# What is the arclength of f(t) = (t+sqrt(lnt),t-sqrtlnt)) on t in [1,e]?

Jan 2, 2018

The arc length between $t = a$ and $t = b$ on the parametric curve $y = f \left(t\right) , x = g \left(t\right)$ is given by
${\int}_{a}^{b} \sqrt{{\left(\frac{\mathrm{dy}}{\mathrm{dt}}\right)}^{2} + {\left(\frac{\mathrm{dx}}{\mathrm{dt}}\right)}^{2}} \setminus \mathrm{dt}$

So, from the question, we have $t$ from $1$ to $e$ and $y = t - \sqrt{\ln \left(t\right)} , x = t + \sqrt{\ln \left(t\right)}$. Then, $\frac{\mathrm{dy}}{\mathrm{dt}} = 1 - \frac{1}{2 t \sqrt{\ln \left(t\right)}}$ and $\frac{\mathrm{dx}}{\mathrm{dt}} = 1 + \frac{1}{2 t \sqrt{\ln \left(t\right)}}$.

Thus, we need to find
${\int}_{1}^{e} \sqrt{{\left(1 - \frac{1}{2 t \sqrt{\ln \left(t\right)}}\right)}^{2} + {\left(1 + \frac{1}{2 t \sqrt{\ln \left(t\right)}}\right)}^{2}} \setminus \mathrm{dx}$
$= {\int}_{1}^{e} \sqrt{2 + \frac{1}{2 {t}^{2} \ln \left(t\right)}} \setminus \mathrm{dx}$

Using numerical approximation methods, we find that the integral is approximately equal to $3$.