Difference between revisions of "009A Sample Final 1, Problem 5"
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|So, we have <math style="vertical-align: -5px">2(40)6=2(50)s'</math>. | |So, we have <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. | + | |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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!Final Answer: | !Final Answer: | ||
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− | | <math>s'=\frac{24}{5} </math>m/s | + | | <math>s'=\frac{24}{5}</math> m/s |
|} | |} | ||
[[009A_Sample_Final_1|'''<u>Return to Sample Exam</u>''']] | [[009A_Sample_Final_1|'''<u>Return to Sample Exam</u>''']] |
Revision as of 11:29, 24 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 |
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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 |