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| | <span class="exam">(b) Find an equation of the tangent line to the curve <math style="vertical-align: -4px">x^3+y^3=6xy</math> at the point <math style="vertical-align: -5px">(3,3)</math>. | | <span class="exam">(b) Find an equation of the tangent line to the curve <math style="vertical-align: -4px">x^3+y^3=6xy</math> at the point <math style="vertical-align: -5px">(3,3)</math>. |
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| | + | [[009A Sample Final 1, Problem 7 Solution|'''<u>Solution</u>''']] |
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| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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| − | !Foundations:
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| − | |'''1.''' What is the result of implicit differentiation of <math style="vertical-align: -4px">xy?</math>
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| − | It would be <math style="vertical-align: -13px">y+x\frac{dy}{dx}</math> by the Product Rule.
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| − | |'''2.''' 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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| − | |'''3.''' What is the slope of the tangent line of a curve?
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| − | The slope is <math style="vertical-align: -13px">m=\frac{dy}{dx}.</math>
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| − | |}
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| | + | [[009A Sample Final 1, Problem 7 Detailed Solution|'''<u>Detailed Solution</u>''']] |
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| − | '''Solution:'''
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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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| − | |Using implicit differentiation on the equation <math style="vertical-align: -4px">x^3+y^3=6xy,</math> we get
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| − | <math>3x^2+3y^2\frac{dy}{dx}=6y+6x\frac{dy}{dx}.</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 move all the <math style="vertical-align: -12px">\frac{dy}{dx}</math> terms to one side of the equation.
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| − | |So, we have
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| − | <math>3x^2-6y=\frac{dy}{dx}(6x-3y^2).</math>
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| − | |We solve to get
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| − | | <math style="vertical-align: -17px">\frac{dy}{dx}=\frac{3x^2-6y}{6x-3y^2}.</math>
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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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| − | |First, we find the slope of the tangent line at the point <math style="vertical-align: -5px">(3,3).</math>
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| − | |We plug <math style="vertical-align: -5px">(3,3)</math> into the formula for <math style="vertical-align: -12px">\frac{dy}{dx}</math> we found in part (a).
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| − | |So, we get
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| − | <math>\begin{array}{rcl}
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| − | \displaystyle{m} & = & \displaystyle{\frac{3(3)^2-6(3)}{6(3)-3(3)^2}}\\
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| − | &&\\
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| − | & = & \displaystyle{-\frac{9}{9}}\\
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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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| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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| − | !Step 2:
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| − | |Now, we have the slope of the tangent line at <math style="vertical-align: -5px">(3,3)</math> and a point.
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| − | |Thus, we can write the equation of the line.
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| − | |So, the equation of the tangent line at <math style="vertical-align: -5px">(3,3)</math> is
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| − | <math>y\,=\,-1(x-3)+3.</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)''' <math>\frac{dy}{dx}=\frac{3x^2-6y}{6x-3y^2}</math>
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| − | | '''(b)''' <math>y=-1(x-3)+3</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>''']] |
