Difference between revisions of "009A Sample Final 1, Problem 8"
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(→3) |
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!Step 1: | !Step 1: | ||
|- | |- | ||
− | |First, we find <math style="vertical-align: | + | |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 '''(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> |
|} | |} | ||
Line 62: | Line 62: | ||
!Step 2: | !Step 2: | ||
|- | |- | ||
− | |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  <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 69: | Line 69: | ||
|- | |- | ||
| | | | ||
− | ::<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: |
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What is the differential of at |
|
Solution:
2
(a)
Step 1: |
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First, we find the differential |
Since we have |
|
Step 2: |
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Now, we plug into the differential from Step 1. |
So, we get |
|
3
(b)
Step 1: |
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First, we find . We have |
Then, we plug this into the differential from part (a). |
So, we have |
|
Step 2: |
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Now, we add the value for to to get an |
approximate value of |
Hence, we have |
|
4
Final Answer: |
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(a) |
(b) |