Variation Models (Cambridge (CIE) IGCSE International Maths: Extended): Revision Note

Consider the following table of data.

Given that $y$ is inversely proportional to $x$, find the values of $p$ and $q$.

As $y$ is inversely proportional to $x$ we can write the equation

$y=\frac{k}{x}$

Substitute the pair of known values into the equation to find $k$

$\begin{array}{rcl}36& =& \frac{k}{1}\\ k& =& 36\end{array}$

So the full equation of the variation model is

$y=\frac{36}{x}$

Substitute in $x=3$ to find $p$

$p=\frac{36}{3}=12$

$\mathit{p}\mathbf{=}\mathbf{12}$

Substitute in $y=9$ to find $q$

$\begin{array}{rcl}9& =& \frac{36}{q}\\ 9q& =& 36\\ q& =& \frac{36}{9}=4\end{array}$

$\mathit{q}\mathbf{=}\mathbf{4}$

Consider the data in the table below.
Decide which out of a linear, quadratic, or cubic variation model would be most suitable for the data, and use this to find an equation for $y$ in terms of $x$.

Start by checking if the model is linear, $y=kx$

Using the first data point to find $k$

$\begin{array}{rcl}20& =& k\times 2\\ k& =& 10\end{array}$

Use this to check the second data point

$$\begin{gathered} y=10x \\ y=10\left(4\right)=40 \end{gathered}$$

Not close to 80, so not linear

Consider if the model is quadratic, $y=k{x}^2$

Use the first data point to find $k$

$\begin{array}{rcl}20& =& k\times 2^2\\ 20& =& 4k\\ k& =& 5\end{array}$

Use this to check the second data point

$\begin{array}{rcl}y& =& 5x^2\\ y& =& 5{\left(4\right)}^2=80\end{array}$

It works for the second point, now check the third

$y=5{\left(6\right)}^2=180$

Also correct for the third point, so this is an appropriate variation model

A quadratic variation model is suitable

Its full equation is $\mathit{y}\mathbf{=}\mathbf{5}{\mathit{x}}^{\mathbf{2}}$

If a cubic model is tested, $y=k{x}^3$, the first point produces the equation $y=\frac{5}{2}{x}^3$ but this does not model the other two data points accurately

Two variation models are suggested for the data in the table below.

Model A: $y=\frac{400}{x^2}$

Model B: $y=\frac{800}{x^3}$

Determine which variation model is more suitable.

Add a row to the table for each of the suggested models

Use the same given values of $x$ (5, 6, 10, 15) to calculate the $y$ values that each model predicts

E.g. To find the $y$ value that model A predicts for $x=5$

$y=\frac{400}{5^2}=16$

Compare the modelled $y$ values to the actual $y$ values

The values for Model A are much closer to the actual values than Model B

Model A is more suitable