A 20 foot ladder is leaning against a wall. The base of the ladder is placed 6 feet from the bottom of the wall. The base of the ladder slips away from the wall at a constant rate of 2 feet/sec. Find the velocity of the top of the ladder as it slider down the wall 3 seconds after it begins to fall.

Answer 1

At a time of t seconds, let the distance the foot is away from the wall be x ft, and let the top of the ladder be y ft above the ground
x^2 + y^2 = 20^2
2x dx/dt + 2y dy/dt = 0
given dx/dt = 2
when t = 3, x = 2(3) = 6ft
when x = 6, 36 + y^2 = 400
y = √(400-36) = 2√91
in 2x dx/dt + 2y dy/dt = 0
12(2) + 4√91 dy/dt = 0
dy/dt = .... , notice the negative indicates y is decreasing, thus the ladder is sliding down

Answer 2

6^2 + y^2 = 20^2,
Y = 19 Ft. = ht. of ladder against wall.
r * T = 19.
r * 3 = 19,
r = 6.36 Ft/s.