Water is leaking out of an inverted conical tank at a rate of 10600.0 cm3/min at the same time that water is being pumped into the tank at a constant rate. The tank has height 14.0 m and the the diameter at the top is 6.0 m. If the water level is rising at a rate of 27.0 cm/min when the height of the water is 1.5 m, find the rate at which water is being pumped into the tank in cubic centimeters per minute.

draw a cross-section. When the water has depth y, the radius of the surface is r = 3/14 y

So, the volume of the water at depth y is

v = 1/3 πr^2y
= π/3 (3/14 y)^2 y
= 3π/196 y^3
dv/dt = 9π/196 y^2 dy/dt
If the water is being pumped in at a rate of k cm^3/min, then

dv/dt = k-10600

dv/dt = dv/dy dy/dt, so

k-10600 = 9π/196 y^2 dy/dt
Plugging in the given values, we have
k-10600 = 9π/196 * 150^2 * 27
k-10600 = 87636
k = 98436 cm^3/min

To find the rate at which water is being pumped into the tank, we can use the concept of related rates. We need to find an equation that relates the rate at which the water level is rising to the rate at which water is being pumped into the tank.

Given:
- The water is leaking out of the tank at a rate of 10600.0 cm3/min.
- The water level is rising at a rate of 27.0 cm/min.
- The tank has a height of 14.0 m and a diameter of 6.0 m.

Let's denote:
- The height of the water in the tank as h (in meters).
- The rate at which water is being pumped into the tank as V (in cm3/min).

We can start by finding the relationship between the height and radius of the tank at any given time.

The tank is a right circular cone, so we can use similar triangles to relate the height and radius.

The ratio of the height to the radius is constant at any given time, so we have:
h/r = H/R, where r is the radius of the tank at height h, R is the radius at the top (given as 6.0 m), and H is the total height of the tank (given as 14.0 m).

Rearranging this equation, we can express r in terms of h:
r = R * h / H

To find the volume of the tank at a given height h, we can use the formula for the volume of a cone:
V = (1/3) * π * r^2 * h

Substituting the expression for r, we get:
V = (1/3) * π * (R * h / H)^2 * h
V = (1/3) * π * R^2 * h^3 / H^2

Now, let's differentiate this equation with respect to time t, since we are interested in the rate of change:
dV/dt = (1/3) * π * R^2 * (3h^2 * dh/dt) / H^2

We are given that dh/dt = 27.0 cm/min, and we need to find dV/dt, the rate at which water is being pumped into the tank.

The leaking water is reducing the volume of the tank at a rate of 10600.0 cm3/min, so we need to subtract this from dV/dt.

Therefore, the final equation becomes:
dV/dt - 10600.0 = (1/3) * π * R^2 * (3h^2 * dh/dt) / H^2

We are given the height when the water level is 1.5 m, so we can substitute h = 1.5 m, H = 14.0 m, R = 6.0 m, and dh/dt = 27.0 cm/min into the equation to find dV/dt.

By solving this equation, we can find the rate at which water is being pumped into the tank in cubic centimeters per minute.