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We are tasked with finding the rate of change of the radius of a cylinder at a specific instant, given the volume equation for a cylinder: \( V = \pi r^2 h \).

### Given:
- Height \( h \) is increasing at a constant rate of \( \frac{dh}{dt} = 2 \, \text{cm/s} \).
- Volume \( V \) is increasing at a rate of \( \frac{dV}{dt} = 1152 \, \text{cm}^3/\text{s} \).
- At the instant where \( h = 3 \, \text{cm} \) and \( V = 403 \, \text{cm}^3 \), we are to find the rate of change of the radius \( \frac{dr}{dt} \).

### Step-by-Step Solution:

1. **Volume Formula:**
   \[
   V = \pi r^2 h
   \]
   Differentiate both sides with respect to time \( t \) using the product rule:
   \[
   \frac{dV}{dt} = \pi \left( 2r \cdot \frac{dr}{dt} \cdot h + r^2 \cdot \frac{dh}{dt} \right)
   \]
   We now have an equation that relates \( \frac{dV}{dt} \), \( \frac{dr}{dt} \), and \( \frac{dh}{dt} \).

2. **Substitute the known values:**
   - \( \frac{dV}{dt} = 1152 \, \text{cm}^3/\text{s} \)
   - \( h = 3 \, \text{cm} \)
   - \( \frac{dh}{dt} = 2 \, \text{cm/s} \)
   - \( V = 403 \, \text{cm}^3 \)

   First, we need to find the value of \( r \) when \( V = 403 \, \text{cm}^3 \) and \( h = 3 \, \text{cm} \). From the volume formula:
   \[
   V = \pi r^2 h \implies 403 = \pi r^2 (3)
   \]
   Solving for \( r \):
   \[
   r^2 = \frac{403}{3\pi} \implies r^2 \approx \frac{403}{9.4248} \approx 42.77
   \]
   \[
   r \approx \sqrt{42.77} \approx 6.54 \, \text{cm}
   \]

3. **Substitute into the differentiated equation:**
   \[
   1152 = \pi \left( 2(6.54) \cdot \frac{dr}{dt} \cdot 3 + (6.54)^2 \cdot 2 \right)
   \]
   Simplify:
   \[
   1152 = \pi \left( 39.24 \cdot \frac{dr}{dt} + 85.57 \cdot 2 \right)
   \]
   \[
   1152 = \pi \left( 39.24 \cdot \frac{dr}{dt} + 171.14 \right)
   \]
   Divide by \( \pi \):
   \[
   \frac{1152}{\pi} = 39.24 \cdot \frac{dr}{dt} + 171.14
   \]
   \[
   366.56 = 39.24 \cdot \frac{dr}{dt} + 171.14
   \]
   Subtract 171.14:
   \[
   195.42 = 39.24 \cdot \frac{dr}{dt}
   \]
   Solve for \( \frac{dr}{dt} \):
   \[
   \frac{dr}{dt} = \frac{195.42}{39.24} \approx 4.98 \, \text{cm/s}
   \]

### Final Answer:
The rate of change of the radius is approximately \( 4.980 \, \text{cm/s} \).

Would you like more details or have any other questions? Here are some related questions to explore:

1. How do you apply the product rule when differentiating the volume of the cylinder?
2. How would the result change if the rate of height increase was different?
3. Can you find the rate of change of the radius when the volume changes but the height stays constant?
4. What would happen if both height and radius changed at different rates?
5. How does changing the value of \( \pi \) in the equation impact the result?

**Tip:** Always ensure units are consistent when dealing with related rates problems to avoid errors in calculation.

The height of a cylinder is increasing at a constant rate of 2 centimeters per second, and the volume is increasing at a rate of 1152 cubic centimeters per second. At the instant when the height of the cylinder is 3 centimeters and the volume is 403 cubic centimeters, what is the rate of change of the radius? The volume of a cylinder can be found with the equation V = πr^2h. Round your answer to three decimal places.

Explore how to calculate the rate of change of the radius of a cylinder given that the height increases by 2 cm/s and the volume by 1152 cm³/s. This problem involves calculus concepts such as related rates and volume differentiation. Ideal for students in Grades 11-12 or AP Calculus.

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