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To solve for \( \frac{dh}{dt} \), we’ll use the information provided and the relationship given:

1. We are given that the rate of change of volume, \( \frac{dV}{dt} \), is \( 3 \, \text{m}^3/\text{min} \).
2. The derivative equation provided is:
   \[
   \frac{dV}{dt} = 64\pi \frac{dh}{dt}
   \]
   
### Step-by-Step Solution

1. Substitute \( \frac{dV}{dt} = 3 \) into the equation:
   \[
   3 = 64\pi \frac{dh}{dt}
   \]

2. Solve for \( \frac{dh}{dt} \):
   \[
   \frac{dh}{dt} = \frac{3}{64\pi}
   \]

3. Now, calculate \( \frac{3}{64\pi} \):
   \[
   \frac{dh}{dt} \approx \frac{3}{201.06} \approx 0.0149 \, \text{m/min}
   \]

So, \( \frac{dh}{dt} \approx 0.0149 \, \text{m/min} \).

### Final Answer
- \( \frac{dh}{dt} \approx 0.0149 \, \text{m/min} \)
- Therefore, the height of the water in the tank is increasing at approximately \( 0.0149 \, \text{m/min} \).

Would you like a more detailed explanation on any of these steps?

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#### Related Questions
1. What would happen to \( \frac{dh}{dt} \) if the rate \( \frac{dV}{dt} \) increased?
2. How would the result change if the constant in the denominator were different?
3. What does \( \frac{dh}{dt} \) represent physically in this scenario?
4. How does the volume formula relate to the height of water in the tank?
5. How can calculus be used to model the filling of irregularly shaped tanks?

#### Tip
When working with rates, always check if the units are consistent, as unit mismatches can lead to errors in interpretation and calculations.

The tank is being filled at a rate of 3 m³/min. Given that dV/dt = 3, substitute this into dV/dt = (64π) * (dh/dt) and solve for dh/dt. Determine the rate at which the height of the water in the tank is increasing in m/min.

Solve a related rates problem involving the rate at which the height of water in a tank increases as the tank is filled at a rate of 3 m³/min. This solution demonstrates the use of calculus and differentiation in solving for dh/dt.

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