# What is the arclength of (9t^3-2t^2+15,-2t^2-12t-1) on t in [-2,3]?

Aug 28, 2016

$\approx 325$

#### Explanation:

Arc length is the time integral of speed, so $s = {\int}_{- 2}^{3} \dot{s} \mathrm{dt}$

and speed $\dot{s} = \sqrt{\vec{v} \cdot \vec{v}}$ where $\vec{v}$ is the velocity vector

$\vec{v} = \frac{d}{\mathrm{dt}} \vec{r}$

$= \frac{d}{\mathrm{dt}} \left(\begin{matrix}9 {t}^{3} - 2 {t}^{2} + 15 \\ - 2 {t}^{2} - 12 t - 1\end{matrix}\right)$

$= \left(\begin{matrix}27 {t}^{2} - 4 t \\ - 4 t - 12\end{matrix}\right)$

So $s = {\int}_{- 2}^{3} \sqrt{\vec{v} \cdot \vec{v}} \mathrm{dt}$

$= {\int}_{- 2}^{3} \sqrt{\left(\begin{matrix}27 {t}^{2} - 4 t \\ - 4 t - 12\end{matrix}\right) \cdot \left(\begin{matrix}27 {t}^{2} - 4 t \\ - 4 t - 12\end{matrix}\right)} \mathrm{dt}$

$= {\int}_{- 2}^{3} \sqrt{{\left(27 {t}^{2} - 4 t\right)}^{2} + {\left(4 t + 12\right)}^{2}} \mathrm{dt}$

$\approx 325$ units according to computer solution