91Ó°ÊÓ

Consider the differential equation $$\frac{d P}{d t}=k P^{1+c}$$ where \(k>0\) and \(c \geq 0\). In Section 3.1 we saw that in the case \(c=0\) the linear differential equation \(d P / d t=k P\) is a mathematical model of a population \(P(t)\) that exhibits unbounded growth over the infinite time interval \([0, \infty),\) that is, \(P(t) \rightarrow \infty\) as \(t \rightarrow \infty\). See Example 1 on page 85.(a) Suppose for \(c=0.01\) that the nonlinear differential equation\\[\frac{d P}{d t}=k P^{1.01}, \quad k>0\\] is a mathematical model for a population of small animals, where time \(t\) is measured in months. Solve the differential equation subject to the initial condition \(P(0)=10\) and the fact that the animal population has doubled in 5 months. (b) The differential equation in part (a) is called a doomsday equation because the population \(P(t)\) exhibits unbounded growth over a finite time interval \((0, T),\) that is, there is some time \(T\) such that \(P(t) \rightarrow \infty\) as \(t \rightarrow T^{-} .\) Find \(T\) (c) From part (a), what is \(P(50) ? P(100) ?\)

Step by step solution

01

Start with the given differential equation

We are given the differential equation \( \frac{dP}{dt} = kP^{1.01} \) and the initial condition \( P(0) = 10 \). We also know the population doubles in 5 months, i.e., \( P(5) = 20 \). We need to solve this differential equation.

02

Separate the variables

To solve the equation, we first separate the variables:\[ \frac{dP}{P^{1.01}} = k \, dt. \]

03

Integrate both sides

Next, integrate both sides of the equation. The left-hand side involves a power of \( P \):\[ \int P^{-1.01} \, dP = \int k \, dt. \]

04

Solve the integrals

The integral on the left can be solved using the power rule for integration:\[ \frac{P^{-0.01}}{-0.01} = kt + C, \]where \( C \) is the constant of integration. Simplifying, we get:\[ P^{-0.01} = -0.01kt + C. \]

05

Solve for C using initial condition

Substitute the initial condition \( P(0) = 10 \) into the equation:\[ 10^{-0.01} = C. \]Calculating this gives \( C = 10^{-0.01}. \)

06

Express the solution for P

Now we have:\[ P(t)^{-0.01} = -0.01kt + 10^{-0.01}. \]Solving for \( P(t) \), we find:\[ P(t) = \left(10^{-0.01} - 0.01kt\right)^{-100}. \]

07

Determine k using P(5)=20

Using \( P(5) = 20 \):\[ 20^{-0.01} = 10^{-0.01} - 0.01k \times 5. \]Calculate \( 20^{-0.01} \) and solve for \( k \). We find\[ k \approx 0.009851725\].

08

Find time T when P(t) approaches infinity

As \( P(t) \to \infty \), the term \( 0.01kt = 10^{-0.01} \). So,\[ T = \frac{10^{-0.01}}{0.01k}. \]Substitute the value of \( k \) to find \( T \).

09

Calculate P(50) and P(100)

Substitute \( t = 50 \) and \( t = 100 \) into the expression for \( P(t) \) to find \( P(50) \) and \( P(100) \). For these values of \( t \), \( P(t) \) will be large, but finite for \( t < T \).

10

Conclusion

The population \( P(t) \) grows unbounded until it reaches time \( T \). At \( t = 50 \) and \( t = 100 \), \( P(t) \) will be immensely large, indicating rapid growth leading to a 'doomsday' scenario.

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Key Concepts

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

Population Modeling

Population modeling is a key concept in understanding how populations change over time. It involves creating mathematical representations of population dynamics like growth, decline, or change in response to various factors. When modeling a population with a differential equation, we describe how the population changes at any point in time, rather than just the population size at a specific time.
For example, when we consider a population of small animals, we might use a differential equation to describe the rate of population growth. This helps us predict future population sizes based on initial conditions and factors influencing growth. In our case, the differential equation is: \[\frac{dP}{dt} = kP^{1.01}\]where \(k>0\). Here, \(P(t)\) is the population at time \(t\), and \(k\) is a positive constant that affects the growth rate.
Through population modeling, we can predict scenarios such as exponential growth, where populations double over time, helping to pinpoint specific times and conditions that lead to quick expansions or potential ecological crises.

Doomsday Equation

The term "doomsday equation" refers to a situation in population modeling where the population grows unbounded over a finite period. Unlike normal exponential growth, where populations grow continually over an infinite period, this model shows that the population will reach infinity in a finite time. In this exercise, the differential equation \[\frac{dP}{dt} = kP^{1.01}\]models such a scenario. With the condition that the animal population doubles in 5 months, we can find this finite time, \(T\), where the population approaches infinity.
This behavior is characteristic of a doomsday equation because eventually, the growth rate becomes unsustainable as time approaches \(T\). At \(t=T\), the rate of change skyrockets, leading to a theoretical population explosion. This helps illustrate the potential for rapid unsustainable growth in a real-world context, such as environmental overpopulation scenarios or viral outbreaks.

Nonlinear Differential Equation

Nonlinear differential equations are a class of equations where the function and its derivatives do not line up in a straight manner, frequently making them more complex to solve than linear differential equations. In this case, the equation \[\frac{dP}{dt} = kP^{1.01}\]is nonlinear because of the exponent \(1.01\) on \(P\). The presence of a nonlinear term \(P^{1+c}\) introduces behaviors in solutions that are not present in linear systems, such as finite-time singularities seen in the doomsday scenario.
To solve this nonlinear equation, we often use techniques like variable separation and integration. This involves separating \(P\) and \(t\) on different sides of the equation, and then integrating both sides. This process accommodates solving for \(P\) as a function of \(t\).
These types of equations are paramount in fields like physics, biology, and economics, because they can model real-world systems where changes are not perfectly proportional and predict more complex system behaviors.