Problem Analysis:

The volume of a cone is given by the formula:

$V = \frac{1}{3} \pi r^2 (10 - r)$

We are tasked to find the rate of change of the volume ($\frac{dV}{dt}$) with respect to time ($t$) in terms of $r$, where the rate of change of the radius ($\frac{dr}{dt}$) is given as 5.

Step-by-Step Solution:

1. Differentiate the volume equation with respect to time $t$: Since $V$ is a function of $r$, and $r$ changes with time, we apply the chain rule:

   $V = \frac{1}{3} \pi r^2 (10 - r)$

   Expand this expression for simplicity:

   $V = \frac{1}{3} \pi (10r^2 - r^3)$

   Differentiate $V$ with respect to $t$:

   $\frac{dV}{dt} = \frac{1}{3} \pi \cdot \frac{d}{dt}(10r^2 - r^3)$

   Using the chain rule:

   $\frac{dV}{dt} = \frac{1}{3} \pi \left( 20r \frac{dr}{dt} - 3r^2 \frac{dr}{dt} \right)$

   Factor out $\frac{dr}{dt}$:

   $\frac{dV}{dt} = \frac{1}{3} \pi \frac{dr}{dt} \left( 20r - 3r^2 \right)$

2. Substitute the given $\frac{dr}{dt} = 5$:

   $\frac{dV}{dt} = \frac{1}{3} \pi (5) \left( 20r - 3r^2 \right)$

   Simplify:

   $\frac{dV}{dt} = \frac{5}{3} \pi \left( 20r - 3r^2 \right)$

3. Final Answer:

   The rate of change of the volume in terms of $r$ is:

   $\frac{dV}{dt} = \frac{100\pi r - 15\pi r^2}{3}$

   Match this result with the given options:

   • Correct choice: (D) $\frac{100\pi r - 15\pi r^2}{3}$

Verification of Units and Logic:

• $\frac{dV}{dt}$: Units of volume change per time.
• Input $r$ and $\frac{dr}{dt}$ have been properly substituted.

Do you have further questions, or would you like an additional explanation? 😊

Related Questions:

1. What happens to $\frac{dV}{dt}$ when $r = 0$?
2. At what value of $r$ does $\frac{dV}{dt} = 0$?
3. What is the maximum possible volume of the cone based on the given formula?
4. If $\frac{dr}{dt}$ is halved, how does $\frac{dV}{dt}$ change?
5. How would the problem change if the cone's height varied with time?

Tip:

Always check if the formula for differentiation requires a product or chain rule when variables depend on time!