In a pristine lake with carrying capacity K and fishing allowed, the logistic differential equation for the population N(t) of fish at time t(in days) is dN/dt= rN(1-(N/K) - H.
H= the fish harvested from the lake each day.
If H= 3/16rK find all of the equilibria for the differential equation.
*I got (1/4) and (3/4).
What will happen to the fish population after a long time if the current population equals half of the carrying capacity?
*Do I solve the differential here?
If the rate of fish harvested is increased from H=3/16 rK fish per day to a larger number, how will the curve that you drew be affected ?
*The graph(dN/dt vs N) I drew is an upside down parabola shape that goes through the x-axis at (1/4) and (3/4)
If H is large enough, then the fish populations is guaranteed to go extinct. Find the smallest value of H for which the fish population is guaranteed to go extinct.
*No idea how to do this question
ummm could you possibly walk me through on how i would solve the DE...?
Thank you for your response!
yesh - why don't we get part B at this point. Solve the DE.
Also, you might google logistic growth to see how the other parameters are used.
To find the equilibria for the differential equation, we can set dN/dt equal to zero and solve for N. In this case, the equation is:
dN/dt = rN(1 - N/K) - H = 0
Substituting H = 3/16rK, we have:
rN(1 - N/K) - 3/16rK = 0
Simplifying further, we get:
N(K - N) - 3/16K = 0
Expanding, we have:
NK - N^2 - 3/16K = 0
Rearranging the equation, we get:
-N^2 + NK - 3/16K = 0
To solve for N, we can factorize the equation:
-(N - 1/4K)(N - 3/4K) = 0
Setting each factor equal to zero:
N - 1/4K = 0 --> N = 1/4K
N - 3/4K = 0 --> N = 3/4K
So the equilibria for the differential equation are N = 1/4K and N = 3/4K.
To determine what happens to the fish population after a long time if the current population equals half of the carrying capacity, we can substitute N = 1/2K into the differential equation and solve:
dN/dt = rN(1 - N/K) - H
Substituting N = 1/2K, we get:
dN/dt = r(1/2)K(1 - (1/2K)/K) - H
Simplifying, we have:
dN/dt = r(1/2)K - H
Since this equation does not depend on N, it means that the population will remain constant over time and will not change.
If the rate of fish harvested is increased from H = 3/16rK fish per day to a larger number, the curve that you drew (the graph of dN/dt vs N) will be affected by shifting downward. This is because a larger value of H represents a greater rate of fish being harvested, leading to a decrease in the population growth rate.
To find the smallest value of H for which the fish population is guaranteed to go extinct, we need to find the point where the fish population reaches zero. This occurs when dN/dt is negative for all values of N.
Since dN/dt = rN(1 - N/K) - H, we can set dN/dt equal to zero and solve for N:
rN(1 - N/K) - H = 0
Simplifying further, we have:
N(K - N) - HK = 0
Expanding, we get:
NK - N^2 - HK = 0
Rearranging, we have:
N^2 - (NK + HK) + 0 = 0
Using the quadratic formula, N will have real roots if the discriminant (b^2 - 4ac) is non-negative:
(NK + HK)^2 - 4(1)(0) >= 0
Simplifying, we get:
N^2K^2 + 2(NHK) + H^2K^2 >= 0
Since K, N, and H are positive values, this inequality is always true. Therefore, there is no smallest value of H for which the fish population is guaranteed to go extinct.