Equation of Cycloid in Cartesian Coordinates

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Theorem

Consider a circle of radius $a$ rolling without slipping along the x-axis of a cartesian coordinate plane.

Consider the point $P$ on the circumference of this circle which is at the origin when its center is on the y-axis.

Consider the cycloid traced out by the point $P$.

Let $\left({x, y}\right)$ be the coordinates of $P$ as it travels over the plane.


The point $P = \left({x, y}\right)$ is described by the equation:

$a \sin^{-1} \left({\dfrac {\sqrt {2 a y - y^2} } a}\right) = \sqrt {2 a y - y^2} + x$


Proof

From Equation of Cycloid, the point $P = \left({x, y}\right)$ is described by the equations:

$x = a \left({\theta - \sin \theta}\right)$
$y = a \left({1 - \cos \theta}\right)$


Expressing $\theta$ and $\sin \theta$ in terms of $y$:

\(\displaystyle \cos \theta\) \(=\) \(\displaystyle 1 - \frac y a\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle \sin \theta\) \(=\) \(\displaystyle \sqrt {1 - \left({1 - \frac y a}\right)^2}\)
\(\displaystyle \) \(=\) \(\displaystyle \sqrt {\frac {2 y} a - \frac {y^2} {a^2} }\)
\(\displaystyle \) \(=\) \(\displaystyle \frac {\sqrt {2 a y - y^2} } a\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle \theta\) \(=\) \(\displaystyle \sin^{-1} \left({\frac {\sqrt {2 a y - y^2} } a}\right)\)


Substituting for $\theta$ and $\sin \theta$ in the expression for $x$:

\(\displaystyle x\) \(=\) \(\displaystyle a \left({\sin^{-1} \left({\frac {\sqrt {2 a y - y^2} } a}\right) - \frac 1 a \sqrt {2 a y - y^2} }\right)\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle a \sin^{-1} \left({\frac {\sqrt {2 a y - y^2} } a}\right)\) \(=\) \(\displaystyle \sqrt {2 a y - y^2} + x\)

$\blacksquare$


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