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Edited...

I have a Cartesian equation of a cycloid: $\arcsin\left(k\sqrt{y(x)}\right) - k\sqrt{y(x)-k^2y(x)^2} + c = x$ where $k$ and $c$ are constants.

How might I parameterize it so that I get the usual parameterizations, i.e. $\begin{align*} x&=r(t-\sin{t})\\ y&=r(1-\cos{t}) \end{align*}$?

Thanks in advance!

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    @jake: Frankly, I've lost interest. You just changed the question for the third time. Why do you think someone should spend time on this version if they can expect it to change a few more times? If you want people to spend time to help you, you should be more careful with their time.2011-08-18

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I'll give you a way to check if something your fiend friend gave you is a cycloid: the Whewell equation (the equation relating tangential angle $\phi$ and arclength $s$) for the cycloid is

$s=k\,\sin\,\phi$

where $k$ is a constant (which is proportional to the radius of the rolling circle). Derive the required expressions for arclength and tangential angle from your parametric equations, and see if the cycloid's Whewell equation holds.

Alternatively, the Cesàro equation (which relates curvature $\kappa$ and arclength) for the cycloid is

$\frac1{\kappa^2}+s^2=c$

where $c$ is a constant (that is also related to the radius of the rolling circle). Substitute the appropriate expressions for the curvature and arclength to verify if you have a cycloid.

Both equations are so-called intrinsic equations; equations that depend only on the nature of the curve, and not its orientation/position in the plane.

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I don't think this is a cycloid, at least not a full cycloid up to a cusp. The cusp of a cycloid has infinite curvature. This curve has finite curvature everywhere.

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    Wait... What about if I throw in a square root over $y(x)$? c.f. Edited question...2011-08-18