Introduction to Related Rates - AP Calculus AB Study Guide

The Core Idea: Introduction to Related Rates

Related rates problems involve finding the rate of change of one quantity by relating it to other quantities whose rates of change are known. The fundamental concept is that multiple variables in a given scenario (such as the radius, height, and volume of a cone) are all changing with respect to a single, often unstated, variable: time.

The process for solving these problems is a direct application of the chain rule and implicit differentiation. By establishing an equation that connects the variables (e.g., a geometric formula), we can differentiate the entire equation with respect to time ($t$). This creates a new equation that relates the rates of change of these variables (e.g., $\frac{dV}{dt}$, $\frac{dr}{dt}$), allowing us to solve for an unknown rate using the known information.

Key Concepts: The Foundation of Related Rates

The AP Calculus AB curriculum does not specify a single "formula" for related rates. Instead, it emphasizes a process rooted in the chain rule. The key idea is to treat every variable in an equation as a function of time, $t$.

When we differentiate an equation with respect to time, we are performing implicit differentiation. For any variable, say $x$, that is a function of time, its derivative is not simply 1, but $\frac{dx}{dt}$. The chain rule must be applied to any function of that variable.

The Chain Rule Applied to Time:

If we have an equation relating variables, such as $A=\pi r^2$, and we differentiate with respect to time $t$, we apply the chain rule to the function of $r$:

1. Differentiate the outer function with respect to $r$: $\frac{d}{dr}[\pi r^2]=2\pi r$

2. Multiply by the derivative of the inner function ($r$) with respect to time: $\frac{dr}{dt}$

Combining these, the differentiation of the full equation with respect to time $t$ is:

$$\frac{d}{dt}[A]=\frac{d}{dt}[\pi r^2]$$

$$\frac{dA}{dt}=2\pi r\cdot \frac{dr}{dt}$$

This new equation relates the rate of change of the area ($\frac{dA}{dt}$) to the rate of change of the radius ($\frac{dr}{dt}$).

Understanding the "With Respect to Time" Concept

In a related rates problem, quantities like distance, volume, area, or angle are changing over time. The phrase "how fast is the volume changing?" is a direct prompt to find the derivative of volume with respect to time, or $\frac{dV}{dt}$. This is the central nuance of the topic.

Consider the simple equation $y=x^3$.

• Standard Differentiation: When asked to find the derivative with respect to $x$, the result is $\frac{dy}{dx}=3x^2$. Here, we treat $x$ as the independent variable.

• Related Rates Differentiation: In a related rates context, both $x$ and $y$ are assumed to be functions of time, $t$. When we differentiate the equation with respect to $t$, we must apply the chain rule to both sides:

  $$\frac{d}{dt}[y]=\frac{d}{dt}[x^3]$$

  $$\frac{dy}{dt}=3x^2\cdot \frac{dx}{dt}$$

The term $\frac{dx}{dt}$ appears because of the chain rule; $x$ is a function of $t$, so its derivative with respect to $t$ is not 1 unless $x=t$. This distinction is the critical step in correctly setting up every related rates problem.

Core Concepts & Rules

• Underlying Relationship: Every related rates problem begins with an equation that connects two or more variables that are changing over time (e.g., a geometric formula, the Pythagorean theorem).

• Implicit Differentiation w.r.t. Time: The core technique is to differentiate this entire equation implicitly with respect to time, $t$.

• The Chain Rule is Essential: Because each variable (e.g., $V$, $r$, $x$, $y$) is treated as a function of time, the chain rule must be applied. The derivative of $x^n$ with respect to $t$ is $n{x}^{n-1}\frac{dx}{dt}$.

• Substitute After Differentiating: Known values for variables (e.g., radius $r=5$) and rates of change (e.g., $\frac{dr}{dt}=2$) should only be substituted into the equation after the differentiation step is complete.

• Solve for the Unknown Rate: After differentiating and substituting, use algebra to solve for the desired rate of change.

Step-by-Step Example 1: A Basic Geometric Application

Problem: The radius $r$ of a circle is increasing at a constant rate of 2 centimeters per second. At what rate is the area $A$ of the circle increasing when the radius is 10 centimeters?

Step 1: Identify Given Information and the Goal

• Rate the radius is increasing: $\frac{dr}{dt}=2\text{ cm/s}$

• The instant in question: $r=10\text{ cm}$

• What we need to find: The rate the area is increasing, $\frac{dA}{dt}$.

Step 2: Write an Equation Relating the Variables

The variables are the area $A$ and the radius $r$. The equation that relates them is the formula for the area of a circle.

$$A=\pi r^2$$

Step 3: Differentiate the Equation with Respect to Time $t$

Apply implicit differentiation and the chain rule to both sides of the equation.

$$\frac{d}{dt}[A]=\frac{d}{dt}[\pi r^2]$$

The derivative of $A$ with respect to $t$ is $\frac{dA}{dt}$. For the right side, $\pi$ is a constant, and we use the chain rule on $r^2$.

$$\frac{dA}{dt}=\pi \cdot (2r)\cdot \frac{dr}{dt}$$

$$\frac{dA}{dt}=2\pi r\frac{dr}{dt}$$

Step 4: Substitute Known Values and Solve

Now, substitute the given values ($r=10$ and $\frac{dr}{dt}=2$) into the derivative equation.

$$\frac{dA}{dt}=2\pi (10)(2)$$

$$\frac{dA}{dt}=40\pi$$

Step 5: State the Final Answer with Units

The rate of change of area is measured in square units per unit of time.

The area of the circle is increasing at a rate of $40\pi$ cm^2/s when the radius is 10 cm.

Step-by-Step Example 2: An Exam-Style Application

Problem: A 13-foot ladder is leaning against a vertical wall. The bottom of the ladder is being pulled away from the wall at a constant rate of 2 feet per second. How fast is the top of the ladder sliding down the wall at the instant the bottom of the ladder is 5 feet from the wall?

Step 1: Identify Given Information and the Goal

• Let $x$ be the distance from the base of the wall to the bottom of the ladder.

• Let $y$ be the distance from the ground to the top of the ladder.

• The length of the ladder is constant: 13 ft.

• Rate the bottom is moving away from the wall: $\frac{dx}{dt}=2\text{ ft/s}$.

• The instant in question: $x=5\text{ ft}$.

• What we need to find: The rate the top is sliding down the wall, $\frac{dy}{dt}$.

Step 2: Write an Equation Relating the Variables

The wall, the ground, and the ladder form a right triangle. The Pythagorean theorem relates $x$, $y$, and the ladder's length (13).

$$x^2+y^2=13^2$$

$$x^2+y^2=169$$

Step 3: Differentiate the Equation with Respect to Time $t$

Differentiate each term with respect to $t$, remembering the chain rule.

$$\frac{d}{dt}[x^2]+\frac{d}{dt}[y^2]=\frac{d}{dt}[169]$$

$$2x\frac{dx}{dt}+2y\frac{dy}{dt}=0$$

Step 4: Find Any Missing Values for the Specific Instant

We need to solve for $\frac{dy}{dt}$, but our differentiated equation includes $x$, $y$, and $\frac{dx}{dt}$. We know $x=5$ and $\frac{dx}{dt}=2$, but we need to find the value of $y$ at the specific instant when $x=5$. We use the original Pythagorean equation.

$$(5{)}^2+y^2=169$$

$$25+y^2=169$$

$$y^2=144⟹y=12\text{ ft}$$

Step 5: Substitute Known Values and Solve for $\frac{dy}{dt}$

Now substitute $x=5$, $y=12$, and $\frac{dx}{dt}=2$ into the differentiated equation.

$$2(5)(2)+2(12)\frac{dy}{dt}=0$$

$$20+24\frac{dy}{dt}=0$$

$$24\frac{dy}{dt}=-20$$

$$\frac{dy}{dt}=-\frac{20}{24}=-\frac{5}{6}$$

Step 6: State the Final Answer with Units

The negative sign indicates that the distance $y$ is decreasing.

The top of the ladder is sliding down the wall at a rate of $\frac{5}{6}$ ft/s at the instant the bottom is 5 feet from the wall.

Using Your Calculator

The process of solving related rates problems is purely analytical. It requires setting up an equation based on the context of the problem, differentiating that equation implicitly with respect to time, and using algebra to solve for an unknown rate.

A calculator is not used to perform the differentiation or setup steps. Its role is limited to performing the final arithmetic calculation if the numbers are complex. For example, in the ladder problem, you would do all the calculus by hand to arrive at $\frac{dy}{dt}=-\frac{20}{24}$. You could then use your calculator to simplify the fraction to $-\frac{5}{6}$ or find its decimal equivalent, $-0.833$. The core calculus work must be done by hand.

AP Exam Quick Hit

Common Question Types

• Geometric Shapes: You are given the rate of change of one dimension of a geometric shape (e.g., radius of a sphere, side length of a cube) and asked to find the rate of change of another property (volume, surface area). Example: Air is being pumped into a spherical balloon at a rate of 100 cm^3/s. How fast is the radius of the balloon increasing when the diameter is 50 cm?

• Pythagorean Theorem: Two objects are moving at right angles to each other, or a ladder is sliding against a wall. You are given the rate of one side of the triangle and asked to find the rate of another side. Example: Car A is traveling west at 50 mph and car B is traveling north at 60 mph. Both are headed for the intersection of the two roads. At what rate are the cars approaching each other when car A is 0.3 mi and car B is 0.4 mi from the intersection?

Common Mistakes

• Substituting a Variable Too Early: The most common mistake is plugging in a value for a variable (e.g., $x=5$) before differentiating the equation. This incorrectly treats a changing quantity as a constant, making its derivative zero and invalidating the entire process.

• Forgetting the Chain Rule: A critical error is to differentiate incorrectly with respect to time. For example, writing that the derivative of $V=\frac{4}{3}\pi r^3$ is $\frac{dV}{dt}=4\pi r^2$ instead of the correct $\frac{dV}{dt}=4\pi r^2\frac{dr}{dt}$. You must include the $\frac{d(\text{variable})}{dt}$ term for every variable you differentiate.

• Sign Errors: Failing to correctly assign a negative sign to a rate for a quantity that is decreasing. If a ladder is sliding down a wall, $\frac{dy}{dt}$ must be negative. If a tank is draining, $\frac{dV}{dt}$ must be negative.

• Forgetting to Solve for an Intermediate Variable: In many problems (like the ladder example), you must use the original equation to solve for a missing variable's value ($y=12$) at the specific instant in time before you can substitute into the differentiated equation.