E2.8 - Direct and inverse proportion (Extended) | Cambridge IGCSE International Mathematics

If an input becomes four times as large, must its output also become four times as large? E2.8 uses direct and inverse proportion to describe several different answers, including square, square-root and cube relationships. You will turn these descriptions into equations and use one known pair of values to find unknown quantities.

Direct proportion

Buying twice as many identical items at a fixed price doubles the cost. If 2 items cost 6 dollars and 4 cost 12 dollars, the cost divided by the number of items stays at 3. The unchanged ratio, rather than simply “both quantities increase”, identifies direct proportion.

Direct proportion

$y$ is directly proportional to a quantity when $y$ divided by that quantity is constant. The symbol $\propto$ means “is proportional to”. For example, $y\propto x$ means $y=kx$ , where $k$ is the constant of proportion.

The quantity after $\propto$ matters. In $y\propto x^2$ , the constant ratio is $y/x^2$ , not $y/x$ . Replacing $\propto$ by an equals sign introduces the multiplier $k$ ; it does not assume $k=1$ .

Direct proportion models

For the positive inputs used here,

Description	Proportional notation	Equation
Directly proportional to $x$	$y\propto x$	$y=kx$
Directly proportional to the square of $x$	$y\propto x^2$	$y=kx^2$
Directly proportional to the square root of $x$	$y\propto\sqrt{x}$	$y=k\sqrt{x}$
Directly proportional to the cube of $x$	$y\propto x^3$	$y=kx^3$

In each row, $k$ stays fixed within that relationship. These models are not on the supplied formula lists: learn them.

For positive $x$ , multiplying $x$ by a factor $r>0$ multiplies these four outputs by $r$ , $r^2$ , $\sqrt r$ and $r^3$ , respectively. Apply the stated power to the input factor before changing the output.

Worked example 1: direct square proportion

Question. $y$ varies directly as the square of $x$ . Write an equation using $k$ , and find the factor by which $y$ changes when $x$ is multiplied by 4.

Plan. Translate “square of $x$ ” into $x^2$ , then square the input factor.

\begin{aligned} y&=kx^2 &&\text{introduce the constant}\\ \text{new }y&=k(4x)^2 &&\text{replace }x\text{ by }4x\\ &=16kx^2 &&\text{square both factors}\\ &=16y \end{aligned}

Answer. $y=kx^2$ ; the output is multiplied by 16.

Check. For $x=2$ , the output is $4k$ . For $x=8$ , it is $64k$ . The input factor is 4 and $64k/(4k)=16$ , for non-zero $k$ .

Keep the same input factor, but change the square relationship to a square-root relationship. That single change makes the output grow much more slowly.

Worked example 2: direct square-root proportion

Question. $y$ varies directly as the square root of $x$ . Write an equation using $k$ , and find the factor by which $y$ changes when $x$ is multiplied by 4.

Plan. Use $y=k\sqrt{x}$ and take the square root of the input factor.

\begin{aligned} y&=k\sqrt{x}\\ \text{new }y&=k\sqrt{4x}\\ &=2k\sqrt{x} &&\text{since }\sqrt4=2\\ &=2y \end{aligned}

Answer. $y=k\sqrt{x}$ ; the output is multiplied by 2.

Check. Moving from $x=4$ to $x=16$ changes the output from $2k$ to $4k$ , which doubles it.

The square root applies to $x$ , not automatically to $k$ . The equation $y=k\sqrt{x}$ follows directly from the words; $y=\sqrt{kx}$ uses a differently defined constant.

Watch out

An increase does not establish direct linear proportion. The rule $y=x+5$ increases with $x$ , but $y/x$ changes: it is 6 at $x=1$ and $7/2$ at $x=2$ . Direct linear proportion requires $y=kx$ , with no added fixed term.

Translate the power before applying the scale factor.

Inverse proportion

Sharing a fixed total of 24 dollars equally gives 12 dollars each for 2 people and 6 dollars each for 4 people. The amount per person decreases as the number of people increases. The product, number of people multiplied by amount each, stays at 24.

Inverse proportion

$y$ is inversely proportional to a quantity when $y$ multiplied by that quantity is constant. Thus $y\propto1/x$ means $y=k/x$ , or equivalently $xy=k$ , with $x\ne0$ .

“Inverse” means take the reciprocal of the stated quantity. Inverse proportion to the square of $x$ uses $1/x^2$ . It does not mean changing $x^2$ into $-x^2$ .

Inverse proportion models

For $x>0$ ,

Description	Proportional notation	Equation	Constant product
Inversely proportional to $x$	$y\propto1/x$	$y=k/x$	$xy=k$
Inversely proportional to the square of $x$	$y\propto1/x^2$	$y=k/x^2$	$x^2y=k$
Inversely proportional to the square root of $x$	$y\propto1/\sqrt{x}$	$y=k/\sqrt{x}$	$y\sqrt{x}=k$
Inversely proportional to the cube of $x$	$y\propto1/x^3$	$y=k/x^3$	$x^3y=k$

These models are not on the supplied formula lists: learn them.

Zero cannot be an input to these reciprocal expressions. The square-root denominator also requires $x>0$ . For positive $k$ , all four outputs decrease as positive $x$ increases, but at different rates.

Worked example 3: inverse linear proportion

Question. $y$ is inversely proportional to $x$ . Write an equation using $k$ , and describe the change in $y$ when $x$ is multiplied by 3.

Plan. Put $x$ in the denominator, so multiplying the denominator by 3 divides the output by 3.

\begin{aligned} y&=\frac{k}{x}\\ \text{new }y&=\frac{k}{3x}\\ &=\frac13\left(\frac{k}{x}\right)\\ &=\frac{y}{3} \end{aligned}

Answer. $y=k/x$ ; $y$ is divided by 3.

Check. With $k=12$ , inputs 2 and 6 give outputs 6 and 2. Both products are 12, and the output has become one third as large.

Change only the power of the input from 1 to 3. Tripling the input now multiplies the denominator by 27.

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E2.8 - Direct and inverse proportion (Extended)

Direct proportion

Inverse proportion

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