Length of Arc of Deltoid

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Theorem

The total length of the arcs of a deltoid constructed within a stator of radius $a$ is given by:

$\mathcal L = \dfrac {16 a} 3$


Proof

Let $H$ be embedded in a cartesian coordinate plane with its center at the origin and one of its cusps positioned at $\left({a, 0}\right)$.


Deltoid.png


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

From Arc Length for Parametric Equations:

$\displaystyle \mathcal L = 3 \int_{\theta \mathop = 0}^{\theta \mathop = 2 \pi/3} \sqrt {\left({\frac{\mathrm d x} {\mathrm d \theta}}\right)^2 + \left({\frac{\mathrm d y} {\mathrm d \theta}}\right)^2} \mathrm d \theta$

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:

\(\displaystyle \frac {\mathrm d x} {\mathrm d \theta}\) \(=\) \(\displaystyle -2 b \sin \theta - 2 b \sin 2 \theta\)
\(\displaystyle \frac {\mathrm d y} {\mathrm d \theta}\) \(=\) \(\displaystyle 2 b \cos \theta - 2 b \cos 2 \theta\)


Thus:

\(\displaystyle \) \(\) \(\displaystyle \left({\frac {\mathrm d x} {\mathrm d \theta} }\right)^2 + \left({\frac {\mathrm d y} {\mathrm d \theta} }\right)^2\)
\(\displaystyle \) \(=\) \(\displaystyle \left({-2 b \sin \theta - 2 b \sin 2 \theta}\right)^2 + \left({2 b \cos \theta - 2 b \cos 2 \theta}\right)^2\)
\(\displaystyle \) \(=\) \(\displaystyle 4 b^2 \left({\left({-\sin \theta - \sin 2 \theta}\right)^2 + \left({\cos \theta - \cos 2 \theta}\right)^2}\right)\)
\(\displaystyle \) \(=\) \(\displaystyle 4 b^2 \left({\sin^2 \theta + 2 \sin \theta \sin 2 \theta + \sin^2 2 \theta + \cos^2 \theta - 2 \cos \theta \cos 2 \theta + \cos^2 2 \theta}\right)\)
\(\displaystyle \) \(=\) \(\displaystyle 4 b^2 \left({2 + 2 \sin \theta \sin 2 \theta - 2 \cos \theta \cos 2 \theta}\right)\) Sum of Squares of Sine and Cosine
\(\displaystyle \) \(=\) \(\displaystyle 8 b^2 \left({1 + \sin \theta \sin 2 \theta - \cos \theta \cos 2 \theta}\right)\)
\(\displaystyle \) \(=\) \(\displaystyle 8 b^2 \left({1 + 2 \sin^2 \theta \cos \theta - \cos \theta \cos 2 \theta}\right)\) Double Angle Formula for Sine
\(\displaystyle \) \(=\) \(\displaystyle 8 b^2 \left({1 + 2 \sin^2 \theta \cos \theta - \cos \theta \left({1 - 2 \sin^2 \theta}\right)}\right)\) Double Angle Formula for Cosine: Corollary 2
\(\displaystyle \) \(=\) \(\displaystyle 8 b^2 \left({1 - \cos \theta + 4 \sin^2 \theta \cos \theta}\right)\) simplifying
\(\displaystyle \) \(=\) \(\displaystyle 8 b^2 \left({1 - \cos \theta + 4 \cos \theta \left({1 - \cos^2 \theta}\right)}\right)\) Sum of Squares of Sine and Cosine
\(\displaystyle \) \(=\) \(\displaystyle 8 b^2 \left({1 - \cos \theta + 4 \cos \theta \left({1 + \cos \theta}\right) \left({1 - \cos \theta}\right)}\right)\) Difference of Two Squares
\(\displaystyle \) \(=\) \(\displaystyle 8 b^2 \left({1 - \cos \theta}\right) \left({1 + 4 \cos \theta \left({1 + \cos \theta}\right)}\right)\)
\(\displaystyle \) \(=\) \(\displaystyle 8 b^2 \left({1 - \cos \theta}\right) \left({1 + 4 \cos \theta + 4 \cos^2 \theta}\right)\)
\(\displaystyle \) \(=\) \(\displaystyle 8 b^2 \left({1 - \cos \theta}\right) \left({1 + 2 \cos \theta}\right)^2\)
\(\displaystyle \) \(=\) \(\displaystyle 8 b^2 \left({2 \sin^2 \frac \theta 2}\right) \left({1 + 2 \cos \theta}\right)^2\) Half Angle Formula for Sine
\(\displaystyle \) \(=\) \(\displaystyle 16 b^2 \sin^2 \frac \theta 2 \left({1 + 2 \cos \theta}\right)^2\)

Thus:

$\sqrt {\left({\dfrac {\mathrm d x} {\mathrm d \theta} }\right)^2 + \left({\dfrac {\mathrm d y} {\mathrm d \theta} }\right)^2} = 4 b \sin \dfrac \theta 2 \left|{1 + 2 \cos \theta}\right|$

In the range $0$ to $2 \pi / 3$, $1 + 2 \cos \theta$ is not less than $0$, and so:

$\displaystyle \mathcal L = 3 \int_0^{2 \pi / 3} 4 b \sin \dfrac \theta 2 \left({1 + 2 \cos \theta}\right) \, \mathrm d \theta$


Put:

$u = \cos \dfrac \theta 2$

so:

$2 \dfrac {\mathrm d u} {\mathrm d \theta} = -\sin \dfrac \theta 2$

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

Then:

\(\displaystyle 1 + 2 \cos \theta\) \(=\) \(\displaystyle 1 + 2 \left({2 \left({\cos \dfrac \theta 2}\right)^2 - 1}\right)\) Half Angle Formula for Cosine
\(\displaystyle \) \(=\) \(\displaystyle 4 u^2 - 1\)


Substituting:

$2 \dfrac {\mathrm d u} {\mathrm d \theta} = -\sin \dfrac \theta 2$

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:

\(\displaystyle \mathcal L\) \(=\) \(\displaystyle 12 b \int_1^{1/2} \left({1 - 4 u^2}\right) 2 \, \mathrm d u\) Integration by Substitution
\(\displaystyle \) \(=\) \(\displaystyle 24 b \left[{u - \frac 4 3 u^3}\right]_1^{1/2}\)
\(\displaystyle \) \(=\) \(\displaystyle 24 b \left({\left({\frac 1 2 - \frac 4 3 \frac 1 {2^3} }\right) - \left({1 - \frac 4 3 }\right) }\right)\)
\(\displaystyle \) \(=\) \(\displaystyle 24 b \left({\frac 1 2 - \frac 1 6 - 1 + \frac 4 3}\right)\)
\(\displaystyle \) \(=\) \(\displaystyle 24 b \left({\frac 1 3 + \frac 1 3}\right)\)
\(\displaystyle \) \(=\) \(\displaystyle 16 b\)
\(\displaystyle \) \(=\) \(\displaystyle \frac {16 a} 3\)

$\blacksquare$


Sources