Stuck on this calculus problem please help

Question

A 15-ft ladder rests against a vertical wall. If the top of the ladder slides down the wall at a rate of 0.33 ft/sec, how fast, in ft/sec, is the bottom of the ladder sliding away from the wall, at the instant when the bottom of the ladder is 9 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). 

Answer 1

let x and y be the usual co-ordinates so that the rate you are looking for is dx/dt.

x(t) = sqrt(225 - y²(t0)

dx/dt = (-2y/sqrt(225-y²)) * dy/dt.

Now plug in the values given in the problem to get the required rate.