In 1995, the moose population in a park was measured to be 5600. By 1999, the population was measured again to be 6500.

Question

In 1995, the moose population in a park was measured to be 5600. By 1999, the population was measured again to be 6500. If the population continues to change linearly, find an equation for the moose population, , as a function of , the years since 1990.

What does your model predict the moose population to be in 2001?

Answer 1

Changing linearly, means the population always changes at the same rate. What is that rate? To determine that, let's first think about the variables. Population is the dependent variable. It changes as time changes. So time is the independent variable. You pick a year, and the population can be determined for that year.

The rate of change (also called the slope of the line on the graph) is the change in dependent variable divided by the change in the independent variable. So in this case, it's the change in moose population divided by the change in time. Or (5600 - 6500) divided by (1995 - 1999) which is -900/-4 = 900/4 = 225. The units of that number are the units of the dependent variable (moose) divided be the units of the independent variable (years), so 225 moose/year or the population changes by 225 moose per year.

In this case, they want us to write an equation considering years since 1990. A linear equation can be written in the form y = mx + b where y is the dependent variable, m is the slope (or rate of change), x is the independent variable, and b is the initial condition when the independent variable equals 0. So, in this case population is the "y" and time (they want use to use "t") is the "x" and m is the rate of change we calculated (225 moose/year) but we need to know what the "b" is, which was the initial population when t = 0 or the population of moose in 1990.

Since the population always changes by 225 moose per year, and since 1990 was 5 years before 1995, we can take the population in 1995 and subtract 5*225. So the population in 1990 was 5600 - 225*5 = 5600 - 1125 = 4475.

Using that, we can write the equation. y = mx + b becomes P = 225t + 4474. One more thing though. They are asking us to write an equation for the moose population (P) as a function of t. To do that, we write it as P(t) = 225t + 4475

Now let's try it to make sure it works. 1995 is 5 years after 1990 and 1999 is 9 years after 1990 so if we plug in 5, we should get the population in 1995 and if we plug in 9, we should get the population in 1999.

225(5) + 4475 = 1125 + 4475 = 5600 (it works)

225(9) + 4475 = 2025 + 4475 = 6500 (it works)