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

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!Step 1:    
 
!Step 1:    
 
|-
 
|-
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|First, we need to calculate <math style="vertical-align: -14px">\frac{dr}{d\theta}</math>.  
+
|First, we need to calculate &nbsp;<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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|Since &nbsp;<math style="vertical-align: -14px">r=\theta,~\frac{dr}{d\theta}=1.</math>
 
|-
 
|-
 
|Using the formula in Foundations, we have  
 
|Using the formula in Foundations, we have  
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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 &nbsp;<math style="vertical-align: -2px">\theta=\tan x.</math> &nbsp; Then, &nbsp;<math style="vertical-align: -1px">d\theta=\sec^2xdx.</math>
 
|-
 
|-
 
|So, the integral becomes  
 
|So, the integral becomes  
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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 &nbsp; <math style="vertical-align: -4px">\theta=\tan x,</math>&nbsp; we have &nbsp;<math style="vertical-align: -1px">x=\tan^{-1}\theta .</math>
 
|-
 
|-
 
|So, we have
 
|So, we have

Revision as of 15:28, 26 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|}

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