The ladder, the ground and the wall form a right triangle; the ladder length (L) is the longest side of this triangle. L^2 = h^2 + x^2, where h represents the height of the point on the wall where the ladder touches the wall, and x represents the distance of the base of the ladder from the wall.
We need dh/dt, which will be negative because the top of the ladder is sliding down the wall.
Starting with h^2 + x^2 = L^2, we differentiate (and subst. known values such as x = 5 feet and 4 ft/sec to find dh/dt. Note that since the ladder length does not change, dL/dt = 0. This leaves us with
dh dx
2h ---- + 2x ----- = 0.
dt dt
Since x^2 + h^2 = 15^2 = 225, h^2 = 225 - (5 ft)^2 = 200, or
200 ft^2 = h^2. Then h = + sqrt(200 ft^2)
Substituting this into the differential equation, above:
2[sqrt(200)] (dh/dt) + 2 (5) (4 ft/sec) = 0. Solve this for the desired quantity, dh/dt:
[sqrt(200)] (dh/dt) + (5)(4) = 0, or
dh/dt = -20 / sqrt(200) = (-1.41 ft / sec) (answer)
This result is negative because the top of the ladder is moving downward.