Arc Length with Parametric Equations — Question 2

Question 2 – Solution

Solution

1. Differentiate the parametric equations: $$x'(t)=1-\cos t, \qquad y'(t)=\sin t.$$

2. Compute and simplify the squared speed: $$\begin{aligned} x'^2+y'^2 &=(1-\cos t)^2+\sin^2t\\ &=1-2\cos t+\cos^2t+\sin^2t\\ &=2-2\cos t =4\sin^2\left(\frac t2\right). \end{aligned}$$ Hence the speed is $$2\left|\sin\left(\frac t2\right)\right|.$$

3. On $0\le t\le2\pi$, we have $0\le t/2\le\pi$, so $\sin(t/2)\ge0$. Thus $$L=2\int_0^{2\pi}\sin\left(\frac t2\right)dt.$$

4. Evaluate the integral: $$L=2\left[-2\cos\left(\frac t2\right)\right]_0^{2\pi} =-4(\cos\pi-\cos0) =\boxed{8}.$$