# What is the arclength of (t/(1+t)^2,-1/t) on t in [2,4]?

Jun 4, 2018

$\approx 0.258144$

#### Explanation:

We have
$x \left(t\right) = \frac{t}{1 + t} ^ 2$
so
$x ' \left(t\right) = {\left(t - 1\right)}^{2} / {\left(1 + t\right)}^{2}$

$y \left(t\right) = - \frac{1}{t}$
so
$y ' \left(t\right) = \frac{1}{t} ^ 4$
and we get the integral

${\int}_{2}^{4} \sqrt{\frac{1}{t} ^ 4 + {\left(t - 1\right)}^{2} / {\left(1 + t\right)}^{6}} \mathrm{dt}$
by a numerical method we get

$\approx 0.258144$