1.11: Optional — The Astroid

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Imagine a ball of radius $a/4$ rolling around the inside of a circle of radius $a\text{.}$ The curve traced by a point $P$ painted on the inner circle (that's the blue curve in the figures below) is called an astroid¹. We shall find its equation.

$astroid1AA.svg$ $astroid1BB.svg$ $astroid1CC.svg$

$astroid1DD.svg$ $astroid1EE.svg$ $astroid1FF.svg$

Define the angles $\theta$ and $\phi$ as in the figure in the left below.

$astroid.svg$ $astroid3.svg$

That is

the vector from the centre, $O\text{,}$ of the circle of radius $a$ to the centre, $Q\text{,}$ of the ball of radius $a/4$ is $\frac{3}{4}a\big(\cos\theta,\sin\theta\big)$ and
the vector from the centre, $Q\text{,}$ of the ball of radius $a/4$ to the point $P$ is $\frac{1}{4}a\big(\cos\phi,-\sin\phi\big)$

As $\theta$ runs from 0 to $\frac{\pi}{2}\text{,}$ the point of contact between the two circles travels through one quarter of the circumference of the circle of radius $a\text{,}$ which is a distance $\frac{1}{4}(2\pi a)\text{,}$ which, in turn, is exactly the circumference of the inner circle. Hence if $\phi=0$ for $\theta=0$ (i.e. if $P$ starts on the $x$-axis), then for $\theta=\frac{\pi}{2}\text{,}$ $P$ is back in contact with the big circle at the north pole of both the inner and outer circles. That is, $\phi=\frac{3\pi}{2}$ when $\theta=\frac{\pi}{2}\text{.}$ (See the figure on the right above.) So $\phi=3\theta$ and $P$ has coordinates

\[ \frac{3}{4}a\big(\cos\theta,\sin\theta\big) +\frac{1}{4}a\big(\cos\phi,-\sin\phi\big) =\frac{a}{4}\big(3\cos\theta+\cos 3\theta,3\sin\theta-\sin 3\theta\big) \nonumber \]

As, recalling your double angle, or even better your triple angle, trig identities,

\[\begin{align*} \cos3\theta&=\cos\theta\cos2\theta-\sin\theta\sin 2\theta\\ &=\cos\theta[\cos^2\theta-\sin^2\theta]-2\sin^2\theta\cos\theta\\ &=\cos\theta[\cos^2\theta-3\sin^2\theta]\\ \sin3\theta&=\sin\theta\cos2\theta+\cos\theta\sin 2\theta\\ &=\sin\theta[\cos^2\theta-\sin^2\theta]+2\sin\theta\cos^2\theta\\ &=\sin\theta[3\cos^2\theta-\sin^2\theta] \end{align*}\]

we have

\[\begin{alignat*}{2} 3\cos\theta+\cos 3\theta &=\cos\theta[3+\cos^2\theta-3\sin^2\theta] & &=\cos\theta[3+\cos^2\theta-3(1-\cos^2\theta)] \\ &=4\cos^3\theta\\ 3\sin\theta-\sin 3\theta &=\sin\theta[3-3\cos^2\theta+\sin^2\theta] & &=\sin\theta[3-3(1-\sin^2\theta)+\sin^2\theta] \\ &=4\sin^3\theta \end{alignat*}\]

and the coordinates of $P$ simplify to

\[ x(\theta)= a\cos^3\theta\qquad y(\theta)=a\sin^3\theta \nonumber \]

Oof! As $\ x^{2/3}+y^{2/3}=a^{2/3}\cos^2\theta+a^{2/3}\sin^2\theta \ ,$ the path traced by $P$ obeys the equation

\[ x^{2/3}+y^{2/3} =a^{2/3} \nonumber \]

which is surprisingly simple, considering what we went through to get here.

There remains the danger that there could exist points $(x,y)$ obeying the equation $x^{2/3}+y^{2/3}=a^{2/3}$ that are not of the form $x= a\cos^3\theta,\ y=a\sin^3\theta$ for any $\theta\text{.}$ That is, there is a danger that the parametrized curve $x= a\cos^3\theta,\ y=a\sin^3\theta$ covers only a portion of $x^{2/3}+y^{2/3}=a^{2/3}\text{.}$ We now show that the parametrized curve $x= a\cos^3\theta,\ y=a\sin^3\theta$ in fact covers all of $x^{2/3}+y^{2/3}=a^{2/3}$ as $\theta$ runs from $0$ to $2\pi\text{.}$

First, observe that $x^{2/3}=\big(\root 3\of x\big)^2\ge 0$ and $y^{2/3}=\big(\root 3\of y\big)^2\ge 0\text{.}$ Hence, if $(x,y)$ obeys $x^{2/3}+y^{2/3}=a^{2/3}\text{,}$ then necessarily $0\le x^{2/3}\le a^{2/3} $ and so $-a\le x\le a\text{.}$ As $\theta$ runs from $0$ to $2\pi\text{,}$ $a\cos^3\theta$ takes all values between $-a$ and $a$ and hence takes all possible values of $x\text{.}$ For each $x\in[-a,a]\text{,}$ $y$ takes two values, namely $\pm{[a^{2/3}-x^{2/3}]}^{3/2}\text{.}$ If $x=a\cos^3\theta_0=a\cos^3(2\pi-\theta_0)\text{,}$ the two corresponding values of $y$ are precisely $a\sin^3\theta_0$ and $-a\sin^3\theta_0=a\sin^3(2\pi-\theta_0)\text{.}$

The name “astroid” comes from the Greek word “aster”, meaning star, with the suffix “oid” meaning “having the shape of”. The curve was first discussed by Johann Bernoulli in 1691–92.