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Chapter 17: Problem 8

Let \(M=10\) and \(r=0.1\) and plot the graphs of $$p(t)=\frac{M p_{0}}{p_{0}+\left(M-p_{0}\right) e^{-r t}}$$ for \(0 \leq t \leq 80\) and a. \(p_{0}=1\) $$\text { b. } p_{0}=12 \quad \text { c. } p_{0}=10$$

Short Answer

Expert verified
Graph the function for three initial conditions to observe different behaviors of \(p(t)\).

Step by step solution

01

Understand the Equation

The function given is the logistic growth model defined as \(p(t)=\frac{M p_{0}}{p_{0}+\left(M-p_{0}\right) e^{-r t}}\). It describes how a population grows over time \(t\) with parameters \(M\), \(r\), and initial population \(p_0\). Here, \(M = 10\) and \(r = 0.1\). We will evaluate this function for different initial conditions \(p_0 = 1, 12, 10\).
02

Setting Up for Plotting

We need to plot the function for these initial conditions while keeping \(t\) in the range \([0, 80]\). This involves calculating the value of \(p(t)\) at various times \(t\) from 0 to 80 for each \(p_0\) value given.
03

Calculate for \(p_0 = 1\)

Substitute \(p_0 = 1\) into the equation: \(p(t) = \frac{10 \times 1}{1+(10-1) e^{-0.1 t}}\). Calculate \(p(t)\) for several values of \(t\) and plot the points.
04

Calculate for \(p_0 = 12\)

Replace \(p_0 = 12\) in the equation: \(p(t) = \frac{10 \times 12}{12+(10-12) e^{-0.1 t}}\). Note that the initial population exceeds the carrying capacity \(M\). Calculate and plot \(p(t)\) for multiple \(t\).
05

Calculate for \(p_0 = 10\)

Substitute \(p_0 = 10\) into the model: \(p(t) = \frac{10 \times 10}{10+(10-10) e^{-0.1 t}} = 10\). As \(p_0 = M\), the population remains constant. Plot \(p(t) = 10\) from \(t=0\) to \(t=80\).
06

Plot the Graphs

Using the calculations from steps 3-5, plot three separate graphs for \(p_0=1\), \(p_0=12\), and \(p_0=10\) with \(t\) on the x-axis and \(p(t)\) on the y-axis. Each graph should depict how \(p(t)\) changes over time for these initial conditions.

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Key Concepts

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

Population Dynamics
Population dynamics is a fascinating branch of biology that studies how populations of organisms, such as animals and plants, change over time. It involves understanding the factors that cause population sizes to increase, decrease, or even stay constant. These changes can occur due to reproduction, death, immigration, and emigration.

The logistic growth model, as seen in the given exercise, is a popular model for simulating population changes. It captures how a population grows in an environment with limited resources, which aligns with real-world scenarios. As resources become scarce, growth slows down, leading to a stable population size, known as the carrying capacity. This model starts with an initial population size (denoted as \( p_0 \)) and predicts how the population will evolve over a specified time frame \( t \).

The model is represented by the equation:

\[p(t) = \frac{M p_{0}}{p_{0} + (M-p_{0}) e^{-r t}}\]

where \( M \) is the carrying capacity and \( r \) is the growth rate. By analyzing how populations behave under various conditions, we can better understand issues like resource allocation and conservation strategy development.
Differential Equations
Differential equations are equations that involve the rates at which quantities change. They are vital in modeling real-world phenomena where change is a factor, such as population dynamics, heat transfer, and motion.

The logistic growth model uses a differential equation to represent how populations grow over time with constraints like a limited carrying capacity. The change in population \( p(t) \) with respect to time \( t \) is described by:

\[p'(t) = rp(t)\left(1 - \frac{p(t)}{M}\right)\]

This equation incorporates the ideas of initial population size, growth rate, and carrying capacity. The term \( \left(1 - \frac{p(t)}{M}\right) \) represents available resources, diminishing as populations approach the carrying capacity. Solving this equation yields the logistic function used in the exercise:

\[p(t) = \frac{M p_{0}}{p_{0} + (M-p_{0}) e^{-r t}}\]

Understanding such equations is essential to predicting how populations will behave over time and assessing the long-term sustainability of an environment.
Carrying Capacity
Carrying capacity, denoted by \( M \) in the logistic growth model, is the maximum population size that an environment can sustainably support. It plays a crucial role in population dynamics as it dictates the point at which a population will stabilize over time.

In the given exercise, the carrying capacity is set to 10. This means that the population will naturally adjust to reach a size where resources are optimally used without depleting the environment.

Using the logistic growth function:

\[p(t) = \frac{M p_{0}}{p_{0} + (M-p_{0}) e^{-r t}}\]

we can see that when the initial population \( p_0 \) is less than \( M \), the population will grow and eventually stabilize at the carrying capacity. Conversely, when \( p_0 \) is initially greater than \( M \), the population will decrease until it reaches the carrying capacity.

This concept highlights the balance between population growth and resource availability, allowing predictions on how populations interact with their environment over time. Evaluating carrying capacity is essential for managing natural resources and ensuring the sustainability of ecosystems.

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Most popular questions from this chapter

The special case of \(y^{\prime}=f(t, y)\) in which \(f(t, y)=F(t)(f\) is independent of \(y)\) has a familiar solution from the Fundamental Theorem of Calculus I. Check by substitution that $$y(t)=y_{a}+\int_{a}^{t} F(x) d x \quad \text { solves } \quad y(a)=y_{a} \quad \text { and } \quad y^{\prime}(t)=F(t)$$ The differential equation $$y(a)=y_{a} \quad y^{\prime}(t)=F(t)$$ has therefore been completely solved. Henceforth we will consider that \(f\) is dependent on \(y\) and possibly also on \(t\).

Continuous infusion of penicillin. Suppose a patient recovering from surgery is to be administered penicillin intravenously at a constant rate of 5 grams per hour. The patient's kidneys will remove penicillin at a rate proportional to the serum penicillin concentration. Let \(P(t)\) be the penicillin concentration \(t\) hours after infusion is begun. Then a simple model of penicillin pharmacokinetics is \(\begin{array}{l}\text { Net Rate of Increase } & \text { Clearance } & \text { Infusion }\end{array}\) $$P^{\prime}(t)=-K \times P(t)+5$$ $$\frac{\mathrm{gm}}{\mathrm{hr}} \quad \frac{1}{\mathrm{hr}} \times \operatorname{gm} \quad \frac{\mathrm{gm}}{\mathrm{hr}}$$ The proportionality constant, \(K,\) must have units \(\frac{1}{\mathrm{hr}}\) in order for the units on the equation to balance. We initially assume that \(K=2.5 \frac{1}{\mathrm{hr}}\) which is in the range of physiological reality. It is reasonable to assume that there was no penicillin in the patient at time \(t=0,\) so that \(P(0)=0\). a. Draw the phase plane for the differential equation $$P(0)=0 \quad P^{\prime}(t)=-2.5 P(t)+5$$ b. Find the equilibrium point of \(P^{\prime}=-2.5 P+5\). c. Is the equilibrium point stable? d. Show that the units of the equilibrium point are grams. e. Suppose the patient's kidneys are impaired and only operating at \(60 \%\) of normal. Then \(K=1.5\) instead of \(2.5 .\) What effect does this have on the equilibrium point.

Exercise 17.4.10 Suppose a marine fish population when not subject to harvest is reasonably modeled by $$u^{\prime}(t)=r \times u(t) \times(1-u(t))$$ with time measured in years. Suppose a harvest procedure is initiated, and that a fraction, \(h,\) of the existing population is harvested every year. The harvest is not a fixed amount each year, but depends on the number of fish available. The growth rate will be the difference between the natural birth-death process and the harvest and may be modeled by $$ u^{\prime}(t)=r \times u(t) \times(1-u(t))-h \times u(t) $$ a. Assume \(h=r\) (the harvest rate equals the low density growth rate) Substitute \(h=r\) in Equation 17.17, and simplify. Show that $$u(t)=\frac{1}{r t+1 / u_{0}} \quad \text { where } \quad u(0)=u_{0}$$ is a solution for this model. What will be the eventual annual fish harvest under this harvest strategy? b. Assume \(h=\frac{3}{4} r\) in Equation 17.17 and simplify. Draw a direction field or phase plane for this model. What will be the eventual annual fish harvest under this harvest strategy? c. Assume \(h=\frac{1}{2} r\) in Equation \(17.17,\) and simplify. Draw a direction field for this model. What will be the eventual annual fish harvest under this harvest strategy? d. Which of the three strategies will provide the largest long term harvest?

Show that the variables are not separable in the equation \(y^{\prime}(t)=t+y .\) That is, there are not two functions, \(g(t)\) and \(h(y),\) which for all \(t\) and \(y \quad t+y=g(t) \times h(y)\) A procedure is to assume two such functions, \(g(t)\) and \(h(y)\) exist and then show that the following equations are incompatible. \(\begin{array}{lllll}t=0 & y=0 & t=0 & y=1 & t=1 & y=0\end{array}\) \(g(0) \times h(0)=0+0=0 \quad g(0) \times h(1)=0+1=1 \quad g(1) \times h(0)=1+0=1\) Show that \(g(0) \times h(0)=0, g(0) \times h(1)=1\) and \(g(1) \times h(0)=1\) are incompatible.

Parameter reduced population models are shown below. In the Ricker equation, find a condition on \(\gamma_{0}\) that will insure that there is a value of \(v\) for which the population is growing the fastest. In the Beverton-Holt equation show that there is no value of \(v\) for which the population is growing the fastest. In the Gompertz equation, find the value of \(v\) for which the population is growing the fastest. $$\text { a. } \quad v^{\prime}=v e^{-v}-\gamma_{0} v \text { Ricker }$$$$\begin{array}{l} \text { b. } \quad v^{\prime}=\frac{v}{1+v} \\\\\text { c. } v^{\prime}=-v \ln (v)\end{array}$$ Beverton-Holt Gompertz The three previous problems have important implications for wildlife management, at least conceptually. Suppose you are managing a wildlife population, salmon, for example, as a renewable resource, and wish to annually harvest as many salmon as possible. If you harvest too severely, the next years spawn will be low, and four years later the harvest will be limited. Your optimum strategy is to maintain the population at the level where the growth is the greatest.
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