## IITJEE Foundation Maths: Variation

DEFINITION:

One quantity A is said to vary directly as another B, when the two quantities depend upon each other in such a manner that if B is changed, A is changed in the same ratio.

NOTE: The word directly is often omitted, and A is said to vary as B.

For instance: if a train a moving at a uniform rate travels 40 miles in 60 minutes, it will travel 20 miles in 30 minutes, 80 miles in 120 minutes, and so on; the distance in each case being increased or diminished in the same ratio as the time. This is expressed by saying that when the velocity is uniform the distance is proportional to the time, or the distance varies as the time.

NOTATION: The symbol $\alpha$ is used to denote variation; so that, $A \alpha B$ is read as “A varies as B.”

Theorem I: If A varies as B, then A is equal to B multiplied by some constant quantity.

Note: If any pair of corresponding values of A and B are known, the constant m can be determined. For instance, if $A=B$, when $B=12$, we have $3=m \times 12$; and $m=\frac{1}{4}$, and $A=\frac{1}{4}B$

DEFINITION: One quantity A is said to vary inversely as another B, when A varies directly as the reciprocal of B.

The following is an illustration of inverse variation: If 6 men do a certain work in 8 hours, 12 men would do the same work in 4 hours; 2 men in 24 hours; and so on. Thus, it appears that when the number of men is increased, the same is proportionately decreased; and vice-versa.

Example 1: The cube root of x varies inversely as the square of y; if $x=8$, when $y=3$; find x when $y=\frac{3}{2}$.

Solution 1: By supposition, $\sqrt[3]{x}=\frac{m}{y^{2}}$, where m is constant. Putting $x=8, y=3$, we have $2=\frac{m}{9}$, so $m=18$, and $\sqrt[3]{x}=\frac{18}{y^{2}}$; hence, by putting $y=\frac{3}{2}$, we obtain $x=512$.

Example 2: The square of the time of a planet’s revolution varies as the cube of its distance from the Sun; find the time of Venus’s revolution, assuming the distances of the Earth and Venus from the Sun to be $91\frac{1}{4}$ and $66$ millions of miles respectively.

Let P be the periodic time measured in days, D the distance in millions of miles; we have

$P^{2} \alpha D^{3}$, or $P^{2}=k \times D^{3}$, where k is some constant.

For the Earth, $365 \times 365 = k \times 91 \frac{1}{4} \times 91 \frac{1}{4} \times 91 \times \frac{1}{4}$

hence, $k=\frac{4 \times 4 \times 4}{3.5}$

so that $P^{2}=\frac{4 \times 4 \times 4}{365}D^{3}$

For Venus, $P^{2}=\frac{4 \times 4 \times 4}{365} \times 66 \times 66 \times 66$

hence, $P= A \times 66 \times \sqrt{\frac{264}{365}} = 264 \times \sqrt{0.7233}$, approximately.

$P^{2}=264 \times 0.85=224.4$.

Hence, the time of revolution is nearly $224\frac{1}{2}$.

DEFINITION: One quantity is said to vary jointly as a number of others, when it varies directly as their product.

Thus, A varies jointly as B and C, when $A=m \times BC$. For instance, the interest on a sum of money varies jointly as the principal, the time, and the rate per cent.

DEFINITION:

A is said to vary directly as B and inversely as C, when A varies as $\frac{B}{C}$.

Theorem:

If A varies as B, when C is constant, and A varies as C when B is constant, then A will vary as BC when both B and C vary.

Proof:

Homework

The following are some illustrations of the theorems stated above:

The amount of work done by a given number of men varies directly as the number of days they work, and the amount of work done in a given time varies directly as the number of men; therefore, when the number of days and the number of men are both variable, the amount of work will vary as the product of the number of men and the number of days.

Again, in plane geometry, the area of a triangle varies directly as its base when the height is constant, and directly as the height when the base is constant; and when both the height and base are variable, the area varies as the product of the numbers representing the height and the base.

Example:

The volume of a right circular cone varies as the square of the radius of the base when the height is constant, and as the height when the base is constant. If the radius of base is 7cm, and the height is 15 cm, the volume is 770 cc, find the height of a cone whose volume is 132 cubic cm, and which stands on a base whose radius is 3cm.

Solution:9

Let h and r denote respectively the height and radius of the base measured in cm.; also let V be the volume in cubic cm.

Then. $V=m \times r^{2} \times h$, where m is constant.

By assumption, $770=m \times 7^{2} \times 15$

hence, $m = \frac{22}{21}$ so that $V=\frac{22}{21}r^{2}h$.

By substituting $V=132, r=3$, we get the following:

$132=\frac{22}{21} \times 9 \times h$

so that $h=14$; and, therefore the height is 14 cm.

Note:

A quantity A can vary jointly as (a product of) more than two variables also as is most often the case in real engineering. Further, the variations may be either direct or inverse. The principle is interesting because of its frequent occurence in physical sciences or engineering. For example, Boyle’s law in chemistry: It is found by experiment that the pressure (P) of a gas varies as the “absolute temperature” (T) (in Kelvin) when its volume (V) is constant and that the pressure varies inversely as the volume when the temperature is constant; that is,

$P \alpha T$, when V is constant and $P \alpha \frac{1}{V}$ when T is constant.

From these results we should expect that, when both t and v are variable, we should have the formula:

$P \alpha \frac{T}{V}$, or $PV=kT$, where k is a constant (based on laws of chemistry). And, by actual experiment this is found to be true.

Example:

The duration of a railway journey varies ditectly as the distance and inversely as the velocity; the velocity varies directly as the square root of the quantity of coal used per kilometer (don’t worry the days of steam engine/coal engine and resulting environmental degradation are over; but this is only a simple engineering application), and inversely as the number of carriages in the train. In a journey of 50 kilometers, in half an hour with 18 carriages 100 kg of coal is required; how much coal will be consumed in a journey of 42 kilometers, in 28 minutes with 16 carriages?

Solution:

Let t be the time expressed in hours; let d be the distance in kilometers; let v be the velocity in kmph; let q be the mass of coal (in kg) used per kilometers; and let c be the number of carriages.

We have $t \alpha \frac{d}{v}$ and $v \alpha \frac{\sqrt{q}}{c}$, and hence, $t \alpha \frac{cd}{\sqrt{q}}$, or $t=\frac{kcd}{\sqrt{q}}$, where k is a constant.

Substituting the values given, we have (since q=2),

$\frac{1}{2} = \frac{k \times 18 \times 50}{\sqrt{2}}$

that is, $k=\frac{1}{\sqrt{2} \times 18 \times 50}$.

Hence, $t=\frac{cd}{\sqrt{2} \times 18 \times 50\sqrt{q}}$

Substituting now the values of t, c, d given in the second part of the question, we have

$\frac{28}{60}=\frac{16 \times 42}{\sqrt{2} \times 18 \times 50 \times \sqrt{q} }$

that is, $\sqrt{q} = 4 \sqrt{2}$, hence $q=32$.

Hence, the quantity of coal is $42 \times 32 = 1344$ kg.

Tutorial problems on Variation:

1. If x varies as y, and $x=8$, when $y=15$, find x when $y=10$
2. If P varies as Q, and $P=7$ when $Q=3$, find P when $Q=2\frac{1}{3}$.
3.  If the square of x varies as the cube of y, and $x=3$, when $y=4$, find the value of y when $x=\frac{1}{\sqrt{3}}$.
4. A varies as B and C jointly; if $A=2$ when $B=\frac{3}{5}$ and $C=\frac{10}{27}$, find C when $A=54$ and $B=3$.
5. If A varies as C, and B varies as C, then $A \pm B$ and $\sqrt{AB}$ will each vary as C.
6. If A varies as BC, then B varies inversely as $\frac{C}{A}$.
7. P varies directly as Q and inversely as R; also $P=\frac{2}{3}$ when $Q=\frac{3}{7}$ and $R=\frac{9}{14}$; find Q when $P=\sqrt{48}$ and $R=\sqrt{75}$.
8. If x varies as y, prove that $x^{2}+y^{2}$ varies as $x^{2}-y^{2}$.
9. If y varies as the sum of two quantities, of which one varies directly as x and the other inversely as x; and if $y=6$ then $x=4$, and $y=3\frac{1}{3}$ when $x=3$, find the equation between x and y.
10. If y is equal to the sum of two quantities one of which varies as x directly, and the other as $x^{2}$ inversely; and, if $y=19$ when $x=2$, or 3; find y in terms of x.
11. If A varies directly as the square root of B and inversely as the cube of C, and if $A=3$, when $B=256$ and $C=2$, find B when $A=24$ and $C = \frac{1}{2}$
12. Given that $x+y$ varies as $x+\frac{1}{x}$, and that $x-y$ varies as $z- \frac{1}{z}$, find the relation between x and z, provided that $z=2$ when $x=3$ and $y=1$.
13. If A varies as B and C jointly, write B varies as $D^{2}$, and C varies inversely as A, show that A varies as D.
14. If y varies as the sum of three quantities of which the first is a constant, the second varies as x, and the third as $x^{2}$; and, if $y=0$ when $x=1$, $y=1$, when $x=2$, and $y=4$ when $x=3$; find y when $x=7$.
15. When a body falls down from rest the distance from the starting point varies as the square of the time it has been falling; if a body falls through 122.6 meters in 5 seconds, how far does it fall in 10 seconds? Also, how far does it fall in the tenth second?
16. Given that the volume of a sphere varies as the cube of its radius, and that when the radius is 3.5 cm, the volume is 176.7 cubic cm; find the volume when the radius is 1.75 cm.
17. The weight of a circular disc varies as the square of the radius when the thickness remains the same; it also varies as the thickness when the radius remains the same. Two discs have their thicknesses in the ratio of 9:8; find the ratio of their radii if the weight of the first is twice that of the second.
18. At a certain regatta, the numbers of races on each day varied jointly as the number of days from the beginning and end of the regatta up to and including the day in question. On three successive days there were respectively 6, 5 and 3 races. Which days were these, and how long did the regatta last?
19. The price of a diamond varies as the square of its weight (mass). Three rings of equal weight, each composed of a diamond set in gold, have values INR a, INR b, INR c, the diamonds in them weighing 3, 4, 5 carats respectively. Show that the value of a diamond of one carat is INR $(\frac{a+c}{2}-b)$, the cost of workmanship being the same for each ring.
20. Two persons are awarded pensions in proportion to the square root of the number of root of the number of years they have served. One has served 9 years longer than the other and receives a pension greater by INR 500. If the length of service of the first had exceeded that of the second by $4\frac{1}{4}$ years their pensions would have been in the proportion of 9:8. How long had they served and what were their respective pensions?
21. The attraction of a planet on the satellites varies directly as the mass (M) of the planet, and inversely as the square of the distance (D); also the square of a satellite’s time of revolution varies directly as the distance and inversely as the force of attraction. If $m_{1}$, $d_{1}$, $t_{1}$ and $m_{2}$, $d_{2}$, $t_{2}$ are simultaneous values of M, D, T respectively, prove that $\frac{m_{1}t_{1}^{2}}{m_{2}t_{2}^{2}} = \frac{d_{1}^{3}}{d_{2}^{3}}$. Hence, find the time of revolution of that moon of Jupiter whose distance is to the distance of our Moon as 35:31, having given that the mass of Jupiter is 343 times that of the Earth, and that the Moon’s period is 27.32 days.
22. The consumption of coal by a locomotive varies as the square of the velocity; when the speed is 32 kmph the consumption of coal per hour is 2 tonnes: if the price of coal is INR 10 per tonne, and the other expenses of the engine be INR 11.25 an hour, find the least cost of a journey of 100 km.

Cheers,

Nalin Pithwa

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