Note. In each example, the illustration shows the space curve. The part of the curve highlighted in red is the part whose length you are asked to find.
Find the length of $\definecolor{answer}{RGB}{0,0,200}\mathbf{r}(t) = \langle \cos t, \sin t, t \rangle$ from $t=-2$ to $t=2$.
Find the length of $\mathbf{v}(t) = \langle 12t,8t^{3/2},3t^2 \rangle$ from $t=0$ to $t=1$.
Find the length of $\mathbf{r}(\theta) = \cos^3\theta\,\mathbf{i} + \sin^3\theta\,\mathbf{j}$ from $\theta=0$ to $\theta=2\pi$.
Find the length of $\mathbf{r}(u) = \langle 1,-u^2,u^3 \rangle$ from $u=0$ to $u=1$.

### Solution

\begin{align*} L &= \int_{-2}^2 \sqrt{\left(\frac{d}{dt}\cos t\right)^2 + \left(\frac{d}{dt}\sin t\right)^2 + \left(\frac{d}{dt}t\right)^2}\,dt \\ &= \int_{-2}^2 \sqrt{(-\sin t)^2 + (\cos t)^2 + (1)^2}\,dt \\ &= \int_{-2}^2 \sqrt{\sin^2 t + \cos^2 t + 1}\,dt = \int_{-2}^2 \sqrt{1 + 1}\,dt = \int_{-2}^2 \sqrt{2}\,dt \\ &= \sqrt{2}\,t\bigg]_{-2}^2 = \color{answer}\mathbf{4\sqrt{2}} \end{align*} \begin{align*} L &= \int_0^1 \left|\mathbf{v}'(t)\right|\,dt \\ &= \int_0^1 \sqrt{(12)^2 + (12t^{1/2})^2 + (6t)^2}\,dt \\ &= \int_0^1 \sqrt{144 + 144t + 36t^2}\,dt = 6\int_0^1 \sqrt{4 + 4t + t^2}\,dt \\ &= 6\int_0^1\sqrt{(2+t)^2}\,dt = 6\int_0^1(2+t)\,dt \\ &= 6\left(2t+\tfrac{1}{2}t^2\right)\bigg]_0^1 = \color{answer}\mathbf{15} \end{align*} Note: the curve is only piecewise smooth. Thus we cannot simply integrate from $0$ to $2\pi$. Instead we integrate from $0$ up to the first corner, which gives us one-quarter of the arc length, and then multiply by $4$. \begin{align*} L &= 4\int_0^{\pi/2}\left|\mathbf{r}'(\theta)\right|\,d\theta \\ &= 4\int_0^{\pi/2}\sqrt{\left(\frac{d}{d\theta}\cos^3\theta\right)^2 + \left(\frac{d}{d\theta}\sin^3\theta\right)^2 + 0^2}\,d\theta \\ &= 4\int_0^{\pi/2}\sqrt{\left(-3\cos^2\theta\sin \theta\right)^2 + \left(3\sin^2\theta\cos \theta\right)^2}\,d\theta = 12\int_0^{\pi/2}\sin \theta\cos \theta\sqrt{\cos^2\theta + \sin^2\theta}\,d\theta \\ &= 6\int_0^{\pi/2}\sin(2\theta)\,d\theta = -3\cos(2\theta)\bigg]_0^{\pi/2} = \color{answer}\mathbf{6} \end{align*} \begin{align*} L&= \int_0^1 \left|\mathbf{r}'(u)\right|\,du = \int_0^1 \sqrt{\left(\frac{d}{du}1\right)^2 + \left(\frac{d}{du}(-u^2)\right)^2 + \left(\frac{d}{du}u^3\right)^2}\,du \\ &= \int_0^1 \sqrt{0 + 4u^2 + 9u^4}\,du = \int_0^1 u\sqrt{4 + 9u^2}\,du \\ &= \frac{1}{18}\left(\frac{2}{3}(4+9u^2)^{3/2}\right)\bigg]_0^1 = \color{answer}\mathbf{\frac{1}{27}\left(13\sqrt{13} - 8\right)} \end{align*}