Modeling Situations with Differential Equations - AP Calculus BC Study Guide

The Core Idea: Modeling Situations with Differential Equations

A differential equation is a powerful mathematical tool that describes the relationship between an unknown function and its derivatives. In essence, it's an equation that contains one or more rates of change. The fundamental purpose of this topic is to translate real-world scenarios, which are often described in terms of how a quantity is changing, into the precise language of mathematics.

Instead of describing what a quantity is, a differential equation describes how it behaves over time or with respect to another variable. For example, we can model population growth not by stating the population at every moment, but by stating that its rate of growth is proportional to its current size. This topic focuses on the critical first step of this modeling process: correctly interpreting a verbal description and constructing the corresponding differential equation. All variables within the model must be clearly defined in the context of the situation.

Key Definitions and Relationships

The process of modeling with differential equations relies on translating specific English phrases into mathematical expressions.

• Differential Equation: An equation involving an unknown function and one or more of its derivatives.

  • Example: $\frac{dy}{dt}=2y+t$

• Rate of Change: The phrase "the rate of change of $y$ with respect to $x$" is a direct instruction to write the derivative $\frac{dy}{dx}$. The variable mentioned first ($y$) is the dependent variable and forms the numerator of the derivative, while the variable mentioned second ($x$) is the independent variable and forms the denominator.

• Proportionality: The phrase "is proportional to" or "is directly proportional to" signifies a multiplicative relationship involving a constant of proportionality, $k$.

  • If a quantity $A$ is proportional to a quantity $B$, we write:

  $$A=kB$$

  • The constant k is a non-zero real number that scales the relationship.

• Common Proportionality Models: Many real-world phenomena are modeled using specific forms of proportionality.

  • Direct Proportionality to the Amount Present: This model is used for things like unrestricted population growth or radioactive decay. If the rate of change of $y$ is proportional to the amount of $y$ present, the equation is:

  $$\frac{dy}{dt}=ky$$

  • Proportionality to a Difference (Newton's Law of Cooling/Heating): This model applies when the rate of change of a quantity is proportional to the difference between that quantity and a constant ambient value. If the rate of change of $y$ is proportional to the difference between $y$ and a constant $A$, the equation is:

  $$\frac{dy}{dt}=k(y-A)$$

Understanding the Language of Rates

The most critical skill in this topic is the careful translation of verbal statements into mathematical symbols. Every part of a descriptive sentence corresponds to a part of the final differential equation.

The core of the process is to deconstruct the sentence. First, identify the quantities that are changing and assign them variables. For example, if a problem discusses the volume of water in a tank over time, you might define $V$ as the volume and $t$ as time.

Next, locate the phrase that describes the rate of change. A statement like "the volume of water is decreasing at a rate..." immediately tells you that you are working with the derivative $\frac{dV}{dt}$. The word "decreasing" implies that this rate is negative.

Finally, identify the relationship that governs the rate. If the sentence continues "...proportional to the square root of the current volume," you translate this to $=k\sqrt{V}$. Combining these pieces gives the differential equation. If the rate is decreasing, you would write $\frac{dV}{dt}=-k\sqrt{V}$ (for a positive $k$) or understand that the resulting $k$ in $\frac{dV}{dt}=k\sqrt{V}$ must be a negative constant. The key is that the structure of the equation is dictated entirely by these translated phrases.

Core Concepts & Rules

• A differential equation is an equation that defines a function by relating it to its own rate of change (its derivative).

• These equations are the mathematical foundation for modeling dynamic real-world systems.

• The phrase "the rate of change of $A$ with respect to $B$" must be translated as the derivative $\frac{dA}{dB}$.

• The statement "is proportional to $Q$" must be translated as a multiplicative relationship, $=kQ$, where $k$ is the constant of proportionality.

• The statement "is inversely proportional to $Q$" translates to $=\frac{k}{Q}$. This is a specific case of the rate being proportional to a function of the quantity.

• When setting up a model, you must explicitly define what each variable ($y$, $t$, $P$, etc.) represents in the context of the problem.

• A common and important model is exponential growth or decay, where a quantity's rate of change is directly proportional to the quantity itself: $\frac{dy}{dt}=ky$.

• Another key model involves a rate of change proportional to the difference between a quantity and a constant, such as in Newton's Law of Cooling: $\frac{dy}{dt}=k(y-A)$.

Step-by-Step Example 1: Basic Application

Problem: The rate of spread of a rumor in a school is jointly proportional to the number of students who have heard the rumor and the number of students who have not. The school has a total of 1200 students. Let $P(t)$ be the number of students who have heard the rumor at time $t$ in hours. Write a differential equation that models this situation.

Step 1: Define Variables and Constants

• Dependent Variable: P = the number of students who have heard the rumor.

• Independent Variable: t = time in hours.

• Total Population: N = 1200 students.

• The number of students who have not heard the rumor is $(1200-P)$.

Step 2: Translate the "Rate of Change"

• The phrase "The rate of spread of a rumor" refers to the rate of change of the number of people who have heard it, $P$, with respect to time, $t$.

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

Step 3: Translate the Proportionality Relationship

• The problem states the rate is "jointly proportional to the number of students who have heard the rumor ($P$) and the number of students who have not (1200 - P)".

• "Jointly proportional to" means proportional to the product of these two quantities.

• This translates to: $=k\cdot P\cdot (1200-P)$.

Step 4: Combine and Finalize the Equation

• Combine the expressions from Step 2 and Step 3 to form the complete differential equation.

$$\frac{dP}{dt}=kP(1200-P)$$

• This is the logistic differential equation, a key model for constrained growth. The constant k would be determined by additional information about the spread of the rumor.

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

Problem: A cylindrical barrel with a radius of 2 feet is being filled with water. The height of the water, $h$, is changing over time $t$ (in minutes). At the same time, water is leaking from a hole in the bottom. The rate of change of the volume of water in the barrel, $\frac{dV}{dt}$, is modeled by the differential equation $\frac{dV}{dt}=5-\sqrt{h}$, where $V$ is the volume in cubic feet. Write a differential equation for the rate of change of the height of the water, $\frac{dh}{dt}$, in terms of $h$.

Step 1: Identify the Goal

• We are given $\frac{dV}{dt}$ and need to find $\frac{dh}{dt}$. This means we need to find a relationship between $V$ and $h$ and then use the chain rule.

Step 2: Establish the Relationship Between Variables

• The problem involves the volume $V$ and height $h$ of water in a cylinder. The formula for the volume of a cylinder is $V=\pi r^{2}h$.

• The radius is given as a constant: r = 2 feet.

• Substitute the constant radius into the volume formula:

$$V=\pi (2{)}^{2}h=4\pi h$$

Step 3: Differentiate the Relationship with Respect to Time

• Take the derivative of the equation $V=4\pi h$ with respect to time, $t$. Remember that both $V$ and $h$ are functions of $t$. We must use implicit differentiation.

$$\frac{d}{dt}(V)=\frac{d}{dt}(4\pi h)$$

$$\frac{dV}{dt}=4\pi \frac{dh}{dt}$$

• This equation now connects the two rates of change.

Step 4: Substitute and Solve for the Desired Rate

• We are given the expression for $\frac{dV}{dt}$ in the problem statement: $\frac{dV}{dt}=5-\sqrt{h}$.

• Substitute this expression into the equation from Step 3:

$$5-\sqrt{h}=4\pi \frac{dh}{dt}$$

• Now, isolate $\frac{dh}{dt}$ to get the final answer.

$$\frac{dh}{dt}=\frac{5-\sqrt{h}}{4\pi }$$

• This is the differential equation that models the rate of change of the height of the water in the barrel.

Using Your Calculator

Topic 7.1 is focused exclusively on translating verbal descriptions and physical situations into mathematical equations. It is a conceptual modeling task that does not involve solving differential equations or analyzing their graphical representations.

Therefore, a calculator is not used for this specific topic. The process of identifying variables, interpreting rates of change, and translating proportionality is purely analytical. All work required for this topic must be done by hand.

AP Exam Quick Hit

Common Question Types

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

  • Example: "The rate of change of a quantity $y$ is inversely proportional to the square of $x$. Which of the following is the differential equation for this relationship?" The answer would be $\frac{dy}{dt}=\frac{k}{x^2}$.

• FRQ Part (a) Setup: The very first part of a multi-part Free Response Question will often be to set up the differential equation that governs the rest of the problem.

  • Example: "A potato is taken from an oven where its temperature is 190°F and left to cool in a kitchen where the temperature is 70°F. The rate of change of the potato's temperature is proportional to the difference between its temperature and the ambient kitchen temperature. If $H$ is the temperature of the potato at time $t$, write a differential equation to model this situation." The answer is $\frac{dH}{dt}=k(H-70)$.

• Related Rates as a Differential Equation: You may be given a geometric scenario and asked to find a differential equation relating the rates of change of different variables, as shown in Example 2 above.

Common Mistakes

• Proportionality vs. Addition: Confusing "is proportional to $y$" ($=ky$) with "is $k$ more than $y$" ($=k+y$). Proportionality is always a multiplicative relationship.

• Forgetting the Constant of Proportionality: A very common error is to omit the constant $k$. Writing $\frac{dy}{dt}=y$ is incorrect if the problem states the rate is proportional to $y$.

• Incorrect Derivative Setup: For "the rate of change of $P$ with respect to $t$," writing $\frac{dt}{dP}$ instead of the correct $\frac{dP}{dt}$. The variable mentioned first is the dependent variable (numerator).

• Sign Errors in Cooling/Decay Models: In a cooling model like $\frac{dH}{dt}=k(H-A)$, if the object's temperature $H$ is greater than the ambient temperature $A$, then $(H-A)$ is positive. Since the object is cooling, $\frac{dH}{dt}$ must be negative. Therefore, the constant $k$ must be negative. Forgetting to consider the physical meaning of the rate can lead to sign errors.

• Misinterpreting "Rate In - Rate Out" Problems: For problems involving tanks or populations where things are both entering and leaving, the overall rate of change is $\frac{dy}{dt}=(\text{rate in})-(\text{rate out})$. A common mistake is to multiply or divide these rates instead of subtracting them.