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

May 1, 2017

$\text{Arclength } \approx 10.601267$

#### Explanation:

$f \left(t\right) = \left({t}^{2} \sqrt{t - 1} , {t}^{2} + t - 1\right)$

Formula for parametric arclength: $\textcolor{b l u e}{L = {\int}_{a}^{b} \sqrt{{\left(\frac{\mathrm{dx}}{\mathrm{dt}}\right)}^{2} + {\left(\frac{\mathrm{dy}}{\mathrm{dt}}\right)}^{2}} \text{ } \mathrm{dt}}$

First, find $f ' \left(t\right)$ by differentiating $f \left(t\right)$

$\frac{\mathrm{dx}}{\mathrm{dt}} = \left({t}^{2}\right) \left(\frac{1}{2} {\left(t - 1\right)}^{- \frac{1}{2}}\right) + \left(2 t\right) {\left(t - 1\right)}^{\frac{1}{2}}$

$\frac{\mathrm{dx}}{\mathrm{dt}} = \frac{{t}^{2}}{2 \sqrt{t - 1}} + 2 t \sqrt{t - 1} = \frac{{t}^{2} + 4 t \left(t - 1\right)}{2 \sqrt{t - 1}}$

$\frac{\mathrm{dy}}{\mathrm{dt}} = 2 t + 1$

Plug in expressions for $\frac{\mathrm{dx}}{\mathrm{dt}}$ and $\frac{\mathrm{dy}}{\mathrm{dt}}$ into the arclength formula:
$L = {\int}_{2}^{3} \sqrt{{\left(\frac{{t}^{2} + 4 t \left(t - 1\right)}{2 \sqrt{t - 1}}\right)}^{2} + {\left(2 t + 1\right)}^{2}} \text{ } \mathrm{dt}$

Use a calculator to evaluate:
$\text{Arclength } \approx 10.601267$