Let's solve the problem step by step.

Given:

• The edges of a cube are expanding at a rate of $\frac{ds}{dt} = 9 \, \text{cm/sec}$, where $s$ is the edge length of the cube.
• We are asked to find the rate of change of the volume, $\frac{dV}{dt}$, when the edge is $s = 4 \, \text{cm}$ and $s = 13 \, \text{cm}$.

Step 1: Formula for the volume of a cube

The volume $V$ of a cube is given by: $V = s^3$ where $s$ is the length of an edge of the cube.

Step 2: Differentiate the volume with respect to time $t$

To find how fast the volume is changing, we differentiate both sides of $V = s^3$ with respect to time $t$ using the chain rule: $\frac{dV}{dt} = 3s^2 \cdot \frac{ds}{dt}$ where $\frac{ds}{dt} = 9 \, \text{cm/sec}$ is the rate at which the edges are expanding.

Step 3: Calculate the rate of change of volume

(a) When $s = 4 \, \text{cm}$:

$\frac{dV}{dt} = 3(4)^2 \cdot 9 = 3 \times 16 \times 9 = 432 \, \text{cm}^3/\text{sec}$

(b) When $s = 13 \, \text{cm}$:

$\frac{dV}{dt} = 3(13)^2 \cdot 9 = 3 \times 169 \times 9 = 4563 \, \text{cm}^3/\text{sec}$

Final Answers:

• (a) The volume is changing at a rate of $432 \, \text{cm}^3/\text{sec}$ when the edge is 4 cm.
• (b) The volume is changing at a rate of $4563 \, \text{cm}^3/\text{sec}$ when the edge is 13 cm.

Let me know if you need more details on any step, or if you have any questions.

Here are 5 related questions:

1. How would the rate of change of volume differ if the edge growth rate was 12 cm/sec instead of 9 cm/sec?
2. What happens to the rate of change of volume as the edge length approaches infinity?
3. How do you compute the surface area change rate if the surface area formula is $6s^2$?
4. Can you derive a general expression for the rate of volume change if the edge length grows at a non-constant rate?
5. How does the rate of change of the diagonal of the cube relate to the change in the edge length?

Tip: When dealing with related rates, always differentiate each part of the equation with respect to time.