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This problem involves a conical reservoir, and we need to determine the rate at which the water level is changing when the water is 3 meters deep, and also the rate at which the radius of the water's surface is changing. Let's break it down into steps.

### Given Information:
- Water flows out at the rate of \( \frac{dV}{dt} = 60 \, \text{m}^3/\text{min} \).
- The cone's base radius \( r = 35 \, \text{m} \) and height \( h = 4 \, \text{m} \).
- The water depth at the point of interest is 3 meters.

### Step 1: Express the volume of the cone
The volume \( V \) of a cone is given by:
\[
V = \frac{1}{3} \pi r^2 h
\]
However, both the radius and height are variables in this case. To express the radius in terms of the height, use the fact that the cone is similar throughout (i.e., the ratio of \( r \) to \( h \) is constant).

From the dimensions of the cone:
\[
\frac{r}{h} = \frac{35}{4}
\]
Thus, \( r = \frac{35}{4} h \).

### Step 2: Substitute for \( r \) in the volume equation
Now, substitute \( r = \frac{35}{4} h \) into the volume formula:
\[
V = \frac{1}{3} \pi \left( \frac{35}{4} h \right)^2 h
\]
Simplify this expression:
\[
V = \frac{1}{3} \pi \left( \frac{1225}{16} \right) h^3 = \frac{1225 \pi}{48} h^3
\]

### Step 3: Differentiate with respect to time
We now differentiate both sides of the volume equation with respect to time \( t \), using the chain rule:
\[
\frac{dV}{dt} = \frac{1225 \pi}{48} \cdot 3h^2 \cdot \frac{dh}{dt}
\]
This simplifies to:
\[
\frac{dV}{dt} = \frac{1225 \pi}{16} h^2 \frac{dh}{dt}
\]

### Step 4: Solve for \( \frac{dh}{dt} \)
We know \( \frac{dV}{dt} = -60 \, \text{m}^3/\text{min} \) (negative because the water is flowing out), so we solve for \( \frac{dh}{dt} \):
\[
-60 = \frac{1225 \pi}{16} h^2 \frac{dh}{dt}
\]
At \( h = 3 \, \text{m} \):
\[
-60 = \frac{1225 \pi}{16} (3)^2 \frac{dh}{dt}
\]
\[
-60 = \frac{1225 \pi}{16} \times 9 \frac{dh}{dt}
\]
\[
-60 = \frac{11025 \pi}{16} \frac{dh}{dt}
\]
\[
\frac{dh}{dt} = \frac{-60 \times 16}{11025 \pi}
\]
\[
\frac{dh}{dt} \approx \frac{-960}{34647.2}
\]
\[
\frac{dh}{dt} \approx -0.0277 \, \text{m/min}
\]
Thus, the water level is dropping at approximately **2.77 cm/min**.

Would you like details on the second part of the problem or further clarification? 

Here are 5 relative questions to explore further:
1. How do you calculate the rate of change of the radius in the second part of the problem?
2. What happens to the flow rate when the water depth changes?
3. How is the concept of similar triangles applied to this type of problem?
4. How would the solution differ if the cone’s height or radius changed?
5. How can this concept be applied to other geometric shapes?

**Tip:** When solving related rates problems, always focus on identifying how one variable depends on another through geometry before differentiating.

Water is flowing at the rate of 60 m^3/min from a shallow concrete conical reservoir of base radius 35 m and height 4 m. At what rate (centimeters per minute) is the water level changing when the water is 3 m deep?

Solve a related rates problem involving water flow from a conical reservoir with a base radius of 35 m and a height of 4 m. Learn how to calculate the rate of change of water level when the depth is 3 meters, with key concepts such as related rates, the volume of a cone, and differentiation.

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