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| | <span class="exam">You may leave your answers in point-slope form. | | <span class="exam">You may leave your answers in point-slope form. |
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| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
| + | <hr> |
| − | !Foundations:
| + | [[009A Sample Final 1, Problem 4 Solution|'''<u>Solution</u>''']] |
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| − | |'''1.''' '''Chain Rule'''
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| − | | <math>\frac{d}{dx}(f(g(x)))=f'(g(x))g'(x)</math>
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| − | |'''2.''' | |
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| − | | <math>\frac{d}{dx}(\cos^{-1}(x))=\frac{-1}{\sqrt{1-x^2}}</math>
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| − | |'''3.''' The equation of the tangent line to <math style="vertical-align: -5px">f(x)</math> at the point <math style="vertical-align: -5px">(a,b)</math> is
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| − | | <math style="vertical-align: -5px">y=m(x-a)+b</math> where <math style="vertical-align: -5px">m=f'(a).</math>
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| − | |}
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| − | '''Solution:''' | + | [[009A Sample Final 1, Problem 4 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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| − | |First, we compute <math style="vertical-align: -13px">\frac{dy}{dx}.</math>
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| − | |Using the Chain Rule, we get
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| − | <math>\begin{array}{rcl}
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| − | \displaystyle{\frac{dy}{dx}} & = & \displaystyle{\frac{-1}{\sqrt{1-(2x)^2}}(2x)'}\\
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| − | &&\\
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| − | & = & \displaystyle{\frac{-2}{\sqrt{1-4x^2}}.}
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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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| − | |To find the equation of the tangent line, we first find the slope of the line.
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| − | |Using <math style="vertical-align: -14px">x_0=\frac{\sqrt{3}}{4}</math> in the formula for <math style="vertical-align: -12px">\frac{dy}{dx}</math> from Step 1, we get
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| − | <math>\begin{array}{rcl}
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| − | \displaystyle{m} & = & \displaystyle{\frac{-2}{\sqrt{1-4(\frac{\sqrt{3}}{4})^2}}}\\
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| − | &&\\
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| − | & = & \displaystyle{\frac{-2}{\sqrt{\frac{1}{4}}}}\\
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| − | &&\\
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| − | & = & \displaystyle{-4.}
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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 3:
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| − | |To get a point on the line, we plug in <math style="vertical-align: -14px">x_0=\frac{\sqrt{3}}{4}</math> into the equation given.
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| − | |So, we have
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| − | <math>\begin{array}{rcl}
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| − | \displaystyle{y_0} & = & \displaystyle{\cos^{-1}\bigg(2\frac{\sqrt{3}}{4}\bigg)}\\
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| − | &&\\
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| − | & = & \displaystyle{\cos^{-1}\bigg(\frac{\sqrt{3}}{2}\bigg)}\\
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| − | &&\\
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| − | & = & \displaystyle{\frac{\pi}{6}.}
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| − | \end{array}</math>
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| − | |Thus, the equation of the tangent line is <math style="vertical-align: -14px">y=-4\bigg(x-\frac{\sqrt{3}}{4}\bigg)+\frac{\pi}{6}.</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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| − | <math>\frac{dy}{dx}=\frac{-2}{\sqrt{1-4x^2}}</math>
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| − | <math>y=-4\bigg(x-\frac{\sqrt{3}}{4}\bigg)+\frac{\pi}{6}</math>
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| − | |}
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| | [[009A_Sample_Final_1|'''<u>Return to Sample Exam</u>''']] | | [[009A_Sample_Final_1|'''<u>Return to Sample Exam</u>''']] |
