Calculus: Exponential Growth

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Differential equations are all over the place. We've seen them and some methods to solve them. But, can they lead to romance? Well, in fact, they really can. Let me just show you how the world works. You have a girl felt figure and then you have a boy felt figure. If they like each other a lot, then they can come together. What can happen is that they can start to reproduce. Then, they can actually start to reproduce other figures, which are no longer felt, but in fact become plastic. This is the way that life really is. Then, when they grow up and they get older and older and older, then they can start to repeat the process, if they so desire. Then you get more and more and more. Welcome to the world of population explosion.
Now, how can you model that? How can you model populations? Well, what's the simplest model you can for this reproductive structure? Now, it sounds like this is maybe really a lecture that should be given in a health class or a hygiene class. But, actually, no. It belongs right in the center of mathematics, because the whole point of mathematics is that it empowers us to understand our world around us in a better, more focused way. So, how can we actually resolve this? Well, let's just think about it. If there were just one person, well then there's not a lot of reproducing going on. Now, if there were two people who decided they wanted to reproduce, well then there would be a little bit of reproducing going on.
So suppose we adopt, as a model, that the change in population will actually be proportional to the amount of people we have. We have a lot of people. We expect a big change in the population in nine months. If we have just a couple of people, then we expect a smaller change in the population. Well, just that model will lead to a very, very accurate predictor of population growth that's used a lot in biology and understanding the population models. This is actually known as "exponential growth."
Let's see if we can figure out why it's called exponential growth. Well, the model, if I write this down, is just the following. The change in population - let's call the change in population at any time P. So, the change in population, which would be , how population is changing at time t. So, the derivative, its change in population with respect to time, let's say that's proportional to the amount of population. So, it equals some number, some constant. I'll call it k, times the current population or = k P. I don't know what this constant is, but it's just some constant, so that we have an equality here. Let's think about what this says. If P is very, very big, then the derivative is big. That means that we have a lot of change. A lot of people, a lot of reproduction. Small population, small amount of reproduction. The method is the same here. Well, it's a differential equation. Remember that k is just a constant number. Don't think of it as a variable. It's fixed. It's like 5 or it's like a or something like that.
OK. So, let's see if we can solve this. If we solve this, what do I see? Well, I can separate this. I can write this as =kdt. I divide both sides by the P and multiply both sides by the dt. It separates. So, since it's separable, I can integrate both of these things separately. Here I see the natural log of the absolute value of P equals the - this is a constant, remember. So, that's just kt, plus some new constant, which I'll call C, constant integration or P=kt +C. Now, that's a great formula. That, of course, is the population growth. I don't need absolute values around it, because the population is the number of people we have. That can never be negative. You can't have -4 people. So, in fact, I can just make these parentheses, if I wanted. Anyway, if you want to solve this for P alone, you've got to undo the natural log. So you can e both sides. I would say here this is just population as a function of time. We're just recording the fact that population depends on time. That would equal this. What is that? Well, this is P(t)=e^kt * (e)^C. Right? Because if you multiply the bases, you just add the powers. So, I get that. So, this is actually the solution. Suppose that our initial population, P at zero, is P[0]. This is the initial population. Then I can actually solve this for the constant C. If I plug this in, what do I see? I see that P[0]=e^k(0). Well, e^0 is just 1. What I see here is just e^C. So, what I see is that e^C is just the initial population. So, what I see here is that P(t)=P[0]e^kt. What that means is that population is growing exponentially. Thus, the little placards exponential growth. If you take this differential equation and solve it, what you actually see is this is the solution, where this is the initial population and this is just the constant required in this exponential growth.
So, let's actually take a look at an example. This model is very accurate in that it does predict accurately what goes on, especially with small things. People, you have to look at a lot of people. Since people are so big, you have to look at a lot of them to see the action happening. But, with small things that are so small, you can see things happening. In fact, a lot of activity, a lot of reproduction goes on in a Petri dish. Here's a Petri dish. You can see we've got two little goldfish. You can think of them as bacteria or something. So, they're going along here. And, suppose that they have tendencies to reproduce. If they have that tendency, let's further hypothesize that we - in fact we start with 1,000. So, this is just going to be 1,000 here. So, let's suppose that we start our initial population at 1,000. Let's further assume that we're told that what happens is that every two hours, the population doubles. So, every two hours, the population doubles. So, what does that mean? It means that if we put this out in an enclosed thing, so that they can be alone, give them a little privacy and if we wait two hours, what do you think happens? If we wait two hours, let's see what happens. There are four of them. They're just going out of here. Now, there are more of them. So, we'd expect that in two hours, we'd see more. If the population doubles every two hours, what's going to happen? Well, the more we have, of course, the more we expect to see in the future. So, that means the faster the growth rate. So, what happens if you wait two more hours? Two hours later, now we've got tons of them here. So, you can see there are eight. You can see what's really going on.
So, how can we figure out a formula for the population? Well, the way to do this is to think about it. We'll say the population is P(t)=P[0]e^kt. That's just some constant yet to be determined. Well, the initial population was given to be 1,000. So, that means that P(t)=1,000e^kt. So, now, how can I figure out what k is? Well, since I know that the population doubles every two hours, what does that tell me? Actually it tells me how many there's going to be two hours later. Let's assume that t is time in hours. Well then, what do I see? What I see here is that if I have 1,000 to begin with, two hours later, I'm going to double and this is going to be 2,000. Now, that one extra piece of data will allow me to figure out what k is. Let me show you how to do that. You just plug in. When I put in t=2, I know that this thing has to be 2,000. So, if I insert that, what I see here is 2,000=1,000e^kt. t is two hours. So, it would be 2,000=1,000e^2k. So now, this is an equation that just has k's in it and numbers. We can actually solve this. In fact, the 1,000 can be divided here. Let's just reduce this to 2=e^2k. I can solve this pretty easily by taking the ln of both sides. The ln2 is just some number equals - this is just going to be 2k or ln2=2k. Therefore, I can solve for k. k= ln2. So, I can actually now insert the ln2. What do I see? I would see that P(t)=1,000e^(tln2/2). That now gives me a formula for population. So, for example, let's just check that and see if it makes sense. Let's plug in and see what we're doing after six hours. So, let's plug in t=6. I see 1,000. How many people are there after six hours? I have P(6)=1,000e^(6ln2/2). What does that equal? Well, 6 divided by 2 is 3. So, I see 3ln2. Is that the answer? Well, we can simplify that a little bit, if we think about it. We have 1,000e. That 3, since it's a multiple of a log, can be brought up as an exponent. So, I could write this as ln2^3. 2^3 is 8. What's e^ln? Well, these are inverse functions, so they kill each other off. e^(ln8) is just the number 8. So, this just equals 1,000 x 8, which equals 8,000. So, the population should be 8,000. Does that make sense? Let's think about it. If every two hours, the population doubles, then after two hours, I'm going to have 2,000. After four hours, I'm going to have double that, which is 4,000. After two more hours, which is going to be six hours, I'm going to have double the 4,000, which is 8,000. So, in fact, that checks. That actually makes complete sense. But, now I have a model for any time t. So, these population models, known also as exponential growth, provides powerful models.
Let's go back and see how our little ameba goldfish are doing. You can see - whoa! They're just actually having a big old party here. So, you can see that population can be modeled, using differential equations. Now, this is a very simple model. You see, in the real world, goldfish die or are eaten. The model gets a little more complicated when you include death into the picture, but they still have this exponential feel and they are delicious. OK. See you next lecture.
Differential Equations
Growth and Decay Problems
Exponential Growth Page [2 of 2]