To find how fast the height of the water is increasing in the tank, we can use the relationship between the volume and height in a cylinder.

1. Given Data:

   • Radius $r = 5$ meters
   • Rate of volume increase $\frac{dV}{dt} = 4$ cubic meters per minute

2. Volume of a Cylinder: The volume $V$ of a cylinder is given by: $V = \pi r^2 h$ where $h$ is the height of the water in the tank.

3. Differentiate with Respect to Time: Since we are looking for $\frac{dh}{dt}$, the rate at which the height of the water is increasing, we can differentiate both sides of the volume formula with respect to time $t$: $\frac{dV}{dt} = \pi r^2 \frac{dh}{dt}$

4. Plugging in the Known Values: Substitute $r = 5$ meters and $\frac{dV}{dt} = 4$ m³/min: $4 = \pi (5)^2 \frac{dh}{dt}$ Simplify the right side: $4 = 25\pi \frac{dh}{dt}$

5. Solve for $\frac{dh}{dt}$: $\frac{dh}{dt} = \frac{4}{25\pi}$ Calculating this value, we get: $\frac{dh}{dt} \approx 0.0509 \text{ m/min}$

Answer

The height of the water is increasing at approximately $0.0509$ meters per minute.

Would you like further details on any part of this solution?

Here are some related questions:

1. What if the radius were doubled? How would that affect the rate of height increase?
2. How would the rate of height increase change if the volume increase was 6 m³/min instead?
3. What is the total time required to fill the tank to a certain height, given this rate?
4. How does this concept apply to other shapes, such as a cone or sphere?
5. What units are typically used for rates like $\frac{dh}{dt}$, and why?

Tip: For problems involving rates of change in geometry, identifying relationships between dimensions is key to setting up correct equations.