Arc Length with Vector Functions — Question 2

Question 2 – Solution

Strategy. Differentiate and look for constant speed.

Step 1: Speed $$\mathbf r'(t)=\left\langle -5\sin t,5\cos t,12\right\rangle.$$ Hence $$\begin{align*} \|\mathbf r'(t)\| &=\sqrt{25\sin^2t+25\cos^2t+144}\\ &=\sqrt{25+144}=13. \end{align*}$$

Step 2: Total length $$L=\int_0^{3\pi}13\,dt =13[t]_0^{3\pi} =\boxed{39\pi}.$$

Step 3: One revolution A complete revolution changes $t$ by $2\pi$. Therefore $$L_{\mathrm{rev}}=\int_t^{t+2\pi}13\,du=\boxed{26\pi}.$$ Because the speed is the same constant $13$ at every parameter value, an interval of width $\Delta t$ always contributes length $13\Delta t$. Thus equal parameter intervals have equal arc lengths.

Verification. The interval $[0,3\pi]$ contains $3/2$ revolutions, and $(3/2)(26\pi)=39\pi$, confirming the total.