Arc Length with Vector Functions — Question 5

Question 5

The upper semicircle $x^2+y^2=4$ is represented by $$\mathbf r(t)=\left\langle 2\cos t,2\sin t,0\right\rangle,\quad 0\le t\le\pi,$$ and also by $$\mathbf R(u)=\left\langle 2\cos(u^3),2\sin(u^3),0\right\rangle,\quad 0\le u\le\sqrt[3]{\pi}.$$ Tasks

1. Compute the length using each parametrization.

2. Explain why the nonconstant speed of $\mathbf R$ does not change the result.

3. Identify the substitution connecting the two integrals.

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.