Question

Gas is being pumped into a spherical balloon at the rate of $30\text{ ft}^3/\text{min}$. Then the rate at which the radius increases when it reaches the value $15\text{ ft}$ is:

• A. $\frac{1}{30\pi}\text{ ft/min}$
• B. $\frac{1}{15\pi}\text{ ft/min}$
• C. $\frac{1}{20\pi}\text{ ft/min}$
• D. $\frac{1}{25\pi}\text{ ft/min}$

Hint:

In related rates problems involving spheres, remember: \[ \frac{dV}{dt}=4\pi r^2\frac{dr}{dt} \] The factor $4\pi r^2$ is actually the surface area of the sphere.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
This problem involves the application of derivatives as a rate of change. We are given the rate of change of the volume of a sphere and asked to find the instantaneous rate of change of its radius.
The relationship between volume (\(V\)) and radius (\(r\)) is governed by the volume formula for a sphere. By differentiating this formula with respect to time (\(t\)), we link the two rates.
Step 2: Key Formula or Approach:
1. Volume of a sphere: \(V = \frac{4}{3} \pi r^3\).
2. Differentiate with respect to time \(t\) using the chain rule:
\[ \frac{dV}{dt} = 4\pi r^2 \frac{dr}{dt} \]
Step 3: Detailed Explanation:
Given information:
Rate of change of volume \(\frac{dV}{dt} = 30\) ft\(^3\)/min.
Instantaneous radius \(r = 15\) ft.
We need to find \(\frac{dr}{dt}\).
Plug the values into the differentiated equation:
\[ 30 = 4\pi (15)^2 \frac{dr}{dt} \]
Calculate the square of 15: \(15^2 = 225\).
\[ 30 = 4\pi (225) \frac{dr}{dt} \]
\[ 30 = 900\pi \frac{dr}{dt} \]
Isolate \(\frac{dr}{dt}\):
\[ \frac{dr}{dt} = \frac{30}{900\pi} \]
Cancel the zeros and simplify the fraction:
\[ \frac{dr}{dt} = \frac{3}{90\pi} = \frac{1}{30\pi} \]
The unit for the rate of change of radius is length per time, which is ft/min.
Step 4: Final Answer:
The radius is increasing at a rate of \(\frac{1}{30\pi}\) ft/min.
Hence, the correct option is (A).