POP

E2.8 Proportion: Direct & Inverse Variation with Powers and Roots

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"Directly proportional" means two quantities grow together; "inversely proportional" means one grows as the other shrinks. This is the algebraic engine behind real contexts like gravitational force, material cost, or braking distance, and it always reduces to the same three-step process: set up the equation, find the constant, then use the equation.

01 — Explanation

How do I set up an equation for direct proportion?

Replace the proportionality symbol $\propto$ with " $=k\times$ ", where $k$ is a fixed constant:

y \propto x^2 \quad\Longrightarrow\quad y = kx^2 \qquad\qquad y \propto \sqrt[3]{x} \quad\Longrightarrow\quad y = k\sqrt[3]{x}

02 — Explanation

How do I set up an equation for inverse proportion?

The variable moves to the denominator instead:

y \propto \frac{1}{x^3} \quad\Longrightarrow\quad y = \frac{k}{x^3} \qquad\qquad y \propto \frac{1}{\sqrt{x}} \quad\Longrightarrow\quad y = \frac{k}{\sqrt{x}}

Exam Tip

The word "inversely" is what sends the variable to the bottom of the fraction. Writing $y=kx^3$ for an inverse relationship is one of the most common marks lost in this topic.

03 — Explanation

How do I find the constant of proportionality, k?

Substitute one known pair of values into your equation and rearrange to make $k$ the subject.

$y$ is inversely proportional to $x^3$ , and $y=2$ when $x=5$ :

y=\frac{k}{x^3} \implies 2=\frac{k}{125} \implies k=250

04 — Explanation

How do I use my equation to find a missing value?

Once $k$ is known, substitute the new given value into the completed equation and solve for whichever variable is missing.

Using $y=\dfrac{250}{x^3}$ , find $y$ when $x=10$ :

y = \frac{250}{10^3} = \frac{250}{1000} = 0.25

Find $x$ when $y=250$ :

250 = \frac{250}{x^3} \implies x^3=1 \implies x=1

05 — Explanation

What effect does doubling x have on y, without finding k at all?

For $y=kx^n$ , doubling $x$ multiplies $y$ by $2^n$ ; for $y=\dfrac{k}{x^n}$ , doubling $x$ divides $y$ by $2^n$ . This shortcut is useful when a question only asks for the effect of a change, not an exact value.

If $y\propto x^2$ and $x$ is tripled, $y$ is multiplied by $3^2=9$ . If $y\propto\dfrac{1}{x^2}$ and $x$ is tripled instead, $y$ is divided by $9$ — it becomes $\dfrac{1}{9}$ of its original value.

Exam Tip

This "scale factor" shortcut only works cleanly for one variable changing by a multiplying factor (e.g. " $x$ is doubled" or " $x$ is halved"). If a question gives an additive change instead (e.g. " $x$ increases by $5$ "), you must find $k$ and substitute properly — the shortcut does not apply.

Worked Example

$y$ is directly proportional to the square root of $x$ . Given that $y=12$ when $x=9$ , find the value of $y$ when $x=25$ .

Write the general equation: $y=k\sqrt{x}$ .
Substitute the known pair: $12 = k\sqrt{9} = 3k \implies k=4$ .
State the full equation: $y=4\sqrt{x}$ .
Substitute $x=25$ : $y = 4\sqrt{25} = 4\times5=20$ .

Exam Tip

Always find and write down the full equation connecting the two variables before substituting the second value — jumping straight to a ratio between the two given situations is a common source of arithmetic errors with roots and powers, and the full equation also earns method marks even if the final arithmetic slips.