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.



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

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