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?

Answer 1

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.

$$\text{ Since triangles ABE and CDE are similar },$$
$$\frac{AB}{CD}=\frac{AE}{CE}$$

$$\Rightarrow \frac{9}{1.8}=\frac{x+y}{y}$$
$$\Rightarrow \frac{x}{y}=\frac{9}{1.8}-1$$
$$\Rightarrow \frac{x}{y}=\frac{7.2}{1.8}$$
$$\Rightarrow x=4y$$
$$\Rightarrow \frac{dy}{dt}=\frac{1}{4}\left(\frac{dx}{dt}\right)$$
$$\Rightarrow \frac{dy}{dt}=\frac{1}{4}\times 2(\because \frac{dx}{dt}=2)$$
$$\Rightarrow \frac{dy}{dt}=0.5m/\sec$$