7.9 Logistic Models with Differential Equations

Logistic models describe growth that speeds up at first, then slows as a quantity approaches its carrying capacity. The logistic differential equation is $\frac{dy}{dt}=ky(a-y)$, and you can find the carrying capacity and the point of fastest growth straight from the equation without solving it. For AP Calculus BC, interpret the model in context before trying to solve the differential equation.

Logistic Growth in AP Calculus BC

In AP Calculus BC, logistic growth usually appears as a differential equation like $\frac{dy}{dt} = ky(M-y)$, where $M$ is the carrying capacity. You do not always need to solve the differential equation. For Topic 7.9, the main job is to interpret the model in context.

The carrying capacity is the nonzero value that makes $\frac{dy}{dt}=0$, and the quantity changes fastest when $y=\frac{M}{2}$. If a problem gives an initial condition, use it to describe whether the quantity starts below or above the carrying capacity and what the long-term behavior means in context.

Why This Matters for the AP Calculus Exam

Logistic growth appears on the AP Calculus BC exam, often as part of differential equations work. The big skill here is interpreting a logistic model in context: you read information about a rate of change directly from the differential equation instead of solving it from scratch. You should be able to find the carrying capacity (the limiting value as time goes to infinity) and the value where the quantity is changing fastest. These questions reward recognizing the form of the equation and reasoning about behavior, which shows up in both multiple-choice and free-response style problems involving rates and long-term behavior.

Key Takeaways

• The logistic model comes from "the rate of change is jointly proportional to the size of the quantity and the difference between the quantity and the carrying capacity": $\frac{dy}{dt} = ky(a - y)$.
• The carrying capacity is the value the quantity approaches as $t \to \infty$, where $\frac{dy}{dt} = 0$. Set the rate equal to zero and solve for the nonzero value.
• The quantity grows fastest when it is at half the carrying capacity, $y = \frac{a}{2}$.
• You can interpret a logistic equation and its initial conditions without solving the differential equation.
• Logistic growth looks exponential when the quantity is small but levels off as it nears the carrying capacity.
• This material is assessed on the BC exam only.

Logistic Models with Differential Equations

The logistic model describes how a population grows over time when growth is self-limiting. The quantity grows proportionally to its size, but the growth slows and stops as the quantity reaches its carrying capacity, the maximum the environment can sustain.

The rate of change of the quantity is jointly proportional to the size of the quantity and the difference between the quantity and the carrying capacity. In symbols, where $y$ is the quantity, $k$ is a positive constant, and $M$ is the carrying capacity:

$$\frac{dy}{dt} = ky(M - y)$$

An equivalent form is:

$$\frac{dy}{dt} = ky\left(1 - \frac{y}{M}\right)$$

The logistic model is used in ecology, biology, and other fields to describe populations of animals, plants, and microorganisms, as well as other quantities that have a carrying capacity. Compared with exponential growth, which increases without bound, logistic growth produces an S-shaped curve that flattens at the carrying capacity.

For the AP Calculus BC exam, you should be able to find the carrying capacity of a logistic differential equation and the value of the dependent variable when it is changing fastest.

How to Use This on the AP Calculus Exam

Problem Solving

Work through an example to see the process.

The population $P(t)$ of germs in a petri dish satisfies the logistic differential equation below, where $t$ is measured in hours and the initial population is 400.

$$\frac{dP}{dt} = 2P\left(5 - \frac{P}{2000}\right)$$

1. What is the carrying capacity of the population?
2. What is the population's size when it is growing the fastest?

Finding the carrying capacity

As $t \to \infty$, $P(t)$ approaches the carrying capacity, so there is a horizontal asymptote there. No matter the initial condition, the population stabilizes at the carrying capacity.

At that value, $\frac{dP}{dt}$ approaches 0. So set the rate equal to zero and solve for $P$:

$$0 = 2P\left(5 - \frac{P}{2000}\right)$$

This product equals 0 in two cases: when $2P = 0$ (which gives $P = 0$) and when $5 - \frac{P}{2000} = 0$.

$$0 = 5 - \frac{P}{2000}$$ $$\frac{P}{2000} = 5$$ $$P = 10000$$

The carrying capacity is 10000.

You can also rewrite the equation in the form $\frac{dy}{dt} = ky(M - y)$, where $M$ is the carrying capacity:

$$\frac{dP}{dt} = \frac{P}{1000}(10000 - P)$$

From this form, the carrying capacity is 10000.

Finding the size at fastest growth

The quantity grows fastest at the vertex of $\frac{dP}{dt}$, which is a downward-opening quadratic in $P$ with zeros at $P = 0$ and at the carrying capacity. The vertex is halfway between the two zeros, so the point of fastest change occurs at:

$$y = \frac{M}{2}$$

Using the carrying capacity of 10,000:

$$y = \frac{10000}{2}$$

The population is growing the fastest at a size of 5000.

Practice Problem

Try this one on your own first.

The population $P(t)$ of bacteria in a petri dish satisfies the logistic differential equation $\frac{dP}{dt} = 4P\left(10 - \frac{P}{2000}\right)$, where $t$ is measured in hours and the initial population is 300.

1. What is the carrying capacity of the population?
2. What is the population's size when it is growing the fastest?

Answers and solutions

As $t \to \infty$, $P(t)$ approaches the carrying capacity and $\frac{dP}{dt}$ approaches 0. Set the rate equal to zero:

$$0 = 4P\left(10 - \frac{P}{2000}\right)$$

This product equals 0 when $4P = 0$ (giving $P = 0$) and when $10 - \frac{P}{2000} = 0$.

$$0 = 10 - \frac{P}{2000}$$ $$\frac{P}{2000} = 10$$ $$P = 20000$$

The carrying capacity is 20000.

For fastest growth, use $y = \frac{M}{2}$ with the carrying capacity of 20,000:

$$y = \frac{20000}{2}$$

The population is growing the fastest at a size of 10000.

Quick reference

Reading the logistic equation $\frac{dy}{dt} = ky(M - y)$:

• The factor $ky$ shows that growth is proportional to the current size.
• The factor $(M - y)$, or equivalently $\left(1 - \frac{y}{M}\right)$, shows that growth slows as $y$ approaches $M$.
• When the starting value is below the carrying capacity ($y_0 < M$), the quantity grows quickly at first, then slows as $y$ nears $M$.
• At $y = M$, growth stops because $\frac{dy}{dt} = 0$.
• As $t \to \infty$, $y$ approaches $M$.
• The quantity changes fastest when $y = \frac{M}{2}$.

Common Misconceptions

• You do not need to solve the differential equation to answer these questions. Carrying capacity and fastest-growth size both come directly from the equation and the model's structure.
• Fastest growth is not at the carrying capacity. Growth slows to zero at $M$. The maximum rate happens at $y = \frac{M}{2}$.
• Carrying capacity is not $P = 0$. Setting the rate to zero gives two values, $P = 0$ and $P = M$. The carrying capacity is the nonzero value the population approaches.
• Watch the form of the equation. A coefficient out front does not change the carrying capacity by itself. Rewrite into the form $ky(M - y)$ to read $M$ correctly, since the value inside the parentheses is what determines where the rate hits zero.
• Logistic is not the same as exponential. Exponential growth keeps increasing, while logistic growth flattens out at the carrying capacity.

zero equilibrium value. In the form $\frac{dy}{dt}=ky(M-y)$, the carrying capacity is $M$.

When is logistic growth fastest?

Logistic growth is fastest when the quantity is half the carrying capacity. If the carrying capacity is $M$, the fastest growth happens at $y=\frac{M}{2}$.

Do you need to solve the logistic differential equation on the AP exam?

Not always. Topic 7.9 focuses on interpreting the logistic equation and initial conditions, including carrying capacity and fastest growth, often without solving the differential equation explicitly.

How is logistic growth different from exponential growth?

Exponential growth keeps increasing without a built-in upper limit. Logistic growth starts similarly when the quantity is small, but the rate slows as the quantity approaches the carrying capacity.

Related AP Calculus Guides

• Unit 7 Overview: Differential Equations
• 7.2 Verifying Solutions for Differential Equations
• 7.3 Sketching Slope Fields
• 7.6 Finding General Solutions Using Separation of Variables
• 7.5 Approximating Solutions Using Euler’s Method
• 7.7 Finding Particular Solutions Using Initial Conditions and Separation of Variables