https://file.aiquickdraw.com/mathbot/file/1737211330940dtmwddp3.jpg

Solve this problem correctly step by step

### 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!

The volume of the cone, given by V = (1/3)πr^2(10 − r), is changing as time goes on. The rate of change of the radius r of the cone with respect to time is 5. In terms of r, what is the rate of change of the volume of the cone with respect to time?

Explore how to compute the rate of change of the volume of a cone, given the radius is changing at a constant rate of 5. Learn about differentiation, the chain rule, and the volume formula in this step-by-step solution. Ideal for Grades 11-12.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation