Water is leaking out of an inverted conical tank at a rate of 13200 cubic centimeters per min at the same time that water is being pumped into the tank at a constant rate. The tank has height 9 meters and the diameter at the top is 7 meters. If the water level is rising at a rate of 28 centimeters per minute when the height of the water is 3 meters, 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 use the principle of related rates. We need to relate the rate at which water is being pumped into the tank to the rate at which the water level is rising.
Let's denote the volume of water in the tank as V, the radius of the water surface as r, and the height of the water in the tank as h.
We are given the following information:
- Water is leaking out of the tank at a rate of 13200 cubic centimeters per minute.
- The water level is rising at a rate of 28 centimeters per minute when the height of the water is 3 meters.
We know that the volume of a cone is given by the formula V = (1/3)πr^2h, where r is the radius of the base and h is the height.
To solve the problem, we'll need to find an expression that relates the rate of change of the water level to the rate of change of the volume of water in the tank.
First, let's find the expression for the radius of the water surface in terms of the height h. Since the diameter of the top of the tank is 7 meters, the radius can be obtained by dividing the diameter by 2, giving us r = (7/2) meters.
To convert the radius from meters to centimeters, multiply it by 100 because 1 meter = 100 centimeters, so r = (7/2) * 100 = 350 centimeters.
Next, we'll differentiate the formula for the volume of a cone with respect to time t using the chain rule:
dV/dt = (1/3)π(2rh * dh/dt + r^2 * dh/dt)
Since we want to find the rate at which water is being pumped into the tank (dV/dt), we can rearrange the equation to solve for it:
dV/dt = (1/3)π(2rh * dh/dt + r^2 * dh/dt) - 13200
Plugging in the known values, we have:
(1/3)π(2 * 350 * h * dh/dt + 350^2 * dh/dt) - 13200 = dV/dt
Now let's substitute the given values into this equation to find the rate at which water is being pumped into the tank.
When the height of the water is 3 meters, the radius of the water surface can be calculated as:
r = (7/2) * 100 = 350 centimeters
Using this value, the equation becomes:
(1/3)π(2 * 350 * 3 * dh/dt + 350^2 * dh/dt) - 13200 = 28
Simplifying the equation, we have:
350 * 3π * dh/dt + 350^2 * π * dh/dt - 39600 = 84π
Now, let's combine like terms and solve for dh/dt:
(1050π + 122500π) * dh/dt = 39600 * π + 84π
Calculating the left-hand side:
(1050π + 122500π) * dh/dt = 123550π * dh/dt
Calculating the right-hand side:
39600π + 84π = 39750π
Now, solve for dh/dt by dividing both sides of the equation by 123550π:
dh/dt = (39750π) / (123550π) ≈ 0.321 centimeters per minute
Therefore, the rate at which water is being pumped into the tank is approximately 0.321 cubic centimeters per minute.