Difference between revisions of "009C Sample Final 3, Problem 7"
Kayla Murray (talk | contribs) |
Kayla Murray (talk | contribs) |
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\end{array}</math> | \end{array}</math> | ||
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| − | |So, <math>\theta=\frac{\pi}{4}.</math> | + | |So, <math style="vertical-align: -15px">\theta=\frac{\pi}{4}.</math> |
|- | |- | ||
| − | |Now, this point in polar is <math>\bigg(1,\frac{\pi}{4}\bigg).</math> | + | |Now, this point in polar is <math style="vertical-align: -15px">\bigg(1,\frac{\pi}{4}\bigg).</math> |
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!Step 2: | !Step 2: | ||
|- | |- | ||
| − | |Now, we plug in <math>\theta=\frac{\pi}{4}</math> into our polar equation. | + | |Now, we plug in <math style="vertical-align: -15px">\theta=\frac{\pi}{4}</math> into our polar equation. |
|- | |- | ||
|We get | |We get | ||
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\end{array}</math> | \end{array}</math> | ||
|- | |- | ||
| − | |So, the point <math>(x,y)</math> belongs to the curve. | + | |So, the point <math style="vertical-align: -5px">(x,y)</math> belongs to the curve. |
|} | |} | ||
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!Step 2: | !Step 2: | ||
|- | |- | ||
| − | |Now, recall from part (a) that the given point in polar coordinates is <math>\bigg(1,\frac{\pi}{4}\bigg).</math> | + | |Now, recall from part (a) that the given point in polar coordinates is <math style="vertical-align: -15px">\bigg(1,\frac{\pi}{4}\bigg).</math> |
|- | |- | ||
|Therefore, the slope of the tangent line at this point is | |Therefore, the slope of the tangent line at this point is | ||
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\end{array}</math> | \end{array}</math> | ||
|- | |- | ||
| − | |Therefore, the equation of the tangent line at the point <math>(x,y)</math> is | + | |Therefore, the equation of the tangent line at the point <math style="vertical-align: -5px">(x,y)</math> is |
|- | |- | ||
| <math>y=-1\bigg(x-\frac{\sqrt{2}}{2}\bigg)+\frac{\sqrt{2}}{2}.</math> | | <math>y=-1\bigg(x-\frac{\sqrt{2}}{2}\bigg)+\frac{\sqrt{2}}{2}.</math> | ||
Revision as of 15:24, 5 March 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+\cos^2(2\theta)}
(a) Show that the point with Cartesian coordinates 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,y)=\bigg(\frac{\sqrt{2}}{2},\frac{\sqrt{2}}{2}\bigg)} belongs to the curve.
(b) Sketch the curve.
(c) In Cartesian coordinates, find the equation of the tangent line at 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 \bigg(\frac{\sqrt{2}}{2},\frac{\sqrt{2}}{2}\bigg).}
| Foundations: |
|---|
| 1. What two pieces of information do you need to write the equation of a line? |
|
You need the slope of the line and a point on the line. |
| 2. 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)
| Step 1: |
|---|
| First, we need to convert this Cartesian point into polar. |
| 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{r} & = & \displaystyle{\sqrt{x^2+y^2}}\\ &&\\ & = & \displaystyle{\sqrt{\frac{2}{4}+\frac{2}{4}}}\\ &&\\ & = & \displaystyle{\sqrt{1}}\\ &&\\ & = & \displaystyle{1.} \end{array}} |
| Also, 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{\tan \theta } & = & \displaystyle{\frac{y}{x}}\\ &&\\ & = & \displaystyle{1.} \end{array}} |
| So, 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 \theta=\frac{\pi}{4}.} |
| Now, this point in polar is 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 \bigg(1,\frac{\pi}{4}\bigg).} |
| Step 2: |
|---|
| Now, we plug in 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 \theta=\frac{\pi}{4}} into our polar equation. |
| 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 \begin{array}{rcl} \displaystyle{r} & = & \displaystyle{1+\cos^2\bigg(\frac{2\pi}{4}\bigg)}\\ &&\\ & = & \displaystyle{1+(0)^2}\\ &&\\ & = & \displaystyle{1.} \end{array}} |
| So, the point 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,y)} belongs to the curve. |
| (b) |
|---|
| Insert graph |
(c)
| 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=1+\cos^2(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}=-4\cos(2\theta)\sin(2\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 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{\frac{dy}{dx}} & = & \displaystyle{\frac{-4\cos(2\theta)\sin(2\theta)\sin\theta+(1+\cos^2(2\theta))\cos\theta}{-4\cos(2\theta)\sin(2\theta)\cos\theta-(1+\cos^2(2\theta))\sin\theta}.}\\ \end{array}} |
| Step 2: |
|---|
| Now, recall from part (a) that the given point in polar coordinates is 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 \bigg(1,\frac{\pi}{4}\bigg).} |
| Therefore, the slope of the tangent line at this point is |
| 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{m} & = & \displaystyle{\frac{-4\cos(\frac{\pi}{2})\sin(\frac{\pi}{2})\sin(\frac{\pi}{4})+(1+\cos^2(\frac{\pi}{2}))\cos(\frac{\pi}{4})}{-4\cos(\frac{\pi}{2})\sin(\frac{\pi}{2})\cos(\frac{\pi}{4})-(1+\cos^2(\frac{\pi}{2}))\sin(\frac{\pi}{4})}}\\ &&\\ & = & \displaystyle{\frac{0+(1)(\frac{\sqrt{2}}{2})}{0-(1)(\frac{\sqrt{2}}{2})}}\\ &&\\ & = & \displaystyle{-1.} \end{array}} |
| Therefore, the equation of the tangent line at the point 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,y)} is |
| 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=-1\bigg(x-\frac{\sqrt{2}}{2}\bigg)+\frac{\sqrt{2}}{2}.} |
| Final Answer: |
|---|
| (a) See above. |
| (b) See above. |
| (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 y=-1\bigg(x-\frac{\sqrt{2}}{2}\bigg)+\frac{\sqrt{2}}{2}} |