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
${\begin{array}{rcl}\displaystyle {L}&=&\displaystyle {{\frac {1}{2}}\sec(\tan ^{-1}(\theta ))\theta +{\frac {1}{2}}\ln |\sec(\tan ^{-1}(\theta ))+\theta |{\bigg |}_{0}^{2\pi }}\\&&\\&=&\displaystyle {{\frac {1}{2}}\sec(\tan ^{-1}(2\pi ))2\pi +{\frac {1}{2}}\ln |\sec(\tan ^{-1}(2\pi ))+2\pi |}\\\end{array}}$

Final Answer:
${\frac {1}{2}}\sec(\tan ^{-1}(2\pi ))2\pi +{\frac {1}{2}}\ln |\sec(\tan ^{-1}(2\pi ))+2\pi |$

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