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

Aug 3, 2017

Approximate arc length via a numerical method is:

$0.79$ (2dp)

#### Explanation:

The arc length of a curve:

$\vec{r} \left(t\right) = \left\langlex \left(t\right) , y \left(t\right)\right\rangle$

Over an interval $\left[a , b\right]$ is given by:

$L = {\int}_{a}^{b} \setminus | | \vec{r} \left(t\right) | | \setminus \mathrm{dt}$
$\setminus \setminus = {\int}_{a}^{b} \setminus \sqrt{x ' {\left(t\right)}^{2} + y ' {\left(t\right)}^{2}} \setminus \mathrm{dt}$

So, for the given curve:

$\vec{r} \left(t\right) = \left\langle\ln \frac{t}{t} , \ln t\right\rangle \setminus \setminus \setminus t \in \left[1 , 2\right]$

Differentiating the components wrt $t$ we get:

$x ' \left(t\right) = \frac{\left(t\right) \left(\frac{d}{\mathrm{dt}} \ln t\right) - \left(\ln t\right) \left(\frac{d}{\mathrm{dt}} t\right)}{{t}^{2}}$
$\text{ } = \frac{\left(t\right) \left(\frac{1}{t}\right) - \left(\ln t\right) \left(1\right)}{{t}^{2}}$
$\text{ } = \frac{1 - \ln t}{t} ^ 2$

$y ' \left(t\right) = \frac{1}{t}$

So, the arc length is given by:

$L = {\int}_{1}^{2} \setminus \sqrt{x ' {\left(t\right)}^{2} + y ' {\left(t\right)}^{2}} \setminus \mathrm{dt}$
$\setminus \setminus = {\int}_{1}^{2} \setminus \sqrt{{\left(\frac{1 - \ln t}{t} ^ 2\right)}^{2} + {\left(\frac{1}{t}\right)}^{2}} \setminus \mathrm{dt}$
$\setminus \setminus = {\int}_{1}^{2} \setminus \sqrt{{\left(\frac{1 - \ln t}{t} ^ 2\right)}^{2} + \frac{1}{t} ^ 2} \setminus \mathrm{dt}$

The integral does not have an elementary antiderivative,and so we evaluate the definite integral al numerically:

$L = 0.7890476183035367 \ldots$