Length of Arc of Cycloid

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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 1

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:

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


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


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

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


Proof 2

Consider the tangent $PQ$ to both the generating circle and the cycloid itself.




$PR = 2 PQ$

In the limit, where $P$ is at the cusp, the tangent $PQ$ is perpendicular to the straight line along which the generating circle rolls.

At this point:

$PQ = 2 a$.

Thus at this point:

$PR = 4 a$

But $4 a$ is half the length of one arc of $C$.

Hence the result.

Also see

Historical Note

The geometric proof of the length of the arc of a cycloid was demonstrated by Christopher Wren in $1658$.