Difference between revisions of "009A Sample Final 1, Problem 8"
Jump to navigation
Jump to search
(Created page with "<span class="exam">Let ::::::<math>y=x^3.</math> <span class="exam">a) Find the differential <math style="vertical-align: -4px">dy</math> of <math style="vertical-align: -...") |
|||
Line 3: | Line 3: | ||
::::::<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>. |
− | <span class="exam">b) Use differentials to find an approximate value for <math style="vertical-align: -2px">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;" |
Revision as of 11:18, 18 April 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 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) |