Problem 49 - College Algebra

Question

In 1991 , the moose population in a park was measured to be \(4,360 .\) By \(1999,\) the population was measured again to be \(5,880\) . Assume the population continues to change linearly. a. Find a formula for the moose population, b. since 1990 . b. What does your model predict the moose population to be in 2003\(?\)

Step-by-Step Solution

Answer

a. The formula is \( P(t) = 190t + 4170 \). b. The population in 2003 is 6,640 moose.

Step 1: Identify the given data

The problem provides measurements for two years: in 1991, the moose population was 4,360, and in 1999, it was 5,880. We will use these data points to find the formula for a linear model of the population growth.

Step 2: Set up the linear equation framework

A linear model can be written in the form \( P(t) = mt + b \), where \( P(t) \) is the population at year \( t \), \( m \) is the slope of the line (rate of change per year), and \( b \) is the y-intercept (population at year 1990). We need to solve for \( m \) and \( b \) using the given data points.

Step 3: Calculate the slope (m)

The slope \( m \) of a line is calculated as the change in population divided by the change in time. Using the points (1991, 4360) and (1999, 5880), we calculate: \[m = \frac{5880 - 4360}{1999 - 1991} = \frac{1520}{8} = 190\]Thus, the population increases by 190 moose per year.

Step 4: Find the y-intercept (b)

Using the point-slope form \( P(t) = mt + b \) and one of the data points, we can solve for \( b \). Using the point (1991, 4360):\[4360 = 190 \times 1 + b \]Thus, \( b = 4360 - 190 = 4170 \). Thus, our model becomes \( P(t) = 190t + 4170 \).

Step 5: Adjust the model for time consistent with the problem

Since the problem asks for the population since 1990, let's redefine \( t = 0 \) for the year 1990. Therefore, \( t = 1 \) for 1991. Our model was already adjusted because we began calculations one year after 1990, so we keep the equation \( P(t) = 190t + 4170 \) valid for this time shift.

Step 6: Predict the population in 2003

Now, using the model \( P(t) = 190t + 4170 \), substitute \( t = 13 \) (since 2003 is 13 years after 1990):\[P(13) = 190 \times 13 + 4170 = 2470 + 4170 = 6640\]The population in 2003 is predicted to be 6,640 moose.

Key Concepts

Linear Equation

A linear equation represents a straight line when graphed, and is often used to model real-world situations where something changes at a constant rate over time. In the context of population growth, this could mean a consistent increase or decrease in population year by year.
For our moose population example, we set up the equation as \( P(t) = mt + b \), where \( P(t) \) is the moose population at year \( t \). Here, \( m \) is the slope representing the yearly change in population, and \( b \) is the y-intercept, or the population at the starting year (in our adjusted example, 1990).
This formulation helps us express population as a function of time, allowing for predictions and better understanding of trends. Understanding linear equations in this way provides a solid mathematical foundation for analyzing consistent changes over time.

Slope Calculation

The slope \( m \) in a linear equation is a critical value that shows how much a population changes each year. To calculate this, you take two data points, say \((x_1, y_1)\) and \((x_2, y_2)\), and use the formula:
\[ m = \frac{y_2 - y_1}{x_2 - x_1} \]
In our moose scenario, we took the years 1991 and 1999 with populations 4,360 and 5,880 respectively. Plugging into the formula, we get:
\[ m = \frac{5880 - 4360}{1999 - 1991} = \frac{1520}{8} = 190 \]
This calculation tells us that the moose population is increasing by 190 each year.
Understanding and calculating slope helps in assessing the rate of change, which is crucial in many predictive and analytical models.

Y-intercept

The y-intercept \( b \) of a linear equation \( P(t) = mt + b \) is the initial value where the line crosses the y-axis. It represents the population size at the very start of the period under consideration.
For our moose model, the y-intercept is found by solving the equation with the slope and one known data point. Using 1991 as an example, where the population was 4,360, we calculate:
\[ 4360 = 190 \times 1 + b \]
Solving this, \( b = 4360 - 190 = 4170 \).
Thus, the y-intercept \( b \) is 4170, indicating that in 1990, the moose population was projected to be 4,170.
The concept of a y-intercept helps us in establishing a starting point for the linear model, contributing to accurate predictions and understanding of data trends.

Predictive Modeling

Predictive modeling involves using mathematical equations to predict future outcomes based on known data. This method applies well to situations where a consistent pattern or trend, like linear growth, is observed.
In our example, once we established the linear growth model \( P(t) = 190t + 4170 \), we used it to predict the population in future years, such as in 2003. To find this prediction, replace \( t \) with 13, since 2003 is 13 years from 1990:
\[ P(13) = 190 \times 13 + 4170 = 6640 \]
This gives us a predicted population of 6,640 moose for that year.
Predictive modeling is a powerful tool for forecasting future trends, facilitating planning, and making informed decisions.