# What is the arc length of f(t)=(1/sqrt(t+t^2),2-8t)  over t in [1,3]?

May 28, 2018

$\approx 16.0071$

#### Explanation:

We hat
$x \left(t\right) = {\left({t}^{2} + t\right)}^{- \frac{1}{2}}$
$x ' \left(t\right) = - \frac{1}{2} \cdot \frac{2 \cdot t + 1}{{t}^{2} + t} ^ \left(\frac{3}{2}\right)$
$y \left(t\right) = 2 - 8 t$
$y ' \left(t\right) = - 8$
So we have to solve
${\int}_{1}^{3} \sqrt{64 + {\left(- \frac{1}{2} \cdot \frac{2 \cdot t + 1}{{t}^{2} + t} ^ \left(\frac{3}{2}\right)\right)}^{2}} \mathrm{dt}$
I have got only a numerical value.