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

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!Step 1:    
 
!Step 1:    
 
|-
 
|-
|
+
|First, we find <math>dx</math>. We have <math>dx=1.9-2=-0.1</math>.
 
|-
 
|-
|
+
|Then, we plug this into the differential from part '''(a)'''.
 
|-
 
|-
|
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|So, we have <math>dy=12(-0.1)=-1.2</math>.
 
|}
 
|}
  
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!Step 2: &nbsp;
 
!Step 2: &nbsp;
 
|-
 
|-
|
+
|Now, we add the value for <math>dy</math> to <math>2^3</math> to get an
|}
+
|-
 
+
|approximate value of <math>1.9^3</math>.
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 
!Step 3: &nbsp;
 
 
|-
 
|-
|
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|Hence, we have
 
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|<math>1.9^3\approx 2^3+-1.2=6.8</math>.
 
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|'''(a)''' <math>dy=12dx</math>
 
|'''(a)''' <math>dy=12dx</math>
 
|-
 
|-
|'''(b)'''   
+
|'''(b)''' <math>6.8</math>  
 
|}
 
|}
 
[[009A_Sample_Final_1|'''<u>Return to Sample Exam</u>''']]
 
[[009A_Sample_Final_1|'''<u>Return to Sample Exam</u>''']]

Revision as of 10:09, 15 February 2016

Let

a) Find the differential of at .

b) Use differentials to find an approximate value for .

Foundations:  

Solution:

(a)

Step 1:  
First, we find the differential .
Since , we have
.
Step 2:  
Now, we plug in into the differential from Step 1.
So, we get
.

(b)

Step 1:  
First, we find . We have .
Then, we plug this into the differential from part (a).
So, we have .
Step 2:  
Now, we add the value for to to get an
approximate value of .
Hence, we have
.
Final Answer:  
(a)
(b)

Return to Sample Exam