Question:
A man 2 metres high walks at a uniform speed of 6 km/h away from a lamp-post 6 metres high. Find the rate at which the length of his shadow increases.
Solution:
Let AB be the lamp post. Let at any time t, the man CD be at a distance of x km from the lamp post and y m be the length of his shadow CE.
Since triangles $A B E$ and $C D E$ are similar,
$\frac{A B}{C D}=\frac{A E}{C E}$
$\Rightarrow \frac{6}{2}=\frac{x+y}{y}$
$\Rightarrow \frac{x}{y}=\frac{6}{2}-1=2$
$\Rightarrow \frac{d y}{d t}=\frac{1}{2} \frac{d x}{d t}$
$\Rightarrow \frac{d y}{d t}=\frac{1}{2}(6)$
$\Rightarrow \frac{d y}{d t}=3 \mathrm{~km} / \mathrm{hr}$