Changes in inverse proportion

When \(y=\dfrac{k}{x^n}\), multiplying \(x\) by a factor \(r\) divides \(y\) by \(r^n\). The multiplier for \(y\) is \(\dfrac{1}{r^n}\).

Why: new \(y=\dfrac{k}{(rx)^n}=\dfrac{1}{r^n}\times\dfrac{k}{x^n}=\dfrac{1}{r^n}\times\text{original}\hspace{0.5em}y\).

Method

1. Convert the change in \(x\) to a multiplier \(r\).
2. Find the multiplier for \(y\): \(\dfrac{1}{r^n}\).
3. Convert back: percentage change in \(y=(\text{multiplier}\hspace{0.5em}-1)\times100\%\).
4. To work backwards from a change in \(y\), solve \(\dfrac{1}{r^n}=\text{multiplier for}\hspace{0.5em}y\) for \(r\).

Example

The number of square tiles, \(N\), needed to cover a floor is inversely proportional to the square of the side length, \(s\) cm, of each tile. The side length is increased by \(25\%\). Find the percentage change in \(N\).

Multiplier for \(s\): \(1.25\)

Multiplier for \(N\): \(\dfrac{1}{1.25^2}=0.64\)

Change: \((0.64-1)\times100\%=-36\%\)

So \(N\) decreases by \(36\%\).

Check: with \(s=20\), \(N=\dfrac{k}{400}\); with \(s=25\), \(N=\dfrac{k}{625}\), and \(\dfrac{400}{625}=0.64\). A common error is to say that a \(25\%\) increase in \(s\) gives a \(25\%\) or \(50\%\) decrease in \(N\).