We are tasked with determining how fast the water level is rising in a hemispherical tank with radius $r = 9$ m, given that water is being pumped in at a rate of $\frac{dV}{dt} = 4 \, \text{m}^3/\text{min}$. Specifically, we need to find $\frac{dh}{dt}$, the rate at which the water level is rising, when the water level $h = 4$ meters.

Key Information

• Radius of the tank: $r = 9 \, \text{m}$
• Inflow rate: $\frac{dV}{dt} = 4 \, \text{m}^3/\text{min}$
• Water height at the moment: $h = 4 \, \text{m}$

The formula for the volume $V$ of the water as a spherical cap of thickness $h$ from a sphere of radius $r$ is given by:

$V = \frac{\pi h^2 (3r - h)}{3}$

Step-by-Step Solution

We will differentiate the volume formula with respect to time $t$ to find $\frac{dh}{dt}$. First, let’s compute $\frac{dV}{dh}$ and then use the chain rule to relate it to $\frac{dV}{dt}$.

The volume formula is:

$V(h) = \frac{\pi h^2 (3r - h)}{3}$

Differentiate this with respect to $h$:

$\frac{dV}{dh} = \frac{\pi}{3} \cdot \left( 2h(3r - h) + h^2(-1) \right)$

Simplify the expression:

$\frac{dV}{dh} = \frac{\pi}{3} \cdot \left( 6rh - 3h^2 \right)$

Substitute $r = 9$:

$\frac{dV}{dh} = \frac{\pi}{3} \cdot \left( 6(9)h - 3h^2 \right)$ $\frac{dV}{dh} = \frac{\pi}{3} \cdot \left( 54h - 3h^2 \right)$ $\frac{dV}{dh} = \frac{\pi}{3} \cdot 3h(18 - h)$ $\frac{dV}{dh} = \pi h (18 - h)$

Now use the chain rule:

$\frac{dV}{dt} = \frac{dV}{dh} \cdot \frac{dh}{dt}$

We know $\frac{dV}{dt} = 4 \, \text{m}^3/\text{min}$. Therefore, at $h = 4$:

$4 = \pi (4)(18 - 4) \cdot \frac{dh}{dt}$ $4 = \pi (4)(14) \cdot \frac{dh}{dt}$ $4 = 56\pi \cdot \frac{dh}{dt}$ $\frac{dh}{dt} = \frac{4}{56\pi}$ $\frac{dh}{dt} = \frac{1}{14\pi} \, \text{m}/\text{min}$

Final Answer:

The water level is rising at a rate of $\frac{1}{14\pi} \, \text{m}/\text{min}$, or approximately $0.0227 \, \text{m}/\text{min}$, when the water level is 4 meters from the bottom.

Do you want further clarifications or details on any part of the solution?

Here are 5 questions for further exploration:

1. How would the rate change if the inflow rate doubled to 8 m³/min?
2. What is the water level rising rate when $h = 6$ meters?
3. How would the formula for $\frac{dh}{dt}$ change for a full sphere instead of a hemisphere?
4. What if the radius of the tank were smaller, say 6 meters, how would the result differ?
5. How can we calculate the time required to fill the tank completely?

Tip: Understanding how to differentiate volume equations with respect to time is key in solving many related rates problems in calculus.