Chapter no 11

Calculus Made Easy

MAxIMA AND MINIMA.

One of the principal uses of the process of differentiating is to find out under what conditions the value of the thing differentiated becomes a maximum, or a minimum. This is often exceedingly important in engi- neering 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

y = x2 − 4x + 7.

By assigning a number of successive values to x, and finding the cor- responding values of y, we can readily see that the equation represents a curve with a minimum.

x	0	1	2	3	4	5
y	7	4	3	4	7	12

These values are plotted in Fig. 26, which shows that y has appar- ently a minimum value of 3, when x is made equal to 2. But are you

sure that the minimum occurs at 2, and not at 21 or at 13 ?

1 2 3 4

Fig. 26.

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

−1

−2

−3

−4

3 4 X

Fig. 27.

Here is another example:

Let y = 3x − x2.

Calculate a few values thus:

x	−1	0	1	2	3	4	5
y	−4	0	2	2	0	−4	−10

Plot these values as in Fig. 27.

It will be evident that there will be a maximum somewhere between x = 1 and x = 2; and the thing looks as if the maximum value of y ought to be about 21 . Try some intermediate values. If x = 11 , y = 2.187; if

4 4

x = 11 , y = 2.25; if x = 1.6, y = 2.24. How can we be sure that 2.25 is the real maximum, or that it occurs exactly when x = 11 ?

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 (78) 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

= 0. 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 x that will make its y a minimum (or a maximum), first differentiate

it, and having done so, write its

as equal to zero, and then solve

for x. Put this particular value of x 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 chap- ter opens, namely

y = x2 − 4x + 7.

Differentiating, we get:

dx = 2x − 4.

Now equate this to zero, thus:

2x − 4 = 0.

Solving this equation for x, we get:

2x = 4,

x = 2.

Now, we know that the maximum (or minimum) will occur exactly when x = 2.

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

y = 22 − (4 × 2) + 7

= 4 − 8 + 7

= 3.

Now look back at Fig. 26, and you will see that the minimum occurs when x = 2, and that this minimum of y = 3.

Try the second example (Fig. 24), which is

y = 3x − x2. dy

Differentiating, = 3 − 2x. dx

Equating to zero,

3 − 2x = 0,

whence x = 11 ;

and putting this value of x into the original equation, we find:

y = 41 − (11 × 11 ),

2 2 2

y = 21 .

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

Now, before we go on to any further cases, we have two remarks to

make. When you are told to equate

to zero, you feel at first (that

is if you have any wits of your own) a kind of resentment, because you

know that

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

= 0,

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 equations 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 x has

when the curve is neither sloping up nor sloping down, that is, at the

particular place where

= 0. So, writing

= 0 does not mean that

it always is = 0; but you write it down as a condition in order to see

how much x will come out if

is to be zero.

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 x that you thereby find is going to give you a maximum value of y or a min- imum value of y. Quite so. It does not of itself discriminate; it finds for you the right value of x but leaves you to find out for yourselves whether the corresponding 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:

y = 4x +

Without stopping to think what curve it corresponds to, differenti- ate it, and equate to zero:

− −

dy = 4 x−2 = 4 1

dx x2

= 0;

whence x = 1 ;

and, inserting this value,

y = 4

will be either a maximum or else a minimum. But which? You will hereafter be told a way, depending upon a second differentiation, (see Chap. xII., p. 109). But at present it is enough if you will simply try any other value of x 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.

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 a 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 be cut into two parts be called n. Then if x is one part, the other will be n − x, and the product will be x(n − x) or nx − x2. So we write y = nx − x2. Now differentiate and equate to zero;

dx = n − 2x = 0

Solving for x, we get

= x.

So now we know that whatever number n 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 = 1 n2.

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

Test Case.

Let us at once apply our knowledge to a case that we can test. Let y = x2 − x;

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

= 2x − 1.

Equating to zero, we get

2x − 1 = 0,

whence 2x = 1,

or x = 1 .

That is to say, when x is made = 1 , the corresponding value of y

will be either a maximum or a minimum. Accordingly, putting x = 1

in the original equation, we get

y = ( 1 )2 − 1 ,

2 2

or y = −1 .

Is this a maximum or a minimum? To test it, try putting x a little bigger than 1 ,—say make x = 0.6. Then

y = (0.6)2 − 0.6 = 0.36 − 0.6 = −0.24,

which is higher up than −0.25; showing that y = −0.25 is a minimum.

Plot the curve for yourself, and verify the calculation.

‌Further Examples.

A most interesting example is afforded by a curve that has both a maximum and a minimum. Its equation is:

y = 1 x3 − 2x2 + 3x + 1.

Now

= x2 − 4x + 3.

−1

−2

−3

−4

2 3 4 5 X

Fig. 28.

Equating to zero, we get the quadratic,

x2 − 4x + 3 = 0;

and solving the quadratic gives us two roots, viz.



x = 3

x = 1.

‌Now, when x = 3, y = 1; and when x = 1, y = 21 . The first of

these is a minimum, the second a maximum.

The curve itself may be plotted (as in Fig. 28) from the values calculated, as below, from the original equation.

−1

−43

A further exercise in maxima and minima is afforded by the follow- ing example:

The equation to a circle of radius r, having its centre C at the point whose coordinates are x = a, y = b, as depicted in Fig. 29, is:

(y − b)2 + (x − a)2 = r2.

Fig. 29.

This may be transformed into

y = √r2 − (x − a)2 + b.

Now we know beforehand, by mere inspection of the figure, that when x = a, y will be either at its maximum value, b + r, or else at its minimum value, b − r. But let us not take advantage of this knowledge; let us set about finding what value of x will make y a maximum or a minimum, by the process of differentiating and equating to zero.

dy 1 1

= √r2 2 × (2a − 2x),

dx 2 − (x − a)

which reduces to

√

dy a − x

dx = r2 − (x − a)2 .

Then the condition for y being maximum or minimum is:

√ = 0.

a − x r2 − (x − a)2

Since no value whatever of x will make the denominator infinite, the only condition to give zero is

x = a.

Inserting this value in the original equation for the circle, we find

y = √r2 + b;

and as the root of r2 is either +r or −r, we have two resulting values of y,



y = b + r

y = b − r.

The first of these is the maximum, at the top; the second the mini- mum, at the bottom.

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 = ax3 + bx + c.

Then

= 3ax2 + b.

Equating this to zero, we get 3ax2 + b = 0,

x2 = −b, and x = r−b, which is impossible.

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.

What are the sides of the rectangle of maximum area inscribed in a circle of radius R?
√

If one side be called x,

the other side = (diagonal)2 − x2;

and as the diagonal of the rectangle is necessarily a diameter, the other side = √4R2 − x2.

Then, area of rectangle S = x√4R2 − x2,

√

dS

dx = x ×

d √4R2 x2

—

+

dx

4R2 − x2 ×

d(x)

.

dx

If you have forgotten how to differentiate √4R2 − x2, here is a hint:

—

write 4R2 x2 = w and y = √w, and seek dy

dw

dw

and ; fight it out,

dx

and only if you can’t get on refer to page 66.

You will get

dS

x √

4R2 − 2x2

× −√

— √

= x               +

dx 4R2 − x2

4R2 x2 =           .

4R2 − x2

For maximum or minimum we must have

—

4R2 2x2

√4R2 − x2 = 0;

that is, 4R2 − 2x2 = 0 and x = R√2.

The other side = √4R2 − 2R2 = R√2; the two sides are equal; the

figure is a square 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.
What is the radius of the opening of a conical vessel the sloping

side of which has a length l when the capacity of the vessel is greatest? If R be the radius and H the corresponding height, H = √l2 − R2.

Volume V = πR2 ×

H

×

= πR2

3

√l2 R2

—

.

3

Proceeding as in the previous problem, we get

× −

dV

= πR2

dR

R

3√l2 − R2

+ 2πR √l2 R2

—

3

− −

2πR(l2 R2) πR3

= 3√l2 − R2 = 0

q

for maximum or minimum.

3

Or, 2πR(l2 − R2) − πR2 = 0, and R = l 2 , for a maximum,

obviously.
Find the maxima and minima of the function
x

y =

4 − x

+ 4 − x.

x

We get

dy (4 − x) − (−x) −x − (4 − x)

= + = 0

dx (4 − x)2 x2

for maximum or minimum; or

4 4

(4 − x)2 − x2 = 0 and x = 2.

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

For

x = 2,

y = 2,

for

x = 1.5,

y = 2.27,

for

x = 2.5,

y = 2.27;

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

function.)
Find the maxima and minima of the function y = √1 + x +

√1 − x. (It will be found instructive to plot the graph.) Differentiating gives at once (see example No. 1, p. 67)

dy 1 1

dx = 2√1 + x − 2√1 − x = 0

for maximum or minimum.

Hence √1 + x = √1 − x and x = 0, the only solution For x = 0, y = 2.

For x = ±0.5, y = 1.932, so this is a maximum.
Find the maxima and minima of the function
x2 − 5

y = .

2x − 4

We have

dy (2x − 4) × 2x − (x2 − 5)2

dx (2x − 4)

2 = 0

=

for maximum or minimum; or

—

2x2 8x + 10

(2x − 4)2 = 0;

or x2 − 4x + 5 = 0; which has for solutions

2

x = 5 ± √−1.

These being imaginary, there is no real value of x for which hence there is neither maximum nor minimum.
Find the maxima and minima of the function
(y − x2)2 = x5.

dy

= 0;

dx

This may be written y = x2

5

± x2 .

dy 5 3

dx = 2x ± 2 x2 = 0 for maximum or minimum;

5 1 5 1

that is, x(2± 2 x2 ) = 0, which is satisfied for x = 0, and for 2± 2 x2 = 0,

25

that is for x = 16 . So there are two solutions.

Taking first x = 0. If x = −0.5, y = 0.25 ± √2 −(.5)5, and if

x = +0.5, y = 0.25 ± √2 (.5)5. 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 curve goes 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

dy

condition for a minimum is satisfied, namely

= 0. It is necessary

dx

therefore always to check by taking one value on either side.

y

0.3

0.2

0.1

0 0.2 0.4 0.6 0.8

1

x

Fig. 30.

25

‌Now, if we take x = 16 = 0.64. If x = 0.64, y = 0.7373 and

y = 0.0819; if x = 0.6, y becomes 0.6389 and 0.0811; and if x = 0.7,

y becomes 0.8996 and 0.0804.

This shows that there are two branches of the curve; the upper one does not pass through a maximum, but the lower one does.
A cylinder whose height is twice the radius of the base is increas- ing in volume, so that all its parts keep always in the same proportion to each other; that is, at 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 the rate of 20 square inches per second; at what

rate is its volume then increasing?

Area = S = 2(πr2) + 2πr × 2r = 6πr2.

Volume = V = πr2 × 2r = 2πr3.

dS dV

= 12πr,

dr dr

= 6πr2,

dS = 12πr dr = 20, dr =

12πr

dV = 6πr2 dr = 6πr2 20 = 10r.

12πr

The volume changes at the rate of 10r cubic inches.

Make other examples for yourself. There are few subjects which offer such a wealth for interesting examples.

‌Exercises IX. (See page 257 for Answers.)‌

What values of x will make y a maximum and a minimum, if
x2

y = ?

x + 1
What value of x will make y a maximum in the equation y =
x

a2 + x2 ?

4
A line of length p is to be cut 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 1 p.
‌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?
Plot the curve corresponding to the equation
10

y = +

x

dy

10

;

8 − x

also find

, and deduce the value of x that will make y a minimum;

dx

and find that minimum value of y.
If y = x5 − 5x, find what values of x will make y a maximum or a minimum.
What is the smallest square that can be inscribed in a given square?
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; (b) whose lateral area is a maximum; (c) whose total area is a maximum.
Inscribe in a sphere, a cylinder (a) whose volume is a maxi- mum; (b) whose lateral area is a maximum; (c) whose total area is a maximum.
A spherical balloon is increasing in volume. If, when its radius is r feet, its volume is increasing at the rate of 4 cubic feet per second, at what rate is its surface then increasing?
Inscribe in a given sphere a cone whose volume is a maximum.
The current C given by a battery of N similar voltaic cells is

n E

C = , where E, R, r, are constants and n is the number of

rn2

R +

cells coupled in series. Find the proportion of n to N for which the

current is greatest.

‌CHAPTER XII.‌

CURVATURE OF CURVES.

Returning to the process of successive differentiation, it may be asked: Why does anybody want to differentiate twice over? We know that when the variable quantities are space and time, by differentiating

twice over we get the acceleration of a moving body, and that in the

geometrical interpretation, as applied to curves,

d2y

means the slope of

the curve. But what can mean in this case? Clearly it means the

dx2

rate (per unit of length x) at which the slope is changing—in brief, it

is a measure of the curvature of the slope.

‌Fig. 31. Fig. 32.

Suppose a slope constant, as in Fig. 31.

Here,

is of constant value.

Suppose, however, a case in which, like Fig. 32, the slope itself is

d dy

getting greater upwards, then

dx , that is,

d2y

dx2 , will be positive.

If the slope is becoming less as you go to the right (as in Fig. 14,

p. 80), or as in Fig. 33, then, even though the curve may be going

d2y

upward, since the change is such as to diminish its slope, its will

dx2

be negative.

Fig. 33.

‌It is now time to initiate you into another secret—how to tell whether the result that 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

d2y

is positive or negative. If

dx2 comes out positive, then you know that

d2y

the value of y which you got was a minimum; but if comes out

dx2

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. 80), or like Fig. 34, where the point of minimum y is marked M , and the curve is concave upwards. To the left of M the slope is downward, that is, negative, and is getting less negative. To the right of M the slope has become upward, and is getting more and more upward. Clearly the change of slope as the

y min.

y max.

‌O

Fig. 34.

d2y

Fig. 35.

curve passes through M is such that is positive, for its operation,

dx2

as x 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. 81), or like Fig. 35, where the curve is convex, and the max- imum point is marked M . In this case, as the curve passes through M

from left to right, its upward slope is converted into a downward or

d2y

negative slope, so that in this case the “slope of the slope” is

dx2

negative.

Go back now to the examples of the last chapter and verify in this

way the conclusions arrived at as to whether in any particular case there is a maximum or a minimum. You will find below a few worked out examples.

Find the maximum or minimum of
1. y = 4x2 − 9x − 6; (b) y = 6 + 9x − 4x2;
  and ascertain if it be a maximum or a minimum in each case.
  
  dy
  
  (a)
  
  dx
  
  8
  
  d2y
  
  = 8x − 9 = 0; x = 11 , and y = −11.065.
  
  dx2 = 8; it is +; hence it is a minimum.
  
  8
  
  dy
  
  (b)
  
  dx
  
  d2y
  
  = 9 − 8x = 0; x = 11 ; and y = +11.065.
  
  dx2 = −8; it is −; hence it is a maximum.
Find the maxima and minima of the function y = x3 − 3x + 16.
dy

dx d2y

= 3x2 − 3 = 0; x2 = 1; and x = ±1.

dx2 = 6x; for x = 1; it is +;

—

hence x = 1 corresponds to a minimum y = 14. For x = −1 it is −; hence x = 1 corresponds to a maximum y = +18.
Find the maxima and minima of y = x − 1 .
x2 + 2

2 2 =

=

dy (x2 + 2) × 1 − (x − 1) × 2x

dx (x + 2)

2x x2 + 2

—

(x2 + 2)2 = 0;

or x2 − 2x − 2 = 0, whose solutions are x = +2.73 and x = −0.73.

d2y

dx2 = −

= −

(x2 + 2)2 × (2x − 2) − (x2 − 2x − 2)(4x3 + 8x) (x2 + 2)4

2 4 .

2x5 − 6x4 − 8x3 − 8x2 − 24x + 8 (x + 2)

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

If we put x = 2.73, the numerator is negative; the maximum, y = 0.183.

If we put x = −0.73, the numerator is positive; the minimum,

y = −0.683.
The expense C of handling the products of a certain factory
varies with the weekly output P according to the relation C = aP +

b

+ d, where a, b, c, d are positive constants. For what output will

c + P

the expense be least?

dC b

dP = a − (c + P )2 = 0 for maximum or minimum;

hence a = (c + P )2 and P = ±r − c.

b b

a

a

As the output cannot be negative, P = +r b − c.

Now

d2C

dP 2 = +

b(2c + 2P ) (c + P )4 ,

a

which is positive for all the values of P ; hence P = +r b −c corresponds

to a minimum.
The total cost per hour C of lighting a building with N lamps of a certain kind is

1000

C = N Cl + EPCe ,

where E is the commercial efficiency (watts per candle),

P is the candle power of each lamp,

t is the average life of each lamp in hours,

Cl = cost of renewal in pence per hour of use,

Ce = cost of energy per 1000 watts per hour.

Moreover, the relation connecting the average life of a lamp with the commercial efficiency at which it is run is approximately t = mEn, where m and n are constants depending on the kind of lamp.

Find the commercial efficiency for which the total cost of lighting will be least.

1000

We have C = N Cl E−n + PCe E ,

dC = PCe dE 1000

nCl −(n+1)

— m E = 0

for maximum or minimum.

En+1 = 1000 × nCl

mPCe

and E = n+r1

1000 × nCl .

mPCe

This is clearly for minimum, since

d2C

dE2 = (n + 1)

nCl E m

−(n+2),

which is positive for a positive value of E.

For a particular type of 16 candle-power lamps, Cl = 17 pence,

Ce = 5 pence; and it was found that m = 10 and n = 3.6.

E = 4r.6

1000 × 3.6 × 17 = 2.6 watts per candle-power.

10 × 16 × 5

‌Exercises X. (You are advised to plot the graph of any numerical example.) (See p. 258 for the Answers.)‌

Find the maxima and minima of
y = x3 + x2 − 10x + 8.

b dy

d2y

—

Given y =

x cx2, find expressions for

, and for

dx2

, also

find the value of x which makes y a maximum or a minimum, and show

whether it is maximum or minimum.

Find how many maxima and how many minima there are in the curve, the equation to which is
x2 x4

y = 1 −

+ ;

2 24

and how many in that of which the equation is

x2 x4 x6

y = 1 − 2 + 24 − 720 .
Find the maxima and minima of
5

y = 2x + 1 + x2 .
Find the maxima and minima of
3

y = x2 + x + 1 .
Find the maxima and minima of
5x

y = .

2 + x2
Find the maxima and minima of
3x x

y = x2 − 3 + 2 + 5.
Divide a number N into two parts in such a way that three times the square of one part plus twice the square of the other part shall be a minimum.
The efficiency u of an electric generator at different values of output x is expressed by the general equation:
x

u = ;

a + bx + cx2

where a is a constant depending chiefly on the energy losses in the iron and c a constant depending chiefly on the resistance of the copper parts. Find an expression for that value of the output at which the efficiency will be a maximum.
Suppose it to be known that consumption of coal by a certain steamer may be represented by the formula y = 0.3+0.001v3; 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 speed 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?
Find the maxima and minima of

6

‌y = ±x√x(10 − x).
Find the maxima and minima of

y = 4x3 − x2 − 2x + 1.

‌CHAPTER XIII.‌

OTHER USEFUL DODGES.

Partial Fractions.

‌We have seen that 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 each 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 appears in a simplified form.

Let us see how to reach this result. Try first the job of adding

two fractions together to form a resultant fraction. Take, for example,

the two fractions

x + 1

and

x − 1

. Every schoolboy can add these

together and find their sum to be

3x + 1

‌x2 − 1 . And in the same 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 every 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 only 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 x is less in the numerator than in

x2 + 2

—

the denominator. If we have to deal with such an expression as x2 1,

—

we can simplify it by division, since it is equivalent to 1 + x2 1; and

—

x2 1 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 x, and no terms in x2, x3, or any other powers of x, 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 denominators of the partial fractions of which we are in search.

3x + 1

Suppose we wish to go back from to the components which

x2 − 1

we know are

x + 1

and

x − 1

. If we did not know what those compo-

nents were we can still prepare the way by writing:

3x + 1 3x + 1

x2 − 1 = (x + 1)(x − 1) = x + 1 + x − 1 ,

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 x at all, and we can call them A, B, C . . . as we please. So, in this case, we have:

3x + 1 A B

x2 − 1 = x + 1 + x − 1 .

If now we perform the addition of these two partial fractions, we

get A(x − 1) + B(x + 1); and this must be equal to 3x + 1 .

(x + 1)(x − 1) (x + 1)(x − 1)

And, as the denominators in these two expressions are the same, the

numerators must be equal, giving us:

3x + 1 = A(x − 1) + B(x + 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. But there is another way out of this difficulty. The equation must be true for all values of x; therefore it must be true for such values of x as will cause x − 1 and x + 1 to become zero, that is for x = 1 and for x = −1 respectively. If we make x = 1, we get 4 = (A×0) +(B ×2), so that B = 2; and if we make x = −1, we get −2 = (A×−2) +(B ×0), so that A = 1. Replacing the A and B of the partial fractions by these

new values, we find them to become done.

x + 1

and

x − 1

; and the thing is

—

4x2 + 2x 14

As a farther example, let us take the fraction x3 + 3x2 − x − 3 . The

denominator becomes zero when x is given the value 1; hence x − 1 is a factor of it, and obviously then the other factor will be x2 + 4x + 3;

and this can again be decomposed into (x + 1)(x + 3). So we may write the fraction thus:

4x2 + 2x − 14 A B C

x3 + 3x2 − x − 3 = x + 1 + x − 1 + x + 3 ,

making three partial factors.

Proceeding as before, we find

4x2 + 2x − 14 = A(x − 1)(x + 3) + B(x + 1)(x + 3) + C(x + 1)(x − 1).

Now, if we make x = 1, we get:

−8 = (A × 0) + B(2 × 4) + (C × 0); that is, B = −1.

If x = −1, we get:

−12 = A(−2 × 2) + (B × 0) + (C × 0); whence A = 3.

If x = −3, we get:

16 = (A × 0) + (B × 0) + C(−2 × −4); whence C = 2.

So then the partial fractions are:

3 1 2

x + 1 − x − 1 + x + 3 ,

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

Case II. If some of the factors of the denominator contain terms in x2, and are not conveniently put into factors, then the corresponding numerator may contain a term in x, as well as a simple number; and hence it becomes necessary to represent this unknown numerator not by the symbol A but by Ax + B; the rest of the calculation being made as before.

− −

x2 3 Try, for instance: (x2 + 1)(x + 1) .

− −

x2 3 Ax + B C

= + ;

(x2 + 1)(x + 1) x2 + 1 x + 1

2 2

−x − 3 = (Ax + B)(x + 1) + C(x + 1).

Putting x = −1, we get −4 = C × 2; and C = −2;

hence −x2 − 3 = (Ax + B)(x + 1) − 2x2 − 2; and x2 − 1 = Ax(x + 1) + B(x + 1).

Putting x = 0, we get −1 = B; hence

x2 − 1 = Ax(x + 1) − x − 1; or x2 + x = Ax(x + 1); and x + 1 = A(x + 1),

so that A = 1, and the partial fractions are:

x − 1

x2 + 1

— x + 1 .

Take as another example the fraction

—

x3 2

(x2 + 1)(x2 + 2) .

We get

x3 − 2

Ax + B

Cx + D

(x2 + 1)(x2 + 2) = x2 + 1 + x2 + 2

(Ax + B)(x2 + 2) + (Cx + D)(x2 + 1)

= (x2 + 1)(x2 + 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 found 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 expressions as those with which we are dealing here, the coefficients of the same powers of x are equal and of same sign.

Hence, since

x3 − 2 = (Ax + B)(x2 + 2) + (Cx + D)(x2 + 1)

= (A + C)x3 + (B + D)x2 + (2A + C)x + 2B + D,

we have 1 = A + C; 0 = B + D (the coefficient of x2 in the left expression being zero); 0 = 2A + C; and −2 = 2B + D. Here are four equations, from which we readily obtain A = −1; B = −2; C = 2;

2(x + 1) x + 2

D = 0; so that the partial fractions are − . This method

x2 + 2 x2 + 1

can always be used; but the method shown first will be found the

quickest in the case of factors in x only.

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

fraction

3x2 2x + 1

—

(x + 1)2(x − 2) we must allow for the possible existence of a

denominator x + 1 as well as (x + 1)2 and (x − 2).

It maybe thought, however, that, since the numerator of the fraction the denominator of which is (x + 1)2 may contain terms in x, we must allow for this in writing Ax + B for its numerator, so that

3x2 − 2x + 1 Ax + B C D

(x + 1)2(x − 2) = (x + 1)2 + x + 1 + x − 2 .

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 three relations connecting them, yet

3x2 − 2x + 1 x − 1 1 1

(x + 1)2(x − 2) = (x + 1)2 + x + 1 + x − 2 .

But if we write

3x2 − 2x + 1 A B C

= + + ,

(x + 1)2(x − 2) (x + 1)2 x + 1 x − 2

we get

3x2 − 2x + 1 = A(x − 2) + B(x + 1)(x − 2) + C(x + 1)2,

which gives C = 1 for x = 2. Replacing C by its value, transposing, gathering like terms and dividing by x − 2, we get −2x = A + B(x + 1), which gives A = −2 for x = −1. Replacing A by its value, we get

2x = −2 + B(x + 1).

Hence B = 2; so that the partial fractions are:

2 2 1

x + 1 − (x + 1)2 + x − 2 ,

instead of

x + 1

+ x − 1

(x + 1)2

x − 2

stated above as being the fractions

—

3x2 2x + 1

from which (x + 1)2(x − 2) was obtained. The mystery is cleared if we

observe that x − 1

(x + 1)2

1 can itself be split into the two fractions x + 1 −

(x + 1)2 , so that the three fractions given are really equivalent to

1 1 2 1 2 2 1

x + 1 + x + 1 − (x + 1)2 + x − 2 = x + 1 − (x + 1)2 + x − 2 ,

which are the 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 a power of a factor of x2 in the denominator, however,

the corresponding numerators must be of the form Ax+B; for example,

3x − 1 = Ax + B

Cx + D E

+ + ,

(2x2 − 1)2(x + 1) (2x2 − 1)2

which gives

2x2 − 1 x + 1

3x − 1 = (Ax + B)(x + 1) + (Cx + D)(x + 1)(2x2 − 1) + E(2x2 − 1)2.

For x = −1, this gives E = −4. Replacing, transposing, collecting like terms, and dividing by x + 1, we get

16x3 − 16x2 + 3 = 2Cx3 + 2Dx2 + x(A − C) + (B − D).

Hence 2C = 16 and C = 8; 2D = −16 and D = −8; A − C = 0 or A − 8 = 0 and A = 8, and finally, B − D = 3 or B = −5. So that we obtain as the partial fractions:

(8x − 5) + 8(x − 1) − 4 .

(2x2 − 1)2 2x2 − 1 x + 1

It is useful to check the results obtained. The simplest way is to replace x 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 quick method is as follows:

4x + 1

Taking, for example, , let x + 1 = z; then x = z − 1.

(x + 1)3

Replacing, we get

4(z − 1) + 1 = 4z − 3 = 4 − 3 .

The partial fractions are, therefore,

4 3

(x + 1)2 − (x + 1)3 .

Application to differentiation. Let it be required to differentiate

—

5 4x

y = ; we have

6x2 + 7x − 3

dx = −

(6x2 + 7x − 3) × 4 + (5 − 4x)(12x + 7) (6x2 + 7x − 3)2

− −

24x2 60x 23

= (6x2 + 7x − 3)2 .

If we split the given expression into

1 2

3x − 1 − 2x + 3 ,

we get, however,

dy 3 4

dx = −(3x − 1)2 + (2x + 3)2 ,

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. 228).

‌Exercises XI. (See page 259 for Answers.) Split into fractions:‌

3x + 5

(1)

(x − 3)(x + 4)

3x + 5

. (2) 3x − 4 .

(x − 1)(x − 2)

x + 1

(3)

x2 + x − 12 . (4)

x2 − 7x + 12 .

(5) x − 8 .

(2x + 3)(3x − 2)

—

x2 3x + 1

(7) .

(x − 1)(x + 2)(x − 3)

5x2 + 7x + 1

(8) .

(6)

x2 13x + 26

—

(x − 2)(x − 3)(x − 4)

(9)

(2x + 1)(3x − 2)(3x + 1)

x3 − 1. (10)

x4 + 1

x3 + 1.

(11)

5x2 + 6x + 4

(x + 1)(x2 + x + 1).

(x − 1)(x − 2)2 .
x
(x2 − 1)(x + 1)

. (14) x + 3 .

(x + 2)2(x − 1)

(15)

(17)

3x2 + 2x + 1

(x + 2)(x2 + x + 1)2 . (16)

—

7x2 + 9x 1

(3x − 2)4 . (18)

5x2 + 8x 12 (x + 4)3 .

—

(x3 − 8)(x − 2).

Differential of an Inverse Function.

Consider the function (see p. 13) y = 3x; it can be expressed in the‌

form x =

; this latter form is called the inverse function to the one 3

originally given.

If y = 3x,

= 3; if x =

dx 1

dy 3

, and we see that

dy 1 dy dx

dx = dx

or dx × dy = 1.

Consider y = 4x2, dy

= 8x; the inverse function is

y 2 dx 1 1 1

x = , and 2

dy = 4√y = 4 × 2x = 8x.

dy dx

Here again

dx × dy = 1.

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

dy dx dy 1

dx × dy = 1 or dx = dx .

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 differ- ential coefficient of the inverse function gives the differential coefficient of the given function itself.

As an example, suppose that we wish to differentiate y = 2

x − 1.

We have seen one way of doing this, by writing u = x − 1, and finding

and

. This gives

dx = −

2x2

x − 1

If we had forgotten how to proceed by this method, or wished to check our result by some other way of obtaining the differential coef-

ficient, or for any other reason we could not use the ordinary method,

we can proceed as follows: The inverse function is x = 1 + y2 .

dx = − 3 × 2y = − 6y ;

hence

dy (1 + y2)2

(1 + y2)2

dy 1

dx = dx

(1 + y2)2

= − 6y = −

3 2

1 + x − 1 3

r 3 = − r 3 .

6 ×

dy x − 1

2x2

x − 1

Let us take as an other example y = √3 θ + 5 .

−3

The inverse function is θ = y3

— 5 or θ = y − 5, and

dθ = −3y−4 = −3√3

(θ + 5)4.

dy 1

It follows that

erwise.

dx = −3√(θ + 5)4 , as might have been found oth-

We shall find this dodge most useful later on; meanwhile you are advised to become familiar with it by verifying by its means the results obtained in Exercises I. (p. 24), Nos. 5, 6, 7; Examples (p. 67), Nos.

1, 2, 4; and Exercises VI. (p. 72), Nos. 1, 2, 3 and 4.

You will surely realize from this chapter and the preceding, that in many respects the calculus is an art rather than a science: an art only to be acquired, as all other arts are, by practice. Hence 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.

Let 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 calcula- tion, according as the calculation is made by what 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 capital, which therefore increases by successive additions.

At simple interest. Consider a concrete case. Let the capital 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 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 cent., he would have

had to hoard for 50 years. It is easy to see that if the value of the yearly 1

interest is

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 at the end of n years, his property will be

, then,

y + n

= 2y.

‌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. 2s.; so that he starts the fourth year with 133. 2s., and so on. It is easy to work it out, and find that at the end of the ten years the total capital will have grown to 259. 7s. 6d. In fact, we see that at the end of each

year, each pound will have earned 1 of a pound, and therefore, if

this is always added on, each year multiplies the capital by 11 ; and if continued for ten years (which will multiply by this factor ten times

over) will multiply the original capital by 2.59374. Let us put this into

symbols. Put y0 for the original capital; n for the fraction added on at

each of the n operations; and yn for the value of the capital at the end of the nth operation. Then

yn = y0

1 n

1 + .

But this mode 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. At the end of half a year it ought to have been at least 105, and it certainly would have been fairer had the interest for the second half of the year been calculated on 105. This would be equivalent to calling it 5% per half-year; with 20 operations, therefore, at each of which the capital is multiplied by 21 . If reckoned this way, by the end of ten years the capital would have grown to 265. 6s. 7d.; for

(1 + 1 )20 = 2.653.

But, even so, the process is still not quite fair; for, by the end of the first month, there will be some interest earned; and a half-yearly reckoning assumes that the capital remains stationary for six months at 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

100

y = £100 1 + 1 100 ;

which works out to 270. 9s. 71 d.

Even this is not final. Let the ten years be divided into 1000 periods, each of 1 of a year; the interest being 1 per cent. for each such period;

then

100

1000

y = £100 1 + 1 1000 ;

which works out to 271. 13s. 10d.

Go even more minutely, and divide the ten years into 10, 000 parts,

each

1 1000

of a year, with interest at 1

of 1 per cent. Then

100

10,000

y = £100 1 + 1 10,000 ;

which amounts to 271. 16s. 31 d.

Finally, it will be seen that what we are trying to find is in reality

the ultimate value of the expression

1 n

1 + , which, as we see, is

greater than 2; and which, as we take n larger and larger, grows closer

and closer to a particular limiting value. However big you make n, the value of this expression grows nearer and nearer to the figure

2.71828 . . .

a number never to be forgotten.

Let us take geometrical illustrations of these things. In Fig. 36,

OP stands for the original value. OT is the whole time during which

the value is growing. It is divided into 10 periods, in each of which

there is an equal step up. Here

is a constant; and if each step up

1 of the original OP , then, by 10 such steps, the height is doubled.

If we had taken 20 steps, each of half the height shown, at the end the

height would still be just doubled. Or n such steps, each of

of the

original height OP , would suffice to double the height. This is the case

of simple interest. Here is 1 growing till it becomes 2.

In Fig. 37, we have the corresponding illustration of the geometrical

progression. Each of the successive ordinates is to be 1 +

n + 1

, that is,

times as high as its predecessor. The steps up are not equal,

O 1 2 3 4 5 6 7 8 9 T

Fig. 36.

because each step up is now

‌

of the ordinate at that part of the curve.

If we had literally 10 steps, with 1 + 1 for the multiplying factor,

the final total would be (1 + 1 )10 or 2.594 times the original 1. But if

only we take n sufficiently large (and the corresponding

sufficiently

small), then the final value be 2.71828.

1 n

1 + to which unity will grow will

2.7182

O 1 2 3 4 5 6 7 8 9 T

Fig. 37.

‌Epsilon. To this mysterious number 2.7182818 etc., the mathemati- cians have assigned as a symbol the Greek letter ϵ (pronounced ep-

silon). All schoolboys know that the Greek letter π (called pi ) stands for 3.141592 etc.; but how many of them know that epsilon means 2.71828? Yet it is an even more important number than π!

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 call a logarithmic rate of grow- ing. Unit logarithmic rate of growth is that rate which in unit time will cause 1 to grow to 2.718281. It might also be called the organic rate of growing: because it is characteristic of organic growth (in certain circumstances) that the increment of the organism in a given time is proportional to the magnitude of the organism itself.

If 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 2.71828 . . . .

A little more about Epsilon. We have seen that we require to know

what value is reached by the expression

1 n

1 + , when n becomes

indefinitely great. Arithmetically, here are 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.

(1 + 1 )2 = 2.25.

(1 + 1 )5 = 2.488.

(1 + 1 )10 = 2.594.

(1 + 1 )20 = 2.653.

100

(1 + 1 )100 = 2.705.

1000

(1 + 1 )1000 = 2.7169.

10,000

(1 + 1 )10,000 = 2.7181.

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

Accordingly, we will avail ourselves of the binomial theorem, and

expand the expression

1 n

‌1 + in that well-known way.

The binomial theorem gives the rule that

(a + b)n

= an

an−1b

+ n 1! + n(n − 1)

an−2b2

Putting a = 1 and b =

+ n(n − 1)(n − 2)

, we get

an−3b3

+ etc.

1 +

1 n

1 n − 1 1 (n − 1)(n − 2)

= 1 + 1 +

+ 1 (n − 1)(n − 2)(n − 3) + etc.

4! n3

Now, if we suppose n to become indefinitely great, say a billion, or a billion billions, then n − 1, n − 2, and n − 3, etc., will all be sensibly equal to n; and then the series becomes

ϵ = 1 + 1 +

1 1

2! 3!

+ etc. . . .

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

	1.000000
dividing by 1	1.000000
dividing by 2	0.500000
dividing by 3	0.166667
dividing by 4	0.041667
dividing by 5	0.008333
dividing by 6	0.001389
dividing by 7	0.000198
dividing by 8	0.000025
dividing by 9	0.000002
Total	2.718281
ϵ is incommensurable with 1, and	resembles	π in being an inter-

minable 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 ex-

pression

1 nx

1 +

, which is the same as ϵx when we make n indefi-

nitely great.

1nx−1

1nx−2

1 2

ϵx = 1nx + nx n + nx(nx − 1) n

1nx−3

1 3

+ nx(nx − 1)(nx − 2) 3! + etc.

1 n2x2 − nx 1 n3x3 − 3n2x2 + 2nx

= 1 + x + 2! ·

n2 + 3! ·

—

x 3x2 2x

n3 + etc.

= 1 + x +

n +

n + n2

+ etc.

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

ϵx = 1 + x + x

x3 x4

+ +

3! 4!

+ etc. . . .

This series is called the exponential series.

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

d(ϵx) dx

= 0 + 1 +

2x 1 · 2

3x2

1 · 2 · 3

4x3

· · ·

1 2 3 4

5x4

1 · 2 · 3 · 4 · 5

+ etc.

or = 1 + x +

1 · 2

1 · 2 · 3

1 · 2 · 3 · 4

+ etc.,

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 x, such that its differential coefficient is the same as itself. Or, is there any expression, involving only powers of x, which is unchanged by differentiation? Accordingly; let us assume as a general expression that

y = A + Bx + Cx2 + Dx3 + Ex4 + etc.,

(in which the coefficients A, B, C, etc. will have to be determined), and differentiate it.

dy = B + 2Cx + 3Dx2 + 4Ex3 + etc. dx

Now, if this new expression is really to be the same as that from

B A

which it was derived, it is clear that A must = B; that C =

= ;

that D =

1 · 2 · 3

; that E =

1 · 2 · 3 · 4

, etc.

2 1 · 2

The law of change is therefore that

y = A

1 + +

1 · 2

1 · 2 · 3

1 · 2 · 3 · 4

+ etc. .

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

y = 1 +

+ 1

1 · 2

1 · 2 · 3

1 · 2 · 3 · 4

+ etc.

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

when x = 1,	y = 2.718281 etc.;	that is, y = ϵ;
when x = 2,	y = (2.718281 etc.)2;	that is, y = ϵ2;
when x = 3,	y = (2.718281 etc.)3;	that is, y = ϵ3;

and therefore

when x = x, y = (2.718281 etc.)x; that is, y = ϵx,

thus finally demonstrating that

ϵx = 1 + x +

1 · 2

+ +

1 · 2 · 3

1 · 2 · 3 · 4

+ etc.

[Note.—How to read exponentials. For the benefit of those who have no tutor at hand it may be of use to state that ϵx is read as “epsilon to the eksth power ;” or some people read it “exponential eks.” So ϵpt is read “epsilon to the pee-teeth-power ” or “exponential pee tee.” Take some similar expressions:—Thus, ϵ−2 is read “epsilon to the minus two power ” or “exponential minus two.” ϵ−ax is read “epsilon to the minus ay-eksth” or “exponential minus ay-eks.”]

Of course it follows that ϵy remains unchanged if differentiated with respect to y. Also ϵax, which is equal to (ϵa)x, will, when differentiated with respect to x, be aϵax, because a is a constant.

Natural or Naperian Logarithms.

Another reason why ϵ is important is because it was made by Napier, the inventor of logarithms, the basis of his system. If y is the value of

ϵx, then x is the logarithm, to the base ϵ, of y. Or, if

y = ϵx,

then x = logϵ y.

The two curves plotted in Figs. 38 and 39 represent these equations.



x	0	0.5	1	1.5	2
y	1	1.65	2.71	4.50	7.39

The points calculated are: For Fig. 38 

0.5 1 1.5

y = ǫ

y	1	2	3	4	8
x	0	0.69	1.10	1.39	2.08

‌For Fig. 39 

x = log

‌Fig. 39. Fig. 38.

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 calcu- lated to base 10 instead of base ϵ, are unfamiliar with the “natural” logarithms, it may be worth while to say a word about them. The or- dinary rule that adding logarithms gives the logarithm of the product still holds good; or

logϵ a + logϵ b = logϵ ab.

Also the rule of powers holds good;

n × logϵ a = logϵ an.

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 0.4343; or

log10 x = 0.4343 × logϵ x,

and conversely, logϵ x = 2.3026 × log10 x.

‌Exponential and Logarithmic Equations.

Now let us try our hands at differentiating certain expressions that contain logarithms or exponentials.

Take the equation:

First transform this into

y = logϵ x.

ϵy = x,

whence, since the differential of ϵy with regard to y is the original function unchanged (see p. 139),

dx = ϵy, dy

A Useful Table of “Naperian Logarithms”

(Also called Natural Logarithms or Hyperbolic Logarithms)

number logϵ number logϵ

1.1

1.2

1.5

1.7

2.0

2.2

2.5

2.7

2.8

3.0

3.5

4.0

4.5

5.0

1.7918

0.0000	6
0.0953	7
0.1823	8
0.4055	9
0.5306	10
0.6931	20
0.7885	50
0.9163	100
0.9933	200
1.0296	500
1.0986	1, 000
1.2528	2, 000
1.3863	5, 000
1.5041	10, 000
1.6094	20, 000

1.9459

2.0794

2.1972

2.3026

2.9957

3.9120

4.6052

5.2983

6.2146

6.9078

7.6009

8.5172

9.2103

9.9035

and, reverting from the inverse to the original function,

dy 1 1 1

dx = dx

= ϵy = x.

Now this is a very curious result. It may be written

d(logϵ x) = x−1. dx

‌Note that x−1 is a result that we could never have got by the rule for differentiating powers. That rule (page 24) is to multiply by the power, and reduce the power by 1. Thus, differentiating x3 gave us 3x2; and differentiating x2 gave 2x1. But differentiating x0 does not give us x−1 or 0 × x−1, because x0 is itself = 1, and is a constant. We shall have to come back to this curious fact that differentiating logϵ x gives

when we reach the chapter on integrating.

Now, try to differentiate

y = logϵ(x + a),

that is ϵy = x + a;

we have d(x + a) = ϵy, since the differential of ϵy remains ϵy. dy

This gives dx = ϵy = x + a; dy

hence, reverting to the original function (see p. 128), we get

dy 1

‌dx = dx

= .

x + a

‌Next try y = log10 x.

First change to natural logarithms by multiplying by the modulus 0.4343. This gives us

y = 0.4343 logϵ x;

whence

dy 0.4343

= .

dx x

The next thing is not quite so simple. Try this:

y = ax.

Taking the logarithm of both sides, we get

logϵ y = x logϵ a,

or x = logϵ y = 1 × log y.

logϵ a logϵ a

Since

logϵ a

is a constant, we get

dx 1 1 1

= ×

= ;

dy logϵ a y ax × log a

hence, reverting to the original function.

dy = 1 = ax log a. dx dx ϵ

We see that, since

dx dy

dx 1 1

1 dy

dy × dx = 1 and

dy = y × log a,

y × dx = logϵ a.

We shall find that whenever we have an expression such as logϵ y =

1 dy

a function of x, we always have

y dx

= the differential coefficient of

the function of x, so that we could have written at once, from logϵ y =

x logϵ a,

1 dy

y dx

= logϵ

a and

= ax logϵ a.

Let us now attempt further examples.

Examples.

y = ϵ−ax. Let −ax = z; then y = ϵz.

Or thus:

dy = ϵz ; dz

dx dx

dy

= −a; hence dx

= −aϵ−ax.

logϵ

1 dy y = −ax; y dx

dy

= −a; dx

= −ay = −aϵ−ax.

x2
y = ϵ 3 . Let

2

x = z; then y = ϵz. 3

Or thus:

dy = ϵz;

dz

dz 2x

= ;

dx 3

dy 2x

=

dx 3

x2

ϵ 3 .

x2 1 dy 2x

dy 2x x2

logϵ y = 3 ;

= ;

y dx 3

= ϵ 3 .

dx 3

2x

y = ϵx+1 .

logϵ

y =

x + 1

1 dy

y dx

= 2(x + 1) − 2x; (x + 1)2

dy 2

hence

dx = (x + 1)2 ϵx+1 .

Check by writing

√

x2+a

x + 1

= z.

2 1

y = ϵ

1 dy

. logϵ y = (x

x

+ a) 2 .

dy

x × ϵ√x2+a

2

=

1

y dx

(x2 + a) 1

and

= .

2

dx (x2 + a) 1

For if (x2

+ a) 2 = u and x2

1

+ a = v, u = v 2 ,

du 1

= 1 ;

dv du x

= 2x; = 1 .

dv 2v 2 dx

dx (x2 + a) 2

Check by writing √x2 + a = z.

y = log(a + x3). Let (a + x3) = z; then y = logϵ z.

dy 1 dz 2 dy 3x2

;

dz z

= 3x ; hence

dx = a + x3 .

y = logϵ{3x2 + √a + x2}. Let 3x2 + √a + x2 = z; then y = logϵ z.

dy 1 dz x

;

dz z

dx = 6x + √x2 + a ;

dy 6x + √x2 + a

x(1 + 6√x2 + a)

dx = 3x2 + √a + x2 = (3x2 + √x2 + a)√x2 + a.

y = (x + 3)2√x − 2.

2

logϵ y = 2 logϵ(x + 3) + 1 logϵ(x − 2).

1 dy

=

y dx

2

+

(x + 3)

1

;

2(x − 2)

dy = (x + 3)2√x − 2 2 + 1 .

dx

2 3 3 2

x + 3

2(x − 2)

y = (x + 3) (x − 2) 3 .

logϵ y = 3 logϵ(x2 + 3) + 2 logϵ(x3 − 2);

1 dy

2x 2

3x2

6x 2x2

y dx = 3 (x2 + 3) + 3 x3 − 2 = x2 + 3 + x3 − 2 .

For if y = logϵ(x2 + 3), let x2 + 3 = z and u = logϵ z.

du 1 dz du 2x

;

dz z

= 2x;

dx = x2 + 3 .

3x2

Similarly, if v = logϵ(x3 − 2),

dx x3 − 2

and

dy 2 3 3

2 6x

2x2

y =

= (x

dx

√2 x2 + a

√3 x3 − a .

+ 3) (x

— 2) 3

x2 + 3 + x3 − 2 .

1 1

log y = log (x2 + a) − log (x3 − a).

ϵ

1 dy

2 ϵ

1 2x

3 ϵ

1 3x2 x x2

y dx = 2 x2 + a − 3 x3 − a = x2 + a − x3 − a

dy √2 x2 + a x x2

and

dx = √3 x3 − a

x2 + a − x3 − a .
y =

1

logϵ x

dy

1

logϵ x × 0 − 1 × x 1

dx = log2 x = −x log2 x.

dy

=

1

z− 2 ;

3

dz

=

1

;

dy

=

1

dz

3

dx

x

dx

3x 3

log2 x
y = √3

log

ϵ

1

x = (log x

z = log

ϵ

1

x; y = z .
y =

1 ax

ax .

ϵ ) 3 . Let ϵ 3

√

logϵ y = ax(logϵ 1 − logϵ ax) = −ax logϵ ax.

1 dy

and

y dx dy

= −ax × ax logϵ

= −

(x × ax+1 logϵ a + a logϵ ax).

1 ax

a − a logϵ

ax.

Try now the following exercises.

‌Exercises XII. (See page 260 for Answers.)‌

Differentiate y = b(ϵax − ϵ−ax).
Find the differential coefficient with respect to t of the expression
u = at2 + 2 logϵ t.
If y = nt, find d(logϵ y).
dt
Show that if y =
1 abx b · logϵ a

dy = abx. dx

,
If w = pvn, find dw .
dv

Differentiate
y = logϵ xn

. (7) y = 3ϵ

— x

x−1 .

(8) y = (3x2 + 1)ϵ−5x. (9) y = logϵ(xa + a).

y = (3x2 − 1)(√x + 1).

×
y = logϵ(x + 3). (12) y = ax xa. x + 3

It was shown by 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 signals s that can be sent per minute can be expressed by the formula
s = ay2 log

1

ϵ y ;

where a is a constant depending on the length and the quality of the materials. Show that if these are given, s will be a maximum if y = 1 ÷ √ϵ.
Find the maximum or minimum of
y = x3 − logϵ x.
Differentiate y = logϵ(axϵx).
Differentiate y = (logϵ ax)3.

The Logarithmic Curve.

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

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

Then when

x = 1, y = bp; x = 2, y = bp2; x = 3, y = bp3, etc.

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 6 ; each ordinate being 6 as high as the preceding one.

5 5

O 1 2 3 4 5 6 X

log b

O 1 2 3 4 5 6 X

log y

‌Fig. 40. Fig. 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 curve, Fig. 41, with values of logϵ y as ordinates, it would be a straight line sloping up by equal steps. In fact, it follows

from the equation, that

logϵ y = logϵ b + x · logϵ p,

whence logϵ y − logϵ b = x · logϵ p.

Now, since logϵ p is a mere number, and may be written as logϵ p = a, it follows that

logϵ b = ax,

and the equation takes the new form

y = bϵax.

The Die-away Curve.

‌If we were to take p as a proper fraction (less than unity), the curve would obviously tend to sink downwards, as in Fig. 42, where each‌

successive ordinate is 3 of the height of the preceding one.

The equation is still

y = bpx;

but since p is less than one, logϵ p will be a negative quantity, and may be written −a; so that p = ϵ−a, and now our equation for the curve takes the form

y = bϵ−ax.

The importance of this expression is that, in the case where the independent variable is time, the equation represents the course of a

1 2 3 4 5 6 X

Fig. 42.

‌great many physical processes 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

θt = θ0ϵ−at;

where θ0 is the original excess of temperature of a hot body over that of its surroundings, θt the excess of temperature at the end of time t, and a is a constant—namely, the constant of decrement, depending on the amount of surface exposed by the body, and on its coefficients of conductivity and emissivity, etc.

A similar formula,

Qt = Q0ϵ−at,

is used to express the charge of an electrified body, originally having a charge Q0, 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 ϵ−at serves 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,

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

is proportional to the height y; the curve becoming flatter

as y grows smaller. In symbols, thus

y = bϵ−ax

or logϵ y = logϵ b − ax logϵ ϵ = logϵ b − ax,

1 dy

and, differentiating,

hence

y dx = −a;

= bϵ−ax × (−a) = −ay;

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

y = bpx;

for then

= bpx × logϵ p.

But logϵ p = −a;

giving us

as before.

dx = y × (−a) = −ay,

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

— t

factor will be written ϵ T ; and it will be seen, by making t = T that the

meaning of T or of 1 is that this is the length of time which it takes for the original quantity (called θ0 or Q0 in the preceding instances) to

die away

th part—that is to 0.3678—of its original value.

The values of ϵx and ϵ−x are continually required in different

branches of physics, and as they are given in very few sets of mathemat- ical tables, some of the values are tabulated on p. 157 for convenience. As an example of the use of this table, suppose there is a hot body cooling, and that at the beginning of the experiment (i.e. when t = 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 its excess

of temperature to fall to

part of 72◦), then we can calculate to what

it will have fallen in any given time t. For instance, let t be 60 minutes.

Then

= 60 ÷ 20 = 3, and we shall have to find the value of ϵ−3,

and then multiply the original 72◦ by this. The table shows that ϵ−3

is 0.0498. So that at the end of 60 minutes the excess of temperature will have fallen to 72◦ × 0.0498 = 3.586◦.

‌x	ϵx	ϵ−x	1 − ϵ−x
0.00	1.0000	1.0000	0.0000
0.10	1.1052	0.9048	0.0952
0.20	1.2214	0.8187	0.1813
0.50	1.6487	0.6065	0.3935
0.75	2.1170	0.4724	0.5276
0.90	2.4596	0.4066	0.5934
1.00	2.7183	0.3679	0.6321
1.10	3.0042	0.3329	0.6671
1.20	3.3201	0.3012	0.6988
1.25	3.4903	0.2865	0.7135
1.50	4.4817	0.2231	0.7769
1.75	5.755	0.1738	0.8262
2.00	7.389	0.1353	0.8647
2.50	12.182	0.0821	0.9179
3.00	20.086	0.0498	0.9502
3.50	33.115	0.0302	0.9698
4.00	54.598	0.0183	0.9817
4.50	90.017	0.0111	0.9889
5.00	148.41	0.0067	0.9933
5.50	244.69	0.0041	0.9959
6.00	403.43	0.00248	0.99752
7.50	1808.04	0.00055	0.99947
10.00	22026.5	0.000045	0.999955

Further Examples.

The strength of an electric current in a conductor at a time t secs. after the application of the electromotive force producing it is given by the expression C = E 1 − ϵ− Rt .

The time constant is

— Rt

If E = 10, R = 1, L = 0.01; then when t is very large the term ϵ L

becomes 1, and C =

= 10; also

= T = 0.01. R

Its value at any time may be written:

C = 10 − 10ϵ

—

0.01 ,

the time-constant being 0.01. This means that it takes 0.01 sec. for the

variable term to fall by

= 0.3678 of its initial value 10ϵ ϵ

—

0.01 = 10.

To find the value of the current when t = 0.001 sec., say,

ϵ−0.1 = 0.9048 (from table).

= 0.1,

It follows that, after 0.001 sec., the variable term is 0.9048 × 10 = 9.048, and the actual current is 10 − 9.048 = 0.952.

Similarly, at the end of 0.1 sec.,

t = 10; ϵ−10 = 0.000045;

the variable term is 10 × 0.000045 = 0.00045, the current being 9.9995.

The intensity I of a beam of light which has passed through a thickness l cm. of some transparent medium is I = I0ϵ−Kl, where I0 is the initial intensity of the beam and K is a “constant of absorption.”

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 cms. of a certain transparent medium, this means that 82 = 100 × ϵ−K×10 or ϵ−10K = 0.82, and from the table one sees that 10K = 0.20 very nearly; hence K = 0.02.

To find the thickness that will reduce the intensity to half its value, one must find the value of l which satisfies the equality 50 = 100×ϵ−0.02l, or 0.5 = ϵ−0.02l. It is found by putting this equation in its logarithmic form, namely,

which gives

log 0.5 = −0.02 × l × log ϵ,

l = −0.3010 = 34.7 centimetres nearly.

−0.02 × 0.4343
The quantity Q of a radio-active substance which has not yet un- dergone transformation is known to be related to the initial quantity Q0 of the substance by the relation Q = Q0ϵ−λt, where λ is a constant and t the time in seconds elapsed since the transformation began.

‌For “Radium A,” if time is expressed in seconds, experiment shows that λ = 3.85 × 10−3. Find the time required for transforming half the substance. (This time is called the “mean life” of the substance.)

We have 0.5 = ϵ−0.00385t.

log 0.5 = −0.00385t × log ϵ;

and t = 3 minutes very nearly.

‌Exercises XIII. (See page 260 for Answers.)‌

— t

Draw the curve y = bϵ T ; where b = 12, T = 8, and t is given
various values from 0 to 20.
If a hot body cools so that in 24 minutes its excess of temperature has fallen to half the initial amount, deduce the time-constant, and find how long it will be in cooling down to 1 per cent. of the original excess.
‌Plot the curve y = 100(1 − ϵ−2t).
The following equations give very similar curves:
ax
1. y =
  
  ;
  
  x + b
  
  — x
2. y = a(1 − ϵ b );
  
  90◦
  
  b
3. y = a arc tan x .
  Draw all three curves, taking a = 100 millimetres; b = 30 millime- tres.
Find the differential coefficient of y with respect to x, if
ϵ .

(a) y = xx; (b) y = (ϵx)x; (c) y = xx
For “Thorium A,” the value of λ is 5; find the “mean life,” that is, the time taken by the transformation of a quantity Q of “Thorium A” equal to half the initial quantity Q0 in the expression
Q = Q0ϵ−λt;

t being in seconds.
A condenser of capacity K = 4 × 10−6, charged to a potential V0 = 20, is discharging through a resistance of 10, 000 ohms. Find the potential V after (a) 0.1 second; (b) 0.01 second; assuming that the fall
— t

of potential follows the rule V = V0ϵ KR .
The charge Q of an electrified insulated metal sphere is reduced from 20 to 16 units in 10 minutes. Find the coefficient µ of leakage, if Q = Q0 × ϵ−µt; Q0 being the initial charge and t being in seconds. Hence find the time taken by half the charge to leak away.
The damping on a telephone line can be ascertained from the re- lation i = i0ϵ−βl, where i is the strength, after t seconds, of a telephonic current of initial strength i0; l is the length of the line in kilometres, and β is a constant. For the Franco-English submarine cable laid in 1910, β = 0.0114. Find the damping at the end of the cable (40 kilometres), and the length along which i is still 8% of the original current (limiting value of very good audition).
The pressure p of the atmosphere at an altitude h kilometres is given by p = p0ϵ−kh; p0 being the pressure at sea-level (760 millimetres).

The pressures at 10, 20 and 50 kilometres being 199.2, 42.2, 0.32 respectively, find k in each case. Using the mean value of k, find the percentage error in each case.
Find the minimum or maximum of y = xx.
1
Find the minimum or maximum of y = xx .
1
Find the minimum or maximum of y = xax .

‌CHAPTER XV.‌

HOW TO DEAL WITH SINES AND COSINES.

Greek letters being usual to denote angles, we will take as the usual letter for any variable angle the letter θ (“theta”).

Let us consider the function

y = sin θ.

dθ

Fig. 43.

What we have to investigate is the value of

‌d(sin θ) dθ

; or, in other

words, if the angle θ 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 θ is the angle. Now, if θ is supposed to increase by the addition to it of the small angle dθ—an element of angle—the height of y, the sine, will be increased by a small element dy. The new height y + dy will be the sine of the new angle θ + dθ, or, stating it as an equation,

y + dy = sin(θ + dθ); and subtracting from this the first equation gives

dy = sin(θ + dθ) − sin θ.

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,

M + N

sin M − sin N = 2 cos 2

sin M − N .

If, then, we put M = θ + dθ for one angle, and N = θ for the other, we may write

dy = 2 cos θ + dθ + θ · sin θ + dθ − θ ,

2 2

or, dy = 2 cos(θ + 1 dθ) · sin 1 dθ.

‌But if we regard dθ as indefinitely small, then in the limit we may neglect 1 dθ by comparison with θ, and may also take sin 1 dθ as being

2 2

the same as 1 dθ. The equation then becomes:

and, finally,

dy = 2 cos θ × 1 dθ;

dy = cos θ · dθ, dy

= cos θ.

dθ

The accompanying curves, Figs. 44 and 45, show, plotted to scale,

‌the values of y = sin θ, and

dθ

of θ.

= cos θ, for the corresponding values

0.5

30◦ 60◦ 90◦

180◦

270◦

360◦

−0.5

−1

Fig. 44.

0.5

30◦ 60◦ 90◦ 120◦ 150◦ 180◦

270◦

‌θ

360◦

−0.5

−1

Fig. 45.

‌Take next the cosine.

−

Let y = cos θ.

Now cos θ = sin π θ .

Therefore

dy = d sin π − θ = cos π − θ × d(−θ),

− × −

dθ

dy = − cos π − θ .

And it follows that

—

= sin θ.

dθ

= cos π θ ( dθ),

Lastly, take the tangent.

Let y = tan θ,

dy = tan(θ + dθ) − tan θ.

Expanding, as shown in books on trigonometry,

tan(θ + dθ) =

tan θ + tan dθ

— ·

;

tan θ tan dθ

tan θ + tan dθ

— ·

whence dy = 1 tan θ tan dθ − tan θ

(1 + tan2 θ) tan dθ

= .

1 − tan θ · tan dθ

Now remember that if dθ is indefinitely diminished, the value of tan dθ becomes identical with dθ, and tan θ · dθ is negligibly small compared with 1, so that the expression reduces to

(1 + tan2 θ) dθ

dy = ,

so that dy = 1 + tan2 θ, dθ

or dy = sec2 θ.

dθ

Collecting these results, we have:

dθ

sin θ

cos θ

tan θ

cos θ

— sin θ

sec2 θ

Sometimes, in mechanical and physical questions, as, for example, in simple harmonic motion and in wave-motions, we have to deal with angles that increase in proportion to the time. Thus, if T be the time of one complete period, or movement round the circle, then, since the angle all round the circle is 2π radians, or 360◦, the amount of angle moved through in time t, will be

θ = 2π

, in radians,

or θ = 360

, in degrees.

If the frequency, or number of periods per second, be denoted by n,

then n =

, and we may then write:

θ = 2πnt.

Then we shall have

y = sin 2πnt.

If, now, we wish to know how the sine varies with respect to time, we must differentiate with respect, not to θ, but to t. For this we must resort to the artifice explained in Chapter Ix., p. 66, and put

dy dy dθ

dθ

Now

dt = dθ · dt .

will obviously be 2πn; so that

dt = cos θ × 2πn

= 2πn · cos 2πnt.

Similarly, it follows that

— ·

d(cos 2πnt)

= 2πn sin 2πnt.

Second Differential Coefficient of Sine or Cosine.

‌We have seen that when sin θ is differentiated with respect to θ it becomes cos θ; and that when cos θ is differentiated with respect to θ it becomes − sin θ; or, in symbols,

d2(sin θ)

dθ2 = − sin θ.

So we have this curious result that we have found a function such that if we differentiate 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 differentiating cos θ gives us − sin θ, and differentiating − sin θ gives us − cos θ; or thus:

d2(cos θ)

dθ2 = − cos θ.

Sines and cosines are the only functions of which the second differ- ential coefficient is equal (and of opposite sign to) the original function.

‌Examples.

With what we have so far learned we can now differentiate expres- sions of a more complex nature.

y = arc sin x.

If y is the arc whose sine is x, then x = sin y.

dx

= cos y.

dy

Passing now from the inverse function to the original one, we get

dy 1

dx = dx

dy

1

= .

cos y

Now cos y = q1 − sin2 y = √1 − x2;

dy 1

hence

dx = √1 − x2 ,

a rather unexpected result.
y = cos3 θ.

This is the same thing as y = (cos θ)3. Let cos θ = v; then y = v3; dy = 3v2.

dv

dv

dθ = − sin θ.

dy = dy × dv = −3 cos2 θ sin θ.

dθ dv dθ
y = sin(x + a).

Let x + a = v; then y = sin v.

dy

= cos v;

dv

dv

= 1 and

dx

dy

= cos(x + a).

dx

y = logϵ sin θ.

Let sin θ = v; y = logϵ v.

dy 1

;

dv v

dy 1

= cos θ;

dθ

y = cot θ =

dθ = sin θ × cos θ = cot θ.

cos θ

.

sin θ

2 2

dy = − sin θ − cos θ dθ sin2 θ

2 2

= −(1 + cot θ) = − cosec θ.
y = tan 3θ.

Let 3θ = v; y = tan v; dy = sec2 v.

dv

dv = 3; dy = 3 sec2 3θ.

dθ dθ
y = √1 + 3 tan2 θ; y = (1 + 3 tan2 θ
Let 3 tan2 θ = v.

1

) 2 .

1

y = (1 + v) 2 ;

dy

=

dv 2

1

√

(see p. 67); 1 + v

dv = 6 tan θ sec2 θ dθ

(for, if tan θ = u,

v = 3u2; dv du

= 6u; du = sec2 θ; dθ

hence dv = 6(tan θ sec2 θ) dθ

dy 6 tan θ sec2 θ

hence

‌dθ = 2√1 + 3 tan2 θ .

y = sin x cos x.

dx = sin x(− sin x) + cos x × cos x

= cos2 x − sin2 x.

‌Exercises XIV. (See page 261 for Answers.)‌

Differentiate the following:
−
1. y = A sin θ π .
  
  2
2. y = sin2 θ; and y = sin 2θ.
3. y = sin3 θ; and y = sin 3θ.
Find the value of θ for which sin θ × cos θ is a maximum.
Differentiate y =
1

cos 2πnt.

2π
If y = sin ax, find dy .
dx
Differentiate y = logϵ cos x.
Differentiate y = 18.2 sin(x + 26◦).
Plot the curve y = 100 sin(θ − 15◦); and show that the slope of the curve at θ = 75◦ is half the maximum slope.
dy
If y = sin θ · sin 2θ, find dθ .
If y = a · tanm(θn), find the differential coefficient of y with respect to θ.
Differentiate y = ϵx sin2 x.
Differentiate the three equations of Exercises xIII. (p. 160), No. 4, and compare their differential coefficients, as to whether they are equal, or nearly equal, for very small values of x, or for very large values of x, or for values of x in the neighbourhood of x = 30.
Differentiate the following:
1. y = sec x. (ii) y = arc cos x.
  
  (iii) y = arc tan x. (iv) y = arc sec x.
  
  (v) y = tan x × √3 sec x.
Differentiate y = sin(2θ + 3)2.3.
Differentiate y = θ3 + 3 sin(θ + 3) − 3sin θ − 3θ.
Find the maximum or minimum of y = θ cos θ.

‌CHAPTER XVI.‌

PARTIAL DIFFERENTIATION.

‌We sometimes come across quantities that are functions of more than one independent variable. Thus, we may find a case where y depends on two other variable quantities, one of which we will call u and the other v. In symbols

y = f (u, v).

Take the simplest concrete case.

Let y = u × v.

What are we to do? If we were to treat v as a constant, and differentiate with respect to u, we should get

dyv = v du;

or if we treat u as a constant, and differentiate with respect to v, we should have:

dyu = u dv.

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 per- formed only partially, that is, has been performed only with respect to one of the independent variables, is to write the differential coefficients with Greek deltas, like ∂, instead of little d. In this way

∂y

= v,

∂u

∂y

= u.

∂v

If we put in these values for v and u respectively, we shall have

dyv dy

∂y



= ∂u du,

∂y

∂v

= dv, 

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

dy =

∂y

du +

∂u

∂y

dv;

∂v

and this is called a total differential. In some books it is written dy =

dy du + dy dv.

Example (1). Find the partial differential coefficients of the expres- sion w = 2ax2 + 3bxy + 4cy3. The answers are:

∂w

= 4ax + 3by.

∂x

∂w







= 3bx + 12cy2.

∂y

The first is obtained by supposing y constant, the second is obtained by supposing x constant; then

dw = (4ax + 3by) dx + (3bx + 12cy2) dy.

Example (2). Let z = xy. Then, treating first y and then x as constant, we get in the usual way

∂z = yxy−1,

∂x

∂z



∂y

= xy × logϵ x,

so that dz = yxy−1 dx + xy logϵ xdy.

Example (3). A cone having height h and radius of base r has volume V = 1 πr2h. If 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



∂V 2π

∂r = 3 rh,

∂V π

∂h 3

r2. 

The variation when both the radius and the height change is given by dV = 2πrh dV + πr2 dh.

‌3 3

Example (4). In the following example F and f denote two ar- bitrary functions of any form whatsoever. For example, they may be sine-functions, or exponentials, or mere algebraic functions of the two

independent variables, t and x. This being understood, let us take the expression

y = F (x + at) + f (x − at),

or, y = F (w) + f (v);

where w = x + at, and v = x − at.

∂y ∂F (w) ∂w ∂f (v) ∂v

Then

∂x = ∂w · ∂x + ∂v · ∂x

= F ′(w) · 1 + f′(v) · 1

(where the figure 1 is simply the coefficient of x in w and v);

and

∂2y

∂x2 = F

′′(w) + f

′′(v).

∂y ∂F (w) ∂w ∂f (v) ∂v

Also

∂t = ∂w · ∂t + ∂v · ∂t

= F ′(w) · a − f′(v)a;

and

∂2y

∂t2 = F

′′(w)a2

+ f ′′

a2;

whence

∂2y

∂t2 = a

2 ∂2y

∂x2 .

This differential equation is of immense importance in mathematical physics.

Maxima and Minima of Functions of two Independent Variables.

‌Example (5). Let us take up again Exercise Ix., p. 107, No. 4.

Let x and y be the length of two of the portions of the string. The third is 30 − (x + y), and the area of the triangle is A =

√ − − −

s(s x)(s y)(s 30 + x + y), where s is the half perimeter, 15, so that A = √15P , where

P = (15 − x)(15 − y)(x + y − 15)

= xy2 + x2y − 15x2 − 15y2 − 45xy + 450x + 450y − 3375.

Clearly A is maximum when P is maximum.

dP =

∂P ∂P

dx +

∂x ∂y

dy.

For a maximum (clearly it will not be a minimum in this case), one must have simultaneously

∂P

= 0 and

∂x

∂P

= 0;

∂y



that is, 2xy − 30x + y2 − 45y + 450 = 0,

2xy − 30y + x2 − 45x + 450 = 0.

An immediate solution is x = y.

If we now introduce this condition in the value of P , we find

P = (15 − x)2(2x − 15) = 2x3 − 75x2 + 900x − 3375.

dP

For maximum or minimum,

dx

= 6x2 − 150x + 900 = 0, which gives

x = 15 or x = 10.

Clearly x = 15 gives minimum area; x = 10 gives the maximum, for

d2P

dx2 = 12x − 150, which is +30 for x = 15 and −30 for x = 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.

The truck is a rectangular box open at the top. Let x be the length

V

and y be the width; then the depth is

. The surface area is S =

xy

2V 2V

xy + + .

x y

∂x

∂y

x2

y2

dS = ∂S dx + ∂S dy = y − 2V dx + x − 2V dy.

For minimum (clearly it won’t be a maximum here),

2V

y − x2 = 0, x −

2V

y2 = 0.

Here also, an immediate solution is x = y, so that S = x2 + 4V ,

x

dS

dx = 2x −

4V

x2 = 0 for minimum, and

x = √3 2V .

‌Exercises XV. (See page 263 for Answers.)‌

3 2

x3 y
1. Differentiate the expression − 2x y − 2y x +
  3 3
  
  with respect
  
  to x alone, and with respect to y alone.
2. Find the partial differential coefficients with respect to x, y
  and z, of the expression
  
  x2yz + xy2z + xyz2 + x2y2z2.
3. Let r2 = (x − a)2 + (y − b)2 + (z − c)2.
  ∂r ∂r ∂r
  
  ∂2r
  
  ∂2r
  
  Find the value of ∂x + ∂y + ∂z . Also find the value of ∂x2 + ∂y2 +
  
  ∂2r
  
  ∂z2 .
4. Find the total differential of y = uv.
5. Find the total differential of y = u3 sin v; of y = (sin x)u; and of
  
  y = logϵ u .
  
  v
6. Verify that the sum of three quantities x, y, z, whose product is a constant k, is maximum when these three quantities are equal.
7. Find the maximum or minimum of the function
  u = x + 2xy + y.
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 great- est volume that can be sent by post (a) in the case of a package of rectangular cross section; (b) in the case of a package of circular cross section.
9. Divide π into 3 parts such that the continued product of their sines may be a maximum or minimum.
10. Find the maximum or minimum of u =
11. Find maximum and minimum of
  ϵx+y
  
  .
  
  xy
  
  u = y + 2x − 2 logϵ y − logϵ x.
12. A telpherage bucket of given capacity has the shape of a hor- izontal isosceles triangular prism with the apex underneath, and the

opposite face open. Find its dimensions in order that the least amount of iron sheet may be used in its construction.

‌CHAPTER XVII.‌

INTEGRATION.

∫

The great secret has already been revealed that this mysterious sym- bol , which is after all only a long S, merely means “the sum of,” or “the sum of all such quantities as.” It therefore resembles that other symbol (the Greek Sigma), which is also a sign of summation. There is this difference, however, in the practice of mathematical men as to

sum of a number of finite quantities, the integral sign

is generally

the use of these signs, that while Σ is generally used t∫o indicate the

∫ ∫

used to indicate the summing up of a vast number of small quantities of indefinitely minute magnitude, mere elements in fact, that go to make up the total required. Thus dy = y, and dx = x.

1,000,000

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 1 of an inch long; or of 100 parts, each part being

100

of an inch long; or of 1, 000, 000 parts, each of which is 1 of

an 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 calculate the bigness of the thing as a whole without reckoning up the sum of a lot of small parts. The process of “integrating ” is to enable us to calculate 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 a lot of separate parts.

Consider the series:

1 + 1 + 1 + 1 + 1 + 1 + 1 + etc.

2 4 8 16 32 64

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 is 2. Think of it, if you like, as a line. Begin with one inch; add a half

1/2

1/4 1/8

Fig. 46.

inch, add a quarter; add an eighth; and so on. If at any point of the operation we stop, there will still be a piece wanting to make up the whole 2 inches; and the piece wanting will always be the same size as the last piece added. Thus, if after having put together 1, 1 , and 1 , we

2 4

stop, there will be 1 wanting. If we go on till we have added 1 , there

will still be 1 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

1024

got to pieces so small that they could not be drawn—that would be after about 10 terms, for the eleventh term is 1 . 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 18th 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 calculus 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.

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 expres- sion 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 case (2) on p. 82. Here we have the simplest of curves, a sloping line with the equation

y = ax + b.

We know that here b represents the initial height of y when x = 0,

and that a, which is the same as

, is the “slope” of the line. The line

Fig. 47.

has a constant slope. All along it the elementary triangles

have the same proportion between height and base. Suppose we were

to take the dx’s, and dy’s of finite magnitude, so that 10 dx’s made up one inch, then there would be ten little triangles like

Now, suppose that we were ordered to reconstruct the “curve,”

starting merely from the information that

= a. 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: 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

= a. And

whether we take the dy’s and dx’s as finite or infinitely small, as they

are all alike, clearly

= a, if we reckon y as the total of all the dy’s, x

and x as the total of all the dx’s. But whereabouts are we to put this

Fig. 48.

‌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 unde- termined. 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 C above O. That is, we have the equation

y = ax + C.

‌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 x grows. In

symbols this is:

= ax.

Or, to give a concrete case, take a = 1 , so that

dy = 1 x. dx

Then we had best begin by calculating a few of the values of the slope at different values of x, and also draw little diagrams of them.

When x = 0, x = 1,

x = 2,

x = 3,

dy = 0,

= 0.2,

= 0.4,

= 0.6,

x = 4,

= 0.8,

x = 5,

= 1.0.

Now try to put the pieces together, setting each so that the middle of its base is the proper distance to the right, and so that they fit together at the corners; thus (Fig. 49). The result is, of course, not a smooth 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 for a perfect curve we ought to take each dx and its corresponding dy infinitesimally small, and infinitely numerous.

Fig. 49.

∫

‌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 1 x · dx, it follows that the whole y will be equal to the sum of

all such bits as 1 x · dx, or, as we should write it, 1 x · dx.

Now if x had been constant, ∫ 1 x · dx would have been the same as 1 x ∫ dx, or 1 x2. But x began by being 0, and increases to the

Fig. 50.

∫

particular value of x at the point P , so that its average value from 0 to that point is 1 x. Hence 1 xdx = 1 x2; or y = 1 x2.

But, as in the previous case, this requires the addition of an unde- termined constant C, because we have not been told at what height above the origin the curve will begin, when x = 0. So we write, as the equation of the curve drawn in Fig. 51,

y = 1 x2 + C.

Fig. 51.

‌Exercises XVI. (See page 264 for Answers.)‌

‌Find the ultimate sum of 2 + 1 + 1 + 1 + 1 + etc.
3 3 6 12 24
Show that the series 1 − 1 + 1 − 1 + 1 − 1 + 1 etc., is convergent,
2 3 4 5 6 7

and find its sum to 8 terms.

x2 x3 x4
If logϵ(1 + x) = x − 2 + 3 − 4 + etc., find logϵ 1.3.
Following a reasoning similar to that explained in this chapter, find y,
dy dy
1. if = 1 x; (b) if
  = cos x.
If

dx 4 dx

= 2x + 3, find y.

‌CHAPTER XVIII.‌

INTEGRATING AS THE REVERSE OF DIFFERENTIATING.

Differentiating is the process by which when y is given us (as a

function of x), we can find .

‌dx

Like every other mathematical operation, the process of differentia-

tion may be reversed; thus, if differentiating y = x4 gives us dy

= 4x3;

if one begins with

= 4x3 one would say that reversing the process

would yield y = x4. But here comes in a curious point. We should get

dy = 4x3 if we had begun with any of the following: x4, or x4 + a, dx

or x4 + c, or x4 with any added constant. So it is clear that in working

backwards from

to y, one must make provision for the possibility

of there being an added constant, the value of which will be unde-

termined until ascertained in some other way. So, if differentiating xn

yields nxn−1, going backwards from dy

= nxn−1 will give us y = xn+C;

where C stands for the yet undetermined possible constant.

Clearly, in dealing with powers of x, 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

working backwards, we get

dy = xn, dx

y =

n + 1

xn+1 + C.

If differentiating the equation y = axn gives us

dy = anxn−1, dx

it is a matter of common sense that beginning with

dy = anxn−1, dx

and reversing the process, will give us

y = axn.

So, when we are dealing with a multiplying constant, we must simply put the constant as a multiplier of the result of the integration.

Thus, if

= 4x2, the reverse process gives us y = 4 x3.

But this is incomplete. For we must remember that if we had started

with

y = axn + C,

where C is any constant quantity whatever, we should equally have found

dy = anxn−1. dx

So, therefore, when we reverse the process we must always remember to add on this undetermined constant, 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

. Hitherto we have as much as

possible kept dy and dx together as a differential coefficient: henceforth

we shall more often have to separate them.

If we begin with a simple case,

dy = x2. dx

We may write this, if we like, as

dy = x2 dx.

Now this is a “differential equation” which informs us that an ele- ment of y is equal to the corresponding element of x multiplied by x2. Now, what we want is the integral; therefore, write down with the proper symbol the instructions to integrate both sides, thus:

∫ dy = ∫ x2 dx.

[Note as to reading integrals: the above would be read thus: “Integral dee-wy equals integral eks-squared dee-eks.”]

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 = ∫ x2 dx.

∫

‌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 dx’s, but all such terms as x2 dx; and this will not be the same as x2 dx, because x2 is not a constant. For some of the dx’s will be multiplied by big values of x2, and some will be multiplied by small values of x2, 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. 189 ante—when dealing with xn is “increase the power by one, and divide by the same number as this increased power.” That is to say, x2 dx will be changed* to 1 x3. Put this into the equation; but don’t forget to add the “constant of integration” C at the end. So we get:

y = 1 x3 + C.

You have actually performed the integration. How easy! Let us try another simple case.

Let

= ax12,

where a is any constant multiplier. Well, we found when differentiat- ing (see p. 27) that any constant factor in the value of y reappeared

* You may ask, what has become of the little dx at the end? Well, remember that it was really part of the differential coefficient, and when changed over to the right-hand side, as in the x2 dx, serves as a reminder that x is the independent variable with respect to which the operation is to be effected; and, as the result of the product being totalled up, the power of x has increased by one. You will soon become familiar with all this.

unchanged in the value of

. 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 = ax12 · dx,

∫ dy = ∫ ax12 · dx,

∫ dy = a ∫ x12 dx,

y = a × 1 x13 + C.

So that is done. How easy!

We begin to realize now that integrating is a process of finding our way back, as compared with differentiating. If ever, during differentiat- ing, we have found any particular expression—in this example ax12—we can find our way back to the y from which it was derived. The contrast between the two processes may be illustrated by the following remark 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 Station 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

= x2 + x3,

then dy = x2 dx + x3 dx.

There is no reason why we should not integrate each term separately: for, as may be seen on p. 34, we found that when we differentiated the sum of two separate functions, the differential coefficient was simply the sum of the two separate differentiations. So, when we work backwards, integrating, the integration will be simply the sum of the two separate integrations.

Our instructions will then be:

∫ dy = ∫ (x2 + x3) dx

= ∫ x2 dx + ∫ x3 dx y = 1 x3 + 1 x4 + C.

3 4

If either of the terms had been a negative quantity, the correspond- ing term in the integral would have also been negative. So that differ- ences 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 = xn + b. dx

This is laughably easy. For you have only to remember that when

you differentiated the expression y = ax, the result was

= a. Hence,

when you work the other way and integrate, the constant reappears

multiplied by x. So we get

dy = xn dx + b · dx,

∫ dy = ∫ xn dx + ∫ b dx,

y =

n + 1

xn+1 + bx + C.

Here are a lot of examples on which to try your newly acquired powers.

Examples.

Given
dx

= 24x11. Find y. Ans. y = 2x12 + C.
Find ∫ (a + b)(x + 1) dx. It is (a + b) ∫ (x + 1) dx

or (a + b) ∫

xdx + ∫ dx

or (a + b)

x2

+ x + C.

2

du 1 2 3
Given
dt = gt2 . Find u. Ans. u = 3 gt2 + C.

dy
dx

= x3 − x2 + x. Find y.

dy = (x3 − x2 + x) dx or

dy = x3 dx − x2 dx + xdx; y = ∫ x3 dx − ∫ x2 dx + ∫ xdx;

4

3

2

and y = 1 x4 − 1 x3 + 1 x2 + C.
Integrate 9.75x2.25 dx. Ans. y = 3x3.25 + C.

All these are easy enough. Let us try another case.

Let

= ax−1.

Proceeding as before, we will write

dy = ax−1 · dx, ∫ dy = a ∫ x−1 dx.

Well, but what is the integral of x−1 dx?

If you look back amongst the results of differentiating x2 and x3 and xn, etc., you will find we never got x−1 from any one of them as the value of dy . We got 3x2 from x3; we got 2x from x2; we got 1

from x1 (that is, from x itself); but we did not get x−1 from x0, for

∫

two very good reasons. First, x0 is simply = 1, and is a constant, and could not have a differential coefficient. Secondly, even if it could be differentiated, its differential coefficient (got by slavishly following the usual rule) would be 0 × x−1, and that multiplication by zero gives it zero value! Therefore when we now come to try to integrate x−1 dx, we see that it does not come in anywhere in the powers of x that are given by the rule:

xn dx = 1 xn+1.

n + 1

It is an exceptional case.

Well; but try again. Look through all the various differentials ob- tained from various functions of x, and try to find amongst them x−1.

‌A sufficient search will show that we actually did get

= x−1 as the

result of differentiating the function y = logϵ x (see p. 145).

Then, of course, since we know that differentiating logϵ x gives us x−1, we know that, by reversing the process, integrating dy = x−1 dx will give us y = logϵ x. But we must not forget the constant factor a 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 = a logϵ x + C.

N.B.—Here note this very remarkable fact, that we could not have integrated in the above case if we had not happened to know the cor- responding differentiation. If no one had found out that differentiating logϵ x gave x−1, we should have been utterly stuck by the problem how to integrate x−1 dx. Indeed it should be frankly admitted that this is one of the curious features of the integral calculus:—that you can’t in- tegrate anything before the reverse process of differentiating 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,

dy = a−x2 , dx

because a−x2 has never yet been found to result from differentiating anything else.

∫

Another simple case.

Find (x + 1)(x + 2) dx.

On looking at the function to be integrated, you remark that it is the product of two different functions of x. You could, you think, integrate (x+1) dx by itself, or (x+2) dx by itself. Of course you could. But what to do with a product? None of the differentiations 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

∫ (x2 + 3x + 2) dx.

And this is the same as

∫ x2 dx + ∫ 3xdx + ∫ 2 dx.

And performing the integrations, we get

1 x3 + 3 x2 + 2x + C.

3 2

Some other Integrals.

‌Now that we know that integration is the reverse of differentiation, we may at once look up the differential coefficients we already know, and see from what functions they were derived. This gives us the following integrals ready made:‌

x−1 (p. 145); ∫ x−1 dx = log x + C.

x + a

1 (p. 145); ∫ 1 dx = log (x + a) + C.

ϵx (p. 139); ∫ ϵx dx = ϵx + C. ϵ−x ∫ ϵ−x dx = −ϵ−x + C

× − ×

1 dy ϵx 0 1 ϵx −

(for if y = −ϵx , dx = − ϵ2x = ϵ ).

sin x (p. 165); ∫ sin xdx = − cos x + C.

cos x (p. 163); ∫ cos xdx = sin x + C.

Also we may deduce the following:

logϵ x; ∫ logϵ xdx = x(logϵ x − 1) + C

dy x

(for if y = x logϵ x − x, dx = x + logϵ x − 1 = logϵ x).

log10 x; ∫ log10 xdx = 0.4343x(logϵ x − 1) + C.

ax (p. 146);

∫ ax

dx =

logϵ a

+ C.

∫

cos ax; cos ax dx = 1 sin ax + C a

(for if y = sin ax,

= a cos ax; hence to get cos ax one must differen-

tiate y =

sin ax). a

sin ax; ∫ sin ax dx = − 1 cos ax + C.

Try also cos2 θ; a little dodge will simplify matters:

cos 2θ = cos2 θ − sin2 θ = 2 cos2 θ − 1;

hence cos2 θ = 1 (cos 2θ + 1),

and ∫ cos2 θ dθ = 1 ∫ (cos 2θ + 1) dθ

= 1 ∫ cos 2θ dθ + 1 ∫ dθ.

sin 2θ

+ C. (See also p. 225).

See also the Table of Standard Forms on pp. 249–251. You should make such a table for yourself, putting in it only the general functions which you have successfully differentiated and integrated. See to it that it grows steadily!

On Double and Triple Integrals.

‌In 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,

∫∫ f (x, y, ) dx dy

means that some function of the variables x and y has to be integrated for each. It does not matter in which order they are done. Thus, take the function x2 + y2. Integrating it with respect to x gives us:

∫ (x2 + y2) dx = 1 x3 + xy2.

Now, integrate this with respect to y:

∫ ( 1 x3 + xy2) dy = 1 x3y + 1 xy3,

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

∫∫ u · dx dy,

where u is some property that depends, at each point, on x and on y. This would then be called a surface-integral. It indicates that the value of all such elements as u · dx · dy (that is to say, of the value of u over a little rectangle dx 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 dimen- sions. Consider any element of volume, the small cube whose dimen- sions are dx dy dz. If the figure of the solid be expressed by the function f (x, y, z), then the whole solid will have the volume-integral,

volume = ∫∫∫ f (x, y, z) · dx · dy · dz.

Naturally, such integrations have to be taken between appropriate lim- its* in each dimension; and the integration cannot be performed unless one knows in what way the boundaries of the surface depend on x, y, and z. If the limits for x are from x1 to x2, those for y from y1 to y2, and those for z from z1 to z2, then clearly we have

∫ z2 ∫ y2 ∫ x2

volume =

f (x, y, z) · dx · dy · dz.

* See p. 206 for integration between limits.

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.

‌Exercises XVII. (See p. 264 for the Answers.)‌

Find ∫ y dx when y2 = 4ax.
∫

∫
Find 3

(x2 + a) dx. (5) Integrate 5x

∫

(4) Find

dx. (3) Find 1 x3 dx. a

2 .

— 7

Find ∫ (4x3 + 3x2 + 2x + 1) dx.
If
dy ax

= +

∫

dx 2

bx2

3

cx3

+

4

; find y.

Find

x2 + a x + a

dx. (9) Find ∫

(x + 3)3

dx.

Find ∫ (x + 2)(x − a) dx.
Find ∫ (√x + √3 x)3a2 dx.
2

3
Find ∫ (sin θ − 1 ) dθ .
Find ∫ cos2 aθ dθ. (14) Find ∫ sin2 θ dθ.

SIMPLE INTEGRATIONS 203

(15) Find ∫ sin2 aθ dθ. (16) Find ∫ ϵ3x dx.

(17) Find ∫ dx . (18) Find ∫ dx .

1 + x

1 − x

‌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.

Q B

Fig. 52.

‌Let AB (Fig. 52) be a curve, the equation to which is known. That is, y in this curve is some known function of x. Think of a piece of the curve from the point P to the point Q.

Let a perpendicular PM be dropped from P , and another QN from the point Q. Then call OM = x1 and ON = x2, and the ordinates PM = y1 and QN = y2. We have thus marked out the area PQNM that lies beneath the piece PQ. The problem is, how can we calculate the value of this area?

The secret of solving this problem is to conceive the area as be- ing divided up into a lot of narrow strips, each of them being of the width dx. The smaller we take dx, the more of them there will be between x1 and x2. 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 strip s. It will be like this: being bounded between two vertical sides,

with a flat bottom dx, and with a slightly curved sloping top. Suppose we take its average height as being y; then, as its width is dx, its area will be y dx. 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 S, meaning surface. The area of one strip will be simply a bit of the whole area, and may therefore be called dS. So we may write

area of 1 strip = dS = y · dx.

If then we add up all the strips, we get

total area S = ∫ dS = ∫ y dx.

So then our finding S depends on whether we can integrate y · dx for the particular case, when we know what the value of y is as a function of x.

∫

For instance, if you were told that for the particular curve in ques- tion y = b + ax2, no doubt you could put that value into the expression and say: then I must find (b + ax2) dx.

‌That is all very well; but a little thought will show you that some- thing more must be done. Because the area 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 be- tween limits. We suppose x to vary, and for the present purpose we do not require any value of x below x1 (that is OM ), nor any value of x above x2 (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 distinguishing 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

∫ x=x2

x=x1

y · dx

will be read: find the integral of y · dx between the inferior limit x1 and the superior limit x2.

Sometimes the thing is written more simply

∫ x2

y · dx.

Well, but how do you find an integral between limits, when you have got these instructions?

Look again at Fig. 52 (p. 204). Suppose we could find the area under the larger piece of curve from A to Q, that is from x = 0 to x = x2, naming the area AQNO. Then, suppose we could find the area under the smaller piece from A to P , that is from x = 0 to x = x1, 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 PQNM , which is what we want. Here we have the clue as to what to do; the definite integral between the two limits is the difference 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:

∫ y dx,

and, as y = b + ax2 is the equation to the curve (Fig. 52),

∫ (b + ax2) dx

is the general integral which we must find.

Doing the integration in question by the rule (p. 193), we get

bx + ax3 + C;

and this will be the whole area from 0 up to any value of x that we may assign.

Therefore, the larger area up to the superior limit x2 will be

bx + ax3 + C;

2 3 2

and the smaller area up to the inferior limit x1 will be

bx + ax3 + C.

1 3 1

Now, subtract the smaller from the larger, and we get for the area S

the value,

area S = b(x − x

1) + 3 (

x3 − x3).

This is the answer we wanted. Let us give some numerical values. Suppose b = 10, a = 0.06, and x2 = 8 and x1 = 6. Then the area S is equal to

10(8 − 6) + 0.06 (83 − 63)

= 20 + 0.02(512 − 216)

= 20 + 0.02 × 296

= 20 + 5.92

= 25.92.

Let us here put down a symbolic way of stating what we have as- certained about limits:

∫

x=x2 x=x1

y dx = y2 − y1,

where y2 is the integrated value of y dx corresponding to x2, and y1 that

corresponding to x1.

All integration between limits requires the difference between two values to be thus found. Also note that, in making the subtraction the added constant C has disappeared.

Examples.

To familiarize ourselves with the process, let us take a case of which we know the answer beforehand. Let us find the area of the triangle (Fig. 53), which has base x = 12 and height y = 4. We know beforehand, from obvious mensuration, that the answer will come 24.

Fig. 53.

‌Now, here we have as the “curve” a sloping line for which the equa- tion is

y =

The area in question will be

∫

x=12

x=0

y · dx =

x=12 x

∫

3 ·

dx.

x=0

Integrating

dx (p. 192), and putting down the value of the general 3

integral in square brackets with the limits marked above and below, we

get

area =

1 1 2

3 · 2 x

x=12

+ C

x=0

x2 x=12

+ C

6 x=0

—

122

144

= = 24. Ans.

Let us satisfy ourselves about this rather surprising dodge of cal- culation, by testing it on a simple example. Get some squared paper, preferably some that is ruled in little squares of one-eighth inch or one-

Fig. 54.

‌tenth inch each way. On this squared paper plot out the graph of the equation,

y =

The values to be plotted will be:

x	0	3	6	9	12
y	0	1	2	3	4

The plot is given in Fig. 54.

Now reckon out the area beneath the curve by counting the little squares below the line, from x = 0 as far as x = 12 on the right. There are 18 whole squares and four triangles, each of which has an area equal to 11 squares; or, in total, 24 squares. Hence 24 is the numerical value

of the integral of limit of x = 12.

dx between the lower limit of x = 0 and the higher 3

As a further exercise, show that the value of the same integral be- tween the limits of x = 3 and x = 15 is 36.

a O

Fig. 55.

Find the area, between limits x = x1 and x = 0, of the curve

y = .

∫

x + a

Area =

x=x1 x=0

y · dx =

x=x1 b

∫

x=0 x + a

= b log (x + a) x1 + C

= b logϵ(x1 + a) − logϵ(0 + a)

= b logϵ

x1 + a. Ans. a

N.B.—Notice that in dealing with definite integrals the constant C

always disappears by subtraction.

Let it be noted that this process of subtracting one part from a larger to find the difference is really a common practice. How do you find the area of a plane ring (Fig. 56), the outer radius of which is r2

Fig. 56.

‌and the inner radius is r1? You know from mensuration that the area of the outer circle is πr2; then you find the area of the inner circle, πr2;

2 1

then you subtract the latter from the former, and find area of ring

= π(r2 − r2); which may be written

2 1

π(r2 + r1)(r2 − r1)

= mean circumference of ring × width of ring.

Here’s another case—that of the die-away curve (p. 153). Find the area between x = 0 and x = a, of the curve (Fig. 57) whose equation is
∫

y = bϵ−x.

x=a

Area = b

x=0

ϵ−x

· dx.

The integration (p. 198) gives

0

= b −ϵ−x a

= b −ϵ−a − (−ϵ−0)

= b(1 − ϵ−a).

Y

b

O

a

X

p

p1

p2

O

v1

v

‌v2

Fig. 57. Fig. 58.
Another example is afforded by the adiabatic curve of a perfect gas, the equation to which is pvn = c, where p stands for pressure, v for volume, and n 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 v2 to volume v1.

∫

Here we have

area =

v=v2

v=v1

cv−n

· dv

= c

1 − n

v1−n

= c 1 (v1−n − v1−n)

0.42

− n

= −c

1 1

—

.

An Exercise.

Prove the ordinary mensuration formula, that the area A of a circle whose radius is R, is equal to πR2.

dr

r

R

Fig. 59.

Consider an elementary zone or annulus of the surface (Fig. 59), of breadth dr, situated at a distance r from the centre. We may consider the entire surface 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 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 r, that is, a length of 2πr. Then we have, as the area of the narrow zone,

dA = 2πr dr.

Hence the area of the whole circle will be:

∫

∫

A = ∫

dA =

r=R r=0

2πr · dr = 2π

r=R r=0

r · dr.

2

Now, the general integral of r · dr is 1 r2. Therefore,

2

r=0

A = 2π 1 r2 r=R;

2

whence A = πR2.

2

or A = 2π 1 R2 − 1 (0)2 ;

Another Exercise.

Let us find the mean ordinate of the positive part of the curve

y = x − x2, which is shown in Fig. 60. To find the mean ordinate, we

Y

M

1/4

N

O

1

Fig. 60.

‌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. At N the ordinate y has zero value; therefore, we must look at the equation and see what value of x will make y = 0. Now, clearly, if x is 0, y will also be 0, the curve passing through the origin O; but also, if x = 1, y = 0; so that x = 1 gives us the position of the point N .

∫

Then the area wanted is

=

x=1 x=0

(x − x2) dx

2

3

0

= 1 − 1 − [0 − 0]

= 1 x2 − 1 x3 1

2 3

6

= 1 .

But the base length is 1.

6

Therefore, the average ordinate of the curve = 1 .

[N.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 x = 0 to x = x1,

∫

is given by the expression,

1

mean y =

x1

x=x1

x=0

y · dx.

One can also find in the same way the surface area of a solid of revolution.

Example.

The curve y = x2 − 5 is revolving about the axis of x. Find the area of the surface generated by the curve between x = 0 and x = 6.

A point on the curve, the ordinate of which is y, describes a circum- ference of length 2πy, and a narrow belt of the surface, of width dx, corresponding to this point, has for area 2πy dx. The total area is

∫

x=6

2π

x=0

y dx = 2π

x=6

∫

x=0

(x2 − 5) dx = 2π

x3 6

0

3 − 5x

= 6.28 × 42 = 263.76.

Areas in Polar Coordinates.

‌When the equation of the boundary of an area is given as a function of the distance r of a point of it from a fixed point O (see Fig. 61) called the pole, and of the angle which r makes with the positive horizontal

θ

O

X

dθ

r

B A

Fig. 61.

‌direction OX, the process just explained can be applied just as easily, with a small modification. Instead of a strip of area, we consider a small triangle OAB, the angle at O being dθ, and we find the sum of all the little triangles making up the required area.

The area of such a small triangle is approximately

r dθ

AB

× r or

× r; hence the portion of the area included between the curve and

two positions of r corresponding to the angles θ1 and θ2 is given by

∫ θ=θ2

1

2

θ=θ1

r2 dθ.

Examples.
1. Find the area of the sector of 1 radian in a circumference of radius a inches.
  ∫
  
  The polar equation of the circumference is evidently r = a. The area is
  
  θ=θ2
  
  1
  
  a2 dθ =
  
  a2 ∫ θ=1
  
  a2
  
  dθ = .
  
  2
  
  θ=θ1
  
  2 θ=0 2
2. Find the area of the first quadrant of the curve (known as “Pas- cal’s Snail”), the polar equation of which is r = a(1 + cos θ).

2 a2(1 + cos θ)2 dθ

∫ θ= π

Area = 1

θ=0

a2 ∫ θ=

= 2 (1 + 2 cos θ + cos2 θ) dθ

θ=0

θ sin 2θ π

= θ + 2 sin θ + +

2 2 4 0

a2(3π + 8)

= .

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.

Examples.

Find the volume of a sphere of radius r.
A thin spherical shell has for volume 4πx2 dx (see Fig. 59, p. 214); summing up all the concentric shells which make up the sphere, we

have

x=r x3 r

volume sphere = ∫ 4πx2 dx = 4π = 4 πr3.

x=0

3 0 3

y

dx y

x

O

x

Fig. 62.

‌We can also proceed as follows: a slice of the sphere, of thickness dx, has for volume πy2 dx (see Fig. 62). Also x and y are related by the expression

y2 = r2 − x2.

∫

x=0

π(r2 − x2) dx

∫ x=r

Hence volume sphere = 2

= 2π

= 2π

∫ x=r

x=0

r2x −

r2 dx − x3 r

3

0

x=r

x=0

=

r3.

4π

3

x2 dx
Find the volume of the solid generated by the revolution of the curve y2 = 6x about the axis of x, between x = 0 and x = 4.

The volume of a strip of the solid is πy2 dx.

∫

x=4 x=4

Hence volume = πy2 dx = 6π xdx

x=0 x=0

= 6π x2 4 = 48π = 150.8.

2 0

On Quadratic Means.

‌In certain branches of physics, particularly in the study of alternat- ing electric currents, it is necessary to be able to calculate the quadratic mean of a variable quantity. By “quadratic mean” is denoted the square root of the mean of the squares of all the values between the limits con- sidered. Other names for the quadratic mean of any quantity are its “virtual” value, or its “r.m.s.” (meaning root-mean-square) value. The French term is valeur efficace. If y is the function under consideration, and the quadratic mean is to be taken between the limits of x = 0 and x = l; then the quadratic mean is expressed as

s2 1 ∫ l y2 dx.

l 0

Examples.

To find the quadratic mean of the function y = ax (Fig. 63).

∫ l

Here the integral is

a2x2 dx, which is 1 a2l3.

3

0

Dividing by l and taking the square root, we have

1

quadratic mean = √3 al.

Y

y

O

X

l

Fig. 63.

2

‌Here the arithmetical mean is 1 al; and the ratio of quadratic to

arithmetical mean (this ratio is called the form-factor ) is

∫
To find the quadratic mean of the function y = xa.
2

√3 = 1.155.

The integral is

x=l

x=0

x2a dx, that is

l2a+1

.

2a + 1

Hence quadratic mean = s2

l2a

.

2a + 1

x

∫

∫
To find the quadratic mean of the function y = a 2 .

The integral is

x=l

x=0

(a 2 ) dx, that is

x=l

x=0

ax dx,

which is

al 1

—

logϵ a

x=l

x=0

s −

al 1

Hence the quadratic mean is 2 .

l logϵ a

‌Exercises XVIII. (See p. 265 for Answers.)‌

Find the area of the curve y = x2 + x − 5 between x = 0 and
x = 6, and the mean ordinates between these limits.
Find the area of the parabola y = 2a√x between x = 0 and x = a. Show that it is two-thirds of the rectangle of the limiting ordinate and of its abscissa.
Find the area of the positive portion of a sine curve and the mean ordinate.
Find the area of the positive portion of the curve y = sin2 x, and find the mean ordinate.
Find the area included between the two branches of the curve
y = x2

5

± x2 from x = 0 to x = 1, also the area of the positive portion

of the lower branch of the curve (see Fig. 30, p. 106).
Find the volume of a cone of radius of base r, and of height h.
Find the area of the curve y = x3 − logϵ x between x = 0 and
x = 1.
Find the volume generated by the curve y = √1 + x2, as it revolves about the axis of x, between x = 0 and x = 4.
Find the volume generated by a sine curve revolving about the axis of x. Find also the area of its surface.
Find the area of the portion of the curve xy = a included be- tween x = 1 and x = a. Find the mean ordinate between these limits.
√
Show that the quadratic mean of the function y = sin x, between 2
the limits of 0 and π radians, is

. Find also the arithmetical mean

2

of the same function between the same limits; and show that the form-

factor is = 1.11.
Find the arithmetical and quadratic means of the function x2 + 3x + 2, from x = 0 to x = 3.
Find the quadratic mean and the arithmetical mean of the func- tion y = A1 sin x + A1 sin 3x.
A certain curve has the equation y = 3.42ϵ0.21x. Find the area included between the curve and the axis of x, from the ordinate at x = 2 to the ordinate at x = 8. Find also the height of the mean ordinate of the curve between these points.
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.
Find the volume generated by the curve y = ±x√x(10 − x)

rotating about the axis of x.

‌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 Calculus 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

∫ u dx = ux − ∫ xdu + C.

∫ ∫

It is useful in some cases that you can’t tackle directly, for it shows that if in any case xdu can be found, then u dx can also be found. The formula can be deduced as follows. From p. 37, we have,

d(ux) = u dx + xdu,

which may be written

u(dx) = d(ux) − xdu,

which by direct integration gives the above expression.

∫

Examples.

Find w · sin w dw.
sin w · dw = − cos w = x.

Write u = w∫ , and for sin w · dw write dx. We shall then have

du = dw, while

Putting these into the formula, we get

∫ w · sin w dw = w(− cos w) − ∫ − cos w dw

= −w cos w + sin w + C.
Find ∫ xϵx dx.

Write u = x, ϵx dx = dv;

then du = dx, v = ϵx,

and ∫ xϵx dx = xϵx − ∫ ϵx dx (by the formula)

= xϵx − ϵx = ϵx(x − 1) + C.
Try ∫ cos2 θ dθ.
‌u = cos θ, cos θ dθ = dv.

Hence du = − sin θ dθ, v = sin θ,

∫ cos2 θ dθ = cos θ sin θ + ∫ sin2 θ dθ

2

= 2 cos θ sin θ + ∫ (1 − cos2 θ) dθ

2

= sin 2θ + ∫ dθ − ∫ cos2 θ dθ.

∫

Hence 2 cos2 θ dθ = sin 2θ + θ

2

and ∫ cos2 θ dθ = sin 2θ + θ + C.

4 2
Find ∫ x2 sin xdx.
Write x2 = u, sin xdx = dv; then du = 2xdx, v = − cos x,

∫ x2 sin xdx = −x2 cos x + 2 ∫ x cos xdx.

Now find ∫ x cos xdx, integrating by parts (as in Example 1 above):

∫ x cos xdx = x sin x + cos x + C.

Hence

∫ x2 sin xdx = −x2 cos x + 2x sin x + 2 cos x + C′

2

2

x ′

= 2 x sin x + cos x 1 − + C .
Find ∫ √1 − x2 dx.

Write u = √1 − x2, dx = dv; xdx

then du = −√1 − x2 (see Chap. Ix., p. 66)

and x = v; so that

∫ √

1 − x2 dx = x

√

1 − x2 +

x2 dx

√

1 − x2

∫

Here we may use a little dodge, for we can write

∫ √

1 − x2 dx =

(1 x2) dx

∫ −

√1 − x2 =

∫

√ −

∫

1 − x2

x2 dx

∫

√

1 − x2

Adding these two last equations, we get rid of have

x2 dx

√1 − x2 , and we

2 ∫ √1 − x2 dx = x√1 − x2 + ∫

√1 − x2 .

Do you remember meeting √1 − x2 ? it is got by differentiating

y = arc sin x (see p. 168); hence its integral is arc sin x, and so

∫ √

1 − x2 dx =

x√1 − x2

+ 1 arc sin x + C.

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. 66. Let us illustrate its application to integration by a few examples.

√3 + xdx.
Let 3 + x = u, dx = du;

u 2 du

3

2

3

2

replace ∫ 1 = 2 u 3 = 2 (3 + x) 3 .

∫
dx  .

ϵx + ϵ−x

ϵ

Let ϵx = u, du = ϵx, and dx = du ;

∫     dx

dx

∫

     du

x

∫    du

∫  du

so that

du

ϵx + ϵ−x =

ϵx(ϵx + ϵ−x) =

1 =

u u +

u

u2 + 1 .

1 + u2 is the result of differentiating arc tan x.

∫ ∫ ∫

Hence the integral is arc tan ϵx.

x2 + 2x + 3

x2 + 2x + 1 + 2

(x + 1)2 + (

2)2

dx = dx = dx √ .

Let x + 1 = u, dx = du;

∫ du du

then the integral becomes

u2 + (√2)2 ; but

u2 + a2 is the result of

differentiating u =

arc tan

a a

Hence one has finally integral.

√2 arc tan

x + 1

√2 for the value of the given

Formulæ of Reduction are special forms applicable chiefly to bino-

mial and trigonometrical expressions that have to be integrated, and have to be reduced into some form of which the integral is known.

Rationalization, and Factorization of Denominator are dodges ap- plicable in special cases, but they do not admit of any short or general explanation. Much practice is needed to become familiar with these preparatory processes.

The following example shows how the process of splitting into partial fractions, which we learned in Chap. xIII., p. 118, can be made use of in integration.

Take again ∫ dx ; if we split 1 into partial frac-

x2 + 2x + 3

tions, this becomes (see p. 230):

1 ∫ dx

x2 + 2x + 3

∫ dx

2√−2 x + 1 − √−2 − x + 1 + √−2

1 x + 1 − √−2

= 2√−2 logϵ x + 1 + √−2 .

Notice that the same integral can be expressed sometimes in more than one way (which are equivalent to one another).

Pitfalls. A beginner is liable to overlook certain points that a prac- tised hand would avoid; such as the use of factors that are equivalent to either zero or infinity, and the occurrence of indeterminate quantities such as 0 . There is no golden rule that will meet every possible case. Nothing but practice and intelligent care will avail. An example of a pitfall which had to be circumvented arose in Chap. xVIII., p. 189, when we came to the problem of integrating x−1 dx.

Triumphs. By triumphs must be understood the successes with which the calculus has been applied to the solution of problems other- wise intractable. Often in the consideration of physical relations one is able to build up an expression for the law governing the interac- tion of the parts or of the forces that govern them, such expression being naturally in the form of a differential equation, that is an equa- tion containing differential coefficients with or without other algebraic quantities. And when such a differential equation has been found, one can get no further until it has been integrated. Generally it is much easier to state the appropriate differential equation than to solve it:— the 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 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

dx a2 − x2

could blossom out into

1 a + x

y = 2a logϵ a − x + C?

yet the latter is the solution of the former.

‌As a last example, let us work out the above together. By partial fractions,

1 1 1

a2 − x2 = 2a(a + x) + 2a(a − x) ,

dy =

2a(a + x)

+ ,

2a(a − x)

y = 1 ∫ dx + ∫ dx

2a a + x a − x

* This means that the actual result of solving it is called its “solution.” But many mathematicians would say, with Professor Forsyth, “every differential equation is considered as solved 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 integrations in the latter can or cannot be expressed in terms of functions already known.”

= 2a (logϵ(a + x) − logϵ(a − x))

1 a + x

= 2a logϵ a − x + C.

Not a very difficult metamorphosis!

There are whole treatises, such as Boole’s Differential Equations, devoted to the subject of thus finding the “solutions” for different orig- inal forms.

‌Exercises XIX. (See p. 266 for Answers.)‌

Find ∫ √a2 − x2 dx. (2) Find ∫ x logϵ xdx.

∫

(3) Find ∫ xa logϵ xdx. (4) Find ∫ ϵx cos ϵx dx.

(5) Find 1 cos(log

x ϵ

∫

(7) Find

∫ (logϵ x)a

x) dx. (6) Find ∫ x2ϵx dx.

∫ −

dx. (8) Find

∫ dx

x log

Find

5x + 1 dx. (10) Find

x2 + x − 2

(x2 3) dx x3 − 7x + 6 .

(11) Find ∫ b dx . (12) Find ∫ 4xdx .

x2 − a2

(13) Find ∫ dx . (14) Find ∫

x4 − 1

√ .

1 − x4

x a − bx2

‌CHAPTER XXI.‌

FINDING SOME SOLUTIONS.

In 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 art. 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

Transposing we have

ay + b

= 0.

bdx = −ay.

Now the mere inspection of this relation tells us that we have got

to do with a case in which

is proportional to y. If we think of the

curve which will represent y as a function of x, 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. 153), and the solution will contain ϵ−x 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 dx to the other. To do this, we must split our usually inseparable companions dy and dx from one another.

dy a

y = − b dx.

Having done the deed, we now can see that both sides have got into

a shape that is integrable, because we recognize

, or

dy, as a differ-

ential that we have met with (p. 143) when differentiating logarithms.

So we may at once write down the instructions to integrate,

∫ dy = ∫ −a dx;

and doing the two integrations, we have:

logϵ y = − b x + logϵ C,

where logϵ C is the yet undetermined constant* of integration. Then,

* We may write down any form of constant as the “constant of integration,” and the form logϵ C is adopted here by preference, because the other terms in this line of equation are, or are treated as logarithms; and it saves complications afterward if the added constant be of the same kind.

delogarizing, we get:

y = Cϵ

— ax,

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 x.

Now, as to the C, its meaning depends on the initial value of y. For if we put x = 0 in order to see what value y then has, we find that this makes y = Cϵ−0; and as ϵ−0 = 1 we see that C is nothing else than the particular value* of y at starting. This we may call y , and so write the solution as

y = y0ϵ− x.

Example 2.

Let us take as an example to solve

ay + b

= g,

where g is a constant. Again, inspecting the equation will suggest,

(1) that somehow or other ϵx will come into the solution, and (2) that if at any part of the curve y becomes either a maximum or a minimum,

so that

= 0, then y will have the value =

. But let us go to work

as before, separating the differentials and trying to transform the thing

* compare what was said about the “constant of integration,” with reference to Fig. 48 on p. 184, and Fig. 51 on p. 187.

into some integrable shape.

bdx = g − ay;

dy = a g − y ;

dx b a

dy a

g y − a

= − b dx.

Now we have done our best to get nothing but y and dy on one side, and nothing but dx on the other. But is the result on the left side integrable?

It is of the same form as the result on p. 145; so, writing the in- structions to integrate, we have:

∫

dyg y − a

= − ∫ a dx;

and, doing the integration, and adding the appropriate constant,

log y − g = −ax + log C;

ϵ a b ϵ

g − ax

whence y − a = Cϵ

b ;

— ax

and finally, y =

which is the solution.

+ Cϵ b , a

If the condition is laid down that y = 0 when x = 0 we can find C; for then the exponential becomes = 1; and we have

0 =

+ C,

or C = −a.

Putting in this value, the solution becomes

y = a (1 − ϵ

— ax).

But further, if x grows indefinitely, y will grow to a maximum; for

when x = ∞, the exponential = 0, giving ymax. = a . Substituting this,

we get finally

— ax

y = ymax.(1 − ϵ ).

This result is also of importance in physical science.

Example 3.

Let ay + b dt = g · sin 2πnt.

We shall find this much less tractable than the preceding. First divide through by b.

dy a

y =

dt b

sin 2πnt. b

Now, as it stands, the left side is not integrable. But it can be made so by the artifice—and this is where skill and practice suggest a

plan—of multiplying all the terms by ϵ b , giving us:

dy at

a at g at

ϵ b

which is the same as

yϵ b =

ϵ b

b b

· sin 2πnt,

dyϵat + y dt

d(ϵ b ) = g ϵat

dt b

· sin 2πnt;

and this being a perfect differential may be integrated thus:—since, if

u = yϵ b ,

du dy a

= ϵ b

dt dt

d(ϵ b )

+ y ,

yϵat = g ∫ ϵat · sin 2πnt · dt + C,

or y = g ϵ− at ∫ ϵat · sin 2πnt · dt + Cϵ− at. [a]

∫

The last term is obviously a term which will die out as t increases, and may be omitted. The trouble now comes in to find the integral that appears as a factor. To tackle this we resort to the device (see p. 224) of integration by parts, the general formula for which is udv =

uv − ∫ vdu. For this purpose write

 u = ϵ b ;

We shall then have

dv = sin 2πnt · dt.

at a

du = ϵ b ×

dt;

 v = −2πn cos 2πnt.

Inserting these, the integral in question becomes:

∫ ϵat · sin 2πnt · dt

2πn

= − 1 · ϵat · cos 2πnt − ∫ − 1 cos 2πnt · ϵat · a dt

2πn

2πnb

= − 1 ϵat cos 2πnt + a ∫ ϵat · cos 2πnt · dt. [b]

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 2πnt;

dv = ϵ b · dt;

whence

du = 2πn · cos 2πnt · dt;



b a

 v = aϵ b

Inserting these, we get

∫ ϵat · sin 2πnt · dt

= b · ϵat · sin 2πnt − 2πnb ∫ ϵat · cos 2πnt · dt. [c] Noting that the final intractable integral in [c] is the same as that

2πnb

in [b], we may eliminate it, by multiplying [b] by

, and multiply-

ing [c] by , and adding them.

2πnb

[d]

The result, when cleared down, is:

ϵ b

· sin 2πnt · dt = ϵ b

∫ a t

at ab · sin 2πnt − 2πnb2 · cos 2πnt

a2 + 4π2n2b2

Inserting this value in [a], we get

y = g a · sin 2πnt − 2πnb · cos 2πnt .

a2 + 4π2n2b2

To simplify still further, let us imagine an angle ϕ such that tan ϕ =

2πnb

Then sin ϕ = √a2 + 4π2n2b2 ,

and cos ϕ = √a2 + 4π2n2b2 .

Substituting these, we get:

cos ϕ · sin 2πnt − sin ϕ · cos 2πnt

y = g

√a2 + 4π2n2b2 ,

which may be written

sin(2πnt − ϕ)

y = g √a2 + 4π2n2b2 ,

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, n the frequency, a the resistance, b the coefficient of self-induction of the circuit, and ϕ is an angle of lag.

Example 4.

Suppose that M dx + N dy = 0.

We could integrate this expression directly, if M were a function of x only, and N a function of y only; but, if both M and N are functions that depend on both x 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

∂U

= M,

∂x



∂U

 ∂y

= N.

And if such a common function exists, then

∂U ∂U

dx + dy

∂x ∂y

is an exact differential (compare p. 172).

Now the test of the matter is this. If the expression is an exact differential, it must be true that

dM dN

= ;

dy dx

for then

which is necessarily true.

d(dU )

dx dy

d(dU )

= ,

dy dx

Take as an illustration the equation

(1 + 3xy) dx + x2 dy = 0.

 dy

Is this an exact differential or not? Apply the test.



d(1 + 3xy)

= 3x,



d(x2)

 dx

= 2x,

which do not agree. Therefore, it is not an exact differential, and the two functions 1 + 3xy and x2 have not come from a common original function.

It is possible in such cases to discover, however, an integrating fac- tor, 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 one 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

(2x + 6x2y) dx + 2x3 dy = 0.



Now apply the test to this.



d(2x + 6x2y)

= 6x2,



d(2x3) dx

= 6x2,

which agrees. Hence this is an exact differential, and may be integrated. Now, if w = 2x3y,

dw = 6x2y dx + 2x3 dy.

Hence ∫ 6x2y dx + ∫ 2x3 dy = w = 2x3y; so that we get U = x2 + 2x3y + C.

Example 5. Let

d2y dt2

+ n2y = 0.

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.

Transposing, we have

d2y

dt2

= −n2y.

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 function—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 = A sin(pt + q). However, let us go to work.

Multiply both sides of the original equation by 2

and integrate,

giving us 2

d2y dy

+ 2x2ydy = 0, and, as

dt2 dt dt

dy 2

d2y dy d dt

dt2

2 =

dy 2

+ n2(y2 − C2) = 0,

C being a constant. Then, taking the square roots,

dy = −n√y2 − C2 and

√C2 − y2

= n · dt.

But it can be shown that (see p. 168)

√C2 − y2 =

d(arc sin

)

C ;

whence, passing from angles to sines,

arc sin C = nt + C1 and y = C sin(nt + C1),

where C1 is a constant angle that comes in by integration.

Or, preferably, this may be written

y = A sin nt + B cos nt, which is the solution.

Example 6.

d2y dt2

— n y = 0.

Here we have obviously 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 (see p. 139), 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

, and integrat-

ing, we get 2

d2y

dy dy

— 2x y

= 0,

dx2 dx

— √

dy 2

and, as 2

d2y

dy d dx

2 2 2

dx2 dx

dy 2

— n (y + c ) = 0,

dy n y2 + c2 = 0, dx

where c is a constant, and

√y2 + c2 = n dx.

Now, if w = logϵ(y + √y2 + c2) = logϵ u,

dw 1 du

y y + √y2 + c2

du u

dy = 1 + √y2 + c2 =

√y2 + c2

and

dy = √y2 + c2 .

Hence, integrating, this gives us

logϵ(y + √y2 + c2) = nx + logϵ C,

y + √y2 + c2 = Cϵnx. (1) Now (y + √y2 + c2) × (−y + √y2 + c2) = c2;

whence −y +

√y2 + c2 =

c2 −nx

. (2)

Subtracting (2) from (1) and dividing by 2, we then have

y =

Cϵnx

1 c2

— 2 C ϵ

−nx,

which is more conveniently written

y = Aϵnx + Bϵ−nx.

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 x increases, and of a second term which dies away as x increases.

Example 7.

d2y dy

Let b dt2 + a dt + gy = 0.

Examination of this expression will show that, if b = 0, 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 the present example is

y = (ϵ−mt)(Aϵnt + Bϵ−nt),

where m = and n = 2b

4b2 − b .

a r a2 g

The steps by which this solution is reached are not given here; they may be found in advanced treatises.

Example 8.

d2y

dt2 = a

2 d2y dx2 .

It was seen (p. 174) that this equation was derived from the original

y = F (x + at) + f (x − 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

d2y

du · dv

= 0,

where u = x + at, and v = x − at, leading to the same general solution.

If we consider a case in which F vanishes, then we have simply

y = f (x − at);

and this merely states that, at the time t = 0, y is a particular func- tion of x, and may be looked upon as denoting that the curve of the relation of y to x has a particular shape. Then any change in the value of t is equivalent simply to an alteration in the origin from which x is reckoned. That is to say, it indicates that, the form of the function being conserved, it is propagated along the x direction with a uniform velocity a; so that whatever the value of the ordinate y at any par- ticular time t0 at any particular point x0, the same value of y will appear at the subsequent time t1 at a point further along, the abscissa of which is x0 +a(t1 −t0). In this case the simplified equation represents the propagation of a wave (of any form) at a uniform speed along the x direction.

If the differential equation had been written

d2y d2y

m dt2 = k dx2 ,

the solution would have been the same, but the velocity of propagation would have had the value

a = k . m

You have now been personally 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 convenient collection of standard forms (see pp. 249–251). In the middle column are set down a number of the functions which most commonly occur. The results of differentiating them are set down on the left; the results of integrating them are set down on the right. May you find them useful!

‌EPILOGUE AND APOLOGUE.‌

It 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 possi- ble 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 puffed 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 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. It would be equally absurd to require general rigid demonstrations to be expounded to beginners in the calculus.

One other thing will the professed mathematicians say about this thoroughly bad and vicious book: that the reason why it is so easy is because the author has 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 calculus by the stupid way in which its teaching is almost al- ways 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 sim- plicities, and to grasp its powerful methods of solving problems, with- out being compelled to toil through the intricate out-of-the-way (and mostly irrelevant) mathematical gymnastics so dear to the unpractical mathematician.

There are amongst young engineers a number on whose ears the adage that what one fool can do, another can, 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.‌

‌dy

→− y −→

∫ y dx

Algebraic.

2x nxn−1

−2

−x

du dv dw dx ± dx ± dx

dv du

u + v

dx dx

du dv

v dx − u dx

v2 du

x a

x ± a ax x2 xn x−1

u ± v ± w uv

u v

1 x2 + C

ax + C

1 x2 ± ax + C

1 ax2 + C

1 x3 + C

xn+1 + C

n + 1

logϵ x + C

∫ u dx ± ∫ v dx ± ∫ w dx

No general form known

ux − ∫ xdu + C

dy dx	→− y −→	∫	y dx
Exponential and Logarithmic.
ϵx x−1 ax logϵ a	ϵx	ϵx + C
	logϵ x	x(logϵ x − 1) + C
	log10 x	0.4343x(logϵ x − 1) + C
	ax	ax	+ C
		logϵ a
Trigonometrical.
cos x — sin x sec2 x	sin x cos x tan x	— cos x + C sin x + C — logϵ cos x + C
Circular (Inverse).
1 √(1 − x2) 1 −√(1 − x2) 1 1 + x2	arc sin x	x · arc sin x + √1 − x2 + C
	arc cos x	x · arc cos x − √1 − x2 + C
	arc tan x	1 2		x2) + C
Hyperbolic.
cosh x sinh x sech2 x	sinh x cosh x tanh x	cosh x + C sinh x + C logϵ cosh x + C

‌dy dx

→− y −→

∫ y dx

Miscellaneous.

−(x + a)2

−(a2 + x2) 3

∓(a ± bx)2

3a2x

−(a2 + x2) 5

a · cos ax

−a · sin ax a · sec2 ax sin 2x

n · sinn−1 x · cos x

cos x

−sin2 x

sin 2x

sin2 x − cos2 x

sin2 x · cos2 x

n · sin mx · cos nx + m · sin nx · cos mx

2a · sin 2ax

−2a · sin 2ax

x + a

√a2 + x2

a ± bx a2

(a2 + x2) 2

sin ax cos ax tan ax sin2 x cos2 x

sinn x

sin x

sin2 x

sin x · cos x

sin mx · sin nx

sin2 ax

cos2 ax

logϵ(x + a) + C

log (x + √a2 + x2) + C

± b logϵ(a ± bx) + C

√a2 + x2 + C

− a cos ax + C

sin ax + C

− a logϵ cos ax + C

x sin 2x

2 − 4 + C

x sin 2x

+ + C

2 4

— cos x sinn−1 x + n − 1 ∫ sinn−2 xdx + C n n

logϵ tan 2 + C

— cotan x + C

logϵ tan x + C

1 cos(m − n)x − 1 cos(m + n)x + C

2 2

x sin 2ax

2 − 4a + C

x sin 2ax

+ + C

2 4a

sin 2x
sin4 x

‌ANSWERS.‌‌

Exercises I. (p. 24.)

(1)	dy dx	12	(2)	dy = −3 x− 5 . 2 dx 2	(3) dy = 2ax(2a−1). dx
(4)	du dt	1.4	(5)	dz = 1 u− 2 . 3 du 3	(6) dy = −5 x− 8 . 3 dx 3
(7)	du dx	8 − 13		dy (8) dx	= 2axa−1.
(9)	dy dx	3 3−q = x q . q		dy (10) dx	= −mx− m+n . n n

= 13x .

= 2.4t .

= −5 x 5 .

‌Exercises II. (p. 31.)

dy 2

dy 3 1

dy − 1

(1)

= 3ax . (2)

dx = 13 × 2 x2 . (3)

= 6x 2 .

(4)

dy 1

dx 2

c 2 x

—

2 . (5)

du an

dz c

zn−1

. (6)

= 2.36t.

πσT

(7) dlt dt

(8)

2DL

= 0.000012 × l0.

= abV b−1, 0.98, 3.00 and 7.47 candle power per volt respectively.

dn 1

πσ3

2DL

(9) = −

rgT , dn = − 1

rgT ,

LD2

πσ

DL2

πσ

dn 1

dσ

= −

r gT , dn = 1

r g .

Rate of change of P when t varies D

(10) = − .

Rate of change of P when D varies t

(11) 2π, 2πr, πl, 2 πrh, 8πr, 4πr2. (12) dD = 0.000012lt .

3 dT π

‌Exercises III. (p. 45.)

x2 x3

(a) 1 + x + +
2 6

x4

+ + . . . (b) 2ax + b. (c) 2x + 2a.

24

(d ) 3x2 + 6ax + 3a2.

(2)

dw

dt = a − bt. (3)

dy

= 2x.

dx

14110x4 − 65404x3 − 2244x2 + 8192x + 1379.
dx = 2y + 8. (6) 185.9022654x2 + 154.36334.

(7) −5 .

(3x + 2)2

ad − bc

(8)

6x4 + 6x3 + 9x2 (1 + x + 2x2)2 .

anx−n−1 + bnxn−1 + 2nx−1

(9)

(cx + d)2 . (10)

(x−n + b)2 .

b + 2ct.
R0(a+2bt), R0

b a + 2√t

R0(a + 2bt)

, −(1 + at + bt2)2 or

R2(a + 2bt)

.

R0
1.4340(0.000014t − 0.001024), −0.00117, −0.00107, −0.00097.

dE k

= b +

dl i

di = −

c + kl

i2 .

17 + 24x; 24.

Exercises IV. (p. 50.)

x2 + 2ax − a

2a(a + 1)
1. 1 + x +
  x2
  
  1 × 2
  
  x3
  
  +
  
  1 × 2 × 3
  
  (2)
  
  ; 1 + x +
  
  (x + a)2 ;
  
  x2
  
  .
  
  1 × 2
  
  (x + a)3 .
2. (Exercises III.):
  d2y
  
  d3y 1 1
  1. (a)
    dx2
    
    = dx3
    
    = 1 + x +
    
    x2 +
    
    x3 + . . ..
    
    2
    
    6
    
    (b) 2a, 0. (c) 2, 0. (d ) 6x + 6a, 6.
  2. −b, 0. (3) 2, 0.
    1. 56440x3 − 196212x2 − 4488x + 8192. 169320x2 − 392424x − 4488.
    2. 2, 0. (6) 371.80453x, 371.80453.
    30 270
    
    (7) , − . (3x + 2)3 (3x + 2)4
    
    (Examples, p. 40):
    
    6a 6a
    
    3a√b
    
    6b√3 a
    
    18b√3 a 3a√b
    
    (1)
    
    b2 x, b2 . (2)
    
    2√x −
    
    x3 ,
    
    x4 − 4√x3 .
    
    2 1.056 2.3232 16
  3. √3 θ8 −
    
    √5 θ11 ,
    
    √5 θ16 − 3√3 θ11 .
  4. 810t4 − 648t3 + 479.52t2 − 139.968t + 26.64. 3240t3 − 1944t2 + 959.04t − 139.968.
  5. 12x + 2, 12. (6) 6x2 − 9x, 12x − 9.
  (7)
  
  4
  
  √θ + √θ5
  
  +
  
  4
  
  √θ7 − √θ3
  
  .
  
  3 1 1 1 15 1
  
  √θ5 − √θ3
  
  —
  
  √θ9 + √θ7
  
  .
  
  3 1 1 15 7 1
  
  8
  
  8
  
  ‌Exercises V. (p. 63.)
64; 147.2; and 0.32 feet per second.
x = a − gt; x¨ = −g. (4) 45.1 feet per second.
1. 12.4 feet per second per second. Yes.
2. Angular velocity = 11.2 radians per second; angular acceleration
  = 9.6 radians per second per second.
3. v = 20.4t2 − 10.8. a = 40.8t. 172.8 in./sec., 122.4 in./sec2.
4. v = 30√3
  
  1
  
  (t − 125)2 , a = −45√3
  
  1
  
  .
  
  (t − 125)5
  
  —
  
  8t
5. v = 0.8 − (4 + t2)2 , a =
6. n = 2, n = 11.
24t2 32

(4 + t2)3 , 0.7926 and 0.00211.

‌Exercises VI. (p. 72.)

x x 1

(1) √x2 + 1. (2) √x2 + a2 . (3) − √ .

2 (a + x)3

—
ax

√(a − x2)3 .
2a2 x2 x3√x2 − a2 .

2
3 x2 8 x (x3 + a) − (x4 + a)
2a (x − a).

3

2
5 y3.

(x4 + a

2

2 9

) 3 (

x3 + a) 3

(x + a)

(1 − θ)√1 − θ2 .

(1)

‌Exercises VII. (p. 74.)

dw 3x2 (3 + 3x3)

dx 27 1 x3 + 1 x6 3

(2)

2 4

1 +

2 + 3x2

dx = −√ √

12x

3 + 4

1 +

2 + 3x2

√

√ √ 2

du x2 √3 + x3

(3)

dx = −,u“

1 +

1 + √3

2#3

‌Exercises VIII. (p. 88.)

1.44.

(4)

= 3x2 + 3; and the numerical values are: 3, 33 , 6, and 15.

±√2.

dy 4 x 1

(6) = −

. Slope is zero where x = 0; and is ∓ 3√2 where x = 1.

dx 9 y

m = 4, n = −3.
Intersections at x = 1, x = −3. Angles 153◦ 26′, 2◦ 28′.
Intersection at x = 3.57, y = 3.50. Angle 16◦ 16′.
3

3

3
x = 1 , y = 21 , b = −5 .

‌Exercises IX. (p. 107.)

Min.: x = 0, y = 0; max.: x = −2, y = −4.
x = a. (4) 25√3 square inches.

dy 10 10

(5) = − + ; x = 4; y = 5.

dx x2 (8 − x)2

Max. for x = −1; min. for x = 1.
Join the middle points of the four sides.
r = 2 R, r = R , no max.
3 2

r2 R

r = R

, r = √ , r = 0.8506R.

3 2

At the rate of
r

square feet per second.

3

r
r = R√8 . (12) n = rNR .

‌Exercises X. (p. 115.)

Max.: x = −2.19, y = 24.19; min.:, x = 1.52, y = −1.38.

dy b

d2y b

(2)

dx = a − 2cx;

dx2 = −2c; x = 2ac (a maximum).

(a) One maximum and two minima.
(b) One maximum. (x = 0; other points unreal.)
Min.: x = 1.71, y = 6.14. (5) Max: x = −.5, y = 4.

Max.: x = 1.414, y = 1.7675. Min.: x = −1.414, y = 1.7675.
r
Max.: x = −3.565, y = 2.12. Min.: x = +3.565, y = 7.88.
0.4N , 0.6N . (9) x = a .

c
1. Speed 8.66 nautical miles per hour. Time taken 115.47 hours. Minimum cost 112. 12s.
2. Max. and min. for x = 7.5, y = ±5.414. (See example no. 10, p. 71.)
  2
  
  3
3. Min.: x = 1 , y = 0.25; max.: x = −1 , y = 1.408.
‌Exercises XI. (p. 127.)

(1)

2

x − 3

1

+

x + 4

. (2)

1

+

x − 1

2

x − 2

. (3)

2

+

x − 3

1

.

x + 4

5 4 19 22

(4) − . (5) − .

x − 4 x − 3 13(2x + 3) 13(3x − 2)

2 4 5

(6) + − .

x − 2 x − 3 x − 4

(7)

1

+

6(x − 1)

7

11

+

15(x + 2)

71

1

.

10(x − 3)

5

(8) + − . 9(3x + 1) 63(3x − 2) 7(2x + 1)

1

3(x − 1)

2x + 1

+ 3(x2 + x + 1)

. (10) x +

3(x + 1)

+ 1 − 2x . 3(x2 − x + 1)

3 2x + 1 1 1 2

(11) + . (12) − + . (x + 1) x2 + x + 1 x − 1 x − 2 (x − 2)2

1 1 1

(13) − + . 4(x − 1) 4(x + 1) 2(x + 1)2

4 4 1

− − . 9(x − 1) 9(x + 2) 3(x + 2)2

1 − x − 1 − 1 .

x + 2 x2 + x + 1 (x2 + x + 1)2

5 32 36

(16) − + .

x + 4 (x + 4)2 (x + 4)3

7 55 73

(17) + + . 9(3x − 2)2 9(3x − 2)3 9(3x − 2)4

1 x

(18) + − . 6(x − 2) 3(x − 2)2 6(x2 + 2x + 4)

‌Exercises XII. (p. 150.)

(1) ab(ϵax + ϵ−ax). (2) 2at + 2 .

(3) logϵ n.

(5) npv

n−1

. (6)

. (7) 3ϵ

x x−1

—

(8) 6xϵ−5x − 5(3x2 + 1)ϵ−5x.

(9)

axa−1 xa + a .

(x − 1)2

6x +

3x2 − 1

1

2 (√x + x)

(3x2 − 1) (√x + 1).
1 − logϵ (x + 3).
(x + 3)2
ax (axa−1 + xa logϵ a). (14) Min.: y = 0.7 for x = 0.694.

1 + x 3

(15) . (16)

(logϵ

x x

ax)2.

Let
T

‌Exercises XIII. (p. 160.)

= x (∴ t = 8x), and use the Table on page 157.
T = 34.627; 159.46 minutes.
Take 2t = x; and use the Table on page 157.

(a) xx (1 + logϵ x); (b) 2x(ϵx)x; (c) ϵxx × xx (1 + logϵ x).
0.14 second. (7) (a) 1.642; (b) 15.58.

µ = 0.00037, 31m 1 .
i is 63.4% of i0, 220 kilometres.
0.133, 0.145, 0.155, mean 0.144; −10.2%, −0.9%, +77.2%.
1

Min. for x =

Max. for x = ϵ.

Min. for x = logϵ a.

‌Exercises XIV. (p. 170.)

(i) dy = A cos θ − π ;
dθ

(ii)

2

dy

= 2 sin θ cos θ = sin 2θ and

dθ

dy

= 2 cos 2θ;

dθ

(iii) dy = 3 sin2 θ cos θ and dy = 3 cos 3θ.

dθ

θ = 45◦ or π

dθ

—

radians. (3) dy = n sin 2πnt.

(4) ax logϵ

a cos ax. (5) cos x = cotan x

sin x

18.2 cos (x + 26◦).
dy
The slope is
dθ

= 100 cos (θ − 15◦), which is a maximum when

(θ −15◦) = 0, or θ = 15◦; the value of the slope being then = 100.

2

When θ = 75◦ the slope is 100 cos(75◦ − 15◦) = 100 cos 60◦ = 100 × 1 = 50.
cos θ sin 2θ + 2 cos 2θ sin θ = 2 sin θ cos2 θ + cos 2θ
= 2 sin θ 3 cos2 θ − 1 .
amnθn−1 tanm−1 (θn) sec2 θn.
ϵx sin2 x + sin 2x ; ϵx sin2 x + 2 sin 2x + 2 cos 2x .

—

dy
(i) =
dx

dy

ab

(x + b)

a

2 ; (ii) b ϵ

x

b ; (iii)

1 ◦ ab

90 × (b2 + x2).
(i)
= sec x tan x;

dx

dy 1

(ii) = −√1 − x2 ;

dx

dy 1

(iii)

dx = 1 + x2 ;

—

dy 1

(iv)

dx = x√x2 − 1;

(v)

dy √3 sec x (3 sec2 x 1)

= .

dx 2
dy = 4.6 (2θ + 3)1.3 cos (2θ + 3)2.3.
dθ

—
dy = 3θ2 + 3 cos (θ + 3) log
dθ ϵ

3 cos θ × 3sin θ + 3θ .
θ = cot θ; θ = ±0.86; is max. for +θ, min. for −θ.

‌Exercises XV. (p. 177.)

x3 − 6x2y − 2y2; 1 − 2x3 − 4xy.
2xyz + y2z + z2y + 2xy2z2; 2xyz + x2z + xz2 + 2x2yz2; 2xyz + x2y + xy2 + 2x2y2z.
1{(x − a) + (y − b) + (z − c)} = (x + y + z) − (a + b + c); 3 .
r r r
dy = vuv−1 du + uv logϵ u dv.
dy = 3 sin vu2 du + u3 cos v dv,

dy = u sin xu−1 cos xdx + (sin x)u logϵ sin xdu,

1 1 1

dy = v u du − logϵ uv2 dv.

Minimum for x = y = −1 .
(a) Length 2 feet, width = depth = 1 foot, vol. = 2 cubic feet.
2

(b) Radius =

π

feet = 7.46 in., length = 2 feet, vol. = 2.54.
All three parts equal; the product is maximum.
Minimum for x = y = 1.
2
Min.: x = 1 and y = 2.
Angle at apex = 90◦; equal sides = length = √3 2V .

‌Exercises XVI. (p. 187.)

(1) 11 . (2) 0.6344. (3) 0.2624.

(a) y = 1 x2 + C; (b) y = sin x + C.
y = x2 + 3x + C.

2

‌Exercises XVII. (p. 202.)

(1)

4√ax 3 3

1

+ C. (2) −x3

+ C. (3)

x4

+ C.

4a

x

3

3

(4) 1

+ ax + C. (5) −2x

5

—

2 + C.
x4 + x3 + x2 + x + C.
ax2

4

bx3

+

9

cx4

+

16

+ C.
x2 + a

x + a = x − a +

2

x2

a2 + a

by division. Therefore the answer is

x + a

2 − ax + (a + a) logϵ(x + a) + C. (See pages 196 and 198.)

x + 3x3 + 4

27 x2 + 27x + C. (10)

x + 2 − a

3 2

x2 − 2ax + C.

2 3 9 4 1 1

(11) a (2x2 + 4 x3 ) + C. (12) −3 cos θ − 6 θ + C.

(13)

θ sin 2aθ

+ + C. (14)

θ sin 2θ

—

+ C.

2 4

(15)

θ sin 2aθ

—

+ C.

2 4a

1 ϵ3x.

3
log(1 + x) + C. (18) − logϵ(1 − x) + C.

‌Exercises XVIII. (p. 221.)

Area = 60; mean ordinate = 10.

3
Area = 2 of a × 2a√a.
2
Area = 2; mean ordinate =
π

= 0.637.
Area = 1.57; mean ordinate = 0.5.
0.572, 0.0476. (6) Volume = πr2 h .

(7) 1.25. (8) 79.4.

Volume = 4.9348; area of surface = 12.57 (from 0 to π). a
a logϵ a, a − 1 logϵ a.

Arithmetical mean = 9.5; quadratic mean = 10.85.

(13) Quadratic mean = √2

A1 + A3; arithmetical mean = 0.

1 √ 2 2

The first involves a somewhat difficult integral, and may be stated thus: By definition the quadratic mean will be

s

∫

1 2π

2π 0

(A1 sin x + A3 sin 3x)2 dx.

Now the integration indicated by

1

3

∫ (A2 sin2 x + 2A1A3 sin x sin 3x + A2 sin2 3x) dx

is more readily obtained if for sin2 x we write
1. − cos 2x.
  
  2
  
  For 2 sin x sin 3x we write cos 2x − cos 4x; and, for sin2 3x,
  1. − cos 6x.
    
    2
    
    Making these substitutions, and integrating, we get (see p. 198)
    
    A
    
    1
    
    2
  2. x −
  sin 2x 2
  
  + A1A3
  
  sin 2x
  
  2 −
  
  sin 4x 4
  
  2
  
  A
  
  3
  
  + 2 x −
  
  sin 6x
  
  .
  
  6
  
  At the lower limit the substitution of 0 for x causes all this to vanish, whilst at the upper limit the substitution of 2π for x gives A2π + A2π. And hence the answer follows.
  
  1 3
  1. Area is 62.6 square units. Mean ordinate is 10.42.
  (16) 436.3. (This solid is pear shaped.)
  
  ‌Exercises XIX. (p. 231.)
  
  (1)
  
  x√a2 − x2
  
  2
  
  + a sin−1 x
  
  2
2. a

+ C. (2)

(logϵ x − 1 ) + C.

xa+1 a + 1

log x − 1

+ C.
sin ϵx + C.
ϵ a + 1
sin(logϵ x) + C. (6) ϵx(x2 − 2x + 2) + C.

(7)

a + 1

(logϵ

x)a+1 + C. (8) logϵ(logϵ x) + C.

2 logϵ(x − 1) + 3 logϵ(x + 2) + C.
1 logϵ(x − 1) + 1 logϵ(x − 2) + 3 logϵ(x + 3) + C.
2 5

b x − a

10

x2 − 1
logϵ + C. (12) logϵ + C. 2a x + a x2 + 1

(13) 1 log 1 + x + 1 arc tan x + C.

− −

4 ϵ 1 − x 2

(14)

√a logϵ

rv2 − b

√a √a bx2

x√a . (Let

= v − u.)

= v; then, in the result, let

You had better differentiate now the answer and work back to the given expression as a check.

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 differentiating 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 suf- fice here to name two: R. G. Blaine’s The Calculus and its Applications, and F. M. Saxelby’s A Course in Practical Mathematics.

GLASGOW: PRINTED AT THE UNIVERSITY PRESS BY ROBERT MACLEHOSE AND CO. LTD.

‌A SELECTION OF

Mathematical Works

An Introduction to the Calculus. Based on Graphical Meth- ods. By Prof. G. A. Gibson, M.A., LL.D. 3s. 6d.

An Elementary Treatise on the Calculus. With Illustra- tions from Geometry, Mechanics, and Physics. By Prof. G. A. Gib- son, M.A., LL.D. 7s. 6d.

Differential Calculus for Beginners. By J. Edwards, M.A.

4s. 6d.

Integral Calculus for Beginners. With an Introduction to the Study of Differential Equations. By Joseph Edwards, M.A. 4s. 6d.

Calculus Made Easy. Being a very-simplest Introduction to

those beautiful Methods of Reckoning which are generally called by the terrifying names of the Differential Calculus and the Integral Calculus. By F. R. S. 2s. net. New Edition, with many Examples.

A First Course in the Differential and Integral Cal-

culus. By Prof. W. F. Osgood, Ph.D. 8s. 6d. net.

Practical Integration for the use of Engineers, etc. By

A. S. Percival, M.A. 2s. 6d. net.

Differential Calculus. With Applications and numerous Exam- ples. An Elementary Treatise by Joseph Edwards, M.A. 14s.

Differential and Integral Calculus for Technical Schools and Colleges. By P. A. Lambert, M.A. 7s. 6d.

Differential and Integral Calculus. With Applica- tions. By Sir A. G. Greenhill, F.R.S. 10s. 6d.

A Treatise on the Integral Calculus and its Ap- plications. By I. Todhunter, F.R.S. 10s. 6d. Key. By

H. St. J. Hunter, M.A. 10s. 6d.

A Treatise on the Differential Calculus and the Elements of the Integral Calculus. With numer- ous Examples. By I. Todhunter, F.R.S. 10s. 6d. Key. By

H. St. J. Hunter, M.A. 10s. 6d.

Ordinary Differential Equations. An Elementary Text-book. By James Morris Page, Ph.D. 6s. 6d.

An Introduction to the Modern Theory of Equations.

By Prof. F. Cajori, Ph.D. 7s. 6d. net.

A Treatise on Differential Equations. By Andrew Rus- sell Forsyth, Sc.D., LL.D. Fourth Edition. 14s. net.

A Short Course on Differential Equations. By Prof. Don- ald F. Campbell, Ph.D. 4s. net.

A Manual of Quaternions. By C. J. Joly, M.A., D.Sc., F.R.S.

10s. net.

The Theory of Determinants in the Historical Order of Development. Vol. I. Part I. General Determinants, up to 1841. Part II. Special Determinants, up to 1841. 17s. net. Vol. II. The Period 1841 to 1860. 17s. net. By T. Muir, M.A., LL.D.,

CATALOGUE

F.R.S.

An Introduction to the Theory of Infinite Series. By

T. J. I’a Bromwich, M.A., F.R.S. 15s. net.

Introduction to the Theory of Fourier’s Series and In- tegrals, and the Mathematical Theory of the Con- duction of Heat. By Prof. H. S. Carslaw, M.A., D.Sc.,

F.R.S.E. 14s. net.

LONDON: MACMILLAN AND CO., LTD.

‌transcriber’s note

The diagrams have been re-created, using accompanying formulas or descriptions from the text where possible.

In chapter XIV, pages 132–159, numerical values of 1 + 1 n, ϵx, and related quantities of British currency have been verified and rounded to the nearest digit.

On page 142 (page 146 in the original), the graphs of the natural logarithm and exponential functions, Figures 38 and 39, have been interchanged to match the surrounding text.

The vertical dashed lines in the natural logarithm graph, Figure 39 (Figure 38 in the original), have been moved to match the data in the corresponding table.

On page 164 (page 167 in the original), the graphs of the sine and cosine functions, Figures 44 and 45, have been interchanged to match the surrounding text.