 Okay let's go ahead and get started. Let's see, it's what's on the agenda for today for this week. Hopefully you guys saw the ELC announcement I sent out which, sorry, there's a little bit of weirdness in the ordering of the homeworks. But this week I want to say it's 4.2 A and B. So I think we did 4.3 last Thursday. But next up will be those two 4.2 A and B. Along with that I've also opened up on Gradescope all of the, essentially all of the assignments for the rest of the semester. Just in case people find themselves with more or less time. So I'll still set the original sort of due dates, you know, when they're due. But you can sort of work ahead as far as you want. If you want to clear out some time, like we didn't, I don't know, the school kind of made fall break not really a thing. And I guess we have some kind of Thanksgiving break coming up. But they've definitely taken some of the breaks out of the semester. And I think they'll be doing that next semester too. Like I think they're removing spring break. So to try and compensate for that, I'm just trying to make it so if you do want to take some kind of break, then you can sort of work around that by just working a few of the work sheets early and just turning them in. What else? Project grading, those should be back pretty soon. Yes, I'm just now getting caught up on grading. Just as an overall effort, I think everyone's doing really well on these. And definitely a lot of improvements since the first project. The only thing I would mention for so we have another project coming up very, very shortly. So I'll either start talking about it today or on Thursday. The only thing I would throw out is that for some people you might want to get started a little bit earlier, just because for the last one we had that kind of rough draft date, or that was originally sort of like the final turn in date. We kind of pushed it back, made that a rough draft date. And I think if people had turned in, like if they're rough drafts, they'd been their final drafts or something like that. I think people would have, I would have had to take off a lot more points on things just because the rubric is sort of set by the department. But yeah, so just that being said, with the next project, try to start earlier if you can. So where you should be at this point is you should have at least looked at the project handout. And then just sort of get an idea of what's going to be on it, sort of what the physical situation is. I would start by going through that sort of introduction section at the beginning, which has sort of a nice summary, and then skip all the way to the end and just go to the example and sort of see what they're working out. And then just kind of go back and forth between the example and the middle part where they're kind of going through the actual situation. So the example is supposed to like parallel situation you have, but it won't be exactly the same. This one will be a little bit trickier because there's not an explanatory video for it. So I just want to make a PSA that, you know, for everything you'll need on this project, we'll either be talking about in class, or it's already on the handout. This one is a little bit more open ended. Something with the last project that wasn't super great is like I did the derivations and the mathematics in class, and very few people included that work, but it was sort of a significant piece, you know, was actually working out those exponential equations using logs and stuff to solve the various things. And a lot of people just said, here's the original equation, here's the solved equation, which is just the solved equation after I worked it out. So try to avoid that. Like you have to like actually include the mathematical write up, you know, how you arrive at these things. And so for this one, we don't have to worry about that so much because there isn't really one solution for these. This is kind of like a modeling exercise, sort of. So the idea is we're going into trigonometric functions now, and, you know, we did sort of linear modeling near the beginning of the semester where you can sort of model things that just increase, you know, at sort of a constant rate over time. But most things in the real world do not do this. You know, at very best, like nothing can increase forever. Like eventually you level off. Some linear models aren't super great for like extrapolating out to various times. But more than that, like most of the things we deal with, you know, if you're in business or even just in biology and the way seasons work, you have a lot of phenomena that are periodic in nature. So, you know, something that's, you know, even just like the temperature throughout the year, right, it's, you know, it peaks in the summer and it kind of dips in the fall and winter and then peaks again in the next summer. So, you know, just like a lot of, a lot of things exhibit this behavior. And so the whole point of this unit is how do we, like what functions do we have to model this behavior? And how do we, you know, how do we build them? It's just kind of a big thing. I think this is kind of useful, like people are thinking about industry or anything like that. This kind of stuff comes up if you're doing any, it's a word that might not be meaningful, but it's a time series analysis is a big thing. If you do any kind of like prediction, modeling, data science, finance kind of work. It's kind of stuff comes up all the time, you know, businesses sales through a year have something called seasonality to them, which is exactly the same thing as the seasonality of temperature. And so, you know, the name of the game there is finding a really nice bunch of cosine waves to, you know, and fitting the parameters kind of like we did with the linear models to find something that sort of nicely models, you know, it's a nice curve to your, your sinusoidal sort of data. But right if you're doing biology, if you're doing, if you're doing physics, this is everything like essentially every, every, every major concept in physics is modeled off of a wave. There's this thing called the wave equation, sort of governance how businesses think most sort of particles work in most forces. And yeah, if you're doing things like biology, of course, you have, you know, things like populations can be sinusoidal. It's, you know, sort of predator prey relationships where, you know, if you have a predator population going up, then the prey population is going down. But then there's not enough prey. So the predator population goes down, the prey goes back up. So there's kind of two different sinusoid, it's governing that relationship. It just, yeah, these are just sort of fundamental functions that show up everywhere. So it's really nice to know how to work with them. And, you know, model some kind of periodicity that way. So that's kind of the whole point of this, this next project is, you know, there'll be some sort of periodic phenomenon that's happening. In this case, it'll be amount of light through the day. So you can imagine, you know, that during the day, if you have like a little slab of concrete or something out in front of your house, you know, at various times, you know, the sun might be directly above it, the sun might be coming in at an angle, or it might be nighttime. And so you can imagine that, you know, the sun is providing some energy to this tiny little square. And that energy is varies in a sinusoidal way. Maybe it's, you know, maxing out when the sun is directly above and all the light rays are sort of hitting the square in the maximum way possible. And it's minimized when the sun is either parallel to the your slab. So it's like off the horizon or something. Or if it's just nighttime, in which case you're getting zero, like all the light rays are missing your little square. So that's the sort of periodic phenomenon that's happening there. And I'll talk about it a little bit in more detail later. But yeah, I think with that, yes, I've make a I just want to make a quick note before we get too far into things. For the project, I've actually already assigned groups. So this one I assigned randomly at ELC, it's on your project three. So I would I would really recommend trying to meet with your partner or partners this week, if you can. I haven't set the final due date for this project yet. But it'll definitely be easier to start earlier. Is that way, if you have questions, either, you know, they'll come up in class, and we'll address them here, or you'll have time to go to office hours, and ask any of the instructors, like clarifications. So yeah, partners on ELC, you go in there, pull their email, put, you know, at uga.edu on the end of their username or whatever, or message them through the group chat or kind of whichever way you have of communicating with other students. And maybe try to set up a meeting for this week, just to talk about, you know, how you both understand the project, how everybody in your group understands it, how much you've read, maybe if you want to start allocating different parts of the analysis, that kind of thing. Yeah, I would start this week. Um, I just looked on ELC on project three, and it doesn't show the partners. Do you know where specifically? Yeah, I'm not sure. Let me make a note of it. And then I will check after class. Okay, thank you. You'll see. Yeah, it might be like, I said them, but they aren't finalized yet or something. So I'll go and see if there's a button to click for it. Okay. So yeah, I'll say some more specifics about like what mathematics is happening in the project later. And try to say, let's go ahead and jump into some of the stuff we've been looking at on pre classes and worksheets. Once I will share my screen. Okay, so right now we're looking at, I guess, trigonometry, I guess is the broad term here. And where we left off last time, I think, was looking at the relationship between, you know, we have a sector or something like that. And it's determined by saying angle and really a radius to right, like we could have radius shooting off to infinity. If we fix a radius, then we get a specific circle. And we had this relationship, you know, the ratio of angles, varies in proportion to the ratio of like the little arc length in your sector to the whole arc length, like the circumference of the circle. I wanted to point out one quick warning here. And a simple simplest way to say it is just to use radians. I guess degrees right come up in engineering applications a lot. But I don't know if for me, it's always tricky because you never know if your is your calculator in radians or or degrees or you need to convert back and forth or how do you check. So I mean, you can totally check if you know some of these like special angles, if you know, cosine of 30 degrees or something. If you put cosine of 30 in your calculator and it comes out as some Oh, yeah. I mean, so you have to know what that looks like in decimal form. So maybe that's a little difficult. But if you put cosine of 30, it'll be some like easily recognizable number. If it's in degrees. And if you put put that in it's in radians, you'll get something that's totally unrecognizable, 30 radians is I was like nine pi or something like that. Pi is around three. It's nine or 10 pi. So that's, you know, 30 degrees kind of facing off this way versus nine or 10 pi, which is facing off in the other way. So it'll be totally different. So the way to get around this, I mean, it's just used to use radians. I think calculators internally do a conversion sort of matter what. So I would just for once and for all set your calculator to radians. And every time you see something in degrees just convert it to radians, at least for this class. All right, this was just because all of these formulas, so radians have this like geometric interpretation, they have these proportionality relationships. And the proportionality relationships are totally different if you're measuring your angles as 360. And remember, we said this was like totally arbitrary, whether it's two pi or 360 or if you call it just one revolution and ask yourself, you know, about half of a revolution or a quarter or whatever. But so you have to pick some convention to go with and radians is the one we'll, we'll run with. And this was like theta r versus theta d. I think this came up on the quiz too, a few times. Sorry, it's just just a matter of this is a common sort of mistake. So I'll try to box that and put the wording sign on it and everything there. So the thing I'm gonna talk about first are areas of sectors. And this should be just for view from last time. But what was the picture here? It's something like this. And we've chosen sort of a reference frame, a coordinate system. Here's my zero. When to start giving these axes, I didn't name them last time, but I'm going to give them a funny name here. So this will be a Y hat. And this will be an X hat. Hat is the little carrot over the letter. And the reason I'm kind of using this notation is because we actually want to start thinking of these as vectors. So I mentioned just the word vectors last time. And I wouldn't expect that you've seen this unless you've maybe taken like a calculus course, or sorry, physics, like an introductory physics course. So vectors were kind of invented for that reason. And to study forces on free bodies and this kind of thing. But they're extremely, extremely useful in mathematics and they're going to be useful for us to simplify some of this discussion about, you know, just relating points in the plane to various trigonometric functions. So the way you think about this is X is just an X hat is just a direction. Y hat is just a direction. And I've kind of chosen them. I've chosen them in such a way that they're orthogonal. So they form a little plane. But you can imagine I, you know, I could choose this on like a sphere or something like that, I could put a little cross on a sphere and just choose some directions and put a little dot to call that the origin of the sphere. I could go out of my yard and put the origin there and then just choose two directions, maybe east is X bar and north is Y bar, but it's totally arbitrary. It could be, you know, west is Y bar and north is X bar. If I just turn by 90 degrees by halves. Yes, you just think of these as like perpendicular or orthogonal directions. So the situation we have here is that we have some kind of sector and we've chosen to measure the sector from like an angle of zero as our reference. And you zoom into this a little bit. And let's say we're on, we have some radius that we have some point out there. So this may or may not be the unit circle. We may not have r equals one. We have our origin O down here. We have some angle here. We'll just call theta. And we want to know, right, so we've picked a radius and we've picked a theta, which means that we've determined a unique sector. And it's the sector kind of swept out by the stuff in green. And just remember like if we didn't pick a radius, we wouldn't have anything to work with yet because right, you know, if the radius was out here instead, this would be a totally different sector and it would have more area. So we couldn't really talk about the, you know, the area. We can't even really talk about a sector really just associated with an angle. You have to sort of choose the angle and choose the radius. And so what'll happen is that we'll have some kind of area of this sector, right? It's just some figure in the plane it encloses some some area. And it's going to be because we think of this as like a function really. This function, if I wanted to code it up on a computer, what information would I have to tell the computer to compute this area? Well, I'd have to tell it something about what's called theta, just the angle. Computer has to know the angle in order to compute this area. And it has to know the radius. And the question is, what is that? So the way I would think about this is that there's the one area we know, namely the area associated to 2 pi and r. And so this is just the area swept out. If you think of an angle of 2 pi, you're just sweeping out the entire circle. And so this is just pi r squared from the back of our recalculus book for the area of a circle of radius r. We probably don't derive it in this class. Okay, so we know the area of one circle. And the whole idea behind all this strict stuff is that somehow knowing something about one circle is kind of enough because we can just sort of scale things up and down appropriately. As long as we know how the proportions are related, then we can compute whatever we want for our new circle. So or for our new sector in this case. So what happens here is that the proportionality relationship is that here's the orange thing. The orange thing is going 2 pi. And that's kind of the orange sector is corresponding to this thing, the full circle. And we want to know something about this angle in green and this sector swept out by this arc length there. And the way I would remember this is that, okay, we have a theta, we have a 2 pi. So kind of doesn't matter how we're measuring the circle. But we know that whatever theta is, we scale it by 2 pi. This kind of scales out whatever choice we made for, you know, for what the measure of a full angle is. So the proportion of this small angle to the proportion of the entire circle is just going to be the same as the proportion of all the area of the small sector to the area of the entire circle. And this thing is area of some of the areas of function depending on theta, depending on the r. And then we know the thing in the denominator. So we get something it's like theta over 2 pi. It's just equal to this area that we're concerned with over pi r squared. And that's the thing that I would remember for this is we kind of, we didn't really have to memorize many formulas to make this work. We just needed to know like one fundamental geometric principle, which was really two. One of them was just how, what does the circle look like? Like what is this well-studied object? Like what properties do we know about that? In particular, we knew the area of one circle. And now, okay, how do we relate our object to the well-studied object while we're using this other principle about proportionality? Where for measuring things in radians, then we get this nice relationship where the ratio of the angles is the same as the ratio of the areas. So I want to throw in one more wording here. Just a common mistake is not drawing a picture. We'll maybe box that one. That is worth keeping in mind pretty much through this entire unit. Really for nearly every, every problem you'll want to practice writing out some kind of diagram that helps you reason about the problem. This is really good practice because then when you get to a question that you don't necessarily know how to do, you have a place to start. You have a first step that you can take towards a solution which is namely, diagramming the situation, drawing it out, getting a picture or schematic that represents things, making it as accurate as you reasonably can. But also, this is good just to do in general, even for problems where you think like, okay, I know how to apply the formula. I don't really need to do anything here. Still draw a picture because you want to like sanity check your work. Oftentimes, I find myself doing something, you know, I'll just rush through it. I won't draw the picture. Like, yeah, I know it. I'll just plug in the angles or whatever. And then something ends up looking weird and then I draw a picture and it's like, oh, of course, the angle. I thought it was in this quadrant, but it's in that quadrant. Just, you know, silly things like that. So pictures, they're kind of important as a way of bookkeeping, I guess, or a way of like catching your mistakes. So I want to add actually one thing to this picture, which is already a little bit complicated. Let me move this over here. And what I just want to say here is that we picked theta and we picked r. And we can actually, I mean, so we actually get a point out of it. We're getting a lot of things out of that choice. So maybe I'll call this point. This little arrow thing is just indicating that it's a vector, right? So it's just an arrow that points somewhere in the plane. And this vector can actually be a function. It can depend on theta, it can depend on r. And all I mean by that is like I chose one theta and r and I got some, you know, you just think of this line segment connecting the origin to your point. It's like a little arrow. So it has a direction and it has a magnitude, like the length of it. And if I chose a different theta, right, I would get a different arrow. And if I chose a different r, I would get a longer arrow. So this is a function. So I guess p bar or p vector is a function that takes two real numbers. So that's like its domain. Here's a notation for this, r2. r2 is saying like you have two copies of the real numbers. And this is the same thing as the Cartesian plane we're used to working with. And what does it, what does it send out? It sends out a vector. This is a more abstract function. The outputs aren't real numbers anymore. They're mathematical objects of some sort. It's like a geometric object now. And what I wanted to point out here is that we have an ocean of polar coordinates. And so, well, okay, let me just write it and I'll explain what it means. So if we try to look at this in terms of coordinates and we ask ourselves, you know, what is what is the x-coordinate of this point and what is the y-coordinate? Maybe I'll make a cleaner picture a little bit later. Then these would be formulas for those coordinates. So for on the unit circle, the x-coordinate would be cosine of theta and the y coordinate would be sine of theta. And if we have an arbitrary radius, well, we just scale that x and y value by r. So anything I wanted to say is that? Okay, so yeah, the point here, this is kind of maybe an aside, since we already have this picture here, is that choosing a theta and an r actually gives us a point in the plane. So we have a geometry, we have a, you know, a sector, and we also have an actual, honest point, namely the end of this little vector that it determines. Okay, so I want to say a little bit about, let's just talk about special angles. And I hinted at this last time that there's, you know, if you go online, actually I wonder if I could even bring it up. That works, yeah. If you just search Google for, you know, unit circle, you're going to get this image right here, which is an absolute mess, it's so complicated. It's just such a pain, why would we ever do this? Yeah, so there's just like way way too much going on here. We definitely don't want to memorize, I don't know what is this, three, six, nine, 12, plus four, so like 16 different angles here, and then each one has two values associated with it, so it's like 32 different things we have to memorize. No thanks, right? The name of the game here is, you know, if you're a mathematician, be lazy, memorize as little as you possibly can. You want to know much more about the process, because that's the thing that's easier to remember, right? Just sort of recalling some sort of rote fact is never great, but if you know maybe like one fundamental fact or like one very minimal set of facts, and you know the process, you can suddenly like generate all of these other things without ever having to memorize them. So I want to to focus on, so I guess we'll start with this, this picture here, something like this, and just thinking of just one quarter of a circle. It's not the greatest circle in the world, but again, this was the Y hat direction, this is the X hat direction. There's my origin that I've chosen, sort of arbitrarily, and I want to think about a few angles here, so let me just mark them with points first, and then I'll label them. So we want the ones that are kind of right side at zero degrees if you like, or 90 degrees, those will be important. We want one exactly between them, so this is going to be 45 degrees, something like that, and then maybe let's also break, that's not going to be exactly half, but let's pick two that are sort of in between, so there's that, and that. So using this notation from earlier, I can actually label these in polar coordinates really easily without having to really worry much about anything. So I need to know that this is, I guess this is pi over six, and I guess to make this bigger, also this one I guess we're thinking of as pi over four, and one up here, this one is pi over three, and then of course we have this thing is just the right angle, and we can label these points in the following way. So if we, if we just imagine, once I haven't told you your radius yet, I've just told you the angles, we can always pick one nice radius if you want, you can just pick a radius of one and ask where do these raise, like at say a pi six, where does this intersect the unit circle, and I get some point, say here this is going to be p bar, and it's going to be, right it's a function, it takes in an angle and a radius, so the angle is pi six, the radius is one, I can also think of this p bar down here, which is zero, angle of zero and a radius of one, this one here, set an angle of pi over four, radius of one, let's hopefully get the idea there, and lastly we have this one here, set pi halves, okay, and using our kind of polar coordinate convention, before we saw that this, if we, so we have just some geometric point in the plane, we want to figure out what its x and y values are, so we can write this as, what we're using the formula is like p bar theta and r is equal to this point, r times cosine of theta in the first coordinate, and r times sine of theta in the second coordinate, I want to box this and mention that it, this is like a really, really key thing to remember for this section is somehow like, if you have an angle, cosine is kind of picking out the horizontal component, and sine is picking out the vertical component, so like you can think of this as like this first coordinate here, I'll erase this in a second, but you can think of what this is, is if you project this down to the x-axis, you get some kind of x value there, so there's maybe some x naught, and you can project it down to the y-axis, you get some y naught, r cosine theta here, this is telling you the length of this segment here in red, and r sine of theta is telling you the length of this segment here in orange, and so you have a segment, you actually have points, you can think of these as actual vectors, if you like, and what happens is like there's a, there's an addition, addition rule for vectors where to add two vectors, you add them as arrows, so if you have two different arrows pointing in two different directions and you want to add them up, you go along the first arrow, and then you pick up your second arrow and put it at the end of the tip of that, and then follow the second one, so that's a little bit, if you haven't seen it before, it's a little bit tricky to understand how that goes, but you can just imagine this red arrow here, if it were up here instead, this is just telling you that this p-bar function decomposes into two components, there's one that goes strictly along the y-hat direction in orange, plus this other red portion that goes strictly along the x-direction in red, so we always have this freedom to like pick up a vector and move it around, some arrow with a fixed length, but it kind of doesn't matter where we put the base of the arrow back for now, and actually I think we can leave, you can leave that on the drawing, that should be fine, but okay, if I wanted to find out what are the x and y coordinates associated to these angles on the unit circle, I can just use this polar coordinate formula, and really the r is one, so it simplifies a lot, so for example here whatever x, y this is, we know that the x-part is cosine of pi over 4, y is sine of pi over 4, and I'll just draw kind of one more just for emphasis here, if we wanted to figure out what the x, y coordinates at this point were, well they'd just be cosine pi over 6 in the x-direction, and then sine of pi over 6 in the y-direction, and so on and so forth for all of these, and we'll write them all out, and so this is actually how we kind of define cosine and sine really, you have some angle, right, we started with some angle like pi fourths, that only determines a ray, so we don't have like a radius yet, but we can just pick a radius and agree that that's always the radius we pick, if we just specify an angle, and like if I want to know what is sine of that angle, okay, I look along this ray, I know sine is going to correspond to a y component somewhere, and I just agree that the somewhere is going one along that ray, so I go along that ray one unit, I look down at the ground, and I see, you know if I'm doing sine, I see how high up I am, and if I'm doing cosine I'm seeing how how horizontally displaced I am, right, so these are the special angles, there's kind of a nice way to remember these, so I will say that this is this is the thing you sort of do have to memorize here, there's kind of no getting around just knowing the values of cosine and sine for these special angles, I don't really know a great way, but I can at least give you a trick that kind of works, so we'll make a little table here, and this is this is honestly just a trick, there's no real reason why it works that I know of, but here we go anyways, you can think of this as just, I guess it's another mnemonic device for memorizing these angles, so you make a table, you put sine for the first column, you put cosine theta for the second column, you list out your special angles in increasing order, zero, let's make some room pi six, pi quarters, pi thirds, and pi halves, table a bit bigger, and okay I'm going to put, so there's going to be an up arrow here, indicate that we're going to be increasing, and down our arrow there to indicate that we'll be decreasing, and I'll just try to write this out and say what the pattern is, so for the first one it's going to be square root of zero over four, the second one is going to be square root of one over four, third one, square root of two over four, importantly this is the square root of the whole quantity, the whole fraction there, and next one will be three over four, square root the whole thing, next one will be four over four, square root the whole thing, okay, and all I really did here was I had square root of something over four and I just increased it from zero, one, two, three, and four, and just if you end up going and checking what these are, this is just zero, this is one half, this is three two over two, this is your three over two, and this one is one, okay, so I mean don't ask me why that works, I don't really know, some mathematical coincidence, but it's kind of a nice trick, so you start with sine, you do square root of something over four and then you just walk up zero, one, two, three, four, and we should ask ourselves, we should actually go back to our picture and ask ourselves does this make sense, did I, was I supposed to be decreasing for cosine, or was I supposed to be increasing, and if we look at this circle, you could imagine, so sine of theta at theta equals zero is just this horizontal line, and let's imagine rotating this this point from an angle of zero to an angle of pi halves, so if I rotate it up, I know that sine is supposed to correspond to the y value, so it's the height of this point, and yeah, that should be increasing, right, if I'm in the first quadrant and, you know, even just on my pen, like, you know, it's going up, right, so that's at least one way to, like, sanity check yourself, so we know that sine at least on zero to pi halves is an increasing function, because it's corresponding to the y coordinate of some point that's getting higher up in the plane, okay, and the way this works for cosine is basically the same thing, just in the reverse order, so in this case, we start, we just really just walk this backwards, start with four over four, three over four, two over four, one over four, zero over four, go back and take square roots of everything, put a line to separate these, and okay, we'll find that this is one, this is root three over two, this is root two over two, this is one half, and this is zero, so there's also kind of a nice pattern here where try and draw some of it in yellow, it's just kind of flipping the order of these things, sorry, this is, it's drawing a kind of messy after a while, it's, yeah, that's just flipping the order, so it's kind of like a little shoelace or something in the middle that connects them down, yeah, so there's some mnemonic device for remembering these, and it works out that way, okay, and then don't worry if you weren't able to copy this down, I'll put these notes up on the website afterwards so you can refer to them if you'd like, but here's sort of the important part of this is why is this enough, why, like why is only having one quadrant enough, so we'll need a picture here, so now we need something from kind of really early in the class, sorry, here's my y hat direction, it's my x hat direction, this is my origin that I've chosen arbitrarily, but I've labeled as oh, okay, so the idea is what if I have some angle, say I'm given this one, maybe this angle is theta equals four pi over three, so we'll do, this is an example, okay, and I've just specified the angle which means I have really just some ray going off to infinity, but if I want I can choose a very special radius of r equals one to get an actual point in the plane, and we knew that in polar coordinates, if we know the angle then we know the point in the plane, so if I had this, this is p bar, it depends on theta, depends on r, we know it's in the x coordinate, it's our cosine of theta, in the y coordinate we know it's our sine of theta, and maybe I should be clear, what are we actually even trying to do here? Given theta equals four pi over three, what are x y coordinates of that point? I guess on the unit circle, so that's what we're trying to figure out, so we're going to use this polar coordinate trick, this tells you that whatever the x coordinate is, it's just the function of r and theta, and it's expressed this way, r is just one for us, and we know theta, so we know that, and this is just equal right to some pair x naught, y naught, that we're sort of used to, on one hand, and then on the other hand, coming from this side on the right, we know r is one, and we know what theta is, but we don't know what cosine of it is, cosine of pi over three, or sorry, four pi over three, and sine of four pi over three. And again, remembering this fact that cosine is the horizontal component, or sort of the component that's like projected along the x-axis or the x-direction, and sine is the vertical, or the one that's projected onto the y-direction, that's going to be pretty important for other stuff, so that's good to remember. But okay, so we know whatever this point is, it's given by cosine of four pi over three, and sine of pi over three, so if we can just find the coordinates of this point, then we're good to go. And so the game you play, when you make this bit smaller, so we have some room, move it over here or something, and let me move... The game we play is this game of essentially, so there's a word for this, you know, there's Euclidean transformation, like SR, or the way you want to think about this, but there are sort of two things we can do to shapes on the plane and end up with shapes that are, you know, like the same from the point of view of geometry. You know, one thing is we can like move them around, we can translate a triangle, and it doesn't really change its angles, it doesn't really change the side lengths. We can rotate, that's another thing you can do, it doesn't really change features of the geometry. But the last thing you can do is you can you can do flips, so you can take something and flip it like a mirror over an axis or over any line, really. So we've seen like a little bit of this, I think, when we're talking about even and odd functions, we were talking about flipping over the y-axis. So I have some point here, and what I want to do is, so here's the goal I guess, so I want to get some, something here in the first quadrant where I know everything, a bit smaller, and so this is going to be some coordinate x1, y1, and if I can figure out what this angle here is, so I need to determine I want to get somehow the angle I have into that first quadrant, so let me just run through it and now try and explain it, hopefully it makes sense. So the first thing we'll do is we have this sort of triangle here, just formed by dropping the perpendicular to the x-axis. We're going to have some angle here, going to put two tick marks on it, we don't necessarily know that it's equal to this first one yet, let's pretend we didn't know anything about this, yeah. So we have some theta here, let's call this thing theta sub 2, so this is whatever the reference angle is, and what I'll do is I will flip it over this axis and then I will just flip it over that axis and maybe I'll label these as like this will be the first step and this will be the second step. Okay so if I do this first flip I get something, let's just copy this picture into a new one so it doesn't get too messy, so what is happening in this first flip? So the first thing is I wanted to choose this reference angle because I'm only concerned with this little triangle, so I do this first flip and I get something that looks like this, it's not going to be the greatest most accurate drawing of all time, but if this had coordinates x0, y0 then the question is what are these coordinates? Well I didn't change the x value at all so it's still x0, but the y value got flipped to a negative sign so that's negative, why not? Okay so that's stage one of this this flip, stage two is now doing this thing and okay this will end up over here, oh and what's important I forgot to mention here is that in this first triangle we just did a flip and it actually preserves this little angle here, the angle between it and the x axis, so the two tick marks there are just indicating that these are literally the same angle, so this is also a theta two and now I'm just going to do this flip around the axis, it's not going to change that angle but it's going to change where it lies, that's the same angle, what are the coordinates of this point? Well I didn't change the y value from my immediately previous step, so this is step two of a process, but I did change the x value, I reflected the x value to negative x naught, okay and the idea here is that now I'm in the first quadrant and I know that these will be, well it's okay, I know that on one hand this is equal to negative x naught, so let me do this, on one hand this new p bar zero p bar naught is equal to this thing that we got from the geometric process flipping, on the other hand we know things about points in the first quadrant, you know it's our cosine of theta, sine of theta, where theta is just this angle here, so there's, this is just the just the x value there, y value there, and if I know theta, so let's see, I have to kind of go back to my original picture at bed, so this was an angle of four pi thirds for the original theta, which means that my theta two, so this is just the reference angle for it, remember I dropped the perpendicular to the axis, if I were above I would have to drop it down, but I want some, some reference angle theta two, well okay, if I've gone around half then that's exactly one pi or three pi over three, and so this additional incremental piece is just one extra pi over three, so that'll be important, this whole determining the reference angle business is not really nice, so we know that this angle here is still pi over three, once we've done these transformations, and so we know that this, since we're still on the unit circle, is cosine pi over three in the x-coordinate, sine of pi over three in the y-coordinate, you know hopefully we know these are you know one of our, one of our special angles, so let's see, we actually know that this is, makes room here, let's see, this is the square root of one over four, if I did the table right, square root of uh, three over four, and okay so that's one half root three over two, so this is just some chain of logic that we're using here, so we've reduced it to something we know about, namely an angle in the first quadrant on the unit circle, it's a special angle where we know the value, and so now we just have to do one last step where we compare this, what we got from the geometry to this, and remember that this down here was what we were originally looking for, x-naught, y-naught, and we just kept track of it how it changed, so we get an equation that looks like, so up here it's negative x-naught, negative y-naught, it's just equal to one half root three over two, and this is the beautiful thing about vectors is that if you have two vectors that are equal, then all of their components are equal, so this is telling me that, let me move this calculation to a new page so you can see it all at once, so this is telling me two things, once that negative x-naught is equal to one half, and negative y-naught is equal to root three over two, and it's telling me about that, those two pieces of information, okay, and then I can do this, and just solve for these, x-naught is equal to negative one half, y-naught is equal to negative root three over two, and so if I know that I can go back up here, and I now know the coordinates at this point, negative one half, negative root three over two, okay, so let me just try to recap this because there's some, I think this gets really difficult otherwise if you try to like actually memorize these, but this is an argument that works almost all the time, so it's a little bit complicated to get it working, but this is like a very useful algorithm for doing this because once you know this, you can do a whole swath of problems and you only have to remember a very, very tiny bit, you don't have to refer to tables or anything, so just a quick recap of how we did that, we started off with the fact that we know sine and cosine for a bunch of special angles in the first quadrant only, and we don't want to memorize any of the things in quadrants two, three or four, so the way we do this is that if we have say an angle in quadrant three, well we want to find what its x and y values are, we flip it up, so we find a reference angle first, so this is dropping the perpendicular, finding this data two, we found in the middle, we do some flips to flip that little triangle made by the reference angle up into the first quadrant, and we just recognize that if we flip from, like if we flip over the x-axis, this is just flipping the sine of the y coordinate, sending y to negative y and negative y to positive y, and then same sort of thing if we flip it over the axis that way in the middle, then we're just flipping the x coordinate, just flipping the sine up or down, and so it sends x to negative x and sends negative x to x, so I start with my coordinates down in whatever quadrant I want, I do the flips to get it up into the first quadrant, and I keep track of the sines, so from third quadrant I had to flip both the x and the y, if I were in second quadrant I would only have to flip the x to get it in the first quadrant, if I were in the fourth quadrant I would only have to flip the y to get it up there, and then I just keep track, I call my original point x not y not, I kind of trace it through this flipping procedure, I know that as I go from this to this, I flip the y value, and if I go from this to this, I flip the x value, and at the end now we're looking at a triangle in the first quadrant with a special angle where we just, we know this, and you get this little equation of setting these two things in red equal to each other, and you just solve, solve that equation. Another way to think about this is that if you know the reference angle, you're basically done, so just ignore signs for a minute, if you just know that the reference angle is pi thirds, you can kind of immediately jump to this step of thinking about this angle of pi thirds in the first quadrant, and you're going to get one half root three over two, but now you just have to go back and correct for the fact that you're in some different quadrant, so you don't necessarily have to do this flipping process as long as you remember, okay, if I'm in quadrant two, now I need to change the sign of the x, if I'm in quadrant three, I need to change the sign of the x and the y, if I'm in quadrant four, I need to change just the y, so that's kind of like why this whole, like why they spend so much time on this reference angle business is because this algorithm is like super good, you don't have to memorize anything once you know it, okay, so I think that's probably all I want to say about that for now, let's say in the last 10 minutes here, so just a little bit about the projects, you pull up notes on that, okay, so this is project three, sorry I'm going to take a quick sip of coffee, but does anybody have any questions about it, or has anybody looked at it yet, and has something they want me to address in the next 10 minutes, okay, sounds like a no, so the question we're looking at here is the following, well maybe I'll draw a diagram first, so here in orange, we have spy model of the giant sun, and it is casting light onto, okay, our planet, okay, so here's equator, north pole, south pole, okay, that's us basically, and we're going to zoom in on some specific place, let's say here, and do some kind of like magnifying glass for what's happening there, so what happens is that we have some kind of like tree or something, and has some leaf sitting on a branch, okay, can I actually draw this, let's see, not the worst, in case you can't tell from heart skills, this is a leaf, this is the earth, this is some point on the earth we're considering, this is zoomed in version of that point, and what's happening is that this leaf is receiving some kind of sunlight, and so these light rays are coming through, but I've drawn it here so you can see they're kind of hitting the leaf at an angle, so hopefully we know all about photosynthesis and all that, so the question we want to find out, so this leaf is like constantly converting the just sort of energy from the sun right into chemical energy internally, and the question we want to answer is how energy hits a leaf over say a 24 hour period, so this is the main question we're trying to do that, okay, but this is kind of a complicated situation for a number of reasons, one of them is that these light rays aren't, okay, so the first one is, is that one it's not the same amount of light every day, so there's a periodic nature right to the sun is, so the earth is rotating around the sun, and the earth is rotating itself, so you have to simplify things somehow, so we're just going to use the fact that you know the sun is out for some number of hours a day, it's casting light on this leaf and it's not out on the other, for you know some amount of time, and so there's some kind of sinusoidal behavior to like the amount of light that's even reaching this leaf, there is another complication though that somehow the amount of energy absorbed depends on the angle that the leaf is sitting with respect to these light rays, so you can imagine that if it's 12 noon and this leaf is perfectly flat and a bunch of light rays are hitting it, you know just totally in a perpendicular way, and this is kind of like maximizing the amount of energy you get, but you can imagine if the light is like you know just tilted off to a degree a little bit, if you want to think of light as like a little particle, like some of the particles are now missing the leaf because they're at an angle, so the leaf is like absorbing slightly less energy at that point, and then if you know the sun is like on the horizon, it's at a completely flat angle, then all the particles are just kind of skipping along the leaf or something or not hitting it at all, so they're parallel to the plane of the leaf, and so it's absorbing like zero energy, this is kind of tricky, we can't just like say oh it's absorbing you know 20 milliwatts every hour and then just add up all of the hours, it's like every hour it's a different intensity of light depending on the angle, so if you want to take the differential integral calculus you'll find that there's a very nice mathematical way to do this that simplifies everything we're doing here, but what we'll do is kind of what comes up in applications a lot is try to approximate this discreetly, so we'll try to like at the end, so one will try to compute some kind of function that tells us depending on what time in the day how much light is hitting that leaf, and we'll try to say okay let's assume that function is constant for like two hours or something like that, we'll just add up how all the light that hits in that two hours, at the end of that two hours we'll recompute the function because we know that the sun's angle is different, we'll get a different energy sort of density hitting it, we'll add that up for another two hours and so on and so forth, so we'll chunk up the day into like 12 parts or something at each little time marker, we'll evaluate this energy function, find out the density, and then add all of that up to get like a total energy input into this thing, let's see what I want to say about this, yeah so that kind of model to keep in mind here is that this is supposed to be like a horizon or something, and this little flat plane is the leaf, and you can imagine the sun is sort of a maybe at some time, kind of acts like a little tangent plane or tangent line here, and so you have, sorry these are supposed to be straight lines, but so you can imagine that it's casting light that way at that time, and at some point it's like directly over and it's casting straight down, and this is kind of like maximizing the amount of energy that the leaf is sort of absorbing because it's sort of all of its surface area is hitting everything at a direct angle, and you know you might imagine that over here, now these are coming in totally parallel to the leaf, and sort of none of that, you know if the leaf was like a solar panel, none of it would be hitting the panel, okay, and the way that we'll sort of model this we'll think of this leaf as having a little vector that points up, and we'll think of the sun, well it'll be coming in at some vector like this, and I'll have to zoom in a little bit, there will be some angle between these two vectors, and the amount of energy that's being absorbed will depend on this angle, so imagine if this angle is zero this is telling you that you're in the maximal situation where they both line up, if this angle is exactly pi halves this is telling you you're in this other situation where they're exactly parallel, so we're going to use some kind of trigonometric function of the angle to model how much energy it's getting over the day, so it's going to be some kind of function just kind of sketch out what it'll be, you can have two axes, and what it's actually going to be, this is very important that it's not x and y, this is theta as a function of t, theta being this this angle here, the angle between these two, sort of the angle associated with the leaf which way it's pointing and the angle that the sun is coming in, and so we know that at like 12, maybe 12 noon or something it maximizes sorry i'll let you go in just one sec here, but and we know it maybe times zero it's zero and maybe time 24, time 24 is just equal to time zero, this is like midnight and we'll want to come up with some function that models, you know, increasing through the day hitting that max and then decreasing, oops and this max what angle is this, this should be some maybe some angle here, so it'll be some function that sort of does this kind of thing and then it also is periodic, so it's going to do the same thing outside of that range, so this won't be a sine function or anything yet, maybe just sort of a linear function or a piecewise linear, and then later on we'll see that the energy ends up being sinusoidal and we'll need to model that too, okay, so definitely take a look at the project handouts, start reading through it, take a look at the example, I'll post up these notes and see you guys on Thursday. Thank you, have a good day. Thank you, you too. You can say hi, say hi. Let's see, Julietta, Parker, did you guys have any questions? No, thank you. Okay, see ya.