Arc Length with Vector Functions — Question 6

Question 6 – Solution

Strategy. Arc length is additive across a corner, even though the derivative changes abruptly there.

Step 1: Join check Both formulas give $$\left\langle 0,2,0\right\rangle$$ at $t=0$ (the second by its right-hand limit), so the path is continuous at $C$.

Step 2: Length of each segment On the first branch, $$\mathbf r'(t)=\left\langle 1,1,0\right\rangle,\qquad \|\mathbf r'(t)\|=\sqrt 2.$$ On the second branch, $$\mathbf r'(t)=\left\langle 1,-1,0\right\rangle,\qquad \|\mathbf r'(t)\|=\sqrt 2.$$ Therefore $$\begin{align*} L&=\int_{-2}^{0}\sqrt 2\,dt+\int_0^2\sqrt 2\,dt\\ &=2\sqrt 2+2\sqrt 2=\boxed{4\sqrt 2}. \end{align*}$$

Step 3: Chord comparison The endpoint distance is $$\|B-A\|=\|\left\langle 4,0,0\right\rangle\|=4.$$ Since $4\sqrt 2>4$, the broken path is longer. Equality would require travel directly along the segment from $A$ to $B$ without the detour through $C$.