1. Lesson 9-5 Logistic Equations

2. Logistic Equation We assume P(t) is constrained by limited resources so : Logistic differential equation for population growth: 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 if P is small dP ----- ≈ kP dt dP ----- ≈ kP (1 – P/K) dt

3. Example 1a Find the solution to the initial value problem dP/dt = 0.01P(1 – P/500) and P(1980) = 200,, where t is given in years since 1980. Hint: Identify P 0 and K; find A. P(t) = K / (1 + A e –kt ) where A = (K – P 0 ) / P 0 Differential Equation: k=0.01; K = 500; A = (500 – 200)/200 = 3/2 P(0)= 200 Equation: t is years since 1980 P(t) = 500 / (1 + (1.5)e -0.01t ) dP ----- ≈ kP (1 – P/K) dt

4. Example 1b Use the equation to find the population in 1985. Equation: t is years since 1980 P(t) = 500 / (1 + (1.5)e -0.01t ) P(5) = 500 / (1 + (1.5)e -0.01(5) ) = 500 /

5. Example 1c When does the population reach 450? Equation: t is years since 1980 P(t) = 500 / (1 + (1.5)e -0.01t ) 450 = 500 / (1 + (1.5)e -0.01t ) (1 + (1.5)e -0.01t ) = 500/450 1.5 e -0.01t = 1/9

6. 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? P = p 0 e ±kt General Equation: decay k < 0; p 0 = 10 grams P(5730)= 5 = 10e 5730k 0.5 = e 5730k ln 0.5 = ln e 5730k = 5730k ln 2 -1 / 5730 = k -(ln 2)/5730 = k = -0.000121 Equation: Use to solve for k P(2000) = 10e 2000(-0.000121) = 7.851 grams

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