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Chapter 8: Problem 15

Write a logistic equation with the following parameter values. Then solve the initial value problem and graph the solution. Let \(r\) be the natural growth rate, \(K\) the carrying capacity, and \(P_{0}\) the initial population. $$r=0.2, K=300, P_{0}=50$$

Short Answer

Expert verified
To summarize, given the parameters \(r\), \(K\), and the initial population \(P_{0}\), we found the logistic equation as: $$\frac{dP}{dt} = 0.2P \left(1-\frac{P}{300}\right)$$ We then solved the initial value problem with the initial condition and found the solution function: $$P(t) = \frac{300(50)}{250(1-e^{-\frac{0.2t}{300}})+50}$$ Graphing the solution \(P(t)\) over the interval \(0 \leq t \leq 100\) will show how the population approaches the carrying capacity \(K=300\) over time.

Step by step solution

01

Write the Logistic Equation with Given Parameters

Using the logistic equation and the given parameter values, we can write the equation as: $$\frac{dP}{dt} = 0.2P \left(1-\frac{P}{300}\right)$$
02

Solve the Initial Value Problem

To solve this differential equation, we need to solve for \(P(t)\) given the initial value \(P_{0}=50\). Let's rewrite the given ODE in the following form $$\frac{dP}{t} = \frac{0.2P(300-P)}{300}$$ We have the pattern of a separable equation: $$\frac{dP}{P(300-P)} = \frac{0.2 dt}{300}$$ Now integrate both sides: $$\int\left(\frac{1}{P}+\frac{1}{300-P}\right)dP = \int\frac{0.2}{300}dt$$ Use partial fraction decomposition on the left-hand side: $$\int\left(\frac{A}{P}+\frac{B}{300-P}\right)dP = A \int\frac{dP}{P} +B \int\frac{dP}{300-P}$$ where \(A\) and \(B\) are constants. By matching coefficients, we find \(A=1\) and \(B=1\). Therefore, the equation becomes: $$\int\frac{dP}{P} + \int\frac{dP}{300-P} = \int\frac{0.2}{300}dt$$ Now integrate: $$\ln |P|-\ln |300-P| = \frac{0.2t}{300} + C$$ Apply exponentiation to eliminate the logarithm: $$\frac{P}{300-P} = e^{\frac{0.2t}{300}+C}$$
03

Use the Initial Condition

Solve for the constant \(C\) using the initial condition \(P_{0}=50\) at \(t=0\): $$\frac{50}{300-50} = e^{C}$$ $$C = \ln\left(\frac{50}{250}\right)$$
04

Find the Solution \(P(t)\)

Rewriting the previous solution we get $$P(t) = \frac{300e^{\frac{0.2t}{300}+\ln(\frac{50}{250})}}{1+e^{\frac{0.2t}{300}+\ln(\frac{50}{250})}}$$ Simplifying this expression gives us the final solution: $$P(t) = \frac{300(50)}{250(1-e^{-\frac{0.2t}{300}})+50}$$
05

Graph the Solution

To graph the solution, plot the function \(P(t)\) over a suitable time interval, say \(0 \leq t \leq 100\). Note that the population will approach the carrying capacity \(K=300\) as time goes on, and observing the graph will help visualize this.

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Key Concepts

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

Natural Growth Rate
In the context of logistic equations, the natural growth rate, represented by the symbol \(r\), is a key parameter. It refers to the rate at which a population grows in an ideal, unlimited environment. For the exercise given, \(r\) is 0.2. This value indicates the population鈥檚 intrinsic growth rate without any restrictions. In reality, the natural growth rate provides insight into how quickly a species would increase if resources were dangerously abundant. It becomes the initial driving force in the logistic model before environmental factors come into play.
Carrying Capacity
Carrying capacity, denoted as \(K\), is a crucial component in understanding population dynamics. It represents the maximum population size that an environment can sustain long-term without degradation. In the logistic equation from the exercise, \(K\) is set at 300. This value influences how the population growth rate slows as the population approaches this threshold. The concept helps to model realistic population growth by showing how resources, disease, and other limiting factors constrain unchecked growth. This ultimately leads the population to stabilize around the carrying capacity value.
Separable Equations
The logistic equation from the exercise is a classic example of a separable differential equation. Separable equations allow us to isolate variables on different sides of the equation, facilitating the process of integrating each side independently. By transforming the equation \(\frac{dP}{P(300-P)} = \frac{0.2}{300}dt\), it becomes possible to integrate the function of \(P\) and \(t\) separately. This method greatly simplifies the problem-solving process. It highlights one of the powerful techniques of calculus where complex equations can be disentangled into more manageable parts.
Initial Value Problem
An initial value problem involves solving a differential equation with a given initial condition. For the logistic equation provided, the initial population \(P_0\) is 50 at time \(t=0\). The initial value problem aspect introduces a real-world starting point to the model, which allows the solution to reflect a specific scenario accurately. By applying this initial condition, you solve for the unknown constant in your solution, tailoring the general solution to fit the particular case described. This is crucial for modeling scenarios that reflect actual conditions.

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Most popular questions from this chapter

A special class of first-order linear equations have the form \(a(t) y^{\prime}(t)+a^{\prime}(t) y(t)=f(t),\) where \(a\) and \(f\) are given functions of \(t.\) Notice that the left side of this equation can be written as the derivative of a product, so the equation has the form $$a(t) y^{\prime}(t)+a^{\prime}(t) y(t)=\frac{d}{d t}(a(t) y(t))=f(t)$$ Therefore, the equation can be solved by integrating both sides with respect to \(t.\) Use this idea to solve the following initial value problems. $$t^{3} y^{\prime}(t)+3 t^{2} y=\frac{1+t}{t}, y(1)=6$$

Plot a direction field for the following differential equation with a graphing utility. Then find the solutions that are constant and determine which initial conditions \(y(0)=A\) lead to solutions that are increasing in time. $$y^{\prime}(t)=0.05(y+1)^{2}(t-1)^{2},|t| \leq 3 \text { and }|y| \leq 3$$

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 and graph the solution to be sure that \(m(0)\) and \(\lim _{t \rightarrow \infty} m(t)\) are correct. A 2000 -L tank is initially filled with a sugar solution with a concentration of \(40 \mathrm{g} / \mathrm{L} .\) A sugar solution with a concentration of \(10 \mathrm{g} / \mathrm{L}\) flows into the tank at a rate of \(10 \mathrm{L} / \mathrm{min} .\) The thoroughly mixed solution is drained from the tank at a rate of \(10 \mathrm{L} / \mathrm{min}\).

The Gompertz growth equation is often used to model the growth of tumors. Let \(M(t)\) be the mass of a tumor at time \(t \geq 0 .\) The relevant initial value problem is $$ \frac{d M}{d t}=-r M \ln \left(\frac{M}{K}\right), M(0)=M_{0} $$ a. Graph the growth rate function \(R(M)=-r M \ln \left(\frac{M}{K}\right)\) (which equals \(M^{\prime}(t)\) ) assuming \(r=1\) and \(K=4 .\) For what values of \(M\) is the growth rate positive? For what value of \(M\) is the growth rate a maximum? b. Solve the initial value problem and graph the solution for \(r=1, K=4,\) and \(M_{0}=1 .\) Describe the growth pattern of the tumor. Is the growth unbounded? If not, what is the limiting size of the tumor? c. In the general solution, what is the meaning of \(K ?\) where \(r\) and \(K\) are positive constants and \(0 < M_{0} < K\)

Widely used models for population growth involve the logistic equation \(P^{\prime}(t)=r P\left(1-\frac{P}{K}\right),\) where \(P(t)\) is the population, for \(t \geq 0,\) and \(r>0\) and \(K>0\) are given constants. a. Verify by substitution that the general solution of the equation is \(P(t)=\frac{K}{1+C e^{-n}},\) where \(C\) is an arbitrary constant. b. Find that value of \(C\) that corresponds to the initial condition \(P(0)=50\). c. Graph the solution for \(P(0)=50, r=0.1,\) and \(K=300\). d. Find \(\lim _{t \rightarrow \infty} P(t)\) and check that the result is consistent with the graph in part (c).
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