Gas is being pumped into a spherical balloon at the rate of `30 ft^(3) min^(-1)`. Then the rate at which the radius increases when it reaches the value 15 ft, is

Options

A. `1/(30 pi) ft min^(-1)`
B. `1/(15 pi) ft min^(-1)`
C. `1/20 ft min^(-1)`
D. `1/15 ft min^(-1)`

Explanation

To solve the problem step by step, we will follow the process of relating the volume of a spherical balloon to its radius and then find the rate of change of the radius with respect to time.

### Step 1: Understand the relationship between volume and radius
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.

### Step 2: Differentiate the volume with respect to time
We need to find the rate of change of the radius with respect to time, \( \frac{dr}{dt} \). We start by differentiating the volume \( V \) with respect to time \( t \):
\[ \frac{dV}{dt} = \frac{d}{dt}\left(\frac{4}{3} \pi r^3\right) \]
Using the chain rule, we have:
\[ \frac{dV}{dt} = \frac{4}{3} \pi \cdot 3r^2 \frac{dr}{dt} = 4\pi r^2 \frac{dr}{dt} \]

### Step 3: Substitute the known rate of volume change
We know that gas is being pumped into the balloon at a rate of \( \frac{dV}{dt} = 30 \, \text{ft}^3/\text{min} \). Therefore, we can set up the equation:
\[ 30 = 4\pi r^2 \frac{dr}{dt} \]

### Step 4: Solve for \( \frac{dr}{dt} \)
Rearranging the equation to solve for \( \frac{dr}{dt} \):
\[ \frac{dr}{dt} = \frac{30}{4\pi r^2} \]

### Step 5: Substitute \( r = 15 \, \text{ft} \)
Now we need to find \( \frac{dr}{dt} \) when the radius \( r \) is 15 ft:
\[ \frac{dr}{dt} = \frac{30}{4\pi (15)^2} \]
Calculating \( (15)^2 \):
\[ (15)^2 = 225 \]
Now substituting this value:
\[ \frac{dr}{dt} = \frac{30}{4\pi \cdot 225} \]
Simplifying:
\[ \frac{dr}{dt} = \frac{30}{900\pi} = \frac{1}{30\pi} \]

### Final Answer
Thus, the rate at which the radius increases when it reaches the value of 15 ft is:
\[ \frac{dr}{dt} = \frac{1}{30\pi} \, \text{ft/min} \]