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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>t_0=\frac{\pi}{4}</math>. |
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| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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| − | !Foundations:
| + | [[009C Sample Final 1, Problem 10 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.''' What is the slope of the tangent line of a parametric curve? | |
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| − | The slope is <math style="vertical-align: -21px">m=\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}.</math>
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| − | '''Solution:''' | + | [[009C Sample Final 1, Problem 10 Detailed Solution|'''<u>Detailed Solution</u>''']] |
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| − | !(a)
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| − | |Insert sketch of curve
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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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| − | |First, we need to find the slope of the tangent line.
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| − | |Since <math style="vertical-align: -14px">\frac{dy}{dt}=-4\sin t</math> and <math style="vertical-align: -14px">\frac{dx}{dt}=3\cos t,</math> we have
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| − | <math>\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}=-\frac{4\sin t}{3\cos t}.</math>
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| − | |So, at <math>t_0=\frac{\pi}{4},</math> the slope of the tangent line is
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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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| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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| − | !Step 2:
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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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| − | |If we plug in <math>t_0=\frac{\pi}{4}</math> into the equations for <math style="vertical-align: -5px">x(t)</math> and <math style="vertical-align: -5px">y(t),</math> we get
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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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| − | <math>y\bigg(\frac{\pi}{4}\bigg)=4\cos\bigg(\frac{\pi}{4}\bigg)=2\sqrt{2}.</math>
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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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| − | |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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| − | <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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| − | | '''(a)''' See above.
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| − | | '''(b)''' <math style="vertical-align: -14px">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>''']] |
