# Find the exact length of the curve (parametrics)?

## Question 43:

May 16, 2018

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

#### Explanation:

Let:

$\left\{\begin{matrix}x = x \left(t\right) \\ y = y \left(t\right)\end{matrix}\right.$

with ${t}_{1} \le t \le {t}_{2}$ be the equation of a curve, the length of the element of the curve is:

$\mathrm{dl} = \sqrt{{\mathrm{dx}}^{2} + {\mathrm{dy}}^{2}} = \sqrt{x ' {\left(t\right)}^{2} + y ' \left(t\right)} \mathrm{dt}$

and so the length is calculated with the integral:

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

In this case (exercise 43):

$\left\{\begin{matrix}x \left(t\right) = t \sin t \\ y \left(t\right) = t \cos t\end{matrix}\right.$

with $0 \le t \le 1$

$\left\{\begin{matrix}x ' \left(t\right) = \sin t + t \cos t \\ y ' \left(t\right) = \cos t - t \sin t\end{matrix}\right.$

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

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

L=int_0^1 sqrt( (sin^2t+cos^2t +t^2(sin^2t+cos^2t))dt

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

Substitute:

$t = \tan \theta$

$\mathrm{dt} = {\sec}^{2} \theta d \theta$

and the limits of integration become $\theta = 0$ and $\theta = \frac{\pi}{4}$:

$L = {\int}_{0}^{\frac{\pi}{4}} {\sec}^{2} \theta \sqrt{1 + {\tan}^{2} \theta} d \theta$

Using the trigonometric identity:

$1 + {\tan}^{2} \theta = {\sec}^{2} \theta$

and considering that for $\theta \in \left[0 , \frac{\pi}{4}\right]$ the secant is positive:

$\sqrt{1 + {\tan}^{2} \theta} = \sec \theta$

so:

$L = {\int}_{0}^{\frac{\pi}{4}} {\sec}^{3} \theta d \theta$

Integrate by parts:

$L = {\int}_{0}^{\frac{\pi}{4}} \sec \theta \cdot {\sec}^{2} \theta d \theta$
$L = {\int}_{0}^{\frac{\pi}{4}} \sec \theta d \left(\tan \theta\right)$
$L = {\left[\sec \theta \tan \theta\right]}_{0}^{\frac{\pi}{4}} - {\int}_{0}^{\frac{\pi}{4}} \tan \theta d \left(\sec \theta\right)$
$L = \sqrt{2} - {\int}_{0}^{\frac{\pi}{4}} {\tan}^{2} \theta \sec \theta d \theta$

$L = \sqrt{2} - {\int}_{0}^{\frac{\pi}{4}} \left({\sec}^{2} \theta - 1\right) \sec \theta d \theta$

$L = \sqrt{2} - {\int}_{0}^{\frac{\pi}{4}} {\sec}^{3} \theta d \theta + {\int}_{0}^{\frac{\pi}{4}} \sec \theta d \theta$

$L = \sqrt{2} - L + {\left[\ln \left\mid \sec \theta + \tan \theta \right\mid\right]}_{0}^{\frac{\pi}{4}}$

$2 L = \sqrt{2} + \ln \left\mid \sqrt{2} + 1 \right\mid$

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