|
|
| Line 5: |
Line 5: |
| | <span class="exam">(b) Use differentials to find an approximate value for <math style="vertical-align: -5px">\tan(0.885).</math> Hint: <math style="vertical-align: -15px">\frac{\pi}{4}\approx 0.785.</math> | | <span class="exam">(b) Use differentials to find an approximate value for <math style="vertical-align: -5px">\tan(0.885).</math> Hint: <math style="vertical-align: -15px">\frac{\pi}{4}\approx 0.785.</math> |
| | | | |
| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
| + | <hr> |
| − | !Foundations:
| + | [[009A Sample Final 3, Problem 10 Solution|'''<u>Solution</u>''']] |
| − | |-
| |
| − | |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: -4px">x=1,</math> the differential is <math style="vertical-align: -4px">dy=2xdx=2dx.</math>
| |
| − | |}
| |
| | | | |
| | | | |
| − | '''Solution:''' | + | [[009A Sample Final 3, Problem 10 Detailed Solution|'''<u>Detailed Solution</u>''']] |
| | | | |
| − | '''(a)'''
| |
| | | | |
| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
| |
| − | !Step 1:
| |
| − | |-
| |
| − | |First, we find the differential <math style="vertical-align: -4px">dy.</math>
| |
| − | |-
| |
| − | |Since <math style="vertical-align: -5px">y=\tan x,</math> we have
| |
| − | |-
| |
| − | |
| |
| − | <math>dy\,=\,\sec^2 x\,dx.</math>
| |
| − | |}
| |
| − |
| |
| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
| |
| − | !Step 2:
| |
| − | |-
| |
| − | |Now, we plug <math style="vertical-align: -15px">x=\frac{\pi}{4}</math> into the differential from Step 1.
| |
| − | |-
| |
| − | |So, we get
| |
| − | |-
| |
| − | |
| |
| − | <math>dy\,=\,\bigg(\sec\bigg(\frac{\pi}{4}\bigg)\bigg)^2\,dx\,=\,2\,dx.</math>
| |
| − | |-
| |
| − | |
| |
| − | |}
| |
| − |
| |
| − | '''(b)'''
| |
| − |
| |
| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
| |
| − | !Step 1:
| |
| − | |-
| |
| − | |First, we find <math style="vertical-align: -1px">dx.</math> We have
| |
| − | |-
| |
| − | | <math style="vertical-align: -1px">dx=0.885-\frac{\pi}{4}\approx 0.885-0.785=0.1.</math>
| |
| − | |-
| |
| − | |Then, we plug this into the differential from part (a).
| |
| − | |-
| |
| − | |So, we have
| |
| − | |-
| |
| − | |
| |
| − | <math>dy\,=\,2(0.1)\,=\,0.2.</math>
| |
| − | |}
| |
| − |
| |
| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
| |
| − | !Step 2:
| |
| − | |-
| |
| − | |Now, we add the value for <math style="vertical-align: -4px">dy</math> to <math style="vertical-align: -16px">\tan\bigg(\frac{\pi}{4}\bigg)</math> to get an
| |
| − | |-
| |
| − | |approximate value of <math style="vertical-align: -5px">\tan(0.885).</math>
| |
| − | |-
| |
| − | |Hence, we have
| |
| − | |-
| |
| − | |
| |
| − | <math>\tan(0.885)\,\approx \, \tan\bigg(\frac{\pi}{4}\bigg)+0.2\,=\,1.2.</math>
| |
| − | |}
| |
| − |
| |
| − |
| |
| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
| |
| − | !Final Answer:
| |
| − | |-
| |
| − | | '''(a)''' <math style="vertical-align: -5px">dy=2\,dx</math>
| |
| − | |-
| |
| − | | '''(b)''' <math style="vertical-align: -1px">1.2</math>
| |
| − | |}
| |
| | [[009A_Sample_Final_3|'''<u>Return to Sample Exam</u>''']] | | [[009A_Sample_Final_3|'''<u>Return to Sample Exam</u>''']] |
