# Equation of Cycloid in Cartesian Coordinates

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

$\blacksquare$