Calculus

A 17 foot ladder is leaning on a 6 foot fence. The base of the ladder is being pulled away from the fence at the rate of 5 feet/minute. How fast is the top of the ladder approaching the ground when the base is 6 from the fence? [Note: The ladder is protruding over the top of the fence.]

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  1. Let:
    h = the height of the fence is 6'.
    x = distance of the base of the ladder from the fence
    t = length of ladder between top of fence and the ground
    = sqrt(h²+x²)
    L = length of ladder
    y = Height of top of ladder from the ground

    From similar triangles,
    h/t = y/L
    or
    y=Lh/t=Lh/sqrt(h²+x²) ..(1)

    dx/dt=5'/min.
    dy/dt=(dy/dx)*(dx/dt) ..(2)

    So differentiate (1) to get dy/dx and substitute in (2).

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