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Chapter 9: Problem 20

Graph the solution to be sure that \(M(0)\) and \(\lim M(t)\) are correct. $$r=0.1, K=500, M_{0}=50$$

Short Answer

Expert verified
Answer: The solution is given by the function \(M(t) = \frac{500}{1 + \frac{450}{50}e^{-0.1t}}\). The graph starts at \(M(0) = 50\) and asymptotically approaches the carrying capacity \(K = 500\), as \(t\) approaches infinity.

Step by step solution

01

Write down the logistic equation

The logistic growth equation can be represented as: $$\frac{dM}{dt} = rM\left(1 - \frac{M}{K}\right)$$ where \(M(t)\) is the population at time \(t\), \(r\) is the growth rate, and \(K\) is the carrying capacity.
02

Separate variables and integrate

Separate the variables and integrate both sides: $$\int\frac{1}{M\left(1 - \frac{M}{K}\right)}dM = \int rdt$$ After performing partial fractions and solving the integrals, we get: $$\ln\left|\frac{M}{K-M}\right| = rt + C$$
03

Solve for \(M(t)\)

To find the function \(M(t)\), raise \(e\) to the power of both sides, and solve for \(M\): $$M(t) = \frac{K}{1 + \frac{K-M_0}{M_0}e^{-rt}}$$
04

Calculate \(M(0)\) and \(\lim M(t)\)

It's given that \(M_0 = 50\), \(r = 0.1\), and \(K = 500\). Plug in these values and find \(M(0)\): $$M(0) = \frac{500}{1 + \frac{500-50}{50}e^{-0.1(0)}} = 50$$ Now, find the limit as \(t\) approaches infinity: $$\lim_{t \to \infty} M(t) = \lim_{t \to \infty} \frac{500}{1 + \frac{450}{50}e^{-0.1t}} = 500$$
05

Graph the solution

To graph the solution, plot the function \(M(t) = \frac{500}{1 + \frac{450}{50}e^{-0.1t}}\). It should start at \(M(0) = 50\) and asymptotically approach the carrying capacity \(K = 500\). Verify these values on the graph, ensuring that they are correct.

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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 tools used for modeling the relationships that exist between changing quantities. They play a crucial role in science, engineering, and economics, linking the rates at which things change to the quantities themselves. In the context of the logistic growth equation, the differential equation represents the rate of change of a population over time.

Specifically, these equations can define the growth of populations or quantities that don't simply increase indefinitely but rather tend towards a maximum limit. The logistic growth equation, characterized as a first-order, non-linear ordinary differential equation, describes how a population grows rapidly at first and then slows down as it approaches a maximum sustainable size, the carrying capacity. Students frequently encounter these kinds of differential equations when analyzing population dynamics, biochemical processes, and resource consumption.
Separation of Variables
Separation of variables is a method for solving differential equations, where we manipulate the equation to isolate the independent and dependent variables on opposite sides. This technique hinges on the idea that if two products equal each other, and each product is a function of a separate variable, then each must be equal to a constant.

In our logistic growth problem, separation of variables allows us to integrate both sides of the equation to find an expression for the population, M(t), given time, t. This approach transforms the problem into one that is manageable and solvable. By breaking down the complex population behavior into integrable parts鈥攐ne showing the dependency on M and the other on t鈥攚e essentially disentangle the intertwined growth factors for easier analysis.
Carrying Capacity
Carrying capacity, denoted by the symbol K, is a concept in ecology and population dynamics that represents the maximum population size that an environment can sustain indefinitely without being degraded. It serves as an ecological ceiling for the growth of any given population.

In the logistic equation, the carrying capacity plays a critical role in controlling the growth rate. Once the population reaches or exceeds this capacity, resources become limited, growth slows down, and the population may fluctuate around K. Understanding carrying capacity is vital for predicting how populations will change over time and for making decisions around conservation, urban planning, and management of natural resources.
Population Dynamics
Population dynamics encompasses the study of how populations of living organisms change over time and space. Factors influencing these changes include birth rates, death rates, immigration, emigration, and the carrying capacity of the environment.

The logistic growth equation is a classic model in population dynamics used to represent how a population grows in a restricted environment. It captures the essence of biological constraints that cause a population to cease exponential growth and stabilize. This balancing act is fundamental to the resilience and sustainability of ecosystems. Through models like the logistic equation, we gain insights into how interventions, environmental changes, or resource depletion can influence the survival and growth of species.

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Most popular questions from this chapter

Solving initial value problems Solve the following initial value problems. $$y^{\prime}(x)=3 x^{2}-3 x^{-4}, y(1)=0$$

Logistic equation for spread of rumors Sociologists model the spread of rumors using logistic equations. The key assumption is that at any given time, a fraction \(y\) of the population, where \(0 \leq y \leq 1,\) knows the rumor, while the remaining fraction \(1-y\) does not. Furthermore, the rumor spreads by interactions between those who know the rumor and those who do not. The number of such interactions is proportional to \(y(1-y) .\) Therefore, the equation that describes the spread of the rumor is \(y^{\prime}(t)=k y(1-y)\) for \(t \geq 0,\) where \(k\) is a positive real number and \(t\) is measured in weeks. The number of people who initially know the rumor is \(y(0)=y_{0},\) where \(0 \leq y_{0} \leq 1\). a. Solve this initial value problem and give the solution in terms of \(k\) and \(y_{0}.\) b. Assume \(k=0.3\) weeks \(^{-1}\) and graph the solution for \(y_{0}=0.1\) and \(y_{0}=0.7.\) c. Describe and interpret the long-term behavior of the rumor function, for any \(0 \leq y_{0} \leq 1\).

For each of the following stirred tank reactions, carry out the following analysis. a. Write an initial value problem for the mass of the substance. b. Solve the initial value problem. A \(1500-\) L tank is initially filled with a solution that contains \(3000 \mathrm{g}\) of salt. A salt solution with a concentration of \(20 \mathrm{g} / \mathrm{L}\) flows into the tank at a rate of \(3 \mathrm{L} / \mathrm{min.}\). The thoroughly mixed solution is drained from the tank at a rate of \(3 \mathrm{L} / \mathrm{min}\).

Consider the general first-order linear equation \(y^{\prime}(t)+a(t) y(t)=f(t) .\) This equation can be solved, in principle, by defining the integrating factor \(p(t)=\exp \left(\int a(t) d t\right) .\) Here is how the integrating factor works. Multiply both sides of the equation by \(p\) (which is always positive) and show that the left side becomes an exact derivative. Therefore, the equation becomes $$p(t)\left(y^{\prime}(t)+a(t) y(t)\right)=\frac{d}{d t}(p(t) y(t))=p(t) f(t)$$ Now integrate both sides of the equation with respect to t to obtain the solution. Use this method to solve the following initial value problems. Begin by computing the required integrating factor. $$y^{\prime}(t)+\frac{2 t}{t^{2}+1} y(t)=1+3 t^{2}, y(1)=4$$

Growth rate functions a. Show that the logistic growth rate function \(f(P)=r P\left(1-\frac{P}{K}\right)\) has a maximum value of \(\frac{r K}{4}\) at the point \(P=\frac{K}{2}\) b. Show that the Gompertz growth rate function \(f(M)=-r M \ln \left(\frac{M}{K}\right)\) has a maximum value of \(\frac{r K}{e}\) at the point \(M=\frac{K}{e}\)
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