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

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<span class="exam"> when 50 (meters) of the string has been let out?
 
<span class="exam"> when 50 (meters) of the string has been let out?
  
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<hr>
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!Foundations: &nbsp;
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[[009A Sample Final 1, Problem 5 Solution|'''<u>Solution</u>''']]
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|'''The Pythagorean Theorem'''
 
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|&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
 
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&nbsp; &nbsp; &nbsp; &nbsp; hypotenuse, we have &nbsp;<math style="vertical-align: -2px">a^2+b^2=c^2.</math>
 
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'''Solution:'''
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[[009A Sample Final 1, Problem 5 Detailed Solution|'''<u>Detailed Solution</u>''']]
  
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!Step 1: &nbsp;
 
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|Insert diagram.
 
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|From the diagram, we have &nbsp;<math style="vertical-align: -3px">30^2+h^2=s^2</math>&nbsp; by the Pythagorean Theorem.
 
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|Taking derivatives, we get
 
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&nbsp; &nbsp; &nbsp; &nbsp; <math>2hh'=2ss'.</math>
 
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!Step 2: &nbsp;
 
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|If &nbsp; <math style="vertical-align: -4px">s=50,</math>&nbsp; then
 
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|&nbsp; &nbsp; &nbsp; &nbsp;<math style="vertical-align: -2px">h=\sqrt{50^2-30^2}=40.</math>
 
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|So, we have
 
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|&nbsp; &nbsp; &nbsp; &nbsp; <math style="vertical-align: -5px">2(40)6=2(50)s'.</math>
 
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|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;
 
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!Final Answer: &nbsp;
 
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|&nbsp; &nbsp; &nbsp; &nbsp;<math style="vertical-align: -14px">s'=\frac{24}{5} \text{ m/s}</math>&nbsp;
 
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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 15:50, 2 December 2017

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?

Solution


Detailed Solution


Return to Sample Exam

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