 This lesson is on how do we measure speed? It is your introduction to differential calculus. In order to do this introduction, we have to look at what is called average velocities and instantaneous velocities. Now the way we're going to look at our average and instantaneous velocities is through a ball bounce lab. And what we're going to do is we're going to determine the velocity of a ball over an interval of time. And then we're going to determine the velocity of the ball at a specific time. So let's go to our ball bounce lab. We are going to do a ball bounce experiment. And this is my assistant, Lynn, who is helping me do this today. I'm holding a ball. She's holding the motion detector, which is hooked up to a calculator, a TI-83 calculator. And that's hooked up to the CBL, which will activate a program on the calculator. And the CBL stands for Calculator-Based Laboratory. So we are going to drop this ball. And what we will eventually see is the graph of the data of the ball bounce. So ready, set, go. What we are going to do now is take some data from the ball bounce lab and use that for our average velocity in our instantaneous velocity. This is what our ball bounce lab looks like. I'm going to zoom in on one of these parabolas. And we're going to work with just one of the parabolas to get average and instantaneous velocity. So let me do that now. Okay, we have zoomed in on one of our parabolas. And it's just an upside-down parabola. And we will have data points. Trace along here, you will see all the different data points. And, of course, we started out at 0.19993 seconds. And we were at a height of 0.00879. And as our time increases, of course, our distance from the ground increases until we reach a max at about 0.6 seconds and a height of around 0.66. This is in meters. And then, of course, the ball comes down. So what we're going to do is make a chart of all these points. And here is our chart of all the different points. And it starts out at the 0.19993, 0.00879. And that parabola ends at 0.95997 for time and distance in meters is 0.0033, where it's almost hitting the ground again. Now, what we are going to find is average velocity over some different points here. So what is average velocity? Well, average velocity is the change in distance over the change in time. That is equal to delta S over delta T. I think you've seen this notation for delta Y over delta X. And, of course, that is equal to the distance S of B, the ending point, minus the distance S of A, the beginning point, over the interval of time, which is B minus A. So let's compute some of these average velocities. So the first one I want to compute is when the first time T equals A, and that is equal to 0.199993, and B is equal to 0.279989. S of B minus S of A over B minus A will be 0.20416 minus 0.00879 over 0.279989 minus 0.199993. And if I calculate that out, it will be approximately equal to 2.442 meters per second. Let's go back to our calculator and see where we're looking at that average value. The first point I picked was right here, 0.199993, and the next one was at 0.27998. So I'm trying to find the speed or velocity between those two points, which is, you almost can see a line being drawn through those two points and finding the slope of that line, and that is what our average velocity is there. I'm going to do another two points, and I'll show you on the calculator first, where they are, and then we'll compute them. This one is from 0.43999, 2.71999. Now you can see I'm on the other side of the maximum part of that curve. So if you drew a line between these two points, it is certainly a different slope than the original line that I had drawn before. And so that average velocity is actually going to be quite different from the original one, and it's going to be less steep, as you can see. Okay, so let's compute that. So this time we are going to do our point at T equals 0.43999 and T equals 0.719988, and then our average velocity will be equal to 0.57338 minus 0.55319 divided by 0.719988 minus 0.43999. And that's going to come out to be approximately 0.86 meters per second. And you can see that's a lot slower than the original one we computed, which was 2.44. The last average velocity we're going to compute is between this point at 0.87998 and then down here at the bottom at 0.959997. Alright, again, it's different from the other two, and of course this one should be negative because it is a negative slope on the line connecting these two. And what I'm trying to have you think of is not only the average rate of change or the average velocity, but slope. So let's compute this. This would go from T is equal to 0.879985 to T is equal to 0.959997. The average velocity equals 0.0033 minus 0.253546 over 0.959997 minus 0.879985. And that's approximately equal to negative 3.128 meters per second. Again, it's negative, and of course it's a pretty high negative compared to the other two numbers, negative three, around negative three meters per second. Well, what is this actually telling us? Well, yeah, they're telling us about average velocities, but only in the sense that at different places on our curve whatever two time values we'll get or pick, we will get a different average velocity. Is it meaningful? Well, it's meaningful in the sense that we know if we're on the left side of our curve, our average velocity should be positive. If we are at the top of the curve, it should be a zero, and if we are on the right hand side of our curve, it should be negative. Something more important than this is instantaneous velocity or instantaneous rate of change, and that's what we will look at next. So what do we want to do? Well, we want to determine the velocity at a point in time. And can we do this? If you know anything about calculus, you know the answer is yes. And if so, how do we do it? Well, we look at a limiting idea, and we'll talk about limits a little bit later. But what we do is take that average change in distance over change in time and limit the change in time to being almost zero. So now, if you think of it being just a very quick change in time, what you get is instantaneous. You can determine your instantaneous rate of change. So what we do is change that time and make it almost the distance between the two times almost zero, and then we look at as a limit. And so that is equal to the limit as B approaches A of your S of B minus S of A over B minus A. Well, how does that work with our problem? Well, let's look at one point. What is the instantaneous velocity at T is equal to 0.4? Well, if we look at our calculator one more time and trace back to where zero is around 0.4, we see we have it at 0.399994. And the next one closest to that to the right is 0.43999. And if we went to the left, it's 0.359985. So from the data, we can only get as close as these two times tell us the difference between our original time. We're going to use the 0.359985 and that 0.399994 as well as the 0.43999 to compute our instantaneous rate of change. So we're going to start out with T is equal to 0.399994 and we're going to go to T is equal to 0.43999. Now if we do our delta S over delta T for this, we are going to get approximately equal to 1.674 meters per second. If we use the other side where T is equal to 0.359985 and T is equal to 0.399994, we get our delta S over delta T is approximately equal to 1.976 meters per second. Well, these numbers are fairly close to each other, but they're not that accurate. So maybe we could use one or the other or we could even take an average of the two and find out that delta S over delta T is equal to 1.674 plus 1.976 over 2 and get an average of around 1.7 or 1.8 for our instantaneous rate of change. That's all well and good, but again our times are pretty far away. This is 0.39 or around 0.4. This is 0.43. There's got to be a way that we can get even closer in because instantaneous rate of change and instantaneous velocity means our delta T has to be really, really, really small. And of course there is a way around this even with this problem. So the next question becomes are there other ways to find more and more exact answer? And of course the answer is yes. Can we get closer in? Yes, we can. Well, how do we go about doing that sort of thing? Well, we have to use on these data points our regression equation. So let's go to our calculator once again to help us with our regression equation. If we want to regression equation on this we need to go to stat calc and since this is a quadratic we'll use number five and my data for this part of the quadratic is in L1 and L3. Let's put that in. Second L1, second L3, and I'm going to paste my regression equation in Y sub 1. So I'll go vars, Y vars, function, function and we'll compute and paste at the same time. And that's my regression equation and you'll see A is equal to negative 4.624, etc. B is equal to 5.425, etc. of 47 and C is equal to negative 0.936, etc. And if I look in my Y1 I will see the equation sitting in there and if I graph, that's the original and this is the regression equation. Now you notice this one doesn't start and stop where we want it to start and stop and the way we do that if you remember from pre-calculus you can divide by X is greater than or equal to the first X coordinate for this set of data which is 0.199993 and the last one X is less than or equal to the last data point for this set which is point for time of course 95997 and we need to enclose all of this in a parenthesis so let's insert parenthesis around everything and if we graph when we see this regression graph we see that it pretty much hits very nicely over the top of the original data points so we really want to once again find out the instantaneous velocity when t is equal to point 4 where are we on our regression equation? Well let's look at this we can look at values in and around 0.4 for time solely from our calculator so if we reset our table to the ask and go to table we want to look for values in and around 0.4 and some of the ones we can look at and we want to get closer and closer to point 4 if I put in point 4 right now you'll see what happens I get a value for the Y's which will eventually when I do my subtraction become 0 so what values could I do? I could do a 0.39 and I can do a 0.399 getting closer to the 0.4 and I could do a 0.3999 so that brings me to the left hand side of my 0.4 and to the right hand side I could do a 0.41 and a 0.401 and a 0.4001 and that gets me closer and closer into point 4 on the right hand side so if I were to make up a table with these values in them here's what it would look like I would have a T and I would have my S of T and I would have a 0.39 and then 0.399 0.3999 and then the 0.4 and then I'll start on the other side 0.4001 the closest one 0.401 and 0.41 0.47585 0.49183 0.49339 0.49356 0.49373 and 0.49528 and the last one 0.51034 so those are I took off the calculator now let's find out what our instantaneous velocity would be around 0.04 so if I take the first set of data points I would have for my first velocity V sub 1 would be equal to the change in y's which is 0.49356 minus 0.47585 over 0.4 minus 0.39 remember this is the value at 0.4 this is the value at 0.39 and when I come up with an answer I come up with approximately 1.771 the second time we're going to use that same value at 0.4 0.49356 and we're going to use the second one at 0.399 which is 0.49183 and again 0.04 for this one minus 0.399 and that's approximately equal to 1.730 if we do it a third time and this time the substitution will be again for s of 0.4 and then we're going to do s of 0.3999 and then our difference will be 0.4 minus 0.3999 we're going to get approximately equal to 1.725 with all that computation and if you notice we're getting closer and closer in the fourth one would be s of 0.4 minus s of 0.4 and that would give us a zero in the numerator and in the denominator thus there would be no value for this one so the next one to do is v sub 5 I'll call this one that's equal to s of 0.4001 minus s of 0.4 over 0.4001 minus 0.4 and that's approximately equal to 1.724 and if I go further out from that v of 6 we have s of 0.401 minus s of 0.4 over 0.401 minus 0.4 we get approximately 1.720 and the last one s of 0.41 minus s of 0.4 over 0.41 minus 0.4 and we get approximately 1.679 and you can see in this one we're going from 1.679 down to the closer value of 1.724 and then from the 1 through 3 we went from 1.771 to 1.725 so the answer for this one will be that our instantaneous velocity is equal to 1.724 or 5 much better answer than we were getting before this concludes your lesson on how do we measure speed