# Length of Arc of Cycloid/Proof 1

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## 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.

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

where, from Equation of Cycloid:

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

we have:

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

Thus:

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

Thus:

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

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

$\blacksquare$