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>dy</math> of <math>y=x^3</math> at <math>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>. |
− | <span class="exam">b) Use differentials to find an approximate value for <math>1.9^3</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;" | ||
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|First, we find the differential <math>dy</math>. | |First, we find the differential <math>dy</math>. | ||
|- | |- | ||
− | |Since <math>y=x^3</math>, we have | + | |Since <math style="vertical-align: -5px">y=x^3</math>, we have |
|- | |- | ||
− | |<math>dy=3x^2dx</math>. | + | | |
+ | ::<math>dy=3x^2dx</math>. | ||
|} | |} | ||
Line 30: | Line 31: | ||
!Step 2: | !Step 2: | ||
|- | |- | ||
− | |Now, we plug in <math>x=2</math> into the differential from Step 1. | + | |Now, we plug in <math style="vertical-align: -1px">x=2</math> into the differential from Step 1. |
|- | |- | ||
|So, we get | |So, we get | ||
|- | |- | ||
− | |<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>dx</math>. We have <math>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)'''. | ||
|- | |- | ||
− | |So, we have <math>dy=12(-0.1)=-1.2</math>. | + | |So, we have |
+ | |- | ||
+ | | | ||
+ | ::<math>dy=12(-0.1)=-1.2</math>. | ||
|} | |} | ||
Line 52: | Line 57: | ||
!Step 2: | !Step 2: | ||
|- | |- | ||
− | |Now, we add the value for <math>dy</math> to <math>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>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 15:37, 22 February 2016
Let
a) Find the differential of at .
b) Use differentials to find an approximate value for .
Foundations: |
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Solution:
(a)
Step 1: |
---|
First, we find the differential . |
Since , we have |
|
Step 2: |
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Now, we plug in into the differential from Step 1. |
So, we get |
|
(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: |
---|
Now, we add the value for to to get an |
approximate value of . |
Hence, we have |
|
Final Answer: |
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(a) |
(b) |