Question

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?

Answer 1

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

Explanation:

First, let's sketch the situation:

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

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

By the properties of similar triangles, we have

`s/15 = (s-m)/6`

`=> 2s = 5s - 5m`

`=>s = 5/3m`

Differentiating with respect to time, we get

`(ds)/dt = 5/3 (dm)/dt = 5/3*5 = 25/3`

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