R10, R13 - Constructing proportional relationships

Turn statements about direct and inverse proportion into equations, then use known values to find the constant. You will also work with powers, roots and two changing quantities, and check whether a proposed model fits its data.

Forming equations and finding the constant (Higher)

Can the same pair of numbers fit both direct and inverse proportion? Suppose $x=5$ and $y=6$ . You will find that this pair supports two very different predictions, so the relationship matters as much as the numbers.

You already know

Direct proportion keeps the ratio $y/x$ constant. Inverse proportion keeps the product $xy$ constant. Substitute a known pair into a given equation, then solve for the unknown number.

Constructing an equation means choosing its form from the words before finding the constant $k$ . The symbol $\propto$ means “is proportional to”. Replacing it with an equals sign requires a constant multiplier.

Direct and inverse proportional equations

\begin{aligned} y\propto x&\quad\Longrightarrow\quad y=kx \\ y\propto\frac1x&\quad\Longrightarrow\quad y=\frac{k}{x}\qquad(x\ne0) \end{aligned}

Use one constant $k$ throughout each relationship. To find $k$ in the direct case, use $y/x$ with a non-zero input; in the inverse case, use $xy$ .

Not on the formula sheet: learn it.

One known pair fixes the constant only after the type of relationship is known.

Worked example 1: a direct relationship

Question. $y$ is directly proportional to $x$ . When $x=5$ , $y=6$ . Find an equation and find $y$ when $x=10$ .

Plan. Use $y=kx$ , substitute the known pair, then use the resulting equation.

\begin{aligned} 6&=5k &&\text{substitute the pair}\\ k&=\frac65=1.2 &&\text{divide by }5\\ y&=1.2x &&\text{insert the constant}\\ y&=1.2(10)=12 &&\text{use the new input} \end{aligned}

Why this form? Direct proportion means that multiplying $x$ by a factor multiplies $y$ by the same factor.

Answer. $y=1.2x$ ; at $x=10$ , $y=12$ .

Check. The input doubled from $5$ to $10$ , and the output doubled from $6$ to $12$ .

Change just the relationship to inverse proportion. Keep the known pair and new input unchanged so you can see what the words change.

Worked example 2: an inverse relationship

Question. $y$ is inversely proportional to $x$ . When $x=5$ , $y=6$ . Find an equation and find $y$ when $x=10$ .

Plan. Use $y=k/x$ , find the constant product, then substitute.

\begin{aligned} 6&=\frac{k}{5} &&\text{substitute the pair}\\ k&=30 &&\text{multiply by }5\\ y&=\frac{30}{x} &&\text{insert the constant}\\ y&=\frac{30}{10}=3 &&\text{use the new input} \end{aligned}

Why multiply by $5$ ? It undoes division by the input in the inverse equation.

Answer. $y=30/x$ ; at $x=10$ , $y=3$ .

Check. Both pairs have product $30$ : $5(6)=10(3)$ .

The opening pair fits both equations, but their predictions are $12$ and $3$ . A single pair alone cannot tell you which relationship applies.

Watch out

Wrong: for the inverse example, $k=6\div5=1.2$ . That finds a constant ratio, which belongs to direct proportion. Correct: $k=6\times5=30$ , so $y=30/x$ . Check by substituting the given pair.

Your turn changes the known output in the inverse example; the relationship and inputs stay the same.

Proportion involving powers (Higher)

If $y$ is proportional to $x^2$ , the squared input takes the place of $x$ in the direct-proportion method.

For example, $y=2x^2$ gives $y=2$ , $8$ and $18$ at $x=1$ , $2$ and $3$ . Doubling $x$ multiplies $x^2$ , and therefore $y$ , by $4$ . Tripling $x$ multiplies $y$ by $9$ .

Proportion to a power

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

For inverse proportion to $x^n$ , use $y=k/x^n$ , with $x\ne0$ . Here the stated power $n$ is part of the relationship; it is not something to guess from one pair.

Not on the formula sheet: learn it.

Start with the square model, then change only direct to inverse proportion.

Worked example 3: finding a squared-input equation

Question. $y\propto x^2$ . When $x=3$ , $y=18$ . Find the equation and the positive value of $x$ when $y=50$ .

Plan. Find $k$ using the squared input, then solve the new equation.

\begin{aligned} 18&=k(3^2)=9k &&\text{square the given input}\\ k&=2 &&\text{divide by }9\\ 50&=2x^2 &&\text{use }y=2x^2\\ x^2&=25 &&\text{divide by }2\\ x&=5 &&\text{take the positive root} \end{aligned}

Why the positive root? The question specifies positive $x$ . Without that restriction, $-5$ also satisfies $x^2=25$ .

Answer. $y=2x^2$ ; the required $x$ is $5$ .

Check. $2(3^2)=18$ and $2(5^2)=50$ .

Change only the proportional relationship to inverse-square. The given pair and target output stay fixed.

Worked example 4: an inverse-square equation

Question. $y\propto1/x^2$ . When $x=3$ , $y=18$ . Find the equation and the positive value of $x$ when $y=50$ .

Plan. Find $k$ from the product $yx^2$ , then rearrange before taking a root.

\begin{aligned} k&=18(3^2)=162 &&\text{multiply by the denominator}\\ y&=\frac{162}{x^2} &&\text{insert the constant}\\ 50x^2&=162 &&\text{multiply by }x^2\\ x^2&=3.24 &&\text{divide by }50\\ x&=1.8 &&\text{positive square root} \end{aligned}

Why is the input smaller now? In inverse-square proportion, a larger positive output requires a smaller positive input.

Answer. $y=162/x^2$ ; $x=1.8$ .

Check. $162/1.8^2=50$ .

The power applies before finding the constant.

Watch out

Wrong: from $18=k(3^2)$ , write $k=18\div3=6$ . This ignores the square. Correct: $k=18\div9=2$ . In inverse-square proportion the constant is instead $yx^2$ , so check which side of the fraction contains the input.

Change the known output in Worked example 4 from $18$ to $32$ .

Proportion involving roots (Higher)

A root can be the whole proportional input too: $y\propto\sqrt{x}$ means $y=k\sqrt{x}$ , where $x\ge0$ for real square roots.

For $y=4\sqrt{x}$ , increasing $x$ from $9$ to $36$ doubles $y$ , from $12$ to $24$ . The input must quadruple to double its square root. Cube-root relationships work in the same way: $y\propto\sqrt[3]{x}$ means $y=k\sqrt[3]{x}$ .

These forms are not on the formula sheet: derive them from the direct-proportion form, using the stated root as its input.

Worked example 5: a square-root relationship

Question. $y\propto\sqrt{x}$ . When $x=9$ , $y=12$ . Find the equation, then find $x$ when $y=28$ .

Plan. Evaluate the given root to find $k$ ; isolate the new root before squaring.

\begin{aligned} 12&=k\sqrt9=3k &&\text{evaluate the root}\\ k&=4 &&\text{divide by }3\\ 28&=4\sqrt{x} &&\text{use the equation}\\ \sqrt{x}&=7 &&\text{divide before squaring}\\ x&=49 &&\text{square both sides} \end{aligned}

Why isolate the root first? Then squaring undoes just the root, without also squaring the multiplier.

Answer. $y=4\sqrt{x}$ ; $x=49$ .

Check. $4\sqrt{49}=4(7)=28$ .

Change only the root from square root to cube root. The known pair and target output stay the same.

Worked example 6: a cube-root relationship

Question. $y\propto\sqrt[3]{x}$ . When $x=9$ , $y=12$ . Find $x$ when $y=28$ , giving an exact answer.

Plan. Keep the constant exact, divide the two output equations, then cube the root ratio.

\begin{aligned} k&=\frac{12}{\sqrt[3]{9}} &&\text{find the constant}\\ \frac{28}{12}&=\frac{\sqrt[3]{x}}{\sqrt[3]{9}} &&\text{cancel the same }k\\ \frac{x}{9}&=\left(\frac73\right)^3=\frac{343}{27} &&\text{cube both sides}\\ x&=\frac{343}{3} &&\text{multiply by }9 \end{aligned}

Why use a ratio? The exact constant cancels, avoiding a rounded cube root.

Answer. $x=343/3$ .

Check. The cube root of $x/9$ is $7/3$ , the required output factor $28/12$ .

A root must stay attached to its intended input.

Watch out

Wrong: from $28=4\sqrt{x}$ , write $x=28^2/4=196$ . Squaring the left side also squares the factor $4$ . Correct: divide first, then square: $x=(28/4)^2=49$ .

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Papers 1, 2 and 3

R10, R13 - Constructing proportional relationships

Forming equations and finding the constant (Higher)

Proportion involving powers (Higher)

Proportion involving roots (Higher)

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