A ladder of 8 m is leaning against a wall begins to slide. If its upper end slides down the wall at a rate of 0.25 m/sec, at what rate is the angle between the ladder and the ground changing when the foot of the ladder is 5 m from the wall?

Answer 1

If the base to the wall is x, and the height up the wall is y, then
x^2+y^2 = 8^2
tanθ = y/x
sec^2θ dθ/dt = (x dy/dt - y dx/dt)/x^2
Now plug in your values for x and y, and find dθ/dt