Let's solve the problem step by step.

Given:

1. Volume of a cube $V = s^3$, where $s$ is the side length of the cube.
2. $\frac{dV}{dt} = -511 \, \text{m}^3/\text{min}$ (volume is decreasing at 511 cubic meters per minute).
3. $V = 186 \, \text{m}^3$ (volume at the instant of interest).
4. Find $\frac{dA}{dt}$, the rate of change of the surface area, where $A = 6s^2$.

Step 1: Relating volume and side length

The volume of a cube is given by: $V = s^3$ Solving for $s$: $s = \sqrt[3]{V}$ At $V = 186 \, \text{m}^3$: $s = \sqrt[3]{186} \approx 5.729 \, \text{m}$

Step 2: Differentiating volume with respect to time

Differentiating $V = s^3$ with respect to $t$: $\frac{dV}{dt} = 3s^2 \frac{ds}{dt}$ Substituting $\frac{dV}{dt} = -511$: $-511 = 3s^2 \frac{ds}{dt}$ At $s \approx 5.729$: $-511 = 3(5.729)^2 \frac{ds}{dt}$ Calculating $s^2$: $5.729^2 \approx 32.826$ $-511 = 3(32.826) \frac{ds}{dt}$ $-511 = 98.478 \frac{ds}{dt}$ $\frac{ds}{dt} = \frac{-511}{98.478} \approx -5.191 \, \text{m/min}$

Step 3: Differentiating surface area with respect to time

The surface area of the cube is: $A = 6s^2$ Differentiating with respect to $t$: $\frac{dA}{dt} = 12s \frac{ds}{dt}$ Substitute $s \approx 5.729$ and $\frac{ds}{dt} \approx -5.191$: $\frac{dA}{dt} = 12(5.729)(-5.191)$ $\frac{dA}{dt} \approx 12(5.729)(-5.191) \approx -356.752 \, \text{m}^2/\text{min}$

Final Answer:

The rate of change of the surface area is approximately: $\boxed{-356.752 \, \text{m}^2/\text{min}}$

Let me know if you need further clarification or additional steps!

5 Related Questions:

1. What is the relationship between surface area and volume for a cube?
2. How does the rate of change of volume relate to the rate of change of side length for a cube?
3. How would the problem change if the cube's volume were increasing instead?
4. How can you compute the side length directly using logarithms for more precision?
5. Can you derive a general formula for $\frac{dA}{dt}$ in terms of $\frac{dV}{dt}$ without calculating $s$?

Tip:

For cubes, always link volume, surface area, and side length using their geometric formulas before differentiating!