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 40 mm? (Round your answer to two decimal places.)


To solve this problem, we need to find the rate at which the volume of the sphere is increasing when the diameter is 40 mm.

### Step 1: Relationship between the volume and the radius of the sphere
The volume \( V \) of a sphere is related to its radius \( r \) by the formula:

\[
V = \frac{4}{3} \pi r^3
\]

### Step 2: Differentiate the volume equation with respect to time
To find how fast the volume is increasing, we differentiate the volume equation with respect to time \( t \), using the chain rule. This gives:

\[
\frac{dV}{dt} = 4 \pi r^2 \frac{dr}{dt}
\]

Here, \( \frac{dV}{dt} \) is the rate at which the volume is increasing (which we are trying to find), and \( \frac{dr}{dt} \) is the rate at which the radius is increasing (given as 5 mm/s).

### Step 3: Find the radius when the diameter is 40 mm
Since the diameter is 40 mm, the radius is:

\[
r = \frac{40}{2} = 20 \text{ mm}
\]

### Step 4: Substitute the known values
Now we can substitute \( r = 20 \) mm and \( \frac{dr}{dt} = 5 \) mm/s into the differentiated equation:

\[
\frac{dV}{dt} = 4 \pi (20)^2 (5)
\]

\[
\frac{dV}{dt} = 4 \pi (400)(5) = 8000 \pi \text{ mm}^3/\text{s}
\]

### Step 5: Compute the value
Now, calculate the numerical value:

\[
\frac{dV}{dt} \approx 8000 \times 3.1416 = 25,132.74 \text{ mm}^3/\text{s}
\]

So, the volume is increasing at a rate of approximately **25,132.74 mm³/s** when the diameter is 40 mm.

Would you like more details on any of these steps, or have any questions?

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Here are 5 related questions for further understanding:
1. How would the result change if the rate of increase of the radius was different?
2. What is the rate of change of the surface area of the sphere at this instant?
3. If the radius is increasing at 5 mm/s, how long would it take for the diameter to double?
4. How would the volume change if the radius was increasing at a constant rate, but the sphere's material was expanding non-uniformly?
5. What is the relationship between the rate of change of radius and the rate of change of volume for other 3D shapes, such as cylinders?

**Tip**: For geometric problems involving rates of change, always remember to relate the variables through their formulas (like volume, surface area) and use the chain rule for differentiation.

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 40 mm? (Round your answer to two decimal places.)

Learn how to calculate the rate at which the volume of a sphere increases when the radius is growing at 5 mm/s, with a detailed step-by-step solution. In this example, we use the sphere volume formula and apply related rates using differentiation, suitable for students in Grades 10-12.

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