Question

A man of height $2$ metres walks at a uniform speed of $6km/h$ away from a lamp post which is $6$ meters high. Find the rate at which the length of his shadow increases. Also find the rate at which the tip of the shadow is moving away from the lamp post.

Answer 1

Let the length of shadow be $DE=y\text{m}$

and let the distance between the man and the lamp post be x meter.

Let AB be the height of lamp post and CD be the height of man.

$\therefore AB=6m\&CD=2m$

Therefore, by similar triangles theorem,

$\frac{AB}{BE}=\frac{CD}{DE}$

$\frac{6}{x+y}=\frac{2}{y}$

$6y=2x+2y$

$\Rightarrow x=2y⟶\left(1\right)$

Differentiating ${eq}^n\left(1\right)$ w.r.t. time, t, we have

$\frac{dx}{dt}=2\frac{dy}{dt}$

$\because \frac{dx}{dt}=6km/hr$

$\therefore 2\frac{dy}{dt}=6$

$\frac{dy}{dt}=3km/hr$

The rate at which the length of his shadow increases is $3km/hr$