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

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|<math>3x^2-6y=\frac{dy}{dx}(6x-3y^2)</math>.
 
|<math>3x^2-6y=\frac{dy}{dx}(6x-3y^2)</math>.
 
|-
 
|-
|We solve for <math>\frac{dy}{dx}=\frac{3x^2-6y}{6x-3y^2}</math>.
+
|We solve to get <math>\frac{dy}{dx}=\frac{3x^2-6y}{6x-3y^2}</math>.
 
|}
 
|}
  
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!Step 1: &nbsp;  
 
!Step 1: &nbsp;  
 
|-
 
|-
|
+
|First, we find the slope of the tangent line at the point <math>(3,3)</math>.
 
|-
 
|-
|
+
|We plug in <math>(3,3)</math> into the formula for <math>\frac{dy}{dx}</math> we found in part '''(a)'''.
 
|-
 
|-
|
+
|So, we get
 +
|-
 +
|<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;
 
|-
 
|-
|
+
|Now, we have the slope of the tangent line at <math>(3,3)</math> and a point.
|}
+
|-
 
+
|Thus, we can write the equation of the line.
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 
!Step 3: &nbsp;
 
 
|-
 
|-
|
+
|So, the equation of the tangent line at <math>(3,3)</math> is
 
|-
 
|-
|
+
|<math>y=-1(x-3)+3</math>.
 
|}
 
|}
  
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!Final Answer: &nbsp;  
 
!Final Answer: &nbsp;  
 
|-
 
|-
|'''(a)'''
+
|'''(a)''' <math>\frac{dy}{dx}=\frac{3x^2-6y}{6x-3y^2}</math>
 
|-
 
|-
|'''(b)'''  
+
|'''(b)''' <math>y=-1(x-3)+3</math>
 
|}
 
|}
 
[[009A_Sample_Final_1|'''<u>Return to Sample Exam</u>''']]
 
[[009A_Sample_Final_1|'''<u>Return to Sample Exam</u>''']]

Revision as of 17:04, 14 February 2016

A curve is defined implicityly by the equation

a) Using implicit differentiation, compute .

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

Foundations:  

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 in into the formula for we found in part (a).
So, we get
.
Step 2:  
Now, we have the slope of the tangent line at and a point.
Thus, we can write the equation of the line.
So, the equation of the tangent line at 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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