Question

A man 2 metres high walks at a uniform speed of 5 km/hr 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. Suppose 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.


$$\begin{gathered} \mathrm{Since}\mathrm{triangles}ABE\mathrm{and}CDE\mathrm{are}\mathrm{similar}, \\ \frac{AB}{CD}=\frac{AE}{CE} \\ \Rightarrow \frac{6}{2}=\frac{x+y}{y} \\ \Rightarrow 3y=x+y \\ \Rightarrow x=2y \\ \Rightarrow \frac{dx}{dt}=2\frac{dy}{dt} \\ \Rightarrow \frac{dy}{dt}=\frac{1}{2}\frac{dx}{dt} \\ \Rightarrow \frac{dy}{dt}=\frac{1}{2}\left(5\right)\left(\frac{dx}{dt}=5\right) \\ \Rightarrow \frac{dy}{dt}=\frac{5}{2}\mathrm{km}/\mathrm{hr} \end{gathered}$$