We are given that the radius $r$ of a sphere is increasing at a constant rate of 6 meters per second, i.e.,

$\frac{dr}{dt} = 6 \, \text{m/s}.$

We are also given the equation for the volume of the sphere:

$V = \frac{4}{3} \pi r^3.$

We are asked to find the rate of change of the volume $\frac{dV}{dt}$ when the radius is $r = 5$ meters.

Step 1: Differentiate the volume formula with respect to time $t$.

Using the chain rule, we differentiate $V = \frac{4}{3} \pi r^3$ with respect to $t$:

$\frac{dV}{dt} = \frac{d}{dt} \left( \frac{4}{3} \pi r^3 \right).$

Applying the chain rule:

$\frac{dV}{dt} = \frac{4}{3} \pi \cdot 3r^2 \cdot \frac{dr}{dt}.$

Simplifying:

$\frac{dV}{dt} = 4 \pi r^2 \cdot \frac{dr}{dt}.$

Step 2: Substitute the known values.

We know $\frac{dr}{dt} = 6 \, \text{m/s}$ and $r = 5 \, \text{m}$. Substitute these values into the equation:

$\frac{dV}{dt} = 4 \pi (5)^2 \cdot 6.$

$\frac{dV}{dt} = 4 \pi \cdot 25 \cdot 6.$

$\frac{dV}{dt} = 600 \pi.$

Step 3: Approximate the value.

Using $\pi \approx 3.1416$:

$\frac{dV}{dt} \approx 600 \cdot 3.1416 = 1884.96 \, \text{m}^3/\text{s}.$

Thus, the rate of change of the volume when the radius is 5 meters is approximately:

$\boxed{1884.961} \, \text{m}^3/\text{s}.$

Additional Questions:

1. How does the rate of change of the volume change if the radius increases at a different rate?
2. What would be the rate of change of the volume when the radius is 10 meters?
3. How would the rate of change of the volume behave if the radius were decreasing instead of increasing?
4. How can you use implicit differentiation to solve similar problems involving other geometric shapes?
5. How would this formula change if the sphere had a different geometric constant?

Tip: Always carefully apply the chain rule when dealing with rates of change of related quantities in calculus.