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Math 104: Calculus14 chapters | 115 lessons | 11 flashcard sets

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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.

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.

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?

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 *P*s on the left-hand side of the equation and all of the *t*s on the right-hand side of the equation. We have (1/*P*)*dP* = 0.05*dt*. If I integrate both sides, I get the natural log (ln) of *P* = 0.05*t* 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*^(0.05*t* + *C*). That's the same as (*e*^(0.05*t*))(*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*^(0.05*t*). 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 1,000*e*^(0.05*t*).

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=1,000*e*^(0.05*t*). Divide both sides by 1,000. Then, take the natural log of both sides. I get ln(2) = 0.05*t*. 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) = 0.05*t*. 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.

This type of exponential growth is the same type of behavior you see in things like bank accounts or stock markets, on average. In a bank account, you'll get a return that is a percent interest based on the amount that you have in the bank. The change in the amount of money that you have in the bank (*dM*) over time (*dt*) equals your interest rate (*r*), per dollar per year, times the number of dollars that you have; let's call that *M*.

This equation looks exactly like the last one. I can separate the variables and integrate, pull down my constant of integration, and I find that the amount of money that I have in the bank equals *C* sub 1 times *e* to my interest rate times time (*M* = (*C* sub 1)(*e*^(*rt*)). What does this mean? Let's say that I start with $10,000 in the bank. If I start with $10,000 in the bank right now, that's at *t*=0, then my constant, *C* sub 1 = 10,000. If the interest rates are 0.9%, or 0.009 for *r*, then after 30 years (*t*=30), I can solve this equation to find out how much money I have in the bank. It's 10,000*e*^(0.009 * 30). That will give me $13,000. That's not very much interest. That's $3,000 interest over 30 years.

But what if I can invest in the stock market and get a return of 5%? If I can get a return of 5%, that means my rate is 0.05. After 30 years, I will have a total of 10,000*e*^(0.05 * 30), which is equal to just under $45,000. Why are these problems so similar? They're similar because they're both **rate problems**. In both of these, the variable that you care about, the population or the amount of money that you have, is changing proportionally to the current value of your variable, like your current population or the current amount of money that you have. You can write both of these in the form *dP*/*dt* = *kP* or *dM*/*dt* = *rM*. They both have the same form. This means that you can solve these by using a separation of variables. When you do that, you find this exponential term, such as *P*=*Ce*^*rt*.

This is this kind of exponential. Now, how high this exponential is, if you graph it, is going to depend on your initial value. How fast it rises depends on the rate. Think of this in terms of your population. How high your population is as a function of time depends in part, at least initially, on how high your population is currently. How fast your population rises depends on how your population is changing, or by what percentage. A higher percentage means it's changing faster and this graph will be steeper. A lower percentage, or a population growth of zero, will give you a flat curve. Both of these are rate problems.

Okay, let's review. Any time you see a **rate problem**, when you have a derivative like *dP*/*dt* on the left side and something proportional to your variable of interest on the right side (*rP*), try using separation of variables and expect a solution that looks like *e*^(*rt*).

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Math 104: Calculus14 chapters | 115 lessons | 11 flashcard sets

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