**Question:**

A man 180 cm tall walks at a rate of 2 m/sec. away, from a source of light that is 9 m above the ground. How fast is the length of his shadow increasing when he is 3 m away from the base of light?

**Solution:**

Let *AB* be the lamp post. Suppose at any time *t*,* *the man *CD* is at a distance *x* km from the lamp post and *y* m is the length of his shadow *CE*.

Since triangles $A B E$ and $C D E$ are similar,

$\frac{A B}{C D}=\frac{A E}{C E}$

$\Rightarrow \frac{9}{1.8}=\frac{x+y}{y}$

$\Rightarrow \frac{x}{y}=\frac{9}{1.8}-1$

$\Rightarrow \frac{x}{y}=\frac{7.2}{1.8}$

$\Rightarrow x=4 y$

$\Rightarrow \frac{d y}{d t}=\frac{1}{4}\left(\frac{d x}{d t}\right)$

$\Rightarrow \frac{d y}{d t}=\frac{1}{4} \times 2$ $\left(\because \frac{d x}{d t}=2\right)$

$\Rightarrow \frac{d y}{d t}=0.5 \mathrm{~m} / \mathrm{sec}$