 Welcome back, everyone. In this video, we're going to continue our discussion of polar coordinates with particular focus on what we call a polar function. What do we mean by a polar function? Well, we have these two coordinates at r, not x and y this time. It's going to be r and theta. So we're going to set r equal to a function of theta. So treating the angle theta as an independent variable, we can see that the variable r will be dependent on that choice by some formulaic relationship, r equals f of theta. So I mostly want to focus about the graphs of polar functions, give you an idea of what things might look like here. And so we're going to switch over to Desmos.com to see exactly what these type of graphs might look like. What you see illustrated on the screen right now would be the function, well, the polar function r equals 2, r equals 2, which according to this, it looks like it's a circle of radius 2. Now, one way to better understand what's going on here is, first of all, graphing calculators, including Desmos, typically have an option of switching to polar coordinates. For Desmos, you just have to click this little button right here, which looks like a target. And so now the axes switched up a little bit. You can see running horizontally the x-axis and the y-axis, but now instead of like the box grid lines, we actually see a bunch of like pizza slices and such. You'll see several concentric circles, which represent different radii. And you'll see a bunch of lines emanating from the origin, which represent different angles such as pi 6, pi thirds, as you can see here. So r graph r equals 2, which is like a constant function for polar coordinates. It's like the equivalent of a horizontal line in Cartesian coordinates. It gives us what would look like a circle. And with a little bit more explanation on what's going on here, play around with the following idea. If we want to draw this, we might start at angle 0. And as we start transversing, you're going to start coloring, coloring, coloring. So theta is allowed to vary, but the radius r is always supposed to equal 2. So as theta continues to get bigger, bigger, bigger, going all the way to 2 pi, you just continue to draw a circle over and over and over again. So that's a very fundamental polar function. And this is an important thing to remember because in Cartesian coordinates, a constant function is a straight line. But in polar coordinates, a constant function is a circle. And that big distinction there will actually, one, it kind of decides why one uses polar coordinates over Cartesian coordinates. And it'll motivate how we do some calculus a little bit later on. Another straight line, so to speak, one needs to consider in polar coordinates would be an equation of the form theta equals pi over 4. In this situation, we just fix theta, so theta's always pi over 4, but r could be whatever it wants. It could be a large radius, a small radius, and so you end up getting this line through the origin. You get the things on the other side as well because you do allow the possibility of a negative radius. This is sort of like the polar equivalent of a vertical line. So what I want to do now is look at some more exotic pictures. Let's zoom in a little bit on this one. So take the graph r equals sine of 2 theta, and you get what is often referred to as a four-petal rose. I'm certainly not a florist of any kind. When I see this picture, I don't see a rose. I mean, I might call it like a four-petal daisy, but, you know, the petals don't quite match up. But you do get something that kind of resembles a flower of some kind. And again, I want to give you some understanding of where does this picture come from. So if we play around with this a little bit, as you start drawing this thing, you start with the origin. Sorry, you start with the angle zero. Sine at zero is equal to zero. So you're going to start at the origin. So then as the angle gets bigger, what happens is your curve will get farther and farther away from the origin. Eventually, at the point when you reach pi force right here, so you're halfway through the first quadrant. Once you reach pi force, you're taking sine of 2 times pi force, that is sine of pi. That's going to equal one, the maximum value that sine could obtain. And so the distance between the origin and the point right now is one. You can see how it intersects this one radius circle right here. And so then as we continue to go down towards pi halves, which would finish off the first quadrant, you're going to get back to zero because you're taking sine of 2 pi over 2, which is just going to be sine of pi. I think I misspoke earlier. Sine obtained at maximum pi halves, pi force times 2 would be pi halves. Sine would be zero at pi. And so then continuing on, when your angle is in the second quadrant, actually what you see drawing is something in the fourth quadrant, excuse me. The reason for that is when you're going from pi halves to pi, sine of 2 theta actually thinks you're going from a different interval. It actually thinks you're going from pi to 2 pi because it's going twice the speed. And in that interval, sine is actually negative. Your r is negative, thus you're going to be graphing on the opposite side of the origin. Then you return to zero. Then it's going to repeat itself here. In the third quadrant, you're going to trace a positive pedal. And then in the fourth quadrant, you're going to trace a negative pedal. And then as sine, of course, is 2 pi periodic, this graph will repeat itself every 2 pi units. So you get this four pedal rose. Some things to mention here that we could adapt it, right? If you switch sine with a cosine here, it doesn't make too much of a difference here. It does change the orientation of the flower a little bit, but be aware. Oops, that doesn't trace it anymore. But this would still give you the same basic picture that we did before. And we kind of rotated things, reflection actually. So we'll switch it back to sine. Some other things we could do is we could put some type of coefficient A in front of the sine. What does that do to it? If we allow A to get bigger, it's actually going to cause the flower to get bigger. If A gets smaller, that actually shrinks it. And so this is the idea of a vertical stretch or compression on a polar function. It makes the radius get bigger or it makes the radius get smaller. If you take a negative one, it's going to reflect back through. You're not going to see any difference because of the symmetry that the graph has. All right. Now, what if you want to tweak the period of sine, right? So do some type of vertical stretch, horizontal stretch or compression. Well, for these polar functions, you get something a little bit different. R equals sine of 3theta. It gives you something that's called a three-pedaled rose. Again, I would say this is either a three-pedaled daisy or honestly, to me, this looks like the flux capacitor. Whatever you want to call it, we now get three pedals showing up this time. And again, if you want to watch the picture as you start tracing this thing, you're going to first trace the first pedal, then you get the second pedal, then you get the third pedal. And this actually happens when you reach pi. My slider right now is about 3.16, which we know is a little bit bigger than pi. When you do the second half from pi to 2pi, it actually repeats itself. So this graph is actually pi periodic. It repeats itself every multiple of pi, not just 2pi, which is a little bit curious. This has to do with the fact that we have an odd number of pedals here. You could switch sine with a cosine and you'll get the same basic picture, slightly different orientation there. You can see your flux capacitor got a little bit turned. And some things that we can mention switching back to sine is if you pick other odd numbers, like if you take sine of 5theta, you'll get five pedals. If you do 7pi, you'll get seven pedals. But if you do something like 2pi or 2theta, you get four pedals that's actually doubled. If you do four, you're going to get eight pedals. If you do six pedals, you get 12 pedals. And so there's a parity thing going on here. If you put a coefficient from a theta that's even, you're going to get double the amount of pedals. If you get one with just an odd number, you're going to get that exact number of pedals. So you can get like 12 pedals, eight pedals. What if we wanted exactly six pedals? How would you accomplish that? Turns out there is a way of doing that. It's sort of the following nice little trick here. Take the equation r squared equals 2 sine 2theta. Like so. This gives what's often referred to as a limiscate, which is like an infinity symbol. Or in this case, we could call this a two-pedaled flower. It does accomplish the fact that we actually get two pedals instead of four, because you're taking the square root of sine of 2theta. It kind of kills off the extra pedals. And so if we take a look at what's going on here, if we draw the picture, as you allow this parameter to get bigger, in the first quadrant, you're going to get the first pedal. In the second quadrant, you're actually going to get nothing. Nothing, nothing, nothing, nothing. The issue here is you're taking the square root of sine, which is negative. So you're actually producing an imaginary output, which we don't see in the graph. Let's see. And that's because, again, since we tweaked the period to be 2theta, sine actually thinks we're on the bottom half of the screen, the third and fourth quadrant. So then when you get to the third quadrant, let's turn this thing off. Here we go. So when you get to the third quadrant, you're going to start tracing again, because the sine of 2theta thinks you're in the first and second quadrant again. And then for the fourth quadrant, you see nothing. And so we can actually get these two pedals by sort of making the other two pedals imaginary. And we could switch this to be like four right here. And so we get four pedals. We could switch it to six. We get four, six pedals right there. And if you did three, you could also get three. This isn't exactly the flux capacitor we had before. If you compare them on the same screen, whoops, that was still six. Go back to three. Like so, you can see that things are a little bit different. The amplitude is a little bit different because one and the amplitudes mostly the difference there. And we can, of course, change the amplitude by adjusting this number in front like an A. We can make this thing get bigger, bigger, bigger, shrink it down to be something like that. Again, the curvature is not exactly the case, but it turns out that using polar functions, we can actually create a variety of different flowers with lots of different pedals we wanted here. Kind of like this. In this situation, if you actually do take a negative, you get this reflection. You can flip the thing upside down. And you can see some fun stuff with these trigonometric graphs right here. Alright, switching back over to our board here. That's going to show us a little bit of polar functions. I want to show you some more in the next video. Stay tuned for that. This actually, though, does end lecture 32. Lecture 33 will continue on this discussion. I'll show you some other graphs of polar functions.