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: -5px">s',</math> we get  <math style="vertical-align: -14px">s'=\frac{24}{5}</math> m/s. | + | |Solving for  <math style="vertical-align: -5px">s',</math> we get  <math style="vertical-align: -14px">s'=\frac{24}{5}</math>  m/s. |
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|- | |- | ||
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| − | :<math>s'=\frac{24}{5}</math> m/s | + | :<math style="vertical-align: -14px">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 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: |
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| Recall: |
| The Pythagorean Theorem: For a right triangle with side lengths , where is the length of the |
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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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|