A cylindrical tank with a radius of 5 meters is being filled with water at a rate of 3 cubic meters per minute. Determine the rate at which the water level is rising in the tank when the water is 4 meters deep.
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To solve this problem, we can use the volume of a cylinder formula: V = πr^2h, where V is the volume, r is the radius, and h is the height of the water in the tank.
We are given that the radius of the tank is 5 meters and the rate at which the water is being filled is 3 cubic meters per minute. Let's call the rate at which the water level is rising dh/dt.
We can differentiate the volume formula with respect to time to relate the rate at which the water level is rising (dh/dt) to the given filling rate (dV/dt = 3):
dV/dt = d(πr^2h)/dt
3 = 2πrh * dh/dt
To find dh/dt, we need to determine the value of rh. We are given that the water is 4 meters deep, so the height (h) is 4 meters. Therefore, rh = 5 * 4 = 20.
Plugging this value into the equation, we have:
3 = 2π * 20 * dh/dt
3 = 40π * dh/dt
Now we can solve for dh/dt by dividing both sides of the equation by 40π:
3 / (40π) = dh/dt
Using an approximation for π = 3.14159, we get:
3 / (40 * 3.14159) ≈ 0.02396 m/min
Therefore, the rate at which the water level is rising in the tank when the water is 4 meters deep is approximately 0.02396 meters per minute.
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