Calculating Rate and Exponential Growth: The Population Dynamics Problem

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  • 0:06 Population Dynamics Problem
  • 2:23 Population Growth Formula
  • 5:46 Other Rate Problems
  • 9:30 Lesson Summary
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Lesson Transcript
Instructor: Jeff Calareso

Jeff teaches high school English, math and other subjects. He has a master's degree in writing and literature.

You know how the world population keeps increasing? It's increasing faster now than it was 100 or 1,000 years ago. In this lesson, learn how differential equations predict this type of exponential growth.

Welcome to Radonville

Both C and e^C are constants
Population Growth Problem

Let's say that you're the mayor of a small town called Radonville, with a population of 1,000. You have been tasked with determining whether or not you need to build a new city hall. Your city hall has to hold all of your residents. Right now, well, let's just say that it's starting to get a little small. To build a new city hall, you need to get a good idea, or a good estimate, of how your population is changing with time. There's no point in building a city hall now if it's going to be too small in five years. You want to build a city hall that might last 30 years or so. To do that, you need to know how your population is changing as a function of time.

Population Dynamics Problem

According to the most recent census, the population is growing at a rate of 5%, and the population is currently at 1,000 people. We can write that as population, or P, equals 1,000 at t equals now; that's t=0. When will the population reach 2,000? When will it reach 5,000? These are the questions you need to answer. And, as the town mayor, you're equipped to do so.

Let's first look at what it means that the population growth is at 5% each year, per person. Because the growth is at 5%, that means that each year, every person that's in the town is responsible for 5% of a new person. If I'm in the town, for every year that I'm there I might create a foot - 5% of a new person. If there are 20 of us, maybe we've created a new person. Each one of us creates 5% of the person, so 20 of us create the whole person. Really, this rate is the number of new people per person, per year. That's a rate per person, per time. To find the population growth as a function of time, we need to multiply this rate per person times the current population. If there are 1,000 people and each one of us is contributing 5% of a person, how many people are we creating each year?

In the population growth problem, the population will be 2,000 in just under 14 years
Population Growth 14 years

Population Growth Formula

If we write this in terms of differentials, we write dP/dt, that's the change in population over time, equals 5%, or 0.05, times P - that's the current population. This is a standard differential equation. We can solve this as for P as a function of t. We're going to use separation of variables. Remember, that's where we're going to get all of the Ps on the left-hand side of the equation and all of the ts on the right-hand side of the equation. We have (1/P)dP = 0.05dt. If I integrate both sides, I get the natural log (ln) of P = 0.05t plus a constant of integration (C). Using what I know about exponentials, I can take e^(ln (P)) and get back P. On the right side, then, I have e^(0.05t + C). That's the same as (e^(0.05t))(e^C). Because C is just a constant, e^C is going to be a constant, so I'm going to call it C sub 1 so that I don't have to write this extra e in here. Okay, so I know that my population is equal to some constant, C sub 1, times e^(0.05t). Can we find this constant C? If we can't find this constant C, then I don't know what the population is at any given point in time. Let's use what we know about the population right now. At this time, at t=0, the population is 1,000. If I plug in P=1,000 and t=0, I can solve for C sub 1. C sub 1 is then 1,000. Overall, I find that the population of Radonville is equal to 1,000e^(0.05t).

When are we going to get to a population of 2,000? Let's plug 2,000 in for population and solve for t. 2,000=1,000e^(0.05t). Divide both sides by 1,000. Then, take the natural log of both sides. I get ln(2) = 0.05t. If I solve for t, I find that the population will be 2,000 in just under 14 years. That's not so bad. Hopefully, I won't be mayor by then. So, if I build a hall that can hold 2,000 people, we should be good for a while. But let's say that we want to prepare for 5,000 people. How long will that hall last? When is our population going to reach 5,000? Let's plug in 5,000. 5,000 divided by 1,000 is 5. Take the natural log of both sides and I get ln(5) = 0.05t. Solve this for t and I get 32.2 years. Okay, so if I build a town hall that can hold 5,000 people, it'll last for 32 years, assuming that our population continues to grow at 5% per year.

Equation for finding the amount of money in the bank
Bank Rate Example

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