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

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!Step 1:    
 
!Step 1:    
 
|-
 
|-
|First, we find <math style="vertical-align: -1px">dx</math>. We have <math style="vertical-align: -1px">dx=1.9-2=-0.1.</math>
+
|First, we find <math style="vertical-align: 0px">dx</math>. We have &thinsp;<math style="vertical-align: -1px">dx=1.9-2=-0.1.</math>
 
|-
 
|-
 
|Then, we plug this into the differential from part '''(a)'''.
 
|Then, we plug this into the differential from part '''(a)'''.
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|-
 
|-
 
|
 
|
::<math>dy=12(-0.1)=-1.2.</math>
+
::<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 style="vertical-align: -4px">dy</math> to <math style="vertical-align: 0px">2^3</math> to get an
+
|Now, we add the value for <math style="vertical-align: -4px">dy</math> to &thinsp;<math style="vertical-align: 0px">2^3</math>&thinsp; to get an
 
|-
 
|-
 
|approximate value of <math style="vertical-align: -1px">1.9^3.</math>
 
|approximate value of <math style="vertical-align: -1px">1.9^3.</math>
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|-
 
|-
 
|
 
|
::<math>1.9^3\approx 2^3+-1.2=6.8.</math>
+
::<math>1.9^3\,\approx \, 2^3+-1.2\,=\,6.8.</math>
 
|}
 
|}
 +
 
== 4 ==
 
== 4 ==
 
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"

Revision as of 13:49, 4 March 2016

Let

a) Find the differential of at .

b) Use differentials to find an approximate value for .

1

Foundations:  
What is the differential of at
Since    the differential is  

Solution:

2

(a)

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

3

(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

4

Final Answer:  
(a)
(b)

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