Difference between revisions of "009C Sample Final 1, Problem 7"

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<span class="exam">A curve is given in polar coordinates by  
 
<span class="exam">A curve is given in polar coordinates by  
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::::::<math>r=1+\sin\theta</math>
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::<math>r=1+\sin\theta</math>
  
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<span class="exam">a) Sketch the curve.
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<span class="exam">(a) Sketch the curve.
  
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<span class="exam">b) Compute <math>y'=\frac{dy}{dx}</math>.
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<span class="exam">(b) Compute &nbsp;<math style="vertical-align: -12px">y'=\frac{dy}{dx}</math>.
  
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<span class="exam">c) Compute <math>y''=\frac{d^2y}{dx^2}</math>.
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<span class="exam">(c) Compute &nbsp;<math style="vertical-align: -12px">y''=\frac{d^2y}{dx^2}</math>.
  
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<hr>
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!Foundations: &nbsp;
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[[009C Sample Final 1, Problem 7 Solution|'''<u>Solution</u>''']]
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'''Solution:'''
 
  
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'''(a)'''
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[[009C Sample Final 1, Problem 7 Detailed Solution|'''<u>Detailed Solution</u>''']]
  
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!Step 1: &nbsp;
 
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|Insert sketch of graph
 
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'''(b)'''
 
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!Step 1: &nbsp;
 
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|First, recall we have
 
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|<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>.
 
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|Since <math>r=1+\sin\theta</math>, <math>\frac{dr}{d\theta}=\cos\theta</math>.
 
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|Hence, <math>y'=\frac{\cos\theta\sin\theta+(1+\sin\theta)\cos\theta}{\cos^2\theta-(1+\sin\theta)\sin\theta}</math>
 
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!Step 2: &nbsp;
 
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|Thus, we have
 
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::<math>\begin{array}{rcl}
 
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\displaystyle{y'} & = & \displaystyle{\frac{2\cos\theta\sin\theta+\cos\theta}{\cos^2\theta-\sin^2\theta-\sin\theta}}\\
 
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&&\\
 
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& = & \displaystyle{\frac{\sin(2\theta)+\cos\theta}{\cos(2\theta)-\sin\theta}}\\
 
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\end{array}</math>
 
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'''(c)'''
 
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!Step 1: &nbsp;
 
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{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 
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!Step 2: &nbsp;
 
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!Final Answer: &nbsp;
 
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|'''(a)''' See '''(a)''' above for the graph.
 
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|'''(b)''' <math>y'=\frac{\sin(2\theta)+\cos\theta}{\cos(2\theta)-\sin\theta}</math>
 
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|'''(c)'''
 
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[[009C_Sample_Final_1|'''<u>Return to Sample Exam</u>''']]
 
[[009C_Sample_Final_1|'''<u>Return to Sample Exam</u>''']]

Latest revision as of 14:23, 3 December 2017

A curve is given in polar coordinates by

(a) Sketch the curve.

(b) Compute  .

(c) Compute  .

Solution


Detailed Solution


Return to Sample Exam

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