To solve this problem, we need to determine the rate at which the radius of a spherical balloon is increasing when the volume is inflating at a constant rate.

Given:

• Rate of change of volume $\frac{dV}{dt} = 11 \, \text{cm}^3/\text{s}$.
• We need to find $\frac{dr}{dt}$ (the rate of change of the radius) when $r = 1$ cm, $r = 10$ cm, and $r = 100$ cm.

Solution Outline:

1. Volume of a sphere: $V = \frac{4}{3} \pi r^3$
2. Differentiate with respect to time: $\frac{dV}{dt} = 4 \pi r^2 \frac{dr}{dt}$
3. Solve for $\frac{dr}{dt}$: $\frac{dr}{dt} = \frac{\frac{dV}{dt}}{4 \pi r^2}$
4. Substitute the given values of $\frac{dV}{dt}$ and $r$ to find $\frac{dr}{dt}$ at each radius.

Let's calculate each one:

1. When $r = 1 \, \text{cm}$:

$\frac{dr}{dt} = \frac{11}{4 \pi (1)^2} = \frac{11}{4 \pi} \approx 0.875 \, \text{cm/s}$

2. When $r = 10 \, \text{cm}$:

$\frac{dr}{dt} = \frac{11}{4 \pi (10)^2} = \frac{11}{400 \pi} \approx 0.00875 \, \text{cm/s}$

3. When $r = 100 \, \text{cm}$:

$\frac{dr}{dt} = \frac{11}{4 \pi (100)^2} = \frac{11}{40000 \pi} \approx 0.0000875 \, \text{cm/s}$

Answers:

• For $r = 1$ cm: $\frac{dr}{dt} \approx 0.875 \, \text{cm/s}$
• For $r = 10$ cm: $\frac{dr}{dt} \approx 0.00875 \, \text{cm/s}$
• For $r = 100$ cm: $\frac{dr}{dt} \approx 0.0000875 \, \text{cm/s}$

Would you like a detailed explanation of each step?

Here are some related questions:

1. What happens to $\frac{dr}{dt}$ as the radius gets larger?
2. How does the rate of radius increase relate to the volume?
3. Can this method be used if the rate of volume increase is not constant?
4. How would the answer change if the units of $\frac{dV}{dt}$ were different?
5. How does the rate of change in surface area compare with the rate of change in volume?

Tip: As the radius increases, $\frac{dr}{dt}$ decreases due to the $r^2$ term in the denominator. This reflects that larger spheres expand slower for a constant volume inflow rate.