Problem 585 - Precalculus

Question

The population of a wildlife habitat is modeled by the equation \(P(t)=\frac{360}{1+6.2 e^{-0.35 t}}\) , where \(t\) is given in years. How many animals were originally transported to the habitat? How many years will it take before the habitat reaches half its capacity?

Step-by-Step Solution

Answer

The initial population is 50 animals, and it takes approximately 5.15 years to reach half capacity.

Step 1: Understanding the Problem

The population model given is \( P(t) = \frac{360}{1 + 6.2 e^{-0.35t}} \). Here, \( P(t) \) represents the population at time \( t \), \( 360 \) is the carrying capacity, and the term \( 6.2 e^{-0.35t} \) modifies the growth rate. We need to find the initial population \( P(0) \) at \( t = 0 \), which represents the animals originally transported, and the time \( t \) when the population is half of the carrying capacity.

Step 2: Calculate the Initial Population

To find the initial population, substitute \( t = 0 \) into the equation. So, we calculate \( P(0) = \frac{360}{1 + 6.2 e^{0}} = \frac{360}{1 + 6.2} = \frac{360}{7.2} \). Solve this to find \( P(0) \).

Step 3: Compute the Result for Initial Population

Calculate \( \frac{360}{7.2} \) to get the initial population: \( P(0) = 50 \). Thus, 50 animals were originally transported to the habitat.

Step 4: Determine Population at Half Capacity

When the population reaches half the carrying capacity, \( P(t) = \frac{360}{2} = 180 \). We need to determine \( t \) such that \( \frac{360}{1 + 6.2 e^{-0.35t}} = 180 \).

Step 5: Solve for Time to Half Capacity

Set the equation \( \frac{360}{1 + 6.2 e^{-0.35t}} = 180 \) and multiply both sides by \( 1 + 6.2 e^{-0.35t} \) to clear the fraction, giving \( 360 = 180(1 + 6.2 e^{-0.35t}) \). Solving for \( e^{-0.35t} \), we get \( 2 = 1 + 6.2 e^{-0.35t} \) and \( 6.2 e^{-0.35t} = 1 \).

Step 6: Simplify and Solve for \( t \)

Divide both sides by 6.2: \( e^{-0.35t} = \frac{1}{6.2} \). Take the natural log of both sides: \( -0.35t = \ln\left(\frac{1}{6.2}\right) \). Solve for \( t \): \( t = -\frac{\ln\left(\frac{1}{6.2}\right)}{0.35} \).

Step 7: Compute Final Time Value

Calculate \( \ln\left(\frac{1}{6.2}\right) \) and then divide by \(-0.35\) to find \( t \). This results in \( t \approx 5.15 \) years.

Key Concepts

Carrying Capacity

The concept of "Carrying Capacity" is essential in understanding population dynamics, especially in ecological settings. Carrying capacity refers to the maximum number of individuals that an environment can sustain over a long period without degrading the habitat. This number is determined by resources such as food, water, and shelter.
In the logistic growth model, like the one presented in the exercise, the population approaches this limiting size as resources become scarce.
For any given environment, managing the population size to match the carrying capacity ensures a sustainable balance, preventing overpopulation and resource depletion.

• The carrying capacity in the logistic equation is often symbolized by "K". In our exercise, this is represented by the number 360.
• It signifies that the maximum population the habitat can support is 360 animals.

Initial Population

The initial population is the starting point of the logistic growth model. It indicates how many individuals were present in the habitat at the beginning, which is essentially when time, denoted as "t", equals 0.

• In the population model given in the original problem, we found this by substituting the value of 0 for "t".
• Mathematically, it is expressed as \( P(0) = \frac{360}{1+6.2}\)

With this substitution, we can calculate the initial population. After simplifying, it results in 50 animals being introduced initially into the habitat.
Actions taken at this stage can significantly influence how the population grows or stabilizes over time.
The initial population sets the stage for how quickly the model will approach its carrying capacity, illustrating the importance of strategic wildlife management.

Exponential Function

An exponential function forms the backbone of population growth models such as the logistic function. In simpler terms, exponential functions relate to processes that grow or decay at a rate proportional to the current size.

• In the logistic growth equation, represented by \(e^{-0.35t}\), the exponential function here describes how the population grows over time.
• The negative exponent reflects the fact that as time progresses, the growth rate slows down.

The model in question modifies this core growth through the term \(6.2 e^{-0.35t}\), adding realism as populations naturally slow their growth as they near their carrying capacity.
Exponential functions are particularly important since they capture the "boom" and "bust" patterns seen in real-world population changes.
Understanding this exponential nature explains why the growth doesn't continue unbounded but instead levels off as it approaches the carrying capacity.