CALCULUS
posted by Anonymous on .
A 25 ft ladder is leaning against a vertical wall. At what rate (with respect to time) is the angle theta between the ground and the ladder changing, if the top of the ladder is sliding down the wall at the rate of r inches per second, at the moment that the top of the ladder is h feet from the ground?
I have no idea how to even start with this problem. Help!

tanθ = h/√(625h^2)
secθ = 25/√(625h^2)
sec^2θ dθ/dt = 625/(625h^2)^3/2 dh/dt
dθ/dt = 1/√(625h^2) dh/dt
so, when dh/dt = r, plug and chug.