If a snowball melts so that its surface area decreases at a rate of6cm2/min, find the rate (in cm/min) at which the diameter decreases when the diameter is 11cm.

To solve this problem, we can use related rates in calculus. Here’s a step-by-step breakdown:

1. **Given Information**:
   - Rate of change of surface area: \(\frac{dA}{dt} = -6 \, \text{cm}^2/\text{min}\) (negative because the surface area is decreasing).
   - Diameter of the snowball at the time we're interested in: \(D = 11 \, \text{cm}\).

2. **Goal**:
   - Find the rate at which the diameter decreases, \(\frac{dD}{dt}\), when \(D = 11 \, \text{cm}\).

3. **Formulas and Relationships**:
   - Surface area \(A\) of a sphere is given by:
     \[
     A = 4 \pi r^2
     \]
     where \(r\) is the radius of the sphere.
   - The diameter \(D\) is twice the radius, so \(D = 2r\), or equivalently \(r = \frac{D}{2}\).
   
4. **Differentiate the Surface Area with Respect to Time**:
   - To relate \(\frac{dA}{dt}\) to \(\frac{dD}{dt}\), we need to differentiate \(A = 4 \pi r^2\) with respect to time \(t\):
     \[
     \frac{dA}{dt} = 4 \pi \cdot 2r \cdot \frac{dr}{dt} = 8 \pi r \cdot \frac{dr}{dt}
     \]
   - Since \(r = \frac{D}{2}\), we have \(\frac{dr}{dt} = \frac{1}{2} \frac{dD}{dt}\).
   - Substitute \(\frac{dr}{dt} = \frac{1}{2} \frac{dD}{dt}\) into the differentiated surface area equation:
     \[
     \frac{dA}{dt} = 8 \pi r \cdot \frac{1}{2} \frac{dD}{dt} = 4 \pi r \cdot \frac{dD}{dt}
     \]

5. **Solve for \(\frac{dD}{dt}\)**:
   - Rearranging for \(\frac{dD}{dt}\), we get:
     \[
     \frac{dD}{dt} = \frac{\frac{dA}{dt}}{4 \pi r}
     \]
   - Now, substitute \(\frac{dA}{dt} = -6 \, \text{cm}^2/\text{min}\) and \(r = \frac{11}{2} = 5.5 \, \text{cm}\):
     \[
     \frac{dD}{dt} = \frac{-6}{4 \pi \cdot 5.5}
     \]
     Simplifying:
     \[
     \frac{dD}{dt} = \frac{-6}{22 \pi} = -\frac{3}{11 \pi} \, \text{cm/min}
     \]

### Answer
The rate at which the diameter decreases when the diameter is 11 cm is:
\[
\frac{dD}{dt} = -\frac{3}{11 \pi} \, \text{cm/min}
\]

Would you like further explanation on any step, or have any other questions?

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**Relative Questions for Further Practice:**

1. How would the rate of change in radius differ from that of the diameter in this scenario?
2. If the surface area decreases at a different rate, how would that affect the diameter's rate of decrease?
3. Can you determine the rate at which the radius decreases when the diameter is 11 cm?
4. How would the problem change if we were given the rate at which the volume decreases instead?
5. What would happen to \(\frac{dD}{dt}\) if the initial diameter was larger?

**Tip:** When working with related rates problems, make sure to write down all known rates and identify which formulas relate them. This helps in setting up the problem correctly before differentiating.

If a snowball melts so that its surface area decreases at a rate of 6 cm^2/min, find the rate (in cm/min) at which the diameter decreases when the diameter is 11 cm.

Explore how to find the rate at which the diameter of a snowball decreases as it melts, given that its surface area is shrinking by 6 cm²/min. This problem utilizes calculus and related rates, making it ideal for advanced high school students in Grades 11-12.

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