If a spherical rain drop evaporates at a rate proportional to its surface area. Form a differential equation indicating the rate of change of the radius of the rain drop.

Answer

To form a differential equation indicating the rate of change of the radius of a spherical raindrop that evaporates at a rate proportional to its surface area, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Problem**: We know that the raindrop is spherical and that it evaporates at a rate proportional to its surface area. This means that the volume of the raindrop decreases over time.

2. **Define the Variables**: Let \( r \) be the radius of the raindrop. The volume \( V \) of a sphere is given by the formula:
\[ V = \frac{4}{3} \pi r^3 \]

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

4. **Express the Rate of Evaporation**: According to the problem, the rate of evaporation (which is the rate of decrease of volume) is proportional to the surface area of the sphere. The surface area \( A \) of a sphere is given by:
\[ A = 4 \pi r^2 \]
Therefore, we can express the rate of evaporation as:
\[ \frac{dV}{dt} = -k \cdot A = -k \cdot 4 \pi r^2 \]
where \( k \) is a positive constant of proportionality.

5. **Set Up the Differential Equation**: Now we can set the two expressions for \( \frac{dV}{dt} \) equal to each other:
\[ 4 \pi r^2 \frac{dr}{dt} = -k \cdot 4 \pi r^2 \]

6. **Cancel Common Terms**: Since \( 4 \pi r^2 \) is common on both sides (and assuming \( r \neq 0 \)), we can cancel it out:
\[ \frac{dr}{dt} = -k \]

### Final Differential Equation:
Thus, the differential equation indicating the rate of change of the radius of the raindrop is:
\[ \frac{dr}{dt} = -k \]