Difference between revisions of "009C Sample Final 3, Problem 7"

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Line 34: Line 34:
 
!Step 1:    
 
!Step 1:    
 
|-
 
|-
−
|
+
|First, we need to convert this Cartesian point into polar.
 +
|-
 +
|We have
 +
|-
 +
|&nbsp; &nbsp; &nbsp; &nbsp; <math>\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}</math>
 +
|-
 +
|Also, we have
 +
|-
 +
|&nbsp; &nbsp; &nbsp; &nbsp; <math>\begin{array}{rcl}
 +
\displaystyle{\tan \theta } & = & \displaystyle{\frac{y}{x}}\\
 +
&&\\
 +
& = & \displaystyle{1.}
 +
\end{array}</math>
 
|-
 
|-
−
|
+
|So, &nbsp;<math>\theta=\frac{\pi}{4}.</math>
 
|-
 
|-
−
|
+
|Now, this point in polar is &nbsp;<math>\bigg(1,\frac{\pi}{4}\bigg).</math>
 
|}
 
|}
  
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!Step 2: &nbsp;
 
!Step 2: &nbsp;
 
|-
 
|-
−
|
+
|Now, we plug in &nbsp;<math>\theta=\frac{\pi}{4}</math>&nbsp; into our polar equation.
 +
|-
 +
|We get
 +
|-
 +
|&nbsp; &nbsp; &nbsp; &nbsp; <math>\begin{array}{rcl}
 +
\displaystyle{r} & = & \displaystyle{1+\cos^2\bigg(\frac{2\pi}{4}\bigg)}\\
 +
&&\\
 +
& = & \displaystyle{1+(0)^2}\\
 +
&&\\
 +
& = & \displaystyle{1.}
 +
\end{array}</math>
 
|-
 
|-
−
|
+
|So, the point &nbsp;<math>(x,y)</math>&nbsp; belongs to the curve.
 
|}
 
|}
  
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{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 
!Step 1: &nbsp;  
 
!Step 1: &nbsp;  
 +
|-
 +
|Since &nbsp;<math style="vertical-align: -5px">r=1+\cos^2(2\theta),</math>
 
|-
 
|-
 
|
 
|
 +
&nbsp; &nbsp; &nbsp; &nbsp;<math>\frac{dr}{d\theta}=-4\cos(2\theta)\sin(2\theta).</math>
 +
|-
 +
|Since
 +
|-
 +
|
 +
&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>
 +
|-
 +
|we have
 
|-
 
|-
 
|
 
|
 +
&nbsp; &nbsp; &nbsp; &nbsp; <math>\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}</math>
 
|-
 
|-
 
|
 
|
Line 74: Line 117:
 
!Step 2: &nbsp;
 
!Step 2: &nbsp;
 
|-
 
|-
−
|
+
|Now, recall from part (a) that the given point in polar coordinates is &nbsp;<math>\bigg(1,\frac{\pi}{4}\bigg).</math>
 +
|-
 +
|Therefore, the slope of the tangent line at this point is
 +
|-
 +
|&nbsp; &nbsp; &nbsp; &nbsp; <math>\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}</math>
 +
|-
 +
|Therefore, the equation of the tangent line at the point &nbsp;<math>(x,y)</math>&nbsp; is
 
|-
 
|-
−
|
+
|&nbsp; &nbsp; &nbsp; &nbsp;<math>y=-1\bigg(x-\frac{\sqrt{2}}{2}\bigg)+\frac{\sqrt{2}}{2}.</math>
 
|}
 
|}
  
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!Final Answer: &nbsp;  
 
!Final Answer: &nbsp;  
 
|-
 
|-
−
|&nbsp;&nbsp; '''(a)'''  
+
|&nbsp; &nbsp; '''(a)''' &nbsp; &nbsp; See above.
 
|-
 
|-
−
|&nbsp;&nbsp; '''(b)'''  
+
|&nbsp; &nbsp; '''(b)''' &nbsp; &nbsp; See above.
 
|-
 
|-
−
|&nbsp;&nbsp; '''(c)'''  
+
|&nbsp; &nbsp; '''(c)''' &nbsp; &nbsp; <math>y=-1\bigg(x-\frac{\sqrt{2}}{2}\bigg)+\frac{\sqrt{2}}{2}</math>
 
|}
 
|}
 
[[009C_Sample_Final_3|'''<u>Return to Sample Exam</u>''']]
 
[[009C_Sample_Final_3|'''<u>Return to Sample Exam</u>''']]

Revision as of 16:21, 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    belongs to the curve.

(b) Sketch the curve.

(c) In Cartesian coordinates, find the equation of the tangent line at  

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}}

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