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> | |
− | + | <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 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: | !Foundations: | ||
|- | |- | ||
− | |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 <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> |
|- | |- | ||
| | | | ||
− | + | Since <math style="vertical-align: -4px">x=1,</math> the differential is <math style="vertical-align: -4px">dy=2xdx=2dx.</math> | |
|} | |} | ||
+ | |||
'''Solution:''' | '''Solution:''' | ||
− | |||
'''(a)''' | '''(a)''' | ||
Line 24: | Line 24: | ||
!Step 1: | !Step 1: | ||
|- | |- | ||
− | |First, we find the differential <math style="vertical-align: -4px">dy.</math> | + | |First, we find the differential <math style="vertical-align: -4px">dy.</math> |
|- | |- | ||
− | |Since <math style="vertical-align: -5px">y=x^3,</math>& | + | |Since <math style="vertical-align: -5px">y=x^3,</math> we have |
|- | |- | ||
| | | | ||
− | + | <math>dy\,=\,3x^2\,dx.</math> | |
|} | |} | ||
Line 35: | Line 35: | ||
!Step 2: | !Step 2: | ||
|- | |- | ||
− | |Now, we plug <math style="vertical-align: 0px">x=2</math>& | + | |Now, we plug <math style="vertical-align: 0px">x=2</math> into the differential from Step 1. |
|- | |- | ||
|So, we get | |So, we get | ||
|- | |- | ||
| | | | ||
− | + | <math>dy\,=\,3(2)^2\,dx\,=\,12\,dx.</math> | |
|} | |} | ||
Line 48: | Line 48: | ||
!Step 1: | !Step 1: | ||
|- | |- | ||
− | |First, we find <math style="vertical-align: 0px">dx</math> | + | |First, we find <math style="vertical-align: 0px">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 | + | |Then, we plug this into the differential from part (a). |
|- | |- | ||
|So, we have | |So, we have | ||
|- | |- | ||
| | | | ||
− | + | <math>dy\,=\,12(-0.1)\,=\,-1.2.</math> | |
|} | |} | ||
Line 61: | Line 63: | ||
!Step 2: | !Step 2: | ||
|- | |- | ||
− | |Now, we add the value for <math style="vertical-align: -4px">dy</math> to & | + | |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 style="vertical-align: -1px">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> | |
|} | |} | ||
+ | |||
{| class="mw-collapsible mw-collapsed" style = "text-align:left;" | {| class="mw-collapsible mw-collapsed" style = "text-align:left;" | ||
!Final Answer: | !Final Answer: | ||
|- | |- | ||
− | | '''(a)''' & | + | | '''(a)''' <math style="vertical-align: -5px">dy=12\,dx</math> |
|- | |- | ||
− | | '''(b)''' & | + | | '''(b)''' <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) |