Difference between revisions of "009C Sample Final 2, Problem 9"

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&nbsp; &nbsp; &nbsp; &nbsp;<math>\begin{array}{rcl}
 
&nbsp; &nbsp; &nbsp; &nbsp;<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(2\theta)\cos\theta}{2\cos(2\theta)\cos \theta-\sin(2\theta)\sin\theta}\bigg)}\\
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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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& = & \displaystyle{\frac{(2\cos(2\theta)\cos \theta-\sin(2\theta)\sin\theta)(-4\sin(2\theta)\sin \theta +2\cos(2\theta)\cos(theta)+2\cos(2\theta)\cos\theta-\sin(2\theta)\sin \theta)}{(2\cos(2\theta)\cos \theta-\sin(2\theta)\sin\theta)^2}}\\
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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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&&\\
 
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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}{(2\cos(2\theta)\cos \theta-\sin(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}{(2\cos(2\theta)\cos \theta-\sin(2\theta)\sin\theta)^2}}\\
 
 
\end{array}</math>
 
\end{array}</math>
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|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;
 
 
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&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>
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&nbsp; &nbsp; &nbsp; &nbsp;<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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|&nbsp; &nbsp; '''(c)''' &nbsp; &nbsp;
 
|&nbsp; &nbsp; '''(c)''' &nbsp; &nbsp;
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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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[[009C_Sample_Final_2|'''<u>Return to Sample Exam</u>''']]
 
[[009C_Sample_Final_2|'''<u>Return to Sample Exam</u>''']]

Revision as of 12:19, 5 March 2017

A curve is given in polar coordinates by

(a) Sketch the curve.

(b) Compute  

(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}.}

Foundations:  
How do you calculate   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'}   for a polar curve  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=f(\theta)?}

       Since   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 x=r\cos(\theta),~y=r\sin(\theta),}   we have

       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}=\frac{\frac{dr}{d\theta}\sin\theta+r\cos\theta}{\frac{dr}{d\theta}\cos\theta-r\sin\theta}.}


Solution:

(a)  
Insert sketch of graph

(b)

Step 1:  
Since  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=\sin(2\theta),}

       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 \frac{dr}{d\theta}=2\cos(2\theta).}

Step 2:  
Since

       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}=\frac{\frac{dr}{d\theta}\sin\theta+r\cos\theta}{\frac{dr}{d\theta}\cos\theta-r\sin\theta},}

we have

        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 \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}}

since
       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 \sin(2\theta)=2\sin\theta\cos\theta,~\cos(2\theta)=\cos^2\theta-\sin^2\theta.}

(c)

Step 1:  
We have   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 \frac{d^2y}{dx^2}=\frac{\frac{dy'}{d\theta}}{\frac{dr}{d\theta}\cos\theta-r\sin\theta}.}
So, first we need to find   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 \frac{dy'}{d\theta}.}
We have

       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 \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}}

Step 2:  
Now, using the resulting formula for   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 \frac{dy'}{d\theta},}   we get

       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 \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)}.}


Final Answer:  
    (a)    See above
    (b)    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{2\cos^2\theta \sin\theta-\sin^3\theta}{\cos^3\theta-2\sin^2\theta\cos\theta}}
    (c)    
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 \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)}}

Return to Sample Exam

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