Length of Arc of Cycloid/Proof 3
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
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Descartes approximated the Cycloid by subtituting a polygon for the generating circle. For example, a hexagon ($n=6$) produces 5 arches or lobes ($n-1$).
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 polygon, 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 |
| \(\ds c\) | \(=\) | \(\ds 2a \sin k \frac {2 \pi} 6\) | chord for hexagon inscribed in circle of radius a, subtending k sides |
| \(\ds l\) | \(=\) | \(\ds \frac {2 \pi} 6 \cdot 2a \sin k \frac {2 \pi} 6\) | length of arch is angle times arm |
| \(\ds L\) | \(=\) | \(\ds \sum \frac {2 \pi} 6 \cdot 2a \sin k \frac {2 \pi} 6\) | total arc, summed over k = n-1 arches |
It is convenient to let the sum range from 0 to 6 (the two extra terms don't contribute anything).
In addition, everywhere $6$ appears we can substitute $n$ for the polygon of $n$ sides.
| \(\ds L\) | \(=\) | \(\ds \sum_{k \mathop = 0}^n \ \ \dfrac {2 \pi} n \cdot 2a \cdot \sin k \dfrac \pi n = 4a \dfrac \pi n \sum_0^n \sin k \theta\) |
where $\theta = \dfrac \pi n$. We use one of Lagrange's trigonometric identities. This can be derived from the Werner Formula for Sine by Sine
| \(\ds 2 \sin A \sin B\) | \(=\) | \(\ds \cos {A - B} - \cos {A + B}\) |
by substituting $A = \sin k \theta$ and $B = \frac 1 2$, and taking the sum of both sides, noting that the infinite sum on the right-hand side is a telescoping series. The identity we need is
| \(\ds \sum_{k \mathop = 0}^n \ \ \sin k \theta\) | \(=\) | \(\ds \frac {\cos \dfrac 1 2 \theta - \cos (n + \dfrac 1 2) \theta} {2 \sin \dfrac 1 2 \theta}\) |
Now let $n \rightarrow \infty$. Since $\theta = \dfrac \pi n$, the angle for the first term in the numerator becomes $\dfrac 1 2 \dfrac \pi n = 0$
so we have for the first term $\cos 0 = 1$.
The second term becomes $\cos 0 + \pi = -1$ so the numerator is just $2$, which cancels the $2$ in the denominator.
Use the Power Series Expansion for Sine Function (since $\pi/n \rightarrow 0$). The denominator is
$\sin \dfrac 1 2 \theta \approx \dfrac 1 2 \dfrac \pi n$
The expression for the total arc length $L$ becomes
| \(\ds L\) | \(=\) | \(\ds 4a \cdot \dfrac \pi n \dfrac {2n} \pi = 8a\) |
$\blacksquare$