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Logistic growth with a threshold: Most species have a survival threshold level, and populations of fewer individuals than the threshold cannot sustain themselves. If the carrying capacity is \(K\) and the threshold level is \(S\), then the logistic equation of change for the population \(N=N(t)\) is $$ \frac{d N}{d t}=-r N\left(1-\frac{N}{S}\right)\left(1-\frac{N}{K}\right) . $$ For Pacific sardines, we may use \(K=2.4\) million tons and \(r=0.338\) per year, as in Example 6.10. Suppose we also know that the survival threshold level for the sardines is \(S=0.8\) million tons. a. Write the equation of change for Pacific sardines under these conditions. b. Make a graph of \(\frac{d N}{d t}\) versus \(N\) and use it to find the equilibrium solutions. How do the equilibrium solutions correspond with \(S\) and \(K\) ? c. For what values of \(N\) is the graph of \(N\) versus \(t\) increasing, and for what values is it decreasing? d. Explain what can be expected to happen to a population of \(0.7\) million tons of sardines. e. At what population level will the population be growing at its fastest?

Short Answer

Expert verified
a. \(\frac{d N}{d t} = -0.338 N (1-\frac{N}{0.8})(1-\frac{N}{2.4})\) b. Equilibriums: 0, 0.8, 2.4 million tons; stable at 0.8 and 2.4. c. Increasing: (0, 0.8), (0.8, 2.4); Decreasing: (2.4, ∞). d. 0.7 million tons will grow towards 0.8 million tons. e. Fastest growth at \(N\approx 1.38\) million tons.

Step by step solution

01

Write the equation of change

Given the parameters for Pacific sardines, substitute into the logistic growth equation: \[-r N\left(1-\frac{N}{S}\right)\left(1-\frac{N}{K}\right)\] where \(r = 0.338\), \(K = 2.4\), and \(S = 0.8\). The equation becomes: \[\frac{d N}{d t} = -0.338 N\left(1-\frac{N}{0.8}\right)\left(1-\frac{N}{2.4}\right).\]
02

Analyze graph for equilibrium solutions

To find equilibrium solutions, set \(\frac{d N}{d t} = 0\): \[-0.338 N\left(1-\frac{N}{0.8}\right)\left(1-\frac{N}{2.4}\right) = 0.\] The solutions occur when: \(N = 0\), \(N = S = 0.8\), and \(N = K = 2.4\). These are the equilibrium solutions. By analyzing the graph of the derivative \(\frac{d N}{d t}\) vs \(N\), we see: - \(N = 0\) is unstable because any small population will grow or increase.- \(N = 0.8\) is stable as perturbations will revert back to equilibrium.- \(N = 2.4\) is stable, maximum carrying capacity.
03

Determine intervals of increase and decrease

The graph of \(N\) versus \(t\) is increasing where \(\frac{d N}{d t} > 0\), and decreasing where \(\frac{d N}{d t} < 0\). Analytically this happens:- Increase between \(0 < N < 0.8\) and \(0.8 < N < 2.4\).- Decrease when \(N > 2.4\).Thus, \(N\) increases towards \(0.8\) and \(2.4\), and decreases after \(2.4\).
04

Analyze population of 0.7 million tons

For \(N = 0.7\), since \(0 < 0.7 < 0.8\), \(\frac{d N}{d t} > 0\) which means the population is subjected to growth. As it is below the threshold \(S = 0.8\), it will tend to grow towards \(0.8\).
05

Find the fastest growth rate

The growth rate is maximized at the vertex of the parabola defined by the growth derivative. Taking the derivative of \(\frac{d N}{d t}\) with respect to \(N\) to locate this point analytically or graphically reveals the fastest growth rate occurs at the middle value between \(S\) and \(K\). Thus, the maximum growth occurs at \(N=\sqrt{S \cdot K} \approx \sqrt{0.8 \cdot 2.4} \approx 1.38\) million tons.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Carrying Capacity
In population dynamics, carrying capacity refers to the maximum population size of a species that a given environment can sustain indefinitely. It is denoted as \(K\). For instance, in our example of Pacific sardines, the carrying capacity is given as \(K = 2.4\) million tons. This means the oceanic environment can support a maximum of 2.4 million tons of sardines without any negative effects from overpopulation. Carrying capacity is influenced by several factors, including resource availability, environmental conditions, and space. When populations reach this size, resources become limited, and growth rates decrease, eventually stabilizing the population. In mathematical models like the logistic growth equation, the carrying capacity shapes the curve representing how the population grows over time.
Population Dynamics
Population dynamics studies how and why population sizes change over time and the effects these changes have on ecosystems. It takes into account factors such as birth rates, death rates, immigration, and emigration. For the Pacific sardines, important factors influencing dynamics include birth rates and natural resource limits.The logistic growth model we use with sardines is a classic example of population dynamics. It expresses how populations grow rapidly when resources are abundant (initial growth phase) and slow down as they near the carrying capacity. In the equation provided, the population \(N\) grows according to the rate \(r = 0.338\), but as \(N\) approaches either the threshold level \(S = 0.8\) or \(K = 2.4\), the growth rate adjusts to reflect available resources and population pressure.Understanding these dynamic processes allows ecologists to predict changes and manage conservation efforts effectively.
Equilibrium Solutions
Equilibrium solutions are points where population changes stabilize, meaning growth and decline pressures balance each other out. In the logistic equation, these points occur when \(\frac{dN}{dt} = 0\), indicating no change in population over time.For our Pacific sardines, the equilibrium points occur at \(N = 0\), \(N = S = 0.8\), and \(N = K = 2.4\). Each of these points represents a stable or unstable state:
  • \(N = 0\): Unsustainable as any small increase leads to growth.
  • \(N = 0.8\): Stable equilibrium as small deviations return to this state.
  • \(N = 2.4\): Stable equilibrium representing the carrying capacity.
At these points of equilibrium, the population does not change unless influenced by significant external factors.
Threshold Level
The threshold level, \(S\), is a critical population size required for a population to sustain itself. For many species, this level represents the minimum population needed to avoid extinction due to factors like genetic diversity loss or societal disruptions.In the case of the sardines, the threshold level is \(S = 0.8\) million tons. Below this threshold, the population \(N\) finds it challenging to grow independently. The logistic growth model highlights how, below this threshold, \(\frac{dN}{dt} > 0\), indicating natural growth as the population tries to reach a stable state.This concept is essential in conservation biology, as it underscores the importance of maintaining populations above this critical level to ensure their long-term viability.

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Most popular questions from this chapter

Fishing for sardines: This is a continuation of Example 6.10. If we take into account an annual fish harvest of \(F\) million tons of fish, then the equation of change for Pacific sardines becomes $$ \frac{d N}{d t}=0.338 N\left(1-\frac{N}{2.4}\right)-F . $$ a. Suppose that there are currently \(1.8\) million tons of Pacific sardines off the California coast and that you are in charge of the commercial fishing fleet. It is your goal to leave the Pacific population of sardines as you found it. That is, you wish to set the fishing level \(F\) so that the biomass of Pacific sardines remains stable. What value of \(F\) will accomplish this? (Hint: You want to choose \(F\) so that the current biomass level of \(1.8\) million tons is an equilibrium solution.) b. For the remainder of this exercise, take the value of \(F\) to be \(0.1\) million tons per year. That is, assume the catch is 100,000 tons per year. i. Make a graph of \(\frac{d N}{d t}\) versus \(N\), and use it to find the equilibrium solutions. ii. For what values of \(N\) will the biomass be increasing? For what values will it be decreasing? iii. On the same graph, sketch all equilibrium solutions and the graphs of \(N\) versus \(t\) for each of the initial populations \(N(0)=0.3\) million tons, \(N(0)=1.0\) million tons, and \(N(0)=2.3\) million tons. \(\rightarrow\) iv. Explain in practical terms what the picture you made in part iii tells you. Include in your explanation the significance of the equilibrium solutions.

The spread of AIDS: The table on the following page shows the cumulative number \(N=N(t)\) of AIDS cases in the United States that have been reported to the Centers for Disease Control and Prevention by the end of the year given. (The source for these data, the U.S. Centers for Disease Control and Prevention in Atlanta, cautions that they are subject to retrospective change.) a. What does \(\frac{d N}{d t}\) mean in practical terms? b. From 1986 to 1992 was \(\frac{d N}{d t}\) ever negative? $$ \begin{array}{|c|c|} \hline t=\text { year } & N=\text { total cases reported } \\ \hline 1986 & 28,711 \\ \hline 1987 & 49,799 \\ \hline 1988 & 80,518 \\ \hline 1989 & 114,113 \\ \hline 1990 & 155,766 \\ \hline 1991 & 199,467 \\ \hline 1992 & 244,939 \\ \hline \end{array} $$

Mileage for an old car: The gas mileage \(M\) that you get on your car depends on its age \(t\) in years. a. Explain the meaning of \(\frac{d M}{d t}\) in practical terms. b. As your car ages and performance degrades, do you expect \(\frac{d M}{d t}\) to be positive or negative?

Equation of change for logistic growth: The logistic growth formula \(N=6.21 /\left(0.035+0.45^{t}\right)\) that we used in Chapter 2 for deer on the George Reserve actually came from the following equation of change: $$ \frac{d N}{d t}=0.8 N\left(1-\frac{N}{177}\right) \text {. } $$ a. Plot the graph of \(\frac{d N}{d t}\) versus \(N\), and use it to find the equilibrium solutions. Explain their physical significance. b. In one plot, sketch the equilibrium solutions and graphs of \(N\) versus \(t\) in each of the two cases \(N(0)=10\) and \(N(0)=225 .\) c. To what starting value for \(N\) does the solution \(N=6.21 /\left(0.035+0.45^{t}\right)\) correspond?

Growth of fish: Let \(w=w(t)\) denote the weight of a fish as a function of its age \(t\). For the North Sea cod, the equation of change $$ \frac{d w}{d t}=2.1 w^{2 / 3}-0.6 w $$ holds. Here \(w\) is measured in pounds and \(t\) in years. b. Make a graph of \(\frac{d w}{d t}\) against \(w\). Include weights up to 45 pounds. c. What is the weight of the cod when it is growing at the greatest rate? d. To what weight does the cod grow?

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