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The problem is about a reservoir shaped like an inverted cone with the following parameters:

- Height of the cone (\( h \)) = 5 m
- Diameter of the top face = 6 m, so the radius (\( r \)) = 3 m
- The water level in the cone is currently 4 m and is lowering at a rate of 0.4 cm/s (which is \( 0.004 \) m/s)

We are asked to:

1. Express the volume \( V \) of water as a function of \( x \), where \( x \) is the height of the water level.
2. Compute \( \frac{dV}{dt} \) at the moment when the water level is at 4 m, considering the rate at which the water level lowers.

### Solution

#### (a) Express \( V \) as a function of \( x \)

The formula for the volume of a cone is:
\[
V = \frac{1}{3} \pi r^2 h
\]
Since the radius and height of the water in the cone change proportionally, we can express \( r \) (radius at water level) in terms of \( x \) (current water height) using similar triangles:

Given the full height of the cone is 5 m and the corresponding radius is 3 m:
\[
\frac{r}{x} = \frac{3}{5} \implies r = \frac{3}{5}x
\]

Now substitute this expression for \( r \) into the volume formula:
\[
V = \frac{1}{3} \pi \left(\frac{3}{5}x\right)^2 x
\]
\[
V = \frac{1}{3} \pi \frac{9}{25} x^3
\]
\[
V = \frac{3\pi}{25} x^3
\]

Thus, the volume \( V \) as a function of \( x \) is:
\[
V(x) = \frac{3\pi}{25} x^3 \quad \text{m}^3
\]

#### (b) Compute \( \frac{dV}{dt} \)

To find \( \frac{dV}{dt} \), we'll use the chain rule:
\[
\frac{dV}{dt} = \frac{dV}{dx} \cdot \frac{dx}{dt}
\]

We already know \( \frac{dx}{dt} \) is the rate at which the water level is lowering, which is \( -0.004 \) m/s (negative because the level is decreasing).

Now, find \( \frac{dV}{dx} \):
\[
V(x) = \frac{3\pi}{25} x^3
\]
\[
\frac{dV}{dx} = \frac{3\pi}{25} \cdot 3x^2 = \frac{9\pi}{25} x^2
\]

When the water level \( x \) is 4 m:
\[
\frac{dV}{dx} = \frac{9\pi}{25} (4)^2 = \frac{9\pi}{25} \cdot 16 = \frac{144\pi}{25} \text{ m}^2
\]

Now, use the chain rule:
\[
\frac{dV}{dt} = \frac{144\pi}{25} \cdot (-0.004)
\]
\[
\frac{dV}{dt} = -\frac{576\pi}{625} \quad \text{m}^3/\text{s}
\]

Calculating the value:
\[
\frac{dV}{dt} \approx -2.89 \times 10^{-3} \; \text{m}^3/\text{s}
\]

### Final Answer:

1. **\( V(x) = \frac{3\pi}{25} x^3 \; \text{m}^3 \)**
2. **\( \frac{dV}{dt} \approx -0.00289 \; \text{m}^3/\text{s} \)**

Do you want any further details or have any questions?

Here are five follow-up questions related to this problem:
1. How would the solution change if the water level were lowering at a different rate?
2. What would be the rate of change of the water volume when the water level is at 2 m?
3. Can you explain how similar triangles were used to derive the relationship between \( r \) and \( x \)?
4. How would the volume equation change if the cone had a different height or diameter?
5. Why is the rate \( \frac{dx}{dt} \) negative in this scenario?

**Tip:** Always check the units of your rates when differentiating; it helps in understanding whether quantities are increasing or decreasing.

A reservoir is shaped like an inverted cone. Its height is 5 m, and the diameter of its top face is 6 m. We need to determine the rate at which the water empties when the water level in the cone is 4 m and is decreasing at a rate of 0.004 m/s.

Learn how to calculate the rate at which water drains from an inverted cone-shaped reservoir, given the water level height and the rate of decrease. This problem involves geometry, volume formulas, and related rates. Suitable for students in Grades 10-12.

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