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

##### 1 Answer
May 13, 2017

$\approx 1.64833$

#### Explanation:

The arc length for a parametric function is:
$L = {\int}_{a}^{b} \left(\sqrt{{\left(\frac{\mathrm{dx}}{\mathrm{dt}}\right)}^{2} + {\left(\frac{\mathrm{dy}}{\mathrm{dt}}\right)}^{2}}\right) \mathrm{dt}$

In order to plug in values into this equation, we need to find $\frac{\mathrm{dx}}{\mathrm{dt}}$ and $\frac{\mathrm{dy}}{\mathrm{dt}}$ by differentiating the given function $\left(x \left(t\right) , y \left(t\right)\right)$:

$x \left(t\right) = {\left(3 t - 2\right)}^{\frac{1}{2}}$

$\frac{\mathrm{dx}}{\mathrm{dt}} = \frac{1}{2} {\left(3 t - 2\right)}^{- \frac{1}{2}} \left(3\right) = \frac{3}{2} {\left(3 t - 2\right)}^{- \frac{1}{2}} = \frac{3}{2 \sqrt{3 t - 2}}$

$y \left(t\right) = {\left(t + 3\right)}^{- \frac{1}{2}}$

$\frac{\mathrm{dy}}{\mathrm{dt}} = \left(- \frac{1}{2}\right) {\left(t + 3\right)}^{- \frac{3}{2}} = \frac{- 1}{2 {\left(t + 3\right)}^{\frac{3}{2}}}$

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

$= {\int}_{1}^{3} \left(\sqrt{\frac{9}{4 \left(3 t - 2\right)} + \frac{1}{4 {\left(t + 3\right)}^{3}}}\right) \mathrm{dt}$

$\approx 1.64833$