Logistic Models

Logistic Growth — The Differential Equation

Part 1 of 7 — From Exponential to Logistic

Why Logistic?

Exponential growth ($dP/dt = kP$) assumes unlimited resources. In reality, populations encounter a carrying capacity $L$ (maximum sustainable population).

The Logistic Differential Equation

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

Behavior Analysis

Key Fact: The growth rate $dP/dt$ is maximized when $P = L/2$. This is the inflection point of the logistic curve.

The Logistic Solution

The general solution to $dP/dt = kP(1 - P/L)$ with $P(0) = P_0$ is:

$\boxed{P(t) = \frac{L}{1 + Ae^{-kt}} \quad \text{where } A = \frac{L - P_0}{P_0}}$

Key Properties of the Solution

Example

A population grows logistically with $L = 1000$, $k = 0.5$, $P_0 = 100$.

$A = (1000 - 100)/100 = 9$

$P(t) = \frac{1000}{1 + 9e^{-0.5t}}$

• $P(0) = 1000/(1 + 9) = 100$ ✓
• $P(\infty) = 1000/(1 + 0) = 1000$ ✓
• Inflection at $P = 500$: $500 = 1000/(1 + 9e^{-0.5t})$ → $t = \ln(9)/0.5 \approx 4.39$

Logistic Fundamentals

Logistic Model Setup

Compute

Summary

• Logistic: $dP/dt = kP(1 - P/L)$
• Solution: $P(t) = L/(1 + Ae^{-kt})$ where $A = (L - P_0)/P_0$
• Max growth at $P = L/2$, rate $= kL/4$
• $P \to L$ as $t \to \infty$

Next: Part 2 — Solving Logistic DEs by Separation of Variables.

Solving Logistic DEs

Part 2 of 7 — Separation of Variables and Partial Fractions

Derivation of the Solution

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

Step 1: Separate variables: $\frac{dP}{P(1 - P/L)} = k\,dt$

Step 2: Partial fractions on the left side: $\frac{1}{P(1 - P/L)} = \frac{L}{P(L - P)} = \frac{1}{P} + \frac{1}{L - P}$

Step 3: Integrate both sides: $\ln|P| - \ln|L - P| = kt + C$

$\ln\left|\frac{P}{L - P}\right| = kt + C$

Step 4: Solve for $P$: $\frac{P}{L - P} = Ae^{kt} \quad \Rightarrow \quad P = \frac{AL e^{kt}}{1 + Ae^{kt}} = \frac{L}{1 + A^{-1}e^{-kt}}$

With initial condition $P(0) = P_0$: $A = P_0/(L - P_0)$, giving $A^{-1} = (L - P_0)/P_0$.

$\boxed{P(t) = \frac{L}{1 + \frac{L - P_0}{P_0}e^{-kt}}}$

Alternative Forms

The logistic equation sometimes appears in different forms:

Example: Non-Standard Form

$dP/dt = P(3 - 0.01P)$

Identify: $a = 3$, $b = 0.01$, so $k = 3$, $L = 3/0.01 = 300$.

Rewrite: $dP/dt = 3P(1 - P/300)$

With $P(0) = 50$: $A = (300 - 50)/50 = 5$

$P(t) = \frac{300}{1 + 5e^{-3t}}$

AP Tip: The AP exam often gives the equation in a non-standard form. Always factor to identify $k$ and $L$.

Identifying Parameters

Solve the Logistic Equation

$dP/dt = 2P(1 - P/100)$, $P(0) = 10$

Carrying Capacity

Summary

• Solve logistic DE via separation of variables + partial fractions
• Solution: $P(t) = L/(1 + Ae^{-kt})$
• Recognize non-standard forms: factor to identify $k$ and $L$
• $dP/dt = aP - bP^2$ → $k = a$, $L = a/b$

Next: Part 3 — Logistic Curve Analysis and Inflection Points.

Logistic Curve Analysis

Part 3 of 7 — Inflection Points, Concavity, and Phase Lines

Inflection Point of the Logistic Curve

The inflection point occurs where $d^2P/dt^2 = 0$ (concavity changes).

Starting from $dP/dt = kP(1 - P/L)$:

$\frac{d^2P}{dt^2} = k\frac{dP}{dt}\left(1 - \frac{2P}{L}\right)$

Setting $d^2P/dt^2 = 0$: since $dP/dt \neq 0$ (away from equilibria), we need:

$1 - \frac{2P}{L} = 0 \quad \Rightarrow \quad \boxed{P = \frac{L}{2}}$

Concavity Regions

Phase Line Analysis

For $dP/dt = kP(1 - P/L)$:

Any initial $P_0 > 0$ → $P(t) \to L$ as $t \to \infty$.

Key Fact: The logistic curve is an S-shape (sigmoid): concave up below $L/2$, concave down above $L/2$.

Finding When the Inflection Occurs

$P(t) = \frac{L}{1 + Ae^{-kt}}$. Set $P = L/2$:

$\frac{L}{2} = \frac{L}{1 + Ae^{-kt}} \quad \Rightarrow \quad 1 + Ae^{-kt} = 2 \quad \Rightarrow \quad e^{-kt} = \frac{1}{A}$

$\boxed{t_{\text{inflection}} = \frac{\ln A}{k}}$

Example

$P(t) = 500/(1 + 4e^{-0.2t})$. $L = 500$, $A = 4$, $k = 0.2$.

Inflection when $P = 250$: $t = \frac{\ln 4}{0.2} = \frac{1.386}{0.2} \approx 6.93$

At $t \approx 6.93$, the population is growing fastest: $dP/dt = kL/4 = 0.2(500)/4 = 25$.

Logistic Curve Shape Summary

Curve Analysis

Concavity Analysis

Inflection Time

Summary

• Inflection at $P = L/2$, time $t = \ln(A)/k$
• Below $L/2$: concave up (accelerating); above: concave down (decelerating)
• Equilibria: $P = 0$ (unstable), $P = L$ (stable)
• S-shaped (sigmoid) curve

Next: Part 4 — Logistic Models in Context (AP Applications).

Logistic Models in Context

Part 4 of 7 — AP Word Problems and Applications

AP Exam Context Problems

The AP exam presents logistic growth in real-world contexts:

Reading a Logistic FRQ

Typical structure:

• "A population of fish in a lake grows at a rate modeled by $dP/dt = 0.4P(1 - P/5000)$."
• Part (a): Find the carrying capacity.
• Part (b): Find the population when growth is fastest.
• Part (c): Find the particular solution given $P(0) = 200$.
• Part (d): When does the population reach 4000?

Example: Complete FRQ

$dP/dt = 0.1P(1 - P/2000)$, $P(0) = 100$.

(a) $L = 2000$

(b) Growth fastest at $P = L/2 = 1000$; rate $= kL/4 = 0.1(2000)/4 = 50$

(c) $A = (2000 - 100)/100 = 19$. $P(t) = \frac{2000}{1 + 19e^{-0.1t}}$

(d) $4000/(1 + 19e^{-0.1t}) = 4000$... wait, that's above $L$. If $P_{\text{target}} = 1500$: $1 + 19e^{-0.1t} = 2000/1500 = 4/3 \quad \Rightarrow \quad e^{-0.1t} = 1/57 \quad \Rightarrow \quad t = \frac{\ln 57}{0.1} \approx 40.4$

AP Tip: Always check units and whether the target population is below $L$. A logistic model never exceeds $L$ (when starting below it).

Euler's Method with Logistic Equations

AP exams sometimes combine Euler's method with logistic growth.

Example: $dP/dt = 0.5P(1 - P/100)$, $P(0) = 20$, $\Delta t = 1$.

$\boxed{P(3) \approx 49.87}$

Notice: $\Delta P$ increases until $P$ passes $L/2 = 50$, then decreases.

Contextual Problems

Word Problem Setup

A population of bacteria grows logistically. At $t = 0$, there are 1000 bacteria. The carrying capacity is 10000, and the initial growth rate is $dP/dt = 450$.

Solve for Time

Summary

• Logistic models appear in population, disease, technology contexts
• Convert non-standard forms to identify $k$ and $L$
• FRQs: find $L$, max rate, solve for particular solutions and specific times
• Can combine with Euler's method

Next: Part 5 — AP Exam Strategies.

AP Exam Strategies — Logistic Growth

Part 5 of 7 — Common Question Patterns

What the AP Tests

Quick Facts to Memorize

$\boxed{\begin{aligned} &\text{Carrying capacity:} \quad L \\ &\text{Max growth at:} \quad P = L/2 \\ &\text{Max rate:} \quad kL/4 \\ &\text{Solution:} \quad P = \frac{L}{1 + Ae^{-kt}} \\ &\text{Inflection time:} \quad t = \frac{\ln A}{k} \end{aligned}}$

Form Recognition

AP Tip: If asked "for what value of $P$ is $d^2P/dt^2 = 0$?", the answer is always $P = L/2$. Don't waste time computing the second derivative.

AP-Style Questions

Quick Identification

Compute A

Summary

• Memorize: $L$, $L/2$, $kL/4$, $A = (L-P_0)/P_0$
• Recognize non-standard forms quickly
• Inflection = $P = L/2$ always
• No need to compute $d^2P/dt^2$ explicitly

Next: Part 6 — Problem-Solving Workshop.

Problem-Solving Workshop

Part 6 of 7 — Mixed Logistic Practice

Work through these problems applying everything you've learned.

Workshop Set A

Complete Analysis

$dP/dt = P(4 - 0.008P)$

Maximum Rate

Workshop Complete

• Practice identifying parameters from non-standard forms
• Apply Euler's method to logistic equations
• Compute carrying capacity, max rate, and inflection points

Next: Part 7 — Comprehensive Review.

Comprehensive Review — Logistic Models

Part 7 of 7 — Final Assessment

Master Reference

$\boxed{P(t) = \frac{L}{1 + \frac{L - P_0}{P_0}e^{-kt}}}$

Review Set A

Review Set B

Full Problem

$dP/dt = 0.6P - 0.001P^2$, $P(0) = 60$.

Final Challenge

Logistic Models — Complete

You've mastered:

• The logistic DE and its solution via separation of variables
• Identifying $k$, $L$, and $A$ from any form
• Inflection points, concavity, and phase line analysis
• Real-world applications and AP exam strategies

$\boxed{\frac{dP}{dt} = kP\left(1 - \frac{P}{L}\right) \quad \Rightarrow \quad P(t) = \frac{L}{1 + Ae^{-kt}}}$