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.