Unit 7.9 – Logistic Models with Differential Equations

AP® Calculus BC ONLY | Limited Growth Models

Why This Matters: Logistic growth is a more realistic model than exponential growth! While exponential models assume unlimited growth, real populations face limits—food, space, resources. The logistic model accounts for these constraints by introducing a carrying capacity. This BC-only topic models everything from population biology to disease spread to product adoption. It's heavily tested on AP® BC exams and shows how calculus models real-world limitations. Master this and you'll understand bounded growth!

🌍 Why Logistic Instead of Exponential?

THE PROBLEM WITH EXPONENTIAL GROWTH

Exponential model \(\frac{dP}{dt} = kP\) assumes:

Unlimited resources
No environmental constraints
Growth continues forever

Reality: Every environment has limits! Populations can't grow infinitely.

The Logistic Solution:

Modify the exponential model to include a limiting factor that slows growth as population approaches a maximum sustainable level called the carrying capacity.

📐 The Logistic Differential Equation

The Logistic Model

THE DIFFERENTIAL EQUATION:

\[ \frac{dP}{dt} = kP\left(1 - \frac{P}{M}\right) \]

Where:

\(P(t)\) = population at time \(t\)
\(k\) = growth rate constant (\(k > 0\))
\(M\) = carrying capacity (maximum sustainable population)
\(\left(1 - \frac{P}{M}\right)\) = limiting factor (slows growth as \(P \to M\))

📝 Key Insight: When \(P\) is small compared to \(M\), \(\left(1 - \frac{P}{M}\right) \approx 1\), so growth is nearly exponential. As \(P \to M\), \(\left(1 - \frac{P}{M}\right) \to 0\), so growth slows to zero!

✅ The Logistic Function (Solution)

General Solution to Logistic DE

THE LOGISTIC FUNCTION:

\[ P(t) = \frac{M}{1 + Ae^{-kt}} \]

Where:

\(M\) = carrying capacity
\(k\) = growth rate
\(A\) = constant determined by initial condition

Finding A from initial condition \(P(0) = P_0\):

\[ P_0 = \frac{M}{1 + A} \quad \Rightarrow \quad A = \frac{M - P_0}{P_0} \]

Alternative Form (with initial condition built in):

\[ P(t) = \frac{M}{1 + \left(\frac{M - P_0}{P_0}\right)e^{-kt}} \]

🔍 Key Features of Logistic Growth

Behavior of the Logistic Curve:

S-shaped curve (sigmoid): Starts slow, speeds up, then levels off
Initial growth nearly exponential: When \(P \ll M\)
Inflection point at \(P = \frac{M}{2}\): Maximum growth rate
Approaches carrying capacity: \(\lim_{t \to \infty} P(t) = M\)
Horizontal asymptote: \(y = M\)
Never exceeds M: If \(P_0 < M\)

The Inflection Point

Where Growth Rate is Maximum:

The population grows fastest when \(P = \frac{M}{2}\)

Proof: Maximum of \(\frac{dP}{dt} = kP\left(1 - \frac{P}{M}\right)\)

Take derivative and set to zero:

\[ \frac{d^2P}{dt^2} = 0 \quad \Rightarrow \quad P = \frac{M}{2} \]

Maximum rate of change:

\[ \left(\frac{dP}{dt}\right)_{\text{max}} = \frac{kM}{4} \]

⚖️ Exponential vs Logistic Growth

Comparing Growth Models
Feature	Exponential	Logistic
Differential Equation	\(\frac{dP}{dt} = kP\)	\(\frac{dP}{dt} = kP\left(1-\frac{P}{M}\right)\)
Solution	\(P = P_0 e^{kt}\)	\(P = \frac{M}{1 + Ae^{-kt}}\)
Long-term behavior	\(P \to \infty\)	\(P \to M\)
Curve shape	J-curve (unlimited)	S-curve (sigmoid)
Inflection point	None	At \(P = M/2\)
Realism	Short-term only	More realistic

📖 Deriving the Logistic Solution

Derivation using Separation of Variables:

Start with:

\[ \frac{dP}{dt} = kP\left(1 - \frac{P}{M}\right) \]

Separate variables:

\[ \frac{1}{P\left(1 - \frac{P}{M}\right)} \, dP = k \, dt \]

Use partial fractions:

\[ \frac{1}{P\left(1 - \frac{P}{M}\right)} = \frac{1}{P} + \frac{1}{M-P} \]

Integrate:

\[ \ln|P| - \ln|M-P| = kt + C \]

\[ \ln\left|\frac{P}{M-P}\right| = kt + C \]

Solve for P:

After algebraic manipulation:

\[ P(t) = \frac{M}{1 + Ae^{-kt}} \]

📊 Comprehensive Examples

Example 1: Complete Logistic Problem

Problem: A population has carrying capacity 1000. Initially there are 100 individuals. After 2 years, the population is 300. Find the logistic model and determine when the population reaches 800.

Given:

\(M = 1000\) (carrying capacity)
\(P_0 = 100\) (at \(t = 0\))
\(P(2) = 300\)

Find A:

\[ A = \frac{M - P_0}{P_0} = \frac{1000 - 100}{100} = 9 \]

Model so far:

\[ P(t) = \frac{1000}{1 + 9e^{-kt}} \]

Find k using \(P(2) = 300\):

\[ 300 = \frac{1000}{1 + 9e^{-2k}} \]

\[ 1 + 9e^{-2k} = \frac{1000}{300} = \frac{10}{3} \]

\[ 9e^{-2k} = \frac{7}{3} \]

\[ e^{-2k} = \frac{7}{27} \]

\[ -2k = \ln\left(\frac{7}{27}\right) \]

\[ k = -\frac{1}{2}\ln\left(\frac{7}{27}\right) \approx 0.674 \]

Complete model:

\[ P(t) = \frac{1000}{1 + 9e^{-0.674t}} \]

Find when \(P = 800\):

\[ 800 = \frac{1000}{1 + 9e^{-0.674t}} \]

\[ 1 + 9e^{-0.674t} = 1.25 \]

\[ e^{-0.674t} = \frac{0.25}{9} = \frac{1}{36} \]

\[ -0.674t = \ln\left(\frac{1}{36}\right) \]

\[ t = \frac{\ln 36}{0.674} \approx 5.32 \text{ years} \]

Example 2: Finding Growth Rate at Specific Population

Problem: For \(\frac{dP}{dt} = 0.5P\left(1 - \frac{P}{200}\right)\), find the rate of growth when \(P = 50\) and when \(P = 100\).

When \(P = 50\):

\[ \frac{dP}{dt} = 0.5(50)\left(1 - \frac{50}{200}\right) = 25\left(\frac{3}{4}\right) = 18.75 \]

When \(P = 100\) (half of carrying capacity):

\[ \frac{dP}{dt} = 0.5(100)\left(1 - \frac{100}{200}\right) = 50\left(\frac{1}{2}\right) = 25 \]

This is the MAXIMUM growth rate (at inflection point)

📋 Logistic Problem-Solving Strategy

Step-by-Step Approach

Systematic Method:

Identify the model: Recognize it's logistic (limited growth mentioned)
Find M: Carrying capacity (often stated directly)
Find \(P_0\): Initial population
Find A: \(A = \frac{M - P_0}{P_0}\)
Find k: Use additional data point
- Substitute known \((t, P)\) into \(P = \frac{M}{1 + Ae^{-kt}}\)
- Solve for \(k\)
Write complete model: \(P(t) = \frac{M}{1 + Ae^{-kt}}\) with values
Answer the question: Evaluate or solve as needed

💡 Essential Tips & Strategies

✅ Success Strategies:

Inflection point: Always at \(P = \frac{M}{2}\) (half of carrying capacity)
Maximum growth rate: Occurs at inflection point
Long-term behavior: \(P \to M\) as \(t \to \infty\)
Finding A is easy: \(A = \frac{M - P_0}{P_0}\)
Finding k requires algebra: Use given data point
Check reasonableness: Does \(P < M\) always?
S-curve visualization: Helps understand behavior

🔥 Key Relationships:

When \(P \ll M\): Growth is nearly exponential
When \(P = M/2\): Growth rate is maximum
When \(P \to M\): Growth rate approaches zero
If \(P_0 > M\): Population decreases to M
Limiting factor: \(\left(1 - \frac{P}{M}\right)\) controls slowdown

❌ Common Mistakes to Avoid

Mistake 1: Confusing logistic with exponential models
Mistake 2: Forgetting that inflection point is at \(P = M/2\)
Mistake 3: Wrong formula for A: it's \(\frac{M-P_0}{P_0}\), not \(\frac{P_0}{M-P_0}\)
Mistake 4: Algebra errors when solving for k
Mistake 5: Not recognizing carrying capacity from context
Mistake 6: Thinking population can exceed M (if starting below)
Mistake 7: Sign errors with logarithms
Mistake 8: Not checking if answer makes sense in context
Mistake 9: Confusing \(\frac{dP}{dt}\) with P
Mistake 10: Not showing work finding k (loses AP® credit)

📝 Practice Problems

Solve these logistic problems:

A lake can support 5000 fish. Initially 500 fish. If \(k = 0.4\), write the model.
For \(\frac{dP}{dt} = 0.6P(1 - P/300)\), at what population is growth fastest?
If \(P(t) = \frac{2000}{1 + 19e^{-0.5t}}\), what is the carrying capacity?
A population grows logistically with \(M = 1000\), \(P_0 = 100\). When does it reach 500?

Answers:

\(P(t) = \frac{5000}{1 + 9e^{-0.4t}}\)
\(P = 150\) (half of 300)
\(M = 2000\)
Need to find k first from additional information

✏️ AP® BC Exam Success Tips

What AP® BC Graders Look For:

Identify the model: State it's logistic growth
Write the DE: \(\frac{dP}{dt} = kP\left(1-\frac{P}{M}\right)\) if asked
Identify M: Clearly state carrying capacity
Show finding A: \(A = \frac{M-P_0}{P_0}\)
Show finding k: Complete algebraic work
Write complete model: With numerical values
Inflection point reasoning: Explain why at \(M/2\)
Interpret in context: What does answer mean?

💯 Exam Strategy:

Read carefully—look for "carrying capacity" or "maximum"
Identify: M, \(P_0\), and any data points
Calculate A immediately: \(\frac{M-P_0}{P_0}\)
Use data point to find k (show all algebra)
Write complete model clearly
For inflection point: state \(P = M/2\)
For max growth rate: evaluate \(\frac{dP}{dt}\) at \(M/2\)
Interpret answer in context of problem

⚡ Quick Reference Guide

LOGISTIC MODEL ESSENTIALS (BC ONLY)

The Differential Equation:

\[ \frac{dP}{dt} = kP\left(1 - \frac{P}{M}\right) \]

The Solution:

\[ P(t) = \frac{M}{1 + Ae^{-kt}} \]

where \(A = \frac{M - P_0}{P_0}\)

Key Features:

Carrying capacity: \(M\) (horizontal asymptote)
Inflection point: \(P = \frac{M}{2}\)
Max growth rate: \(\frac{kM}{4}\) (at \(P = M/2\))
Long-term: \(\lim_{t \to \infty} P(t) = M\)
Shape: S-curve (sigmoid)

Master Logistic Models! (BC ONLY) The logistic differential equation \(\frac{dP}{dt} = kP\left(1-\frac{P}{M}\right)\) models limited growth with carrying capacity \(M\). Unlike exponential growth (unlimited), logistic growth has an S-shaped curve that levels off at \(M\). The solution is \(P(t) = \frac{M}{1 + Ae^{-kt}}\) where \(A = \frac{M-P_0}{P_0}\). Key features: (1) Inflection point at \(P = M/2\) where growth is fastest, (2) Maximum growth rate of \(\frac{kM}{4}\), (3) Approaches carrying capacity as \(t \to \infty\), (4) S-shaped (sigmoid) curve. To solve problems: find M (carrying capacity), \(P_0\) (initial), calculate \(A = \frac{M-P_0}{P_0}\), use data point to find k, write complete model. The limiting factor \(\left(1-\frac{P}{M}\right)\) slows growth as population approaches capacity. When \(P \ll M\), growth is nearly exponential; when \(P \to M\), growth approaches zero. Applications: population biology, disease spread, product adoption. This BC-only topic appears on most BC exams—know the formulas cold! 🎯✨