We are asked to find the rate at which the volume of a cube increases, given that the edge of the cube is increasing at a rate of 3 cm/s and the edge length is 10 cm at the specific moment.

Step 1: Volume of a Cube

The volume $V$ of a cube is given by the formula:

$V = s^3$

where $s$ is the edge length of the cube.

Step 2: Rate of Change of Volume

We need to find how fast the volume is changing with respect to time, i.e., $\frac{dV}{dt}$. To do this, we can differentiate the volume formula with respect to time $t$.

$\frac{dV}{dt} = \frac{d}{dt}(s^3)$

Using the chain rule:

$\frac{dV}{dt} = 3s^2 \frac{ds}{dt}$

Step 3: Substituting Known Values

We are given:

• $\frac{ds}{dt} = 3 \, \text{cm/s}$ (the rate at which the edge is increasing),
• $s = 10 \, \text{cm}$ (the edge length at the moment of interest).

Substituting these values into the equation:

$\frac{dV}{dt} = 3(10)^2 \times 3$

$\frac{dV}{dt} = 3 \times 100 \times 3$

$\frac{dV}{dt} = 900 \, \text{cm}^3/\text{s}$

Final Answer:

The volume of the cube is increasing at a rate of $900 \, \text{cm}^3/\text{s}$ when the edge of the cube is 10 cm.

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Would you like further clarification on any of the steps? Or do you have another related question in mind?

Here are some related questions for further exploration:

1. How do you find the rate of change of surface area when the edge length is increasing?
2. What happens to the rate of change of volume if the edge length increases at a different rate?
3. How would this calculation change if the cube's edge length was decreasing?
4. How can the rate of change of volume be used in real-world applications like manufacturing or fluid dynamics?
5. Can you use the same method to find the rate of change of the volume of other three-dimensional objects?

Tip: Always ensure the units of all quantities are consistent before substituting into formulas!