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 1 answerLast reply by: Professor Selhorst-JonesMon Jan 30, 2017 11:45 AMPost by Hana Abid on January 30 at 07:56:14 AMHi, how would you plot (-1,-pi/4)? thanks 1 answerLast reply by: Professor Selhorst-JonesThu Feb 12, 2015 1:29 AMPost by Kishor Pant on February 10, 2015You are a great teacher! Thank you! :)

### 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 counter-clockwise from where we used to have the positive x-axis, like we measured angles in the unit circle.)
In polar coordinates, we give points as the ordered pair (r, θ).
• 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:     r2 = x2 + y2           tanθ = y/x
Converting to polar can be tricky, though, even with the above formulas. We can't tell if r is positive or negative from its equation. Worse, the function tan−1 can only output from −[(π)/2] to [(π)/2] (this is a problem because we have to use it to find θ). To get around these limitations, make sure to always draw a picture when converting from rectangular to polar coordinates. The picture does not need to be extremely accurate, just clear enough for you to see which quadrant the point is in and help give you a sense of what angle to expect.

### Polar Coordinates

Plot each of the below points (given in polar coordinates).
 ⎛⎝ 2 , π6 ⎞⎠ ⎛⎝ 1 , 3π4 ⎞⎠ ⎛⎝ 3 , 5π3 ⎞⎠
• 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 x-axis, 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.
Plot each of the below points (given in polar coordinates).
 ⎛⎝ −2 , π6 ⎞⎠ ⎛⎝ 3 , − π2 ⎞⎠ ⎛⎝ −1 , − 2π3 ⎞⎠
• 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 x-axis, 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 x-axis; (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.
Plot each of the below points (given in polar coordinates).
 ⎛⎝ 2.5 , π3 ⎞⎠ ( 1 , 3.1 )            ( −3 , 0.9 )
• 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 x-axis, 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 r-values 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:
 π6 ≈ 0.52 π4 ≈ 0.79 π3 ≈ 1.05 π2 ≈ 1.57
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.
 Radians:   θ = 0.9     ⇒     Degrees:   θ = 0.9 2π ·360°    ≈     51.57°
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.]
In polar coordinates, name the point below. Then give two alternate ways to name it, where one of the ways has r < 0.
• 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:
 θ   is equivalent to   θ+k·(2π),  where k is any integer
This allows us to create a whole bevy of equivalent polar coordinate points:
 ( 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 half-rotation). 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.
( 3,  π),    ( 3,  −π),    ( −3,  0 ) [Note: There are an infinite number of ways to answer this question, because there are an infinite number of ways to name the location in polar coordinates. The three points above just represent a few of the possible ways to name the point. See the steps to the problem for more information.]
In polar coordinates, name the point below. Then give two alternate ways to name it, where one of the ways has r < 0.
• 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, 5π3 ⎞⎠ .
• 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:
 θ   is equivalent to   θ+k·(2π),  where k is any integer
This allows us to create a whole slew of equivalent polar coordinate points:
 ⎛⎝ 2, 5π3 ⎞⎠ ≡ ⎛⎝ 2, 11π3 ⎞⎠ ≡ ⎛⎝ 2,  − π3 ⎞⎠ ≡ ⎛⎝ 2,  − 7π3 ⎞⎠ ≡ ⎛⎝ 2, 65π3 ⎞⎠ ≡     …
• 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 half-rotation). 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.
( 2,  [(5π)/3] ),    ( 2,  −[(π)/3] ),    ( −2,  [(2π)/3] ) [Note: There are an infinite number of ways to answer this question, because there are an infinite number of ways to name the location in polar coordinates. The three points above just represent a few of the possible ways to name the point. See the steps to the problem for more information.]
In polar coordinates, name the point below. Then give two alternate ways to name it.
• 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.
(0, 0),    (0,  [(9π)/2] ),    (0, −4.7 ) [Note: There are an infinite number of ways to answer this question, because there are an infinite number of ways to name the location in polar coordinates. The three points above just represent a few of the possible ways to name the point. See the steps to the problem for more information.]
Convert the point from polar coordinates to rectangular coordinates: (4, [(5π)/6] ).
• 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 ⎛⎝ 5π6 ⎞⎠ ⎢⎢ y = 4sin ⎛⎝ 5π6 ⎞⎠

 x = 4 ⎛⎝ − √3 2 ⎞⎠ ⎢⎢ y = 4 ⎛⎝ 1 2 ⎞⎠

 x = −2√3 ⎢⎢ y = 2
Thus, we now have our rectangular coordinates: (x, y)     ⇒     (−2√3,  2).
(−2√3,  2)
Convert the point from polar coordinates to rectangular coordinates: (−8, 7.2 ). [Give your answer to three decimal places of accuracy.]
• 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)

 x = −4.867 ⎢⎢ y = −6.349
Thus, we now have our rectangular coordinates: (x, y)     ⇒     (−4.867,  −6.349).
(−4.867,  −6.349)
Convert the point from rectangular coordinates to polar coordinates: (−5, −5). Give the polar coordinates such that r > 0 and 0 ≤ θ < 2π.
• 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
 r2 = x2 + y2                      tanθ = y x
Finding r is easy, we simply solve by plugging in to the identity:
 r2 = (−5)2 + (−5)2     ⇒     r = √ 50 ⇒     r = 5√2
Working out θ is a little more difficult, though, as we will see:
 tanθ = −5 −5 ⇒     tanθ = 1     ⇒     θ = tan−1(1)     ⇒     θ = π4
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.
• 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:
 θ = π4 + π    = 5π4
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.
(5√2, [(5π)/4] )
Convert the point from rectangular coordinates to polar coordinates: (3, −4). Give the polar coordinates such that r > 0 and 0 ≤ θ < 2π.
• 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
 r2 = x2 + y2                      tanθ = y x
Finding r is easy, we simply solve by plugging in to the identity:
 r2 = (3)2 + (−4)2     ⇒     r = √ 25 ⇒     r = 5
Working out θ is a little more difficult, though, as we will see:
 tanθ = −4 3 ⇒     θ = tan−1 ⎛⎝ −4 3 ⎞⎠ ⇒     θ ≈ −0.927
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.
• 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:
 θ = −0.927 + 2π   ≈     5.356
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.
(5,  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.

### Polar Coordinates

Lecture Slides are screen-captured 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 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 Counter-Clockwise 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

### Transcription: Polar Coordinates

Hi--welcome 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 y--0008

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 origin--0033

the distance from the center, and the angle that it is on--what 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 origin--0067

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 x-axis.0098

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.0102

On the unit circle, we always measured angles by going counterclockwise from that positive x-axis.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 seeing--we 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 x-axis, what we normally used to call our positive x-axis, 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 coordinates--they 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 coordinates--why 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 circles--concentric 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 coordinates--the 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 this--if we want to avoid being able to accidentally loop and show up at the same angle--0682

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 positive--what 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 spinning--the same thing, just like normal--counterclockwise 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 work--why 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 coordinates--how 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 diagram--through 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 triangle--what 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

r2 = x2 + y2 because of the Pythagorean theorem.0992

The hypotenuse, squared, is equal to the other two legs, squared and added together.0997

So, we have r2 = x2 + y2.1002

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 r2 = x2 + y2 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 r2 = x2 + y2.1062

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

r2 = x2 + y2: 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 accurate--it 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, either--more 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, 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 middle--the 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

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 points--some 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 way--that 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 + 2π--what 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

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 dot--where 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

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

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

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 this--sketching, really quickly, about where that would end up showing.1868

8--a distance of 8; 5π/6...well, we end up spinning to somewhere out here for 5π/6 and then a distance of 8.1877

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 x-axis 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 x-value and a negative y-value.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√2/2--basically 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 r2 = x2 + y2 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 r2 = x2 + y2:2234

so our -3 and 3√3 are x and y: r2 is equal to (-3)2 + (3√3)2.2240

So, we have r2 = +9 +...3 squared is 9; √3 squared is 3; 9 times 3 is 27.2252

So, we have r2 = 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 distance--that doesn't seem so problematic.2275

Well, let's try using the tan(θ) = y/x.2277

Great--let'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 blindly--try to rely on some formula--chances 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 polar--why 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 check--it 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 smiley-face; 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 easily--a 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 r2 is equal to...I'll take the square root of both sides...2587

and eventually, we work this all out: 172 + 192...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 pictures--pictures 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 differently--I 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 round-off 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 round-off 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 round-off error.2845

This checks out, because we knew that we would have some rounding-off 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 right--2856

like you are on a test, or you are confused about how this stuff is working out--2859

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 later--goodbye!2885