Problem 83 - Precalculus Mathematics for Calculus

Question

Fish Population \(A\) small lake is stocked with a certain species of fish. The fish population is modeled by the function $$P=\frac{10}{1+4 e^{-0.8 t}}$$ where \(P\) is the number of fish in thousands and \(t\) is measured in years since the lake was stocked. (a) Find the fish population after 3 years. (b) After how many years will the fish population reach 5000 fish?

Step-by-Step Solution

Answer

(a) After 3 years, the population is 7340 fish. (b) The population will reach 5000 fish after approximately 3.46 years.

Step 1: Understand the Function

We have the fish population function given by \( P = \frac{10}{1+4 e^{-0.8 t}} \), where \( P \) represents the population in thousands and \( t \) is the number of years since the fish were stocked. We need to calculate \( P \) after a specific number of years and find \( t \) when \( P = 5 \) (since 5000 fish is 5 in thousands).

Step 2: Calculate Population at 3 Years

Substitute \( t = 3 \) into the population equation: \( P = \frac{10}{1 + 4 e^{-0.8 \times 3}} \). Evaluate the exponent in the denominator first: \( e^{-2.4} \). Use a calculator to get \( e^{-2.4} \approx 0.0907 \). Now, calculate: \( 1 + 4 \times 0.0907 = 1.3628 \). The population \( P \) is therefore \( \frac{10}{1.3628} \approx 7.34 \).

Step 3: Solve for Time when Population is 5000 Fish

Set \( P = 5 \) in the equation and solve for \( t \): \( 5 = \frac{10}{1 + 4 e^{-0.8 t}} \). Rearrange to isolate the exponential term: \( 1 + 4 e^{-0.8 t} = 2 \) then \( 4 e^{-0.8 t} = 1 \). Divide by 4: \( e^{-0.8 t} = 0.25 \). Take the natural logarithm of both sides: \( -0.8 t = \ln(0.25) \). From this, calculate \( t = \frac{\ln(0.25)}{-0.8} \approx 3.46 \).

Key Concepts

Exponential Functions

Exponential functions are powerful mathematical models used to describe a wide range of real-world scenarios. An exponential function has the form \( f(t) = ae^{kt} \), where \( a \) is the initial amount, \( e \) is the base of the natural logarithms, and \( k \) is a constant that determines the rate of growth or decay.
In these functions, the rate of change is proportional to the current value. As time progresses, the value of an exponential function can grow or shrink dramatically.
These functions are particularly useful in modeling situations where growth accelerates rapidly, such as populations, radioactive decay, and interest in banking, making them vital tools in various scientific fields.

Population Growth

Population growth is often modeled using exponential functions due to their ability to mathematically depict rapid growth. The function \( P = \frac{10}{1+4 e^{-0.8 t}} \) is an excellent example of how exponential growth and decay can be factored into a model.
In our model, \( P \) represents the number of fish (in thousands) in a lake over time \( t \).

• The numerator "10" signifies the carrying capacity or maximum sustainable population.
• The variable \( t \) is the time in years, showing how long the population has been growing.
• The exponential term \( e^{-0.8t} \) indicates how quickly the population reaches equilibrium.

As \( t \) increases, the exponential function declines towards zero, causing the denominator to approach one, and \( P \) approaches the carrying capacity. This behavior mimics how natural populations stabilize due to environmental limits.

Natural Logarithm

The natural logarithm, denoted as \( \ln \), is the logarithm to the base \( e \), where \( e \) is approximately 2.71828. It's a key component in solving problems involving exponential equations because it is the inverse operation of exponentiation with base \( e \).
In the problem, we used the natural logarithm to solve for the time \( t \) when the fish population reached 5000:

• By setting \( P = 5 \), we simplified the equation to \( e^{-0.8 t} = 0.25 \).
• Taking the natural logarithm of both sides, we have \( -0.8 t = \ln(0.25) \).
• Solve for \( t \) by dividing through by \(-0.8\).

Understanding how to work with natural logarithms is crucial for solving exponential growth or decay problems, as seen in our fish population model.

Problem Solving Steps

Breaking down complex problems into smaller, manageable steps can greatly enhance understanding and accuracy in calculations. Let's revisit how this methodical approach is applied in our exercise:

• **Understand the problem:** Familiarize yourself with the given equation and what each variable represents.
• **Calculate specific values:** For instance, substituting \( t = 3 \) into the model to find the population at 3 years was the first step.
• **Solve for unknowns:** When determining the time for a population of 5000, we set \( P = 5 \) and solved analytically to find \( t \).
• **Use mathematical tools:** Utilize calculators or software for accurate computations, especially for exponential and logarithmic values.

Each step helps in focusing on a part of the problem, ensuring clarity. This approach avoids errors that can arise when trying to solve problems in one go.