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Polar Equations & Functions

  • We can set up equations and functions using r and θ exactly the same way as we did for x and y. Generally, θ is the independent variable (like x was), while r is the dependent variable (like y was).
  • We always assume that θ is in radians ([(π)/2],  [(7π)/4], etc.).
  • We graph polar equations/functions in the same way we graph "normal" stuff: plug in values, plot points, and connect with curves to make a graph. Since θ is the independent variable, we plug in some value for it, then see what distance r we get out at that angle.
  • Just like graphing with rectangular equations, you don't need to plot a huge number of points-merely enough to sketch the graph. Many polar equations involve trigonometric functions. The "interesting" points are when the trig function produces a zero or an extreme value (sin, cos ⇒ ±1). Figure out where these interesting values will occur and use them to help plot the graph.
  • If at anytime you're unsure how the graph will behave, just plot more points. The easiest way to work out your uncertainty is by calculating more points.
  • Occasionally you will see an equation that only uses one variable: that's fine! It means that variable is fixed, while the other can change freely.
  • Sometimes we'll want to convert an entire equation or function from polar to rectangular, or vice-versa. We can do so with the same conversion formulas we figured out and used in the previous lesson. Since these formulas were based off any x, y, r, θ, they work fine for substituting in equations.
    • Polar ⇒ Rectangular:    x = r cosθ              y = rsinθ
    • Rectangular ⇒ Polar:     r2 = x2 + y2           tanθ = y/x
  • If you have access to a graphing calculator, it's great to try graphing some polar equations with it. It's a new way of looking at graphing, so it helps just to play around. For more information, check out the appendix on graphing calculators.

Polar Equations & Functions

Fill in the table of values for the polar function below.
r(θ) = 3+cos(2θ)                   
θ
r
0
π

4
π

2

4
π

4

2

4
  • Filling in a table of values for a polar function is the same as doing it for a normal function. The input value is θ for the function, so we plug in some value for θ, then we see what output value we get for r.
  • For example, to find r(0), we plug in like normal:
    r(0) = 3 + cos(2 ·0)     =     3 + cos(0)     =     3+1     =     4
  • Repeat the above process for each value of θ in the table:
    θ
    r
    0
    4
    π

    4
    3
    π

    2
    2

    4
    3
    π
    4

    4
    3

    2
    2

    4
    3
    4
    [Notice that the values of r repeat after θ = π. This is because cos(2θ) has a period of π. Still, the values in the table after that are important because the θ values are distinctly different. When it comes time to draw polar graphs, remember that the θ value is just as key to plotting a point as the r value.]
θ
r
0
4
[(π)/4]
3
[(π)/2]
2
[(3π)/4]
3
π
4
[(5π)/4]
3
[(3π)/2]
2
[(7π)/4]
3
4
Below is the graph of r(θ) = 3cos(5 θ). Find an interval of θ where the graph is only drawn out once. (That is, the graph does not get re-traced as later values for θ wind up creating matching locations to those already drawn in by earlier values for θ.)
  • In polar coordinates, it's possible (and common) for numerically different points to give the same location. For example, all of the below points are equivalent:
    (7, π)    ≡     (7, 3π)    ≡     (−7, 0)
    [If this idea is unfamiliar to you, make sure to check out the previous lesson on the idea of polar coordinates. Being comfortable with this idea and how polar coordinates are plotted is critical to understanding polar equations/functions.]
  • We are looking for when the equation r=3 cos(5θ) begins to repeat locations. While it will never repeat the same coordinate numerically, it will eventually create equivalent coordinates, and the equation will re-trace the locations the graph has covered in previous points. Let us arbitrarily choose a starting angle of θ = 0. We could start before that, but 0 is a nice, easy place to start. With that in mind, notice that we will be pointing back in the same direction at θ = 2π. Thus, if we get the same r value for θ = 0 and θ = 2π, we see that the graph will be repeating then. Plug in to check:
    r(0) = 3cos(5 ·0)    =   3       
           r(2π) = 3cos(5 ·2π)    =   3
    Furthermore, if we continue onward from θ = 2π, all the values will be the exact same as if we had started at θ = 0 because of the periodic nature of the trig function. Thus, the graph is repeating after 2π. However, there is one more thing to keep in mind...
  • It is also possible to have duplicate locations from different coordinates if the r is negative in one of them while the θ points in the opposite direction. For example, (7, π)  ≡  (−7,  2π). This means that, assuming we start at θ = 0, there is another, earlier opportunity for repetition: θ = π. If that causes the function to spit out a negative version of the r, they will be giving equivalent locations:
    r(0) = 3cos(5 ·0)    =   3       
           r(π) = 3cos(5 ·π)    =   −3
    Thus we see that, yes, r(0) and r(π) graph to the exact same location. Furthermore, if we look at how r(θ) will behave for angles starting at θ = π, we see that it will produce the exact opposite (negative) to what it produced when starting at θ = 0. [This happens because θ = π catches cos(5θ) in the exact middle of a period interval, so everything is flipped to the opposite from there on out.] Thus, we see that when we start at θ = 0, the first repetition of the function begins at θ = π. This means we can create the entire graph of the function by just using the interval  θ: [0,  π).
  • Finally, if you found all the above a little difficult to follow, don't worry! Parametric coordinates are confusing when they're new to you, and parametric equations/functions are even tougher the first time around. A great way to help yourself understand what's happening is to use a graphing calculator. Enter in the function to be graphed (don't forget to give it an interval of θ to graph, most graphing calculators will require you to enter one), then try out the trace function. Trace the graph using the calculator and get a better understanding of how it's working. Feel free to play around with the functions you're looking at too! Adjust some numbers, try other trig functions, and just generally explore. The easiest way to get comfortable with polar equations is by exploring them, and a graphing calculator makes it possible to see a lot of different things very quickly.
θ: [0,  π)
Graph the polar equation:   r = sin(θ).
  • Unless you're already extremely familiar with a certain type of equation, it's almost always necessary to create a table of values to help see how the function graphs. Make sure to get enough values so that you understand what's going on in the equation and can comfortably graph it. If you're ever unsure about what it will look like, just plot more points until it becomes clear to you.
  • We might create a table of values like the one below:
    θ
    r
    0
    0
    π

    4
    0.71
    π

    2
    1

    4
    0.71
    π
    0

    4
    −0.71

    2
    −1

    4
    −0.71
    0
  • Once you have enough points to feel comfortable with how the equation works, you're ready to start plotting. With polar equations, it's very important to keep in mind that the r value changes based on the θ involved. Thus, as the angle you are drawing on rotates, the distance from the origin varies. Pay careful attention to this as you connect the points with curves: don't just go directly from point to point, make sure to fulfill the changing θ-direction as r grows or shrinks (or goes negative!). For this problem in specific, notice how sin(θ) hits its maximum at θ = [(π)/2]. It starts at 0, then increases to 1 as it rotates, then comes back down to 0. The graph is then re-traced by r becoming negative, telling us to go in the opposite direction to the θ angle, causing us to cover the same graph again.
Graph the polar equation θ = − [(π)/6].
  • Notice that this polar equation has no r term in it whatsoever. The only thing that has a restriction on it is the θ term.
  • This means that θ is restricted to only being −[(π)/6], but that r can be anything at all (since r does not appear anywhere, the equation puts no restrictions on it). Thus, we set the angle of θ = −[(π)/6] while r is allowed to run the interval of (−∞, ∞). This is similar to a rectangular equation like x=5. We set x as 5, then y is allowed to run anywhere, so we get a vertical line out of that equation.
  • When drawing in the graph, don't forget that negative values for r cause them to go "backwards". This means θ sets an angle to point in, and the various possible r values fill out the forward and backward directions for that angle, creating a line.
Graph the polar equation  r=√{θ} with the restriction that θ ≤ 4 π.
  • Begin by noticing that while there is an explicit upper restriction on what θ can be (the problem says θ ≤ 4 π), there is also an implied lower restriction. The equation we're working with is r=√{θ}: if θ < 0, then the equation won't make any sense, because we can't take the square root of negative. Thus, the only domain we have to consider for this problem is θ: [0, 4π].
  • Once we know where we're looking in terms of θ, we approach the problem like other graphing problems. Create a table of values so you have a sense of how r grows and changes and so you can later plot points. Using a calculator to find approximate values for r, we have:
    θ
    r
    0
    0
    π

    2
    1.25
    π
    1.77

    2
    2.17
    2.51

    2
    2.80
    3.07

    2
    3.32
    3.54
    As always, if you're unsure about how the graph works or would like to have more plots to point, just calculate some extra points until you feel comfortable with what you have.
  • Once you have enough points to feel comfortable with how the equation works, you're ready to start plotting. With polar equations, it's very important to keep in mind that the r value changes based on the θ involved. Thus, as the angle you are drawing on rotates, the distance from the origin varies. Pay careful attention to this as you connect the points with curves: don't just go directly from point to point, make sure to fulfill the changing θ-direction as r grows or shrinks (or goes negative!). For this problem in specific, notice that r=√{θ} grows very quickly at first, but slows down the larger it gets. This means that the changes when it first starts out at θ = 0 will be especially quick, so it would probably be useful to get a few extra points for those early θ-values:
    θ
    r
    π

    6
    0.72
    π

    4
    0.88
    π

    3
    1.02
    This goes to show just how quickly r grows for early values of θ. Although r=0 when θ = 0, it shoots out very quickly for the first tiny bit of turning. As θ turns more, the growth rate of r slows down.
Graph the polar function  r(θ) = 1+4sin(2θ).
  • Unless you're already extremely familiar with a certain type of equation, it's almost always necessary to create a table of values to help see how the function graphs. Make sure to get enough values so that you understand what's going on in the equation and can comfortably graph it. If you're ever unsure about what it will look like, just plot more points until it becomes clear to you. When figuring out what values of θ to use in your table, think in terms of which values will be "interesting". For example, the most "interesting" values for cos(x) are x=0, [(π)/2], π, [(3π)/2]. These values are extremely important because they make up the zeros, maximum, and minimum for cos(x). Along these lines, think of what values for θ will be "interesting" for sin(2θ).
  • Because sin(2θ) effectively goes twice as "fast" as sin(θ), this causes all the interesting values to occur on intervals of [(π)/4]. We will see our zeros, maximums, and minimums all fall on the below values:
    θ = 0,    π

    4
    ,    π

    2
    ,   

    4
    ,   …
    Furthermore, notice that sin(2θ) will start repeating outputs after π since π is its period. We still have to care about the θ-values after π because, even though the r-values will repeat, the θ-values will indicate new angles and thus create unique locations.
    θ
    r
    0
    1
    π

    4
    5
    π

    2
    1

    4
    −3
    π
    1

    4
    5

    2
    1

    4
    −3
    1
    As always, if you're unsure about how the graph works or would like to have more points to plot, just calculate some extra points until you feel comfortable with what you have.
  • Once you have enough points to feel comfortable with how the equation works, you're ready to start plotting. With polar equations, it's very important to keep in mind that the r value changes based on the θ involved. Thus, as the angle you are drawing on rotates, the distance from the origin varies. Pay careful attention to this as you connect the points with curves: don't just go directly from point to point, make sure to fulfill the changing θ-direction as r grows or shrinks (or goes negative!). For this problem in specific, notice that we get the exact same r values, in the same order, for θ from [0, π] as we do for [π, 2 π]. This, combined with how the angle is spinning, causes the graph to mirror around the origin. Noticing this mirroring effect can make it easier to draw in the graph. As you get really skilled, you can even notice this sort of mirroring before ever making the table, allowing you to create graphs much more quickly but without losing any quality.
Convert the equation from polar to rectangular.
r = cos(θ)
  • We can convert from polar coordinates to rectangular coordinates through the use of the following identities:
    x = rcosθ,       y = rsinθ,        r2 = x2 + y2,       tanθ = y

    x
    When using them, it's almost always a good idea to swap one entire side of the identity for the other entire side. For example, modifying the first identity to [x/r] = cosθ so you can swap out cosθ for the fraction [x/r] is almost never a good idea-you want to convert to one coordinate type, not use a mix of them.
  • With this idea in mind, we see that it's difficult to directly swap in an identity for the equation r = cos(θ): neither side directly fits one of the identities. However, we see that we're close to an identity. If we had r2 or r cosθ, we could match up to an identity. We can achieve both of these by multiplying each side of the equation by r:
    r = cos(θ)     ⇒     r ·r = r·cos(θ)     ⇒     r2 = r cos(θ)
  • With this new formation of the original equation, it's now quite easy to swap based on identities:
    r2 = r cos(θ)     ⇒     (x2 + y2) = (x)     ⇒     x2 + y2 = x
  • At this point, we're done: we've achieved a rectangular equation. However, if we want to understand it even better, we can go one more step and put this into a form we're used to-a circle:
    x2 + y2 = x     ⇒     x2 − x + y2 = 0     ⇒     x2 − x + 1

    4
    + y2 = 1

    4
        ⇒    
    x− 1

    2

    2

     
    + y2 =
    1

    2

    2

     
    If we're familiar with conic sections, we see that this is the graph of a circle centered at ([1/2], 0) with a radius of [1/2]: exactly what we would have gotten by graphing the original polar equation.
x2 + y2 = x, or, equivalently, (x−[1/2])2 + y2 = ([1/2] )2
Convert the equation from polar to rectangular, then solve for y.
r= −1

1+sinθ
  • We can convert from polar coordinates to rectangular coordinates through the use of the following identities:
    x = rcosθ,       y = rsinθ,        r2 = x2 + y2,       tanθ = y

    x
    When using them, it's almost always a good idea to swap one entire side of the identity for the other entire side. For example, modifying the first identity to [x/r] = cosθ so you can swap out cosθ for the fraction [x/r] is almost never a good idea-you want to convert to one coordinate type, not use a mix of them.
  • Looking at the equation, we see that there's nothing we can currently swap out with an identity. However, having that fraction on the right side isn't helping things. Let's get rid of that by multiplying both sides by the denominator of 1+sinθ:
    r= −1

    1+sinθ
        ⇒     r (1+sinθ) = −1     ⇒     r + r sinθ = −1
    Now we see that we can apply the identity y = rsinθ:
    r + r sinθ = −1     ⇒     r + y = −1
  • At this point, we still need to convert that r into something rectangular. The problem is that the only identity that uses just r is based on r2, which we don't currently have. However, we can get r2 to appear if we move the y to the other side, then square both sides:
    r+ y = −1     ⇒     r = −1 − y     ⇒     r2 = (−1−y)2
    Now that r2 has appeared, we can apply the identity r2 = x2 + y2:
    r2 = (−1−y)2     ⇒     x2 + y2 = (−1−y)2
  • We now have a rectangular equation, but the problem told us to also solve for y, so we need to get y alone on one side. Start off by expanding:
    x2 + y2 = (−1−y)2     ⇒     x2 + y2 = 1 +2y + y2
    We can subtract y2 on both sides, then get the y alone:
    x2 + y2 = 1 +2y + y2     ⇒     x2 = 1+2y     ⇒     x2 − 1 = 2y     ⇒     1

    2
    x2 1

    2
    = y
y = [1/2] x2 − [1/2]
Convert the rectangular equation to polar form.
y = x
  • We can convert from polar coordinates to rectangular coordinates through the use of the following identities:
    x = rcosθ,       y = rsinθ,        r2 = x2 + y2,       tanθ = y

    x
    When using them, it's almost always a good idea to swap one entire side of the identity for the other entire side. For example, modifying the first identity to [x/r] = cosθ so you can swap out cosθ for the fraction [x/r] is almost never a good idea-you want to convert to one coordinate type, not use a mix of them.
  • We can apply the identities right from the start because they have x and y alone in them. Swap out based on the identities:
    y = x     ⇒     r sinθ = r cosθ
  • At this point, we've technically completed the problem, since we have an equation in polar form. Still, it might be nice to simplify it in case we wanted to graph it. In that case, start off by canceling out the r's on each side;
    r sinθ = r cosθ    ⇒     sinθ = cosθ
    Next, we can combine the trig functions into a single tangent function by dividing cosθ on both sides:
    sinθ = cosθ    ⇒     sinθ

    cosθ
    = 1     ⇒     tanθ = 1
    Finally, to make it super easy to graph, we can take the inverse tangent of both sides to just get θ alone:
    tanθ = 1     ⇒     θ = tan−1 ( 1)     ⇒     θ = π

    4
    This makes sense! The rectangular equation y=x is just a straight line that goes up at an angle of 45° (because it has a slope of 1). The polar equation gives us the exact same line, just like it should.
θ = [(π)/4]
Convert the rectangular equation to polar form, then solve for r.
x2−9 = 6y
  • We can convert from polar coordinates to rectangular coordinates through the use of the following identities:
    x = rcosθ,       y = rsinθ,        r2 = x2 + y2,       tanθ = y

    x
    When using them, it's almost always a good idea to swap one entire side of the identity for the other entire side. For example, modifying the first identity to [x/r] = cosθ so you can swap out cosθ for the fraction [x/r] is almost never a good idea-you want to convert to one coordinate type, not use a mix of them.
  • This is a very tricky problem. Our knee-jerk reaction would be to swap out x and y based on the identities, as below:
    x2−9 = 6y     ⇒     (r cosθ)2 − 9 = 6 (rsinθ)     ⇒     r2 cos2 θ− 9 = 6 ·r sinθ
    However, it turns out that it's actually pretty difficult to solve for r from the above equation. Give it a try, and you'll quickly notice how tough it is to pin down r. [It is possible, but you have to use the identity sin2θ+ cos2 θ = 1 to change the cos2 to a sin2, then also do something similar to what we do below.] Instead, it will help us a bit to take a different approach we might not have initially thought of...
  • Notice that we have x2 on the left. One of the identities we have is r2 = x2 + y2, so alternatively, if we had y2 show up, we could use that identity. Let's try that instead:
    x2 − 9 = 6y     ⇒     x2 + y2 − 9 = y2 + 6y     ⇒     r2 − 9 = y2 + 6y
    At this point, we might try moving the 9 to the other side:
    r2 − 9 = y2 + 6y     ⇒     r2 = y2 + 6y + 9
    Which then might cause us to realize we can factor it:
    r2 = y2 + 6y + 9     ⇒     r2 = (y+3)2
  • Our goal is to eventually get r alone, so since we now have something squared on both sides, we might as well take the square root. Remember that taking a square root causes a `±' to show up.
    r2 = (y+3)2     ⇒    

     

    r2
     
    = ±

     

    (y+3)2
     
        ⇒     r = ±(y+3)
    We still need to fully convert to polar, so switch out the y:
    r = ±(y+3)     ⇒     r = ±(rsinθ+ 3)
    Remember, ± means that there is both a + version and − version, so we can split the above into its two versions:
    r = rsinθ+ 3              
        r = −(rsinθ+3)     ⇒     r = −rsinθ− 3
  • Finally, we have an equation that is entirely polar and it's not too difficult to solve for r. To do so, we need to get all the r's on one side, then pull out the r, then put everything else on the other side.
    r=rsinθ+ 3       
           r = −rsinθ− 3

    r − r sinθ = 3       
           r + r sinθ = −3

    r(1−sinθ) = 3       
           r(1+sinθ) = −3

    r = 3

    1−sinθ
          
           r = −3

    1+sinθ
    At this point, we have a polar equation and we've solved for r. There is a little bit of strangeness, though...  How can we have two different solutions? Each of those equations is distinct from the other, but somehow they are both conversions of the same rectangular equation? That's weird.
  • It turns out, there's an explanation. If you graph either of them, you get the exact same graph, which is pictured below. Each of the equations produces the same parabola that the original rectangular equation gave, they just do it from different starting locations. The graph of r = [3/(1−sinθ)] "starts" from the positive x-axis at 3, then works up to the right, until eventually flipping to the left side. The other one starts differently: r = [(−3)/(1+sinθ)] "starts" from the negative x-axis at −3, then works down at first, eventually coming up on the right side, and then also flipping like the other. The fact that we can express the original rectangular equation as two different polar equations is based on how we draw polar coordinates. We can name the same location with different sets of polar coordinates, and these two equations correspond to the different ways we can name the points. We can name with positive ("forward") r or we can name them with negative ("backward") r, which is the two types that come out of the equations.
r = [3/(1−sinθ)]     or     r = [(−3)/(1+sinθ)] [See the last step for a discussion of why the rectangular equation can be turned into two different polar equations.]

*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 Equations & Functions

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
  • Equations and Functions 1:16
    • Independent Variable
    • Dependent Variable
    • Examples
    • Always Assume That θ Is In Radians
  • Graphing in Polar Coordinates 3:29
    • Graph is the Same Way We Graph 'Normal' Stuff
    • Example
  • Graphing in Polar - Example, Cont. 6:45
  • Tips for Graphing 9:23
    • Notice Patterns
    • Repetition
  • Graphing Equations of One Variable 14:39
  • Converting Coordinate Types 16:16
    • Use the Same Conversion Formulas From the Previous Lesson
  • Interesting Graphs 17:48
    • Example 1
    • Example 2
  • Graphing Calculators, Yay! 19:07
    • Plot Random Things, Alter Equations You Understand, Get a Sense for How Polar Stuff Works
    • Check Out the Appendix
  • Example 1 21:36
  • Example 2 28:13
  • Example 3 34:24
  • Example 4 35:52

Transcription: Polar Equations & Functions

Hi--welcome back to Educator.com.0000

Today, we are going to talk about polar equations and functions.0002

In the previous lesson, we introduced a new way to look at location in the plane, polar coordinates.0005

By now, we have seen thousands of graphs (probably literally), where x and y are in relationship with each other.0010

As the horizontal changes, the vertical changes somehow.0018

Either x and y are both in an equation, and we graph the equation, and we figure out all of the solutions to that equation;0020

we graph it; or y is a function of x, and we graph the function--as x changes, how will y respond?0025

We can do the exact same thing with polar coordinates: the variables r and θ can be put in some sort of relationship,0031

and then we can graph the resulting polar coordinates.0037

Before you watch the lesson, make sure you watch the previous lesson (before watching this one).0039

You really, really need to understand polar coordinates on their own.0045

You have to be able to understand how polar coordinates make locations on the plane before any of this stuff is going to really make sense.0048

So, if you are having difficulty with working through polar equations, but you don't have a really good understanding0056

of polar coordinates yet, that is the thing--you really want to work on polar coordinates.0060

Make sure you have watched the previous lesson before watching this one,0063

and that you are comfortable with polar coordinates before you try to work on polar equations and functions.0065

And if you can make sense of how polar coordinates work, polar equations and functions probably won't actually be that much harder.0070

All right, let's get started.0074

We can set up equations and functions using r and θ, exactly the same way as we did for x and y.0077

Generally, θ is going to be the independent variable, like x was.0082

Our x was allowed to change and vary around, and then r will be our dependent variable, in the same way that y was.0086

So, x was allowed to change around, and y responded to x's changes.0094

Similarly, here it is going to be θ that will be allowed to change around, and our distance r will respond to the angle that we are at.0098

Here are some examples; we could have an equation r = 1 + 2cos(θ) (we plug in a θ, and it tells us an r),0106

or a function r(θ) = 3sin(2θ), which is the same thing as...we plug in a θ, and it gives us what the r-value will be.0113

So, while r could be the independent, and θ the dependent, such relationships are pretty uncommon in polar equations and functions.0121

We normally think in terms of how length changes based on the angle--0132

if we go to this angle, what length it will be at--and not the opposite way,0136

just like when we are doing a rectangular graph, we normally think in terms of "for this horizontal location,0140

what height will I be at? For this horizontal location, what height will I be at?" and not0145

"For this height, what horizontal location will I be at?"0150

We don't normally think of height and then horizontal; we think of horizontal, and then height,0152

because x is the independent, and y is the dependent, just like here θ is the independent and r is the dependent.0156

We always, always assume that θ is in radians.0164

Whenever we are looking at our θ, it is assumed that θ is going to be in radians.0169

That is numbers like π/2, 7π/4, decimal things, etc.0173

It could be explicitly put in degrees, but that would be extremely rare.0177

Always, always, always use radians, unless you are being told explicitly otherwise, that this thing is in degrees.0182

It is extremely uncommon: you almost never see something like that.0189

I can't even think of one time I have seen it; so just don't expect that to happen--expect to be using radians when you are working with polar equations and graphs.0193

So, always think in terms of radians: you are plugging in radian values, and you are plotting with radian values on your angles.0202

All right, how do we graph in polar coordinates?0210

We graph polar equations and functions pretty much the same way that we graphed normal stuff.0212

You plug in some value for your independent (in this case, θ).0217

You plot the point that gets put out; and then you connect the whole thing with curves to make it into a graph.0220

So, θ is the independent variable; we plug in some value for it, and then we see what distance r would get out.0226

Over here, let's plot some points: if we are looking at the equation r = 1 + 2cos(θ), then r is going to come out once we plug in some θ.0231

So, if we plug in 0, well, 1 + 2cos(0)...cos(0) is 1, so 2 times 1 is 2, and 1 + 2 is 3.0239

So, we now have the point...not (0,3), but (3,0); that is one little confusing thing--0248

the fact that it is not (x,y); it is now (r,θ); so our independent variable is actually the thing that comes second.0255

So, don't let that confuse you.0261

Our distance out is 3, and our θ is going to be an angle of 0.0263

So, we end up getting this point right here; we have an angle of 0; we are at 0 above the starting location.0269

We haven't moved at all, and we are out on the third circle out; so we are at (3,0).0276

Next, π/4: if we plug in π/4, 1 + 2cos(π/4)...cosine of π/4 is √2/2; 2(√2/2) would be √2;0282

so that means that we get 1 + √2 out of this.0293

1 + √2...we figure that out with a calculator, so we can actually plot something down.0296

That is approximately 2.41; so at this point, we are at angle π/4, so we are on this arc sector line right here.0300

Notice, it is broken into eight pieces, so each one of them is going to be π/4, because up here is π/2.0309

So, we are at the line of π/4 angle; and then we go out 2.41: (2.41,π/4) is the point that we get out of this.0315

So, we are at 2.41 out, somewhere that looks around 2.4, a little more than 2, but even more less than 3.0328

So, we get this point right here; OK.0336

The same thing if we plug in π/2: if we plug in π/2, 1 + 2cos(π/2)...cosine of π/2 is 0, so we just get 1.0339

So, we have the point (1,π/2); and that gets us this point here.0348

So now, we think about how these things are going to connect through curves.0354

If we are really confused about it, we could just plot down more points.0357

We could put down π/6 and π/3, as well.0360

And if we want even more points, we could continue to plot down more and more points.0362

But we would probably be able to get a pretty good sense with just π/6 and π/3.0366

But we can even just figure this out by thinking, "Well, how is it curving--what is happening?"0370

So, as the angle goes up, notice that, because it is cosine of θ, as this goes to larger and larger values,0373

0 to π/4 to π/2, cosine shrinks down and down.0381

So, this portion of our equation, the 2cos(θ), will get smaller and smaller as cos(θ) gets smaller and smaller.0384

So, as we get larger and larger, it shrinks down to ultimately this value of 1 right here.0391

We go up, and as it gets farther out, it shrinks down; it shrinks down; it cuts through here; it shrinks down; it shrinks down.0396

And we get something like that curve.0403

We can see it precisely here, now drawn by a computer: r = 1 + 2cos(θ).0405

At this point, we just keep plotting more points.0411

Here is a new, interesting one to think about: if we plotted 3π/4, well, that is 1 + 2cos(3π/4).0413

What does 3π/4 come out to be? That is 1 + 2(-√2/2), so that gets us 1 - √2.0421

We approximate that with a calculator; we get -0.41, which means that we have the point (-0.41,3π/4).0430

So, here is our 3π/4; it is right here; this is the 3π/4 location.0440

We are on this line; but we have -0.41, so we are going in the opposite direction, and we are here for our point.0447

We plug in π; 2 times cosine of π; what is cos(π)? It is -1.0455

So, 2 times -1 get us -2; 1 - 2 gets us -1; so we have gotten the point (-1,π) out of this.0462

So, at the angle of π (here is angle π), we are going to be going opposite the length of 1, and so we are here, as well.0470

So, once again, it is continuing to drop down; this part here is continuing to get smaller and smaller, until eventually it becomes negative.0478

It actually continues out to this curve here, and then it cuts through.0485

It will always have to cut through the origin, because if it gets a 0 in the r, then it has to go through the origin,0488

because distant 0 from the origin, distant 0 from the pole, means at the pole; cut through here.0493

And then, curve up to here; and that is what we have so far.0500

It is computer-drawn there, so it is a little more accurate; but we basically just keep plotting points.0503

If we plug in 5π/4, we see that that is at -0.41; so at 5π/4, that would be this angle here.0508

So, we are in the opposite now; we are over here at 3π/2; we have length 1, so at 3π/2, we are here; length is 1 in that direction.0515

7π/4: we are at a length of approximately 2.41, so at 7π/4...we are at 2.41 from here.0523

And we continue this curving; we can also see at this point, perhaps, that is going to end up being symmetric.0529

If we looked at the entire table of values, we would see that there is some symmetry going on in the way these distances are coming out,0534

and that we are going to end up seeing the top part...this curve so far will happen and just sort of flip over;0539

and we will end up getting it out like this.0547

And finally, here is a computer-drawn version that is better than my slightly-imperfect drawing.0553

And that is what we end up getting out of this.0559

You plot points; you work it out; you graph the whole thing.0560

Just like graphing with rectangular equations, you don't need to plot a huge number of points.0564

It won't hurt; if you plot more points, it is not going to hurt.0568

But you really only have to plot enough so that you can sketch the graph.0571

Which points should you plot--what are the useful points to plot--what are the interesting points to plot?0575

Many polar equations involve trigonometric functions; the interesting points are when the trigonometric function produces a zero,0580

when we plug some θ into that trigonometric function and it puts out 0,0588

or when it puts out an extreme value (for sine and cosine, that is positive or negative 1).0592

Plugging in those values for our θ lets us see what the most extreme values are that our function is going to make,0597

which helps us have an understanding of how the whole thing is behaving.0602

More points will just make it smoother and easier to make the curves.0605

But those are the most important ones of all; so you definitely want to make sure that you are plotting those.0608

If we want to try to plot these extreme values and these 0's, we want to figure out where these interesting values will occur.0613

How are we going to get to it? Trigonometric functions tend to have a pattern.0618

We are used to working with sine and cosine; so we want to think in terms of sine--0621

we have to plug in 0 or π/2 or π; if it is sin(θ), we have to plug in 0 or π/2 or π or 3π/2...0625

any of these would end up getting an interesting number.0633

Sin(0) is 0; sin(π/2) is 1; sin(π) is 0; sin(3π/2) is -1; we are used to working with that from all of our work in trigonometry.0635

If we go to something else, though, like cos(5θ), well, it is not just θ anymore; it is 5 times that.0644

So, if it were cosine of something, we would want that something to be 0, and then π/2, and then π, and then 3π/2, and then 2π, and so on.0651

But in this case, it is 5 times θ; so that has to be taken into account in how our thing works.0658

So, the 0 is still just going to be the same; if we plug in a 0, 5 times 0 still gets a 0.0663

But if we wanted to try to figure out π/2 = 5θ, what does our θ have to go in to get that interesting first value of π/2?0669

We divide both sides by 5, and we would get π/10 = θ.0677

π/10 = θ, because if we plug in π/10, 5 times π/10 is π/2; cos(π/2) gets us 0.0682

So, that is our next interesting thing.0688

Similarly, it is going to continue on this pattern of π/10 being the interval here.0690

π/5 (I said that the wrong way...) 5 times π/5 would get us π; cos(π) is -1, an interesting value.0695

The cosine of 5 times 3π/10 would get us cos(3π/2), which is 0, an interesting value.0704

So, we are thinking in terms of what we have to plug in here, total, to get 0, π/2, π, 3π/2, 2π, the normal, interesting values.0711

But then, we have to pay attention to the fact that it is not just θ; it is something interacting with θ.0720

So, we have to pay attention to what the numbers are going to be.0725

And that helps us figure out what numbers we want to plot in when we are trying to graph.0727

Finally, what if it was 2θ + 1? In this case, θ = 0 won't even show up, because we have to figure out what would make this something.0732

What would make 2θ + 1 turn into 0?0739

Well, if we plug in -1/2 for θ, 2 times -1/2 gets us -1; -1 + 1 gets us 0.0742

So, if we plug in -1/2 for our θ, we get out a 0 from it.0752

What if we plug in θ = π/4 - 1/2?0756

Well, then we have 2θ + 1; so 2(π/4) - 1/2 gets us π/2 - 1 + 1; so we get π/2 out of this.0759

Let's work that one out: if we had 2θ + 1 = π/2, we would have to work out how to get to this.0768

2θ = π/2 - 1...θ...if we are trying to get to this value of π/2,0777

if we want this whole something here to come out to be π/2, the θ will have to end up being π/4 - 1/2.0784

And that is where we are getting it.0792

It is the same thing if we want this whole thing to be π; we end up needing π/2 - 1/2.0793

We want the whole thing to be π/2; we need to plug in 3π/4 - 1/2 for our angle θ.0798

Notice that, in each one of these, we are stepping up by π/4 each time.0804

Over here, we were stepping up by π/5 each time.0808

Once you start to notice the pattern, the pattern will normally continue.0811

But it will depend on the specific circumstances of what is inside of your trigonometric function.0814

Beyond looking for these interesting points, you might notice repetition in trigonometric functions leading to symmetries in the graph.0820

Depending on the way the trigonometric functions are set up, you might notice that all of the same stuff is going to happen here.0825

And because of the way that my angle is going to work, there is going to end up being a symmetry occurring in the graph.0830

If you notice the symmetry, if you see that a symmetry will certainly occur, just use that to make graphing easier.0835

You won't have to plot all of those points, because you will see0839

that it is just going to end up doing effectively the same thing, but reversed or flipped in some way.0842

If at any time you are unsure how the graph will behave, just plot more points; that is the easiest way to be sure of what is going to happen of all.0846

If you are uncertain about what will come next, calculate a bunch of points.0854

Just drop them in and connect the curve through those points.0859

The more points you have down, the easier it will be to see how the curve works.0861

After time, you will develop a sensibility; you will get an intuition for how these curves are going to come out--how they are going to look.0865

You won't have to plot as many points.0871

But when you are just starting out, if you are uncertain about a graph, plot more points, and you will have a good idea of where to go.0873

All right, occasionally you will see an equation that only uses one variable, where there is just one variable there.0879

A lot of students get scared; there is no reason to get scared.0884

It is totally fine to have one variable; it just means that that variable that you said is fixed, and the other one can change freely.0887

For example, if we had r = 2, then that means our distance is always 2.0893

But θ--does θ show up in r = 2? θ doesn't show up at all.0898

So, θ can be anything over here; in our red one over here, θ can be anything.0902

So, if θ can be anything, then that means we set r = 2; so it is going to always be on this length of 2, this distance of 2 from the center.0908

But our θ can end up going to any spin at all--it can be anything.0916

So, as θ is allowed to spin positive or negative, it is going to always end up being stuck on the circle.0921

So, we end up drawing this perfect red circle at a distance of 2 from the origin.0926

A similar thing is going on over here with θ = π/3; if θ = π/3, we never mentioned r in there.0932

We never mentioned the distance, so that means r can be anything.0937

Since r can be anything, we have to be on the angle of π/3.0941

But r could be any positive thing, so it could go out forever this way.0947

And r could be any negative thing, so it could go out in the opposite direction.0950

So, with θ = π/3, with θ at a fixed value, we end up creating a line.0953

With r at a fixed value, like r = 2, we end up creating a circle, because either with r (we fix r) we have fixed a distance,0959

but we are able to go to any angle (so we are making a circle); or if we fix an angle,0966

and we are allowed to go to any distance, we have drawn a line.0970

And that is what is going on if we just fix a single variable.0972

Converting coordinate types: sometimes, we will want to convert an entire equation or function from polar to rectangular, or vice versa.0976

We can do so with the same conversion formulas we figured out and used in the previous lesson.0983

So, when we figured out these formulas, they were based off any x, y, r, and θ.0987

We didn't say what x, y, r, and θ had to be; so this is going to work for converting the variables in equations.0992

They can be used to convert from one variable type to the other variable, to convert from (x,y) to (r,θ) or (r,θ) to (x,y).0998

So, previously we had x = rcos(θ), y = rsin(θ) for one set.1004

And then, the other set of formulas was r2 = x2 + y2; tan(θ) = y/x.1009

Pretty much any way that you can see these working out to allow you to swap variables around,1014

so you can get to the kind of variables you want--go ahead and use it.1018

These are the formulas that will allow you to convert from one type of equation to another type of equation,1021

to switch from polar to rectangular or rectangular to polar.1026

And if you ever forget any of these formulas, if you forget, you can re-derive them by drawing this picture right here,1029

because we know that r is always the hypotenuse, and we know that x and y are the horizontal and vertical;1036

and we know that θ has to be the angle inside of the triangle.1043

So, we can figure out all of these equations: x = rcos(θ); y = rsin(θ);1046

r2 = x2 + y2, tan(θ) = y/x,1051

just through basic trigonometry, because we know that (right down here) it is a right triangle.1054

So, you can just use basic trigonometry and re-derive this if, in the middle of an important exam, they all disappear from your head.1058

You can just draw a picture and do basic trigonometry; it is not too hard.1063

All right, polar equations allow us to make really interesting graphs--that is to say, crazy, bizarre, cool, strange graphs--1068

graphs that look nothing like the kind of graphs that we are used to with rectangular equations.1074

They allow us to make really, really interesting stuff very easily.1080

For example, look at this red one right here: could you imagine figuring out any way to graph that with rectangular x and y,1083

to make an x and y equation that could make this flower-looking thing?1089

That looks so unlike what we are used to graphing, but it only takes 2sin(6θ) + 0.5.1094

We are able to say that incredible thing, where it has these large petals and these small petals,1101

and a bunch of them repeating in this kind-of symmetrical pattern--with very, very little work.1105

Very, very little writing is required to be able to make this incredibly detailed picture.1110

The same thing over here in the blue picture: we are able to get this weird, squished thing that doesn't really look like anything specific.1114

But it is a picture, and it is not very hard to write out.1120

Once again, 2cos(θ) - sin(5θ)--we are able to get this very strange-looking picture,1124

that there would be no easy way for us to create a graph with x and y coordinates that we would be used to using here.1130

But in polar graphs, it is pretty easy to do.1136

We can make really interesting things, things that we were really not used to seeing before, with not that long of an equation.1139

Polar equations and functions are really a new way of thinking about graphing.1148

As such, it is a great time to use a graphing calculator.1151

This is the best time for graphing calculators: plot random equations in there.1155

Alter equations that you already understand, and just get a sense for how polar stuff works by playing around with graphs on a graphing calculator.1159

If you want more information, check out the appendix.1166

There is an appendix to the course entirely on graphing calculators--how to use graphing calculators,1168

what good graphing calculators are, and even if you don't own one, and you are not going to buy one--1172

absolutely, for sure--there are free options out there.1176

So, there is lots of cool stuff where you can go on the Web really quickly.1179

And in five minutes, you can be graphing polar stuff; probably in one minute, you can be graphing polar stuff.1183

And you would be able to get that without having to spend any money on a graphing calculator.1188

And just playing around with this stuff is going to help you understand polar graphs massively.1192

Trying different things, playing around, putting in things, looking at how changing one number here changed the whole thing--1199

just being able to see that incredible speed of responsiveness of a graphing calculator,1205

being able to change immediately when you do something...1210

You don't have to take all of the time to plot it carefully, because that goes so slowly; it is hard to realize what is going on.1212

But if you change one variable in a graphing calculator, and it creates a new graph,1217

you will be able to gain this really beautiful intuition of how polar graphs work--how a polar equation creates the graph associated with it.1220

I really, really, strongly recommend: if you have a graphing calculator, and even if you don't have a graphing calculator,1228

check out the appendix; I will talk about lots of places where you can get free ones,1233

where you can just go and play with them right now, immediately.1236

And you will be able to get a really good understanding of how polar graphs work with not that much effort.1239

Just playing around for 10 or 15 minutes will give you such a better understanding of polar graphs, if you find them difficult, even in the slightest.1244

Also, I want to just point out one really important thing.1250

When you are using a graphing calculator, pay attention to the interval that θ has.1253

Normally, they are going to start your θ being from 0 to 2π.1256

And that is going to often be enough to give you the entire graph; but it won't always.1261

It might not show you everything; so if it doesn't show you everything you need to see--1265

if there is some part missing (you might want to expand it, and you might not even realize that there is something missing)--1270

you might want to just try increasing your interval to -6π, +6π, or -10 to +10.1274

And just see if that changes the graph.1280

If it doesn't change the graph, then you know that a 0 to 2π interval is enough.1282

But if it does change the graph, that tells you that you need to think about how big your interval needs to be, to be able to see the whole graph.1285

And maybe it won't even be possible to graph the entire interval all at once on one graph, like we are going to see in Example 2.1290

All right, we are ready for some examples.1296

First, let's graph the function r(θ) = 3sin(2θ).1297

We plug in θ's here; it gives some value here, and that tells us what our r is going to be.1302

It is just plugging in a number and getting something out of it; and that tells us our r.1307

We plug in some value for θ; we get r.1312

Let's figure out some values for this; let's figure out how often we need to do this.1314

Well, we have 2θ as our thing in here: 2θ...if we were solving for the very first interesting point,1319

which would normally occur at π/2 (0 is an interesting point, but we can be certain that 0 is already going to show up,1324

and that one doesn't tell us as much), then that is going to be θ = π/4.1330

So, we are going to have interesting points occur at an interval of every π/4 that we go out.1334

Let's plug in π/4 to figure this out.1339

We will plug in θ's; we are going to make a whole bunch of them.1342

And here are our r's; and if we plug in 0, 3sin(0)...sin(0) is 0, so we are just going to get 0 in here.1350

If we plug in π/4 (we figured out that our first interesting point was π/4), 3sin(2π/4)...2π/4 is π/2; sin(π/2) is 1; 3 times 1 is 3.1362

So, we get +3 over here for our r.1372

At this point, we can draw a graph; so we can start plotting some things.1374

We know that we are going to have to at least get out to a distance of 3.1378

And I will tell you that it actually is only going to go out to a total distance of 3, at the maximum.1381

So, we will have circles every 3; so here is our first distance circle...1385

Pardon me if my circles aren't absolutely perfect; I am but a human.1392

The second distance circle; and our third distance circle...OK.1396

And then, let's see: since we know that we are going to be based off of π/4, let's cut angles at π/4, as well; cool.1416

So now, we can plot some points: at an angle of 0, we go out 0, so our first point is just at the pole, on what we used to call the origin.1428

At an angle of π/4, this one right here, we are a distance of 3: 1, 2, 3 distance out.1438

Next, let's try π/2, the next π/4 forwards, the next interesting point.1445

We plug in sin(2π/2); 2π/2 is π; sin(π) is 0, so we end up getting 3(0) is 0.1450

By the time we get back to π/2, this angle here, we are back down to 0; what does that look like?1458

As π/4 goes to π/2, it goes down; from 0 up to π/2, we increase to 3, and then we decrease back down to 0.1464

We increase to 3; and then we decrease down to 0; from 0 to π/2, we go up to 3, and then down to 0.1473

So, as we spin counterclockwise, our thing increases; we touch that, and we spin back down.1481

We get smaller and smaller, back down to 0; so that is the first part of our graph.1491

Let's see what else is going to happen here: if we plug in 3π/4, sin(2(3π/4))...2(3π/4) is 3π/2;1495

sin(3π/2) is -1; 3 times -1 is -3; so at an angle of 3π/4, which is here, we are at -3.1506

So, we are going to go in the opposite direction; we are going to go opposite.1515

We were going this way, but we are now going to go opposite, because it is -3.1518

So, we are out 1, 2, 3 here; and then, at π, the next interesting place, sin(2π) is 0;1521

2 times π is 2π, so sine of 2π is 0; so we get 0 once again here.1528

And so, it is going to do the same thing, where, as it goes from π/2 to π, it becomes...1532

here is 0; here is π/2; here is π; so for the first part, it went up to 3, and then it went back down to 0,1538

when it got to π/2; and now it is going to go down to -3, and then it is going to go up to 0.1544

So, we are seeing it go up, and then down, and then negative, and then back to 0.1550

So, for this part, it is going more and more negative; so we end up seeing it curve out like this as it gets to larger and larger angles.1554

We see it spin this way; OK.1561

Next, we have 5π/4; at 5π/4, we plug that in; 2 times 5π/4 is going to be 10π/4, which is the same thing as 2π +...1567

well, let's just write this out, because that way it is a little less confusing.1581

5π/4 times 2 is equal to 10π/4; and look, that is the same thing as 8π/4 + 2π/4.1583

So, that is the same thing as 2π + π/2; everything that we have here is just going to end up being the same thing as this.1599

The π/4 here will match up to the 5π/4 here; so we are going to end up getting 3.1608

With this idea in mind, we could actually realize, really quickly, that π/2 here is going to match to 6π/4, which is 3π/2.1613

So, as we go that extra π forward, because we have this 2 here, it is going to end up repeating everything, as well.1622

Our angles are going to be new and different and interesting, but the r's will end up repeating.1627

We are going to see a repeat of this part here; it is just going to repeat here.1631

So, 3π/2 gets us 0; 7π/4 will get us a -3; and at 2π, we will be right back where we started at 0.1635

So, it will end up getting back to where we started.1644

Let's plot 5π/4; at an angle of 5π/4, we are at this part; we go out a distance of 3, so we are here.1647

So, we just got back to the origin when we were at π; the same thing--it curves out and grows larger when it gets out to it.1653

And then, it gets smaller and smaller at this point; it drops back down.1659

We are starting to see some symmetry; it is like there are petals here.1662

"Petals" is a way of talking about this.1665

At 7π/4, we are here; but once again, it is -3, so we go in the opposite direction.1667

We go out to here, and there we go--they should be perfectly symmetrical, if the graph was absolutely perfect.1675

If it was drawn by a computer, they would end up being perfectly symmetrical petals.1684

But we can get a pretty good sense of what is going on, even drawing it by hand.1687

The second example: Graph the equation r = 1.5θ/2, where θ is between -2π and 2π; it goes from -2π to 2π.1692

At this one, let's start...we will graph...well, we really have no idea of what this is going to end up being.1702

We have 1.5r = 1.5θ/2; we are dividing our θ by 2.1708

So, we know that we are going to want 0 in there, because it is right in between -2π and +2π.1719

And we want to go all the way down to -2π and all the way up to 2π.1724

So, let's do this by π/2; π/2 will be easy to graph, as well, since they are the cross-axis that we normally have in there.1727

A distance of r...if we plug in θ at 0 first, well, 1.5 raised to the 0--that is easy.1734

We always end up getting 1; if you raise any number to the 0, you end up getting 1 out of it.1741

Next, though, this is a little bit more difficult: if we plug in π/2, then what do we get out of this?1746

Well, we will have to use a calculator; so here is how we do the first one.1752

If we were to try to figure out 1.5, raised to the π/2, over 2...we have π/2, over 2...1754

well, that is going to be the same thing as 1.5 to the π/4.1762

We might not be able to figure this out by hand, but we could put this into a calculator: 3.14...1767

1.57 would be half of that; so it is just going to be plugging in those numbers and getting an approximate value.1772

So, this comes out to be approximately...using a calculator, we get 1.37 out of it.1779

Sorry, I meant to say 1.57, because 3.14/2 would be equal to 1.57.1784

But then, this is 3.14/4, so that is 1.57/2; the point is that we could work it out with a calculator1790

and get an approximate decimal value for what that ends up being.1799

That comes out to be 1.37; we do the same thing for each one of these.1802

We plug in π; so it is 1.5 to the π/2 or 1.5 to the approximately 1.57.1806

We raise that...our calculator gives us that that is approximately 1.89.1813

3π/2...our calculator gives us that that gives us approximately 2.60...2π...we get approximately 3.57.1818

What if we went the other way? -π/2...raising that up there...1.5-π/2 divided by 2, so 1.5-π/4...1829

we get approximately 0.72; raising it to the -π/2...we get 0.53; raising it to the -3π/2, over 2, gets us 0.38.1837

And finally, at -2π, we get 0.28.1852

So notice; at 0, if we take 0, and we go up with 0, if we go up, the value gets larger and larger.1857

We have it getting larger and larger faster and faster, because remember: this is an exponential function.1864

So, it will get faster and faster growing as we go to larger and larger values of θ.1869

As we go to negative values of θ, though, it gets smaller and smaller, because it is an exponential function.1874

Once again, we are seeing the tail part get really down close to that x-axis.1877

So, we can draw this out: the most extreme value that we get out to is 3.57.1882

So, I will set our most extreme circle at a distance of 4.1889

These single cross bars will be enough, because we are only concerned1893

if π/2 is the only real reference angle we have going on here; so that will be OK.1896

So, here is a distance of 1 circle; here is a distance of 2 circle; here is a distance of 3 circle; here is a distance of 4 circle.1901

Oops, that got a little bit out of hand; here is a distance of 4 circle; that is better.1916

OK, that is a pretty reasonable setup for our axes.1929

At this point, we can plot some points: at 0, at an angle of 0, we are 1 out; so here is our first point.1932

At an angle of π/2, going straight up, we are at 1.37; so we are almost to halfway out.1940

At π, this angle here, we are at 1.89, getting pretty close to that distance of 2.1949

At 3π/2, we are at 2.60, a little bit over halfway between the 2 and the 3 ring.1956

At 2π, we are at 3.57, a little bit over halfway between the 3 and the 4 ring.1962

We have this...as we go to a larger and larger angle, the distance out increases.1968

We see it spiraling out, the farther out it gets.1973

It continues to spiral out this way; so this is what we see as the angle gets larger and larger--it spirals out.1977

What if the angle goes negative? Well, at -π/2, remember, this is -π/2,1986

because it is talking about going clockwise instead; so at -π/2, we have 0.72.1991

0.72 is around here; at -π, that is here, because remember: it is clockwise now;1997

we are at 0.53; at -3π/2 (that is this one here), we are at 0.38; at -2π, we are at 0.28.2005

So, we end up seeing it continue to spiral in and in and in and in and in.2018

-2π to 2π: that is what is stopping this from continuing out forever.2024

If this was allowed to keep going forever, we would see it spiral off way out forever.2027

If this was allowed to continue forever, we would see it spiral into the center more and more and more and more.2032

So, that -2π to 2π--that is why we have to have an interval set--because sometimes,2036

if we don't set an interval, we could just keep going forever.2041

All right, and that is why we have it set in our graphing calculator, if you are using a graphing calculator.2044

You have to pay attention to the interval, because sometimes it will end up cutting off parts of the graph that you want to see.2047

You wouldn't see that part where it gets to continue to spin in if you didn't have a larger than 0 as the bottom of your interval.2052

You have to go to -2π, -4π, -10π...to really get a sense of just how much that spirals into the center.2057

All right, the third example: convert the equation from polar to rectangular, then solve for y in terms of x.2064

What were all of our equations? We have x = rcos(θ) as one of our formulas for changing.2070

y = rsin(θ) is another formula; r2 = x2 + y2 is another; and tan(θ) = y/x.2076

So in this case, we see right away that 6r2 times cos2(θ)...2086

we are not quite sure about that part, but here is our sin(θ), and here is our sin(θ).2091

So, we can swap them out; that is 3y = 7.2096

What about this part here--what about r squared, cosine squared, θ?2099

Well, we realize that that sounds an awful lot like r cosine θ; so how can we get rcos(θ) to show up there?2103

We maybe think, "Oh, well, there is r2; there is cosine squared; we can pull that squared out, and we could write this as 6rcos(θ)."2109

And then, that whole thing is squared; and then, minus 3y = 7.2117

At this point, we can swap out x = rcos(θ); we have 6 times x2 - 3y = 7.2122

Add 3y; subtract 7 on both sides; so we have 6x2 - 7 = 3y.2131

We can divide by 3 on both sides; we could write this as y = 6x2 - 7, and that divided by 3.2137

We have managed to convert this to a rectangular x and y, and it is now in the form y = stuff involving x.2145

What if we are doing the reverse? We are going from rectangular to polar.2153

Last time, we went from polar to rectangular; now we are going the other way.2157

Solve for r in terms of θ; once again, we have the same x = rcos(θ), y = rsin(θ);2160

r2 = x2 + y2; and tan(θ) = y/x.2169

In this case, we see that that is nice; we have rcos(θ), rsin(θ)...we have y; we have x.2176

So, we can just swap those out directly; we can swap them out for what we have here.2182

We swap them out; y is rsin(θ)...equals 2 times...x is rcos(θ), so 2 times rcos(θ), plus 3.2187

At this point, we were told to solve for r in terms of θ, so we need to get our r's on one side, so we can get just r by itself.2199

We move the 2rcos(θ) over by subtracting it, so we have rsin(θ), now minus 2rcos(θ), from both sides;2207

that equals...+3 is still left over here; now, notice: we have an r here and an r here, so we can use the distributive property in reverse;2214

we pull that r out, and we have sin(θ) - 2cos(θ) = 3.2222

We now divide that out; so we have just r; so r = 3/(sin(θ) - 2cos(θ)).2232

That might be a little surprising; that seems like a fairly complicated thing, if it is just going to give us a line.2243

But some things...polar is better at graphing certain kinds of pictures, and rectangular is graphing other kinds of pictures.2248

So, it depends on things; rectangular is great for graphing lines; polar is not as great at graphing lines.2255

And you might be surprised that that would even end up coming out to be a line.2259

Try plugging it into a graphing calculator, and you will see that that ends up giving us y = 2x + 3.2263

It is just another way of graphing it.2267

You might have a little bit of difficulty seeing why that ends up giving it.2269

If you think about it, this bottom part here is going to sort of work as an asymptote,2272

as it approaches the same angle that this line is based on, which is why it is going to shoot off infinitely in both the top and the bottom.2276

So, think about that for a while; try graphing it.2283

Just in general, try to play around with graphing as many polar functions as you can.2285

It will really give you such a great sense of how the stuff is working if you just play with it for a while on a graphing calculator.2289

All right, we will see you at Educator.com later--goodbye!2294