Length of Arc of Nephroid

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Theorem

The total length of the arcs of a nephroid constructed around a stator of radius $a$ is given by:

$\LL = 12 a$


Proof

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


Nephroid.png


We have that $\LL$ is $2$ times the length of one arc of the nephroid.

From Arc Length for Parametric Equations:

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

where, from Equation of Nephroid:

$\begin{cases} x & = 3 b \cos \theta - b \cos 3 \theta \\ y & = 3 b \sin \theta - b \sin 3 \theta \end{cases}$


We have:

\(\displaystyle \frac {\d x} {\d \theta}\) \(=\) \(\displaystyle -3 b \sin \theta + 3 b \sin 3 \theta\)
\(\displaystyle \frac {\d y} {\d \theta}\) \(=\) \(\displaystyle 3 b \cos \theta - 3 b \cos 3 \theta\)


Thus:

\(\displaystyle \) \(\) \(\displaystyle \paren {\frac {\d x} {\d \theta} }^2 + \paren {\frac {\d y} {\d \theta} }^2\)
\(\displaystyle \) \(=\) \(\displaystyle \paren {-3 b \sin \theta + 3 b \sin 3 \theta}^2 + \paren {3 b \cos \theta - 3 b \cos 3 \theta}^2\)
\(\displaystyle \) \(=\) \(\displaystyle 9 b^2 \paren {\paren {-\sin \theta + \sin 3 \theta}^2 + \paren {\cos \theta - \cos 3 \theta}^2}\)
\(\displaystyle \) \(=\) \(\displaystyle 9 b^2 \paren {\sin^2 \theta - 2 \sin \theta \sin 3 \theta + \sin^2 3 \theta + \cos^2 \theta - 2 \cos \theta \cos 3 \theta + \cos^2 3 \theta}\) Square of Difference
\(\displaystyle \) \(=\) \(\displaystyle 9 b^2 \paren {2 - 2 \sin \theta \sin 3 \theta - 2 \cos \theta \cos 3 \theta}\) Sum of Squares of Sine and Cosine
\(\displaystyle \) \(=\) \(\displaystyle 18 b^2 \paren {1 - \paren {\sin \theta \sin 3 \theta + \cos \theta \cos 3 \theta} }\)
\(\displaystyle \) \(=\) \(\displaystyle 18 b^2 \paren {1 - \cos 2 \theta}\) Cosine of Difference
\(\displaystyle \) \(=\) \(\displaystyle 18 b^2 \paren {2 \sin^2 \theta}\) Square of Sine
\(\displaystyle \) \(=\) \(\displaystyle 36 b^2 \sin^2 \theta\) simplifying

Thus:

$\sqrt {\paren {\dfrac {\d x} {\d \theta} }^2 + \paren {\dfrac {\d y} {\d \theta} }^2} = 6 b \sin \theta$


So:

\(\displaystyle \LL\) \(=\) \(\displaystyle 2 \int_0^\pi 6 b \sin \theta \rd \theta\)
\(\displaystyle \) \(=\) \(\displaystyle 12 b \int_0^\pi \sin \theta \rd \theta\)
\(\displaystyle \) \(=\) \(\displaystyle 12 b \bigintlimits {-\cos \theta} 0 \pi\)
\(\displaystyle \) \(=\) \(\displaystyle 12 b \paren {-\cos \pi - \paren {-\cos 0} }\)
\(\displaystyle \) \(=\) \(\displaystyle 12 b \paren {-\paren {-1} - \paren {-1} }\)
\(\displaystyle \) \(=\) \(\displaystyle 24 b\)
\(\displaystyle \) \(=\) \(\displaystyle 12 a\)

$\blacksquare$


Sources