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A quantity \(P\) satisfies the differential equation \(\frac{d P}{d t}=k P\left(1-\frac{P}{100}\right)\) Sketch approximate solutions satisfying each of the following initial conditions: (a) \(\quad P_{0}=8\) (b) \(\quad P_{0}=70\) (c) \(\quad P_{0}=125\)

Short Answer

Expert verified
The solutions show logistic growth or decay toward the carrying capacity of 100, depending on the initial value.

Step by step solution

01

Introduction to the Logistic Equation

The given differential equation \( \frac{dP}{dt} = kP(1 - \frac{P}{100}) \) resembles the logistic growth model. This type of equation models how a population \( P \) grows considering both the growth rate \( k \) and a carrying capacity, which in this case is 100.
02

Equilibrium Points

To find equilibrium points, set \( \frac{dP}{dt} = 0 \). This yields the solutions \( P = 0 \) and \( P = 100 \) as equilibrium points. Thus, values below 100 result in growth, while values above 100 indicate a decrease back toward the carrying capacity.
03

Sketch for Initial Condition P_0 = 8

For \( P_0 = 8 \), \( P \) is below the carrying capacity (100). The solution curve starts from 8 and asymptotically approaches the carrying capacity (100) as \( t \to \infty \), showing logistic growth.
04

Sketch for Initial Condition P_0 = 70

For \( P_0 = 70 \), \( P \) is still below carrying capacity. The solution curve starts from 70 and increases, though slower than starting further from 100, as it approaches the carrying capacity (100).
05

Sketch for Initial Condition P_0 = 125

For \( P_0 = 125 \), \( P \) is above the carrying capacity. This results in \( P \) decreasing toward the carrying capacity. The curve starts at 125 and decreases towards 100 over time.

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Key Concepts

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

Differential Equation
A differential equation involves an unknown function and its derivatives. It describes the rate of change of a quantity concerning another variable. In the context of the logistic growth equation, it models how a population evolves over time.

The given differential equation is \( \frac{dP}{dt} = kP \left(1 - \frac{P}{100}\right) \). Here, \( P \) represents the population, and \( t \) is time.
  • The term \( \frac{dP}{dt} \) stands for the rate of change of the population \( P \) with respect to time.
  • \( k \) is a constant that represents the growth rate. It dictates how quickly the population grows.
  • The equation considers the self-limiting nature of population growth by including the term \( 1 - \frac{P}{100} \).
This logistic growth captures the essence that as the population approaches a certain level (in this case, 100), its rate of growth reduces gradually due to limited resources or other factors.
Equilibrium Points
Equilibrium points in differential equations are where the rate of change is zero. These points indicate stable populations or scenarios where the population remains constant.

To find equilibrium points in the logistic equation, set \( \frac{dP}{dt} = 0 \). Thus, solving

\[ kP \left(1 - \frac{P}{100}\right) = 0 \]
leads to two possibilities:
  • \( P = 0 \)
  • \( P = 100 \)
At these points, the population does not change over time.

- \( P = 0 \) represents a scenario where no individuals exist, and remains an unchanging equilibrium if starting from zero.
- \( P = 100 \) is the carrying capacity, where the population remains stable.

Any deviation from these points will lead the population to adjust back towards equilibrium, reflecting real-world dynamics where populations naturally stabilize under constrained resources.
Population Dynamics
Population dynamics describe how populations change over time. In the logistic growth model, the dynamics are reflected through the logistic function.

Initially, when the population \( P \) is much smaller than the carrying capacity, growth is exponential.
  • Population grows fast due to less competition for resources.
  • It is represented by the term \( kP \) in the equation.

As \( P \) nears 100, growth slows due to increased competition and limited resources. This is modulated by \( 1 - \frac{P}{100} \), effectively reducing the rate \( \frac{dP}{dt} \).

When \( P \) exceeds 100 due to overshooting, the negative growth rate pulls it back towards 100. This showcases a regulatory mechanism ensuring population stability.

These dynamics give rise to the characteristic S-shaped logistic curve. It demonstrates how real populations adjust over time.
Carrying Capacity
Carrying capacity is the maximum population size an environment can sustain indefinitely without degrading. It is a crucial component of the logistic growth model.

In our model, the carrying capacity is determined to be 100.
  • 91Ó°ÊÓ such as food, habitat, and other essentials constrain this capacity.
  • It signifies a balance between population size and the resources available.

As the population nears carrying capacity:
  • Growth slows down as resources become scarcer.
  • The logistic equation reflects this slowdown, eventually flattening out.

When \( P \) is below the carrying capacity, populations tend to grow. If \( P \) is above, they decrease back to this level.

This concept underlines the importance of environmental limits in population studies. It describes how ecosystems maintain stability and prevent overexploitation.

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Most popular questions from this chapter

(a) Sketch the slope field for \(y^{\prime}=-y / x\) (b) Sketch several solution curves. (c) Solve the differential equation analytically.

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 \(g(0)=1\) and \(g(x)\) is increasing for \(x \geq 0,\) then \(f(x)\) is also increasing for \(x \geq 0.\)

According to an article in The New York Times,' ' pigweed has acquired resistance to the weedkiller Roundup. Let \(N\) be the number of acres, in millions, where Roundup-resistant pigweed is found. Suppose the relative growth rate, \((1 / N) d N / d t,\) was \(15 \%\) when \(N=5\) and \(14.5 \%\) when \(N=10 .\) Assuming the relative growth rate is a linear function of \(N,\) write a differential equation to model \(N\) as a function of time, and predict how many acres will eventually be afflicted before the spread of Roundup-resistant pigweed halts.

The population data from another experiment on yeast by the ecologist G. F. Gause is given. $$\begin{array}{l|c|c|c|c|c|c|c}\hline \text { Time (hours) } & 0 & 13 & 32 & 56 & 77 & 101 & 125 \\\\\hline \text { Yeast pop } & 1.00 & 1.70 & 2.73 & 4.87 & 5.67 & 5.80 & 5.83 \\\\\hline\end{array}$$ (a) Do you think the population is growing exponentially or logistically? Give reasons for your answer. (b) Estimate the value of \(k\) (for either model) from the first two pieces of data. If you chose a logistic model in part (a), estimate the carrying capacity, \(L,\) from the data. (c) Sketch the data and the approximate growth curve given by the parameters you estimated.

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 $$\begin{array}{l} \frac{d x}{d t}=-a y \\ \frac{d y}{d t}=-b x \quad a, b>0 \end{array}$$. (a) For two armies of strengths \(x\) and \(y\) fighting a conventional battle governed by Lanchester's differential equations, write a differential equation involving \(d y / d x\) and the constants of attrition \(a\) and \(b\) (b) Solve the differential equation and hence show that the equation of the phase trajectory is $$a y^{2}-b x^{2}=C$$ for some constant \(C .\) This equation is called Lanchester's square law. The value of \(C\) depends on the initial sizes of the two armies.

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