# Rate of Change of Cartesian Coordinates of Cycloid

## Theorem

Let a circle $C$ of radius $a$ roll without slipping along the x-axis of a cartesian coordinate plane at a constant speed such that the center moves with a velocity $\mathbf v_0$ in the direction of increasing $x$.

Consider a point $P$ on the circumference of this circle.

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

Then the rate of change of $x$ and $y$ can be expresssed as:

 $\displaystyle \frac {\mathrm d x} {\mathrm d t}$ $=$ $\displaystyle \mathbf v_0 \left({1 - \cos \theta}\right)$ $\displaystyle \frac {\mathrm d y} {\mathrm d t}$ $=$ $\displaystyle \mathbf v_0 \sin \theta$

where $\theta$ is the angle turned by $C$ after time $t$.

## Proof

Let the center of $C$ be $O$.

Without loss of generality, let $P$ be at the origin at time $t = t_0$.

By definition, $P$ traces out a cycloid.

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

$(1): \quad \begin{cases} x & = a \left({\theta - \sin \theta}\right) \\ y & = a \left({1 - \cos \theta}\right) \end{cases}$

Let $\left({x_c, y_c}\right)$ be the coordinates of $O$ at time $t$.

We have that:

$y_c$ is constant: $y_c = a$
$x_c = \mathbf v_0 t$

as the acceleration of $O$ is zero.

But $x_c$ is equal to the length of the arc of $C$ that has rolled along the $x$-axis.

$x_c = a \theta$

and so:

 $(2):\quad$ $\displaystyle \theta$ $=$ $\displaystyle \frac {\mathbf v_0 t} a$ $(3):\quad$ $\displaystyle \leadsto \ \$ $\displaystyle \frac {\mathrm d \theta} {\mathrm d t}$ $=$ $\displaystyle \frac {\mathbf v_0} a$

Thus:

 $\displaystyle x$ $=$ $\displaystyle a \left({\frac {\mathbf v_0 t} a - \sin \theta}\right)$ substituting for $\theta$ from $(2)$ $\displaystyle x$ $=$ $\displaystyle \mathbf v_0 t - a \sin \theta$ $\displaystyle \leadsto \ \$ $\displaystyle \frac {\mathrm d x} {\mathrm d t}$ $=$ $\displaystyle \mathbf v_0 - a \cos \theta \frac {\mathrm d \theta} {\mathrm d t}$ Chain Rule $\displaystyle$ $=$ $\displaystyle \mathbf v_0 - a \cos \theta \frac {\mathbf v_0} a$ from $(3)$ $\displaystyle$ $=$ $\displaystyle \mathbf v_0 \left({1 - \cos \theta}\right)$

and:

 $\displaystyle y$ $=$ $\displaystyle a \left({1 - \cos \theta}\right)$ $\displaystyle \leadsto \ \$ $\displaystyle \frac {\mathrm d y} {\mathrm d t}$ $=$ $\displaystyle a \sin \theta \frac {\mathrm d \theta} {\mathrm d t}$ Chain Rule $\displaystyle$ $=$ $\displaystyle a \sin \theta \frac {\mathbf v_0} a$ from $(3)$ $\displaystyle$ $=$ $\displaystyle \mathbf v_0 \sin \theta$

$\blacksquare$