Question

a street light is at the top of a 16.5 ft. tall pole. a man 5.5 ft tall walks away from the pole with a speed of 5.5 feet/sec along a straight path. how fast is the tip of his shadow moving when he is 34 feet from the pole?
hint: draw a picture and use similar triangles.
your answer: □ft/sec

Explanation:

Step1: Set up variables and similar triangles

Let \(x\) be the distance of the man from the pole, and \(y\) be the distance from the tip of the shadow to the pole.
By similar triangles, \(\frac{5.5}{16.5}=\frac{y - x}{y}\).
Simplify \(\frac{5.5}{16.5}=\frac{1}{3}\), so \(3(y - x)=y\), which gives \(3y-3x = y\), and then \(2y = 3x\), or \(y=\frac{3}{2}x\).

Step2: Differentiate with respect to time

Differentiate both sides with respect to time \(t\).
Using the chain - rule, \(\frac{dy}{dt}=\frac{3}{2}\frac{dx}{dt}\).
We know that \(\frac{dx}{dt}=5.5\) ft/sec.

Answer:

Since \(\frac{dy}{dt}=\frac{3}{2}\times5.5 = 8.25\) ft/sec.
The tip of his shadow is moving at a rate of \(8.25\) ft/sec.