# What is the arc length of the curve given by r(t)= (-t,2t^2,-t^3) on  t in [3,4]?

Aug 24, 2016

see below

#### Explanation:

I'm assuming that the $t$ is indeed time, so conveniently for arc length we have

$s = {\int}_{3}^{4} \dot{s} \setminus \mathrm{dt}$

$= {\int}_{3}^{4} \left\mid \vec{v} \right\mid \setminus \mathrm{dt}$

$= {\int}_{3}^{4} \sqrt{\vec{v} \cdot \vec{v}} \setminus \mathrm{dt}$

with $\vec{v} = \frac{d \vec{r}}{\mathrm{dt}} = \left(\begin{matrix}- 1 \\ 4 t \\ - 3 {t}^{2}\end{matrix}\right)$

$\implies {\int}_{3}^{4} \sqrt{1 + 16 {t}^{2} + 9 {t}^{4}} \setminus \mathrm{dt} \approx 39.6 \text{ units}$

the actual integration was done by computer.