Let's break this problem down and set up the relations we need to solve it.

Step 1: Volume of a cone

The container is in the shape of an inverted right circular cone. The volume $V$ of a cone is given by the formula: $V = \frac{1}{3} \pi r^2 h$ where:

• $r$ is the radius of the base of the cone,
• $h$ is the height (or depth in this context),
• $\pi$ is the mathematical constant Pi (approximately 3.1416).

However, the radius of the water level decreases as the water level drops. Since the cone is similar to the full cone, the radius $r$ of the water surface at any depth $h$ is proportional to the depth $h$. For the full cone, when $h = 16$ cm, $r = 4$ cm. Therefore, we can express the radius $r$ as a function of $h$ using the proportionality: $\frac{r}{h} = \frac{4}{16} = \frac{1}{4}$ Thus, at any depth $h$, the radius $r$ of the water is: $r = \frac{h}{4}$

Step 2: Volume as a function of depth

Now substitute this expression for $r$ into the formula for the volume: $V = \frac{1}{3} \pi \left(\frac{h}{4}\right)^2 h = \frac{1}{3} \pi \frac{h^3}{16}$ This simplifies to: $V(h) = \frac{\pi h^3}{48}$ This gives the volume of water in the container as a function of the depth $h$.

Step 3: Rate of change of volume

The water is being pumped out at a constant rate of 60 cm³/s. This means that the volume $V$ is decreasing over time at a rate of: $\frac{dV}{dt} = -60 \, \text{cm}^3/\text{s}$ We can use the chain rule to relate the rate of change of volume to the rate of change of depth $h$: $\frac{dV}{dt} = \frac{dV}{dh} \cdot \frac{dh}{dt}$

First, find $\frac{dV}{dh}$ by differentiating the volume function $V(h) = \frac{\pi h^3}{48}$: $\frac{dV}{dh} = \frac{3\pi h^2}{48} = \frac{\pi h^2}{16}$

Now, substitute into the chain rule: $-60 = \frac{\pi h^2}{16} \cdot \frac{dh}{dt}$ Solving for $\frac{dh}{dt}$, we get: $\frac{dh}{dt} = \frac{-60 \times 16}{\pi h^2} = \frac{-960}{\pi h^2}$ This gives the rate of change of the water depth $h$ with respect to time $t$.

Step 4: Initial conditions and further analysis

• At $t = 0$, the container is fully filled, meaning $h = 16 \, \text{cm}$.
• The relationship between volume, depth, and time could be further analyzed by integrating the rate of change of $h$ with respect to time.

Let me know if you'd like to explore more detailed steps in solving for $h(t)$ or specific questions about this setup!

Follow-up questions:

1. How long does it take to completely empty the container?
2. What is the depth of the water after 30 seconds?
3. How does the rate at which the depth decreases change as the water level drops?
4. How would the problem change if the cone had a different base radius or height?
5. Could we solve for the time when the water depth reaches a specific value (e.g., 8 cm)?

Tip:

When working with related rates, always express all quantities in terms of a single variable before differentiating to avoid errors.