1. ## Related Rates Problem

A runner sprints around a circular track of radius 100 m at a constant speed of 7 m/s. The runner's friend is standing at a distance 200 m from the center of the track. How fast is the distance between the friends changing when the distance between them is 200 m.

I attached a sketch of the problem so you can visualize what I'm doing.

$l=r\theta$

$\frac{dl}{dt}=r\frac{d\theta}{dt}$

$7=r\frac{d\theta}{dt}$

$\frac{d\theta}{dt}=\frac{7}{100}$

I decided to use the law of cosines, $s$ is the distance between them:

$s^2=(100)^2+(200)^2-2(200)(100)\cos(\theta)$

Differentiating with respect to $t$

$2s\frac{ds}{dt}=40,000\sin(\theta)\frac{d\theta}{d t}$

$\frac{ds}{dt}=\frac{20,000\sin(\theta)}{2s}\frac{d \theta}{dt}$

When the $s=200$

$40,000=10,000+40,000-40,000\cos(\theta)$

$cos(\theta)=\frac{1}{4}\rightarrow \sin(\theta)=\frac{\sqrt{15}}{4}$

So no I plug everything into the equation for the time derivative of s:

$\frac{ds}{dt}=\frac{(20,000)(\sqrt{15})(7)}{4(100) }$

$=\frac{7\sqrt{15}}{8}$

This is very close to the answer in my book which is $\frac{7}{4}\sqrt{15}$

I can't seem to find where I want wrong.

2. It's where you go from $2s\frac{ds}{dt}=,$ to $\frac{ds}{dt}=.$ You've missed a cancellation by 2.