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Assume that the water level is rising at a rate of 0.3m/min when it is 2m

User Realistic
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The rate of change of the volume of water is 0.8π cubic meters per minute.

Given:

Height of the conical tank (h) = 3m

Radius of the conical tank at the top (r) = 2m

Rate of change of water level (dh/dt) = 0.3m/min

Water level (h) = 2m

We need to find the rate of change of the volume of water (dV/dt).

The volume of a cone is given by:

V = (1/3)πr²h

We can differentiate both sides of the equation to find the rate of change of the volume of water:

dV/dt = (1/3)πr²dh/dt

Plugging in the given values:

dV/dt = (1/3)π(2m)²(0.3m/min) = 0.8π m³/min

Therefore, the rate of change of the volume of water is 0.8π cubic meters per minute.

Assume that the water level is rising at a rate of 0.3m/min when it is 2m-example-1
User Naaman
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