Question

A person whose height is $6\mathrm{feet}$ is walking away from the base of a street light along a straight path at a rate of $4$ feet per second.

If the height of a street light is $15\mathrm{feet}$, what is the rate at which a person’s shadow is lengthening?

Answer 1

Find the rate of change in length of shadow:

Let the distance from the pole be $x$ and the length of the shadow be $y$
Using similar triangle property

Given : Height of a person $=6\mathrm{feet}$

Walking at speed of $4\mathrm{feet}/\sec$

$\therefore \frac{\mathrm{dx}}{\mathrm{dt}}=4\mathrm{feet}/\sec$

Height of a streetlight $=15\mathrm{feet}$
$$\begin{gathered} \therefore \frac{y}{6}=\frac{x+y}{15} \\ \Rightarrow 9y=6x \\ \Rightarrow y=\frac{2x}{3}...\left(\mathrm{i}\right) \end{gathered}$$
Differentiate equation $\left(\mathrm{i}\right)$ with respect to time $\left(t\right)$, we get
$\Rightarrow \frac{dy}{dt}=\frac{2}{3}\frac{dx}{dt}...\left(\mathrm{ii}\right)$
$\Rightarrow \frac{dy}{dt}=\frac{2}{3}\times 4$ $\mathbf{[}\mathbf{\because }\frac{\mathbf{d}\mathbf{x}}{\mathbf{d}\mathbf{t}}\mathbf{=}\mathbf{4}\mathbf{}\mathbf{]}$
$=\frac{8}{3}\mathrm{feet}/\sec$

Hence, the rate at which a person's shadow is lengthening is $\frac{8}{3}\mathrm{feet}/\sec$.