A runner runs along a straight course at a speed of 4m/s. A camera is located on the ground 20m from the course (at the nearest point). At what rate is the camera rotating when the runner is 15m beyond the closest point to the course? Give your answer in radians.
So I know this is a related rates problem. I'm just not sure what I need to relate. I found that tan(x)=15/20 so that x would be .6435. Where do I go from here? Thanks!
Best Answer
Assume the runner is going to the right along the $x$-axis. Let $C$ be the position of the camera, let $N$ be the nearest point on the course ($x$-axis) to the camera. Let $R$ be the position of the runner. Note that $R$ is changing, while $C$ and $N$ are fixed.
Then $\triangle CNR$ is right-angled at $N$. Let $\theta=\angle RCN$. Draw a labelled picture.
As the runner runs, $\theta$ changes, since the camera is tracking her. We are asked how fast the camera is rotating. Thus we want $\dfrac{d\theta}{dt}$.
Let $x=NR$. Then $\dfrac{dx}{dt}$ is how fast the runner is running. We know $\dfrac{dx}{dt}$.
In order to exploit our knowledge about $\dfrac{dx}{dt}$ to find $\dfrac{d\theta}{dt}$, we will need a relationship between $x$ and $\theta$.
Since $CN=20$ we have $$\frac{x}{20}=\tan\theta.$$ Differentiate. We get $$\frac{1}{20}\frac{dx}{dt}=(\sec^2\theta)\frac{d\theta}{dt}$$ (we have used the Chain Rule for the calculation on the right-hand side).
Now use the fact that we know $\dfrac{dx}{dt}$, and that when $x=15$ we can find $\sec^2\theta$, to finish. It is better not to bother with a calculator.
Remark: We could have written the relationship as $x=20\tan\theta$. Or else we could have written $\theta=\tan^{-1}\left(\frac{x}{20}\right)$. Usually, once we have a nice simple-looking relationship, it is best to differentiate right away.