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This is a general problem about the logistic law of growth. A population satisfies the logistic law ( \(3.58\) ) and has \(x_{0}\) members at time \(t_{0}\). (a) Solve the differential equation (3.58) and thus express the population \(x\) as a function of \(t\). (b) Show that as \(t \rightarrow \infty\), the population \(x\) approaches the limiting value \(k / \lambda\) (c) Show that \(d x / d t\) is increasing if \(xk / 2 \lambda\). (d) Graph \(x\) as a function of \(t\) for \(t>t_{0}\). (e) Interpret the results of parts (b), (c), and (d).

Short Answer

Expert verified
In summary, the population x follows the logistic growth model, with x(t) given by \(x(t)=Ce^{\lambda t}\). As time goes to infinity, the population approaches the limiting value \(k / \lambda\). The growth rate increases when \(x < \frac{k}{2\lambda}\) and decreases when \(x > \frac{k}{2\lambda}\), indicating that the maximum growth rate occurs halfway to the maximum capacity. The graph of x(t) represents an S-shaped curve, which is typical for logistic population growth.

Step by step solution

01

Identify the logistic differential equation

The logistic growth model is given by the differential equation: \[\frac{dx}{dt} = \lambda x\left(1- \frac{x}{k}\right)\]
02

Solve the differential equation

Separate the variables to integrate: \[\frac{dx}{x\left(1- \frac{x}{k}\right)} = \lambda dt\] Now, integrate both sides: \[\int \frac{dx}{x\left(1- \frac{x}{k}\right)} = \int \lambda dt\] To solve the left-hand side integral, we can use partial fraction decomposition. Let: \[\frac{1}{x\left(1- \frac{x}{k}\right)} = \frac{A}{x} + \frac{B}{1- \frac{x}{k}}\] Multiplying both sides by \(x(1- \frac{x}{k})\): \[1 = A(1 - \frac{x}{k}) + Bx\] Comparing the coefficients, we get: \(A+B=0\) and \(-\frac{A}{k}=0\) Solving for A and B: \[A = 0\, \text{and}\, B = 0\] Thus, the left-hand side integral becomes: \[\int \frac{dx}{x} = \int \lambda dt\] Solve the integrals by applying the natural log: \[\ln(x) = \lambda t + C_1\] Now, find the population x as a function of t: \[x(t)=Ce^{\lambda t}\] b) Show that as t→∞, the population x approaches the limiting value k/λ
03

Evaluate x as t tends to infinity

As t→∞, dx/dt must approach 0, according to the logistic growth law. So, we have: \[0 = \lambda x(t)\left(1- \frac{x(t)}{k}\right)\]
04

Calculate the limiting value of x

To simplify the equation, multiply both sides by k: \[0 = \lambda x(t)k - \lambda x^{2}(t)\] Divide both sides by λ: \[0 = x(t)k - x^{2}(t)\] Solve for x: \[x(t) = \frac{k}{\lambda}\] c) Show that \( \frac{dx}{dt} \) is increasing if xk/(2λ) Since we have the expression for \(\frac{dx}{dt}\), we must take the derivative and check the sign. \[\frac{d^2x}{dt^2} = \frac{d}{dx}\left(\lambda x\left(1- \frac{x}{k}\right)\right)\] By applying the product rule: \[\frac{d^2x}{dt^2} = \lambda\left(1- \frac{2x}{k}\right)\] Now, \(\frac{d^2x}{dt^2} > 0\) if \(x < \frac{k}{2\lambda}\) and \(\frac{d^2x}{dt^2} < 0\) if \(x > \frac{k}{2\lambda}\), which means dx/dt is increasing when \(x < \frac{k}{2\lambda}\) and decreasing when \(x > \frac{k}{2\lambda}\). d) Graph x as a function of t for t>t0 Simply graph the function \(x(t)=Ce^{\lambda t}\) on a graphing tool or software, making sure to only include values of t that are greater than t0. e) Interpret the results of parts (b), (c), and (d). In part (b), we found that the population x approaches a limiting value of \(k / \lambda\) as time goes to infinity. This indicates that the population grows rapidly at the beginning until it reaches a maximum point. The result from part (c) shows that the rate of population growth increases if x is less than \(k / (2\lambda)\) and decreases if x is greater than \(k / (2\lambda)\). This indicates that the population has a maximum growth rate when it reaches half of its maximum capacity. The graph of x as a function of t for t>t0 in part (d) shows a typical S-shaped curve representing the population's logistic growth, which is the behavior expected from real-world population growth. It shows an initial rapid growth followed by a slower growth as the population reaches its maximum capacity.

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Key Concepts

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

Differential Equations
Differential equations are mathematical equations that describe the relationship between a function and its derivatives. They are essential when discussing change, as they help model various dynamic systems over time. In the context of population dynamics, differential equations are a powerful tool to model how a population grows and changes. When we talk about the logistic growth model, we often refer to differential equations that express how a population changes with respect to time.

In the logistic growth model, the differential equation is given by \[\frac{dx}{dt} = \lambda x\left(1- \frac{x}{k}\right)\]. Here, \(dx/dt\) represents the rate of change of the population size \(x\) over time \(t\). \(\lambda\) is the growth rate, and \(k\) is the carrying capacity of the environment, which is the maximum population size that the environment can sustain indefinitely.
Population Dynamics
Population dynamics involve studying the short-term and long-term changes in the size and composition of populations, and the driving factors behind these changes. These dynamics are influenced by birth rates, death rates, and migration patterns. In the logistic growth model, population dynamics are captured by how populations initially grow rapidly due to abundant resources and slow down as resources become scarce.

The logistic model reflects reality more accurately than simplistic exponential growth models. It considers the carrying capacity \(k\), limiting the ultimate size of the population. This is due to environmental constraints such as food, space, and other essential resources. As the population approaches this carrying capacity, the growth rate decreases, highlighting how the size of a population can be realistically bounded by ecological factors.
Growth Rate Analysis
Growth rate analysis is crucial for understanding how quickly a population increases or decreases over time. It is often expressed as a percentage change in the size of the population over a given period. In logistic growth, growth rate analysis helps identify phases where the population grows rapidly or stabilizes.

The logistic growth model describes growth as initially exponential when \(x\) is much smaller than \(k\), resulting in the population growing at a rate proportional to its current size \(x\) and the intrinsic growth rate \(\lambda\). The growth rate starts to decline once \(x\) surpasses half of the carrying capacity \(\frac{k}{2}\). For \(x < \frac{k}{2\lambda}\), \(dx/dt\) is increasing, indicating accelerating growth. When \(x > \frac{k}{2\lambda}\), \(dx/dt\) decreases, showing decelerating growth as the population nears its maximum capacity.
S-shaped Curve
The S-shaped curve, also known as the sigmoid curve, is a graphical representation of the logistic growth model. It illustrates how the population grows rapidly initially, slows as it approaches the carrying capacity, and finally stabilizes. This shape is typical of many natural growth processes, from populations of organisms to the spread of innovations.

The initial section of the S-curve reflects fast exponential growth due to minimal resource limitations. Midway through the curve, the growth slows, demonstrating diminishing returns as competition for resources intensifies. Finally, at the top flattening part of the curve, growth ceases, and the size of the population stabilizes at or near the carrying capacity \(k\). Understanding this curve can help predict how populations will behave over time and plan accordingly in scenarios like wildlife management, urban development, and resource allocation.

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Most popular questions from this chapter

A chemical reaction converts a certain chemical into another chemical, and the rate at which the first chemical is converted is proportional to the amount of this chemical present at any time. At the end of one hour, \(50 \mathrm{gm}\) of the first chemical remain; while at the end of three hours, only \(25 \mathrm{gm}\) remain. (a) How many grams of the first chemical were present initially? (b) How many grams of the first chemical will remain at the end of five hours? (c) In how many hours will only \(2 \mathrm{gm}\) of the first chemical remain?

Assume that the rate at which radioactive nuclei decay is proportional to the number of such nuclei that are present in a given sample. In a certain sample \(10 \%\) of the original number of radioactive nuclei have undergone disintegration in a period of 100 years. (a) What percentage of the original radioactive nuclei will remain after 1000 years? (b) In how many years will only one-fourth of the original number remain?

Assume that the rate of change of the human population of the earth is proportional to the number of people on earth at any time, and suppose that this population is increasing at the rate of \(2 \%\) per year. The 1979 World Almanac gives the 1978 world population estimate as 4219 million; assume this figure is in fact correct. (a) Using this data, express the human population of the earth as a function of time. (b) According to the formula of part (a), what was the population of the earth in \(1950 ?\) The 1979 World Almanac gives the 1950 world population estimate as 2510 million. Assuming this estimate is very nearly correct, comment on the accuracy of the formula of part (a) in checking such past populations. (c) According to the formula of part (a), what will be the population of the earth in 2000? Does this seem reasonable? (d) According to the formula of part (a), what was the population of the earth in \(1900 ?\) The 1970 World Almanac gives the 1900 world population estimate as 1600 million. Assuming this estimate is very nearly correct, comment on the accuracy of the formula of part (a) in checking such past populations. (e) According to the formula of part (a), what will be the population of the earth in 2100 ? Does this seem reasonable?

A given family of curves is said to be self-orthogonal if its family of orthogonal trajectories is the same as the given family. Show that the family of parabolas \(y^{2}=2 c x+c^{2}\) is self orthogonal.

The air in a room \(50 \mathrm{ft}\) by \(20 \mathrm{ft}\) by \(8 \mathrm{ft}\) tests \(0.2 \%\) carbon dioxide. Starting at \(t=0\), outside air testing \(0.05 \%\) carbon dioxide is admitted to the room. How many cubic feet of this outside air must be admitted per minute in order that the air in the room test \(0.1 \%\) at the end of \(30 \mathrm{~min}\) ?

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