Difference between revisions of "022 Sample Final A, Problem 4"

From Grad Wiki
Jump to navigation Jump to search
Line 8: Line 8:
 
|When we use implicit differentiation, we combine the chain rule with the fact that <math style="vertical-align: -18%">y</math> is a function of <math style="vertical-align: 0%">x</math>, and could really be written as <math style="vertical-align: -21%">y(x).</math> Because of this, the derivative of <math style="vertical-align:-17%">y^3</math> with respect to <math style="vertical-align: 0%">x</math> requires the chain rule, so  
 
|When we use implicit differentiation, we combine the chain rule with the fact that <math style="vertical-align: -18%">y</math> is a function of <math style="vertical-align: 0%">x</math>, and could really be written as <math style="vertical-align: -21%">y(x).</math> Because of this, the derivative of <math style="vertical-align:-17%">y^3</math> with respect to <math style="vertical-align: 0%">x</math> requires the chain rule, so  
 
|-
 
|-
−
|&nbsp;&nbsp;&nbsp;&nbsp; <math>\frac{d}{dx}\left(y^{3}\right)=3y^{2}\cdot\frac{dy}{dt}.</math>
+
|&nbsp;&nbsp;&nbsp;&nbsp; <math>\frac{d}{dx}\left(y^{3}\right)=3y^{2}\cdot\frac{dy}{dx}.</math>
 +
|-
 +
|For this problem we also need to use the product rule.
 
|}
 
|}
  
−
&nbsp;'''Solution:'''
+
'''Solution:'''
  
 
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 
!Step 1: &nbsp;
 
!Step 1: &nbsp;
 
|-
 
|-
−
|First, we differentiate each term separately with respect to <math style="vertical-align: 0%">x</math> to find that&thinsp; <math style="vertical-align: -18%">x^{3}-y^{3}-y=x</math> &thinsp;differentiates implicitly to
+
|First, we differentiate each term separately with respect to <math style="vertical-align: 0%">x</math> and apply the product rule on the right hand side to find that&thinsp; <math style="vertical-align: -18%">x + y = x^3y^3</math> &thinsp;differentiates implicitly to
 
|-
 
|-
−
|&nbsp;&nbsp;&nbsp;&nbsp; <math>3x^{2}-3y^{2}\cdot\frac{dy}{dx}-\frac{dy}{dx}=1</math>.
+
|
 +
::<math>1 + \frac{dy}{dx} = 3x^2y^3 + 3x^3y^2 \cdot \frac{dy}{dx}</math>.
 
|}
 
|}
 +
 
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 
!Step 2: &nbsp;
 
!Step 2: &nbsp;
 
|-
 
|-
−
|Since they don't ask for a general expression of <math style="vertical-align:-24%">dy/dx</math>, but rather a particular value at a particular point, we can plug in the values <math style="vertical-align:-5%">x=1</math> and <math style="vertical-align:-18%">y=0</math> &thinsp;to find
+
|Now we need to solve for <math>\frac{dy}{dx}</math> and doing so we find that <math>\frac{dy}dx} = \frac{3x^2y^3 - 1}{1 - 3x^3y^2}</math>
−
|-
 
−
|&nbsp;&nbsp;&nbsp;&nbsp; <math>3(1)^{2}-3(0)^{2}\cdot\frac{dy}{dx}-\frac{dy}{dx}=1,</math>
 
−
|-
 
−
|which is equivalent to <math style="vertical-align:-60%">3-\frac{dy}{dx}=1</math>. This solves to <math style="vertical-align:-24%">dy/dx=2.</math>
 
 
|}
 
|}
  
Line 33: Line 33:
 
!Final Answer: &nbsp;
 
!Final Answer: &nbsp;
 
|-
 
|-
−
|&nbsp;&nbsp;&nbsp;&nbsp; <math style="vertical-align:-24%">dy/dx=2.</math>
+
|<math>\frac{dy}dx} = \frac{3x^2y^3 - 1}{1 - 3x^3y^2}</math>
 
|}
 
|}
 
[[022_Exam_1_Sample_A|'''<u>Return to Sample Exam</u>''']]
 
[[022_Exam_1_Sample_A|'''<u>Return to Sample Exam</u>''']]

Revision as of 13:16, 30 May 2015


Use implicit differentiation to find

Foundations:  
When we use implicit differentiation, we combine the chain rule with the fact that 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} is a function of 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} , and could really be written as 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(x).} Because of this, the derivative of 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^3} with respect to 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} requires the chain rule, so
     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{d}{dx}\left(y^{3}\right)=3y^{2}\cdot\frac{dy}{dx}.}
For this problem we also need to use the product rule.

Solution:

Step 1:  
First, we differentiate each term separately with respect to 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} and apply the product rule on the right hand side to find that  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 + y = x^3y^3}  differentiates implicitly to
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 1 + \frac{dy}{dx} = 3x^2y^3 + 3x^3y^2 \cdot \frac{dy}{dx}} .
Step 2:  
Now we need to solve 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}} and doing so we find that 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^2y^3 - 1}{1 - 3x^3y^2}}
Final Answer:  
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^2y^3 - 1}{1 - 3x^3y^2}}

Return to Sample Exam

Retrieved from "https://gradwiki.math.ucr.edu/index.php?title=022_Sample_Final_A,_Problem_4&oldid=846"