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

Revision as of 17:10, 2 December 2017

A curve is given in polar coordinates by

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle r=1+\sin\theta}

(a) Sketch the curve.

(b) Compute Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle y'=\frac{dy}{dx}} .

(c) Compute Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle y''=\frac{d^2y}{dx^2}} .

Solution

Detailed Solution

Return to Sample Exam

@@ Line 8: / Line 8: @@
 <span class="exam">(c) Compute &nbsp; <math style="vertical-align: -12px">y''=\frac{d^2y}{dx^2}</math>.
-{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
+<hr>
-!Foundations: &nbsp;
+[[009C Sample Final 1, Problem 7 Solution|'''<u>Solution</u>''']]
-|-
-|How do you calculate &nbsp; <math style="vertical-align: -5px">y'</math> &nbsp; for a polar curve &nbsp;<math style="vertical-align: -5px">r=f(\theta)?</math>
-|-
-|
-&nbsp; &nbsp; &nbsp; &nbsp;Since &nbsp; <math style="vertical-align: -5px">x=r\cos(\theta),~y=r\sin(\theta),</math>&nbsp; we have
-|-
-|
-&nbsp; &nbsp; &nbsp; &nbsp;<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:'''
+[[009C Sample Final 1, Problem 7 Detailed Solution|'''<u>Detailed Solution</u>''']]
-{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
-!(a) &nbsp;
-|-
-|Insert sketch of graph
-|}
-'''(b)'''
-{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
-!Step 1: &nbsp;
-|-
-|First, recall we have
-|-
-|
-&nbsp; &nbsp; &nbsp; &nbsp;<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 &nbsp;<math style="vertical-align: -4px">r=1+\sin\theta,</math>
-|-
-|
-&nbsp; &nbsp; &nbsp; &nbsp;<math>\frac{dr}{d\theta}=\cos\theta.</math>
-|-
-|Hence,
-|-
-|
-&nbsp; &nbsp; &nbsp; &nbsp;<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: &nbsp;
-|-
-|Thus, we have
-&nbsp; &nbsp; &nbsp; &nbsp;<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: &nbsp;
-|-
-|We have &nbsp; <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 &nbsp; <math>\frac{dy'}{d\theta}.</math>
-|-
-|We have
-|-
-|
-&nbsp; &nbsp; &nbsp; &nbsp;<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 &nbsp;<math style="vertical-align: -2px">\sin^2\theta+\cos^2\theta=1</math>&nbsp; and &nbsp;<math style="vertical-align: -5px">2\cos^2(2\theta)+2\sin^2(2\theta)=2.</math>&nbsp;
-|}
-{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
-!Step 2: &nbsp;
-|-
-| Now, using the resulting formula for &nbsp; <math>\frac{dy'}{d\theta},</math>&nbsp; we get
-|-
-|
-&nbsp; &nbsp; &nbsp; &nbsp;<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: &nbsp;
-|-
-|&nbsp;&nbsp; '''(a)''' &nbsp; &nbsp; See above.
-|-
-|&nbsp;&nbsp; '''(b)''' &nbsp; &nbsp; <math>\frac{\sin(2\theta)+\cos\theta}{\cos(2\theta)-\sin\theta}</math>
-|-
-|&nbsp;&nbsp; '''(c)''' &nbsp; &nbsp; <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|'''<u>Return to Sample Exam</u>''']]

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

Revision as of 17:10, 2 December 2017

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