To determine the rate of change of the volume of a sphere, we use the formula for the volume of a sphere:

$V = \frac{4}{3} \pi r^3$

Step 1: Differentiate the volume with respect to time

Using the chain rule: $\frac{dV}{dt} = 4\pi r^2 \frac{dr}{dt}$

Here:

• $r$ is the radius of the sphere.
• $\frac{dr}{dt}$ is the rate of change of the radius.
• $\frac{dV}{dt}$ is the rate of change of the volume.

Step 2: Plug in the known values

At the instant when $r = 99$ inches and $\frac{dr}{dt} = 2$ inches per minute: $\frac{dV}{dt} = 4\pi (99)^2 (2)$

Step 3: Simplify the expression

First, calculate $99^2$: $99^2 = 9801$

Now substitute: $\frac{dV}{dt} = 4\pi (9801)(2) = 8\pi (9801)$

Finally, compute: $\frac{dV}{dt} = 78408\pi \, \text{cubic inches per minute.}$

Step 4: Approximate the result (if needed)

Using $\pi \approx 3.1416$: $\frac{dV}{dt} \approx 78408 \times 3.1416 = 246,407.01 \, \text{cubic inches per minute.}$

Final Answer:

The rate of change of the volume of the sphere is: $\boxed{78408\pi \, \text{cubic inches per minute} \quad \text{or approximately } 246,407.01 \, \text{cubic inches per minute.}}$

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Related Questions:

1. What is the relationship between surface area and rate of change of volume?
2. How does the rate of change of the radius affect the rate of change of the volume?
3. How would the result change if $\frac{dr}{dt}$ were a variable instead of a constant?
4. What is the formula for the surface area of a sphere, and how does its rate of change relate to this problem?
5. If the radius doubles, what happens to the volume and its rate of change?

Tip:

When solving problems involving rates of change, always identify which variables are constants and which are changing with respect to time.