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009C Sample Final 1, Problem 9 - Revision history
2026-04-22T13:54:08Z
Revision history for this page on the wiki
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https://wiki.math.ucr.edu/index.php?title=009C_Sample_Final_1,_Problem_9&diff=1412&oldid=prev
MathAdmin at 00:16, 8 June 2016
2016-06-08T00:16:24Z
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<td colspan="2" style="background-color: #fff; color: #202122; text-align: center;">Revision as of 00:16, 8 June 2016</td>
</tr><tr><td colspan="2" class="diff-lineno" id="mw-diff-left-l58" >Line 58:</td>
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<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>!Step 3: &nbsp;</div></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>!Step 3: &nbsp;</div></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>|-</div></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>|-</div></td></tr>
<tr><td class='diff-marker'>−</td><td style="color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #ffe49c; vertical-align: top; white-space: pre-wrap;"><div>|Since <math style="vertical-align: -<del class="diffchange diffchange-inline">1px</del>">\theta=\tan x,</math> we have <math style="vertical-align: -1px">x=\tan^{-1}\theta .</math></div></td><td class='diff-marker'>+</td><td style="color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div>|Since <math style="vertical-align: -<ins class="diffchange diffchange-inline">4px</ins>">\theta=\tan x,</math> we have <math style="vertical-align: -1px">x=\tan^{-1}\theta .</math></div></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>|-</div></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>|-</div></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>|So, we have</div></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>|So, we have</div></td></tr>
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MathAdmin
https://wiki.math.ucr.edu/index.php?title=009C_Sample_Final_1,_Problem_9&diff=1406&oldid=prev
MathAdmin at 20:07, 13 May 2016
2016-05-13T20:07:04Z
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<td colspan="2" style="background-color: #fff; color: #202122; text-align: center;">Revision as of 20:07, 13 May 2016</td>
</tr><tr><td colspan="2" class="diff-lineno" id="mw-diff-left-l1" >Line 1:</td>
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<tr><td colspan="2"> </td><td class='diff-marker'>+</td><td style="color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div><ins style="font-weight: bold; text-decoration: none;">[[File:009C_SF1_9_GP.jpg|right|400px]]</ins></div></td></tr>
<tr><td colspan="2"> </td><td class='diff-marker'>+</td><td style="color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div><ins style="font-weight: bold; text-decoration: none;"></ins></div></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div><span class="exam">A curve is given in polar coordinates by </div></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div><span class="exam">A curve is given in polar coordinates by </div></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>::::::<span class="exam"><math>r=\theta</math></div></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>::::::<span class="exam"><math>r=\theta</math></div></td></tr>
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MathAdmin
https://wiki.math.ucr.edu/index.php?title=009C_Sample_Final_1,_Problem_9&diff=1393&oldid=prev
MathAdmin: Created page with "<span class="exam">A curve is given in polar coordinates by ::::::<span class="exam"><math>r=\theta</math> ::::::<span class="exam"><math>0\leq \theta \leq 2\pi</math> <span..."
2016-04-19T01:29:12Z
<p>Created page with "<span class="exam">A curve is given in polar coordinates by ::::::<span class="exam"><math>r=\theta</math> ::::::<span class="exam"><math>0\leq \theta \leq 2\pi</math> <span..."</p>
<p><b>New page</b></p><div><span class="exam">A curve is given in polar coordinates by <br />
::::::<span class="exam"><math>r=\theta</math><br />
::::::<span class="exam"><math>0\leq \theta \leq 2\pi</math><br />
<br />
<span class="exam">Find the length of the curve.<br />
<br />
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"<br />
!Foundations: &nbsp; <br />
|-<br />
|'''1.''' The formula for the arc length <math style="vertical-align: 0px">L</math> of a polar curve <math style="vertical-align: -5px">r=f(\theta)</math> with <math style="vertical-align: -4px">\alpha_1\leq \theta \leq \alpha_2</math> is <br />
|-<br />
|<br />
::<math>L=\int_{\alpha_1}^{\alpha_2} \sqrt{r^2+\bigg(\frac{dr}{d\theta}\bigg)^2}d\theta.</math><br />
|-<br />
|'''2.''' How would you integrate <math style="vertical-align: -14px">\int \sqrt{1+x^2}~dx?</math><br />
|-<br />
|<br />
::You could use trig substitution and let <math style="vertical-align: -1px">x=\tan \theta .</math><br />
|-<br />
|'''3.''' Recall that <math>\int \sec^3x~dx=\frac{1}{2}\sec x \tan x +\frac{1}{2}\ln|\sec x +\tan x|+C.</math><br />
|}<br />
<br />
'''Solution:'''<br />
<br />
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"<br />
!Step 1: &nbsp; <br />
|-<br />
|First, we need to calculate <math style="vertical-align: -14px">\frac{dr}{d\theta}</math>. <br />
|-<br />
|Since <math style="vertical-align: -14px">r=\theta,~\frac{dr}{d\theta}=1.</math><br />
|-<br />
|Using the formula in Foundations, we have <br />
|-<br />
|<br />
::<math>L=\int_0^{2\pi}\sqrt{\theta^2+1}d\theta.</math><br />
|}<br />
<br />
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"<br />
!Step 2: &nbsp;<br />
|-<br />
|Now, we proceed using trig substitution. Let <math style="vertical-align: -2px">\theta=\tan x.</math> Then, <math style="vertical-align: -1px">d\theta=\sec^2xdx.</math><br />
|-<br />
|So, the integral becomes <br />
|-<br />
|<br />
::<math>\begin{array}{rcl}<br />
\displaystyle{L} & = & \displaystyle{\int_{\theta=0}^{\theta=2\pi}\sqrt{\tan^2x+1}\sec^2xdx}\\<br />
&&\\<br />
& = & \displaystyle{\int_{\theta=0}^{\theta=2\pi}\sec^3xdx}\\<br />
&&\\<br />
& = & \displaystyle{\frac{1}{2}\sec x \tan x +\frac{1}{2}\ln|\sec x +\tan x|\bigg|_{\theta=0}^{\theta=2\pi}.}\\<br />
\end{array}</math><br />
|}<br />
<br />
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"<br />
!Step 3: &nbsp;<br />
|-<br />
|Since <math style="vertical-align: -1px">\theta=\tan x,</math> we have <math style="vertical-align: -1px">x=\tan^{-1}\theta .</math><br />
|-<br />
|So, we have<br />
|-<br />
|<br />
::<math>\begin{array}{rcl}<br />
\displaystyle{L} & = & \displaystyle{\frac{1}{2}\sec (\tan^{-1}(\theta)) \theta +\frac{1}{2}\ln|\sec (\tan^{-1}(\theta)) +\theta|\bigg|_{0}^{2\pi}}\\<br />
&&\\<br />
& = & \displaystyle{\frac{1}{2}\sec(\tan^{-1}(2\pi))2\pi+\frac{1}{2}\ln|\sec(\tan^{-1}(2\pi))+2\pi|.}\\<br />
\end{array}</math><br />
|}<br />
<br />
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"<br />
!Final Answer: &nbsp; <br />
|-<br />
|&nbsp;&nbsp; <math>\frac{1}{2}\sec(\tan^{-1}(2\pi))2\pi+\frac{1}{2}\ln|\sec(\tan^{-1}(2\pi))+2\pi|</math><br />
|}<br />
[[009C_Sample_Final_1|'''<u>Return to Sample Exam</u>''']]</div>
MathAdmin