A spherical ice ball is melting at the rate of `100 pi cm^(3)` /min. The rate at which its radius is decreasing, when its radius is 15 cm, is

Options

A. `(1)/(9) ` cm/min
B. `(1)/(9 pi)` cm/min
C. `(1)/(18)` cm/min
D. `(1)/(36)` cm/min

Answer

To solve the problem, we need to find the rate at which the radius of a spherical ice ball is decreasing when its radius is 15 cm, given that the volume of the ice ball is decreasing at a rate of \(100\pi \, \text{cm}^3/\text{min}\).

### Step-by-Step Solution:

1. **Understand the Volume of a Sphere**:
The volume \(V\) of a sphere is given by the formula:
\[ V = \frac{4}{3} \pi r^3 \]

2. **Differentiate the Volume with Respect to Time**:
We need to find the rate of change of volume with respect to time, \( \frac{dV}{dt} \). Using the chain rule, we differentiate \(V\):
\[ \frac{dV}{dt} = \frac{d}{dt} \left( \frac{4}{3} \pi r^3 \right) = 4\pi r^2 \frac{dr}{dt} \]

3. **Set Up the Equation**:
We know from the problem that the volume is decreasing at a rate of \(100\pi \, \text{cm}^3/\text{min}\). Therefore, we can write:
\[ \frac{dV}{dt} = -100\pi \]
Substituting this into our differentiated equation gives:
\[ -100\pi = 4\pi r^2 \frac{dr}{dt} \]

4. **Simplify the Equation**:
We can divide both sides by \(\pi\) (since \(\pi \neq 0\)):
\[ -100 = 4r^2 \frac{dr}{dt} \]

5. **Substitute the Radius**:
We need to find \(\frac{dr}{dt}\) when \(r = 15 \, \text{cm}\):
\[ -100 = 4(15^2) \frac{dr}{dt} \]
Calculating \(15^2\):
\[ 15^2 = 225 \]
So, substituting this back gives:
\[ -100 = 4 \times 225 \frac{dr}{dt} \]
Simplifying further:
\[ -100 = 900 \frac{dr}{dt} \]

6. **Solve for \(\frac{dr}{dt}\)**:
Now, we can solve for \(\frac{dr}{dt}\):
\[ \frac{dr}{dt} = \frac{-100}{900} = -\frac{1}{9} \, \text{cm/min} \]

7. **Conclusion**:
The rate at which the radius is decreasing when the radius is 15 cm is:
\[ \frac{dr}{dt} = -\frac{1}{9} \, \text{cm/min} \]
Since we are interested in the rate of decrease, we can express the answer as:
\[ \text{Rate of decrease of radius} = \frac{1}{9} \, \text{cm/min} \]