In 1993, the moose population in a park was measured to be 5000. By 1996, the population was measured again to be 5150.

Question

In 1993, the moose population in a park was measured to be 5000. By 1996, the population was measured again to be 5150. If the population continues to change linearly:

A.) Find a formula for the moose population, , in terms of , the years since 1990.

P(t)=

B.) What does your model predict the moose population to be in 2003?

Answer 1

To "change linearly" means that the population will change by the same amount over equal durations of time. For example, if the population changed by 100 over 2 years, then we can assume the population will continue to change by 100 the next 2 years.

In this case, we're told that, in 1993, the population was 5000, and after 3 years, in 1996, the population increased by 150, yielding 5150 moose. This means that after another 3 years, in 1999, the population will continue to increase by 150, yielding 5300 moose.

That's what it means to change linearly, but how do we write this more mathematically, in the formula their asking for? What we want to do is create a rule (or function) so that if we're given t, the number of years since 1990, we can calculate P, the population.

So far, we know that if t = 3, the year is 1993, and the population is 5000. We also know that every 3 years, the population increases by 150. This also means if we go backwards in time 3 years, then the population decreases by 150. We can then deduce that at t = 0, or 1990 which is 3 years before 1993, the population was 150 less, so 5000 - 150 = 4850.

But now, what if t = 1? How do we find the population between intervals of 3 years? Well, because it changes linearly, the difference in population from 0-1, 1-2, and 2-3 years should all be the same. That means we have to chop up our 3 year difference into 3 equal parts. Effectively, what we're doing is finding a constant ratio that represents the change in population over the change in time. So the ratio for this problem is 150 moose : 3 years, or 50 moose/yr (150 / 3 = 50). If we multiply this ratio by how many years have passed, we will get the increase in population.

With this, we've figured out one part of the rule to calculate the population: dP = 50t, where dP is the change in population. Using it, we can partially solve B) by plugging in 13 for t (2003 is 13 years passed 1990): dP = 50(13) = 650. But is that the total population? No, just the change. To get the total, we have to add that to the original population, which, at 1990 was 4850. Adding gives us 650 + 4850 = 5500. So we can predict the moose population in 2003 is 5500 moose.

But we still have to complete our formula for part A. We found that the change in population dP could be represented by dP = 50t. What did we have to do to get the total population? Just add 4850, the initial population! So to get our final formula, we find P(t) = 50t + 4850.

Does this look familiar? This is y = mx + b, the slope-intercept form of a line! The reason we call this change linear is because it looks like a line when you graph it. Hopefully that helps your understanding of the problem, as well as give insight into the connection between things in math.

Answer 2

"normally" you'd do this problem with an exponential growth rate,

but the problem says "linear"

so, that means (0, 5000) and (3, 5150) as two points on a line

with the equation (5150-5000)/3= 150/3 = 50 = slope = y-5000)/(x)= 50

y = 50x + 5000

or P = 50t + 5000 is the equation with t measured in years since 1990

P(t) = 50t + 5000

P(13) = 50(13) + 5000

= 650 + 5000 = 5650

P(13) = 5650 in year 2003 (2003-1990 = 13 = t)