 Okay, so I'll just let you all please the whole class sitting and waiting, so let me just remind you what we're doing is we're talking about polar graphs, and so this means that we have something describing, so polar coordinates, we have, we give the angle, we give the distance from the origin to give a plate, we give the point, and well it's useful sometimes to be able to invert one to the other, which we can do, this is just a review, I'm just talking about a few people are still finishing, we used the fact, the definition of the cosine and the sine to relate r, x, y, theta, whatever, r is the cosine of theta, what am I doing? x is r cosine of theta from the picture, y is r sine of theta, r squared is x squared plus y squared, and theta is the r cosine of x. So we can switch back and forth from polar to rectangular, usually when you're using polar coordinates you don't even want to think in polar or rectangular, just like if you're speaking Spanish, it's better if you just think in Spanish rather than trying to think English and translate. So the power of polar coordinates is a much better way to describe that are hard if not impossible to describe in terms of rectangular coordinates, so sort of better to just naturally think polar. Can I stop this quicker question? Anybody still working on it? Okay, so we'll now pause, see how people are doing, if anybody has a clue. Okay, so almost everybody got this right, well more than half of you got this right, you can't see this chart unless you have really good eyes because they didn't go over the screen yet. I guess like, let me try and go over the screen first to see where it looks like when I go over the screen. So you can see that most of you know that the right answer is one plus the sign. The way to figure this out, since I'm giving you a choice, is to just test some points. So not to try and convert one to the other, although you can do that, but just think about what's going on here. When the angle is zero degrees, well, we go through some positive number, and so that already, all of these go through zero one. Well, this one goes through zero two in rectangular. So all of these are sort of a good candidate, but now think about what happens to the sign as the, or the cosine. What happens to the cosine as the angle increases, is that the cosine decreases. And so if we went, and the cosine of pi over two, or straight up, is zero. So that would mean that if it were one plus the cosine, this should come in. It should be closer here, closer here than it is here. So that rules this one out. One minus the sign as the angle increases. This should go down to zero because the sign of pi over two is one. So I would have one minus zero, one minus one, which is zero. That means that when I look this way, I should be hitting the origin. So if it were one minus the sign, I would want to do something like that. So that rules this guy out. This guy is still in the running because this goes from one to two. And it increases. If you look here at this angle, this distance gets longer and longer as I go up. And then as I continue, the sign is going to decrease down to negative one. So a decrease back down to the graph of the sign looks like this. So the radius will decrease back down to zero when I'm looking this way. So I should be symmetric here. I should decrease back down to the same distance here, but looking the other way. Again, this one is matching. This one is not doing the right thing. And then as I continue looking more, it should decrease because now I'm in the negative signs. I should get radii less than one down here. So all of these radii are less than one. So this one is really the only one that's viable out of these choices. So that's sort of a way that you can look at it, compare it to a known equation, and say, yeah, that's reasonable. So if you make a graph, I don't know. What I want to do for the next couple of minutes is actually show you. So plotting polar graphs just like plotting any kind of graph is a little bit, I'm a little bit of this. Sorry. Is a little bit, I don't know, tedious to do by hand and computers are very good at this sort of thing. So for example, if I just did the one that's on the floor there. So if I want to plot on plus the sign of T, there it is. It looks kind of like what I drew. And again, you just think about what's going on with the relationship with R. Now essentially what I want to do for the next, I don't know, five minutes or so is just play with this. Give you some sense of what other sorts of things we can see when we mess with polar graphs. And then I want to turn to actually doing some calculus with polar graphs. So instead of changing this T, well you can't see here but I'm tracing out more than 2 pi here. I'm just running over the same graph every 2 pi. So for example, if I just went up the first half from 0 to pi then I just get the top half of this business and then from pi to pi over 2, I mean to 2 pi, I pick up the bottom bit. So it's just going around and around. There's a little touch in there. But the way that I've put this is very sensitive to this 1 because the sign is 1, is negative 1 here, which means that it will just touch the origin at some place. So if I increase this constant here a little bit from 1 plus the sign to say, I don't know, 3 halves plus the sign, what do you think is going to happen? No one has a clue? Go ahead, you're making a fixed conclusion. It'll be followed towards the top. What will happen on the bottom? No clue? Anybody have any kind of intuition? You could be wrong. It's okay. It'll get further out. So it won't touch here. Right? Because I'm increasing the radius. So I'm going to think of this as taking the circle, r equals 1, and then I'm adding sort of a sign curve onto it so it bows out because I'm going to sign out and then bows back in because of the sign. So if I, you know, because the sign here, the radius is 1, but I'm moving inward by 1, so it's just touching. So if I change this from 1 to 3 half, a couple of people said it's going to get further away and it's not going to touch, and I'll get a little lump on the bottom rather than a peak. What if I made it less than 1? So if I went instead of 3 halves, I went to 1 half. Now what's going to happen? The top will come in closer. What will happen at the bottom? It'll come up higher. So do you think I will see something? You know, I can just do it if that's my mind. You think, so it'll come in lower and then here it's going to come up higher. Is it going to come up higher like this, you think? What's going to happen? It'll make you move. Exactly. So if I make it a half, then I get a loop, like that. It makes a little loopy loop and then goes away again. Again, if you think of this as being taking the circle and then moving the radius in and out as I go, then this makes some amount of sense. You've got to work on it a little bit to see if it will make a loop. What if I increase the angle so that the angle goes faster than the angle? Does that make any sense? No. Let's try this again. So instead of that graph, let's try a different one. So instead of doing that, so let's just, we talked about this one last time, r equals sine of theta, just gives us a circle like that. What if, instead of being the sine of theta, I make it the sine of 2 theta? So let's think about, so let's remember here, when r is sine theta, this actually sweeps out this circle. Let me first do the first quarter. So when theta goes from 0 to pi over 2, I get the one side of the circle. And then when I go from pi over 2 to pi, I get the next side of the circle. And then as I go from pi to pi over 2 to 3 pi over 2, then I get the same side of the circle but looking backwards. So when I go from pi to 3 pi over 2, this is a little harder to see. The angle is down here, and we're going backwards because the sine is negative in this and the lower two quadrants. The radius is negative, so I'm going backwards. And looking this way, my distance is back there. Okay? So again, we're a little, if I, if I just, so let's just go with the whole 2 pi business, 0. So this is traversing the circle twice, once in the first and second quadrant and then the second time with the negative r when theta is in the third and fourth quadrant. So for every pi, it goes around the circle once. What if I just multiply a constant here? What's going to happen? It'll get bigger. So all I get is a bigger circle. Nothing interesting there. You probably can't even see that it goes through 3 now instead of 1. What if I put my constant inside the angle instead? So instead of looking at sine of pi or sine of theta, I'm going to look at twice the angle. Now something very different will happen. In the first quadrant, let's just look in the first quadrant to get an idea of what's going on. So I'm just going to go from 0 to pi over 2. Well, when theta goes from 0 to pi over 2, what happens to the sine of theta? It goes from 0 to 0 passing through 1. I'm sorry. When theta goes from 0 to pi over 2, the sine of 2 theta goes from 0 up to 1 and then back down again. So that means that instead of being at its maximum angle here, as soon as you get to the maximum here, so instead of going out and in in the first quarter of a turn, I'm going to go out and in the first quarter of a turn. I said quarter twice. Instead of in the first half turn, I go all the way out and all the way in. I knew it in a quarter of a turn. So I'll have something that would look kind of like a circle, but I squash it all over here. So it gives me a little loop over there. Now if I continue, now I'm looking in this quadrant, this is my angle here. As my angle ranges from here to here, what is sine of twice that angle to? Will it be positive or negative? Did I ask this as a quicker question? Sure. She said sure. Okay. So my question is sine of 2 theta. When theta is here, will it be in... So theta is between pi over 2 and pi is the graph in A. A, first quadrant. D, second quadrant. C, third quadrant. D, fourth quadrant. Roth. I'd like you to answer this quickly because we have to get somewhere. It's channel 26 or 49. Okay, so I'm going to give you 15 seconds more. I don't know if you can make a random guess. Or don't. Okay, so what is the game? I'm going to stop this now. Stop. Are you too late? So it seems that nobody thinks it's in Roth's quadrant. It's either in the second quadrant or the fourth quadrant. So let's think about which it would be. So theta is between pi over 2 and pi. And so the sine... So that means 2 theta is between pi and 2 pi. So when 2 theta is between pi and 2 pi the sine is positive or negative. Negative. So that means that when I look over here my radius is negative. So I'm down here. So we'll be in the fourth quadrant. And let's just check that. I'll let you take a picture and see if I got it wrong. So here if we go from pi over 2 let me get rid of some of this jumpless plug on your screen. Pi over 2. You can't quite see that that's the fourth quadrant so let's include the first part as well. And maybe you can see that's the fourth quadrant. It's down below. Because our angle is looking this way but our radius is negative so we have to back up. It's behind us. And as I continue when I go from pi to 3 pi over 2 I'm hearing this quadrant again I'm looking negative so it will be behind. And so if we do the whole business well let's just look at the next section. So I go from pi to pi times 3 over 2 I'm sorry, it's positive again. Sorry, so the radius is positive I'm thinking straight because it's the sine not the cosine. It's positive again and then in the next quadrant it's negative again. So what's happening during the curve in the first quadrant it's here then when I'm looking over here r is negative so I get this and then when I'm looking over here r is positive again I'll get this and then when I'm looking over here r is negative again I get this a little clover leaf so let me just do the whole thing out for 2 pi 4 leaf I usually call it 4 leaf rose but I need it inside. Okay and just to go a little further than that if we go from 2 pi to 4 pi it will double the number of leaves and if we go 8 pi then I'll keep on here we'll go to 2 pi and then I'll get a whole bunch of leaves there so something like that so let's go back to something you can see like 4 pi we get twice as many leaves as the doubling of the angle I'm sorry I'm writing J pi writing T and meaning theta yeah this is why I'm confused I could write theta here but you know theta is the type of theta so here and the same business happens with the cosine if you think about it for a minute it's just going to turn it a little okay but something a little bit funny happens if this angle is odd instead so let's look at 2 pi just so we can see so for 2 pi we get 4 leaves on the rose but if I go to 3 pi 3 theta I keep saying pi thank you if I go to 3 theta instead of getting 6 leaves I only get 3 so for odd numbers I get the same if you think about this this is because 0 to 4 pi divides nicely by 2 but by 3 it doesn't so here I go over this same thing twice if you just think about how the angle varies so for odd odd multiples I get the same number of leaves and for even I get twice as many don't memorize this it's stupid to memorize this just realize that there can be some surprises here and if I add a little constant to it say 1 plus and this will pull it out a little bit it will come in and touch several times and just to emphasize that it's coming in and touching let's make it 1.2 plus so it comes in and then it goes back out again so here I'm taking the circle and I'm moving in and out as I go around the circle if we go to an even number back to 4 pi you can see what's going on with that loop difference that we got before sometimes it's coming in making it go back sometimes not it's the radius where the length is less than 1 like 0.4 then you can see how the double loops are changing this is where it's positive and then this is where it's negative and then it's positive and then it's negative and then it's positive and then it's negative the big loops are where it's positive in this case and the little loops are where it's negative and of course you get really crazy things if we don't put an integer here like I don't know 4 7 oops if I put 4 7 here I need a little more than 2 pi let's do let's go a lot so we get really nice pictures that make like little mandalas and things this is fairly complicated because what's happening well let's do 1 7 instead so you can see that we get little variations from going in and out and in and out and all around so we can make all sorts of crazy stuff happen I would encourage you to play with this with something that can graph polar graphs I really have to be wrong because the point is I want you to be able to talk about the area calculating the area here doing calculus with these pictures because when this isn't a calculus class this is not a pre-calculus class so this is not a study of functions so these are good to play with but the real point is I want to introduce now some calculus so I've done with this I'm going to turn it off so one of the things one of the things that calculus is useful for is taking derivatives so if I have some function like let me just drive this way r equals capital theta I can certainly take the derivative of r and this tells me something what does it tell me so what does prime of theta equals zero mean what is this telling me is happening with the graphs is the derivative of zero the derivative of r with respect to theta it reaches either a max or a min or maybe a flat spot so it means that f of theta is a critical point which is it's a max a min or maybe an inflection point so in terms of a polar graph this is telling us so again remember we're talking about the radius it's either at a max or a min I'm assuming we're in the first part or maybe some kind of inflection point where it's going out flat is out and then continues perpendicular to the radius so this tells you the same kind of thing also if the derivative is infinite it means it's parallel to the radius if f prime of theta so if it's infinite then this means that the graph is moving parallel to the radius here's my angle that I'm looking at and the graph is going to be like this kind of peak like that something coming tangent to the radius usually it's going to be some kind of peak but it could also do something like this something like that so I don't really want to focus a whole lot on that because really the focus of this class is not on differential calculus although this is a context in which you're thinking about derivatives slightly differently so this is just to give you some sense of how the derivative is useful in other contexts you just have to change how you're thinking about it a little bit in fact polar coordinates are an example of something more general called parametric equations usually the focus of parametric equations at Stony Brook anyway is done why does it seem like they don't have all the lights because they aren't usually the focus on parametric equations at Stony Brook is done in calculus 3 in math 203 or math 307 or AMS 2, 6, 3 I don't remember the number does anyone know calculus 3 for applied matter whatever it is I used to know it, I knew it anyway the focus in parametric equations so parametric equations is how it's useful to describe curves in space and really this is part of the focus of calculus 3 so I don't want to do a lot of this but now let's turn to the idea of area because this is really somewhat relevant to what we've been doing so again thinking that I have some curve it doesn't really matter what the curve looks like too much and I want to understand so I want to understand how I might find the area inside such a curve of course one way is to convert it to a rectangular do the integral in rectangular and then say there's the area that way pretty much sucks because a lot of times you can't do the conversion or you get something horrible but here we can just sort of think polar and think about obviously it should have something to do with the integral but we don't want to just integrate f of r I mean r equals f of theta because that will not give us actually what we want so let's think about let's say maybe this looks like a circle let's do something a little more complicated like that so I'll just do the exact this example that I just did r equals well I'll just do it as f of theta but you can think of f of theta as being the sine of 2 theta and how would I get the area of this loop so I would do the same kind of thing that we do to get the area under the curve like this we chop it up into little bits but instead of my bits being dx my bits will be d theta so I want to cover this in little sectors which sweep out the graphs where the angle here is some little fraction of theta d theta and I want to think about what is the area of this so I want the area of such a little slice like this so the area here would be the integral from theta equals let's just call it a 2 theta equals b and let me not write down what it is let's just say the area of the little wedge d theta does this make sense so if we figure out how to calculate what the area of the little wedge is for a tiny angle d theta then we can figure out the area of the whole thing by adding up the areas of the infinitesimal wedges that is integrating d theta for little wedges in the same sense that we found volumes by adding up the volumes of infinitesimal little disks or cylinders or whatever and the same way that we found work by adding up the amount of work over an infinitesimal time unit or distance unit so it's the same idea that we've been doing already we figure out what the area of the slice is and then we integrate that area to get a whole thing okay well I guess the d theta is in here otherwise it would get involved it's just the area of these where this is d theta so let's figure out what the area of such a slice would look like well just like when we were doing all of the other things we just assumed that the function is pretty much constant over that little slice alright so here's my polar curve I'm trying a big version of d theta I'm going to assume that the function is pretty much constant this is the number our doesn't change so this distance here and I want the area of this piece of pie the piece of pie is going to get smaller and smaller and smaller because you don't want to start with four people and we say oh let's have some pie and now I decided to put one pie then I'm going to make everybody in this room right you all get about a hundredth of a pie it's a pretty small number and then you get to like and then we all get it okay okay so here we have a little slice of pie and we want to figure out what's the area of this slice of pie well it's a piece of a big circle right it's just a big I have a circle a radius r what's the area of this now you're jumping ahead the whole circle what's the area of the circle of radius r better know that that's pie r squared you're not going to have to go back to third grade okay so this is pie r squared but I don't have the circle all the way around the circle is 2 pie I only have d theta around the circle so I only have a d theta for the circle so the area of the d theta of the circle is pie r squared divided by 2 pie times d theta the amount that I have is d theta of 2 pie of the whole circle times pie r squared which is the pie is cancelled and you give me one half r squared d theta that's my little area element so now as I go back I do the idea that I want to integrate from a to b the area of a wedge well r squared is f of theta r is f of theta so the area of a little wedge is that so what I want to integrate is just I lost my pie no I did lose my pie that was not good it's just one half of the function squared d theta so that means that if my goal say is to calculate the area of one so if I want the area of equal sine 2 theta so one thing I need to think about is a sine 2 theta well my picture is bad already I want the area inside this clover and so this means it's just going to be well I start here and I sweep out a whole clover in sort of a funny order but if I go theta from zero to pie then I pick up the whole thing I want here the whole thing is picked up from theta to pie what am I doing wrong no, two pops what's wrong here I want two pops so or I can just think of one leaf this is four times the area of this and I get this as theta goes from zero to pie over two so anyway it's the integral of the integral of of sine 2 theta square that's the area of this but I want the whole thing so I want four of those from in the right from zero to pie over two which is also the same as the area from zero to two pie without multiplying by four that's enough okay so that gives me the area let me not do that integral so let me remind you how to do that integral so this is a sine squared so this means that you use the half angle formula to turn it into a cosine of four theta right people remember how to do that or you ask your computer no do it no do it two minutes to do it don't do it so I'm hearing don't do it let me just give you the hint we use the fact do it jump jump kill yourself do it we use the fact that sine squared of theta is one half one minus the cosine of two theta right so in this case instead of having a theta here we have a two theta here so I have sine squared of two theta is one half one minus the cosine of four theta and so this integral so this integral so I want to do two zero and pi over two sine squared theta d theta becomes one half and that gives me well let's just do a two so one half one minus the cosine of four theta d theta and so this part this part the half of this is gone forget it so the integral of one is theta minus zero pi over two integral of cosine four theta let me just say it the integral of minus cosine four theta is sine four theta but I need to divide by four because it's integral evaluated from zero pi over two so at zero this is zero or at pi over two this is a quarter of the sine of four times pi over two that's two pi so that's zero so that just gives me pi over two and at zero this is zero so I'm going to screw it up that's what the answer is so let me just point out if you really insist on Monday I will do another problem like this but on Monday we're supposed to start a new aspect of this course which is on infinite series so on Monday and all of your refugees from 127 have just showed up jump jump, slip your wrist let me see if you're really feeling suicidal but okay so I will see you Monday