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Full text of "The calculus for beginners"

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 MARUZEN-KABUSHIKI-KAISHA 



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 Co-ordinates. 

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 co-ordinate 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 speed-time formula . . . 205 253 
Speed-time 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 

203-213 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 CO-ORDINATES 

283289 Meaning of polar co-ordinates .... 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. 



1-3] 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 



1-2 



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 9-75 

(4) I 4-69 

(5) 1 2-30 

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 37-52 

(5) 1 36-80 

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 2-30 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 2-625 ins., and 

in the same way if we assumed uniform speeds of 39 and 
37'52 ins./sec., we should get 2-44 and 2-34 ins. respectively. 

These distances 3, 2'625, 2-44, 2-34 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 36-8 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 
36-8 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 quarter-mile 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 quarter-mile 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 1-25 125 

5 55-75 111-5 

2 20-776 103-88 

1 10-142 10142 

01 0-992402 99-2402 

001 0-099024002 99-024002 

0001 0-009900240002 99-00240002 

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 + 2-4 + -02 or 101-42.] 

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, 3-2, 3-1, 
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 (3-j) 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. 



5-8 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 = -. = = = 111-50, 
Increase in t 'O 



12 CALCULUS FOR BEGINNERS [CH. I 

and we say that 111-50 is the average rate of increase of s 
per unit increase of t between t = 4 and t = 4-5; for if the 
body were moving uniformly in such a way that 

an increase of -5 in t produced an increase of 55-75 in s, 
then 1 would produce 111-50 in s. 

Similarly 101-42 is the average rate of increase of s per unit 
increase of t between t= 4 and t = 4-1, for if s were increased by 
10-142 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 

111-50 is the average rate of increase of s with respect 
to t between t = 4 and t = 4-5, meaning that as t increases from 
4 to 4*5, s increases 111-50 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 + 4A-2A 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, * = 50-2A 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 



8-10] 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.] 



10-13] 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-- - = =2-1 (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 1-001" 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 =1-001 is 2-001. 

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 



= 1-9. 



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* 



13-16] 



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 


40-5 


40 + 7* 














1200 


p 


30 


26-67 


28-57 


29-27 


29-63 




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 = 40-5 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 4-figure 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, 
>> P-s^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), 

15-10 1 
and gradient = ^- lo =2 ; 

coordinates of P 2 are ( 2, 4) and of Q 2 ( 6, 2), 

2-4 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 

.. . 9-5 



20-23] 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 2-4. 
Gradient of PQ 4 or average rate of increase of y per unit 

increase of x between a; 1 and x= 1*2 is 2-2. 
Gradient of PQ 5 or average rate of increase of y per unit 

increase of x between x = 1 and x = 1 ! is 2-1. 
Gradient of PG^ or average rate of increase of y per unit 

increase of x between x 1 and x = 1-01 is 2-01. 
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. 




2h-h 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? 



= 2-A, 



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. 



23-25] 



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 y-units 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 y-unit 
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, 3-5, 3-1, 3-01, 3 + fe, 2-9, 3-fc. 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;=3-5, 3-1, 3-01, 3 + ft, 2-9, 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=2-m 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 y-unit. 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, 4-5, 4-2. 

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 space-time graph 
is drawn, the gradient of the curve at any point gives the 
speed at the instant corresponding to that point. 

Similarly if the speed-time 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 space-time 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 space-time 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 -l-3333 101 -1-01 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) 9-30 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 = l-Qlx. 

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 



30-32] 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 
5-67 


0152 
11-43 


0038 -00095 -00015 
22-90 45-83 114-6 


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(l-cos0) 




Fig. 18. 






AM=r(l-cos20) 
= 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 
3-9924 



1 

3-9997 



30' 

3-99992 



10' 

3-999992 



3-999999... 



44 



CALCULUS FOR BEGINNERS 



[CH. I 



35. (4) In Fig. 19, let XOY be a triangle right-angled 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 


1-0013 
6 


1-00103 
4 


1-0008 
3 


1-0006 
2 


1-0005 
1 


Chora AP 

LAOP 

in degrees 


Arc^P 


0872665 


0698132 


0523599 


0349066 


0174533 


Chord AP 


0872388 


0697990 


0523538 


0349048 


0174530 


AxoAP 


1-0003 


1-0002 


1-0001 


1-00005 


1-00002 



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 half-second 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 

41-44] MEANING OF -7- 51 

43. We have found that the speed at the end of t seconds 
is (3-J-6* 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 



45-47] -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 

* = 6-2, i.e. Increase in y = 6*2 times increase in x 

= 6-2x -2 = 1-24. 

i.e. if the side be originally 3" and increase by -2" the area 
increases by 1-24 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 3-02 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 x-rate 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. 



48-50] 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 1-2 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 

joV-1200 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 59-7 Ibs./sq. in. 

[Actually the decrease is '2985.] 



51-53] 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 = 2n-r . 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 2-2 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). 



55-59] 



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 (3-02, 9-1204). 
/. RQ= -1204 and RQ' = 6 x-02 = -12. 




Fig. 24. 

The error in taking RQ = RQ' is about ^ / . 

o 

If PR = -001 then Q is the point (3-001, 9-006001). 
.'. 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). 



59-61] 



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 y-unit. 

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 2-01 2 be taken as 4-04 ; 
(ii) if7'03 2 49-42; 
(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. 



62-64] 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. 



64-66] 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 space-time 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 speed-time 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 speed-time 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. 



06-68] 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 



G9-71] 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 30-301. 

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 


3-5 


42-875 


5 


15-875 


31-750 


3-2 


32-768 


2 


5-768 


28-84 


3-1 


29-791 


1 


2-791 


27-91 


3-01 


27-270901 


01 


270901 


27-0901 



3-001 27-027009001 -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- 



Ax-f- 





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 text-books 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 n-w +><> 



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 



73-76] 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 s-1 

(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 



1-P-l -331 = 331 

wvi 

-3-0301, 



1-1-1 ~ -1 ' 
1-OP-l -030301 



1-01 - 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-\ W-1 ~ W-l 

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. 



1-1- 1 1-0741-1 -0741 

e -s- TTT r = 1-1-1 ^r - ' 741j 

1 -01 s - 1 1-00749-1 _ -00749 _ 
1-01-1" 1-01-1 -01 ' 

and in this case = ='75. 
4 

(3) Let i be negative = - m where m is positive. 

1-1 

" 



" 2-1 ~ a-1 "T^T """*' a-1 ' 

z m 1 
Since m is positive, Lt. -- = m, and Lt. z m = 1. 

-*! z ~ 1 s-fl 

z n 1 
.*. Lt. - 7- = m or w. 

*i -! 

M**-l -9535-1 

ag - : -- " 465 



99504- 1_ 



1-01-1 -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 2-1 

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 



80-83] 



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 



6-2 



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 A-y, 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- = -, l--y- + -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=x3-3x, 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;3-f-4x2-3 what is -j-1 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'2x-3 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 = 83x-162. 

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. co-ordinates of T are ( 0, -r- ) or OT = -: , but ON =7-. 
\ 4a/ 4a 4a 



/. OT = ON. 



Fig. 34. 
A 



In (2) put rr-0. .'. y = +2a, i.e. the co-ordinates 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 semi-circle on TG as diameter. 

But this semi-circle 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 y-axis 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. 



88-90] 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 nf---T : 

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.) 



90-93] 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, speed-time and acceleration-time graphs. 

93. We get interesting special cases when f(x) is of the 
1st or 2nd degree. 

e.g. if /(a) = 2o: 2 -3a:-t-U 

/ = 4*-3> 
and f" (x) - 4. J 



96 



CALCULUS FOR BEGINNERS 



[CH. Ill 



The graphs of y=f(x), y=f'(x), y-f"(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 speed-time graph is a straight line and the 
acceleration-time 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 = 3-4 + 2t 2 + t 3 find the speed and acceleration at the end 
of 3 seconds. 

Draw the space-time, speed-time and acceleration-time 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 (ar-1) (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 co-ordinates. 

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 co-ordinates of the points T, G where they 
meet the y-axis. 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 + t-5t 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 co-ordinates of the points at which the tangent is parallel 
to the x-axis. 



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 x-axis in T, G, and the y-axis 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 



V-RC+L.*?. 



where R = 0-5, 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 10-6 02 + 3 x io-7 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*-2x3-x2 + 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 straight-line 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 x-axis 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 rr-axis. 

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 co-ordinates 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 mid-point 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. 1-5) 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. 



98-100] 



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 
x-scale each time, the points where y =f (x) cuts the cc-axis 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 x-direction, 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 -/ ' 



100-102] 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(x-l)(x-l)(x-2). 

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 = (2-x)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 1-5 

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? + 2-n-xti) 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 -- -= ./ - = 3-17, 
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 



= + 4-899... 

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 ] 



103-105] MAXIMA AND MINIMA 11 

Put = x* 



.*. y is a minimum when x= 3. [Why minimum?] 
.'. least distance is after 3 hrs, 



A r 



75-9x - > 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* + 2-n-xh. 

The error consists in saying that 

-- = 2-n-x + 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 one-half 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 + 0-OOlV 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 semi-circle. 
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 right-angled 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=CE-C2r. 

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 candle-power 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 E-e=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. 



107-109] 



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 



1-8 



IE 



Y-L 



^ 



w 



'@L 



1-2 



^ 



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.) 



111-113] 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 (x-l) (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 
ce-axis, and there i 1 ? not a turning point on y=f(x) corresponding 



113-116] 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 text-books. 



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 mid-point 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 co-ordinates 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 (x-2). 

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 co-ordinates. 

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 = 2-irr . 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.] 



118-120] 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 18-5", 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 =l-6, a = -00388, b= -000000587, find the approximate change in R 
when the temperature rises from 100 to 101. 



120-122] 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 v-axis 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 p-axis 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 Cross-Section (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 = --jrp-j-7-xd 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=32-2.] 



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. 



14-2 



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 = 3-01. 
Calling the expression y (or) we have 



= 662, /'(3)=1136. 
.'./(8-01) = 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. 



124-127] 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 =2-0087. 

3. Find approximately the value of Gx 3 - 7* 2 + 2x - 9 when x = 3-002. 

4 

4. Find approximately the value of 7a; 2 -3x + - when x=2-01. 

5. Find approximately the value of 

2z 3 - 9* 2 + 12s- 3 when x= (i) 2-005, (ii) 1-995. 
In each case find also the accurate value. 

6. Find approximately the value of 

3a;*-16aj3 + 30a;2-24a; + 5 when x = (i) 2-01, (ii) 1-01. 
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, /' (3-01) = 1151-68. 

So that 

/(3-01)-/( 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 

/(3-005) = 1143-8200, 
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/' (3-2). 

a. If / (x) = x 3 - 6x - 5, shew that 

/ (1-4) -/(I) =-0-656, 

4/' (1-4) =-0-048, 
4/' (1-205) =-0-65757. 

3. If/(a;)=4-4a+16a;-7, shew that 

ft /(3-2)-/ (3) = 3-9856, (ii) / (4) -/ (3) = 43, 

2/'(3)=3-2, /'(3) = 16, 

2/' (3-2) = 4-8384, /' (4) = 80, 

2/' (3-102) = 3-9852. /' (3-539) = 43-003. 



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 n-1 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. 



128-130] 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. 



10-2 



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 gradient-law 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 co-ordinates (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 gradient-law is -^- = 1 - 4r + 2 . 
dx 

In what point and at what angle does this curve cut the y-axis? 

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> = 60-32f. 

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 co-ordinates 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 mid-point 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 co-ordinates 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 x-axis horizontal, and the unit along 
each axis 1 foot. 
Find: 

(i) The co-ordinates 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*-2x3-x 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" 



x-scale unit 2", y-scale 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(x-l) z between x=0 and x=S. Find the 
co-ordinates 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 - 2-5x + 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 co-ordinates 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 co-ordinates 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 x-scalo 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 -6x-8y + 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 co-ordinates 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 13-6. 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 2-J? 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. 



135-137] 



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 = BK-AH, 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, 4-1, 4-2, ... 13-9, 14. 
.'. the ordinates are 

12, 12-3, 12-6, ...417, 42. 
The width of each rectangle is !. 
.'. sum of inside rectangles 

= -1 [12 + 12-3 + 12-6 + ... + 41-7] 100 terms 
= .! x 100 x 53-7 = 268 . 5> 
2 

and sum of outside rectangles 

= -l[12-3+12-6+...+42] 
= . lx 100^54-3 = 2?1 . 5 

The difference between these is 3, the area of a rectangle, 
height 30 and width !. 

The area ABKH lies between 268-5 and 271-5. 



166 CALCULUS FOR BEGINNERS [CH. VII 

Similarly shew that if AB be divided into 1000 equal parts 
the area ABKH lies between 269-85 and 270-15. 
We have two sets of numbers: 

255, 268-5, 269-85,... 
285, 271-5, 270-15,... 
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, ...,42-SA, 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 a-scale and the 
y-scale 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 oj-unit and the y-unit. 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 a-unit = -3" and y-unit = -05", 

unit of area = '015 sq. ins., 
and area ABKH = 270 x -015 = 4-05 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 ar-unit is J" and the y-unit 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 ar-unit is 3" and the y-unit "? 

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 ar-axis 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. 



= 294-24 
= 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 , x-0, 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 



142-144] 



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 j-j. -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, 



145-148] 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, ...(42-3A), 

being the values of y or 3x when x = 4, 4 + A, 4 + 2A, ... (14 A)], 
or more shortly 

x=U-h 

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=14-fc 

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=n-h 

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 



148-151] 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 
J-i a 


13. 


P X 


JjS^/x 


14. 
16. 


/8 
Jo 

r 

\ Idx. 

J-3 


15. 
17. 


./ 30 v1 ' 4 

(Y3x3-5x2 + 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 
one-third and two-thirds 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 one-quarter and three-quarters 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 co-ordinates 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 co-ordinates 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 + 
J-a L 



2 3 J-a 

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}. 



155-157] 



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 ic-axis 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 x-axis. 

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 

J-a 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 oj-axis and one above. 




Fig. 84. 



190 CALCULUS FOR BEGINNERS [CH. VII 

r5 r~ %y3 y^~\ > 

Now I (4 + 3o: - or) dx -\ 4x+ - 

j-a 4 * J -s 



2 
. 21-. 133 



The co-ordinates 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 (l-x)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 x-axis 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 x-axis and the maximum and minimum ordinates. 

6. Draw the curve y*=x(x-l) 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 h-k, 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 = h-k, x=h + k and y=x 3 is the 
same as that bounded by j/ = 0, x h-k, 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 x-axis. 



192 CALCULUS FOR BEGINNERS [CH. VII 

12. Trace the curve 



and find the areas of the two portions bounded by the curve and the x-axis. 

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 two-thirds 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 two-thirds 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 l-a; 



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 ar-axis 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 

3af-4a? + 6x*-7x+ 8 when a: = 3-01. 
/' (*) = 15x 4 -l2x*+ 12* - 7, 

/"(*) = 60^-240; +12. 
= 662, /'(3) = 1136, /"(3)=1560, 



/. /(3-01) =/(S) + -01 /' (3) + f" (3) app. 



= 662 + 1 136 x -01 + x -0001 

2 

= 673-4380 app. 
[Actually /(3-01) - 673-438264503.] 



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 zig-zag boundary 
AB' is 

and MM' = 4-2. 

So that the mean height of this figure is 

2 fl 2 + 1 '2 2 + . . . + 5 2 ] I 2 + 1 -2 2 + . . . + 5 3 



4-2 21 

= the Arithmetic mean of the 21 ordinates from AM to BN inclusive. 



202 CALCULUS FOR BEGINNERS [CH. VII 

[This is 10^ = 10-47.] 

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, 2-2, 2-4, 2-6, ...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, !!, 1-2, ... 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 a-axis 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 
_ 2-2 3 + 2-4 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 z-axis. 

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 +... (n-1) terms] =-(-l) n(2-l) 



= 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 



168-170] DISTANCE FROM SPEED-TIME 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 



170-174] DISTANCE FROM SPEED-TIME FORMULA 209 

(0, 3A 2 , 3(2/t) 2 ... being the values of v or 3t 2 when = 0, h, 

2h } 
***j /> 

t=12-h =12-fc 

or shortly 2 3 2 . h or 2 v . A. 

t=o t=o 

t=12-A 

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 SPEED-TIME 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 speed-time graph. 

175. Generally. If the speed be given as a function of 
the time, say v=f(t), and the speed-time 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 speed-time and space-time graphs be drawn, the number 
of units of length in the difference between two ordinates of the 
space-time graph is equal to the number of units of area in the 
corresponding portion of the speed-time 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 speed-time 
curve, the time-axis 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=112-32, 

rt>A = ["ll2-16< 1 T= 112x3-16 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. 



175-177] SPEED-TIME AND SPACE-TIME GRAPHS 



213 



If we draw the graph v=112-32< 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 space-time 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 17-h 

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. 



177-179] 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 

x-8 aj + Aoi-8 . 
between - Aaj and = Aa; in. Ibs. wt. 

'-i 2i 

05-8 A . x + Ace 8 A 

i.e. AW lies between Ace and = Ace in. Ibs. wt. 

2i '2i 

AW ,. cc-8 , 05 + Aa;-8 . 

.*. lies between ^ and in. IDS. wt. 

Aa; 2 2 

dw x-8 _ 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 x-8 
,\ = - approximately. 



216 CALCULUS FOR BEGINNERS [CH. VIII 

Now when x = 18, W = 0. 

18 2 
.'. = 4. 18 + c, 



x 2 
/. W=^-4x-9. 

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 

r-y.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=25-h x8 =25-A 

Or shortly 2 h . ^ or 2 hT. 

z=18 x=18 

=25-fc 

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 



179-182] 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 anti-differentiation. 
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 
anti-differentiation 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 

5-10 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 


257-14 


245-45 


180 


234-78 


225 




216 


207-69 




200 


192-86 




186-21 




450 


1094 ; 13 


871-00 




4376-52 






1742-00 






450 





6568-52 
Area = ^ x 6568-5 

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 



2-2 



2-4- 2-6 

Fig. 101. 



2-8 



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, cross-section 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 right-angled triangle 
about one of the sides containing the right angle. 

A sphere is generated by the rotation of a semi-circle 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. 



186-188] 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 PMNQ-2Acc 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;. 



188-191] 



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 y-axis. 



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.] 



191-193] 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 

J-a 



~5 - ij 
er - x*dx = 



or 



I y? \/a a x*dx= -5- . 

J-a 



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 x-axis, (ii) the 

y-axis. 

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 co-ordinates 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 co-ordinates 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. 



lltiiizf-i~ 


3?^rES3| 


m 4-lji -! ! 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 \ 
. . Co-ordinates 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. 



16-2 



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 co-ordinates 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 x-axis, 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 mid-points 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 co-ordinates 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 

i-e 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 sc-axis and the ordinates x = a and x = b. This 
mean value is 



Ja 



f(x) . dx 



b-a 
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 0-2- . 

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, ... (b-h), 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(b-h) +f(b) 
n+l ' 



200, 201] MEAN VALUES 251 

hf(a) + hf(a + h) + ... + hf(b- h) + hf(b) 

(b-a) + 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 two-thirds of the final speed. 

8. A stone falls freely from rest. Shew that the mean kinetic energy 
with respect to the time is two-thirds 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 x-axis in A, B and 
the y-axis in C. 

Find the areas AOC, BOG. 

2. The parabola j/ 2 = 6x forms a paraboloid of revolution by revolving 
about the x-axis. 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(5-x) 2 between x= - 1 and x=6. 

Find the area bounded by the x-axis 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 x-axis 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 x-axis. Calculate the maximum and minimum ordinates and the 
gradient at each of the points where the curve cuts the x-axis. 

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 = 4x-x 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 s-axis 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) (x-1) (x-4). 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 x-axis 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;i-2=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 co-ordinates 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= 20-61 when #=5 find a and n. 
Let the curve rotate about the s-axis 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 x-axis. 

If your x-unit is 5" and your y-unit 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 x-axis 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 x-axis is - hk. 

o 

Prove that the radius of gyration about the x-axis of a lamina shaped 

/3~ / 3 3 \ 

like this area is k ./ = and that the co-ordinates 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 cigar-shaped 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) (x-3) from x= -4 to x=4. 

If it cuts the x-axis in A, B, C reading from the left and the y-axis 
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-&x-12 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 

17-a 



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 x-axis. 

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 (v-73.) 

/. 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.] 



204-206] 



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' 
1-0 57 18' 
1-1 G3 02' 
1-2 68 45' 
1-3 74 29' 
1-4 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 y--8776x=-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 
= -3-1520-1-8471 
= -4-9991. 
If a = c.M. of (-144) 

y = - 4 cos 36 - 3 sin 36 
= -3-2360-1-7634 
= -4-9994. 

.*. - 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 { - - } , ir-tan" 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 y-sinnx 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 x-axis 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 x-axis 
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 

y-l-Ay 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. 



210-212] 



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 = l-2 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 y-8ecx, 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 45-1. 

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 = 14-Gtan0. 

.'. Total time taken if he lands at P is 
/6sec0 14-6 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 = ~ =6-801, 
and 



~V7 2x/7 2- 2- 
[Shew that this is a minimum value.] 

.'. He must land 6-804 miles from N and his time will be 
4-823 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 lamp-post 
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. 



214-216] 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 Ax--0, 
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 



216-219] 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 . 



219-221] 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 ' .T-i + 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 



223-225] 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 



1-cc 



, 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 

7-3 ~ * 

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 : 



/l-x* i-z s /I -sin 2 x 

l - /v/1 5' 8 - xJa 2 -**. 3. A /, r-s-- 

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 co-ordinates (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, = 40-32*. 

at at 

i.e. x is increasing 30 times as fast as t, and y is increasing 
(40 _ 32<) times as fast as t. 

40-32* . 
. . y is increasing ^ times as fast as x, 



dy _ 40 - 



30 



dy Idx 
~dtdi' 



From the given equations we might eliminate t. The first 
o-ives 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 

40-32* 

= ~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=a8m0-bcos$, find 2, 



230, 231] x AND y IN TERMS OF THIRD VARIABLE 297 

x <i 

4. Shew that the co-ordinates of any point of the parabola y may 
be written in the form x = am, y = am 2 . 

Find -/-, (i) from-j-^ 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 co-ordinates of a point on a cycloid are given by x = a (9 - sin 0), 
y=o(l-cos0). 

Prove -j^ = cotjj. 

dx 2 

8. The co-ordinates of a point on an epicycloid are given by 

x = (a + b) cos 0-6 cos J-T ' 

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 co-ordinates 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 co-ordinates 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 ar-axis 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 y-axis, 

(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. 



234-237] INVERSE FUNCTIONS 301 

235. If y=x>, x = Jy. 

Here y is said to be a single-valued function of x, i.e. given 
any value of x, there is only one corresponding value of y, but x 
is a double-valued 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 single-valued 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 co-ordi- 

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 many-valued 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 T-JT 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 


1-19 + 


5592 


5934 


0592 


0698 


1-18- 


5299 


5585 


0299 


0349 


1-17 


5150381 


5410.521 


0150381 


0174533 


1-16 


5075384 


5323254 


0075384 


0087266 


1-157 




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 = tan-ix, 

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. 



240-242] TAN-IS 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 = Qy-x- 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 
8-12 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^ + S-i, 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 7-figure tables. With 4-figure 

2' 01 1 
tables the best we can do is to say that ^ = '7 approximately. 

With 7-figure 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 = 5-1, 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 = 7-1, 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 *=7-01, 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. 



24G-248] 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 , -/-- 1-61 x 5 X approximately. 

Q.3C 

3 . If y = 2-5*, ^ = 0-92 x 2 -5* approximately. 

(UB 

4. If y = 2-7*, ^' = 0-99 x 2-7* approximately. 

5 . If y = 2-8*, - = 1-03 x 2'8* approximately. 

248. The results of Exs. LXXV. 4, 5 suggest that there is a 

value of n between 2'7 and 2-8 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 2-718281828. 

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 = 2-7183*, 

^=2-7183 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 
2-5 -92 
2-7 -99 
2-8 1-03 

3 MO 
5 1-61 



249-252] 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 2-3026, each being correct to 5 significant 
figures. 



316 



CALCULUS FOR BEGINNERS 



[CH. XI 



We thus have 

log e n = Iog 10 n x 2-3026 and Iog 10 n = log, n x -43429, 
e.g. log, 2 = Iog 10 2 x 2-3026 

= 30100x2-3026 
= 6931. 

Graphic treatment. 
253. Fig. 124 shews the graph of y = 2*. 



70 



60 



50 



30 



20 



10 






Fig. 124. 



252-254] GRAPHIC TREATMENT 317 

Now if we keep the same axes and make a suitable change in 
the cc-scale, 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 jc-scale 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 1-58B . 

.', if the ce-scale 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 oj-axis 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 



254-257 



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 or-scale 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 2-75. 



320 



CALCULUS FOB BEGIMNERS 



[CH. XI 



4 



- 



2 Scale for 2 X 
1 4 

e 



257-259] 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 

2-1 

2-01 

2-001 
[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 sea-mile 
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 




1-2 1-4 



1-6 



1-8 2 2-2 
Fig. 131. 



2-4 2-6 K2-8 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 + 1-01 



and the summation is easily effected by means of a table of 
reciprocals. 

Using 4-figure tables [the numbers to be added are very 
conveniently placed for addition] we get '01 x 95-8596. 

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= 2-8. 

Area OA"B"R < 1 '032839 and > 1-026410, 
a good approximation is 1*029624. 
.'. e is between 2 '7 and 2*8. 

Try 2-75. 

Area < 1-014789 and > 1-008425, 

a good approximation is 1-011607. 

Try 2-72. 

Area < 1-003800 and > -997476, 

a good approximation is 1-000638 

Try 2-71. 

Area < 1-000110 and > -993800, 

a good approximation is '996955. 

.'. e < 2-72 and > 2-71 and apparently nearer to 2 - 72. 

If we use a table of reciprocals which gives more figures and 
divide the area into strips '001 aj-units wide, we shall be able to 
shew that 

e> 2-718 and <2-719. 



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 co-ordinates of a point on the circle x 2 +y 2 =a 2 mav t e 
taken to be (acos0, asinfl), so the co-ordinates 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 =-2-5 and a; =2-5. 

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 = 2-5. 

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.] 



261-263] 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 

l-a; "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 g|if 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 half-yearly, the increase in capital 

during any half-year is -Q5x x - where x is the capital at the 

Z 

beginning of the half-year. 



263-265] 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 = 20-2. 

And j2 = ^*> 

'* A = 20 10ge l2 

= -017. 
/, $ = 20-2e--<>. 



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 = 2100e--oooo4A. 

(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, 2-25, 2-37, 2-44, 2-594, 2-692, 2-705, 2-717. 

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 fly-wheel, 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(v-850), where t; ft./sec. is the speed and & a constant 
depending on the form and weight of the shell. If V stands for (w-850), 

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 1-5 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 1-5 oz. per cubic inch. Find the law connecting the cross-section 
with the distance from the end if the tensile stress is everywhere 
400 Ibs./sq. in. (Fig. 132.) \y sq. ins. is area of cross-section 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) g-4,=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 = Aa-2z, \ - 4y = 3e- x . 

17. Shew that - | + 2p-p + g'?/:=r ) where p, q, r are constant?, is the 

uX (tX 

same as 5-5+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 IX-XI. 

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 cos0-2sm0 + 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 y-axis 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 x-axis. 

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) (2-3x2)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 -L-d-r, (iii) j\os(l-x)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 r-a 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 14-75 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 
x-axis in G, PO=PG. 

sin nx 



(ii) Shew that y= - 



7*2-1 



is a solution of -r-^ + V = sin nx. 

dx* 

a. Given log, 6 = 1-7918, find log.,6-1. 

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 x-axis and the ordinates x a, x = 2a. 

Find also the co-ordinates 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 mid-point 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 2-rrt 

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 x-axis 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 x-axis at 
tune t = 0'2. Find area contained between x-axis and the path from t = to 
t=0-5. 

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 x-axis 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 mid-point. 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 x-axis. 



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 foot-poundals be the 
kinetic energy, shew that 

dk , dk 

-j-=ma and = ma.r. 
d dt 



267] MISCELLANEOUS EXAMPLES 357 

4. If a piston-rod drives a crank of length r through a connecting-rod 
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 P-side 
of AB is u ft. /sec. and on the Q-side 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 connecting-rod 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 
connecting-rod is 

* .'l2i " RP, 

lcos<(> I 

where P is the end of the crank and R the point where the connecting-rod 
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 (m-n) 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 x-axis and the ordinate 



267] MISCELLANEOUS EXAMPLES 359 

Divide the base into 10 equal parts. Shew that the area lies between 
919 and 1-08. 

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 5f-6^ 

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 co-ordinates of any point on a cycloid being given in the form 

x=a (0-sin 0), 

y = a(i-coa 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 space-average and the time-average 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 cross-section 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? + 9x-l, 

then / / (*) = 3* a -12a; + 9=3(a;-l)(a:-3). 

Make a table of values : 

x 01234 

/() -1 3 1-1 3 

/'(*) 90-309 

If we draw the graph (Fig. 133) the roots appear to lie 
between (i) and -2, (ii) 2-3 and 2-4, (iii) 3-5 and 3-6. 

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 6-72 




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 
/'() 7-60 7-49 

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 aj-axis, 
therefore we might say that the graph of y =f(x) cuts the OS-axis 
at some intermediate point. 




Fig. 135. 

It may happen that y =f(x) cuts the cc-axis 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. 



270-273] 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 x-3, 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 .r-axis in 
R, the point A where the curve cuts the aj-axis 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 2-50. 

["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 2-34 and 2-50.] 



273, 274] EQUATIONS 369 

To get a closer appi-oximation, use the points P', Q' corre- 
sponding to a; = 2-33 and as = 2'50 exactly as we have just used 
P and Q. 

/(2-33) = 0-045937, /(2'5) = - 0-375, 

/'(2-33) = -2-6733. 
Equation P'T' y - -045937 = - 2-6733 (x - 2-33). 



/. atT' * = 2-33 + -. = 2-33 + (-017+). 



Equation P'Q' y - -Q45937 = - ' 42Q ? 37 ( x - 2-33) 

= - 2-4761 (x- 2-33). 
. 04^007 
/. at R' x= 2-33 + ^f. = 2-33 + (-019 -). 

.'. the root lies between 2-347 and 2-349. 
If we want a still closer approximation we use the points P", 
Q" corresponding to x= 2-347 and 05 = 2'349. 

/(2-S47) = 0-000781923, /(2-S49) =-0-004491451, 
/' (2-347) =-2-638773. 
Equation P"T" 

y - -000781923 = - 2-638773 (* - 2-347). 



Equation P"Q" 

y- -000781923 = - 2-636687 (x - 2-347). 
.*. at R" 



.'. the root lies between 2-3472963 and 2-3472966. 

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 2-3 
and 2-4. 

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. 



274-277] 



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 
co-ordinates 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". 

24-2 



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 5-figure 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] = 1-68200 (+ -000005). 
Q 




Pig. 139. 
Equation QU is 

y- -01914 = 1-68200 (a;- [43]). 

01 Q14- * 
/. atu aj = 

Equation PQ is 



-- 01914 = 



02924 _r 43 .-,v 
02924. 

ol745 <*-[ 43 ]> 

l-6756(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 1-1 and 1-6. 

2. 4z 6 -24a; 3 + 2 = has a root near 3. Shew that 2-7 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 3-3 and 3 -6. 

6. x a -Bx 2 + 2x-5 = has a root near 2-9. Shew that it is between 
2-9041 and 2'9042. 

6. Shew that x s -2x-5 = 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.n-1 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 


tanh-i- 






\u y 


- tanh" 1 - 


a 




(x'ca 9 ) 


1 


a a 


a + x 






a 2 -x 2 


1 o + x 


2 ge a-xj 


Or 2a 10g a-x 



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 x-a 


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 u-i 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?*jcfl-x*.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[x-2 = z]. 12. 

13. \ ^dx. 14. /- 

J 2a-x 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 right-hand 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 (sin-cosa:), 
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(3-2a:)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 co-ordinates 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 one-half 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 half-a-minute and after half-an-hour. 
[0 = 82.] 



25-2 



CHAPTER XIV 

POLAR CO-ORDINATES 

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 co-ordinates " 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 Co-ordinates." 

284. O is a fixed point (fig. 140), OX a fixed line through 

p 



Fig. 140. 



283-287] POLAR CO-OTCDINATES 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 
co-ordinates " 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 
co-ordinates 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 co-ordinates 
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 CO-ORDINATES 



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 co-ordinates. 

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 CO-ORDINATES 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 co-ordinates 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) 39-39 m./hr. (ii) 57-77 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.; 50-77 ft./sec. A's is best. 22ft.; 23Jft.; 
25-4 ft. A's. 

2. [All in chs./min.] 31, 39, 57, 52, 59, 58, 62, 62. (i) 59, (ii) 62. 

3. 50, 57-36, 42-26, 54-46, 45 -40, 51-50, 48-48 cms. 8-832 -006, 9-288 -006, 
8-92-01, 9'20-01, 9-00-03, 9-12-03 cms./sec. Between 8-97 and 
9-12 cms./sec. 

4. 173-21, 142-81, 214-45, 153'99, 196-26, 166-43, 180-40 cms. 36-480-012, 
49'488-012, 38-44-02, 46'10-02, 40-68-06, 43-14-06 cms./sec. 
Between 41-4 and 42-3 cins./sec. 



EXERCISES III. p. 9. 

L. (99- 24A + 2A2) ft./sec., ( 99 + ^ ft./sec. 



2. 42, 68, 54-25, 46-72, 44-33, 42-2303, 42-023003 ft. 26, 24-5, 23-6, 23-3, 
23 03, 23-003 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. { 

4-5 
Six 
364-5 


>. (i) 67T. 

4-2 

70-56T 
296-352 


(ii) 2*-. (iii) 3. 

4-1 

67-247T 
275-684 


3 ' 3 *" 

4-01 
64-3204jr 
257-924804 


3 *" 3 " 
4+7t 
(64 + 32ft + 4 7(2) n 


3 

p 


2 + 4871 + 47^ 


3 
17 201-76 


196-84 


3 

192-4804 19 


(1) 8 -, 3 -, 3 .., 3 . 


3 *"' 3 



(ii) 367T, 347T, 32-87T, 32-47T, 32-04ir, (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, 12-25, 9-61, 9-0601, 9 + 6/i + ft 2 , 8-41, 9-6/ + / t 2 ; 
6-5, 6-1, 6-01, 6 + h, 5-9, Q-h; 6; tan-i6=80 32'. 

2. 6. 3. tan-il-2=5012'. 4. 12 + 6/j + 7i2. 12. 
5. 12 + ft 2 ; 12 + 6(ji-m) + (jt2 +nm+m 2). e . 0,3,27,48. 
7. tan- 1 4-8 =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 = 100-32t. (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. ; 6-4 ft. ; 2-44 o/ . -64 ft. ; -249 o/ . 

3. 4 ft./sec. 2 4. 13-6 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. 22-G3sq. 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; 14-03; about -21 o/ . 5. (i) -0025. (ii) -00184. (iii) -2. (iv) 2-5. 

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 = 60x-80. 

4. 1-728 c. ins./sec. 8. 9ir = 28-27 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. 

| x-t, - 3x-, j x~S, - i , l-8x-8, -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 0-22 

^ 9 0-3 09 

dx 



2. 4.T 

a: -1 1 23 
y 6-3-2-36 

$^ -24 24 

dx 

3. Gx 2 -18a; + 12 = 6(x-l) (x-2); 

x 0123 
y -3 2 1 6 

12 12 



- - - -nr and > ' - or (1 '" 77 ' ~ ' 38 ^ and (-423> ' 385) - 

EXERCISES XIX. p. 90. 



5. 3*-y-2 = 0, rr + 3y-4 = 0; T,o<(0, -2), G (4, 0), g 

80 

6. 9Gz-y-144=0, a; + 96y -4610 = 0; x 

S. 3x-y-4 = 0| 3a; + 7/-4=Oi 

a; + 3i/-28 = 0j" ' *-3y -28 = 0) " 

9. ^ = 40x-57, 



11. y = 7x-4, x + ly-12, -=- sq. units. 



EXERCISES XX. p. 92. 



3. 5+1,1-8415,0. 



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 y-2c = 0\ Area of triangle formed by the tangent at 

1 I ; 2. any point and the axes of co-ordinates is 
c ~ constant. 

" J ; (0, - 9), (0, 9) ; 18, s . 
ar + 6y-57 = 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, 14-03, -21 o/o . 8. y = 144x-400. (0, 5), (-1, 0), (2, -27). 
9. (i) 1--. (ii) 1-n. (iii) n. (iv) -. 



10 - (^ T = >0693 ins./sec. (ii) = -0173 ins./sec. 
1UU 1UU 

11. 2-65. 12. Average speed = n times speed at end. 

15. (25-8)a, -6 -70ix 10- 6 ,2 -68x10-*. 16. 1-00896. 



ANSWERS 401 



17. 0-172. 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 1-20 1-80 2-50 3-30 4'08 4-73 5-25 5-63 5-88 6'00 

^ -25 -35 -45 -55 -65 "75 '85 '72 -58 -45 -32 -18 -05 

dx 

25. 3z-2M= s . 26. 92-5 Ibs. wt. 27. 27 ft./sec. 

o 

28. 305 radians/sec. 120 radians/sec. 2 



(i) -4-Slbs. wt., (ii) -4-2 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. (-, 3|J 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- . -~- =108-6 c. ins. 

O O iJ ^y t>7T 

9. 2r=ft=. / . 1O. 9-42 inches. 11. 4/150 = 5-31 knots. 
12 = 3-183 acres. 13. Total height = width = -. =8-40 feet. 

7T + 4 

14. A = M55 ft., 2 ^=1-633 ft. 15. ~pft.j 

sees. 



2^i cos ft 9 cos /3 

17. (i) 2 c. ft. (ii) -=2-546 c. ft. (iii) = '849 c. ft. 

7T OTT 

18. 7. 18s. 9<L [Ht. = J side of base.] 10. 4-8 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= 6-06, Z=10 y = 10' 5, volume = 330 c. ft. 

26. 66-1. 27. 860-9 sq. ft. [A = tx/2]. 28. ^ = 3-224 c. ft 
20. 5. 30. R=r. 31. 100. 

32. a=900; 5 = 60,000; m= -404-8; 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 = 4-58 (when x =4/15). 

5. (0, 0) max., f~y , - ? (y Yl or [1-71, - 7-25] 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. 45-24 sq. ins. 3. 362 c. ins. 4. 197 ft., 9'8 ft 

6. (i) 19-6 c. ft. (ii) 7-07 c. ft. 6. irr 2 coto.Ar. 

7. (i) 46-5 sq. ins. (ii) 430 o. ins. 8. R (a + 2bt), -0064. 



EXERCISES XXVII. p. 139. 

I. 1-5 <V . 3. Decrease 1 %. 1O. 6-08 ft. /sec., -025. 

II. 2-214, -0115. 13. 1-1 % decrease. 



EXERCISES XXVIII. p. 143. 

a. 37-2088. 3. 96-244. 4. 24-24. 
6. (i) 1. (ii) 1. 1-00007525, 1-00007475. 
6. (i) -3. (ii) -2. - 2-99D39197, -2-00000397. 

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/ = + 4x-l. 11. = 



ANSWERS 405 

EXERCISES XXXHI. p. 150. 

; 2 . a. (i) y = Zx-1x*. (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. (60-162)ft. 3. 186 ft./sec.; 404ft. 






3. s=(3 + t 2 + t + 5. (i) 5 ft., 1 ft./sec. (ii) 97^ft-> 48|ft./seo. 
6. = 






MISCELLANEOUS EXAMPLES ON CHAPTERS I-VI. 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 3-016 sq. ins./sec. 
(u) (a) -04ir or -1257 c. ins./sec. (ii) (b) 5-16-rr or 18-096 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. = 15-9 ft. and side of square = 

7T 

(not a true max.). Area = 796 sq. ft. 

Least when side of square = diameter of circle= -- - ft. = 8-97 ft. (true 

TT-fO 

min.). Area = - 5 = 224 sq.ft. 

7T + O 

. =. 



2. Max. 142 when x= -2 ; Min. - 2055 when a; =11. 

3. (i) 150ft./sec. (ii) 58-4 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. (16-90) when ,= 

at 

... 160-56J28. ..... , ..-., 

Mm. - ^^ ( - 5-Oo) when x= (2-097). 

iif 



3. 



/I 32\ 
\3' 27/ 



4. -^- = 1-41 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 (27-146, 21J) and (100-954, 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. 3-91C. 2. Q, -385^; y = x-l; y=-x + l. 

4. 1-09 x 10< ft. ibs. wt. ; 3740 Ibs. ft. /sec. ; 60 Ibs. wt. 5. 18-18. 

G. 

1. 1080 ft. 3. Lengths of pieces are - - , - - inches, or '56a, '44a 



H. 

. (x-2)2(4x + l); 6 (x -2) (2x- 1) ; 6(4x-5). 

(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) -8-2 4 

/'(x) -27 4 5 13 

/"(x) 54 12 -6 30 

/'"(x) -54 -30 -6 18 42 

4. J below centre. 5. (3-99, 2-26) ; (1*05, 4-46). 

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. 4-5 %. 

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.; 4-875 sq. ins. 2. 4; 6 sq. ins. 

3. (i) 285, 385. (ii) 328'35, 338-35. 

(iii) |(200-30A+/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 +2ar-16. (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. 4-032. 12. ^ + 2c. 13. L 14. 1164*. 

O J 

15. -0695. 16. 56. 17. 614^ J. 18. 2. 

19. (i) 278. (ii) 225-9875. 2O. (i) 56. (ii) 56. 21. 2-431. 

EXERCISES XL. p. 183. 

1. 2|. (ii) 5|. 2. (i) 4. (ii) 12. 3. (i) g. (ii) ^. 

k-h 



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) 4046-41. (2) (ii) 3381-04. 
(3) (i) 3382-91. (3) (ii) 3355-04. 

EXERCISES XLIL p. 19tt 

2. 1-4. 8. |. 4. 14. 6. A. 

6. a = 0, b= -8, c=6. 4. 

7. a = h (h?-k*), 6=A;2-3/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, 18-4. a. . 3. -69315, -6989. 4. 170H- 

EXERCISES XL VII. p. 204. 
1. 0. 3. (i) 6. (ii) 0. 3. 3J, 2-75. 4. 4J. 

6. |a*. 6. 11J. 

EXERCISES XL VIII. p. 206. 
(i) 112J, 112-325, '155%. (ii) 71*, 71-465, -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. 1-19 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. 6-71 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=376-99 c. ins. 3. irr 2 Ac. ins. 3. ^=67-02 c. ins. 

o o 



4. ~(3r-fc). 6- (i) 261ir=820c. ins. (ii) ir= 2060 c. ins. 

o o 

qo_. 
6. (i) 16?r = 50-27. (ii) 247r = 75-40. 7. =33-51. 

8. ^. 0. 2 = 2-84 = l-862. 

10. =5-x*. 



11. (i) 120jr = 37G-99. (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)M-jg . 6. -Ma 2 . 



8. (i). Wr. 9. . 10. 2-025, 1-62, 1-35 ft. 



11. M-. Ift. M. . 13. 85-7 ft. Ibs. wt. 

^ A 



90 

14. (i) ^M6. (ii)|Ma 3 . 16. 6-34M. 



EXERCISES LIII. p. 244. 



.. (7-71,85.34). 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. M-t-c'. 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 mid-points of the parallel sides 
... o4-37> ft . ft (a + 3fc) + e(2a + 4fc) j, 

I 1 ) ST . (") r-; STT nr- rrv below the side a. 

l 'a--2b 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 (3-75). 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; 42-04, -28-56; 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. ; 46-5 Ibs. wt./sq. ft. 6. 



D. 

I. 30; 4; 26. 2. (i) 20J. (ii) 20J. 3. 10 inches. 

4. 3| m. 6. 8-713 x 108 gm. cm.2 ; 1-075 x 10 9 ergs. 



CALCULUS FOB BEGINNERS 



E. 
2. 135*- = 424-1 c. ft. x /22 7 5=4'743 ft. 



3. tan-i^- = 4ll'; 0-122; (1-61, 1-12). 

4. ^ (2aR - a2) ft. Ibs. wt. ; ,^ ft. Ibs. wt. ; 35 : 32. 



(R + a) 
P. 



1. (i) g*' + c. (ii) 37 J. 
3. -4514, 2-374 ; 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) 291-6T=916. (iv) |, 



'7 o.ft. 



6. 360 ft. Ibs. wt. 



J. 



, 



12 



9. = 3-82ins./min. 3. 439 ft. Ibs. wt. 






6. 



r+o 



aib3. w t. r 



ANSWERS 415 



EXERCISES LVII. p. 268. 



3. 2, -1-444, -3-560, 0-232, -3598, 0, V!3 = 3-606, -^13= -3-606. 
y = 1x-\, j/ + -144.r=4-748, y + 3-598* = 7'77. 
y = 3-606* + 3 -545, y = 3-606. 

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) 0-4598. (ii) 0-8415. (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 (1-5708, 0-3927). 2O. . 



21. [Add an arbitrary constant.] 

... l/cos5x \ .... l/cosSa; 



sin5x 
-5- 

cos(p-g)x 



p-q 



\ 

)' 

1 fsin(p + q)x sin (y-g)A 

V P + q ~Y-q )' 

sin(p-g)x\ 

P-g /' 



416 CALCULUS FOR BEGINNERS 

22. -5150. 23. -96 sq. ina. 24. 6. 

25. -791. 26. - 1-745 ft./sec., - 4-685 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, 150-75, 1-523, 0-381, 0-095 ft./sec. ; 
i, |, 5-53, 547, 2189, 8754 ft./sec. 



EXERCISES LIX. p. 277. 

BOA nv 

96 



1. 15-535 ft. 2. -1396 ft./sec. 3. 35-S 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. ^ y-cos^/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. j-coajx. 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 n-1 (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 x-3sinx)= -3sin3x. 

51. cosw.^. 52. 3 sin 2 u. cos w-^*. 53. 2u.^. 

ax da; dx 



ANSWERS 419 



64. *u-l 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&nnx-x. 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(cosx-xsinx). 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; 4-3 

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 xl-9459. 
7. [Add an arbitrary constant.] 

(i) e*. (ii) \e**. (iii) le**. (iv) \e*\ (v) e tanz . 

O A 






(vi) -e*. (vii) y- s=7*x-5139. 
le ' 

8. -e 2 = 47-209; 2-36 sq. ins. 1O. -0-237. 



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 ) 0-5 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 x-x. (xv) + 2x + log e x. (xvi) - + 3x-7log 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 =1-0986. (ii) \ log e ^= 0-1062. (iii) ~ log. 3 
(iv) i log. 2 = 0-3466. (v) log. (^2 + 1) = -8813. 

IB 



10 . , or (1-8484, 0-21875). 

11. ^2=0-6931. 12. 3030. 

13. Max. value (-ll + 301og < 3) = 10-98, when *=1. 



Min. value ( - 27 -f 30 log, 5) = 10-64, when x = 3. 



ANSWERS 423 

EXERCISES LXXXI. p. 334. 



3.34.1 



(37s 4 + 81* 3 + 24*2 + x-3). 

(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.; 1-304 sees. 2. 81-1 Ibs. wt. ; ^AOP=0644'. 

3. After 6-575 minutes. 

4. h=- j; (i) 2 mins., (ii) 3 mins., (iii) 3-32 mins. 

6. v - 850 = (u - 850) e~ w . [ u ft. /sec. is speed when t = 0.] 

6. 48-634 ft. [=50(1 --).] 

J -32_ 

7. y = -e 12800 . [j/ sq. ins. is cross-section 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/ = e-2*(Acosx+B sinx). 
(iii) y = e-2* (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_ j-2 1 



1. -at ; a. 2. ^; -500. 3. -.70"; -87 ft. 



ANSWERS 425 

D. 
2-2* 2 1 

1. (l) - -r. (ll) = :. 

2* (1 +*)*(!-*)* 

2. (2, 1), (-1, 1); 2x-3j/-l=0, x-3y + 4=0. 3. ^13-2 = 1-606. 

4. -8933r. 6. (l-c-"" 2 ); 14-2. 

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 =63-496. 

P. 

1. T-g ; TTg, - 6 cos 2 2x sin 2ar, e 01 (a sin fa; + 6 cos bx), 4 sec 4x. 

PN 2 

2. c; PG= - . 3. 25-1 ft./sec., 24-1 ft./sec., 18-85 ft./seo. 

c 

0. ?rG cos a ft. Ibs. 

G. 



(iv) 4sec 2 4a;. (v) e-3* (1 - 3*). (vi) 



a. 18 1 

2. -^=. 4. (i) 22 (ii) 0-3054. (iii) 0-2456. (iv) 5. 

v/ ^ & w 

(v) 0-0429. (vi) 0-3466. 
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.J-4 

(v) - ^. (vi) 6sec23*tan3*. (vii) - - 

(1 - x 2 )? V a 2 - a; 2 

a. -0-3690 and -1-7602 Minimum. 
0-3690 and 1-7602 Maximum. 

3. 38-2 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 

-3-715 ft. /sec. 
4. 6-28 ft. Ibs. wt., 6-98 ft. Ibs. wt. 

J. 



a. 1-8083. a. c2('l+Vl = c 2 x 1-1353. 

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 ; (l-513a, l'003a). 

L. 

8. 2, 2, -|. a. 02(^/2 -l)=0-4142a. 

4. ( 6, J ; ft. /sec. making 210 with OX. 

(3-96) ft./sec.2 making - 13 54' with OX. 
6. 59-3 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'; - = 0-3183. 



3. 11-16 ft. - from A. 8-85 ft. from B. 



4. A = 100^13 = 360 -6. g = tan- 1 l-5=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. 3-165a 3 . 

O. 





(iii) (12a^ + 18* + 5) (6*2 + 3). (iv) y cos a: or e" n!B . cos *. 

& ^T- ? - (vi ) 55-5- < vii ) 

4. 20 log. 3 = 21-97 ; (7-28, 1-52) ; 1-84. 

/\ ^ /* \ ^ \ -* /% -^ 

6 ' (l) T' (ll) T* (m) ' (W) 2' 



P. 

I. (i) x 2 cos x + 2x sin *. (ii) 2xe 2a! (a; + l). 

(iii) ^(aco&Zbx-ZbcoBbxsinbx). (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. - 1-482 ft./sec. ; 6-187 ft./sec. 2 

4. (i) |. (ii) -0243. (iii) 0. (iv) 0. 



S. 
a. IJins./sec.; radians/sec. 4. e5 = l-649. 

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-^ = 0-2146. (v)^ 

3. 8ins./min.; 3*175 ins./min. 

3am(2-7ft3) 3a(l-2m 3 ) m(2-m 

*' 






6 . 7 . 82ins . 



EXERCISES LXXXIV. p. 373. 

e. 2-095. 7. -0-315, 0-446, 1-069. 8. 0-4142. . 0-85138. 
1O. 12-03. 11. 11-34 ins. 12. 1-1839. 13. 1-80, -1-13. 

14. 4-167 ins. 15. 28 23 / . 1. 1-166. 17. 2-289. 

18. 0-657. 19. 4-536. SO. 3-945. 



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 coth-i (2x - 3) or log e ^| . 

a/ ~~ A 

2 a; 
(ii) - 2 tanh- 1 (2x - 3) or log e = . 

3 1 

19. ^logj^ + lJ + tan-Jx. 3O. - - log e cos (era; + b). 

2 1 / 

31. --cos s a; + cosx. 33. --*/2-3a: 2 . 



as. _t a nh-i- + lo g (3-x2. 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*z-l. 3. 



1 . _ 

4. a; tan- 1 x -jrlog^l-l- a; 2 ). . a;sin~ 1 a;+ Jl-x\ 6. - x cos a; + sin x. 



a; -^rn 8in(7K+n)a: ) 

+ - ( x cos (m - ?t)ar -- sin (TO - n) x } I . 
m-n\ TO-n ' 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. ~- + 2x-2log 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. x-o 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 ' -3-2x)(27 + 30,). 17. 



18. l ogs seo + a 19. - tanh-i - + log e (3 - x^J. 



.. / a\ ~, < i 

+ 241og,(x-2). 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. 1-568. 26. 0-430. 27. -0-148. 

a a a 

28. |3. 29. 0-0266, 3O. 0-430 f^l . 

oo I 2ob I 



81. 5tan-i = 0-2145. 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"* 

41-77 m./hr.; 49-9998 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 2-5, at Intervals of !. 



X 


ex 


e~* 


cosh x 


sinh x 


tanh x 





1-000 


1-000 


1-000 








1 


1-105 


905 


1-005 


100 


100 


2 


1-221 


819 


1-020 


201 


197 


3 


1-350 


741 


1-045 


305 


291 


4 


1-492 


670 


1-081 


411 


380 


5 


1-649 


607 


1-128 


521 


462 


6 


1-822 


549 


1-185 


637 


537 


7 


2-014 


497 


1-255 


759 


604 


8 


2-226 


449 


1-337 


888 


664 


9 


2-460 


407 


1-433 


1-027 


716 


10 


2-718 


368 


1-543 


1-175 


762 


1-1 


3-004 


333 


1-669 


1-336 


801 


12 


3-320 


301 


1-811 


1-509 


834 


1-3 


3-669 


273 


1-971 


1-698 


862 


14 


4-055 


247 


2-151 


1-904 


885 


15 


4-482 


223 


2-352 


2-129 


905 


1-6 


4-953 


202 


2-577 


2-376 


922 


1-7 


5-474 


183 


2-828 


2-646 


935 


1-8 


6-050 


165 


3-107 


2-942 


947 


1-9 


6-686 


150 


3-418 


3-268 


956 


20 


7-389 


135 


3-762 


3-627 


964 


2-1 


8-166 


122 


4-144 


4-022 


970 


2-2 


9-025 


111 


4-568 


4-457 


976 


23 


9-974 


100 


5-037 


4-937 


980 


24 


11-023 


091 


5-557 


5-466 


984 


25 


12-182 


082 


6-132 


6-050 


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 


1-030 


1-065 


3 


1-099 


1-131 


1-163 


1-194 


1-224 


1-253 


1-281 


1-308 


1-335 


1-361 


4 


1-386 


1-411 


1-435 


1-459 


1-482 


1-504 


1-526 


1-548 


1-569 


1-589 


5 


1-609 


1-629 


1-649 


1-668 


1-686 


1-705 


1-723 


1-740 


1-758 


1-775 


6 


1-792 


1-808 


1-825 


1-841 


1-856 


1-872 


1-887 


1-902 


1-917 


1-932 


7 


1-946 


1-960 


1-974 


1-988 


2-001 


2-015 


2-028 


2-041 


2-054 


2-067 


8 


2-079 


2-092 


2-104 


2-116 


2-128 


2-140 


2-152 


2-163 


2-175 


2-186 


9 


2-197 


2-208 


2-219 


2-230 


2-241 


2-251 


2-262 


2-272 


2-282 


2-293 



log 10=2-303, log 102=4-605, log 103=6'908. 



M. C. 



INDEX 

The numbers refer to pages. 

Acceleration 19 

,, from gradient of speed-time graph 33, 67, 68 

Approximate equality, test of 39 

,, solution of equations 362-373 

Approximations 136-144 

to f(x + h) 141, 196-198 
Arbitrary constants 145-147, 173, 174, 181 
Area of plane curve 160-192 
by summation 161, 165, 166 

from 5rW(*) 170, 177 

OfX 

approximate rules for 184-186 

sign of 187-190 

,, ; , from ordinates of integral curve 193-196 

Area traced by moving ordinate 169-177 
Atmospheric pressure as function of height 337 
Average speed 2 

,, gradient 22, 23 

rate of increase 12, 15, 16 

ordinate 199-203 

values 250-253 

Centre of gravity 238-244 

,, ,, 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 334-339 
Cycloid 297 

area of 299 

Definite integral 179, 180 

,, approximation to value of 192, 193 

represented by area 192, 219 

Derived curves 94-97 
Differential coefficient 51 

,, coefficients, higher 93 
equations, Exs. on 342-344 
Differentiation, meaning of 51 

from first principles 47-71 

of a* 69, 70 
* 75-79 
kx n 80-82 

x* + c 80, 82, 83 

sin x 261-265, 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; 272-275, 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 278-282 

,, quotient 283, 284 

,, ,, function of function 285-290 

,, ,, inverse functions 301-307 

,, ,, 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 40-45 
dt , As . 
Jt and At 48 - 5 

f and 55-57 
dv Av 



INDEX 437 



and 52-54, 58-65 
fix Ax 

da 

-t- as speed 48 

^ ae gradient 58-61, 85, 86 
dx 

dy 

~- as rate of increase 52, 53 
dx 

~ as ratio of time-rate of increase of w to time-rate of increase of x 54 
dx 

^ , sign of 66, 71-73, 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 speed-time formula 205-210, 221 
,, graph 211-213 

e, definition of 313 

e, approximation to value of 319, 328-330 

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 362-373 

Exs. on differential 342-344 

Error, relative 139 
Errors, approximation to value of small 136-144 

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 285-290 
/ (x + h) =/ (x) + hj ' (x) approximately 141 

/ (x + h) =f (x) + Jif (x) + ^ h*f" (x) approximately 196-198 
Fundamental notions 1-46 

Geometry, Applications of Differential Calculus to 87-90 

Gradient 19-29 

uniform 19-21 
average 22-23 
at a point 23-26, 38 
,, representing rate of increase 33, 34 
of space-time graph represents speed 33 
of speed- time graph represents acceleration G3 

Graphical solurion of equations 362-366 

Gyration, radius of 234 

Harmonic motion, simple 269 
Higher differential coefficients 93 
Hyperbolic functions 330-333 
logarithms 315 

Implicit functions, differentiation of 308, 309 
Inertia, moments of 233-237 

, theorem of perpendicular axes 237 

parallel axes 245-247 

Inflexion, points of 127-133 
Integral curve 193-196 

definite 179, 180, 209 
indefinite 180, 181 
Integration 160-172 

,, meaning of 1GO-162 

,, as a process of summation 219 

> anti-differentiation 219 

of - 323 

x 

list of standard results in 376, 377 
some methods of 377-383 

by change of variable 299 

examples of 220-226 

Inverse functions, differentiation of 301-307 

operation, the fgiven ^ find y\ 145-152 



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 106-133 

,, tests for 108, 110, 123 

,, Exs. on 112-117 

Mean ordinate 199-203 

,, values 250-253 
Moments of inertia 233-237 

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 88-90 

Parallel axes, theorem of 245-247 

Perpendicular axes, theorem of 236, 237 

Points of inflexion 127-133 

Polar co-ordinates 388-393 

Product, differentiation of 278-282 

Quadrant of circle, centre of gravity of 243 
Quotient, differentiation of 283, 284 

Eadius of gyration 234 

Bate of increase- 10-17, 38 
,, ,, ,, meaning of negative 13, 16, 17 
., ,, ,, ,, ,, zero 13 

,, ,, represented by gradient 33, 34 

Katio of continually diminishing quantities 41-46 



44-0 INDEX 

Rectangle, moment of inertia of 233, 234 
Revolution, solids of 228-232 

sec a:, differentiation of 274 
Shape of curve 99-101 
Sign of area 187-190 

Sign of ^ and ^ 56, 71-73, 99 
ax dx* 

Simpson's rule 185, 186 

sinar, differentiation of 261-265 

Solids of revolution, volume of 228-232 

Space-time graph 33, 35, 36, 66, 211, 213 

Speed 1-10 

,, uniform 1 

average 2 

,, at an instant 3-10, 37 

given by gradient of space-time graph 33, 66, 67 
Speed-time 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 261-277 

,, ,, inverse 304-307 

Turning points 107-111, 129-132 

Uniform gradient 20, 21 
,, speed 1 

Volume of frustum of cone 229, 230 

sphere 231, 232 
Volumes of solids of revolution 228-232 
Volume strain 140 

Work done in stretching elastic string 213-218 
,, by expanding gas 224-226 

Zero speed 13 



CAMBRIDGE \ PRINTED BY JOHN CLAY, M.A. AT THE UNIVERSITY PRESS. 



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