Problem 83 - Algebra and Trigonometry

Question

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) 7331 fish after 3 years; (b) 1.73 years to reach 5000 fish.

Step 1: Understand the Problem

We are given the function modeling the fish population: \(P=\frac{10}{1+4 e^{-0.8 t}}\). This function represents the fish population (in thousands) as a function of time \(t\) in years. We need to find \(P\) when \(t=3\) and find \(t\) when \(P=5\) (since 5000 fish mean \(P=5\) in thousands).

Step 2: Calculate Population After 3 Years

Substitute \(t=3\) into the function: \(P=\frac{10}{1+4 e^{-0.8 \times 3}}\). First, calculate the exponent: \(-0.8 \times 3 = -2.4\). Then, find \(e^{-2.4}\) and substitute the value to find \(P\).

Step 3: Compute for \(e^{-2.4}\)

Using a calculator, compute \(e^{-2.4}\) which is approximately \(0.090717953\).

Step 4: Solve for \(P\) with \(t=3\)

Plug \(e^{-2.4} = 0.090717953\) into the equation: \(P = \frac{10}{1 + 4 \times 0.090717953} \approx \frac{10}{1 + 0.362871812}\). Simplify the expression to get \(P\approx 7.331\). Thus, the fish population after 3 years is approximately 7331.

Step 5: Set Up Equation for Part (b)

Set \(P = 5\) in the equation: \(5 = \frac{10}{1+4 e^{-0.8 t}}\). Begin solving for \(t\) by multiplying both sides by the denominator to clear the fraction: \(5(1+4 e^{-0.8 t}) = 10\).

Step 6: Simplify and Solve for \(e^{-0.8 t}\)

Distribute \(5\) over \(1+4 e^{-0.8 t}\) to get \(5 + 20 e^{-0.8 t} = 10\). Then, subtract 5 from both sides: \(20 e^{-0.8 t} = 5\). Divide by 20 to isolate the exponential term: \(e^{-0.8 t} = 0.25\).

Step 7: Solve the Exponential Equation

Take the natural logarithm of both sides: \(\ln(e^{-0.8 t}) = \ln(0.25)\). Using the property \(\ln(e^x) = x\), simplify to \(-0.8t = \ln(0.25)\).

Step 8: Calculate \(t\)

Solve \(-0.8t = \ln(0.25)\) for \(t\) by dividing both sides by \(-0.8\). Calculate \(t = \frac{\ln(0.25)}{-0.8}\), where \(\ln(0.25)\approx -1.386294361\). Thus, \(t \approx 1.732868\).

Step 9: Conclusion

The answer to (a) is that the fish population is approximately 7331 after 3 years, and for (b), the fish population will reach 5000 fish after approximately 1.73 years.

Key Concepts

Fish Population Model

The fish population model is a mathematical representation used to understand how the number of fish in a specific environment changes over time. In this particular case, the population of fish in a lake is represented by the function:\[P=\frac{10}{1+4 e^{-0.8 t}}\]Here, \(P\) represents the fish population in thousands, and \(t\) symbolizes time in years since the lake was stocked. This model is crucial for managing fish populations, ensuring ecological balance, and making informed environmental decisions. By inputting different values for \(t\), researchers can predict future population sizes, helping to optimize conservation efforts and fishery management practices.

Exponential Functions

Exponential functions are mathematical expressions where a constant base is raised to a variable exponent. In the fish population model, the exponent is represented as \(e^{-0.8t}\). The constant \(e\) is known as Euler's number, approximately equal to 2.71828, and it is a fundamental base for natural logarithms.

Exponential functions display rapid growth or decay. In population models, they often describe situations where growth rate is proportional to the current amount; for example, how quickly fish populations can grow. Understanding exponential functions is essential because it allows us to predict how populations will change over time and to notice trends such as possible overpopulation or extinction.

Population Dynamics

Population dynamics is the study of how and why population sizes change over time in natural systems. It considers a variety of factors, such as birth rates, death rates, immigration, and emigration. In the context of the exponential growth model, the dynamics of fish populations are shaped by the initial stocking rate of fish, the habitat conditions, and natural growth constraints.

For example, when using our fish population model, we can predict how a small lake's fish population is expected to evolve over a number of years based on the mathematical expression provided. Effective management of population dynamics ensures that ecosystems remain healthy and sustainable, preventing issues like depletion or overstocking which could lead to habitat damage.

Logarithmic Equations

Logarithmic equations involve the use of logarithms, which are the inverse operations of exponentiation. In the fish population problem, logarithms are used to solve for the time \(t\) when the exponential equation is set up.

The process involves isolating the exponential term and taking the natural logarithm of both sides, allowing the exponent to be expressed as a linear equation. For example:- Start with an exponential equation: \(e^{-0.8t} = 0.25\)- Apply the natural logarithm to both sides: \(\ln(e^{-0.8t}) = \ln(0.25)\)- Simplify using the rule \(\ln(e^x) = x\) to find \(-0.8t = \ln(0.25)\)- Solve for \(t\) by dividing by \(-0.8\)Mastering logarithmic equations allows one to flexibly switch between exponential growth and linear understanding, aiding in solving population dynamics problems efficiently.