For more information, please see full course syllabus of Pre Calculus
For more information, please see full course syllabus of Pre Calculus
Polar Coordinates
 Previously, whenever we've talked about the location of a point on the plane, we've described its horizontal and vertical distance from the origin: x and y. These are rectangular coordinates.
 Polar coordinates give us a new way to describe the location of a point. Instead of using horizontal and vertical components, we can talk about the point's distance from the origin and the angle it is on.
 We plot points with polar coordinates using two things:
 r: the distance of the point from the origin (from here on, we will call the origin the pole).
 θ: the angle of the point. (We measure this counterclockwise from where we used to have the positive xaxis, like we measured angles in the unit circle.)
 It is normally assumed that θ is in the unit of radians. (Degrees are used occasionally, but much less often.)
 Visualizing polar coordinates can be really difficult the first few times you try to use them. Make sure to watch the video to see how we can visualize what polar coordinates mean.
 Unlike rectangular coordinates, there are multiple ways to name the same point in polar coordinates. For any angle θ, we can make an equivalent angle by adding or subtracting a multiple of 2π (360^{°}).
 If r is negative, we go in the direction opposite the one that the angle θ points out. This also means there is another alternative way to name the same point: add π to θ and make r negative.
 We can convert between coordinate systems if we want:
 Polar ⇒ Rectangular: x = r cosθ y = rsinθ
 Rectangular ⇒ Polar: r^{2} = x^{2} + y^{2} tanθ = y/x
Polar Coordinates

 Start off by drawing in some "polar graph paper". It helps to have a series of concentric circles to help you judge distance from the center, and it similarly helps to have light lines drawn in to help you see major angles (like every [(π)/6] / 30^{°}). See the video lesson for more of an explanation and to see what it looks like if you're unfamiliar with the idea.
 When plotting a point (r, θ) in polar coordinates, there are two ways to approach it: A) Go out a length of r along the xaxis, then rotate by θ. B) Imagine aiming in the direction determined by θ, then go out a distance of r. Either one of these methods to visualize what's happening work fine. Think in terms of the one that makes more sense to you.
 Make sure you're comfortable with measuring angles in radians. They will come up a lot when working with polar coordinates, so it's important that they make sense and feel natural to you. If you have difficulty knowing where a given angle in radians will lie, keep a unit circle diagram near you and reference it as you work.

 Start off by drawing in some "polar graph paper". It helps to have a series of concentric circles to help you judge distance from the center, and it similarly helps to have light lines drawn in to help you see major angles (like every [(π)/6] / 30^{°}). See the video lesson for more of an explanation and to see what it looks like if you're unfamiliar with the idea.
 When plotting a point (r, θ) in polar coordinates, there are two ways to approach it: A) Go out a length of r along the xaxis, then rotate by θ. B) Imagine aiming in the direction determined by θ, then go out a distance of r. Either one of these methods to visualize what's happening work fine. Think in terms of the one that makes more sense to you.
 Normally we think of r as being a positive number. When r is positive we (A) go right on the xaxis; (B) go in the direction pointed out by θ. However, when r is a negative number we go in the opposite direction instead.
 Normally we think of θ as being a positive number, causing the angle to "spin" counterclockwise. However, when θ is negative it causes the angle to go in the opposite direction: clockwise.
 When both r and θ are negative, just do both of the above things together.

 Start off by drawing in some "polar graph paper". It helps to have a series of concentric circles to help you judge distance from the center, and it similarly helps to have light lines drawn in to help you see major angles (like every [(π)/6] / 30^{°}). See the video lesson for more of an explanation and to see what it looks like if you're unfamiliar with the idea.
 When plotting a point (r, θ) in polar coordinates, there are two ways to approach it: A) Go out a length of r along the xaxis, then rotate by θ. B) Imagine aiming in the direction determined by θ, then go out a distance of r. Either one of these methods to visualize what's happening work fine. Think in terms of the one that makes more sense to you.
 For rvalues that are not integers, just plot the point so that it is appropriately placed between the correct "distance circles". For the point ( 2.5 , [(π)/3] ), we see that the point should be plotted half way between the circles indicating r=2 and r=3.
 The hardest thing to deal with are θ angles that are not "nice" fractional multiples of π. An angle in radians does not necessarily have to show π in it, it just makes it more difficult to get a sense of where it lies if it is in decimal instead. To figure out where to place such an angle, use other common radian locations as references. For the point ( 1 , 3.1 ), we're trying to figure out where θ = 3.1 is located. Notice that 3.1 is very close to π ≈ 3.14. Thus, we can place the angle θ = 3.1 as being just a hair above the angle of π, which we know from our work with the unit circle.
 We do the same thing for the point ( −3 , 0.9 ), it's just a little more difficult. We need to figure out the location of θ = 0.9. As reference, we can use a calculator to find other common radian locations in decimal form:
Thus, we see that θ = 0.9 is roughly 40% of the way from [(π)/4] to [(π)/3]. Alternatively, we could convert θ = 0.9 into degrees, since degrees are sometimes easier to visualize and plot.π 6≈ 0.52 π 4≈ 0.79 π 3≈ 1.05 π 2≈ 1.57
In any case, once we roughly know where the angle lies, we can plot the point. [Don't forget to notice the negative r in ( −3 , 0.9 ), as it will cause the point to appear in the opposite direction.]Radians: θ = 0.9 ⇒ Degrees: θ = 0.9 2π·360^{°} ≈ 51.57^{°}
 Let's begin by naming it in the most natural form: where r > 0 and 0 ≤ θ < 2π.
Looking at the point, we see that it is a distance of 3 from the origin: r=3. We also see that the angle is pointing directly to the left, that is θ = π. Putting this together, we get
( 3, π).  Notice that because radians are based on a circle, they repeat the same direction for every 2π of "spin". That means, for any angle θ, the angle of θ+ 2π points in the same direction.
This is true also if we spin an angle of −2π (a full rotation clockwise), so we have that θ and θ− 2π are equivalent as well.
Finally, it doesn't matter how many rotations we make, so as long as we add or subtract a whole multiple of 2π, we continue to point in the same direction:
This allows us to create a whole bevy of equivalent polar coordinate points:θ is equivalent to θ+k·(2π), where k is any integer ( 3, π) ≡ ( 3, 3π) ≡ ( 3, −π) ≡ ( 3, −5π) ≡ ( 3, 47π) ≡ …  To create an equivalent point where r < 0 requires a little bit more thought. Remember, a negative value for r causes the point to go in the opposite direction to where the angle is aimed. Furthermore, notice that we can aim any angle in the opposite direction to where it is currently pointing by adding π to the angle (since π is a halfrotation). Thus, if we make both our r negative and increase the angle θ by π, we wind up at the same location: the two opposites cancel each other out and we get back to where we started. By this logic, we have that ( −3, 2π) is equivalent to ( 3, π). Furthermore, if we wanted to make even more equivalent points, we know from the above that we can add or subtract any whole multiple of 2π to this alternate point's angle and get another equivalent point.
 Let's begin by naming it in the most natural form: where r > 0 and 0 ≤ θ < 2π.
Looking at the point, we see that it is a distance of 2 from the origin: r=2. We also see that the angle lines in the "polar graph paper" split the circle into a total of twelve sectors. That means each angle line "turns" an angle of [(π)/6]. Therefore, we have the angle is θ = [(5π)/3].
⎛
⎝2, 5π 3⎞
⎠.  Notice that because radians are based on a circle, they repeat the same direction for every 2π of "spin". That means, for any angle θ, the angle of θ+ 2π points in the same direction.
This is true also if we spin an angle of −2π (a full rotation clockwise), so we have that θ and θ− 2π are equivalent as well.
Finally, it doesn't matter how many rotations we make, so as long as we add or subtract a whole multiple of 2π, we continue to point in the same direction:
This allows us to create a whole slew of equivalent polar coordinate points:θ is equivalent to θ+k·(2π), where k is any integer ⎛
⎝2, 5π 3⎞
⎠≡ ⎛
⎝2, 11π 3⎞
⎠≡ ⎛
⎝2, − π 3⎞
⎠≡ ⎛
⎝2, − 7π 3⎞
⎠≡ ⎛
⎝2, 65π 3⎞
⎠≡ …  To create an equivalent point where r < 0 requires a little bit more thought. Remember, a negative value for r causes the point to go in the opposite direction to where the angle is aimed. Furthermore, notice that we can aim any angle in the opposite direction to where it is currently pointing by adding π to the angle (since π is a halfrotation). Thus, if we make both our r negative and increase the angle θ by π, we wind up at the same location: the two opposites cancel each other out and we get back to where we started. By this logic, we have that ( −2, [(8π)/3] ) is equivalent to ( 2, [(5π)/3] ). Furthermore, if we wanted to make even more equivalent points, we know from the above that we can add or subtract any whole multiple of 2π to this alternate point's angle and get another equivalent point.
 This question is a little bit of a trick question compared to the previous two problems. Notice what the point's distance from the origin is.
 Once we realize that the point has r=0 because it has no distance from the origin, we realize that the angle does not matter. It has absolutely no importance what angle winds up being used, since every direction gives the same location if we go out absolutely no distance in that direction.
 Thus, we can name the point as (0, θ) for absolutely any value of θ we choose. Any number at all is fine for the angle, so long as r=0.
 We can convert from polar to rectangular coordinates with the following conversion identities:
x = rcosθ y = rsinθ  Converting to rectangular coordinates is quite simple: just plug in to the above identities. Whatever we get out tells us the rectangular coordinates.
 Plug in:
x = 4cos ⎛
⎝5π 6⎞
⎠⎢
⎢y = 4sin ⎛
⎝5π 6⎞
⎠x = 4 ⎛
⎝− √3 2⎞
⎠⎢
⎢y = 4 ⎛
⎝1 2⎞
⎠
Thus, we now have our rectangular coordinates: (x, y) ⇒ (−2√3, 2).x = −2√3 ⎢
⎢y = 2
 We can convert from polar to rectangular coordinates with the following conversion identities:
x = rcosθ y = rsinθ  Converting to rectangular coordinates is quite simple: just plug in to the above identities. Whatever we get out tells us the rectangular coordinates. Don't worry about the fact that θ is not a "standard" radian angle, the method will still work exactly the same.
 Plug in [don't forget that θ is in radians, even though π does not appear]:
x = −8cos(7.2) ⎢
⎢y = −8sin(7.2) x = −8·(0.60835) ⎢
⎢y = −8 ·(0.79367)
Thus, we now have our rectangular coordinates: (x, y) ⇒ (−4.867, −6.349).x = −4.867 ⎢
⎢y = −6.349
 When converting from rectangular coordinates to polar, make sure to start off by drawing a quick sketch to help you see what's going on:
 From the lesson, we have the identities
Finding r is easy, we simply solve by plugging in to the identity:r^{2} = x^{2} + y^{2} tanθ = y x
Working out θ is a little more difficult, though, as we will see:r^{2} = (−5)^{2} + (−5)^{2} ⇒ r =
√
50⇒ r = 5√2
However!, notice that if r=5√2, we can not have θ = [(π)/4] due to where the point is located. This is why drawing the picture at the beginning is so important: it allows us to easily check our work.tanθ = −5 −5⇒ tanθ = 1 ⇒ θ = tan^{−1}(1) ⇒ θ = π 4  So what's going on here? Inverse tangent (tan^{−1}) can only give results of (−[(π)/2], [(π)/2]). Looking at our picture, though, we see that θ must be greater than [(π)/2]. Thus, to get to the other side of the circle, we must add π to the angle we found above. Therefore we have:
Combined with the r we found previously, we have the point (5√2, [(5π)/4]). We can check this against the picture we drew at the beginning, and we see that the polar coordinates are fairly reasonable and match up with the rectangular coordinates we started with.θ = π 4+ π = 5π 4
 When converting from rectangular coordinates to polar, make sure to start off by drawing a quick sketch to help you see what's going on:
 From the lesson, we have the identities
Finding r is easy, we simply solve by plugging in to the identity:r^{2} = x^{2} + y^{2} tanθ = y x
Working out θ is a little more difficult, though, as we will see:r^{2} = (3)^{2} + (−4)^{2} ⇒ r =
√
25⇒ r = 5
However!, notice that the problem requires us to have 0 ≤ θ < 2π. Instead, the angle we got was if we went clockwise (instead of counterclockwise) to the direction of the point.tanθ = −4 3⇒ θ = tan^{−1} ⎛
⎝−4 3⎞
⎠⇒ θ ≈ −0.927  So what's going on here? Inverse tangent (tan^{−1}) can only give results of (−[(π)/2], [(π)/2]). Looking at our picture, though, we see that θ is equivalent to the angle we found above, it's just going clockwise (−) instead of counterclockwise (+). Thus, to force it to go in the normal, counterclockwise direction we can add 2π, which will result in a positive θ that goes in the same direction:
Combined with the r we found previously, we have the point (5, 5.356). We can check this against the picture we drew at the beginning, and we see that the polar coordinates are fairly reasonable and match up with the rectangular coordinates we started with.θ = −0.927 + 2π ≈ 5.356
*These practice questions are only helpful when you work on them offline on a piece of paper and then use the solution steps function to check your answer.
Answer
Polar Coordinates
Lecture Slides are screencaptured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.
 Intro
 Introduction
 Plotting Points with Polar Coordinates
 The Distance of the Point from the Origin
 The Angle of the Point
 Give Points as the Ordered Pair (r,θ)
 Visualizing Plotting in Polar Coordinates
 First Way We Can Plot
 Second Way We Can Plot
 First, We'll Look at Visualizing r, Then θ
 Rotate the Length CounterClockwise by θ
 Alternatively, We Can Visualize θ, Then r
 'Polar Graph Paper'
 Horizontal and Vertical Tick Marks Are Not Useful for Polar
 Use Concentric Circles to Helps Up See Distance From the Pole
 Can Use Arc Sectors to See Angles
 Multiple Ways to Name a Point
 Negative Values for r
 If r Is Negative, We Go In The Direction Opposite the One That The Angle θ Points Out
 Another Way to Name the Same Point: Add π to θ and Make r Negative
 Converting Between Rectangular and Polar
 Rectangular Way to Name
 Polar Way to Name
 The Rectangular System Must Have a Right Angle Because It's Based on a Rectangle
 Connect Both Systems Through Basic Trigonometry
 Equation to Convert From Polar to Rectangular Coordinate Systems
 Equation to Convert From Rectangular to Polar Coordinate Systems
 Converting to Rectangular is Easy
 Converting to Polar is a Bit Trickier
 Draw Pictures
 Example 1
 Example 2
 Example 3
 Example 4
 Example 5
 Intro 0:00
 Introduction 0:04
 Polar Coordinates Give Us a Way To Describe the Location of a Point
 Polar Equations and Functions
 Plotting Points with Polar Coordinates 1:06
 The Distance of the Point from the Origin
 The Angle of the Point
 Give Points as the Ordered Pair (r,θ)
 Visualizing Plotting in Polar Coordinates 2:32
 First Way We Can Plot
 Second Way We Can Plot
 First, We'll Look at Visualizing r, Then θ
 Rotate the Length CounterClockwise by θ
 Alternatively, We Can Visualize θ, Then r
 'Polar Graph Paper' 6:17
 Horizontal and Vertical Tick Marks Are Not Useful for Polar
 Use Concentric Circles to Helps Up See Distance From the Pole
 Can Use Arc Sectors to See Angles
 Multiple Ways to Name a Point 9:17
 Examples
 For Any Angle θ, We Can Make an Equivalent Angle
 Negative Values for r 11:58
 If r Is Negative, We Go In The Direction Opposite the One That The Angle θ Points Out
 Another Way to Name the Same Point: Add π to θ and Make r Negative
 Converting Between Rectangular and Polar 14:37
 Rectangular Way to Name
 Polar Way to Name
 The Rectangular System Must Have a Right Angle Because It's Based on a Rectangle
 Connect Both Systems Through Basic Trigonometry
 Equation to Convert From Polar to Rectangular Coordinate Systems
 Equation to Convert From Rectangular to Polar Coordinate Systems
 Converting to Rectangular is Easy
 Converting to Polar is a Bit Trickier
 Draw Pictures 18:55
 Example 1 19:50
 Example 2 25:17
 Example 3 31:05
 Example 4 35:56
 Example 5 41:49
Precalculus with Limits Online Course
Transcription: Polar Coordinates
Hiwelcome back to Educator.com.0000
Today, we are going to talk about polar coordinates.0002
Previously, whenever we have talked about the location of a point on the plane,0005
we have described its horizontal and vertical distance from the origin, x and y0008
how much we go out horizontally and how much we go out vertically.0013
We call them rectangular coordinates because, if we look at a horizontal and a vertical put together,0016
we are just drawing out a rectangle on the plane; so they are called rectangular coordinates.0020
Now, we are going to look at a totally new way to talk about coordinates; we are going to talk about polar coordinates.0026
This gives us a new way to describe the location of a point.0031
Instead of using horizontal and vertical components, we can talk about the point's distance from the origin0033
the distance from the center, and the angle that it is onwhat angle the point is on.0038
This gives us a whole new way to talk about location in the plane.0043
Using it, we can create graphs like we have never seen before.0046
We will explore these in the next lesson, Polar Equations and Functions.0049
For now, though, let's work on a strong understanding of just what polar coordinates are and how they work.0053
We really have to have a good understanding of how polar coordinates work, so that it is an intuitive thing,0057
before we will really be able to use it in equations and functions; so let's get that learned in this lesson.0061
All right, we plot points with polar coordinates using two things: r, the distance of the point from the origin0067
how far we are from the origin (and from here on, we are going to call the origin the pole; the center of the graph0073
will be called the pole now, since we are talking about polar coordinates; and in a little while,0078
we will see why it makes sense to be talking about the pole; and there is sort of vaguely a connection0083
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).0087
All right, next we have θ, the angle that the point occurs on.0092
We measure this counterclockwise from where we used to have the positive xaxis.0098
Our positive xaxis used to go out to the right, and we measure counterclockwise from that, just like we measured angles in the unit circle.0102
On the unit circle, we always measured angles by going counterclockwise from that positive xaxis.0110
We always spun counterclockwise; so it is a lot like when we worked with the unit circle in that manner.0118
We give points as the ordered pair (r,θ); it is distance first, then angle(distance,angle).0123
It is normally assumed that θ is going to be in the unit of radians.0131
Degrees are used occasionally in polar coordinates, but much, much less often.0135
Mostly, unless you see it really explicitly shown otherwise (there is a degree symbol or something else0140
to make us definitely sure that we are talking about degrees), just assume that it is radians,0145
because that will be the usual thing we are talking about when we are using polar coordinates.0148
All right, how can we visualize polar coordinates?0152
When we plot in rectangular coordinates, we usually think of a point (x,y) in one of two ways:0154
the "go left/right by x," how much we go horizontally, and then how much we go vertically, "go up/down by y."0159
We do these two things, and we get to some location.0167
Alternatively, we can do the "go up/down by y" first; we can do that vertical motion first, and then our horizontal motion.0170
But we end up getting to the same location.0179
Either way is fine; both go to the same place.0181
Likewise, there are two ways to visualize polar coordinates; so let's see both of them.0184
First, we will look at visualizing r, then θ.0189
The first thing we do: we can go out a length of r from the pole, what we used to call the origin,0193
the center of the graph, directly to the right.0198
So, if we have...here is our center, where I am holding this ruler.0201
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.0206
We go out a distance of r to the right from that pole.0213
The next step is that we rotate the length counterclockwise by whatever our angle θ is.0217
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.0222
This distance here is r; that is what we are ending up seeingwe have just put out some distance, and then we spin the distance.0233
And combining these two things, we arrive at some point (r,θ), some distance, comma, angle.0240
Alternatively, we can visualize θ, then r; so first, we rotate counterclockwise by θ, and we create an imaginary line at that angle.0246
We spin from that right xaxis, what we normally used to call our positive xaxis, but now it is just going right from the center.0255
We spin up some angle θ; so what we will have is this little stub.0262
We spin up some little stub; and so, going off in this direction is now going off in the angle θ.0266
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).0271
We are already at some angle θ, this imaginary line right here; and then we just end up going out the distance of r.0281
We go out our distance r along that direction, and we end up getting out to (r,θ).0290
We spin, and then we go out to the distance.0296
Either of these two ways is just fine for visualizing polar coordinatesthey both work perfectly reasonably.0299
Whichever one makes a lot more sense to you, that is the way I would recommend visualizing it.0305
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.0308
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.0315
Also, now that we have some sense of how polar coordinates work, visualizing it,0321
we can also see why we call them polar coordinateswhy the idea of a pole makes sense.0324
Imagine if you were standing on the North Pole.0328
If we wanted to talk about any location on Earth, well, we could talk about it as:0330
you stand on the pole; you face in some direction (you just choose a direction, arbitrarily, to face in);0333
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.0339
Some place in South Africa: you just spin some angle, and you walk down to it.0346
Some place in Uruguay: you just spin to some angle, and you walk down to it.0350
Some place in America: you just spin to some angle, and you walk down to it.0354
Wherever you are going to go to, it is starting at some pole; you spin around that pole, and then you walk a distance.0358
Or, alternatively, you can walk a distance, and then spin around (effectively) to some different longitude.0364
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.0369
All right, polar graph paper: normally, when we graph rectangular coordinates,0376
we think in terms of these tick marks on the axes: 1, 2, 3, 1, 2, 3, positive 1, positive 2, positive 3,0382
1, 2, 3...to help us see horizontal and vertical distance.0391
We can easily see that, at this point here, we have a distance of 2 horizontally and a distance of 2 vertically.0395
But horizontal and vertical tick marks aren't so useful for polar.0402
How far is the distance from here to here, based on these tick marks here and here?0406
We can't figure out how far we are from the center, how far we are from the pole,0411
based on these locations of these tick marks for horizontal and vertical tick marks.0415
So, for polar, we need a new way of marking our graph paper so that we can see things more easily.0419
We do this with concentric circles; we use concentric circles to help us see distance from the pole.0427
The first circle (at least in this graph of concentric circlesconcentric just means one, and then another around that,0432
and another around that, and so on and so on)this first one would be at a distance of 1.0437
Anything on this circle is a distance of 1 from the pole.0441
This next one: anything on this circle is a distance of 2 from the pole.0446
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.0451
It is like horizontal and vertical tick marks, but instead, it is for how far you are from the center of the graph.0457
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 coordinatesthe distance from the center.0463
It works really well to think in terms of these concentric circles.0472
Similarly, we need some way to be able to think about the angle θ.0475
We use as a reference...we can talk about the angle θ with arc sectors; that is lines coming out of the origin.0478
In this, we have lines coming out of the origin here and here and here, and we continue around in this way out.0483
We see that it is cut a total of 12 times; so an entire revolution would be 2π;0491
so an entire, cut...2π divided by 12 times that it has been cut would be π/6 for each one.0497
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.0503
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.0512
And the concentric circles let us easily see our distance from the center.0520
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.0523
Now, we don't necessarily have to format our graph with concentric circles and arc sectors,0529
just like we don't have to use tick marks when we are using rectangular coordinates.0534
We can make a graph that doesn't have any tick marks on it, and still probably understand what is going on.0538
But it often makes it easier; it will make it easier to graph things, so it is useful to have it.0542
It is not absolutely necessary, and there will be some times where it is not worth it for us to put them down.0547
But if we are trying to be really careful with the graph, it is a good idea to sketch these in first,0551
so we can be really accurate when we are drawing it out.0554
Something that is special about the polar coordinates is that, unlike rectangular coordinates, there are multiple ways to name the same point.0558
There are multiple ways to name a single point in polar coordinates.0565
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).0569
They are both going to end up being out at a distance of 3 on this third circle here.0578
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?0585
Well, think about this: π/4 ends up being here, but we can break 9π/4 into 2π + π/4.0589
So, 9π/4 is just the same thing as 2π + π/4.0599
What we have here is that 9π/4 is effectively spinning the first 2π; it is making an entire revolution,0602
and then it is doing the last π/4 to end up being on this angle; and then, it puts the same distance out.0608
Our first one is π/4 at a distance of 3; but then, the next one is 9π/4...0614
so it just makes an entire revolution for the first 2π, but then that last part of 9π/4,0620
after that 2π, will just be an additional π/4; so it ends up being at the same angle.0626
We end up getting an equivalent point for polar coordinates; so this is something special about polar coordinates.0630
We can end up lapping any given θ away; any θ where we can talk about a specific angle...0635
we can lap it by adding 2π or adding multiples of 2π.0640
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.0644
You add 2π; you just do one loop counterclockwise.0654
If you add 4π, you do two loops counterclockwise; if you add 2π, you do a loop clockwise.0658
However you end up adding a multiple of 2π, whether it is adding or subtracting it,0665
you end up just getting back to where you started,0669
so it has no real effect, just like when we worked with the unit circle.0671
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.0675
If we want to avoid thisif we want to avoid being able to accidentally loop and show up at the same angle0682
we can restrict our θ to be between 0 and 2π, so it is allowed to be 0, inclusive,0688
but not allowed to get up to 2π, since the 0 and the 2π match up.0694
But often, we won't actually make this restriction; we very often won't use this restriction.0698
It sometimes can be used, but very often, we will be allowed to go larger than 2π.0703
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.0707
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.0713
There is also another idea that we can talk about: we can talk about negative values for r.0719
So, when we talked about the idea of r, we just talked about r representing how far out we are.0723
But we never required it to be positive.0728
If it is positive, that makes sense: we go to some angle, and then we go out by r.0731
But we need a way to interpret negative values for r, because what if our r wasn't positivewhat if we ended up having a negative value for r?0735
If we have a negative value for r, we do the following.0742
If r is negative, we go in the direction opposite the one that angle θ points out.0745
For example, in this one here, we have (2,3π/4); so the first thing that we do is spin the angle to 3π/4.0753
And that goes off in this direction here.0762
But then, we have 2; so instead of going off in this direction, we go in the opposite direction, down to this way.0765
So, the negative says to go in the direction opposite that which the angle does.0774
It is like we are spinning to some angle, but then we end up going in the opposite direction, like this.0778
We spin to some angle, but then we end up spinning in the opposite direction by whatever our r's absolute value is.0784
Alternatively, this is the way of thinking "spin, then go out the distance"; if you really prefer thinking0790
in terms of "distance out, then spin," you can think of r as going left.0795
So, here would be a 2: r = 2 would go like this.0800
And then, we just end up spinningthe same thing, just like normalcounterclockwise to 3π/4.0807
So, as opposed to positive r coming out from our pole like this, we end up having r come out the other way.0812
And then, we just end up spinning, like usual, to whatever our point ends up being.0820
This means that there is yet another alternative way to name the same point.0825
We can add π to the θ and then make r negative.0829
Why does this workwhy does this end up giving the same point?0833
Well, imagine that we had some point that we normally got to, with some θ and some r distance out.0836
Well, if we add π to θ, we end up spinning to the opposite direction.0841
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.0846
π to θ puts us in the opposite direction, but negative on r puts us in the opposite direction.0852
Opposite opposite means that we are back where we started; so combining these two things0856
means that we have yet another way to express the same point.0860
So, that is something we really have to keep in mind.0864
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.0865
We have to think about if this could be the same thing as something else.0873
How about converting between rectangular and polar?0878
At this point, for a point in the plane, we now have two ways to name it.0880
We can talk about the rectangular (x,y) coordinates, how far we go out horizontally and how far we go out vertically.0883
But we can also talk about the polar coordinateshow far we go out in distance from the pole and how much we spin that angle:0889
the distance out we go, r, and then the angle that we spin.0897
So, how can we convert between the two coordinate systems?0901
They both end up calling out the exact same point up here; so how can we convert between the two?0903
Well, let's just layer both of them down simultaneously.0909
If we look at both systems mapping the same point, we see that they are both mapping the same point.0911
So now, we just want to see how they relate to each other through this diagramthrough this picture.0915
Another thing to point out is that, since it is a rectangular coordinate system,0920
we know that it has to have a right angle in the corner, because it is based on a rectangle.0924
A horizontal portion and a vertical portion means that where they meet, they have to have a right angle right there.0928
So, we end up seeing a right angle in the triangle we have made with r, x, and y.0933
Great; a right trianglewhat we have here is something that we can work with using basic trigonometry.0938
Looking at both systems simultaneously, we can connect through basic trigonometry.0944
This triangle right here has an angle in the corner and has some right angle in the other corner.0948
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.0956
It is good stuff that we have been learning since geometry.0964
So, let's start working through these: we can relate θ to our other information:0968
cos(θ) would be equal to the adjacent, divided by the hypotenuse; so that gets us cos(θ) = x/r.0972
sin(θ) is going to be equal to the opposite, divided by the hypotenuse; so that gets us sin(θ) = y/r.0982
r^{2} = x^{2} + y^{2} because of the Pythagorean theorem.0992
The hypotenuse, squared, is equal to the other two legs, squared and added together.0997
So, we have r^{2} = x^{2} + y^{2}.1002
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.1005
Great; so we can use the following equations to convert between coordinate systems.1015
We figured out where they are coming from; so at this point, we can go from polar to rectangular.1019
We had cos(θ) = x/r; we just multiply both sides, and we get x = rcos(θ); similarly, y = rsin(θ).1024
On the other direction, rectangular to polar, we had r^{2} = x^{2} + y^{2} and tan(θ) = y/x.1033
Converting to rectangular is easy; if we want to convert to rectangular,1040
all we have to do is plug in r and θ, and we will automatically get the right x and y.1044
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.1049
That part is pretty easy; converting to polar, though, is a bit trickier.1057
We can't tell if r is positive or negative, because it is r^{2} = x^{2} + y^{2}.1062
So, r could be negative; r could be positive; x could be positive; y could be negative.1069
The squareds are going to end up turning everything into things that look positive.1072
So, we can't tell if it ends up being positive or negative from its equation.1076
Worse, the function tan^{1} can only output from π/2 to π/2.1079
Remember: the arctan, the tan^{1} function, only outputs on this side of the unit circle.1085
That is one of the things that we talked about when we learned that in trigonometry.1092
So, tan^{1} can only output from π/2 to π/2, and this is a problem,1095
because since we have tan(θ) here, if we want to figure out what just θ is,1099
we are going to have to use tan^{1} on the way to figuring out what θ is.1103
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.1108
That means that we will end up being restricted to just seeing one side.1115
So, converting to polar is this kind of difficult thing.1118
r^{2} = x^{2} + y^{2}: we don't know if r should be positive or negative from that information.1121
tan(θ) = y/x: we don't really have a great way to talk about it.1125
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.1129
How do we deal with this? To get around these limitations, make sure you always, always1134
(I am serious about this) draw a picture whenever you are converting from rectangular to polar coordinates.1139
So, pictures are the way we are going to solve this issue.1146
The picture does not need to be extremely accurateit doesn't need to be a perfect picture.1148
But it needs to be clear enough for you to see which quadrant the point you are talking about is in.1152
And it will give you a sense of what angle to expect, a sense of what things should come out to be.1157
It will be a sanity check that lets you see, "Oh, my answer is possibly right" or "my answer is obviously, clearly wrong."1162
It lets you have some idea of what is going on; and we will see how useful this is in Example 4 and Example 5.1169
Of course, it wouldn't hurt to draw a picture when you are converting to rectangular, eithermore pictures are never a bad thing.1175
But it is not quite as necessary; still, a visual aid always helps, so I would recommend:1180
the first couple of times you do this, draw a picture; it will help you see what is going on1184
and see the relationship between rectangular and polar coordinates.1187
All right, we are ready to talk about some examples.1190
All right, plot the points below: we have this nice diagram with concentric circles and lines (arc sectors) to cut things up.1194
So, let's just mark out what all of our arc sectors are here, first.1202
We see that there is a total of 12 pieces that it has been cut up into.1206
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/6,1209
which is just 5π/6; 6π/6, which will just be π; 7π/6, which is 7π/6,1221
8π/6, which is 4π/3; 9π/6, which will be 3π/2; 10π/6, which will be 5π/3; 11π/6;1227
and we now wrap to 2π, so we are back where we started, at 0.1240
OK, with that in mind, we can see how to plot points; let's starting plotting some points.1244
The first point is (2,π/3); so if we are on 2, we will be on the concentric circle representing a length of 2.1248
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.1255
The next one we will mark in blue: (1,225°): like I said, degrees are sometimes used in polar coordinates.1269
They are pretty rare, but we can deal with them; we can work between radians and degrees when we need to.1277
But mostly, we will end up sticking with radians.1281
But we do have to know how to deal with it.1283
Where would 225 degrees appear? Well, 180 degrees gets us to π.1285
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.1290
So, we are going to just cut this arc sector right down the middlethe other arc sector is 30 degrees, so we can split down the middle with this one.1299
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.1306
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.1314
5π/4 means that we are going to go to π; we spin to π here, and then we go an additional π/4.1323
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.1331
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.1340
The next one: (2.5,7π/6); 7π/6: we mark out that that one will be this angle right here.1348
So, here we are on this one; what is 2.5?1358
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.1360
Next, we will go back to red, but this time we will mark it with a star after we get the point down.1370
(4,3π/4): the first thing we do is spin to 3π/4.1375
Here is π/2; and then, here is another π/4; so we are on the same angle as when we figured out 5π/4.1380
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.1387
4 means we are going to go, not out on this angle, but in the opposite direction.1395
Negatives tell us to go in the opposite direction; so we are going out 4, so 4 out in the opposite direction.1401
We are on the fourth circle; we are here, and we will mark that with a big star to help us see the difference.1407
The final point: (3,1.47): this is probably the hardest point of all.1415
Since it doesn't have a degree symbol, we know that 1.47 is in radians.1420
1.47 radians is an anglebut what is 1.47 radians?1425
I don't know how to do very well with a decimal form of radians.1431
I am used to π/2, π/3, π/6, and things like that; so how do we deal with 1.47 radians?1433
Well, one way is to just get some reference pointssome reference ideas.1440
What does π mean in decimal? What does π/2 mean in decimal?1444
Well, π is about the same thing as 3.14; π/2 is about the same thing as 1.57.1448
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/2.1458
If we want to have an even better idea of what it is, we could divide 1.47, the angle we are going to,1467
and then divide it by π/2, which ends up being about 1.57.1477
When we plug that into a calculator, that comes out to be around .94, which means we are 94% of the way to π/2.1481
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.1490
But we can also think of it as a ratio, so we can get this decimal, which is basically a percent.1496
We can think that we are 94% of the way from 0 to π/2.1500
94% of the waythat means we are going to be very close to it.1505
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.1508
The next example: In polar coordinates, name each point below.1517
Then give two alternate ways to name them, where one of them will have r < 0.1521
First, we see that this is at what distance? Well, here is the 1 circle and the 2 circle.1527
We do have to make sure that the scale for our circles ends up being 1, 2, 3.1531
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.1537
We could have 20, 40, 60 be the scale on our tick marks.1544
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.1547
So here, indeed, is the 5; so we have r = 5.1553
We are at a distance of r = 5; what is the angle that we are at?1558
Well, we are on this, and we can see that it is cut into π/6; so this is π/3.1561
So, we have θ = π/3; so our first, most basic point, the one we would normally call out, will be (5,π/3):1566
a distance of 5 from the center, at an angle, a counterclockwise spin angle, of π/3.1575
Oops, I accidentally meant to write π/3; I didn't mean to write π/6 the second time.1580
We are at π/3 there; now remember: we can talk about any of these points in multiple ways,1584
because if we spin an additional 2π, we end up just getting to the same place.1590
Here, if we are at this place, and then we just spin another 2π, we are still at the same place.1593
So, we can just add an entire rotation onto it; so we will take θ = π/3;1598
and now, we can just add: so π/3 + 2πwhat does that end up being?1603
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).1608
Alternatively, we could also, if we want to get r less than 0 (the last part of this problem1620
requires us to get the less than 0 value for our r, because we can always go in the opposite direction)...1625
what if we had spun to the opposite direction?1630
If we had spun to the opposite direction, well, this would be a total angle for θ of 4π/3.1632
So, at opposite angles, opposite θ ends up being 4π/3.1644
So, the opposite r: well, normally our r is equal to 5.1650
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,1655
so we can go back in the direction we want to go), that would be 5.1660
Put those two things together, and we have the point (5,4π/3).1665
We have our normal canonical way, the way we would usually talk about this, (5,π/3).1674
But then, there are other things, where we have just done an additional rotation (5,7π/3),1680
or when we spin in the opposite direction, and then go in the opposite opposite direction by having that negative value for r.1684
All right, next: let's look at the blue dotwhere would this end up being?1693
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.1700
What angle is it at? Well, we see that it is along this angle here, which splits the entire quadrant evenly.1710
You split quadrants on π/4; to get from here to here would be π, so an additional π/4 would be 5π/4.1720
So, our normal θ would be equal to 5π/4.1728
The normal way to talk about this point would be (3,5π/4); great.1734
However, we can also talk about this in different ways.1741
Remember: we can go any addition of spins around, so any number of 2π's added to the θ.1743
And we will still end up being at the same place.1749
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.1750
Whatever the multiple is, is how many times we spin around.1758
So, if we want to spin around 5 times, we would add 10π, 5 times 2π.1760
We can take 5π/4, and we can add 10π to that; what does that end up coming out to be?1764
That ends up being 45π/4; so we ended up spinning to the 5π/4,1770
and then we just count 1, 2, 3, 4, 5...and we land back right where we had been previously.1776
We have another way of talking about this point as the same distance, 3, comma...1782
but now we have a totally different angle for it, 45π/4.1786
Great; and our last point, where r is less than 0, where we end up going in the opposite direction:1791
well, the easiest way to call this out would be: what would be the opposite direction?1800
Well, we are splitting the quadrant this way; we could talk about this as being π/4.1803
So, that would be π/4 here, and then, if this is the direction we normally go, we are going the opposite direction.1808
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.1815
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/4.1825
So, there are two different ways of talking about it; both are perfectly fine.1833
π/4...and then the opposite r...well, normally our r is 3, so if we are going to have the opposite r,1836
once we are going the opposite direction in our angle, we want to have 3.1845
We put these two things together, and we end up getting (3,π/4).1849
We have our angle pointing us in the opposite direction; but then, the negative in our r tells us to walk the opposite direction.1855
Opposite opposite means that we get to the point that we wanted to be at.1861
All right, the third example: Convert the polar coordinates to rectangular.1864
Our first one is (8,5π/6) for this one; let's start by just drawing thissketching, really quickly, about where that would end up showing.1868
8a distance of 8; 5π/6...well, we end up spinning to somewhere out here for 5π/6 and then a distance of 8.1877
We are at something like that; we don't know for sure what the point is going to be from that drawing.1884
But we do at least have a sanity check that whatever we end up getting, it should be in the second quadrant.1891
We know that our y should come out to be positive, and we know that our x should come out to be negative.1896
It is a negative x and a positive y if we are going to be in this quadrant right here.1900
So, if we are going to be in that quadrant, we have some sanity check to help us out here.1904
Now, what are the formulas? Remember: the formulas for figuring out our rectangular coordinates1908
from polar are x = rcos(θ) and y = rsin(θ).1913
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...1920
You can draw it out and figure it out.1926
All right, let's work through this: here is our r, and here is our θ.1928
That means that x is equal to 8 times cosine of 5π/6.1933
We work with the unit circle; we remember that cosine of 5π/6...8 times cos(5π/6) is √3 over 2.1940
We are going to end up getting 4√3 as our value for x.1951
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.1955
8sin(θ) is 5π/6; remember, we are on the unit circle, so now we have 8 times sine of 5π/6.1963
Looking on the unit circle, we get positive 1/2; it isn't below the xaxis yet; so 8 times 1/2...we get positive 4, so our y is equal to +4.1972
So, putting these two pieces of information together, we end up getting the point (4√3,4).1982
If we end up looking against the point that we figured out here, that seems to makes sense.1989
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).1992
So, this seems pretty reasonable; it checks out for at least a quick sanity check.2000
All right, the next one: (5,7π/4): let's draw what this would end up looking like.2004
7π/4 is going to spin clockwise, as opposed to counterclockwise; so we spin to 7π/4.2012
That would put us at the same angle as having spun positive 5π/4, somewhere out here.2017
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.2021
We are going to be at some point out here at 5.2029
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.2032
We should have a negative xvalue and a negative yvalue.2038
The same basic thing to figure this out: x = rcos(θ): our r is 5; our cosine is 7π/4.2041
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/4,2052
as we can see from our picture (another reason why pictures are useful).2061
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 2.2064
And we have x = 5√2/2basically the same as if we are working through 5sin(7π/4).2076
We work that through; once again, we have 7π/4 as an equivalent to having written just π/4.2092
So, y is equal to 5sin(π/4); we also could remember that, since we are splitting the quadrant,2097
it is going to end up just being the same thing as it was for cosine.2104
y = 5 times √2/2, just like we got for cosine, because we are splitting that first quadrant.2108
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).2115
And if we were to compare that to the picture that we originally drew,2129
just to give us a quick idea of what is going on, yes, that seems perfectly reasonable.2133
(5√2/2,5√2/2): that is in the third quadrant, because both our x and our y are negative.2137
And we see that this point here splits the x and the y pretty evenly, which makes sense,2143
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.2147
So, it passes the sanity check; we have a pretty good check to see that we are probably right.2152
All right, next, Example 4: Convert the rectangular coordinates to polar, and give your answers with r greater than 0,2156
and 0 less than or equal to θ, which is less than 2π.2162
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.2165
It is a fairly normal restriction here that we might have in this sort of thing.2173
It is not necessary, but it doesn't really hurt us or make this problem much harder to do.2176
The first thing we want to do: we always, always, always want to do this.2181
When we are converting from rectangular to polar, always, always begin by drawing a picture.2184
So, the first thing we do is draw a picture.2190
It doesn't have to be absolutely perfect, but we want to have some sense of what is going on here.2192
And also, we will see how drawing a picture can be really useful for figuring out the numbers.2195
So, we have 3 horizontally and then 3√3 vertically.2199
Here is 3 out this way and 3√3 out this way.2206
OK, so what were our formulas? They were r^{2} = x^{2} + y^{2} and tan(θ) = y/x.2211
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.2222
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.2228
But if we had just used them blindly, then we would end up having r^{2} = x^{2} + y^{2}:2234
so our 3 and 3√3 are x and y: r^{2} is equal to (3)^{2} + (3√3)^{2}.2240
So, we have r^{2} = +9 +...3 squared is 9; √3 squared is 3; 9 times 3 is 27.2252
So, we have r^{2} = 36; take the square root of both sides,2260
and we know that we can't have a negative r, so we know that it has to be positive,2265
even though we get that ± technically; but generally we will leave it as positive; so r = 6; great.2269
We have figured out our r distancethat doesn't seem so problematic.2275
Well, let's try using the tan(θ) = y/x.2277
Greatlet's toss it in: tan(θ) = our y, 3√3, over our x, 3; OK.2281
We see that the 3's will cancel, and we will end up having √3; great.2290
We take the arctan of both sides; you punch that into a calculator, and you end up getting π/3.2295
Wait, what? That doesn't make sense!how is that possible?2303
How can we get π/3? If we go back and look at our picture, this is our θ right here.2308
This here is not the case; that does not work out; that doesn't make sense; what is going on here?2314
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.2319
Arctan talks about this side of the unit circle; but the point that we want is over here on this side.2326
So, arctan can't get us to our point; so here is what we do instead.2332
We end up looking at this as just a normal triangle.2335
If it was just a normal triangle, then this would be a length of 3.2339
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.2343
And this would be 3√3; and now, we could talk about this angle in here as being α.2346
Then, tan(α) would be equal to the opposite, 3√3, divided by the adjacent, 3; so we would get tan(α) = √3...2352
we take the inverse tangent of both sides; tan^{1}(√3) is π/3, so we get π/3 from this.2364
However, this is α, not θ; what we want is θ, and what we have here is the α inside of it.2372
But how can we figure out what θ is?2379
Well, if we were to draw in the axes, the θ we want is this one here.2380
We have figured out what α is; well, how do θ and α connect together?2386
Well, this here would be π in terms of the angle.2389
So, what we have is: we have that θ + α is equal to π for this specific picture.2394
And now, how these two things will relate is depending on the picture we are dealing with.2402
So, we really have to keep our wits about us and think about what we are looking at.2405
If we just try to do it blindlytry to rely on some formulachances are that we are more likely to make mistakes2408
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.2413
But if you end up making a picture and thinking about how this angle connects,2417
and paying attention to the unit circle, it is not that hard to do at all.2420
So, θ + α = π; we just figured out that α is equal to π/3.2423
So, we have θ + π/3 = π, which means that our θ must be 2π/3.2428
And that makes total sense over here, since that is 2π/3, and inside of the triangle is π/3.2437
We add those together, and we would have the entire top arc from here to here.2441
That makes perfect sense; so we now have...here is our θ.2447
Figuring out the r still didn't cause any problems for us; that is our 6.2452
And so, we have the point; we put these two pieces of information together.2456
And the distance is 6, and the angle is 2π/3.2461
And I hope this makes clear why it is so, so important to figure out how these things work,2465
if we are converting from rectangular to polarwhy it is so important to draw a picture.2471
If we don't draw a picture, we are going to get lost.2475
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.2478
So, it is (6,2π/3).2482
Also, if we were to draw that as a picture, where would we get?2484
Well, 2π/3 would end up being out here, and a distance of 6 is about here; compare that to what we got here.2487
That is pretty good; it seems to pass the sanity checkit seems like we probably got the answer right; great.2494
Convert the rectangular coordinates to polar, and give your answers with r > 0 and 0 ≤ θ < 2π.2499
It is basically the same problem as previous, but now we have (17,19), new points.2504
The first thing we do: we always draw a picture.2509
Now, at this point, we understand how useful a picture is.2511
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.2515
17 would be out here; 19 would be down here; this is a distance of 19.2522
It is 19 as a coordinate, but we can also just treat it as being a distance of 19.2531
Here is a distance of 17: our r is going to be this side of the triangle.2535
Remember: it is a right triangle, and we know that...not θ; θ is not inside of the triangle;2540
θ was the counterclockwise spin, like this; this is our θ; what we have inside of the triangle2546
is some other angle; I am deciding to call it α; you can call it smileyface; you could call it any symbol that you want to use.2552
α is an easy one for me to draw, so I am calling it α inside of the triangle.2557
We can figure out r pretty easily; it is just a right triangle; we can figure out α pretty easilya basic trigonometric function, tan(θ).2562
And then, we can use θ and α's interaction together to figure out what θ has to be.2568
So, first, let's figure out what our r is.2573
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.2577
We work that out; so we have r^{2} is equal to...I'll take the square root of both sides...2587
and eventually, we work this all out: 17^{2} + 19^{2}...and it simplifies to r = 5√26.2594
And you could probably get away with putting that in a decimal form if you wanted; but exactly what it is, is r = 5√26.2603
Next, let's figure out what α has to be.2609
Well, since α is in here, we see that tan(α) is equal to...what is the opposite?2612
The opposite, away, opposite from that angle, is 19; so 19 divided by...the adjacent to that angle is 17,2617
so divided by 17; we take the arctan of both sides; α is equal to the arctan of 19/17.2626
Now, that doesn't come out to be any friendly, nice form.2635
That is not a really nice number to have to deal with.2638
But if we punch it into a calculator, it will end up coming out to be approximately 0.84.2641
If we wanted to have it absolutely perfectly, we would have to leave it as tan^{1}(19/17).2647
But we can probably have a decimal approximation; so α is approximately equal to 0.842653
will probably do, for our purposes of answering this question.2658
Now, how do we get to the actual answer? Well, we have to figure out how θ and α relate.2660
Well, if we did a spin here, we could do one entire revolution: one entire revolution would be 2π.2664
So, we see that θ, combined with α...if we spin most of the way,2671
and then we have the angle that we do in the triangle, well, that comes out to be one entire revolution.2675
One entire revolution is 2π; so that means that θ + α, in this case, comes out to be 2π.2679
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.2687
There is not just a simple formula here; it is thinking and using picturespictures and thinking.2691
Without thinking, nothing will end up working.2695
θ + α = 2π; what is our α? We have that our α is 0.84.2698
So, θ + 0.84 = 2π; actually, let's write it a little bit differentlyI prefer to write it as θ = 2π  α.2702
Then, we will substitute in: so we have θ = 2π  0.84, which gives us that θ is approximately equal to...2714
plug in 2π; 2 times 3.14, minus 0.84; we get approximately 5.44 for our value here.2723
That means we could plot the point as being 5√26.2731
Our r value over here of 5√26, and comma...5.44...sorry, θ comes out to be 5.44.2737
So, 5.44: that is what we end up figuring out as our approximate value.2749
If we wanted it precisely, we could leave it as 2π  tan^{1}(19/17).2753
But we will probably end up being OK with just having this approximate value right here at 5.44.2758
Now, we didn't do this in the last problem, but I want you to see that you can also check your work here.2765
If you are not quite sure that it all came out right, we can check our work; we can go to polar,2769
but then, since it is so easy to convert from polar back to rectangular, we just plug them in.2774
Remember: to convert to rectangular, x = rcos(θ); y = rsin(θ).2777
So, I will put in the middle this checking box; if we want to do a check, well, x = rcos(θ),2785
so that means x is equal to...what is our r? Our r is 5√26.2793
So, 5√26 times cosine of 5.44: we plug that into a calculator, and we end up getting x = 16.96.2798
What did we have originally? We originally had 17.2812
And remember: we ended up having some roundoff error, because we had to round,2815
because it was tan^{1}(19/17), so we rounded to 0.84.2818
So, that makes sense, that we are ending up seeing 16.96 compared to 17.2822
It is just roundoff error, but basically, our answer is correct, so it checks out.2826
The same thing for y: y is equal to r times sin(θ); our r here is 5√26, and our sine is 5.44.2830
We plug that into a calculator, and we end up getting 19.04.2840
Compare that, once again, to 19: the only thing we are seeing here is a slight bit of roundoff error.2845
This checks out, because we knew that we would have some roundingoff error,2849
because we had rounded it, as opposed to using what it was precisely.2852
So, if you are ever in a situation where it really, really matters that you end up getting this right2856
like you are on a test, or you are confused about how this stuff is working out2859
just end up checking it; it is so easy to convert from polar to rectangular.2862
The hard part is converting from rectangular to polar.2866
So, if you are in polar, you might as well convert back to rectangular at the end,2869
and check and make sure that you got the answer right,2871
especially if it really ends up mattering, like in a situation where you are taking a test or an exam of some sort.2873
All right, so now that we have a good understanding of how polar coordinates work,2878
we are ready to talk about polar equations and functions and really get into graphing this stuff.2881
All right, we will see you at Educator.com latergoodbye!2885
1 answer
Last reply by: Professor SelhorstJones
Thu Feb 12, 2015 1:29 AM
Post by Kishor Pant on February 10, 2015
You are a great teacher! Thank you! :)