 Hi, everyone. Here we're going to talk about rates of change by using equations. So let's take a look at an example. Suppose we have a ball that is thrown vertically upward from an initial height of 30 feet and an initial velocity of 55 feet per second. The equation of motion is s equals negative 16t squared plus 55t plus 30, where t is the elapsed time since the ball was thrown. And s is the distance from the ground. So we're going to talk about both average and instantaneous rate of change. And we'll start with average velocity. So we want to calculate the average velocity between t equals 0 and t equals 1 second. So what that means is that if we were to take a look at the motion of this ball in between t equals 0 and t equals 1, what is its average rate of change over that course of time? So you might remember that average velocity is really a slope calculation. We can use the symbol v bar, and average velocity is always change in position over change in time. So in this case, we need to calculate the position at 1 second minus the position at 0 seconds and divide that by 1 minus 0. So I'll allow you to substitute the values in. You should be getting 69 and 30. So for this first answer, we get 39 feet per second. An easy way to figure out the units of measure, go back to that slope calculation. In the numerator, its position, which we measure here in feet over time, which in this case, we measure in seconds. That's why it's feet per second. Now let's talk about the fact that this came out as a positive value. Since we have a positive 39, that tells us that the motion of the ball is still upward in between 0 and 1 seconds. Let's take a look at the graph really quick. And that's what the graph looks like. So here's t equals 0. Here's t equals 1. And you can see that it is going upward over that period of time. That's why the average velocity came out positive. So let's take a look at another one. Let's consider in between t equals 2 and 3 seconds. So once again, in order to do our calculations, we need to do the position at 3 minus the position at 2 over 3 minus 2. Once again, I'll let you substitute in. Hopefully you're getting 51 when you substitute in 3 and 76 when you substitute in 2. So that gives us a negative 25. Once again, the units of measure are feet per second. Now the fact it came out negative though, that tells us that in between 2 and 3 seconds, the ball is now going downward. So let's take a look again at the graph. So here's 2 seconds. Here's 3 seconds. And notice on that period of time, the ball is on its downswing. That's why the average velocity came out negative. So let's now consider the instantaneous velocity. And the first one we're going to look at is the instantaneous velocity at 3 seconds. So while we're here on the graph, here's 3 seconds about, right about there. So what we want to try to estimate is pretty much how fast is the ball going at that particular point in time? It would be helpful if we could do a slow calculation. So let's maybe try to work with that angle. So now that we're finding instantaneous velocity, we can call it just plain v. Plain v typically indicates an instantaneous velocity v bar is the average velocity. Now notice how it says estimate, and compare that to the directions you saw with average velocity. All we can really do at this point in time is estimate the instantaneous velocity. As a result, we need to make sure we denote that and how we write that up. So we do need to use an approximately equal to symbol. You might remember there's two approximately equal to symbols in mathematics. There's the equals sign with the dot over it. That's the one I tend to use most often. But of course, you probably know the squealies better, which is fine. You can use either one. So let's try to do this as a slow calculation. So you might remember back when we did limits. And we talked about a delta neighborhood around the x value that we were approaching. And we talked about that delta neighborhood as being a really teeny, tiny area around that value we're approaching. We're going to use that same concept here. So let's think about maybe choosing t values pretty close to 3 seconds. And we're going to pick one that's a little bit above 3 seconds and one that's a little bit below 3 seconds. And we do want to go equidistant on either side of 3. So let's maybe go a tenth above and a tenth below. So let's maybe do the position at 3.1 seconds minus the position at 2.9 over 3.1 minus 2.9. So we do need to substitute the 3.1 and the 2.9 into that position function. So this is where we can let the calculator do all the work for us. So let's go make sure you have under y equals your equation. Please take a second to do that. If you don't have it there, it could be because you will want to practice this that I'm going to show you. And once you have it there, go to your quit screen. Basically, you're going to use function notation until the calculator to do the position function at 3.1 minus the position function at 2.9. So in order to do that, hit your vars button. Go across the top to y vars and into function and pull up y1 and in parentheses next to that, you're going to put 3.1. It looks just like f of x notation. Now we're going to do minus and the same thing for 2.9. So vars across the top to y vars and then in parentheses 2.9. So this is the numerator for our slope calculation. And we get negative 8.2. Now we need to divide that by 0.2 which is the difference between 3.1 and 2.9. So in this case, we got negative 41. Now just because we're doing an instantaneous velocity now, does not mean that the units of measure change. They remain feet per second. So let's think about what this means. This is telling us that at three seconds, the ball is going downward, that's what the negative tells us, at a rate of 41 feet per second. Those of you who have had physics might remember the absolute value of velocity is what gives you the speed of the object. Your speed is basically your velocity regardless of direction. The negative really just tells us the direction that the ball is traveling at that point in time. So let's do another one. Let's try it one and a half seconds. And we're going to use the same idea of choosing t values surrounding 1.5 seconds. Don't forget your approximately equal to. So let's stick to using a tenth above and a tenth below that seemed to work well before. Now if you were to be doing this on your own, you definitely can use a smaller delta value. Here I'm using a delta value of a tenth. You very much could use 0.05, 0.01. The more granularity you have in making that delta value smaller, you typically get a more accurate answer. But it's really up to you what you want to use. Don't go any larger than a tenth. Usually your results come out better this way. So once again, we're going to let the calculator do the number crunching for us. So we're going to do that vars and across the top to Y vars. And we're going to have the calculator evaluate the function at 1.6 minus at 1.4. So once again, this is the numerator from our slope calculation, and we need to divide that by 0.2, which is the 1.6 minus 1.4, and we get seven. Now notice this time the answer's positive, so think about what that means. That's telling us that at one and a half seconds the object, this ball, is still traveling upward because we got a positive velocity. And if you take a look at the graph, so here would be one and a half seconds right here. Notice it's almost at the very, very top, but it is still a positive slope there. It is still traveling upward. Over here at three seconds, notice how that's sort of on the downside of the ball coming down on this side of the parabola. So let's talk about some other things, maybe. Let's see if we can figure out when it is the ball hits the ground. If you take a look at the graph, it looks like it's hitting ground pretty close to four, but not quite four. So let's see if we can figure out a way to find that algebraically. Well, if the ball's hitting the ground, that means its height is zero. So an easy way to do this would be to set the quadratic equation, the position function, equal to zero and solve it. So since we have a quadratic equal to zero, of course I'm sure you remember the quadratic formula, but no, we are not going to make you do that by hand for heaven's sakes. Hopefully by now you have the calculator program that will do it for you. So let's go back to the calculator. If you hit your program button, now I probably have a lot more programs than you do, but you want the one that you probably have called quadrat. Nope, there it is. So when you run the program, it's going to ask you for the coefficients, assuming of course your quadratic is of the form ax squared plus bx plus c. So in this case, a, which is the first question mark, is a negative 16, b is 55, you just enter them in order, c is 30, hit enter, and of course we can't have a negative time. So the positive one, 3.916, is how long it takes for the ball to hit the ground. And remember we had noticed from the graph that it was pretty close to four, which obviously you just found it is. Remember that we always round our decimals to thousands in AP calculus. So we would have 3.916 seconds when it hits the ground. So that's one more question to think about. Can we determine the maximum height of the ball and when it is at that maximum height? Well, this is something you might have done in algebra two or pre-calculus. Let's go back to the graph. So obviously we can see when it hits the maximum, so we can just go ahead and find that. So let's trace to the top right about there. Let's run our maximum program. Go a little to the left, little to the right. So there's our maximum. So think about what these values are telling us. The X value is the time at which it hits the maximum height. So that would be at 1.719 seconds. And the Y value is the height itself, the maximum height. So in this case it's 77.266 feet.