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.

From Arc Length for Parametric Equations:

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

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.

$\blacksquare$


Sources

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