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On a UIL Calculator Applications test (A Texas public high school calculator-based math competition), I've come across a problem that I can only remember the problem, not the numbers.

A man is standing on the center of a platform of radius $R$, which makes one full revolution every $P$ (for period) seconds. He begins to walk in a straight line at a velocity of $V$, but his path is curved by the spinning of the platform. What total distance is covered by the man?

The speed of a point on the edge of the platform is given by $2{\pi}R/P$. If the platform speed was $0$, then the distance would simply be $R$, however this is no the case. During the contest (under a very tight time constraint), that his distance covering per second would be what he covers at at distance of $R/2$. The time it takes to cross the platform would be $R/V$, and his speed at $R/2$ would be half of the maximum speed, which results as ${\pi}R/P$. The distance covered due to spinning would be $({\pi}R^2)/(2P)$, added to the original distance gave a final answer of $({\pi}R^2)/(2P) + R/2$.

Did I take the most efficient approach to this problem, if even correct? Maybe there is an application to an Archimedean Spiral? Any information would be greatly appreciated.

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We have $r(t) = Vt, \theta(t) = \omega t$, with $\omega = {2 \pi \over P}$

The position is given by $x(t) = Vt( \cos \omega t, \sin \omega t ) $ and so $\dot{x}(t) = V [ ( \cos \omega t, \sin \omega t ) + \omega t ( -\sin \omega t, \cos \omega t ) ]$ and $\|\dot{x}(t)\| = V \sqrt{1+(\omega t)^2}$ and $L = V \int_0^{R \over V} \sqrt{1+(\omega t)^2} dt $.

This gives $L = {V \over 2\omega} \left[ \operatorname{arcsinh} ({\omega R \over V}) + {\omega R \over V} \sqrt{({\omega R \over V})^2+1} \right] $.

Replacing $\omega$ gives: $L = {VP \over 4 \pi} \left[ \operatorname{arcsinh} ({2 \pi R \over VP}) + {2 \pi R \over VP} \sqrt{({2 \pi R \over VP})^2+1} \right] $.

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  • $\begingroup$ That looks familiar to the formula for the length of an Archimedean Spiral. Is that the application? $\endgroup$ – Joe Doe Apr 2 '16 at 0:28
  • $\begingroup$ I don't understand your question. What application are you referring to? We have $\operatorname{arcsinh} x = \log(x+\sqrt{x^2+1})$, so yes it is the same as the length of an Archimedean Spiral. $\endgroup$ – copper.hat Apr 2 '16 at 4:50

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