009C Sample Final 1, Problem 8 - Revision history

Kayla Murray at 01:12, 3 December 2017

2017-12-03T01:12:24Z

2017-02-26T23:23:05Z

2017-02-26T19:09:56Z

2017-02-26T00:11:09Z

2017-02-26T00:10:57Z

2017-02-26T00:10:05Z

2016-04-19T01:39:46Z

2016-02-29T18:42:44Z

2016-02-24T18:32:45Z

2016-02-23T22:51:13Z

@@ Line 7: / Line 7: @@
 <span class="exam">(b) Find the area enclosed by the curve.
-{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
+<hr>
-!Foundations: &nbsp;
+[[009C Sample Final 1, Problem 8 Solution|'''<u>Solution</u>''']]
-'''Solution:'''
+[[009C Sample Final 1, Problem 8 Detailed Solution|'''<u>Detailed Solution</u>''']]
 [[009C_Sample_Final_1|'''<u>Return to Sample Exam</u>''']]

@@ Line 10: / Line 10: @@
 !Foundations: &nbsp;
 |-
-|The area under a polar curve <math style="vertical-align: -5px">r=f(\theta)</math> is given by
+|The area under a polar curve &nbsp; <math style="vertical-align: -5px">r=f(\theta)</math>&nbsp; is given by
 |-
+|
-&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>
+&nbsp; &nbsp; &nbsp; &nbsp;<math>\int_{\alpha_1}^{\alpha_2} \frac{1}{2}r^2~d\theta</math>&nbsp; for appropriate values of &nbsp;<math>\alpha_1,\alpha_2.</math>
 |}
@@ Line 39: / Line 39: @@
 !Step 2: &nbsp;
 |-
-|Using the double angle formula for <math style="vertical-align: -5px">\sin(2\theta),</math> we have
+|Using the double angle formula for &nbsp;<math style="vertical-align: -5px">\sin(2\theta),</math>&nbsp; we have
 |-
+|
@@ Line 72: / Line 72: @@
 !Final Answer: &nbsp;
 |-
-|&nbsp;&nbsp; '''(a)''' &nbsp; &nbsp; See Step 1 above.
+|&nbsp;&nbsp; '''(a)''' &nbsp; &nbsp; See above.
 |-
 |&nbsp;&nbsp; '''(b)''' &nbsp; &nbsp; <math>\frac{3\pi}{2}</math>
 |}
 [[009C_Sample_Final_1|'''<u>Return to Sample Exam</u>''']]

@@ Line 18: / Line 18: @@
 '''Solution:'''
 {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
-!Step 1: &nbsp;
+!(a) &nbsp;
 |-
 |Insert sketch
 |}
 '''(b)'''

← Older revision		Revision as of 00:11, 26 February 2017
Line 6:		Line 6:

	<span class="exam">(b) Find the area enclosed by the curve.		<span class="exam">(b) Find the area enclosed by the curve.
−

	{\| class="mw-collapsible mw-collapsed" style = "text-align:left;"		{\| class="mw-collapsible mw-collapsed" style = "text-align:left;"

@@ Line 37: / Line 37: @@
 |-
+|
-::<math>2\int_{-\frac{\pi}{4}}^{\frac{3\pi}{4}}\frac{1}{2}(1+\sin (2\theta)^2)~d\theta.</math>
+&nbsp; &nbsp; &nbsp; &nbsp;<math>2\int_{-\frac{\pi}{4}}^{\frac{3\pi}{4}}\frac{1}{2}(1+\sin (2\theta)^2)~d\theta.</math>
 |}
@@ Line 46: / Line 46: @@
 |-
+|
-::<math>\begin{array}{rcl}
+&nbsp; &nbsp; &nbsp; &nbsp;<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} \\
 &&\\
@@ Line 63: / Line 63: @@
 |-
+|
-::<math>\begin{array}{rcl}
+&nbsp; &nbsp; &nbsp; &nbsp;<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]}\\
 &&\\

@@ Line 1: / Line 1: @@
 <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.
@@ Line 14: / Line 14: @@
 |-
+|
-::<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:'''
@@ Line 70: / Line 71: @@
 \end{array}</math>
 |}
 {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 !Final Answer: &nbsp;
 |-
-|&nbsp;&nbsp; '''(a)''' See Step 1 above.
+|&nbsp;&nbsp; '''(a)''' &nbsp; &nbsp; See Step 1 above.
 |-
-|&nbsp;&nbsp; '''(b)''' <math>\frac{3\pi}{2}</math>
+|&nbsp;&nbsp; '''(b)''' &nbsp; &nbsp; <math>\frac{3\pi}{2}</math>
 |}
 [[009C_Sample_Final_1|'''<u>Return to Sample Exam</u>''']]

@@ Line 14: / Line 14: @@
 |-
+|
-::<math>\int_{\alpha_1}^{\alpha_2} \frac{1}{2}r^2~d\theta</math> for appropriate values of <math>\alpha_1,\alpha_2</math>.
+::<math>\int_{\alpha_1}^{\alpha_2} \frac{1}{2}r^2~d\theta</math> for appropriate values of <math>\alpha_1,\alpha_2.</math>
 |}
@@ Line 36: / Line 36: @@
 |-
+|
-::<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>
 |}
@@ Line 44: / Line 42: @@
 !Step 2: &nbsp;
 |-
-|Using the double angle formula for <math style="vertical-align: -5px">\sin(2\theta)</math>, we have
+|Using the double angle formula for <math style="vertical-align: -5px">\sin(2\theta),</math> we have
 |-
+|
@@ Line 54: / Line 52: @@
 & = & \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}}}\\
+& = & \displaystyle{\frac{3}{2}\theta-\cos(2\theta)-\frac{\sin(4\theta)}{8}\bigg|_{-\frac{\pi}{4}}^{\frac{3\pi}{4}}.}\\
 \end{array}</math>
 |}
@@ Line 69: / Line 67: @@
 & = & \displaystyle{\frac{9\pi}{8}+\frac{3\pi}{8}}\\
 &&\\
-& = & \displaystyle{\frac{3\pi}{2}}\\
+& = & \displaystyle{\frac{3\pi}{2}.}\\
 \end{array}</math>
 |}