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Biologists stocked a lake with 400 fish and estimated the carrying capacity (the maximal population for the fish of that species in that lake) to be \(10,000 .\) The number of fish tripled in the first year. (a) Assuming that the size of the fish population satisfies the logistic equation, find an expression for the size of the population after \(t\) years. (b) How long will it take for the population to increase to 5000\(?\)

Short Answer

Expert verified
Fish population: \( P(t) = \frac{10,000}{1+24e^{-4.277t}} \). It takes about 0.742 years to reach 5000 fish.

Step by step solution

01

Define the Logistic Equation

The logistic equation for a population is given by: \[ \frac{dP}{dt} = rP\left(1 - \frac{P}{K}\right) \]where \(P(t)\) is the population at time \(t\), \(r\) is the intrinsic growth rate, and \(K\) is the carrying capacity. For this lake, \(K = 10,000\).
02

Determine Initial Conditions

Initially, \(P(0) = 400\). After one year, the number of fish tripled, giving us \(P(1) = 1200\). We will use these initial conditions to find \(r\).
03

Solving for Growth Rate \(r\)

The solution to the logistic equation is: \[ P(t) = \frac{K}{1 + \left(\frac{K-P_0}{P_0}\right)e^{-rt}} \]Given \(P(1) = 1200\), substitute \(P_0 = 400\) and \(K = 10,000\):\[ 1200 = \frac{10,000}{1 + \left(\frac{10,000-400}{400}\right)e^{-r}} \] We need to solve this equation to find \(r\).
04

Solving the Equation for \(r\)

Rearrange and simplify the equation to solve for \(r\): \[ 1200 = \frac{10,000}{1 + 24e^{-r}} \] This leads to: \[ 1 + 24e^{-r} = \frac{10,000}{1200} \] \[ 24e^{-r} = \frac{10,000}{1200} - 1 \] \[ e^{-r} = \frac{1}{24} \left(\frac{10,000}{1200} - 1\right) \] \[ e^{-r} = \frac{1}{24} \cdot \frac{400}{1200} \] \[ e^{-r} \approx \frac{1}{72} \]Solving for \(r\): \[ r \approx -\ln\left(\frac{1}{72}\right) \approx \ln(72) \approx 4.277 \].
05

Expression for Population \(P(t)\)

Now that we have \(r\), substitute it back into the logistic function:\[ P(t) = \frac{10,000}{1 + \left(\frac{10,000-400}{400}\right)e^{-4.277t}} \]This is the expression for the population after \(t\) years.
06

Calculating Time for Population to Reach 5000

We set \(P(t) = 5000\) to find the time \(t\):\[ 5000 = \frac{10,000}{1 + 24e^{-4.277t}} \]\[ 1 + 24e^{-4.277t} = 2 \]\[ 24e^{-4.277t} = 1 \]\[ e^{-4.277t} = \frac{1}{24} \]\[ -4.277t = \ln\left(\frac{1}{24}\right) \]\[ t = -\frac{1}{4.277} \cdot \ln\left(\frac{1}{24}\right) \approx \frac{1}{4.277} \cdot 3.178 \approx 0.742 \]Thus, it takes approximately 0.742 years, or about 9 months, to reach 5000 fish.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Carrying Capacity
In any ecosystem, there is a maximum number of individuals that can be sustained sustainably over time. This limit is known as the carrying capacity, denoted by \(K\). The carrying capacity is determined by the amount of resources, such as food, water, and shelter, available in the environment. For the fish in our lake, the carrying capacity \(K\) is defined as 10,000 fish. By knowing this value, biologists can predict how populations will grow and stabilize in the ecosystem.

Carrying capacity serves multiple purposes in population modeling:
  • It sets an upper limit to how large a population can grow.
  • Helps in managing wildlife and conservation efforts.
  • Indicates the balance between resource availability and consumption.
Understanding carrying capacity is crucial for maintaining biodiversity and ensuring a stable ecosystem.
Growth Rate
The growth rate in a population model indicates how quickly a population increases over a given timeframe. In the logistic equation, this is represented by the variable \(r\), known as the intrinsic growth rate. The value of \(r\) significantly impacts how fast a population approaches its carrying capacity.

In our fish population example, the initial number of fish tripled in the first year, suggesting a high growth rate. We calculated \(r\) to find it was approximately 4.277. This value showcases a rapid increase from the initial population of 400.

The intrinsic growth rate \(r\) plays a vital role in population dynamics:
  • It determines the speed of population growth or decline.
  • Higher \(r\) values indicate faster growth and vice versa.
  • It is critical for making predictions about population sustainability and management.
By understanding \(r\), conservationists and biologists can better plan for population challenges.
Population Model
A population model is a mathematical framework used to simulate how populations change over time. The logistic equation used in our exercise is a common population model that describes limited growth, where the population stabilizes at the carrying capacity. The equation is given by:

\[ \frac{dP}{dt} = rP\left(1 - \frac{P}{K}\right) \]

Here, \(P(t)\) is the population at time \(t\), \(r\) is the intrinsic growth rate, and \(K\) is the carrying capacity. The logistic model accounts for the rapid initial population growth that later slows as it approaches the carrying capacity.

Key characteristics of the logistic model include:
  • Incorporates both growth potential and environmental limits.
  • Provides a sigmoid curve that reflects realistic growth scenarios.
  • Widely used for modeling populations with self-limiting growth.
The application of population models like the logistic equation is essential for understanding real-world ecological dynamics and guiding wildlife management.

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