# Thread: inverse variation

1. ## inverse variation

The illumination produced by a light source varies inversely as the square of the distance from the source. If the illumination produced 4m from a light source is 75 units, find the illumination produced 8 meters from the same source.

I understand variations but inverse variations mess me up...

2. Hello,

Just translate the text !

If i is the illumination and d the distance, we have, with k a constant of proportionnality :

$\displaystyle i = \frac{k}{d^2}$

You've been given d=4 and i=75. So if you want, you can try to calculate k. But it's not necessary :

You want the light i' produced at a distance of 8 = d'. d'=2d.

So $\displaystyle i' = \frac{k}{d'^2} = \frac{k}{(2d)^2} = \frac{1}{4} \frac{k}{d^2}= \frac{i}{4}$

3. I still don't understand how to work through the equation you gave me. In my math book, the answer is 18.75 units..how did they come to this answer?

4. I think it's just a simple proportion.

$\displaystyle \begin{array}{lll} 75&\longrightarrow&4^2\\ x&\longrightarrow&8^2 \end{array}$

$\displaystyle 75\cdot 16 = x \cdot 64$

$\displaystyle x = \frac{75\cdot 16}{64} = \frac{75}{4} = 18.75$