2 - 8x - 24y + 4 = 0, show that when -f = 0, d 2 y 4 x + y = 1 and — =f = x — t . Hence find the maximum and J dx &-5y minimum values of y. 8. Determine the smallest positive value of x at which a point of inflexion occurs on the graph of y = 3 e 2x cos(2x - 3). 9. If y 3 = 6xy -x 3 - 1 , prove that -f- = % _ ~ and that the maximum value of y occurs where x 3 = 8 + 2\/l4 and the minimum value where x 3 =8-2Vl4. 247 Differentiation Applications 2 10. For the curve y = e x sin x, express -j- in the form Ae x cos(x + a) and show that the points of inflexion occur at x = -j + kit for any integral value of k. 1 1 . Find the turning points and points of inflexion on the following curves, and, in each case, sketch the graph. (i) y = 2x 3 - 5x 2 + Ax - 1 ,... x(x-l) (iii) y = x + sin x (Take x and y scales as multiples of 77.) 12. Find the values of x at which points of inflexion occur on the following curves. (i) y = e~* 2 (ii) y = z 2x {2x 2 + 2x + 1) (iii) y=x* -\0x 2 +7jc + 4 13. The signalling range (x) of a submarine cable is proportional to r 2 In ( -J, where -r is the ratio of the radii of the conductor and cable. Find the value of/- for maximum range. 14. The power transmitted by a belt drive is proportional to Tv - — , where v = speed of the belt, T = tension on the driving side, and w = weight per unit length of belt. Find the speed at which the transmitted power is a maximum. 15. A right circular cone has a given curved surface A. Show that, when its volume is a maximum, the ratio of the height to the base radius isV2: 1. 16. The motion of a particle performing damped vibrations is given by y = e~ f sin 2t,y being the displacement from its mean position at time t . Show that y is a maximum when t = ^ tan" 1 (2) and determine this maximum displacement to three significant figures. 17. The cross-section of an open channel is a trapezium with base 6 cm and sloping sides each 10 cm wide. Calculate the width across the open top so that the cross-sectional area of the channel shall be a maximum. 248 Programme 8 18. The velocity (v) of a piston is related to the angular velocity (oj) of the crank by the relationship v = cor \ sin 6 + — sin 20 I where r = length of crank and / = length of connecting rod. Find the first positive value of for which v is a maximum, for the case when l = 4r. 19. A right circular cone of base radius r, has a total surface area S and volume V. Prove that 9V 2 = r 2 (S 2 - 2tt/- 2 S). If S is constant, prove that the vertical angle (0) of the cone for maximum volume is given by = 2 sin -1 (jj. d x dx 20. Show that the equation 4tj + 4jjt- + y?x = is satisfied by x = (At + B) e" M '' 2 , where A and B are arbitrary constants. If dx x = and t = C when t = 0, find A and B and show that the at ■ 2C 2 maximum value of x is — and that this occurs when t = — . lie li 249 Programme 9 PARTIAL DIFFERENTIATION PART1 Programme 9 1 Partial differentiation The volume V of a cylinder of radius r and height h is given by V = nr 2 h i.e. V depends on two quantities, the values of r and ft. If we keep r constant and increase the height h, the volume V will increase. In these circumstances, we can consider the differential coef- ficient of V with respect to h - but only if r is kept constant. i.e. dV dh is written r constant 3V dh Notice the new type of 'delta'. We already know the meaning of By dy 9V 3V ■=— and -j— . Now we have a new one, — r-. — r-is called the partial differential Bx dx dh dh ^ J coefficient of V with respect to h and implies that for our present purpose, the value of r is considered as being kept constant DDDaDnaDDnnDannunDDnaDnDnnnnDDDDDnnnnn ,, „ » .3V V = ' o v ■nr 2 h. To find — we differentiate the given expression, taking all 3V : AT" symbols except V and h as being constant .'. K ^- = 7ir i \ ah Of course, we could have considered h as being kept constant, in which case, a change in r would also produce a change in V. We can therefore talk about ~ which simply means that we now differentiate V = -nr 2 h dr with respect to r, taking all symbols except V and r as being constant for the time being. ^ v :. Y- =Tr2rh = 2nrh dr In the statement, V = irr 2 h, V is expressed as a function of two variables, r and h. It therefore has two partial differential coefficients, one with respect to and one with respect to 251 Partial Differentiation 1 One with respect to r\ one with respect to h DDnnDDDDDDDDDDDDQDnODnDQaDDDDDDDDDDnDO Another Example r- Let us consider the area of the curved surface of the cylinder. 7 A = 2-nrh A is a function of r and A, so we can ~ , dA , 9A nnd— and — dr dh 3A To find-^— we differentiate the expression for A with respect to r, keep- ing all other symbols constant. 9A To find — we differentiate the expression for A with respect to h, keep- ing all other symbols constant. So, if A= 2irrh, then-r— = and -tt- = dr dh A = 2irrh JjA dr = 2irh and 9A dh = 2irr ODDDDnUDDDDDDDDDDDDDDDDDUDDDDDDDDnnDDO Of course, we are not restricted to-the mensuration of the cylinder. The same will happen with any function which is a function of two independent variables. For example, consider z = x 2 y 3 . Here z is a function of x and v. We can therefore find-r- and-r- dz bx hy (i) To find ^- , differentiate w.r.t. x, regarding^ as a constant. ■'■i =2 ^ 3= ^l dz (ii) To find^-, differentiate w.r.t. y, regarding x as a constant. ~=x 2 3y 2 = 3x 2 y 2 dy Partial differentiation is easy! For we regard every independent variable, except the one with respect to which we are differentiating, as being for the time being 252 Programme 9 constant □□□DnonnnnnnnnooQannDDnnonQaaoaQnonoon Here are one or two examples: Example 1. u=x 2 +xy+y du dx (i) To find ^ , we regards as being constant. Partial diff. w.r.t. x oix 2 = 2x " " " ''- " xy = y (y is a constant factor) " " " y 2 =0 (y 2 is a constant term) du „ , — = 2x+y (ii) To find -£■ , we regard x as being constant. Partial diff. w.r.t. y of x 2 =0 (x 2 is a constant term) " " " " " xy = x (x is a constant factor) — =x + 2y by Another example on frame 6. Example 2. z=x 3 +y 3 ~2xy — = 3x 2 + - 4xy = 3x 2 - Axy dx - = + 3v 2 - 2x 2 = 3y 2 - 2x 2 by i And it is all just as easy as that. Example 3. z = (2x - y) (x + 3j)) This is a product, and the usual product rule applies except that we keep y constant when finding — , and x constant when finding — f 5 = (2x -y) (1 + 0) + {x + 3y) (2 - 0) dx -Ix-y + 2x + 6y - 4x + 5y f=(2x-y)(0 + 3) + (x + 3y)(0-l) = 6x-3y - x-3y = 5x-6y Here is one for you to do. If z = {Ax - 2y) (3x + Sy), find -^ and ^ Find the results and then turn on to frame 7. 253 Partial Differentiation 1 Results: |£ = 24x + 14y bx ^ = 14jc-20j ^ For z = (4x- 2y) (3x + 5y), i.e. product . 9z bx 9z 9>> = (4x - 2y) (3 + 0) + (3x + Sy) (4 - 0) = 1 2x - 6y + \2x + 20y = 24a: + 14y = (4* - 2y) (0 + 5) + (3x + Sy) (0 - 2) = 20x - lQy - 6x - lOy = 14x-20y There we are. Now what about this one? Example* U z=^LZZ t find |^ and ^ x +y bx by Applying the quotient rule, we have az_ fr+jO(2-0)-(2x-jQ(l+0) . bx (x+y) 2 aM by~ (x+yf ~(x+y) 2 3y (x+yf That was not difficult. Now you do this one: T - 5x +y _ 9z 9z If z = jf-, find -5- and -*- x - ly ox dy When you have finished, on to the next frame. bz -lly bx (x - 2y) 2 bz _ llx 3y (x-2^) 2 Here is the working: (i) To find— , we regard jy as being constant. 9z bx , (x-2y)(S+0)-(Sx+y)(l-0) 5s- 10y-5s-j ;_ —1 !>■ 9z (x-2y? - Oi,^ (*-2y) : 8 (ii) To find-^ , we regard x as being constant. y . 9z (s-2y)(0 + l)-(5s+>Q(0-2) " by (x - 2yf _ x-2y + \0x + 2y 1 Ijc (s-2>>) 2 (s-2y) a In practice, we do not write down the zeros that occur in the working, but that is how we think. Let us do one more example, so turn on to the next frame. 254 Programme 9 Example 5. If z = sin(3x + 2y) find t- and r- Here we have what is clearly a 'function of a function'. So we apply the usual procedure, except to remember that when we are finding dz (i) -^- , we treaty as constant, and (ii) -r- , we treat x as constant. dy Here goes then. 9z = = cos(3x + 2y) X 3 = 3 cos(3x + 2y) |[ = cos(3x + 2y) X ^ (3x + 2y) I 2 = cos(3x + 2y) X ^- (3x + 2y) dy dy = cos (3* + 2^X2 = 2 cos (3x + 2y) There it is. So in partial differentiation, we can apply all the ordinary rules of normal differentiation, except that we regard the independent variables other than the one we are using, as being for the time being 10 constant DaanDnnnnnDDDnDnnDnannDannaaDDDnaDnann Fine. Now here is a short exercise for you to do by way of revision Exercise In each of the following cases, find ^- and ^- 1 . 2 = 4x 2 + 3xy + Sy 2 2. z = (3jc + 2y) (4x - 5y) 3. z = tan(3* + 4y) sin(3x + 2y) 4. z = — - — xy Finish them all, then turn to frame 11 for the results. 255 Partial Differentiation 1 Here are the answers: 1 . z = Ax 2 + 3xy + Sy 2 ^~=8x + 3y ^=3x + lOy dx L. by — 2. z = (3x + 2y) (Ax - 5y) ^=24x-7y ^=-7x-20y Ox — ay 11 3. z = tan(3x + Ay) ¥ = 3 sec 2 (3x + Ay) ^ = 4 sec 2 (3x + Ay) dx ay 4. z = _ sin(3x + xy bz 3x cos(3x + 2v)- sin(3x + 2v) + 2v>- sin (3.x bx x 2 y bz 2y_ cos(3x + 2y) by xy 2 DDnDnDDnnnaanQaDDDnnaDDnnnnDaDaDnDnana If you have got all the answers correct, turn straight on to frame 15. If you have not got all these answers, or are at all uncertain, move to frame 12. Let us work through these examples in detail. 1 . z = Ax 2 + 3xy + 5y 2 |2 bz To find — , regard y as a constant. .'. ~ = 8x + 3y + 0, i.e. 8jc + 3y :. j = 8x + 3y Similarly, regarding x as constant, Y = + 3x + 10y, i.e. 3x + lOy .*. r 2 = 3x + 10)> 2. z = (3x + 2j>) (4x - 5>0 Product rule. || = (3x + 2^)(4) + (4x-5^)(3) = 12x + 8.y + 12* - \5y = 2Ax-ly ^•=(3x + 2y)(-5) + (4*-5j/)(2) = -15* - 10y + 8* - 10>- = -lx - 20)' Turn o/i /or ?fte solutions to Nos. 3 and 4. 256 Programme 9 13 3. z = tan(3x + Ay) Y = sec 2 (3x + Ay) (3) = 3 sec 2 (3x + Ay) Y = sec 2 (3x + Ay) (4) = 4 sec 2 (3x + Ay) 4 2 = sin(3x + 2y ) xy dz _ xy cos(3x + 2y) (3) -- sin(3x + 2y) (y) bx x 2 y 2 _ 3x cos(3x + 2y) - sin(3x + 2y) x 2 y dz Now have another go at finding r- in the same way. Then check it with frame 14. 14 Here it is: z = _ sin(3x + 2y) xy dz -xy cos(3x + 2y).{2) - sin (3* + 2y).(x) by x 2 y 2 -2y cos(3x + 2y)-sin(3x + 2y) xy That should have cleared up any troubles. This business of partial differentiation is perfectly straightforward. All you have to remember is that for the time being, all the independent variables except the one you are using are kept, constant — and behave like constant factors or constant terms according to their positions. On you go now to frame 15 and continue the programme. 257 Partial Differentiation 1 Right. Now let us move on a step. Consider z = 3x 2 + Axy - Sy 1 Then — = 6x + Ay and — = Ax - lOj dx dy bz The expression — = 6x + Ay is itself a function of x and y. We could therefore find its partial differential coefficients with respect to x or to y. (i) If we differentiate it partially w.r.t. x, we get: 3/2 l ^2 r- I -^ J and this is written r-j (much like an ordinary second differential coefficient, but with the partial 9) . 9 2 z 9 ,, , . , , This is called the second partial differential coefficient of z with respect tox. (ii) If we differentiate partially w.r.t. y, we get: 9 fi z \ a *u- ■ •♦♦ 3 * z -r— { tt- } and this is written r — r- by \bx) by. ox Note that the operation now being performed is given by the left-hand of the two symbols in the denominator. 15 b 2 z 9 (3z\ i)L A , 1 . So we have this: z = 3x 2 + Axy - Sy 2 — = 6x + Ay — = Ax-10y dx by 2=6 bx ^- A by.bx Of course, we could carry out similar steps with the expression for -^- on the right. This would give us: 2 - - 10 dy bx.by KT ♦ ♦!, ♦ ° Z ° \ 0Z \ 2 Z Note that ^ — ;- means ^- -r— } so - — — means , ay. ax oy\ox) ox. by 16 258 Programme 9 17 a 2 z a Ldz r—r- means -r-{ — DDDnDDDnDnnnDnDDnnnDnDDnnnnnDDnnDDnnDD Collecting our previous results together then, we have z = 3x\ + 4xy - 5y 2 ,— = 6x + 4 y ■; dx SfTy' 4 *- 1 * '^ 2 z f 9.T *>= 4 dy.dx a = 4 a.x.a.y as, that ^ = a 2 z We see, in this case, that . . . . ay. ax dx.dy There are then, two first differential coefficients, and four second differential coefficients, though the last two seem to have the same value. Here is one for you to do. 3z Bz 3 2 z 3 2 z 9 2 z 3 2 z If z = 5x 3 + 3x y + 4y 3 , find — , — , t-t , -r-j , t — r- , r — r- ' dx by tor' dy dx.dy dy.dx When you have completed all that, turn to frame 18. 18 Here are the results: z = 5x 3 + 3x 2 y + Ay 3 ,'f^T = \5x 2 +6xy / ; dx i, a 2 z a* 2 * J!* dy.dx ^=3x 2 +12y 2 '.' dy ^2 = 30x + 6y = 6x Again in this example also, we see that oy a 2 z ti = 24 y = 6x dx.dy 3 2 z 3 2 z . Now do this one. dy.dx dx.dy ' It looks more complicated, but is done in just the same way. Do not rush at it; take your time and all will be well. Here it is. Find all the first and second partial differential coefficients of z =x.cosy—y.cosx. Then to frame 19. 259 Partial Differentiation 1 Check your results with these. 1 Q z = x cosy-y.cosx When differentiating w.r.t. x,y is constant (and therefore cos.y also) " y,x " " ( " " COS* " ) So we get : " y,x" ( " " cos. bz . — = cos v + y.smx dx 3z -r— = — x.sin y — cos by b 2 z d 2 z -^^-x.cosy a 2 z . A . - — r- = -sin v + sin x by. ox b 2 z . + . -■ . - sin v + sin a: bx.dy A A ■ ° 2Z ° 2Z And again, — — =r— r - by.bx bx.by In fact this will always be so for the functions you are likely to meet, so that there are really three different second partial diff. coeffts. (and not ~2 four). In practice, if you have found -r — ^— it is a useful check to find -\2 ' ■? — r— separately. They should give the same result, of course. What about this one? 2 If V = ln(x 2 + y 2 ), prove that-|J +-|J = This merely entails finding the two second partial diff. coeffts. and sub- stituting them in the left-hand side of the statement. So here goes : V = \n(x 2 +y 2 ) 3V 1 „ 2x 2x = - bx (jc 2 + y 2 ) x 2 +y 2 3 2 V = {x 2 +y 2 )2-2x.2x bx 2 (x 2 +y 2 ) 2 = 2x 2 +2y 2 -4x 2 = 2y 2 - 2x 2 (i) (x 2 +y 2 ) 2 (x 2 +y 2 ) 2 a 2 v Now you find-7— y in the same way and hence prove the given identity. by When you are ready, turn on to frame 21. 20 260 Programme 9 01 9 2 V 2y 2 ~2x 2 £ I We had found that r-r = ;*V~; — 5^ 9x z (x + .y ) So making a fresh start from V = ln(x 2 + y 2 ), we get 9y = _i_, 2v= 2y 9j x 2 + j 2 ^ x 2 +/ 9 2 V^( * 2 +y 2 )2-2y.2y by 2 (x 2 +y 2 ) 2 _ 2x 2 + 2y 2 - 4y 2 _ 2x 2 - 2y 2 (ii) (x 2 +J 2 ) 2 (x 2 +/) 2 Substituting now the two results in the identity, gives JlY 3 2 V = 2y 2 -2x 2 2x 2 - 2y 2 9x 2 by* (x 2 +y 2 ) 2 (x 2 +y 2 ) 2 2y 2 - 2x z + 2x 2 - 2y z _ (x 2 +j 2 ) 2 "^ Afew o« to frame 22. O O Here is another kind of example that you should see. Example 1. If V =/(x 2 +y 2 ), show that x — -y — = Here we are told that Visa function of (x 2 + y 2 ) but the precise nature of the function is not given. However, we can treat this as a 'function of a function' and write/'(x 2 +y 2 )to represent the diff. coefft. of the func- tion w.r.t. its own combined variable (x 2 +y 2 ). ■•• ^=/V + /)X"|(* 2 +J' 2 )=/V +J 2 )-2* =/V +y 2 )-^ (x 2 +y 2 ) =/ V +y 2 ).2y •'• XTr-y¥ =x -f'<- x * +y 2 )-2y-y-f'(.x 2 +y 2 )-2x ay ox = 2xy.f(x 2 +y 2 )-2xy.f'(x 2 +y 2 ) = Let us have another one of that kind on the next frame. 261 Partial Differentiation 1 Example 2. If z = /[£] , sho w that x b I + y 1- = £. «J 9>- Much the same as before. 3z 9x =/'Bl@=/'H(^)=-^/'H 9y y Ur by\xl J \x> x x 1 \xl x \xi x \x) = And one for you, just to get your hand in. If V = f(ax + £>>), show that b ^ - a ^— = dx oy When you have done it, check your working against that on frame 24. Here is the working; this is how it goes. SIX- V=f(ax+by) ,<£=f ( ax + by).-±(ax + by) = f'(ax +by).a = a.f'(ax + by) ... (i) — = f'(ax + by) — (ax + by) = f'(ax + by).b = b.f'(ax + by) ....(ii) " b a* ~ a dy = ab -f' ( ax + b y)~ ab - f'( ax + b y) = Turn on to frame 25. 262 Programme 9 ^JJ So to sum up so far. Partial differentiation is easy, no matter how complicated the expres- sion to be differentiated may seem. To differentiate partially w.r.t. x, all independent variables other than x are constant for the time being. To differentiate partially w.r.t. y, all independent variables other thanj are constant for the time being. So that, if z is a function oix and 7, i.e. if z = f(x,y), we can find bz_ dx bz by b 2 z dx 2 b 2 z ay 2 b 2 z b 2 z by.bx bx.by b 2 z _ b 2 z And also: by. bx bx. by Now for a few revision examples. Revision Exercise 1 . Find all first and second partial differential coefficients for each of 26 the following functions. (i) z = 3x 2 + 2xy + 4y 2 (ii) z = sin xy (111) z = - v ' x-y 2. Ifz = ln(e x +e^), show that ^ + ^-=1. v bx by 3 . If z = x.fixy), express x- — y — in its simplest form. When you have finished check with the solutions on frame 27. 263 Partial Differentiation 1 Results 1. (i) z = 3x 2 +2xy+4y 2 9 2 z „ 9 2 z _„ 27 9y.9x 9x.9y (ii) z = sin xy dz dz -r~ = ycosxy — =xcosxy 9x 3y 9 2 z , . 9 2 z , . ■^2 = -y 2 sin xy -^ = ~x 2 sin xy g— ^ = j>(--x sin xy ) + cos xy -^-^ = x(-y sin xy) + cos xy = cos xy — xy sin xy = cos xy - xy sin xy ...., _x+y (iu) z = - x-y 9z = (x-y)l-(x+y)l = -2y 9x (x-y) 2 (*~y) 2 9z _ (x-y)l-(x+y)(-l) 2x 9y (*~y) 2 (x-y) 2 -^=(-2v) ( " 2) - -&- 9x 2 <■ y) (x~yf (x-y) 3 ' By 2 ^{x-yf { U (x-y) 3 9 2 z ^ (x - y) 2 (-2) - (~2y)2?x - y) (-1) 3y.9x (x-y) 4 _ -2(x-y) 2 -4y(x-y) (x-yf = -2 4y {x-yf {x-yf _ -2x + 2y ~ 4y _ -2x - 2y (x-y) 3 (x-y) 3 /continued Programme 9 b 2 z _ (x -y) 2 (2)-2x.2 (x -y)l Continuation of frame 27. bx. by (x -y) A _ 2(x-y) 2 -4x(x-y) (x-y)* _ _2 4x (x-yf (x-y) 3 ^ 2x-2y-4x _ -2x - 2y (x-y) 3 (x-y) 3 2. z = ln(e* + e^) bz 1 bz 1 bx e x + &y " by e* + e^ 9z bz__ e x &y .*y bx by e* + e^ e* + eJ' . e* + e^ e^ + e-V 9z 9z _ 9.x: by 3. z = x/(xy) -|=*./'(x>0-* x~ ~y-^y =x2 yf'( x y) +x f( x y)- x2 yf'( x y) bz bz _ , N _ That was a pretty good revision test. Do not be unduly worried if you made a slip or two in your working. Try to avoid doing so, of course, but you are doing fine. Now on to the next part of the programme. Turn on to frame 28. 265 Partial Differentiation 1 So far we have been concerned with the technique of partial differenti- £ Q ation. Now let us look at one of its applications. Small increments If we return to the volume of the cylinder with which we started this programme, we have once again that V = nr 2 h. We have seen that we can 9V 9V find-^- with/i constant, and -57- with br an - nr r constant. or bh Now let us see what we get if r and h both change simultaneously. If r becomes r + Br, and h becomes h + oh, let V become V + 5 V. Then the new volume is given by V + SV = n(r + hrf(h + bh) = Ti(r 2 +2r.dr + 8r 2 )(h+8h) = Ti(r 2 h + 2rhdr + hdr 2 + r 2 8h + Irbrbh + br 2 .bh) Subtract V = nr 2 h from each side, giving bV=n(2rh.br+ h.br 2 + r 2 bh\ Irbrbh + br 2 .bh) —n(2rhbr + r 2 .bh) since r and h are small and all the remain- ing terms are of a higher degree of smallness. .". 5V === 2-nrhbr + Trr 2 bh bV -— br +— bh br bh Let us now do a numerical example to see how it all works out. On to frame 29. 2( Programme 9 29 30 Example. A cylinder has dimensions r = 5 cm, h = 10 cm. Find the approximate increase in volume when r increases by 0-2 cm and h decreases by 01 cm. Well now, V = -rrr 2 h — =2nrh — = nr or dh In this case, when r = 5 cm, /i = 10 cm, ~— = 2tt5.10= IOOtt ~ = Trr 2 =n5 2 = 25n dr dh br = 0-2 and 6/? =-01 (minus because h is decreasing) .. 5V- — .br + rT-5/j 5V=100rr(0-2) + 257r(-01) = 207T-2-57T= 17-57T .'. 5V^ 54-96 cm 3 i.e. the volume increases by 54-96 cubic centimetres. Just like that! This kind of result applies not only to the volume of a cylinder, but to any function of two independent variables. Example. If z is a function of x and y, i.e. z =f(x,y) and if x andy increase by small amounts 8x and by, the increase dz will also be relatively small. If we expand dz in powers of 8x and Sy, we get 5z = A5x + B by + higher powers of bx and 5j>, where A and B are functions of x andy. If y remains constant, so that by = 0, then bz = A5jc + higher powers of 6* .'. — - = A. So that if bx ->• 0, this becomes A = t— 5* ox Similarly, if x remains constant, making by -> gives B = — .". bz = — bx + — by + higher powers of very small y quantities which can be ignored. 5z - — 5x + — 5 v 3jc dy 261 Partial Differentiation 1 So, if z=f(x,y) . dz dz bz = -r- bx + -r- 5 V ox oy This is the key to all the forthcoming applications and will be quoted over and over again. The result is quite general and a similar result applies for a function of three independent variables e.g. If z=f(x,y,w) then bz = — bx + — by + — 5vv ox oy ow If we remember the rule for a function of two independent variables, we can easily extend it when necessary. Here it is once again: If z =f(x,y) then bz = — 5x + — by ox oy Copy this result into your record book in a prominent position, such as it deserves! 31 Now for an example or two. a a V j£ Example 1. If I =— and V = 250 volts and R'= 50 ohms, find the change in I resulting from an increase of 1 volt in V and an increase of 0-5 ohm in R. I=/(V,R) .". 6I=|^5V+|i5R 9V R Md 9R R 2 R 5V R 2 :. 5I = ^5V-— 2 §R So when R= 50, V = 250, 5V= 1, and 5R = 0-5, 6I = 5>-I> 5 ) = J-_ J_ 50 20 = 0-02 -0-05 =-0-03 i.e. I decreases by 0-03 amperes. 268 Programme 9 33 Here is another example. 3 Example 2. My = -73 , find the percentage increase in y , when w increases by 2 per cent, s decreases by 3 per cent and d increases by 1 per cent. Notice that, in this case, j is a function of three variables, u>, s and d. The formula therefore becomes: dw ds dd We have *2L = *L • & =— ■ & = -^ Weh3Ve dw 7' 95 d 4 ' dd d 5 •• &y=-p&w+—j4 8s+ - d5 8d „ Now then, what are the values of 5vv, 6x and 8d1 2—3 1 Is it true to say that 5w = Too ; 5x = T00' 5 ^l00 ? If not, why not? Next frame. Q/l No. It is not correct. For 8w is not — of a unit, but 2 per cent of w, ie - 6w -Too ofw -Too Similarly, 8s = — of s = -^ and 5d = y^- . Now that we have cleared that point up, we can continue with the problem. „ s 3 /2w\ , 3ws 2 (-3s\ 4ws 3 ( d \ by = d 4 \ioo) d 4 \l00> d s mo' _ ws 3 1 2 \ ws 3 t 9 \ _ ws? 1 4 \ ~ d 4 VlOO/ d 4 llOO/ d 4 \100/ .wsM_2 9___4_] d 4 1 100 100 100) ^{-jk} = " •11 per cent ofj> i.e. y decreases by 1 1 per cent Remember that where the increment of w is given as 2 per cent, it is not 2 2 jp— of a unit, but y^j of w, and the symbol w must be included. Turn on to frame 35. 269 Partial Differentiation 1 Now here is one for you to do. Exercise P = w 2 hd. If errors of up to 1% (plus or minus) are possible in the measured values of w, h and d, find the maximum possible percentage error in the calculated value of P. This is very much like the last example, so you will be able to deal with it without any trouble. Work it right through and then turn on to frame 36 and check your result. 35 ? = w 2 hd. :. 8? = ^-.8w + ^-.8h + ^.8d bw oh od dP „ , . 9P 2 , 3P j. ow oh od 5P = Iwhd.ow + w 2 d.8h + w*h.8d Now 5w = ±r ^, 6A = ± I g 5 , Sd = ±^ 5P=2wM( ± ^) + w^( ± ^) + w^( ± ^) _ + 2w 2 hd ( w 2 dh + w 2 hd ~ 100 " 100 " 100 The greatest possible error in P will occur when the signs are chosen so that they are all of the same kind, i.e. all plus or all minus. If they were mixed, they would tend to cancel each other out. 36 • SP = ± w 2 hd\— + — + — > = + p(— a \l00 100 100) M00 .'. Maximum possible error in P is 4% of P 00/ Finally, here is one last example for you to do. Work right through it and then check your results with those on frame 37. Exercise. The two sides forming the right -angle of a right-angled triangle are denoted by a and b. The hypotenuse is h. If there are possible errors of ± 0-5% in measuring a and b, find the maximum possible error in calculating (i) the area of the triangle and (ii) the length of h. 270 Programme 9 37 Results: (i) 5A=l%ofA (ii) hh =0-5% of h DDnnnnnDDDnnnnnnnnnnnannnDDnnnannnnnDD Here is the working in detail: r-\ a a - b <- . 9A . ^ 9A s 9A_^ 9A _a . _ _a_ s , _ + fe 9a~2 ; 96 2 ; M '200' ° ~ 200 (ii) ~ -2 |_200 200 ;. 5A= l%of A = ±A 100 h = ^(a 2 +b 2 )Ha 2 +b 2 ) -, dh dh hh =— 5a ■+ ^r ofc Also a = ± 200' ft = + 200 6 a J a \ & / , 6 \ •' 5/! "vV+fe 5 )'^ 20 ° ' V(« 2 + & 2 A" 200 ' a 2 +b 2 = + 200V(« 2+ ^ 2 ) = ^ ^ 2 + t 2 )-±^( h ) :. hh = 0-5% of h That brings us to the end of this particular programme. We shall meet partial differentiation again in a later programme when we shall consider some more of its applications. But for the time being, there remains only the Test Exercise on the next frame. Take your time over the questions; do them carefully. So on now to frame 38. Ill Partial Differentiation 1 Text Exercise - IX Q 8 Answer all questions. 1 . Find all first and second partial differential coefficients of the following: (i) z = 4x 3 - 5xy 2 + 3y 3 (ii) z = cos(2x + 3y) (iii) Z = e(* 2- J' 2 ) (iv) z=x 2 sin(2x + 3y) 2. (i) If V = x 1 + y 2 + z 2 , express in its simplest form av av av X— +V-T- +Zt- dx ' dy dz (ii) If z =/(x + a>») + F(x - aj), find-ri and Tl and hence P rove that -r-2 = a -Tl by 2 dx* E 2 3. The power P dissipated in a resistor is given by P = jt . If E = 200 volts and R = 8 ohms, find the change in P resulting from a drop of 5 volts in E and an increase of 0-2 ohm in R. _l 4. If = fcHLV 2 , where k is a constant, and there are possible errors of + 1% in measuring H, L and V, find the maximum possible error in the calculated value of 8. That's it. 272 Programme 9 Further Problems - IX L lfi %r^j. showthat^| +J -|=-2z(l +Z ). 3x 2 dy 2 2. Prove that, if V = ln(x 2 + j> 2 ), then-f-j +^4=0. 3. If z = sin(3;c + 2y), verify that 3 f-f - 2 f4 = 6z. dy 2 2 dx 2 4. lfu = -^lIlL_ showtiatx ^ + ^i +z ^ sQ ( X i +y i +z y * x & ** 5. Show that the equation — + r-f = 0, is satisfied by z = lnV(x 2 +y)+^tan- 1 (^) 6. If z = e x (x cosy ~y siny), show that^-| + r-§ = 0. a* 2 ay 7. If m = (1 + at) sinh (5* - 2y), verify that 4ff + 20r^ + 25|^ a* 2 a*.a.y ay 2 8. Ifz =/(-), show that a a 2 z ^ „ a 2 z ^ . a 2 z n 9. If z = (* + >")■/(), where/is an arbitrary function, show that 9z 3z X Tx+ y Ty =Z Eh 3 10. In the formula D = • _ 2 , h is given as 0-1 ± 0-002 and v as 0-3 ± 0-02. Express the approximate maximum error in D in terms ofE. a 2 1 1 . The formula z = ~~; ; ; is used to calculate z from observed x z + y* -a 1 values ofx and y. If jc'andjy have the same percentage error p, show that the percentage error in z is approximately ~2p(l + z). 273 Partial Differentiation 1 12. In a balanced bridge circuit, Rj = R2R3/R4 . If R 2 , R3, R4, have known tolerances of±x%,±y%,±z% respectively, determine the maximum percentage error in Rj , expressed in terms of x, y and z. 13. The deflection y at the centre of a circular plate suspended at the edge and uniformly loaded is given hy y = — -5— , where w = total load, d = diameter of plate, f = thickness and k is a constant. Calculate the approximate percentage change in y if w is increased by 3%, d is decreased by 2Vi% and t is increased by 4%. 14. The coefficient of rigidity (w) of a wire of length (L) and uniform AL diameter (d) is given by n = -r% , where A is a constant. If errors of ± 0-25% and ± 1% are possible in measuring L and d respectively, determine the maximum percentage error in the calculated value of n. 15. If k/k = (TlT o y.p/760, show that the change in k due to small changes of a% in T and b% in p is approximately (na + b)%. 1 6. The deflection y at the centre of a rod is known to be given by kwl 3 y ~~~74 , where k is a constant. If w increases by 2%, / by 3%, and d decreases by 2%, find the percentage increase in y. 17. The displacement^ of a point on a vibrating stretched string, at a distance x from one end, at time t, is given by 3f 2_c 'dx 2 T)X Show that one solution of this equation is y = A sin — .sm(pt ■*■ a), where A, p, c and a are constants. 18. If.y = A sin(px +«) cos(qt + b), find the error iny due to small errors 8x and 6f in x and t respectively. 19. Show that = Ae~ kt ' 2 sin pt cos que, satisfies the equation 3 2 1 f 3 2 ^ , 9<&1 . . , t . t 2 2 2 k 2 ^ = ^3^ +k a? " • P rovlded thatp =cV "T- ^ a2w a2\/ a2\/ 3 2 V 3 2 V 3 2 V bx 2 + 9^ 2 3z 2 20. Show that (i) the equation ^-5- + ^y + ^7 = is satisfied by 1 3 2 V 3 2 V V = 77^5 5";, and that (ii) the equation — ~T + "r~T = is v(* + y +z ) 9 * ^ satisfied by V = tanrV-^'V 274 Programme 10 PARTIAL DIFFERENTIATION PART 2 Programme 10 1 Partial differentiation In the first part of the programme on partial differentiation, we estab- lished a result which, we said, would be the foundation of most of the applications of partial differentiation to follow. You surely remember it: it went like this: If z is a function of two independent variables, x and y, i.e. if z=f(x,y), then E dz bz bz = — 8x + -r- by ox oy We were able to use it, just as it stands, to work out certain problems on small increments, errors and tolerances. It is also the key to much of the work of this programme, so copy it down into your record book, thus: If z = f(x,y), then 6z =-^8* + g- &y If z =f{x,y), then bz = -^bx+j by In this expression,— and— are the partial differential coefficients of z with respect to x and y respectively, and you will remember that to find bz (i) — , we differentiate the function z w.r.t. x, keeping all independent ox variables other than x, for the time being, bz (ii) — , we differentiate the function z w.r.t. y, keeping all independent by variables other than j>, for the time being, 277 Partial Differentiation 2 constant constant DDnDDDDDanDDnaDDnnnnnDDnDaanDnDDDnnnDa An example, just to remind you: If z = x 3 + 4x 2 y - 3y 3 (y is constant) bz then -£- = 3x 2 + 8xy - and —- = + Ax 2 - 9y 2 by (x is constant) In practice, of course, we do not write down the zero terms. Before we tackle any further applications, we must be expert at find- ing partial differential coefficients, so with the reminder above, have a go at this one: (1) Ifz = tan(x 2 -j> 2 ),find-^-and|^ ox ay When you have finished it, check with the next frame. tor and ^ = 2x sec 2 (;c 2 -y 2 ); j- = -2y sec 2 (x 2 - y 2 ) z = tan(x 2 -y 2 ) ^. = sec 2 (x 2 -y 2 )X-^(x 2 -y 2 ) - sec 2 (x 2 -y 2 ) (2x) = 2x sec 2 (x 2 - y 2 ) h-=sec 2 (x 2 -y 2 )X-^(x 2 -y 2 ) = sec 2 (x 2 -j 2 ) (-2)0 = ~2y sec 2 (x 2 -y 2 ) That was easy enough. Now do this one: d 2 z <Pz b 2 z bx 2 ' by 2 Finish them all. Then turn on to frame 5 and check your results. (2) Ifz = e 2 *~ 3 >', find |-|, . 2 , a a <kr d)^ 3x3y 278 Programme 10 Here are the results in detail: bz z = e 2X — 3}> ■ ¥± - Q 2X -3y 9 = 7 e 2 * ~ 3y bx — bz_ by = e 2X - 3J » ( _ 3) = - 3e 2X-3y ^=2.e 2 ^- 3 >'.2 = 4.e«- 3 >' by b 2 z bx.by ^ 2 =-3.e 2 *-W(-3) = 9.e 2 *-^ = -3.e 2X ~ 3y .2=-6.e 2X - 3y All correct? You remember, too, that in the 'mixed' second partial diff. coefft., the order of differentiating does not matter. So in this case, since d 2 z s 2X-3V .i. d 2 z r— — = ~6.e 2 * iy , then — — = dx. dy by. dx b 2 z b 2 z -6.e 2x ~ 3y bx.by by.bx DnnnnDnDnDnDnDnnDDnnnnnnnnnnnnnaDnnann Well now, before we move on to new work, see what you make of these. Find all the first and second partial differential coefficients of the following: (i) z = x sin y (ii) z = (x+y)\n(xy) When you have found all the diff. coefficients, check your work with the solutions in the next frame. 279 Partial Differentiation 2 Here they are. Check your results carefully. (i) z = x sin y . bz bz ~r- = sin y — = x cos y - bx ""' 3y 9iz a 2 z b 2 z 2-0 r-2=-A:sin>' = cosj - — — =COSJ 3j. bx bx. by (ii) z=(jc+y)ln(^) - 3* v " + y) — .x+ln(j xy ~ {x 2 + y) + l .y vj,) x 1 Ul^J 9z , V) = ^j^ + Wxy) .b 2 Z X "bx 2 ■K--X- y x 1 X -4 X y xy x-y X 2 b 2 z _y dy 2 ~ jf-x-r 1 2 ' _.y-* b 2 z 1 by. bx x , 1 11 xy x y r~ _y +x xy b 2 z 1 3x by y xy ' y x _x +y xy I 280 Programme 1 8 Well now, that was just by way of warming up with work you have done before. Let us now move on to the next section of this programme. Rates-of-change problems Let us consider a cylinder of radius r and height h as before. Then the volume is given by V = nr 2 h av 9V ■r— = 2trrh and — = Tir dr dh Since V is a function of r and h, we also know that ov --r— .Or + —.0H (Here it is, popping up again!) ,,,,... c 5V 3V fir 3V Sh Now divide both sides by of: — - — .^j+ g^ §7 Then iffi,^o,-|?- %,% ■* I'l? "*§ > but the P artial di f f erential of af of at ot at coefficients, which do not contain Sf, will remain unchanged. dV So our result now becomes — — = dt dV = 3V.fr dVdh dt dr dt dh'dt This result is really the key to problems of the kind we are about to consider. If we know the rate at which r and h are changing, we can now find the corresponding rate of change of V. Like this: Example 1. The radius of a cylinder increases at the rate of 0-2 cm/sec while the height decreases at the rate of 0-5 cm/sec. Find the rate at which the volume is changing at the instant when r - 8 cm and h = 12 cm. WARNING: The first inclination is to draw a diagram and to put in the given values for its dimensions, i.e. r = 8 cm, h = 12 cm. This we must NOT do, for the radius and height are changing and the given values are instan- taneous values only. Therefore on the diagram we keep the symbols r and h to indicate that they are variables. 281 Partial Differentiation 2 Here it is then: V = nr 2 h 6V=— dr + — 5h br bh ■ ^Y = 5Y dL+W dh " dt br 'dt bh dt 10 — = 2i\rh\ — =ttr l br bh dV. dt . , dr ___ 2 dh dt dt Now at the instant we are considering r = 8, h = 12, — =0-2, — = -0-5 (minus since h is decreasing) So you can now substitute these values in the last statement and finish off the calculation, giving dV = dt 4^=20-1 cm 3 /sec dt for dV. dt dt dt : 27r8.12.(0-2) + 7r64(-0-5) ■■ 38-4-n - 32ir ■64tt= 20-lcm 3 /sec. Now another one. Example 2. In the right-angled triangle shown, x is increasing at 2 cm/ sec while y is decreasing at 3 cm/sec. Calculate the rate at which z is changing when x = 5 cm and v = 3 cm. The first thing to do, of course, is to express z in terms of x and _y. That is not difficult. z = 11 282 Programme 10 12 Z = V(* 2 -/) DnDDnnnnnDDnnDDnnannaannnDDDnDannDannn z : ^{x i -y*) = (x 1 -y 2 $ In this case . . dz _ , 9z t " dr 3x"df 3/df £-W>"Wvc?^) (The key to the whole business) f4(^-^W) = -J dz dx V(* 2 -J' 2 ) J' dy dr"V(x 2 -^ 2 )'^ j(x 2 -y 2 ydt So far so good. Now for the numerical values dx =7 dy__. dt ' * x = 5, y = 3, ^-" = 2, -^ = -3 dz. dr' Finish it off, then move to frame 13. 13 for we have — =4-75 cm/sec at dz >(2) ^(-3) dr V(5 2 -3 2 ) w V(5 2 -3 2 )' 5(2) 3(3) 10^9 19_._, , = _i_i-+ _i_Z = — + -=—= 4-75 cm/sec 4 4 4 4 4 .'. Side z increases at the rate of 4-75 cm/sec Now here is Example 3. The total surface area Sofa cone of base radius r and per- pendicular height h is given by S = Trr 2 +irrs/(r 2 + h 2 ) If r and h are each increasing at the rate of 0-25 cm/ sec, find the rate at which S is increasing at the instant when r = 3 cm and h = 4 cm. Do that one entirely on your own. Take your time : there is no need to hurry. Be quite sure that each step you write down is correct. Then turn to frame 14 and check your result. 283 Partial Differentiation 2 Solution. Here it is in detail. S = nr 2 + nr^r 2 + h 2 ) = nr 2 + -nrir 2 + h 2 $ ^JS,, JS-, . dS bSdrbSdh OS = -r-.br +—.8h .. — = ^-.tt+^T-X dr dh dt dr dt dh dt (i) |^ = litr + Ttr.^r 2 + h 2 )^(2r) + ir(r 2 + h 2 f When r = 3 and h = 4, dr 5 5 5 /••\ <*S_ 1, , ,2\ll/ii\ urn (u)-^= rrr^ 2 + fc 2 ) 2 (2fc) " y / ( ^J ?) = *3A = 127T 5 5 Also we are given that-— = 0-25 and— = 0-25 a? dt . dS_64Tt _1 12tt 1 " dr 5 '4 5 '4 _ 167T + 37[ = 197T 5 5 5 3-87T= ll-93cm 2 /sec 14 So there we are. Rates-of-change problems are all very much the same. What you must remember is simply this: j C (i) The basic statement If z=f(x,y) then bz = -r-.bx +-^-.by (i) (ii) Divide this result by 8t and make 8t -*■ 0. This converts the result into the form for rates-of-change problems: dz bz dx bz dy ,..^ — = — — + — — *■ (ii) dt bxdt bydt v ; The second result follows directly from the first. Make a note of both of these in your record book for future reference. Then for the next part of the work, turn on to frame 1 6. 284 Programme 10 16 Partial differentiation can also be used with advantage in finding differential coefficients of implicit functions. For example, suppose we are required to find an expression for-j- when we are given that x 2 + 2xy + y 3 = 0. We can set about it in this way: Let z stand for the function of x and>>, i.e. z = x 2 + 2xy +y 3 . Again we use the basic relationship 5z = — 8x +— 8y. If we divide both sides by 8x, we get 8z _ dz dz 5y 8x dx dy' 8x xt f t .« dz dz , dz dy Now, ifSx-*0, -3-= _+_-.-£. ox dx by dx If we now find expressions for — and r- , we shall be quite a way towards finding-^- (which you see at the end of the expression). In this particular example, — = and — = 17 z = x 2 + 2xy + y 3 f x -2x + 2y;f y -7x + 3y 2 Substituting these in our previous result gives us f-(2x + *0 + (2* + 3,')g dz If only we knew — , we could rearrange this result and obtain an expres- sion for-r ■ So where can we find out something about -3- ? dx dx Refer back to the beginning of the problem. We have used z to stand for x 2 + 2xy + y 3 and we were told initially that x 2 + 2xy + y 3 =0. dz Therefore z = 0, i.e. z is a constant (in this case zero) and hence— = 0. :. = (2x + 2y) + (2x + 3y 2 )? y From this we can find-f-. So finish it off. dx dx On to frame 18. dx 285 Partial Differentiation 2 dy__2x + 2y 1 dx 2x + 3y 2 j 18 DnnnDnDDnDnnaaDDDDDDnDnDnDnDnnDDnannDD This is almost a routine that always works. In general, we have — If /(x,;0 = 0, find-^ Let z =f(x,y) then 8z = r^ Sx + y dy. Divide by 8x and make 5x -> 0, in which case dz_dzdz dy dx dx dy'dx But z = (constant) /. f = :. = ^ + ^ ^ A 9.x dy'dx dy dz ,dz The easiest form to remember is the one that comes direct from the basic result _ _dz dz oz -— ox +— 8y dx dy Divide by 8x, etc. dz _ dz dz dy j dz _ ) dx dx dy'dx \ dx j Make a note of this result. Now for one or two examples. Example 1. If e*y +x +y = 1, evaluate-^ at (0,0). The function can be |9 written e^y +x + y— 1 =0. Letz = e^+x+j-l 5z = ^.8x + ^.8y :. <** = ** +** dy dx dy ' dx dx dy'dx Bu tz = .-.-^=0 :. ^ = -(Z£^±i dx dx lx.e*.y + 1 Atx = 0,j; = 0,-^ = -4 = -l :.^ = -l dx 1 dx All very easy so long as you can find partial differential coefficients correctly. On to frame 20. 286 Programme 10 20 d± Nowhere is: Example 2. If xy + sin y = 2, find ^ Let z = xy + sin j> - 2 = to dy But z = Here is one for you to do: dz _dz dz dy dx dx dy'dx dz dz — =y; — = x + cosy dx dy f =0 dx dz , ^dy :. =f = y + (x + cos y)-f dx * v dx . dy _ -y dx x + cos y dy. Example 3. Find an expression for -jf- when x tan y = y sin x. Do it all on your own. Then check your working with that in frame 21. 21 dy tan y - y cos x dx x sec^-sinx Did you get that? If so, go straight on to frame 22. If not, here is the working below. Follow it through and see where you have gone astray! x tan y = y sin x .'. x tan y - y sin x = Let z =x tany-y sin* = dz c , dz r 9a: dy bz_ dx dz = dz dz_ dy dx dx dy'dx dz_ dy ~- = tan y - y cos x ; -^7; = x sec 2 y - sin x But z =0 ;. -^ = (tan y - y cos x) + (x sec 2 y - sin x) -^ dy __ tany-y cosx dx: x sec 2 j> - sin x On now to frame 22. 287 Partial Differentiation 2 Right. Now here is just one more for you to do. They are really very much the same. Example 4. If e* + y = x 2 y 2 , find an expression for-r- e x+ y-x 7 y i = 0. Letz = e x+ y-x 2 y 2 = . dz - , Bz 5z= — 5x +-r-5^ 9x 9.y dz _ dz 9z <iy etc 9.x 9j'dx So continue with the good work and finish it off, finally getting that dy_ 22 dx Then move to frame 23. dy_ 2xy 2 -e x+ y dx e x+y-2x 2 y For = e x+ y-x 2 v 2 = z = e |*=e* + ?-2xy 2 ; ^=e x+ y~2x 2 y dx ' dy ■ dz ^ = ( e x + y- 2x yi) + (e x + y-2x 2 y)%- dy dx dx But z = dz_. dx 23 . dy. (e* + y - 2xy 2 ) dx (e* + y - 2x 2 y) : dy._ dx . 2xy 2 " e x+y - e x+y -2x 2 y That is how they are all done. Now on to frame 24. 288 Programme 10 24 There is one more process that you must know how to tackle. Change of variables If z is a function of x and>>, i.e. z = f(x,y), and* andy are themselves functions of two other variables u and v, then z is also a function of u and v. We may therefore need to find— and — . How do we go about it? bu bv *=/( x,y) :.8z = f x 8x + f y 8y Divide both sides by 8u hz _ dz 8x bz by 8u bx' 8u by' 8u If v is kept constant for and — becomes ~. 8u bu and Next frame. the time being, then-r - when 8u -> becomes — 8u du . bz _ bz bx bz by bu bx' bu by' bu bz _bz bx bz by bv bx'bv by' bv rNuie Liicbc 25 Here is an example on this work. If z = x 2 +y 2 , where x = r cos 9 andy =r sin 20, find -r- and-r^ bz _ bz bx bz by br bx' br by' br and Now, bz _bz bx bz by bd~bx'bO by'bd bz „ bz „ •r- = 2x rr- = 2y bx by !^=cos9 |* = sin 29 br br bz — = 2x cos 9 + 2v sin 29 And — =-r sin and ~ = 2r cos 29 ■. %; = 2x(-r sin 0) + 2^(2/- cos 20) 00 — = 4 ^r cos 20 - 2 XT- sin And in these two results, the symbols x andy can be replaced by r cos and r sin 20 respectively. 289 Partial Differentiation 2 One more example. Ifz = e x y where x =ln(« +v)and.y = sin(u-v), find -r^ and— . Wehave £ *|* + |E.|V = y.e*^. +x.e x >'.cos(M- v) u+v - e xy) y +x.cos(u-v) (u + v v 26 and bz _ bz bx bz by bv bx'bv by'bv = v.e^. + x.e x y\-cos(u-v)\ u+v { j = e*.W— — xcos(u — v)\ [u+v j Now move on to frame 27. Here is one for you to do on your own. All that it entails is to find the various partial differential coefficients and to substitute them in the established results. dz _ dz bx_ bz_ by bu dx' bu by' bu bz _bz bx bz by bv bx' bv by' bv So you do this one: Ifz = sinQc +y), where x = u 1 + v 2 andy = 2uv, find— and — The method is the same as before. When you have completed the work, check with the results in frame 28. 27 290 Programme 10 28 z = sin(jc + y); x=u 2 +v 2 ; y = 2uv — = cos(x +y) ; -r- = cos(x + y) dx by -r- = 2u -^ = 2v du ow 9z _ 9z 9x 9z 9y 8« 9a;' 3m 9,y' du = cos(x +>>).2m + cos(x +y).2v = 2(» + v) cos(x + y) Also a £ = 9 i^ + 9£^ 9v dx' dv dy' 9v — = 2v ; -r L = 2u 9v 9v 9z — = cos(x +y).2v + cos(x +y).2u av = 2(m + v) cos (x + >Q You have now reached the end of this programme and know quite a bit about partial differentiation. We have established some important results during the work, so let us list them once more. 29 1 . Small increments z=f(x,y) 5z = |£ 6jc+ j£ 5> , (i) 2. Rates of change dz _bz_ dx fa d£ ,..> dt dx'dt dy'dt W 3 . Implicit functions — - — + — Q (—\ dx dx dy'dx ^ ' 4 . Change of variables dz_ _ 9z dx 9z dy du dx' du dy'du dz _dz dx 9z dy dv dx' dv dy' dv All that now remains is the Test Exercise, so turn on to frame 30 and work through it carefully at your own speed. The questions are just like those you have been doing quite successfully. 291 Partial Differentiation 2 Test Exercise - X Answer all the questions. Take your time over them and work care- fully. dy 1 . Use partial differentiation to determine expressions for — in the following cases: (i) x 3 +y 3 ~2x 2 y = (ii) e* cos y = tV sin x (iii) sin 2 jc - 5 sin x cos y + tan y = 2. The base radius of a cone, r, is decreasing at the rate of 0- 1 cm/sec while the perpendicular height, h, is increasing at the rate of 0-2 cm/sec. Find the rate at which the volume, V, is changing when r = 2 cm and h = 3 cm. 3. If z = Ixy - 2x 2 y and x is increasing at 2 cm/sec determine at what rate y must be changing in order that z shall be neither increasing nor decreasing at the instant when x = 3cm and y=\ cm. 9z 9z 4. If z = x 4 + 2x 2 y + y 3 and x = r cos 6 and y = r sin 9 , find — and r-: in their simplest forms. 30 292 Programme 10 Further Problems - X 1 . If F = f(x,y) where x = e" cos v and y = e" sin v, show that 3F 3F 3F , 3F 3F 3F -r- -x-r- + y-r- and t-=-.} ; -t~" +x ~Z~ ou dx by bv bx by 2. Given that z = x 3 + y 3 and x 2 +y 2 = 1 , determine an expression for — - in terms of x and y. dx 3. Ifz=/(^jO = 0,showthat-^=-!r/^r. The curves 2.y 2 + 3x - 8 = v " dx dx' dy J and x 3 + 2xy 3 +3y— 1 =0 intersect at the point (2, -1). Find the tangent of the angle between the tangents to the curves at this point. prove that 2 J t=(x 2 -y 2 ){tf"(t) + 3f'(t)} 4. If u = (x 2 —y 2 )f(t) where t = xy and/ denotes an arbitrary function, bx.by 5. If V = xyj(x 2 +y 2 ) 2 andx = rcos0,j> =/• sinfl, show that 3 2 V I9V 1 9?V_ n dr 2 r br V 30 2 ~° 6. If u ~f(x,y) where x = r 2 - s 2 andy = 2rs, prove that 7. If/= F(x,j>) and* =re e and j> = re e , prove that „ 3/ 3/3/ ., „ 3/ 3/ 3/ 2* IT =, "-r + ^ and 2^-a ^a ™ 3.x 3r 30 3y 3r 30 8. If z = x ln(x 2 + j» 2 ) - 2y tan" 1 (— ) verify that 9z 3z x— +y—=z + 2x dx dy 9. By means of partial differentiation, determine -f- in each of the following cases. (i) xy + 2y-x = 4 ( "\*L _?£_"3 (ii) x 3 y 2 - 2x 2 y + 3xy 2 - &xy = 5 ^ x + y 293 Partial Differentiation 2 10. If z = 3xy -y 2, + (y 2 ~ Txfl 2 , verify that (i) 9 2 z b 2 z , ^ . n;\ &L&L-( 32 2 \ 2 .... d'z b 2 z _ ( b 2 z \* , and that (11) -^ . ^ = [ ^y j bx. by by. bx' bx 2 'by 2 ^ dx. by. H- If /= //7— =T^ j.,showthat>'-/ = (j:-3')-f 12. If z = x.f {^-\+ F(A prove that ,.s 9z 9z c/^X /■••> 2 9 2 z .o 9 2 z L 2 9 2z _ n 13. If z = e k ( r ~ x \ where k is a constant, andr 2 = x 2 +y 2 , prove ,.. /dzf /9z\ 2 _,_,, , 9z _ n .... b 2 z b 2 z _, bz _kz 14. If z = /(* - 2y) + F(3x + j), where/and F are arbitrary functions, ~2 ^2 ti2 and if-r-2 + a t — r + b -r—z = 0, find the values of a and b. bx bx.by by 2 15. Ifz =xyl(x 2 + y 2 ) 2 , verify that0 +0 = 0. 16. If sin 2 *- 5 sin* cos.y + tanj> = 0, find-pby using partial differentiation. 17. Find -~ by partial differentiation, when x tan y = y sin x. 18. IfV = tan _1 -^^L prove that ,., 9V 9V „ ,... 9 2 V a 2 v n (i) x—+y— =0, (n) r-T + n" = w 9x by bx 1 by 2 19. Prove that, if z = 2xy + *•/(-) then 20. (i) Find— given that x 2 }' + sin xy = dv (ii) Find-^ given that x sin xy = 1 294 Programme 11 SERIES PART1 Programme 1 1 1 Series A series, Wj ,w 2 ,« 3 ... is a sequence of terms each of which is formed according to some definite pattern. e.g. 1,3,5,7,... is a series (the next term would be 9) 2, 6, 18, 54, . . . is a series (the next term would be 3 X 54, i.e. 162) 1 2 , - 2 2 , 3 2 , -4 2 , . . . is a series (the next term would be 5 2 ) but 1,-5, 37, 6, . . . is not a series since the terms are not formed to a regular pattern and one cannot assess the next term. A finite series contains only a finite number of terms. An infinite series is unending. So which of the following constitutes a finite series: (i) All the natural numbers, i.e. 1, 2, 3, . . . etc. (ii) The page numbers of a book, (iii) The telephone numbers in a telephone directory. The page numbers of a book Correct. Since they are in regular sequence and terminate at the last page. (The natural numbers form an infinite series, since they never come to an end: the telephone numbers are finite in number, but do not form a regular sequence, so they do not form a series at all.) nnnDnDnnqnnnDnnnnnnnnnDanDDDnnnnDnnDan We shall indicate the terms of a series as follows: «i will represent the first term, u 2 the second term, « 3 the third term, etc. so that u r will represent the r th term, and u r + i the (r + l) th term, etc. Also the sum of the first 5 terms will be indicated by S 5 . So the sum of the first n terms will be stated as 297 Series 1 naDDnnnDnnnnnDDnnnannnnnnnDnnnnnnnnaan You will already be familiar with two special kinds of series which have many applications. These are (i) arithmetic series and (ii) geometric series. Just by way of revision, however, we will first review the important results relating to these two series. 1 . Arithmetic series (or arithmetic progression) denoted by A.P. An example of an A.P. is the series 2,5,8,11,14, You will note that each term can be written from the previous term by simply adding on a constant value 3. This regular increment is called the common difference and is found by selecting any term and subtract- ing from it the previous term e.g. 11-8 = 3; 5-2 = 3; etc. Move on to the next frame. The general arithmetic series can therefore be written: a, a+d, a + 2d, a + 3d, a + Ad, (i) where a = first term and d = common difference. You will remember that (i) the « th term = a+{n-\)d (ii) (ii) the sum of the first n terms is given by S„=|(2a+ I ^ r Td) (iii) Make a note of these three items in your record book. By way of warming up, find the sum of the first 20 terms of the series : 10,6,2,-2,-6, . . . etc. Then turn to frame 5. 298 Programme 11 S,o=-560 Since, for the series 10, 6, 2, -2, -6, . . . etc. a = 10 and d = 2-6 = -4 S„ =|(2fl+'n-l'd) :. S a 20 (20 + 19 [-4]) = 10(20 - 76) = 10(-56) = - 560 OQaaDOoaaaDnoQaaaaaaoaaaoDaoaQDaaaanoa Here is another example: If the 7 th term of an A.P. is 22 and the 12 th term is 37, find the series. We know 7th term = 22 .\ a + 6d = 22 ' , 5<2=15 :. d = 3 and 12th term = 37 ;. a +lltf=37j . fl = 4 So the series is 4, 7, 10, 13, 16, ... etc. Here is one for you to do: The 6 th term of an A.P. is -5 and the 10*h term is -21 . Find the sum of the first 30 terms. since: S 3O =-1290 6 th term = -5 .'. a + 5d = -5 10 th term = -21 .\ a + 9d = -21 4c? = -16 :. d- 15 a= 15, 6? = -4, « = 30, S n =—(2a+n~ Id) 30 .-. S 3O = y(30 + 29[-4]) = 15(30- 116)= 15(-86) = -1290 Arithmetic mean We are sometimes required to find the arith. mean of two numbers, P and Q. This means.that we have to insert a number A between P and Q, so that P, A and Q form an A.P. A-? = d and Q-A = g? P + Q :. A-P = Q-A 2A = P + Q .. A = — -^ The arithmetic mean of two numbers, then, is simply their average. There- fore, the arithmetic mean of 23 and 58 is 299 Series 1 The arithmetic mean of 23 and 58 is 40-5 If we are required to insert 3 arithmetic means between two given numbers, P and Q, it means that we have to supply three numbers, A, B, C between P and Q, so that P, A, B, C, Q form an A.P. Example. Insert 3 arithmetic means between 8 and 18. Let the means be denoted by A, B, C. Then 8, A, B, C, 18 form an A.P. First term, a = 8. fifth term = a + \d = 18 4d=10 :. d = 2-5 Required arith. means are 10-5, 13, 15-5 Now, you find five arithmetic means between 1 2 and 21 -6. Then turn to frame 8. <z = 8 a + 4d = 1 A = 8 + 2-5 = 10-5 B = 8 + 5 = 13 C = 8 + 7-5 = 15-5 Required arith. means: |13-6, 15-2, 16-8, 18-4, 20 Here is the working: Let the 5 arith. means be A, B, C, D, E. Then 12, A, B, C, D, E, 21-6 form an A.P. .'. a= 12; fl + 6d = 21-6 :. 6d = 9-6 :. d = 1 -6 Then A = 12 + 1-6 = 13-6 A =13-6 B = 12 + 3-2= 15-2 B = 15-2 C = 12 + 4-8= 16-8 C = 16-8 0=12 + 6-4=18-4 D=18-4 E = 12 + 8-0 = 20-0 E=20. So that is that! Once you have done one, the others are just like it. Now we will see how much you remember about Geometric Series. So, on to frame 9. 8 300 Programme 11 2. Geometric series (Geometric progression) denoted by G.P. An example of a G.P. is the series: 1,3,9,27,81, ... etc. Here you see that any term can be written from the previous term by multiplying it by a constant factor 3. This constant factor is called the common ratio and is found by selecting any term and dividing it by the previous one. e.g. 27^-9 = 3; 9-^3 = 3; etc. A G.P. therefore has the form: a, ar, ar 2 , ar 3 , ar 4 , etc. where a = first term, r = common ratio. So in the geometric series 5, -10, 20, -40, etc. the common ratio, r, is 10 20 -10 = -2 The general geometric series is therefore : a, ar, ar 2 , ar 3 , ar 4 , etc. (iv) (v) and you will remember that (i) the « th term =ar n ' x (ii) the sum of the first n terms is given by c a(l-r») ,~ S„= \-/ ( V1 > Make a note of these items in your record book. So, now you can do this one: For the series 8, 4, 2, 1 , \ , ... etc., find the sum of the first 8 terms. Then on to frame 11. 301 Series 1 15 Since, for the series 8,4,2,1, ... etc. " 8 ' r 4"2* Sn 1^7" 11 ••■ s s _ 8(1 256 } _16.255_255_ if 1-i 256 16 ^— Now here is another example. If the 5th term of a G.P. is 162 and the 8 th term is 4374, find the series. We have 5 th term = 162 .". a.r"' = 162 8 th term = 4374 .'. a.r 1 = 4374 ar n 4374 ar" 162 :. r* = 27 :. r = 3 a = 2 12 for a/- 4 = 162; ar 7 = 4374 and r = 3 162 .-. a.3" = 162 :. a = -rrr :. a = 2 81 .'. The series is: 2, 6, 18, 54, . . . etc. Of course, now that we know the values of a and r, we could calculate the value of any term or the sum of a given number of terms. For this same series, find (i) the 10th term (ii) the sum of the first 10 terms. When you have finished, turn to frame 13. 302 Programme 11 13 a = 2; r = 3 (i) 10th term = ar 9 =2.3 9 = 2(19683) = 39366 ,.., „ fl (l-r 10 ). 2(1- (u) Sio \_ r j -3 10 ) -3 _ 2(1- 59049) _ -2 59048 Geometric mean The geometric mean of two given numbers P and Q is a number A such that P, A and Q form a G.P. A . Q_ p = 'and--r :.£=? :. A 2 = PQ A = V(PQ) P A So the geometric mean of 2 numbers is the square root of their product. Therefore, the geom. mean of 4 and 25 is 14 A = V(4X25) = VlOO = 10 nnnnDDDnnnannnnnnnDnnnDannnnnnDnnDnnnn To insert 3 G.M's between two given numbers, P and Q means to insert 3 numbers, A, B, C, such that P, A, B, C, Q form a G.P. Example. Insert 4 geometric means between 5 and 1215. Let the means be A, B, C, D. Then 5, A, B, C, D, 1215 form a G.P. i.e. a = 5 and ar s = 1215 • r s = 1211 = 243 :. r = 3 The required geometric means are: 15,45,135,405 /. A= 5.3 = 15 B = 5.9 = 45 C =5.27= 135 D = 5. 81 =405 Now here is one for you to do: Insert two geometric means between 5 and 8-64. Then on to frame 15. 303 Series 1 Required geometric means are 6-0, 7-2 15 For, let the means be A and B. Then 5, A, B, 8-64 form a G.P. .'. a = 5; :. ar 3 = 8-64; .\ r 3 = 1 -728; .'. r = 1 -2 A = 5.1-2 = 6 1 Required means are B= 5. 1-44 = 7-20 6-0 and 7-2 Arithmetic and geometric series are, of course, special kinds of series. There are other special series that are worth knowing. These consist of the series of the powers of the natural numbers. So let us look at these in the next frame. 16 Series of powers of the natural numbers n 1 . The series 1+2 + 3+4 + 5 + .. .+n etc. = 2/-. i This series, you will see, is an example of an A.P., where a = 1 and d= 1. The sum of the first n terms is given by: £/• =1+2 + 3+4 + 5 + . ..+« i =-(2a + «- 1 a) = " _ w(w + l) i 2 So, the sum of the first 100 natural numbers is Then on to frame 1 7. 304 Programme 11 17 100 2/- = 5050 i for , = iopiion =50(101)=5050 DnnaDDDnnaDDDDDDnnDnnDDnDDnnaDDDDDDDnD 2. That was easy enough. Now let us look at this one: To establish the result for the sum of n terms of the series 1 2 + 2 2 + 3 2 + 4 2 + 5 2 + . . . +n 2 ., we make use of the identity (w + l) 3 = n 3 + 3n 2 + 3n + 1 We write this as 3 3-32^1^1 (« + 1) -n = 3« + 3« + 1 Replacing n by « - 1 , we get n 3 -(n- l) 3 = 3(n - l) 2 + 3(« - 1) + 1 and again (« - l) 3 - (h - 2) 3 = 3(n - 2) 2 + 3(« - 2) + 1 and (« - 2) 3 -(« - 3) 3 = 3(n - 3) 2 + 3(n - 3) + 1 Continuing like this, we should eventually arrive at: 3 3 -2 3 = 3.2 2 + 3.2+1 2 3 -l 3 = 3.1 2 + 3.1 + 1 If we now add all these results together, we find on the left-hand side that all the terms disappear except the first and the last. (n + l) 3 - l 3 = 3{« 2 + (n - l) 2 + (#1 - 2) 2 + . . . + 2 2 + l 2 ) + 3{n+(n-l) + («-2) + ... + 2 + lJ +«(1) n n = 3.S> 2 + 3Sr + « 1 1 :. n 3 + 3n 2 + 3n + y- ¥= 32r 2 + 3Z,r + n = 3£r 2 + 3 "*■" * ^ + « 11 1 2 :. n 3 + 3« 2 + 2« = 32r 2 +y(« 2 + «) .-. 2« 3 +6« 2 +4n=6Sr 2 +3n 2 + 3n 1 n 6Zr 2 =2n 3 + 3n 2 +m 1 . £ ra _ fi(/i + l)(2n + l ) 1 6 So, the sum of the first 12 terms of the series l 2 + 2 2 +3 2 + . . . is 305 Series 1 g r a. "(" + l)(2» + l ) 18 1 " 6 »,, 12(13] 1(25) = 26(25); 650 i 6 3. The sum of the cubes of the natural numbers is found in much the same way. This time, we use the identity (n + 1)* = n 4 + 4n 3 + 6n 2 + An + 1 We rewrite it as before (n + l) 4 -« 4 =4« 3 4 6n 2 + 4n + l If we now do the same trick as before and replace n by (n - 1) over and over again, and finally total up the results we get the result Note in passing that S n [n \\ Let us collect together these last three results. Here they are: 1. Sr . (vu) i I „ " , n(n + l)(2n + l) , .... 2. 2r 2 =— i4 (vm) 1 o 3.|,3={i!^l)} 2 (ix) These are handy results, so copy them into your record book. Now turn on to frame 20 and we can see an example of the use of these results. 19 306 Programme 11 £(1 Example: Find the sum of the series 2 n(3 + 2n) n = 1 s s = 2 n(3 + 2w) = 2 (3w l i = 2 3« + 2 2n 2 i i = 32n + 22« 2 i i _ 3.5.6 2. 5.6.11 ~T~ 6 = 45+ 110 = 155 + 2« 2 ) It is just a question of using the established results. Here is one for you to do in the same manner. 4 Find the sum of the series 2 (2« + « 3 ) n = \ 21 S 4 =2(2n+« 3 ) i 4 4 = 22« + 2n 3 i i _ 2.4.5 (±5) 2 2 [ 2 = 20+ 100 120 Remember Sum of first n natural numbers = -!— — — ' 2 Sum of squares of first n natural numbers = Z^H. '^ n ' Sum of cubes of first n natural numbers = _ j n(n + iy 2 307 Series 1 Infinite series So far, we have been concerned with a finite number of terms of a given series. When we are dealing with the sum of an infinite number of terms of a series, we must be careful about the steps we take. Example: Consider the infinite series 1 + 5 + 3 + g + . . . This we recognize as a G.P. in which a = 1 and r = \. The sum of the first n terms is therefore given by Now if n is very large, 2" will be very large and therefore — will be 1 very small. In fact, as n -+ °°, ► 0. The sum of all the terms in this 2" infinite series is therefore given by Soo = the limiting value of S„ as n -»•«>. i.e. Soo = Lt {S„} = 2(1 - 0) = 2 This result means that we can make the sum of the series as near to the value 2 as we please by taking a sufficiently large number of terms. Next frame. 22 This is not always possible with an infinite series, for in the case of an o*l A.P. things are very different. ^3 Consider the infinite series 1+3+5 + 7 + .. . This is an A.P. in which a = 1 and d = 2. Then S„ = j(2a +n - \.d) = j(2 +n - 1.2) = |(2 + 2«-2) S„ >n Of course, in this case, if n is large then the value of S„ is very large. In fact, if n -*■ °° , then S„ -> °° , which is not a definite numerical value and of little use to us. This always happens with an A.P.: if we try to find the "sum to infinity", we invariably obtain + °° or - °° as the result, depending on the actual series. Turn on now to frame 24. 308 Programme 11 £j\ In the previous two frames, we made two important points. (i) We cannot evaluate the sum of an infinite number of terms of an A.P. because the result is always infinite. (ii) We can sometimes evaluate the sum of an infinite number of terms of a G.P. since, for such a series, S„ = ^ and provided |r| <1 , then n->» r «^0.In that case S«, = - \_ r = jTTp ie - s °° ~ i^r So, find the 'sum to infinity' of the series 20 + 4 + 0-8 + 016 + 0-032 + as 25 5^=25 For 20 + 4 + 0-8 + 0-16 + 0-032 + .. . n 8 i a = 20; /•=-^ = 0-2=-r 4 5 5 QDDODDDOODDDDDDDDDDDPDDDDDaODDDDDDDDOD Limiting values In this programme, we have already seen that we have sometimes to determine the limiting value of S„ as n -» °°. Before we leave this topic, let us look a little further into the process of finding limiting values. One or two examples will suffice. So turn on to frame 26. 309 Series 1 Example 1. To find the limiting value of as'n ->■ °° We cannot just substitute n = °° in the expression and simplify the result, since °° is not an ordinary number and does not obey the normal rules. So we do it this way: rp — - = a — 4r (dividing top and bottom by n) 2n - 7 2 - i/n .. . t (5n + 3\ T .. 5 + 3/n Limit \ - } = Limit - — =y- „->oc\2»-7/ „_„«, 2-7/n Now when n-+°°, 3/n^-O and 7/n^>-0 5n+3 _ 5 + 3/w = 5+0 = 5 "„Voo2«-7 „ioo2-7/rt 2-0 2 c c c We can always deal with fractions of the form—, -3 , ~l , etc., for when ' n n n n -> °°, each of these tends to zero, which is a precise value. Let us try another example. On to the next frame then. 26 2n 2 + 4/1 — 3 Example 2. To find the limiting value of 2 _ as n -» <*>. First of all, we divide top and bottom by the highest power of n which is involved, in this case n 2 . In 2 + An - 3 = 2 + 4/n - 3//i 2 5« 2 -6n + l 5-6/H + l/n 2 . u 2» 2 + 4w - 3 _ Lt 2 + 4//i - 3//i 2 " „->oo5n 2 -6n + 1 „->-oo 5~6//i + 1//J 2 2+0-0 2 27 5-0+0 5 „3-2 Example 3. To find Lt 3 - n~*-oo zn ~y on <+ In this case, the first thing is to Turn on to frame 28. 310 Programme 11 28 Right. So we get Divide top and bottom by n 3 n 3 - + 3n 2 -4 n 3 - 1 - 2/r 3 2n 3 It 2 + 3/; 2 7 2 - 4/n 3 2n 3 + 3n - 4 Finish it off. Then move on to frame 29. 29 aDDDnDanDnnDDDannnnnDDaaannnDDnnnaDDDD Convergent and divergent series A series in which the sum (S„) of n terms of the series tends to a definite value, as n -* °°, is called a convergent series. If S n does not tend to a definite value as n -»■ °°, the series is said to be divergent. Example: Consider the G.P. l + o + g + 27 + oi+--- «(1 -r") We know that for a G.P., S„ = _ — - so in this case since a = 1 and r = t, we have : '<'-^> '-f" 3 ■4 2 3 ;(•-«) 1 3 .\ As «-»•<» ,— ->-0 :. Lt S„=- ~> n-t-ao £ The sum of n terms of this series tends to the definite value ■=■ as n -* °°. It is therefore a series. (convergent/divergent) 311 Series 1 convergent 30 If S n tends to a definite value as n -*■ °°, the series is convergent. If S„ does not tend to a definite value as n -> °°, the series is divergent. DnnnnDDDnnDnnnnnaDDDDDDnnDDDnDnnDDDnnn Here is another series. Let us investigate this one. 1+3+9 + 27 + 81+... This is also a G.P. with a = 1 and r = 3. .a(l -/■")_ 1(1 -3") _ 1-3" ••• s w l-r 1-3 3"- 1 2 Of course , when n -*■ °°, 3 " ->• °° also . Lt S„ = °° (which is not a definite numerical value) So in this case, the series is divergent We can make use of infinite series only when they are convergent and it is necessary, therefore, to have some means of testing whether or not a given series is, in fact, convergent. Of course, we could determine the limiting value of S„ as n -» °°. as we did in the examples a moment ago, and this would tell us directly whether the series in question tended to a definite value (i.e. was convergent) or not. That is the fundamental test, but unfortunately, it is not always easy to find a formula for S„ and we have therefore to find a test for con- vergence which uses the terms themselves. Remember the notation for series in general. We shall denote the terms by u 1 + u 2 + u 3 + w 4 + . . . So now turn on to frame 32. 31 312 Programme 11 j £ Tests for convergence Test I. A series cannot be convergent unless its terms ultimately tend to zero, i.e. unless Lt u n = 0. If Lt u n /0, the series is divergent. „-*oo This is almost just common sense, for if the sum is to approach some definite value as the value of n increases, the numerical value of the individual terms must diminish. For example, we have already seen that (i) the series 1 + 3 + o" + 27 + ^i + • • ■ converges, while (ii) the series 1+3 + 9 + 27 + 81+... diverges. So what would you say about the series 1+ i + l + L + l + I + 9 1 2 3 4 5 6 •■• ■ Just by looking at it, do you think this series converges or diverges? O «J Most likely you said that the series converges since it was clear that the numerical value of the terms decreases as n increases. If so, I am afraid you were wrong, for we shall show later that, in fact, the series 1 + -J + -J + ^ + -J + ■ • • diverges. It was rather a trick question, but be very clear about what the rule states. It says: A series cannot be convergent unless its terms ultimately tend to zero, i.e. Lt u„ = 0. It does not say that if the terms tend to zero, then the series is convergent. In fact, it is quite possible for the terms to tend to zero without the series converging - as in the example stated. In practice, then, we use the rule in the following form: If Lt u n = 0, the series may converge or diverge and we must test further. If Lt u n /0, we can be sure that the series diverges. Make a note of these two statements. 313 Series 1 Before we leave the series 1 +i + 4- + T + F + r+---+-+-- • 2 3 4 5 6 n here is the proof that, although Lt u„ = 0, the series does, in fact, diverge. " °° We can, of course, if we wish, group the terms as follows: 34 l + WhMH'M and {1 + 1 + 1 + 1}>{± + 1 + 1 + Ij> 2 etC - So that S„>l+^-+y+7+3+y+-.- This is not a definite numerical value, so the series is divergent 35 The best we can get from Test 1, is that a series may converge. We must therefore apply a further test. Test 2. The comparison test A series of positive terms is convergent if its terms are less than the corresponding terms of a positive series which is known to be convergent. Similarly, the series is divergent if its terms are greater than the correspond- ing terms of a series which is known to be divergent. An example or two will show how we apply this particular test. So turn on to the next frame. 314 Programme 11 36 Example. To test the series we can compare it with the series which is known to converge. If we compare corresponding terms after the first two terms, we see that— 3 <- 3 ; — 4 <- 4 ; and so on for all further terms, so that, after the first two terms, the terms of the first series are each less than the corres- ponding terms of the series known to converge. The first series also, therefore, 37 converges The difficulty with the comparison test is knowing which convergent series to use as a standard. A useful series for this purpose is this one: It can be shown that (i) ifp>l, the series converges (ii) ifp < 1, the series diverges 00 1 , n = i n Does it converge or diverge? 315 Series 1 Converge since the series 2- 2 is the series Z-„ withp > 1 38 DDDDnnDDDDDDDOODDDDDDODDnnnDDDDDDDDODD Let us look at another example. To test tne series ^ + J^ + JL + A + ... If we take our standard series Tp + 2P + lP + 4P + Jp + 6P + " " when p = 2, we get T 2+ "2 2+ I 2+ 4 2+ 5 2+ 6 2 + --' which we know to converge. J_<1. _L<J_. J_<J_ 1.2 l 2 ' 2.3 2 2 ' 3.4 3 2 Each term of the given series is less than the corresponding term in the series known to converge. Therefore But — <-r 2 ; ^<T2> T7 <_ ^ etc - The given series converges 39 DDDDDnDODDDDDDODDODDDDnDDDDDDDDDDODDDD It is not always easy to devise a suitable comparison series, so we look for yet another test to apply, and here it is; Test 3. D'Alembert's ratio test for positive terms Let Mj + u 2 + u 3 + w 4 + . . . + u n + . . . be a series of positive terms. Find expressions for u n and u n + 1 , i.e. the nth term and the (n + l)th term, and form the ratio — — . Determine the limiting value of this ratio u n as n -* °° . If Lt — - — < 1 , the series converges " > 1 , the series diverges " = 1 , the series may converge or diverge and the test gives us no definite information. Copy out D'Alembert's ratio test into your record book. Then on to frame 40. 316 Programme 11 Af\ Here it is again: D'Alembert's ratio test for positive terms If Lt ""•*•' „-><» u n < 1, the series converges > 1 , the series diverges = 1 , the result is inconclusive. □DDODnnoDQnnaDQODDnanaoonaannnDoannDaa 13 5 7 Example: To test the series- +T + "j 2 +~z% + . . . We first of all decide on the pattern of the terms and hence write down the nth term. In this case u„ ='^|fA The (n + l) th term will then be the same with n replaced by (n + 1) _2n + l i.e. u„ + i rw - u n + 1 _ 2n + 1 2"' 1 _ 1 In + 1 ~£7" T^b-I 2'2n-l We now have to find the limiting value of this ratio as n -*■ °° . From our previous work on limiting values, we know that the next step, then, is to divide top and bottom by 41 Divide top and bottom by n cn Tt «» + !_ Tt 1 2« + l_ 1 2+1/n SO Lt Lt - .x r - Lt - , „^.oo U„ „->-oo2 2h-1 „-*oo / 2 l/« -I 112= ! "2"2-0 2 Since, in this case. Lt - "" + 1 < 1 , we know that the given series is convergent. nnnDnnDnnDDDDDaaDDnnDDDDnnaanDnnnnnnnD Let us do another one in the same way. Example: Apply D'Alembert's ratio test to the series _L + 2 + 3 + 4 + 5 + 2 3 4 5 6 First of all, we must find an expression for u n . In this series, u n = 317 Series 1 H*H- U n = « + 1 42 Then k„ + 1 is found by simply replacing n by (n + 1). -Ill Un + i- n + 1 n+ 1 _ w 2 + In + 1 So that u„ n + 2' n n 2 + 2n We now have to find Lt "" + 1 and in order to do that we must divide „->oo U n top and bottom, in this case, by 43 T u n + i T , n 2 + 2n + l_ ,. f 1 + 2/n + \\n 2 Lt — Lt >, . n — - Lt n^oo U„ „X» n I + 2n n ^oo 1 + 2/« 1 +0 + Lt "rt + 1 _ 1+0 1 ► OO Uy\ 1 , which is inconclusive and which merely tells us that the series may be convergent or divergent. So where do we go from there? We have, of course, forgotten about Test 1, which states that (i) if Lt u n = 0, the series may be convergent rc-yoo (ii) if Lt u n f 0, the series is certainly divergent In our present series, u n = . .'. Lt u n = Lt Lt 1 n + 1 „^»oo 1 + 1/n 1 This is not zero. Therefore the series is divergent. □□nanannQnononnQnanQnnoDnnDnoonDnnonoo Now you do this one entirely on your own: Test the series I + 2 + l 2 + A 3 + 1 4 + 5 6 7 8 9 When you have finished, check your result with that in frame 44. 318 Programme 11 44 Here is the solution in detail: see if you agree with it. 1 2 2 2 2 3 2 4 — + - + — + — + — + 5 6 7 8 9 ■■• " 4+H* 2 n "« +1 5+n . u n+l 2" 4+« «« 5 + n 2"" 1 The power 2"~ : cancels with the power 2 n to leave a single factor 2. • "» + i = 2 ( 4 + ") w„ 5 +n ■ Lt J^i= Lt 1«*±»>« Lt 2(4Aill) n ->oo « K n->°o 5 + ft n ->oo 5/w + 1 = 2(0+1 ) 0+1 • Lt i^li = 2 And since the limiting value is > 1 , we know the series is 45 divergent Series in general. Absolute convergence So far, we have considered series with positive terms only. Some series consist of alternate positive and negative terms. Example: the series 1— ~ + ~ — ~ + ... is in fact convergent while the series ^ + o + q + 2 + --- * s divergent. If u n denotes the n m term of a series in general, it may well be positive or negative. But \u n \ , or 'mod u n ' denotes the numerical value of u n , so that if u x + u 2 + u 3 + w 4 + . . . is a series of mixed terms, i.e. some positive, some negative, then the series \ui |+ \u 2 \ + \u 3 1+ |w 4 1+ . . . will be a series of positive terms. So if S«„ = 1 - 3 + 5 - 7 + 9 - . . . Then 2|w„| = 319 Series 1 2|w„| = l +3 + 5 + 7 + 9 + . 46 DnDnDnDnnnDanDnnannnnanDDDDnanannnnnDa Note: If a series 2m„ is convergent, then the series T,\u n \ may very well not be convergent, as in the example stated in the previous frame. But if 2|w„| is found to be convergent, we can be sure that 2«„ is convergent. If 2|«„| converges, the series 2u„ is said to be absolutely convergent. If 2|w„| is not convergent, but £m„ does converge, then Y,u n is said to be conditionally convergent. So, if Sh„ =1 ~7 + "? _ 4 + ? _ ■ • ■ converges and 2|w^| = 1 +2 + 3 + T + T + • • • diverges then Sw„ is convergent. (absolutely or conditionally) conditionally Example: Find the range of values of x for which the following series is absolutely convergent. jl _ .*i +-^- - -Z— + * s - 2.5 3.5 2 4.5 3 5.5 4 6.5 s ■" x" I | x n+1 '""' = (n + 1)5" ; r" + 1 l = (« + 2)5" + 1 . |"« + i| x" + 1 (n + 1)5" "" l (« + 2)5 n + 1 x n - 4" + _ *0 + i/«) = 5(« + 2) 5(1 + 2/«) Lt |"« + „-*ooi u„ I 5 For absolute convergence Lt M-^l < l . ;. Series convergent when|-|< 1, i.e. for|x|<5. On to frame 48. 47 320 Programme 11 A O You have now reached the end of this programme, except for the test "** exercise which follows in frame 49. Before you work through it, here is a summary of the topics we have covered. Read through it carefully: it will refresh your memory of what we have been doing. Revision Sheet 1. Arithmetic series: a, a + d, a + 2d, a + 3d, u„=a + (n-\)d S„ =y(2fl +7F r T.d) 2. Geometric series: a, ar, ar 2 , ar 3 , ar u n = ar n ~ l S„ a _„,*-. c -a(l-r») If|r|<l, Soo-^ 3. Powers of natural numbers'- 2 i n . .3 -\ nin + 1 ) i 2 4. Infinite series: S n = u t + u 2 + u 3 + u^ + . . . + u„ + . . . If Lt S„ is a definite value, series is convergent If " is not a definite value, series is divergent. 5 . Tests for convergence : (1) If Lt u n = 0, the series may be convergent If " /o, the series is certainly divergent. (2) Comparison test - Useful standard series Jp + Jp + 3? + 4> + 5? + • ■ • TzP ' " For p > 1 , series converges: for p < 1, series diverges. (3) D'Alembert's ratio test for positive terms. If Lt " n + 1 < 1 , series converges. „->-oo M„ " > 1 , series diverges. " =1, inconclusive. (4) For general series (i) If 2|m„ I converges, 2w M is absolutely convergent, (ii) If £|m„ I diverges, but 2w„ converges, then 2h„ is conditionally convergent. Afow you are ready for the Test Exercise so turn to frame 49. 321 Series 1 Test Exercise - XI Answer all the questions. Take your time over them and work carefully. 1. The 3rd term of an A.P. is 34 and the 17th term is -8. Find the sum of the first 20 terms. 2. For the series 1 , 1 -2, 1 44, find the 6th term and the sum of the first 10 terms. 8 3. Evaluate 2 n(3 + In + n 2 ). « = i 4. Determine whether each of the following series is convergent. 49 2.3 3.4 4.5 5.6 1 2 2 3 2 4 2" (ii)f 2 4+f^ + --^ + -- 0") "« "TTT?" (iv) u n~-y 5. Find the range of values of x for which each of the following series is convergent or divergent. x 2 x 3 x 4 (i)l+x + |+f ] +- ! + x x^ x^ x (li) D + T3 + T4 + 43 + (iii) 2 ^— r- x" 322 Programme 11 Further Problems - XI 1 . Find the sum of n terms of the series S„ = l 2 + 3 2 +5 2 + ... + (2«-l) 2 2. Find the sum to n terms of 1 + 3 .5 + 7 + 1.2.3 2.3.4 3.4.5 4.5.6 " 3. Sum to n terms, the series 1.3.5+2.4.6 + 3.5.7 + .. . 4. Evaluate the following: (i) £r(>- + 3) (ii)£(r+l> 5 . Find the sum to infinity of the series 3 6. For the series S -i + 5_5. . (-If- 1 5 . find an expression for S„, the sum of the first n terms. Also, if the series converges, find the sum to infinity. 7. Find the limiting values of ,.. 3x 2 + 5x - 4 (l) 5x 2 -x+7 ^ X ^°° x 2 + 5x-4 2x 2 - 3x + 1 (») o^2_^„^, as x -* °° 8. Determine whether each of the following series converges or diverges. 00 n °° n (iii) f^TT < iv ) f^TTT)! 323 Series 1 9. Find the range of values of x for which the series JL + *L + Y " 27 125 '" (2n + l) 3 "" is absolutely convergent. 10. Show that the series 1+ r2 + S + ^ + --- is absolutely convergent when-1 <x < +1. 1 1 . Determine the range of values of x for which the following series is convergent 2 Y 3 v 4 Jz— +— + x + x + 1.2.3 2.3.4 3.4.5 4.5.6 " - ' 12. Find the range of values of x for convergence for the series ^ 2V A 3 V A 4V x X + -2T + ^T + — + --' 13. Investigate the convergence of the series Otitis + fr--- for * >0 14. Show that the following series is convergent 2 + 3I + 4 1_ + 5I 2 + 2"4 3"4 2 4"4 3 ■■• 15. Prove that VT + V^ + V^ + V4" + --- isdivergent and that 1111 T 2 '2 T+ ~3 5 + 4* + •■• 1S convergent. 16. Determine whether each of the following series is convergent or divergent. (l) Z 2n(2» + 1) (U) 2 TT7^ ,...,. v n ,. , v 3ra + 1 (lll) S V(4^TT) (1V) S ^^2 324 Programme 11 17. Show that the series 2x 3x 2 4x 3 1 +— + — + — + . . . is convergent if -5 < x < 5 and for no other values of x. 18. Investigate the convergence of 3 7 15 31 (ii) \2 +: h + TT^i2* + --- 19. Find the range of values of x for which the following series is convergent. 0-2 ) (x_-2) 2 + ix-2) 3 + + (x-2f 1 2 3 n - ' ' 20. If« r =r(2r+ l) + 2'' +1 , find the value of E« r i 325 Programme 12 SERIES PART 2 Programme 12 1 Power series Introduction: In the first programme (No. 1 1) on series, we saw how important it is to know something of the convergence properties of any infinite series we may wish to use and to appreciate the conditions in which the series is valid. This is very important, since it is often convenient to represent a function as a series of ascending powers of the variable. This, in fact, is just how a computer finds the value of the sine of a given angle. Instead of storing the whole of the mathematical tables, it sums up the terms of a series representing the sine of an angle. That is just one example. There are many occasions when we have need to express a function of x as an infinite series of powers of x. It is not at all difficult to express a function in this way, as you will soon see in this programme. So make a start and turn on to frame 2. Suppose we wish to express sine x as a series of ascending powers of x. The series will be of the form sin x = a + bx + ex 2 + dx 3 + ex 4 + . . . where a, b, c, etc., are constant coefficients, i.e. numerical factors of some kind. Notice that we have used the 'equivalent' sign and not the usual 'equals' sign. The statement is not an equation: it is an identity. The right-hand side does not equal the left-hand side: the R.H.S. is the L.H.S. expressed in a different form and the expression is therefore true for any value of x that we like to substitute. Can you pick out an identity from these? (x + 4) 2 = 3x 2 - 2x + 1 (2* + l) 2 =4x 2 +4x-3 (x + 2) 2 = x 2 + 4x + 4 When you have decided, move on to frame 3. 327 Series 2 (x + 2) 2 =X 2 +4x+4 Correct. This is the only identity of the three, since it is the only one in which the R.H.S. is the L.H.S. written in a different form. Right. Now back to our series: sin x = a + bx + ex 2 + dx 3 + ex 4 + . . . To establish the series, we have to find the values of the constant coeffi- cients a, b, c,d, etc. Suppose we substitute x = on both sides. Then sinO=a + + + + 0+... and since sin = 0, we immediately get the value of a. a = a = Now can we substitute some other value for x, which will make all the terms disappear except the second? If we could, we should then find the value of b. Unfortunately, we cannot find any such substitution, so what is the next step? Here is the series once again: sin x = a + bx + ex 2 + dx 3 + ex 4 + . . . and so far we know that a = 0. The key to the whole business is simply this: Differentiate both sides with respect to x. On the left, we get cos x. On the right the terms are simply powers of x, so we get cos* = 328 Programme 12 cosx = b + c.2x + d.3x 2 + e.4x 3 + . . This is still an identity, so we can substitute in it any value for x we like. Notice that the a has now disappeared from the scene and that the constant term at the beginning of the expression is now b. So what do you suggest that we substitute in the identity as it now stands, in order that all the terms except the first shall vanish? We substitute* = again. Substitute x = again Right: for then all the terms will disappear except the first and we shall be able to find b. Putx = cos* = b + c.2x + d.3x 2 + e.4x 3 + . . . :. cos 0=1=6 + + + + + ... :. b = 1 So far, so good. We have found the values of a and b. To find c and d and all the rest, we merely repeat the process over and over again at each successive stage. i.e. Differentiate both sides with respect to x and substitute 329 Series 2 substitute x = So we now get this, from the beginning: sin x = a + bx + ex 2 + dx 3 + ex* +fx s + . . . Put x = 0. :. sin = = a + + + + . . . .". g = ( Diff. cos x = b+ c.2x + d.3x 2 + eAx 3 +/.5x 4 . . . I Put x = 0. .'. cos = 1 = b + + + + . . . .'. b = 1 [ Diff. -sinx = c.2+d.3.2x +e.43x 2 +f.5Ax 3 . . 1 Putjc = 0. .'. -sin0 = 0= C.2 + + 0+... ■'. c = [Diff. -cosx= d.3.2.1 + e.43.2x +f.5A3x 2 . I Putx = 0. :. -cos0=-l= c?.3!+0 + + . .. :.d=-^ J And again, sin x = e.4.3.2.1 +f.5A3.2x + . . . [ Put x = 0. A sin = = e.4! +0 + 0+ :. e = I Once more. cosx= /.5.4.3.2.1 +... I Putx = 0. V. cos0= 1 = /.5!+0+ ■■ f=\i etc. etc. All that now remains is to put these values for the constant coeffi- cients back into the original series. sinx = 0+ I.jc + 0.x 2 + -|, x 3 +0.x 4 + ^x 5 + . . . ^ s X , X i.e. sinx =x-— , +-r-j - Now we have obtained the first few terms of an infinite series represent- ing the function sin x, and you can see how the terms are likely to proceed. Write down the first six terms of the series for sin x. When you have done so, turn on to frame 8. 330 Programme 12 8 x 3 , x 5 X 1 , x 9 x 11 Provided we can differentiate a given function over and over again, and find the values of the derivatives when we put x = 0, then this method would enable us to express any function as a series of ascending powers of x. However, it entails a considerable amount of writing, so we now establish a general form of such a series, which can be applied to most functions with very much less effort. This general series is known as Maclaurin 's series. So turn on to frame 9 and we will find out all about it. Maclaurin 's series: To establish the series, we repeat the process of the previous example, but work with a general function, /(x), instead of sin x. The first differential coefficient of/(x) will be denoted by fix); the second by f"{x); the third byf'"(x); and so on. Here it is then: Let f(x) =a+bx+cx 2 + dx 3 + ex 4 +fx 5 + . . . Putx = 0. Then/(0) = a + + + 0+. . . :. a=fj0). i.e. a = the value of the function with x put equal to 0. Diff. f'(x) = b + c.2x + d.3x 2 + e.4x 3 +f.5x 4 + . . . Putx = :.f'(0)= b + + + ... :. b=f'(0) Diff. /"(*)= c.2.1 + d.3.2x + eA3x 2 +f.SAx 3 . . . Put jc = ."./"(O) = c.2! + + + . . . :. c ,/"(0) 2! Now go on and find d and e, remembering that we denote |{/"W). by fix) and <k{ f '" {x) } by fiv{x) > etc - So, d= and e = 331 Series 2 , f"(o) /i v (o) Here it is. We had: /"(*) = c.2.1 + d.3.2x+eA3x 2 +f.5Ax 3 + . . . rDiff. •'. /'"W= rf.3.2.1+e.4.3.2x+/.5.4.3* 2 + ... / ; '"(0) Putx = :. /'"(0)= d.3!+0 + 0... .'. rf- 3! (Diff. .'. /iV( X ): e A3. 2.1 +f.5A3.2x + . . 1 Put a: = :. /^(O) ■■ e.4! +0 + + .. . etc. etc. _ /*(()) 4! 10 So a-/(0); b=f'(0); c= ~^—\ d=—^— ; e=-^— ;... Now, in just the same way as we did with our series for sin x, we put the expressions for a,b,c, . . . etc., back into the original series and get: /(*) = f"(Q) f'"(G) fix) =/(0) +f'(0).x +-^F.* 2 + 7 -^.x 3 + . . 3! and this is usually written as f(pc) =/(0) +*./'(0) +J/"(0) +|J./'"C0) ... . I This is Maclaurin 's series and important! Notice how tidy each term is. The term in x 2 is divided by 2! and multiplied by /"(0) " " " x 3 " " " 3! " " " /'"(0) " " " a: 4 " " " 4! " " " / iv (0) Copy the series into your record book for future reference. Then on to frame 12. 11 Programme 12 12 13 Maclaurin's series fix) =/(0) +*. /'(O) +f?./"(0) +ff--/"'(0) + . . . DnDDDDDDDDOnDDDDDDDnDDODDDDDDDDDDDDDDD Now we will use Maclaurin's series to find a series for sinh;*. We have to find the successive differential coefficients of sinh x and put x = in each. Here goes, then: /(jc) = sinhx /(0) = sinhO = fix) = cosh x /'(0) = cosh = 1 f'\x) = sinh x /"(0) = sinh = f'ix) = cosh x /'"(0) = cosh = 1 f iy ix) = sinh x /» v (0) = sinh = / v (x) = cosh x / v (0) = cosh = 1 etc. :. sinh* =^x.l+j£<6) + f^.(l) +J^<6) + fJ.(l)+... 7wrn oh ?o /rame 13. JC 3 JC JC 7 sinh x=x+-^ + T\ + nJ + - Now let us find a series for ln(l + x) in just the same way. /(x) = ln(l+x) A /(0) = /,( * )= rb =(1+ * T1 •'• / ' (0) = /■••(*) = -(i + *T 2 = ( Y7^2 •• /"(0) = r"w = 2(i +xj 3 = (T 77)3 ■"■ /'"(o) = /*(*)= "3.2(1 + ^" 4 = -(n|)4 •• / iv (0) = /v(x) = 4.3.2(l+^)- s = (T ^y 5 .-./ v (0) = You complete the work. Evaluate the differentials when x = 0, remembering that In 1 = 0, and substitute back into Maclaurin's series to obtain the series for ln(l + x). So, ln(l +x) = 333 Series 2 /(0) = lnl=0; /'(0)=1; /»'(0) = -l; /'"(O) = 2; / iv (0) = -3!; / v (0) = 4!; ... Also f{x) =/(0) + x/'(0) + fJ/"(0) +f , /'"(O) + . . . ln(l+x) = 0+x.l+f*(-i)+^( 2 ) + £(-3!) + ... 14 4! Y 2 3 4 „5 ln(l+x)=x-f + f-±-+Z-- Note that in this series, the denominators are the natural numbers, not factorials! Another example in frame 15. 15 Example: Expand sin 2 * as a series of ascending powers of x. Maclaurin's series: f{x) =/(0) +x.f'(0)+^.fm +f-'-/'"(0) + • ■ • :. f(x) = sin 2 * /(0) = /'(*) = 2 sin x cos x = sin 2x /'(0) = /"(*) = 2 cos 2x /"(0) = /-'"(*) =-4 sin 2x /"'(0) = / iv (*) = / iv (0) = There we are! Finish it off: find the first three non-vanishing terms of the series. Then move on to frame 16. 334 Programme 12 16 sin * K 3 45* For /(*) = sin 2 * /'(*) = 2 sin x cos * = sin 2x f"(x) = 2 cos 2x /'"(*) = -4 sm 2x / 1V (*) = -8 cos 2x / v (*) = 16sin2x / vi (*) = 32 cos 2* ■'• /(0) = /. /'(0) = .-. /"(0) = 2 •'• /'"(0) = /. / iv (0) = -8 .-. / v (0) = /. / vi (0) = 32 etc. ..3 /(*) =/(0) + *./'(0) +|j ./-"(O) +fy./-"-'(0) + ■ • • sin 2 * = + *(0) +§J (2) +fj (0) + f-, (-8) + f-| (0) + jj (32) sin 2 *=* 2 -y +^- 17 Next we will find the series for tan x. This is a little heavier but the method is always the same. Move to frame 1 7. Series for tan x f(x) = tan x :. /(0) = /. f'{x) = sec 2 * /. /'(0) = 1 .-. fix) = 2 sec 2 * tan x :. /"(0) = /. /'"(*) = 2 sec 4 * + 4 sec 2 * tan 2 * .'. /'"(0) = 2 = 2 sec 4 * + 4(1 + tan 2 *) tan 2 * = 2 sec 4 * + 4 tan 2 * + 4 tan 4 * .'. / iv (*) = 8 sec 4 * tan* + 8 tan* sec 2 * + 16 tan 3 * sec 2 * = 8(1 + t 2 ) 2 t + 8r(l + t 2 ) + 16f 3 (l + t 2 ) = 8(1 + 2f 2 + f)t + 8t + 8t 3 + 16r 3 + I6t 5 = I6t + 40t 3 + 24t s :. / iv (0) = /. / v (*) = 16 sec 2 * + 120? 2 .sec 2 * + 120f 4 sec 2 * _ :. /v(0) = 16 . . tan * = 335 Series 2 :. tanx = x +•=- + tt + Standard series By Maclaurin's series, we can build up a list of series representing many of the common functions - we have already found series for sin x, sinhx and ln(l + x). To find a series for cos x, we could apply the same technique all over again. However, let us be crafty about it. Suppose we take the series for sin x and differentiate both sides with respect to x just once, we get 3 5 7 X , X X, smx = x - tt+c7~ 7~r + • • • 18 Din. cos x = l - wf + ft ~~ jC etc. ~2\ 4l - 6T In the same way, we can obtain the series for coshx. We already know that 3 5 7 sinh x=x + T-f+ry + 7T + so if we differentiate both sides we shall establish a series for cosh x. What do we get? We get: Diff. giving: 3 5 7 9 X j X , X j_X sinh x = -x + ^T + "rf + yr + gT + i i . jX i ja , IX . yX coshx- 1 +2|- + JT + 7!~ + 9T x 2 x 4 x 6 x 8 coshx = l +2! + 4J + 6T + 8T + Let us pause at this point and take stock of the series we have obtained. We will make a list of them, so turn on to frame 20. 19 336 Programme 12 20 Summary Here are the standard series that we have established so far. X , X X' .XI IT sinx *"3T + 5T - 7!" + "9f- • • " cosjc = i-fr + 4T _ fr + tr" ■ m tan* =x+j + ff- + • .. IV x*x 5 sinh x =jc + t7 + -f7 + ^7+ . coshx=l+|j- + |f+|y+|y 2 3 4 X , x X m(i + x) = x-f + f-f + f... Make a note of these six series in your record book. Then turn on to frame 21. V VI VII 21 The binomial series By the same method, we can apply Maclaurin's series to obtain a power series for (1 + xY 1 . Here it is: f{x) = {\+xT /(0)=1 f'(x) = n.(\+x)"- 1 f'{0) = n f"(x) = 7i(h - 1) .(1 + x)"- 2 /"(0) = n(n - 1) f'"{x) = n(n - 1) (« - 2).(1 + x) n ' 3 /"'(0) = «(n - 1) (n - 2) / iv (x) = n(«-l)(n-2)(n-3).(l +x)"^ / iv (0) = «(«-l)(n-2)(n-3) etc. etc. General Maclaurin's series: /(*) =/(0) + x.fXO) + f?/"(0) + f,/"'(0) . . . Therefore, in this case, ( 1 + x)" = 1 + x« + |j w(n - 1 ) + |y "(" ~ ! ) (" - 2) . . . n(n-l) 2 «(«-l)(«-2) 3 WTTI (l+x)" = l +nx+ 2\ IT Add this result to your list of series in your record book. Then, by replacing x wherever it occurs by (-x), determine the series for (1 -xf. When finished, turn to frame 22. 337 Series 2 DoaQnpnanaaoQananonDDaanaDnnaaaanDDOQD Now we will work through another example. Here it is: Example: To find a series for tan" 1 *. As before, we need to know the successive differential coefficients in order to insert them in Maclaurin's series. f(x) = tan" 1 * and /'(*) 1 1 +x .2 If we differentiate again, we get /"(*) = - ( y^r )2 , after which the work- ing becomes rather heavy, so let us be crafty and see if we can avoid unnecessary work. We have/(x) = tan" 1 * and A*) = j~2 =(1 + *T- If we now expand (1 +* 2 )" 1 as a binomial series, we shall have a series of powers of* from which we can easily find the higher differential coefficients. So see how it works out in the next frame. To find a series for tan l x fix) = tan x x :. /(o) = Z U 1.2 1.2.3 ■■• = 1 -x 2 +x 4 -x 6 +x 8 ~. . . /'(0)=1 •"• f"(x) = - 2x + 4x 3 - 6x 5 + 8x 7 - ... /"(0) = ••• f'"(x) = - 2 + 1 2* 2 - 30a: 4 + 56xr 6 - . . . /'"(0) = -2 :. f iv (x) = 24x-l 20a: 3 + 336jc s - . . . /'"(()) = .'. f y (x) = 24 - 360tc 2 + 1680a: 4 - . . . /v(0) = 24 etc. :. tan'jc =/(0) + x./'(0) +f?/"(0) + §?/' "(0) + . . .. Substituting the values for the derivatives, gives us that tan" 1 * = Then on to frame 24. 338 Programme 12 24 tan" 1 * = + x{\) + §J (0) +fj (-2) +£ (0) + f? (24) . . . X -1 A; X X tan x = x— -=- +■? — =- + This is also a useful series, so make a note of it. DDDDDDDnODODDDDDDDDDODDDDQDDDDDDDDDDDD Another series which you already know quite well is the series for e*. Do you remember how it goes? Here it is anyway. e* = 1 +x+ x T j+h + X + . . . 2! ' 3! ' 4! and if we simply replace x by (— x), we obtain the series for e" XI ■rX = l-x+-- _*_ + * 2! 3! 4!'-' So now we have quite a few. Add the last two to your list. And then on to the next frame. XII Examples: Once we have established these standard series, we can of course, combine them as necessary. 25 Example 1. Find the first three terms of the series for e*.ln(l +x). We know that and that 2 3 4 1+ * + 2! + 3! + 4T + ' ln(l+x)=x-|- +f"-f- + e*.ln(l +x)- X X X 1+x + 2! + 3T + 4T 2 3 Now we have to multiply these series together. There is no constant term in the second series, so the lowest power of x in the product will be x itself. This can only be formed by multiplying the 1 in the first series by the x in the second. The* 2 term is found by multiplying 1 XI ~y\ and X X x x 3 The x 3 term is found by multiplying 1 X ~- and x XI - ~- I and j X x 3 3 3 T ~2~ 2 ~3 x and so on. 339 Series 2 •\ e*.ln(l+jc)=x+y +j +... ^" i 3 X j_X It is not at all difficult, provided you are careful to avoid missing any of the products of the terms. DaanDDDDDDDnDnnanGnnDDnannDDDnnnDDannn Here is one for you to do in the same way: Example 2. Find the first four terms of the series for e* sinh x. Take your time over it: then check your working with that in frame 27. Here is the solution. Look through it carefully to see if you agree with the result. 27 *x 1+ * + il + 3T + 4T + 3 5 7 sinh * = * + §] + fi + 77 + e*.sinhx= 1+ * + f! + IT---r + f! + 5T + 3 S X , X Lowest power is x Term in* = l.x =x " "x 2 =x.x=x 2 x i. 3! + 2 ,.x xy 6 + 2 ) 3 x?.x? = 3/I.I \_ 2x 3 31 2! x x 3 3 H v X , X — 4 X X . ** I ■ n I ,X ~~~ X 3 ^3 ./! |i v 4 3! + 3T » "^4 =v .ir'_ L ^ 4/'l + L\_^C \6 6J 3 •7 y 3 y 4 e*.sinhx =x +x 2 + =- +4- + . 77ze7-e we are. A^ow turn on to frame 28. 340 Programme 12 28 Approximate values This is a very obvious application of series and you will surely have done some examples on this topic some time in the past. Here is just an example or two to refresh your memory. Example 1. Evaluate \/l-02 correct to 5 decimal places. 1-02= 1 +0-02 Vl-02 = (1 +0-02) 1 / 2 = 1 + ±(0-02) + ±jj- (0-02) 2 + \ 23 l (0-02) 2 . . . = 1 + 001 - 1 (0-0004) + ^(0000008)-. . . = 1 + 001 - 0-00005 + 0-0000005 . . . = 1-010001-0000050 = 1 009951 :. Vl 02=1 00995 Note that whenever we substitute a value for x in any one of the standard series, we must be satisfied that the substitution value for x is within the range of values of x for which the series is valid. The present series for ( 1 + x) n is valid for | x | < 1 , so we are safe enough on this occasion. Here is one for you to do. Example 2. Evaluate tan" 1 01 correct to 4 decimal places. Complete the working and then check with the next frame. 29 tan" 1 01 =00997 3 5 7 tari~ 1 x=x--^ +T -7j +... .,-!„, n , 0001 000001 00000001 .. tan '0-1 =01 — + ^ . . . = 0-1 - 0-00033 + 0000002 - . . . = 0-0997 We will now consider a further use for series, so turn now to frame 30. 341 Series 2 Limiting values — Indeterminate forms In Part I of this programme on series, we had occasion to find the limiting value of -£±i as n ->■ °°. Sometimes, we have to find the limiting U n o value of a function of x when x -> 0, or perhaps when x -> a. T . f.x 2 + 5.x -14} + 0-14 14 7 e.g. Lim -2- 30 x ->-oU 5jc+ 8 ) 0-0+ 8 8 4 That is easy enough, but suppose we have to find ' x 2 + 5x-\4 Lim \ ■, ,. x-+2\ x 5x + 6 Puttings = 2 in the function, gives-; — - — -= — and what is the value „ 4 — 1U + O U of "o ? Is it zero? Is it 1? Is it indeterminate? When you have decided, turn on to frame 31. . — as it stands, is indeterminate We can sometimes, however, use our knowledge of series to help us out of the difficulty. Let us consider an example or two. Example 1. Find the Lim { => — If we just substitute* = in the function, we get the result —which is indeterminate. So how do we proceed? x 3 2x s Well, we already know that tan x = x + ~ +■!-=-+... So if we replace tan x by its series in the given function, we get 31 Lim | 3 — } = Lim x-+o { x j x ^ - 3 2x s ^+f+ff +...)-/ x-*0 \ x ; x ^q = Lim l-k+^r + l + *L\ 1 = 1 ^0l3 15 ■•■/ 3 Lim | 3 — } = :r — and the iob is done! x + I x 3 ) 3 J Move on to frame 32 for another example. 342 Programme 12 32 Example 2. To find Lim ! x^-0 { x Direct substitution of x = gives — — which is— again. So we will express sinhx by its series, which is sinhx = (If you do not remember, you will find it in your list of standard series which you have been compiling. Look it up.) Then on to frame 33. 33 sinh x= x + y, + 71 + 7T + So 3 S 7 Lim/^^Lim 1 3! 5! 7! x^O x^-0 _ I X 2 X 4 = 1 +0 + + . .. =1 ■ Lim (^) = 1 x->0 \ x Now, in very much the same way, you find Lim <. , x->0 I * Work it through: then check your result with that in the next frame. 34 !sin x — 2~ 1 = 1 Here is the working: Lim x-*0 = Lim x^Q 2-X + 2*. x 3 45 x 2 2x 4 -SPol 1 "!^ .'. Lim (sin 2 x I I * J Here is one more for you to do in like manner. „ , „ . f sinh x - x Find Lim Then on to frame 35. m - ► V = 1 343 Series 2 Lim x-+0 sinh x - x Here is the working in detail: 3 5 7 sinh x=x+-+o+Ti + 5! 7! 35 . sir ihx - ■x / 3 5 7 3! 5! 7! • •■-/ x 3 _ 1 3! X 2 x 4 Lim x-+0 sinhx-x) 1 * 3 1 fl x 2 = Lim t, + 77 + x h>o(3! 5! x 4 7! + - 1 1 3! 6 Lim 1 x-*(H sinhx-x) 1 x 3 J 6 So there you are: they are all done the same way. (i) Express the given function in terms of power series (ii) Simplify the function as far as possible (iii) Then determine the limiting value — which should now be possible. DnDnDDnDannnnanDnDDnnnnnnnanDannrjDDnnn Of course, there may well be occasions when direct substitution gives the indeterminate form — and when we do not know the series expansion of the function concerned. What are we going to do then? All is not lost! — for we do in fact have another method of finding limiting values which, in many cases, is quicker than the series method. It all depends upon the application of a rule which we must first establish, so turn to the next frame for details thereof. 344 Programme 12 36 L 'Hopital's rule for finding limiting values. fix) Suppose we have to find the limiting value of a function F(x) = -7—, at x = a, when direct substitution of x = a gives the indeterminate form 0_ i.e. at x = a, f{x) - and g(x) = 0. If we represent the circumstances graphically, the diagram would look like this: — Note that at x = a, both of the graphs y = f(x) and y = g(x) cross the x-axis, so that at x = a, f(x) = ( ~-9 {x) and #00 = < At a point K, i.e. x = (a + h), KP =/(fl + h) and KQ = g(a + h) f{a + h) _ KP g( a +fc) KQ Now divide top and bottom by AK fja+ji) = KP/AK = tan PAK g(a + h) KQ/AK tan QAK Now Lim fix). T . f(a+h) T . tan PAK Lim ; . ,( = Lim : rk) x^agix) h -> g(a+h) h .+Q tan QAK £» r/(*). i e the limiting value of -~ as x -» a (at which the function value by #0) direct substitution gives^) is given by the ratio of the differential coeffi- cients of numerator and denominator at x = a (provided, of course, that both/'(a) and g'(a) are not zero themselves)! ... Lim (®Ul) =Lim (^] ., Lim (/^)U T; „//M x-y a \gix) '■ Lim 1 , , . x^a[g(x) This is known us I 'Hopital's rule and is extremely useful for finding limiting values when the differential coefficients of the numerator and denominator can easily be found. Copy the rule into your record book. Now we will use it. 345 Series 2 x^a\g(x)f x-+a [g(x)j 37 ix Example 1. To find Lim { - + x 2 -x-\ l( x 2 + 2x - 3 Note first that if we substitute* = 1, we get the indeterminate form tt. Therefore we will apply 1'Hopital's rule. We therefore differentiate numerator and denominator separately (not as a quotient). (3x 2 + 2x - 1 Lim ( x 3 + 1 ( x 2 + 2x - 3 : Lim 2x + 2 3+2-1 4 , Lim 2x-3 2 + 2 4 = 1 and that is all there is to it! Let us do another example, so, on to the next frame. 38 {cosh x — €^ We first of all try direct substitution, but we find that this leads us to the result -— — , i.e. —which is indeterminate. Therefore, apply 1'Hopital's rule x-*a\g(x)) x^a\g(x)) i.e. differentiate top and bottom separately and substitute the given value of x in the differential coefficients. . T . lcoshx~e x \ T . Isinhx — e x • • Lim { J = Lim { : x-±o{ x J x -+ \ 1 0-1 = -1 Now you can do this oneT Determine - Umf cosh x-e x \__, x-*0\ x f l A- sin 3x Lim I .. x^o\ x*+4x 346 Programme 12 39 I j*{^K The working is simply this s q , so we apply r 2x - 3 cos 3x Direct substitution gives x, so we apply l'Hopital's rule which gives T . [ x 2 - sin 3x \ , . i Lim — J-— — = Lim x ^ol x 2 +4x ) x -* [ 2x + 4 = ~ 3 = 3 + 4 4 WARNING: l'Hopital's rule applies only when the indeterminate form arises. If the limiting value can be found by direct substitution, the rule will not work. An example will soon show this. „ ■* T- (x 2 +4x~3\ Consider Lim { — ? — - — ) x -y 2 \ 5 ~2x ) By direct substitution, the limiting value = = 9. By l'Hopital's rule Lim \ * I '**„_, J J = Lim { ^ '^ }= -4. As you will see, these results ►2 1 5-2* J x ^i\ -2 do not agree. Before using l'Hopital's rule, therefore, you must satisfy yourself that direct substitution gives the indeterminate form q. If it does, you may use the rule, but not otherwise. 40 Let us look at another example {x — sin x 2 X 0-0 By direct substitution, limiting value = -— — - -. Apply l'Hopital's rule: T . fx-sinx) T . (l-cosxl We now find, with some horror, that substituting x = in the differen- tial coefficients, again produces the indeterminate form ~ . So what do you {1 — COS X \ — , L (without bringing in the use of 2x | series)? Any ideas? We 347 Series 2 We apply the rule a second time. Correct, for our immediate problem now is to find Urn | ^ 0SX 1. If we do that, we get: Lim x - sin x i\ -cos x = Lim { ►0 x- ) x ^. { 2x \- J v j First stage Second stage 2x L imI ^U = o . T . (jc-sinxl ■ ■ Lim i 5 / : x->0\ x ) So now we have the rule complete: For limiting values when the indeterminate form (i.e. k) exists, apply l'Hopital's rule Um m--ujm x-+a\g(xj) x ->a\g(x)) and continue to do so until a stage is reached where either the numerator and/or the denominator is not zero. Next frame. 41 Just one more example to illustrate the point. Example: Determine Lim \ 5 } x->0\ * J Direct substitution gives — — , i.e. — . (indeterminate) T . f sinh x — sin x Lim -3 x-»0( X 3x 2 T . I sinh x + sin x ■■ Limj 2 (cosh* -cosx) 1-1 Lim ( r— j J, gives — — =- x ^. { 3x 2 )' b + = 1 + 1 = 1_ 6 3 gives = Lim x->0 ( /coshx + cosx Lim x I sinh x -sinx n ■ ol Note that we apply l'Hopital's rule again and again until we reach the stage where the numerator or the denominator (or both) is not zero. We shall then arrive at a definite limiting value of the function. Turn on to frame 43. 42 348 Programme 12 *t«J Here are three Revision Examples for you to do. Work through all of them and then check your working with the results set out in the next frame. They are all straightforward and easy, so do noi peep at the official solutions before you have done them all. Determine (i) Lim { *' ~f_ ^ 3 ) (ii) Lim(^^) (hi) Lirn/ * C0S y sin * | x^-olsin x-x ) x -*q\ x 3 J 44 Solutions: (0 jft J 4^-5x 4 +l 3 } (Substitution gives g) T . j 3x 2 - 4x + 4 ) 3 , =^1 8^-5 rr l . T . (x 3 - 2x 2 + Ax - 3 ) , ■• ^t 4^-5, + i r .... T . (tanx — x\ ,_, , . .0. (11) Lim { / (Substitution gives — ) x ^-q |sin x-x) a = Lim I r \ (still gives -) ^^olcosx-1) 5 y T . (2sec 2 xtanx) , ... = Lim { : / (and again!) x -*o\ -sinx J (2 sec 2 x sec 2 * + 4 sec 2 x tan 2 x \ 2 + = Lim = — : — = -2 x^-oX -cos* J -1 , Lim( t -^^) = -2 ...... fxcosx-sinx) . (k (111) Lim ^ 5 — - J (Substitution gives — ) T . ( ~x sin x + cos x — cos x = Lim { —j x ->ol 3x 2 T . (-sinx) T . /-cosx) 1 = Lim < — ^ = Lim x^0{ 3 * i x^o\ 3 / 3 . . J x cos x - sin x ) _ 1 x-*ol * / 3 Next frame. 349 Series 2 Let us look at another useful series: Taylor's series. Maclaurin's series /(x) =/(0) + *./'(0) + |j /"(0) + . . . expresses a Y f y = /"(a:) function in terms of its differential coefficients at x = 0, i.e. at the point K. 45 At P,/(/0 =/(0) + h.f'(0) + ^ /»(0) + ^f'"{0) . . . If we now move the _y-axis a ?/v= x a umts t0 t j ie j e f t> t jj e e q Ua tion of the curve relative to the new axes now becomes Fih + a) y = F(a+ x) and the value at K is now F(a) At P, F(a + h) = F(a) + /z. F'(a) + §y F"(a) +fy F'"(a) + . . . This is, in fact, a general series and holds good when a and h are both variables. If we write a = x in this result, we obtain f{x + h)=f(x) + h.fXx) +|j /"(*) + §■*/'»(*) + . . . which is the usual form of Taylor's series. Maclaurin's series and Taylor's series are very much alike in some _ g^ respects. In fact, Maclaurin's series is really a special case of Taylor's. 4d Maclaurin's 2 v 3 series: f(x) = /(0) +x.f(0) + fj/"(0) + fr/"'(0) + . . . Taylor's h i ,3 scnes: f(x + h)= f{x) + h .f'{x) + § , /"(*) + % f'"(x) + ... Copy the two series down together: it will help you learn them. 350 Programme 12 Jti Example 1. Show that, if h is small, then t* _ h xh 2 tarf \x + h) = tan *x + r—— 2 ~ n , 2-> 2 approximately. DDDaDDDDDDDDDnDnnDDDDaDDDDaDnDnDDnDDan Taylor's series states f{x + A) =/(*) + A./'(x) +^r/"W + §t/'"(*) • • • where f{x) is the function obtained by putting h = in the function f(x + h). In this case then, /(*) = tan 1 x. •••A*)—, -d /"W=- (T ^ )2 Putting these expressions back into the series, we have -i,l h 2 2x tan" 1 (pc + h) = tan l x + h. j-^ 2 - ^r^r^js + • • • h xh 2 48 l+x 2 (l+x 2 ) 2 approx. Why are we justified in omitting the terms that follow? The following terms contain higher powers of h which, by definition, is small. These terms will therefore be very small. Example 2. Express sin (x + h) as a series of powers of h and evaluate sin 44° correct to 5 decimal places. sin(x + h) =f(x) + h.p{x) +y, /"(*) +|j /'"(*) + ■ ■ • f(x) = sin x; fix) = cos x ; f"(x) = - sin x; f"'(x) = -cos x; / iv (x) = sin x; etc. h 2 . h 3 :. sin(x + h) = sin x + h cos x -— sin x - — cos x + . . . sin 44° = sin(45°-l°) = sin (—-0-01745) and sin- = cos-r= ,- /.sin 44 = pr 1 +h~j- - g- +. . . A = -0-01745 = ^{l-0.01745- -^^ + °-^f^ 3 + = Ml - 0-01745 - 0-0001523 + 0-0000009 . . = 0-7071 (0-982399) = 0-69466 351 Series 2 You have now reached the end of the programme, except for the test TrO exercise which follows. The questions are all straightforward and you will have no trouble with them. Work through all the questions at your own speed. There is no need to hurry. Test Exercise— XII 1. State Maclaurin's series. 2. Find the first 4 non-zero terms in the expansion of cos 2 *. 3. Find the first 3 non-zero terms in the series for sec x. 3 5 7 _ jY X X 4. Show that tan 1 x = x — — + - — =-+... 5. Assuming the series for e x and tan x, determine the series for e*.tan;>c up to and including the term in x 4 . 6. Evaluate Vl -05 correct to 5 significant figures. 7. Find (i) Limf 1 " 251 "^^^ V x-+ol 5x 2 .... T . (tan x .tan l x~x 2 \ (11) Lim -e x^Ol x I ..... T . Ix-sm.r (111) Lim ( v x ^.q[x- tan* 8. Expand cos(x + h) as a series of powers of h and hence evaluate cos 31° correct to 5 decimal places. You are now ready to start the next programme. 352 Programme 12 Further Problems— XII X 2 x 4 X 6 1 . Prove that cosx=l-yj+-jy-^r+... and that the series is valid for all values of x. Deduce the power series for sin 2 x and show that, if x is small, sin 2 x-x 2 cosx I x 2 . x . 7 = 6 + 360 a PP roxlmatel y- 2. Apply Maclaurin's series to establish a series for ln(l + x). If 1 +x = — , show that 2 3 (b 2 ~a 2 )l2ab=x-^ +*--... Hence show that, if b is nearly equal to a, then (b 2 -a 2 )j2ab exceeds ln(— ]by approximately (b -a) 3 /6a 3 . „ ^ . ,.. T . f 1 - 2 sin 2 * - cos 3 * 1 3. Evaluate (l) Lim { ;— 3 J ,... T . /sinx-xcosjcl ..... T . f tan jc- sin x \ (iv) lU*^) ( V ) Lim(^l^) 4. Write down the expansions of (i) cos x and (ii) ——,and hence show that cosx , x 2 x 3 , 13* 4 — =1-*+---+^-... 5. State the series for ln(l + x) and the range of values of x for which it is valid. Assuming the series for sin x and for cos x, find the series for ln /sinx \ and m( - cos x ^ as far as the term in yA Hence show that, if x is small, tan x is approximately equal to x.e x 6. Use Maclaurin's series to obtain the expansion of e x and of cos x in ascending powers of x and hence determine (e x +e x -2 Lim \ >-o(2 cos x ->-n I ^ cos 2jc— 2 353 Series 2 x — 3 . 7. Find the first four terms in the expansion of . _ s 2 , 2 \ m ascending powers of x. 8. Write down the series for ln(l + x) in ascending powers of x and state the conditions for convergence. If a and b are small compared with x, show that ln(x+fl)-lnjc=| (l+^){ln(x + 6)-lnx) 9. Find the value of k for which the expansion of (l+toOO+fF'lnO- 1 -*) contains no term in x 2 . .., T . ( sinh x - tanh x 10. Evaluate (i) Lim\ 5 .... T . / lnx\ ..... T . f x + sins \ (11) Lim 1 — : (111) Lim — j— — 11. If u r and « A .i indicate the r th term and the (/• - l) th term respectively of the expansion of (1 + x) n , determine an expression, in its simplest form, for the ratio ^—. Hence show that in the binomial expansion of (1 + 0-03) 12 , the r th term is less than one-tenth of the (r - l) th term if r > 4. Use the expansion to evaluate (1 -03) 12 correct to three places of decimals. 12. By the use of Maclaurin's series, show that . -, x ix , sin x = x + t + 7^" + • • 6 40 Assuming the series for e x , obtain the expansion of e x sin -1 *, up to and including the term in x 4 . Hence show that, when x is small, the graph of j = e x suf 1 * approximates to the parabola y = x 2 + x. 13. By application of Maclaurin's series, determine the first two non- vanishing terms of a series for In cos x. Express (1 + cos 8) in terms of cos 9/2 and show that, if 8 is small, ln(l + cos 0) = In 2-— -— approximately. 354 Programme 12 14. If x is small, show that (i) — 1 + x+ — 1 -x \ 2 V{ (l+3* 2 )e* } ^ 3* 2_5x 2 *■ 1 -* 2 8 15. Prove that x* U e*=T "" 2 12 ~ 720 + """ lU e* + l ~2 _ 4 + 48~' •• 16. Find (i) Lim( S -£K^], (ii) Lim f «"" *- 1 ~ * ) , 17. Find the first three terms in the expansion of sinhx/.-ln(l +x) x 2 (l +x) 3 18. The field strength of a magnet (H) at a point on the axis, distance x from its centre, is given by H= N|f 1 1 2l\(x-[) 2 (x + l) 2 j where 2/ = length of magnet and M = moment. Show that, if / is 2M very small compared with x, then H — — r. x 19. Expand [ln(l + x)] 2 in powers of jc up to and including the term in x 4 . Hence determine whether cos 2x + [ln(l + x)] 2 has a maximum value, minimum value, or point of inflexion at x = 0. 20. If / is the length of a circular arc, a is the length of the chord of the whole arc, and b is the length of the chord of half the arc, show that (i) a = 2r sin — and (ii) b = 2r sin — where r is the radius of the 2r v ' Ar circle. By expanding sin r- and sin —as series, show that / = — - — 2r Ar ' 3 approximately. 355 Programme 13 INTEGRATION PART1 Programme 13 1 Introduction You are already familiar with the basic principles of integration and have had plenty of practice at some time in the past. However, that was some time ago, so let us first of all brush up our ideas of the fundamentals. Integration is the reverse of differentiation. In differentiation, we start with a function and proceed to find its differential coefficient. In integra- tion, we start with the differential coefficient and have to work back to find the function from which it has been derived. e.g. — (x 3 + 5) = 3x 2 . Therefore it is true, in this ease, to say that the integral of 3x 2 , with respect to x, is the function from which it came, i.e. 1 3x 2 dx = x 3 +5. However, if we had to find 1 3x 2 dx, without know- ing the past history of the function, we should have no indication of the size of the constant term involved, since all trace of it is lost in the differ- ential coefficient. All we can do is to indicate the constant term by a symbol, e.g. C. So, in general, 1 3x 2 dx = x 3 + C Although we cannot determine the value of this constant of integration without extra information about the function, it is vitally important that we should always include it in our results. There are just one or two occasions when we are permitted to leave it out, not because it is not there, but because in some prescribed situation, it will cancel out in sub- sequent working. Such occasions, however, are very rare and, in general, the constant of integration must be included in the result. If you omit the constant of integration, your work will be slovenly and, furthermore, it will be completely wrong! So, do not forget the constant of integration. 1. Standard integrals Every differential coefficient, when written in reverse, gives us an integral, e.g. ^ (sin x) = cos x .'. I cos x dx = sin x + C dx It follows then that our list of standard differential coefficients will form the basis of a list of standard integrals — sometimes slightly modified to give a neater expression. 357 Integration 1 Here is a list of basic differential coefficients and the basic integrals that go with them: 1. £ (x") = nx"' 1 dx 2 1^4 3. f-(e*) = e* dx 4. ^-( e kx^ = ke kx dx 5. — (a x ) = a x In a dx 6. — (cosx) = -sinx d , ■ x 7. — (sinx) = cos* d 9 8. — (tan x) = sec^x dx 9. -r- (coshx) = sinhx 10. t- (sinh x) = cosh x 12. ^(cos- 1 ^)^^-^ ■■ 13. ^(tan" 1 *)^^- 14. ^(rinh- 1 *)'^!) •• 15. ^(cosh- 1 *)= > ^ rT) :. 16. |(tanh-x)= r ^ 2 - ••• DDDDDDDDDnDDDnnDnD Spend a little time copying this as a reference list. x"dx=^ +C {^ d ) — dx = In x + C x dx = e x + C a x dx = , — + C In a sin x dx = - cos x + C cos x dx = sin x + C sec 2 x dx = tan x + C sinh x dx = cosh x + C cosh x dx = sinh x + C 1 V(i-* 2 ) dx = s i n »x + C -1 V(l-x 2 ) dx = cos l x + C 1 +x^ dx=tan 'x + C r— -,^ = sinh 1 x + C V(* a + V(* a -0 dx = cosh J x + C 1 ■dx = tanh^ + C 1-x' DnDnDDDDnnnDnDnDnDDD list carefully into your record book 358 Programme 13 Here is a second look at the last six results, which are less familiar to you than the others. \j5=?f = sin ' lx + c \^Tif sinh ' lx + c \jy^f = cos ~ lx + c \j^ Notice (i) How alike the two sets are in shape, (ii) Where the small, but all important, differences occur. On to frame 4. -dx = cosh 1 x + C dx = tanh" 1 * + C Now cover up the lists you have just copied down and complete the following. (i) («) (iii) (iv) (v; e sx dx= ... x 1 dx = \Jx dx = sinxdx = .. 2 sinh x dx : (vi) \-dx = ii)f -JL_ JVO-* 2 (vii (viii) I 5 X dx = ■ dx = ■ (ix) [ 1 Hx : (x) V(* 2 -i) 1 \l + x'< dx = When you have finished them all, check your results with those given in the next frame. 359 Integration 1 Here they are (i) y sx dx = 6 —+C (vi)l|-dx = 51nJC + C (ii) L- 7 dx = ~ + C (vii) [ * dx = sin" 1 * + C (iii) y/x dx = J x 1 / 2 dx (viii) [ 5 x dx= Jl +c = 2 T +C (iv) J sin * tfx = -cos x + C (i x ) f , *_ rf x = cosh" ^ + ( (v) I 2 sinh x dx = 2 cosh x + C (x) I — — 2 rfx = tan" 1 * + C All correct? - or nearly so? At the moment, these are fresh in your mind, but have a look at your list of standard integrals whenever you have a few minutes to spare. It will help you to remember them. Now move on to frame 6. 2. Functions of a linear function of x We are very often required to integrate functions like those in the standard list, but where x is replaced by a linear function of x, e.g. I (5x - 4) 6 dx, which is very much like x 6 dx except that jc is replaced by (5x - 4). If we put z to stand for (5x - 4), the integral becomes z 6 dx and before we can complete the operation, we must change the variable, thus !«•*,- J 6 dx z° —dz dz Now — can be found from the substitution z = 5x - 4 for — = 5 there- at dx dx 1 fore — = - and the integral becomes dz 5 ^dx^pz^^dz-^dz 4f + C Finally, we must express z in terms of the original variable, x, so that \(5x-4) 6 dx= 1 360 Programme 13 \ (5x-4) 6 dx = (5x-4) 7 + i 5.7 ( 5x-4) 7 ~" 35~ f x 7 The corresponding standard integral is \x 6 dx = =- + C. We see, there- fore, that when x is replaced by (5x - 4), the 'power' rule still applies, i.e. (5x - 4) replaces the single x in the result, so long as we also divide by the coefficient ofx, in this case 5. (Vdx^+C :. f(5x-4) 6 dJC = ( ^li )7 + C This will always happen when we integrate functions of a linear function ofx. ;.g. [ e x dx = [e x + C :. j e 3x + 4 dx = ' + C i.e. (3x + 4> replaces x in tne integral, then (3x + 4) " " " " result, provided we also divide by the coefficient ofx. Similarly, since Icos x dx = sin x + C, then icos (2x + 5) dx = Similarly, f ,„ « ., sin(2x + 5) „ cos(2x + 5) dx = — i-^ 7 + C I sec 2 x S* 1 dx = tan x + C dx = In x + C ■f 2 . , tan 4x _ sec 4x dx = — — + C sin x dx = —cos x + C e x dx = e x + C •■•J __ cosh(3-4x) _ . _ cos 3x sin 3x dx = - — r — + C e«dx=— + C 4 So if a linear function ofx replaces the single x in the standard integral, the same linear function ofx replaces the single x in the result, so long as we also remember to 361 Integration 1 divide by the coefficient of x Now you can do these quite happily — and do not forget the constants of integration! 1 . \{2x - If dx 2. \ cos (7* + 2) dx i (2x)- dx 3.\e 'dx 4. \ sinh Ix dx dx' 7. lsec 2 (3* + l)rf;c 8. f sm(2x-5)dx 9.\ cosh(l +4x)dx 10. f 3 5x dx Finish them all, then move on to frame 10 and check your results. Here are the results: J 2 - 4 8 2.^co S (7* + 2) t fr = Sin(7 * 7 +2 ) + C 3. U 5 * + 4 dx= e -y-+C 4.fsinh7x^ = C 2t^ + C J 4x + 3 4 8.{sin(2x-5)^ = - COS(2 ^" 5) + C 9.[c sh(l +4jC )^ = sinh(1 4 +4 ^ + C 10. 3 5X dx- + C 5 In 3 II Now we can start the next section of the programme. So turn on to frame 11. 362 Programme 13 11 3 . Integrals of the form \ £& dx and I/O) ./'(*) dx . r i a- ^ Consider the integral V 2 _ , dx. This is not one of our standard integrals, so how shall we tackle it? This is an example of a type of integral which is very easy to deal with but which depends largely on how keen your wits are. You will notice that if we differentiate the denominator, we obtain the expression in the numerator. So, let z stand for the denominator, i.e.z = x 2 + 3.x -5 /. — = 2x + 3 .'. dz = (2x + 3)dx dx The given integral can then be written in terms of z. f ( 2x + 3> > dx = [ d _L and we know that f -dz = In z + C J x 2 + 3x - 5 ) z Jz = In z + C If we now put back what z stands for in terms of x, we get (2x + 3) J x 2 + 3x - 5 dx = . 12 J,-^ 5 *-ln(,» + ax-5) + C Any integral, in which the numerator is the differential coefficient of the denominator, will be of the kind ^Y dx = M/C*) } + c - -|^- dx is of the form p since j|(x 3 - 4) = 3x 2 , i.e. the differ- ential coefficient of the denominator appears as the numerator. Therefore, we can say at once, without any further working I 3x2 <2x=ln(x 3 -4) + C x 3 -4 Similarly, f-p^ dx = 2 f^TT^* = 2 ln (* 3 " 4) + C x 2 and 1—5 — - dx = 363 Integration 1 ^^^to-'jW-O + c 13 We shall always get this log form of the result, then, whenever the numerator is the differential coefficient of the denominator, or is a multiple or sub-multiple of it. (I cos X cot xdx = \ - dx and since we know that cos x is the Jsinx differential coefficient of sin x, then fcosx In the same way, I f cos X I cot x dx = I - — dx = In sin x + C J Jsinx J tan x dx = \ dx J cos a; J tan x dx = \ dx = - I d* ] cos x J cos x -In cos x + CJ 14 Whenever we are confronted by an integral in the form of a quotient, our first reaction is to see whether the numerator is the differential coeffi- cient of the denominator. If so, the result is simply the log. of the denominator. [ 4x-8 '■ g -)x 2 -4x+i dx = 364 Programme 13 15 f ^-8 f .2x-4 Here you are: complete the following: ax 21n(x 2 -4;c + 5) + C , . sec x , 1 . I dx = tan* 2 f 2x + . Z -Jx 2 +4x- _ fsinhx 3. 1 — : — ax = J coshx 4 -)x 2 -6x + 2' dx = . 16 Here are the results: check yours. f sec jc 1 . I dx = In tan x + C I tan x " J x 2 + 4x - 1 dx = ln(x 2 + 4je-l) + C 3. I !^LiL <& = in cosh x + C J cosh* I* jc-3 ' J x 2 - 6x + - dx=^\n(x 2 -6x+2) + C Now turn on to frame 1 7. 365 Integration 1 In very much the same way, we sometimes have integrals such as I tan x. sec 2 * dx This, of course, is not a quotient but a product. Nevertheless we notice that one function (soc 2 *) of the product is the differential coefficient of the other function (tan x). If we put z = tan x, then dz = sec 2 * ctx and the integral can then be 17 f z 2 written I z dz which gives y + C ■■■I' , , tan 2 * „ an x. sec * dx = „ + C Here, then, we have a product where one factor is the differential coeffi- cient of the other. We could write it as I tan x. d(tan x) f z 2 This is just like I z dz which gives — + C ,|,a„,.s«A^-J, tun jc tan x . sec 2 * dx = \ tan x . <i(tan *) = — - — + C On to the next frame. 18 Here is another example of the same kind: I sin* .cos* dx = lsin*.<i(sin*) i.e. like \zdz~ —x— + C The only thing you have to spot is that one factor of the product is the differential coefficient of the other, or is some multiple of it. = fln*.rf(ln*) = (1 -^ )2 + C Example 1. \ ^-^- dx = \ In x. - dx 2 T 1 {* sin x 1 Example 2. I -r zdx = I sin" 1 *. 7— ,.<. JV0-* 2 ) J V(i-*) = 1 sin~ 1 *.d(siri" 1 *) .(an 1 *) 2 f ~~ T ~ Example 3. I sinh * . cosh * dx = 366 Programme 13 19 I sinh x . cosh x dx = I sinh x. d(smh x) sinh 2 * Now here is a short revision exercise for you to do. Finish all four and then check your results with those in the next frame. 2x + 3 x 2 + 3x - 7 ' i.\ 2 ^:_\ dx 2.\^^ z dx r cosx J 1 + sin x 2. Ux 2 +7x-4)(2x + l)dx 4 -\x^ 2 - dx Results: Notice that the top is exactly the diff. 20 fc w f 2x + 3 Notice that the top is exact J x 2 + 3x - 7 C oefft. of the bottom, i.e. 1 f 2x + 3 _ [ d{x 2 + 3x-7 = ln(x 2 + 3x-7) + C } J* cosx _ fi i(l + sir " J 1 + sin x J 1 + sin = ln(l +sinx) + C^ sin x) sin* 3. Ux 2 + Ix - 4) (2x + 7) cfcc = Jo 2 + 7x - 4).d(x 2 + Ix - 4) = (x 2 +7x-4) 2 2 , r tf , 4f 3x 7 4 -\x^7 dX= 3)x^ 2 - dx = jln(x 3 -7) + C Always be prepared for these types of integrals. They are often missed, but very easy if y.ou spot them. Now on to the next part of the work that starts in frame 21 . 367 Integration 1 21 4 Integration of products — integration by parts » ■ We often need to integrate a product where either function is not the differential coefficient of the other. For example, in the case of x 2 . In x dx, In x is not the differential coefficient of x 2 x 2 " " " " " " In x so in situations like this, we have to find some other method of dealing with the integral. Let us establish the rule for such cases. If u and v are functions of x, then we know that d dv du dx ( dx dx Now integrate both sides with respect to x. On the left, we get back to the function from which we started. du , dx dx uv = \u j- dx + l v and rearranging the terms, we have f dv , [ du 1 u -- dx = uv - \v — dx J dx J dx On the left-hand side, we have a product of two factors to integrate. One factor is chosen as the function u: the other is thought of as being the differential coefficient of some function v. To find v, of course, we must integrate this particular factor separately. Then, knowing u and v we can substitute in the right-hand side and so complete the routine. You will notice that we finish up with another product to integrate on the end of the line, but, unless we are very unfortunate, this product will be easier to tackle than the original one. This then is the key to the routine: I dV A u — dx = uv — dx du , v — dx dx For convenience, this can be memorized as ■ du \udv = uv—\vt In this form it is easier to remember, but the previous line gives its mean- ing in detail. This method is called integration by parts. 368 Programme 13 22 So \ u — dx - uv — \v — dx J dx J dx i.e. \udv = uv — \ v du \udv = uv — \ v c Copy these results into your record book. You will soon learn them. Now for one or two examples involving integration by parts. Examplel. \x 2 .lnxdx The two factors are x 2 and In x, and we have to decide which to take as u and which as dv. If we choose x 2 to be u and In x to be dv, then we , I In x i shall have to integrate In x in order to find v. Unfortunately, In x dx is not in our basic list of standard integrals, therefore we must allocate u and dv the other way round, i.e. let In x = u and x 2 = dv. x 2 . In x dx = In xl — ) - — 1 x 3 .— dx. Notice that we can tidy up the writing of the second integral by writing the constant factors involved, outside the integral. :. \x 2 hix dx = \nx(—\ ~-\x 3 .-dx = —lnx-^\x 2 dx x 3 ■ 1 * 3 ^ * 3 (, 1 \ r, = 3- to *-3-3 +C= 3 ta *"3 +C Note that if one of the factors of the product to be integrated is a log term, this must be chosen as (u or dv) 23 Example 2. \x 2 e 3x dx Let u - x 2 and dv = e 3x Then I x 2 e 3x dx = x 2 { e -^)-~[e 3x xdx ) x 2 .e 3x 2( (e 3X \ lf 3Xj ). * 2 e 3x 2x e 3x 2 e 3x „ X — I - -\e ix dx) = — = — rr- +-x.-z- +C 3 3\ \ 3/3)1 3 9 9' 3 On to frame 24. 369 Integration 1 In Example 1 we saw that if one of the factors is a log function, that log function must be taken as u. In Example 2 we saw that, provided there is no log term present, the power of x is taken as u. (By the way, this method holds good only for positive whole-number powers of x. For other powers, a different method must be applied.) So which of the two factors should we choose to be u in each of the following cases? , (i) Ix.lnxefo (ii) ix 3 . sinxdx 24 f» In x.lnx dx, u = In x x 3 sin x dx, u=x 3 J 25 Right. Now for a third example. Example 3. \ e 3x sin x dx. Here we have neither a log factor nor a power of x. Let us try putting u = e 3x and dv = sin x. .'. e 3X sin x dx = e 3 *(-cos x) + 3 \ cos x. e 3x dx = -e 3x cos x + 3 I e 3x cos x dx and it looks as though we are back where we started. However, let us write I for the integral I e 3x sin x dx I = -e 3x cos x + 3 e 3x sin x - 9 1 Then, treating this as a simple equation, we get 101 = e 3X (3 sin x - cos x) + Cj e 3x I = jY7 (3 sin x - cos x) + C Whenever we integrate functions of the form e kx sin x or e ** cos x, we get similar types of results after applying the rule twice. Turn on to frame 26. 370 Programme 13 J Jj The three examples we have considered enable us to form a priority order for u: (i) lnx (ii) x n (iii) e kx i.e. If one factor is a log function, that must be taken as V. If there is no log function but a power of x, that becomes '«'. If there is neither a log function nor a power of x, then the exponen- tial function is taken as V. Remembering the priority order will save a lot of false starts. So which would you choose as '«' in the following cases (i) I* 4COS2 ^' U ~~ (ii)l x*e 3x dx, u= (iii) \ x 3 ln(x + 4)dx, u= W j e "c„ ste<ft , .- 27 wJ (i) \ x cos 2xdx, u=x (ii) J x 4 e 3x dx, u=x A (iii) f x 3 ln(jc + 4)dx, u = \n(x + 4) (iv) fe 2X cos4:cdx, « = e 2x Right. Now look at this one. e sx sin 3x dx J' Following our rule for priority for u, in this case, we should put u = 371 / Integration 1 \ e sx sin 3x dx .. u = e Correct. Make a note of that priority list for u in your record book. Then go ahead and determine the integral given above. When you have finished, check your working with that set out in the next frame. 28 e 5x sin 3x dx = — j - sin 3x ~ cos 3x } + C Here is the working. Follow it through. r <l X { cos 3x\ 5 f e sin 3xdx = e cas.lx.e dx e sx cos 3x 29 + \{<"^)-%^>***\ e sx cos 3x 5 ,;v . . 25 T I = r + - e 5X sin 3x - — I 34,e«/5 . , '1 _ q" I — - sm 3x - cos 3x ) + Ci 3e 5 */5 I = -r— ■ j - sin 3x - cos 3* }+ C There you are. Now do these in much the same way. Finish them both before turning on to the next frame. (i) \ x In x dx 00 J*""* 372 V Programme 13 30 31 Solutions: (i) \x\nxdx = ]nx(—j-j\x 2 .-dx x 2 kix 1 , , = ~T~ i} xdx _ x 2 lnx _ 1 x 2 "2 2"2 4 2 {lnx-l} + C 00 [x 3 e 2 *dx=x 3 ( e -^)-||"e 2 *x 2 dx x 3 e 2X 3x 2 e 2x | 3x g 2 * 3g" |C 2 4 4 4 2 !?( 3_3z! 3x_3_' 2 1* 2 + 2 4; That is all there is to it. You can now deal with the integration of products. The next section of the programme begins in frame 31, so turn on now and continue the good work. 5. Integration by partial fractions f x + 1 Suppose we have I 2 _ dx. Clearly this is not one of our standard types, and the numerator is not the differential coefficient of the denominator. So how do we go about this one? In such a case as this, we first of all express the rather cumbersome algebraic fraction in terms of its partial fractions , i.e. a number of simpler algebraic fractions which we shall most likely be able to integrate separately without difficulty. x+1 . . + , a 3 2 can, in fact, be expressed as —^- — -— : $X i z. X Z, X l ■'■ \ ~i — ^ \ dx = — - dx - 1 dx Jx 2 -3x + 2 Jx-2 Jx-1 373 Integration 1 3]n(x-2)-21n(.x-l) + C The method, of course, hinges on one's being able to express the given function in terms of its partial fractions. The rules of partial fractions are as follows: (i) The numerator of the given function must be of lower degree than that of the denominator. If it is not, then first of all divide out by long division. (ii) Factorizethe denominator into its prime factors. This is important, since the factors obtained determine the shape of the partial fractions. (iii) A linear factor (ax + b) gives a partial fraction of the form — — 7 A R (iv) Factors (ax + b) 2 give partial fractions r + , T\2 v ' v /or ax + b (ax + b) A R C (v) Factors (ax + b) 3 give p.f.'s + , p.-, + , tm v ' v > *> v ax + b (ax + b) 2 (ax + b) 3 A v + R (vi) A quadratic factor (ax 2 + bx + c) gives a p.f. — 5 — ; — — v ' ^ v /or- ax j. + j }X + c Copy down this list of rules into your record book for reference. It will be well worth it. Then on to the next frame. Now for some examples Example 1 x + 1 x+ 1 A B ■ + 32 )me examples. f * +i dx 33 • J x 2 - 3x + 2 aX x 2 -3x + 2 (x-l)(x~2) x-l x-2 Multiply both sides by the denominator (x - 1) (x - 2). x+ 1 = A(x-2) + B(;c-l) This is an identity and true for any value of x we like to substitute. Where possible, choose a value of x which will make one of the brackets zero. Let (x — 1) = 0, i.e. substitute x = 1 /. 2 = A(-1) + B(0) :. A =-2 Let (x — 2) = 0, i.e. substitute x = 2 :. 3 = A(0) + B(1) .'. B = 3 So the integral can now be written 374 V Programme 13 34 2 X r> ~^ dx = dx - dx Jx -3x+2 Jx-2 Jx-1 Now the rest is easy. x+1 — — ; — — dx = 3 dx - 2 dx Jx -3x + 2 Jx-2 Jx-1 x'-3x + 2~" "Jx — 2 "Jx-1 3 ln(x-2)-2 ln(x-l) + C (Do not forget the constant of integration!) And now another one. Example 2. To determine dx (x + 1) (x-1) 2 Numerator = 2nd degree; denominator = 3rd degree. Rule 1 is satisfied. Denominator already factorized into its prime factors. Rule 2 is satisfied. x 2 = _A_ JB_ + C (x + 1) (x-1) 2 x + 1 x-1 (x-1) 2 x 2 = A(x - l) 2 + B(x + 1) (x - 1) + C(x + 1) Clear the denominators Put(x-1) = 0, i.e.x= 1 1 = A(0) + B(0) + C(2) • C=i .. c 2 Put (x + 1) = 0, i.e. x = -1 .-. 1 = A(4) + B(0) + C(0) •'. A = ± When the crafty substitution has come to an end, we can find the remain- ing constants (in this case, just B) by equating coefficients. Choose the highest power involved, i.e. x 2 in this example. [x 2 ] .". 1 = A + B .\ B = 1 - A = 1 ■ i T 1 1 (x+i)(x-iy 3 + — . 1 1 4 'x+1 4 'x-1 1_ x-1 ■■ , 7w ^^2 dx =- 1 -— dx + -\ t dx + - (x - J (x+1) (x-1) 2 4Jx + 1 4Jx-l 2J V \_ V(x-1) 2 dx + ^-|(x- l)~ 2 <ix 35 [J (x+1) (x-1) 2 dx = iln(x+l) + 7ln(*-l)- 1 2(x-l) + C Example 3. To determine ■ f * 2 + l Ine J iT&? dx Rules 1 and 2 of partial fractions are satisfied. The next stage is to write down the form of the partial fractions. x 2 + l (x + 2) 3 375 Integration 1 X 2 + 1 = _A_ _B C (jc + 2) 3 x + 2 + (x + 2) 2 (* + 2) 3 36 Now clear the denominators by multiplying both sides by (x + 2) 3 . So we get * 2 + l = jc 2 + 1 = A(x + 2) 2 + B(x + 2) + C 37 We now put (x + 2) = 0, i.e. x = —2 .'. 4 + 1 = A(0) + B(0) + C :. C = 5 There are no other brackets in this identity so we now equate coeffi- ( cients, starting with the highest power involved, i.e. x 2 . What does that give us? x 2 + 1 = A(x + 2) 2 + B(x + 2) + C. C = 5 [x 2 ] :. l=A :. A = 1 We now go to the other extreme and equate the lowest power involved, i.e. the constant terms (or absolute terms) on each side. [C.T.] :. 1 =4A+2B + C .-.1=4 +2B+5 :. 2B = -8 .'. B = -4 x 2 + 1 1 (x + 2) 3 x + 2 (x + 2f (x + 2) 3 ■ f * 2 + l "J + 2) 3 dx = 38 7 - x 2 + l (x + 2T 1 M 5 (x + 2)' 2 + c Now for another example, turn on to frame 40. 39 376 Programme 13 40 Example 4. To find C x 2 ](*-2)(x 2 + 1) dx In this example, we have a quadratic factor which will not factorize any further - x 2 A + Rx + C " (x-2)(x 2 + l) x-2 ' x 2 +1 :. x 2 = A(x 2 + 1) + (je - 2) (Bx + C) Put(x-2) = 0, i.e.* = 2 .-. 4 = A(5) + Equate coefficients [x 2 1 = A + B .\ B = 1 - A = 1 - F [C.T.] = A-2C .'. C = A/2 -5 B = C = 4 1 + 5* + 5 (x-2)(x 2 + l) 5 'x-2 x 2 + l | 2 1 5'x-2 ' 5'x 2 + 1 + 5'x 2 + 1 4 11 x 1 (x~2)(x 2 + l) (ix : 41 i (x - 2) (x- — dx = j ln(x - 2) + - ln(x 2 + 1) + jtan" 1 * + C Here is one for you to do on your own. 1 Example 5. Determine r 4x 2 + Jx(2x- 1) : dx Rules 1 and 2 are satisfied, and the form of the partial fractions will be 4x 2 + l = A _B_ C x(2x-l) 2 x + 2x-l (2x-l) 2 Off you go then. When you have finished it completely, turn on to frame 42. 377 Integration 1 J. 4x 2 + l , . 2 , „ 2 ax = In x - 7 + C x{2x-\f 2x-\ Check through your working in detail. Ax 2 + 1 A , = - + B C x(2x-l) 2 x 2x-\ (2x-l) 2 /. 4x 2 + 1 = A(2x-l) 2 + Bx(2x-l) + Cx Put(2jt-l) = 0,i.e.x = 1/2 :. 2 = A(0) + B(0) + j C = 4 [x 2 ] 4 = 4A + 2B :. 2A + B = 2 \ A= 1 [C.T.] 1=A J B = . 4x 2 + 1 1 4 = - + -. . , '" x(2jc-1) 2 x (2x-l) 2 . .4.(2x-l)T 1 4 . r ]nx + — _ / + C 42 - to *-^-i tc Afove on to frame 43. We have done quite a number of integrals of one type or another in 4«J our work so far. We have covered: 1 . the basic standard integrals, 2. functions of a linear function of x, 3. integrals in which one part is the differential coefficient of the other part, 4. integration by parts, i.e. integration of products, ' 5. integration by partial fractions. Before we finish this part of the programme on integration, let us look particularly at some types of integrals involving trig, functions. So, on we go to frame 44. 378 Programme 13 *X^T 6. Integration of trigonometrical functions (a) Powers of sin x and of cos x (i) We already know that 1 sinxdx = -cos* + C h xdx ' sln ' tc (ii) To integrate sin 2 * and cos 2 *, we express the function in terms of the cosine of the double angle. cos 2x = 1 - 2 sin 2 * and cos 2* = 2 cos 2 * - 1 .'. sin 2 * = - (1 - cos 2x) and cos 2 * = - (1 + cos 2*) .". 1 sin 2 * dx = - \ (1 - cos 2x) dx - - - — - — + C .-. [ cos 2 *d* = M(l+cos2*)tf*=|+ S -^ + C Notice how* nearly alike these two results are. One must be careful to distinguish between them, so make a note of them in your record book for future reference. Then move on to frame 45. _ _ (iii) To integrate sin 3 * and cos 3 *. *f U To integrate sin 3 *, we release one of the factors sin * from the power and convert the remaining sin 2 * into (1 - cos 2 *), thus: I sin 3 * dx = \ sin 2 * .sin * dx = (1 - cos 2 *) sin * dx = I sin * dx - I cos 2 * . sin * dx „ ^ . cos 3 * , n = -COS * + rz- + C We do not normally remember this as a standard result, but we certainly do remember the method by which we can find I sin 3 * dx when necessary. So, in a similar way, you can now find cos 3 * dx. When you have done it, turn on to frame 46. 379 Integration 1 For: I 3 j • sin 3 x cos x dx = sin x — + C : l cos xdx- ^^.[^.co.**,^-*.',)**,*, = \ cos x dx - I sin 2 * . cos x& = sin x =— + C Now what about this one? (iv) To integrate sin 4 * and cos 4 x. jsin^=j(sinV^=J 1—2 cos 2x + cos 2 2jc 2^2 ^ _ I (1 ~ cos 2xf dx f cos 2 x = 5 (1 + cos 2x) dx N.B. cos 2 2x:=^(l + cos 4a:) If 11 = - (1 - 2 cos 2x +- + - cos 4x)dx H- 2 cos 2x + - cos 4x) dx _ 1 (3x _ 2 sin 2x 1 sin 4x | 4\2 2 2' 4 J 8 4 Remember not this result, but the method. Now you find cos 4 jc dx in much the same way 3x sin 2x sin Ax _ 32 f 4 . 3x sin 2x sin Ax „ The working is very much like that of the last example. cos 2.x) 2 ■I°- J cos 4 x dx = I (cos 2 x) 2 dx = ■ dx 2 cos 2x + cos 2 2jc) dx -tf (1 + 2 cos 2x + — + - cos Ax) dx ,i i jl ^ \ j 1 (3* . „ sin Ax] _, + 2 cos 2x + -. cos 4x)c?x = -I - + sin 2x + \+ C On to the next frame 3x sin 2x sin 4x 8 4 32 46 47 380 Programme 13 48 (v) To integrate sin 5 * and cos 5 * We can integrate sin 5 * in very much the same way as we found the integral of sin 3 *. j sin 5 * dx = sin in 4 *.sm*d*=J(l-cos 2 *) 2 sin*d* (1-2 cos 2 * + cos 4 *) sin * dx s^xdx^x.^xdx^x^xdx 2 cos 3 * cos 5 * _, = -COS * + — r — + C Similarly, i 2 = I (1-2 sin 2 * + sin 4 *) cos * dx = sin*- cos * dx — 2 1 sin 2 * . cos * dx + | sin 4 * . cos * dx 2 sin 3 * . sin 5 * 3 + t-" + c 49 Note the method, but do not try to memorize these results. Some- times we need to integrate higher powers of sin * and cos * than those we have considered. In those cases, we make use of a different approach which we shall deal with in due course. (b) Products of sines and cosines Finally, while we are dealing with the integrals of trig, functions, let us consider one further type. Here is an example: I sin 4*. cos 2xdx To determine this, we make use of the identity 2 sin A cos B = sin (A + B) + sin(A - B) .'. sin 4* . cos 2* = ~ (2 sin 4* cos 2*) = 2 sin (4* + 2*) + sin (4* - 2*)| = ~jsin 6* + sin 2* J /. sin 4* cos 2* dx = U(sin 6* + sin 2x)dx = - ?°|6* _ cos2* + c 381 Integration 1 This type of integral means, of course, that you must know your trig, identities. Do they need polishing up? Now is the chance to revise some of them, anyway. There are four identities very like the one we have just used. 2 sin A cos B = sin (A + B) + sin (A - B) 2 cos A sin B = sin (A + B) - sin (A - B) 2 cos A cos B = cos(A + B) + cos(A - B) 2 sin A sin B =^sim(A - B) - cos (A + B) Remember that the compound angles are interchanged in the last line. These are important and very useful, so copy them down into your record book and learn them. Now move to frame 51. Now another example of the same kind Example: 1 cos 5x sin 3x dx 50 51 = z 1(2 cos 5x sin 3x) dx = -\hm(5x + 3x)~ sin(5x- 3x)\dx ■■ -l jsin 8x - sin 2x\dx 1 /' cos 8x cos 2x , _, ^2tT + — J+C cos 2x cos 8jc 16 + C And now here is one for you to do: ;cos6*cos4^ = Off you go. Finish it, then turn on to frame 52. I- 382 Programme 13 52 f, , sin lOx sin 2x _, cos 6x cos 4x dx = — — — + — - — + L For cos 6x cos Ax dx = - 1 2 cos 6x cos 4x dx -if cos 1 Ojc + cos 2x\dx if sin lOx , sin 2x\ 2( + 10 2 sin lOx sin 2x + C 20 + + C Well, there you are. They are all done in the same basic way. Here is one last one for you to do. Take care! i sin 5x sin x dx ■ This will use the last of our four trig, identities, the one in which the compound angles are interchanged, so do not get caught. When you have finished, move on to frame 53. 53 Well, here it is, worked out in detail. Check your result. sin 5x sin x dx : — \2 sin 5x sinx dx = -\lcos(5x - x) - cos(5x + x)\dx = — \ (cos 4x— cos 6x\dx _ 1 f sin Ax _ sin 6x 2{ 4 sin Ax 6 sin 6x 12 + C + C DnDDDDnDDDnDnDnDnannDDDDDnDDaDDDDDaDDa This brings us to the end of Part 1 of the programme on integration, except for the Test Exercise which follows in the next frame. Before you work the exercise, look back through the notes you have made in your record book, and brush up any points on which you are not perfectly clear. When you are ready, turn on to the next frame. 383 Integration 1 Here is the Test Exercise on the work you have been doing in this pro- gramme. The integrals are all quite straightforward so you will have no trouble with them. Take your time: there is no need to hurry — and no extra marks for speed! Test Exercise — XIII Answer all the questions. Determine the following integrals: 54 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. e cos x sm x tf x \nx , —j— ax V* tan 2 * dx x" 1 sin 2x dx e~ 3x cos 2x dx sin 5 * dx cos 4 x dx Ax + 2 dx x 2 +x + 5 *VQ +x 2 )dx 2x-l dx : 2 - 8x + 1 5 ■ 2x 2 + x + 1 (*-l)(x 2 + l) dx 12. I sin 5x cos 3x dx You are now ready to start Part 2 of the programme on integration. 384 Programme 1 3 Further Problems - XIII Determine the following: f 3 X 2 L J (x-l)(x 2 +x + l)^ ,. f sin 2x , 3. I ,— t~ ax J 1 + cos x 5. 1 x sin 2 * dx J ° 7 ' J (x-l)(x 2 +x+l) ^ f 2x 2 +x + 1 y -J(*-l)(x 2 +l) . f"x 2 («- f T/2 Jo C?X 11 x)Pdx,forp>0 13. | sin 5a; cos 3x dx x 2 -2x ■i: (2x+l)(x 2 + l) dx 17-1 a: 2 sin 2 xdx dx 19 f * x 2 ) 21 f 8-x J (x - 2) 2 (x + ■j: n f 2x + 3 J (x-4)(5x + 2) f 5x 2 + 1 lx - 2 J (x + 5)(x 2 +9) „ >/2 . 5 3 , 23. 1 sin x cos x ax 25 . \ sin co? cos 2cot dt 27. sin 7x cos 5x dx •I o 4. f fl/2 xV-x 2 )- 3 / 2 rfx 10 12 14 x 2 sin x dx J(x 2 +x+l) 3 / 2 ^ — — dx |X + 1 frr I (w — x) cos x dx J o f 4x 2 -7x + 13 J(x-2)(x 2 + l)^ (* sin' 1 ; 16 ! : 18- x tan _1 xdx J° 20. j xV(l + x 2 )dx ■I* I" J . tan 2 x f dx ' J \Jx 2 + 4x + I 22. 1 e 2X cos 4x dx i 24. | " * e 2e cos 30 dd 26. | tan 2 x sec 2 x dx 28. x-1 9x 2 -18x+ 17 385 Integration 1 3l -\^~[ dx 32 - [x 2 \n(l+x 2 )dx 3 3 f cosg -sin0 34 r i-sinfl Jcosfl+sinfl J -^FF^ 35 " J (*-l)(**-2)(* + 3) dX 36 - J o (1 + S cos"x) 2 d * 37. | 2 (x-\f]nxdx 38. [ 4 *' ~* + „ J l J *(* + 4 [ x 3 +x + J x 4 +x 2 ■4) 39. | " _*~'J dx 40. If L— + Rj = E, where L, R and E are constants, and it is known that dt i = at t = 0, show that U 2 j: (E/-R/ 2 )A = y TVofe. Some of the integrals above are definite integrals, so here is a reminder. In 1 f(x)dx, the values of a and b are called the //m/'fs of the integral. rb , I f(x)dx, Ja lf\f(x)dx=F(x) + C then[f(x)dx=[F(x)] x = b - [F(x)] } 386 1 Programme 14 INTEGRATION PART 2 Programme 14 1 I. Consider the integral f dZ JZ 2 -A 2 From our work in Part 1 of this programme on integration, you will recognize that the denominator can be factorized and that the function can therefore be expressed in its partial fractions. 1 = 1 _ P , Q Z 2 -A 2 (Z-A)(Z + A) Z-A Z + A where P and Q are constants. .-. 1 =P(Z + A) + Q(Z-A) :. 1 = P(2A) + Q(0) :. P = Put Z = A Put Z = -A 2A 1 = P(0) + Q(-2A) • Q = 2A 1 1 1 1 1 Z 2 -A 2 2AZ-A 2A'Z + A 1 I 1 •'Jz 2 -A 2C?Z "2aJz dZ ~k\zh dZ I: ;j^dZ=i.ln(Z-A)-\.ln(Z + A) + C ■A" 2A' 2A" k--m^ / This is the first of nine standard results which we are going to establish in this programme. They are useful to remember since the standard results will remove the need to work each example in detail, as you will see. 1 We have Ji ■A-' dZ = 2A ta (i~ — 1 + C ; Jz^r6 dZ= jz^ dZ = HzTT} +c (Note that 5 can be written as the square of its own square root.) So jz^ z = 2-Hfrx) + c ® Copy this result into your record book and move on to frame 3. 389 Integration 2 We had So therefore: J dZ .Vln Z 2 -A 2 2A Z + A Z-A + C dZ Z 2 -25 dZ 72 _ 7 r rfz f jj JZ 2 -25 Jz 2 - f_rfz_ r , Jz 2 -7 JZ 2 - 5 2 tfZ 1 /Z-5 + c 277-Hl^ 1+C 4 Z 2 -7 " 1Z 2 -(V7) 2 DDDnnDDDDDDDDDDDDDDnDDDanDDnnnDDDDDDnD Now what about this one? I 1 x 2 + 4x + 2 6?X At first sight, this seems to have little to do with the standard result, or to the examples we have done so far. However, let us re-write the denominator, thus: x 2 + 4x + 2 = x 2 + Ax +2. (Nobody will argue with that!) Now we complete the square with the first two terms, by adding on the square of half the coefficient of x. x 2 + 4x + 2 = x 2 + 4x + 2 2 +2 and of course we must subtract an equal amount, i.e. 4, to keep the identity true . .". x 2 + 4x + 2 = x 2 + 4x + 2 2 + 2 - 4 (jc + 2) 2 -2 So 1 x 2 + 4x + 2 Turn on to frame 5. dx can be written dx 390 Programme 14 J x 2 + Ax + 2 dx j (jc + 2) 2 2 dx Then we can express the constant 2 as the square of its own square root. 1 dx i(x 1 dx x 2 + Ax + 2 ""* j(x + 2) 2 - (V2) 2 . ' You will see that we have re-written the given integral in the form I 7 2 _ .2" dZ where, in this case, Z = (x + 2) and A. = sjl. Now the standard result was h 1 ._ 1 . JZ-A. _ - 2 £?Z = — ln{ - , , } + C Z 2 -A 2 "^ 2A"'|Z + A, Substituting our expressions for Z and A in this result, gives J x 2 + Ax + 2 * C= ](jc + 2) 2 -(V2) 2 dX 2^2 |x + 2 + V2 Once we have found our particular expressions for Z and A, all that remains is to substitute these expressions in the standard result. On now to frame 6. Here is another example. 1 x 2 + 6x + 4 dx First complete the square with the first two terms of the given denominator and subtract an equal amount. x 2 + 6x + A = x 2 + 6x +4 = x 2 + 6x + 3 2 +4-9 = (x + 3) 2 - 5 = (x + 3) 2 - (V5) 2 S ° J* 2 + 6x + 4 d H(x + 3) 2 -(V5) 2CbC 391 Integration 2 7 i x 2 + 6x + 4 dx ■ 1 , JT + 3-V5 2V5 X + 3+V5 nDDnaaDDannnDaDDDDDnnDaDnDnDnnaDDnDDDa And another on your own: Find )x 2 -10x + lo Men you have finished, move on to frame 8. dx J; l 10* + 18 dx = 1 ( X-5-V7 ', , r 2V^ kl U-5 + V7 } + C 8 For: x 2 - 10x + 18 = x 2 -10x +18 = x 2 -10;c + 5 2 + 18-25 = (x-5) 2 -7 = (x-5) 2 -(V7) 2 L 2 -lto + 18 dX = 277 ^f^W?} + C Now on to frame 9. Now what about this one? 'fi 1 dx '5x 2 - 2x - 4 ' In order to complete the square, as we have done before, the coeffi- cient of x must be 1. Therefore, in the denominator, we must first of all take out a factor 5 to reduce the second degree term to a single x 2 . Sx 2 -2x-A aX 5 1 x 2 -\x- 4_ 5 ■ dx Now we can proceed as in the previous examples. ^ ^ 5 X 5 -*'-MiM- 1 25 2i 25 44m 5jc 2 - 2x - 4 dx = (Remember the factor 1/5 in the front!) 392 Programme 14 10 11 1 1 , L , f 5x-l-V21 , . _ 5^ 2 -2^-4 dX ~2V21 |5x-l+V21 f L Here is the working: follow it through. 5x 2 - 2x - 4 dx - (*-iFWJ dx 1 _5_ 1|L [ x-l/S-V21/5 \ S'ly/ll U-1/5+V21/5/ hfcM 5x-l-\/2l + C 2V21 \5x-\ +V21 DDnDnDnDQDnnnnoDDDDDDnnDDannQDnDnDDDnD II. Now, in very much the same way, let us establish the second standard result by considering f dZ Ja 2 -: This looks rather like the last one and can be determined again by partial fractions. Work through it on your own and determine the general result. Then turn on to frame 11 and check your working. \ dZ 1 , IA + Z1 _ F^ = 2A ln A^Z +C For: Put Z = A PutZ=-A 1 1 = = _L_ + _Q_ A 2 -Z 2 (A-Z)(A + Z) A-Z A + Z .'. 1 =P(A + Z) + Q(A-Z) :. P^ ] .-. 1 =P(2A) + Q(0) .'. 1 = P(0) + Q(2A) Ja^Z 2 dZ = TaJaTz dZ + 2A Ja^Z dZ 2A Q = 2A IA 2 -Z 2 -^.ln(A + Z)-^.ln(A-Z) + C 2A' A-Z •00 Copy this second standard form into your record book and compare it with the first result. They are very much alike. Turn to frame 12. 393 Integration 2 So we have: y dZ -A 2 1 2A '»{1t-aH u dZ l -z 2 1 2A ta NH 12 Note how nearly alike these two results are. Now for some examples on the second standard form. Example l.\^dx=)-^ 2 dx=\)n^Yc Example 2. J j^p dx = J y^h^dx = ^ ln{^ Example 3. Ir^ — 2 dx= + C Example 4. \r. 1 . (\/3 + x ) „ 13 6x dx We complete the square in the denominator as before, but we must be careful of the signs — and, do not forget, the coefficient of x 2 must be 1 . So we do it like this: 3 + 6x - x 2 - (x 1 - 6x ) Note that we put the x 2 term and the x term inside brackets with a minus sign outside. Naturally, the 6x becomes - 6x inside the brackets. Now we can complete the square inside the brackets and add on a similar amount outside the brackets (since everything inside the brackets is negative). So 3 + 6x~x 2 = 3-(x 2 -6x + 3 2 ) + 9 = 12-(x-3) 2 = (2V3) 2 -(x-3) 2 In this case, then, A = 2\/3 and Z = (x - 3) 1 . f 1 •'■ j3 + 6x-x 2 ^ Ji (2V3) 2 -(x-3) : dx Finish it off 394 Programme 14 14 1 , (2V3+A--3; ^ 473 ln 2V3-x + 3 ^ + C DQnnanDDaaaaDDDDDDDDDDannnnnoaDDDaaDaa Here is another example of the same type. Example 5. I _ 2 dx First of all we carry out the 'completing the square' routine. 9 - 4x - x 1 = 9 - (x 2 + 4x ) = 9 - (x 2 + 4x + 2 2 ) + 4 = 13-(x+2) 2 = (Vl3) 2 -(x + 2) 2 In this case, A = \j\ 3 and Z = (x + 2) Now we know that I - "^ So that, in this example f— 1 - J9~4x-x 2A' 2 dx '■ „-Aln(^l) + C A-Z 15 ±-m Ml +x + h + c 2\/l3 K/13-X-2 DaDODDDDDaaODDOOaDODDDDDDaDDOnonDDDDDQ 1 Example 6. k dx 4x - 2x 2 Remember that we must first remove the factor 2 from the denominator to reduce the coefficient ofx 2 to 1. '''J 5 + 4x-2x 2 ^ 2J5 + . dx Now we proceed as before. + 2x-x^ | + 2x-x 2 =^-(x 2 -2x ) 5 '2 5 '2 7 '2 1 (x 2 -2x+l 2 )+l (x-1) 2 = (V3-5) 2 -(x-l) 2 . dx = . " J5+4x-2x 2 (Do not forget the factor 2 we took out of the denominator.) 395 Integration 2 i ,JV3!±£zi) + 4V3-5 (V3-5-X+ 1 16 DDDnnDDDaanDnnnDDDnDDanDDannaanDDaDDna Right. Now just one more. Example 7. Determine J 6 - 6x - 5x- dx. What is the first thing to do? Reduce the coefficient of x 2 to 1, i.e. take out a factor 5 from the denominator. Correct. Let us do it then. )6-6x-Sx 2dX 5j6_6 v _ y j 5 5 X dx Now you can complete the square as usual and finish it off. Then move to frame 18. dx 1 K/39 + 5* + 3| + 6-6x-5x 2UX 2V39 \/39-5x-3 For: k- 6x — 5x 2 5 Jf- 6 J-* 2 dx 5 5 5-5*"* 2 4-^ + f- So that A Now ■■ ^y and Z = (x + |) -g-H)' ^ 22 dZ ■>m+ c f 1 rf = I J_ i K/39/5+X + 3/5 ) j6-6jc-x 2fi!X 5'2\/39_ m \V39/5-*-3/5/ Now turn to frame 19. ln K^ + 3 l 2V39 1V39 - 5* - 3 17 18 396 Programme 14 19 By way of revision, cover up your notes and complete the following. Do not work out the integrals in detail; just quote the results. 00 I A dZ Z 2 -A 2 dZ Check your results with frame 20. 20 DDDDnDDDDDDDDnaannDDDDDDDnnnnDDDnnnnnn III. Now for the third standard form. dZ Consider f Z 2 +A 2 Here, the denominator will not factorize,so we cannot apply the rules of partial fractions. We now turn to substitution, i.e. we try to find a sub- stitution for Z which will enable us to write the integral in a form which we already know how to tackle. Suppose we put Z = A tan 9 . Then Z 2 + A 2 = A 2 tan 2 + A 2 = A 2 (l + tan 2 0) = A 2 sec 2 dZ Also The integral now becomes dZ — = A sec 2 i.e. dZ = A sec 2 6 dd do \z?TK* dZ= \ 75 jt .A sec 2 dd A 2 sec^fl ■a dd 1 = T-0 + C A This is a nice simple result, but we cannot leave it like that, for is a variable we introduced in the working. We must express 8 in terms of the original variable Z. Z = A tan 6 , ■'■ -r = tan i A ♦ -i z = tan A I Z 2 +A 2 ^ Z i tan 1ll +c (iu) Add this one to your growing list of standard forms. 397 Integration 2 \ 1 z> + A^ z= i tan ii +c Example 1. -5 — — dx = —z — -= dx = — tan 'f - } + C J x 2 + 16 J x 2 + 4 2 4 |4j £-x«mpte2. ^ x 2 + l Q x + 3Q dx As usual, we complete the square in the denominator x 2 + I0x + 30 = x 2 + \0x + 30 = x 2 + 10x + 5 2 + 30-25 = (x + 5) 2 + 5 = (x + 5) 2 + (V5) 2 •'• J x 2 + 10x + 30 ^ = J (x + 5) 2 + (V5) 2 dX 1 -Jjc + 5 V5" 1 " 7T 21 22 DnnDDDnnDnnDDDnDDDannannDnnnnannDnDnnD Once you know the standard form, you can find the expressions for Z and A in any example and then substitute these in the result. Here you are; do this one on your own: Example 3. Determine ( L_ J2x 2 + 12x + 32 dx Take your time over it. Remember all the rules we have used and then you cannot go wrong. When you have completed it, turn to frame 23 and check your working. 398 Programme 14 23 1 2x 2 + 1 2x + 32 dx 2-Jl tan'(^} + . Check your working. f 1 _l[ \__ J 2x 2 + 12;c + 32 2Jx 2 + 6jc + 12.x + 32""" 2Jx 2 + 6jt+ 16 x 2 + 6x + 16 = x 2 + 6x + 16 = x 2 + 6x + 3 2 + 16-9 dx So Z = (x + 3) and A = \Jl JZ 2 +A 2 . f 1 = (x + 3) 2 + 7 = (x + 3) 2 +(V7) 2 ,z4tan-|l + C " J2x 2 + 12* + 32 Afow move to frame 24. dx '■ k4tan"fe- 3 + 2'^ ian IVT _ _ IV. Let us now consider a different integral 24 r i We clearly cannot employ partial fractions, because of the root sign. So we must find a suitable substitution. Put Z = A sin 6 Then A 2 - Z 2 = A 2 - A 2 sin 2 = A 2 (l - sin 2 0) = A 2 cos 2 V(A 2 -Z 2 ) = Acosfl dZ Also dd = A cos ( dZ = A cos 6. d6 So the integral becomes Ivt^i 2 )^ 1 Acos6.de A cos 6 Expressing 6 in terms of the original variable Z = A sin i . . sin a = — A 1 = \d6- t ™-i z + C (iv) - J^A a -Z') dZ = Silfl lxj +C This is our next standard form, so add it to the list in your record book. Then move on to frame 25. 399 Integration 2 \ V(A 2 -z 2 y dZ = sin ■(!}- Example 1. Example 2. As usual Jvc^M i VC5 2 -* 2 )' dx dx = sin If) + C J V(3-2x-x 2 )' 3~2x-x 2 = 3-(x 2 + 2x ) = 3-(x 2 + 2x + 1 2 )+ 1 = 4-(jc+ l) 2 = 2 2 - (x + l) 2 So, in this case, A = 2 and Z = (x + 1) 1 -dx Similarly, j V(3-2x~x 2 ) " A J V{2 2 - (x + l) 2 } .-if* +11. = sin { —z— + i dx J. sinM -y J + C 1 dx = For: ^5-J-x') dx=sin "T'i 3 ' +c 5 - 4x - x 2 = 5 - (x 2 + 4x ) = 5-(x 2 +4x + 2 2 ) + 4 = 9 - (x + 2) 2 = 3 2 -(x + 2) 2 _ . _ x /x + 2 1 = sin { — ^ — J + C Now this one : 1 V(5-4x:-x 2 ) 2-, dx = sin 3 hne J\ 25 26 Example 4. Determine >,. . _ . - — __ 2 . dx. Before we can complete the square, we must reduce the coefficient of x 2 to 1, i.e. we must divide the expression 14- 12x — 2x 2 by 2, but note that this becomes \/2 when brought outside the root sign. N dx = H. J V(14-12x-2* 2 ) V2jV(7-6x:-x 2 ) Now finish that as in the last example. dx 400 Programme 14 27 For: j^-iL-^vi^i^r)^ i : dx : -If V(14 - 1 2x - 2X 2 ) aX V2 J V(7 - 6x -x 2 ) 7 - 6x - x 2 = 7 - (x 2 + 6x ) = 7 - (x 2 + 6x + 3 2 ) + 9 = 16-(x + 3) 2 dx ■(x + 3) 2 So A = 4 and Z = (* + 3) •"■I I 1 vtA a -z a ) dZ=sin " 1 {i} +c 1 dx = VCH-12X-2JC 2 ) \/2 1 . -Jx + 3\ + C 0% q V. Let us now look at the next standard integral in the same way. To determine I ;,„ 2 - 2 y Again we try to find a convenient substitu- tion for Z, but no trig, substitution converts the function into a form that we can manage. We therefore have to turn to the hyperbolic identities and put Z = A sinh 8 . Then Z 2 + A 2 = A 2 sinh 2 + A 2 = A 2 (sinh 2 + 1) Remember cosh 2 - sinh 2 = 1 ■'. cosh 2 6 = sinh 2 + 1 l 2 Also So /. Z 2 + A 2 = A 2 cosh 2 :. V(Z 2 + A 2 ) = A cosh 6 — = A cosh 6 .'. dZ = A cosh 6 . dd da But JV(Z 2 +A 2 f J sinh -J 1 A cosh 6 .A coshd dd=\de =6 +C i< Z = A sinh 6 .'. sinh # = — .". 6 = sinh A c?Z V(Z 2 +A 2 ) = sinh , ID '(I)- W Copy this result into your record book for future reference. Then J vfr a + 4)' dx = 401 Integration 2 iv^W* =sinlfl {!) + c 29 DnDDnnnDDaDnDDnDDDnDnnnnnDnnoDnaanDDDD Once again, all we have to do is to find the expressions for Z and A in any particular example and substitute in the standard form. Now you can do this one all on your own. 1 dx Determine J ^ + 5x TT2) ' Complete the working: then check with frame 30. Uf+L + nf = sinh_1 { ^M ' + c Here is the working set out in detail: x 2 + 5x+ \2 = x 2 + 5x + 12 :x 2 + 5x+(|) 2 +12 ~4^" So that Z = x + r and A = ^-— - \yjix 2 +.5x+12) ( x+ i) d , = rinh -i|_j +c = smh- 1 {^) + C Now do one more. 30 IV(2x 2 +8x+15) dx : 402 Programme 14 31 Here is the working: I V2 sinh {-7T-] +c ng: 1 *,- x [ l -4 /(2x x : 2 +8*+15) J V2J n /^2 + 4;c + 4x + y = x 2 + 4x + j = x 2 + 4x + 2 2 + 1 ^ = (* + 2) 2 +j dx = (x + 2) 2 +U- So that Z = (jc + 2) and A = / ^ •'•JV(2x 2 +8 1 ,,-ik 2 — 5)^ = V2 Slnh j-jT" 2 - + c Fme. M)w on fo /rame 52. 1 . , ( x + 2)V2 ; - smh -^y-^C Now we will establish another standard result. 32 v,. Consider j^^ The substitution here is to put Z = A cosh 9. Z 2 - A 2 = A 2 cosh 2 - A 2 = A 2 (cosh 2 - 1) = A 2 sinh 2 .-. V(Z 2 -A 2 )= Asinhfl Also Z = A cosh 6 :. dZ = A sinh d d0 ■ f dZ =[-*- ■■ JV(Z 2 -A 2 ) J A sin! JV(Z 2 -A 2 ) Z = A cosh 6 . f dZ sinh0 .Asinh0d0 =\ d e = < + c cosh<?=| :. = cosh- 1 j|| + C JV(Z 2 -A 2 )' = cosh 11} + C .(vi) This makes the sixth standard result we have established. Add it to your list- Then move on to frame 33. 403 Integration 2 Example 1 Example 2. I V(Z 2 -A 2 ) ID 33 = cosh" 1 r + C V^-9)' J\/(* 2 + 6x + dx = cosh" '(!) + c i) dx = You can do that one on your own. The method is the same as before: just complete the square and find out what Z and A are in this case and then substitute in the standard result. J V(* 2+ 6* + - ) ^ = cosh-(U|] + C Here it is: x 2 + 6x + 1 = x 2 + 6x +1 = x 2 + 6x + 3 2 + 1 - 9 = (x + 3) 2 - 8 = (x + 3) 2 - (2V2) 2 So that Z = (x + 3) and A = 2\/2 1 ■'•Jv(* 2 +6x + l) dx= Jv/{(* + dx 34 3) 2 -(2V2) 2 }' = cosh-^} + C Let us now collect together the results we have established so far so that we can compare them. So turn on to frame 35. 404 Programme 14 35 Here are our standard forms so far, with the method indicated in each case. 1 ■ I 72 _ ,2 = o~a ' n J 7 . a I + ^ Partial fractions r dz 1 f z ) 3 " Jz^TA 2= A tan \t) + C PutZ = Atan0 4 - JvlA^) =Sin "{!) +C P"tZ = Asin0 5- X z2+ a ^sinh" 1 !- +C Put Z = A sinh 6 6. J ^ Z 2^ Z A 2x = cosh"' HI + C PutZ = Acosh0 Note that the first three make one group (without square roots). Note that the second three make a group with the square roots in the denominators. You should make an effort to memorize these six results for you will be expected to know them and to be able to quote them and use them in various examples. m g% You will remember that in the programme on hyperbolic functions, J h we obtained the result sinhT 1 * = ln/x + \J(x 2 + 1)} •■■HM7(S*')} Similarly °-^}-F^/ This means that the results of standard integrals 5 and 6 can be expressed either as inverse hyperbolic functions or in log form according to the needs of the exercise. Turn on now to frame 37. 405 Integration 2 The remaining three standard integrals in our list are: Q"7 7. f V(A 2 - Z 2 \dZ 8. f V(Z 2 + A 2 ).dZ 9. j V(Z 2 - A 2 ).dZ In each case, the appropriate substitution is the same as with the corresponding integral in which the same expression occurred in the denominator. i.e. for J y/(A 2 -Z 2 ).dZ put Z = A sin 6 W(Z 2 +A 2 ).dZ " Z = Asinh0 V(Z 2 -A 2 ). dZ " Z = Acosh0 Making these substitutions, gives the following results. jV(A 2 -Z 2 ).,Z = f{s^|) + Z -^l^l>} + C (vh) J V(Z 2 + A>Z = f 2 |sinh- ( I) + Z -^^} + C (viii) JV(Z 2 - A 2 ). dZ = f 2 {W^ 2) - cosh- (f)} + C (ix) These results are more complicated and difficult to remember but the method of using them is much the same as before. Copy them down. Let us see how the first of these results is obtained. V(A 2 -Z?).dZ Put Z = A sin 6 :. A 2 - Z 2 = A 2 - A 2 sin 2 = A 2 (l - sin 2 0) = A 2 cos 2 .". V(A 2 -Z 2 ) = A cos 6 Also dZ = A cos d dB f V(A 2 - Z 2 ). dZ = j A cos 6 . A cos 6 d% = A 2 [cos 2 d9 = A 2 [f + ^V = f^+~ £^} + C Now sin 6 = | and cos 2 = 1 - |* = J ~^ :. cos 6 = ^^L^l The other two ar e proved in a similar manner. Now on to frame 39. 406 38 Programme 14 39 Here is an example yf(x 2 +4x+13).dx First of all complete the square and find Z and A as before. Right. Do that. 40 41 x 2 +4x+ 13=(x + 2) 2 + 3 2 So that, in this case Z=x + 2 and A = 3 This is of the form .-. f V(* 2 + 4* + I3).dx = f V{(* + 2) 2 + 3 2 } .dx So, substituting our expressions for Z and A, we get f V(* 2 + 4x + 13).dx = J V(* 2 + 4, + 13). *c = |(smh-f-f 2 ) + ( ^2)V(^4, + 13) j + c We see then that the use of any of these standard forms merely involves completing the square as we have done on many occasions, finding the expressions for Z and A, and substituting these in the appropriate result. This means that you can now tackle a wide range of integrals which were beyond your ability before you worked through this programme. DDDDDDnnnDDDDDDDDDDnnDnDDDaDDnDnnDanq n Now, by way of revision, without looking at your notes, complete the following: r , f dZ o J dZ (iii) A 2 -Z 2 ' dZ Z 2 +A 2 407 Integration 2 42 And now the second group: dZ V(A 2 -Z 2 ) dZ V(Z 2 + A 2 )" dZ V(Z 2 -A 2 )" dZ V(A 2 -Z 2 )= Sm lAi +C >BU dZ 2 ,-smh 1 {-\+C V(Z 2 + A 2 ) dZ U -JZ\ „ v(z 2 "^) =cosh Ur c 43 You will not have remembered the third group, but here they are again. Take another look at them. I VCZ-AVZ'fP-^^'-cosh-djUc Notice that the square root in the result is the same root as that in the integral in each case. DaDDnnaDnnDDDnnaDnnDDDDDanDDDDnDDDDDDD That ends that particular section of the programme, but there are other integrals that require substitution of some kind, so we will now deal with one or two of these. Turn on to frame 44. 408 Programme 14 44 Integrals of the form J- 1 b sin 2 * + c cos 2 * dx k 1 COS X 2 — dx, which is different from any we Example 1. Consider have had before. It is certainly not one of the standard forms. The key to the method is to substitute t = tan x in the integral. Of course, tan x is not mentioned in the integral, but if tan x = t , we can soon find corresponding expressions for sin x and cos x. Draw a sketch diagram, thus: t Vd + f 2 ) _ _JL Vd + f 2 ) tan x = t . . sin x '■ COS X ... dt Also, since t = tan x, — = sec dx . dx '■x = 1 + tan 2 x = 1 + t 2 1 . , _ dt ..dx— , , 1 +f 2 Then 3 + cos 2 * = 3 + dt 1 + t' 1 _ 3 + 3f 2 + 1 = 4 + 3f 2 1 +r 2 1+r 1+r 8 So the integral now becomes: (* L_ ,, _ f 1 + f 2 <ft j3 + cos 2 x j4 + 3f 2 '1 +f 2 = j4T37 df = 3j4~77 2 df and from what we have done in the earlier part of this programme, this is 45 ljiV'^ tt "-'feH Finally, since t = tan x, we can return to the original variable and obtain f 1 , 1 -,|V3.tanjcl _ — r dx = -JZ tan J v — + C J 3 + cos^ 2-y/3 ( 2 J T«r« f o frame 46. 409 Integration 2 The method is the same for all integrals of the type 1 46 i a + b sin 2 x + c cos 2 * dx In practice, some of the coefficients may be zero and those terms missing from the function. But the routine remains the same. Use the substitution / = tan x. That is all there is to it. From the diagram we get sinx : cosx = sin x : W(i + t 2 ) cosx = Vd + f 2 ) We also have to change the variable. dt t = tan x .'. r 1 = sec 2 * = 1 + tan 2 jc = 1 + t 1 dx 1 dx _ di ~ 1 + t 2 dx = . 47 dx = _ dt l+t 2 48 Armed with these substitutions we can deal with any integral of the present type. This does not give us a standard result, but provides us with a standard method. We will work through another example in the next frame, but first of all, what were those substitutions? sin x = cosx : 410 Programme 14 49 smx y(i7?) cosx yur?) Right. Now for an example Example 2. Determ ,ine j- 1 2 sin 2 * + 4 cos * dx dt Using the substitution above, and that dx = ~~~.\ , we have „ • 2 , 2 2f 2 ,4 2 f 2 +4 2 sm 2 * + 4 cos * = — 2 + — 2 = y^ r i _ f i + 1 2 jt_ " J 2 sin 2 * + 4 cos 2 * J It 2 + 4 ' 1 + t = 2-R-2 * 50 272-14^ and since t = tan *, we can return to the original variable, so that 1 I 2 sin 2 * + 4 cos 2 * * = 2>-f7rH Now here is one for you to do on your own. Remember the substitutions: t = tan * sin * cos* : t dx Vd+/ 2 ) 1 V(i+f 2 ) = dt V0 + ' 2 ) Right, then here it is: Example 3. 2 cos 2 * + 1 dx = Work it right through to the end and then check your result and your working with that in the next frame. 411 Integration 2 \T^kTi dx = ^ t:in ~ l t^f} +c Here is the working: •-I 2 cos 2 * + 1 = 2 " l+t 2 3+t 2 'l+t 2 1 dx = f 1 +r 2 cos 2 x + 1 J 3 + r + 1 = 2 + l + r l +r 2 -J 1 j 1, -,/ (\ 3T? df = V3 tan WSJ + c 51 1 -Jtanx) So whenever we have an integral of this type, with sin 2 * and/or cos 2 * in the denominator, the key to the whole business is to make the substitu- tion t = Let us now consider the integral t = tan * ~,-„i 1 1 ■ 4 cos * dx This is clearly not one of the last type, for the trig, function in the denominator is cos* and not cos 2 *. In fact, this is an example of a further group of integrals that we are going to cover in this programme. In general they are of the form 1 dx, i.e. sines and cosines in the denominator but not f_ J a + b sin * + c cos * squared. So turn on to frame 53 and we will start to find out something about these integrals. 52 412 53 Programme 14 Integrals of the type J- 1 b sin x + c cos x dx The key this time is to substitute t = tan j X X From this, we can find corresponding expressions for sin -~ and cos x from a simple diagram as before, but it also means that we must express sin x and cos x in terms of the trig, ratios of the half -angle — so it will entail a little more work, but only a little, so do not give up. It is a lot easier than it sounds. First of all let us establish the substitutions in detail. t sin 2V(l+r 2 ) x _ . . cos ~ = 2 V(l+/ 2 ) sin x = 2 sin § cos f = 2.^.^-^ ^ cos x = cos — — sin — : , x 2 ,x r 2 1 f- l + r l + 1 2 1-f 2 1 + f 2 Also, since f = tan f , f- = \ sec 2 f = kl + tan 2 £) 2' dx rff : 2 1+f 2 2 2 1+f 2 2 2 V So we have: If t = tan ~ dx sin x _ 2rff ~l+f 2 2f 1+f 2 cos* : dx = 1-f 2 1+f 2 2<ft 1+f 2 It is worth remembering these substitutions for use in examples. So copy them down into your record book for future reference. Then we shall be ready to use them. On to frame 54. 413 Integration 2 f dx J 5 + 4 cos x Example 1 Using the substitution t = tan - we have (l-t 2 ) 5+4cosx = 5+4 v : -J 5 + 5t 2 + 4 - 4r 2 1 +? 2 9 + r l+r 2 • f ^ = f 1 + f 2 2<ft =2 [ dt ■'J5 + 4cosx J 9 + t 1 ",1 + f 2 J9 + / 2 Here is another . Example 2. |— — J3sinx + 4cosx x Using the substitution ? = tan— 3 sin x + 4 cos x : 6r , 4(l-f') l +? 2 TTF 4 + 6/ - 4f 2 1 + f 2 f <fr f ltt ; 2A " j3sinx + 4cosx J 4 + 6? -4f 2 '1 + f 2 }2 + 3t-2t 2 2 h+h-t 2 dt 54 55 Now complete the square in the denominator as we were doing earlier in the programme and finish it off. Then on to frame 56. 414 Programme 14 56 I f 1 + 2 tan x/2 1 5 -Jn \4-2 tan x/2 j For l+^-r 2 = l-<> 2 = l-(f 2 25 :f + ) 2x 9 (irw M-(-ir (!f-B)* Integral "it 1,(1+2*1 ^ 1 , / 1 + 2 tan x/2 5 ( 4 + 2 tan x/2 1 And here is one more for you, all on your own. Finish it: then check your working with that in the next frame. Here it is. 1 Example 3. ■■JiT sin x — cos x dx = 57 InL-jSL^CC 1 + tan x/2 Here is the working. 1 + sin x - cos x - 1 + , , - : i - 1 + ? 2 + 2f - 1 + r 2 = 2(r 2 + t) 1 +t' l+f T = f l+{2 2dt -f 1 1, J2(r 2 + f)'i+f 2 )t 2 +t dt t l + 1) dt 1 + H 1 1 + tan x/2J 415 Integration 2 You have now reached the end of this programme except for the Test Exercise which follows. Before you work through the questions, brush up any parts of the programme about which you are not perfectly clear. Look back through the programme if you want to do so. There is no hurry. Your success is all that matters. When you are ready, work all the questions in the Test Exercise. The integrals in the Test are just like those we have been doing in the pro- gramme, so you will find them quite straightforward. Test Exercise — XIV Determine the following: 58 10. l dx V(49-x 2 )' dx x 1 + 3x - 5 dx 2x 2 +8x + 9 1 V(3x 2 + 16) dX dx 9-Sx-x 2 sj{\-x-x 2 ).dx dx V(5x 2 + 10;c-16) dx 1 + 2 sin 2 * dx 2 cos x + 3 sin x sec x dx You are now ready for your next programme. Well done! 416 Programme 14 Further Problems - XIV Determine the following: 1. 13 15 29 (Put x = a tan d) - dx o r dx 9 r dx J x 2 + I2x+ 15 J 8-12*-* 3 f dx f x-8 J x 2 + 14* + 60 Jx 2 +4;t + l ■ J V(* 2 + 12x + 48) J V(17-14x -x 2 ) 7 f *c o f 6x-5 J V(* 2 + 16* + 36) J V0c 2 -12jc + 52) aJC 9 f ^ 10 f" /2 ^ J 2 + cos jc J o 4 sin 2 * + 5 cos 2 x ii f dx .. f 3x 3 - 4x 2 + 3x ., J x 2 + 5x + 5 J x 1 + 1 4 dx f ^* ., (* dx J vtx 2 -4x-21) J 4 sin 2 * + 9 cos 2 * i7. f - . d \ *-V /r- J 3 sin x - 4 cos x J V 2 —x dx (Put x = 2 sin 2 0) 19. f rf^Kdx 20. fr^^-dx J v(l~* ) J 2- cosx f x 2 -x + 14 f dx J (x + 2)(x 2 + 4) * J 5 + 4 cos 2 x 23 -Jv#^)* 24 "jv(2x 2 -t + 5) 9 c f 4 dx f _ . dfl J t V{(* + 2) (4-x)} i6 ' J 2 sin 2 - cos 2 27 -lv(?^^)^ 28.jVd5-2x-x 2 d x) f a dx in f a 2 dx J (« 2 + x 2 ) 2 J (x + a)(x 2 + 2a 2 ) 417 Programme 15 REDUCTION FORMULAE Programme 15 Iln an earlier programme on integration, we dealt with the method of integration by parts, and you have had plenty of practice in that since that time. You remember that it can be stated thus: I u dv = u v - \v du So just to refresh your memory, do this one to start with. [x 2 e x dx = When you have finished, move on to frame 2. J" x 2 e x dx=e x [x 2 -2x + 2] +C Here is the working, so that you can check your solution. { X 2 e x dx=x 2 (e x )-2[e x xdx = x 2 e x ~2[x(e x )-^e x dx] = x 2 e x ~2xe x + 2e x +C = e x [x 2 - 2x + 2] + C On to frame 3. Now let us try the same thing with this one — J f x n e x dx=x n (e*) - n je x x"' 1 dx = x n e x -n[e x x"- l dx Now you will see that the integral on the right, i.e. e* x n ~ l dx, is of exactly the same form as the one we started with, i.e. I e* x" dx, except for the fact that n has now been replaced by (n~l) Then, if we denote I x" e* dx by I„ we can denote [x"' 1 e x dx by \ n _ t So our result can be written [x n e x dx=x n e x -n fc* x"' 1 dx l n =x n e x - Then on to frame 4. 419 Reduction Formulae tl fl — 1 This relationship is called a reduction formula since it expresses an integral in n in terms of the same integral in (n-\). Here it is again. If I„=jx" 'e*dx then I„ =x n ex - n.l„ *n Make a note of this result in your record book, since we shall be using it in the examples that follow. Then to frame 5. Example Consider \x 2 e* dx This is, of course, the case of I„ = \x n e* dx in which n = 2. We know that I„ = x" e* - n I„. x applies to this integral, so , putting n = 2, we get I 2 =;c 2 e*-2.Ii and then Ii =x 1 e* - l.I Now we can easily evaluate I in the normal manner - l = \x°e x dx = \\e x dx = \e x dx = e x +C So l 2 =x 2 e x -2.li and Ii =xe x -e* + Q :. \ 2 =x 2 e x -2xe x +2e x +C = e x [x 2 -2x + 2]+C And that is it. Once you have established the reduction formula for a particular type of integral, its use is very simple. In just the same way, using the same reduction formula, determine the integral \x i e x dx. Then check with the next frame. 420 Programme 15 I x 3 e x dx = e x [x 3 -3x 2 +6x-6]+C Here is the working. Check yours. I„ = x" e x - n l n _ ! n = 3 n = 2 n= 1 I 3 = x 3 e*-3.I 2 I 2 = x 2 e* -2.1 l and In Ii =xe x - l.I 0= \x li e x dx = \e x dx = e x + C :. l3=* 3 e*-3.i 2 = x 3 e x - 3x 2 e x +6.1 1 = jc 3 e*-3x 2 e* + 6xe*-6e* +C = e* [x 3 - 3jc 2 + 6x - 6] + C Now move on to frame 7. 1 Let: us now find a reduction formula for the integral I x" cos x dx. l n = \x n cos x dx = x"(sinx)-n \smxx"' 1 dx = x" sin* -« at"' 1 sinxdx. Note that this is wor a reduction formula yet, since the integral on the right is not of the same form as that of the original integral. So let us apply the integration-by-parts routine again. :-n [x"- 1 sinx dx ■ x sin x — n . 8 I„ = x n sin x + n x" 1 cos x - n («-l) j x" 2 cos x dx Now you will see that the integral x"~ 2 cos x dx is the same as the integral jVcos, dx, with n replaced by 421 Reduction Formulae n-2 9 i.e. I„ = x n sin x + n x" 1 cos x - n(n - 1) I„_ 2 So this is the reduction formula for I„ = \x" cos x dx Copy the result down in your record book and then use it to find J x 2 cos x dx. First of all, put n = 2 in the result, which then gives I 2 -x sin* + 2x cos x-2A. I 10 Now And so 1,-jx^osxdx-joosxdx-sinx.C, Ii = x 2 sin x + 2x cos x - 2 sin x + C Now you know what it is all about, how about this one? Find a reduction formula for jc" sin x dx. Apply the integration-by-parts routine: it is very much like the last one. When you have finished, move on to frame 1 1. \ n = —x" cos x + nx" sin x - n(n-\) l n 11 For: I sin x dx n =Jx"si = x"(-cos;c) + h cos**"" 1 dx = -x" cos x + nix" 1 (sin x) - (n - 1) sin x x"~ 2 dx \ .'. I„ = -x" cos x + n x"' 1 sin x - n(n - 1) I„. 2 Make a note of the result, and then let us find x 3 sin x dx. Putting n = 3, I 3 = -x 3 cos x + 3 x 2 sin x - 3.2. Ii and then Ii = \x sin x dx Find this and then finish the result - then on to frame 12. 422 Programme 15 12 li = -x cos x + sin x + C So that I 3 = -x cos x + 3 x sin x - 6 1 1 I3 I 3 = -x 3 cos x + 3 x 2 sin x + 6x cos x — 6 sin x + C Note that a reduction formula can be repeated until the value of n decreases to n = 1 or n = 0, when the final integral is determined by normal methods. Now move on to frame 13 for the next section of the work. I J Let us now see what complications there are when the integral has limits. Example. To determine I x n cos x dx. J Now we have already established that, if I„ = x n cos x dx, then I„ = x" sin x + n x"' 1 cos x - n(n - 1) l n _ 2 If we now define I„ = x" cos x dx, all we have to do is to apply the J limits to the calculated terms on the right-hand side of our result I* = x" -n(«-l)I„_ = [0 + n tt"- 1 (-1)] -[0 + 0] -n(n - 1) I„_ 2 .-. l n =-n^' l -n(n-l)l n _ 2 This, of course, often simplifies the reduction formula and is much quicker than obtaining the complete general result and then having to substitute the limits. Use the result above to evaluate First put n = 4, giving I 4 = I X cosx J dx. 14 I 4 = -4tt 3 - 4.3. 1, Now put n ~ 2 to find I 2 , which is I 2 = 423 Reduction Formulae I 2 =-2.7r-2.1.I and I = So we have and o = x° cos x dx = Jo Jo cos x dx - | sin jc L = I 4 = -4ir 3 — 12 I 2 I 2 =-2n U = 15 x 4 cos x dx = I 4 = -47r 3 + 247r Now here is one for you to do in very much the same way. Evaluate x s cos x dx. J Work it right through and then check your working with frame 1 7. Working: and I s = -5tt 4 + 60tt 2 - 240 I, l„=-nir n - l -n(n-l)l n _ 2 :. I 5 =-5tt 4 -5.4. I 3 I 3 =-3tt 2 -3.2. I x (•7T I - n ("it I x cos x dx = x(sin jc) — I sin x r L n Jo J = [0-0] cos^: -cos x : (ix = (-!)-(!) = -2 :. I s =-5tt 4 -2OI3 I 3 = -3tt 2 - 6(-2) :. I, = -5tt 4 + 60tt 2 - 240 Turn on to frame 18. 16 17 424 Programme 15 "xdx. 1 8 Reduction f or mulae for (i) sin"* dx and (ii) cos n x (i) sin"* dx. Let I„ = J sin"* dx = I sin" -1 * . sin * d* = sin" -1 * . d(-cos x) Then, integration by parts, gives I„ = sin" -1 * . (-cos x) + (n - 1 ) j cos x . sin"" 2 * . cos x dx = -sin" -1 * . cos x + (n - 1) f cos 2 * . sin" -2 * dx = -sin"" 1 * . cos * + (« - 1) f ( 1 - sin 2 *) sin" -2 * dx = -sin" _1 *.cos* + (n- l)|fsin"- 2 *d*- | sin"*d*| .-. I„ = -sin"" 1 * . cos * + (n - 1) I„_ 2 - (« - 1) I„ Now bring the last term over to the left-hand side, and we have nl n = -sin" -1 * . cos * + (n - 1) I So finally, if /„ = f sin"* <&r, I„ = -i sin"" 1 * . cos * + ^ I„_ 2 Make a note of this result, and then use it to find j sin 6 * dx 19 I 6 --?sin s *.cos*-^sin 3 *..cos*- T 2 sin* cos* + f£ + C 5* 16 16 For _ 1 • s 16 - — A sin * tu!i - , c T T' • *4- 16 =-?sin 5 * cos*+-r-. I 4 U = -4 sin 3 * cos * + 7 . I 2 • I2 =_ 2 ^^ cos * + 2 ■ !<>■ Io = jdx = x + C • T 1-5 J .. I 6 = -gSirr* COS* + , — j-sin 3 * cos * + -T-. Io = -7 sin 5 * cos * _ 24 ^n 3 * cos * + "e 1 . * -~ sin* cos* + :r + C = -g sin 5 * cos * -24 sin 3 * cos * - j2 sin * cos * + ~ + C 425 Reduction Formulae 20 (ii) I co^x dx. Let I„ = I cos"* dx = I cos" -1 * cos* dx = I cos" -1 * d(sinx) = cos" -1 *, sin x - (n - 1) J sin x . cos"" 2 * (-sin x) dx = cos" -1 *. sin x + (n - 1) I sin 2 * cos"" 2 * dx = cos"" 1 *, sin* + (« — 1) I (1 - cos 2 *) cos"" 2 * dx Now finish it off, so that L, = I„ =— cos *.sin*+ . " n n n-2 For I„ = cos" l x . sin * + (« - 1) I„_ 2 - (« - 1) I„ n l n = cos"" 1 *. sin* + (« - 1) l n _ 2 . 1 n-\ . n-l , .. I„ = — cos '*.sin* + l„_, " n n " 2 Add this result to your list and then apply it to find I cos 5 * dx When you have finished it, move to frame 22. \ 1 A rt cos 5 * dx = —cos 4 * sin * + tt cos 2 * sin * + -rr sin * + C Here it is: I 5 =— cos 4 * sin* +— 1 3 21 22 I 3 =— cos z * sin* +-li And I x = cos*dx = sin* + Ci .'. I 5 =— cos x sin* + — 1 2 • J. ' — cos * sin * +~ sin * + C On to frame 23. =— cos 4 * sin* + — cos 2 * sin* + — sin* + C 426 Programme 15 £m U The integrals I sin"x dx and cos"x dx with limits x = and x = tt/2, give some interesting and useful results. We already know the reduction formula I sin"xdx = I„ = — sin"' 1 x . cos x + 1„ , Inserting the limits In 1 --sin" 'x cosx r."- n tt/2 w-1 T 1 M J o = [ -o] + ^iL "-2 I =-^-1 n-2 And if y'ou do the same with the reduction formula for I cos^x dx, you get exactly the same result. _ r ,*W2 . „ . . rW2 So for /•t/2 f* Ji : sin x dx and I Jo Jo dx and | cos n xdx, we have «-l- I„ I„ Also — (i) If n is even, the formula eventually reduces to I rn/2 tt/2 i.e. ldx = [x] = 7T/2 :. I = 7r/2 Jo o (ii) If n is odd, the formula eventually reduces to Ii /•w/2 r ,jt/2 i.e. sinx dx = l-cosxj = _ (~1) •'■ Ii - 1 Jo o ("n/2 So now, all on your own, evaluate snrx dx. What do you get? ■J. 24 i =1 1 i = - ls - 5 ■ 3 • 1 - , c 15 For i5- ? .u and we know that Ii = 1 15 5-3" 1 15 •jt/2 In the same way, find I cos x dx. J Then to frame 25. 427 Reduction Formulae T _ J« 16 ~ — 5jr 32 25 For I 6 ~.I 4 u = -.I andl =y 5 3 1 n 5-n 6 ' 4 " 2 ' 2 32 Note that all the natural numbers from n down to 1 appear alternately on the bottom or top of the expression. In fact , if we start writing the numbers with the value of n on the bottom, we can obtain the result with very little working. (n-1) (n-3) (n-5) n (n- 2) (« - 4) If n is odd, the factors end with 1 on the bottom 6.4.2 ■ etc. e-g- and that is all there is to it. 7.5.3.1 If n is even, the factor 1 comes on top and then we add the factor ir/2 e-g- 7.5.3.1 2 .6.4.2 So (i) sin 4 * dx - cos 5 a: dx = and ( »f sin 4 * dx = — , \ cos 5 x dx- 15 26 This result for evaluating sin"x dx or cos"x dx between the limits x = and x = 7r/2, is known as Wallis's formula. It is well worth remembering, so make a few notes on it. Then onto frame 27 for a further example. 428 27 Here is another example on the same theme. Example. Evaluate I sin 5 * cos 2 x dx. •* o We can write Programme 15 .1,12 rn/2 I sin 5 * cos 2 * dx = I sin s *(l - sin x) dx Jo J o = I (sin 5 x - sin x) dx = Is -I, Finish it off. 28 105 4.2 = J5_ 6.4.2 = 16 5 5.3.1 15 ' 7 7.5.3.1 35 ■" 5 7 15 35 105 29 All that now remains is the Test Exercise. The examples are all very straightforward and should cause no difficulty. Before you work the exercise, look back through your notes and revise any points on which you are not absolutely certain: there should not be many. On then to frame 30. 429 Reduction Formulae Test Exercise— XV Work through all the questions. Take your time over the exercise: there are no prizes for speed! Here they are then. 1 . If I„ = \x n e 2x dx, show that 30 l. If I„ = [*" _ x n e 2x n l " 2 2 ' "- 1 and hence evaluate (x 3 e 2x dx. J * nil r W2 2. Evaluate (i) I sin 2 * cos 6 * dx •' o r nil (ii) I sin 4 jc cos 5 * dx J o 3. By the substitution x = a sin 6, determine f %V - J o .2> 3/2 x 2 ) dx 4. By writing tan"* as tan" 2 *. (sec 2 * — 1), obtain a reduction formula for tan"* dx. • tt/4 J Hence show that \„ = tan"* dx = r - I „ 0*' 5 . By the substitution * = sin 2 8 , determine a reduction formula for the integral J* 5/2 (l-*) 3 ^* Hence evaluate J'- 5/2,, \3/l , ( 1 - *) dx 430 Programme 15 Further Problems- XV /•f/2 1 . If I„ = I x cos"x dx, when n > 1 , show that J _ «(» - 1) _ x " „2 l n-2 i 2. Establish a reduction formula for I sin"x dx in the form r sin"x < T 1-72-1 . n ~ 1 T I„ = - — sin x cos x + 1 n and hence determine I sin 7 x dx. n-2 Yl e ax dx, show that I„ = — . I„ , . Hence evaluate a n ' 1 tine J 3. If I„ = [ x n Jo f x 9 e" 2 *dx 4. If I„ = f" e - * sin"x dx, show that I„ = "^" 7 ^ I J o n v,2 n 2 + 1 "-2" 5. If I„ = I x" sinx dx, prove that, for« ^>2, J o Hence evaluate I 3 and I 4 . 6. If I„ = x" e* dx, obtain a reduction formula for I„ in terms of I n _ 1 and hence determine x 4 e* dx. 7. If I = sec"x dx, prove that 1 n -2 I„ = , tan x sec" 2 x + -L, , (« > 2) n - 1 n - 1 ""2 v r »7r/6 Hence evaluate | sec 8 x dx. 431 Reduction Formulae ir/2 8. If I„ = I e~ x cos n x dx, where n >2, prove that J (i) I„ = 1 - n \ e x sin x cos" *x dx /I (ii) (n 2 + 1) I„ = 1 + «(« - 1) I„_ 2 cv. ♦», ♦ i - 263-144e-" /2 Show that I 6 = 7^- ozy 9. If I„ = l(x 2 + a 2 )" dx, show that 10. If I„ = I cot"x dx,(n>\), show that cot"" 1 * _ j l "~ (n - 1) "" 2 Hence determine I 6 . 11. If I„ = (lnx)" dx, show that I„=x(lnx)"-«.I„_ 1 Hence find I (In x) 3 dx. 12. If I„ = I cosh"x dx, prove that I„ =— cosh" _1 x sinhx + I n _ 2 Hence evaluate I cosh 3 x dx, where a = cosh" 1 (\/2). J 432 Programme 16 INTEGRATION APPLICATIONS PART1 Programme 16 1 fte). We now look at some of the applica- tions to which integration can be put. Some you already know from earlier work: others will be new to you. So let us start with one you first met long ago. Areas under curves To find the area bounded by the curve y = fix), the x-axis and the ordinates at x = a and x = b There is, of course, no mensuration formula for this, since its shape depends on the function /(x). Do you remember how you established the method for finding this area? Move on to frame 2. y = Hx). Let us revise this, for the same principles are applied in many other cases. Let P(x, y) be a point on the curve y -J\x) and let A x denote the area under the curve measured from some point away to the left of the diagram. The point Q, near to P, will have co-ordinates (x + bx, y + 8y) and the area is increased by the extent of the shaded strip. Denote this by§A x . If we 'square off the strip at the level of P, then we can say that the area of the strip is approximately equal to that of the rectangle (omitting PQR). i.e. area of strip = 6 A x — Turn to frame 3. 435 Integration Applications 1 SA X n^ydx Therefore, -r-^ &x -y i.e. the total area of the strip divided by the width, 8x, of the strip gives approximately the value y. The area above the rectangle represents the error in our stated approximation, but if we reduce the width of the strips, the total error is very much reduced. Sx- ff we continue this process and make 8x -> 0, then in the end the error will vanish, and, at the same time,-r-^ ox dA x 5A Y dA r Sx dx 4 Correct. So we have-j-^ =y (no longer an approximation) Y UX y=f{x) ■ A x =J y dx =jf{x) dx A x = F(x) + C and this represents the area under the curve up to the point P. Note that, as it stands, this result would not give us a numerical value for the area, because we do not know from what point the measurement of the area began (somewhere off to the left of the figure). Nevertheless, we can make good use of the result, so turn on now to frame 5. 436 Programme 1 6 ■ \y dx gives the area up to the point ?(x, y). J Y y-K*L So: (i) If we substitute x = b, we have the area up to the point L ■I' i.e. Aj = I y dx with x = b. (ii) If we substitute x = a, we have the area up to the point K ■f i.e. Aa = \y dx with x = a. If we now subtract the second result from the first, we have the area Y y=f(x) under the curve between the ordinates at x = a and x = b. i.e.h=\ydx _ -\ydx , This is written and the boundary values a and b are called the limits of the integral. Remember: the higher limit goes at the top. ] _, , ,, , ,. .. A .. , ^ } That seems logical, the lower limit goes at the bottom. I ° So, the area under the curves = fix) between x = 1 and x = 5 is written -5 A = | y dx. <\\ Si is milarly, the written A = area under the curves = f(x) between x = -5 and x = On to frame -1 6. 437 Integration Applications 1 ■0 y dx Let us do a simple example. Find the area under the curves = x 2 + 2x + 1 between x = 1 and x = 2. A = f y dx = j (x 2 + 2x + 1) dx -\ *** + x + ( 2 1 = ^+4+2+C - Y+ 1 + 1 + c (putting x = 2) (putting x = 1 ) = 8§+C - 4 +c = 6v units 2 A/ofe: When we have limits to substitute, the constant of integration appears in each bracket and will therefore always disappear. In practice therefore, we may leave out the constant of integration when we have limits, since we know it will always vanish in the next line of working. Now you do this one: Find the area under the curve y = 3x 2 + Ax - 5 between x = 1 and jc = 3. Then move on to frame 7. A = 32 units 2 For r> X 2 + 4x - 5)d> c = x 3 + Z K 2 -5x 27+ 18-15 - 1 + 2-5 30 - -2 = 32 units 2 Definite integrals An integral with limits is called a definite integral. With a definite integral, the constant of integration may be omitted, not because it is not there, but because On to frame 8. 438 Programme 1 6 8 ...it occurs in both brackets and disappears in subsequent working. So, to evaluate a definite integral (i) Integrate the function (omitting the constant of integration) and enclose within square brackets with the limits at the right-hand end. (ii) Substitute the upper limit, (iii) Substitute the lower limit, (iv) Subtract the second result from the first result. j: y dx- F(*) F(6)-F(a) Now, you evaluate this one. \ 4e 2 * dx~ Here it is: 5-166 I 4e 2x dx = 4 = 2 = 2 e-e -i-i-- 2 -1-1 1 e-~2 e = 5-166 Now, what about this one: :dx. rn/2 2: I xcosxi ^ 0- First of all, forget about the limits. I jc cosxdx = When you have done that part, turn to frame 10. 439 Integration Applications 1 = x sinx + cosx + C ,«7r/2 ■'■}' x cos x dx = x sinx + cosx 77/2 You finish it off. 10 !-> , jr/2 for | x cos x dx = f J x sin x + cos x IT /2 [i*°. - 0+ 1 J -■§-' If you can integrate the given function, the rest is easy. So move to the next frame and work one or two on your own. Exercise Evaluate: (1) f (2x - 3) 4 dx (4) x 2 In x rfx When you have finished them all, check your results with the solutions given in the next frame. 11 12 440 Programme 16 13 Solutions (1) (2) (3) J 2 (2x-3) 4 dx = V* 3)5 1 = — (rn 5 -(-n s 10 j l 10 ( l ; *• u '■(.)-(-.)} 44 5 — 10 ln(;c + 5) Jo 10 f 3 dx = lnl0-ln5 = lny=ln2 ! .13 , 3 tan " 3J (tan" 1 ^-(tan" 1 [-1]) ~JL— (--'Tl - — _4 \ 4 'J "i. (4) x 2 lnxcfx = lnx(— )-— \x 3 — dx x 3 \nx x 3 + C x 2 Inx dx ■ 1 On to frame 14. x 3 lnx _xj 9 3 „3 „3 (i-?)-(«-5) Win very many practical applications we shall be using definite integrals, so let us practise a few more. Do these: 'Jo ( 5 ) I , ^ m „2, dx 2 1 + cos X (6) I ^ G?X (7) x 2 sin x , J Finish them off and then check with the next frame. ■ dx 441 Integration Applications 1 Solutions ,„ r' 2 sin2x , ( 5 ) i ^ 2 dx = J 1 + cos 2 x -In (1 + cos 2 *) tt/2 = -In (1 + 0) - -ln(l + 1) = -In 1 + In 2 = ln2 (<5) \**«-*^-\** = xe x -e x + C ■■•/:■ e* dx e*(*-l)] = e 2 - = e^ (7) x 2 sin x dx = x 2 (-cos x) + 2 1 cos x dx = -x 2 cos x + 2 | x(sin x) - J sin x dx = -x 2 cos x + 2x sin x + 2 cos x + C x 2 sin x dx = (2-x 2 ) cosx + 2x sinx (2-7T 2 )(-l) + 7T 2 -2~2 = 7r 2 -4 2 + JVow move on to frame 16. 15 Before we move on to the next piece of work, here is just one more example for you to do on areas. Example. Find the area bounded by the curve y -x 2 - 6x + 5, the x-axis, and the ordinates at x = 1 and x = 3. Work it through and then turn on to frame 1 7. 16 442 Programme 16 17 A = — 5 =- units 2 Here is the working: r3 A = y dx = (x 2 - 6x + 5) dx ■■ ^ - 3x 2 + 5x = (9-27+15)-(j-3 + 5) = (-3)-(2|) = -5junits 2 If you are concerned about the negative sign of the result, let us sketch the graph of the function. Here it is: Y y =x -6x + 5 We find that between the limits we are given, the area lies below the x-axis. For such an area, y is negative .'. ySx is negative .'. 6 A is negative .'. A is negative. So remember, Areas below the x-axis are negative. Next frame. 18 The danger comes when we are integrating between limits and part of the area is above the x-axis and part below it. In that case, the integral will give; the algebraic sum of the area, i.e. the negative area will partly or wholly cancel out the positive area. If this is likely to happen, sketch the curve and perform the integration in two parts. Now turn to frame 19. 443 Integration Applications 1 Parametric equations Example. A curve has parametric equations x = at 2 ,y = 2 at. Find the area bounded by the curve, the x-axis, and the ordinates at t = 1 and t=2. ,& We know that A = I y dx where a and b are the limits or boundary J a values of the variable. Replacing y by 2 at, gives *b A = I 2 at dx ■■( 2 at J a but we cannot integrate a function of t with respect to x directly. We therefore have to change the variable of the integral and we do it thus We are given x = at 2 .'. -?- = 2at .'. dx =■ 2at dt We now have A = j 2at.2at dt=\ 4 a 2 t 2 dt = Finish it off. A-J|«.- 2 1 2 dt = 4a 2 = 4a 2 u3 J2 (8_l\_28a 2 (3 3| = J_ The method is always the same — (i) Express x and>» in terms of the parameter, (ii) Change the variable, (iii) Insert limits of the parameter. Example. If x = a sin 6 , y = b cos 6 , find the area under the curve between 9 = and 6 = ir. A = I y dx = i b cos . a cos & . dd x = a sin dx=a cos d0 J a JO 19 20 COS 2 Q?0 444 Programme 1 6 21 A = nab For •> dd-- = ab ~0 sin 2d' L2 4 J IT ■nab J2_ ~ 2 = ab ± = Now do this one on your own: Example. If x = - sin , y = 1 - cos , find the area under the curve between = and 9 = n. When you have finished it, move on to frame 22. 22 Working: A = y units'' -J. A = I y dx = \ (1 ~ cos 6) (1- cos 6)d6 Jo = \ (l-2cos0 + cos 2 0)d0 JO -a -> • n , g , sin 20 = -2 sin +:r+— - — ^ = (1 -cos0) x = (0- sin 0) djc = (l-cos0)d0 377 .2. 3?r .. 7 = y units' 445 Integration Applications 1 Mean values To find the mean height of the students in a class, we could measure their individual heights, total the results and divide by the number of subjects. That is, in such cases, the mean value is simply the average of the separate values we were considering. To find the mean value of a continuous function, however, requires further consideration. 23 When we set out to find the mean value of the function y = f(x) between x = a and x = b, we are no longer talking about separate items but a quantity which is continuously changing from* = a to x = b. If we estimate the mean height of the figure in the diagram, over the given range, we are selecting a value M such that the part of the figure cut off would fill in the space below. y - fix) In other words, the area of the figure between x = a and x = b is shared out equally along the base line of the figure to produce the rectangle. /. M = Area A Base line b — a :. M = J—[ b y dx So, to find the mean value of a function between two limits, find the area under the curve between those limits and divide by On to frame 24. 446 Programme 16 24 length of the base line So it is really an application of areas. Example. To find the mean value of y = 3x 2 + 4x + 1 between jc = -1 and x = 2. M=- -a J „ y dx 2, 2ZTZ[\ \ ( 3 * 2 + 4x + 1) dx x 3 + 2x 2 + x (8 + 8 + 2) - (-1 + 2 - 1) It = 6 :. M = 6 Here is one for you: Example. Find the mean value of y = 3 sin 5r + 2 cos 3r between r = and t = it. Check your result with frame 25. ' 25 Here is the working in full: M = I (3 sin 5? + 2 cos 3?) dt 77 -3 cos 5/^ 2 sin 3Z" 5 3 7T ■i{ -3 cos Sn 2 sin 3n L 5 3 - [-H = MM) ^1 6 5tl 447 Integration Applications 1 R.M.S. values The phrase 'r.m.s. value of y' stands for 'the square root of the mean value of the squares ofy' between some stated limits. Example. If we are asked to find the r.m.s. value of y = x 2 + 3 between x = 1 and x = 3 , we have — r.m.s. =>/(Mean value of^ 2 between x = 1 and* = 3) .". (r.m.s.) 2 = Mean value ofy 2 between x = 1 and x - 3 .3 26 K y 2 dx (r.m.s.) 2 ^J (x 4 + 2x 3 + 9x 243 + 54 + 27 = — {129-6- 11-2 + 6x 2 + 9) dx 13 1 _-+2 + 9 48-6 + 81-11-2 118-4 =59-2 r.m.s. =V59-2 = 7-694 /. r.m.s. = 7-69 So, in words, the r.m.s. value ofy between x = a and x = b means (Write it out) Then to the next frame. 21 448 Programme 16 28 '.. the square root of the mean value of the squares of >> between x = a and x = b J There are three distinct steps: (1) Square the given function. (2) Find the mean value of the result over the interval given. (3) Take the square root of the mean value. So here is one for you to do: Example. Find the r.m.s. value of y = 400 sin 2007T? between t = and 1 t = 100 When you have the result, move on to frame 29. 29 See if you agree with this — y 2 = 160000 sin 2 2007tf = 160000.-|(1 - cos 400irf) = 80000(1 -cos4007rr) '1/100 i r i/ioo :. (r.m.s.) 2 = t - - 80000 (1 - cos 4007rr) dt = 100.80000 t- sin 4007T? 400?r 1/100 = 8.10 6 _m-°_ = 8.10 4 .'. r.m.s. = V(8.1 4 ) = 200 v/2 = 282.8 Now on to frame 30. 449 Integration Applications 1 Before we come to the end of this particular programme, let us think back once again to the beginning of the work. We were, of course, considering the area bounded by the curve .y =/(*), the x-axis, and the ordinates at* =a and* = b. v = f(x) We found that 30 ■r J a y dx Let us look at the figure again. Y y=f{oc) Xi y If P is the point (x, y) then the area of the strip 5 A is given by 8A^y.8x If we divide the complete figure up into a series of such strips, then the total area is given approximately by the sum of the areas of these strips. 8i 2 = 'the sum of all terms like..' i.e. A = sum of the strips between x = a and x : x = b i.e. A - 2 y.&x x = a The error in our approximation is caused by ignoring the area over each rectangle. But if the strips are made narrower, this error progressively decreases and, at the same time, the number of strips required to cover the figure increases. Finally, when 8x -*■ 0, A = sum of an infinite number of minutely thin rectangles f b x = b .'. A = l ydx= 2 y .8x whenx -> J a * = fl It is sometimes convenient, therefore, to regard integration as a summing up process of an infinite number of minutely small quantities each of which is too small to exist alone. We shall make use of this idea at a later date. Next frame. 450 Programme 1 6 31 Summary Sheet 1 . Areas under curves Y y = Hx) f J a A = l y dx a b Areas below the x-axis are negative. 2. Definite integrals A definite integral is an integral with limits. r b r l b ydx = F(x) = F(6) - F(a) 3. Parametric equations x=/(0, j = F(r) rx 2 y dx = J X! F(t)dx = \ m-^dt Xi J t x 4. Mean va/ues 5. R.M.S. values I 1 c b vl = y dx b -a 1 (r.m. ^ a 6. Integration as a summing process x = b rb When 5jc-*0, 2 .y.5x=| j/& x = a All that now remains is the Test Exercise set out in the next frame. Before you work through it, be sure there is nothing that you wish to brush up. It is all very straightforward, so take your time. On then to frame 32. 451 Integration Applications 1 Test Exercise— XVI j £ Work all the questions. 1 . Find the area bounded by the curves jy = 3 e 2x andy = 3 e~ x and the ordinates at x = 1 and x = 2. 2. The parametric equations of a curve are 77 7T y = 2 sin — f, x = 2 + 2/ - 2 cos — ? "^10 10 Find the area under the curve between t = and t = 10. 3. Find the mean value of y = t—s between x = -- 2 - x - 3x 3 and* = +-. 4. Calculate the r.m.s. value of/' = 20 + 100 sin 1007it between r = andf= 1/50. 5. If /' = I sin cot and v = L— + Rz, find the mean value of the product 2tt vi between t = and f = ■— . to 6. If/' = 300 sin 100-rrf + I, and the r.m.s. value of i between t-0 and t = 002 is 250, determine the value of I. 452 Programme 16 Further Problems-XVI 1 . Find the mean height of the curve y - 3x 2 + 5x - 7 above the x-axis between x = -2 and x = 3. 2. Find the r.m.s. value of /' = cos x + sin x over the range x = to 3tt 3. Determine the area of one arch of the cycloid x = 6 - sin 6 , y = 1 - cos 6 , i.e. find the area of the plane figure bounded by the curve and the x-axis between = and 9 = 2it. 4. Find the area enclosed by the curves jy = sin x andy = sin 2x, between x = and x = 7r/3. 5. If/' = 0-2 sin lOnt + 0-01 sin 30nt, find the mean value of/' between ? = 0andf = 0-2. 6. If i = i x sin p/ + i 2 sin 2p£, show that the mean value of i 2 over a period is i(/'! + z' 2 ). 7. Sketch the curves j> = 4e* andj> = 9 sinhx, and show that they intersect when x = In 3. Find the area bounded by the two curves and the y -axis. 8. If v = v sincof and /' = i sin(utf - a), find the mean value of vi between f = and t = — . CO E 9. If/' = — + I sincof, where E, R, I, co are constants, find the r.m.s. 2-n value of/' over the range t = to t = — . CO 10. The parametric equations of a curve are x = a cos 2 / sin/, j =a cosr sin 2 / Show that the area enclosed by the curve between / = and f = — 2 na 2 is— units 2 453 Integration Applications 1 11. Find the area bounded by the curve (1 ~x 2 )y = (x - 2) (x - 3), the x-axis and the ordinate s at x = 2 and x = 3. 12. Find the area enclosed by the curve a(a - x) y = x 3 , the x-axis and the line 2x = a. 13. Prove that the area bounded by the curve y = tanh x and the straight line y=\ between x = and x = <*>, is In 2. 14. Prove that the curve defined by x = cos 3 1, y = 2 sin 3 f , encloses an 3"' -j. 2 area — units . 4 15. Find the mean value of y - x e~ x ' a between x = and x = a. ( 16. A plane figure is bounded by the curves 2y = x 2 and x 3 y = 16, the ? x-axis and the ordinate at x = 4. Calculate the area enclosed. J 17. Find the area of the loop of the curve y 2 = x 4 (4 + x). 18. If i" = i! sin(cor + a) + I 2 sin(2cor + j3), where Ii , I 2 , u, a, and $ are constants, find the r.m.s. value of/ over a period, i.e. from / = 2tt to t = — . CO 19. Show that the area enclosed by the curve x = a (2t - sin 2t), y = 2a sin 2 f , and the x-axis between t = and r = n is 37ra 2 units 2 . 20. A plane figure is bounded by the curves y = 1/x 2 ,y = e*' 2 - 3 and the lines x = 1 andx = 2. Determine the extent of the area of the figure. 454 Programme 17 INTEGRATION APPLICATIONS PART 2 Programme 1 7 1 Introduction In the previous programme, we saw how integration could be used (a) to calculate areas under plane curves, (b) to find mean values of functions, (c) to find r.m.s. values of functions. We are now going to deal with a few more applications of integration: with some of these you will already be familiar and the work will serve as revision; others may be new to you. Anyway, let us make a start, so move on to frame 2. Volumes of solids of revolution If the plane figure bounded by the curve y = f{x), the x-axis, and the ordinates at x = a and x = b, rotates through a complete revolution about the x-axis, it will generate a solid symmetrical about OX. 'f(x) Let V be the volume of the solid generated. To find V, let us first consider a thin strip of the original plane figure. y=f(x) S^' y __L_ 8x The volume generated by the strip - the volume generated by the rectangle. i.e. 5V^ 457 Integration Applications 2 8V^.iry 2 .dx Correct, since the solid generated is a flat cylinder. If we divide the whole plane figure up into a number of such strips, each will contribute its own flat disc with volume Try 2 . 8x. fix) Y %-'''' A ^* fill x --XLiJJ^ Y, ~-^ x = b . .'. Total volume, V— 2H vy.bx x = a The error in the approximation is due to the areas above the rectangles, which cause the step formation in the solid. However, if bx -*■ 0, the error disappears, so that finally V = -f V = | ity 1 . dx a This is a standard result, which you have doubtless seen many times before, so make a note of it in your record book and move on to frame 5. Here is an example: Example. Find the volume generated when the plane figure bounded by 77 y = 5 cos 2x, the x-axis, and ordinates at x = and x=— rotates about the x-axis through a complete revolution. We have : -tt/4 p7r/4 \ iry 2 .dx = 25ir \ < Jo Jo cos 2 2x dx Express this in terms of the double angle (i.e. Ax) and finish it off. Then turn on to frame 6. 458 Programme 1 7 25-rr 2 units For: -f ■ n/4 r- rr/4 y 2 dx = 25tt I cos 2 2x dx Jo = -r- (1 + co$4x)dx z Jo cos 20 = 2cos 2 0- 1 cos 2 = i(l + cos 20) 257T : 2 x + sin4x -{i +0 }_{ + Q }] = 2|i! unit , = 25tt 2 Now what about this one? Example. The parametric equations of a curve are x = 3f 2 ,_y = 3t - t 2 . Find the volume generated when the plane figure bounded by the curve, the x-axis and the ordinates corresponding to t = and t = 2, rotates about the x-axis. [Remember to change the variable of the integral!] Work it right through and then check with the next frame. V = 49-62?r= 156 units 3 Here is the solution. Follow it through. .b V = v = iry 2 dx Tr(3t-t 2 ) 2 dx a t= 2 f-0 wf (9t 2 -6t 3 +t*)6tdt Jo 2 (9t 3 -6f 4 +f s )df * = 3t 2 , y = 3r - r x = 3t 2 dx = 6f dt = 6tt r Jo = 6*| f 9f 6t s 5 + 6. = 6tt 36-384+ 10-67 = 6n = 6tt(8-27) = 49-62tt = 156units : So they are all done in very much the same way. Turn on now to frame 8. 46-67-384 "3 459 Integration Applications 2 Here is a slightly different example. Example. Find the volume generated when the plane figure bounded by the curve .y = x 2 + 5, the x-axis, and the ordinates x = 1 and x = 3, rotates about the y-axis through a complete revolution. Note that this time the figure rotates about the axis of y. Y x* +5 f(x) Half of the solid formed, is shown in the right-hand diagram. We have rb V = 7r j> 2 dx refers to rotation J a In all such cases, we build up the integral from first no standard formula for this case. about the x-axis, principles. To see how we go about this, move on to frame 9. Here it is: note the general method. Y If we rotate an elementary strip PQ, we can say — Vol. generated by the strip — vol. generated by rectangle (i.e. hollow thin cylinder) .'. 5 V — area of cross section X circumference 6 V — ybx . 2nx — 2nxy 8x For all such strips between x = 1 andx = 3 V^S5V^*f 2nxy.bx x= 1 As usual, if 8x -> 0, the error disappears and we finally obtain V = 2 | nxy dx € Since y -x 1 + 5, we can now substitute ioiy and finish the calculation. Do that, and then on to the next frame. 8 460 Programme 1 7 10 V = 80rr units 3 Here is the working: check yours •3 V = = 2-nxy dx = 2n I x(x 2 = 2tt J 1 + 5)d* (x 3 + 5x) dx 2ff| t +; 2 = 2tt = 2tt = 2tt [{T + f) 3 1 4 2 '80 40 4 + 2 20 + 20 = 807T units 3 Whenever we have a problem not covered by our standard results, we build up the integral from first principles. 11 This last result is often required, so let us write it out again. The volume generated wnen the plane figure bounded by the curve y = f{x), the x-axis and the ordinates x = a and x = b rotates completely about the y-axis is given by : V = 2-n I xy dx f J a Copy this into your record book for future reference. Then on to frame 12, where we will deal with another application of integration. 461 Integration Applications 2 Centroid of a plane figure The position of the centroid of a plane figure depends not only on the extent of the area but also on how the area is distributed. It is very much like the idea of the centre of gravity of a thin plate , but we cannot call it a centre of gravity, since a plane figure has no mass. We can find its position, however, by taking an elementary strip and then taking moments (i) about OY to find x, and (ii) about OX to find y. No doubt, you remember the results. Here they are: y=f(x) x = b Ax— Y2 x.ybx x = a 12 x = b y A y- 12 w 5x x = a z Which give if \\ y 2 dx ' a J a y dx y dx Add these to your list of results. Now let us do one example. Here goes. Find the position of the centroid of the figure bounded by y = e 2X , the x-axis, the y -axis, and the ordinate at x = 1. First, to find* 2 xy dx f 2 . i: y dx - Ii We evaluate the two integrals quite separately, so let x - Then I J h dx = 13 462 Programme 1 7 14 ii = 3e 4 + 1 For: -£- \e 2X dx x e 2x e 2X ~\ 2 -C-3-H) 3e" 1 = 3 e 4 + 1 4 4 4 2 ■r Jo Similarly, I 2 = \ e 2x dx which gives I 2 = 15 e 4 -l For: So, therefore I 2 = e 2X dx = J 2 _e 4 lj 4 -l n 2 2 2 I i _3V+iv-2_ * " i 2 " 4 X e 4 - 1 16 3c= 1-523 _ = 3e 4 + 1 = 3(54-60) + 1 = 163-8+ 1 = 164-8 X 2(e 4 -l) 2(54-60-1) 109-2-1 108-2 /. x = 1 -523 Now we have to find.y D _y 2 dx I, Note that the < J J o y dx denominator is the * 2 same as before. I 3 =^\ y 2 dx = 'o 463 Integration Applications 2 i. = 5-[«* - 1] .'-Mk + i] 17 -A 1 -if-.] 7 2 c?x = y 12 I 2 (e 8 -l) dX=^r 1 ,„ 4 . „ 1 J o Jc 1) ^(e 4 + l) = ^(54-60+1) 55-60 = 13 9 So the results are: 3c = 1-523; y = 13-9 Now do this one on your own in just the same way . Example. Find the position of the centroid of the figure bounded by the curve y = 5 sin 2x, the x-axis, and the ordinates at x = and x = - 6 (First of all find x and check your result before going on to findjO x = 0-3424 18 • ?r/6 (• 7l/6 Ii = \ xy dx = 5 I x sin 2x c?x Jrr/6 (• 7r/( xy dx = 5 I Jo C T (-cos 2x ) J f^ 6 , " = 5 x K r — + t I cos 2x dx = 5 = 5 = 5 x cos2x sin 2x 1 "^ -* 2 4 _ZL 1 i + Vl 6 '2 "2 8 V 3 _ jr_ " 8 24 V3_?r' 2 6 ■I» : Also' I 2 = I 5 sin 2x dx = 5 _ 5 * = 4 5 COS 2 2x~ C 6 5 [ ~~2[ H V: 5 tt" 2 6. 4. '5' V3 tt" 2 6 _5_ 4 = 0-8660-0-5236 :. x = 0-3424 Do you agree with that? If so, push on and findjy. When you have finished, move on to frame 1 9. 464 Programme 1 7 19 Here is the working in detail ■W6 y = 1-542 I, j pw/6 = ^l ±(\-cos4x)dx 2 Jo 2 25 sin 2 2x dx W6 25 T sin 4x : 7 *~~. tt/6 Therefore .25 4 .25 ' 4 .25 ' 4 .25 ' 4 TT_^ sin(27r/3) 6 4 . 6 8 0-5236-0-2153 .277 .11 \/3 0-3083 L = 25(007708) = 1-927 _ = I3 = L927 = (± 927)4 = 1542 20 5/4 5 So the final results are x = 0-342, >>= 1-542 Now to frame 20. Here is another application of integration not very different from the last. Centre of gravity of a solid of revolution To find the position of the centre of gravity of the solid formed when the plane figure bounded by the curves =f(x), the x -axis, and the ordinates at x = a and x = b rotates about the x-axis. y = fix) If we take elementary discs and sum the moments of volume (or mass) about OY, we can calculate X. P b \ xy 2 dx This gives x = — y 2 dx What about j? Clearly,^ = 465 Integration Applications 2 7 = 21 Correct, since the solid generated is symmetrical about OX and therefore the centre of gravity lies on this axis, i.e.y = 0. So we have to find only 3c, using >b xy 2 dx a -Ii b I 2 y 2 dx f and we proceed in much the same way as we did for centroids. Do this example, all on your own: Example. Find the position of the centre of gravity of the solid formed when the plane figure bounded by the curve x 2 + y 2 = 16, the x-axis, and the ordinates x = 1 and x = 3 rotates about the x-axis. When you have finished, move to frame 22. Check your working. x= 1-89, y = Ii=f x(\6-x 2 )dx = \ (16x-x 3 )dx = -(™-i)-(«4) 8*' "J ■i: A) \~ 4. 64-20 = 44 ;. I, =44 (\6~x 2 )dx- 16x-| 48 -9)-(.6-i) .. x = 23^ :. I 2 = 23^ _ i! _ 44 ~3 U 132 1 70 70 1-89 So x= 1-89, y = They are all done in the same manner. Now for something that may be new to you. Turn on to frame 23. 22 466 Programme 1 7 O O Lengths of curves To find the length of the arc of the curve y = f(x) between x = a and x = b. Y y - ri b/ 8 F s ^^ ! y a b X -• — x— ■- 8x I— Let P be the point (x, y) and Q a point on the curve near to P. Let 5s = length of the small arc PQ. Then IWy+tof .'.[g^§! V Sx ds Iffix-0 S -/{1 + Make a note of this result. Then on to the next frame. (£)') ■-Vnth O /I Example. Find the length of the curve y 2 = x 3 between x = and x = 4. 10V10-1 27 27 31-62- 1 = £ (30-62) = 907 units That is all there is to it. Now here is one for you: x Example. Find the length of the curve y = 1 cosh — between x = —l and* = 2. Finish it, then turn to frame 25. 467 Integration Applications 2 s = 3-015 units Here is the working set out. 25 y = 10 cosh — s : r)V -\ 2 •'• ! 'lM° osh '^}'' fc "I, CO!h to = 10 [sinh 0-2 - sinh (-Ol)j sinh(-;c) = -sinh x = 10[sinh 0-2+ sinh 0l] = 10[0-2013 + 01002] = 10 [ 0-3015 ] = 3 015 units Now to frame 26. Lengths of curves — parametric equations Instead of changing the variable of the integral as we have done before £ (J when the curve is defined in terms of parametric equations, we establish a special form of the result which saves a deal of working when we use it. Here it is. _\ 2^ Let^=/(f),* = F(r) s y As before -L (6s) 2 & (5x) 2 + (Sy) 2 Divide by (5r) 2 s2 ,R^2 ,£,,,2 ■■Q^hm If St -*• 0, this becomes /dsy dt) ~\dti \dt> ■i=y((f) % (f) f,'::,V((f)'if)V 77iis is a very useful result. Make a note of it in your book and then turn on to the next frame. .. s : 468 Programme 1 7 £ / Example. Find the length of the curve x = 2 cos 3 8,y = 2 sin 3 between the points corresponding to 6 = and = tt/2. dx We have -rz = 6 cos 2 (- sin 0) = -6 cos 2 sin do -75= 6 sin 2 cos0 ■'• ® + (S) = 36 cos4(5 sin20 + 36 sin4e cos20 = 36 sin 2 cos 2 (cos 2 + sin 2 0) = 36 sin 2 cos 2 •■•y((f) 2+ (sn =6sin0cos9=3sin2e /•tt/2 :. s = 3 sin 20 d0 J = Finish it off. 28 Is = 3 units •tt/2 For we had s = I 3 sin 20 <20 f*/2 J - ,[-f!]' - 3 [(-5H-i)>i=!=. ]n/2 It is all very straightforward and not at all difficult. Just take care not to make any silly slips that would wreck the results. Here is one for you to do in much the same way. Example. Find the length of the curve x = 5 (2? - sin 2t),y = 10 sin 2 1 between t = and t = it. When you have completed it, turn on to frame 29. 469 Integration Applications 2 s = 40 units 29 For: x = 5(2r - sin 2t), y = 10 sin 2 ? dx :. j- = 5(2 - 2 cos 2r) = 10(1 - cos 2r) -r = 20 sin t cos r = 10 sin 2f. {—) +(&\ = 100(1 - 2 cos 2f + cos 2 2f) + 100 sin 2 2? = 1 00( 1—2 cos It + cos 2 2f + sin 2 2f) = 200(1 -cos 2t) Butcos2r= 1 -2 sin 2 ? = 400 sin 2 1 ■V!(£W1 -">■*' /• 7T J 20 sin t dt = 20 -cos f Next frame. 20 (l)-(-l) : 40 units So, for the lengths of curves, there are two forms: (i)* = PV( 1+ @V when ' = F (*) ^-rVKi/^V 9 forparametnc fii equations. Just check that you have made a note of these in your record book. Now turn on to frame 31 and we will consider a further application of integration. This will be the last for this programme. 30 470 Programme 1 7 31 Surfaces of revolution If an arc of a curve rotates about an axis, it will generate a surface. Let us take the general case. Find the area of the surface generated when the arc of the curve y = f(x) between x = Xi and x = x 2 rotates about the x-axis through a complete revolution. y = fix) y Dividing by 8x, gives and if 8x -»■ 0, — - 2ny — ox ox dk „ ds — = 2Try — dx dx If we rotate a small element of arc 5s units long, it will generate a thin band of area 5 A. ThenSA — Iny. 8s 8s Now we have previously seen that — = s / ( 1 + {~jz) ■■■£WM£)'l So that A = 471 Integration Applications 2 H>V( 1+ ©> This is another standard result, so copy it down into your record book. Then on to the next frame. Here is an example requiring the last result. Example. Find the area generated when the arc of the parabola y 2 = &c between x = and x = 2 rotates about the x-axis. We have y 2 - i ■■>-^ ■••£->*«-* •■•(&'-§ .,,+(£)' = ,,2,.£ii \ ax/ x x A = \ 2ir 2\/2 x^ J {^A dx Jo * 1 = 4>/2 77 (x + 2)^ Jo x* F dx 32 33 Finish it off: then move on. A= 19-577 = 61 -3 units 2 34 For we had A=<H/2. r* i 7T (X + 2P Jo F dx = V2.7T (* + 2) 3/2' = 8V2_7T 3 3/2 (8)-(2>/2) 8V2-4 8jr 3 19-5tt =61-3 units 2 7-312 Now continue the good work by moving on to frame 35. 472 Programme 1 7 j Jj Surfaces of revolution — parametric equations We have already seen that if we rotate a small arc 8s, the area 6 A of the thin band generated is given by 8A^2iry.5s If we divide by 86, we get and if 6 6 -> , this becomes SA^_ 8s Je- 2ny -Te We already have established in our work on lengths of curves that ,2 /J, A 2\ ■'■a-PW|(S)**(§)V J 8] This is a special form of the result for use when the curve is defined as a pair of parametric equations. On to frame 36. O C Example. Find the area generated when the curve x = a{6 - sin 0), J (J J* = a ~ cos 6) between 6 = and 6 = n, rotates about the jr-axis through a complete revolution. Here ^=a(l-cos0) :. (^) = <z 2 (l -2 cos Q + cos 2 0) ■' (ID" + (dflf =fl2(1 ~2 costf +cos 2 + sin 2 0) = 2a 2 (l-cos0) But cos 6 = 1 -2 sin 2 - $-*• •■■(S) , - > -' , « v2 ■7(( = 4« ,! snr — Finish the integral and so find the area of the surface generated. 473 Integration Applications 2 ,/{£>' ♦©>"*i *-j>^«sw* = 2tt I a(\ - cos 6). 2a sin-^.dd Jo l ■2-ni" «(2 sin 2 -|). 2a sin-| d6 (1- cos 2 !) sin -|. d6 = %m 2 = 8ra 2 = 8to 2 = 8ra 2 = 8ra 2 /(' sm -r- - COS . 2 cos 3 0/2 -2cos I+ 3 61 • °\m ]" Jo (0)-(-2 + 2/3) 4/3 32wa 2 ., 2 — - — units Here is one final one for you to do. Example. Find the surface area generated when the arc of the curve y = 3f 2 , x = 3t - f 3 between t = and f = 1 rotates about OX through 2n radians. When you have finished - next frame. Here it is in full. y = 3t 2 :.% = 6t * dt , m 2 = \dtl 36f 2 x = 3t-t 3 .-. J^ = 3 -3r 2 =3(1 -r 2 ) :. (f^ =9(1 -2r 2 + f) (w) + ("f ) 2 = 9 " 18 ' 2 + 9f4 + 36 ' 2 x =9+18f 2 +9f 4 =9(1 + f 2 ) 2 /. A = f 27r3r 2 V9(l + r 2 ) 2 .dr J = 1873-f t 2 (l+t 2 )dt=l8TT\ (t 2 +t*)dt Jo Jo 1 -18,I 1+T 37 38 .-■-r-Hi"-^"* 1 474 Programme 1 7 39 Rules of Pappus There are two useful rules worth knowing which can well be included with this stage of the work. In fact we have used them already in our work just by common sense. Here they are: 1 . If an arc of a plane curve rotates about an axis in its plane, the area of the surface generated is equal to the length of the line multiplied by the distance travelled by its centroid. 2. If a plane figure rotates about an axis in its plane, the volume generated is equal to the area of the figure multiplied by the distance travelled by its centroid. You can see how much alike they are. By the way, there is just one proviso in using the rules of Pappus: the axis of rotation must not cut the rotating arc or plane figure. So copy the rules down into your record book. You may need to refer to them at some future time. Now on to frame 40. 40 Revision Summary 1 . Volumes of solids of revolution (a) about x-axis y = fix) = IT) * n V = | ny i dx a Parametric equations V = (b) about y-axis f 2 2 dx dd 0) (ii) y = fix) "J J a V = I 2irxy dx (iii) 475 Integration Applications 2 2. Centroids of plane figures Y y c rT 1 ■ i y = f{x) «« - X- -1 X j: xy dx — _* a x =- ! J 2 c?X (iv) (v) 3. Centres of gravity of solids of revolution y=fix) f x = — r •» a 7=o xy 2 cfr (vi) y 2 dx 4. Lengths of curves y=Ax) Parametric equations 5. Surfaces of revolution Parametric equations r /('♦(&> « A = f ! 2 Vv /(l + (£)), A (ix) *-i;;^(f)'^> « 10. ,4M f/zaf now remains is the Test Exercise in frame 41, so when you are ready, turn on and work through it. 476 Programme 1 7 41 Test Exercise XVII The problems are all straightforward so you should have no trouble with them. Work steadily: take your time. Do all the questions. Off you go. 1 . Find the position of the centroid of the plane figure bounded by the curves = 4 -x 2 and the two axes of reference. 2. The curve/ 2 = x(l -x) 2 between* = andx = 1 rotates about the x-axis through 2ir radians. Find the position of the centre of gravity of the solid so formed. 3. If x = a(6 - sin0), y = a{\ - cos 6), find the volume generated when the plane figure bounded by the curve, the x-axis, and the ordinates at 0=0 and 6 = 2tt, rotates about the x-axis through a complete revolution. 4. Find the length of the curve Axy = x 2 + 4 between x = 1 and x = e. 5. The arc of the catenary y - 5 cosh-rbetweenx = andx = 5 rotates about OX. Find the area of the surface so generated. 6. Find the length of the curve x = 5 (cos Q + 8 sin 8), y = 5 (sin 6 - 8 cos 8) between 8 = and 8 = it/2. 1 . The parametric equations of a curve are x = e l sin t,y =e l cos t. If the arc of this curve between t = and t = 7r/2 rotates through a complete revolution about the x-axis, calculate the area of the surface generated. Now you are all ready for the next programme. Well done, keep it up! 477 Integration Applications 2 Further Problems-XVII .2 J 1. Find the length of the curve 7 ~\~\ +2 ln ~x) between x = and x = 2- x 2 For the catenary y = 5 cosh-?-, calculate (i) the length of arc of the curve between x = and x = 2. (ii) the surface area generated when this arc rotates about the x-axis through a complete revolution. 3. The plane figure bounded by the parabola y 2 = Aax, the x-axis and the ordinate at x = a, is rotated through a complete revolution about the line x = -a. Find the volume of the solid generated. 4. A plane figure is enclosed by the parabola y 2 = Ax and the liney = 2x. Determine (i) the position of the centroid of the figure, and (ii) the centre of gravity of the solid formed when the plane figure rotates completely about the x-axis. 5. The area bounded byy 2 x = 4a 2 (2a -x), the x-axis and the ordinates x = a, x = 2a, is rotated through a complete revolution about the x-axis. Show that the volume generated is 4™ 3 (2 In 2 - 1). 6. Find the length of the curve x 2/3 + y 2 ^ = 4 between x = and 7 . Find the length of the arc of the curve 6xy = x* + 3, between x = 1 and* = 2. 8. A solid is formed by the rotation about the y-axis of the area bounded by the y-axis, the lines y = -5 andj> = 4, and an arc of the curve 2x 2 -y 2 =8. Given that the volume of the solid is — j^ , find the distance of the centre of gravity from the x-axis. 9. The line .y = x - 1 is a tangent to the curves = x 3 - 5x 2 + 8x - 4 at x = 1 and cuts the curve again at x = 3. Find the x coordinate of the centroid of the plane figure so formed. 478 Programme 1 7 10. Find by integration, the area of the minor segment of the circle x 2 +y 2 = 4 cut off by the limy = 1. If this plane figure rotates about the jc-axis through 2tt radians, calculate the volume of the solid generated and hence obtain the distance of the centroid of the minor segment from the x-axis. 1 1 . If the parametric equations of a curve are x = 3a cos 9 - a cos 39, y = 3a sin 9 ~a sin 39, show that the length of arc between points corresponding to 6 = and 6 = (pis, 6a(l ~ cos <p). 12. A curve is defined by the parametric equations x = 9 - sin 9 , y = 1 - cos 6 (i) Determine the length of the curve between 9 = and 6 = 2n. (ii) If the arc in (i) rotates through a complete revolution about the x-axis, determine the area of the surface generated, (hi) Deduce the distance of the centroid of the arc from the x-axis. 13. Find the length of the curves = coshx between x = andx = 1. Show that the area of the surface of revolution obtained by rotating the arc through four right-angles about thej>-axis is — *■ -' units. 14. A parabolic reflector is formed by revolving the arc of the parabola y 2 = 4ax from x = to x = h about the x-axis. If the diameter of the reflector is 21, show that the area of the reflecting surface is ~\(?+Ah^-A 1 5 . A segment of a sphere has a base radius r and maximum height h. Prove that its volume is~T"{ h 2 + 3r 2 16. A groove, semi-circular in section and i cm deep, is turned in a solid cylindrical shaft of diameter 6 cm. Find the volume of material removed and the surface area of the groove. 17. Prove that the length of arc of the parabola y 2 = Aax, between the points where y = and y = 2a, is a\ \/2 + ln(l + >/2)] This arc is rotated about the x-axis through 27T radians. Find the area of the surface generated. Hence find the distance of the centroid of the arc from the line y = 0. 479 Integration Applications 2 18. A cylindrical hole of length la is bored centrally through a sphere. Prove that the volume of material remaining is -^— ■ 19. Prove that the centre of gravity of the zone of a thin uniform spherical shell, cut off by two parallel planes is halfway between the centres of the two circular end sections. 20. Sketch the curve lay 2 = x(x - a) 2 , when a > 0. Show that dv 3x~a , . , , -f- = ±~ ,,. — 7 and hence prove that the perimeter of the loop is 4a/ \/3 units. 480 Programme 18 INTEGRATION APPLICATIONS PART 3 Programme 18 1 1 . Moments of inertia The amount of work that an object of mass m, moving with velocity v, will do against a resistance before coming to rest, depends on the values of these two quantities: its mass and its velocity. The store of energy possessed by the object, due to its movement, is called its kinetic energy, and it can be shown experimentally that the kinetic energy of a moving object is proportional (i) to its mass, and (ii) to the square of its velocity That is, K.E. ccmv 2 .'. K.E. = kmv 2 and if standard units of mass and velocity are used, the value of the constant Aris j. .'. K.E. =\mv 2 No doubt, you have met and It is important, so make a used that result elsewhere, note of it. 3 K.E. =\mv 2 In many applications in engineering, we are concerned with objects that are rotating — wheels, cams, shafts, armatures, etc. — and we often refer to their movement in terms of 'revolutions per second'. Each particle of the rotating object, however, has a linear velocity, and so has its own store of K.E. — and it is the K.E. of rotating objects that we are concerned with in this part of the programme. So turn on to frame 4. 483 Integration Applications 3 Let us first consider a single particle P of mass m rotating about an axis X with constant angular velocity co radians per second. This means that the angle 6 at the centre is increasing at the rate of co radians/ per second. Of course, the linear velocity of P, v cm/s, depends upon two quantities (i) the angular velocity (co rad/s) and also (ii) how far P is from the centre 1 radian To generate an angle of 1 radian in a second, P must move round the circle a distance equal to 1 radius length, i.e. r (cm). If 6 is increasing at 1 rad/s, P is moving at r cm/s, " " " " " 2 " Pis moving at 2r cm/s, " " " " " 3 " P is moving at 3r cm/s, etc. So, in general, if 9 is increasing at co rad/s, P is moving at cor cm/s. Therefore, if the angular velocity of P is co rad/s, the linear velocity, v, of Pis v = cor We have already established that the kinetic energy of an object of mass m moving with velocity v is given by K.E. = 484 Programme 18 K.E.=-i mv So, for our rotating particle, we have K.E. = ii = \m{ix>r) 2 = \m(x> 2 r 2 and changing the order of the factors we can write K.E. = \u> 2 .mr 2 where to - the angular velocity of the particle P about the axis (rad/s) m = mass of P r = distance of P from the axis of rotation Make a note of that result: we shall certainly need that again. 8 2 oj . mr If we now have a whole system of particles, all rotating about XX with the same angular velocity go rad/s, each particle contributes its own store of energy. *m 2 -»m 4 K.Ej =\ cj 2 . Mj rj 2 K.E 2 = K.E 3 = K.E 4 = 485 Integration Applications 3 K.Ej 2 CO m 1 r x 2 K.E 2 = \(j0 2 m 2 r 2 2 K.E 3 _ 1 . ,2 - "J CO m 3 r 3 2 K.E 4 2 CO m 4 r 4 2 So that, the total energy of the system (or solid object) is given by K.E. = K.Ej + K.E 2 + K.E 3 + K.E 4 + . . . = •2- co 2 . m^^+f co 2 . m 2 r 2 2 + 2 co 2 . m 3 r 3 2 + . . . K.E. = 2-2- co 2 . m a- 2 K.E. =4-co 2 .£mr 2 This is another result to note. (since co is a constant) 10 This result is the product of two distinct factors: (i) \ co 2 can be varied by speeding up or slowing down the rate of rotation, but (ii) ~Lm r 2 is a property of the rotating object. It depends on the total mass but also on where that mass is distributed in relation to the axis XX. It is a physical property of the object and is called its second moment of mass, or its moment of inertia (denoted by the symbol I). .'. I = ~Lm r 2 (for all the particles) Example: For the system of particles shown, find its moment of inertia about the axis XX. 1 kg. 4kg* 1m 2m 3 kg 2m 1 = 486 Programme 18 11 1 = 21 kgm 2 Since hEmr 2 = 2.3% 1.1 +3.2"+ 4.2" Move on to frame 12. # + 1 + X +/# =2f^jn /£ ^ 12 2. Radius of gyration x 13 ■ r \ • T>, — r 2 -*/77g -j[ ®(M) r, ./n 3 If we imagine the total mass M of the system arranged at a distance k from the axis, so that the K.E. of M would be the same as the total K.E. of the distributed particles, •m 4 then ^co 2 .Mk 2 =\ co 2 .Zmr 2 :. Mk 2 = -£mr 2 and k is called the radius of gyration of the object about the particular axis of rotation. So, we have l = Zmr 2 : Mk 2 = l 1 = moment of inertia (or second moment of mass) k= radius of gyration about the given axis. Now let us apply some of these results, so on you go to frame 13. Example 1. To find the moment of inertia (I) and the radius of gyration (k) of a uniform thin rod about an axis through one end perpendi- cular to the length of the rod. x b P Q VA Let p = mass per unit length of rod Mass of element PQ = p.Sx. Sjc .'. Second moment of mass of PQ about XX = mass X (distance) 2 = p.Sx. x 2 = px 2 .8x. ■'■ Total second moment for all such elements can be written 1=2= --M 487 Integration Applications 3 I a px 2 .bx 14 The approximation sign is included since x is the distance up to the left-hand side of the element PQ. But, if bx -> 0, this becomes -3"" I px .dx = p <2£- 3 ■ t = ££? 3 3 Now, to find k, we shall use Mk 2 - I, so we must first determine the total mass M. Since p = mass per unit length of rod, and the rod is a units long, the total mass, M = M =ap Uk 2 = \ ■■• ">■*' = 7 .. ^ 3 ,3 :. k = yft Now for another: pa- a I= T and /c= v - Example 2. Find I for a rectangular plate about an axis through its c.j parallel to one side, as shown. X Let p = mass per unit area of plate. Mass of strip PQ = b.bx.p Second moment of mass of strip about XX — b bx p.x 2 x sx (i.e. mass X distance 2 ) Total second moment for all strips covering the figure 15 I- 2 x = Programme 18 16 17 i x = d/2 2 bpx 2 .8x Did you remember the limits? So now, if 5x^0, •d/2 1 = bpx 2 .dx = bp ■d/2 *'{(£)-(-£) -"3J- d/2 ■d/2 d 3 \)_bpd 3 12 12 Mtf 2 and since the total mass M = bdp, I = — - :. I M 3 p _ Mrf 2 12 12 This is a useful standard result for a rectangular plate, so make a note of it for future use. Here is an example, very much like the last, for you to do. Example 3. Find I for a rectangular plate, 20 cm X 10 cm, of mass 2 kg, about an axis 5 cm from one 20-cm side as shown. -• 10cm — -i I,5cm ( P Q J X 20 cm Take a strip parallel to the axis and argue as before . Note that, in this case, p= i^o = 25o = - 01 i.e. p = 0-01 kg/cm 2 Finish it off and then turn on to the next frame. 489 Integration Applications 3 I = 217 kg cm 2 Here is the working in full: p = 0-01 kg cm 2 Area of strip = 20. 8x .'. Mass of strip = 20.8x.p .'. 2nd moment of mass of strip about XX ^20.Sjc.p.x 2 x = 15 Total 2nd moment of mass = I — 2 20 p x 2 .bx. x= 5 •15 If 5x^0, 20 px 2 .dx = 20 p 15 J5 20 p 3375-125 = ^3250U = 3 100 650 = 217 kg cm 2 Now, for the same problem, find the value of k. 18 k= 104 cm 19 for Mfc 2 =I and M = 2 kg /. 2/c 2 = 217 108-5 /. fe = VlQ8-5 = 104 cm Normally, then, we find I this way: (i) Take an elementary strip parallel to the axis of rotation at a distance x from it. (ii) Form an expression for its second moment of mass about the axis, (iii) Sum for all such strips. (iv) Convert to integral form and evaluate. It is just as easy as that! 490 20 Programme - 3. Parallel axes theorem If I is known' about an axis through the e.g. of the object, we can easily write down the value of I about any other axis parallel to the first and a known distance from it. A Let G be the centre of gravity of the object Let m = mass of the strip PQ Then I G = Unix 2 and I AB = 2m(x + I) 2 /. I AB = Sm(x 2 + 2/x + I 2 ) = Zmx 2 + S2mx/+ "Lml 2 = Unix 2 + 2l2,mx + l 2 Y,m (since / is a constant) Now, I,mx 2 = and 2m = 21 Zmx 2 =l G ; Sw = M Right. In the middle term we have Eroc. This equals 0, since the axis XX by definition passes through the e.g. of the solid. In our previous result, then, Smx 2 = I G ; Xmx = 0; Im = M and substituting these in, we get I AB =I G +M/ 2 Thus, if we know I G , we can obtain I AB by simply adding on the product of the total mass X square of the distance of transfer. This result is important: make a note of it in your book. 491 Integration Applications 3 Example 1. To find I about the axis AB for the rectangular plate shown O Q below. A tLtL -3 cm- 5 cm Total mass = 3 kg We have : Ud 2 3.16 A , , j g = T 2 = TT =4kgcm I AB =I G+ M/ 2 = 4 + 3.25 = 4 + 75 = 79 kg cm 2 ,2 As easy as that! Next frame. :. I AB = 79 kg cm^ You do this one: 0*» Example 2. A metal door, 40 cm X 60 cm, has a mass of 8 kg and is fc«J hinged along one 60-cm side. Here is the figure : f 40 cm H Calculate J~ (i) I about XX, the axis through the e.g. (ii) I about the line of hinge, AB. (iii) k about AB. 60 cm Find all three results: then turn on to frame 24 and check your working. 492 Programme 18 24 I X x = 1067 kg cm 2 ; I AB =4267 kg cm 2 ; k AB =23-1 cm Solutions: (i) 25 (ii) I ab = Ig +M/2 = 1067 + 8. 20 2 = 1067 + 3200 = 4267 kg cm 2 (iii) Mk 2 = \ Ah :. 8k 2 = 4267 :. k 2 = 533-4 :. fc = 23-1 cm If you made any slips, be sure to clear up any difficulties. Then move on to the next example. Let us now consider wheels, cams, etc. — basically rotating discs. To find the moment of inertia of a circular plate about an axis through its centre, perpendicular to the plane of the plate. If we take a slice across the disc as an elementary strip, we are faced with the difficulty that all points in the strip are not at the same distance from the axis. We therefore take a circular strip as shown. Mass of strip — 2irx.8x.p (p = mass per unit area of plate) .'. 2nd moment of strip about ZZ — 493 Integration Applications 3 2-nx.bx.p.x 7 .'. 2nd moment of strip about ZZ = 2npx 3 .8x .'. Total 2nd moment for all such circular strips about ZZ, is given by 26 h ~ 2 2npx 3 .8x x = = I 2npx 3 .dx = JO - 2-np _ 2npr 4 _ -nr 4 p 4 2 _M.r 2 2 lf8x-+0, Total mass, M = ur 2 p This is another standard result, so note it down. Next frame. Jo I. _ w *p - M.r 2 2 2 27 Example 1. Find the radius of gyration of a metal disc of radius 6 cm and total mass 0-5 kg. We know that, for a circular disc, M.r 2 1 7 =-^-^> and, of course, MA: 2 =I z 2 so off you go and find the value oik. fc = 4-24cm 28 . M.r 2 0-5.36 . , 2 I z =— = — ^— = 9 kg cm 2 M/fc 2 =I .'. i^ 2 = 9 /. £ 2 = 18 .'. fc = 4-24 cm They are all done in very much the same way. Turn to frame 29. 494 Programme 18 yD 4 Perpendicular axes theorem (for thin plates) Let Smbea small mass at P. Then I x — 28m.y 2 X P ^\ ^z o J X Let ZZ be the axis perpendicular to both XX and YY. Then I z = S5m.(OP) 2 = S6m.(;c 2 + y 2 ) = XSm.y 2 + 25ot.jc 2 •'■ 1 Z= I X +I Y .". If we know the second moment about two perpendicular axes in the plane of the plate, the second moment about a third axis, perpendicular to both (through the point of intersection) is given by !z = I X + l Y And that is another result to note. 30 To find I for a circular disc about a diameter as axis. We have already established that _ 7rr 4 p _ M.r 2 z_ 2 ~ 2 Let XX and YY be two diameters perpendicular to each other. Then we know But all diameters are identical I x + I Y = I z M.r 2 M.r 2 M.r 2 •'• Iy _ Iv •'■ 2 Iv — .. Iv - . A X _1 Y .'. For a circular disc: Make a note of these too. _ OT- 4 p _ M.r 2 _ w 4 p _ M.r 2 L Z ~2 T and x ~4~ 4 495 Integration Applications 3 Example. Find I for a circular disc, 40 cm diameter, and of mass 12 kg, (i) about the normal axis (Z axis), (ii) about a diameter as axis, (iii) about a tangent as axis. Work it through on your own. When you have obtained (ii) you can find (iii) by applying the parallel axes theorem. Then check with the next frame. 31 For: I z = 2400 kg cm 2 ; I x = 1 200 kg cm 2 ; I T = 6000 kg cm 2 (0 h M.r 2 = 12.20 2 2 2 = 2400 kg cm 2 32 M r 2 1 (ii) I x ~ =^I Z = 1200kg cm 2 (iii) I x = 1 200 kg cm 2 By the parallel axes theorem I T = I X +M/ 2 = 1200 + 12.20 2 = 1200 + 4800 = 6000 kg cm 2 In the course of our work, we have established a number of important results, so, at this point, let us collect them together, so that we can see them as a whole. On then to the next frame. 496 Programme 18 33 Useful standard results, so far. 1. I = 2mr 2 ; M.fc 2 = I 2. Rectangular plate (p = mass/unit area) - d GT 3. Grculardisc 4. Parallel axes theorem A I AB =I G +M/ 2 lr bd 3 p _M.d 2 12 12 h = ■nr 4 p _ M.r 2 2 ~ 2 _ 7ir 4 p _ M.r 2 5 . Perpendicular axes theorem z I Z = I X +I Y These standard results cover a large number of problems, but some- times it is better to build up expressions in particular cases from first principles. Let us see an example using that method. 497 Integration Applications 3 Example 1. Find I for the hollow shaft shown, about its natural axis. Density of material = 0-008 kg/cm 3 . 34 ffi — y m 40cm- First consider a thin shell, distance x from the axis Mass of shell -2irx.Sx.40p. kg ;. 2nd mt. about XX -2irx.Sx.40p.x 2 -80irpx 3 .8x x = % :. Total 2ndmt. S 8>Qirpx 3 .8x x = 4 Now, if Sx -* 0, 1 = and finish it off, then check with the next frame. 1= 1931 kg cm 2 For c 8 I = 80ttp x 3 dx=&0irp 35 807TP [64 2 -16 2 ] = 207rp . 48. 80 = 20tt .48.80.0-008 = 614-4tt= 1931 kg cm 2 Here is another: Example 2. Find I and k for the solid cone shown, about its natural axis of symmetry. First take an elementary disc at distance x from the origin. For this disc, OX is the normal axis, so *x = Then sum for all the discs, etc. Finish it off. 498 Programme 18 36 I x =256ttp ; fc = 2-19cm Solution: For elementary disc: lx = iry 4 .bxp Total I * = 10 TT!, 4 ny hxp x = l If 5x^0, Jo z l Jo Now, from the figure, the slope of the generating line is 4/10. Ax ' = I0 ■■■'*-?j:°(sj* wp 016' 10 Tip 0-0256 10 5 " L 5- = ;rp0-0256. 10" =256ttp Now we proceed to find k. -r * i ,, 1 ,2,„ 1607rp Total mass = M = -7r4 lOp = — — Uk 2 =\ . 160 -np , , ~ c ^ .. — 5— - fc =256 7rp • 7,2 _ 3.256.7rp " 160.7TP _ 3.64 40 = 4-8 .'. fc=v / 4-8 = 2-19cm Turn now to frame 37. 1 I 499 Integration Applications 3 5. Second moments of area In the theory of bending of beams, the expression Xar 2 , relating to the cross-section of the beam, has to be evaluated. This expression is called the second moment of area of the section and although it has nothing to do with kinetic energy of rotation, the mathematics involved is clearly very much akin to that for moments of inertia, i.e. I,mr 2 . Indeed, all the results we have obtained for thin plates, could apply to plane figures, provided always that 'mass' is replaced by 'area'. In fact, the mathematical processes are so nearly alike that the same symbol (I) is used in practice both for moment of inertia and for second moment of area. 37 Moments of inertia I = Emr 2 Mfc 2 = I Second moments of area I = Zar 2 kk 2 = \ 38 Rectangular plate _ bd 3 p G 12 M.d 2 12 Rectangle Ic= T2 = A J d_ 2 12 Grcular plate 17 = 4 nr p M.r 2 2 4 M.r 2 . 7Tf 4 p Circle iz=- ! x% A.r 2 2 Trr 4 A.r 2 Parallel axes theorem — applies to both. I AB =I C +A/ 2 Perpendicular axes theorem — applies to thin plates and plane figures only. Turn on. h W+U 500 Programme 18 JU There is really nothing new about this: all we do is replace 'mass' by 'area'. Example 1. Find the second moment of area of a rectangle about an axis through one corner perpendicular to the plane of the figure. T _bd 3 6. 4 3 . 4 l K}-T2 = ~12 -32cm By the parallel axes theorem, I x 40 for Also I x = 128 cm 4 I x = 32 + 24.2 2 = 32 + 24.4 = 32 + 96= 128 cm 4 r b(T Irs=- = 41 for I RS = 72 cm 4 i - 4- 6 3 _ 4 Irs ~~12 _72cm :. I Y = 42 I Y = 288 cm 4 For, again by the parallel axes theorem, I Y = 72 + 24.3 2 = 72 + 216 = 288 cm 4 So we have therefore: I x = 128 cm 4 and I Y = 288 cm 4 .'. I z (which is perpendicular to both I x and I Y ) = 501 Integration Applications 3 I z =416cm" □ aaanDonanncioaanaanQnnnanaQaanaQQOooaa When the plane figure is bounded by an analytical curve, we proceed in much the same way. Example 2. Find the second moment of area of the plane figure bounded by the curve y = x 2 + 3, the x-axis, and the ordinates at x = 1 and x = 3, about the y-axis. Y x 2 + 3. 43 If 5jc-> 0, Finish it off. x y dx '■ Are; i of strip PQ =y.8x ". 2nd mt. of strip about OY = y.8x.x 2 = x 2 .y.8x • h'~ x = 3 ± 2 x 2 ybx x = l I Y = 74-4 units 4 for iy = (•3 (-3 \ x 2 (x 2 + 3) dx = \ (x 4 + 3x 2 ) dx 44 ^ +x 3 5 = (f t27 )-(i + 1 ) 242 + 26 = 48-4 + 26 = 74-4 units 4 Note: Had we been asked to find I x , we should take second moment of /V\ 2 X = 3 y 3 the strip about OX, i.e. y8x\^~) ; sum for all strips 2 — 8x; and then evaluate the integral. Now, one further example, so turn on to the next frame. 502 Programme 18 45 Example 3. For the triangle PQR shown, find the second moment of area and k about an axis AB through the vertex and parallel to the base. P First consider an elementary strip. Area of strip =x.Sy ■'. 2nd mt. of strip about AB = x.by.y 2 = xy 2 .8y .'. Total 2nd mt. about AB for all such strips y = S ^ 2 xy 2 .by y = If by -> 0, AB f JO xy 2 dy We must now write x in terms of y — and we can obtain this from the figure by similar triangles. Finish the work off, so that I AB = 46 I = 250 cm 4 ; k = 3-536 < For we have 5 8 Sy 'AB r Jo 5 8 xy 2 dy=^\ y 3 dy=- ^[5-0]=-(625) = 2jW Also, total area, A = -r- = 20 cm 2 :. kk 2 = I .'. 20k 2 = 250 :. k 2 = \2-5 .'. k = 3-536 cm Next frame. 503 Integration Applications 3 Composite figures If a figure is made up of a number of standard figures whose individual second moments about a given axis are I 1; I 2 , I 3 , etc., then the second moment of the composite figure about the same axis is simply the sum of Ij.Ij.Ia.etc. Similarly, if a figure whose second moment about a given axis is I 2 is removed from a larger figure with second moment ^ about the same axis, the second moment of the remaining figure is I = I 2 — 1 2 . Now for something new. 47 6. Centres of pressure Pressure at a point P depth z below the surface of a liquid. s s *7777i 48 If we have a perfect liquid, the pressure at P, i.e. the thrust on unit area at P, is due to the weight of the column of liquid of height z above it. Pressure at P = p = wz, where w = weight of unit volume of the liquid. Also, the pressure at P operates equally in all directions. s s \ / JL Note that, in our considerations, we shall ignore the atmospheric pressure which is also acting on the surface of the liquid. ''t< P =, The pressure, then, at any point in a liquid is proportional to the of the point below the surface. 504 Programme 18 49 depth Total thrust on a vertical plate immersed in liquid s s "y hz~ P Q n Consider a thin strip at a depth z below the surface of the liquid. d z Pressure at P = wz. .'. Thrust on strip PQ —wz (area of strip) — w.z.a.Sz Then the total thrust on the whole plate If Sz -* 0, total thrust rd 2 z=d 2 — 2 awzbz z = d i awzdz= 50 a w 2 d 2 2 -d l 2 for: total thrust = aw This can be written T\d d\ 2 aw T dJ-dS Total thrust =~{d 2 -d l ) (d 2 + d t ) = W a{d 2 -d x )(^) Now,( 2 — ij is the depth halfway down the plate, i.e. it indicates the depth of the centre of gravity of the plate. Denote this by z. Then, total thrust = wa(d 2 -di)z = a(d 2 — d l )wz. Also a(d 2 ~dx) is the total area of the plate. So we finally obtain the fact that total thrust = area of plate X pressure at the e.g. of the plate. In fact, this result applies whatever the shape of the plate, so copy the result down for future use. On to the next frame. 505 Integration Applications 3 Total thrust = area of plate X pressure at the e.g. of plate So, if w is the weight per unit volume of liquid, determine the total thrust on the following plates, immersed as shown. 0) £ (ii) £ 51 So, thrust (i) = and thrust (ii) - thrust (i) = 336 w : thrust (ii) = 180 w For, in each case, total thrust = area of surface X pressure at the e.g. 0) 4^ 7 cm t 6 cm G»— L 1 ■* 8cm 52 Area = 6 X 8 = 48 cm 2 Pressure at G = 7 w Total thrust = 48.7 w = 336w 10X6 , n 2 Area = — - — = 30 cm Pressure at G = 6 w Total thrust = 30. 6 w = 180w On to the next frame. 506 Programme 1 8 53 ^ tne P late is not vertical > but inclined at an an 8 le ^ t0 { ^ e horizontal, the rule still holds good. imple: S s s S \~ 1 2 IT t , . /%30» * G« < 1 < ^ \^^^ ^0^5 Depth of G=d l +-sin30° = cf 1 +- Pressure at G = (di + 7 V Total area = ab .'. Total thrust = 54 ab{d x +-)w Remember this general rule enables us to calculate the total thrust on an immersed surface in almost any set of circumstances. So make a note of it: total thrust = area of surface X pressure at the e.g. Then on to frame 55. 507 Integration Applications 3 Depth of centre of pressure s 55 The pressure on an immersed plate increases with depth and we have seen how to find the total thrust T on the plate. The resultant of these forces is a single force equal to the total thrust, T, in magnitude and acting at a point Z called the centre of pressure of the plate. Let f denote the depth of the centre of pressure. To find I we take moments of forces about the axis where the plane of the plate cuts the surface of the liquid. Let us consider our same rectangu- lar plate again. r tz JL >///;//;;;;; ;;;/;;;;;;; dz The area of the strip PQ : a.8z 56 The pressure at the level of PQ - zw 57 So the thrust on the strip PQ : 508 Programme 18 58 a.hz.w.z i.e. awzSz The moment of this thrust about the axis in the surface is therefore = a wz 8z. z = awz 2 , 5z So that the sum of the moments of thrusts on all such strips 59 a 2 2 awz 2 8z Now, if §z -> 0, the sum of the moments of thrusts = \ * awz 2 .dz Also, the total thrust on the whole plate = .. rd 2 60 awz dz Right. Now the total thrust X z = sum of moments of all individual thrusts. ' d i - [d 2 i awzdzX z = \ awz dz Therefore, we have .'. Total thrust X z = w\ az 2 Jrfi dz ~ wl = wl wkk 2 Awl total thrust .'. z = k 7 ~z Make a note of that and then turn on. 509 Integration Applications 3 So we have these two important results: (i) The total thrust on a submerged surface = total area of face X pressure at its centroid (depth z) (ii) The resultant thrust acts at the centre of pressure, the depth of k 2 which, z, is given by z = — . 61 Now for an example on this. Example J. For a vertical rectangular dam, 40m X 20m, the top edge of the dam coincides with the surface level. Find the depth of the centre of pressure. 62 I 7 /////////////// •+ 40m - — »■ B i 2 t C t E o CM \ In this case,z = 10m. To find k 2 about AB ^A<i 2 = 40.20.400 = 80 000 4 Ic " 12 12 3 m Iab = Ic+a/2 = 80|00 + 800 .100 j. (80 000) A/c 2 =I :. k l = - 4 80 000 400 800 _ k 2 400 40 . _ _, z = _=-—= — = 13-33 m z 3.10 3 Note that, in this case, (i) the centroid is half-way down the rectangle, but (ii) the centre of pressure is two-thirds of the way down the rectangle. 510 Programme 18 63 Here is one for you. Example 2. An outlet from a storage tank is closed by a circular cover hung vertically. The diameter of the cover = 1 m and the top of the cover S s is 2-5 m below the surface of the ? liquid. Determine the depth of the I centre of pressure of the cover. 2-5 m Q 1 m —L Work completely through it: then check your working with the next frame. 64 3m e z = 3-02m For AB k 2 =- ~z AB We have: (i) Depth of centroid = z = 3 m (ii) To find k 2 about AB I, X AB At 4 ■ JL 64 2 *(4) 2 .(4) 2 7T 64 + A.3 2 -£♦•<«'•' ■n 9ir 64 4 1457T 64 145tt 4. 'n A 64 = 145 1 = 145 16 "3 48 .145 " 16 = 3-02m DannDnnnnDnnnnDDDnnDnnDnnnDDnnnnDDnnnn And that brings us to the end of this piece of work. Before you work through the Test Exercise, check down the revision sheet that follows in frame 65 and brush up any part of the programme about which you may not be absolutely clear. Then, when you are ready, turn on to the Test Exercise. 511 Integration Applications 3 Revision Sheet 1 . SECOND MOMENTS Mts. of Inertia (i) I = 2mr 2 Mfc 2 = I (ii) Rectangular plate: = bd 3 p _U.d 2 12 12 lr 0) (ii) Rectangle: 2nd Mts. of Area l = I,ar 2 Afc 2 =I lc = 12 A.J 2 12 65 (iii) Circular disc: nr 4 p = Mr 2 2 2 M.r 2 lz= J E _ 7rr 4 p (iv) Parallel axes theorem: I AB =I G +M/ 2 (iii) Circle: I 7 nr '' 2 A^ 2 - H 4 - A -^" 2 Ix " 4 ~ 4 I AB =I C +A/ 2 (v) Perpendicular axes theorem (thin plates and plane figures only): I Z =I X +I Y 2. CENTRES OF PRESSURE (i) Pressure at depth z = wz (w> = weight of unit volume of liquid) (ii) Total thrust on plane surface = area of surface X pressure at the centroid. (iii) Depth of centre of pressure (z): Total thrust X z = sum of moments of distributed thrust = k 2 z =— z where k = radius of gyration of figure about axis in surface of liquid, z = depth of centroid. Note: The magnitude of the total thrust = (area X pressure at the centroid) but it acts through the centre of pressure. DnnDnnDDnDDnnanDDnnannnnnDDnnnannDnnnD Now for the Test Exercise, on to frame 66. 512 Programme 18 66 Work through all the questions in theTest Exercise. They are very much like those we have been doing, so will cause you no difficulty: there are no tricks. Take your time and work carefully. Test Exercise — XVIII 1 . (i) Find the moment of inertia of a rectangular plate, of sides a and b, about an axis through the mid-point of the plate and perpendicular to the plane of the plate, (ii) Hence find also the moment of inertia about an axis parallel to the first axis and passing through one corner of the plate, (iii) Find the radius of gyration about the second axis. 2. Show that the radius of gyration of a thin rod of length / about an axis through its centre and perpendicular to the rod is — j . An equilateral triangle ABC is made of three identical thin rods each of length /. Find the radius of gyration of the triangle about an axis through A, perpendicular to the plane of ABC. 3. A plane figure is bounded by the curve xy = 4, the x-axis, and the ordinates at x = 2 and x = 4. Calculate the square of the radius of gyration of the figure (i) about OX, and (ii) about O Y. 4. Prove that the radius of gyration of a uniform solid cone with base radius r about its natural axis is /rjr . 5. An equilateral triangular plate is immersed in water vertically with one edge in the surface. If the length of each side is a, find the total thrust on the plate and the depth of the centre of pressure. 513 Integration Applications 3 Further Problems - XVIII 1 . A plane figure is enclosed by the curve y = a sin x and the x-axis between x = and x = n. Show that the radius of gyration of the figure about the x-axis is ~- . 2. A length of thin uniform wire of mass M is made into a circle of radius a. Find the moment of inertia of the wire about a diameter as axis. A solid cylinder of mass M has a length / and radius r. Show that its moment of inertia about a diameter of the base is M ■ r 2 I 2 7 + 3_ 4. Show that the moment of inertia of a solid sphere of radius r and 2 mass M, about a diameter as axis, is— Mr 2 . 5. Prove that, if k is the radius of gyration of an object about an axis through its centre of gravity, and kj is the radius of gyration about another axis parallel to the first and at a distance / from it, then 6. A plane figure is bounded by the parabola y 2 = Aax, the x-axis and the ordinate x = c. Find the radius of gyration of the figure (i) about the x-axis, and (ii) about thej-axis. 7. Prove that the moment of inertia of a hollow cylinder of length /, with inner and outer radii r and R respectively, and total mass M, about its natural axis, is given by I =\M (R 2 + r 2 ). 8. Show that the depth of the centre of pressure of a vertical triangle with one side in the surface is-^, if/? is the perpendicular height of the triangle. 9. Calculate the second moment of area of a square of side a about a diagonal as axis. 1 0. Find the moment of inertia of a solid cone of mass M and base radius r and height h, about a diameter of the base as axis. Find also the radius of gyration. 514 Programme 18 11. A thin plate in the form of a trapezium with parallel sides of length a and b, distance d apart, is immersed vertically in water with the side ef length a in the surface. Prove that the depth of the centre of pressure (z) is given by = _ d(a + 3b) Z 2(a + 2b) 12. Find the second moment of area of an ellipse about its major axis. 13. A square plate of side a is immersed vertically in water with its upper side horizontal and at a depth d below the surface. Prove that the 2 centre of pressure is at a distance — — - below the centre of the 6(a + 2d) square . 14. Find the total thrust and the depth of the centre of pressure when a semicircle of radius a is immersed vertically in liquid with its diameter in the surface. 15. A plane figure is bounded by the curve y = e x , the x-axis, the j>-axis and the ordinate at x = 1 . Calculate the radius of gyration of the figure (i) about OX as axis, and (ii) about OY as axis. 16. A vertical darn is a parabolic segment of width 12 m and maximum depth 4 m at the centre. If the water reaches the top of the dam, find the total thrust on the face. 17. A circle of diameter 6 cm is removed from the centre of a rectangle measuring 10 cm by 16 cm. For the figure that remains, calculate the radius of gyration about one 10-cm side as axis. 18. Prove that the moment of inertia of a thin hollow spherical shell of 2 mass M and radius r, about a diameter as axis is -Mr 2 . 19. A semicircular plate of radius a is immersed vertically in water, with its diameter horizontal and the centre of the arc just touching the surface. Find the depth of the centre of pressure. 20. A thin plate of uniform thickness and total mass M, is bounded by the curve y = c cosh—, the x -axis, the y-axis, and the ordinate x = a. c Show that the moment of inertia of the plate about the _y-axis is Ma 2 - 2oz coth~ ^-U 2c 2 ' 515 Programme 19 APPROXIMATE INTEGRATION Programme 19 1 Introduction In previous programmes, we have seen how to deal with various types of integral, but there are still some integrals that look simple enough, but which cannot be determined by any of the standard methods we have studied. For instance, I xe x dx can be evaluated by the method of integration u Jo by parts. What do you get? 1 xe x dx ■ 1-We for: I x e x dx : x(e x ) J dx e x (x-l) = ^H)-e°(-l)=l-iVe That was easy enough, and this method depends, of course, on the fact that on each application of the routine, the power of x decreases by 1, until it disappears, leaving I e x dx to be completed without difficulty. But suppose we try to evaluate x T e x dx by the same method. The J process now breaks down. Work through it and see if you can decide why. When you have come to a conclusion, move on to the next frame. Reducing the power of x by 1 at each application of the method, will never give x° , i.e. the power of* will never disappear and so the resulting integral will always be a product. For we get: jc 2 e x dx- x*(e x )\ -4 e x x 2 dx JO L Jo J and in the process, we have hopped over x°. So here is a complication. The present programme will show you how to deal with this and similar integrals that do not fit in to our normal patterns. So on, then, to frame 4. 517 Approximate Integration Approximate integration First of all, the results we shall get will be approximate in value, but like many other 'approximate' methods in mathematics, this does not imply that they are 'rough and ready' and of little significance. The word 'approximate' in this context simply means that the numerical value cannot be completely defined, but that we can state the value to as many decimal places as we wish. e.g. To say x = \/3 is exact, but to say x = 1 -732 is an approximate result since, in fact, -y/3 has a value 1-7321 . . . with an infinite number of decimal places. Let us not be worried, then, by approximate values: we use them whenever we quote a result correct to a stated number of decimal places, or significant figures. tt = 34-; 77 = 3-142 : tt = 3-14159 are all values approximate We note, of course, that an approximate value can be made nearer and nearer to the real value by taking a larger number of decimal places — and that usually means more work! Evaluation of definite integrals is often required in science and engineer- ing problems: a numerical approximation of the result is then quite satisfactory. Let us see two methods that we can apply when the standard routines fail. On to frame 6. 518 Programme 19 Method 1 . By series Consider the integral 1 1 x 1 e x dx, J o which we have already seen cannot be evaluated by the normal means. We have to convert this into some other form that we can deal with. Now we know that 2 3 4 v y x e -l + x + f !+ f !+ - + .. '^ x^e x dx=r x^^+x+~+j ] +...]dx .f* x 1 e x dx = P x 1 jl ,5/2 -.7/2 Jc l/2 +x 3/2 + i_ + ±_ + ...| c?x Now these are simply powers of x, so, on the next line, we have 1 = ~2x 3 ' 2 2x 5 ' 2 2x l/2 2x 9 ' 2 [ 3 + 5 + 7.2 + 9.6 + r 2x 3 ^ 2 2x s ' 2 x 1 ' 2 x 9 ' 2 1 L 3 + 5 + 7 + 27 + '--_ i "2" 1 T To ease the calculation, take out the factor x 2 I = 2x 2x 2 x 3 x 4 x 5 X { 3 5 7 27 132 JO = JJl + -2_ + J_ + _!_ + V213 4.5 8.7 16.27 - V- > 0-3333 + 0-1000 + 0-0179 + 0-0023 + 0-0002 = V f{ 0-4537 = 1-414(0-2269) = 0-3207 All we do is to express the function as a series and integrate the powers of x one at a time . Let us see another example, so turn on to frame 8. 519 Approximate Integration Here is another. ■/ ,2ln(l+jc). To evaluate — i — ? — ax V* 8 First we expand ln(l + x) as a power series. Do you remember what it is? ln(l +x) = 2 3 4 5 x x x* x , ln(l +*) = *-- + T ~-r +- + 2 3 4 5 .2 v 3 v 4 v 5 ln(l + x) -i I x' . x" x" , x \Jx x 'i \ x — — + — — — + - * 2 3 4 5 v 3/2 v 5/2 y 7/2 9/2 i/2_£_ ,±_ _^ + _ 2 3 4 5 'ln(l +x) tfx 1 r*/2 V* 3 5 21 18 So that, applying the limits, we get 2 ln(l +x) V* dx '■ * 3 5 21 18 " " ifl_JL + i _ l . l 'Jo l V2 13 20 84 288 880 2496'"'] : 0-7071 ( 0-3333 - 0-0500 + 0-0119- 0-0035 + 0-0011-0-0004.. .] = 0-7071 (0-2924) = 0-2067 10 Here is one for you to do in very much the same way. Evaluate \ y/x.cosx dx JO Complete the working and then check your result with that given in the next frame. 520 Programme 1 9 11 0-531 to 3 decimal places Solution: , X 2 X A X 6 X B . , l/2 jc s/2 X 9 ' 2 x 13 ' 2 . . \/X COS X = X ' + V 2 24 720 .'. I \/x cos x dx '■ J 2x 3/2 x 1 ' 2 ^x ll < 2 x ls/2 3 7 + 132 5400 2_J_ J 1_ 3 7 132 5400 -j: = 0-6667 - 0-1429 + 0-007576 - 0-000185 + . . = 0-531 to 3 dec. pi. Check carefully if you made a slip. Then on to frame 12. 12 The method, then, is really very simple, providing the function can readily be expressed in the form of a series. But we must use this method with caution. Remember that we are dealing with infinite series which are valid only for values of x for which the series converges. In many cases, if the limits are less than 1 we are safe, but with limits greater than 1 we must be extra careful. For instance, f 4 1 the integral — — - 3 dx would give a divergent series when the limits J2 l +x were substituted. So what tricks can we employ in a case such as this? On to the next frame, and we will find out. 521 Approximate Integration To evaluate f 4 1 J 2l+- ,dx 13 We first of all take out the factor x 3 from the denominator 1 1 1 U^hV 1 +x 3 x s U . ,, x 3 + 1 This is better, for if x 3 is going to be greater than 1 when we substitute 1 the limits," i will be x less than 1 Right. So in this form we can expand without further trouble. "J/ l=\ 4 x- 3 l\--,+~- 1 - q +...\dx {/•{ \-X ij rX b ~X*+...\dX lit x *-x " + X + . . Adx Now finish it off. 14 forI = ft 0-088 to 3 decimal places ■x- 6 +x' 9 -x- i2 + ...\dx ' 2 + 5 8 + 11 " 1 1 1 1 2x 2 5x 5 Sx a llx 1 -i 4 2 1 1 -^ + 1 32 5120 524288 1 15 8 160 2048 ' ' = - 0-03125 + 0-00020 - 0-00000 + 0-12500 - 0-00625 + 0-00049 = 0-12569-0-03750 = 0-08819 = 0-088 to 3 dec. pi. 522 Programme 19 16 Method 2. By Simpson 's rule Integration by series is rather tedious and cannot always be applied, so let us start afresh and try to discover some other method of obtaining the approximate value of a definite integral. We know, of course, that integration can be used to calculate the area under a curve y = f(x) between two given points x = a and x = b. Y = f ydx = \ J a J a f(x) dx So, if only we could find the area A by some other means, this would give us the numerical value of the integral we have to evaluate. There are various practical ways of doing this and the one we shall choose is to apply Simpson's rule. So on to frame 1 7. Simpson 's rule To find the area under the curve y =f(x) between x = a and x = b. 4 5 17 y = ft,x) (a) Divide the figure into any even number (n) of equal-width strips (width s) (b) Number and measure each ordinate: yi,y 2 ,yz, ■ ■ ■ , yn + i ■ The number of ordinates will be one more than the number of strips. (c) The area A of the figure is then given by: (F + L) + 4E + 2R Where s = width of each strip, F + L = sum of the first and last ordinates, 4E = 4 X the sum of the even-numbered ordinates, 2R= 2 X the sum of the remaining odd-numbered ordinates. Note that each ordinate is used once — and only once. Make a note of this result in your record book for future reference. 523 Approximate Integration s (F + L) + 4E + 2R J The symbols themselves remind you of what they represent. Example: To evaluate \ y dx for the function y = f(x), the graph of which is shown. Y ^4 5 6 7 q _ 1 j s - 6 r To find I y dx J2 If we take 8 strips, then s =■ — — = n = ;r- s = :r o o 2 2 Suppose we find the lengths of the ordinates to be as follows: Ord. No. Length 12 3 4 5 6 7-5 8-2 10-3 11-5 12-4 12-8 7 8 9 12-3 11-7 11-5 Then we have F + L= 7-5 + 11-5= 19 4E= 4(8-2 + 11-5 + 12-8 + 1 1-7) = 4(44-2) = 176-8 2R= 2(10-3 + 12-4+ 12-3) = 2(35) = 70 So that A--y [19+ 176-8 + 70] = -7- [265 -8] =44-3 :. A = 44-3 units 2 6 6 ■'■J 2 /(x)dx^44-3 The accuracy of the result depends on the number of strips into which we divide the figure. A larger number of thinner strips gives a more accurate result. Simpson's rule is important: it is well worth remembering. Here it is again: write it out, but replace the query marks with the appropriate coefficients. A-^ (F + L) + ?E + ?R 18 524- Programme 19 19 20 ^f (F + L)+4E + 2R In practice, we do not have to plot the curve in order to measure the ordinates. We calculate them at regular intervals. Here is an example. r»/3 . Example: To evaluate I V sm x dx, using six intervals. Jo (a) Find the value of s: n 1 3-0 n . s = — = — (=10 intervals) 6 lo (b) Calculate the values of y(i.e. Vsin x) at intervals of 7r/18 between x = (lower limit) and x = 77/3 (upper limit), and set your work out in the form of the table below. X sinx Vsin* (0°) 0-0000 0-0000 77/18(10°) 0-1736 0-4166 tt/9 (20°) 0-3420 Leave the right-hand side of 77/6 (30°) 0-5000 your page blank for the 2;r/9 (40°) moment. 57r/18(50°) tt/3 (60°) Copy and complete the table as shown on the left-hand side above. Here it is: check your results so far. (0 00 (iii) X sinx Vsin x F+L E R (0°) 77/18(10°) rr/9 (20°) tt/6 (30°) 277/9 (40°) 577/18(50°) 77/3 (60°) 00000 0-1736 0-3420 0-5000 0-6428 0-7660 0-8660 0-0000 - 0-4166 - 0-5848 - 0-7071 - 0-8016 - 0-8752 - 0-9306 - _ -- -_ , -- - " ' Now form three more columns on the right-hand side, headed as shown, and transfer the final results across as indicated. This will automatically sort out the ordinates into their correct groups. Then on to frame 21. 525 Approximate Integration (i) (ii) (iii) F + L E R Note that O-OOOO., (a) You start in column 1 04166 (b) You then zig-zag down the "0-5848 two right-hand columns 0-707 K' (c) You finish back in column 1 . .0-8752 .-'0-8016 0-9306' Now total up each of the three columns. Your results should be: Now (a) Multiply column (ii) by 4 so as to give 4E, (b) Multiply column (iii) by 2 so as to give 2R, (c) Transfer the result in columns (ii) and (iii) to column (i) and total column (i) to obtain (F + L) + 4E + 2R. Now do that. This gives: F+L- 4E- 2R- (F + L) + 4E + 2R . s F + L 0-9306 7-9956 2-7728 1 1 -6990 R 1 -9989 4 7-9956 1-3864 2 2-7728 21 The formula is A - 1 [(F + L) + 4E + 2R] so to find A we simply need to multiply our last result by 4. Remember s = n/l&. So now you can finish it off. I \/smxdx= 22 23 526 Programme 19 24 0-681 For: A-| [(F + L) + 4E + 2R] - E ^ [11-6990] ^tt/54 [11-6990] ^0-6806 •f tt/3 Vsinxdx —0-681 Before we do another example, let us see the last solution complete. To evaluate I Vsin x dx by Simpson's rule, using 6 intervals. Jo , n s = ILLJ^. = 7r /i8 (=10° intervals) 6 X sin x Vsinx F + L E R (0°) 00000 0-0000 0-0000 7i7l8(10°) 0-1736 0-4166 0-4166 tt/9 (20°) 0-3420 0-5848 0-5848 tt/6 (30°) 0-5000 0-7071 0-7071 2?r/9 (40°) 0-6428 0-8016 0-8016 5tt/18(50°) 0-7660 0-8752 0-8752 tt/3 (60°) 0-8660 0-9306 0-9306 F+L >■ 0-9306 1-9989 1-3864 4E 2R t- 4E + 2R > 7-9956 2-7728 4 2 7-9956 2-7728 (F + L) 11-6990 I-| [(F + L) + 4E + 2R] ^- [11-6990] = 0-6806 •tt/3 Vsinx dx —0-681 Now we will tackle example 2 and set it out in much the same way. Turn to frame 25. 527 Approximate Integration •1-0 Example 2. To evaluate I VO + x 3 )dx, using 8 intervals. J 0-2 First of all, find the value of s in this case. 25 o-i For s = l-0-0-2 _0- 8 8 26 o-i 0-1 Now write the column headings required to build up the function values. What will they be on this occasion? X X 3 1 +x 3 Vd +* 3 ) F + L E R Right. So your table will look like this, with x ranging from 0-2 to 1-0. x x 3 1 +x 3 V(l+x 3 ) F + L E R 0-2 0-008 1-008 1-0039 0-3 0-027 1-027 1-0134 04 0-064 0-5 0-125 0-6 0-216 0-7 0-343 0-8 0-9 1-0 F + L > 4E ► 2R > 4 2 (F + L) + 4E + 2 U ) Copy down and complete the table above and finish off the working to evaluate 1 y/( 1 + x 3 )dx . J 0-2 Check with the next frame. _ 27 528 Programme 19 28 ■1-0 V(l +x 3 )<±c = 0-911 J 0-2 X X 3 1 + x 3 V(i+* 3 ) F + L E R 0-2 0-008 1-008 1 -0039 1 -0039 0-3 0-027 1-027 1-0134 1-0134 0-4 0-064 1-064 1-0316 1-0316 0-5 0-125 1-125 1-0607 1-0607 0-6 0-216 1-216 1-1027 1-1027 0-7 0-343 1-343 1-1589 1-1589 0-8 0-512 1-512 1-2296 1-2296 0-9 0-729 1-729 1-3149 1-3149 1-0 1-000 2-000 1-4142 1-4142 F + L 4E 2R 4E + 2R 2-4181 4-5479 3-3639 2 (F + L) + 18-1916 6-7278 4 18-1916 6-7278 y 27-3375 I = | [(F + L) + 4E + 2R] There it is. Next frame = ^y [27-3375] =^ [2-73375] =0-9113 .'. f y/(l+x 3 )dx* 0-911 J 0-2 " f_ g* Here is another one: let us work through it together. £«J Example 3. Using Simpson's rule with 8 intervals, evaluate V ydx, where the values of y at regular intervals of x are given. •" 1 1-0 1-25 1-50 1-75 2-00 2-25 2-50 2-75 3-00 2-45 2-80 3-44 4-20 4-33 3-97 3-12 2-38 1-80 If these function values are to be used as they stand, they must satisfy the requirements for Simpson's rule, which are: (i) the function values must be spaced at intervals of x, and (ii) there must be an number of strips and therefore an number of ordinates. 529 Approximate Integration regular; even; odd These conditions are satisfied in this case, so we can go ahead and evaluate the integral. In fact, the working will be a good deal easier for we are told the function values and there is no need to build them up as we had to do before. In this example, s= For 5 = - 1 2 -=4=0-25 0-25 30 31 Off you go, then. Set out your table and evaluate the integral defined by the values given in frame 29. When you have finished, move on to frame 32 to check your working. 6-62 I=| [(F + L) + 4E + 2R] = ^ [79-43] = 6-62 0-25 X y F + L E R 1-0 2-45 2-45 1-25 2-80 2-80 1-50 3-44 3-44 1-75 4-20 4-20 2-00 4-33 4-33 2-25 3-97 3-97 2-50 3-12 3-12 2-75 2-38 2-38 3-00 1-80 1-80 F + L^- 4-25 13-35 10-89 4E-^ 53-40 21-78 4 2 2R-h> 53-40 21-78 (F + L) + 4E + 2R— ► 79-43 [79-43] ■J; y dx —6-62 32 530 Programme 1 9 33 Here is one further example. Example 4. A pin moves along a straight guide so that its velocity v (cm/s) when it is a distance x (cm) from the beginning of the guide at time t (s), is as given in the table below. '00 v (cm/s) 0-5 1-0 1-5 2-0 2-5 3-0 3-5 4-0 4-00 7-94 11-68 14-97 17-39 18-25 16-08 dx v =— .. x dt Apply Simpson's rule, using 8 intervals, to find the approximate total distance travelled by the pin between t = and t = 4. We must first interpret the problem, thus: •4 vdt 10 and since we are given values of the function v at regular intervals of t, and there is an even number of intervals, then we are all set to apply Simpson's rule. Complete the problem, then, entirely on your own. When you have finished it, check with frame 34. 34 46-5 cm t V F + L E R 0-00 0-00 0-5 4-00 4-00 1-0 7-94 7-94 1-5 11-68 11-68 2-0 14-97 14-97 2-5 17-39 17-39 3-0 18-25 18-25 3-5 16-08 16-08 4-0 0-00 0-00 F + L-* 0-00 49-15 41-16 4E-»- 2R— ► h L) + 4E + 2R — ► 196-60 82-32 4 2 196-60 82-32 (F- 278-92 x=|[(F + L ) + 4E + 2 R] and s = 0-5 .'. X = •^[278-92] =46-49 .\ T otal distance —46-5 cm 531 Approximate Integration Proof of Simpson's rule Q |" So far, we have been using Simpson's rule, but we have not seen how it is ** ** established. You are not likely to be asked to prove it, but in case you are interested here is one proof. Y Divide into an even number of strips (2k) of equal width (s). Let the ordinatesbe>' 1 ,_y 2 ,j 3 ,...>' 2 „ + i. Take OX and OY as axes in the position shown. Then A = {~s,y 1 ); B = (0,y 2 ); C = {s,y 3 ) Let the curve through A, B, C be represented by y =a + bx + ex 2 y t =a+b(-s) + cs 2 (i) yj = a (ii) y 3 = a + bs + cs 2 (iii) (iii) - (i) y 3 -y x = 2bs :. b = — (y 3 -y{) (i) + (iii) - 2(ii) yi + y 3 -2y 2 = 2 cs 2 :. c=A (y, - 2y 2 +y 3 ) 2s 1 Let A t = area of the first pair of strips. Aj = I y dx — \ (a + bx + ex 2 ) dx ■ 1>*1 -s _]-s „3 bx 2 ex 3 ax+ — + ~ -2as+~- ~2sy 2 +-| .—. - 2 (>, ~2y 2 +y 3 ) -f (6y 2 +y l -2y 2 +y 3 )^±(y 1 + 4y 2 +y 3 ) So &i-j(y 1 +4y 2 +y 3 ) Similarly A 2 -| (y 3 + 4y 4 + j/ s ) A 3 --f Os +4y 6 +^ 7 ) A n -| 2n -i +4j 2 „ +Ji„ + V ) Total area A = Aj + A 2 + A 3 + + A„ . Oi +^2» + i) + 4(y 2 + j> 4 + . . . + >> 2 „) + 2(y 3 +.y s + A = I [(F + L) + 4E + 2R] A^- • A 3 + yin-l) On to frame 36. 532 Programme 19 36 We have almost reached the end of the programme, except for the usual Test Exercise that awaits you. Before we turn to that, let us revise once again the requirements for applying Simpson's rule. (a) The figure is divided into an even number of strips of equal width x. There will therefore be an odd number of ordinates or function values, including both boundary values. [b (b) The value of the definite integral I f{x)dx is given by the numerical value of the area under the curve .y ~f(x) between x = a and x = b I = A-| [(F + L) + 4E + 2R] where s = width of strip (or interval), F + L = sum of the first and last ordinates, 4E = 4 X sum of the even-numbered ordinates, 2R = 2 X sum of remaining odd-numbered ordinates. (c) A practical hint to finish with: Always set your work out in the form of a table, as we have done in the examples. It prevents your making slips in method and calculation, and enables you to check without difficulty. Now for the Test Exercise. The problems are similar to those we have been considering in the programme, so you will find them quite straight- forward. On then to frame 37. 533 Approximate Integration Test Exercise — XIX Work through all the questions in the exercise. Set the solutions out neatly. Take your time: it is very easy to make numerical slips with work of this kind. 1 . Express sin x as a power series and hence evaluate 37 sin x dx to 3 places of decimals. r 0-2 2. Evaluate \ x l e 2x dx correct to 3 decimal places. J 0-1 3. The values of a function^ =f(x) at stated values of x are given below. X 2-0 2-5 3-0 3-5 4-0 4-5 5-0 5-5 6-0 y 3-50 6-20 7-22 6-80 5-74 5-03 6-21 8-72 11-10 Using Simpson's rule, with 8 intervals, find an approximate value r 6 of \ y dx. J2 fir/2 4. Evaluate I Vcos 9 d6, using 6 intervals. JO frr/2 5. Find an approximate value of I vU ~~ 0-5 sm 2 d)d8 using Simpson's rule with 6 intervals. ° Now you are ready for the next programme. 534 Programme 1 9 Further Problems - XIX Si \/(l ~x 2 )dx (i) by direct integration, (ii) by expanding as a power series, (iii) by Simpson's rule (8 intervals). 2. State the series for ln(l + x) and for ln(l — x) and hence obtain a . , , ,1 +x] series tor In 1 -xY 0-3 n + x Evaluate \ ln| _ \dx, correct to 3 decimal places. 3. In each of the following cases, apply Simpson's rule (6 intervals) to obtain an approximate value of the integral. f v l 2 dx r n J- WJ HTZTx (b)J o (5-4co,*)* dfl {C) )o Vd-isin 2 ^) 4. The coordinates of a point on a curve are given below. X 1 2 3 4 5 6 7 8 y 4 5-9 7-0 6-4 4-8 34 2-5 1-7 1 The plane figure bounded by the curve, the x-axis and the ordinates at x = and x = 8, rotates through a complete revolution about the x-axis. Use Simpson's rule (8 intervals) to obtain an approximate value of the volume generated. 5. The perimeter of an ellipse with parametric equations x = 3 cos 8 , y = 2 sin 6 , is 2 V2 \ (1 3 - 5 cos 20) T dd . Evaluate this integral JO using. Simpson's rule with 6 intervals. 2 6. Calculate the area bounded by the curve y = e x , the x-axis, and the ordinates at x = and x = 1 . Use Simpson's rule with 6 intervals. 535 Approximate Integration 7. The voltage of a supply at regular intervals of 0-01 s, over a half- cycle, is found to be: 0, 19-5, 35, 45, 40-5, 25, 20-5, 29, 27, 12-5, 0. By Simpson's rule (10 intervals) find the r.m.s. value of the voltage over the half-cycle. 8. Show that the length of arc of the curve x = 3d — 4 sin 6 , 7 = 3 — 4 cos 9, between 0=0 and 6 = 2-7T, is given by the integral r 2n \ V(25 - 24 cos 9)dd . Evaluate the integral, using Simpson's rule JO with 8 intervals. i 9. Obtain the first four terms of the expansion of (1 + x 3 } 1 and use them to determine the approximate value of I >/(l +x 3 )dx, correct to three decimal places. 10. Establish the integral in its simplest form representing the length of the curve y =\ sin 6 between 0=0 and 6 - tt/2. Apply Simpson's rule, using 6 intervals, to find an approximate value of this integral. 1 1 . Determine the first four non-zero terms of the series for tan -1 * and hence evaluate \ \Jx.Un l x dx correct to 3 decimal places. JO 12. Evaluate, correct to three decimal places, (i) \ y/x.cos x dx, (ii) I y/x. sin x dx. JO Jo r»/2 13. Evaluate \ V(2-5 - 1-5 cos 26)dd by Simpson's rule, using 6 Jo intervals. [1 J. 14. Determine the approximate value of \ (4 + jc 4 ) 2 dx JO (i) by first expanding the expression in powers of x, (ii) by applying Simpson's rule, using 4 intervals. In each case, give the result to 2 places of decimals. 536 Programme 20 POLAR CO-ORDINATES SYSTEM Programme 20 1 Introduction to polar co-ordinates We already know that there are two main ways in which the position of a point in a plane can be represented. (i) by Cartesian co-ordinates, i.e. (x,y) (ii) by polar co-ordinates, i.e. (r, 8). .The relationship between the two systems can be seen from a diagram. T (x,y) (r,6) For instance, x and y can be expressed in terms of rand 6 . '■ r cos 6 ; y - r sin ( Or, working in the reverse direction, the co-ordinates r and 6 can be found if we know the values of x and y. r = y/(x i +y 2 y,e=^ri 1 (^) This is just by way of revision. We first met polar co-ordinates in an earlier programme on complex numbers. In this programme, we are going to direct a little more attention to the polar co-ordinates system and its applications. First of all, an easy example or two to warm up. Example 1. Express in polar co-ordinates the position (-5, 2). Important hint: always draw a diagram; it will enable you to see which quadrant you are dealing with and prevent your making an initial slip. p Remember that 6 is measured from -js the positive OX direction. In this case, the polar co-ordinates of P are 539 Polar Co-ordinates System (5-385, 158°12') (i) r 2 = 2 2 + 5 2 = 4 + 25 = 29 .-. r = \/29 = 5-385 . (ii) tan E =|=0-4 /. E = 21°48' :. e = 158°12' Position of Pis (5-385, 158°1 2') A sketch diagram will help you to check that is in the correct quadrant. Example 2. Express (4, -3) in polar co-ordinates. Draw a sketch and you cannot go wrong! When you are ready, move to frame 5. (5,323°8') (i) ,-* = 3 2 + 4 2 = 25 :. r = 5 (ii) tan E =|=0-75 :. E = 36°52' (4,-3) = (5,323°8') .-. 6 = 323°8' Example 3. Express in polar co-ordinates (-2, -3). Finish it off and then move to frame 6. 3-606, 236°19' Check your result . 8 ^ (i) r 2 = 2 2 +3 2 = 4 + 9= 13 x r = Vl3 = 3-606 (ii) tan E =|=l-5 /. E = 56°19' .-. e =236°19' (-2,-3) = (3-606,236°19') Of course, conversion in the opposite direction is just a matter of evaluat- ing x = r cos 6 and y = r sin 6 . Here is an example. Example 4. Express (5, 124°) in Cartesian co-ordinates. Do that, and then mov e on to frame 7. 540 Programme 20 Working (-2-796,4-145) (i) x = 5 cos 124° = -5 cos 56° = -5 (0-5592) =-2-7960 (ii) y = 5 sin 124° = 5 sin 56° = 5(0-8290) = 4-1450 .-.(5, 124°) = (-2-796, 4-145) That was all very easy. Now, on to the next frame. 8 Polar curves In Cartesian co-ordinates, the equation of a curve is given as the general relationship between x and_y, i.e. y =f(x). Similarly, in the polar co-ordinate system, the equation of a curve is given in the form r = f{&). We can then take spot values for 6 , calculate the corresponding values of r, plot r against 8 , and join the points up with a smooth curve to obtain the graph of r = f(8). Example 1. To plot the polar graph of r = 2 sin 8 between 0=0 and = 277. We take values of 8 at convenient intervals and build up a table of values giving the corresponding values of r. 6° 30 60 90 120 150 180 sin 8 0-5 0-866 1 0-866 0-5 r=2sind 10 1-732 2 1-732 1-0 e° 210 240 270 300 330 360 sin 8 r = 2 sin 8 Complete the table, being careful of signs. When you have finished, turn on to frame 9. 541 Polar Co-ordinates System Here is the complete table. sin 6 r = 2 sin 30 0-5 1-0 60 0-866 1-732 90 1 2 120 0-866 1-732 150 0-5 1-0 180 e° sin 6 2 sin 6 210 -0-5 -1-0 240 270 -0-866 -1 -1-732 -2 300 330 360 -0-866 -0-5 -1-732 -1-0 180' (i) We choose a linear scale for r and indicate it along the initial line. (ii) The value of r is then laid off along each direction in turn, points plotted, and finally joined up with a smooth curve. The resulting graph is as shown above. Note that when we are dealing with the 210° direction, the value of r is negative (-1) and this distance is therefore laid off in the reverse direc- tion which once again brings us to the point A. So for values of 6 between 6 = 1 80° and 6 = 360°, r is negative and the first circle is retraced exactly. The graph, therefore, looks like one circle, but consists, in fact, of two circles, one on top of the other. Now, in the same way, you can plot the graph of r = 2 sin 2 . Compile a table of values at 30° intervals between 6=0° and 6 and proceed as we did above. Take a little time over it. When you have finished, move on to frame 10. 360° 542 Programme 20 10 Here is the result in detail. 6 30 60 90 120 150 180 sin 9 sin 2 0-5 0-25 0-866 0-75 1 1 0-866 0-75 0-5 0-25 r - = 2 sin 2 0-5 1-5 22 1-5 0-5 e 210 240 270 300 330 360 sin 6 sin 2 -0-5 0-25 -0-866 0-75 -1 1 -0-866 0-75 -0-5 0-25 r= 2sin 2 0-5 1-5 2 1-5 0-5 This time, r is always positive and so there are, in fact, two distinct loops. Now on to the next frame. 11 Standard polar curves Polar curves can always be plotted from sample points as we have done above. However, it is often useful to know something of the shape of the curve without the rather tedious task of plotting points in detail. In the next few frames, we will look at some of the more common polar curves. So on to frame 12. 543 Polar Co-ordinates System Typical polar curves 1 . r = a sin i 3. r = a cos i >. r = a sin 4. r = a cos 2 150' 12 There are some more interesting polar curves worth seeing, so turn on to frame 13. 544 Programme 20 13 9. r = a(\ +.cos0) 10. r = a(l + 2cos0) 11. r 2 = a 2 cos 20 12. r = «0 135°\ '45° Sketch these 12 standard curves in your record book. They are quite common in use and worth remembering. Then on to the next frame. The graphs of r = a + b cos give three interesting results, according to the relative values of a and b. L 14 (i) If a = b, we get (ii) If a < b , we get — -> (iii) If a > b, we get — (cardioid) (re-entrant loop) (no cusp or re-entrant loop) So sketch the graphs of the following. Do not compile tables of values, (i) r = 2 + 2 cos (iii) r = 1 + 2 cos (ii) /■ = 5 + 3 cos (iv) r = 2 + cos 545 Polar Co-ordinates System Here they are. See how closely you agree. (i) r = 2 + 2 cos 8 (a = b) I (ii) r = 5 + 3 cos6 (a>b) 15 (iii) r = 1 + 2 cos 8 (a < b) (iv) r = 2 + cos 8 (a>b) j If you have slipped up with any of them, it would be worth while to plot a few points to confirm how the curve goes. On to frame 16. To find the area of the plane figure bounded by the polar curve r = /(#) and the radius vectors at 8 =8 X and 8 =8 2 . 16 r=fW p(/-+sr,e+se) <-,e> Area of sector OPQ = 5 A — \r(r + 8r) sin 36 ■ 8A „ , , ^ r . sin 58 fr(r + 8r)- " 86 JC . . . 5A dA . „ sin 50 If 50 -*■ ► — 5y ->■ ■ — ' 88 dd ' ' 88 Next frame. 88 546 Programme 20 17 sin 88 dA_ d8 \r{r + <$)\ =\r 2 J »i /. A= | ~ \r 2 d8 Example 1. To find the area enclosed by the curve r = 5 sin 8 and the radius vectors at 8 = and 8 = tt/3. r = 5sinff "tt/3 A=| Vd0 r.,3 Jo "/3 25 . J ;. A = ^f ' ~ 4(1 - cos 20) c?0 J sin 2 d0 Finish it off. 18 A = ^ A 4 ~ir V3" .3 4_ = 3-84 For: 25 f */ 3 A = 4 (1 - cos 28) , 25 4 sin 29 2 jt/3 25 7T sin 2tt/3 4 3 2 25 4 ~77 V3" .3 4_ = 3-8 A= 3-84 to 2 decimal places Now this one: Example 2. Find the area enclosed by the curve r = 1 + cos and the radius vectors at 8 = and = 7r/2. First of all, what does the curve look like? 547 Polar Co-ordinates System 19 Right. So now calculate the value of A between = and 6 - it/2. When you have finished, move on to frame 20. A = ^+l =2-178 20 For: -rr/2 fn/2 A = i| r 2 dd = M (1 + 2 cos 6 + cos 2 0) d6 (•tt/2 fn Jo Jo ^ . n 6 sin20"W 2 + 2 sin0 + J + — T" t j(f + 2 + )-(0) .'. A=^ + l =2-178 o So the area of a polar sector is easy enough to obtain. It is simply r>e 2 A = Make a note of this general result in your record book, if you have not already done so. Next frame. Example 3. Find the total area enclosed by the curve r = 2 cos 3d . *)\ Notice that no limits are given, so we had better sketch the curve to see *■ ■ what is implied. This was in fact one of the standard polar curves that we listed earlier in this programme. Do you remember how it goes? If not, refer to your notes: it should be there. Then on to frame 22. 548 Programme 20 22 Since we are dealing with r = 2 cos 36, r will become zero when cos 3d = 0, i.e. when 3d = ir/2, i.e. when 6 = 7r/6. We see that the figure consists of 3 equal loops, so that the total area, A, is given by A = 3 (area of one loop) = 6 (area between 0=0 and 6 = 77/6.) rrr/6 Ctt/6 A = 6 \r 1 dd=3\ 4cos 2 36d6 JO Jo 23 ■n units since p»r/6 \{\ + cos 66) dd = 6 L + sin60l 7T/6 = 77 units 2 Now here is one for you to do on your own. Example 4. Find the area enclosed by one loop of the curve r = a sin 26 . First sketch the graph. 24 Arguing as before, r = when a sin 26 = 0, i.e. sin 26 = 0, i.e. 26 = 0, so that 26 = 0, tt, 2ti, etc. .'. 6 =0,77/2,tt, etc. So the integral denoting the area of the loop in the first quadrant will be A = 549 Polar Co-ordinates System < 2 , JO r 2 dd 25 Correct. Now go ahead and calculate the area. A = 7ra 2 /8 units 2 Here is the working: check yours •n/2 26 2 Jo rd6 -2) l sin 2 26 dd a 2 ri 2 = - (1 - cos 46) dd Now on to frame 27. sin 40 tt/2 na . , = —r units To find the volume generated when the plane figure bounded by r=f(d) and the radius vectors at 6 = 6 X and 6 = 6 2 rotates about the initial line. \e = e 2 p(c + Sr, 6 + 88) D X If we regard the elementary sector OPQ as approximately equal to the 2r A OPQ, then the centroid C is distance —from 0. ■ 1, 27 Area OPQ -jr(r + 8r) sin 86 We have: Volume generated when OPQ rotates about OX = 5 V .'. 5V = area OPQ X distance travelled by its centroid (Pappus) = \r(r + 8r) sin 86. 2n CD 1 2 = ~r(r + 8r) sin 86 .2-nr^r sin 6 = -j itr 2 (r + 8r) sin 86 . sin 6 :.f e =l«r\r + 8r) dV Then when 50 -*■ 0, — = at* sin 55 . 50 sin 6 550 Programme 20 28 dY 2 3 ■ a -T- =-z-irr° sin 6 ax 3 and .-. V = 29 C°2 1 V= -r-jrr 3 sin (9 d0 J 8l Correct. This is another standard result, so add it to your notes. Then move to the next frame for an example. 30 Example 1. Find the volume of the solid formed when the plane figure bounded by r = 2 sin 6 and the radius vectors at = and 8 = n/2, rotates about the initial line. Well now, V fTT/2 2 = Jo 3* /•7T/ Jo r 3 sin dd 7r.(2 sin9) 3 . sine dd (*7T Jo W2 16 7T sin 4 dfl Since the limits are between and 7r/2, we can use Wallis's formula for this. (Remember?) So V= 31 V = -n- 2 units 3 For V ' 3 J 167T 3.1 77 ~3~'4.2 '2 sin 4 6> d6> . - = 7T units Example 2. Find the volume of the solid formed when the plane figure bounded by r = 2a cos 6 and the radius vectors at 8 = and = 7r/2, rotates about the initial line. Do that one entirely on your own. When you have finished it, move on to the next frame. 551 Polar Co-ordinates System v = - units For •tt/2 9 V = | — .tt./- 3 sin ^ (ifi" and r = 2a cos -/ f tt/2 2 I 9 . sin c?0 cos 9 (- sm 0) <ro \6tra z cos 4 e W2 167T0 3 Jo V = — — units So far, then, we have had 9 2 , „ (i) A re 2 J B t _f 92 J 9j 9 2 9 (ii) V = | -ttt?- 3 sine d6 ■ Check that you have noted these results in your book. To find the length of arc of the polar curve r =f(6), between 6 =9 t and 9 = 6 2 . \e=e 2 -r = fW) 32 33 5s 2 or 2 With the usual figure 5s 2 - r 2 . 50 2 + dr 2 :. — 2 - r 2 + —. ■- se 2 66 1 If 66+0, (| r-^ii ■■■svMS) 1 )--- s = 552 Programme 20 34 Example 1. Find the length of arc of the spiral r = ae 36 from 6 = to = 2tt. Now, dr 2 dO- 2> c2tt ' 2 e 6e +9a 2 e 6e lCtoV .. s 35 = £V10| C 67T. Since 2 " /in 39 w fl VlOa 3 ^"Vio/^.j As you can see, the method is very much the same every time. It is merely a question of substituting in the standard result, and, as usual, a knowledge of the shape of the polar curves is a very great help. Here is our last result again. ■0HS)> Make a note of it: add it to the list. 36 Now here is an example for you to do. Example 2. Find the length of the cardioid r = a(\ + cos 8) between 6 = and 6 = n. Finish it completely, and then check with the next frame. 553 Polar Co-ordinates System s = Aa units Here is the working: r = a(l + cos0) 20 .. — = -a sin 6 ad •'• r% + (to) =a2 { 1 +2cos6 + cos20 + sin20 } = a 2 h + 2 cos fl}= 2a 2 (1 + cos 6) Now cos 6 can be re-written as ( 2 cos 2 ^ - l) ■'■J {•"&)>»■ .'. s =1 2a COSrrdd Jo l = Aa [l - 0] = 4a units ,2 iL 2 COS' 2 sin ~ Next frame. 37 Let us pause a moment and think back. So far we have established J Q three useful results relating to polar curves. Without looking back in this programme, or at your notes, complete the following. If /■=/((?), (i) A = 00 V=.... (iii) s= To see how well you have got on, turn on to frame 39. 554 Programme 20 39 A = T 2 \ r 2 dd V= 2 ^.n.r 3 sind dd -i:7i"^)> If you were uncertain of any of them, be sure to revise that particular result now. When you are ready, move on to the next section of the programme. 40 Finally, we come to this topic. To find the area of the surface generated when the arc of the curve r = f(d) between 6 = d t and 6 = 9 2 rotates about the initial line. Once again, we refer to our usual figure. If the elementary arc PQ rotates about OX, then, by the theorem of Pappus, the surface generated, 5S, is given by (length of arc) X (distance travelled by its centroid). :. 5S-5s. 27rPL-Ss.27rrsin0 .. --2*rsm6- From our previous work, we know that — — /{ r 2 + (~Tq) } so that 6S -o • , 7T — Z7TT sm I 00 And now, if 88 -*■ 0, — =2flrsin do This is also an important result, so add it to your list. 555 Polar Co-ordinates System S=J^2 W rsinfl/(r» + (|))d» This looks a little more involved, but the method of attack is much the same. An example will show. Example 1. Find the surface area generated when the arc of the curve r = 5(1 + cos 0) between = and 6 = it, rotates completely about the initial line. dr Now, r=5(l+cos0) .'. — = -5sin0 \Jd) *«% - 41 50(1 +cos0) for r 2 +(j) = 25(1 + 2 cos d + cos 2 + sin 2 0) = 25(2 + 2 cos 6) = 50(1 +cos0) We would like to express this as a square, since we have to take its root, so we now write cos d in terms of its half angle. ■•-^1/ =50(1 + 2 cos' | -1) = 100 cos 2 - ' idr\ 2 > 42 •VMS)]- 10 cos- So the formula in this case now becomes S= 556 Programme 20 43 C n ft = 2tt.5(1 + cose) sin 0.10 cos j: -dd Jo 2 •»/: S = IOOtt (1 + cos 0) sin cos ^ dd 2 We can make this more convenient if we express (1 + cos 0) and sin ( a in terms of-. What do we get? 44 S = 400 ■j; cor- sin- dd S= 100? cos0) sin cos- dd 2 2' 7T 2 cos 2 - 2 sin ^ cos^-, cos-dd JO •j: ^ I sin ^ -i Now the differential coefficient of cos- is { j^- \ T Jo = 1007 = 400? cos* - sin - dd . :. S=-800 7 cos4 ff"~r^ Finish it off. 45 S = 1 60 7T units Since S = -800 Trf" cos 4 |f- ^ dd -800tt 'COS -800tt [0-1] S = 1607T units And finally, here is one for you to do. Example 2. Find the area of the surface generated when the arc of the curve r = ae e between 0=0 and d = 7r/2 rotates about the initial line. Finish it completely and then check with the next frame. 557 Polar Co-ordinates System S = ^.7ra 2 (2e" + 1) For, we have: And, in this case, fW2 S = \ 27rrsin0 Jo >&» e ■ dr r = ae .. — =ae dO ,dr- 2 2 e 2e +a 2 e 2f) = 2a 2 e 26 iar\ fir/2 s = Jo = 2y/2na 2 [ n '~ e 26 sin d6 2nae e smd.s/2ae 6 dd 7T-/2 Let I = U 2 " sin 6 dd = e 2e (- cos d) + 2 fcos e 26 d6 e 26 cos 6» + 2 [e 26 sin Q - 2 [sin 6 e 26 dd 20 l=-e 2 ° cos9 + 2e 2H sin 0-41 .". 51 = e 20 \2 sin0 -cos i e 26 I = — } 2 sin 6 - cos i ( : :. S = 2V2.7T.fl 2 - e 28 — { 2 sin d - cos i r/2 46 = 2v^L£ 2 Lir (2 -0)- 1(0-1) S = ^%^- 2 (2^ + l) units 2 We are almost at the end, but before we finish the programme, let us collect our results together. So turn on to frame 47. 558 Programme 20 47 Revision Sheet Polar curves — applications. 1. Area K=\ d2 \r 2 dB J 6i 2. Volume V= 2 \ it r 3 sin d dd 3. Length of arc -r./^^f 4. Surface of revolution S= 27rrsin0 /v^i'Tn))^ It is important to know these. The detailed working will depend on the particular form of the function r=f{6), but, as you have seen, the method of approach is mainly consistent. The Test Exercise now remains to be worked. First brush up any points on which you are not perfectly clear; then, when you are ready, turn on to the next frame. 559 Polar Co-ordinates System Test Exercise — XX Answer all the questions. They are quite straightforward: there are no tricks. But take your time and work carefully. 1. Calculate the area enclosed by the curve rd 2 - 4 and the radius vectors at 6 = -a\2 and 9 = it. 2. Sketch the polar curves: (i) r = 2 sin 6 (ii) r = 5 cos 2 . (iii) r = sin 20 (iv) r = 1 + cos (v) r = 1 + 3 cos 6 (vi) r = 3 + cos 6 3. The plane figure bounded by the curve r = 2 + cos 6 and the radius vectors at - and 9 = it, rotates about the initial line through a complete revolution. Determine the volume of the solid generated. 4. Find the length of the polar curve r = 4 sin 2 -= between (5 = and 6 =7r. 5. Find the area of the surface generated when the arc of the curve r = a(l - cos 9) between 6=0 and 9 = it, rotates about the initial line. That completes the work on polar curves. You are now ready for the next programme. 48 560 Programme 20 Further Problems - XX 1 . Sketch the curve r = cos 2 0. Find (i) the area of one loop and (ii) the volume of the solid formed by rotating it about the initial line. 3 1 1 2. Show that sin 4 = --— cos 28 +—cos40. Hence find the area o z 8 bounded by the curve r = 4 sin 2 and the radius vectors at 8 = and 6 = n. a 3. Find the area of the plane figure enclosed by the curve r = a sec 2 (— ) and the radius vectors at 8 = and 6 = 7r/2. 4. Determine the area bounded by the curve r = 2 sin 8 + 3 cos 8 and the radius vectors at 6 = and 6 = 7t/2. 2 5. Find the area enclosed by the curve r = — and the radius 1 + cos 26 vectors at 8 = and 8 = tt/4. 6. Plot the graph of r = 1 + 2 cos at intervals of 30° and show that it consists of a small loop within a larger loop. The area between the two loops is rotated about the initial line through two right -angles. Find the volume generated. 7. Find the volume generated when the plane figure enclosed by the curve r~2a sin 2 [— (between 0=0 and 6 - n, rotates around the initial line . 8. The plane figure bounded by the cardioid r = 2a{\ + cos 6) and the parabola r(l + cos 8) = 2a rotates around the initial line. Show that the volume generated is 18;ra 3 . 9. Find the length of the arc of the curve r = a cos 3 (— | between 0=0 and 8 = 3n. 10. Find the length of the arc of the curve r = 3 sin 8 +4 cos 8 between 6 = and d = ■nil. 561 Polar Co-ordinates System 1 1 . Find the length of the spiral r = ad between 6=0 and 6 = 2n. — J and calculate its total length. 13. Show that the length of arc of the curve r = a cos 2 6 between 0=0 and 6 = tt/2 is a [2\/3 + ln(2 + y/3)] l(2s/3). 14. Find the length of the spiral r = ae be between 6=0 and 9 = d x , and the area swept out by the radius vector between these two limits. 15. Find the area of the surface generated when the arc of the curve r 2 = a 2 cos 26 between 6=0 and 6 = rr/4, rotates about the initial line. 562 Programme 21 MULTIPLE INTEGRALS Programme 21 4 Summation in two directions Let us consider the rectangle bounded by the straight lines, x = r, x = s, y = k,y = m, as shown. Y P y i' i i l 1 i i So i i °! i i ■• x — Hsxh- ■ X Then the area of the shaded element, 6a = 5a = by. bx If we add together all the elements of area, like 6a, to form the vertical strip PQ, then 6 A, the area of the strip, can be expressed as 6A= 5A y = m = 2 y = k by bx Did you remember to include the limits? Note that during this summation in the ^-direction, bx is constant. Y t If we now sum all the strips across the figure from x = r to x = s, we shall obtain the total area of the rectangle, A. .'. A= 2 (all vertical strips like PQ) x = r y = m x = s = 2 x = r 2 by.bx Removing the brackets, this becomes x = s y = m A= 2 2 by.bx. x = r y = k If now by -*■ and bx -*■ 0, the finite summations become integrals, so the expression becomes A = 565 Multiple Integrals -r i J x = r J y = m y = k dy dx To evaluate this expression, we start from the inside and work outwards. ; dy\ dx=\ X = s x = r y = m dx y = k -r J X = 1 (m - k) dx and since m and k are constants, this gives A : A = (m - k) . (s - r) 4 for A = (m - k) x = (m-k) A = {m - k) . (s - r) which we know is correct, for it is merely A = length X breadth. That may seem a tedious way to find the area of a rectangle, but we have done it to introduce the method we are going to use. First we define an element of area by ■ 8x. Then we sum in the j> -direction to obtain the area of a Finally, we sum the result in the x-direction to obtain the area of the vertical strip; whole figure We could have worked slightly differently: Y 8y i i t y c ! ! &° i i D \ " 1 j , " — x Asxr- ■ X As before 5a = 8x. by. If we sum the elements in the x-direction this time, we get the area 5Ai of the horizontal strip -CD :. 5A, 566 Programme 21 x = s SAi = 2 8xj. by x = r 8^ x Now sum the strips vertically and D we obtain once again the area of the whole rectangle. y ~ m Ai = 2 (all horizontal strips like CD) y = k y = m I x = s 2 2 6jc. 5y y = & { x = r As before, if we now remove the brackets and consider what this becomes when bx -> and by -* 0, we get Ai = 8 n y = m >> = * dx.dy To evaluate this we start from the centre -y = m A, = >> = * <ix > dy Complete the working to find Ai awe? tfien move on to frame 9. Ai =(s-r).(m- k) For /•^ ~ "<r -is f'" Ai = * dy = (s - r) dy = (s - r) J > = fc L J'" J/t .'. Ai = (s - r) . (m - k) which is the same result as before. So the order in which we carry out our two summations appears not to matter. Remember (i) We work from the inside integral. (ii) We integrate w.r.t. x when the limits are values of x. (hi) We integrate w.r.t. y when the limits are values of.y. Turn to the next frame. 567 Multiple Integrals Double integrals The expression 1 ■ yi yi J X2 10 fix, y) dx dy is called a double integral Xi (for obvious reasons!) and indicates that (i) f[x, y) is first integrated with respect to x (regarding y as being constant) between the limits x = x v and x - x 2 , (ii) the result is then integrated with respect to y between the limits y=yi ■&ndy=y 2 - Example 1 2 r 4 Evaluate 1=1 (x + 2y) dx dy So (x + 2y) is first integrated w.r.t. x between x = 2 and x = 4, with y regarded as constant for the time being. < : f 4 ; 1 (jc + 2y) dx \ dy. -i: V j +2xy_ 4 ■ dy 2 = [ j(8 + 8^)-(2 + 4^)J dy = 1 (6 + 4y)dy = Finish it off. J 1 For 1= 12 6y + 2y 2 ~j 2 11 I = f (6 + 4 y)dy = 1 1 Here is another. = (1 2 + 8) - (6 + 2) = 20 - 8 =12 Example 2 Evaluate 1=1 I x 2 y dx dy J 1 J Do this one on your own. Remember to start with x 2 y dx with J y constant. Finish the double integral co mpl etely an d then turn on to frame 12. 568 Programme ^ 12 1=13-5 Check your working: "2 r3 -J J ■J x 2 y dx dy = J J x 2 y dx dy x = 3 x = ^ (9^)dv = 9f~ 2 = 18-4-5 = 13-5 Now do this one in just the same way. Example 3 Evaluate I = j f (3 + sin 0) d6 dr When you have finished, check with the next frame. 13 I = 3tt + 2 Here it is: I=f (" (3 + sm6)dedr. ■i: 3d - cos d ■n dr = f /(3ir+l)-(-l)}tfr =J (3v + 2)dr = (3w + 2)/|* = ( > + 2) (2-1) = 3u + 2 On to the next frame. 569 Multiple Integrals Triple integrals. Sometimes we have to deal with expressions such as •b r d r f ro r a /•/ 1= f(x,y,z)dx.dy-dz •'a Jc Je but the rules are as before. Start with the innermost integral and work outwards. Jb i »d ; » / a \ J c \Je ■<D ■<2) ■Q> f(x,y,z)dx- dy dz All symbols are regarded as constant for the time being, except the one variable with respect to which the stage of integration is taking place. So try this one on your own straight away. Example 1. Evaluate 1=1 I (x + 2y-z)dx. dy. dz Jl J-lJ I = -8 Did you manage it first time? Here is the working in detail. 1 = 1 I I (x + 2y-z)dx.dy.dz J 1 J— lJ -ft ~ + 2xy - xz dy.dz = j j (2 + Ay - 2z) dy. dz = f = (" {(2 + 2-2 2 )-(-2 + 2 + 2 2 ))rfz =f(4- 3 2y + 2y 2 - 2yz ■3 1 dz 4z)dz 4z - 2z 2 = (12-18)-(4-2) =- 8 f2f3 pi Example 2. Evaluate (p 2 + q 2 -r 2 )dp.dq.dr ■J iJ qJq When you have finished it, turn on to frame 16. 14 15 570 Programme Zx 16 1 = 3 For 1=1 ( (p 2 +q 2 ~r 2 )dpdqdr 0^0 + pq 2 -pr 2 -in -J =1 dq dr H -^ (1 +9-3r 2 )dr IQr-r 3 J 1 = 12-9 = 3 ;(20 ~8) -(10-1) It is all very easy if you take it steadily, step by step. Now two quickies for revision: 4 i>lx Evaluate (i) dy dx, (ii) 2ydydx. J 1 J 3 J Jl Finish them both and then move on to the next frame. 17 (i) 1 = 2; (ii) 1=1$ Here they are. =4-2=2 .y c?x = | (5-3)dJC = f 2cfr = 2x lii) 1= lydydx =\ y 2 dx = \ (9x 2 -l)dx J oJ i J oL J * Jo 3x -jc = 192-4= 15 And finally, do this one. 1 = 1 I (3x 2 - 4) dx dy = u 571 Multiple Integrals Check the working. 1=15 = f f (3x 2 -4r)dxdy J QJ l -J =1 x 3 - 4x dfy (8-8)-(l-4)Uy 3rfy = 3y. >15 Now let us see a few applications of multiple integrals. Move on then to the next frame. Applications 4X Example 1. Find the area bounded by y = -=, the ^r-axis and the ordinate at x - 5 . 5. Area of element = by. bx y ~y\ .'. Area of strip 2 Sj> . 5x >> = The sum of all such strips across the figure gives us x= 5 ( y =yi A^ 2 2 5^.6jc x = | ^ = x = 5 y = y\ — 2 2 hy.hx x = y = Now, if by -* and fix -> 0, then J oJ But y x 4x = ^ dx = \ yidx J oL Jo Jo Fini's/i ir off. SoA = 18 19 572 Programme 21 20 A= 10 units 2 For A=| ^rdx 1 2x 2 = 10 Right. Now what about this one? Example 2. Find the area under the curve y - 4 sin — between x = — and x = 7r, by double integral method. Y / = 4 sin ^ Steps as before. Area of element = dy.Bx Area of vertical strip y -y\ 2 5j.5x >> = * = t \ y -yi A- 2 2 Sj.Sx x = tt/31 >> = Sx Total area of figure: If hy -* and 8x -+ 0, then A = Jir/3J0 dyabr : Complete it, remembering that y\ = 4 sin: 21 For you get A = 4a/3 units 3 fir ryi fw r -ij'i T f " A= rf}>rfx = y dx=\ y J*/3J JW3L J° J t/3 •r i dx 4 sin^ cfrc : o X -8 cos— = (-8 cos tt/2) - (-8 cos rr/6) = 0-8.^|= 4V3units 2 Now for a rather more worthwhile example - on to frame 22. 573 Multiple Integrals Example 3. Find the area enclosed by the curves ji 2 =9xand>'2 =^ First we must find the points of intersection. For that,ji = y 2 ■ • 9x = i_ ;. x = o orx 3 = 729, i.e. x = 9 81 So we have a diagram like this: l2 As usual, Area of element = dy. 8x Area of strip PQ L dy-8x y = y% Y *-y *1 """""/! /? =9;r h (mi ~^**^ ' ' -r—X H \*~ 9 X Sx Summing all strips between x = and x = 9, ^ = b=y 2 J * = .y =j>2 IfSj>->-0and6;c-+0, r9 pi J oJv 2 dy dx Now finish it off, remembering that y x 2 = 9x and .y 2 22 Here it is. A = 27 units 2 A= dydxA J J v 2 J - 3~l9 2jc 3 / 2 -M 27 ^i dx yi 23 = 54 - 27 = 27 units 2 Now for a different one. So turn on to the next frame. 1 574 Programme 21 24 Double integrals can conveniently be used for finding other values besides areas. Example 4. Find the second moment of area of a rectangle 6 cm X 4 cm about an axis through one corner perpendicular to the plane of the figure. z 4,. Second moment of element P about OZ ^ 8a (OP) 2 ^8y.8x.(x 2 + y 2 ) Total second moment about OZ I**! 6 y i 4 (x 2 +y 2 )dydx If 6x -* and by -* 0, this becomes I= ( (pc*+y 2 )dydx 25 For: f (* 2+ : )Jo Now complete the workingj = ... I = 416 cm 4 J JoJo y 2 )dydx = *♦* dx J 4x 2 +f U* 4x 3 , 64x" = 288+ 128 = 416 cm'* about one 5 cm side as axis. Complete it and then twmonjofrome^___ :== ___ :r _ : ^ ::z ^ 575 Multiple Integrals I = 45 cm 4 Here it is: check through the working. Y Area of element = 8a = 8y. 8x Second moment of area of 8a about OX = 8ay 2 = y 2 8y 8x y = 3 Second moment of strip — 2 y . 8y.8x y = x = 5 y = 3 Second moment of whole figure - 2 2 y .8y.8x x=0 y=0 U8y-*Oand8x^O -n 3 - J oj y 2 dy dx ■: I = 4r 5 ryi _3_ 3 r 5 r -| dx = 9dx = 9x J - I = 45 cm 4 On to frame 2 7. 26 Now a short revision exercise. Finish both integrals, before turning on to the next frame. Here they are. Revision Evaluate the following: (0 f t (y 2 -xy)dydx (ii) f [ (x 2 +y 2 )dydx. When you have finished both, turn on. ZJ 576 Programme 21 28 (i) 1 = 9^; (ii) 1=16 Here they are in detail. •2 (-3 (0 I J J 1 (y 2 ~ xy) dy dx J c > 3 _xy 2 ' dx (-f)-(l-f))- 17|-8«4 -2x 2 (ii) Jo Ji Jo (*• ♦!H* 1 4)1 * ^ T dx ■r.( =i 3 7jc^ 3 3 3 = 9 + 7=16 Now on to frame 29. 29 Alternative notation Sometimes, double integrals are written in a slightly different way. -3 ,2 For example, the last double integral 1=1 (x 2 + y 2 ) dy dx could i\y have been written HI (x 2 +y 2 )dy The key now is that we start working from the right-hand side integral and gradually work back towards the front. Of course, we get the same result and the working is identical. Let us have an example or two, to get used to this notation. Move on then to frame 30. 577 Multiple Integrals Example 1. •tt/2 = dx 5 cos d dd Jo Jo = | dx J 5 sin0 V2 = f dx Jo 5 = \ 5dx-- Jo = 10 5x It is all very easy, once you have seen the method. You try this one. » 6 p 7r/2 Example 2. Evaluate I = \ dy\ 4 sin Zx dx po rail = dy\ J 3 Jo 1 = 4 Here it is. jt/2 4 sin 3x dx 3 •6 = | dy •6 -4 cos 3x~ tt/2 I >k(-!)H >f 4y 3 = (8)-(4) = 4 Now do these two. Example 3. Example 4. ih: (x-x 2 )d> •iy dy\ (x-y)dx (Take care with the second one) When you have finished them both, turn on to the next frame. 30 31 578 Programme 21 32 Results-: Example 3. Example 4 Next frame. Ex.3. I = -4-5, Ex.4. I = £ = I dx I (x - x 2 Jo J o ■f Jc )rfy xy - x 2 y l JO = dx(x-x 2 ) = \ (x-x 2 )dx JO Jo 2 3 -f-9-4* 1 = •2y >2 (• ^ 1 J j; •2 x = 2y x = y 6 ■1_±=2 6 6 6 j 3 Now, by wa y °f revision, evaluate these. f4r2j> (i) (2x + 3y)dxdy J OJ y (ii) f dxf (2y-5*)dy. PV/ienjow /lave completed both of them, turn on to frame 34. 579 Multiple Integrals Working (i) 128, (ii) -54-5 (i) I ■J.! (2x + 3y) dx dy x 2 + 3xy x = 2y x = y dy -\ (4y 2 +6y 2 )-(y 2 +3y 2 )\dy \0y 2 -4y 6y 3 J 6y 2 dy 2y 3 = 128 ■A (ii) I = f dxl (2y-5x)dy = dx ly 2 - 5xy J 1 L J y = Q = f dxlx-5x V2 f (x - 5x zl2 ) dx = - - 2x 512 2 . = (8-64)-(^-2) = -56+ 1-5 = -54-5 34 So it is just a question of being able to recognize and to interpret the two notations. Now let us look at one or two further examples of the use of multiple integrals. Turn on then to frame 35. 580 Programme 21 35 Example To find the area of the plane figure bounded by the polar curve r =f(9), and the radius vectors at 6 = 1 and 6 = 8 2 . e = e. Small arc of a circle of radius r, subtending an angle 86 at the centre. .'. arc = r. 50 We proceed very much as before. Area of element — 8r,.r86 r= r t Area of thin sector — 2 8r.r86 r=0 e = e 2 Total area - 2 (all such thin sectors) e = 0! = e 2 ( r = r< -2 2 r.8r.8d = 0! ( r=0 Then if 60 -^ and 5r-*0, = 2 r=ri ^2 2 r.8r.86 " ~- 0i r= ?2 /•'■l A = | I r.dr.dd r»2 r r\ J 01 J Finish it off. 36 The working continues: A = i.e. in general, _r 62 J ©I dd r 1 d6 Which is the result we have met before. Let us work an actual example of this, so turn on to frame 37. 581 Multiple Integrals Example. By the use of double integrals, find the area enclosed by the 0"7 polar curve r - 4(1 + cos 6) and the radius vectors at = and 6 = rr. ' J # r = 4 (1 + cos 8) 6 = 7r /■ = n 6 = r = A f J 0. rdrdd .dd i r n * dd Butr t = /(0) = 4(1 + cos 6>) .-. A=f 8(1 +cos0) 2 dd J 8(1 +2 cos6 + cos 2 0)^0 1 For A= I2n units 2 A = i (i + 2cose + cos 2 e)^e „ . „ 6 sin 20 + 2 sin 8 +T+—T - 2 4 J -8( W+ f)-(0) = 87r + 4tt = 12ir units 2 Now let us deal with volumes by the same method, so move on to the next frame. 38 582 Programme 21 j Jj Determination of volumes by multiple integrals Surface z, = f {x, y) Element of volume 6v = bx . by . bz. Summing the elements up the column, we have 5V c = r S Zl 5jc.5^.5z z = If we now sum the columns between y =y 1 and y = y 2 , we obtain the volume of the slice. y = v 2 z = zi bV s = S 2 Sx.S.y.5z .y =>>i z = Then, summing all slices between x = x l and x = x 2 , we have the total volume. x = x 2 y = y-i z = 2i V= 2 2 2 Sx.Sy.Sz x = Xi x = Xl z = Then, as usual, if fix -* 0, 6_y ->• and 62 ~* •*2 /"JV2 /«Zl dx.dy. dz The result this time is a triple integral, but the development is very much the same as in out previous examples. Let us see this in operation in the following examples. Next frame. 583 Multiple Integrals Example 1. A solid is enclosed by the plane z = 0, the planesx- l,x- 4, ^Ifl y = 2 , y = 5 and the surface z = x + >\ Find the volume of the solid. ** First of all, what does the figure look like? The plane z = is the x-y plane and the plane x = 1 is positioned thus: z Plane x = \ Working on the same lines, draw a sketch of the vertical sides. The figure so far now looks like this: z If we now mark in the calculated heights at each point of intersection (z = x+y), we get z This is just preparing the problem, so that we can see how to develop the integral. For the calculation stage, turn on to the next frame^ 41 584 Programme 21 42 Volume of element - 8x . 8y . dz z = (x + v) Volume of column -Sx.Sy 2 s z r = , r , y ~ 5 z = x + y Volume of slice - 8x 2 by 2 5z ^=2 z=0 JC ~ 4 V = S 7 — 1" + V Volume of total solid - 2 5a: 2 5 v 2 8z jc = 1 >>= 2 z = Then, as usual, if 8x -> 0, 6^ -> 0, 5z -»• 0, this becomes /•4 f5 /•x+j V= dx dy dz J 1 J 2 Jo And this you can now finish off without any trouble. (With this form of notation, start at the right-hand end. Remember?) SoV= 43 IV =■ 54 units" 4 |«5 rx+y M [S dx\ dy\ dz = \ dx\ dy(x+y) 1 •> 2 J J 1 J 2 = dx\ (x+y)dy=\ dx xy+ :} ~ J 1 J 2 J I L l J2 -^[sx^-Tx^Y^x^dx '3£ 2 + 21x 2 2 J 2 3x 2 + 2 be 4{(48 + 84)-(3 + 21)j=4[l32-24J= 54units 3 585 Multiple Integrals Example 2. Find the volume of the solid bounded by the planes, z = 0,x= l,x = 2,y = -l,y = 1 and the surface z = x 2 + y 2 . In the light of the last example, can you conjure up a mental picture of what this solid looks like? As before it will give rise to a triple integral. 44 -I! 41 dy x 2 + y 2 Evaluate this and so find V. V = . dz V = ^units 3 For we have: rl rl rx 2 +y 2 V = \ dx \ dy\ dz J 1 J -i Jo = [ dx( dy(x 2 +y 2 ) -jH^i', -j' {(•'♦*)-(-*•-*)}* *ii{ 2x '+?} dx 2 3 X 3 +x 2 1 = -|{(8 + 2)-(l + l)| 16 .. 3 = -5- units Next frame. 45 586 Programme 21 46 That brings us almost to the end of this programme. In our work on multiple integrals, we have been developing a form of approach rather than compiling a catalogue of formulae. There is little therefore that we can list by way of revision on this occasion, except perhaps to remind you, once again, of the two forms of notation. Remember: pd pb (i) For integrals written f(x, y) dx.dy, work from the centre J cJ a outwards. (ii) For integrals written dy J fix, y) dx work from the right-hand side. Now there is the Test Exercise to follow. Before working through it, turn back into the programme and revise any points on which you are not perfectly clear. If you have followed all the directions you will have no trouble with the test. So when you are ready, move on to the Test Exercise. 587 Multiple Integrals Test Exercise— XXI Answer all questions. They are all quite straightforward and should cause you no trouble. 1. Evaluate (i) (y 3 ~xy) dy dx (ii) I dx I' * (x -y) dy, where y ,, = vV - x 2 ) (n) dx\ (x JO Jo 2. Determine „ ,, . ,, '■'j + 2 |* tt/3 (i) (2cos0-3sin30)d0.dr j* ^3 + 2 J" IT; Jo J ■offT J 2J 1 J (iii) I dz I cfx (j: Jo J 1 Jo ■4 p2 p4 (ii) 1 xy(z + 2) dx dy dz + y + z)dy 3. The line y = 2x and the parabola y 2 = \6x intersect at x = 4. Find by a double integral, the area enclosed by y = 2x, y 2 = 16x and the ordinate at x = 1 . 4. A triangle is bounded by the x-axis, the line y = 2x and the ordinate at x = 4. Build up a double integral representing the second moment of area of this triangle about the x-axis and evaluate the integral. 5. Form a double integral to represent the area of the plane figure bounded by the polar curve r = 3 + 2 cos d and the radius vectors at 6=0 and 9 = n/2, and evaluate it. 6. A solid is enclosed by the planesz = 0,y = l,y = 3,x = 0,x = 3, and the surface z = x 2 + xy. Calculate the volume of the solid. That's it! 47 588 Programme 21 Further Problems- XXI • 77 /• COS e 1. Evaluate I I r sin ddrdd f n /«co J oJ 2 ' " I *f r 3 (9-r 2 ) drdd J J j*l rlx + 2 3. " II G?y dx J_2J^: 2 +4x [ a f>b pc 4. " (x 2 +/)d;ciydz J oJ oJ • 7T j»7r/2 (.r - 2 sin 6 dx dd dip J ojo Jo 6. Find the area bounded by the curve 7 = x 2 and the line y =x + 2. 7. Find the area of the polar figure enclosed by the circle r = 2 and the cardioid r = 2(1 + cos 6). 8. Evaluate I dxl dy xy 2 zdz J J 1 J l 9. " i dx[ (x 2 +y 2 )dy f 1 /• 7r/' Jo Jo .tt/4 10. " I dr\ rcos 2 dde 11. Determine the area bounded by the curves x =y 2 and x = 2y ~y 2 . 12. Express as a double integral, the area contained by one loop of the curve r = 2 cos 30 and evaluate the integral. .tt/2 /•tan" 1 (2) M |.7r// /»tan (i) ft 13. Evaluate x sin y dx dy dz J J tt/4 Jo /•tt /»4 cos z /•, J J J ( • 4 cosz pj(l6 —y 2 ) 14. Evaluate | | | y dx dy dz 589 Multiple Integrals 15 . A plane figure is bounded by the polar curve r = a(l + cos 0) between = and = ti, and the initial line OA. Express as a double integral the first moment of area of the figure about A and evaluate the ^ 2 integral. If the area of the figure is known to be — — units 2 , find the distance (h) of the centroid of the figure from OA. 16. Using double integrals, find (i) the area and (ii) the second moment about OX of the plane figure bounded by the x-axis and that part of x 2 y 2 the ellipse -j- + rj = 1 which lies above OX. Find also the position of the centroid. 17. The base of a solid is the plane figure in the ;cy-plane bounded by x = 0, x = 2,y = x, and y = x 2 + 1 . The sides are vertical and the top is the surface z=x 2 + y 2 . Calculate the volume of the solid so formed. 18. A solid consists of vertical sides standing on the plane figure enclosed by x = 0, x = b, y = a andy = c. The top is the surface z = xy. Find the volume of the solid so defined. 19. Show that the area outside the circle r- a and inside the circle r = 2a cos 9 is given by •7r/3 /-2a cos 6 A = 2| rdrdd frr/i pla J J a Evaluate the integral. 20. A rectangular block is bounded by the co-ordinate planes of reference and by the planes x = 3,y = 4, z = 2. Its density at any point is numerically equal to the square of its distance from the origin. Find the total mass of the solid. 590 Programme 22 FIRST ORDER DIFFERENTIAL EQUATIONS Programme 22 1 Introduction A differential equation is a relationship between an independent variable,*, a dependent variable,^, and one or more differential coefficients of y with respect to x. 2 dy , ■ r, e.g. x ~f~ + y smx = Differential equations represent dynamic relationships, i.e. quantities that change, and are thus frequently occurring in scientific and engineering problems. The order of a differential equation is given by the highest derivative involved in the equation. x -j- -y 2 = is an equation of the 1st order xy~i -y 2 sinx = " " " " " 2nd " dll- y dy + e 4x = „ „ „ „ >, 3rd » dx 3 J dx So that-r^ + 2~ + 10 y - sin 2x is an equation of the order. dx 2 dx second Since in the equation j{ + 2— + 10 y = sin 2x, the highest derivative involved is -r-? 1 „.* £y dx 2 Similarly, (i) x-*- =y 2 + 1 is a order equation (iij cos 2 x-^ -V-y =1 is a order equation (iii) -r^j - 3 -J- + 2y = x 2 is a order equation (iv) (y 3 + l)j ~xy 2 = .xis a order equation On to frame 3. 593 First Order Differential Equations (i) first, (ii) first, (iii) second, (iv) first. Formation of differential equations Differential equations may be formed in practice from a consideration of the physical problems to which they refer. Mathematically, they can occur when arbitrary constants are eliminated from a given function. Here are a few examples: Example 1. Consider y - A sin x + B cos x, where A and B are two arbitrary constants. If we differentiate, we get -f- = A cos x — B sin x dx and dx 2 ■■ -A sin x - B cos x which is identical to the original equation, but with the sign changed. i.e. <ry . d 2 y A dx' dx A y = This is a differential equation of the order. second Example 2. Form a differential equation from the function y = x + '■ 4 We have y =x + — =x + Ax 1 x :.^ = l-Ax- 2 = l-4 dx x 2 From the given equation, — = y -x ■'■ A = x(y—x) ■ dZ = i x(y-x) " dx x 2 = 1 _y~x _x~y+x _2x~y :.x*L=2x-y dx This is an equation of the order. 594 Programme 22 first Now one more. Example 3. Form the cliff, equation for y = A x 2 +Bx. We have y = Ax 2 + Bx (i) (ii) . d 2 y dx dx z ■2K (m) A ~2d? Substitute for 2A in (ii) -~- = x -A + B ax ax dx dx Substituting for A and B in (i), we have v=x 2 i€z +x (dz_ tfy\ y X 2dx 2 X \dx X dxV dx i x 2 d 2 y dy 2 d 2 y "2 dx 2 X dx X ■ - " dx 2 d£ " y X dx 2'dx 2 2 d 2 y and this is an equation of the order. second If we collect our last few results together, we have: d 2 y y = A sin x + B cos x gives the equation — ^ + y = (2nd order) y = Ax 2 + Bx ii -if dy d 2 y -■ o- A v = x + — x (1st order) If we were to investigate the following, we should also find that y=X tc-2-dT 2 ( 2ndorder ) x& -2x-y dx y = Axe* gives the diff. equation x-j--y{\ +x) = (1st order) y = Ae~ 4 * + Be' *2. >dy. d £ + 10 jg + 24y=0 (2nd order) Some of the functions give 1st order equations: some give 2nd order equations. Now look at these five results and see if you can find any distinguishing features in the functions which decide whether we obtain a 1st order equation or a 2nd order equation in any particular case. When you have come to a conclusion, turn on to frame 7. 595 First Order Differential Equations A function with 1 arbitrary constant gives a 1st order equation. " " 2 arbitrary constants " " 2nd order " Correct, and in the same way, A function with 3 arbitrary constants would give a 3rd order equation. So, without working each out in detail, we can say that (i) y = e~ 2x (A + Bx) would give a diff. equation of order. (u) >' = A^y " " " " " " " - „3* (iii) y = e (A cos 3x + B sin 3x) (i) 2nd, (ii) 1st, (iii) 2nd since (i) and (iii) each have 2 arbitrary constants, while (ii) has only 1 arbitrary constant. Similarly, (0 *'■£+* = constants. (i) x 2 — + y = 1 is derived from a function having arbitrary (ii) cos 2 x-r- = l-y " " a function having arbitrary dx j2. constants. (iii) -r-y + 4 -*- + y = e 2x " a function having arbitrary constants. 8 596 Programme 22 10 (i) 1, (ii) 1, (iii) 2 So from all this, the following rule emerges: A 1st order diff. equation is derived from a function having 1 arbitrary constant. A 2nd " " " " " " " " " 2 arbitrary constants. An nth order differential equation is derived from a function having n arbitrary constants. Copy this last statement into your record book. It is important to remember this rule and we shall make use of it at various times in the future. Then on to frame 10. Solution of differential equations To solve a differential equation, we have to find the function for which the equation is true. This means that we have to manipulate the equation so as to eliminate all the differential coefficients and leave a relationship between y andx The rest of this particular programme is devoted to the various methods of solving first order differential equations. Second order equations will be dealt with in a subsequent programme. So, for the first method, turn on to frame 11. 597 First Order Differential Equations Method 1 By direct integration dy. If the equation can be arranged in the form-j- = f(x), then the equation can be solved by simple integration. dy 2 Example 1. -4- = 3x dx Then 6x + 5 = 1 (3x 2 - 6x + 5) dx = x 3 - 3x 2 + 5x + C i.e. y=x 3 ~3x 2 +5x + C As always, of course, the constant of integration must be included. Here it provides the one arbitrary constant which we always get when solving a first order differential equation. Example 2. Solve In this case, dx ^ = 5x 2 + 4 - dx x So, y = 11 y- 5x 3 + 4 In x + C As you already know from your work on integration, the value of C cannot be determined unless further information about the function is given. In its present form, the function is called the general solution (or primitive) of the given equation. If we are told the value of .y for a given value of x, C can be evaluated and the result is then a particular solution of the equation. Example 3. Find the particular solution of the equation e* ■—■ = 4, given that y = 3 when* =0. First re-write the equation in the form-p= -^ = 4e' x 12 Then y -\ 4e~ x dx = -4e' x +C Knowing that when* = 0,y = 3, we can evaluate C in this case, so that the required particular solution is y = 598 Programme 22 13 y = -4e x + 7 Method 2 By separating the variables If the given equation is of the form-f^ =f(x, y), the variable y on the right-hand side, prevents solving by direct integration. We therefore have to devise some other method of solution. Let us consider equations of the form-^ =f(x) .F(y) and of the form —— = pjr-r ,i.e. equations in which the right-hand side can be expressed as products or quotients of functions of x or of y. A few examples will show how we proceed. Example 1. Solve — ^ We can re-write this as (y + 1) — = 2x dx dx y+\ Now integrate both sides with respect to x \(y+l)^ x dx=\2xdx i.e. \{y + 1) dy = fzx dx v 2 and this gives ^— + y =x 2 + C mn Example 2. Solve -^ = (1 + x ) (1 + y) 1 d y - 1 + 1 +y dx Integrate both sides with respect to x \ \hd\ dX = J + X) dX •'• JlTy ^ ={(1 + X) dx \n(l+y) = x+Y + C The method depends on our being able to express the given equation in the form F(y). -j- =f(x). If this can be done, the rest is then easy, for we have JF(y)®-dx=jf(x)dx :. ^F(y)dy = \f(x)dx and we then continue as in the examples. Let us see another example, so turn on to frame 15. 599 First Order Differential Equations Example 3. Solve £z = 1_Z (j) IE dx 2+x I «J This can be written as — _ Si = _L_ Integrate both sides with respect to x \iryfx dX= i2T-x dx (ii) /. ln(l +^) = ln(2+x) + C It is convenient to write the constant C as the logarithm of some other constant A , , . , In (1 +>0 = In (2 + x) + In A :. 1 + y = A (2+x) Note: We can, in practice, get from the given equation (i) to the form of the equation in (ii) by a simple routine, thus: dy _ 1 + y dx 2+x First, multiply across by the dx dy = -z — — dx y 2 + x Now collect the '^-factor' with the dy on the left, i.e. divide by (1 + y) 1 a - 1 A 7— dy = r— — dx \+y 2+x Finally, add the integral signs JTT-y dy= S^x dx and then continue as before. This is purely a routine which enables us to sort out the equation algebraically, the whole of the work being done in one line. Notice, how- ever, that the R.H.S. of the given equation must first be expressed as 'x-factors' and '_y-f actors'. Now for another example, using this routine. 2 j. „,,2 Example 4. Solve -f-= -^ — ^— y dy _ y + xy dx x 2 y - x First express the R.H.S. in 'x-factors' and '^-factors' dy_ y 2 (l+x) dx x l (y-l) Now re-arrange the equation so that we have the 'y-factors' and dy on the L.H.S. and the 'x -factors' and dx on the R.H.S. So we get 600 Programme 22 16 v- 1 , 1 + x .,„ We now add the integral signs \ y 7^ dy= \ l -^ dx and complete the solution JM*-Jk ^r Here is another. Example 5. Solve Re-arranging, we have ■'• In y + y x = In x - x 1 + C .". lny+~=\nx-j+C dy_ y 2 -1 y 2 -i <fy=* dx —5 — ~dy = — dx y 2 - 1 J x ■■ jyrzi <* =j> Which gives 17 1 v- 1 r-ln*— r = lnjc + C 2 y + 1 :. ln^-4 = 21nx + lnA Ax 2 y + i y-1 =Ax 2 (y+ 1) You see they are all done in the same way. Now here is one for you to do: ci dy _ x 2 + 1 Example 6. Solve xy - — r First of all, re-arrange the equation into the form ¥{y)dy=f{x)dx i.e. arrange the '^-factors' and dy on the L.H.S. and the 'x-factors' and cfrontheR.H.S. What do you get? 601 First Order Differential Equations yiy +\)dy = x 2 + 1 dx for xy dy _ x 2 + 1 dx y + 1 .'. xy dy ="- x 2 +1 y +i dx :. y{y + 1) dy x 2 + 1 dx So we now have j(y 2 +y)dy=^(x + ±)dx Now finish it off, then move on to the next frame. iH'.4 +lni - + c 18 19 aannnnDDDDDaDa.DDDDnaDonnannDDnnDDDDDDa Provided that the R.H.S. of the equation -j- = f(x, y) can be separated into 'x-factors' and '^-factors', the equation can be solved by the method of separating the variables. Now do this one entirely on your own. Example 6. Solve dy x-f- ~y + xy dx When you have finished it completely, turn to frame 20 and check your solution. 602 Programme 22 £(J Here is the result. Follow it through carefully, even if your own answer is correct. dy , dy ,. , . x^ = y + xy .. X -S-=y(l + x) xdy = y{\ + x)dx :.& = ^dx y x .'. \ny = \nx + x + C At this stage, we have eliminated the differential coefficients and so we have solved the equation. However, we can express the result in a neater form, thus: \ny-\nx = x + C :. In y_ = e t + c = e x e c %■' + c Now e c is a constant; call it A. Next frame. .'. — = Ae x :.y = Axe x x — 21 This final example looks more complicated, but it is solved in just the same way. We go through the same steps as before. Here it is. Example 7. Solve y tan x-j- = (4 +y 2 ) see 2 * First separate the variables, i.e. arrange the '_y-factors' and dy on one side and the 'x-factors' and dx on the other. So we get 22 y 4+y sec 2 x , ? dy = tln-x- dx Adding the integral signs, we get r y , f sec 2 x , in— 5 dy = dx 4+y 2 ' J tanjc Now determine the integrals, so that we have 603 First Order Differential Equations -±ln(4+,y 2 ) = lntanx + C This result can now be simplified into: In (4 + y 2 ) = 2 In tan x + In A (expressing the .". 4+y 2 = A tan x .'. y 2 = A tan 2 *- 4 constant 2C as In A) 23 So there we are. Provided we can factorize the equation in the way we have indicated, solution by separating the variables is not at all difficult. So now for a short revision exercise to wind up this part of the programme. Move on to frame 24. Revision Exercise Work all the exercise before checking your results. Find the general solutions of the following equations: dx x dy 24 l. dx ■■(y + 2)(x+l) i dy cos x-r = v + 3 dx dy -dx- = Xy ~ y sin* dy _ 1 +y'~dx~ cosx When you have finished them all, turn to frame 25 and check your solutions. 604 Programme 22 25 Solutions 1. 3. dx x dy — dx x .'. In y = In x + C = In x + In A .'. y - Ax dx (y + 2) (x + l) JVh'H ( * +1)dx /. ln(y + 2)=— +x + C cos^x-p- = v + 3 ax "J' — T- C?X COS X 1^ + 3 ' sec 2 x dx In + 3) = tan x + C ^7 • dy , ,, ■■•I> = S<*- l)dx /. lnj;=— -* + C sinx <iy 1 + /dx cos* (^ dy JS212L dx J l +y J smx .'. ln(l +y) =lnsin;c + C = In sin x + In A 1 + y = A sin x ■'. y = A sin x — 1 nnnnnnnQnnnooanaaQaQaaaaaaaaaDDaaaaoaa If you are quite happy about those, we can start the next part of the programme, so turn on now to frame 26. 605 r First Order Differential Equations. Method 3 Homogeneous equations - by substituting y = vx ?fi dy x + 3y tU Here is an equation: —j- = — r — This looks simple enough, but we find that we cannot express the R.H.S. in the form of 'x-factors' and '_y-factors', so we cannot solve by the method of separating the variables. In this case we make the substitution y = vx, where v is a function ofx. So y = vx Differentiate with respect to x (using the product rule). • dy , _,_ dv dv dx dx dx „ _, _ x + 3vx _ 1 + 3v Also -dv =\ —dx x 2x 2 dv 1 + 3v The equation now becomes v + x— = — r — . dv 1 + 3v dx 2 1 + 3v - 2v _ 1 +v 2 2 . dv_ \ + v_ " * dx' 2 The given equation is now expressed in terms of v and x, and in this form we find that we can solve by separating the variables. Here goes: .-. 2 In (1 + v) = In x + C = In x + In A (1 +vf = Ax But y = vx :. v=J^j :. (l+jf=Ax which gives (x + y) 2 = Ax 3 jVofe — - x + 3y is an example of a homogeneous diff. equation, dx 2x This is determined by the fact that the total degree in x and y for each of the terms involved is the same (in this case, of degree 1). The key to solving every homogeneous equation is to substitute^ = vx where v is a function ofx. This converts the equation into a form in which we can solve by separating the variables. Let us work another example, so turn on to frame 27. 606 Programme 22 07 Example 2. Solve dy _x 2 + y 2 dx xy -^ Here, all terms on the R.H.S. are of degree 2, i.e. the equate. „ homogeneous. :. We substitute^ = vx (where v is a function of x) tion is and The equation now becomes ■ dy dv ■■ ~y~ =v + x—- dx dx x 2 +y 2 _x 2 +v 2 x 2 1 xy V + X i vx V dv l+v 2 dx v dv l+v 2 dx v -v l+v 2 - -v 2 dv " " dx v Now you can separate the variables and get the result in terms of v and x. Off you go: when you have finished, move to frame 28. 28 — = In x + C for Wi dx v 2 lnx + C All that now remains is to express v back in terms of x and y. The substitution we used was y - vx Vn 2 - v=L x " 2\x) Now, what about this one? Example 3. Solve y lnx + C 2 = 2 x 2 (In x + C) dy_ 2xy + 3y 2 dx x 2 + 2xy Is this a homogeneous equation? If you think so, what are your reasons? When you have decided, turn on to frame 29. 607 First Order Differential Equations Yes, because the degree of each term is the same Correct. They are all, of course, of degree 2. So we now make the substitution, y = 29 y = vx, where v is a function of x Right. That is the key to the whole process. dy _ 2xy + 3y 2 dx x 2 + 2xy So express each side of the equation in terms of v and x. dx 2xy + 3y 2 _ x 2 + 2xy and When you have finished, move on to the next frame. 30 dy , dv ~-=v + x-t- dx dx 2xy + 1y 2 _ 2vx 2 + 3v 2 x 2 = 2v_+3v^ x 2 +2xy x 2 + 2vx 2 1 + 2v 31 So that dv 2v + 3v 2 v + x— _ dx 1 + 2v Now take the single v over to the R.H.S. and simplify, giving dv X Tx = 608 Programme 22 32 dv 2v + 3v 2 X dx l+2v v 2v + 3v 2 -v- -2v 2 1 + 2v dv v + v 2 "* cfcc 1 + 2v Now you can separate the variables, giving 33 — ; — 2 dv =\ — dx J V + V 2 J X Integrating both sides, we can now obtain the solution in terms of v and x. What do you get? 34 ln(v + v 2 ) = lnx + C = In x + lnA .'. V + v 2 = Ax We have almost finished the solution. All that remains is to express v back in terms of x and y. Remember the substitution was v = vx, so that v =— x So finish it off. Then move on. 35 X y +y 2 = AX 3 for v + v 2 = Ax and v =— ~+ y ~Y=Ax X X xy +y 2 = Ax 3 And that is all there is to it. Turn to frame 36. 609 First Order Differential Equations Here is the solution of the last equation, all in one piece. Follow it through again. T , dy_ 2xy + 3y 2 Tosolve d*~ x 2 +2xy This is homogeneous, all terms of degree 2. Put y = vx dy dv dx dx 2xy + 3y 2 = 2vx 2 + 3v 2 x 2 = 2vJJ3v_ 2 x 2 + 2xy ~ x 2 + 2vx 2 1 + 2v . , dv 2v + 3v 2 .. V +X-r-= , . ~ dx 1 + 2v dv 2v + 3v 2 v = y dx 1 + 2v 2v + 3v 2 -v-2v 2 1 + 2v .. x dv v + v 2 dx 1 + 2v — ■ — 2 dv = —dx V + V ■fc In (y + v 2 ) = In x + C = In x + In A v + v 2 = Ax But y = vx :.2- + X x 2 = Ajc •'• xy + j' 2 = Ax 3 36 Now, in the same way, you do this one. Take your time and be sure that you understand each step. Example 4. Solve (x 2 + y 2 )-^ = xy When you have completely finished it, turn to frame 37 and check your solution. 610 Programme 22 37 Here is the solution in full. (x + y ) dx xy .. dx x2+y2 d ♦ . dy _, dv Put y = vx ■■—f = v+x—r- dx dx and xy _ vx 2 v x 2 +y 2 x 2 +v 2 x 2 1 +v 2 , dv v v + x— =- dx 1 + v l dv dx 1+v 2 But dx 1 + v* 1 + v 2 ■ — 5 — dv = - — dx J v 3 Jx :. J (v~ 3 +— )dv = -lnx + C -v~ 2 .'■ -r— + In v = -In x + In A In v + In x + In K = r-5 In Kvx = -t~t 2v* v=£ :. \nKy = ^r x 2r 2y 2 \nKy=x 2 This is one form of the solution: there are of course other ways of expressing it . Now for a short revision exercise on this part of the work, move on to frame 38. Solve the following: l . ( x - y) — = x + y Revision Exercise e: 1 (v — m] 2. 2x 2 ^=x 2 + y 2 3. (x * +xy) f x=xy - y 2 When you have finished all three, turn on and check your results. 611 First Order Differential Equations The solution of equation 1 can be written as tan _1 J^}= In A + lnx +yln j 1 +^ 39 Did you get that? If so, move straight on to frame 40. If not, check your working with the following. 1. Put , ,dy . dy x+y dx dx x-y . dy , dv y = vx ..-= v+x - . ^ dv 1 +v • ■ v + x T = 1 dx 1 -v dv 1+v x-r--~. v = dx l-v x +y _ 1 + v x-y l-v 1 + v-v + v 2 1+v 2 1 -v l-v ••• \\t? dv -\i dx •■• f (rb - tt?} dv - lnx + c .". tan l v- — ln(l + v 2 ) = lnx + In A But v= y - :. tan'M^-UlnA + lnx+^-lnO +^) This result can, in fact, be simplified further. Now on to frame 40. Equation 2 gives the solution 2x x-y ^ln x + C If you agree, move straight on to frame 41 . Otherwise, follow through the working. Here it is. 2x 2 dy^^i ^„ 2 . dy _ x +y dx ■ dy- Put y = vx •'• -f = V + X , J dx dx ■ + dv^ l+v 2 dx 2 ■x' +y* dv dx 2x 2 u.^2 2~ x 2 + y 2 _ x 2 +v i x i _ 1 +v^ 2x 2 2x 2 2 dv = 1 +v 2 _ = 1 - 2v + v 2 = (v-1) 2 X dx 2 V 2 '2 •'•j(^T) 2dv= B dx :. -2 r=lnx + C v-1 But „=Z and r-=- x 1 -v On to frame 41. = lnx + C 2x x-y lnx + C 40 612 Programme 22 - - - - 41 One form of the result for equation 3 is xy = A e x ' y Follow through the working and check yours. 3. (x'+xy^^xy-y* :. d J= Xy 2 ' y2 "dx ' * dx x 2 +xy . dy ^ dv xy -y Put y = vx ■■ -f = v + x — ; \ / J dx dx x + x. l 7 2 2 2 vx -v x v - V v x 2 +vx 2 1 + V dv v-v 2 .. v+x — = dx \+v dv v~v 2 v - v 2 - v - v 2 -2 v 2 X — = — v = : = dx 1 + v 1 + v 1 + v • • 2 dv -\ — dx J v 2 J X ^ 2+ >"=-j> :. lnv-- = -21nx + C. Let C = In A V In v + 2 In x = In A + — V lnRx 2 }=lnA + * :. xy = Ae x fy -. Now move to the next frame. 42 Method 4 Linear equations - use of integrating factor Consider the equation -j- + Sy = e 2x This is clearly an equation of the first order, but different from those we have dealt with so far. In fact, none of our previous methods could be used to solve this one, so we have to find a further method of attack. In this case, we begin by multiplying both sides by e sx . This gives dy + dx _j, 5e 5*= e 2*. e S* We now find that the L.H.S. is, in fact, the differential coefficient of y.e sx =!'■«" Now, of course, the rest is easy. Integrate both sides w.r.t. x. S* dx = V + C y.e sx -J- y = 613 First Order Differential Equations e 2x 43 Did you forget to divide the C by the e sx ? It is a common error so watch out for it. □DnnannDnnQQDQnnnnannnnannnaaannoonnnD The equation we have just solved is an example of a set of equations of the form 4^ + Py = Q, where P and Q are functions of x (or constants). dx This equation is called a linear equation of the first order and to solve any such equation, we multiply both sides by an integrating factor which is always e^ e dx . This converts the L.H.S. into a complete differential coefficient. In our last example,-^- + 5y = e 2x , P = 5. .-.J P dx = 5x and the integrating factor was therefore e sx . Note that in determining Pdx, we do not include a constant of integration. This omission is purely for convenience, for a constant of integration here would in fact give a constant factor on both sides of the equation, which would subsequently cancel. This is one of the rare occasions when we do not write down the constant of integration. So : To solve a differential equation of the form dx where P and Q are constants orfunctionsofx, multiply both sides by ■ <■ ! ?dx the integrating factor e This is important, so copy this rule down into your record book. Then move on to frame 44. dy Example 1. To solve —-y-x. If we compare this with-f^ + Py = Q, we see that in this case 44 P = -l andQ = :c f pdx j i. lys e and her ? dx =-x and the integrating factor is therefore rp dx The integrating factor is always e and here P = -1 614 Programme 22 45 We therefore multiply both sides by e x . •'• e x ~-ye x =xe x dx -~r\e x y\=xe x .'. y e x = \x e x dx The R.H.S. integral can now be determined by integrating by parts. y e' x = x (-e' x ) + J e* dx = - x e -x - e ~* + c :. y = -x - 1 + C e* :■ y =Ce x -x-\ The whole method really depends on (i) being able to find the integrating factor, (ii) being able to deal with the integral that emerges on the R.H.S. Let us consider the general case. ay A C Consider -?- + P y = Q where P and Q are functions of x. Integrating fp^v dv fPdx (?dx i?dx factor, IF =-e* ?dx :.=f.e ) +Pye =Qe J dx JP dx „ „ w „ ^ iiW W lt V,! J, C ,^P"-].Q^* Integrate both sides with respect to x ye J = I Q e . dx This result looks far more complicated than it really is. If we indicate the integrating factor by IF, this result becomes , IF -/q. Wax and, in fact, we remember it in that way. So, the solution of an equation of the form -^ + Vy = Q (where P and Q are functions of x) is given by y. IF = I Q . IF ate, where IF = e dx Copy this into your record book. Then turn to frame 47. 615 First Order Differential Equations So if we have the equation -r-+ 3y = sin x dx dy dx + Py = Q then in this case (0 P= 00 ?dx= (iii) IF; 47 (0 P=3; (ii) ?dx = 3x; (iii) IF = e a 48 nanDnannnnDDDnnnDDnDnnDDnnDnDDnnnnDann Before we work through any further examples, let us establish a very useful piece of simplification, which we can make good use of when we are finding integrating factors. We want to simplify e to F , where F is a function ofx. Let y = e ln F Then, by the very definition of a logarithm, ln_y = In F :. y = F .". F = e ln F i.e. e ln F = F This means that e^ (function) = f unct i n. e tox =x e hlsinx = smx e lntanhx =t<mhx e ln(x 2 ) = Always! s Similarly, what about e k ln F ? If the log in the index is multiplied by any external coefficient, this coefficient must be taken inside the log as a power. e-g- and e 2 1nx =g ln(jr) =y p- g 3 In sin x = gin (sin x) - sin 3 x e -lnx = e ln(^ 1 ) = x -i =J_ -l ln x - 49 616 Programme 22 50 51 l for "2 Inx = e ln (x~ 2 ) =x -2 -1 So here is the rule once again: e ln F = F Make a note of this rule in your record book. Then on to frame 51. Now let us see how we can apply this result to our working. Example 2. Solve dy , 3 x~ + y =x 3 ax First we divide through by x to reduce the first term to a single — dx dy A 2 i.e. -y- + — . y=x 2 dx x Compare with Jy_ dx + ?y = Q P = - and Q = x 2 x ^ IF = e^ P ^^ \Pdx=[jdx = lnx .-. iF = e inx =x ;.\F=x The solution is y. IF = j Q.IF dx so y x = \ x 2 . x dx Move to frame 52. 4 4 x 3 dx=*-+C :. xy=*- + C 4 '4 52 Example 3. Solve Compare with dy dx + y cotx = cos x !♦»-<> P = cot x Q = cos x COS JC P dx = I cot x dx = I — c?x = In sin * = \ cot x dx = - J J smx IF = e lnsin * = sinx .y.IF = >x=js Q.IFdx .'. _y sin x = j sin x cos x dx - sm x \ q _ sin x y r— + C cosec x Now here is another. Example 4. Solve (x + 1) -^ + y = (x + 1 ) 2 The first thing is to 617 First Order Differential Equations Divide through by (x + 1) dy Correct, since we must reduce the coefficient of-— to 1 . dy 1 _,_ . dx -~ + . . y=x + 1 Compare with In this case dx x + 1 dx 1 x + 1 and Q = x + 1 Now determine the integrating factor, which simplifies to IF IF = x+ 1 for .-. iF = e ln (* +1 ) = (x+ 1) The solution is always y . IF = I Q . IF dx and we know that, in this case, IF = x + 1 and Q = x + 1 . So finish off the solution and then move on to frame 55. y _(*+0 2 + x+l Here is the solution in detail: y .(x+ 1)= (x+ l)(x+ \)dx = f(je + l) 2 dx _(x+l) 2 + C 3 x+l •• 7 Now let us do another one. Example 5. Solve x^-5y=x 1 .dy. dx In this case, P= Q = 53 54 55 618 56 P--;Q = *« for if Compare with . dy 5 6 dx x So the integrating factor, IF = 57 for IF = $Pdx So the solution is J p *-Ji y C x * J 2 dx = -5 hue x J = -c + C 58 .y=^ + c* 5 Programme 22 Did you remember to multiply the C by x s ? DDaaaDDDnaDDDnDDnDnanQDDDDannaanDnDDDn Fine. Now you do this one entirely on your own. Example 6. Solve ( 1 - x 2 ) ^ - xv = 1 H%e« jow have finished it, turn to frame 59. 619 First Order Differential Equations W(l ~x 2 ) = sin J x + C Here is the working in detail. Follow it through. • & - x - j ■■ dx \-x 2 - y ~l-x 2 Now •'• y 4 y.lF = \Q.lFdx V(i-^ 2 ) = |— ^V(i-^).dx i jVO-* 2 ) j>V(l _;>c2 ) = sin" 1 * + C dx = sin ' x + C Now on to frame 60. 59 In practically all the examples so far, we have been concerned with finding the general solutions. If further information is available, of course, particular solutions can be obtained. Here is one final example for you to do. Example 7. Solve the equation {x~2)f x -y = ( x -2f given that y = 10 when x = 4. Off you go then. It is quite straightforward When you have finished it, turn on to frame 61 and check your solution. 60 620 Programme 22 61 Here it is: 2 y = (x - 2) 3 + 6(jc - 2) (x-2)£-,-(*-2)' J—J^ dx = -ln(x-2) . IF = e -ln (x- 2) = e lnj(x- 2) * j = (jf _ 2) -l 1 x-2 (*-2) = (x - 2) dx _(x-2) 2 dx + C •'• y ^ + C(x C(x - 2) ... General solution. When* = 4, ^=10 10 = |+C2 :. 2C = 6 :. C = 3 .-. 2^ = (x-2) 3 +6(x-2) gi ~ Finally, for this part of the programme, here is a short revision exercise. (J ^ Revision Exercise Solve the following: 1. J^ + 3y = e* x dx 2. d y j. x-~ + y =x sinx dx 3. tanx-r- + y = secx dx Wor/fc through them all: then check your results with those given in frame 63. 621 First Order Differential Equations Results: 63 1. p 4x (IF = e 3x ) 2. xy = sin x - x cos x + C (IF=*) 3. y sin x = x + C (IF = sin x) DDDnannnnnnnnnnanannnnaaanDnDnnnnDnDDa There is just one other type of equation that we must consider. Here is an example: let us see how it differs from those we have already dealt with. To solve -r- +— . y = x y 2 dx x Note that if it were not for the factor^ 2 on th& right-hand side, this equation would be of the form -p + ?y = Q that we know of old. To see how we deal with this new kind of equation, we will consider the general form, so move on to frame 64. 64 Bernoulli's equation. Equations of the form , ,,♦,/-•) where, as before, P and Q are functions of x (or constants). The trick is the same every time: (i) Divide both sides by y" . This gives -"igl+P/-" =Q (ii) Now put z =y 1 ~ r y dx dz so that, differentiating — = 622 Programme 22 65 So we have dz ,, , - n dy ■— =(1 -n)y n -f dx dx f x + ?y = Qy» dz (i) (ii) Putz=y-" sothat^l=(l-«)j;-"^ If we now multiply (ii) by (1 - n) we shall convert the first term into dz dx ' (i-»)7" ! -^ + (i-")P/-"=(i-")Q Remembering that z = y l " and that -j- = (1 ~ «) v n -f- , this last line can now be written dz_ dx + PlZ = Q! with Pi andQi functions of x. This we can now solve by use of an integrating factor in the normal way. Finally, having found z, we convert back to y using z = y 1 '" . Let us see this routine in operation - so on to frame 66. 66 dy A -T- + — y = x j dx x' (i) Divide through by y 2 , giving Example 1. Solve 67 _ 2 dy , 1 _! y -f- + — y =x dx x (ii) Now put z= j 1 ", i.e. in this case z =y 1 2 =y -i z=j> _dz _ _ - 2 dy dx J dx (iii) Multiply through the equation by (-1), to make the first dz term dx' -2 dy 1 -i _ -y -r--~y =~x dx x dz 1 dz so that — z = -x which is of the form— + P z = Q so that you can CUC Ji (XX, now solve the equation by the normal integrating factor method. What do you get? When you have done it, move on to the next frame. 623 First Order Differential Equations y = {Cx-x 2 y x Check the working: dz 1 — z = -x ax x IF=e $?dx \?dx= \--dx = -\nx .-. IF=e -ln^= e ln ^~ 1 >=x- 1 =- x z.IF = \ Q.lFdx :. z- = \-x.— dx \dx=-x + C M z = Cx-x 2 z=y' x :.—=Cx-x 2 :.y=(Cx-x 2 T But Right! Here is another. Example 2. Solve x 2 y -x 3 -— =y i cosx First of all, we must re-write this in the form-^- + Fy = Q y n So, what do we do? 68 Divide both sides by (~x 3 ) _dy_ dx 1 --■y = y* cos x X 3 giving Now divide by the power of y on the R.H.S., giving _ 4 dy _ 1 - 3 cosx y ~dx T y 'x s ~ Next we make the substitution z =y l " which, in this example, is z=y l -*=y~ 3 -3 J ■ dz ..z=y and .. — - : dx 69 70 624 Programme 22 71 *1 = _ 3 -*& dx y dx If we now multiply the equation by (-3) to make the first term into — - , we have dx -»-<%<?-'' 3 cosx dz 3 _ 3 cosx i£ Tx + " x z= ^r~ This you can now solve to find z and so back to y. Finish it off and then check with the next frame. 72 / = 3 sin x + C For: IF = e IPdx dz 3 3 cos x —r+—.z = — 3 — dx x x hH- -dx = 3 In x IF=e 3lnx =e ln0c 3 ) =;c 3 Z.IF =/ QIF dx , f 3 c osx 3 , '■ x "]ir~ x dx ■J .'. zx But, in this example, z ~y~ = 3 cos x dx 3 sin x + C .-.^5- =3sinx + C 3 sin x + C Let us look at the complete solution as a whole, so on to frame 13. 625 First Order Differential Equations Here it is: To solve 2 x y ■ . dy_ dx -4 dy y dx _ r 3dy X dx 1 1 -3 = y cosx _y 4 cosx cosx x 3 Put z = =/■ "" =y l "4 =J ,-3 . dz _ 4 ..— =-3y ■ dx 4v dx Equation becomes -3y^ dy_ + l dx x' , 3 cosx ■ y x 3 i.e. dL + h dx x' 3 cosx Z= x 3 l F=e SPdx P dx = 1 — <2x = 3 In x ■. IF = gSMX; = e m(* 3 ) =JC 3 .'. z ;x 'f cosx s X J =.[3 cos x dx :. z x 3 = 3 sinx + C But z = =y~ 3 . x 3 -y 3 ■■■y 3 = 3 sinx + C x 3 3 sin x + C 73 They are all done in the same way. Once you know the trick, the rest is very straightforward . On to the next frame. Here is one for you to do entirely on your own. 7/1 Example 3. Solve 1y - 3 ^ = y* e 3x Work through the same steps as before. When you have finished, check your working with the solution in frame 75. 626 Programme 22 75 f 5 ~e 5x e 2x + A Solution in detail: •' dx 3 y 3 „- 4 ^ -2 v- 3 = - ,3X y "-an' Put Z=>' 1 " 4 = >>~ 3 . dz _ _ 4 a> ■■~r 3y j dx dx Multiplying through by (-3), the equation becomes ^l + 2v- 3 =e 3x -3y- dx dz i.e. -3- + 2z = e J ax IF = e JPd* Jpdx-J 2 dx = 2jc .'. IF = e 2 .. K = f e 3x e 2x dx = \ 7.X = e 3X e 2X dx = g 5X dx + C Butz=>> 3 . e 2x ■ v 3 e 5x + A 5 5e 2x •'■ y e 5 *+A On to frame 76. _ Finally, one further example for you, just to be sure. #U Example 4. Solve y -2x^- = x(x + \)y 3 First re-write the equation in standard form j^ + ?y-Qy" This gives 77 d>_ 1 dx 2x ' y (x + l)y 3 2 Now off you go and complete the solution. When you have finished, check with the working in frame 78. 627 First Order Differential Equations 6x 2x* + 3x 2 + A Solution: dy _ 1 „__ (* + i)y 3 2 1 _, dx ' 2x ' y y dx 2x ■y (x+l) Put Equation becomes 2= yl-3 =y -2 . dz_ _ _ - 3 dy dx dx -2y-^ + L y -2 ={x+l) dx x' v ' dz a. ! l.e.-r- +— . z = jc + 1 dx x • i F =e ta *=jc 2. IF = Q.IFcfr .'. z x = \(x + 1) x dx "J = (x 2 + x) G?.X x 3 JC 2 . X _ 2x 3 + 3x 2 + A ■ M 2 6 fa ■■ y 2xr 3 + 3x 2 + A 78 Butz=/" DDDnnDnDnnnnnaDODDannnDDnnDDnnDnnnnnnD There we are. You have now reached the end of this programme, except for the Test Exercise that follows. Before you tackle it, however, read down the Revision Sheet presented in the next frame. It will remind you of the main points that we have covered in this programme on first order differential equations. Turn on then to frame 79. 628 Programme 22 79 Revision Sheet 1 . The order of a differential equation is given by the highest derivative present. An equation of order n is derived from a function containing n arbitrary constants. 2. Solution of first order differential equations. (a) By direct integration: —7--f(x) gives y =Jf(x)dx (b) By separating the variables: F(y). -f- =f(x) gives F0>) dy = I fix) dx (c) Homogeneous equations: Substitute y = vx , dv _ „, , gives v + x—= Hv) dx (d) Linear equations: -r- + Py = Q Integrating factor, IF = e-' p dx and remember that e to F = F gives y IF=\Q.IF dx (e) Bernoulli's equation:— j- + Py = Qy n Divide by y n : then put z =y 1 ~ n Reduces to type (d) above. DDODODDDDDDDODODOODDDDDDDDDDDDDDDDDDDD If there is any section of the work about which you are not perfectly clear, turn back to that part of the programme and go through it again. Otherwise, turn on now to the Test Exercise in frame 80. 629 x First Order Differential Equations The questions in the test exercise are similar to the equations you have been solving in the programme. They cover all the methods, but are quite straightforward. Do not hurry: take your time and work carefully and you will find no difficulty with them. Test Exercise— XXII Solve the following differential equations: 1. x^=x 2 +2x-3 dx 2. (1 +,)'-£= l + „» 3. ® + 2y=e 3 * dx a dy 2 4. x-i--y =x dx 5. x 2 -r = x 3 sin3x + 4 dx 6. x cos v-^-— sin v = dx 7. (x 3 + xy 2 )^- = 2 y 3 dx 8. (x 2 -\)^ + 2xy=x dx 80 4h t 9. — f- + y tanh x = 2 sinh x dx dy -j 10. x—f--2y=xcosx dx 11. ^ + ^y 3 dx x 12. x d f+3y=x 2 y 2 dx 630 Programme 22 Further Problems-XXlI Solve the following equations. I. Separating the variables 1. ^"3)^ = 4, 2. (l + x 3 )j^-=x 2 y given that* = 1 when y = 2. 3. x 3 +(j> + l) 2 -g = 4. cosy + (1 + e~ x ) siny^- = 0, given that y = tt/4 when x = 0. 5. x 2 (y+l)+j> 2 (x-l)£=0 II. Homogeneous equations 6 (2v - x)-^ = 2x + y, given that y = 3 when x = 2. y J dx 7. (xy+y 2 )Hx 2 -xy)^ = 8. (x 3 +y 3 ) = 3xy 2 ^ 9. y-3x + (4y + 3x)^=0 10. (x 3 +3xy 2 )^=y 3 +3x 2 y III. Integrating factor 11. x^-j=x 3 +3x 2 -2x ax „ dV 12. — r-+ y tanx = sinx dx 13 x izl-y=x 3 cosx, given that y- when x = 7T. ax 14. (1 + x 2 ) ^f- + 3x>> = 5x, given that y = 2 when x = 1 . 15 j£l + v cot x = 5 e cosx , given that >> = -4 when x = ff/2. 631 First Order Differential Equations IV. Transformations. Make the given substitutions and work in much the same way as for first order homogeneous equations. 16. (3x + 3y-4)=^-=-(x+y) Putx+y = v 17. (y-xy 2 ) = (x+x*y)® ?nty = V - 18. (x-y-\) + (4y + x-l)^-=0 Putv = x-1 19. (3y-7x + 7) + (7y-3x + 3)^-=0 Putv=x-1 20. y(xy+i)+x(l+xy+x 2 y 2 )-^-=0 Put^=— V. Bernoulli's equation 21. f x+ y=xy> 22. %*y-f* 23. 2^+y=y\x-i) 24. -£-- 2y tanx = y 2 tan 2 * 25. — ; - + y tan x - y 3 sec 4 x dx J VI. Miscellaneous. Choose the appropriate method in each case. 26. (l-x 2 )^=l+xy 27. xy^-(l+x)s/(y 2 ~\) = 28. 0c 2 -2jcy + 5j; 2 ) = (;t 2 + 2xy+y 2 ) d £- 29. —r-~y cot jc = .y 2 sec 2 x, given >■ =-1 when* = 77/4. 30. y + (x 2 -4x)^=0 632 I Programme 22 VII. Further examples 31. Solve the equation-^- -y tanx = cos* - 2x sinx, given that >» = when x = tt/6. 32. Find the general solution of the equation dy _ 2xy +y 2 dx x 2 + 2xy 33. Find the general solution of (1 + x 2 )-j- = x(l + y 2 ). 34. Solve the equation x-j- + 2y = 3x - 1 , given that y = 1 when* = 2. 35. Solve x 2 —?-=y 2 - xy-==, given that y = 1 when* = 1. 36. Solve -?- = e 3x ~ 2y , given that y = when x = 0. 37. Find the particular solution of-p + — : j> = sin 2x, such that y = 2 when x = 7f/4. 38 . Find the general solution of y 2 + x 2 ~j- = xy 39. Obtain the general solution of the equation 2xy-$=x 2 -y 2 dx 40. By substituting z = x - 2y, solve the equation dy _ x - 2y + 1 dx 2x -Ay given that y = 1 when x = 1 . 41. Find the general solution of (1 -x 3 )— +x 2 y =x 2 (l -x 3 ) dx K '' 42. Solve—- +— = sin x, given that y = at x = ■nil. dx x 43. Solve-^- + x + xy 2 = 0, given_y = when x = 1 . 633 First Order Differential Equations 44. Determine the general solution of the equation dy (I 2x \ = _±_ dx \x l-x 2 j y l-x 2 45. Solve (l+* 2 )-^ + *>> = (l + * 2 ) 3/2 46. Solve x(l + y 2 ) -y{\ + x 2 )^ = 0, given y = 2 at x = 0. 47 . Solve 2 _ 2 .-£ = 1 , given /• = when = 7r/4. 48. Solve -3~ +^ cot jc = cos x, given thatj = when x = 0. 49. Use the substitution j =—, where v is a function of at only, to transform the equation 4y . y 2 -r-+- =xy 2 dx x into a differential equation in v and x. Hence find y in terms ofx 50. The rate of decay of a radio-active substance is proportional to the amount A remaining at any instant. If A = A at t = 0, prove that, if the time taken for the amount of the substance to becomeyAo is T, then A = A e~ (t ln 2)/T . Prove also that the time taken for the amount remaining to be reduced to^j A is 4-32 T. 634 Programme 23 SECOND ORDER DIFFERENTIAL EQUATIONS Programme 23 J Many practical problems in engineering give rise to second order differential equations of the form where a, b, c are constant coefficients and/(x) is a given function of x. By the end of this programme you will have no difficulty with equations of this type . Let us first take the case where f(x) = 0, so that the equation becomes Let y = u and y = v (where u and v are functions of x) be two solutions of the equation. ,-, , n d l u , ,du _ _ ~ a ±? + b fc + CU ~° d 2 v dv dx 2 dx and a -r-2+b—+cv = Adding these two lines together, we get N °» £ <« + ') =f + 1 » d i? <" + "> -0 + S- ,te " fore ,te equation can be written a-3-2 (u + v) + fe /•(« + v) + c(« + v) = which is our original equation withy replaced by (« + v). d 2 y dy _ n i.e. If j> = w and j> = v are solutions of the equation a^-£ + *^ + 0> "" u > so also isy = m + v. This is an important result and we shall be referring to it later, so make a note of it in your record book. Turn onto frame 2. 637 Second Order Differential Equations Our equation was a-rr + b-^-+ cy = 0. If a = 0, we get the first order equation of the same family b-r + cy = i.e. -r-+ ky = where k =t dx 7 dx b Solving this by the method of separating the variables, we have &=- ky ;.[*L = -[ kdx dx J y J which gives In >> =-kx + c :. y = e kx + c = e kx .e c = Ae kx (since e c is a constant) i.e. y = Ae kx If we write the symbol m for —k, the solution is y = A e m * In the same way,^ = Ae mx will be a solution of the second order equation a—\ + b-^-+ cy = 0, if it satisfies this equation. dx dx Now, if y = Ae n dy. dx = Am e p 2 =Am*e mx dx z and substituting these expressions for the differential coefficients in the left-hand side of the equation, we get On to frame 4. 638 Programme 23 aAm 2 e mx +bAme mx +cAe mx = Right. So dividing both sides by Ae mx , we obtain am 2 + bm + c = which is a quadratic equation giving two values for m. Let us call these m-m^ and m = m 2 i.e. y = Ae m i x and;; = Be" 1 ^ are two solutions of the given equation. Now we have already seen that if _y = u and y = v are two solutions so also is y = u + v. .'• lfy = A e miX and y = B e m ^ x are solutions, so also is y = Ae m i x +Be m * x Note that this contains the necessary two arbitrary constants for a second order differential equation, so there can be no further solution. Move to frame 5. The solution, then, of a— 4" + b-^-+ cy = is seen to be ax dx v = Ae m i x + Be m i x where A and B are two arbitrary constants and m, and m 2 are the roots of the quadratic equation am 2 + bm + c = 0. This quadratic equation is called the auxiliary equation and is obtained directly from the equation a—^ + b-f-+ cy = 0, by writing m 2 for ^-\, (XX CtX QX m for-^-. 1 for v. dx Example: For the equation 2-—^ + 5 -p+ 6j> = 0, the auxiliary equation is 2m 2 + 5m + 6 = 0. In the same way, for the equation— \ + 3-p+ 2y = 0, the auxiliary equation is Then on to frame 6. 639 Second Order Differential Equations m 2 + 3m + 2 = Since the auxiliary equation is always a quadratic equation, the values of m can be determined in the usual way. e.g. if m 2 + 3m + 2 = (m + 1) (m + 2) = :. m=-l and w = -2 .". the solution of-p£ + 3^- + 2y = is j> = A e~* + B e 2x In the same way, if the auxiliary equation were m 2 + Am - 5 = 0, this factorizes into (m + 5) (w - 1) = giving m = 1 or -5, and in this case the solution would be y = A er + B e The type of solution we get, depends on the roots of the auxiliary equation. (i) Real and different roots Example J. —^ + 5-f^-+ 6y = dx l dx J Auxiliary equation: m 2 + Sm + 6 = •'. (m + 2) (m + 3) = .'. m = -2 or m = -3 .'. Solution is y = A e~ 2x + B e" 3x Example 2. -A~ 1~+ 12v = Auxiliary equation : m 2 - 7x + 1 2 = (/w - 3) (w - 4) = -'. w = 3 or w = 4 So the solution is Turn to frame 8. 640 Programme 23 8 y = Ae 3x +Be 4 Here you are. Do this one. Solve the equation -j-=j + 3 -j- — 1 Oy = d 2 y , 3 dy dx 2 dx 9 When you have finished, move on to frame 9. y = Ae 2x + Be Now consider the next case. (ii) Real and equal roots to the auxiliary equation. Let us take d 2 y dx dx , + 6± + 9y = 0. The auxiliary equation is: m 2 + 6m + 9 = .'. (m + 3) (m + 3) = .'. m = -3 (twice) If m, = -3 and m 2 = -3 then these would give the solution y = A e" 3x + B e~ 3x and their two terms would combine to give y = Ce~ 3x . But every second order differential equation has two arbitrary constants, so there must be another term containing a second constant. In fact, it can be shown that y = Kx e~ 3x also satisfies the equation, so that the complete general solution is of the formj' = Ae' 3x + Bxe~ 3x i.e. y =e' 3X (A + Bx) In general, if the auxiliary equation has real and equal roots, giving m= m x (twice), the solution of the differential equation is y =e m i x (A + Bx) Make a note of this general statement and then turn on to frame 10. 641 Second Order Differential Equations Here is an example: J II Example 1. Solve -—7 + 4 -~ + 4v = dx L dx Auxiliary equation: m 2 +4m + 4 = Q (m + 2) (m + 2) = .\ m = -2 (twice) The solution is: _y = e' 2x (A + Bx) Here is another: Example 2. Solved + 1 o4^+ 25 y = dx 1 dx 7 Auxiliary equation : m 2 + 10m + 25 = (m + 5) 2 = :. m=-5 (twice) 7 =. e~ 5 *(A + Bx) Now here is one for you to do : Solve 4^ + 8^+16^ = When you have done it, move on to frame 11. y = <f 4 *(A + Bx) Since if J0 +8 £ +16j; = o the auxiliary equation is m 2 + 8w + 16 = .'. (w + 4) 2 = .". m = -4 (twice) ■•■ j = e~ 4 *(A + Bjc) So, for real and different roots m = m x and m~m 2 the solution is y = Ae m i x +Be m * x and for rea/ a«<? equal roots m-m x (twice) the solution is y=e m i x (A+Bx) Just find the values of m from the auxiliary equation and then substitute these values in the appropriate form of the result. Move to frame 12. 11 642 Programme 23 |Z ( iji ) Complex roots to the auxiliary equation. Now let us see what we get when the roots of the auxiliary equation are complex. Suppose m = a±$, i.e. m x = a + jj3 and m 2 = a- j)3. Then the solution would be of the form = Ce ax .e ii3x +De ax .e- i < 3x = e aX {Cei< 3x +Dc*| Now from our previous work on complex numbers, we know that e ]X = cos x + j sin x e~ iX = cos x— j sin x e i$x = C os jSx; +j sin j3x and that e"-" 3 * = cos 0X - j sin fix Our solution above can therefore be written y = e a *{C(cos fix + j sin fix) + D(cos fix - j sin |3jc)} = e ax {(C + D) cos |3x + j(C - D) sin fix} y = e ax {A cos (3x + B sin |3x} where A = C + D B=j(C-D) .". If m = a ± j|3, the solution can be written in the form y = e ax {A cos (3x + B sin fix} Example: If m = - 2 + j3 , then .y = e' 2X {A cos 3x + B sin 3x} Similarly, if m = 5 + j2, then 7 = 643 Second Order Differential Equations y = e sx [A cos 2x + B sin 2x] Here is one of the same kind: Solve Auxiliary equation: d^ + 4 ^ +9 ' = ° m 2 + Am + 9 = .. m _ -4±V(16-36) -4±V-20 _ -4 ± 2JV5 -2±jV5 In this case a = -2 and /3 = \/S Solution is: y = e' 2x (A cos \/5x + B sin y/5x) Now you can solve this one : When you have finished it, move on to frame 14. y = e x (A cos 3x + B sin 3x) Just check your working: ax* dx Auxiliary equation: m 2 - 2m + 10 = 2 ± V(4 - 40) m '■ _2±V-36 l±j3 y = e x (A cos 3x + B sin 3*) Jwrfl fo /rame 75. 13 14 644 Programme 23 1 3 Here is a summary of the work so far. Equations of the form a-rp + b ^-+ cy = Auxiliary equation: am 2 +bm + c = Q (i) Roots real and different m = m x and m = m 2 Solution is j = Ae w ' Jt + Be m ^ (ii) Real and equal roots m = m^ (twice) Solution is y = e m i x (A + foe) (iii) Complex roots m = a±]j5 Solution is y = e ax (A cos fix + B sin /3x) In each case, we simply solve the auxiliary equation to establish the values of m and substitute in the appropriate form of the result. On to frame 16. 1 fi Equations of the form-pr ± n 2 y = cf 2 y dy _ n Let us now consider the special case of the equation a—r^r b—+ cy = when b = 0. -.£♦■»- -0 + f-« and this can be written as-p£ ± n 2 y = to cover the two cases when the coefficient of y is positive or negative. f n I f^ + „2y = 0, ™ 2 +n 2 =0 .'• w 2 =-« 2 /. m = ±]n (This is like m - a ± j|3, when a = and )3 = n) :. y = A cos «x + B sin nx (ii ) lf^-« 2 y = 0, m 2 -« 2 =0 :. w 2 =« z :. m = ±n dx :. y = Ce nx +De' nx This last result can be written in another form which is sometimes more convenient, so turn on to the next frame and we will see what it is. 645 / Second Order Differential Equations You will remember from your work on hyperbolic functions that coshrar =- z e nx _ e -nx sinh nx = :. e nx ~ e nx = 2 sinh nx Adding these two results: 2 e nx = 2 cosh nx + 2 sinh nx .'. e nx = cosh nx + sinh nx Similarly, by subtracting: e nx = cosh nx - sinh nx Therefore, the solution of our equation, y = Ce nx + De~" x , can be written y = C(cosh nx + sinh nx) + D(cosh nx - sinh nx) = (C + D) cosh nx + (C - D) sinh nx i.e. y = A cosh nx + B sinh nx Note. In this form the two results are very much alike: d 2 y (0 -j-r + n 2 y = y = A cos nx + B sin nx d 2 y 00 -f^-n 2 y~ y = A cosh nx + B sinh nx Make a note of these results in your record book. Then, next frame. Here are some examples: d^y dx 2 d 2 y Example 1. -~r^j+l6y = .'. m 2 =-16 .'. m = ±j4 ■'. jy = A cos 4x + B sin 4x dx 2 d 2 v Example 2. —^ - 3y = .'. m 2 = 3 .'. m = ± \/3 Similarly Example 3. — ^ + 5y = cfor 77zen f «r« oh ro frame 1 9. y = A cosh \/3x + B sinh \/3x 17 18 646 \ Programme 23 19 y = A cos y/5x + B sin \j5x And now this one: d 2 y Example 4. — -=§■ — 4y = .'. m = 4 :. m = ±2 y = 20 j' = A cosh 2x + B sinh 2x Now before we go on to the next section of the programme, here is a revision exercise on what we have covered so far. The questions are set out in the next frame. Work them all before checking your results. So on you go to frame 21. 21 ™ ■ Revision Exercise Solve the following: 1. d 2 y ,~,dy 2. d 2 y 3. d 2 y,^dy dS +2 dx- 3y -° 4. 2^ + 4^+3^ = dx l dx 5. d 2 y „ dx For the answers, turn to frame 22. 647 Second Order Differential Equations Results 1 . y = e 6X (A + Bx) 2. j> = A cos \Jlx + B sin V?* 3. >> = Ae* + Be" 3x _ x x 4. j> = e x (A cos /— + B sin jt) 5. _y = A cosh 3x + B sinh 3x By now, we are ready for the next section of the programme, so turn on to frame 23. 22 So far we have considered equations of the form T T| a — K + b -r-+ cy = f(x) for the case where f(x) = dx dx Iff(x) = 0, then am 2 + bm + c = giving m = m l and m= m 2 and the solution is in general j> = Ae™ 1 * + Be™ 2 *. In the equation a-t-+ fe-r-+ c^ =/(*), the substitution y = Ae miX + Be™ 2 * would make the left-hand side zero. Therefore, there must be a further term in the solution which will make the L.H.S. equal to f(x) and not zero. The complete solution will therefore be of the form y = A e m i* + B e™ 2 * + X, where X is the extra function yet to be found. y = Ae miX + Be m2X is called the complementary function (C.F.) y = X(a function of x)" " " particular integral (P. I.) Note that the complete general solution is given by general solution = complementary function + particular integral Our main problem at this stage is how are we to find the particular integral for any given equation? This is what we are now going to deal with. So on then to frame 24. 648 Programme 23 24 d 2 y dy To solve an equation a —^j + b -f- + cy = f(x) (i) The complementary function is obtained by solving the equation with/(x) = 0, as in the previous part of this programme. This will give one of the following types of solution: (i) y = Ae m i x +Be m * x (ii) y = e m i x (A + Bx) (iii) y = e ax (A cos fix + B sin fix) (iv) y = A cos nx + B sin nx (v) y = A cosh «x + B sinh nx (ii) The particular integral is found by assuming the general form of the function on the right-hand side of the given equation, substituting this in the equation, and equating coefficients. An example will make this clear: Example: Solve 44 - 5 ^ + 6v = x 2 dx* dx ' (i) To find the C.F. solve L.H.S. = 0, i.e. m 2 - 5m + 6 = •'. (m - 2) (m - 3) = .\ m = 2 or m = 3 .'. Complementary function is y = Ae 2x +Be 3x (i) (ii) To find the P.I. we assume the general form of the R.H.S. which is a second degree function. Let y = Cx 2 + Dx + E. ^=2C* + Dand^- dx dx* Substituting these in the given equation, we get 2C - 5(2Cx + D>+ 6{Cx 2 + Dx + E) = x 2 2C - IOCjc - 5D + 6Cx 2 + 6Dx + 6E = x 2 6Cx 2 + (6D - 10C)x + (2C - 5D + 6E) = x 2 Equating coefficients of powers of x, we have [x 2 ] 6C=1 /. C = i [x] 6D-10C = :. 6D = i£ = | ■'• D = TS" [CT] 2C-5D + 6E = :. 6E=f§-§-=±§ /. E = ^ x 2 5x 19 .'. Particular integral is y = -r- + -yo + Too (ii) Complete general solution = C.F. + P.I. General solution is y = A e 2 * +Be 3x +~- + t§- + r?L O I 5 lUo This frame is quite important, since all equations of this type are solved in this way. On to frame 25. Then -J- = 2Cx + D and -^ = 2C 649 Second Order Differential Equations We have seen that to find the particular integral, we assume the general £ jj form of the function on the R.H.S. of the equation and determine the values of the constants by substitution in the whole equation and equat- ing coefficients. These will be useful: If f{x) = k fix) = kx fix) = kx 2 fix) = k sin x or k cos x fix) = k sinh x or k cosh x fix)=e kx Assume y - C y = Cx + D y = Cx 2 +Dx + E y = C cos x + D sin x y = C cosh x + D sinh * " y = Ce kx ikely to meet at this stage. This list will cover all the cases you are '. So if the function on the R.H.S. of the equation is/(x) = 2x 2 + 5, you would take as the assumed P.I., y : y = Cx 2 +Dx + E 26 Correct, since the assumed P.I. will be the general form of the second degree function. What would you take as the assumed P.I. in each of the following cases: 1. f(x) = 2x-3 2. fix) = e sx 3. fix) = sin Ax 4. /(x) = 3 - 5x 2 5. fix) = 21 6. fix) = 5 cosh 4-x When you have decided all six, check your answers with those in frame 27. 650 Programme 23 27 Answers 1. /(x) = 2jc-3 2. /(*)=e 5 * 3. /(x) = sin 4x 4. /(*) = 3 - 5x 2 5. /(x) = 27 6. /(jc) = 5 cosh4x P.I. is of the form y = Cx + D Ce 5 * y = C cos 4x + D sin 4jc 7 = Cx 2 + Dx + E j> = C cosh 4x + D sinh 4* All correct? If you have made a slip with any one of them, be sure that you understand where and why your result was incorrect before moving on. Next frame. 28 Let us work through a few examples. Here is the first. d 2 v Example 1. Solve ^~* - 5~-+ 6v dx 2 dx J ■-24 (i) C.F. Solve L.H.S. = /. m 2 - 5m + 6 = .'. (m-2)(m-3) = :. m = 2 and m = 3 .". y = Ae 2X + Be 3x 0) (ii) P.I. f(x) = 24, i.e. a constant. Assume^ = C Then 4Z = and ^ = dx (ii) dx Substituting in the given equation 0-5(0) + 6C = 24 C = 4 .". PJ.is y = 4 General solution is y = C.F. + P.I. i.e. y = Ae 2X +Be 3 * + 4 : C.F. P.I. Now another: d 2 y dy Example 2. Solve —4 - 5 -f- + 6v = 2 sin 4x dx 1 dx (i) C.F. This will be the same as in the last example, since the L.H.S. of this equation is the same. i.e. y = Ae 2X + Be 3 * (ii) P.I. The general form of the P.I. in this case will be 651 Second Order Differential Equations y = C cos Ax + D sin Ax Note: Although the R.H.S. is/(x) = 2 sin Ax, it is necessary to include the full general function y = C cos Ax + D sin Ax since in finding the differential coefficients the cosine term will also give rise to sin Ax. So we have y = C cos Ax + D sin Ax dy — = -4C sin Ax + 4D cos Ax d 2 y ^~2 - -1 6C cos Ax - 1 6D sin Ax We now substitute these expressions in the L.H.S. of the equation and by equating coefficients, find the values of C and D. Away you go then. Complete the job and then move on to frame 30. 29 C_ 25' D_ ~25 ; >" ~ 25 ( 2 cos 4x - sin 4x) Here is the working: -1 6C cos Ax - 16D sin Ax + 20C sin Ax - 20D cos Ax + 6C cos Ax + 6D sin Ax = 2 sin Ax (20C -.1 OD) sin Ax - (1 OC + 20D) cos Ax = 2 sin Ax 20C-10D = 2 40C-20D = 4) 50C = 4 :. C = ^ 10C + 20D = 10C + 20D = 0, 25 D=- In this case the P.I. is y=j^ (2 cos Ax - sin 4*) The C.F. was y = Ae 2x + Be 3x The general solution is y = Ae 2X +Be 3x +^j(2 cos4*-sin Ax) 25 30 652 Programme 23 J | Here is an example we can work through together. First we have to find the C.F. To do this we solve the equation 32 §♦ »£♦ 49,-0 dx dx Correct. So start off by writing down the auxiliary equation, which is 33 m 2 + 14w + 49 = This gives (m + 7) (m + 7) = 0, i.e.m=-7 (twice). /. The C.F. is y = e lx (A + Bx) 0) Now for the P.I. To find this w the given equation, i.e. we assun e take the general form of the R.H.S. of 34 y = Ce sx Blight. So we now differentiate dy_ dx twice , whi A d and — d: :h gives us K 1 653 Second Order Differential Equations *l=5Ce«: 4^=25C e - dx dx 1 The equation now becomes 25Ce SJC + 14.5Ce 5JC + 49Ce 5X = 4e 5x Dividing through by e sx : 25C + 70C + 49C = 4 1 144C = 4 -C = ^ The P.I. is y = -^7- 36 So there we are. The C.F. is y = e lx (A + Bx) e sx and the P.I. is y = -57- and the complete general solution is therefore . (ii) 35 y = e lx (A + Bjc) + -^ Correct, for in every case, the general solution is the sum of the complementary function and the particular integral. Here is another. Solve 4^ + 6^-+ 10v = 2sin 2x dx' dx (i) To find C.F. solve L.H.S. = :. m 2 + 6m + 10 = -6 ±7(36-40) = -6± V-4 _ 2 ~ 2 y = e~ 3x (A cos x + B sin x) m '■ 3± (ii) To find P.I. assume the general form of the R.H.S. i.e. y = On to frame 37. (i) 36 654 Programme 23 37 y = C cos 2x + D sin 2x Do not forget that we have to include the cosine term as well as the sine term, since that will also give sin 2x when the differential coefficients are found. As usual, we now differentiate twice and substitute in the given equation — — j + 6 -j- + 1 Oj = 2 sin 2x and equate coefficients of sin 2x and of cos 2x. Off you go then. Find the P.I. on your own. When you have finished, check your result with that in frame 38 38 y = TF (sin 2x - 2 cos 2x) For if y = C cos 2x + D sin 2x dv -f-= -2C sin 2x + 2D cos 2x dx . £y " dx 2 = -4C cos 2x - 4D sin 2x Substituting in the equation gives — 4C cos 2x - 4D sin 2x— 1 2C sin 2x + 1 2D cos 2x + IOC cos 2x + 10D sin 2x = 2 sin 2x (6C + 12D ) cos 2x + (6D - 12C) sin 2x = 2 sin 2x 6C + 12D = /. C = -2D 6D-12C = 2 /. 6D + 24D=2 .\ 30D = 2 °-rV C 15 P.I. is y = r-F (sin 2x - 2 cos 2x) So the C.F. is j = e" 3x (A cos x + B sin x) and the P.I. is y = Tr(sin 2jc - 2 cos 2x) The complete general solution is therefore y = (ii) 655 Second Order Differential Equations y = e 3X (A cos x + B sin x) + jf (sin 2x - 2 cos 2x) 39 Before we do another example, list what you would assume for the P.I. in an equation when the R.H.S. function was (1) f(x) = 3 cos 4x (2) f{x) = 7e lx (3) f(x) = 3 sinh x (4) f(x)=2x 2 -l (5) f(x) = x + 2e x Jot down all five results before turning to frame 40 to check your answers. (1) y = C cos4x + Dsin4x (2) y = Ce lx (3) y = C cosh x + D sinh x (4) y = Cx 2 + Dx + E (5) .y = Cx + D + Ee* 40 Note that in (5) we use the general form of both the terms. General form for x is Cx + D " " e x isEe* .'. The general form of x + e x isy = Cx + D + Ee x Now do this one all on your own. Solve ax ax Do not forget: find (i) the C.F. and (ii) the P.I. Then the general solution isj/ = C.F. + P.I. Off you go. When you have finished completely, turn to frame 41. 656 Programme 23 41 7 = Ae* + Be 2 * + ^ (2x 2 + 6x + 7) Here is the solution in detail. dx dx J (i) C.F. m 2 - 3m + 2 = :. (m - 1) (m - 2) = :. m = 1 or 2 :. y = Ae x +Be 2x (i) (ii) P.I. y = Cx 2 + Dx + E ax d 2 v •'• -TT = 2C dx 2C - 3(2Cx + D) + 2(Cx 2 + Dx + E) = jc 2 2Cx 2 + (2D - 6C)x + (2C - 3D + 2E) = x 2 2C=1 /. C=-i 2D-6C = /. D=3C :. D=| 2C - 3D + 2E = /. 2E = 3D - 2C =-|- 1 =j :. E = -|- A P.I. is y=Y + Y + 4 = 4( 2x2 + 6x + 1 ) 00 General solution: y = Ae x + Be 2x + \ (2x 2 + 6x + 7) Next frame. . 4 42 Particular solutions. The last result was^ = Ae x + Be 2 * + x(2* 2 +6x + 7) and as with all second order differential equations, this contains two arbitrary constants A and B. These can be evaluated when the appropriate extra information is provided. e.g. In this example, we might have been told that at x = 0, y = ~r and dy_ = 5_ dx V It is important to note that the values of A and B can be found only from the complete general solution and not from the C.F. as soon as you obtain it. This is a common error so do not be caught by it. Get the com- plete general solution before substituting to find A and B. In this case, we are told that when x = 0, y = -r, so inserting these values gives Turn on to frame 43. 657 Second Order Differential Equations A + B = -l 43 For: 3 7 A + B = -l We are also told that when x = 0,-r~ = -~, so we must first differentiate dx I the general solution, y = Ae x + Be" + \{2x 2 + 6x + 7) to obtain an expression for dy_ dx' So, dx dx 2 44 Now we are given that when x = 0. dx 2 :. | = A + 2B + 1 .'. A + 2B = 1 So we have A + B = -1 and A + 2B = 1 and these simultaneous equations give: A= ; B=. Then on to frame 45. 658 Programme 23 45 A = -3;B = 2 Substituting these values in the general solution y = Ae x + Be 2 *+^-(2x 2 + 6x + 7) gives the particular solution y = 2e 2X -3e x + ^ (2x 2 + 6x + 7) And here is one for you, all on your own. Solve the equation — -^ + 4-^- + 5y = 13e 3x given that when ax cix n _ 5 , dy _ 1 x = 0, v -tt and -7-- ~. 2 dx I Remember: (i) Find the C.F.; (ii) Find the P.I.; (iii) The general solution is.y = C.F. + P.I.; (iv) Finally insert the given conditions to obtain the particular solution. When you have finished, check with the solution in frame 46. 46 y = e 2X (2 cos x + 3 sin x) +- For: &*%+w (i) C.F. m 2 + 4m + 5 = :. m = " 4±V(16 " 20 ) = ZilE w = -2 ± j :. y = e 2X (A cos x + B sin x) (0 (ii) P.I. 7 = Ce 3 * ■ ■& = 3Ce 3 *, Q = 9Ct ax dx .-. 9Ce 3x + 12Ce 3 * + 5Ce 3 3X 26C= 13 :. C = | .'. P.I.is>> = 13e 3 * g3X 2 (ii) General solution y = e 2X (A cos x + B sin x) + 4>— ; x = 0, y =■ '• 2 dy dx 5 = A+i /. A=2 y = e~ 2x (2 cos x + B sin x) + — -^ = e 2X (-2 sin x + B cos x) - 2e 2 * (2 cos x + B sin x) + 3e 3 ^ = 1 • I dx 2 "2 •'■ Particular solution is " ~ "~~ 2x x = 0, ^ = ^ :. ^ = B - 4 + ^ /. B = 3 y = e x (2 cos x + 3 sin x) + -y- 659 Second Order Differential Equations Since the C.F. makes the L.H.S. = 0, it is pointless to use as a P.I. a *\§ term already contained in the C.F. If this occurs, multiply the assumed P.I. by x and proceed as before. If this too is already included in the C.F., multiply by a further x and proceed as usual. Example: Solve -^ - 2 &-- &y = 3 e 2x (i) C.F. m 2 - 2m - 8 = .\ (m + 2) (m - 4) = .'. m = -2 or 4 y = Ae« x + Be~ 2x (i) (ii) P.I. The general form of the R.H.S. is Ce~ 2x , but this term in e 2x is already contained in the C.F. Assume y = Cxe~ 2x , and continue as usual. y = Cxe 2x = Cx(-2e- 2x ) + Ce 2x = Ce _2X (l - 2x) dx ^ = Ce 2X (-2) - 2Ce" 2 * (1 - 2x) = Ce~ 2 * (4jc - 4) Substituting in the given equation, we get Ce 2x {Ax - 4) - 2.Ce" 2 * (1 - 2x) - 8Cxe" 2X = 3e 2X (4C + 4C - 8C)a: - 4C - 2C = 3 -6C = 3 :. C = -i P.I. is y=-^xe~ 2x (ii) — 2JC General solution y = Ae 4X + Be" 2x -^V~ So remember, if the general form of the R.H.S. is already included in the C.F., multiply the assumed general form of the P.I. by x and continue as before. Here is one final example for you to work. Solve £y + *L- 2 y = e x dx dx Finish it off and then turn to frame 48. 660 Programme 23 48 y = Ae x +Be- 2X +^- Here is the working: To solve (i) C.F. m 2 + m - 2 = dx 2 dx (m-l)(m + 2) = :. m = 1 or -2 .'. y = Ae x + Be 2x (i) (ii) P.I. Take >> = Ce*. But this is already included in the C.F. Therefore, assume j> = Cxe* . Then dx dll dx Cxe? +Ce x = Ce x (x + \) f =Ce* +Cxe x + Ce* = Ce*(x + 2) Ce* (x + 2) + Ce*(x + 1) - 2C*e* = e* C(x + 2) + C(x+ l)-2Cx= 1 3C = 1 :. C = j P.I. is y .xe* Oi) and so the general solution is y = Ae x +-Qe 2x +Zf- You are now almost at the end of this programme. Before you work through the Test Exercise, however, look down the revision sheet given in frame 49. It lists the main points that we have established during this programme, and you may find it very useful. So on now to frame 49. 661 Second Order Differential Equations Revision Sheet d y dy 1 . Solution of equations of the form a — 3 + b —+ cy = f(x) 2. Auxiliary equation: am 2 + bm + c = 3. Types of solutions: (a) Real and different roots m = m^ and m= m 2 y = Ae m 1 x +Be m 2 x (b) Real and equal roots rn^m^ (twice) y = e m i x (A + Bx) (c) Complex roots m = a ± j/3 y = e ax (A cos 0* + B sin 0x) 4. Equations of the form -j-t+ « 2 v = j = A cos «jc + B sin «x <i 2 y 5. Equations of the form -72 ~n 2 y - y = A cosh nx + B sinh wx 6. General solution j> = complementary function + particular integral 7. (i) To find C.F. solve a^f + 6 ^ + cy = rf 2 _y dy_ dx 2 dx 49 (ii) To find P.I. assume the general form of the R.H.S. Note: If the general form of the R.H.S. is already included in the C.F., multiply by x and proceed as before, etc. Determine the complete general solution before substituting to find the values of the arbitrary constants A and B. Now all that remains is the Test Exercise, so on to frame 50. 662 Programme 23 50 The Test Exercise contains eight differential equations for you to solve, similar to those we have dealt with in the programme. They are quite straightforward, so you should have no difficulty with them. Set your work out neatly and take your time: this will help you to avoid making unnecessary slips. Test Exercise - XXIII Solve the following: dx 2 dx ' 2. g-4y»10^ 4. j^ + 25y = 5x 2 +x 6. -Hr + 4-r-+ 5v = 2e 2X , given that at x = 0,y = 1 and -r-=-2. dx L dx dx dx 2 dx 1. 3^ r i-2 : ±-y = 2x-3 4^-6^ + 8y=8e™ dx 2 dx 663 Second Order Differential Equations Further Problems - XXIII Solve the following equations: '■ *£-'£-*— 2 &-«£♦»-*"♦■« d 2 y „ dy 5 ' 5? + S" - ?V = 2 cosh 2« d 2 y 10. ^2 - 9y = e 3x + sin 3x 11. For a horizontal cantilever of length I, with load w per unit length, the equation of bending is d x 2 { dy dx in terms of x. Hence find the value of y when x = l. where E, I, w and Z are constants. If y = and =£ = at x = find y 664 Programme 23 12. Solve the equation ' 2 - 1 + 4 — + 3jc = e" £? X . „ G?X . „ _ -3/ given that at r = 0, x = ~ and-^ = -2. df 2 ' ' dt dt 2 3 — + 2x = sin f 13. Obtain the general solution of the equation d 2 y „dy r dF + 4 dF + 5y = 6smt and determine the amplitude and frequency of the steady-state function. 14. Solve the equation d 2 x dx dt 2 dt dx given that at t = 0, x = and -r- = 1 . d 2 y dv 15. Solve -j-j + 3 — - + 2y = 3 sin*, given that when jc = 0,>> =-0-9 and-^ = -0-7. dx 16. Obtain the general solution of the equation d 2 y , 6 Q[y dx 2 dx y + 6-7-+ 10y = 50x 2 + 2 — +2x = 85sin3r 17. Solve the equation g? 2 x . dx d? 2 dt dx given that when t = 0, x = and— = -20. Show that the values of t for stationary values of the steady-state solution are the roots of 6 tan 3? = 7. d 2 y 18. Solve the equation-— ^-= 3 sinx-4-.y, given that y = at x = and dy that— = 1 at x = rr/2. Find the maximum value of y in the interval 0<x<tt. 665 Second Order Differential Equations 19. A mass suspended from a spring performs vertical oscillations and the displacement x (cm) of the mass at time t (s) is given by 2 dr If x = \ and -r = when t = 0, determine the period and amplitude 6 dt of the oscillations. 20. The equation of motion of a body performing damped forced vibra- tions is— r+ 5 -r + 6x = cos t. Solve this equation, given that x = 0-1 dt dt dx and— = when t = 0. Write the steady -state solution in the form dt K sin (f + a). 666 Programme 24 OPERATOR D METHODS Programme 24 1 Operator D d d j£ (sin x) = cos x d , , du dv dx dx dx These results, and others like them, you have seen and used many times in the past in your work on differentiation. The symbol — of course, can have no numerical value of its own, nor can it exist alone. It merely indicates the process or operation of finding the differential coefficient of the function to which it is attached, and as such it is called an operator. For example, — (e sx ) denotes that we are carrying out the operation of finding the differential coefficient of e sx with respect to x, which in fact gives us — (e 5JC ) = i^ =5e5x a, fd\ 2 d 2 -. . . J ,_ , Also, j — } , or T-5 as it is written, denotes that the same operation is to be carried out twice - so obtaining the second differential coefficient of the function that follows. Of course, there is nothing magic about the symbol — . We could use any symbol to denote the same process and, for convenience, we do, in fact, often use the letter D to indicate the same operation. i.e. D = — o . dy dx So that — can be written Dv. dx ' and D(sin x) = cos x D(e**) =ke kx D(x 2 + 6x - 5) = 2x + 6 etc., etc. So that D(sinhx) = Turn to frame 3. 669 Operator D Methods D(sinhx) = coshx Similarly, D(tanx) = sec 2 *, D(lnx) = 1 D(cosh 5jc) = 5 sinh 5x. Naturally, all the rules of differentiation still hold good. e.g. D(x 2 sin x) = x 2 cos x + 2x sin x (product rule) and similarly, by the quotient rule, D sin 5x x+ 1 D sin 5x x+ 1 (x + 1)5 cos 5x — sin 5x (x + iy In the same way, D 2 {x 3 } = D{D (x 3 )} = D{3x 2 } = 6x. So: The symbol D denotes the first differential coefficient, D 2 " " second " D 3 " " third and, if n is a positive integer, D" denotes Correct. the » th differential coefficient (i) D 2 (3 sin x + cos Ax) = D(3 cos x - 4 sin Ax) = -3 sin x — 1 6 cos Ax (ii) D 2 (5x 4 -7jc 2 + 3) = D(2Qx 3 -14;t) = 60x 2 -14 All very easy: it just means that we are using a different symbol to repre- sent the same operators of old. D (e 2X + 5 sin 3x) = 2e 2x + 1 5 cos 3x 45 sin 3x 135 cos 3x D 2 (e 2X + 5 sin 3x) = Ae 2X D 3 (e 2x +5 sin3x)=8e 2X Here are some for you to do. Find (i) D (Ae 5x - 2 cos 3jc) = (ii) D 2 (sinh 5x + cosh 3x) = (iii) D 3 (5x 4 - 3x 3 + lx 2 + 2x - 1) = When you have finished, turn to frame 6. etc. 670 Programme 24 (0 20e 5 * + 6 sin 3x (") 25 sinh 5x + 9 cosh 3x (in) 120x- -18 The special advantage of using a single letter as an operator is that it can be manipulated algebraically. Example 1. (D + 4){sin x) = D{sin x) + 4 sin x = cos x + 4 sin x i.e. we just multiply out in the usual way. Example 2. (D + 3) 2 {sin x) = (D 2 + 6D + 9) {sin x] = D 2 {sin x} + 6D{sin x} + 9 sin x D (sin x) = cos x D 2 (sin x) = -sinx = -sin x + 6 cos x + 9 sin x Similarly (D - 3) {cos 2x} = —2 sin 2x - 3 cos 2x For Similarly, (D - 3){cos 2x} = D{cos 2x} - 3 cos 2x = —2 sin 2x — 3 cos 2x (i) (D + 4){e 3x }=D{e 3 *}+4e = 3e 3 * +4e 3X 3X _ t^3X 1e 5 (ii) (D 2 -5D + 4){x 2 + 4x- 1} = 2 - 5 {2x + 4) + 4(x 2 + Ax - 1) = 2-10x-20 + 4x 2 + 16x-4 = 4x 2 + 6x - 22 D (x 2 +4x-l)= 2x + 4 D 2 (jc 2 +4x- 1) = 2 Now you determine this one: (D 2 - 7D + 3){sin 3x + 2 cos 3x} = When you are satisfied with your result, turn on to frame 8. 671 Operator D Methods 36 sin 3x - 33 cos 3x 8 Since D (sin 3x + 2 cos 3x) = 3 cos 3x - 6 sin 3* and D 2 (sin 3x + 2 cos 3jc) = -9 sin 3x - 1 8 cos 3x :. (D 2 - 7D + 3) {sin 3x + 2 cos 3x) = -9 sin 3x - 18 cos 3x - 21 cos 3.x + 42 sin 3x + 3 sin 3x + 6 cos 3x = 36 sin 3a: - 33 cos 3x Remember that the operator can be manipulated algebraically if required. Here is one more: (D 2 + 5D + 4){5e 2X } = 90e 2 Since (D 2 + 5D + 4){5e 2 *} = D 2 {Se 2x } + 5D{5e 2x } +*{5e 2X } ■Now D{5e 2x } = 10e 2x and D 2 {5e 2x } = 20e 2x (D 2 + 5D + 4){5e 2x } = 20e 2x + 50e 2x + 20e 2x = 90e 2x or we could have said: (D 2 + 5D + 4){5e 2x '} = (D + 4) (D + 1){5<? 2X } = (D + 4){l0e 2X + 5e 2X } = (D + 4){l5e 2x ) = 30e 2x +60e 2X = 90e 2x On now to the next frame. b/_ Programme 24 I (J The inverse operator -rr We define the inverse operator - as being one, the effect of which is cancelled out when operated upon by the operator D. That is, the inverse operator - is the reverse of the operator D, and since D indicates the pro- cess of differentiation, then — indicates the process of 11 integration 1 Right, though our definition of — is a little more precise than that. Here it is: Definition: The inverse operator - denotes integration with respect to x, omitting the arbitrary constant of integration. .g. — {sinx} D l ; 3 COS X 3X M 12 Similarly, and D X } 5 jr{sinh 3x + cosh 2x} cosh 3x sinh 2x hH + lnjc Therefore, we have that (i) the operator D indicates the operation of (ii) " D Turn on to frame 13. 673 Operator D Methods D denotes differentiation D integration 13 Of course, ^-5 =(^) and ^{/(x)} therefore indicates the result of function /(x) twice with respect to x, the arb ation being omitted. e.g. & x2 + 5x -^ = h{T + 5 -T- 4x } integrating the function /(x) twice with respect to x, the arbitrary con- stants of integration being omitted. ' 12 + 6 4x 2 " 2 x Sx ' 12 +_ 6 tx + ^-2x 2 Note that the constant of integration is omitted at each stage of integration. j So -r-2 {sin 3x - 2 cosx} = 2 cosx - sin 3x 14 Since 1 , „ „ 1 1 cos 3x , . -j~2 { sin 3x - 2 cosjc} = — j 2sinx sin 3x -JT2 {sin 3jc - 2 cos x } = 2 cos x + 2 cosx sin 3x D Here is a short exercise. Work all the following and then check your results with those in frame 1 5 . (i) D (sin 5x + cos 2x) = (ii) D(jc 2 e 3x ) Oii) 5 ( 2x 2 + 5 +- (iv) — (cosh 3x) (v) ^(3x 2 + sin2x) = When you have completed all five, move on to frame 15. 674 Programme 24 15 Here are the results in detail. (i) D (sin 5x + cos 2x) = 5 cos 5x - 2 sin 2x (ii) D(x 2 e 3X ) = x 2 3e 3x + 2xe 3x 3X n„2 = e ix (3a: 2 + 2x) 3 (iii) i(2x 2 +5+J) =2^ +5 x + 2\nx (iv) I (cosh 3*) =^|3£ (v) i I (3x 2 + sin2x) =I^3-^2!|£) Tow must have got those right, so on now to frame 16. | Q Before we can really enjoy the benefits of using the operator D, we have to note three very important theorems, which we shall find most useful a little later when we come to solve differential equations by operator D methods. Let us look at the first. Theorem I F(D){e ax }=e ax F(a) (I) where a is a constant, real or complex D{e ax } = ae <" D 2 !^} =a 2 e ax ■■■ (D 2 + D){e^} = a 2 e ax + a e™ = e^ia 2 + a) Note that the result is the original expression with D replaced by a. This applies to any function of D operating on e"* . Example 1. (D 2 + 2D - 3) {e **} = e ax (a 2 + la - 3) This sort of thing works every time: the e ax comes through to the front and the function of D becomes the same function of a, i.e. D is replaced by a. So (D 2 -5){e 2 *} = Turn to frame 1 7. 675 Operator D Methods (D 2 -5){e 2X } = -e 2 17 Similarly, (2D 2 + 5D- 2) {e 3 *} = e 3 *(2.9 + 5.3 - 2) = e 3x (18 + 15-2) = 31 e 3X The rule applies whatever function of D is operating on e ax . e.g. _i_{ e «}= e s*._i_ = i£_ e -§- K2 D-2 2 e.g. D 2 + 3 1 {e 3x } 5-2 3 2 e 3 '9 + 3 12 i ( e -2X\ = „-2X • D 2 -4D-1 ( ' (~2) 2 - 1 ■4(-2)-l „-2X 4 + 8-1 11 So (D 2 - 5D + 4) {e 4x } for (D 2 -5D + 4){e 4 *}=e«(4 2 -5.4 + 4) = e 4x (16-20+4) = Right, and in the same way, 1 for D 2 + 6D - 2 [e 3X j 25 D 2 + 6D - 2 {e 3x }=< 1 •9+ 18-2 18 19 25 Fine. Turn on now to frame 20. 676 Programme 24 £\j Just for practice, work the following: (i) (D 2 +4D-3){<? 2 *} = (iii) (D 2 - 7D + 2) {W 2 } = <*> D'-3D-2 ^> = ^ (D - 3) (D + 4) ^ e ' = When you have finished, check your results with those in the next frame. 2X t\m Results (i) (D 2 +4D-3){e 2 *}=e 2 *(4 + 8-3) = 9e (li) ^{ e - }=e -3*_J_ = JI (iii) (D 2 - 7D + 2) {e*/ 2 } = e x ' 2 (I - \ + 2) (") 5*^=2 <«">='"■ 25=1^2 Cv^) ! (^-x ) = -x 1 y ' (D - 3) (D + 4) ** ' • (-1 " 3) (-1 + 4) = -x 1 6 (-4) (3) = _£l 12_ All correct? Turn on now then to the next part of the programme that starts in frame 22. 677 Operator D Methods Theorem II £■ £ F(D){^ x V;=e^F(D + fl ){V} (II) where a is a constant, real or complex, and V is a function of x. Consider (D 2 + D + 5) {e^V} D{e ax V} = e ax D{v}+ae ax V = e ax [D{V}+aV] D 2 {e ax V} =e ax [D 2 {V} +aD{V}] +ae ax [D{V] +a V] = e ax [D 2 {V} + 2aD{v}+a 2 V] Therefore (D 2 + D + 5){e ax V} = e** [D 2 {V} + 2«D{V} + a 2 V] + e^ [D{V} + a V] + 5e a *V = e ax [(D 2 + 2Da + a 2 ){V} + (D + b){V} + 5 V] = e' zx [(D + a) 2 +(D + a) + 5]{V} which is the original function of D with D replaced by (D + a). So, for a function of D operating on {e™ V}, where V is a function of x, the e ax comes through to the front and the function of D becomes the same function of (D + a) operating on V. F(D){e ax V}=e ax F(D + a){V} An example or two will make this clear. (1) (D + 4){e 3x x 2 } In this case, a = 3 and V = x 2 = e 3x {(D + 3) + 4}{x 2 } = e 3x (D + 7){x 2 }=e 3x (2x + 7x 2 ) = (7x 2 + 2x)e 3x (2) (D 2 +2D-3){e 2 *sinx} = e 2X [(D + 2) 2 + 2(D + 2) - 3] . {sin x] = e 2X (D 2 +4D + 4+2D + 4-3) {sin*} = e 2X (D 2 + 6D + 5)(sinx} f D(sinjc) = cos x = e 2X [4sinx +6 cosx] } D 2 (sin x) = -sin x And, in much the same way, (3) (D 2 - 5){e sx cos 2x} = 678 Programme 24 23 4e sx (4 cos 2x - 5 sin 2s) for: (D 2 -5){e 5 *cos2s} = e s *[(D + 5) 2 -5] .{cos 2s} = e"[D 2 +10D + 25-5] {cos 2s} = e sx [D 2 + 10D + 20] {cos 2s} D(cos2s)=-2sin2x sx, D 2 (cos2s)=-4cos2s = e x (-4 cos 2x - 20 sin 2x + 20 cos 2x) = 4 e sx (4 cos 2jc - 5 sin 2s) Now here is another: 1 D 2 -8D + 16 - „4X {e 4x s 2 } 1 U 2 } (D + 4) 2 -8(D + 4)+ 16 ' 1 D 2 + 8D+ 16-8D-32 + 16 D 2 {x 2 } = e 4x ^{x 2 } h^-Y 12 D : (x 2 )-- 12 Now this one: they are all done the same way. (D 2 -3D + 4){e~*cos3s} The first step is to 24 (i) bring the e x through to the front (ii) replace D by (D-l) Right, so we get (D 2 - 3D + 4) {e x cos 3s} = e x [(D-l) 2 -3(D-l) + 4].{cos3s} = e x (D 2 - 2D + 1 - 3D + 3 + 4). (cos 3s} = e~ x (D 2 -5D + 8){cos3s} i D(cos 3s)=-3 sin 3s \ D 2 (cos 3s) =-9 cos 3s When you have sorted that out, turn on to frame 25. 679 Operator D Methods e x (\5 sin 3x - cos 3x) 25 Now let us look at this one. 1 D 2 + 4D + 5 - „"2X {x 3 e' 2x } Here a = -2 and V = jc 3 1 (D - 2) 2 + 4(D - 2) + 5 {* 3 } - „ _ 2X e 2x lx 3 \ D 2 -4D+4+4D-8+5 l ; 1 D 2 + 1 VI and we are now faced with the problem of how to deal with 2 , i t* 3 -' Remember that operators behave algebraically. -9 v A D 2 + l ■{x 3 } = e 2x {\ + D 2 )- 1 {x 3 } and (1 + D 2 ) : can be expanded by the binomial theorem. • (1 + D 2 )- 1 = (1 + D 2 )- 1 - D 2 + D 4 - D 6 + . 26 e' 2x (l + D 2 )" 1 {x 3 } = e- 2X (l-D 2 + D 4 -D 6 + ...).{x 3 } = e 2X (x 3 - 6x + - . . . ) = e 2x {x 3 - 6x) Here is another. D(jc 3 ) = 3x 2 D 2 (x 3 ) = 6jc D 3 (x 3 ) = 6 D 4 (x 3 ) = etc. Note we take out the factor 3 to reduce the denominator to the form (1 +w) On to frame 27. 680 Programme 24 27 M Similarly 1 1 1 2 {* 4 } = -i (l+?+F+f ! + ... ){x4} F/kmA z'r off. Then move on to frame 28. 28 --/V» 2 (* + 6a: 2 + 6) Right. So far we have seen the use of the first two theorems. Theorem! F(D){e ax } = Theorem II F(D){e ax v} = Check your results with the next frame. 29 F(D){e ax } = e ax F(a) FCD^e"* V} = e fl * F(D + fl){V} Now for Theorem III Theorem III cos ax F(D a ){^?J = F(- fl a ) ,SinflX cos ax (III) If a function of D 2 is operating on sin ax or on cos ax (or both) the sin ax or the cos ax is unchanged and D 2 is everywhere replaced by(-a 2 ). Note that this applies only to D 2 and not to D. Example 1. (D 2 + 5){sin Ax} = (-16 + 5) sin 4* = -1 1 sin Ax Just as easy as that! Example 2. p2 _ 3 {cos 2*}'= -~^ cos 2x = ~ cos 2x Example 3. 2+ { sin 3x + cos 3x] = — — - (sin 3x + cos 3x) Example 4. (2D 2 - 1 ) {sin x } = : — c" (sin 3x + cos 3x) 681 Operator D Methods -3 sin x for (2D 2 -l){sinx}= [2(-l) - 1] {sinx} = -3 sin x If the value of a differs in two terms, each term is operated on separately. e.g. ="2 — r- {sin 2x + cos 3x} = D 2 + 2 ( sin 2x ) + d^TT^ 05 3x ^ = ^-^ {sin 2x} + 3^- { cos 3x } sin 2x cos 3x 30 So therefore D^ -{sinx + cos4x} Here it is: sin x cos 4x 6 21 D 2 -5 {sin x + cos 4x} -^j {sin x} + ^—^ { cos 4x) 1 1 _ 1-5 {sinx}+q^{cos4x} sin x _ cos 4x 6 21 (I) («) 31 Here are those three theorems again: Theorem I F(D){e"> = e ax F(a) Theorem II F(D){e»V} = e flX F(D+a){V} Be sure to copy these down into your record book. You will certainly be using them quite a lot from now on. We have now reached the stage where we can use this operator D to our advantage, so turn now to frame 32. 682 Programme 24 32 Solution of differential equations by operator D methods The reason why we have studied the operator D is mainly that we can now use these methods to help us solve differential equations. You will remember from your previous programme that the general solution of a second order differential equation with constant coeffi- cients, consists of two distinct parts. general solution = complementary function + particular integral. (i) The C.F. was easily found by solving the auxiliary equation, obtained from the given equation by writing m 2 for -7-7, m for ---, and 1 for v dx dx This gave a quadratic equation, the type of roots determining the shape of the C.F. (a) Roots real and different y = Ae OTl * + Be™ 2 * (b) Roots real and equal y = e miX (A + Bx) (c) Roots complex y = e ax (A cos fix + B sin fix) (ii) The P.I. has up to now been found by 33 . . . assuming the general form of the function /(xt) on the R.H.S., substituting in the given equation and determining the constants involved by equating coefficients. In using operator D methods, the C.F. is found from the auxiliary equation as before, but we now have a useful way of finding the P.I. A few examples will show how we go about it. Example 1. ^4 + 4^-+ 3y = e 2x dx z dx ' (i) C.F. m 2 + 4m + 3 = :. (m + 1) (m + 3) = /. w=-lor-3. y = A e x + B e~ 3x (ii) P.I. First write the equation in terms of the operator D D 2 y + 4Dy + 3y = e 2X (D 2 +4D + 3> =e 2x and, applying theorem I, we get y = 683 Operator D Methods 15 34 for y = e' 1 4 + 8 + 3 15 So C.F. is and P.I. is y = Ae x + Be- „2X y- 15 So the complete general solution is y = .... ,-3X , £1 y = Ae x + Be' ix +~j 35 Correct. Notice how automatic it all is when using the operator D. Here is another. Solve dy , 6 dy [ dx 2 dx 9y = e 5 (i) First find the C.F. which is y- y = e' 3x (A + Bjc) since m 2 + 6m + 9 = :.(m + 3) 2 = :.m=-3 (twice) y = e~ 3x (A + Bx) (ii) To find the P.I., write the equation in operator D form 36 D 2 y + 6Dy + 9y = e sx (D 2 + 6D + 9)y =e 5X y : l and by theorem I General solution is y = e- D 2 + 6D + 9 1 {e 5x } 25 + 30 + 9 64 C.F. is y = e~ 3x (A + Bx) sx P.I.is y^ y = On to frame 37. 684 Programme 24 37 y = e 3x (A + Bx) + ,P — Now that you see how it works, solve this one in the same way. Solve d y , .dy , , dS +4 dx + 5y = e ' (i) C.F. m 2 + Am + 5 = .". am -4±V(16-20) 38 y = e 2X (A cos jc + B sin x) (ii) Now for the P.I. D 2 y + 4Dy + 5y = e x :. (D 2 + 4D + 5)j> = <f* y Now finish it off and obtain the complete general solution. When you have it, move on to frame 39. 39 y = e 2X (A cos x + B sin x) + y- for the P.I. is 1 e x a=-1 = e" 1-4+5 2 -2X/ i.e. 7 : .'. General solution is y = e *(A cos x + B sin x) + -~- Now here is one for you to do all on your own. Solve &♦'&♦>*-*» When you have finished it, turn on to frame 40 and check your result. 685 Operator D Methods y = Ae 3x + Be « + e_ 40 Since (i) C.F. m 2 + 1m + 12 = .'. (m + 3)(m + 4) = /. m = -3 or -4 y = Ae~ 3x + Be~* x (ii) P.I. D 2 .y + 7D.y + I2y=5e 2x (D 2 + 7D+ I2)y =5e 2x y = 5e y~- 1 1 D 2 + 7D+ 12 {5e 2x } General solution: 4+ 14+ 12 y = Ae~ 3x + Be" 4;c + 30 6 7 o[y 5 Now if we were told that at x = 0, y = -g and— = - y, we could differen- tiate and substitute, and find the values of A and B. So off you go and find the particular solution for these given conditions. Then on to frame 41. y = 2e~ 3x -e^ x ^\ 41 for x = 0, .y=-g- ir A + B 4 It: :. A + B = 1 2e 2X x = 0. dx -3Ae~ 3x -4Be-« x + -|=-3A-4B + | /. 3A + 4B = 2 3A + 4B = 2 3A + 3B = 3 .'. Particular solution is B = -l, A = 2 y = 2e- 3x -e' 4x +^ So (i) the C.F. is found from the auxiliary equation as before, (ii) the P.I. is found by applying operator D methods to the original equation. Now turn on to frame 42. 686 Programme 24 42 Now what about this one? 4^ + 3 ^-+2y = sin 2x dx z dx Solve (i) C.F. m 2 + 3m + 2 = .". (m + 1) (m + 2) = m = -1 or -2 j = Ae Jt +Br (ii) P.I. (D 2 + 3D + 2)y = sin 2x 1 y= W73DT2 {sin2x ^ By theorem III we can replace D 2 by -a 2 , i.e. in this case by -4, but the rule says nothing about replacing D by anything. -4 + 3D + 2 ■ 1 '3D -2 { sin 2xj Now comes the trick! If we multiply top and bottom of the function of D by (3D + 2) we get y = 43 3D + 2 , . „ , ■y = 9D 2_ 4 {sin2x} Correct, and we can now apply theorem III again to the D 2 in the denominator, giving: _ 3D + 2, . „ x 3D + 2 , . „ , y ~ ^36^4 t sln 2 *} =^4o-{sin 2x} Now the rest is easy, for D(sin 2x) = 2 cos 2x ■'■ y = - 4Q (6 cos 2x + 2 sin 2x) 1 20 i-e. y = — ^ (3 cos 2x + sin 2x) So C.F. is P.I. is .'. General solution is y = Ae x +Be 2x y = — ^7j (3 cos 2x + sin 2x) y = Ae x + Be 2X - -^ (3 cos 2x + sin 2*) Note that when we were faced with ^ {sin 2xj, we multiplied top and bottom by (3D+2) to give the difference of two squares on the bottom, so that we could then apply theorem III again. Remember that move: it is very useful. Now on to frame 44. 687 Operator D Methods Here is another example. Solve 4^ + 10^-+ 25y = 3 cos Ax dx 1 dx (i) Find the C.F. You do that. y- 44 y = e' sx {A + Bx) Since m 2 + 10m + 25 = .'. (m + 5) 2 = .'. m = -5 (twice) Lp- y = e -s*(A + Bx) (ii) Now for the P.I. (D 2 + 10D + 25)y = 3 cos 4x y= tf + l l 0D + 25 {3cOs4x ) Now apply theorem III, which gives us on the next line I y = 45 l -16 + 10D + 25 {3 cos 4xj since, in this case, a = 4 .'. -a 2 =-16 .'. D 2 is replaced by -1 6. Simplifying the result gives j; = TodT9 {3cos4a:} Now then, what do we do next? When you have decided, turn on to frame 47. 46 688 Programme 24 47 We multiply top and bottom by (10D - 9) Correct — in order to give D 2 in the denominator. So we have y = 10D-9 (10D + 9)(10D-9) 10D-9 {3 cos4x} 100D 2 -81 We can now apply theorem III, giving y = {3 cos Ax} 1 1 48 Here it is: y = jggj (120 sin Ax + 27 cos Ax) _ 10D-9 f „ A , '~100D 2 -81 {3C ° S4 * } _ 10D-9 f „ . , "-1600-81 t3cos4x} 168 j-(10D-9) {3cos4x} D(3 cos Ax) = -12 sin Ax tt^y (-120 sin Ax - 27 cos Ax) y = tttt (1 20 sin Ax + 27 cos Ax) So C.F.: y = e sx {A + Bx) 1 P.I.: y = - 1681 Therefore, the general solution is y = ■ Now turn on to frame 49. (120 sin Ax + 27 cos Ax) 689 Operator D Methods y = e' 5x (A + Bx) + 1681 (120 sin Ax + 27 cos Ax) Let us look at the complete solution. Here it is: To solve dx 2 dx 2Sy = 3 cos Ax (i) C.F. m 2 + 10m + 25 = :. (m + 5) 2 = :. m = -5 (twice) .". y = e~ sx (A + Bx) (ii) P.I. (D 2 + 10D + 25)7 = 3 cos 4x 1 y = y D 2 + 10D + 25 1 (3 cos 4x} -16+ 10D + 25 1 {3 cos 4x} 10D + 9 { 3 cos 4x] ioor?-8i {3cos4x} : -1600-81 {3cOs4 ^ 1 y- 1681 1 (-120 sin 4jc- 27 cos Ax) 1681 Therefore, the general solution is (120sin4jt + 27 cos Ax) y = e' 5X (A + Bx) + 1 1681 (120sin4jt + 27cos4x) That is it. Now you can do this one in very much the same way. Solve 44-4^+13^ = 2sin3x dx 2 dx 49 Find the complete general solution and then check your solution with that given in the next frame. 690 Programme 24 ■J II Here is the solution in detail. 44~4^-+ 13y = 2 sin 3x dx 1 dx J (i) C.F. m 2 -4m + 13 = ■ m = 4±V(16-52) _ 4 + V-36 2 .". j> = e 2 *(A cos 3jc + B sin 3x) (ii) P.I. (D 2 - 4D + 1 3)y = 2 sin 3x > ,= -F^rTFTTT7{ 2sin3 ^} 2±j3 \2 sin 3x} D 2 -4D+ 13 1 -9-4D+ 13 = 4(1-D) {2sin3x} - 2 _L_/ • i \ _1 1+D f . . , ~2- l-D 2 ^ Sm ' - 1 1+D / • o > " 2- 1 - (-9) ( sm3x J = ^(1 + D){sin3x} y = 2fj (sin 3.x + 3 cos 3x) General solution is y = e 2x (A cos 3x + B sin 3*) + ~q (sin 3a: + 3 cos 3x) Now let us consider the following example. Solve -73-6-7^-+ 5y = e 2X sin 3x dx 1 dx (i) First find the C.F. in the usual way. This comes to y = On to frame 51. 691 Operator D Methods y = Ae* + Be sx Since m 2 - 6m + 5 = .\ (m - 1) (m - 5) = .'. m = 1 or 5 .'• y ~ A e* + B e 5x Now for the P.I. (D 2 - 6D + 5)j> = e 2x sin 3* 7 = D 2 -6D + 5 {e2Arsin3x ^ This requires an application of theorem II F(D) {e ax V} = e ax F(D + a) {V } Here a = 2 V = sin 3x So the e 2 * comes through to the front and the function of D becomes the same function of (D + a), i.e. (D + 2), and operates on V, i.e. sin 3x 51 y = e' (D + 2) 2 - 6(D + 2) + 5 {sin 3x} 6 D 2 +4D + 4-6D-12 + 5^ S ' n3;C ^ 1 D 2 -2D-3 Now, applying theorem III, gives y = — 7 {sin 3x] y = e2X - - 9 -2D-3 ^ 3x ^ y = e 2X -2D- 12 y = -- {sin 3x} „2X 1 1 2 D + 6 Now what? Multiply top and bottom by . 2 'D + 6 [sin 3x} {sin 3x] Right. D-6 €1 2 D-6 ■• y = - So the P.I. is finally y D-6 ( . x D-6 r . x IX ^ (D-6) {sin 3x) v = = _£ 2;c _e_ 52 53 692 Programme 24 54 y = -t— (cos 3x - 2 sin 3*) So C.F.: y~- --Ae x + Be 5x P.I.: y- 2X 30 l cos 3x - -2 sin 3x) .'. General solution: Ae* + Be 5x e 2X + lo (cos 3x — 2 sin 3x) This is an example of the use of theorem II. Usually, we hope to be able to solve the given equation by using theorems I or III, but where this is not possible, we have to make use of theorem II. Let us work through another example. d 2 y Solve dx '2~y = x 2 e x (i) Find the C.F. What do you make it? y = 55 y = Ae* + Be : since m — I = Now for the P.I. m 1 .'. m = 1 or— 1 (D 2 -l)y=x 2 e x 1 Applying theorem II, the e x comes through to the front, giving i D 2 + 2D + 1 — U 2 } 1 1 " 'D'D + 2 * x 1 1 {x 2 } {x 2 } 2 ' D ' 1 + D/2 Now expand (1 + D/2) -1 as a binomial series, and we get y- On to frame 56. £. 1 I 2 D ■[- ■){* 2 } 693 Operator D Methods _e x 1 /, D D 2 \, 56 But D{x 2 } = 2x; D 2 (x 2 }=2; D 3 {* 3 } = etc. •'• y = ^--R{ x2 - x + k 1 2 Df " ' 2J and since ^ denotes integration, omitting the constant of integration, then e x / \ r =-( ) .-f(f-f^) 57 So the general solution is j = Ae* + Be~* + §-(§- -y- + §) Now here is one for you to do on your own. Tackle it in the same way. d 2 y_ dy Solve , 2 ~6^-+9y = x 3 e 3x dx z dx Find the complete general solution and then check with the next frame. y = e 3x (A + Bx+^) 58 (i) C.F. (ii) P.I. y = e 3x (A + Bx) 1 y D 2 - 6D + 9 {x 3 e 3X } (D + 3) z -6(D + 3) + 9^ 3 ^ D 2 + 6D + 9-6D-18 + 9^ 3 ^ ' Dl4 '20 y=- x 5 e 3x 20 .'. General solution is y = e 3x (A + Bx)+ : y = e 3x U + Bx+—) \ 20/ Now move on to frame 597 20 694 Programme 24 59 Special cases By now, we have covered the general methods t^at enable us to solve the vast majority of second order differential equations with constant coefficients. There are still, however, a few tricks that are useful when the normal methods break down. Let us see one or two in the following examples. Example 1. |^T + 4 ^T + 3 >> = 5 (i) C.F. m 2 + Am + 3 = /. (m + 1) (m + 3) = .'. m = -1 or -3 .". y = A e x + B e' 3x (ii) P.I. (D 2 + 4D + 3)y = 5 y = D 2 + 4D + 3 ^ This poses a problem, for none of the three theorems specifically applies to the case when/(x) is a constant. Have you any ideas as to how we can make progress? When you have thought about it, turn on to frame 60. 60 Wehave y = p2 + ^ + 3 {5} The trick is to introduce a factor e ox with the constant 5 and since e ox = g o = j ^ Ulis W jn not a j ter its va i U e. So we have: We can now apply theorem I to the function. The e ox comes through to the front, the function of D becoming the same function of a which, in this case, is 0. ox 1 { 5 } y e 0+0+3 1 J = e ox . 3 and since e ox = 1 , 5 y = i So the general solution is: y = Ae' x + Be~ 3 * +^ * j- n»-ix .. 5 Now for another. Turn on to frame 61. 695 Operator D Methods Here is another example. C j d 2 y dy Example 2. -7-5- + 2— =5 (i) C.F. m 2 + 2m = .'. m(m + 2) = :. m = or -2 :. ^ = Ae ox + Be' 2X .'. j> = A + B e' 2X (ii) P.I. (D 2 + 2D).y = 5 y= WT2D {5} If we try the same trick again, i.e. introduce a factor e ox and apply theorem I, we get y = y = l f0r ■ y = D 2 +2D ^ 5 ^ becomes ■ y = D 2 +2D ^ 5e( " C ^ j = e ox — — —{5} which is infinite ! So our first trick breaks down in this case. However, let us try another approach. y D 2 + 2D { {5} D(D + 2) 1 1 1 {5} D ' (D + 2) Now introduce the e ox factor and apply only the operator 2 y = ^- 1 1 D ' (D + 2) {5e ox } = * e ox l {5} D + 2 1 ; = 5 ^(5) since e ox = 1 y= M which is y = 62 696 Programme 24 63 _5* y ~ — since —denotes integration (with the constant of integration omitted). Note that we can apply the operators one at a time if we so wish. TheC.F. was y = A + Be' 2x The P.I. was found thus: look at it again. (D 2 + 2D)y = 5 ' = rFT2D {5) D D + 2 1 ' 1 1 {5e *} D'D + 2 = rj' e °*o72 {5} b y theoremI -i Mi)) -m General solution is 5* y = A + Be- 2X +^- Now here is another one. Let us work through it together. On to frame 64. C VI Example 3. —-^ - 16y = e 4 (i) C.F. m 2 -16 = :.m 2 = l6 :.m = ±4 y = Ae 4X +Be~ 4x (ii) P.I. (D 2 -16)y = e 4x V = —^ \e 4x } Theorem I applied to this breaks down, giving ^ again. .'. Introduce a factor 1 with the e 4x We now apply theorem II and on the next line we get y = Turn to frame 65. 697 Operator D Methods y 1 (D + 4) 2 -16 ■{1} 65 i.e. the e 4 * comes through to the front and the function of D becomes the same function of (D + 4). Then y- l D 2 + 8D+ 16-16 (1} 1 1 "{1} D'D + 8 The function 1 can now be replaced by e ox and we can apply theorem I to the second operator 1 D + 8 which then gives us y- y e D'D + 8 1 ; = e 4x 4 e ox — D + 8 D18 66 (since e ox = 1) since — denotes integration. So we have: General solution AX ■ y = e* x ■ 67 C.F. P.I. y = Ae y=- Ae" +Be" « + Be -« + £|_ Notice this trick then of introducing a factor 1 or e ox as required, so that we can use theorem I or II as appropriate. There remains one further piece of work that can be very useful in the solution of differential equations, so turn on to frame 68 and we will see what it is all about. 698 Programme 24 68 Consider d 2 y dx f +4y = 3sin2x (i) C.F. w 2 + 4 = :.m 2 =-4 :. m = +J2 >> = A cos 2x + B sin 2x (ii) P.I. (D 2 + 4)y = 3 sin 2x -> ; = 5T74-{3sin2x} The constant factor 3 can be brought to the front to simplify the work. y = 3 -^T^jUm2x} If we now apply theorem III (since we are operating on a sine term) we get 6 y= 69 y = 3 'I4T4 ^ sin 2x ^ = 3 • h ^ sin 2 *} and theorem III breaks down since it produces the factor -A-. Our immediate problem therefore is what to do in a case like this. Let us think back to some previous work. From an earlier programme on complex numbers, you will remember that e* 6 = cos0 +j sin 6 so that cos 6 = the real part of ei 6 , written^ {e' e } and sin 6 = the imaginary part of e> 6 , written J{e* 6 } . In our example, we could write sin 2x=J{ } 699 Operator D Methods sin 2x =./{e J2X } 70 So we can work this way: 1 y = 3. D 2 +4 •{sin 2x} = 3. 1 D 2 + 4 /{ £ J2X} 3./ 1 D 2 +4 •{e j2x } Theorem I now gives y = 3J ei™ . , : ^ 2 \ A = 3je> 2x 1 Q2f+4 -4 + 4 = 3 Je> 2x 4- so this does not get us very far. Since this does not work, we now introduce a factor 1 and try theorem II. ,-3/^ei".!}- y = 3/e J2 1 (D+J2)' TT^ 71 .'. j = 3ie j2 * 1 D 2 +j4D-4 + 4 (1) = 3 5^ j2 *^. 1 DD+J4 1 {e ox } putting e w for 1 . = 3jM 2 *jj.e°* 1 0+J4 ,j«l|l' theorem I on second operator. = 3ie-^ = 3^«.£ - T^^(cos 2x + j sin 2x) writing e y2X back into 1 its trig. form. [x cos 2x . „ \ — + x sm 2x) J ; = 4/^ sin 2x-j x cos2x) /. 3; = - 3x cos 2a; That seems rather lengthy, but we have set it out in detail to show every step. It is really quite straightforward and a very useful method. So finally we have „ ^ a ->ir>--i r> t 3x cos 2x C.F. y = A cos 2x + B sm 2x P.I. y = General solution y = A cos 2x + B sin 2x — 3x cos 2x Look through the last example again and then solve this following equation in much the same way. d 2 y Solve ^-T+9>> = cos3x When you have finished, turn on to frame 72 and check your result. 700 Programme 24 72 Solution: y = A cos 3x + B sin 3x + xsm3x Here are the steps in detail: (i) You will have had no trouble with the complementary function y = A cos 3x + B sin 3x (ii) Now for the particular integral:- (D 2 + 9) y = cos 3x •'• y = tf~^ ( cos 3x ) Theorem HI breaks down. Therefore use cos 3x + j sin 3x = e^ x i.e. cos 3x =0t { e i 3x } y D 2 + 9 l ' Theorem I breaks down. Therefore introduce a factor 1 and use theorem II. y =0l , ! { e )3X ,} (D+j3) 2 +9 l ; D 2 +j6D-9 + 9 1 ^ ' e i3xL 1 D' (D+j6) { e OXj £ 0X =1 \y Operate on e ox with the second operator — — — - using theorem I (D+j6) y D e j6 =®eV xX -[-\ =0> e i 3x ~ D\j6j j6 =^ -rr (cos 3x + j sin 3jc) writing e )3x back in V tr ig- form. -*r- (cos 3x + j sin 3x) J - oi> I "J* cos 3* jc sin 3x j 6 + 6 * sin 3x 7 = 6 Then, combining the C.F. and the P.I. we have the general solution y = A cos 3x + B sin 3x + x sm 3 * Aote. These special methods come to your aid when the usual ones break down, so remember them for future reference Turn to frame 73. 701 Operator D Methods You have now completed this programme on the use of operator D f J methods for solving second order differential equations. All that remains is the Test Exercise, but before you tackle that, here is a brief summary of the items we have covered. Summary Sheet 1. OperatorD D=£; D'-jJ; D» =£ 2. Inverse operator - = l ... dx, omitting the constant of integration. 3. Theorem I F(D) { e™ } = e ax .F(a) 4. Theorem II F(D) {e ax V} = e ax F(D + a) {v} 2 . ] sin ax \ . 2x ( sin ax 5. Theorem III F(D 2 ) = F(-a 2 ) , I cos ax] 1 cos ax 6. General solution y = complementary function + particular integral 7. Other useful items (where appropriate) (i) Introduction of a factor 1 or e ox (ii) Use of e* 9 = cos 6 +jsin0 i.e. cos0=^?{e j9 } sing =J { e J e } Revise any part of the programme that you feel needs brushing up before working through the Test Exercise. When you are ready, turn on to the next frame and solve the equations given in the exercise. They are all straightforward and similar to those you have been doing in the programme, so you will have no difficulty with them. On to frame 74. 702 Programme 24 /£} Work through the whole of the exercise below. Take your time and work carefully. The equations are just like those we have been dealing with in the programme: there are no tricks to catch you out. So off you go. Test Exercise - XXIV Solve the following equations: 1. &♦*£♦*-•" 2. £♦«&♦*-"« 3. d 2 y . dy ^ + 4^ + 3j = cos3* 4. d 2 y dv -j-T - 4-7-+ 5/ = sin 4x dx dx 5 &*%+*-**>* 7. ^U+>> = 3e* + 5e 2 * GOT 8 ' ^ + 6 ^ +8 J = 2sinAr + sin3x d 2 v 9. ^f+25j = sin5* Well done. 3X 703 Operator D Methods given at x = 0, y = 2 and Dy = 0. Further Problems - XXIV Wore: Where hyperbolic functions occur, replace them by their corresponding exponential expressions. Employ operator-D methods throughout. ananQnDnQOQQnaQDnaaaononDDnnonQnDaoano Solve the following equations by the use of the operator D. 1. D 2 y + 2Dy-3.y = 4e _3X 2. D 2 y + 3Dy + 2y=xe' x 3. D 2 y+y = sir\x 4. D 2 y-2Dy +y = sin x + x 2 5. D 2 y-3Dy + 2y = -4e x sinhx — 6. D 2 y-5D.y + 6y = e 3x 7. D 2 y - 5Dy + 6y = e 4 * sin 3x 8. D 2 j> + 4Dy + 5y = x + cos 2x 9. D 2 y + 2Dy + 5y = 11 cos 2x 10. D 2 j> + 4Dy + 5y = 8 cos x 11. D 2 y + 2flDy + a 2 y = x 2 e~ a * 12. D 2 j + Dy +y = xe x + e x sinx 13. D 2 y-6Dy + 9y = e 3x +e' 3 * 14. D 2 y + 4Dy + 4y = cosh 2x 15. D 2 j> + 6Dy + 9y = e~ 3 * cosh 3x 16. D 2 y-Dy-6y=xe 3x 17. D 2 y + 4Dy + 5y = 8cos 2 x 18. D 2 y + 2Dy + 5y = 34 sin x cos x 19. 2D 2 y +Dy-y = e x sin 2x 20. D 2 y + 2Dy + 5y = x + e x cos 3x 21. D 2 j-2D>' + 4.y = e* sin3x 22. D 2 y - 4Dy + 4y = e 2x 23 . D V - 9y = cosh 3x + x 2 24. D 2 y + 3Dy + 2y = e x cos x 25. D 2 y + 2Dj + 2.y = x 2 e~* 704 ANSWERS Answers ANSWERS Test Exercise I (page 32) 1- (0 -j, (ii) j, (iii) 1, (iv) -1 2. (i) 29-J2, (ii) -j2, (iii) 111+J56, (iv) 1 +j2 3. (i) 5-831 1S9°3', (ii) 6-708 jl53°26', (iii) 6-403 |231°24' 4. (i) -3-5355(1+ j), (ii) 3-464 -j2 5. x= 10-5, 7 = 4-3 6. (i) 10ei°- 6so , (ii) 10 e-J 0650 ; 2-303 +J0-650, 2-303 -J0-650 7. je Further Problems I (page 33) 1. (i) 115+J133, (ii) 2-52+J0-64, (iii) cos 2x +j sin 2x 2. (22-j75)/41 3. 0-35+J0-17 4. 0-7,0-9 5. -24-4+J22-8 6. 1-2 + jl-6 8. x=18,y=l 9. a=2,b = -2Q 10. x = ±2,y = ±3l2 12. a=l-5,Z> = -2-5 13. V2e j2 ' 3562 14. 2-6 16. R=(R 2 C 3 -R 1 C 4 )/C 4 ; L = R 2 R 4 C 3 18. E = (1811 +jll24)/34 20. 2+j3,-2+j3 707 Answers Test Exercise II (page 67) 1. 5-831 l210°58' 2. (i) -1-827 +J0-813, (ii) 3-993 -J3-009 3. (i) 36 |_19T\ (ii) 4 |53° 4. 8 [75° 5. 2 |88°, 2 1 208°, 2 [328°; p.r. = 2 [328° 2 1 208' 6. sin 40 = 4 sin cos - 8 sin 3 cos 7. cos 4 = r L [cos 40 + 4 cos 20 + 6] 16 8. (i) x 2 +y 2 -8x + 7 = Further Problems II (page 68) 1. x = 0-27, >> = 0-53 2. -3+jV3; -J2V3 3. 3-606 |56°19' , 2-236 |296°34' ; 121-3 -J358-4; 378eJ 1 ' 244 4. 1-336(12T, [99^ , lilll. [243.°. 1315° ) 1-336 (eJ°' 4712 , e^ 1 ' 7279 , e-i 2 ' 9845 , e -J 2 - 0420 ; e -Jo-78S4^ 708 Answers 5. 2173 +J0-899, 2-351 e-J 0392 6. V2(l + j), V2(-l + j), V2(-l -j), V2(l -j) 7. 1 [36^, 1 [108°, 1 [180°, 1 [252°, 1 [324°; e J°-«83 8. x = -4andx = 2±j3464 9. 1 1102°18 ', 1 1 222° 18' , 1 1342°18' ; 0-953 - jO-304 11. 1 -401 ( )58°22' , 1 130°22' , |202°22' , [274°22\ |346°22 '); p.r.= l-36-j0-33= 1-401 «-*•«" 12. -0-36 +J0-55, -1-64 -J2-55 13. -je, i.e. -J2-718 14. sin 70 = Is - 56s 3 + 1 12s 5 - 68s 7 (s = sin 0) 15. 22 [10- 15 cos 2x + 6 cos 4.x -cos 6x] 16. x 2 + y 2 + ^x + 4 = 0; centre (-X'°)> radius 8 / 3 17. x 2 +7 2 -(l+V3)Ar-(l+V3)>-+V3 = 0, centre 1 , radius \/2 18. 19. 20. f l+s/3 1+V3 v 2 ' 2 x 2 +y 2 = 16 (i) 2x 2 + 2^ 2 -x - 1 = 0, (ii) x 2 + y 2 + 2x + ly = (i) x 1 +y 2 - 4x = 0, (ii) x 2 +y 2 + x - 2 = 22. (i)y = 3, (ii) x 2 +y 2 ■4k 2 Test Exercise HI (page 97) 1. 67-25 2. 19-40 709 Answers 4. -COth A 5. (i) 1-2125, (ii) ±0-6931 6. x = 0-3466 7. (i) y = 224, (ii) x = ±48-12 8. sinx cosh y — j cos x sinh_y Further Problems III (page 98) 2. x = 0,x = 0-549 5. (i) 0-9731, (ii) 1-317 7. (i) 0-9895 +J0-2498, (ii) 0-3210 + jO-3455 10. x = 0, x=-jln2 12. x = 0-3677 or -1-0986 14. 1-528 +J0-427 18. 1-007 Test Exercise IV (page 135) 1. (a) 4, (b) 18 2. Equations not independent 3. x = 3, y = -2, z = -1 4. A; =3 or -25 5. jc = 3, 1-654, -6-654 710 Answers Further Problems IV (page 136) 1. (i) 144, (ii) 2. (i) 0, (ii) 666 3. x = 5, y = 4, z =-2 4. x = 2-5,y = 3, z = -4 5. x = 2, y= 1-5, z = -3-5 6. 4or-14 7. 5 or -2-7 8. (a) 0or±V2, (b) (a -b)(b -c)(c -a) (a + b + c) 9. x= 1 OTX = S±y/34 10. jc = -1-5 11. -2(a-b)(b-c)(c-a)(a + b+c) 12. / 2 =5-2 13. (a + Z>+c) 2 (a-fc)(6- c )(c-a) 14. 2 or -16/3 15. (x-y)(y-z)(z-x)(x+y+z) 16. x = -3or±V3 17 Y _ (2M, +M 2 )W ' M!(Mi+2M 2 ) 18. i x =0, z 2 =2, / 3 =3 'in a - 7 n 1 If 20. e-^or — Test Exercise V (page 167) 1 . 2i - 5/, -4* + /, 2/' + 4/; AB = V29, BC = Vl 7, CA = V20 2. (i) -8, (ii) -2i-7/-18* 3. (0-2308, 0-3077, 0-9230) 4. (i) 6, = 82°44'; (ii) 47-05, 6 = 19°3l' 711 Answers Further Problems V (page 168) 1. OG = 5(l(K +2/) 2 - V5o (3>4>5): vk* 1 ' 2 '- 3 * - 8005 ' 3. Moduli: V74, 3Vl0, 2n/46; D.C's: ,-^(3,7,-4), ^0,-5,-8), ^(6,-2,12); Sum=10/ 4. 8, I7i-7j+2k, = 66°36' 5. (i) -7, (ii) 7(i-/-fc), (iii) cos 0= -0-5 6. cos = -0-4768 7. (i) 7, 5; - 3/ - k; (ii) 8, 1 1/ + 1 8/ - 1 9k 8 - VTll /+ Vl^ / + V^ fc;sine = () - 997 9. -£-, ^0; 5 ' " 2 Vl3' Vl3' V30' >/30' V30 10. 6V5; 3^5, 3^5, 3^5- 11. (i) 0, = 90°; (ii) 68-53, (-0-1459, -0-5982, -0-7879) 12. 4* - 5/ + 1 lfc; 9^2 ( 4 > -5- H) 13. (i) ; + 3/ - 7k, (ii) -4z + / + 2k (iii) 1 3 (i + 2/ + k), (iv)^-|0 + 2/+fc) Test Exercise VI (page 191) 1. (i) 2sec 2 2x, (ii) 30(5x + 3) 5 , (iii) sinh2x, (iv) (x 2 -?x- 3 l)lnl0' W- 3tan3 ^ 712 Answers (vi) 1 2 sin 2 Ax cos Ax, (vii) e 2x (3 cos 3x + 2 sin 3x), 4* e sin* (vm) -ornr- (1X) ^cos2x 4 + cotx--+2tan2x a: 2. 3_ _25 4' 64 2 . z.,,2 3 3 * + Ay 3y 2 + 8xy 4. tan|, l/(l2sinyCos 3 -|) Further Problems VI (page 192) 1- (i) - x, (ii) secx, (iii) 4 cos 4 * sin 3 *- 3 cos 2 * sin 5 * 2. (i) x sinx 1 + cos* l x . , sinx — + cot x + • X 1 + COS X , (") -Ax 1-x 4 4. 7 2 -2jcj 5. (i) 5 sin 2*^*, (ii)-^-, (iiO^l 2 x 6. (i) 2x cos 2 x -2x 2 sin* cosx, (ii) — - , _ 2 , x l-x 2 (iii) g In x (*-D 3 -4, -42 2 + 1 x In x x - 1 12. -^j,- & -f-,x 2 + y 2 -2y = 14. -tan i 1 3a sin 6 cos 4 15. -cot 3 0; -cot 2 cosec 5 ( 713 Answers Test Exercise VII (page 217) 1. = 37°46' 2. 16y + 5* = 94, 5y = 16x-76 3. _y = ;c 4. >- = 2-598x - 3-849 5. R = 477; C: (-470, 50-2) 6. R = 5-59; C: (-3-5, 2-75) Further Problems VII (page 218) 1. 20y = 125* -363; y = 2x 2. y + 2x = 2; 2y=x + 4; x=l,y = 3. ]CCQsfl t j ' sin ^l; 5j>=13tan0.x-144sin0; ON.OT=144 4. ^-^; 3^ + 5x=14 x + 3y 5. R = >>i 2 /c 6. 5>> + Sx = 43 7. a 2 cos 3 f sinr 8. *lz*!;* a 9. (i) y=x;y = -x, (ii)R = V2, (iu) (1,-1) 10. (i) R = -6-25; C: (0,-2-25) (ii) R=l; C:(2,0) (iii) R = -ll-68; C: (12-26, -6-5) 11. R = -0-177 14. R= 2-744 17. p = t; (h, k) = (cos t, sin t) 18. R = -10-54, C: (11,-3-33) 20. (i)j> = ± -£, (iii) R = 0-5 714 Answers Test Exercise VIII (page 246) 1. (i) 130°, (ii) -37° 9 / ; \ -3 r .x -1 cos -1 * W V(-9* a -12*-3)' W xVO-^ 2 ) * 2 ' 2* 2 . „ , -i/jc\ ,. N -3 (iii)^ + 2*tan^), (iv)^^ W VCcos^+1)' (V1) 1 - 25 x 2 3 - (0 ^max^lOatoc"!; j min = 6 at * = 3; P of I at (2, 8) (iii) y max =e' 1 = 0-3679 at* = 1; P of I at (2, 0-271) Further Problems VIII (page 247) 1. (i) 1, (ii) 2V(1-* 2 ) 3 - (i) vxd + 2 4*)- (ii) irb '11 250 4- 0) (t>-w)> 00 (-0-25, -4-375) 3' -'min 5 - J ; max =0at ^=^;>'min = 4at*=l 6 - ^max at;,c = 2 ; ^min at:,c = 3 ; P°fIat*=V6 16 5 7 16 4 11 7 " >'max = T at;<: = ~T;> ; min =0at ^ := l X= 1-5 10. | =V 2.^cos(* + j) 11- (0 >W at(|, ^) , y min at (1 , 0); P of I at (|, 1) (ii) 7max a t (2 - V2, 3 - 2V2); y min at (2 + y/2, 3 + 2V2) (iii) Pof I at (n7r, mi) 715 Answers 12. (i) ±0-7071, (ii) 0, (iii) ± 1-29 13. 0-606 14. V3w 16. jw = °- 514 17. 1746 cm 18. d = lT 20. A = C, B = Test Exercise IX (page 272) 1. (i) ^ = \2x 2 -Sy 2 |£=-io^ + 9j 2 9 x oy ^ = 24* *~ Z dx 2 24 * oy o 2 z _ 9 2 z ^-24* ££ = -10x+18y 9J.3X 9x.9y (ii) |^- = -2sin(2x + 3>0 -! L=-3sin (2x + 3> ; ) ox oy ^|=-4cos(2x + 3>) f-f=-9cos(2x+3j0 ox 1 y ' oy d * Z -6cos(2x + 3.y) £-?- = -6 cos(2x + 3y) 9.y.9 x 9*.9.y /■■■\ ^ z -i _x: 2 -v 2 9z „ y 2 -v 2 (m) ^ = 2*e* > T^- 7 ** ~- 2 = 2 ex 2 -? 2 (2x 2 + 1) |4 = 2 e^ 2 -^ 2 {2y 2 - 1) 9x" v ' oy 1 o 2 z „ x 2 -v 2 o 2 z = -4xy e* ^ — — = -4xy e *-y oy.ox ox.oy (iv) ~ = 2x 2 cos(2x + 3^) + 2x sin(2x + 3j>) i-§ = (2 - 4jc 2 ) sin(2x + 3y) + 8x cos (2x + 3y) 716 Answers by.bx = -6x 2 sin(2x + 3y) + 6x cos(2x + 3y) —- = 3x 2 cos(2x + 3v) dy i 2 3 2 = -9x 2 sin(2x + 3y) d'z = -6x 2 sin(2x + 3y) + 6x cos(2x + 3y) dx.dy 2. (i) 2V 3. P decreases 375 W 4. ± 2-5 % Further Problems IX (page 273) 10. + 105EX 10~ s approx. 12. ±(x+y + z)% 13. y decreases by 19% approx. W ±4-25% 16\ 19% 18. 5X= y {&x. p cot (px + a) - St. q tan (qt + b)} Test Exercise X (page 292) ... 4ry - 3x 2 e* cos y —e^ cos jc '• W 3^ 2 -2x 2 ' (U) <? sxny + e? sinjc 5 cos x cos 7 — 2 sin x cos jc 5 sinx sin_y + sec 2 _y 2. V decreases at 0-419 cm 3 /s 3. y decreases at 1 -524 cm/s 717 Answers 4. p- = (4x 3 +4xy) cosB + (2x 2 +3y 2 ) sind or |§ = r { (2x 2 + 3y 2 ) cos - (4x 3 + 4xy) sin 6 } de Further Problems X (page 293) 2. 3x 2 - 3xy 3. tan 6 = 17/6 = 2-8333 « ,-s l-J' /-N 8y - 3y 2 + 4xy - 3*V ,.... y 5 3 14. fl =-- 6 =-~ , cos x (5 cos y - 2 sin x) f£ ' 5 sin* siny + sec 2 j H 5& , _ v cos x - tan y $ : 17. — 2 r-^- *sv * sec j - sin x |g Test Exercise XI (page 322) || 1. 230 P 2. 2-488, 25-945 3. 1812 4. (i) convergent, (ii) divergent, (iii) divergent, (iv) convergent 5. (i) convergent for all values of x. (ii) convergent for - 1 ^ x <C 1 (iii) convergent for - 1 <C jc<C 1 718 Answers Further Problems XI (page 323) 1. f(4« 2 -l) n(3n + 1) ' 4(m + !)(« + 2) 3. j(n+l)(« + 4)(n + 5) 4. (i) j(/i + 1) (« + 5), (ii) 1(« 2 + 3n) (n 2 + 3« + 4) 5. 2 •»«+ 1 ■ 1:S = 3 , ^,.m (l+ t^i K .» 7. (i) 0-6, (ii) 0-5 8. (i) diverges, (ii) diverges, (iii) converges, (iv) converges 9. -1<jc<1 11. -l^x^l 12. All values of x 13. -i<:x<:i 16. (i) convergent, (ii) divergent, (iii) divergent, (iv) divergent 18. (i) convergent, (ii) convergent 19. l^x<:3 20. -|(« + l)(4« + 5) + 2" + 2 -4 Test Exercise XII (page 352) 2 1. /(*)=/(0)+x/'(0) + f r /"(0) + 2. 1-x 2 +^- -^C + 2!' 45 719 Answers X 2 5x* 5. x + x 2 +^+- 6. 1-0247 12 1 7 - ® ~10' ^"9"' (iii ) _ 2~ 8. 0-85719 Further Problems XII (page 353) 111 1 3- 0) " jq, (ii) 3-, (iii) y, (iv) "g, (v) 2 3 5x \lx 2 13x 3 2 2 4 4 2 9 - 3 10. (i) -\, (ii) |, (iii) 2 11. ("- y + 2 )* ; 1-426 r- 1 x 2 x* 13. lncosx = -- yj~ — 16. (i) -|", (ii)y 17. l-y + 8x 2 llx 4 19. x 2 -x 3 + ~ry; max. at x = 720 Answers Test Exercise XIII (page 384) 1. - e C0S *+C 2. 2Vx(lnx-2) + C 3. tanx-x + C x sin 2x _ x 2 cos 2* cos 2x 2 2 +_ 4~ +C 2e~ 3:x: f 3 ) 5. — pr— J sin 2*- — cos2x! +C , , 2 cos 3 x cos 5 jc _, 6. -COSOC+ — — +C „ 3* sin 2x sin 4x 8 4 32 +L J. 21n(x 2 + x + 5) + C 10. -|ln(x-5)-jln(x-3) + C 11. 21n(x-l) + tan _1 x + C 1? _ / cos 8* cos2jc \ I 16 4 / L Further Problems XIII (page 385) 1. ln{A(x-l)(x 2 +x+ 1)}+C - I 3. -In (1 + cos 2 *) + C 4 I_I 721 Answers 6 - c (* 2 +* + i) l/2 7. -|ln(*-l)-jln(* 2 + *+l) + C 8. y-* + ln(* + 1) + C 9. 2 In (* - 1) + tan" 1 * + C 10. 2 -2«" +3 11. (p + l)(p + 2)(p + 3) 12. 31n(*-2)+-jln(* 2 + l)-5tan _1 jc + C 13. 1 2 14. (sin" 1 *)* 2 ^ 15. £(2 In 3 -ir) 16. 7T 2 -4 17. 7r 3 rr 6 4 18. 7T 1 4 2 19. tan _1 * + C * 20. j(i + * 2 ) 3/2 + c i\ In Cv + n-1n (v-V\ 2 22. ^(e 2w - 1) = 53-45 23 ± ZJ - 24 722 Answers 24. ^( 3e -/'-2 3co 26. ^L? + C 27. i-l n (x-4)-^ln(5x + 2) + C 28. ln(jc + 2) + C 29. 2 In (* + 5) + 1 In (x 2 + 9) - jtan" 1 (j\ + C 30. ln(9x 2 - 18* + 17) 1 / 18 +C 31. 2x 2 +ln{(x 2 -l)/(x 2 + 1)} + C 32. lJ3x 2 ln(l +x 2 )-2x 3 +6x-6tan^x + C 33. In (cos 6 + sin 6) + C 34. tan 6 -seed +C 20 35. ^ln(x-l)+yln(x-2)-^ln(x + 3) + C -1 37. |ln2- 1 | 31nx+yln(jc 2 + 4) - jtaif'd) + C 38 39. lnx-tan'x—T+C 723 Answers Test Exercise XIV (page 416) 1. sm^yj + C 3. /ftan" 1 {(^ + 2)V2}+C 4. ^-sinh- , (-^) + < 1 V3 - > f^H 6 ik(^) + ^ ii) ^ i -*-* a)+c V5 7 VT cosh 1v(tl7i)) +C 8. ^-tan" 1 (V3tanx) + C V3 9 _Li [ Vl3-3 + 2tanx/2 \ + c ' >Jn n lVl3 + 3-2tanjc/2 ' 10 . ta (H^2j + c (1 -tanx/2 j Further Problems XIV (page 417) L 2V2l ln (^6 + V2l") +C 1 , (7sJ\\+x + 6\ ^ 2 47n ln (2Vn^^) +c 3 . J_ tan -.(*+Z\ + c Vl l w 1 1 / 4. 4ln(* 2 + 4* + 16)-^tan-(^) + C 724 Answers 7. co*- (£±1) + C 8. 6V(^ 2 -12jc + 52) + 31 sinlW^pWc 9. ^^(4^(£)} +C 10. Vl£ =0 .3 511 20 11. 7r-ln{- Vl + C V5 l2x + 5+V5) 3x 2 12. -y -4x + 4taiT 1 x + C 13. -£±1 V(3-2x-^ 2 ) + 2sm 1 (^j 1 ) +C 14. 7T 15. cosh^^-^ + C 16. ^■tan _1 |-|tan^| +C 17. l ln ptanx-l ^ 5 | tanx + ? ' 18. |-l 19. 3sin _1 x-V(l-* 2 ) + C 20. TjtaiT 1 |V3tanYf-x + C 725 Answers 21. |ln(;c + 2)-|ln(jc 2 + 4) + tan- 1 (|)+ C 22 - 3V5" t3n (3") +C 23. V(* 2 + 9) + 2 ln{x + V(* 2 + 9)} + C 24 - ^ C ° sh " 1 (^) + C 25 -T 26 . 1 ln (V2tan -H + c 2V2 \\/2tan0 + lJ 27. V(* 2 + 2x + 10) + 2 sinh" 1 (ill) + C 28. 8 sin ' (^) + *y* V(15 - 2x -x 2 ) + C 29. ^(* + 2) 30. -L ttt ->(^)4ln{^if4) + C 3v/2 W2/ 6 lx 2 +2a 2 J Test Exercise XV (page 430) x 3 _ 3x 2 3x _ : 2 4 4 8 L e w|^ 3 _3x 2 + 3^_|l + c 2. ... 57T ,.... 8 « 256 > (U) 31? 3. 2a 7 35 4. I = * , tan"" 1 > " n-\ 5 3tt 256 726 Answers Further Problems XV (page 431) 1 , fi 6 , 8 , 16 2. --s c - — s^c-— s'c-— c + Ci where I s=sinx \ c = cos x 2835 5. I 3 =^--6; I 4 = j-12tt + 24 6. I„=x"e x -nl„_ 1 ; I 4 = e* (x 4 -4x 3 + 12x 2 - 24* + 24) 7 1328V3 2835 .„ . cot 5 x cot 3 X 10. I 6 =-— ^— + — — -cotx-x + C 11. I 3 =x{(lnx) 3 -3(lnx) 2 +61nx-6}+C Test Exercise XVI (page 452) 1. 70-12 2. — +2tt= 31-75 ■n 3. -|ln6 = 2-688 4. 73-485 5. I RI2 6. 132-3 727 Answers Further Problems XVI (page 453) 1. 24 2. 1 3. 3tt «• i 5. 7. 2 »■ ? ■ >• JH 9. /i^+7 12 11. ln(2 n .3" 6 )-l 12. a 2 (ln2-|) 15. a(l-2e _1 ) 16. 2-83 17. 39-01 18. ^ 0? +13) 20. 1-361 Test Exercise XVII (page 477) 1. (0-75, 1-6) 2. (3-1, 0) 3. 5tt 2 a 3 4. <-l±l 728 Answers 5. 70-35 n 6. 5.7T 2 8 7 e"-2 Further Problems XVII (page 478) L f 6 + T ln2 2. (i) 2-054, (i) 66-28 - 64jr£ 3 15 4. (i) (0-4, 1), (ii) (0-5, 0) 6. 24 7. 8. 9. 17 12 19 20 U 5 10. A = 2-457, V=47rV3, J=l-409 12. (i) 8, (ii) ^, (iii)-i 13. 1-175 16. V= 25-4 cm 3 , A = 46-65 cm 2 17. S =15-31 a 2 , y= \-062a 729 Answers Test Exercise XVIII (page 513) 1. (i)I z =^(*W), (ii)I AB =fV+* 2 ), *» ■/ + b 2 3 2. '■jr 3. (l) ln4' (n) ln2 5. ^-\ 0433 a 2. JM*» 6. (i) /4ac V 5 ' 9. a 4 12 Further Problems XVIII (page 514) /3c 2 12. Hi' 4 , . 2u>a 3 3;ra 14 - T--16 15. (i)i-V(^ 2 +e + l), (ii) yjfy 16. 51-2w 17. 9-46 cm (157T-32) a 4(3vr-4) 730 Answers Test Exercise XIX (page 534) 1. 0-946 2. 0-926 3. 26-7 4. 1188 5. 1-351 Further Problems XIX (page 535) 1. 0-478 2. 0-091 3. (i) 0-6, (ii) 6-682, (iii) 1-854 4. 560 5. 15-86 6. 0-747 7. 28-4 8. 28-92 9. 0-508 V2 r W2 10 - J y/(9 + cos2B).dd; 4-99 J Y" "V* "Y* 11. tan 1 x=jc-j+y-y; 0-076 12. (i) 0-5314, (ii) 0-364 13. 2-422 14. 2-05 731 Answers Test Exercise XX (page 560) 56 377 3 1- ~„3 (ii) r = 5 cos 2 (iii) r = sin 2 Q (iv) r = 1 + cos I (v) r = 1 + 3 cos ( (vi) r = 3 + cos I 4Chr 3 4. 8 327Tfl 2 732 Answers Further Problems XX (page 561) 16' v 21 2. 3tt 3. 4 a 2 3 4. £♦' 5. 2 3 6. 20rr 3 7. 87Tfl 3 3 9. 3ir a 2 10. 5n 2 11. 21-25 a 3na U - 2 14. |{V(fc 2 + 1)} O fie i - 1) ; £ (e^i - 1) 15. 7ra 2 (2-V2) Test Exercise XXI (page 588) 1. (i) 0, (ii) 2 -f 2. (0-1, (ii) 168, (iii) ¥■ 733 Answers 3. 13-67 4. 170-67 5. M? + 6 4 6. 54 Further Problems XXI (page 589) 1. I 2. 243 n 2 3. 4-5 . abc ,, 2 , 2\ 4. — (b 2 +c 2 ) 5. - 6. 4-5 7. 7T + 8 8. 26 22 *• T n.1 7r/6 (• 2 cos 30 ^ (■ ir/6 f 2 = 2 J J 12. A = 2\ \ rdrd6=- 13 - 47r !v^"V5 734 Answers 64 14. ^(3tt-4) r n ffl(l + cos0) 4fl 3 16 ^ 15. M= r 2 sind drdd=i-; h=-^- JoJo 3 9tt 16. (i) —-nab, (ii) —irab 3 : centroid lO, — j 17. 19-56 18. ^(c 2 -« 2 ) 2 19. |-(2tt + 3V3) 20. 232 Test Exercise XXII (page 630) x 2 1. y=Y +2x-3 1nx + C 2. tan"V = C- 1 1 + x 3X 3. ^ = j-+Ce- 2x 4. jk=jc 2 + Cx c x cos 3x sin 3.x 4 ^ 5. , = —_+__- - + c 6. sinj' = Ax 7. ,y 2 -x 2 = Ax 2 .y 8. ;y(* 2 -l)=y + C C 9. v = cosh a; + — : — cosh a: 735 Answers 10. y=x 2 (sinx + C) 11. xy 2 (Cx + 2)=l 12. y=l/(Cx 3 +x 2 ) Further Problems XXII (page 631) 1. jeV=Aey 2. y 3 =4(l+x 3 ) 3. 3x 4 + 4(y + l) 3 = A 4. (1 + e*)secy = 2y/2 5. jc 2 + >> 2 +2x-2j> + 21n(x-l) + 21nO + 1) = A 6. ^ 2 -xy-x 2 + 1 =0 7. xy = A^ x 8. x 3 -2^ 3 =Ac 9. A(x - 2yf (3x + 2j) 3 = 1 10. (x 2 - y 2 ) 2 = Axy 11. 2y=x 3 +6x 2 ~4x\nx + Ax 12. _y = cos* (A + In secx) 13. y = x(l +x sinx + cosx) 14. (3^-5)(l+x 2 ) 3/2 = 2V2 15. jsiruc + 5e cosx = l 16. x + 3y + 2 In (x + y - 2) = A 17. x = Aye xy 18. ln{47 2 +0-l) 2 }+tan- 1 .(^ 1 19. 0"* + l) 2 0+*-l) 5 =A 20. 2jc 2 ^ 2 In y - 2xy - 1 = Ax 2 .y 2 736 Answers 21. ~^ = 2x+ l + Ce 2 * 22. 7=^" 23. j 2 (x + C e*) = 1 24. sec 2 x _ n tan 3 x J 3 25. cos 2 x=^ 2 (C-2tan^) 26. jV(1 -x 2 ) = A + siif';c 27. x + In Ajc = V(y 2 - 1) 2x 2 Ax 28. ln(x-^) = A+ * 29. 7 (y-x) 2 j>-x \/2 sin 2x 2(cos x - sjl) 30. (*-4)>> 4 =Ax 31. 7 = a: cosx- — sec* 32. (x-jO 3 -Axy = 33. 2tan" 1 ^ = ln(l+x 2 ) + A 34. 2x 2 y = 2x 3 -x 2 -4 x-y 35. y = e x 36. 3<? 2 >' = 2e 3 * + l 37. Axy = sin 2jc - 2x cos 2x + 27T - 1 38. y = k e yl x 39. x 3 '-3jcy 2 =A 40. x 2 -4xy + 4^ 2 +2x-3 = 41. j(1-jc 3 )- 1 / 3 =-^(1-x 3 ) 2 / 3 +C 737 Answers 42. xy + x cosx-sinx+ 1 = 43. 2tan _1 .y=l-jc 2 x 2 +C 44. j> = 2x(l-x 2 ) 3 45. >>V(l+* 2 ) = x + 3- + C 46. l+/=A(l+x 2 ) 47. sin 2 0(a 2 -r 2 ) = |- 48. j=ysinx 49. 7 = * x(A-x) Test Exercise XXIII (page 663) 1. y = Ae' x +Be 2x -4 2. y = Ae 2x +Be' 2x +2e 3x 3. y = e x (A + Bx) + e 2x 4. ^ = Acos5x + Bsin5x+y|j(25x 2 +5x-2) 5. _y = e* (A + Bx) + 2cosx 6. y = e~ 2x (2-cosx) 7. j = Ae x +Be' x ^ -2x + 7 8. y = Ae M +B e 4x + Ax e AX Further Problems XXIII (page 664) 1. y = Ae* x +Be- x/2 -j- 2. .y = e 3x (A + Rx) + 6x + 6 3. ^ = 4cos4x-2sin4x + Ae 2x + Be 3 * 738 Answers 4. y = e x (Ax + B) + | — x 2 e x 1X y p ~2X 5. y = Ae x + Be- 2x + e -^-- X -~ 2X 6. y = e 3x (A cos x + B sin x) + 2 - ~- 7. y = e 2x (A + Bx) + j+ |-sin 2x 8. 7 = Ae x + Be 3x +|<3x + 4)-e 2;c 9. j = e*(Acos2x + Bsin2jc)+^ + ^- -L 10. j; = Ae 3x +Be- 3x --isin3x+|jce^ lo 6 2 - - w/1 8EI. 1 1L ^ = 2rii^ 2 - to+6 ' 2 }^- w/ ' 12. x=^(l-f) e - 3 ' 13. y = e' 2t (A cos t + B sin t) - -(cos f - sin 0; ,. j 0\/2 1 amplitude——, frequency — 4 2tt 14. Jf = --e f +y-e 2f +^(sinf + 3cos0 15. 7 = e -2^- e -^ + — (sinx-3cosx) 16. .y = e~ 3 *(Acos;c + Bsinjc) + 5;c-3 17. x = e~ f (6 cos t + 7 sin f) - 6 cos 3f - 7 sin 3f 1 9 18. 7 = sinx-ysin2x:; y max = 1-299 atx=^ 19. 1 = ^ = 0.6413; A =| 20. x= — (e" 3r -e" 2f + cos f + sinf}; 10 idv state: x = ■ 10 739 Steady state: x = ^— sin (t + — ) 10 V 4/ 3. y = Ae x + Be 3X - — (cos 3x - 2 sin 3x) Answers Test Exercise XXIV (page 703) e 4 * 1 . y = Ae' x +Be 2x + — 2. y = e' 2X (A + Bx) + 5e- 3x _1_ 30' 4. j> = e 2x (A cos x + B sin x) + ^r (1 6 cos Ax - 1 1 sin 4x) 5 . >> = e x (A cos x + B sin x) - — (8 cos 2x - sin 2x) 3e* 7. 7 = A cosjc + B sin a: + -r— +e 8. _y = Ae~ 2x + Be' 4x -— {6 cosx - 7 sinx} 85 9. 7 = A cos 5x + B sin 5x xe 3x 10. >> = Ae" x +Be 3x +-r— - rrr {18 cos 3x + sin 3x] x cos 5* 10 Further Problems XXIV (page 704) 1. y = Ae* + Be"" -xe~ 3x 2. j = Ae- x + Be- 2x +e- x fy-x) 3. y = A cosx + B sin* — — cosx 4. y=e x (A + Bx)+—cosx+x 2 +4x + 6 5. >>= 1 +e 2x (l-2x) 6. j> = Ae 2x + Be 3x + x e 3x 7. y = A e 2x + B e 3x - e 4x (9 cos 3x + 7 sin 3x)/130 740 Answers 8. y = e 2x (A cosx + B sin x) +j- ~ + (8 sin 2x + cos 2x)/65 9. y = e' x (A cos 2x + B sin 2x) + cos 2x + 4 sin 2x 10. y = e" 2 * (A cos x + B sin x) + cos x + sin x 11. j> = e- a *(^ + A + Bjc) 12. V3 . „ . V3 y = e x > 2 ( A cos ^ * + B sin ^- xj + i- (* - 1 ) - —(3 cos x - 2 sin jc) V 2 /> 3 * p~ 3X 13. > , = e 3JC (A + &c)+^y-+|^- P 2 * Y 2 P~ 2X 14. 7 = e -"(A + Bx)+l T +i^ 15. ^ = e" 3x (A + Bx)+-i(l+e- 6x ) lo 16. j = A e 3x + B e~ 2 * + e 3x (5x 2 - 2jc)/50 17. y = e~ 2x (A cos x + B sin x) + — + — (8 sin 2x + cos 2x) 18. y = e' x (A cos 2x + B sin 2x) + sin 2x - 4 cos 2x 19. j = A W 2 + B <f* - e* (3 sin 2jc + 5 cos 2x)/68 20. y =e~ x (A cos 2x + B sin 2x) +|- ^ - e ~* C0S 3 * 21. ^ = e JC (Acos v / 3x + B sin 73^-^4^ 6 22. y = e 2x (A + Bx+Y) 23. J ,« e »x( A+ ^) + e -ax(B-^)- 8 4. (9x a +2) 24. y = e~* | A + y sin jc - y cos x) + B e~ 2 * 25. .y = e~*(A cos* + Bsinx +x 3 -6x) 741 INDEX INDEX (References given are page numbers) Absolute convergence, 319 Angle between vectors, 163 Approximate integration, 517 by series, 519 by Simpson's rule, 523 Approximate values, 341 Areas by double integrals, 581 Areas enclosed by polar curves, 546 \reas under curves, 435 Argand diagram, 19 Arithmetic means, 299 Arithmetic series, 298 Bernoulli's equation, 622 Binomial series, 337 Centre of gravity, 465 Centre of pressure, 504 Centroid of a plane figure, 462 Complementary function, 648 Complex numbers, 4 addition and subtraction, 4 conjugate, 8 De Moivre's theorem, 50 division, 12 equal, 14 exponential form, 27 graphical representation, 17 logarithm of, 29 multiplication, 6 polar form, 22, 37 principal root, 55 roots of, 5 1 Consistency of equations, 1 26 Convergence, 311 absolute, 319 tests for, 313 Curvature, 206 centre of, 208, 213 radius of, 207 743 D'Alembert's ratio test, 317 Definite integrals, 438 De Moivre's theorem, 50 Determinants, 101 evaluation, 110 properties, 130 solution of equations, 105, 114 third order, 109 Differentiation, 171 function of a function, 173 implicit functions, 185, 285 inverse hyperbolic functions, 229 inverse trig, functions, 226 logarithmic, 180 parametric equations, 187 products, 177 quotients, 178 Differentiation applications, 195 curvature, 206 tangents and normals, 200 Differential equations, 593 direct integration, 598 first order, separating the variables, 599 homogeneous, 606 integrating factor, 613 Bernoulli's equation, 622 second order linear, 637 solution by operator-D, 683 Differentiation, partial, 251, 277 Direction cosines, 156 Direction ratios, 165 Double integrals, 565 Equation of a straight line, 195 Expansion of sin nd and cos nd, 57 Expansion of sin"0 and cos"6), 59 Exponential form of a complex number, 27 First order differential equations, 593 Bernoulli's equation, 622 by direct integration, 598 Index homogeneous, 606 integrating factor, 613 variables separable, 599 Function of a function, 173 Geometric means, 303 Geometric series, 301 Homogeneous differential equations, 606 Hyperbolic functions, 73 definitions, 74 evaluation, 83 graphs of, 77 inverse, 84 log. form of the inverse, 87 series for, 75 Hyperbolic identities, 89 /', definition, 1 powers of, 2 Lengths of curves, 467, 552 Limiting values, 309, 342 l'Hopitai's rule, 345 Loci problems, 61 Logarithm of a complex number, 29 Logarithmic differentiation, 180 • Maclaurin's series, 331 Maximum and minimum values, 235 Mean values, 446 Moment of inertia, 483 Normal to a curve, 200 Operator-D methods, 669 Identities, trigonometric/hyperbolic, 89 Implicit functions, 185, 285 Indeterminate forms, 342 Inertia, moment of, 483 Infinite series, 308 Integrals, basic forms, 358 definite, 438 I — dx mdff(x).f(x)dx, 363 J fix) J linear functions, 360 standard forms, 389 Integrating factor, 613 Integration, 357, 389 partial fractions, 373 by parts, 368 powers of sin x and of cos x, 379 products of sines and cosines, 381 reduction formulae, 419 substitutions, 389 as a summation, 450 by t = tan x, 409 by t = tan — ,413 y 2 Inverse hyperbolic functions, 84 log. form, 87 Inverse operator — , 673 Inverse trig, functions, 223 inverse operator — .673 in solution of differential equations, 683 Theorem I, 675 Theorem II, 678 Theorem III, 681 Pappus, theorem of, 475 Parallel axes theorem, 491 Parametric equations, 187, 211, 444, 468,473 Partial differentiation, 251 change of variables, 289 rates of change, 281 small finite increments, 266 Partial fractions, 373 Particular integral, 649 Perpendicular axes theorem, 495 Points of inflexion, 240 Polar co-ordinates, 539 Polar curves, 541 areas enclosed by, 546 lengths of arc, 552 surfaces generated, 555 volumes of revolution, 550 Polar form of a complex number, 22 Power series, 327 744 Powers of natural numbers, series of, 304 Properties of determinants, 130 Radius of curvature, 207 Radius of gyration, 487 Rates of change, 281 Reduction formulae, 419 R.M.S. values, 448 Roots of a complex number, 51 Scalar product of vectors, 157 Second moment of area, 500 Second order differential equations, 637 Separating the variables, 599 Series, 297 approximate values by, 341 arithmetic, 298 binomial, 337 convergence and divergence, 311 geometric, 301 infinite, 308 Maclaurin's, 331 powers of natural numbers, 304 standard, 336 Taylor's, 350 Simpson's rule, 523 proof of, 532 Simultaneous equations, consistency, 126 solution by determinants, 105 Index Small finite increments, 266 Standard integrals, 358 Straight line, 195 Summation in two directions, 565 Surfaces of revolution, 471, 555 Tangent to a curve, 200 - Taylor's series, 350 Trigonometric and hyperbolic identities, 93 Triple integrals, 570 Turning points, 235 Unit vectors, 152 Vectors, 141 addition and subtraction, 144 angle between vectors, 163 components, 147 direction cosines, 156 direction ratios, 165 equal, 143 representation, 142 scalar product, 157 in space, 154 in terms of unit vectors, 152 vector product, 159 "Volumes of revolution, 457, 550 Volumes by triple integrals, 583 Wallis's formula, 428 745 STROUD: ENGINEERING MATHEMATICS This book provides y complete one-year course in mathematics tav means of an integrated series of programmes together with extensive exercises, and is destined *o< use by undergraduates during th* first year of engineering degree studies and for National Diploma and Certificate courses. The course consists of 24 programmes devised as weekiv assignments of work. Each programme contains a number of worked examples through wKich the student is guided with a gradual vvithdrawal of support as the topic is mastered, nrtd concludes with a criterion test relating to the techniques covered in that programme. There are also exercises for further practice and probiem solving and a full range of answers is provided. The work has been designed to be equally suitable for class use w individual study. All the .programmes have been subjected to rigorous validatt in procedures and havo been proven highly successful. r *