https://file.aiquickdraw.com/mathbot/file/1728396295288janssnwo.jpg

Let's break down and solve the problem in the image.

### Given:
- Water is pouring into a **conical tank**.
- **Height (h) of the cone** = 10 m.
- **Radius (r) of the base** = 4 m.
- Water is poured at a rate of **6 m³/min** (i.e., \(\frac{dV}{dt} = 6 \, \text{m}^3/\text{min}\)).
  
### Part (a):
We need to find the rate at which the **water level (h)** is rising when the water level is **5 m**.

#### Step 1: Formula for the volume of a cone
The volume of a cone is given by:
\[
V = \frac{1}{3} \pi r^2 h
\]
However, \(r\) and \(h\) are not independent. They maintain a ratio that matches the dimensions of the cone.

#### Step 2: Relating radius and height
From the given dimensions of the cone, we know:
\[
\frac{r}{h} = \frac{4}{10} = \frac{2}{5} \implies r = \frac{2}{5}h
\]
Now, substitute \(r = \frac{2}{5}h\) into the volume formula to express \(V\) in terms of \(h\) only.

\[
V = \frac{1}{3} \pi \left(\frac{2}{5}h\right)^2 h = \frac{1}{3} \pi \frac{4}{25} h^3 = \frac{4\pi}{75} h^3
\]

#### Step 3: Differentiate volume with respect to time
Now, differentiate both sides of the volume equation with respect to time \(t\):

\[
\frac{dV}{dt} = \frac{4\pi}{75} \cdot 3h^2 \frac{dh}{dt}
\]
Simplify:
\[
\frac{dV}{dt} = \frac{4\pi}{25} h^2 \frac{dh}{dt}
\]

#### Step 4: Substitute known values
We are given that \(\frac{dV}{dt} = 6 \, \text{m}^3/\text{min}\) and \(h = 5 \, \text{m}\). Plug these into the equation:

\[
6 = \frac{4\pi}{25} (5)^2 \frac{dh}{dt}
\]
\[
6 = \frac{4\pi}{25} \cdot 25 \cdot \frac{dh}{dt}
\]
\[
6 = 4\pi \cdot \frac{dh}{dt}
\]
\[
\frac{dh}{dt} = \frac{6}{4\pi} = \frac{3}{2\pi} \, \text{m/min}
\]

So, the water level is rising at a rate of \(\frac{3}{2\pi} \, \text{m/min} \approx 0.48 \, \text{m/min}\) when the water level is 5 m.

---

### Part (b):
As time passes, what happens to the rate at which the water level rises?

#### Explanation:
The rate of rise of the water level depends on the height \(h\). From the equation \(\frac{dV}{dt} = \frac{4\pi}{25} h^2 \frac{dh}{dt}\), we can observe that the relationship between \(\frac{dh}{dt}\) and \(h\) is inversely proportional to \(h^2\). As the height \(h\) increases, \(h^2\) increases, which in turn reduces \(\frac{dh}{dt}\) for a constant inflow rate \(\frac{dV}{dt}\). 

Thus, **as time passes and the water level rises, the rate at which the water level rises decreases**.

---

Would you like further details or have any questions about this solution?  
Here are some related questions for practice:

1. What would the rate of the water level be when the height is 7 m?
2. How would the rate change if the water poured in at a different rate, say 8 m³/min?
3. What happens if the cone's height-to-radius ratio changes, and how does it affect the rise of the water level?
4. Can you find the total time it takes for the tank to fill up?
5. What would happen to the rate of rise if the tank were cylindrical instead of conical?

**Tip:** The relationship between radius and height in a cone simplifies complex problems like this by reducing the number of variables!

Water pours into a conical tank of height 10 m and radius 4 m at a rate of 6 m³/min. At what rate is the water level rising when the level is 5 m high? As time passes, what happens to the rate at which the water level rises?

Solve a related rates problem where water pours into a conical tank at 6 m³/min. Learn how to calculate the rate of water level rise when the height is 5 meters using calculus and geometric relationships. Includes step-by-step explanations of key concepts and formulas, suitable for Grades 11-12.

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