009A Sample Final 1, Problem 8 - Revision history

MathAdmin at 16:14, 10 April 2017

2017-04-10T16:14:22Z

MathAdmin at 22:15, 18 April 2016

2016-04-18T22:15:57Z

MathAdmin at 19:18, 18 April 2016

2016-04-18T19:18:05Z

MathAdmin: Created page with "Let :::::: $y=x^3.$ a) Find the differential $dy$ of
2016-03-05T07:10:42Z

Created page with "Let ::::::<math>y=x^3.</math> a) Find the differential <math style="vertical-align: -4px">dy</math> of <math style="vertical-align: -..."

New page
Let

::::::<math>y=x^3.</math>

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>.

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;"
!Foundations:  
|-
|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:'''

'''(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=x^3,</math>  we have
|-
|
::<math>dy\,=\,3x^2\,dx.</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
|-
|Now, we plug <math style="vertical-align: 0px">x=2</math>  into the differential from Step 1.
|-
|So, we get
|-
|
::<math>dy\,=\,3(2)^2\,dx\,=\,12\,dx.</math>
|}

'''(b)'''

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
|-
|First, we find <math style="vertical-align: 0px">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)'''.
|-
|So, we have
|-
|
::<math>dy\,=\,12(-0.1)\,=\,-1.2.</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: 0px">2^3</math>  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>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Final Answer:  
|-
|'''(a)'''  <math style="vertical-align: -5px">dy=12\,dx</math>
|-
|'''(b)'''  <math style="vertical-align: -1px">6.8</math>
|}
[[009A_Sample_Final_1|'''Return to Sample Exam''']]

@@ Line 3: / Line 3: @@
 ::::::<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;"

@@ 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>.
 {| 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>
+|What is the differential  &nbsp;<math style="vertical-align: -4px">dy</math>&nbsp; of &nbsp;<math style="vertical-align: -4px">y=x^2</math>&nbsp; at &nbsp;<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>
+&nbsp; &nbsp; &nbsp; &nbsp; Since &nbsp;<math style="vertical-align: -4px">x=1,</math>&nbsp; the differential is &nbsp;<math style="vertical-align: -4px">dy=2xdx=2dx.</math>
 |}
 '''Solution:'''
 '''(a)'''
@@ Line 24: / Line 24: @@
 !Step 1: &nbsp;
 |-
-|First, we find the differential <math style="vertical-align: -4px">dy.</math>
+|First, we find the differential &nbsp;<math style="vertical-align: -4px">dy.</math>
 |-
-|Since <math style="vertical-align: -5px">y=x^3,</math>&thinsp; we have
+|Since &nbsp;<math style="vertical-align: -5px">y=x^3,</math>&nbsp; we have
 |-
+|
-::<math>dy\,=\,3x^2\,dx.</math>
+&nbsp; &nbsp; &nbsp; &nbsp;<math>dy\,=\,3x^2\,dx.</math>
 |}
@@ Line 35: / Line 35: @@
 !Step 2: &nbsp;
 |-
-|Now, we plug <math style="vertical-align: 0px">x=2</math>&thinsp; into the differential from Step 1.
+|Now, we plug &nbsp;<math style="vertical-align: 0px">x=2</math>&nbsp; into the differential from Step 1.
 |-
 |So, we get
 |-
+|
-::<math>dy\,=\,3(2)^2\,dx\,=\,12\,dx.</math>
+&nbsp; &nbsp; &nbsp; &nbsp;<math>dy\,=\,3(2)^2\,dx\,=\,12\,dx.</math>
 |}
@@ Line 48: / Line 48: @@
 !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>
+|First, we find &nbsp;<math style="vertical-align: 0px">dx.</math>&nbsp;  We have
 |-
-|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>
+&nbsp; &nbsp; &nbsp; &nbsp;<math>dy\,=\,12(-0.1)\,=\,-1.2.</math>
 |}
@@ Line 61: / Line 63: @@
 !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
+|Now, we add the value for &nbsp;<math style="vertical-align: -4px">dy</math>&nbsp; to &nbsp;<math style="vertical-align: 0px">2^3</math>&nbsp; to get an
 |-
-|approximate value of <math style="vertical-align: -1px">1.9^3.</math>
+|approximate value of &nbsp;<math style="vertical-align: -1px">1.9^3.</math>
 |-
 |Hence, we have
 |-
+|
-::<math>1.9^3\,\approx \, 2^3+-1.2\,=\,6.8.</math>
+&nbsp; &nbsp; &nbsp; &nbsp;<math>1.9^3\,\approx \, 2^3+-1.2\,=\,6.8.</math>
 |}
 {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 !Final Answer: &nbsp;
 |-
-|&nbsp;&nbsp; '''(a)''' &thinsp;<math style="vertical-align: -5px">dy=12\,dx</math>
+|&nbsp; &nbsp; '''(a)''' &nbsp; &nbsp; <math style="vertical-align: -5px">dy=12\,dx</math>
 |-
-|&nbsp;&nbsp; '''(b)''' &thinsp;<math style="vertical-align: -1px">6.8</math>
+|&nbsp; &nbsp; '''(b)''' &nbsp; &nbsp; <math style="vertical-align: -1px">6.8</math>
 |}
 [[009A_Sample_Final_1|'''<u>Return to Sample Exam</u>''']]