# How do you find the length of the curve x=3t-t^3, y=3t^2, where 0<=t<=sqrt(3) ?

Aug 15, 2014

$6 \sqrt{3}$.

#### Explanation:

The answer is $6 \sqrt{3}$.

The arclength of a parametric curve can be found using the formula: $L = {\int}_{{t}_{i}}^{{t}_{f}} \sqrt{{\left(\frac{\mathrm{dx}}{\mathrm{dt}}\right)}^{2} + {\left(\frac{\mathrm{dy}}{\mathrm{dt}}\right)}^{2}} \mathrm{dt}$. Since $x$ and $y$ are perpendicular, it's not difficult to see why this computes the arclength.

It isn't very different from the arclength of a regular function: $L = {\int}_{a}^{b} \sqrt{1 + {\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}^{2}} \mathrm{dx}$. If you need the derivation of the parametric formula, please ask it as a separate question.

We find the 2 derivatives:
$\frac{\mathrm{dx}}{\mathrm{dt}} = 3 - 3 {t}^{2}$
$\frac{\mathrm{dy}}{\mathrm{dt}} = 6 t$

And we substitute these into the integral:
$L = {\int}_{0}^{\sqrt{3}} \sqrt{{\left(3 - 3 {t}^{2}\right)}^{2} + {\left(6 t\right)}^{2}} \mathrm{dt}$

And solve:
$= {\int}_{0}^{\sqrt{3}} \sqrt{9 - 18 {t}^{2} + 9 {t}^{4} + 36 {t}^{2}} \mathrm{dt}$
$= {\int}_{0}^{\sqrt{3}} \sqrt{9 + 18 {t}^{2} + 9 {t}^{4}} \mathrm{dt}$
$= {\int}_{0}^{\sqrt{3}} \sqrt{{\left(3 + 3 {t}^{2}\right)}^{2}} \mathrm{dt}$
$= {\int}_{0}^{\sqrt{3}} \left(3 + 3 {t}^{2}\right) \mathrm{dt}$
$= 3 t + {t}^{3} {|}_{0}^{\sqrt{3}}$
$= 3 \sqrt{3} + 3 \sqrt{3}$
$= 6 \sqrt{3}$

Be aware that arclength usually has a difficult function to integrate. Most integrable functions look like the above where a binomial is squared and adding the two terms will flip the sign of the binomial.