Problem 391 - Precalculus

Question

For the following exercises, use a graphing calculator and this scenario: the population of a fish farm in \(t\) years is modeled by the equation \(P(t)=\frac{1000}{1+9 e^{-0.6 t}}\) To the nearest tenth, what is the doubling time for the fish population?

Step-by-Step Solution

Answer

The doubling time is approximately 1.4 years.

Step 1: Identify Initial Population

To find the initial population, calculate \(P(0)\). In the given equation, substitute \(t = 0\):\[ P(0) = \frac{1000}{1 + 9e^{-0.6 \cdot 0}} = \frac{1000}{1 + 9 \cdot 1} = \frac{1000}{10} = 100.\] Thus, the initial population is 100 fish.

Step 2: Determine Double Population

The goal is to find out when the population doubles. If the initial population is 100, the doubled population is \(2 \times 100 = 200\).

Step 3: Set Equation for Double Population

Set the equation equal to 200 and solve for \(t\):\[ 200 = \frac{1000}{1 + 9e^{-0.6t}}.\]

Step 4: Solve for Exponential Term

Rearrange the equation to isolate the exponential term:\[ 1 + 9e^{-0.6t} = \frac{1000}{200} = 5,\] which simplifies to \[ 9e^{-0.6t} = 5 - 1 = 4.\] Then, divide both sides by 9:\[ e^{-0.6t} = \frac{4}{9}.\]

Step 5: Solve for t Using Natural Logarithm

Take the natural logarithm of both sides to solve for \(t\):\[ -0.6t = \ln\left(\frac{4}{9}\right),\]\[ t = -\frac{1}{0.6} \ln\left(\frac{4}{9}\right).\] Use a calculator to compute the value, \[ t \approx -\frac{1}{0.6} \times (-0.81093) \approx 1.35.\]

Step 6: Round to Tenth

Round the answer to the nearest tenth. Therefore, the doubling time \(t\) is approximately 1.4 years.

Key Concepts

Population Modeling

Population modeling is a technique used to predict the growth or decline of a population over time. In this exercise, we're looking at the fish population in a farm. The population at time \(t\) is given by the function \(P(t) = \frac{1000}{1 + 9e^{-0.6t}}\). This is known as a logistic growth model, where:

• The numerator, 1000, represents the maximum carrying capacity, or the largest number of fish the environment can support.
• The denominator involves an exponential decay term, \(9e^{-0.6t}\), which indicates how the population approaches the carrying capacity over time.

This model helps predict when resource limitations will slow population growth, eventually stabilizing it. Initially, the population grows rapidly, but as time progresses, the growth rate decreases and the population levels off.

Doubling Time

Doubling time is the time it takes for a population to double in size. For exponential growth, it can be found using the natural logarithm.
In this exercise, we start with an initial population of 100 fish. We need to find out how many years it takes for this population to grow to 200 fish.
To do this, we set our model equation equal to 200, representing the doubled population amount, and solve for \(t\).

• The process involves rearranging the equation to isolate the \(e^{-0.6t}\) term.
• By taking the natural logarithm, we can solve for \(t\) to find the doubling time.

In this context, doubling time is an important measure because it gives insight into how quickly a population can grow under specific conditions.

Natural Logarithm

The natural logarithm, denoted as \(\ln(x)\), is the logarithm to the base \(e\), where \(e\) is an irrational constant approximately equal to 2.71828.
It is a fundamental concept in understanding exponential growth and decay, often used in population modeling.
In solving the exercise, we needed to take the natural logarithm of both sides of the equation \(e^{-0.6t} = \frac{4}{9}\).

• Applying \(\ln\) allows us to bring down the exponent \(-0.6t\) to solve for \(t\).
• This property of logarithms is particularly useful for converting exponential equations into linear ones, making them easier to solve.

Understanding the natural logarithm is essential for effectively analyzing and solving real-world problems involving exponential change.