Equation of Astroid/Parametric Form

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Theorem

Let $H$ be the astroid generated by the epicycle $C_1$ of radius $b$ rolling without slipping around the inside of a deferent $C_2$ of radius $a = 4 b$.

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

Let $P$ be a point on the circumference of $C_1$.

Let $C_1$ be initially positioned so that $P$ is its point of tangency to $C_2$, located at point $A = \tuple {a, 0}$ on the $x$-axis.


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 parametric equation:

$\begin{cases}

x & = a \cos^3 \theta \\ y & = a \sin^3 \theta \end{cases}$ where $\theta$ is the angle between the $x$-axis and the line joining the origin to the center of $C_1$.


Proof

By definition, an astroid is a hypocycloid with $4$ cusps.

Astroid.png


By Equation of Hypocycloid, the equation of $H$ is given by:

$\begin{cases}

x & = \paren {a - b} \cos \theta + b \map \cos {\paren {\dfrac {a - b} b} \theta} \\ y & = \paren {a - b} \sin \theta - b \map \sin {\paren {\dfrac {a - b} b} \theta} \end{cases}$

From Number of Cusps of Hypocycloid from Integral Ratio of Circle Radii, this can be generated by a epicycle $C_1$ of radius $\dfrac 1 4$ the radius of the deferent.

Thus $a = 4 b$ and the equation of $H$ is now given by:

$\begin{cases}

x & = 3 b \cos \theta + b \cos 3 \theta \\ y & = 3 b \sin \theta - b \sin 3 \theta \end{cases}$


From Triple Angle Formula for Cosine:

$\cos 3 \theta = 4 \cos^3 \theta - 3 \cos \theta$

and from Triple Angle Formula for Sine:

$\sin 3 \theta = 3 \sin \theta - 4 \sin^3 \theta$


Thus $H$ can be expressed as:

$\begin{cases}

x & = 4 b \cos^3 \theta = a \cos^3 \theta \\ y & = 4 b \sin^3 \theta = a \sin^3 \theta \end{cases}$

$\blacksquare$


Sources

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