idahogie's Comments

I haven't checked your math, deadskin, but I can see that I made the error that you describe. So I'm inclined to believe you on the extra credit question. Good work.
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Deadskin's right. The rabbit can maintain a position opposite the agent while he stays 1/4 of the way to shore. From there it's a race - the rabbit goes (3/4)R at speed x and the agent goes (pi)R at speed 4x. Time for the rabbit is distance/speed, or (3/4)(R/x). Time for the agent is (pi/4)(R/x). Pi/4 is about 0.79, which is longer than than the rabbit, which is 0.75. Go, bunny, go!

For the maximum agent speed, set (3/4)(R/x) = (pi/max)(R/x). Solve for max = (4/3)(pi)
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  • Member Since 2012/08/11


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