How to compute the arc length of a parametric curve?

**Derivation of the formula:** The formula comes from approximating the curve by small line segments. For a small change $\Delta t$, the horizontal displacement is approximately $x'(t)\Delta t$ and the vertical displacement is approximately $y'(t)\Delta t$. By the Pythagorean theorem, the segment length is approximately: $$\Delta L \approx \sqrt{(x'(t)\Delta t)^2 + (y'(t)\Delta t)^2} = \sqrt{x'(t)^2 + y'(t)^2} \, \Delta t$$ Summing and taking the limit as $\Delta t \ o 0$ gives the integral. **Application to the Cycloid:** $x = t - \sin t$, $y = 1 - \cos t$ \begin{align*} \frac{dx}{dt} &= 1 - \cos t \\ \frac{dy}{dt} &= \sin t \end{align*} \begin{align*} \left(\frac{dx}{dt}\ ight)^2 + \left(\frac{dy}{dt}\ ight)^2 &= (1 - \cos t)^2 + \sin^2 t \\ &= 1 - 2\cos t + \cos^2 t + \sin^2 t \\ &= 2 - 2\cos t = 4\sin^2\left(\frac{t}{2}\ ight) \end{align*} Since $\sqrt{4\sin^2(t/2)} = 2|\sin(t/2)|$, and $\sin(t/2) \geq 0$ for $t \in [0, 2\pi]$: $$L = \int_0^{2\pi} 2\sin\left(\frac{t}{2}\ ight) \, dt = 2\left[-2\cos\left(\frac{t}{2}\ ight)\ ight]_0^{2\pi} = -4(\cos\pi - \cos 0) = -4(-1 - 1) = 8$$ So one arch of the cycloid has length $8$ units.