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Suppose that water is pouring into a swimming pool in the shape of a right circular cylinder at a constant rate of 3 cubic feet per minute. If the pool has radius 6 feet and height 10 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 6 feet?

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BOB

  1. at a time of t minutes, let the height of the water be h ft
    V = π(6^2)(h) = 36π h
    dV/dt = 36π dh/dt

    subbing in our values
    3 = 36π dh/dt
    dh/dt = 1/(12π) ft/min

    Did you notice that the rate was constant (it did not depend on what h was) ?

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    Reiny

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