🎯⭐ INTERACTIVE LESSON

Logistic Models

Learn step-by-step with interactive practice!

← Back to Standard Lesson

Logistic Models - Complete Interactive Lesson

Part 1: The Logistic Differential Equation

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

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

Parameter	Meaning
$P$	Population at time $t$
$k$	Growth rate constant
$L$	Carrying capacity
$kP$

Behavior Analysis

Condition	$1 - P/L$	$dP/dt$	Population...
$P \ll L$

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:

Logistic Fundamentals

Logistic Model Setup

Compute

Summary

Logistic: $dP/dt = kP(1 - P/L)$
Solution: $P(t) = L/(1 + Ae^{-kt})$ where

Part 2: The Solution

Solving Logistic DEs

Part 2 of 7 — Separation of Variables and Partial Fractions

Derivation of the Solution

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

Part 3: Inflection Point

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 $d P / d t = k P (1 - P / L)$ :

Part 4: Analyzing Logistic Problems

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:

Context	$P$ represents	$L$ represents
Population biology	Number of organisms	Environment capacity
Disease spread	Number infected	Total susceptible population
Technology adoption	Number of users	Market size
Rumors	People who heard	Total community

Reading a Logistic FRQ

Typical structure:

Part 5: Logistic vs Exponential

AP Exam Strategies — Logistic Growth

Part 5 of 7 — Common Question Patterns

What the AP Tests

Concept	How it's tested	Point value
Identify $L$	"What is the carrying capacity?"	1
Max growth rate	"When is growth fastest?" or "Find max $dP/dt$ "	1–2
Solve for $P(t)$

Part 6: Practice 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

Part 7 — Comprehensive Review.

Part 7: Final Assessment

Comprehensive Review — Logistic Models

Part 7 of 7 — Final Assessment

Master Reference

Concept	Formula
Logistic DE	$dP/dt = kP(1 - P/L)$
Solution

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

Key Properties of the Solution

Property	Value/Behavior
$P(0)$	$P_0$
$\lim_{t \to \infty} P(t)$	$L$
Inflection point	$P = L/2$
Max growth rate	$kL/4$
Shape	S-curve (sigmoid)

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 $P = 500$ : $500 = 1000/(1 + 9e^{-0.5t})$ → $t = \ln$

Max growth at

P = L/2

, rate

= kL/4

P \to L

t \to \infty

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

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:

Form	Equivalent standard form	$k$	$L$
$dP/dt = kP(1 - P/L)$	Standard	$k$	$L$
$dP/dt = P(a - bP)$	$= aP(1 - P/(a/b))$
$dP/dt = rP(L - P)/L$	Same as standard	$r$	$L$
$dP/dt = rP - rP^2/L$	Expanded standard	$r$

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 $L$
$dP/dt = aP - bP^2$ → $k = a$ , $L = a /$

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

d P / d t = k P (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:

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

Region	$1 - 2P/L$	Concavity	Growth behavior
$P < L/2$	$> 0$	Concave UP	Growth accelerating
$P = L/2$	$= 0$	Inflection	Growth rate maximum
$L/2 < P < L$	$< 0$	Concave DOWN	Growth decelerating

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

Equilibrium	Stability	Type
$P = 0$	Unstable	Repelling
$P = L$	Stable	Attracting

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

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

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

Feature	Value
Initial growth	Nearly exponential ( $\approx kP$ )
Inflection point	$P = L/2$ , $t = \ln(A)/k$

Concavity Analysis

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).

"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

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

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

$t$	$P$	$f = 0.5P(1-P/100)$

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

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 $dP/dt = 450$ .

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.

Quick Facts to Memorize

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

Given equation	$k$	$L$
$dP/dt = 0.3P(1 - P/400)$	$0.3$	$400$
$dP/dt = P(5 - 0.01P)$	$5$	$500$
$dP/dt = 2P - P^2/250$	$2$	$500$

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

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.

P (t) = L / (1 + A e^{- k t})

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

Full Problem

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

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

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

t = ln (9) /0.5 \approx 4.39

Inflection time:t=klnA

0	20	$0.5(20)(0.8) = 8$	8	28
1	28	$0.5(28)(0.72) = 10.08$	10.08	38.08
2	38.08	$0.5(38.08)(0.6192) = 11.79$	11.79	49.87

Logistic Models

Key Properties of the Solution

Example

Alternative Forms

Example: Non-Standard Form

Summary

Concavity Regions

Phase Line Analysis

Finding When the Inflection Occurs

Example

Logistic Curve Shape Summary

Summary

Example: Complete FRQ

Euler's Method with Logistic Equations

Summary

Quick Facts to Memorize

Form Recognition

Summary

Logistic Models — Complete