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Logistic Growth & Phase Line
AP Calculus BC · Unit 7 Differential Equations
Learning Objectives
Analyze dP/dt = kP(1 − P/K)
Identify carrying capacity K
Locate max growth rate at P = K/2
Read the Phase Line (equilibria & stability)
Equation Properties
ODE:
dP/dt = kP(1 − P/K)
Growth Rate vs P:
Parabola opening downward.
Roots at P=0 and P=K.
Vertex at P = K/2 (Max Growth).
Phase Line:
P=0 is unstable, P=K is stable.
Tags
logistic
differential equation
phase line
Growth Rate k
0.5
Capacity K_max
1000
Initial Pop P₀
100
Max Growth Rate Location
P =
500
(K/2)
Equilibrium Points
P=0 (Unstable)
P=
1000
(Stable)
Logistic Solution P(t)
P(t) = K / (1 + [(K−P₀)/P₀] e⁻ᵏᵗ)
Top Graph:
Solution curve P(t) over time. Notice the inflection point at P = K/2.
Bottom Graph:
Growth rate dP/dt vs P. Note the parabolic shape confirming dP/dt is maximum exactly halfway to capacity.