Logistic Models with Differential Equations (BC ONLY) - AP Calculus BC Study Guide

The Core Idea: Logistic Models with Differential Equations (BC ONLY)

In many real-world scenarios, population growth is not unlimited. While exponential models describe unrestricted growth, logistic models are used when there are constraints, such as limited resources or space. This constraint is known as the carrying capacity, which represents the maximum sustainable population. The logistic model describes a population that initially grows rapidly (similar to an exponential model) but whose growth rate slows down as the population approaches the carrying capacity, eventually leveling off.

The logistic differential equation mathematically defines this relationship, where the rate of change of the population is proportional to both the current population size and the difference between the carrying capacity and the current population. This topic explores how to analyze this differential equation to understand the system's behavior and how to solve it to find an explicit function for the population over time.

Key Formulas

The logistic model is defined by a specific type of differential equation and its corresponding solution.

The Logistic Differential Equation

The rate of change of a population $P$ with respect to time $t$ is given by:

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

or in an equivalent factored form:

$$\frac{dP}{dt}=kP\left(1-\frac{P}{L}\right)$$

• $P$ is the population at time $t$.

• $L$ is the carrying capacity, the maximum value that $P$ can sustain.

• $k$ is a positive constant of proportionality.

The General Solution

The solution to the logistic differential equation $\frac{dP}{dt}=kP(L-P)$ gives the population $P$ as a function of time $t$. The general solution is:

$$P(t)=\frac{L}{1+C{e}^{-Lkt}}$$

• $L$ and $k$ are the same constants from the differential equation.

• $C$ is a constant determined by an initial condition (e.g., the population at $t=0$).

Point of Fastest Growth

The population grows fastest when it reaches exactly half of the carrying capacity.

• Population at fastest growth:$P=\frac{L}{2}$

This point corresponds to the maximum value of the rate $\frac{dP}{dt}$.

Understanding the Rate of Change

A crucial aspect of the logistic model is understanding how the growth rate, $\frac{dP}{dt}$, changes as the population $P$ changes. The differential equation $\frac{dP}{dt}=kP(L-P)$ shows that the rate of change is a quadratic function of $P$.

If we consider the function $f(P)=kP(L-P)=kLP-k{P}^2$, this is a downward-opening parabola with roots at $P=0$ and $P=L$. The vertex of this parabola represents the maximum value of the function, which corresponds to the maximum rate of population growth. The vertex of a parabola $a{x}^2+bx+c$ occurs at $x=-b/(2a)$. For our function $f(P)=-k{P}^2+kLP$, the maximum occurs at:

$$P=-\frac{kL}{2(-k)}=\frac{L}{2}$$

This confirms that the population grows fastest when $P=L/2$. At this point, the solution curve $P(t)$ has a point of inflection. For $P<L/2$, the growth rate is increasing (the $P(t)$ graph is concave up). For $P>L/2$, the growth rate is decreasing (the $P(t)$ graph is concave down).

Core Concepts & Rules

• Logistic Differential Equation: A population that follows a logistic model is described by the differential equation $\frac{dP}{dt}=kP(L-P)$.

• Carrying Capacity: The constant $L$ in the logistic equation is the carrying capacity, representing the long-term limit of the population. As $t\to \infty$, $P(t)\to L$.

• Maximum Growth Rate: The population's growth rate is at its maximum when the population is exactly half the carrying capacity, i.e., at $P=L/2$.

• Solving the Equation: The logistic differential equation can be solved analytically using the method of separation of variables, which requires partial fraction decomposition to perform the integration.

• The Solution Form: The solution to the logistic differential equation is always of the form $P(t)=\frac{L}{1+C{e}^{-Lkt}}$. Knowing this form can be a shortcut to finding a particular solution if $L$ and $k$ are known.

Step-by-Step Example 1: Analyzing a Logistic Model

A population of fish is modeled by the logistic differential equation $\frac{dP}{dt}=0.001P(500-P)$, where $P$ is the number of fish and $t$ is the time in years.

(a) What is the carrying capacity of the fish population?

(b) What is the population size when the population is growing fastest?

(c) What is the rate of change of the population when it is growing fastest?

Solution

Step 1: Identify the form of the equation and the constants.

The equation is given as $\frac{dP}{dt}=0.001P(500-P)$. This matches the standard form $\frac{dP}{dt}=kP(L-P)$.

By comparing the two forms, we can identify:

• $k=0.001$

• $L=500$

(a) Determine the carrying capacity.

The carrying capacity is the value of $L$.

• Carrying Capacity:$L=500$ fish.

(b) Determine the population when growth is fastest.

The population grows fastest when $P=L/2$.

• $P=\frac{500}{2}=250$ fish.

(c) Determine the value of the fastest growth rate.

To find the rate of change when the population is growing fastest, substitute $P=250$ back into the differential equation $\frac{dP}{dt}$.

• $\frac{dP}{dt}{|}_{P=250}=0.001(250)(500-250)$

• $\frac{dP}{dt}=0.001(250)(250)$

• $\frac{dP}{dt}=0.001(62500)$

• $\frac{dP}{dt}=62.5$

The fastest growth rate is 62.5 fish per year.

Step-by-Step Example 2: Solving a Logistic Differential Equation

Find the particular solution to the differential equation $\frac{dy}{dt}=\frac{y}{8}(6-y)$ with the initial condition $y(0)=1$.

Solution

Step 1: Separate the variables.

First, rewrite the equation in the standard form $\frac{dy}{dt}=ky(L-y)$.

$\frac{dy}{dt}=\frac{1}{8}y(6-y)$. Here, $k=1/8$ and $L=6$.

To solve, we separate variables $y$ and $t$.

$$\frac{1}{y(6-y)}dy=\frac{1}{8}dt$$

Step 2: Use partial fraction decomposition for the left side.

We need to decompose $\frac{1}{y(6-y)}$.

$$\frac{1}{y(6-y)}=\frac{A}{y}+\frac{B}{6-y}$$

Multiplying by $y(6-y)$ gives $1=A(6-y)+By$.

• If $y=0$, then $1=A(6)⟹A=1/6$.

• If $y=6$, then $1=B(6)⟹B=1/6$.

So, the decomposition is $\frac{1}{6y}+\frac{1}{6(6-y)}$.

Step 3: Integrate both sides of the separated equation.

$$\int \left(\frac{1/6}{y}+\frac{1/6}{6-y}\right)dy=\int \frac{1}{8}dt$$

$$\frac{1}{6}\int \left(\frac{1}{y}+\frac{1}{6-y}\right)dy=\frac{1}{8}\int dt$$

$$\frac{1}{6}(\ln |y|-\ln |6-y|)=\frac{1}{8}t+C_1$$

Note the minus sign from the integral of $\frac{1}{6-y}$ due to the chain rule.

Step 4: Simplify and solve for $y$ in terms of $t$.

$$\ln \left|\frac{y}{6-y}\right|=6\left(\frac{1}{8}t+C_1\right)=\frac{3}{4}t+6C_1$$

Exponentiate both sides:

$$\left|\frac{y}{6-y}\right|=e^{\frac{3}{4}t+6C_1}=e^{6C_1}{e}^{\frac{3}{4}t}$$

Let $C=\pm e^{6C_1}$ to absorb the absolute value and the constant.

$$\frac{y}{6-y}=C{e}^{\frac{3}{4}t}$$

Step 5: Use the initial condition $y(0)=1$ to find $C$.

$$\frac{1}{6-1}=C{e}^{\frac{3}{4}(0)}⟹\frac{1}{5}=C\cdot 1⟹C=\frac{1}{5}$$

Our equation is now $\frac{y}{6-y}=\frac{1}{5}{e}^{\frac{3}{4}t}$.

Step 6: Perform the final algebra to isolate $y(t)$.

$$5y=(6-y)e^{\frac{3}{4}t}$$

$$5y=6e^{\frac{3}{4}t}-y{e}^{\frac{3}{4}t}$$

$$5y+y{e}^{\frac{3}{4}t}=6e^{\frac{3}{4}t}$$

$$y(5+e^{\frac{3}{4}t})=6e^{\frac{3}{4}t}$$

$$y(t)=\frac{6e^{\frac{3}{4}t}}{5+e^{\frac{3}{4}t}}$$

To match the standard solution form, divide the numerator and denominator by $e^{\frac{3}{4}t}$:

$$y(t)=\frac{6}{5e^{-\frac{3}{4}t}+1}$$

This matches the form $P(t)=\frac{L}{1+C{e}^{-Lkt}}$ with $L=6$, $C=5$, and $-Lkt=-3/4t$.

Check: $-6(1/8)t=-6/8t=-3/4t$. The solution is correct.

Using Your Calculator

Solving a logistic differential equation is an analytical process requiring separation of variables and partial fractions. A calculator is not used to find the symbolic solution. However, it can be a powerful tool for verification and analysis.

1. Verifying the Solution:

Once you have found the particular solution, such as $y(t)=\frac{6}{1+5e^{-0.75t}}$, you can graph this function on your calculator.

• Check if the initial condition is met: Does the graph pass through $(0,1)$?

• Check the carrying capacity: Does the graph have a horizontal asymptote at $y=L=6$?

• Check the point of fastest growth: Does the graph have an inflection point at $y=L/2=3$?

2. Visualizing the Slope Field:

You can use your calculator's slope field feature to visualize the logistic model.

• Enter the differential equation, e.g., $dy/dx=(y/8)(6-y)$.

• Graph the slope field. You should see slopes that are small near $y=0$ and $y=6$, and steepest around $y=3$.

• You can then graph your particular solution on top of the slope field to see if it "follows" the field lines, providing a strong visual confirmation of your answer.

AP Exam Quick Hit

Common Question Types

• Analysis from the Differential Equation: Given $\frac{dP}{dt}=0.02P(1000-P)$, you will be asked to state the carrying capacity ($L=1000$) and the population value where growth is fastest ($P=L/2=500$).

• Solving from an Initial Condition: You will be given a logistic differential equation and an initial value, like $\frac{dP}{dt}=kP(L-P)$ with $P(0)=P_0$, and be required to find the particular solution $P(t)$. This tests the full separation of variables and partial fractions procedure.

• Interpretation from the Solution: Given a solution like $P(t)=\frac{200}{1+4e^{-0.5t}}$, you may be asked to find the carrying capacity (the numerator, $L=200$), the initial population ($P(0)=\frac{200}{1+4}=40$), or the limit as $t\to \infty$ (which is the carrying capacity, 200).

Common Mistakes

• Confusing Growth Rate and Population: Students often confuse the population value where growth is fastest ($P=L/2$) with the maximum growth rate itself (the value of $\frac{dP}{dt}$ at $P=L/2$). Read the question carefully to see which is being asked.

• Partial Fraction Errors: Simple arithmetic or algebraic errors when solving for the constants $A$ and $B$ in the partial fraction decomposition are very common. Double-check this step.

• Integration Errors: Forgetting the negative sign when integrating a term like $\frac{1}{L-P}$. The integral is $-\ln |L-P|$, not $\ln |L-P|$.

• Algebraic Manipulation: The final steps of solving for $P(t)$ after integrating and finding the constant $C$ can be tricky. Be careful and methodical when isolating $P$.

• Identifying Constants: When given the form $\frac{dP}{dt}=kP(1-P/L)$, students sometimes mistakenly identify $L$ as $1/L$ or misidentify $k$. Ensure you match the equation to the correct standard form before pulling out constants.