 So, yeah, we'll just settle down. OK, it's good to see you all. And welcome back to parametric equations. I want to begin by revisiting what I talked about at the end of last lecture, because I didn't have enough time to explain it in detail. I talked about arc lengths of curves. And I wrote a formula for the arc length. And I wanted to tell you a little more about how this formula is derived. So we look at a piece of a curve on a plane, which is given in parametric form. And we look at the segment of this curve for t between alpha and beta. So this point would correspond to t equal alpha, and this point would correspond to t equal beta. Remember, we discussed last time that in the case of areas, it was important to keep track of which endpoint was which. And so we would have to go, when we compute the area, we have to go from left to right. And it just might happen that the value of t on the left would be larger than the value of t on the right. So then we end up with an integral where the limits are kind of unusual. The lower limit is larger than the upper limit. But it's no problem, you know that if you have such an integral, it's just minus of the integral in which the limits are switched, they're reversed. So there is this subtle point when you do areas. When you do arc lengths, there is no such subtlety. It actually doesn't matter. You could put here t equal beta or t equal alpha. Either way, you're going to get a good formula for it. So how do we find the arc length of this curve? The idea, which in fact you will find in many places in this course and in all of calculus, is to approximate the curve by a union of small segments like this. So you break it into many, many pieces. And then each of those segments, you want to approximate by an interval, by a segment of a straight line. So that is a crucial idea of all of calculus. Because you remember we discussed last time that lines are the simplest curves. So whenever you can approximate a curve, a general curve by a segment of a line, you're doing well. You are able to obtain good results. It's not a good idea to approximate the entire curve by a straight line segment, because you see that it looks very different from this curve. And the length of this would certainly not be the same as the length of this curve. And in fact, the curve could be much worse and something could be much more wiggly like this. So certainly the length does depend on the shape of the curve. However, when we break into small pieces, each piece can be successfully approximated by a straight line segment. You know how these days, you have these advertisements, sublime advertising. I almost feel like, yeah, we should get product placements and make some money on this. Nice watch, especially at this time, time of financial crisis, I think university can do really well given that we have hopefully a very large worldwide audience for this. Because what else people want to watch at home, but multivariable calculus comes to think about it. And not just before they go to sleep. So anyway, going back to this, let's just blow up one small segment of this curve. So it will look something like this. That's just one of those little guys. And now we approximate it by a straight line. And so then we approximate the length of this segment by the length of the straight line segment. And that we can easily compute from the Pythagoras theorem. This would be the displacement delta f, displacement in x. This would be displacement in y. And this would be delta l. Now I want to say that this is for a particular segment. And let's say the segments will be numbered. There will be segments numbered from 1 to n, where n is some large number, say 1,000. On this picture, it's about 10 or so. But you want to break it into many pieces. So this would be really delta xi and delta yi and delta li. And delta li you find by Pythagoras theorem to be delta xi squared plus delta yi squared, squared. And now, so this is nice. But I would like to rewrite this in a slightly more convenient form. You will see why it's more convenient. I want to divide and multiply by delta ti, where delta ti is the range of our parameter t from this point to this point. So what I do is I divide by this. And I also multiply by this. So here I have a square root, but in each square I divide by delta ti. So the net result is that I'm dividing by delta ti. It's like I'm dividing by delta ti. But then at the same time, I'm multiplying by delta ti. So the result is the same. But this form will be more convenient for us. And now the arc length of our curve, which is the sum of the arc length of these little segments of the curve, can be approximated by the sum of the arc length of the straight line segments. So we are going to get the sum of this delta li with i going from 1 to n. This is to say that we have delta l1, the length of the first segment, delta l2, the length of the second segment, and so on up to the length of the nth segment, delta ln. And I hope you remember this notation from previous calculus courses. This means the summation of all those pieces. And so now I continue the formula and I substitute this expression here. So what I get is a sum from 1 to n. And here I put the square root of delta xi divided by delta ti squared plus delta yi delta ti squared delta ti. And now the point is, so you see this expression, which it starts approximating the length of the curve, the finer the partition becomes, the more number n capital n becomes. So for 1,000 pieces, you'll get a good approximation. For a million pieces, you'll get even better approximation. So in the limit, when n goes to infinity, this will actually give us the arc length of the curve. And this sum is called the partial sum. And the integral is defined as a limit of this kind of partial sum. And so if you remember how the integral in one variable is defined, and this is really going to be an integral in one variable, because it's just a summation in one variable. So this is going to be precisely what we call the integral from t equal alpha to t equal beta, which are the endpoints of this expression in which you substitute instead of delta x delta t, you substitute dx dt. This becomes well-approximated in the limit, becomes equal to, I mean, it is well-approximated for finite n, but becomes equal to dx dt, the derivative of x with respect to t. And likewise, this one becomes, perhaps I should do it like this, and this one becomes dy dt. So this whole thing becomes dx dt squared plus dy dt squared dt. And that's the formula which I wrote down at the end of last lecture. I explained it very quickly, so I skipped these intermediate steps about summation and taking a limit. But that's the formula we get by doing this calculation. If you will, we can write it down even more precisely remembering that x is equal to a function f of t and y is equal to the function g of t. So we can write it as the integral from alpha to beta square root of f prime of t squared plus g prime of t squared dt. So that's the formula. And this is a very representative example for many other things which we will study in this course. Because oftentimes, we will try to approximate things, we will try to approximate various quantities for curvy objects like curves or surfaces by sums of the same kind of quantities for straight objects like straight line intervals or squares or parallelograms and things like that. And inevitably, we end up with an expression like this where you have a summation over all pieces in the partition where you have some expression involving your parameter, in this case t, times delta ti. And under good circumstances, and in this course, we don't really discuss the sort of the subtle points here about what are the conditions on the functions f and g. But let's just say good circumstances which are known and well understood, all the functions which we will study in this course will satisfy those conditions. Under those conditions, the sum in the limit when n goes to infinity, when partition becomes more and more fine becomes an integral like this. And it's very easy to read off the expression of the integral from the expression of this partial sum, as you can see in this example. So this is a good guiding principle for many other things that come later. For example, even in this chapter, we discuss also the area of surface of revolution and also in Cartesian coordinates and polar coordinates and the idea is always the same. So I'm not going to explain it each time. I'm going to sort of stress it here in this particular example. And I want to say that in those other cases, it's going to work in a very similar way. OK. But as always in math or in other subjects, there are two, there are somehow two parts of the story. The first part of the story is to get the formula. And here I give you an intuitive derivation of the formula. I give you a rough explanation of why this formula is true. But once you derive this formula, you can sort of forget about this derivation and you can just use it. So when you do homework exercises from a practical point of view, you don't need to remember every step of this derivation. I think it is worthwhile to really to understand, to think about it and to understand how this thing work. Not only for this particular example, but for other problems or questions which will come next, which will come in the future. But at the end of the day, when you do homework, all you need to know is the formula. And you need to know how to work with this formula. So in this case, it's very easy to work with the formula because normally, you are going to be given the parameterization. Or maybe you have to find it yourselves. But once you have the parameterization, you have the formula, you just plug things in, and you get an integral. What kind of integral? Well, you get a single integral, an integral one variable. And so you get something which was the subject matter of the single variable calculus. So at this point, you have to remember all the tricks and rules that you learned in Math 1a1b on how to compute integrals in one variable. This is not the subject of this course. In other words, we already use that information in this course. So what we've done here is we have reduced the problem which arises in multivariable calculus, a problem which involves two variables, x and y. It is a curve on the plane. And we want to calculate this arc leg. But what we have done is we have reduced the problem to a question in single variable calculus. And the question really is to calculate an integral in one variable. And how do you calculate our integral? Well, at this point, you don't need to know anything about multivariable calculus. You switch back to the single variable calculus, and you have at your disposal all kinds of methods and tools and tricks that you've learned before. And so this is a good time to refresh your memory on this, to look back. In your notes or in the book and to see how to compute those integrals. But this will be assumed. We are not going to dwell on this too much. There are some standard methods like change of variables, like integration by parts, and things like that. So you have to also remember the anti-derivatives of various trigonometric functions, for example. And there are special functions like exponential function, logarithm, and things like that. And so you just take that toolbox of single variable calculus and you apply it here. So in a way, in this course, and in the homework and in the test, it is not our goal to test your knowledge necessarily on one B. In other words, I'm not going to give you very hard integrals to work on, but still the basic methods you have to know. And even if it's a simple integral, you still need to know some basic methods to be able to compute it. All right, so that's about arc length. And next, we have one more calculation, one more integral expression, which is not for the length, but it's for surface area. Namely, surface area of the surface of a revolution surface, which means that we look at the surface obtained by rotating x-axis or y-axis. Let's start, let's look at first at the case of x-axis. So here, again, we have, let's say we have some curve, a segment of a curve between some points here, A and B, say. And we rotate it about the x-axis. So what we get is, let me actually make a slightly, let me make a slightly better picture for you. And so when we rotate, each point on the curve makes a circle. Like for example this, this point makes a circle. And then this point also makes a smaller circle because it's closer. And the curve itself appears many times, right? So it looks something like this. And then here also you have a circle, and here you have a circle, and so on. So you get a kind of a cylindrical looking picture. In fact, you would get a cylinder if your curve were a straight line or segment of a straight line parallel to the x-axis. So the simplest, let me erase it again. The simplest example would be if you take this and then the surface is just like this. So this is a kind of a kind of a wiggly, kind of like a vase if you want to think about it this way, which is turned on its side. So you want to compute the area of this object and a cylinder is one example. Another example is actually sphere because you can think of a sphere as a surface of revolution over half a circle because if you start with half a circle which has the end points precisely on the x-axis when you rotate it, you're going to get a sphere. So this is kind of a generalization of a sphere and a cylinder at the same time. So it's kind of very useful to know what the surface area is. For instance, this way you can derive formula for the area of a sphere, which is very useful. And so here the idea, again, there are two parts. One is derivation of the formula, which kind of proceeds in a very similar way as here. And the second is using the formula. Once you get the formula, all you need to do is to substitute the information you are given and then use tools and methods from single variable calculus. So what is the formula? The formula looks like this. It is the integral from alpha to beta, two pi y. And then the familiar expression, which we have there for the arc length. And where does this formula come from? Well, it comes in exactly the same way. Just like here, I have to break everything into small pieces. And so a small piece would be, say, I break on the curve into small pieces. And then for each small piece, I'm going to end up with a little cylinder on the surface, which looks like this. It almost looks like a cylinder. And so I will approximate the area of the cylinder by taking the product of the arc length of this curve. In other words, sort of the height of the cylinder. And two pi times the radius of the circle, which is the circumference of the circle, as we discussed last time. So this part comes from the circumference from the circumference of the circle. And this part comes from the arc length of the length of the segment of the curve. So the formula is not at all surprising. It's just the formula realizes very simple fact that if you have a cylinder, then the area of the cylinder is just going to be equal to two pi times the radius of the cylinder, which in this case, the radius is y. That's why you get two pi y times the length of the side. The length of the side. And the length of the side is the arc length that we talked about. So it's given by this formula. That's how you get this. And once you get it, then you substitute f and g for the x and y and you get a single variable integral. Any questions about this? Yes? Do you have to add the sides? Okay, good question. It depends on what is being asked. Let me, first of all, let me repeat the question. The question is whether in calculating this, we have to add the areas of the top and the bottom of this, right? That's the question. So it depends on what is asked. If you are asked to calculate sort of the entire, that if you are told that you have to look at the figure, which includes both the surface of revolution and the top and the bottom. And you're asked to calculate the area of the whole thing, then you have to take the sum of three terms. One is given by this integral. And that is strictly speaking, the area of revolution. The area of revolution by itself does not include top and bottom. But if you're asked, you can add those two pieces as well. Any other questions? Yes? I'm sorry? What if you rotated something beside the axis, right? That's a very good question. So here I talked about rotating about the x-axis, but in principle, we could rotate. Well, the next level would be to look at, rotate around the y-axis, right? But the formula would be very similar. It would be instead of two pi y, we'll get two pi x. But now the question is, suppose that it's rotated about a different axis, which is neither of these two, but some other axis like this. Well, in this case, the formula could also be adapted. And, but to really do justice to this, you have to know the general rules for changing variables under linear transformations. That's a subject of minus 54. So in this course, we're not going to focus on this kind of questions, rotation around lines other than the coordinate axis. But in principle, you could, and the way, roughly speaking, you do it is by making a transformation of the whole picture, by rotating the picture, so that that line becomes one of the coordinate axis, and then doing the calculation using the formula that we got. There wasn't one more question. You are, did you still have a question? Same question? Okay, good. Okay, one more. Why does this formula use the, both of them, this formula in a special case when x is t and y is equal to f of t, which is, as we discussed, the case of graph of a function y equals f of x. This formula becomes the old formula, which we had in a single variable calculus. So it is really a generalization. Okay, so let's move on to the next subject. And the next subject is polar coordinates. Now, up to now, we have studied, up to now, we have studied various questions about curves, and in all of this discussion, our initial point was a certain parametrization of this curve. In other words, an expression for both x and y, coordinates of points, which are on this curve, as functions of an auxiliary parameter t. So what are x and y here? X and y refer to the coordinates of the point. So we are using here a way to parametrize points on the plane by pairs of numbers, x and y, by their coordinates. And as I said already at the first lecture, you should think of this as a way of addressing those points. You can think of this as a unique address of this point amongst all the points on the plane. So the system of coordinates, x and y, is called the Cartesian coordinate system in honor of French mathematician philosopher Descartes. But in fact, there are other systems of coordinates which in many situations are more convenient and more useful than the Cartesian system of coordinates. And the typical example of a different coordinate system is a polar coordinate system, which we are going to talk about now. Okay? So what is a polar coordinate system? A polar coordinate system is a different way to assign an address to a given point on the plane. And it is defined by different rule. So what is this rule? Let me do it here. So the rule is instead of projecting a point onto the x and y coordinates, I still draw those coordinates just because it's a tribute to the fact how deeply entrenched this Cartesian coordinate system is in our minds. Because I kind of said it's like formatting, think about it as a formatting a disk. The plane doesn't have any coordinate system, but I kind of like to draw it to just indicate that we are viewing it as a plane on which we are going to draw curves and do various mathematical calculations. However, given a point now, we are not going to assign to it an address by dropping perpendicular lines onto x and y axis the way we did before. But instead, we will measure different characteristics. Namely, we'll measure the distance to the origin. Okay? And so we will call this R. We'll call this R. Now, if we just measure this, that's not going to give this point a unique address because there are many points on the circle for which the distance to the origin is equal to R. In fact, we know precisely what the set is. This is a circle of radius R. So there are way too many. What we are striving to do on the other hand is to find a way to assign to each point a unique address. So just measuring this by itself is not going to help us. We need additional information. And what gives us the additional information which already uniquely determines the point is the angle which this segment connecting the origin and our point makes with the x axis. Let's call this angle theta. So now, you see, as I said, the set of all points which have distance R to the origin is the entire circle. But within that circle, there is only one point for which the angle is going to be a particular angle theta. So now we pin down this point in a unique way once we know these two numbers, R and theta. And these are called the polar coordinates. So there are several questions here. First of all, why do we need another coordinate system to begin with? Why can't we be satisfied with the original coordinate system, the Cartesian coordinate system? And the answer to this is that oftentimes if you use the traditional, the Cartesian coordinate system, you end up, for example, with various types of integrals. And these integrals are going to be single variable integrals like this for arc length or surface area. And sometimes they're just too hard. And even if you had an A plus on a one B, you won't be able to get a number out of it. So even if you apply all the tricks, you're still unable to solve it. And so oftentimes there is a, you should try a different approach. And so oftentimes the same quantity can be expressed as a different type of integral, right? And the way to get to a different type of integral is to use a different coordinate system. And polar coordinate turns out to be very convenient in many cases. In many cases it simplifies the answer. It simplifies the kind of integrals that we get. In fact, just a couple of minutes ago we discussed the question of rotating around the different line. You see, if we were to try to tackle this question by using just the X and Y coordinates, we would get nowhere. It would be very difficult. What I mean to say is that suppose that we were asked to rotate a curve not around this axis but around this axis, you know, it's a legitimate question. And so the answer to this question, the correct answer is to realize that in fact, in addition to this coordinate system X, Y that I kind of draw without thinking on the board has as much right to exist as a different coordinate system which is obtained by rotating this one by a small angle in which we'll have two different axis which I draw with pink chalk. So this one I call X prime and this one I would call Y prime. And once you translate it to this coordinate system, the question becomes exactly identical to the question we've discussed now, right? So this already makes you appreciate the fact that first of all, there is not a unique coordinate system on the plane. So it's an illusion that there is a unique coordinate system because the way I draw it, I draw the horizontal line sort of parallel to the floor, right? But if I kind of tilt my body a little bit then it will be like this and also tilt the floor, right? So it's not a good reason. And so this tells you already there is a whole variety of coordinates that you can get by rotating. All of those coordinate systems though have the same flavor. They are all Cartesian coordinate systems. They just rotated one from with respect to another by a certain angle. This one is sort of radically different but this is already a good illustration that you should not be stuck with a particular coordinate system that oftentimes to get an answer or to get a good solution or to get a better approach to your problem. It is advantageous to try a different coordinate system. So here we try something which is different and the advantage of this is that equations of curves simplify of certain curve, curve simplify when you use this coordinate system. And of course the curve for which the equation simplifies is the circle. Circle. Circle I recall can be parameterized using the traditional Cartesian coordinate system in the following way. We write X is cosine T and Y is sine T. That's not too bad but see we are using two trigonometric functions. By the way, this is for a circle of radius one but if you want a circle of radius R, let's call R capital to distinguish it from the other one. So the circle of radius R. For example, a circle of radius five would have X, five cosine T and Y, five sine T. That's not too bad but this is some trigonometric functions which are not kind of not elementary functions. On the other hand, in polar coordinate system the equation we could try to write equations for this curve as well, for the circle as well using this new coordinate system and the equation will simply be R equals R. So this is in polar coordinates. So it's the same curve, the circle but represented in two different coordinate systems. Here we use cosine and sine, here we use nothing. It's just R equals R is equal to R capital. Keep in mind that here the small R and the big R play completely different roles. This R is one of the two polar coordinates. This is a coordinate like X or Y and this R is a number. Because I'm asking you to write an equation for the circle of radius R. So for example, it's a number which could be equal to any number you want, like five, for example. So in this case, the equation would read just R equals five. So it's this type of equations that we get when looking at the circle from the point of view of polar coordinates. And surely this is a much simpler equation than this one. It doesn't involve sine or cosine. It doesn't even involve any function of anything but a constant function five or R in general. So that's a good illustration of the advantages of this coordinate system. The equations for some important curves like the circle simplify. And for this reason also, various integrals that you get if you try to calculate the arc length, the surface area and so on, will also simplify it. So that's the first point. Why do we need this coordinate system? It's useful in applications. The second question that we can ask is how to convert one coordinate system into another? Because let's say, okay, I convinced you that this is a very useful coordinate system. And suppose you are given a curve in the Cartesian coordinates and you would like to translate this into the polar coordinate system. Or maybe conversely, you are given something in a polar coordinate system, you want to translate back into Cartesian. So you need some tools, a kind of a dictionary, how to go from one coordinate system to the other. And that's actually done in a very straightforward way. You just have to express the coordinates X and Y in terms of R and theta. And conversely, you have to express R and theta in terms of X and Y. Once you do that, you have a dictionary which will enable you to go between these two coordinate systems very easily. So the dictionary is very simple. For this, we need to remember how the X and Y coordinates are obtained. So this is X and this is Y. And so now we see very clearly how to find what X and Y are. Because X and Y are two sides of this triangle in which one of the angles is 90 degrees, or pi over two, the right angle. And another angle is theta. So for such a triangle, we can find the lengths of the sides by taking the lengths of the long side and multiplying by cosine and sine of theta. So therefore, we get this formula. X is R cosine theta and Y is R sine theta. So that's it. What does it mean? It means that if you are given a point represented in polar coordinates by two numbers, R and theta, you can find the X, Y coordinates of this point. For example, let's say theta is, let's look at this example. Let's say theta is pi over four, 45 degrees, and R is two, okay? So this is two, this is pi over four, right? So then you can ask what is X and Y? So let's say two and pi over four. You have to take two times cosine theta, and cosine theta is one over square root of two. So it's going to be two over square root of two. And likewise, in this case, sine and cosine are the same. So you end up with square root of two, square root of two. A very simple rule. So that's one way. What about the other way if you want to go from X, Y coordinates to R theta coordinates? Well, in this case, we are given the two sides like this, and we need to find the lengths of the long side, and we also need to find the angle. And of course, we can do that because we can use Pythagoras's theory for the first one. So we get R is a square root of X squared plus Y squared and theta. To find theta, we have to find what we can find is the tangent of theta. The tangent of theta is going to be the ratio between this side and this side. So instead of writing what theta is, I'll just write the tangent of theta, and that's Y over X. And once you know the tangent of theta, you should be able to find what theta is. So at this point, actually, we have to address one question which I kind of swept under the rug, which is what are the possible, what is the possible range for R and theta? What are the possible values for R and theta? So because if we wanna study, if we wanna use it as a bona fide coordinate system, we better know what are the possible values of this coordinates. So what are the possible values and theta? Before answering this, let's ask what are the possible values of X and Y? This is also a legitimate question. We never asked it because it was sort of a given that both X and Y can take arbitrary values. When I say arbitrary, any real number. For minus infinity to plus infinity. For X and likewise for Y. Because we assume that when we talk about the plane, we talk about the infinite plane, not just this blackboard, but the infinite plane, which is obtained by extending this blackboard in all possible directions. So X and Y, the ranges for X and Y are from minus infinity to plus infinity. Not so for the polar coordinates, right? For one thing, R was defined as a distance from the origin. And the distance is a positive number. Or more precisely, it's a non-negative number. It could be zero or positive. So R in this definition is greater than or equal to zero. What about theta? Theta is the angle. And we know that the angle goes from zero to two pi. So it is wise to say that actually theta takes values between zero and two pi. Because if we start looking at theta, if we allow theta greater than two pi or less than zero, what will happen is that we'll sort of get double billing. We'll get different ways of representing the same point. Because we'll be, the same point, for example, be pi over four, but also two pi plus pi over four, or four pi plus pi over four, and so on. And when I was telling you about this coordinate system, one of the important properties that we wanted it to satisfy was that it gives us a unique address to a given point. So if we allow theta to take arbitrary real values, there will be infinitely many ways to represent the same point. Because we could always then shift theta by two pi, and we'll get the same point. So that's why it's better to specify the range of theta as being from zero to two pi. If we want to be pedantic, we should also realize that actually two pi is like zero. And so once you get back, once you get to two pi, it's like you're back at zero. So you don't want to use the same angle twice, even if it's just one angle. So strictly speaking, it has to be from zero, greater than or equal to zero, but less than two pi. So this is how it's defined, how it should be defined in order to get sort of an ambiguous answer. And even then, we actually do have a small ambiguity. I kind of put it in brackets because it's really kind of a subtle point which we are not going to dwell on too much. But there is a small ambiguity. There is a small ambiguity. If r is equal to zero, if r is equal to zero, we are actually at the origin. And so theta becomes meaningless. We can't really say at what angle our point is with respect to the origin because it is the origin. So if r is equal to zero, unfortunately my board doesn't look so good. So my zero looks like theta now. I mean zero really. If r is zero, then theta is not determined. So there is a subtle point that when I promised you that polar coordinate system assigns a unique address to each point, it's not exactly speaking true. It does assign a unique address to all points except the origin. So the origin, the address will be r equals zero and theta could be anything. So there are too many addresses that you assign to that point. But because it's only one point, we're not really going to worry about it too much. So I'm just mentioning it to you so that you are not sort of startled by this when you realize it. It is a fact, but it's not going to be a big problem for us just because it's only one point which is potentially problematic. Now there is kind of a more important point here which is that in fact, we are going to allow, we are going to allow in our calculations, we are going to allow more values for r and theta. We will allow, and this is sort of a technical thing. What I write here is how the polar coordinates are defined strictly speaking. If we really want to define an unambiguous, say for this origin, unambiguous coordinate system. But in our calculations, it will be very convenient to have a certain rule and to allow points with negative r. So this is a rule of convenience really. The rule of convenience is that if you have negative r, if r is negative, we will allow negative r with the following interpretation, negative values of r. Instead of explaining it in words, I'll just draw a picture because it's much easier to see that once than to hear it 10 times. And many of you probably already know that, who have already tried to do the homework exercises for this chapter, for this section. The rule is like this, that if you have a point like this r theta, then it's mirror image. It's the point which is obtained on the other side of the slide, will be called minus r theta minus r theta plus pi. So we'll have, we'll follow this rule that negative r. So if r is positive, you have this point. But then the point with negative r is going to be a point for which lies on the opposite end of the slide, which is symmetric to this point. Good, I'm glad. So okay, correct me, what should I write? Right, that's right, exactly. So if r is negative, then let's say you're given point minus 10 and then pi over four. So where would we plot it? So if it was 10 pi over four, clearly it would be somewhere here, right? But if it is, so the way I wrote it is not a good way. So let's just, let me just give an example using a particular values of r and theta. So this is a representation for r theta, if r is positive, let's put it this way. If r is positive, this is a representation, okay? And if r is negative, then the representation if r is negative, then it will be minus r theta plus r. Like this. So that's what I meant in this picture, but it was a little bit ambiguous. So in particular, if you have 10 pi over four, it will be here. But minus 10 pi over four will be here. This is going to be the point pi over four, which will be the same as 10 and three pi, pi plus pi over four, okay? It's like this. Yeah, when you write it, it's not clear. When I say minus r, do I mean the original one or the minus of the minus of the negative one? I mean minus of the negative one. So it's not the same r as this one. If r is positive, r theta is just formed like this. And if r is negative, let's just do it like this. If r is negative, then you take, in other words here, you take the length r. And if r is negative, you take the length negative r, which will be now positive, and you take the angle, which is theta plus pi. I think it's clear now, okay? You have to be careful when you try to explain it. But we'll see it now in the calculation, so you will see better what I'm talking about. All right, so let's see what we can do now with this coordinate system. Everything all right? Are you getting tired? I guess Thursday afternoon? Are you all excited about the long weekend? Yeah, well, we still have half an hour, but okay, let me take two minutes. But afterwards, you will. So let me, actually, let's take a two minute break just to relax a little bit. I want to tell you something. It doesn't mean that you should start all the talk, which I still have to listen to me. This is my time. So I want to tell you about an article I read today in Forbes magazine. Not that I read Forbes magazine regularly. Don't get the wrong idea. I saw it mentioned on the blog and the link to this article, and that's why I read it. It's available electronically. And it's a very interesting article about string theory and about the controversy about string theory. And I'm interested in this because it's not very far from my research interests, which involve quantum field theory and mathematical aspects of quantum field theory. But I like this article because remember on the first day when I said that people think of mathematics and science as a kind of something which has been written in stone and hasn't changed for many years, which in fact is not true. And this article is very good illustration and I even copied a sentence from this article from the very beginning, which I think is written very nicely. So he writes, the author writes, lay people tend to regard science as the lofty temple inhibited by serene, spoke like wise men. Working scientists though will tell you, it's more like a stock market full of fads and fashions, booms and busts. So I like this analogy, not so much, I don't like it if it's not, the meaning is not to say that it's as speculative as stock market. But I think it's a good illustration of the fact that it is really a live organism which is constantly changing like the stock market. So in that sense, I think it's a good analogy and because it's Forbes magazine, of course they use that analogy, that makes a lot of sense. Just like stock market goes up and down and things change, likewise in science, mathematics and physics, things also go up and down and certain things become fashionable and then they fall out of fashion. And at the end of the day, it's the fundamentals, which you really care about, not down the speculation and not sort of all the artificial things. So I think that it's really cool and I really like this analogy with the stock market and I really recommend this article. It's very short and it sort of talks about one of the most interesting ideas of the last maybe 30 years in science, the string theory. So I'm going to put a link to this article on the B-space, on our B-space page. And if you come across any article about math or science, which you would like to share with other students, please send me the link and I'll post it as well, okay? All right, now back to boring stuff. I'm kidding, it's not boring at all. And maybe another day we'll take a small break and I'll tell you a little bit about what string theory really is. Okay, but now let's go back to polar coordinates. So let's look at various curves and how to represent those curves using polar coordinates. So we already talked about the circle. Circle of radius R is represented by the equation R equals R. Where R is one of the two polar coordinates and this is the radius. So here's another example. A line passing through the origin with angle theta to the x-axis. Also given by a very simple equation, namely now theta is a constant, angle theta zero, say. What I mean to say is that this line, this line, which forms angle theta zero to the x-axis, is described by the equation theta equals theta zero. In the first equation, R equals R, theta is arbitrary. Well, not exactly arbitrary as we saw, it should really be between zero and two pi. But in this equation, R is arbitrary, both positive and negative. So in fact, this is the reason why we introduced this rule, which may look strange at first sight. Because remember, as I said at the beginning, strictly speaking, R should be non-negative. If we follow the definition of polar coordinates in which R is just a distance, so it has to be non-negative. But if we were to adopt this point of view, then the equation theta equals theta zero would actually correspond not to this entire line, but only to half a line, a kind of array which goes from the origin to infinity. Which is fine, it's a fine geometric object on its own right. But if we adopt this rule, if we allow R to be negative, and we plot points with negative R's the way we just discussed, then not only half a line will be represented in this equation, which would correspond to positive R, but also the entire line, the second half a line will correspond to negative values of R. So this is the advantage of this rule that if we follow this rule, then we have a nice representation by an equation of the entire line like this, and not just half a line, okay? Okay, what else can we learn from equations with polar coordinates? So I want to look at a couple of more complicated examples. I'm sorry? What is? Arbitrary, why is it arbitrary? This, the question is why did I say R is arbitrary? Because when I write this equation, R is nowhere to be found in this equation, right? So the meaning of this equation is we look at all points on the plane whose theta coordinate is fixed. It is equal to theta zero. Theta zero could be any number between zero and two pi, like pi over four, pi over three, whatever, whatever you want, any real number between zero and two pi. So theta is fixed, but R is arbitrary. When I write this equation, because this equation does not involve R, this equation means that R can take arbitrary values within the allowed range. The allowed range initially was stipulated to be R greater than zero, maybe R greater than or equal to zero. But eventually we decided that we'll allow negative values of R. So that's why, well actually maybe it's better to write it like this. And on this line, we actually see the part corresponding to R greater than zero, the part corresponding to R less than zero, and there is one point which corresponds to R equals zero. Likewise, when I write this equation, when I write this equation, what I'm saying is that I look at all points on the plane for which the R coordinate is fixed. It's equal to some number capital R, which could be five, 10, 13, whatever you want. But because theta does not appear in the equation, it means that theta is arbitrary within the allowed range of theta. And what is allowed range? The allowed range here is from zero to two pi. So that's the meaning of this. Okay, now let's go back to, let's look at more complicated examples. So three, R equals cosine theta. So again, a perfectly legitimate equation involving the two polar coordinates, R and theta. In other words, we are looking at all points on the plane for which, for whose R and theta coordinates are constrained by this relation. It's like writing in the case of Cartesian coordinates, writing an equation x squared plus y squared equals one. It's one equation involving our two variables x and y. Likewise, this is also one equation involving our two variables, R and theta. We have now switched to the polar coordinate system. So the question you can be asked is to draw this. What does it, to see what this curve represents? And in this case, actually, it's easier to understand what this curve represents by switching back to the Cartesian coordinates. This is not to say that this equation is useless. In fact, on the contrary, you can then use this equation to compute various things about the curve as opposed to the equation in Cartesian coordinates. But first, we wanna visualize it. Like, what does it mean? And here, we have to go look back at the dictionary. At the formulas expressing the polar coordinates in terms of the Cartesian coordinates. And so we could just write R is square root of x squared plus y squared equals cosine theta. And then, but we know tangent theta, so it's a little bit, so it's not clear how do we, how do we get a good formula? Okay, so it doesn't work. So let's look instead at the geometry of this picture. So then, let's see what cosine is, cosine theta. What is it in terms of x and y? Well, it's better to say not in terms of x and y, but in terms of x and r, right? Because this length is x, this length is y, and this length is r. So the cosine theta is the ratio of this side to this side. That's why cosine theta is x divided by r. Now let's substitute this in this formula. So we get r equals x divided by r. And now we can multiply both sides by r. So we get r squared equals x. And now is a good time to express r squared in terms of x and y. Because remember, r squared is x squared plus y squared. So we substitute here. And so the result is x squared plus y squared equals x. And this is already much more manageable. Let's take x to the left-hand side and let's complete the square. We can write x squared minus two times one-half, two times x times one-half plus one-half squared. And then, so I introduce an additional term, one-half squared, which is one-quarter. On this side, and to compensate for this, I also introduce it on the right-hand side so that I get the true quality. And of course I shouldn't forget y squared as well. You see, so I just put two additional terms on the left and right-hand side, which is one-half squared. But if I do that, then the first three terms combined into a square, the first three terms give you x minus one-half squared plus y squared equals one-half squared. And then the result, the end result is already an equation, which is familiar. The end result is x minus, let me use a different chalk. So we get x minus one-half squared plus y squared equals, okay, yes? Okay. This was by way of simply making things easier for you, but I guess maybe I didn't achieve that goal. Two times one-half is one, right? So that's why I wanted to use the formula, a plus b squared is equal to a squared plus two a b plus b squared. And I had x, and I want to have something which is twice. So that's why I wrote x is two times x times one-half. Does that make sense? And then I added, and then I saw that, so that means x is a and b is one-half, and then I put b squared, which is one-half squared, which I also introduced on the other side. Completing squared, that's right, that's right. That's all it is. So I'm not cheating, it's all legitimate. But yeah, but please check, because sometimes I could make a mistake. It works out. So that's the equation we get. And now we can draw it because you see, of course, we are more familiar, we are more familiar with the curve x squared plus y squared is one-half squared. That curve is a circle, of course, of radius one-half. Let me make it bigger. One-half. What's the difference between this equation and this equation? We just shift by one-half, right? So what does it mean geometrically? Geometrically, it means that we shift it to the right by one-half. Why to the right? Because, for example, here the point with x equals zero will correspond to the point with x equals one-half so that this whole thing will become zero. So shifting x by minus one-half means shifting everything to the right by one-half. And when we shift everything to the right by one-half, it means that the center of the circle which used to be the origin now becomes this point, whereas this point now becomes this point. And so it's actually a circle which looks like this, okay? So this is the circle which is represented by this equation, which is kind of neat because if you wanna write it in Cartesian coordinates, you get this equation, which is not too bad, but certainly this looks much more canonical in some sense. And so for the purposes of some calculations, it could be very useful. Okay, what else can we do? Let's do a small variation on this problem. So that would be number four. A small variation would be r equals cosine of two theta. So see, now we can't really get away with a simple formula for it because here we use the fact that cosine theta has such a nice expression. You can still try cosine theta is cosine squared of theta minus sine squared of theta. So then you can rewrite cosine squared and sine squared, but you'll get something really complicated. So when you get to a situation like this where you can't really rewrite it in a nice way by using the Cartesian coordinate system, you need to try to just kind of understand qualitatively what this curve looks like. And to do that, we should look at this equation as if r and theta were Cartesian coordinates in some other world. r and theta are Cartesian coordinates, right? So I'll kind of, in a pink world, in a pink world it will be like Cartesian coordinates, right? So we'll just draw theta and r and we'll draw this graph in r and theta, and then we'll try to see what it means in our world, which is like a yellow world for now, okay? So what this is is just, it's a cosine function except the period of the cosine function is now shrunk to pi instead of two pi, right? Because already one theta is pi, you get cosine of two pi, and so you're back to square one, back to zero. So cosine looks like this. At zero it's one, and then it becomes zero at pi over two, then it becomes negative one, then it becomes zero at three pi over two, and then it goes back to one at two pi, and then it continues like this, right? But now we have to realize that normally if it were just cosine function, normally if it were cosine theta, this would be pi over two, but because it's two theta, it's going to be pi over four, and this will be pi over, normally it would be pi, but now it's pi over two, and this is like three pi over four, this is pi, and then it continues like this. So for example, at five pi over four is going to reach zero again and so on. And now we would like to plot this on the plane where r and theta are polar coordinates, at least qualitatively to get sort of qualitative understanding of what it looks like. So we just have to see what the points here correspond to on this picture. The first point is a point when theta is zero and r is one, right? So it's this point in which r is one and theta is zero. Let's plot it on this graph now, on this plane. So theta is zero, which means that we are on the x-axis. All points on the x-axis have theta equals zero, right? Because I remember theta is supposed to be the angle with the x-axis. So if theta is zero, it means you are on the x-axis. So you are on the x-axis, and the distance is one. So you start here. That's your point, or even let's make it nicer, let's start here. That's our point, that's one. What happens next? What happens next is we are increasing the angle from zero to pi over four. What is pi over four? Pi over four is sort of this bisector. That's pi over four. So when we reach pi over four, r becomes zero, r becomes zero. And r equals zero, no matter what theta is, is as we discuss the origin. So we start here and then the angle should increase. We should go from theta equals zero to theta equals pi over four. But at the same time, r is going to decrease until it becomes zero. So this is going to look like this, roughly. Because what else can it look like? You see, that's the point. I mean, it's a qualitative picture. I'm not insisting that, I know exactly. See, there is no other way to represent it by this. If you try to write it in terms of Cartesian coordinates, you're getting nowhere. You're going to get a very complicated expression, which is not going to help you. So we can only study it qualitatively and that's what it looks like. And now we continue. But see what happens now is that r is a theta, sorry, theta goes from pi over four to pi over two, but r becomes negative. That's where our rule becomes really handy because if we were, how should I say, too rigid and said no, r has to be positive and we don't accept any negative values, we would have to say that this part is not acceptable. We can only look at the picture in this range and this range and so on, but not here. And what this would mean is that we would not be able to draw some, represent some very nice pictures by using polar coordinates. But because we've been flexible and we said we will accept negative rs but following a particular rule which we discussed, then we'll be able to actually draw the whole thing. And so what this is going to look like is that now pi is supposed to go from pi over four to pi over two, but r becomes negative. r becoming negative means that we take the absolute value of r but we shift the angle by pi. So that means that in the next segment on this segment, we look at theta from pi over four plus pi to pi over two plus pi. Right? So where is that? This is pi over four. Pi over two, pi over four plus pi is this, right? And pi over two plus pi is this. So it's going, and it's going to go in such a way that r is going to go to one again. Well, it's to negative one, but we have to take the absolute values. So it's going to be like that. Now it becomes positive. Oh, sorry, it still stays negative and it continues to three pi over four. And that's like this. So see, this segment is this. This segment is this. This segment is this. And I think now you can probably already guess what is going to look like. It's going to be like this. That's right, like this. Then it goes like this, huh? It's not so bad, huh? All right, make it a little bit nice. Like this. A good thing to chew on over the long weekend. So there is one more thing which we need to discuss, which is the formula for the surface area for polar curves. But it's fairly straightforward. I'll just write the answer. Hold on, hold on. We still have three minutes, okay? So the formula for the area, which is one half r squared d theta. And that's for a picture, which you get by taking a segment, by taking a sector on the plane, which is bounded by the curve and by the line theta equals alpha and theta equals beta. The way you derive this formula is exactly the same. What you need to know is the same as the method which we use to understand arc length and surface areas for surfaces of revolution. And the method is to break it into small sectors and evaluate the approximate area of which small sector. And that's something you can read about in the book. It's very straightforward. Now I want to end with two announcements. Two announcements. The first announcement is that, as you know, the first homework assignment is due on Wednesday because Monday is Labor Day. You turn your homework in at your sections, to your GSI, to your TAs, all right, on Wednesdays. Now, on Wednesday night, after all the sections are over, I will post the solutions to all the homework problems in this set on the B-space page online, okay? That's the first announcement. And the second announcement about my office hours, normally my office hours are supposed to be in my office in Evans Hall. But I think we're lucky that there is nobody, at least in the first two lectures, after the first two lectures, there was nobody here. So we'll just hold office hours here, in this room from five to six thirty, all right? So I let you go. Have a good weekend.