Snowball melts, surface area decreases at 9 cm^2/min, find rate at which diameter decreases when diameter is 11 cm

To solve this problem, we'll need to use related rates. Let's break it down step by step:

### Step 1: Formula for the Surface Area of a Sphere
The surface area \(A\) of a sphere is given by the formula:
\[
A = 4\pi r^2
\]
where \(r\) is the radius of the sphere.

### Step 2: Relationship Between Diameter and Radius
The diameter \(d\) is related to the radius by:
\[
d = 2r
\]

### Step 3: Differentiate the Surface Area
We are given that the surface area decreases at a rate of \( \frac{dA}{dt} = -9 \, \text{cm}^2/\text{min} \).

We want to find the rate at which the diameter is decreasing, i.e., \( \frac{dd}{dt} \), when the diameter is 11 cm. 

### Step 4: Use Chain Rule
To connect the rate of change of surface area to the rate of change of the diameter, we first differentiate the surface area with respect to time \(t\):
\[
\frac{dA}{dt} = \frac{d}{dt} \left( 4\pi r^2 \right) = 8\pi r \frac{dr}{dt}
\]

Since \(r = \frac{d}{2}\), we also know:
\[
\frac{dr}{dt} = \frac{1}{2} \frac{dd}{dt}
\]

### Step 5: Solve for \( \frac{dd}{dt} \)
We can now solve for \( \frac{dd}{dt} \). Substituting \(r = \frac{d}{2} = \frac{11}{2} = 5.5 \, \text{cm}\) and using \( \frac{dA}{dt} = -9 \, \text{cm}^2/\text{min} \):

\[
-9 = 8\pi (5.5) \frac{dr}{dt}
\]

Solving for \( \frac{dr}{dt} \):
\[
\frac{dr}{dt} = \frac{-9}{8\pi(5.5)} = \frac{-9}{44\pi}
\]

Finally, since \( \frac{dr}{dt} = \frac{1}{2} \frac{dd}{dt} \), we multiply by 2 to find \( \frac{dd}{dt} \):

\[
\frac{dd}{dt} = 2 \times \frac{-9}{44\pi} = \frac{-18}{44\pi} = \frac{-9}{22\pi} \, \text{cm/min}
\]

### Conclusion
The diameter of the snowball is decreasing at a rate of \( \frac{9}{22\pi} \, \text{cm/min} \) when the diameter is 11 cm.

Would you like a further explanation or more details?

Here are 5 follow-up questions:
1. How is the chain rule used in related rates problems?
2. How does the surface area change impact the rate of diameter decrease?
3. What would happen if the snowball's surface area decreased at a faster rate?
4. Can this method be applied to other geometric shapes?
5. How would the answer change if we were given the volume instead of the surface area?

**Tip**: In related rates problems, always identify what rates are given and what rates need to be found before differentiating.

Learn how to calculate the rate at which a snowball's diameter decreases when its surface area is shrinking at 9 cm²/min. This related rates problem involves geometry, differentiation, and the chain rule, and is suited for high school students in Grades 11-12.

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