# Length of Arc of Astroid

## 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.

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

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

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 $\ds$ $=$ $\ds 6 a \frac {-\paren {-1} + 1} 2$ $\ds$ $=$ $\ds 6 a$

$\blacksquare$