 Now we'll introduce some calculus through the instantaneous rate of change. The problem is that average rates of change aren't necessarily all that informative. For example, suppose I have a function that gives the height of a rock in meters t seconds after it's released. We want to find and interpret the average rate of change of h of t over the interval 0 to 2. So remember that the average rate of change is found by finding the ratio, the quotient, of the difference of the function values over the length of the interval. So we'll find our function values at 2 and at 0, and substitute these in, making sure we take care of the unit appropriately. And we get an average rate of change of 0 meters per second. So we might interpret this as meaning the height changed by an average of 0 meters every second during that 2 second interval. And so this raises a new question. If the average change in height was 0 meters per second, was the rock not moving? The problem here is that our interval is very long, and it might not capture what's really going on. And so we can try to use a shorter interval. For example, we might want to look at the average rate of change of height over the interval between 0 and 1. So we'll compute that average rate of change of height, and find that it has an average rate of change of height of 5 meters every second. Or maybe we'll take a look at the average rate of change of height over the interval between 0 and 0.1. And here we find it's 9.5 meters every second. What about the average rate of change of height over the interval between 0 and 0.01? And that'll work out to be 9.95 meters per second. Or we could even take a look at the average rate of change of height over the interval between 0 and 0.001. And that works out to be 9.995 meters every second. And we might make the following observation. As the length of the interval gets closer to 0, the average rate of change gets closer to 10 meters every second. At this point, a good question to ask yourself is, self, where have I seen this before? Now ordinarily, you have to scan through the entirety of your human existence to answer that question. But because this is coming up very early in a calculus course, you should suspect that this has something to do with some of our previously introduced concepts. And in fact, it does. Remember, our concept of a limit is that given some function and some value, the limit of our function as x approaches a is some number that our function gets close to. And it appears that this average rate of change seems to be getting close to the value of about 10 meters per second. So there's some sort of limiting process going on here. And this suggests our notion of the instantaneous rate of change. The instantaneous rate of change at x equals a is the limit of the average rate of change as the length of an interval including a goes to 0. Now, that may seem a little bit nebulous. What do we mean about an interval including a? And so here's an idea that's worth remembering. An easy way to get an interval including a is to begin or end the interval at a. So let's talk a little bit of grammar. The average rate of change is always over some interval. And we might express this in a couple of different ways. We might say from some place x equals a to some other place x equals b, or we might say something like during the interval a less than or equal to x less than or equal to b. In contrast, the instantaneous rate of change is always at a single point at x equals a when x is equal to a. And as with average rate of change, problems that seek to find the instantaneous rate of change might not use any of the words instantaneous, rate, or change. So it's important to be able to recognize when you're looking for an instantaneous rate of change without seeing any of those keywords. Let's see if we can find an instantaneous rate of change. Well, we've already collected some information about the average rate of change. And the one extra important thing to observe is that all of our intervals include zero. So if we're going to use this information to find an instantaneous rate of change, it'll tell us something about the instantaneous rate of change at t equal to zero. So let's go ahead and collect our data. The interval length is either 1 second, 0.1 second, 0.01, or 0.001 seconds. And the corresponding rates of change are 5, 9.5, 9.95, 9.995. And so we make our observation. As the interval length goes to zero, the rate of change appears to approach 10 meters per second. Finally, let's interpret our answer and figure out what it's telling us. So to interpret a rate of change, what we should do is to identify the quantity that is changing and when the rate is applicable. So in this case, the height is the thing that's changing. And since the rate of change is positive, we can say that the height is increasing. Now, when is this rate of change applicable? Well, the rate of change was computed for intervals that included t equals zero, when t is the number of seconds after the rock was thrown. So t equals zero is when the rock was thrown. And so we might interpret our finding as follows. When the rock was thrown, its height was increasing by 10 meters per second.