Length of Arc of Astroid

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Theorem

The total length of the arcs of an astroid constructed within a deferent of radius $a$ is given by:

$\LL = 6 a$


Proof

Let $H$ be embedded in a cartesian plane with its center at the origin and its cusps positioned on the axes.


Astroid.png


We have that $\LL$ is $4$ times the length of one arc of the astroid.

From Arc Length for Parametric Equations:

$\ds \LL = 4 \int_{\theta \mathop = 0}^{\theta \mathop = \pi/2} \sqrt {\paren {\frac {\d x} {\d \theta} }^2 + \paren {\frac {\d y} {\d \theta} }^2} \rd \theta$

where, from Equation of Astroid:

$\begin{cases}

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


We have:

\(\ds \frac {\d x} {\d \theta}\) \(=\) \(\ds -3 a \cos^2 \theta \sin \theta\)
\(\ds \frac {\d y} {\d \theta}\) \(=\) \(\ds 3 a \sin^2 \theta \cos \theta\)


Thus:

\(\ds \sqrt {\paren {\frac {\d x} {\d \theta} }^2 + \paren {\frac {\d y} {\d \theta} }^2}\) \(=\) \(\ds \sqrt {9 a^2 \paren {\sin^4 \theta \cos^2 \theta + \cos^4 \theta \sin^2 \theta} }\)
\(\ds \) \(=\) \(\ds 3 a \sqrt {\sin^2 \theta \cos^2 \theta \paren {\sin^2 \theta + \cos^2 \theta} }\)
\(\ds \) \(=\) \(\ds 3 a \sqrt {\sin^2 \theta \cos^2 \theta}\) Sum of Squares of Sine and Cosine
\(\ds \) \(=\) \(\ds 3 a \sin \theta \cos \theta\)
\(\ds \) \(=\) \(\ds \frac {3 a \sin 2 \theta} 2\) Double Angle Formula for Sine


Thus:

\(\ds \LL\) \(=\) \(\ds 4 \frac {3 a} 2 \int_0^{\pi / 2} \sin 2 \theta \rd \theta\)
\(\ds \) \(=\) \(\ds 6 a \intlimits {\frac {-\cos 2 \theta} 2} 0 {\pi / 2}\) Primitive of $\sin a x$
\(\ds \) \(=\) \(\ds 6 a \paren {-\frac {\cos \pi} 2 + \frac {\cos 0} 2}\) evaluating limits of integration
\(\ds \) \(=\) \(\ds 6 a \frac {-\paren {-1} + 1} 2\)
\(\ds \) \(=\) \(\ds 6 a\)

$\blacksquare$


Sources

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