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.

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.

A’b’c’ are the coordinates to the new location of the transformed triangle abc. find the distance between the coordinates points to use for your calculations.