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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?

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Closely related: A lady and a monster –  t.b. May 19 '11 at 19:54
Think about Archimedes's spiral. –  soandos May 19 '11 at 19:54
Maybe changing the viewpoint will help: Relative to the rabbit, the fox is moving in a straight line... –  Aryabhata May 19 '11 at 20:03
@Aryabhatta, Indeed. Your can write down the following: $x(t) = R(t)y(t)$, then $|x'(t)| = |R'(t)y(t) + R(t)y'(t)|$, but I'm not sure how this can help. –  Phonon May 19 '11 at 20:06
I don't think it is Archimedes' spiral. The increase in radius is not linear with angle. –  Ross Millikan May 19 '11 at 20:11
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4 Answers

up vote 7 down vote accepted

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 $r=R \sin\phi$)

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Thanks! = ) Didn't realize that it could be integrated this easily. –  Phonon May 19 '11 at 20:28
You can go slightly further than this, and conclude that while the rabbit has run a quarter circle, the fox has run half a smaller circle (half the radius) –  Henry May 20 '11 at 0:59
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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.

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Tried that. That makes the most sense. One of them is angular velocity, the other one is velocity in direction of radius vector, so they are neatly perpendicular, however I just can't manipulate the numbers so that I get a single solid answer. –  Phonon May 19 '11 at 20:10
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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.

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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.

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If the fox is not so smart, he is a factor $\pi/2 \approx 1.5$ slower -> still not so bad. –  Fabian May 19 '11 at 20:22
"it is always between the center of the circle and the rabbit". Cool solution, but not to this problem = ) –  Phonon May 19 '11 at 20:25
Yeah, I just noticed that, too. Oh well, I guess there was a reason for it being in a Calculus text book after all. –  Ben Hocking May 19 '11 at 20:28
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