Question

The top of a 15-foot ladder leans against a wall, and the bottom of the ladder slides away from the wall at a rate of 4 ft/sec. find the velocity of the top of the ladder at time t = 1 sec if the bottom of the ladder is 5 feet away from the wall at time t = 0.

Answer 1

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.