If a snowball melts so that its surface area decreases at a rate of 6 cm2/min, find the rate (in cm/min) at which the diameter decreases when the diameter is 11 cm. (Round your answer to three decimal places.)


We are given that the surface area of the snowball is decreasing at a rate of \( \frac{dA}{dt} = -6 \, \text{cm}^2/\text{min} \), and we are asked to find the rate at which the diameter is decreasing, i.e., \( \frac{dD}{dt} \), when the diameter \( D = 11 \, \text{cm} \).

### Step 1: Relating Surface Area and Diameter
The 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 related to the radius by \( D = 2r \), so \( r = \frac{D}{2} \).

### Step 2: Differentiate with Respect to Time
We differentiate the surface area formula with respect to time \( t \):
\[
\frac{dA}{dt} = \frac{d}{dt} \left( 4 \pi r^2 \right) = 8 \pi r \frac{dr}{dt}
\]
Now, we need to express the rate of change of the radius \( \frac{dr}{dt} \) in terms of the rate of change of the diameter. Since \( r = \frac{D}{2} \), we have:
\[
\frac{dr}{dt} = \frac{1}{2} \frac{dD}{dt}
\]
Substitute this into the derivative of the surface area:
\[
\frac{dA}{dt} = 8 \pi r \left( \frac{1}{2} \frac{dD}{dt} \right) = 4 \pi r \frac{dD}{dt}
\]

### Step 3: Solve for \( \frac{dD}{dt} \)
We are given that \( \frac{dA}{dt} = -6 \, \text{cm}^2/\text{min} \), and we need to find \( \frac{dD}{dt} \) when \( D = 11 \, \text{cm} \). First, find \( r \) when \( D = 11 \):
\[
r = \frac{11}{2} = 5.5 \, \text{cm}
\]
Now substitute \( r = 5.5 \) and \( \frac{dA}{dt} = -6 \) into the equation:
\[
-6 = 4 \pi (5.5) \frac{dD}{dt}
\]
Simplify and solve for \( \frac{dD}{dt} \):
\[
-6 = 22 \pi \frac{dD}{dt}
\]
\[
\frac{dD}{dt} = \frac{-6}{22 \pi} \approx \frac{-6}{69.115} \approx -0.087 \, \text{cm/min}
\]

### Final Answer:
The diameter is decreasing at a rate of approximately \( \boxed{-0.087} \, \text{cm/min} \) when the diameter is 11 cm.

Would you like any more details or clarifications?

Here are some related questions:
1. How does the volume of the snowball change with time?
2. What happens to the rate of diameter change if the surface area decreases at a faster rate?
3. Can you find the rate of change of the radius at any given time?
4. How is the volume change related to the diameter change?
5. What is the relationship between surface area and volume in this context?

**Tip**: Always check units when solving rate-related problems to ensure consistency across the equations.

If a snowball melts so that its surface area decreases at a rate of 6 cm2/min, find the rate (in cm/min) at which the diameter decreases when the diameter is 11 cm. (Round your answer to three decimal places.)

Learn how to calculate the rate at which the diameter of a snowball decreases as it melts, given that its surface area decreases at 6 cm2/min. This problem involves related rates, surface area of spheres, and differentiation techniques. Suitable for students in Grades 11-12.

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