# 15-16 Movement of shadow from light at eye level

**Problem 15**

A light at eye level stands 20 ft from a house and 15 ft from the path leading from the house to the street. A man walks along the path at 6 ft per sec. How fast does his shadow move along the wall when he is 5 ft from the house?

**Solution 15**

$\dfrac{x}{y} = \dfrac{x + 15}{20}$

$20x = xy + 15y$

$(20 - y)x = 15y$

$x = \dfrac{15y}{20 - y}$

$\dfrac{dx}{dt} = \dfrac{(20 - y)\left(15\dfrac{dy}{dt}\right) - 15y\left( -\dfrac{dy}{dt} \right)}{(20 - y)^2}$

$\dfrac{dx}{dt} = \dfrac{15(20 - y) + 15y}{(20 - y)^2}\,\dfrac{dy}{dt}$

$\dfrac{dx}{dt} = \dfrac{300}{(20 - y)^2}\,\dfrac{dy}{dt}$

when y = 5 ft

$\dfrac{dx}{dt} = \dfrac{300}{(20 - 5)^2}\,(6)$

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

**Problem 16**

In Problem 15, when the man is 5 ft from the house, find the time-rate of change of that portion of his shadow which lies on the ground.

**Solution 16**

$x^2 + y^2 = s^2$

$2x \dfrac{dx}{dt} + 2y \dfrac{dy}{dt} = 2s \dfrac{ds}{dt}$

$x \dfrac{dx}{dt} + y \dfrac{dy}{dt} = s \dfrac{ds}{dt}$

dx/dt = 8 ft/sec and

x = 15(5)/(20 - 5) = 5 ft, then

s = √(x

^{2}+ y

^{2}) = √(5

^{2}+ 5

^{2}) = 5√2 ft

Thus,

$5(8) + 5(6) = 5\sqrt{2}\dfrac{ds}{dt}$

$\dfrac{ds}{dt} = 9.9 \, \text{ ft/sec}$ *answer*

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