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| | <span class="exam">A curve is given in polar parametrically by | | <span class="exam">A curve is given in polar parametrically by |
| − | ::::::<span class="exam"><math>x(t)=3\sin t</math>
| + | ::<span class="exam"><math>x(t)=3\sin t</math> |
| − | ::::::<span class="exam"><math>y(t)=4\cos t</math>
| + | ::<span class="exam"><math>y(t)=4\cos t</math> |
| − | ::::::<span class="exam"><math>0\leq t \leq 2\pi</math>
| + | ::<span class="exam"><math>0\leq t \leq 2\pi</math> |
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| − | <span class="exam">a) Sketch the curve. | + | <span class="exam">(a) Sketch the curve. |
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| − | <span class="exam">b) Compute the equation of the tangent line at <math>t_0=\frac{\pi}{4}</math>. | + | <span class="exam">(b) Compute the equation of the tangent line at <math style="vertical-align: -14px">t_0=\frac{\pi}{4}</math>. |
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| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
| + | <hr> |
| − | !Foundations:
| + | [[009C Sample Final 1, Problem 10 Solution|'''<u>Solution</u>''']] |
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| − | |Review tangent lines of polar curves
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| − | |} | |
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| − | '''Solution:'''
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| − | '''(a)''' | + | [[009C Sample Final 1, Problem 10 Detailed Solution|'''<u>Detailed Solution</u>''']] |
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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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| − | |Insert sketch of curve
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| − | |}
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| − | '''(b)'''
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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 find the slope of the tangent line.
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| − | |-
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| − | |Since <math>\frac{dy}{dt}=-4\sin t</math> and <math>\frac{dx}{dt}=3\cos t</math>,
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| − | |-
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| − | |we have <math>\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}=\frac{-4\sin t}{3\cos t}</math>.
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| − | |-
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| − | |So, at <math>t_0=\frac{\pi}{4}</math>, the slope of the tangent line is
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| − | |-
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| − | |<math>m=\frac{-4\sin\bigg(\frac{\pi}{4}\bigg)}{3\cos\bigg(\frac{\pi}{4}\bigg)}=\frac{-4}{3}</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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| − | |Since we have the slope of the tangent line, we just need a find a point on the line in order to write the equation.
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| − | |-
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| − | |If we plug in <math>t_0=\frac{\pi}{4}</math> into the equations for <math>x(t)</math> and <math>y(t)</math>, we get
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| − | |-
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| − | |<math>x\bigg(\frac{\pi}{4}\bigg)=3\sin\bigg(\frac{\pi}{4}\bigg)=\frac{3\sqrt{2}}{2}</math> and
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| − | |-
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| − | |<math>y\bigg(\frac{\pi}{4}\bigg)=4\cos\bigg(\frac{\pi}{4}\bigg)=2\sqrt{2}</math>.
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| − | |-
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| − | |Thus, the point <math>\bigg(\frac{3\sqrt{2}}{2},2\sqrt{2}\bigg)</math> is on the tangent line.
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| − | |}
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| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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| − | !Step 3:
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| − | |-
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| − | |Using the point found in Step 2, the equation of the tangent line at <math>t_0=\frac{\pi}{4}</math> is
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| − | |-
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| − | |<math>y=\frac{-4}{3}\bigg(x-\frac{3\sqrt{2}}{2}\bigg)+2\sqrt{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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| − | |-
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| − | |'''(a)''' See '''(a)''' above for the graph.
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| − | |-
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| − | |'''(b)''' <math>y=\frac{-4}{3}\bigg(x-\frac{3\sqrt{2}}{2}\bigg)+2\sqrt{2}</math>
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| − | |}
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| | [[009C_Sample_Final_1|'''<u>Return to Sample Exam</u>''']] | | [[009C_Sample_Final_1|'''<u>Return to Sample Exam</u>''']] |
