# 17-18 Rate of shadow in the wall of a building

**Problem 17**

A light is placed on the ground 30 ft from a building. A man 6 ft tall walks from the light toward the building at the rate of 5 ft/sec. Find the rate at which the length of his shadow is changing when he is 15 ft from the building.

**Solution 17**

$\dfrac{y}{30} = \dfrac{6}{x}$

$y = \dfrac{180}{x}$

$\dfrac{dy}{dt} = \dfrac{-180\dfrac{dx}{dt}}{x^2}$

when x = 30 - 15 = 15 ft

$\dfrac{dy}{dt} = \dfrac{-180(5)}{15^2}$

$\dfrac{dy}{dt} = -4 \, \text{ ft/sec}$ *answer*

The negative sign in the answer indicates that the length of the shadow is shortening.

**Problem 18**

Solve Problem 17, if the light is 10 ft above the ground.

**Solution 18**

$\dfrac{10 - y}{30} = \dfrac{4}{x}$

$y = 10 - \dfrac{120}{x}$

$\dfrac{dy}{dt} = - \dfrac{-120\dfrac{dx}{dt}}{x^2}$

$\dfrac{dy}{dt} = \dfrac{120}{x^2}\,\dfrac{dx}{dt}$

when x = 30 - 15 = 15 ft

$\dfrac{dy}{dt} = \dfrac{120}{15^2}\,(5)$

$\dfrac{dy}{dt} = \dfrac{8}{3} \, \text{ ft/sec}$ *answer*