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    • Thread Starter

    Im trying to find the distance traced out by a point on a wheel's circumference over one revolution where the wheel is rolling on a horizontal x axis with constant velocity.

    the original question did not give the parametric form of the cycloid so I had to derive it. I started with a point on the top of the wheel (call this point A). suppose the wheel has radius a and at time T has turned through an angle t where t is the angle between A and the upward vertical. through this i deduced that the parametric form of the cycloid must be


    I then used the formula used to find the arc length of a curve ((dy/dt)^2+(dy/dt)^2)^1/2 dt. putting the limits as 2pi and 0 (as im trying to find the arc length traced during 1 revolution). However, the final integral yields 0 and I do not think I have done anything wrong. furthermore, if I change the limits to pi and 0 and double the integral I get at the end (instead of integrating between 2pi and 0) I obtain the correct answer. If I model the cycloid from a particle at the bottom of the wheel at time 0, then the problem does not exist and I obtain the same answer as if I integrated the previous integral between pi and 0 then doubled my answer.

    does anyone know why this is? does it have something to do with the fact that the curve is not continous at t=pi? (the curve has infinite gradient there, imagine two circles touching each other through a common and vertical tangent).

    Did you at some point decide something like \sqrt{2+2 \cos 2t} = \sqrt{4 \cos^2 t} = \cos t?

    If so, the last step is actually wrong; cos^2 t is always positive, so \sqrt(cos^2 t) is positive too, and you need |cos t| instead of cos t after you take the square root.
    • Thread Starter

    Ah I see thanks. So it's ok to integrate arc lengths over a discontinuous(can't think of a better word to describe that region of infinite gradient) region? I recall normal integration doesn't work over discontinuous regions.

    This was a poor question(STEP III 1997) in my opinion since if you modelled the wheel's motion from a particle at the bottom of a circle you'd get away with the right answer(since you'd have to integrate sin(t/2) which is non negative from 0 to 2pi) without being careful about the modulus sign (and the Mark scheme itself suggested the candidate would get away with it as there's no mention of the modulus thing on it).

    There's no discontinuity in the cycloid, there's discontinuity in its differential. One might say that the cycloid belongs to C^0 but not C^1, where the class C^k contains function whose k-th derivative is continuous.

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