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zeesbrat3

  • one year ago

A 25-ft ladder rests against a vertical wall. If the bottom of the ladder slides away from the wall at a rate of 0.18 ft/sec, how fast, in ft/sec, is the top of the ladder sliding down the wall, at the instant when the bottom of the ladder is 20 ft from the wall? Answer with 2 decimal places. Type your answer in the space below. If your answer is a number less than 1, place a leading "0" before the decimal point (ex: 0.35).

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  1. zeesbrat3
    • one year ago
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    Please help

  2. zepdrix
    • one year ago
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    |dw:1436663055042:dw|So they want us to figure out y' when x=20.

  3. zepdrix
    • one year ago
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    Setup your Pythagorean Identity here :)\[\Large\rm x^2+y^2=z^2\]The length of the ladder is the only thing that is `not changing`. It's being held constant, so you can plug that value in if you want.\[\Large\rm x^2+y^2=25^2\] Oh oh oh, you'll need the length y near the end of the problem. Might make sense to deal with that now.\[\Large\rm 20^2+y^2=25^2\qquad \text{solve for y}\]Then back to this setup,\[\Large\rm x^2+y^2=25^2\]Differentiate with respect to `time` and then simply plug in all of your stuff and solve for y' ! :)

  4. zeesbrat3
    • one year ago
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    Thank you! I think I have it from here :)

  5. zepdrix
    • one year ago
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    cool :3

  6. anonymous
    • one year ago
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    Refer to the calculation from Mathematica 9:

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