Arc Length with Vector Functions — Question 8

Question 8 – Solution

Strategy. Factor $t^2$ from the squared speed and use $t\ge 0$ before integrating.

Step 1: Speed $$\mathbf r'(t)=\left\langle 2t,3t^2,0\right\rangle.$$ Hence $$\begin{align*} \|\mathbf r'(t)\| &=\sqrt{4t^2+9t^4}\\ &=\sqrt{t^2(4+9t^2)}\\ &=|t|\sqrt{4+9t^2}. \end{align*}$$ On $[0,2]$, $|t|=t$, so the speed is $t\sqrt{4+9t^2}$.

Step 2: Integrate Let $u=4+9t^2$, so $du=18t\,dt$: $$\begin{align*} L&=\int_0^2t\sqrt{4+9t^2}\,dt\\ &=\frac 1{18}\int_4^{40}u^{1/2}\,du\\ &=\frac 1{18}\left[\frac 23u^{3/2}\right]_4^{40}\\ &=\frac 1{27}(40^{3/2}-4^{3/2})\\ &=\boxed{\frac{80\sqrt{10}-8}{27}}. \end{align*}$$

Step 3: Endpoint regularity Although $\mathbf r'(0)=\mathbf 0$, the speed $t\sqrt{4+9t^2}$ is continuous at $0$ and equals zero there. A continuous integrand on a closed interval defines an ordinary proper integral; no singularity is present.