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We are asked to determine how fast the height of the water column is rising when the water is 9 meters deep in a cylindrical tank, given that:

- Water is being pumped into the tank at a rate of \(15 \, \text{m}^3/\text{min}\).
- The height \(h\) of the water column is always four times the radius \(r\) of the column, i.e., \( h = 4r \).
- The volume of a cylinder is given by \( V = \pi r^2 h \).

### Step 1: Express everything in terms of \(h\)
We are told that \( h = 4r \), so the radius \(r\) can be written as:
\[
r = \frac{h}{4}
\]

Now, substitute this into the volume formula \( V = \pi r^2 h \):
\[
V = \pi \left( \frac{h}{4} \right)^2 h
\]
\[
V = \pi \left( \frac{h^2}{16} \right) h = \frac{\pi h^3}{16}
\]

### Step 2: Differentiate with respect to time
We now differentiate the volume \(V\) with respect to time \(t\) to relate the rate of change of height \( \frac{dh}{dt} \) to the rate of change of volume \( \frac{dV}{dt} \):
\[
\frac{dV}{dt} = \frac{d}{dt} \left( \frac{\pi h^3}{16} \right)
\]
\[
\frac{dV}{dt} = \frac{3\pi h^2}{16} \cdot \frac{dh}{dt}
\]

### Step 3: Plug in the given values
We are given that \( \frac{dV}{dt} = 15 \, \text{m}^3/\text{min} \) and \( h = 9 \, \text{m} \). Substituting these into the differentiated equation:
\[
15 = \frac{3\pi (9)^2}{16} \cdot \frac{dh}{dt}
\]
\[
15 = \frac{3\pi \cdot 81}{16} \cdot \frac{dh}{dt}
\]
\[
15 = \frac{243\pi}{16} \cdot \frac{dh}{dt}
\]

### Step 4: Solve for \( \frac{dh}{dt} \)
Now, solve for \( \frac{dh}{dt} \):
\[
\frac{dh}{dt} = \frac{15 \cdot 16}{243\pi}
\]
\[
\frac{dh}{dt} = \frac{240}{243\pi}
\]
\[
\frac{dh}{dt} \approx \frac{240}{763.407} \approx 0.314 \, \text{m/min}
\]

Thus, the height of the water column is rising at approximately **0.314 meters per minute** when the water is 9 meters deep.

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Do you have any questions or would you like more details?

Here are five related questions to expand on this topic:
1. How would the result change if the radius was always three times the height instead of four?
2. What is the physical significance of the rate of volume change \( \frac{dV}{dt} \) in this context?
3. How would you approach the problem if the relationship between height and radius was not linear?
4. What would happen to the rate of height increase if the pumping rate were doubled?
5. How could this concept be applied to real-world engineering problems involving fluid dynamics?

**Tip**: Understanding how to relate variables through differentiation is key to solving many real-world rate of change problems!

Water is being pumped into a cylindrical tank at a rate of 15 m^3 per minute, forming a cylindrical water column. At all times, the height of the water column is four times the radius of the column. How fast is the height of the water column rising when the water is 9 meters deep? The volume of a cylinder with radius r and height h is given by V = πr^2h.

Solve a related rates problem involving a cylindrical tank filled at a rate of 15 m^3/min. The height of the water column is four times the radius. Learn how fast the height is rising when the water is 9 meters deep using implicit differentiation.

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