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

Jul 9, 2018

$\approx 0.4287742274$

#### Explanation:

We have

$x \left(t\right) = \ln \left(t + 3\right)$ then

$x ' \left(t\right) = \frac{1}{t + 3}$

$y \left(t\right) = \ln \frac{t}{t}$

then

$y ' \left(t\right) = \frac{\frac{1}{t} \cdot t - \ln \left(t\right)}{t} ^ 2$

$y ' \left(t\right) = \frac{1 - \ln \left(t\right)}{t} ^ 2$
and we have to integrate
${\int}_{1}^{2} \sqrt{{\left(\frac{1}{t + 3}\right)}^{2} + {\left(\frac{1 - \ln \left(t\right)}{t} ^ 2\right)}^{2}} \mathrm{dt}$
by a numerical method we get

$\approx 0.4287742274$