Difference between revisions of "009A Sample Final 1, Problem 5"

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|Recall:
 
|Recall:
 
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|'''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  
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|'''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  
 
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|-
 
|
 
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|Insert diagram.
 
|Insert diagram.
 
|-
 
|-
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|From the diagram, we have <math style="vertical-align: -2px">30^2+h^2=s^2</math> by the Pythagorean Theorem.
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|From the diagram, we have <math style="vertical-align: -3px">30^2+h^2=s^2</math> by the Pythagorean Theorem.
 
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|-
 
|Taking derivatives, we get  
 
|Taking derivatives, we get  
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!Step 2: &nbsp;
 
!Step 2: &nbsp;
 
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|If <math style="vertical-align: -1px">s=50,</math> then <math style="vertical-align: -3px">h=\sqrt{50^2-30^2}=40.</math>
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|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>
 
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|So, we have <math style="vertical-align: -5px">2(40)6=2(50)s'.</math>
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|So, we have&thinsp; <math style="vertical-align: -5px">2(40)6=2(50)s'.</math>
 
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|Solving for <math style="vertical-align: 0px">s',</math> we get <math style="vertical-align: -14px">s'=\frac{24}{5}</math> m/s.
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|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.
 
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!Final Answer: &nbsp;  
 
!Final Answer: &nbsp;  
 
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| <math>s'=\frac{24}{5}</math> m/s
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:<math>s'=\frac{24}{5}</math> m/s
 
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[[009A_Sample_Final_1|'''<u>Return to Sample Exam</u>''']]
 
[[009A_Sample_Final_1|'''<u>Return to Sample Exam</u>''']]

Revision as of 11:49, 4 March 2016

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?

Foundations:  
Recall:
The Pythagorean Theorem: For a right triangle with side lengths , where is the length of the
hypotenuse, we have

Solution:

Step 1:  
Insert diagram.
From the diagram, we have by the Pythagorean Theorem.
Taking derivatives, we get
Step 2:  
If  then 
So, we have 
Solving for  we get  m/s.
Final Answer:  
m/s

Return to Sample Exam

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