I have literally posted the complete assignment information, can I please have some help with these two problems?

MAT 375 Module Two Guided Activity: (Continuous) Dynamical Systems Our textbook has a series of Exercises and Examples dealing with a model of whales. The textbook starts off with relatively simpler models and then adds features. This Guided Activity is part of Exercise 5 in Section 4.4 of your textbook. Exercise 5 is based on Example 4.2 which is in turn based on Exercise 1 in Section 2.4. The equations for the rates of change for B (B') and for F (F') are 3' = gB com/0B =rBB 3'65 13 aBF B+cB KB = 0.053 m 1 L 1081.err B + 3,000 150,000 and 0.08F 1 L 1083;: F + 15,000 400,000 These equations for B' and F' are the dynamical system we wish to use. Step 4: Solve the model. To find the equilibrium points, we solve the steady-state system in B and F (with both rates, 3' and F', being 0), which is a system of two variables with two unknowns. o =.053 w 1- L 10'8BF B + 3,000 150,000 0=.08F 1; 10_SBF F + 15,000 400,000 It is often very difficult/ impossible to find closed form solutions to these types of problems, i.e. to be able to write a formula for B(t) and Flt). Instead, we can use software such as Sage worksheets in CoCalc to calculate specific values of B(t) and F(t) for example equilibrium points. Step 43: Software information setting up a CoCalc / SageMath worksheet This is how to set up a solution using a SageMath worksheet in CoCalc. 1. Log into www.cocalc.com creating an account if you don't have one yet. 2. In Projects, click on "Create New Project". Give it a title such as \"Module 2 Guided Activity" Click on \"Create Project\" When the project appears in the project list, click on the project name When you get the message \"No files found\# Set up the variables and parameters B,F = var('B, F') rB=.05 CB=3000 KB=150000 IF=.08 CF=15000 KF=400000 alpha=10^(-8) compB=alpha*B*F compF=alpha*B*F The next few lines set up the equations for B' and F', first calculating gs and gF., continuing to follow the set-up described in Step 3. # Calculate the derivatives gB=rB*B*(B-CB)/(B+CB)*(1-B/KB) Bprime=gB-compB gF=rF*F*(F-CF)/(F+CF)*(1-F/KF) Fprime=gF-compF Now that the functions are set up, we can use CoCalc / Sage to solve the equations B' = 0 and F' = 0. Note that the script requires a double equals sign, or Bprime==0 and Fprime==0. Typing in Bprime=0 with one equals sign will replace the value of Bprime with 0. # Solve the equations solve((Bprime==0,Fprime==0), B,F) Run After entering this entire Sage script into CoCalc, we click Run and get# Set up the variables and parameters B, F = var('B, F' ) rB=. 05 CB=3000 KB=150000 rF=. 08 CF=15000 KF=400000 alpha=10^ (-8) 10 compB=alpha*B*F 11 compF=alpha*B*F 12 13 # Calculate the derivatives 14 gB=rB*B*(B-CB) / (B+CB) *(1-B/KB) 15 Bprime=gB-compB 16 gF=rF*F*(F-CF)/ (F+CF)*(1-F/KF) 17 Fprime=gF-compF 18 19 # Solve the equations 20 solve( (Bprime==0, Fprime==0) , B, F) [ [B == 0, F == 0], [B == 3000, F == 0], [B == 150000, F == 0], [B == 0, F == 400000], [B == 0, F == 15000], [B == 3018. 440677966102, F == 15011.76470588235], [B == 149513.0, F == 15594.99074074074], [B == 137698. 3636363636, F == 392568.0], [B == 3535. 178571428572, F == 399809.5]] The output marked with a green line is the solution to the "solve" command. It lists 9 solutions B = 0, F = 0 . B = 3,000, F = 0 . B = 150,000, F = 0 Etc. Can the two species coexist? Out of the 9 solutions, in 5 of them either B is 0 or F = 0. In those equilibrium points, there are either no blue whales or no fin whales so that the two types of whales do not coexist. We have 4 other equilibrium points where both B and F are not zero. Weneed to analyze whether any of these equilibrium points are stable. If one or more is stable, then the two species can coexist. This completes Step 4: We now know the equilibrium points. Solution to Question b: CoCalc / Sage shows that there are 9 equilibrium points. In one of those, both species are extinct (B = 0, F = 0). In 4 of them, one of the species is extinct (B = O or F = 0). In the remaining 4, the two species could possibly coexist. The following are some vector fields drawn using CoCaIc / Sage. The script begins right after the previous \"solve" step. if Draw the vector field for the entire range from B = 0 to B = 15,000, F = 0 to F = 400,000. plot_vector_field((Bprime(B,F),Fprime(B,F)),(B,-5,150005),(F,-5,400005)) # Draw the vector field for the entire range from B = 0 to B = 15,666, F = e to F = 466,099. plot_vector_fie1d((Bprime(B, F), Fprime(B, F) ), (B, -5, 159065), (F, -5,490695)) i f a ' ' \\ \\\\\\\\\\\\\\\\\\\\\\\\\\\\ * ' ' ' \\ \\\\\\\\\\\\\\\\\\\\\\\\\\\\ ' ' ' ' ' \\ \\\\\\\\\\\\\\\\\\\\\\\\' ' ' ' ' ' \\ \\\\\\\\\\\\\\\\\\\\\\\\\" ' ' ' ' \\ \\\\\\\\\\\\\\\\\\\\\\\\\" ' ' ' ' \\ \\\\\\\\\\\\\\\\\\\\\\\\' ' ' ' ' ' \\ \\\\\\\\\\\\\\\\\\\\\\\\' ' ' ' ' ' \\\\\\\\\\\\\\\\\\\\\\\\\\' ' ' ' ' ' \\ \\\\\\\\\\\\\\\\\\\\\\\\' " ' ' ' '\\\\\\\\\\\\\\\\\\\\\\\\' ' ' ' ' ' ' \\\\\\\\\\\\\\\\\\\\\\\\' ' ' ' ' ' \\ \\\\\\\\\\\\\\\\\\\\\\\\' ' ' ' ' ' \\ \\\\\\\\\\\\\\\\\\\\\\\\' ' ' ' ' ' \\ \\\\\\\\\\\\\\\\\\\\\\\\' ' ' ' ' ' \\ \\\\\\\\\\\\\\\\\\\\\\~u ' ' ' ' ' \\\\\\\\\\\\\\'ms\"' ' ' ' ' {'5'va .v'i'W/l'4' - In this plot, the xaxis is associated with the B variable (the blue whale population) and the y-axis is associated with the F variable (the fin whale population). You can see that if the initial population is zero on either axis, then it stays zero, but othenNise, the vectors point toward both reaching their expected maximum population. However, this plot of the vector field is not detailed enough to analyze each of the 9 equilibrium points found when solving the steady state system. For that analysis, we need to zoom in on a small region around each equilibrium point. The following shows the vector field around 4 of the 9 equilibrium points. You can plot the others to verify the summary table below. We can examine the vector field around B = 150000 and F =0 by plotting B values between 140,000 and 160,000 and F values between -500 and 500. This is the case when the blue whales are at their maximum and the fin whales are extinct. We get the following vector field: if Draw the vector field around B = 150000 and F = 0 if Use the range from B = 140,000 to B = 160,000, F = -500 to F = 500 plot_vector_field((Bprime(B,F),Fprime(B,F)),(B,140000,150000),(F,500,500)) # Draw the vector field around B = 150000 and F = 0 # Use the range from B = 140,000 to B = 160,000, F = -500 to F = 500 plot_vector_field((Bprime(B, F), Fprime(B, F) ), (B, 140000, 160000), (F, -500, 500) ) 400 200 -200 -400 140000 145000 150000 155000 160000 We see that this is a stable equilibrium point since the arrows point towards the equilibrium point in all directions. We can examine the vector field around B = 3018 and F =15,002 by plotting B values between 2900 and 3100 and F values between 14,000 and 16,000. This is the first coexistence equilibrium point found above. We get the following vector field: # Draw the vector field around B = 3018 and F =15,002 # Use the range from B = 2,900 to B = 3,100, F = 14,000 to F = 16,000plot_vector_field((Bprime(B,F), Fprime(B,F)),(B,2900,3100), (F,14000,16000)) # Draw the vector field around B = 3018 and F =15,002 # Use the range from B = 2,900 to B = 3, 100, F = 14,000 to F = 16,000 plot_vector_field( (Bprime(B, F) , Fprime (B, F) ), (B, 2900, 3100), (F, 14000, 16000) ) 16000 11117 HILL .1 117 15500 15000 14500 14000 - 2900 2950 3000 3050 3100 We see that this is an unstable equilibrium point since the arrows point away from the equilibrium point in some (actually all) directions. We can examine the vector field around B = 149,513 and F = 15,595 by plotting B values between 140,000 and 160,000 and F values between 10,000 and 20,000. We get the following vector field: # Draw the vector field around B = 149,513 and F = 15,595# Use the range from B = 140,000 to B = 160,000, F = 10,000 to F = 20,000 plot_vector_field((Bprime(B,F),Fprime(B,F)),(B,140000,160000),(F,10000,20000)) # Draw the vector field around B = 149,513 and F = 15,595 # Use the range from B = 140,000 to B = 160,000, F = 10,000 to F = 20,000 plot_vector_field( (Bprime(B, F), Fprime(B, F) ), (B, 140000, 160000), (F, 10000, 20000) ) 20000 18000 16000 1011111 1 1 1111/1/ 14000 12000 10000 - 140000 145000 150000 155000 160000 We see that this is an unstable equilibrium point since although the arrows in the horizontal (B) direction point towards the equilibrium point, the arrows in the vertical direction (F) point away.Section 2.4 Exercise 1. Ecologists use the following model to represent the growth process of two competing species, x and y dx [ x] =r1x 1 0ilxy dt K1 dy y =r 1 (x x dt 2% K2] 2 y The variables x and y represent the number in each population; the parameters r; represent the intrinsic growth rates of each species; K; represents the maximum sustainable population in the absence of competition; and a,- represents the effects of competition. Studies of the blue whale and fin whale populations have determined the following parameter values (t in years) Blue Fin r 0.05 0.08 K 150,000 400,000 01 10" (8) 10"(8) (The Exercise continues by asking several questions about population levels.) Instructor's Note: Instead ofx and y, we will use the variable B for blue whales and F for fin whales. It's easier to remember! We can examine the vector field around B = 137,698 and F =392,568 by plotting B values between 130,000 and 150,000 and F values between 380,000 and 420,000. We get the following vector field: if Draw the vector field around B = 137,698 and F = 392,568 if Use the range from B = 130,000 to B = 150,000, F = 380,000 to F = 420,000 plot_vector_field((Bprime(B,F),Fprime(B,F)),(B,130000,150000),(F,380000,420000)) # Draw the vector field around B = 137, 698 and F = 392,568 # Use the range from B = 130,000 to B = 150,000, F = 380,000 to F = 420,000 plot_vector_field((Bprime(B, F), Fprime(B, F) ), (B, 130000, 150000), (F, 380000, 420000)) 420000 - 415000 410000 405000 400000 395000 AA AAAA A 390000 385000 380000 130000 135000 140000 145000 150000 We see that this is a stable equilibrium point since the arrows in all directions point towards the equilibrium point. We can continue and draw the vector fields around the other 5 equilibrium points. A summary of the equilibrium points is in the table below.Blue Whale Population (B) Fin Whale Population (F) Stable or Unstable? 0 0 Stable 3,000 0 Unstable 150,000 0 Stable 0 400,000 Stable 0 15,000 Unstable 3,018 15,012 Unstable 149,513 15,595 Unstable 137,698 392,568 Stable 3,535 399,809 Unstable There is only one stable equilibrium point where blue whales and fin whales coexist. We have solved the system using graphical, tabular, and algorithmic methods, so we move on to Step 5: Answering our initial question. Step 5: Answer the question. Our initial question was \"Can the two species coexist?\" From our solution, we see that it is possible for them to coexist, assuming that both species exceed their minimum survival level for the species. Solution to Question c: Let us examine the vector field close to B = 5,000 and F = 70,000. # Draw the vector field around B = 5,000 and F = 70,000 WNA # Use the range from B = 3,000 to B = 7,000, F = 50,000 to F = 150,000 plot_vector_field( (Bprime(B, F) , Fprime(B, F) ), (B, 3000, 7000), (F, 50000, 150000) ) 1.4e5 1.2e5 1e5 8e4 - 6e4 3000 3500 4000 4500 5000 5500 6000 6500 7000 We see that the arrows point towards increases in both B and F, suggesting that neither species will become extinct. There are four possibilities when they coexist, but due to stability, the whale populations will converge to 137,698 blue whales and 392,568 fin whales - the only stable coexistence equilibrium point.Part 2 (Independent Practice) Complete Problem 6 from section 4.4 of your textbook. Your solution should be similar in scope to the example in this Module Two Guided Activity document, taking care to account for all steps in the five-step model. You are welcome to first try solving the textbook problem on your own, but I am also going to include some steps and hints to help walk you through this portion as well in case you get stuck. Note: There are multiple ways to label these variables. 0 A classical mathematical way is to let x1 be the number of blue whales and x; be the number of fin whales. The textbook uses this. 0 For a specific application, it is often easier to name variables to be easier to remember. In our scripts, we use B for the number of blue whales and F for the number of fin whales. Section 4.4 Exercise 6. Reconsider the whale problem of Example 4.2, and assume that a = 10'\". In this problem we will investigate the effects of harvesting on the two whale populations. Assume that a level of effort E boat days will result in the annual harvest of qu1 blue whales and quZ fin whales, where the parameter q (catchability) is assumed to be approximately 105. (a) Under what conditions can both species continue to coexist in the presence of harvesting? Use the five-step method, and model as a dynamical system in steady state. (b) Draw the vector field for this problem, assuming that the conditions identified in part (a) are satisfied. (c) Find the minimum level of effort required to reduce the fin whale population to its current level of around 70,000 whales. Assume that we started out with 150,000 blue whales and 400,000 fin whales before mankind began to harvest them. (cl) Describe what would happen to the two populations if harvesting were allowed to continue at the level of effort identified in part (c). Draw the vector field in this case. This is the situation which lead the IWC to call for an international ban on whaling. Task A: Complete Part (a). Here are some hints. The two equations for x1' (i.e. B') and x2' (i.e. F') now need to add a harvesting factor, subtracting q Ex1 (i.e. q E B) from x1' (i.e. B') and q E x2 (i.e. q E F) from x2' (i.e. F'). B' = 0.05 B (1 - B / 150,000) - 10-8 BF - 10 5 EB F' = 0.08 F (1 - F / 400,000) - 10-8 BF - 10-5 EF To find equilibrium points, we need to solve B' = 0 and F' = 0. We can write B' = B [ .05 (1 - B / 150,000) - 10-8 F - 10 5 6 ] so that if B' = 0, then either B = 0 or .05 (1 - B / 150,000) - 10-8 F- 10" E = 0. We can get a second equation in the same way from F' = 0. The two equations to solve are 0.05 / 150,000 B + 10-8 F = 0.05 - 10-5 E 0.08 / 400,000 F - 10-8 B = 0.08 - 10-5 E This is a system of linear equations in B and F, so it can be solved exactly (in terms of E). If we keep E as an undefined quantity in a Sage script in CoCalc, CoCalc / SageMath can solve these equations in terms of a variable (not value) E! We can use the following script, setting E as a variable instead of as a parameter. B,F,E = var('B, F, E ') rB=.05 KB=150000 rF=.08 KF=400000 gB=rB*B*(1-B/KB)\f# Include a harvesting factor E B, F, E = var( 'B, F, E ' ) rB=. 05 KB=150000 rF= . 08 6 KF=400000 gB=rB*B*(1-B/KB) 8 gF=rF*F*(1-F/KF) alpha=10^ (-8) 10 compB=alpha*B*F 11 compF=alpha*B*F 12 Bprime=gB-compB 13 Fprime=gF-compF 14 solve( (Bprime==0, Fprime==0) , B, F) 15 q=10^(-5) 16 harvestB=q*E*B 17 harvestF=q*[*F 18 Bharvprime=Bprime-harvestB 19 Fharvprime=Fprime-harvestF solve( (Bharvprime==0, Fharvprime==0) , B, F) [[B == 0, F == 0], [B == 150000, F == 0], [B == 0, F == 400000], [B == (276000000/1997), F == (785000000/1997) ]] [ [B == 0, F == 0], [B == -30*E + 150000, F == 0], [B == 0, F == -50*E + 400000], [B == -57000/1997*E + 276000000/1997, F == -97000/1997*E + 785000000/1997]] The stable coexistence equilibrium point that contains the variable E is the last one listed above: B = -57,000/1997 E + 276,000,000 / 1997 F =-97,000/1997 E + 785,000,000 / 1997 Now use the fact that B > 0 and F > 0 to create two inequalities above; solve each inequality for E. This will give us two bounds for E (one for each inequality). What are the two bounds? Thinking about what these bounds mean, which of the two values will be our upper bound for E for coexisting populations?Task B: Complete Part (b). Here are some hints. Let's try two values for E, E = 3,000 and E = 7,000. The value of E = 3,000 should lead to an equilibrium point where the two species coexist. The value of E = 7,000 should not lead to any equilibrium points where the two species coexist (based on the bounds from Part a / Task A). We can use the following Sage script in CoCalc for the E = 3,000 case. To analyze the E = 7,000 case, you only need to change the one line E=3000 to E=7000. B,F = var('B, F ') rB=.05 KB=150000 rF=.08 KF=400000 gB=rB*B*(1-B/KB) gF=rF*F*(1-F/KF) alpha=10^(-8) compB=alpha*B*F compF=alpha*B*F Bprime=gB-compB Fprime=gF-compF solve((Bprime==0,Fprime==0), B,F) (Instruction for the student - click on Run after typing in this line) q=10^(-5) E=3000 harvestB=q*[*B harvestF=q*[*F Bharvprime=Bprime-harvestB Fharvprime=Fprime-harvestF solve((Bharvprime ==0, Fharvprime ==0),B, F) (Instruction for the student - click on Run after typing in this line)To complete the analysis, we can continue as follows: (Instruction for the student enter equilibrium values from the previous step). Enter these two numbers below. EquilB=Equi|ibrium value for B EquilF=Equi|ibrium value for F EquilB.n(),Equi|F.n() (Instruction for the student choose a range for B, say LowB to HighB, that contains EquilB. Choose a range for F, say LowF to HighF, that contains EquilF). Enter these four numbers below. plot_vector_field((Bharvprime, Fharvprime),(B,LowB,HighB), (F, LowF,HighF)) (Instruction for the student click on Run after typing in this line. This vector eld will show you whether the equilibrium point is stable or not. The difference between this script and the one above is that E is no longer a variable but a specified parameter. To display the calculations along the way (for solve((Bprime==0,Fprime==0),B,F) and solve((Bharvprime ==0, Fharvprime ==0),B,F)), you need to input the script just to that point and click \"Run" without typing in all the script. If you type in the entire script and then click "Run\Example 4.2. The blue whale and fin whale are two similar species that inhabit the same areas. Hence, they are thought to compete. The intrinsic growth rate of each species is estimated at 5% per year for the blue whale and 8% per year for the fin whale. The environmental carrying capacity (the maximum number of whales that the environment can support) is estimated at 150,000 blues and 400,000 fins. The extent to which the whales compete is unknown. In the last 100 years intense harvesting has reduced the whale population to around 5,000 blues and 70,000 fins. Will the blue whale become extinct? Section 4.4 Exercise 5. In the whale problem of Example 4.2 we used a logistic model of population growth, where the growth rate of population P in the absence of interspecies competition is p g(P) rP[1- E) In this problem we will be using a more complex model P-c P MP) _rP[P+c][1 E] in which the parameter c represents a minimum viable population level below which the growth rate is negative. Assume that the minimum viable population level is 3,000 for blue whales and 15,000 for fin whales. a) Can the two species of whales coexist? Use the five-step method, and model as a dynamical system in steady state. b) Sketch the vector field for this model. Classify each equilibrium point as stable or unstable. c) Assuming that there are currently 5,000 blue whales and 70,000 fin whales, what does this model predict about the future of the two populations. d) Suppose that we have underestimated the minimum viable population for the blue whale, and that it is actually closer to 10,000. Now what happens to the two species? Instructor's Note: This can be summarized as Blue Fin 0.05 0.08 150,000 400,000 10" (-8) 10" (-8) 3,000 15,000 Part 1, Instruction, provides 0 A CoCalc / SageMath based solution for Exercise 5 parts a, b, and c in Section 4.4. Part 2, Independent Practice, consists of o A CoCaIc / SageMath based solution for Exercise 6 parts a, b, and c in Section 4.4. Part 1 (Instruction) In this module, we are going to investigate dynamical systems. It is important to build on what we have done before, so we are going to work through an example problem together, emphasizing the key concepts, and then you will build upon this on your own. In Module One, we analyzed discrete time dynamical systems. They are called discrete because time is measured in discrete increments (month 1, month 2, month 3, etc.). In Module Two, we analyze a continuous dynamical system (also called a gal dynamical system) where time is measured as a real number. Solutions to these continuous time models often use differential equations instead of spreadsheet recursions. Section 4.2 in the text discusses continuous dynamical systems while section 4.3 discusses discrete dynamical systems. In the readings, Example 4.2 was an elementary population dynamics problem. Read the problem and example carefully before moving forward with this activity. We are going to use the same scenario, but extend it together, and then you will extend in a different way on your own. Sce na rio In the whale problem of Example 4.2, we used a logistic model of population growth where the growth rate of population, P, in the absence of interspecies competition is 9(P) = rp 1 - - This logistic model builds in a maximum population K. Applied to blue whales (B) and fin whales (F), we get two equations in our logistic model B 9B = B B 1 - KB 9F = IF F |1 F KE To that, we added an interspecies competition factor resulting in a logistic model where the two variables impact each other. Using interspecies competition factors CB and CF with CB = CF = Q B F B' = gB -CB F' = gF - CF we get equations for the growth rates B' and F' of B'=rB B|1 B - a BF KB F'= IF F |1 - F -a BF KF In this Guided Activity, we will start with a more complex model for each species, a model that not just includes a maximum population K but also a minimum population c. The following growth function includes a maximum value K for the population and a minimum value c._ P'3 -5 g(P)_rP[P+c][1 K] We can add the interspecies competition factor to this model, getting the two growth equations In this Guided Activity, we will be using a more complex model Pc P atP)rPlp..ll1-;) Here the parameter c represents a minimum viable population level below which the growth rate is negative. Assume that the minimum viable population levels are 3,000 for blue whales and 15,000 for fin whales. Assume that a :10s for the interspecies competition factor. a) Can the two species of whales coexist? Use the five-step method and model this scenario as a dynamical system in steady state to determine whether there are any possible coexistence equilibrium points. b) Sketch the vector field for this model. Classify each equilibrium point as stable or unstable. Conclude whether the two species can or cannot coexist. c) Assuming there are currently 5,000 blue whales and 70,000 fin whales, what does this model predict about the future of these populations? Solution to Question a: Step 1: Determine what question is to be answered. In this problem, we have a few things that the problem is asking for. This is a bit more formulaic than most real-world problems, but it does guide us in determining what can be answered here. First, we must determine whether or not the species of whales can coexist. To coexist, it means both species survive long term. Essentially, we are trying to determine whether one species will go extinct in the presence of the other. We are given initial values for the whale population and minimum capacity for survival. So our problem is... Can the two whale species coexist? Our state variables for this equation are: B= Population of blue whales F= Population of fin whales State variables are the key variables for the system, which can often be thought of as the measurable variables. Here we are investigating population, so it makes sense that the state variables are the population of the two types of whales. There are some other variables that are important, but not necessarily state variables. gs: Growth rate of blue whale population (per year) gF= Growth rate of fin whale population (per year) CB: Competition factor for blue whale population (per year) CF: Competition factor for fin whale population (per year) Using the same values for the parameters values from Exercise 5 in Section 4.4, the assumptions for our model are B , B 93 = _05 3 1 _ _ B + 3,000 150,000 F 15,000 F g, = .08F 1 F + 15,000 400,000 CB=CF=CBF BBQFEO Next we move on to Step 2: Choose your modeling technique. Step 2: Choose your modeling technique Here, initially, we want to model the situation as a steady-state problem. Steady-state problems imply that the derivative of the total population function is zero, i.e. that the population is not changing. When we are investigating the total population change at any point, common sense tells us that the population change will be the growth minus the competition factor, if we do not account for whale death rates. In this model, we are not accounting for death rates, although if we were, that competition factor would be integrated into the death rate and would be subtracted from the growth rate. Steady state will determine at what point these population change rates are not changing, so it gives us an equilibrium point. Note that we are just using this steady-state modeling approach to determine equilibrium points. Once we have found the equilibrium points, we will next model the system as a dynamical system, which allows us to use tools such as vector fields to determine the long-term behavior of the system. So you can essentially consider steady-state systems a specific case of the dynamical system. Now that we have decided our approach to modeling this system, we next move to Step 3: Formulate the model. Step 3: Formulate the model We are modeling this as a dynamical system (x1, x2) or (x, y), where our x1 (or x) = B and x; (or y) = F. Our system space would then be S = { (B, F): B 2 0, F 2 0}. The equations in the initial variables B and F and their associated parameters r, c, and K would be: I"); = 0.05 Cs = 3,000 KB = 150,000 rF = 0.08 CF = 15,000 KF = 400,000 Because this model uses the letter c for both the minimum viable population levels and the competition factors, we will use comps and compp here instead for the two competition factors. compa=aB F compp=a BF 11:108 Consider Guided Activity 2, Part 2, Task A: After finding the stable equilibrium point, we have that B = 57,000/1997*E + 276,000,000/1997. Using this and the fact that B > 0, set up an inequality, and solve for E. E is less than what value? Round your answer to the nearest whole number. Your Answer: Q Answer Consider Guided Activity 2, Part 2, Task A: After finding the stable equilibrium point, we have that F = 97,000/1997*E + 785,000,000/1997. Using this and the fact that F > 0, set up an inequality, and solve for E. E is less than what value? Round your answer to the nearest whole number. Your