Problem 36
Question
Give an example of: A system of differential equations for two populations \(X\) and \(Y\) such that \(Y\) needs \(X\) to survive and \(X\) is indifferent to \(Y\) and thrives on its own. Let \(x\) represent the size of the \(X\) population and \(y\) represent the size of the \(Y\) population.
Step-by-Step Solution
Verified Answer
The system is: \(\frac{dx}{dt} = ax\); \(\frac{dy}{dt} = bxy - cy\).
1Step 1: Determine the Relationship Between Populations
First, identify the nature of the interaction between the two populations. According to the problem, population \(Y\) relies on \(X\) to survive, which suggests a commensal relationship for \(Y\). Conversely, \(X\) is indifferent to \(Y\) and would thrive on its own.
2Step 2: Formulate the Differential Equation for Population X
Since population \(X\) thrives on its own, it can be modeled as an exponential growth equation. Thus, use the differential equation for \(X\) as: \(\frac{dx}{dt} = ax\), where \(a\) is the rate constant for growth.
3Step 3: Formulate the Differential Equation for Population Y
Population \(Y\) needs population \(X\) to survive, which can be modeled as a dependence on \(X\). Therefore, we can write its differential equation as: \(\frac{dy}{dt} = bxy - cy\), where \(b\) represents the dependence of \(Y\) on \(X\), and \(c\) represents the decay rate of \(Y\) without \(X\).
Key Concepts
Population DynamicsExponential GrowthCommensal Relationship
Population Dynamics
Population dynamics involves understanding how and why populations change over time. It considers factors like birth rates, death rates, and the impact of interactions between different populations. In population dynamics, mathematical models such as differential equations are often used to predict how populations will evolve.
In this particular exercise, we have two populations, \(X\) and \(Y\). Population \(Y\) needs \(X\) to survive, indicating a dependency. Hence, \(Y\)'s growth and survival are directly influenced by the presence of \(X\). On the other hand, population \(X\) is self-sufficient and remains unaffected by \(Y\). This unique interaction impacts how these populations will be represented and understood mathematically.
Overall, studying population dynamics helps us comprehend complex ecological relationships and predict how species might fare in changing environments.
In this particular exercise, we have two populations, \(X\) and \(Y\). Population \(Y\) needs \(X\) to survive, indicating a dependency. Hence, \(Y\)'s growth and survival are directly influenced by the presence of \(X\). On the other hand, population \(X\) is self-sufficient and remains unaffected by \(Y\). This unique interaction impacts how these populations will be represented and understood mathematically.
Overall, studying population dynamics helps us comprehend complex ecological relationships and predict how species might fare in changing environments.
Exponential Growth
Exponential growth is a fundamental concept in population dynamics that occurs when the rate of growth of a population is proportional to its current size. When a population experiences exponential growth, it means that as the population increases, it grows faster each period.
For population \(X\), the growth is modeled using the differential equation \(\frac{dx}{dt} = ax\). This equation conveys that the rate of change in population \(X\) is directly proportional to its current size, \(x\), with \(a\) representing the growth rate constant.
Exponential growth can lead to rapid increases in population size, which can be unsustainable if resources become limited. However, in the context of the exercise, \(X\) thrives on its own, suggesting it has access to abundant resources.
For population \(X\), the growth is modeled using the differential equation \(\frac{dx}{dt} = ax\). This equation conveys that the rate of change in population \(X\) is directly proportional to its current size, \(x\), with \(a\) representing the growth rate constant.
Exponential growth can lead to rapid increases in population size, which can be unsustainable if resources become limited. However, in the context of the exercise, \(X\) thrives on its own, suggesting it has access to abundant resources.
Commensal Relationship
A commensal relationship describes an interaction where one species benefits while the other remains unaffected. In this exercise, population \(Y\) has a commensal relationship with population \(X\).
Population \(Y\), which benefits from \(X\), requires \(X\) for its survival and can be modeled by the differential equation \(\frac{dy}{dt} = bxy - cy\). Here, \(bxy\) represents the growth rate influenced by dependency on \(X\), while \(-cy\) accounts for \(Y\)'s natural decay rate in the absence of \(X\).
This relationship highlights an important ecological interaction where \(Y\) relies on \(X\) without impacting \(X\)'s population dynamics. Understanding such relationships aids in ecological modeling and conservation efforts.
Population \(Y\), which benefits from \(X\), requires \(X\) for its survival and can be modeled by the differential equation \(\frac{dy}{dt} = bxy - cy\). Here, \(bxy\) represents the growth rate influenced by dependency on \(X\), while \(-cy\) accounts for \(Y\)'s natural decay rate in the absence of \(X\).
This relationship highlights an important ecological interaction where \(Y\) relies on \(X\) without impacting \(X\)'s population dynamics. Understanding such relationships aids in ecological modeling and conservation efforts.
Other exercises in this chapter
Problem 35
Give an example of: A differential equation that has a logarithmic function as a solution.
View solution Problem 35
Solve the differential equations in Problems \(34-43 .\) Assume \(a, b,\) and \(k\) are nonzero constants. $$\frac{d Q}{d t}-\frac{Q}{k}=0$$
View solution Problem 36
Give an example of: A differential equation for any quantity which grows in two ways simultaneously: on its own at a rate proportional to the cube root of the a
View solution Problem 36
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
View solution