A bag is tied to the top of a 5m ladder resting against a vertical wall. Suppose the ladder begins sliding down the wall in such a way that the foot of the ladder is moving away from the wall. How fast is the bag descending at the instant the foot of the ladder is 4m from the wall and foot is moving away at the rate 2 m/sec.

Answer 1

when the foot of the ladder is x m from the wall, and the top of the ladder is y m up the wall,
x^2+y^2 = 25
x dx/dt + y dy/dt = 0
So now plug in your numbers to find dy/dt