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

May 31, 2016

Approximately 68.8711...

#### Explanation:

Arclength is defined by the integral

${\int}_{a}^{b} \sqrt{{\left(\frac{\mathrm{dx}}{\mathrm{dt}}\right)}^{2} + {\left(\frac{\mathrm{dy}}{\mathrm{dt}}\right)}^{2}} \mathrm{dt}$

In this case, $x = t - {t}^{3}$, $y = t + 1$, $a = - 2$, and $b = 4$. Plugging these values into the formula gives

${\int}_{-} {2}^{4} \sqrt{{\left(1 - 3 {t}^{2}\right)}^{2} + {\left(1\right)}^{2}} \mathrm{dt}$

Now simplifying gives

${\int}_{-} {2}^{4} \sqrt{9 {t}^{4} - 6 {t}^{2} + 2} \mathrm{dt} \approx 68.8711 \ldots$