math - calculus

Suppose that water is pouring into a swimming pool in the shape of a right circular cylinder at a constant rate of 9 cubic feet per minute. If the pool has radius 3 feet and height 11 feet, what is the rate of change of the height of the water in the pool when the depth of the water in the pool is 9 feet?

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  1. Let the height of water be h ft

    Volume of water = π(3^2)h = 9πh
    dV/dt = 18π dh/dt
    9/(18π) = dh/dt
    dh/dt = 1/(2π) , a constant no matter what the height is

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    posted by Reiny
  2. The above answer is incorrect..

    everything is right except v=9pi(h)
    differentiate both sides. gives you...
    dv/dt=9pi dh/dt

    9/9pi= dh/dt.
    dh/dt= 1/pi

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    posted by Tom

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