Difference between revisions of "009C Sample Final 1, Problem 7"
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::::::<math>r=1+\sin\theta</math> | ::::::<math>r=1+\sin\theta</math> | ||
| − | <span class="exam">a) Sketch the curve. | + | ::<span class="exam">a) Sketch the curve. |
| − | <span class="exam">b) Compute <math style="vertical-align: -12px">y'=\frac{dy}{dx}</math>. | + | ::<span class="exam">b) Compute <math style="vertical-align: -12px">y'=\frac{dy}{dx}</math>. |
| − | <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>. |
{| class="mw-collapsible mw-collapsed" style = "text-align:left;" | {| class="mw-collapsible mw-collapsed" style = "text-align:left;" | ||
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::<math>\frac{dr}{d\theta}=\cos\theta.</math> | ::<math>\frac{dr}{d\theta}=\cos\theta.</math> | ||
|- | |- | ||
| − | |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> | + | |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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!Final Answer: | !Final Answer: | ||
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| − | |'''(a)''' See Step 1 above for the graph. | + | | '''(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> |
|- | |- | ||
| − | |'''(c)''' <math>\frac{3-3\sin\theta\cos(2\theta)+3\sin(2\theta)\cos\theta}{(\cos(2\theta)-\sin\theta)^3}</math> | + | | '''(c)''' <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>''']] | [[009C_Sample_Final_1|'''<u>Return to Sample Exam</u>''']] | ||
Revision as of 18:38, 18 April 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: |
|---|
| 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)?} |
|
|
Solution:
(a)
| Step 1: |
|---|
| Insert sketch of graph |
(b)
| Step 1: |
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| First, recall we have |
|
| 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,} |
|
| Hence, |
|
| Step 2: |
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Thus, we have
|
(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 |
|
| 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 |
|
| 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}} |