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Chapter 11: Problem 14

Four pairs of species are given, with descriptions of how they interact. I. Bees/flowers: each needs the other to survive II. Owls/trees: owls need trees but trees are indifferent III. Elk/buffalo: in competition and would do fine alone IV. Fox/hare: fox eats the hare and needs it to survive Match each system of differential equations with a species pair, and indicate which species is \(x\) and which is \(y.\) (a) \(\frac{d x}{d t}=-0.2 x+0.03 x y\) \(\frac{d y}{d t}=0.4 y-0.08 x y\) (b) \(\frac{d x}{d t}=0.18 x\) \(\frac{d y}{d t}=-0.4 y+0.3 x y\) (c) \(\frac{d x}{d t}=-0.6 x+0.18 x y\) \(\frac{d y}{d t}=-0.1 y+0.09 x y\) (d) Write a possible system of differential equations for the species pair that does not have a match.

Short Answer

Expert verified
(a) Bees/flowers, (b) Fox/hare, (c) Elk/buffalo, (d) Owls/trees.

Step by step solution

01

Analyze System (a)

The system (a) is \(\frac{dx}{dt}=-0.2x+0.03xy\) and \(\frac{dy}{dt}=0.4y-0.08xy\). This set of equations indicates a **mutualistic** relationship identified by the positive interaction terms \(+0.03xy\) and \(-0.08xy\). The interaction term suggests both species benefit when they interact, which matches the **bees/flowers** pair, where both species need each other to survive. Assign \(x\) to bees and \(y\) to flowers.
02

Analyze System (b)

The system (b) is \( \frac{dx}{dt}=0.18x \) and \( \frac{dy}{dt}=-0.4y+0.3xy \). Here, \(x\) grows naturally and \(y\) has a predatory or competitive relationship with \(x\). The positive interaction term \(+0.3xy\) indicates that \(y\) benefits from \(x\). This corresponds to a **predator-prey** relationship, particularly the **fox/hare** pair. Assign \(x\) to hare and \(y\) to fox.
03

Analyze System (c)

The system (c) is \( \frac{dx}{dt}=-0.6x+0.18xy \) and \( \frac{dy}{dt}=-0.1y+0.09xy \). Both equations have negative intrinsic growth and positive interaction terms, indicating **competition**. Since elks and buffaloes are competitive and can survive independently, this system resonates with the **elk/buffalo** pair. Assign \(x\) to elk and \(y\) to buffalo.
04

Determine matching system for owls/trees (System (d))

The owls need trees, but trees are indifferent to owls. To model this with differential equations, consider: \( \frac{dx}{dt}=-0.3x+0.05xy \) (trees, \(x\), are constant but available for use) and \( \frac{dy}{dt}=0.5y-0.04y \) (owls, \(y\), need trees, \(x\), to survive but do not impact trees' growth). Assign \(x\) to trees and \(y\) to owls.

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Key Concepts

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

Mutualistic Relationship
In biological systems, a mutualistic relationship is where both species involved derive benefits from their interaction with each other. This positive interaction is crucial to their survival and overall health. For example, consider bees and flowers. Bees get nectar from the flowers, which they use to make food, while flowers receive the service of pollination, which is essential for their reproduction.

Mathematically, this relationship can be modeled using differential equations that capture the benefits each species obtains from the interaction. In system (a) mentioned earlier, the positive interaction terms, such as - "+0.03xy" and - "+0.08xy", indicate how the presence of one species positively affects the growth rate of the other.

It is important to recognize that without mutualistic relationships, many ecosystems would struggle because the interconnectedness is key to the survival of various species. Understanding these equations helps ecologists forecast changes in species populations over time.
Predator-Prey Models
Predator-prey models are used to describe the interactions between two species where one species, the predator, hunts and consumes the other species, the prey. This relationship is fundamental for understanding population dynamics.

System (b) in the original exercise provides an example of this model, representing the fox and hare relationship. The interaction term - "+0.3xy" indicates that the growth of the fox population depends on the availability of hares, while the natural growth rate of hares is denoted by another term in the equation. The foxes need the hares for survival, but this has an upper limit based on the availability of prey.

These models can give insights into how populations oscillate over time. When prey numbers increase, predator numbers may also rise due to more readily available food, potentially leading to a decrease in prey numbers, which subsequently affects the predators as well.
Competition in Species
Competition occurs when two species vie for the same resources in an ecosystem. This competition influences both species' survival rates, often negatively impacting their growth.

In the differential equation system (c), both elks and buffaloes are depicted as being in competition. Each species has terms like - "-0.6x" for elks and - "-0.1y" for buffaloes. These terms show that their populations inherently decline without interaction. Yet, through competition, positive terms like - "+0.18xy", emerge when they interact, suggesting that shared resources create a competitive balance.

Through this modeling, ecologists can predict how species populations interact and fluctuate based on resource availability and other environmental factors.
Species Interaction Modeling
Understanding species interactions is crucial for ecologists. Differential equations provide a powerful tool to model these interactions, leading to better predictions regarding ecosystem dynamics.

System (d) modeled the relationship between owls and trees, demonstrating a unidirectional dependency. Trees, represented as a resource, are constants ( - "-0.3x + 0.05xy" ), meaning their growth isn't affected by the presence of owls. Owls, however, rely on trees for survival ( - "0.5y - 0.04y" ), showcasing a dependent interaction.

This modeling allows us to explore essential ecological scenarios where one species depends heavily on another while the latter remains unaffected. By understanding how such dependencies work, we can better appreciate the intricacy of species interactions and manage ecosystems more sustainably.

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Most popular questions from this chapter

In Problems \(52-54\), explain what is wrong with the statement. Separating variables in \(d y / d x=e^{x+y}\) gives \(-e^{y} d y=\) \(e^{x} d x\)

Is the statement true or false? Assume that \(y=f(x)\) is a solution to the equation \(d y / d x=g(x) .\) If the statement is true, explain how you know. If the statement is false, give a counterexample. If \(g(x)\) is even, then so is \(f(x).\)

When people smoke, carbon monoxide is released into the air. In a room of volume \(60 \mathrm{m}^{3},\) air containing \(5 \%\) carbon monoxide is introduced at a rate of \(0.002 \mathrm{m}^{3} / \mathrm{min}\) (This means that \(5 \%\) of the volume of the incoming air is carbon monoxide.) The carbon monoxide mixes immediately with the rest of the air, and the mixture leaves the room at the same rate as it enters. (a) Write a differential equation for \(c(t),\) the concentration of carbon monoxide at time \(t,\) in minutes. (b) Solve the differential equation, assuming there is no carbon monoxide in the room initially. (c) What happens to the value of \(c(t)\) in the long run?

In the 1930 s, the Soviet ecologist G. F. Gause \(^{22}\) studied the population growth of yeast. Fit a logistic curve, \(d P / d t=k P(1-P / L),\) to his data below using the method outlined below. $$\begin{array}{l|c|c|c|c|c|c|c}\hline \text { Time (hours) } & 0 & 10 & 18 & 23 & 34 & 42 & 47 \\ \hline \text { Yeast pop } & 0.37 & 8.87 & 10.66 & 12.50 & 13.27 & 12.87 & 12.70 \\\\\hline\end{array}$$(a) Plot the data and use it to estimate (by eye) the carrying capacity, \(L\) (b) Use the first two pieces of data in the table and your value for \(L\) to estimate \(k\) (c) On the same axes as the data points, use your values for \(k\) and \(L\) to sketch the solution curve \(P=\frac{L}{1+A e^{-k t}} \quad\) where \(\quad A=\frac{L-P_{0}}{P_{0}}\)

Consider a conflict between two armies of \(x\) and \(y\) soldiers, respectively. During World War I, F. W. Lanchester assumed that if both armies are fighting a conventional battle within sight of one another, the rate at which soldiers in one army are put out of action (killed or wounded) is proportional to the amount of fire the other army can concentrate on them, which is in turn proportional to the number of soldiers in the opposing army. Thus Lanchester assumed that if there are no reinforcements and \(t\) represents time since the start of the battle, then \(x\) and \(y\) obey the differential equations $$\begin{array}{l} \frac{d x}{d t}=-a y \\ \frac{d y}{d t}=-b x \quad a, b>0 \end{array}$$. Near the end of World War II a fierce battle took place between US and Japanese troops over the island of Iwo Jima, off the coast of Japan. Applying Lanchester's analysis to this battle, with \(x\) representing the number of US troops and \(y\) the number of Japanese troops, it has been estimated \(^{31}\) that \(a=0.05\) and \(b=0.01\) (a) Using these values for \(a\) and \(b\) and ignoring reinforcements, write a differential equation involving \(d y / d x\) and sketch its slope field. (b) Assuming that the initial strength of the US forces was 54,000 and that of the Japanese was 21,500 draw the trajectory which describes the battle. What outcome is predicted? (That is, which side do the differential equations predict will win?) (c) Would knowing that the US in fact had 19,000 reinforcements, while the Japanese had none, alter the outcome predicted?
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