# What is the arclength of (t-2t^3,4t+1) on t in [-2,4]?

Jul 19, 2017

Approximate arc length via a numerical method is:

$146.64$ (2dp)

#### Explanation:

The arc length of a curve:

 vec(r) (t) = << f(t), g(t)) >>

Over an interval $\left[a , b\right]$ is given by:

$L = {\int}_{a}^{b} \setminus | | \vec{r} \left(t\right) | | \setminus \mathrm{dt}$
$\setminus \setminus = {\int}_{a}^{b} \setminus \sqrt{f ' {\left(t\right)}^{2} + g ' {\left(t\right)}^{2}} \setminus \mathrm{dt}$

So, for the given curve:

$\vec{r} \left(t\right) = \left(t - 2 {t}^{3} , 4 t + 1\right) \setminus \setminus \setminus t \in \left[- 2 , 4\right]$

the arc length is given by:

$L = {\int}_{- 2}^{4} \setminus \sqrt{{\left(\frac{d}{\mathrm{dt}} \left(t - 2 {t}^{3}\right)\right)}^{2} + {\left(\frac{d}{\mathrm{dt}} \left(4 t + 1\right)\right)}^{2}} \setminus \mathrm{dt}$
$\setminus \setminus = {\int}_{- 2}^{4} \setminus \sqrt{{\left(1 - 6 {t}^{2}\right)}^{2} + {\left(4\right)}^{2}} \setminus \mathrm{dt}$
$\setminus \setminus = {\int}_{- 2}^{4} \setminus \sqrt{1 - 12 {t}^{2} + 36 {t}^{4} + 16} \setminus \mathrm{dt}$
$\setminus \setminus = {\int}_{- 2}^{4} \setminus \sqrt{36 {t}^{4} - 12 {t}^{2} + 17} \setminus \mathrm{dt}$

The integral does not have an elementary antiderivative,and so we evaluate the definite integral al numerically:

$L = 146.64637228 \ldots$