 This lesson is a continuation of our first lesson on how do we measure speed? However, in this lesson, we will take our concept of velocity and speed into the concept of slopes. Slopes of tangent lines, slopes of secant lines. But before we do this, we have to find out why we are here. What is this calculus all about? Well, we are beginning to develop one piece of the calculus when we talk about slopes or rates of change. And that piece is called derivatives. And derivatives deal with differentiation. You're going to hear that term a lot, differentiation. And what happens in calculus, you take your static mathematics, what you have learned in previous classes, and you make it rates of change or active. So what we did is a static way becomes active now in our instantaneous rates of change. In our instantaneous rates of change really are derivatives. We look at those not only through velocity, but slopes of tangent lines. And that's what we'll be talking about right now. The other piece of calculus is called integrals or integration. And that deals with antiderivatives, which means having a derivative and going backwards. And also it deals with area under curve. That allows us to do all kinds of problems that deal with area. Let's go back to our ball bounce graph and talk about slopes of secant lines. That's the first thing we want to look at. Now a slope of a secant line means we get a slope from one point to another point on a graph like that. And so this would be a secant line. And you'll see that this slope is very positive and just goes from one point to another one. If I do one that's like this, well this slope is also positive, a little more positive or more steep than the original one I drew. And then I can even do one starting at this point and going there, which is even steeper. I can have multiple slopes of secant lines. I could also do one from that point back here to a point on that part of the graph. Again, a very steep slope for my secant line. And remember, we don't only want the line itself, we are thinking about slopes. So as we keep drawing those secant lines, we can see that as we get closer and closer in to the point in question, that secant line can indeed become a tangent line. So our instantaneous rate of change here brings the secant line to a tangent line. So how do we determine these? The same way we determined our instantaneous velocities and our average velocities. For secant lines, we are doing averages and for tangent lines, we are doing instantaneous. So if we were to compute the slope between 0.5 and 0.6, we would say the slope is equal to function at 0.6 minus the function at 0.5 over 0.6 minus 0.5. If we went to our calculator and computed this, and this is our original graph, and if I did a table on that, I can put in 0.5 and 0.6 and get those two values and compute from those two values what the average rate of change is, slope of the secant line is. So that's equal to 0.65357 minus 0.61981 all over the 0.6 minus the 0.5. And I get something that's approximately equal to 0.3376. So if I tried to find the slope of the secant line for my other values, it would be 0.59483 minus 0.65357 and this time it's 0.7 minus 0.6 and I get approximately negative 0.5874. Now again, these are slopes of secant lines. If I indeed wanted to find the slope of the tangent line using these, I would try to average these two together if this is all the information I had. But as you well know, we can get closer and closer into the value if we are indeed looking for slopes of tangent lines. So let's go to the next one and determine what the slope of the tangent line is at 0.6. Well in this one, instead of going 0.6 to 0.7 or 0.6 to 0.5, what we are going to do is make the table x in our f of x and we'll start at 0.59, get closer and closer into 0.6, 0.599, 0.5999. And again we do this because we are looking for the trend. And then on the other side of 0.6 we have 0.6001, 0.601, and 0.61. Let me put in my f of x's which is 0.65436 and the last one is 0.65186. And you notice they are very, very close in y values to what would be at 0.6. Now remember, each time I compute the slope I'm going to compute it with 0.6 and the number for x that I'm using and the first case would be 0.59. So when I do that slope, it's going to be negative 0.079. And then the next one would be negative 0.12 and the next one would be negative 0.1. With this one we can't do it. This one is negative 0.1 and you notice these two are exactly the same and the trend from left to right to that original 0.6 brings you from negative 0.079 to negative 0.1 and then here we also have the negative 0.1. So we know we are in the ballpark of what we need to find for the slope of the tangent line. And then this one is negative 0.13 and then this one is negative 0.171. So now we can actually state that the slope of the tangent line at x is equal to 0.6 is negative 0.1 and that we have figured out by coming closer and closer in to our values. This is just one way to figure out what that slope of the tangent line is. Eventually we'll learn other ways. Well how do we use this? That becomes the next question. So here's a problem. Given the function s of t is equal to negative 16t squared plus 55t plus 30, graph the function on your calculator and then determine the following. We're going to graph the function on our calculator first. So let's go to our calculator and this is what the graph looks like. Now we want to find the average velocity first of all between t equals 0 and t equals 1. So we have to start again computing with our table at 0 and at 1 and get those values. We are going to find the average velocity between t equals 0 and t equals 1. So that is equal to we'll just call it average velocity is equal to 69 and 30. So it's 69 minus 30 over the differential which is 1. So that is equal to 39. The instantaneous velocity at t is equal to 1.71875. There's a nice way to do this on your calculator. Go to second draw and number five is tangent and we just have to place it where we want it. And I want it at 1.71875. Okay it will draw the tangent line and then give you even the equation for it. And this time I get a 0 in front of x because 0 is slope. So the slope of this is a 0. The next question we have is the instantaneous velocity at t equals 3. Well I can go back to my calculator again and go to second draw again number five and this time I want it at 3 and it will actually draw it at 3 and show me what the slope is. This slope is negative 41. So negative 41 is written in and then the instantaneous velocity at t is equal to 1.5. Let's look at that on the calculator and of course that's a positive 1 and it's positive 7. Let's go back and put that answer on. And then we want the average velocity between t equals 2 and t equals 3. So that again the difference is 1 and I just have to go to my table and get 2 and 3 in there and 76 and 51. So let's compute that 51 minus 76 over 1 which is negative 25. The next question would be okay we've computed these but let's put them in increasing order. So we have the first one which is negative 41 and that is c. The next one is negative 25 and that is e. The third one is 0 and that's b. The fourth one is 7 and that's d and of course the last one is a which is 39. Well that's all well and good. Yes we can sit there and compute these and then put them in order but what if we don't really know the numbers to compute and we are just given a graph how would we do that? Well let's just look at this. Our original graph looked like this. Our original a was between 0 and 1 second so that's sitting about here. The next one at b was sitting right up there so that was b. Our c was at three seconds and that was down in here. Our d was at 1.5 seconds and that was in here so that's d and e connected two points which was in here. So now we can just compare slopes and we can see that a and d are positive slopes and a is much higher than d so a and d being the only two we have to have that order of a being the highest and d being just less than that. b is 0, e and c are negative so we'll put the b in that spot and now we just have to compare c and e and we see the slope on e is less negative than the slope on c so we put those two in order and we get the same order as we did before by computing only this time we are able to do it just by looking at slopes. Another topic we need to consider in this lesson is limits and we're going to be looking at limits in very different ways. Last time we looked at them graphically this time we're going to do numerical calculations and eventually we will do the algebraic way of finding limits. So here's the problem given the function determine the limit at x is equal to zero and our function is f of x is equal to three plus x quantity squared minus nine over x. Okay different representations for this we could say the limit as x approaches zero that's what that arrow means of three plus x quantity squared minus nine over x or we can even change the variable because it doesn't matter what variable we have we can say the limit as h approaches zero of three plus h quantity squared minus nine over h. Just like when we were computing slopes of tangent lines in order to compute a limit numerically we have to get closer and closer into the limiting value in this case of zero. So we will use the numbers in and around zero. Well here's a graph of our function and you notice it looks like a line and when I try to evaluate it at x is equal to zero I don't get any value for y. So let's do our computations again when we get closer and closer to zero we won't have a value at zero but closer and closer in we can compute these numbers. And if I put a negative one in and you do the same thing for these as you do for slopes of tangent lines we get for our function five point nine and if I put a negative zero point zero one we get five point nine nine if I put a negative zero point zero zero one I get five point nine nine nine. You can almost see that the limiting value is going to be six and in this one if we put a point zero zero one in there we get six point oh oh one and zero point zero one six point oh one and if we put a point one and we get a six point one so on this one as we get closer and closer in to the zero we know that our limiting value is six and if we go back to our graph we see if we just count up this is one two three four five and that point is just about at six so this is how you would compute limits using numerical calculations this concludes our lesson on how do we measure speed continued