Logistic Models

Part 1 of 7 — The Logistic Differential Equation

The Model

Logistic Basics 🎯

Key Takeaways — Part 1

Logistic: $dP/dt = kP(1 - P/L)$. Max growth at $P = L/2$.

Logistic Models

Part 2 of 7 — The Solution

General Solution

P(t) = rac{L}{1 + Ae^{-kt}}

where A = rac{L - P_0}{P_0}.

Behavior

• As $t o infty$: $P o L$
• $P(t)$ is always between $P_0$ and $L$ (assuming $0 < P_0 < L$)
• S-shaped (sigmoid) curve

Solution 🎯

Key Takeaways — Part 2

$P(t) = L/(1 + Ae^{-kt})$ where $A = (L-P_0)/P_0$. Always approaches $L$.

Logistic Models

Part 3 of 7 — Inflection Point

The Inflection Point

The growth rate changes from increasing to decreasing at $P = L/2$.

To verify: rac{d^2P}{dt^2} = 0 when $P = L/2$.

rac{dP}{dt} = kP - rac{kP^2}{L}

rac{d^2P}{dt^2} = krac{dP}{dt} - rac{2kP}{L}rac{dP}{dt} = rac{dP}{dt}left(k - rac{2kP}{L} ight)

Set to zero (with $dP/dt eq 0$): $k - 2kP/L = 0 implies P = L/2$ ✓

Inflection 🎯

Key Takeaways — Part 3

Inflection at $P = L/2$: growth changes from accelerating to decelerating.

Logistic Models

Part 4 of 7 — Analyzing Logistic Problems

From the DE

rac{dP}{dt} = 0.1P(1 - P/2000), $P(0) = 200$

• Carrying capacity: $L = 2000$
• Growth constant: $k = 0.1$
• Max growth rate at $P = 1000$
• Max $dP/dt = 0.1 cdot 1000 cdot (1 - 1000/2000) = 0.1 cdot 1000 cdot 0.5 = 50$

Reading the DE

$kP(1 - P/L) = kP - kP^2/L$: if given as $0.1P - 0.00005P^2$, then $k = 0.1$ and $k/L = 0.00005$ so $L = 2000$.

Analysis 🎯

Key Takeaways — Part 4

Factor the DE to identify $k$ and $L$. $kP - (k/L)P^2 = kP(1-P/L)$.

Logistic Models

Part 5 of 7 — Logistic vs Exponential

Comparison

Compare 🎯

Key Takeaways — Part 5

Logistic starts exponential, then levels off at $L$. More realistic.

Logistic Models

Part 6 of 7 — Practice Workshop

Workshop 🎯

Workshop Complete!

Logistic Models — Review

Part 7 of 7 — Final Assessment

Final 🎯

Logistic Models — Complete! ✅