Silvanus P. Thompson - Calculus Made Easy-Johnston Press (2008).pdf

CALCULUS MADE EASY

LIE

THE MACMILLAN COMPANY
EM YORK « MOSTON + CHICAGO . DALLAS
“ATLANTA + SAN FRANCISCO
MACMILLAN AND CO. Linsen
LONDON + BOMDAY CALCUTTA - MADRAS
‘THE MACMILLAN COMPANY
OF CANADA, Lixitso
oxosto

CALCULUS MADE EASY:

BEING A VERYSIMPLESF INTRODUCTION TO

THOSE BEAUTIFUL METHODS OF RECKONING

WHICH ARE GENERALLY CALLED BY THE
TERRIFYING NAMES OF THE

DIFFERENTIAL CALCULUS

AND THE

INTEGRAL CALCULUS,

By
SILVANUS P. THOMPSON, FRS,

SECOND EDITION, ENLARGED

Noto Bork
THE MACMILLAN COMPANY

All righes reserned

“es

Fred in the United States of America

‘What one fool can do, another ean.
Láncient Kimian, Proverdy

PREFACE TO THE SECOND EDITION.

Tue surprising success of this work has led the
author to add a considerable number of worked
examples and exercises, Advantage hes also been
taken to enlarge certain parts where experience
showed that further explanations would be nseful.

"he author acknowledges with gratitude many
valuable suggestions and letters received from teachers,
students, and—crities

October, 1914

u.
tL

XIL.
X.
XIV.

XV.
XVI.
XVI.

CONTENTS.

Prowse» + + - - + = =

To DELIVER You FROM THE PRELIMINARE TER-
POR

Os Drerenewr Dronees or SMALLNESS ==
ON RuLATIVE Growings - + - + =
BIMPLEST Cases = + + -
Nexr Stacz, Waar 10 DO WITH CONSTANTS -
Sons, Drrrzrances, PRODUCTS, AND QUOZIENTS
Svconssive Durrereyiation - +, + +
Wars Time VARIES + + + + +
Inrnoovcısa 4 Usurun Done + + +
Geowwraicar Maanine OP DIFFERENTIATION -
Maxma AND Moma + + + +
CUBVATURE OF Cunves + + + + +
Omase Useruz Dovces - + + + 0 =
Ox rave Courouxp Isterest AND mue Law or
ORGANICO GROWTH - + + + =
How 10 DEAL WITE Sixes AND Costas + +
PARTIAL DiPFERENTIATION + + + +

Invecration + + ee ee

pen
x

18
28

28521

76
93
12
121

134
165
175
182

XVIII.
xix.

XXL
XXIL
XXUL

CONTENTS

INTEGRATISG as rae Reverse or DIFFERENTIATING 191

Ox Frvpina Angas sy INTEGMATING =~
Dovers, Prreanis, AND THIUMPUS + +
FINDING SOME SOLUTIONS + + + + -
A LITILE MORE ABOUT CURYATURE OF CURVES -

How ro Fınp run Laxora or ax Año on u
o

EPILOGUE ann APOLOGUE- + + + +
Table of Standard Forms - - + +
Answers to Exercises - + + - *

206
226
234
949

285

289

PROLOGUE.

ConsmERING how many fools can caleulate, it is
surprising that it should be thought either a difficult
or a tedious task for any other fool to learn how to
master the same tricks.

Somo calenlus-tricks are quite easy. Some are
enormously difficult. ‘The fools who write the text-
books of advanced mathematics—and they are mostly
clever fools—seldom take the trouble to show you how
easy the easy calculations are. On the contrary, they
seera to desire to impress you with their tremendous
cleverness by going about it in the most difficult way.

Being myself a remarkably stupid fellow, I have
had to unteach myself the difficulties, and now beg
to present to my fellow fools the parts that are not
hard. Master these thoroughly, and the rest will
follow. What one fool can do, another can.

CHAPTER L

TO DELIVER YOU FROM THE PRELIMINARY
TERRORS.

Tue preliminary terror, which chokes off most fifth-
form boys from even attempting to learn how to
calculate, can be abolished once for all by simply stating
what is the meaning—in common-sense terms—of the
two prineipal symbols that are used in calculating.

These dreadful symbols are:

(1) 2 which merely moans “a little bit of.”

Thus de means a little bit of ©; or du means a
little bit of « Ordinary mathematicians think it
more polite to say “an element of,” instead of “ a little
bit of.” Just as you please. But you will find that

these little bits (or elements) may be considered to be
indefinitely small.

e) | which is merely a long S, and may be called
G£ you like) “the sum of”

Thus {de means the sum of all the little bits
of 2; or fae means the sum of all the little bits
of t. Ordinary mathematicians call this symbol “the

on à

2 CALCULUS MADE EASY

integral of” Now any fool can see that if & is
considered as made up of a lot of little bits, each of
which is called da, if you add them all up together
you get the sum of all the das, (which is the same
thing as the whole of 2). The word “integral” simply
means “the whole.” If you think of the duration
of time for one hour, you may (if you like) think of
it as cut up into 3600 little bits called seconds. ‘The
whole of the 3600 little bits added up together make
one hour,

When you see an expression that begins with this
terrifying symbol, you will henceforth know that it
is put there merely to give you instructions that you
are now to perform the operation (if you can) of
totalling up all the little bits that are indicated by
the symbols that follow.

That's all.

CHAPTER IL
ON DIFFERENT DEGREES OF SMALLNESS.

We shall find that in our processes of calculation we
have to deal with small quantities of various degrees
of smallness.

We shall have also to learn under what circumstances
we may consider small quantitics to be so minute
that we may omit them from consideration. Every-
thing depends upon relative minuteness.

Before we fix any rules let us think of some
familiar cases, There are 60 minutes in the hour,
24 hours in the day, 7 days in the week. There are
therefore 1440 minutes in the day and 10080 minutes
in the week,

Obviously 1 minute is a very small quantity of
time compared with a whole week. Indeed, our
forefathers considered it small as compared with an
hour, and called it “one mindte” meaning a minute
fraction—namely one sixtieth—of an hour. When
they came to require still smaller subdivisions of time,
they divided each minute into 60 still smaller purts,
which, in Queen Elizabeth's days, they called “second
minites” (i.e. small quantities of the second order of
minuteness), Nowadays we call these small quantities

4 CALCULUS MADE EASY

of the second order of smallness “seconds.” But few
people know why they are so called. °

Now if one minute is so small as compared with a
whole day, how much smaller by comparison is one
second !

Again, think of a farthing as compared with a sove-
reign: it is worth only a little more than 1945 part.
A farthing more or less is of precious little importance
compared with a sovereign’ it may certainly be re-
garded as a small quantity. But compare a furthing
with £1000: relatively to this greater sum, the
farthing is of no more importance than zu of a
farthing would be to a sovereign, Even a golden
sovereign is relatively a negligible quantity in the
wealth of a millionaire.

Now if we fix upon any numerical fraction as
constituting the proportion which for any purpose
we call relatively small, we can easily state other
fractions of a higher degree of smallness. Thus if,
for the purpose of time, yy be called a small fraction,
then gy of gy (being a small fraction of a small
fraction) may be regarded as a small quantity of the
second. order of smaliness.*

Or, if for any purpose we were to take 1 per cent.
(ia, 130) as a small fraction, then 1 per cent. of
1 per cont. (ie, 25) would be a small fraction

19.000,
of the second order of smallness; and would

7090,00

+ The mathomaticians talk about the second order of “ magnitude”
(e grestoess) when they really mean second onder of saline

is very confusing to beginners.

DIFFERENT DEGREES OF SMALLNESS 5

be a small fraction of the third order of smallness,
being 1 per cent. of 1 per cent, of 1 per cent,

Lastly, suppose that for some very precise purpose
we should regard 32575 as “small” Thus, if a
first-rate chronometer is not to lose or gain more than
half a minute in a year, it must keep time with an
aceuracy of 1 part in 1,051,200. Now if, for such a
purpose, we regard zz 555 (or one millionth) as a
small quantity, then of that fa,

taras (or aa TE a small
Guantity of the second order of amallness, and may
be utterly disregarded, by comparison,

Then we seo that the smaller a small quantity itself
is, the more negligible does the corresponding small
quantity of the second order become, Hence we
know that in all cases we are justified in neglecting
the small quantities of the second—or third (or
higher)—orders, if only we take the small quantity
of the first order small enough in itself.

But ib must be remembered that small quantities,
if they occur in our expressions as factors multiplied
by some other factor, may become important if the
other factor is itself large. vena farthing becomes
important if only it is multiplied by a few hundred.

Now in the caleulus we write die for a little bit
of @ These things such as da, and du, and dy, are
called “differentials,” the differential of m, or of 1,
or of y, as the case may be. [You read them as
des-cks, oc dee-you, or dee-wy.| IE de be a small bit
of a and relatively small of itself, it does not follow

6 CALCULUS MADE EASY

that such quantities as «0» dx, or ad, or atda are
negligible. But dexde would be negligible, being a
small quantity of the second order.

A very simple example will serve as illustration,

Let us think of = as a quantity that can grow by
a small amount so as to become x+dz, where da is
the small increment added by growth. The square
of this is a*4+20-de+(def. The second term is
not negligible because it is a first-order quantity;
while the third term is of the second order of small-
ness, being a bit of a bit of a. Thus if we took
des to mean numerically, say, x of æ, then the second
term would be 4% of a*, whereas the third term would
be yes ola”. This last term is clearly less important
than the second. But if we go further and take
dz to mean only vos of w, then the second term
will be yo of a, while the third term will be
only dr of 2%

x

Fig. 1.

Geometrically this may be depicted as follows:
Draw a square (Fig. 1) the side of which wo will
take to represent =. Now suppose the square to
grow by having a bit de added to its size each

DIFFERENT DEGREES OF SMALLNESS 7

way. The enlarged square is made up of the original
square a2, the two rectangles at the top and on the
right, each of which is of area ada (or together
22 de), and the little square at the top right-hand
corner which is (dw, In Fig. 2 we have taken dev as

x de
Pe de 2
= (dx)
= Le
edo
x de
Fro. 2. Fro. 3,

quite a big fraction of «—about }. But suppose we
had taken it only —jy—about the thickness of an
inked line drawn with a fine pen. ‘Then the little
corner square will have an ares of only „tz of af,
and be practically invisible. Clearly (de)? is negligible
if only we consider the increment dw to be itself
small enough.

Let us consider a simile.

Supposo a millionaire were to say to his secretary:
next week I will give you a small fraction of any
money that comes in to me. Suppose that the
secretary were to say to his boy: I will give you a
small fraction of what I get. Suppose the fraction
in each case to be +3 part. Now if Mr. Millionaire
received during the next week £1000, the secretary

8 CALCULUS MADE EASY

would receive £10 and the boy 2 shilling, Ten
pounds would be a small quantity compared with
£1000; but two shillings is a small small quantity
indeed, of a very secondary order. But what would
be the disproportion if the fraction, instead of being
dm had been settled at zus part? Then, while
Mr. Millionaire got his £1000, Mr. Secretary would
get only £1, and the boy less than one farthing!

The witty Dean Swift * once wrote:

“So, Nat'ralists observe, a Flea

“Tlath smaller Fleas that on him prey.
“And these have smaller Fleas to bite em,
“And so proceed ad infinitum.”

An ox might worry about a flea of ordinary
sıze—a small creature of the first order of smallness
But he would probably not trouble himself about a
flea’s flea; being of the second order of smaliness, it
would be negligible, Bven a gross of fleas’ fleas
would not be of much account to the ox.

"On Perry = Rhapsody (p, 20), printed 1783—usually misquoted,

CHAPTER UL
ON RELATIVE GROWINGS.

Aux through the calculus we are dealing with quan-
tities that are growing, and with rates of growth.
We classify all quantities into two classes: constants
and variables. Those which we regard as of fixed
value, and call constants, we generally denote alge-
braically by letters from the beginning of the
alphabet, such as a, 6, or e; while those which we
consider as capable of growing, or (as mathematicians
say) of “varying,” we denote by letters from the end
of the alphabet, such as æ, y, £, u, v, W, or sometimes 4,

Moreover, we are usually dealing with more than
one variable at once, and thinking of the way in
which one variable depends on the other: for instance,
we think of the way in which the height reuched
by a projectile depends on the time of attaining that
height. Or, we are asked to consider a rectangle of
given area, and to enquire how any increase in the
length of it will compel a corresponding decrease in
the breadth of it. Or,we think of the way in which
any variation in the slope of a ladder will cause the
height that it reaches, to vary.

Suppose we have got two such variables that

10 CALCULUS MADE EASY

depend one on the other, An alteration in one will
bring about an alteration in the other, because of this
dependence. Let us call one of the variables æ, and
the other that depends on it y.

Suppose we make x to vary, that is to say, we
either alter it or imagine it to be altered, by adding
to it a bit which we call da. We are thus causing @
to become æ-+dæ. Then, because w has been altered,
y will have altered also, and will have become y+dy.
Here the bit dy may be in some eases positive, in
others negative; and it won't (except very rarely) be
the same size as dx.

Take to examples.

(1) Let @ and y be respectively the base and the
height of a right-angled triongle (Fig. 4), of which

Ad

x dx
Fo. 4.

the slope of the other side is fixed at 30%. If we
suppose this triangle to expand and yet keep its
angles the same as at first, then, when the base grows
so as to become a+de, the height becomes y+dy.
Here, increasing @ results in an increase of y. The
little triangle, the height of which is dy, and the baso

ON RELATIVE GROWINGS n
of which is da, is similar to the original triangle; and
it is obvious that the value of the ratio

same as that of the ratio % As the angle is 30° it
will be seen that here

ue

en
173

(2) Lot æ represent, in Fig. 5, the horizontal dis-
tance, from a wall, of the bottom end of a ladder,

B

om
x

Fie, 5,

AB, of fixed length; and let y be the height it
reaches up the wall. Now y clearly depends on a
It is easy to sce that, if we pull the bottom end A a
bit furthor from the wall, the top ead B will come
down a little lower. Let us state this in scientific
language. If we increase @ to &+dx, then y will
become y—dy; that is, when æ receives a positive

12 CALCULUS MADE EASY

increment, the increment which results to y i
negative.

Yes, but how much? Suppose the ladder was so
long that when the bottom end A was 19 inches from
tho wall the top end B reached just 15 feet from the
ground. Now, if you were to pull the bottom end
out 1 inch more, how much would the top end come
down? Put it all into inches: #=19 inches, y=180
inches. Now the increment of a which we call da,
is 1 inch: or #+de=20 inches,

How much will y be diminished? The new height
will be y—dy. If we work out the height by Euclid
L 47, then we shall be able to find how much dy will
be. The length of the ladder is

A/(150PF(19P= 181 inches,
Clearly then, the new height, which is y—dy, will be
such that
(y— dy) = (181° —(20) = 82761 — 400 = 32861,
y dy = 32801 = 17989 inches.
Now y is 180, so that dy is 180—179-89—0-11 inch,

So we see that making dz an increase of 1 inch
has resulted in making dy a decrease of 0-11 inch.

And the ratio of dy to das may be stated thus:

dy, 1
de Y

Tt is also easy to see that (except in one particular
position) dy will be of a different size from dar,

Now right through the differential caleulus we
are hunting, hunting, hunting for a curions thing,

ON RELATIVE GROWINGS 13

a mere ratio, namely, the proportion which dy
bears to de when both of them are indefinitely
small.

It should be noted here that we can only find

this ratio du when y and © are related to each

other in some way, so that whenever a varies y does
vary also. For instance, in the first example just
taken, if the base æ of the triangle be made longer,
the height y of the triangle becomes greater also,
and in the second example, if the distance a: of the
foot of the ladder from the wall be made to increase,
the height y veached by the ladder decreases in a
corresponding manner, slowly at first, but more and
more rapidly as a becomes greater, In these cases
the relation between «w and y is perfectly detinite,

it can be expressed mathematically, being Y

e tan 30°
and a*+y?=]? (where 1 is the length of the Indder)
respectively, and e has the meaning we found in
each case.

If, while æ is, as before, the distance of the foot
of the ladder from the wall, y is, instead of the
height reached, the horizontal length of the wall, or
tho number of brieks in it, or the number of years
sineo it was built, any change in æ would naturally

cause no change whatever in y; in this ease 4Y has

no meaning whatever, and it is not possible to find

14 CALCULUS MADE EASY

an expression for it. Whenever we use differentials
de, dy, dz, ete, the existence of some kind of
relation between a y, x ete, is implied, and this
relation is called a “function” in x, y. 2, ete.; the
two expressions given above, for instance, namely

= =tan 30° and a? +y"

Such expressions contain implicitly (that is, contain
without distinetly showing it) the means of expressing
either a in terms of y or y in terms of m, and for
this reason they are called implicit functions in
wand y; they.can be respectively put into the forms

4, are functions of & and y

y=atan30 or a=—Y
and VF or x=

These ia expressions state explicitly (hat is, die-
tinctly) the value of a in terms of y, or of y in terms
of z, and they are for this reason called explicit
functions of @ or y. For example a*43=2y-7 is
an implicit function in # and y; it may bo written

a
y 22 (explicit function of 2) or 4/27 10

(explicit function of y) We see that an explicit
function in a, y, # ete, is simply something the
value of which changes when à, y, =. ete, are
changing, either one at the time or several together.
Because of this, the value of the explicit function is
called the dependent variable, as it depends on the
valué of the other variable quantities in the function;

ON RELATIVE GROWINGS 15

these other variables are called the independent
variables because their value is not determined from
the value assumed by the function. For example,
if u=2*sin 0, x and 6 are the independent variables,
and # is the dependent variable.

Sometimes the exact relation between several
quantities a, y, % either is not known or it is not
convenient to state it; it is only known, or con-
venient to state, that there is some sort of relation
between these variables, so that one cannot alter
either a or y or x singly without affecting the other
quantities; tho existence of a function in a, y, 2 is
then indicated by the notation F(a, y, 2) (implicit
function) or by w= Fy, 2), y= F(a, 2) or z= F(a, y)
(explicit function). Sometimes the letter for ¢ is used
instead of F, so that y=F(a), y=f(w) and y=9(a)
all mean the same thing, namely, that the value of
y depends on tho value of @ in some way which is
not stated,

We call the ratio E “the differential coeficient of

y with respect to «.” It is a solemn scientific name
for this very simple thing. But we are not going
to be frightened by solemn names, when the things
themselves aro so easy. Instead of being frightened
we will simply pronounce a brief curse on tho
stupidity of giving long crack-jaw names; and, having
relieved our minds, will go on to the simple thing
dy

itself, namely the ratio

16 CALCULUS MADE EASY

In ordinary algebra which you learned at school,
you were always hunting after some unknown
quantity which you called & or y; or. sometimes
there were two unknown quantities to be hunted
for simultaneously. You have now to learn to go
hunting in a new way; the fox being now neither
æ nor y. Instead of this you have to hunt for this

curious cub called a The process of Änding the
value of BU is called “diferentiating:” But, remember,
what is wanted is the value of this ratio when both
dy and dx are themselves indefinitely small. The
true value of the differential coefficient is that to which
it approximates in the limiting case when each of
them is considered as infinitesimally minute.

Let us now learn how to go in quest of a

ON RELATIVE GROWINGS Y

NOTE TO CHAPTER III.
How to read Diferentials.

It will never do to fall into the schoolboy error of
thinking that de means d times æ, for d is not a
factor—it means “an clement of” or “a bit of”
whatever follows, One reads de thus: “ deo-eks”

In caso the reader has no one to guide him in such
matters it may here be simply said that one reads
differential coefficients in the following way. The
differential coefficient,

se is read “dee-wy by dee-chs,” or “dee-wy over
dec-ehs?
So also = is read “ dee-you by dee-tee.”

Second differential coefficients will be met with
later on. They are like this:
er which is read “ dee-two-wy over dee-cks-squared,”

and it means that the operation of differentiating y
with respect to æ has been (or has to be) performed
twice over.

Another way of indicating that a function has been
differentiated is by putting an accent to the symbol of
the function, ‘Thus if y=F(2), which meons that y
is some unspecified function of æ (see p. 14), we may
write (m) instead of EE similarly, Pa;

de y, Pa)

will mean that the original function F(a) has been
differentiated twice over with respect to a
eus a

CHAPTER IV.
SIMPLEST CASES.

Now let us see how, on fitst principles, we can
differentiate some simple algebraical expression.

Case 1.

Let us begin with the simple expression y=a%,
Now remember that the fundamental notion abet
the calculos is the idea of growing. Mathematicians
call it varying. Now as y and a? are equal to one
another, it is clear that if 2 grows, a? will also grow.
And if 22 grows, then y will also grow. What we
have got to find out is the proportion between the
growing of y and the growing of a. In other words
our task is to find out the ratio between dy and da,

or, in brief, to find the value of 2

Let @, then, grow a little bit bigger and become
*+de; similarly, y will grow a bit bigger and will
become y+dy. Then, clearly, it will still be true
that the enlarged y will be equal to the square of the
enlarged @. Writing this down, we have:

ytdy=(e+ de).

Doing the squaring we get:

y+dy=0420-de+(doA

SIMPLEST CASES 19

What does (dw)? mean? Remember that da meant
a bit—a little bit—of # Then (de)? will mean a little
bit of a little bit of #; that is, as explained above
(p. 4), it is a small quantity of the second order
of smallness. It may therefore be discarded as quite
inconsiderable in comparison with the other terms
Leaving it out, we then have:

ytdy=x*+ 2a de.

Now y=2"; so let us subtract this from the equa-
tion and we have left

dy=20-de,
Dividing across by da, we find

dy
de

Now this* is what we set out to find. The ratio of
the growing of y to the growing of æ is, in the case
before us, found to be 22.

*N.B.—Thia ratio SE ia the remit of diferentiating y with
respect to =. Differentiating means finding the differential eo-
efficient, Supposo we had some other function of +, as, for
example, u=T2t+3, Then if wo wore told to diferentiato thin
with respect to 2, we should havo to dnd dE, or, what is tho same
thing, H+
timo was the independent variable (900 p. 18), much as thia:
9=b+3a@. Then, if we wore told to differentiate it, that means wo
oust find ito differential cocificiont with respect to &. So that then

A 2% that is dE +300)
inese would be to try to find 2Y, that ie, to fin .
our busi 1d be to try Gf, that i, 10 ünd 204100)

On the other hand, we may have a case in which

20 CALCULUS MADE EASY

Numerical example,

Suppose &=100 and -. y=10,000. "Then let # grow
till it becomes 101 (that is, let dx=1). Then the
enlarged y will be 101 x 101=10,201, But if we agree
that we may ignore small quantities of the second
order, 1 may be rejected as compared with 10,000; so
we may round off the enlarged y to 10,200. y has
grown from 10,000 to 10,200; the bit added on is dy,
which is therefore 200.

24.290 900, According to the algebra-working

=2n, And so

of the previous paragraph, we find 5,

it is; for &= 100 and 2= 200.
But, you will say, we neglected a whole unit.
Well, try again, making de a still smaller bit.
Try do=yy. Then ©+d0=1001, and

(2+d2)=1001 x 100-1 =10,020°01,

Now the last figure 1 is only one-millionth part of
the 10,000, and is utterly negligible; so we may
take 10,020 without the little deal at the end.

And this makes dy=20; and a 29 = 200, which
is still the same as 2a,

Case 2.

Try differentiating y=a in the saine way.

We let y grow to y+dy, while » grows to 2+de
Then we have

ytdy=(@+day,

SIMPLEST CASES a

Doing the eubing we obtain

ytdy =a? + 8x2 d+ 3x (da) + (day.

Now we know that we may neglect small quantities
of the second and third orders ; since, when dy and da
are both made indefinitely small, (de)? and (de)
will become indefinitely smaller by comparison. So,
regarding them as negligible, we have left:

yt dy =a? +32 de.

But y=2°; and, subtracting this, we have:

dy=32* dn,

ay _y
and dn

Case 3.
Try differentiating y=at, Starting as before by
letting both y and a grow a bit, we have:
ytdy =(a+day.
Working out the raising to the fourth power, we get,
yr dy =a + dan +6 da + dale) + (dey.
Then, striking ont the terms containing all the
higher powers of dz, as being negligible by com-
parison, we have
+ dy = + dde.
Subtracting the original y=0*, we have left
dy= hd,

AY_ da
and te

22 CALCULUS MADE EASY

Now all these cases are quite easy, Let us collect:
the results to see if we can infer any general rule,
Put them in two columns, the values of y in one
and the corresponding values found for $2 in the
other: thus

Just look at theso results: the operation of differen-
tinting appears to have had the effect of diminishing
the power of a by 1 (for example in the last case
reducing 2 to a), and at the same time multiplying
by a number (the same number in fact which originally
appeared as the power). Now, when you have once
seen this, you might easily conjecture how the others
will run. You would expect that differentiating af
would give 52%, or differentiating a would give 64%.
IE you hesitate, try ono of these, and see whether
the conjecture comes right.

Try y=añ.

Then y-+dy=(+dx)

=a + Sutde-+ 100 (dep + 100 doy
+5x(de + do).

Neglecting all the terms containing small quantities

of the higher orders, we have left
yt dy=aP+5atde,

SIMPLEST CASES 23

and subtracting y=a* leaves us
dy Sade,

whenco

5a, exactly as we supposed,

Following out logically our observation, we should
conclude that if we want to deal with any higher
power,—call it #—we could tackle it in the same
way.

Let y=ar,
then, we should expect to find that

na,

For example, let n=8, then y=a*; and differs
entiating it would give 44 =807,

And, indeed, the rule that differentiating a” gives as
the result na“ is true for all cases where n is a
whole number and positive. [Expanding (æ-+ dx)" by
the binomial theorem will at once show this.) But
the question whether it is true for cases where m
has negative or fractional values requires further
consideration.

Case of a negative power.
Let y=a-*, Then proceed as before:
ytdy=(a+dx)*

(14%)

24 CALCULUS MADE EASY

Expanding this by the binomial theorem (see p. 141),
we get

no fie]
20 due + 30 (da? do (depto.
So, neglecting the small quantities of higher orders
of smallness, we have:
y+dy=0"*-20"de,
Subtracting the original y=a-%, we find
dy= 22 de,
dae

And this is still in accordance with the rule inferred
above,

Case of a fractional power.
Let y=, Then, as before,
3
vray orde)
1 de _1 (de)
(ER terms with higher
+9 Te E aja terne with, bleh
Subtracting the original y= a}, and neglecting higher
powers we have left:

_l dw dy,
dy= ata ¿o? de,

and Dies. This agrees with the general rule,

=NVe+

SIMPLEST CASES 25

Summary. Let us see how far we have got. Wo
have arrived at the following rule: To differentiate
a”, multiply it by the power and reduce the power by
one, so giving us na”! as the result.

Exercises I. (See p. 288 for Answers.)
Differentiate the following:

(D yaa O yet
(8) y=a™ (4) wat
6) zu (6) y= Vas
m ual (8) y=228.
CFE COPENES

Fou have now learned how to differentiate powers
ofa. How easy it is!

CHAPTER V.
NEXT STAGE, WHAT TO DO WITH CONSTANTS.

IN our equations we have regarded æ as growing,
and as a result of æ being made to grow y also
changed its value and grew. We usually think of æ
as a quantity that we can vary; and, regarding the
variation of & as a sort of cause, we consider the re-
sulting variation of y as an effect, In other words, we
regard the value of y as depending on that of a. Both
wand y are variables, but æ is the one that we operate
upon, and y is the “dependent variable” In all the
preceding chapter we have been trying to find out
rules for the proportion which the dependent variation
in y bears to the variation independently made in x.

One next step is to find out what effect on the
process of differentiating is caused by the presence of
constants, that is, of numbers which don’t change
when a or y changes its value.

Added Constants.

Let us begin with some simple case of an added
constant, thus:

Let y=0%45,
Just as before, let us suppose æ to grow to æ-+d and
y to grow to y+dy.

WHAT TO DO WITH CONSTANTS 22

Then: y+dy=(a-+daP+5
= a+ 3xtde+3x(dx)+(daP+5.
Neglecting the small quantities of higher orders, this
becomes y+dy=0*+32*-de+5.
Subtract the original y=a*+5, and we have left:
dy=3x%de,

So the 5 has quite disappeared. It added nothing
to the growth of a, and docs not enter into the
differential coefficient. If we had put 7, or 700, or
any other number, instead of 5, it would have dis-
appeared. So if we take the letter a, or b, or € to
represent any constant, it will simply disappear when
we differentiate.

If the additional constant had been of negative value,
such as —5 or —b, it would equally have disappeared,

Multiplied Constants.
Take as a simple experiment this case:

Let y=7a%.
Then on proceding as before we get:
Yrdy=Ka+day

{+ 20-da+(de)}
=Tat + 140-de+ (da)
Then, subtracting the original y=7a*, and neglecting
the last term, we have
dy= Made.

dy
Gia lia.

28

CALCULUS MADE EASY

Let us illustrate this example by working out the

graphs of the equations y=7e? and au

14, by

assigning to æ a set of successive values; 0,1,2, 3, ete,

and finding the corresponding values of y and of 42.

These values we tabulate as follows:

efo]+fa[ofefof fs

y [o [7 [2863 [18/15] 7 | 98 | ca

glo | 14 ss ao [56 | 70 | -14| ss | =«2
| =

Y

200

a Ne les

1
1
i
1
tot
,
tt
“ar 0 TE SA

Fre, 6.— Graph of y=70%.

Gi ay
Fis. 6a—Graph of aa,

Now plob these values to some convenient scale,
and we obtain the two curves, Figs. 6 and 6a,

WHAT TO DO WITH CONSTANTS 29

Carefully compare the two figures, and verify by
inspection that the height of the ordinate of the
derived eurve, Fig. 6a, is proportional to the slope of
the original curve,* Fig. 6, at the corresponding value
of @ To the left of the origin, where the original
curve slopes negatively (that is, downward from left
to right) the corresponding ordinates of the derived
curve are negative,

Now, if we look back at p. 19, we shall sec that
simply differentiating a* gives us 2x. So that the
differential coeflicient of 72% is just 7 times as big as
that of 2% If we had taken 82* the differential
coefficient would have come out eight times as great
as that of at. If we put y=aé, we shall get

lt we had begun with y=au", we should have had
Dax nav: So that any mere multiplication by
a constant reappears as a mere multiplication when
the thing is differentiated. And, what is true about
multiplicetion is equally true about division: for if,
in the example above, we had taken as the constant +
instead of 7, we should have had the same + come
out in the result after differentiation,

“See p. 77 about slopes of curven

30 CALCULUS MADE EASY

Some Further Examples.

The following further examples, fully worked out,
will enable you to master completely the process of
differentiation as applied to ordinary algebraical ex-
pressions, and enable you to work out by yourself the
examples given at the end of this chapter.

3 is an added constant and vanishes (see p. 26).

We may then write at once

The term 1/a vanishes, being an added constant;

and as au/z, in the index form, is written azi, we

have d
YY ax lio?
daga xat,
dy a
or Y,
de 2/0

GE aytbu=by- a+ (cry BE,
find the differential coefficient of y with respect to a
As a rule an expression of this kind will need a
little more knowledge than we have acquired so far:

WHAT TO DO WITH CONSTANTS 31

it is, however, always worth while to try whether the
expression can be put in a simpler form.

First we must try to bring it into the form y=some
expression involving a: only.

‘The expression may be written

(«dy + (at d)a=@et+y) BE
Squaring, we get
(abr + (a+ bya? +Ua+b)(a— Day
=(0+y420y Na? 0),

which simplifies to

(AAA) a;
a [(a—-bP-(a*—b1P=[(07-0)—(a+bP Jar,
that is 2b G-a)y’=—2b(b+a)a*;

hence

(4) The volume of a cylinder of radius r and height
his given by the formula = rh. Find the rate of
variation of volume with the radius when r=55 in
and h=20 in. If r=h, find the dimensions of the
eylinder so that a change of 1 in. in radius causes a
change of 400 cub. in. in the volume.
The rate of variation of V with regard to r is
av
d
If 755 in. and h=20 in. this becomes 6908, It
means that a change of radius of 1 inch will causo a
change of volume of 6908 cub. inch. This can be
easily verified, for the volumes with r=5 and 7=6

=2rrh.

32 CALCULUS MADE EASY

are 1570 cub, in. and 22608 cub. in. respectively, and
22608— 1570 = 6908,

Also, if
gah, Vom and ro

dr

(6) The reading 9 of a Féry's Radiation pyrometer
is related to the Centigrade temperature ¢ of the
observed body by the relation

8 y
6) ’
where 6, is the reading corresponding to a known tem»
perature t, of the observed body.

Compare the sensitiveness of the pyrometer at
temperatures 800° C., 1000° ©., 1200” C,, given that it
read 25 when the temperature was 1000° C.

The sensitiveness is the rate of variation of the

reading with the temperature, that is 22, The formula,
a di
may be written

and we have
do _ 1008 e
dt 1000 T0,000,000,000°
When £=800, 1000 and 1200, we get 00513, 04
and 01728 respectively.
The sensitiveness is approximately doubled from
800° to 1000”, and becomes three-quarters as great
again up to 1200°

WHAT TO DO WITH CONSTANTS 33
Exercises IT, (Sce p. 288 for Answers.)
Differentiate the following:

@) y=1824-e,
3) ys leat +. () y-dat.

ar—1
01 EZ

(6) y=1-188 422

Make up somo other examples for yourself, and try
your hand at differentiating them.

(1) IE 4, and 7, be the lengths of a rod of iron at
the temperatures # C. and 0°C. respectively, then
1,=1,(1+0'0000122), Find the change of length of the
rod per degree Centigrade.

(8) It has been found that if ¢ be the candle power
of an incandescent electric lamp, and V be the voltage.
e=aV?, where a and D are constants.

Find the rate of change of the candle power with
the voltage, and calculate the change of candle power
per volt at 80, 100 and 120 volts in the case of alamp
for which a=05 x 10-1 and b=

(9) The frequency m of vibration of a string of
diameter D, ‘cagth L and specific gravity o, stretched
with a force 7, is given by
ı HT
"=D N zo"
Find the rate of change of the frequency when D, L,
y and 7 are varied singly.
CE c

84 CALCULUS MADE EASY

(10) The greatest external pressure P which a tube

esn support without collapsing is given by
2Exé
P(e

where E and y are constants, ¢ is the thickness of the
tube and D is its diameter. (This formula assumes
that 4¢ is small compared to D.)

Compare the rate at which P varies for a small
change of thickness and for a small change of diameter
taking place separately.

(11) Find, from first principles, the rate at which
the following vary with respect to a change in
radius :

(a) the circumference of a circle of radius 7;

€) the area of a circle of radius 1;

(©) the lateral area of a cone of slant dimension 7;

(d) the volume of a cone of radius r und height à ;

(e) the area of a sphere of radius 7;

(f) the volume of a sphere of radius r.

(12) The length Z of an iron rod at the temperature
T being given by L=1,[14+0000012(T—t)), where I,
is the length at the temperature ¢, find the rate of
variation of the diameter D of an iron tyre suitable
for being shrunk on a wheel, when the temperature
T varies.

CHAPTER VI.

SUMS, DIFFERENCES, PRODUCTS, AND
QUOTIENTS.

Wu have learned how to differentiate simple alge-
braical functions such as fte or au“, and wo have
now to consider how to tackle the sum of two or
more functions,
For instance, let
y= (a +e) +(aat+b);
what will its 4% he? How are we to go to work

on this new job?

The answer to this question is quite simple: just
differentiate them, one after the other, thus:

d
de

If you have any doubt whether this is right, try
a more general case, working it by first principles.
And this is the way.

Let y=u+v, where u is any function of +, and
v any other function of æ. Then, letting æ increase
to xtde, y will increase to y+dy; and u wil)
increase to u+du; and e to v+du.

=2044ax%. (Ans)

36 CALCULUS MADE EASY

And we shall have:
y+dy=u+du+v+do.
Subtracting the original y=u+v, we get
dy=du+do,
and dividing through by da, we get: -
dy _du de,
da da” da
This justifies the procedure, You differentiate each
function separately and add the results. So if now
wo take the example of the preecding paragraph, and
put in the values of the two functions, we shall have,
using the notation shown (p. 17),
dy to AUD)
da de
= oe Haas,
exactly as before,
If there were three functions of , which we may
call u, v and w, so that

y=u+o+w;
dy_du do do
“in dede det ade

As for the rule about subtraction, it follows at once;
for if the function v had itself had a negative sign, its
differential coefficient would also be negative; so that
by differentiating

y=u-e,

dy_du_do
we should get La

SUMS, DIFFERENCES, PRODUCTS 37

But when we como to do with Products, the thing
is not quite so simple.
Suppose we were asked to differentiate the expression
y=(a*+ €) x (aat+b),
what are we to do? The result will certainly not
be 2x dax; for it is easy to see that neither ex aa,
nor ax b, would have been taken into that product.
Now there are two ways in which we may go
to work,
First way. Do the multiplying first, and, having
worked it out, then differentiate. .
‘Accordingly, we multiply together a°+c and aat-+6,
This gives a+ acu! + ba°-+ be.
Now differentiate, and we gob:
DU at + Aaa? + 2b.
Second way. Go back to first principles, and
consider the equation
y=uxw;
where u is one function of a, and e is any other
function of æ. Then, if æ grows to be #-+da; and
y to y+dy; and w becomes ud; and v becomes
+de, we shall havi
y+dy=(u+du)x (0+do)
-v+udo+o-du+ dude.
Now du+ dv is a small quantity of the second order
of smallness, and therefore in the limit may be
discarded, leaving
y+dy=u-w+udo+o da.

38 CALCULUS MADE EASY

Then, subtracting the original y=w-v, we have lefé
dy=wde+vedu;
and, dividing through by dar, we get the result:
dy_ do, du
oe =4 dat a
This shows that our instructions will be as follows:
To differentiate the product of two functions, multiply
each function by the differential coefficient of the
other, and add together the two products so obtained.
You should note that this process amounts to the
following: Treat a as constant while you differen-
tiate »; then treat vas constant while you differentiate

#3 and the whole differential cooficient SU will be

the sum of the results of these two treatments,

Now, having found this rule, apply it to the
conercto example which was considered above,

We want to differentiate the product

(a? +e) x(ax*+b).

Call (a’+0)=u; and (axt+b)=0,

Then, by the general rule just established, we
may write:

tt as last)
de

240) (ao! +b)

esa + (act! +b)
an’ + 4aca* +2ax5+ 2bw,

À Gax?-+4acu® +20,

exactly as before.

d(x +e)
de

QUOTIENTS 39

Lestly, we have to differentiate quotients,

3
Think of this example, yee In such a ease

it is no use to try to work out the division beforehand,
because æ+a will not divide into da*+e, neither
have they any common factor. So there is nothing
for it but to go back to first principles, and find a
rule,

So we will put y

where u and v are two different functions of the
independent variable a Then, when « becomes
u+da, y will become y+dy; and u will become
u+du; and e will become v+de. So then
utdu
ae
Now perform the algebraie division, thus:
v+dv|u+du E que um
de
ur

ude

du

v
du „du: de
ae ae

40 CALCULUS MADE EASY

As both these remainders are small quantities of
the second order, they may be neglected, and the
division may stop here, sinee any further remainders
would be of still smaller magnitudes.

So we have got:

y+dy

which may be written
v-du-u-do
A

Now subtract the original y=“, and we have left:

dy ut,

Fi

du_ de

whenee dy, E da
da

This gives us our instructions as to how to differ-
entiate a quotient of two functions, Multiply the
divisor function by the differential coefficient of
the dividend function; then multiply the dividend
function by the differential coeficient of the divisor
Function; and subtract the latter product from the
former. Lastly, divide the difference by the square of
the divisor function.

Going back to our example y=
write bte;
and ata

DIFFERENTIATION 4

Then
rer Gas+e ete)
— 777 —“*
+a) +e)(2x)

dy _3bas+5abat — 2e
DRE EURE. (Answer)
The working out of quotients is often tedious, but
there is nothing difficult about it.

Some further examples fully worked out aro given
hereafter.

(1) Differentiate

= peg Zo +
Being a constant, % vanishes, and we have

Ea,

=1; so we get:

pa
(2) Differentiste y=2a/bae— a _ 2d.

E
Putting w in the index form, we get

y= a/b 03 Ja 2a) db.
Now

A SS

d 3/8
or, AN REZA

42 CALCULUS MADE EASY

(8) Differentiate BEN. y

This may be written: ¿=180 449-927,
The 27° vanishes, and we have

18% 3x0 o;

de
&__ 26s 0886:
or, $ rot,
or, de_088_12
, Je Je

(4) Differentiate v=(3P—1:2t41Y.

A direct way of doing this will be explained Jater
(eee p. 67); but we can nevertheless manage it now
without any difficulty.

Developing the cube, we get
p= QTE — 32-4 + DOG 233280413320 361415
hence

de_
de
(5) Differentiate y=(2x—3)(0-+1)%

Un EDEN ER

=(22— [eye y n)

1624 — 1624 + 159:848* — 69-9840? + 26:648—38,

ter te)

=20+1)[(20-3)+(0+1)]]=2(0+1B7-2%
or, more simply, multiply out and then differentiate,

LLFFERENTIATION 43
(6) Differentiate y
dy _ os sd
Ge osas

=05[0*+(0—3)x322]=2 "4508,
Same remarks as for preceding example.
a Nig.)
(D Diterentinte = (+1) (Vö+Zg)-
This way be written
Ve

dore ya +6)

or Helo on »

=(04+0-900 40) +0 +0 NEL)
E E)

= 1 A. 1
Ze)

This, again, could be obtained more simply by
multiplying the two factors first, and differentiating
res, This no, however; always posible:
sve, for instance, p. 173, example 8, in which the
mule for differentiating u produet must be used.

(8) Differentiate y=

ee er
L+an/a are

+).
(+axt+ax)

e CALCULUS MADE EASY

an at
(9) Differentiate y= Ep

dy (@+1)2a—a®x 2e 2e
de GIF FI

(10) Difforentiate y=4

atat

In the indexed form, ES
ae

dy ta aa array _a- trat

CS aabt”

dy a
ps de ala

Se A LI-a ve
AL) Differentiate 0=, Ya

af
Now Linie,
l+ath

40 _(1+ath)(— gat)—G—at) x4
a G+atiÿ

_ Y

Sra VAT
(12) A reservoir of square cross-section has sides
sloping at an angle of 45° with the vertical. Ihe side

DIFFERENTIATION 45

of the-bottom is 200 feet. Find an expression for the
quantity pouring in or out when the depth of water
varies by 1 foot; hence find, in gallons, the quantity
withdrawn hourly when the depth is reduced from
14 to 10 feet in 24 hours.

The volume of a frustum of pyramid of height ZZ,

and of bases A and a, is V=2(44a+,/Aa). It is

easily seen that, the slope being 45°, if the depth be
h, the length of the side of the square surface of the
water is 20042 feet, so that the volume of water is

" L200:+(200+2%)-+ 200(200 +21]

= 40,000 + 400K? +- ze.

40,000 +800%-+4%*=eubic feet per foot of depth

variation. The mean level from 14 to 10 feet is
av
dh
Gallons per hour corresponding to a change of depth

4x 50178625 _ 52,267 gallons.

12 feet, when h=12 50,176 cubic feet,

of 4 ft. in 24 hours =

(13) The absolute pressure, in atmospheres, P, of
saturated steam at the temperature °C. is given by

: 4040
Dulong as being P= (4048)
80°. Find the rate of variation of the pressure with

the temperature at 100°C,

as long as £ is above

46 CALCULUS MADE EASY

Expand the numerator by the binomial theorem
(see p. 141).

Pp (40545 404 10 x 40104. 10 x 4088
+5x 40048);

hence

Pl

de ~ 337, 324.x10°

404+ 20 x 40% +30 x 40% + 20 x 408 458%),
when #=100 this becomes 0036 atmosphere per
degree Centigrade change of temperature.

Exercises III. (See the Answers on p. 289.)
(1) Differentiate

@) If w=at—4b8, find de.
(8) Find the differential coefficient of
y=(at =D x(0/—D.

entiate
y=(1970— 342°) x (7+ 2207 — 8329),

@ D

(5) Tf o=(y+3)x(y+5), find de.
ly
(6) Differentiate y=1'3709@ x (11264452022).

DIFFERENTIATION 47
Find the differential coefficients of

_2æ+3 44204 30%
Verre ©) = Let
_aath _ Ca
OU ced GO y= EE

(11) The temperature ¢ of the filament of an in-
candescent electric lamp is connected to the current
passing through the lamp by the relation

C=atbt+et.

Find an expression giving the variation of the
current corresponding to a variation of temperature.

(12) The following formulae have been proposed to
express the relation between the electric resistance R
of a wire at the temperature #° C.. and the resistance
R, of that same wire at 0° Centigrade, a and D being
constants. R=R,(1+at+be).

RoR (i tat+b/d.
R=R,(1+at+be) 1.

Find the rate of variation of the resistance with
regard to temperature as given by each of these
formulae.

(13) The electromotive-force Æ of a certain type of
standard cell has heen found to vary with the tom-
perature ¢ according to the relation

B= 1-4340[1 — 0-000814(¢~ 15)
+0000007(£—15)] volts,

Find the change of electromotive-foree per degree,
at 15°, 20° and 25°.

48 CALCULUS MADE EASY

(14) The electromotive-foree necessary to maintain
an electric ure of length 7 with a eurent of intensity
à has been found by Mrs. Ayrton to be

Baa El

where a, b, e, k are constants,
Find an expression for the variation of the electro-

motive force (a) with regard to the length of the are;
(b) with regard to the strength of the current,

CHAPTER VII.
SUCCESSIVE DIFFERENTIATION.

Ler us try the effect of repeating several times over
the operation of differentiating a function (see p. 14).
Begin with a concrete case.

Let

= 20a,
Third differentiation, 5x4x 0% = 602%,
Fourth differentiation, 5x4x3x 22 =120m.
Fifth differentiation, 5x 4x 3x 2x1=120.
Sixth differentiation, =0.

There is a certain notation, with which we are
already acquainted (sce p. 15), used by some writers,
that is very convenient. This is to employ the
general symbol fiz) for any function of a Here
the symbol J( ) is read as “function of,” without
saying what purticular function is meant. So the
statement y=/(e) merely tells us that y is a function
of a, it may be a or de®, or cosas or any other com-
plicated function of a»

The corresponding symbol for the differential co-

efficient is f'(æ), which is simpler to write than

This is called the “derived function” of a,
QuE D

dy,
da

50 CALCULUS MADE EASY

Suppose we differentiate over again, we shall get
the “second derived function” or second differential
coeflicient, which is denoted by f(x); and so on.

Now lot us generalize.

Let yafla)=ar.

First differentiation, F(æ)=na"-1

Second differentiation, (2) =n(n— 1)an=2,

Third differentiation, f”"(@)=n(n—1)(n—2)a"-%

Fourth differentiation,

Fan a Hager
eto, ete.

But this is not the only way of indicating successive

differentiations. For,

if the original function bo y=f(x);

once differentiating gives a Os

twice differentiating gives =2-

and this is more conveniently written as

ay
y"
more usually 5%, @Y Similarly, we may write as the
result of thrice 7 ocentiating, = fa).

SUCCESSIVE DIFFERENTIATION 51

Examples,
Now let us try y=/(0)=710'4350 104202

Uf (0) 2280410522041,

= =f" (w)=84024 210-1,

=f"(w)= 1080-421,
FY = f°" (@) 2168,
ty f(a) =0.
Ina similar manner if y= $(o)=80(2*—4),
ik ain 1]=3(802—4),

¿E =3x ba= 18%,

$"@)- wy Tha 1s,

¢"@)= Len

Exercises IV. (See page 28) for Answers)

Fina 4 and ¿2 for the following expressions :
. „ta
@ oe ate ton aorta
E a
@ y= yt AAN
(4) Find the 2nd and 3rd derived functions in
the Exercises LIL. (p. 46), No. 1 to No.7, and in the
Examples given (p. 41), No. 1 to No. 7.

CHAPTER VII.
WHEN TIME VARIFS.

Som of the most important problems of the calculus
are those where time is the independent variable, and
we have to think about the values of some other
quantity that varies when the time varies, Some
things grow larger as time goes on; some other things
grow smaller. ‘The distance that a train has travelled
from its starting place goes on ever increasing as time
gocs on. Trees grow taller as the years go by.
Which is growing at the greater rate; a plant 12
inches high which in one month becomes 14 inches
high, or a tree 12 feet high which in a year becomes
14 fect high?

In this chapter we are going to make much use
of the word rate. Nothing to do with poor-rate, or
police-rate (except that even here the word suggests
a proportion—a ratio—so many pence in the pound).
Nothing to do even with birth-rate or death-rate,
though these words suggest so many births or deaths
per thousand of the population. When a motor-car
whizzes by us, we say: Whut a terrific rate! When
a spendthrift is flinging about his money, we remark
that that young man is living at a prodigious rate.

WHEN TIME VARIES 53

What do we mean by rate? In both these cases we
are making a mental comparison of something that is
happening, and the length of time that it takes to
happen, If the motor-car flies past us going 10 yards
per second, a simple bit of mental arithmetic will
show us that this is equivalent—while it lasts—to a
rate of 600 yards per minuto, or over 20 miles per
hour.

Now in what sense is it true that a speed of
10 yards per second is the sume as 600 yards
per miaute? Ten yards is not the same as 600 yards,
nor is one second the same thing as one minute.
What we mean by saying that the rate is the same,
is this: that the proportion borne between distance
passed over and time taken to pass over it, is the
same in both enses.

Take another example. A man may have only
a few pounds in his possession, and yet be able to
spend money at ihe rate of millions a year—provided
he goes on spending money at that rate for a few
minutes only. Suppose you hand a shilling over
the counter to pay for some goods; and suppose the
operation lasts exactly one second. Then, daring
that brief operation, you are parting with your money
at the rate of 1 shilling per second, which is the
same rate as £3 per minute, or £180 per hour, or
£4320 per day, or £1,576,800 per year! If you have
£10 in your pocket, you can go on spending money
at the rate of a million a year for just 5} minutes.

Tt is seid that Sandy had not been in London

54 CALCULUS MADE EASY

above five minutes when “bang went saxpence” If
he were to spend money at that rate throughout a.
day of 12 hours, he would be spending 6 shillings an
hour, or £3. 12s, per day, or £21, 12s, a week, not
counting the Sawbath.

Now try to put some of these ideas into differential
notation.

Let y in this case stand for money, and let £ stand
for time.

If you are spending money, and the amount you
spend in a short tise dt be called dy, the rate of

spending it will be ©; or, as regards saving, with a

minus sign, as — U,
2

not an inerement. But money is not a good example
for the calculus, because it generally comes and goes
by jumps, not by a continuous flow—you may earn
£200 a year, but it does not keep running in all
day long in a thin stream; it comes in only weekly,
or monthly, or quarterly, in lumps: and your ex-
penditure also goes out in sudden payments.

A more apt illustration of the idea of a rate is
furnished by the speed of a moving body. From
London (Euston station) to Liverpool is 200 miles
It a train leaves London at 7 o'clock, and reaches
Liverpool ab 11 o'clock, you know that, since it has
travelled 200 miles in 4 hours, its average rate must
have been 50 miles per hour; because #22=49, Here
you are really making a mental comparison between

because then dy is a devrement,

WHEN TIME VARIES 55

the distance passed over and the time taken to pass
over it. You are dividing one by the other. If y is
the whole distance, and ¢ the whole time, clearly thé

average rato is E Now the speed was not actually

constant all the way: at starting, and during the
slowing up at the end of the journey, the speed was
Jess. Probubly at some part, when running down-
hill, the speed was over 60 miles an hour. If, during
any particular element of time dt, the corresponding
element of distance passed over was dy, then at that
dy
ES

part of the journey the speed was The rate at

which ono quantity (in the present instance, distance)
is changing in relation to the other quantity (in this
caso, time) is properly expressed, then, by stating the
differential coefficient of one with respect to the other,
A velocity, scientifically expressed, is the rate at which
a very small distance in any given direction is being
passed over; and may therefore be written
dy
de

But if the velocity e is not uniform, then it must
be either increasing or else decreasing. The rate at
which a velocity is increasing is called the acceleration,
If a moving body is, at any particular instant, gaining
an additional velocity de in an element of time dé,
then the acceleration « at that instant may be written
dv
a

w=

56 CALCULUS MADE EASY

but dv is itselt a(

|, Hence we may put

and this is usually written a= 7;

or the acceleration is the second differential coefficient
of the distance, with respect to time, Acceleration is
expressed as a change of velocity in unit time, for
instance, as being so many feet per second per second;
the notation used being feet-+second®.

When a railway train has just begun to move, its
velocity y is small; but it is rapidly gaining speed—it
ig being hurried Lup or accelerated, by the effort of the
engine. So its oe is large. When it has got up ita
top speed it is no longer being accelerated, so that
then EY has fallen to zero. But when it nears its

stopping place its speed begins to slow down; may,
indeed, slow down very quickly if the brakes are put
on, and during this Period of Aclara or slackening
of pace, the value of © = that is, of D 2 X, will be negative,

To accelerate a mass m requires the continuous
application of force. The force necessary to accelerate
a mass is proportional to the mass, and it is also
proportional to the acceleration which is being im-
parted. Hence we may write for the force 4 the

expression fama;

WHEN TIME VARLES 57

nn de,

or In
a

or fan

The product of a mass by the speed at which it is
going is called its momentum, and is in symbols mo.
If we differentiate momentum with respect to time

we shall get EP) for the rate of chango of mo-

mentum. But, since m is a constant quantity, this
may be written 292, which we see above is the same
as f That is to say, force may be expressed either
as mass times acceleration, or as rate of change of
momentum.

‘Again, if a force is employed to move something
(against an equal and opposite counter-foree), it does
work; and the amount of work done is measured by
the product of the force into the distance (in its
own direction) through which its point of application
moves forward. So if a force f moves forward
through a length y, the work done (which we may
call 20) will bo ska

where we take / as a constant force. If the force
varies at different parts of the range y, then we must
find an expression for its value from point to point.
If / be the force along the small element of length
dy, the amount of work done will be fxdy. But as
dy is only an clement of length, only an clement of

58 CALCULUS MADE EASY

work will be done. If we write w for work, then an

element of work will be dw; and we havo
dw=fxdy;

which may be written

awe sai
or dw= ml. dy
or do=m dy.

Further, we may transpose the expression and write

This gives us yet a third definition of force; that
if it is being used to produce a displacement in any
direction, the foree (in that direction) is equal to the
rate at which work is being done per unit of length
in that direction. In this last sentence the word
rate is clearly not used in its time-sense, but in its
meuning as ratio or proportion.

Sir Isaac Newton, who was (along with Leibnitz)
an inventor of the methods of the calculus, regarded
all quantities that wore varying as flowing; and the
ratio which we nowadays call the differential co-
efficient he regarded as the rate of flowing, or the
fluxion of the quantity in question. He did not use
the notation of the dy and de, and dt (this was due
to Leibnitz), but had instead a notation of his own.
If y was o quantity that varied, or “flowed,” then his
symbol for its rate of variation (or “fluxion”} was

WHEN TIME VARIES 58

9 TE w was the variable, then its fluxion was called
&. The dot over the letter indicated that it had been
differentiated. But this notation docs not tell us
what is the independent variable with respect to
whick the differentiation has been effected. When

dy
we see GE ;
respect tot If we see a we know that y is to be

we know that y is to be differentiated with

differentiated with respect to æ. But if we see merely
Y. we caunot tell without looking at the context
whether this is to mean 42 or U or À, or what is
the other variable. So, therefore, this fiuxional no-
tation is less informing than the differential notation,
and has in consequence largely dropped ont of use.
But its simplicity gives it an advantage if only we
will agree to use it for those cases exclnsively where
time is the independent vera In that case Au will
mean Y and % will mean = and & will mean ve
Adopting this fuxional notation we may write the
mechanical equations considered in the paragraphs
above, as follows:
distaneo
velocity
acceleration
force
work

60 CALCULUS MADE EASY

Hoamples.

(1) A body moves so that the distance æ (in feet),
which it travels from a certain point O, is given by
the relation 2=02#+4104, where £ is the time in
seconds elapsed since a certain instant, Find the
velocity and acceleration 5 seconds after the body
began to move, and also find the corresponding values
when the distance covered is 100 feet, Find also
the average velocity during the first 10 seconds of
its motion. (Suppose distances and motion to the
right to be positive.)

Now w=02F +104,

7204 = constant,

When £=0, =10% and p=0, ‘The body suswd
from a point 104 feet to the right of the point O:
and the time was reckoned from the instant the
body started.

When ¢

v= a eo; and a

v=04x5=2 H./scc.; a=04 ft./sect,

When #=100,100=02F+4104, or @=448,
and t= 21-17 sec.; v= 04 x 2117 = 8-468 1t./sec,

When t= 10,

distance travelled = 0-2 x 104 10:4—104=20 fé.

Average velocity = 3g =2 ft./soc.

(It is the same velocity as the velocity at the middle
of the interval, £=5; for, the acceleration being con-
stant, the velocity las varied uniformly from zero
when t=0 to 4 ft/sec. when ¢=10.)

WHEN TIME VARIES 61

(2) In the above problem let us suppose
w= 024324104.

>
0443; a8 72

Li =04= constant.
When t=0, x=104 and y ft./sec., the time is
reckoned from the instant at which the body passed a
point 10-4 it. from the point O, its velocity being then
already 8 ft./sce. ‘Lo find the time elapsed since it began
moving, let v=0; then 04¢4+3=0, t= —
The body began moving 75 sec. before time was
begun to be observed; 5 seconds after this gives
t= —25 and v=U'4x —254+3=2 ft/sec.
When a=100 ft,
100=0:2P+36+10:4; or #+154—448=0;
hence t=1495 sec, 0=04x1495+3=898 1t./sec.
To find the distance travelled during the 10 first
seconds of the motion one must know how far the
body was from the point O when it started.
When ¢
æ=02x(—75)—3 x754+104= — 085 ft,
that is 0:85 ft, to the left of the point O.
Now, when t=2%5,
w= 02K 2543 x 25410-4= 19-15,
So, in 10 sceonds, the distance travelled was
19°15 4-085 = 20 ft. and
the average velocity

2 ft/sec.

(3) Consider a similar problem when the distance
is given by a=02P-31+104, Then v=044-3,
constant. When ¢=0, a= 104 as before, and

62 CALCULUS MADE EASY

v= —3; so that the body was moving in the direction
opposite to its motion in the previous cases. As the
acceleration is positive, however, we see that this
velocity will decrease as time goes on, until it becomes
zero, when v=0 or 0'4¢—3=0; or t=75 sec. After
this, the velocity becomes positive; and 5 seconds
after the body started, t= 125, and
v=04x125—-3=2 ft /sco.
When 2=100,
100= 02? 31+104, or £—151—448=0,

and t=2995; v=04x2995-3=898 ft/sec.

When vis zero, @«=02x75°-3x75+104= — 085,
informing us that the body moves back to 0:85 it,
beyond the point O before it stops. Ten seconds later

4=175 and 2=02x175-3x175+104=1915.
The distance travelled ='$5+1915=200, and the
average velocity is again 2 ft./sec.

(4) Consider yet another problem of the same sort
with 2=026-38 4104; v=06t—6t; a=12-6,
‘The acceleration is no more constant.

When t=9, æ=104, 9=0, a=—6. The body is
at rest, bus just ready to move with a negative
acceleration, that is to gain a velocity towards the
point O.

(5) If we have #=0-2¢—3t+10-4, then v=060—3,
and a=12,

When ¿=0, =104; v=~8; a=ü.

‘The body is moving towards the point O with

WHEN TIME VARIES 63

a velocity of 3 ft/sec, and just at that instant the
velocity is uniform.

We see that the conditions of the motion can always
be at once ascertained from the time-distance equation
and its first and second derived functions. In the
last two cases the mean velocity during the first
10 seconds and the velocity 5 seconds after the start
will no more be the same, becanse the velocity is not
increasing uniformly, the acceleration being no longer
constant.

(6) The angle 9 (in radians) turned through by a
wheel is given by 0=3+2t—0-18, where £ is the
time in seconds from a certain instant; find the
angular velocity « and the angular acceleration a,
(a) after 1 second; (b) after it has performed one
revolution. At what time is it at rest, and how many
revolutions has it performed up to that instant?

Writing for the acceleration

de ‚38, de
072034, a= 5= Fe = 062

When ¢=0,0=8; w=2 rad.fsec.; a=

When t=1,

—03=17 rad /see. ; a = —0°6 rad./see?,
This is a retardation; the wheel is slowing down,
After 1 revolution

0=27=628; 628=3 42-018.
By plotting the graph, 0=3+2(—016, we can get
the value or values of t for which @=628; these
are 2:11 and 303 (there is a third negative value).

.=

6 CALOULUS MADE EASY
When t=2-11,

—184=066 rad./see, 5
—1-27 rad./see?.

28; = 2—-2754 = —0754 rad /sec.;
a= — 1-82 rad /sec?.

The velocity is reversed. The wheel is evidently
at rest between these two instants; it is at rest when
w=, that is when 0=2-03t%, or when £=258 sec,
it has performed

"
2 A O 1288 1.025 revolutions

Exercises V. (Ses page 290 for Answers.)
a) It ysatbe-tet; find Y ana OY

dy A 4
Ans. 7, = t+ ses Fs

= 2b + 128,

(2) A body falling freely in space describes in ¢
seconds a space s, in feet, expressed by the equation
s=168, Draw a curve showing the relation between
sand £. Also determine the velocity of the body at
the following times from its being let drop: ¢=2
seconds; t=4'6 seconds; ¢= 0-01 second.

(3) If x=at 2; find & and &.

(4) If a body move according to the law

g= 12454628,
find its velocity when t=4 seconds; s being in feet.

WHEN TIME VARIES 65

(5) Find the acceleration of the body mentioned in
the preceding example. Is the acceleration the same
for all values of £?

(8) The angle @ (in radians) turned through by
a revolving wheel is connected with the time £ (in
seconds) that has elapsed since starting, by the law

0=21-321+480,

Find the angular velocity (in radians per second) of
that wheel when 1} seconds have elapsed. Find also
its angular acceleration,

(7) A slider moves so that, during the first part of
its motion, its distance s in inches from its starting
point is given by the expression

8=688—108¢; t being in seconds.

Find the expression for the velocity and the accelera-
tion at any time; and hence find the velocity aud the.
acceleration after 3 seconds,

(8) The motion of « rising balloon is such that its
height %, in miles, is given at any instant by the
expression k=05+ y /(—125; £ being in seconds.

Find an expression for the velocity and the accelera-
tion at any time. Draw curves to show the variation
of height, velocity and acceleration during the first
ten minutes of the ascent.

(9) A stone is thrown downwards into water and
its depth p in metres at any instant ¢ seconds after
reaching the surface of the water is given by the
expression 4

ppt 08

am. E

66 CALCULUS MADE EASY

Find an expression for the velocity and the accelera-
tion at any time. Find the velocity and acccloration
after 10 seconds.

(10) A body moves in such a way that the spaces
described in the time ¢ from starting is given by
=i", where n is a constant, Find the vaine of m
when the velocity is doubled from the 5th to the 10th
second ; find it also when the velocity is numerically
equal to the acceleration at the end of the 10th second.

CHAPTER IX.
INTRODUCING A USEFUL DODGE,

Someries one is stumped by finding that the ex-
pression to be differentiated is too complicated to
tackle directly.

Thus, the equation

y=(e+ayt

is awkward to a beginner

Now the dodge to turn the difficulty is this: Write
some symbol, such as 1, for the expression 2*+a;
then the equation becomes

ya,

which you can easily manage; for
3
Hein,
u Pi u

‘Then tackle the expression

u=xita,
and differentiate it with respect to a
du

=2r.

68 CALCULUS MADE EASY

‘Then all that remains is plain sailing;

dy
for de
that is, ito
= Matta? x 2e
=30(0*+a2);

and so the trick is dono,

By and bye, when you have learned how to deal
with sines, and cosines, and exponentials, you will
find this dodge of increasing usefulness.

Example.
Let us practise this dodge on a few examples.

(1) Differentiate y=VJate.

Let a+o=w
dur. yey, Werte 3
Zr yews y te? Hat)
dy _dy du _ 1

da” Au” de Bate

e A ae
(2) Differentiate "Jere
Let ata?=u.

INTRODUCING A USEFUL DODGE 69
(8) Differentiate y= (m—nect+ BY.

Let m-net+pat=u.

ce na pnt;

u: Beane,
vu; q sara,

dy sa
= mamon 2) (ent sper
(4) Differentiate y= IE
Let w=a5—
de di en
Gases yout; = ey,

dy_dy du_ 3°
di di Je
(5) Differentiate y= N
a-ak
Ata
aj 1
ee
de THe ie

(We.may also write y=(1—2)'(14+2)* and differs
ontiate as a product.)

Write thisas y=

70 CALCULUS MADE EASY

Proceeding as in exercise (1) above, we get

daa 1, a Ute 1
de Ever de 2V1+x
Hence
dy__ +a wt
de une Mae
Yiz_.
EVE © AF

1
(Maia?

(6) Difierentiate y= =

We may write this
yaa,
201407740 UA an)
Differentiating (L+4?) 4, as shown in exercise (2)
above, we get

dfa+2y À]
de

80 that

du. Va
EE JA) ara

INTRODUCING A USEFUL DODGE 71
(7) Differentiate y=(&+va +2 ta).
Let a+ rates.
gen ee
yaw; and 4 Eat NA
Now act and (244 a)=w

do_ 3; Ru
tl; veut; pur

ren
du 2,
Hence de +975
dy_dy an
dede da
= TT a) 2241
3@+Va Feta) arten

18) Differentiate y= ere | NE =
We got

a N
er ei ta,

dy os er er PRICE SUN
de (+ Haedo

12 CALCULUS MADE EASY
RN

du__1 ¿de

+ E nm

ut; ¿ot Dem
Liat.
jet ayi

Let w=(a?+a%)t and 2=(a’+2°)

wat; Belt, Mom

Laa yt

dy e

>. á
a tan 3@— at os

Fat
ee Wen HOTT rer)
(9) Differentiate y" with respect to yP.

nn ns
ay) ar
(10) Find the first and second differential coefficients

of yaa.
dy _e damn, Vado,
de db de b
Let [(a—a)a]!=u and let (a—2)e =; then u= ww
du 1,4 1 1 _
do 2 (axe

INTRODUCING A USEFUL DODGE 73

Pa m.

x ae du___a-2xr

= gn Ba dat)b(a 22)
ey Dada 8) Sea
dat ours ea om yw

Ba 12a

(We shall need these two last differential coefficients
later on, See Ex, X. No. 11.)

Exercises VI. (See page 291 for Answers)

Differentiate the following:

2) y=Sa Fah

a+ a
OI ay

#4 CALCULUS MADE EASY

(8) Differentiate y* with respect to y
JA

(9) Differentiate y=

The process can be extended to three or more
" dy _dy „de „dv
differential coefficients, so that da de® d* To

Examples.
(1) If 2=Bet; iy Mers fina YY

We have
14. den

>

28
"evito? Bert
ee a=r4l
do _Tx(5u—-6), de_gpyy. d_ 1
GV di 0H Te
Tx(See—6)(82-+2)

de_ 3
Hence E E

an expression in which a must be replaced by its
value, and ¢ by its value in terms of 6.

Ban, „SIE.
= rg; and 9

INTRODUCING A USEFUL DODGH 75

We get o
0-30 we and He

1

TANTE

dg
So that SP =<,
Replace now first w, then @ by its value.

Exercises VII.
You can now successfully try the following, (See

page 291 for Answers.)

() Hunts; 0=8 (0-402); and w=, find de,

na

5 — 1
= Bat. 2: gea/1 q =>
Or y=3at+J/2: an Ey; ando Je

find
fina Zu

CHAPTER X.
GEOMETRICAL MEANING OF DIFFERENTIATION.

Ir is useful to consider what geometrical meaning can
bo given to the differential coefficient.

In the first place, any function of =, such, for
example, as a, or /z, or a+b, can be plotted as
a curve; and nowadays every schoolboy is familiar
with the process of curve-plotting.

Fa.

Let PQR, in Fig. 7, be a portion of a curve plotted
with respect to the axes of coordinates OX and OY.
Consider any point Q on this curve, where the
abscissa of the point is æ and its ordinate is y.
Now observe how y changes when w is varied. If 2

MEANING OF DIFFERENTIATION 77

is made to increase by a small increment de, to the
right, it will bo observed that y also (in this particular
curve) increases by a small increment dy (because this
particular curve happens to be an ascending curve).
‘Then the ratio of dy to dz is a measure of the degree
to which the curve is sloping up between the two
points Q and 7, As a matter of fact, it can be seen
on the figure that the curve between Q and 7 has
many different slopes, so that we cannot very well
speak of the slope of the curve between Q and 7. If,
however, Q and T are so near each other that the
small portion QT of the curve is practically straight,

then it is true to say that the ratio dy is the slope of

the curve along Q7. The straight line Q7 produced
on either side touches the eurve along the portion QT
only, and if this portion is indefinitely small, the
straight line will touch the curve at practically
one point only, and be therefore a tangent to the
curve,

This tangent to the curve has evidently the same

slope as QT, so that ©

is the slope of the tangent to

the curve at the point Q for which the valuo of Ed is
found. .

We have seen that the short expression “the slope
of a curve” has no precise meaning, because a. curve
has so many slopes—in fact, every small portion of a
curve has a different slope. “The slope of a curve at
a point” is. however, a perfectly defined thing; it is

18 CALCULUS MADE EASY

the slope of a very small portion of the curve situated
just at that point; and we have scon that this is the
Sumo as “the slope of the tangent to the curve at that
point.”

Observe that dr is a short step to the right, and
dy the corresponding short step upwards. These
steps must be considered as short as possible—in fact
indefinitely short,—though in diagrams we have to
represent them by bits that are not infinitesimally
small, otherwise they could not be seen.

We shall hereafter muke considerable use of this

circumstance that Ÿ represents the slope of the curve
at any point.

Y]

Fie, &

If a curve is sloping up at 45° at a particular point,
as in Fig. 8, dy and de will be equal, and the value

of Wat,

MEANING OF DIFFERENTIATION 7%
If the curve slopes up steeper than 45° (Fig, 9)

ps will be greater than 1.

de
Y

C2

Aa
ta
Lt
or x o

Fre, de ie. 10,

If the curvo slopes up very gently, as in Fig. 10,
4Y Wil be a fraction smaller than 1.

For a horizontal line, or a horizontal place in a
curve, dy=0, and therefore Woo,

Vio, la.

If a curve slopes downward, as in Fig. 11, dy will
be a step down, and must therefore be reckoned of

80 CaLCULUS MADE EASY
dy

negative value; hence $% will have negative sigo
aiso.

If the “curve” happens to be a straight line, like
that in Fig: 12, the valuo of @ will be tho same at

all points along it. In other words its slope is constant.

é x

Fig, 12,
LE a curve is one that turns more upwards as it

goes along to the right, the values of 42 will become

C2

Fro. 18,

greater and greater with the increasing steepness, as
in Fig. 18,

MEANING OF DIFFERENTIATION &
E a curve is one that gets flatter and flatter as it

goes along, the values of oe will become smaller and

smaller as the flatter part is reached, as in Fig, 14

Y]

1
Hi
A
{
‘
A
}

1

t
o x
Yo. 14. Fis. 15,

IE a curve first descends, and then goes up again,
as in Fig. 15, presentinz a concavity upwards, then
clearly a will first be negativo, with diminishing
values as the curve flattens, then will be zero at the
point where the bottom of the trough of the curve is

reached; and from this point onward dy will have

positive values that go on inercasing. Tn such a case y
is said to pass through a minimum. ‘The minimum
value of y is not necessarily the sinallest value of y.
it is that value of yy corresponding to the bottom of
the trough; for instance, in Fig. 28 (p. 101), the
value of y corresponding to the bottom of the trough
is 1, while y takes elsewhere values which are smaller
than this, The characteristic of a minimum is that

y must increase on either side of it.
om. F

32 CALCULUS MADE EASY

N.B—For the particular value of æ that maxes

dy
aan

If a curve first ascends and then descends, the

y = minimum, the value of

values of Me will be positive at first; then zero, as
th; summit is reached; then negative, as the curve
alcpes downwards, as in Fig. 16. In this case y is
said to pass through a mazimum, but the maximum
value of y is nob necessarily the greatest value of y.
In Fig, 28, the maximum of y is 23, but this is by no
means the greatest value y can have at some other
point of the curve,

Fig, 16, Fra, 17,

W.B.—For the particular value of æ that makes

y a macimum, the value of ¿2 0.

If a curve has the particular form of Fig. 17, the
values of 4 will always be positive; but there will
be one particular place where the slope is least steep,
where the value of du will be a minimum; thet 18,

less than it is ab any other part of the curve.

MEANING OF DIFFERENTIATION 83

dy
de
will be negative in the upper part, and positive in the
lower part; while at the noso of the eurve where it

IE a curve has the form of Fig. 18, the value of

becomes actually perpendicular, the value of 92 will

be infinitely great.

Y

x

dy
de
steepness of a curve at any point, let us turn to some
of the equations which we have already learned how
to differentiate.

Now that we understand that measures the

(1) As tho simplest case take this:
yaar.

It is plotted out in Fig. 19, using equal seales
for w and y. If we put 2=0, then the corresponding
ordinate will be y=; that is to say, the “curve”
crosses the y-axis at tho height 5. From here it

84 CALCULUS MADE EASY

ascends ab 45°; for whatever values wo give to «to
the right, we have an equal y to ascend. The line
has a gradient of 1 in 1.

Now differentiate y=æ+5, by the rules we have

already learned (pp. 22 and 26 ante), and we get u- 1.

The slope of tho line is such that for every little
step dev to the right, we go an equal little step dy
upward, And this slope is constant—always the
same slope.

Fig, 19. Fra, 20,

(2) Take another ease:

y=00+b,
We know that this curve, like the preceding one, will
start from a height 8 on the y-axis. But before we
draw the curve, let us find its slope by differentiating;

a. The slope will be constant, at

an angle, the tangent of which is here called a. Let
as assign to a some numerical value—say.}. Then we
must give it such a slope that it ascends 1 in 3; or

MEANING OF DIFFERENTIATION 85

dar will be 3 times as great as dy; as magnified in
Fig. 21. So, draw the line in Fig. 20 at this slope.

Fi. 21.

(8) Now for a slightly harder case.

Let

Again the eneve will start on the y-axis at a height
6 above the orig’

Now differentiate, [If you have forgotten, turn
back to p. 26; or, rather, don’t turn back, but think
out the differentiation.]

dy
Fi 2a.

Y|

o! =
Fic. 22,

This shows that the steepness will not be constant:
it increases as a increases. At the starting point P,

86 CALCULUS MADE EASY

where æ=0, the curve (Fig. 22) has no steepness
—that is, it is level. On the left of the origin, where

2 has negative values, 42 will also have negative

de
values, or will descend from left to right, as in the
Figure.

Let us illustrate this by working out a particular
instance. Taking the equation

y= ets,
and differentiating it, we get

dy _
de

Now assign a few successive values, say from 0 to
5, to a; and calculate the corresponding values of y
by the first equation; and of qu from the second

equation, Tabulating results, we have:

æ lo
vloa
dy

2|o

Then plot them out in two eurves, Figs. 23 and 24
in Fig. 23 plotting the values of y against those of @,

and in Fig, 24 those of ae against those of æ For

MEANING OF DIFFERENTIATION 87

any assigned value of a, the height of the ordinate
in the second curve is proportional to the slope of the
first curve.

rio, 23 Pia. 24,

If a curve comes to a sudden cusp, as in Fig. 25,
the slope at that point suddenly changes from a slope

Fis. 25.

upward to a slope downward. In that case ES will

clearly undergo an abrupt change from a positive to
a negative value.

88 CALCULUS MADE EASY

The following examples show further applications
of the principles just explained.
(A) Find the slope of the tangent to the eurve

1
33 +3

y=

at the point where a=—1. Find the angle which this
tangent makes with the curve y=2a*+2,

‘The slope of the tangent is the slope of the curve at
the point where they touch one another (see p. 77);
that is, it is the

1 dy 1.
gaa and for w= —1, 9Y=-5, which is the
slope of the tangent and of the curve at that point,
The tangent, being a straight line, has for equation

of the curve for that point. Here

y=an-+b,and ite slope is (La, honce a= À. Also
1 lar >

if w=—1, y=g y+3=24; and as the tangent

passes by this point, the coordinates of the point must
satisfy the equation of the tangent, namely

1
—¿o+b,

Y
so that 2}= —}x(—1)-+5 and D=2; the equation of
the tangent is therefore y= jor 2

Now, when two curves meet, the intersection being
a point common to both curves, its coordinates must
satisfy the equation of cach one of the two curves;

MEANING OF DIFFERENTIATION 38

that is, it must be a solution of the system of simul-
taneous equations formed by coupling together the
equations of the curves. Hore the curves meet one
another at points given by the solution of
{ y= +2,
y=-ja+2 or 2+2=—-}0+2;

that is, 2204 )=0.

This equation has for its solutions «= 0 and = —}
The slope of the curve y=2a*+2 at any point is

CE

For the point where = 0, this slope is zero; the curve
is horizontal. For the point where

1 dy
“+ &
hence the curve at that point slopes downwards to
the right at such an angle @ with the horizontal that
tan 921; that is, at 45° to the horizontal.

The slope of the straight line is —} ; that is. it slopes
downwards to the right and makes with the horizontal
an angle ¢ such that tan =}; that is, an angle of
26° 34, It follows that at the first point the curve
cuts the straight line at an angle of 26° 34/, while at
the second it cuts it at an angle of 45°— 26° 34’ =18" 26.

1;

(5) A straight line is to be drawn, through a point
whose coordinates are &=2, y==1, as tangent to the
curve y=2*—524+6, Find the coordinates of the
point of contact.

90 CALCULUS MADE EASY

The slope of the tangent must be the same as the
a of the curve; that is, 22-5.

The equation of the straight line is y=a2+D, and

as it is satisfied for the values @=2, y=—1, then
“tg U

-1=4X2+b; also, ite ZU

The & and the y of the point of contact must also
satisfy both the equation of the tangent and the
equation of the curve.

We have then

=20-5.

5046,
y=ax+b,
—1=2a+6,
a — 5,

four equations in a, b, x, y.

Equations (1) and (ii) give a—5a-+6=an+b.
Replacing a and b by their value in this, we get
e—b0+6=(20—5)0—1- 2205),
which simplifies to 2*—40+3=0, the solutions of
which are: 2=3 and @=1. Replacing in (i), we get
y=0 and y=2 respectively ; the two points of contact

are then 2=1, y=2; and &=5, y=0.

Note-—In all exercises dealing with curves, students
will find it extremely instructive to verify the dedue-
tions obtained by actually plotting the curves.

MEANING OF DIFFERENTIATION 9

Exercises VIII. (See page 291 for Answers.)

(1) Plot the curve y=jat-5, using a scale of
millimetres. Measure at points corresponding to
different values of 2, the angle of its slope.

Find, by differentiating the equation, the expression
for slope; and see, from a Table of Natural l'engents,
whether this agrees with the measured angle,

(2) Find what will be the slope of the curve

y=012%*-2,
at the particular point that has as abscissa æ= 2.

(3) If y=(w—a@)(w—b), show that at the particular

point of the curvo where 92.0, a will have the value
dia+b).
(4) Find the du of the equation y=x+ 3%; and

ealeulate the numerical values of qu for the points

corresponding to 2=0, 2-4, a=1,

(5) In the curve to which the equation is a? +y*=4,
find the values of æ at those points where the slope=1.

(6) Find the slope, at any point, of the curve whose
Ep

equation is & 4/1; and give the numerical value

of the slope at the place where #0, and at that
where æ=1.

(7) The equation of a tangent to the curve
y=5—20+052, being of the form y=matn, where
m and m are constunts, find the value of m and n if

92 CALCULUS MADE EASY

the point where the tangent touches the curve has
æ=2 for abscissa,

(8) At what angle do the two curves

y=350'+2 and y=a%—5a+95
cut one another ?

(9) Tangents to the curve y= + 4/35 2 are drawn.
at points for which &=3 and w= 4, the value of y being
positive. Find the coordinates of the point of inter-
section of the tangents and their mutual inclination.

(10) A straight line y=2x—b touches a curve
y=Su!+2 at one point. What are the coordinates
of the point of contact, and what is the value of 4?

CHAPTER XL
MAXIMA AND MINIMA.

A quawtirY which varies continuously is said ta
pass by (or through) a maximum or minimum value
when, in the courso of its variation, the immediately
preceding and following values are both smaller or
greater, respectively, than the value referred to. An
infinitely great value is therefore not a maximum
vaine.

One of the principal uses
of the process of differen-
tiating is to find out under
what conditions the value
of the thing differentiated
becomes a maximum, or &
minimam. This is often ex-
ceedingly important in en-
gineering questions, where
it is most desirable to
know what conditions will make the cost of working
a minimum, or will make the efficiency a maximum,

Now, to begin with a concrete case, let us take the
equation yaa het.

By assigning a number of successive values to =,
and finding the corresponding values of y, we can

H
H
3

x

Fro. 26,

y CALCULUS MADE EASY

readily see that the equation represents a curve with
@ minimum,

SOHEIEIEIE
|

ylrlalslalrlu

These values are plotted in Fig. 26, which shows
that y has apparently a minimum value of 3, when &
is made equal to 2. But are you sure that the
minimum occurs at 2, and not at 2} or at 131

Of course it would be possible with any algebraic
expression to work out a lot of values, and in this
way arrive gradually at tho particular value that
may be a maximum or a minimum.

Here is another exampl
Let y=8n—2%.
Calculate a few values thus:

MAXIMA AND MINIMA E

Plot these values as in Fig. 27.

It will be evident that there will be a maximum
somewhere between #=1 and &=2; and the thing
tools as if the maximum value of y ought to be
about 2}. Try some intermediate values, If æ=1},
y=2187; if o=14 y=225; if 0-16, y=224
How can we be sure that 2:25 is the real maximum,
or that it oceurs exactly when #=1}?

Now it may sound like juggling to be assured that
there is a way by which one can arrive straight at a
maximum (or minimum) value without making a lot of
preliminary trials or guesses. And that way depends
on differentiating. Look back to'an earlier page (81) for
the remarks about Figs. 14 and 15, and you will see
that whenever a curve gets either to its maximum

or to its minimum height, at that point its au,

Now this gives us the clue to the dodge that is
wanted. When there is put before you an equation,
and you want to find that value of that will make
its y a minimum (or a maximum), first diferentiate

it, and having done so, write its 5 as equal to zero,

and then solve for #. Pub this particular value of =
into the original equation, and you will then get the
required value of y. This process is commonly called
“equating to zero.”
To see how simply it works, take the example with
which this chapter opens, namely
yao 4047.

96 CALCULUS MADE EASY
Differentiating, we get:
YY on 4,

de
Now equate this to zero, thus:
20—4=0,
Solving this equation for æ, we get:
20=4,

æ=2.

Now, we know that the maximum (or minimum)
will occur exactly when +

Putting the value &=2 into the original equation,
we get

Now look back at Fig 26, and you will see that the
minimum oceurs when #=2, and that this minimum
of y=3.
‘Try the second example (Fig. 24), which is
y=3e—a
Differentiating, W320,

Equating to zero,

3-22=0,
whenco e=1);
and putting this value of a into the original equation,
weänd: y=44- (4x1,

This gives us exactly the information as to which
the method of trying a lot of values left us uncertain,

MAXIMA AND MINIMA 97

Now, before we go on to any farther cases, we have
two remarks to make, When you are told to equate

we to zero, you feel at first (that is if you have any
wits of your own) a kind of resentment, because you
know that @U has all sorts of different values at

different parts of the curve, according to whether it
is sloping up or down, So, when you are suddenly
told to write d

Yo,

you resent it, and feel inclined to say that it can't be
true, Now you will have to understand the essential
difference between "an equation,” and “an equation
of condition.” Ordinarily you are dealing with equa-
tions that are true in themselves, but, on occasions,
of which the present are examples, you have to write
down equations that are not necessarily true, but are
only true if certain conditions are to be fulfilled ; and
you write them down in order, by solving them, to
find the conditions which make them true Now we
want to find the particular value that « has when
the curve is neither sloping up nor sloping down, that
is, at the particular place where (2
Mano does not mean that it always is =0; but you

So, writing

write it down as a condition in order to see how
much a will come out if Y is to be zero,

ce &

98 CALCULUS MADE EASY

‘The second remark is one which (if you have any
wits of your own) you will probably have already
made: namely, that this much-belauded process of
equating to zero entirely fails to tell you whether
the = that you thereby find is going to give you
a maximum value of y or a minimum value of y.
Quite so. It does not of itself discriminate; it finds
for you tho right value of æ but leaves you to find
out for yourselves whether the rorresponding y is a
maximum or a minimum. Of course, if you have
plotted the curve, you know already which it will be.
For instance, take the equation :

1
yet

Without stopping to think what curve it corre-
sponds to, differentiate it, and equate to zero:

Wagon
whence w=;
and, inserting this value,

y=4

will be either a maximum or else a minimum. But
which? You will hercafter be told a way, depending
upon a second differentiation, (see Chap. XIL, p. 112)
But at present it is enough if you will simply try
any other value of æ differing a little from the one
found, and see whether with this altered value the
corresponding value of y is less or greater than that
already found.

MAXIMA AND MINIMA 99

Try another simple problem in maxima and minima,
Suppose you were asked to divide any number into
two parts, such that the product was a maximum?
How would you set about it if you did not know
the trick of equating to zero? I suppose you could
worry it out by the rule of try, try, try again, Let
60 be the number. You can try cutting it into two
parts, and multiplying them together. ‘Thus, 50 times
10 is 500; 52 times 8 is 416; 40 times 20 is 800; 45
times 15 is 675; 30 times 30 is 900. This looks like
2 maximum: try varying it 31 times 29 is 899,
which is not so good; and 32 times 28 is 896, which
is worse. So it seems that the biggest product will
be got by dividing into two equal halves,

Now see what the calculus tells you Let the
number to bo ent into two parts be called m. Then
if x is one part, the other will be n— x, and the product
will be e(n—a) or ne—a®. So we write y=na—a%
Now differentiate and equate to zero;

U _n-20=0.

Solving for æ, we get ga

So now we know that whatever number # may be,
we must divide it into two equal parts if the product
of the parts is to be a maximum; and the value of
that maximum product will always be =n.

This is a very useful rule, and appliec to any number
of factors, so that if m+n+p=a constant number,
mxnxp is a maximum when m=n=p.

100 CALCULUS MADE EASY

Test Case.

Let us at once apply our knowledge to a case that
we can test.

Let y=0—2;
and let us find whether this function has a maximum
or minimum ; and if so, test whether it is a maximum
or a minimum.

Differentiating, we get

Gl=2n~L
Equating to zero, we get
2x—-1=0,
whence 2n=1,
or ah

That is to say, when x is made=}, the corresponding
value of y will be either a maximum or a minimum.
Accordingly, putting æ= } in the original equation, we
Bet yo Gib
or y=-2

Is this a maximum or a minimum? To test it, try
putting & a littie bigger than 4,—say make #=0€
Then y=(06)—06=036-=06=-0:24,
which is higher up than —025; showing that

=-—025 is a minimum.

Plot the curve for yourself, and verify the cal-
culation,

MAXIMA AND MINIMA 101

Further Examples.

A most interesting example is afforded by a curve
that has both a maximum and a minimum. Its
equation is: y=10-20 43041

Now du dut 3.

Kio, 28,
Equating to zero, we get the quadratic,
w—4¢4+3=0;
and solving the quadratic gives us two roots, viz

Now, when #=8,
The first of these is
maximum,

The curve itself may be plotted (as in Fig. 28)

1; and when w=1, y=2h
a minimum, the second a

102 CALCULUS MADE EASY

from the values calculated, as below, from the
original equation.

A further exereise in maxima ond minima is
afforded by the following example:

‘The equation to a circle of radius +, having its
centre O at the point whose coordinates are 2=a,
y=b, as depicted in Fig, 29, is:

G-D+(@-a)=

This may be transformed into

y=NP-@—aF tb

a
Fre, 29,

Now we know beforohand, by mere inspection of
the figure, that when 2=a, y will be either at its
maximum value, d-Fr, or else at its minimum
value, 6—7. But let us not take advantage of this

MAXIMA AND MINIMA 103

Knowledge; let us set about finding what value
of æ will make y a maximum or a minimum, by the
process of differentiating and equating to zero.

gy 1 CN
do Ira CO
which reduces to
Wan
Laser

Then the condition for y being maximum or
minimum is:

a-2 _g
Nr-@-a

Since no value whatever of y will make the de-

nominator infinite, the ‘only condition to give zero is
æ=a.

Inserting this value in the original equation for
the circle, we find u
NR+8;
and as the root of 7? is either +r or —r, we have
two resulting values of y,

iis

The first of these is the maximum, at the top;
the second the minimum, at the bottom.

104 CALCULUS MADE EASY

If the curve is such that there is no place that is a
maximum or minimum, the process of equating to
zero will yield an impossible result, For instance:

Let y=a+bet+e
D _ ou.
Ten Gan Baa +b

Equating this to zero, we get 8aa°+-b=0, ang,
SEE which is impossible, eupposs
and y= vg which is impossible, supposing « and &
to have the same sign.

‘Therefore y has no maximum nor minimum,

A few more worked examples will enable you to
thoroughly master this most interesting and useful
application of the calculus.

(1) What are the sides of the rectangle of maximum
area inscribed in a circle of radius 2?

Jf one side be called =,

the other side = (diagonal? —a?;

and as the diagonal of the rectangle is necessarily a
diameter, the other side=/ER*— 2%,

Then, area of rectangle S=a/ TP =a,
a a AO ST 42.

1£ you have forgotten how to differentiate /IR añ

here is a hint: write 4R'-a%=w and y=./w, and
de dw a

seek Zz and 973 fight it out, and only if you can't

get on refer to page 67.

MAXIMA AND MINIMA 105

You will get
as æ oe _ HR 22
Te

For maximum or minimun we must have
4R°- 2a
SER

that is, 4R*—2a%=0 and a= R4/2.
The other side= VAR IR R/2; the two sides
are equal; the figure is a syuare the side of which is
equal to the diagonal of the square constructed on the

radius, In this case it is, of course, a maximum with
which we are dealing.

=0,

(2) What is the radius of the opening of a conical
vessel the sloping side of which has a length Z when
the capacity of the vessel is greatest?

ER be the radius and HT the corresponding height,
H=NFZR.

JR
Volume V=rrex Hex VER.
Proceeding as in the previous problem, we get
2rR
HN PR

EZ
for maximum or minimum.
Or, 27R(P—R*)—R*=0, and R= 14/5, for a maxi-
mum, obviously.

106 CALCULUS MADE EASY

€) Find the maxima and minima of the function
=

a
NES
We get
dy _(4-2)-(-0), ee)
de Gay tr

for maximum or minimum or

4

¿320 and 2=2.

(ar

There is only one value, hence only one maximum
or minimum,

it is therefore a minimum. (It is instructive to plot
the graph of the function.)

(4) Find the maxima and minima of the function
y=J/i\+a+/1—2. (It will be found instructive to
plot the graph.)

Differentiating gives at once (see example No 1,
p. 68)

dy 1___ 1 9

de 2/42 21a

for maximum or minimum.
Hence /T+o=w/1=w% and x=0, the only solution.
Foræ=0, y
Fora==0%, y

MAXIMA AND MINIMA 107

(5) Find the maxima and minima of the function

YE
We Snes
_ (2-4) x 2p (a —5)2_
a ~~ Ge
for maximum or minimum ; or
2a?
¢
2°-40+5=0; which has for solutions

e=t4 VI.

‘These being imaginary, there is no real value of =

for which Y- ; hence there is neither maximum nor
minimum,

(6) Find the maxima and minima of the function
Ga) =.
This may be written y=a*+aé,
d

de 20 +g0*=0 for maximum or minimum;
that is, &(2+#æl)=0, which is satisfied for 2=0,
and for 2+§a%=0, that is for æ=}2 So there are
two solutions.

Taking first &=0. If x2=-05,y=0254/ (5,
and if æ=+05, y=0254x/05% On one side y is
imaginary ; that is, there is no value of y that can be
represented by a graph; the latter is therefore entirely
on the right side of the axis of y (see Fig. 30),

On plotting the graph it will be found that the

108 CALCULUS MADE EASY

eurve gocs to the origin, as if there were a minimum
there; but instead of continuing beyond, as it should
do for a minimum, it retraces its steps (forming what
is called a“cusp”). There is no minimum, therefore,
although the condition for a minimum is satistied,

namely E O. It is necessary therefore always to

check by taking one value on either side.
Y

os]

O 02 OF 06 0810 E
Fic. 30,

Now, if we take a: 064 Ifx=064,y=07873
and y=0:0819; if a =0'6, y becomes 06389 and 00811;
and if =0°7, y becomes 0°8996 and 00804.

This shows that there are two branches of the curve,
the upper one does not pass through a maximum, but
the lower one does.

(7) A eylinder whose height is twice the radius of
the base is increasing in volume, so that all its parte

MAXIMA AND MINIMA 109

keep always in the same proportion to each other;
that is, ab any instant, the cylinder is similar to the
original cylinder. When the radius of the base is
r feet, the surface area is increasing at tho rate of
20 square inches per second; at what rate per second
is its volume then increasing?
Area=S=2(r1%)+ 201 x2=670%
Volume=V= ._ X= ri.

as dr_ 20
Sas Mn;
POP UE rn
Lu and

ay _

Le

‘The volume changes ut the rate of 10r cubic inches
per second.

Make other examples for yourself. ‘Chere are few
subjects which offer such a wealth for interesting
examples,

Exercises IX. (See page 292 for Answers.)
(1) What values of & will make y a maximum

and a minimum, if y="?
(2) What value of æ will make y a maximum in

the equation

10 CALCULUS MADE EASY

(8) A lino of length p is to be eut up into 4 parts
and put together as a rectangle. Show that the area
of the rectangle will be a maximum if each of its
sides is equal to tp.

(4) A piece of string 30 inches long has its two
ends joined together and is stretched by 3 pegs so
as to form a triangle. What is the largest triangular
area that can be enclosed by the string?

(Hint: Apply last three lines of p. 99.)

(5) Plot the curve corresponding to the equation

also find 4% and deduce the value of æ that will

make y a minimum; and find that minimum value
of y.

(6) If y=2%—50, find what values of a will make
y a maximum or a minimum.

(7) What is the smallest square that can be in-
scribed in a given square?

(8) Inscribe in a given cone, the height of which
is equal to the radius of the base, a cylinder
(a) whose volume is a maximum; (6) whose lateral
area is a maximum; (c) whose total area is a
maximum.

(9) Inseribe in a sphere, a eylindor (a) whose
volume is a maximum; (b) whose lateral area is a
maximum ; (c) whose total area is a maximum,

MAXIMA AND MINIMA m

(10) A spherical balloon is increasing in volumo,
If, when its radius is r feet, its volume is increasing
at the rate of 4 cubie fect per second, at what rate is
its surface then increasing?

(11) Inscribe in a given sphere a cone whose volume
is maximum.

(12) The current O given by a battery of N similar

voltaie cells is O=22,, where E, R,x, are constante

ty
and # is the number of cells coupled in series. Find
the proportion of a to N for which the current is
greatest,

CHAPTER XIL
CURVATURE OF CURVES.

PerurNING to the process of successive differentia-
tion, it may be asked: Why does anybody want to
differentiate twice over? We know that when the
variable quantities are space and time, by differ-
entinting twice over we geb the acceleration of a
moving body, and that in the geometrical interpreta-

Fre, A. Fra. 32,

tion, as applied to curves, © means the slope of the
curve, But what can $4 mean in this caso? Clearly

it means the rate (per unit of length x) at which the
slope is changing—in brief, it is an indication of the
manner im which the slope of the portion of curve
considered. varies, that is, whether the slope of the
curve increases or decreases when æ increases, or, in

CUKVATURE OF CURVES 13

other words, whether the curve curves up or down
towards the right
Suppose a slope constant, as in Fig, 31.

Here, SU is of constant value.

Suppose, however, a case in which, like Fig, 32,
the slope itself is getting greater upwards; then

A) dl

Er the slope is becoming less as you go to the
right (as in Fig. 14, p. 81), or as in Fig. 33, then,
even though the curve may be
going upward, since the change
is such as to diminish its slope,

Y, will be positive.

its FY will be negative,

ES is now time to initiate
you into another secret —how —.
to tell whether the result that mae
you get by “equating to zero” +

is a maximum or a minimum. The trick is this: After
you have differentiated {so as to get the expression
Which you equate to zero), you then differentiate a
second time, and look whether the result of the second

differentiation is positive or negative. IE FU comes

ont positive, then you, know that the value of y
which you grt was a minimum; bat it GP, comes
OM. H

114 CALCULUS MADE EASY

out negative, then the value of y which you got must
be a maximum. That's the rule.

The reason of it ought to be quite evident. Think
of any curve that has a minimum point in it, like
Fig. 15 (p. 81), or like Fig. 34, where the point of
minimum y is marked M, and the curve is concave
upwards. To the left of H the slope is downward,
that is, negative, and is getting less negative. To the
right of M the slope has become upward, and is

Y

=
Fic. 34. Fie, 85.

getting more and more upward. Clearly the change
of slope as the curve passes through M is such that
gu is positive, for its operation, as æ increases toward.
the right, is to convert a downward slope into an
upward one.

Similarly, consider any curve that has a maximum
point in it, like Fig. 16 (p. 82), or like Fig, 35, where
the curve is convex, and the maximum point is
marked Jf. In this case, as the curve passes through
M from left to right, its upward slope is converted

CURVATURE OF CURVES 15

into a downward or negative slope, so that in this
By
das

Go back now to the examples of the lest chapter
and verify in this way the conclusions arrived at as to
whether in any particular case there is a maximum
ora minimum. You will find below a few worked
out examples.

case the “slope of the slope” ZH is negative,

(1) Find the maximum or minimum of

(a) y-4at-98-6; (0) y=64 90-402;
and ascertain if i be a maximum or a minimum in
each case.

=80—9=0; w=1}; and y=—11-065.

de

Hs; it is + ; hence it is a minimum.

® 2 =9-80=0; w=14; and y=+11-065.

—8; it is —; hence it is a maximum.

dee

(2) Find the maxima and minima of the function
y=0*-30+16.

UY 95230; af=1; =
2802-30; at=1; and a=].

dy

das

6x; for @=1; itis +;

hence @=1 corresponds to a minimum y=14. For
@=—1 it is —; hence = —1 corresponds to a maxi-
mum y= +18.

116 CALCULUS MADE EASY

(3) Find the maxima and minima of y

dy _(@+2)x1—(w-1)x 2a _w—aP-+2

de + Ay
or 22-22 2=0, whose solutions are &= +273 and
m= ~073.

(oP + 2) x (2 — 2) — (0? in 2) (408 + Bar)
FOF

The denominator is always positive, so it is sufficient
to ascertain the sign of the numerator.

IE we put x=278, the numerator is negative; the
maximum, y =0:183,

If we put == —078, the numerator is positive; the
minimum, y= —0-683.

(4) The expense O of handling the products of a
certain factory varies with the weekly output P
é
o+P
a, b, e, d are positive constants. For what output

will the expense be least ?
aC 4-2 =0 for maximum or minimum;
dp c+PF ’

b ib
hence a= py and P= taba

according to the relation C=aP+

+d, where

As the output cannot be negative, P= + be

CURVATURE OF CURVES” Wr

EC _,bCer2P)
dP? (c+ Py”
which is positive for all the values of P; hence

Now

5 GE
P=+ Y —e corresponds to a minimum.

(5) The total cost per hour C of lighting a building
with N lamps of a certain kind is
G, EPC,
C= ro HS Sum)
where His the commercial efficiency (watts per candle),
P is the candle power of each lamp,
t is the average life of cach lamp in hours,
C,=cost of renewal in pence per hour of use,
C,=cost of energy per 1000 watts per hour.
Moreover, the relation connecting the average hie
of a lamp with the commercial efficiency at which it
is run is approximately (= m", where m and n are
constants depending on the kind of lamp.
Kind the commercial efficiency for which the total
cost of lighting will be least.

We have CeN(C gr FE 1),
e (PO, Cigar

for ms or minimum.

1000 x nO; "+ [1000 x nO,
he A PA A
E" mPO, and E a

118 CALCULUS MADE EASY
This is clearly for minimum, since
PO RO, nm.
rie,

which is positive for a positive value of E.

For a particular type of 16 candle-power laraps,
O/=17 pence, C,=5 pence; and it was found that
m=10 and x

2-6 watts per candle-power.

Exercises X. (You are advised to plot the graph
of any numerical example.) (See p. 292 for the
Answers.)
(1) Find the maxima and minima of
y=0"+0'—1024+8.
(2) Given y= Lo—e, find expressions for LY, and

a
for 2: also find the value of a which makes y a

maximam or a minimum, and show whether it is
maximum or minimum.

(3) Find how many maxima and how many minima
there are in the curve, the equation to which is

yA;
CES

and how many in that of which the equation is

RAR
vl gr

CURVATURE OF CURVES 118

(4) Find the maxima and minima of

. y=20+14+ 5
(5) Find the maxima und minima of
a
UO FT
(6) Find the maxima and minima of

fe
Y= ga

(7) Find the maxima and minima of
8a

æ
y= +5 +5

E

(8) Divide a number N into two parts in such a
way that three times the square of one part plus
twice the square or the other part shall be e
minimum,

(9) The efficiency u of an electric generator at
different values of output 2 is expressed by the
general equation :

2
“abate?

where « is a constant depending chiefly on the energy
losses in the iron and e a constant depending chiefly
on the resistance of the copper parts, Find an ex
pression for that value of the output ab which tho
efficiency will be a maximum.

120 CALCULUS MADE EASY

(10) Suppose it to be known that consumption of
coal by a certain steamer may be represented by the
formula y=0'3-+-0-001v"; where y is the number of
tons of coal burned per hour and v is the speed
expressed in nautical miles per hour. The cost of
wages, interest on capital, and depreciation of that
ship are together equal, per hour, to the cost of
1 ton of coal, What spced will make the total cost
of a voyage of 1000 nautical miles a minimum?
And, if coal costs 10 shillings per ton, what will that
minimum cost of the voyage amount to?

(11) Find the maxima and minima of

vera.

(12) Find the maxims and minima of
ya tea 2 +1

CHAPTER XHE
OTHER USEFUL DODGES.
Partial Fractions,

We have seen thet when we differentiate a fraction
we have to perform a rather complicated operation;
and, if the fraction is not itself a simple one, the result
is bound to be a complicated expression. If we could
split the fraction into two or more simpler fractions
such that their sum is equivalent to the original
fraction, we could then proceed by differentiating
sach of these simpler expressions, And the result of
differentiating would be the sum of two (or more)
differentials, each one of which is relatively simple;
while the final expression, though of course it will be
the same as that which could be obtained without
resorting to this dodge, is thus obtained with much
less effort and appoars in a simplified form.

Let us see how to reach this result, Try first the
job of adding two fractions together to form & resultant
fraction, Take, for example, the two fractions peat
and + Every schoolboy can add these together

and find their sum to be SCH. And in the same

122 CALCULUS MADE EASY

way he can add together three or more fractions
Now this process can certainly be reversed : that is to
say that, if this last expression were given, it is certain
that it can somehow be split back again into its
original components or partial fractions. Only we do
not know in overy case that may be presented to us
how we can so split it. In order to find this out
we shall consider a simple case at first. But it is
important to bear in mind that all which follows
applies ouly to what are called “proper” algebraic
fractions, meaning fractions like the above, which have
the numerator of a lesser degree than the denominator ;
that is, those in which the highest index of a is less
in the numerator thon in the denominator. If we

. . +2
have to deal with such an expression as PL, we can

simaplity it by division, since it is equivalent to
3
; and

A is a proper algebraic fraction

to which the operation of splitting into partial fractions
can be applied, as explained hereafter.

Case I. If we perform many additions of two or
more fractions the denominators of which contain only
terms in æ, and no terms in a2, a, or any other powers
of æ, we always find that the denominator of the final
resulting fraction is the product of the denominators
of the fractions which were added to form the result,
It follows that by factorizing the denominator of this
final fraction, we can find every one of the denomina-
tors of the partial fractions of which we are in search.

OTHER USEFUL DODGES 123

Suppose we wish to go back from SCH to the
components which we know are ty and =". If

we did not know what those components were we can
still prepare the way by writing:
A Sol re eee
a ae atar
leaving blank the places for the numerators until we
know what to put there. We always may assume the
sign between the partial fractions to be plus, since, if
it be minus, we shall simply find the corresponding
numerator to be negative. Now, since the partial
fractions are proper fractions, the numerators are
mere numbers without @ at all, and we can call them
A, B,C... as we please. So, in this ease, we have:
Setl_ A , B.
al Rita
If, now, we perform the addition of these two

A sons we vot AED E BAD. 7
partial fractions, we got SE ASAS: and this

must bo equal to 7;

Da And, as the de-

nominators in these two expressions are the same,
the numerators must be equal, giving us:
30+1=4(2-1)+B(0+1)

Now, this is an equation with two unknown

quantities, and it would seem that we need another
equation before we can solve them and find A and B

124 CALCULUS MADE EASY

Bub there is another way out of this difficulty. The
equation must be true for all values of a; therefore
it must be true for such values of a as will causo
2—1 and a-+1 to become zero, that is for a=1 and
for w=—1 respectively. If we make æ=1, we get
4=(Ax0)+(BX2), so that B=2; and if we make
w= —1, we geb —2=(4 x —2+(BX0), so that A=1.
Replacing the A and B of the partial we by

these new values, we find them to become sn and
Zar; and the thing is done,
As a farther example, let us take the fraction

The denominator becomes zero when

x is given the value 1; henee »—1 is a factor of it,
and obviously then the other factor will be 2°+-4a-+3;
and this can again be decomposed into (a+ 1)(@+8)
So we may write the fraction thus:
4420-14 A BO
Baer a+ 343
making three partial factors.
Proceeding as before, we find
4a? 4-22-14 A(w—1)(w+8) + Blw+1)(@43)
+€(a@+1)(@—-1).
Now, if we make æ=1, we get:
—8=(4 x0)+Bi2x4)+(Cx0); that is, B=—L.
If w= —1, we get
~12=A(—2x2)+(Bx0)+(Cx 0); whence À =3.

OTHER USEFUL DODGES 125

If 2= —3, we get:

16=(A x0)+(Bx0)+C(—2x —4); whence O=2

So then the partial fractions are:

3 1 2
241 2-1 ++ rg

which is far easier to differentiate with respect to ar
than the complicated expression from which it is
derived.

Case II. If sume of the factors of the denominator
contain terms in a”, and are not conveniently put
into factors, then the corresponding numerator may
contain a term in =, as well as a simple number, and
hence it becomes necossary to represent this unknown
numerator not by the symbol A but by 4a+B; the
rest of the calculation being made as before.

A
Tay, for instance: (¿a FA ee y
8-3 _AxtB, O,
E

a ~3=(Aw+ Bla 1)4 C+).
Putting == —1, we get —4=02; and C=—2;
hence —a*—3=(Aa+ B)(w+1)—20*-25
and = Aa(a+1)+ Ble+1).
Putting #=0, we get —1=B;
hence

21e A+ Dr; or a +e=Aa(et);
and æ+i= d+ Ld.

126 CALCULUS MADE EASY
so that 4 =1, and the partial fractions are:

2-1 2
SF ET
Take as another example the fraction
2-2
(CAC
We get
4-2 Ax+B, Ox+D

FL Tarte
w+ BY a+ 2)+(Cn+ D)(a2-41),
(41) +2)

In this case the determination of A, B, C, D is not
so easy. It will be simpler to proceed as follows:
Since the given fraction and the fraction founá by
adding the partial fractions are equal, and have
identical denominators, the numerators must also be
identically the same. In such a case, and for such
algebraical exproesions as those with which we are
dealing here, the coefficients of the same powers of æ
are equal and of same sign,

Hence, since
w—2=(Aw+ B)(a*+2)+(Cu+D)(a2+1)

=(A+C)a?+(B+D)a*+ (244 0)04+2B 4D,
we have 1=A+C; 0=B+D (the coefficient of a?
in the left expression being zero); 0=24 +0; and
—2=2B+D. Here are four equations, from which
we readily obtain A=-1; B=-2; (=2; D=2;
2w+1)_ 242
A FT

€

so that the partial fractions are

OTHER USEFUL DODGES 127

This method can always be used; but the method
shown first will be found the quickest in the case of
factors in æ only.

Case III. When among the factors of the denomi«
nator there are some which are raised to some power,
one must allow for the possible existence of partial
fractions having for denominator the several powers
of that factor up to the highest. For instance, in

splitting the fraction we must allow for

(041) )
the possible existence of a denominator «+1 as well
as (e+1) and (2-2).

It may be thought, however, that, since the numerator
of the fraction the denominator of which is (4-1)
may contain terms in q, we must allow for this in
writing Aw+ B for its numerator, so that

B#-%e41 A+B, O, D
e+ ly ett
If, however, we try to find A, B, C and D in this case,
we fail, because we get four unknowns; and we have
only tires relations connecting them, yeb

ft Dae 1
a —3) “atada
But if we write

32041 _ A , BO

(De CFI a
we get
30-204 1=4(0—2)+ B(041D(2-2)+C(04+15

128 CALCULUS MADE EASY

which gives @=1 for =2 Replacing C by its value,
transposing, gathering like terms, and dividing by
2-2, we got —20= A +B(w-+1), which gives A = —2
ior@=—1. Replacing A by its value, we get

20=-24B(0+1)
Hence B=2; so that the partial fractions are:

=.» y Tr
æ+1 pr a
21
instead of q + Ca pasear
Br

the fractions from which was oblainei

are à pa
The mystery is cleared if we observe that 777 > oa

itself be split into the two fractions ——

1
ari
that the three fractions given are really equivalent to

12 2 1
SI E A is MN E av
which are tho partial fractions obtained.

We see that it is sufficient to allow for one numerical
term in each numerator, and that we always get the
ultimate partial fractions,

When there is u power of a factor of a in the
denominator, however, the corresponding numerators
must be of the form Ax+B; for example,

O AwtB , Ont D, E
GDF) Get BL tae

OTHER USEFUL DODGES 129
which gives
30-1=(424+BX(0+1)
+(Ox+ DNa+ 1) Cat -1)+ Er -1%,
For w= —1, this gives E=—4. Replacing, trans-
posing, collecting like terns, and dividing by æ+1,
we get
168-1902 +8 = 2028 +2D2"+0(A —C)+(B—D).
Henee 20=16 and O=8;
á—C=00r A—5=0and A=5; and finally, B— D
So that we obtain as the partial fractions:
e.
el
It is useful to check the results obtained. The
simplest way is to replace a by a single value, say
+1, both in the given expression and in the partial
fractions obtained.
Whenever the denominator contains but a power of
a single factor, a very ce un is as follows:

king, fe > > 1=
gone or example, en let a+1=2; then
Replacing, we get
4(e-1)+1_ 4-38 4 3
RES -5-3

‘The partial fractions are, therefore,

gl rt
CNET)
1

oun

130 CALCULUS MADE EASY
Applying this to differentiation, let it be required
to differentiate y= ES ; we have

dy _ Gat+%a-3)x4 212047)
ate à
_ 2a? — 60% — 23

Apra"
If we split the given expression into

we get, however,
dy 3

4
TE Garay
which is really the same result as above split into
partial fractions, But the splitting, if done after
differentiating, is more complicated, as will easily be
seen, When we shall deal with the integration of
such expressions, we shall find the splitting into
partial fractions a precious auxiliary (see p. 280).

Exercises XI. (See page 293 for Answers.)
Split into fractions :

Ba 3a—4
Va 9 ea
3045 atl
O Beene O prep iy
a — 189426

æ—8
Dares 9 ars

OTHER USEFUL DODGES 131

e—30+1
Varas

+1

a EE

54604 >
Manera Mina

A
O aer

Ba +2041
05) ar

Ta?+90—1
a) a

Differential of an Inverse Function.
Consider the function (see p. 14) y=8ar; it can be
this latter form is called

expressed in the form a
the inverse function to the one originally given.
de 1 and we seo thas

132 CALCULUS MADE BASY
dy de
de dy

It can be shown that for all functions which can be
put into the inverse form, one can always write

Bu io Y
y

Here again x

a,

dy

It follows that, being given a function, if it be
easier to differentiate the inverse function, this may
be done, and the reciprocal of the differential coefficient
of the inverse function gives the differential coefficient
of the given function itself.

As an example, suppose that we wish to differentiate
2

1. We have seen one way of doing this,

y
by writing u=2-1, and finding Y and de This

gives

JE we had forgotten how to proceed by this method,
or wished to check our result by some other way of
obtaining the differential coefficient, or for any other
reason we could not use the ordinary method, we coulé

proceed as follows: The inverse function is = 7.5.
de__ 8x2y _
dy (yr

OTHER USEFUL DODGES 133

_1 a Y

A a

1

Let us take, as another example, y = eee

The inverse function is 0-5 or 6=y-°—5, and

By =—3 OEA
dy

It follows that 40

been found otherwise.

We shall find this dodge most useful Inter on;
meanwhile you are advised to become familiar with
it by verifying by its means the results obtained in
Exercises I. (p. 25), Nos. 5, 6, 7; Examples (p. 68),
Nos. 1, 2, 4; and Exercises VL (p. 78), Nos. 1, 2, 3
and 4

ara as might have

You will surely realize from this chapter and the
preceding, that in many respeets the calculus is an
art rather than a science: an art only to be acquired,
as all other arts are, hy practice, Henee you should
work many examples, and set yourself other examples,
to see if you can work them out, until the various
artifices become familiar by use,

CHAPTER XIV.

ON TRUE COMPOUND INTEREST AND THE
LAW OF ORGANIC GROWTH.

Ler there be a quantity growing in such a way that
the increment of its growth, during a given time,
shall always be proportional to its own magnitude.
This resembles the process of reckoning interest on
money at some fixed rate; for the bigger the capital,
the bigger the amount of interest on it in a given
time.

Now we must distinguish clearly between two
cases, in our calculation, according as the caleulation
is made by whut the arithmetic books call “simple
interest,” or by what they call “compound interest.”
For in the former case the capital remains fixed,
while in the latter the interest is added to the eap-
ital, which therefore increases by successive additions,

(1) At simple interest, Consider a concrete case.
Let the capitol at start be £100, and let the rate
of interest be 10 per cent. per annum. Then the
increment to the owner of the capital will be £10
every year, Let him go on drawing his interest
every year, and hoard it by putting it by in a

ON TRUE COMPOUND INTEREST 135

stocking, or locking it up in his safe. Then, if he
goes on for 10 years, by the end of that time he will
have received 10 increments of £10 each, or £100,
making, with the original £100, a total of £200 in all.
His property will have doubled itself in 10 years.
If the rate of interest had been 5 per cent, he would
have had to hoard for 20 years to double his property.
If it had been only 2 per cont, he would have had
to hoard for 50 years. It is easy to see that if the

value of the yearly interest is z of the capital, he

must go on hoarding for n years in order to double
his property.
Or, if y be the original capital, and the yearly

interest is 2 then, at the end of m years, his property
will be yen ny.

(2) At compound interest, As before, let the owner
begin with a capital of £100, earning interest at the
rate of 10 per cent. per annum; but, instead of
hoarding the interest, let it be added to the capital
each year, so that the capital grows year by year,
Then, at the end of one year, the capital will have
grown to £110; and in the second year (still at 10%)
this will earn £11 interest. He will start the third
year with £121, and the interest on that will-be
£12. 28,5 so that he starts the fourth year with
£183. 2s,, and so on. It is easy to work it out, and
find that at the end of the ten years the total capital

136. CALCULUS MADE EASY

will have grown to £259, 73. 6d. In fact, we see that
at the end of each year, each pound will have earned
25 of a pound, and therefore, if this is always added
on, each year multiplies the capital by 43; and it
continued for ten years (which will multiply by this
factor ten times over) the original capital will be
multiplied by 250375. Let us put this into symbols,

Put yy for the originel capital; À for the fraction

added on at each of the operations ; and yn for the
value.of the capital at the end of the 1% operation,
Then n=0w(1+).
But this modo of reckoning compound interest once
a year, is really not quite fair; for even during the
first year the £100 ought to have been growing, Ab
the end of half a year it ought to have been at least
£105, and it certainly would have been fairer had
tho interest for the second half of the year been
calculated on £105, This would be equivalent to
calling it 3% per half-year; with 20 operations, there-
fore, at each of which the capital is multiplied by 34.
If reckoned this way, by the end of ten years the
capital would have grown to £265, 89,; for
(lt gy)? = 2654
But, even so, the process is still not quite fair; for,
by tho end of the first month, there will be some
interest carned ; and a half-yearly reckoning assumes
that the capital remains stationary for six months at

ON TRUE COMPOUND INTEREST 137

a time. Suppose we divided the year into 10 parts,
and reckon a one-per-cent, interest for each tenth of
the year, We now have 100 operations lasting over
the ten years; or

Yu= £100 (14,0);
which works out to £270. 8s,

Even this is ot final. Let the ten years be divided
into 1000 periods, each of rÿ of a year; the interest
being py per cent. for each such period; then

Yn= £100 (ro):
which works out to £271. 14s, 24d.

Go even more minutely, and divide the ten years
into 10,000 parts, each yyy of a year, with interest
at dy of 1 per cent. Then

Ya = L100 (14000;

T9008
which amounts to £271. 16s, 4d.
Finally, it will be seen that what we are trying to

find is in reality the ultimate value of the expression

(142), which, as we see, is greater than 2; and

which, as we take m larger and larger, grows closer
and closer to a particular limiting value. However
big you make m, the value of this expression grows
nearer and nearer to the figure

271828 ...

a number never to de forgotten.
Let us take geometrical illustrations of these things.
In Fig. 36, OP stands for the original value, OF is

138 CALCULUS MADE EASY

the whole time during which the value is growing,
It is divided into 10 periods, in each of which there is
an equal stop up. Here 4Y is a constant; and if each
step up is 24 of the original OP, then, by 10 such
steps, the height isdoubled, If we had taken 20 steps,

each of half the height shown, at the end the height
would still be just doubled. Or m such steps, each

ot à of the original height OP, would suffice to

double the height. This is the case of simple interost,
Here is 1 growing till it becomes 2.

In Fig. 87, we have the corresponding illustration of
the geometrical progression. Each of the successive
n+1

rn
its predecessor. The steps up are not equal, because

each step up is now À of the ordinate af that part of

the curve. If we had literally 10 steps, with (1+)
for the multiplying factor, the final total would be

ordinates is to be 14, that is, ET timos as high as

ON TRUE COMPOUND INTEREST 139

(+) or 2593 times the original 1. But if only
we take sufficiently large (and the corresponding

Loa 1»
à sufficiently small), then the final value (142) to
which unity will grow will be 271828,

Epsilon. To this mysterious number 2-7182818
ete., the mathematicians have assigned as a symbol
the Greek letter e (pronounced epsilon) or the English
letter e. AJl schoolboys know that the Greek letter m
(called pi) stands for 3-141592 ete.; but how many of
them know that epsilon means 271828? Yetitisan
even more important number than m!

What, then, is epsilon?

Suppose we were to let 1 grow at simple interest
till it became 2; then, if at the same nominal rate of
interest, and for the same time, we were to let 1 grow
at true compound interest, instead of simple, it would
grow to the value epsilon.

This process of growing proportionately, at every
instant, to the magnitude at that instant, some people

149 CALCULUS MADE EASY

call a logarithmic rate of growing. Unit logarithmie
rate of growth is that rate which in unit time will
cause 1 to grow to 2718281. Ib might also be
called the organic rate of growing: because it is
characteristic of organic growth (in certain eircum-
stances) that the increment of the organism in a
given time is proportional to the magnitude of the
organism itself,

IE we take 100 per cent, as the unit of rate,
and any fixed period as the unit of time, then the
result of letting 1 grow arithmetically at unit rate,
for unit time, will be 2, while the result of letting 1
grow logarithmically at unit rate, for the same time,
will be 271828... ,

A little more about Epsilon. We have seen that
we require to know what value is reached by the
expression (1+1)", when m becomes indefinitely
great. Arithmetically, here ate tabulated a lot of
values (which anybody can calculate out by the help

of an ordinary table of logarithms) got by assuming
n=2; n=5; n= 10; and so on, up to n= 10,000,

d+ =225.
(Y =2:480.
o ;
A+)

At

Uta
(yp hag) = 27182

THE LAW OF ORGANIC GROWTH 141

It is, however, worth while to find another way of
ealeulating this immensely important figure.

Accordingly, we will avail ourselves of the binomial
theorem, and expand the expression (142) in that
well-known way. ”

The binomial theorem gives the rule that

1 we gab
®

A Cored pee ee

1 (n=1)(n—2)(n—8)
ta = +ete.

Putting a=1 and b=

Now, if we suppose to become indefinitely great,
say a billion, or a billion billions, then 1-1, 1-2,
and n—8, ete, will all be sensibly equal to m; and
then the series becomes

ent pt tele

By taking this rapidly convergent series to as
many terms as we please, we ean work out the sum to
any desired point of accuracy, Here is the working
for ten terms:

142 CALCULUS MADE EASY

1:000000
dividing hy 1 1000000
dividing by 2 0:500000
dividing by 3 0160667
dividing by 4 0041667
dividing by 5 — 0008833
dividing by 6 0001389
dividing by 7 0000198
dividing by 8 0.000025
dividing by 9 0000002
Total 2'718281
€ is incommensurable with 1, and resembles + in
being on interminable non-recurrent decimal,
The Exponential Series, We shall have need of yet
another series,
Let us, again making use of the binomial theorem,
expand the expression (1+4)", which is the same

as e when we make m indefinitely great.

ma a( Y ar (DY
E at noes @

eal" +nw

+no(no—1)no—2) «roto.

1 nenn
> + A

—+ ete

1 nana + 2n0
+

THE LAW OF ORGANIC GROWTH 143

But, when m is made indefinitely great, this
simplifies down to the following:

€=14+0+ oo.

Etat i

This series is called the exponential serie

‘The great reason why ¢ is regarded of importance
is that e* possesses a property, not possessed by any
other function of æ, that when you differentiate it
tts value remains unchanged; or, in other words, its
differential coefficient is the same as itself. This can
be instantly seen hy differentiating it with respect
to a, thus:

REIF TER
which is exactly the same as the original series.

Now we might have gone to work the other way,
and said: Go to; let us find a function of =, such
that its differential coefficient is the same as itself.
Ox, is there any expression, involving only powers

144 CALCULUS MADE EASY
of a, which is unchanged by differentiation? Accord»
ingly, let us asswme as a general expression that

y= At Bet Cxt+Dxi+Eat+ete,
(in which the coefficients A, B, O, ete. will have to be
determined), and differentiate it.

di
de

-B4+2Cx43D2*4+4Ex*+eto,

Now, if this new expression is really to be the same
as that from which it was derived, it is clear that
c A

B_A

A must =B; that 0-5 =775; that D=

that B= P= E

A
Tage e
The law of change is therefore that
ree).

If, now, we take A=1 for the sako of further
simplicity, we have

2
y

Differentiating it any number of times will give
always the same series over again.

If, now, we take the particular case of A=1, and
evaluate the series, we shall get simply
2718281 ete; thatis, y
18281 ete}; thatis, yee;
2718281 ete): thabis, y=e;

when 9=1, y

when æ

THE LAW OF ORGANIC GROWTH 145

and therefore

when a=a, y=(2718281 ete); that is, y=e",
thus finally demonstrating that

wat æ a4
eltitistrasttegate

[Nore.—Howtoreud exponentials, For the benefit
of those who have no tutor at hand it may be of use
to state that e is read as “epsilon to the eksth power ;”
or some people read it “exponential els.” So er is
read “epsilon to the pee-teeth-power” or “ exponential
pee tee” Take some similar expressions :—Thus, e~? is
read “epsilon to the minus tv power” or “exponential
minus two.” e- is read “epsilon to the minus
ay-eksth” or “ exponential minus ay-eks."]

Of course it follows that e’ remains unchanged if
differentiated with respect to y. Also e”, which is
equal to (e*)*, will, when differentiated with respect
to æ, be ae“, because @ is a constant,

Natural or Napieriam Logarithms.

Another reason why e is important is because it
was made by Napier, the inventor of logarithms, the
basis of his system. If y is the value of ¢% then a
is the logarithm, to the base e, of y. Or, if

y=e,
then w=log.y.
The two curves plotted in Figs. 38 and 39 represent

these equations.
CALE x

146 CALCULUS MADE EASY
‘The points calculated are:

vera as (| 212128 | [os [2 |
y | 1 2:65 [971 [so] rco

For Fis, 39 2 1 2 3 + be
x | o jos9|1:10|1:39/2:08

Y Y
pa 8
7 7.
6 6
5 5
4 4
3 3
2 2
2 1
O o!

1
}
H
H
t
H
H
2

Fis, 89.

It will be seen that, though the calculations yield
different points for plotting, yet the result is identical,
‘The two equations really mean the same thing,

As many persons who use ordinary logarithms,
which are calculated to base 10 instead of base e, are
unfamiliar with the “natural ” logarithms, it may be
worth while to say a word about them, ‘The ordinary
rule that adding logarithms gives the logarithm of
the product still holds good ; or

log. a+ log. b=log, ab.
Also the rule of powers holds good ;
nx log. a=10g, a”,

THE LAW OF ORGANIC GROWTH 147

But as 10 is no longer the basis, one cannot multiply
by 100 or 1000 by merely adding 2 or 3 to the
index. One can change the natural logarithm to
the ordinary logarithm simply by multiplying it by
04343; or log,æ=04348x log, a,
and conversely, log, x = 2/3026 x log,» x.

A Userun Tanto or “ NAPIERIAN LOGARITHMS ”
(Also called Natural Logarithms or Hyperbolic Logarithms),

Number Lor, Sonder
1 | 00000 6
0.0963 7
01823 8
15 | 04055 9
17 | 05306 10
20 | 06031 20
22 | 07885 50
25 | 09163 100
27 | 09933 200
28 | 10206 500
30 | 10988 1,000
35 |1:2528 2,000
40 | 19803 5.000
45 | 15041 10,000
50 | 16098 20,000

Exponential and Logarithmic Equations.
Now let us try our hands at differentiating certain
expressions that contain logarithms or exponentials.
Take the equation:
y=log, w.
First transform this into
van,

145 CALCULUS MADE EASY

whence, since the differential of el witn regard to y is
the original function vase (see p. 143)

dea,

% nae
and, reverting from the inverse to the original funo-
tion,

Now this is a very curious result. It may be
written dog, a)
dn

Note that a? is a result that we could never have
got by the rule for differentiating powers. That rule
(page 25) is to multiply by the power, and reduce the
power by L Thus, differentiating af gave us Ba*;
and differentiating a® gave 2a. But differentiating
a gives us 0X3 , because 2° is itself =1, and
is a constant. We shall have to come back to this
curious fact that differentiating log. gives us

À when we reach the chapter on integrating.
Now, try to differentiate
y=log.(a+a),
that is Sota;
.. since the differential of es

THE LAW OF ORGANIC GROWTH 149

This gives yrezata;

hence, reverting to the original function (see p. 181),
we get

Next try Y= logy Y.

First change to natural logarithms by multiplying
by the modulus 04343. ‘This gives us
y= 04343 log.n;
dy _ 0
de

whence

The next thing is not quite so simple. Try thi
y=e.
‘Taking the logarithm of both sides, we get
log.y =a log.a,

or „ey_ 1
log.a logía

xlogey.
Since ;-1 is & constant, we got
log.a
dæ_ 1 1 oa
dy Toga y &xlopa’
hence, reverting to u original function,

d;
Lu

de

=a" x log.a,

150 CALCULUS MADE EASY
We see that, since

de dy 1 ong Moy 1, OL,

dy de dy y Toga y xi

We shall find that whenever we have an expression
such as log.y=a function of æ, we always have

=log.a.

à de he differential coefficient of the function cf a,

so that we could have written at once, from
log.y=2 log. a,
1 dy

FA iog.a end a a log,a.

Let us now attempt further examples,

Examples.

(1) y=er™ Let —aa=e; then y=e.
Wie, @__ U _ _ grs,
E E a; hence GY = —ae-*,

Or thus:

.1dy
log.y= — ax; y dz

e a
@ yao. Let Faz; then ya.

THE LAW OF ORGANIC GROWTH 151

@ yao

_ 22 1dy_%+1)=20,
BEN NC

dy 2 5

hence E Eger
eco! 20)

Check by writing „==

4) y= FF log.y= (+0)
æ dy ox er
art) da (a)
(Por if (2? +a)H=u and a’+a=v, w=04,

du_ 1 do, du_ o )

de Ww? de ‘de (ta).
Check by writing Vatta=z.
(5) y=log{a+a) Let (a42*)=2; then y=l0g.%

dy 1. delos dy_ 32
daz) a hone Goa

(6) y=log.{380°+Jata%}. Let Bett /ateaaj
then y=log.z.
1, de 2,
de de Jeeta

æ

ay TERA ABE.
de Bt Nate (+ aa

152 CALCULUS MADE EASY

MD y-(@+ 3/22
ro Les
1dy_ 2
= arar aa

(yla a Aaa

€) hear
log.y=3log. (a? +3) -+§ log. 2);
1dy_¿ 2e ,2 3% __ 0m , et
yde PESTE B28 BD

(For if w=log.(a*+8), lot a?+3=5 and u=loge

du_1, de_y,. du_ 20

de de O A

Et dv _ Sa

Similarly, if v=log,(2°—2), 2 and

a 2 2

Opa

verein
dl 20 180 gt

dy_S#+a
«a À ya =

THE LAW OF ORGANIC GROWTH 153

1
(10) a

1
ay log, o-1x, 1
da Lg ul

(1D) y=Togz=(logæ). Let z=log.x; y=2%,
Wil, de 1. dy_ 1
de 3 * de ©’ de 30 logéz

Le
on (3)
log.y = — aa log,a*= —aa?, log.a.

= 2aw. iog a

OR

‘Try now the following exercises.

2eca!- Joga.

Exercises XII, (See page 294 for Answers.)
(1) Lifferentiate y= b(e"—e-*),
(2s Find the differential coefficient with respect ta
t of the expression’ u=al+2log.t.

(8) If y=nt, find og)

SS
Al
x

(4) Show that Bun ea?

du,
de

(5) E w=po", find

154 CALCULUS MADE EASY

Differentiate .
log. a”. (D y=30 rs.
(8) y=(@a?-+1)e-™, (9) y=log.(w*+a).
(10) y=(3a*—D(/34D»

(12) y=a* x24

(13) It was shown hy Lord Kelvin that the speed of
signalling through a submarine cable depends on the
value of the ratio of the external diameter of the core
to the diameter of the enclosed copper wire. If this
ratio is called y, then the number of sigrials s that can
be sent per minute can be expressed by the formula
i,

s=ay' log,
where @ is a constant depending on the length and
the quality of the materials. Show that if these are

given, 8 will be a maximum if y=1+4/6.
(14) Find the maximum or minimum of
y=a—log.a.

(15) Differentiate y =l0g.(aze”).
(16) Differentiate y=(log.az),

Tho Logarithmic Curve.

Let us return to the curve which has its successive
ordinates in geometrical progression, such as that
represented by the equation y=bp*.

We can see, by putting æ=0, that b is the initial
height of y.

‘Then when

weal, y=bp; 2=2 y=bp?; 2-3, y=bp ete.

THE LAW OF ORGANIC GROWTH 155

‘Also, we see that p is the numerical value of the
ratio between the height of any ordinate and that of
the next preceding it. In Fig. 40, we have taken p
as $; each ordinate being $ as high as the preceding

one,

Fw. 40. Pro. 41.
If two successive ordinates are related together
thus in a constant ratio, their logarithms will have a
constant difference; so that, if we should plot out
a new curvo, Fig. 41, with values of log, y as ordinates,
it would be a straight line sloping up by equal steps.
In fact, it follows from the equation, that
log. y=log.b+2+log.p,
whenco logy —log.b =a log.p.
Now, since log,p is a mere number, and may be
written as log.p=a, it follows that
hog. =ar,
and the equation takes the new form
yaber.

156 CALCULUS MADE EASY

The Die-away Curve.

If we were to take p as a proper fraction (less than,

unity), the curve would obviously tend to sink down-

wards, as in Fig. 42, where each successive ordinate
is $ of the height of the preceding one, .

‘The equation is still
yal:
Y

y

F

© sex

ia. à

but since p is less than one, log.p will be a negative

quantity, and may be written —a; so that p=e"s,

and now our equation for the curve takes the form
yobe™,

The importance of this expression is that, in the
case where the independent variable is time, the
equation represents the course of @ grest many
physical processos in which something is gradually
dying away. Thus, the cooling of a hot body is
represented (in Newton’s celebrated “law of cooling”)
by the equation @,=@ue-";

THE LAW OF ORGANIC GROWTIL 157

where 6, is the original excess of temperature of a
hot body over that of its sunoundings, @, the excess
of temperature at-the end of time ¢, and a is a con-
stant—namely, the constant of decrement, depending
on the amount of surface exposed by the body, and
on its coefficients of conductivity and emissivity,
ete,

A similar formula,

Q= Qe,

is used to express the charge of an electrified body,
originally having a charge Q,, which is leaking away
with a constant of decrement a; which constant
depends in this case on the capacity of the body and
on the resistance of the leakage-path.

Oscillations given to a flexible spring die out after
a time; and the dying-out of the amplitude of the
motion may be expressed in a similar way.

In fact e-* sorves as a die-away factor for all
those phenomena in which the rate of decrease
is proportional to the magnitude of that which is

decreasing; or where, in our usual symbols, u is

proportional at every moment to the value that y has

at that moment. For we have only to inspect the
curve, Fig. 42 above, to see that, at every part of it,

the slope Fa is proportional to the height y; the
curve becoming flatter as y grows smaller. In sym-

bols, thus ne

158 CALCULUS MADE EASY

or log.y=log.b—awlog.e=l0g.b—az,
’ .. lé .

and, diorentiating, | = a;

hence be x(—a)= 09;

or, in words, the slope of the curve is downward, and
proportional to y and to the constant a.

We should have got the same result if we had
taken the equation in the form

= bps
d
for then = px log.p.
But log. p=—a;
ivi OY AER
giving us Ia) = ay,

as before.

The Time-constant. In the expression for the “ die-
away factor” e-*, the quantity a is the reciprocal of
another quantity known as “ the time-constant,” which
we may denote by the symbol 7. Then the die-away

factor will be written eE; and it will be seen, by
making £=7 that the meaning of 7 (or of 2) is that
this is the length of time which it takes for the original
quantity (called 6, or Q, in the preceding instances)
to die away to 1th part—thut is to 0:3678—of its
original value.

THE LAW OF ORGANIC GROWTH 158

The values of e* and e-* are continually required
in different branches of physics, and as they are given
in very fow sets of mathematical tables, some of tho
values are tabulated here for convenience.

æ € e We
000 1.0000 10000 00000
010 1.1052 09048 00952
020 12214 08187 01813
050 16487 0:8065 03935
075 21170 09724 05276
090 2:4596 01068 5
100 27188 03679 06321
110 30042 03329 06671
120 3:3201 03012 06988
125 34903 02865 07135
150 44817 02231 07769
175 5754 01738 08962
200 7:389 01353 08647
250 12183 00821 09179
300 20-085 00498 09502
3:50 33115 00302 0.9698
400 54598 00183 09817
450 90017 on 09889
500 14841 00067 09933
5:50 244-69 00041 09939
600 40343 000248 099752
750 1808-04 0-00053 099947

[re 220265 0000045 0999955

As an example of the use of this table, suppose
there is a hot body cooling, and that at the beginning

160 CALCULUS MADE EASY

of the experiment (ü.e. when £=0) it is 72° hotter than
the surrounding objects, and if the time-constant of its
cooling is 20 minutes (that is, if it takes 20 minutes

for ite excess of temperature to fall to 4 part of 72°)

then we can calculate to what it will have fallen in
any sven time & For instance, let £ be 60 minutes.

Then poe 20=3, and we shall have to find the

value de e-, and then multiply the original 72° by
this, The table shows that ¢~* is 00498. So that
at the end of 60 minutes the excess of temperature
will have fallen to 72° x 00498 — 3"

588%.

Further Examples.

(1) The strength of an electric current in a con-
ductor at a time £ secs. after the application of the
electromotive force producing it is given by the ex-

=
pression e=z{1 3).

The time constant is D,

If B=10,R=1, L=001; then when ¢is very large

the term 1—e" becomes 1, and C= = 10; also

L 9
Re T=00.
Its value at any time may be written:
0=10-106°8,

THE LAW OF ORGANIC GROWTH 161

the time-constant being 001. This means that it
takes 001 sec. for the variable term to fall to

0
203678 of its initial value 106700 =10.

To find the value of the current when £=0'001 sec,
say, Z=01, €91=0:9048 (from table).

It follows that, after 0001 see, the variable term
is 09048x10=9048, and the actual current is
10—9-048=0:952.

Similarly, sé the end of O1 sec,

€ _10: 10 = 0 .
77105 e-10=0000045 ;

the variable term is 10 x 0000045
being 99995.

(2) The intensity 7 of u beam of light which has
passed through a thickness J cın. of some transparent
medium is =J,e~™, where 1, is the initial intensity
of the beam and Æ is a “constant of aksorption.”

This constant is usually found by experiments If
it be found, for instance, that a beam of light has
its intensity diminished by 18% in passing through
10 ems, of a certain transparent medium, this meuns
that 82=100xe"%%0 or e“W%=082, and from the
table one sees that 10K=020 very nearly; hence
K=002.

To find the thickness that will reduce the intensity
to half its value, one must find the value of Z which
satisfies the equality 50=100xe-°, or 056-0

ome L

"00045, the current

162 CALCULUS MADE EASY

It is found by putting this equation in its logarithmic
form, namely,
log 05 = — 002 x 2x log e,
which gives _
¿10990
002 x 04348

(8) The quantity Q of a radio-active substance
which has not yet undergone transformation is known
to be related to the initial quantity Q, of the sub-
stance by the relation Q= @Q,e-, where À is a constant
and ¢ the time in seconds elapsed since the trans-
formation began.

For “Radium A,” if time is expressed in seconds,
experiment shows that A=385 x 10-3, Find the time
required for transforming half the substance. (This
time is called the “mean life” of the substance.)

We have OS evo,

log 0°5 = — 000385 x log e;
minutes very nearly.

=345 centimetres nearly.

and

Exercises XIII. (See page 294 for Answers.)

(1) Draw the curve y=be"?; where b=12, T=8,
and ¢ is given various values from 0 to 20.

(2) If a hot body cools so that in 24 minutes its
excess of temperature has falien to half the initial
amount, deduco the time-constant, and find how Jong
it will be in cooling down to 1 per cent. of the original
excess,

THE LAW OF ORGANIC GROWTH 163

€) Plot the curve y=100(1—e-*).
(4) The following equations give very similar curves:
oy;
10?
Gi) yale");
a 2
Gi) y= A aretan (à

Draw all three curves, taking a=100 millimetres;
b=30 millimetres.

(5) Find the differential coefficient of y with respect
tom (a) year year Our

(6) Kor “Thorium A,” the value of A is 5; find the
“mean life,” that is, the time taken by the trans-
formation of a quantity Q of “Thorium A” equal to
half the initial quantity Q, in the expression

we Q= Que;
t being in seconds,

(7) A condenser of capacity K=4x10-9, charged
to a potential V,=20, is discharging through a resist-
ance of 10,000 ohms. Find the potential P after (a) 0
second; (b) 0-01 second; assuming that the fall of
potential follows the rule V= Vie”.

(8) The charge @ of an electrified insulated metal
sphere is reduced from 20 to 16 units in 10 minutes,
Find the coefficient x of leakage, if Q-Q,xe"; €,
being the initial charge and £ being in seconds, Hence
find the time taken by half the charge to leak away.

164 CALCULUS MADE EASY

(9) The damping on a telephone line can be ascer+
tained from the relation i=éje-*, where à is the
strength, after £ seconds, of a telephonic current of
initial strength à; 2 is the length of the line in kilo-
metres, and 8 is a constant... For the Franco-English
submarine cable laid in 1910, 8=00114. Find the
damping at the end of the cable (40 kilometres), and
the length slong which é is still 8% of the original
current (limiting value of very good audition).

(10) The pressure p of the atmosphere at an altitude
h kilometres is given by p=pye-™; p, being the
pressure at sea-level (760 millimetres).

‘The pressures at 10,20 and 50 kilometres being
1992, 422, 032 millimetres respectively, find Æ in
each case. Using the mean value of £, find the per-
centage error in each case.

(11) Find the minimum or maximum of y=0%,
1
(12) Find the minimum or maximum of y=a#.

1
(13) Find the minimum or maximmn of y-xa#.

CHAPTER XV.
HOW TO DEAL WITH SINES AND COSINES,

Greux letters being usual to denote angles, we will
take as the usual letter for any variable anglo the
letter 0 (* theta”).
Let us consider the function
y=sin0.

21

AY

Fis. 43,

What we have to investigateis the value of 888);

or, in other words, if the angle @ varies, we have to
find the relation between the increment of the sine
and the increment of the angle, both increments being
indefinitely small in themselves, Examine Fig. 43,
wherein, if the radius of the circle is unity, the height
of y is the sine, and 6 is the angle. Now, if D is

166 CALCULUS MADE EASY

supposed to increase by the addition to it of the
small angle d@—an element of angle—the height
of y, the sine, will be inereased by a small element dy.
The new height y+dy will be the sine of the new
angle 6-+d8, or, stating it as an equation,
y+dy=sin(0+d0);
and subtracting from this the first equation gives
dy=sin (0+d0)—sin 8.

‘The quantity on the right-hand side is the difference
between two sines, and books on trigonometry tell
us how to work this out. For they tell us that if
M and N are two different angles,
peed vain EN,

If, then, we pub ee for one angle, and
N=6 for the other, we may write

9+40+0, sin 2+49—0
in

sin M—sin N=2 cos —

dy= 2008 0+

or, dy=2 nn sin Jd.

But if we regard dé as indefinitely small, then in
the limit we may neglect 146 by comparison with 6,
and may also take sin ¿d9 as being the same as 348.
The equation then becomos:

dy=2e0s 0 x 3d0;
dy=c0s 0: d8,

and, finally, % =c08 8.

SINES AND COSINES 167

The accompanying curves, Figs. 44 and 45, show,

plotted to scale, the values of y=sin 8, and ÍY
for the corresponding values of 6.

cos 8,

Fo, 46,

163 CALCULUS MADE EASY

Take next the cosine.
Let y=c0s0,

Now cos @=sin (5-9).
Therefore
dy=a(sin(Z—6)) =c0s (5-8) xa(—6),

=cos 5-0) x(-d0),

dy_ m
ar G z }
And it follows that

—sin 6.

Lastly, take the tangent.
Let y=tand
_sing
cos 6
The differential cocftciont of sing is 46), ang

the differential coefficient of cos is eme),

Apply-
ing the rule given on page 40 for differentiating a
quotient of two functions, we got

SINES AND COSINES 768

0886+ sin’
"cost
at.
* coro"
FY _ soc
a geo,
Collecting these results, we haves
o | &
sin | cos0
cos | si. 8
tand | el

Sometimes, in mechanical and physical questions,
us, for example, in simple harmonic motion and in
wave-motions, we have to deal with angles that in-
crease in proportion to the time, Thus, if 7 be the
time of one complete period, or movement round the
circle, then, since the angle all round the circle is 27
radians, or 860°, the amount of angle moved through
in time £, will be

O= 20, in radians,

170 CALCULUS MADE EASY
If the frequency, or number of periods per second,

be denoted by m, then n=>,, and we may then write:

7
0=2mnt.
Then we shall have
y=sin ant.

If, now, we wish to know how the sine varies with
respect to time, we must differentiate with respect, not
to £, but to & For this we must resort to the artifice
explained in Chapter IX. p. 67, and put

Y. dy de

ar

Now = will “hue be 21; so that

D eos 0x 2a

= 2rn cos Irak.
Similarly, it follows that
eos 2ant) _ _ oan sin dan.
de “

Second Differential Coeficient of Sine or Cosine.

We have seen that when sin 8 is differentiated with
respect to 6 it becomes cos 9; and that when cos 6 is
differentiated with respect to O it becomes —sin 6;
or, in symbols,

d'in 8) _

= sin 8.

SINES AND COSINES m

So we have this curious result that we have found
a function such that if we difforentiate it twice over,
we get the same thing from which we started, but
with the sign changed from + to —.

The same thing is true for the cosine; for differen-
tinting cos 0 gives us —sinO, and differentiating
—sin 6 gives us —cos 0; or thus:

pa —cos 6.

Sines und cosines are the only functions of which
the seoond differential coefficient is equal and of
opposite sign to the original function,

Examples.

With what we have so far learned we can now
differentiate expressions of a more complex nature.
(1) y=aresin a
If y is the are whose sine is &, then @=siny.
de
Tyne

Passing now from the inverse function to the original
one, we get

7]

Now cosy=n/1—sinty=/1—e 3 +
h dy_ 1

ence

de Ji

a rather unexpected result.

172 CALCULUS MADE EASY

(2) y=cos*0.
This is the same thing as y =(cos 0).
Let cos 6=03 then y="; nn.

= sind,

—3cos’@sin 8.

Let z+a=v; then y=sine.
WU =esso; Bat and
(4) y=log,sin 0
Let sin 0=0; y=log.v.
dy de
duo’ de
dy __1 =
d0 mag * 05 0 = cot 6,

= cos 8;

cos 6
©) y=cot 0

a
= -(1+cot?0)= — ease
(6), y=tan 30.
La 30=v; y=tane; YY _ seedy,
mere “do
de EU 3000
26 der 30.

SINES AND COSINES 173
CD y=V 143 tans; y

=(1+3tan?0jh

Let 3taud=o.
= ‘A
va ue TA CP 68);
de_ 2
756 tan 0 sec 0
(or, if tan =,
v=3u; 6u;

due;
hence 6 tan 9 sec*0);

2
hanes 6 tan O sec?0

(8) y==sin x08 as

ee
de sin «(—sin 2) +e08 2 x cos 2

Beorcises XIV. (See page 295 for Answers)
(1) Differentiate the following:
. alo”
6) y=Asin (0-3).
Gi) y=sin?0; and y=sin 20.
(ii) y=sinS0; and y=sin 30.

(2) Find the value of 4 for which sing xcos6 is a
maximum.

(8) Differentiate y + cos 2rnt.

174 CALCULUS MADE EASY

(4) IE y=sin as, find tu.

(5) Differentiate y =log.cos a.
(6) Differentiate y=182 in (a+ 26°).
(7) Plot the curve y=100 sin (9—15"); and show

that the slope of the curve at 0=75” is half the
maximum slope.

eos ana ae
(8) If y=sin 0-sin 26, find 9%.
(9) If y=a-tan”(9%), find the differential coefficient
of y with respect to 0.
(10) Differentiate y
(11) Differentiate the three equations of Exercises
XUL (p. 163), No. 4, and compare their differential
coefficients, as to whether they are equal, or nearly
equal, for very small values of a, or for very large
values of «, or for values of & in the neighbourhood
of 2=30.
(12) Differentiate the following :
a)
Gi)
Gi)
(iv) y=aresec a.
©) y=tanax/Tseew.
(13) Differentiate y=sin (26+3)"*.
(14) Differentiate y =6"+3 sin (043) 3110089,
(5) Find the maximum or minimum of y—6 cos 8.

*sinta.

CHAPTER XVI.
PARTIAL DIFFERENTIATION.

We sometimes come across quantities that are fune-
tions of more than one independent variable. ‘Thus,
we may find a ease where y depends on two other
variable quantities, one of which we will call # and
the other ». Tn symbols

vu)
Take the simplest concrete case.
Let y=ux»,

What are we to do? If we were to treat v as a

constant, and differentiate with respect to u, we

should get dy.=vdu;

or if we treat u as a constant, and differentiate with

respect to ©, we should hav
dy,=udv.

The little letters here put as subscripts are to show
which quantity has been taken as constant in the
operation,

Another way of indicating that the differentiation
has been performed only partially, that is, has been
performed only with respect to one of the independent

176 CALCULUS MADE EASY

variables, is to write the differential coefficients with
Greek deltas, like 2, instead of little d. In this way

2y

Ou

2

ow

If we put in these values for » and w respectively,
we shall have

dy,

dy.= u de,

v

Us

ey
By

which are partial differentials,

But, if you think of it, you will observe that the
total variation of y depends on both these things at
the same time, That is to say, if both are varying,
the real dy ought to be written

Y dn Y dy;
dy= 3 dur do;
and this is called a total differential. In some books

ah te oot _ (dy dy
it is written dy= (9%) du+ (32) do.
Eaample (1), Find the partial differential co-
efficients of the expression «= 2aa®-+Sbay-+4ey’,
The answers are:
ED
bgp + By.

ew
5. =3b04+12ey?
ay + 12ey!

PARTIAL DIFFERENTIATION 177

The first is obtained by supposing y constant, the
second is obtained by supposing æ constant; ther
dw = (a+ Bby)dee-+ (Bbae + 12ey?)dy.
Beample (2). Let e=a%. Then, treating first y
and then a as constant, we get in the usual way

vr, |

so that de ya" "td a log.a dy.

Example (3). A cone having height % and radins
of base r, has volume V=3r7°h. I its height remains
constant, while r changes, the ratio of change of
volume, with respect to radius, is different from ratio
of change of volume with respect to height which
would occur if the height were varied and the radius
kept constant, for

E
|

dr
a 3
The variation when both the radius and the height
Qn as

rh dr + rt.

Example (4). In the following example F and f
denote two arbitrary functions of any form whatso-
ever. For example, they may be sine-functions, or

exponentials, or mere algebraic functions of the two
oun. m

change is given by dV=

178 CALCULUS MADE EASY
independent variables, ¢ and æ This being under
stood, let us take the expression
y= F(a+at)+f(w—at),

cr, y=F(m+ Ao):
where w=a+at, and v=2-at.
dy _ 2F(r), din | lv) de
daw de! d de

= Fwy. LE feoy-1
(where the figure 1 is simply the coefficient of æ in
wand n);

and WEL).

Then

#6 dé dw dit dé

=F(w)-a—f (a;

and —P(w)a+f (0);
&y_ by,
whence de Ce

This differential equation is of immense importance
in mathematical physics. (See also page 247.)

Maxima and Minima of Functions of two
Independent Variables.
Example (5). Let us take up again Exercise IX,
p.110, No. 4
Let x and y be the length of two of the portions of
the string. The third is 30-~(w+-y), and the area of the

PARTIAL DIFFERENTIATION 179

triangle is A=4/s(s—28—y}(s—30+%+7) where
8 is the half perimeter, 15, so that A =w/J5P, where
P=(15—2)(13—y)(a+y—15)
y ay — 150° — 159 — Asay + 450% +450y~ 3375.
Clearly A is maximum when P is maximum.
ap E
7
For a maximum (clearly it will not be a minimum in
this case), ono must have ed
Lo and $” 5
that is, 2xy—30x%+y*—45y +450:
acy —30y + a —450+450
An immediate solution is a=y.
IE we now introduce this condition in the value
of P, we find
P=(15- 222 15)= 248 — 7522+ 9002 — 8375,
For maximum or minimum, D Ga? -1502 +000 = 0,
which gives &=15 or «=10,
Clearly æ=15 gives zero area; æ=10 gives the

oP
dæe+— dy
ety Ue

à EP à Sn à
maximum, for = 12x—150, which is +30 for

æ=15 and —30 for »=10.

Example (6). Find the dimensions of an ordinary
railway coal truck with rectangular ends, so that,
for a given volume V the area of sides and floor
together is as small as possible.

180 CALCULUS MADE EASY

The truck is a rectangular box open at the top.
Let w be the length and y be the width; then tho

LT . 24,27.
depth ig 7 The surface aren is S=0y+ +="

Say ly) de+(@-%) dp

For minimum (clearly it won't be a maximum here),

2 9 14
mer
Here also, an immediate solution is 2=y, so that
4V dS 4v nt
PAE
S=e4+ = da Le: 0 for minimum, and

æ=4#/2V.

Exercises XV. (See page 206 for Answers.)
(1) Differentiate the expression À —22%y—2yt0+Y
with respect to @ alone, and with respect to y alone.
(2) Find the partial differential coefficients with
respect to 2, y and 2, of the expression
yz ays aye pas
(8) Let (oa + yb Her

' Or, Or, Or,
Find the value of 57+ 5745; Also find the value
Sr Pr, Or
of tot

(4) Find the total differential of y=u".

PARTIAL DIFFERENTIATION 181

(5) Find the total differential of y=utsinv; of
y=(in a); and of y= LE,

(6) Verify that the sum of three quantities æ, y, 2,
whose product is a constant k, is minimum when
these three quantities are equal.

(7) Find the maximum or minitaum of the function

U= D+ ay ty.

(8) The post-office regulations state that no parcel
is to be of such a size that its length plus its girth
exceeds 6 feet, What is the greatest volume that
can be sent by post (a) in the case of a package of
rectangular cross section; (2) in the ease of a package
of circular cross section,

(9) Divide + into 3 parts such that the continued
product of their sines may be a maximun or minimum.

ta
(10) Find the maximum or minimum of Mr

(11) Find maximum and minimum of
u=y-+2a—2log.y—log.a.

(12) A telpherage bucket of given capacity has
the shape of a horizontal isosceles triangular prism
with the apex underneath, and the opposite face open.
Find its dimensions in order that the least amount
ol iron sheet may be used in its construction,

CHAPTER XVIL
INTEGRATION,

Tax great secret has already been revealed that this
mysterious symbol |, which is after all only a long $,
merely means “the'sum of,” or “the sum of all such
quantities as.” It therefore resembles that other
symbol Y (the Greek Sigma), which is also a sign
of summation. There is this difference, however, in
the practice of mathematical men as to the use of
these signs, that while 3 is generally used to indicate
the sum of a number of finite quantities, the integral
sign ji is generally used to indieate the summing up
of u vast number of small quantities of indefinitely
minute magnitude, mere elements in fact, that go
to make up the total required, Thus fay=». and
doma. :

Any one can understand how the whole of anything
can be conceived of as made up of a lot of little bits;
and the smaller the bits the more of them there will
be, Thus, a line one inch long. may be conceived as
made up of 10 pieces, each y of an inch long; or
of 100 parts, each part being yy of an inch long;

INTEGRATION 183
or of 1,000,000 parts, each of which is 77355, of an

1976008
inch long; or, pushing the thought to the limits of
conceivability, it may be regarded as made up of
an infinite number of elements each of which is
infinitesimally small,

Yes, you will say, but what is the use of thinking
of anything that way? Why not think of it straight
off, as a whole? The simple reason is that there are
a vast number of cases in which one cannot caleulate
the bigness of the thing as a whole without reckoning
up the sum of a lot of email parts. The process of
“ integrating” is to enable us to ealeulate totals that
otherwise we should be unable to estimate directly.

Let us first take one or two simple cases to
familiarize ourselves with this notion of summing
up x lot of separate parts,

Consider the series;

LEE Het dr tete,

Here each member of the series is formed by taking
it half the value of the preceding. What is the value
of the total if we could go on to an infinite number
of terms? Every schoolboy knows that the answer
ie 2, Think of it, if you like, as a line. Begin with

Ye 18
Fro 46.

one inch; add a half inch; add a quarter; add an
eighth; and so on, I at any point of the operation

184 CALCULUS MADE EASY

we stop, there will still be a piece wanting to make
up the whole 2 inches; and the piece wanting will
always be the sume size as the last pieco added,
‘Thus, if after having put together 1, $, and }, we stop,
there will be } wanting. If we go on till we have
added 2, there will still be ¿7 wanting. The
remainder needed will always be equal to the last
term added. By an infinite number of operations
only should we reach the actual 2 inches. Practically
we should reach it when we got to pieces so small
that they could not be drawn—that would be after
about 10 terms, for the eleventh term is y If we
want to go so far that not even a Whitworth’s
measuring machine would detect it, we should merely
have to go to about 20 terms, A microscope would
not show even the 18% term! So the infinite number
of operations is no such dreadful thing after all.
The integral is simply the whole lot, But, as we
shall see, there are cases in which the integral
caleulus enables us to get at the exact total that
there would be as the result of an infinite number
of operations. In such cases the integral calculus
gives us a rapid and easy way of getting at a result
that would otherwise require an interminable lot of
elaborate working out. So we had best lose no time
in learning how to integrate,

INTEGRATION 185

Slopes of Curves, and the Curves themselves,

Let us make a little preliminary enquiry about the
slopes of curves. For we have seen that differentiating
a curve means finding an expression for its slope (or
for its slopes at different points). Can we perform
the reverse process of reconstructing the whole curve
if the slope (or slopes) are prescribed for us?

Go back to caso (2) on p.84 Here we have the
simplest of curves, a sloping line with the equation

y=ax+b.

re. 47

We know that here 5 represents the initial height
dy

of y when 2=0, and that a, which is the same as

is the “slope” of the line, The line has a constant
dopo. All along dé che dlementany til 8
En

have the same proportion between height and base.
Suppose we were to take the da’s and dys of finite

186 CALCULUS MADE EASY

magnitude, so that 10 da's made up one inch, then
there would be ten little triangles like

A4d444d44444

Now, suppose that we were ordered to reconstruct
the “curve,” starting merely from the information

that Yona, What could we do? Still taking the

little d’s as of finite size, we could draw 10 of them,
all with the same slope, and then put them together,
end to end, like this:

Fie, 48,

And, as the slope is the same for all, they would join
to make, as in Fig. 48, a sloping line sloping with the

correct slope Ya, And whether we take the dy’s
and dx’s as finite or infinitely small, as they are all

INTEGRATION 181

alike, clearly g =a, if we reckon y as the total of

all the dy’s, and & as the total of all the da's. But
whereabouts are we to put this sloping line? Are
we to start at the origin O, or higher up? As the
only information we have is as to the slope, we are
without any instructions as to the particular height
above O; in fact the initial height is undetermined.
The slope will be the same, whatever the initial height.
Let us therefore make a shot at what may be wanted,
and start the sloping line at a height © above O.
‘That is, we have the equation
y=00+0.

It becomes evident now that in this case the added
constant means the particular value that y has when
x=0.

Now let us take a harder case, that of a line, the
slope of which is not constant, but turns up more and
more. Let us assume that the upward slope gets
greater and greater in proportion as = grows, In
symbols this is:

YY a
da

Or, to give a concrete case, take a=4, so that

¿Lam

‘Then we had best begin by ealeulating 2 few of
the values of the slope at différent values of x, and
also draw little dingrams of them.

188 CALCULUS MADE EASY

ans, W-10.
de

DANA 1]

Now try to put the pieces together, setting each so
that the middie of its base is the proper distance to
the right, and so that they ft together at the corners;
thus (Fig. 49). ‘he result is, of course, not a smooth

x

wie, 48.

curve: but it is an approximation to one. If we had
taken bits half as long, and twice as numerous, like
Fig, 50, we should have a better approximation. But *

INTEGRATION 189

for a perfect curve we ought to take each dx and its
corresponding dy infinitesimally small, and infinitely
numerous.

Y Pp

abe

s x

Then, how much ought the value of any y to be?
Clearly, at any point P of the curve, the value of
y will be the sum of all the little dy/s from 0 up to

that level, that is to say, [dy=y. And as each dy is

equal to 3a- de, it follows that the whole y will be
equal to the sum of all such bits as 10- do, or, as we

should write it, fre. da,
Now if & had been constant, fre-ae would have

been the same as tofda, or ¿al But æ began by

being 0, and increases to the particular value of æ at
the point P, so that its average value from 0 to that

point is }=. Hence [jede at; or y= yea",

But,as in the previous case, this requires the addition
of an undetermined constant C, because we have not

190 CALCULUS MADE £ASY

been told at what height above the origin the curve
will begin, when @=0. So we write, as the equation
of the curve drawn in Fig. 51,

ya tee +O.

Y

Exercises XVI. (See page 296 for Answers.)
(1) Find the ultimate sum of 3-+3+3 454204 ete,

(2) Show that the series 1-34+}—}4}-3+} ote,
is convergent, and find its sum to 8 terms.

at at at
(8) If log.(1+a)= 0-5 5 reto, find log, 13.

(4) Following a reasoning similar to that explained
in this chapter, find y,

e if Wa; @) it Wcose,
on Y &Y = 20-48, find y.

CHAPTER XVHL

INTEGRATING AS THE REVERSE OF
DIFFERENTIATING.
DIrrORENITATING is the process by which when y is

given us (as a function of ©), we can find a.

Like every other mathematical operation, the
process of differentiation may be reversed. ‘Thus, if
differentiating y=2* gives us Ds, then, if one

begins with a. 4, one would say that reversing the
process would yield y=a4, But here comes in a curious
point, We should get 4 if we had begun with

any of the following: æ or ata, or at+e, or at
with any added constant. So it is clear that in

working backwards from Y to y, ono must make

provision for the possibility of thore being an added
constant, the valuo of which will be undetermined

192 CALCULUS MADE EASY

until ascertained in some other way. So, if diffère
entiating a” yields na*-}, going backwards from

Dunes will give us y=a?-+O; whore C stands

for the yet undetermined possible constant.

Clearly, in dealing with powers of a, the rule for
working backwards will be: Increase the power by 1,
then divide by that increased power, and add the
undetermined constant.

So, in the case where

LE
de?

working backwards, we get

si 1 1.
yea

If differentiating the equation y=aw* gives us
U _ ange
Fenn,

it is a matter of common sense that beginning with
Y =anar-ı,
and reversing the process, will give us
y=aar,
So, when we are dealing with a multiplying constant,

we must simply put the constant as a multiplier of
the result of the integration.

HOW TO INTEGRATE 193

Thus, if Pa, the reverse process gives us
yas

But this is incomplete. For we must remember
that if we had started with

y=aat+0,
where (is any constant quantity whatever, we should
equally have found
UY ama-ı,
manana,

So, therefore, when we reverse the process we must
always remember to add on this undetermined eon-
stant, even if we do not yet know what its value
will be,

This process, the reverse of differentiating, is called
‘integrating; for it consists in finding the value of
the whole quantity y when you are given only an
expression for dy or for 4% Hlitherto we have as

much as possible kept dy and de together as a dif-
ferential coofficient: henceforth we shall more often
have to separate them,

Hf we begin with a simple case,

dr
We may write this, if we like, as
dy= cde.

Now this is a “ differential equation” which informs
us that an element of y is equal to the corresponding
element of æ multiplied by a. Now, what we want

eus. N

194 CALCULUS MADE EASY

is the integral ; therefore, write down with the proper
symbol the instructions to integrate both sides, thus:

Jay=[arae.

[Note as to reading integrals: the above would be
read thus:
“Integral dee-wy equals integral eks-squared dee-eke”]

We haven't yet integrated: we have only written
down instructions to integrate—if we can. Let us
try. Plenty of other fools can do it—why not we
also? The left-hand side is simplicity itself. The

sum of all the bits of y is the same thing as y itself,
So we may at once put:

y= [ado

But when we come to the right-hand side of the
equation we must remember that what we have got
to sum up together is not all the de's, but all such
terms as afdo; and this will not be the same as

af da, because a” is not a constant. For some of the

dés will be multiplied by big values of 2°, and some
will be multiplied by small values of 4°, according to
what x happens to be. So we must bethink ourselves
as to what we know about this process of integration
being the reverse of differentiation, Now, our rule
for this reversed process—see p. 191 ante—when
dealing with a is “inerease the power by one, and
divide by the same number as this increased power.”

HOW TO INTEGRATE 195

That is to say, ae will be changed* to $a% Pub
this into the equation; but don't forget to add the
“constant of integration” C at the end. So we get:
u=ya + C.
You have actually performed the integration. How
easy |
Let us try another simple case.

dy _
Let Ea,

where a is any constant multiplier. Well, we found
when differentiating (see p. 29) that any constant
factor in the value of y reappeared unchanged in the

value of wy In the reversed process of integrating.
it will therefore also reappear in the value of y. So
we may go to work as before, thus:

dy=aat.de,

fa = foo «de,

fey = aferas,
y=ax 40,
So that is done, How easy!

You may ask: whut has become of the little dz at the end?

Well, remeuiber that it was really part of the differential coefficient,

and when changed over to the fighthand side, as in the ade,

serves as a reminder that z is tho independent variablo with respoot

to which the operation ie to be effected; and, as the result of the

Product being totaled up, the power of = has increased hy one
ou will soon become familiar with all this.

196 CALCULUS MADE EASY

We begin to realize now that integrating is a
process of finding our way back, as compared with
differentiating, If ever, during differentiating, we
have found any particular expression—in this example
ame can find our way back to the y from which
it was derived. The contrast between the two pro-
cosses may be illustrated by the following illustration
due to a well-known teacher. IF a stranger were set
down in Trafalgar Square, and told to find his way to
Euston Station, he might find the task hopeless. But
if he had previously been personally conducted from
Euston Stution to Trafalgar Square, it would be
comparatively easy to him to find his way back to
Euston Station.

Integration of the Sum or Difference of two

Functions,
dy _
Let Grete
then dy=sde+ ade.

There is no reason why we should not integrate
each term separately: for, as may be seen on p. 35,
we found that when we differentiated the sum of two
separate functions, the differential cocflicient was
simply the sum of the two separate differentiations.
So, when we work backwards, intograting, the integra-
tion will be simply the sum of the two separate
integrations.

HOW TO INTEGRATE 197
Our instructions will then be:
fay= [er+ando
= foracs foro
y=04 q Co
If either of the terms had been a negative quantity,
the corresponding term in the integral would have
also been negative, So that differences are as readily
dealt with as sums,
How to deal with Constant Terms.
Suppose there is in the expression to be integrated
a constant term—such as this:
dy _
Gand.

This is laughably easy. For you have only to
remember that when you differentiated the expression

y=az, the result was Wag, Henco, when you work

the other way and integrate, the constant reappears
multiplied by a. So we get
dy=adr+b dx,

Jou= [rans [oae,

1
y= aia

Here are a lot of examples on which to try your
newly acquired powers.

198 CALCULUS MADE EASY
Examples.

(1) Given LY 240, Findy Ans, y=209+0.

€) Find (arm Lis (a+b)((e+1)de

or (at) frde+{ae] or (a40)(S+2) +0.
6) oe Mega. Findu Ans u=igth+é.
ou =0-2+w. Find y.

y Parada ot

dead tar: y= [edo [odes odo;

and y= tar jar O
(5) Integrate 9:752%%dw, Ans. y

All these are easy enough. Let us try another casa
Let
d

Proceeding as before, we will write
dy=ax de, fay=afesae

Well, but what is the integral of ade?
TE you look back amongst the results of differen-
tiating 2* and a and a”, ete, you will find wo never

got ="! from any one of them as the value of ar.
We got 32% from a’; we got 2 from a; we got 1
from a? (that is, from æ itself); but we did not get
2-1 from af, for two very good reasons. First, a! is
simply =1, and is a constant, and could not have

EASIEST INTEGRATIONS 199

a differential coefficient. Secondly, even if it could
bo differentiated, its differential coefficient (gob by
slavishly following the usual rule) would be 0x «74,
and that multiplication by zero gives it zero value!
Therefore when we now come to try to integrate
æ-ide, we sce that it does not come in anywhere
in the powers of æ that are given by the rule:
1
fra nes.

It is an exceptional case.

Well; but try again. Look through all the various
differentials obtained from various functions of +, and
try to find amongst them wv", A sufficient search

will show that we actually did get Y 21 as the

result of differontiating the function y=loga@ (see
p. 148).

Then, of course, since wo know that differentiating
log.æ gives us æ-1, we know that, by reversing the
process, integrating dy=a-tde will give us y =log.w.
But we must not forget the constant factor @ that
was given, nor must we omit to add the undetermined
constant of integration. ‘This then gives us as the
solution to the present problem,

y=alogæ+ 0.

N.B.—Here note this very remarkable fact, that we
could not have integrated in the above ease if we had
not happened to know the corresponding differentia-
tion. If no one had found out that differentiating
loge gave a", wo should have been utterly stuck by

200 CALCULUS MADE EASY

the problem how to integrate &-1dæ. Indeed it should
bo frankly admitted that this is one of the curious
features of the integral caleulus:—that you can’t
integrate anything before the reverse process of differ-
entiating something else has yielded that expression
which you want to integrate. No one, even to-day,
is able to find the general integral of the expression,

because a" has never yet been found to result from
differentiating anything else.

Another simple case,

Find ferne + Dan.

On looking at the function to be integrated, you
remark that it is the product of two different functions
of ©. You could, you think, integrate (-+1)da by
itself, or (@-+2)da by itself, OF course you could.
But what to do with a product? None of the differ-
entiations you have learned have yielded you for the
differential coefficient a product like this. Failing
such, the simplest thing is to multiply up the two
functions, and then integrate. This gives us

flettact2 a
And this is the same as
erdo+ [sede [ade
And performing the integrations, we get
10420242040.

SOME OTHER INTEGRALS 201

Some other Integrals.

Now that we know that integration is the reverse
of differentiation, we may at once look up the differ-
ential coefficients we already know, and see from
what functions they were derived. This gives us the
following integrals ready made :

x”: (p.148); ford A

0.109);

ata dx =log.(2+a)+0,

lira
e (p.143): feto =e+0.

er fe =-e-=+0
ss 1 dy ex0—i1xe
(for if y= a: e” nen

sina (p. 168); jsnoan = cos a+,

cosa (p. 166); feos ada =sinæ+C.
Also we may deduce the following:

loge: flog, ardor er(log.ar— +0

(for it y=alog.a—a, WE 4 10g,0—1 =log.2)

202 CALCULUS MADE EASY

log; flosioe de =043482008,2—1+0.

. py ‘as a
ao @ 100); fear “gat
cos az; feosazdo=2sinaz+0
’ a

(for it y=sinaz, ¿L=acosam; hence to get cosaz
one must differentiate y=t sin ax).

sina; fsinardo=—Leosar+0.

Try also costó; a little dodge will simplify matters:
cos 29=cos'0—sin*9=2cost0—1;
hence cos? 8=4 (008 2041),

and foostadd=3[ (cos 20-+1)a0
=1feos20ae+ 3fao

dc. (Seo also p. 227.)

Seo also the Table of Standard Forms on pp.286, 287.
You should make such a table for yourself, putting
in ib only the general functions which you have
successfully differentiated and integrated. See to it
that it grows steadily !

DOUBLE INTEGRALS 203

On Double and Triple Integrals,

Án many cases it is necessary to integrate some
expression for two or more variables contained in it;
and in that case the sign of integration appears more
than once. Thus,

[free ware

means that some function of the variables © and y
has to be integrated for each. It does not matter in
which order they are done. Thus, take the function
ai+y. Integrating it with respect to & gives us:

forty acaer tay.
Now, integrate this with respect to y:

Jar -+ay)dy = ray,

to which of course a constant is to be added. If we
had reversed the order of the operations, the result
would have been the same.

In dealing with areas of surfaces and of solids, we
have often to integrate both for length and breadth,
and thus have integrals of the form

if u-dedy,

where a is some property that depends, at each point,
on æ and on y. This would then be called a surface-
integral. It indicates that the value of all such

204 CALCULUS MADE EASY

elements as u- de» dy (that is to say, of the value of w
over a little rectangle der long and dy broad) has to be
summed up over the whole length and whole breadth.

Similarly in the case of solids, where we deal with
three dimensions. Consider any clement of volume,
the small cube whose dimensions are de dy dz If
the figuro of the solid be expressed by the function
F(a, y, 2), then the whole solid will have the volume-
integral,

volume = re y,2)-de-dy-de.

Naturally, such integrations have to be taken be-
tween appropriate limits* in each dimension; and the
integration cannot be performed unless one knows in
what way the boundaries of the surface depend on
a, y, and x If the limits for æ are from a, to 2,
those for y from y, to Yo, and those for # from =,
to 2, then clearly we have

volume= IS 1,2): de-dy-de.

There are of course plenty of complicated and
difficult cases; but, in general, it is quite easy to
see the significance of the symbols where they are
intended to indicate that a certain integration has to
be performed over a given surface, or throughout a
given solid space.

* See p. 208 for intogration between limita.

SIMPLE INTEGRATIONS 205
Exercises XVII. (Sec p. 297 for the Answers.)
(1) Find |y dar when y! =4a2.

(2) Find (Far. (8) Find [lasdo.
(4) Find ((2%+a)de. (5) Integrate Bark.

(6) Find |(4a*+ 32:°+ 20+ Ida.

(1) TE dy oe e find y.
7
(8) Find 16 +2) (9) Find Jessa.

(10) Find |(w+2\a—a)dz.
(11) Find fiver Yaseas.

(12) Find [isin o—y@.

(13) Find [eostaa dé. (14) Find (sin*0 48.

(15) Find ET de. (16) Find |*-dx.

@7) Find FE (18) Find f2

CHAPTER XIX.
ON FINDING AREAS BY INTEGRATING.

‘One use of the integral calculus is to enable us to
ascertain the values of areas bounded by curves,
Let us try to get at the subject bit by bit,

Y

Fue, 82

Let AB (Fig. 52) be a curve, the equation to which
is known. That is, y in this curve is some known
function of m. Think of a piece of the curve from
the point P to the point Q.

Let a perpendicular PM be droppod from P, and
another QN from the point @. Then call OM=a,
and ON =a, and the ordinates PM=y, and QN =»
Wo have thus marked out the area PONM that lies

FINDING AREAS BY INTEGRATING 207

beneath the piece PQ. The problem is, how ean we
ealeulate the value of this area?

The secret of solving this problem is to conccive
the area as being divided up into a lot of narrow
strips, each of them being of the width de. The
smaller we take da, the more of them there will be
between a, and.ay. Now, the whole area is clearly
equal to the sum of the areas of all such strips. Our
business will then be to discover an expression for
the area of any one narrow strip, and to integrate it
so as to add together all the strips Now think of
any one of the strips It will be like this:
being bounded between two vertical sides, with
a flat bottom dz, and with a slightly curved
sloping top. Suppose we take its average
height as being y; then, as its width is de, its
area will be yde. And seeing that we may
take the width as narrow as we please, if we
only take it narrow enough its average height will be
the same as the height at the middle of it. Now
let us call the unknown value of the whole area
8, meaning surface The area of one strip will be
simply a bit of the whole ares, and may therefore
be called dS. So we may write

area of 1 strip=dS=y- de.
If then we add up all the strips, we get
total area S= | dS=|yas.

So then our finding S depends on whether we can

208 CALCULUS MADE EASY

integrate y- dx for the particular case, when we know
what the value of y is as a function of 2.

For instance, if you were told that for the particular
curve in question y=0-+as%, no doubt you could put
that value into the expression and say: then I must
find ‘i (b+a2) do.

That is all very well; but a little thought will show
you that something more must be done. Because the
arca we are trying to find is not the area under the
whole length of the curve, but only the area limited
on the left by PM, and on the right by QN, it follows
that we must do something to define our area between
those limits!

This introduces us to a new notion, namely that of
integrating between limits. We suppose & to vary,
and for the present purpose we do not require any
value of w below æ, (that is OM), nor any value of
& above a, (that is ON), When an integral is to be
thus defined between two limits, we call the lower
of the two values the inferior limit, and the upper
value the superior limit. Any integral so limited
we designate as a definite integral, by way of dis-
tinguishing it from a general integral to which no
limits are assigned.

In the symbols which give instructions to integrate,
the limits are marked by putting them at the top
and bottom respectively of the sign of integration,
‘Thus the instruction

FINDING AREAS BY INTEGRATING 209

will bo read: find the integral of y- de between the
inferior limit æ, and the superior limit ay.

Somotimes the thing is written more simply

Poy de,
a
Well, but how do you find an integral between limits,
when you have got these instructions ?

Look again at Fig. 52 (p. 206). Suppose we could
find the area under the larger piece of curve from
A to Q, that is from 2=0 to w=a, naming the area
AQNO. Then, suppose we could find the area under
the smaller piece from A to P, that is from æ=0 to
æ=,, namely the area APMO. If then we were to
subtract the smaller area from the larger, we should
have left as a remainder the area PONM, which is
what we want. Here we have the clue as to what
to do; the definite integral between the two limits is
the différence between the integral worked out for
the superior limit and the integral worked out for the
lower limit.

Let us then go ahead. First, find the general
integral thus: es

and, as y=b-+ ax is the equation to the curve (Fig, 52),
foraxax

is the general integral which we must find.
Doing the integration in question by the rule

‘p. 196), 11
BIO weg 1840;

eus o

210 CALCULUS MADE EASY

and this will be the whole arca from 0 up to any
value of æ that we may assign.
‘Therefore, the larger area up to the superior limit
ill di
ANR da + Ga + 0;
and the smaller area up to the inferior limit a, will be
da + ga +0.

Now, subtract the smaller from the larger, and we
get for the area S the value,

area Saba, 0) + Sasa).

This is the answer we wanted. Let us give some
numerical values. Suppose 6=10, a=006, and 2,=8
and #,=6. Then the area S is equal to

o
106-0466)
=204+002(512—216)
=20+0°02 x 296
=20+592
=2592.

Let us here put down a symbolic way of stating
what we have ascertained about limits:

==
f yde=Y Y
mao

where y, is the integrated value of ydw correspondin
to a, and y, that corresponding to a.

FINDING AREAS BY INTEGRATING 211

All integration between limits requires the differ-
ence between two values to be thus found. Also note
that, in making the subtraction the added constant O
has disappeared. :

Examples.

(1) To familiarize ourselves with the process, let us
take a case of which we know the answer beforehand.
Let us find tho area of the triangle (Fig, 53), which

Fic. 58.

has base @=12 and height y=4. We know before-
hand, from obvious mensuration, that the answer will
come 24,

Now, here we have as the “curve” a sloping line
for which the equation is

e
y=y

‘The area in question will be

eo pant
I | Sd.
0 zo

Integrating Gd (p 194), and putting down the

212 CALCULUS MADE EASY

value of the general integral in square brackets with
the limits marked above and below, we get

orea=[4-5 20]
[Eso a
-[F+e]- E +0]

144
== =24 Ans.

Note that, in dealing with definite integrals, the
constant C always disappears by subtraction.
Let us satisfy ourselves about this rather sur-
Y prising dodge of caleula
tion, by testing it on
a simple example. Get
some squared paper, pre-
ferably some that is
ruled in little squares of
Fis. Dh one-eighth inch or ane-
tenth inch each way. On this squared paper plot
out the graph of this equation,

DT
q

Seer

‘Lhe values to be plotted will be:
= JelsIs Is I»

vol:

‘Che plot is given in Fig. 54

FINDING AREAS BY INTEGRATING 213

Now reckon out the arca beneath the curve by
counting the Little squares below the line, from æ=0
as far as æ=12 on the right. There are 18 whole
squares and four triangles, each of which has an area
equal to 1} squares; or, in total, 24 squares. Hence

24 is the numerical value of the integral of gan

between the lower limit of #=0 and the higher limit
of æ=12.

As a further exercise, show that the value of the
same integral between the limits of 2=3 and a=15
is 36.

(2) Find the arca, between limits æ=œ, and 2=0,

of the curvo y=

ata
Y

POT x

Pio. 55,

| van at
o Je

214 CALCULUS MADE EASY

=> Diog.(o-+0)+ c

=bllog. (2, +a)+C—log.(0+0)— 0]

At Ans.
a

=

=blog,

Let it be noted that this process of subtracting one
part from a larger to fiad the difference is really a
common practice, How do you find the area of a

ie, 58.

plane ring (Fig. 56), the outer radius cf which is r,
and the inner radius is 1,1 You know from men-
suration that the area of the outer circle is mr; then
you find the area, of the inner circle, wr,*; then you
subtract the latter from the former, and find area of
ring="(r2—r/); which may be written
r(r Hr (m7)

= mean cireumference of ring x width of ring.

(8) Here's another case—that of the die-away curve

FINDING AREAS BY INTEGRATING 215

(p. 156). Find the area between »=0 and «=a, of
the curve (Fig. 57) whose equation is

y=be-=.
Area=0f"e-*-de,

‘The integration (p. 201) gives

@ La j
ë
Fra, 87. Eso, 08,

(4) Another example is afforded by the adiabatic
eurve of a perfect gas, the equation to which is
pw"=c, where p stands for pressure, w for volume,
and 2 is of the value 1:42 (Fig, 58).

Find the area under the curve (which is proportional
to the work done in suddenly compressing the gas)
from volume 2, to volume %;.

216 CALCULUS MADE EASY

Here we have
area = Fa »-de

AT

An Exercise,

Prove the ordinary mensuration formula, that the
area A of a circle whose radius is 12, is equal to TR?

Consider an elementary zone or annulus of the
surface (Fig. 59), of breadth dr, situated at a distance

Fig, 50.

r from the centre. We may consider the entire sur-
face as consisting of such narrow zones, and the
whole area A will simply be the integral of all
such elementary zones from centre to margin, that is,
integrated from r=0 to r=R.

We have therefore to find an expression for the

FINDING AREAS BY INTEGRATING 217°

elementary area dA of the narrow zone. Think of
it as a strip of breadth dr, and of a length that is
the periphery of the circle of radius 7, that is, n
length of Zrw. Then we have, as tho area of the
narrow zone, dA=2rrdr.

Hence the area of the whole circle will be:
i ae
A=[a4 = feras tar.
E KA

Now, the general integral of r- dr is jr, Therefore,

A= E
or A=2r[4R—4(0)];
whence Anal.

Another Exercise.

Let us find the mean ordinate of the positive part
of the curve y=w=af, which is shown in Fig. 60.

LA M

N

7
Fis. 60,

To find the mean ordinate, we shall have to find the
area of the piece OMN, and then divide it by the
length of the base ON. But before we can find
the area we must ascertain the length of the base,
so as to know up to what limit we are to integrate,

218 CALCULUS MADE EASY

At N the ordinate y has zero value; therefore, we
must look at the equation and see what value of a
will make y=0. Now, clearly, if a is 0, y will also be
0, the curve passing through the origin O; but also,
if a=1,y=0: so that z=1 gives us the position of
the point N.

Then the ares, wanted is

ON ES
= fear -[ 10-40] 21-17-1007.

But the base length is 1.

‘Therefore, the average ordinate of the eurve=3.

[V.B.—It will be a pretty and simple exercise in
maxima and minima to find by differentiation what
is the height of the maximum ordinate. It must be
greater than the average]

The mean ordinate of any curve, over a range from
@=0 to æ=æ,, is given by the expression,

1 (ea
mn yo [Tran

If the mean ordinate be required over a distance not
beginning at the origin but beginning at a point
distant æ, from the origin and ending at a point
distant æ, from the origin, the value will be

mean y=
2

FINDING AREAS BY INTEGRATING 219
Areas in Polar Coordinates.

When the equation of the boundary of an area is
given as a function of the distance 7 of a point of it
from a fixed point O (see Fig. 61) called the pole, and.

BA

x
Fro. 61.

of the angle which y makes with the positive hori-
zontal direction OX, the process just explained can
be applied just as easily, with a small modification.
Instead of a, strip of ares, we considera small triangle
OAB, the angle at O being dé, and we find the sum
of alí the Mille telaaghs making up the sequined
area,

The area of such a small triangle is approximately
AB xr or xr; hence the portion of the area
included between the curve and two positions of #
corresponding to the angles A, and 6, is given by

220 CALCULUS MADE EASY

Examples,

(1) Find the area of the sector of 1 radian in a
circumference of radius a inch.

‘The polar equation of the circumferenee is evidently
r=a. The area is

ei pret
AI ar | =.
Pr a
(2) Find the area of the first quadrant of the curve
(known as “Pascal’s Snail”), the polar equation of
which is =a(1+c08 8)

a(1-+cos OY 40

2
A *(1+2c089+cos"9)d0

-£ F[o+2sina + 45 =

_ (348)
PAD

Volumes by Integration,

What we have done with the area of a little strip
of a surface, we can, of course, just as easily do with
the volume of a little strip of a solid. We can add
up all the little strips that make up the total solid,
and find its volume, just as we have added up all the
small little bits that made up an area to find the final
area of the figure operated upon,

FINDING AREAS BY INTEGRATING 221
Examples.

(1) Find the volume of a sphere of radius r.

A thin spherical shell has for volume 4ræ*dx (see

Fig. 59, p. 216); summing up all the concentric shells
which make up the sphere, we have

[Zamarao= 1%

EN
Y

We can also proceed as follows: a slice of the

sphere, of thickness da, hes for volume y'da (see

Fig. 62). Also æ and y are related by the expression
Para.

volume sphere

Fra. 62.

Hence volume sphere =

(2) Find the volume of the solid generated by the
revolution of the curve yi=6x about the axis of a
between w= 0 and w= 4.

202 CALCULUS MADE EASY
‘The volume of a slice of the solid is ry*do.
mn le
are
Hence volume Kr de= 6) ode

=

On Quadratic Means.
In certain branches of physics, particularly in the
study of alternating electric currents, it is necessary
to be able to calculate the quadratic mean of à
variable quantity. By “quadratic mean” is denoted
the square root of the mean of the squares of all the
values between the limits considered. Other names
for the quadratic mean of any quantity are its
“virtual” value, or its “RMS” (meaning root-mean-
square) value. The French term is valeur efficace. IE
y is the function under consideration, and the quad-
ratic mean is to be taken between the limits of æ=0
and w=1; then the quadratic mean is expressed as

Vij ae
Examples.

(1) To find the quadratic mean of the function
y=ae (Fig. 63).
Here the integral is | aa*do,

which is Ja".
Dividing by 7 and taking the square root, we have

487 = 1508.

" i
quadratic mean} ah

FINDING AREAS BY INTEGRATING 223

Here the arithmetical mean is ¿al; and the ratio
of quadratic to arithmetical mean (this ratio is called

the form-factor) is Be 1155.
Y]
y
ö x
fo. os

(2) To find the quadratic mean of the function y= at

The inte vis (7% san that is LA
o integral is | da, that is ea

. wi

Hence qundratic mean 4/, fe

(8) To find the quadratic mean of the function y =a%

The integral is

(ai) dex, thet is f

a ya
= Loa lo
1

whien is De
log. @

de,
©

Hencs the quadratic mean is

224 CALCULUS MADE EASY

Evercises XVIIL (See p. 297 for Answers.)

(1) Find the area of the curve y==*%4+0=5 be-
tween @=0 and @=6, und the mean ordinate between
these limits,

(2) Find the arca of the parabola y= 2ax/2 between
æ=0 and a=a, Show that it is two-thirds of the
rectangle of the limiting ordinate and of its abscissa.

(3) Find the area of the positive portion of a sine
curve and the mean ordinate,

(4) Find the area of the portion of the curve y=sin?æ
from 0° to 150', and find the mean ordinate,

(5) Find the area included between the two branches

ot the curve y=a*+a? from æ=0 to æ=1, also the
area of the positive portion of the lower branch of
the curve (see Fig. 30, p. 108).

(6) Find the volume of a cone of radius of base 7,
and of height A.

(7) Find the area of the curve y=a'—log.a be-
tween 2=0 and w= 1.

(8) Find the volume generated by the curve
y= 1+2%, as it revolves about the axis of a, be-
tween a=0 and a=4

(9) Find the volume generated by a sine curve
revolving about the axis of =.

(10) Find the area of the portion of the curve
wy =a included between #=1 and a=a. Find the
mean ordinate between these limits,

FINDING AREAS BY INTEGRATING 225

(11) Show that the quadratic mean of the function
in w, between the limits of © and 7 radians, is

a Find also the arithmetical mean of the same

funetion between the same limits; and show that the
form-factor is =1:13.

(12) Find the arithmetical and quadratic means of
the function 2*43x+2, from #=0 to 2=3.

(13) Find the quadratic mean and the arithmetical
mean of the function y =A, sin «+4, sin 3a.

(14) A certain curve has the equation y=3'42e"™™.
Find the area included between the curve and the
axis of a, from the ordinate at a=2 to the ordinate
at w=8, Find also the height of the mean ordinate
of the curve between these points.

(15) Show that the radius of a circle, the area of
which is twice the area of a polar diagram, is equal
to the quadratic mean of all the values of r for that
polar diagram.

(16) Find the volume generated by the curve

y= +2 2002) rotating about the axis of a

CHAPTER XX.
DODGES, PITFALLS, AND TRIUMPHS.

Dodges, A great part of the labour of integrating
things consists in licking them into some shape that
can be integrated. The books—and by this is meant
the serious books—on the Integral Culculus are full
of plans and methods and dodges and artifices for
this kind of work. The following are a few of
them.

Integration by Parts. This name is given to a
dodge, the formula for which is

fucto=ua— [adu+.

16 is useful in some cases that you can’t tackle
directly, for it shows that if in any case fra can
be found, then [ud can also be found. ‘The formula
can be deduced as follows. From p. 38, we have,

dua) =ude+edu,
which may be written
nd = dux)— edu,
‘which by direct integration gives the above expression,

DODGES, PITFALLS, AND TRIUMPHS 227

Examples.

©) Find fw-sin ew du.

Write u=w, and for sin w- dw write de, We shall
then have du= dis, while [sin w- dn= eos w=.

Putting these into the formula, we get

Joo-sin waw= wc) - [-eos din
= —weos w+ sin w+0.

(2) Find [re de.

Write
then
and feed se | ‘dx (by the formula)

=00-e=e(9-1)+0.

@ Try feosa do.

Hence
fooste do=cos asin + [sine do
EE + [i1—costoyao
= 22 as fevode.
Honce 2fcost00"" 2840
and feoste aa = 882848 +6.

228 CALCULUS MADE EASY

(4) Find |2*sin ada.

Write a sinada=dv;
then du=2x dr, v= —cosm,

sin à dx = —a* cos a+ 2|acosade,

i
Now find fecoseede, integrating by parte (as in

Example 1 above):

ar cos w dav =a sin 2-4 008 2+ 0.

Hence

a sin a du= — 2° cos e+ 2x sin e+2 cos 040"
2

-22sino+0s2(1-%) |+0.

(5) Find [VI= de.

Write u=V 12, de=de;

then du= — FERS (see Chap. IX, p. 07)

and æ=v; so that
imac star
Here we may use a little dodge, for we can write

fumer [GD MES

Adding these two last equations, we get rid of
2

and we have

af Viper 1e +

da
VIE

DODGES, PITFALLS, AND TRIUMPHS 229
de

Do you remember meeting 1 it is got by

differentiating y=awe sin a (see p. 171); hence its in-
tegral is arc sin a, and so

fr Pap ZE +4 aresino+ 0.

You can try now some exercises by yourself; you
will find some at the end of this chapter.

Substitution. This is the same dodge as explained
in Chap. IX, p. 67. Let us illustrate its application
to integration by a few examples.

@ [uz Fa de.
Let Sta=u, de=du;
replace futdu=sut=3+0))

a fete

Let =u, da, and daa,
du f du,

=f = du
are Bee +e*)

ner

lute

so that je

du

Tap is the result of differentiating are tan u

Hence the integral is are tan e%.

de de
Sat aaa

230 CALCULUS MADE EASY

Let whlen de=du;
then the integral becomes [
E

the result of differentiating À aro tan 2.

1 etat

Fi

Hence one has finally

for the value

of the given integral,

Formule of Reduction ar» special forms applicable
chiefly to binomial and trigonometrical expressions
that have to be integrated, and bave to be reduced
into some form of which the integral is known.

Rationalization, and Factorization of Denominator
are dodges applicable in special cuses, but they do not
admit of any short or general explanation. Much
practice is needed to become familiar with these pre-
paratory processes.

‘The following example shows how the process of
splitting into partial fractions, which we learned in
Chap, KILL, p 122, con bo made use of in integration,

Take again

5 if we split

+ F5
into partial fractions, this becomes (see p. 282):

57=l de Lt
212 d+ 14/22

Notice that the same integral can be expressed

DODGES, PITFALLS, AND TRIUMPHS 231

sometimes in more than one way (which are equivalent
to one another).

Pitfalls, A beginner is liable to overlook certain
points that a practised hand would avoid; such as
the use of factors that are equivalent to either zero or
infinity, and the oceurrenco of indeterminate quantities
such as $. ‘There is no golden rule that will meet
every possible ease, Nothing but practice and intelli-
gent care will avail. An example of a pitfall which
had to be circumvented arose in Chap. XVIIL, p. 199,
when we came to the problem of integrating =": da.

Triwmphs. By triumphs must be understood the
successes with which the calculus has been applied to
the solution of problems otherwise intractable. Often
in the consideration of physical relations one is able
to build up an expression for the law governing the
interaction of the parts or of the forces that govern
them, such expression being naturally in the form of
a differential equation, that is an equation containing
differential coefficients with or without other algebraic
quantities, And when such differential equation
has been found, one can get no further until it has
been integrated. Generally itis much easier to state
the appropriate differential equation than to solve it:
tho real trouble begins then only when one wants to
integrate, unless indeed the equation is seen to possess
some standard form of which the integral is known,
and then the triumph is easy. The equation which
results from integrating a differential equation is

232 CALCULUS MADE EASY

called * its “solution”; and it is quite astonishing
how in many cases the solution looks as if it had no
relation to the differential equation of which it is
the integrated form. The solution often seems as
different from the original expression as a butterfly
does from the caterpillar that it was. Who would
have supposed that such an innocent thing as
dy_ 1
de aa?
could blossom out into
atx
+ ot
yet the latter is the solution of the former.
Asa last example, let us work out the above together
By partial fractions,
1 1 + 1
Fé a aaa)

y= 0
Y= Fy 8s,

de
= Galata) Talay

rl i)

=p (log (a+@)log.(a—2))

1, at
=p tr
Pre
* This means that the actual result of solving it is called its

“solution.” But many mathemeticians would say, with Professor
Forsyth Severy'diftrsntal equation se considera! an soled when,
‘the value of the dependent variable is expressed as a function of
the independent variable by means either of known functions, or of
integrals, whether the integratione in the latter can or cannot be
expressed in terms of fanetions already known.”

DODGES, PITFALLS, AND TRIUMPHS 233

Not a very difficult metamorphosis!
There are whole treatises, such as Boole’s Diféren-

tial Equations, devoted to the subject of thus finding
the “solutions” for different original forms.

Exercises XIX. (See p. 298 for Answers)
(1) Find fe Ed. (2) Find (olog.o do.

(8) Find je loge dz, (4) Find fecose da,
(5) Find fi cos(log,w)da. (6) Find Fe

(7) Find [ESPA (8) Fina [ sty
(9) Find 14240. (10) Find popa zo,
don de

(12) Find

Ea

a

(14) Find at
aa

CHAPTER XXI
FINDING SOLUTIONS,

Is this chapter we go to work finding solutions to
some important differential equations, using for this
purpose the processes shown in the preceding chapters.

‘The beginner, who now knows how easy most of
those processes are in themselves, will here begin to
realize that integration-is an ort. As in all arts, so
in this, facility can be acquired only by diligent and
regular practice. He who would attain that facility
must work out examples, and more examples, and yet
more examples, such as are found abundantly in all
the regular treatises on the Calculus. Our purpose
here must be to afford the briefest introduction to
serious work.

Example 1. Find the solution of the differential
equation ay +620,

Transposing we have

FINDING SOLUTIONS 235

Now the mere inspection of this relation tells us

that we have got to do with a caso in which a is

proportional to y. If we think of the curve which
will represent y as a function of a, it will be such
that its slope at any point will be proportional to
the ordinate at that point, and will be a negative
slope if y is positive. So obviously the curve will
be a die-away curve (p. 156), and the solution will
contain e"* as a factor. But, without presuming on
this bit of sagacity, let us go to work.

As both y and dy occur in the equation and on
opposite sides, we can do nothing until we get both
y and dy to one side, and de to the other. To do
this, we must split our usually inseparable companions
dy and des from one another.

dy _ a
ay ¿da

Having done the deed, we now can see that both
sides have got into a shape that is integrable, because
we recognize Y, or je as a differential that we
have met with (p. 147) when differentiating logarithms.
So we may at once write down the instructions to
integrate, p

a
e] ¿ao
and doing the two integrations, we have:

logy

Gu+log, O,

236 CALCULUS MADE EASY

where log, O is the yet undetermined constant * of
integration. ‘Then, delogatizing, we get:
y=Ce*,

which is the solution required. Now, this solution
looks quite unlike the original differential equation
from which it was constructed: yet to an expert
mathematician they both convey the same information
as to the way in which y depends on a.

Now, as to the O, its meaning depends on the
initial value of y. For if we put #=0 in order to
see what value y then has, we find that this makes
y=Ce-%; and as e-9=1, we see that O is nothing else
than the particular valuet of y at starting. This we
may call yy, and so write the solution as

Ye

Example 2.
Let us take as an example to solve

ay +0 dl =p,

where g is a constant. Again, inspecting the equation
will suggest, (1) that somehow or other & will come
into the solution, and (2) that if at any part of the

“We may write down any form of constant as the “constant of
integration,” and the form loge © is adopted here by preference,
because the other terms in this line of equation are, or are treat
as logurithins; und it saves complications aftorward if the added
constant be of the same kind,

Compare what was said about the “constant of integration,”
with reference to Fig, 48 on , 187, and Fig. 51 on p. 190.

FINDING SOLUTIONS 237

curve y becomes either a maximum or a minimum, 80
that 2 =o, then y will have the value=2, But lot
us go to work as before, separating the differentials
and trying to transform the thing into some in-
tegrable shape.

Ce
BI

Now we have done our best to get nothing but y
and dy on one side, and nothing but de on the other,
But is the result on the left side intograble ?

It is of the same form as the result on p. 148;
writing the instructions to integrate, we have:

A A
5 pens

u:

and, doing the integration, and adding the appropriate
constant,

log. (y-2) = — Fotos. 0;

whence Ice”,
a

and finally, valtac®”.
which is the solution.

238 CALCULUS MADE EASY

If the condition is laid down that y=0 when 2=0
we can find C'; for then the exponential becomes =1;
and we have

eel.
0=,+6

or 0=-2.
a

Putting in this value, the solution becomes

But further, if æ grows indefinitely, y will grow to
a maximum; for when æ=, the exponential=0,

giving Yous. =

2 Substituting this, we get finally

Y= Yan (14).
This result is also of importance in physical science.

Example 8.
dy at
Let ay+bGf=g- sin 2rnt.
We shall find this much less tractable than the
preceding. First divide through by d,

de, Li
+ Fy =F sin Dent.

Now, as it stands, the left side is not integrable,
But it can be made so by the artifice—and this is

FINDING SOLUTIONS 239

where skill and practice suggest a plan—of multiplying
all the terms by el, giving us:
ay + Gye! Hai sin ant,

which is the same as

Yea, yee 1-98 sin aunts
and bis bing poco differential may be integrated

yale)

se y= fet fe sin 2ent- dt 06

The last term is obviously a term which will die
out as £ increases, and may be omitted. * The trouble
‘now comes in to find the integral that appears as 6
factor, To tackle this we resort to the device (see
p. 226) of integration by parts, the genoral formula for

which is fudo=wo— fou. For this purpose write

uae
dv = sin 2rnt- di,
We shall then have

240 CALCULUS MADE EASY

fe

Inserting these, the integral in question becomes:

in Qarnt de

1 4 1 a
ed eos Innt- [55 cos 2nd "zu
1% a
pe
game Colette,

‘The last integral is still irreducible. To evade the
difficulty, repeat the integration by parts of the left
side, but treating it in the reverse way by writing:

u=sin 2ants
due". de;

du
whence

Inserting these, we get

fet cos 2rnt de [8]

re -cos rnt-dt;

in 2rnt—

68. cos Int dt. [0]

Noting that the final intractable integral in [c] is
the same as that in [3], we may eliminate it by

roping Ge] by 29, and multiplying [0] by

y and adding them.

FINDING SOLUTIONS 21
The result, when cleared down, is:

sin 2and- dt

+ fab «sin 2ant—2rnb? cos Bent
dene oll
Inserting this value in [A], we get
_ fa: tene Bun on 2th
y=g) Der
To simplify still further, let us imagine an angle $
such that tong =?

. 2rnb
Then sin $= Trae
a
sd cud apa

Substituting these, we ats

008 008 trat
we nn
Which may be written
genug)
re
which is the solution desired.

This is indeed none other than the equation of an
alternating electric current, where g represents the
amplitude of the electromotive force, # the frequency,
a the resistance, b the coefficient of self-induction of
the circuit, and ¢ is an angle of lag,

one. Q

242 CALCULUS MADE EASY

Example 4,
Suppose that Mdz+Ndy=0.

We could integrate this expression directly, if M
were a function of « only, and N a function of y
only; but, if both 4 and N are functions that depend
on both æ and y, how are we to integrate it? Is it
itself an exact differential? That is: have M and N
each been formed by partial differentiation from some
common function U, or not? If they have, then

eu.
CR

M,

And if such a common function exists, then
SU, SU
ae

is an exact differential (compare p. 175).

Now the test of the matter is this, If the expression
is an exact differential, it must be true that
aM _ aN,
dy da’
UdV)_dKav)
for then de
which is necessarily true.
Take as an illustration the equation

(1+30yde+a?dy=0.

FINDING SOLUTIONS 243

Is this an exact differential or not? Apply the
test,
dA +32) _ go,
dy

ds) _
de m

which do not agree. Therefore, it is not an exact
differential, and the two functions 14+32y and a?
have not coma from a common original function,

It is possible in such cases to discover, however, un
integrating factor, that is to say, a factor such that
if both are multiplied by this factor, the expression
will become an exact differential. There is no ono
rule for discovering such an integrating factor; but
experience will usually suggest one. In. the present
instance 2x will act as such. Multiplying by 2x, we

get (a+ G.x"y) dae + 2a%dy =0.
Now apply the test to this.
[aout a,
Y

which agrees. Hence this is an exact differential, and
may be integrated. Now, if w=2x%y,
dw=6x'y de+20 dy.
Hence foo dn fre ay= y
so that we get U=sa+2y+0.

244 CALCULUS MADE EASY

Ecample 5. Lot PL enty=o.

In this case we have a differential equation of the
second degree, in which y appears in the form of
a second differential coefficient, as well as in person,

i
Transposing, we have vy = nly.

It appears from this that we have to do with a
function such that its second differential coefficient is
proportional to itself, but with reversed sign, In
Chapter XV. we found that there was such a func-
tion—namely, the sine (or the cosine also) which
possessed this property. So, without further ado,
we may infer that the solution will be of the form
y=Asin(pt+q). However, let us go to work.

Multiply both sides of the original equation by 24%

and integrate, giving us ate D + ony Y =0, a as
any
al
a
ada dy _ Gi (ey Hg 020,

O being a constant. OO The taking the square roots,
dy _

But it can be shown that (see p. 171)
ve sin Y,

Alaresind,

JO dy

whenee, passing from angles to sines,

aresin D 1040, and y=Csin(nt+ Ci),

FINDING SOLUTIONS 245

whereC, is a constantangle that comes inby integration,
Or, preferably, this may be written
y=Asinnt+Beosnt, which is the solution.

ey

Example 6. ny =0.

Here we have say to deal with a function y
which is such that its second differential coefficient is
proportional to itself. The only function we know
that has this property is the exponential function
(eee p. 143), and we may be certain therefore that the
solution of the equation will be of that form.
Proceeding as before, by multiplying through by
2% Py dy d
D and integrating, we get 274 72 —2nty22=0,
«ey
dde SS

apre

and,as 2

where e is a constant, and 7

y
Now, if w=log.(yt/P+2)=log. 1
de_1 du_ RON Zn;
du a dy EN Nro

and

Hence, integrating, thi
log. (y +! næ+ log. C;

yt y+ e)= Cem.

Now (ytVP+E)x(— yt Vo)
whence yr,

246 CALCULUS MADE EASY
Subtracting (2) from (1) and dividing by 2, ws

then have om] em

soo"

which is more corn written
ya ASE Be

Or, the solution, which at first sight does not look
as if it had anything to do with the original equation,
shows that y consists of two terms, one of which
grows logarithmically as æ increases, while the other
term dies away as = increases.

Example. y
æ D =
Let are +a=0.

Examination of this expression will show that, if
) it has the form of Example 1, the solution of
which was a negative exponential. On the other
hand, if a=0, its form becomes the same as that of
Example 6, the solution of which is the sum of a
positive and a negative exponential. It is therefore
not very surprising to find that the solution of thr
present example is

yale \(det + Be”

where m= and n=

The steps by which this solution is reached are not
given here; they may be found in advanced treatises

FINDING SOLUTIONS 247
Example 8.

It was seen (p. 177) that this equation was derived

from the original
y=F(0+at)4f(0—at),

where F' and f were any arbitrary functions of t.

Another way of dealing with it is to transform it
by a change of variables into

Y o,

Tu
where u=c-+at, and v=æ— at, leading to the same
general solution, If we consider a case in which
F vanishes, then we have simply

y-fa-at);
and this merely states that, at the time t=0, y is a
particular function of «, and may be looked upon as
denoting that the curve of the relation of y to a: has
a particular shape. Then any change in the value
of ¢ is equivalent simply to an alteration in the origin
from which w is reckoned. That is to say, it indicates
that, the form of the function being conserved, it is
propagated along the « direction with a uniform
velocity «a; so that whatever the value of the
ordinate y at any particular timo t, at any particular
point a, the same value of y will appear at the sub-
sequent time £, at a point further along, the abscissa
of which is æ,+a(t—t,) In this case the simplified

248 CALCULUS MADE EASY

* equation represents the propagation of a wave (of any
form) at a uniform speed along the æ direction.
If the differential equation had been written

the solution would have been the same, but the
velocity of propagation would have had the value

CHAPTER XXI.

A LITTLE MORE ABOUT CURVATURE
OF CURVES.

In Chapter XII. we have learned how wo can find out
which way a curve is curved, that is, whether it
curves upwards or downwards towards the right.
This gave us no indication whatever as to how much
the curve is curved, or, in other words, what is its
curvature.

By curvature of a line, we mean the amount of
bending or deflection taking place along a certain
length of the line, say along portion of the line the
length of which is one unit of length (the same unit
which is used to measure the radius, whether it be
one inch, one foot, or any other unit). For instance,
consider two cirenlar paths of centre O and O' and of
equal lengths AB, A'B' {see Fig. 64). When passing
from A to B along the are AB of the first one, one
changes one’s direction from AP to BQ, since at A
one faces in the direction AP and at B one faces in
the direction BQ. In other words, in walking from A
to B one unconsciously turns round through the angle
PCQ, which is equal to the angle AOB. Similarly,

250 CALCULUS MADE EASY

in passing from A’ to B', along the are A'B of
equal length to AB, on the second path, one turns
round through the angle P’C’Q, which is equal to the
angle A’0'B’, obviously greater than the correspond-

Fra, 64,

ing angle AOB. The second path bends therefore
more than the first for an equal length.

This fact is expressed by saying that the curvature
of the second path is greater than that of the first
one. The larger the circle, the lesser the bending,
that is the lesser the curvature. If the radius of the
first circle is 2, 3, 4, ... ete. times greater than the
radius of the second, then the angle of bending or
defection along an are of unit length will be 2, 3,
4, ... ete. times less on the first circle than on the
second, that is, it will be 4, 4, +, ... ete. of the bending
or deflection along the are of same length on the
second circle. In other words, the curvature of the

CURVATURE OF CURVES 251

first circle will be 3, 3, 4, ... ete. of that of the second
circle, We see that, as the radius becomes 2, 3, 4, ..

ete. times greater, the curvature becomes 2, 3, 4, ...
eto. times smaller, and this is expressed by saying that
the curvature of a circle is inversely proportional to
the radius of the civele, or

1
curvature =1 x He

where & is a constant. It is agreed to take &=1, so
that 1
curvature = PUT
always.
If the radius becomes indefinitely great, the curva-

1
ture becomes infinity

nator of a fraction is indefinitely large, the value of
the fraction is indefinitely small. For this reason
mathematicians sometimes consider a straight line as
an are of circle of infinite radius, or zero curvature.
In the case of a circle, which is perfectly symnetri-
cal and uniform, so that the curvature is the same at
every point of its circumference, the above method of
expressing the curvature is perfectly definite. In the
case of any other curve, however, the curvature is
not the same at different points, and it may differ
considerably even for two points fairly close to one
another, It would not then be accurate to take the
amount of bending or deflection between two points
as a measure of the curvature of the are between

zero, since when the denomi-

A

252 CALCULUS MADE EASY

these points, unless this are is very small, in fact,
unless it is indefinitely small.

If then we consider a very small are such as À B (sea
Fig, 65), and if we draw such a circle that an are AB

of this circle coincides with the are AB of the curve
more closely then would be the case with any other
circle, then the curvature of this circle may be taken
as the curvature of the are AB of the curve. The
smaller the are AB, the easier it will be to find a
circle an aro of which most nearly coincides with the
are AB of the curve. When A and B are very near
one another, so that AB is so small so that the length
ds of the are AB is practically negligible, then the
coincidence of the two ares, of cirele and of curve,
may be considered as being practically perfect, and
the curvature of the curve at the point A (or B),

+

CURVATURE OF CURVES 253

being then the same as the curvature of the circle,
will be expressed by the reciprocal of the radius of

E + 1
this circle, that is, by a>,
measuring curvature, explained above.

Now, at first, you may think that, if AB is very
small, then the cirele must be very small also. A little
thinking will, however, cause you to perceive that it is
by no means necessarily so, and that the circle may
have any size, according to the amount of bending of
the curve along this very small are AB. In fact,
if the curve is almost Aut af that point, the cirele will
be extremely large. This circle is called the circle of
curvature, or the osoulating circle at the point con-
sidered. Its radius is the radius of curvature of the
curve at that particular point.

If the aro AB is represented by de and the angle
AOB by d0, then, if r is the radius of curvature,

de_1
>

according to our way of

ds=rd9 or

The secant AB makes with the axis OX the angle
0, and it will be seen from the small triangle ABC

that dy tanó. When AB is indefinitely small, so

that B practically eoineides with A, the lino AB
becomes a tangent to the curve at the point 4
(or B).

Now, tan € depends on the position of the point A
(or B, which is supposed to nearly coincide with it),

254 CALCULUS MADE EASY

that is, it depends on a, or, in other words, tan @ is
“a function” of a.

Differentiating with regard to © to get the slope
(see p. 112), wo get

a)
de) dan de d0_ 1 dd
de de dé da cos bd
(see p. 168);
hence CA
de 02
But GE cos 6, and for 2 one may write a,
therefore Ey
1_d0_d9 _de_ ¿dy de.
LS =
dee se

F
but sec@=./T-+ tan"; hence

and finally,

The muraerator, being a square root, may have tho
sign + or the sign —. One must select for it the
sume sign as the denominator, so as to have y positive
always, as a negative radius would have no meaning,

CURVATURE OF CURVES 255

It has been shown (Chapter XII), that if LY is

positive, the curve is convex downwards, while if
FY is negative, the curvo is concave downwards, IE

Fu =0, the radius of curvature is infinitely great,
that is, the corresponding portion of the curve is a bit
of straight line. This necessarily happens whenever
a curve gradually changes from being convex to con-
cave to the axis of a or vice versa. The point where
this occurs is called a point of inflenion.

The centre of the circle of curvature is called the
centre of curvature. If its coordinates are 4, Y, then
the equation of the circle is (see p. 102)

(ay;
hence (wa, )dw+2(y—y,)dy =0,

and ayy) Lo. ay

Why did we differentiate? To get rid of the con-
stant y. This leaves but two unknown constants =,
and y,; differentiate again ; you shall get rid of one of
them. This last differentiation is not quite as easy as
it seems; let us do it together; we have:

256 CALCULUS MADE EASY

the numerator of the second term is a product; hence
differentiating it gives

ae
ow te) ae dude) 2 (y (de

80 that the result of iene (Dis
dys! a
1444) + 20;
from this we at once get
ays?
ny te)
dy

dai

Replacing in (1), we get

@-a)tyy-y- —
ze

and finally,

2, and y, give the position of the centre of curvature,
‘Tho use of these formulae will be best seen by care-
fully going through a fow worked-out examples.

Example 1. Find the radius of curvature and the
coordinates of the centre of curvature of the curve
y=2a'—a+8 at the point ==0.

CURVATURE OF CURVES 257

We have sat, FY ma,
a+
oe 1+ (5%) } _ da
3 AAA

des
when #=0; this becomes
army}

Hf my, yy, are the coordinates of the centre f curva
tuve then

dy, (aay
_ +) (4e—1) 0 + 40-1)
ao ee — Ge=D + de

Y
de
ao br _1
~ 4 2

when 2=0, y=3, so that

=34

dy
+ (by 1+ (40-1 14-1?
i Alene et Aa Se a Tas

da

Plot the curve and draw the circle, it is both in.
teresting and instructive. The values can be checked
easily, as since when #=0, y=3, here

ate (pS Pr? or StF Ste 50= 707%

ome, R

238 CALCULUS MADE EASY

Example 2, Find the radius of curvature and the
position of the centre of curvature of the curve
y'=ma at the point for which y=0.

Here
henee
Sa (yy { 3
Lee tite it Gotm},
Ey a Smt
det Tui

taking the — sign at the numerator, so as to have r

positivo.

+

Since, when y=0, »=0, we get r=
y estr

m,
2
Also, if ary, y, are the coordinates of the centre,

By) te

ay mi
Px a
aot PEM 2,

when w=, then m=z

CURVATURE OF CURVES 259

‘dy? m
ei) o
or ne

when 2=0, y,=0.

Example 3. Show that the circle is a curve of
constant curvature.

If a, y, are the coordinates of the centre, and R
is the radius, the equation of the cirele in rectangular
coordinates is
(@—a¥+(y- y= BP;
this is easily put into the form
y Eo +4, = {Rea} Y yo
To differentiate, let R°—(@—a,)'=0} then

A di
von Bait, 20,

Le HR 2(@—a,)
| EY
{R—@—a,3*

Differentiate again ; using the rule for differentiation
of a fraction, we get

(Rea a) lea

Ey La
de R-@-a}

260 CALCULUS MADE EASY

(it is always a good plan to write out the whole
expression in this way when dealing with a compli-
cated expression) ; this simplifies to
2 tae yy at
ey E a
ae RG)
re
a AP

hence
AIRE
= = E

de Pa A
the radius of curvature is constant and equal to the
radius of the circle,

Example 4. Find the radius and the centre of cur-
vature of the curvo y=a*—2a*4-a—1 at points where
2=0, #=05 and 2=10. Find also the position of
the point of inflexion of the curve.

dy 379
Here SU —8a"—4e-+1,

- das
den 4

Ba — Aa +1) (1 HE de D
6a—4

1 um er A

e =.—

CURVATURE OF CURVES 261
When 2=0, y=—1,

707, 2=0+3=03, y=

Plot the ourve, mark the point #=0, y=—L, take
two points on either side about half an inch away and
construct georactrically the circle passing by the three
points; measure the radius and the coordinates of
the centre, and compare with the above results. Ona
diagram, the scale of which was 2 inches =unit length,
the construction gave a circle for which r=072,
&=047, y= -153, a very fair agreement.
When &=05, y= —0875,

{+ 02594

Pr

=109,

—T

205 A =033,

yo 0875 41%. 190.

The diagram gave r=0%8, ,~033, y =-—183,
When #=1, y=-1,

O
os,
ant UN,
nee -05.

The diagram gave r=037, 2,096, y =—044,

262 CALCULUS MADE EASY

At the point of inflexion $40, 6x—4=0, and
= 3; hence y=09%,

Example 5, Wind the vedias and contre of eurva-

tare of the curve y= ¿4 lat the point for
which &=0. (This curve is called the catenary, as
a hanging chain affects the same slope exactly.) The
equation of the curve may be written
yagedgess
then (sce p. 150 Examples),
dy a lia

Fa,
Similarly
dy_1
de 2a
itt
=
since e* =.

when #=0, en ‘ae #0;

2
@
hence r=

CURVATURE OF CURVES 263

The radius of curvature at the vertex is equal to
the constant a.

Also aj=0- Do,
a
140
nays ata=20.
a

Yon are now sufficiently familiar with this type of
problem to work out the following exereises by your-
sel. You are advised to check your answers by

careful plotting of the curve and construction of the
circle of curvature, as explained in Example 4.

Exercises XX, (For Answers see p. 299.)

(1) Find the radius of curvature and the position
of tho centre of curvature of the curve y=e* at the
point for which @=0.

(2) Find the radius and the centre of curvature of
the curve y=2(3—1) at the point for which æ=2.

(3) Find the point or points of curvature unity in
the curve y=a.

(4) Find the radius and the centre of curvature of
the curve wy =m, at the point for which a= „/m.

(5) Find the radius and the centre of curvature of
the curve y?=4aw at the point for which 2=0.

(6) Find the radius and the centre of eurvature of

264 CALCULUS MADE EASY

the curve y=a5 at the points for which w= +09 and
also a@=0.

(7) Find the radius of curvature and the coordi-
nates of the centre of curvature of the curve

y=—-242

at the two points for which 2=0 and æ=1, re-
spectively. Find also the maximum or minimum
value of y. Verify graphically all your results,

(8) Find the radius of curvature and the eoordi-
nates of the centre of curvature of the curve

y=wW-2-1
at the points for which «= —2, 2=0, and æ=1.
(9) Find the coordinates of the point or points of
inflexion of tho curve y= a? +a°-+1.

(10) Find the radius of curvature and the coordi-

nates of the centre of curvature of the curve
y=(40—e—3
at the points for which w=12, w=2 and æ=25.
‘What is this curve?

(11) Find the radius and the centre of curvature of
the curve y=="—30*420+1 at the points for which
2=0, 2=+15. Find also the position of the point
of inflexion.
~ (12) Find the radius and centre of curvature of

the curve y=sin0 at the points for which 9=% and
6=5. Find the position of the point of inflexion.

OURVATURE OF CURVES 265

(13) Draw a circle of radius 8, the centre of which
has for its coordinates æ=1, y=0. Deduce the
equation of such a circle from first principles (seo
p- 202). Find by calculation the radius of curvature
and the coordinates of the centre of curvature for
several suitable points, as accurately as possible, and
verify that you get the known values.

(14) Find the radius and contre of curvature of the
curve y=cos@ at the points for which @=0, 07

x
and 0-3: .
(15) Find the radius of curvature and the centre of

. a
curvature of the ellipse +

=1 at the points for
which 2=0 and at the points for which y=0,

CHAPTER XXIIL

HOW TO FIND THE LENGTH OF AN ARC ON
A CURVE.

SINCE an are on any curve is made up of a lot of
little bits of straight lines joined end to end, if we
could add all these little bits, we would get the
length of the arc. But we have seen that to add a
lot of little bits together is precisely what is called

Y Ns
x vr
aly
= ¡a
¿A E
¡An
Ir
E
o “ LE) o “ x
(a) (5)
Fre, 66.

integration, so that it is likely that, since we know
how to integrate, we can find also the length of an

LENGTH OF AN ARC ON A CURVE 267

are on any curve, provided that the equation of the
curve is such that it lend itself to integration.

If MN is an are on any curve, the length ¢ of
which is required (see Fig. 66a), if we call “a little
bib” of the are ds, then we see at once that

(ds) = (da) + (dy),

or either

Bei) or ds~ 14 (By de.

Now the aro MN is made up of the sum of all the
httle bits ds between M and N, that is, between a
and a or between y, and y, so that we get either

sf” 14 WY ae or om a+ (gy) an

That is all!

The second integral is useful when there are several
points of the eurve corresponding to the given values
of «(asin Fig 668). In this case the integral between
a, and a, leaves a doubt as to the exact portion of the
curve, the length of which is required. It may be
ST, instead of MN, or SQ; by integrating between
y, and y, the uncertainty is removed, and in this case
one should use the second integral.

IE instead of æ and y coordinates,—or Cartesian
coordinates, as they are named from the French
mathematician Descartes, who invented them—we
have r and 6 coordinates (or polar coordinates, sec
p.219); then, if MN be a small are of length ds on

268 CALCULUS MADE EASY

any enrve, the length s of which is required (see
Fig. 67), O being the pole, then the distance ON will
generally differ from OM by a small amount dr. If
the small angle MON 15 called d@, then, the polar
coordinates of the point M being 6 and r, those of N
are (9-+d0) and (r+dr) Let MP be perpendicular

to ON, and let OR=OM; then RN=dr, and this
is very nearly the same as PN, as long as dO 1s
a very small angle. Also RM=rd8, and RM is
very nearly equal to PAL, and the are MN is very
neatly equal to tho chord MN. In fact we can write
PN=dr, PM=rd0, and aro MN =chord MN with
out appreciable error, so that we have.

(ds? =(chord MNY=PN°+ PU? = di?-+-1°de?.

Dividing by dé? we get (26) = 224 CAE honce

LENGTH OF AN ARC ON A CURVE 269

hence, since the length s is made up of the sum of all
the little bits ds, between values of 9=0, and 0=0,

we have
e % dr‘
fa fe
ha)
We can proceed at once to work out a fow examples.
Example 2. ‘The equation of a circle, the centre of
which is at the origin—or intersection of the axis of

æ with the axis of y—is a*4y!=r*; find the length
of an are of one quadrant.

dy
= 1a and 2y dy= — 2x de, so that = —%,
y aw and 2y dy so that =,

lola Jrs
ler;

‘The length we want—one quadrant—extends from
a point for which #=0 to another point for which

hence

z=r. We express this by writing
ee ee
alar)"
or, more simply, by writing
sl re
Va)"

the 0 and r to the right of the sign of integration

270 CALCULUS MADE EASY

merely meaning that the integration is only to be
performed on a portion of the curve, namely that
between 2=0, x=7, as we have seen (p. 210).

Here is a fresh integral for you! Can you manage
it?

On ts 171 we have differentiated y=are (sin æ) and

found À Tf you have tried all sorts of

2 75 By
variations of the given examples (as you ought to
have done), you perhaps tried to differentiate some-

thing like y=a are (sin 2), which gave

‚0 WE A

de aa Na”
that is, just the same expression as the one we have
to integrate here.

Honce 6 Ze ge rare (end) +0, © being a

constant.
As the integration is only to be made between
2=0 and æ=r, we write

“fa leo oe
proceeding thon as explained in Example (1), p.211,
we get s=rare(sin? 7)+0-rare (sin D- 6

or s=rXH

LENGTH OF AN ARC ON A CURVE 271

since are(sin 1) is 90° or 5 and are(sin 0) is zero, and
the constant C disappears, as has been shown.
The length of the quadrant is therefore 7, and

the length of the cireumference, being four times this,

Example 2. Find the length of the are AB between
@,=2 and a, =5, in the circumference x*+y*= 6? (seo
Fig. 68).

Fro. 68.

Here, proceeding as in previous example,

o=[raresin (2 8) ) +0] =[Saresin(2) +0]
=6[are QE aresin(? JE 6 (0:9850— 03397)
=38718 inches (the ares being expressed in radians),

272 CALCULUS MADE EASY

It is always well to check results obtained by a
new and yet unfamiliar method. This is easy, for

3 and cos BOX=§;
hence AOX=10°32, BOX =33"34,
and AOX—BOX = AOB=36°58'

36.9667

58

radian =0-6451 radian=38706 inches,

the discrepancy being merely due to the fact that the
last decimal in logarithmic and trigonometrical tables
is only approximate.

Example 3. Find the length of an arc of the curve

jj

[las
=} ysrerte 2 ade,

ILE EF
1, so that | 24e +

we can replace 2 by 2xe'=2xe:"2; then

LENGTH OF AN ARC ON A CURVE 273

and s=§(e-2)-

Example 4. A curve is such that the length of the
tangent at any point P (see dig. 69) from P to the
Y]

Fra. 69,

intorsection 7’ of the tangent with a fixed lino AB is
a constant length a. Find an expression for an arc
of this curve, —which is called the tractrix,—and find
the length, when a=3, between the ordinates y=a
and y=

eux 8

274 CALCULUS MADE EASY

We shall take the fixed line for the axis of a,
The point D, with DO=a, is a point on the curve,
which must be tangent to OD at D. We take OD
as the exis of y; AB and OD are what are called
axes of symmetry, that is the curve is symmetrical
about them; PT=a, PN=y, ON=2.

If we consider a small portion ds of the curve, ab

P, then sin@=
slopes downwards to the right, see p. 79).

Y (minus because the curve

de__2 q ¿2 = af,
Hence ay ds= a and ad,
that is, = —alog.y+ 9

When æ=0, s=
and C=a log.a.

, y=a, so that 0=—alog,a+0,

It follows that s=alog.a—alog.y=a lg >
When a=3, s between y=a and y=1 is therefore
o
$=3 [18.3] =3 (log, 1—leg, 3) =3 x (0-1:0986)
Ñ
296 or 3296,
as the sign — refers merely to the direction in which
the length was measured, from D to P, or from P
to D.
Note that this result has been obtained without a
knowledge of the equation of the curve, This is

sometimes possible. In order to get the length of an
are between two points given by their abseissae, how-

LENGTH OF AN ARC ON A CURVE 275

ever, it is necessary to know the equation of the curve;
this is easily obtained as follows:

hence

The integration will give us a relation between æ
and y, which is the equation of the curve .

(VE FAY __ os dy ydy
AE et
To integrate
_ _ 1, dy__1_ y
E ar
dy__ de,
so that y Ta
The integral becomes — Ss To integrate
N 1
this let Va T=0— ar, that io,
ela? ave
and 0=2v dv—2azdv—2aw de,

from which de =? = do, so that, replacing, we get

v—az 1
Sin J ao “ar
1(d__1
19-208

276 CALCULUS MADE EASY

dy Ligg 2+: A
PRE Er a MoE +O.
Now, for

y dy = À = YY.
145 lot am VE then dem U;

hence

so that J

=— [a= = Ja
We have then, finally,
analog NEE JE,
When 2=0, y=a, so that 0=alog.1—0+€, and

C=0; the equation of the tractrix is therefore

Tf a=3, as before, and if the length of the are from
æ=0 to æ=1 is required, it is not an casy matter to
calculate the value of y corresponding to any given
numerical value of a. It is, however, easy to find
graphcally an approximation as near the correct
value as we desire, when we are given the value of
a as follows:

Plot the graph, giving suitable values to y, say 3,
2,155, 1. From this graph, find what values of y
correspond to the two given values of a determining
the are, the length of which is needed, as accurately
35 the scale of the wraph allows. Tor @=0, y=3 nf

LENGTH OF AN ARC ON A CURVE 277%

course; suppose that for 2=1 you find y=172 on
the graph. This is only approximate, Now plot
again, on as large a seale as possible, taking only
three values of y, 1:6, 17, 18, On this second graph,
which is nearly, but not quite a straight line, you
will be probably able to read any value of y correct
to three places of decimals, and this is sufficient for
our purpose. We find from the graph that y=1723
corresponds to a=1. Then

ges ay
= 3] log, = =3| log.~
avg], [e- 5]
=3(log. 1741 —0)=1:66.

If we wanted a more accurate value of y we could
plot a third graph, taking for values of y 1°722, 1723,
1724, ...; this would give us, correct to five places of
decimals, the value of y corresponding to a=1, and
so on, till the required accuracy is reached.

Example 5. Find the length of an are of the
logarithmie spiral y=e* between 9=0 and 6— 1 radian.

Do you remember differentiating y=e*? Tt is an
easy one to remember, for it remains always the same
whatever is done to it: B-e (see p.143).

dr
do
If we reverse the process and integrate [exo we

Here, since r=e,

=r.

get back to r+C, the constant C being always intro-

278 CALCULUS MADE EASY

duced by such a process, as we have seen in Chap
XVIL
It follows that

le) Ja [eras
=2[rd0=/3[ed9=v/2(0+0)

Integrating between the two given values 9=0
and 6=1 we get

[+ (2 Ja=[Vhe+0),
= 20 —1/20=/2(c—1)
=141x1713= inches,
since r=e%=1 inch when 0=0.
Example 6, Find the length of an aro of the
logarithmic spiral r=? between @=0 and 6=6,.
As we have just seen,

2 0-0 [a0]=/2(091)-

Example T. As a lost example let us work fully a
case leading to a typical integration which will be
found useful for several of the exercises found at tho
end of this chapter. Let us find the expression for

the length of an are of the curve IES

Mean, [VERE da.

LENGTH OF AN ARC ON A CURVE 27%

Integrate this by parts: let

u=v ita? and de=dv;
Aedo

Y= d du=
then æ=v and du Visa

by the method of differentiation explained in Chap. IX,
Since fra uw ~fo du (see p. 226), we have
fur Fata de= oT Fra | ee. qa)
Also, we can write

a, (tata?) de.
[Mr 2 EI 5

hence
a dee ade
Fdo ta [Pdo
(vera air Kier a
Adding (1) and (2) we get
Tota i Trea. de
FF Fdo TEE Hit (3)
de

+

Remains to integrate

A =; for this purpose,
let VIF =v—ae; then
L4arat=v*—2ave+a@at or 1=0'—2ave.
Differentiating this, to get rid of the constant, we gef:
0=2v dv—2av da—2aw dv or avde= = andes
—ax) do
av

4 (o
that is de=
obtain ira Ta a"

[come al (SE 1 fae} 1
avira a) var) alo

; replacing in Es

Logo;

280 CALCULUS MADE EASY
de

hence ala + T+ aa),
Replacing in (3) and dividing by 2 we get, finally,

se [ur de

TERA flog (a0 TERR,

which can easily be calculated between any given
limits,

You ought now to be able to attempt with sucecas
the following exercises. You will find it interesting
as well as instructive to plot the eurves and verify
your results by measurement where possible.

The integration is usually of the kind shown on
p. 225, Ex. (5), or p. 229, Ex. (1), or p. 278, Ex. (7).

Exercises XXI. (For Answers, see p. 300.)

(1) Find the length of the line y=3æ+2 between
the two points for which #=1 and #=4.

(2) Find the length of the line y=an+b between
the two points for which a =a? and æ= —1.

(3) Find the length of the curve y=3æ? between
the two points for which 2=0 and »=1,

(4) Find the length of the curve y=a between the
two points for which 2=0 and æ=2.

(5) Find the length of the curve y=ma? between

the two points for which 2=0 and on

LENGTH OF AN ARC ON A CURVE 281
(6) Find the length of the curves r=acos@ and
r=asin 8 between 6=6, and 0=6).
(7) Find the length of the curve r=a secó.

(8) Find the length of the are of the curve y'=4aw
between 2 =0 and w=a.
(9) Find the length of the are of the eurve

1)

between 2=0 and 2=4.

(10) Find the length of the are of the curve y=e*
between @=0 and a=1.

(Note. This curve is in rectangular coordinates,
and is not the same curve as the logarithmic spiral
y= which is in polar coordinates. ‘The two cqua-
tions are shnilar, but the curves are quite different.)

(11) A curve is such that the coordinates of a point
on it aro a:=a(8--sin 6) and y=«(1—cos 8), 8 being
a certain angle which varies between 0 and 2x. Find
the length of the curve. (It is called a eyeloid.)

(12) Find the length of an are of the curve y*=ma

between @=0 and w=

(13) Find the expression for the length of an are of
a
2.
(14) Find the length of the curve y°=84 between
the two points for which w=] and 2=2.

the curve y=

282 CALCULUS MADE EASY

(15) Find the length of the curve y#-+a3=a? be
tween #=0 and a=a.

(16) Find the length of the curve r=a(1- cos €)
between 9=0 and @=7.

You have now been yersonally conducted over the
frontiers into the enchanted land. And in order that
you may have a handy reference to the principal
results, the author, in bidding you farewell, begs to
present you with a passport in the shape of a con-
venient collection of standard forms (see pp. 286, 287).
In the middle column are seb down a number of the
functions which most commonly occur. The results
of differentiating them aro set down on the left; the
results of integrating them are set down on the right,
May you find them useful |

EPILOGUE AND APOLOGUE.

Ir may be confidently assumed that when this
tractate “Calculus made Easy” falls into the hands
of the professional mathematicians, they will (if not
too lazy) rise up as one man, and damn it as being a
thoroughly bad book. Of that there can be, from
their point of view, no possible manner of doubt
whatever. It commits several most grievous and
deplorable errors.

First, it shows how ridiculously easy most of the
operations of the calculus really are.

Secondly, it gives away so many trade secrets. By
showing you that what one fool can do, other fools
can do also, it lets you see that these mathematical
swells, who pride themselves on having mastered such
an awfully difficult subject as the calculus, have no
such great reason to be pufled up. They like you to
think how terribly difficult it is, and don't want that
superstition to be rudely dissipated.

Thirdly, among the dreadful things they will say
about “So Easy ” is this: that there is an utter failure
on the part of the author to demonstrate with rigid

284 CALCULUS MADE EASY

and satisfactory completeness the validity of sundry
methods which he has presented in simple fashion,
and has even dared to use in solving problems! But
why should he not? You don’t forbid the use of
a watch to every person who does not know how to
make one? You don't object to the musician playing
on a violin that he has not himself constructed. You
don’t teach the rules of syntax to children until they
have already become fluent in the use of speech. Tt
would be equally absurd to require general rigid
demonstrations to be expounded to beginners in the
caleulus.

One other thing will the professed mathematicians
say about this thoroughly bad and vicious book : that
the reason why it is so easy is bocause the author hos
left out all the things that are really difficult. And
the ghastly fact about this accusation is that—it
is true! That is, indeed, why the book has been
written—written for the legion of innocents who have
hitherto been deterred from acquiring the elements of
the calenlus by the stupid way in which its teaching
is almost always presented, Any subject can be made
repulsive by presenting it bristling with difficulties.
The aim of this book is to enable beginners to learn
its language, to acquire familiarity with its endearing
simplicities, and to grasp its powerful methods of
solving problems, without being compelled to toil
through the intricate out-of-the-way (und mostly
irrelevant) mathematical gymnastics so dear to the
‘unpractical mathematieian.

EPILOGUE AND APOLOGUE 285

There are amongst young engineers a number on
whose ears the adage that what one fool can do,
another cam, may fall with a familiar sound, They
are earnestly requested not to give the author
away, nor to tell the mathematicians what a fool
he really is

TABLE OF STANDARD FORMS

4
eo ——uv— [var |
Algebraic.
1 æ dette
o a ax+C
1 ata Hô a+ C
a ax das+0
See a 4+
a ers
na” a Eres lod 1+C
as a logex+ 0
¿9 „dee
tte utotw Jude Juda: [wde
ae +0 = uv No general form known
du, de
as
m # No general form known
= u -fadu+O
} Exponential and Logarithmic.
' e e e+e
at logece æQog.æ-1)+C
04343 x 207? logya US 1+C
alogea e mate
Trigonometrical.
cose sing 0004 O
ina cosa sing+C
sectas tana: —log.cosx+C
Circular (Inverse).
arcsino arena Tl
rer} Are cos 1 æ-arccosæ-#l-xi4+ O
| us tana | w-aretana—plog(1-+a%)}4+C

r = = 2
Do y —> f yde
Eyperbolio.
cosh as sinhæ coshao+ O
sink cosh a sinhæ+C
sechtar tanh a logecoshe+O |
‘Miscellaneous.
1
ra loge(æ+a)+C
æ
<< log. (æ+Var+af)+ C
ri Ze a)
poe + +
ua) 3, l08.(at5a)+0
Bar LE
rat Vars
ac aze Loos aut 0
san Y sin an+C
asta tanda — Diag. 00s a+ 0
in 2
sin 200 site +0
ein? costar sc
sin + cos sine A [sin tude +O]
Jose logetan 5 +0
sinto 2
sin du —cotano+ O
logetan a+ 0
| }cos(m - nj - Feos(m4n)a+C]
Sosa | wer gie, c
—20. sin 2ax cota sin Sang

ANSWERS,

Exercises I. (p. 25)
Ouanna,
© Set aps, © Bobet © au Bek,
mW, © Pesar,

© BT. (10) Y RE,

Exereises IL. (p. 33.)
0) Besant. © ars xg — (9) Wear
a, (6) Es =2:30t.

ae _,
CD f= 0000012 x ly.

(e) Lab Vi, 098, 300 and 7:47 candle power per volt

respectively.
du__ 1 dn__ 1, gT
ob N

dn__ 1 [GT dn_1 fg
DNS ar

ANSWERS 289
(10) Rate of change of P when $ varies_ D,
Rate of change of P when D varios &

(1) 2e ar, ml, Zur, Bar, dart, as E 2

Exercises III. (p. 46.)

CROSS CE CE

24
(0) 8x4 ban + daR.

o Waa, oz

(1) 1411070 —G5L0LwS—22441*4 819204 1379.

fo) ee 2448. (G) 185-9022654.0% 4 15436984.
bi, Gar $6304 8208

Mm fx? ® (+0 +2)

(o) ‚ad=be (10) GRE bnew,

(cut dy?
(11) b+ 208,
12 ‚nlar 2), aid rare),
(12) Rıla+2bt) Rdarb;) E Bas
(13) 1-4840(0-000014¢—0-001024), —0-00117, —0-00107, —0-00097.

+0

Exercises IV. (p- 51.)

(1) 74210; 2.

a
+2

rd 5
O Let
(4) (Exercises ITT):
oO @ ne MO bal tne G) 2a, 0.
© 2,0. @ bx+6a, 8.
eat. 1

a
Lans
vers

290 ANSWERS

(2) -b, 0. (3) 2,0.
(4) 6644028 — 1962122 — 44880048192,
1693202 — 3924242 — 4488.
(5 2,0. (6) 371804532, 371:80403,
a _ mo
Gx+9 Garat
rmnpler pa:
ES O
2 1056 299888 16.
Fr

m eb¥a 18bYa_ Ban,

(4) 8108 — 64884479520 — 139-0688 + 2664.
32408 — 19448? +959-04¢ ~ 130-968,

(6) 12042, 12. (6) Ga? 9m, 12094
BE Er,
ee)

Exercises V. (p.64)
(2) 64; 147-2; and 0:32 feet: per second.
(3) #=a-gt; &=-9. (4) 451 feet per second.
(5) 124 feet per second per second. Yes.
(6) Angular velocity=112 radians per second; angular ac.
eeleration=9'6 radians per second per second.
(1) v=2048—108. a=408f. 1728 in./sec, 1224 in./seot.

a
Ov VG)

ET a

© v=08- any an ern > 0-7928 and 0-00211.
(10) n=2, n=1L.

ANSWERS 29%
Exercises VL (p.73)

E e
Via Sas o -
EE
9-2 Oo
jota) (4) 2afe—a),
© (40 rra? © exer
(8) iy €) CONTA

Exercises VIL. (p. 76.)
2
CEE
NEN sn (V8 +414 V8 +52)"
ph _ He _
du

Exercises VIII. (p. 91.)

@

(2) 144,
(1) Y 3040; and the numerical values are:
3, 3, 6, and 15.
(6) 402.
(o W=-42 slope is zero whore æ=0; and is +7

y
where &=1.

( m=4, n=-2.
(8) Intersections at a=
(9) Intersection at x:
110) w= 3, y=2},

292 ANSWERS

Exercises IX. (p. 109.)
(1) Min.: &=0, y=0; max: 2=-2, y
@) a=a. (4) 2543 square inch

AA AR
CEE Ot ni
(6) Max. for x= — 1; min, for æ=1.
(7) Join the mide points of the four sides,

@ r=3R, eE no max,

@) r= Bas, ir r=08506 R.
(10) At the rate of 8 7 Square feet per second,

as)

Exercises X. (p.118)
© Max: w= -219, y=2419; min. : 22158, y=—198,

(8) (a) One maximum and two minima.
(b) One maximusn, (e=0; other points unreal.)

T1, y (6) Max: = 5,
418, y

Tals, y= 1707.

3505, y=212

= +3965, y=788.
aN, 01 = afte
(8) Ot, 00N. (9) e Wi

iles per hour. “Lime taken 115'47 hours
120

(20) Speed 8:56 nautical
Minimum cost.

ANSWERS 293

(12) Max. and min. for &=75, y>45414 (See example
no, 10, p. 72)
(12) Min.: =$, y=025; max. : &= 4, y=1408.

Exercises XI. (p. 190)

Orte gr

1 i 2
Mater
orton SE

Teer) Wa) TERN)

1 2041 2 1-2

Mayta O try

3 , et 11, 2
U) zittert ara

1 1 $
2) gery AREA

AA
MD ID A

5 A 1 1
5) Ss ni

EEE
08) a ter

7 73
E ae O

1 1 =
O EEE

294 ANSWERS

Exercises XIL (p.153;

a) abler+c=). CEE @) logar.
n
6) apura, OF
DÉS (8) ee
ae tt 120/21
re (10) we
a A (12) az(aat-L400logaa).
(14) Min. : y=07 for x:=0604.
(15) te. (16) À Cog. cae

Exercises KIL. (p. 162)
© Lot few (
(2) T=34627 ; 15946 minutes.

(2) Take 2t=25 and use the Table on page 159.

6) @ a. +logen)s (1) Bl); (0 ka (Loge).
(6) 014 second. (D (a) 1'642; (6) 1558,

(8) 1000087, 314.

(9) £ is 63:47 of day 22155 kilometers,

(10) Working as accurately as possible with a table of four-dgure
logarithms, #=0'1330, 0-1445, 0-1563, mean-01446;
percentage errors :—102%, practically nil, 4710 %.

(1) Min. for a=!. (12) Max. for z=e.

(29) Min. for as=logua-

= 8x), and uso the Table on page 15%

ANSWERS 295
Exercisos XIV. (p 173°
aw DA co (0-2);
60 asin 6cos é=sin20 and Y 2000205
0 SY =Bein?9c060 and GH —3.c0030.

€) 0=45° or Trndians, (8) wn -nsinient.

sino,

(4) a log.a cos añ. © ©) sx

(6) 18200 (:04-26°).

= — tana.

(7) The slope is HY =100000 (0167) which is a maximum

when (9-15*)=0, or 6=15" ; the value of the slope
being then=100. When 9=75° the slope is
100 cos(75° — 15°)=100 cos 60° = 100 x $=50.

(8) cos bain 29-+2.cos 20 sin G=2sin O(cos*9+c0320)
=2sin G(3 cos? @— 1).

©) amor tanm-1(9r) sector,
(10) e(sintar+sin 2); (sino +2 ein 22-42 c08 2).

ab
ray

me o 6 Es ED gy
ay @ We scoxtan es
Gi) Le Gi) BL
CEE Greco =D,

a Bog 6(20-+3)'*cos 284372,

296 ANSWERS

ay Y SY 84200 (04+3)- log. 3(cos Ox 320 +30)
(15) cos = 2086: y= 2050; ismax. for +0, min. for-@

Exercises XV. (p. 180.)

QD) tay 2Y% 3-20 day.
0) Seys+ pete y +2;
Boy oa + act + DU à

eoye tasty ay + ae.
O He-a}q-0+6-0- Ha dr), 2,

(4) dy=vw-du+urlog ado.

(5) dy=3sin vudu+u5cosu do,
nain 0 toos de + (in Top ei da,

dyn) Ldu-togu dau.

(7) Minimum for a=y=—4.
(8) (2) Length 2 feet, width=depth=1 foot, vol.=2 cubie

(&) Badius=2 foct=1:46 in, lengths? feot, vol= 254
(9) All three parts equal ; the product is maximun,

(10) Minimum =é for æ=y=1.
:307 for a=}, yo.

(11) Minimum
(12) Angle at apex=90° ; equal sides=length= YF
Exercises XVI, (p. 190.)
(3) 02094,

a1. €) voran
(4) (a) y=la!+C 5 @ y=sinx+C.

6) yart+3n+C.

ANSWERS 297
Exercises XVIL (p. 205.)

a Stig aja wire
(0) + 004+ (5) 2040.
© rara O. ERAS
® Erbe gp ete by division, Therefore the answer
is% a+ (a"+a)loge(ot a) +C.
(800 pages 199 and 201,)
OT aer (0 À
(1) aaa q) + O. (12) —$00s 0-30+C.
(13) 8212208 , q, ay 2-22, 0.
(as) fc a9 390.

(17) dog(1+2) +0. (18) log, (L-@)+C.

Exercises XVIIL (p.294)
(1) Area=40; mean ordinate=10,
€) Area=3 of ax 2aVa.
(8) Area=2; mean ordinato= 20697.
(4) Area= 1:57; mean ordinate=0"5,
©) 0973, 00478. (6) Volume=meh,

(D 1:25, @ 98
(9) Volume = 4-945 ; (trom 0 to m).

og
00) alog.a, =“ log.

(12) Arithmetical mean =v ; quaaranc mean =10'85,

298 ANSWERS

(13) Quadratio mean= JADE Az; arithmetical mean =0,

‘The first involves a somewhat difficult integral, and may be
stated thus: By definition the quadratie mean will be

NE [Asin + À soin 30d.

Now the integration indicated by
farias ‚Assin a: sin 30 +A sin?3x) de
is more readily obtained if for sin! we write
T= cos 2e
2
For 2sin æ sin 3% we write cos 2a: — cos das ; and, for sin*30s,
1 — cos 6a
Lee,
Making these substitutions, and integrating, we get (see
Oa 202)
sin 22 ‘sin 2% e 4x Ag ff, sin 6a"
Alo) (eto ant Ala te)

At the ER limit the ace of 0 in causes all
this to vanish, whilst at the upper limit the substitu:
tion of 27 for a gives Ayır+ Ag’, And hence the
answer follows,

(14) Area is 626 square units. Mean ordinate is 1042,

(16) 4362. (This solid is pear shaped.)

Exercises XIX. (p.233)

WEE Css. (oa pro

yo“

e leo Jo mera
©) sin (log, a) +C. (6) e@®-20+2)+0.

ANSWERS 299

o ay (og, + O. (8) log, (log,æ)+C.
(9) Blog, (2-1) + Blog, (@+2)+ 0.

(10) log, (@-1)-+ Log, (a 2)-+ y log, (ar+3)+ Co

u =
ay ña mt e. as) log ETO.
(13) Hog EE +4 arc tan a+ O,

(9 Jong NE

a)

You had better differentiate now the answer and work
back to the given expression as a check.

(Let Law; then, in the result,

Exercises XX. (p. 263)

{D r=W3, «=
@) w= 40388, y

2 m8 (2) r=283, a,
47. (4) r=,

3
(©) 7-20, %,=20+30, ne when 0=0, 21-20, yy =0
a

(6) When 2=0, r=y, =infinity,
When w= +09, 36, 2 =—221, y= +201.
When x= -09, 36, = +221, y= - 201.

(7) When æ=0, P=141, a,
When 2=1, 7=1%41, 24=0, Y
Minimw 5.

(8) For = 9, r=1129, 1098, y= — 272,
For x=0, =, =Y, = infinity.

For v=1, 067, y =-017.

(9) w= -033, y:

(10) r=1, &=2, y=0 for all points, A circle,
(11) When æ=0, 7=186, 2,=167, y¿=017.
"59, y, 098.
, yy=1 for zero curvature.

0.

300 ANSWERS
x =
(12) When 6=3, r=1, 2-7, i= 0.
When 0-7, 7-2598, 12285, = - 141,

4
(14) When 9=0, r=1, 2=0, yı=0.
When

598, 22,=0"7146, y, =- 1:41.

When =yı=ininity.

ar: CRE A where &=0, r=É, a =0, u

Exercise XXI. (p. 280)

G) s=948, @ s=(ta%j# =.
2
® a= [ Jide =[ TFT tog, Ge + Tr],
=404.

(6) s=a(0,-0). (Ms

@s-[Nırlae and s-av5+ate (HN?)

es

PFI +4 log, (001) +W2-1+ Tj and
3=680, ag
Y
aa “he Ke a v-;
in the second; this leads to
a Ting Ta ta

in the first and

a) 8-20 sin ao and a=85,

ANSWERS au
3) san ig, (url) ana

8 + loge (1 + Ds
8 NO
CRE) .

a9 a> f NTFS de. Let 1+180=2, express 8 in termnof à

and integrate between the values of # corresponding
toxw=1and x=2 8=527.

as) sE

Every earnest student is exhorted to manufacture more
examples for himself at every stage, so as to test his powers.
When integrating he can always test his answer by differ-
entiating it, to see whether he gets back the expression from
which he started.

There are lots of books which give examples for practice.
‘It will suffice here to name two: R. G. Blaine’s The Calculus

and its Applications, and F. ML Saxelby's À Course in Practioat
Mathematics.

(16) 4a.

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