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

Nov 12, 2016

In parametric form, the Arc Length is given by:

$L = {\int}_{\alpha}^{\beta} \sqrt{{\left(\frac{\mathrm{dx}}{\mathrm{dt}}\right)}^{2} + {\left(\frac{\mathrm{dy}}{\mathrm{dt}}\right)}^{2}} \mathrm{dt}$

We have:

$\left\{\begin{matrix}x \left(t\right) = 3 {t}^{2} & \implies \frac{\mathrm{dx}}{\mathrm{dt}} = 6 t \\ y \left(t\right) = {t}^{4} - t & \implies \frac{\mathrm{dy}}{\mathrm{dt}} = 4 {t}^{3} - 1\end{matrix}\right.$

So the Arc Length is given by;

$L = {\int}_{-} {4}^{1} \sqrt{{\left(6 t\right)}^{2} + {\left(4 {t}^{3} - 1\right)}^{2}} \mathrm{dt}$
$\therefore L = {\int}_{-} {4}^{1} \sqrt{36 {t}^{2} + 16 {t}^{6} - 8 {t}^{3} + 1} \mathrm{dt}$

And this is as far as we can get analytically as there is no elementary anti-derivative and we would have to resort to a numerical solution to find the value of the definite integral