To solve this problem, we need to relate the volume and surface area of the cube, and then find how the rate of change of the volume affects the rate of change of the surface area.

Step 1: Relationships between volume and side length

Let the side length of the cube be $s$. The volume $V$ and surface area $A$ of the cube are given by:

• Volume: $V = s^3$
• Surface area: $A = 6s^2$

We are given that the volume is decreasing at a constant rate: $\frac{dV}{dt} = -1699 \text{ cubic meters per second}$

We are also given that at the instant of interest, the volume is: $V = 629 \text{ cubic meters}$

Step 2: Differentiate the volume and surface area equations

To find the rate of change of the surface area, we will differentiate both the volume and surface area equations with respect to time $t$.

1. Differentiating $V = s^3$: $\frac{dV}{dt} = 3s^2 \frac{ds}{dt}$

2. Differentiating $A = 6s^2$: $\frac{dA}{dt} = 12s \frac{ds}{dt}$

Step 3: Solve for $\frac{ds}{dt}$

From the first equation, we can solve for $\frac{ds}{dt}$: $\frac{dV}{dt} = 3s^2 \frac{ds}{dt}$ $-1699 = 3s^2 \frac{ds}{dt}$ $\frac{ds}{dt} = \frac{-1699}{3s^2}$

Step 4: Find $s$ when $V = 629$

Since $V = s^3$, we can solve for $s$: $s^3 = 629$ $s = \sqrt[3]{629}$

Let’s calculate $s$:

$s \approx 8.559 \text{ meters}$

Step 5: Calculate $\frac{ds}{dt}$

Now that we have $s$, we can substitute it into the equation for $\frac{ds}{dt}$: $\frac{ds}{dt} = \frac{-1699}{3(8.559)^2}$

Let’s calculate $\frac{ds}{dt}$:

$\frac{ds}{dt} \approx \frac{-1699}{3 \times 73.236} = \frac{-1699}{219.708}$ $\frac{ds}{dt} \approx -7.735 \text{ meters per second}$

Step 6: Calculate $\frac{dA}{dt}$

Finally, we substitute $s$ and $\frac{ds}{dt}$ into the equation for $\frac{dA}{dt}$: $\frac{dA}{dt} = 12s \frac{ds}{dt}$ $\frac{dA}{dt} = 12(8.559)(-7.735)$ $\frac{dA}{dt} \approx -794.825 \text{ square meters per second}$

Final Answer:

The rate of change of the surface area of the cube is approximately: $\boxed{-794.825} \text{ square meters per second}.$

Let me know if you'd like further details or clarifications!

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Here are 5 related questions:

1. How would the solution change if the volume were increasing instead of decreasing?
2. What is the relationship between the side length and surface area in terms of rates of change?
3. How does the surface area change if the volume remains constant?
4. Can the rate of volume change ever directly equal the rate of surface area change?
5. How would this problem change if we were working with a rectangular prism instead of a cube?

Tip: Remember that volume changes scale with the cube of the side length, while surface area changes scale with the square of the side length.