Difference between revisions of "009C Sample Final 1, Problem 8"

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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  
::::::<math>r=1+\sin 2\theta</math>
+
::<math>r=1+\sin 2\theta</math>
::::::<math>0\leq \theta \leq 2\pi</math>
+
::<math>0\leq \theta \leq 2\pi</math>
  
::<span class="exam">a) Sketch the curve.
+
<span class="exam">(a) Sketch the curve.
  
::<span class="exam">b) Find the area enclosed by the curve.
+
<span class="exam">(b) Find the area enclosed by the curve.
  
  
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|-
 
|
 
|
::<math>\int_{\alpha_1}^{\alpha_2} \frac{1}{2}r^2~d\theta</math> for appropriate values of <math>\alpha_1,\alpha_2.</math>
+
&nbsp; &nbsp; &nbsp; &nbsp;<math>\int_{\alpha_1}^{\alpha_2} \frac{1}{2}r^2~d\theta</math> for appropriate values of <math>\alpha_1,\alpha_2.</math>
 
|}
 
|}
 +
  
 
'''Solution:'''
 
'''Solution:'''
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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: &nbsp;  
 
!Final Answer: &nbsp;  
 
|-
 
|-
|&nbsp;&nbsp; '''(a)''' See Step 1 above.
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|&nbsp;&nbsp; '''(a)''' &nbsp; &nbsp; See Step 1 above.
 
|-
 
|-
|&nbsp;&nbsp; '''(b)''' <math>\frac{3\pi}{2}</math>
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|&nbsp;&nbsp; '''(b)''' &nbsp; &nbsp; <math>\frac{3\pi}{2}</math>
 
|}
 
|}
 
[[009C_Sample_Final_1|'''<u>Return to Sample Exam</u>''']]
 
[[009C_Sample_Final_1|'''<u>Return to Sample Exam</u>''']]

Revision as of 16:10, 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=1+\sin 2\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}

(a) Sketch the curve.

(b) Find the area enclosed by the curve.


Foundations:  
The area under 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)} is given 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 \int_{\alpha_1}^{\alpha_2} \frac{1}{2}r^2~d\theta} for appropriate values of 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,\alpha_2.}


Solution:

(a)

Step 1:  
Insert sketch


(b)

Step 1:  
Since the graph has symmetry (as seen in the graph), the area of the curve 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 2\int_{-\frac{\pi}{4}}^{\frac{3\pi}{4}}\frac{1}{2}(1+\sin (2\theta)^2)~d\theta.}
Step 2:  
Using the double angle formula for 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 \sin(2\theta),} 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{2\int_{-\frac{\pi}{4}}^{\frac{3\pi}{4}}\frac{1}{2}(1+\sin (2\theta))^2~d\theta} & = & \displaystyle{\int_{-\frac{\pi}{4}}^{\frac{3\pi}{4}}1+2\sin(2\theta)+\sin^2(2\theta)~d\theta} \\ &&\\ & = & \displaystyle{\int_{-\frac{\pi}{4}}^{\frac{3\pi}{4}}1+2\sin(2\theta)+\frac{1-\cos(4\theta)}{2}~d\theta}\\ &&\\ & = & \displaystyle{\int_{-\frac{\pi}{4}}^{\frac{3\pi}{4}}\frac{3}{2}+2\sin(2\theta)-\frac{\cos(4\theta)}{2}~d\theta}\\ &&\\ & = & \displaystyle{\frac{3}{2}\theta-\cos(2\theta)-\frac{\sin(4\theta)}{8}\bigg|_{-\frac{\pi}{4}}^{\frac{3\pi}{4}}.}\\ \end{array}}
Step 3:  
Lastly, we evaluate to get
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{\frac{3}{2}\theta-\cos(2\theta)-\frac{\sin(4\theta)}{8}\bigg|_{-\frac{\pi}{4}}^{\frac{3\pi}{4}}} & = &\displaystyle{\frac{3}{2}\bigg(\frac{3\pi}{4}\bigg)-\cos\bigg(\frac{3\pi}{2}\bigg)-\frac{\sin(3\pi)}{8}-\bigg[\frac{3}{2}\bigg(-\frac{\pi}{4}\bigg)-\cos\bigg(-\frac{\pi}{2}\bigg)-\frac{\sin(-\pi)}{8}\bigg]}\\ &&\\ & = & \displaystyle{\frac{9\pi}{8}+\frac{3\pi}{8}}\\ &&\\ & = & \displaystyle{\frac{3\pi}{2}.}\\ \end{array}}


Final Answer:  
   (a)     See Step 1 above.
   (b)     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{3\pi}{2}}

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