1 / 7

Lesson 9-5

Lesson 9-5. Logistic Equations. Logistic Equation. dP ----- ≈ kP dt. if P is small. We assume P(t) is constrained by limited resources so : Logistic differential equation for population growth:

sancha
Download Presentation

Lesson 9-5

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Lesson 9-5 Logistic Equations

  2. Logistic Equation dP ----- ≈ kP dt if P is small 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 – P0) / P0 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 P0 and K; find A. P(t) = K / (1 + A e–kt) where A = (K – P0) / P0 Equation: P(0)= 200 t is years since 1980 dP ----- ≈ kP (1 – P/K) dt Differential Equation: k=0.01; K = 500; A = (500 – 200)/200 = 3/2 P(t) = 500 / (1 + (1.5)e-0.01t )

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

  5. Example 1c When does the population reach 450? P(t) = 500 / (1 + (1.5)e-0.01t ) t is years since 1980 Equation: 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 = p0e±kt General Equation: decay k < 0; p0 = 10 grams P(5730)= 5 = 10e5730k 0.5 = e5730k ln 0.5 = ln e5730k = 5730k ln 2-1 / 5730 = k -(ln 2)/5730 = k = -0.000121 Use to solve for k Equation: P(2000) = 10e2000(-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

More Related