Arc Length with Parametric Equations — Question 3

Question 3 – Solution

Solution

1. Apply the product rule to each coordinate: $$x'(t)=\cos t-t\sin t, \qquad y'(t)=\sin t+t\cos t.$$

2. Square and add: $$\begin{aligned} x'^2+y'^2 &=(\cos t-t\sin t)^2+(\sin t+t\cos t)^2\\ &=\cos^2t+\sin^2t+t^2(\sin^2t+\cos^2t)\\ &\quad-2t\sin t\cos t+2t\sin t\cos t\\ &=1+t^2. \end{aligned}$$ The cross terms cancel exactly.

3. Therefore, the speed is $\sqrt{1+t^2}$, and the exact length is $$\boxed{L=\int_0^{2\pi}\sqrt{1+t^2}\,dt}.$$

4. If an evaluated form is desired, use $$\int\sqrt{1+t^2}\,dt =\frac12\left(t\sqrt{1+t^2}+\operatorname{arsinh}t\right),$$ giving $$L=\pi\sqrt{1+4\pi^2}+\frac12\operatorname{arsinh}(2\pi).$$