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>dy</math> of <math>y=x^3</math> at <math>x=2</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">b) Use differentials to find an approximate value for <math>1.9^3</math>.
+
<span class="exam">b) Use differentials to find an approximate value for <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;"
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|First, we find the differential <math>dy</math>.
 
|First, we find the differential <math>dy</math>.
 
|-
 
|-
|Since <math>y=x^3</math>, we have
+
|Since <math style="vertical-align: -5px">y=x^3</math>, we have
 
|-
 
|-
|<math>dy=3x^2dx</math>.
+
|
 +
::<math>dy=3x^2dx</math>.
 
|}
 
|}
  
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!Step 2: &nbsp;
 
!Step 2: &nbsp;
 
|-
 
|-
|Now, we plug in <math>x=2</math> into the differential from Step 1.
+
|Now, we plug in <math style="vertical-align: -1px">x=2</math> into the differential from Step 1.
 
|-
 
|-
 
|So, we get  
 
|So, we get  
 
|-
 
|-
|<math>dy=3(2)^2dx=12dx</math>.
+
|
 +
::<math>dy=3(2)^2dx=12dx</math>.
 
|}
 
|}
  
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!Step 1: &nbsp;  
 
!Step 1: &nbsp;  
 
|-
 
|-
|First, we find <math>dx</math>. We have <math>dx=1.9-2=-0.1</math>.
+
|First, we find <math style="vertical-align: -1px">dx</math>. We have <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 <math>dy=12(-0.1)=-1.2</math>.
+
|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
+
|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
 
|-
 
|-
|approximate value of <math>1.9^3</math>.
+
|approximate value of <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>.
+
|
 +
::<math>1.9^3\approx 2^3+-1.2=6.8</math>.
 
|}
 
|}
  

Revision as of 15:37, 22 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)

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