Asked by Anonymous

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?

Answers

Answered by Steve
If the base of the ladder is x away from the wall, and the top of the ladder is at height y, then

x^2+y^2 = 25^2

x dx/dt + y dy/dt = 0

So, figure y when x=20, and plug in dx/dt=0.18 to find dy/dt.
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