THE CALCULUS
FOR BEGINNERS
CAMBRIDGE UNIVERSITY PRESS
ilonUon FETTER LANE, E.G.
C. F. CLAY, MANAGER
Binburgb: io, PRINCES STREET
Berlin: A. ASHER AND CO.
ILfipMs: F. A. BROCKHAUS
fhto Boris: G. P. PUTNAM'S SONS
Bombay nnt) Calcutta: MACMILLAN AND Co., LTD.
Toronto: J. M. DENT AND SONS, LTD.
THE MARUZENKABUSHIKIKAISHA
All rights reserved
THE CALCULUS
FOR BEGINNERS
v
J/W^MERCER, M.A.
Head of the Mathematical Department, Royal Naval College, Dartmouth
Author of Trigonometry for beginners
Cambridge :
at the University Press
1914
First Edition, 1910.
Reprinted, 1912, 1914.
PREFACE
IN writing this short course, I have been guided by my
conviction that it is much more important for the be
ginner to understand clearly what the processes of the
Calculus mean, and what it can do for him, than to acquire
facility in performing its operations or a wide acquaintance
with them. I had much rather that a boy, confronted with
a problem, should, after analysing it, be able to say " If I
could differentiate (or integrate) this function of x, I could
solve the problem " than that he should be able to perform
the operation without seeing its bearing on the problem.
The book is intended primarily for those who are or will
be interested in the applications of the Calculus to Physics
and Engineering.
I believe that boys, especially on the Science side of a
School, could profitably begin the study of the subject at a
much earlier age than is usual. I believe, too, that even
those who are to be Mathematicians would do well to go
through some such course as is here indicated before pro
ceeding to the more formal treatises, for comparatively young
students gain by approaching a subject inductively before
studying it deductively. Even if they are possessed of
mathematical ability, it is good that they should have a
thorough comprehension of the goal in particular cases
VI PREFACE
before undertaking general proofs, and should be reminded by
a preliminary course depending on more laborious methods, of
the advantage of replacing " mere counting " by mathematical
processes.
As I have already indicated, no great prominence is given
to methods of differentiating and integrating complicated
expressions, in fact, x n is the only function whose differential
coefficient is required in the first 250 pages. Later, sin a;,
cosic, etc., e? and log a; are dealt with, and the Engineer
needs little beyond these.
Before reaching " ~ ," I have devoted many pages some
may think too many to the ideas of rate of change and of
the limiting value of the ratio of two continually diminish
ing quantities. Unless the ground is cleared in this way,
a boy is apt to begin and continue indefinitely his study of
the subject in a state of mental fog. I do not believe that
anything is gained by hurrying over the early stages. On
the contrary, I believe that the ultimate result of reducing
the time spent on them is a very great loss of power.
When cc n has been differentiated, the result is applied
to problems of maxima and minima, geometrical properties
of curves, approximations, etc. Then the inverse problem
"Given ~, find y" is discussed, and this leads up to the
CLOD
Integral Calculus and its application to Areas, Volumes,
Work, etc. These should not be looked upon as so many
detached problems each requiring its own particular rule,
but if the meaning of integration is properly understood,
they will all be seen to depend on one general principle.
This will explain why I have given so much space to the
discussion of a few typical problems ab initio.
After this, rules are investigated for differentiating
PREFACE Vll
sin a;, cosx, tana?, etc., e*, \og e x, products, quotients, etc.,
and more problems are dealt with of the same nature as
before but requiring a knowledge of these rules.
Chapters XII. XIV. form a sort of appendix with short
notices of the application of the Calculus to the Solution of
Equations, Methods of Integration and Polar Coordinates.
No great amount of previous mathematical knowledge
is assumed. A boy is supposed to know his Elementary
Algebra and Trigonometry and to have some slight acquaint
ance with the coordinate geometry of the straight line.
He should be able to write down the equation of a line
through a given point with a given gradient and should know
the relation between the gradients of perpendicular lines.
He is also expected to have had some practice in drawing
graphs from their equations and to know what these graphs
mean. Many of the illustrations given are arithmetical,
algebraical and geometrical, but in addition to these many
will be found drawn from the scientific knowledge usually at
the command of boys of this age.
Several of the examples are intended to lead up to the
ideas afterwards presented in the text, and in the more
difficult ones, especially in some which will be found in the
miscellaneous sets, the boy of mathematical ability will find
plenty of scope for his skill in the manipulation of symbols.
It would probably be well to omit the section on Points
of Inflexion (pp. 127 135) on a first reading It is found
there, because there did not seem to be a more convenient
place for it.
I take this opportunity of thanking Professor Hobson of
Cambridge for some notes on the earlier chapters, which he
sent to me after seeing the manuscript, but this must not
be taken to mean that anything I have written has his
sanction.
Vlll PREFACE
I am very grateful also to Mr A. W. Siddons of Harrow,
for some criticisms and suggestions which he made after
wading through most of the original manuscript.
The subject has been taught here during the last few
years to boys of 16 on the lines of this book and I am much
indebted to my colleagues who have helped in various ways
during the passage of the book through the press.
Thanks are also due to Professor Lamb of Manchester for
permission to reprint two tables from his "Infinitesimal
Calculus," to Professor Ewing, Director of Naval Education,
for permission to use questions set to Naval Officers, and
to the Controller of His Majesty's Stationery Office for
permission to include some questions set at Examinations
by the Board of Education and the Civil Service Com
missioners.
J. W. MERCER.
EOYAL NAVAL COLLEGE,
DARTMOUTH,
June 1910.
CONTENTS
CHAPTER I
FUNDAMENTAL NOTIONS
SECTIONS PAGES
1 37 Uniform speed, average speed, speed at an instant . 1 46
Rate of increase.
Function.
Uniform gradient, average gradient, gradient at a
point.
Tangent at a point of a curve.
Rate of increase as gradient.
Test of approximate equality.
Ratio of continually diminishing quantities.
Ess. I. VII.
DIFFERENTIATION FROM FIRST PRINCIPLES
38 71 Notation. Meaning of A, etc 47 74
Meaning of = . Its connection with r .
dt A*
dy f .
f as a rate of increase.
dx
(%/U
f as ratio of time rates of increase of x and y.
dx y
f as gradient
(HP
X CONTENTS
SECTIONS PAGES
Formal definition of / and summary of process of
cLx
finding it.
Speed and acceleration as gradients.
Differentiation of # 3 .
Geometrical illustration.
Sign of g.
Exs. VIII. XIII.
CHAPTER III
DIFFERENTIATION OF X n
7295 Notation 75101
Differentation of x n (i) with, (ii) without
Binomial Theorem.
Differentiation of kx n , a? n +c.
Graphical and kinematical illustrations.
Use of j in drawing a curve from its equation.
Applications to Geometry.
Some properties of the parabola.
The functional notation.
Higher differential coefficients.
Derived curves.
Relation between the signs of ~ , ~^ and tho
shape of a curve.
Exs. XIV. XXII.
Miscellaneous Examples on Chapter III. . 102105
CHAPTER IV
MAXIMA AND MINIMA
96117 Turning points of a curve .... 106135
Tests for discriminating between maximum and
minimum.
Points of inflexion.
Exs. XXIII. XXV.
CONTENTS XI
CHAPTER V
SMALL ERRORS AND APPROXIMATIONS
SECTIONS PAGES
118127 The relation Ay = j . tuc approximately . . 136144
Relative error.
The relation f(x + h)=f(x) + hf'(x} approxi
mately.
Exs. XXVI. XXIX.
CHAPTER VI
THE INVERSE OPERATION
128133 Given ^, find y ...... 145152
CLX
 =
gives a family of curves.
/
Exs. XXX. XXXIV.
Miscellaneous Examples on Chapters T. VI. . 153 159
CHAPTER VII
INTEGRAL CALCULUS. AREAS OF PLANE CURVES. MEAN ORDINATE
134 167 Areas by summation 160 204
Meaning of integration.
n
Meaning of / f(x) . dx.
J a,
Definite and indefinite integrals.
Approximate rules for Area : Trapezoidal and
Simpson's rules.
Sign of Area.
Approximation to value of definite integral.
Xll CONTENTS
SECTIONS PAGES
Area represented by difference of ordinates of
integral curve.
The relation f(x + A) =/(*) + hf(x) + } ^f"(x)
A
approximately.
Mean ordinate.
Exs. XXXV. XLVII.
FURTHER APPLICATIONS OF THE INTEGRAL CALCULUS
168 202 Distance from speedtime formula . . . 205 253
Speedtime and space time graphs.
Work done in stretching string.
Aspects of integration.
Resultant thrust on immersed area.
Work done by expanding gas.
Volume of solids of revolution (frustum of cone,
sphere).
Moment of inertia of rectangle.
Moment of inertia of circular disc.
Radius of gyration.
Theorems of perpendicular and parallel axes.
Centre of gravity (parabola, paraboloid of
revolution, quadrant of circle).
Centre of pressure.
Mean values.
Mean speed with respect to (i) time, (ii) distance.
Exs. XLVIII. LVI.
Miscellaneous Examples on Chapters VII. and
VIII. . 254260
CHAPTER IX
DIFFERENTIATION OF TRIGONOMETRICAL RATIOS
203213 Exs. LVII. LIX 261277
CONTENTS Xlll
CHAPTER X
HARDER DIFFERENTIATION
SECTIONS PAGES
214 244 Differentiation of a product of two or more
functions 278310
Differentiation of a quotient.
Differentiation of a function of a function.
r when y and x are given as functions of a
third variable.
Integration sometimes simplified by the use of
a third variable.
Differentiation of inverse functions.
Differentiation of inverse trigonometrical func
tions.
Differentiation of implicit functions.
Exs. LX. LXXIV.
CHAPTER XI
DIFFERENTIATION OF n x
245267 Ify=* ^ = N.y 311344
Approximate values of N for different values
of n.
Definition of e.
Relation between the logarithms of the same
number to different bases.
Graphic treatment.
Differentiation of \og e x, log a #.
Approximation to value of e.
Hyperbolic functions.
Logarithmic differentiation.
Compound Interest Law.
/ j\n
e as Lt ( 1 +  ) .
*oo \ */
Exs. LXXV. LXXXIII.
Miscellaneous Examples on Chapters IX. XI. 345361
xiv CONTENTS
CHAPTER XII
APPROXIMATE SOLUTION OF EQUATIONS
SECTIONS PAGES
268277 Graphical solution 362374
Successive approximation by calculus.
Exs. LXXXIV.
CHAPTER XIII
METHODS OP INTEGRATION
278282 List of standard forms 375384
r j r
The formula I $ (u) . ^ . dx = I <f> (u) , du.
Integration by substitution. 1
Integration by parts.
Exs. LXXXV. LXXXVIL
Miscellaneous Exercises on Chapter XIII. . 385 387
CHAPTER XIV
POLAR COORDINATES
283289 Meaning of polar coordinates .... 388394
Angle between tangent to a curve and the
radius vector.
Areas.
Exs. LXXXVIII., LXXXIX.
ANSWERS 395432
Table of exponential and hyperbolic functions
of numbers from to 2'5 at intervals of '1.
Table of logarithms to base e 433
INDEX . . 435440
CHAPTER I
FUNDAMENTAL NOTIONS
Speed
1. Uniform, speed. To measure the speed of a moving
body, we must know the distance it travels in some definite time.
The speed is said to be uniform if equal distances are travelled
in equal times, however small those equal times may be. Thus if
a train has a uniform speed of 60 miles an hour, it will travel
30 miles in every half hour, 10 miles in every 10 mins., 1 mile in
every min., 88 ft. in every sec., 8 '8 ft. in every ! sec., and so on,
and we could state the speed as 60 miles/hour, or 1 mile/minute,
or 88 ft. /sec., etc.
We can deduce any one of these expressions for the speed
from any one of the relations between distance and time :
e.g. 60 miles an hour is 60 x 5280 ft. in 60 x 60 sees.
60 x 5280 ,,
or ~n R?T ft  m 1 sec>
60 x 60
or 88 ft. /sec.,
8 *8
8*8 ft. in 1 sec. is^p ft. in 1 sec.
or 88 ft./sec.,
and generally if a body moving with uniform speed travels s ft.
o
in an interval of t sees.,  will have the same value whatever be
I
M. C. I
2 CALCULUS FOR BEGINNERS [CH. I
the duration of the interval and at whatever stage of the journey
c
it be taken, and in this case  ft./sec. is said to be the speed of
the body at every instant.
2. Average speed. Suppose a train travels 60 miles in a
certain hour, but that we do not know that its speed has been
uniform during the hour, we can say that its average speed
during the hour is 88 ft./sec., meaning that if it kept up a
uniform speed of 88 ft. /sec. for an hour, i.e. travelled 88 feet in
every separate second, the distance covered in the hour would be
60 miles, the same as the distance actually covered. Actually it
might cover 70 ft. in one second, 90 ft. in another, and so on,
provided that the total distance was 60 miles.
If a body travels 3 ft. in j sec., we say that its average
speed during this quarter of a second is 12 ft./sec., meaning
that if a body had a uniform speed of 12 ft./sec., it would in
sec. travel 3 ft, the same as the distance actually travelled.
The distance travelled in any other quarter of a second would
not as a rule be 3 feet, unless the motion were uniform ; suppose
for example that in the next quarter of a second the distance
travelled were 5 feet, then the average speed during this next
quarter of a second would be 20 ft./sec. and during the complete
half second 16 ft./sec.
Generally, if a body travels s ft. in t sees., its average speed
during this interval of t sees, is
 ft./sec.,
and as a rule the value of  will depend on the duration of the
t
interval and on the stage of the journey at which it is taken.
13] SPEED AT AN INSTANT 3
EXERCISES. I.
1. A train leaves Bristol at 4.0 p.m. and arrives at Exeter, 75J miles
distant, at 5.55 p.m. Find its average speed (i) in miles per hour, (ii) in
feet per secohd.
2. A train travels 40 miles in the first hour, 35 in the second, 45 in the
third, 50 in the fourth and 40 in the fifth. Find in miles per hour its
average speed for the whole journey.
3. A stone falls 1 ft. in  sec. Find its average speed during this
interval (i) in miles per hour, (ii) in yards per minute.
4. A body travels 0035 ins. in 02 of a second. Find in ft./sec. its
average speed during this interval.
6. A body travels 00063 ft. in 001 of a second. Find in miles/hour its
average speed during this interval.
3. Speed at an instant. Suppose we are studying the motion
of some moving body and that we are able to measure the dis
tances it travels in certain small intervals of time after passing
some fixed mark. In our first experiment we will suppose that
we get the distance travelled in one second. The next experiment
we will suppose to be made with improved apparatus which will
enable us to find the distance travelled in half a second, and
so on, each experiment being a more refined one than its pre
decessor.
Tabulate the results of the experiments as follows :
Time in sees.
Distance in ins.
(1)
1
48
1
(2)
2
24
1
(3)
4
12
1
/ A\
G
*
8
(5)
1
16
3
12
4 CALCULUS FOR BEGINNERS [CH. I
In this case we should probably conclude that the body was
moving with a uniform speed of 48 ins./sec.
If however the results were
(1) 1 48
( 2 ) 5 21
(3) I 975
(4) I 469
(5) 1 230
it is obvious that the speed is not uniform.
What shall we say then is the speed of the body as it passes
the fixed point?
The average speeds during the intervals used in the different
experiments are as shewn in the following table :
Interval in sees. Average speed in ins.jsec,
(1) 48
(2) \ 42
(3) \ 39
(4) I 3752
(5) 1 3680
These results are all different. Now suppose we took the
result of the first experiment and assumed that the body moved
for the whole second with a uniform speed of 48 ins./sec. and
from this calculated how far it would go in ^ sec., we should
get 3 ins., which is by no means the same as 230 ins., the actual
distance travelled. If, however, we used the result of the second
experiment, and assumed that the body moved for half a second
3] SPEED AT AN INSTANT 5
with a uniform speed of 42 ins./sec., and calculated from this the
distance it would go in y^ sec., we should get 2625 ins., and
in the same way if we assumed uniform speeds of 39 and
37'52 ins./sec., we should get 244 and 234 ins. respectively.
These distances 3, 2'625, 244, 234 ins. approximate more
and more to the true distance 2 3 ins. If we tried to deduce
the distance travelled in ^ sec. we should get a better approxi
Oa
mation by using the average speed 368 ins. /sec. than by using
any of the others, but the result 1*15 ins. would not be correct.
Still, if the apparatus at our disposal did not permit us to
measure the distance travelled in any shorter period of time
than y^ sec., we should content ourselves by saying that the
speed of the body as it passed the given point was approximately
368 ins./sec.
If we could improve our apparatus and measure the distance
travelled in ^TWC sec., say 361 ins., we should say that the speed
of the body as it passed the given point was still more nearly
36 ! ins./sec. If from this we calculated the distance travelled
in any shorter interval, say sec., our result 036 1 ins. would
be nearer the truth than a result obtained by using any of the
other average speeds.
But however refined our measurements, we could never by a
calculation of this kind obtain the speed of the body as it passed
the given point. We calculate each time the average speed of
the body during a short interval following the instant in question
and the effect of improving our apparatus is that we are able to
shorten the interval for which the average speed is calculated.
We cannot say that the body moves for any interval of time,
however short, with uniform speed, but the shorter the interval
of time for which we make this assumption the smaller is the
error made in the calculated distance for a still shorter interval.
CALCULUS FOR BEGINNERS
[CH. I
EXERCISES. U.
1. Three men A, B, C begin to observe the motion of a train just as it
reaches a quartermile post. A finds that in the next second it goes 44 ft.,
B that it takes 4 sees, for the whole train (210 ft. long) to pass him, C that
it takes 26 sees, to reach the next quartermile post. Find in ft./sec. what
each would obtain for the average speed of the train. What is the best
available result for the speed of the train as it passes the first post ? If each
assumes that the speed he obtains is uniform over the time cousidered, and
uses it to calculate the distance travelled by the train in the next half second
after leaving the post, what would their results be, and which would be
nearest to the truth? [Nearest tenth of a foot.]
2. A train leaves the terminus A and passes through the stations
B, C, D, etc. at the times indicated in the table. The second column
gives distances from A. [80 chains = 1 mile.]
A
D
C
D
E
F
G
H
I
Find the average speed between the stations in chains (nearest chain) per
minute. What would you guess to be the approximate speed as the train
passed through (i) F, (ii) H ?
n.
ch.
00
Time oj passing
5.20
1
43
5.24
2
41
5.26
3
4
4
81J
04
5.27J
5.284
G
8
34*
5.34
8
12
5.35
Fig 1.
4] SPEED AT AN INSTANT 7
3. A point P starts from A (Fig. 1) and moves round a circle of radius
100 cms. with uniform speed, making a complete revolution in 1 min. PM
is perpendicular to OA. What is OM when L AOP is (i) 60, (ii) 55, (iii) 65,
(iv) 57, (v) 63, (vi) 59, (vii) 61? What is the average speed (cms. /sec.)
of M as L AOP increases (i) from 55 to 60, (ii) from 60 to 65, (iii) from 57
to 60, (iv) from 60 to 63, (v) from 59 to 60, (vi) from 60 to 61? What is
approximately the speed of M when / AOP = 60?
4. A point P starts from A and moves round a circle of radius 100 cms.
with uniform speed, making a complete revolution in 1 min.' OP meets the
tangent at A in T. What is AT when L AOP is (i) 60, (ii) 55, (iii) 65,
(iv) 57, (v) 63, (vi) 59, (vii) 61? What is the average speed (cms./sec.)
of T as L AOP increases (i) from 55 to 60, (ii) from 60 to 65, (iii) from
57 to 60, (iv) from 60 to 63, (v) from 59 to 60, (vi) from 60 to 61 ?
What is approximately the speed of T when L AOP = 60?
4. Now suppose the distance travelled in a given time is not
determined by experiment but given by an algebraical formula, say
s = 5 + 3t + 2t3,
s ft. being the distance from some fixed point at the end of t sees,
measured from some fixed instant, and suppose we want the speed
at the end of 4 sees.
If in the formula we put t = 4, we get s = 145 ; i.e. the distance
from the fixed point at the end of 4 sees, from the fixed instant
is 145 ft.
Similarly the distance from the fixed point at the end of
5 sees, from the fixed instant is 270 ft.
Reckoning from the end of the fourth second, i.e. from the
instant at which we are to find the speed, we see that the
distance travelled in the next second is 125 ft. and the average
speed during this second is 125 ft. /sec. Now calculate the
distance travelled in 4^ sees, from the beginning. It is 200f ft.,
and the distance described in the ~ sec. after the end of the
fourth second is 55f ft. .'. the average speed during this ^ sec.
is 55^= or 111^ ft. /sec.
Similarly find the distances and average speeds for 2, !, '01,
8 CALCULUS FOR BEGINNERS [CH. I
001, 0001 of a second after the end of the fourth second.
The results should be as tabulated.
Time in sees. Distance in ft. Average speed in ft. I sec.
1 125 125
5 5575 1115
2 20776 10388
1 10142 10142
01 0992402 992402
001 0099024002 99024002
0001 0009900240002 9900240002
This table could be continued to any extent, and it appears
that the average speed as we diminish the interval of time over
which it is calculated is always greater than 99 ft./sec., ap
proaches more and more to 99 ft./sec., and apparently could be
made as near to 99 ft./sec. as we like if we extend the table far
enough, i.e. if we make the interval of time small enough.
To make quite sure of this let us consider th'e distance
travelled in (4 + h) sees., where h can eventually be as small
a fraction as we please. The distance is
5 + 3 (4 + h) + 2 (4 + h) 3 ft. or 145 + 99A + 24& 2 + 2A ft,
i.e. the distance travelled in the interval h sees, after the end of
the fourth second is (99A + 24A 2 + 2A 3 ) ft. .'. the average speed
during this interval is
99A+24/i 2 +2A 3
r or (99 + 24A + 2A 2 ) ft. /sec.
[Notice that this result includes all the results in the table,
e.g. if we put h= 1 we get 99 + 24 + 02 or 10142.]
As h is made smaller and smaller (247i + 2/i 2 ) becomes smaller
and smaller and can be made as small as we please if h be made
small enough. In other words the average speed during the
interval h can be made as near to 99 ft./sec. as we please if we
take the interval short enough.
4] SPEED AT AN INSTANT 9
This, so to speak, ideal speed of 99 ft./sec. is called the
speed of the body at the end of 4 sees.
As this is an extremely important idea, it may be well to
summarise the steps by which we arrived at our result.
We found the distance travelled by the body in a short
interval immediately following the end of the fourth second and
calculated the average speed during this interval. We then did
the same thing with a shorter interval, then with a shorter, and
so on, and found that our average speeds continually approached
99 ft./sec. and that if we took our interval short enough, we
could make the average speed come as near to 99 ft./sec. as we
wished.
This limiting value of the average speed which can be
approached as nearly as we please by taking the interval
short enough, is called the speed of the body at the end
of 4 sees.
EXERCISES. III.
1. If = 5 + 3f + 2t 3 , find the average speed during the interval A sees,
which immediately precedes the end of the fourth second. Shew that it is
less than 99 ft./sec., continually approaches 99 ft./sec. as h is made smaller
and can be brought as near to 99 ft./sec. as we please if we make h small
enough.
Also find the average speed during the interval of h seconds which
extends from ^ sees, before to  sees, after the end of the fourth second, and
j a
shew that this average speed also can be brought as near to 99 ft./sec. as we
please if we make h small enough.
2. If s = 5t + 3t 2 find the distances travelled in 3, 4, 3'5, 32, 31,
3  01, 3'001 sees, from the beginning. Hence find the average speeds
during the intervals 1, 5, 2, 1, 01, 001 sees, immediately following the
end of the third second. Also find the distance travelled in (3 + h) sees,
and the average speed during the interval h sees, after the end of the third
second.
What is the speed at the end of the third second ?
10 CALCULUS FOR BEGINNERS [CH. I
Find also the average speed during (i) the interval of h seconds
immediately preceding the end of the third second, (ii) during the interval
of h seconds which extends from the end of ( 3   j sees, to the end of
3 + o ) secs > (iii) during the interval between the end of (3j) sees, and
(
the end of (3 + ) sees., and shew in each case that by making the interval
short enough we can bring the average speed as near as we please to
23 ft./sec.
3. In Qu. 2 find (i) the average speed during the fourth second,
(ii) the speeds at the end of 3, 4, 3 sees., (iii) the mean (half the sum)
of the speeds at the end of 3 and 4 sees.
4. If s S + 9t 3 , find (i) the average speed during the fourth second,
(ii) the speed at the end of 3J sees., (iii) the mean of the speeds at the end
of 3 and 4 sees.
6. If s = t*'6t + 5, find the distance gone in 4 sees, and in (4 + /<) sees.
Hence get the speed at the end of the fourth second.
5. From Exs. 1 and 2 we see that the speed at a given
instant may be deduced from the average speed during a short
interval, (i) immediately following, or (ii) immediately preceding,
or (iii) including, the instant in question and may be defined as
that speed which any of these average speeds can be made to
approach as nearly as we please if the interval be made short
enough. See 28.
Rate of Increase.
6. The question of speed might be presented somewhat
differently, as shewn in the following examples.
Suppose a train to travel uniformly between two stations
distant 15 and 17 miles from the terminus, and to pass those
stations at 5.20 and 5.23.
Let s feet be the distance of the train from some fixed point
of the path, say the terminus, at the end of t sees, from some
fixed instant, say 5 o'clock.
58 J RATE OF INCREASE 11
Then as t increases from 20 x 60 to 23 x 60
a 15 x 5280 to 17 x 5280,
i.e. as t increases by 3 x 60, s increases by 2 x 5280,
1 2x528Q or 582
3x60 *'
or 5S is the rate of increase of s per unit increase of t.
[With our earlier terminology we should say the speed was
58 ft/sec.]
, Increase in a
Notice that this 58 is = .
Increase in t
7. It will avoid confusion if we remember in cases like this
where we are dealing with quantities of two different kinds,
viz. : distance and time, that a and t are numbers, s being the
number of feet in the distance and t the number of seconds in
the time.
Increase in s
When we say T :  = 58, we are stating, not that
Increase in t
58 is the result of dividing 2 x 5280 feet by 3 x 60 seconds, but
that it is the result of dividing 2 x 5280 by 3 x 60; 2 x 5280
being the number of feet and 3 x 60 the number of seconds.
The statement in 2 that if a body travels s feet in t seconds
o
its average speed during the interval is  ft. /sec. is equivalent to
this : If the number of feet in the distance travelled be divided
by the number of seconds in the time taken, the resulting
number is the number of ft./sec. in the average speed.
, Increase in s .
8. If the motion is not unitorm. i.e. it the ratio =  :  is
Increase in t
not constant, as in 4, we might state our results as follows:
When I is increased by 5, a is increased by 55'75.
Increase in a 55*75 ,,, Kr ,
The ratio = . = = = 11150,
Increase in t 'O
12 CALCULUS FOR BEGINNERS [CH. I
and we say that 11150 is the average rate of increase of s
per unit increase of t between t = 4 and t = 45; for if the
body were moving uniformly in such a way that
an increase of 5 in t produced an increase of 5575 in s,
then 1 would produce 11150 in s.
Similarly 10142 is the average rate of increase of s per unit
increase of t between t= 4 and t = 41, for if s were increased by
10142 in every 1 sec. s would be increased by 101 '42 in 1 sec.
and so on.
99 + 24A + 2h 2 is the average rate of increase of s per unit
increase of t between t= 4 and t = 4 + h and 99 is said to be the
rate of increase of s per unit increase of t when t = 4.
Or we might say
11150 is the average rate of increase of s with respect
to t between t = 4 and t = 45, meaning that as t increases from
4 to 4*5, s increases 11150 times as much, and 99 is the rate of
increase of s with respect to t when t = 4, this 99 being the
number towards which our succession of average rates continually
tends.
9. Suppose the formula connecting s and t to be
s = 20t  2t 2 ,
and to fix our ideas we will suppose that s is measured to the
right.
If < = 4, 8 = 4:8: if t = 4: + h, s = 48 + 4A2A 2 and proceeding
as before we find that the speed at the end of 4 seconds is
4 ft./sec. to the right.
If <=5, s = 50: if t = 5+h, * = 502A 2 . If we go through
the usual steps we say that in the interval of A seconds which
follows the end of the 5th second, the body moves  2k" 2 ft. to
the right and its average speed in the interval is 2h ft./sec.
The significance of the negative sign is easily seen. At the
end of 5 seconds, the distance of the body to the right of the
810] RATE OF INCREASE 13
starting point is 50 feet: at the end of (5 + h) seconds it is
(50 2h?) ft., i.e. in the interval of h sees, the body has moved
2& 2 ft. to the left and the average speed is 2h ft./sec. to the
left.
If t 5 h, s = 50 2/i 2 and in the interval h sees, before the
end of the 5th second, the body has moved 2A 2 ft. to the right
and its average speed is 2h ft./sec. to the right.
This average speed can be made as small as we like if h is
made small enough, i.e. if the interval be made short enough.
During any interval, however short, immediately before the
end of the 5th second the body is moving to the right and during
any interval however small immediately after the end of the 5th
second, it is moving to the left and in each case the average
speed during the interval can be made as small as we please if
we make the interval short enough. We say then, that at the
end of 5 seconds the speed of the body is zero and that the
direction of motion changes at the end of 5 seconds.
10. If t = 7, s = 42 : if t = 7 + h, s = 42  8h  2h\
Proceeding as before we say that in the interval of h seconds
following the end of the 7th second the body moves 8A 2h? ft.
to the right and its average speed in that direction is ( 8 2/t)
ft./sec. and the speed at the end of 7 sees, is  8 ft./sec.
The negative sign is again easily explained.
At the end of 7 sees, the body is 42 ft. to the right of the
starting point. At the end of (7 + h) sees, it is 42 8h 2A 2 ft.
to the right, i.e. in the interval of h sees, it has moved (8h + 2h 2 )
ft. to the left, the average speed in this direction being (8 + 2h)
ft./sec. and the speed at the end of 7 sees, being 8 ft./sec. to the
left.
A negative speed is thus seen to indicate that the body is
moving in the opposite direction to that in which s is measured.
Or we might say
As t increases from 7 to 7 +/*, s changes from 42 to 42 8h 2h\
14. CALCULUS FOR BEGINNERS [CH. I
Or, as t increases by h, s increases by Sh  1h*
or decreases by Sh + 2h?,
and the average rate of increase of s per unit increase of t is
 8  2h,
or the average rate of decrease of s per unit increase of t is 8 + 2h.
These two statements mean the same thing; in fact, to say
that the rate of increase of s with respect to t is negative
simply means that s is decreasing as t is increasing.
11. Function. When two quantities are connected in such
a way that to each value of the first there corresponds a definite
value of the second, the second quantity is said to be a function
of the first,
e.g. in the case considered in 4, s is a function of t the
time being given we can calculate the distance of the body from
the fixed point. The relation between s and t is shewn in the
form of an equation
s = 5 + St + 2t 3 ,
[In 3 the relation was exhibited by means of a table of
pairs of corresponding values.]
and we have seen how to find
(i) The average rate of increase of s per unit increase of t
or the average rate of increase of s with respect to t between two
r~ Ti . Increase in s ~]
specified values of t. It is .= : .
Increase in t
(ii) The rate of increase of s per unit increase of I or the
rate of increase of s with respect to t for a specified value of t.
rr , . , , Increase in s
lit is the value which the ratio =^ = can be made to
Increase in t
approach as nearly as we please if the increase in t be made
small enough.]
1013] RATE OF INCREASE 15
12. The area of a square is a function of the length of the
side. If A sq. ins. is the area of a square whose side is x ins.
A = x 2 .
Now suppose we have a square plate whose side is 1" and that it
is expanding under the action of heat so that it remains a square.
The original area is 1 sq. in., i.e. the original value of A= 1.
If the side be increased to 1*1" the area increases to 1*21 ins., i.e.
if x be increased by !, A is increased by 21,
Increase in A 21
T  = =21 (see 7),
Increase in a; '1
and this is the average rate of increase of A per unit increase of
x between x = 1 and x= I'l. .
Similarly if the side be increased from 1" to 1001" the area
is increased from 1 sq. in. to 1 '002001 sq. ins., and the average
rate of increase of A per unit increase of x between x = 1 and
x =1001 is 2001.
If the increase in the side be made smaller and smaller this
average rate of increase can be brought as near to 2 as we please,
for if x is increased from 1 to 1 + h, A is increased from 1 to
1 + 2h + h?
Increase in A 2h + h? .
and : = j  2 + h,
increase in a? h
which can be made as near to 2 as we please if h is made small
enough.
And we say that when x = 1, the rate of increase of A per
unit increase of x, or the rate of increase of A with
respect to x, is 2.
13. If the plate be supposed to be cooling and contracting
we shall find that
if x is diminished by '1, A is diminished by 19, or we might
say
16
CALCULUS FOR BEGINNERS
[CH. I
if x is increased by *1, A is increased by '19, so that
or we might say
Decrease in A '19
Decrease in a? '1
Increase in A '19
= 19.
and
Increase in a? !
Similarly if x changes from 1 to 1  h
A 1 to 1  2h + .
Decrease in A 2A A 2
or
Decrease in a; h
Increase in A 2k + h?
= 2*
Increase in a? h
and as before, when x = 1, the rate of increase of A with respect
to x is 2.
14. It thus appears that if the rate of increase of one
quantity with respect to another is positive, the two
quantities are increasing or decreasing together, but if
the rate is negative, one is increasing while the other
is decreasing.
15. The pressure of a given mass of gas at constant tempera
ture is a function of the volume.
Let V c. ins. denote the volume of a given mass of gas and
p Ibs. per sq. in. its pressure, and suppose the following table
given
P
V
30
40
40
30
48
25
50
24
i.e.
As V decreases from 40 to 24, p increases from 30 to 50,
Increase of p 20 5
Decrease of V 16 ~ 4*
1316]
RATE OF INCREASE
17
or  is the average rate of increase of p per unit decrease of V
between V = 40 and V = 24.
Similarly
g
v is the average rate of increase of p per unit decrease of V
5
o
or = is the average rate of increase of p with respect to V
D
between V = 30 and V = 25.
16. Now suppose the relation between p and V given by
the formula pM = 1 200.
We can by calculation form the following table :
V
40
45
42
41
405
40 + 7*
1200
p
30
2667
2857
2927
2963
40 + /i
and the average rates of increase of p with respect to V between
(i) V = 40 and V = 45
(ii) V = 40 and V = 42
(iii) V =. 40 and V = 41
(iv) V = 40 and V = 405 J
will be found to be respectively '67, *71, *73, '74 and
what we call the rate of increase of p with respect to V when
V = 40 is the quantity towards which this succession of numbers
continually tends as the increase in V is made smaller.
To make quite certain what this limit is, find the average
rate of increase of p with respect to V between V = 40 and
30
V = 40 + h. It is JJT j ; and by making h small enough we
30
can make this as near or 75 as we like.
40
Thus 75 is the rate at which p is increasing with
respect to v when v = 40.
M. c. 2
18 CALCULUS FOR BEGINNERS [CH. 1
EXERCISES. IV.
1. In the following table A sq. ins. is the area of a segment of height
h ins. in a circle of diameter 1000 ins.
h 100 101 102 103 104
A 40875 41477 42081 42687 43296
Find the average rate of increase of A with respect to h between the
following values of h :
(i) 100 and 104; (ii) 100 and 103; (iii) 100 and 102; (iv) 100 and 101.
2. Use tables to find the average rate of increase of sin with respect
to 6 (0 being the number of radians in the angle) (i) between = 5061 and
0=5236, (ii) between 0='5236 and 6 = 5411.
With 4figure tables what is the best value you can get tor the rate of in
crease of sin0 with respect to when 6 = z1
o
3. A square is expanding in such a way that it remains a square.
Find the rate of increase of the number of sq. ins. in the area per unit
increase of the number of sq. ins. in the side when the side of the square is
(i) 3", (ii) a".
4. If a cubical block expands so as to remain cubical, find the rate of
increase of the number of cubic inches in the volume with respect to the
number of inches in the edge when the edge ia
(i) 1"; (ii)5"; (iii) a".
6. If a circle expands so as to remain circular, and if the radius,
circumference and area are respectively r ins., C ins. and A sq. ins., find the
rates of increase of
(i) A with respect to r,
(ii) C r,
(iii) A C.
when r=3.
6. The radius of a spherical bubble is increasing at the rate of 1 cm.
per second. At a certain instant its radius is 4 cms. What will its radius be
I, 5, 2, !, 01, h sees, after this? In each case find the surface and
volume.
17] GRADIENT 19
Find during each interval
(ij the average rate of increase of the number of cubic cms. in the
volume with respect to the number of cms. in the radius ;
(ii) the average rate of increase of the number of sq. cms. in the
surface with respect to the number of cms. in the radius ;
(iii) the average increase per sec. of the volume ;
(iv) ,, ,, ,, surface.
Also find at the end of 4 sees.
(v) the rate of increase of the number of c.c. in the volume per unit
increase of the number of cms. in the radius ;
(vi) the rate of increase of the number of sq. cms. in the surface per
unit increase of the number of cms. in the radius ;
(vii) the rate of increase of the volume per sec. ;
(viii) ,, surface .
7. The speed of a body (v ft. /sec.) at the end of t sees, is given by the
formula v=3 + 4t 2 .
Find the average rate of increase of v with respect to t between
(i) t=2andt=3; (ii) f=2 and t = 2 + fc,
and deduce the rate of increase of v with respect to t when t = 2.
[Note. The first result is 20 and this means that in 1 sec. the speed has
increased 20 ft. /sec. or that the average acceleration is 20 ft./sec 2 . The last
result gives the acceleration at the end of 2 sees.]
Gradient.
17. Uniform gradient. Fig. 2 shows a portion of a
straight line Kl_
P, Q are any two points on the line.
PM, Q.N are the ordinates of P, Q and PR is parallel to OX.
The A PQR has always the same shape wherever we take
RQ
P and Q. In particular the ratio is constant.
PR
This ratio is called the gradient of the line.
It is the tangent of the angle RPQ or the tangent of the angle
which KL makes with OX.
22
20
CALCULUS FOR BEGINNERS
[CH. 1
18. We may suppose that we pass from P to Q by two steps,
(i) a horizontal step PR, (ii) a vertical step RQ. The gradient is
Fig. 2.
the ratio of the vertical step to the horizontal step, and is the
same whatever the lengih of the horizontal step and from
whatever point we start.
Following the usual convention we call a horizontal step
positive if it is to the right and a vertical step positive if it is
upwards; so that the gradient is positive if both steps are
positive or both negative, and the gradient is negative if the
steps are of opposite signs.
e.g. in Fig. 2
Horizontal step P^ = + 10, Vertical step RjQj = + 5,
>> Ps^2 = ~^ : > ;> n R a Q 3 = 2,
and the gradient of the line is + jr .
18, 19]
UNIFORM GRADIENT
21
In Fig. 3
Horizontal step P^ = + 10, Vertical step R J Q 1 =  15,
f22 = ~ 4, II 11 RjQa = + 6,
3
and the gradient of the line is ^ .
Fig. 3.
19. Expressing this in the language of coordinates we maj
say that if (a^) (x.gj^) be the coordinates of any two points on
the line, the gradient is this being a fraction which has the
#2 a a
same value whatever the coordinates of the points chosen.
e.g. in Fig. 2
coordinates of P l are (10, 10) and of QI (20, 15),
1510 1
and gradient = ^ lo =2 ;
coordinates of P 2 are ( 2, 4) and of Q 2 ( 6, 2),
24 1
and gradient __=.
22 CALCULUS FOR BEGINNERS [CH. 1
20. Another way of looking at the question is as follows :
In Fig. 2 if we take a horizontal step + 1 [P 3 R 3 ], the vertical
step is +^r [R 3 Q 3 ] ; so that we may say that the gradient gives
2i
the vertical ascent per unit length of horizontal advance, or the
increase of y per unit increase of x.
21. Example. Shew that with our definition the gradient
of y = mx + n is m.
Let (#1^1) (#22/2) be any two points on the line,
.'. 2/ 2 = mx 2 + n >
and 2/1= I m&\ + n ;
' 2/2  /i = * 0*2  i),
3/2 &!
or m is the gradient of the line.
22. Average gradient. If P, Q bo two points on a path
RQ
which is not straight as in Fig. 4, is called the average
PR
gradient of the path between P and Q; e.g. the average
/>
gradient between P and Q is ^. This is the gradient of the line
y
PQ, and we may say that if we move from P along the path so as
to make a horizontal step PR, the vertical step RQ is the same as
if we had moved along the line PQ which has a uniform
gradient <r .
y
The average gradient will depend on the length of the
horizontal step and on the point from which we start, e.g. the
1 fi
average gradient between P and Qj is y^r and between P 2 and Qj
i y
.. . 95
2023] AVERAGE GRADIENT 23
Generally if (x^) (#2^2) ^ e the coordinates of two points
P, Q on a curve the average gradient between P and Q is
and this will be different for different positions of P and Q,
i.e. for different values of the coordinates.
o/r 4  rb trr 5
Fig. 4.
Gradient at a point of a curve.
23. Figure 5 shews a portion of the curve
P is the point (1, 1), Qj is (3, 9).
Let PQj be joined and produced indefinitely both ways.
Draw the ordinates PM, Q 1 N l ,and draw PF^ parallel to OX,
24 CALCULUS FOR BEGINNERS
Y
[CH. I
7
Fig. 5.
4 X
23] GRADIENT AT A POINT 25
Then as x increases from 1 to 3,
y 1 to 9 >
RiQ, Increase in y 8
i.e.  = f .  =  = 4 = average rate of increase of
PRj Increase in a; 2
y per unit increase in x between x= 1 and x = 3.
This is tan RjPQj or tan XK X P or the tangent of the angle
which PQ a makes with OX and is called the gradient of the
chord PQj.
[.Notice. If PR 2 = 1 and RaSj be drawn parallel to OY meeting
PQj in Su then R 2 Sj = 4.]
Now suppose the line PQj to rotate about P until it cuts the
curve in Q 2 corresponding to x = 2.
We shall get in an exactly similar way tan R 2 PQ 2 or
tan XK 2 P or the gradient of PG^ or the average rate of
increase of y per unit increase of x between x = 1 and x = 2 is
j or 3.
Let the line continue to rotate and cut the curve successively
in Q 3 , 04, Q 5 , Qg corresponding to a3 = l'4, 1'2, I'l, I'Ol, and we
get
Gradient of PQ 3 or average rate of increase of y per unit
increase of x between x 1 and x = 1*4 is 24.
Gradient of PQ 4 or average rate of increase of y per unit
increase of x between a; 1 and x= 1*2 is 22.
Gradient of PQ 5 or average rate of increase of y per unit
increase of x between x = 1 and x = 1 ! is 21.
Gradient of PG^ or average rate of increase of y per unit
increase of x between x 1 and x = 101 is 201.
If we continue this process making the increase in x and
therefore at the same time the increase in y smaller and smaller
or bringing the point Q nearer and nearer to P, the gradient of
the rotating line comes gradually nearer to 2, is always greater
than 2, but can be made as near to 2 as we please by bringing Q
near enough to P.
26
CALCULUS FOR BEGINNERS
[CH. I
To make quite sure of this let the line cut the curve in Q for
which x = 1 + h and /. y = 1 + 2A + A 9 (Fig. 6).
Increase in y 2h + h 2
Then gradient of PQ = ^  = 5 = 2 + h,
Increase in a; n
and by making h small enough we can make this as near 2 as
we like.
2hh 2
Fig. 6.
If we continue the rotation, the line will cut the curve again
on the other side of P, say in a point for which x = 1 h and
(Fig. 7).
Then gradient of PQ =
Decrease in y 2A  h?
= 2A,
Decrease in a; A
i.e. the gradient of a chord joining P to a point Q for which
x < 1 is less than 2, but can be made as near to 2 as we like, by
bringing Q near enough to P.
Thus if a chord through P cut the curve in a point to the
right of P, however near to P, its gradient > 2 ; if the chord cut
the curve in a poiut to the left of P, its gradient < 2 but in either
case the gradient can be made as near to 2 as we please by
bringing the second point near enough to P.
The gradient of the curve at P is said to be 2.
It is the same thing that we have previously called the rate of
increase of y per unit increase of x t when x=l.
2325]
GRADIENT AT A POINT
27
24. The line through P (PT) whose gradient is 2 is called
the tangent at P.
This line does not cut the curve again in a point near P, for
if it did its gradient would not be 2.
In fact PT is not one of the series of chords PQ 15 PQ 2 , PQ 3 ,
etc. for we have seen that none of these chords has a gradient 2,
but by bringing Q near enough to P we can bring PQ as near to
coincidence with PT as we like, i.e. we can make the angle TPQ as
small as we like.
This is expressed shortly by saying that the tangent PT is
the limiting form of the chord PQ when Q is brought indefinitely
near to P.
If PT (Fig. 8) is the tangent at P and PR = 1 then if RT be
drawn parallel to OY
RT = 2.
Fig. 8.
25. In these figures we have taken the same scale along
both axes and what we have called the gradient of a line is
the tangent of the angle which it makes with OX.
28
CALCULUS FOB BEGINNERS
[CH. I
. Actual length of R,Q,
e.g. the gradient of PQ, is . 2__ _t _ l or tan PKjX.
Actual length of PRj
If the scales are not the same the gradient, being
Increase in y . Number of yunits in
Increase in x Number of a>units in PRj '
and this is not the tangent of PK X X.
e.g. if y = y? be drawn with the scales shewn in Fig. 9, the
R O
gradient may be written provided it is understood that RjQj
PRj
is to be measured by means of the scale marked on OY, and PRj
by means of the scale marked on OX.
Thus gradient =  = 4 as before but now 4 is not tan RjPQj in
Fig. 9. It is the tangent of the angle that we should get in place
of RjPQj if we altered the scales so that the oMinit and the yunit
were of the same length, i.e. it is tan R 1 PQ 1 in Fig. 5.
25] GRADIENT 29
We may look upon Fig. 9 as a deformed copy of Fig. 5 and
consider either that horizontal distances are correctly represented
and vertical distances shortened in the ratio 1 to 2 or that
vertical distances are correctly represented and horizontal
distances lengthened in the ratio 2 to 1.
The tangent of the actual angle F^PQi that we see in Fig. 9 is
not 4 but 2.
EXERCISES. V.
1. If y=x*, find y when x=3, 35, 31, 301, 3 + fe, 29, 3fc. Hence
get the gradients of the chords joining the point on the curve y=x% at which
a;=3 to the points at which a;=35, 31, 301, 3 + ft, 29, 3 A.
What is the gradient of the curve at the point (3, 9) ?
Draw the graph of y=x% between x = I and x = 4 [taking " as the unit
for both x and y~\. Through (3, 9) draw the line whose gradient is 6. It
should be a tangent to the curve. What angle does it make with OX ?
2. What is the gradient of the chord joining the points where
a:=3/j and x=3 + h. Shew that by making h smaller and smaller we
can bring this chord as near as we please to coincidence with the tangent
at (3, 9).
3. Draw y=x z between x = l and x = 4 taking 1" as the unit along OX
and 2" as the unit along OY. As in Qu. 1 draw through (3, 9) the line
whose gradient is 6. It should be a tangent. What angle does it make
with OX ?
4. Find the gradient of the chord joining the point on the curve y=x s
at which a; =2 to the point at which x = 2 + h.
Hence get the gradient at the point (2, 8).
5. Find the gradient of the chord joining the points where x= 2  h and
x = 2 + h on the curve y = x 3 . Shew that by making h smaller and smaller we
can bring the chord as near as we please to coincidence with the tangent
at (2, 8). Similarly for the chord joining the points where x=2m and
6. Find the gradient of y=x* at the points where x = Q, 1, 3^4.
7. In the curve y=x 3 you have found the values of y at the points
where x = 0, 1, 2, 3, 4 and also the gradients at these points. Make use
30
CALCULUS FOR BEGINNERS
[CH. I
of these ten facts to draw the curve from x=0 to x=4. Take 1" as the
a:unit and 1" as the yunit. What angle does the tangent at (4, 64) make
with OX ?
8. Find the gradient of the curve y =
at the points where
9. Find the gradient of y=4x + 3 at the points where x=0, 1, 2.
10. If y = x 3 (4  x) find the values of y when x = I and when x = 1 + 7.
Use your result to find ' ' the gradient of the curve y = x 3 (4  x) at the point
where x=l," explaining carefully what the phrase means. Similarly find the
gradient where x=3.
Using these results and the values of y when x =0, 1, 2, 3, 4 draw the
graph between x = Q and a; = 4.
26. In curves with which we shall deal, the limiting position
of the chord PQ when Q is brought indefinitely near to P will be
the same on whichever side of P we take Q, and will be the
Fig. 10.
same as the limiting position of the chord RS (P lying between
R and S) when R and S are each brought indefinitely near to P.
[See Figs. 10, 11.]
26]
TANGENT
31
Fig. 11.
If a curve had a sharp point at P as in Fig. ] 2, it would not
be true that the limiting position of the chord PQ is the same on
whichever side of P we take Q.
In fact there are two tangents at P, viz., PT 1} PT a , one to
each branch.
Fig. 12.
32
CALCULUS FOB BEGINNERS
[CH. I
125
4;
Fig. 13.
26, 27] RATE OF INCREASE AS GRADIENT 33
27. If one quantity be a function of another, the relation
may be shewn analytically by an equation or geometrically by
a graph.
e.g. in 4 a is a function of t, this fact being expressed in
the form of an equation
s = 5 + 3t + 2t 3 .
This relation may be exhibited graphically.
In Fig. 13, the portion of the graph between t = 3 and t = 5 is
shewn, P, Q 1? GLj, Q 3 being the points on the graph corresponding
to* = 4, 6, 45, 42.
The number of units of length in F^Qj is the increase in s as
t increases from 4 to 5 [or it is the number of feet travelled
between the ends of the 4th and 5th seconds].
The increase in t is the number of units of length in PR lf
Increase in s
Thus = ; or what we have called the average rate
Increase in t
of increase of s with respect to t between t = 4 and t = 5 is the
gradient of the chord PQ^
Similarly the gradients of PQ 2 and PQ 3 give the average rates
of increase of s with respect to t between t = 4 and t = 4'5 and
between t = 4 and t = 4*2.
The limit to which this average rate of increase of with
respect to t tends as the increase of t is diminished more and
more, or what we have called the rate of increase of s with
respect to t when t = 4 is the limit to which the gradient of the
chord PQ tends as Q is brought nearer and nearer to P, i.e. the
gradient of the curve at P.
In this particular case, then, when the spacetime graph
is drawn, the gradient of the curve at any point gives the
speed at the instant corresponding to that point.
Similarly if the speedtime graph is drawn, the gradient
of the curve at any point gives the acceleration at the
instant corresponding to that point.
M. C. 3
34
CALCULUS FOR BEGINNERS
[CH. I
Similarly, if we draw the graph pM = 1200 the gradient of PQ
(3'33\
J is the average rate of increase of p with
29
27
43
Fig. 14.
respect to V between V = 40 and V = 45. The gradient of the
curve at P is the rate of increase of p with respect to v
when v = 40.
27, 28]
APPARENT EXCEPTION
35
28. There are cases in which the statement in 5 appears
not to be true.
e.g., suppose a marble falls vertically from a height of
36 feet. It strikes the ground with a speed of 48 ft./sec., and
will rebound with reduced speed, say 40 ft./sec.
Fig. 15 is the spacetime graph, the abscissae representing
the number of seconds from the instant when it begins to fall
and the ordinates the height above the ground in feet.
Fig. 15.
The speed 48 ft. /sec. is the limit of the average speed taken
over a short interval preceding the instant when it strikes the
ground, and the speed 40 ft./sec. is the limit of the average speed
taken over a short interval following the instant when it leaves
the ground.
32
36
CALCULUS FOR BEGINNERS
[CH. I
If we take these two instants to be the same, we seem to
have two different gradients at A, or in other words to have two
different speeds at the instant of impact according as we consider
the limit of the average speed over a short interval preceding or
over a short interval following the instant.
Fig. 16.
28, 29] RATE OF INCREASE AS GRADIENT 37
As a matter of fact the two instants are not the same and the
marble is in contact with the ground for some short time during
which there is compression followed by recovery.
Suppose for instance that the time of contact is T ^ 7 sec.
Fig. 16 shews roughly what the true spacetime graph is like.
It is a portion of the graph, the horizontal and vertical scales
being each magnified 100 times as compared with Fig. 15.
A corresponds to the instant when the marble strikes the
ground and B to the instant when it leaves the ground.
It will now be seen that the statement in 5 still applies to
the speeds at the instants corresponding to A and B.
EXERCISES. VI.
1. Draw pretty accurately between x =3 and a: =5 the graph of y=log 10 ar.
Draw by eye the tangent at the point corresponding to ar=4.
What is the rate of increase of Iog 10 a; per unit increase in x when x=4 ?
2. A body is moving in a straight line and its distances (s ft.) from a
fixed point in the line at the end of (t sees.) are given by the following table
t 1 2 3 4 5
it 7 22 45 76 115'
Draw as accurate a graph as possible and from it get the speed at the end of
3 sees.
3. Get the results of Exs. TV. 5, p. 18, graphically.
29. The processes by which we arrived at our results in
the preceding illustrations were in principle the same.
In finding the speed of a moving body at a given instant, we
calculated the average speed during a short interval of time
starting from the instant in question. We then found that as
the interval over which the average speed was calculated was
made shorter and shorter, the average speed continually
approached some definite speed from which we could make it
38 CALCULUS FOB BEGINNERS [CH. I
differ by as little as we pleased by making the interval short
enough. This speed we called the speed of the body at the given
instant.
In finding the gradient of a curve at a given point we
calculated the gradient of a line passing through the given point
and cutting the curve again in a neighbouring point. We then
found that as the neighbouring point was moved nearer and
nearer to the given point, this gradient continually approached
some definite value from which we could make it differ by as
little as we pleased by making the points close enough together.
This value we called the gradient of the curve at the given point.
In finding the rate at which y is changing with respect to x
for a given, value of x, we made a small change in x, calculated
the corresponding change in y and found the ratio of the change
in y to the change in x. We then found that as the change in x
was made smaller and smaller, this ratio continually approached
some definite value from which we could make it differ by as
little as we pleased by making the change in x small enough.
This value we called the rate at which y was changing with
respect to x for the given value of x.
30. Our final result was in each case obtained by finding the
ratio of two numbers each of which was supposed to become
less than any assigned number however small.
Now it might at first sight appear that if two quantities are
constantly decreasing they must become more and more nearly
equal. This is because we have a wrong idea of the test of
approximate equality ; we think the two quantities must approach
to equality because their difference becomes smaller and smaller.
In estimating whether two quantities are nearly equal or not,
we must look not merely at the difference between them, but at
the ratio which this difference bears to the quantities compared.
For example, two lengths of 3 and 4 inches are not so nearly
equal as two lengths of 100 and 101 inches, for though the actual
difference is the same in both cases, viz. : 1 inch, this difference
29, 30] TEST OF APPROXIMATE EQUALITY 39
is ^ of the smaller length in the first case, but only ^7 of the
smaller length in the second case. Lengths of 1000 and 1003 inches
are still nearer to equality, for though the actual difference is 3"
and therefore greater than the difference in either of the other
two cases, this difference is only 1fif)ft of the smaller length, and
3 J_
1000 < 100*
The point is that there are different standards of size in the
three cases. The same difference of length is of less importance
when we are dealing with 1000 inches than when we are dealing
with only 3 or 4.
An error of 1 foot made in measuring the length of a room
would be considerable, but the same error made in finding the
height of a mountain would be of very little account.
A loss of 200 would be serious for a man with an income
of 300, but of little moment to a man with an income of
100,000.
An error of 10 million miles in the Sun's distance (about 90
millions of miles) would be considerable, but the same error
would be inappreciable in the distance of Sirius (more than
500,000 times that of the Sun).
A carpenter would neglect an error of ^ of an inch, in
fact he would not detect it, but in cases where it was necessary
to work to . nnnn of an inch, it would be a gross error.
We have no right to say that all microscopic creatures are of
the same length, because they are all so small as to be invisible
to the naked eye. In comparing two such creatures, we must,
so to speak, think microscopically, i.e. use a microscopic standard
of length. It is just as true that a creature ^7; of a millimetre
40 CALCULUS FOE BEGINNERS [CH. I
in length is twice as long as a creature r^r of a millimetre in
length as that a man 6 ft. in height is twice as high as a child of
3 ft. in height
To say that the difference between two quantities is great or
small is meaningless. The terms are purely relative and both
may be used of the same quantity in different cases. The
important thing is the ratio of the difference to the quantities
compared.
The test of approximate equality might be put in another
way :
4 l3333 101 101 1003 _1. OS
3~ 33 ' TOO" 01 ' 1000 3 '
and we say that the last pair of numbers are more nearly equal
than either of the other pairs, because their ratio is nearer to 1
than eillier of the other ratios.
EXERCISES. VII.
Which of the following pairs are most and which least nearly equal ?
1. (a) 7" and 8" ; (b) 9 ft. Sins, and 11 ft. ; (c) 0054" and 0062".
a. (a) 93,000,000 miles and 92,900,000 miles.
(b) 930 inches and 9'29 inches.
(c) 93,000,000 miles and 93,100,000 miles.
(d) 930 yds. and 931 yds.
3. (a) x and y ; (b) Wx and W*y ; (c) JL and JL ;
where y = lQlx.
31. We shall now give a few geometrical illustrations to
shew that if x and y are two quantities which continually
decrease towards zero and become zero (or vanish) together, the
ratio of a: to y may
(1) increase continually, i.e. it may be greater after
a decrease than it was before, still greater after a further
decrease and so on, becoming greater than any number that can
3032] RATIO OF CONTINUALLY DIMINISHING QUANTITIES 41
be named, however large, if the decrease in x and y be carried
far enough ;
or (2) decrease continually, i.e. it may be less after a
decrease than it was before, still less after a further decrease and
so on, becoming less than any fraction that can be named,
however small, if the decrease in x and y be carried far enough ;
or (3) increase or decrease towards a limit, i.e. con
tinually approach some value which it never reaches, but to
which it can be brought as near as we like if the decrease of
x and y be carried far enough ;
or (4) always remain the same.
32. (1) Fig. 17 shews a circle centre O. OA, OB are two
radii at right angles. P is a point
which is supposed to move along the Q
circumference nearer and nearer to A.
PM is perpendicular to OA and AP is
produced to meet OB produced in Q.
As P approaches A, PM and MA
both diminish without limit, i.e. assign
any length however small, we can find
a position of P so that PM or MA shall
be less than that length. If we sup
pose P to be at A, there is no triangle
PMA at all, or as we may put it, PM
and MA vanish together, but we have
no right to say that PM and MA are
all the time tending to become equal.
For every position of P which
is distinct from A, PMA is similar to
PM QO
A QOA, i.e. = .
MA OA
But as P approaches A, Q recedes
further and further from O, i.e. QO increases without limit,
42 CALCULUS FOR BEGINNERS [cil. I
QO PM
and OA remains the same, .'. increases without limit, i.e. 
OA MA
increases without limit, i.e. state any number however great, PM
may be made to contain MA more than that number of times
before P reaches A.
Suppose the radius of the circle to be 1" and call L AOP 6
Then PM = sin 6 ins. and MA = (1  cos 0} ins. We can form this
table :
20
10
5
Q1O 1 O
" *
PM
3420
1736
0872
0436 01745
MA
PM
MA
0603
567
0152
1143
0038 00095 00015
2290 4583 1146
6
3<X
i<y
4'
2'
PM
0087265
0029089
0011636
0005818
MA
PM
MA
0000381
229
0000042
688
0000007
1719
0000002
3438 etc.
sin
TN.B. Since , 7 .=cot ~ the numbers in the last line .can
L 1  cos 6 2
be written down from a table of cotangents.]
Suppose our eyes incapable of seeing any length less than
th of an inch ; then when L AOP is about 2, MA is
1000
invisible but PM is more than = of an inch.
zo .
Again, when the angle is about 4', PM is just visible, but we
should have to use a microscope magnifying about 1700 times to
make MA visible.
33. (2) If in the same figure we consider , the ratio
can be made less than any fraction we like to assign, however
small, before P reaches A.
32 34] RATIO OF CONTINUALLY DIMINISHING QUANTITIES 4*
34. (3) In Fig. 18, OA is a fixed radius of a circle, centre O.
P, Q. are two points on the cir
cumference, so that Q is the raid
point of the arc AP; PM, QN are
perpendicular to OA.
As P approaches A, Q ap
proaches A, and AM, AN both
diminish without limit. If we
suppose P to have arrived at A,
Q will also be at A, and AM, AN
will have ceased to exist. But
we must not say that AM, AN are
all the time tending to equality.
Call L AOQ 0, then L AOP = 20.
If the radius of the circle be r,
AN=r(lcos0)
Fig. 18.
AM=r(lcos20)
= 2rsin 8
&
AM
AN
Now as 6 diminishes cos ^ increases and can be made as near
m
to 1 as we please if & be made small enough.
Thus is a continually increasing ratio whose limit is 4,
AM
i.e. is always less than 4, but state a number whose defect
from 4 is as little as we please, we can place P so that
shall be equal to this number.
AN
10
5
39924
1
39997
30'
399992
10'
3999992
3999999...
44
CALCULUS FOR BEGINNERS
[CH. I
35. (4) In Fig. 19, let XOY be a triangle rightangled at X,
such that OX = 2XY,
and let a point P move along YO towards O. Draw PM perpen
dicular to OX.
Kg. 19.
Then as P approaches O, PM and OM both diminish without
limit. If we suppose P to have reached O the A PMO has ceased
to exist. Bub for every position of P distinct from O, however
near to it
OM OX
~MP ~ XY ~ "'
OM
i.e. is a constant ratio, however small OM and MP are.
MP
36. Of course, it may happen, that, if two quantities
continually approach zero and vanish together, their ratio
continually approaches the limit 1, but it must be borne in mind
that this is an exceptional case.
The following illustration is an important example of this
case.
In Fig. 20, if the point P move along the circumference of a
circle towards A, the arc PA and the chord PA diminish together
and can each be made less than any assigned length, however small,
35, 36] RATIO OF CONTINUALLY DIMINISHING QUANTITIES 45
if P be brought near enough to A ; but we have no right to con
3iT*C AP
elude from this that the ratio : , continually approaches 1.
chord AP
Fig. 20.
As a matter of fact it does, as may be seen from the table
below.
LAOP
in degrees
10
9
8
7
6 3
Arc^P
1745329
1570796
1396263
1221730
1047198
Chord AP
1743114
1569182
1395130
1220970
1046720
AioAP
10013
6
100103
4
10008
3
10006
2
10005
1
Chora AP
LAOP
in degrees
Arc^P
0872665
0698132
0523599
0349066
0174533
Chord AP
0872388
0697990
0523538
0349048
0174530
AxoAP
10003
10002
10001
100005
100002
Chord AP
46
CALCULUS FOR BEGINNERS
[CH. I
37. Another difficulty which often occurs is this :
How can a quantity continually decrease without eventually
disappearing altogether ?
In Fig. 21, OAB is a quadrant of a circle, centre O.
BX is the tangent at B.
B T T' T" T'"
M M'M" A
Fig. 21.
Suppose T to travel along *BX, at any rate whatever, say 1000
miles an hour, and in every position of T, join OT cutting the
circumference in P. It is obvious that as T travels along BX, P
moves nearer to A, but P will never reach A, however far T be
supposed to travel along BX.
The lengths PM, MA considered in 32 can be supposed to go
on decreasing for ever without actually disappearing and it has
PM
been shewn that during the motion of P towards A, will
MA
increase without limit.
CHAPTEE II
DIFFERENTIATION FROM FIRST PRINCIPLES
38. WE shall now go through some of the investigations
of Chapter I in a slightly different manner, using the notation of
the subject.
39. We have
s = 5 + 3t f 2 3 ,
giving the distance s feet travelled in t seconds.
If we take a slightly greater value of t, say t + At, we shall
obtain a slightly different value of s, say s + As.
[A is merely shorthand for " an ' increment ' or ' increase ' in
the value of t," A being the Greek equivalent of capital D.
A by itself means nothing, and A2 must be considered as one
symbol; in fact, Ai takes the place of A in 4.]
There will be the same relation between (s + As) and (t + A<)
as between s and t ;
i.e. s + As = 5 + 3 (t + A<) + 2 (t + A<) 3
= 5 + 3* + 2^ + (3 + 6 2 ) A< + Qt (A) 3 f 2 (A*) 8 ,
.*. As = (3 + W) A* + Qt (A) 2 + 2 (A<) 3 .
This gives the extra distance travelled in the extra interval
of time A.
48 CALCULUS FOB BEGINNERS [CH. II
[e.g. If we want the distance travelled in the halfsecond following the
end of the 3rd second, we put t=3, At=.
A
/. =? = (3 + 6* 2 ) + 6* . A* + 2
Since As feet is the distance described in the interval A sees
As
.". gives the average speed during this interval in ft./sec.
Av
As A is made smaller and smaller Qt . A + 2 (A) a becomes
smaller and smaller and can be made as small as we please if A
is made small enough.
i.e. can be made as near to (3 + 6< 2 ) as we please if Ai be
A
made small enough.
As
The limit to which continually approaches and from
which we can make it differ by as little as we please by taking
A small enough is called y .
dt
/Jo (7$
Thus 5 = 3 + 6^ and 3 ft. /sec. is the speed of the body at the
end of t seconds.
\V. definition of speed at an instant 4.]
40. Notice that if s = 5 + 3t + 2
As
(i) = (3 + 6< 2 ) + 6* . A< + 2 (A*) 2 is an accurate statement
whatever the value of A, e.g. take =4, A = ^ and we get
99 + 12+^ or 111 ft./sec. as the average speed during the
interval of second immediately after the end of the 4th second.
39, 40] DIFFERENTIATION FROM FIRST PRINCIPLES 49
As
(ii) = 3 + 6^ 2 is approximately true if A be small and
t\t
becomes more and more nearly true as A is made smaller, but
there is no value of A2 however small for which the statement is
accurately true.
As
[See table in 4 where the third column gives for
LA 6
different values of AZ when t = 4 and in this case 3 + Qt" 2 = 99.]
In other words, the average speed during a small interval A
following the end of t seconds is approximately (3 + 6^ 2 ) ft./sec.
and this is more nearly true, the shorter we make the interval
A&
e.g. if we want approximately the distance described my^th
of a second after the end of the 3rd second we have
As = (3 + 6 2 ) A
approximately, meaning that if A be small the distance described
in A2 seconds immediately following the end of t seconds is nearly
(3 + 6 2 ) At feet.
Put t = 3 and A=y^ and we have approximately as the
57
distance required in our special case, yr or 57 ft.
[Shew that the distance is really 571802 ft.]
This gives an error of about '3 / o .
If we used the formula to give the distance in ycnnr sec. we
should get '057 ft. instead of 057018002 ft., an error of about
03 / an d so on.
ds
(iii) j = 3 + 6f 2 is accurately true. It is in fact a convenient
of*
As
way of saying that the unattainable limit which is trying to
reach is 3 t 6^, or  r stands for the ideal value which ~ is
at t\t
striving to attain.
M. c. 4
50 CALCULUS FOR BEGINNERS [CH. II
41. ds and dt have no separate meanings, dt does not stand
for a very small increment in t, nor ds for a small increment iii s,
for we have just seen that however small such increments are
made, their ratio is never 3 + 6Z 2 .
ds
TT then must be treated as a whole, and simply means
dt
the limit to which is constantly tending as At and with
At
As
it As becomes smaller and smaller. 7 can be brought as
At
ds
near as we please to ? if we make At small enough,
dt
ds
does not mean the result of dividing ds by dt as
dt
means the result of dividing As by At. The fractional
At
ds
form T7 is retained merely to remind us of the fraction
dt
of which it is the limiting value.
42. It is a common error to say that r is the value of 
dt A<
when As and A are each zero. If this statement be examined, it
will be seen to have no meaning.
As
Let us go back to the stage at which was obtained.
ttv
We had As = (3 + 6* 2 ) A + 6t . (A*) 2 + 2 . (A<) 3 .
We then divided both sides of the equation by A.
Now if A = we have no right to do this.
[It is certainly true that 8x0 = 3x0 but we cannot divide
both sides by and deduce 8 = 3. Similarly if a;xO = 7/xOwe
cannot conclude that x =y, though if x x T^= y x y^ it is true
that x = y.]
We should in fact be trying to estimate the speed of the body
from the fact that it had travelled feet in sees.
ds
4144] MEANING OF 7 51
43. We have found that the speed at the end of t seconds
is (3J6* 2 ) ft/sec.
The result obtained in 4 is merely a particular case of
this, for when = 4, 3 +6f 2 = 99 and our formula enables us to
write down the speed at any instant, e.g. to get the speed at the
end of 10 seconds, put t 10 and we get 603 ft. /sec.
EXERCISES. VIII.
1. Obtain a formula giving the speed at the end of t seconds of a body
whose distance (s feet) from a fixed point at the end of t seconds is given by
(i) 8 = 16(2, (ii) 8 = lOOt 16*2,
(iii) s=5 + t 3 , (iv) s = 5t + 9,
and in each case find the speeds at the end of 3, 5, 17 seconds.
2. If 8 = 16^2, what is the speed at the end of 2 seconds? Assuming
this speed to remain constant for the next tenth of a second, how far would
the body move, and what percentage error would be made in taking this as
the true distance ? Do the same for a hundredth of a second.
3. If the speed of a body increase by 12 ft./sec. in 3 seconds, what is the
average acceleration during the 3 seconds ?
4. If the speed at the end of 4 seconds is 78 ft./sec. and at the end of 9
seconds 146 ft./sec., what is the average acceleration between the end of the
4th and the end of the 9th second ? [v. Ex. 7, p. 19.]
6. If the speed at the end of t seconds is v ft./sec. and at the end of
(t + At) seconds, (v + Av) ft./sec. , what is the average acceleration during the
interval At? By what symbol should the acceleration at the end of t seconds
be denoted ?
6. In Ex. 1, find the accelerations at the end of 3, 5, 17 seconds.
7. With the formula 8 = 5 + t 3 what is approximately the distance
described in of a second immediately following the end of the 3rd second?
DU
What is approximately the increase of speed during this interval ?
ds
44. DEFINITION, rr is called the differential coefficient
dt
of s with respect to t, and "to differentiate a with respect to t"
means to find 7 .
at
42
52
CALCULUS FOR BEGINNERS
[CH. II
ds
45 Instead of saying that if s = 3t + 2t 3 then 3= 3 + G^ 2 we
0V
might say
dt
= 3 + Gf. This form of stating the result
will sometimes be found convenient.
46. Consider a square plate whose side is x ins. and let its
area be y sq. ins. so that y = a?.
Suppose the side increased to (x+ Aa:) ins., and as a result let
the area be increased to (y + Ay) sq. ins.
.'. y + Ay = (x + Aa;) 2 ,
.'. Ay = 2xAa + (Aa;) 2 .
i.e. the increase in area due to an increase Aa; in the side is
2.rAa; + (Aa;) 2 sq. ins.
[In Fig. 22 each of the shaded rectangles has area x&x and
the square in the top corner has area (Aa;) 2 .]
Increase in number of sq. ins. in area Aw
^ . _ n  = _ 2.
Increase in number of ins. in side Ax
dii
and as before we deduce ?= 2x.
ax
_ 2. = 2x + Aa;
Ax
>*) 2
x
x
Fig. 22
4547] r AS A RATE OF INCREASE 53
CLOS
47. As before notice that if y = a?
AV
(i) = 2o? + Ax is an accurate statement whatever the
Ax
value of Ax, e.g. if x = 3, and Ax = 2,
Ay
* = 62, i.e. Increase in y = 6*2 times increase in x
= 62x 2 = 124.
i.e. if the side be originally 3" and increase by 2" the area
increases by 124 sq. in.
A?/
(ii) ~ = 2x is approximately true and becomes more nearly
1.+OC
true as Ax is made smaller.
e.g. if x = 3, Ay = 6Ax nearly, i.e. if a square have a side of
3 ins. and a small increase be made in the side, the increase in
the number of sq. ins. in the area will be nearly 6 times the
increase in the number of ins. in the side.
Suppose for example the side increases from 3 to 302 ins.
the area increases approximately by *12 sq. ins.
Actually the increase is "1204 sq. in. [0004 sq. in. is the
area of the small square in Fig. 22.]
fjft
(iii) ~ = 2x is accurately true, and is simply a convenient
tUB
way of saying that the approximate statement in (ii),
Increase in y
^ = 2x,
Increase in x
can be brought as near to the truth as we please by making the
increase in x small enough. Taking again the special case when
x = 3, ~ = 6, this might be stated thus :
ux
When x = 3, the rate at which y is increasing compared with
x is 6, or the xrate of increase of y is 6, or the rate of increase of
y per unit increase of x is 6.
54 CALCULUS FOR BEGINNERS [CH. II
48. If we introduce the idea of time into the question and
suppose the side of the square to be steadily growing, then in
the same time in which Aic ins. is added to the side, Ay sq. ins.
Aw
is added to the area and ~=6'2 tells us that y increases 6'2
Aa?
times as much as x increases in the same time. If we call the
time in which the change takes place A< sees, we have
i.e. average rate of increase of y during this interval = 6'2 times
the average rate of increase of x during the interval.
In the general case
At/ Ay
If A, and therefore at the same time Arc and Ay, be made
smaller and smaller, ~ and can be made as near as we please
A A
dx , dy , Ay dy
to T and ? respectively, as near as we please to ? , and
dt dt Ao? dx
2o? + Aa; as near as we please to 2x.
dx
l
But 37 and = are what we call the rates at which y and x
at at
respectively are increasing at the end of t sees.
di/ dij flit*
Thus ~ = 2x being equivalent to j = 2oj . j may be looked
CtOC ut Cut
upon as telling us that y is increasing 2x times as fast as x.
di/
In the special case when x = 3, f=, and this tells us that
Cww
at the particular instant when the side, of the square is 3 inches
y is increasing 6 times as fast as x, or the number of square
inches in the area is increasing 6 times as fast as the number of
inches in the side.
4850] DIFFERENTIATION FROM FIRST PRINCIPLES 55
49. Example. The side of a square increases uniformly at
the rate of ! inch per second. At what rate is the area growing
when the side of the square is 6 inches ?
If y sq. ins. is the area of a square of side x ins.
x
' dx **
Ul/
and when x = 6, ^=12.
i.e. at the instant when the side of the square is 6 ins. y is
increasing 12 times as fast as a?, or the number of square inches
in the area is increasing 12 times as fast as the number of inches
in the side.
But x is increasing '1 per sec.
/. y ,, 12 x ! or 12 per sec.
i.e. the area is increasing at the rate of 1*2 sq. in. per sec.
50. Suppose the pressure (p Ibs./sq. in.) and the volume
(V cub. ins.) of a given mass of gas are connected by the relation
joV1200 or j = ^p.
(1) If we take a slightly different value of V, V + AV we shall
get a slightly different value of p, p + Ap such that
1200
1200 1200 1200AV
A <v\
(2)
P V + AV V (V + AV)V'
Ap 1200
AV (V + AV)V'
As AV is made smaller and smaller this comes nearer and
. 1200 1200 ...
nearer to and can be made as near to ^ as we like it
AV be made small enough.
dv
 . dp_ 1200
\/ ' "XT, vl" '
56 CALCULUS FOR BEGINNERS [CH. II
51. The meaning of the negative sign is simply this :
As V increases p decreases, so that AV and &p have opposite
signs, therefore ^ is negative.
We might read the final result either
The rate of increase of p per unit increase of V = 5,
or the rate of decrease of p per unit increase of V = f ^ ,
or p is decreasing 5 times as fast as V is increasing.
52. As before, notice, if pV = 1200
(i) ^ = AvW * s an accurate statement whatever the
values of V and AV.
e.g. if V = 40, and AV = 5
Ajp_ 1200 _2
AV~ 40 x 45~~3'
2
Le. decrease in p = ^ (increase in V) = 3.
9
.... Ap 1200 .
(n) ^ = is approximately true and becomes more
nearly true the smaller we make AV.
e.g. if V = 20, &p =  3 . AV nearly.
i.e. if the volume be 20 cub. ins. and a small increase be made
in the volume, the decrease in the number of Ibs./sq. in. in the
pressure will be nearly 3 times the increase in the number of
cub. ins. in the volume.
Suppose for example the volume increases from 20 to 20'1
cub. ft., the pressure decreases approximately 4 3 Ibs./sq. in., i.t).
from 60 to 597 Ibs./sq. in.
[Actually the decrease is '2985.]
5153] DIFFERENTIATION FROM FIRST PRINCIPLES 57
.... dp 1200.
(111) j  j is accurately true and is merely a convenient
A/?
way of saying that ~ can be brought as near as we please to
if AV be made small enouh.
,o
dv
is a formula giving the rate at which p is changing with respect
to V for any assigned value of V.
i.e. when the volume is 10 cubic feet,
. increasing 1 . (decreasing
p is . . } 12 times as fast as V is < .
decreasing J (.i
(.increasing
EXERCISES. IX.
1. A eq. ins. is the area of a circle of radius r ins.
What is the relation between A and r ?
If r be increased to r + Ar, find the corresponding increase AA in A, and
dA
prove as above that r=2?rr.
This tells us that approximately AA = 2nr . Ar.
Interpret this as a rule for the approximate area of a thin circular ring,
and apply it to find approximately the area of a circular ring whose inner
radius is 3 ft. and thickness ! inch.
2. Shew that the percentage error made in taking 2irr. Ar for the area
50. Ar
of the ring is less than  .
What is approximately your percentage error in Ex. 1 ?
3. The radius of a circle is increasing at the rate of 1 mm. per second.
At what rate is the area increasing when the radius is (i) 1 mm., (ii) 1 cm.,
(iii) 20 cms. ?
4. What will be the approximate increase in area during the next tenth
of a second after the radius reaches 20 cms. ?
58
CALCULUS FOR BEGINNERS
[CH. II
6. The side of a square is measured and found to be 8 inches. If an
error of '01 inch is made in measuring the side, find approximately the error
in the calculated area.
6. The area of a circle grows at the rate of 2 sq. his. per second. At
what rate is the radius growing
(i) when the radius is 5 inches,
(ii) when the area is 20 sq. ins. ?
7. If pv=80, find the rate of increase of p with respect to v when
t;=(i) 10, (ii) 80.
8. OX, OY are two lines at right angles. A and B are points in OX,
OY respectively such that the area of the triangle OAB is always 10 square
inches. If A moves along OX at a constant speed of O'l inch per second,
what is the speed of B (i) when OA = 8", (ii) when OB = S"?
54. Let P be a point whose coordinates are (x, y) on the
curve y = a? and let Q be a point on the curve near to P.
Its coordinates will differ slightly from those of P and may be
denoted by (a; + Aa;, y + Ay). (Fig. 23.)
Fig. 23.
54, 55] j AS GRADIENT 59
Then PR = Aa;, RQ = Ay.
.*. tan RPQ or tan XKP1 Aw r
r, I" = T [ V  n te On Sca H 25 ]'
or gradient or PQ ) Aa: L
Now y + Ay = (a; + Ace) 2
= a; 2 + 2a;Aa; + (Ax) 2 ,
.'. Ay = 2x . Ax + (Ax)*,
.'. t^=2a; + Aa;;
Ace
i.e. gradient of PQ= 2a; + Aa;.
Now suppose Q to move along the curve towards P, so that
Aa; and Ay become smaller and smaller, then by bringing Q near
Aw
enough to P, we can make  as near to 2x as we please.
Ace
Av
The limit which ^ continually approaches and from
which it can be made to differ by as little as we please if
dv
we make Ax small enough is called ^ .
(iij
Thus in this case ~= 2cc, and this is called the gradient of
CL3C
the curve at the point P.
55. Notice, \iy = a?
A?/
(i)  = 2o3 + Aa; is an accurate statement whatever the value
Ace
of Ace, e.g. take x = 1, Ace= 2, and we have 22 as the gradient of
the chord joining the point for which x = I to the point for which
1 ?/
(ii) ~ = 2cc is approximately true if Ace is small and becomes
AdB
more and more nearly true as Ace is made smaller, but there is no
value of Aa; however small for which = 2a;, in other words if
Ace
60 CALCULUS FOR BEGINNERS [CH. II
we draw a chord through P cutting the curve in a point Q as near
to P as we please, the gradient of this chord will never be exactly
2x, but can be made as near to 2x as we please if we bring Q
/\ 1 1
near enough to P. For instance if x = 3, ^ = 6 nearly, i.e. the
Ax
gradient of a chord joining (3, 9) to a neighbouring point on the
curve is nearly 6 and becomes nearer to 6 the closer this
neighbouring point is to (3, 9).
(iii) ~ = 2o; is accurately true and is merely a convenient
dx
A?/
way of saying that the unattainable limit which 2 is trying to
reach is 2x. In fact 2x is the gradient of a line through P
towards which the chord PQ is continually tending as Q approaches
P, but with which it can never be made to coincide however near
Q is to P.
This line towards which PQ continually tends is called the
tangent at P.
dv
Thus ^ gives the gradient of the tangent to the curve
at P, or as it is sometimes called, the gradient of the curve at P.
dv
56. The result ~ = 2x may be looked upon as a
QiZ
formula giving the gradient at any point of the curve.
Thus, if 05 = 3, ^=6, i.e. the gradient of the curve at the point
(3, 9) is 6.
57. Notice that although ~ is not the gradient of any
ctx
chord through P however near the other end of the chord may
be to p, yet it is the gradient of an actual line through P,
namely the line which we call the tangent to the curve at P.
58. We can now see better the significance of the approxi
mation in (ii).
5559]
dy
f AS GRADIENT
dx
61
Let P be the point (3, 9) and let Q be a point on the curve
near to P. (Fig. 24.)
Let PT the tangent at P cut QR in Q'.
Then the statement is that the gradient of PQ is nearly 6, in
RO RQ'
other words is nearly 6 ; but since  = 6, this is equivalent
RP RP
to saying that RQ is nearly the same as RQ'.
e.g. suppose PR = '02 then Q is the point (302, 91204).
/. RQ= 1204 and RQ' = 6 x02 = 12.
Fig. 24.
The error in taking RQ = RQ' is about ^ / .
o
If PR = 001 then Q is the point (3001, 9006001).
.'. RQ = 006001 and RQ' = 6 x 001 = 006.
The error is about ^ / o .
59. A caution similar to that issued in 41 applies here
also, dy and dx have no separate meanings : they do not stand
for small changes in length, for we have seen that however small
the increments in x and y are supposed to be, their ratio is never
exactly 2x.
"We must not say that is the value of ^ when Aa; and
dx AOJ
Ay
Ay are each zero. The whole process by which we obtained ~
62 CALCULUS FOR BEGINNERS [CH. II
becomes unintelligible if we imagine affairs pushed so to speak
to the limit. The whole argument depends on the existence of a
triangle PQR, which even though every side is out of the range
of visibility can be supposed magnified so as to become visible,
e.g. [with unit 1" along each axis] if x = 1 and Ao;= 0001 then
Ay '00020001 and if we used a microscope magnifying 1000
times the triangle PQR would appear as a triangle in which
PR = !" and RQ= 20001". But if we push Q to absolute coinci
dence with P, no amount of magnifying will separate the points
P, Q, R.
60. If the equation of a curve is given as y = a?, we may
look upon this as a formula for finding the value of y corre
sponding to any assigned value of x, or for finding the ordinate
corresponding to any given abscissa. In other words it is a
formula by means of which we may obtain the positions of any
number of points on the curve.
dij
j = 2x is a formula giving the gradient corresponding to any
given abscissa.
e.g. if x = 5, y= 25 and ^ = 10.
ax
i.e. (5, 25) is the point on the curve whose abscissa is 5 and
the gradient at this point is 10.
61. In drawing the curve y = y? it is a great help to make
use of both formulae.
Suppose we draw it between a? = 3 and x 3, we have the
following table
x 3 2 1 1 2 3
y 9 4 10149
g 6 4 2 2 4 6
Plot the points
(3,9) (2,4) (1,1) (0,0) (1,1) (2,4) (3,9).
5961]
dy
 AS GRADIENT
dx
63
Through these points draw lines whose gradients are 6, 4,
2, 0, 2, 4, 6 respectively. These lines then are to be tangents
to the curve (Fig. 25).
If we try to draw a curve passing through these 7 points and
touching these 7 lines, we shall find that the run of the curve is
pretty well determined.
64 CALCULUS FOR BEGINNERS [CH. II
EXERCISES. X.
1. Use the result just obtained to get the gradient of y = x 2 at the
points where
ar=3, 2, 1, 0, 1, 2, 3.
Draw the graph j/=x 2 , and through the points where
x=3, 2, 1, 0, 1, 2, 3
draw lines with gradients just found. They should be tangents to your
1"
curve. Take 1" as the a;unit and  as the yunit.
m
2. Find the gradient of j/ = x 3 at the points where
x=B, 1, 0, 5.
Shew that the gradient of t/=x 3 is never negative. Draw a figure.
3. Find the gradient of y=x*4x at the points where x = 0, 1, ^4, 2.
Draw a figure.
4. If y=x 2 , what is p when x = 7? Suppose we move along the curve
(tvG
from the point where a; = 7 to the point where x = 7'03, what is the gradient
of the chord joining these points ? What would be the percentage error if we
took the value of ^ just obtained as the gradient ?
else
6. Shew that the percentage error when 2x is taken instead of 2x + Ax
or 2x . Ax instead of 2x . Ax + (Ax) 2 is less than  and if x 2 + 2x Ax be
/Ax\ 2
taken instead of (x + Ax) 2 the percentage error < 100 ( I .
Use these results to find approximately the p.c. error
(i) if 201 2 be taken as 404 ;
(ii) if7'03 2 4942;
(iii) if the gradient of the chord joining the point where x = 5 to the
point where x = 5'02 be taken as 10;
(iv) if RQ in Fig. 24 be taken as equal to RQ', P being the point
(10, 100) and PR = 5.
6. Find (i)
l*O (/ \*V
62. The process by which we obtained j in 39 ; j in
cCt else
(I'D
46 and 54, j in 50 should be carefully studied, as it is
the same in all cases.
6264] DIFFERENTIATION FROM FIRST PRINCIPLES 65
Taking the first case : We had s given as a function of t
[s = 5 + 3t + 2t s ].
(1) We gave t a small increment (Atf) and found the
corresponding increment in s
As
(2) We found the ratio .
A^
As
(3) We found the limit to which tended as A< was made
A
smaller and smaller, and from which it could be made to differ
by as little as we please by taking A< small enough and this limit
ds
63. Generally if a quantity y be a function of another
quantity x, a small change, Ax, in x, will produce in y
a corresponding small change which we call Ay, and in
Av
the cases with which we have to deal tends continually
Ax
to some limit from which it can be made to differ by as
little as we please by making Ax small enough. This
dv
limit is denoted by ~ , and in the Differential Calculus the
dZ
ul/
fundamental problem is to determine  for different forms of
ax
the relation connecting y and x.
dy
64. p gives the rate of increase of y per unit in
crease of x or the rate of increase of y with respect to x
for any value of x. It tells us that y is increasing so
many times as fast as x.
In the special case when y and x are replaced by s and t with
ds
the usual meanings, y gives the speed at a given instant.
Ct'U
dv
Similarly j gives the acceleration at a given instant.
at
M. a 5
66
CALCULUS FOR BEGINNERS
[CH. II
If y and x are the coordinates of a point on a curve whose
equation is the given equation connecting y and x, = gives the
cLx
gradient of the curve at any point.
65. If we draw the graph of
[Fig. 26 shews it drawn between t = and t = 3.J
70
60
50
40
30
s
20
Fig. 26.
6466] SPEED AND ACCELERATION AS GRADIENTS 67
ds
the gradient at any point corresponding to time t will be j
citi
or 3 + 6* J .
ds
But r gives the speed at the end of time t.
Cut
Thus we see that if the spacetime graph be drawn the speed
at any instant is given by the gradient of the graph at the point
corresponding to that instant. (See 27.)
e.g. in the figure Q is the point corresponding to t = 2, s = 27.
The gradient of the tangent at Q is 27. The speed at the end
of 2 seconds is 27 ft./sec.
66. Similarly if we draw the speedtime graph
the gradient at any point is [shew that this is 12<] and =
dt dt
gives the acceleration in ft./sec. a
V
40
Fig. 27.
2 3
t
52
68 CALCULUS FOR BEGINNERS [CH. II
Thus if the speedtime graph be drawn the acceleration at
any instant is given by the gradient of the graph at the point
corresponding to that instant.
e.g. Q' is the point corresponding to t= 2, v= 27.
The gradient of the tangent is 24. The acceleration at the
end of 2 sees, is 24 ft./sec. 2
67. Generally if y is given as a function of x [or y=f(x),
f(x) being algebraic shorthand for a function of x\ and if the
graph of y=f(x) be drawn, the rate at which y is increasing
compared with x for any value of x is the gradient of the graph
at the point corresponding to this value of a?.
Thus if we could draw accurately the graph of y f (x) and
also draw accurately the tangent at a given point we should be
cty
able to read off the value of  for the value of x corresponding
MX
to that point.
EXERCISES. XL
[To be done as accurately as possible by drawing tangents ; v. 67.]
1. Draw the graph of y = */25  x 2 or a: 2 + j/ 2 = 25.
It is a circle centre at the origin, radius 5.
What is ^ when x = 1, 3.  4 ?
ax
3. Draw the graph of j/= between x = and a; =5.
What is ^ when x=l, 2 ?
dx
6?77
68. Example. Find from first principles j when y = x*.
CX/3C
Let x receive a small increment Aa; and let the resulting
increment in y be Ay.
0668] DIFFERENTIATION FROM FIRST PRINCIPLES 69
Then y + Ay = (x + Aa?) 3
= a? + 3a; 2 . Aa: + 3a; (Ai) 2 + (Aa:) 3
and */ <&>
.'. Ay = 3 2 . Ax + 3a; (Aa;) a + (Aa:) 3 .
.'. ^ = 3a; 2 + 3 . A + (A) 2 .
A
Now if Aaj be made smaller and smaller, each of the last two
terms on the right can be made as small as we please by taking
Aa; small enough.
A?/
.'.  can be made as near to 3a? as we please by making Aa?
small enough.
EXERCISES. XIL
1. Find p if y = k . x 3 where k is any constant.
dx
Make use of the result of Ex. 1 to solve the remaining questions.
2. If a body is travelling ia a straight line and its distance from a fixed
point in the line at the end of t seconds is given by s = 4t 3 , where s is the
number of feet, find the speed at the end of 3 seconds.
3. If the graph y = 5x? be drawn, what is the gradient at the point
(2, 40) and what is the equation of the tangent at that point ?
4. If the edge of a cube be increasing at the rate of '001 of an inch per
second, find at what rate the volume is increasing when the edge is 2 feet.
6. If the radius of a sphere be increasing at the rate of 3 inches per
second, at what rate is the volume increasing when the radius is 3 feet ?
6. If the edge of a cube be measured and found to be 8 inches and if an
error of 5^ inch has been made, what is the approximate error in the
U
calculated volume?
7. If an error of ! / is made in measuring the radius of a sphere, find
approximately the perceutage error in the calculated volume.
70
CALCULUS FOR BEGINNERS
[CH. II
69. The result obtained in 68 that if y a?
Ay = 3ar>Aa; + 3x (A*) 2 + (Ax) 3
may be geometrically illustrated as follows.
The edge of a cube is x ins. and is increased by Ax ins. On
inspection of Fig. 28 the volume added will be seen to be made
Fig. 28.
up of 3 slabs each'of volume x 2 Ax, 3 bars each of volume x (Asc) 8
and a cubical block of volume (Ax) 3 .
/\/y*
The ratio of a bar to a slab is . and the ratio of the small
x
A 3?
block to a bar is .
x
G971] DIFFERENTIATION FROM FIRST PRINCIPLES 7l
When we say that the increase in volume is approximately
3o; 2 Aa; we omit the bars and the small block. Suppose for example
the edge of the original cube is 10 ins. and that it is increased by
Ao; 1
1 in. [i.e. x= 10, Aa; = 1, so that = ^. .1
a; 100 J
= 300 x 1 = 30,
(Az) 3 =001.
So that we take 30 cub. ins. as the increase in volume instead
of 30301.
70. Another illustration of the fact that if y = a?, then
dy 3a*
"1  C/iA/
ax
If x = 3, y = 27.
Take an increased value of x and find the increased value of y
as shewn in the table :
x + Ax
y + Aj,
Ax
Ay
Ay
Ax
35
42875
5
15875
31750
32
32768
2
5768
2884
31
29791
1
2791
2791
301
27270901
01
270901
270901
3001 27027009001 001 27009001 27*009001
Thus as A* is made smaller  continually approaches 3x*
and can be made as near to it as we like if we take Aa; small
enough.
71. Sign of ~ . In 8 50 r? turned out to be negative
dx cN
because p decreased when V increased.
Generally : Suppose y = any function of x.
CALCULUS FOR BEGINNERS
[CH. II
Ay
If  is positive A.r and Ay have the same sign ; i.e. an
increase in x produces an increase in y and a decrease in x
produces a decrease in y.
Suppose the graph of y =f(x) be drawn, P being the point
(x, y) and Q (a; + Ace, y + Ay). Then if Ax and Ay are both
positive P and Q will be placed as in (1), if both negative as
in (2). (Fig. 29.)
Ay
Axf
rig. 29.
In each case the chord PQ has a positive gradient.
But if A* be + and Ay , P and Q will be placed as in (3), if
Aa; and Ay +, as in (4).
In each case the chord PQ has a negative gradient.
dy
So, if is positive for a given value of x, x and y are
71]
dy
SIGN OF f
das
73
dv
both increasing or both decreasing, but if  is negative,
u,Jx
x is increasing and y decreasing or vice versa.
di/
In the graph if ~ is positive at the point (x, y) the gradient
(KB
of the tangent is positive and the curve in the neighbourhood of
ftV
P is shaped like (5) or (6), but if ~ is negative, the shape is like
(7) or (8). (Fig. 30.)
Fig. 30.
74 CALCULUS FOR BEGINNERS [CH. II
EXERCISES. XIII.
1. Find from first principles ^ when y =  .
Illustrate your result by making a table like that in 70. [Use a table
of reciprocals.]
2. Find from first principles ~ when y = x +  .
What is ~ when x=l and when x = 2 ?
dx
Draw the graph of y=x +  between x =  and x=3, and illustrate your
X Ji
answer by reference to this graph.
3. What is the equation of the tangent at the point (2, 2J) on the
curve y = x +  ?
x
ds
4. Find from first principles when s= 3 + 2i + 1 2 .
(tt
If a body move in a straight line, so that its distance from a fixed point
at the end of t sees, is given by s = 3 + 2t + t 2 , find its speed at the end of
(i) 3 seconds, (ii) 10 seconds.
6. Find from first principles ^ :
CiX
(i) ify=p, (ii) ify=2*3+3.
Find the equations of the tangent and normal to each of the curves
7
y = j and j/=2x 3 + 3 at the point corresponding to x = l.
6. Find from first principles ^ if y = *Jx~.
["Use JxTh ^J^*) =
L Jx + h+Jx v/ar + A+x J
7. Find from first principles p when j/ = fc (a constant).
~
8. From first principles, shew that if y =#' then ^ = x
dx o
r n . , , . Ay 2x + Ax "I
[Put y*=x> and get J  %2+% Ay + (Aj/)2 .J
9. From first principles, shew that if a; 2 +u 2 = a 2 , then ^=
dx
T . Aw 2z + Aa; ~1 _ .
Get =    . Interpret this result g ;ometrically.
CHAPTER III
DIFFERENTIATION OF a*
(ll/
72. IN future instead of saying that " 2 is the quantity which
! continually approaches as Ace continually approaches zero, and
from which we can make it differ by as little as we please by
making Aas small enough " we say that " ^ is the limit of 
(or Lt.  j when Ax is indefinitely diminished," but it must be
constantly borne in mind that this is merely an abbreviated form
of the longer statement.
73. A sign which has been recently introduced into
mathematical textbooks enables us to compress this phrase still
further. The sign is * . z * a means that z continually
approaches a and can be brought as near to a as we please.
Thus in 68 we have ^= 3x 2 + 3x. Ax + (Ax) 2 and we may say
Ax
Ay
that as Ax ^ 0,  * Sx 2 , meaning that as Ax approaches zero,
 continually approaches Sx 2 and can be made as near to 3x 2 as*
76 CALCULUS FOR BEGINNERS [CH. Ill
we please if Aa; be brought near enough to zero. The statements
in inverted commas in 72 may be conveniently written
rf y_ Lt *y
~T~ *** A 
Aa;0 Ax
A?/
so that when we have found that as Ax  0, ~ * Bar 2 we may
say ^=3ar>.
J oa?
74. If y = x, find S.
Let as receive a small increment Aa; and let the corresponding
increment in y be Ay.
Then y + A^ = (03 + '&x) n ,
and y = a;",
.'. Ay = (a; + Aa?) n  aj n .
[Now it is proved in algebra that if h < x
(x + h) n = x n + nx n ~ l h + n ( n ~ l ) x nw +><>
for all values of n : also that starting from any term and taking
any number of terms however large, the sum never exceeds a
finite limit.]
.'. A?/= nx n ~ l (Aa;) + . ^af~ 2 (Aa;) 2 + terms containing
2i
higher powers of (Aa;)
= nx n ~ l Aa? + (Ax) 2 x L, where L is a finite quantity.
.*. ^ = nx" 1 + L . Aa;.
Aa;
AV
.*. as Aas * 0, ^  naf*.
Aa;
or = nx n ~ a .
dx
7376] DIFFERENTIATION OF x n 77
75. The proof of the Binomial Theorem for indices Avhich
are not positive integers is a very difficult piece of mathematics,
which will probably never be met with by many readers of this
= x n , =
which does not depend on this theorem, such a proof is subjoined.
book. For those who prefer a proof that if y = x n , = nx
76. We shall first establish the following theorem :
For all values of n, positive and negative, integral and
z n 1
fractional,' as z approaches nearer and nearer to 1,
z 1
approaches nearer and nearer to n and can be made as near to n
as we like if z be brought near enough to 1, or shortly :
t> n 1
T z L
lit.   ~n.
2^1 s1
(1) Let n be a positive integer.
Then by actual division
z n l
  = z n  l + z n * + z n  3 +...+ z + 1.
z 1
There are n terms on the right, each of which can be made as
near to 1 as we please if z is brought near enough to 1.
z 1
or
1Pl 331 = 331
wvi
30301,
111 ~ 1 '
1OPl 030301
101  1 01
etc., and in this case n = 3.
tn
(2) Let n be a positive fraction, say n =  where p and q are
positive integers.
78 CALCULUS FOR BEGINNERS [CH. Ill
p
Also let zP = u, .'. 3 = u q and z = u p .
Z\ W1 ~ Wl
As a * 1, so does w.
Since ^ and g are positive integers,
, W P 1 ,
. . as *!,  ?*j9 and
a w l p
.. Lib.  r = or n.
11 1 107411 0741
e s TTT r = 111 ^r  ' 741j
1 01 s  1 1007491 _ 00749 _
1011" 1011 01 '
and in this case = ='75.
4
(3) Let i be negative =  m where m is positive.
11
"
" 21 ~ a1 "T^T """*' a1 '
z m 1
Since m is positive, Lt.  = m, and Lt. z m = 1.
*! z ~ 1 sfl
z n 1
.*. Lt.  7 = m or w.
*i !
M**l 95351
ag  :  " 465
99504 1_
1011 01
and in this case n =  =  5.
76, 77] DIFFERENTIATION OF X n 79
77. Now with the notation of 74,
Ay _ (x + Aa;) re  x n
Aa; (x + Aa;)  x
/
(
x
x + Ate
( x
~n 1
Z 1
n )
H
)
X 4
where z stands for
x
Now as Ace  0, or z > 1 and the limit of ~, is
x 21
therefore n.
. f. % ._!
Ijt. = nx ,
5 = nx"
dx
This is a most important formula and includes all the results
we have obtained hitherto.
e.g. H,.<* = W.
EXERCISES. XIV.
1. Write down ~ when
dx
dy
dx
= sf>, x 20 , x, a; 3 , ^ar, ., x 1 *, x*, x, 5,
2. In the above piece of work ( 74 or 77) what difference would it
make if y = kx n where k is a constant, i.e. some number independent of x?
3. What difference would it make if y = x n + c where c is a constant?
80 CALCULUS FOR BEGINNERS [CH. Ill
78. The results of the last two examples are very important,
(i) Taking y = kx n
we should have y + Ay k (x + &x) n .
.'. Ay = [(x + Ax') rt  x n ].
i.e. the value of Ay in this case is k times its value in the
case when y = x n .
The rest of the work is the same to the end, except that this
factor k remains, and we get eventually
^ = knx" 1 .
dx
(ii) Taking y = x n + e,
we should have y + Ay = (03 + Aa;) n + c.
.*. Ay = (x + Ace) 71 x"
exactly as when y x n .
dy
.'. in this case ~ = nx"" 1 .
dx
EXERCISES. XV.
en j/=3x 5 , 2x, ,
ax x
I. Write down ^ when j/=3x 5 , 2x, , 5Jx, 2a;2 + 3,  8,
 1
2. Write down if =t 2 + 5.
di o
a. Write down  if pv* >4 = 500.
dv
4. Write down if R= k (1 + a0) where k and a are constants.
79. Graphical illustration of the result of 78 (i).
Draw on the same sheet, using the same scales
y = a; 2 ...(l)andy=3 2 ...(2). [Fig. 31.]
In (1)^ = 2*. In (2)^ = to
^ ' dx ^ ' dx
78/79]
DIFFERENTIATION OF kx n
81
So that at the point (2, 4) on (1), the gradient of the curve
is 4 and at the point (2, 12) on (2), the gradient of the curve is 12.
i.e. the gradient of (2) at any point is 3 times the gradient
of (1) at the point with the same abscissa, i.e. at the point
vertically under it.
II. 0.
82 CALCULUS FOR BEGINNERS [CH. Ill
EXERCISES. XVI.
1. Shew that the tangents at (2, 4) and (2, 12) to the curves y=x z and
y = 3x 2 meet OX in the same point.
2. Shew that if a tangent be drawn to y = x 2 at the point (c, c 2 ) and a
tangent to y kx 2 at the point (c, kc 2 ) the gradient of the second tangent is k
times that of the first and that the two tangents meet OX in the same point.
3. Make and prove similar statements about y x 3 and y = kx* and
generally about y = x n and y = kx n .
80. Kinematical illustration. Suppose that two bodies
are moving in the same straight line and that the distance of
the first from a fixed point in the line at the end of a given time
is determined by the equation s = fi, and that the distance of the
second body from the same point is given by s = 3t 2 ; then the
speed of the first at any instant is 3 times that of the second at
the same instant ; in fact, in any interval of time however small,
the second body moves 3 times as far as the first.
81. Graphical illustration of the result of 78 (ii).
Draw on the same sheet, using the same scales [Fig. 32]
2/=ic 2 ...(l) and y = x*+2. ..(2),
then in each case ^ = '2x,
dx
so that at the point (2, 4) on (1) the gradient is 4 and at the
point (2, 6) on (2) the gradient is 4.
i.e. the gradient at two corresponding points is the same.
This is obvious from the fact that (2) may be obtained by
sliding (1) bodily parallel to OY through a distance 2.
82. Kinematical illustration. Suppose that two bodies
are moving in the same straight line and that their distances
from a fixed point in the line at the end of a given time are
8083]
DIFFERENTIATION OF X n + C
given by the equations s = t 2 and s = t 2 + 2 respectively, then their
speeds are the same at every instant, for the second body is
always the same distance, 2 feet, ahead of the first.
12
10
QY 2 3
Fig. 32.
83. We found in 39 that if s = 5 + 3t + 2 s ,
ds
dt
fjo fiQ /7Q
Now if s = 5,^ = 0; if s = St, 33 = 8; if s=W, ^=
(tt Ctt (Aft
ds d(5) d(3t) d(2f)
so that in this case ^  ~i + \ ' + \ ' .
dt dt dt dt
62
84 CALCULUS FOR BEGINNERS [CH. Ill
Generally if y = u + v where u and v stand for two functions
of x whose differential coefficients we know,
dy du dv
dx dx dx '
for if x receive a small increment ACT, there will be a resulting
small increment in u which we may call Aw, and a small
increment in v which we may call Ay, and Ay, the increment
in y, is equal to the sum of these.
In fact y + Ay = u + AM + v + Aw,
so that Ay = AM + Aw.
A'// A?t i\v
^ y. _j_ ^
Ao; A# Aa: '
If Aos be made smaller and smaller  and  can be made
A Aa;
to differ by as little as we please from = and 7 respectively.
ax dx
Ay du dv .
. .  can be made to diner from y + j by as little as we
Ax ax ax '
please; ie.
dy _ du dv
dx dx dx '
dy du dv dw
Similarly if y = u + v + w+ ... , j = , ly + 5 + ... .
dx dx dx dx
EXERCISES. XVII.
1. Write down ^ (i) if j/=3 + 2a; + 4x2, (ii) if y = 5 + \.
uX ' X X
2. Writedown^ if *= 5(3 Jt2 + i.
(it '2
3. Draw on the same sheet the graphs of y = 3, j/ = 2a;, y = 4a; 2 , and
j/ = 3 + 2x + 4a; 2 , and verify that at any point [say where x=l] the gradient
of the fourth is equal to the sum of the gradients of the other three.
4.
If y = ox 2 + bx + c, shew that f  \ = 4aj/ + fc 2  4ac.
83, 84]
DIFFERENTIATION OF X n
85
84. Example. Find (i) the gradients of the tangents to
the curve
y = 2a; 2 3o;l
at the points where x = 1, 0, 1, 2 ;
(ii) at what point the tangent is parallel to OX.
Also find y when x = 1, 0, 1, 2 and make use of all this
information to draw the graph of y = 2X 2 3x 1 between x = 1
and x = 2.
y = 2x* 3x 1.
.'. we
X
~ l
1
2
y
4
1
2
1
dy
dx
7
3
1
5
Also the tangent is parallel to OX when the gradient is 0,
3
i.e. when x = r, and for this value of x, y  2^.
Plot the points ( 1, 4) (0,  1) (1,  2) (2, 1) ( ,  2^ and
through these points draw lines whose gradients are 7, 3,
1, 5, respectively.
Then our curve must go through all these points and touch
all these lines. [Fig. 33.J
EXERCISES. XVIII.
1. If v=x33x, what is ? Find the values of y and  corre
dx Q*
spending to x 2, 1, 0, 1, 2, and draw the curve y = x 3 3x between
x= 2 and x = 2.
3. If i/=a;*4a;3f4x23 what is j1 Find the values of y and ^
when x= 1, 0, 1, 2, 3, and draw the curve y = x*  4z 3 + 4*2  3 between
x=  1 and x = 3.
86
CALCULUS FOR BEGINNERS
[CH. Ill
3. If y = 2a: 3 9*2 + 12* 3 find ^.
Also find the values of y and ^ when x = 0, 1, 2, 3.
From these data, draw the graph of y = 2x 3 $x + l'2x3 between a;=0
and x=3.
4. Find the points on y =x*  3x 2 + 2x at which the gradient is zero and
draw the graph between x=0 and x=3.
Fig. 33.
85, 86] APPLICATION TO GEOMETRY 87
85. In the following examples we make use of the following
theorem :
If a line whose gradient is m be drawn through the point
(h, k") its equation is
y k = m (x  h).
That this is so is easily seen, for
(1) The equation represents a straight line since it is of the
first degree in x and y.
(2) It is satisfied when x = h and y = k and therefore passes
through (h, k).
(3) Its gradient is m since it may be written
y mx + (k mh).
e.g. the line through (3, 5) whose gradient is 4 is
y  5 = 4 (x  3) or y = 4x  7.
86. Ex. 1. Find the equations of the tangent and normal
(perpendicular to the tangent) at the point for which x = 3 on the
curve y ~ 2x + 3x 3 .
We have ^ = 2 + 9a; 2 .
ax
Whena;=3,y = 87 and ^ = 83.
ax
.'. the tangent is the line through (3, 87) whose gradient
is 83.
.'. its equation is y 87 = 83 (x 3)
or y = 83x162.
The normal is perpendicular to the tangent.
.'. its gradient is ^ ,
oo
and its equation y  87 = ^ (x  3)
or x+83y = 7224.
88 CALCULUS FOR BEGINNERS [CH. Ill
or 2
Ex. 2. In the curve y=~, P is the point where x = h.
Find the equations of the tangent and normal at P.
3C fLfJ 3f*
Wehavey = , '.^=77.
4a ax 2a
W , dy h
.. when a: = ft, y=r and ~ = .
la dx 2a
. h
ls ~
(h^\
h, J whose gradient
., ,. n? h . ..
. . its equation is y  = (x  h),
or
2a
or
The normal is the line through (h,  J whose gradient
2a
'. its equation is y = = (x  Ji),
^rr. 3. From the results of Ex. 2 we can deduce some
a; 2
interesting properties of the curve y = . (Fig. 34.)
4:C&
First we notice that the curve passes through the origin, and
is symmetrical about OY, i.e. the same value of y is obtained by
putting x = h as by putting x h. Also y is always positive.
/ 7t 2 \
Let P be the point f 7t, : ) , and let the tangent and normal
at P meet OY in T and G and let PN be perpendicular to OY.
86]
APPLICATION TO GEOMETRY
h*
80
In (1) put x = 0, .*. y
1U>
f A 2 \ A a h*
i.e. coordinates of T are ( 0, r ) or OT = : , but ON =7.
\ 4a/ 4a 4a
/. OT = ON.
Fig. 34.
A
In (2) put rr0. .'. y = +2a, i.e. the coordinates of G
4#
are ( 0, / + 2a] or OG = j + 2a, but ON = .
\ 4a / 4a 4 a
.'. NG = 2 a, i.e. NG is the same for all positions of P.
90 CALCULUS FOR BEGINNERS [CH. Ill
Again if S is the point (0, a),
h?
ST  SO + OT = a + 7 ,
4a
h?
and SG =OG  OS = a + j .
4a
/. ST = SG,
ie. S is the centre of the semicircle on TG as diameter.
But this semicircle passes through P, since TPG is a right
angle.
.'. ST = SG = SP.
EXERCISES XIX.
1. Taking the unit as ^ along each axis and a = l, what is the
m
equation of the curve in Exs. 2 and 3, 86? Draw the curve from x=  4
to x= + 4.
Let P be the point at which x=3a.
Take OT = ON and verify that TP is the tangent at P.
Take NG = 2a and verify that GP is perpendicular to TP.
2. If P be any point on j/= , S the point (0, a), KK' the line y=  a,
PM perpendicular to KK', prove SP=PM.
[This curve is called a parabola, S is its focus and KK' its directrix.]
9. In your figure to Question 1 shew the focus and directrix, and verify
SP=PM.
x z
4. If the tangent at any point P on y= meets the directrix in Z,
Qct
shew that PSZ is a right angle.
6. Find the equations of the tangent and normal at the poiut (1, 1) to
the curve y = x 3 . If the tangent meet OX and OY in T and ( respectively
and the normal meet OX and OY in G and g respectively, find the co
ordinates of T, t, G, g.
6. Find the equations of the tangent and normal at the point where
x = 2 on the curve t/ = 3a; 4 : also at the point where x=3 on the curve
86, 87] THE FUNCTIONAL NOTATION 91
/ h s \ x 3
7. P is the point f h,  1 on the curve y = %. PN and Pn are perpen
dicular to OX, OY. The tangent at P meets OX, OY in T, t and the
normal meets OX, OY in G, g. Prove
OT = ON, Ot = 2On, ON. NG=3.O7i 2 , Ore.n</ =
o 8
8. Find the equations of the tangent and normal at each of the points
on the curve y 2 =x 3 where x = 4. Draw a figure.
8. Find the equations of the tangent and normal to the curve
at the point where x= 3.
10. The equation of a curve is x + y = a%x3.
Find the gradient of the tangent at the point where x l&a. Find where
this tangent meets the line x+y 0.
11. Find the equations of the tangent and normal at the point where
.T = 2 on the curve y = 3x + a; 2 .
If the point be P and if the tangent and normal meet the yaxis in T and
G respectively find the area of ATPG.
The functional notation.
87. The fact that y is a function of x is sometimes expressed
in the form y=f(x), e.g. if y y?~ Ix + 8, f(x) is x 7x + 8.
/"(2) means the result of substituting 2 for x in this expression,
i.e. /(2) = 2 2 7x2+8 2.
Similarly /(0)=8, /(3) = 38,
f(2p) = 4p 2  14jt> + 8, f(z> + 2) = (z 2 + 2) 2  7 (s 2 + 2) + 8
= z 4  3z 2  2, etc.
Other notations sometimes used are F (x), < (x), etc.
It is very important to bear in mind thaty^a?) does not stand
for the product of x and some quantity f, but is really an
instruction written in shorthand to perform a certain operation
or series of operations on x.
Thus in the above example where f(x) stands for .r 2 7x + 8,
the symbol/ placed before a number means square the number,
subtract 7 times the number and add 8.
92 CALCULUS FOR BEGINNERS [CH. Ill
EXERCISES. XX.
2. If 0(x) = So? 2*2 + , find 0(6), <f>(y + l).
cc
3. Itf(x)=x + sinx [x being in radians] find/f Y/(1),/(0).
88. If y=f(x), the differential coefficient of y with respect
to x may be written ^ or 7 or /' (a;).
rfcc ox
Thus if ^Sa^
f (x) is itself a function of or, and we may therefore evaluate
such expressions as
/' (3), /')),/'(* + 3), etc.
e.g. /'(3) = 9x3 2 18x3+7
= 34.
With our usual notation we should say that this was the
d'U
value of r when x = 3.
ax
Thus we may say that the gradient at any point of the
curve y =f(x) is /' (a:), or that the gradient at the point for
which x = a isf (a).
Similarly if s=f(t),
ds
j may be written y (<),
and we may say that /' (t) gives the speed at the end of time t.
8890] HIGHER DIFFERENTIAL COEFFICIENTS 93
Higher differential coefficients.
89. If y =/() then or/' (x) is also a function of x and
may therefore itself be differentiated.
The result of this differentiation may be written
dx df'(x)
dx dx
d?y
These are abbreviated into  and/" (x) respectively.
e.g. if y =/() = 2x 4 3x 3 + 2x,
90. Just as p or/' (x) tells us the rate of increase of y orf(x)
CvX
d*y
with respect to x, so 7^ or /" (x) tells us the rate of increase of
DNv
, or/' (x) with respect to x.
dx
Considered in relation to the graph y =/(*), ~ or/' (05) gives
cLx
d?u
the gradient at any point on the graph ; ~ or/" (a;) gives the
doc?
rate at which the gradient is increasing with respect to x.
If in the last illustration we put x = 2, we get
/(2) = 2x2 4 3x2 s + 2x2
= 12.
/' (2) = 8 x 2 3  9 x 2 2 + 2
= 30.
/"(2)=24x2 2 18 x2
= 60.
CALCULUS FOR BEGINNERS
[CH. Ill
i.e. at the point (2, 12) on the curve y = 2o^  Sx 3 + 2x, y is
increasing 30 times as fast as x, or the gradient of the curve is
30, and the gradient is increasing 60 times as fast as x. (Fig. 35.)
30 '" ztjZ^^d:
20 ;;;;li;ii;lil nfT :
10 1 i iiiiiiii
o:;; l f~2~^.^
60!; jijjijijj
; ;;!
40:::
20;:; :;i ! ill!!!!!!';;
jiii jiji;  3^
100 ::
:::::::::::::::::! JCTjto'^^i
HJ:;;;  i:;;;;;!
O 1 i 4 2 3
igs. 35, 36, 37.
Or we may say that at the point (2, 30) on the curve
= Bar* 
which is called the first derived curve of y = 2x*  3x? + 2x, the
gradient is 60. (Fig. 36.)
9093] DERIVED CURVES 95
91. If we draw the graphs
y =/(),
y=f'(*\
y =f" (*)>
the number of units in the ordinate of y =/' (x) corresponding to
any value of x gives the gradient at the corresponding point on
y =f(x), and the number of units in the ordinate of y =f" (x)
gives the gradient at the corresponding point on y =f (x}.
e.g., looking at Figs. 35, 36, 37 which give the graphs of
y =f(x), y =f (x), y f" (x) for the particular case when
f(x)= 2x 4 3x s + 2x,
P, P', P" are the points corresponding to x 2. The ordinate of
P" is 60 units and 60 is the gradient at P'. The ordinate of P' is
30 imits and 30 is the gradient at P.
ds
92. If s=f(t), j or f (t) (sometimes written s) gives the
ut
speed at end of time t.
cti}
If we call this v we have seen [v. Ex. (5) p. 51] that y gives
ttt
the acceleration at end of time t.
dv . cPs .
<Ps
Thus j^ or f" (t) [sometimes written s] gives the acceleration
ut
at end of time t.
The graphs of f(t), f (t), f" (t) are respectively the space
time, speedtime and accelerationtime graphs.
93. We get interesting special cases when f(x) is of the
1st or 2nd degree.
e.g. if /(a) = 2o: 2 3a:tU
/ = 4*3>
and f" (x)  4. J
96
CALCULUS FOR BEGINNERS
[CH. Ill
The graphs of y=f(x), y=f'(x), yf"(x) are shewn in
Fig. 38.
y f ( x ) is a straight line, and the gradient at every point is
the same.
y =f" (a;) is a straight line, the ordinate at every point being
the same, i.e. it is a straight line parallel to OX.
1 2
Fig. 38.
In the general case,
if f( x } =a y ? + bx + c, where a, b, c are constant,
f (x) = 2ax + b t
and f" (x) = 2a>
93] DERIVED CURVES 97
If s =f(t) = at 2 + bt + c,
v =/' (t) = 2at + b,
and a =/" (t) = 2a.
i.e. if s be a quadratic function of t, the acceleration is
constant.
In this case the speedtime graph is a straight line and the
accelerationtime graph a straight line parallel to OX.
If f(x) = 4:X + 3,
/'(a) = 4,
/"(*)= a
The graphs are shewn in Fig. 39.
y =f(x) is a straight line,
y =f (<) is a straight line parallel to OX,
y ~f" (x) is OX.
Generally if f(x) = ax + b t where a and b are constant,
/ (x) = a,
/*(*)<>.
If s =f(t) = at + b,
v=f'(t) = a,
a=f"(t) = 0.
i.e. if s be a linear function of t the speed is constant and the
acceleration zero.
EXERCISES. XXI.
1 . If / (*) = 2z +  , find /' (x) and /" (x) .
X
Alsofind/(2),/'(2),/"(2).
Draw the graphs y=f(x), y = f (x), y=f"(x) between z=l and
*= + !.
Make a statement like that in 91, taking x ^ .
2. If = 34 + 2t 2 + t 3 find the speed and acceleration at the end
of 3 seconds.
Draw the spacetime, speedtime and accelerationtime graphs.
M. C. 1
98
CALCULUS *OR BEGINNERS
[CH. Ill
fjm
94. We have seen (71) that if ^ is positive, y is increasing
dy .
as x increases, but that if f is negative y is decreasing as x
increases.
Q. .. . ... dx) . dy . .
Similarly if ^ or   is positive,  is increasing as x
increases, but if ~ is negative, ~ is decreasing as a? increases.
94]
dy d?v
SIGNS OF / AND ~
dx da?
99
Now if between P and Q a curve has a shape as in Fig. 40 (a)
it is clear that as x increases from OM to ON, y increases and also
Ul/
the gradient or ~ increases. [The positive directions of the x
CvCC
and y axes are supposed to be as usual 'right' and 'up' respectively.]
(a)
(b)
(c)
(d)
Fig. 40.
i.e. for all points between P and Q
dy . . (Fy .
? is + and ~ is + .
dx dy?
72
100 CALCULUS FOR BEGINNERS [CH. Ill
Shew similarly that corresponding to a shape like Fig. 40 (6)
dy . , o? 2 v .
f is + and r^ is .
ax ax
For Fig. 40 (c) ^ is  and ^ is + .
dx dx*
For Fig. 40 (d) ^ is  and ^ is  .
dx dx
95. e.g. consider the curve y = 2x* 9x 2 + l'2x 3.
We have ^. = Qx 2  1 8* + 1 2
ax
= 6 (ar1) (x 2),
and ^=12a;18.
do: 2
For all values of x between 2 and 3,  is + and ^ is + .
dx dx 2
.'. between x = 2 and x = 3 the curve is of the form shewn
in Fig. 40 (a).
For all values of x between and 1, ^ is + and . is ,
dx dx 2
.'. between x = and x = 1 the curve is of the form shewn
in Fig. 40 (6).
For all values of x between 1A and 2, ~ is and ~ is + .
dx dx 2
.'. between xl^ and x= 2 the curve is of the form shewn
in Fig. 40 (c).
For all values of x between 1 and 1 1, ^ is and ~ z is .
CvSO CL3&
.'. between x= 1 and x=\^ the curve is of the form shewn
in Fig. 40 (d).
[See Fig. 41.]
04, 95]
du d 2 y
SIGNS OF f AND ~,
doc dx*
Fig. 41.
101
EXERCISES. XXII.
What is the form of the following curves ?
: (i) between x = 2 and x=3,
(ii) x = and x = l.
: (i) a;=l and x = 2,
(ii) x= 1 and x= 2,
(iii) a; = 3anda; = 4.
(i) when a; is positive,
(ii) ,, negative.
a. =
102 CALCULUS FOR BEGINNERS [CH. Ill
MISCELLANEOUS EXAMPLES ON CHAPTER III.
1. Find from first principles
(i) ^ify = x^. (ii) p if pv z =k (a constant).
a. Find the equations of the tangent and normal to the curve y = 
at each of the points (2, ^ ) , (5, = j and in each case find the area of the
triangle formed by the tangent and the axes of coordinates.
3. Do the same for the point ( c,  ) . State a geometrical property of
\ c J
the tangent at any point of the curve.
4. P is the point (3, 9) on the curve y=x 2 . Find the equations of the
tangent and normal at P, and the coordinates of the points T, G where they
meet the yaxis. If PN is perpendicular to the ?/axis, find the lengths of
NT and NG.
6. A body moves in a straight line in such a way that its distance (s ft.)
from a fixed point in the line at the end of t seconds is given by
Find its distance from the fixed point, its speed and acceleration at the end
of 5, 6, 7 seconds.
6. The same if =3 + t5t 2 + 7t 3 .
7. In the curve j/ = 2a;+3rr 2 , find the gradient at the point P where
x = 2.
Find also the gradient of the chord joining P to Q, Q being the point
where x =2 01.
The tangent at P and the line through P parallel to OX meet the ordinate
of Q in T and R respectively. Find the percentage error made in using RT
instead of RQ. Explain the connection between the statements RT = RQ
approximately and Ay = ~.Ax approximately.
ax
8. Find the equation of the tangent to the curve
y = 3x*  4x 8  12x 2 + 5,
at the point where x 3.
Also find the coordinates of the points at which the tangent is parallel
to the xaxis.
95] MISCELLANEOUS EXAMPLES 103
9. P is the point on y = cx n (c a constant) at which x = h. The tangent
and normal at P meet the xaxis in T, G, and the yaxis in t, g. PN, Pn
are perpendicular to OX, OY. Find
.. OT ., Ot .. ON.NG . On. ?i0
* (m) ' (lv)
ON oS* 5S
10. A trough is in the shape of a right prism with its ends equilateral
triangles. The length of the trough is 10 feet. It contains water, which
leaks at the rate of 1 cubic foot per minute. Find in ins. /sec. the rate at
which the level is sinking when the depth of water is (i) 3 inches, (ii) 1 foot.
11. A current C of electricity is changing according to the law
C
where t is seconds.
The voltage V is such that
VRC+L.*?.
where R = 05, and L = 0'01.
Find V when t=2.
12. A man runs a given total distance in such a way that the time,
t sees. , he takes to run any part, s feet, of the distance is given by the equation
t=as n , a and n being constants. Compare his average speed over the whole
distance with his speed at the end.
13. A body of constant mass m is moving with variable speed v.
JLf
If K is its kinetic energy and M its momentum shew that = M.
State what your units are.
14. If pV=k (a constant) shew that Tn = r>
andifpV n =&, ^ = ~ n U'
16. The coefficient of cubical expansion of a substance at temperature
6 is the rate of increase of volume per unit increase of temperature.
The volume (V c.c.) of a gram of water at 9 C. is given by
where a=8'38x!0 9 .
rfV
Find and hence get the coefficient of cubical expansion of water at
da
C. and at 20 C.
104 CALCULUS FOB BEGINNERS [CH. Ill
16. The specific heat of a substance at temperature 6 is the rate of
increase of Q per unit increase in 0, where Q is the number of heat units
required to raise the temperature of 1 gm. from some standard temperature
to0.
The total heat required to raise the temperature of 1 gm. of water from
to C. is given by
Q= e + 2 x 106 02 + 3 x io7 03.
Find the specific heat of water at 80 C.
17. For diamond the formula is
Q= 09470 + 0004970 2  0000001203.
Find the specific heat of diamond at 80 C.
18. If y = 2x*2x3x2 + l find y when x=l,   , 0, 1, 2.
Also find T when x=  1, 2 and find what values of x make ; = 0.
dx dx
Using all this information draw the graph of
between x=l and a; = 2.
19. In a certain case of straightline motion, the number of feet in
the distance from a fixed point, at a given instant t seconds after the start, is
given by the formula
Calculate
(i) The average speed between the end of the 2nd and the end of the
4th second.
(ii) The mean of the speeds at the instants 2 and 4 seconds after
the start.
(iii) The speed 3 seconds after the start.
2O. Find the equations of the tangent and normal to y = x 3 at the point
(2, 8).
If P is the point (2, 8) and if the tangent and normal meet the xaxis in
T and G and PN is perpendicular to OX, find the lengths NT, NG.
(h 2 \ x 2
h, 1 on the parabola y= ; PT is the tangent
at P. PM is perpendicular to the line y = a. S is the point (0, a).
Prove SM is perpendicular to PT and meets it on the rraxis.
22. A vessel containing water is in the form of an inverted hollow cone
with vertical angle 90. If the depth of water be x feet, what is the volume
of water ?
If water flows in at the rate of 1 cubic foot per minute, at what rate is
the level of water rising when the depth is 2 feet ?
21.
95] MISCELLANEOUS EXAMPLES 105
23. Find the coordinates of the point of intersection of the tangents to
y = So; 2 + 5x  7 at the points where x = 2 and a; = 5.
Verify that this point and the midpoint of the chord joining the two
points of contact lie on a line parallel to OY.
24. The shape of the vertical section of a hill is given (approximately)
by the graph of y = 05x 2 + 25x from x = Q to x=6 and by the graph of
15
froma? = 6 to x = 12.
Calculate the ordinates when a; = l, 2, 3, ... 12 and the gradient at each
of the corresponding points. Shew that when x = 6 the gradient is the same
for each part of the hill. Draw the figure taking  as the unit each way.
m
25. Find the equation of that tangent to the parabola y = Bx 2 which is
parallel to 'Ax  2y = 7.
26. If s = 3t  7t 2 + 16t 3 , s being the number of feet and t the number of
seconds, find the formula for the acceleration at the end of t seconds. If
8 Ibs. is the mass of the moving body, what is the force acting on it at the
end of 4 seconds ?
27. A body is moving in a straight line and its distance (.s feet) from a
fixed point at the end of t seconds is given by the following table :
t 1 2 3 4 5 6
s 7 22 45 76 115 162
Find graphically the speed at the end of 5 seconds.
28. A wheel in t seconds rotates through (5t + 4( 3 ) radians from some
standard position. Find its angular velocity and acceleration after 5
seconds.
29. The side walls of a truck are vertical and its section by a plane
parallel to the side walls is a trapezium, upper side 14 feet, lower side 10
feet, depth 3 feet. The width of the truck is 4 feet. It is originally full of
sand (sp. gr. 15) which leaks out in such a way that the depth of sand
diminishes at a constant rate of 1 inch per minute. Find the mass of sand
in the truck after t seconds. If the truck is made to move in a straight line
with a uniform speed of 15 miles an hour, what is the momentum of the sand
in the truck at the end of t seconds ? What is the resultant force on the
truck at the end of t seconds ? [Kemember force is given by rate of change
of momentum.]
Find the force at the end of (i) 5 minutes, (ii) 20 minutes.
30. If
di/
find graphically the value of 2 when x =
CHAPTER IV
MAXIMA AND MINIMA
96. PLOT roughly the curve y = 2ac?  $a? + 12x  3 between
x = and x = 3.
You will get a curve as in Fig. 42.
s
: 
Fig. 42.
d/u .
Between A and B, ^ is positive,
CHB
B and C negative,
C and D ,, positive again,
96, 97] MAXIMA AND MINIMA 107
or in other words as we advance from left to right, y increases
up to B and then begins to decrease ; as we come to C, y goes
on decreasing and then begins to increase. (71.)
Points like B and C are called turning points and at B
y is said to have a maximum value, and at C a minimum
value.
Notice, that although we call the value of y at B a maximum,
yet this is not by any means the greatest value y can have.
A maximum point is one where the ordinate (representing the
function) stops increasing and begins to decrease; in other words
the value of y at a maximum point is greater than the value of y
at neighbouring points on either side, and similarly at a minimum
point the value of y is less than the value of y at neighbouring
points on either side.
97. Now suppose we wish to find the exact position of these
turning points B and C.
At a turning point the tangent to the curve is parallel to OX,
i.e. the gradient of the curve is zero, or
.
ax
dv
.'. to find the turning points, we get ~ and find the
dv
values of x which make ^ = 0.
dx
e.g. in the case above
:. f = when x = I or 2.
dx
When x = 1, y  2 ; when x = 2, y = 1.
.'. (1, 2) and (2, 1) are the turning points.
108 CALCULUS FOR BEGINNERS [CH. IV
98. In order to find whether a turning point corresponds to
a maximum or minimum we may either draw a rough graph or
proceed as follows :
when x 1, y = 2 ;
when x= 1 + h, y = 2(l + A) 3  9 (1 + A) 2 + 12 (1 + A)  3
= 2  3A 2 + 2A 3 .
Now if h is a small fraction, 27i 3 is small compared with 3A 2 ,
and the sign of 3A 2 + 2A 3 is that of the first term, and is there
fore negative whether h is positive or negative.
Thus when x=\ t y = 2 ; but if x is slightly greater or less
than 1, y < 2. .'. x = 1 corresponds to a maximum value of y.
Similarly, shew that when x ~ 2, y = 1 ; but if x is slightly
greater or less than 2, y> 1.
.'. x = 2 corresponds to a minimum value of y.
99. There is an exceptional case which we shall consider
later. Suppose
y = So/  IQx 3 + SOar*  24a + 5.
ax
= 12 (!)( !)(* 2),
and ? = when as = 1 or 2.
aa5
Ifoj = l, y = 2.
If x= 1 + A, y =  2  47i 3 + 3A 4 .
If A is a small positive fraction, y < 2.
If A is a small negative fraction, y > 2.
Thus when x = 1, y 2.
If x is slightly greater than 1, y < 2.
If as is slightly less than 1, y> 2.
So that here there is neither maximum nor minimum.
The graph is shewn in the first part of Fig. 56, p. 130, and
the point P is called a point of inflexion.
98100]
MAXIMA AND MINIMA
109
100. We may state the results of 97 :
"The values of x which make f (x) a maximum or
minimum are roots of f'(x) = 0."
In the case considered
/ (x) = 2a?  9x 2 + I2x  3,
Fig. 43.
110
CALCULUS FOR BEGINNERS
[CH. IV
If we plot the curves y =f(x) and y =/' (x) taking the same
xscale each time, the points where y =f (x) cuts the ccaxis will
correspond to turning points on y =f(x). (Fig. 43.)
Another Test for Maxima and Minima.
101. It is evident from Fig. 44 that if we advance from
dv
left to right, i.e. in the positive xdirection, f'(x) or p
changes from + to as we pass through a maximum
point and from to + as we pass through a minimum
point. (See 71.)
orffx)
a maximum
a mnimum
or f(x)
Fig. 44.
Consider the example in 96.
/ (x) = Sec 8 9^+ 12* 3,
= 6 (x !)( 2).
The values of x, which make/' (a;) = 0, are 1 and 2.
If x is slightly less than \,f'(x) is + for x 1 is 
and x 2 is / '
If x is slightly more than 1, /' (x) is  for x  1 is
05
'I is +]
'.  2 is / '
100102] MAXIMA AND MINIMA 111
This may be abbreviated conveniently :
If =!, /=()()
If =!+, /'(*) = (+)()
.'. as we pass from left to right through the point (1, 2)
/' (x) changes from + to .
.'. (1, 2) is a maximum point.
Similarly shew that (2, 1) is a minimum point.
102. If we apply this test to the case in 99,
where f(x) = 3x*  16x* + 30a; a  24a + 5,
we have f'(x) = l2(xl)(xl)(x2).
The values of x which make /' (a;) = are 1 and 2.
If x=\ /'() = ()()()=)
If x=I + , / = ( + )( + )() = /'
i.e. f(x) does not change sign "as we pass through the point
(1, 2), and this point is neither a maximum nor a minimum.
EXERCISES. XXIII.
1. Find the turning points of the curve y = x 3  Bx and plot enough of
the curve to shew these points.
2. Shew that y=x 3 + 3x has no turning points. Plot it between x=  2
and x = + 2. Also plot y =/' (x).
3. Find the turning points of the curve
4. Find the turning points of the curve
and plot between x=2 and x=+2.
Also plot y =f (x) between the same limits.
5. Shew that y=x*+2x z + l has only one turning point and plot
between x =  2 and x = + 2. Plot y =/' (x).
9. Find the turning points on
Plot between x=  2 and x= +2. Plot y=f (x).
7. Shew that y=x +  has two turning points and that the maximum
SB
point is lower than the minimum.
112 CALCULUS FOR BEGINNERS [CH. IV
8. Find the turning point of y = 2z 2 3a; + 5. Is it a maximum or a
minimum point?
9. Shew that y = a.r 2 + bx + c has always one turning point and that it
is a maximum or minimum according as a is negative or positive.
10. Shew that y = ax z + bx* + ex + d has two turning points or none.
103. Examples. (1) The stiffness of a rectangular beam
varies as bcP where b is the breadth and d the depth. Find
the breadth and depth of the stiffest beam the cross section of
which has a perimeter of 4 feet.
Let the depth be x feet.
.'. the breadth is (2 a;) feet, and the beam will be stiffest
when (2 a;) 3? is greatest. Call this y.
y = (2x)x*
= 2** x\
:. ^ = 6** 4**.
dx
For a maximum or minimum value of y
.
dx
3
.'. x = or = .
A
x = obviously corresponds to a zero value of y.
*jc =  ives a maximum.
* It is clear that there must be a maximum between x = and x = 2, for
y = when x = or 2 and is positive for intermediate values of x, or we may
say
ft.
dx~
When *=,
when = + ,
.*. x = g gives a maximum value of y.
103J
MAXIMA AND MINIMA
113
Thus the stiffest beam is 1 ft. deep and ft. broad.
Fig. 45 shews the graph of y = (2 x)x 3 between x and
1 15
Fig. 45.
(2) The volume of an open cylindrical canister is to be 100
cubic inches. Find the most economical dimensions, i.e. the
dimensions which give the least surface area.
Let the radius of base be x ins. and the height h ins.
Then surface area = (TTO? + 2nxti) sq. ins.
We want the least value of this.
At present the surface involves two variables x and h but we
are iven
7,_ 100
Surface area =7rjc 2 + 2x .
200
100
M. C.
114 CALCULUS FOR BEGINNERS
Calling this y, we have
[CH. IV
dx
.
a?
dy
For a maximum or minimum value of y, f = 0.
<ttB
and
_
Sr
ioo
TT
100
To6
3J / o 1 1 j
100 TT 3 /lQO
=  x  = ./  = 317,
v 10QS V *
350
300
250
20(
150
100
50
1
2345
Fig. 46.
The corresponding value of y is 30 \/!OTT = 94*6. [Shew that
this is a minimum value.]
.*. The canister has the radius of the base and the height
each 3'17 ft. and 94*6 sq. ins. of material are required to
make it.
103] MAXIMA AND MINIMA
[That x  h can be seen as follows:
X J 100 f " x=Jl \
Fig. 46 shews the graph of
y=ira
115
200
x
(3) Find two factors of 24 whose sum is a minimum.
24
Let the factors be x and and call their sum y.
x y
Fig. 47.
82
116 CALCULUS FOR BEGINNERS [OH. IV
24
Thus y x + ,
oc
__
* dx~ ar"
/. ^ = o whenar> = 24.
ax
is a maximum or minimum when
= + 4899...
If a: = 4'899, sum of factors = 9'798... .'. Sum is a minimum
If x 5, other factor is 4 '8, sum = 9'8j when x = \/24.
Similarly shew that sum is a maximum when
Thus the required factors are 4 '899, 4 '899.
Fig. 47 shews the graph of
24
104. The following example illustrates a useful device.
A ship B is 75 miles due east of a ship A. B sails west at
9 knots and A south at 12 knots.
Find when their distance apart is least. Find also what the
east distance is.
P, Q are positions after x hrs. [Fig. 48.]
AP = 75  9cc = 3 (25  3x) miles,
AQ = 12o; =3(4tt),
/. PQ 2 = 9 [(25  3ic) 2 + (4o;) 2 ]
= 9 [625 150Z+250 2 ]
103105] MAXIMA AND MINIMA 11
Put = x*
.*. y is a minimum when x= 3. [Why minimum?]
.'. least distance is after 3 hrs,
A r
759x  > p
I2x
Fig. 48.
When <r = 3, PQ 2 = 225xl6,
.'. PQ=60.
Instead of finding when PQ. is a minimum we found when
PQ 2 was a minimum. If we had expressed PQ in terms of a?, we
should have 15^25 6a; + y? and we do not yet know how to
differentiate this.
105. Notice that we have to express the quantity, of which
we wish to find the maximum or minimum value, as a function
of one unknown.
In some cases it is first of all necessary to express it in terms
of two or more unknowns, and then by means of given conditions
to express all these unknowns in terms of one of them.
118 CALCULFS FOR BEGINNERS [CH. IV
Thus in the second example the surface was written first
as a function of x and h, then by means of the given relation,
?rx 2 A= 100, we were able to express h in terms of x, and so get
the surface as a function of x only.
106. A common error in questions of maximum and
minimum values arises from a confusion between variables and
constants. Take, for instance, the second example of 103.
If y sq. ins. is the surface area
y = irx* + 2nxh.
The error consists in saying that
 = 2nx + 2irh.
ax
This would only be right if h were constant, i.e. if A remained
the same when x changed, which is obviously not the case, for
since the volume remains the same, h must increase when x
decreases and decrease when x increases. In fact h is a function
of x, being connected with x by the equation
100
or h = y,
7TC 3
and this value must be substituted for h before we proceed to get
dy
dx'
EXERCISES. XXIV.
1. Find the number such that when it is added to its reciprocal the
sum is a minimum.
2. The sum of the square and the reciprocal of a number is to be a
minimum.
Find the number. Draw a graph shewing how this sum changes with
the number (positive numbers only).
3. Divide a number a into two parts so that three times tbe square of
one part plus four times the square of the other shall be a minimum.
105, 106] EXERCISES 119
4. Find the dimensions of the rectangle of greatest area with a perimeter
of 10 feet.
Draw a graph shewing how the area changes with the length of the
rectangle.
6. Prove that the greatest rectangle that can be inscribed in a given
circle is a square.
6. A rectangle PQRS is inscribed in a triangle so that P lies in AB,
Q in AC and R, S in BC. If BC = a, QR=a; and the altitude of the
triangle = h, shew that PQ = a ( 1   J . Shew that the rectangle of maximum
area has its height onehalf that of the triangle.
7. Find the dimensions of the greatest cylinder that can be inscribed in
a cone of radius r and height h, the planes of the bases of the cone and
cylinder being the same.
8. The surface of a hollow cylinder without top is 100 sq. ins. Find
the maximum volume.
9. The volume of a solid cylinder is V cubic ins. Find its dimensions
if the total surface is a minimum.
10. A sheet of tin is 6 feet by 4 feet. A square is cut out of each
corner, and a tank made by bending up the projecting portions. Find the
side of the squares cut out so that the volume of the tank may be a
maximum.
11. The number of tons of coal consumed per hour by a certain ship
is 0'3 + 0OOlV 3 where V knots is the speed. For a voyage of 1000 miles at
V knots, find the total consumption of coal. Find for what speed the coal
consumption is least.
Do the same for 100 miles.
12. A running track is in the form of a rectangle ABCD with semi
circles on AB, CD. If the perimeter is a quarter of a mile find the
maximum area.
13. A window is in the form of a rectangle surmounted by a semicircle.
If the perimeter is 30 feet, find the dimensions so that the greatest possible
amount of light may be admitted.
14. The strength (i.e. resistance to breaking) of a rectangular beam
varies as bd z where b is breadth and d depth. Find the breadth and depth of
the strongest beam that can be cut from a circular log of diameter 2 feet.
120 CALCULUS FOR BEGINNERS [CH. IV
15. A shot is projected in vacuo with velocity ft./sec. in a direction
making an angle a with the horizontal. Its height above the ground at the
end of t seconds is tu ein o  ^ gt 2 . Find the greatest height and the time of
i
reaching it.
16. A shot is projected with velocity u ft./sec. in a direction making an
angle a with the horizontal. An inclined plane making an angle /3 with the
horizontal passes through the point of projection. The distance of the
shot from this plane at the end of t sees, is
tu sin (a  ft)   g cos . t 2 .
2i
Find its greatest distance from the plane and the time of reaching it.
17. By the G.P.O. regulations the combined length and girth of a
parcel must not exceed 6 feet. Find the volume of the greatest parcel that
can be sent
(i) in the shape of a rectangular solid with square ends,
(ii) in the shape of a cylinder,
(iii) in the shape of a cone.
18. An open tank whose base is a square has to contain 1000 cubic feet
of water. Find the least cost of lining it with lead at 4d. per sq. ft.
19. A man has 108 square feet of iron which he is going to make into
a rectangular tank, lidless, and such that the depth is to be double the
breadth. What will be the length, breadth and depth of the tank con
structed under these conditions which will hold the greatest volume of
water?
20. Find the circular cylinder of largest volume which can be cut from
a sphere of radius 6 inches, the plane ends being perpendicular to the axis.
21. Find the volume of the greatest cylinder that can be inscribed in
a sphere of radius a.
What is the ratio of the height of the cylinder to the diameter of the
base?
22. Find the angle of the circular sector of greatest area that can be
made having a given perimeter a.
23. A dynamo is in two parts whose weights are x and y. The cost of
the machine is (r/ + 4a:). The usefulness of the machine is proportional to
(a; 2 + 3xy). Find the values of x and y for maximum usefulness in a dynamo
worth 10.
106] EXERCISES 121
24. The cost per hour of driving a steamer through water varies as the
cube of the speed relative to the water. Shew that the most economical speed
3
against a current of V miles/hour is ^ V. [If C is cost per mile, find when
31
^ is a maximum.]
C
25. Given 200 sq. feet of canvas, find the greatest conical tent that can
be made out of it. [Get V 2 in terms of r and I (slant height).]
26. Given a circular sheet of paper. Find the angle of the sector
which must be cut out so that the remainder may be folded to give a conical
vessel of maximum volume.
27. Find the least area of canvas that can be used to construct a conical
tent whose cubical capacity is 800 cubic feet.
28. Find the volume of the greatest right cone that can be described by
the revolution about a side of a rightangled triangle of hypotenuse 2 feet.
29. There are n voltaic cells each of E.M.F. E volts and internal
resistance r ohms; x cells are arranged in series and  rows in parallel. The
current C amperes that the battery will send through an external resistance
R is given by
Given n=20, R = 25, r = 2,
find how many cells should be arranged in series to give the maximum current.
30. A battery, internal resistance r ohms and E.M.F. E, sends a current
through an external resistance R.
The power W given to the external circuit is given by
RF2
W
~(R + r)2*
Given E and r find what the external resistance must be so that the power
may be the greatest possible.
31. The power W given to an external circuit by a generator of internal
resistance r ohms and E.M.F. E when the current is C amperes is given by
W=CEC2r.
If E=60, r=3,
find C so that W may be a maximum.
122 CALCULUS FOR BEGINNERS [CH. IV
33. The annual cost of giving a certain amount of electric light to a
certain town, the voltage being V and the candlepower of each lamp C, is
found to be
A=a + y for electric energy,
o
_ 7 nV , ,
and B = ^ i   for lamp renewals.
C T
The following figures are known when C is 10.
V 100 200
A 1500 1200
B 300 500
Find a, b, m, n. If C=20 what value of V will give the minimum total
cost?
33. Shew that the Mechanical Power of a motor is greatest when the
efficiency is 50 / .
[Mec. Power = e.Ca; EfBciency=pj Ee=Ca. Ra. E and Ra are con
stants.]
34. A given weight W hangs from a point B of a straight horizontal
lever ABC which can turn about A. The lever weighs w Ibs. per foot length.
If the length of AB is given, say a feet, find the length of lever for which
the requisite vertical supporting force at C will be a minimum.
35. A piece of wire of length 6 feet is to be cut into six portions, two
equal of one length and four equal of another. The two former are each
bent into the form of a circle and these are held in parallel planes and
fastened together by the four remaining pieces which are perpendicular to
the planes of the circles. The whole thus forms a model of a cylinder.
Calculate the lengths into which the wire must be divided so as to produce a
cylinder of maximum volume.
36. In measuring electric current by a tangent galvanometer the per
centage error due to a given small error in the reading is proportional to
tan x + cot x. Find the value of x for which this is a minimum. [Put
37. A log is in the form of a frustum of a cone, 20 feet long, the
diameters of the ends being 2 feet and 1 foot. Find the length of the greatest
beam of square section that can be cut from it.
Do the same if the length is I feet and the diameters of the ends a and b
feet.
What is the relation between a and b if the length of the greatest beam
is the same as the length of the log ?
107] EXERCISES 123
38. Find the length of the shortest line which will divide a given triangle
into two parts of equal area.
39. An electric current flows round a coil of radius a.
A small magnet is placed with its axis on the line perpendicular to the
plane of the coil through its centre.
If x is the distance of the magnet from the plane of the coil, the force
exerted on it by the current is proportional to
Find x so that the force may be a minimum. [Put a; 2 + a 2 =j/.]
4O. A tank standing on the ground is kept full of water to a depth a ft.
Water issues from a small aperture at a depth ft ft. below the surface with
velocity *J%gh ft. /sec.
Find ft in order that the water may strike the ground at the greatest
possible distance from the tank.
Another way of distinguishing between Maximum
and Minimum.
107. If f(x) is a function of x which is increasing as x
increases, or decreasing a? x decreases, f'(x)is positive [71],
but if f(x) is decreasing as x increases, or increasing as x
decreases, f'(x) is negative. [Fig. 49.]
increases
f (x) increasing as X increases f (x) decreasing as X
f (x) +
Fig. 49.
124
CALCULUS FOR BEGINNERS
[CH. IV
Now as we pass through a maximum point /' (x) is decreasing
as x increases.
In Fig. 50 (a) the gradient of the tangent (1) is positive; that
of (2) is positive and less than that of (1) and so on, that of (4) is
zero, that of (5) is negative, that of (6) is negative and numerically
greater than that of (5) and so on, so that all the way f (x) is
algebraically decreasing.
Similarly from Fig. 50 (6) we see that as we pass through
a minimum pointy (x) increases as x increases, for we start with
a large negative gradient \j and gradually increase up to a large
positive gradient s**
2 3
Ax*
Fig. 50.
But by the above if f (x) increases as x increases, f" (x) is
positive, and \if (x) decreases as x increases/" (x) is negative.
/. If f(x) is a maximum, /' (x)  and f " (x) is negative.
minimum, /' (x) = and f" (x) is positive.
107109]
MAXIMA AND MINIMA
125
We see then that where the curve is concave down
wards r^ is , but where it is concave upwards ~ is +.
ax* dx 2
(See 94.) Fig. 51. [Positive directions of axes as usual; v. p. 99.]
Sign of l!l or f ( x )
p ig. 51.
108. Consider the example in 96.
/() = 2or J 9x 2 i 12a;  3,
= 6 (a; 2 3a;2),
/"(*)= 6 (2* 3).
The values of x which make/' (x) are 1 and 2.
If x =l,f"(x) is.
If * = 2,/'is+.
.'. x = 1 gives a maximum value of f(x) \
and x 2 gives a minimum value of f(x) ) '
[See Fig. 41, p. 101.]
109. If we apply this test to the case in 99,
where f( x ) 3* 4 16& 8 + 30ic 2  24a? f 5,
The values of x which make/' (a;) = are 1 and 2.
If x 2,/" (x) is +, .'. f(x) is a minimum,
but if a;= I,/" (x) = 0, and the test fails.
126
CALCULUS FOB BEGINNERS
[CH. IV
Fig. 52.
110. In the case of 108, putting
we have
and
/' (x) = 6 (a?  3x + 2),
If we draw the graphs of y =/(#), y =/' (x), and y =f" (x)
(Fig. 53) we get:
When = 1 or 2,
f(x) is maximum or minimum,
i,e.
y =f (x) cuts the %axis.
110, 111] MAXIMA AND MINIMA
When x 1,
127
and
f (x) is a minimum,
/' = 0,
y =f" (x) cuts the a;axis.
Kg. 53.
Points of inflexion.
111. A point like P where the gradient is a maximum or
minimum is called a point of inflexion.
128
CALCULUS FOlt BEGINNERS
[CH. IV
The portion BPC of the curve is shewn magnified in Fig. 54.
Between B and P the gradient diminishes (algebraically)
steadily from to  1 *5.
la
18
IE
YL
^
w
'@L
12
^
a
Fig. 54.
Fig. 55.
Between P and C it increases (algebraically) from 1 *5 to 0.
The gradient is a minimum at P and the curve crosses the
tangent there.
If we suppose a point to advance along the curve from B to C,
and a line to be drawn from any point K. parallel to the tangent
at each position, this line swings in a clockwise direction until
the moving point reaches P, and then back again in a counter
clockwise direction. (Fig. 55.)
111113] POINTS OF INFLEXION 129
The tangent at a point like P is sometimes called a stationary
tangent.
Notice (v. Fig. 41, p. 101) that from A to P the curve is concave
downwards and ~ is : from P to D the curve is concave
dy?
j fy
upwards and ~ 2 is +.
At P where we change from one kind of bending to the other
112. If we call a point where a curve crosses the axis a
zero point we see
(1) To a turning point on y = f (x) corresponds a zero
point on y = f ' (x).
(2) To a point of inflexion on y = f(x) corresponds
a turning point on y = f ' (x) and a zero point on
y = f"(x).
113. In the case of 109 :
f(x) = 3x 4  IGo 3 + SOar*  24* + 5,
/' () = 12 (y?  4ic 2 + 5x  2) = 12 (x  1) 3 (x  2),
f"(x) = 12 (3x*8x+ 5) = 12 (xl) (Sec 5).
When x= 2, f( x ) i & a minimum,
/=<>,
f"(x)is+.
When x = 1,
f(x) is neither a maximum nor a minimum [i.e. there
is a point of inflexion],
/' (x) = and is a maximum,
/"<)= a
There is also another point of inflexion on y=f(x) corre
5
spending to x = 5 .
^
M. C. 9
130 CALCULUS FOR BEGINNERS
This makes f" (x) = 0,
f (x) a minimum.
[CH. IV
114. Fig. 56 shews the gra
y=.
y=
y=*
3hs of
flrb
f (T^
r<).
,fc'! 3 ^3QX?i& : 3!I.' \\
: jljjj jjjjj 1;
;; i;i;:;;;:;;;];;i;;;;J!;;;;
'' / \ ' ' '
; ;;  ;; 
fr ;rL BfV p ]
:;l;;;;;;;!:i;;;:
I ii 21%* i jijiiii
Notice when x= 1, f (x)  0, but y=f (x) does not cross the
ceaxis, and there i 1 ? not a turning point on y=f(x) corresponding
113116] POINTS OF INFLEXION 131
to this value of x. f" (%) = Q, and y=f" (x) crosses the a>axis,
arid we have a turning point on y =/' (x) and a point of inflexion
on y =/(*).
115. We have seen that if a be a value of x which makes
f (x) = Q, it may happen that /() is not a maximum or mini
mum value of /(a:), and that usually if/"(a) = 0, it will not,
because then as a rule f (x) does not change sign as x passes
through the value a.
It may happen, however, that even if/" (a) = O,/' (a;) changes
sign as x passes through the value a, in which case we get a
turning point corresponding to x = a.
Similarly if b be a value of x which makes/" (x) = Q, it may
happen that/' (6) is not a max. or a min. value of /'(a?) and that
usually if /"' (6) = 0, it will not, because then, as a rule /" (x)
does not change sign as x passes through the value b.
116. Summary. If a is a value of x which makes f'(x) =
and if /" (a) ^fe 0, there is a turning point corresponding to x a
and the value of f(x) is a max. or iniii. according as/" (a;) is
or +.
If/' (a) = and also/" (a) = we cannot tell without further
investigation whether there is a turning point or not. If /' (x)
changes sign as x passes through the value a, there will be a
turning point and f(x) will be a max. or min. according as /' (x)
changes from + to or from to +.
If b be a value of x which makes /" (a?) = and if /'" (b) J=
there is a point of inflexion corresponding to x = b. If /" (b) =
and also /'" (b) = we cannot tell without further investigation
whether there is a point of inflexion or not. If /" (x) changes
sign as x passes through the value 6, there is a point of inflexion.
e.g. if y=x*(x+\],
we have f(x) = x 5 + x 4 ,
f (x)
92
132 CALCULUS FOR BEGINNERS [CH. IV
Now/' (a;) = 0, when x = or '8.
When as =  8, /" (x) is  .
.'. x = '8 gives a maximum value of f(x).
When x = 0, /" (a;) = 0.
But f'(x)
.'. !* = <)*, /'(*) = (
.*. x = gives a minimum value of /(a:).
Again /" (a;) = 4x 2 (5* + 3).
.'. /" (a;) = when x = or  6,
/'" (a) = 60* 2 + 24* = llx (5x + 2).
When x =  6, /'" (x) = () () = +.
.'. there is a point of inflexion when x =  6 and the gradient
is a minimum.
[Of course we could have seen this without using f" (x), for
if =6,/"(*) = <+)() = J
if =  6 +,/"(*) = (+)(+) = 4,
.'. /' (a;) is a minimum and there is a point of inflexion.]
When aj = 0,/"'() = 0.
In this case, if x = , f" (x) = (+) (+) = + ;
if x = +,/"(*) = (+)(+) = +.
.*. there is not a point of inflexion. [See Fig. 57.]
117. If we go on and calculate /"" (a;), etc. we get
/"" (a?) = 1 20x + 24 = + 24 when x = 0,
and generally, it will be found that if f (a:) = when x = c, and
if the first of the derived functions of x which does not vanish
when x = c is an even one, i.e. if it is one of the set/" (a;), / IT (x)
etc., x = c gives a turning point, but if the first one which does
uot vanish when x = c is an odd one, i.e. if it is one of the set
/'" (*),/ v (x) etc., x= c does not give a turning point.
For instance in the above/"" (a;) was the first derived function
which did not vanish when x = 0. For a proof, the reader must
refer to more advanced textbooks.
116, 117] MAXIMA AND MINIMA
133
134 CALCULUS FOR BEGINNERS [CH. IV
EXERCISES. XXV.
1. Find the maximum and minimum values of
2a; 3 3x 2 36a; + 10.
Also find the point of inflexion on
and shew that it is the midpoint of the line joining the turning points.
2. Find the maximum and minimum values of
3. In the curve y=x*2x 3 , shew that there is a minimum point where
3
x  and a point of inflexion where x = 0.
m
Find the coordinates of the other point of inflexion.
Draw the graph between x=  1 and x =3.
a; 2 5
4. Find a minimum value of ^ + 2 + 5 and shew that there is no
o x
maximum.
6. Shew that there are two turning points on
y=x 6 (x2).
How many points of inflexion are there? Draw a sketch between
0;= 1 and x=2.
6. If ^ (x + 1) 3 (x  2)* shew that x=  1 gives a minimum value of y,
fljp
but that x = 2 gives neither a max. nor a rnin.
7. If /(a;)=a;*8x3 + 24a: 2 32a;
shew that/' (2)=0 and/" (2)=0.
Is the point corresponding to x=2 a turning point or a point of inflexion?
Draw a rough sketch of y = f(x) from x=  1 to x 5.
8. In the curve
find the turning points and the point of inflexion and shew that the point of
inflexion is midway between the turning points.
If the point of inflexion be taken as origin, the axes being parallel to the
original axes, shew that the equation of the curve is
y=2x*24:x,
and that the curve is symmetrical in opposite quadrants.
117] EXERCISES 135
. Shew that y = ax 3 + bx* f ex + d
has always one point of inflexion and find its coordinates.
Shew also that if the curve has two turning points, the point of inflexion
is midway between them.
Shew that the equation of the curve referred to parallel axes through the
point of inflexion is
and that the curve is symmetrical in opposite quadrants.
10. Shew that y = ax* + bx 3 + ex 2 + dx + e
has two points of inflexion or none according as 36 2  8ac is + or .
11. Draw on a large scale the graph of y = 2x 4  3x? + 2 x between a; =
and 2 = 1.
CHAPTER V
SMALL ERRORS AND APPROXIMATIONS
118. WE have already had a few examples of the application
of the Differential Calculus to this kind of problem :
y being a given function of x, what will be the approximate
change in y, due to a small change in x 1
A?y
We make use of the fact that is approximately equal to
JT, the approximation being better as Aa; is diminished. The
CM?
approximation may be put in the form
dv
Ay = . Ax approximately.
e.g. if A sq. ins. is the area of a circle of radius r ins.
A = 71^.
Now suppose a small error, which we may call Ar, has been
made in measuring r, there will be a corresponding small error
in A, which we may call AA, and we have seen that
AA dA
= approximately,
Ar dr J
i.e. AA = 2irr . Ar approximately.
Suppose the radius was measured as 10 inches and that there
was an error of *1 inch.
Then we have r = 10, Ar = !.
.'. AA = 27T x 10 x 1 approximately
= 2ir approximately,
ie. the error in the area is approximately 6*28 sq. ins.
[Shew that actually the error is 6'31 sq. ins.]
118120] SMALL ERRORS AND APPROXIMATIONS 137
If the error in the radius was 01" we should get for the error
in the area 6284 sq. ins. instead of '6287.
119. The result AA = 2irr. Ar approximately, admits of a simple
geometrical interpretation which is important. It tells us that
if the radius (r) of a circle be increased or decreased by a small
amount (A?) the increase or decrease in area is approximately
2rrr . Ar or
Area of thin circular ring is approximately circum
ference of either boundary x breadth.
e.g. Suppose we have a ring with inner radius 20" and
breadth 06", the area is approximately
2ir x 20 x 06 = 2 Air= 7'54 sq. ins. approximately.
[Shew that area is actually 2'4036n= 7 '55 sq. ins.]
120. We might get this result from first principles thus.
Area of ring, inner and outer radii a and b,
= Tr(& 2 a 3 )
= IT (b + a) (b a) = ( 2ir . ) t> where t is the breadth
= Circumference of concentric circle lying midway
between the two circumferences x breadth. [Fig. 58.]
Fig. 58.
138 CALCULUS FOE BEGINNERS [CH. V
As the breadth is made less, the circumference of this
intermediate circle comes nearer to the circumference of either
boundary and can be brought as near to it as we please by
making the breadth small en'ough.
EXERCISES. XXVI.
1. The radius (r) of a sphere is increased by a small length Ar. Shew
that the increase in volume is approximately 4ar 2 . Ar.
State this as a formula for the approximate volume of a thin spherical
shell.
2. The radius of a sphere Is 6". Find approximately the diminution
in volume if the radius is diminished by ^ ".
3. Find approximately the volume of a thin spherical shell, internal
radius 1 foot, thickness 2". [Shew tbat your result is less than 2/ in
error.]
4. A stone is thrown at an angle of 40 to the horizontal with a
velocity of 80 ft./sec. Find the range R = and the approximate
I Jf _J
increase in the range if the velocity is changed to 82 ft./sec., the angle of
projection remaining unchanged.
6. A cylindrical well is said to be 25 feet deep and 6 feet in diameter.
Find the error in the calculated volume if there is an error of (i) 1" in the
diameter, (ii) 3" in the depth.
6. The radius of the base of a cone is r and its vertical angle 2a. Find
the approximate increase in volume due to a small increase Ar in the radius,
the vertical angle remaining constant. Hence shew that the volume of a
conical shell, internal radius r, thickness t, t being small, is approximately
rrlt where I is the slant height.
7. The radius of a sphere is found by measurement to be 185", with a
possible error of !". Find the consequent errors possible in (i) the surface
area, (ii) the volume, as calculated from this measurement.
8. R = R (l + at + & 2 ) is a formula for the electrical resistance of a
metal, R being the resistance at C. and t C. the temperature.
Find the rate of increase of R per unit increase of t.
If R =l6, a = 00388, b= 000000587, find the approximate change in R
when the temperature rises from 100 to 101.
120122] RELATIVE ERROR 139
121. Since A = nr 2
and A A = 2irr . Ar approximately,
AA Ar
7T =3 'T'
AA
or the ratio of the error in the area to the original area, is
called the relative error in the area.
Thus the relative error in the area is approximately twice the
relative error in the radius.
If, for example, the radius increases by 1 / o , the area will
increase approximately by 2 / o ,
Ar 1 AA 2
for T = loo' " "A" = 100 a pp roximatel y
EXERCISES. XXVII.
1. If V c. ins. is the volume of a cube of edge x ins. prove
AV Az
rj = 3 . approximately.
V X
Hence find approximately the percentage increase in volume due to an
increase of *5 / in the edge.
2. Prove that the relative increase in the volume of a sphere is
approximately 3 times the relative increase in the radius.
3. If pv= k, prove = approximately.
If the pressure increase 1/ , what is the approximate change in the
volume ?
4. Ifpv l *=k, prove =  1'4 approximately.
6. If y = x n , prove =n . approximately.
"
122. Elasticity of volume. Suppose v c. ins. to be the
volume of unit mass of a fluid and p Ibs./sq. in. the pressure, p
being some given function of v.
If p be increased to p + AJL>, v will become v + Av (Aw being,
140
CALCULUS FOR BEGINNERS
[CH. V
of course, negative, since an increase in pressure will produce a
decrease in volume).
Aw will be the diminution of volume.
Aw
(the ratio of this diminution to the original volume) is
called the volume strain.
The ratio of A/> (the increase of pressure required to produce
this) to the volume strain is v . and the limit of this when
Av
Aw is indefinitely diminished, i.e. V ==, is called the elasticity
of volume or the bulk modulus of the fluid.
6. If pv constant, prove that the elasticity of volume is p.
7. If j>v n = constant, prove that the elasticity of volume is np.
8. If p be given as a function of v, say p=f(v) and the graph p=f(v)
be drawn, the vaxis being horizontal, shew that if P be a point on the
graph corresponding to any given value of v and MQ be drawn through the
foot of the ordinate of P parallel to the tangent at P, meeting the paxis in
Q, then OQ will give the elasticity of volume for this value of v. [Fig. 59.]
Fig. 59.
9. A formula giving the deflection (D) of a beam in terms of the
length (I), the load (W), the Moment of Inertia of the CrossSection (I) and
WJ 3 1
Young's modulus (E) is D = x .
122, 123] SMALL ERRORS AND APPROXIMATIONS 141
Shew that if a small error be made in I, the resulting percentage error in
D is approximately 3 times the percentage error in I.
10. The velocity of discharge (v ft. /sec.) of water through a long pipe is
LJ
given by u 1>87 = jrpj7xd 1 * where H ft. ~ head of water, d ft. = diameter of
'UUU4L
pipe, L ft. = length of pipe.
Find the velocity of discharge when the head is 35 feet, the length of
pipe 1 mile and its diameter 18 inches.
If a small error Ad is made in estimating the diameter, prove that the
error (Av) in the velocity =  x  x Ad approximately.
If the error in the diameter is ", find Av.
11. If t sees, is the time of oscillation of a pendulum of length I feet
i. [0=322.]
Shew that the relative error in t is approximately half that in I.
Calculate t when Z=4 and find approximately the error in t corresponding
to an error of ~ " in the length.
m
12. If pv l '% = k and the pressure increase 2/ , what is approximately
the percentage change in volume ?
123. = r approximately,
Ax ax
i.e. if x is increased by Ace, y is increased by
Ace x J? approximately.
Suppose y =/(), then j =f'(x), and the above statement is
ctx
equivalent to this :
If x is increased by a small quantity A, then f(x) is increased
by hf'(x) approximately. \h taking the place of Ax and/'(x)
J.1 Or
dx
f (x + h) = f (x) + hf (x) approximately.
142
CALCULUS FOR BEGINNERS
[CH. V
124. Let P be the point (a;, y) on the curve y =f(x) and let
Q be a neighbouring point whose abscissa ON is x + h. [Fig. 60.]
M N
Fig. 60.
Then
NQ =f(x + h),
MP =f(x).
:. if PR is parallel to OX, RQ =f(x + h) f(x).
If PT is the tangent at P, the gradient of PT is/' (x) and
RT = hf'(x),
so that in taking f(x + h) /(*) as equal to hf (x) we take PQ as
coincident with PT, in other words we consider the portion PQ of
the curve as straight.
125. Ex. Find the value of
32*  4X 3 + 6a: 2  7x + 8 when x = 301.
Calling the expression y (or) we have
= 662, /'(3)=1136.
.'./(801) = 662 + 1136 x 01 approximately
= 673 '36 approximately.
126. Notice that if /' (a;) = 0, i.e. if the tangent at P is
parallel to OX, this approximation will not shew any difference
between f(x) and f(x + h). See Exs. 5 and 6 below.
124127] SMALL ERRORS AND APPROXIMATIONS 143
EXERCISES. XXVIII.
1. Shew that if f(x}=ax + b where a and b are constants the statement
f(x + h)=f(x)+hf'(x)
is accurately true for all values of h.
2. Find approximately the value of 5x 2 + 4x + 9 when x =20087.
3. Find approximately the value of Gx 3  7* 2 + 2x  9 when x = 3002.
4
4. Find approximately the value of 7a; 2 3x +  when x=201.
5. Find approximately the value of
2z 3  9* 2 + 12s 3 when x= (i) 2005, (ii) 1995.
In each case find also the accurate value.
6. Find approximately the value of
3a;*16aj3 + 30a;224a; + 5 when x = (i) 201, (ii) 101.
In each case find the accurate value.
127. The approximation f (x + h) =f (x) 4 hf'(x) is, as we
have seen, equivalent to the statement that the gradient of PQ
[Fig. 61] is that of the tangent at P. If we suppose P and Q to
be such that f (x) and f" (x) do not change sign between P and
Q, so that between these points the curve has one of the shapes
shewn on p. 99, it is clear that the gradient of PQ is between the
gradients of the tangents at P and Q and equal to the gradient
at some intermediate point. Thus the quantity [represented by
RQ] added to f(x) to make f(x + h} is between hf (x) [RT] and
hf'(x + h) [RUj and is actually hf'(x+Qh) where 6 is some
proper fraction, so that (x f 6h} is between x and (x + h).
e.g. in 125 /' (3) = 1136, /' (301) = 115168.
So that
/(301)/( 3 ) is between 01 x/'(3) and 01 x/'(3'01),
i.e. between 11 '36 and 11 '52.
Actually it is H'43826..., and as a matter of fact
/(3005) = 11438200,
so that in this case f(x + h) f(x) = hf (x + ^h).
[Correct to 3 decimal places.]
The more rapidly f (x) is changing, the less reliable will
f(x) + hf'(x) be as an approximation tof(x + h).
144
CALCULUS FOR BEGINNERS
[CH. V
Fig. 61.
EXERCISES. XXIX.
1 . If / (x) = ax* + Ix + c, shew that / (x + h) =f (x) + hf ( x f U).
In particular, if/(a;) = 7a; 2 lla;+15, shew that/(3'4)=/(3) + 4/' (32).
a. If / (x) = x 3  6x  5, shew that
/ (14) /(I) =0656,
4/' (14) =0048,
4/' (1205) =065757.
3. If/(a;)=44a+16a;7, shew that
ft /(32)/ (3) = 39856, (ii) / (4) / (3) = 43,
2/'(3)=32, /'(3) = 16,
2/' (32) = 48384, /' (4) = 80,
2/' (3102) = 39852. /' (3539) = 43003.
CHAPTER VI
THE INVERSE OPERATION
128. GIVEN , find y.
CwB
di/
We know that if y = c w , ~f = a; TO ~ 1 .
cKE
ttW
Can we say that if ~ nx n1 then y = x n t Not unless we are
CwB
certain that x n is the only function whose differential coefficient
is nx n \ As a matter of fact we have seen that if y = x n + any
d'u
constant whatever,  = nx n ~\ .". if we are given that
Cnv
j = wa3 n ~ 1 , we can only conclude that y = JB W + some constant, and
if no further information is given " some constant " may be
replaced by "any constant whatever." As a rule sufficient
information is given to enable us to fix the constant for the
particular case in question.
In practice f will not present itself in such a convenient
fUD
form as rac"" 1 , e.g. we might have the value of ~ given not as
Gfos
5x* but as ar 4 or 3x*. To deal with such cases we have only to
U'l/
remember that if y = k . x n where k is a constant, ^ = k. nx n ~\
o o
Thus 3x 4 = .5oj 4 and therefore arises from differentiating = , X s .
O
M. C. 10
146 CALCULUS FOR BEGINNERS [CH. VI
Remembering what was said about the addition of an arbitrary
constant, we see that if ^ = 3x* then y = =x 5 + c where c is some
dx o
constant.
Similarly if ^ = Sjx, i.e. 5o?, we know that x* must come
from differentiating x% + ', i.e. a;^ : but if y = x*. ^=H#
dx 2
EXERCISES. XXX.
1. Write down the values of y corresponding to the following values
dm
(i) So; 2 . (ii) S* 3 . (iii) ^. (iv) 7. (v) 5x~%, ( v i)  x n .
(vii) ^. (viii)^. (ix) i. (x) *i (xi) 2*3 + 3*. (xii)  2 + 4.
(xiii) 7* 2 + 3* + 5 . (xiv)
X**
a. What is /(*) if /'(*) =
(i) A (ii) 8a?* + l. (iii) 2
d 2 ^
129. If T^ is given there will be two arbitrary constants
QBP
in the value of y, as the following example will shew :
Given ^ = 2* + 3, find y.
Since ^~ means = where z stands for ~ , we have
da? dx dx
.'. z = x? + 3x + a
where a is any constant.
128130] THE INVERSE OPERATION 147
(*?/
i.e. ~ = a? + 3x + a.
ax
x 3 3x 2
" y=3+ 2 +ax+b
where b is any constant.
EXERCISES. XXXI.
1. Find the values of y corresponding to the following values of 3^
dx^
(i) 3x2. (ii) . (iii) 7. (iv) ^ + 5. (v) ^. (vi) V*.
2 . If /" (x) = 3x  8, find / (x).
3. If /"(*) = 5, find f(x).
4. Iff" (x) = 2x 3 3x + l, find/{a;).
130. The following examples will shew how to fix the value
of the arbitrary constant in special cases.
CtAJ
Ex. 1. Given that ^ = 3x, and that t/=5 when a? = 2, fiud
flHD
y in terms of x.
3
We have y = x z + c.
c must be chosen so that this shall be satisfied when
x = 2 and y = 5.
102
148 CALCULUS FOR BEGINNERS [CH. VI
EXERCISES. XXXII.
du
1. Given r^ = a; 2 + l and y = 3 when a; = 3, express y in terms of a;.
2. Given =v + ^ and p = 5 when v = 2, express^ in terms of v.
3. Given ^ =N /a; and y = 7 when a; =4, find y in terms of x.
dA
4. Given ==5y + 4 and A = 6 when y = l, express A in terms of y.
6. Given p = (a; + l) (a; + 2) andy=12 when a; =3, express y in terms
of a;.
6. Given 5 = (2t  3) 2 ands = 52 when t = 5, express s in terms of t.
7. Given = 4irr 2 and V = 367r when r=3, find V in terms of r.
8. Given J? = and y = 8 when a; =27, find y in terms of x.
ax *jx
9. The speed of a body at the end of t sees, is given by the formula
v=u + at where u and a are constants. Find the relation between s and t,
given s=0 when t=0.
<Z 2 w
10. Given j^=3x; also y = ll when a; =2, and y = 47 when a; =4,
find y.
^32
~I n vii . cuovs u ^ v T i_i^u *L> v c^ii^ ~7"~ 
dx 2 djj
i:d y.
11. Given j=32; also y=0 when x = 0, and ^ = 5 when a; = 0,
131. jE'x. 2. The gradient of a curve at the point (x, y) is
2a; + 3, and the curve passes through (1, 2). Find its equation.
We have ^ = 2x + 3.
dx
.'. y = a; 2 + 3a; + c
[where c is some constant to be fixed].
131, 132] THE INVERSE OPERATION
This has to be satisfied by x= 1, y  2.
149
.'. the equation of the curve is
y = x 2 + 3x  2.
132. If we had not been given the second piece of in
formation we should have had the equation of the curve
y = a? + 3a; + e
where c is any constant whatever.
Fig. 02.
150 CALCULUS FOR BEGINNERS [CH. VI
For different values of c this really represents a whole family
of curves got by taking
y = a? + 3*
and sliding it through any distance parallel to Oy.
is that particular member of the family which goes through
(1, 2). (Fig. 62.)
"We might say that ^ = 2x + 3 represents a family of curves,
(tx
viz. that obtained by drawing y = a? + 3x + c for different values
of c.
EXERCISES XXXIII.
1. Find the equation of a curve whose gradient at any point (x, y) is
3  4 and which passes through (2, 3).
a. Find the equations of the curves with the same gradientlaw as in
Ex. 1 passing through (i) (0, 0), (ii) (3, 1).
3. Draw the 3 curves of Exs. 1 and 2.
*. The gradient of a certain curve at any point (x, y) is 2x + l, and the
curve passes through the point (1, 5). Find the equation of the curve and
the equation of the tangent at the point (1, 5).
6. The gradient of a certain curve at the point (x, y) is 2a; 2 + 3x  7, and
the curve passes through (1, 2). Find its equation and also the equation of
the normal at the point (1, 2).
6. The gradient of a curve at the point (x, y) is 3x  5. Shew that it
has a minimum point and find its coordinates (i) if the curve passes through
the origin, (ii) if the curve passes through (  3, 2).
7. Find the equation of a curve which passes through the point (3, 5)
and whose gradientlaw is ^ = 1  4r + 2 .
dx
In what point and at what angle does this curve cut the yaxis?
133. Ex. 3. A point is moving in a straight line so that its
speed at the end of t sees, is (3t + 2^) ft/sec, and its distance from
a fixed point in the line at the end of 2 seconds is 20 feet. Find
its distance from the fixed point at the end of t sees.
132, 133] THE INVERSE OPERATION 151
If the required distance is s ft. we have
at
3
[where c is a constant to be determined].
Now we are told that when t = 2, s = 20.
Eoj. 4. A point is moving in a straight line so that its
acceleration at the end of t seconds is (3 + 4f) ft./sec. a At the end
of 2 seconds its speed is 16 ft. per second and its distance from a
fixed point in the line is 20^ feet. Find its distance from the
fixed point at the end of t seconds.
Let the required distance be s ft. and let its speed at the end
of t sees, be v ft./sec.
We have ^ = 3 + 4*.
at
:. v = a + 3 + 2 a ,
where a is some constant to be determined.
Now when t = 2, v 16.
.'. 16 = a + 6+8. .'. a = 2.
/. v = 2 + 3< + 2< 2 .
i.e. ^=2 + 3<+2<?.
at
3 2
;P + ~1>*,
*t O
where 6 is some constant to be determined.
152 CALCULUS FOR BEGINNERS [CH. VI
Now when t = 2, s = 20 .
/. 20 = b + 4 + 6 + 5J. .'. 6 = 5.
EXERCISES. XXXIV.
1. The speed of a body moving in a straight line is given hy
t> = 6032f.
Its distance from a fixed point O in the line at the end of 1 second is 44 feet,
find its distance from O at the end of t seconds.
2. The acceleration of a body moving in a straight line is 32 ft. /sec. 2
Its speed at the end of 2 seconds is 154 ft. sec. and its distance from a fixed
point O in the line is 234 feet. Find its speed and distance from O at the
end of 3 seconds.
3. The acceleration of a body moving in a straight line at the end of
t seconds is (2 + 3t) ft./sec. 2 Its distances from a fixed point O in the line at
the end of 1 and 2 seconds are respectively 7 ft. and 15 ft. ; find s in terms
of t and get the distance from O and the speed of the body (i) when t=0,
(ii) when t=5.
4. A body moves in a straight line with constant acceleration a ft./sec. 2
Its initial speed is M ft./sec. If v ft./sec. is its speed at the end of t sees.,
and a ft. its distance from the initial position at the end of t sees., prove
and s=
6. If J = g^ express p as a function of v, given that p = 18  95 when
20.
6. If ~ = ax + b express y as a function of x, given the following pairs
of corresponding values of x and y :
a; 1 2
y 3 5 9.
133] MISCELLANEOUS EXAMPLES 153
MISCELLANEOUS EXAMPLES ON CHAPTERS I VI.
A.
1. From first principles find ^ when
dx
Also write down the equation of the tangent to the curve represented by
y = 3x 3 + 2x + 1 at the point where x= 1.
2. A body starts from rest and moves in a straight line in such a way
that its speed after t seconds is (16t  4t 2 ) ft. /sec. Find the distance described
during the fourth second.
3. The strength of a rectangular beam varies as fed 2 where b is the
breadth and d the depth. Find the depth and breadth of the strongest
rectangular beam of perimeter 3 feet.
4. The radius of a spherical bubble grows at the rate of  01 inch per
second. At what rate is (i) the surface, (ii) the volume increasing when the
radius is (a) 1 inch, (b) 1 foot?
6. The volume of a cylindrical tin canister closed at both ends is 300 c.c.
Find the most economical dimensions.
B.
1. (i) Find from first principles ~ when y=*
tlx x
(ii) Find the equations of the tangent and normal to the curve
xy = l at the point P, where x = 5.
(iii) If the tangent meets the axis in T and t, shew that PT = P*.
2. Writedown (i) ~ when y = 5*Jx + / .
dx y x
d\/ lex*
(ii) if V =  (R x), where R is constant.
3. Water is poured at the rate of 5 cubic feet per minute into a vessel
in the shape of a hollow cone with a vertical angle of 90. When the vessel
contains 12 c. ft., at what rate is the depth increasing ?
154 CALCULUS FOR BEGINNERS [CH. VI
4. The sum of the perimeters of two equal squares and a circle is
100 feet. When is the sum of the areas least, and when is it greatest ?
Are these really maximum and minimum values ?
6. There is a certain curve such that the gradient at any point (x, y)
is Bx. Find its equation, given that it passes through the origin.
C.
1. If j/ = 3# 2 + 2, obtain from first principles the differential coefficient
of y with respect to x, and shew with a diagram the geometrical meaning of
your result. For what value of y is this differential coefficient equal to
unity ?
2. If y = 2x 3 27x*132x + 2
find its maximum and minimum values.
3. A lamp is 60 feet above the ground. A stone is let drop from a
point at the same level as the lamp and 20 feet away from it. Find the
speed of the shadow of the stone on the ground (i) after 1 second, (ii) when
it has fallen 30 feet. [0=32.]
4. Find the coordinates of the point of intersection of the tangents to
j/ = 3a; 2 +5j;7 at the points where x=2 and x=5, and the equation of the
line joining this point of intersection to the midpoint of the chord joining
the points of contact.
6. The graph of/' (x) is the parabola
y + 2 = 3 (x 1)2.
What information does this give respecting .; (a;) ?
What is the gradient at the point on y =f(x) where x=l 1
D.
1. A body is thrown into the air with a velocity of 100 ft. /sec. at a
certain elevation. If the resistance of the air be neglected, the horizontal
distance from the point of projection (a; feet) and the height above the ground
(y feet) after t seconds are given by the equations x = 80t, y = 60t  16t 2 .
Find the horizontal and vertical velocities after 3 seconds, and the magni
tude and the direction of their resultant. By eliminating t between the
3 1
given equations, get the equation of the path in the form y = j x  ^ x, and
find the direction of the tangent to this path when x = 240.
a. Find the maximum and minimum values of
(*!)(* 3).
133J MISCELLANEOUS EXAMPLES 155
3. Find the equations of the tangents to the curve
at the points where the curve crosses the axes and also the coordinates of
the point of inflexion.
4. The volume of a sphere is V cuhio inches and the surface S square
inches. Shew that V = ^ . S i
6VT
The surface of a sphere is given as 100 square inches, and from this the
volume is calculated. If the surface is really 101 square inches, find the
approximate error in the calculated volume.
dy_
curve goes through the point (1, 1). Find its equation.
5. The gradient of a certain curve obeys the law ^=4a; 2 +3 and the
clx
E.
1. The surface of a sphere is given and the volume calculated from it.
If there is an error of '4 / in the surface, find approximately the resulting
percentage error in the calculated volume.
2. An open tank with a square base and vertical sides is to have a
capacity of 4000 cubic feet. Find the dimensions so that the cost of lining
it with lead may be a minimum.
3. The equation of the path of a projectile thrown in vacuo with a
x 2
velocity of 64 ft./sec. at an angle of 45 to the horizon is y x ^^ , the
I2o
origin being the point of projection, the xaxis horizontal, and the unit along
each axis 1 foot.
Find:
(i) The coordinates of the highest point,
(ii) The equation of the tangent at the point for which a; =32.
(iii) Where the direction of motion makes an angle 30 with the
horizon.
4. If y=2x*2x3x z + l, find y when x = 1, ,0,1,2.
Also find ^ and the values of j when x  1, 2.
ax dx
Also find what values of * make j^=0.
dx
156 CALCULUS FOR BEGINNERS [CH. VI
Using all this information, draw the graph of
y = 2x 4  2x 3  x 2 + 1 between x =  1 and x = 2.
, 1"
xscale unit 2", yscale unit ^ .
6. In a certain case of straight line motion, the velocity in feet per
second, at a given instant t seconds after the start, is given by the formula
Calculate :
(i) The average velocity between the end of 2 seconds and the end of
4 seconds.
(ii) The arithmetic mean of the velocities at the inatauts 2 and
4 seconds after the start.
(iii) The velocity 3 seconds after the start.
1. If v be the volume of a given mass of water at C., and v its
volume at 6 C., Hallstrom's formula is
for values of between and 30, where
a =000057577,
6 = 0000075601,
c= 00000003509.
What is the temperature of maximum density ?
2. Draw the graph of y^=x(xl) z between x=0 and x=S. Find the
coordinates of the turning points and the equations of the tangents at (1, 0).
3. The normal at a point of y 3 =x meets OX in G. Prove that the
minimum value of OG is 7698.
4. A body of mass 20 Ibs. is moving in a straight line so that
s feet being its displacement from some standard position at the end of
t seconds. Find its kinetic energy and momentum and the force acting on
it at the end of 4 seconds.
6. Find approximately the value of 16  25x + 47x 2 , when x = '997.
133] MISCELLANEOUS EXAMPLES 157
G.
1. In a certain case of straight line motion the acceleration is
proportional to the time which has elapsed since the moving body passed a
fixed point A. Find the distance of the body from A after 10 seconds,
given that its speed at A is 8 ft. /sec. and its distance from A after 1 second
is 9 ft.
2. If y = ax 2 + bx where a and 6 are constants, prove
rttfir .*%*
dx 2 dx
3. A piece of wire, a inches long, is cut into two parts, one of which
is bent into the form of a square and the other into that of a circle. Find
where it must be cut so that the sum of the areas enclosed may be a
minimum and shew that the diameter of the circle is equal to the side of the
square.
4. P is the point on the curve y=zP whose coordinates are (h, ft 3 ).
The tangent and normal to the curve at P meet the axis of y in T and G
respectively, and PN is the perpendicular from P on the y axis. Prove that
OT=2ON, and that as P changes its position on the curve, NG varies
inversely as N P.
6. The pressure and volume of a gas are connected by the relation
pv 8 = constant. When the volume is 20 cubic feet the pressure is 30 Ibs.
per sq. ft. Find the elasticity of volume when the volume is 40 cubic feet.
H.
1. If/() = (* + !) (as 2)3, find /'(*),/"(*), and/'" (x).
Find the values of x which make
(i) / (*) = 0, (ii) /'(*) = 0, (iii) /"(*) = 0, (iv) /'" (x) = 0.
Find the coordinates of the maximum and minimum points on
y=f(x), y =/'(*), y =/"(*),
and the points of inflexion on y=f(x) and y=f (x).
Also calculate the values off(x),f (x\, f" (x), /'" (x), when x=  1, 0,
1, 2, 3.
Using the information you have now collected draw on squared paper the
curves
y =/(*), y =/'(*). y =/"(*). y =/'"(*)
using the same xscalo for all.
158 CALCULUS FOR BEGINNERS [CH. VI
2. A conical vessel has vertical angle 60. Water is flowing into it at
the rate of 1 cubic foot per minute. Shew that when the water is 1 foot
deep the level is rising at about 19 inch per second.
3. The tangent at P to the curve aj/ 2 =x 3 meets OX in T and OY in t.
PN is the ordinate of P.
Prove OT = i ON and Ot= PN.
O a
4. A solid sphere radius r floats just immersed in water. Find where
the sphere must be cut by a horizontal plane, so that if the top part be
removed the thrust on the plane surface of the remainder may be a
maximum.
6. Draw the circle x z + y z 6x8y + 21=0, and by drawing find the
points on it at which p =  .
1. A vertical line moves with uniform speed of 1" per second, so that
its lower end always lies in the axis and its upper end on the parabola
y=x%. [The unit being 1" each way.] At what rate is the length of the
line growing when it is 10" from the origin?
3.2
3. P, Q are any two points on the parabola y=. V is the mid
ft0
point of the chord PQ and T is the point of intersection of the tangents at
P and Q.. Shew that TV is parallel to OY. [Take coordinates of P
(,,) a of a (,.?).]
3. ABCD is a rectangle. On AB, on the side remote from CD, an
equilateral triangle ABE is drawn. If AB=a;" and BC=y" write down
(i) the perimeter, (ii) the area of the figure EADCB.
If the perimeter is 33 inches, what is the maximum area of the figure?
4. A man 6 ft. high walks at a uniform rate of 5 ft./sec. away from a
lamp 10 ft. high. Find (i) the rate at which the length of his shadow grows,
(ii) the rate at which the end of his shadow moves.
6. The formula for the velocity of sound is
_ /Volume Elasticity
Density
[If the volume is in c.c., the pressure in dynes per sq. cm. and the density
in gms. per c.c., this gives the velocity in cms. /sec.]
133] MISCELLANEOUS EXAMPLES 159
Find the velocity of sound in air, at C. , the height of the mercury
barometer being 76 cms., given that density of air under these conditions is
001293 gms. and that the sp. gr. of mercury is 136. The pressure and
volume of the air are supposed to be connected by the formula
p V 1 ' = const.
J.
1. A body moves in a straight line in such a way that its acceleration
t seconds after it has passed a fixed point O on the line is (3t + 2t 3 ) ft. /sec. 2
Find its distance from O after 5 seconds, given that its speed when it
passes O is 10 ft./sec.
2. A uniform sphere diameter 2 T 1 B inches is turned down uniformly to
diameter 2J? inches. Find approximately the percentage by which its
weight is reduced. [Use no tables.]
3. Shew that the expression 1 + 12%x 3 increases in value as x
increases from 2 to +2 and after that diminishes as x increases.
4. In a certain curve the subtangent at any point is always equal to
3 times the abscissa of the point. Express this as a relation between the
(x, y) and ~ of the point. [If the tangent at P meet OX in T and PN is
the ordinate of P, TN is the subtangent.]
Shew that y = 2x* fulfils the above condition.
6. A steamer has to go 20 miles up a river which is flowing at 3 miles
an hour. The resistance of the water is proportional to the square of the
relative velocity and the coal consumed in a given time is proportional to the
product of this resistance into the distance the steamer has moved relative to
water. If the cost of the coal is to be made a minimum, find the speeds
of the steamer relative to shore and relative to water and the time taken on
the journey.
CHAPTER VII
INTEGRAL CALCULUS. AREAS UNDER PLANE
CURVES. MEAN ORDINATE
134. Integration. Suppose we want the area of the figure
bounded by HK part of the curve y=f(x), the a?axis, and the
ordinates AH, BK corresponding to x = a, x = b.
Fig. 63.
Divide AB into any number of equal parts and complete the
rectangles as shewn in Fig. 63.
134, 135] AREA BY SUMMATION 161
We get an approximation to the area we require by taking it
as equal to the sum of these rectangles. This is too small by the
sum of the blank portions HOP, PDQ, etc.
If AB were divided into twice a.s many parts we should get a
better approximation to the area, the sum of the 10 rectangles
in Fig. 64 being nearer to the area sought than the sum of the
5 rectangles in Fig. 63 [the sum of the two blank portions
between H and P < the portion HOP in the previous figure].
A L M
A L M
Fig. 65.
Fig. 66.
By dividing AB into more and more parts we obtain a series
of closer and closer approximations to the area sought.
135. If we complete rectangles as in Ficr. 65, we get an
approximation to the area required by taking the sum of these
external rectangles. This is too large by the sum of the black
portions HC'P etc.
If AB were divided into twice as many parts we should get a
better approximation, the sum of the 10 rectangles in Fig. 66
being nearer to the area sought than the sum of the 5 rectangles
in Ficr. 65.
M. a
11
162
CALCULUS FOR BEGINNERS
[CH. VII
[The sum of the two small pieces between H and P in Fig. 66
< the small piece HC'P in Fig. 65.]
By dividing AB into more and more parts we obtain a series
of closer and closer approximations to the area sought
136. Combining Figs. 63 and 65 into one (Fig. 67) we see
that the difference between the sum of the outside and the sum
of the inside rectangles is the sum of the rectangles CC', DD', EE',
FF', = rectangle G F'.
If we increase the number of parts into which AB is divided
the area of GF' diminishes, for its height remains constant and
its breadth diminishes, and by making the number of divisions
large enough, we can make the area of GF' as small as we please.
i.e. by taking a sufficiently large number of divisions, we can
make the difference between the sum of the outside rectangles
and the sum of the inside rectangles as small as we please ; and
since the area sought lies between these two sums, it is the limit
towards which the sum of outside or inside rectangles tends as
the number of divisions is indefinitely increased, and the breadth
of each division consequently indetinitely diminished.
The determination of this limit is called Integration.
135137]
INTEGRATION
163
137. Note. If the shape of the curve is like that shewn
in Figs. 68 and 69, the sum of the first set of rectangles is too
large and the sum of the second set too small, otherwise the
argument is the same and the area sought is the limit towards
which the sum of either set tends as the number of divisions is
indefinitely increased.
\
A B
Fig. 68;
A B
Fig. 69.
If the shape is as in Figs. 70 and 71, we cannot say with
certainty of the sum of either set that it is too great or too small,
C'
Fig. 71.
112
164
CALCULUS FOR BEGINNERS
[CH. VII
for in each case some of the strips are greater and some less than
the corresponding portions of the required area.
By dividing the area as in Fig. 72, we can apply the theorem
to each part separately.
Y
W Op 45 10 14j 5 X
Pig. 72. Pig. 73.
138. To make the principle clear we will take an easy
special case.
HK is a portion of the line y = 3. (Fig. 73.)
AH, BK are the ordinates corresponding to x = 4 and x = 14.
Divide AB into 10 equal parts and complete the rectangles as
shewn.
The different ordinates are the values of y corresponding to
x = 4, 5, ... 14.
137, 138] AREA BY SUMMATION 165
Now y = 3ox Hence the ordinates are 12, 15, 18 ... 42.
The width of each rectangle = 1.
.*. sum of inside rectangles
= 12 + 15 + 18 + ... + 39 (10 terms)
Sum of outside rectangles
= 15 + 18 + ... + 42(10 terms)
The difference between these is 30, the area of the rectangle
GK. [Height 30 = BKAH, width 1.]
The area ABKH lies between 255 and 285.
Now suppose AB divided into 100 equal parts.
The different ordinates will be the values of y corresponding
toa? = 4, 41, 42, ... 139, 14.
.'. the ordinates are
12, 123, 126, ...417, 42.
The width of each rectangle is !.
.'. sum of inside rectangles
= 1 [12 + 123 + 126 + ... + 417] 100 terms
= .! x 100 x 537 = 268 . 5>
2
and sum of outside rectangles
= l[123+126+...+42]
= . lx 100^543 = 2?1 . 5
The difference between these is 3, the area of a rectangle,
height 30 and width !.
The area ABKH lies between 2685 and 2715.
166 CALCULUS FOR BEGINNERS [CH. VII
Similarly shew that if AB be divided into 1000 equal parts
the area ABKH lies between 26985 and 27015.
We have two sets of numbers:
255, 2685, 26985,...
285, 2715, 27015,...
apparently both continually approaching a limit 270.
To make quite sure of this, suppose AB divided into n equal
parts each = h, so that nh = 10.
The ordinates are the values of y corresponding to
a? = 4, 4+A, 4 + 2A, ..., 14 A, 14.
Hence the ordinates are
12, 12 + 3A, 12 + GA, ...,42SA, 42.
The width of each rectangle is h.
.*. sum of inside rectangles
= A[12 + (12 + 3A) + (12 + 6A) + ... + (42  37*)] (n terms)
and sum of outside rectangles
= h [(12 + 3A) + (12 + 6A) + ... + 42]
.
m
The difference is 30/t the area of a rectangle height 30,
width A.
The area ABKH lies between 270  15h and 270 + 15h.
Now, as the number of strips is indefinitely increased, i.e. as
A  0, each of the areas (1) and (2) can be made as near as we
please to 270.
.'. the area ABKH which is the limit of either sum is 270.
Since the limit of the sum of inside rectangles when A *
is the same as the limit of the sum of outside rectangles, we
need only calculate one of the sums.
138, 139]
UNITS
167
We may say
either sum of inside rectangles = 270 15 A,
.'. area reqd. = Limit of this when h  0, = 270 ;
or sum of outside rectangles = 270 + 157*,
.'. area reqd. = Limit of this when h * 0, = 270.
139. Note on units. In Fig. 73 the ascale and the
yscale are the same, and the unit of area is the small square
which is shaded, whose side is the unit of length.
The area ABKH is 270 times this unit.
/. if the unit be 1" each way, the unit of area is 01 sq. in.
and area ABKH = 270 x 01 = 2'7 sq. ins.
If however the scales are different as in Fig. 74 the unit of
area is no longer a square but a rectangle whose sides are equal
respectively to the ojunit and the yunit. The area ABKH is 270
times this rectangle.
Y
40
30
20
10
2 3 4 5 6 7 8 9 10 11 12 13 14
Fig. 74.
If aunit = 3" and yunit = 05",
unit of area = '015 sq. ins.,
and area ABKH = 270 x 015 = 405 sq. ins.
X
168 CALCULUS FOR BEGINNERS [CH. VII
EXERCISES. XXXV.
1. Find by the method of 138 the area bounded by
If the unit is fo" each way what is the area in sq. ins.?
What is it if the arunit is J" and the yunit T V I
2. Find the area bounded by
y = 3x + 2, o;=0, y = B, y = 7.
[Divide into strips parallel to Ox.]
What is the area in sq. ins. if the arunit is 3" and the yunit "?
3. P is the point (10,100) on the curve y = x z . PN is the ordinate of P
and O is the origin.
Find the area OPN bounded by ON, NP and the curve.
(i) Divide ON into 10 equal parts,
(ii) ,, ,,100
(iii) ,, ,, n ,, each h,
and in every case find the sum of the inside and outside rectangles, using
the formula
..
b
(iv) Find the limit of each of the last results as h * 0.
4. Find the area OPN bounded by y^x 3 , the araxis and the ordinate
x=3.
Divide ON into n equal parts and use the formula
140. We shall now solve the problem of 138 by adopting
an entirely different point of view.
P is any point (a;, y) on the line y = 3x. (Fig. 75.)
An ordinate is supposed to advance from the position AH
corresponding to x = 4, to the position BK corresponding to #= 14,
its length continually changing in such a way that its upper end
140J
INTEGRATION
169
is always in the line HK. When the ordinate reaches MP, it has
swept out a certain area which is obviously a function of x, i.e. it
is determined when x is known and changes as a? changes. Call
this area A.
7
A MN
Fig. 75.
"We will consider the change in the area due to a short
advance of the ordinate.
Let Q be a point (a; + Ao;, y + Ay) on the line and complete
the rectangles as in the figure.
Then the area PMNQ is the increment in the area A, due to
an. increment AOJ in a;, and may therefore be denoted by AA.
170 CALCULUS FOR BEGINNERS [CH. Vil
Now PMNQ lies between PMNR and SMNQ.
i.e. AA lies between y . Ace and (y + Ay) . Ace.
.*. lies between y and (y + Ay).
As Arc, and with it Ay * 0, (y + Ay) * y.
dA
 = y = 3x.
.". A is a function of x which when differentiated gives 3x.
3
where c is some constant to be determined.
Now since A stands for the area traced out by an ordinate
starting from AH, A = when the ordiuate coincides with AH, i.e.
when x = 4.
= 24.
We want the value of A corresponding to x = 14.
= 29424
= 270.
3
Notice 294 is the value of x 2 when x= 14
3
and 24 is the value of =a? when x = 4,
and the area sought is the difference of these.
140, 141]
INTEGRATION
171
EXERCISES. XXXVI.
To be solved by the method of 140.
1. Find the area bounded by y = 3x, y = 0, a? =3, x = 9.
3. Find the area bounded by y = 3x, y = 0, x = 0, x=9.
3. Find the area bounded by y = Bx, y = 0, x = 0, x=3.
4. Find the area bounded by t/ =  x + 2, y = 0, x = 5, a:=10.
IB
5. Find the area bounded by y = 5x + 7, J/ = 0, x = Q, x=5.
6. Find the area bounded by y = Bx + 2, aj = 0, y = 3, j/ = 7.
7. Find the area bounded by y = 4x 3, a:=0, y=0, y = ll.
141. Example. Find the area OBK bounded by y = 0, o?=10
and part of y = x*.
As before let M P be the ordinate of the point P (x, y) and
PMNQ a strip of width A. [Fig. 76.]
MN , X
Fig. 76.
Let A be the area OMP, then AA lies between ar'Aa; and
x + Aa;' 2 Aa;.
172 CALCULUS FOR BEGINNERS [CH. VII
This time our moving ordinate which is supposed to trace out
the area starts from O where x 0.
.*. when x = 0, A = 0.
r 3
' A
~3'
10*
,\ area OBK = ^ = 333^.
EXERCISES. XXXVIL
1. Find the area bounded by y=x z , y = 0, x = B, a: = 14.
2. Find the area in the first quadrant bounded by
y = x 2 , x0, y = l, y = 4.
3. Find the area bounded by y=x 5 , y=0, x = 2, x=5.
4. Find the area bounded by y = 2x 2 + Bx + 1, y = 0, x = 3, x = 7.
5. Find the area bounded by y=x 3 + x, y = 0, x=3, a: = 11.
6. Find the area bounded by y = x 3 , x=Q, y = S, j/ = 27.
7. Find the area bounded by y = x 3 , y = Q, x=a, x = b.
142. Notice that in leading up to the equation
//A
~ = 3x in 140
ax
we could have supposed the moving ordinate to start from any
position. The fact that it was supposed to start from the
position AH was used in the determination of the constant c in
3
A = H a; 2 + c.
4i
3
The equation A  ^ a? 24
a
is a formula for the area traced out by an ordinate which starts
from AH and moves into any other position.
141, 142]
INTEGRATION
173
Suppose now that the moving ordinate started from A'H' (a; = 2).
[Fig. 11.}
I
OA'A B X
Fig. 77.
We should get in the same way
as a formula for the area traced out by an ordinate which starts
from A'H' and moves into any other position.
Thus
area A'H'HA = . 4 2  6,
area A'H'KB =  . 14 2 6.
2
  4 a
2 '
as before, the constant 6 disappearing in the subtraction.
Therefore if we are finding the area between two given
ordinates x  4, a; =14 we can deduce at once from
dx
174 CALCULUS FOR BEGINNERS [CH. VII
that area between x= 4 and x= 14
rs i 14
or as it may be written ^ or ;
and the same result would be obtained if we wrote for
3 3
where c is any constant whatever.
3
In this particular case, A = ^ ic 2 is a formula for the area traced
2i
out by an ordinate which starts from the origin, so that when we
T3 ~l 14 T3 I 14
= UarM e<#6
L^ 1 4 I ^ _J 4
say
we are merely saying that
AHKB = OKB  OHA = A'H'KB  A'H'HA.
143. In Ex. xxxvn (4), the result is 309 34, 309^ and 34
2X 3 3JC 2
being the values of ~ + ^ + x when x = 7 and x =3 respec
O L
tively.
2a; 3 3JC 2
j + ^r + a; gives the area traced out by an ordinate starting
o a
from OL
309^ is area OLKB ; 34^ is area OLHA. [Fig. 78.]
We miht have said
295 being area A'H'KB and 21^ the area A'H'HA, where A'H'
is the ordinate x = 2, the constant 13 being chosen so that
5 + jf + a; 13^ = when x = 2.
o 2
142144]
INTEGRATION
175
HO
120
100
80
60
40
20
01234S678
Fig. 78.
& a 4 b* , a 4
144. In Ex. xxxvn (7) the result is jj. j and 4
bein<* the values of ^ when x = b and * = respectively. We
e 4
fa; 4 "! 6
might write the result A = T meaning
f A 1 f QtA
Value of  when x = b  Value of ^ when a  aj
and the work might be stated shortly thus :
^ = .'
being a function [regardless of arbitrary constant] which when
differentiated gives a?.
176 CALCULUS FOR BEGINNERS [Ctf. VTT
145. Notice in Ex. (4), instead of finding the values of
when x= 7 and when x = 3 we might more conveniently say
= 274f.
146. Generally. If we want the area bounded by y =f(x}
[any function of x], y = 0, x = a, x = b, we have, making the same
construction as in 140 :
Area of strip PMNQ lies between yAo; and (y + Ay) A#.
[Fig. 79.]
A MN. B X
Fig. 79.
i.e. AA is between yAa; and (y + Ay) Ase.
.'. is between y and y + Ay ;
but as A, and with it Ay,  0, y + Ay  y,
145148] INTEGRATION 177
.'. A is some function of x which when differentiated gives
/(#). Suppose <f> (x) to be such a function,
.'. A = d> (x) + c.
L*^
Now A = when x = a, .'. c =$(), .'. A = < (a;) < (a),
.*. area between AH and BK = <f> (b)  (a) = \<f> (x)~
147. Notice that PMNR is the area which would be traced
out by the moving ordinate between M and N if it kept through
out the length which it has at M.
SMNQ is the area which would be traced out if it kept
throughout the length which it has at N.
d\
Notice also that so far as finding is concerned, it is suffi
ax
cient to say
AA =f(x) . Ao: approximately,
AA
.*. =f(x) approximately,
it being understood that the statement
AA f(x) . Aos approximately
means that AA is betweeny(a;) . Aa; and/(o; + Ace) . AiK.
EXERCISES. XXXVIII.
Find formulae for the area traced out by an ordinate of y = 5x s + 2 which
starts from the position given by (i) # = 0, (ii) x=3, (Hi) x= 2 and moves
into any other position. Deduce from each of these formulae the area
traced out by an ordinate which moves from x=5 to =10.
148. Returning to the investigation in 138, p. 166, each
term in the series (1) such as (12 + 6h)h is the area of a rect
angular strip.
(12 + Qh) is the value of y or 3x when x = 4 + 2h.
M. C. 12
178 CALCULUS FOR BEGINNERS [CH. VII
The series is
12/i + (12 + 3A) h + (12 + 6A) A + ... + (42  3A) h ;
[12, 12 + 3A, 12 + 6A, ...(423A),
being the values of y or 3x when x = 4, 4 + A, 4 + 2A, ... (14 A)],
or more shortly
x=Uh
2 3x . h,
a=4
meaning "Take values of x, starting at 4 and increasing by
h at a time up to (14 A) ; for each value of x, find the corre
sponding value of 3x', multiply each of these values by A, and
add all the results."
The actual area sought is the limit of this sum when A~*0.
X=li z=14fc
2 3x . h would mean the same as 2 3a:.A with the
z=4 z=4
addition of an extra strip 42h (42 being the value of 3x when
03= 14) and this can be made as small as we please by taking A
small enough.
x=i4 x=nh
i.e. the limits when A * of 2 3a? . A and 2 3x . h are
x=4 x=4
the same.
a =14
.*. we may say that the area ABKH is Lt 2 3.rA.
h*~o 1=4
149. Generally if the curve is y=f(x), adopting the
usual notation as in 146, the area of the rectangular strip
PMNR is 2/A.r, and the sum of the areas of all such strips between
=6
AH and BK may be denoted by 2 yAo?, meaning : take values of
x=a
x starting at a and increasing Aa; at a time up to b ; for each
value of x get the corresponding value of y from y=f(x);
multiply each value of y by Arc and add all the results.
The area of the figure bounded by HA, AB, BK and the curve
is the limit to which this sum tends as Ace * and this is
148151] INTEGRATION 179
rb r r
denoted by / ydx I being simply a lengthened form of the
J a LJ
letter si.
rb x=b
150. Thus I ydx is defined to be Lt 2 yAx, and
.'a Ax*Ox=a
we have seen that in order to get this limit, we must find a
function of x which when differentiated gives y, and subtract
the value of this function when x = a from its value when x = b.
where = f(x).
dx
/i4
e.g. j[ (8)dte
that is to say instead of finding the area by a process of direct
summation, we find the rate at which the area increases with
respect to x and from this deduce the area as a function of x.
In other words instead of finding A from
A= ( f(x)dx= Lt 2 f(x) Aaj,
a summation which in most cases it would be impossible or at
any rate inconvenient to effect, we find A from
and we say that
rb
]ft
()
Ja
where </> (x) is a function of x such that
d(fr(x) ,. .
TT : t/
151. Notice that just as dy and dx in the expression j
have no separate meanings, but the form ~ is preserved to
122
180 CALCULUS FOR BEGINNERS [CH. VII
remind us of of which ^ is the limit, so dx in the expression
*Au? CISC
/14 [14
I (3#) dx has no separate meaning, but the form I (So;) dx is
x=u
preserved to remind us of 2 (3a;) Ax of which it is the limit.
x=4
x=14
Az represents a definite length and 2 (3x) Ase represents the sum
x=4
of a finite number of rectangles.
L
14
(3x) eta stands for the limit of this sum as Ax * 0.
4
fit
152. An expression like I (3a;) dx is called a definite
J4
integral, 4 and 14 being called the limits of the integral.
It is read "Integral of (3a;) dx between 4 and 14."
We have seen that in order to find the value of this we must
first discover a function of x which when differentiated gives
(80).
This function is written I (3a;) dx and is read " Integral
(3) dx."
Thus
/1
J
An expression like I (3x) dx is called an indefinite integral.
y = l(3x)dx,
are merely different forms of the same statement. When the
151, 152] DEFINITE AND INDEFINITE INTEGRAL 181
value of an indefinite integral has been found, it should always
be checked by differentiation.
// 1 9 \ ^<y4 9 o 9
I / o K \ r "^ $ **
Check :
d^ /3jB_ 4 _ 2 f 2 \_3 23 ^ / 1\
dx\ 4 3 x J 4' 3'2'' \ x~)
i 2
= 3x 3 x + 2 .
The first step in the evaluation of a definite integral is the
determination of the indefinite integral*.
EXERCISES. XXXIX.
Write down the values of the following indefinite integrals and in each
case check by differentiation ;
1. 2x*dx. 2.
. dx.
J
. I (ax 2 + bx + c)dx. 4. /  ? .
J J ^v*
5. (5t + 6t 2 t)d. 6.
. \ldx. 8. f Ux*~
dx.
*'
* Notice that the indefinite integral always contains an arbitrary
constant, but in the subtraction which forms part of the process of
evaluating the definite integral, this constant disappears and may therefore
be omitted, so that, instead of saying
_
we may say
/
,
182 CALCULUS FOR BEGINNERS [CH. VII
Hence write down the values of the following definite integrals :
10.
n
h
11.
/'('^ + ?) fc
12.
ri
Ji a
13.
P X
JjS^/x
14.
16.
/8
Jo
r
\ Idx.
J3
15.
17.
./ 30 v1 ' 4
(Y3x35x2 + 2x
J 3\
f 2 / l \
i \
_LW
o*^/
18.
^ 1
Find the values of the following:
z=7
19. 2 2x 2 . Ax (i) when Ax = l, (ii) when Ax= 05.
X=5
20. 2 7. Ax (i) when Ax=l, (ii) when Ax = '01.
z=3
Z=2 / 1 \
21. 2 I x + ^ Ax when Ax ='2.
=i V 2 /
Compare your results with those of Exs. 10, 16, 18.
153. The process of finding an area may now be written
down as follows : (taking again the example of 140)
Area of strip = (3aj) Ax app.
{z=u
.*. Sum of strips = 2 (3a;) Aw app. and area required is the
x=i
limit of this sum when Ax * 0. !
/M
.*. Area required = I (3x) dx
Ji
After a little practice the part in brackets {} may be omitted.
153] INTEGRATION 183
EXERCISES. XL.
1. P is the point (2, 4) on the parabola y=x i . PM and PN are perp.
to OX and OY.
Find (i) area OMP, (ii) area ON P.
Verify that their sum is the area of the rectangle OMPN.
2. The same where P is (2, 8) on y = x 3 .
f h z \ x 2
3. P is the point ( h, J on the parabola y . PM and PN are
perpendicular to OX and OY.
Find separately areas OMP, ONP and shew that they are respectively
onethird and twothirds of the rect. OMPN.
/ / t 3\ x 3
4. P is the point I h, 5 ) on the curve y = 5 , PM, PN are perp. to OX,
V a* I a 2
OY.
Prove that OMP, ONP are respectively onequarter and threequarters of
the rect. OMPN.
/ h n \ x n
5. P is the point I h,  r I on the curve y==, . PM, PN are per
\ a" 1 / a*
pendicular to OX, OY. Prove OMP= ^ rect. OMPN.
n + 1
6. Find the area bounded by
= 0, x=2, x = 7.
7. Find the area bounded by
c, y=0, x=p, x=p.
8. The coordinates of H, K, L are respectively
(a, h) (a, k) (0, I). [Fig. 80.]
A curve whose equation is yp + qx + rx 2 passes through H, L, K.
Find p, q, r in terms of a, h, k, I.
9. In Qu. 8, if HA, KB are the ordinates of H, K, prove that the area
184
CALCULUS FOB BEGINNERS
[CH. VII
154. The result of Exs. XL. (9) leads to a very important
rule called Simpson's Rule for finding approximately the area of
any figure.
A B X
Fig. 80.
Suppose we want the area bounded by the curved line
PQRSTUV, the ordinates AP, GV, and AG perp. to AP and GV.
[Fig. 81.]
Q R
A B C D E F G
Fig. 81.
We can get a first approximation as follows :
Divide AG into any number of equal parts and draw ordinates
through the points of division as in the figure. Join PQ, QR etc.
154, 155] APPROXIMATE RULES FOR AREA 185
Then if AP = A n BQ = A 2 , e *c. and AB = BC =etc. =a we have
Area of trapezium ABQP =  (Aj + A 3 ).
Area of trapezium BCRQ=  (/* 2 + A 8 ) and so on.
Area of trapezium FGVU = ^ (A 6 + hj).
.'. by addition, Area of figure is approximately
d
or generally, for any number of equidistant ordinates
Area is approximately ^. Distance between consecutive
ordinates x {sum of extreme ordinates + twice sum of
intermediate ordinates}.
This is the Trapezoidal rule and its accuracy is obviously
increased by taking more and more strips.
Notice that this result is obtained by substituting for the
curved boundary, a boundary composed of straight lines through
pairs of consecutive points.
In other words the portion of the boundary between two
consecutive points is replaced by a line whose equation is of the
first degree or of the form y =p + qx.
155. Now if we suppose the portion of the boundary passing
through 3 consecutive points to be replaced by a curve whose
equation is of the second degree of the form y = p + qx + rx*, we
shall get a closer approximation.
Considering the portion PQR.
Take B as origin and axes along BC and BQ.
Then coordinates of P, Q, R are respectively
( a, Aj) (0, A 2 ) (a, A 8 )
186 CALCULUS FOR BEGINNERS [CH. VII
and if y p + qx + ne 2 passes through these three points, we have
A! = p qa> + ra?
whence
3 = p + qa + ra
h 3 fir,
Now area ACRP bounded by PA, AC, CR and y =p + qx + rx z
/ of
( p + qx + rar) dx  px +
Ja L
2 3 Ja
2ra 8
~3~*
Substituting the values of p and r, this becomes
2/4 2 a + (^i  2^ + ^3) = o (^i + 4A 2 + 7< 3 ),
and this is taken as being approximately the area of the portion
PACR of the given figure.
If there is an even number of strips as in the figure, we shall
get in exactly the same way
Area RCET = ^ (h 3 + 4A 4 + h 6 ) approx.
ci
and Area TEGV  ~(h 6 + 4ft 6 + h,)
3
Adding we get
Area of figure = ^ {/t x + h 7 + 4 (h 2 + h 4 + h 6 ) + 2 (h 3 + /i & )} approx.
and similarly for any even number of strips.
We thus have Simpson's rule :
Divide the area into an even number of strips of equal width
by an odd number of ordinates ; the area is approximately
7j . Width of a strip x {sum of extreme ordinates + twice sum
3
of other odd ordinates t 4 times sum of even ordinates}.
155157]
SIGN OF AREA
187
EXERCISES. XLI.
1. Find by integration the area bounded by y=0, x=2, #=12 and the
curve y = x 3 .
3. Find the approximate area in (1) by using 3 ordinatea and applying
(i) the trapezoidal rule, (ii) Simpson's rule.
3. Do the same as in (2) using 11 ordinates.
4. Do the same as in (1), (2) and (3) for the area bounded by y = 0,
x = 2, x = 7 and y = x*.
156. Note on sign of area. If we advance from left to
right, i.e. if Ao; is positive, y&x representing the area of our
typical strip is positive if y is positive, i.e. if the strip lies
above the icaxis and negative if the strip lies below the a:axis
[Fig. 82.]
I
+ y  y + y 
LX
+ Ao:
+ AX
Ao:
y
kx
\
+ y&x
y*x
y&x
+
*
~~ 1
J
}
I
Fig. 82.
If we advance from right to left, i.e. if Aw is negative, y&x is
positive if the strip lies below and negative if the strip lies above
the xaxis.
a?
157. Thus if the curve be y =  2 and P, Q be respectively the
a
f of
points ( a,  a), (a, a) [Fig. 83], / z dx will give the area PMO
Ja a
and will be negative since we are advancing from left to right
188
CALCULUS FOR BEGINNERS
[CH. VII
and the area is below the axis, i.e. during the whole motion of
the ordinate from MP to O, Ax is + and y is .
/;
a
Fig. 83.
z dx will give the area OMP and will be positive since
we are advancing from right to left and the area is below the
axis, i.e. Ace is and y is .
f a y? [ y?
Similarly I , dx will be positive and I dx will be
h ' Ja o?
negative.
/ y?
~2 dx we shall get zero, which simply means
a a
that the moving ordinate in passing from MP to NQ sweeps out
two numerically equal areas of opposite signs.
If we want the actual area shaded, we must find one of the
portions PMO, ONQ separately.
.'. Shaded area =  .
157, 158] SIGN OF AREA
158. As another example, consider the curve
189
between x=  2 and x= + 5. [Fig. 84.]
The area bounded by the curve, the a:axis, and the extreme
ordinates is seen from the figure to consist of 3 parts, two below
the ojaxis and one above.
Fig. 84.
190 CALCULUS FOR BEGINNERS [CH. VII
r5 r~ %y3 y^~\ >
Now I (4 + 3o:  or) dx \ 4x+ 
ja 4 * J s
2
. 21. 133
The coordinates of the points R and S are (1, 0) and (4, 0).
~
f 1
2
i.e. actual area of portion PMR = 2.
This is area RAS.
r5
i.e. actual area of portion SNQ = 2.
i.e. the definite integral gives the algebraic sum of the 3 areas
PMR, RAS, SNQ, or the excess of the actual area PMR over the
sum of the actual areas of RAS and SNQ.
EXERCISES. XLII.
f2
1. Shew that I (lx)dx=0.
Jo
Draw a figure to explain the result.
/I
(x 3 3x*+2x)dx=0.
Draw a figure to explain the result.
It the curve y = x 3  Sx 2 + 2x meet the xaxis in O, B, D (in order from
left to right), and the ordinates at O and D meet the line y =  ? in E, F,
what is the area of the figure bounded by the curve OE, EF, FD?
158J INTEGRATION 191
/+!
3. Find I (x 2 x 3 )dx and interpret the result.
4. Find the area bounded by
y =0^6x2 + 9*4 5
the xaxis and the maximum and minimum ordinates.
6. Draw the curve y*=x(xl) 2 between x=0 and x=2 and find the
area of the loop.
6. P, Q, R are the points on y = x 3 at which z=0, 2, 4.
Find a, b, c BO that the parabola
may pass through P, Q, R.
Draw the two curves on as large a scale as possible and shew that the two
closed portions contained between the curves are equal in area. Also find
the area of each portion.
7. P, Q, R are the points on y = x 3 at which x hk, h, h + k,
Find a, b, c so that the parabola
may pass through P, Q, R.
Shew that the area bounded by y = 0, x = hk, x=h + k and y=x 3 is the
same as that bounded by j/ = 0, x hk, x=h + k and this parabola.
8. If the area bounded by y = x 3 , y = and any two ordinates be found
by Simpson's Rule, what does (7) tell you about the result?
9. What is the area bounded by
y=p + qx + rx* + sx3, y=0, x=k, x+k?
Notice that your result is independent of 3 and s.
10. If P, Q, R be the points
(a, ftj) (i\ fc 2 ) (a, fcs),
and if the curve y=p + qx + rx z + sx s passes through P, Q, R, shew that p,
r have the same values as in 155 but that q, s are indeterminate.
Find the area bounded by this cubic, t/ = 0, and the ordinates at P and R.
Shew that it is the same as that obtained in 155.
1 1 . Draw roughly y  (.T + 1 ) (2  x) .
Find the area of the part above the xaxis.
192 CALCULUS FOR BEGINNERS [CH. VII
12. Trace the curve
and find the areas of the two portions bounded by the curve and the xaxis.
13. Find the area common to the two curves
y z =iax and a: 2 =4ay.
14. Find the area bounded by y%=x 3 and x=2.
15. P, Q are the points on the parabola y = 5 + 3x  2z 2 where x= and
x1^. Find the area bounded by the chord PQ and the portion of the
curve above it. If the tangent to the curve which is parallel to PQ meet the
ordinates of P, Q in M, N, shew that the area just obtained is twothirds of
the parallelogram PQNM.
16. P, R, Q are three points on
y = ax* + bx + c
corresponding to x=h, x = 0, x = h.
The tangent at R meets the ordinates at P, Q in M, N. Shew that this
tangent is parallel to PQ and that area PRQ bounded by the curve and the
chord PQ is twothirds of the parallelogram PQNM.
159. The fact that any definite integral may be interpreted
as an area can be used to find sometimes accurately, sometimes
approximately the value of a given definite integral.
ra . _
Ex. Find I *Ja?x>dx.
Jo
At present we do not know how to find the indefinite integral
V a 2 x 1 dx.
But if we draw the graph of y = \/a 2 or 2 between x = and
x = a we get a quadrant of a circle, centre at the origin, radius a.
[Fig. 85.]
The given definite integral is the area of this.
159, 160]
APPROXIMATE INTEGRATION
103
Even if we did not recognise the curve as part of a circle we
could find several points on it, i.e. calculate the lengths of a
number of ordinates and apply Simpson's Kule to find the
approximate area.
Fig. 85.
EXERCISES. XLIII.
1. Shew from a figure that
= 2

J 
3. Find approximately by Simpson's rule [use 11 ordinates]
f 3 1
3. Similarly find I y
J 2 L +
160. If we draw
and
dx.
x
dx and
__1
1
= 4^ +
[3 1
I ,
y 2 la;
dx.
(which may be called the integral curve of t/ = ^ + 2 j between
the same values of x (say 2 and 12), the number of units of length
in the difference of the extreme ordinates in the second graph is
M. 0, 13
194
CALCULUS FOR BEGINNERS
[CH. VII
equal to the number of units of area in the figure bounded
by the curve, the araxis and the extreme ordinates in the first
graph. [Fig. 86.]
S::
GO
40
20
**
20
40
Fig. 86.
Or No. of units of length in gk (or g'k') (55) = No. of units of
area in AHKB and generally if the second graph be y=f(x) and
the first y /' (x) a corresponding result will be true.
160]
195
Notice that it does not matter which curve of the family
we use.
e.g. in the figure hk is y   y? + 2x (c = 0),
h'k'isy = jx> + 2x'20 (c = 20).
The ordinate of ti = (the ordinate of h)  20
and the ordinate of k'  (the ordinate of k)  20.
So that difference between ordinates of h' and k' difference
between ordinates of h and k.
132
196 CALCULUS FOR BEGINNERS [CH. VII
161. As another example take the graphs of
y = 6a; 2 18aj+ 12 [/'(*)] and y = 2a 8 9a? + 12+c [/()].
[Fig. 87.]
No. of units of area in STLP = No. of units of length in
Ip ts, i.e. in up or in lp'ts, i.e. in u'p'l2^.
No. of units of area in PLQ = No. of units of length in
mq lp, i.e. in vq or in mq' lp\ i.e. in v'q' = l.
No. of units traced out as ordinate moves from L to R = No.
of units of length in nr Ip or in nr  lp' = 0.
i.e. area PLQ = area QRK.
No. of units traced out as ordinate moves from Q to R = No.
of units of length in nr mq or in nr mq' = vq or v q' = 1.
i.e. area QRK = 1 unit.
162. We saw on p. 142 that if h is small
to a first approximation.
We can now obtain a closer approximation.
As a first example suppose
so that
/'(*) = 4* + 3.
In this case y =f (x) is a straight line. [Fig. 88.]
Now, No. of units of length in nq mp = No. of units of area
in PMNQ.
The gradient of PQ is/" (x).
.'. RQL=hf"(x).
.'. No. of units of area in PMNQ
161, 162]
INTEGRAL CURVES
197
/. f(x + h) f(x) = hf (x) + \ Vf" (x)
or
This is accurately true in this case and in every case where
f(x) is of the form
aa? + bx + c,
since in all such cases y =f (a;) is a straight line.
y=f(x)
f(x+h)
f(x)
rn n
y=f(x)
?B
R
hf"(x)
If'
x
M N
Fig. 88.
198 CALCULUS FOR BEGINNERS [CH. VII
EXERCISES. XLIV.
Verify f(x + h) =f (x) + hf (x) + i fc/' (*),
(i) when / (x) =
(ii)
(iii)
163. If y =f (x) is not a straight line, we have as before
f(x + h) f(x) = No. of units of area in PMNQ
= No. of units of area in PMNT approximately
[Fig. 89]
i.e. f(x + h) =f(x) + hf (x) +  tff" (x) approximately.
2
Notice that in this case we take the arc of y =f (x) as being
approximately a straight line, whereas in obtaining the first
approximation we took the arc of y =f(x) as being approxi
mately straight
e.g. taking the example on p. 143.
Find approximately the value of
3af4a? + 6x*7x+ 8 when a: = 301.
/' (*) = 15x 4 l2x*+ 12*  7,
/"(*) = 60^240; +12.
= 662, /'(3) = 1136, /"(3)=1560,
/. /(301) =/(S) + 01 /' (3) + f" (3) app.
= 662 + 1 136 x 01 + x 0001
2
= 6734380 app.
[Actually /(301)  673438264503.]
MEAN ORDINATE
199
163, 164]
164. Mean or average ordinate. Suppose AB to be a
portion of the curve y f(x\ A, B being the points corresponding
to x = a and x = b. [Fig. 90.]
5
I
M
Fig. 89.
Fig. 90.
The mean ordinate of the curve between x  a and x = b is the
height of the rectangle with base MN equal in area to the figure
AMNB.
It is in fact that ordinate of constant length which in
moving from M to N traces out the same area that the variable
ordinate of the curve does in moving from M to N.
[Compare idea of average speed p. 2.]
200
CALCULUS FOR BEGINNERS
[CH. VII
165. Suppose for example, we want the mean ordinate of
the curve y = x 2 between x = \ and x = 5. [Fig. 91.]
and
T 2 3 4 5
Fig. 91.
[* > 7 f^l 5 124
AreaAMNB=/ x*dx= ^ = ~3~
.*. Mean ordinate = 5  = = lOi.
6 x 4 6
If CD be the line y = 10,
Area CMND = area AMNB.
EXERCISES. XLV.
1. Find the mean ordinate of the curve y x 2 between x=4 and a; = 10.
Find also the arithmetic mean of the two ordinates a;=4, a; = 10 and the
midway ordinate corresponding to a; = 7.
a. Find the mean ordinate of the line j/ = 3x + 5 between x = 3 and
Find also the arithmetic mean of the two ordinates x = 3 and x = 15 and
the midway ordinate.
165, 166]
MEAN ORDINATE
201
3. Prove by calculus and geometrically that the mean ordinate of
y=mx + n between x = a and x=b,
the arithmetic mean of the ordinates x = a, x = b and the midway ordinate
are all equal.
4. Find the mean ordinate of y=x 3 (i) between x = 3 and x=7,
(ii) between x = a and x=b.
6. Find the mean ordinate of y=x z 4x + 3 between x=l and x=5.
6. Find the mean ordinate of y = x% between x=0 and x=+3, also
between x=  3 and x= +3 and explain why these results are the same.
166. If we divide the area into a number of strips (say 20)
of equal width (2), and complete the inside rectangles in the
usual way with an extra one on the right [Fig. 92], the area of
"1 2 3 4 5
Fig. 92.
the figure bounded by AM, MN', N'B' and the zigzag boundary
AB' is
and MM' = 42.
So that the mean height of this figure is
2 fl 2 + 1 '2 2 + . . . + 5 2 ] I 2 + 1 2 2 + . . . + 5 3
42 21
= the Arithmetic mean of the 21 ordinates from AM to BN inclusive.
202 CALCULUS FOR BEGINNERS [CH. VII
[This is 10^ = 1047.]
If we take 100 strips the Arithmetic mean of the 101
ordinates is 1043.
The more strips we take the nearer does the Arithmetic mean
approach to 10^.
This is only what might be expected, for we have shewn that
by increasing the number of strips the area AMN'B' can be
brought as near to the area bounded by AM, MN, NB and the
curve BA as we please.
The value 10^ may thus be called the mean or average value
of a; 2 between x = 1 and x = 5, and it is the limit to which the
Arithmetic mean of a number of equidistant ordinates the first
of which is AM and the last BN, approaches as the number of
ordinates is indefinitely increased.
EXERCISES. XLVI.
1. Find the mean value of x 2 + 3x between x=2 and z=4.
Find also the Arithmetic mean of 11 values of x 2 + Bx corresponding to
x = 2, 22, 24, 26, ...4.
2. Find the mean value of , between x = 1 and a; =4.
x e
3. Use Simpson's rule to find approximately the mean value of 
between x = l and x = 2. (11 ordinates.)
Also use a table of reciprocals to find the Arithmetic mean of 11 values of
 corresponding to o; = l, !!, 12, ... 2.
*C
4. Find the mean value of z 3 + 2x z + 1 between x = 2 and x = 7.
167. If part of the curve y =J (x) is below the aaxis the
corresponding portion of area must be reckoned negative.
Thus to get the mean ordinate of
y = y?
between x = 2 and a? = 3.
166, 167]
We have
MEAN ORDINATE
208
65
and this gives actual area ONB  actual area OMA.
65 13
. . Mean ordmate = = =  = 3 i .
4x5 4
If CD be y = ^,
Area CMND = actual area ONB actual area OMA [Fig. 93]
or
= algebraic sum of areas ONB, OMA.
Fig. 93.
If we take 26 equidistant ordinates distance apart 2 the
Arithmetic mean will be
(_ 2) 3 + ( 1 8) 3 + ( 1 6) 3 +...( 2) 3 + O 3 + (2) 3 + (4) 3 + . . . + 3 8
~2Q
_ 22 3 + 24 3 +...3 8 _ 91 _
~~26~ ~26~ *
and as before the greater the number of ordinates we take the
nearer will the Arithmetic mean be to 3f .
204 CALCULUS FOR BEGINNERS [CH. VII
EXERCISES. XLVII.
I. Find the mean ordinate of y = 2x + 6 between x=  11 and x=5.
Draw a figure to explain the result.
3. Find the mean ordinate of y=a?
(i) between and 3,
(ii) between  3 and 3.
Draw a figure.
3. Find the mean value of 4 + 3x  x z between x =  2 and x = 3.
Find also the Arithmetic mean of 11 equidistant ordinates.
4. Find the mean height with respect to x of that portion of
y = 4 + 3a;x 2
which lies above the zaxis.
6. Find the mean value with respect to the abscissa of the square of the
ordinate of a semicircle of radius a.
6. Find the mean value of 2x 3  3a; a  36z f 30 between x= 3 and
CHAPTER VIII
FURTHER APPLICATIONS OF THE INTEGRAL CALCULUS
168. So far we have only considered the Integral Calculus
in its application to the areas of plane curves.
We shall now shew how similar methods may be applied to
other problems.
Ex. 1. The speed of a body in ft. /sec. at the end of t seconds
is given by
V = 3t2.
Find the distance travelled in 12 seconds from rest.
Suppose the whole time 12 sees, to be divided into n equal
intervals each h sees., so that nh=l'2, and suppose the speed to
remain constant during each interval and equal to the speed at
the beginning of the interval.
The speeds at the beginning of successive intervals are
0, 3A 2 , 3 (2hy, ... 3 (n^lhf,
and the total distance described on our assumption would be
h [0 + 3# + 3 . (2A) 2 + ... + 3 (n^\hy\
01J
2 2 + 3 2 +... (n1) terms] =(l) n(2l)
= 1728  216h + 6h2 (= aj.
206 CALCULUS FOR BEGINNERS [CH. VIII
Now suppose the speed to remain constant during each inter
val and equal to the speed at the end of the interval. The total
distance would be
h [3/t 2 + 3 . (2&) 2 + . . . + 3 (w/i) 2 ]
= 3h 3 [P + 2 2 + 3 2 + ... n terms]
o 73
=  . n (n + 1) (2n + 1)
= 1728 + 216h + 6h2 (= S 2 ).
Now the actual distance (S) lies between Sj and S. 2 . But by
diminishing h indefinitely we can make Sj and S 2 each as near to
1728 as we like.
.'. S is the limit of either Sj or S 2 as h*Q, and S= 1728.
Notice that it is unnecessary to find both S, and 83. We can
say
either S 1 = 1728  216A+ 6A 2 , .'. S=1728,
or S=1728 + 216A + 6/t 2 , .'. S=1728.
EXERCISES. XLVm.
If v=5 + 7t find by this method the distance travelled,
(i) in the first 5 seconds,
(ii) in the 10th second.
Also in each case find what the distance would be if we supposed the time
to be divided up into intervals of 01 sec. and the speed to remain constant
during each interval and equal to the speed at the beginning of the interval.
Find by how much per cent, the distance calculated on this assumption
differs from the actual distance.
169. The method employed in 168 corresponds exactly to
that employed in the determination of an area in 138. There
we had a continually changing ordinate (y) whose length was given
as a function of the abscissa (a;). Here we have a continually
changing speed (v) whose magnitude is given as a function of the
time (t). There we got an approximation to the area traced out
168170] DISTANCE FROM SPEEDTIME FORMULA 207
by the moving ordinate by dividing the distance travelled by it
into small pieces and supposing the ordinate to remain the same
length as it passed from end to end of each small piece : we then
added the areas of the thin rectangles so obtained and finally found
the limit of the sum of these areas as the number of them was
indefinitely increased and the breadth of each consequently
indefinitely diminished.
Here we get an approximation to the distance travelled by
the body by dividing the whole time into small intervals and
supposing the speed to remain constant during each interval ; we
then add the small distances so obtained and finally find the
limit of the sum of these distances, as the number of them is
indefinitely increased and the length of each interval consequently
indefinitely diminished.
170. We shall now approach the problem from a different
point of view corresponding to that adopted in 140, in finding
the area traced out by a moving ordinate.
Let s ft. be the distance described in t sees, and s + As ft. be
the distance described in t + A sees.
The speed at the beginning of the interval A is 3t* f t./sec. and
at the end 3 (t + A) 2 .
/. As lies between 3< 2 . A< and 3 (t + A*) 2 A,
As
.. lies between 3i! 2 and 3 (t + A) a .
L\t
But as At ^ 0, (t + A) 2 ^ < 2 .
/. s = t 3 + c where c is some constant.
ds
* [So far as finding T is concerned we might simply say
CLt
As = 3t 2 . A app.
/. ?=8A v. 147.]
at
208 CALCULUS FOR BEGINNERS [CH. VIII
Since we are finding the distance the body has travelled from
its position when t = 0, we have s = when t 0,
.'. = + c or c = 0.
.'. Distance in 12 sees. = 12 3 = 1728 ft.
Notice that this is
EXERCISES. XLIX.
1. Find by this method the distance travelled in the 5th second when
r=3t 2 [ft. sec. units].
2. Find the distance travelled between the ends of the 3rd and 10th
seconds when w=2t 3 + 3t + 5.
171. Generally if v =f(t) and we want the distance between
the end of a sees, and the end of b sees., we say As lies between
f (t) . A< andy( + A2) . A and as before we deduce
Let < (t) be a function of t which when differentiated gives
/<*)
.*. s = < (t) + c.
Now s = when t = a.
i.e. * = < (t} <f> (a),
and the distance required = <j> (b) </> (a) = (t) \ .
172. Each term in the first series in 168 is the distance
travelled in a time h, the speed being supposed constant through
out the interval. The series is
170174] DISTANCE FROM SPEEDTIME FORMULA 209
(0, 3A 2 , 3(2/t) 2 ... being the values of v or 3t 2 when = 0, h,
2h }
***j />
t=12h =12fc
or shortly 2 3 2 . h or 2 v . A.
t=o t=o
t=12A
As in 148 we may shew that the limits of 2 3 3 . A and
<=o
e=i2
2 3Z 2 . h when A * are the same.
t=o
.'. we may say that the actual distance required is the limit
=12
of 2 3Z 2 . h when h is indefinitely diminished.
<=o
173. Generally if v =f(t}, the distance described in a small
interval A following the end of t seconds is approximately v. A
and the sum of all such distances between the end of a seconds
and the end of b seconds may be denoted approximately by
<=b
2 v&t.
t=a
The actual distance is the limit of this when A * and this
is denoted by
/ vdt or [ f(t}. dt,
Ja. J a
and we have seen that this is <j>(b) <f> (a) or < (t) \
e.g. In our problem ( 168, 170)
distance  I W . dt = [f]  1728.
Jo L Jo
174. This problem can be reduced to the problem of finding
an area as follows :
The relation v  3 2 can be exhibited in the form of a graph.
[Fig. 94.]
M. C. 14
210
CALCULUS FOR BEGINNERS
[CH. VIII
Distances along the horizontal axis represent seconds of time
and distances along the vertical axis represent speed in ft./sec.
OM represents t seconds, i.e. OM contains t horizontal units of
length.
MP represents a speed of 3 2 ft./sec., i.e. MP contains 3< J
vertical units of length.
ON represents (t+ A) seconds and NQ 3 (t + A) 8 ft./sec.
The number of feet described in the interval A on our first
hypothesis is v A, and this is the number of units of area in the
rectangle PMNR.
The number of feet described in 12 seconds will be the number
of units of area in all strips like PMNR between O and the
ordinate corresponding to t = l2.
t=12
In other words 2 vA< may be looked upon either as the
t=o
number of feet described in all the intervals or as the number of
174, 175] DISTANCE FROM SPEEDTIME GRAPH 211
units of area in all the strips and the limit of this when A< 
f 12
(which we call I vdt) is either the number of feet described in
JO
12 seconds or the number of units of area in the curvilinear
figure OAB where AB is the ordinate 03= 12.
General statement for speedtime graph.
175. Generally. If the speed be given as a function of
the time, say v=f(t), and the speedtime graph be drawn, the
400
300
V
200
2000
1500
S
1000
5
2 4 6 8 10 12
Fig. 95.
142
212 CALCULUS FOR BEGINNERS [CH. VIII
number of units of area bounded by the curve, v = Q, t = a, t b,
gives the number of units of length in the distance described
between the end of a seconds and the end of b seconds.
If the speedtime and spacetime graphs be drawn, the number
of units of length in the difference between two ordinates of the
spacetime graph is equal to the number of units of area in the
corresponding portion of the speedtime graph.
e.g. number of units of area in OPM is number of units of
length in mp or m'p. [Fig. 95.]
i.e. the number of units of area lying between the speedtime
curve, the timeaxis and the ordinates t = a, t = b is the number
of units of length in the distance travelled between the end of a
seconds and the end of b seconds.
Another example.
176. If w=11232,
rt>A = ["ll216< 1 T= 112x316 x 21=0.
The meaning of this is not hard to find.
From t = 2 to t = 3 J, v is positive.
From t = 3 to t = 5, v is negative.
t=5
So that of the terms which go to make up 2 v&t, some are
t=2
positive and some negative and our result tells us that the sums
of positive and negative terms are numerically equal.
As a matter of fact our formula gives the speed at any time
of a body projected vertically with speed 112 ft./sec.
From t = 2 to = 3J the body ascends and from t = 3^ to t = 5 it
descends an equal distance, so that its distance from the starting
point, or what we may call the effective distance described, is
zero.
If we want the actual distance described we must integrate
from 2 to 3.
175177] SPEEDTIME AND SPACETIME GRAPHS
213
If we draw the graph v=11232< we get a straight line
cutting y  where x = 3. From x = 2 to x = 3 the area is
above the a;axis and from x = 3 to x = 5 below and the two
portions of area are numerically equal but of opposite sign.
[Fig. 96.]
100
50
50
200
150
I
'100
50
123456
t
Fig. 96.
The corresponding ordinates in the spacetime graph are
equal.
177. Ex. 2. Find the work done in stretching an elastic
string from a length of 18 to a length of 25 inches, given that the
natural length is 8 inches and that it has a length of 14 inches
when sustaining a pull of 3 Ibs. wt.
214 CALCULUS FOR BEGINNERS [CH. VIII
By Hooke's law if the pull be T Ibs. wt. and the stretch s
inches,
T = ms (where m is a constant).
Now when T = 3, s = 6,
.. T = S.
Divide the total extension of 7 inches into n equal parts, each h
inches, so that nh=7 and suppose that as the string was stretched
from 18 to (18 + A) inches the pull remained the same as it was at
18 inches and that as it was stretched from (18 + h) to (18 + 2A)
the pull remained the same as at (18 + h), and so on.
The pulls corresponding to lengths
18, 18+ A, 18 + 2A, ...25 A,
h 17h
are 5, 5 + ^, o + n, ... ^ ,
and the work done on our assumption would be
/_ h\ 17 /i~l .
h 5 + 1 5 + 5 J + (5 + h) + ...  in. Ibs. wt.
L V z ' ^ J
_nh 27 h
27 1 , , 27 , 7, /189 7h\ .
=  r . nh .nh.h = r . 7 rh[ 7 I in. Ibs. wt. = W,.
4 4 4 \ 4 4 /
If we supposed that as the string was stretched from 18 to
(18 + A) inches the pull was the same as at 18 + h, and so on, we
should get as the work done
189 7A\ .
] in. IDS. wt. = Vj.
Now the actual work done lies between Wj and W 2 but each of
189
these can be made as near j as we like if h be made small enough.
.'. Actual work (W in. Ibs. wt.) is the limit of either W x or W,
when h  and is 47 j in. Ibs. wt.
177179] WORK DONE IN STRETCHING STRING 215
Notice that we need not calculate both Wj and W 2 . We can
say
.,, u , 189 7A . 189
either Wj  j , . . W = ,
189 77* 189
w 2 =  T  + T , .. w= r .
178. The method employed here corresponds to that
employed in 138 and 168. We shall now solve the problem
by a method corresponding to that employed in 140 and 170.
179. Let W in. Ibs. wt. be the work done in stretching the
string from 18 ins. to x ins. and (W + AW) in. Ibs. wt. the work
done in stretching it from 18 ins. to (x + Ace) ins.
ic 8
The pull corresponding to length x ins. is Ibs. wt.
2
The pull corresponding to length (x + Ax) ins. is
C+ Ace
Ibs. wt.
.". the work done in stretching from x ins. to (x 4 Aa;) ins. lies
x8 aj + Aoi8 .
between  Aaj and = Aa; in. Ibs. wt.
'i 2i
058 A . x + Ace 8 A
i.e. AW lies between Ace and = Ace in. Ibs. wt.
2i '2i
AW ,. cc8 , 05 + Aa;8 .
.*. lies between ^ and in. IDS. wt.
Aa; 2 2
dw x8 _ x . %
" dS~2~2~
a?
:. W = r  4ce + c.
4
dW
* So far as finding r is concerned it is sufficient to ay
ax
Q
AW = ^ Aa; approximately.
2
AW x8
,\ =  approximately.
216 CALCULUS FOR BEGINNERS [CH. VIII
Now when x = 18, W = 0.
18 2
.'. = 4. 18 + c,
x 2
/. W=^4x9.
625
.*. work done in stretching to 25 inches= j
= 56^9
= 47 in. Ibs. wt.
yS,
Notice that 564 and 9 are the values of 7 4a3 when x = 25
4
and 18 respectively, and
ry.1 i 25
work done =  4cc .
Jtt
180. Each term in the series in 177 is the work done in
stretching the string through a small distance A, the pull being
supposed to remain constant throughout this small stretch.
The series is h 5 + ( 5 +  ) + (5 + K) + . . . in. Ibs. wt.
L V */ J
(5, 5 + ^ , 5 + h ... being the values of T or ^ , when x = 1 8.
2i 2i
18 + A, 18 + 2^...).
x=25h x8 =25A
Or shortly 2 h . ^ or 2 hT.
z=18 x=18
=25fc
As in 148 we may shew that the limits of 2 AT and
x=18
*=25
2 AT when h * are the same.
=18
.'. we may say that the actual distance required is the limit
a;=25
of 2 AT when A * 0.
z=18
179182] WORK 217
181. Generally if T=f(x) gives the pull in terms of the
length, the work done in stretching from length x to length
(a; + Ace) is approximately T . Aa; and the sum of all such works
done as the length increases from a to b may be denoted
approximately by
x=b
2 T . Aa
x=a
The actual work done is the limit of this when Ax + and
this is denoted by
rb /&
I T dx or ; f (x) . dx,
J a J a
and we have seen that this is <f> (b) <j> (a) or </> (x)\ where
dx ~'
e.g. In our problem
rx 8 Vx* ~\ w
_ . dx = \ 4* = 471 in. Ibs. wt.
2 L 4 JM
182. This can also be reduced to finding the area of a curve
as follows.
Tdx where T is a given function of x,
^ x ~ 8
in this case ^ .
m
x 8
This relation T= ^ can be exhibited in the form of a
graph. [Fig. 97.]
Distances along the horizontal axis represent length of string
in ins., and distances along the vertical axis represent pull of
string in Ibs. wt.
In this case the graph is a straight line.
OM represents x ins., i.e. OM contains x horizontal units, MP
represents a pull of T ( = ^ J Ibs. wt.,i.e. MP contains T ( = ^ J
vertical units.
218
CALCULUS FOR BEGINNERS
[CH. VIII
ON represents (x + Ace) ins. and NQ a pull of
x + Ace i
T + AT =
Ibs. wt.
2
4*
20
B
10
15 A MN 25
Fig. 97.
The work done on our first hypothesis as the string is stretched
from length x to length x + Ao; is T . Ao; in. Ibs. wt. and this is
the number of units of area in the rectangle PMNR. The work
done in stretching the string from 18 to 25 ins. will be the
number of units of area in all the strips like PMNR between A
and B.
z=25
In other words S T . Aa; may be looked upon either as the
2=18
number of in. Ibs. wt. of work done on our first assumption or
as the number of units of area in all strips like PMNR, and the
(26
limit to which this tends as Aa;0 which we call I T dx is
J18
either the number of in. Ibs. wt. of work done or the number of
units of area in the figure ABKH.
182, 183] ASPECTS OF INTEGRATION 219
183. The following statements should now be intelligible.
(1) Integration is a process of summation.
If f(x) be any function of x, the values which f(x) takes as x
increases from a to b by equal increments h are
/(a), f(a + h),f(a + 2h)....
The limit of the sum h ./(a) + hf(a + h) + ... hf(b) when h *0
x=b
or as it may be briefly written the limit of S f(x) . Aa; when
x=a
rt>
Ace * is written I /(a?) . dx and the finding of this limit is
Ja
called integration.
(2) Integration is a process of antidifferentiation.
To find the limit of the sum just mentioned it is necessary to
discover a function of x, say <f> (x), such that ^ ' =f(x), i.e. to
dx
find a function (f> (x) which when differentiated gives /().
This process of finding < () which may be called a process of
antidifferentiation is also called integration.
If
(3) Any definite integral may be interpreted as an
area.
f(x) . dx is the area bounded by the curve y =f(x), the
axis and the ordinates x = a and x = 6.
Therefore, even if we cannot find the value of the definite
integral by the ordinary methods, that is to say if we cannot dis
cover a function < (x) such that *V =f( x }i we can nn d its
CtOb
value approximately by drawing the curve y=f(tt) between x = a
220
CALCULUS FOR BEGINNERS
[CH. VIII
and x = b and applying some one of the approximate methods
previously mentioned for finding the area.
184. In any problem requiring the Integral Calculus for its
solution it is advisable to keep in mind what we should do if we
wanted an approximate solution.
Ex. 1 . Find the area bounded by y = cc 3 , y = 0, x = 3, x = 1.
We have to find the area traced out by an ordinate which
moves from the position AH to the position BK, its length changing
continually in accordance with the law y = a?. [Fig. 98.]
To get an approximate area we should divide AB into a num
ber of equal parts like MN, and suppose the ordinate to keep the
same length MP as it moved from M to N.
The sum of all such rectangles as MR would be an approxima
tion to the area required, and we have seen that the actual area
required is the limit of the sum of such rectangles when their
width is indefinitely diminished.
We have seen that so far as the value of this limit is concerned
it is immaterial whether we suppose the ordinate in its passage
183, 184] EXAMPLES OF INTEGRATION 221
from M to N to preserve its initial length M P or its final length
NQ, or of course any intermediate length.
The work may be stated thus :
Suppose the area divided into a number of strips by lines
parallel to OY.
Let PMNQ be a typical strip, MP being the ordinate corre
sponding to abscissa x. .'. MP = x 3 and area PMNQ =
x=1
approximately and sum of strips = 5 a? 3 Ace approximately.
z=3
x s dx
Ex. 2. The speed (v ft./sec.) of a body at the end of t seconds
is given by
v = 6t+17t 2 .
Find the distance travelled between the end of the. 2nd and
the end of the 5th second.
Here we have to find the distance travelled in a certain time
by a body whose speed continually changes in accordance with
the law v = 6t + 17t 2 .
To get an approximate result we should divide the whole time
into a number of small intervals and suppose the speed to keep
throughout each interval the value which it has at the beginning
of the interval. We should then calculate the distance travelled
in each interval on this assumption and the sum of all the distances
would be an approximation to the required distance.
The actual distance is the limit of this sum when the duration
of each interval is indefinitely diminished. We have seen that
so far as the value of this limit is concerned it is immaterial
whether we suppose the speed throughout an interval to remain
222 CALCULUS FOR BEGINNERS [OH. VIII
the same as it was at the beginning or at the end of the interval,
or indeed at any intermediate instant.
The work may be stated thus :
Suppose the whole time divided into a number of small
intervals.
The interval A following the end of t seconds may be taken
as the typical interval.
The speed at the beginning of this interval is
(6J+17* 2 ) ft/sec.
.*. Distance travelled in this interval = (6f+ 17 2 ) A feet
approximately and the total distance is 2 (Qt+ 17< 2 ) A approxi
mately.
/B
Actually, distance is / (6< + Iltydt
Ja
= 63 + 663
= 726 feet.
Ex. 3. A tank 6 feet deep is filled with water. Find the
magnitude of the resultant thrust on an end which is 4 feet wide.
The pressure at any point is proportional to the depth and is
therefore continually changing as we pass downwards from top to
bottom of the tank.
We should get an approximation to the required thrust by
dividing the area into a number of thin strips by horizontal lines,
and supposing the pressure to remain constant over each strip. If
PQNM is such a strip we suppose the pressure not to change as
we pass from the level PM to the level QN but to keep throughout
the value that it has at the level PM. [Fig. 99.]
184]
EXAMPLES OF INTEGRATION
223
We should then calculate the thrust on the atrip PQN M on this
assumption, and the sum of the thrusts on all the strips would be
an approximation to the required result.
4'
X 1
Fig. 99.
The actual resultant thrust is the limit of this sum when the
width of each strip is indefinitely diminished. The limit would
be the same if we supposed the pressure at every point of PQNM
to be the same as it is at the level QN or at any level intermediate
between PM and QN.
The work may be stated thus :
Suppose the end divided into thin strips by horizontal lines.
PQNM is such a strip, the depth of P being x feet and
PQ = Ax feet.
The pressure at the level PM = wt. of x cubic feet of water per
sq. foot= 62'5a; Ibs. wt./sq. ft.
The area of the strip = 4Aa? sq. ft.
.*. Thrust on strip = 62'5a? x 4Aa; Ibs wt. approximately
= 250ajAo3 Ibs. wt. approximately.
=6
.'. Total thrust on end = 2 SSOceAa? Ibs. wt. approximately.
*=o
224
CALCULUS FOR BEGINNERS
[CH. VIII
Actually,
thrust =
= 250
ra:
= 4500 Ibs. wt.
Ex. 4. Find the work done by a gas in expanding from 2 to
3 cubic feet, the pressure and volume being connected by the law
pv = constant, given that the pressure is 2 160 Ibs. wt./sq. ft. when
the volume is  cubic foot.
4
Let p Ibs. wt./sq. ft. be the pressure]
v cub. ft. the volume / *
.'. pv = 21 60x^ = 540.
Suppose the gas contained in a cylinder closed at one end and
fitted with a piston free to slide along the cylinder.
Call the area of the piston a sq. ft. [Fig. 100.]
R 5
P/Q
v
Ay
a
Fig. 100.
When the volume of the gas increases from v to v + &v the
piston moves feet and the force on the piston at the beginning
Ctr
of this small movement is . a Ibs. wt.
v
184] FURTHER APPLICATIONS 225
.*. Work done by gas when volume increases from v to v + Av
is
510 Aw,, .,
. a . ft.lbs. wt. approximately
= . Aw ft.lbs. wt. approximately,
and the total work done in expanding from 2 to 3 cubic feet is
= 3 540
2   Av app.,
=2
r540
. dv.
v
At present we do not know how to find the value of the
indefinite integral I  dv, but we can obtain an approximate value
for the definite integral by looking upon it as an area and applying
Simpson's rule. [Fig. 101.]
Taking 11 ordinates, we have
Extreme
Even
Odd
270
25714
24545
180
23478
225
216
20769
200
19286
18621
450
1094 ; 13
87100
437652
174200
450
656852
Area = ^ x 65685
U
= 219 nearly.
.'. Work done = 219 ft.lbs. wt.
M. 0. 15
226
CALCULUS FOR BEGINNERS
[CH. VIII
[We shall see later, Exs. LXXIX. 6, that this is correct to three
figures.]
300
200
100
22
24 26
Fig. 101.
28
Ex. 5. If in the last example the pressure and volume wero
connected by a law of the form pv n = K (constant), say pv 1 ' 13 = K, wo
should get in exactly the same way
"['*>*
Work done
where
Now
2160 540

._
13 (2' 13 3' 1
= 5 1 JJL
e 13 18^~
= ^{7631 7239}
= 163 ft.lbs. wt.
FURTHER APPLICATIONS 227
EXERCISES. L.
1. The speed of a body (v ft./sec.) at the end of * sees, from a fixed
instant is given by
[where u and a are constants].
Shew that the distance travelled in these t seconds ia
ut+ iat2.
Shew also that u is the speed at the fixed instant and that the accelera
tion is constant and equal to a.
2. If v = 3t + 7t 2 , find the space passed over between the beginning of the
3rd and the end of the 5th second.
3. If a = 3< + 7< 2 [ ft./sec. 2 is acceleration], find the increase of speed
between the end of the 4th and the end of the 8th second.
Also, if the speed when t = Q is 20 ft./sec., find the speed at the end of the
4th second.
Find a formula for the speed at the end of t seconds and the distance
passed over between the end of the 4th and the end of the 8th second.
4. Find the work done in the expansion of a quantity of steam from
2 cubic feet at 4000 Ibs. per sq. ft. pressure to 8 cubic feet. The steam
expands so as to satisfy the law pv " 9 = constant.
5. The pressure (p Ibs. w f .. per sq. ft.) and the volume (v cubic feet) of
a gas are connected by the law pv*'' l = constant.
When the volume is 40 cubic feet the pressure is 100 Ibs. wt. per sq. ft.
Find the work done in compressing the gas from 40 to 35 cubic feet.
6. Find the resultant thrust on a rectangular area 6 ft. by 4 ft. im
mersed in water with its long sides vertical and the upper side 3 feet below
the surface.
7. Find the resultant thrust on a triangular plate immersed in water
with its plane vertical, its vertex A in the surface, and its base BC
horizontal.
BC=4 ft. and the perpendicular from A to BC = 3 ft.
8. Same with BC in the surface.
9. Find the work done in stretching an elastic string from length a ft.
to length b ft. given that the natural length is I ft. and the pull when it is
stretched to twice its natural length is k Ibs. wt.
152
228 CALCULUS FOR BEGINNERS [CH. VIII
10. A ship of 600 tons displacement goes a distance of 900 feet after
steam has been shut off. Supposing that the resistance of the water varies as
the square of the distance which it has to go before coming to rest and that
the initial resistance is 15 Ibs. wt. per ton, find the work done before the
ship comes to rest.
This will he the initial Kinetic Energy of the ship. Hence find the
initial speed in knots.
11. A body of mass 5 Ibs. moves in a straight line under the action of
a force F Ibs. wt. which obeys the law
Given that the body starts from rest, i.e. v=0 when t = 0, find (i) the speed
at the end of 5 seconds, (ii) the distance travelled in 5 seconds.
12. A body of mass m Ibs. moves in a straight line so that its accelera
tion (a ft./sec. 2 ) is given by a=  w 2 x, where w is a constant and x ft. is the
distance from a fixed point O in the line.
Find the work done against the force as the body moves from O to a
distance r.
If the speed at distance r is zero, what is the speed at O ?
13. Supposing the force of the earth's attraction at points above the
surface to vary inversely as the square of the distance from the earth's
centre, find the work done in moving a body of mass m Ibs. from the earth's
surface to a height of 4000 miles. [Take earth's radius =4000 miles.]
14. Find the weight of a rod AB 10 feet long, crosssection 1 sq. in.,
whose density changes from point to point in such a way that it is propor
tional to the distance from one end A, given that the density at B is 1 Ib. per
cubic inch.
Volumes of solids of revolution.
185. If a plane area be made to rotate about any line in its
plane a solid of revolution is generated. Familiar instances of
such solids are the cylinder, cone and sphere.
A cylinder is generated by the rotation of a rectangle about
one side.
185, 186]
VOLUMES OF REVOLUTION
229
A cone is generated by the rotation of a rightangled triangle
about one of the sides containing the right angle.
A sphere is generated by the rotation of a semicircle about
the diameter.
186. To find the volume of a solid of revolution.
oc
Ex. 1. OK is the line y=^'> AH > BK are tne ordinates jc = 2,
a
x = 9. The area AHKB (Fig. 102) is rotated through a complete
revolution about OX generating a frustum of a cone (Fig. 103).
Required the volume of this solid.
H
6" A
Pig. 102.
M N
Fig. 102 a.
Suppose the area AHKB divided into strips by lines parallel to
Qy and the inner rectangles completed as in Fig. 102 a.
If the area composed of these rectangles rotate about OX we
shall get a solid as in Fig. 103 a consisting of a number of cylinders,
and just as the sum of the rectangles in Fig. 102 a can be made as
230
CALCULUS FOR BEGINNERS
[CH. VIII
near as we like to the area AH KB in Fig. 102, if the strips are made
thin enough, so the sum of the cylinders in Fig. 103 a can be
made as near as we like to the volume required (Fig. 103).
Fig. 103.
Fig. 103 a.
186]
VOLUMES OF REVOLUTION
231
Now if OM=#and
= y =  and the volume of
m
the typical slice in Fig. 103 a formed by the revolution of PMNR
y?
= TT. M P 2 . M N = TT . y 2 . Ao; = TT . . Aa;.
4
x=9 yS
.*. Sum of slices in Fig. 103 a = S IT Aa; approximately,
*=2 4
and actual volume of solid in Fig. 103
721
Ex. 2. To find volume of sphere, radius a.
The sphere may be considered as generated by the rotation of
the area ABA' about OX. [Fig. 104.]
Suppose the area split into strips by lines parallel to OY.
PMNQ is such a strip.
232 CALCULUS FOR BEGINNERS [CH. VIII
When the rotation about A'A takes place this strip will
generate a slice of the sphere.
Volume of slice = TT . PM 2 . MN approximately
= Try Ace approximately
= TT (or x") Aa; approximately.
(v. 147.)
/a
TT (a 2 a; 2 ) d
a
2a 3
EXERCISES. LL
1. Find the volume of a cone radius of base 6", height 10".
2. Find the volume of a cone radius of base r", height h".
3. In a sphere of radius 6" find the volume of a cap of height 2".
4. In a sphere of radius r" find the volume of a cap of height h".
6. In a sphere of radius 10" find the volume of a slice contained between
two parallel planes at distances 2" and 5" from the centre (i) on the same
side, (ii) on opposite sides of the centre.
6. Find the volumes of the solids formed by the revolution of
^+T = 1 > (i) about OX, (ii) about OY.
y 4
[These solids are called respectively oblate and prolate spheroids.]
7. Find the volume of that portion of the first solid in Qu. 6 which is
cut off by planes perpendicular to the axis of rotation through the points
(1,0), (2,0).
8. Find the volume of the solid formed by the rotation about OY of that
x z
portion of the parabola y = which lies between the origin and y=k. Shew
that it is half the volume of the circumscribed cylinder.
This solid is called a paraboloid of revolution.
186188] MOMENTS OF INERTIA 233
9. Trace roughly the curve 36?/ 2 = x (4  a:) 2 .
Find the area of the loop and the volume obtained by revolving the loop
about OX.
10. Find the equation of the parabolic arc with axis parallel to axis of
y which passes through the points ( 2, 3), (0, 5), (2, 3).
Find the volume formed by rotating the area bounded by this arc and the
extreme ordinates about OX.
11. Find the volume of the solid formed by the rotation (i) about OX,
(ii) about OY of that part of xy = 20 which lies between x = '2 and x=5.
12. Shew that the volume formed by the rotation about OX of that
portion of y =f (x) [where / (x) denotes any function of x] which lies between
[b
x=a and x = b is I wy 2 .dx.
Ja
Moments of inertia.
187. Definition. If a particle of mass m be placed
at a distance r from a fixed axis, its moment of inertia
about that axis is mr 2 .
If a number of particles m^, m a , m^ ... be placed at distances
r n r 2> r s ..."from a fixed axis, the moment of inertia of all these
masses about the fixed axis is
m^Ti + m. 2 r + m 3 r^ + ... or shortly 2 mr 3 .
If instead of a series of disconnected particles we have a
continuous body, we split it up into strips or slices as when we
were finding areas or volumes, find the sum of the moments of
inertia of all the pieces and the limit to which this sum tends as
the pieces are indefinitely diminished.
188. Ex. Find the moment of inertia about a short side
of a thin plate of length 3 feet and breadth 2 feet, the mass
being 20 Ibs.
XX' is the axis about which the M.I. is required. [Fig. 105.]
PMNQ is a thin strip. AP = o?feet. PQ=Ao?.
.'. area PMNQ2Acc square feet.
20, 20
.'. mass PMNQ = (2A)~ A# Ibs.,
234
CALCULUS FOR BEGINNERS
[CH. VIII
arid the distance of any point in the strip from XX' lies between
x and x + Ao:.
X
pa
x'
MN
Fig. 105.
20
.'. The M.I. of the strip = ^ Ace . as 2 approximately.
(v. 146.)
.*. M. I. of rectangle
20, r20, s 20x27 ., .
/20
= Jo T
189. Radius of gyration. If the M. i. of a body of mass M
about an axis be M& 3 , k is called the radius of gyration about
this axis.
In this example since the mass is 20 Ibs., k* = 3, Le. radius of
gyration = \/3 feet.
190. Sometimes a more convenient method of splitting into
elements than by lines parallel to an axis may be found.
e.g. Find the moment of inertia of a circular disc radius a
about an axis through the centre perpendicular to its plane.
We naturally seek for a set of points at the same distance
from the axis and this suggests splitting the disc into thin con
centric rings.
Such a ring is shewn in figure 106.
Let OP = x, PQ = Ao;.
188191]
MOMENTS OF INERTIA
235
Then area of ring= 2irx&x approximately, and if m be mass
of unit area
Mass of ring = 'Zirxm&.x approximately,
M. i. of ring = ^TTxrn &x x a; 2 approximately,
.'. M.I.
f
of disc = /
Jo
Fig. 106.
Now if M be mass of disc, M = rnira*.
Ma 2
/. M. I. Of dlSC = ~ .
/a 2 at
10n = V 2=J1'
Radius of gyrat
191. Suppose we try to find the M.I. of a circular disc
radius a about a diameter.
Take the diameter as the yaxis.
236
CALCULUS FOR BEGINNERS
[CH. VIII
Here points equidistant from the axis lie on a parallel line
and this suggests division into strips by lines parallel to OY.
[Fig. 107.]
Y
Fig. 107. Fig. 108.
Let PpqQ. be such a strip, and let OM=a?, MN=Aaj,
= y(Wa 2 x*).
Mass of strip = 2y&xxm approximately
= 2m >Ja? y? . A.
M. I. of strip = 2m \fa tf&xxa? approximately,
/a _
M. I. of disc = / 2mx 2 Va 2  a?dx.
J a
Here we have a function which at present we do not know
how to interate.
and
192. We can avoid the difficulty by making use of the
following important theorem.
If OX, OY are two perpendicular axes in the plane of a lamina
and OZ is a third axis perpendicular to the plane, and if the
moments of inertia about OX, OY, OZ are respectively l z , !, I,,
then
Suppose a mass m situated at P in the plane XOY. [Fig. 108.]
191193] MOMENTS OF INERTIA 237
Draw PM, PN perpendicular to OX, OY and join OP.
Then OP is perpendicular to OZ.
M. i. of mass about OX  m . PM a .
M.I. of mass about OY = m . PN a .
M. I. of mass about OZ = m . PO 3 .
Bub PO 2 = PM 2 + PN 2 .
.'. M.I. of mass about OZ =sum of M. i.'s about OX and OY.
This is true for any number of masses in the plane XOY whence
the theorem follows.
193. To apply this to the present case.
Take an axis OZ perpendicular to the plane of the disc. We
ft /I 2
have just found that l a =  t ~ ( 190),
/ t I I I
But by symmetry \ x = \ y .
_ Ma 2
a; ~ y ^ ^M
[N.B. This shews that
mira? . a 9
f
Ja
~5  ij
er  x*dx =
or
I y? \/a a x*dx= 5 .
Ja
EXERCISES. LI!/
1. Find the moment of inertia of a uniform rod 10 feet long, mass
20 Ibs., about an axis perpendicular to its length (a) through one end,
(b) through a point 3 ieet from one end.
2. Find the M.I. about AB of a rectangular plate ABCD in which
AB=a ft., BC = & ft., the mass being M Ibs.
238 CALCULUS FOR BEGINNERS [CH. VIII
3. Find the si. i. of the same plate about a line parallel to AB and DC
and midway between them.
4. Find the M. i. of the same plate about a line perpendicular to its
plane (i) through A, (ii) through the centre.
5. Find the M. i. of a sphere, radius a, mass M, about a diameter.
6. Find the radius of gyration about OY (i) of a lamina in the form of
x 2
part of the parabola y = bounded by the curve and y = k, (ii) of the solid
formed by the revolution of this area about OY.
7. Find the M. i. of a rod AB length a ft. about a line through O per
pendicular to the plane OAB. O is on right bisector of A B at distance p.
8. Find the radius of gyration about the axis of (i) a hollow cone,
(ii) a solid cone (radius r, height h).
9. Find the moment of inertia of a triangular lamina ABC of mass
m Ibs. about the side BC, given BC = 2 ft. and the perpendicular from A to
BC=3 ft. [First find the length of a line parallel to BC at a distance x ft.
from it and terminated by the other sides.]
10. Find the radius of gyration about each side of a triangular lamina,
mass M, sides 4, 5, 6 feet.
11. Find the M. i. of a solid cylinder about the axis of figure.
12. Find the M.I. of a hollow cylinder, internal and external radii R
and r, about the axis of figure.
13. A grindstone, radius 1 foot, mass 100 Ibs., is rotating about its axis
at 100 revolutions per minute. Find its kinetic energy.
14. Find the moment of inertia about the axis of revolution of the
x 2 t/ 3
solid formed by the revolution of 2 + r2 = l about (i) the xaxis, (ii) the
yaxis.
15. Find the moment of inertia about its axis of a frustum of a cone,
the radii of the ends being 3 ft. and 4 ft. and the thickness 5 feet.
Centre of gravity.
194. If we have a number of particles of masses m n m z , m^ &c.
situated in a plane at points whose coordinates referred to some
fixed axes in the plane are (a^t/j), (xg/J), (x 3 y 3 ) &c., and if x and y
194] CENTilE OF GRAVITY 239
are the coordinates of the centre of gravity of the masses
[Fig. 109],
then
and
^ +
m. A x z +
or
2m
y=^
[If we call m^ the moment of the mass m^ about OY, these
results merely express the fact:
8a
7a
6a
5a
4a
3a
2a
m,
Fig. 109.
a 2a 3a
Fig. 110.
Moment about OY (or OX) of sum of masses supposed collected
at centre of gravity = sum of moments about OY (or OX) of the
separate masses, for this is equivalent to
240 CALCULUS FOR BEGINNERS [CH. VIII
Now suppose we are dealing with a continuous body instead
of a set of disconnected particles, we proceed as when we required
the area or volume, that is to say, suppose the body split up into
suitable strips or slices.
195. The method will be best seen by examples.
x 2
Ex. 1. HK is a portion of the curve y = ; AH and BK are
the ordinates x = a, x = 3a. [Fig. 110.]
Find the centre of gravity of the area ABKH.
Suppose the area divided into strips like PMNQ by lines
parallel to OY.
x*
Let OM = x, MN = A*, MP = y = .
Let m be the mass of unit area.
Area of strip = y&x approximately, (v. 146.)
Mass of strip = my&x approximately.
Moment of mass about OY = myAa; x x approximately.
Moment of mass about OX = my Ace x ^ approximately.
Sum of masses of strips = 2 my&x approximately.
x a
x=3a
Sum of moments about OY = 2 mxy&x approximately.
z=3a ? ,2
Sum of moments about OX = 2 m^ Ax approximately.
x=a "
rSa f3<L /yi2 O
.*. Mass of ABKH = I mydx = \ m dx = ^ma 3 .
Jo, Jet
/**** /"* ftlSu
Moment of mass about OY= I mxydx= I dx=2Qma?.
Ja Ja a
fZa y2
Moment of mass about OX = / m ^ dx
Ja "
=/'
Ja
mx* . 121
194, 195]
CENTRE OF GRAVITY
'241
20m 3 30
= T3 a
~26
T""*
121
. "5"" 363
y ~~^~ I30 a
a;'
Ex. 2. HK is a portion of the curve y = ; CH and DK are
the lines y = a, y = 9a. [Fig. 111.]
Find the centre of gravity of the solid formed by a complete
revolution of the area CHKD about OY.
lltiiizfi~
3?^rES3
m 4lji ! ! i i i i
^ ::
ill
jiii 1
6a
N :
Mg
0> a
Fig. 1
2a 3a
11.
M. C.
16
242 CALCULUS FOR BEGINNERS [CH. VIII
Divide the area into strips like PMNQ by lines parallel to OX.
Let OM = y, MN = Ay and MP = :(= Jay}.
When the area is rotated about OY this strip will generate a
slice of the solid which will be approximately a cylinder, radius of
base MP, height MN.
Volume of slice = TT . MP 2 . MN = ira; 2 Ay approximately
= Tray . Ay approximately.
. . If m be mass of unit volume,
Mass of slice = miray . Ay approximately.
Moment of slice about OX = miray&y x y approximately
y 2 Ay approximately.
mira x 80a 2
, ,, I" 90 rwiTray 2 ! 90
.*. Mass of solid = J miray ay = =
Ja * Ja
r 90 PmTray 3 ] 90 nura x 728a 8
Moment about OX = I nnray^dy = 5^ =  
Ja L 3 J a
728
= 57
728
91
and by symmetry 5 = 0.
/ 91 \
. . Coordinates of centre of gravity are 1 0, TF ) .
196. Suppose we try to find the centre of gravity of the
area of a quadrant of a circle, say OAB, radius a. [Fig. 112.]
Area of strip PMNQ = yAa; or va 2 & 2 Aa: approximately.
Mass of strip  m va 2 y? Ax approximately.
Moment of strip about OY = in va 2 a; 2 A;r x x approximately.
195, 196]
.'. Mass of OAB
Moment about OY
CENTRE OF GRAVITY
243
Fig. 112.
Here we have apparently two integrals neither of which we
can evaluate.
The value of the fir?t is however easily obtained, for it is ^
of mass of whole disc = . mira?.
4
[~N.B. This shews that I* N/a 2 ~^o? dx = ~ .1
To avoid the difficulty of evaluating the second integral con
sider the moment of the strip about OX.
It is
i/
my&x x ^ approximately
2i
in , A
= jr y* Aa; approximately
in
= (a?~ a; 2 ) Aa? approximately.
162
244 CALCULUS FOR BEGINNERS [CH. VIII
f a m
,*. Moment of whole area about OX =1 ^ (a z x 3 ) dx
and it is obvious from symmetry that the moment about OY is
the same. [This tells us that
x va a y? dx s a 3 .!
o J
ma"
3 " 4a
__2 ~~ O_
ftnra OTT
EXERCISES. LIII.
1. OP is the line y = kx. P is the point on it whose abscissa is a.
Find by integration the coordinates of the o. G. of the triangle OMP
where M P is the ordinate of P.
2. Find the distance from the base BC of the c. o. of a triangle ABC
whose altitude is h.
x 2
3. Find the c. o. of that portion of the parabola y= cut off by the
line y = k.
4. Find the c. G. of a cone. [Take your origin at the vertex and suppose
the cone generated by the revolution of part of the line y = kx about the axis
of*.]
6. Find the c.o. of the area bounded by the xaxis, the ordinates a; = 3,
x= 10 and the curve y = 3a; 2  2x  1.
6. Find the c. G. of the trapezium whose corners are (a, 0), (o, 0),
(P, 9), (~P, q)
7. Find the c. G. of a semicircle, radius a.
8. Find the c. G. of a hemisphere, radius a.
9. The lengths of the parallel sides of a trapezium are a, b and the
distance between them is h.
Shew that the c. G. lies on the line joining the midpoints of the parallel
sides and that its distance from the side a is ^ .  r . [First find the
3 a + o
length of a line parallel to the side a distance x from it.] What does your
result give you if o=0 ?
196, 197]
CENTRE OF GRAVITY
245
1O. Shew that the coordinates of the o. o. of a uniform lamina bounded
by y = 0, 3 = a, x=b and the curve y=f(x) are given by
If the lamina is uniform this result is independent of the density and the
point is sometimes called the Centre of area of the figure bounded by y = 0,
x = a, x=b and the curve y =f (x).
11. A lamina, area A sq. ft., is totally immersed in water with its plane
vertical and its o. a. at a depth h ft. below the surface. Shew that the
resultant thrust on it = to Aft Ibs. wt. where w Ibs. is the weight of a cubic
foot of water.
197. (i) The moment of inertia of a lamina about any axis
in its plane is equal to the moment of inertia about a parallel
axis through the centre of gravity + the moment of inertia about
the first axis of the whole mass supposed concentrated at the
centre of gravity.
Let particles of masses m^, m^ <fec. lie in the plane of the paper
at points Pj, P s , &c., and let G be their o. Q.
H
B
Fig. 113.
Suppose we want the M.I. of these masses about AB and let
HG be a parallel axis through G.
246
CALCULUS FOR BEGINNERS
[CH. VIII
Let Xi, x 2 , &c. be the distances of the masses from HG, and
let h be the distance between the two axes.
M. i. of mass m l about AB = m l (x^ Kf =
and similarly for each of the other masses.
.". M. i. of all the masses about AB = 2 (my?) 2h (%mx)
and M.I. GH=2 l (mx?).
Also since G is the c. a. of the masses ^7 r =
.*. M. I. about AB = M. I. about GH + h?. (Sm)
= M.I. about GH + M.I. about AB of the whole
mass (2m) supposed placed at G.
Hence the theorem.
(ii) Now let O be any point in the plane of the lamina,
G its C. G. ; OX, OY any two axes at right angles through O;
GX', GY' two parallel axes through G.
K X
Fig. 114.
Then, if \ x be the M.I. about OX, \ x ' about GX', &c.,
we have I. = I.' + Mb 2 ,
\ y = \y + Ma 2 .
i.e.
197, 198] THEOREM OF PARALLEL AXES
247
where \ a and I,' are M.I. about axes through O and G perpendicular
to the plane of the lamina.
(iii) Now let AB be parallel to the plane of the lamina, GH
a parallel axis through G.
Fig. 115.
Then P^ 3 X* + h? 2ha^ cos 0, where 6 is the angle between
the plane ABHG and the plane of the lamina.
. . 2 (mPQ 2 ) = 2 (ma: 2 ) + /* 2 2 (m)  2 h cos 02, (mx),
and as before 2 (mx) = 0.
So that M.I. of masses about AB = M. I. about GH + M.J. about
AB of 2 (m) placed at G.
198. Hence the moment of inertia of a plane lamina
about any axis in its plane, parallel to its plane, or
perpendicular to its plane = the moment of inertia about
a parallel axis through the centre of gravity + the moment
of inertia about the original axis of the whole mass
supposed concentrated at the centre of gravity.
248 CALCULUS FOR BEGINNERS [CH. VIII
Ex. The M. I. of a circular disc, mass M, radius a, about an
axis through the centre O, perpendicular to its plane, is
Ma 2
2 '
If we want the M. I. about an axis through A a point on the
circumference perpendicular to the plane, the theorem tells us
that it is
Ma 2
s+ Ma*
a
(Ma 2 being the M.I. about A of the whole mass supposed concen
trated at the centre of gravity O).
EXERCISES. LIV.
1. Find the radius of gyration of a circular lamina about a tangent.
2. Find the M.I. of a circular lamina, radius a, about a line perpendicular
to its plane and meeting it in a point distance c from the centre.
3. Find the M.I. of a circular lamina, radius a, about a line in its plane
distance c from the centre.
4. Find the M.I. of a thin circular hoop, radiug a, about an axis
perpendicular to its plane through a point in the rim.
6. Find the H.I. of a circular lamina about a line parallel to its plane at
a distance c from the centre.
6. Find the M.I. of a solid cylinder about a line through the centre
perpendicular to the axis of figure.
7. Find the radius of gyration of a solid cone about a line through the
vertex perpendicular to the axis.
Centre of pressure.
199. In Ex. 3, 184 we found the resultant thrust on a
rectangular area immersed in water.
Suppose we wish to find through what point of the plate this
resultant pressure acts. [This point is called the centre of
pressure.]
Our approximate solution would be the finding of the line of
action of the resultant of the thrusts on the strips into which the
rectangle is supposed to be divided.
198, 199] CENTRE OF PRESSURE 249
The principle of which we make use is that if we take
moments about any line, the moment of the resultant of any
number of forces = the sum of the moments of the separate
forces.
Now it is obvious that the centre of pressure lies on the
vertical line of symmetry. It therefore remains to find its
distance from AB.
Take moments about AB.
The thrust on our typical strip PQNM = 250* A,w Ibs. wt.
approximately and may be taken to act at the centre of the rect
angle PQNM.
.\ Moment of this thrust about AB = 250* Aa; x x Ibs. ft.
approximately and the sum of the moments of all the thrusts
ie F250 ~1 6
= I 250^. dx = ~y? = 250 x 72 Ibs. ft.
/ L ^ I o
The resultant thrust we have seen to be
f
J
T250 ~l 6
250xdx= \..a?\ = 250 x 18 Ibs. wt.
L ^ Jo
If X ft. be the distance below AB of the centre of pressure the
moment of this resultant thrust
= 250 x 18 x X Ibs. ft.
.. 250 x 18xX = 250x72.
. . X = 4.
.. Centre of pressure is 4 ft. below the surface.
EXERCISES. LV.
1. Find the centre of pressure of a rectangle a feet by b feet, placed in
water with its plane vertical and one of the a feet sides (i) in the surface,
(ii) c feet below the surface.
2. Find the centre of pressure of a triangle, base a feet, height h feet,
placed in water with its plane vertical and (i) its vertex, (ii) its base in the
surface, (iii) its vertex c feet and its base (c + h) feet below the surface,
(iv) its base c feet and its vertex (c + h) feet below the surface.
3. The parallel sides of a trapezium are a, b and the distance between
them is h. It is placed in water with its plane vertical ; find the centre of
pressure if the side a is (i) in the surface, (ii) c ft. below the surface.
250 CALCULUS FOR BEGINNERS [CH. VIII
Mean values.
200. We have already ( 1 64) defined the mean value of f(x)
between x = a and x = b as being the height of a rectangle whose
base is (b a) and whose area is equal to that bounded by the
curve y =f(x), the scaxis and the ordinates x = a and x = b. This
mean value is
Ja
f(x) . dx
ba
and we might take this as the definition of the mean value.
It will be instructive to obtain this result by a method which
does not involve the idea of area.
In ordinary arithmetic the mean or average of a set of
quantities is obtained by dividing the sum of the quantities by
their number [i.e. it is the Arithmetic mean of the quantities].
Thus if 5 men have heights 5 ft. 8 ins., 5 ft. 6 ins., 6 ft., 6 ft.
2 ins. and 5 ft. 2 ins. the average height is
b' 8" + 5' 6" + 6' + 6' 2" + 5' 2" 28' 6"
p D 02 .
5 o
The combined height of 5 men each of height 5 ft. 8 ins. is
the same as that of the given 5 men.
Now suppose we have a function of x, /(a?), and suppose x to
pass from the value a to the value b by n equal steps e^ch = A so
that nh = b a.
Successive values of x are
a, a + h, a+2h, ... (bh), b,
and the corresponding values of f(x) are
/(a), /(a + A), /(a + 2A), . . ./(b  A), /(&),
(n + 1) in number.
The Arithmetic mean of these values is
/() +f(a + h) +f(a + 2A) + ... +f(bh) +f(b)
n+l '
200, 201] MEAN VALUES 251
hf(a) + hf(a + h) + ... + hf(b h) + hf(b)
(ba) + h '
since nh b a.
We define the mean value of f(x) between a and b as the
limit of this when h * 0, i.e. when the number of steps from a to
b is indefinitely increased. We have seen that the numerator can
be written
"?/<) **,
x=o
where Aw takes the place of h and the limit of this when
AOJ  is
rt>
f(x) dx.
Ja
Also the limit of the denominator is b a.
.'. The mean value of f(x) between a and b is
L
b
f(x) . dx
b a
201. In questions of mean value it is important to make
clear what the independent variable is.
The following example will illustrate what is meant.
A stone falls to the ground from the top of a tower 400 feet
high. Find its mean speed. [g= 32.]
We may think of the speed as a function of either (i) the
time, or (ii) the distance fallen.
If we take (i) we suppose the whole time of fall divided into
n equal parts, the speeds at the ends of these equal intervals of
time calculated, and the Arithmetic mean of them obtained.
Our mean speed is the limiting value of this Arithmetic mean
when n is indefinitely increased.
It is easily found that 5 sees, is the time of falling and that
7> = 32< is the formula which expresses v as a function of .
252 CALCULUS FOR BEGINNERS [CH. VIII
.'. The mean speed on the first assumption is
Cvdt /*
Jo _ Jo
32*.<ft
o _ o Qr . f. I
5 5  =  5  = 8O it. /sec.
If we take (ii) we suppose the whole distance, 400 feet,
divided into n equal parts, the speeds at the points of division
calculated and the Arithmetic mean obtained.
Our mean speed is the limiting value of this Arithmetic mean
when n is indefinitely increased.
The formula connecting v and s is v* = 64s.
.'. The mean speed on the second assumption is
202. Ex. 1. Find the average speed between the end of
the 2nd and the end of the 5th second of a body whose speed
changes according to the law v = Qt + 17 &.
i.e. find the mean value of Qt + 17 2 between t = 2 and t = 5.
This is  _.  = = 242 ft./sec.
Z o
/6
Since I (6< + 17 2 ) rf< or 726 ft, is the distance travelled in the
J2
3 seconds between the end of the 2nd and the end of the 5th
second, this speed of 242 ft./sec. is the uniform speed with which
the body would have to move to cover the same distance in the
same time (v. definition of average speed in Chap. L).
Ex. 2. The pressure and volume of a gas are connected by
the law pv = 540, p being in Ibs. wt./sq. ft. and v in cub. ft.
Find the mean pressure as the gas expands from 2 to 3 cub. ft.
i.e. find the mean value of  between v = 2 and v = 3.
v
201, 202] MEAN VALUES 253
/ 3 540
I . dv
This is h " = 219 Ibs.wt. /sq.ft. [v. Ex. 4, 184.]
o 2i
This is the uniform pressure which would have to act on the
piston to do the same amount of work that is actually done.
EXERCISES. LVL
1. Find the mean sectional area of a sphere supposed cut by a series of
equidistant parallel planes.
Interpret the result geometrically.
2. Find the mean sectional area of a cone supposed cut by a series of
planes parallel to the base. [Radius of base r, height ft.]
3. If v = u + at where u and a are constants, shew that the average speed
between the end of ti seconds and the end of t% seconds is equal to half the
sum of the speeds at these instants.
4. If v=u + at + bt* where u, a, b are constants, find the average speed
between the end of ^ seconds and the end of t% seconds.
Also find half the sum of the speeds at these instants.
O. The pressure and volume of a gas are connected by the law
pv l  2 = constant.
If the pressure is 400 Ibs. per sq. ft. when the volume is 3 cub. ft., find the
mean pressure as the gas expands from 2 to 4 cub. ft.
6. Find the mean gradient (i) with respect to x, (ii) with respect to y,
oty = x z , between the points (2, 4), (5, 25).
7. A stone falls freely from rest; shew that the mean speed with respect
to the time is half the final speed, but the mean speed with respect to the
distance is twothirds of the final speed.
8. A stone falls freely from rest. Shew that the mean kinetic energy
with respect to the time is twothirds of the mean kinetic energy with
respect to the distance fallen.
254 CALCULUS FOB BEGINNERS [CH. Vlll
MISCELLANEOUS EXAMPLES ON CHAPTERS
VII. AND VIIL
1. Draw roughly the curve y = 6  4x  x z cutting the xaxis in A, B and
the yaxis in C.
Find the areas AOC, BOG.
2. The parabola j/ 2 = 6x forms a paraboloid of revolution by revolving
about the xaxis. Find the volume of a segment of this paraboloid of length
10 units, and shew that it is half the volume of the circumscribing cylinder.
3. Find the mean value of 10x 2  X s between x =3 and x= 5. Explain
your result graphically.
4. A point starts with initial velocity u feet per second, and moves
along a straight path with a constant acceleration a feet per second per
second. Prove by integration that its displacement, a, at any time t sees.
after the start is given by
/ i \
feet.
6. ABC is a triangle having a right angle at C ; BC is horizontal, and
A, which is vertically above BC, is in the surface of a liquid ; find the centre
of pressure.
If the triangle were turned about BC until A were vertically below C, find
where the centre of pressure would now be.
B.
1. Write down the values of
J (2x 2 
3. Sketch the curve t/=x(5x) 2 between x=  1 and x=6.
Find the area bounded by the xaxis and that portion of the curve which
lies between x=0 and x = 5.
Find also the maximum ordinate and the area of the circumscribed
rectangle.
3. The equation of an ellipse is x 2 + 4j/ 2 = 4. Shew that the volume
generated by the rotation of this ellipse about the xaxis is half the volume
generated by its rotation about the t/axis.
202J MISCELLANEOUS EXAMPLES 255
4. Find the abscissae of the point in which the curve y=
cuts the xaxis. Calculate the maximum and minimum ordinates and the
gradient at each of the points where the curve cuts the xaxis.
Sketch the curve roughly between the extreme points in which it cuts the
axis. Find the areas of the two portions in the 2nd and 4th quadrants.
6. Find the Moment of Inertia of a circular disc, radius a, about a line
parallel to its plane at a distance c from its centre.
Hence find the Moment of Inertia of a solid cone about a line through
its C.Q. perpendicular to the axis.
0.
1. Consider only the part of the curve y = 4xx s , for which neither x
nor y is negative.
Make a rough sketch of this part.
Find the equations to the tangents at the points where it cuts the axis
of x.
Find the length of its greatest ordinate.
Find the area between the curve and the axis of x.
2. A solid is formed by the revolution about the saxis of the curve
y2 = a + bx + ex 2 + dx 3 .
Find volume of solid between x=h and x=h and prove that this
volume is given by Simpson's rule
where Aj and A 3 are the areas of the end sections, A2 of the middle section
(corresponding to a:=0).
3. Find the moment of inertia of a uniform thin triangular lamina
ABC about the side BC.
4. A quantity of steam expands so as to follow the law pv' 6 = constant.
Find the work done in expanding from 1 to 4 cub. ft., given that when the
volume is 2 cub. ft. the pressure is 50 Ibs. per sq. ft.
Also find the mean pressure during this expansion.
5. A square side a is placed in water with a corner in the surface and
the diagonal through that corner vertical. Find the depth of the centre of
pressure.
256 CALCULUS FOR BEGINNERS [CH. VIII
(V
J o
D.
dx
1. Find
2. Draw roughly y=(a; + 2) (x1) (x4). If the points in which this
cuts the a;axis be, in order from the left, A, B, C, find the areas bounded by
the xaxis and (i) that portion of the curve between A and B, (ii) that portion
between B and C.
3. Two parallel planes cut a sphere in circles whose diameters are 12
inches and 10^/3 inches respectively, the perpendicular distance between the
planes being 13 inches. Find the radius of the sphere and shew that the
planes cut off spherical segments whose volumes are in the ratio 112 : 625.
4. An isosceles triangle 10 metres in altitude is immersed in water
with its base in the surface and vertex pointing downwards. Find the depth
of its centre of pressure supposing the pressure at the surface to be that due
to a column of water 10 m. high.
6. Find the moment of inertia about its axis of a grindstone 120 cm.
in diameter, 20 cm. thick, whose density is 2*14 gms. per c.c.
Also find its kinetic energy at 15 revolutions per minute.
E.
fs
1. Shew that (x 2  4x + 3) dx = 0.
Jo
Draw a figure to explain the result.
2. A rectangle 5 ft. by 3 ft. revolves about a straight line parallel to
the longer side and 3 ft. from the nearest one. Find the volume generated
and the radius of gyration of the solid so formed about the axis of
revolution.
3. Indicate in a diagram the curves
7/a;i2=2 and j/x 1>5 =2.
At what angle do these curves intersect each other ? Find the area of the
closed figure bounded by these curves and the ordinate x = 2.
Find the coordinates of the centre of gravity of this area.
202] MISCELLANEOUS EXAMPLES 257
4. Assuming that above the surface the force of the earth's attraction
varies inversely as the square of the distance from the centre and below the
surface directly as the distance, find the work done in moving a mass of
m Ibs. from a feet below the surface to the surface, calling the earth's radius
R feet ; also the work done in moving the mass from the surface to a feet
above.
Find the ratio of the work done in moving a mass from 1000 miles
below the surface to the surface, to that done in moving it from the surface
to 1000 miles above, taking the earth's radius as 4000 miles.
6. A zone of a sphere has thickness t. A cylinder has the same
thickness and the area of its base equal to the Arithmetic mean of the
areas of the ends of the zone. Shew that volume of the zone = the volume
of cylinder + the volume of sphere, diameter t.
F.
1. Find the values of
(i) I3x%.dx, (ii)
2. A curve has to be drawn through the points (0, 0), (1, 4), (2, 1),
(3, 5). Assuming the equation to be of the form y ax 3 + bx" + ex + d deter
mine a, b, c, d and find by integration the area of the curve between the
extreme ordinates and the axis of x.
The area comes out to be that of a rectangle on the same base with height
2. Draw a figure shewing the curve and this rectangle, and shew that the
result can be obtained from symmetry.
3. There is a curve y = ax n .
If y = 2'34 when x=2 and y= 2061 when #=5 find a and n.
Let the curve rotate about the saxis forming a surface of revolution.
Find the volume between the sections at x = 2 and x = 5.
4. AB is a rod I ft. long and O a point in the rod a ft. from A.
Find M. i. of the rod about an axis through O perpendicular to rod.
Find O so that this M.I. is a minimum.
5. A zone of a sphere has thickness t, A cylinder has the same
thickness and the diameter of its base equal to the diameter of the section
midway between the ends of the zone. Shew that the volume of the zone
= the volume of the cylinder the volume of a hemisphere, diameter .
M. C. 17
258 CALCULUS FOR BEGINNERS [CH. VIII
fc
G.
1. Trace roughly y = x x 3 from x = to x = l.
Find the area between it and the xaxis.
If your xunit is 5" and your yunit 10" what is the actual area in square
inches ?
What is the mean height of this portion of the curve ?
Find also the area bounded by the curve, the xaxis and the maximum
ordinate between x = and x=l.
2. A hemisphere 1 foot in radius has to be divided into two equal parts
by a plane parallel to the base. Prove that if the distance of the plane from
the centre is x feet, then x is a root of the equation
x 3 3x + l = 0.
3. Prove that the area contained between the parabola y s =4ax, the
axis of y and the line through the point (h, k) on the parabola parallel to
the xaxis is  hk.
o
Prove that the radius of gyration about the xaxis of a lamina shaped
/3~ / 3 3 \
like this area is k ./ = and that the coordinates of the c.o. are ( ^ h,  k \ .
4. A torque T Ib. ft. acts on a shaft and is proportional to the angle
turned through (T = fc0).
Shew that the work done in turning the shaft through an angle a radians
is  Tjtt ft. Ibs. where Tj is the value of the torque when the shaft has
2i
turned through an angle a.
6. A cigarshaped strut, 40 ft. long, is of diameter ( 10  ^ n ) inches at
a section x feet from the middle. Find its total volume.
1. Trace the curve 5y = (x + 3) (x 1) (x3) from x= 4 to x=4.
If it cuts the xaxis in A, B, C reading from the left and the yaxis
in D, find the areas of the figures bounded by the curve and (i) OA, OD,
(ii) OB, OD, (iii) BC.
Also find the mean value of y between x=  4 and z=4.
202] MISCELLANEOUS EXAMPLES 259
2. Find
(i) The area comprised between the a;axis and that portion of
y=6x&x12 which lies below it.
(ii) The c.o. of the area.
(iii) The volume generated by the revolution of this area about OX.
(iv) The C.G. of this volume.
3. Find the equation of a parabola, whose axis is along the axis of y,
and which passes through the three points (  3, 2), (0, 2), (3, 2).
Use your result to find the volume of a cask, height 6 feet, equal circular
ends of radii 2 feet, radius of the middle section 2 feet, and such that the
curve of a vertical section is a parabola.
4. Find the radius of gyration of a uniform solid hemisphere about
a diameter of the base.
6. A body mass 10 Ibs. moves in a straight line in such a way that its
acceleration at any instant varies as its distance from a fixed point O in the
line and is directed towards O.
At a distance of 1 foot from O the acceleration is 64 ft. /sec. 2 Find the
work done in bringing the body to O from a distance of 6 feet.
L
1. Shew that the value of
I ft.
(5 + mx) dx
i"
a
is the same for all values of m. Interpret this result geometrically.
Do the same for
2. The bounding radii of a sector of a circle of radius a include an
acute angle 0.
If this sector revolves about one of its bounding radii, shew that the
volume of the spherical sector thus generated is
8
17a
260
CALCULUS FOR BEGINN 7 ERS
[CH. VIII
3. The section of a girder is of the form shewn in the accompanying
figure (Fig. 116). It is 3" high, the greatest breadth is 2" and the breadth
of the straight part in the middle which is 1" high is 1".
Fig. 116.
The section may thus be regarded as made up of a rectangle 3" x 1" with
4 semicircles 1" in diameter.
Find the radius of gyration of the section about the axis AB through the
centre perpendicular to the greatest length.
4. Shew that I x t jd 2 x 2 dx=0.
J
6. Find the average length of the ordinates to the parabola y*=8x
erected at equidistant intervals from x Q to x=6.
J.
1. Find the volume formed by the revolution about OX of that portion
of the curve x + y = 2x*, for which x and y are both positive.
3. A hemispherical bowl radius 9" is being filled with water at the
rate of half a cubic foot per minute. Find the rate at which the depth is
increasing when the water is 6 inches deep.
3. A solid iron cylinder, diameter 6 inches, length 2 feet, rotates about
an axis along a diameter of one end. If it performs 100 revolutions per
minute, find its K. E. , given that a cubic inch of iron weighs '28 Ibs.
4. Find the area enclosed between y z =x 5 and x 2 =y s , and the volume
generated by the revolution of this area about the xaxis.
6. If the radius of the earth is r feet and a body of mass m Ibs. fall from
a height a feet to the earth, find the work done on it by the earth's attraction
and shew that if a be very large compared with r the work done is approxi
mately mr ft. Ibs. wt.
CHAPTER IX
DIFFERENTIATION OF TRIGONOMETRICAL RATIOS
203. WE have so far confined ourselves to problems
depending on the differentiation and integration of x n .
We shall now investigate rules for dealing with other
functions.
Differential coefficient of sin x.
204. Let y  sin x, where x is the number of radians in the
angle.
Then with the usual notation
y + Ay = sin (x + Aa;).
.". Ay = sin (x + Aa;)  sin x
= 2 cos ( x +  ) sin ^ .
\ * / J
/ AccX . Ao;
2 cos ( x + . ] sm ~^
V 2 / 2
=
Aa; Aic
. Aa;
262
CALCULUS FOR BEGINNERS
[CH. IX
Now, as Aa; * 0, cos (a; + r ) * cos x and  * 1.
\ L J Aa;
T (v73.)
/. The limit of when Aa; * is cos x.
Ax
i.e.
dy
dx
= COS X.
Illustrations of this result.
205. (1)
X
Angle in = No. of y
degrees radians in / = sin x
40 6981317 6427876
x + Ax
y + Ay
Ax
Ay
Ay
Ax
41
7155850
6560590
0174533
0132714
760
40 30'
7068583
6494480
0087266
0066604
763
40 10'
7010406
6450132
0029089
0022256
7651
40 6'
6995861
6439011
0014544
0011135
7656
40 1'
6984226
6430104
0002909
0002228
7659
And cos 40 ='7660.
Thus we see that as the increase in x is made smaller,
increase in sin x
*. : approaches the value cos x. What we proved in
increase in x
204 is that it can be brought as near to cos a? as we like if the
increase in x is made small enough.
206. (2) Draw the graph of y = sin x (where x is C.M.). At
a few points, say x Q, x=7, x=l, a=l'3 draw lines whose
gradient is cos x and verify that they are tangents to the graph.
[Fig. 117.]
204206]
DIFFERENTIATION OF SIN X
263
Angle
x in deg. &c.
1 5 44'
2 11 28'
3 17 11'
4 22 55'
5 28 39'
6 34 23'
7 40 06'
8 45 50'
9 51 34'
10 57 18'
11 G3 02'
12 68 45'
13 74 29'
14 80 13'
i : ^A
llill

y <ty
_ . j = COSX
1
100
199
295
389
479
565
644 765
717
783
842 540
891
932
964 268
985
T  'I' y/X^  
11 ii ;
iffc>F
iiiijiiiiiiiHIil^M
'//"
;S::?::j:j!::   ' ' ^ p
/ k 4 r o . .4 c e* 
i j
r .0 .0 i.n 4.1 1.0 i.o <.ii
Fig. 117.
264
CALCULUS FOR BEGINNERS
[CH. IX
Geometrical proof.
207. P, Q are points on a circle radius r such that
AOP x radians,
POQ = Ax radians. [Fig. 118.]
OS bisects L POQ.
Fig. 118.
MP
y = sin x = ,
r
A
y + Ay = sm x + Aa: = ,
r
RQ
and
Now
PQ
r
. Ay _ RQ _ RQ PQ
' Ate ~ ^ ~ PQ
PQ
L. PQR = z. ACS = X +
PQ
Aw
T
207, 208] DIFFERENTIATION OF SIN X 265
[The arms of L PQR are perpendicular to those of L. AOS.]
Ay / AoA PQ
. . r^ = cos (x + ^ } . = .
A /y X V /
V * / PQ
/ At\ PQ
As Aa; * 0, cos ( x + ^ ) * cos x and ^  1.
V 2* / PQ
. dy = ( ,
" da;
It is very important to remember that x is the number of
radians in the angle.
208. Ex. 1. To find the gradient of the curve y = sina:
at the point where x = *5. We have
dy
~ = cos x.
ax
.'. Gradient where x= '5 is cos *5, i.e. cos(*5 radians)
= cos 28 39'
= 8776.
To find the equation of the tangent at this point we have
when x = 5
y = sin 5
= sin 28 39'
= 4795.
/. We want the equation of the line through (5, 4795) whose
gradient is (8776).
This is y  4795 = 8776 (*  5),
or y8776x=0407.
Ex. 2. If y = sina;,
dy
f = cos x.
ax
.'. Ay = cos x . Aa; approximately.
266 CALCULUS FOR BEGINNERS L CH  Ix
e.g. if x = [c. M. of 30], sin x = '5, cos x = '8660 and
Ay = '8660 x Ax approximately.
Suppose Ace = c. M. of 1 = 0175,
then Ay = 8660 x 0175 approximately
= '0152 approximately,
i.e. if the angle increases from 30 to 31 the increase in the
sine is approximately 0152, or
sin 31 = '5152 approximately.
[Actually sin 31" = *5150 correct to 4 figures.]
This might be stated thus :
f(x + h) =f(x) + hf (x) approximately.
Here f (x) = sin x\
.*. f (x) = cos x) '
Also x = ?, h = 0175.
/. sin 31 = sin 30 + 0175 x cos *
D
= 5000 + 0175 x 8660, &c.
Ex. 3. Find maximum and minimum values of
4 cos x + 3 sin x.
Let y = 4 cos as + 3 sin x.
.; ^=4sino;+3cosa;. [Ex. 1, Exs. LVII.]
ax
dy ^
If y is a maximum or minimum, j = 0.
.'. tan x = . .
4
i.e. a: = c.M. of 36 52' 143 OS'
216 52' 323 08'
&c. &c.
208, 209] DIFFERENTIATION OF SIN X 267
Take for example C = C.M. of ( 143 08'),
y = 4 cos ( 143 08') + 3 sin ( 143 08')
=  4 cos 36 52'  3 sin 36 52'
4 ?_
If # = C.M. of (142)
y =  4 cos 38"  3 sin 38
= 3152018471
= 49991.
If a = c.M. of (144)
y =  4 cos 36  3 sin 36
= 3236017634
= 49994.
.*.  5 corresponding to x = C.M. of ( 143 08') is a minimum
value.
Or we might say, since
~ = 4 sin x + 3 cos x %
ax
.'. . = 4 cos x 3 sin x.
dx*
and when x c.M. of 143 08', cos x and sin x are both negative.
.'. ~. is positive,
ClZF
and a; = C.M. of 143 08' gives a minimum value of y.
209. {This particular kind of problem is more easily solved
by ordinary Trigonometry, for
4 cos x + 3 sin x = 5 sin (x + a),
4
where a is the acute angle whose tangent is 5, i.e. 53 8', and
268 CALCULUS FOR BEGINNERS [CH. IX
5 sin (x + a) always lies between  5 and + 5, having the value  5
when
sin (a; + a) = 1,
i.e. x + 53 8' = 270, 630, ..., 90, 450, ...,
and having the value + 5 when sin (a; + a)  + 1 ,
i.e. x +53 8' = 90, 450, ..., 270, 630.}
EXERCISES. LVII.
dy
1. Prove that if y=cosx, =  sin x.
dx
Illustrate as in 205, 206.
2. Get  =cos x by taking it as
sin/; AaA _._ f Aa;
Lt ^
3. Find the gradients of the curve
y = 2 sin x + 3 cos x
n o / o\ Q
at the points where x = 0, 1, 2,  , , tan' 1  , i&n~ l {   } , irtan" 1 ^.
O O O \ &j t
Draw the graph of y = 2 sin x + 3 cos* from x=v to X=T, and check
your results.
Find the equations of the tangents to y = 2 sin x + '6 cos x at the points
where x=0, 1, ^, tan" 1 ! 5),
\ /
dy
4. If ysinnx where n is a constant, prove =ncosnx, and if
dy
y = cosnz, =nsinnx.
dx
dy
A. Write down ~ for the following values of y :
dx
(i) cos 5x, (ii) 3 sin 2x, (Hi) j sin 4x   cos Bx,
(iv) osinnx + 6 cosna:, where o, 6, n are constants.
6. The radius of a circle starts in the position OA and rotates with
uniform angular velocity a> radians/sec. OP is its position at the end of t
seconds. OB is the radius perpendicular to OA and PM, PN are perpen
dicular to OA, OB. If the radius is a, what are OM (x) and ON (y)?
209] DIFFERENTIATION OF SIN # AND COS X 269
What is the speed of M at the end of t seconds ?
What is the acceleration of M at the end of t seconds ?
What is the speed of N at the end of t seconds?
What is the acceleration of N at the end of t seconds ?
In each case shew that the acceleration varies as the distance from O
[you should get M's acceleration =  <a 2 x], and that M oscillates backwards
and forwards from A to A' and back, performing a complete oscillation in
O
seconds. M is said to have Simple Harmonic Motion.
w
7. If a; increases at a uniform rate, shew that the rate of increase of
sin x decreases as x increases from to  , i.e. that sin x increases faster
l
when x is near zero and more slowly when x is near .
31
[Look at the difference columns in the table of Natural Sines.]
8. Given sin 60 ='8660 and cos 60 = 5000, find approximately sin 61,
sin 60 30', cos 61, cos 60 30'.
0. Find r^, (i) if y = sin a;, (ii) if y = cos x.
In each case shew that ~^ + y = 0.
1O. Find j^, (i) if y = acosnx, (ii) if y = a sin nx.
d 2 y
In each case shew that ~ > + n 2 y = 0.
ax
il. Find maximum and minimum values of
(i) 2 sin x + 3 cos x, (ii) 4 cos x  5 sin x.
What is (i) I sin a;, da;, (ii) \ co&x.dxt
13. Find the area bounded by the a;axis and that portion of y = s'm x
which lies between x = and x = rr.
Draw a figure and be quite clear what the answer means, e.g. if 1" repre
sents ^ along OX and 1" represents 1 along OY, what is the area in square
2
inches ?
ri fi
Find (i) / sin x . dx, (ii) I cos x . dx,
Jo Jo
12.
14
f2 f2
(iii) I sin x . dx, (iv) I cos x . dx.
Jo Jo
270 CALCULUS FOR BEGINNERS [CH. IX
/2ir r2ir
sin x . dx, (ii) I cos x . dx,
o jo
Sir Sir
/2 /" 2
sin x . dx, (iv) / cos x . dx.
2 2
Draw figures to explain the results.
16. What is (i) I sin nx . dx, (ii) Icosnx.dx?
[v. Ex. 4.]
17. Prove and draw figures to illustrate the following results :
fir [2 [Sir
(i) I sin x . dx = 2 I sin x . dx = I sin x . dx
Jo Jo Jo
/"~2~
= 21 sin x . dx,
Jo
cos a; .
o
f2 /"
(ii) I sin x .dx I
Jo Jo
/n fZir rSn
cos a; . dx = 0= / cos x. dx =/ cosx.dx,
o Jo Jo
(iv) ' r sinx.dx = 2  sin 2x . dx = 3 I siia.3x.dx.
Jo Jo Jo
18. Write down
(i) / sin 2x . dx, (ii) I cos 2x . dx,
(iii) I (3 sin 2x + 4 cos 3x) dx, (iv) I (1 + cos 2x) dx,
/a /""
cos 2x . dx, (vi) I cos 2x . dx,
~2
(vii) I sin x cos x . dx [remember sin 2x = 2 sin x cos x],
(viii) I sin 2 x . dx [cos 2x = 1  2 sin' 2 x],
(ix) Fcoa 2 x . dx [cos 2x = 2 cos 2 x  1],
209] SIN X AND COS X 271
/IT 'T
sin 2 x . dx, (xi) I cos 2 x . dx.
o Jo
What is the sum of (viii) and (ix) ? How could you foretell this result
without finding the separate integrals ?
19. Find the C.G. of the area bounded by the xaxis and that portion of
y = sinx that lies between a; = and x = ir.
[If you try to find x by the ordinary method, you will meet with an inte
gral which you do not know how to evaluate. How can you evade the
difficulty?]
[v
20. What is the value of I x sin x . dx ?
Jo
21. Find
(i) I sin Bx cos Ix . dx, (v) I sin px cos qx . dx,
(ii) I cos Bx sin Ix . dx, (vi) I cospx cos qx . dx,
(in) I cos Bx cos 2x . dx, (vii) I sin px sin qx . dx.
I
(iv) I sin Bx sin 2x . dx,
22. In Ex. 2, 208, what would you get if you used the approximation
23. The area of a triangle is calculated from the statement that two
sides are 2 feet and 3 feet and the included angle 40. If the sides are
measured correctly, but an error of 10' is made in measuring the angle, find
approximately in square inches the error in the area.
Use no tables. You are given sin 40 = 6428 ; cos 40= '7660.
24. Find the area bounded by the curve y = B sin x + 2 cos x, the xaxis
and the ordinates x = and x = ir.
25. Find the mean value of sin x between x = and x= .
4 o
Check approximately by calculating ^ [sin 45 + sin 46 + . . . + sin 60J.
26. The distance ( feet) of a particle moving in a straight line, from a
fixed point in the line, at the end of t seconds is given by s = acos t + b sin t.
Shew that the acceleration is  s ft./sec>
Find the speed and acceleration at the end of one second, if a = 4, 6 = 3.
27. Find the mean value of the ordinate of a semicircle when the
ordinates are erected at equal intervals (i) along the arc, (ii) along the
diameter.
Draw figures shewing 15 ordinates in each case.
272 CALCULUS FOR BEGINNERS [CH. J]
Differential coefficient of tanx.
210. With same notation as in 201
y tau a?,
y + Ay = tan (x + Ax).
.". Ay = Ian (a; + Ax) tan x
sin (x + Ax) sin x
cos (a; + Ax) cos x
_ sin Aa;
cos (a; + Ax) cos x '
. Ay sin Aa; 1 1
A* Aa; * cos (x + Aa;) ' cos x '
. sin Ax 1 1
Vti hen Aa;  0, 1 and  . 
Ax cos (a; + Ax) cos x
dv 1
/. ~ = ., = sec 2 x (= 1 + tan 3 x).
dx cos^x
211. Illustration of this result.
Angle in a; y
degrees =C.H. of angle =tanx
30 5235988 5773503
ylAy Aa; Ay
31
5410521
6008606
30 30'
5323254
5890450
30 12'
5270894
5820139
30 6'
5253441
5796797
30 1'
5238897
5777382
Fill up the blank columns and shew that the last column comes
nearer and nearer to sec 2 30.
210212]
TAN a;
273
Geometrical proof.
212. As in 207, L AOP = x radians, L. AOQ = x + Ax radians.
OP, OQ meet the tangent at A in p, q. [Fig. 119.]
Fig. 119.
qr, QR are perpendicular to OP.
V = tan x = .
r
y + Ay = tan (as + A) =
and
sr. c.
274 CALCULUS FOR BEGINNERS _CH. IX
Ay pq
. sec x QR
QR 'p^
Now = = = sec (x + A.C).
QR OQ OA
.Ay . , QR
. .  = sec x . sec (a; + Aa:) . ^^ .
Aa; PQ
QR
As Arc * 0, sec (x + Aa;) * sec x and ^ * 1.
PQ
..
dx
EXERCISES. LVIII.
1. Draw y=tan or from x = to x = l2 and verify that tbe gradient is
sec 2 a; at a few points.
3. Draw a sketch of y = tan x from x =  2ir to x = + 2?r. How can you
see that is positive for all values of * ?
dx
dy
3. (i) If y = cotx, prove =cosec 2 x.
dx
dy
(n) If y8ecx, prove see x tans.
dx
dy
(iii) If y = cosec x, prove =  cosec x cot x.
dx
4. If y = t&nnx, prove ^ = n sec 2 nx, and find p (i) if y = cotna;, (ii) if
;, (iii) if y = cosec na.
6. Write down :
(i) I sec 2 x . dx, (ii) I cosec 2 x . dx,
(iii) I secxta.nx.dx, (iv) I cosec x cot a; . dx,
(v) I (1+ tan 2 x) dx, (vi)  (l + cot 2 x) dx,
212, 213] TAN x 275
. ... [ sino; . ... f cosx
(vii) I y dx, (vni) / . dx,
' J cos 2 x ' J sin 2 a;
(ix) I sec 2 3x . dar, (x) \ia.vtx.dx,
(xi) I tan 2 3 . dx.
6. If i/ = sec x + tan x t
shew that
.
y dx
1. Knowing tan 45= 1, and see 45= /s /2, find approximately tan 451.
8. AOB is a quadrant of a circle, centre O. BX is the tangent at B.
A point T travels along BX and in every position of T, TO is joined
cutting the circumference in P.
If / BOP = 0, shew that the
speed of P = speed of T x cos 2 0.
If the speed of T be uniform 5000 ft. /sec. find the speed of P when
6 = (i) 45, (ii) 60, (iii) 80, (iv) 89, (v) 89$, (vi) 89. (v. 37.)
Also if OB be 10 ft. and OP have a uniform angular velocity of 1 radian
per minute, find in ft./sec. the speed of T for the same values of 6.
9. An electric current is measured by a tangent galvanometer, the
current being proportional to the tangent of the deflection. If the deflection
is read as 45 and an error of 1% is made in reading it, shew that the
percentage error in the current is approximately 5 .
a
10. If x is the deflection in a tangent galvanometer and a given small
error is made in reading the deflection, shew that the percentage error in the
current is proportional to (taux + cotx).
213. Ex. A man in a boat 6 miles from shore wishes to
reach a village 14 miles distant from the point of the shore
nearest to him. He can walk 4 miles an hour and row 3 miles
an hour. Where should he land in order to reach the village in
the least possible time ? What will this least time be ?
[The shore is supposed straight.]
182
276
CALCULUS FOR BEGINNERS
[CIL IX
B is the boat. V the village. NV the shore. BN perpen
dicular to NV. P any point in NV. Call the angle NBP 0.
[Fig. 120.]
V
B 6m N
Fig. 120.
Then BP = 6 sec 6,
and PV = 14Gtan0.
.'. Total time taken if he lands at P is
/6sec0 146 tan0
Call this t, so that
3 7
t = 2 sec 6  tan B + .
. dt_ _
" d6~
For a minimum value of t t
dt
dO
= 0.
(3 \
2 tau B  H sec \ = 0.
.'. 2
sin 6
cos 6 2 cos
= 0,
since sec ^ 0.
213] DIFFERENTIATION OF TRIGONOMETRICAL RATIOS 277
3
.*. sin = r.
4
.*. sec = j= ,
v7
and tan = ^= .
\/7
I Q
/. NP = 6tan0 = ~ =6801,
and
~V7 2x/7 2 2
[Shew that this is a minimum value.]
.'. He must land 6804 miles from N and his time will be
4823 hours.
EXERCISES. LIX.
1. A wall 6 feet high runs parallel to and 5 feet from another wall.
Find the length of the shortest ladder that will reach from the ground to the
second wall over the first.
2. An elastic string has one end fixed at A and the other B moves along
OY perpendicular to OA.
If OA=2 feet, find at what rate the other end is moving when / OAB = 60,
supposing the angle OAB to be increasing at a uniform rate of 1 per second.
3. With the same string as in Ex. 2, find the rate at which ^OAB is
increasing when L OAB = 60, supposing B to be moving at 1 inch per
second.
4. A man 6 feet high walks at 6 feet per second away from a lamppost
10 feet high.
Find the rate at which his shadow is increasing. If 6 be the angle made
with the ground by the line joining the top of his head to the top of the lamp
post when he is x feet from the post, prove =4 cot &. Hence find the rate
at which 9 is decreasing when he is 8 feet from the lamp.
CHAPTER X
PRODUCT. QUOTIENT. FUNCTION OF FUNCTION.
INVERSE FUNCTION. IMPLICIT FUNCTIONS
Differential coefficient of a product.
214. SUPPOSE y = a? sin x.
Here y is the product of two functions, viz. a? and sin x, each
of which we can differentiate separately.
y + Ay = (x + Aa;) 2 sin (a; + Aa;).
.'. Ay = or 1 {sin (x + Ace)  sin a;} + 2x Aa: . sin (a; + Aa;)
+ (Ate) 2 sin (a; + Aa;).
. Ay sin (a; + Aa;)  sin x
. . ~ = a, 2 . a ' + 2aj sm (a; + Aa;) H Aa; . sin (a; + Aa?).
sin(a; + Aa;)sinaj
Now as Aa;  0, * * cos x, sin (a; + Aa;)  sin a?,
OdB
and Aa?*0.
.*. jr = x* cos x+2x sin a;.
cc
215. We might have arranged this as follows, so as to shew
more clearly the significance of the different components of the
result.
214216] DIFFERENTIATION OF PRODUCT 279
Ay = (x + Ax) 2 sin (x + Ax)  (a; + A*) 2 sin x + (x + Aw) 8 sin x  or 2 sin x,
where the term (x + Aa;) 2 sin x has been added and subtracted as
a link between (x + Aa;) 2 sin (x f Ax) and a 2 sin x.
. Ay . sin (x + Ax) sin x (x + Ax) 2  a?
.'. = (a; 4 Aa;) 1   + sin a ^ ^
Ax Ax Ax
= A (x + Ax) 2 + B . sin x, say.
is Ax0,
Also (x + Ax) 2 x' J .
d since . d.x? .
Now, as Ax ^ 0, A  ; . i.e. cosx. and B * . , i.e. 2x.
dx dx
dv
.'. ^ = x 2 cos x + 2x sin x,
c?(ar s sinx) rfsinx rfx a
or * ; = x 3 . ; H sin x . ^ .
dx dx dx
EXERCISES. LX.
1. Go through the work again, using as a link
a; 2 sin (a; + Ax).
a. Find in the same way p ,
(i) when y = (2x + 3) cos a;,
(ii) when y = x 3 ia,nx,
216. In the general case let y = uv, where u and v are
functions of x.
If x receive a small increment Aa 1 , this will produce small
increments AM, Aw in u and v and these will produce a small
increment Ay in y.
Thus y + ky = (u + AM) (v + Aw)
= MV + u . A + w . AM + A?t . Aw.
.*. Ay = (M + AM) Av + v A?*.
Aw \ Av Aw
.*. r^ = (M + AM) . r + v . .
Ace Ao: AJC
280 CALCULUS FOR BEGINNERS [C1I. X
Av dv AM du
As Ace * 0, (u + A?n  u,   = and   y .
Aw cte Aaj do
. dy_ dy du
" dx ~~ dx dx '
217. Notice that this result may be stated as follows :
(i) Differentiate uv as if u were constant. This gives u 
^ ' dx
(ii) Differentiate uv as if v were constant. This gives v . y .
Qu6
(iii) Add these together.
If we divide both sides by y or uv, we get
1 dy _ 1 du 1 dv
y' dx u' dx v' dx'
218. Ex. 1. If
Then by the formula
which agrees with what we get by writing
y = x 13
and differentiating in the ordinary way.
Ex. 2. If y = cos x . tan x,
dij
~ = cos x . sec 2 x 4 tan x ( sin x)
dx
1 sin 2 x
cos a; cos x
cos 2 a;
= COS X,
cos as
which agrees with what we get by writing
T sin of!
y = sin x\ = cos x x .
L cos xj
216219] DIFFERENTIATION OF PRODUCT 281
Ex. 3. If y sin 2 x = sin x . sin x,
dy
f sin x . cos x + sin a; . cos x
ax
= 2 sin x cos x.
EXERCISES. LXL
Find ~ for the following values of y :
I. x sin x. 2. a 2 tanar.
3. (3x 2 + 2.r + l) (2.r + 3), (i) by treating as a product, (ii) by multiplying
out.
4. 2 sin x cos x. [Cf. Exs. LVII. 4.]
3 __A
5. x 2 " x a; % (i) by treating as a product, (ii) by writing as one term.
6. tan x . cosec x, (i) by treating as a product, (ii) by writing it sec x.
7. 4s 6 sin or. 8. 5x 2 cos3#. 9. cos 2 x.
1O. (i) cos 2 5x, (ii) sin 2 5x.
What is the sum of these two results ? How can you see that this must
be so without actually finding them ?
sin x
II. Jxcosx, 12. jj
x*
Product of more than two functions.
219. Suppose y = uvw, where u, v, w are all functions of x.
uv is a function of x. Call it z.
.'. y = zw.
dy dw dz
:. By our rule ~ = z . 7 + w . r :
aa3 ax ax
but since z = uv,
dz dv du
(!.< dx dx '
dii dw f dv du\
. . f = uv . = h w I u = + v =]
dx ax \ ax dx/
du dv dw
282 CALCULUS FOB BEGINNERS [CH. X
Ex. Prove similarly that if y is the product of four functions,
say uvtvz, then
dy du dv dw dz
3^. = VWZ = + WZU 3 + ZUV 3 + UVW 3 .
dx dx dx dx dx
220. If we divide both sides of the result in 219 by y or
uvw t we get
1 dy 1 du 1 dv 1 dw
? i i
y' dx u' dx v' dx w' dx'
and similarly if y uvwz,
I dy 1 du I dv 1 div 1 dz
y' dx u' dx v' dx w' dx z' dx'
[Another way of proving these results will be seen later, 263.]
1 dy _ 1 2 3 4
y' dx x+l 2x + 3 3a; + 5 4x + 7'
2 3 4 "I
j. _i_ 4. I
EXERCISES. LXII.
(i) by treating y as the product of z 2 , (x + 1) , (x + 3),
(ii) by treating y as the product of a; 2 and a; 2
(iii) by taking y = x 4 + 4s 3 + 3x 2 .
(i) by treating y as the product of (x + 1), (x + 2), ,
2
(ii) by taking y = x + 3 + .
X
(y
d. If y = :r 2 sins cos x, find .
219221] DIFFERENTIATION OF QUOTIENT 283
4. If y =
6. If u is any function of x, find
... d 8 .... du 3 .... du* . du
' (11) ' (U1) m terms of " and '
Differential coefficient of a quotient.
sin x
221. Suppose y = j
sn
sin (x + Ax) sin x
a; 2 sin (a; + Ax) (a; + A.*) 2 sin x
a? (x + A*) 2
_ ar'lsin (a;+ Ax) sina:}sina; {(a; + Acc) 2 x 2 }
a 2 (a; + A*) 2 ~"'
where a? sin x has been added and subtracted as a link between
a? sin (a; + Aa;) and (a; + Ax) 2 sin x.
. Ay Aa; 2  B sin x
sin (a? + Aa;) sin x
where A stands tor ,
Aa;
(x + Ax) 2  a?
and B stands for J  '  .
Ax
Now the limits of A and B as Ax*0 are cosx and 2x
respectively [the differential coefficients of sin x and x 2 ], and the
limit of (a; + Ax) 2 is a?.
dy x 2 cos x 2x sin x x cos x 2 sin x
" dx x* x 8
rf(sinx) d(a?)
ar. ^j ^sinx. V~^
dx dx
284 CALCULUS FOR BEGINNERS [CU. X
222. Generally, if y = ,
u + Ait
u + AM
v + Aw
v (v + Aw)
AM At?
. Ay Ax Ax
Ax v (v + Av) '
and as Ax  the limits of , , (v + Av) are = , 7 and r
Ax Aa; ax ax
respectively.
du dv
. dy _ dx ~ dx
" dx~ v a ""'
Ex. 1. Take y = ^ = ,
/r 1 ?/ ^ / >*" v vR'** 4.1' v Qo'2
Lvf/ JtO A M OiX' T*O A i/^O
_48x 9 _16
~ 9^" ~ 3 ^ '
4
which agrees with what we get by taking y = x*.
o
sinx
x.2. y =
dy cos x x cos x sin x ( sin x) 1
v ' _ ' __ ^^^^^^ COf* '*
j  s  s set  X,
ax cos x cos x
which agrees with what we have previously determined as. the
differential coefficient of tan x.
222, 223] DIFFERENTIATION OF FUNCTION OF FUNCTION 285
EXERCISES. LXIII.
Find / for the following values of y:
x + 1 x
x + 2 ' .Ti + 1 '
sin x r in
3. (i) quotient, (11) sin ex.
x L tfj
4. : [(i) quotient, (ii) x cosec xl
sm x
6.
1 sin x 1
1 f sin x ' ' cos x "
O.  , where u is any function of x.
Function of a function.
223. The only functions of x which we have yet dealt with
are powers and roots of x, sin x, cos x, &c. and products of these,
and we have shewn how to find the rate at which any one of these
changes with respect to x.
For instance we have
d sin x
which we may read :
The differential coefficient of the sine of any angle with
respect to the angle itself = the cosine of the angle,
or, the rate of change of the sine of an angle per unit increase
in the angle = cosine of the angle, it being understood that by
" angle " we mean " number of radians in the angle."
We may write the above
d . sin (angle)
d . angle
= cosine (angle).
286
Thus
CALCULUS FOR BEGINNEIiS
d . sin 4:X
 = cos 4or,
[CH. X
= cos
&c.
Similarly, from the result
d 
dx
c . sn a;
;  : 
a . sin x
Write down
d.(y?
1.
7.
EXERCISES. LXIV.
d.st? '
d . tan 4x
d . cos 2 (4 x + 7)
d . cos (4as + 7)
'
d.afi
d.x*'
d.sec 6 *
d . sec x '
3.
d .
d. sin x
8.
1O.
d.(2sinx + 3 cos xfi
d . (2 sin x + 3 cos z) "
224. Now suppose we have a function y = sin3, and we
want j . We do not get cos 3a;, for the fact corresponding to
d sin x L d sin 3x d sin 3x
 =cosa is not ;  =cos3a;, but .. . = cos 3x, i.e.
dx dx d(3x)
sin 3cc is increasing cos 3x times as fast as 3x,
but
dx
223225] DIFFERENTIATION OF FUNCTION OF FUNCTION 287
i.e. 3x is increasing 3 times as fast as x.
.'. combining these two statements
sin 3o5 is increasing 3 cos 3x times as fast as x,
d . sin 3x
or = = 3 cos 3x
dx
td . sin 3o; _ d . sin 3a; d . 3afl
dx d,3x dx J '
225. As another example take y = v/tana;.
If y = *Jx, we know that ~ ^ , but we must not say that
if y = \/tan x, ~ = . , for the statement corresponding to
: 1 cwtan x 1
= T= is not
but
7 T^. AO XHJU ' = .
dx Zjx dx 2 N/ tan x'
c?vtan x 1
tana? 2\/tana;'
i.e. vtana; is increasing . times as fast as tana?.
2 v tan x
dtanx
Now j = seo j a;,
i.e. tan x is increasing sec 2 a; times as fast as x.
i . . . sec 2 aj
. . v tan a; is increasing . times as tast as a;,
2v tan x
d \/tanx sec 2 x
or
c
c/vtan a; c? \/tana: </ tan a;~
cfe .
288 CALCULUS FOR BEGINNERS [CH. X
EXERCISES. LXV.
By similar reasoning find ^ for the following values of y:
1. cos 5x. 2. tan 3.T. 3. sin 3 x [i.e. (sin a:) 3 ].
4. sin(z 2 + 3). 6. sin *fx. G.
7. (3x 2 + 5x + l). 8.
226. We might have presented the argument of 224
somewhat differently.
y = sin M, where u = 3x.
Now suppose y to be any function of u, and u any function
of x,
If x receive a small increment AOJ, this will produce a small
increment AM in M, and this increment in u will produce a small
increment Ay in y.
Ay Ay AM
however small Aa;, AM, Ay are.
. Ay Ay AM c/y ty du
But asArc^O =*,  ,  +, ^ , 7 respectively.
Aas AM Aa; ax du ax
5z=^
dx ~ du ' dx '
But in this case ^ = cos M and = = 3.
an ax
.'. = 3 cos M
cw;
= 3 cos 3a?.
2.26, 227]
FUNCTION OF A FUNCTION
289
Ex. 1. y = su\ 3 x.
Put y v? where u = sin x.
dy du
j = ow 2 , i = cos x.
du dx
.'. ~ 3
ax
= 3 sin 2 x cos x.
This might be written
d (sin 3 x) d (sin 8 x) d (sin a;)
dx d (sin a;) fo;
3 sin 2 x x cos x.
Ex. 2. y = A/3aT+l.
Put = iju where u = 3x + 4.
nnn o i i / <V CW v
227. Similarly ^ = /. . .
^a;. 1. Suppose y = \/sin (403 2 + 6x).
We may say y \/w where u = sin v, where v = 4* 2 + 6a5.
dy \ du dv
^
du
j
dv
= cos v, y =
dx
6.
.  =
dx
= x cos v x (80; + 6)
w
3 + 3) cos (4o; 2 + 6x)
\/sin (4a; 2 + 6.7;)
M. 0.
19
290 CALCULUS FOR BEGINNERS [CH. X
dy d vsin (4x 2 + 6r) d&in (4ce 2 + 603)
for ^ =
L dx d sin (4x 2 + 6*) d(*ia? + 6x) . dx
= . x cos (4a; 2 + Qx] x (&e + 6).
2 Vsin (4aj + 6a;)
Ex. 2. y = sin 4 (3x + 4).
Put y = u* where u  sin (3# + 4).
Put u = sin v, v = 3x + 4.
dy , du do
~ = 4w , 7 = cos v.  3.
du dv dx
. dy
f~ 4w x cos wxo
 12 sin 3 (Sx + 4) cos (3.x + 4),
tfX "T" T: j Cv Sill ( tjQG ~r 4: ) tt ( OiK *T~ rr )
= 4 sin 3 (3* + 4) x cos (3x + 4) x 3.
EXERCISES. LXVI.
Find J for the following values of y :
1. v'sin x. 2. sin /. 3. sin 4 x.
4. sin 4a;. 6. sin (x 4 ). 6. fcin" x.
1. sin nx. 8. rasinx. 9. cos n x.
1O. cosnx. 11. sin(4r + 5). 12. (3 sin x + 4 cos x) 2 .
13. tan 2 x. 14. sec 2 x.
[Why are the results of (13) and (14) the same?]
15. (a sin x + b cos x) n . 16. tan n x. 17. sec n ar.
18. *Jt&nx. 19. tan ,Jx. 2O.
21. tx* + 3z + 1. 22.
227]
23.
29.
32.
35.
(6* + 1)20.
1
24.
27.
3O.
33.
36.
(a* + b) n .
cos (sin x).
sin x.
sin 8 (4* + 5).
25.
28.
31.
34.
37.
(3* 2 + 2* + 1) 10
ZJax+b.
(* 2 + l)6'
sin (cos *).
1
jSx* + 2x + 1.
sin (ax + 6).
sin" (ax + b).
njain. (ax + b).
38. tan" (ax + 6).
4O. ^/sin (2* 2 + 5* + 6).
42. (a sin 8 x + b cos 3 *) 5 .
44. cos^Sx.
46. sin (a + to").
48. sin m (a; n ).
6O. 4 cos 3 a;  3 cos *.
[In Exs. 51 60, u denotes any function of x.]
51. sinu. 62. sin 3 w. 53. u 2 .
39. sec" (a* + 6).
41. (a sin 3 as + 6 cos 3 *).
43. *Js sin 2 x  4 cos 2 *.
46. tan s (x 2 ).
47. */5 tan 2x + 3 sec2 2ar.
49. 3 sin x  4 sin 3 x.
64. M n .
56.
66. .
tt
67.
68. cos n M. 59. sin ^i. 6O. *Ja? + u?.
Write down the values of / ydx corresponding to the following values of y :
61.
sin 2 x cos *.
62.
sin" x cos x.
63.
cos n x sin a;.
64.
sinna;.
66.
cos nx.
66.
sec 2 nx.
67.
tan 2 nx.
68.
sin (ax + b).
69.
cos (a* + 6).
7O.
(ax+b)*.
71.
*(* 2 + 4)6.
72.
sin *.
73.
sec 5 x . tan x.
74.
2x+B
75.
2* + 3
192
292
CALCULUS FOR BEGINNERS
[CH. X
228. By using the methods of 218, 219, 222, 226 we may
differentiate more complicated expressions.
Ex. 1. If
We have y = Ju, where u =
du = 2
1cc
, du 2
and =
dx (1 + a;) 2
1
[ 222]
dx
Ex. 2. If
* 4
, ,
> find
method, y = , where u = x Jx* 4, v = v a; 2 1.
v
du dv
dy _ dx dx
dx v*
du
To get  r we have u  x x vo; 2 4.
ax
du
.*. rr var* 4 + x x
a05
4
4 +
2a; 2 4
4 v a; 2 4
and
x. *Jx?  4.
a; 2 !
/a.2_4
2 no? method, y = an*, where u = ^/ ^.
3x ~~ X
T
ox
228, 229] MISCELLANEOUS EXPRESSIONS 293
x 2 4
Now u = Jv, where v =
.
r 1
(>4J* (>!)*"
jEfo 3. y = sin 3 2#.cos 4 5x,
i.e. y = uv, where u = sin 8 2x and v = cos 4 5x.
~r~ == V j W = ,
dx dx dx
du d.s\n 3 2
dx d sin 2x d.2x ' dx
= 3 sin 2 2x . cos 2x . 2
= 6 sin 2 2x cos 2.
0B
Similarly = 20 cos 3 5x . sin 5a;.
CttiC
du
.*. j = 6 sin 2 2o; cos 2x cos 4 5ic 20 sin 3 2x cos 3 5as sin 5x
dx
= 2 sin 2 2x cos 3 5a? (3 cos 2x cos 5cc 10 sin 2a; sin 5cc)
= sin 2 2cc cos 3 5x (13 cos 7x  7 cos 3x).
229. Notice that we could differentiate quotients without
the special rule.
Take as an example the one first worked out
sing
y= ^
We could put this in the form
y x~ z x sin x,
and treat it as a product.
294 CALCULUS FOK BEGINNERS [CH. X
dy
.*. f = ar a x cos x + ( 2ar 3 ) sin x
_ cos x 2 sin x
x cos x 2 sin x
73 ~ *
tiff
n R ' n *
Ex. 2. v = ,
cos a:'
\'Q could put this in the form
y = sin x x (cos a;)" 1
= u x v,
where u = sin x and = (cos a;)" 1 .
We know j = cos x.
dx
dv d (cos a;)" 1 d (cos a;)
and  1 = A x 1 '
dx d (cos x) dx
= ( 5) (sin x)
\ cos 2 x) v
COS 2 X '
dy dv du
MOW ^=14 + v
ax ax ax
sin a; 1
= sin x x H x cos x
cos 2 x cos a;
= tan 2 x + 1
= sec 2 x.
EXERCISES. LXVIL
Find  for the following values of y :
/lx* iz s /I sin 2 x
l  /v/1 5' 8  xJa 2 **. 3. A /, rs
V U* 2 V 1 + sin 2 a:
230] MISCELLANEOUS EXPRESSIONS
4. x^cos^x. 6.
6. f.r 2 + x + 1)4 / 2 . T s _ 3^2 + 4)2. 7.
295
8.
1O. sin 3 a; . sin 3s.
0.
12. 2 cos 3 5a;  3 sin 4 2z.
14. sin n xcosnx. 15.
16. sin" 1 w cos" v.
11. x (x z + 4) ^z 2  4.
13. sin" x . sin nar.
^", where M and v are functions of x.
. /
\/
17
smJ5a: .,
5 + cos 5o;
18.
19.
ao . /.
\/ ^
230. Sometimes y and x are each given as the function ot
a third variable.
e.g. the coordinates (measured horizontally and vertically) at
the end of t seconds, of a body projected with a velocity whose
horizontal and vertical components are 30 and 40 ft./sec., are
.. = 30, = 4032*.
at at
i.e. x is increasing 30 times as fast as t, and y is increasing
(40 _ 32<) times as fast as t.
4032* .
. . y is increasing ^ times as fast as x,
dy _ 40 
30
dy Idx
~dtdi'
From the given equations we might eliminate t. The first
oives t q^: , and substituting in the second, we get
296 CALCULUS FOR BEGINNERS [CH. X
. dy _ 4 32a?
"dx = 3~900
_ 4 32 x 3Qt
~3 900
4032*
= ~r as before,
uU
but in many cases the elimination of t would be inconvenient if
not practically impossible.
The general case.
231. Suppose x and y are given in terms of a third
variable t.
So that *=/(<) and y = <(<)
Suppose u receives a small increment A. This will produce
small increments A and Ay in x and y.
Now Ay _ Ay /A*
A^~ Sr/A*'
however small A, Aa:, Ay are, but as A2, and with it Ay and
Ay Ay Aa? dy dy dx
Ace, 0,  .  . ^ gr i 5 ,  rr respectively.
' Ao; ' A< ' A< dx' dt' dt
y _ y /^
' dx" dt/dt '
EXERCISES. LXVIIL
1. If =3t + 4t 2 anda; = 2 + l, find  in two ways.
dx
2. If z = 4 + 3t + 2j3and?/ = 7 + t 3 , find .
(US
3. If a; = acos5 + tsin0 and y=a8m0bcos$, find 2,
230, 231] x AND y IN TERMS OF THIRD VARIABLE 297
x <i
4. Shew that the coordinates of any point of the parabola y may
be written in the form x = am, y = am 2 .
Find /, (i) fromj^ and = , (ii) fromw = .
dx dm dm x ' a *
and shew that the results agree.
6. Shew that whatever 6 is, x=a cos 0, y = a sin 9 satisfy the eqno.tion
Prove / =  cot =   .
dx y
Interpret this result geometrically.
Shew that the equation of the tangent at the point a cos o, a sin a is
x cos a + y sin a = a.
6. Shew that whatever is, x = a cos 0, y = b sin satisfy the equation
dy b Wx
Prove 3^= cot = 5 .
dx a a?y
Hence shew that the equation of the tangent at the point (a cos a, b sin a)
is
x y .
 cos a + f sin a=l.
a o
7. The coordinates of a point on a cycloid are given by x = a (9  sin 0),
y=o(lcos0).
Prove j^ = cotjj.
dx 2
8. The coordinates of a point on an epicycloid are given by
x = (a + b) cos 06 cos JT '
y = (a + 6) sin b sin
dy , a + 26
Prove ~ = tan 0.
da; 26
298
CALCULUS FOR BEGINNERS
[CH. X
9. A body is projected from O with velocity ft./sec. in a direction
making an angle a with the horizontal. Taking as axes the horizontal and
vertical through O in the plane of projection, the coordinates of the body at
the end of t seconds are ( tu cos a, tu sin a  5 gt z j . Prove that at the end
of t seconds the direction of motion makes an angle tan" 1 ( ) with
\ u cos a /
the horizontal.
Hence find the time of reaching the greatest height.
232. The same problem might be presented in another form,
c^sin x
Ex. Fin'd
We have
da?
i.e. find = where u = sin x and v = x*.
dv
du
dv
du /dv
dx/ dx
cosx
Illustration of the meaning of this result.
Angle
50
X = C.M.
8726646
x+ Ax
51
8901179
50 30'
8813913
50 10'
8755735
50 05'
8741191
and
Find:
a
m. \
a?
sinx
664572
7660444
(x + Ax) 3
sin (x + Ax)
705251
7771460
684709
7716246
671242
7679110
667901
7669785
cos a;
2(
B13.
3x2
EXERCISES. LXIX.
d (tan x)
A (?in x)
0111016
040679
0055802
020137
0018666
006670
0009341
003329
= 2729
= 2771
= 2798
= 2806
3.
5.
d (sin 2x)
d(sinx)
d sin x
d cos x '
232, 233] x AND y IN TERMS OF THIRD VARIABLE 299
233. The device of expressing x and y in terms of some
third variable is sometimes useful in integration.
Take the example in 191 which we failed to do directly
because we could not evaluate
a? \/a 2  # 2 dx.
a
If however we work in terms of ( AOP) we have,
x = a cos 6.
.". Aa; =  a sin . A0 app.
and y = a sin 6.
[Aa; and A0 are of opposite signs.]
.'. Area of strip PpqQ. = 2 2 sin 2 6 . A0 app.
and M. I. = 2ma 4 sin 2 6 cos 2 6 A0 app.
/. M. I. of disc = I 9 2ma 4 sin 2 6 cos 2 6dO
Jo
ma 4 [* . n . 7 . ma 4 /"",,
= I sm 2 20 . dO =  \ (1 cos 4(9) dB
* JQ * Jo
EXERCISES. LXX.
1. Find the ares of a quadrant of
yfi yZ
[Put x=acos 0, y = 5sin 8.]
3. The coordinates of a point on a cycloid are given by x=a (0  sin 6),
y = a (1  cos 8), where 6 is the angle turned through by the rolling circle.
Find the area between the araxis and the portion of the curve traced out
in one complete revolution of the rolling circle.
300 CALCULUS FOR BEGIN NEKS [CH. X
x 2 w 2
3. Find the M.I. of the ellipse 5 + TO = !
a 2 6 2
(i) about the 2 axis,
(ii) ahout the yaxis,
(iii) about a line through the origin perpendicular to the plane of
the ellipse.
4. Find the resultant thrust on a circle, radius a ft., just immersed in
water with its plane vertical. Also find the centre of pressure.
5. Find the c.o. of a quadrant of an ellipse.
Inverse functions.
234. If y is a function of x, then a? is a function of y ; for
instance if y = x + 3, x = ^r . and generally if y =f(x), x is
some function of y, say x = <j> (y).
In the above example f(x) is 4ce + 3 and < (y) is .
Notice that /{< (x)} =f ~ = * . + 3 = x,
(v. 87.)
_ Q
and <j> {/(x)} = < (4* + 3) = = x,
or the two operations denoted by y and < neutralise one another.
f(x) and < (a;) are called inverse functions. Sometimes < (x)
is written y*" 1 (a:).
e.g. sin x and sin" 1 x are inverse functions
sin (sin" 1 x) = x and sin" 1 (sin x) = x.
EXERCISES. LXXI.
1. If y= j , find x in terms of y.
a.
Verify /{/I (*)}=*,
and =x.
234237] INVERSE FUNCTIONS 301
235. If y=x>, x = Jy.
Here y is said to be a singlevalued function of x, i.e. given
any value of x, there is only one corresponding value of y, but x
is a doublevalued function of y, i.e. given any value of y, there
are two corresponding values of x.
e.g. if x = 3, y = 9 ; but if y = 9, x = 3 or  3.
In this case/(cc) is y?, and < (y) or/ 1 (y) is >Jy and one of
the values of \f(x)} is x.
Differentiation of inverse functions.
236. If
dtl
From the first of these relations we get ~ = 4, and from the
ctx
, dx 1 1
second 7 =  = .
dy 4 dy
dx
EXERCISES. LXXII.
Verify that ~ x . =1 in each of the following cases :
1. 3z + 47/ = 7. 2. y = x s . 3. .r?/ = *. 4. ysl.
43? i
237. We shall now prove the general theorem
dy dx
a x _ = 1.
dx dy
j\'l ' Z\T*
The relation x  = 1 is always true however small Aa; and
Ace Ay
Ay are. But by diminishing Ace and Ay indefinitely we can bring
Aw , A* dy , dx
and as near as we like to  and respectively.
Ao; Ay dx dy
, dy dx _.
~j~ x ~j~ := i
ax ay
This formula is of great use in differentiating inverse functions,
for it often happens that 5 is easier to get than ~ .
302 CALCULUS FOR BEGINNERS [CH. X
238. So long as y is a singlevalued function of x, and x
of y, this formula presents no difficulty.
Suppose y = y?, so that x = vy.
mu dy dx 1
Then f = 2x and r = =.
dx dy 2 V y '
the sign being the same as in the line above.
.. r dx
i.e. if x = +*Jy. ; =
dy
.. i dx
,f .._v?. 5
. , , . dx 1
so that in each case r = TT ,
ay 2
dy dx
and thus T^.^S:!.
c??/
Suppose for instance we want the value of ^ when x = 3
[and therefore y = 9].
We have ^ = 2a;,
dx
dy
so that when x = 3, ~ =Q.
dx
d'U
Similarly, when a; = 3, ^ Q.
CHP
Suppose however we take x = + vy, then when y = 9, a; = + 3,
dx 1
and i = +
<&c 1
so that when y = 9. =  7; ,
dy 6
1
^ corresponding to the case when x = 3
and ^ corresponding to the case when x 3
238, 239]
INVERSE FUNCTIONS
303
.". If in each case we are dealing with the same values of
both x and y t
dy dx
x = 1.
dx dy
239. If we draw the graph of y = x z , then x = 3 gives y = 9
and determines one point P x [Fig. 121],
^ gives tan XTjP,,
dx .
gives tan Y^P Jt
Fig. 121.
304 CALCULUS FOR BEGINNERS [CH. X
and tan XT^ . tan ^t 1 P 1 = 1 since the angles are complementary :
but if y = 9, x = 3, and two points P x and P 2 are determined:
dx
gives tan XTjPj or tan XT 3 P 2 according as we take x = 3 or 3 and
dx
j gives tan Y^Pj or tan YtjP 2 (negative angle).
ay
Now tan XTjPj x tan Y^Pj =
tan XT a P 2 x tan Y 2 P 2 =
but tan XTj?! x tan Y< 2 P 2 = 1
and tan XT,P, x tan
=:!}
i.e. ~ * T = 1 if we deal with the same point throughout
dx dy
(either Pj or P,), but if we take ? with reference to the coordi
CBB
(iy
nates of P, and with reference to those of P, we do not get 1.
dy
The inverse trigonometrical ratios.
dii
240. Given y = sin" 1 x, to get ~ we have
CL3C
x sin y.
/. j = cos y = +\/l  a?.
dy
The ambiguity of sign is accounted for as follows :
If we take y = sin" 1 x to mean that y is some angle whose sine
is x, then y is a manyvalued function of x : for any value of x
between 1 and I, say 866, there is an infinite number of
values of y, viz. :
T 27r TJT Sir 4ir 5ir lOir UTT
3' T' ~3~' "3 '"' ~T' ~T' ~~3~' ~~3~""
/"/ay
and for some of these ~ is +, for others (see Fig. 122).
'
239, 240]
SIN" 1 X
305
If we agree to understand by sin l x the angle between 
7T
and + jr whose sine is x, then a glance at the figure shews that.
z
~ is always positive. [Fig. 122.]
Fig. 122.
If y = sin" 1 x, we take
dy 1
M. 0.
20
306 CALCULUS FOR BEGINNERS [CH. X
x y = sin" 1 a; = o. M. of acute angle whose sine is '5000.
5000
5235988
x + Ax
y + Ay
As
Ay
Ay
As
5878
6283
0878
1047
119 +
5592
5934
0592
0698
118
5299
5585
0299
0349
117
5150381
5410.521
0150381
0174533
116
5075384
5323254
0075384
0087266
1157
1 1
/4
A/ ./!;
J33 = 1'166.
V/S
241. Similarly taking cos" 1 x to mean the angle between
and IT whose cosine is x, we can prove that if y = cos" 1 x
dy _ 1
dx = ~
Notice that since whatever x is,
. , TT
sin" 1 x + cos' 1 x = ,
d.sin^x d.cos^x _
' 'dF "dF
242. If y = tanix,
x = tan y.
dx
.*. f = sec 3 y = 1 + tan 2 y  1 + a?.
dy
 2l JL
' dx l + x2'
Usually tan" 1 x is taken to mean the angle between = and
2i
+  whose tangent is x, but the result just obtained shews that
2
even though we make no such restriction, but take tan" 1 x to
mean any angle whose tangent is *, ~ is always the same.
240242] TANIS 307
The graph shews that  is always positive. [Fig. 123.]
Fig. 123.
EXERCISES. LXXIII.
dy 1
1. If y = cot" 1 x, prove p =
(b
3. Ify=sec~ 1 a?, prove ^ =
^ L
8. If y=cosec~ 1 *, prove r= 
202
308 CALCULUS FOR BEGINNERS [CH. X
Implicit functions.
243. It often happens that the relation between x and y is
in such a form that it is impossible or undesirable to find y in
terms of x.
e.g. we might have
/2 = ............... (1).
6R?y
We can get without finding y in terms of x.
ClX
We must remember that
dy* = dy*^dy = 2 ^
dx dy'dx dx'
Differentiate both sides of (1) with respect to x.
... ( x _ 6y + 4) + (4* + y + 1) = 0.
. dy 4x + y + 1
" dx " x  6y + 4 '
In this particular case we could have solved (1) as a quadratic
in y.
Rearranging we get
V  ( + 4) y  (2x 2 + a;  2) = 0.
(x + 4) N/(o;+'
*
(a; + 4) NSSar* + 2Qa;8
6
Taking the positive sign we have
50x + 20
dy _ 1 (
fa ~ 6 I
25*+ 10
243, 244] IMPLICIT FUNCTIONS 309
Now 25^ + 20a;8 = Qyx 4.
... ^=1/1 , 25a;+10l
dx 6 1 6y  as 4J
4o3 + y + 1
6y x 4 '
which agrees with the previous result. Similarly when we take
the negative sign.
244. Ex. Find the equation of the tangent at the point
(2, 3) on the curve
We have 4x 4 (x^ + y) + Uy ^= 0.
\ ttuC / W'iC
4a;
/. Gradient at (2, 3) =
fe 4x 14
812 2
.'. Equation of tangent is y 3 = y= (a; 2),
2x  17y + 47 = 0.
EXERCISES. LXXIV.
1. If x 2 + y 2 = a 2 , find ~ ; and find the equations of the tangents at the
ctx
3
points where x=a.
fl'fj __ __
Also find ~ by taking y = Jo?  x 3 .
ctx
a. if^ + Si, find ^.
a a o* dx
3. If a;?/ = ft 2 , shew that / =  y  . Hence shew that if the tangent at
ax x
P meets the axes in Q and R,
PO = PQ=PR.
310 CALCULUS FOR BEGINNERS [CH. X
4. If *V= **! shew that ^ =  ^ .
ax ox
Verify by writing y = .
dy
5. Find 2 ^ ax 2 + 2^a;r/ + 6t/ 2 =l ; a, ft. b being constants.
6. Verify that (2, 1) is a point on the curve
3x 2  4xy + 2j/2 _ Q x + i y _ 3 = Q,
and find the equation of the tangent to the curve at this point.
7. Find j if 2cosx + 3cosy = 4.
8. If x 2 + j/ 2  3x + lOy  15 = 0, find / . Use this to find the equations of
the tangent and normal at (4, 11) to the circle a; 2 + y 2 3a; + 10j/15 = 0.
Verify that the normal passes through the centre.
9. Find i
10. If x m y n =a m+*
shew that = ^.
ax nx
11. If Bin m x cos* y = const.
dy m
shew that ~ = cot x cot y.
dx n
CHAPTER XI
DIFFERENTIATION OF * e* log,*. HYPERBOLIC FUNC
TIONS. LOGARITHMIC DIFFERENTIATION. COMPOUND
INTEREST LAW. SOME DIFFERENTIAL EQUATIONS
245. HERE n is constant and the index x is variable.
With the usual notation put
y = w",
then y + Ay = ri**** = n* . n** t
.. &y = n*(n* x \\
Ay , n Aa! l
and T^ = w H 
Ao, 1 Ax
n* 3 * 1
Now Lt   is obviously a function of n and is in
Ax ^o Aa;
dependent of x.
.*. we have = N . n* or Ny, where N is a constant, ie. it is
ax
the same for all powers of the same number n.
246. We can find an approximate value for this constant in
particular cases, e.g. if n = 2,
.
Ax Ax
By logarithms we get
2' 2 l
= 743,
2
2" 1 !
= 718,
^*
^~
312 CALCULUS FOR BEGINNERS [CH. XI
The last two were got by using 7figure tables. With 4figure
2' 01 1
tables the best we can do is to say that ^ = '7 approximately.
With 7figure tables it is impossible to distinguish between
2' 0001 1
.'. we may say that if y = 2 Z ,
^= 693 x 2* approximately.
247. It is important to realise that this number is '693
because we are dealing with powers of 2.
Take the case when x = 5, y = 2 5 ,
if x = 51, y = 2"  2 5 x 2' 1 ,
Increase in y _ 2 8 ' 1  2 8 _ p 1
Increase in a; '1 _ !
Similarly if x = 5 01, y = 2 8l " = 2 s x 2' 01 ,
Increase in y _ _ 6 ~2' 01 1~
and T :  * x rpj I ,
Increase in a; \_ '01 J
and so on.
Take the case when x = 7, y = 2 7 ,
if x = 71, 2 / = 2=2 7 x2 1 ,
Increase in y 2 7  2 7 f2 "  1~
and :  =  T = 2 7 x .
Increase in x L * J
Similarly if *=701, y = 2 7 ' 01  2 7 x 2 fll
Increase in )/ r2 01 l~l
and :  = 2 7 x ,
Increase in a: L '01 J
and so on.
The quantities in square brackets are the same whatever
power of 2 we start from, and 693 is the limiting value to which
2' 1  1 2' 01  1
the series of fractions ^ , >Q . etc. continually approach
as the index * 0.
24G248] e 313
If we are dealing with powers of 3, MO takes the place of
693 [this being the limiting value to which the series of fractions
3'i  1 3' 01  1
q > .Qi etc. continually approach as the index * 0.]
And similarly for powers of other numbers.
EXERCISES. LXXV.
As in 246 shew that
1. If y = 3 z , ^=1'10 x C* approximately.
CLX
3. If y = 5 X , / 161 x 5 X approximately.
Q.3C
3 . If y = 25*, ^ = 092 x 2 5* approximately.
(UB
4. If y = 27*, ^' = 099 x 27* approximately.
5 . If y = 28*,  = 103 x 2'8* approximately.
248. The results of Exs. LXXV. 4, 5 suggest that there is a
value of n between 2'7 and 28 such that if y = n y> , J^=] *n x .
dtOts
This value of n is called e. Its value can be calculated to any
degree of accuracy required, by methods which have no place in
this book. Correct to 9 decimal places it is 2718281828.
dv
e is defined to be such that if y = 6*, then ~ = y.
cue
So far as numerical results are concerned, it will be suffi
cient for us to know that it is 2*7183 correct to 5 significant
figures.
EXERCISES. LXXVI.
2 7183 ooi_l
Shew that approximately ^TT = 1, and hence that if y = 27183*,
^=27183 a: approximately.
00
314 CALCULUS FOR BEGINNERS [CH. XI
249. We can now express the constant N in 245 in terms
of n and e.
d.P
First, since i = e ,
CHB
d.e d.ef d.ax
it follows that ; = =  . = = a . d.
ax a .ax ax
Now by the definition of a logarithm
n = d*',
.'. if y = n,
dy
i.e. f = iogt n x ,*.
So that N is log e n.
250. We have thus two very important results:
if y = e*, g = y,
if y = n",
where stands for the number 2718281828. ...
251. The results in Exs. LXXV. therefore give us the
following approximate values for logarithms to the base e.
No. log. to base e
2 693
25 92
27 99
28 103
3 MO
5 161
249252] e 315
252. Tables have been made giving logarithms of numbers
to the base e [called natural or Hyperbolic, or Napierian loga
rithms, the last from the inventor John Napier of Merchiston
15501617].
If these are not available, the logarithm of any number to
the base e can be obtained from the logarithm to the base 10 by
using the formula
Iog 10 n
log e n=^ .
log
This is a particular case of a general theorem, viz.
For let Iog 6 n = x and Iog 6 a = y,
:. n=b x ...(l) and a = 6*.. .(2). .
To get log a n we want n as a power of a.
i
(2) gives b = av y
f l\a> x
/. from (1) n= \av) = av,
x log b n
i.e. log a n =  ==.
y Iog 6 a
The special case when n = b is important. This gives
As we shall be mainly concerned with the bases e and 10 we
take the particular results got by substituting e and 10 for
a and 6.
Thus
and log, 10 =
Iog 10 e[i.e. log,, 27183...] is 43429,
and its reciprocal is 23026, each being correct to 5 significant
figures.
316
CALCULUS FOR BEGINNERS
[CH. XI
We thus have
log e n = Iog 10 n x 23026 and Iog 10 n = log, n x 43429,
e.g. log, 2 = Iog 10 2 x 23026
= 30100x23026
= 6931.
Graphic treatment.
253. Fig. 124 shews the graph of y = 2*.
70
60
50
30
20
10
Fig. 124.
252254] GRAPHIC TREATMENT 317
Now if we keep the same axes and make a suitable change in
the ccscale, this can be made to stand for the graph of y = n x ,
where n is any number whatever.
e.g. 4 = 2 2 ,
so that 4 X = 2*,
4 s = 2 6 ,
4 1 ' 5 = 2 s etc.,
the index when a number is expressed as a power of 4 being half
the index when it is expressed as a power of 2. Therefore if the
a;scale be as indicated in the second line, the graph is that of
Similarly since 8 = 2 s and 8* = 2 s *,
the graph will be that of y = 8 X if the jcscale be as indicated in
the third line.
Any number can be expressed as a power of 2, for instance
3=10' 4771 and 2 = 10' S01 ,
4771
/. 3 = 2' 3U1 = 2 158B .
.', if the cescale be as indicated in the fourth line, the graph
will be that of y = 3 X .
And so on for any value of n.
The ojaxis is in each case OX.
254. We shall now prove an important geometrical property
of the curve y = n x [Fig. 1 25].
P is the point (a, n a ),
Q is the point (a + h, n a+h ),
RN _ NP
PS~ SQ'
. RN_ n a I
" h ~n a+h n a w*l*
h
318
CALCULUS FOR BEGINNERS
[CH. XI
and is independent of a, i.e. of the position of P on the curve
but depends only on the length of the horizontal step PS.
R N M
Fig. 125.
EXERCISES. LXXVIL
Verify this property on the graph of y = 2*
(i) by joining the points for which x = 2 and a: =3 and also those for
which a; =4 and x = 5.
(ii) by joining the points for which x = and x = 2, also those for which
x=2 and x=4t, also those for which x = 3 and x=5.
255. Now let A*0 and we get the property that in the
curve y = n x the subtangent is the same at every point.
i.e. if NP be the ordinate of P and if the tangent at P meet
OX in T, then TN is the same for all positions of P on the curve.
[Fig. 126.]
Now if P be the point (x, y) and if TN = t, we have
y = t x gradient of TP
^y^. w here t is constant
ax t
254257
GEAPHIC TREATMENT
319
256. There is one of the curves y = n x for which t=\ t in
other words on some scale or other TN will be of unit length.
Y1
Let this particular curve be called y e",
Then as before e is defined by the property that
257. The numerical value of e can be found approximately
from the graph as follows.
Let A be the point on the graph the tangent at which goes
through O. [This point can be found fairly accurately with a
straight edge.] Then if AB be the ordinate of A and we choose
our orscale so that OB is unit length the graph will be y = d e ,
and moreover BA gives the value of y corresponding to # = 1,
i.e. the number of y units in BA is e.
From Fig. 124 we can see that e is something between 2*5
and 3.
From Fig. 127 which is part of Fig. 124 drawn to a larger
scale we see that e is between 2 '7 and 275.
320
CALCULUS FOB BEGIMNERS
[CH. XI
4

2 Scale for 2 X
1 4
e
257259] LOG e o; 323
1O. Prove that y^e' 3 * and y=e 3x sin2x touch each other at the
points corresponding to x= , ^,  , etc., and find the gradient where
11. Prove that the curve y=e" 3x sin 2x has turning points where
a IT a a .'Sir a
, etc.,
and points of inflexion where
 ir Sir
where a^tan" 1 .
13. On the same sheet plot between x=0 and x=2ir t
(i) y = e~' 3x , (ii) .y = s'm2x t (iii) y = e' 3z sin 2a;. [Fig. 129.]
Differentiation of
258. If y = \og e x,
x = e y ,
dx
This is a most important result. We deduce that
being the only power of x we have hitherto been unable to
x
integrate.
259. Ex, 1. If y = Iog 10 x, we use the fact that
Iog 10 x = log e x x 43429,
dy _ 43429
dx~ x
21 a
324
CALCULUS FOR BEGINNERS
[CH. XI
4
p"
\
7
Z?j
&7
f^
^
N<NJ
y
259] LOG. x 325
Ex. 2. If y = log e sina;,
dy d \og e sin x d sin x
dx rfsina; dx
1
.cosx
sin x
cot a:.
EXERCISES. LXXIX.
1. Illustrate the fact that if y = log,, x, ~ = , by filling op the following
table :
x y=log.x
2
x+Ax y + Ay Ax Ay
OdB
21
201
2001
[If you have no table of logs to base e, yon must say
log, 2 = Iog 10 2 x 2 3026. ]
Also make a similar table for x =6.
Illustrate also by reference to the graph of y=log.x. [Pig. 130.]
a. Find for the following values of y :
dx
(i) log. 3*, (ii) log,*', (iii) log. (3* + 5),
(iv) log cos x, (v) log. tan x, \ (vi) log. (3x 2 + ox + 1),
(vii)* log. sin 2x, (viii) log s x, (ix) log. tan ,
(x) log. (sec x + tan x), (xi) log, tan ( ^ +  j ,
(xii) log.^Tl, (xiii) log.{
(xiv) log.(a; + Va; 2 + a 2 ). (**) log. (x +
(xvi) log ( X 2 < a 2), (xvii) I
(Z J? 3/ ~~ fit
(xvjii) x log. x, (xix) x 2 log. x, (xx) log. (ax + b),
(xxi) log. (ax 2 + fex + c), (xxii) log. u, where u is any function of x.
326
CALCULUS FOR BEGINNERS
[CH. XI
3. Write down / ydx for the following values of y :
(iv) tan x,
(vii)
M
(xiii)
(xvi)
(rvii)
x+~2
(v) cosec x,
1
ax + b ' V
1
(vi) sec x,
(viii)
log e x, (xv)
[Get this equal to a; + 3 5. I
Fig. 130.
ft I /! nooi
4. Shew that /  dx = I  dx= dx,
J \x J 3 a; J so x
/b 1 f kb 1 h
da;= / dx=log,.
a * J ka 3 ' "
259 j LOG, ab 32?
6. Find the area bounded by
(i) xy = l, y = 0, x = %, x = 9,
(ii) xy = l, y = 0, x = 3, a: = 18.
6. Find the work done by a gas in expanding from 2 to 3 cubic feet, the
pressure and volume being connected by the law pv = const., given that the
pressure is 2160 Ibs. wt./sq. ft. when the volume is J c ft. [v. Ex. 4, 184].
7. In the last example find the work required to compress the gas from
(i) 30 cubic feet to 20 cubic feet,
(ii) 3 cubic inches to 2 cubic inches.
8. Evaluate
(o r dx
f 2 dX
J o >jx* + 9 '
IT
fs
(iv) / tan x dx,
4
IT
/I
cosec xdx.
4
ft. Find the c. G. of the area contained between xy = l, y = 0, x=l, x=4.
10. Find the C.G. of the area contained between x 2 j/ = l, y = 0, x = 1, x = 4.
11. Find the mean value of  between x = l and x = 2. [Cp. Exs.
XLVI. 3.]
13. P, Q are two points on the same meridian whose latitudes are x,
x + Ax (radians).
Shew that arc PQ=fc. Ax nautical miles, where k= .
7T
On a Mercator chart, taking as unit the distance representing a seamile
at the equator, the distance corresponding to PQ will be k . Ax multiplied
by some factor between sec a; and sec (x + Ax), and the whole distance from
the equator to a point B, latitude X, will be represented on the Mercator
chart by a distance / k sec x . dx.
Jo
Find the distance on a Mercator chart from the equator to a spot in
lat. 45 and compare with what you find in the table of meridional parts.
13. For values of x >  1
2y =x*  I2x + 30 log e (x + 2).
Find the maximum and minimum values of y, and evaluate them numeric
ally.
328
CALCULUS FOR BEGINNERS
[CH. XI
260. Another method of approximating to the value of e.
If * = lo.*i
and
r* 1
I cte = log e &
Jo, x
Put 6 = e and a = 1 and we have
e=
Draw the graph of y . Then the area bounded by the
3?
curve, the ovaxis and the ordinates x = 1 and a; = e is 1 unit of
area, i.e. it is equal to the area of the square OPQR [Fig. 131].
6
12 14
16
18 2 22
Fig. 131.
24 26 K28 3 X
If we can place an ordinate KL so that the^area OKLR = area
OPQR or area PKLS = area SQR the value of x given by OK is e.
Now it is obvious from the figure that e lies between 2'6 and
3 [mere counting of whole square inches is sufficient to settle
this].
260 j APPROXIMATION TO VALUE OF 6 329
We shall get a very good approximation to the area OABR by
drawing rectangles as in Figs. 68, 69, p. 163, each of width 01,
and taking the arithmetic mean of the sums of inside and outside
rectangles. [When x is positive, / is negative, and ^ positive,
cfcc CLXJ
so that the curve is like Fig. 40 (c), 94.]
The sum of the outside rectangles is
ni ri i i i
1 + 101
and the summation is easily effected by means of a table of
reciprocals.
Using 4figure tables [the numbers to be added are very
conveniently placed for addition] we get '01 x 958596.
The sum of the inside rectangles < the sum of the outside
rectangles by
.'. Area OABR < 958596 and > 952442,
a good approximation will be '955519 [as a matter of fact the
first 5 figures are right],
Now try OA'B'R where A'B' is x = 27.
Sum of outside rectangles will be greater than in previous
case by
* b ? 37810 '
and is therefore 996406.
The sum of the inside rectangles is less than this by
01 l r '006296.
.*. Area OA'B'R < '996406 and > 990110,
a good approximation is 993258.
330 CALCULUS FOR BEGINNERS [CH. XI
Now try OA"B"R where A"B" is x= 28.
Area OA"B"R < 1 '032839 and > 1026410,
a good approximation is 1*029624.
.'. e is between 2 '7 and 2*8.
Try 275.
Area < 1014789 and > 1008425,
a good approximation is 1011607.
Try 272.
Area < 1003800 and > 997476,
a good approximation is 1000638
Try 271.
Area < 1000110 and > 993800,
a good approximation is '996955.
.'. e < 272 and > 271 and apparently nearer to 2  72.
If we use a table of reciprocals which gives more figures and
divide the area into strips '001 ajunits wide, we shall be able to
shew that
e> 2718 and <2719.
Hyperbolic Functions.
A* c ft"***
261. DEFINITIONS. = is called the hyperbolic sine
A
gX 1 Q X
of x (written sinh x) ; ^ is called the hyperbolic cosine
px ^ ft""*" sinh iK
of x (written cosh x) ; x  x or  ^ is called the hyper
bolic tangent of x (written tanhx);  r is cotho;, r is
sech x % and . . is cosech x.
sinh a;
260, 261] HYPERBOLIC FUNCTIONS 331
EXERCISES. LXXX.
1 . Prove cosh 2 x  sinh 2 x = 1 .
[Just as the coordinates of a point on the circle x 2 +y 2 =a 2 mav t e
taken to be (acos0, asinfl), so the coordinates of any point on the
rectangular hyperbola x 2 y* = a z may be taken to be (a cosh 0, a sinh 0).]
2. Shew that cosh x is never less than 1, by writing it
3. Shew that sinh (  x) =  sinh x and cosh (  x) = cosh x.
4. On the same sheet draw (i) y = e x , (ii) y=e~ x , (ui) y = sinhz,
(iv) j/ = cosh.T between x =25 and a; =25.
6. Shew that tanh x lies between  1 and + 1.
Write it 1 
Shew also that tanh (  x) =  tanh x and draw the graph of y = tanh x
between x=2  5 and x = 25.
6. Shew that cosh 2* = cosh 2 x + sinh 2 x
= 2 cosh 2 a; 1
= l + 2sinh 2 o;
sinh 2x=2 sinh x cosh .c.
dcoshx
=inhx,
8. If y = A cosh x + B sinh x, shew that y^  n%y = 0.
9. What are
(i) / sinh a; dx, (ii) Icoshxdx?
10. Shew that
x sinh 2x
"2 + ~4~ '
+ 8i ^ X . [v.Ex.6.]

332 CALCULUS FOR BEGINNERS [CH. XI
11. If y=sinli~ 1 x[i.e. a;=sinhw], prove =
and if 
a
[Remember cosh y is necessarily positive.]
12. If y = sinh" 1 x, prove = x + *Jx* + 1 , i.e. sinh^x is the flame as
log (a; + >Jx* + 1). Similarly shew that sinh" 1  is the same as
log.
[Compare results of Exs. LXXIX. 2 (xiv) and Exs. LXXT. 11.]
13. If y=coli 1 x(a;>l), prove =
and if y=s
14. If y^cosh" 1 ^, prove e'=a: N /a:2_i j j >e cosh" 1 * is the same as
log,, (a; V* 2  1 ) or Io 3 (* + v/* 2  1), since
_
Similarly shew that cosh 1  is the same as log  .
[Compare results of Exs. LXXIX. 2 (zv) and Ezs. LXXX. 13.]
16. If y= taut 1 x(x 2 <l), prove d Z=_i_;
and If y = tanli" 1  (z 2 < o 2 ), prove = 2 ^ a .
16. If y=coth~ 1 x or tanh" 1 (x 2 > Improve =  5  ;
and if y = coth" 1  or tanh" 1  (a; 2 > a 2 ), prove =  5   .
17. If j/ = tanh~* x (x 3 <l), prove e 2 "= 1  , i.e. tanh" 1 * is the same as
X 3C
n lSe I  Similarly tanh" 1  (a; 2 <a 8 ) is the same as ^ log  .
Also if y=coth~ 1 (a; 8 >a 2 ), prove y = log  .
[Compare results of Exs. LXXIX. 2 (xvi and xvii) and Exs. LXXX. 15, 16.]
261263] LOGARITHMIC DIFFERENTIATION 333
262. Note. If y = sinh" 1 x, the quadratic for e v gives
e* = x \/ar a + 1,
but since e v is essentially positive, e v must be x+ *Jx* + 1 and
there is no ambiguity.
If y = cosh' 1 x,e v = x \Ac 2 1 and both values are admissible,
i.e. for any value of x (> 1) there are two equal and opposite
values of y such that cosh y = x. If we take the positive value,
~ will be , ; if we take the negative value, ~ will be
dx Vcc 2 ! <*"
. . These facts will bo readily seen from the graphs of
vie 2 1
y = cosh" 1 x and y = sinh" 1 x.
If we agree that cosh' 1 a; shall mean the positive value,
i.e. if we take cosh" 1 a; as being log e (x+ va^ 1) we may say
, d cosh" 1 x 1
1 4~ 00 / 1 + *C
If v = tanh~ 1 a;(ar ! <l), 6^ =   and e = + v /  , but as
la; "VI x
before, e v must be positive, so that
/I +x
6 =vr^
, I +x
and =
In this case therefore there is no ambiguity.
Logarithmic Differentiation.
263. The process of differentiating a product is often
simplified by taking logarithms before differentiating.
, dy .. (a; + 3) 3 (2a;l)
Ex. I. Find gif y = ^ (3 ^ 2)5 >.
We have
log y = 2 log (x + 3) + 3 log (2z  1)  5 log (3a: + 2),
1 dy _ 2 6 15
' ~
334 CALCULUS FOB BEGINNER'S [CH. XI
dy _ / 2 6 15 \
* ~~ ~
(7a;77)(a;
(3x + 2) 6
Ex. 2. If y = mw3 where u, v, w, z are functions of x,
log ?/ = log u + log v + log w + log z.
I dy I du I dv 1 c?w 1 dz
_ _ _ _ _ i _ _ i _ _ t _ _
y ' dx u ' dx v ' dx w' dx z ' dx '
dy _ __ du dv dw dz
a
(Cp. 219.)
~ = VWZ . 7= + WZU . rr + ZUV . r + UVW . j .
ax ax dx dx dx
EXERCISES. LXXXI.
Find ~ if
da
1. y =
a. y =
6. y=x t . 6. y=x x . 7. y = (sin a;) 00 '" 5 .
The Compound Interest Law.
264. Suppose money put out to compound interest at 5 "/
and suppose in the first instance that interest is payable yearly.
Then if x be the capital at the beginning of any year, the
increase in capital during the year is '05x.
If interest be payable halfyearly, the increase in capital
during any halfyear is Q5x x  where x is the capital at the
Z
beginning of the halfyear.
263265] COMPOUND INTEREST LAW 335
If interest be payable every th of a year, the increase in
n
capital during any period is 'Q5x x where x is the capital at
the beginning of the period.
Now suppose the increase in capital to be continuous and not
to increase suddenly at the end of stated periods.
If x be the capital at the end of t years, the increase (As;)
during the next interval A< is between
05x.A< and 05 (x + Ax) . A*,
Ai
. .   is between Q5x and 05 (a; + Ace) ;
Cu
**
and generally if interest is at r/
= r?)f) = kx (where k = Int. on 1 for 1 yr. at r/ ),
dx
i.e. x is a function of t such that y is proportional to x.
Civ
i.e. a; is a function of the form Ae w .
Now when t = 0, x = P [P being original capital].
.'. P = Ae = A.
1
.'. x = Pe** = Pe 100.
265. If y is a function of x such that
dx
y and x are said to be connected by a Compound Interest Law.
Notice if x increases in A.P., y increases in G.P., for suppose x
takes successively values
a, a + 6, a +26, ...,
336 CALCULUS FOR BEGINNERS [CH. XI
the corresponding values of y are
and these form a G.P. with common ratio e**.
Notice also that A is the value of y when x = 0.
Examples of the Compound Interest Law.
266. (1) Newton's Law of Cooling. Under certain
conditions the rate of fall of temperature of a cooling body
is proportional to the excess of its temperature above that of
surrounding bodies.
i.e. if be this excess of temperature at the end of t seconds,
A0 _ , n ("the temperature of surrounding bodies']
dt ~~ being supposed to remain constant
where k is some constant. [The sign is to indicate that
decreases as t increases.]
From this we get = Ae" 6 *,
where A is some constant.
The constants A and k can be determined experimentally by
making two observations.
e.g. Suppose when t=W, 0=17"
and when t = 30, = 12J'
then 17 = Ae~ l
12 = Ae 30 *J
17 s
A 2 
12'
.. A = 202.
And j2 = ^*>
'* A = 20 10ge l2
= 017.
/, $ = 202e<>.
265, 266] COMPOUND INTEREST LAW * 337
(2) Atmospheric pressure. Let the pressure be p Ibs. per
sq. ft. at a height h ft. above the earth's surface, and let (p + &p)
be the pressure at height h + A/t [Ajo is negative if A/i is positive].
Also let wlbs. be the weight of a cubic foot of air at pressure p
and (w + Aw) its weight at pressure (p + A/)).
Then Ap is the weight of A h cubic feet of air whose average
density is between w and w + Aw.
i.e. Ap is between wA/i and (w + Aw) AA,
dp = _ w
Now if the temperature be supposed constant w varies as p
by Boyle's Law, or w = w , where p Ibs./sq. ft. is the pressure
and w Ibs./c. ft. the density at the surface.
dh P(,
and when A = 0, p=p , .'. A=JO O .
_w^
' P = Po e *>
e.g. if the pressure at the earth's surface be 2100 Ibs./sq. foot
and the density of air at the surface be '08 Ibs./cubic foot, and
if the temperature be supposed the same from the surface to
height A,
p = 2100eoooo4A.
(3) A body is moving in a straight line and the retardation
at any instant = kv ft./sec. a , where k is constant and v ft./sec. the
speed at that instant.
If v ft./sec. is the speed when < = find formulae for
(i) the speed at the end of t sees.,
(ii) the distance travelled in t sees.
M. o. 22
338 CALCULUS FOR BEGINNERS [CH. XI
We have y = kv.
at
where A is constant ; and v = v when t 0, .*. A = v
:. v = v e w , (i)
k
and s = when t 0, .". c = y
267. We have found in 264 that the amount at the end
of t years of P at Compound Interest, the increase in capital
being supposed continuous, is
Pe** where ^ = rKA
This was obtained by consideration of the increase of capital
in a small time A, which led to the differential equation
dx 7
We shall now consider the question from a different point of
view.
Suppose interest paid at intervals of th of a year. Then if
x be the capital at the beginning of any interval, the capital at
/ &\
the end will be x ( 1 +  ) .
\ n)
266, 267] COMPOUND INTEREST LAW 339
Thus we have Original capital = P.
. *. Capital at end of first interval = P{ 1 +  ) .
\ nj
' 2nd  ( 1+ s)( I + s)
(it*
,, 3rd
etc.
t years
The amount when the increase in capital is supposed continuous
is the limit of this when x is increased indefinitely.
(k\ nt
1 +  ) when n is increased indefinitely
, ja,
1 > O
Special cases.
>fc=land=l, IA1+
*\
To illustrate the meaning of this last result calculate by
loarithms
(1 \50 / 1 NlOO / 1 \1000
1 + 50J ' ( l+ TOO/ ' ( l + TOOOJ '
You should get
2, 225, 237, 244, 2594, 2692, 2705, 2717.
The theorem is that as n is increased we get nearer and
nearer to e (2'7 182818...) and can get as near to e as we like by
taking n large enough.
222
CALCULUS FOR BEGINNERS [CH. XI
EXERCISES. LXXXII.
1. A body starts 2 feet from O and moves so that when it is x feet
from O its speed ia 3a; feet/second.
Find a formula giving its distance from O at the end of t seconds.
Find the distance travelled in 3 seconds and the time taken to travel
100 feet.
a. A rope passes round a drum, centre O. A and B being the points
where the rope leaves the drum.
P is a point on the drum such that L AOP = radians.
The tension of the rope varies according to the law
r=^iT where u. is the coefficient of friction.
act
If TO is the tension at A and TI at B, shew that TjsToe'*" where a is the
angle AOB.
If n='7, T =20 (Ibs.wt.), arcAB = lfoot and diameter of drum = 1 foot,
find Tj. Also find at what point the tension is 40 Ibs. wt.
3. A cistern full of water has a leak in the bottom. Assuming that the
rate at which the water escapes is proportional to the pressure, shew that
the rate at which the height of water in the cistern diminishes is proportional
to the height
Ddx . 
ue. 3 *J.
If the height of the cistern is 10 feet, and the level falls 1 foot in the
first minute, when will the cistern be half empty ?
4. Suppose a flywheel, moment of inertia I Ibs. ft. 2 , rotating in a fluid
which produces a resisting torque cw Ib. ft., where u is the angular velocity;
shew that the angular velocity at the end of time t is given by
W=Wo t *i
where k is, & constant, and UQ is the initial angular velocity, and find k in
terms of c and I.
If the wheel is initially rotating at 200 revolutions a minute, and after
a minute at 100 revolutions a minute, after what time will the number of
revolutions per minute be (i) 50, (ii) 25, (iii) 20 ?
267]
341
5. Mallock's formula for the retardation due to air resistance of a
projectile is k(v850), where t; ft./sec. is the speed and & a constant
depending on the form and weight of the shell. If V stands for (w850),
dV
shew that the acceleration is r
at
Hence get a formula giving the speed at the end of t seconds.
6. A body is moving in a straight line and the retardation at any instant
is 0'03 ft./sec. 2 , where v ft. /sec. is the speed at that instant. If the initial
speed is 15 ft./sec., find the distance of the body at the end of 2 minutes
from its position when t=0 and shew that it can never reach a point 50ft.
from that position.
7 A rod hanging vertically carries a weight of 100 Ibs. The weight of
the rod is 15 oz. per cubic inch. Find the law connecting the crosssection
with the distance from the end if the tensile stress is everywhere
400 Ibs./sq. in. (Fig. 132.) \y sq. ins. is area of crosssection x ins. from
lower end.]
400
342 CALCULUS FOR BEGINNERS [CH. XI
EXERCISES. LXXXIII.
1. If y = A cos nx + B sin nx, where A and B are any constants whatever,
shew that
_
dx 2
8. If 7/ = Ae n *+ Be'** or if y = A cosh nx+ B sinhnx, shew that
*.
dx 2
3. Write down solutions (each involving two arbitrary constants) of
, S + 4,0. (u) g4,=0.
4. If 5 + 4w = and if, when z=0, y = 5 and = 4, find in terms
ax ax
of x.
6. If =^4y = and if, when z=0, t/ = 5 and ^=4, find y in terms
uX CtJC
of x.
6. A body moves in a straight line and ft. is its distance from a fixed
point O in the line at the end of t seconds. The acceleration is 9s ft./sec. 2
towards O.
When t=0, the body is 15ft. to the right of O and when = ^,it is 20ft.
to the right of O.
Shew that s = 15cos3 + 20sin 3t
3
= 25 sin (3t + 0) , where 6= tan' 1 ^ .
Hence shew that the body oscillates backwards and forwards between
two points 25 feet on either side of O, and that the time of a complete
oscillation is seconds.
O
What is the speed of the body
(i) when t = 0, (ii) when it passes through O?
7. Ify=ze~ ix , where z is a function of x, shew that
and hence that if z A cos 2x + B sin 2x,
or y = e~ 2x (Acos2x + B sin 2x),
267] DIFFERENTIAL EQUATIONS 343
8. If y ze^ x , where z is & function of x, shew that
and hence that if z = Ae& + Be ~ % x or y = Ae 31 + Be 2 *,
d?y _ dy .
cA 5 + *y= Q '
9. If y =ze~ Sx , where z is a function of x, shew that
3+Z+*
and hence that if = Az + B or y = e~** (Ax + B),
1O. If y = ze~P x , where z is a function of x and j>, q are constants,
shew that
If ? +2p ~ +qy = 0, what value should 2; have
da; 2 do;
(i) ifg>p 2 , (ii) if2<p 2 , (iii) ifg=
11. By the device indicated in Exs. 7 10 get solutions of
* +**

1 2 . Find A so that if y = Ae 2 *,  5y = Se 22 .
d 2 ?/
13. Find A so that if y = A sin 2x, ^ + 5y = 3 sin 2s.
14. Find A and B so that if y = A sin 6ar + B cos 6x,
f + 4 ^
das 2 dx
344 CALCULUS FOR BEGINNERS [CH. XI
1. If y = xe**, shew that ^  5 ^ + 6y = e 3 *.
CuC* (IX
!. Find A so that if y = Aa2z, \  4y = 3e x .
17. Shew that   + 2pp + g'?/:=r ) where p, q, r are constant?, is the
uX (tX
same as 55+20 r + g=0, where M is (11 ); and hence that the solutions
ax 2 dx \ qj
of ^ + 2p j= + qy =r are got by adding the constant quantity  to those of
dx GLX q
18. Write down solutions of
267] MISCELLANEOUS EXAMPLES 345
MISCELLANEOUS EXAMPLES ON CHAPTERS IXXI.
A.
1. A ladder 20 feet long has one end on the ground and the other in
contact with a vertical wall. The lower end slips along the ground. Shew
that when the foot of the ladder is 16 feet away from the wall the upper end
is moving 1 times as fast as the foot.
IT
I. Evaluate I (7 cos02sm0 + l)d0,
}dx.
3. A point moves in a straight line according to the law r =
Prove that the distance described from t=0 until it comes to rest is
6 feet.
4. A variable torque produced by a couple of moment M sin Ib. ft.
acts on a shaft, M being constant and 9 the angle turned through by the
shaft. Find the work done in turning the shaft from 0=0 to G=ir.
' 6. Find the mean value of tan x between
TT 2ir
x=7 and x= 7r .
b 9
Check your result by finding the Arithmetic mean of
tan 30, tan 31, ... tan 40.
B.
f?V
1. Find j* in the following cases :
x
_
(i) y = sm 2 3a;, (ii) y = x a cos 4x, (iii) j^log^
(iv) y = s*sinfx+V (v) x*2xy 3y2=0.
346 CALCULUS FOR BEGINNERS [CH. XI
L
2. Write down the values of
(i) / e ' 3* dx, (ii) I (2 sin 3* + 4 cos 6x) dx,
dx
m I 4
J o
3. A rod 10 feet long elides with its ends A, B on two lines at right
angles which intersect in O.
If A has a uniform speed of 3 ft./sec., find the speed and acceleration of
B when A is 8 ft. from O.
4. If y =xe~ mx (m positive), find the maximum value of y.
6. O is the centre of a circle, radius 2a. From a point P inside the
circle, PB and PC are drawn to the circumference each of length a. If
OP=x, shew that for the figure PBC (bounded by PB, PC and arc BC) to
have maximum area
0.
1. If
x
find ^ and find the length of the tangent intercepted between the curve and
dx
the yaxis for the point where x=h.
9. Find the area bounded by the curve
y = sin 2 , j/=0, ar=0, *=1,
I
(i) by integration, (ii) by Simpson's rule (11 ordinates). Draw a figure.
3. The height of a tower is calculated from the observation of the
elevation 6 at a distance d ft.
If d = 200 and = 35, find the consequent error in the calculated
height due to (i) an error of 1 inch in measuring d, (ii) an error of 10' in
measuring 0.
4. A chord divides a circle into two segments of heights hi, h^. If A
is the area of either segment and the angle which its arc subtends at the
centre, prove
267] MISCELLANEOUS EXAMPLES 347
6. In a simple horizontal engine a is the length of the crank and I
the length of the connecting rod.
Shew that when the crank makes an angle B with the line of dead centres
the angular velocity of the connecting rod is
a cos
where u is the angular velocity of the crank.
D.
... _,. dy ,
1. (i) Find when y = lo
x j
a > Find
3. Find the intersections of
y=^2 with y=lt
and find the equations of the normals at these points.
3 . Sketch roughly the curve y = 3 coax 2 sin x between # = 0andx=:.
39
Find the area bounded by the axes and that part of the curve which lies
between =0 and the point where the curve first cuts the xaxis.
4. The blade of a fan consists of a uniform circular disc, centre O, from
which a small portion has been cut away by a chord AB equal to the radius r.
Find the distance of the c.o. from AB.
6. The thickness x at the distance r from the centre of a disc of varying
thickness is given by x=be~ ar . Find the volume of the disc, if its radius
is R. What is it in the special case when R = 4, 6 =5, a= '08?
E.
1. Find f'(x) and f"(x) for the following values of / (x) :
(i) xlog e x, (ii) e*sina;, (iii)  , (iv) sin**.
348 CALCULUS FOR BEGINNERS [CH. XI
2. Draw a triangle BAG, the angle A being about 120, and produce
AC to D. Let AD represent the plan of a door hinged at A, which is being
shut into a position along BA produced by a rod BC turned round the fixed
point B by a spring and sliding on the door at C.
If the angular velocity of the door is u, find the velocity with which the
end C of the rod is sliding along the door at any instant.
3. If a;=acos 3 0, y = asin 3 0,
express j= and 3^ as functions of 0.
dx dx 2
4. c is the distance between the centres of two spheres, radii a and b.
Find a point in the line joining the centres so that the total spherical surface
visible from it shall be a maximum.
[Surface of segment of sphere = 2wr h, where r is radius of sphere and /
height of segment.]
6. The half water line section of a small boat is bounded by the curve
t ranging from to 10.
Determine the area of the water line section.
P.
1. Differentiate with respect to x.
=  % , cos 3 2x, * sin bx, log. tan ( 2x 4 . } .
x*x+l \ 4/
If 8=be nt +ce~ nt ,
where b, c, n are constants, shew that the acceleration is proportional to the
displacement.
3. In the catenary
or y=ccosh ;
shew that the length of the perpendicular let fall from N, the foot of the
ordinate PN, upon the tangent at P is of constant length.
Also if the normal at P meet the axis of x in G, shew that PG varies
as PN.
267] MISCELLANEOUS EXAMPLES 349
3. A stick 3 feet long rotates about one end in a vertical plane with
uniform angular velocity, making one complete revolution in each second.
A light is 6 feet vertically above the centre of rotation and casts a shadow
of the stick on the floor which is 6 feet below the centre. Find the speed
of the end of the shadow when the stick makes an angle of (i) 30, (ii) 60,
(iii) 90 with the downward vertical.
4. Shew that the radius of gyration of a triangular lamina about an
axis through its o.o. perpendicular to its plane is
6
5. A torque acts on a shaft. When the shaft has turned through an
angle the torque is G sin sin (0  a) lb.ft., where G and a are constant.
Find the work done in a complete revolution.
G.
1. Find 3^ for the following values of y :
dx
(i) 2 i ^ f (ii) (23x2)4, (iii) log. (2***),
(iv) tan4x, (v) xe" 3 *, (vi) sin
2. A lighted candle is raised above a horizontal table. Find the height
for maximum illumination of the surface at a given point, whose distance
from the point vertically under the candle is a feet.
[If A is the flame, P the point, 6 the angle between AP and the normal
to the table at P, the intensity of illumination varies directly as cos 6 and
inversely as AP 2 .]
3. The roof of a house is inclined at an angle 60 to the horizontal and
the height of the wall is h feet. A ladder whose length is I feet rests with
one end on the ground at a distance y feet from the foot of the wall and the
other end on the roof at a distance x feet from the top of the wall. If the
inclination of the ladder to the ground is , prove ^ =    ^ tan 6.
dx &
350 CALCULUS FOR BEGINNERS [CH. XI
4. Evaluate the following integrals :
(i) J ***<?*, (ii) P Ldr, (iii) j\os(lx)dx,
6
(iv) I 1 x (1  a?)* dx, (v) /V**d* f (vi)
6. A thin uniform rod OA, 6 feet long, swings in a vertical plane about
a horizontal axis through one end O. Express the velocity of a point in the
rod distant r feet from O at the instant when the rod makes an angle & with
the vertical, in terms of . Hence write down an approximate expression
at
for the kinetic energy of a small portion, length Ar, of the rod, and fiud by
integration the total kinetic energy of the rod at this instant. If the rod
falls from the horizontal position, shew that when OA is vertical, the velocity
of A is 24 feet/sec, nearly.
H.
1. Find ~ for the following values of y :
(1)

(2) r (3) (2x 3) (3x2^ + 1) (4x3 2x2 3),
(4) tan 5*. (5) ^
(6) sec 2 3x, (7) * (a 2 + x 2 ) Ja^Hfi.
2. Find the maximum and minimum values of
sin x sin 2x between x = and x=2ir.
Sketch the curve y = sin x  sin 2x.
3. OA, AB are the crank and connecting rod of a steam engine,
OA rotating about O and B sliding along a fixed line through O. If
OA = a, AB = c, OB=r, / BOA = 0,
dr ar sin 6
express cos0 in terms of r, a, c and shew that J7 .= .
dO ra cos
O A = 3^", A B = 12", / O A B = 90, angular velocity of O A = 100 revolutions
per minute; find velocity of B.
2.67] MISCELLANEOUS EXAMPLES 351
4. The motion of the needle of a galvanometer is given by the equation
6 = &e~^ t sin3f, where is the angle in radians made by the needle with the
zero position at the end of t seconds. Find the angular velocity of the needle
at time t and shew that the extreme excursions to the right and left of the
zero position occur at intervals of  seconds and that the angles corre
o
IT
spending to these extreme excursions form a a. P. of common ratio  e 6.
irx
5. The equation of a curve is y = b sin 2 . Find the mean height of
that portion for which x lies between and a.
I.
(2x ir\
T + 4 )'
dy , d?y
find  and r4,
dx dx 2
and shew that T^~ 6 ^ + 7ry = 
dx 2 dx 9 J
2. A crank CP revolves uniformly n times a minute and the other end
A of a rod PA moves to and fro in the straight line AC.
If CP=aft., PA = 2aft., CA=a;ft., shew that
(1) sin = 2 sin 0,
(2) x = a cos 6 + 2a cos $,
where is L ACP and <j> is / CAP.
Hence get ^ and = in terms of and if>.
dt dt
What are the angular velocity of AP and the speed of A when n=40,
o = l, 0=39?
3. On the same sheet trace the curves
between a; = and x = l, using as large a scale as possible. Find the slope of
each at the points where x = 0, !, '2, 4, 6, '8, 1.
352 CALCULUS FOR BEGINNERS [CH. XI
4. Find the work done in slowly compressing a gas from a volume of
10 c. ins. to 6 c. ins., the initial pressure being atmospheric and the
temperature being kept constant. Take atmospheric pressure to be 1475 Ibs.
per sq. inch. [pv = const.]
If the compression takes place under adiabatic conditions, i.e. so rapidly
that no heat is lost, find the work done. [In this case pv l ' tl = constant.]
5. Draw a figure to shew that
Hence prove
 w w
f 2 . , fa 1 [2 TT
sm^xax= I cos^x.ax=^ l.ax = .
JO JO 2jo 4
I. (i) If
prove ^ = *.
ax y
Hence prove that if the normal at any point P on x 2 t/ 2 =a 2 meet the
xaxis in G, PO=PG.
sin nx
(ii) Shew that y= 
7*21
is a solution of r^ + V = sin nx.
dx*
a. Given log, 6 = 17918, find log.,61.
3. A chain hangs in the form given by the equation
\ v*
c J or 7/ = ccosh,
the axis of y being vertical.
Find the area of the figure enclosed between the curve and the chord
which joins the lowest point of the curve to the point where x=2c.
4. Find the C.G. of a circular segment whose arc subtends an angle 2a
at the centre of a circle of radius r.
Find also the C.G. of a sector of angle 2a and of an arc of angle 2a.
5. The sides of a triangular lamina are 3, 4, 5 feet. Find the radius of
gyration about the axis of revolution of the solid formed by the revolution of
the triangle about the longest side.
267] MISCELLANEOUS EXAMPLES 353
E.
1. If p = ab m + ne where a, 6, m, n are constants [De la Boche's
formula connecting vapour pressure and absolute temperature], shew that
dp _ mp log b
d0
2. If the equation of a curve is xy+y + x=Q, find the equations of the
tangent and normal at the point where a; =4.
3. P moves in a straight line AB and O is a fixed point. Shew that
the angular velocity of P about O at any instant is  , where r ft. is the
length OP, v ft./sec. the velocity of P at the instant, and 6 the angle between
PO and AB.
4. If a plane area [A sq. ft.] revolve about an axis in its plane which
does not intersect it, the volume generated is Al c. ft., where I ft. is the
length of the path traced by the o.o. of the area. [Theorem of Pappus, end
of 3rd century A.D.]
6. Find the area included between the curve
x 2 y = x 3 + a 3 ,
the xaxis and the ordinates x a, x = 2a.
Find also the coordinates of the C.G. of this area.
L.
1. If
where A is a constant, prove
8. If (*  a)* is & factor of / (a;), shew that (x  a)"* is a factor of /' (x).
Hence given that 3x 3  1x*  8x + 20 =
has two equal roots, solve the equation.
3. OA, OB are the bounding radii of a quadrant of a circle, PQRS is a
rectangle having one corner (P) in OA, one (Q.) in OB and two (R, S) in the
arc AB. Find the maximum area of this rectangle. [Work in terms of
/COR, C being the midpoint of arc AB.]
M. C. 23
354 CALCULUS FOR BEGINNERS [CH XI
4. A point in a certain mechanism moves so that its position at the end
of t seconds is given by
irt 2rrt
x + 4 =3 cos + cos ,
o o
. irt
J/ = sm T .
Find the position, velocity and acceleration of the point when t = 2.
[Distances are in feet.]
6. The xaxis is taken along the axis of the cylinder of an engine sc
that the piston moves between x=0 and a; = 10 [unit 1 inch]. The thrust on
the piston is 81 Ibs. wt. from x=Q to * = 6 and 1000.E 1 ' 4 Ibs. wt. from x G
to a; =10. Find the work done in one stroke in ft. Ibs. wt.
Also draw the graph shewing the relation between the thrust and x, and
get the work done by counting squares in the appropriate area.
M.
1 . Find the differential coefficients with respect to x of
2. A point moves in a plane so that at end of t sees, its coordinates are
x = t, j/ = sin2jrt, the unit being 1". Draw the path from t = to t = 2.
Find the angle the direction of motion of the point makes with the xaxis at
tune t = 0'2. Find area contained between xaxis and the path from t = to
t=05.
3. A and B are two sources of heat, 20 feet apart. The intensity of A
is twice that of B. Find the position between them where the combined
heating effect is least.
4. If V
at
and C = 1008in600t,
R being 2 and L being O'OOS, find V in the form
5. A point is moving over a straight line of length 2a with simple
harmonic motion, period T seconds. Find its speed and acceleration at a
time t seconds from rest. If V be the maximum value of v, shew that
r
Jc
267] MISCELLANEOUS EXAMPLES 355
N.
1. Find equations of tangent and normal to ^ + ^ = 1 at the point (h, k).
Find the distances from the origin at which this tangent and normal
cut the xaxis and the product of these distances.
2. A ladder has one end on the ground and the other projects over a
vertical wall 10 feet high. The lower end slides along the ground with
uniform speed 1 foot per second. Find the angular velocity of the ladder
when its foot is 10 feet from the wall, the motion being supposed to take
place in a plane perpendicular to the wall.
3. A straight road runs along the edge of a common and a person on
the common, at a distance of 1 mile from the nearest point (A) of the road,
wishes to go to a distant place (C) on the road in the least time possible.
If his rates of walking on common and road are 4 and 5 miles an hour
respectively, prove that he must strike the road at a point B, 1$ miles
from A.
IT
n fi
4. Evaluate (i) cos (3  2x) dx, (ii) sin x. cos 2 x dx,
Jo JO
It TT
/"2 /"30 /"'1047
(iii) ' sm*xGOs 2 xdx, (iv) sinx.dx, (v) xdx.
Jo Jo Jo
Account for the fact that the last two results are approximately equal.
6. CA is a radius of a circle, O its midpoint. OB perpendicular to
CA meets the circumference in B. The head of a shell is formed by the
Devolution of the arc AB about BO. Find its volume if CA = 2a.
[If P be any point on the arc, work in terms of / ACP (=0).]
O.
1. Find ^ if
dx
(i)
(ii)
(iii) ?/ = 3u 2 7u + 2 where u=2x s + 3x + 2,
(iv) sin x = log, y.
. d(3w 2 7w + ll) , .. dx .... dsinx
Fmd W    
232
356
CALCULUS FOR BEGINNERS
[CH. XI
2. (i) If y=xlogx, shew that y is least when x = ~.
6
(ii) If y= j , shew that y is greatest when x=e.
3. (i) Shew that v is acceleration.
at
(ii) If v 2 =a + fc, prove that the acceleration is constant.
(iii) If t> 2 =as 2 + 6s + c, prove that the acceleration varies as the distance
from a certain fixed point in the line of motion.
(iv) If s = ~vt, prove that the acceleration is constant.
S3
4. A curve whose equation is xy = constant passes through the point
(4, 5). Find the area of the figure bounded by the curve, the axis of x and
the two ordinates x =4, x = 12.
Find also the position of the centre of gravity of this area and the radius
of gyration of the area about the xaxis.
Find the mean value between 0=0 and 0= of
a
(i) sin 6,
(iv)
(ii) cos 6,
(v) cos 2 6.
(iii) sin 6 cos 6,
P.
1. Find 3^ when y is
ax
(i) x 2 sin x,
tana:
(ii) a;^ 2 *,
(v) ^a; sin x.
(iv) x ,
(vii) ^/sin >Jx.
(iii) e * cos 2 6x,
'l + cosa;
3. A line is drawn through (1, 3) cutting the axes in A and B
respectively, so that OA . OB is a minimum. Find the equation of the line.
3. A body mass m Ibs. is moving in a straight line. If at the end of
t sees, its displacement from some standard position is < ft., its speed
v ft. /sec. and its acceleration a ft. /sec. 2 , and if k footpoundals be the
kinetic energy, shew that
dk , dk
j=ma and = ma.r.
d dt
267] MISCELLANEOUS EXAMPLES 357
4. If a pistonrod drives a crank of length r through a connectingrod
of length I, prove that its acceleration at the end of the stroke is u*r ( 1  ) ,
where u is the uniform angular velocity of the crank.
6. A cart weighing 200 Ibs. containing 400 Ibs. of sand, ascends a
straight hill rising 40 feet vertically altogether. The sand is assumed to
be thrown out uniformly so that the cart reaches the top empty. Find
by integration the whole work done against gravity in the ascent.
Q.
1. Give equations of tangent to x 2 + 2y*=5 at each of the points where
2/=i.
2. AB is a straight line ; P, Q two given points on opposite sides of it,
PM, QN are perpendicular to AB. PM =p ft., QN =q ft., MN =a ft. R is
any point in AB such that MR=a; ft. If the speed of light on the Pside
of AB is u ft. /sec. and on the Qside v ft. /sec., find the time light would take
to travel from P to Q by the path PRQ. If R' be the position of R for
which this time is the least and if PR', QR' make angles 0, <f> with the
normal to AB, prove
sin u
3. P is any point (h, Jc) on the curve y = be a . PN is the ordinateof P, PT
and PG the tangent and normal meeting OX in T and Q.
ft 2
Prove NT constant and NG = .
a
4. In a tangent galvanometer the current is proportional to the tangent
of the angle of deflection (c = fctan &). Find the approximate error in the
inferred value of the current if an error A0 is made in reading the deflection.
Shew that the relative error in the current ( J is approximately '
and for a given error in the deflection is least when 0=45.
6. Prove that the area common to the two parabolas
is ^(a + b)Jab,
and find the o. o. of this area.
358 CALCULUS FOE BEGINNERS [CH. XI
R.
1. The equation s=ae~ M sin=
gives the displacement (s feet) of the end of a stiff spring from the position
of equilibrium at the end of t seconds.
If a = 10, T = '9, b ='75, find the speed and acceleration of the end of the
spring at the end of 5 seconds.
V 2 V  12
2. Shew that rr~3 g ^ has a minimum value when V = 6 and that
it has no maximum value.
3. In a simple horizontal engine the crank is r feet and the connectiug
rod I feet long. At any instant the crank and connectingrod make angles 6
and <j> respectively with the line of dead centres. If the angular velocity of
the crank be constant and equal to w, shew that the angular velocity of the
connectingrod is
* .'l2i " RP,
lcos<(> I
where P is the end of the crank and R the point where the connectingrod
meets the perpendicular to the line of dead centres through the centre of the
crank circle.
a v
fZ [4
4. Find (i)  sin 3x cos 2x dx, (ii) I cos 3x sin 2x dx,
(iii) I cos Bx cos 2x dx, (iv) I sin 3* sin 2x dx ;
Jo jo
and shew that if m and n are whole numbers
r
sin mx cos nx dx =
/O
2m
/sii
according as (m + n) and (mn) are both even or both odd.
 2 2
Shew also that
fir frr
I sin mx sin nx dx = I cos mx cos nx dx = 0.
Jo Jo
6. Draw roughly y = sin* from x = Q to x = ^ .
Consider the area bounded by the curve, the xaxis and the ordinate
267] MISCELLANEOUS EXAMPLES 359
Divide the base into 10 equal parts. Shew that the area lies between
919 and 108.
Divide the base into n equal parts each ~ or h.
&n
Shew that the area lies between
ft . if . fif h\ h . IT . ITT h\
sm^sm^.J and _sm 3 n (  + ) .
sin  am 
Then shew that when h is indefinitely diminished the limit of each of
these is
2 sin 2 T or 1.
4
IT
f*
Compare with what you get by finding I . sin xdx in the ordinary way.
[Assume sin a + sin (a + /3) + sin (a + 2) + . . . to n terms
' J
s.
1. (i) If
shew that
(ii) If q =
shew that 5f6^
at* at
2. OX, OY are two lines at right angles to each other. A, B are two
points 5 feet apart in a rod which is constrained to move so that A is always
in OX and B in OY. Supposing A to move along OX at a uniform rate of
1 inch per second, find the rate of motion of B along OY and the angular
velocity of AB at the instant when OB is 3 ft.
3. The coordinates of any point on a cycloid being given in the form
x=a (0sin 0),
y = a(icoa 6),
shew that the whole area bounded by one complete arch and the line on
which the generating circle rolls is 3ira 2 .
360 CALCULUS FOR BEGINNERS [CH. XI
4. A submarine cable consists of a circular core surrounded by a con
centric circular covering. The speed of signalling through it varies as ^
x*
where x is the ratio of the radius of the covering to that of the core.
Find x so that the speed may be a maximum.
5. Find the centre of pressure of a circular area (radius r) with its
plane vertical and its centre at a distance c below the surface (or).
T.
C /v*3 i y8
1. Find
m I x * +a * dx
(ii) 1 cot xdx,
ir
f*
(iv) I tun 2 a; da;.
() J 2*
It
[2
(Hi) I sin 2 2xdx,
(v)
[0
2. Water is poured at a constant rate into a conical glass, which is filled
in 2 minutes, the height of the cone being 12 inches. At what rate in
inches per minute is the surface of water rising (1) when the glass is filled
to half its height, (2) when half the liquid has been poured in ?
3. The equation to the Folium of Descartes is
Shew that any point given by
3am
lies on the curve.
_.. , du , dx dy
Find r^ and , and hence / .
am am ax
Find the equations to the tangents at the points where m=  and m = l.
Draw the curve carefully for positive values of m taking a=l. [Unit 1".]
Find j in the above otherwise.
dx
267] MISCELLANEOUS EXAMPLES 361
4. If = A sin n + B cos n, where A, B, n are constants, find the
maximum kinetic energy, the spaceaverage and the timeaverage of the
kinetic energy for one journey between the extreme positions and shew
that the three results are proportional to 6, 4, 3.
6. The crosssection of an I girder has the following dimensions :
Top Flange 6x2 ins.
Bottom Flange 15 x 2 ins.
Web 18 x 2 ins.
Find the radius of gyration of this section about an axis in its plane
through its C.G. perpendicular to the web.
CHAPTER XII
APPROXIMATE SOLUTION OF EQUATIONS
268. SUPPOSE we wish to solve
x3 _ 6x 2 + 9x  1 = 0.
The ordinary graphical method is to draw the graph of
and find where it cuts the a;axis.
The degree of accuracy attained will depend on the care
expended in drawing the graph and on the size of the scales used.
We have seen that a knowledge of the gradients at different
points is a great help in drawing the graph.
Put f (x) =a?Ga? + 9xl,
then / / (*) = 3* a 12a; + 9=3(a;l)(a:3).
Make a table of values :
x 01234
/() 1 3 11 3
/'(*) 90309
If we draw the graph (Fig. 133) the roots appear to lie
between (i) and 2, (ii) 23 and 24, (iii) 35 and 36.
269. If we want better approximations, it is better to
work at each root separately.
268, 269]
EQUATIONS
3(53
Take the first root, that is the one lying between and 2.
x 2
/(*)  1 568
/' (x) 9 672
Fig. 133.
P, Q (Fig. 134) are the points (0,  1) and (2, 568). PS, QT
are the lines through these points with gradients 9 and 6 '7 2,
i.e. they are the tangents at P and Q.
It is evident from the figure without actually drawing the
curve that it cuts the axis very nearly where x = '12.
We find /(12) = 004672.
If we want something nearer, the figure suggests trying 13.
/(13) = 070797.
364 CALCULUS FOB BEGINNERS
.*. the root is between 12 and '13.
x 12 13
f(x)  00467 07080
/'() 760 749
P' is (12,  00467), Q' is (13, 07080).
[CH. XII
Fig. 134.
It will be found, even on a large scale drawing (Fig. 135),
difficult to distinguish between P'Q' and the tangents at P'
and Q'.
1205 seems the best approximation.
As a matter of fact /(1 205) =  0009 nearly.
269, 270]
EQUATIONS
365
270. In the example just worked out /(O) = 1, /(I) = 3,
i.e. the point on the graph of y =f(x) corresponding to x =
is below, while that corresponding to x = 1 is above the ajaxis,
therefore we might say that the graph of y =f(x) cuts the OSaxis
at some intermediate point.
Fig. 135.
It may happen that y =f(x) cuts the ccaxis at more than one
point between the points corresponding to je = and x = l, e.g.
referring to Fig. 133,
and there are 3 roots between and 4.
and there is 1 root between 1 and 3.
366
CALCULUS FOR BEGINNERS
[CH. XII
Also if / (x) has the same sign for two different values of x,
say a and b, it does not follow that there is no root between
a and b ; for instance, taking the same figure, /(2) = l,/(4) = 3
and there are 2 roots between 2 and 4_
Iff (a) and/ (6) have opposite signs, there is an odd number
of roots between a and b. If /(a) and/(6) have the same sign,
there is no root or an even number of roots between a and b.
271. There is an exceptional case if f (x) becomes infinite
between a and b.
e.g. suppose /(#)
Fig. 136.
270273] EQUATIONS 367
Here /( 5) =  175,
but there is no root between 5 and  5.
A glance at Fig. 136 will shew the reason of this.
When x = 0, y is infinite.
When x is very small and negative, y is very large and
negative.
When x is very small and positive, y is very large and
positive.
so that as x passes through the value 0, y suddenly changes from
a very large negative value to a very large positive value.
It thus appears that \i.f(a) andy*(6) have opposite signs we
must take care that/(a;) does not become infinite for some value
of x between a and b, otherwise our inference that there is a root
between a and b may be wrong.
272. If f(a) and f(b) have opposite signs and if moreover
f (x) and f" (x) do not change sign between x = a and x = b, then
between x=a and x = b the graph of y=f(x) will have one of
the shapes shewn in 94, and will cut the o;axis only once
between x = a and x = 6.
273. A method of continual approximation to the value of
a root will now be considered which does not require the actual
drawing of the graph.
Taking the same equation as before, we have
/' (x) = 3a; 2  1 2* + 9 > = 3 (x  1 ) (x  3),
/ = 6(*2).
Now/(2)=l,/(3) = l.
368
CALCULUS FOR BEGINNERS
[CH. XII
Also from o? = 2 to x3, f'(x) is negative and f"(x) is
positive.
.'. the curve y =/() between x = 2 and x 3 is like Fig. 137,
and there is only one root between 2 and 3.
M
Fig. 137.
If PT is the tangent at P and the chord PQ cuts the .raxis in
R, the point A where the curve cuts the ajaxis is between
T and R.
Now/' (2) =3.
.'. equation of PT is y 1 = 3 (x  2).
Putting y = 0, x = 2 + . [This is OT where O is the origin.]
This value of x [2J] is a better approximation to the root
than 2.
2
Again gradient of PQ is  y , and equation of PQ is
Putting y = 0, x = 2 + J. [This is OR.]
This value of x [2] is a better approximation to the root
than 3.
.'. we know that the root is between 2J and 2J and therefore
between 2 33 and 250.
["We must be careful here not to overstate the smaller or
to understate the larger value, for instance we cannot safely say
that the root is between 234 and 250.]
273, 274] EQUATIONS 369
To get a closer appioximation, use the points P', Q' corre
sponding to a; = 233 and as = 2'50 exactly as we have just used
P and Q.
/(233) = 0045937, /(2'5) =  0375,
/'(233) = 26733.
Equation P'T' y  045937 =  26733 (x  233).
/. atT' * = 233 + . = 233 + (017+).
Equation P'Q' y  Q45937 =  ' 42Q ? 37 ( x  233)
=  24761 (x 233).
. 04^007
/. at R' x= 233 + ^f. = 233 + (019 ).
.'. the root lies between 2347 and 2349.
If we want a still closer approximation we use the points P",
Q" corresponding to x= 2347 and 05 = 2'349.
/(2S47) = 0000781923, /(2S49) =0004491451,
/' (2347) =2638773.
Equation P"T"
y  000781923 =  2638773 (*  2347).
Equation P"Q"
y 000781923 =  2636687 (x  2347).
.*. at R"
.'. the root lies between 23472963 and 23472966.
274. We might have " closed in " on the root more rapidly
by drawing first the portion of the graph of
x z  6x 2
M. c. 24
370 CALCULUS FOR BEGINNERS [CM. XII
which lies between x = 2 and x = 3 from the data
This would shew quite clearly that the root lies between 23
and 24.
In working out the smallest root, we might proceed graphically
as far as the stage reached in 269. Fig. 135 seems to shew
that the root is between 1205 and '1207. Also/' (x) is positive
and /" (x) negative so that the curve is like Fig. 40 (6), 94.
Proceeding on the lines of 273, we find that the root lies
between 12061474 and 12061477.
275. Sometimes it may not be so easy as in the example
just considered to decide whether f (x) keeps the same sign
between the values of x chosen, and it may be useful, to argue
as follows:
Between x = 2 and x = 3, f" (x) is clearly positive.
i.e. f (x) continually increases j but/' (2) = 3 and /' (3) = 0.
.'. /' (x) is negative all the way.
276. The rough figure (137) should be drawn, so that there
shall be no doubt at which end of the arc the tangent ought
to be drawn.
The following example will shew what may happen if we
neglect this precaution.
Suppose f(x) = Saj 8 80; + 1 "j
f'(x) = Qx*8 \.
f"(x) = lSx )
Here /(I) =4, /(2) = 9,
and between x = 1 and x = 2, /' (x) and /" (x) do not change
sign ; they are both positive.
.*. there is 1 root between 1 and 2.
274277]
EQUATIONS
371
Now if we start from P (the point on y =f(x) corresponding
to x = 1), we have since y (1) = 1 :
Equation of tangent PT is y + 4 = 1 (x 1).
.". at T, x = 5.
But 5 is not a better approximation to the root than 1.
Since f (x) and f" (x) are both positive between x = 1 and
x = 2 the curve is of the form shewn in Fig. 138 and a glance
at the figure will shew that we should have worked from Q
(corresponding to x = 2).
Q
Fig. 138.
U is nearer to A than N is.
T is not nearer to A than M is.
Of course with a curve of this form it may happen that T
comes nearer to A than M does, but on the other hand it may
not, whereas U must be nearer to A than N is.
What we ought to do in this case is to consider 2 as the first
approximation to the root and get a better one by finding the
coordinates of U.
277. As another example take the equation coso?=o; and
let it be required to find the least positive root.
An inspection of the tables shews that x is between [42] and
[43] where [42] stands for the C.M. of 42".
242
372
CALCULUS FOB, BEGINNERS
[CH. XII
If f( x ) is a: cos x }
f (x) = 1 + sin x I .
f" (x)  cos x }
"Working with 5figure tables
/[42] = 73304  74314 =  01010 ( 00001) j
/[43] = 75049  73135 = + 01914 (+ 00001)} '
Also between x = [42] and x = [43]/' (x) and/" (x) are both
positive and the curve is like Fig. 139.
/' [43] = 168200 (+ 000005).
Q
Pig. 139.
Equation QU is
y 01914 = 168200 (a; [43]).
01 Q14 *
/. atu aj =
Equation PQ is
 01914 =
02924 _r 43 .,v
02924.
ol745 <*[ 43 ]>
l6756(a;[43 J).
.'. atR
* We must be careful to understate the value of the first of these
fractions, and it will be safe to take it as ^ !^r . Similarly the value
I*o82005
of the other fraction should be overstated.
277] EQUATIONS 373
.'. x lies between 73906 and 73912, or since 01137 and
01143 each differ from [39'j by less than [!'], we may say x is
approximately the C.M. of 42 21'.
As a matter of fact for this value of x
f(x) = 73914  73904 = 0001.
EXERCISES. LXXXIV.
1. Shew that a: 3 a: 1=0 has one and only one root between 1 and 2,
and then shew by the method of 273 that this root lies between 11 and 16.
2. 4z 6 24a; 3 + 2 = has a root near 3. Shew that 27 is a better
approximation.
3. x*  2x  1 = has a root between  1 and 0. Shew that  3 is a
better approximation than and ? than 1.
4. 3x*  40z 3 + 1200z  2900 = has a root between 3 and 4. Shew that
it is between 33 and 3 6.
6. x a Bx 2 + 2x5 = has a root near 29. Shew that it is between
29041 and 2'9042.
6. Shew that x s 2x5 = Q has a root between 2 and 3, and find its
value correct to 3 decimal places.
7. Find all three roots of 20x 3 24x 2 +3=0 correct to 3 decimal places.
8. Shew that x 3  5x + 2=0 has one and only one root between and 1,
and find its value correct to 4 decimal places.
9. Shew that 3x 3 =x + l has only one real root and find its value
correct to 5 decimal places.
10. Shew that x 3  30x 2 + 2600 = has a root between 12 and 13 and
find its value correct to 2 decimal places. [First write for x, y + 10.]
11. Shew that if a floating sphere of radius r" and specific gravity *,
sink in water to a depth x", then
x 3 
If the sphere be of wood (sp. gr. 6) and its radius be 10", to what depth
will it sink? [Correct to nearest hundredth of an inch.]
374 CALCULUS FOR BEGINNERS [CH. XII
12. x 3  Wx^ + iOx  35=0 has a root between 1 and 2. Find its value
correct to 4 decimal places.
13. Find the real roots of x*3x 5 = 0, each correct to 3 significant
figures.
14. A plane is drawn parallel to the base of a hemisphere of 1 foot
radius, dividing it into two parts of equal volume. Find its distance from
the centre correct to the nearest thousandth of an inch.
15. x sin x =02 is satisfied if x is the C.M. of some angle near 30.
Find the angle so that your error does not exceed 1 minute.
16. Solve approximately t&nx = 2x between x = and x = ^. Your
2
answer must not be more than 001 in error.
17. Solve tan x=   x between and IT correct to 3 decimal places.
2
18. Shew from a graph that the line By = 2x + 1 cuts the curve y = tan x
roughly where x =  .
Find the value of x at the point of intersection correct to 3 decimal
places.
19. Solve the equation log g a; = a: correct to 3 decimal places.
O
SO. Sketch the shape of the curve y = x 2 sinx: suppose x to range from
OtOTT.
Determine approximately the maximum ordinate within this range.
CHAPTER XIII
SOME METHODS OF INTEGRATION
Further examples in integration.
278. OUR success in evaluating integrals of the form ]f(x) dx
has so far depended on our ability to recognise f(x) as the result
of differentiating some standard function,
e.g. we know that
d (tan x)
7 = sec 2 x,
dx
and hence that
f
sec 2 x dx = tan x + c.
It will be convenient at this stage to collect the most impor
tant standard results. They have all been worked out either in
the text or in the exercises. It is understood that an arbitrary
constant should be added to every function in the column headed
z dx.
When a more complicated expression is to be integrated, our
object will be to reduce it to one or other of these standard forms.
A few of the simpler methods of effecting such reduction are
given in this chapter, and it will be seen that many functions
yield to these methods of treatment, but no rule can be given for
the integration of any given function.
370
CALCULUS FOR BEGINNERS
[CH. XIII
y
dx
2
/ zdx
x
X*
nx* 1
except when
11= 1
T n +l
n+1
sinx
cosx
cosx
sinx
cosx
sinx
sinx
cosx
tanx
Bec 2 x
sec 2 x
tanx
cotx
 cosec 2 x
cosec 2 x
cotx
secx
sec x tan x
sec x tan x
secx
cosec x
 cosec x cot x
cosec x cot x
cosec x
Bin" 1 
a
1
1
, x
sin 1 
a
V/a 2 * 2
v/a 2  x 2
tan" 1 x
a
1
ta.n1 X
a
a 2 + x 2
a 2 + x 2
a a
e*
*
e*
t*
log,x
1
1
x
log,*
sinhx
coshx
coshx
sinhx
coshx
sinhx
sinhx
coshx
tauhx
sech 2 x
sech 2 x
tanhx
sinh" 1 
sinh 1 
a
1
1
a
or locr
*&+*
Vi 2 ^
x + Jx^ + a?
be a J
cosh" 1 
cosh" 1 
1 ^2,1^
1
a
orlop a; + V a;2  2
^A ;
Va % a2
x+>v/x 2 a 2
a
( >1 J
( 1 x
tanhi
\u y
 tanh" 1 
a
(x'ca 9 )
1
a a
a + x
a 2 x 2
1 o + x
2 ge axj
Or 2a 10g ax
278, 279] METHODS OF INTEGRATION
377
y
dy
dx
Izdx
X \
1
coth" 1 
a a
or tanh" 1  V
x
a 2 1
1
1 , .a
or tanh" 1 
a x
x 2  a 2 *
x' 2  a 2
,11 **!
(x3>a>)
1 ln x + a
2o &8 xa
log e g i n x
loge cos x
loge (sec x + tan x) 1
cotx
tanx
cotx
tanx
loge s i Q x
logeCosx or loge secx
i log, (sec x + tan x)
or loge tan (T + O)
secx
secx
 or log e * an ( 7 + H )
log e (cosec x + cot x) ~j
/  log e (cosec x + cot x)
or  log tan  I
cosec x
cosec x
J x
Or loge * an 9
\
279. Corresponding to the formula of differentiation
df(u) df(u) du ., . du
V \ f %f \ / Qp J. Inl \
we have the formula of integration
for since
,
and
du
=f ( u ),
ff ().*. dx =/() and /() = I/' (u) du,
or
where <^> (u) stands for/' (u).
That is to say v . dx occurring under the integral sign may
dx
be replaced by du.
378 CALCULUS FOR BEGINNERS [CH. XIII
Ex. 1. [sin 3 x cos x dx = I sin 3 x ^ . dx
~ I sin 3 x d (sin x) = Iu 3 du [where u is sin a;]
u* sin* x
4 4
Everything here depends on noticing that the expression to
be integrated (sin 3 x cos x) is the product of a function of sin x
(viz. sin 3 x) and cos x, which is the differential coefficient of sin x.
Generally if the expression, f(x), to be integrated can be
written in the form <f> (u) . r we can say that
dx
\f(x).dx=\<t>(u).du,
and it may happen that I < (w) . du is easier to get than
ff(x).dx.
Ex. 2. Find ( , 2a; + 3 dx.
J >Ja? + 3x + 5
Here we notice that (2a? + 3) is ' ^ and thus
dx
(2x + 3)dx may be replaced by d (ar* + 3x + 5) and the integral
may be written
or I j where u is (x 2 + 3x + 5),
which is 2/Ju or 2\Ae a + 3x + 5.
Ex. 3. lsiii s x.dx= Isin 4 a;. sino;c&c = f  (1  cos 2 a) 2 d (
in 2x2^ / 2 1 1
= J^ M >^ = ( M 3 W+ 5 W
cos a
where = cos x
2 1
= cos x + = cos 3 x   cos 5 x.
3 5
279] METHODS OF INTEGRATION 379
dx
Ex. 4.
The expression under the >J
11
T2'
and
j. i du
. . integral is
 j 
where u stands for
1 ui u * i, i 6 * 1
= fo sinh J T= = rr smh" 1 j=
v** /ll 'v
V 12
or
EXERCISES. LXXXV.*
Find I zdx for the following values of z :
1. sin x cos 8 x. 2. xsina; 2 . 3.
1
6 s ^ =. 6.
1 gdn'as
7. sin (log e a;). 8. . O.
l0 ' 9^^ 2<9 )' 4^9 (^> 9 )' "' 9+1F3
13. cos 3 x. 14. tan 8 x sec 2 x. 15. tan 3 x.
* In this and following Exs. results should be checked by differentiation.
380 CALCULUS FOE BEGINNERS [CH. XIII
X6. *j2a + x. 17 sin 3 a; cos 2 ar.
18. _   R . [(i) x*3x + 2pos. (ii) x*  3x + 2 neg.]
2 3x + 2
19 3x + l 20. tan(aa; + 6). 21. sina;cos2a;.
1
as. . 26.
cos 5 a;
Integration by Substitution.
280. This depends on the same principle as the last
method, viz. that
Ex. 1. Find \*Jd i a?.dx.
Put x = a sin 0; then v a 2 x 2 = a cos 0, and
efcc
so that I v 2 # 2 cfo; becomes I a cos . d (a sin 0),
= I a cos . a cos c?0
x
280] INTEGRATION BY SUBSTITUTION 381
Ex. 2. Find \x /2x + 5 . dx.
z 5 (i oc
Put 2x + 5 = z, .*. x = , and =  ,
'Jj dz 2t
so that I x v 2x + 5 . dx = I
Jz.^
=  f(a* 
4 5
EXERCISES. LXXXVI.
Find:
3 
5. I x*Ja*x*.dx. 6. I o?*jcflx*.dx.
1. I . a . da;. 8. I >/a a + o; 2 .da;[a;=osmh0].
y v x + ") /y
. / Vx  a 2 rfz [a; = a cosh tfl. 1O. / / . , ^*
J J x*Ja?x 2
11. I  dx[x2 = z]. 12.
13. \ ^dx. 14. /
J 2ax j ,
15. By means of the substitution tan x = sinh <j>, or seca; = cosh0,
shew that Isec 3 xdx= /cosh 2 (f>.d<t>, and hence evaluate Isec 3 a;dx. (See
Exs. LXXXVII. 15.)
382 CALCULUS FOB BEGINNERS [CH. XIII
Integration by Parts.
281. This is an inversion of the rule for differentiating a
product.
d (x sin x)
Ex.  ; as sin x + xcosx (1).
ax
If one of the members of the righthand side can be integrated
so can the other.
(1) may be written
d (x sin x) d cos x
. = + X COS X.
dx ax
d (x sin x + cos x)
/. x cos x = = ,
dx
or I x cos x dx = x sin x + cos x.
Similarly by differentiating x cos x get I x sin x . dx,
and x log x get I log x . dx.
282. The work can be arranged more conveniently.
,. , d(uv) dv du
We have  \ ' = u = + v 7 ,
dx dx dx
.'. uv = \udv + Ivdu,
or hi d v = u v I v du,
and it may happen that Ivdu is easier to get than ludv.
Ex. 1. \xco&xdx= I x d (sin a:) = x sin x l&inxdx
= x sin x + cos x.
281, 282] INTEGRATION BY PABTS 383
We might have said
/ x cos xdx = lcosx.d(~}
y? [a?
= ^ cos x I d (cos x)
A J Z
= cos x + I sin x . dx,
f 1$
but this leads to an integral I ~ sin xdx which is no easier to get
than the original one / x cos x dx.
Ex. 2. I log e xdx = x \og e x ~ I X( % (lge x )
f 1 ,
*:xlog e x Jx.ax
= X loge X X.
EXERCISES. LXXXVIL
Find:
1. I x log e x dx. 6. I x sin xdx.
2. I xe x dx. 7. Ix sin mx cos nx dx.
3. I x z e x dx. 8. / a; 3 log e o;<ZaJ.
I
4. I tan" 1 xdx. 0.
x 3
6. I sin' 1 xdx 1O. I x sec* xdx.
384 CALCULUS FOE BEGINNERS [CH. XIII
11. Shew that
/ e? sin x dx = e x cos x + I f* cos x dx,
and I e?cosxdx=e x Binx I e'sinxdx.
Hence get I e x siuxdx = e ae (sincosa:),
and I e* cos x dx=^e" (sin* + cos x).
12. Similarly get
I e^sin^x + ^dx and I e * cos (bx + c) dx.
13. Shew that
I siu 6 xdx=  sin* cos a; + I 4 sin 3 x cos 2 * dx.
Hence 5 I sin 6 x dx =  sin 4 x cos x + 4 I sin 3 x dx.
Similarly shew
n I ain n xdx=  sin"" 1 x cos x + (n  1) I sin n ~ 2 a;. dx,
and n / cos n x dx = cos* 1 " 1 x sin x + (n  1) I cos n ~ 2 x . dx.
14. Shew that
a it
/sm n xdx= / si
n J o
ir r
/"a /a
sin 7 xdx and J sit
J o ; o
Hence find / sin 7 x dx and  sin 8 x dx.
16. Shew that
sec 3 x dx = sec x tana; / tan 2 # sec a da?.
Hence 2 / sec s xda;=secxtana;+ / secai.d*.
I sec 3 x dx = sec x tan a? / i
! / sec s xda;=secxtanx+ / sec a;, i
282]
EXERCISES IN INTEGRATION
385
MISCELLANEOUS EXERCISES ON CHAPTER XIII.
Find I z dx for the following values of z :
7 1
m log e ax cos .T
a ' 1 + sino;"
C 9 .....
10
a; 2 5
2 v /a;
13. xaiuBx.
16. ar(32a:)i
14. x^sinBx. 15. (3 2s;)*.
22. sec a;.
\
Evaluate the following definite integrals:
25.
28.
27. r ;;.
29. ein 6 xdx.
31.
.
2
3O. I cos 8 x dx.
32
11. 0.
J o 4 + 5 cos a;
25
386 CALCULUS FOR BEGINNERS [dl. XIII
83. O is the lowest point of the chain of a suspension bridge, P any
other point on the chain whose coordinates are (x, y) referred to horizontal
and vertical axes through O. If the resultant load on the portion OP of the
chain is wx Ibs. (w being constant) and if T, T Ibs. wt. are the pulls
a P and O respectively, shew that
dy _ w
Tx = T x '
and hence get the equation of the curve of the chain.
/ w*
Also shew that T = T ^J 1 + ^ a a; 2 .
34. A body rr.nss m Ibs. starts with a speed of u ft. /sec. and moves
horizontally, the resistance to motion being Kt> 2 poundals, where K is
a constant, and v ft./sec. its speed at any instant. Shew that the speed at
the end of t seconds is  ^ ft./sec.
35. Assume that the resistance to the movement of a ship through the
water is of the form (a 2 + 6 2 t; 2 ) Ibs. wt., where v is the speed and a, b are
constants, m Ibs. being the mass of the ship. If the engines are stopped,
find a formula for the time in which the speed falls to onehalf of its
original amount u.
36. A body of mass m Ibs. falls from rest vertically in a medium in
which the resistance is K 2 Ibs. wt. where K is constant and v ft./sec. is the
speed.
Shew that
o
wliere Vl=
Hence get v in terms of t and t>i , and shew that
r i 2 , / , gt\
= log ( cosh ) .
9 \ vij
87. A block slides in a straight line OA. The resistance to its
movement when it is at a distance x ft. from O is R Ibs. wt.
a 3 f a; 2
where OA is a ft.
Find the work done in pushing the block from A to O.
282] EXERCISES IN INTEGRATION 387
38. A train is moving on the level, the power exerted by the engine
being constant and the total resistance proportional to the speed. Shew
that ^ =   bv, where a and 6 are constants, and hence that a  6u 2 = Ae~ 2bt ,
at v
where A is a constant.
If the constant power is 400 H.P. and if the resistance is 12 Ibs. wt. per
ton at 40 miles an hour, and if the mass of the train be 200 tons, find the
values of the constants a, b, A ; t being reckoned from the instant at which
the speed is 40 miles an hour. Shew that the speed can never reach 50 miles
an hour. Find the speed after halfaminute and after halfanhour.
[0 = 82.]
252
CHAPTER XIV
POLAR COORDINATES
283. WE have hitherto considered the position of a point in
a plane as determined by its distances from two fixed axes.
These distances are the " Cartesian coordinates " of the point.
There is another system in which the position of a point is
determined by a distance and an angle. The position is obviously
determined if we know the length, direction and sense of the line
joining the point to some fixed point. For instance if A is a
given point and we say that B is 100 yards N. 30 E. from A, B
is determined. This is the principle of " Polar Coordinates."
284. O is a fixed point (fig. 140), OX a fixed line through
p
Fig. 140.
283287] POLAR COOTCDINATES 389
O. The position of P is known if we are given the length OP and
the angle, XOP, through which a line must swing to get from the
position OX to the position OP. These are called the "polar
coordinates " of P, and are denoted by (r, 0). O is called the
pole and OX the "initial line."
285. There is an infinite number of ways of writing the
coordinates of a point P, e.g. if XOP = 30 and OP is 3 inches,
taking 1" as the unit of length we may give the coordinates
of P as
(3, 30), (3, 390), (3,  330) etc.
286. If r is negative the radius bounding the angle must
be produced backwards through O, e.g. if the point is ( 3, 210),
first draw OP' so that XOP'= 210, then produce P'O backwards
through O so that OP is 3". This brings us to the same point P
as in 285.
EXERCISES. LXXXVIII.
1. Taking 1" as the unit of length, plot the points (3, 40), (3, 140),
(  3, 220), (3, 320), (  3,  50), (3,  500), (  3,  1000).
2. Draw the loci :
(i).r=a, (ii) 0=40, (iii) r=acos0, (iv) rcos0 = a, (v) r=asin0,
(vi) rsin 0=a.
287. Figure 141 shews part of the curve r=f(0), P is the
point (r, 0) and Q the point (7 + Ar, 6 + A0), PS is perpendicular
to OQ, and R is the point where a circle, centre O radius OP,
cuts OGL
.'. POQ = A0 and RQ = An
QS QR + RS QR RS , ,
JMow cot PQR = = = + and by sufficiently
SP SP SP SP J J
RS
diminishing A0 we can make as small as we like ( 32) ; also
SP
390
CALCULUS FOR BEGINNERS
[CH. XIV
=  = ^ . f. and by sufficiently diminishing A0
SP r sin Ad r. A<9 sin A0
1 dr
we can brine this as near to  . =7. as we please.
r dO
Fig. 141.
.'. if < be the angle between OP and the tangent at P (fig.
r
r 'dr'
1 dr
142) cot = . g or
r du
Ex. If
Fig. 142.
r = a (1 cos 0),
dr
287, 288]
POLAR COORDINATES
891
1 dr
sin 6
6
2'
e.g. it 6 = 90, <f> = 45 ; if $ = 1 20,
[See fig. 143.]
= 60 ; it 6= 60, <= 30.
These facts may be found useful in drawing the curve, just as
the knowledge of the gradient was a help in drawing a curve
whose equation was given in Cartesian coordinates.
288. Areas.
Suppose we want the area bounded by HK, part of the curve
r =/(#), and the radii OH, OK.
392
CALCULUS FOR BEGINNERS
[CH. XIV
Divide Z HOK into any number of equal parts and with O as
centre describe circular arcs as shewn in fig. 144.
Fig. 144.
The area required lies between the sum of the internal sectors
OHC, OPD, OQE, ... and the sum of the external sectors OPC',
OQD', ....
The difference between these sums is the sum of the areas
CHC'P, DPD'Q ..., and this is the same as the area GK. By
sufficiently diminishing the angle of the sectors or increasing the
number of the sectors, we can make this area GK as small as we
like, and thus the area required is the limit of the sum of either
the internal or external sectors as the number of sectors is
indefinitely increased or the angle of the sectors indefinitely
diminished [cp. 136].
If P be any point (r, ff) on the arc HK and if Q be (r + Ar,
6 + A0), the area of the sector POD is ^ r 3 . A0, and area
e=/s
HOK = Lt 2
A0H.0 6=a
where a and ft are the angles XOH, XOK respectively.
rft i
 ^
J a *
288, 289] POLAR COORDINATES 393
289. Ex. If the curve be r = a (1  cos 6) [Fig. 143],
area OAB
3
f*l
= / 
J TT 2
( z Vi o /> 1 + cos 2#\
= /  ( 1  2 cos 4 J o?0
= 1505a 2 .
EXERCISES. LXXXIX.*
1. Find values of r corresponding to 0=0, 20, 40, 60, 80, 90, ... 360 ;
if r=a (1 cos0).
Also find the values of corresponding to = 0, 60, 90, 120, 150, 180.
Draw the graph of r=a (1  cos 6), using these values. (See Fig. 143.)
2. Find tan <j> if
(i)r=asin0, (ii)r=acos0, (iii) r cos a, (iv)rsin0 = a.
Interpret the results geometrically.
3. Draw graphs of
(i) r = asin20,
(ii) r=acos20,
(iii) r=asinC0,
(iv) r=acos30,
(v) r sin 29= a,
(vi) r sin 30 = a,
., a
(vn)  = l + cos0,
' r
.. a . 1
(vm)  = l + cos0,
(ix)  = 1 + 2 cos
(x> 2= a 2 cos 20,
(xi) r=3 + 2cos0.
4. Find the area enclosed by r = a cos 0.
* Paper specially ruled for polar coordinates can be obtained. Lines are
drawn through O (see Fig. 143) at intervals of 5, and concentric circles are
described at small equal distances.
394 CALCULUS FOR BEGINNERS [CH. XIV
6. Find the area of one loop oir=a sin 20.
6. Find the area of a loop of r 2 =a 2 cos 20.
7. Draw the curve r=a0. (The spiral of Archimedes.)
8. In the curve r = ae ke , shew that <f> is constant.
[This is called the equiangular or logarithmic spiral.]
9. Draw the curves (i) r=ae , (ii) r=ae .
10. Shew that r=6e e may be written r = c e+log < 6 and hence that
r 6e e is r=e 6 turned round the pole through a certain angle.
Generally shew that r=oe fctf , r=le* e represent the same spiral in different
positions.
ANSWERS
EXERCISES I. p. 3.
1. (i) 3939 m./hr. (ii) 5777 ft./sec. 2. 42 m./hr.
3. (i) 2^ m./hr. (ii) 80 yds./min.
4. 0146 ft./sec. . 43 m./hr.
EXERCISES II. p. 6.
1. 44 ft./sec.; 461 ft./sec.; 5077 ft./sec. A's is best. 22ft.; 23Jft.;
254 ft. A's.
2. [All in chs./min.] 31, 39, 57, 52, 59, 58, 62, 62. (i) 59, (ii) 62.
3. 50, 5736, 4226, 5446, 45 40, 5150, 4848 cms. 8832 006, 9288 006,
89201, 9'2001, 90003, 91203 cms./sec. Between 897 and
912 cms./sec.
4. 17321, 14281, 21445, 153'99, 19626, 16643, 18040 cms. 36480012,
49'488012, 384402, 46'1002, 406806, 431406 cms./sec.
Between 414 and 423 cins./sec.
EXERCISES III. p. 9.
L. (99 24A + 2A2) ft./sec., ( 99 + ^ ft./sec.
2. 42, 68, 5425, 4672, 4433, 422303, 42023003 ft. 26, 245, 236, 233,
23 03, 23003 ft./sec. (42 + 23ft + 3fc2) ft. ; (23 + 3/i) ft./sec.; 23 ft./sec.;
(23  3/z) ft./sec. ; 23 ft./sec. ; {23 + 3 (  m)} ft. /sec.
3. (i) 26 ft./sec. (ii) 23, 29, 26 ft./sec. (iii) 26 ft./sec.
4. (i) 333 ft./sec. (ii) 330J ft./sec. (iii) 337 ft./sec.
. 249 ft., (249 + 253/t + 96/i 2 + 16/i 3 + fc 4 ) ft., 253 ft./seo.
396
CALCULUS FOR BEGINNERS
EXERCISES IV. p. 18.
1. (i) 605J. (ii) 604. (iii) 603. (iv) 602. 2. '87 ; '86 ; about 87.
3. 6; 2a.
6. r
S
V
m 244 r 2
4. 3; 75; 3c
4 5
64* 1007T
256 500
[2. {
45
Six
3645
>. (i) 67T.
42
7056T
296352
(ii) 2*. (iii) 3.
41
67247T
275684
3 ' 3 *"
401
643204jr
257924804
3 *" 3 "
4+7t
(64 + 32ft + 4 7(2) n
3
p
2 + 4871 + 47^
3
17 20176
19684
3
1924804 19
(1) 8 , 3 , 3 .., 3 .
3 *"' 3
(ii) 367T, 347T, 3287T, 3247T, 3204ir, (32 + 4/z) x.
244
(iii) 5~T etc. c.c./sec. [see (i)]. (iv) 36ir etc. sq. cms./sec. [see (ii)].
o
(v) 64*. (vi) 327T. (vii) 64* c.c./sec. (viii) 32?r sq. cms./sec.
7. (i) 20. (ii) 16 + 4/1. (iii) 16.
EXERCISES V. p. 29.
1. 9, 1225, 961, 90601, 9 + 6/i + ft 2 , 841, 96/ + / t 2 ;
65, 61, 601, 6 + h, 59, Qh; 6; tani6=80 32'.
2. 6. 3. tanil2=5012'. 4. 12 + 6/j + 7i2. 12.
5. 12 + ft 2 ; 12 + 6(jim) + (jt2 +nm+m 2). e . 0,3,27,48.
7. tan 1 48 =78 14'. a. 8,0. 9. 4; 4; 4.
10. 3; 3 + 8A + 6fc 2 fc 4 ; 8; 0.
1. 109.
EXERCISES VI. p. 37.
2. 27 ft./sec.
EXERCISES VUL p. 5L
(i) =32f. (ii) r = 10032t. (iii) u = 3t 2 . (iv) v = 5.
(i) 96, 160, 544. (ii) 4,  60,  444. (iii) 27, 75 867.
(iv) 5, 5, 5 (ft./sec).
ANSWERS 397
2. 64 ft./sec. ; 64 ft. ; 244 o/ . 64 ft. ; 249 o/ .
3. 4 ft./sec. 2 4. 136 ft./sec. 2 5. ^ ft./sec. 2 , ~.
6. (i) 32, 32, 32. (ii)  32,  32,  32. (iii) 18, 30, 102. (iv) 0, 0,
(ft./sec. 2 ).
7. 54ft.; 36 ft./sec.
EXERCISES IX. p. 57.
1. 22G3sq. ins. 2. 14%.
3. (i) 2ir, (ii) 207T, (iii) 400jr (sq. mm./sec.). 4. 407r sq. mm.
5. 16sq. ins. 6. (i) = '064 ins. /sec. (ii) ./_L= '126 ins./sec.
7. (i) 8. (ii) Jr. 8. (i) ^= 03125 ins./sec. (ii) 32 ins./sec.
o(J Oa
EXEEOISES X. p. 64.
!. 6, 4, 2, 0,2, 4, 6. 2. 27,3,0,75. 3. 4,0,12,28.
4. 14; 1403; about 21 o/ . 5. (i) 0025. (ii) 00184. (iii) 2. (iv) 25.
9. (i) 3 to. (ii)8. (iii)  3 .
EXERCISES XL p. C3.
EXERCISES XII. p. 69.
1. 3kx*. 2. 108 ft./sec. 3. 60; y = 60x80.
4. 1728 c. ins./sec. 8. 9ir = 2827 c. ft./seo. 6. 9 '6 c. ins.
7. 3<Y .
398 CALCULUS FOR BEGINNERS
EXERCISES XIIL p. 74.
i 2. 1 0: . a. 3x
x a ' x ' 4
4. 2 + 2t; 8 ft. /sec.; 22 ft./seo.
/\ ^^ /' *\ /* Q
EXEKCISES XIV. p. 79.
 xt,  3x, j x~S,  i , l8x8, A*", 1, 0,
7 5 ^ ! * 2 *
 fa7 '* ' ' X ' ""
2.  = kn**i. 3. =
ax ax
EXERCISES XV. p. 80.
3. gt. 3. ^5. . a&.
EXERCISES XVII. p. 84.
1. (i) 2 + 8x, (ii) ^ 2 l. 2. 15t 2 t.
ANSWERS 399
EXERCISES XVIII. p. 85.
1. 3x*3; x 2 1 12
y 2 2 022
^ 9 03 09
dx
2. 4.T
a: 1 1 23
y 63236
$^ 24 24
dx
3. Gx 2 18a; + 12 = 6(xl) (x2);
x 0123
y 3 2 1 6
12 12
   nr and > '  or (1 '" 77 ' ~ ' 38 ^ and (423> ' 385) 
EXERCISES XIX. p. 90.
5. 3*y2 = 0, rr + 3y4 = 0; T,o<(0, 2), G (4, 0), g
80
6. 9Gzy144=0, a; + 96y 4610 = 0; x
S. 3xy4 = 0 3a; + 7/4=Oi
a; + 3i/28 = 0j" ' *3y 28 = 0) "
9. ^ = 40x57,
11. y = 7x4, x + ly12, = sq. units.
EXERCISES XX. p. 92.
3. 5+1,18415,0.
400 CALCULUS FOR BEGINNERS
EXERCISES XXI. p. 97.
. 2^;  3 ; 4, If, *. 2. 35 ft. /sec.; 22 ft./sec
X X
EXERCISES XXII. p. 101.
Letters refer to figures on p. 99.
1. (i) a. (ii) c. a. (i) b. (ii) a. (iii) d. 3. (i) b. (ii) a.
MISCELLANEOUS EXAMPLES ON CHAPTER III. p. 102.
.. 2 _i 2ft 2
1. (i) gX S. (h) __ =  J: .
2.
3. x + c z y2c = 0\ Area of triangle formed by the tangent at
1 I ; 2. any point and the axes of coordinates is
c ~ constant.
" J ; (0,  9), (0, 9) ; 18, s .
ar + 6y57 = 0j 2
6. t 5 6 7 6. t 5 6 7
s 207 301 413 s 758 1341 2166
v 85 103 121 v 476 697 960
a 18 18 18 a 200 242 284
7. 14, 1403, 21 o/o . 8. y = 144x400. (0, 5), (1, 0), (2, 27).
9. (i) 1. (ii) 1n. (iii) n. (iv) .
10  (^ T = >0693 ins./sec. (ii) = 0173 ins./sec.
1UU 1UU
11. 265. 12. Average speed = n times speed at end.
15. (258)a, 6 70ix 10 6 ,2 68x10*. 16. 100896.
ANSWERS 401
17. 0172. 13. a; 1^01
4
12 36
da;
IS. (i) 115 ft./scc. (ii) 123 ft./sec. (iii) 111 ft./sec.
, _ 19r Ifil 2 3
; ;96> "  or '955 ins./niin. 23. (Si, 40J).
24. x 1 2 3 4 5 6 7 8 9 10 11 12
t/ 30 70 120 180 250 330 4'08 473 525 563 588 6'00
^ 25 35 45 55 65 "75 '85 '72 58 45 32 18 05
dx
25. 3z2M= s . 26. 925 Ibs. wt. 27. 27 ft./sec.
o
28. 305 radians/sec. 120 radians/sec. 2
(i) 4Slbs. wt., (ii) 42 Ibs. \vt.
30. L
EXERCISES XXIII. p. 111.
/2 11\ / 3 83\
1. (1, 2)min., (1, 2) max. 3. (^, ^J min., (^  , j J
4. (  1, 0) min., (0, 1) max., (1, 0) min. 6. (0, 1) min.
6. (  1, 0) min., (0, 5) max., (2,  27) min.
7. (  1,  2) max., (1, 2) min. 8. (, 3J minimum;
M. c.
402 CALCULUS FOR BEGINNERS
EXERCISES XXIV. p. 118.
1. 1. 3. 4/5 = 7937. 3. a, a. 4. 1\ ft. by 2Jft.
7. Radius ^ r , height 5. 8. 5 . ~ =1086 c. ins.
O O iJ ^y t>7T
9. 2r=ft=. / . 1O. 942 inches. 11. 4/150 = 531 knots.
12 = 3183 acres. 13. Total height = width = . =840 feet.
7T + 4
14. A = M55 ft., 2 ^=1633 ft. 15. ~pft.j
sees.
2^i cos ft 9 cos /3
17. (i) 2 c. ft. (ii) =2546 c. ft. (iii) = '849 c. ft.
7T OTT
18. 7. 18s. 9<L [Ht. = J side of base.] 10. 48 ft., 3 ft., 6 ft.
2O. rad.=2 N /6 ins., ht. = 4,/3 ins., vol. =96^^3=522 c. ins.
a i. J_ (=5774) of vol. of sphere; 1 : /2 or 7071 : 1.
v^
15 50
22. 2 radians =114 36'. 23. , 
25. r= 10 \32= 606, Z=10 y = 10' 5, volume = 330 c. ft.
26. 661. 27. 8609 sq. ft. [A = tx/2]. 28. ^ = 3224 c. ft
20. 5. 30. R=r. 31. 100.
32. a=900; 5 = 60,000; m= 4048; n = 733'6; V=358.
34 . /'^L. 35. 4 of 6 ins. each and 2 of 2 feet each.
36. 45. 37. 13i feet; ___. . 6 = a .
y\ a ffl
38. A/26C . sin 5 , where A is the least angle. 3O.  .
ANSWERS 403
EXERCISES XXV. p. 134.
1. 71 (whena=3) min., 54 (when a:= 2) max.; (> ~ 8 i)
a. 12J ( when x =  J Min., 44 (when x = 4) Max.
3. (1, 1). 4. 2 + 2 ./I = 458 (when x =4/15).
5. (0, 0) max., f~y ,  ? (y Yl or [171,  725] min. 1 point of inflexion
7. A minimum point.
8. (3, 47) Min., (  1, 17) Max., (1,  15) point of inflexion.
EXERCISES XXVI. p. 138.
a. 4524 sq. ins. 3. 362 c. ins. 4. 197 ft., 9'8 ft
6. (i) 196 c. ft. (ii) 707 c. ft. 6. irr 2 coto.Ar.
7. (i) 465 sq. ins. (ii) 430 o. ins. 8. R (a + 2bt), 0064.
EXERCISES XXVII. p. 139.
I. 15 <V . 3. Decrease 1 %. 1O. 608 ft. /sec., 025.
II. 2214, 0115. 13. 11 % decrease.
EXERCISES XXVIII. p. 143.
a. 372088. 3. 96244. 4. 2424.
6. (i) 1. (ii) 1. 100007525, 100007475.
6. (i) 3. (ii) 2.  299D39197, 200000397.
20 2
404 CALCULUS FOB BEGINNERS
EXERCISES XXX. p. 146.
[Add an arbitrary constant in each case.]
1. (i) a*. (ii)**. (Hi) \. (iv) 7*. (v) 10A (
(vii) jp. (viii)jal () g*. (x) gg*^. (a) ^ + ^
(xii) +4z. (xiii) o;3 + a; 2 . (xiv)
EXERCISES XXXI. p. 147.
[o and 6 are arbitrary constants.]
x* ..5 ... 7
1 4 s
( v ) ^z +ax + b  (vi) ** + *+ *
2.
EXERCISES XXXI. p. 147.
(iv) Q + x 2 + ax + 6.
4 5
(v) jg+ax + o. (vi) ^ x^ + ax + o.
x^ 5 x^ x"
 4x 2 + ax + o. 3.  x 2 + ax + o. 4. ^  ^
2i & JLO i
EXERCISES XXXH. p. 148.
,. ,i + J. a. ,
. , ,. V = r3. 8. y = ,
O j>
10. t/ = + 4xl. 11. =
ANSWERS 405
EXERCISES XXXHI. p. 150.
; 2 . a. (i) y = Zx1x*. (ii) y = 28 + 3x  2x.
4. y =
5 25\ .... /5 92\
3,  T ). (11) (3,  T ).
'
*3; (0, 7); 8 8'.
o
EXERCISES XXXIV. p. 152.
1. (60162)ft. 3. 186 ft./sec.; 404ft.
3. s=(3 + t 2 + t + 5. (i) 5 ft., 1 ft./sec. (ii) 97^ft> 48ft./seo.
6. =
MISCELLANEOUS EXAMPLES ON CHAPTERS IVI. p. 153.
A.
1. ^=9x2 + 2. y = lla;5. 3. 6 ft. 3. 1 ft., ft.
4. (i) (a) 08w or 2513 sq. ins./seo. (i) (6) 96^ or 3016 sq. ins./sec.
(u) (a) 04ir or 1257 c. ins./sec. (ii) (b) 516rr or 18096 c. ins./sec.
3 /]50
6. Height = diameter =2 * / = 7'26 cms. Surface = 248 sq. cms.
V 7T
B.
(i) 2 (ii)
406 CALCULUS FOR BEGINNERS
4. Greatest when radius of circle = ft. = 159 ft. and side of square =
7T
(not a true max.). Area = 796 sq. ft.
Least when side of square = diameter of circle=   ft. = 897 ft. (true
TTfO
min.). Area =  5 = 224 sq.ft.
7T + O
. =.
2. Max. 142 when x= 2 ; Min.  2055 when a; =11.
3. (i) 150ft./sec. (ii) 584 ft./sec. 4. (3$, 400, s=3J.
5. f(x) = x*3x 2 + x + C} 2.
D.
1. 80 ft./sec.,  36 ft. /sec., 87'72 ft./seo. at angle 65 46' with downward
vertical.
a . Max. (1690) when ,=
at
... 16056J28. ..... , ...,
Mm.  ^^ (  5Oo) when x= (2097).
iif
3.
/I 32\
\3' 27/
4. ^ = 141 c. ins. 6. y = x 3 + 3x^.
V*
E.
1. 6 o/ . 2. 20 ft. by 20 ft. by 10 ft.
3. (i) (64,32). (ii) *2j/ + 16=0.
(iii) [J64 (l=F JgV y] or (27146, 21J) and (100954, 21J).
x 1 i 1 2
l). 12; 36; i, 0,1.
y 4 i o 13
6. (i) 46 ft./sec. (ii) 49 ft./sec. (iii) 45 ft./sec.
ANSWERS 407
F.
1. 391C. 2. Q, 385^; y = xl; y=x + l.
4. 109 x 10< ft. ibs. wt. ; 3740 Ibs. ft. /sec. ; 60 Ibs. wt. 5. 1818.
G.
1. 1080 ft. 3. Lengths of pieces are   ,   inches, or '56a, '44a
H.
. (x2)2(4x + l); 6 (x 2) (2x 1) ; 6(4x5).
(i) x=lor2. (ii) x=jor2. (iii) x  or 2. (iv) x = j.
1 2187X
JMax. pt. on y =f (x) ^ , j .
iMin. pt. on y=/' (a;) (2, 0).
Min. pt. on y =/" (x) ^ ,  ^J .
. of infl. on y =f(x] (2, 0), Q ,  g) .
Pts
x 1 0123
f(x) 82 4
/'(x) 27 4 5 13
/"(x) 54 12 6 30
/'"(x) 54 30 6 18 42
4. J below centre. 5. (399, 226) ; (1*05, 446).
o
I.
1. 20ins./sec. 3. (i) (3x + 2y) ins. (ii) f X y + ^ sq. ins.
4. (i) 7ift./sec. (ii) 12^ ft./sec. 5. 3 '32 x 10* cms./seo.
408 CALCULUS FOB BEGINNERS
J.
1. 425ft. a. 45 %.
3. is + from x=  2 to x=2 and  when x > 2.
dx
V_
dy
dx
4. j = 3z. 6. 1J m./hr., 4J m./hr.,
EXERCISES XXXV. p. 168.
1. 97J; 975 sq. ins.; 4875 sq. ins. 2. 4; 6 sq. ins.
3. (i) 285, 385. (ii) 328'35, 33835.
(iii) (20030A+/i2), jj (200 + 307 + ft2). (iv) 333^.
. 20*. [inside rects.{^ 2 , outside
EXERCISES XXXVI. p. 171.
1. 108. a. 121J. 3. 13J. 4. 28f.
6. 97$. 6. 4. 7. 23 J.
EXERCISES XXXVII. p. 172.
1. 905J. a. 4f. 3. 152J. 4. 274J.
6. 3696. 6. 48^. * &4 " 4 .
EXERCISES XXXVIII. p. 177.
(i) A = x4 + 2x. (ii) A = a:4 + 2a;
(iii) A=a; 4 +2ar16. (iv) 11728.
ANSWERS . 409
EXERCISES XXXIX. p. 181.
[In 1 9 an arbitrary constant should be added.]
2 . 2x __. 3. *3 + + c*. 4. ..
6 . ^. 7. 7*
. 8. ^* 2 ~. 1O. 223J.
11. 4032. 12. ^ + 2c. 13. L 14. 1164*.
O J
15. 0695. 16. 56. 17. 614^ J. 18. 2.
19. (i) 278. (ii) 2259875. 2O. (i) 56. (ii) 56. 21. 2431.
EXERCISES XL. p. 183.
1. 2. (ii) 5. 2. (i) 4. (ii) 12. 3. (i) g. (ii) ^.
kh
6. 211i. 7.  + 2 C2 ,. 8. p = l, q =
EXERCISES XLL p. 187.
1. 5180. 2. (i) 6055. (ii) 5180. 3. (i) 5215. (ii) 5180.
4. (1) 3355. (2) (i) 404641. (2) (ii) 338104.
(3) (i) 338291. (3) (ii) 335504.
EXERCISES XLIL p. 19tt
2. 14. 8. . 4. 14. 6. A.
6. a = 0, b= 8, c=6. 4.
7. a = h (h?k*), 6=A;23/i2, c=3A. Each area =
O. 2pk + ~. 10. 1(^ + 4^+ AS) "
12. 8j\, 1J. 13. <& 14. ' 2 = 4<
410 CALCULUS FOR BEGINNERS
EXERCISES XLIII. p. 193.
a. 69315. 3. 1419, 2027.
EXERCISES XLV. p. 200.
1. 52, 58, 49. a. 32, 32, 32.
4 . (i ) Ho. (ii) ('^H''**). 6. 14. .. 3.
EXERCISES XLVL p. 202.
1. 18J, 184. a. . 3. 69315, 6989. 4. 170H
EXERCISES XL VII. p. 204.
1. 0. 3. (i) 6. (ii) 0. 3. 3J, 275. 4. 4J.
6. a*. 6. 11J.
EXERCISES XL VIII. p. 206.
(i) 112J, 112325, '155%. (ii) 71*, 71465, 05%.
EXERCISES XLIX. p. 208.
1. 61 ft. 2. 5131 ft.
EXERCISES L. p. 227.
3/2 7/8
3. 304 ft. 3. 1117i ft./sec. 193ift./sec. = ^1 + 11 + 20, 2544ft.
2t o
4. 119 x 10* ft. Ibs. wt. 6. 560 ft. Ibs. wt. 6. 9000 Ibs. wt.
7. 750 Ibs. wt. 8. 375 Ibs. wt. O. * ( b ~ ) f^ + g _ {] ft lbs wt .
I \_ i J
10. 2,700,000 ft. Ibs. wt. 671 knots. 11. 192 ft./sec. 346 ft.
771/,)2y2
13.   ft. lbs. wt., ur ft./sec. 13. 10,560,000m It. Ibs. wt.
9g
14. 60 lbs.
ANSWERS
EXERCISES LI. p. 232.
1. 1207r=37699 c. ins. 3. irr 2 Ac. ins. 3. ^=6702 c. ins.
o o
4. ~(3rfc). 6 (i) 261ir=820c. ins. (ii) ir= 2060 c. ins.
o o
qo_.
6. (i) 16?r = 5027. (ii) 247r = 7540. 7. =3351.
8. ^. 0. 2 = 284 = l862.
10. =5x*.
11. (i) 120jr = 37G99. (ii) 60ir=188'5.
EXERCISES LII. p. 237.
1. (a) 666 Ib. ft. 2 (6) 246$ Ib. ft. 2. M ^. 3. M ~ .
O XA
r\ /i r\ R /i iv/i 2
4. (i) M g . (n)Mjg . 6. Ma 2 .
8. (i). Wr. 9. . 10. 2025, 162, 135 ft.
11. M. Ift. M. . 13. 857 ft. Ibs. wt.
^ A
90
14. (i) ^M6. (ii)Ma 3 . 16. 634M.
EXERCISES LIII. p. 244.
.. (771,85.34). 8.
.
/
412 CALCULUS FOR BEGINNERS
a
7. from centre, on a radius perpendicular to diameter.
oTT
8. Q a from centre, on a radius perpendicular to base.
8
EXERCISES LIV. p. 248.
2. M 2 +c2. 8.
. 2Ma 6. Mtc'. 6.
EXERCISES LV. p. 249.
1. On vertical line of symmetry (i) 5, (ii) '= .  below the top side.
O tC T O
3 1
2. On the median drawn to the base a (i) T ft below, (ii) h ahove,
. 2c + ft ft
3. On the line joining the midpoints of the parallel sides
... o437> ft . ft (a + 3fc) + e(2a + 4fc) j,
I 1 ) ST . (") r; STT nr rrv below the side a.
l 'a2b 2 ' ft (a + 26) + c (3a + 36) 2
EXERCISES LVt p. 253.
2
1.  Tra 8 . [Area of base of cylinder, same height and volume.]
4.
5. 420 Ibs. wt./sq. ft. 6. (i) 7. (ii) 7f .
ANSWERS 413
MISCELLANEOUS EXAMPLES ON CHAPTERS VII
AND VIII. p. 254.
A.
I. 2 ^f^(38.41); 2 ^^ 2 (375). a. 300,. 3. 95*.
o
6. r of the way from A down the median.
5 of the way from A up the median,
o
B.
2 Q 3 2s 1
. =a?x
32
500 /101*\ /Q016\
(18if); ^(92if). 3. 3;^.
4. 5, 0, 4; 4204, 2856; 45, 20, 36; 135&, 74.
3. M  (where A is the altitude from A) = ^ s (s  a) (s  6) (s  c).
4. 139 ft. Ibs. wt. ; 465 Ibs. wt./sq. ft. 6.
D.
I. 30; 4; 26. 2. (i) 20J. (ii) 20J. 3. 10 inches.
4. 3 m. 6. 8713 x 108 gm. cm.2 ; 1075 x 10 9 ergs.
CALCULUS FOB BEGINNERS
E.
2. 135* = 4241 c. ft. x /22 7 5=4'743 ft.
3. tani^ = 4ll'; 0122; (161, 112).
4. ^ (2aR  a2) ft. Ibs. wt. ; ,^ ft. Ibs. wt. ; 35 : 32.
(R + a)
P.
1. (i) g*' + c. (ii) 37 J.
3. 4514, 2374 ; 1160.
7 O1 73
I.*, ?. 0;?4.
4. ~(P
G.
1. i, 12isq. ins.; , 2iins.; ^, 6H sq. ins. 6. 7r =
H.
1. 7{, IS, H; .
2. (i)27. (ii).y (lii) 2916T=916. (iv) ,
'7 o.ft.
6. 360 ft. Ibs. wt.
J.
,
12
9. = 382ins./min. 3. 439 ft. Ibs. wt.
6.
r+o
aib3. w t. r
ANSWERS 415
EXERCISES LVII. p. 268.
3. 2, 1444, 3560, 0232, 3598, 0, V!3 = 3606, ^13= 3606.
y = 1x\, j/ + 144.r=4748, y + 3598* = 7'77.
y = 3606* + 3 545, y = 3606.
5. (i) 5 sin 5*. (ii) 6 cos 2*. (iii) cos 4a; + sin 3x.
(iv) n(acosnx & sin no;).
6. a cos art, a sin art, aaisinart, aw 2 cos art, aw cos art, aw 2 sin art.
9. (i) sinx. (ii) cos a:. 1O. (i) a>i 2 cosnx. (ii) an 2 sinnx.
11. (i) ^13, ^13. (ii) ^41, ^41.
4
12. (i) cosz + c. (ii) sinx + c. 13. 2,  = 1'273 sq. ina.
14. (i) 04598. (ii) 08415. (iii) 1. (iv) 1.
15. (i) 0. (ii) 0. (iii) 0. (iv) 2.
16. (i) COS7UC + C. (ii) sinnx + c.
n n
18. [An arbitrary constant to be added in each case except those
marked *.]
1 1 34
(i) cos2a;. (ii) siu2x. (iii)  ^ cos 2x +  sin 3ar.
(iv) ar + sin2x. (v) 0*. (vi) 0*. (vii)  Z cos2ar.
2i 4
.. x sin2 . x sin2a; . . IT* , ., T*
( T1U ) _ _ ^ . ()  + __. ( X )  . (xi) .
19. , or (15708, 03927). 2O. .
21. [Add an arbitrary constant.]
... l/cos5x \ .... l/cosSa;
sin5x
5
cos(pg)x
pq
\
)'
1 fsin(p + q)x sin (yg)A
V P + q ~Yq )'
sin(pg)x\
Pg /'
416 CALCULUS FOR BEGINNERS
22. 5150. 23. 96 sq. ina. 24. 6.
25. 791. 26.  1745 ft./sec.,  4685 ft./sec.
27. (i)  = 6366r. (ii) ^=7854r.
7T 4
EXERCISES LVIII. p. 274.
4. (i) n cosec 2 n*. (ii) + n sec nx tan nx. (iii)  ;i cosec nx cot n.r.
6. An arbitrary constant to be added in every case.
(i) tan*, (ii) cot a;, (iii) sec a;, (iv) cosec*. (v) tan*.
(vi)  cot x. (vii) sec x. (viii)  cosec a;. (ix)  tan 3*.
(x) tan x  x. (xi)  tan 3*  x.
o
B. 2500, 1250, 15075, 1523, 0381, 0095 ft./sec. ;
i, , 553, 547, 2189, 8754 ft./sec.
EXERCISES LIX. p. 277.
BOA nv
96
1. 15535 ft. 2. 1396 ft./sec. 3. 35S mins./sec. or radians/sec.
4. 9 ft./sec.; 17 11' per sec. or ^ radian/sec.
EXERCISES LX. p. 279.
3. (i) (2* + 3)sin* + 2cos*. (ii) * 3 sec 2 * + 3x 2 tan x.
EXERCISES LXI. p. 281.
1. *cos* + sin*. a. * 2 sec 2 * + 2* tan *. 3.
7
4. 2 cos 2*. 5. T*~*T. e. sec* tan*.
7. 4* 6 cos* + 20* 4 sin*. 8. 15* 2 sin 3* +10* cos 3*.
9. 2 cos* sin*. 1O. (i)  10 cos 5* sin 5*. (ii) 10 sin 5* cos 5*.
cos x cos * 2 sin *
ANSWERS 417
EXERCISES LXII. p. 282.
2
1. 4r 3 + 12.r 2 + 6. 2. 1=. 3. x sin 2x + a: 2 cos 2ar.
4. 3 (4x
EXERCISES LXIII. p. 285.
1  a; 2 x cos x  sin x
2 ' "'
Bin x x cos a;
sin 2 * (x 2 
2 cos a; sin a;
tj m _ O , ^
(1 + sina;) 2 ' ' cos 2 ar '
EXERCISES LXIV. p. 286.
4. sec 2 4x.
2 N /x 2 + a; + l'
'. 2cos(4x + 7). 8.
5 ji
IO. (2 sinx + 3 cosx)*.
EXERCISES LXV. p. 288.
1. 5sin5x. 2. 3sec 2 3x. 3. 3 sin 2 x cos a?.
1
4. 2xcos(x 2 + 3). 6. ^ ycos^/x. 6. 14(2x + 3).
7. 4(6a; + 5)(3x 2 + 5x + l) 3 . 8. 21 sin 2 (7x + 5) cos (Ix + 5).
EXERCISES LXVI. p. 290.
lf C08X . a. jcoajx. 3. 4 sin 3 x cos x.
2^/sina: 2 v a;
4. 4cos4x. 6. 4x 3 cos(a; 4 ). 6. n sin" 1 x cos x.
M. C.
27
418 CALCULUS FOR BEGINNERS
7. ncosnx. 8. ncosa?. 9. nco8 n ~ 1 xsinx.
1O. nsinnar. 11. 4cos(4x + 5).
12. 2 (3 sin x + 4 cos x) (3 cos x 4 sin a;). 13. 2 tan x sec 2 .
14. 2 sec 2 a; tan x. 15. n (o sin x +b cos x)* 1 " 1 (a cos x b sin z).
sec 2 *
16. ni&n n ~ l x sec 2 *. 17. nsec B x tanx. 18.
2 A^ tan j;
1 la 8* + 3
21.
23. 120 (6x+ ip. 24.
26.  .s rj. 27. na(ax + b)*~ 1 .
(x 2 +l) 8
28. r a(ax + ft) . 29.  sin x. cos (cos x). 3O.  cos x. sin (sin x).
3
a cos (ax + b)
34. acos(ax + &). 35. v ' .
36. 12sin 2 (4a; + 5)cos(4:c + 5). 37. wasin" 1 (aa; + 6)cos (ax + b).
38. na tan n1 (ax + b) sec 2 (aa: + 6). SO. na sec" (ax + b) tan (ax + b).
(ix + 5) cos (2z 2 + 5x + 6)
4O. s =^. 41. 3 sin x cos x (a sm x  b cos a;).
42. 15 sin as cos a; (a sin x  b cos a;) (a sin 3 x + b cos 3 a;) 4 .
7 sin x cos a;
43. 44.  lo cos* 3a; sm 3x.
^3 sin 2 x  4 cos 2 x
45. 6a;tan 2 (x2) S ec 2 (x 2 ). 46. n6x n icos (a + 6x).
5 sec 2 2x + 6 sec 2 2x tan 2.T
47.  . . 48. mnx n ~ l sin" 1 " 1 (.c n ) cos (x n ).
49. 3 cos x  12 sin 2 x cos x = 3 (4 cos 3 x  3 cos &) = 3 cos 3x.
50. 12cos 2 xsinx + 3sinx = 3(4sin 3 x3sinx)= 3sin3x.
51. cosw.^. 52. 3 sin 2 u. cos w^*. 53. 2u.^.
ax da; dx
ANSWERS 419
64. *ul 65. *.. 56. .
ax 2*Ju dx it 2 da;
cos u du du
67. . . 68.  n cos"" 1 w . sin 3 .
2^/sinu dx *
59. CO.,/.. 60. ,==..
V" V **" 2 o
An arbitrary constant to be added from 61 75.
61. 5 sin 3 *. 62.  rain** 1 *. 63.  rcos w+1 z.
n + 1 n + 1
64. cosnx. 66. siuna;. 66. tan war.
n n n
67. t&nnxx. 68. cos (ax + 6). 69. sin (ax + 6).
n a a ^
70.
73. ^ec^x. 74. . 76.
EXERCISES LXVII. p. 294.
2z a 2  2x 2 2 sin x
1. ; r. 2. . 3. 
4. 2xcosx(cosxxsinx). 6. (2x + 1) 6 (3x + 4) 4 (7
6
7 . ^ 8 .
( a 2_a;2)t 2*Jx
9. (ax + 6)"*i (ex f d)" 1 {(w + n) acx + mad + nbc}.
10. 8 sin 2 x sin 4x. 11. 4 (^ ~ x2 ~ 4 ) .
 15 cos 2 5x sin 5x  12 sin 3 2x cos 2x
12.   . 13.
^2 cos 3 5x  3 sin* 2as
272
420 CALCULUS FOR BEGINNERS
,,,_i _i f du dv\
14. n sin"" 1 x cos (n + 1) x. 15. u m l v n 1 1 mv +nu ] .
V (IX CLxJ
16. sin" 1 " 1 u cos"" 1 u ( wi cos M cos v n sin w sin v 1 .
V &3/ (tXJ
5 ^/sin 5x (15 cos 5x + cos IPs + 2) __!L__ f* j. /;
(du du\
r ^~ M ^j
v 2 I
EXERCISES LXVIH. p. 296.
t 2 _ cos 9 + b sin g
' ~
usina
4. 2?w. 9.  seconds.
EXERCISES LXIX. p. 298.
5 2cos2j; 43
1. 5x8. 2. sec 3 x. 3.  . 4. ; s. 6. cotz.
3 coso; bt + 2
EXERCISES LXX. p. 299.
M& Ma 2 M
 ; 8  "' ~
4. 7ra 3 io [to = weight of 1 c. ft. of water] ; z ft. below centre.
6.
EXERCISES LXXI. p. 300.
ANSWERS 421
EXERCISES LXXIV. p. 309.
3
o.
EXERCISES LXXVIII. p. 321.
1
. (i) W. (ii) I
(i) Se 3 *. (ii) 2ff.*. (iii) 2e 2 *+3. (iv) ae ***. (v)
(vi) e x (x + 'i). (vii) e z (x 2 + 2a;). ('iii) e* z (kx n + nx n ~*
(ix) e* (sin x + cos x). (x) e 2z (2 sin 3x + 3 cos 3x).
(xi) e aa; {a sin (6ar + c) + 6 cos (fix + c)}.
(xii) e~ 3* {  3 sin 2x + 2 cos 2o; } . (xiii)
,. du
(xiv) e. . (xv)
(xvi) e3*(3sin*5a; + 20sin35xcos5x). (xvii) 7 a: log e 7=7 :B xl9459.
7. [Add an arbitrary constant.]
(i) e*. (ii) \e**. (iii) le**. (iv) \e*\ (v) e tanz .
O A
(vi) e*. (vii) y s=7*x5139.
le '
8. e 2 = 47209; 236 sq. ins. 1O. 0237.
EXERCISES LXXIX. p. 325.
13 3
2. (i) . (ii) . (iii) 5 =. (iv) tanas, (v) 2cosec2x.
X X OX "T*
6a; + 5 .. 910 F 1 H
( V1 ) 05 5 r ( V11 ) 2cot2x. (vui) = : I.
3z 2 + 5x + l v ' v ' x L log e 3j
422 CALCULUS FOR BEGINNERS
(ix) cosecx. (x) secx. (xi) secx. (xii) ^ ' ^
2ax + b .. 1 du
(XX)  r . (XXl)  5  r  . (XXll)  . j .
v ' z 2 6a; c dx
3. [An arbitrary constant should be added in each case.]
(i) log e s. (ii) log e x or log e jx. (iii) log (2* + 3).
(iv)  log e cos x or log e sec x. (v) log tan ^
(vi) log e tan + or log e (seca;+tana:). (vii)
(viii) log(ax + b). (ix) log e (x 2 + 4*  7). (x)
a &
(xi) log.^ + ^^). (xii)lo g< . (xiii)
a:
(xiv) xlog e xx. (xv) + 2x + log e x. (xvi)  + 3x7log e (x + 2).
(xvii) j + * 2  a; + 5 log e (x 3).
6. (i) and (ii) log <J 6=l'792. 6. 219 ft. Ibs. wt. 7. 219ft. Ibs. wt.
8. (i) log. 3 =10986. (ii) \ log e ^= 01062. (iii) ~ log. 3
(iv) i log. 2 = 03466. (v) log. (^2 + 1) = 8813.
IB
10 . , or (18484, 021875).
11. ^2=06931. 12. 3030.
13. Max. value (ll + 301og < 3) = 1098, when *=1.
Min. value (  27 f 30 log, 5) = 1064, when x = 3.
ANSWERS 423
EXERCISES LXXXI. p. 334.
3.34.1
(37s 4 + 81* 3 + 24*2 + x3).
(96*6 + 24x  19* 2 + 32* + 12).
/cos 2 a; \
7. (Binx)c** ( sm x . log e sin x J .
EXERCISES LXXXII. p. 340.
1. o;=2e3; 16206ft.; 1304 sees. 2. 811 Ibs. wt. ; ^AOP=0644'.
3. After 6575 minutes.
4. h= j; (i) 2 mins., (ii) 3 mins., (iii) 332 mins.
6. v  850 = (u  850) e~ w . [ u ft. /sec. is speed when t = 0.]
6. 48634 ft. [=50(1 ).]
J 32_
7. y = e 12800 . [j/ sq. ins. is crosssection x ins. from the lower end.]
EXERCISES LXXXIII. p. 342.
8. (i) y = A cos 2z + B sin 2x.
(ii) y = Ae 2iE +Bc 2a: or A' cosh 2a; + B' sinh 2o5.
7 3
4. ?/ = 5cos2a; + 26in2a;. 6. j/ = 
2
6. (i) 60 ft. /sec. (ii) 75 ft./seo.
424 CALCULUS FOR BEGINNERS
* '* V
10. (i)
(iii) Ax+B.
11. (i) y = ke*+Be* x . (ii) j/ = e2*(Acosx+B sinx).
(iii) y = e2* (Ax + B). (iv) y = Ae** + Be*.
(v) y = e Aeos*+Bsin. (vi) y = e* x (*x+ B).
76 3
12. 3. 13. 3. 14. g,  6l . 16. j.
/ /3 /3 \ 11
18. (i) y = Ae*+B 3a! + 2. (ii) y = e& ( A cos ^ x + B sin ^ a; J + y
(iii) y = e
MISCELLANEOUS EXAMPLES ON CHAPTERS IX XI.
p. 344.
A.
2. 4i 2 + 7 = 5 ' 149 ;  + log, 2 = 2 193. 4. 2Mft.lba.
v^
6. 7033; 7034.
B
1. (i) 6 sin Bx cos 3x. (ii) ^x~^ (8x sin 4a; + cos4a;). (iii) tan.
(iv) e
22
(ii) cos3x + sin6x + c. (iii) 1.
3. 4 ft./sec. ; 4 ft./sec. 2 ; both towards O if A is moving away from O.
4. when x = .
me m
0.
fl 2_ j2 1
1. at ; a. 2. ^; 500. 3. .70"; 87 ft.
ANSWERS 425
D.
22* 2 1
1. (l)  r. (ll) = :.
2* (1 +*)*(!*)*
2. (2, 1), (1, 1); 2x3j/l=0, x3y + 4=0. 3. ^132 = 1606.
4. 8933r. 6. (lc"" 2 ); 142.
E.
1. (i) log,,a; + l, . (ii) e x (sino:+cosa;), 2e*coacB,
x
... xcosx sinx ,
(iv) 4 sin 3 a; cos x, 4 sin 2 a; (3 cos 2 x  sin 2 x).
2.  6 tan C . w (where & is CA and C the angle ACB).
4. 7 : <: from centre of first sphere, ; c from centre of second.
a^ + 67
6. 2 + 2jr =63496.
P.
1. Tg ; TTg,  6 cos 2 2x sin 2ar, e 01 (a sin fa; + 6 cos bx), 4 sec 4x.
PN 2
2. c; PG=  . 3. 251 ft./sec., 241 ft./sec., 1885 ft./seo.
c
0. ?rG cos a ft. Ibs.
G.
(iv) 4sec 2 4a;. (v) e3* (1  3*). (vi)
a. 18 1
2. ^=. 4. (i) 22 (ii) 03054. (iii) 02456. (iv) 5.
v/ ^ & w
(v) 00429. (vi) 03466.
5. 
426 CALCULUS FOR BEGINNERS
H.
(iii) 144x6 _ 280s* + 168x 3 120.r2 + 78o; 15. (iv) 5 sec 2 5*.
1 /j4 _L /j2 T 2 _ 4.J4
(v)  ^. (vi) 6sec23*tan3*. (vii)  
(1  x 2 )? V a 2  a; 2
a. 03690 and 17602 Minimum.
03690 and 17602 Maximum.
3. 382 ins./sec. 4. be'* ( 3cos3 jjsin3n . 6. 5
\ J / ^
dd> irn cos Ar airn s <
a. j7 = ^ . ; 3=  57T r^ > 1>714 radians/sec. ;
dt 60 cos</> dt, 30 co
3715 ft. /sec.
4. 628 ft. Ibs. wt., 698 ft. Ibs. wt.
J.
a. 18083. a. c2('l+Vl = c 2 x 11353.
V V
2r sin 3 o 2r sin a * ? sin a ,
4. ^j : ; ; from centre.
3 (a  sin a cos a) 3a a
K.
6. 2a 2 ; (l513a, l'003a).
L.
8. 2, 2, . a. 02(^/2 l)=04142a.
4. ( 6, J ; ft. /sec. making 210 with OX.
(396) ft./sec.2 making  13 54' with OX.
6. 593 ft. Ibs. wt.
ANSWERS 427
K/7
1. 6tan3xsec3x, x (1 + 2 log 4x),
2 (5*+3) (2x2 + 7)" (170x2+72x4 175), 3x(a2+*2)i,
a. 62 45';  = 03183.
3. 1116 ft.  from A. 885 ft. from B.
4. A = 100^13 = 360 6. g = tan 1 l5=c.M. of 5619'.
6. aw sin wt, aw 2 COB w.
N. .
to to ^.fo/.^fl M. 2 . a *~ b
2 + 2~ ' 2 2  IK 2 ~ z ' ~>T
2. ^Q radians/sec. 4. (i) 350. (ii) . (iii) ^.
(iv) 0055. (v) 0055.
6. 3165a 3 .
O.
(iii) (12a^ + 18* + 5) (6*2 + 3). (iv) y cos a: or e" n!B . cos *.
& ^T ?  (vi ) 555 < vii )
4. 20 log. 3 = 2197 ; (728, 152) ; 184.
/\ ^ /* \ ^ \ * /% ^
6 ' (l) T' (ll) T* (m) ' (W) 2'
P.
I. (i) x 2 cos x + 2x sin *. (ii) 2xe 2a! (a; + l).
(iii) ^(aco&ZbxZbcoBbxsinbx). (iv)  ^ .
. sin x + x cos x ... 1 . x .... cosjx
00 / , (vi o sm o ( vu ) , ='
2 ^/a; sm x 4
a. 3x + = 6. 6. 16000 ft. Ibs. wt.
428 CALCULUS FOR BEGINNERS
Q
B.
1.  1482 ft./sec. ; 6187 ft./sec. 2
4. (i) . (ii) 0243. (iii) 0. (iv) 0.
S.
a. IJins./sec.; radians/sec. 4. e5 = l649.
oo
r 2
6. 7 below centre.
4c
T.
x 3 a 3
1 . (i) g + log e a; + c. (ii) log a sin x + c or log e A sin at.
(iii) J. (iv) 1^ = 02146. (v)^
3. 8ins./min.; 3*175 ins./min.
3am(27ft3) 3a(l2m 3 ) m(2m
*'
6 . 7 . 82ins .
EXERCISES LXXXIV. p. 373.
e. 2095. 7. 0315, 0446, 1069. 8. 04142. . 085138.
1O. 1203. 11. 1134 ins. 12. 11839. 13. 180, 113.
14. 4167 ins. 15. 28 23 / . 1. 1166. 17. 2289.
18. 0657. 19. 4536. SO. 3945.
ANSWERS 429
EXERCISES LXXXV. p. 379.
[Add an arbitrary constant in each case.]
1.  5 cos 8 x. 3. cosz 2 . 3. log e (x 2 + 3x + 5).
O a
10 2x + 3
6. 21og e (jc 2 + 3a;+5) +  7T  r tan 1  Trr . 7. cos (log,
8. e 8 " 8 . 9. log. (1 + log.*).
1. , ,2x 1 .
10.
2a;
11. " T coth r or log . 13.
13 sin x  5 sin 3 x. 14. 5 tan 6 ar. 16. s tan 2 x + log., cos x.
O U &
2 . ,s cos 3 a; cos 5 a;
16. (2a + a;)*. 17. a~ + 5
1 8. (i)  2 cothi (2x  3) or log e ^ .
a/ ~~ A
2 a;
(ii)  2 tanh 1 (2x  3) or log e = .
3 1
19. ^logj^ + lJ + tanJx. 3O.   log e cos (era; + b).
2 1 /
31. cos s a; + cosx. 33. */23a: 2 .
as. _t a nhi + lo g (3x2. 34.
38. 2 tanh l Jx. 3.
30. log. 2.^* + (log. a:) 2 .
430 CALCULUS FOR BEGINNERS
EXERCISES LXXXVI. p. 381.
1 f jZ
 2^3 V a
ax
\.
' "
6 (* 3) '
2ax ~ 1 ~ , , x ax
6. 
15
8. ^f
x 1
14. sinh" 1 . 15.  [log, (sec a; + tan x) + tan a; sec x]
(i J
EXERCISES LXXXVII. p. 383.
. 2. e*zl. 3.
1 . _
4. a; tan 1 x jrlog^ll a; 2 ). . a;sin~ 1 a;+ Jlx\ 6.  x cos a; + sin x.
a; ^rn 8in(7K+n)a: )
+  ( x cos (m  ?t)ar  sin (TO  n) x } I .
mn\ TOn ' JJ
B. 
13. g p. e"* [a sin (fee + c) 6 cos (for + c)];
jp p . c * [6 sin (bx + c) + a cos (bx + c)].
16 357T F6 . 4 . 2 7.5.3.1 iT\
35' 256' l_7.5.8 ; 8.6.4.2'2_T
ANSWERS 431
MISCELLANEOUS EXAMPLES ON CHAPTER XIII. p. 385.
3. ~ + 2x2log e (x + l). 4. tan 1 ^.
5. log e a.log e x + (log e x)*. e. log, (1 + sin*).
3
. , , 1 ,
8mh or log
8. ^V^+sM* 3  2 " 2 )
1O. xo 11. x^5(
O JO
12. sin 4 2ar. 13. ^x cos 3a; + 5 sin 3a5.
o o J
122 8
14. ~a; 2 cos3a; + xsin3a5+2=cos3a;. 15.  (3
16 ' 32x)(27 + 30,). 17.
18. l ogs seo + a 19.  tanhi  + log e (3  x^J.
.. / a\ ~, < i
+ 241og,(x2). 21. log, t=log e tan.
22.  log,, tan I  I =log. tan ( r +  } . 23. = sin~ 1  .
\4 2/ \4 a/ a
24.  = tan J . 25. 1568. 26. 0430. 27. 0148.
a a a
28. 3. 29. 00266, 3O. 0430 f^l .
oo I 2ob I
81. 5tani = 02145. 82. log,2=0'2310. 33. y = .x^.
432 CALCULUS FOR BEGINNERS
36.   tan 1 a " 2 . 36. v = vi tanh . 37. ^^ ft. Ibs. wt.
38. If v ft./sec. is speed, and t sees, time,
_110 9 A _198
~~T~' '~3080' ' "86"*
4177 m./hr.; 499998 m./hr.
EXERCISES LXXXIX. p. 393.
2. (i) tan0. (ii) cote, (iii) cotfl. (iv) tantf.
ira a ira s a 3
1O. r=ae is the same as r=e
*
r=6e is the same as r=e
Table of the Exponential and Hyperbolic Func
tions of Numbers from O to 25, at Intervals of !.
X
ex
e~*
cosh x
sinh x
tanh x
1000
1000
1000
1
1105
905
1005
100
100
2
1221
819
1020
201
197
3
1350
741
1045
305
291
4
1492
670
1081
411
380
5
1649
607
1128
521
462
6
1822
549
1185
637
537
7
2014
497
1255
759
604
8
2226
449
1337
888
664
9
2460
407
1433
1027
716
10
2718
368
1543
1175
762
11
3004
333
1669
1336
801
12
3320
301
1811
1509
834
13
3669
273
1971
1698
862
14
4055
247
2151
1904
885
15
4482
223
2352
2129
905
16
4953
202
2577
2376
922
17
5474
183
2828
2646
935
18
6050
165
3107
2942
947
19
6686
150
3418
3268
956
20
7389
135
3762
3627
964
21
8166
122
4144
4022
970
22
9025
111
4568
4457
976
23
9974
100
5037
4937
980
24
11023
091
5557
5466
984
25
12182
082
6132
6050
987
Table of Logarithms to Base e.
o
1
2
3
4
5
6
7
8
9
I
095
182
262
336
405
470
531
588
642
2
693
742
788
833
875
916
956
993
1030
1065
3
1099
1131
1163
1194
1224
1253
1281
1308
1335
1361
4
1386
1411
1435
1459
1482
1504
1526
1548
1569
1589
5
1609
1629
1649
1668
1686
1705
1723
1740
1758
1775
6
1792
1808
1825
1841
1856
1872
1887
1902
1917
1932
7
1946
1960
1974
1988
2001
2015
2028
2041
2054
2067
8
2079
2092
2104
2116
2128
2140
2152
2163
2175
2186
9
2197
2208
2219
2230
2241
2251
2262
2272
2282
2293
log 10=2303, log 102=4605, log 103=6'908.
M. C.
INDEX
The numbers refer to pages.
Acceleration 19
,, from gradient of speedtime graph 33, 67, 68
Approximate equality, test of 39
,, solution of equations 362373
Approximations 136144
to f(x + h) 141, 196198
Arbitrary constants 145147, 173, 174, 181
Area of plane curve 160192
by summation 161, 165, 166
from 5rW(*) 170, 177
OfX
approximate rules for 184186
sign of 187190
,, ; , from ordinates of integral curve 193196
Area traced by moving ordinate 169177
Atmospheric pressure as function of height 337
Average speed 2
,, gradient 22, 23
rate of increase 12, 15, 16
ordinate 199203
values 250253
Centre of gravity 238244
,, ,, of plane area 240
,, ,, ,, solid of revolution 241
,, ,, ,, quadrant of circle 243, 244
,, ,, ,, quadrant of ellipse 300
Centre of pressure 248, 249
,, of circle 300
,, ,, rectangle 249
,, triangle 249
436 INDEX
Centre of pressure of trapezium 249
Circular ring, thin 137
disc, moment of inertia of 235
Compound Interest Law 334339
Cycloid 297
area of 299
Definite integral 179, 180
,, approximation to value of 192, 193
represented by area 192, 219
Derived curves 9497
Differential coefficient 51
,, coefficients, higher 93
equations, Exs. on 342344
Differentiation, meaning of 51
from first principles 4771
of a* 69, 70
* 7579
kx n 8082
x* + c 80, 82, 83
sin x 261265, sin 1 x 304, 305, sinh x 331, sinh 1 x
332, 333
,, cos a; 268, cos 1 * 306, cosh a; 331, cosh" 1 a; 332, 333
tana; 272275, tan 1 a; 306, tanha; 331, tanh 1 x 332
cot a: 274, cot" 1 a; 307, coth 1 x 332
sec a; 274, sec" 1 a; 307
,, ,, coseca: 274, cosec^a: 307
,, sum 84
,, product 278282
,, quotient 283, 284
,, ,, function of function 285290
,, ,, inverse functions 301307
,, ,, implicit functions 308, 309
,, when two variables are each given as functions of a third
295, 296
of rf 311, 314
,, c* 313, 319
log,* 323; log a a; 323
list of standard results in 376, 377
Diminishing quantities, ratio of continually 4045
dt , As .
Jt and At 48  5
f and 5557
dv Av
INDEX 437
and 5254, 5865
fix Ax
da
t as speed 48
^ ae gradient 5861, 85, 86
dx
dy
~ as rate of increase 52, 53
dx
~ as ratio of timerate of increase of w to timerate of increase of x 54
dx
^ , sign of 66, 7173, 99
d3j
dy and dx no meaning when separated 61, 62, 180
dy d z y
Z. and , sign of, in relation to shape of curve 99
dx dx i
At, meaning of 47
Distance from speedtime formula 205210, 221
,, graph 211213
e, definition of 313
e, approximation to value of 319, 328330
e as Lt fl + V 339
e x , differentiation of 313, 319
Elasticity of volume 139
graphical representation of 140
Ellipse, moment of inertia of 300
centre of gravity of quadrant of 300
Equality, approximate 39
Equations, approximate solution of 362373
Exs. on differential 342344
Error, relative 139
Errors, approximation to value of small 136144
Family of curves given by ^=/(x) 149
dx
Frustum of cone, volume of 229, 230
Function 14
represented graphically 33
Functional notation 91, 92
438 INDEX
Function of function differentiated 285290
/ (x + h) =/ (x) + hj ' (x) approximately 141
/ (x + h) =f (x) + Jif (x) + ^ h*f" (x) approximately 196198
Fundamental notions 146
Geometry, Applications of Differential Calculus to 8790
Gradient 1929
uniform 1921
average 2223
at a point 2326, 38
,, representing rate of increase 33, 34
of spacetime graph represents speed 33
of speed time graph represents acceleration G3
Graphical solurion of equations 362366
Gyration, radius of 234
Harmonic motion, simple 269
Higher differential coefficients 93
Hyperbolic functions 330333
logarithms 315
Implicit functions, differentiation of 308, 309
Inertia, moments of 233237
, theorem of perpendicular axes 237
parallel axes 245247
Inflexion, points of 127133
Integral curve 193196
definite 179, 180, 209
indefinite 180, 181
Integration 160172
,, meaning of 1GO162
,, as a process of summation 219
> antidifferentiation 219
of  323
x
list of standard results in 376, 377
some methods of 377383
by change of variable 299
examples of 220226
Inverse functions, differentiation of 301307
operation, the fgiven ^ find y\ 145152
INDEX 439
Limit, limiting value 9, 30, 38, 41, 48, 75, 162
Limits of definite integral 180
Logarithms, Napierian 314, 315
,, of same number to different bases 315
log e x, differentiation of 323
Logarithmic differentiation 333
Maxima and minima 106133
,, tests for 108, 110, 123
,, Exs. on 112117
Mean ordinate 199203
,, values 250253
Moments of inertia 233237
of rectangle 233, 234
,, circular disc 235, 236
,, ellipse 300
Napierian logarithms 315
Natural 315
n x , differentiation of 311
,, graph of 316
Negative speed 13
,, rate of increase 16, 17
Newton's law of cooling 336
Pappus, theorem of 353
Parabola 8890
Parallel axes, theorem of 245247
Perpendicular axes, theorem of 236, 237
Points of inflexion 127133
Polar coordinates 388393
Product, differentiation of 278282
Quadrant of circle, centre of gravity of 243
Quotient, differentiation of 283, 284
Eadius of gyration 234
Bate of increase 1017, 38
,, ,, ,, meaning of negative 13, 16, 17
., ,, ,, ,, ,, zero 13
,, ,, represented by gradient 33, 34
Katio of continually diminishing quantities 4146
440 INDEX
Rectangle, moment of inertia of 233, 234
Revolution, solids of 228232
sec a:, differentiation of 274
Shape of curve 99101
Sign of area 187190
Sign of ^ and ^ 56, 7173, 99
ax dx*
Simpson's rule 185, 186
sinar, differentiation of 261265
Solids of revolution, volume of 228232
Spacetime graph 33, 35, 36, 66, 211, 213
Speed 110
,, uniform 1
average 2
,, at an instant 310, 37
given by gradient of spacetime graph 33, 66, 67
Speedtime graph 33, 67, 211, 213
Sphere, volume of 231, 232
Square plate, expanding 15, 52, 53
Standard forms in differentiation and integration 376, 377
Stationary tangent 129
Tables 433
Tangent at a point of a curve 27
tanx, differentiation of 272, 273
Thrust on immersed area 223
Trapezoidal rule 184, 185
Trigonometrical ratios, differentiation of 261277
,, ,, inverse 304307
Turning points 107111, 129132
Uniform gradient 20, 21
,, speed 1
Volume of frustum of cone 229, 230
sphere 231, 232
Volumes of solids of revolution 228232
Volume strain 140
Work done in stretching elastic string 213218
,, by expanding gas 224226
Zero speed 13
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