Problem 26
Question
Bird Population The population of a certain species of bird is limited by the type of habitat required for nesting. The population behaves according to the logistic growth model $$n(t)=\frac{5600}{0.5+27.5 e^{-0.044 t}}$$ where \(t\) is measured in years. (a) Find the initial bird population. (b) Draw a graph of the function \(n(t)\) (c) What size does the population approach as time goes on?
Step-by-Step Solution
Verified Answer
(a) 200; (b) S-shaped logistics curve; (c) 11200.
1Step 1: Find the Initial Population
To find the initial bird population, we need to evaluate the function \( n(t) \) at \( t = 0 \). Substitute \( t = 0 \) into the logistic growth model: \[ n(0) = \frac{5600}{0.5 + 27.5 e^{-0.044 \times 0}} \] Since \( e^0 = 1 \), the expression simplifies to: \[ n(0) = \frac{5600}{0.5 + 27.5 \times 1} = \frac{5600}{28} \] Calculating \( \frac{5600}{28} \), we get \( n(0) = 200 \). So the initial bird population is 200.
2Step 2: Draw the Graph of the Function
To draw the graph of the function \( n(t) = \frac{5600}{0.5 + 27.5 e^{-0.044t}} \), create a table of values by choosing different \( t \) values and computing \( n(t) \). Example points might include \( t = 0, 10, 20, 30, \ldots \). Plot these points on a graph, with \( t \) on the x-axis and \( n(t) \) on the y-axis. The graph should show an S-shaped curve characteristic of logistic growth, starting at an initial value of 200 and leveling off as \( t \) increases.
3Step 3: Determine the Limit of Population Growth
To find out what size the population approaches as time goes on, calculate the limiting value of \( n(t) \) as \( t \to \infty \). When \( t \to \infty \), the term \( e^{-0.044t} \to 0 \). Substituting into the expression gives us: \[ n(\infty) = \frac{5600}{0.5 + 27.5 \cdot 0} = \frac{5600}{0.5} \] Calculating \( \frac{5600}{0.5} \), we find that \( n(\infty) = 11200 \). The population approaches a size of 11200.
Key Concepts
Initial PopulationGraphing FunctionsPopulation Limit
Initial Population
Understanding the initial population involves evaluating the bird population at the start, specifically when time \( t \) equals zero. To find this, substitute \( t = 0 \) into the logistic growth model:
- The equation given is \( n(t) = \frac{5600}{0.5 + 27.5 e^{-0.044 t}} \).
- When \( t = 0 \), the term \( e^{-0.044 \times 0} = 1 \) (because any number raised to the power of zero is 1).
- Thus, the equation simplifies to \( n(0) = \frac{5600}{0.5 + 27.5 \times 1} = \frac{5600}{28} \).
- Performing the division results in \( n(0) = 200 \).
Graphing Functions
To visualize the bird population over time, we use graphing. The function \( n(t) = \frac{5600}{0.5 + 27.5 e^{-0.044t}} \) needs to be plotted over various time points:
- Choose sample points for \( t \) such as \( t = 0, 10, 20, 30, 40 \), and so on.
- Calculate \( n(t) \) for each of these \( t \) values to get corresponding \( n(t) \) values.
- Plot these \( t \) and \( n(t) \) pairs on a graph with \( t \) on the x-axis (years) and \( n(t) \) on the y-axis (populations).
- It starts off slowly, representing the early exponential growth phase.
- The growth becomes rapid as the population rises.
- Eventually, it levels off, showing the effect of limiting factors (such as habitat).
Population Limit
The population limit in a logistic growth model is the maximum population size that the environment can sustain. To find this limit, consider what happens as \( t \, \rightarrow \, \infty \):
- For large \( t \), the term \( e^{-0.044t} \rightarrow 0 \), simplifying the function to \( n(t) = \frac{5600}{0.5} \).
- So, \( n(t) \) approaches \( \frac{5600}{0.5} = 11200 \) as \( t \to \infty \).
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