 So let's take a look at an example of finding the average and instantaneous rates of change. So the population of a country is given by P of t equals 52e to power 0.02t million persons t years after 2000. And we want to find out how rapidly was the population growing between 2010 and 2015, and how rapidly was the population growing in 2017. So if we take this question apart, we see that the first rate of change is over the interval between 2010 and 2015. So remember that when we looked at the grammar of change, the average rate of change is always over some interval from somewhere to somewhere. And that's exactly what we seem to have. And so we need to know the population in 2010 and the population in 2015. Since t is the number of years after 2000, then the population in 2010 is the population when t is equal to 10. So we'll substitute t equals 10 into our function and evaluate it. And we get a value of 63.513 that's in millions of persons. Because this is an average rate of change, we also need the population in 2015 when t is equal to 15. Now let's record our answer. In 2010 the population was 63.513 million, and in 2015 the population was 70.193 million for an average rate of change between 2010 and 2015 of n minus beginning over how long that change took. Since that change took place between 2010 and 2015, it was over 5 years. Now the numerical value is 1.336. However, the number without units is meaningless. So we do have to include the units. Both of these numerator values are measured in millions of persons. Meanwhile, the 5 in the denominator is a number of years. And so the units will be millions of persons per year. Now can we determine how rapidly was the population growing in 2017? Now the second rate of change is at a specific instant 2017. And this means we're looking at an instantaneous rate of change in 2017 where t is equal to 17. So remember the instantaneous rate of change is the limit of the average rate of change as the length of the interval goes to zero. Now we can approximate this using an average rate of change between t equals 17 and a very close by value of t. So we find that at t equals 17, our population is, then let's use a close by value of t, maybe 17.001, and we find that our population at 17.001 is approximately, and we can summarize our results. Using the population at t equals 17 and t equals 17.001, we find the rate of change of population in 2017 to be about n minus beginning over how long it took. And that works out to be about 1.46 million persons per year. Now because we're dealing with actual numbers here, and we can't record an infinite number of decimal places, we do have to be concerned about rounding. Unfortunately, rounding is a very complicated issue. And will generally be dealt with in more detail in courses in the sciences. However, for our purposes, it is important to keep in mind the following ideas. When calculating instantaneous rates of change numerically, it's important to be aware of what's called round off error. And the idea is that every time you round a number, you introduce an inaccuracy. And what this means is that you shouldn't round a number until you are done with it until you've reached the very end of the problem. And then, and only then, can you round your final answers. To see why this might be a problem, let's see what happens. If we rounded our numbers to two decimal places, we'd have the following. At t equals 17, p of 17 is approximately. And at t equals 17.001, p of 17.001 is approximately. Since our instantaneous rate of change would work out to be, but that's completely wrong. And again, the reason that it's wrong is that we've rounded off and then use these rounded numbers. So how many decimal places should you keep? There's no good answer to that without going into a lot of details over rounding. But as a general rule, keep as many decimal places as you can until you get to the very end of the problem.