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

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::::::<math>x^3+y^3=6xy</math>
 
::::::<math>x^3+y^3=6xy</math>
  
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<span class="exam">a) Using implicit differentiation, compute <math>\frac{dy}{dx}</math>.
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<span class="exam">a) Using implicit differentiation, compute <math style="vertical-align: -12px">\frac{dy}{dx}</math>.
  
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<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>.
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<span class="exam">b) Find an equation of the tangent line to the curve <math style="vertical-align: -4px">x^3+y^3=6xy</math> at the point <math style="vertical-align: -4px">(3,3)</math>.
  
 
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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!Step 1: &nbsp;  
 
!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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|Using implicit differentiation on the equation <math style="vertical-align: -4px">x^3+y^3=6xy</math>, we get
 
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|-
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|<math>3x^2+3y^2\frac{dy}{dx}=6y+6x\frac{dy}{dx}</math>.
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|
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::<math>3x^2+3y^2\frac{dy}{dx}=6y+6x\frac{dy}{dx}</math>.
 
|}
 
|}
  
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!Step 2: &nbsp;
 
!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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|Now, we move all the <math style="vertical-align: -12px">\frac{dy}{dx}</math> terms to one side of the equation.
 
|-
 
|-
 
|So, we have
 
|So, we have
 
|-
 
|-
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|<math>3x^2-6y=\frac{dy}{dx}(6x-3y^2)</math>.
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|
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::<math>3x^2-6y=\frac{dy}{dx}(6x-3y^2)</math>.
 
|-
 
|-
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|We solve to get <math>\frac{dy}{dx}=\frac{3x^2-6y}{6x-3y^2}</math>.
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|We solve to get <math style="vertical-align: -12px">\frac{dy}{dx}=\frac{3x^2-6y}{6x-3y^2}</math>.
 
|}
 
|}
  
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!Step 1: &nbsp;  
 
!Step 1: &nbsp;  
 
|-
 
|-
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|First, we find the slope of the tangent line at the point <math>(3,3)</math>.
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|First, we find the slope of the tangent line at the point <math style="vertical-align: -4px">(3,3)</math>.
 
|-
 
|-
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|We plug in <math>(3,3)</math> into the formula for <math>\frac{dy}{dx}</math> we found in part '''(a)'''.
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|We plug in <math style="vertical-align: -4px">(3,3)</math> into the formula for <math style="vertical-align: -12px">\frac{dy}{dx}</math> we found in part '''(a)'''.
 
|-
 
|-
 
|So, we get
 
|So, we get
 
|-
 
|-
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|<math>m=\frac{3(3)^2-6(3)}{6(3)-3(3)^2}=\frac{9}{-9}=-1</math>.
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|
 +
::<math>m=\frac{3(3)^2-6(3)}{6(3)-3(3)^2}=\frac{9}{-9}=-1</math>.
 
|}
 
|}
  
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!Step 2: &nbsp;
 
!Step 2: &nbsp;
 
|-
 
|-
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|Now, we have the slope of the tangent line at <math>(3,3)</math> and a point.  
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|Now, we have the slope of the tangent line at <math style="vertical-align: -4px">(3,3)</math> and a point.  
 
|-
 
|-
 
|Thus, we can write the equation of the line.
 
|Thus, we can write the equation of the line.
 
|-
 
|-
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|So, the equation of the tangent line at <math>(3,3)</math> is  
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|So, the equation of the tangent line at <math style="vertical-align: -4px">(3,3)</math> is  
 
|-
 
|-
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|<math>y=-1(x-3)+3</math>.
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|
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::<math>y=-1(x-3)+3</math>.
 
|}
 
|}
  

Revision as of 14:28, 22 February 2016

A curve is defined implicityly by the equation

a) Using implicit differentiation, compute 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 \frac{dy}{dx}} .

b) Find an equation of the tangent line to the curve 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+y^3=6xy} 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 (3,3)} .

Foundations:  

Solution:

(a)

Step 1:  
Using implicit differentiation on the equation 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+y^3=6xy} , 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 3x^2+3y^2\frac{dy}{dx}=6y+6x\frac{dy}{dx}} .
Step 2:  
Now, we move all the 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 \frac{dy}{dx}} terms to one side of the equation.
So, we have
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^2-6y=\frac{dy}{dx}(6x-3y^2)} .
We solve to 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 \frac{dy}{dx}=\frac{3x^2-6y}{6x-3y^2}} .

(b)

Step 1:  
First, we find 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 (3,3)} .
We plug in 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 (3,3)} into the formula 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 \frac{dy}{dx}} we found in part (a).
So, 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 m=\frac{3(3)^2-6(3)}{6(3)-3(3)^2}=\frac{9}{-9}=-1} .
Step 2:  
Now, we have the slope of the tangent line at 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 (3,3)} and a point.
Thus, we can write the equation of the line.
So, the equation of the tangent line at 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 (3,3)} 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(x-3)+3} .
Final Answer:  
(a) 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 \frac{dy}{dx}=\frac{3x^2-6y}{6x-3y^2}}
(b) 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(x-3)+3}

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