Rate of Change of Cartesian Coordinates of Cycloid

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Theorem

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


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

\(\ds \frac {\d x} {\d t}\) \(=\) \(\ds \mathbf v_0 \paren {1 - \cos \theta}\)
\(\ds \frac {\d y} {\d t}\) \(=\) \(\ds \mathbf v_0 \sin \theta\)

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, $P = \tuple {x, y}$ is described by:

$(1): \quad \begin{cases} x & = a \paren {\theta - \sin \theta} \\

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

Let $\tuple {x_c, y_c}$ be the coordinates of $O$ at time $t$.

We have that:

$y_c$ is constant: $y_c = a$

From Body under Constant Acceleration: Velocity after Time

$x_c = \mathbf v_0 t$

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.


By Arc Length of Sector:

$x_c = a \theta$

and so:

\(\text {(2)}: \quad\) \(\ds \theta\) \(=\) \(\ds \frac {\mathbf v_0 t} a\)
\(\text {(3)}: \quad\) \(\ds \leadsto \ \ \) \(\ds \frac {\d \theta} {\d t}\) \(=\) \(\ds \frac {\mathbf v_0} a\)


Thus:

\(\ds x\) \(=\) \(\ds a \paren {\frac {\mathbf v_0 t} a - \sin \theta}\) substituting for $\theta$ from $(2)$
\(\ds x\) \(=\) \(\ds \mathbf v_0 t - a \sin \theta\)
\(\ds \leadsto \ \ \) \(\ds \frac {\d x} {\d t}\) \(=\) \(\ds \mathbf v_0 - a \cos \theta \frac {\d \theta} {\d t}\) Chain Rule for Derivatives
\(\ds \) \(=\) \(\ds \mathbf v_0 - a \cos \theta \frac {\mathbf v_0} a\) from $(3)$
\(\ds \) \(=\) \(\ds \mathbf v_0 \paren {1 - \cos \theta}\)


and:

\(\ds y\) \(=\) \(\ds a \paren {1 - \cos \theta}\)
\(\ds \leadsto \ \ \) \(\ds \frac {\d y} {\d t}\) \(=\) \(\ds a \sin \theta \frac {\d \theta} {\d t}\) Chain Rule for Derivatives
\(\ds \) \(=\) \(\ds a \sin \theta \frac {\mathbf v_0} a\) from $(3)$
\(\ds \) \(=\) \(\ds \mathbf v_0 \sin \theta\)

$\blacksquare$


Sources

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