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    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

    (asint+at),(a+acost)

    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).
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    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.
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    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).
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    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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