# What is the arclength of f(t)=(3t^2-1, 3t^3-t) on t∈[-1/sqrt(3),1/sqrt(3)]?

Jun 23, 2018

$\frac{4 \sqrt{3}}{3}$

#### Explanation:

For the given curve, we have

$x \left(t\right) = 3 {t}^{2} - 1 , q \quad y \left(t\right) = 3 {t}^{3} - t$

Hence

$\frac{\mathrm{dx}}{\mathrm{dt}} = 6 t , q \quad \frac{\mathrm{dy}}{\mathrm{dt}} = 9 {t}^{2} - 1$

Thus, the infinitesimal arc length is

$\mathrm{ds} = \sqrt{{\left(\frac{\mathrm{dx}}{\mathrm{dt}}\right)}^{2} + {\left(\frac{\mathrm{dy}}{\mathrm{dt}}\right)}^{2}} \mathrm{dt}$
$q \quad = \sqrt{36 {t}^{2} + {\left(9 {t}^{2} - 1\right)}^{2}} \mathrm{dt}$
$q \quad = \sqrt{{\left(9 {t}^{2} + 1\right)}^{2}} \mathrm{dt}$
$q \quad = \left(9 {t}^{2} + 1\right) \mathrm{dt}$

Thus, the arc length required is

${\int}_{- \frac{1}{\sqrt{3}}}^{+ \frac{1}{\sqrt{3}}} \left(9 {t}^{2} + 1\right) \mathrm{dt} = {\left(3 {t}^{3} + t\right)}_{- \frac{1}{\sqrt{3}}}^{+ \frac{1}{\sqrt{3}}}$

$q \quad q \quad = {\left\{t \left(3 {t}^{2} + 1\right)\right\}}_{- \frac{1}{\sqrt{3}}}^{+ \frac{1}{\sqrt{3}}} = \frac{4}{\sqrt{3}}$