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

May 30, 2018

$\approx 0.431873$

#### Explanation:

We have
$x \left(t\right) = \sqrt{t}$
then
$x ' \left(t\right) = \frac{1}{2 \cdot \sqrt{t}}$
$y \left(t\right) = \frac{1}{\sqrt{{t}^{2} + 3}}$

then
y'(t)=-t/sqrt(t^2+3)^3
So we get the integral
${\int}_{1}^{2} \sqrt{\frac{1}{4 \cdot t} + {t}^{2} / {\left(3 + {t}^{2}\right)}^{3}} \mathrm{dt}$
I have found only a numerical value:

$\approx 0.431873$