Difference between revisions of "009A Sample Final 1, Problem 8"

Latest revision as of 15:57, 2 December 2017

Let

y=x^{3}.

(a) Find the differential $dy$ of $y=x^{3}$ at $x=2$ .

(b) Use differentials to find an approximate value for $1.9^{3}$ .

Solution

Detailed Solution

Return to Sample Exam

@@ Line 1: / Line 1: @@
 <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 &nbsp;<math style="vertical-align: -4px">dy</math>&nbsp; of &nbsp;<math style="vertical-align: -4px">y=x^3</math>&nbsp; at &nbsp;<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 &nbsp;<math style="vertical-align: -2px">1.9^3</math>.
-== 1 ==
-{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
-!Foundations: &nbsp;
-|-
-|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 &thinsp;<math style="vertical-align: -4px">x=1,</math>&thinsp; the differential is &thinsp;<math style="vertical-align: -4px">dy=2xdx=2dx.</math>
-|}
-'''Solution:'''
+<hr>
+[[009A Sample Final 1, Problem 8 Solution|'''<u>Solution</u>''']]
-== 2 ==
-'''(a)'''
-{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
+[[009A Sample Final 1, Problem 8 Detailed Solution|'''<u>Detailed Solution</u>''']]
-!Step 1: &nbsp;
-|-
-|First, we find the differential <math style="vertical-align: -4px">dy.</math>
-|-
-|Since <math style="vertical-align: -5px">y=x^3,</math>&thinsp; we have
-|-
-|
-::<math>dy\,=\,3x^2\,dx.</math>
-|}
-{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
-!Step 2: &nbsp;
-|-
-|Now, we plug <math style="vertical-align: 0px">x=2</math>&thinsp; into the differential from Step 1.
-|-
-|So, we get
-|-
-|
-::<math>dy\,=\,3(2)^2\,dx\,=\,12\,dx.</math>
-|}
-== 3 ==
-'''(b)'''
-{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
-!Step 1: &nbsp;
-|-
-|First, we find <math style="vertical-align: 0px">dx</math>. We have &thinsp;<math style="vertical-align: -1px">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>
-|}
-{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
-!Step 2: &nbsp;
-|-
-|Now, we add the value for <math style="vertical-align: -4px">dy</math> to &thinsp;<math style="vertical-align: 0px">2^3</math>&thinsp; to get an
-|-
-|approximate value of <math style="vertical-align: -1px">1.9^3.</math>
-|-
-|Hence, we have
-|-
-|
-::<math>1.9^3\,\approx \, 2^3+-1.2\,=\,6.8.</math>
-|}
-== 4 ==
-{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
-!Final Answer: &nbsp;
-|-
-|'''(a)''' <math>dy=12dx</math>
-|-
-|'''(b)''' <math>6.8</math>
-|}
 [[009A_Sample_Final_1|'''<u>Return to Sample Exam</u>''']]

Difference between revisions of "009A Sample Final 1, Problem 8"

Latest revision as of 15:57, 2 December 2017

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