01 P 70 2 Phase Portrait 18 a The solution follows the logistic growth solution

01 p 70 2 phase portrait 18 a the solution follows

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01( P - 70) 2 ) Phase Portrait 18. a. The solution follows the logistic growth solution seen in 1. d. The solution is F ( t ) = 10 , 000 50 + 150 e - 0 . 4 t . b. This is a standard logistic growth model, so the equilibria are F e = 0 and 200 (thousand). Below is a sketch of the function with the phase portrait. The equilibrium F e = 0 is unstable, while the carrying capacity, F e = 200 (thousand), is a stable equilibrium. -50 0 50 100 150 200 250 -15 -10 -5 0 5 10 15 20 25 > > > > < < F 0 . 4 F ( 1 - F 200 ) Phase Portrait -50 0 50 100 150 200 250 -30 -25 -20 -15 -10 -5 0 5 10 < > > < < < F 0 . 4 F ( 1 - F 200 ) - 15 Phase Portrait Problem 18. b Problem 18. c c. With harvesting, the right hand side of the differential equation is written 0 . 4 F 1 - F 200 - 15 = - 0 . 002 F 2 + 0 . 4 F - 15 = - 0 . 002( F - 50)( F - 150) . It follows that the equilibria are F e = 50 and 150 (thousand). Above is a sketch of the function with the phase portrait. The equilibrium F e = 50 (thousand) is the critical number of fish needed
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to avoid extinction and this equilibrium is unstable. The carrying capacity, F e = 150 (thousand), is a stable equilibrium.
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