We are given that the edges of a cube are expanding at a rate of 8 cm per second, and we need to determine how fast the volume is changing at different edge lengths.

Step 1: Volume of a Cube

The volume $V$ of a cube with edge length $x$ is given by: $V = x^3$

Step 2: Differentiate the Volume with Respect to Time

To find the rate at which the volume is changing with respect to time, we differentiate both sides of the volume equation with respect to $t$: $\frac{dV}{dt} = 3x^2 \cdot \frac{dx}{dt}$ Here, $\frac{dV}{dt}$ is the rate of change of volume, and $\frac{dx}{dt}$ is the rate of change of the edge length (given as 8 cm/s).

Step 3: Use the Given Information

We are given that $\frac{dx}{dt} = 8$ cm/s.

Now we will solve the two parts of the problem:

(a) When the edge length is 3 cm

Substitute $x = 3$ cm and $\frac{dx}{dt} = 8$ cm/s into the formula: $\frac{dV}{dt} = 3(3)^2 \cdot 8 = 3 \times 9 \times 8 = 216 \text{ cm}^3/\text{s}$ Thus, the volume is changing at a rate of 216 cm³/s when the edge length is 3 cm.

(b) When the edge length is 15 cm

Substitute $x = 15$ cm and $\frac{dx}{dt} = 8$ cm/s into the formula: $\frac{dV}{dt} = 3(15)^2 \cdot 8 = 3 \times 225 \times 8 = 5400 \text{ cm}^3/\text{s}$ Thus, the volume is changing at a rate of 5400 cm³/s when the edge length is 15 cm.

Would you like more details or have any questions?

Related Questions:

1. What is the general formula for the rate of change of volume for any geometric shape?
2. How does the volume change compare between small and large edge lengths?
3. If the rate of edge expansion doubles, how will the volume change rate be affected?
4. Can you explain the concept of related rates in calculus using this problem?
5. What happens to the volume rate if the edges expand at different rates?

Tip:

When solving related rates problems, always write down what each variable represents and differentiate with respect to time using the chain rule.