Question

an inverted cylindrical cone, 44 ft deep and 22 ft across at the top, is being filled with water at a rate of 11 ft3/min. at what rate is the water rising in the tank when the depth of the water is:

Answer 1

The rate of the water rising in the tank when the depth of the water is 1 ft is 56.05 ft/min.

Explain the term rate of change?

• The term "rate of change" (ROC) describes the rate at which something changes over time.

The rate of change of volume is-

dV/dt = 11 ft3/min.

Volume of cylindrical cone = (1/3)πr²h

As we're trying to find dh/dt, we need to calculate the volume within terms of just one variable, h.

Diameter =  22 ft

Thus, radius is 11 ft

By using height and radius of a water in the tank, we may calculate h using similar ratios.

11/44 = r/h

1/4 = r/h

r= h/4

Put into volume;

V = (1/3)π(h/4)h²

V = (1/3)(π)h³ / (16)

dV/dt = [(1/3)(3)(π)(h)² / 16](dh/dt)

V = [(π)(h)² / 16](dh/dt)

Solve for (dh/dt) ;

dh/dt = (dV/dt)(16) / [((π)(h)²]

Now, the change in volume is dV/dt = 11 ft³/min

dh/dt = (11 ft³/min)(16) / [((π)(h)²]

For h = 1 ft,

dh/dt = (11 ft³/min)(16) / [((π)(1 ft)²]

dh/dt ≈ 56.05 ft/min

Thus, the rate of the water rising in the tank when the depth of the water is 1 ft is 56.05 ft/min.

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