## 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

2. zepdrix

|dw:1436663055042:dw|So they want us to figure out y' when x=20.

3. zepdrix

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

Thank you! I think I have it from here :)

5. zepdrix

cool :3

6. anonymous

Refer to the calculation from Mathematica 9: