MathAdmin at 20:02, 13 May 2016

2016-05-13T20:02:01Z

MathAdmin: Created page with "A curve is given in polar coordinates by :::::: $r=1+\sin\theta$ ::a) Sketch the curve. ::b) Compute
2016-04-19T01:26:29Z

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

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

::a) Sketch the curve.

::b) Compute <math style="vertical-align: -12px">y'=\frac{dy}{dx}</math>.

::c) Compute <math style="vertical-align: -12px">y''=\frac{d^2y}{dx^2}</math>.

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Foundations:  
|-
|How do you calculate <math style="vertical-align: -5px">y'</math> for a polar curve <math style="vertical-align: -5px">r=f(\theta)?</math>
|-
|
::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>
|}

'''Solution:'''

'''(a)'''

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

'''(b)'''

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
|-
|First, recall 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>
|-
|Since <math style="vertical-align: -2px">r=1+\sin\theta,</math>
|-
|
::<math>\frac{dr}{d\theta}=\cos\theta.</math>
|-
|Hence,
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::<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>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
|-
|Thus, we have
::<math>\begin{array}{rcl}
\displaystyle{y'} & = & \displaystyle{\frac{2\cos\theta\sin\theta+\cos\theta}{\cos^2\theta-\sin^2\theta-\sin\theta}}\\
&&\\
& = & \displaystyle{\frac{\sin(2\theta)+\cos\theta}{\cos(2\theta)-\sin\theta}.}\\
\end{array}</math>
|}

'''(c)'''

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
|-
|We have <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>
|-
|We have
|-
|
::<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{(\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(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{3-3\sin\theta\cos(2\theta)+3\sin(2\theta)\cos\theta}{(\cos(2\theta)-\sin\theta)^2}}\\
\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>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
|-
| Now, using the resulting formula for <math>\frac{dy'}{d\theta},</math> we get
|-
|
::<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>
|-
|
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Final Answer:  
|-
|   '''(a)''' See Step 1 above for the graph.
|-
|   '''(b)''' <math>\frac{\sin(2\theta)+\cos\theta}{\cos(2\theta)-\sin\theta}</math>
|-
|   '''(c)''' <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|'''Return to Sample Exam''']]

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 !Step 1: &nbsp;
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-|Insert sketch of graph
+|[[File:009C_SF1_7_GP.jpg|center|400px]]
 |}