Arc Length with Vector Functions — Question 5

Question 5 – Solution

Strategy. Calculate both speed integrals, taking care that $u\ge 0$ when simplifying an absolute value.

Step 1: Standard parametrization $$\mathbf r'(t)=\left\langle -2\sin t,2\cos t,0\right\rangle,\qquad \|\mathbf r'(t)\|=2.$$ Hence $$L_r=\int_0^\pi 2\,dt=\boxed{2\pi}.$$

Step 2: Cubic reparametrization $$\mathbf R'(u)=\left\langle -6u^2\sin(u^3),6u^2\cos(u^3),0\right\rangle.$$ Because $u\ge 0$, $$\|\mathbf R'(u)\|=6u^2.$$ Therefore $$L_R=\int_0^{\sqrt[3]\pi}6u^2\,du =\left[2u^3\right]_0^{\sqrt[3]\pi} =\boxed{2\pi}.$$

Step 3: Invariance The substitution $t=u^3$ gives $dt=3u^2du$, so $$2\,dt=6u^2du.$$ The second parametrization changes the rate at which points are visited, but it is increasing and covers the same semicircle once. The speed factor exactly compensates for the parameter distortion.