Length of Arc of Deltoid

Theorem
The total length of the arcs of a deltoid constructed within a deferent of radius $a$ is given by:
 * $\LL = \dfrac {16 a} 3$

Proof
Let $H$ be embedded in a cartesian plane with its center at the origin and one of its cusps positioned at $\tuple {a, 0}$.


 * Deltoid.png

We have that $\LL$ is $3$ times the length of one arc of the deltoid.

From Arc Length for Parametric Equations:


 * $\ds \LL = 3 \int_{\theta \mathop = 0}^{\theta \mathop = 2 \pi/3} \sqrt {\paren {\frac {\d x} {\d \theta} }^2 + \paren {\frac {\d y} {\d \theta} }^2} \rd \theta$

where, from Equation of Deltoid:
 * $\begin{cases}

x & = 2 b \cos \theta + b \cos 2 \theta \\ y & = 2 b \sin \theta - b \sin 2 \theta \end{cases}$

We have:

Thus:

Thus:
 * $\sqrt {\paren {\dfrac {\d x} {\d \theta} }^2 + \paren {\dfrac {\d y} {\d \theta} }^2} = 4 b \sin \dfrac \theta 2 \size {1 + 2 \cos \theta}$

In the range $0$ to $2 \pi / 3$, $1 + 2 \cos \theta$ is not less than $0$, and so:
 * $\ds \LL = 3 \int_0^{2 \pi / 3} 4 b \sin \dfrac \theta 2 \paren {1 + 2 \cos \theta} \rd \theta$

Put:
 * $u = \cos \dfrac \theta 2$

so:
 * $2 \dfrac {\d u} {\d \theta} = -\sin \dfrac \theta 2$

As $\theta$ increases from $0$ to $\dfrac {2 \pi} 3$, $u$ decreases from $1$ to $\dfrac 1 2$.

Then:

Substituting:
 * $2 \dfrac {\d u} {\d \theta} = -\sin \dfrac \theta 2$

and the limits of integration:
 * $u = 1$ for $\theta = 0$
 * $u = \dfrac 1 2$ for $\theta = \dfrac {2 \pi} 3$

we obtain, after simplifying the sign: