Find the arc length of one arch of the cycloid x = a(t

Find the arc length of one arch of the cycloid $x = a(t - \sin t)$, $y = a(1 - \cos t)$.

We compute the arc length of one full arch, $t\in[0,2\pi]$, using the same half-angle trick as the cardioid.

Step 1: Compute derivatives

$\frac{dx}{dt}=a(1-\cos t)$, $\frac{dy}{dt}=a\sin t$.

Step 2: Simplify the integrand

\[\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2 = a^2\left[(1-\cos t)^2+\sin^2t\right] = a^2[2-2\cos t] = 2a^2(1-\cos t)\]

Step 3: Use the half-angle identity

Since $1-\cos t=2\sin^2\frac t2$ and $\sin\frac t2\ge0$ on $[0,2\pi]$:

\[\sqrt{2a^2(1-\cos t)} = 2a\sin\frac t2\]

Step 4: Integrate

With $u=\frac t2$: $L=\int_0^{2\pi}2a\sin\frac t2\,dt = 4a\int_0^\pi\sin u\,du = 4a\cdot2 = 8a$.

\[\boxed{L = 8a}\]

Step 5: Geometric interpretation

One arch spans a horizontal base of $2\pi a$ (the rolling circle's circumference); the arc length $8a$ is exactly four times the circle's radius, and equals the diameter of the rolling circle times $4$.

Summary

\[\boxed{L = 8a}\]

Key insight: The cycloid's arc length ($8a$) matches the cardioid's ($8a$) exactly, both reducing to the same half-angle-identity integral — a coincidence that reflects a deeper kinship between rolling-circle curves and cusped polar curves. This was among the first successful arc-length computations in the early history of calculus (Christopher Wren, 1658).