CALCULUS RELATED RATE PROBLEM. PLEASE HELP??

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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?

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Answer 1 · 2016-09-25T17:33:59

The tip of the shadow is moving at a speed of $\frac{25}{3} = 8 . ¯ 3 \frac{ft}{s}$ $25/3 = 8.bar(3)"ft"/"s"$

Explanation:

First, let's sketch the situation:

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

As derivatives are rates of change, we can rewrite our goal as trying to find $\frac{d s}{d t}$ $(ds)/dt$ given $\frac{d m}{d t} = 5$ $(dm)/dt = 5$ and $m = 40$ $m=40$ .

By the properties of similar triangles, we have

$\frac{s}{15} = \frac{s - m}{6}$ $s/15 = (s-m)/6$

$\Rightarrow 2 s = 5 s - 5 m$ $=> 2s = 5s - 5m$

$\Rightarrow s = \frac{5}{3} m$ $=>s = 5/3m$

Differentiating with respect to time, we get

$\frac{d s}{d t} = \frac{5}{3} \frac{d m}{d t} = \frac{5}{3} \cdot 5 = \frac{25}{3}$ $(ds)/dt = 5/3 (dm)/dt = 5/3*5 = 25/3$

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

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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

Explanation:

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