Water is leaking out of an inverted conical tank at a rate of 6 liters per minute. At
the same time that water is being pumped into the tank at a constant rate. The tank has a
height of 18 meters and the diameter at the top is 12 meters. If the water level is rising at a
rate of 15 centimeters per minute when the height of the water is 2 meters, find the rate at
which water is being pumped into the tank in milliliters per minute.

To solve this problem, we need to apply the concepts of related rates in calculus. Here’s how we can approach it:

### Step 1: Understand the problem and the given values

1. **Tank shape**: Inverted conical tank.
2. **Height of the tank** (\( H \)) = 18 meters.
3. **Diameter of the tank at the top** = 12 meters, so the radius (\( R \)) = 6 meters.
4. **Rate at which water is leaking** = 6 liters per minute.
5. **Water level rising rate** (\( \frac{dh}{dt} \)) = 15 cm/min = 0.15 meters/min.
6. **Current height of the water** (\( h \)) = 2 meters.
7. **We need to find** the rate at which water is being pumped into the tank, in milliliters per minute.

### Step 2: Set up the relationship between the variables

The volume \( V \) of water in the conical tank is given by the formula:

\[
V = \frac{1}{3} \pi r^2 h
\]

where \( r \) is the radius of the water surface and \( h \) is the height of the water.

### Step 3: Relate the radius \( r \) to the height \( h \)

Using similar triangles, since the tank’s full height and radius are 18 m and 6 m respectively, the relationship between \( r \) and \( h \) is:

\[
\frac{r}{h} = \frac{6}{18} = \frac{1}{3}
\]

So,

\[
r = \frac{h}{3}
\]

### Step 4: Substitute \( r \) into the volume formula

Substituting \( r = \frac{h}{3} \) into the volume formula:

\[
V = \frac{1}{3} \pi \left(\frac{h}{3}\right)^2 h = \frac{1}{3} \pi \frac{h^3}{9} = \frac{\pi h^3}{27}
\]

### Step 5: Differentiate the volume with respect to time

Differentiate \( V = \frac{\pi h^3}{27} \) with respect to time \( t \):

\[
\frac{dV}{dt} = \frac{\pi}{27} \cdot 3h^2 \cdot \frac{dh}{dt} = \frac{\pi h^2}{9} \cdot \frac{dh}{dt}
\]

### Step 6: Plug in the given values

Now, substitute \( h = 2 \) meters and \( \frac{dh}{dt} = 0.15 \) meters/min into the differentiated equation:

\[
\frac{dV}{dt} = \frac{\pi (2)^2}{9} \times 0.15
\]

\[
\frac{dV}{dt} = \frac{\pi \times 4}{9} \times 0.15 = \frac{0.6\pi}{9} \text{ cubic meters per minute}
\]

Convert cubic meters to liters (since \(1 \text{ cubic meter} = 1000 \text{ liters}\)):

\[
\frac{dV}{dt} = \frac{0.6\pi \times 1000}{9} \text{ liters per minute}
\]

\[
\frac{dV}{dt} = \frac{600\pi}{9} \approx 209.44 \text{ liters per minute}
\]

### Step 7: Account for the water being leaked

Let \( R \) be the rate at which water is pumped into the tank.

Since the tank is leaking water at 6 liters per minute, the actual rate of water being pumped in \( R \) is:

\[
R = \frac{dV}{dt} + 6 \text{ liters per minute}
\]

\[
R \approx 209.44 + 6 \approx 215.44 \text{ liters per minute}
\]

### Step 8: Convert to milliliters per minute

Since \(1 \text{ liter} = 1000 \text{ milliliters}\):

\[
R \approx 215.44 \times 1000 \approx 215,440 \text{ milliliters per minute}
\]

### Final Answer:

The rate at which water is being pumped into the tank is approximately **215,440 milliliters per minute**.

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Would you like to dive deeper into any part of this solution or explore related topics?

Here are some follow-up questions to consider:

1. How would the problem change if the tank were cylindrical?
2. What would happen if the rate of water leaking varied with time?
3. Can you solve for the rate if the water level is halfway up the tank?
4. What if the radius-to-height ratio of the tank was different?
5. How does the shape of the tank affect the rising water level?
6. Could you solve this problem using numerical methods?
7. What happens if the tank starts with some initial water level?
8. How does the pumping rate change if the water level is very low?

**Tip:** Always double-check unit conversions, especially when dealing with multiple units like meters, liters, and milliliters.

Learn how to solve a related rates problem involving an inverted conical tank leaking water and being pumped in at a constant rate. Calculate the rate of water being pumped into the tank given specific dimensions and rates. Suitable for advanced high school students and calculus learners.

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