MathAdmin at 17:18, 24 May 2016

2016-05-24T17:18:19Z

MathAdmin at 20:05, 13 May 2016

2016-05-13T20:05:10Z

MathAdmin: Created page with "A curve is given in polar coordinates by :::::: $r=1+\sin 2\theta$ :::::: $0\leq \theta \leq 2\pi$ ::a) Sketch the..."

2016-04-19T01:27:51Z

Created page with "A curve is given in polar coordinates by ::::::<math>r=1+\sin 2\theta</math> ::::::<math>0\leq \theta \leq 2\pi</math> ::a) Sketch the..."

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A curve is given in polar coordinates by
::::::<math>r=1+\sin 2\theta</math>
::::::<math>0\leq \theta \leq 2\pi</math>

::a) Sketch the curve.

::b) Find the area enclosed by the curve.

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Foundations:  
|-
|The area under a polar curve <math style="vertical-align: -5px">r=f(\theta)</math> is given by
|-
|
::<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:'''

'''(a)'''

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
|-
|Insert sketch
|}

'''(b)'''

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
|-
|Since the graph has symmetry (as seen in the graph), the area of the curve is
|-
|
::<math>2\int_{-\frac{\pi}{4}}^{\frac{3\pi}{4}}\frac{1}{2}(1+\sin (2\theta)^2)~d\theta.</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
|-
|Using the double angle formula for <math style="vertical-align: -5px">\sin(2\theta),</math> we have
|-
|
::<math>\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}</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 3:  
|-
|Lastly, we evaluate to get
|-
|
::<math>\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}</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Final Answer:  
|-
|   '''(a)''' See Step 1 above.
|-
|   '''(b)''' <math>\frac{3\pi}{2}</math>
|}
[[009C_Sample_Final_1|'''Return to Sample Exam''']]

@@ Line 24: / Line 24: @@
 !Step 1: &nbsp;
 |-
-|Insert sketch
+|[[File:009C_SF1_8.jpg |right|450px]]
 |}

@@ Line 33: / Line 33: @@
 !Step 1: &nbsp;
 |-
-|Since the graph has symmetry (as seen in the graph), the area of the curve is
+|Since the graph has symmetry (as seen in the previous image), the area of the curve is
 |-
+|
-::<math>2\int_{-\frac{\pi}{4}}^{\frac{3\pi}{4}}\frac{1}{2}(1+\sin (2\theta)^2)~d\theta.</math>
+::<math>2\int_{-\frac{\pi}{4}}^{\frac{3\pi}{4}}\frac{1}{2}(1+\sin (2\theta))^2~d\theta.</math>
 |}

009C Sample Final 1, Problem 8 - Revision history

MathAdmin at 17:18, 24 May 2016

MathAdmin at 20:05, 13 May 2016

MathAdmin: Created page with "A curve is given in polar coordinates by ::::::r=1+\sin 2\theta ::::::0\leq \theta \leq 2\pi ::a) Sketch the..."

MathAdmin: Created page with "A curve is given in polar coordinates by :::::: $r=1+\sin 2\theta$ :::::: $0\leq \theta \leq 2\pi$ ::a) Sketch the..."