AP Calculus AB 4.4 Intro to Related Rates: Cylinder Formula

In **AP Calculus AB**, section 4.4 covers **Related Rates**, which includes problems involving the **cylinder formula**. Here's an outline of how this topic might be introduced, focusing on how to solve related rates problems using the volume and surface area of a cylinder.

### The Cylinder Formula in Related Rates

When working with cylinders in related rates problems, we usually deal with the following key quantities:

- **Volume of a cylinder**: \( V = \pi r^2 h \), where \( r \) is the radius of the base, and \( h \) is the height of the cylinder.
- **Surface area of a cylinder**: \( A = 2\pi r h + 2\pi r^2 \), which includes both the lateral surface area and the areas of the two circular bases.

In related rates problems, you are often given information about how one of these quantities is changing with respect to time, and your goal is to determine the rate of change of another quantity. For example, you might be asked to find how fast the height of a cylinder is changing as water is being poured into it, or how fast the radius is expanding when the volume is increasing at a certain rate.

### Differentiating the Cylinder Formula with Respect to Time

Related rates problems typically involve implicit differentiation with respect to time \( t \), since the variables (like \( r \), \( h \), or \( V \)) change over time. Let’s focus on the volume formula for a cylinder:

\[
V = \pi r^2 h
\]

To find how the volume or height changes over time, you differentiate both sides with respect to \( t \). Since \( r \), \( h \), and \( V \) can all be functions of time, we apply the **product rule** and the **chain rule**.

Differentiating \( V = \pi r^2 h \) with respect to time:

\[
\frac{dV}{dt} = \pi \left( 2r \frac{dr}{dt} \cdot h + r^2 \frac{dh}{dt} \right)
\]

This equation tells us how the volume is changing with respect to time, depending on both the rate of change of the radius \( \frac{dr}{dt} \) and the rate of change of the height \( \frac{dh}{dt} \).

### Key Steps for Solving Related Rates Problems with Cylinders

1. **Identify the variables**: Determine what quantities are changing with respect to time (e.g., the radius, height, or volume) and identify what you're solving for.

2. **Write the equation**: Use the cylinder formulas for volume or surface area to write an equation that relates the variables.

3. **Differentiate the equation with respect to time**: Apply implicit differentiation, remembering that each variable is a function of time. Use the product rule and chain rule where necessary.

4. **Substitute known values**: Once you have the differentiated equation, substitute the values you're given (such as the rate at which the volume is increasing or the radius at a specific moment) and solve for the unknown rate.

### Example Problem: Water Filling a Cylinder

#### Problem Statement:
Water is being poured into a cylinder at a rate of 10 cubic meters per minute. The radius of the cylinder is 2 meters, and the height of the water is increasing. Find the rate at which the height of the water is rising when the water is 5 meters high.

#### Step-by-Step Solution:

1. **Given Information**:
- \( \frac{dV}{dt} = 10 \, \text{m}^3/\text{min} \)
- \( r = 2 \, \text{m} \) (radius is constant)
- Find \( \frac{dh}{dt} \) when \( h = 5 \, \text{m} \)

2. **Use the Volume Formula**:
The volume of the cylinder is \( V = \pi r^2 h \).

3. **Differentiate with Respect to Time**:
\[
\frac{dV}{dt} = \pi r^2 \frac{dh}{dt}
\]
Since \( r = 2 \, \text{m} \) is constant, we can plug it into the equation:
\[
\frac{dV}{dt} = \pi (2^2) \frac{dh}{dt} = 4\pi \frac{dh}{dt}
\]

4. **Substitute Known Values**:
We know \( \frac{dV}{dt} = 10 \, \text{m}^3/\text{min} \), so:
\[
10 = 4\pi \frac{dh}{dt}
\]

5. **Solve for \( \frac{dh}{dt} \)**:
\[
\frac{dh}{dt} = \frac{10}{4\pi} = \frac{5}{2\pi} \, \text{m/min} \approx 0.796 \, \text{m/min}
\]

So, the height of the water is rising at approximately 0.796 meters per minute when the height is 5 meters.

### Conclusion

Related rates problems involving cylinders often focus on the volume or surface area changing over time. By differentiating the appropriate formula and applying the chain rule, you can relate how different quantities (like radius, height, or volume) change with respect to time. These problems help develop an understanding of how rates of change in calculus are applied to real-world scenarios.

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Nick Perich
Norristown Area High School
Norristown Area School District
Norristown, Pa

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