 Hello, and welcome back. In today's screencast, we're going to use data from a real-world periodic phenomena to try to develop a sinusoidal function that can be used to model that periodic phenomena. So for example, in the case that we're going to be looking at, we're going to be looking at the number of daylight hours on the first of each month for Edinburgh, Scotland. The key here is that one thing we're going to be doing is to, what we say, model the data. So in some sense, our mathematical equation will not be a perfect fit for the data, but we'll should very closely approximate the data. And in fact, by how we look at the data, different people may look at it slightly differently and end up with slightly different equations. So again, in this situation, there is no one correct answer. But if we do follow the principles that we have learned about sinusoidal functions, we should always have sinusoidal equations that are very close to this. So what we're going to do to actually get data that we can plot is to change the months into numbers, and we'll do the usual thing, one for January, two for February, and so forth until we finish up, of course, with 11 and 12 for November and December. And again, there are different ways to do this, one of which is to use technology in particular to develop a, to get a sine regression equation, and we can do that, for example, on a TI-84 calculator. But today, we're going to learn how to model this and develop a sinusoidal equation by hand. And what we'll do is use very much the same principles that we used when we developed a sinusoidal equation when we were given the actual graph of the sinusoid. And in particular, what we're going to do is focus on a low point and a high point. And for this one, we'll use these two. So let's say that this low point has coordinates t1 and y1, and this high point has coordinates t2 and y2. And remember, that then allows us to determine many of the characteristics of the sine curve that we can determine the parameters a, b, c, and d. And in particular, what we're interested in, of course, is the amplitude, which is basically this distance here. That's probably more important to remember than any particular formula. And again, you can see it's kind of half of the difference between the high point and low point. So we can say two times the amplitude is actually y2 minus y1. And so we can use that then to determine the amplitude or the value for a in our equation. The other thing we can do then, of course, is to use that information to get the vertical shift for our sinusoid. And that's basically going to be the high point y2 minus the amplitude. And again, there are different ways of doing that. In particular, another way is sometimes to use y1 plus y2 divided by 2. And that will give you the same value, and you can get the amplitude from that. The other thing that the high point and low point get us is information about the period. And in particular, we know that 1 half the period is basically going to be the horizontal distance from a high point to a low point. So the way I've got this labeled, this will be t2 minus t1. And again, we can use that then to determine the period and from that determine the coefficient b in our equation. And of course, the last thing we have to do is get the phase shift. And if we focus on a sine function, the way I would get a phase shift might be to focus on this point right here, where it crosses the axis of the sinusoid. And we can see, and the way I've got this set up, this is a quarter of a period back to the left of the high point, or a quarter of a period to the right of the low point. So again, we can get that phase shift by using that general principle. In this case, we could write that as t2 minus a quarter of the period. So I've got to fix a little something here. So we'll write that it's a quarter of a period. So let's get to work on this. Use the fact that we can go back a quarter of a period from. So remember January, February, March, April, and so forth are 1, 2, 3, 4, and so forth. And we're going to do something fairly simple to start with. We're basically going to use this as the high point and this as the low point. So our high point will have coordinates 7 for July and 17.48 for the number of hours of daylight. And the low point will have coordinates 1, 7.08. So we can now use that information to determine our values for a, b, c, and d. And in particular, we can see that in this case, we do have a period of 12 months. It is an annual thing. And we can also see that the difference between the coordinate, the first coordinate for the high point and low point is 6, which tells us that the period is 12. So we do get 2 pi over b equals 12. And if we solve that for b, we get 2 pi equals 12b. Or b is going to be pi divided by 6. 2 divided by 12, we reduce to 1, 6. So there we have our value for b. The amplitude, of course, or in this case, actually 2 times the amplitude, is going to be the 17.48, the y-coordinate of the high point minus 7.08. And if we solve that then for the amplitude, we get an amplitude of 5.2. And then we can get the vertical shift by taking that high point, 17.48, and subtracting the amplitude of 5.2, which gives us a vertical shift of 12.28. And then, of course, the last thing we have to do is to get the phase shift. And again, what we're going to do if we go back to do here, and we can see the high point occurs at 7. So if I take the phase shift, we'll get 7 minus a quarter of a period. And remember, the period is 12, so a quarter of the period is 3. And if we take 7 minus 3, we get 4. And now we actually have an equation that we can completely write down. 5.2, the amplitude times the sine of pi over 6, the value for b, times t minus 4, phase shift, plus 12.28. And what would be good to do here is, again, and you can do this on your calculator. If you enter the data, you can get the scatter plot for the data. And then you can superimpose upon that the graph of this equation. And you might want to take some time now to try to do that. If you do, this is basically what you would get. And as you can see, if we look at the data, it looks like the curve is a little bit too far to the right. So what we're going to try to do is move that to the left a little bit. And there's a couple of ways of doing that. One is simply to adjust the phase shift. We're going to do something a little more substantial than that. In particular, what we're going to do is think about the fact that what we know about days of the year, and we had put the high point on July 1st, the maximum number of hours of sunlight. We know that that is not the case. The high point actually occurs around June 21st or so at the summer solstice. And so what we're going to do is put the high point at about month 6.7 instead of 7. And we might make it, if we look at how those two points up in here look together, the high point then might actually be a little bit higher than the 17.48. So we're just going to try 17.50. And if we use that as the high point, our low point is going to be half a period back. So if we look at 6.7 minus 6, that comes out to be 0.7. So the low point will have a coordinate of 0.7. And that, again, corresponds to the fact that the minimum number of hours of sunlight does not occur on January 1st. And again, we're going to make that just slightly lower than January 1st, and make it 7.06. And that's the type of information we're going to use now to try to modify this graph or the equation just a little. So here's a summary of kind of what we have done. The maximum number of hours of sunlight occurs about 10 days earlier. And the minimum, the sunlight does not occur on January 1st. So what we had just talked about, we're going to try a maximum of 17.5 hours and a minimum of 7.06 and have the maximum occur at t equal to 6.7, which is relatively close to June 21st. So here's, again, the data we're going to work with. And what we're going to try to do now is use that data in much the same way that we had before. Notice that the period is still 12. So that will give us the same value for b that we had before. So that'll be b equals pi over 6. And the amplitude now is going to be, we're going to look at the maximum of 17.5. If we look at 17.5 minus 7.06, what we end up with is 10.44. And so what we're going to have is an amplitude will be half of that, or 5.22. So you can see we've changed the amplitude just a little bit. And now we're going to go through our usual procedure here. The vertical shift will be the value of the maximum, 17.5 minus the amplitude of 5.22. And that, again, comes out to be 12.28. Okay, and then, of course, the last thing we need is the phase shift. And again, we're going to use the same principle we had before. We're going to go a quarter of a period back from where the maximum occurs. The maximum occurs at 6.7. And a quarter of a period is 1 fourth times 12. And so 1 fourth times 12 is 3, and we get a phase shift of 3.7. So we now have an equation of 5.22 sine, again, pi over 6, times t minus 3.7 plus 12.28. And again, if we've already got the data entered into our calculator, we can now take out the other equation and sketch the graph of this equation. It should get something that looks like that. And you can see that is quite a good fit of the data. Not a perfect fit, but a very good fit. And we can use this equation then to model the number of hours of daylight. In Edinburgh, Scotland. In the next screencast, we're going to look at how to compare this to what we would get using a sine regression function on something like the TI-84 calculator. That's all. I will see you at that next screencast.