Optional — The Astroid

Section 1.11 Optional — The Astroid

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 astroid

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.

. We shall find its equation.

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

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

\begin{equation*} \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) \end{equation*}

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

\begin{equation*} x(\theta)= a\cos^3\theta\qquad y(\theta)=a\sin^3\theta \end{equation*}

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

\begin{equation*} x^{2/3}+y^{2/3} =a^{2/3} \end{equation*}

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 \(-a\sin^3\theta_0=a\sin^3(2\pi-\theta_0)\text{.}\)

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