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| | <span class="exam">(c) Compute <math style="vertical-align: -14px">y''=\frac{d^2y}{dx^2}.</math> | | <span class="exam">(c) Compute <math style="vertical-align: -14px">y''=\frac{d^2y}{dx^2}.</math> |
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| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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| − | !Foundations:
| + | [[009C Sample Final 2, Problem 9 Solution|'''<u>Solution</u>''']] |
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| − | |How do you calculate <math style="vertical-align: -5px">y'</math> for a polar curve <math style="vertical-align: -5px">r=f(\theta)?</math>
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| − | Since <math style="vertical-align: -5px">x=r\cos(\theta),~y=r\sin(\theta),</math> 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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| − | '''Solution:''' | + | [[009C Sample Final 2, Problem 9 Detailed Solution|'''<u>Detailed Solution</u>''']] |
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| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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| − | !(a)
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| − | |Insert sketch of graph
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| − | '''(b)'''
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| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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| − | !Step 1:
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| − | |Since <math style="vertical-align: -5px">r=\sin(2\theta),</math>
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| − | <math>\frac{dr}{d\theta}=2\cos(2\theta).</math>
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| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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| − | !Step 2:
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| − | |Since
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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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| − | |we have
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| − | <math>\begin{array}{rcl}
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| − | \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}}\\
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| − | &&\\
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| − | & = & \displaystyle{\frac{2\cos^2\theta \sin\theta-\sin^3\theta}{\cos^3\theta-2\sin^2\theta\cos\theta}}
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| − | \end{array}</math>
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| − | |since
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| − | | <math>\sin(2\theta)=2\sin\theta\cos\theta,~\cos(2\theta)=\cos^2\theta-\sin^2\theta.</math>
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| − | '''(c)'''
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| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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| − | !Step 1:
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| − | |We have <math>\frac{d^2y}{dx^2}=\frac{\frac{dy'}{d\theta}}{\frac{dr}{d\theta}\cos\theta-r\sin\theta}.</math>
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| − | |So, first we need to find <math>\frac{dy'}{d\theta}.</math>
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| − | |We have
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| − | <math>\begin{array}{rcl}
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| − | \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)}\\
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| − | &&\\
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| − | & = & \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}.}
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| − | \end{array}</math>
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| − | |}
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| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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| − | !Step 2:
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| − | |Now, using the resulting formula for <math>\frac{dy'}{d\theta},</math> we get
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| − | <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>
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| − | |}
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| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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| − | !Final Answer:
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| − | | '''(a)''' See above
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| − | | '''(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>
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| − | | '''(c)'''
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| − | |<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>
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| − | |}
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| | [[009C_Sample_Final_2|'''<u>Return to Sample Exam</u>''']] | | [[009C_Sample_Final_2|'''<u>Return to Sample Exam</u>''']] |
