 Welcome to the GVSU Calculus Screencasts. In this edition we're going to talk about a differential equation to model exponential growth. Populations of things like people or viruses typically spread exponentially at first if there's no quarantine to contain them or no immunization for them. In this screencast we'll uncover a differential equation that can be used to model the growth of a virus. We need to start with some initial assumptions. We'll assume that a virus grows at a rate of 11% per day. And we'll assume that we're going to start with a virus population of 100. So pause the video for a moment and consider the following question. We have a population of 100 and the virus grows by 11% per day. What would the population be after day one? Resume the video when you're ready. Well, if we have 100 virus and the growth rate is 11%, there will be an additional 11% of 100 or 11 virus the next day. Add that to the original 100 virus and we'd end up with 111 virus after day one. Now we have 111 virus. So pause the video and answer the same kind of question. With 111 virus and an 11% growth rate, what will the virus population be after two days? Resume the video when you're ready. With an assumption of a virus population of 111 and a growth rate of 11%, then there will be an additional 11% of 111 virus or about 12.21 virus that we add the next day for a total population of 111 after one day plus 12 virus the next day or 123 virus. And we're not going to count fractions of virus. Now this example illustrates an important idea that the rate of growth of the virus population depends on how many virus we have. The larger the population of virus, assuming an 11% growth rate, the faster the population is going to grow. In other words, as the population increases, so does the rate of growth of the population. And this should be expected. When you have a large population, there are more individuals to reproduce. And in general, when we're modeling population growth, we often make the assumption that the rate of growth of the population is proportional to the population. To make this more clear from a mathematical perspective, we'll want some notation. So let's let p equal p of t be the population of the virus at time t. So p is a function of time. The differential equation that we're going to build to model this population growth is going to tell us how the population changes over time. So we're going to need a function that tells us how the population changes. Pause the video for a moment and write down such a function. And then resume the video when you're ready. Now we know a function that tells us the rate of change of a function. So the function that's going to tell us how the population changes is the derivative dp dt. Recall that we're making the assumption that our virus population grows at a rate proportional to the population that's present. And two quantities are proportional to each other if one is a non-zero constant multiple of the other. So since the rate of change of the population is proportional to the population, we conclude that dp dt is a non-zero constant multiple of p. And this gives us a differential equation. dp dt equals kp, where k is some constant that models the growth of our virus population. And this constant k in this differential equation, dp dt equals kp, tells us how fast the population is growing. Now in our virus population example, we're assuming a growth rate of 11%, so the value of k is 0.11. And that gives us a differential equation, dp dt equals 0.11 p, that represents the growth of our virus population. Now it's worthwhile to think about what kind of function satisfies differential equation dp dt equals 0.11 p. Recall that the chain rule tells us that if we differentiate a constant times e to the 0.11 t, we get 0.11 times that constant times e to the 0.11 t. And this works for any constant c. So the function p of t equals some constant times e to the 0.11 t, has the property that the derivative of p of t is just 0.11 times p. All of this tells us that our virus population can be modeled with an exponential function of the form some constant times e to the 0.11 t. All we need to do now is figure out what's the value of that constant c. If we start with an initial population of 100 virus, then notice that means that p of 0 is 100, where p of t is the population at time t. If we use our function p of t equals c times e to the 0.11 t, then p of 0 is c times e to the 0, e to the 0 is 1, so p of 0 is c, that makes c equal to 100. And so our model for our population growth of our virus is 100 e to the 0.11 t at time t. Now if you evaluate this function p at 1, you get about 111, and if you evaluated it at 2, you get about 124. So this solution p actually does a pretty good job of approximating our earlier calculations. To summarize, as our example demonstrates, an exponential function y of t equals c e to the kt is a solution to a differential equation of the form dy dt equals k times y. And this means that any function that satisfies a differential equation of the form dy dt equals k times y is said to grow exponentially, and grow if k is positive and decay if k is negative. That concludes our screencast on differential equations modeling exponential growth. We hope to see you back again soon.