Water is leaking out of an inverted conical tank at a rate of 0.0109 {\rm m}^3{\rm /min}. At the same time water is being pumped into the tank at a constant rate. The tank has height 9 meters and the diameter at the top is 4.5 meters. If the water level is rising at a rate of 0.24 m/min when the height of the water is 4 meters, find the rate at which water is being pumped into the tank. Must be accurate to fifth decimal place

To solve this problem, we'll need to use related rates. Let's denote the radius at any given height h as r(h). We are given that the tank is inverted and has a height of 9 meters and a diameter (which is twice the radius) at the top of 4.5 meters.

First, let's find the formula for the radius of the cone at any given height. From similar triangles, we can write:

r(h) / h = r / 9,

where r is the radius at the top of the cone (when h = 9). Rearranging this equation, we find:

r(h) = r * h / 9,

where r represents the radius at the top.

Next, we are given that the height of the water, h, is changing with time. We'll denote the rate of change of height with respect to time as dh/dt.

We are also given that the water is leaking out of the tank at a rate of 0.0109 m^3/min. This means that the volume of water in the tank is decreasing over time. We'll denote the rate of change of volume with respect to time as -dV/dt (negative because the volume is decreasing).

The volume V of the cone is given by the formula:

V(h) = (1/3) * π * r(h)^2 * h.

Differentiating this formula with respect to time t, we have:

dV / dt = (1/3) * π * [2 * r(h) * (dr/dt) * h + r(h)^2 * (dh/dt)].

We know that the rate of change of volume, dV/dt, is -0.0109 m^3/min, and we are given that dh/dt = 0.24 m/min when h = 4. Plugging in these values, we can solve for dr/dt.

-0.0109 = (1/3) * π * [2 * r(4) * dr/dt * 4 + r(4)^2 * 0.24].

We need to find r(4) to complete the equation. Plugging in h = 4 into the equation r(h) = r * h / 9, we find:

r(4) = (4 * r) / 9.

Substituting this back into the previous equation, we have:

-0.0109 = (1/3) * π * [2 * (4 * r) / 9 * dr/dt * 4 + [(4 * r) / 9]^2 * 0.24].

Simplifying this equation, we can solve for dr/dt.

Now we need to find the value of r. We're given that the diameter at the top is 4.5 meters, which means the radius r at the top is 4.5 / 2 = 2.25 meters.

Substituting r = 2.25 into the equation and solving for dr/dt will give us the rate at which water is being pumped into the tank.