From cartesian to polar coordinates

Let’s look now in the other direction. How do you go from Cartesian to polar coordinates? Well, here’s a Cartesian plane, and x, y are the Cartesian coordinates. And I’m going to place the pole in the polar axis in the canonical configuration along the positive x-axis. How do you find the polar coordinates? Well, the r is pretty simple because it’s the distance from the origin to x, y, and we know how to express that. Here it is. So that’s r in terms of x, y. As for theta, well, the angle theta, if you look at that same triangle that was already useful to us, you see that the tangent of theta is equal to y over x, opposite over adjacent.

Therefore, theta is the arctangent of y over x. That’s a simple formula. But careful, there’s two things about it. The arctangent has values in minus pi over 2, pi over 2. So if theta is not being considered in that range, this won’t work Furthermore, as you can see, the formula for theta is not going to apply if x is 0. So really, this is a good, useful formula if you’re in the positive half plane– that is, the half plane where x is strictly positive. These formulas work, and they give you an answer. But you’re going to need different formulas if you’re considering other parts of the plane.

In fact, there’s no single nice formula that will work everywhere in the plane. Let’s see why. We’ll get some idea of why by considering this. If I give you a point that looks like it’s in the first quadrant, for instance, generally speaking it corresponds to an angle theta between 0 and pi over 2, say. If I now give you a point that seems to be in the fourth quadrant, well, it corresponds to theta between minus pi over 2 and 0. As you know, negative theta just means you measure out theta by going clockwise rather than counterclockwise. Now, consider a point in the third quadrant. It’s just a point in which I’ve gone around clockwise a little further.

The theta would be between minus pi and minus pi over 2. Similarly, the point in the second quadrant is where I kept going counterclockwise a little further from my very first point, and so the theta is between pi over 2 and pi. But now, what happens if I give you a point that is on the negative x-axis, so to speak? Is it of the form r, pi or r, minus pi? If you consider that it came from above, then it’s of the form r, pi. Theta is pi. If you consider that it comes from below, then the theta would naturally be minus pi.

Now, of course, it’s a little bit the same thing because pi and minus pi differ by 2 pi, but you see that there is going to be a definition problem along this negative x-axis and whether you should take pi or minus pi. You’ll need a convention, and there are different ones. And furthermore, the choice you make might depend on the context of the problem you’re considering. So the answer is both possibilities might be useful. There’s even some interest in allowing negative values of r, as we’ll see in a few minutes. To begin with, though, let’s look at a typical problem involving polar coordinates. We have a point whose polar coordinates are specified.

We want to find the Cartesian coordinates, and then we want to find all other possibilities for the polar coordinates of the same point. Well, we measure out the angle pi over 3 in the usual way from the polar axis counterclockwise, and then we go a distance 2 along that angle and we reach our point, which has polar coordinates 2 pi over 3. How do we find the Cartesian coordinates? Well, it’s the formulas from before– r cos theta, r sin theta. They give us the values of x and y, and we find that the Cartesian coordinates are 1 and the square root of 3. Now, what are the other polar coordinates for this very same point?

Well, clearly, we could change the angle pi over 3, the theta here, by adding a multiple of 2 pi, say. That means it would just go around counterclockwise once and wind up at the same point. Similarly, it would be possible to subtract a multiple of 2 pi. That just means we go around clockwise once, and we wind up at the same point. So more generally speaking, the angle pi over 3 could be replaced by pi over 3 plus any integer multiple of 2 pi. So the k here would be positive or negative integer a point in Z. Is that all the possibilities? Well, there’s one other that arises, and it’s the following.

Suppose we add an angle of pi to pi over 3. That brings us in theory to what we’ve called the antipodal point. Now, the antipodal point is not where we want to go, at least not if we measure distance positively from the pole. But what if we were to measure distance negatively from that antipodal point and minus 2 would bring us back to the point we started with? In other words, we’re ready to consider that another polar coordinate for our initial point is the point r equals minus 2 with the angle theta equal to 4 pi over 3. This will have us wind up at the same point.

And then, the 4 pi over 3 could be modified by a multiple of 2 pi. And so we come to all the possible polar coordinates of the given initial point that we had. That’s quite a few. Now, we’ll see in a moment why r being negative is sometimes a necessary thing in some applications. First, though, let’s look at one of the standard applications of polar coordinates– namely, sketching the graph of certain level curves that are naturally described in terms of those coordinates. Consider the graph of the level curve r equals 1 plus cos theta. We want to know what all of those points look like when you find them in the plane.

Well, we can start with theta equals 0, and you can see that r is going to be 1 plus 1– namely, 2. So the point 2, 0 in polar coordinates is a point on our graph. You can also see that the value of r is never going to become negative here, so that complication does not arise because the cosine is never smaller than minus 1. So r is going to be positive. Now, let’s start from our initial point 2, 0 and increase the value of theta. When theta reaches pi over 2, the cosine is equal to 0, and you can see that r is going to equal 1.

As theta continues increasing from pi over 2 to pi, then r is going to attain the value 0. Why? Because cosine pi equals minus 1. So then, you’re going to be at the pole, the point 0, pi. As you continue to increase theta from the value pi, you see that the cosine of pi plus something is going to equal the cosine of pi minus the something. That’s an identity we know. And therefore, you’re going to duplicate the previous values but for bigger values of the angle. And finally, you’re going to wind up at the point where you started when theta is equal to 2 pi. So your graph is going to have this shape.

By the way, because this shape is thought of as resembling the shape of a heart, this graph– famous graph– is called the cardioid, from the Latin for heart. Now, let’s modify the example a little bit. I’ve multiplied the cosine theta term by 2. That’s going to make a difference because now the r is going to become negative, as you can see. Now, it’s true that you’ll start off with r being positive. When theta is 0, r is going to equal 3. So the point 3, 0 is on our graph. And you’re going to keep increasing theta. And you can see that, when theta hits the value 2 pi over 3, the cosine there is minus one half.

r will equal 0. In other words, that’s a point that’s going to get you at the pole. And now, as you keep increasing theta a little bit, r will become negative. And this is a very useful thing in this application because that’s what enables you to describe the inner loop in this graph, that r becomes negative after theta passes the value 2 pi over 3. You describe the inner loop, and then eventually you come back to the pole at 4 pi over 3. And then, you eventually come back to your starting point when theta is 2 pi.

The famous graph that you see here, by the way, is called a limacon because it presumably represents or looks like a snail shell. Now, polar coordinates will enable you to put your graphing application to work and generate some really nice pictures. For example, if you look at the graph of the function, or the level curve r equals a cosine 2 theta, where a is a positive number, that will be the graph of a four-petaled rose as it’s called– rather nice, n’est-ce pas? Or, if you look at the level curve r equals a theta, you will get one of the most famous curves of antiquity, the spiral studied by Archimedes.

Or, if you look at the graph of the level curve r equals a e to the b theta, a and b positive parameters, you will get a famous curve discovered by Johann Bernoulli, Euler’s tutor when Euler was a university student, and it’s called a logarithmic spiral. It has remarkable properties of self-similarity when you look at any scale of this graph. And Bernoulli was so taken with it that he wanted it to be engraved on his gravestone, but they made a mistake and they put the spiral of Archimedes instead. The moral of the story? Well, life can be difficult, but mathematics is beautiful. And in your life, I wish you a wealth of beautiful mathematics.