Calculus: A ladder 15ft long rests against a vertical wall.?

Question

A ladder 15ft long rests against a vertical wall. If the bottom of the ladder slides away from the wall @ a rate of 3 ft/s, how fast is the top of the ladder sliding down the wall when the bottom of the ladder is 6ft from the wall.

Answer 1

Let x be the distance along the floor from the wall to the foot of the ladder, and let y be the distance from the floor to the top of the ladder. If you draw a diagram of this, you have a right triangle whose hypotenuse is the ladder. The ladder is 15 ft long, so x and y are related by

x² + y² = 15²

Take derivative with respect to t of both sides:

(2x)(dx/dt) + (2y)(dy/dt) = 0 so

dy/dt = -(x/y)(dx/dt)

You're given values for dx/dt and x; use the right-triangle relation between x and y to determine what y is at that time, and use these to evaluate dy/dt.

Answer 2

h^2+x^2 = 15^2, where h is the height and x is the horizontal leg.
Differentiate with respect to time t,
2hh' + 2xx' = 0
h' = -xx'/h = -6(3)/sqrt(15^2-6^2) = -18/sqrt(189) ft/s

Answer 3

It's doesn't matter. OSHA will consider this ladder unsafe.