We are tasked with finding the rate at which water is being pumped into an inverted conical tank. To solve this, we will apply the principles of related rates and use the given information to find the unknown rate.

Step 1: Understand the geometry of the conical tank

The tank is an inverted cone with:

• Height of the cone: $h = 6$ meters = 600 cm (since 1 meter = 100 cm)
• Diameter of the top of the cone: $d = 4$ meters = 400 cm (so the radius $r = 200$ cm)

At any time, the water inside the cone has a height $h_w$ and a radius $r_w$, forming a smaller cone that is similar to the overall cone. The relationship between the radius $r_w$ of the water and its height $h_w$ follows from the similarity of triangles:

$\frac{r_w}{h_w} = \frac{200}{600} = \frac{1}{3}$

Thus, the radius of the water is related to the height by:

$r_w = \frac{1}{3} h_w$

Step 2: Volume of water in the cone

The volume $V$ of water in the tank is given by the formula for the volume of a cone:

$V = \frac{1}{3} \pi r_w^2 h_w$

Substitute $r_w = \frac{1}{3} h_w$ into this equation:

$V = \frac{1}{3} \pi \left( \frac{1}{3} h_w \right)^2 h_w$

$V = \frac{1}{3} \pi \frac{1}{9} h_w^3$

$V = \frac{\pi}{27} h_w^3$

Step 3: Differentiate with respect to time $t$

Next, differentiate both sides of the volume equation with respect to time $t$:

$\frac{dV}{dt} = \frac{\pi}{27} \cdot 3 h_w^2 \frac{dh_w}{dt}$

$\frac{dV}{dt} = \frac{\pi}{9} h_w^2 \frac{dh_w}{dt}$

Step 4: Plug in the known values

We are given the following:

• The water is leaking out at a rate of $6,500$ cm³/min, so $\frac{dV_{leak}}{dt} = -6,500$ cm³/min.
• The water level is rising at a rate of $\frac{dh_w}{dt} = 20$ cm/min when the height of the water is $h_w = 200$ cm (since $2$ meters = $200$ cm).
• We need to find the rate at which water is being pumped into the tank, $\frac{dV_{in}}{dt}$.

We know the total rate of change of volume $\frac{dV}{dt}$ is the rate at which water is being pumped in minus the rate at which it is leaking out:

$\frac{dV}{dt} = \frac{dV_{in}}{dt} - \frac{dV_{leak}}{dt}$

Substitute the known values into the volume equation:

$\frac{dV}{dt} = \frac{\pi}{9} (200)^2 (20)$

$\frac{dV}{dt} = \frac{\pi}{9} \cdot 40,000 \cdot 20$

$\frac{dV}{dt} = \frac{\pi}{9} \cdot 800,000$

$\frac{dV}{dt} = \frac{800,000 \pi}{9} \approx 279,252 \, \text{cm}^3/\text{min}$

Step 5: Solve for the rate of water being pumped in

We know:

$\frac{dV}{dt} = \frac{dV_{in}}{dt} - 6,500$

So,

$279,252 = \frac{dV_{in}}{dt} - 6,500$

Solving for $\frac{dV_{in}}{dt}$:

$\frac{dV_{in}}{dt} = 279,252 + 6,500 = 285,752 \, \text{cm}^3/\text{min}$

Final Answer:

The rate at which water is being pumped into the tank is approximately 285,752 cm³/min.

Would you like further details or have any questions? Here are some follow-up questions you might find interesting:

1. How would the solution change if the water level was rising at a different rate?
2. What happens if the tank is leaking more water?
3. How would the solution differ if the shape of the tank was a cylinder instead of a cone?
4. How do you calculate the volume of a cone at a given height and radius?
5. Can the water's volume ever exceed the capacity of the tank? If so, how would that affect the solution?

Tip: When working with related rates problems, always draw a diagram and define variables clearly. Use similar triangles when dealing with cones or pyramids to simplify the relationships between the dimensions.