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

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|First, recall we have
 
|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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|
 +
::<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>,  
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|Since <math style="vertical-align: -2px">r=1+\sin\theta</math>,  
 
|-
 
|-
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|<math>\frac{dr}{d\theta}=\cos\theta</math>.
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|
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::<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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|Hence, <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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\end{array}</math>
 
\end{array}</math>
 
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|-
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|since <math>\sin^2\theta+\cos^2\theta=1</math> and <math>2\cos^2(2\theta)+2\sin^2(2\theta)=2</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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| Now, using the resulting formula for <math>\frac{dy'}{d\theta}</math>, we get  
 
| 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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|
 +
::<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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!Final Answer: &nbsp;  
 
!Final Answer: &nbsp;  
 
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|'''(a)''' See '''(a)''' above for the graph.
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|'''(a)''' See Step 1 above for the graph.
 
|-
 
|-
 
|'''(b)''' <math>\frac{\sin(2\theta)+\cos\theta}{\cos(2\theta)-\sin\theta}</math>
 
|'''(b)''' <math>\frac{\sin(2\theta)+\cos\theta}{\cos(2\theta)-\sin\theta}</math>

Revision as of 13:47, 22 February 2016

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}} .

Foundations:  
Review derivatives in polar coordinates

Solution:

(a)

Step 1:  
Insert sketch of graph


(b)

Step 1:  
First, recall 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}} .
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=1+\sin\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}=\cos\theta} .
Hence, 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{\cos\theta\sin\theta+(1+\sin\theta)\cos\theta}{\cos^2\theta-(1+\sin\theta)\sin\theta}}
Step 2:  
Thus, 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\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}}

(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{\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}}
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+\cos^2\theta=1} and 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 2\cos^2(2\theta)+2\sin^2(2\theta)=2} .
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{3-3\sin\theta\cos(2\theta)+3\sin(2\theta)\cos\theta}{(\cos(2\theta)-\sin\theta)^3}} .
Final Answer:  
(a) See Step 1 above for the graph.
(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 \frac{\sin(2\theta)+\cos\theta}{\cos(2\theta)-\sin\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{3-3\sin\theta\cos(2\theta)+3\sin(2\theta)\cos\theta}{(\cos(2\theta)-\sin\theta)^3}}

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