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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—one showing the dependency on M and the other on t—we 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

For the following separable equations, carry out the indicated analysis. a. Find the general solution of the equation. b. Find the value of the arbitrary constant associated with each initial condition. (Each initial condition requires a different constant.) c. Use the graph of the general solution that is provided to sketch the solution curve for each initial condition. $$\begin{aligned}&e^{-y / 2} y^{\prime}(x)=4 x \sin x^{2}-x ; y(0)=0,\\\&y(0)=\ln \left(\frac{1}{4}\right), y(\sqrt{\frac{\pi}{2}})=0\end{aligned}$$

Explain why or why not Determine whether the following statements are true and give an explanation or counterexample. a. If the growth rate function for a population model is positive, then the population is increasing. b. The solution of a stirred tank initial value problem always approaches a constant as \(t \rightarrow \infty\) c. In the predator-prey models discussed in this section, if the initial predator population is zero and the initial prey population is positive, then the prey population increases without bound.

Is the differential equation that describes a stirred tank reaction (as developed in this section) linear or nonlinear? What is its order?

What is a carrying capacity? Mathematically, how does it appear on the graph of a population function?

Solve the differential equation for Newton's Law of Cooling to find the temperature function in the following cases. Then answer any additional questions. A glass of milk is moved from a refrigerator with a temperature of \(5^{\circ} \mathrm{C}\) to a room with a temperature of \(20^{\circ} \mathrm{C}\). One minute later the milk has warmed to a temperature of \(7^{\circ} \mathrm{C}\). After how many minutes does the milk have a temperature that is \(90 \%\) of the ambient temperature?

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