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| | <span class="exam">(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> | | <span class="exam">(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> |
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| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
| + | <hr> |
| − | !Foundations:
| + | [[009C Sample Final 3, Problem 7 Solution|'''<u>Solution</u>''']] |
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| − | |'''1.''' What two pieces of information do you need to write the equation of a line? | |
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| − | You need the slope of the line and a point on the line.
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| − | |'''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>
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| − | Since <math style="vertical-align: -5px">x=r\cos(\theta),~y=r\sin(\theta),</math> 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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| − | |}
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| − | '''Solution:''' | + | [[009C Sample Final 3, Problem 7 Detailed Solution|'''<u>Detailed Solution</u>''']] |
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| − | '''(a)'''
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| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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| − | !Step 1:
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| − | |-
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| − | |First, we need to convert this Cartesian point into polar.
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| − | |We have
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| − | | <math>\begin{array}{rcl}
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| − | \displaystyle{r} & = & \displaystyle{\sqrt{x^2+y^2}}\\
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| − | &&\\
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| − | & = & \displaystyle{\sqrt{\frac{2}{4}+\frac{2}{4}}}\\
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| − | &&\\
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| − | & = & \displaystyle{\sqrt{1}}\\
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| − | &&\\
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| − | & = & \displaystyle{1.}
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| − | \end{array}</math>
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| − | |Also, we have
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| − | | <math>\begin{array}{rcl}
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| − | \displaystyle{\tan \theta } & = & \displaystyle{\frac{y}{x}}\\
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| − | &&\\
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| − | & = & \displaystyle{1.}
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| − | \end{array}</math>
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| − | |-
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| − | |So, <math style="vertical-align: -15px">\theta=\frac{\pi}{4}.</math>
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| − | |-
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| − | |Now, this point in polar is <math style="vertical-align: -15px">\bigg(1,\frac{\pi}{4}\bigg).</math>
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| − | |}
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| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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| − | !Step 2:
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| − | |Now, we plug in <math style="vertical-align: -15px">\theta=\frac{\pi}{4}</math> into our polar equation.
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| − | |We get
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| − | | <math>\begin{array}{rcl}
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| − | \displaystyle{r} & = & \displaystyle{1+\cos^2\bigg(\frac{2\pi}{4}\bigg)}\\
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| − | &&\\
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| − | & = & \displaystyle{1+(0)^2}\\
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| − | &&\\
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| − | & = & \displaystyle{1.}
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| − | \end{array}</math>
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| − | |-
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| − | |So, the point <math style="vertical-align: -5px">(x,y)</math> belongs to the curve.
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| − | |}
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| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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| − | !(b)
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| − | |Insert graph
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| − | |}
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| − | '''(c)'''
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| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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| − | !Step 1:
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| − | |Since <math style="vertical-align: -5px">r=1+\cos^2(2\theta),</math>
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| − | <math>\frac{dr}{d\theta}=-4\cos(2\theta)\sin(2\theta).</math>
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| − | |-
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| − | |Since
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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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| − | |-
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| − | |we have
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| − | <math>\begin{array}{rcl}
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| − | \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}.}\\
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| − | \end{array}</math>
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| − | |}
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| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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| − | !Step 2:
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| − | |-
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| − | |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>
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| − | |-
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| − | |Therefore, the slope of the tangent line at this point is
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| − | |-
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| − | | <math>\begin{array}{rcl}
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| − | \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})}}\\
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| − | &&\\
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| − | & = & \displaystyle{\frac{0+(1)(\frac{\sqrt{2}}{2})}{0-(1)(\frac{\sqrt{2}}{2})}}\\
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| − | &&\\
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| − | & = & \displaystyle{-1.}
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| − | \end{array}</math>
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| − | |Therefore, the equation of the tangent line at the point <math style="vertical-align: -5px">(x,y)</math> is
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| − | |-
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| − | | <math>y=-1\bigg(x-\frac{\sqrt{2}}{2}\bigg)+\frac{\sqrt{2}}{2}.</math>
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| − | |}
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| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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| − | !Final Answer:
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| − | | '''(a)''' See above.
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| − | | '''(b)''' See above.
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| − | | '''(c)''' <math>y=-1\bigg(x-\frac{\sqrt{2}}{2}\bigg)+\frac{\sqrt{2}}{2}</math>
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| − | |}
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| | [[009C_Sample_Final_3|'''<u>Return to Sample Exam</u>''']] | | [[009C_Sample_Final_3|'''<u>Return to Sample Exam</u>''']] |
