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>
  
<span class="exam">(a) Using implicit differentiation, compute &thinsp;<math style="vertical-align: -12px">\frac{dy}{dx}</math>.
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<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 style="vertical-align: -4px">x^3+y^3=6xy</math> at the point <math style="vertical-align: -5px">(3,3)</math>.
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<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>.
  
 
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 
!Foundations: &nbsp;  
 
!Foundations: &nbsp;  
 
|-
 
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|'''1.''' What is the result of implicit differentiation of <math style="vertical-align: -4px">xy?</math>
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|'''1.''' What is the result of implicit differentiation of &nbsp;<math style="vertical-align: -4px">xy?</math>
 
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&nbsp; &nbsp; &nbsp; &nbsp; It would be&thinsp; <math style="vertical-align: -13px">y+x\frac{dy}{dx}</math>&thinsp; by the Product Rule.
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&nbsp; &nbsp; &nbsp; &nbsp; It would be &nbsp;<math style="vertical-align: -13px">y+x\frac{dy}{dx}</math>&nbsp; by the Product Rule.
 
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|'''2.''' What two pieces of information do you need to write the equation of a line?
 
|'''2.''' What two pieces of information do you need to write the equation of a line?
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&nbsp; &nbsp; &nbsp; &nbsp; The slope is&thinsp; <math style="vertical-align: -13px">m=\frac{dy}{dx}.</math>
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&nbsp; &nbsp; &nbsp; &nbsp; The slope is &nbsp;<math style="vertical-align: -13px">m=\frac{dy}{dx}.</math>
 
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!Step 1: &nbsp;  
 
!Step 1: &nbsp;  
 
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|Using implicit differentiation on the equation&nbsp; <math style="vertical-align: -4px">x^3+y^3=6xy,</math> &nbsp; we get
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|Using implicit differentiation on the equation &nbsp;<math style="vertical-align: -4px">x^3+y^3=6xy,</math>&nbsp; we get
 
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|First, we find the slope of the tangent line at the point &nbsp;<math style="vertical-align: -5px">(3,3).</math>
 
|First, we find the slope of the tangent line at the point &nbsp;<math style="vertical-align: -5px">(3,3).</math>
 
|-
 
|-
|We plug  <math style="vertical-align: -5px">(3,3)</math>&thinsp; into the formula for &nbsp;<math style="vertical-align: -12px">\frac{dy}{dx}</math>&nbsp; we found in part (a).
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|We plug  &nbsp;<math style="vertical-align: -5px">(3,3)</math>&nbsp; into the formula for &nbsp;<math style="vertical-align: -12px">\frac{dy}{dx}</math>&nbsp; we found in part (a).
 
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|So, we get
 
|So, we get
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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 style="vertical-align: -5px">(3,3)</math>&nbsp; and a point.  
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|Now, we have the slope of the tangent line at &nbsp;<math style="vertical-align: -5px">(3,3)</math>&nbsp; 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 style="vertical-align: -5px">(3,3)</math>&nbsp; is  
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|So, the equation of the tangent line at &nbsp;<math style="vertical-align: -5px">(3,3)</math>&nbsp; is  
 
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Revision as of 08:29, 27 February 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  .

Foundations:  
1. What is the result of implicit differentiation of  

        It would be    by the Product Rule.

2. What two pieces of information do you need to write the equation of a line?

        You need the slope of the line and a point on the line.

3. What is the slope of the tangent line of a curve?

        The slope is  


Solution:

(a)

Step 1:  
Using implicit differentiation on the equation    we get

       

Step 2:  
Now, we move all the    terms to one side of the equation.
So, we have

       

We solve to get  

(b)

Step 1:  
First, we find the slope of the tangent line at the point  
We plug    into the formula for    we found in part (a).
So, we get

       

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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