Variations

A straight line through the origin (0,0) shows a direct variation.

When one quantity increases as another decreases at the same rate, they are inversely proportional.

For example:

• The more workers used to dig a field, the less time it takes.
• The time taken for a journey is inversely proportional to the speed.

This is written as:

y ∝ 1/x => y = k/x

Example 3

The volume V of a gas is inversely proportional to the pressure P. If V = 0.8 m³ when P = 250 kg/m², find:

(a) the formula for V in terms of P (b) V when P = 1000 kg/m²

Solution

(a) V = k/P → 0.8 = k/250 → k = 200

So, V = 200/P

(b) V = 200/1000 = 0.2 m³

Example 4

Given y ∝ 1/x, and y = 15 when x = 8.

Find the equation, and use it to find:

(a) y when x = 10 (b) x when y = 3

Solution

y = k/x → 15 = k/8 → k = 120 So, y = 120/x

(a) y = 120/10 = 12 (b) 3 = 120/x → x = 40

The graph of y against x is a curve that falls as x increases.

Sometimes, a quantity is proportional to a power of another quantity.

For example:

The area A of a circle is proportional to the square of its radius r. => A ∝ r² → A = kr²

Example 5

The mass m of a metal sphere is proportional to the cube of its radius r. If m = 42 kg when r = 10 cm, find:

(a) the formula for m (b) the radius when m = 5.25 kg

Solution

(a) m = kr³ → 42 = k × 1000 → k = 0.042

So, m = 0.042r³

(b) 5.25 = 0.042r³ → r³ = 125 → r = 5 cm

Example 6

M ∝ N² and M = 2.7 when N = 0.3

Find:

(a) M in terms of N (b) M when N = 1.5 (c) N when M = 0.3

Solution

(a) M = kN² → 2.7 = k × 0.09 → k = 30

So, M = 30N²

(b) M = 30 × 1.5² = 67.5 (c) 0.3 = 30N² → N² = 0.01 → N = 0.1

When a quantity varies as the product of two or more others, it is said to vary jointly with them.

Example 7

Volume V varies jointly with temperature T and inversely with pressure P. If V = 0.5 m³ when T = 300 K and P = 80 kg/m², find the formula for V in terms of T and P.

Solution

V = kT/P 0.5 = k × 300 / 80 → k = 0.5 × 80 / 300 = 0.133 So, V = 0.133T/P

Example 8

m varies jointly with p and q. If m = 15 when p = 12 and q = 5, find m in terms of p and q. Then find m when p = 3 and q = 28.

Solution

m = kpq → 15 = k × 12 × 5 → k = 1/4 So, m = (1/4)pq When p = 3, q = 28 → m = (1/4) × 3 × 28 = 21

Example 9

A cylinder has radius 3 cm and volume 10 cm³. If the radius increases to 4 cm and the height remains unchanged, what happens to the volume?

Solution

Volume ∝ r² → Old volume ∝ 3² = 9 New volume ∝ 4² = 16 Ratio = 16:9 New volume = (16/9) × 10 = 17.78 cm³

Example 10

A pyramid has a square base. The height is decreased by 10% and volume remains constant. What percentage increase in the side of the base offsets the decrease in height?

Solution

Volume of a pyramid ∝ base area × height Let side be s, new height = 0.9h To maintain volume: s² × 0.9h = original s² × h So, s² must increase by 1/0.9 = 1.111 → 11.1% increase in area √1.111 = 1.054 → side must increase by 5.4%