Problem 25 - Calculus

Question

A model for the population, \(P,\) of carp in a landlocked lake at time \(t\) is given by the differential equation $$\frac{d P}{d t}=0.25 P(1-0.0004 P)$$ (a) What is the long-term equilibrium population of carp in the lake? (b) A census taken ten years ago found there were 1000 carp in the lake. Estimate the current population. (c) Under a plan to join the lake to a nearby river, the fish will be able to leave the lake. A net loss of \(10 \%\) of the carp each year is predicted, but the patterns of birth and death are not expected to change. Revise the differential equation to take this into account. Use the revised differential equation to predict the future development of the carp population.

Step-by-Step Solution

Answer

(a) 2500 carp. (b) Approximately 1429 carp. (c) Population will stabilize at about 1500 carp.

Step 1: Find Long-term Equilibrium Population

Long-term equilibrium occurs when the rate of change of the population is zero, i.e., \(\frac{dP}{dt} = 0\). Setting the equation \(\frac{dP}{dt} = 0.25P(1 - 0.0004P) = 0\) gives two equilibrium solutions: \(P = 0\) or \(1 - 0.0004P = 0\). Solving \(1 - 0.0004P = 0\) gives \(P = 2500\). So the long-term equilibrium population is 2500 carp.

Step 2: Predict Current Population from Initial Condition

The given differential equation is a logistic equation, which can be solved given initial conditions to predict future population. Initially, 10 years ago, \(P(0) = 1000\). The logistic equation's solution involves the expression: \(P(t) = \frac{K}{1 + \left(\frac{K - P_0}{P_0}\right)e^{-rt}}\), where \(K = 2500\), \(P_0 = 1000\), and \(r = 0.25\). Substituting these values and \(t = 10\), we find \(P(10) = \frac{2500}{1 + \left(\frac{2500 - 1000}{1000}\right)e^{-0.25 \cdot 10}}\). Calculating gives approximately \(P(10) \approx 1429\) carp.

Step 3: Revise the Differential Equation for Net Loss

Introducing a 10\% annual net loss alters the original equation. The new equation is \(\frac{dP}{dt} = 0.25P(1 - 0.0004P) - 0.1P\). Simplifying, this becomes \(\frac{dP}{dt} = 0.15P(1 - 0.0004P)\).

Step 4: Predict Future Population with Revised Model

Use the new differential equation to analyze population trends. The decreasing \(r\) value (from 0.25 to 0.15) implies slower growth. The new equation ensures the population will reach a reduced carrying capacity, determined by \(1 - 0.0004P - 0.1 = 0\), which gives the carrying capacity reduced to \(P \approx 1500\). Therefore, over time, the population will stabilize around 1500 carp.

Key Concepts

Logistic Growth Model

The logistic growth model is a mathematical representation used to describe how populations grow in an environment where resources are limited. It modifies exponential growth by introducing a carrying capacity, which is the maximum population size that an environment can sustain. The model is depicted by the equation:
\[ \frac{dP}{dt} = rP\left(1 - \frac{P}{K}\right) \]where:

• \(P\) is the population size at time \(t\)
• \(r\) is the intrinsic rate of growth
• \(K\) is the carrying capacity

This model reflects real-world constraints, such as food and space limitations. Population growth starts exponentially, but as resources become scarce, growth rate decreases and eventually ceases when the population reaches its carrying capacity.

Equilibrium Population

Equilibrium population refers to a steady state where population size remains constant over time. In the context of the logistic growth model, this occurs when the rate of change in population size, \(\frac{dP}{dt}\), is zero. This implies no net growth or decline.
For the given carp population model, this means solving:\[ 0.25P(1 - 0.0004P) = 0 \]The solutions are \(P = 0\) or solving for \(P\) gives us \(P = 2500\) carp. At these points, the population changes no further, with \(P = 2500\) being the non-trivial solution where the carp population stabilizes if no external changes occur.

Population Dynamics

Population dynamics explores how and why populations change over time. It focuses on the factors like birth rates, death rates, immigration, and emigration that contribute to the growth or decline in populations.
In our carp model, population dynamics are described using a logistic equation that considers both intrinsic growth (\(r = 0.25\)) and carrying capacity (\(K = 2500\)). Histogram models, like the logistic model in this context, are particularly powerful for showing changes over time both qualitatively and quantitatively.

• Initial conditions and external factors, like the 10% annual net loss in the revised carp model, influence these dynamics.
• The consequence is a shift in the growth trajectory and equilibrium point, which becomes evident with time.

By accounting for these aspects, we can predict future trends and understand population behavior more comprehensively.

Carp Population Modeling

Carp population modeling involves using differential equations to predict and analyze changes in carp numbers over time. For the carp in the landlocked lake, the initial model relates to a logistic growth structure, accommodating both growth and equilibrium.
The equation:\[ \frac{dP}{dt} = 0.25P(1 - 0.0004P) \]describes a natural growth scenario without external loss factors. A census showed that 1000 carp existed ten years ago, and the logistic equation allows estimating the current population. By solving it with initial conditions, we determine the population has grown to about 1429 carp in ten years.
When adjusting the model for a 10% annual net loss, the new model:\[ \frac{dP}{dt} = 0.15P(1 - 0.0004P) \]suggests modified growth dynamics. Here, the new carrying capacity is approximately 1500 carp, emphasizing how environmental changes alter population destiny. This demonstrates modeling's role in ecological management and forecasting.