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### Problem Summary
- We have an inverted conical reservoir with a height of \(4 \, \text{m}\) and a top diameter of \(6 \, \text{m}\) (so a radius of \(3 \, \text{m}\)).
- We are given:
  - Height (\(h\)) of the cone = \(4 \, \text{m}\).
  - Radius (\(r\)) at the top = \(3 \, \text{m}\).
  - The water level is lowering at a rate of \(5 \, \text{cm/s} = 0.05 \, \text{m/s}\) when the water depth (\(x\)) is \(2 \, \text{m}\).
- We need to:
  1. Express the volume \(V\) of water as a function of \(x\) (height of water level).
  2. Calculate \( \frac{dV}{dt} \) when \(x = 2 \, \text{m}\).

---

### Part (a): Volume \(V\) as a Function of \(x\)

For a cone, the volume \(V\) is given by:
\[
V = \frac{1}{3} \pi r^2 h
\]
However, since the water level is decreasing, the radius of the water surface is proportional to the water height \(x\).

#### Setting up the relationship between \(r\) and \(x\):
Using similar triangles:
\[
\frac{r}{h} = \frac{3}{4} \Rightarrow r = \frac{3}{4}x
\]

Substitute \(r = \frac{3}{4}x\) into the volume formula:
\[
V = \frac{1}{3} \pi \left(\frac{3}{4}x\right)^2 x
\]
Simplifying this:
\[
V = \frac{1}{3} \pi \frac{9}{16} x^3 = \frac{3\pi}{16} x^3
\]

So, **the volume \(V\) as a function of \(x\) is**:
\[
V = \frac{3\pi}{16} x^3
\]

---

### Part (b): Compute \(\frac{dV}{dt}\) when \(x = 2 \, \text{m}\)

To find \(\frac{dV}{dt}\), we differentiate \(V = \frac{3\pi}{16} x^3\) with respect to time \(t\).

1. **Differentiate \(V\) with respect to \(t\):**
   \[
   \frac{dV}{dt} = \frac{3\pi}{16} \cdot 3x^2 \cdot \frac{dx}{dt} = \frac{9\pi}{16} x^2 \cdot \frac{dx}{dt}
   \]

2. **Substitute \(x = 2\) and \(\frac{dx}{dt} = -0.05 \, \text{m/s}\):**
   \[
   \frac{dV}{dt} = \frac{9\pi}{16} \cdot (2)^2 \cdot (-0.05)
   \]
   Simplifying this:
   \[
   \frac{dV}{dt} = \frac{9\pi}{16} \cdot 4 \cdot (-0.05)
   \]
   \[
   \frac{dV}{dt} = \frac{36\pi}{16} \cdot (-0.05)
   \]
   \[
   \frac{dV}{dt} = -\frac{36\pi}{320}
   \]
   Approximating this:
   \[
   \frac{dV}{dt} \approx -0.353 \, \text{m}^3/\text{s}
   \]

---

### Final Answers
1. **(a) \( V = \frac{3\pi}{16} x^3 \)**
2. **(b) \( \frac{dV}{dt} \approx -0.353 \, \text{m}^3/\text{s} \)**

Would you like further details or have any questions?

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### Related Questions
1. How would the solution change if the rate of water level decrease was different?
2. How can we verify that the cone's geometry scaling was applied correctly?
3. What would happen to \(\frac{dV}{dt}\) if the cone was taller but with the same top radius?
4. Can we solve similar problems for shapes other than cones?
5. What if the radius at the top was unknown? How would it affect the equation setup?

### Tip
In related rates problems, always identify relationships between variables and use similar triangles for scaling dimensions within geometric shapes.

A reservoir is shaped like an inverted cone with a height of 4 m and a top diameter of 6 m. It empties its water with a faucet at the bottom. We want to determine the rate at which the water empties from the cone when the water level in the cone is 2 m, given that the water level is lowering at 5 cm/s. Let x be the water level (in m), V the volume of water (in m³), and t the time (in s).

Solve a related rates problem involving an inverted conical reservoir with a height of 4 m and a diameter of 6 m. Calculate the rate at which water volume changes as the water level decreases, using calculus and geometric scaling. Suitable for college-level calculus students.

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