Question

A light hangs in 15 feet directly above a straight walk on which a man is 6 feet tall is walking. How fast is the end of the man's shadow travelling when he is walking away from the light at the rate of 3 miles per hour?

Answer 1

Let |DF|=h – man height, |BC|=H – light height, V=3 miles/hour – walking rate, |AF|=S – shadow size.

|DE|=V$\times$ t – horizontal distance from light on time t.

Triangles ADF and DBE are similar.

So $\frac{S}{h}$=$\frac{|DE|}{H-h}$ /(H-h) 

$\frac{S}{h}$=$\frac{(V\times t)}{(H-h)}$   

S= $\frac{(V\times t\times h)}{(H-h)}$

From last expression or differentiating this expression by t shadow size rate Sh = $\frac{(V\times H)}{(H-h)}$

=$\frac{(3\times6)}{(15-6)}$ = 2 miles/hour.   

The rate of the end of the man's shadow is the velocity of point A

Sa = $\frac{(V\times H)}{(H-h)}$ +V = $\frac{(V\times H)}{(H-h)}$ = $\frac{(3\times5)}{(15-5)}$ (15-6) = 5 miles/hour.

Answer

The rate of the end of the man's shadow Sa = 5 miles/hour.