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

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Shadow on the wall with light on the ground

By similar triangle:
$\dfrac{y}{30} = \dfrac{6}{x}$

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

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

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

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By similar triangle:
$\dfrac{10 - y}{30} = \dfrac{4}{x}$

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

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

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

when x = 30 - 15 = 15 ft
$\dfrac{dy}{dt} = \dfrac{120}{15^2}\,(5)$

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

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