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6 March, 08:45

A cup has the shape of a right circular cone. The height of the cup is 12 cm, and the radius of the opening is 3 cm. Water is poured into the cup at a constant rate of 32 cm / sec. What is the rate at which the water level is rising when the depth of water in the cup is 5cm?

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  1. 6 March, 09:03
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    about 6.52 cm/s

    Step-by-step explanation:

    The radius of the cone is 3/12 = 1/4 of its height, so the radius of the water surface at the time of interest is (5 cm) / 4 = 1.25 cm.

    The area of the water's surface is ...

    A = πr² = π (5/4 cm) ² = 25π/16 cm²

    The rate of change of depth multiplied by this area will give the rate of change of volume.

    dV/dt = (25π/16 cm²) (dh/dt)

    dh/dt = (32 cm³/s) / (25π/16 cm²) = 512 / (25π) cm/s ≈ 6.52 cm/s

    The water is rising at the rate of about 6.52 cm/s.
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