Maximum Rate of Change of Y Coordinate of Astroid

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Theorem

Let $C_1$ and $C_2$ be the epicycle and deferent 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:

\(\ds \dfrac {\d y} {\d t}\) \(=\) \(\ds \dfrac {\d y} {\d \theta} \dfrac {\d \theta} {\d t}\) Chain Rule for Derivatives
\(\ds \) \(=\) \(\ds 3 a \omega \sin^2 \theta \cos \theta\) Power Rule for Derivatives, Derivative of Sine Function, Chain Rule for Derivatives


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:

\(\ds \dfrac {\d^2 y} {\d t^2}\) \(=\) \(\ds \map {\dfrac \d {\d \theta} } {3 a \omega \sin^2 \theta \cos \theta} \dfrac {\d \theta} {\d t}\) Chain Rule for Derivatives
\(\ds \) \(=\) \(\ds 3 a \omega^2 \paren {2 \sin \theta \cos^2 \theta - \sin^3 \theta}\) Product Rule for Derivatives and others


Hence:

\(\ds \dfrac {\d^2 y} {\d t^2}\) \(=\) \(\ds 0\)
\(\ds \leadsto \ \ \) \(\ds 3 a \omega^2 \paren {2 \sin \theta \cos^2 \theta - \sin^3 \theta}\) \(=\) \(\ds 0\)
\(\ds \leadsto \ \ \) \(\ds 2 \sin \theta \cos^2 \theta\) \(=\) \(\ds \sin^3 \theta\)


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:

\(\ds 2 \cos^2 \theta\) \(=\) \(\ds \sin^2 \theta\)
\(\ds \leadsto \ \ \) \(\ds \tan^2 \theta\) \(=\) \(\ds 2\) Tangent is Sine divided by Cosine
\(\ds \leadsto \ \ \) \(\ds \tan \theta\) \(=\) \(\ds \sqrt 2\)


We have:

\(\ds x\) \(=\) \(\ds a \cos^3 \theta\)
\(\ds y\) \(=\) \(\ds a \sin^3 \theta\)
\(\ds \leadsto \ \ \) \(\ds \frac y x\) \(=\) \(\ds \tan^3 \theta\) Tangent is Sine divided by Cosine
\(\ds \) \(=\) \(\ds 2^{3/2}\)
\(\ds \leadsto \ \ \) \(\ds y\) \(=\) \(\ds 2^{3/2} x\)


From Equation of Astroid: Cartesian Form:

$x^{2/3} + y^{2/3} = a^{2/3}$

Hence:

\(\ds x^{2/3} + \paren {2^{3/2} x}^{2/3}\) \(=\) \(\ds a^{2/3}\)
\(\ds \leadsto \ \ \) \(\ds x^{2/3} \paren {1 + 2}\) \(=\) \(\ds a^{2/3}\) simplifying
\(\ds \leadsto \ \ \) \(\ds x\) \(=\) \(\ds \paren {\frac {a^{2/3} } 3}^{3/2}\)
\(\ds \leadsto \ \ \) \(\ds x\) \(=\) \(\ds a \paren {\frac 1 3}^{3/2}\)


Similarly:

\(\ds \frac y x\) \(=\) \(\ds 2^{3/2}\)
\(\ds \leadsto \ \ \) \(\ds x\) \(=\) \(\ds \frac 1 {2^{3/2} } y\)
\(\ds \paren {\frac 1 {2^{3/2} } y}^{2/3} + y^{2/3}\) \(=\) \(\ds a^{2/3}\)
\(\ds \leadsto \ \ \) \(\ds y^{2/3} \paren {1 + \frac 1 2}\) \(=\) \(\ds a^{2/3}\) simplifying
\(\ds \leadsto \ \ \) \(\ds y^{2/3}\) \(=\) \(\ds \frac 2 3 a^{2/3}\) simplifying
\(\ds \leadsto \ \ \) \(\ds y\) \(=\) \(\ds \paren {2 \frac {a^{2/3} } 3}^{3/2}\)
\(\ds \leadsto \ \ \) \(\ds x\) \(=\) \(\ds a \paren {\frac 2 3}^{3/2}\)

$\blacksquare$


Sources