Lesson 9 5 Logistic Equations Logistic Equation We

Lesson 9 -5 Logistic Equations

Logistic Equation We assume P(t) is constrained by limited resources so : d. P ----- ≈ k. P if P is small dt Logistic differential equation for population growth: d. P ----dt ≈ k. P (1 – P/K) where small k is proportional constant of growth and big K is the population carry capacity (via resource restrictions) so if P > K the population decreases and if P < K the population increases Its solution is P(t) = K / (1 + A e–kt) where A = (K – P 0) / P 0

Example 1 a Find the solution to the initial value problem d. P/dt = 0. 01 P(1 – P/500) and P(1980) = 200, , where t is given in years since 1980. Hint: Identify P 0 and K; find A. Equation: P(t) = K / (1 + A e–kt) where A = (K – P 0) / P 0 P(0)= 200 Differential Equation: d. P ----dt ≈ k. P (1 – P/K) t is years since 1980 k=0. 01; K = 500; A = (500 – 200)/200 = 3/2 P(t) = 500 / (1 + (1. 5)e-0. 01 t )

Example 1 b Use the equation to find the population in 1985. Equation: P(t) = 500 / (1 + (1. 5)e-0. 01 t ) P(5) = 500 / (1 + (1. 5)e-0. 01(5) ) = 500 / t is years since 1980

Example 1 c When does the population reach 450? Equation: P(t) = 500 / (1 + (1. 5)e-0. 01 t ) 450 = 500 / (1 + (1. 5)e-0. 01 t ) = 500/450 1. 5 e-0. 01 t = 1/9 t is years since 1980

Example 2 A biologist stocks a shrimp farm pond with 1000 shrimp. The number of shrimp double in one year and the pond has a carrying capacity of 10, 000. How long does it take the population to reach 99% of the pond’s capacity? General Equation: decay k < 0; p 0 = 10 grams P = p 0 e±kt Equation: Use to solve for k P(5730)= 5 = 10 e 5730 k 0. 5 = e 5730 k ln 0. 5 = ln e 5730 k = 5730 k ln 2 -1 / 5730 = k -(ln 2)/5730 = k = -0. 000121 P(2000) = 10 e 2000(-0. 000121) = 7. 851 grams

Summary & Homework • Summary: – Logistics Equations are another differential equation – Carrying capacity makes them unique • Homework: – pg 629 -631: 5, 6, 8