Chapter no 5

Calculus Made Easy

NExT STAGE. WHAT TO DO WITH CONSTANTS.

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

Our next step is to find out what effect on the process of differenti- ating is caused by the presence of constants, that is, of numbers which don’t change when x or y change their values.

‌Added Constants.

Let us begin with some simple case of an added constant, thus: Let y = x3 + 5.

Just as before, let us suppose x to grow to x+dx and y to grow to y+dy.

Then: y + dy = (x + dx)3 + 5

= x3 + 3x2 dx + 3x(dx)2 + (dx)3 + 5.

Neglecting the small quantities of higher orders, this becomes

y + dy = x3 + 3x2 · dx + 5.

Subtract the original y = x3 + 5, and we have left:

dy = 3x2 dx.

dy = 3x2. dx

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

If the additional constant had been of negative value, such as

−5 or −b, it would equally have disappeared.

Multiplied Constants.

Take as a simple experiment this case: Let y = 7x2.

Then on proceeding as before we get:

y + dy = 7(x + dx)2

= 7{x2 + 2x · dx + (dx)2}

= 7x2 + 14x · dx + 7(dx)2.

Then, subtracting the original y = 7x2, and neglecting the last term, we have

dy = 14x · dx. dy

= 14x.

Let us illustrate this example by working out the graphs of the

equations y = 7x2 and dy

= 14x, by assigning to x a set of successive

values, 0, 1, 2, 3, etc., and finding the corresponding values of y and

of .

These values we tabulate as follows:

−1

−2

−3

112

175

−14

−28

−42

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

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

Now if we look back at p. 18, we shall see that simply differenti- ating x2 gives us 2x. So that the differential coefficient of 7x2 is just

* See p. 76 about slopes of curves.

150

100

−3 −2 −1

2 3 4

−50

y dx

200

150

100

−3 −2 −1 0 1 2 3 4 5

Fig. 6.—Graph of y = 7x2.

‌Fig. 6a.—Graph of dy = 14x.

7 times as big as that of x2. If we had taken 8x2, the differential coeffi- cient would have come out eight times as great as that of x2. If we put y = ax2, we shall get

dx = a × 2x.

If we had begun with y = axn, we should have had dy

= a × nxn−1.

So that any mere multiplication by a constant reappears as a mere

multiplication when the thing is differentiated. And, what is true about multiplication is equally true about division: for if, in the example above, we had taken as the constant 1 instead of 7, we should have had

the same 1 come out in the result after differentiation.

Some Further Examples.

The following further examples, fully worked out, will enable you to master completely the process of differentiation as applied to ordinary

algebraical expressions, and enable you to work out by yourself the examples given at the end of this chapter.

x5 3

Differentiate y =

7 − 5 .

is an added constant and vanishes (see p. 25). 5

We may then write at once

× ×

dy = 1 5 x5−1, dx

or dy = 5 x4.

dx 7

—

Differentiate y = a√x 1√a.

2

The term 1√a vanishes, being an added constant; and as a√x, in 2

1

the index form, is written ax 2 , we have

dy 1 1 −1 a − 1

dx = a × 2 × x2 = 2 × x 2 ,

dy a

or dx = 2√x.
If ay + bx = by − ax + (x + y)√a2 − b2,

find the differential coefficient of y with respect to x.

As a rule an expression of this kind will need a little more knowledge than we have acquired so far; it is, however, always worth while to try whether the expression can be put in a simpler form.

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

The expression may be written

(a − b)y + (a + b)x = (x + y)√a2 − b2.

Squaring, we get

(a − b)2y2 + (a + b)2x2 + 2(a + b)(a − b)xy = (x2 + y2 + 2xy)(a2 − b2),

which simplifies to

(a − b)2y2 + (a + b)2x2 = x2(a2 − b2) + y2(a2 − b2); or [(a − b)2 − (a2 − b2)]y2 = [(a2 − b2) − (a + b)2]x2, that is 2b(b − a)y2 = −2b(b + a)x2;

hence y = r a + bx and dy = r a + b .

a − b dx a − b
The volume of a cylinder of radius r and height h is given by the formula V = πr2h. Find the rate of variation of volume with the radius when r = 5.5 in. and h = 20 in. If r = h, find the dimensions of the cylinder so that a change of 1 in. in radius causes a change of 400 cub. in. in the volume.

The rate of variation of V with regard to r is

dV

= 2πrh.

dr

If r = 5.5 in. and h = 20 in. this becomes 690.8. It means that a change of radius of 1 inch will cause a change of volume of

690.8 cub. inch. This can be easily verified, for the volumes with

r = 5 and r = 6 are 1570 cub. in. and 2260.8 cub. in. respectively, and

2260.8 − 1570 = 690.8.

Also, if

dV

r = h,

dr

= 2πr2 = 400 and r = h = 400 = 7.98 in.

r

2π
The reading θ of a F´ery’s Radiation pyrometer is related to the Centigrade temperature t of the observed body by the relation

θ t 4

= ,

θ1 t1

where θ1 is the reading corresponding to a known temperature t1 of the observed body.

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

The sensitiveness is the rate of variation of the reading with the

dθ

temperature, that is

. The formula may be written

and we have

θ1

θ = 4

4 25t4

= 10004 ,

dθ 100t3 t3

dt = 10004 = 10, 000, 000, 000 .

dθ

When t = 800, 1000 and 1200, we get

respectively.

= 0.0512, 0.1 and 0.1728

The sensitiveness is approximately doubled from 800◦ to 1000◦, and becomes three-quarters as great again up to 1200◦.

‌Exercises II. (See p. 252 for Answers.) Differentiate the following:‌

y = ax3

+ 6. (2) y = 13x2 − c.

1 1 1 1

(3) y = 12x2 + c 2 . (4) y = c 2 x2 .

u =

azn 1

—

.

c
y = 1.18t2 + 22.4.

Make up some other examples for yourself, and try your hand at differentiating them.
If lt and l0 be the lengths of a rod of iron at the temperatures t◦ C. and 0◦ C. respectively, then lt = l0(1+0.000012t). Find the change of length of the rod per degree Centigrade.
It has been found that if c be the candle power of an incandes- cent electric lamp, and V be the voltage, c = aV b, where a and b are constants.
Find the rate of change of the candle power with the voltage, and calculate the change of candle power per volt at 80, 100 and 120 volts in the case of a lamp for which a = 0.5 × 10−10 and b = 6.
The frequency n of vibration of a string of diameter D, length L
and specific gravity σ, stretched with a force T , is given by

DL

πσ

n = 1 rgT .

Find the rate of change of the frequency when D, L, σ and T are varied singly.
The greatest external pressure P which a tube can support with- out collapsing is given by
2E t3

P =

1 − σ2

D3 ,

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

Compare the rate at which P varies for a small change of thickness and for a small change of diameter taking place separately.
Find, from first principles, the rate at which the following vary with respect to a change in radius:
1. the circumference of a circle of radius r;
2. the area of a circle of radius r;
3. the lateral area of a cone of slant dimension l; (d ) the volume of a cone of radius r and height h;
  (e) the area of a sphere of radius r;
  
  (f ) the volume of a sphere of radius r.
The length L of an iron rod at the temperature T being given by L = lt 1 + 0.000012(T − t) , where lt is the length at the temperature t, find the rate of variation of the diameter D of an iron tyre suitable for being shrunk on a wheel, when the temperature T varies.