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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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In what sense is this "a Cartesian equation of a cycloid"? Assuming that $c$ is a constant, it's just an equation for $x$ which, depending on the function $a(x)$, can have any number of solutions for $x$; how does that describe a cycloid? – joriki Aug 18 '11 at 6:15
@joriki: Indeed, thanks for the valid point. What if I let $a(x) = k\sqrt{y(x)}$ where $k$ is a constant? I shall edit the post. – jake Aug 18 '11 at 6:32
@joriki: Ignore the sqrt... I really meant let $a(x) = ky(x)$... – jake Aug 18 '11 at 6:41
It really can't be a cycloid; you have a $y$ outside your square root in the implicit Cartesian equation... – J. M. Aug 18 '11 at 9:31
@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. – joriki Aug 18 '11 at 11:08

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


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


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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Thanks, joriki. You are right that it is not a full cycloid. But neither is the Cartesian equation for a cycloid obtained by combining the parametric equations of a cycloid, – jake Aug 18 '11 at 9:16
@jake: I don't understand the connection (assuming that "But neither" is intended to imply a connection) between this and what I wrote. Also, if it's not a full cycloid, perhaps you could say a bit more about how you got it and what part of a cycloid you believe it represents? – joriki Aug 18 '11 at 9:24
This isn't a hw question, but rather something my friend (or rather, fiend) set me. Now I think he's just pulling my leg...! Thanks anyway :) – jake Aug 18 '11 at 10:02
Wait... What about if I throw in a square root over $y(x)$? c.f. Edited question... – jake Aug 18 '11 at 10:16

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