Water is leaking out of an inverted right conical tank at a rate of 0.0115 m^3/min. At the same time water is being pumped into the tank at a constant rate. The tank has height 11 meters and the diameter at the top is 7 meters. If the water level is rising at a rate of 0.17 m/min when the height of the water is 2.5 meters, find the rate at which water is being pumped into the tank.

To find the rate at which water is being pumped into the tank, we need to use the concept of related rates. We can start by setting up a few variables and equations to represent the given information.

Let's denote:
- V as the volume of water in the tank (which is a function of time, t).
- h as the height of the water in the tank at time t.

Given information:
- The tank has a height of 11 meters.
- The diameter at the top is 7 meters, so the radius (r) is 7/2 = 3.5 meters.
- The water level is rising at a rate of 0.17 m/min when the height of the water is 2.5 meters.
- Water is leaking out of the tank at a rate of 0.0115 m^3/min.

We know that the volume of a cone is given by V = (1/3)πr^2h, where r is the radius and h is the height. We can differentiate this equation with respect to time t:

dV/dt = (1/3)πr^2 * dh/dt.

Now, we are given that the water level is rising at a rate of 0.17 m/min when the height of the water is 2.5 meters. This means dh/dt = 0.17 m/min when h = 2.5 meters.

So, substituting the given values into the equation above, we have:

dV/dt = (1/3)π(3.5)^2 * 0.17 m/min.

We also know that water is leaking out of the tank at a rate of 0.0115 m^3/min. Therefore, the rate at which water is being pumped into the tank should be the sum of the rate at which the water level is rising and the rate at which water is leaking out:

Rate of water being pumped into the tank = 0.0115 m^3/min + dV/dt.

Now, we can substitute the value of dV/dt that we calculated earlier to find the rate at which water is being pumped into the tank.