35-36 Time Rates: Lengthening of shadow and movement of its tip in 3D space

Problem 35
An arc light hangs at the height of 30 ft above the center of a street 60 ft wide. A man 6 ft tall walks along the sidewalk at the rate of 4 ft/sec. How fast is his shadow lengthening when he is 40 ft up the street?

Solution 35

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From the figure:
$x = \sqrt{(4t)^2 + 30^2}$

$x = \sqrt{16t^2 + 900}$

$\dfrac{s}{6} = \dfrac{x}{24}$

$s = \frac{1}{4}x$

$s = \frac{1}{4}\sqrt{16t^2 + 900}$

$\dfrac{ds}{dt} = \dfrac{1}{4}\left( \dfrac{32t}{2\sqrt{16t^2 + 900}} \right)$

$\dfrac{ds}{dt} = \dfrac{4t}{\sqrt{16t^2 + 900}}$

when 4t = 40; t = 10 sec
$\dfrac{ds}{dt} = \dfrac{4(10)}{\sqrt{16(10^2) + 900}}$

$\dfrac{ds}{dt} = 0.8 \, \text{ ft/sec}$ answer

Problem 36
In Problem 35, how fast is the tip of the shadow moving?

Solution 36

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Triangle LAB,
$\dfrac{x}{6} = \dfrac{x + 30}{30}$

$5x = x + 30$

$x = 7.5 \,\, \text{ ft }$

Triangle ABC,
$\dfrac{s}{x + 30} = \dfrac{4t}{30}$

$\dfrac{s}{7.5 + 30} = \dfrac{2t}{15}$

$s = 5t $

$\dfrac{ds}{dt} = 5 \, \text{ ft/sec}$ answer

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