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 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">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>. | ||
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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: -1px">x=1</math> | + | ::Since <math style="vertical-align: -1px">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>dy</math> | + | |First, we find the differential <math>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^2dx</math> | + | ::<math>dy=3x^2dx.</math> |
|} | |} | ||
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|- | |- | ||
| | | | ||
| − | ::<math>dy=3(2)^2dx=12dx</math> | + | ::<math>dy=3(2)^2dx=12dx.</math> |
|} | |} | ||
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!Step 1: | !Step 1: | ||
|- | |- | ||
| − | |First, we find <math style="vertical-align: -1px">dx</math>. We have <math style="vertical-align: -1px">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)'''. | ||
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|- | |- | ||
| | | | ||
| − | ::<math>dy=12(-0.1)=-1.2</math> | + | ::<math>dy=12(-0.1)=-1.2.</math> |
|} | |} | ||
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|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> |
|- | |- | ||
|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 12:44, 1 March 2016
Let
a) Find the differential of at .
b) Use differentials to find an approximate value for .
| Foundations: |
|---|
| What is the differential of at |
|
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) |
