Problem 7 - Active Calculus

Question

The population of a species of fish in a lake is \(P(t)\) where \(P\) is measured in thousands of fish and \(t\) is measured in months. The growth of the population is described by the differential equation $$ \frac{d P}{d t}=f(P)=P(6-P) $$ a. Sketch a graph of \(f(P)=P(6-P)\) and use it to determine the equilibrium solutions and whether they are stable or unstable. Write a complete sentence that describes the long-term behavior of the fish population. b. Suppose now that the owners of the lake allow fishers to remove 1000 fish from the lake every month (remember that \(P(t)\) is measured in thousands of fish). Modify the differential equation to take this into account. Sketch the new graph of \(d P / d t\) versus P. Determine the new equilibrium solutions and decide whether they are stable or unstable. c. Given the situation in part (b), give a description of the long-term behavior of the fish population. d. Suppose that fishermen remove \(h\) thousand fish per month. How is the differential equation modified? e. What is the largest number of fish that can be removed per month without eliminating the fish population? If fish are removed at this maximum rate, what is the eventual population of fish?

Step-by-Step Solution

Answer

a. Equilibrium points are P=0 (unstable), P=6 (stable). b. New equilibria: approximately P ≈ 0.343 (unstable), 3.343 (stable). c. Long-term population ≈ 3.343 thousand fish. d. Modified equation: \( \frac{dP}{dt} = P(6-P) - h \). e. Maximum removal rate: 9 thousand fish per month; eventual population: 3 thousand fish.

Step 1: Graph the Function

Plot the graph of the function, which is defined as: $$ f(P) = P(6-P) $$ Note that this is a quadratic function that opens downward. Determine the roots by solving the equation: $$ P(6-P) = 0 $$ Solving this yields equilibrium points at: $$ P = 0 \text{ and } P = 6 $$

Step 2: Determine Stability

Examine the sign of the derivative, $$ f(P) = P(6-P) $$,- If $$ P < 0 $$ or $$ P > 6 $$, then $$ f(P) < 0 $$ which signifies a decline in population.- If $$ 0 < P < 6 $$, then $$ f(P) > 0 $$ which signifies a growth in population.Hence, P = 0 is unstable and P = 6 is stable.

Step 3: Long-term Behavior

The fish population will eventually stabilize at the equilibrium point where P = 6 thousand fish.

Step 4: Modify Differential Equation with Fishing

When fishers remove 1000 fish per month, the new differential equation becomes: $$ \frac{d P}{d t} = P(6-P) - 1 $$

Step 5: Graph the Modified Function

Graph the new function: $$ g(P) = P(6 - P) - 1 $$ To find the new equilibrium points, solve: $$ P(6-P) - 1 = 0 \Rightarrow P^2 - 6P + 1 = 0 $$ Use the quadratic formula to obtain: $$ P = \frac{6 \pm \sqrt{36 - 4}}{2} = 3 \pm 2\sqrt{2} $$

Step 6: Determine Stability of New Equilibrium Points

Determine the sign of the derivative around the new equilibrium points $$ P = 3 - 2\sqrt{2} $$ and $$ P = 3 + 2\sqrt{2} $$:- If $$ P < 3 - 2\sqrt{2} $$ or $$ P > 3 + 2\sqrt{2} $$, population declines.- If $$ 3 - 2\sqrt{2} < P < 3 + 2\sqrt{2} $$, population increases.Thus, P = 3 - 2√2 is unstable and P = 3 + 2√2 is stable.

Step 7: Long-term Behavior with Fishing

With the new removal rate, the fish population will stabilize at approximately P = 3 + 2√2 thousand fish.

Step 8: Introduce Fishing Rate h

Modify the differential equation to include an arbitrary removal rate: $$ \frac{dP}{dt} = P(6-P) - h $$

Step 9: Determine Maximum Removal Rate

Equilibrium occurs when $$ P(6-P) = h $$ Solving $$ P^2 - 6P + h = 0 $$ gives: $$ P = \frac{6 \pm \sqrt{36 - 4h}}{2} $$ For a real solution, ensure: $$ 36 - 4h \geq 0 \Rightarrow h \leq 9 \text{ thousand fish/month} $$

Step 10: Stabilized Population at Maximum Rate

At maximum rate, the fish population stabilizes as: $$ P = \frac{6}{2} = 3 \text{ thousand fish} $$

Key Concepts

equilibrium solutions

In the context of population dynamics, equilibrium solutions refer to the population sizes where the growth rate of the population is zero. These are the points where the population remains constant if reached. For the given problem, the differential equation is:
\[ \frac{dP}{dt} = P(6 - P) \] To find the equilibrium solutions, we solve for when the derivative \[ \frac{dP}{dt} = 0 \] This gives us the equation:
\[ P(6 - P) = 0 \] Solving this, we find that P equals either 0 or 6. Thus, the equilibrium solutions are P=0 (a population of zero) and P=6 thousand fish. P=0 represents extinction, while P=6 represents the carrying capacity of the lake.

stability analysis

Stability analysis involves determining whether small disturbances around equilibrium solutions will die out or amplify. To perform this, we examine the sign of the growth function around the equilibrium points.
For the given differential equation: \[ \frac{dP}{dt} = P(6 - P) \]
- When P < 0 or P > 6, \[ P(6 - P) \] is negative. This means the population decreases.
- When 0 < P < 6, \[ P(6 - P) \] is positive. This means the population increases.

From this analysis, P=0 is an unstable equilibrium because any small positive disturbance leads to population growth. P=6 is a stable equilibrium because any small disturbance brings the population back to 6 thousand fish.

long-term behavior of populations

The long-term behavior of populations under this model can be predicted based on the stability of equilibrium points:
- In the absence of fishing, if the population starts anywhere between 0 and 6 thousand fish, it will ultimately move towards 6 thousand fish, stabilizing at this point.
- If the population starts above 6 thousand, it will decrease back to this stable equilibrium point.
Conversely, if the population starts at 0 (extinction point), it will remain at 0. Thus, without intervention, the population tends to stabilize around the lake's carrying capacity of 6 thousand fish over the long term.

modified differential equations

When fishing is introduced, removing a consistent number of fish each month, the differential equation needs modification:
Suppose fishermen remove 1000 fish per month. The modified equation becomes:
\[ \frac{dP}{dt} = P(6 - P) - 1 \] Here the term '-1' accounts for the removal of 1 thousand fish per month.
Equilibrium solutions are determined by setting \[ \frac{dP}{dt} = 0 \] \[ P(6 - P) - 1 = 0 \] Solving the quadratic equation, we find two new equilibrium points:
\[ P = 3 \] \[ \text{and} \] \[ P = 3 + 2\theta{2} \] Stability analysis shows that P = 3 - 2√2 is unstable and P=3 + 2√2 is stable. The long-term behavior now tends towards approximately 3.83 thousand fish if the fishery management removes 1000 fish per month.

maximum sustainable yield

The maximum sustainable yield (MSY) is the largest number of fish that can be harvested regularly without leading to population decline in the long run. To find the MSY, the modified differential equation is set to:
\[ \frac{dP}{dt} = P (6 - P) - h \] Solving for the equilibrium points, we get:
\[ \text{For real solutions to exist}, \] \[ P^2 - 6P + h \] must equal 0 and the discriminant \[ (6^2 - 4h) \] must be non-negative:
\[ 36 - 4h \] \[ must be \] greater than or equal to 0.
Solving this inequality, we see that the maximum rate of sustainable fish removal h is:
\[ h = 9 \] At this removal rate, the population stabilizes at:
\[ P = 3 \] This means, removing 9 thousand fish per month, the fish population will stabilize at 3 thousand fish, representing the maximum sustainable yield without leading to population collapse.