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

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<span class="exam">A curve is given in polar coordinates by  
 
<span class="exam">A curve is given in polar coordinates by  
::::::<span class="exam"><math>r=\theta</math>
+
::<span class="exam"><math>r=\theta</math>
::::::<span class="exam"><math>0\leq \theta \leq 2\pi</math>
+
::<span class="exam"><math>0\leq \theta \leq 2\pi</math>
  
 
<span class="exam">Find the length of the curve.
 
<span class="exam">Find the length of the curve.
  
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<hr>
!Foundations: &nbsp;
+
[[009C Sample Final 1, Problem 9 Solution|'''<u>Solution</u>''']]
|-
 
|The formula for the arc length <math>L</math> of a polar curve <math>r=f(\theta)</math> with <math>\alpha_1\leq \theta \leq \alpha_2</math> is
 
|-
 
|<math>L=\int_{\alpha_1}^{\alpha_2} \sqrt{r^2+\bigg(\frac{dr}{d\theta}\bigg)^2}d\theta</math>.
 
|}
 
  
'''Solution:'''
 
  
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[[009C Sample Final 1, Problem 9 Detailed Solution|'''<u>Detailed Solution</u>''']]
!Step 1: &nbsp;
 
|-
 
|First, we need to calculate <math>\frac{dr}{d\theta}</math>. Since <math>r=\theta,~\frac{dr}{d\theta}=1</math>.
 
|-
 
|Using the formula in Foundations, we have
 
|-
 
|<math>L=\int_0^{2\pi}\sqrt{\theta^2+1}d\theta</math>.
 
|}
 
  
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!Step 2: &nbsp;
 
|-
 
|Now, we proceed using trig substitution. Let <math>\theta=\tan x</math>. Then, <math>d\theta=\sec^2xdx</math>.
 
|-
 
|So, the integral becomes
 
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|<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>.
 
|}
 
  
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!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>.
 
|-
 
|Thus, <math>L=\frac{1}{2}\sec(\tan^{-1}(2\pi))2\pi+\frac{1}{2}\ln|\sec(\tan^{-1}(2\pi))+2\pi|</math>.
 
|}
 
 
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!Final Answer: &nbsp;
 
|-
 
|<math>L=\frac{1}{2}\sec(\tan^{-1}(2\pi))2\pi+\frac{1}{2}\ln|\sec(\tan^{-1}(2\pi))+2\pi|</math>
 
|}
 
 
[[009C_Sample_Final_1|'''<u>Return to Sample Exam</u>''']]
 
[[009C_Sample_Final_1|'''<u>Return to Sample Exam</u>''']]

Latest revision as of 18:14, 2 December 2017

A curve is given in polar coordinates by

Find the length of the curve.


Solution


Detailed Solution


Return to Sample Exam