# Find the arc length for the following question?

May 7, 2018

$2$

#### Explanation:

For $x = x \left(y\right)$, arc length is $s = {\int}_{{y}_{1}}^{{y}_{2}} \setminus \sqrt{1 + {\left(x '\right)}^{2}} \setminus \mathrm{dy}$

From Fundamental Theorem of Calculus:

• $x \left(y\right) = {\int}_{1}^{y} \setminus \sqrt{{\sec}^{4} t - 1} \setminus \mathrm{dt} \implies x ' = \sqrt{{\sec}^{4} y - 1}$

So:

$s = {\int}_{- \frac{\pi}{4}}^{\frac{\pi}{4}} \setminus \sqrt{1 + {\left(\sqrt{{\sec}^{4} y - 1}\right)}^{2}} \setminus \mathrm{dy}$

$= {\int}_{- \frac{\pi}{4}}^{\frac{\pi}{4}} \setminus {\sec}^{2} y \setminus \mathrm{dy}$

$= {\int}_{- \frac{\pi}{4}}^{\frac{\pi}{4}} \setminus {\left(\tan y\right)}^{'} \setminus \mathrm{dy}$

$= {\left(\setminus \tan y\right)}_{- \frac{\pi}{4}}^{\frac{\pi}{4}} = 2$