Problem 40 - Applied Calculus

Question

A fishery stocks a pond with 1000 young trout. The number of trout \(t\) years later is given by \(P(t)=1000 e^{-0.5 t}\) (a) How many trout are left after six months? After 1 year? (b) Find \(P(3)\) and interpret it in terms of trout. (c) At what time are there 100 trout left? (d) Graph the number of trout against time, and describe how the population is changing. What might be causing this?

Step-by-Step Solution

Answer

(a) ~778 trout after 6 months, ~607 after 1 year. (b) ~223 trout after 3 years. (c) ~4.61 years for 100 trout remaining. (d) Population declines rapidly over time due to natural decay or fishing.

Step 1: Calculate Trout After Six Months

To calculate the number of trout after six months, substitute \( t = 0.5 \) into the formula \( P(t) = 1000e^{-0.5t} \). \[ P(0.5) = 1000e^{-0.5 \times 0.5} = 1000e^{-0.25} \] Using a calculator, find \( e^{-0.25} \) and multiply by 1000 to get \( P(0.5) \).

Step 2: Calculate Trout After 1 Year

Substitute \( t = 1 \) into the formula \( P(t) = 1000e^{-0.5t} \). \[ P(1) = 1000e^{-0.5 \times 1} = 1000e^{-0.5} \] Find \( e^{-0.5} \) using a calculator and multiply by 1000 to calculate \( P(1) \).

Step 3: Calculate P(3) for Three-Year Trout Population

Substitute \( t = 3 \) to find \( P(3) \) using \( P(t) = 1000e^{-0.5t} \). \[ P(3) = 1000e^{-0.5 \times 3} = 1000e^{-1.5} \] Calculate \( e^{-1.5} \) and multiply by 1000 to determine \( P(3) \). Interpreting this, the result represents the number of trout left after three years.

Step 4: Solve for Time When 100 Trout Are Left

Set \( P(t) = 100 \) and solve for \( t \):\[ 100 = 1000e^{-0.5t} \] Divide both sides by 1000:\[ 0.1 = e^{-0.5t} \] Take the natural logarithm of both sides:\[ \ln(0.1) = -0.5t \] Solve for \( t \): \[ t = \frac{\ln(0.1)}{-0.5} \] Using a calculator, find the value of \( t \).

Step 5: Graph and Analyze Trout Population Dynamics

Create a graph with time \( t \) on the x-axis and \( P(t) \) on the y-axis. Plot the equation \( P(t) = 1000e^{-0.5t} \) to visualize the exponential decay over time. The graph should show a rapid decline initially, leveling out as time progresses. This decreasing pattern suggests that either the trout are being fished quickly or there are high natural mortality rates within the pond.

Key Concepts

Exponential Function

An exponential function is a mathematical expression where a constant base is raised to the power of a variable. In the context of the fishery problem, the function used is \( P(t) = 1000e^{-0.5t} \). Here, the constant base is \( e \), which is approximately 2.71828.
This base is used because the function involves exponential decay, rather than growth. The exponent \(-0.5t\) dictates the rate at which the trout population decreases over time.

• \( e \) is a special mathematical constant known as Euler's number.
• \(-0.5t\) represents continuous decline, leading to a multiplicative decrease in the population over time.
• As \( t \) increases, \( P(t) \) decreases, indicative of fewer trout over time.

Population Modeling

Population modeling involves using mathematical equations to predict how a population changes over time. In the case of the trout pond, the exponential function \( P(t) = 1000e^{-0.5t} \) models the population. This serves as a realistic framework to understand population decline.
The model assumes the trout die off following a specific exponential rule, represented by \(-0.5t\). This decline could be due to various factors:

• Fishing activities removing trout from the pond.
• High natural mortality perhaps due to unsuitable living conditions.

It is vital for managers of the fishery to monitor and adjust their practices based on the outcomes from such models. By understanding these factors, they can take corrective actions to maintain a sustainable population.

Natural Logarithm

The natural logarithm, denoted by \( \ln \), is the inverse function of the exponential function. It helps solve equations where the variable is an exponent, such as determining when only 100 trout are left in the pond.
To find this time, the equation \( P(t) = 100 \) is set. This resolves as \( 100 = 1000e^{-0.5t} \). After simplification and using the natural logarithm:

• Divide by 1000 to isolate the exponential expression: \( 0.1 = e^{-0.5t} \).
• Apply \( \ln \) to both sides: \( \ln(0.1) = -0.5t \).
• Solve for \( t \) using: \( t = \frac{\ln(0.1)}{-0.5} \).

Using a calculator, \( \ln(0.1) \) gives a negative result, which fits the decline context as \( t \) will be positive, indicating it is a future time point where this critical level of 100 trout is reached. This illustrates the powerful role of logarithms in population modeling and analysis.