# 009C Sample Final 1, Problem 9

A curve is given in polar coordinates by

$r=\theta$ $0\leq \theta \leq 2\pi$ Find the length of the curve.

Foundations:
1. The formula for the arc length $L$ of a polar curve $r=f(\theta )$ with $\alpha _{1}\leq \theta \leq \alpha _{2}$ is
$L=\int _{\alpha _{1}}^{\alpha _{2}}{\sqrt {r^{2}+{\bigg (}{\frac {dr}{d\theta }}{\bigg )}^{2}}}d\theta .$ 2. How would you integrate $\int {\sqrt {1+x^{2}}}~dx?$ You could use trig substitution and let $x=\tan \theta .$ 3. Recall that $\int \sec ^{3}x~dx={\frac {1}{2}}\sec x\tan x+{\frac {1}{2}}\ln |\sec x+\tan x|+C.$ Solution:

Step 1:
First, we need to calculate ${\frac {dr}{d\theta }}$ .
Since $r=\theta ,~{\frac {dr}{d\theta }}=1.$ Using the formula in Foundations, we have
$L=\int _{0}^{2\pi }{\sqrt {\theta ^{2}+1}}d\theta .$ Step 2:
Now, we proceed using trig substitution. Let $\theta =\tan x.$ Then, $d\theta =\sec ^{2}xdx.$ So, the integral becomes
${\begin{array}{rcl}\displaystyle {L}&=&\displaystyle {\int _{\theta =0}^{\theta =2\pi }{\sqrt {\tan ^{2}x+1}}\sec ^{2}xdx}\\&&\\&=&\displaystyle {\int _{\theta =0}^{\theta =2\pi }\sec ^{3}xdx}\\&&\\&=&\displaystyle {{\frac {1}{2}}\sec x\tan x+{\frac {1}{2}}\ln |\sec x+\tan x|{\bigg |}_{\theta =0}^{\theta =2\pi }.}\\\end{array}}$ Step 3:
Since $\theta =\tan x,$ we have $x=\tan ^{-1}\theta .$ So, we have
${\begin{array}{rcl}\displaystyle {L}&=&\displaystyle {{\frac {1}{2}}\sec(\tan ^{-1}(\theta ))\theta +{\frac {1}{2}}\ln |\sec(\tan ^{-1}(\theta ))+\theta |{\bigg |}_{0}^{2\pi }}\\&&\\&=&\displaystyle {{\frac {1}{2}}\sec(\tan ^{-1}(2\pi ))2\pi +{\frac {1}{2}}\ln |\sec(\tan ^{-1}(2\pi ))+2\pi |.}\\\end{array}}$ ${\frac {1}{2}}\sec(\tan ^{-1}(2\pi ))2\pi +{\frac {1}{2}}\ln |\sec(\tan ^{-1}(2\pi ))+2\pi |$ 