MathAdmin at 17:04, 12 March 2017

2017-03-12T17:04:59Z

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

2017-03-12T16:57:19Z

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

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A curve is given in polar coordinates by
::<math>r=\sin(2\theta).</math>

(a) Sketch the curve.

(b) Compute  <math style="vertical-align: -14px">y'=\frac{dy}{dx}.</math>

(c) Compute  <math style="vertical-align: -14px">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:'''

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

'''(b)'''

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
|-
|Since  <math style="vertical-align: -5px">r=\sin(2\theta),</math>
|-
|
       <math>\frac{dr}{d\theta}=2\cos(2\theta).</math>
|-
|
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
|-
|Since
|-
|
       <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>
|-
|we have
|-
|
        <math>\begin{array}{rcl}
\displaystyle{y'} & = & \displaystyle{\frac{2\cos(2\theta)\sin\theta+\sin(2\theta)\cos\theta}{2\cos(2\theta)\cos \theta-\sin(2\theta)\sin\theta}}\\
&&\\
& = & \displaystyle{\frac{2\cos^2\theta \sin\theta-\sin^3\theta}{\cos^3\theta-2\sin^2\theta\cos\theta}}
\end{array}</math>
|-
|since
|-
|       <math>\sin(2\theta)=2\sin\theta\cos\theta,~\cos(2\theta)=\cos^2\theta-\sin^2\theta.</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{2\cos^2\theta \sin\theta-\sin^3\theta}{\cos^3\theta-2\sin^2\theta\cos\theta}\bigg)}\\
&&\\
& = & \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>
|}

{| 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{(\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>
|-
|
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Final Answer:  
|-
|    '''(a)'''    See above
|-
|    '''(b)'''    <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>
|-
|    '''(c)'''    
|-
|<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>
|}
[[009C_Sample_Final_2|'''Return to Sample Exam''']]

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 !(a) &nbsp;
 |-
-|Insert sketch of graph
+|&nbsp;
 |}

009C Sample Final 2, Problem 9 - Revision history

MathAdmin at 17:04, 12 March 2017

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

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