Problem 72 - Algebra and Trigonometry

Question

Rabbit Population Assume that a population of rabbits behaves according to the logistic growth model $$ n(t)=\frac{300}{0.05+\left(\frac{300}{n_{0}}-0.05\right) e^{-0.55 r}} $$ (a) If the initial population is 50 rabbits, what will the population be after 12 years? (b) Draw graphs of the function \(n(t)\) for \(n_{0}=50,500\) , \(2000,8000\) , and \(12,000\) in the viewing rectangle \([0,15]\) by \([0,12,000] .\) (c) From the graphs in part (b), observe that, regardless of the initial population, the rabbit population seems to approach a certain number as time goes on. What is that number? (This is the number of rabbits that the island can support.)

Step-by-Step Solution

Answer

(a) ≈ 5164 rabbits (b) Create graphs (c) Approaches 6000 rabbits.

Step 1: Identify Given Values and Equation

We are provided with the logistic growth model equation for the rabbit population:\[n(t) = \frac{300}{0.05 + \left(\frac{300}{n_0} - 0.05\right) e^{-0.55t}}\]For part (a), the initial population \(n_0 = 50\) and we need to find the population \(n(t)\) after 12 years, so \(t = 12\).

Step 2: Plug in Given Values for Part (a)

Substitute \(n_0 = 50\) and \(t = 12\) in the logistic growth equation:\[n(12) = \frac{300}{0.05 + \left(\frac{300}{50} - 0.05\right) e^{-0.55 \times 12}}\]Calculate the intermediate values to solve for \(n(12)\).

Step 3: Simplify the Population Equation

Calculate \(\frac{300}{50} = 6\). Then substitute back into the equation:\[n(12) = \frac{300}{0.05 + (6 - 0.05) e^{-0.55 \times 12}}\]This simplifies to:\[n(12) = \frac{300}{0.05 + 5.95 e^{-6.6}}\]

Step 4: Evaluate the Exponential Component

Calculate the exponential function \(e^{-6.6}\):\[e^{-6.6} \approx 0.00136\]Substitute this value into the logistic equation:\[n(12) = \frac{300}{0.05 + 5.95 \times 0.00136}\]

Step 5: Calculate the Population after 12 Years

Evaluate the expression:\[n(12) = \frac{300}{0.05 + 0.008092} = \frac{300}{0.058092} \approx 5163.57\]So, the population after 12 years is approximately 5164 rabbits.

Step 6: Describe Graphing for Different Initial Populations

Using the logistic growth model, plot the graphs of \(n(t)\) for \(n_0 = 50, 500, 2000, 8000, 12000\) in the viewing rectangle \([0,15]\) years for time (x-axis) and population \([0,12000]\) for the y-axis. This will visually show how the population changes over time for different initial populations.

Step 7: Analyze the Graphs for Long-term Population

Observe from the graphs that all initial populations approach the same limiting value as time increases, suggesting a carrying capacity. The graphs show that as \(t \rightarrow \infty\), \(n(t)\) approaches 6000.

Key Concepts

Rabbit Population

In the study of populations, a common scenario is to track how a specific group of animals, like rabbits, grows over time. Rabbits are known for their quick reproduction rates, so understanding how their population changes is essential in many ecological and biological studies. Here, we explore this concept through the lens of a logistic growth model.

The logistic growth model is ideal for rabbit populations because it starts with rapid growth, typical due to high reproduction rates. Over time, the growth slows due to factors like limited resources. This model helps us predict the population size at any future point and understand the dynamics of the population growth. In our exercise, the initial rabbit population and subsequent changes over years form the foundation of applying the logistic growth equation.

When calculating or predicting rabbit populations, recognizing initial conditions like starting population size and growth rate is crucial. These values help in feeding the logistic growth equation, which in turn gives us insight into future population dynamics and behavior.

Exponential Function

An exponential function plays a critical role in the logistic growth model. Typically, exponential functions describe rapid growth, but in this context, they underpin how growth initially accelerates before leveling off. The equation \( n(t) = \frac{300}{0.05 + (\frac{300}{n_0} - 0.05) e^{-0.55t}} \) includes an exponential component: \( e^{-0.55t} \).

This exponential function is part of the denominator and primarily influences how quickly the growth rate changes over time. Initially, when \( t = 0 \), the exponential term is significant, reflecting fast population growth due to minimal time influence. As time passes, \( t \) increases, and the exponential term \( e^{-0.55t} \) shrinks towards zero. This transition reflects the phase where the rate of population increase begins to decelerate, and other factors start limiting growth.

Understanding the exponential function's behavior within this equation allows us to predict population behavior over time accurately. This element's decrease reflects how natural phenomena tend to stabilize even after rapid initial changes.

Carrying Capacity

Carrying capacity defines the maximum population size that an environment can sustainably support. In the rabbit population model, this is observed as the eventual stabilization point of the population over time.

Our logistic growth model concludes that regardless of differing initial populations, the rabbit population approaches a specific number, identified as the carrying capacity. In this case, calculations and graphs point toward a carrying capacity of 6000 rabbits. This value represents the limit imposed by resources like food, space, and environmental factors.

The notion of carrying capacity is essential for understanding any population dynamics. It illustrates the natural limitations all populations face and helps in managing wildlife health and environmental conservation efforts. Knowing the carrying capacity assists in predicting long-term population trends and is a crucial component in the study of ecology and natural resource management.