To find the rate at which the surface area of a sphere is increasing, we can use the relationship between the volume and surface area of a sphere.

1. Volume of a sphere $V$ is given by: $V = \frac{4}{3} \pi r^3$ where $r$ is the radius of the sphere.

2. Surface area of a sphere $S$ is given by: $S = 4 \pi r^2$

Step 1: Differentiate the Volume with Respect to Time

The rate of change of volume with respect to time $\frac{dV}{dt}$ is given as $8 \, \text{cm}^3/\text{s}$.

Differentiating $V = \frac{4}{3} \pi r^3$ with respect to $t$: $\frac{dV}{dt} = 4 \pi r^2 \frac{dr}{dt}$

Since we know $\frac{dV}{dt} = 8$, we can solve for $\frac{dr}{dt}$: $8 = 4 \pi (12)^2 \frac{dr}{dt}$ $8 = 4 \pi \cdot 144 \cdot \frac{dr}{dt}$ $8 = 576 \pi \frac{dr}{dt}$ $\frac{dr}{dt} = \frac{8}{576 \pi} = \frac{1}{72 \pi} \, \text{cm/s}$

Step 2: Differentiate the Surface Area with Respect to Time

Now, we need to find $\frac{dS}{dt}$, the rate at which the surface area is increasing.

Differentiating $S = 4 \pi r^2$ with respect to $t$: $\frac{dS}{dt} = 8 \pi r \frac{dr}{dt}$

Substitute $r = 12$ and $\frac{dr}{dt} = \frac{1}{72 \pi}$: $\frac{dS}{dt} = 8 \pi \cdot 12 \cdot \frac{1}{72 \pi}$ $\frac{dS}{dt} = 8 \cdot 12 \cdot \frac{1}{72}$ $\frac{dS}{dt} = \frac{96}{72}$ $\frac{dS}{dt} = \frac{4}{3} \, \text{cm}^2/\text{s}$

Answer

The rate at which the surface area of the sphere is increasing when the radius is 12 cm is: $\frac{4}{3} \, \text{cm}^2/\text{s}$

Would you like further details or have any questions?

Here are some additional questions for deeper exploration:

1. How would the rate of surface area change if the radius were different?
2. What happens to $\frac{dS}{dt}$ as the radius grows larger?
3. Can this approach be used for other shapes like cylinders or cones?
4. How does the volume rate of change affect the surface area rate in general?
5. What are the practical implications of this calculation in physics or engineering?

Tip: Remember that the chain rule is often key in problems where rates of related quantities change over time.