Question

A ladder leans against a vertical wall and the top of the ladder is sliding down the wall at a
constant rate of 1/2 ft/sec. At the moment when the top of the ladder is 16 feet above the
ground, the bottom of the ladder is sliding away from the wall (horizontally) at the rate of
2/3 ft/sec. At what rate will the bottom of the ladder be sliding away from the wall when the
top of the ladder is 12 feet above the ground?

Answer 1

if the ladder base is x feet from the wall and reaches y feet high and the ladder length is a,

x^2+y^2 = a^2
2x dx/dt + 2y dy/dt = 0
we are told that dy/dt = -1/2
so, when y=16,

2x (2/3) + 2(16)(-1/2) = 0
x=12
so, x^2+y^2 = 12^2+16^2 = 20^2

when y=12, x=16, and we have

2(16) dx/dt + 2(12)(-1/2) = 0
dx/dt = 3/8