Problem 450 - Precalculus

Question

For the following exercises, use this scenario: The population \(P\) of an endangered species habitat for wolves is modeled by the function \(P(x)=\frac{558}{1+54.8 e^{-0.462 x}},\) where \(x\) is given in years. Use the intersect feature to approximate the number of years it will take before the population of the habitat reaches half its carrying capacity.

Step-by-Step Solution

Answer

It will take approximately 12 years for the population to reach half its carrying capacity.

Step 1: Understand the Problem

We need to find the number of years it takes for the wolf population, modeled by the function \( P(x) = \frac{558}{1+54.8 e^{-0.462 x}} \), to reach half its carrying capacity. The carrying capacity is the maximum population that the habitat can support.

Step 2: Determine Carrying Capacity

The carrying capacity of the model is the population value as \( x \to \infty \). In this case, as \( e^{-0.462x} \to 0 \), the value becomes \( \frac{558}{1} = 558 \). Thus, the carrying capacity is 558 wolves.

Step 3: Calculate Half of Carrying Capacity

Half of the carrying capacity is \( \frac{558}{2} = 279 \). We need to determine when the population \( P(x) \) is equal to 279.

Step 4: Set Up Equation

Set up the equation where the population is equal to half of the carrying capacity: \( \frac{558}{1 + 54.8 e^{-0.462 x}} = 279 \).

Step 5: Solve for \( x \)

Multiply both sides by \( 1 + 54.8 e^{-0.462 x} \) to get \( 558 = 279(1 + 54.8 e^{-0.462 x}) \). Simplifying, this becomes \( 558 = 279 + 279 \times 54.8 e^{-0.462 x} \), leading to \( 279 \times 54.8 e^{-0.462 x} = 279 \).

Step 6: Simplify to Find e^{-0.462x}

Divide both sides by 279: \( 54.8 e^{-0.462 x} = 1 \). Then solve for \( e^{-0.462 x} \): \( e^{-0.462 x} = \frac{1}{54.8} \).

Step 7: Solve the Exponential Equation

Take the natural logarithm of both sides to solve for \( x \): \( -0.462 x = \ln \left(\frac{1}{54.8}\right) \).

Step 8: Calculate \( x \)

Calculate \( x \) by rearranging the equation: \( x = \frac{-\ln(\frac{1}{54.8})}{0.462} \). Use a calculator to find \( x \approx 11.8 \).

Step 9: Final Solution

Round the value to the nearest whole number as it is unrealistic to consider a fraction of a year: \( x \approx 12 \). This means it will take approximately 12 years.

Key Concepts

Carrying Capacity

In population modeling, carrying capacity refers to the maximum number of individuals that a particular environment can sustainably support. It is a key concept in ecology and helps in understanding the limits of population growth. In the context of our given problem, carrying capacity is the term used in the logistic growth model to signify this biological limit.
The equation given in the ORIGINAL EXERCISE models the population of wolves in terms of years, with the carrying capacity denoted by the number 558 in the function's denominator. This means that the habitat can sustain up to 558 wolves without causing environmental degradation.

• As the population approaches the carrying capacity, the growth rate decreases.
• This results in a leveling off of the population number over time.
• Understanding carrying capacity helps in wildlife management and conservation efforts by predicting population sustainability.

Hence, the carrying capacity provides a realistic limit to how large the population of wolves can grow in this habitat.

Exponential Function

An exponential function is a mathematical expression where a constant base is raised to the power of a variable exponent. It's crucial in modeling dynamic systems like population growth. In the logistic growth model for the wolf population, the exponential function appears as the term inside the denominator: \(e^{-0.462x}\).
Here, \(e\) signifies Euler's number, a fundamental mathematical constant approximately equal to 2.71828. This function models the growth rate, which is initially rapid when resources are abundant. However, it exponentially declines as the population nears its carrying capacity.

• The base \(e\) represents continuous growth or decay, making it suitable for modeling real-world processes.
• In population terms, the negative exponent \(-0.462x\) indicates a damping effect, slowing growth as time passes.
• This characteristic ensures that the population does not exceed available resources and stabilizes over time.

Understanding the exponential part of the logistic growth model is key to predicting population trends over time.

Population Modeling

Population modeling involves creating mathematical frameworks to predict changes in population size over time. These models help us understand the dynamics of species' populations in varying environmental conditions. The given wolf population model is an example of logistic growth, where the growth rate is restricted as resources become limited.
The model used here includes:

• Initial exponential growth at lower population sizes.
• A gradual reduction in growth rate as the population nears its carrying capacity.
• An eventual stabilization of the population size around the carrying capacity.

The logistic equation in this context is:\[ P(x) = \frac{558}{1 + 54.8 e^{-0.462 x}} \]This equation effectively captures the rapid early growth and subsequent stabilization. Population modeling is vital in conservation biology, aiding in the development of strategies for species protection and resource management.

Natural Logarithm

The natural logarithm, known as \(\ln\), is the inverse function of the exponential function involving Euler's number \(e\). It's especially useful in solving equations where the unknown is an exponent, such as in our population model equation.
In step 7 of the original solution, the natural logarithm is applied to both sides of the equation:\[ -0.462 x = \ln \left(\frac{1}{54.8}\right) \]This transformation allows us to isolate \(x\) by linearizing the equation, leading to a straightforward solution. Applying the natural logarithm here means:

• We convert the multiplicative form of growth into an additive one, simplifying complex exponential relationships.
• This technique is vital in manipulating and solving equations involving exponential growth or decay.
• It's especially helpful in determining the time needed for certain population shifts.

Understanding how the natural logarithm functions help simplify otherwise intricate exponential expressions into understandable linear relations.