Difference between revisions of "009A Sample Final 1, Problem 8"
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!Step 1: | !Step 1: | ||
|- | |- | ||
− | | | + | |First, we find <math>dx</math>. We have <math>dx=1.9-2=-0.1</math>. |
|- | |- | ||
− | | | + | |Then, we plug this into the differential from part '''(a)'''. |
|- | |- | ||
− | | | + | |So, we have <math>dy=12(-0.1)=-1.2</math>. |
|} | |} | ||
Line 52: | Line 52: | ||
!Step 2: | !Step 2: | ||
|- | |- | ||
− | | | + | |Now, we add the value for <math>dy</math> to <math>2^3</math> to get an |
− | | | + | |- |
− | + | |approximate value of <math>1.9^3</math>. | |
− | |||
− | |||
|- | |- | ||
− | | | + | |Hence, we have |
|- | |- | ||
− | | | + | |<math>1.9^3\approx 2^3+-1.2=6.8</math>. |
|} | |} | ||
Line 68: | Line 66: | ||
|'''(a)''' <math>dy=12dx</math> | |'''(a)''' <math>dy=12dx</math> | ||
|- | |- | ||
− | |'''(b)''' | + | |'''(b)''' <math>6.8</math> |
|} | |} | ||
[[009A_Sample_Final_1|'''<u>Return to Sample Exam</u>''']] | [[009A_Sample_Final_1|'''<u>Return to Sample Exam</u>''']] |
Revision as of 10:09, 15 February 2016
Let
a) Find the differential of at .
b) Use differentials to find an approximate value for .
Foundations: |
---|
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) |