Final answer:
The volume of a cube changes at 288 cm³/s when each edge is 4 cm, and at 1458 cm³/s when each edge is 9 cm, calculated by taking the derivative of the volume with respect to time.
Step-by-step explanation:
The student is asking about rate of change in the context of a geometric problem. Specifically, they are interested in how the volume of a cube changes as its edges are expanding at a constant rate. The cube's edge length (s) is changing, and so we will use the formula for the volume of a cube V = s³ to find how fast the volume is changing over time.
To solve this, we take the derivative of the volume with respect to time (dv/dt) using the chain rule:
dV/dt = 3s²(ds/dt)
Given that ds/dt (the rate at which the side of the cube is expanding) is 6 cm/s, we plug in the values of s for each scenario:
- When s = 4 cm:
dV/dt = 3(4 cm)²(6 cm/s) = 3(16 cm²)(6 cm/s) = 288 cm³/s - When s = 9 cm:
dV/dt = 3(9 cm)²(6 cm/s) = 3(81 cm²)(6 cm/s) = 1458 cm³/s
Answers:
- The volume is changing at a rate of 288 cm³/s when each edge is 4 cm.
- The volume is changing at a rate of 1458 cm³/s when each edge is 9 cm.