Math
posted by Connie on .
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?

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