The radius of a sphere is increasing at a rate of 5 mm/s. How fast is the volume increasing (in mm3/s) when the diameter is 60 mm? (Round your answer to two decimal places.)

To determine how fast the volume of a sphere is increasing when the diameter is 60 mm, we need to use related rates in calculus.

First, we identify the variables and their relationships:
- Let \( r \) be the radius of the sphere in mm.
- Let \( V \) be the volume of the sphere in mm³.
- The relationship between the volume and the radius of a sphere is given by the formula: 
  \[
  V = \frac{4}{3} \pi r^3
  \]
- Given: The radius \( r \) is increasing at a rate of \( \frac{dr}{dt} = 5 \text{ mm/s} \).

We need to find how fast the volume \( V \) is increasing, i.e., \( \frac{dV}{dt} \), when the diameter is 60 mm.

First, convert the diameter to the radius:
\[
r = \frac{\text{diameter}}{2} = \frac{60 \text{ mm}}{2} = 30 \text{ mm}
\]

Next, we differentiate the volume formula with respect to time \( t \):
\[
V = \frac{4}{3} \pi r^3
\]
Taking the derivative with respect to \( t \):
\[
\frac{dV}{dt} = \frac{d}{dt} \left( \frac{4}{3} \pi r^3 \right) = 4 \pi r^2 \frac{dr}{dt}
\]

Now, we substitute \( r = 30 \text{ mm} \) and \( \frac{dr}{dt} = 5 \text{ mm/s} \) into the differentiated formula:
\[
\frac{dV}{dt} = 4 \pi (30)^2 (5)
\]

Calculate the value:
\[
\frac{dV}{dt} = 4 \pi (900) (5) = 18000 \pi
\]

Finally, round to two decimal places:
\[
\frac{dV}{dt} \approx 18000 \times 3.14 = 56520 \text{ mm}^3/\text{s}
\]

Therefore, the volume of the sphere is increasing at approximately \( 56520 \text{ mm}^3/\text{s} \) when the diameter is 60 mm.

Would you like more details or have any questions?

Here are 8 relative questions that expand on this information:

1. How do you find the rate of change of the surface area of a sphere?
2. What is the relationship between the radius and the diameter of a sphere?
3. How does the rate of change of the radius affect the volume of a sphere?
4. How would the volume change if the radius was increasing at a different rate?
5. What are some practical applications of related rates in real-life scenarios?
6. How do you differentiate more complex volume formulas with respect to time?
7. What is the significance of using \(\pi\) in the volume formula for a sphere?
8. How would you calculate the rate of change of the volume if the diameter was given instead of the radius?

**Tip:** When dealing with related rates problems, always identify the variables, their rates of change, and how they are related through differentiation.

Learn how to calculate the rate at which the volume of a sphere increases when the diameter is 60mm. Detailed calculus-based solution using the volume formula and related rates concept. Suitable for advanced high school students.

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