The volume of the spherical ball is increasing at the rate of `4pi` cc/sec. Find the rate at which the radius and the surface area are changing when the volume is `288pi` cc.

Answer

To solve the problem step by step, we will follow the given information and apply the necessary mathematical formulas.

### Step 1: Identify the given information
We are given:
- The rate of change of volume, \( \frac{dV}{dt} = 4\pi \) cc/sec.
- The volume at a specific moment, \( V = 288\pi \) cc.

### Step 2: Use the volume formula for a sphere
The formula for the volume \( V \) of a sphere in terms of its radius \( r \) is:
\[ V = \frac{4}{3} \pi r^3 \]

### Step 3: Set up the equation for the given volume
Substituting the given volume into the formula:
\[ 288\pi = \frac{4}{3} \pi r^3 \]
We can simplify this equation by dividing both sides by \( \pi \):
\[ 288 = \frac{4}{3} r^3 \]

### Step 4: Solve for \( r^3 \)
Multiply both sides by \( \frac{3}{4} \):
\[ r^3 = 288 \times \frac{3}{4} = 216 \]

### Step 5: Find the radius \( r \)
Taking the cube root of both sides:
\[ r = \sqrt[3]{216} = 6 \text{ cm} \]

### Step 6: Differentiate the volume with respect to time
Now, we differentiate the volume formula with respect to time \( t \):
\[ \frac{dV}{dt} = 4\pi r^2 \frac{dr}{dt} \]

### Step 7: Substitute known values into the differentiated equation
We know \( \frac{dV}{dt} = 4\pi \) and \( r = 6 \):
\[ 4\pi = 4\pi (6^2) \frac{dr}{dt} \]
This simplifies to:
\[ 4\pi = 4\pi (36) \frac{dr}{dt} \]
Dividing both sides by \( 4\pi \):
\[ 1 = 36 \frac{dr}{dt} \]

### Step 8: Solve for \( \frac{dr}{dt} \)
\[ \frac{dr}{dt} = \frac{1}{36} \text{ cm/sec} \]

### Step 9: Find the surface area \( S \) of the sphere
The formula for the surface area \( S \) of a sphere is:
\[ S = 4\pi r^2 \]

### Step 10: Differentiate the surface area with respect to time
Differentiating with respect to \( t \):
\[ \frac{dS}{dt} = 8\pi r \frac{dr}{dt} \]

### Step 11: Substitute known values into the differentiated surface area equation
Substituting \( r = 6 \) and \( \frac{dr}{dt} = \frac{1}{36} \):
\[ \frac{dS}{dt} = 8\pi (6) \left(\frac{1}{36}\right) \]
This simplifies to:
\[ \frac{dS}{dt} = 8\pi \cdot \frac{6}{36} = 8\pi \cdot \frac{1}{6} = \frac{4\pi}{3} \text{ cm}^2/\text{sec} \]

### Final Answers
- The rate at which the radius is changing: \( \frac{dr}{dt} = \frac{1}{36} \) cm/sec
- The rate at which the surface area is changing: \( \frac{dS}{dt} = \frac{4\pi}{3} \) cm²/sec