A qualitative look at a fish population Example. A population of fish is being monitored in a lake. The fish population at any time t we'll call p(t), where p is measured in thousands of kilograms and t is measured in weeks. Then suppose p(t) satisfies the DE (fishpop) p' = p -0.2p² - 0.7, where p' = dp/dt. You'll see in a later lesson how such an equation can be determined. For now, you'll do some qualitative analysis of the equation and see what you can predict about the growth or decline of the population using only the DE itself, without knowing the solutions. Fill in the details of the following steps: 1. First, notice that the population doesn't change if p' = 0. The values of p at which this happens are called the equilibrium points. Find these equilibrium points to 1 decimal place accuracy. (Do this by setting the right-hand side of equation (fishpop) equal to zero and then solving the resulting quadratic equation for p.) These equilibrium values are constant solutions of the DE, that is, they are constant functions of time which are solutions. (Can you verify this?) 2. Now you'll learn how to draw a phase line plot of the solutions. A phase line plot is a 1- dimensional plot of a p-axis that shows where the solutions are increasing and decreasing. Use the p-axis shown below, and mark the equilibrium positions on it; you should have found these to be: p₁ = 0.8 and P₂ = 4.2. 0 Pi Р In between the equilibrium values, the population p is increasing if p' is positive, or decreasing if p' is negative. For the values of p for which p'>0, draw small arrows on the phase line pointing to the right, and for p' <0, point the arrows to the left. Some of the arrows have been drawn for you; fill in the rest. (Hint: To figure out where p' is positive or negative, either plug some p values into the right-hand side of (fishpop), or sketch a graph of p' vs. p).
A qualitative look at a fish population Example. A population of fish is being monitored in a lake. The fish population at any time t we'll call p(t), where p is measured in thousands of kilograms and t is measured in weeks. Then suppose p(t) satisfies the DE (fishpop) p' = p -0.2p² - 0.7, where p' = dp/dt. You'll see in a later lesson how such an equation can be determined. For now, you'll do some qualitative analysis of the equation and see what you can predict about the growth or decline of the population using only the DE itself, without knowing the solutions. Fill in the details of the following steps: 1. First, notice that the population doesn't change if p' = 0. The values of p at which this happens are called the equilibrium points. Find these equilibrium points to 1 decimal place accuracy. (Do this by setting the right-hand side of equation (fishpop) equal to zero and then solving the resulting quadratic equation for p.) These equilibrium values are constant solutions of the DE, that is, they are constant functions of time which are solutions. (Can you verify this?) 2. Now you'll learn how to draw a phase line plot of the solutions. A phase line plot is a 1- dimensional plot of a p-axis that shows where the solutions are increasing and decreasing. Use the p-axis shown below, and mark the equilibrium positions on it; you should have found these to be: p₁ = 0.8 and P₂ = 4.2. 0 Pi Р In between the equilibrium values, the population p is increasing if p' is positive, or decreasing if p' is negative. For the values of p for which p'>0, draw small arrows on the phase line pointing to the right, and for p' <0, point the arrows to the left. Some of the arrows have been drawn for you; fill in the rest. (Hint: To figure out where p' is positive or negative, either plug some p values into the right-hand side of (fishpop), or sketch a graph of p' vs. p).
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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![### A Qualitative Look at a Fish Population
**Example.** A population of fish is being monitored in a lake. The fish population at any time \( t \) is called \( p(t) \), where \( p \) is measured in thousands of kilograms and \( t \) is measured in weeks. Suppose \( p(t) \) satisfies the differential equation (DE):
\[
p' = p - 0.2p^2 - 0.7,
\]
where \( p' = \frac{dp}{dt} \).
You'll see in a later lesson how such an equation can be determined. For now, you'll do some qualitative analysis of the equation and see what you can predict about the growth or decline of the population using only the DE itself, without knowing the solutions. Fill in the details of the following steps:
1. **Equilibrium Points:**
- Notice that the population doesn’t change if \( p' = 0 \). The values of \( p \) at which this happens are called the equilibrium points. Find these equilibrium points to 1 decimal place accuracy. (Do this by setting the right-hand side of equation (\( \text{fishpop} \)) equal to zero and then solving the resulting quadratic equation for \( p \).)
- These equilibrium values are constant solutions of the DE, that is, they are constant functions of time which are solutions. (Can you verify this?)
2. **Phase Line Plot:**
- Learn how to draw a phase line plot of the solutions. A phase line plot is a 1-dimensional plot of a \( p \)-axis that shows where the solutions are increasing and decreasing. Use the \( p \)-axis shown below, and mark the equilibrium positions on it; you should have found these to be: \( p_1 = 0.8 \) and \( p_2 = 4.2 \).
```
<---------- -----------> <----------
0 p₁ p₂
0.8 4.2
```
- In between the equilibrium values, the population \( p \) is increasing if \( p' \) is positive, or decreasing if \( p' \) is negative. For the values of \( p \) for which \( p' > 0 \), draw small arrows on the phase line pointing to the right, and for \( p'](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F15c022aa-6711-4b29-ab7f-4e5d70ca82f7%2F2f84366a-918a-47ab-86d7-7f5d11585d97%2Frl5eg9cb_processed.png&w=3840&q=75)
Transcribed Image Text:### A Qualitative Look at a Fish Population
**Example.** A population of fish is being monitored in a lake. The fish population at any time \( t \) is called \( p(t) \), where \( p \) is measured in thousands of kilograms and \( t \) is measured in weeks. Suppose \( p(t) \) satisfies the differential equation (DE):
\[
p' = p - 0.2p^2 - 0.7,
\]
where \( p' = \frac{dp}{dt} \).
You'll see in a later lesson how such an equation can be determined. For now, you'll do some qualitative analysis of the equation and see what you can predict about the growth or decline of the population using only the DE itself, without knowing the solutions. Fill in the details of the following steps:
1. **Equilibrium Points:**
- Notice that the population doesn’t change if \( p' = 0 \). The values of \( p \) at which this happens are called the equilibrium points. Find these equilibrium points to 1 decimal place accuracy. (Do this by setting the right-hand side of equation (\( \text{fishpop} \)) equal to zero and then solving the resulting quadratic equation for \( p \).)
- These equilibrium values are constant solutions of the DE, that is, they are constant functions of time which are solutions. (Can you verify this?)
2. **Phase Line Plot:**
- Learn how to draw a phase line plot of the solutions. A phase line plot is a 1-dimensional plot of a \( p \)-axis that shows where the solutions are increasing and decreasing. Use the \( p \)-axis shown below, and mark the equilibrium positions on it; you should have found these to be: \( p_1 = 0.8 \) and \( p_2 = 4.2 \).
```
<---------- -----------> <----------
0 p₁ p₂
0.8 4.2
```
- In between the equilibrium values, the population \( p \) is increasing if \( p' \) is positive, or decreasing if \( p' \) is negative. For the values of \( p \) for which \( p' > 0 \), draw small arrows on the phase line pointing to the right, and for \( p'
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![**Title: Phase Line Analysis and Differential Equations**
**Text Explanation:**
In between the equilibrium values, the population \( p \) is increasing if \( p' \) is positive, or decreasing if \( p' \) is negative. For the values of \( p \) for which \( p' > 0 \), draw small arrows on the phase line pointing to the right, and for \( p' < 0 \), point the arrows to the left. Some of the arrows have been drawn for you; fill in the rest.
*(Hint: To figure out where \( p' \) is positive or negative, either plug some \( p \) values into the right-hand side of \( \text{fishpop} \), or sketch a graph of \( p' \) vs. \( p \).)*
**Tasks:**
- **Task 1:** Suppose that the lake is initially stocked with 1200 kg of fish. Translate this statement into an initial condition for the differential equation (DE). Over the next few weeks, what will the population do — grow? die out? remain the same?
- **Task 2:** Same question as in Task 1, but suppose the initial population is 5000 kg of fish.
- **Task 3:** Draw a phase line plot for the DE:
\[
\frac{dy}{dt} = 4.1y - y^2 - 4.2
\]
**Graph/Diagram Explanation:**
The task involves creating a phase line plot, which is a graphical tool used to determine the stability of equilibrium points in differential equations. It shows the direction of change of the population \( p \) at different points on the line.
1. **Equilibrium Points:** Calculate where \( \frac{dp}{dt} = 0 \), as these points are equilibrium solutions.
2. **Arrow Directions:**
- If \( p' > 0 \), draw arrows pointing to the right to indicate growth.
- If \( p' < 0 \), draw arrows pointing to the left to indicate decline.
3. **Analysis:** Use the phase line to interpret how initial conditions affect the long-term behavior of the fish population.
This study provides a visual understanding of how differential equations can predict changes in real-world scenarios like fish population dynamics in a lake.](https://content.bartleby.com/qna-images/question/15c022aa-6711-4b29-ab7f-4e5d70ca82f7/4cc9071c-1b5a-4795-a81f-4e800d2a3a99/ujnuswl_processed.png)
Transcribed Image Text:**Title: Phase Line Analysis and Differential Equations**
**Text Explanation:**
In between the equilibrium values, the population \( p \) is increasing if \( p' \) is positive, or decreasing if \( p' \) is negative. For the values of \( p \) for which \( p' > 0 \), draw small arrows on the phase line pointing to the right, and for \( p' < 0 \), point the arrows to the left. Some of the arrows have been drawn for you; fill in the rest.
*(Hint: To figure out where \( p' \) is positive or negative, either plug some \( p \) values into the right-hand side of \( \text{fishpop} \), or sketch a graph of \( p' \) vs. \( p \).)*
**Tasks:**
- **Task 1:** Suppose that the lake is initially stocked with 1200 kg of fish. Translate this statement into an initial condition for the differential equation (DE). Over the next few weeks, what will the population do — grow? die out? remain the same?
- **Task 2:** Same question as in Task 1, but suppose the initial population is 5000 kg of fish.
- **Task 3:** Draw a phase line plot for the DE:
\[
\frac{dy}{dt} = 4.1y - y^2 - 4.2
\]
**Graph/Diagram Explanation:**
The task involves creating a phase line plot, which is a graphical tool used to determine the stability of equilibrium points in differential equations. It shows the direction of change of the population \( p \) at different points on the line.
1. **Equilibrium Points:** Calculate where \( \frac{dp}{dt} = 0 \), as these points are equilibrium solutions.
2. **Arrow Directions:**
- If \( p' > 0 \), draw arrows pointing to the right to indicate growth.
- If \( p' < 0 \), draw arrows pointing to the left to indicate decline.
3. **Analysis:** Use the phase line to interpret how initial conditions affect the long-term behavior of the fish population.
This study provides a visual understanding of how differential equations can predict changes in real-world scenarios like fish population dynamics in a lake.
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