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

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(Created page with "<br> <span style="font-size:135%"><font face=Times Roman> 4. Find an equation for the tangent line to the function  <math style="vertical-align: -13%;">-x^{3}-2xy+y^{3}=-...")
 
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|&nbsp;&nbsp;&nbsp;&nbsp; <math>2\cdot y+2x\cdot y' = 2y +2x\cdot y'.</math>
 
|&nbsp;&nbsp;&nbsp;&nbsp; <math>2\cdot y+2x\cdot y' = 2y +2x\cdot y'.</math>
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|We use implicit differentiation on our original equation to find
 
|We use implicit differentiation on our original equation to find
 
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|<br>&nbsp;&nbsp;&nbsp;&nbsp; <math>-3x^{2}-2y-2x\cdot y'+3y^{2}\cdot y'=0.</math>
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|&nbsp;&nbsp;&nbsp;&nbsp; <math>-3x^{2}-2y-2x\cdot y'+3y^{2}\cdot y'=0.</math>
 
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|<br>From here, I would immediately plug in (1,1) to find ''y'' ':
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|From here, I would immediately plug in (1,1) to find ''y'' ':
 
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|<br>&nbsp;&nbsp;&nbsp;&nbsp; <math style="vertical-align: -20%">-3-2-2y'+3y'=0</math>, or <math style="vertical-align: -20%">y' = 5.</math>
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|&nbsp;&nbsp;&nbsp;&nbsp; <math style="vertical-align: -20%">-3-2-2y'+3y'=0</math>, or <math style="vertical-align: -20%">y' = 5.</math>
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!Writing the Equation of the Tangent Line: &nbsp;
 
!Writing the Equation of the Tangent Line: &nbsp;
 
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|Now, we simply plug our values of ''x'' = ''y'' = 1 and ''m'' = 5 into the point-slope form to find the tangent line through (1,1) is <math style="vertical-align: -22%">y-1=5(x-1)</math>, or in slope-intercept form <math style="vertical-align: -20%">y=5x-4</math>.
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|Now, we simply plug our values of ''x'' = ''y'' = 1 and ''m'' = 5 into the point-slope form to find the tangent line through (1,1) is
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|&nbsp;&nbsp;&nbsp;&nbsp; <math>y-1=5(x-1),</math>
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|or in slope-intercept form
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|&nbsp;&nbsp;&nbsp;&nbsp; <math>y=5x-4.</math>
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[[009A_Sample_Final_A|'''<u>Return to Sample Exam</u>''']]
 
[[009A_Sample_Final_A|'''<u>Return to Sample Exam</u>''']]

Revision as of 10:19, 25 March 2015


4. Find an equation for the tangent line to the function  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 -x^{3}-2xy+y^{3}=-1} 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 (1,1)} .

Foundations:  
Since only two variables are present, we are going to differentiate everything with respect to x in order to find an expression for the slope, m = y ' = dy/dx. Then we can use the point-slope equation form 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-y_{1} = m(x-x_{1})} at the point to find the equation of the tangent line.
Note that implicit differentiation will require the product rule and the chain rule. In particular, differentiating 2xy must be treated as
     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 (2x)\cdot (y),}
which has as a derivative
     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 2\cdot y+2x\cdot y' = 2y +2x\cdot y'.}

Finding the slope:  
We use implicit differentiation on our original equation to find
     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}-2y-2x\cdot y'+3y^{2}\cdot y'=0.}
From here, I would immediately plug in (1,1) to find y ':
     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 -3-2-2y'+3y'=0} , or 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' = 5.}

Writing the Equation of the Tangent Line:  
Now, we simply plug our values of x = y = 1 and m = 5 into the point-slope form to find the tangent line through (1,1) 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-1=5(x-1),}
or in slope-intercept form
     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=5x-4.}

Return to Sample Exam

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