Fun calculus problem I can't seem to solve 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?
 A: For this problem it is advisable to introduce polar coordinates ($r$ and $\phi$). The rabbit runs at speed $v$ on a perfect circle with radius $R$. Therefore, $\phi =vt/R$.  As the fox stays between the center and the rabbit, it is at the same $\phi$. The fox's speed (which is constant) has two component (prime denotes the derivative with respect to time)
$$ v = \sqrt{r'^2+ r^2 \phi'^2}.$$
From the knowledge of $\phi'= v/R$ and the fact that $v$ is constant, we can deduce $$r' = v\sqrt{1 - \frac{r^2}{R^2}}.$$
This differential equation can be solve by separating the variables. The time $T$ it takes to reach the rabbit is given by
$$T=\int_0^R \frac{dr}{v \sqrt{1 - \frac{r^2}{R^2}}} = \frac{\pi R}{2v}.$$
(the last integral has been solve by substituting $r=R \sin\phi$)
A: Hint:  the rabbit moves at constant angular velocity of $\frac{v}{r}$.  If we let the fox's radius be R, we have that $(R')^2+(\frac{Rv}{r})^2=v^2$, decomposing the fox's speed into radial and tangential pieces.
A: 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.
A: Assuming a smart fox (and aren't they all), calculus is not required. If the fox anticipated the rabbit's path, and took the shortest path, then it has to travel a distance r. With speed v, the time is simply r/v. (I.e., the fox can travel in a straight line from the center of the circle to the point in the circle the rabbit will be when it gets there.)
Edit to add: note that this does not solve the requested problem because the fox will not always be between the rabbit and the center of the circle, as required.
