a ladder 20 ft. long rests against a vertical wall.If the bottom of a ladder slides away from the wall at a rate of 4 ft./sec, how fast is the top of the ladder sliding down the wall when the bottom of the ladder is 12 ft. from the wall?

Question

x^2 +y^2 = 20^2

x^2 + y^2 = 400 Pythagoras Theorem

differentiate w.r.t to t

2x dx/dt + 2y dy/dt = 0

x dx/dt = -y dy/dt

now when x = 12, dx/dt = -4 (sliding away from the wall)

y = sqrt (400 - 144) = 16 ft

insert these in the differential eq. above to get dy/dt

(12)(-4) = -16dy/dt

dy/dt = 3 ft/sec

Answer 1

According to your answer of a positive dy/dt, the laddder would be sliding UP the wall.

Since the ladder is sliding away from the wall along he ground, dx/dt should be +4 ft/sec
which would make dy/dt = -3 ft/sec, showing that y is decreasing, thus sliding DOWN the wall as expected.

Your number-crunching is ok