Problem 69 - Algebra and Trigonometry

Question

Logistic Growth Animal populations are not capable of unrestricted growth because of limited habitat and food supplies. Under such conditions the population follows a logistic growth model $$ P(t)=\frac{d}{1+k e^{-c t}} $$ where \(c, d,\) and \(k\) arc positive constants. For a certain fish population in a small pond \(d=1200, k=11, c=0.2,\) and \(t\) is measured in years. The fish were introduced into the pond at time \(t=0\) . (a) How many fish were originally put in the pond? (b) Find the population after \(10,20,\) and 30 years. (c) Evaluate \(P(t)\) for large values of \(t\) . What value does the population approach as \(t \rightarrow \infty\) Does the graph shown confirm your calculations?

Step-by-Step Solution

Answer

(a) 100 fish, (b) 483 at 10 years, 999 at 20 years, 1168 at 30 years, (c) approaches 1200 fish as \(t \to \infty\).

Step 1: Determine initial population (t=0)

To find the initial number of fish introduced into the pond, substitute \(t=0\) into the logistic growth equation:\[ P(0) = \frac{1200}{1 + 11e^{-0.2 \times 0}} = \frac{1200}{1 + 11} = \frac{1200}{12} = 100.\]Therefore, originally, 100 fish were introduced into the pond.

Step 2: Calculate population at t=10

Substitute \(t=10\) into the logistic model to find the population after 10 years:\[P(10) = \frac{1200}{1 + 11e^{-0.2 \times 10}} = \frac{1200}{1 + 11e^{-2}}.\]Calculate \(e^{-2}\) approximately as 0.135:\[P(10) = \frac{1200}{1 + 11 imes 0.135} = \frac{1200}{2.485} ≈ 483.22.\]Thus, the population after 10 years is approximately 483 fish.

Step 3: Calculate population at t=20

Substitute \(t=20\) into the logistic equation:\[P(20) = \frac{1200}{1 + 11e^{-0.2 \times 20}} = \frac{1200}{1 + 11e^{-4}}.\]Calculate \(e^{-4}\) approximately as 0.0183:\[P(20) = \frac{1200}{1 + 11 imes 0.0183} = \frac{1200}{1.2013} ≈ 998.92.\]The population after 20 years is approximately 999 fish.

Step 4: Calculate population at t=30

Substitute \(t=30\) into the equation:\[P(30) = \frac{1200}{1 + 11e^{-0.2 \times 30}} = \frac{1200}{1 + 11e^{-6}}.\]Calculate \(e^{-6}\) approximately as 0.0025:\[P(30) = \frac{1200}{1 + 11 imes 0.0025} = \frac{1200}{1.0275} ≈ 1167.54.\]The population after 30 years is approximately 1168 fish.

Step 5: Evaluate population as t approaches infinity

As \(t\) becomes very large, the term \(e^{-ct}\) approaches zero. Therefore, the equation becomes:\[P(t) = \frac{1200}{1 + 11 \times 0} = \frac{1200}{1} = 1200.\]Thus, the population approaches 1200 as \(t \to \infty\). This value represents the carrying capacity of the environment for the fish population.

Key Concepts

Animal Populations

Animal populations in ecological systems usually cannot grow indefinitely. This is because they are constrained by environmental factors that include available resources like food and space. Understanding how these populations change over time is essential in ecology. In a small pond, for example, fish populations can initially grow rapidly, but their growth rate will slow as they near the environment's limits.
When studying animal populations, scientists often use mathematical models to predict changes over time. These models help forecast how many individuals there might be in some future period based on current conditions and historical data. For instance, if 100 fish are introduced into a pond, researchers would use these models to estimate how that population might change in the coming years.
Using the logistic growth model allows ecologists to incorporate the practical constraints faced by populations in nature. Let's explore how population dynamics unfold based on such models.

Population Dynamics

Population dynamics describe how populations change in size and composition over time. These changes are driven by various factors, including birth rates, death rates, and migration.

• Birth and death rates: A balance between these rates determines growth or decline in population numbers.
• Migration: Populations can grow or shrink depending on the movement of individuals into or out of a given area.

The logistic growth model is a practical tool in understanding population dynamics. Unlike simple exponential growth that assumes unlimited resources, logistic growth considers that resources are finite. This aligns closer with real-world scenarios, where limitations like food supply and living space result in a carrying capacity—beyond which the population cannot sustainably grow.
In mathematical terms, a population follows the logistic growth model when it changes according to the equation:\[P(t)=\frac{d}{1+k e^{-c t}}\]The constants in this equation (\(c, d, k\)) relate to the specific circumstances of the population under study. For instance, if you substitute different time values (\(t\)), you'll find the projected population at that time. Over time, the population will approach a maximum size determined by the carrying capacity.

Carrying Capacity

Carrying capacity is a crucial concept when analyzing population dynamics. It refers to the maximum number of individuals that an environment can support sustainably.
The carrying capacity takes into account the availability of resources like food, water, and shelter, as well as other ecological constraints. For example, a pond's carrying capacity for fish will depend on factors such as oxygen levels, availability of food, and suitable habitat conditions.
In logistic growth, the carrying capacity is represented by the variable \(d\) in the equation:\[P(t)=\frac{d}{1+k e^{-c t}}\]As time advances towards infinity, the exponential component diminishes, allowing the population to approach the carrying capacity value. This is illustrated in the example of a fish population in a pond, which moves toward a stable maximum as time progresses.
Understanding carrying capacity helps ecologists and wildlife managers make informed decisions about conservation strategies and natural resource management, ensuring that populations remain healthy and ecosystems remain balanced.