# Length of Arc of Nephroid

## Theorem

The total length of the arcs of a nephroid constructed around a deferent 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 $\tuple {\pm a, 0}$. We have that $\LL$ is $2$ times the length of one arc of the nephroid.

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

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:

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

Thus:

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

Thus:

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

So:

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

$\blacksquare$