All edges of a cube are expanding at a rate of 7 centimeters per second. (a) How fast is the volume changing (in \( \mathrm{cm}^{3} / \mathrm{sec} \) ) when each edge is 5 centimeter(s)? \( \square \) \( \mathrm{cm}^{3} / \mathrm{sec} \) (b) How fast is the volume changing (in \( \mathrm{cm}^{3} / \mathrm{sec} \) ) when each edge is 12 centineters? \( \square \) \( \mathrm{cm}^{3} / \mathrm{sec} \)

To find out how fast the volume of a cube is changing as its edges expand, we can use the formula for the volume of a cube \( V = s^3 \), where \( s \) is the length of each edge. ### (a) When each edge is 5 cm: We know that the volume \( V \) changes with respect to time as the edge length \( s \) changes: \[ \frac{dV}{dt} = \frac{dV}{ds} \cdot \frac{ds}{dt} \] First, we calculate \( \frac{dV}{ds} = 3s^2 \). If \( s = 5 \) cm, then: \[ \frac{dV}{ds} = 3(5)^2 = 3 \times 25 = 75 \, \text{cm}^2 \] Given that \( \frac{ds}{dt} = 7 \, \text{cm/sec} \): \[ \frac{dV}{dt} = 75 \, \text{cm}^2 \cdot 7 \, \text{cm/sec} = 525 \, \text{cm}^3/\text{sec} \] So, the volume is changing at a rate of **525 cm³/sec** when each edge is 5 cm. ### (b) When each edge is 12 cm: Using the same approach, we find: \[ \frac{dV}{ds} = 3s^2 \] For \( s = 12 \) cm: \[ \frac{dV}{ds} = 3(12)^2 = 3 \times 144 = 432 \, \text{cm}^2 \] Still using \( \frac{ds}{dt} = 7 \, \text{cm/sec} \): \[ \frac{dV}{dt} = 432 \, \text{cm}^2 \cdot 7 \, \text{cm/sec} = 3024 \, \text{cm}^3/\text{sec} \] Thus, the volume is changing at a rate of **3024 cm³/sec** when each edge is 12 cm.