Problem 44 - Calculus and its applications

Question

A lake is stocked with 400 rainbow trout. The size of the lake, the availability of food, and the number of other fish restrict population growth to a limiting value of 2500 trout. The population of trout in the lake after time \(t\), in months, is approximated by $$ P(t)=\frac{2500}{1+5.25 e^{-0.32 t}} $$ a) Find the population after 0 months, 1 month, 5 months, 10 months, 15 months, and 20 months. b) Find the rate of change, \(P^{\prime}(t)\). c) Sketch a graph of the function.

Step-by-Step Solution

Answer

a) Populations: 400, 566, 1029, 1693, 2194, 2404. b) Rate of change: \( P'(t) = -\frac{2500 \times 1.68 e^{-0.32t}}{(1 + 5.25 e^{-0.32 t})^2} \).

Step 1: Understand the Function

The population of trout in the lake is modeled by the function \( P(t) = \frac{2500}{1 + 5.25 e^{-0.32 t}} \). This is a logistic growth model, where 2500 is the carrying capacity of the population, and the initial population is influenced by the denominator value at \( t = 0 \).

Step 2: Calculate Population at Specified Times

To find the population at different times, substitute \( t = 0, 1, 5, 10, 15, 20 \) into the function: - \( t = 0 \): \( P(0) = \frac{2500}{1 + 5.25 e^0} = \frac{2500}{1 + 5.25} = \frac{2500}{6.25} = 400 \). - \( t = 1 \): \( P(1) \approx \frac{2500}{1 + 5.25 e^{-0.32}} \approx 566 \). - \( t = 5 \): \( P(5) \approx \frac{2500}{1 + 5.25 e^{-1.6}} \approx 1029 \). - \( t = 10 \): \( P(10) \approx \frac{2500}{1 + 5.25 e^{-3.2}} \approx 1693 \). - \( t = 15 \): \( P(15) \approx \frac{2500}{1 + 5.25 e^{-4.8}} \approx 2194 \). - \( t = 20 \): \( P(20) \approx \frac{2500}{1 + 5.25 e^{-6.4}} \approx 2404 \).

Step 3: Differentiate to Find Rate of Change

To find the rate of change, differentiate \( P(t) \). This involves using the quotient rule \( P'(t) = \frac{d}{dt}\left(\frac{2500}{1 + 5.25 e^{-0.32t}}\right)\): Given \( u = 2500 \) and \( v = 1 + 5.25 e^{-0.32 t} \), \( \frac{du}{dt} = 0 \) and \( \frac{dv}{dt} = -5.25 \times -0.32 e^{-0.32t} = 1.68 e^{-0.32t} \). Apply the rule: \( P'(t) = \frac{0 \cdot v - u \cdot \frac{dv}{dt}}{v^2} = -\frac{2500 \times 1.68 e^{-0.32t}}{(1 + 5.25 e^{-0.32 t})^2} \).

Key Concepts

Population Dynamics

Population dynamics involves understanding how and why populations of organisms change over time. In ecosystems, various factors such as available resources, environmental conditions, and competition influence these changes. For example, when a new population, like trout in a lake, is introduced, it initially grows slowly.
As the population becomes established and resources are abundant, growth accelerates. However, as the population nears its maximum sustainable size, or carrying capacity, growth slows. This pattern of growth can be seen in many natural populations and is often visualized using models like the logistic growth model.

• Initially slow population growth as the species adapts to the environment.
• Explosive growth when resources are plentiful and competition is low.
• Stabilization as resources become limited and the population reaches carrying capacity.

Carrying Capacity

Carrying capacity refers to the maximum number of individuals an environment can support sustainably. It depends on various factors like food supply, habitat space, water availability, and the presence of predators or competitors.
In the example of the stocked lake, we see that the carrying capacity is 2500 trout. This figure is derived from considerations of the lake's size and resources that can support the trout population without causing overuse or depletion.

• Defined by the availability of food and habitat space.
• Natural limit preventing perpetual population growth.
• Determines sustainable population size over time.

As populations reach carrying capacity, they tend to stabilize around this figure due to decreased birth rates and increased mortality rates. Changes in environmental conditions or resources can adjust the carrying capacity over time.

Differentiation in Calculus

Differentiation in calculus is a mathematical process used to determine the rate at which a function is changing at any given point. In the context of population modeling, differentiation helps us find how quickly a population is growing or shrinking at a specific time.

For the trout population problem, differentiation involves calculating the derivative of the logistic growth function. This provides the rate of change of the population's size over time, expressed as \( P'(t) \). This derivative tells us how fast the population is increasing or decreasing at any moment.

• Used to calculate the slope or rate of change of a function.
• Essential for finding how quickly populations grow relative to time.
• Involves applying rules like the product, quotient, or chain rules depending on the function form.

Understanding \( P'(t) \) in this context helps predict future population trends and manage resource allocation for sustainable practices.

Population Modeling

Population modeling uses mathematical functions to represent and predict how populations will change over time. A key tool in studying population dynamics, these models help us understand complex interactions within ecosystems. The logistic growth model is widely used due to its ability to realistically portray populations' growth phases.

Models like \( P(t) = \frac{2500}{1 + 5.25 e^{-0.32 t}} \) reflect real-world constraints such as carrying capacity, showcasing a sigmoidal growth curve. Initially, there is exponential growth that later stabilizes as it approaches carrying capacity.

• Predicts future population sizes based on current data.
• Helps in understanding impacts of different factors on growth rates.
• Essential for planning conservation and resource management efforts.

Using models, ecologists and resource managers can simulate different scenarios, such as changes in carrying capacity or resource availability, to make informed decisions about conservation strategies.