Math Problem Statement

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.)

Solution

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 rr be the radius of the sphere in mm.
  • Let VV 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=43πr3V = \frac{4}{3} \pi r^3
  • Given: The radius rr is increasing at a rate of drdt=5 mm/s\frac{dr}{dt} = 5 \text{ mm/s}.

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

First, convert the diameter to the radius: r=diameter2=60 mm2=30 mmr = \frac{\text{diameter}}{2} = \frac{60 \text{ mm}}{2} = 30 \text{ mm}

Next, we differentiate the volume formula with respect to time tt: V=43πr3V = \frac{4}{3} \pi r^3 Taking the derivative with respect to tt: dVdt=ddt(43πr3)=4πr2drdt\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 mmr = 30 \text{ mm} and drdt=5 mm/s\frac{dr}{dt} = 5 \text{ mm/s} into the differentiated formula: dVdt=4π(30)2(5)\frac{dV}{dt} = 4 \pi (30)^2 (5)

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

Finally, round to two decimal places: dVdt18000×3.14=56520 mm3/s\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 mm3/s56520 \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 π\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.

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Math Problem Analysis

Mathematical Concepts

Calculus
Related Rates

Formulas

Volume of a sphere: V = (4/3) * π * r^3

Theorems

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Suitable Grade Level

Advanced High School