Suppose a species of fish in a particular lake has a population that is modeled by the logistic population model with growth rate k, carrying capacity N, and time t measured in years.
Suppose the growth-rate parameter k=.3 and the carrying capacity N=2500. Suppose P(0)=2500.
(a) If 100 fish are harvested each year, what does the model predict for the long-term behavior of the fish population?
(b) What if one-third of the fish are harvested?
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so the logistic equation is dp/dt=.3(1-P/2500)*P
and for (a) you subtract 100
and for (b) you subtract P/3.
I can kinda guess at the answers but i'm not sure how to show my work...
a) The model predicts that the long-term behavior of the fish population will decrease over time, eventually reaching a steady-state population of approximately 2,400 fish.
b) The model predicts that the long-term behavior of the fish population will decrease over time, eventually reaching a steady-state population of approximately 2,333 fish.
To show your work for predicting the long-term behavior of the fish population using the logistic population model, you need to solve the differential equation dp/dt = 0.3(1 - P/2500)P, where P represents the population of fish at time t.
(a) If 100 fish are harvested each year, the model predicts that the population will decrease by 100 every year. Mathematically, this can be represented by subtracting 100 from the differential equation:
dp/dt = 0.3(1 - P/2500)P - 100
To determine the long-term behavior, you need to find the equilibrium solution of the differential equation. In this case, the equilibrium solution occurs when dp/dt = 0, meaning the population is neither increasing nor decreasing.
Setting dp/dt = 0 and solving for P gives:
0 = 0.3(1 - P/2500)P - 100
Simplifying the equation, you get:
0 = (1 - P/2500)P - 333.33
0 = P - P^2/2500 - 333.33
Combining like terms, you have:
P^2 - 2500P - 333.33 = 0
To solve this quadratic equation, you can use the quadratic formula:
P = (-b ± √(b^2 - 4ac))/(2a)
In this case, a = 1, b = -2500, and c = -333.33.
Plugging these values into the quadratic formula, you can calculate the potential equilibrium population values.
(b) If one-third of the fish are harvested, you need to subtract P/3 from the differential equation:
dp/dt = 0.3(1 - P/2500)P - P/3
To find the long-term behavior, you repeat the same process as in part (a) by finding the equilibrium solution of the differential equation. Set dp/dt = 0 and solve for P to obtain the equilibrium population values.
To answer these questions, we will use the given logistic equation dp/dt = 0.3(1 - P/2500) * P, where P represents the population of fish.
(a) If 100 fish are harvested each year, we can modify the equation by subtracting 100 from the population change per year. The new equation becomes dp/dt = 0.3(1 - P/2500) * P - 100.
To determine the long-term behavior of the fish population, we need to find the equilibrium points of the population (where the population remains constant). These occur when dp/dt = 0.
Setting dp/dt = 0, we get 0 = 0.3(1 - P/2500) * P - 100.
Simplifying the equation, we have 0 = 0.3P - P^2/2500 - 100.
Rearranging the equation, we get P^2 - 750P + 75000 = 0.
Solving this quadratic equation for P, we find two solutions: P ≈ 0 and P ≈ 750.
Since P cannot be negative, the only viable equilibrium point is P ≈ 750.
Therefore, the long-term behavior of the fish population when 100 fish are harvested each year is expected to stabilize around 750 fish.
(b) If one-third of the fish are harvested each year, we modify the equation by subtracting P/3 from the population change per year. The new equation becomes dp/dt = 0.3(1 - P/2500) * P - P/3.
Setting dp/dt = 0, we get 0 = 0.3(1 - P/2500) * P - P/3.
Simplifying the equation, we have 0 = 0.3P - P^2/2500 - P/3.
Rearranging the equation, we get 3P^2 - 2250P + 250000 = 0.
Solving this quadratic equation for P, we find two solutions: P ≈ 0 and P ≈ 750.
Since P cannot be negative, the only viable equilibrium point is P ≈ 750.
Therefore, the long-term behavior of the fish population when one-third of the fish are harvested each year is also expected to stabilize around 750 fish.
In summary:
(a) When 100 fish are harvested each year, the long-term behavior of the fish population is expected to stabilize around 750 fish.
(b) When one-third of the fish are harvested each year, the long-term behavior of the fish population is also expected to stabilize around 750 fish.