Difference between revisions of "009A Sample Final 1, Problem 8"
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!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> | |
|} | |} | ||
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!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> | |
|} | |} | ||
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!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> | |
|} | |} | ||
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!Step 1: | !Step 1: | ||
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| − | |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 61: | ||
!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> | ||
| Line 68: | Line 68: | ||
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| | | | ||
| − | + | <math>1.9^3\,\approx \, 2^3+-1.2\,=\,6.8.</math> | |
|} | |} | ||
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!Final Answer: | !Final Answer: | ||
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| − | |'''(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>''']] | ||
Revision as of 17:59, 25 February 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) |