Difference between revisions of "009A Sample Final 1, Problem 8"

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<span class="exam">Let  
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<span class="exam">Let
  
::::::<math>y=x^3</math>
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::<math>y=x^3.</math>
  
<span class="exam">a) Find the differential <math>dy</math> of <math>y=x^3</math> at <math>x=2</math>.
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<span class="exam">(a) Find the differential &nbsp;<math style="vertical-align: -4px">dy</math>&nbsp; of &nbsp;<math style="vertical-align: -4px">y=x^3</math>&nbsp; at &nbsp;<math style="vertical-align: 0px">x=2</math>.
  
<span class="exam">b) Use differentials to find an approximate value for <math>1.9^3</math>.
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<span class="exam">(b) Use differentials to find an approximate value for &nbsp;<math style="vertical-align: -2px">1.9^3</math>.
  
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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<hr>
!Foundations: &nbsp;
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[[009A Sample Final 1, Problem 8 Solution|'''<u>Solution</u>''']]
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'''Solution:'''
 
  
'''(a)'''
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[[009A Sample Final 1, Problem 8 Detailed Solution|'''<u>Detailed Solution</u>''']]
  
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 
!Step 1: &nbsp;
 
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|First, we find the differential <math>dy</math>.
 
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|Since <math>y=x^3</math>, we have
 
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|<math>dy=3x^2dx</math>.
 
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{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 
!Step 2: &nbsp;
 
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|Now, we plug in <math>x=2</math> into the differential from Step 1.
 
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|So, we get
 
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|<math>dy=3(2)^2dx=12dx</math>.
 
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'''(b)'''
 
 
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 
!Step 1: &nbsp;
 
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{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 
!Step 2: &nbsp;
 
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{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 
!Step 3: &nbsp;
 
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{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 
!Final Answer: &nbsp;
 
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|'''(a)''' <math>dy=12dx</math>
 
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|'''(b)''' 
 
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[[009A_Sample_Final_1|'''<u>Return to Sample Exam</u>''']]
 
[[009A_Sample_Final_1|'''<u>Return to Sample Exam</u>''']]

Latest revision as of 15:57, 2 December 2017

Let

(a) Find the differential    of    at  .

(b) Use differentials to find an approximate value for  .


Solution


Detailed Solution


Return to Sample Exam