QOD #21 — Solution
Solution
The path of the bugs describe four congruent logarithmic spirals which meet at the center of the square beach. At any given instant the four bugs form the corners of a square which shrinks and rotates as the bugs move closer together. The path of each pursuer will therefore at all times be perpendicular to the path of the pursued. This tells us that as Zahra, for example, approaches Stas, there is no component in Stas's motion which carries him toward or away from Zahra. Consequently, Zahra will reach Stas in the same time that it would take if Stas had remained stationary. The length of each spiral path will be the same as the side of the square beach: 100 meters.
Incidentally, it's too bad Stas forgot his calculator, so all four of them were doomed to actually use their minds in order to solve the homework set.
Comments
- The reasoning is actually very sound, though those who still feel uncomfortable can dip into the differential equations
- Incidentally, this year at Caltech, I had forgot this reasoning for why they should meet at the center, and inspired by my differential equations class, I tried to solve it. Unfortunately, I calculated that the bugs would never actually meet at the center, but instead spiral endlessly in a smaller and smaller square. And of course, I thought I was right. And doubly unfortunately, the exact same problem popped up on my final exam. I solved the differential equation again, and I once again got the same answer, that meeting at the center was a limit after an infinite time. So I lost most of the points on that question.
The moral of the story? Always double-check when you convert between Cartesian coordinates and polar coordinates when dealing with derivatives (and integrals).
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