 A spherical snowball is melting so that its diameter is decreasing at rate of 0.4 cm/min. At what is the rate is the volume of the snowball changing when the diameter is 13 cm?
The volume is changing at a rate of  cm3/min.

To find the rate at which the volume of a spherical snowball is changing, we can use the formula for the volume of a sphere:

\[
V = \frac{4}{3} \pi r^3
\]

where \( r \) is the radius of the sphere. The relationship between the diameter \( d \) and the radius is \( r = \frac{d}{2} \). Given that the diameter is decreasing at a rate of 0.4 cm/min, we can express this rate in terms of the radius.

1. **Find the radius when the diameter is 13 cm**:
   \[
   d = 13 \text{ cm} \implies r = \frac{d}{2} = \frac{13}{2} = 6.5 \text{ cm}
   \]

2. **Relate the change in diameter to the change in radius**:
   Since the diameter is decreasing, the rate of change of the radius is:
   \[
   \frac{dr}{dt} = \frac{1}{2} \frac{dd}{dt} = \frac{1}{2} (-0.4) = -0.2 \text{ cm/min}
   \]

3. **Differentiate the volume formula with respect to time**:
   Using the chain rule:
   \[
   \frac{dV}{dt} = \frac{dV}{dr} \cdot \frac{dr}{dt}
   \]

   First, compute \(\frac{dV}{dr}\):
   \[
   \frac{dV}{dr} = 4\pi r^2
   \]

4. **Substituting values**:
   At \( r = 6.5 \text{ cm} \):
   \[
   \frac{dV}{dr} = 4\pi (6.5)^2 = 4\pi (42.25) = 169\pi \text{ cm}^2
   \]

5. **Now substitute everything back**:
   \[
   \frac{dV}{dt} = 169\pi \cdot (-0.2) = -33.8\pi \text{ cm}^3/\text{min}
   \]

6. **Calculate the numerical value**:
   \[
   \frac{dV}{dt} \approx -33.8 \times 3.14 \approx -106.7 \text{ cm}^3/\text{min}
   \]

Thus, the volume of the snowball is changing at a rate of approximately \(-106.7 \, \text{cm}^3/\text{min}\) when the diameter is 13 cm.

### Summary:
- The volume is changing at a rate of \(-106.7 \, \text{cm}^3/\text{min}\).

Do you have any questions about this solution or would you like more details? Here are some related questions you might consider:

1. What would happen to the volume if the melting rate changed?
2. How does the rate of volume change compare at different diameters?
3. What is the significance of the negative sign in the rate of volume change?
4. Can we derive a formula for the rate of volume change in terms of diameter instead of radius?
5. How does this concept apply to other geometrical shapes? 

**Tip:** Understanding the relationship between volume and radius is key in solving similar problems involving spheres and other 3D shapes.

A spherical snowball is melting so that its diameter is decreasing at rate of 0.4 cm/min. At what is the rate is the volume of the snowball changing when the diameter is 13 cm?

Discover how to determine the rate of change of volume for a spherical snowball melting at a rate of 0.4 cm/min in diameter. At a diameter of 13 cm, the volume changes at approximately -106.7 cm³/min. This solution involves calculus concepts and the formula for the volume of a sphere.

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