Problem 441
Question
For the following exercises, use this scenario: The population \(P\) of a koi pond over \(x\) months is modeled by the function \(P(x)=\frac{68}{1+16 e^{-0.28 x}}\). Graph the population model to show the population over a span of 3 years.
Step-by-Step Solution
Verified Answer
Graph \( P(x) \) from \( x = 0 \) to 36; the population starts at 4 and approaches 68.
1Step 1: Understanding the Function
The given function is a population model for a koi pond over time, expressed as \( P(x) = \frac{68}{1 + 16 e^{-0.28 x}} \). This formula is a logistic growth model, where \( P(x) \) denotes the population at month \( x \).
2Step 2: Determine the Range for Graphing
Since we need to graph the population over 3 years, and 1 year has 12 months, the total period is 36 months. Thus, we will graph \( P(x) \) for \( x \) ranging from 0 to 36 months.
3Step 3: Identify Important Points on the Graph
To effectively graph the function, identify key points such as the initial population (at \( x = 0 \)) and the behavior as \( x \) increases. At \( x = 0 \), \( P(0) = \frac{68}{1 + 16 \times e^0} = \frac{68}{17} = 4 \). For large values of \( x \), \( e^{-0.28x} \to 0 \), which implies \( P(x) \to 68 \).
4Step 4: Plot the Graph
Using graphing software or graph paper, plot the graph for \( x \) ranging from 0 to 36. Start with the point at \( x = 0 \), where \( P(x) = 4 \), and continue plotting points to capture the logistic curve progression from the initial value to approaching the carrying capacity (68). The graph should show a curve that starts at a small population, grows rapidly for a period, and then levels off as it approaches 68.
5Step 5: Verify and Analyze the Graph
Check that the graph reflects logistic growth; it should be an S-shaped curve. Ensure that the curve starts from approximately \( P(0) = 4 \) and gradually approaches 68 as \( x \) increases to 36 months. This confirms that the population model accurately depicts the population behavior over three years.
Key Concepts
Population ModelingGraphing FunctionsAsymptotic Behavior
Population Modeling
A logistics growth model is an excellent tool for understanding how populations grow over time. In this scenario, we model the population of a koi pond. Population modeling involves using mathematical functions to predict how a population of organisms changes. These changes can be due to factors like birth rates, death rates, and migration.
This specific logistic growth model, \( P(x) = \frac{68}{1 + 16 e^{-0.28 x}} \), describes how the koi fish population starts small and increases to a larger stable number. Logistic models are practical because they show how populations grow rapidly at first, then slow as they approach a maximum limit. This limit is called the carrying capacity.
In the model, several things are at play:
This specific logistic growth model, \( P(x) = \frac{68}{1 + 16 e^{-0.28 x}} \), describes how the koi fish population starts small and increases to a larger stable number. Logistic models are practical because they show how populations grow rapidly at first, then slow as they approach a maximum limit. This limit is called the carrying capacity.
In the model, several things are at play:
- The carrying capacity: the maximum population size the environment (pond) can support, which is 68 koi fish in this case.
- The initial population: the starting number of koi, calculated at 4 when \( x = 0 \).
- The rate of growth: the speed at which the population grows, determined by the constant \( 0.28 \) in the exponential part of the function.
Graphing Functions
Graphing the logistic function helps visualize how the koi population changes over the designated time frame, which is 3 years or 36 months.
When graphing functions like these:
Graphing functions provides intuitive insights into how quickly populations change over time, helping identify phases where interventions or additional resources may be necessary.
When graphing functions like these:
- Start by identifying the range of the independent variable, \( x \). Here, \( x \) ranges from 0 to 36 months.
- Calculate key points to plot; \( P(0) = 4 \) is the initial population and \( P \) approaches 68 as \( x \) increases.
- Use graphing software or graph paper to visualize these points.
Graphing functions provides intuitive insights into how quickly populations change over time, helping identify phases where interventions or additional resources may be necessary.
Asymptotic Behavior
Asymptotic behavior describes trends of a function as the input values become very large or very small. In the context of the logistic growth model for the koi pond, it refers to how the population approaches the carrying capacity over time.
As the months go on and \( x \) becomes significantly large, \( e^{-0.28x} \) approaches zero. This means the denominator of the fraction \( 1 + 16 e^{-0.28x} \) becomes close to 1, simplifying \( P(x) \) to no greater than 68.
The function approaches an asymptote at 68, indicating that even as \( x \) increases indefinitely, the population won’t exceed this number under normal circumstances. This is because
As the months go on and \( x \) becomes significantly large, \( e^{-0.28x} \) approaches zero. This means the denominator of the fraction \( 1 + 16 e^{-0.28x} \) becomes close to 1, simplifying \( P(x) \) to no greater than 68.
The function approaches an asymptote at 68, indicating that even as \( x \) increases indefinitely, the population won’t exceed this number under normal circumstances. This is because
- the logistic model captures the law of diminishing returns, where increases in population growth become negligible as it approaches the environmental limit,
- external factors such as food availability and space, that prevent the population from growing beyond this threshold.
Other exercises in this chapter
Problem 438
A logarithmic model is given by the equation \(h(p)=67.682-5.792 \ln (p) .\) To the nearest hundredth, for what value of \(p\) does \(h(p)=62 ?\)
View solution Problem 439
A logistic moden by the equation \(P(t)=\frac{90}{1+5 e^{-0.42 t}}\) To the nearest hundredth, for what value of \(t\) does \(P(t)=45 ?\)
View solution Problem 442
For the following exercises, use this scenario: The population \(P\) of a koi pond over \(x\) months is modeled by the function \(P(x)=\frac{68}{1+16 e^{-0.28 x
View solution Problem 443
For the following exercises, use this scenario: The population \(P\) of a koi pond over \(x\) months is modeled by the function \(P(x)=\frac{68}{1+16 e^{-0.28 x
View solution