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Calculus!

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A 16 ft. ladder leans against a wall, with the top being 12 ft. from the wall when time t=0. The top is sliding down at a constant rate of 4 ft./second. Given that the top of the ladder touches the ground at 3 seconds, what is the velocity of the bottom of the ladder when t=3?

  • Calculus! -

    I suspect a typo somewhere. When the ladder is lying flat at t=3, the base is no longer moving.

  • Calculus! -

    Make a sketch:

    x^2 + y^2 = 256
    when t=0 , y=12, x = √112
    when t = 1, y = 8 , x = √192
    when t = 2 , y = 4 , x = √240
    when t = 3, y = 0, x = √256 = 16

    given dy/dt = -4 ft/s

    2x dx/dt + 2y dy/dt = 0

    dx/dt = -y (dy/dt) / x
    = 0/16 = 0

    check:
    let t=1
    y = 8, y = √192
    dx/dt = -8(-4)/√192 = appr 2.3

    let t = 2
    y = 4, x = √240
    dx/dt = -4(-4)/√240 = appr 1.03

    slowing up ....

    let t = 2.99 sec
    y = 12 - 2.99(4) = .04 , x = √255.9984
    dx/dt = -.04(-4)/√255.9984 = appr .01 (close to zero)

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