Difference between revisions of "009A Sample Final 1, Problem 5"
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Kayla Murray (talk | contribs) |
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!Step 1: | !Step 1: | ||
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− | | | + | |Insert diagram. |
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− | | | + | |From the diagram, we have <math>30^2+h^2=s^2</math> by the Pythagorean Theorem. |
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− | | | + | |Taking derivatives, we get |
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− | | | + | |<math>2hh'=2ss'</math>. |
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!Step 2: | !Step 2: | ||
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− | | | + | |If <math>s=50</math>, then <math>h=\sqrt{50^2-30^2}=40</math>. |
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− | | | + | |So, we have <math>2(40)6=2(50)s'</math>. |
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− | | | + | |Solving for <math>s'</math>, we get <math>s'=\frac{24}{5} </math>m/s. |
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!Final Answer: | !Final Answer: | ||
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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 11:30, 15 February 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: |
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Solution:
Step 1: |
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Insert diagram. |
From the diagram, we have by the Pythagorean Theorem. |
Taking derivatives, we get |
. |
Step 2: |
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If , then . |
So, we have . |
Solving for , we get m/s. |
Final Answer: |
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m/s |