# Length of Arc of Cycloid/Proof 1

## Theorem

Let $C$ be a cycloid generated by the equations:

$x = a \paren {\theta - \sin \theta}$
$y = a \paren {1 - \cos \theta}$

Then the length of one arc of the cycloid is $8 a$.

## Proof

Let $L$ be the length of one arc of the cycloid.

$\displaystyle L = \int_0^{2 \pi} \sqrt {\paren {\frac {\d x} {\d \theta} }^2 + \paren {\frac {\d y} {\d \theta} }^2} \rd \theta$

where, from Equation of Cycloid:

$x = a \paren {\theta - \sin \theta}$
$y = a \paren {1 - \cos \theta}$

we have:

 $\displaystyle \frac {\d x} {\d \theta}$ $=$ $\displaystyle a \paren {1 - \cos \theta}$ $\displaystyle \frac {\d y} {\d \theta}$ $=$ $\displaystyle a \sin \theta$

Thus:

 $\displaystyle \paren {\frac {\d x} {\d \theta} }^2 + \paren {\frac {\d y} {\d \theta} }^2$ $=$ $\displaystyle a^2 \paren {\paren {1 - \cos \theta}^2 + \sin^2 \theta}$ $\displaystyle$ $=$ $\displaystyle a^2 \paren {1 - 2 \cos \theta + \cos^2 \theta + \sin^2 \theta}$ $\displaystyle$ $=$ $\displaystyle 2 a^2 \paren {1 - \cos \theta}$ $\displaystyle$ $=$ $\displaystyle 4 a^2 \map {\sin^2} {\theta / 2}$ Half Angle Formula for Sine

Thus:

 $\displaystyle L$ $=$ $\displaystyle \int_0^{2 \pi} 2 a \, \map \sin {\theta / 2} \rd \theta$ $\displaystyle$ $=$ $\displaystyle \bigintlimits {-4 a \, \map \cos {\theta / 2} } 0 {2 \pi}$ $\displaystyle$ $=$ $\displaystyle 8 a$

So $L = 8 a$ where $a$ is the radius of the generating circle.

$\blacksquare$