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Featured Proof
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
Descartes approximated the cycloid by substituting a polygon for the generating circle.
For example, a hexagon (where $n = 6$) produces $5$ arches (that is, $n - 1$).
Each arch is generated by turning through the external angle of the polygon.
For the hexagon, this is $\dfrac {2 \pi} 6$.
The arches have arms of different lengths.
The first is simply the side of the hexagon.
The arm of the second arch is the chord which forms a triangle with two adjacent sides of the hexagon.
The arm of the third arch is the chord which forms a trapezium with three adjacent sides of the hexagon, etc.
We can obtain the lengths of those sides by considering the angles that subtend them.
The central angle subtending a single side of the hexagon is:
| \(\ds \theta\) | \(=\) | \(\ds \frac {2 \pi} 6\) | central angle for hexagon | ||||||||||||
| \(\ds \theta\) | \(=\) | \(\ds k \frac {2 \pi} 6\) | angle subtending k sides | ||||||||||||
| \(\ds c\) | \(=\) | \(\ds 2 a \map \sin {k \times \frac 1 2 \times \frac {2 \pi} 6}\) | chord for hexagon inscribed in circle of radius a, subtending k sides and Inscribed Angle Theorem | ||||||||||||
| \(\ds l\) | \(=\) | \(\ds \frac {2 \pi} 6 \cdot 2 a \map \sin {\frac {k \pi} 6}\) | Arc Length of Sector | ||||||||||||
| \(\ds L\) | \(=\) | \(\ds \sum_{k \mathop = 0}^6 \frac {2 \pi} 6 \cdot 2 a \map \sin {\frac {k \pi} 6}\) | total arc, summed over $k = n - 1$ arches |
Let the sum range from $0$ to $6$ (the two extra terms contribute nothing).
In addition, everywhere $6$ appears we can substitute $n$ for the polygon of $n$ sides.
| \(\ds L\) | \(=\) | \(\ds \sum_{k \mathop = 0}^n \frac {2 \pi} n \cdot 2 a \cdot \map \sin {\frac {k \pi} n }\) | |||||||||||||
| \(\ds \) | \(=\) | \(\ds 4 a \frac \pi n \sum_{k \mathop = 0}^n \map \sin {k \theta}\) |
where $\theta = \dfrac \pi n$.
From Lagrange's Sine Identity, we have:
| \(\ds \sum_{k \mathop = 0}^n \map \sin {k \theta}\) | \(=\) | \(\ds \frac {\map \cos {\frac 1 2 \theta} - \map \cos {n \theta + \frac 1 2 \theta} } {2 \map \sin {\frac 1 2 \theta} }\) | |||||||||||||
| \(\ds \leadsto \ \ \) | \(\ds \sum_{k \mathop = 0}^n \map \sin {\frac {k \pi} n}\) | \(=\) | \(\ds \frac {\map \cos {\frac \pi {2 n} } - \map \cos {\pi + \frac \pi {2 n} } } {2 \map \sin {\frac \pi {2 n} } }\) | $\theta = \dfrac \pi n$ |
Now let $n \rightarrow \infty$.
| \(\ds \sum_{k \mathop = 0}^\infty \map \sin {\frac {k \pi} n}\) | \(=\) | \(\ds \lim_{n \mathop \to \infty} \frac {\map \cos {\frac \pi {2 n} } - \map \cos {\pi + \frac \pi {2 n} } } {2 \map \sin {\frac \pi {2 n} } }\) | |||||||||||||
| \(\ds \) | \(=\) | \(\ds \frac {\map \cos 0 + \map \cos 0} {2 \map \sin {\frac \pi {2 n} } }\) | Cosine of Angle plus Straight Angle | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \frac 2 {2 \times \frac \pi {2 n } }\) | Cosine of $0 \degrees$ and Limit of $\dfrac {\sin x} x$ at $0$ | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \frac {2 n} \pi\) |
Therefore, the expression for the total arc length $L$ becomes:
| \(\ds L\) | \(=\) | \(\ds 4a \cdot \dfrac \pi n \dfrac {2n} \pi = 8a\) |
$\blacksquare$
