Simple Explanation of Polar Coordinates & How to Use Them

Añadido: 2026-05-10

[00:00:00]What we're talking about now, we're not talking about any integrals, we're not talking about derivatives, we're not talking about anything other than in this section talking about what a polar coordinate is, okay? So, remember that a Cartesian coordinate or a Cartesian coordinate is the one that you've learned about all your life. You have two numbers, X and Y, that define that.

[00:00:20]So, if this is your your graph, you know, that you've grown up knowing all the time, it's the XY graph, right?

[00:00:28]And I have a point here, okay? And when you learned about this point, you were taught that this point can be represented by two numbers, X {comma} Y.

[00:00:37]Okay, now you're very comfortable with this idea now because you've been doing it for so long. But when you were first introduced to this, this was a pretty far out concept, okay? That you would take a point, you'd put it on a graph, and that location of that point would be the X coordinate, and it would you'd put a comma here, and then you'd also put the Y coordinate. And those are just numbers. Could be 3 {comma} 4, 2 {comma} 5, -1 {comma} 2, whatever it is. Two numbers to represent a point in a plane.

[00:01:06]And as you know, when you get into talking about three-dimensional functions and three-dimensional space, which is the real world we live in, when you get off to calculus three next semester, you're going to have a third dimension there, so it'll be X and Y and Z. But for now, everything we're talking about is only in a plane, okay?

[00:01:22]Cuz that's that makes it easier.

[00:01:23]>> [snorts] >> Okay, so here are points, these are Cartesian points. Those are the ones you already know about. And you write equations, equations in Cartesian coordinate systems are just they're very simply the same sort of thing. You have X and Y.

[00:01:41]All right, on a graph. And then, you know, if you have a if you have some sort of function, you know, it would go up and come down or whatever it whatever it does, some smooth continuous function, okay? And we say that Y is equal to f of X. Now, this is all review here, okay? This this vertical direction here, Y, that shapes the what this graph actually looks like, is a function of X.

[00:02:01]You take a value of X, you stick it in this function, you get back the value of Y, okay? So, [snorts] we say here that X is the independent variable.

[00:02:12]And that's because that's the number you actually plug into the function, so it's independent. The dependent variable is Y because it's dependent on whatever value of X you have, and also on on the function that you have. Okay, so these are Cartesian coordinates, okay? You have a when you want to plot one on a graph, you go over X number of units, and you go up Y number of units, and it takes two numbers to uniquely define any point in that plane, okay? Now, polar coordinates polar coordinates also take two numbers to to identify a point in the plane, okay? So, it's two numbers, so it's really not that different.

[00:02:48]You're still going to have a number {comma} and then another number to represent that point.

[00:02:53]But in the Cartesian space, I mean in the polar space it's going to look a little bit different, obviously. So, you're going to have you're still going to have a you know, a plane like this, and you can still call this X, and you can still call this Y, just for reference, okay?

[00:03:07]Just for reference. But if I have a point here, you know, and notice up above I called that point X {comma} Y.

[00:03:14]This point would be labeled by or defined by two by two numbers, and those numbers are going to be R {comma} theta, okay? R {comma} theta. What does that mean? So, I can define this point instead of marching over so many units and then marching up so many units, those being the numbers, I can define this point as a radius R, that's just the length of some arrow that I describe, and I can also define it by the angle that this this radius vector or this radius line segment here forms with the X axis.

[00:03:48]So, you see, it's two numbers, okay?

[00:03:50]It's two numbers. And by the way, it's always going to take two numbers to uniquely define a point in a two-dimensional plane, cuz you only have two dimensions in that plane, so no matter how you define your coordinate system, you're always going to need two numbers. When you have three-dimensional space, you're always going to need three numbers, and so on, okay?

[00:04:06]So, in this coordinate system, and I'll just go ahead and write it down just so you know, from the polar representation of coordinates and polar coordinates, okay? Every point in this plane can be defined by two numbers. The first number is the the distance from the origin, basically. So, wherever this point is, it's the distance to the origin, that's what we call R. And then the second number is the angle that that vector makes here with the point, okay? So, you could see that if you had a point anywhere, you could have another point over here, well, I could define let's that's too similar. Let's make it way over here. I could have a point over here like this.

[00:04:42]And I could have a R that would be the length of this vector.

[00:04:46]>> [snorts] >> And I and I have an angle that this thing forms with the X axis. So, the angle would be this giant open angle here.

[00:04:52]Theta is always in referenced to the X axis. So, if I have a a point way down here, the angle is going to be all the way over here, you see? So, that's how you have to do it. That's what that's what polar coordinates are, okay? So, it's referenced to the X axis. Those are for positive angles, okay? Now, you can always have you can also have representation of negative angles as well, okay? If I wanted to have if I wanted to use a negative angle, I could also do that, okay? So, it would be R, and then I would have a negative angle, okay? So, the angle can be expressed either by going all the way around to that place or by just using a negative angle. So, these are all positive directions, and these are all negative directions in the angle, okay? So, two numbers, not that terribly different.

[00:05:35]It's just a little bit different way of looking at it. Now, why do you think you would do that, by the way? Why do you think we would ever use polar coordinates, okay? The reason is usually you're using a polar coordinate system when you have something circular that you're studying. And and gosh, there's there's millions of applications for it. I mean, you could be studying, let's say, the physics of a of the surface of a drum, okay? Or maybe a raindrop dropping in a puddle of water and the waves that come out from the center. See, you have circular symmetry like that. So, anytime you have circular symmetry, anything circular where where you know, the equation of a circle in in Cartesian coordinates is kind of ugly, but in polar coordinates is very simple, and because of this nice symmetry, you can have easier and simpler integrals. So, yeah, you have to have a little overhead learning what this stuff means, polar coordinates and polar equations and polar integrals.

[00:06:22]We're going to get to all that stuff.

[00:06:23]But once you know it, it really makes the math a lot easier than actually trying to do these integrations and these other things in Cartesian coordinates, okay? So, let me show you one more thing. Here's an example, an additional example, of polar coordinates, okay? Let's say I have an XY plane for reference, okay? And let's say my point is actually up here just like this, okay? And this point I'm going to say is it has it is a distance of one from the origin, okay? And the angle here, theta, is equal to pi over 4.

[00:07:02]So, you should know pi over 4 is 45 degrees. So, all I'm saying is at 45 degrees from the X axis, one unit away, there is a point. So, it's two numbers, so it's my polar representation. So, the polar representation okay, of this point is just what we talked about before. It is R {comma} theta. So, it's 1 {comma} pi over 4. So, that's how you would write it down in your test or your homework, that's how you would see it in your book. It would be the radius {comma} and then the angle. That's how you write it down, okay? Just like it's always X {comma} Y in the books, you know, for Cartesian, it's always R {comma} theta, okay?

[00:07:38]Now, let me ask you this. This is the location of the point at pi over 4, okay? Now, you should know from your trig and your introductory calculus and all of that that you know, in trig in radian angle measure, you can continue on past pi over 4, and you can go all the way around another full circle and arrive back at the same point. And one full circle one full circle in trig is always 2 pi radians. Any circle you make is always 2 pi radians, okay? So, in theory, I could add another 2 pi radians to this angle, and I would end up in exactly the same place.

[00:08:12]So, what would that look like, okay? So, what I'm trying to say is that this representation is equally valid equally valid representation at plus or minus 2 pi angles.

[00:08:34]Now, what am I going to what do I mean by that? Let me switch colors and show you what I mean by that, okay? This is the polar representation of this point that we already talked about. Now, I can also say that that's going to equal pi over 4, which is my angle, plus 2 pi.

[00:08:49]Okay? So, it's the same radius, the same radius, and it's the same starting angle. But what I'm saying is if I add 2 pi, I come back to exactly the same spot, so it's really the same point, okay? So, if I were going to simplify this, if I were going to simplify this, it would be 1 {comma} pi over 4, and how would I write 2 pi in order to add these fractions together? It would be it would be 8 pi divided by 4.

[00:09:15]8 pi divided by 4. I chose this way because look, 8 pi divided by 4 is equal to 2 pi. I wrote it this way because I want to add these fractions. I want to add these fractions, okay? So, that would be equal to 1 {comma} 8 pi plus pi is 9 pi divided by 4. 9 pi divided by 4. So, what am I trying to say here? I'm trying to say that this polar representation, which is 1 radius of 1, 9 pi over 4 radians, is exactly the same point. It represents exactly the same point as this one. The angles look different, but the only difference between them is that I've gone an entire circle around again. So, it's really exactly the same thing. So, any polar representation you have, you can always add or subtract 2 pi. Uh in fact, you can add or subtract increments of 2 pi, and you'll always get you'll get a different looking answer, but you'll get it'll represent the same point. Okay?

[00:10:07]Uh I can take I can take this initial point that I had, pi over 4, okay? And instead of adding 2 pi, I can also add 4 pi to it. Because think about it, 2 pi goes around once, and then another 2 pi goes around again. So, I can add 4 pi to it, and what would that look like? It would be 1, pi over 4. And I'm going to add to it I'm going to add uh 4 pi, but that's going to be 16 pi divided by 4. I'm I'm writing 4 pi as 16 pi over 4 because I want to have a common denominator here.

[00:10:38]So, 16 pi over 4, when you do this division, is equal to 4 pi, okay? So, what would that be? That would be 1, uh pi plus 16 pi is 17 pi divided by 4, okay? 17 pi divided by 4, okay? So, this representation 1, 17 pi over 4 for the angle, this representation 1, 9 pi over 4, and the original representation, they look different, but they're exactly the same thing. They're they have different angles, but because the angles are just 2 pi going around the circle 2 pi, which is a full circle, they actually describe the same thing. So, and the same thing goes for subtracting 2 pi and subtracting 4 pi and subtracting uh 6 pi and so on and so on.

[00:11:20]So, I'm adding them here, but you could just as just as legally subtract 2 pi. So, instead of adding 2 pi, I just subtract 2 pi, okay? And I get back to where I started from. So, any polar point you have, you can always add to any multiple of 2 pi, 2 pi, 4 pi, 6 pi, 8 pi, so on.

[00:11:37]You can also subtract them the same way I'm doing here, and you'll get different points, but they're going to all represent the same thing, okay? So, the the long and short of it is uh the thing that I want you to take home is that polar representation of points have infinite number of representation uh representations in polar coordinates, and they're all equally spaced by 2 pi radians. So, for any point you define, uh and that's different than rectangular. In rectangular, it's x, y, and that's it. In polar, it's r, theta, but there's actually infinite variations that I can write, and they all represent the same point because the angles can be can be uh manipulated like this.

[00:12:17]Okay, so we've introduced the concept of the polar uh representation of a point, okay? We've done that. So, we know we have this Cartesian thing, x, y, and we know we have this polar thing, r, theta.

[00:12:28]We know that they're two different ways of writing exactly the same thing, which is just simply trying to describe the location of a point in space. That's all it's trying to do. It's just different ways of doing it, okay? So, naturally, uh it's it's it's something that you were would like to be able to do to be able to convert between the two representations, right? I mean, somebody uh some engineer might be working in x, y, and you might get his work and say, "Well, that's that's nice, but it would really be a lot simpler if we just worked in polar coordinates." So, you'd be able to like to be able to take his work or her work, okay? And convert it into polar coordinates, and uh and then you and then continue on from there and and do your own work. And likewise, you might be working in polar coordinates, and then somebody comes back and says, "Well, that's nice, but I need to work in Cartesian for whatever reason, and it would be nice if you could convert all of his work into Cartesian." And it should be possible to do this because they're just two different ways of saying the same thing. So, it's just like converting from miles to feet or centimeters to kilometers. You can do it. It's just you have to know how to convert them. So, that's what we're going to be able to do here. We're going to learn how to convert these things.

[00:13:32]Okay. So, to go from I'm going to do the easier one first. To go from They're both actually pretty easy, but one of them's actually a little bit easier. From polar and just to remind you, that's r, theta, okay? Uh to Cartesian Okay? Cartesian, which is x, y, just to remind you. To go from polar coordinates to Cartesian coordinates. So, I'm given an r, theta, and I would like to convert to x, y.

[00:14:06]Uh I'm going to go ahead and write it down. I'm going to show you why. The x coordinate Let me I'll go ahead and change colors here just to make it clear. The x coordinate is going to equal r times cosine of theta.

[00:14:20]Cosine of theta.

[00:14:22]And the y coordinate is equal to r times sine of theta. That's all you have to do to convert from polar coordinates to rectangular coordinates or Cartesian coordinates. Because if I'm given a polar value, okay? And I'm trying to convert it here, I know what r is, so I know that the distance from the origin, and I know the angle, which is just the angle uh from the x-axis, either positive or negative, okay? And so, I know what r is, I know what theta is, I just plug it in here, and I'll get an x, and I'll plug it in here, and I'll get a y, and that's the corresponding x and y. Now, why does this work? Why does this work this way?

[00:14:56]Well, if you remember back, let me just draw a quick little picture here to the side.

[00:15:01]Okay, if here's x and y, and here's the point I'm interested in, okay? And obviously, it's going to have a radius here and an angle here, okay?

[00:15:10]Uh well, what you're doing here, you see, this is r and theta that I'm given, and I'd like to convert to Cartesian coordinates, okay?

[00:15:16]So, what I've claimed here is I'm telling you that this x value okay? This x value of this point is equal to r times cosine of theta. So, I'm saying this is r times cosine of theta, okay? And it's this way just because you know, when you think back to your trig, you know, or you should know, if you don't already know, that the cosine function, what it does is it's you know, forgetting about this, the cosine of something, when you march around the unit circle and you start taking the cosine of something, like cosine of zero, you know it's equal to one because you're looking at the projection of that unit circle radius onto the x-axis. That's what you're doing. So, as you take angles around uh the origin, all you're doing is you're looking at the projection along the x-axis. And then you're talking about a unit circle, so in that case, the radius is always equal to one. So, in this case, you're just generalizing it, and you're saying, "Well, the radius is no longer equal to one. The radius is is whatever it is because I'm talking about arbitrary point." So, I'm looking at the projection of this thing onto the x-axis. So, it's still cosine theta, but I'm just multiplying it, scaling it by r.

[00:16:18]Okay, that hopefully is very clear to you, but if it's not, you could also just use your regular trig here.

[00:16:24]Uh some people like to write it this way. Cosine of of uh I keep writing a theta here. Cosine of theta, you could say cosine of theta is equal to the adjacent uh the adjacent value over the hypotenuse, okay? The hypotenuse is r, right? So, you could say r uh here. So, if you were to solve this thing for the adjacent value, which is this, which is what you're trying to find for the value of x, it would be r times cosine of theta, okay? So, it's really just triangle trig is all it is.

[00:16:52]That's all these things are right here.

[00:16:54]And so, by analogy, the same exact thing is true here. This value, which is the y value of this point, is equal to r times sine of theta because you're looking at the projection of this onto the y-axis. And it's the same thing as before. The sine function, you should know in general, is the projection of that unit circle onto the y-axis. It's just here, it's not a unit circle. It's it's a arbitrary length r, depending on where the point is. So, you're doing the same thing. It's sine of theta, and you're multiplying by r because you're scaling it, okay? So, this is the easy one. All you do is you take your r and your theta that's given to you in uh in the problem, which is the numbers that represent the point in polar coordinates, plug them in here, you get an x, you get a y, it's very simple, bulletproof, and you have your Cartesian coordinates. Now, the next one is a little bit a little bit more challenging, but it really isn't that much more challenging.

[00:17:42]Okay? So, let me go ahead and write this down.

[00:17:45]To go from Okay? Cartesian And just for a reminder, that's x, y, to polar.

[00:18:00]And as a reminder, that's r, theta. To If you're given x and y, and you'd like to go to r, theta, I'm going to go ahead and write it down in all its glory, and then we'll talk about it, okay? The way you would do that is you just have to remember the following.

[00:18:14]r squared is equal to x squared plus y squared. We'll talk about this in a second. And the tangent of theta is equal to y over x. This is how you convert, okay? So, I'm given x and y here, and I'm going to plug in x squared plus y squared, I'm going to get r squared, and I just take the square root of that, and I'll get r. So, that's how I arrive at r if I'm converting to polar. And then, I know what y and x are, so I divide them like this, and that's equal to the tangent of theta.

[00:18:43]So, if I take the arc tan or the inverse tan, then I will get the theta here that I need in the polar coordinates, okay?

[00:18:49]Now, why does this work?

[00:18:51]Okay, I'm going to draw really basically the same picture right over here.

[00:18:54]Okay, here's your thing here, your x and y, and here is some point right here.

[00:18:59]So, there's an angle theta involved, and there's r, okay? And what you're saying here is let's talk about the first part here. This is the easier part to understand. r squared, which is this distance squared, is equal to x squared plus y squared. That is nothing more than the distance formula, okay?

[00:19:14]Remember what the distance formula is?

[00:19:15]You know, you might have learned that in algebra. The distance formula is equal to in any the distance between any two points in in the rectangular system is the square root of delta x squared plus delta y squared. The difference between the points in x and the difference between the points in y, square them, add them, take the square root. This is exactly the same thing. It's just I didn't write the square root here. I'm writing it as r squared is equal to x squared plus y squared. Because you're always measuring this from the origin, that's just the definition. The point you have here is always the the length r is always from the origin, so delta x and delta y are just going to be equal to the values of x and y of the point, okay? You square them, uh you add them up, you take the square root, and you'll get R, okay? Uh you'll get R, okay? And another way to think about it, if you if you want to think about it, is also this is the equation of a circle.

[00:20:01]In uh this is the equation of a circle, if you remember back to algebra, centered at the origin, x squared plus y squared is r squared. R is the radius, and that's exactly what's going on here.

[00:20:10]So, I'm telling you two different ways to remember. This is pretty easy. You take You plug in your x and your y, and you get your r back, okay? You take the square root and get your r back. This is also pretty easy, because if you look here, okay? This is If this is a triangle, this is the side of a triangle, and this is the side of a triangle. Now, just the definition of the tangent uh with a right triangle here, cuz remember this is a right triangle, tangent of theta is equal to opposite over adjacent.

[00:20:34]Opposite is y, cuz this is y, and adjacent is x. So, all I've written here is is the exact definition of what the tangent is in terms of a triangle.

[00:20:43]Tangent of theta is opposite over adjacent. That's all it is. The only reason that this one is a little bit tricky is you really have to be careful about your quadrants, okay?

[00:20:51]Uh and we'll see it when we get to the problems, but basically, when you're given an x and you're given a y, okay?

[00:20:57]Remember back to the very first section on the on the inverse tangent function.

[00:21:00]So, what you're going to end up doing is you're going to divide x by y, and you're going to take the arc tangent to get the angle back. Well, arc tangent, okay? Arc tangent only returns returns angles uh in the following range.

[00:21:20]Theta from negative pi over two to pi over two.

[00:21:27]Okay? It only returns So, in other words, if you go to your calculator and you put any any very any ratio of numbers here, and you take the arc tangent, it's only going to return values in this range. Even if the angle actually lies in another quadrant.

[00:21:39]That's just You can go back to the inverse tangent and the inverse um trigonometric function section in the in the DVD and read and kind of watch and see why that's the case. But, the arc tangent in a computer, in a calculator, or whatever, is only going to return these values here. And these values, by the way, if this is x and y, negative pi over two is down here, positive pi over two is up here. So, it's really going to only return values in this half of the plane, okay? So, when I divide y by x, okay? If I get a positive number, y divided y divided by x is a positive number, the angle it's going to return back is always going to be in this quadrant, because it's positive. And if if whatever I divide by here ends up being negative, I'm always going to get an angle down here, okay? But, depending upon the actual values of x and the actual angle might be over here, and the only way to figure that out is going to be to actually look at the x and y values and figure out what quadrant you're in. So, that may sound a little complicated, but we'll walk through it several examples, and I'll show you it's not that hard. It's just the main thing to remember here is when you divide y by x and you take the arc tangent in your calculator, or if you do it, you know, in your head, uh just make sure the angle's right based on the values of x and y. That's all I'm trying to say here. You got to make sure, cuz your calculator's only going to return this, your actual angle may be over here, depending on the values of x and y.

[00:22:54]Okay. You can do this stuff all day, but it really is helpful just to work some problems. So, let's do that now.

[00:22:59]Okay, what we're going to do is we're going to plot the point in these problems. We're going to plot the point, okay? Which is going to be a uh polar point, okay? And we're also going to find another equivalent uh polar point, polar representation of the point. Okay? And all I'm trying to say here is is, you know, we already talked about there's infinite number of ways to write a polar point. You can keep adding 2 pi uh around and around and around, or you can subtract 2 pi, and you're going to get another representation equally valid. That's all I'm trying to say there. So, let's go ahead and do it.

[00:23:42]Let's go ahead and say that we have a point. These are polar points, 1 comma pi over two. And you were asked to plot that, okay? So, how would you do it?

[00:23:54]Well, here is xy, okay? The first number is the radius, and the second number is the angle, pi over two. Pi over two radians is right up here on the y axis, and the radius was said to be one, so this point is exactly right here.

[00:24:12]Okay? So, the point is 1 comma pi over two. And the reason it's up there is because the radius is one, so I'm labeling this radius as one, and it's up here at an angle of pi over two. So, I only took two numbers to plot that point. Now, if I wanted to find another representation of that, all I got to do uh is add 2 pi to that, right? Or I can subtract 2 pi. I can do lots of things, but I'm just going to go ahead and add 2 pi. So, if I wanted to do that, it would be 1 comma pi over two plus 2 pi.

[00:24:39]And I wanted to go ahead and add that up, it would be 1 comma Let's go ahead and add this up. Pi over two plus Now, I'm going to write this in terms of 4 pi over two, because I want to be able to add these fractions. So, 2 pi is 4 pi over two. I just rewrote it so I could add them up, okay? And this is going to be equal to 1 comma pi plus 4 pi is 5 pi over two. So, the question just said find another representation. It didn't say find them all, because you can't find them all.

[00:25:08]You can You know, you'll you'll you'll be going on and on forever. Now, look here. 5 pi over two. How would you plot that if you if you were given this?

[00:25:14]Well, here's pi over two, 2 pi over two, 3 pi over two, 4 pi over two, 5 pi over two. So, you see it's the same point.

[00:25:22]It's the same place. That's why they're they're valid. So, these two representations are exactly uh equal exactly equivalent, I should say. No Neither is is is better than the other.

[00:25:31]Okay?

[00:25:32]So, what if I were given a point that's 4 comma negative 2 pi over three, and I wanted to plot that point.

[00:25:40]And that's a polar uh the polar representation of a point. Well, if I draw my coordinate axis here, x and y, uh then basically, I need to start looking at negative 2 pi over three.

[00:25:53]Now, remember, uh any multiple of pi over three is is going to be like 60 degrees, okay? So, and it's negative, so we're going to start going this way. So, this is roughly 60 degrees. That's negative uh 1 pi over three, and then this one over here will be negative 2 pi over three. So, it's going to be along basically along this direction here, and the distance uh is four, so you go 1 2 3 4 units out along this vector here. So, this is the point right there. So, the angle is negative pi over three, negative 2 pi over three, and it's four units out. That's what you're doing right there. The angle the uh the point there is 4 comma negative 2 pi over three. And if you wanted to find an equivalent version of that of this point, easiest way to do it is just to add 2 pi. 4 comma negative 2 pi over three, and we can just add 2 pi to it.

[00:26:40]It's easiest thing to do.

[00:26:42]Okay? So, it's going to be 4 comma negative 2 pi over three plus Now, I want to write this 2 pi over three in terms of something Now, I want to write this 2 pi in terms of something uh right here that I can uh add together. So, I write it with a 6 pi over three, which if you do this division is equal to 2 pi. I just wrote it that way so I could have this common denominator. And so, the final answer would be 4 comma negative 2 pi over three plus 6 pi there is going to give you a 4 pi over three.

[00:27:16]4 pi over three. And so, the answer would be 4 comma 4 pi over three, uh which is the you know, which is the equivalent form of this. So, these are representing the same point, you know?

[00:27:26]And looking back at my drawing here, I mean, it's a little bit misleading. It's I'm doing this all freehand. I mean, pi over three is roughly 60 degrees, which is which is here. So, negative pi over three is here. Negative 2 pi over three is probably going to be something more more along the lines of this, instead of kind of flatter like that. So, just in case you're wondering why I drew it that way, it's just because I'm looking at the board at an angle like that.

[00:27:47]Now, just a minute ago, we were looking at polar points, and we were plotting them on the graph just so you can get a feel for that. And we were also finding the equivalent representations by adding or subtracting 2 pi or 4 pi, or whatever you want to do to get an equivalent representation. Now, we're going to practice on converting from one coordinate system to another. And first, we're going to start it with problems where I'm giving you the polar representation, r comma theta, and then we're converting to Cartesian. And you'll see that it really isn't all that hard, okay? So, if I had given you a point square root of two comma pi over four. This is the radius. It's just a number. This is the distance from the origin, and this is pi over four, which is 45 degrees, okay? So, if I wanted to draw that, just to give you kind of a flavor for what it looks like, okay?

[00:28:31]This is x. This is y. It's 45 degrees, which is pi over four. And this point, I would just define it uh to be uh square root of two distance from the origin, okay? Whatever that is in your calculator. I think it's about 1.4. So, this would be square root of two comma pi over four. So, it's exactly uh unique representation of this point right here, okay?

[00:28:53]All right. And this would be a distance of square root of two from the origin, okay? Now, if I want to do that, if I want to convert to find the x and y values here, remember remember back from the previous uh board that x is equal to r cosine of theta, and y is equal to r times sine of theta.

[00:29:11]Uh you really should burn these in your head if you haven't already. You're going to use them so much. Uh it's really useful just to go ahead and remember them. So, the x coordinate is going to equal r, which is square root of two, times cosine of theta, which is pi over four.

[00:29:24]So, it's going to be square root of two times What is cosine pi over four? You should remember that that's equal to square root of two over two. Okay?

[00:29:32]So, x is going to equal uh I should I should say uh over here, square root of two times square root of two is square root of four over two, and because I multiply two times two under the radical, and square root of four is two over two. So, obviously X is going to be equal to one. That's the X value, and Y is going to equal to square root of two times the sine of the angle, which is pi over four. And in this case, it's going to give you the exactly the same thing.

[00:30:01]Square root of two times sine of pi over four is also square root of two over two. So, we've already done all this uh stuff, and we can say that Y is also equal to one.

[00:30:11]Okay? So, this representation here uh this representation here at which is square root of two as a radius at pi over four is exactly equivalent to saying that X is equal to one and Y is equal to one. So, if I were to go off on a sheet of graph paper and mark a point one {comma} one, okay, and then measure the distance to the origin, I would get square root of two, uh which is a a little bit greater than one. If you take the square root of two, it'll be greater than one. That kind of makes sense from this picture. It should be greater than one, okay? So, you see, converting from polar to Cartesian is easy. You just plug it in there, as long as you know how to take the sine and the cosine. Uh that's all you have to do, and it's it's pretty bulletproof, okay?

[00:30:52]Let's do another one.

[00:30:54]Okay, let's say I was given a polar representation, which is 1.5 {comma} uh 3 pi over two.

[00:31:02]3 pi over two.

[00:31:04]Well, X is equal to R cosine of theta, which is equal to R, 1.5 times cosine of 3 pi over two.

[00:31:15]Okay? What is the cosine of 3 pi over two?

[00:31:19]Uh well, cosine 3 pi over two, uh if this is your coordinate system, this is pi over two, 2 pi over two, 3 pi over two.

[00:31:27]So, the cosine of that is zero cuz it lies along the X axis, right?

[00:31:34]Okay, so obviously X is going to be equal to zero.

[00:31:38]X is equal to zero, okay?

[00:31:40]Now, what would Y equal to? Y equals R times the sine of theta.

[00:31:46]So, it's 1.5 times the sine of 3 pi over two, and we just saw that that if this is your axis, this is pi over two, 2 pi over two, 3 pi over two. So, it's down here. So, uh this is going to be 1.5 times -1 because the sine of 3 pi over two is -1.

[00:32:03]Okay?

[00:32:04]And then that would simply mean that Y is going to equal -1.5.

[00:32:13]Okay? -1.5. [snorts] So, if I were going to write this down and just draw a picture of it, okay?

[00:32:19]X and Y, okay? X and Y. Uh what I would find here, let's just plot these these coordinates here. Now, I know what X and Y are. X = 0, okay, here's uh -1, here's -2.

[00:32:33]What I'm saying here is that zero {comma} -1.5, X {comma} Y, that's where that point is in the rectangular coordinates. Now, also let's look at the polar representation. It's 1.5 is the distance from the origin. That's exactly matching up with this 1.5, and the angle is 3 pi over two. Pi over two, 2 pi over two, 3 pi over two. So, you see, they represent exactly the same point.

[00:32:55]Now, given the final example problem of this type, if it were and I'm doing a conversion from four {comma} -7 pi over six, which is the polar representation, okay? Then, if I wanted to find the X coordinate, it's going to be uh it's going to be R times cosine of theta, which is R, which is four, times cosine of -7 pi over six. Now, you really have to to know what you know, how to do your trig to find your your sine and your cosine to be able to do that because if you look here, where is -7 pi over six?

[00:33:29]Okay, well, pi over six is kind of like 30°. That's how I remember it. So, here's roughly 30°. So, it's -pi over six, -2 pi over six, -3 pi over six, -4 pi over six, -5 pi over six, -6 pi over six, which is -pi, -7 pi over six. So, this is roughly over here. So, it's about 30° up uh from there.

[00:33:51]-7 pi over six is up there in that quadrant. Why do you need to know that?

[00:33:55]Because when you take this cosine, you're going to have to know that. Four times, what is the cosine of this? -7 pi over six. Well, cosine of of regular old pi over six is square root of three over two, but because we're in this quadrant, you're you're projecting on a negative cosine, it's going to be negative square root of three over two. You have to know how to count in radians around the circle, and you have to know how to do your trig functions. Uh and if you don't know, you should go pick up the trig and precalculus tutor to refresh on that. Um but you have to be able to do that uh kind of in your sleep in order to survive this. So, X is going to be equal to negative 4 / 2 is 2 times the square root of three. Okay? X is equal to -2 times the square root of three. That's factor number one. And then Y is going to be equal to R times the sine of theta.

[00:34:46]And R is again equal to four, and we have sine of -7 pi over six. Now, again, the angle is in the same place. Sine of regular old pi over six, if it were over here, is equal to, because it's sine of 30°, roughly, is 1/2. So, sine of pi over six is 1/2, but you're in this quadrant over here, and so you're projecting against a positive Y. So, it's going to be the same thing, positive 1/2. So, Y is going to be equal to four times positive 1/2. You have to know where this thing lies in the XY plane to be able to figure out if uh your sine or your cosine is going to be negative or positive. So, Y is going to be equal to two.

[00:35:24]Okay? So, the angle or the uh I should say the Cartesian representation is at -2 times the square root of three {comma} two. If you were to plot that, then you would find uh exactly where that point lies, and this representation is exactly the same as this. Uh it's just written a little bit differently.

[00:35:43]Okay, now just a few minutes ago, we did several problems where we were given the polar representation. We put into our uh relations for X and Y, and we were able to arrive at the X and Y coordinates for the polar representation of that function. So, we're going from polar to uh rectangular, polar to Cartesian here just a minute ago. Now, we're going to do the opposite. I'm going to give you the uh rectangular version, the Cartesian version, and I'm going to ask you to give me the polar representation.

[00:36:09]So, that's what we're going to do next.

[00:36:10]So, if I had given you a rectangular uh point at two times the square root of three {comma} -2.

[00:36:21]These is This is just an X value and a Y value. That's all that is, okay?

[00:36:24]Now, the way you do that, remember, to find the R, which is the the distance from the origin, it's R squared is equal to X squared plus Y squared, okay?

[00:36:37]R squared is equal to X squared plus Y squared, okay?

[00:36:40]So, R squared is equal to What do I have here? I have two times the square root of three squared plus -2 squared.

[00:36:51]Now, what do I have in here? The two squared is going to give me four, and the square root of three squared is just going to give me three.

[00:36:58]Okay?

[00:36:59]And -2 squared is just going to give me positive four. So, R squared is equal to 12 plus four, uh which is going to give me 16, okay?

[00:37:09]And then to actually find what R is, which is what I'm going to need, I just take the square root of 16, which is four.

[00:37:15]So, R is equal to four. So, that's I'm halfway home. I'm done. Now, one thing I want to point out to you here is that anytime you take the square root of a of a both sides of an equation, you really should have a plus or minus here. So, this would be plus or minus four, but uh you know, since you're talking about a distance from the origin, you can't have a negative number there. So, it's one of those solutions you throw out. So, if you're one of these people that's used to putting plus or minus when you take a radical over something, that's right, but you just throw out the negative because you can never have a negative uh distance when you're talking about polar coordinates. It's always a positive number from the origin is is the short answer, okay?

[00:37:48]>> [snorts] >> So, that is R. That's halfway home, literally halfway home. Now, what we have to do is we have to find the angle.

[00:37:54]The way you do that is you remember that it's tangent of theta is equal to Y over X, okay? So, tangent of theta is equal to Y was -2, and X was two times the square root of three.

[00:38:10]Okay?

[00:38:11]So, what you can say, tangent of theta is going to be equal to -1 square root of three because you just cancel the twos, okay? So, all I have to do now is find out what this angle is, okay? Find out what that angle is, and and the way you do that is you take the arc tangent. You could put it in your calculator, but really you need to be uh come proficient at just doing this on your own. So, two things you need to do uh do here. You need to find out what the angle is that's going to give you this, and then you have to double-check your answer against the quadrant that you know that this point is in. Now, just from even without doing this, what quadrant is this guy in? I mean, you know, if this is XY, okay? I have a positive number for X.

[00:38:51]It's going to be somewhere over here, two times the square root of three, and -2. That's somewhere over here. So, this point is going to be somewhere over in this quadrant down down here, okay? The the point is going to be somewhere in that quadrant down over there, uh and you just want to make sure that whatever angle you get is going to give you uh that is the right answer.

[00:39:08]Okay, it's going to give you the the right the right quadrant. And there are going to be some cases when your calculator will not give you the right quadrant. So, that's why you have to be careful.

[00:39:15]So, let's just say uh let's do you you basically the way you do these is you start marching around the unit circle and plugging in values until you find the right value.

[00:39:24]Uh so, let's just say if theta is equal to -pi over six, what will we have? Okay?

[00:39:31]If I did that, remember, tangent is equal to sine over cosine, so it would be sine of -pi over six, if this were the angle, if this were the correct angle, over the cosine of -pi over 6.

[00:39:43]Now, let me evaluate that. What is sin of -π / 6? -π / 6 uh roughly right here.

[00:39:50]Uh so, it would be uh -1/2.

[00:39:53]Because it's basically sin of 30° down here below projecting on the negative here. So, it's negative uh 1/2. And then on the bottom, cosine of the very same angle, cosine of -π / 6, it's projecting on the positive x axis, so it's going to be positive. And you should remember from your trig it's square root of 3 over 2. Okay? Now, the twos cancel with the twos, right? So, what if this were the correct angle, the uh what I would get here by taking the tangent, sin over cosine is a tangent, is -1 over square root of 3.

[00:40:23]So, what I'm trying to say here is I shortcut a couple things because if I'm doing this, you know, and I hadn't worked it before, and I know that tangent of a number is equal to -1 over square root of 3, I'm going to start picking values of of theta, plugging them in to the tangent, and seeing when I get something that agrees. So, if theta is -π / 6, we found that it does in fact give us uh what we said it should give us, okay? Now, this this could be the right answer for theta, but the other thing we have to make sure is does it agree? Okay? Does the Does the quadrant agree? Okay?

[00:40:55]Look at this. Where would this angle be?

[00:40:57]Negative -π / 6. Well, positive angles are this way, negative angles are this way. π / 6 is roughly 30°, so that does tell me that it's right there. So, it's in the right quadrant. You see, this angle -π / 6 lies down here. It exactly agrees with the quadrant that that the point is actually in, x {comma} y. So, because they agree, this is your theta, and this is your r, okay? So, you would say 4 {comma} -π / 6 will be the answer to that uh question.

[00:41:25]And we're going to get to some problems in a minute where maybe your first guess at the angle doesn't actually uh or isn't exactly right, uh even though it may look like it's right, and I'm going to show you how to identify that.

[00:41:43]So, let's do that here.

[00:41:45]Let's say your point is -2 {comma} square root of 3 {comma} 2.

[00:41:51]That is your rectangular point. Now, I want to draw your attention to something. Let's go back and look at the previous problem. In this problem, x was 2 {comma} square root of 3, y was -2. In this problem, it's the same numbers, but now I've got negative for x and positive for y. I'm choosing this problem on purpose because I'm trying to illustrate something to you. You see, it's the same numbers, it's just that one uh the negative is in a different place. Uh over here, y was negative, and over here, x is negative, but the numbers are the same. So, we're going to go through the math in a minute, but because of the way we have defined r as x² + y², I think you can convince yourself, we'll do it in a second, I'm not shortcutting you here, you're going to get the same value of r, okay? So, that part's okay.

[00:42:33]And then when you come down to this tangent, you're going to be dividing y by x, you're going to be dividing the same numbers, it's just that in this case, a different the different a different one is negative than in the previous problem. But because you're dividing them, you're still going to get -1 over square root of 3, okay? So, you're going to get uh you're you're going to be tempted to calculate the same angles, all I'm trying to get at here. Uh but you're going to see as we go through it that this angle is actually not correct for this problem.

[00:42:59]It's correct for this problem, but not for this problem, even though really the math is almost the same. So, let's do that. That's sort of a prelude more than anything else.

[00:43:07]To find r, it's r² is equal to x² + y².

[00:43:13]Okay? x² is going to be this guy, which is -2 * the square root of 3 squared plus 2 squared.

[00:43:22]So, r² is equal to -2 squared is 4 * square root of 3 squared is just 3 cuz I dropped the radical. Plus 2 * 2 is 4.

[00:43:33]So, r² is going to be 12 + 4, which is 16, and so r is equal to 4 when I take the square root. So, you see, we did get the same radius. Okay, we did get the same radius.

[00:43:44]Uh and that's because these problems are very similar. Now, to find the angle, we just remember that the tangent of theta is equal to y over x.

[00:43:52]So, the tangent of theta uh is going to be equal to y, which is 2, over x, which is -2 * the square root of 3.

[00:44:01]Okay? So, it looks incredibly similar, it's just that in the last problem, the minus sign was up here instead of down here. But in both cases, I end up concluding that the tangent of this angle, whatever this angle is, is got to be equal to -1 over square root of 3, okay? -1 over square root of 3.

[00:44:17]So, the way you find these angles is you start marching around the unit circle calculating the tangents until you get the right value here. That's how you find the angle. But I'm cautioning you that when you do that, you just need to make sure that the angle you picked is in the right quadrant for the problem that you have. So, we know from looking at our point that we know what quadrant it's in. Negative x, somewhere over here, let's say, positive y, which is somewhere up here.

[00:44:44]So, this point is actually up here. In the previous problem, the point was down here. You see why that was the case? The reason is because uh one of the points is one of the coordinates in x and y, one of the x's were in the last problem, uh x was positive, y was negative. In this problem, x was negative, y was positive. So, it kind of maps it to the other quadrant.

[00:45:06]But when you divide y by x, as long as one or the other is negative, you're still going to get the same uh value here uh that the tangent of theta is equal to -1/3. So, the way if you didn't have the benefit of this lecture in front of you, you might just uh put this in your calculator, and remember that the that the tangent that the arc tangent in your calculator is only going to return values from -π / 2 to π / 2.

[00:45:31]So, that's -π / 2 to π / 2. So, it's only going to return angles over here, and it's going to give you an angle over here uh for this problem because here you have x is positive and y is negative. So, the tangent of the theta is going to give you a negative right there, and it's going to it's going to give you an answer in this quadrant, which is what the other problem did, and that problem's not exactly right. So, just going through the motion cuz I want to make it complete, if you didn't have the benefit of this lecture, you might say, "Well, maybe theta is equal to -π / 6. Maybe that's the case, okay?" Let's see if that's right. Then you would say, "Okay, well, tangent is sin over cosine, so sin of -π / 6 over cosine of -π / 6 is equal to what? Sin of -π / 6, we already did this stuff before.

[00:46:18]Uh sin of π over -π / 6 is -1/2, and we just did this before. Uh cosine of -π / 6 is like cosine of -30° is uh positive square root of 3 over 2.

[00:46:31]Okay? And we get excited here because we see the twos cancel, and this is going to equal -1 over square root of 3. So, you see, you might do this, and you might say, "Oh, look, I found the answer because I found an angle so that if I take the tangent of this angle, I actually get back the -1 over square root of 3. So, I'm done. I'm I'm finished. I'm I've completed the problem. The answer is 4 {comma} uh and the angle would be -π / 6." But look at what you've done here. You found an angle here, -π / 6. But we know from the original coordinates that it doesn't lie over here. It lies way up here. So, if you calculate an angle, uh or if you dump it in your calculator, and you're going to get an angle in one of these quadrants here, uh and and if you know by looking at the point, and you should always look at the point, that the actual uh point is in the opposite quadrant, all you have to do is take the angle you found, which does give you the right tangent, by the way, okay? And just add π radians to it. You don't want to add 2π radians because that would just map it all the way back to where you started. You want to take it here, and you want to add π to it, which is going to map it in the quadrant you want.

[00:47:37]That is the short story there. So, all you have to do uh is you'll say theta is going to equal -π / 6 + π. You're just going to add π to it, and that's going to map it in the right quadrant. So, it's going to be -π / 6 plus, and I got to write π as 6π / 6 because I need a common denominator here. So, then theta is going to equal -π uh plus 6π is going to be 5π / 6. So, this is the angle, and this is the radius.

[00:48:10]Okay. So, 5π / 6. Now, does that make sense to you here? 5π / 6, this is roughly 30°.

[00:48:17]π / uh 6, 2π / 6, 3π / 6, 4π / 6, 5π / 6.

[00:48:23]And my drawing's not exactly to scale, but it's it's it's it's roughly right, okay? So, short answer is, when you're doing this part where you're converting from rectangular to polar, getting the radius is easy.

[00:48:36]x² + y² uh gets your value of r, but when you take the ratio of y to x and you calculate something, and you're taking that arc tangent, go ahead and find an angle that works, but when you find that angle, always compare what you found, the quadrant that it's in, to the actual quadrant of the point that you have, and if they're in the same quadrant, like in the previous problem, you got the right angle. But if the quadrant is mapped on the other side like this, then you're going to have to go back and add π to it or subtract π to from it, it's the same thing, to get the proper angle. If you put this angle on your test, it will be wrong because this is totally wrong point.

[00:49:13]Okay, now the final problem is, what if I give you a rectangular uh point, and that's going to be uh negative square root of 2 {comma} negative square root of 2. This is rectangular, x {comma} y, and I want to find the polar representation. Well, the first thing you do is you try to find r.

[00:49:32]So, r² is equal to x² + y². That's just the distance formula.

[00:49:37]So, this is equal to negative square root of 2 squared plus negative square root of 2 squared because here's y, okay? So, r squared is equal to negative times negative gives me positive. Square root of 2 squared is just positive 2 plus same thing here, positive 2. r squared is equal to 4 and then r is equal to 2, taking the square root of both sides. So, I found my radius.

[00:50:03]Finding the radius is always really simple to do. You're always going to be able to find that really quickly. Now, what we need to do is find the angle.

[00:50:10]So, remember the tangent of theta is equal to y over x, okay?

[00:50:16]So, the tangent of theta is equal to y, which is negative square root of 2 over x, which is negative square root of 2.

[00:50:25]Okay? So, the tangent of theta is equal to 1.

[00:50:30]Okay, the tangent of theta is equal to 1. So, all I have to do now is go find an angle. Okay, go find an angle uh where uh the tangent of that angle gives me 1. And I think I have an angle in my head right now that I think may be the right angle, let's find out. Let's say if uh let me change colors here just to make it clear.

[00:50:53]Let me say and this just just popped in my head, let's just say theta is equal to pi over 4.

[00:50:59]Let's see if that's the right angle, okay? Tangent is equal to sine over cosine. So, sine of pi over 4 over cosine of pi over 4. Okay? Let's see what this is. Sine of pi over 4, you should remember, is the square root of 2 over 2.

[00:51:15]And the cosine of pi over 4, you should remember, is the square root of 2 over 2. So, divide these out, you get 1. So, I have found an angle, pi over 4, so that if I take the tangent of it, which is sine over cosine, what I get in the end is 1. So, I really think I've got the answer because tangent of this angle is 1. If I put this in my calculator, tangent of that angle, I will get 1 back. So, I'm tempted to circle this, but the final step is always to go back and look at the quadrant.

[00:51:42]Okay? So, here is your coordinate axis, okay? Now, look at the point I have.

[00:51:47]It's negative square root of 2, which is somewhere over here, negative, and that comma negative square root of 2, which is somewhere over here, negative. So, I'm over in this quadrant here somewhere, okay?

[00:51:57]Now, look at the angle I found. I found a positive pi over 4, which is squarely over here.

[00:52:03]Now, if you put uh if you put 1 in your calculator and took the arc tangent or the inverse tangent, you're going to get pi over 4 because remember, calculators and computers are only going to return angles over on this side of the uh on the that side of the plane there, okay?

[00:52:17]So, I got pi over 4 because that was my first guess, but the actual point is nowhere close to that. Uh it's actually pi radians away. So, I need to add pi to mark to map the angle I found all the way over here.

[00:52:30]Okay, so to find the answer, all I have to do, say theta is equal to pi over 4 plus pi. I'm just going to take it and I'm going to add pi to it.

[00:52:40]So, that's pi over 4 plus and I'm going to have to make this pi 4 pi over 4 in order to add them cuz I want a common denominator. So, the angle is going to be pi plus 4 pi is 5 pi over 4 and that is the angle. And to check yourself, let's look at it. 5 pi over 4. Let's start counting. Here's 1 pi over 4, 2 pi over 4, 3 pi over 4, 4 pi over 4, 5 pi over 4, right here. That's the quadrant that you're in. So, we do a quadrant check with the angle we found versus that. And if you just want to double-check yourself, I mean, if you look at the at the sine and the cosine, if you were to plug it in here, okay? What is if I want to take the tangent of it, what is the sine of 5 pi over 4 divided by the cosine of 5 pi over 4? What is that going to be equal to? Well, the sine of 5 pi over 4, which we know is down here, the sine's going to be negative. So, it's going to be negative square root of 2 over 2.

[00:53:39]And the cosine again is going to be mapping it over here, so it's also going to be negative.

[00:53:44]Negative square root of 2 over 2.

[00:53:46]Negative divided by negative gives me a positive, so this is going to give me a positive 1. So, what I've found here is the sine over cosine, which is the tangent, of this angle that I have found is equal to 1, okay? I'm proving it to you. I'm showing you that this angle also has a tangent equal to 1 and it's in the proper quadrant. So, you have to be on the lookout for that. So, in this section, we have done a good uh introduction to polar coordinates and comparison to the rectangular coordinates that you've been using pretty much forever. There are certain applications, very common applications.

[00:54:19]You're going to have to get comfortable with polar coordinates, so don't don't uh uh you know, don't just breeze through this and think, "Well, I probably won't use this." This polar stuff is actually very, very important and you'll probably use it uh all throughout your school and definitely beyond school, too. Okay, so basically consists of two numbers, just like x and y, except it's r {comma} theta. A distance from the origin is r, theta is just the angle uh from the uh from the axis. And you can have many different representations because you can continue adding 2 pi or subtracting 2 pi or 4 pi or 6 pi uh to get different angles.

[00:54:52]We also learned how to convert polar to rectangular and rectangular to polar, back and forth, and we have shown several problems to do that here. So, those are uh central skills that you're going to need to tackle the remaining sections in this class and also in your uh in your furthering your education and also in your career. Learn anything at mathandscience.com.