# What is the arc length of the curve given by x = t^2 and y= 2t^2 = 1, for  1<t<3?

Oct 12, 2017

The arc length is $8 \sqrt{5}$

#### Explanation:

For this problem, we will use this formula:

${\int}_{1}^{3} \sqrt{{\left(\frac{\mathrm{dx}}{\mathrm{dt}}\right)}^{2} + {\left(\frac{\mathrm{dy}}{\mathrm{dt}}\right)}^{2}} \mathrm{dt}$

First, let's find the derivatives of $x$ and $y$ with respect to $t$:

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

$\frac{\mathrm{dy}}{\mathrm{dt}} = \frac{d}{\mathrm{dt}} \left(2 {t}^{2}\right) = 4 t$

Now we can plug these into the original formula:

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

${\int}_{1}^{3} \sqrt{4 {t}^{2} + 16 {t}^{2}} \mathrm{dt}$

${\int}_{1}^{3} \sqrt{20 {t}^{2}} \mathrm{dt}$

${\int}_{1}^{3} \sqrt{20} t \mathrm{dt}$

${\int}_{1}^{3} 2 \sqrt{5} t \mathrm{dt}$

$\sqrt{5} {\int}_{1}^{3} 2 t \mathrm{dt}$

$\sqrt{5} {\left[{t}^{2}\right]}_{1}^{3}$

$\sqrt{5} \left({3}^{2} - {1}^{2}\right)$

$8 \sqrt{5}$