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

Not a full solution, but...

#### Explanation:

This function is a parametric curve $f \left(t\right) = \left(x \left(t\right) , y \left(t\right)\right)$, where
$x \left(t\right) = \ln \frac{t}{t}$ and $y \left(t\right) = \ln \left(t + 2\right)$.

To measure its length we sould first use Pythagorean theorem to find an element of lenght $\mathrm{ds}$ (a curve is approximated with a series of small, infinitesimal segments)
$\mathrm{ds} = \sqrt{{\mathrm{dx}}^{2} + {\mathrm{dy}}^{2}}$

Now because $x$ and $y$ are functions of $t$ we use chain rule
$\mathrm{ds} = \sqrt{{\left(\frac{\mathrm{dx}}{\mathrm{dt}}\right)}^{2} {\mathrm{dt}}^{2} + {\left(\frac{\mathrm{dy}}{\mathrm{dt}}\right)}^{2} {\mathrm{dt}}^{2}} = \sqrt{{\left(\frac{\mathrm{dx}}{\mathrm{dt}}\right)}^{2} + {\left(\frac{\mathrm{dy}}{\mathrm{dt}}\right)}^{2}} \mathrm{dt}$

The total lenght for $t \in \left[1 , e\right]$ is an integral

$s = {\int}_{\text{curve}} \mathrm{ds} = {\int}_{1}^{e} \sqrt{{\left(\frac{\mathrm{dx}}{\mathrm{dt}}\right)}^{2} + {\left(\frac{\mathrm{dy}}{\mathrm{dt}}\right)}^{2}} \mathrm{dt}$

Let's find $\frac{\mathrm{dx}}{\mathrm{dt}}$ and $\frac{\mathrm{dy}}{\mathrm{dt}}$ first.

$x \left(t\right) = \ln t \cdot \frac{1}{t}$

By product rule
$\frac{\mathrm{dx}}{\mathrm{dt}} = \frac{1}{t} \cdot \frac{1}{t} + \ln t \cdot \frac{1}{t} ^ 2 = \frac{\ln t + 1}{t} ^ 2$

$y \left(t\right) = \ln \left(t + 2\right) t$

By chain rule
$\frac{\mathrm{dy}}{\mathrm{dt}} = \frac{1}{t + 2} \cdot 1$

By plugging this in, we get

$s = {\int}_{1}^{e} \sqrt{{\left(\frac{\ln t + 1}{t} ^ 2\right)}^{2} + {\left(\frac{1}{t + 2}\right)}^{2}} \mathrm{dt}$
$s = {\int}_{1}^{e} \sqrt{{\left(\ln t + 1\right)}^{2} / {t}^{4} + \frac{1}{t + 2} ^ 2} \mathrm{dt}$
Common denominator
$s = {\int}_{1}^{e} \frac{\sqrt{{\left(t + 2\right)}^{2} {\left(\ln t + 1\right)}^{2} + {t}^{4}}}{{t}^{2} \left(t + 2\right)} \mathrm{dt}$
$s = {\int}_{1}^{e} \frac{\sqrt{{\left(t + 2\right)}^{2} {\ln}^{2} \left(e t\right) + {t}^{4}}}{{t}^{2} \left(t + 2\right)} \mathrm{dt}$