To solve this problem, we need to determine the rate of change of the water depth when the water depth is $h = 4 \, \text{ft}$. The water is draining from an inverted conical tank, and we are given that the volume of water is decreasing at a rate of $\frac{dV}{dt} = -6 \, \text{ft}^3/\text{s}$.

Step 1: Relate the volume of water in the tank to its depth.

The volume $V$ of a cone is given by:

$V = \frac{1}{3} \pi r^2 h$

where $r$ is the radius of the surface of the water and $h$ is the height (or depth) of the water.

Step 2: Express $r$ in terms of $h$ using similar triangles.

Since the tank is an inverted cone, the ratio of the radius to the height remains constant. The original tank has a height of 8 feet and a radius of 4 feet. Using similar triangles, we have:

$\frac{r}{h} = \frac{4}{8} = \frac{1}{2}$

Thus, the radius $r$ of the water at any height $h$ is:

$r = \frac{h}{2}$

Step 3: Substitute $r = \frac{h}{2}$ into the volume formula.

Now, substitute $r = \frac{h}{2}$ into the volume formula:

$V = \frac{1}{3} \pi \left( \frac{h}{2} \right)^2 h = \frac{1}{3} \pi \cdot \frac{h^2}{4} \cdot h = \frac{\pi}{12} h^3$

So the volume of water as a function of the height $h$ is:

$V = \frac{\pi}{12} h^3$

Step 4: Differentiate the volume with respect to time.

We want to find the rate of change of the height of the water, $\frac{dh}{dt}$. To do this, we differentiate the volume $V$ with respect to time $t$:

$\frac{dV}{dt} = \frac{\pi}{12} \cdot 3h^2 \cdot \frac{dh}{dt}$

Simplifying:

$\frac{dV}{dt} = \frac{\pi}{4} h^2 \frac{dh}{dt}$

Step 5: Solve for $\frac{dh}{dt}$.

We are given that $\frac{dV}{dt} = -6 \, \text{ft}^3/\text{s}$ and we want to find $\frac{dh}{dt}$ when $h = 4 \, \text{ft}$. Substituting these values into the equation:

$-6 = \frac{\pi}{4} (4)^2 \frac{dh}{dt}$

Simplifying:

$-6 = \frac{\pi}{4} \cdot 16 \cdot \frac{dh}{dt}$

$-6 = 4\pi \cdot \frac{dh}{dt}$

$\frac{dh}{dt} = \frac{-6}{4\pi} = \frac{-3}{2\pi} \, \text{ft/s}$

Final Answer:

The rate of change of the water depth when the water depth is 4 feet is approximately:

$\frac{dh}{dt} \approx \frac{-3}{2\pi} \approx -0.477 \, \text{ft/s}$

This means that the water depth is decreasing at a rate of approximately $0.477 \, \text{ft/s}$.

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Would you like to dive into more details, or do you have any further questions?

Here are 5 related questions to deepen your understanding:

1. How does the relationship between height and radius change if the cone’s dimensions are different?
2. What happens to the rate of change $\frac{dh}{dt}$ if the draining rate $\frac{dV}{dt}$ changes?
3. Can you derive a similar formula for a cylindrical tank instead of a conical one?
4. What is the physical meaning of a negative $\frac{dh}{dt}$?
5. How would the problem change if the tank were draining faster, say at $10 \, \text{ft}^3/\text{s}$?

Tip: When working with similar triangles, always look for proportional relationships between the dimensions; they often help simplify the variables involved.