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| | <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>. |
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| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
| + | <hr> |
| − | !Foundations:
| + | [[009C Sample Final 1, Problem 7 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 1, Problem 7 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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| − | |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 style="vertical-align: -4px">r=1+\sin\theta,</math>
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| − | <math>\frac{dr}{d\theta}=\cos\theta.</math>
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| − | |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>
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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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| − | |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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| − | |}
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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{\sin(2\theta)+\cos\theta}{\cos(2\theta)-\sin\theta}\bigg)}\\
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| − | &&\\
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| − | & = & \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}}\\
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| − | &&\\
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| − | & = & \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}}\\
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| − | &&\\
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| − | & = & \displaystyle{\frac{3-3\sin\theta\cos(2\theta)+3\sin(2\theta)\cos\theta}{(\cos(2\theta)-\sin\theta)^2}}\\
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| − | \end{array}</math>
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| − | |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>
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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{3-3\sin\theta\cos(2\theta)+3\sin(2\theta)\cos\theta}{(\cos(2\theta)-\sin\theta)^3}.</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>\frac{\sin(2\theta)+\cos\theta}{\cos(2\theta)-\sin\theta}</math>
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| − | | '''(c)''' <math>\frac{3-3\sin\theta\cos(2\theta)+3\sin(2\theta)\cos\theta}{(\cos(2\theta)-\sin\theta)^3}</math>
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| − | |}
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| | [[009C_Sample_Final_1|'''<u>Return to Sample Exam</u>''']] | | [[009C_Sample_Final_1|'''<u>Return to Sample Exam</u>''']] |
