A horizontal disk diameter 3 inches rotates once every 15 seconds. An insect starts at the southernmost point of the disk facing due north. Always facing due north, it crawls over the disk at 1 inch per second. Where does it again reach the edge of the disk?

**Solution**

Answer: at the northernmost point of the disk.

Take polar coordinates with r = 3/2, θ = 0 at the start. The equations of motion are dr/dt = - cos θ, r dθ/dt = 2rπ/15 + sin θ.

Differentiating the second equation: (dr/dt) (dθ/dt) + r d^{2}θ/dt^{2} = (2π/15) dr/dt + (dθ/dt) cos θ. Substituting from the first equation, 2 (dr/dt) (dθ/dt) + r d^{2}θ/dt^{2} = (2π/15) dr/dt. Multiplying through by r and integrating wrt t, we get r^{2} dθ/dt = (π/15) r^{2} + C, for some constant C. At t = 0, r = 3/2 and dθ/dt = 2π/15, so C = 3π/20. Thus r^{2} dθ/dt = (π/15) r^{2} + 3π/20. But the original equation gives r^{2} dθ/dt = 2r^{2}π/15 + r sin θ. Hence π r^{2}/15 + r sin θ = 3π/20 (**). Hence if r = ±3/2, we have sin θ = 0 and hence θ = 0 or π.

That is not quite enough to show that the insect reaches the edge again at θ = π. But we can treat (**) as a quadratic in r and solve to get r^{2} = (sin^{2}θ + (π/5)^{2})^{1/2} - sin θ. This shows that r first decreases, but then increases again to ±1 at θ = π. We can rule out -1 because r^{2} is always positive, and r starts positive. So by continuity it must always remain positive. Thus the insect next reaches the edge at the northenmost point of the disk.

*Many thanks to Melih Ozlem for pointing out an error in the original solution.*

© John Scholes

jscholes@kalva.demon.co.uk

5 Mar 2002

Last corrected/updated 9 July 03