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In Exercises you are given a logistic equation with one or more initial conditions.(a) Determine the carrying capacity and intrinsic rate. (b) Sketch the graph of \(\frac{d N}{d t}\) versus \(N\) in an \(N z\) -plane. (c) In the \(t N\) -plane, plot the constant solutions and place a dashed line where the concavity of certain solutions may change. (d) Sketch the solution curve corresponding to each given initial condition. $$d N / d t=.3 N(100-N), N(0)=25$$

Short Answer

Expert verified
Carrying capacity is 100, intrinsic rate is 0.3. The graph of \( \frac{dN}{dt} \) vs. \( N \) is an inverted U. Solution curve starting from \( N(0)=25 \) approaches \( N = 100 \).

Step by step solution

01

Identify the logistic equation

The given logistic differential equation is \ \( \frac{dN}{dt} = 0.3N(100 - N) \). Here, the function describes how the population \( N \) changes over time.
02

Determine the carrying capacity and intrinsic rate

The logistic equation is of the form \ \( \frac{dN}{dt} = rN(K - N) \), where \( r \) is the intrinsic growth rate and \( K \) is the carrying capacity. By comparing with the given equation, we find: Intrinsic rate \( r = 0.3 \), Carrying capacity \( K = 100 \).
03

Sketch \( \frac{dN}{dt} \) versus \( N \)

Plotting \( \frac{dN}{dt} \) versus \( N \), we know \( \frac{dN}{dt} = 0.3N(100 - N) \). This represents a parabola opening downward with roots at \( N = 0 \) and \( N = 100 \). The maximum is reached at \( N = 50 \). The graph looks like an inverted U-shape.
04

Plot constant solutions in the \( tN \)-plane

In the \( tN \)-plane, the constant solutions are \ \( N = 0 \) and \( N = 100 \). Draw horizontal lines to represent these equilibrium solutions. The solution curve will start at \( N(0) = 25 \) and approach \( N = 100 \) over time.
05

Identify concavity changes

The inflection point or point where concavity changes is where the second derivative changes sign. This typically occurs around \( N = 50 \), as this is the midpoint of the carrying capacity. Dashed vertical lines at \( N = 50 \) indicate changes in concavity.
06

Sketch the solution curve for the initial condition

Starting from the initial condition \( N(0) = 25 \), the solution curve increases monotonically towards the upper equilibrium \( N = 100 \). The curve is increasing and concave up until \( N = 50 \), and concave down afterwards until it stabilizes at \( N = 100 \).

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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, the carrying capacity, denoted by \( K \), is the maximum population size of a species that the environment can sustain indefinitely. This means that given food, habitat, water, and other environmental necessities, the population will stabilize at this number.
For the logistic equation provided \( \frac{dN}{dt}=0.3N(100-N) \), the carrying capacity \( K \) can be identified as 100. This shows that the population cannot exceed 100 due to environmental constraints.
Understanding carrying capacity is essential as it sheds light on the limitations faced by populations and the balance required for sustainable existence.
Intrinsic Rate
The intrinsic rate of increase, represented by \( r \), is the rate at which the population grows when it is far below the carrying capacity and resources are abundant. It is a measure of the population's potential growth under ideal conditions.
In the given logistic differential equation \( \frac{dN}{dt}=0.3N(100-N) \), the intrinsic rate \( r \) is 0.3. This value determines how fast the population can grow when starting from a low population size.
Higher intrinsic rates result in faster population growth, whereas lower rates indicate slower growth. This rate is fundamental in predicting how quickly a population can bounce back from low numbers.
Population Dynamics
Population dynamics studies the changes in population size and composition over time. The logistic equation \( \frac{dN}{dt}=0.3N(100-N) \) is a classic model used to describe population growth, considering both the intrinsic rate and carrying capacity.
Initially, when the population is small, growth is approximately exponential governed by the intrinsic rate. As the population increases, resources become limited, and growth slows down.
Eventually, growth halts when the population reaches the carrying capacity, creating an S-shaped curve known as the logistic growth curve. This model is realistic as it considers environmental limits and intraspecific competition.
Differential Equations
Differential equations like the logistic equation \( \frac{dN}{dt}=0.3N(100-N) \) are mathematical tools used to describe how a quantity changes with respect to another, often time.
In this equation, \( \frac{dN}{dt} \) represents the rate of change of the population \( N \) over time \( t \).
Solving this differential equation provides insights into how the population evolves. By integrating the equation, we can get a function \( N(t) \) that predicts future population sizes based on initial conditions.
This process involves finding equilibrium points, stability analysis, and sometimes phase diagrams, making it a crucial concept in mathematical biology.
Solution Curve
The solution curve of a differential equation like \( \frac{dN}{dt}=0.3N(100-N) \) illustrates how the population evolves over time, given an initial condition.
For the initial condition \( N(0)=25 \), the curve starts at 25 and gradually rises as the population grows.
Initially, the population increases quickly due to the high growth rate. As the population approaches 50, growth slows down. This point, where the growth rate is highest, is the inflection point. Concavity changes here, and after this point, the curve flattens as it approaches the carrying capacity of 100.
Plotting these solution curves helps visualize the dynamics and predict long-term behavior. The analysis aids in understanding how initial conditions affect population trends.

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Most popular questions from this chapter

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