The ladder, the ground and the wall form a right triangle; the ladder length (L) is the longest side of this triangle. Â 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. Â Note that since the ladder length does not change, dL/dt = 0. Â This leaves us with
   dh      dx
2h ---- + 2x -----Â = 0.
    dt      dt
Since x^2 + h^2 = 15^2 = 225, Â Â Â Â Â h^2 = 225 - (5 ft)^2 = 200, or
200 ft^2 = h^2. Â Then h = + sqrt(200 ft^2)Â
Substituting this into the differential equation, above:
2[sqrt(200)] (dh/dt) + 2 (5) (4 ft/sec) = 0. Â 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.
