# Length of Arc of Nephroid

## Theorem

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

$\mathcal L = 12 a$

## Proof

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

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

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

where, from Equation of 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:

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

Thus:

 $\displaystyle$  $\displaystyle \left({\frac {\mathrm d x} {\mathrm d \theta} }\right)^2 + \left({\frac {\mathrm d y} {\mathrm d \theta} }\right)^2$ $\displaystyle$ $=$ $\displaystyle \left({-3 b \sin \theta + 3 b \sin 3 \theta}\right)^2 + \left({3 b \cos \theta - 3 b \cos 3 \theta}\right)^2$ $\displaystyle$ $=$ $\displaystyle 9 b^2 \left({\left({-\sin \theta + \sin 3 \theta}\right)^2 + \left({\cos \theta - \cos 3 \theta}\right)^2}\right)$ $\displaystyle$ $=$ $\displaystyle 9 b^2 \left({\sin^2 \theta - 2 \sin \theta \sin 3 \theta + \sin^2 3 \theta + \cos^2 \theta - 2 \cos \theta \cos 3 \theta + \cos^2 3 \theta}\right)$ $\displaystyle$ $=$ $\displaystyle 9 b^2 \left({2 - 2 \sin \theta \sin 3 \theta - 2 \cos \theta \cos 3 \theta}\right)$ Sum of Squares of Sine and Cosine $\displaystyle$ $=$ $\displaystyle 18 b^2 \left({1 - \sin \theta \sin 3 \theta - \cos \theta \cos 3 \theta}\right)$ $\displaystyle$ $=$ $\displaystyle 18 b^2 \left({1 - \sin \theta \left({3 \sin \theta - 4 \sin^3 \theta}\right) - \cos \theta \cos 2 \theta}\right)$ Triple Angle Formula for Sine $\displaystyle$ $=$ $\displaystyle 18 b^2 \left({1 - \sin \theta \left({3 \sin \theta - 4 \sin^3 \theta}\right) - \cos \theta \left({4 \cos^3 \theta - 3 \cos \theta}\right)}\right)$ Triple Angle Formula for Cosine $\displaystyle$ $=$ $\displaystyle 18 b^2 \left({1 - 3 \sin^2 \theta + 4 \sin^4 \theta - 4 \cos^4 \theta + 3 \cos^2 \theta}\right)$ simplifying $\displaystyle$ $=$ $\displaystyle 18 b^2 \left({1 + 3 \cos 2 \theta + 4 \sin^4 \theta - 4 \cos^4 \theta}\right)$ Double Angle Formula for Cosine $\displaystyle$ $=$ $\displaystyle 18 b^2 \left({1 + 3 \cos 2 \theta + \dfrac {3 - 4 \cos 2 x + \cos 4 x} 2 - 4 \cos^4 \theta}\right)$ Power Reduction Formula for $\sin^4$ $\displaystyle$ $=$ $\displaystyle 18 b^2 \left({1 + 3 \cos 2 \theta + \dfrac {3 - 4 \cos 2 \theta + \cos 4 \theta} 2 - \dfrac {3 + 4 \cos 2 \theta + \cos 4 \theta} 2}\right)$ Power Reduction Formula for $\cos^4$ $\displaystyle$ $=$ $\displaystyle 18 b^2 \left({1 - \cos 2 \theta}\right)$ simplifying $\displaystyle$ $=$ $\displaystyle 18 b^2 \left({2 \sin^2 \theta}\right)$ Square of Sine $\displaystyle$ $=$ $\displaystyle 36 b^2 \sin^2 \theta$ simplifying

Thus:

$\sqrt {\left({\dfrac {\mathrm d x} {\mathrm d \theta} }\right)^2 + \left({\dfrac {\mathrm d y} {\mathrm d \theta} }\right)^2} = 6 b \sin \theta$

So:

 $\displaystyle \mathcal L$ $=$ $\displaystyle 2 \int_0^\pi 6 b \sin \theta \, \mathrm d \theta$ $\displaystyle$ $=$ $\displaystyle 12 b \int_0^\pi \sin \theta \, \mathrm d \theta$ $\displaystyle$ $=$ $\displaystyle 12 b \Big[{-\cos \theta}\Big]_0^\pi$ $\displaystyle$ $=$ $\displaystyle 12 b \left({-\cos \pi - \left({-\cos 0}\right)}\right)$ $\displaystyle$ $=$ $\displaystyle 12 b \left({- \left({-1}\right) - \left({-1}\right)}\right)$ $\displaystyle$ $=$ $\displaystyle 24 b$ $\displaystyle$ $=$ $\displaystyle 12 a$

$\blacksquare$