Water is leaking out of an inverted conical tank at a rate of 11400 cm^3/min at the same time that water is being pumped into the tank at a constant rate. The tank has height 8.0m and the the diameter at the top is 3.5m. If the water level is rising at a rate of 19.0cm/min when the height of the water is 4.5m , find the rate at which water is being pumped into the tank in cubic centimeters per minute.
To find the rate at which water is being pumped into the tank, we can first calculate the rate at which the volume of water inside the tank is changing. Then, we can determine the difference between the rate at which water is leaking and the rate of change of the volume inside the tank.
Step 1: Find the volume of the tank at the given height
The tank has the shape of an inverted cone, which means its volume can be calculated using the formula:
V = (1/3) * π * r^2 * h
where V is the volume, π is a mathematical constant approximately equal to 3.14159, r is the radius of the tank's cross-section, and h is the height of the water inside the tank.
Given that the diameter at the top is 3.5m, the radius (r) is half of that: r = 3.5/2 = 1.75m.
The height of the water inside the tank is given as 4.5m.
Using these values, we can calculate the volume (V1) when the height is 4.5m.
V1 = (1/3) * π * (1.75^2) * 4.5
V1 ≈ 16.313 m^3 (cubic meters)
Step 2: Find the rate of change of the volume inside the tank
We are given that the water level is rising at a rate of 19.0cm/min. This means that the height of the water inside the tank is changing at this rate. We need to convert this rate from cm/min to m/min before we can use it in the volume formula.
19.0cm/min = 19.0/100 m/min = 0.19 m/min
Step 3: Find the rate at which water is being pumped into the tank
Let's assume the rate at which water is being pumped into the tank is P cm^3/min.
The volume inside the tank is changing because water is being pumped in as well as leaking out, so we can set up the following equation:
P - 11400 = (dV/dt)
where (dV/dt) is the rate of change of the volume inside the tank, which we can calculate using the height rate of change (0.19 m/min).
Substituting the values, we get:
P - 11400 = (dV/dt)
P - 11400 = (d/dt) [(1/3) * π * (1.75^2) * h]
P - 11400 = (1/3) * π * (1.75^2) * (0.19)
Now, we can substitute the value of P by the expression (11400 + dV/dt) and solve for dV/dt.
11400 + dV/dt - 11400 = (1/3) * π * (1.75^2) * (0.19)
dV/dt = (1/3) * π * (1.75^2) * (0.19)
dV/dt ≈ 0.934 m^3/min
Step 4: Convert the rate of change of volume from m^3/min to cm^3/min
To find the rate at which water is being pumped into the tank in cubic centimeters per minute, we need to convert the rate of change of volume from m^3/min to cm^3/min.
1 m^3 = 1,000,000 cm^3
So, 0.934 m^3/min = 0.934 * 1,000,000 cm^3/min
Therefore, the rate at which water is being pumped into the tank is approximately 934,000 cm^3/min.