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

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!Step 1:    
 
!Step 1:    
 
|-
 
|-
−
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|We use implicit differentiation to find the derivative of the given curve.
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|-
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|Using the product and chain rule, we get
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|-
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|&nbsp; &nbsp; &nbsp; &nbsp; <math>6x+xy'+y+2yy'=5.</math>
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|-
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|We rearrange the terms and solve for &nbsp;<math style="vertical-align: -5px">y'.</math>
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|-
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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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|-
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|&nbsp; &nbsp; &nbsp; &nbsp; <math>y'=\frac{5-6x-y}{x+2y}.</math>
 
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|}
  
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!Step 2: &nbsp;
 
!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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|-
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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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|-
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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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|-
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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;  
 
!Final Answer: &nbsp;  
 
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|&nbsp; &nbsp; &nbsp; &nbsp;<math>f(x)=-\frac{1}{3}(x-1)-2.</math>
 
|}
 
|}
 
[[009A_Sample_Final_2|'''<u>Return to Sample Exam</u>''']]
 
[[009A_Sample_Final_2|'''<u>Return to Sample Exam</u>''']]

Revision as of 17:03, 7 March 2017

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

  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)}
Foundations:  
The equation of the tangent line to  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 f(x)}   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 (a,b)}   is
        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 y=m(x-a)+b}   where  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 m=f'(a).}


Solution:

Step 1:  
We use implicit differentiation to find the derivative of the given curve.
Using the product and chain rule, we get
        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 6x+xy'+y+2yy'=5.}
We rearrange the terms and solve for  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 y'.}
Therefore,
        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 xy'+2yy'=5-6x-y}
and
        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 y'=\frac{5-6x-y}{x+2y}.}
Step 2:  
Therefore, the slope of the tangent line 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)}   is
        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 \begin{array}{rcl} \displaystyle{m} & = & \displaystyle{\frac{5-6(1)-(-2)}{1-4}}\\ &&\\ & = & \displaystyle{\frac{1}{-3}.} \end{array}}
Hence, the equation of the tangent line to the curve 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)}   is
       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 f(x)=-\frac{1}{3}(x-1)-2.}


Final Answer:  
       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 f(x)=-\frac{1}{3}(x-1)-2.}

Return to Sample Exam

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