Water is leaking out of an inverted conical tank at a rate of 13200.0 cubic centimeters per min at the same time that water is being pumped into the tank at a constant rate. The tank has height 14.0 meters and the diameter at the top is 3.5 meters. If the water level is rising at a rate of 21.0 centimeters per minute when the height of the water is 4.0 meters, find the rate at which water is being pumped into the tank in cubic centimeters per minute.
To solve this problem, we can use the related rates concept from calculus.
First, let's denote:
- V as the volume of water in the tank (in cubic centimeters)
- h as the height of the water in the tank (in meters)
- r as the radius of the water surface in the tank (in meters)
- dh/dt as the rate at which the height of the water is changing (in centimeters per minute)
- dV/dt as the rate at which the volume of water is changing (in cubic centimeters per minute)
We are given the following information:
- dh/dt = 21.0 cm/min (the rate at which the water level is rising)
- h = 4.0 m (the current height of the water)
- r = (3.5 m) / 2 = 1.75 m (the radius of the top of the tank)
- V = (1/3) * π * r^2 * h (the volume of a cone formula)
We need to find dV/dt, the rate at which the volume of water is changing.
To find dV/dt, we need to relate it to the given rates and variables by using the chain rule of calculus.
1. First, let's differentiate the volume equation with respect to t (time):
dV/dt = (dV/dh) * (dh/dt)
2. Now, let's find dV/dh, the derivative of the volume equation with respect to h:
dV/dh = (d/dh)[(1/3) * π * r^2 * h]
= (1/3) * π * r^2 * (d/dh)[h]
= (1/3) * π * r^2
3. Plug in the values for r and solve for dV/dh:
dV/dh = (1/3) * π * (1.75 m)^2
≈ 9.62 m^2π
4. Finally, substitute the value of dV/dh and dh/dt into the equation from step 1 to find dV/dt:
dV/dt = (9.62 m^2π) * (21.0 cm/min)
≈ 6405.23 cm^3/min
Therefore, the rate at which water is being pumped into the tank is approximately 6405.23 cubic centimeters per minute.