# Length of Arc of Cycloid

## 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 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:

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

\(\ds \frac {\d y} {\d \theta}\) | \(=\) | \(\ds a \sin \theta\) |

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.

$\blacksquare$

## Proof 2

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

By

Work In ProgressIn particular: still to be doneYou can help $\mathsf{Pr} \infty \mathsf{fWiki}$ by completing it.To discuss this page in more detail, feel free to use the talk page.When this work has been completed, you may remove this instance of `{{WIP}}` from the code. |

:

- $PR = 2 PQ$

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.

This needs considerable tedious hard slog to complete it.To discuss this page in more detail, feel free to use the talk page.When this work has been completed, you may remove this instance of `{{Finish}}` from the code.If you would welcome a second opinion as to whether your work is correct, add a call to `{{Proofread}}` the page. |

## Also see

## Historical Note

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

## Sources

- 1926: Henry Ernest Dudeney:
*Modern Puzzles*... (previous) ... (next): Solutions: $209$. -- A Wheel Fallacy - 1968: Henry Ernest Dudeney:
*536 Puzzles & Curious Problems*... (previous) ... (next): Answers: $293$. A Wheel Fallacy - 1968: Murray R. Spiegel:
*Mathematical Handbook of Formulas and Tables*... (previous) ... (next): $\S 11$: Special Plane Curves: Cycloid: $11.6$ - 1972: George F. Simmons:
*Differential Equations*... (previous) ... (next): $1$: The Nature of Differential Equations: $\S 6$: The Brachistochrone. Fermat and the Bernoullis