To find the rate at which the volume $V$ of a sphere is changing with respect to time ($\frac{dV}{dt}$) when $r = 5$ inches and $\frac{dr}{dt} = 12$ in/min, we use the formula for the volume of a sphere:

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

Step 1: Differentiate $V$ with respect to time $t$

Using the chain rule:

$\frac{dV}{dt} = \frac{d}{dt} \left( \frac{4}{3} \pi r^3 \right) = 4 \pi r^2 \frac{dr}{dt}$

Step 2: Substitute the known values

• $r = 5$ in
• $\frac{dr}{dt} = 12$ in/min

$\frac{dV}{dt} = 4 \pi (5)^2 (12)$

Step 3: Perform the calculations

1. $5^2 = 25$
2. $4 \cdot 25 = 100$
3. $100 \cdot 12 = 1200$

$\frac{dV}{dt} = 1200 \pi \, \text{in}^3/\text{min}$

Final Answer:

The rate at which the volume is changing with respect to time is:

$\boxed{1200 \pi \, \text{in}^3/\text{min}}$

This is approximately $3769.91 \, \text{in}^3/\text{min}$ if you need a decimal value.

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Do you want a detailed explanation of the differentiation process or a graph to visualize the rate of change? Here are 5 related questions:

1. What is the formula for surface area of a sphere and how does it relate to changes in $r$?
2. How does the rate of change of the radius affect the surface area of the sphere?
3. What happens to $\frac{dV}{dt}$ if $\frac{dr}{dt}$ doubles?
4. Could we solve a similar problem for a cylinder? How would the process differ?
5. How does $\frac{dV}{dt}$ change as $r$ increases?

Tip: Always check your units when dealing with rates; mismatched units can lead to incorrect results!