[The volume of a cone of height $h$ and base radius $r$ is $\frac{1}{3}\pi r^2 h$.] A hollow right circular cone of base radius 5 cm and height 20 cm has a small opening at the base. Initially, the cone is fully filled with water. Due to the leak from the base, water flows out at a constant rate of 2 $\text{cm}^3/\text{s}$.

1. Identify the geometry of the water volume

The volume of water $V$ is the volume of the entire cone minus the volume of the empty cone at the top.
Let the height of the empty cone at the top be $h_{top} = 20 - x$.

2. Use similar triangles to find the radius of the empty top cone

Let the radius of the empty top cone be $r_{top}$.
By similar triangles,
$\frac{r_{top}}{5} = \frac{20 - x}{20}$
$r_{top} = \frac{5(20 - x)}{20} = \frac{20 - x}{4}$

3. Calculate the volume of the water

Volume of entire cone = $\frac{1}{3}\pi (5)^2 (20) = \frac{500\pi}{3}$
Volume of empty top cone = $\frac{1}{3}\pi (r_{top})^2 (h_{top}) = \frac{1}{3}\pi \left(\frac{20 - x}{4}\right)^2 (20 - x)$
$= \frac{1}{3}\pi \frac{(20 - x)^2}{16} (20 - x) = \frac{\pi}{48}(20 - x)^3$
Therefore, $V = \frac{500\pi}{3} - \frac{\pi}{48}(20 - x)^3$ (Shown)

1. Differentiate the volume equation with respect to $x$ to find $\frac{dV}{dx}$.

Given $V = \frac{500\pi}{3} - \frac{\pi}{48}(20 - x)^3$.
$\frac{dV}{dx} = -\frac{\pi}{48} \cdot 3(20 - x)^2 \cdot (-1)$
$\frac{dV}{dx} = \frac{\pi}{16}(20 - x)^2$

2. Set up the connected rates of change equation.

We are given $\frac{dV}{dt} = -2$ (since water flows out).
By the chain rule:
$\frac{dV}{dt} = \frac{dV}{dx} \times \frac{dx}{dt}$

3. Substitute $x = 8$ and $\frac{dV}{dt} = -2$ to find $\frac{dx}{dt}$.

When $x = 8$:
$\frac{dV}{dx} = \frac{\pi}{16}(20 - 8)^2 = \frac{\pi}{16}(144) = 9\pi$
Substitute into the chain rule equation:
$-2 = 9\pi \times \frac{dx}{dt}$
$\frac{dx}{dt} = -\frac{2}{9\pi}$