TITLE: Fun calculus problem I can't seem to solve
QUESTION [8 upvotes]: I've recently picked up a math book I haven't read since college (highly recommended reading by the way!). I was reviewing multi-dimentional derivatives and such, and I stumbled upon a problem I've been trying to solve for two days, and I can't get it out of my head, so please help me out! = )

Problem (from memory):
There is a rabbit that runs in a perfect circle of radius $r$ with a constant speed $v$. A fox chases the rabbit, starting from the center of the circle and also moves with a constant speed $v$ such that it is always between the center of the circle and the rabbit. How long will it take for the fox to catch the rabbit?

I tried using the fact that $|x'(t)| = |y'(t)| = v$ where $x(t)$ is fox's position and $y(t)$ is rabbit's position, and that $x'(t)x''(t) = 0$ because of constant speed restriction, but I'm still failing to find a solution.
Anyone feel like attacking this one?

REPLY [2 votes]: It is easiest to solve the problem with polar coordinates.  WLOG we may set $r=1$ so that the rabbit's angle at time $t$ is $vt$.  Everything can be rescaled for different values of $r$.  If we let $r(t)$ be the distance between the fox and the center of the circle at time $t$, his $(r,\theta)$ position at time $t$ is $(r(t),vt)$, which in $(x,y)$ coordinates is $(r(t)\cos(vt),r(t)\sin(t))$.  The velocity vector is then $r'(t)(\cos(vt),\sin(vt))+vr(t)(\sin(vt),-\cos(vt))$, and so if the speed is $v$, then we must have $v^2=r'(t)^2+v^2r(t)^2$, or $r'=v\sqrt{1-r^2}$.  
The solution to this differential equation, subject to the initial condition that $r(0)=0$ is $r(t)=\sin(vt)$, which takes the value $1$ when $vt=\pi/2$, and so the fox will catch the rabbit when the rabbit is one quarter of the way around the circle.