To solve this problem, let's go through the steps outlined in the feedback:

1. Understanding Surface Area: The surface area $A$ of a sphere with radius $r$ is given by: $A = 4\pi r^2$

2. Differentiating with Respect to Time: Since the surface area $A$ changes over time as the snowball melts, we need to find $\frac{dA}{dt}$ and relate it to $\frac{dr}{dt}$.

   Differentiating $A = 4\pi r^2$ with respect to $t$, we get: $\frac{dA}{dt} = 4\pi \cdot 2r \cdot \frac{dr}{dt} = 8\pi r \cdot \frac{dr}{dt}$

3. Using the Given Rate of Change: We're told that the rate at which the surface area decreases is $\frac{dA}{dt} = -2 \, \text{cm}^2/\text{min}$ (negative because the area is decreasing).

   So, we substitute $\frac{dA}{dt} = -2$ and solve for $\frac{dr}{dt}$: $-2 = 8\pi r \cdot \frac{dr}{dt}$ $\frac{dr}{dt} = \frac{-2}{8\pi r} = \frac{-1}{4\pi r}$

4. Finding the Rate of Change of the Diameter: The diameter $D$ of the sphere is $D = 2r$, so the rate of change of the diameter with respect to time $\frac{dD}{dt}$ is: $\frac{dD}{dt} = 2 \cdot \frac{dr}{dt} = 2 \cdot \frac{-1}{4\pi r} = \frac{-1}{2\pi r}$

5. Evaluating at $r = 4$ cm: Since the diameter is 8 cm, the radius $r$ is 4 cm. Substitute $r = 4$ into the equation: $\frac{dD}{dt} = \frac{-1}{2\pi \cdot 4} = \frac{-1}{8\pi}$

6. Final Answer: Calculating $\frac{-1}{8\pi}$ to three decimal places, $\frac{dD}{dt} \approx -0.040 \, \text{cm/min}$

So, the rate at which the diameter decreases is approximately $-0.040 \, \text{cm/min}$.

Would you like further details on any step, or do you have additional questions?

Here are five related questions to deepen your understanding:

1. How would this approach change if we were given the rate of change of the volume instead of the surface area?
2. What is the physical significance of the negative sign in $\frac{dD}{dt}$?
3. How does the rate of change of the radius relate to the rate of change of the diameter in general?
4. Could we solve for $\frac{dD}{dt}$ directly without first finding $\frac{dr}{dt}$?
5. How would the rate of diameter change if the snowball melted faster (e.g., if $\frac{dA}{dt}$ was -5 cm²/min)?

Tip: Always keep track of units when differentiating physical quantities with respect to time; this helps to verify the validity of your final answer.