Answer:
285,753 cm³/min
Step-by-step explanation:
The rate of change of volume is the product of the water's surface area and its rate of change of depth.
At a depth of 2 m, the water has filled 1/3 of the 6 m depth of the tank. So, the radius at the water's surface will be 1/3 of the tank's radius of 2 m. The water's surface area is ...
A = πr² = π(2/3 m)² = 4π/9 m²
The rate of change of depth is 0.2 m/min, so the volume of water is increasing at the rate ...
dV/dt = (0.20 m/min)(4π/9 m²) = 8π/90 m³/min ≈ 279253 cm³/min
This change in volume is the difference between the rate at which water is being pumped in and the rate at which it is leaking out:
2.8×10⁵ cm³/min = (input rate) - 6500 cm³/min
Adding 6500 cm³/min to the equation, we get ...
input rate ≈ 285,753 cm³/min
