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>
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!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>
 
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'''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}
 
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\displaystyle{m} & = & \displaystyle{\frac{5-6(1)-(-2)}{1-4}}\\
 
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&&\\
 
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& = & \displaystyle{-\frac{1}{3}.}
 
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\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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!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.

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 3x^2+xy+y^2=5}   at the point  Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle (1,-2)}

Solution


Detailed Solution


Return to Sample Exam

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