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>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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| − | + | <math>L=\int_{\alpha_1}^{\alpha_2} \sqrt{r^2+\bigg(\frac{dr}{d\theta}\bigg)^2}d\theta.</math> | |
|- | |- | ||
|'''2.''' How would you integrate <math style="vertical-align: -14px">\int \sqrt{1+x^2}~dx?</math> | |'''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> | |
|- | |- | ||
|'''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> | |'''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:''' | '''Solution:''' | ||
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| − | + | <math>L=\int_0^{2\pi}\sqrt{\theta^2+1}d\theta.</math> | |
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| − | + | <math>\begin{array}{rcl} | |
\displaystyle{L} & = & \displaystyle{\int_{\theta=0}^{\theta=2\pi}\sqrt{\tan^2x+1}\sec^2xdx}\\ | \displaystyle{L} & = & \displaystyle{\int_{\theta=0}^{\theta=2\pi}\sqrt{\tan^2x+1}\sec^2xdx}\\ | ||
&&\\ | &&\\ | ||
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| − | + | <math>\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{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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\end{array}</math> | \end{array}</math> | ||
|} | |} | ||
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{| class="mw-collapsible mw-collapsed" style = "text-align:left;" | {| class="mw-collapsible mw-collapsed" style = "text-align:left;" | ||
!Final Answer: | !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> | + | | <math>\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>''']] | ||
Revision as of 16:14, 25 February 2017
A curve is given in polar coordinates by
- 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 r=\theta}
- 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 0\leq \theta \leq 2\pi}
Find the length of the curve.
| Foundations: |
|---|
| 1. The formula for the arc length 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} of a polar curve 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 r=f(\theta)} with 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 \alpha_1\leq \theta \leq \alpha_2} is |
|
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=\int_{\alpha_1}^{\alpha_2} \sqrt{r^2+\bigg(\frac{dr}{d\theta}\bigg)^2}d\theta.} |
| 2. How would you integrate 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 \int \sqrt{1+x^2}~dx?} |
|
You could use trig substitution and let 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 \theta .} |
| 3. Recall that 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 \int \sec^3x~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 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{dr}{d\theta}} . |
| Since 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 r=\theta,~\frac{dr}{d\theta}=1.} |
| Using the formula in Foundations, 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=\int_0^{2\pi}\sqrt{\theta^2+1}d\theta.} |
| Step 2: |
|---|
| Now, we proceed using trig substitution. Let 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 \theta=\tan x.} Then, 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 d\theta=\sec^2xdx.} |
| So, the integral becomes |
|
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{\int_{\theta=0}^{\theta=2\pi}\sqrt{\tan^2x+1}\sec^2xdx}\\ &&\\ & = & \displaystyle{\int_{\theta=0}^{\theta=2\pi}\sec^3xdx}\\ &&\\ & = & \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 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 \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|} |