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Math

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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 -

    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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