Direct and inverse proportion

In a nutshell

Proportion describes a relationship between two sets of quantities which are related by the constant ratio, $k$ . The symbol $\propto$ means "is proportional to".

Direct proportion

Two directly proportional quantities are related by a constant $k$ such that if $y \propto x^n$ then $y = kx^n$ . This means, as $y$ increases, $x^n$ increases whilst maintaining a constant ratio of $k$ between their quantities.

Note: Linear models show proportion where $(y-c) \propto x$  with the constant ratio being the gradient, $m$ .

Example 1

When $y= 45$ , $x=3$ . Given that $y$ is proportional to $x^2$ , find the value of $y$  when $x=2$ .

Represent the relationship of $y$ and $x$ as an equation.

$\begin{aligned} y &\propto x^2 \\ y&=kx^2 \end{aligned}$

Find the value of the constant $k$ by inserting the known values.

$\begin{aligned}(45) &= k(3)^2 \\ 45 &=9 k \\ 5&=k \end{aligned}$

To find $y$ when $x=2$ , substitute $x=2$ into $y=kx^2$ , where $k=5$ .

$\begin{aligned}y&= (5)(2)^2 \\ y&=(5)(4) \\ y &= 20 \end{aligned}$ 

Therefore, when $\underline{ x =2, \ y=20}$ .

Inverse proportion

Two inversely proportional quantities are also related by the constant ratio, $k$ , but such that if $y \propto \dfrac 1{x^n}$ then $y = \dfrac{k}{x^n}$ . This means as $y$ increases, $x^n$ decreases and vice versa whilst maintaining a constant ratio of $k$ between their quantities.

Example 2

When $y= 1$ , $x=1$ . Given that $y$ is inversely proportional to $x^2+1$ , find the value of $y$  when $x=3$ .

Represent the relationship of $y$ and $x$ as an equation.

$\begin{aligned} y &\propto \dfrac{1}{x^2 +1} \\ \\ y &= \dfrac{k}{x^2 +1}\end{aligned}$

Find the value of the constant $k$ .

$\begin{aligned}(1) &= \dfrac {k}{(1)^2+1} \\ \\ 1 &=\dfrac {k}{2} \\ \\ 2&=k \end{aligned}$

Substitute $x=3$ into the equation $y = \dfrac{2}{x^2 +1}$ .

$\begin{aligned}y&= \dfrac{2}{(3)^2+1} \\ \\ y&= \dfrac{2}{10}\\ \\ y &= \dfrac15 = 0.2 \end{aligned}$ 

Therefore, when $\underline{ x =3, \ y=0.2}$ .

Sketching graphs of proportion

After relating two proportional quantities by a constant $k$ , it is possible to form an equation in order to sketch a graph .

procedure

$1.$	Create an equation relating two directly or inversely proportional quantities.
$2.$	Insert given values and solve for the constant $k$ .
$3.$	Write an equation only in terms of $x$ and $y$ .
$4.$	Find three properties of the curve: the asymptotes, the $y\text-intercept$ and where the curve crosses the $x$ axis.
$5.$	Use the information to mark the relevant points and sketch the curve.

Note: Graphs of variables which are inversely proportional will often produce asymptotes. An asymptote is a line that gets closer and closer to a curve but never touches it.

Example 3

Given that $y$ is inversely proportional to $x+3$ , when $y=2$ , $x=7$ . Sketch a graph to represent this relationship.

Create an equation to relate the quantities.

$\begin{aligned} y &\propto \dfrac{1}{x+3} \\ \\ y &= \dfrac{k}{x+3} \end{aligned}$ 

Solve for constant $k$ .

$\begin{aligned} (2) &= \dfrac {k}{(7)+3} \\ \\ 2&= \dfrac {k}{10} \\ \\ 20 &=k \end{aligned}$ 

Write an equation in terms of $x$ and $y$ .

$y = \dfrac{20}{x+3}$ 

Find any asymptotes. Vertical asymptotes occur when the denominator here equals zero.

$\begin{aligned}0 &= x+3 \\ x&=-3\\ \\ y&=0 \end{aligned}$ 

Find the $y\text-intercept$ by setting $x=0$ .

$y = \dfrac{20}{(0)+3} = \dfrac{20}3$ 

Note: In some cases there isn't a $y\text-intercept$ . In this case there will be no value for $y$ when $x=0$ . Likely this will mean there is an asymptote on the $y$ axis.

Use the information gathered to sketch the curve.

Maths; Sketching graphs; KS5 Year 12; Direct and inverse proportion