Difference between revisions of "009A Sample Final 1, Problem 7"

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<span class="exam">A curve is defined implicityly by the equation
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<span class="exam">A curve is defined implicitly by the equation
  
::::::<math>x^3+y^3=6xy</math>
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::<math>x^3+y^3=6xy.</math>
  
<span class="exam">a) Using implicit differentiation, compute <math>\frac{dy}{dx}</math>.
+
<span class="exam">(a) Using implicit differentiation, compute &nbsp;<math style="vertical-align: -12px">\frac{dy}{dx}</math>.
  
<span class="exam">b) Find an equation of the tangent line to the curve <math>x^3+y^3=6xy</math> at the point <math>(3,3)</math>.
+
<span class="exam">(b) Find an equation of the tangent line to the curve &nbsp;<math style="vertical-align: -4px">x^3+y^3=6xy</math>&nbsp; at the point &nbsp;<math style="vertical-align: -5px">(3,3)</math>.
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<hr>
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[[009A Sample Final 1, Problem 7 Solution|'''<u>Solution</u>''']]
  
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!Foundations: &nbsp;
 
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'''Solution:'''
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[[009A Sample Final 1, Problem 7 Detailed Solution|'''<u>Detailed Solution</u>''']]
  
'''(a)'''
 
  
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 
!Step 1: &nbsp;
 
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|Using implicit differentiation on the equation <math>x^3+y^3=6xy</math>, we get
 
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|<math>3x^2+3y^2\frac{dy}{dx}=6y+6x\frac{dy}{dx}</math>.
 
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{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 
!Step 2: &nbsp;
 
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|Now, we move all the <math>\frac{dy}{dx}</math> terms to one side of the equation.
 
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|So, we have
 
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|<math>3x^2-6y=\frac{dy}{dx}(6x-3y^2)</math>.
 
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|We solve for <math>\frac{dy}{dx}=\frac{3x^2-6y}{6x-3y^2}</math>.
 
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'''(b)'''
 
 
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!Step 1: &nbsp;
 
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!Step 2: &nbsp;
 
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{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 
!Step 3: &nbsp;
 
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!Final Answer: &nbsp;
 
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|'''(a)'''
 
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|'''(b)'''
 
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[[009A_Sample_Final_1|'''<u>Return to Sample Exam</u>''']]
 
[[009A_Sample_Final_1|'''<u>Return to Sample Exam</u>''']]

Latest revision as of 15:55, 2 December 2017

A curve is defined implicitly by the equation

(a) Using implicit differentiation, compute  .

(b) Find an equation of the tangent line to the curve    at the point  .


Solution


Detailed Solution


Return to Sample Exam