Population Growth

6.5 Logistic Growth Model 100 80 60 Bears 40 20 0 20 40 Years 60 80 100 Greg Kelly, Hanford High School, Richland, Washington We have used the exponential growth equation to represent population growth. y  y0ekt The exponential growth equation occurs when the rate of growth is proportional to the amount present. If we use P to represent the population, the differential equation becomes: dP dt  kP The constant k is called the relative growth rate. dP / dt k P  The population growth model becomes: P  Pekt 0 However, real-life populations do not increase forever. There is some limiting factor such as food, living space or waste disposal. There is a maximum population, or carrying capacity, M. A more realistic model is the logistic growth model where growth rate is proportional to both the amount present ( P) and the fraction of the carrying capacity that remains: M P M  The equation then becomes: dP  M P  kP   dt  M  Our book writes it this way: Logistics Differential Equation dP k  P  M  P dt M We can solve this differential equation to find the logistics growth model.  Logistics Differential Equation dP k  P  M  P dt M 1 k dP  dt P  M  P M 1 1 1  k  P  M  P  dP  M dt M  ln P  ln  M  P  kt  C 1 A B   P  M  P P M  P 1  A M  P  BP Partial Fractions 1  AM  AP  BP 1  AM 1 A M 0   AP  BP AP  BP A B 1 B M P ln  kt  C M P  Logistics Differential Equation dP k  P  M  P dt M 1 k dP  dt P  M  P M 1 1 1  k  P  M  P  dP  M dt M  ln P  ln  M  P  kt  C P kt C e M P M  P kt C e P M 1  ekt C P M  1  ekt C P  P ln  kt  C M P Logistics Differential Equation P kt C e M P M  P kt C e P M 1  ekt C P M  1  ekt C P M P 1 ekt C P M 1 eC  ekt C Let A  e M P 1  Aekt  Logistics Growth Model M P  kt 1  Ae  Example: Logistic Growth Model Ten grizzly bears were introduced to a national park 10 years ago. There are 23 bears in the park at the present time. The park can support a maximum of 100 bears. Assuming a logistic growth model, when will the bear population reach 50? 75? 100?  Ten grizzly bears were introduced to a national park 10 years ago. There are 23 bears in the park at the present time. The park can support a maximum of 100 bears. Assuming a logistic growth model, when will the bear population reach 50? 75? 100? M P 1  Aekt M  100 P  10 0 P  23 10  M P  kt 1  Ae 100 10  0 1 Ae 100 10  1 A 10 10 A  100 10 A  90 A9 M  100 P  10 0 P  23 10 At time zero, the population is 10. 100 P  kt 1  9e  M P  kt 1  Ae M  100 P  10 0 P  23 10 100 23  1 9ek10 100 10k 1  9e  23 9e e 10k 100 P  kt 1  9e After 10 years, the population is 23. 10k  0.988913 k  0.098891 77  23 10k  0.371981 100 P 1 9e0.1t  100 P 0.1t 1 9e We can graph this equation and use “trace” to find the solutions. Bears 100 80 60 40 20 0 20 40 Years 60 80 100 y=50 at 22 years y=75 at 33 years y=100 at 75 years p