Problem 21
Question
Suppose that a population, whose size at time \(t\) is given by \(N(t)\), grows according to $$\frac{d N}{d t}=\frac{1}{100} N^{2}, \quad \text { with } N(0)=10$$ (a) Solve \((8.43)\). (b) Graph \(N(t)\) as a function of \(t\) for \(0 \leq t<10\). What happens as \(t \rightarrow 10 ?\) Explain in words what this means.
Step-by-Step Solution
Verified Answer
The population grows towards infinity as \( t \to 10 \), showing a mathematical blow-up.
1Step 1: Separate Variables
We start with the differential equation given \( \frac{d N}{d t} = \frac{1}{100} N^{2} \). To solve this, first separate variables by dividing both sides by \( N^2 \) and multiplying by \( dt \): \[ \frac{dN}{N^2} = \frac{1}{100} dt \].
2Step 2: Integrate Both Sides
Now, integrate both sides of the equation. The integral of \( \frac{dN}{N^2} \) is \( -\frac{1}{N} \), and the integral of \( \frac{1}{100} dt \) is \( \frac{1}{100} t + C \), where \( C \) is the constant of integration. Therefore, \[ -\frac{1}{N} = \frac{1}{100} t + C \].
3Step 3: Solve for C using Initial Condition
Use the initial condition \( N(0) = 10 \) to find \( C \). Substitute \( t = 0 \) and \( N = 10 \) into the equation: \[ -\frac{1}{10} = C \]. So, \( C = -\frac{1}{10} \).
4Step 4: Solve for N as a Function of t
Substitute \( C = -\frac{1}{10} \) back into the equation: \[ -\frac{1}{N} = \frac{1}{100} t - \frac{1}{10} \]. Solve for \( N \): \[ \frac{1}{N} = -\left( \frac{1}{100} t - \frac{1}{10} \right) \]. This simplifies to \[ N = \frac{1}{\frac{1}{10} - \frac{1}{100} t} \].
5Step 5: Analyze the Graph and Behavior as t Approaches 10
Plot the function \( N(t) = \frac{1}{\frac{1}{10} - \frac{1}{100} t} \) for \( 0 \leq t < 10 \). As \( t \to 10 \), \( \frac{1}{10} - \frac{1}{100}t \to 0 \) from positive values, and \( N(t) \to \infty \). This indicates that the population size becomes infinitely large, implying a vertical asymptote at \( t = 10 \).
6Step 6: Explain the Meaning
The growth implies that as time approaches 10 units, the population undergoes a rapid acceleration in growth and reaches an unbounded size. This phenomenon is known as a "mathematical blow-up" where the model predicts the population can no longer be sustained under this growth law.
Key Concepts
Population GrowthSeparation of VariablesIntegrationInitial Value Problem
Population Growth
Population growth is a common scenario modeled by differential equations. In this problem, the population size at time \( t \) is described by \( N(t) \), and it increases according to a specific growth law given by the differential equation \( \frac{dN}{dt} = \frac{1}{100}N^2 \). This equation suggests that the growth rate depends on the square of the population size, indicating non-linear growth.
As the population grows, the rate at which it grows also increases, making this an example of exponential or super-exponential growth. The larger the population, the faster it grows, potentially leading to a situation where the population size becomes incredibly large in a short period. Understanding how populations grow under different scenarios helps us appreciate real-world biological and ecological systems.
This type of growth model is useful for studying populations that reproduce rapidly, like algae blooms or bacterial colonies, where the resource availability and environmental factors play crucial roles.
As the population grows, the rate at which it grows also increases, making this an example of exponential or super-exponential growth. The larger the population, the faster it grows, potentially leading to a situation where the population size becomes incredibly large in a short period. Understanding how populations grow under different scenarios helps us appreciate real-world biological and ecological systems.
This type of growth model is useful for studying populations that reproduce rapidly, like algae blooms or bacterial colonies, where the resource availability and environmental factors play crucial roles.
Separation of Variables
Separation of Variables is a common technique for solving differential equations, especially when dealing with equations that can be rearranged to isolate different functions of the variables you're solving for. Our equation \( \frac{dN}{dt} = \frac{1}{100} N^2 \) is an example where separation of variables can be applied.
By moving all terms involving \( N \) to one side and all terms involving \( t \) to the other, the equation is rearranged as \( \frac{dN}{N^2} = \frac{1}{100} dt \). This step simplifies the integration process by creating two separate integrals that can be independently solved.
Through this method, integration becomes straightforward, allowing for a clearer path to finding the solution to the differential equation.
By moving all terms involving \( N \) to one side and all terms involving \( t \) to the other, the equation is rearranged as \( \frac{dN}{N^2} = \frac{1}{100} dt \). This step simplifies the integration process by creating two separate integrals that can be independently solved.
- Move \( dt \) to the other side: multiply both sides by \( dt \) and divide by \( N^2 \).
- End with two separate expressions, each involving only one variable.
Through this method, integration becomes straightforward, allowing for a clearer path to finding the solution to the differential equation.
Integration
Integration is the process used to solve the equations separated in the previous step. By integrating each side of \( \frac{dN}{N^2} = \frac{1}{100} dt \), we can find a relationship between \( N \) and \( t \).
The integral of \( \frac{dN}{N^2} \) results in \( -\frac{1}{N} \), and for \( \frac{1}{100} dt \), it becomes \( \frac{1}{100}t + C \), where \( C \) is the constant of integration. This integration step is crucial as it transforms a differential equation into an algebraic one, making it easier to handle.
The process of integrating involves finding antiderivatives.
Integration turns a problem of rates into a tangible expression involving our quantities of interest, bridging the gap between calculus and practical applications.
The integral of \( \frac{dN}{N^2} \) results in \( -\frac{1}{N} \), and for \( \frac{1}{100} dt \), it becomes \( \frac{1}{100}t + C \), where \( C \) is the constant of integration. This integration step is crucial as it transforms a differential equation into an algebraic one, making it easier to handle.
The process of integrating involves finding antiderivatives.
- For \( \frac{dN}{N^2} \): You restructure it to \( N^{-2} dN \) and find the integral.
- For \( \frac{1}{100} dt \): Recognize it as a linear function of \( t \), yielding a simple constant term.
Integration turns a problem of rates into a tangible expression involving our quantities of interest, bridging the gap between calculus and practical applications.
Initial Value Problem
An Initial Value Problem (IVP) involves a differential equation accompanied by an initial condition that specifies the value of the unknown function at a particular point. In the context of our problem, the initial condition given is \( N(0) = 10 \).
Using this initial value, we can determine the constant \( C \) from our integrated expression. Substituting \( t = 0 \) and \( N = 10 \) into the equation \( -\frac{1}{10} = \frac{1}{100}(0) + C \) gives \( C = -\frac{1}{10} \). This step is crucial as it customizes the solution to uniquely fit the scenario described by the original problem.
Solving an IVP ensures that the resulting function \( N(t) \) accurately represents the real-life situation we are modeling from its beginning state, displaying how populations evolve from set starting conditions.
Using this initial value, we can determine the constant \( C \) from our integrated expression. Substituting \( t = 0 \) and \( N = 10 \) into the equation \( -\frac{1}{10} = \frac{1}{100}(0) + C \) gives \( C = -\frac{1}{10} \). This step is crucial as it customizes the solution to uniquely fit the scenario described by the original problem.
- Substitute initial values to the equation obtained after integration.
- Solve for \( C \) to find the specific solution fitting the initial condition.
Solving an IVP ensures that the resulting function \( N(t) \) accurately represents the real-life situation we are modeling from its beginning state, displaying how populations evolve from set starting conditions.
Other exercises in this chapter
Problem 20
Assume that \(W(t)\) denotes the amount of radioactive material in a substance at time \(t\). Radioactive decay is then described by the differential equation $
View solution Problem 21
Denote by \(p=p(t)\) the fraction of occupied patches in a metapopulation model, and assume that $$\frac{d p}{d t}=0.5 p(1-p)-1.5 p \quad \text { for } t \geq 0
View solution Problem 22
Denote by \(p=p(t)\) the fraction of occupied patches in a metapopulation model, and assume that $$\frac{d p}{d t}=c p(1-p)-p^{2} \quad \text { for } t \geq 0$$
View solution Problem 22
Denote by \(L(t)\) the length of a fish at time \(t\), and assume that the fish grows according to the von Bertalanffy equation $$\frac{d L}{d t}=k(34-L(t)) \qu
View solution