CALCULUS RELATED RATE PROBLEM. PLEASE HELP??

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 5ft/s along a straight path. How fast is the tip of the shadow moving when he is 40 ft from the pole?

1 Answer
Sep 25, 2016

The tip of the shadow is moving at a speed of 25/3 = 8.bar(3)"ft"/"s"253=8.¯3fts

Explanation:

First, let's sketch the situation:

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In the above image, mm is the distance from the pole to the man, and ss is the distance from the pole to the tip of the man's shadow. Our goal is to find the rate of change of ss with respect to time given that rate of change of mm with respect to time is 5"ft"/"s"5fts and m=40"ft"m=40ft

As derivatives are rates of change, we can rewrite our goal as trying to find (ds)/dtdsdt given (dm)/dt = 5dmdt=5 and m=40m=40.

By the properties of similar triangles, we have

s/15 = (s-m)/6s15=sm6

=> 2s = 5s - 5m2s=5s5m

=>s = 5/3ms=53m

Differentiating with respect to time, we get

(ds)/dt = 5/3 (dm)/dt = 5/3*5 = 25/3dsdt=53dmdt=535=253

As it so happens, the rate of change of the tip of the shadow with respect to time is independent of the value of mm, and our final result is that the tip of the shadow is moving at a speed of 25/3 = 8.bar(3)"ft"/"s"253=8.¯3fts