Difference between revisions of "009C Sample Final 1, Problem 9"

Revision as of 13:07, 9 February 2016

A curve is given in polar coordinates by

r=\theta

0\leq \theta \leq 2\pi

Find the length of the curve.

Foundations:
The formula for the arc length $L$ of a polar curve $r=f(\theta )$ with $\alpha _{1}\leq \theta \leq \alpha _{2}$ is
Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle L=\int _{\alpha _{1}}^{\alpha _{2}}{\sqrt {r^{2}+{\bigg (}{\frac {dr}{d\theta }}{\bigg )}^{2}}}d\theta } .

Solution:

Step 1:
First, we need to calculate ${\frac {dr}{d\theta }}$ . Since Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle r=\theta ,~{\frac {dr}{d\theta }}=1} .
Using the formula in Foundations, we have
Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle L=\int _{0}^{2\pi }{\sqrt {\theta ^{2}+1}}d\theta } .

Step 2:
Now, we proceed using trig substitution. Let Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle \theta =\tan x} . Then, $d\theta =\sec ^{2}xdx$ .
So, the integral becomes
$L=\int _{\theta =0}^{\theta =2\pi }{\sqrt {\tan ^{2}x+1}}\sec ^{2}xdx=\int _{\theta =0}^{\theta =2\pi }\sec ^{3}xdx$ .
We integrate to get Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle L={\frac {1}{2}}\sec x\tan x+{\frac {1}{2}}\ln \|\sec x+\tan x\|{\bigg \|}_{\theta =0}^{\theta =2\pi }} .

Step 3:
Since Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle \theta =\tan x} , we have Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\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 L=\frac{1}{2}\sec (\tan^{-1}(\theta)) \theta +\frac{1}{2}\ln\|\sec (\tan^{-1}(\theta)) +\theta\|\bigg\|_{0}^{2\pi}} .

Final Answer:

@@ Line 28: / Line 28: @@
 !Step 2: &nbsp;
 |-
-|Now, we proceed using trig substitution. Let <math>\theta=\tan x</math>. Then, <math>d\theta=\sec^2xdx</math>
+|Now, we proceed using trig substitution. Let <math>\theta=\tan x</math>. Then, <math>d\theta=\sec^2xdx</math>.
+|-
+|So, the integral becomes
+|-
+|<math>L=\int_{\theta=0}^{\theta=2\pi}\sqrt{\tan^2x+1}\sec^2xdx=\int_{\theta=0}^{\theta=2\pi}\sec^3xdx</math>.
+|-
+|We integrate to get <math>L=\frac{1}{2}\sec x \tan x +\frac{1}{2}\ln|\sec x +\tan x|\bigg|_{\theta=0}^{\theta=2\pi}</math>.
 |}
+{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
+!Step 3: &nbsp;
+|-
+|Since <math>\theta=\tan x</math>, we have <math>x=\tan^{-1}\theta</math>.
+|-
+|So, we have
+|-
+|<math>L=\frac{1}{2}\sec (\tan^{-1}(\theta)) \theta +\frac{1}{2}\ln|\sec (\tan^{-1}(\theta)) +\theta|\bigg|_{0}^{2\pi}</math>.
+|-
+|
+|}
 {| class="mw-collapsible mw-collapsed" style = "text-align:left;"