Proportional Relationships

Direct and inverse proportion, finding constants, and graph shapes

SPEC 2.7 • GRAPHS AND FUNCTIONS

What is Proportion?

A proportional relationship describes how one quantity changes in relation to another.

Direct Proportion

When one quantity increases, the other increases at a constant rate.

y ∝ x

"y is proportional to x"

Inverse Proportion

When one quantity increases, the other decreases at a constant rate.

y ∝ 1/x

"y is inversely proportional to x"

Key Concept

The symbol ∝ means "is proportional to". To convert to an equation, we introduce a constant of proportionality, k.

Direct Proportion

Definition

If y ∝ xⁿ, then y = kxⁿ for some constant k > 0.

y \propto x^{n} \Rightarrow y = k x^{n}

Common forms of direct proportion:

ProportionEquationExample

y ∝ xy = kxDistance ∝ Time (constant speed)

y ∝ x²y = kx²Area ∝ (Radius)²

y ∝ x³y = kx³Volume ∝ (Length)³

y ∝ √xy = k√xPeriod ∝ √(Length) for pendulum

Finding the Constant k

To find the constant of proportionality:

Method

Write the equation: Convert y ∝ f(x) to y = k·f(x)

Substitute known values: Use given (x, y) pair

Solve for k: Rearrange to find k

Write complete equation: y = (value of k)·f(x)

Tip

Once you have k, you can find y for any value of x, or find x for any value of y.

Worked Example01

Given that y is directly proportional to x, and y = 12 when x = 4, find:

(a) the value of k

(b) y when x = 7

(c) x when y = 21

1

y ∝ x means y = kx

2

Substitute y = 12, x = 4: 12 = k × 4

3

(a) k = 12 ÷ 4 = 3

4

So y = 3x

5

(b) When x = 7: y = 3 × 7 = 21

6

(c) When y = 21: 21 = 3x, so x = 7

Worked Example02

y is directly proportional to x². Given that y = 32 when x = 4, find the value of y when x = 5.

1

y ∝ x² means y = kx²

2

Substitute y = 32, x = 4: 32 = k × 4² = 16k

3

k = 32 ÷ 16 = 2

4

So y = 2x²

5

When x = 5: y = 2 × 5² = 2 × 25 = 50

Inverse Proportion

Definition

If y ∝ 1/xⁿ, then y = k/xⁿ for some constant k > 0.

y \propto \frac{1}{x ^{n}} \Rightarrow y = \frac{k}{x ^{n}}

Common forms of inverse proportion:

ProportionEquationExample

y ∝ 1/xy = k/xTime ∝ 1/Speed (fixed distance)

y ∝ 1/x²y = k/x²Gravitational force ∝ 1/r²

y ∝ 1/√xy = k/√xVarious physics applications

Important

With inverse proportion, as x increases, y decreases (and vice versa). The product xy = k remains constant for y ∝ 1/x.

Worked Example03

y is inversely proportional to x. Given that y = 10 when x = 3, find the value of y when x = 6.

1

y ∝ 1/x means y = k/x

2

Substitute y = 10, x = 3: 10 = k/3

3

k = 10 × 3 = 30

4

So y = 30/x

5

When x = 6: y = 30/6 = 5

Note: When x doubled (3 → 6), y halved (10 → 5). This confirms inverse proportion!

Worked Example04

y is directly proportional to √x. Given that y = 6 when x = 4, find the value of x when y = 15.

1

y ∝ √x means y = k√x

2

Substitute y = 6, x = 4: 6 = k × √4 = k × 2

3

k = 6 ÷ 2 = 3

4

So y = 3√x

5

When y = 15: 15 = 3√x

6

√x = 5, so x = 5² = 25

Graphs of Proportion

Key Point

All direct proportion graphs pass through the origin (0, 0).

Inverse proportion graphs (y ∝ 1/xⁿ) have asymptotes at x = 0 and y = 0.

Worked Example05

y is inversely proportional to x². If x is doubled, find the percentage change in y.

1

y ∝ 1/x² means y = k/x²

2

Original: y₁ = k/x²

3

New (x doubled): y₂ = k/(2x)² = k/4x² = ¼ × (k/x²) = ¼y₁

4

Change = y₂ − y₁ = ¼y₁ − y₁ = −¾y₁

5

Percentage change = (−¾y₁/y₁) × 100% = −75%

y decreases by 75% (or y becomes ¼ of its original value)

Combined Proportion

Sometimes a quantity depends on more than one variable:

Combined Proportion

If z ∝ x²/y, then z = kx²/y for some constant k.

z \propto \frac{x ^{2}}{y} \Rightarrow z = \frac{k x ^{2}}{y}

Examples

z ∝ xy → z = kxy

z ∝ x/y → z = kx/y

z ∝ x²y³ → z = kx²y³

z ∝ √x/y² → z = k√x/y²

Method

1

Write equation with k

2

Substitute ALL known values

3

Solve for k

4

Use equation to find unknowns

Worked Example06

z is directly proportional to x² and inversely proportional to y. Given that z = 27 when x = 3 and y = 2, find z when x = 6 and y = 4.

1

z ∝ x²/y means z = kx²/y

2

Substitute z = 27, x = 3, y = 2: 27 = k × 3²/2 = 9k/2

3

27 = 9k/2, so 54 = 9k, k = 6

4

So z = 6x²/y

5

When x = 6, y = 4: z = 6 × 36/4 = 216/4 = 54

Real-World Applications

Proportional relationships appear throughout science and engineering:

ContextRelationshipEquation

Gravitational forceF ∝ 1/r²F = Gm₁m₂/r²

Light intensityI ∝ 1/d²I = k/d²

Kinetic energyE ∝ v²E = ½mv²

Pendulum periodT ∝ √LT = 2π√(L/g)

Pressure (gas)P ∝ 1/VPV = constant

Exam Tip

In physics problems, proportionality relationships let you compare two situations without knowing the exact constant k.

Worked Example07

The gravitational force F between two objects is inversely proportional to the square of the distance r between them. If the force is F₀ when the distance is r₀, find the force when the distance is 3r₀.

1

F ∝ 1/r² means F = k/r²

2

Original: F₀ = k/r₀², so k = F₀r₀²

3

New distance: r = 3r₀

4

New force: F = k/(3r₀)² = k/9r₀²

5

Substitute k = F₀r₀²: F = F₀r₀²/9r₀² = F₀/9

The force is 1/9 of the original force.

Practice

Question 1: Given y ∝ x³ and y = 54 when x = 3, find y when x = 2.

Step 1: y = kx³

Step 2: 54 = k × 3³ = 27k, so k = 2

Step 3: y = 2x³

Step 4: When x = 2: y = 2 × 8 = 16

Practice

Question 2: P is inversely proportional to √V. Given P = 8 when V = 9, find V when P = 6.

Step 1: P = k/√V

Step 2: 8 = k/√9 = k/3, so k = 24

Step 3: P = 24/√V

Step 4: When P = 6: 6 = 24/√V

Step 5: √V = 24/6 = 4, so V = 16

Practice

Question 3: The intensity I of light is inversely proportional to d². If I = 200 when d = 3, find I when d = 5.

Step 1: I = k/d²

Step 2: 200 = k/3² = k/9, so k = 1800

Step 3: I = 1800/d²

Step 4: When d = 5: I = 1800/25 = 72

Summary

Key Concepts

TypeSymbolEquationGraph

Directy ∝ xy = kxLine through O

Direct (squared)y ∝ x²y = kx²Parabola through O

Inversey ∝ 1/xy = k/xHyperbola

Inverse (squared)y ∝ 1/x²y = k/x²Steeper hyperbola

Method

Convert ∝ to equation with constant k

Substitute known values to find k

Use equation to find unknown values