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

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<span class="exam">Let   
 
<span class="exam">Let   
  
::::::<math>y=x^3.</math>
+
::<math>y=x^3.</math>
  
::<span class="exam">a) Find the differential <math style="vertical-align: -4px">dy</math> of <math style="vertical-align: -4px">y=x^3</math> at <math style="vertical-align: 0px">x=2</math>.
+
<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 style="vertical-align: -2px">1.9^3</math>.
+
<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;"
 
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 
!Foundations: &nbsp;  
 
!Foundations: &nbsp;  
 
|-
 
|-
|What is the differential <math style="vertical-align: -4px">dy</math> of <math style="vertical-align: -4px">y=x^2</math> at <math style="vertical-align: -1px">x=1?</math>
+
|What is the differential &nbsp;<math style="vertical-align: -4px">dy</math>&nbsp; of &nbsp;<math style="vertical-align: -4px">y=x^2</math>&nbsp; at &nbsp;<math style="vertical-align: -1px">x=1?</math>
 
|-
 
|-
 
|
 
|
::Since &thinsp;<math style="vertical-align: -4px">x=1,</math>&thinsp; the differential is &thinsp;<math style="vertical-align: -4px">dy=2xdx=2dx.</math>
+
&nbsp; &nbsp; &nbsp; &nbsp; Since &nbsp;<math style="vertical-align: -4px">x=1,</math>&nbsp; the differential is &nbsp;<math style="vertical-align: -4px">dy=2xdx=2dx.</math>
 
|}
 
|}
 +
  
 
'''Solution:'''
 
'''Solution:'''
 
  
 
'''(a)'''
 
'''(a)'''
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!Step 1: &nbsp;  
 
!Step 1: &nbsp;  
 
|-
 
|-
|First, we find the differential <math style="vertical-align: -4px">dy.</math>
+
|First, we find the differential &nbsp;<math style="vertical-align: -4px">dy.</math>
 
|-
 
|-
|Since <math style="vertical-align: -5px">y=x^3,</math>&thinsp; we have
+
|Since &nbsp;<math style="vertical-align: -5px">y=x^3,</math>&nbsp; we have
 
|-
 
|-
 
|
 
|
::<math>dy\,=\,3x^2\,dx.</math>
+
&nbsp; &nbsp; &nbsp; &nbsp;<math>dy\,=\,3x^2\,dx.</math>
 
|}
 
|}
  
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!Step 2: &nbsp;
 
!Step 2: &nbsp;
 
|-
 
|-
|Now, we plug <math style="vertical-align: 0px">x=2</math>&thinsp; into the differential from Step 1.
+
|Now, we plug &nbsp;<math style="vertical-align: 0px">x=2</math>&nbsp; into the differential from Step 1.
 
|-
 
|-
 
|So, we get  
 
|So, we get  
 
|-
 
|-
 
|
 
|
::<math>dy\,=\,3(2)^2\,dx\,=\,12\,dx.</math>
+
&nbsp; &nbsp; &nbsp; &nbsp;<math>dy\,=\,3(2)^2\,dx\,=\,12\,dx.</math>
 
|}
 
|}
  
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!Step 1: &nbsp;  
 
!Step 1: &nbsp;  
 
|-
 
|-
|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>
+
|First, we find &nbsp;<math style="vertical-align: 0px">dx.</math>&nbsp;  We have  
 +
|-
 +
|&nbsp; &nbsp; &nbsp; &nbsp;<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).
 
|-
 
|-
 
|So, we have  
 
|So, we have  
 
|-
 
|-
 
|
 
|
::<math>dy\,=\,12(-0.1)\,=\,-1.2.</math>
+
&nbsp; &nbsp; &nbsp; &nbsp;<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 &thinsp;<math style="vertical-align: 0px">2^3</math>&thinsp; to get an
+
|Now, we add the value for &nbsp;<math style="vertical-align: -4px">dy</math>&nbsp; to &nbsp;<math style="vertical-align: 0px">2^3</math>&nbsp; to get an
 
|-
 
|-
|approximate value of <math style="vertical-align: -1px">1.9^3.</math>
+
|approximate value of &nbsp;<math style="vertical-align: -1px">1.9^3.</math>
 
|-
 
|-
 
|Hence, we have  
 
|Hence, we have  
 
|-
 
|-
 
|
 
|
::<math>1.9^3\,\approx \, 2^3+-1.2\,=\,6.8.</math>
+
&nbsp; &nbsp; &nbsp; &nbsp;<math>1.9^3\,\approx \, 2^3+-1.2\,=\,6.8.</math>
 
|}
 
|}
 +
  
 
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 
!Final Answer: &nbsp;  
 
!Final Answer: &nbsp;  
 
|-
 
|-
|'''(a)''' &thinsp;<math style="vertical-align: -5px">dy=12\,dx</math>
+
|&nbsp; &nbsp; '''(a)''' &nbsp; &nbsp; <math style="vertical-align: -5px">dy=12\,dx</math>
 
|-
 
|-
|'''(b)''' &thinsp;<math style="vertical-align: -1px">6.8</math>   
+
|&nbsp; &nbsp; '''(b)''' &nbsp; &nbsp; <math style="vertical-align: -1px">6.8</math>   
 
|}
 
|}
 
[[009A_Sample_Final_1|'''<u>Return to Sample Exam</u>''']]
 
[[009A_Sample_Final_1|'''<u>Return to Sample Exam</u>''']]

Latest revision as of 08:14, 10 April 2017

Let

(a) Find the differential    of    at  .

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

Foundations:  
What is the differential    of    at  

        Since    the differential is  


Solution:

(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

       

(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