Lecture on Exponential Growth and Decay in Population Dynamics

Key Equation

• Equation: ( \frac{dP}{dt} = kP )
  • Describes population growth rate proportional to the population size.
  • Variables:
    • ( \frac{dP}{dt} ): Growth rate of the population at a given time.
    • ( k ): Relative growth rate.
    • ( P ): Size of the population at time ( t ).

Conceptual Understanding

• Example: Bacteria doubling
  • If initial bacteria population grows at 50 cells/hr, doubling the population leads to growth at 100 cells/hr.
  • Growth rate increases proportionally with population size.

Deriving the General Formula

• Start with ( \frac{dP}{dt} = kP ).
• Separate variables: ( \frac{1}{P} dP = k dt ).
• Integrate both sides:
  • ( \ln P = kt + C ).
• Exponentiate to solve for ( P ):
  • ( P = e^{kt+C} ).
  • Simplify: ( P = Ce^{kt} ), where ( C ) is a constant.
• Initial population: ( P_0 = C ).
• General formula: ( P(t) = P_0 e^{kt} ).

Application Example

• Rabbit population on an island, starting year: 2000
• Determine Relative Growth Rate:
  • Use initial condition: ( P_0 = 1500 ) (population in 2000).
  • Second data point (2001): ( P(1) = 1577 ).
  • Solve for ( k ):
    • ( \frac{1577}{1500} = e^k )
    • Natural log gives: ( k = \ln(1.0513) \approx 0.05 ).
• General Formula: ( P(t) = 1500e^{0.05t} ).
• Relative Growth Rate: 5% annually, continuously compounded.

Estimating Future Populations

• Estimate Population in 2010:
  • ( t = 10 ) (since 2000): ( P(10) = 1500e^{0.5} \approx 2473 ) rabbits.

Population Doubling Time

• Double from 1500 to 3000:
  • Set ( P(t) = 3000 ).
  • Solve: ( 2 = e^{0.05t} ).
  • ( \ln 2 = 0.05t ) implies ( t = \frac{\ln 2}{0.05} \approx 13.86 ) years.

The lecture concludes with methods and applications of exponential growth, particularly emphasizing the derivation and practical use of the exponential growth formula in predicting population changes.