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an inverted pyramid is being filled with water at a constant rate of 60 cubic centimeters per second. the pyramid, at the top, has the shape of a square with sides of length 2 cm, and the height is 10 cm. find the rate at which the water level is rising when the water level is 8 cm.

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Answer: he volume of a pyramid is 1/3 Bh, where B is the area of the base and h is the height. Since the base is square, B = s2, where s is the side of the base. So the volume is 1/3 s2h.

We know that the full pyramid has a height of 11 cm and a side of the base is 8 cm. Also, no matter what the height of the water is, the surface of the water will form a square which has the same ratio to the depth of the water as a side of the top of the pyramid does to the full height of the pyramid. So that means that no matter what the depth of the water, s = 8/11 h.

So that means that the formula for the volume of the pyramid formed by the water is V = 1/3 (8/11 h)2 h = 64/363 h3.

Differentiate this as a function of time to get the rate of change of the volume of water:

V = 64/363 h3

dV/dt = 3 (64/363) h2 dh/dt = 192/363 h2 dh/dt

But we know that the volume is changing at a rate of 65 cc / sec. So at any point in time,

65 = 192/363 h2 dh/dt

65 * 363 / 192 = h2 dh/dt

23595 / 192 = h2 dh/dt

(23595 / 192) / h2 = dh/dt

When the water level (h) is 8, that means:

(23595 / 192) / 64 = dh/dt

1.92 = dh/dt

So the water level is rising at the rate of 1.92 cm per second when the water level is 8 cm.

Explanation:

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