MathAdmin: Created page with "A curve is given in polar coordinates by :: $r=1+\cos^2(2\theta)$ (a) Show that the point with Cartesian coordinates <..."

2017-03-19T19:26:36Z

Created page with "A curve is given in polar coordinates by ::<math>r=1+\cos^2(2\theta)</math> (a) Show that the point with Cartesian coordinates <..."

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A curve is given in polar coordinates by

::<math>r=1+\cos^2(2\theta)</math>

(a) Show that the point with Cartesian coordinates  <math style="vertical-align: -15px">(x,y)=\bigg(\frac{\sqrt{2}}{2},\frac{\sqrt{2}}{2}\bigg)</math>  belongs to the curve.

(b) Sketch the curve.

(c) In Cartesian coordinates, find the equation of the tangent line at  <math style="vertical-align: -15px">\bigg(\frac{\sqrt{2}}{2},\frac{\sqrt{2}}{2}\bigg).</math>

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!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   <math style="vertical-align: -5px">y'</math>   for a polar curve  <math style="vertical-align: -5px">r=f(\theta)?</math>
|-
|
       Since   <math style="vertical-align: -5px">x=r\cos(\theta),~y=r\sin(\theta),</math>  we have
|-
|
       <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>
|}

'''Solution:'''

'''(a)'''

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
|-
|First, we need to convert this Cartesian point into polar.
|-
|We have
|-
|        <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
|-
|        <math>\begin{array}{rcl}
\displaystyle{\tan \theta } & = & \displaystyle{\frac{y}{x}}\\
&&\\
& = & \displaystyle{1.}
\end{array}</math>
|-
|So,  <math style="vertical-align: -15px">\theta=\frac{\pi}{4}.</math>
|-
|Now, this point in polar is  <math style="vertical-align: -15px">\bigg(1,\frac{\pi}{4}\bigg).</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
|-
|Now, we plug in  <math style="vertical-align: -15px">\theta=\frac{\pi}{4}</math>  into our polar equation.
|-
|We get
|-
|        <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  <math style="vertical-align: -5px">(x,y)</math>  belongs to the curve.
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!(b)  
|-
| 
|-
|[[File:9CSF3_7.jpg |center|441px]]
|}
'''(c)'''

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
|-
|Since  <math style="vertical-align: -5px">r=1+\cos^2(2\theta),</math>
|-
|
       <math>\frac{dr}{d\theta}=-4\cos(2\theta)\sin(2\theta).</math>
|-
|Since
|-
|
       <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
|-
|
        <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>
|-
|
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
|-
|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
|-
|        <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  <math style="vertical-align: -5px">(x,y)</math>  is
|-
|       <math>y=-1\bigg(x-\frac{\sqrt{2}}{2}\bigg)+\frac{\sqrt{2}}{2}.</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Final Answer:  
|-
|    '''(a)'''     See above.
|-
|    '''(b)'''     See above.
|-
|    '''(c)'''     <math>y=-1\bigg(x-\frac{\sqrt{2}}{2}\bigg)+\frac{\sqrt{2}}{2}</math>
|}
[[009C_Sample_Final_3|'''Return to Sample Exam''']]

009C Sample Final 3, Problem 7 - Revision history

MathAdmin: Created page with "A curve is given in polar coordinates by ::r=1+\cos^2(2\theta) (a) Show that the point with Cartesian coordinates <..."

MathAdmin: Created page with "A curve is given in polar coordinates by :: $r=1+\cos^2(2\theta)$ (a) Show that the point with Cartesian coordinates <..."