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The problem describes an inverted cylindrical cone, 60 meters deep and with a diameter of 30 meters, being filled with water at a constant rate of \(20 \, \text{m}^3/\text{min}\). You are asked to determine how fast the water level is rising when the water is at depths of 1 meter, 10 meters, and 59 meters.

### Step-by-step approach:

1. **Volume of a cone:**
   The volume \(V\) of a cone is given by:
   \[
   V = \frac{1}{3} \pi r^2 h
   \]
   where \(r\) is the radius of the water surface, and \(h\) is the depth of the water.

2. **Relation between \(r\) and \(h\):**
   The problem specifies the cone is 60 meters deep, and the top diameter is 30 meters. Therefore, the radius of the top of the cone is \(r_{\text{max}} = 15 \, \text{m}\). Since the radius increases linearly with height, we have:
   \[
   \frac{r}{h} = \frac{15}{60} = \frac{1}{4}
   \]
   So, \(r = \frac{h}{4}\).

3. **Volume in terms of \(h\):**
   Substituting \(r = \frac{h}{4}\) into the volume formula:
   \[
   V = \frac{1}{3} \pi \left(\frac{h}{4}\right)^2 h = \frac{1}{3} \pi \frac{h^3}{16} = \frac{\pi h^3}{48}
   \]

4. **Differentiate volume with respect to time:**
   We are given the rate of change of volume, \(\frac{dV}{dt} = 20 \, \text{m}^3/\text{min}\), and we need to find \(\frac{dh}{dt}\), the rate at which the water level is rising. Differentiate \(V = \frac{\pi h^3}{48}\) with respect to time \(t\):
   \[
   \frac{dV}{dt} = \frac{\pi}{48} \cdot 3h^2 \frac{dh}{dt}
   \]
   Simplifying:
   \[
   \frac{dV}{dt} = \frac{\pi h^2}{16} \frac{dh}{dt}
   \]
   Now solve for \(\frac{dh}{dt}\):
   \[
   \frac{dh}{dt} = \frac{16}{\pi h^2} \frac{dV}{dt}
   \]
   Substituting \(\frac{dV}{dt} = 20\):
   \[
   \frac{dh}{dt} = \frac{320}{\pi h^2}
   \]

5. **Calculate \(\frac{dh}{dt}\) for each value of \(h\):**
   - For \(h = 1 \, \text{m}\):
     \[
     \frac{dh}{dt} = \frac{320}{\pi (1)^2} = \frac{320}{\pi} \approx 101.86 \, \text{m/min}
     \]
   - For \(h = 10 \, \text{m}\):
     \[
     \frac{dh}{dt} = \frac{320}{\pi (10)^2} = \frac{320}{100\pi} \approx 1.02 \, \text{m/min}
     \]
   - For \(h = 59 \, \text{m}\):
     \[
     \frac{dh}{dt} = \frac{320}{\pi (59)^2} = \frac{320}{3481\pi} \approx 0.0293 \, \text{m/min}
     \]

### Answers:
- When the depth is 1 meter, the water level is rising at approximately **101.86 m/min**.
- When the depth is 10 meters, the water level is rising at approximately **1.02 m/min**.
- When the depth is 59 meters, the water level is rising at approximately **0.0293 m/min**.

The units of the answers are in **meters per minute** \((\text{m/min})\).

Would you like more details on any of these steps? Here are some related questions you might consider:

1. What is the relation between the radius and height of the cone?
2. How does the rate of water level rise change as the depth increases?
3. What would happen to the rate if the cone were wider or narrower?
4. How would you approach this problem if the volume flow rate were not constant?
5. Could this method be used to calculate other shapes, such as a cylinder?

**Tip**: Always ensure that the relationship between the variables is correctly substituted before differentiating in related rates problems.

An inverted cylindrical cone, 60 m deep and 30 m in diameter across the top, is being filled with water at a constant rate of 20 m³/min. At what rate is the water rising in the tank when the depth of the water is 1 meter, 10 meters, and 59 meters?

This problem involves an inverted cone being filled with water at a constant rate of 20 m³/min. Learn how to calculate how fast the water level is rising when the water is at depths of 1 meter, 10 meters, and 59 meters. This step-by-step solution covers conical volume, calculus, and related rates concepts.

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