WEBVTT mathematics/pre-calculus/selhorst-jones
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Hi--welcome back to Educator.com.
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Today, we are going to talk about polar coordinates.
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Previously, whenever we have talked about the location of a point on the plane,
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we have described its horizontal and vertical distance from the origin, x and y--
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how much we go out horizontally and how much we go out vertically.
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We call them rectangular coordinates because, if we look at a horizontal and a vertical put together,
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we are just drawing out a rectangle on the plane; so they are called rectangular coordinates.
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Now, we are going to look at a totally new way to talk about coordinates; we are going to talk about polar coordinates.
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This gives us a new way to describe the location of a point.
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Instead of using horizontal and vertical components, we can talk about the point's distance from the origin--
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the distance from the center, and the angle that it is on--what angle the point is on.
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This gives us a whole new way to talk about location in the plane.
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Using it, we can create graphs like we have never seen before.
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We will explore these in the next lesson, Polar Equations and Functions.
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For now, though, let's work on a strong understanding of just what polar coordinates are and how they work.
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We really have to have a good understanding of how polar coordinates work, so that it is an intuitive thing,
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before we will really be able to use it in equations and functions; so let's get that learned in this lesson.
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All right, we plot points with polar coordinates using two things: r, the distance of the point from the origin--
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how far we are from the origin (and from here on, we are going to call the origin the pole; the center of the graph
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will be called the pole now, since we are talking about polar coordinates; and in a little while,
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we will see why it makes sense to be talking about the pole; and there is sort of vaguely a connection
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between north pole and south pole, like we normally hear the word "pole" show up when we are talking about the poles of the earth).
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All right, next we have θ, the angle that the point occurs on.
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We measure this counterclockwise from where we used to have the positive x-axis.
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Our positive x-axis used to go out to the right, and we measure counterclockwise from that, just like we measured angles in the unit circle.
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On the unit circle, we always measured angles by going counterclockwise from that positive x-axis.
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We always spun counterclockwise; so it is a lot like when we worked with the unit circle in that manner.
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We give points as the ordered pair (r,θ); it is distance first, then angle--(distance,angle).
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It is normally assumed that θ is going to be in the unit of radians.
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Degrees are used occasionally in polar coordinates, but much, much less often.
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Mostly, unless you see it really explicitly shown otherwise (there is a degree symbol or something else
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to make us definitely sure that we are talking about degrees), just assume that it is radians,
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because that will be the usual thing we are talking about when we are using polar coordinates.
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All right, how can we visualize polar coordinates?
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When we plot in rectangular coordinates, we usually think of a point (x,y) in one of two ways:
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the "go left/right by x," how much we go horizontally, and then how much we go vertically, "go up/down by y."
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We do these two things, and we get to some location.
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Alternatively, we can do the "go up/down by y" first; we can do that vertical motion first, and then our horizontal motion.
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But we end up getting to the same location.
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Either way is fine; both go to the same place.
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Likewise, there are two ways to visualize polar coordinates; so let's see both of them.
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First, we will look at visualizing r, then θ.
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The first thing we do: we can go out a length of r from the pole, what we used to call the origin,
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the center of the graph, directly to the right.
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So, if we have...here is our center, where I am holding this ruler.
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Here is our center; and what we do is have some distance r that we then go out to the right along from that pole.
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We go out a distance of r to the right from that pole.
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The next step is that we rotate the length counterclockwise by whatever our angle θ is.
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We spin that length that we just put out by that angle; we spin out that r by that angle, and so, we end up having the same distance r here.
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This distance here is r; that is what we are ending up seeing--we have just put out some distance, and then we spin the distance.
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And combining these two things, we arrive at some point (r,θ), some distance, comma, angle.
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Alternatively, we can visualize θ, then r; so first, we rotate counterclockwise by θ, and we create an imaginary line at that angle.
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We spin from that right x-axis, what we normally used to call our positive x-axis, but now it is just going right from the center.
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We spin up some angle θ; so what we will have is this little stub.
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We spin up some little stub; and so, going off in this direction is now going off in the angle θ.
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Next, we apply going out a distance r; so we then go out in that direction some distance r, and we achieve our point (r,θ).
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We are already at some angle θ, this imaginary line right here; and then we just end up going out the distance of r.
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We go out our distance r along that direction, and we end up getting out to (r,θ).
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We spin, and then we go out to the distance.
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Either of these two ways is just fine for visualizing polar coordinates--they both work perfectly reasonably.
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Whichever one makes a lot more sense to you, that is the way I would recommend visualizing it.
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For me personally, I prefer this method; I think it is easier to think in terms of what you spin to, and then how much you go out to.
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But if you think it is easier to think of going out the distance, then spinning, then fine; go ahead and use whatever makes sense to you.
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Also, now that we have some sense of how polar coordinates work, visualizing it,
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we can also see why we call them polar coordinates--why the idea of a pole makes sense.
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Imagine if you were standing on the North Pole.
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If we wanted to talk about any location on Earth, well, we could talk about it as:
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you stand on the pole; you face in some direction (you just choose a direction, arbitrarily, to face in);
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and then we can talk about every location on Earth as being how far you spin, and then how far you walk down to get to that location.
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Some place in South Africa: you just spin some angle, and you walk down to it.
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Some place in Uruguay: you just spin to some angle, and you walk down to it.
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Some place in America: you just spin to some angle, and you walk down to it.
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Wherever you are going to go to, it is starting at some pole; you spin around that pole, and then you walk a distance.
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Or, alternatively, you can walk a distance, and then spin around (effectively) to some different longitude.
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It depends on how you want to approach it; but I think it makes a little more sense to do spin, then walk, and so I tend to visualize it that way.
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All right, polar graph paper: normally, when we graph rectangular coordinates,
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we think in terms of these tick marks on the axes: 1, 2, 3, -1, -2, -3, positive 1, positive 2, positive 3,
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-1, -2, -3...to help us see horizontal and vertical distance.
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We can easily see that, at this point here, we have a distance of 2 horizontally and a distance of 2 vertically.
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But horizontal and vertical tick marks aren't so useful for polar.
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How far is the distance from here to here, based on these tick marks here and here?
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We can't figure out how far we are from the center, how far we are from the pole,
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based on these locations of these tick marks for horizontal and vertical tick marks.
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So, for polar, we need a new way of marking our graph paper so that we can see things more easily.
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We do this with concentric circles; we use concentric circles to help us see distance from the pole.
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The first circle (at least in this graph of concentric circles--concentric just means one, and then another around that,
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and another around that, and so on and so on)--this first one would be at a distance of 1.
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Anything on this circle is a distance of 1 from the pole.
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This next one: anything on this circle is a distance of 2 from the pole.
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That way, we can see easily how far you are by just seeing which circle you are on, or which circle you are closer to.
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It is like horizontal and vertical tick marks, but instead, it is for how far you are from the center of the graph.
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It is a new way of talking about it, which works really, really well for the r of (r,θ), for the r in the polar coordinates--the distance from the center.
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It works really well to think in terms of these concentric circles.
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Similarly, we need some way to be able to think about the angle θ.
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We use as a reference...we can talk about the angle θ with arc sectors; that is lines coming out of the origin.
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In this, we have lines coming out of the origin here and here and here, and we continue around in this way out.
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We see that it is cut a total of 12 times; so an entire revolution would be 2π;
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so an entire, cut...2π divided by 12 times that it has been cut would be π/6 for each one.
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So, here is π/6, and then π/3 (2π/6); 3π/6 would be π/2, and so on and so on and so on.
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So, we can see the angles here; cutting it up into these arc sectors lets us see that we are on this angle here, this angle here, or this angle here.
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And the concentric circles let us easily see our distance from the center.
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So, this works really well as a way to talk about polar graph paper, a way to easily see where we are on a polar graph.
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Now, we don't necessarily have to format our graph with concentric circles and arc sectors,
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just like we don't have to use tick marks when we are using rectangular coordinates.
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We can make a graph that doesn't have any tick marks on it, and still probably understand what is going on.
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But it often makes it easier; it will make it easier to graph things, so it is useful to have it.
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It is not absolutely necessary, and there will be some times where it is not worth it for us to put them down.
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But if we are trying to be really careful with the graph, it is a good idea to sketch these in first,
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so we can be really accurate when we are drawing it out.
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Something that is special about the polar coordinates is that, unlike rectangular coordinates, there are multiple ways to name the same point.
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There are multiple ways to name a single point in polar coordinates.
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So, what do I mean about this? Well, notice that the point (3,π/4) ends up giving us the exact same location as (3,9π/4).
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They are both going to end up being out at a distance of 3 on this third circle here.
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So, we can see that they are both going to end up being the same distance out; but why are they on the same angle?
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Well, think about this: π/4 ends up being here, but we can break 9π/4 into 2π + π/4.
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So, 9π/4 is just the same thing as 2π + π/4.
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What we have here is that 9π/4 is effectively spinning the first 2π; it is making an entire revolution,
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and then it is doing the last π/4 to end up being on this angle; and then, it puts the same distance out.
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Our first one is π/4 at a distance of 3; but then, the next one is 9π/4...
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so it just makes an entire revolution for the first 2π, but then that last part of 9π/4,
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after that 2π, will just be an additional π/4; so it ends up being at the same angle.
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We end up getting an equivalent point for polar coordinates; so this is something special about polar coordinates.
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We can end up lapping any given θ away; any θ where we can talk about a specific angle...
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we can lap it by adding 2π or adding multiples of 2π.
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This means that, for any angle θ that we give in polar coordinates, we can make an equivalent angle by just adding or subtracting a multiple of 2π.
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You add 2π; you just do one loop counterclockwise.
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If you add 4π, you do two loops counterclockwise; if you add -2π, you do a loop clockwise.
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However you end up adding a multiple of 2π, whether it is adding or subtracting it,
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you end up just getting back to where you started,
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so it has no real effect, just like when we worked with the unit circle.
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2π and 0 are the same angle on the unit circle; so adding or subtracting 2π has no effect on our location in terms of angle.
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If we want to avoid this--if we want to avoid being able to accidentally loop and show up at the same angle--
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we can restrict our θ to be between 0 and 2π, so it is allowed to be 0, inclusive,
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but not allowed to get up to 2π, since the 0 and the 2π match up.
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But often, we won't actually make this restriction; we very often won't use this restriction.
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It sometimes can be used, but very often, we will be allowed to go larger than 2π.
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And we are allowed to end up looping multiple times and then landing on the angle that we are actually going to end up using.
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We can have this restriction, but very often it won't be put on, so don't think that it is just going to always show up.
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There is also another idea that we can talk about: we can talk about negative values for r.
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So, when we talked about the idea of r, we just talked about r representing how far out we are.
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But we never required it to be positive.
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If it is positive, that makes sense: we go to some angle, and then we go out by r.
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But we need a way to interpret negative values for r, because what if our r wasn't positive--what if we ended up having a negative value for r?
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If we have a negative value for r, we do the following.
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If r is negative, we go in the direction opposite the one that angle θ points out.
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For example, in this one here, we have (-2,3π/4); so the first thing that we do is spin the angle to 3π/4.
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And that goes off in this direction here.
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But then, we have -2; so instead of going off in this direction, we go in the opposite direction, down to this way.
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So, the negative says to go in the direction opposite that which the angle does.
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It is like we are spinning to some angle, but then we end up going in the opposite direction, like this.
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We spin to some angle, but then we end up spinning in the opposite direction by whatever our r's absolute value is.
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Alternatively, this is the way of thinking "spin, then go out the distance"; if you really prefer thinking
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in terms of "distance out, then spin," you can think of r as going left.
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So, here would be a -2: r = -2 would go like this.
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And then, we just end up spinning--the same thing, just like normal--counterclockwise to 3π/4.
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So, as opposed to positive r coming out from our pole like this, we end up having -r come out the other way.
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And then, we just end up spinning, like usual, to whatever our point ends up being.
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This means that there is yet another alternative way to name the same point.
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We can add π to the θ and then make r negative.
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Why does this work--why does this end up giving the same point?
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Well, imagine that we had some point that we normally got to, with some θ and some r distance out.
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Well, if we add π to θ, we end up spinning to the opposite direction.
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But then, if we make r negative, we push back in the opposite direction once again; so we end up getting back to our original thing.
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π to θ puts us in the opposite direction, but negative on r puts us in the opposite direction.
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Opposite opposite means that we are back where we started; so combining these two things
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means that we have yet another way to express the same point.
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So, that is something we really have to keep in mind.
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When we are working with polar coordinates, there is not just one way to call a point if it is simplified, like there is when we are working with rectangular coordinates.
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We have to think about if this could be the same thing as something else.
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How about converting between rectangular and polar?
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At this point, for a point in the plane, we now have two ways to name it.
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We can talk about the rectangular (x,y) coordinates, how far we go out horizontally and how far we go out vertically.
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But we can also talk about the polar coordinates--how far we go out in distance from the pole and how much we spin that angle:
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the distance out we go, r, and then the angle that we spin.
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So, how can we convert between the two coordinate systems?
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They both end up calling out the exact same point up here; so how can we convert between the two?
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Well, let's just layer both of them down simultaneously.
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If we look at both systems mapping the same point, we see that they are both mapping the same point.
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So now, we just want to see how they relate to each other through this diagram--through this picture.
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Another thing to point out is that, since it is a rectangular coordinate system,
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we know that it has to have a right angle in the corner, because it is based on a rectangle.
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A horizontal portion and a vertical portion means that where they meet, they have to have a right angle right there.
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So, we end up seeing a right angle in the triangle we have made with r, x, and y.
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Great; a right triangle--what we have here is something that we can work with using basic trigonometry.
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Looking at both systems simultaneously, we can connect through basic trigonometry.
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This triangle right here has an angle in the corner and has some right angle in the other corner.
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It has the angle θ in this corner; it has the right angle in this corner; so this is a perfect chance to use basic trigonometry.
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It is good stuff that we have been learning since geometry.
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So, let's start working through these: we can relate θ to our other information:
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cos(θ) would be equal to the adjacent, divided by the hypotenuse; so that gets us cos(θ) = x/r.
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sin(θ) is going to be equal to the opposite, divided by the hypotenuse; so that gets us sin(θ) = y/r.
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r² = x² + y² because of the Pythagorean theorem.
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The hypotenuse, squared, is equal to the other two legs, squared and added together.
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So, we have r² = x² + y².
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And finally, tan(θ): if we take the tangent of θ, then that is the opposite over the adjacent, so that means tan(θ) = y/x.
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Great; so we can use the following equations to convert between coordinate systems.
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We figured out where they are coming from; so at this point, we can go from polar to rectangular.
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We had cos(θ) = x/r; we just multiply both sides, and we get x = rcos(θ); similarly, y = rsin(θ).
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On the other direction, rectangular to polar, we had r² = x² + y² and tan(θ) = y/x.
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Converting to rectangular is easy; if we want to convert to rectangular,
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all we have to do is plug in r and θ, and we will automatically get the right x and y.
00:17:29.400 --> 00:17:37.300
You just plug in r; you plug in θ; you work with the numbers, and you end up getting out what your horizontal x is and what your vertical y is.
00:17:37.300 --> 00:17:42.000
That part is pretty easy; converting to polar, though, is a bit trickier.
00:17:42.000 --> 00:17:49.000
We can't tell if r is positive or negative, because it is r² = x² + y².
00:17:49.000 --> 00:17:52.600
So, r could be negative; r could be positive; x could be positive; y could be negative.
00:17:52.600 --> 00:17:55.900
The squareds are going to end up turning everything into things that look positive.
00:17:55.900 --> 00:17:59.200
So, we can't tell if it ends up being positive or negative from its equation.
00:17:59.200 --> 00:18:05.300
Worse, the function tan^-1 can only output from -π/2 to π/2.
00:18:05.300 --> 00:18:11.800
Remember: the arctan, the tan^-1 function, only outputs on this side of the unit circle.
00:18:11.800 --> 00:18:15.400
That is one of the things that we talked about when we learned that in trigonometry.
00:18:15.400 --> 00:18:19.000
So, tan^-1 can only output from -π/2 to π/2, and this is a problem,
00:18:19.000 --> 00:18:23.500
because since we have tan(θ) here, if we want to figure out what just θ is,
00:18:23.500 --> 00:18:28.100
we are going to have to use tan^-1 on the way to figuring out what θ is.
00:18:28.100 --> 00:18:34.900
If we are going to solve tan(θ) and get to θ on its own, we need to use tan^-1; we need to use arctan on both sides.
00:18:34.900 --> 00:18:37.900
That means that we will end up being restricted to just seeing one side.
00:18:37.900 --> 00:18:41.300
So, converting to polar is this kind of difficult thing.
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r² = x² + y²: we don't know if r should be positive or negative from that information.
00:18:45.600 --> 00:18:49.100
tan(θ) = y/x: we don't really have a great way to talk about it.
00:18:49.100 --> 00:18:54.300
If it is in the very first quadrant, it is easy; but if it is in any of the other three, things get a little bit trickier.
00:18:54.300 --> 00:18:58.700
How do we deal with this? To get around these limitations, make sure you always, always
00:18:58.700 --> 00:19:05.700
(I am serious about this) draw a picture whenever you are converting from rectangular to polar coordinates.
00:19:05.700 --> 00:19:08.400
So, pictures are the way we are going to solve this issue.
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The picture does not need to be extremely accurate--it doesn't need to be a perfect picture.
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But it needs to be clear enough for you to see which quadrant the point you are talking about is in.
00:19:17.200 --> 00:19:22.500
And it will give you a sense of what angle to expect, a sense of what things should come out to be.
00:19:22.500 --> 00:19:29.500
It will be a sanity check that lets you see, "Oh, my answer is possibly right" or "my answer is obviously, clearly wrong."
00:19:29.500 --> 00:19:34.700
It lets you have some idea of what is going on; and we will see how useful this is in Example 4 and Example 5.
00:19:34.700 --> 00:19:39.700
Of course, it wouldn't hurt to draw a picture when you are converting to rectangular, either--more pictures are never a bad thing.
00:19:39.700 --> 00:19:43.700
But it is not quite as necessary; still, a visual aid always helps, so I would recommend:
00:19:43.700 --> 00:19:47.000
the first couple of times you do this, draw a picture; it will help you see what is going on
00:19:47.000 --> 00:19:50.200
and see the relationship between rectangular and polar coordinates.
00:19:50.200 --> 00:19:53.700
All right, we are ready to talk about some examples.
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All right, plot the points below: we have this nice diagram with concentric circles and lines (arc sectors) to cut things up.
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So, let's just mark out what all of our arc sectors are here, first.
00:20:05.900 --> 00:20:08.900
We see that there is a total of 12 pieces that it has been cut up into.
00:20:08.900 --> 00:20:20.800
It means that our first angle will be 0; here is π/6, 2π/6, so π/3, 3π/6 is π/2, 4π/6, which will be 2π/3, 5π/6,
00:20:20.800 --> 00:20:27.300
which is just 5π/6; 6π/6, which will just be π; 7π/6, which is 7π/6,
00:20:27.300 --> 00:20:40.500
8π/6, which is 4π/3; 9π/6, which will be 3π/2; 10π/6, which will be 5π/3; 11π/6;
00:20:40.500 --> 00:20:44.000
and we now wrap to 2π, so we are back where we started, at 0.
00:20:44.000 --> 00:20:48.600
OK, with that in mind, we can see how to plot points; let's starting plotting some points.
00:20:48.600 --> 00:20:55.000
The first point is (2,π/3); so if we are on 2, we will be on the concentric circle representing a length of 2.
00:20:55.000 --> 00:21:09.000
We go out to angle π/3; that is this one right here; so distance 2 from the center and angle of π/3 means that that is our point, right there.
00:21:09.000 --> 00:21:16.800
The next one we will mark in blue: (1,225°): like I said, degrees are sometimes used in polar coordinates.
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They are pretty rare, but we can deal with them; we can work between radians and degrees when we need to.
00:21:20.900 --> 00:21:23.300
But mostly, we will end up sticking with radians.
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But we do have to know how to deal with it.
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Where would 225 degrees appear? Well, 180 degrees gets us to π.
00:21:29.900 --> 00:21:38.700
π is at 180 degrees, and then another 145 degrees will get us to 225 degrees; another 45 degrees would be along this angle right here.
00:21:38.700 --> 00:21:45.900
So, we are going to just cut this arc sector right down the middle--the other arc sector is 30 degrees, so we can split down the middle with this one.
00:21:45.900 --> 00:21:53.800
We are at a distance of 1, so we are going to end up being here on this concentric circle; and we have that point right there.
00:21:53.800 --> 00:22:03.300
The next point (in green): (3,-5π/4): if it is negative on our θ, that means that instead of going counterclockwise, we will go clockwise.
00:22:03.300 --> 00:22:11.100
-5π/4 means that we are going to go to π; we spin to π here, and then we go an additional -π/4.
00:22:11.100 --> 00:22:20.100
So, we spun -π; that is clockwise π; and then we spin an additional -π/4, so that puts us splitting this arc sector in the middle.
00:22:20.100 --> 00:22:28.400
We end up having a distance of 3 away from the center: 1, 2, 3...the third circle out; and here is our point, right here.
00:22:28.400 --> 00:22:37.800
The next one: (2.5,7π/6); 7π/6: we mark out that that one will be this angle right here.
00:22:37.800 --> 00:22:40.600
So, here we are on this one; what is 2.5?
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Well, 2.5 is just going to be halfway between 2 and 3; halfway between 2 and 3 looks to be right around here to me, and that is our point.
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Next, we will go back to red, but this time we will mark it with a star after we get the point down.
00:22:55.500 --> 00:23:00.600
(-4,3π/4): the first thing we do is spin to 3π/4.
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Here is π/2; and then, here is another π/4; so we are on the same angle as when we figured out -5π/4.
00:23:07.500 --> 00:23:15.500
That makes sense: -5π/4 and 3π/4 are together there, making a total...meeting in the middle of 2π...one way of thinking about it.
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-4 means we are going to go, not out on this angle, but in the opposite direction.
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Negatives tell us to go in the opposite direction; so we are going out -4, so 4 out in the opposite direction.
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We are on the fourth circle; we are here, and we will mark that with a big star to help us see the difference.
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The final point: (3,1.47): this is probably the hardest point of all.
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Since it doesn't have a degree symbol, we know that 1.47 is in radians.
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1.47 radians is an angle--but what is 1.47 radians?
00:23:50.800 --> 00:23:53.500
I don't know how to do very well with a decimal form of radians.
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I am used to π/2, π/3, π/6, and things like that; so how do we deal with 1.47 radians?
00:23:59.700 --> 00:24:04.600
Well, one way is to just get some reference points--some reference ideas.
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What does π mean in decimal? What does π/2 mean in decimal?
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Well, π is about the same thing as 3.14; π/2 is about the same thing as 1.57.
00:24:18.200 --> 00:24:26.900
So, that means that π/2...if it is at 1.57, and we are going to 1.47, that means that we are going to be most of the way to having made it to π/2.
00:24:26.900 --> 00:24:37.000
If we want to have an even better idea of what it is, we could divide 1.47, the angle we are going to,
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and then divide it by π/2, which ends up being about 1.57.
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When we plug that into a calculator, that comes out to be around .94, which means we are 94% of the way to π/2.
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If that is confusing, just think in terms of 1.57 being here; I am going to 1.47, so it is going to be roughly most of the way.
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But we can also think of it as a ratio, so we can get this decimal, which is basically a percent.
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We can think that we are 94% of the way from 0 to π/2.
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94% of the way--that means we are going to be very close to it.
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We are at a distance of 3, so we are pretty close to being right on π/2; so call it about here, and that is our last point; great.
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The next example: In polar coordinates, name each point below.
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Then give two alternate ways to name them, where one of them will have r < 0.
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First, we see that this is at what distance? Well, here is the 1 circle and the 2 circle.
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We do have to make sure that the scale for our circles ends up being 1, 2, 3.
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We could have the scale be 5, 10, 15, just as we have had tick marks on the axis be more than just 1 for each tick mark.
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We could have 20, 40, 60 be the scale on our tick marks.
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But in this case, we see that 2 does match up; the second circle does match up to it a 2 distance; the fourth circle matches up to a 4 distance.
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So here, indeed, is the 5; so we have r = 5.
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We are at a distance of r = 5; what is the angle that we are at?
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Well, we are on this, and we can see that it is cut into π/6; so this is π/3.
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So, we have θ = π/3; so our first, most basic point, the one we would normally call out, will be (5,π/3):
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a distance of 5 from the center, at an angle, a counterclockwise spin angle, of π/3.
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Oops, I accidentally meant to write π/3; I didn't mean to write π/6 the second time.
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We are at π/3 there; now remember: we can talk about any of these points in multiple ways,
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because if we spin an additional 2π, we end up just getting to the same place.
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Here, if we are at this place, and then we just spin another 2π, we are still at the same place.
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So, we can just add an entire rotation onto it; so we will take θ = π/3;
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and now, we can just add: so π/3 + 2π--what does that end up being?
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That ends up being 6π/3 + π/3, or 7π/3; so we can make another point to call this out with as (5,7π/3).
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Alternatively, we could also, if we want to get r less than 0 (the last part of this problem
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requires us to get the less than 0 value for our r, because we can always go in the opposite direction)...
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what if we had spun to the opposite direction?
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If we had spun to the opposite direction, well, this would be a total angle for θ of 4π/3.
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So, at opposite angles, opposite θ ends up being 4π/3.
00:27:30.200 --> 00:27:35.600
So, the opposite r: well, normally our r is equal to 5.
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So, if we are going to go to the opposite direction (we want to have an opposite, and then another opposite stacked on top of it,
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so we can go back in the direction we want to go), that would be -5.
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Put those two things together, and we have the point (-5,4π/3).
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We have our normal canonical way, the way we would usually talk about this, (5,π/3).
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But then, there are other things, where we have just done an additional rotation (5,7π/3),
00:28:04.500 --> 00:28:13.100
or when we spin in the opposite direction, and then go in the opposite opposite direction by having that negative value for r.
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All right, next: let's look at the blue dot--where would this end up being?
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Well, we see that at a distance, the circle is 1, 2, 3; so we have r = 3 for our normal way of talking about it.
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What angle is it at? Well, we see that it is along this angle here, which splits the entire quadrant evenly.
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You split quadrants on π/4; to get from here to here would be π, so an additional π/4 would be 5π/4.
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So, our normal θ would be equal to 5π/4.
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The normal way to talk about this point would be (3,5π/4); great.
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However, we can also talk about this in different ways.
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Remember: we can go any addition of spins around, so any number of 2π's added to the θ.
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And we will still end up being at the same place.
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So, if our normal θ is equal to 5π/4, we can add any multiple of 2π, and that is just that many times that we spin around.
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Whatever the multiple is, is how many times we spin around.
00:29:20.200 --> 00:29:24.300
So, if we want to spin around 5 times, we would add 10π, 5 times 2π.
00:29:24.300 --> 00:29:30.500
We can take 5π/4, and we can add 10π to that; what does that end up coming out to be?
00:29:30.500 --> 00:29:36.300
That ends up being 45π/4; so we ended up spinning to the 5π/4,
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and then we just count 1, 2, 3, 4, 5...and we land back right where we had been previously.
00:29:42.400 --> 00:29:46.300
We have another way of talking about this point as the same distance, 3, comma...
00:29:46.300 --> 00:29:50.900
but now we have a totally different angle for it, 45π/4.
00:29:50.900 --> 00:29:59.900
Great; and our last point, where r is less than 0, where we end up going in the opposite direction:
00:29:59.900 --> 00:30:02.700
well, the easiest way to call this out would be: what would be the opposite direction?
00:30:02.700 --> 00:30:08.300
Well, we are splitting the quadrant this way; we could talk about this as being π/4.
00:30:08.300 --> 00:30:15.300
So, that would be π/4 here, and then, if this is the direction we normally go, we are going the opposite direction.
00:30:15.300 --> 00:30:24.900
The opposite θ is π/4; we can see that 5π/4 minus π or plus π ends up getting us to an opposite direction.
00:30:24.900 --> 00:30:33.100
So, 5π/4 - π gets us π/4; or if we had added π, we would be at 9π/4, but 9π/4 is still the same direction as π/4.
00:30:33.100 --> 00:30:36.100
So, there are two different ways of talking about it; both are perfectly fine.
00:30:36.100 --> 00:30:45.000
π/4...and then the opposite r...well, normally our r is 3, so if we are going to have the opposite r,
00:30:45.000 --> 00:30:49.200
once we are going the opposite direction in our angle, we want to have -3.
00:30:49.200 --> 00:30:55.500
We put these two things together, and we end up getting (-3,π/4).
00:30:55.500 --> 00:31:01.200
We have our angle pointing us in the opposite direction; but then, the negative in our r tells us to walk the opposite direction.
00:31:01.200 --> 00:31:04.300
Opposite opposite means that we get to the point that we wanted to be at.
00:31:04.300 --> 00:31:08.500
All right, the third example: Convert the polar coordinates to rectangular.
00:31:08.500 --> 00:31:16.900
Our first one is (8,5π/6) for this one; let's start by just drawing this--sketching, really quickly, about where that would end up showing.
00:31:16.900 --> 00:31:24.400
8--a distance of 8; 5π/6...well, we end up spinning to somewhere out here for 5π/6 and then a distance of 8.
00:31:24.400 --> 00:31:31.300
We are at something like that; we don't know for sure what the point is going to be from that drawing.
00:31:31.300 --> 00:31:36.300
But we do at least have a sanity check that whatever we end up getting, it should be in the second quadrant.
00:31:36.300 --> 00:31:40.300
We know that our y should come out to be positive, and we know that our x should come out to be negative.
00:31:40.300 --> 00:31:44.200
It is a negative x and a positive y if we are going to be in this quadrant right here.
00:31:44.200 --> 00:31:47.700
So, if we are going to be in that quadrant, we have some sanity check to help us out here.
00:31:47.700 --> 00:31:52.900
Now, what are the formulas? Remember: the formulas for figuring out our rectangular coordinates
00:31:52.900 --> 00:31:59.900
from polar are x = rcos(θ) and y = rsin(θ).
00:31:59.900 --> 00:32:06.500
If you ever end up forgetting these, you can pretty easily just figure them out by drawing that triangle, the x, y, r, θ in the corner...
00:32:06.500 --> 00:32:07.700
You can draw it out and figure it out.
00:32:07.700 --> 00:32:12.900
All right, let's work through this: here is our r, and here is our θ.
00:32:12.900 --> 00:32:20.100
That means that x is equal to 8 times cosine of 5π/6.
00:32:20.100 --> 00:32:30.800
We work with the unit circle; we remember that cosine of 5π/6...8 times cos(5π/6) is -√3 over 2.
00:32:30.800 --> 00:32:35.600
We are going to end up getting -4√3 as our value for x.
00:32:35.600 --> 00:32:43.500
OK, the other one, y: y is going to be equal to r, sine...what is our r? Our r is 8; we might as well write that in.
00:32:43.500 --> 00:32:52.400
8sin(θ) is 5π/6; remember, we are on the unit circle, so now we have 8 times sine of 5π/6.
00:32:52.400 --> 00:33:01.700
Looking on the unit circle, we get positive 1/2; it isn't below the x-axis yet; so 8 times 1/2...we get positive 4, so our y is equal to +4.
00:33:01.700 --> 00:33:09.100
So, putting these two pieces of information together, we end up getting the point (-4√3,4).
00:33:09.100 --> 00:33:12.600
If we end up looking against the point that we figured out here, that seems to makes sense.
00:33:12.600 --> 00:33:19.800
That checks out, because we see that it ends up having more x distance than it should have y distance, based on (8,5π/6).
00:33:19.800 --> 00:33:24.200
So, this seems pretty reasonable; it checks out for at least a quick sanity check.
00:33:24.200 --> 00:33:31.800
All right, the next one: (-5,-7π/4): let's draw what this would end up looking like.
00:33:31.800 --> 00:33:37.200
-7π/4 is going to spin clockwise, as opposed to counterclockwise; so we spin to -7π/4.
00:33:37.200 --> 00:33:41.200
That would put us at the same angle as having spun positive 5π/4, somewhere out here.
00:33:41.200 --> 00:33:48.700
But instead, we are going to -5; -5 will actually go...as opposed to going in the normal direction of this way, we will end up going in the opposite direction.
00:33:48.700 --> 00:33:52.200
We are going to be at some point out here at -5.
00:33:52.200 --> 00:33:58.000
All right, so we see how that works; so we have some sense that we should at least end up being in the third quadrant.
00:33:58.000 --> 00:34:00.900
We should have a negative x-value and a negative y-value.
00:34:00.900 --> 00:34:11.700
The same basic thing to figure this out: x = rcos(θ): our r is -5; our cosine is -7π/4.
00:34:11.700 --> 00:34:21.300
If we are not sure how to take a cosine of -7π/4, we could remember that -7π/4 is equivalent to have written it as just π/4,
00:34:21.300 --> 00:34:24.400
as we can see from our picture (another reason why pictures are useful).
00:34:24.400 --> 00:34:36.400
So, we could write this as x = -5cos(π/4); the cosine of π/4...we check the unit circle; we get x = -5 times √2 over 2.
00:34:36.400 --> 00:34:51.900
And we have x = -5√2/2--basically the same as if we are working through -5sin(-7π/4).
00:34:51.900 --> 00:34:56.800
We work that through; once again, we have -7π/4 as an equivalent to having written just π/4.
00:34:56.800 --> 00:35:04.300
So, y is equal to -5sin(π/4); we also could remember that, since we are splitting the quadrant,
00:35:04.300 --> 00:35:07.900
it is going to end up just being the same thing as it was for cosine.
00:35:07.900 --> 00:35:15.300
y = -5 times √2/2, just like we got for cosine, because we are splitting that first quadrant.
00:35:15.300 --> 00:35:29.500
And so, we get y = -5√2/2; combining those two pieces of information together, we now have the point (-5√2/2,-5√2/2).
00:35:29.500 --> 00:35:33.000
And if we were to compare that to the picture that we originally drew,
00:35:33.000 --> 00:35:37.200
just to give us a quick idea of what is going on, yes, that seems perfectly reasonable.
00:35:37.200 --> 00:35:43.100
(-5√2/2,-5√2/2): that is in the third quadrant, because both our x and our y are negative.
00:35:43.100 --> 00:35:47.500
And we see that this point here splits the x and the y pretty evenly, which makes sense,
00:35:47.500 --> 00:35:52.000
because we are based off of splitting the quadrant; so it makes sense that our x and y values will end up being the same.
00:35:52.000 --> 00:35:56.100
So, it passes the sanity check; we have a pretty good check to see that we are probably right.
00:35:56.100 --> 00:36:01.900
All right, next, Example 4: Convert the rectangular coordinates to polar, and give your answers with r greater than 0,
00:36:01.900 --> 00:36:04.700
and 0 less than or equal to θ, which is less than 2π.
00:36:04.700 --> 00:36:13.600
So, this is a restriction to just require us to put our answers out in that normal form of positive r and θ's that aren't crazily large or negative.
00:36:13.600 --> 00:36:16.500
It is a fairly normal restriction here that we might have in this sort of thing.
00:36:16.500 --> 00:36:20.700
It is not necessary, but it doesn't really hurt us or make this problem much harder to do.
00:36:20.700 --> 00:36:24.400
The first thing we want to do: we always, always, always want to do this.
00:36:24.400 --> 00:36:29.700
When we are converting from rectangular to polar, always, always begin by drawing a picture.
00:36:29.700 --> 00:36:32.300
So, the first thing we do is draw a picture.
00:36:32.300 --> 00:36:35.400
It doesn't have to be absolutely perfect, but we want to have some sense of what is going on here.
00:36:35.400 --> 00:36:39.100
And also, we will see how drawing a picture can be really useful for figuring out the numbers.
00:36:39.100 --> 00:36:46.100
So, we have -3 horizontally and then 3√3 vertically.
00:36:46.100 --> 00:36:51.300
Here is -3 out this way and 3√3 out this way.
00:36:51.300 --> 00:37:02.500
OK, so what were our formulas? They were r² = x² + y² and tan(θ) = y/x.
00:37:02.500 --> 00:37:07.800
If you remember, earlier, when we talked about these, I warned about how it is dangerous to just use them blindly, without thinking about them.
00:37:07.800 --> 00:37:14.500
We are going to now use them blindly, just so we can see how we have to be careful and why it is so important that we pay attention to this picture.
00:37:14.500 --> 00:37:20.500
But if we had just used them blindly, then we would end up having r² = x² + y²:
00:37:20.500 --> 00:37:32.100
so our -3 and 3√3 are x and y: r² is equal to (-3)² + (3√3)².
00:37:32.100 --> 00:37:40.000
So, we have r² = +9 +...3 squared is 9; √3 squared is 3; 9 times 3 is 27.
00:37:40.000 --> 00:37:44.800
So, we have r² = 36; take the square root of both sides,
00:37:44.800 --> 00:37:49.200
and we know that we can't have a negative r, so we know that it has to be positive,
00:37:49.200 --> 00:37:55.000
even though we get that ± technically; but generally we will leave it as positive; so r = 6; great.
00:37:55.000 --> 00:37:57.500
We have figured out our r distance--that doesn't seem so problematic.
00:37:57.500 --> 00:38:00.900
Well, let's try using the tan(θ) = y/x.
00:38:00.900 --> 00:38:10.400
Great--let's toss it in: tan(θ) = our y, 3√3, over our x, -3; OK.
00:38:10.400 --> 00:38:14.800
We see that the 3's will cancel, and we will end up having -√3; great.
00:38:14.800 --> 00:38:23.200
We take the arctan of both sides; you punch that into a calculator, and you end up getting -π/3.
00:38:23.200 --> 00:38:28.000
Wait, what? That doesn't make sense!--how is that possible?
00:38:28.000 --> 00:38:34.300
How can we get -π/3? If we go back and look at our picture, this is our θ right here.
00:38:34.300 --> 00:38:39.500
This here is not the case; that does not work out; that doesn't make sense; what is going on here?
00:38:39.500 --> 00:38:46.000
Well, this is why it is so important: we have to be talking about a θ that is in the part that arctan cannot talk about.
00:38:46.000 --> 00:38:51.800
Arctan talks about this side of the unit circle; but the point that we want is over here on this side.
00:38:51.800 --> 00:38:55.400
So, arctan can't get us to our point; so here is what we do instead.
00:38:55.400 --> 00:38:59.600
We end up looking at this as just a normal triangle.
00:38:59.600 --> 00:39:03.100
If it was just a normal triangle, then this would be a length of 3.
00:39:03.100 --> 00:39:06.600
We don't have to worry about -3, because we are not talking about a coordinate; we are talking about the length of the side.
00:39:06.600 --> 00:39:12.000
And this would be 3√3; and now, we could talk about this angle in here as being α.
00:39:12.000 --> 00:39:23.900
Then, tan(α) would be equal to the opposite, 3√3, divided by the adjacent, 3; so we would get tan(α) = √3...
00:39:23.900 --> 00:39:32.100
we take the inverse tangent of both sides; tan^-1(√3) is π/3, so we get π/3 from this.
00:39:32.100 --> 00:39:38.700
However, this is α, not θ; what we want is θ, and what we have here is the α inside of it.
00:39:38.700 --> 00:39:40.500
But how can we figure out what θ is?
00:39:40.500 --> 00:39:46.200
Well, if we were to draw in the axes, the θ we want is this one here.
00:39:46.200 --> 00:39:49.400
We have figured out what α is; well, how do θ and α connect together?
00:39:49.400 --> 00:39:53.900
Well, this here would be π in terms of the angle.
00:39:53.900 --> 00:40:01.800
So, what we have is: we have that θ + α is equal to π for this specific picture.
00:40:01.800 --> 00:40:05.300
And now, how these two things will relate is depending on the picture we are dealing with.
00:40:05.300 --> 00:40:08.500
So, we really have to keep our wits about us and think about what we are looking at.
00:40:08.500 --> 00:40:13.500
If we just try to do it blindly--try to rely on some formula--chances are that we are more likely to make mistakes
00:40:13.500 --> 00:40:17.000
than to actually be able to figure it out, because there are a lot of things to have to memorize, because there are so many special cases.
00:40:17.000 --> 00:40:20.100
But if you end up making a picture and thinking about how this angle connects,
00:40:20.100 --> 00:40:23.100
and paying attention to the unit circle, it is not that hard to do at all.
00:40:23.100 --> 00:40:27.800
So, θ + α = π; we just figured out that α is equal to π/3.
00:40:27.800 --> 00:40:37.000
So, we have θ + π/3 = π, which means that our θ must be 2π/3.
00:40:37.000 --> 00:40:41.600
And that makes total sense over here, since that is 2π/3, and inside of the triangle is π/3.
00:40:41.600 --> 00:40:47.500
We add those together, and we would have the entire top arc from here to here.
00:40:47.500 --> 00:40:52.300
That makes perfect sense; so we now have...here is our θ.
00:40:52.300 --> 00:40:56.000
Figuring out the r still didn't cause any problems for us; that is our 6.
00:40:56.000 --> 00:41:00.700
And so, we have the point; we put these two pieces of information together.
00:41:00.700 --> 00:41:05.600
And the distance is 6, and the angle is 2π/3.
00:41:05.600 --> 00:41:11.600
And I hope this makes clear why it is so, so important to figure out how these things work,
00:41:11.600 --> 00:41:15.300
if we are converting from rectangular to polar--why it is so important to draw a picture.
00:41:15.300 --> 00:41:17.900
If we don't draw a picture, we are going to get lost.
00:41:17.900 --> 00:41:22.400
You really have to draw a picture; it doesn't have to be complicated, but we need that picture to help us figure this out.
00:41:22.400 --> 00:41:24.600
So, it is (6,2π/3).
00:41:24.600 --> 00:41:27.400
Also, if we were to draw that as a picture, where would we get?
00:41:27.400 --> 00:41:33.800
Well, 2π/3 would end up being out here, and a distance of 6 is about here; compare that to what we got here.
00:41:33.800 --> 00:41:39.100
That is pretty good; it seems to pass the sanity check--it seems like we probably got the answer right; great.
00:41:39.100 --> 00:41:44.500
Convert the rectangular coordinates to polar, and give your answers with r > 0 and 0 ≤ θ < 2π.
00:41:44.500 --> 00:41:49.000
It is basically the same problem as previous, but now we have (17,-19), new points.
00:41:49.000 --> 00:41:51.200
The first thing we do: we always draw a picture.
00:41:51.200 --> 00:41:55.100
Now, at this point, we understand how useful a picture is.
00:41:55.100 --> 00:42:02.000
So, we are going to draw a huge picture, because now we can just work on the triangle inside of it, now that we see how this thing works.
00:42:02.000 --> 00:42:11.400
17 would be out here; -19 would be down here; this is a distance of 19.
00:42:11.400 --> 00:42:15.400
It is -19 as a coordinate, but we can also just treat it as being a distance of 19.
00:42:15.400 --> 00:42:20.100
Here is a distance of 17: our r is going to be this side of the triangle.
00:42:20.100 --> 00:42:26.300
Remember: it is a right triangle, and we know that...not θ; θ is not inside of the triangle;
00:42:26.300 --> 00:42:32.400
θ was the counterclockwise spin, like this; this is our θ; what we have inside of the triangle
00:42:32.400 --> 00:42:37.600
is some other angle; I am deciding to call it α; you can call it smiley-face; you could call it any symbol that you want to use.
00:42:37.600 --> 00:42:41.700
α is an easy one for me to draw, so I am calling it α inside of the triangle.
00:42:41.700 --> 00:42:48.500
We can figure out r pretty easily; it is just a right triangle; we can figure out α pretty easily--a basic trigonometric function, tan(θ).
00:42:48.500 --> 00:42:52.700
And then, we can use θ and α's interaction together to figure out what θ has to be.
00:42:52.700 --> 00:42:57.300
So, first, let's figure out what our r is.
00:42:57.300 --> 00:43:07.300
We can see that we have r here; the hypotenuse squared is equal to the other two legs squared, so it is 17 squared plus 19 squared.
00:43:07.300 --> 00:43:14.300
We work that out; so we have r² is equal to...I'll take the square root of both sides...
00:43:14.300 --> 00:43:23.000
and eventually, we work this all out: 17² + 19²...and it simplifies to r = 5√26.
00:43:23.000 --> 00:43:29.600
And you could probably get away with putting that in a decimal form if you wanted; but exactly what it is, is r = 5√26.
00:43:29.600 --> 00:43:32.500
Next, let's figure out what α has to be.
00:43:32.500 --> 00:43:37.600
Well, since α is in here, we see that tan(α) is equal to...what is the opposite?
00:43:37.600 --> 00:43:45.700
The opposite, away, opposite from that angle, is 19; so 19 divided by...the adjacent to that angle is 17,
00:43:45.700 --> 00:43:55.200
so divided by 17; we take the arctan of both sides; α is equal to the arctan of 19/17.
00:43:55.200 --> 00:43:58.600
Now, that doesn't come out to be any friendly, nice form.
00:43:58.600 --> 00:44:00.900
That is not a really nice number to have to deal with.
00:44:00.900 --> 00:44:07.400
But if we punch it into a calculator, it will end up coming out to be approximately 0.84.
00:44:07.400 --> 00:44:13.600
If we wanted to have it absolutely perfectly, we would have to leave it as tan^-1(19/17).
00:44:13.600 --> 00:44:17.700
But we can probably have a decimal approximation; so α is approximately equal to 0.84
00:44:17.700 --> 00:44:20.100
will probably do, for our purposes of answering this question.
00:44:20.100 --> 00:44:24.300
Now, how do we get to the actual answer? Well, we have to figure out how θ and α relate.
00:44:24.300 --> 00:44:30.700
Well, if we did a spin here, we could do one entire revolution: one entire revolution would be 2π.
00:44:30.700 --> 00:44:35.300
So, we see that θ, combined with α...if we spin most of the way,
00:44:35.300 --> 00:44:39.400
and then we have the angle that we do in the triangle, well, that comes out to be one entire revolution.
00:44:39.400 --> 00:44:47.000
One entire revolution is 2π; so that means that θ + α, in this case, comes out to be 2π.
00:44:47.000 --> 00:44:51.200
As we saw in the previous one, it came out to be π; so we really have to pay attention to the picture we are looking at.
00:44:51.200 --> 00:44:55.300
There is not just a simple formula here; it is thinking and using pictures--pictures and thinking.
00:44:55.300 --> 00:44:57.800
Without thinking, nothing will end up working.
00:44:57.800 --> 00:45:02.500
θ + α = 2π; what is our α? We have that our α is 0.84.
00:45:02.500 --> 00:45:13.700
So, θ + 0.84 = 2π; actually, let's write it a little bit differently--I prefer to write it as θ = 2π - α.
00:45:13.700 --> 00:45:23.000
Then, we will substitute in: so we have θ = 2π - 0.84, which gives us that θ is approximately equal to...
00:45:23.000 --> 00:45:31.600
plug in 2π; 2 times 3.14, minus 0.84; we get approximately 5.44 for our value here.
00:45:31.600 --> 00:45:37.200
That means we could plot the point as being 5√26.
00:45:37.200 --> 00:45:49.200
Our r value over here of 5√26, and comma...5.44...sorry, θ comes out to be 5.44.
00:45:49.200 --> 00:45:53.300
So, 5.44: that is what we end up figuring out as our approximate value.
00:45:53.300 --> 00:45:58.600
If we wanted it precisely, we could leave it as 2π - tan^-1(19/17).
00:45:58.600 --> 00:46:04.900
But we will probably end up being OK with just having this approximate value right here at 5.44.
00:46:04.900 --> 00:46:09.000
Now, we didn't do this in the last problem, but I want you to see that you can also check your work here.
00:46:09.000 --> 00:46:13.800
If you are not quite sure that it all came out right, we can check our work; we can go to polar,
00:46:13.800 --> 00:46:17.400
but then, since it is so easy to convert from polar back to rectangular, we just plug them in.
00:46:17.400 --> 00:46:25.200
Remember: to convert to rectangular, x = rcos(θ); y = rsin(θ).
00:46:25.200 --> 00:46:33.600
So, I will put in the middle this checking box; if we want to do a check, well, x = rcos(θ),
00:46:33.600 --> 00:46:38.200
so that means x is equal to...what is our r? Our r is 5√26.
00:46:38.200 --> 00:46:51.900
So, 5√26 times cosine of 5.44: we plug that into a calculator, and we end up getting x = 16.96.
00:46:51.900 --> 00:46:54.800
What did we have originally? We originally had 17.
00:46:54.800 --> 00:46:58.200
And remember: we ended up having some round-off error, because we had to round,
00:46:58.200 --> 00:47:02.000
because it was tan^-1(19/17), so we rounded to 0.84.
00:47:02.000 --> 00:47:05.900
So, that makes sense, that we are ending up seeing 16.96 compared to 17.
00:47:05.900 --> 00:47:09.900
It is just round-off error, but basically, our answer is correct, so it checks out.
00:47:09.900 --> 00:47:20.300
The same thing for y: y is equal to r times sin(θ); our r here is 5√26, and our sine is 5.44.
00:47:20.300 --> 00:47:24.800
We plug that into a calculator, and we end up getting -19.04.
00:47:24.800 --> 00:47:29.100
Compare that, once again, to -19: the only thing we are seeing here is a slight bit of round-off error.
00:47:29.100 --> 00:47:32.000
This checks out, because we knew that we would have some rounding-off error,
00:47:32.000 --> 00:47:35.700
because we had rounded it, as opposed to using what it was precisely.
00:47:35.700 --> 00:47:39.300
So, if you are ever in a situation where it really, really matters that you end up getting this right--
00:47:39.300 --> 00:47:42.200
like you are on a test, or you are confused about how this stuff is working out--
00:47:42.200 --> 00:47:46.600
just end up checking it; it is so easy to convert from polar to rectangular.
00:47:46.600 --> 00:47:49.100
The hard part is converting from rectangular to polar.
00:47:49.100 --> 00:47:51.600
So, if you are in polar, you might as well convert back to rectangular at the end,
00:47:51.600 --> 00:47:53.600
and check and make sure that you got the answer right,
00:47:53.600 --> 00:47:57.900
especially if it really ends up mattering, like in a situation where you are taking a test or an exam of some sort.
00:47:57.900 --> 00:48:01.100
All right, so now that we have a good understanding of how polar coordinates work,
00:48:01.100 --> 00:48:04.800
we are ready to talk about polar equations and functions and really get into graphing this stuff.
00:48:04.800 --> 00:48:07.000
All right, we will see you at Educator.com later--goodbye!