# A street light is mounted at the top of a 15-ft-tall pole. A man 6 ft tall walks away from the pole with a speed of 4 ft/s along a straight path. How fast is the tip of his shadow moving when he is 50 ft from the pole?

Dec 22, 2017

$\frac{20}{3} \frac{f t}{s}$

#### Explanation: in this diagram, x is the distance from the man to the pole, and y is the distance from the tip of the man's shadow to the pole. i assume the man and pole are standing straight up, which means the 2 triangles are similar.
by similarity, $\frac{y - x}{y} = \frac{6}{15}$
$15 \left(y - x\right) = 6 y$
$15 y - 15 x = 6 y$
$9 y = 15 x$
$y = \frac{5}{3} x$

differentiate both sides with respect to $t$ or time.
$\frac{\mathrm{dy}}{\mathrm{dt}} = \frac{5}{3} \frac{\mathrm{dx}}{\mathrm{dt}}$

you know $\frac{\mathrm{dx}}{\mathrm{dt}} = 4 \frac{f t}{s}$ because the man is walking that speed away from the pole. you want to find $\frac{\mathrm{dy}}{\mathrm{dt}}$, how fast the tip of the shadow is moving.

that means $\frac{\mathrm{dy}}{\mathrm{dt}} = \frac{5}{3} \cdot 4 \frac{f t}{s} = \frac{20}{3} \frac{f t}{s}$

actually, the man's distance from the pole doesn't matter since only his speed affects how fast his shadow moves.