009A Sample Final 1, Problem 5 - Revision history

Kayla Murray at 15:05, 3 December 2017

2017-12-03T15:05:38Z

Kayla Murray at 22:50, 2 December 2017

2017-12-02T22:50:24Z

Kayla Murray at 20:16, 18 March 2017

2017-03-18T20:16:24Z

Kayla Murray at 16:22, 27 February 2017

2017-02-27T16:22:24Z

Kayla Murray at 00:41, 26 February 2017

2017-02-26T00:41:49Z

Kayla Murray at 01:47, 19 February 2017

2017-02-19T01:47:37Z

Grad at 04:31, 5 March 2016

2016-03-05T04:31:14Z

Grad at 18:49, 4 March 2016

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Kayla Murray at 19:32, 29 February 2016

2016-02-29T19:32:26Z

Kayla Murray at 00:18, 25 February 2016

2016-02-25T00:18:03Z

← Older revision		Revision as of 15:05, 3 December 2017
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−	<span class="exam"> A kite 30 (meters) above the ground moves horizontally at a speed of 6 (m/s). At what rate is the length of the string increasing	+	<span class="exam"> A kite 30 (meters) above the ground moves horizontally at a speed of 6 (m/s). At what rate is the length of the string increasing when 50 (meters) of the string has been let out?
−
−	~~<span class="exam">~~ when 50 (meters) of the string has been let out?

	<hr>		<hr>

@@ Line 3: / Line 3: @@
 <span class="exam"> when 50 (meters) of the string has been let out?
-{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
+<hr>
-!Foundations: &nbsp;
+[[009A Sample Final 1, Problem 5 Solution|'''<u>Solution</u>''']]
-'''Solution:'''
+[[009A Sample Final 1, Problem 5 Detailed Solution|'''<u>Detailed Solution</u>''']]
 [[009A_Sample_Final_1|'''<u>Return to Sample Exam</u>''']]

@@ Line 33: / Line 33: @@
 !Step 2: &nbsp;
 |-
-|If &nbsp; <math style="vertical-align: -4px">s=50,</math>&nbsp; then &nbsp;<math style="vertical-align: -2px">h=\sqrt{50^2-30^2}=40.</math>
+|If &nbsp; <math style="vertical-align: -4px">s=50,</math>&nbsp; then
 |-
-|So, we have &nbsp; <math style="vertical-align: -5px">2(40)6=2(50)s'.</math>
+|&nbsp; &nbsp; &nbsp; &nbsp;<math style="vertical-align: -2px">h=\sqrt{50^2-30^2}=40.</math>
 |-
-|Solving for &nbsp; <math style="vertical-align: -5px">s',</math>&nbsp;  we get &nbsp; <math style="vertical-align: -14px">s'=\frac{24}{5}</math> &nbsp; m/s.
+|So, we have
 |}
@@ Line 44: / Line 48: @@
 !Final Answer: &nbsp;
 |-
-|&nbsp; &nbsp; &nbsp; &nbsp;<math style="vertical-align: -14px">s'=\frac{24}{5}</math>&nbsp; m/s
+|&nbsp; &nbsp; &nbsp; &nbsp;<math style="vertical-align: -14px">s'=\frac{24}{5} \text{ m/s}</math>&nbsp;
 |}
 [[009A_Sample_Final_1|'''<u>Return to Sample Exam</u>''']]

@@ Line 8: / Line 8: @@
 |'''The Pythagorean Theorem'''
 |-
-|&nbsp; &nbsp; &nbsp; &nbsp; For a right triangle with side lengths <math style="vertical-align: -4px">a,b,c</math>, where <math style="vertical-align: 0px">c</math> is the length of the
+|&nbsp; &nbsp; &nbsp; &nbsp; For a right triangle with side lengths &nbsp;<math style="vertical-align: -4px">a,b,c</math>&nbsp; where &nbsp;<math style="vertical-align: 0px">c</math>&nbsp; is the length of the
 |-
+|
-&nbsp; &nbsp; &nbsp; &nbsp; hypotenuse, we have <math style="vertical-align: -2px">a^2+b^2=c^2.</math>
+&nbsp; &nbsp; &nbsp; &nbsp; hypotenuse, we have &nbsp;<math style="vertical-align: -2px">a^2+b^2=c^2.</math>
 |}
@@ Line 22: / Line 22: @@
 |Insert diagram.
 |-
-|From the diagram, we have <math style="vertical-align: -3px">30^2+h^2=s^2</math> by the Pythagorean Theorem.
+|From the diagram, we have &nbsp;<math style="vertical-align: -3px">30^2+h^2=s^2</math>&nbsp; by the Pythagorean Theorem.
 |-
 |Taking derivatives, we get
@@ Line 33: / Line 33: @@
 !Step 2: &nbsp;
 |-
-|If &nbsp; <math style="vertical-align: -4px">s=50,</math> then&thinsp; <math style="vertical-align: -2px">h=\sqrt{50^2-30^2}=40.</math>
+|If &nbsp; <math style="vertical-align: -4px">s=50,</math>&nbsp; then &nbsp;<math style="vertical-align: -2px">h=\sqrt{50^2-30^2}=40.</math>
 |-
 |So, we have &nbsp; <math style="vertical-align: -5px">2(40)6=2(50)s'.</math>

@@ Line 6: / Line 6: @@
 !Foundations: &nbsp;
 |-
-|Recall:
+|'''The Pythagorean Theorem'''
 |-
-|'''The Pythagorean Theorem:''' For a right triangle with side lengths <math style="vertical-align: -4px">a,b,c</math>, where <math style="vertical-align: 0px">c</math> is the length of the
+|&nbsp; &nbsp; &nbsp; &nbsp; For a right triangle with side lengths <math style="vertical-align: -4px">a,b,c</math>, where <math style="vertical-align: 0px">c</math> is the length of the
 |-
+|
-::hypotenuse, we have <math style="vertical-align: -2px">a^2+b^2=c^2.</math>
+&nbsp; &nbsp; &nbsp; &nbsp; hypotenuse, we have <math style="vertical-align: -2px">a^2+b^2=c^2.</math>
 |}
@@ Line 27: / Line 27: @@
 |-
+|
-::<math>2hh'=2ss'.</math>
+&nbsp; &nbsp; &nbsp; &nbsp; <math>2hh'=2ss'.</math>
 |}
@@ Line 33: / Line 33: @@
 !Step 2: &nbsp;
 |-
-|If&thinsp; <math style="vertical-align: -4px">s=50,</math> then&thinsp; <math style="vertical-align: -2px">h=\sqrt{50^2-30^2}=40.</math>
+|If &nbsp; <math style="vertical-align: -4px">s=50,</math> then&thinsp; <math style="vertical-align: -2px">h=\sqrt{50^2-30^2}=40.</math>
 |-
-|So, we have&thinsp; <math style="vertical-align: -5px">2(40)6=2(50)s'.</math>
+|So, we have &nbsp; <math style="vertical-align: -5px">2(40)6=2(50)s'.</math>
 |-
-|Solving for&thinsp; <math style="vertical-align: -5px">s',</math> we get&thinsp; <math style="vertical-align: -14px">s'=\frac{24}{5}</math>&thinsp; m/s.
+|Solving for &nbsp; <math style="vertical-align: -5px">s',</math>&nbsp;  we get &nbsp; <math style="vertical-align: -14px">s'=\frac{24}{5}</math> &nbsp; m/s.
 |}
@@ Line 44: / Line 44: @@
 !Final Answer: &nbsp;
 |-
+|&nbsp; &nbsp; &nbsp; &nbsp;<math style="vertical-align: -14px">s'=\frac{24}{5}</math>&nbsp; m/s
 |}
 [[009A_Sample_Final_1|'''<u>Return to Sample Exam</u>''']]

@@ Line 36: / Line 36: @@
 |So, we have&thinsp; <math style="vertical-align: -5px">2(40)6=2(50)s'.</math>
 |-
-|Solving for&thinsp; <math style="vertical-align: -5px">s',</math> we get&thinsp; <math style="vertical-align: -14px">s'=\frac{24}{5}</math> m/s.
+|Solving for&thinsp; <math style="vertical-align: -5px">s',</math> we get&thinsp; <math style="vertical-align: -14px">s'=\frac{24}{5}</math>&thinsp; m/s.
 |}
@@ Line 43: / Line 43: @@
 |-
+|
-:<math>s'=\frac{24}{5}</math> m/s
+:<math style="vertical-align: -14px">s'=\frac{24}{5}</math>&thinsp; m/s
 |}
 [[009A_Sample_Final_1|'''<u>Return to Sample Exam</u>''']]

@@ Line 11: / Line 11: @@
 |-
+|
-::hypotenuse, we have <math style="vertical-align: -2px">a^2+b^2=c^2</math>.
+::hypotenuse, we have <math style="vertical-align: -2px">a^2+b^2=c^2.</math>
 |}
@@ Line 26: / Line 26: @@
 |-
+|
-::<math>2hh'=2ss'</math>.
+::<math>2hh'=2ss'.</math>
 |}
@@ Line 32: / Line 32: @@
 !Step 2: &nbsp;
 |-
-|If <math style="vertical-align: -1px">s=50</math>, then <math style="vertical-align: -3px">h=\sqrt{50^2-30^2}=40</math>.
+|If <math style="vertical-align: -1px">s=50,</math> then <math style="vertical-align: -3px">h=\sqrt{50^2-30^2}=40.</math>
 |-
-|So, we have <math style="vertical-align: -5px">2(40)6=2(50)s'</math>.
+|So, we have <math style="vertical-align: -5px">2(40)6=2(50)s'.</math>
 |-
-|Solving for <math style="vertical-align: 0px">s'</math>, we get <math style="vertical-align: -14px">s'=\frac{24}{5}</math> m/s.
+|Solving for <math style="vertical-align: 0px">s',</math> we get <math style="vertical-align: -14px">s'=\frac{24}{5}</math> m/s.
 |}