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 1

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.