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Sagot :
To find the rate at which the water level is rising, we can use the formula for the volume of a cylinder: V = πr^2h, where r is the radius and h is the height of the water.
Given that the radius is 1 and the length is 2, the volume V = π(1)^2(2) = 2π.
Since water is being pumped in at a rate of m^3 per minute, the rate of change of volume with respect to time is dm/dt = m.
To find the rate at which the water level is rising when the water is m deep, we can differentiate the volume formula with respect to time t: dV/dt = πr^2 dh/dt.
Substitute the known values:
2π = π(1)^2 dh/dt,
2 = dh/dt.
Therefore, the rate at which the water level is rising when the water is m deep is 2 units per minute.
Given that the radius is 1 and the length is 2, the volume V = π(1)^2(2) = 2π.
Since water is being pumped in at a rate of m^3 per minute, the rate of change of volume with respect to time is dm/dt = m.
To find the rate at which the water level is rising when the water is m deep, we can differentiate the volume formula with respect to time t: dV/dt = πr^2 dh/dt.
Substitute the known values:
2π = π(1)^2 dh/dt,
2 = dh/dt.
Therefore, the rate at which the water level is rising when the water is m deep is 2 units per minute.
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