009A Sample Final 1, Problem 5 - Revision history

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2016-03-05T07:16:12Z

MathAdmin: Created page with " 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 whe..."

2016-03-05T07:09:13Z

Created page with " 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 whe..."

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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?

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Foundations:  
|-
|Recall:
|-
|'''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
|-
|
::hypotenuse, we have <math style="vertical-align: -2px">a^2+b^2=c^2.</math>
|}

'''Solution:'''

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
|-
|Insert diagram.
|-
|From the diagram, we have <math style="vertical-align: -3px">30^2+h^2=s^2</math> by the Pythagorean Theorem.
|-
|Taking derivatives, we get
|-
|
::<math>2hh'=2ss'.</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
|-
|If  <math style="vertical-align: -4px">s=50,</math> then  <math style="vertical-align: -2px">h=\sqrt{50^2-30^2}=40.</math>
|-
|So, we have  <math style="vertical-align: -5px">2(40)6=2(50)s'.</math>
|-
|Solving for  <math style="vertical-align: -5px">s',</math> we get  <math style="vertical-align: -14px">s'=\frac{24}{5}</math>  m/s.
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Final Answer:  
|-
|
:<math style="vertical-align: -14px">s'=\frac{24}{5}</math>  m/s
|}
[[009A_Sample_Final_1|'''Return to Sample Exam''']]

@@ Line 56: / Line 56: @@
 |-
+|
-:<math style="vertical-align: -14px">s'=\frac{24}{5}</math>&thinsp; m/s
+&nbsp;&nbsp; <math style="vertical-align: -14px">s'=\frac{24}{5}</math>&thinsp; m/s
 |}
 [[009A_Sample_Final_1|'''<u>Return to Sample Exam</u>''']]

@@ Line 6: / Line 6: @@
 !Foundations: &nbsp;
 |-
-|Recall:
+|'''The Pythagorean Theorem'''
 |-
+|&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
 |-
+|
-::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; hypotenuse, we have &nbsp;<math style="vertical-align: -2px">a^2+b^2=c^2.</math>
 |}
 '''Solution:'''
@@ Line 25: / Line 22: @@
 |[[File:9AF_5_GP.png|center|550px]]
 |-
-|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
 |-
+|
-::<math>2hh'=2ss'.</math>
+&nbsp; &nbsp; &nbsp; &nbsp; <math>2hh'=2ss'.</math>
 |}
@@ Line 36: / Line 33: @@
 !Step 2: &nbsp;
 |-
-|If&thinsp; <math style="vertical-align: -4px">s=50,</math> then
+|If &nbsp; <math style="vertical-align: -4px">s=50,</math>&nbsp; then
 |-
-|So, we have
+|&nbsp; &nbsp; &nbsp; &nbsp;<math style="vertical-align: -2px">h=\sqrt{50^2-30^2}=40.</math>
 |-
+|So, we have
 |-
-|Solving for&thinsp; <math style="vertical-align: -5px">s',</math> we get
+|&nbsp; &nbsp; &nbsp; &nbsp; <math style="vertical-align: -5px">2(40)6=2(50)s'.</math>
 |-
+|Solving for &nbsp; <math style="vertical-align: -5px">s',</math>&nbsp;  we get &nbsp; <math style="vertical-align: -14px">s'=\frac{24}{5} \text{ m/s.}</math> &nbsp;
 |}
 {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 !Final Answer: &nbsp;
 |-
+|&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: @@
 |Recall:
 |-
 |'''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
 |-
+|
@@ Line 32: / Line 36: @@
 !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&thinsp; <math style="vertical-align: -4px">s=50,</math> then
 |-
-|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>&thinsp; m/s.
+|So, we have
 |}

@@ Line 19: / Line 19: @@
 !Step 1: &nbsp;
 |-
-|Insert diagram.
+|[[File:9AF_5_GP.png|center|550px]]
 |-
 |From the diagram, we have <math style="vertical-align: -3px">30^2+h^2=s^2</math> by the Pythagorean Theorem.