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

May 27, 2018

$0.326586$

#### Explanation:

We have
$x \left(t\right) = \frac{t}{\sqrt{t - 1}}$
x'(t)=1/sqrt(t-1)-t/(2*(t-1)^(3/2)
$y \left(t\right) = \frac{t}{{t}^{2} - 1}$
then
$y ' \left(t\right) = \frac{1}{{t}^{2} - 1} - 2 \cdot {t}^{2} / {\left({t}^{2} - 1\right)}^{2}$
and our integral will be
${\int}_{2}^{3} \sqrt{{\left(t - 2\right)}^{2} / \left(4 \cdot {\left(t - 1\right)}^{3}\right) + {\left({t}^{2} + 1\right)}^{2} / \left({\left(t - 1\right)}^{4} \cdot {\left(t + 1\right)}^{4}\right)} \mathrm{dt}$
which i can only solve by a numerical method.