To find how fast the radius of the balloon is increasing when the radius is 5 inches, we can use related rates in calculus.

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

2. Given Information:

   • The rate of change of the volume with respect to time is $\frac{dV}{dt} = 10$ cubic inches per minute.
   • We want to find $\frac{dr}{dt}$ when $r = 5$ inches.

3. Differentiate the Volume with Respect to Time: Using the chain rule, we differentiate both sides with respect to time $t$: $\frac{dV}{dt} = 4\pi r^2 \frac{dr}{dt}$

4. Substituting the Known Values: Plugging in the known values:

   • $V = 10$ cubic inches per minute
   • $r = 5$ inches

   We get: $10 = 4\pi (5^2) \frac{dr}{dt}$ $10 = 100\pi \frac{dr}{dt}$

5. Solve for $\frac{dr}{dt}$: $\frac{dr}{dt} = \frac{10}{100\pi} = \frac{1}{10\pi}$

6. Numerical Value: Approximating $\pi$ as 3.14: $\frac{dr}{dt} \approx \frac{1}{31.4} \approx 0.0318 \text{ inches per minute}$

Thus, the radius of the balloon is increasing at approximately 0.0318 inches per minute when the radius is 5 inches.

Questions for Further Exploration:

1. What would happen to the rate of change of the radius if the inflation rate increases to 20 cubic inches per minute?
2. How would the rate of increase of the radius change if the radius was initially 10 inches?
3. Can you explain how the formula for the volume of a sphere is derived?
4. What other shapes can you apply similar related rates concepts to?
5. How does temperature affect the volume of a gas in a balloon?

Tip:

When dealing with related rates, always identify what is changing, write down the relevant formulas, and differentiate with respect to time carefully.