Logistic Growth - Biology

1. Fundamental Concepts

Definition: Logistic growth is a model that describes how a population grows in a limited environment, characterized by an initial exponential growth followed by a leveling off as the population approaches the carrying capacity of the environment.
Carrying Capacity: The maximum population size that the environment can sustain indefinitely, given the food, habitat, water, and other necessities available in the environment.
Logistic Growth Equation: The logistic growth equation is given by $$\frac{dN}{dt} = rN \left(1 - \frac{N}{K}\right)$$ where $N$ is the population size, $r$ is the intrinsic growth rate, and $K$ is the carrying capacity.

2. Key Concepts

Basic Rule: The logistic growth curve is S-shaped (sigmoidal), reflecting the initial rapid growth followed by a slowdown as the population approaches the carrying capacity.

Differential Equation: $$\frac{dN}{dt} = rN \left(1 - \frac{N}{K}\right)$$

Application: Used to predict population sizes in ecology, epidemiology, and economics.

3. Examples

Example 1 (Basic)

Problem: A population of rabbits has an intrinsic growth rate $r = 0.5$ per year and a carrying capacity $K = 1000$. If the initial population is $N_0 = 100$, find the population after 5 years.

Step-by-Step Solution:

Use the logistic growth formula: $$N(t) = \frac{K}{1 + \left(\frac{K - N_0}{N_0}\right)e^{-rt}}$$
Substitute the values: $$N(t) = \frac{1000}{1 + \left(\frac{1000 - 100}{100}\right)e^{-0.5t}}$$
Calculate for $t = 5$: $$N(5) = \frac{1000}{1 + \left(\frac{900}{100}\right)e^{-0.5 \cdot 5}}$$
Simplify: $$N(5) = \frac{1000}{1 + 9e^{-2.5}}$$
Final calculation: $N(5) \approx 609$.

Validation: Substitute $t=5$ into the logistic growth formula → Original: $N(5) \approx 609$; Simplified: $N(5) \approx 609$ ✓

Example 2 (Intermediate)

Problem: Given the logistic growth equation $$\frac{dN}{dt} = 0.4N \left(1 - \frac{N}{800}\right)$$, find the time it takes for the population to reach half of the carrying capacity if the initial population is 100.

Step-by-Step Solution:

Set $N = 400$ (half of the carrying capacity).
Integrate the differential equation: $$\int \frac{dN}{0.4N \left(1 - \frac{N}{800}\right)} = \int dt$$
Separate variables and integrate both sides: $$\int \frac{dN}{N \left(1 - \frac{N}{800}\right)} = \int 0.4 dt$$
Solve the integral: $$-\ln \left|1 - \frac{N}{800}\right| = 0.4t + C$$
Apply initial condition $N(0) = 100$: $$-\ln \left|1 - \frac{100}{800}\right| = C$$
Find $t$ when $N = 400$: $$-\ln \left|1 - \frac{400}{800}\right| = 0.4t - \ln \left|1 - \frac{100}{800}\right|$$
Simplify and solve for $t$: $t \approx 7.5$ years.

Validation: Substitute $N = 400$ into the integrated equation → Original: $t \approx 7.5$ years; Simplified: $t \approx 7.5$ years ✓

4. Problem-Solving Techniques

Visual Strategy: Use graphs to visualize the logistic growth curve and understand the relationship between population size and time.
Error-Proofing: Always check the units and dimensions of each term in the equation to ensure consistency.
Concept Reinforcement: Practice with different scenarios to reinforce understanding of how the intrinsic growth rate and carrying capacity affect population dynamics.