Water is leaking out of an inverted conical tank at a rate of 12700 cubic centimeters per min at the same time that water is being pumped into the tank at a constant rate. The tank has height 10 meters and the diameter at the top is 5 meters. If the water level is rising at a rate of 22 centimeters per minute when the height of the water is 5 meters, find the rate at which water is being pumped into the tank in cubic centimeters per minute. \[ \frac{\mathrm{cm}^{3}}{\mathrm{~min}} \]

Step 1 ：The volume of the water in the tank changes by 12700 cubic centimeters each minute, while the height of the water in the tank rises by 22 centimeters each minute. The volume of a cone is \(\frac{1}{3}\pi r^2h\). If we look at the changes in volume and height in just one minute, we can solve for the rate at which water is being pumped into the tank.

Step 2 ：First, we need to express the radius in terms of the height. Since the height and diameter of the tank are in a 2:1 ratio, we have \(r = \frac{h}{2}\).

Step 3 ：Substitute \(r = \frac{h}{2}\) into the volume formula, we get \(V = \frac{1}{3}\pi (\frac{h}{2})^2h = \frac{1}{12}\pi h^3\).

Step 4 ：Then, we differentiate both sides with respect to time \(t\), we get \(\frac{dV}{dt} = \frac{1}{4}\pi h^2\frac{dh}{dt}\).

Step 5 ：Substitute \(h = 5\) meters and \(\frac{dh}{dt} = 22\) centimeters per minute into the above equation, we get \(\frac{dV}{dt} = \frac{1}{4}\pi (5)^2(22) = 1375\pi\) cubic centimeters per minute.

Step 6 ：Since the water is leaking out of the tank at a rate of 12700 cubic centimeters per minute, the rate at which water is being pumped into the tank is \(1375\pi + 12700 = 12700 + 4319.46891 = 17019.46891\) cubic centimeters per minute.

Step 7 ：Since the rate at which water is being pumped into the tank must be positive, the final answer is \(\boxed{17019.46891}\) cubic centimeters per minute.