009A Sample Final 1, Problem 8 - Revision history

Kayla Murray at 22:57, 2 December 2017

2017-12-02T22:57:30Z

Kayla Murray at 20:22, 18 March 2017

2017-03-18T20:22:18Z

Kayla Murray at 16:32, 27 February 2017

2017-02-27T16:32:06Z

Kayla Murray at 00:59, 26 February 2017

2017-02-26T00:59:54Z

Kayla Murray at 01:49, 19 February 2017

2017-02-19T01:49:09Z

Kayla Murray at 01:44, 19 February 2017

2017-02-19T01:44:46Z

Grad at 20:51, 4 March 2016

2016-03-04T20:51:36Z

Grad: /* 4 */

2016-03-04T20:51:08Z

Grad: /* 4 */

2016-03-04T20:49:54Z

Grad: /* 3 */

2016-03-04T20:49:24Z

@@ Line 7: / Line 7: @@
 <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;"
+<hr>
-!Foundations: &nbsp;
+[[009A Sample Final 1, Problem 8 Solution|'''<u>Solution</u>''']]
-'''Solution:'''
+[[009A Sample Final 1, Problem 8 Detailed Solution|'''<u>Detailed Solution</u>''']]
 [[009A_Sample_Final_1|'''<u>Return to Sample Exam</u>''']]

@@ Line 48: / Line 48: @@
 !Step 1: &nbsp;
 |-
-|First, we find &nbsp;<math style="vertical-align: 0px">dx.</math>&nbsp;  We have &nbsp;<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).

@@ 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 &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>&nbsp; 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>
 |-
+|
@@ Line 24: / Line 24: @@
 !Step 1: &nbsp;
 |-
-|First, we find the differential &nbsp; <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>&nbsp; we have
+|Since &nbsp;<math style="vertical-align: -5px">y=x^3,</math>&nbsp; we have
 |-
+|
@@ Line 35: / Line 35: @@
 !Step 2: &nbsp;
 |-
-|Now, we plug <math style="vertical-align: 0px">x=2</math>&nbsp; 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
@@ Line 48: / Line 48: @@
 !Step 1: &nbsp;
 |-
-|First, we find <math style="vertical-align: 0px">dx.</math>&nbsp;  We have &nbsp;<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 &nbsp;<math style="vertical-align: -1px">dx=1.9-2=-0.1.</math>
 |-
 |Then, we plug this into the differential from part (a).
@@ Line 61: / Line 61: @@
 !Step 2: &nbsp;
 |-
-|Now, we add the value for <math style="vertical-align: -4px">dy</math>&nbsp; to &nbsp;<math style="vertical-align: 0px">2^3</math>&nbsp; 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

@@ Line 10: / Line 10: @@
 !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  <math style="vertical-align: -4px">dy</math>&nbsp; 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>
+&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>
 |}
@@ 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 <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 <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 <math style="vertical-align: 0px">dx.</math>&nbsp;  We have &nbsp;<math style="vertical-align: -1px">dx=1.9-2=-0.1.</math>
 |-
-|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 61: @@
 !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 <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>
@@ Line 68: / Line 68: @@
 |-
+|
-::<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>
 |}
@@ Line 75: / Line 75: @@
 !Final Answer: &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>
 |-
-|'''(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>''']]

@@ Line 15: / Line 15: @@
 ::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:'''
 '''(a)'''
@@ Line 70: / Line 70: @@
 ::<math>1.9^3\,\approx \, 2^3+-1.2\,=\,6.8.</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 <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 76: / Line 76: @@
 !Final Answer: &nbsp;
 |-
-|'''(a)''' <math>dy=12dx</math>
+|'''(a)''' &thinsp;<math>dy=12dx</math>
 |-
-|'''(b)''' <math>6.8</math>
+|'''(b)''' &thinsp;<math>6.8</math>
 |}
 [[009A_Sample_Final_1|'''<u>Return to Sample Exam</u>''']]