009C Sample Final 2, Problem 9 - Revision history

Kayla Murray at 01:36, 3 December 2017

2017-12-03T01:36:16Z

Kayla Murray at 20:19, 5 March 2017

2017-03-05T20:19:23Z

Kayla Murray at 20:08, 5 March 2017

2017-03-05T20:08:29Z

Kayla Murray at 04:52, 5 March 2017

2017-03-05T04:52:53Z

Kayla Murray at 04:51, 5 March 2017

2017-03-05T04:51:37Z

Kayla Murray at 04:50, 5 March 2017

2017-03-05T04:50:48Z

Kayla Murray at 04:38, 5 March 2017

2017-03-05T04:38:23Z

Kayla Murray at 01:16, 5 March 2017

2017-03-05T01:16:05Z

Kayla Murray at 00:50, 5 March 2017

2017-03-05T00:50:44Z

Kayla Murray at 00:46, 5 March 2017

2017-03-05T00:46:09Z

@@ Line 8: / Line 8: @@
 <span class="exam">(c) Compute &nbsp;<math style="vertical-align: -14px">y''=\frac{d^2y}{dx^2}.</math>
-{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
+<hr>
-!Foundations: &nbsp;
+[[009C Sample Final 2, Problem 9 Solution|'''<u>Solution</u>''']]
-'''Solution:'''
+[[009C Sample Final 2, Problem 9 Detailed Solution|'''<u>Detailed Solution</u>''']]
 [[009C_Sample_Final_2|'''<u>Return to Sample Exam</u>''']]

@@ Line 77: / Line 77: @@
+|
 &nbsp; &nbsp; &nbsp; &nbsp;<math>\begin{array}{rcl}
-\displaystyle{\frac{dy'}{d\theta}} & = & \displaystyle{\frac{d}{d\theta}\bigg(\frac{2\cos(2\theta)\sin\theta+\sin(2\theta)\cos\theta}{2\cos(2\theta)\cos \theta-\sin(2\theta)\sin\theta}\bigg)}\\
+\displaystyle{\frac{dy'}{d\theta}} & = & \displaystyle{\frac{d}{d\theta}\bigg(\frac{2\cos^2\theta \sin\theta-\sin^3\theta}{\cos^3\theta-2\sin^2\theta\cos\theta}\bigg)}\\
 &&\\
-& = & \displaystyle{\frac{(2\cos(2\theta)\cos \theta-\sin(2\theta)\sin\theta)(-4\sin(2\theta)\sin \theta +2\cos(2\theta)\cos(theta)+2\cos(2\theta)\cos\theta-\sin(2\theta)\sin \theta)}{(2\cos(2\theta)\cos \theta-\sin(2\theta)\sin\theta)^2}}\\
+& = & \displaystyle{\frac{(\cos^3\theta-2\sin^2\theta\cos\theta)(-4\cos\theta\sin^2\theta+2\cos^3\theta-3\sin^2\theta\cos\theta)-(2\cos^2\theta\sin\theta-\sin^3\theta)(-3\cos^2\theta\sin\theta-4\sin \theta\cos^2\theta+2\sin^3\theta)}{(\cos^3\theta-2\sin^2\theta\cos\theta)^2}.}
 \end{array}</math>
 |}
@@ Line 95: / Line 89: @@
 |-
+|
-&nbsp; &nbsp; &nbsp; &nbsp;<math>\frac{d^2y}{dx^2}=\frac{3-3\sin\theta\cos(2\theta)+3\sin(2\theta)\cos\theta}{(\cos(2\theta)-\sin\theta)^3}.</math>
+&nbsp; &nbsp; &nbsp; &nbsp;<math>\frac{d^2y}{dx^2}=\frac{(\cos^3\theta-2\sin^2\theta\cos\theta)(-4\cos\theta\sin^2\theta+2\cos^3\theta-3\sin^2\theta\cos\theta)-(2\cos^2\theta\sin\theta-\sin^3\theta)(-3\cos^2\theta\sin\theta-4\sin \theta\cos^2\theta+2\sin^3\theta)}{(\cos^3\theta-2\sin^2\theta\cos\theta)^2(2\cos(2\theta)\cos \theta-\sin(2\theta)\sin\theta)}.</math>
 |-
+|
@@ Line 109: / Line 103: @@
 |-
 |&nbsp; &nbsp; '''(c)''' &nbsp; &nbsp;
 |}
 [[009C_Sample_Final_2|'''<u>Return to Sample Exam</u>''']]

@@ Line 53: / Line 53: @@
 |-
+|
-&nbsp; &nbsp; &nbsp; &nbsp;<math style="vertical-align: -18px">y'=\frac{2\cos(2\theta)\sin\theta+\sin(2\theta)\cos\theta}{2\cos(2\theta)\cos \theta-\sin(2\theta)\sin\theta}.</math>
+&nbsp; &nbsp; &nbsp; &nbsp; <math>\begin{array}{rcl}
 |}
@@ Line 69: / Line 77: @@
+|
 &nbsp; &nbsp; &nbsp; &nbsp;<math>\begin{array}{rcl}
-\displaystyle{\frac{dy'}{d\theta}} & = & \displaystyle{\frac{d}{d\theta}\bigg(\frac{\sin(2\theta)+\cos\theta}{\cos(2\theta)-\sin\theta}\bigg)}\\
+\displaystyle{\frac{dy'}{d\theta}} & = & \displaystyle{\frac{d}{d\theta}\bigg(\frac{2\cos(2\theta)\sin\theta+\sin(2\theta)\cos\theta}{2\cos(2\theta)\cos \theta-\sin(2\theta)\sin\theta}\bigg)}\\
 &&\\
-& = & \displaystyle{\frac{(\cos(2\theta)-\sin\theta)(2\cos(2\theta)-\sin\theta)-(\sin(2\theta)+\cos\theta)(-2\sin(2\theta)-\cos\theta)}{(\cos(2\theta)-\sin\theta)^2}}\\
+& = & \displaystyle{\frac{(2\cos(2\theta)\cos \theta-\sin(2\theta)\sin\theta)(-4\sin(2\theta)\sin \theta +2\cos(2\theta)\cos(theta)+2\cos(2\theta)\cos\theta-\sin(2\theta)\sin \theta)}{(2\cos(2\theta)\cos \theta-\sin(2\theta)\sin\theta)^2}}\\
 &&\\
-& = & \displaystyle{\frac{2\cos^2(2\theta)+2\sin^2(2\theta)-3\sin\theta\cos(2\theta)+3\sin(2\theta)\cos\theta+\sin^2\theta+\cos^2\theta}{(\cos(2\theta)-\sin\theta)^2}}\\
+& = & \displaystyle{\frac{2\cos^2(2\theta)+2\sin^2(2\theta)-3\sin\theta\cos(2\theta)+3\sin(2\theta)\cos\theta+\sin^2\theta+\cos^2\theta}{(2\cos(2\theta)\cos \theta-\sin(2\theta)\sin\theta)^2}}\\
 &&\\
-& = & \displaystyle{\frac{3-3\sin\theta\cos(2\theta)+3\sin(2\theta)\cos\theta}{(\cos(2\theta)-\sin\theta)^2}}\\
+& = & \displaystyle{\frac{3-3\sin\theta\cos(2\theta)+3\sin(2\theta)\cos\theta}{(2\cos(2\theta)\cos \theta-\sin(2\theta)\sin\theta)^2}}\\
 \end{array}</math>
 |-
@@ Line 98: / Line 106: @@
 |&nbsp; &nbsp; '''(a)''' &nbsp; &nbsp;See above
 |-
-|&nbsp; &nbsp; '''(b)''' &nbsp; &nbsp;<math style="vertical-align: -18px">y'=\frac{2\cos(2\theta)\sin\theta+\sin(2\theta)\cos\theta}{2\cos(2\theta)\cos \theta-\sin(2\theta)\sin\theta}.</math>
+|&nbsp; &nbsp; '''(b)''' &nbsp; &nbsp;<math style="vertical-align: -18px">y'=\frac{2\cos^2\theta \sin\theta-\sin^3\theta}{\cos^3\theta-2\sin^2\theta\cos\theta}</math>
 |-
 |&nbsp; &nbsp; '''(c)''' &nbsp; &nbsp;
 |}
 [[009C_Sample_Final_2|'''<u>Return to Sample Exam</u>''']]

@@ Line 61: / Line 61: @@
 !Step 1: &nbsp;
 |-
+|We have &nbsp; <math>\frac{d^2y}{dx^2}=\frac{\frac{dy'}{d\theta}}{\frac{dr}{d\theta}\cos\theta-r\sin\theta}.</math>
 |-
+|
 |-
+|since &nbsp;<math style="vertical-align: -2px">\sin^2\theta+\cos^2\theta=1</math>&nbsp; and &nbsp;<math style="vertical-align: -5px">2\cos^2(2\theta)+2\sin^2(2\theta)=2.</math>&nbsp;
 |}
 {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 !Step 2: &nbsp;
 |-
+|
 |-
+|

@@ Line 34: / Line 34: @@
 !Step 1: &nbsp;
 |-
-|Since &nbsp;<math style="vertical-align: -4px">r=\sin(2\theta),</math>
+|Since &nbsp;<math style="vertical-align: -5px">r=\sin(2\theta),</math>
 |-
+|

@@ Line 4: / Line 4: @@
 <span class="exam">(a) Sketch the curve.
-<span class="exam">(b) Compute &nbsp;<math>y'=\frac{dy}{dx}.</math>
+<span class="exam">(b) Compute &nbsp;<math style="vertical-align: -14px">y'=\frac{dy}{dx}.</math>
-<span class="exam">(c) Compute &nbsp;<math>y''=\frac{d^2y}{dx^2}.</math>
+<span class="exam">(c) Compute &nbsp;<math style="vertical-align: -14px">y''=\frac{d^2y}{dx^2}.</math>
 {| class="mw-collapsible mw-collapsed" style = "text-align:left;"

@@ Line 21: / Line 21: @@
+|
 |}
 '''Solution:'''
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+|
 |}
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@@ Line 70: / Line 92: @@
 |-
 |&nbsp;&nbsp; '''(b)'''
 |}
 [[009C_Sample_Final_2|'''<u>Return to Sample Exam</u>''']]

@@ Line 1: / Line 1: @@
 <span class="exam">A curve is given in polar coordinates by
-::::::<span class="exam"><math>r=\sin(2\theta).</math>
+::<span class="exam"><math>r=\sin(2\theta).</math>
-::<span class="exam">a) Sketch the curve.
+<span class="exam">(a) Sketch the curve.
-::<span class="exam">b) Compute <math>y'=\frac{dy}{dx}.</math>
+<span class="exam">(b) Compute &nbsp;<math>y'=\frac{dy}{dx}.</math>
-::<span class="exam">c) Compute <math>y''=\frac{d^2y}{dx^2}.</math>
+<span class="exam">(c) Compute &nbsp;<math>y''=\frac{d^2y}{dx^2}.</math>
 {| class="mw-collapsible mw-collapsed" style = "text-align:left;"