|
|
| Line 1: |
Line 1: |
| | <span class="exam"> Calculate the equation of the tangent line to the curve defined by <math style="vertical-align: -4px">x^3+y^3=2xy</math> at the point, <math style="vertical-align: -5px">(1,1).</math> | | <span class="exam"> Calculate the equation of the tangent line to the curve defined by <math style="vertical-align: -4px">x^3+y^3=2xy</math> at the point, <math style="vertical-align: -5px">(1,1).</math> |
| | + | <hr> |
| | + | [[009A Sample Final 3, Problem 5 Solution|'''<u>Solution</u>''']] |
| | | | |
| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
| |
| − | !Foundations:
| |
| − | |-
| |
| − | |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
| |
| − | |-
| |
| − | | <math style="vertical-align: -5px">y=m(x-a)+b</math> where <math style="vertical-align: -5px">m=f'(a).</math>
| |
| − | |}
| |
| | | | |
| | + | [[009A Sample Final 3, Problem 5 Detailed Solution|'''<u>Detailed Solution</u>''']] |
| | | | |
| − | '''Solution:'''
| |
| | | | |
| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
| |
| − | !Step 1:
| |
| − | |-
| |
| − | |We use implicit differentiation to find the derivative of the given curve.
| |
| − | |-
| |
| − | |Using the product and chain rule, we get
| |
| − | |-
| |
| − | | <math>3x^2+3y^2y'=2y+2xy'.</math>
| |
| − | |-
| |
| − | |We rearrange the terms and solve for <math style="vertical-align: -5px">y'.</math>
| |
| − | |-
| |
| − | |Therefore,
| |
| − | |-
| |
| − | | <math>3x^2-2y=2xy'-3y^2y'</math>
| |
| − | |-
| |
| − | |and
| |
| − | |-
| |
| − | | <math>y'=\frac{3x^2-2y}{2x-3y^2}.</math>
| |
| − | |}
| |
| − |
| |
| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
| |
| − | !Step 2:
| |
| − | |-
| |
| − | |Therefore, the slope of the tangent line at the point <math style="vertical-align: -5px">(1,1)</math> is
| |
| − | |-
| |
| − | | <math>\begin{array}{rcl}
| |
| − | \displaystyle{m} & = & \displaystyle{\frac{3(1)^2-2(1)}{2(1)-3(1)^2}}\\
| |
| − | &&\\
| |
| − | & = & \displaystyle{\frac{3-2}{2-3}}\\
| |
| − | &&\\
| |
| − | & = & \displaystyle{-1.}
| |
| − | \end{array}</math>
| |
| − | |-
| |
| − | |Hence, the equation of the tangent line to the curve at the point <math style="vertical-align: -5px">(1,1)</math> is
| |
| − | |-
| |
| − | | <math>y=-1(x-1)+1.</math>
| |
| − | |}
| |
| − |
| |
| − |
| |
| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
| |
| − | !Final Answer:
| |
| − | |-
| |
| − | | <math>y=-1(x-1)+1</math>
| |
| − | |}
| |
| | [[009A_Sample_Final_3|'''<u>Return to Sample Exam</u>''']] | | [[009A_Sample_Final_3|'''<u>Return to Sample Exam</u>''']] |
