Difference between revisions of "009A Sample Final 1, Problem 8"
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!Step 1: | !Step 1: | ||
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| − | |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> | ||
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|- | |- | ||
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| − | ::<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: |
|---|
| What is the differential of at |
|
Solution:
2
(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 |
|
|
3
(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 |
|
|
4
| Final Answer: |
|---|
| (a) |
| (b) |