AP Calculus AB TOPIC 4.4 Introduction to Related Rates

### **Learning Objective: CHA-3.D**
The goal of **Learning Objective CHA-3.D** is to teach students how to **calculate related rates** in applied contexts, which involves differentiating multiple variables that are interconnected and changing with respect to time or another independent variable. In **related rates problems**, the relationship between the variables is typically given by a geometric or physical formula, and the task is to find how one variable's rate of change affects another's.

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### **Essential Knowledge:**

#### **CHA-3.D.1**
The **chain rule** is fundamental to solving related rates problems. The chain rule allows us to differentiate variables that are functions of an independent variable, often time (\( t \)). In related rates problems, we typically express the rate of change of one quantity in terms of another by applying the chain rule.

**Example**:
If \( y = f(x) \) and \( x \) is a function of time \( t \), then by the chain rule:
\[
\frac{dy}{dt} = \frac{dy}{dx} \cdot \frac{dx}{dt}
\]
This equation shows how the rate of change of \( y \) with respect to \( t \) depends on both the rate of change of \( y \) with respect to \( x \), and the rate of change of \( x \) with respect to \( t \).

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#### **CHA-3.D.2**
In addition to the chain rule, other differentiation rules such as the **product rule** and the **quotient rule** may also be necessary to differentiate more complex expressions where two or more functions are multiplied or divided.

**Example (Product Rule)**:
In a problem where two quantities \( u \) and \( v \) are functions of time, their product \( uv \) must be differentiated using the product rule:
\[
\frac{d}{dt}(uv) = u \frac{dv}{dt} + v \frac{du}{dt}
\]

**Example (Quotient Rule)**:
In a problem where two quantities \( u \) and \( v \) (functions of time) are divided, the quotient rule is applied:
\[
\frac{d}{dt}\left( \frac{u}{v} \right) = \frac{v \frac{du}{dt} - u \frac{dv}{dt}}{v^2}
\]

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### **Putting It All Together in Applied Contexts:**

In applied related rates problems, we are given relationships between variables (e.g., geometric formulas like the volume of a sphere or cylinder), and the goal is to find how one variable's rate of change affects the others. The chain rule helps connect these rates, and the product or quotient rules help with more complex relationships.

#### **Example Problem:**
A balloon is being inflated, and its radius is increasing at a rate of 2 cm per second. How fast is the volume of the balloon increasing when the radius is 5 cm?

1. **Formula**: The volume \( V \) of a sphere is \( V = \frac{4}{3} \pi r^3 \).
2. **Differentiate**: Use the chain rule to differentiate the volume with respect to time \( t \), knowing \( r \) changes with \( t \):
\[
\frac{dV}{dt} = 4\pi r^2 \cdot \frac{dr}{dt}
\]
3. **Plug in values**: Given \( \frac{dr}{dt} = 2 \, \text{cm/s} \) and \( r = 5 \, \text{cm} \), substitute into the equation to find \( \frac{dV}{dt} \).

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Nick Perich
Norristown Area High School
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