Problem 32 - 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) After 6 months: approximately 778 trout; after 1 year: approximately 607 trout. (b) \(P(3)\) is around 223 trout; 223 trout remain after 3 years. (c) 100 trout remain at approximately 4.61 years. (d) The trout population decreases exponentially due to natural factors.

Step 1: Evaluate P(0.5) for six months

Given the formula for trout population over time \(P(t) = 1000 e^{-0.5t}\), we start with calculating the number of trout after six months. Since six months is \(0.5\) years, substitute \(t = 0.5\) into the equation: \[ P(0.5) = 1000 e^{-0.5 imes 0.5} = 1000 e^{-0.25} \] Calculate the value of \(P(0.5)\) using a calculator to find the number of fish left after six months.

Step 2: Evaluate P(1) for 1 year

Next, we calculate the trout population after 1 year by substituting \(t = 1\) in the equation: \[ P(1) = 1000 e^{-0.5 imes 1} = 1000 e^{-0.5} \] Use a calculator to find the value of \(P(1)\) to obtain the number of fish left after 1 year.

Step 3: Evaluate P(3) to interpret population

We find the population after 3 years by substituting \(t = 3\): \[ P(3) = 1000 e^{-0.5 imes 3} = 1000 e^{-1.5} \] Calculate this value to determine the number of trout remaining after 3 years. Interpret this result as the number of trout left in the pond three years after stocking.

Step 4: Solve P(t) = 100 for 100 trout left

To find when there are 100 trout left, set \(P(t)\) equal to 100 and solve for \(t\): \[ 100 = 1000 e^{-0.5t} \] Divide both sides by 1000: \[ 0.1 = e^{-0.5t} \] Take the natural logarithm of both sides: \[ ext{ln}(0.1) = -0.5t \] Solve for \(t\) by dividing by \(-0.5\): \[ t = \frac{ ext{ln}(0.1)}{-0.5} \] Calculate \(t\) to find when the population reaches 100 trout.

Step 5: Graphing and Analysis

Graph the function \(P(t) = 1000 e^{-0.5t}\) using a graphing tool or calculator for a range of \(t\) values, such as from 0 to 10 years. Observe that the number of trout decreases exponentially over time. The exponential decay depicts a decrease in population likely due to environmental factors such as predation, disease, or lack of resources.

Key Concepts

Trout Population

The concept of trout population refers to the number of trout found in a specific area, such as a pond. This population can change over time due to various environmental factors. In our exercise, the trout population is modeled to understand how many fish remain in a pond after being initially stocked.
The population is originally 1000 trout. Over time, this number changes based on factors such as birth rates, death rates, and environmental conditions, including food availability and predation.
In the scenario presented, the trout population is decreasing over time, which is a representation of exponential decay. This means that the number of trout decreases at a rate proportional to its current value, leading to a rapid decline in population if not managed.

Population Modeling

Population modeling is the process of using mathematical equations to represent how populations change over time. It helps in understanding trends and predicting future population sizes. In the case of our trout population, an exponential decay model is used. This model is presented by the equation: \( P(t) = 1000 e^{-0.5t} \).

• \(P(t)\) represents the population at time \(t\).
• The term 1000 is the initial number of trout in the pond.
• The constant \(-0.5\) is a decay rate, indicating how quickly the population decreases over time.

Such models are crucial for fisheries management. They allow for informed decision-making regarding stocking amounts and times, and help determine appropriate management actions to maintain a stable fish population.

Logarithmic Equations

Logarithmic equations are pivotal in solving problems where exponential growth or decay is involved. They help 'reverse' or solve equations that involve exponential functions, allowing us to find unknowns.In our trout population exercise, we encounter a situation where we need to determine when the population will reach 100 trout. The equation \( P(t) = 100 \) is used, and through substitution and rearrangement, we arrive at \( e^{-0.5t} = 0.1 \).
Taking the natural logarithm, \( \ln(0.1) = -0.5t \), simplifies finding \(t\). Solving for \(t\), using \( t = \frac{\ln(0.1)}{-0.5} \), provides us with the specific time when only 100 trout remain in the pond.
Understanding logarithmic equations is essential when working with exponential models and real-world contexts where rates of change are not constant but rather proportional to the current state.