Step 1: Understanding the Concept:
This problem falls under "Related Rates" in differential calculus. We have a physical quantity (volume) changing at a known rate over time, and we need to find how another related quantity (radius) changes at the same instant.
As the balloon is a sphere, the relationship between its volume and radius is governed by geometric formulas.
Key Formula or Approach:
The volume \( V \) of a sphere with radius \( r \) is given by:
\[ V = \frac{4}{3}\pi r^3 \]
We need to find the derivative of volume with respect to time, \( \frac{dV}{dt} \), which relates to the derivative of the radius with respect to time, \( \frac{dr}{dt} \).
Step 2: Detailed Explanation:
Differentiating both sides of the volume equation with respect to time \( t \) using the chain rule:
\[ \frac{dV}{dt} = \frac{d}{dt}\left( \frac{4}{3}\pi r^3 \right) \]
\[ \frac{dV}{dt} = \frac{4}{3}\pi \cdot 3r^2 \cdot \frac{dr}{dt} \]
Simplifying the constants:
\[ \frac{dV}{dt} = 4\pi r^2 \frac{dr}{dt} \]
Note that \( 4\pi r^2 \) is the formula for the Surface Area (\( S \)) of the sphere. This means the rate of change of volume is the surface area multiplied by the rate of change of the radius.
The problem provides the following values:
\( \frac{dV}{dt} = 30 \, \text{ft}^3/\text{min} \)
\( r = 15 \, \text{ft} \)
Plug these into our differentiated equation:
\[ 30 = 4\pi (15)^2 \cdot \frac{dr}{dt} \]
Calculate \( 15^2 \):
\[ 30 = 4\pi (225) \cdot \frac{dr}{dt} \]
\[ 30 = 900\pi \cdot \frac{dr}{dt} \]
Solving for \( \frac{dr}{dt} \):
\[ \frac{dr}{dt} = \frac{30}{900\pi} \]
Reducing the fraction by dividing both numerator and denominator by 30:
\[ \frac{dr}{dt} = \frac{1}{30\pi} \, \text{ft}/\text{min} \]
Step 3: Final Answer:
The rate of increase of the radius is \( \frac{1}{30\pi} \, \text{ft}/\text{min} \).