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Population density. It is often convenient to measure population abundance (size) as a population density (number of animals per unit area). What difference does it make to the population equations? To find out, let \(n(t)=N(t) / A\), where \(A\) is the fixed area where the population resides. Given the population logistic equation $$ \frac{d N}{d t}=r N\left(1-\frac{N}{K}\right) $$ what is the differential equation for the density \(n(t) ?\)

Short Answer

Expert verified
The differential equation for the density \(n(t)\) is \(\frac{dn}{dt} = r n \left(1 - \frac{nA}{K}\right)\).

Step by step solution

01

Express Population in Terms of Density

We start by expressing the total population \(N\) in terms of the population density \(n(t)\) and the area \(A\). Since \(n(t) = \frac{N(t)}{A}\), we can rewrite this as \(N(t) = n(t) \cdot A\). This relationship helps us substitute into the logistic growth equation for \(N(t)\).
02

Differentiate Population with Respect to Time

Differentiate \(N(t) = n(t) \cdot A\) with respect to time. This gives: \[ \frac{dN}{dt} = A \cdot \frac{dn}{dt} \] This equation expresses the rate of change of the total population \(N\) in terms of \(n\) and the area \(A\).
03

Substitute into Logistic Growth Equation

Substitute the differentiated form into the logistic equation: \[ A \cdot \frac{dn}{dt} = r(N) \left(1 - \frac{N}{K}\right) \] Substituting \(N = n \cdot A\), the equation becomes: \[ A \cdot \frac{dn}{dt} = r (n \cdot A) \left(1 - \frac{n \cdot A}{K}\right) \] Simplifying gives us \(A \cdot \frac{dn}{dt} = r n A (1 - \frac{nA}{K})\).
04

Simplify the Equation

Since \(A\) is a constant, we can divide both sides of the equation by \(A\): \[ \frac{dn}{dt} = r n \left(1 - \frac{nA}{K}\right) \] This equation now describes the rate of change of population density \(n(t)\) over time, within the area \(A\). The logistic growth equation for density looks similar but scaled by the area.

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Key Concepts

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

Logistic Growth Equation
The logistic growth equation is a fundamental concept used to model how populations grow under limited resources. This model is not linear, as it takes into account the carrying capacity, which is the maximum population size that the environment can sustain. The original equation is \( \frac{dN}{dt} = rN\left(1-\frac{N}{K}\right) \), where:
  • \( \frac{dN}{dt} \) represents the rate of change of the population over time.
  • \( r \) is the intrinsic growth rate, indicating how fast the population can grow.
  • \( N \) is the current population size.
  • \( K \) is the carrying capacity.
As the population gets closer to \( K \), the growth slows down, representing the saturation effect in a real-world scenario. By considering both individual growth rate and environmental constraints, the logistic growth model provides a realistic way to predict population dynamics. Its main feature is the S-shaped curve it produces, capturing how populations stabilize when resources become more scarce.
Changing the representation of a population from raw numbers to density involves integrating the surface area \( A \), which adjusts the logistic equation appropriately for spatial constraints.
Differential Equations
Differential equations are mathematical expressions that describe how a quantity changes over time. They are crucial for modeling dynamic processes, such as population growth, in a precise manner.
In the logistic growth model, the equation \( \frac{dN}{dt} = rN\left(1-\frac{N}{K}\right) \) is such a differential equation. It demonstrates how populations change over time considering intrinsic factors like growth rate \( r \) and extrinsic constraints like carrying capacity \( K \).
The exercise involved modifying the equation to account for population density: \( \frac{dN}{dt} = A \cdot \frac{dn}{dt} \). Here, \( n(t) \) is the population density, or population per unit area. This transformation results in a new equation \( \frac{dn}{dt} = r n \left(1 - \frac{nA}{K}\right) \), allowing us to analyze how populations behave in a constrained space with respect to density rather than number. This insight tells us how the total area \( A \) affects population dynamics.
Population Dynamics
Population dynamics is the field of study concerned with how populations change over time and space. These changes are influenced by factors such as birth and death rates, immigration and emigration, and environmental limitations.
Using a density-based approach helps scientists and ecologists assess how population changes relate to available resources or spatial constraints. With the formula derived in the exercise \( \frac{dn}{dt} = r n \left(1 - \frac{nA}{K}\right) \), the focus shifts from total population to density, allowing a more nuanced understanding.
  • **Density dependence:** As density increases, competition for resources can limit growth.
  • **Habitat size:** The larger the area \( A \), the more individuals it can support, impacting the speed of growth.
  • **Carrying capacity:** A measure of how many individuals can thrive in given conditions, \( K \) serves as a threshold for sustainable population size.
This density perspective highlights how environment and resources directly affect population sustainability and growth patterns, providing essential insights for ecological management and conservation.

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Most popular questions from this chapter

Finding equilibrium solutions. For the following discrete population models find all the equilibrium solutions by setting \(N_{k+1}=N_{k}=N .\) Determine the stability of each of the equilibrium populations. (a) \(N_{k+1}=5 N_{k}\) (b) \(N_{k+1}=0.8 N_{k}-0.1 N_{k}^{2}\)

Stability of 2-cycles. Consider the discrete logistic equation (with \(K=1\) ) $$ X_{n+1}=X_{n}+r X_{n}\left(1-X_{n}\right) $$ (a) Show that every second term in the sequence \(X_{0}, X_{1}, X_{2}, \ldots\) satisfies the difference equation $$ \begin{gathered} X_{n+2}=\left(1+2 r+r^{2}\right) X_{n}-\left(2 r+3 r^{2}+r^{3}\right) X_{n}^{2} \\ +\left(2 r^{2}+2 r^{3}\right) X_{n}^{3}-r^{3} X_{n}^{4} \end{gathered} $$ (b) For equilibrium solutions, let \(S=X_{n+2}=X_{n}\) and obtain a quartic equation (that is, an equation with the unknown raised to the fourth power, at most). Explain why \(S=0\) and \(S=1\) must be solutions of this quartic equation. Hence show that the other two solutions are $$ S=\frac{(2+r) \pm \sqrt{r^{2}-4}}{2 r} $$ [Note: Comparing with Figure \(3.11(r=2.2)\) we see that these two values of the two nonzero equilibrium solutions are the values between which the population oscillates in a 2 -cycle. Furthermore, when \(r\) increases to where these two equilibrium solutions become unstable, this corresponds to where the 2-cycle changes to a 4-cycle.]

Allee effect model. Consider the population model $$ \frac{d N}{d t}=r N(N-b)\left(1-\frac{N}{K}\right) $$ where \(r\) is the intrinsic growth rate, \(K\) is the carrying capacity and \(b\) is a positive constant where \(b

Density-dependent births. Many animal populations have decreasing per-capita birth rates when the population density increases, as well as increasing per- capita death rates. Suppose the density-dependent per-capita birth rate \(B(X)\) and density-dependent death rate \(A(X)\) are given by $$ B(X)=\beta-(\beta-\alpha) \delta \frac{X}{K}, \quad A(X)=\alpha+(\beta-\alpha)(1-\delta) \frac{X}{K} $$ where \(K\) is the population carrying capacity, \(\beta\) is the intrinsic per- capita birth rate, \(\alpha\) is the intrinsic per-capita death rate and \(\delta\), where \(0 \leq \delta \leq 1\), is a parameter describing the extent that density dependence is expressed in births or deaths. Show that this still gives rise to the standard logistic differential equation $$ \frac{d X}{d t}=r X\left(1-\frac{X}{K}\right) $$

Fishing with quotas. In view of the potentially disastrous effects of overfishing causing a population to become extinct, some governments impose quotas that vary depending on estimates of the population at the current time. One harvesting model that takes this into account is $$ \frac{d X}{d t}=r X\left(1-\frac{X}{K}\right)-h X $$ (a) Show that the only non-zero equilibrium population is $$ X_{e}=K\left(1-\frac{h}{r}\right) $$ (b) At what critical harvesting rate can extinction occur? Although extinction can occur with this model, as the harvesting parameter \(h\) increases towards the critical value the equilibrium population tends to zero. This contrasts with the constant harvesting model in Section \(3.3\) and Exercise \(3.7\), where a sudden population crash (from a large population to extinction) can occur as the harvesting rate increases beyond a critical value.

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