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A hemispherical tank with a radius of 9 m9 m is filled from an inflow pipe at a rate of 4 m cubed divided by min4 m3/min ​(see figure). How fast is the water level rising when the water level is 4 m4 m from the bottom of the​ tank? (Hint: The volume of a cap of thickness h sliced from a sphere of radius r is StartFraction pi h squared left parenthesis 3 r minus h right parenthesis Over 3 EndFraction
πh2(3r−h)
3​.)

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.

A hemispherical tank with a radius of 9 m is filled from an inflow pipe at a rate of 4 m³/min. How fast is the water level rising when the water level is 4 m from the bottom of the tank? (Hint: The volume of a cap of thickness h sliced from a sphere of radius r is V = (πh²(3r−h))/3.)

Solve a related rates problem involving a hemispherical tank with a radius of 9 m, where water is filled at a rate of 4 m³/min. Learn how to calculate the rate at which the water level is rising when the water level is 4 m from the bottom.

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