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| | <span class="exam">Find the length of the curve. | | <span class="exam">Find the length of the curve. |
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| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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| − | !Foundations:
| + | [[009C Sample Final 1, Problem 9 Solution|'''<u>Solution</u>''']] |
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| − | |'''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
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| − | <math>L=\int_{\alpha_1}^{\alpha_2} \sqrt{r^2+\bigg(\frac{dr}{d\theta}\bigg)^2}d\theta.</math>
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| − | |'''2.''' How would you integrate <math style="vertical-align: -14px">\int \sqrt{1+x^2}~dx?</math> | |
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| − | You could use trig substitution and let <math style="vertical-align: -1px">x=\tan \theta .</math>
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| − | |'''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>
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| − | '''Solution:''' | + | [[009C Sample Final 1, Problem 9 Detailed Solution|'''<u>Detailed Solution</u>''']] |
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| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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| − | !Step 1:
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| − | |First, we need to calculate <math style="vertical-align: -14px">\frac{dr}{d\theta}</math>.
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| − | |Since <math style="vertical-align: -14px">r=\theta,~\frac{dr}{d\theta}=1.</math>
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| − | |Using the formula in Foundations, we have
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| − | <math>L=\int_0^{2\pi}\sqrt{\theta^2+1}d\theta.</math>
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| − | |}
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| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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| − | !Step 2:
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| − | |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>
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| − | |So, the integral becomes
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| − | <math>\begin{array}{rcl}
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| − | \displaystyle{L} & = & \displaystyle{\int_{\theta=0}^{\theta=2\pi}\sqrt{\tan^2x+1}\sec^2xdx}\\
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| − | &&\\
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| − | & = & \displaystyle{\int_{\theta=0}^{\theta=2\pi}\sec^3xdx}\\
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| − | &&\\
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| − | & = & \displaystyle{\frac{1}{2}\sec x \tan x +\frac{1}{2}\ln|\sec x +\tan x|\bigg|_{\theta=0}^{\theta=2\pi}.}\\
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| − | \end{array}</math>
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| − | |}
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| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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| − | !Step 3:
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| − | |Since <math style="vertical-align: -4px">\theta=\tan x,</math> we have <math style="vertical-align: -1px">x=\tan^{-1}\theta .</math>
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| − | |So, we have
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| − | <math>\begin{array}{rcl}
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| − | \displaystyle{L} & = & \displaystyle{\frac{1}{2}\sec (\tan^{-1}(\theta)) \theta +\frac{1}{2}\ln|\sec (\tan^{-1}(\theta)) +\theta|\bigg|_{0}^{2\pi}}\\
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| − | &&\\
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| − | & = & \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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| − | \end{array}</math>
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| − | |}
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| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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| − | !Final Answer:
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| − | | <math>\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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| − | |}
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| | [[009C_Sample_Final_1|'''<u>Return to Sample Exam</u>''']] | | [[009C_Sample_Final_1|'''<u>Return to Sample Exam</u>''']] |
