Posted by **Connie** on Sunday, December 9, 2012 at 6:21pm.

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?

- Math -
**Steve**, Sunday, December 9, 2012 at 7:52pm
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

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