Maximum Rate of Change of Y Coordinate of Astroid

Theorem
Let $C_1$ and $C_2$ be the rotor and stator respectively of an astroid $H$.

Let $C_2$ be embedded in a cartesian plane with its center $O$ located at the origin.

Let the center $C$ of $C_1$ move at a constant angular velocity $\omega$ around the center of $C_2$.

Let $P$ be the point on the circumference of $C_1$ whose locus is $H$.

Let $C_1$ be positioned at time $t = 0$ so that $P$ its point of tangency to $C_2$, located on the $x$-axis.

Let $\theta$ be the angle made by $OC$ to the $x$-axis at time $t$.

Then the maximum rate of change of the $y$ coordinate of $P$ in the first quadrant occurs when $P$ is at the point where:
 * $x = a \paren {\dfrac 1 3}^{3/2}$
 * $y = a \paren {\dfrac 2 3}^{3/2}$

Proof

 * Astroid.png

The rate of change of $\theta$ is given by:
 * $\omega = \dfrac {\d \theta} {\d t}$

From Equation of Astroid: Parametric Form, the point $P = \tuple {x, y}$ is described by the parametric equation:
 * $\begin {cases}

x & = a \cos^3 \theta \\ y & = a \sin^3 \theta \end{cases}$

The rate of change of $y$ is given by:

By Derivative at Maximum or Minimum, when $\dfrac {\d y} {\d t}$ is at a maximum:
 * $\dfrac {\d^2 y} {\d t^2} = 0$

Thus:

Hence:

We can assume $\sin \theta \ne 0$ because in that case $\theta = 0$ and so $\dfrac {\d y} {\d t} = 0$.

Thus when $\sin \theta = 0$, $y$ is not a maximum.

So we can divide by $\sin \theta$ to give:

We have:

From Equation of Astroid: Cartesian Form:
 * $x^{2/3} + y^{2/3} = a^{2/3}$

Hence:

Similarly: