Answer to Question #278242 in Calculus for Rolan

Question

A light hangs 15 feet directly above a straight walk on which man 6 feet tall is walking. How fast (in ft/sec) is the end of the man’s shadow travelling when he is walking away from the light at rate of 3 mile 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*t – horizontal distance from light on time t.

Triangles ADF and DBE are similar.

So S/h=|DE|/(H-h)  =>  S/h=V*t/(H-h)  => S=V*t*h/(H-h)  

From last expression or differentiating this expression by t shadow size rate Sh = V*h/(H-h) = 3*6/(15-6) = 2 miles/hour.   

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

Sa = V*h/(H-h)+V = V*H/(H-h) = 3*15/(15-6) = 5 miles/hour.

Answer

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