Step 1: Understanding the Concept:
This problem involves related rates in calculus. We are given the rate of change of volume and asked to find the rate of change of the radius at a specific moment.
Since the balloon is spherical, we must use the geometry of a sphere.
Step 2: Key Formula or Approach:
The volume \(V\) of a sphere with radius \(r\) is given by:
\[ V = \frac{4}{3} \pi r^3 \]
To find the relationship between the rates, we differentiate this equation with respect to time \(t\):
\[ \frac{dV}{dt} = \frac{4}{3} \pi \cdot (3r^2) \frac{dr}{dt} \]
Simplifying:
\[ \frac{dV}{dt} = 4 \pi r^2 \frac{dr}{dt} \]
(Note: \(4 \pi r^2\) is the surface area of the sphere).
Step 3: Detailed Explanation:
Given:
Rate of volume increase \(\frac{dV}{dt} = 30 \, \text{ft}^3/\text{min}\).
Radius at the given instant \(r = 15 \, \text{ft}\).
We need to find \(\frac{dr}{dt}\).
Substitute the known values into the differentiated equation:
\[ 30 = 4 \pi (15)^2 \cdot \frac{dr}{dt} \]
First, calculate \(15^2\):
\[ 15^2 = 225 \]
Now substitute:
\[ 30 = 4 \pi (225) \cdot \frac{dr}{dt} \]
\[ 30 = 900 \pi \cdot \frac{dr}{dt} \]
Solve for \(\frac{dr}{dt}\):
\[ \frac{dr}{dt} = \frac{30}{900 \pi} \]
Cancel the common factor of 30:
\[ \frac{dr}{dt} = \frac{1}{30 \pi} \, \text{ft/min} \]
Check the logic: As the balloon gets bigger, the same volume of gas added leads to a smaller increase in radius because it must be spread over a much larger surface area. Since the surface area at \(r=15\) is quite large (\(900\pi\)), the radial increase should be very small, which \(\frac{1}{30\pi} \approx 0.01\) is.
Step 4: Final Answer:
The rate of increase of the radius is \(\frac{1}{30\pi} \, \text{ft/min}\).