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PROGRAMMES AND PROBLEMS
PROGRAMMES AND PROBuEMS
PROGRAMMES AND PROBLEMS
PROGRAMMES AND PROBLEMS
PROGRAMMES AND PROBLEMS
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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 photolithography 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 selfstudy, 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 Coordinates Systems
Introduction to polar coordinates 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 twentyfour 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 0sin 2
= 12 sin 2
= 2 cos 2  1
, na _ 2tan0
tan 20  j^^
xin
(4) Letfl= ;. sin = 2sincos
ra 2 ■ 2^
cos = cos^ —  sin 2 —
= l2sin'f
.2
= 2cos 2 ^ 1
2 tan §
tan0=
2
1tan^
(5) sin C + sin D = 2 sin —  — cos
. n . „ . C+D . CD
sin C  sin D = 2 cos —  — sin
n,. r. , C + D CD
cos C + cos D = 2 cos — — cos
2 2
p. „ . . C + D . CD
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 jaxis: y = mx + c
(ii) Passing through (x j, .yj): ^^i=w(xXi)
(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 /•: (xh) 2 + (yk) 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: xct, 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(8156) _ 9±V25 9 + 5
X 4 4 " 4
• v _4 14
" *" 4 0r ~T
:. x =1 or 35
That was straightforward 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=j2646
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(36100) _ 6 ± V(~64)
5* tx + i u  x To To
: ' x= ^JcT " x = °' 6 * j0 ' 8
.'. x = 06+j08 or x = 06j08
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 = lj=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)
,xl^3 ±j 5
i.e. * = 3+j5 or x = 3j5
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) + (3j2). 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)(2j5) = 4+j72+j5 = (42)+j(7 + 5)
= 2+jl2
So, in general, (a + ]b) + (c + )d) = (a + c) + j(b + d)
Now you do this one:
(5+j7) + (3j4)(6j3)=
7
8
2+j6
Programme 1
since (5+j7) + (3j4)(6j3)
= 5+j7 + 3j46+j3
= (5+36)+j(74 + 3)
= 2+j6
Now you do these in just the same way:
(i) (6+j5)(4j3) + (2j7) =
and (ii) (3+j5)(5j4)(2j3) =
10
(i)4+j (ii)jl2
Here is the working:
(i) (6+j5)(4j3) + (2j7)
= 6+J54+J3 + 2J7
= (64 + 2)+j(5+37)
= 4+j
(ii) £+j5)(5j4)(2j3)
= 3+J55+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 lefthand terms
(ii) the two inner terms
(iii) the two outer terms
14
(iv) the two righthand terms
11
(3 + J4) (2 + j5)
■ .2.
"3"
6+j8 +jl5+j 2 20
6+J2320 (since j 2 =l)
14+J23
Ukewise, (4j5)(3 +]2)
12
22 j7
for: (4j5)(3+j2)=12jl5+j8j 2 10
= 12J7 + 10 (j 2 =l)
= 22j7
If the expression contains more than two factors, we multiply the
factors together in stages:
(3+j4)(2j5)(lj2)
= (6+j8jl5j 2 20)(lj2)
= (6j7 + 20)(lj2)
= (26j7)(lj2)
Finish it off.
Programme 1
13
12J59
for: (26j7)(lj2)
= 26j7j52+j 2 14
= 26 j59 14 = 12J59
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)(5j8)=
14
89
Here it is:
(5 + j8) (5  j8) = 25 +j40j40j 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 (5j8)
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) (ab) = 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) (4j3)(4+j3)
(ii) (4+j7)(4j7)
(hi) (fl+j6)(aj6)
(iv) (xiy)(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) (4J3) (4+j3) = 4 2 j 2 3 2 = 16 + 9 =
(ii) (4+j7)(4j7) = 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
(3j5)(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) (5j9)(2j6) + (3j4)
(ii) (6j3)(2+j5)(6j2)
(iii) (4J3) 2
(iv) (5j4)(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) (5j9)(2j6) + (3j4)
= 5j92+j6 + 3j4
= (52 + 3)+j(694) =
(ii) (6j3)(2+j5)(6j2)
= (12j6+j30j 2 15)(6~j2)
= (27+j24)(6j2) ,_
= 162 +J144J54 + 48
(iii) (4J3) 2 = 16J249
6J7
210+J90
7J24
(iv) (5j4)(5+j4)
= 25 j 2 16 = 25 + 16 =
41
Programme 1
Required factor is the conjugate of the given complex number.
(4j3)(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=1.67jl.33
But how do we manage with
3 3 J 3
7j4 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.
7j4 = (7j4)(4j3) = 28j3712 _ 16J37
4+J3 (4+j3)(4j3) 16 + 9 25
ifj=0.64jl48
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.
4i5
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:
4j5 _ (4j5)(lj2) = 4J1310
1+J2 (l+j2)(lj2) 1+4
— 6 — j 1 3 6 .13
= l2j26
Now here is one for you to do:
Simplify
3 +j2
1J3
When you have done it, move on to the next frame.
25
Result
03+jll
3+j2 _ (3+,j2)(l+j3) _ 3+jll6
1J3 (lj3)(l+j3) 1+9
= 3+jll = _ Q . 3+jl , 1
10
naDnannDDDnDnnnnnDDDannnnDDDnnnnDDDnnD
Now do these in the same way:
(i)
4J5
(ii)
3+j5
2j v " _/ 5J3
(2+j3)(lj2)
(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 4p_(4j5)(2+j)_8j6 + 5
1 K> 2j (2j)(2+j) 4 + 1
13 j6 _
26J12
(ii)
(iii)
3+j5 _ (3+j5)(5+j3) = 15+J3415
5j3 (5j3)(5+j3) 25+9
(2+j3)(lj2)_2j+6_8j
(3+J4) 3+J4 3+J4
_ (8j)(3j4)
(3+j4)(3j4)
. • _24j354_20j35
9+16
25
= 08J14
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
Rearranging terms, we get
a +}b = c + \d
ac=](db)
In this last statement, the quantity on the lefthand side is entirely real,
while that on the righthand 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)
ac = 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 = 6j3, thena = and b = .
29
a = 6
and
b=~3
Be careful to include the sign!
aDannnanDnnnnDDnnDnnnDannnDnnnaDnnnDDn
Now what about this one?
If (a + b)+)(ab)= 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
ab = 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 = 45; 6 = 25
31
For
a + 6=7
ab = 2
2a = 9 .'. a = 45
26 = 5 :. 6 = 25
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 Xaxis represents real numbers.
XX! is therefore called the real axis.
(ii) The scale on the Yaxis 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 =4j3 (iv) z 4 =^4j5
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 coordinates.
The real part corresponds to the xvalue.
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 = lj6
Then
OP = 2! + 2 2
OQ=Z! + z 2 z 3
3j
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=^=O75
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 = 3606 (cos 56°19' + j sin 56°19')
Here is the working
z = 2 + j3 = /(cos + j sin 0)
r i =4 + 9=13 r = 3606
tan0 == 15
= 56°19'
z = 3606 (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, thenz =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 = 5385 and from above
6 = 158°12'
• The full polar form is z = 5385 (cos 158°12' + j sin 158°12') and this
can be shortened to z = 5385 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 = 075 /. £ = 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(08192 + jO5736)
z = 40960 +J38680
Now you do this one o
Express in the forma + )b, 4(cos 65° + j sin 65 )
49
z = 16904 +J36252
for z = 4(cos 65°+j sin 65> 4(04226 + J09063) = 16904 + J36252
nDDDDanDDDDnDDDClQDDQnDDDDDaaanDDDDDDDD
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(08660j05)
= l732j
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 =38300 +J32140
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(07660 + jO6428)
= 38300 +J32140
□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 = 6403
tan E = 08 :. E = 38°40'
• e = l41°20'
5 + j4 = 6403(cos 141°20' + j sin 141°20') = 6403 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(0500 J0866)
Turn to frame 52.
1500J2598
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 = 642eJ 1  57
lnz = ln642+jl57
= 18594 +J157
and the result is once again a complex number.
And if z = 38eJ 0236 , then In z =
29
Complex numbers 1
lnz = ln38jO236 :
13350J0236
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(cos0jsin0) = reJ e
7. Logarithm of a complex number
z = rei 6 :. lnz = lnr + j0
or if z = re~i e .'. Inz = lnrj0
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) (4j7)(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) 4J5
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(*.>') = 148+j62
6. Express in exponential form
0) z, = lo37°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)
.... (2j3)(3+j2) ,..., cos3x + jsin3x
K} (4j3) 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(4j6), 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)(3j4) = 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. Uzx+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(zz) = 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 = 6403(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 =6403
tan E = j =
■. E = 38°40'
In this case, 8 = 180° +£ = 218°40'
So z = 5 j4 = 6403(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. 6403(cos 218°40' + j sin 218°40') is
4
38
Programme 2
6403 l218°40'
nnnuuDDononnnnaoaaaaaaaaanaoaaauaaaaaa
322°15'
105 c
Correct. Likewise:
5 7 2(cos 322° 1 5 ' + j sin 322° 1 5 ') is written 572
5(cosl05°+jsinl05°) " " 5
34(cos + jsing) " " 34
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 = 075 /. E = 36°52'
6 = 360°  36°52' = 323°8'
First the diagram.
Y
4j3 = 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°
5 160"
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 )
Rearranging 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(0i0 2 )]
18
divide the r's and subtract the angle
DDDDDDDnDDDDDDDDDDDDDDDDnDDaDnnDDDDaDD
That is correct.
eg.
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 = 4472
8 tan E = 05 :. £ = 26°34'
x, —4 ■ o~ ^< ■'• 6 = 153°26'
z =4 + j2 = 4472(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(0866 +j05) = 1732 +j
(ii) 5(cos57°jsin57°)
= 5(05446 J08387)
= 27230 J41935
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 ^ )
= l71(cos75°+jsin75°)
zj = 171  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 = 171 I 75°
z 2 = 171 [195°
z 3 = l71 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) = 02158
Taking antilogs, A = 1 644
The first of the 5 fifth roots is therefore, z t = 1 .64460°
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 =
1644  60°
X, ^
*2 =
■ 644  136°
1644204°
*3
*5
*4 =
1644 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 = (0845 1) = 02113 and A = 1627
z x = 1 627  20°
The other roots will be separated by intervals of — = 90°
Therefore the four fourth roots are —
zj = 1627 20°_
z 3 = 1627  200°
And once again, draw an Argand diagram to illustrate these roots.
z 2 =1627 1 110°
z 4 = l627[ 290°
55
Complex numbers 2
Y
\ J
*i 0\ X
Zi = 1627 J 20
2z = 1627 J110
*3 = 1 627 200 °
z 4 = 1627 [290°
37
aDDDDnnDDDnnDnnannDDDDDaDnnnnnanDDDaDa
And in this example, the principal fourth root is
38
Principal root: z x = 1627 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 = 1817 1 200°
z 3 = 1817 320°
N^O"
*2
*3
Principal root : Zj =
■817 320°
<l
u
□□□□□□□□□□□□□□DDnnnnnnnQnnannnoanaonQO
Here is the working.
z = 6 I 240°
Zi =63
240"
1817 80°
•5zrr\°
Interval between roots = — — = 120°
Therefore the roots are:
1817 180
z 2 =1817 1200
z 3 = 1817 1 320°
The principal root is the root nearest to the positive OX axis. In this case,
then, the principal root is z 3 = 1817 ] 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 righthand 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 s3cs 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 s3cs 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 03 cos0
sin 30 =3 sin04sin 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 08cos 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 (lc 2 ) + (lc 2 ) 2
c 4  6c 2 + 6c 4 + 1  2c 2 + c 4
8c 4 8c 2 + 1
= 8 cos 4 08cos 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 rewrite 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 464 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 jj
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
zl=x+]yl =(xl)+]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 \ ^/[{xlf+y 2 ]
= 2
■Wl(x\) 2 +y 2 ]
■ (^ + i) 2 +y 2 =
" (xl) 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
;
(xiy^ry
(x+l) 2 +y 2 =4J(xl) 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
z1 =x+iyl =(xl) + iy
:. zl =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)z4 = 3 (ii) arg(z + 2)=5
67
Complex numbers 2
Further ProblemsII
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 3j
2. In the Argand diagram, the origin is the centre of an equilateral
triangle and one vertex of the triangle is the point 3 + jy/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—ii . 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 nonreal 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.
z2
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 +zj2J
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
z1
: 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 rightangled 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 rightangled
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 xaxis
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 xaxis.
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 midpoints, 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 xaxis at C, i.e. P : .
Note that on the left of the origin, BP is negative and is therefore
placed below the xaxis.
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 1275 = 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 1275.
This is 3579 .". A = 3579 (as easy as that!)
So e 1 " 275 =3579 and e" 1 " 275 =^ = 02794
:. sinh 1275 =K3579 0279)
= 1(3300)= 165
:. sinh 1275 = 165
In the same way, you now find the value of cosh 2156.
When finished, move on to frame 18.
18
cosh 2156 = 4377
cosh
□ □□DDDDnDDDDDDDDDnDDDDDnnnnDDDDDDDDnnD
Here is the working:
Example 2. cosh 2156 =\(e 2 ' 1S6 + e' 2 " 156 )
Let A = e 2 " 156 .'. In A= 2156 :. A = 8637 and  = 01158
:. cosh 2 156 = £(8637 + 01 16)
= (8753) = 4377
:. cosh 2156 = 4377
Right, one more. Find the value of tanh 1 27.
When you have finished, move on to frame 19.
83
Hyperbolic Functions
tanh 127 = 08539
19
DDnaDnnDDDDDnnanDanDnaDDDnnDnDnDQDnnnD
Working: e i27_ e i27
Example 3. tanh 127 :
e   ■ r e
Let A=e 127 ;. In A =127 .\ A = 3561 and \ = 02808
A
, , „„ 35610281 _ 3280 O" 5159
•'• tanh l  21 ~ 3561 +0281 = 3842 Q'5845
19314
tanh 127 = 08539 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 1475, i.e. to find the value of x such that xll
sinhx= 1475.
Here it is: sinh x = 1 475 .'. %(e x  e x ) = 1 475
.. e*  \ = 2950
Multiplying both sides by e x : (e x f  1 = 295(e*)
(e x ) 2 295(e*)l=0
This is a quadratic equation and can be solved as usual, giving
&x = 295±V(295 2 +4) = 295 ± V(8703 + 4)
2 2
= 295 ±y/l 2703 = 295+3564
2 2
_ 6514 0614 _
= — j— or 7y— = 3257 or 0307
But e* is always positive for real values of x. Therefore the only real solu
tion is given by e* = 3257.
:. x = In 3257=11809
.. x= 11809
Exercise 2.
Now you find cosh" 1 2364 in the same way.
84
Programme 3
21
cosh 1 2364 = ±1507
naaannnnDnannaDDnaannDDnDaDnnnnaaaaDDD
For: To evaluate costf 1 2364, let x = cosh"" 1 2364
.'. cosh* = 2364
+ e'
= 2364
:. e* +  Y = 4728
(e x ) 2 4728(e*)+l =0
e* = 4728±V(22364) Vlg . 36 = ^
= i(4728 ± 4285) = i(9013) or ^(0443)
e*= 45065 or 02215
.'. x = In 45065 or In 02215
= 15056 or 24926 i.e. 15074
x =±1507
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" 0623.
Let x = tanh" 1 0623 .' tanh x = 0623
Dnnnannn
• e c =06^
e x + e~ x
.. e * e *=0623(e*+e x )
:. (1 0623) e*=(l + 0623) e
X
0377 e* = 1623 e' x
1623
e*
02103
■ ( s x\2 1623
"0377
T5763
2) 06340
:. e x = 2075
03170
:. x = In 2075 = 07299
:. tanh" 1 0623 = 0730
Now one for you to do on your own. Evaluate sinrf
1 05
85
Hyperbolic Functions
sinh" 1 05 = 04810
23
aQnaaonanaanannaaaaQDqnoQQDQDQDOODODOn
Check your working.
Let x = sinh" 1 05 .'. sinh x = 05
1
p x _ p x
e e = 05 :. e*
= 1
/. (e*) 2  1 = e*
(e*) 2 (e*)l=0
a *_ 1±V(1+4)_1±V5
32361
or
2
12361
2 "* 2
= 16181 or 06181
.. x = ln 16181 =04810
sinh" 1 05 = 04810
And just one more! Evaluate tanh" 1 075.
tanhT 1 075 = 09731
e* =06181
gives no real
value of x.
24
□QnQQnnnnonnaonannnnQnnooonoQannaanaQn
Let
x = tanh" 1 075 .'. tanh x = 075
■ = 075
e x  t x = 075(e* + e"*)
(l075)e*=(l + 075)e"*
025 e* = l75e*
175
(e x Y
025
= 7
e* = ±V7 = ±26458
But remember that e x cannot be negative for real values of x.
Therefore e* = 26458 is the only real solution.
.'. x = In 26458 = 09731
tanh" 1 075 = 09731
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
lx
1 +x
1
,1 1, fl +x
y = t<mh l x=$\n\^—^
tanlf 1 05=1 In j^)
= iln3=i(10986) =05493
26
And similarly, tanh ' (06) :
tanh" 1 (06) =06932
nnDDDnnDnnannDnDnnnaDnnnnnDnDDnnnnDnnn
For, tanh" 1 * = jln{ t~z — 
= \ In 025
= i(26137)
= i(l3863)
25 09163
10 23026
26137
= 06932
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 =xsj(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 = xy/(x 2  1)
Both these results are positive, since \J(x 2 ~ \)<x.
u 1 i x\J(x 2 \)
However, 775 —  = 775 —  . }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 thejyaxis (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
■iir^
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) = rr —
These, by the way, are pronounced (i) coth, (ii) sheck and (iii) cosheck
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 +eJ 9 =
38
2 cos 6
So that.
cos 8
e i e + eJ 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 eJ 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 = 18 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 1532 (ii) cosh 1 125
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 coshpr
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" '075, (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 coordinates 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 = 0215 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
= 98=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
eg
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
= 125 =
□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
= 4.35.2= 1210 =
= 7.36.4 = 2124 =
= 2(3)4.1 =64 =
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 = ij2 11 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 xterms 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
= 5.43.2 = 206= 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
8557 = 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 =
\3x2y+ 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+3y14 =
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(4028)3(3214) + 2(1610)
= 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+ 14060 = 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 + 15272 = 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 lefthand 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 lefthand 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 + 12632=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 87
7 3 1
4 6 9
= 7(1 8 48) + 3(9 32) 1(6 8)
= 7(30) + 3(23)l(2)
= 21069 + 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
ci 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 xterms
A 2 = " " " " " " " jterms
A 3 = " " " " " " " zterms
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 + 515 = 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(54)
1 2 13
= 3(43)+l(37)4(9)
25 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 + 2y3z+ 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 + 16077 = 63
To find A , omit the constant terms
••• Ao =
115
Determinants
for
Ao=21
Ao =
2 1 5
1 1 1
4 23
=
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 + 730 = 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+2y3z 3 =
2x y i 11 =0
3x + 2y + z + 5=0
 Ay + 2z + 8 = 
+ 5y3z+ 2 = \
J
3x
x + 5y
5x + 3y  z+ 6
Find y.
Find* andz.
2x2y 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
16 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 = 21 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 + 12915= 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)
= 11021 =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 + 5y3z + 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 .xterms.
/. 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(306)+8(5 + 9)
= 4(16) 2(24) + 8(4)
= 6448 + 32 = 9648 = 48
:. Aj = 48
(ii) To find A 3 , form the determinant of coefficients omitting the zterms.
■■• A,
A 3 =
34 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)
= 7216176.
= 72192 = 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 + 5644
= 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.
2x2y z3 =
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 y4 =
(0
2x + 3y~8 =
(ii)
x y4=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 lefthand side of (i), we obtain
3x j4 = 324= 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 + 25 =
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.
fliAi ~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(103)5(l12)
= 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 = .\ fc16 = .\ 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) =
262fc 2 llfc9+ 14fc15 =
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 48
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 kbt
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.
(1X1) 2 3
5 1
1(215) = 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
34 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
34 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 jc4 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) [(x4)(x+ l)4] =0
:. (x + 2) (x 2  3x  8) =
=
x + 2 = or jc 2 3x8 =
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 2x
,2
:. (x+4)(2xx 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 =
x2y + 2z+ 3=0
3x+ y + z 10 =
3. Use determinants to solve completely
*3.y+4z5 =
2x+ y+ z3 =
[4x + 3y + 5zl=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
4x5v + 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)v5 =
(\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(k2)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 >rM 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(xa) + 2xz =
k(ya) + 2yz =
k(za)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 98 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 northeasterly 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.
35
The vector quantity AB is referred
to as
AB or a
The magnitude of the vector
quantity is written ABl , ora~,
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 equallength 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 ABCB + CDED
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
ABCB + CDED
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 + BCDCAD = 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 + CLML =
(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 + CLML= 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 midpoints 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 + (AGAG) + (HBHB)
= 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
= 4AL3AL + 3LCLCAL
= 4AL4AL+3LCLC
= 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 lefthand 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 = APAP=0.
In the same way, (QB + QD) =
27
QB + QD =
Since Q is the mid point of BD .'. QD = QB
.'. QB + QD = QBQB =
.'. AB + AD + CB + CD = 4PQ + +
= 4PQ
28
Programme 5
Here is one more.
Example 4.
Prove by vectors that the line joining the midpoints of two sides of a
triangle is parallel to the third side and half its length.
A
Let D and E be the midpoints 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+iAC
= 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/ + 3/3z2;
= 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 21)/
= 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 = OBOA
= (5/ 2/) (3/ + 5/) = 2i 7/
On to frame 35.
33
34
Vectors in space z
The axes of reference are defined by
the 'righthand' rule.
OX, OY, OZ form a righthanded set
if rotation from OX to OY takes a
"7 righthanded corkscrew action along
the positive direction of OZ.
35
Similarly, rotation from OY to OZ gives righthand 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 =
= 5385
PQ =4i + 3j+2k
pq= V(4 2 +3 2 +2 2 )
= V(16 + 9 + 4) = V29=5385
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
cr 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=3i2j+6k
Then to frame 39.
7=3i2j + 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 = aii + 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+ aib 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  64
:. 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 righthanded 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 righthanded 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
34 2
2 51
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 coordinates (/, m, ri)
and P' " " (l',m',ri)
{mm')
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 = 22(//' +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 )
= 22cos0 (ii) \unitvectors j
So, from (i) and (ii), we have:
(PP') 2 = 22(//' +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] = [05, 03, 04]
and [l',m',ri] = [025, 06, 02]
the angle between the vectors is 6 =
163
Vectors
6= IT
for, we have
cos 6 = 11' + mm + nri
= (05) (025) + (03) (06) + (04) (02)
= 0125 + 018  008
= 0308  008 = 0225
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 = 4i3j + 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 02414
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
righthanded 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 righthanded 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 = (4x5) 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 = (4x5) 6 ^ = 6(4x5) 5 .4 = 24(4x5)
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(34cosjc) 7== — (4smi)
& 34cosx 34 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 (2xl)
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 (2xl) ^ = (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 «(I21nx)
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 notebook.
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 twofactor 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 righthand 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
_44x 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
..— ^ rr 42tanh2x
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 2x4tan4x
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 lefthand 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
:. yf=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
:.(3y6)g = 22*
■ dy _ 22x _ 1 x
dx 2y  6 _>>  3
^ v rf,ir> (y3)(l)(lx)^
~, « J' _ d l x  dx
Then —
28
dx 2 dx\y3) iy~3) 2
Oy)(lx)^
_ dx
(y3) 2
d 2 y. (32)(l3)2 _ l(4) .
dx 2 (23) 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 ry
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 thatf =£■ —
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
• • —rn = 4 COS t. = 4
dx cos t
••■gi
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
. 23?
1 +t
3 + 2?
dx
It
dy
dt
(1+Q (3) (2 3Q
(T+Tp
(1+0 (2) (3 +20
1+?
dx 33/
2 + 3?
*
(i
+o 2
dy.
2 + 2?
32?
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 66 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 — ^jjj
(ii)^ = lnjxV(lx 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 = 3cos0cos 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 jyaxis
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(xx 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.y3 = 2x\0
y = 2x7
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 = i2x
For
y(S) = 2(x2)
.'• 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
yyi =m(xx 1 )
It is well worth remembering.
DnDDnnnDnnnDDaDDDDDDDnnDaDnnDnnnDanDnD
So for one last time:
If a point P has coordinates (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)
= 2x8
.'. y=2x5
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>x5 and 3y = x + 2.
(ii) 2y = x  5 and 7 = 6  x
(iii) y3x2 = and 3.y+;c + 9 = 0.
(iv) 5yx = 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 :. mmifl 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— /"
2x8
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(jc3)
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 + 62= 10, i.e.m= 10
(^1
\dx( x = 1
Passes through P, i.e. x =1,^ = 0.
y~y\ = wOcXj) gives j>0 = 10(xl)
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: y0= — (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),^= 128 + 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)
:.y5=±{x2)
7y35=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 findp 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)
7y14 = 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:
Jxl)
8j16 = 7x7
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=31x62 :. 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)33f _3 + 3t  3t
1 +f " dt (1 + f) 2 (1+0 (1 +0 :
r . dy _(1 +t)2tt 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
(x2)
3y~4 = 8xL6 :. 3j; = 8;c12 (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 — = 05; y = co
6
05
Tangent is y  ^ = 2(jc  ~) .'. 2y  \ = Ax + 2
.". 2y + Ax = 3 (Tangent)
Slope of normal = y • Line passes through (05, 05)
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 + 8x16 =
8 ± V(64 + 64) _ 8 ± y/128
X 2 2
8 ±11314 3314 19314
= ~ = — « — or ■ —
x = 1 657 or [9655] Not a real point of
intersection.
When jc = 1657, y 2 = 8(1657) = 13256, y = 3641
Coordinates of P are x = 1657, y = 3641
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 3641 0910
tan 0! =1099 /. di = 47°42'
(ii) Similarly for x 2 + y 2 = 1 6
^ = _x = _F65_7
dx y 3641
tan d 2 =04551 /. 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
= 04551
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 coordinates 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 = 2828 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£.. ,($.
dx" dx 2
{l + l}3/2 _ 2 3/2 _2x/2_V2
6 6 6 3
■'. R = 0471 units
Now you do this one:
Find the radius of curvature of the curve y 2 =^ at the point (l,4)
When you have finished, check with the solution oh frame 32.
31
210
Programme 7
32
R = 521 units
Here is the solution in full.
y =
dy_ = ^_ . dy _ 3x 2
" y dx 4 " dx Sy
 At V'V>dx 4 " \dx) 16
dy = 3x^_ . d 2 y =
dx Sy " dx 2
8y(6x)3:c 2 8
dx
64 y 2
R =
1 +
At(l,I),
2 > 3/2
1 \ d 2 y _ 24  24.f _ 24  18 _ 3
dx'
16
16
(dy\ 2 } 312 f. _9J 3/2 {251 3/2
W > _ r + 16) _ \16J _ 8 125 _ 1
dx 2
64
25 _5_
24 24
•'• R= 521 units
33
Of course, the equation of the curve could be an implicit function, as
in the last example, or a pair of parametric equations.
e.g. \fx = sin0 andy = 1  cos 0, find Rwhen0 = 60° =
* = 0sin0 •§=lcosfl > l dy = dy_dB_
d Y I dx dd ' dx
y  1  cos 8 /. ^ = sin I
. dv . „ 1 _ sing
dx 1  cos 1  cos
At = 60°, sin 8 =^j, cos0=i; ^ = ^
dx
_d i sin0 \ = d_ ( sing 1 dd_
dx\lcos0) dd \lcosd\dx
_ (1  cos 0) cos 8  sin . sin 6 1
(1  cos 8) 2 1  cos 8
_ cos 0 cos 2 8 sin 2 6 _ cos  1 = 1
(1  cos 0) 3
:. At = 60°,$ =
" (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 coordinates of the centre
of curvature of the curve y = _„ at the point (2,3).
3x
1
dy (3jc) M)(114*)(1) _ 12 + 4jc+U4x ,
±c~ (3^x7 (3x) 2 (3x) 2
dx 1
W
d 2 y_d
2
dx
=^H3^}=2(3x)(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 =jat 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 rewrite 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 = 2795, i.e. 2795 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«Mi..f4
A*g£i~')n~'}£r.i,s
. d 2 y_ 1
"" dx 2 2
R= dV " i Wi
dx 2 2
_25n/5_5\/5_ 5 (22361)
8 4 4
= lM805 = _ 27951
4
R =2795
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') = 08944
Xi = 2 cos 60° = 2. — = 1
7i = sin'
 ^H
and you have already proved that R = 2795.
What then are the coordinates of the centre of curvature?
Calculate them and when you have finished, move on to the next frame.
Results:
For:
and
/z=025; fc=l75
43
h=\ (2795) (04472)
= 1 1250
h = 025
k = 075 + (2795) (08944)
= 075250
k = \15
04464
T6505
00969
04464
19515
03979
Therefore, the centre of curvature is the point (025 , 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 coordinates 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 75 = 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, 06) 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 coordinates 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 xaxis at the points T and N respectively, show that
ON.OT is constant, O being the origin of coordinates.
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 jaxis 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 midpoint
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 coordinates of the centre of curvature.
10. In each of the following cases, find the radius of curvature and the
coordinates of the centre of curvature for the point stated.
(1)^4=1 at (0,4)
(ii) y 2 = Ax x 2  3 at x = 25
(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 coordinates 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 coordinates.
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 coordinates (h,k) of the centre of curvature in terms
off.
219
Differentiation Applications 1
18. Find the radius of curvature and the coordinates of the centre of
curvature of the curves = 3 lnx, at the point where it meets the
xaxis.
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 05 = the angle whose sine is the value 05
= 30°
There are, of course, many angles whose sine is 05, e.g. 30°, 1 50°, 390°,
510°, 750°, 870°, .. .. etc., so would it not be true to write that
sin 1 05 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 05 is the numerically smallest angle
(measured between 0° and 180°, or 0° and 180°) whose sine is 05.
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 05, 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 05 = 30°
and no other angle!
~X to 180°
to 180°
Similarly, if sin 8 = 07071 , 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 = 07071 .\ 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 07071 =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 08 1 92.
From the cosine tables, we find one angle whose cosine is 08192 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 08192 = 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 (0866)
Work through it carefully and then check your result with that on frame 6.
cos" 1 (0866) = 150°
For we have
cos £' = 0866 .'. £=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 (0866)= 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(lx 4 )
5 ''■• ta  1 (f)S* a 7p^)i^^(f)
2V{iV
= 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 ••■^ = rT—
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
>,= ltanh 2 ,= lx 2 :.%^
£ tanhx U^
Now here are the results, all together, so that we can compare them.
1 (iv)
Ijsinhxj = V(x2 + i}
d_
dx
1
(v)
{ COSh_lx jV(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 j32x
. 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 169jc 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
jtanh" (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^tanCsinh*) 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 ii 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 Sbend.
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', 75 is positive
for PofL
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 PofI) 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 jy
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 x2 = :. (x2)(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 =41=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 PofI 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 righthand bend to a left
hand bend, or from a lefthand bend to a righthand bend.
32
The point C we considered is, of course, a PofI, but it is not essential at
a PofI 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 PofI 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 75 = 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 PofI
d 2 y
the graph of ^
crosses the xaxis.
In the
case
of no PofI, 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 PofI,
try =
dP
with
change of sign
This last phrase is allimportant.
DGnnnnDnDnnnanDnnDnnnnnDDDnnDDnnnnDDDD
Example 1. Find the points of inflexion, if any, on the graph of the function I
y^jTx + 5.
(i) Diff. twice. d £=x 2 x2, 4$= 2x  1
For PofI,
d 2 y_
dx
= 0, with change of sign. .'. 2x  1 = .'. x = —
If there is a PofI, 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 = l2al
= 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 righthand
curve changes to the lefthand 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 Pof1, rj = 0, with change of sign.
:. 60x 2 (xl) = :. 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 (al)
= (+)(+)() = negative J
No sign change.
No PofI.
(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 +al)
= (+)(+)(+) = positive
► Change in sign.
:. PofI.
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 (06428), (ii) tan 1 (07536).
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/(lx 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 coordinates of the point of inflexion on the curves
(i).y = (x2) 2 (x7)
(ii) y = 4x 3 +3x 2 18x9
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 =82Vl4.
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(xl)
(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 crosssection 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 crosssectional 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. — ris 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 = 2nrh
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, thenr— = and tt =
dr dh
A = 2irrh
JjA
dr
= 2irh
and
9A
dh
= 2irr
ODDDDnUDDDDDDDDDDDDDDDDDUDDDDDDDDnnDDO
Of course, we are not restricted tothe 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 findr andr
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
Ixy + 2x + 6y  4x + 5y
f=(2xy)(0 + 3) + (x + 3y)(0l)
= 6x3y  x3y = 5x6y
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
^ = 14jc20j
^
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 = 14x20y
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(20)(2xjQ(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 (x2^) 2
Here is the working:
(i) To find— , we regard jy as being constant.
9z
bx
, (x2y)(S+0)(Sx+y)(l0)
5s 10y5sj ;_ —1 !>■
9z
(x2y?
 Oi,^
(*2y) :
8
(ii) To find^ , we regard x as being constant.
y . 9z (s2y)(0 + l)(5s+>Q(02)
" by (x  2yf
_ x2y + \0x + 2y 1 Ijc
(s2>>) 2 (s2y) 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)
^=24x7y ^=7x20y
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) + (4x5^)(3)
= 12x + 8.y + 12*  \5y = 2Axly
^•=(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 rj (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 lefthand
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 — = Ax10y
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 — , — , tt , rj , 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 cosyy.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 thatJ +J =
This merely entails finding the two second partial diff. coeffts. and sub
stituting them in the lefthand 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 )22x.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 find7— 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 rr = ;*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 )22y.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
•'• XTry¥ =x f'< x * +y 2 )2yyf'(.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 ' xy
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 = 
xy
9z = (xy)l(x+y)l = 2y
9x (xy) 2 (*~y) 2
9z _ (xy)l(x+y)(l) 2x
9y (*~y) 2 (xy) 2
^=(2v) ( " 2)  &
9x 2 <■ y) (x~yf (xy) 3 '
By 2 ^{xyf { U (xy) 3
9 2 z ^ (x  y) 2 (2)  (~2y)2?x  y) (1)
3y.9x (xy) 4
_ 2(xy) 2 4y(xy)
(xyf
= 2 4y
{xyf {xyf
_ 2x + 2y ~ 4y _ 2x  2y
(xy) 3 (xy) 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(xy) 2 4x(xy)
(xy)*
_ _2 4x
(xyf (xy) 3
^ 2x2y4x _ 2x  2y
(xy) 3 (xy) 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^ + eV
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 === 2nrhbr + 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 02 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 = 02 and 6/? =01 (minus because h is
decreasing)
.. 5V — .br + rT5/j
5V=100rr(02) + 257r(01)
= 207T257T= 1757T
.'. 5V^ 5496 cm 3
i.e. the volume increases by 5496 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 05 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 = 05,
6I = 5>I> 5 )
= J_ J_
50 20
= 002 005 =003
i.e. I decreases by 003 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 rightangled triangle
are denoted by a and b. The hypotenuse is h. If there are possible errors
of ± 05% 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 =05% 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 = 05% 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— +VT +Zt
dx ' dy dz
(ii) If z =/(x + a>») + F(x  aj), findri and Tl and hence P rove
that r2 = 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 02 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 ), thenfj +^4=0.
3. If z = sin(3;c + 2y), verify that 3 ff  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 — + rf = 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 01 ± 0002 and v as
03 ± 002. 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
± 025% 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 2X3y
^=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 2X3V .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
20 r2=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
■KX y x 1
X 4 X
y xy
xy
X 2
b 2 z _y
dy 2 ~
jfxr 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.
Ratesofchange 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 02 cm/sec while the
height decreases at the rate of 05 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, — =02, — = 05 (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^=201 cm 3 /sec
dt
for
dV.
dt
dt dt
: 27r8.12.(02) + 7r64(05)
■■ 384n  32ir
■64tt= 20lcm 3 /sec.
Now another one.
Example 2.
In the rightangled 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
— =475 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 = — + =—= 475 cm/sec
4 4 4 4 4
.'. Side z increases at the rate of 475 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 025 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+^TX
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— = 025 and— = 025
a? dt
. dS_64Tt _1 12tt 1
" dr 5 '4 5 '4
_ 167T + 37[ = 197T
5 5 5
387T= ll93cm 2 /sec
14
So there we are. Ratesofchange 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 ratesofchange 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 forj
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 forr ■ 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 findf. 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+jl 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 tanyy 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 __ tanyy 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 forr
e x+ yx 7 y i = 0. Letz = e x+ yx 2 y 2 =
. dz  , Bz
5z= — 5x +r5^
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+y2x 2 y
For
= e x+ yx 2 v 2 =
z = e
*=e* + ?2xy 2 ; ^=e x+ y~2x 2 y
dx ' dy
■ dz
^ = ( e x + y 2x yi) + (e x + y2x 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, thenr  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 andr^
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(uv), find r^ and— .
Wehave £ ** + E.V
= y.e*^. +x.e x >'.cos(M v)
u+v
 e xy) y +x.cos(uv)
(u + v v
26
and
bz _ bz bx bz by
bv bx'bv by'bv
= v.e^. + x.e x y\cos(uv)\
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 02 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 xr + yr 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 + 2yx = 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 ifr2 + 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, findpby 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) rT + 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. 118 = 3; 52 = 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 = 26 = 4
S„ =(2fl+'nl'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 QA = g?
P + Q
:. AP = QA 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 405
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 = 25
Required arith. means are
105, 13, 155
Now, you find five arithmetic means between 1 2 and 21 6.
Then turn to frame 8.
<z = 8
a + 4d = 1
A
= 8 + 25
= 105
B
= 8 + 5
= 13
C
= 8 + 75
= 155
Required arith. means: 136, 152, 168, 184, 20
Here is the working:
Let the 5 arith. means be A, B, C, D, E.
Then 12, A, B, C, D, E, 216 form an A.P.
.'. a= 12; fl + 6d = 216
:. 6d = 96 :. d = 1 6
Then A = 12 + 16 = 136 A =136
B = 12 + 32= 152 B = 152
C = 12 + 48= 168 C = 168
0=12 + 64=184 D=184
E = 12 + 80 = 200 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(lr») ,~
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
1i 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 (lr 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 = 'andr
:.£=? :. 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 864.
Then on to frame 15.
303
Series 1
Required geometric means are 60, 72
15
For, let the means be A and B.
Then 5, A, B, 864 form a G.P.
.'. a = 5; :. ar 3 = 864; .\ r 3 = 1 728; .'. r = 1 2
A = 5.12 = 6 1 Required means are
B= 5. 144 = 720 60 and 72
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 3 332^1^1
(« + 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 lefthand 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+(nl) + («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 + 08 + 016 + 0032 +
as
25
5^=25
For 20 + 4 + 08 + 016 + 0032 + .. .
n 8 i
a = 20; /•=^ = 02=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/ „_„«, 27/n
Now when n+°°, 3/n^O and 7/n^>0
5n+3 _ 5 + 3/w = 5+0 = 5
"„Voo2«7 „ioo27/rt 20 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 56/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+00 2
27
50+0 5
„32
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„=
~> ntao £
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") _ 13"
••• s w
lr 13
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^bI 2'2nl
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 2h1 „*oo / 2 l/«
I 112= !
"2"20 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
rcyoo
(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 2w„ =
319
Series 1
2w„ = 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
2w„ 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 2w„ 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 2w^ = 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. forx<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(lr»)
Ifr<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 2m„ 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 + 5x4
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 when1 <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.
02 ) (x_2) 2 + ix2) 3 + + (x2f
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 righthand side does not equal the lefthand 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 +4x3
(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— , +rj 
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)=xf + 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 nonvanishing 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 + Tf+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 = ifr + 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) = xf + ff + 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)(n2)(n3).(l +x)"^ / iv (0) = «(«l)(n2)(n3)
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(nl) 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) = 24xl 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
= lx+
_*_ + *
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! + ITr + 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 + .
77ze7e 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 \/l02 correct to 5 decimal places.
102= 1 +002
Vl02 = (1 +002) 1 / 2
= 1 + ±(002) + ±jj (002) 2 + \ 23 l (002) 2 . . .
= 1 + 001  1 (00004) + ^(0000008). . .
= 1 + 001  000005 + 00000005 . . .
= 10100010000050
= 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 '01 =01 — + ^ . . .
= 01  000033 + 0000002  . . .
= 00997
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} + 014 14 7
e.g. Lim 2
30
x >oU 5jc+ 8 ) 00+ 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 lk+^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
2X + 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
sinhxx)
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
sinhxx) 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 xaxis, 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
xy 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+21 4 ,
Lim
2x3
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
01
= 1
Now you can do this oneT
Determine
 Umf cosh xe 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 52* 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
00
By direct substitution, limiting value = — —  .
Apply l'Hopital's rule:
T . fxsinx) T . (lcosxl
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 . (jcsinxl
■ ■ 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 mujm
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) 11
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 xx ) 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 xx) a
= Lim I r \ (still gives )
^^olcosx1) 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
...... fxcosxsinx) . (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 _yaxis 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 (—001745) and sin = cosr= ,
/.sin 44 = pr 1 +h~j  g +. . . A = 001745
= ^{l0.01745 ^^ + °^f^ 3 +
= Ml  001745  00001523 + 00000009 . .
= 07071 (0982399) = 069466
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 nonzero terms in the expansion of cos 2 *.
3. Find the first 3 nonzero 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 . Ixsm.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=lyj+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 xx 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 . /sinxxcosjcl ..... 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 + 003) 12 , the r th term is less than onetenth 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= Nf 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. (tanhx)= 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(lx 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
1x'
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)ldx = 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
\(5x4) 6 dx=
1
360
Programme 13
\
(5x4) 6 dx
= (5x4) 7 + i
5.7
( 5x4) 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(5x4) 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(34x) _
. _ 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(2x5)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(2x5)^ =  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(,» + ax5) + 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, fp^ dx = 2 f^TT^* = 2 ln (* 3 " 4) + C
x 2
and 1—5 —  dx =
363
Integration 1
^^^to'jWO + 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 submultiple 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.
[ 4x8
'■ g )x 2 4x+i
dx =
364
Programme 13
15 f ^8 f .2x4
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 + 4jel) + C
3. I !^LiL <& = in cosh x + C
J cosh*
I* jc3
' 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 +7x4)(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 +
3x7
= ln(x 2 + 3x7) + 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 +7x4) 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 lefthand 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 righthand 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 *33 +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 wholenumber 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(—jj\x 2 .dx
x 2 kix 1 , ,
= ~T~ i} xdx
_ x 2 lnx _ 1 x 2
"2 2"2
4 2 {lnxl} + 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 Jx2 Jx1
373
Integration 1
3]n(x2)21n(.xl) + 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 (xl)(x~2) xl x2
Multiply both sides by the denominator (x  1) (x  2).
x+ 1 = A(x2) + B(;cl)
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 Jx2 Jx1
Now the rest is easy.
x+1
— — ; — — dx = 3 dx  2 dx
Jx 3x + 2 Jx2 Jx1
x'3x + 2~" "Jx — 2 "Jx1
3 ln(x2)2 ln(xl) + C (Do not forget the constant of integration!)
And now another one.
Example 2. To determine
dx
(x + 1) (x1) 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) (x1) 2 x + 1 x1 (x1) 2
x 2 = A(x  l) 2 + B(x + 1) (x  1) + C(x + 1)
Clear the denominators
Put(x1) = 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)(xiy
3
+ — .
1
1
4 'x+1 4 'x1
1_
x1
■■ , 7w ^^2 dx = 1 — dx + \ t dx +  (x 
J (x+1) (x1) 2 4Jx + 1 4Jxl 2J V
\_
V(x1) 2
dx + ^(x l)~ 2 <ix
35 [J
(x+1) (x1)
2 dx = iln(x+l) + 7ln(*l)
1
2(xl)
+ 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
" (x2)(x 2 + l) x2 ' x 2 +1
:. x 2 = A(x 2 + 1) + (je  2) (Bx + C)
Put(x2) = 0, i.e.* = 2
.. 4 = A(5) +
Equate coefficients
[x 2
1 = A + B .\ B = 1  A = 1  F
[C.T.] = A2C .'. C = A/2
5
B =
C =
4 1 + 5* + 5
(x2)(x 2 + l) 5 'x2 x 2 + l
 2 1
5'x2 ' 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(2xl) 2 x + 2xl (2xl) 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(2xl) 2 x 2x\ (2xl) 2
/. 4x 2 + 1 = A(2xl) 2 + Bx(2xl) + Cx
Put(2jtl) = 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(2jc1) 2 x (2xl) 2
. .4.(2xl)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(lcos 2 *) 2 sin*d*
(12 cos 2 * + cos 4 *) sin * dx
s^xdx^x.^xdx^x^xdx
2 cos 3 * cos 5 * _,
= COS * + — r — + C
Similarly, i 2
= I (12 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
2xl
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 (xl)(x 2 +x + l)^
,. f sin 2x ,
3. I ,— t~ ax
J 1 + cos x
5. 1 x sin 2 * dx
J °
7 ' J (xl)(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
171 a: 2 sin 2 xdx
dx
19 f *
x 2 )
21
f 8x
J (x  2) 2 (x +
■j:
n
f 2x + 3
J (x4)(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 xVx 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(x2)(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.
x1
9x 2 18x+ 17
385
Integration 1
3l \^~[ dx 32  [x 2 \n(l+x 2 )dx
3 3 f cosg sin0 34 r isinfl
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 (ZA)(Z + A) ZA Z + A
where P and Q are constants.
.. 1 =P(Z + A) + Q(ZA)
:. 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 2AZA 2A'Z + A
1 I 1
•'Jz 2 A 2C?Z "2aJz
dZ ~k\zh dZ
I:
;j^dZ=i.ln(ZA)\.ln(Z + A) + C
■A"
2A'
2A"
km^
/
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 = 2Hfrx) + 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
ZA
+ 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 /Z5
+ c
277Hl^ 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 rewrite 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 rewritten 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 . JZA. _
 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 +49
= (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 + 3V5
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 ( X5V7 ', , r
2V^ kl U5 + V7 } + C
8
For:
x 2  10x + 18 = x 2 10x +18
= x 2 10;c + 5 2 + 1825
= (x5) 2 7
= (x5) 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 2xA 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 5xlV21 , . _
5^ 2 2^4 dX ~2V21 5xl+V21 f L
Here is the working: follow it through.
5x 2  2x  4
dx 
(*iFWJ
dx
1 _5_ 1L [ xl/SV21/5 \
S'ly/ll U1/5+V21/5/
hfcM
5xl\/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 (AZ)(A + Z) AZ A + Z
.'. 1 =P(A + Z) + Q(AZ)
:. 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(AZ) + C
2A'
AZ
•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
'»{1taH
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(x3) 2
= (2V3) 2 (x3) 2
In this case, then, A = 2\/3 and Z = (x  3)
1 . f 1
•'■ j3 + 6xx 2 ^ Ji
(2V3) 2 (x3) :
dx
Finish it off
394
Programme 14
14
1 , (2V3+A3; ^
473 ln 2V3x + 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 
J9~4xx
2A'
2 dx '■
„Aln(^l) + C
AZ
15
±m Ml +x + h + c
2\/l3 K/13X2
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 + 4x2x 2 ^ 2J5 + .
dx
Now we proceed as before.
+ 2xx^
 + 2xx 2 =^(x 2 2x )
5
'2
5
'2
7
'2
1
(x 2 2x+l 2 )+l
(x1) 2
= (V35) 2 (xl) 2
. dx = .
" J5+4x2x 2
(Do not forget the factor 2 we took out of the denominator.)
395
Integration 2
i ,JV3!±£zi) +
4V35 (V35X+ 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.
)66xSx 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 +
66x5x 2UX 2V39 \/395x3
For:
k
6x — 5x 2 5
Jf 6 J* 2
dx
5 5
55*"* 2 4^ + f
So that A
Now
■■ ^y and Z = (x + )
gH)'
^
22 dZ
■>m+ c
f 1 rf = I J_ i K/39/5+X + 3/5 )
j66jcx 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
= T0 + 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 + 3025
= (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
2Jl
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 + 169
dx
So Z = (x + 3) and A = \Jl
JZ 2 +A 2
. f 1
= (x + 3) 2 + 7
= (x + 3) 2 +(V7) 2
,z4tanl + 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(32xx 2 )'
3~2xx 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(32x~x 2 ) " A J V{2 2  (x + l) 2 }
.if* +11.
= sin { —z— +
i
dx
J.
sinM y J + C
1
dx =
For:
^5Jx') 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(54x: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(1412x2* 2 ) V2jV(76x: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 =
VCH12X2JC 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 jjT"
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 = (smhff 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
VCZAVZ'fP^^'coshdjUc
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^ 2y/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
= 2R2 *
50
27214^
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
1f 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 =
1f 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
(lt 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(lf')
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 + 3t2t 2
2 h+ht 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 \42 tan x/2 j
For
l+^r 2 = l<> 2
= l(f 2
25
:f +
)
2x 9
(irw
M(ir
(!fB)*
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
InLjSL^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(49x 2 )'
dx
x 1 + 3x  5
dx
2x 2 +8x + 9
1
V(3x 2 + 16) dX
dx
9Sxx 2
sj{\xx 2 ).dx
dx
V(5x 2 + 10;c16)
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 812**
3 f dx f x8
J x 2 + 14* + 60 Jx 2 +4;t + l
■ J V(* 2 + 12x + 48) J V(1714x x 2 )
7 f *c o f 6x5
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 4x21) 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) (4x)} i6 ' J 2 sin 2  cos 2
27 lv(?^^)^ 28.jVd52xx 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 +6x6]+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 integrationbyparts 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
n2
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 x2A. I
10
Now
And so
1,jx^osxdxjoosxdxsinx.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 integrationbyparts 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 righthand 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(nl)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 =2.7r2.1.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(nl)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
= [00]
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 lefthand 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 15 J
.. I 6 = gSirr* COS* + ,
— jsin 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
n2
For I„ = cos" l x . sin * + («  1) I„_ 2  («  1) I„
n l n = cos"" 1 *. sin* + («  1) l n _ 2 .
1 n\ . nl ,
.. 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 w1
T 1 M
J o
= [ o] + ^iL
"2
I =^1
n2
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 = lcosxj = _ (~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 53" 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 5n
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.
(n1)
(n3) (n5)
n (n 2) («  4)
If n is odd, the factors end with 1 on the bottom
6.4.2
■ etc.
eg
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
eg
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 n2 i
2. Establish a reduction formula for I sin"x dx in the form
r sin"x <
T 1721 . n ~ 1 T
I„ =  — sin x cos x + 1
n
and hence determine I sin 7 x dx.
n2
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  263144e" /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 xaxis 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
coordinates (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 havej^ =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
yK*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+ 1815

1 + 25
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 righthand
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:
5166
I
4e 2x dx = 4
= 2
= 2
ee
ii
2
11
1
e~2
e
= 5166
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 (2x3) 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
= lnl0ln5 = 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 =
(2x 2 ) cosx + 2x sinx
(27T 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
xaxis, 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
= (927+15)(j3 + 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
xaxis.
For such an area, y is negative
.'. ySx is negative
.'. 6 A is negative .'. A is negative.
So remember,
Areas below the xaxis are negative.
Next frame.
18
The danger comes when we are integrating between limits and part of
the area is above the xaxis 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 xaxis, 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.
AJ«.
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
= \ (l2cos0 + cos 2 0)d0
JO
a > • n , g , sin 20
= 2 sin +:r+—  —
^ = (1 cos0)
x = (0 sin 0)
djc = (lcos0)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
= — {1296 112
+ 6x 2 + 9) dx
13
1
_+2 + 9
486 + 81112
1184 =592
r.m.s. =V592 = 7694 /. r.m.s. = 769
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 xaxis, 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 xaxis 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 100rrf + I, and the r.m.s. value of i between t0
and t = 002 is 250, determine the value of I.
452
Programme 16
Further ProblemsXVI
1 . Find the mean height of the curve y  3x 2 + 5x  7 above the
xaxis 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 xaxis 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/' = 02 sin lOnt + 001 sin 30nt, find the mean value of/' between
? = 0andf = 02.
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.
2n
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
xaxis 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 xaxis 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 ?
xaxis 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 xaxis 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 xaxis, and
the ordinates at x = a and x = b, rotates through a complete revolution
about the xaxis, 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 xaxis, and ordinates at x = and x=— rotates about
the xaxis 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
25rr 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 }] = 2i! 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 xaxis and the ordinates corresponding to t = and t = 2, rotates
about the xaxis. [Remember to change the variable of the integral!]
Work it right through and then check with the next frame.
V = 4962?r= 156 units 3
Here is the solution. Follow it through.
.b
V =
v =
iry 2 dx
Tr(3tt 2 ) 2 dx
a
t= 2
f0
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
36384+ 1067
= 6n
= 6tt(827) = 4962tt = 156units :
So they are all done in very much the same way.
Turn on now to frame 8.
4667384
"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 xaxis, and the ordinates x = 1 and x = 3, rotates
about the yaxis 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 righthand 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 xaxis,
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 =
= 2nxy 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 xaxis and the ordinates x = a and x = b rotates completely
about the yaxis is given by :
V = 2n 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 xaxis, 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
C3H)
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+iv2_
* " i 2 " 4 X e 4  1
16
3c= 1523
_ = 3e 4 + 1 = 3(5460) + 1 = 1638+ 1 = 1648
X 2(e 4 l) 2(54601) 10921 1082
/. 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) = ^(5460+1)
5560
= 13 9
So the results are:
3c = 1523; y = 139
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 xaxis, and the ordinates at x = and x = 
6
(First of all find x and check your result before going on to findjO
x = 03424
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
= 0866005236 :. x = 03424
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 = 1542
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
0523602153
.277 .11 \/3
03083
L
= 25(007708) = 1927
_ = I3 = L927 = (± 927)4 = 1542
20
5/4 5
So the final results are
x = 0342, >>= 1542
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 xaxis.
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 xaxis, and
the ordinates x = 1 and x = 3 rotates about the xaxis.
When you have finished, move to frame 22.
Check your working.
x= 189, y =
Ii=f x(\6x 2 )dx = \ (16xx 3 )dx =
(™i)(«4)
8*' "J
■i:
A) \~ 4.
6420 = 44 ;. I, =44
(\6~x 2 )dx
16x
48
9)(.6i)
.. x
= 23^ :. I 2 = 23^
_ i! _ 44 ~3
U
132
1 70 70
189
So x= 189, 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
Iffix0 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.
10V101
27
27
3162 1
= £ (3062) = 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 = 3015 units
Here is the working set out.
25
y = 10 cosh — s :
r)V
\ 2
•'• ! 'lM° osh '^}'' fc "I, CO!h to
= 10 [sinh 02  sinh (Ol)j sinh(;c) = sinh x
= 10[sinh 02+ sinh 0l]
= 10[02013 + 01002]
= 10 [ 03015 ] = 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 [(5Hi)>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 xaxis 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 xaxis.
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= 19577 = 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)
8V24
8jr
3
195tt =613 units 2
7312
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\
■'■aPW(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 jraxis
through a complete revolution.
Here ^=a(lcos0) :. (^) = <z 2 (l 2 cos Q + cos 2 0)
■' (ID" + (dflf =fl2(1 ~2 costf +cos 2 + sin 2 0)
= 2a 2 (lcos0) 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
■2ni" «(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 = 3tt 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
= 1873f t 2 (l+t 2 )dt=l8TT\ (t 2 +t*)dt
Jo Jo
1
18,I 1+T
37
38
.■rHi"^"* 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 xaxis
y = fix)
= IT)
* n
V =  ny i dx
a
Parametric equations V =
(b) about yaxis
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
xaxis 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 xaxis, and the ordinates at
0=0 and 6 = 2tt, rotates about the xaxis 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 coshrbetweenx = 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 xaxis, 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 ProblemsXVII
.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 xaxis
through a complete revolution.
3. The plane figure bounded by the parabola y 2 = Aax, the xaxis 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 xaxis.
5. The area bounded byy 2 x = 4a 2 (2a x), the xaxis and the ordinates
x = a, x = 2a, is rotated through a complete revolution about the
xaxis. 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 yaxis of the area bounded
by the yaxis, 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 xaxis.
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 jcaxis through 2tt radians, calculate the volume of the
solid generated and hence obtain the distance of the centroid of the
minor segment from the xaxis.
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
xaxis, determine the area of the surface generated,
(hi) Deduce the distance of the centroid of the arc from the xaxis.
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 rightangles 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 xaxis. 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, semicircular 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 xaxis 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. = 22 co 2 . m a 2
K.E. =4co 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
lefthand 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 20cm 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 = 001 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 = 001 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
3375125
= ^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
1085
/. fe = VlQ85 = 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 60cm 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 =231 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 = 5334 :. fc = 231 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
2nx.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
 2np
_ 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 05 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 = 424cm
28
. M.r 2 05.36 . , 2
I z =— = — ^— = 9 kg cm 2
M/fc 2 =I .'. i^ 2 = 9 /. £ 2 = 18
.'. fc = 424 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 = 0008 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 .48.80.0008
= 6144tt= 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 = 219cm
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 00256
10 5 "
L 5
= ;rp00256. 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
= 48
.'. fc=v / 48 = 219cm
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
crosssection 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 xaxis, and the ordinates at x = 1 and x = 3,
about the yaxis.
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 = 744 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 = 484 + 26 = 744 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 = 3536 <
For we have
5 8
Sy
'AB
r
Jo
5 8
xy 2 dy=^\ y 3 dy=
^[50]=(625) = 2jW
Also, total area, A = r = 20 cm 2
:. kk 2 = I .'. 20k 2 = 250
:. k 2 = \25
.'. k = 3536 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
dJdS
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 = 8000 + 800 .100
j. (80 000)
A/c 2 =I :. k l = 
4 80 000 400
800
_ k 2 400 40 . _ _,
z = _=—= — = 1333 m
z 3.10 3
Note that, in this case,
(i) the centroid is halfway down the rectangle,
but (ii) the centre of pressure is twothirds 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 25 m below the surface of the
? liquid. Determine the depth of the
I centre of pressure of the cover.
25 m
Q
1 m
—L
Work completely through it: then check your working with the next
frame.
64
3m
e
z = 302m
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
= 302m
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 midpoint 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 xaxis, 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 xaxis
between x = and x = n. Show that the radius of gyration of the
figure about the xaxis 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 xaxis and
the ordinate x = c. Find the radius of gyration of the figure
(i) about the xaxis, and (ii) about thejaxis.
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 xaxis, 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 10cm 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 yaxis, and the ordinate x = a.
c
Show that the moment of inertia of the plate about the _yaxis 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 ■
1We
for:
I
x e x dx :
x(e x )
J
dx
e x (xl)
= ^H)e°(l)=liVe
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 17321 . . . 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 = 3142 : tt = 314159
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 > 03333 + 01000 + 00179 + 00023 + 00002
= V f{ 04537
= 1414(02269)
= 03207
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'"']
: 07071 ( 03333  00500 + 00119 00035
+ 0001100004.. .]
= 07071 (02924)
= 02067
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
0531 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:
= 06667  01429 + 0007576  0000185 + . .
= 0531 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
0088 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 ' '
=  003125 + 000020  000000 + 012500  000625 + 000049
= 012569003750
= 008819
= 0088 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 equalwidth 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 evennumbered ordinates,
2R= 2 X the sum of the remaining oddnumbered 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
75 82 103 115 124 128
7 8 9
123 117 115
Then we have
F + L= 75 + 115= 19
4E= 4(82 + 115 + 128 + 1 17) = 4(442) = 1768
2R= 2(103 + 124+ 123) = 2(35) = 70
So that
Ay [19+ 1768 + 70]
= 7 [265 8] =443 :. A = 443 units 2
6 6
■'■J 2 /(x)dx^443
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 30 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°)
00000
00000
77/18(10°)
01736
04166
tt/9 (20°)
03420
Leave the righthand side of
77/6 (30°)
05000
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 lefthand 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
01736
03420
05000
06428
07660
08660
00000 
04166 
05848 
07071 
08016 
08752 
09306 
_ 
_
, 
 " '
Now form three more columns on the righthand 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
OOOOO.,
(a) You start in column 1
04166
(b) You then zigzag down the
"05848
two righthand columns
0707 K'
(c) You finish back in column 1 .
.08752
.'08016
09306'
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
09306
79956
27728
1 1 6990
R
1 9989
4
79956
13864
2
27728
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
0681
For:
A [(F + L) + 4E + 2R]
 E ^ [116990]
^tt/54 [116990]
^06806
•f
tt/3
Vsinxdx —0681
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
00000
00000
7i7l8(10°)
01736
04166
04166
tt/9 (20°)
03420
05848
05848
tt/6 (30°)
05000
07071
07071
2?r/9 (40°)
06428
08016
08016
5tt/18(50°)
07660
08752
08752
tt/3 (60°)
08660
09306
09306
F+L
>■
09306
19989
13864
4E
2R
t 4E + 2R
>
79956
27728
4
2
79956
27728
(F + L)
116990
I [(F + L) + 4E + 2R]
^ [116990]
= 06806
•tt/3
Vsinx dx —0681
Now we will tackle example 2 and set it out in much the same way.
Turn to frame 25.
527
Approximate Integration
•10
Example 2. To evaluate I VO + x 3 )dx, using 8 intervals.
J 02
First of all, find the value of s in this case.
25
oi
For s =
l002 _0
8 8
26
oi
01
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 02 to 10.
x
x 3
1 +x 3
V(l+x 3 )
F + L
E
R
02
0008
1008
10039
03
0027
1027
10134
04
0064
05
0125
06
0216
07
0343
08
09
10
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 02
Check with the next frame. _
27
528
Programme 19
28
■10
V(l +x 3 )<±c = 0911
J 02
X
X 3
1 + x 3
V(i+* 3 )
F + L
E
R
02
0008
1008
1 0039
1 0039
03
0027
1027
10134
10134
04
0064
1064
10316
10316
05
0125
1125
10607
10607
06
0216
1216
11027
11027
07
0343
1343
11589
11589
08
0512
1512
12296
12296
09
0729
1729
13149
13149
10
1000
2000
14142
14142
F + L
4E
2R
4E + 2R
24181
45479
33639
2
(F + L) +
181916
67278
4
181916
67278
y
273375
I =  [(F + L) + 4E + 2R]
There it is. Next frame
= ^y [273375] =^ [273375] =09113
.'. f y/(l+x 3 )dx* 0911
J 02 "
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
10 125 150 175 200 225 250 275 300
245 280 344 420 433 397 312 238 180
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=025
025
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.
662
I= [(F + L) + 4E + 2R]
= ^ [7943] = 662
025
X
y
F + L
E
R
10
245
245
125
280
280
150
344
344
175
420
420
200
433
433
225
397
397
250
312
312
275
238
238
300
180
180
F + L^
425
1335
1089
4E^
5340
2178
4
2
2Rh>
5340
2178
(F + L) + 4E + 2R— ►
7943
[7943]
■J;
y dx —662
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)
05 10 15 20 25 30 35 40
400 794 1168 1497 1739 1825 1608
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
465 cm
t
V
F + L
E
R
000
000
05
400
400
10
794
794
15
1168
1168
20
1497
1497
25
1739
1739
30
1825
1825
35
1608
1608
40
000
000
F + L*
000
4915
4116
4E»
2R— ►
h L) + 4E + 2R — ►
19660
8232
4
2
19660
8232
(F
27892
x=[(F + L
) + 4E + 2
R] and s = 05
.'. X
= •^[27892] =4649 .\ T
otal distance —465 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 &ij(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
+ yinl)
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 evennumbered ordinates,
2R = 2 X sum of remaining oddnumbered 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 02
2. Evaluate \ x l e 2x dx correct to 3 decimal places.
J 01
3. The values of a function^ =f(x) at stated values of x are given below.
X
20 25 30 35 40 45 50
55
60
y
350 620 722 680 574 503 621
872
1110
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 ~~ 05 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
03 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 (54co,*)* dfl
{C) )o Vdisin 2 ^)
4. The coordinates of a point on a curve are given below.
X
1
2
3
4
5
6
7
8
y
4
59
70
64
48
34
25
17
1
The plane figure bounded by the curve, the xaxis and the ordinates
at x = and x = 8, rotates through a complete revolution about the
xaxis. 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 xaxis, 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 001 s, over a half
cycle, is found to be: 0, 195, 35, 45, 405, 25, 205, 29, 27,
125, 0. By Simpson's rule (10 intervals) find the r.m.s. value of the
voltage over the halfcycle.
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 = 27T, 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 nonzero 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(25  15 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 COORDINATES SYSTEM
Programme 20
1
Introduction to polar coordinates
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 coordinates, i.e. (x,y)
(ii) by polar coordinates, 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 coordinates 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 coordinates in an
earlier programme on complex numbers. In this programme, we are going
to direct a little more attention to the polar coordinates system and its
applications.
First of all, an easy example or two to warm up.
Example 1. Express in polar coordinates 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 coordinates of P are
539
Polar Coordinates System
(5385, 158°12')
(i) r 2 = 2 2 + 5 2 = 4 + 25 = 29
.. r = \/29 = 5385 .
(ii) tan E ==04 /. E = 21°48'
:. e = 158°12'
Position of Pis (5385, 158°1 2')
A sketch diagram will help you to check that is in the correct quadrant.
Example 2. Express (4, 3) in polar coordinates. 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 ==075 :. E = 36°52'
(4,3) = (5,323°8')
.. 6 = 323°8'
Example 3. Express in polar coordinates (2, 3).
Finish it off and then move to frame 6.
3606, 236°19'
Check your result . 8 ^
(i) r 2 = 2 2 +3 2 = 4 + 9= 13
x r = Vl3 = 3606
(ii) tan E ==l5 /. E = 56°19'
.. e =236°19'
(2,3) = (3606,236°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 coordinates.
Do that, and then mov e on to frame 7.
540
Programme 20
Working
(2796,4145)
(i) x = 5 cos 124° = 5 cos 56°
= 5 (05592) =27960
(ii) y = 5 sin 124° = 5 sin 56°
= 5(08290) = 41450
..(5, 124°) = (2796, 4145)
That was all very easy.
Now, on to the next frame.
8
Polar curves
In Cartesian coordinates, the equation of a curve is given as the general
relationship between x and_y, i.e. y =f(x).
Similarly, in the polar coordinate 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
05
0866
1
0866
05
r=2sind
10
1732
2
1732
10
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 Coordinates System
Here is the complete table.
sin 6
r = 2 sin
30
05
10
60
0866
1732
90
1
2
120
0866
1732
150
05
10
180
e°
sin 6
2 sin 6
210
05
10
240 270
0866 1
1732 2
300 330 360
0866 05
1732 10
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
05
025
0866
075
1
1
0866
075
05
025
r 
= 2 sin 2
05
15
22
15
05
e
210
240
270
300
330
360
sin 6
sin 2
05
025
0866
075
1
1
0866
075
05
025
r= 2sin 2
05
15
2
15
05
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 Coordinates 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)
(reentrant loop)
(no cusp or reentrant
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 Coordinates 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_
= 384
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_
= 38
A= 384 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 Coordinates 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 =2178
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 =2178
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 Coordinates 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 .2nr^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 =zirr° sin 6
ax 3
and
.. V =
29
C°2 1
V= rjrr 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 Coordinates 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 Coordinates 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 rewritten 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).
:. 5S5s. 27rPLSs.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 Coordinates 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^, cosdd
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
[01]
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 Coordinates 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 041
.". 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(01)
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 Coordinates 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 Coordinates 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
= (mk)
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 xdirection 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
xdirection 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 =(sr).(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=135
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
= 1845 = 135
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) (21)
= 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.dydz
•'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 + 2yz)dx. dy. dz
Jl JlJ
I = 8
Did you manage it first time? Here is the working in detail.
1 = 1 I I (x + 2yz)dx.dy.dz
J 1 J— lJ
ft
~ + 2xy  xz
dy.dz
= j j (2 + Ay  2z) dy. dz = f
= (" {(2 + 22 2 )(2 + 2 + 2 2 ))rfz =f(4
3
2y + 2y 2  2yz
■3
1
dz
4z)dz
4z  2z 2
= (1218)(42) = 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 +93r 2 )dr
IQrr 3
J 1
= 129 = 3
;(20 ~8) (101)
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.
=42=2
.y
c?x
=  (53)dJC = f 2cfr =
2x
lii) 1= lydydx =\ y 2 dx = \ (9x 2 l)dx
J oJ i J oL J * Jo
3x jc
= 1924= 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
(88)(l4)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 ^raxis 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)
= 08.^= 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 dy8x
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)(lf))
178«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 righthand 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:
(xx 2 )d>
•iy
dy\ (xy)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 = 45, Ex.4. I = £
= I dx I (x  x 2
Jo J o
■f
Jc
)rfy
xy  x 2 y
l JO
= dx(xx 2 ) = \ (xx 2 )dx
JO Jo
2 3
f94*
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 (2y5*)dy.
PV/ienjow /lave completed both of them, turn on to frame 34.
579
Multiple Integrals
Working
(i) 128, (ii) 545
(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 (2y5x)dy
= dx ly 2  5xy
J 1 L J y = Q
= f dxlx5x V2
f
(x  5x zl2 ) dx =
  2x 512
2 .
= (864)(^2)
= 56+ 15 = 545
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 = yi 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 xy 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 righthand 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[l3224J= 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 righthand
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) (2cos03sin30)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 xaxis, 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 xaxis 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 (9r 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 xaxis 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 ;cyplane 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 coordinate 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 thatr^ + 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^ Vy =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
:.^ = lAx 2 = l4
dx x 2
From the given equation, — = y x ■'■ A = x(y—x)
■ dZ = i x(yx)
" dx x 2
= 1
_y~x _x~y+x _2x~y
:.x*L=2xy
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 tc2dT 2 ( 2ndorder )
x& 2xy
dx
y = Axe*
gives the diff. equation xjy{\ +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 xr = ly " " a function having arbitrary
dx
j2.
constants.
(iii) ry + 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 formj = 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 rewrite the equation in the formp= ^ = 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 formf^ =f(x, y), the variable y on the
righthand 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
—— = pjrr ,i.e. equations in which the righthand 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 rewrite 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= i2Tx 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
JTTy 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
'xfactors' and '_yf 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 'xfactors' and '^factors'
dy_ y 2 (l+x)
dx x l (yl)
Now rearrange the equation so that we have the 'yfactors' 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
Rearranging, we have
■'• In y + y x = In x  x 1 + C
.". lny+~=\nxj+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
rln*— r = lnjc + C
2 y + 1
:. ln^4 = 21nx + lnA
Ax 2
y + i
y1 =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, rearrange the equation into the form
¥{y)dy=f{x)dx
i.e. arrange the '^factors' and dy on the L.H.S. and the 'xfactors' 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 'xfactors' 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
xf ~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 xj = (4 +y 2 ) see 2 *
First separate the variables, i.e. arrange the '_yfactors' and dy on one
side and the 'xfactors' and dx on the other.
So we get
22
y
4+y
sec 2 x ,
? dy = tlnx 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 xr = 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^xp = 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 'xfactors' and '_yfactors', 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 + xt
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 +Xr= , . ~
dx 1 + 2v
dv 2v + 3v 2
v = y
dx 1 + 2v
2v + 3v 2 v2v 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 = r5
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 xy
. dy , dv
y = vx ..= v+x 
. ^ dv 1 +v
• ■ v + x T = 1
dx 1 v
dv 1+v
xr~. v =
dx lv
x +y _ 1 + v
xy lv
1 + vv + v 2 1+v 2
1 v
lv
••• \\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
xy
^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 = (v1) 2
X dx 2 V 2 '2
•'•j(^T) 2dv= B
dx :. 2 r=lnx + C
v1
But „=Z and r=
x 1 v
On to frame 41.
= lnx + C
2x
xy
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^^xyy* :. 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 vv 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 —yx.
If we compare this withf^ + 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.
eg
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 ~lx 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)
(x2)£,(*2)'
J—J^
dx = ln(x2)
. IF = e ln (x 2) = e lnj(x 2) * j = (jf _ 2) l
1
x2
(*2)
= (x  2) dx
_(x2) 2
dx
+ C
•'• y
^ + C(x
C(x  2) ... General solution.
When* = 4, ^=10
10 = +C2 :. 2C = 6 :. C = 3
.. 2^ = (x2) 3 +6(x2)
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.
tanxr + 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& righthand 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 = {Cxx 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 = Cxx 2
z=y' x :.—=Cxx 2 :.y=(Cxx 2 T
But
Right! Here is another.
Example 2. Solve x 2 y x 3 — =y i cosx
First of all, we must rewrite 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*
JpdxJ
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 rewrite the equation in standard form j^ + ?yQy"
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= yl3 =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 = \(x + 1) x dx
"J
= (x 2 + x) G?.X
x 3 JC 2
. X
_ 2x 3 + 3x 2 + A
■ M 2
6
fa
■■ y
2xr 3 + 3x 2 + A
78
Butz=/"
DDDnnDnDnnnnnaDODDannnDDnnDDnnDnnnnnnD
There we are. You have now reached the end of this programme,
except for the Test Exercise that follows. Before you tackle it, however,
read down the Revision Sheet presented in the next frame. It will
remind you of the main points that we have covered in this programme
on first order differential equations.
Turn on then to frame 79.
628
Programme 22
79
Revision Sheet
1 . The order of a differential equation is given by the highest derivative
present.
An equation of order n is derived from a function containing
n arbitrary constants.
2. Solution of first order differential equations.
(a) By direct integration: —7f(x)
gives y =Jf(x)dx
(b) By separating the variables: F(y). f =f(x)
gives F0>) dy = I fix) dx
(c) Homogeneous equations: Substitute y = vx
, dv _ „, ,
gives v + x—= Hv)
dx
(d) Linear equations: r + Py = Q
Integrating factor, IF = e' p dx
and remember that e to F = F
gives y IF=\Q.IF dx
(e) Bernoulli's equation:— j + Py = Qy n
Divide by y n : then put z =y 1 ~ n
Reduces to type (d) above.
DDODODDDDDDDODODOODDDDDDDDDDDDDDDDDDDD
If there is any section of the work about which you are not perfectly
clear, turn back to that part of the programme and go through it again.
Otherwise, turn on now to the Test Exercise in frame 80.
629
x
First Order Differential Equations
The questions in the test exercise are similar to the equations you
have been solving in the programme. They cover all the methods, but are
quite straightforward.
Do not hurry: take your time and work carefully and you will find
no difficulty with them.
Test Exercise— XXII
Solve the following differential equations:
1. x^=x 2 +2x3
dx
2. (1 +,)'£= l + „»
3. ® + 2y=e 3 *
dx
a dy 2
4. xiy =x
dx
5. x 2 r = x 3 sin3x + 4
dx
6. x cos v^— sin v =
dx
7. (x 3 + xy 2 )^ = 2 y 3
dx
8. (x 2 \)^ + 2xy=x
dx
80
4h t
9. — f + y tanh x = 2 sinh x
dx
dy j
10. x—f2y=xcosx
dx
11. ^ + ^y 3
dx x
12. x d f+3y=x 2 y 2
dx
630
Programme 22
Further ProblemsXXlI
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 (xl)£=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. y3x + (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 izly=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 + 3y4)=^=(x+y) Putx+y = v
17. (yxy 2 ) = (x+x*y)® ?nty = V 
18. (xy\) + (4y + xl)^=0 Putv = x1
19. (3y7x + 7) + (7y3x + 3)^=0 Putv=x1
20. y(xy+i)+x(l+xy+x 2 y 2 )^=0 Put^=—
V. Bernoulli's equation
21. f x+ y=xy>
22. %*yf*
23. 2^+y=y\xi)
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. (lx 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 xj + 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 ofp + — : 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 lx 2 j y lx 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 radioactive 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 432 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 r2+b—+cv =
Adding these two lines together, we get
N °» £ <« + ') =f + 1 » d i? <" + "> 0 + S ,te " fore ,te
equation can be written
a32 (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 arr + b^+ cy = 0. If a = 0, we get the first order
equation of the same family
br + 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
lefthand 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
mm^ 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—^ + bf+ 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— \ + 3p+ 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 ofp£ + 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. —^ + 5f^+ 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 mm 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(CD)
.". If m = a ± j3, 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(1636) 4±V20
_ 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±V36
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 arp + 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 formpr ± 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 asp£ ± 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 ± j3, 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 jr + 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 at+ fer+ c^ =/(*), the substitution
y = Ae miX + Be™ 2 * would make the lefthand 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 righthand 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] 6D10C = :. 6D = i£ =  ■'• D = TS"
[CT] 2C5D + 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) = 2x3
2. fix) = e sx
3. fix) = sin Ax
4. /(x) = 3  5x 2
5. fix) = 21
6. fix) = 5 cosh 4x
When you have decided all six, check your answers with those in frame 27.
650
Programme 23
27
Answers
1. /(x) = 2jc3
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 =
.'. (m2)(m3) = :. 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
05(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
20C10D = 2 40C20D = 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(3640) = 6± V4 _
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
6D12C = 2 /. 6D + 24D=2 .\ 30D = 2
°rV
C 15
P.I. is y = rF (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
2D6C = /. 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
(ml)(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 jt+ « 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. g4y»10^
4. j^ + 25y = 5x 2 +x
6. Hr + 4r+ 5v = 2e 2X , given that at x = 0,y = 1 and r=2.
dx L dx dx
dx 2 dx
1. 3^ r i2 : ±y = 2x3
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 steadystate
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 jj + 3 —  + 2y = 3 sin*, given that when jc = 0,>> =09
and^ = 07.
dx
16. Obtain the general solution of the equation
d 2 y , 6 Q[y
dx 2 dx
y + 67+ 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 steadystate solution are the roots of
6 tan 3? = 7.
d 2 y
18. Solve the equation— ^= 3 sinx4.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 = 01
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 T5 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)
= 210x20 + 4x 2 + 16x4
= 4x 2 + 6x  22
D (x 2 +4xl)= 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 r2 {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 + 152)
= 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
D2
2
e.g.
D 2 + 3
1
{e 3x }
52 3
2 e 3
'9 + 3 12
i ( e 2X\ = „2X
• D 2 4D1 ( ' (~2) 2 
1
■4(2)l
„2X
4 + 81 11
So (D 2  5D + 4) {e 4x }
for (D 2 5D + 4){e 4 *}=e«(4 2 5.4 + 4)
= e 4x (1620+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+ 182
18
19
25
Fine. Turn on now to frame 20.
676
Programme 24
£\j Just for practice, work the following:
(i) (D 2 +4D3){<? 2 *} =
(iii) (D 2  7D + 2) {W 2 } =
<*> D'3D2 ^> =
^ (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 +4D3){e 2 *}=e 2 *(4 + 83) = 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 +2D3){e 2 *sinx}
= e 2X [(D + 2) 2 + 2(D + 2)  3] . {sin x]
= e 2X (D 2 +4D + 4+2D + 43) {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 + 255] {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+ 168D32 + 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 (Dl)
Right, so we get
(D 2  3D + 4) {e x cos 3s}
= e x [(Dl) 2 3(Dl) + 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+4D8+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 (lD 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
_ 15 {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 77, 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=lor3.
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(1620)
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"
14+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 +
=3A4B +  /. 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
ie. 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 =
10D9
(10D + 9)(10D9)
10D9
{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)
_ 10D9 f „ A ,
'~100D 2 81 {3C ° S4 * }
_ 10D9 f „ . ,
"160081 t3cos4x}
168
j(10D9) {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}
: 160081 {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
444^+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(1652)
_ 4 + V36
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
94D+ 13
= 4(1D) {2sin3x}
 2 _L_/ • i \
_1 1+D f . . ,
~2 lD 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 7367^+ 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 + 46D12 + 5^ S ' n3;C ^
1
D 2 2D3
Now, applying theorem III, gives
y =
— 7 {sin 3x]
y = e2X   9 2D3 ^ 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.
D6
€1
2
D6
■• y = 
So the P.I. is finally
y
D6 ( . x
D6 r . x
IX
^ (D6) {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(ff^)
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 + 96D18 + 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. 75 + 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+ 1616
(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 +j4D4 + 4
(1)
= 3 5^ j2 *^. 1
DD+J4
1
{e ox } putting e w for 1 .
= 3jM 2 *jj.e°*
1
0+J4
,j«ll'
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 2xj 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 +j6D9 + 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
jT  47+ 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 operatorD methods throughout.
ananQnDnQOQQnaQDnaaaononDDnnonQnDaoano
Solve the following equations by the use of the operator D.
1. D 2 y + 2Dy3.y = 4e _3X
2. D 2 y + 3Dy + 2y=xe' x
3. D 2 y+y = sir\x
4. D 2 y2Dy +y = sin x + x 2
5. D 2 y3Dy + 2y = 4e x sinhx —
6. D 2 y5D.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 y6Dy + 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 yDy6y=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 +Dyy = e x sin 2x
20. D 2 y + 2Dy + 5y = x + e x cos 3x
21. D 2 j2D>' + 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) 29J2, (ii) j2, (iii) 111+J56, (iv) 1 +j2
3. (i) 5831 1S9°3', (ii) 6708 jl53°26', (iii) 6403 231°24'
4. (i) 35355(1+ j), (ii) 3464 j2
5. x= 105, 7 = 43
6. (i) 10ei° 6so , (ii) 10 eJ 0650 ; 2303 +J0650, 2303 J0650
7. je
Further Problems I (page 33)
1. (i) 115+J133, (ii) 252+J064, (iii) cos 2x +j sin 2x
2. (22j75)/41
3. 035+J017
4. 07,09
5. 244+J228
6. 12 + jl6
8. x=18,y=l
9. a=2,b = 2Q
10. x = ±2,y = ±3l2
12. a=l5,Z> = 25
13. V2e j2 ' 3562
14. 26
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. 5831 l210°58'
2. (i) 1827 +J0813, (ii) 3993 J3009
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 = 027, >> = 053
2. 3+jV3; J2V3
3. 3606 56°19' , 2236 296°34' ; 1213 J3584; 378eJ 1 ' 244
4. 1336(12T, [99^ , lilll. [243.°. 1315° )
1336 (eJ°' 4712 , e^ 1 ' 7279 , ei 2 ' 9845 , e J 2  0420 ; e Jo78S4^
708
Answers
5. 2173 +J0899, 2351 eJ 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' ; 0953  jO304
11. 1 401 ( )58°22' , 1 130°22' , 202°22' , [274°22\ 346°22 ');
p.r.= l36j033= 1401 «*•«"
12. 036 +J055, 164 J255
13. je, i.e. J2718
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. 6725
2. 1940
709
Answers
4. COth A
5. (i) 12125, (ii) ±06931
6. x = 03466
7. (i) y = 224, (ii) x = ±4812
8. sinx cosh y — j cos x sinh_y
Further Problems III (page 98)
2. x = 0,x = 0549
5. (i) 09731, (ii) 1317
7. (i) 09895 +J02498, (ii) 03210 + jO3455
10. x = 0, x=jln2
12. x = 03677 or 10986
14. 1528 +J0427
18. 1007
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, 1654, 6654
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 = 25,y = 3, z = 4
5. x = 2, y= 15, z = 35
6. 4or14
7. 5 or 27
8. (a) 0or±V2, (b) (a b)(b c)(c a) (a + b + c)
9. x= 1 OTX = S±y/34
10. jc = 15
11. 2(ab)(bc)(ca)(a + b+c)
12. / 2 =52
13. (a + Z>+c) 2 (afc)(6 c )(ca)
14. 2 or 16/3
15. (xy)(yz)(zx)(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) 2i7/18*
3. (02308, 03077, 09230)
4. (i) 6, = 82°44'; (ii) 4705, 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, I7i7j+2k, = 66°36'
5. (i) 7, (ii) 7(i/fc), (iii) cos 0= 05
6. cos = 04768
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) 6853, (01459, 05982, 07879)
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
1x 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 lx 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 = 16x76
3. _y = ;c
4. > = 2598x  3849
5. R = 477; C: (470, 502)
6. R = 559; C: (35, 275)
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.x144sin0; 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 = 625; C: (0,225)
(ii) R=l; C:(2,0)
(iii) R = ll68; C: (1226, 65)
11. R = 0177
14. R= 2744
17. p = t; (h, k) = (cos t, sin t)
18. R = 1054, C: (11,333)
20. (i)j> = ± £, (iii) R = 05
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 = 03679 at* = 1; P of I at (2, 0271)
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 (025, 4375)
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= 15
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) ±07071, (ii) 0, (iii) ± 129
13. 0606
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>) ff=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. ± 25 %
Further Problems IX (page 273)
10. + 105EX 10~ s approx.
12. ±(x+y + z)%
13. y decreases by 19% approx.
W ±425%
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 0419 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 = 28333
« ,s lJ' /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. 2488, 25945
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) 06, (ii) 05
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. 1x 2 +^ ^C +
2!'
45
719
Answers
X 2 5x*
5. x + x 2 +^+
6. 10247
12 1
7  ® ~10' ^"9"' (iii ) _ 2~
8. 085719
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 )* ; 1426
r 1
x 2 x*
13. lncosx =  yj~ —
16. (i) ", (ii)y
17. ly + 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(lnx2) + C
3. tanxx + 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(x5)jln(x3) + C
11. 21n(xl) + tan _1 x + C
1? _ / cos 8* cos2jc \
I 16 4 / L
Further Problems XIII (page 385)
1. ln{A(xl)(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 + n1n (vV\
2
22. ^(e 2w  1) = 5345
23 ±
ZJ  24
722
Answers
24. ^( 3e /'2
3co
26. ^L? + C
27. il n (x4)^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 +6x6tan^x + C
33. In (cos 6 + sin 6) + C
34. tan 6 seed +C
20
35. ^ln(xl)+yln(x2)^ln(x + 3) + C
1
37. ln2 1 
31nx+yln(jc 2 + 4)  jtaif'd) + C
38
39. lnxtan'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 [ Vl33 + 2tanx/2 \ + c
' >Jn n lVl3 + 32tanjc/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. 7rln{ Vl + C
V5 l2x + 5+V5)
3x 2
12. y 4x + 4taiT 1 x + C
13. £±1 V(32x^ 2 ) + 2sm 1 (^j 1 ) +C
14. 7T
15. cosh^^^ + C
16. ^■tan _1 tan^ +C
17. l ln ptanxl ^
5  tanx + ? '
18. l
19. 3sin _1 xV(l* 2 ) + C
20. TjtaiT 1 V3tanYfx + 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 = j12tt + 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 =— ^— + — — cotxx + C
11. I 3 =x{(lnx) 3 3(lnx) 2 +61nx6}+C
Test Exercise XVI (page 452)
1. 7012
2. — +2tt= 3175
■n
3. ln6 = 2688
4.
73485
5.
I RI2
6.
1323
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(l2e _1 )
16. 283
17. 3901
18. ^ 0? +13)
20. 1361
Test Exercise XVII (page 477)
1. (075, 16)
2. (31, 0)
3. 5tt 2 a 3
4. <l±l
728
Answers
5.
7035 n
6.
5.7T 2
8
7
e"2
Further Problems XVII (page 478)
L f 6 + T ln2
2. (i) 2054, (i) 6628
 64jr£ 3
15
4. (i) (04, 1), (ii) (05, 0)
6. 24
7.
8.
9.
17
12
19
20
U
5
10. A = 2457, V=47rV3, J=l409
12. (i) 8, (ii) ^, (iii)i
13. 1175
16. V= 254 cm 3 , A = 4665 cm 2
17. S =1531 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  T16
15. (i)iV(^ 2 +e + l), (ii) yjfy
16. 512w
17. 946 cm
(157T32) a
4(3vr4)
730
Answers
Test Exercise XIX (page 534)
1. 0946
2. 0926
3. 267
4. 1188
5. 1351
Further Problems XIX (page 535)
1. 0478
2. 0091
3. (i) 06, (ii) 6682, (iii) 1854
4. 560
5. 1586
6. 0747
7. 284
8. 2892
9. 0508
V2 r W2
10  J y/(9 + cos2B).dd; 499
J
Y" "V* "Y*
11. tan 1 x=jcj+yy; 0076
12. (i) 05314, (ii) 0364
13. 2422
14. 205
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.
2125 a
3na
U  2
14. {V(fc 2 + 1)} O fie i  1) ; £ (e^i  1)
15. 7ra 2 (2V2)
Test Exercise XXI (page 588)
1. (i) 0, (ii) 2 f
2. (01, (ii) 168, (iii) ¥■
733
Answers
3. 1367
4. 17067
5. M? + 6
4
6. 54
Further Problems XXI (page 589)
1. I
2.
243 n
2
3. 45
. abc ,, 2 , 2\
4. — (b 2 +c 2 )
5. 
6. 45
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. ^(3tt4)
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. 1956
18. ^(c 2 « 2 )
2
19. (2tt + 3V3)
20. 232
Test Exercise XXII (page 630)
x 2
1. y=Y +2x3 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 +2x2j> + 21n(xl) + 21nO + 1) = A
6. ^ 2 xyx 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 +0l) 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 (C2tan^)
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
(yx) 2 j>x
\/2 sin 2x
2(cos x  sjl)
30. (*4)>> 4 =Ax
31. 7 = a: cosx — sec*
32. (xjO 3 Axy =
33. 2tan" 1 ^ = ln(l+x 2 ) + A
34. 2x 2 y = 2x 3 x 2 4
xy
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 +2x3 =
41. j(1jc 3 ) 1 / 3 =^(1x 3 ) 2 / 3 +C
737
Answers
42. xy + x cosxsinx+ 1 =
43. 2tan _1 .y=ljc 2
x 2 +C
44. j> =
2x(lx 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(Ax)
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+yj(25x 2 +5x2)
5. _y = e* (A + Bx) + 2cosx
6. y = e~ 2x (2cosx)
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. ^ = 4cos4x2sin4x + 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=^(lf) 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 +ye 2f +^(sinf + 3cos0
15. 7 = e 2^ e ^ + — (sinx3cosx)
16. .y = e~ 3 *(Acos;c + Bsinjc) + 5;c3
17. x = e~ f (6 cos t + 7 sin f)  6 cos 3f  7 sin 3f
1 9
18. 7 = sinxysin2x:; y max = 1299 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 fyx)
3. y = A cosx + B sin* — — cosx
4. y=e x (A + Bx)+—cosx+x 2 +4x + 6
5. >>= 1 +e 2x (l2x)
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 operatorD, 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
OperatorD 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 coordinates, 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 oneyear 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
*