MathAdmin at 01:37, 15 September 2017

2017-09-15T01:37:30Z

MathAdmin: Created page with "Use implicit differentiation to find an equation of the tangent line to the curve at the given point. ::
2017-04-10T16:22:37Z

Created page with "Use implicit differentiation to find an equation of the tangent line to the curve at the given point. ::<math style="vertical-align: -4p..."

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Use implicit differentiation to find an equation of the tangent line to the curve at the given point.

::<math style="vertical-align: -4px">3x^2+xy+y^2=5</math>  at the point  <math style="vertical-align: -5px">(1,-2)</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>6x+xy'+y+2yy'=5.</math>
|-
|We rearrange the terms and solve for  <math style="vertical-align: -5px">y'.</math>
|-
|Therefore,
|-
|        <math>xy'+2yy'=5-6x-y</math>
|-
|and
|-
|        <math>y'=\frac{5-6x-y}{x+2y}.</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,-2)</math>  is
|-
|        <math>\begin{array}{rcl}
\displaystyle{m} & = & \displaystyle{\frac{5-6(1)-(-2)}{1-4}}\\
&&\\
& = & \displaystyle{-\frac{1}{3}.}
\end{array}</math>
|-
|Hence, the equation of the tangent line to the curve at the point  <math style="vertical-align: -5px">(1,-2)</math>  is
|-
|       <math>y=-\frac{1}{3}(x-1)-2.</math>
|-
|
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Final Answer:  
|-
|       <math>y=-\frac{1}{3}(x-1)-2.</math>
|}
[[009A_Sample_Final_2|'''Return to Sample Exam''']]

@@ Line 21: / Line 21: @@
 |Using the product and chain rule, we get
 |-
-|&nbsp; &nbsp; &nbsp; &nbsp; <math>6x+xy'+y+2yy'=5.</math>
+|&nbsp; &nbsp; &nbsp; &nbsp; <math>6x+xy'+y+2yy'=0.</math>
 |-
 |We rearrange the terms and solve for &nbsp;<math style="vertical-align: -5px">y'.</math>
@@ Line 27: / Line 27: @@
 |Therefore,
 |-
-|&nbsp; &nbsp; &nbsp; &nbsp; <math>xy'+2yy'=5-6x-y</math>
+|&nbsp; &nbsp; &nbsp; &nbsp; <math>xy'+2yy'=-6x-y</math>
 |-
 |and
 |-
-|&nbsp; &nbsp; &nbsp; &nbsp; <math>y'=\frac{5-6x-y}{x+2y}.</math>
+|&nbsp; &nbsp; &nbsp; &nbsp; <math>y'=\frac{-6x-y}{x+2y}.</math>
 |}
@@ Line 40: / Line 40: @@
 |-
 |&nbsp; &nbsp; &nbsp; &nbsp; <math>\begin{array}{rcl}
-\displaystyle{m} & = & \displaystyle{\frac{5-6(1)-(-2)}{1-4}}\\
+\displaystyle{m} & = & \displaystyle{\frac{-6(1)-(-2)}{1-4}}\\
 &&\\
-& = & \displaystyle{-\frac{1}{3}.}
+& = & \displaystyle{\frac{4}{3}.}
 \end{array}</math>
 |-
 |Hence, the equation of the tangent line to the curve at the point &nbsp;<math style="vertical-align: -5px">(1,-2)</math>&nbsp; is
 |-
-|&nbsp; &nbsp; &nbsp; &nbsp;<math>y=-\frac{1}{3}(x-1)-2.</math>
+|&nbsp; &nbsp; &nbsp; &nbsp;<math>y=\frac{4}{3}(x-1)-2.</math>
 |-
+|
@@ Line 56: / Line 56: @@
 !Final Answer: &nbsp;
 |-
-|&nbsp; &nbsp; &nbsp; &nbsp;<math>y=-\frac{1}{3}(x-1)-2.</math>
+|&nbsp; &nbsp; &nbsp; &nbsp;<math>y=\frac{4}{3}(x-1)-2.</math>
 |}
 [[009A_Sample_Final_2|'''<u>Return to Sample Exam</u>''']]