Difference between revisions of "009A Sample Final 2, Problem 4"

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::<span class="exam"><math style="vertical-align: -4px">3x^2+xy+y^2=5</math>&nbsp; at the point &nbsp;<math style="vertical-align: -5px">(1,-2)</math>
 
::<span class="exam"><math style="vertical-align: -4px">3x^2+xy+y^2=5</math>&nbsp; at the point &nbsp;<math style="vertical-align: -5px">(1,-2)</math>
  
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<hr>
!Foundations: &nbsp;
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[[009A Sample Final 2, Problem 4 Solution|'''<u>Solution</u>''']]
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|The equation of the tangent line to &nbsp;<math style="vertical-align: -5px">f(x)</math>&nbsp; at the point &nbsp;<math style="vertical-align: -5px">(a,b)</math>&nbsp; is
 
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|&nbsp; &nbsp; &nbsp; &nbsp; <math style="vertical-align: -5px">y=m(x-a)+b</math>&nbsp; where &nbsp;<math style="vertical-align: -5px">m=f'(a).</math>
 
|}
 
  
  
'''Solution:'''
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[[009A Sample Final 2, Problem 4 Detailed Solution|'''<u>Detailed Solution</u>''']]
  
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!Step 1: &nbsp;
 
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|We use implicit differentiation to find the derivative of the given curve.
 
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|Using the product and chain rule, we get
 
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|&nbsp; &nbsp; &nbsp; &nbsp; <math>6x+xy'+y+2yy'=5.</math>
 
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|We rearrange the terms and solve for &nbsp;<math style="vertical-align: -5px">y'.</math>
 
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|Therefore,
 
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|&nbsp; &nbsp; &nbsp; &nbsp; <math>xy'+2yy'=5-6x-y</math>
 
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|and
 
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|&nbsp; &nbsp; &nbsp; &nbsp; <math>y'=\frac{5-6x-y}{x+2y}.</math>
 
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!Step 2: &nbsp;
 
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|Therefore, the slope of the tangent line at the point &nbsp;<math style="vertical-align: -5px">(1,-2)</math>&nbsp; is
 
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|&nbsp; &nbsp; &nbsp; &nbsp; <math>\begin{array}{rcl}
 
\displaystyle{m} & = & \displaystyle{\frac{5-6(1)-(-2)}{1-4}}\\
 
&&\\
 
& = & \displaystyle{\frac{1}{-3}.}
 
\end{array}</math>
 
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|Hence, the equation of the tangent line to the curve at the point &nbsp;<math style="vertical-align: -5px">(1,-2)</math>&nbsp; is
 
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|&nbsp; &nbsp; &nbsp; &nbsp;<math>f(x)=-\frac{1}{3}(x-1)-2.</math>
 
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|
 
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!Final Answer: &nbsp;
 
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|&nbsp; &nbsp; &nbsp; &nbsp;<math>f(x)=-\frac{1}{3}(x-1)-2.</math>
 
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[[009A_Sample_Final_2|'''<u>Return to Sample Exam</u>''']]
 
[[009A_Sample_Final_2|'''<u>Return to Sample Exam</u>''']]

Latest revision as of 10:13, 1 December 2017

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

  at the point  

Solution


Detailed Solution


Return to Sample Exam