Answer
The distance between the friends is changing at a rate of $~~6.78~m/s$
Work Step by Step
Let $a$ be the distance from the friend to the center of the circle.
Let $b$ be the distance from the runner to the center of the circle.
Let $c$ be the distance between the two friends.
Note that these three lines form a triangle.
Let $\theta$ be the angle which subtends side $c$.
We can find $\theta$:
$c^2 = a^2+b^2-2ab~cos~\theta$
$2ab~cos~\theta = a^2+b^2-c^2$
$cos~\theta = \frac{a^2+b^2-c^2}{2ab}$
$\theta = arccos(\frac{a^2+b^2-c^2}{2ab})$
$\theta = arccos(\frac{200^2+100^2-200^2}{(2)(200)(100)})$
$\theta = arccos(0.25)$
$\theta = 1.318~rad$
We can find the time it takes the runner to complete one lap:
$T = \frac{(2\pi)(100~m)}{7~m/s} = 89.76~s$
We can find $\frac{d\theta}{dt}$:
$\frac{d\theta}{dt} = \frac{(2\pi)~rad}{89.76~s} = 0.07~rad/s$
We can find $\frac{dc}{dt}$:
$2c~\frac{dc}{dt} = 2a~\frac{da}{dt}+2b~\frac{db}{dt}-2b~cos~\theta~\frac{da}{dt}-2a~cos~\theta~\frac{db}{dt}+2ab~sin~\theta~\frac{d\theta}{dt}$
$\frac{dc}{dt} = \frac{1}{c}~[a~(0)+b~(0)-b~cos~\theta~(0)-a~cos~\theta~(0)+ab~sin~\theta~\frac{d\theta}{dt}]$
$\frac{dc}{dt} = \frac{1}{c}~(ab~sin~\theta~\frac{d\theta}{dt})$
$\frac{dc}{dt} = \frac{(200)(100)~(sin~1.318~rad)(0.07~rad/s)}{200}$
$\frac{dc}{dt} = 6.78~m/s$
The distance between the friends is changing at a rate of $~~6.78~m/s$
