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.

Answer 1

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.

$$\begin{gathered} \mathrm{Since}\mathrm{triangles}ABE\mathrm{and}CDE\mathrm{are}\mathrm{similar}, \\ \frac{AB}{CD}=\frac{AE}{CE} \end{gathered}$$

$$\begin{gathered} \Rightarrow \frac{6}{2}=\frac{x+y}{y} \\ \Rightarrow \frac{x}{y}=\frac{6}{2}-1=2 \\ \Rightarrow \frac{dy}{dt}=\frac{1}{2}\frac{dx}{dt} \\ \Rightarrow \frac{dy}{dt}=\frac{1}{2}\left(6\right) \\ \Rightarrow \frac{dy}{dt}=3\mathrm{km}/\mathrm{hr} \end{gathered}$$