# Question 1a66a

Jan 6, 2017

Arc Length = ln (sqrt(2)+1 )) = 0.8813137... 

#### Explanation:

The Arc Length for a Curve $y = f \left(x\right)$ is given by

$L = {\int}_{\alpha}^{\beta} \sqrt{1 + {\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}^{2}} \setminus \mathrm{dx} = {\int}_{\alpha}^{\beta} \sqrt{1 + {\left(f ' \left(x\right)\right)}^{2}} \setminus \mathrm{dx}$

So in this problem we have

$y = \ln \left(\sec x\right) \implies \frac{\mathrm{dy}}{\mathrm{dx}} = \tan x$

So the Arc Length is;

$L = {\int}_{0}^{\frac{\pi}{4}} \sqrt{1 + {\tan}^{2} x} \setminus \mathrm{dx}$
$\setminus \setminus \setminus = {\int}_{0}^{\frac{\pi}{4}} \sqrt{{\sec}^{2} x} \setminus \mathrm{dx}$
$\setminus \setminus \setminus = {\int}_{0}^{\frac{\pi}{4}} \sec x \setminus \mathrm{dx}$
$\setminus \setminus \setminus = {\left[\ln | \sec x + \tan x |\right]}_{0}^{\frac{\pi}{4}}$
$\setminus \setminus \setminus = \ln | \sec \left(\frac{\pi}{4}\right) + \tan \left(\frac{\pi}{4}\right) | - \ln | \sec \left(0\right) + \tan \left(0\right) |$
$\setminus \setminus \setminus = \ln | \sqrt{2} + 1 | - \ln | 1 + 0 |$
$\setminus \setminus \setminus = \ln \left(\sqrt{2} + 1\right) - \ln 1$
 \ \ \=ln (sqrt(2)+1 )) #
$\setminus \setminus \setminus = 0.8813137 \ldots$