Logistic Growth – AP Calculus BC

Learning Objectives

The Logistic Equation: $\frac{dP}{dt} = kP(1 - \frac{P}{K})$ describes a population that grows exponentially at first but hits a defined limit (Carrying Capacity).

Equilibrium Solutions: Since $\frac{dP}{dt}$ depends only on $P$, there are two horizontal asymptotes: $P=0$ and $P=K$.

The Inflection Point: The growth rate $\frac{dP}{dt}$ is MAXIMIZED at exactly $P = \frac{K}{2}$! The population curve shifts from curving up to curving down at this threshold.

Tags

Diff EQLogistic GrowthPartial FractionsCarrying Capacity

DRAG P₀ UP/DOWN TO CHANGE INITIAL POPULATION

System Parameters

Carrying Cap ($K$) = 1000

Growth Rate ($k$) = 0.50

Interactive Analysis ($P_0$)

$\frac{dP}{dt} = kP(1 - \frac{P}{K})$

Initial Pop $P_0$ 200

TARGET INFLECTION ANALYSIS

Max Rate at $P =$ 500

Inflection Time $t =$ 2.77

LIMIT BEHAVIOR

$\lim_{t \to \infty} P(t) =$ 1000

Behavior: Drag $P_0$ above the carrying capacity $K$ and watch! The population cannot sustain itself, resulting in a negative derivative $\frac{dP}{dt} < 0$ that forces the population to exponentially decay down back to the $K$ asymptote!