Equation of Cycloid in Cartesian Coordinates

Theorem

Consider a circle of radius $a$ rolling without slipping along the $x$-axis of a cartesian 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 $\tuple {x, y}$ be the coordinates of $P$ as it travels over the plane.

The point $P = \tuple {x, y}$ is described by the equation:

$a \sin^{-1} \paren {\dfrac {\sqrt {2 a y - y^2} } a} = \sqrt {2 a y - y^2} + x$

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

$x = a \paren {\theta - \sin \theta}$

$y = a \paren {1 - \cos \theta}$

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

$\ds \cos \theta$

$=$

$\ds 1 - \frac y a$

$\ds \leadsto \ \ $

$\ds \sin \theta$

$=$

$\ds \sqrt {1 - \paren {1 - \frac y a}^2}$

$\ds $

$=$

$\ds \sqrt {\frac {2 y} a - \frac {y^2} {a^2} }$

$\ds $

$=$

$\ds \frac {\sqrt {2 a y - y^2} } a$

$\ds \leadsto \ \ $

$\ds \theta$

$=$

$\ds \sin^{-1} \paren {\frac {\sqrt {2 a y - y^2} } a}$

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

$\ds x$

$=$

$\ds a \paren {\sin^{-1} \paren {\frac {\sqrt {2 a y - y^2} } a} - \frac 1 a \sqrt {2 a y - y^2} }$

$\ds \leadsto \ \ $

$\ds a \sin^{-1} \paren {\frac {\sqrt {2 a y - y^2} } a}$

$=$

$\ds \sqrt {2 a y - y^2} + x$

$\blacksquare$

1992: George F. Simmons: Calculus Gems ... (previous) ... (next): Chapter $\text {B}.21$: The Cycloid: Problem $1$