Modeling Situations with Differential Equations - AP Calculus AB Study Guide

The Core Idea: Modeling Situations with Differential Equations

A differential equation is a powerful mathematical tool that describes the relationship between a function and its derivatives. The core idea of this topic is to translate real-world scenarios, particularly those involving a rate of change, into the precise language of differential equations. Instead of starting with a function and finding its rate of change (differentiation), or starting with a rate of change and finding the original function (integration), we begin with a verbal description of how a quantity changes and construct an equation that models this dynamic relationship.

This process is foundational for modeling phenomena in science, economics, and engineering. For instance, we can describe population growth, the cooling of an object, or the decay of a radioactive substance by stating how the rate of change of the quantity depends on the quantity itself. The primary skill is not to solve these equations, but to accurately interpret a given situation and write the corresponding differential equation, carefully defining all variables involved.

Key Definitions and Interpretations

A differential equation is an equation that contains an unknown function and one or more of its derivatives. The primary interpretation used in modeling is that of the derivative as a rate of change.

• Differential Equation: An equation involving a function and its derivatives.

  • Example: $\frac{dy}{dt}=2y$ relates the rate of change of a function $y$ to the value of the function $y$ itself.

• The Derivative as a Rate of Change: The expression $\frac{dy}{dx}$ represents the instantaneous rate of change of the quantity $y$ with respect to the quantity $x$. In many modeling problems, the independent variable is time, $t$, so the rate of change is expressed as $\frac{dy}{dt}$.

Understanding the Language of Rates

Translating verbal descriptions into mathematical equations is the key skill for this topic. Certain phrases correspond directly to mathematical operations. It is critical to identify the variables and the relationship between them.

• "The rate of change of Y...": This phrase translates directly to the derivative of Y. If the rate is with respect to time $t$, this becomes $\frac{dY}{dt}$.

• "...is..." or "...is equal to...": This translates to the equals sign, $=$.

• "...proportional to X...": This means "is a constant multiple of X." It translates to $=kX$, where $k$ is the constant of proportionality.

• "...inversely proportional to X...": This means "is a constant multiple of the reciprocal of X." It translates to $=\frac{k}{X}$.

When building a model, always define your variables first. For example, let $P(t)$ be the population at time $t$. Then, "the rate of change of the population" is $\frac{dP}{dt}$.

Core Concepts & Rules

• A differential equation is a mathematical model that relates a function to its own rate of change.

• The primary purpose of a differential equation in this context is to represent a real-world situation where a rate is described verbally.

• The expression $\frac{dy}{dx}$ must be interpreted as the instantaneous rate of change of the quantity $y$ with respect to the quantity $x$.

• To write a differential equation from a verbal description, you must first identify the quantities involved, define variables for them, and then translate the described relationship into a mathematical equation.

• Pay close attention to key phrases like "rate of change," "proportional to," and "inversely proportional to," as they dictate the structure of the resulting equation.

• The context of the problem is crucial for defining variables and determining the signs of terms (e.g., a rate of decrease implies a negative term).

Step-by-Step Example 1: Direct Proportionality

Problem: The rate of growth of a bacterial culture is directly proportional to the number of bacteria present. Write a differential equation that models this situation.

Step 1: Define the Variables

First, identify the quantities involved and assign variables.

• Let $B(t)$ represent the number of bacteria at time $t$.

• Let $t$ represent time.

Step 2: Identify the Rate of Change

The phrase "The rate of growth of a bacterial culture" describes how the number of bacteria, $B$, changes with respect to time, $t$.

• This translates to the derivative: $\frac{dB}{dt}$.

Step 3: Translate the Relationship

The phrase "is directly proportional to the number of bacteria present" describes the relationship between the rate and the variable $B$.

• "is" translates to $=$.

• "directly proportional to the number of bacteria present ($B$)" translates to $kB$, where $k$ is the constant of proportionality.

Step 4: Combine and Write the Equation

Combine the translated parts to form the complete differential equation.

$$\frac{dB}{dt}=kB$$

This equation models the described situation. Since it describes growth, we know that $k>0$.

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

Problem: A water tank is being filled at a constant rate of 5 gallons per minute. At the same time, water is leaking out of the tank at a rate proportional to the square root of the volume of water in the tank. Write a differential equation for the volume of water, $V$, in the tank at time $t$.

Step 1: Define the Variables

• Let $V(t)$ be the volume of water in the tank (in gallons) at time $t$ (in minutes).

• Let $t$ be the time in minutes.

Step 2: Identify the Overall Rate of Change

The problem describes two competing processes: water coming in and water going out. The overall rate of change of the volume, $\frac{dV}{dt}$, will be the difference between the rate in and the rate out.

$$\frac{dV}{dt}=(\text{Rate In})-(\text{Rate Out})$$

Step 3: Translate Each Component Rate

• Rate In: "A water tank is being filled at a constant rate of 5 gallons per minute." This is a direct statement.

  • Rate In $=5$.

• Rate Out: "water is leaking out... at a rate proportional to the square root of the volume of water."

  • "proportional to" means we use a constant of proportionality, $k$.

  • "the square root of the volume" is $\sqrt{V}$.

  • So, Rate Out $=k\sqrt{V}$. Since this is a rate of removal (leaking out), it contributes negatively to the overall change, and the constant $k$ must be positive.

Step 4: Combine and Write the Equation

Substitute the expressions for the rates into the overall rate equation.

$$\frac{dV}{dt}=5-k\sqrt{V}$$

This is the differential equation that models the change in the volume of water in the tank.

Using Your Calculator

This topic is purely analytical and conceptual. The task is to translate a verbal description into a mathematical equation. A graphing calculator is not used for this process. The skills required are reading comprehension, identification of key mathematical phrases, and correct formulation of an equation.

While a calculator might be used in later topics to analyze a given differential equation (for example, by graphing its slope field), it does not play a role in the initial modeling and setup process covered in this topic.

AP Exam Quick Hit

Common Question Types

• Multiple-Choice Translation: You will be given a one-sentence description of a rate and asked to select the matching differential equation from a list of options.

  • Example: "The rate of change of a quantity $y$ with respect to $x$ is inversely proportional to the square of $y$. Which of the following is the differential equation that describes this relationship?"

  • Answer Choices: (A) $\frac{dy}{dx}=k{y}^2$ (B) $\frac{dy}{dx}=\frac{k}{y}$ (C) $\frac{dy}{dx}=\frac{k}{y^2}$ (D) $\frac{dy}{dx}=k\sqrt{y}$

• FRQ Part (a) Setup: The first part of a Free Response Question will often describe a real-world scenario and ask you to write, but not solve, a differential equation that models it.

  • Example: "The rate at which the number of people in a park, $P$, is changing is given by the difference between the rate at which people enter, $E(t)$, and the rate at which they leave, $L(t)$. Write a differential equation for $P(t)$."

  • Answer:$\frac{dP}{dt}=E(t)-L(t)$.

Common Mistakes

• Forgetting the Constant of Proportionality: Writing $\frac{dy}{dt}=y$ when the problem states "proportional to $y$." The correct form is $\frac{dy}{dt}=ky$.

• Confusing Proportionality Types: Mistaking "directly proportional to $y$" ($ky$) for "inversely proportional to $y$" ($\frac{k}{y}$).

• Incorrect Signs for Rates: In a "Rate In - Rate Out" model, students may incorrectly add the rates or assign the wrong sign to a rate. Remember that a quantity that is decreasing, decaying, or being removed corresponds to a negative term in the rate of change equation.

• Using the Wrong Variables: Writing the derivative with respect to the wrong variable (e.g., using $\frac{dy}{dx}$ when the independent variable is time, $t$). Always check what the rate is "with respect to."

• Mixing Up the Function and its Derivative: Placing the derivative expression on the wrong side of the equation. The phrase "The rate of change of Y..." always sets up the $\frac{dY}{dt}$ side of the equation.