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 Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x=\tan^{-1}\theta .}
So, we have
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \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:
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \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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