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

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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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::<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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|'''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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::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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|'''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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!Step 1: &nbsp;  
 
!Step 1: &nbsp;  
 
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|First, we need to calculate <math style="vertical-align: -14px">\frac{dr}{d\theta}</math>. Since <math style="vertical-align: -14px">r=\theta,~\frac{dr}{d\theta}=1</math>.
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|First, we need to calculate <math style="vertical-align: -14px">\frac{dr}{d\theta}</math>. 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  
 
|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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::<math>L=\int_0^{2\pi}\sqrt{\theta^2+1}d\theta.</math>
 
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!Step 2: &nbsp;
 
!Step 2: &nbsp;
 
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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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|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  
 
|So, the integral becomes  
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& = & \displaystyle{\int_{\theta=0}^{\theta=2\pi}\sec^3xdx}\\
 
& = & \displaystyle{\int_{\theta=0}^{\theta=2\pi}\sec^3xdx}\\
 
&&\\
 
&&\\
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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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& = & \displaystyle{\frac{1}{2}\sec x \tan x +\frac{1}{2}\ln|\sec x +\tan x|\bigg|_{\theta=0}^{\theta=2\pi}.}\\
 
\end{array}</math>
 
\end{array}</math>
 
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!Step 3: &nbsp;
 
!Step 3: &nbsp;
 
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|Since <math style="vertical-align: -1px">\theta=\tan x</math>, we have <math style="vertical-align: -1px">x=\tan^{-1}\theta</math>.
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|Since <math style="vertical-align: -1px">\theta=\tan x,</math> we have <math style="vertical-align: -1px">x=\tan^{-1}\theta .</math>
 
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|So, we have
 
|So, we have
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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}}\\
 
\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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& = & \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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& = & \displaystyle{\frac{1}{2}\sec(\tan^{-1}(2\pi))2\pi+\frac{1}{2}\ln|\sec(\tan^{-1}(2\pi))+2\pi|.}\\
 
\end{array}</math>
 
\end{array}</math>
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Revision as of 11:44, 29 February 2016

A curve is given in polar coordinates by

Find the length of the curve.

Foundations:  
1. The formula for the arc length of a polar curve with is
2. How would you integrate
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|}

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