Q.249. Application of differentiation

Question 249.
A light is at the top of a pole whose height is 14 feet. A boy 5 feet tall is walking away from the pole at
3 ft/sec. At what rate is the length of the boy's shadow increasing when he is 25 feet from the pole?

Answer 249.
Refer to the figure as under.

PL is the pole and BY is the boy.
x = length of the shadow of the boy and
y = distance of the boy from the pole.

Δs PLO and BYO are similar
=> (x+y)/14 = x/5 = (x+y-x)/(14-5) = y/9
 => x = (5/9)y
=> rate of increase in the length of the shadow,
dx/dt = (5/9) dy/dt, where dy/dt = speed of the boy
=> dx/dt = (5/9) * 3 = (5/3) ft/s.

 [Note that the rate of increase of length of the shadow of the boy is independent of his distance from the pole.]

Link to YA!