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

May 9, 2017

$\text{Arc Length } \approx .863746$

#### Explanation:

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

Differentiate using the chain rule:
$\frac{\mathrm{dx}}{\mathrm{dt}} = \frac{1}{2} {\left(t + 2\right)}^{- \frac{1}{2}} = \frac{1}{2 \sqrt{t + 2}}$

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

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

Simplify a bit:
$= {\int}_{1}^{3} \left(\sqrt{\frac{1}{4 \left(t + 2\right)} + \frac{1}{t} ^ 4}\right) \mathrm{dt}$

$= {\int}_{1}^{3} \left(\sqrt{\frac{{t}^{4} + 4 t + 8}{4 {t}^{4} \left(t + 2\right)}}\right) \mathrm{dt}$

Plug this into a calculator to get

$\text{Arc Length } \approx .863746$