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Water runs into a conical tank (pointy part facing downward) at the rate of 12 ft^3/minute. The tank stands down and has a height of 12 ft. and a base radius of 6 ft. How fast is the water level rising when the water is 10 ft deep?

The tank is an inverted cone with height of 12 ft and radius of 6 ft.

The water in the tank is the shape of a cone with height h, radius r, and volume V.

dV/dt = 12 ft³/minute

The water and tank are similar cones, so r = h/2.

V = ⅓πr²h = πh³/12

dV/dh = πh²/4

dV/dt = dV/dh · dh/dt

12 = πh²/4 · dh/dt

dh/dt = 48/(πh²)

When h = 10, dh/dt = 48/π10² ≅ 0.153 ft/minute.

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