# Related rates, problem with an absolute answer

A round shaped athletic track (centre in a point C) has a radius of $50.0$ $mts$. An athlete runs around the track and a referee is going to time from a starting point A. If the athlete is in a point B, 'how fast does the area of the triangle ABC increases when the central angle is $\pi/4$?

I started by stating that $h=r\sin \alpha$ and $b=r$

Then I said that $A={r^2 \over 2}\sin \alpha$

Implicitly deriving...

$${dA \over dt}={r^2 \over 2} \cos \alpha {d\alpha \over dt}$$ Replacing:

$${dA \over dt}=625 \sqrt2 {d\alpha \over dt}$$

But I don't know how to get ${d\alpha \over dt}$? I even tried $l=r*\alpha$

$${dl \over dt} = r{d\alpha \over dt}$$

And then replacing again:

$${dA \over dt}={25 \over 2} \sqrt2 {dl \over dt}$$

But then again I have no way out of there.

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## 1 Answer

You can’t really say more than $${dA \over dt}=625 \sqrt2 ~\frac{d\alpha}{dt}\tag{1}$$ unless you have some information on the runner’s speed. The most you can do is rewrite $(1)$ directly in terms of his speed. If he’s running at a speed of $v$ metres per second, then $$v=r\frac{d\alpha}{dt}=50\frac{d\alpha}{dt}\;,$$ and $$\frac{dA}{dt}=\frac{625\sqrt2}{50}v=\frac{25\sqrt2 v}2\;.\tag{2}$$ Although they say essentially the same thing, I slightly prefer $(2)$ to $(1)$, since linear units are a little more familiar in this context than angular units.

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Why did you write that $v=r {d\alpha \over dt}?$ – ChairOTP Nov 13 '12 at 20:07
@ChairOTP: Because it’s true. The arc of the circle that subtends an angle $\alpha$ has length $r\alpha$, and $v$ is the rate of change of that length. – Brian M. Scott Nov 13 '12 at 20:13
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