# Related Rates calculus problem

For some reason I keep getting this question wrong.

Suppose a 6 feet tall man is walking away from a 15 foot tall lamp post at 5ft/s. What is the rate at which the man's shadow is moving when he is 40 ft from the lamp post.

Here is what I did: $\dfrac{dx}{dt} = 5$ The man's speed at which he is walking. We want $\dfrac{ds}{dt}$. And we can use the chain rule to get: $\dfrac{dx}{dt} = \dfrac{dx}{ds}\cdot \dfrac{ds}{dt}$ Equation (1)

Let $s$ be the position of the tip of the man's shadow. Then $x$ and $s$ are related by similar triangles:

$\dfrac{15}{x+s} = \dfrac{6}{s} \iff 15s = 6x + 6s \iff \frac{3}{2}s = x$

Now we have: $\dfrac{dx}{dt}$ and $x$ in terms of $s$. $\dfrac{dx}{ds} =\dfrac{3}{2}$

Evaluating Equation (1): $5 = \dfrac{3}{2}\cdot \dfrac{ds}{dt}$ and hence $\dfrac{10}{3} = \dfrac{ds}{dt}$

The answer is supposed to be $\dfrac{25}{3}$ where did I go wrong?

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## 1 Answer

At any given time the tip of the shadow is $x+s$ ft away from the lamp post. Therefore the speed value you are looking for is $$v=\frac{d(x+s)}{dt}=\frac{dx}{dt}+\frac{ds}{dt}=\frac{dx}{dt}+\frac{ds}{dx}\frac{dx}{dt}=\frac{dx}{dt}+\frac{\alpha}{1-\alpha}\frac{dx}{dt}=\frac{1}{1-\alpha}\frac{dx}{dt}$$ where $\alpha = \frac{h}{H}$ is the ratio of the heights of the man and the lamp post. Hence $$v=\frac{1}{1-\frac{2}{5}}5=\frac{25}{3}$$

Since the motion is uniform I do not see the relevance of the initial condition here.

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It looks great, but could you tell me where I went wrong? I would also like to learn from my mistakes. –  CodeKingPlusPlus Oct 29 '12 at 0:24
effectively, you evaluated the speed of the shadow relatively to the walking man. It takes adding the speed of the man himself to get the answer –  Valentin Oct 29 '12 at 0:45
$\dfrac{dx}{ds}$ is the rate of change in $x$ with respect to $s$ correct? –  CodeKingPlusPlus Oct 29 '12 at 3:14