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

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|So, we have&thinsp; <math style="vertical-align: -5px">2(40)6=2(50)s'.</math>
 
|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.
 
|}
 
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|-
 
|-
 
|
 
|
:<math>s'=\frac{24}{5}</math> m/s
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:<math style="vertical-align: -14px">s'=\frac{24}{5}</math>&thinsp; 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 21:31, 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