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Full text of "Engineering Mathematics 1st Ed"



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K A STROUD 

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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 
*