To solve the problem from the image, we need to determine how fast the surface area of a cube is changing given that the cube's volume is increasing at a rate of $3 \text{ cm}^3/\text{s}$, and each edge of the cube is $17 \text{ cm}$.

Step 1: Set Up the Relationships

1. Volume of the cube: $V = s^3$ where $s$ is the side length of the cube.

2. Surface area of the cube: $A = 6s^2$

Step 2: Use Derivatives to Relate the Rates of Change

We know the rate of change of volume, $\frac{dV}{dt}$, is $3 \text{ cm}^3/\text{s}$. We need to find the rate of change of surface area, $\frac{dA}{dt}$.

Differentiate the volume: $\frac{dV}{dt} = 3s^2 \cdot \frac{ds}{dt}$ Rearranging to solve for $\frac{ds}{dt}$: $\frac{ds}{dt} = \frac{\frac{dV}{dt}}{3s^2} = \frac{3}{3 \cdot 17^2} = \frac{3}{867}$

Differentiate the surface area: $\frac{dA}{dt} = 12s \cdot \frac{ds}{dt}$

Step 3: Calculate $\frac{dA}{dt}$

Plugging in $s = 17$ and $\frac{ds}{dt} = \frac{3}{867}$: $\frac{dA}{dt} = 12 \cdot 17 \cdot \frac{3}{867} = \frac{612}{867} \approx 0.706 \text{ cm}^2/\text{s}$

Thus, the surface area of the cube is changing at approximately $0.706 \text{ cm}^2/\text{s}$ when each edge is 17 cm.

Would you like any further details or clarification on this solution? Here are some follow-up questions you might consider:

1. What would happen to the rate of change of surface area if the rate of volume expansion were different?
2. How does the rate of change of the surface area change with respect to different side lengths?
3. Can you determine the time it takes for the volume of the cube to double?
4. How would you solve this problem if the cube were shrinking instead of expanding?
5. Could you apply similar principles to find how the volume or surface area changes for other shapes, like a sphere or cylinder?

Tip: When dealing with related rates problems, always express all variables in terms of a single variable before differentiating if possible. This simplifies the process and minimizes errors.