# Find arc length given x=t\sint, y=t\cost and 0\let\le1?

## I got up to $\setminus {\int}_{0}^{1} \setminus \sqrt{1 + {t}^{2}} \mathrm{dt}$ using the parametric arc length formula \color(indianred)(\int_a^b\sqrt((\dotx)^2+(\doty)^2)dt, but a solution I found online changes the bounds from $\setminus {\int}_{0}^{1}$ to \color(palevioletred)(\int_0^(\tan^-1(1))... I assume the 0 is because $\setminus {\tan}^{-} 1 \left(0\right) = 0$, but why arctangent? Is it because $t = \setminus \tan \setminus \theta$ from trigonometric substitution?

May 28, 2018

$L = \frac{1}{2} \left(\sqrt{2} + \ln \left(1 + \sqrt{2}\right)\right)$ units.

#### Explanation:

$x = t \sin t$
$x ' = \sin t + t \cos t$

$y = t \cos t$
$y ' = \cos t - t \sin t$

Arc length is given by:

$L = {\int}_{0}^{1} \sqrt{{\left(\sin t + t \cos t\right)}^{2} + {\left(\cos t - t \sin t\right)}^{2}} \mathrm{dt}$

Expand and simplify:

$L = {\int}_{0}^{1} \sqrt{1 + {t}^{2}} \mathrm{dt}$

Apply the substitution $t = \tan \theta$:

$L = {\int}_{0}^{{\tan}^{- 1} \left(1\right)} {\sec}^{3} \theta d \theta$

This is a known integral. If you do not have it memorized look it up in a table of integrals or apply integration by parts:

$L = \frac{1}{2} {\left[\sec \theta \tan \theta + \ln | \sec \theta + \tan \theta |\right]}_{0}^{{\tan}^{- 1} \left(1\right)}$

Insert the limits of integration:

$L = \frac{1}{2} \left(\sqrt{2} + \ln \left(1 + \sqrt{2}\right)\right)$