009A Sample Final 3, Problem 5 - Revision history

MathAdmin: Created page with " Calculate the equation of the tangent line to the curve defined by $x^3+y^3=2xy$ at the point,
2017-04-10T16:30:16Z

Created page with "<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, <m..."

New page
<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>

{| 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>
|}

'''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>''']]