5.02a, 5.02b, 6.02a - Proportion with powers and roots | OCR GCSE Mathematics

A quantity can be proportional to a square, a cube or a root rather than to the input itself. Build the correct equation, find its constant and track how powers change scale factors, while keeping the allowed values in view.

This lesson is for Higher tier.

Direct proportion to powers

If doubling a length quadruples a quantity, what should happen when the length triples? The answer can be nine times as much: proportionality to a square changes the scale factor itself.

Start with $y=3x^2$ . When $x=2$ , $y=12$ ; when $x=4$ , $y=48$ . Doubling $x$ has multiplied $x^2$ by four, so it has multiplied $y$ by four. It is $y/x^2$ , rather than $y/x$ , that stays constant.

Direct proportion to a power

y\propto x^n\quad\Longrightarrow\quad y=kx^n

Find $k=y/x^n$ from a known pair with $x^n\ne0$ . For a square, use $x^2$ ; for a cube, use $x^3$ . These proportionality equations are not supplied on the 2027 formula sheets.

For a nonzero scale factor $a$ , replacing $x$ by $ax$ multiplies $x^n$ , and hence $y$ , by $a^n$ , provided the powers are defined and the new input stays in the allowed domain. Thus doubling an input gives a factor of four for a square law and eight for a cube law. Use the actual power in the question.

Worked example 1: a square law

Question. $y$ is directly proportional to $x^2$ . When $x=3$ , $y=54$ . Find $y$ when $x=6$ .

Plan. Find the constant by dividing by the square, then use the new input.

\begin{aligned} 54&=k\times3^2 &&\text{known pair}\\ k&=54\div9=6 &&\text{divide by the square}\\ y&=6x^2 &&\text{particular equation}\\ y&=6\times6^2=216 &&\text{new input} \end{aligned}

Answer. $y=216$ .

Check. The input doubled, so the output should be four times 54: $4\times54=216$ .

Keep the square law and change which quantity is unknown. Reversing a square can produce two possible inputs.

Worked example 2: reversing a square law

Question. $y$ is directly proportional to $x^2$ , with $y=54$ when $x=3$ . Find all real values of $x$ when $y=150$ .

Plan. Use the same constant, isolate $x^2$ , then consider both signs.

\begin{aligned} k&=54\div3^2=6 &&\text{known constant}\\ 150&=6x^2 &&\text{substitute the output}\\ x^2&=25 &&\text{divide by 6}\\ x&=5\text{ or }-5 &&\text{both have square 25} \end{aligned}

Answer. $x=5$ or $x=-5$ .

Check. Both $6(5)^2$ and $6(-5)^2$ equal 150. A length context would require only the positive value; this question asks for all real values.

For a cube law, the procedure is the same but the reverse operation is a cube root. For example, $z=2t^3$ and $z=250$ give $t^3=125$ , so $t=5$ . There is no second real value $-5$ : its cube is $-125$ . In a square law $y=kx^2$ with nonzero $k$ , a real input requires $y/k\ge0$ .

Watch out

From $y\propto x^2$ , writing $y=kx$ loses the square. From $x^2=25$ , writing only $x=5$ loses a possible solution unless the context excludes negative inputs.

Reverse a new square law and retain all allowed inputs.

Inverse proportion to powers

Inverse proportion to a square keeps $x^2y$ fixed, so a change in the input has a stronger effect than in ordinary inverse proportion.

If $y=36/x^2$ , then doubling $x$ makes the denominator four times as large and makes $y$ one quarter as large. Inverse proportion to a cube uses $x^3$ instead; doubling a positive input makes the output one eighth as large.

Inverse proportion to a power

y\propto\frac1{x^n}\quad\Longrightarrow\quad y=\frac{k}{x^n},\qquad x^n\ne0

Find $k=yx^n$ . Multiplying the input by a nonzero factor $a$ divides the output by $a^n$ , provided the new input stays in the domain. These equations are not supplied on the 2027 formula sheets.

The power belongs in the denominator before you substitute any values. Finding $k$ is then a multiplication, because you are undoing division by that power.

Worked example 3: inverse proportion to a square

Question. $y$ is inversely proportional to $x^2$ . When $x=2$ , $y=18$ . Find $y$ when $x=6$ .

Plan. Write the inverse-square equation, find its constant, then substitute.

\begin{aligned} 18&=\frac{k}{2^2} &&\text{known pair}\\ k&=18\times4=72 &&\text{multiply by the square}\\ y&=\frac{72}{6^2}=2 &&\text{new input} \end{aligned}

Answer. $y=2$ .

Check. The input tripled, so the output is divided by $3^2=9$ : $18\div9=2$ .

Change only the specified power from a square to a cube. The known pair and the new input stay the same; the constant must be recalculated for the new law.

Worked example 4: inverse proportion to a cube

Question. $y$ is inversely proportional to $x^3$ . When $x=2$ , $y=18$ . Find $y$ when $x=6$ .

Plan. Use cubes in both the constant calculation and the new denominator.

\begin{aligned} 18&=\frac{k}{2^3} &&\text{new inverse-cube law}\\ k&=18\times8=144 &&\text{multiply by the cube}\\ y&=\frac{144}{6^3}=\frac23 &&\text{new input} \end{aligned}

Answer. $y=\frac23$ .

Check. The input tripled, so the output is divided by $3^3=27$ : $18\div27=\frac23$ .

To find the input instead, first isolate its power. For example, $y=80/x^2$ and $y=5$ give $5x^2=80$ , then $x^2=16$ , so $x=4$ or $-4$ . Retain only values allowed by the context, and never include zero in an inverse-power denominator.

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5.02a, 5.02b, 6.02a - Proportion with powers and roots

Direct proportion to powers

Inverse proportion to powers

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