009C Sample Final 1, Problem 7 - Revision history

Kayla Murray at 21:23, 3 December 2017

2017-12-03T21:23:35Z

Kayla Murray at 01:10, 3 December 2017

2017-12-03T01:10:07Z

Kayla Murray at 23:23, 26 February 2017

2017-02-26T23:23:44Z

Kayla Murray at 23:23, 26 February 2017

2017-02-26T23:23:28Z

Kayla Murray at 23:21, 26 February 2017

2017-02-26T23:21:31Z

Kayla Murray at 23:21, 26 February 2017

2017-02-26T23:21:00Z

Kayla Murray at 19:09, 26 February 2017

2017-02-26T19:09:22Z

Kayla Murray at 00:09, 26 February 2017

2017-02-26T00:09:10Z

Kayla Murray at 00:07, 26 February 2017

2017-02-26T00:07:02Z

Kayla Murray at 01:38, 19 April 2016

2016-04-19T01:38:19Z

@@ Line 4: / Line 4: @@
 <span class="exam">(a) Sketch the curve.
-<span class="exam">(b) Compute &nbsp; <math style="vertical-align: -12px">y'=\frac{dy}{dx}</math>.
+<span class="exam">(b) Compute &nbsp;<math style="vertical-align: -12px">y'=\frac{dy}{dx}</math>.
-<span class="exam">(c) Compute &nbsp; <math style="vertical-align: -12px">y''=\frac{d^2y}{dx^2}</math>.
+<span class="exam">(c) Compute &nbsp;<math style="vertical-align: -12px">y''=\frac{d^2y}{dx^2}</math>.
 <hr>

@@ Line 8: / Line 8: @@
 <span class="exam">(c) Compute &nbsp; <math style="vertical-align: -12px">y''=\frac{d^2y}{dx^2}</math>.
-{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
+<hr>
-!Foundations: &nbsp;
+[[009C Sample Final 1, Problem 7 Solution|'''<u>Solution</u>''']]
-'''Solution:'''
+[[009C Sample Final 1, Problem 7 Detailed Solution|'''<u>Detailed Solution</u>''']]
 [[009C_Sample_Final_1|'''<u>Return to Sample Exam</u>''']]

@@ Line 4: / Line 4: @@
 <span class="exam">(a) Sketch the curve.
-<span class="exam">(b) Compute <math style="vertical-align: -12px">y'=\frac{dy}{dx}</math>.
+<span class="exam">(b) Compute &nbsp; <math style="vertical-align: -12px">y'=\frac{dy}{dx}</math>.
-<span class="exam">(c) Compute <math style="vertical-align: -12px">y''=\frac{d^2y}{dx^2}</math>.
+<span class="exam">(c) Compute &nbsp; <math style="vertical-align: -12px">y''=\frac{d^2y}{dx^2}</math>.
 {| class="mw-collapsible mw-collapsed" style = "text-align:left;"

@@ Line 101: / Line 101: @@
 !Final Answer: &nbsp;
 |-
-|&nbsp;&nbsp; '''(a)''' &nbsp; &nbsp; See above for the graph.
+|&nbsp;&nbsp; '''(a)''' &nbsp; &nbsp; See above.
 |-
 |&nbsp;&nbsp; '''(b)''' &nbsp; &nbsp; <math>\frac{\sin(2\theta)+\cos\theta}{\cos(2\theta)-\sin\theta}</math>

@@ Line 101: / Line 101: @@
 !Final Answer: &nbsp;
 |-
-|&nbsp;&nbsp; '''(a)''' &nbsp; &nbsp; See (a) above for the graph.
+|&nbsp;&nbsp; '''(a)''' &nbsp; &nbsp; See above for the graph.
 |-
 |&nbsp;&nbsp; '''(b)''' &nbsp; &nbsp; <math>\frac{\sin(2\theta)+\cos\theta}{\cos(2\theta)-\sin\theta}</math>

@@ Line 11: / Line 11: @@
 !Foundations: &nbsp;
 |-
-|How do you calculate <math style="vertical-align: -5px">y'</math> for a polar curve <math style="vertical-align: -5px">r=f(\theta)?</math>
+|How do you calculate &nbsp; <math style="vertical-align: -5px">y'</math> &nbsp; for a polar curve &nbsp;<math style="vertical-align: -5px">r=f(\theta)?</math>
 |-
+|
-&nbsp; &nbsp; &nbsp; &nbsp;Since <math style="vertical-align: -5px">x=r\cos(\theta),~y=r\sin(\theta),</math> we have
+&nbsp; &nbsp; &nbsp; &nbsp;Since &nbsp; <math style="vertical-align: -5px">x=r\cos(\theta),~y=r\sin(\theta),</math>&nbsp; we have
 |-
+|
@@ Line 39: / Line 39: @@
 &nbsp; &nbsp; &nbsp; &nbsp;<math>y'=\frac{dy}{dx}=\frac{\frac{dr}{d\theta}\sin\theta+r\cos\theta}{\frac{dr}{d\theta}\cos\theta-r\sin\theta}.</math>
 |-
-|Since <math style="vertical-align: -2px">r=1+\sin\theta,</math>
+|Since &nbsp;<math style="vertical-align: -4px">r=1+\sin\theta,</math>
 |-
+|
@@ Line 66: / Line 66: @@
 !Step 1: &nbsp;
 |-
-|We have <math>\frac{d^2y}{dx^2}=\frac{\frac{dy'}{d\theta}}{\frac{dr}{d\theta}\cos\theta-r\sin\theta}.</math>
+|We have &nbsp; <math>\frac{d^2y}{dx^2}=\frac{\frac{dy'}{d\theta}}{\frac{dr}{d\theta}\cos\theta-r\sin\theta}.</math>
 |-
-|So, first we need to find <math>\frac{dy'}{d\theta}.</math>
+|So, first we need to find &nbsp; <math>\frac{dy'}{d\theta}.</math>
 |-
 |We have
@@ Line 83: / Line 83: @@
 \end{array}</math>
 |-
-|since <math style="vertical-align: -2px">\sin^2\theta+\cos^2\theta=1</math> and <math style="vertical-align: -5px">2\cos^2(2\theta)+2\sin^2(2\theta)=2.</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;
 |}
@@ Line 89: / Line 89: @@
 !Step 2: &nbsp;
 |-
-| Now, using the resulting formula for <math>\frac{dy'}{d\theta},</math> we get
+| Now, using the resulting formula for &nbsp; <math>\frac{dy'}{d\theta},</math>&nbsp; we get
 |-
+|
@@ Line 101: / Line 101: @@
 !Final Answer: &nbsp;
 |-
-|&nbsp;&nbsp; '''(a)''' &nbsp; &nbsp; See Step 1 above for the graph.
+|&nbsp;&nbsp; '''(a)''' &nbsp; &nbsp; See (a) above for the graph.
 |-
 |&nbsp;&nbsp; '''(b)''' &nbsp; &nbsp; <math>\frac{\sin(2\theta)+\cos\theta}{\cos(2\theta)-\sin\theta}</math>

@@ Line 40: / Line 40: @@
 |-
+|
-::<math>y'=\frac{dy}{dx}=\frac{\frac{dr}{d\theta}\sin\theta+r\cos\theta}{\frac{dr}{d\theta}\cos\theta-r\sin\theta}.</math>
+&nbsp; &nbsp; &nbsp; &nbsp;<math>y'=\frac{dy}{dx}=\frac{\frac{dr}{d\theta}\sin\theta+r\cos\theta}{\frac{dr}{d\theta}\cos\theta-r\sin\theta}.</math>
 |-
 |Since <math style="vertical-align: -2px">r=1+\sin\theta,</math>
 |-
+|
-::<math>\frac{dr}{d\theta}=\cos\theta.</math>
+&nbsp; &nbsp; &nbsp; &nbsp;<math>\frac{dr}{d\theta}=\cos\theta.</math>
 |-
 |Hence,
 |-
+|
-::<math style="vertical-align: -18px">y'=\frac{\cos\theta\sin\theta+(1+\sin\theta)\cos\theta}{\cos^2\theta-(1+\sin\theta)\sin\theta}.</math>
+&nbsp; &nbsp; &nbsp; &nbsp;<math style="vertical-align: -18px">y'=\frac{\cos\theta\sin\theta+(1+\sin\theta)\cos\theta}{\cos^2\theta-(1+\sin\theta)\sin\theta}.</math>
 |}
@@ Line 57: / Line 57: @@
 |-
 |Thus, we have
-::<math>\begin{array}{rcl}
+&nbsp; &nbsp; &nbsp; &nbsp;<math>\begin{array}{rcl}
 \displaystyle{y'} & = & \displaystyle{\frac{2\cos\theta\sin\theta+\cos\theta}{\cos^2\theta-\sin^2\theta-\sin\theta}}\\
 &&\\
@@ Line 76: / Line 76: @@
 |-
+|
-::<math>\begin{array}{rcl}
+&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)}\\
 &&\\
@@ Line 95: / Line 95: @@
 |-
+|
-::<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{3-3\sin\theta\cos(2\theta)+3\sin(2\theta)\cos\theta}{(\cos(2\theta)-\sin\theta)^3}.</math>
 |-
+|

@@ Line 1: / Line 1: @@
 <span class="exam">A curve is given in polar coordinates by
-::::::<math>r=1+\sin\theta</math>
+::<math>r=1+\sin\theta</math>
-::<span class="exam">a) Sketch the curve.
+<span class="exam">(a) Sketch the curve.
-::<span class="exam">b) Compute <math style="vertical-align: -12px">y'=\frac{dy}{dx}</math>.
+<span class="exam">(b) Compute <math style="vertical-align: -12px">y'=\frac{dy}{dx}</math>.
-::<span class="exam">c) Compute <math style="vertical-align: -12px">y''=\frac{d^2y}{dx^2}</math>.
+<span class="exam">(c) Compute <math style="vertical-align: -12px">y''=\frac{d^2y}{dx^2}</math>.
 {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
@@ Line 14: / Line 14: @@
 |-
+|
-::Since <math style="vertical-align: -5px">x=r\cos(\theta),~y=r\sin(\theta),</math> we have
+&nbsp; &nbsp; &nbsp; &nbsp;Since <math style="vertical-align: -5px">x=r\cos(\theta),~y=r\sin(\theta),</math> we have
 |-
+|
-::<math>y'=\frac{dy}{dx}=\frac{\frac{dr}{d\theta}\sin\theta+r\cos\theta}{\frac{dr}{d\theta}\cos\theta-r\sin\theta}.</math>
+&nbsp; &nbsp; &nbsp; &nbsp;<math>y'=\frac{dy}{dx}=\frac{\frac{dr}{d\theta}\sin\theta+r\cos\theta}{\frac{dr}{d\theta}\cos\theta-r\sin\theta}.</math>
 |}
 '''Solution:'''
@@ Line 98: / Line 99: @@
+|
 |}
 {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 !Final Answer: &nbsp;
 |-
-|&nbsp;&nbsp; '''(a)''' See Step 1 above for the graph.
+|&nbsp;&nbsp; '''(a)''' &nbsp; &nbsp; See Step 1 above for the graph.
 |-
-|&nbsp;&nbsp; '''(b)''' <math>\frac{\sin(2\theta)+\cos\theta}{\cos(2\theta)-\sin\theta}</math>
+|&nbsp;&nbsp; '''(b)''' &nbsp; &nbsp; <math>\frac{\sin(2\theta)+\cos\theta}{\cos(2\theta)-\sin\theta}</math>
 |-
-|&nbsp;&nbsp; '''(c)''' <math>\frac{3-3\sin\theta\cos(2\theta)+3\sin(2\theta)\cos\theta}{(\cos(2\theta)-\sin\theta)^3}</math>
+|&nbsp;&nbsp; '''(c)''' &nbsp; &nbsp; <math>\frac{3-3\sin\theta\cos(2\theta)+3\sin(2\theta)\cos\theta}{(\cos(2\theta)-\sin\theta)^3}</math>
 |}
 [[009C_Sample_Final_1|'''<u>Return to Sample Exam</u>''']]