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

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::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&thinsp; <math style="vertical-align: -13px">y+x\frac{dy}{dx}</math>&thinsp; 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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|-
 
|
 
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::You need the slope of the line and a point on the line.
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&nbsp; &nbsp; &nbsp; &nbsp; You need the slope of the line and a point on the line.
 
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|'''3.''' What is the slope of the tangent line of a curve?
 
|'''3.''' What is the slope of the tangent line of a curve?
 
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|-
 
|
 
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::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&thinsp; <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&thinsp; <math style="vertical-align: -4px">x^3+y^3=6xy,</math> 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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::<math>3x^2+3y^2\frac{dy}{dx}=6y+6x\frac{dy}{dx}.</math>
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&nbsp; &nbsp; &nbsp; &nbsp; <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 &thinsp;<math style="vertical-align: -12px">\frac{dy}{dx}</math>&thinsp; terms to one side of the equation.
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|Now, we move all the &nbsp;<math style="vertical-align: -12px">\frac{dy}{dx}</math>&nbsp; terms to one side of the equation.
 
|-
 
|-
 
|So, we have
 
|So, we have
 
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|-
 
|
 
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::<math>3x^2-6y=\frac{dy}{dx}(6x-3y^2).</math>
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&nbsp; &nbsp; &nbsp; &nbsp;<math>3x^2-6y=\frac{dy}{dx}(6x-3y^2).</math>
 
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|We solve to get &nbsp;<math style="vertical-align: -17px">\frac{dy}{dx}=\frac{3x^2-6y}{6x-3y^2}.</math>
 
|We solve to get &nbsp;<math style="vertical-align: -17px">\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 &thinsp;<math style="vertical-align: -5px">(3,3).</math>
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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>
 
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|We plug  <math style="vertical-align: -5px">(3,3)</math>&thinsp; into the formula for &thinsp;<math style="vertical-align: -12px">\frac{dy}{dx}</math>&thinsp; we found in part '''(a)'''.
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|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).
 
|-
 
|-
 
|So, we get
 
|So, we get
 
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|-
 
|
 
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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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&nbsp; &nbsp; &nbsp; &nbsp;<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 style="vertical-align: -5px">(3,3)</math>&thinsp; and a point.  
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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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|-
 
|Thus, we can write the equation of the line.
 
|Thus, we can write the equation of the line.
 
|-
 
|-
|So, the equation of the tangent line at <math style="vertical-align: -5px">(3,3)</math>&thinsp; is  
+
|So, the equation of the tangent line at <math style="vertical-align: -5px">(3,3)</math>&nbsp; is  
 
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|-
 
|
 
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::<math>y\,=\,-1(x-3)+3.</math>
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&nbsp; &nbsp; &nbsp; &nbsp; <math>y\,=\,-1(x-3)+3.</math>
 
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!Final Answer: &nbsp;  
 
!Final Answer: &nbsp;  
 
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|'''(a)'''&thinsp; <math>\frac{dy}{dx}=\frac{3x^2-6y}{6x-3y^2}</math>
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|&nbsp; &nbsp; '''(a)'''&nbsp; &nbsp; <math>\frac{dy}{dx}=\frac{3x^2-6y}{6x-3y^2}</math>
 
|-
 
|-
|'''(b)'''&thinsp; <math>y=-1(x-3)+3</math>
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|&nbsp; &nbsp; '''(b)'''&nbsp; &nbsp; <math>y=-1(x-3)+3</math>
 
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|}
 
[[009A_Sample_Final_1|'''<u>Return to Sample Exam</u>''']]
 
[[009A_Sample_Final_1|'''<u>Return to Sample Exam</u>''']]

Revision as of 16:53, 25 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 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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