# Graphing Polar Equations, Test for Symmetry & 4 Examples Corrected

BAM!!! Mr. Tarrou In this lesson we are going
to take a very detailed in depth look at graph of polar equations. If you are just looking
for a little piece of information, I have a whole bunch of links in the description
of this lesson if you just want to skip ahead to maybe just look at graphing Rose curves.
In this full length lesson, we are going to be looking at or identifying the three types
of symmetry that we can expect to see in our graph by some type of algebraic test to our
equations. There are three types of symmetry to test for. We are going to be looking at
the special types of graphs. What is the polar graph going to look like, just by looking
at the equations? Can we by just looking at the equation, identify if it is going to be
a circle, a limacon, a rose curve, of finally a lemniscate. I always say that wrong…lemniscate.
Our last pages here is just our examples being done. First, by just looking at the equation,
can we identify any symmetries that will be in the graph. That will save us some points.
Maybe we can only plot half the points maybe instead of all of them. We notice as we are
plotting, we are kind of getting a picture, we know there is going to be some kind symmetry,
and we can stop earlier and kind of just mirror it and be done a lot quicker. These are going
to be tedious graphs if you are doing them completely by hand. We will be using a little
bit of aid from a calculator. Symmetry of polar graphs. They can be symmetric to the
polar axis, the horizontal axis. Don’t call it the x axis, because we are not in the rectangular
coordinate system. Of course I am assuming we have already talked about graphing just
simple polar coordinate. Where we rotate the pointer if you will and if after the rotation
theta, if the radius is positive we go towards the direction of the pointer and if the radius
is negative then we will go in the opposite direction of the pointer. So this is not just
plotting individual coordinates, that is already assumed to be understood. I will have a link
here if you are not sure how to graph polar coordinates. This is about the graphs. If
you can replace theta with negative theta, or replace (r,theta) with (-r,pi-theta), some
textbooks include this extra check and some don’t and it is OR you don’t have to do both,
if that yields an equivalent equation then your graph will be symmetric to the polar
axis. If you know that you can maybe plot fewer or less points and get your graph quicker.
If your equation only involves cosine, then it will be symmetric to the polar axis. Now
by symmetry to a polar axis, that is symmetry to a line so we are talking about a reflection.
So if we have a point in quadrant one, say here, then you can be guaranteed that if you
reflect or fold that graph over the polar axis, this line right here, then you will
land on a point another point on the graph. Now, when you are graphing polar coordinates
you are allowed to have a negative radius, right? So, if you again have that pointer
starting off at zero degrees radians or zero degrees. Did I just say zero degrees radians?
Zero radians or zero degrees. If you rotate that pointer in a direction which is opposite
to where your point is, now opposite means that we are looking at a difference between
this angle and this angle of some kind of… Well, I guess you could talk about multiple
rotations, but lets just say that they are separated by basically 180 degrees or pi radians,
then you could just simply backwards. Now, all of these thetas that I am drawing remember
too these are all reference angles which are equal to the number of degrees or radians
between the polar axis and well, if we were drawing angles, the terminals sides of the
angles. So, if this were say pi/6 radians, then it could be a reference angle over here
of pi/6 radians, or pi minus pi/6 which would be 5pi/6 radians. If you could then go backwards,
with that opposite radius value, that negative radius value, you could land on that point
as well. So that is where this negative r, pi minus theta alternate substitution comes
in for checking for if you have any symmetry to the polar axis. Now, on our first example
here and it will be the only simple example for testing for symmetry until I get to the
graphs and I will kind of build it into those problems. We are going to check for symmetry
for the equation r equals three plus two times cosine theta. I kind of already gave you a
cheater note. I said if your equation only involves cosine, you will have symmetry to
the polar axis. But assuming I didn’t give you that little short cut tip, you can have
other polar equations be symmetric to the polar axis, otherwise if that is all there
was I would not need to be teaching you some kind of check, right. Ok, so if replace theta
with negative theta and it simplifies to get back to the original equation, or take out
r and replace it with negative r and take out theta and replace it with pi minus theta,
if that also simplifies to the original equation, both checks don’t have to pass, but that one
substitution worked then we could be guaranteed to have symmetry to the polar axis. So, I
have taken out our theta and replaced it with negative theta. But, the cosine of negative
theta is the same as the cosine of theta because the cosine function is an even function. Hopefully
you have already studied that in your trig class. Even functions, and you studied the
definition of even functions back in algebra2 and precalculus. If you plug in opposite numbers
you get the same answer. The cosine of pi/6 is the square root of three over two, and
the cosine of negative pi/6 is also square root of three over two. So, just by the fact
that cosine is even we are already back to our original equation and thus are guaranteed
to have symmetry to the polar axis. Now if I take out r and replace it with negative
r and the out theta and replace it with pi minus theta, we then use the difference identity
of cosine to expand and then start to simplify. The cosine of theta is negative one and the
sine of, did I just say theta? I did. The cosine of pi is negative one and the sine
of pi is zero. You see here as we continue to simplify we get r, we have to divide everything
by negative one, we get r is equal to negative three plus two times the cosine of theta.
That does not look the same as this. We again have those special difference between plotting
polar coordinates as opposed to rectangular coordinates where one point in a polar system
can have many different ways, an infinite number of ways, to getting to that one point,
where as in a rectangular system it is unique. So, while this equation does not look the
same as this, they actually yield the same graph. Now of course you don’t know that,
we are just starting off. So I am just going to say this. When you are testing for symmetry,
if this is the only test that we did, the algebra would show that, well not that there
is no symmetry to the polar axis, just that we are not guaranteed. The three checks that
I am giving you, and oh by the way you would not have to do both of these…as soon as
we did one check and it worked we could have stopped, if the algebra check fails it is
not that symmetry does not exist, we are just not guaranteed that it exists. You might still
notice a symmetry as you develop your graph, it is just that you had to plot more points
maybe to see that symmetry start to happen as opposed to an algebraic check that just
told you immediately and thus saved you some time. Respect to the line of theta is equal
to pi over two. Well that is again theta is equal to pi over two. Hopefully in an earlier
section you noticed that, let’s see… If I want to graph theta equals pi over two.
I don’t care what the radius is, all I care is that every point on that graph has a rotation
of pi over two radians or 90 degrees. So, if I start from zero and rotate pi/2 radians,
or if I rotate 90degrees, and then I don’t care what r is then that is this line right
here. Every single point, what color have I not used yet, every single point on this
line has a rotation of pi over two radians or ninety degrees. But this equation, it is
sort of weird that I am giving you an equation of a polar graph in a section where we are
introducing graphing. So, I am just trying to explain this. I don’t care how far each
of these points are from the pole, there is no value of r in this equation, just theta.
They have to have a rotation of 90 degrees or pi over two radians. So we have a symmetry
to this line. That means that we have a reflection. Every point to the right I guess you could
say is going to have a reflection to the left. Or, if you fold the graph on that line they
are going to line up with another point on the graph. That symmetry, that reflection
over the line theta is equal to pi/2, that can be identified ahead before you actually
start making the graph by doing the following. If you replace (r,theta) with (r,pi-theta)
or (-r,-theta), this is going to be the one that we use the most in the examples of our
lesson, if you can plug those in, simplify, and get an equivalent equation then you are
guaranteed to have symmetry to the line theta is equal to pi over two. Any equation, polar
equation, that just involves sine will have symmetry to the line theta is equal to pi/2.
There is one more check to symmetry. Before I get to the next board, again I am going
to say one more time, if any of these algebraic checks fail, you may still have that symmetry.
You are just going to have to plot more points to see it your graph. nanananana The third
type of symmetry that we can check for algebraically and we are going to be concerned with in this
lesson, is symmetry with respect to the pole. If you can replace (r,theta) with (r,pi plus
theta) or (-r,theta) and get an equivalent equation, then you are guaranteed to have
symmetry to the pole. Now, what does symmetry to the pole look like? Well, it looks like
we have this point here in quadrant one. It does need to be in quadrant one. But, if the
graph is symmetric to the pole, then you can either look at it two different ways. You
can look at it as a reflection over the line, the polar axis. So we can reflect that line
over the polar axis and get it down here. Then reflect it again over the line theta
equals pi over two. You can see two reflections to get to the original point to another point
which is symmetric to the pole. Or, I always just say that it is a rotation of 180 degrees,
or pi radians, around the pole. So, if I take this graph…if I take any kind of graph and
I rotate it pi radians about the pole, about the center, if it was a rectangular system
about the origin, and the graph lines up on itself again, then you have that symmetry
to the pole. With this rotation idea, if we rotate some kind of radians and let’s say
it is a reference angle like…I don’t know…maybe pi over four radians, then if I go over here
at pi and I rotate the same angle of pi over four radians keeping the r value the same
you would end up again with that pi radian rotation. Or that other point where if you
reflected about the polar axis and about the line of theta is equal to pi over two you
would end up on that same point. Or you keep your pointer going in the same direction,
right? Your pointer indicating of course your angle of rotation that you do first and then
you plot the point. Well, instead of stepping out a positive r units for the radius, you
can just simply go in the opposite direction and have negative r…theta. One more time,
if any of these symmetry tests fail, you may still see the graph show one or more of these
symmetries when you are done. These tests again, these algebraic tests of these symmetries,
is if they pass then you are guaranteed that type of symmetry and that can save you time.
You can reduce the number of points you have to plot to finish your graph. Now these special
types of graphs. The first special type of graph…. you are looking for the fact that
it may be a circle. If you see r is equal to a which is any kind of constant like three
or five or it could be negative two, but some value of a, it could be a decimal as well
like negative two point three, but if you have r is equal to some kind of coefficient…
some kind of constant times the cosine of theta it is a circle. Now notice there is
no b or c. There is no variable in front of theta with this general form, in this equation.
If you see specifically a times the cosine of theta, one theta, then you are looking
at a circle. If you have some kind of 2theta, 3theta situation going on in there, then it
is not going to be a circle. It is going to be some other type of polar graph that we
are going to be talking about shortly. Now if it is r is equal to a times cosine of theta,
and cosine being an even function, you are going to have that symmetry about the polar
axis. The distance from, the diameter basically of the circle passing through (in this form)
passing from the pole to the opposite side of the circle, that entire distance of the
diameter is going to be a. If you have r is equal to the a times the sine of one theta,
then you have again a circle. So, any kind of constant, coefficient of one on theta,
all you have going on there is no plus two or something…you know there is no plus one
or something like that which is some other type of graph, it is just a times the sine
of theta, then you are looking at a circle. Again passing through the pole, I think I
said origin instead of pole over here, but it is passing through the pole. It is going
to be symmetric to theta equals pi/2, because the sine function is odd. It is going to be
symmetric to the line theta is equal to theta over two. Again, that entire diameter is going
to be a. By the way, and this will become really nice when we are graphing the roses…our
examples there, to solve for the maximum value of r, the maximum distance any point on your
graph will be away from the pole, to solve for that maximum absolute value of r remember
that the sine of b*theta, (Now we don’t have b*theta here, this is just in general terms
of the entire lesson) the sine of some kind of angle measure and the cosine of some kind
of angle measure… Well, sine and cosine is that wave right? It only goes down to a
minimum value of negative one and a positive value of positive one. So, the range of sine
and cosine is only between negative one and one. Ultimately if there is some kind of coefficient
out front of course it is going to increase/decrease that value. Like if this three times the cosine
of theta, then farthest distance away from the pole is going to be three units away because
one times three is equal to three. If it was negative three, well we have the absolute
value wrapped around the r. Ok, now next type of graph that we are going to be taking a
look at, and we are going to do an example of graphing. This special type of graph is
a limacon. nanananana… After circles our second form of polar graphs is limacon. They
are going to be symmetrical to the polar axis when they are in the form of r is equal to
a plus or minus b times the cosine of theta. They are symmetric to the line theta is equal
to pi over two if they are in the form of r is equal to a plus or minus b times the
sine of theta. Now these are, I mean if an equation is in exactly in this form, it cannot
deviate. A and b can be pretty much any value, but we have coefficients of one on the theta
in the functions of cosine and sine. That is how you recognize it. If the equation looks
like this, it is going to be a limacon. There are four different types of limacons as well.
Even though I am telling you this is going to be symmetric to the polar axis and this
is symmetric to the line theta is equal to pi/2, when we get to this example (We are
going to do one of these examples) am going to go through those three symmetry checks
that I showed you at the beginning of the lesson. Now, if you take the constant of a
and you divide it by the coefficient of b and you get a value that is less than one,
then you are going to be looking at a limacon which has an inner loop. Our example is going
to be a limacon with an inner loop that we are going to be doing after this scene. If
you divide, if you take a and divide by b, and you get a ratio which is equal to one,
then you are going to be looking at a limacon which is specifically a cartiod. This kind
of comes to a point of you will. It starts to, the inner loop sort of starts I guess,
but it fails to finish. You have a little point, a little sharp bend at the pole. Now
you just simply have a dimple as you can see here if a divided by b is greater than one
and less than two. And if you take a divide b and you get value which is greater than
two, then your limacon is going to be defined as being convex. There is no place where it
caves in, so it is convex instead of concave. Now up here, I kind of ran out of room when
started to draw my first polar grid. You definitely want to go the internet, I will put link right
here of some website that you can go to and print a piece of paper with probably about
six polar grids on them. You don’t want to make these from scratch every time like I
am doing. I am identifying the angles here are really the angles about the unit circle.
You don’t have to use the angle measures around the unit circle to make a polar graph. Indeed,
when we get to roses, when there is a lot of petals, if you were to make those by plotting
points you would want to make those increments much smaller than they are on the unit circle.
But, there is not a tremendous amount of crazy movement going on with these four types of
limacons, so when you do these particular graphs, when you make the table of values
you can do them by hand and maybe count every…. Well, if you are doing them by hand you can
put the angles that are from the unit circle into your table. Or, and I will be showing
you in a pop up window in our lesson, that you can put your calculator, you can use the
table feature in your calculator and set it up to work with increments of 15 degrees or
pi over 12 radians. Then you can skip some of these because this is going to go from,
if this is the unit circle angles, from here to here is pi/6 which is 30 degrees, and then
you get to pi/4 or 45 degrees and it starts to count by 15 or pi/12. That is why you want
to set your calculator, your table, to go by 15’s or by pi/12 because that is going
to hit all the values around the unit circle. Now let’s get to that first example right
now. BAM! nananana… Example number one. r is equal to one minus three times cosine
theta. Test for symmetry. Well is it symmetric to the polar axis? Can I plug in (r,- theta)
in place of (r,theta), or really just substituting theta with negative theta. If I do that, we
are going to get r is equal to one minus cosine, excuse me, three times cosine of negative
theta. Because again cosine is an even function, we get one minus three times cosine of theta.
So, yes we plugged in negative theta and we got right back to where we started from so
it is going to be symmetric to the polar axis. Is it going to be symmetric to the line theta
is equal to pi over two. In that test I need to replace r and theta with negative r and
negative theta. Well, again the even nature of cosine is going to allow that negative
theta to just be cancelled out and keep it as negative three or minus three cosine theta.
But there is no way to get rid of the negative r and simplify it back to what it was in the
first place. So, that does not mean that this is not going to have symmetry to the line
theta equals pi over two, we just are not positive. We can already see here if I let
r become negative theta, excuse me if I replace r with negative r, there is no way of canceling
that either and get us back to where we started from. So, we are guaranteed with our three
algebraic checks that we are going to have symmetry to the polar axis. As far as the
line theta is equal to pi over two and the pole, it may have symmetry to those items
and it may not. We are not sure. If it just happens to show up in the graph, then all
the better. What is the maximum radius? Well again, that… Now you might be looking at
these problems, and they might have other trig functions besides sine and cosine. But
most of these that we deal with in our lesson will. So, the maximum cosine is ever going
to be is positive one. One minus three times positive one is one minus three which is negative
two. The smallest value that cosine of theta can take on as an answer is negative one.
So, one minus three times negative one is going to be one plus three which is positive
four. So the maximum the radius is ever going to be, or the maximum value for the absolute
value of r is going to be four. At what values or measurements of theta is that maximum value
of r going to occur? This particular step is really convenient when you are graphing
roses, but let’s go ahead and do it now for this limacon. We have four is equal to one
minus three times cosine theta. We are going to subtract both sides by one and get three
is equal to negative three times cosine theta. Divide both sides by negative one, excuse
me I am thinking of the answer, by negative three and get negative one is equal to the
cosine of theta. Now to get that theta out of the cosine function we need to take the
inverse cosine of both sides of the equation. I am also going to flip this around. We have
theta is equal to the inverse cosine of negative one. If you know your unit circle, you don’t
need a calculator for that. What angle measure in radians is going to give you a cosine value
of negative one? Of course cosine of theta is equal to x over r. That is going to be
pi. So on that graph over there we are guaranteed, or we just simply know at this point that
we are going to have a point of (4,pi). So, if this is of course is zero radians, pi/2
going to give you a point right here. So at least we know one point now that make up our
polar graph. Now to complete the table, I don’t think your teacher… Well it depends
on your teacher. Again, we can put these into just a unit circle type thing. Why am I stopping
this table at pi? Well, it symmetric to the polar axis. So if I work through a rotation
of zero to pi, I believe I am going to have a nice set of points that I can reflect over
the polar axis and finish my graph. I think that is all we are going to have to do. At
zero, the cosine of zero is one, and of course that is going to be one minus three times
one which is negative two. Then we can put in pi/6. The cosine of pi/6 is the square
root of three over two. The square root of three over two times three and then that is
subtracted from one, right off the top of my head is going to be approximately equal
to negative 1.6. Now plugging in pi over four, the cosine of pi/4 is square root of two over
two, multiply that by three and then subtract that product from one and you get approximately
negative 1.12. Now, I am not putting my answers in exact form because that is going to be
quite difficult for plotting on a polar grid. While I am calling this a graph and we are
going to have quite a few points to make it up, I don’t want to be graphing…I don’t
know five square root of three over two. Where exactly is that? So, I am getting decimal
approximations and that means that, I really can just use a calculator to get these values.
So let me show you how that looks on the TI-NSPIRE. So, you would like to graph a polar equation
with your graphing calculator. Well this is what it is going to look like on ti-nspire
whether it is CAS or not. We are going to toggle between the scratch pad and go to the
calculator. Nope, I don’t want the calculator. We are going to the graphing calculator portion.
We are going to hit menu, then go down and select three for graphing and edit entry.
That is where you can tell it that you want to graph a polar equation. So, we are going
to be graphing a lexicon with an inner loop. That equation is going to be one minus three
times, sometimes you can get away with not hitting that times button and you can just
let it be assumed to be multiplication but sometimes the calculator does not like that.
So one minus three cosine of theta. Where are you going to go for theta? One place is
to come down here where the pi symbol is, open up that window, arrow over to theta,
select it, and of course that is what we want to graph. One minus three times the cosine
of theta. You can change the step of the graph here if you find that this is necessary. You
can decrease the interval that theta is going to graph in. Like say that some graphs you
can get away with, like a rose with an odd number of petals only if you were going from
zero to 3.14. And then some of the more complicated polar graphs you actually want to extend that
beyond 6.28 to get the entire graph. But, this is going to be good with what we are
working with. We are going to go ahead and hit enter. There is our limacon with an inner
loop. Now calculator tutorial is actually part of a full length lesson about graphing
polar equations and we were interested in getting the values off the tables. So, we
are going to hit menu, we are going to go down to table, and I could have just hit seven,
we are going to do a hit split screen. Right here I am going to hit one. Now I have a table
already set up but I wanted to count by specifically by pi/12 radians or effectively 15 degrees.
So, I am going to go to menu, go to table, we are going to edit the table setting, we
are going to start that at zero just for convenience and we are going to count by pi over twelve.
So, pi enter divide that by twelve, and now we have a table of values starting at zero
of approximately negative 1.9, at 2pi/12 or pi/6 radians we have a radius of negative
1.6 approximately, at 3pi/12 or pi/4 we have a radius of negative 1.12 approximately, and
so on and so on counting by increments of p/12 radians or every 15 degrees. Now, if
once you are done getting information off this table that you like, you can turn it
off. Now, if you select the graph portion of your split screen and you hit menu you
might find it quite difficult to figure out how to get rid of that table. But, if you
select the table portion of your split screen, hit menu, go down to table, and you can remove
it. And that is how you graph a polar equation with your inspire calculator and create or
look at the table of values if you need it. Now, back to our full length lesson or if
this is just the tutorial, the short tutorial you watched, thank you for watching! Ok, so
as you saw from that demonstration of the use of the TI-NSPIRE, you might of noticed
that I made the graph to create the table. But, just to make sure that we are comfortable
with plotting points. Let’s just walk through a couple of these. We have a rotation of zero
and a radius measure of negative two, so with my pointer if you will… Again with plotting
points, I like to think of it as a pointer on the pole, you rotate first and then you
move. Well the pointer is pointing to zero and then we move back two units because the
radius is negative two. Then we rotate that pointer to pi over six and then move back
approximate, you know, somewhere around 1.6 approximately so… I don’t know, somewhere
around here. Then at pi/4, so our pointer is looking at pi over four, we go back 1.12
units. Obviously this is going to be a little bit more than negative one. Pi over three,
that is negative one-half, so pi/3 and negative one half. I almost hit the wrong line there.
At pi over two we have a radius of one. So, pi/2 and we step out a length of one. At 2pi/3
we have a positive radius of 2.5. At one, 2pi/3 we go out 2.5 units. At 3p/4 we go out
to just past three, so 1..2..3…just past that. And at finally, where are we at? At
5pi/6 we are a little bit past 3.5 so somewhere around there. And then at pi we are at four,
so if we look, and that we already had that. So, if you look at what we plotted so far.
Zero radians and negative two, and then wrapping around we have
half of our…uh…cartiod, half of our lemicon
just going from zero to pi. Remember we had that relation, that symmetry to the polar
axis. So, I am going to grab another color just to show you the reflection. Let’s see
here. That point reflected over the polar axis again would be just past 3.5. Then we
were just past three. And, this reflected down is somewhere in the middle at…. What
was that? Um, I lost my place here. Let’s see, how about… Oh yes, at 2pi/3 we were
at we were at, at 2pi/3 three we were at two and a half so reflecting that down to two
and half radiuses length over the polar axis. Now, instead of looking at the numbers, I
am just going to keep the pattern going on up here. This is reflected down, this is reflected
down, this is reflected down, this is reflected down. Here we had at pi over two, we are going
to reflect that down to match this radius of one reflected down over the polar axis.
This is just going to reflect over and just be itself. This is going to reflect up now.
This point is going to reflect up. Try not to loose your track here. That is the other
half of your limacon with an inner loop. So that is the end of our first example. Let’s
move on to a different type of polar graph which is called roses. Ah yes, the rose curve.
One of the prettiest curves you will make when making your polar graphs. nanananana…
Aren’t you glad I was wearing a t-shirt that said that I am right 98 percent, who cares
about the other three percent. Well, this is one of those three percents. I made a couple
of small corrections through our lesson so far, but a nice student slash viewer pointed
out to me that my labels on my three petal curves in this introduction scene to graphing
rose curves were flip-flopped. That is kind of a big mistake that is sort of graphically
screaming at you on the screen. So, let’s take two on this introduction to rose curves
and see if we can do a better job. The rose curves again r is equal to a sine of n times
theta and r is equal to a cosine of n times theta. Both the sine and cosine functions
make rose curves. It is the first general form we are looking that we are able to have
a coefficient in front of theta other than just one. And a cannot equal zero because
a is going to be the overall maximum length of these petals. If a was zero you would be
just looking at a point or a rose with petals of zero length. Ok, so if n is an even number,
then there are 2n petals. If n is odd, then there are a n number of petals. The maximum
radius to the polar graph is going to take on is going to be the absolute value of a.
And this is where the mistake was. Some simple examples of what these are going to look like.
We are going to do two real examples with all the nice points and table of values and
what not. r is equal to a cosine of two theta. That coefficient of n is even, so you are
going to have twice the number of petals. Or, r is equal to a cosine of three theta
and it might look something like this. Here we have r is equal to a times the sine of
four theta giving us eight petals. Here we have another three petal rose based of of
sine. I just didn’t want to try to graph a five petal rose. Now one thing to just reiterate
one more time in our lesson, if you have polar equation only involving the cosine function
it is going to be symmetric to the pole. So, if you see…if I fold this over the pole
the top half of this graph, if drawn accurately, is going to perfectly line up with the bottom
half. With these even numbers of petals you have a lot more symmetry going on. But it
is really nice with the odd number of petals to make sure that your sketch is actually
coming out correctly. And sine has symmetry to the line theta is equal to pi over two.
So, reflecting that over the line theta is equal to pi over two you can see the graph
line up on itself. That symmetry is thankfully there. You know, sometimes I just make careless
mistakes. I try to catch them all, but that is what allowed the viewer to catch on to
the fact that I had made a mistake in my lesson. So, thank you:) Set the equation equal to
the maximum r if it is odd or r and negative r if n is even to solve for theta to find
the location of the petals. See, there is a lot of motion in these graphs. Look how
much the radius is changing as I go around just this simple three petal rose. All of
that movement, all of that change that we are seeing with our radius as we rotate around
with theta means that if you are just going to plot rose curves with a table of values,
you are going to need a lot of points. So, hopefully we can set up our equations here
and find out theta, find the angles at which we can go out our maximum r value and place
the end points of our petals to get a rough sketch of what the rose curve is going to
look like and then if you want to add a few points you can. I was running out of room,
so this is purple because we are really going to start up here. To make a really accurate
plot, you need more than just the endpoints of these pedals if you are required to do
more than that. You want to start with n times theta to start building your t-table. You
can see that I am just pulling values or angles off of the unit circle. So we have zero, pi/6,
pi/4, and so on. Make that list and then go through and divide them all by n. So if n
is just a simple two, but that…well I put that down but it gives you four petals. There
is a lot of movements with that. So we would have an angle of zero, an angle of pi over
twelve, pi over eight, an angle of pi over six, and angle of pi over four. I mean that
is a lot of angles to just get to pi/4 radians. So you can see how long that list or that
t-table would need to be to plot some points to get a good really accurate graph of your
polar equation. So, again hopefully we can just highlight the endpoints of our petals
to get our sketch. Ok, and as you make this list of theta, I lost track of what I was
going to say, you need to let theta go from greater than or equal to zero up to pi if
n is odd. To take care of the double the number of petals that you get when n is even, you
would need to let go from greater than or equal to zero up to less than 2pi if n is
even to again get all those petals. So, now we have got our correction done, let’s get
back to our regularly scheduled program. BAM! So, from our algebraic tests of plugging in
negative theta, (-r, – theta), and finally negative r, we know that this graph is going
to be symmetric about the line theta is equal to pi/2. That may allow us, if we were doing
a t-table to do this graph, to maybe stop the t-table a little short or just do it in
pieces so that we can get the entire picture of the graph. But, we are going to focus on
just finding the endpoints of the petals. It failed the symmetry to the polar axis and
the pole. But again, that does not mean that it will not be symmetric, it just means that
we are not guaranteed for that to occur. Ok, now each of these petals are going to have
a maximum length of four. So, we are going to just identify the tips of these petals
again. We are going to set four equal to this equation. We are going to divide by four.
Now we are going to get that sine function away from the four theta by using the inverse
sine function. This is a little bit of a review of solving trig equations. Now we are just
start going through this. We are going to keep going looking for values/angles that
give us sine values of one. Where do we stop? Well, because this n is even, we need our
final answer of theta to go from zero to 2pi to make sure that we get the endpoints of
all eight of those petals. So, let’s see here. We are going to do four theta and we are going
to…. Let’s see here. This is the inverse sine of one. So where is sine equal to one?
That is going to be pi over two, and two pi over two, three, four, five pi over two. Five,
six, seven, eight, nine pi over two. I hope you realize I am just trying to move my arm
around the unit circle to give you a visual. We are we at…. Nine pi over two, now nine,
ten, eleven, twelve, thirteen pi over two. Finally thirteen, fourteen, fifteen, sixteen,
seventeen pi over two. Now I am showing you the hand movements just give you a visual,
but i am just adding the numerator here by four over and over again. There is a pattern
to it of course. We are just cycling around a unit circle. Ok, and that list keeps going
but do I need more to that list or do I need to stop there? We are now going to divide
everything by four and we get theta, again we are dividing by four or multiplying by
one over four, so one over two times one over four is going to be one over eight. So we
have pi over eight. We also have 5pi/8, 9pi/8, 13pi/8, and 17pi/8. Now if I were to extend
that by one more value, you can see my numerator is just getting greater by four, my next point
would be 21pi/8. Well, 21 divided by 8 is bigger than two so we have moved past or beyond
the 2pi that we need. Also, if you seventeen pi over eight and you subtract 2pi you are
going to realize that actually we can get rid of that as well because 17pi/8… Not
only…actually what? I really really went too far because 17 divided by 8 is also bigger
than 2. So, I really went way too far. If I take 17/8, sorry about that, that is bigger
than 2pi, but if I take 17pi/8 and I subtract 2pi… I am looking for a coterminal angle.
Coterminal angles are angles that are separated by 360 degrees or 2pi. If I take this and
find a common denominator, that is going to be…CHALK DOWN… 17pi/8 minus 16pi/8 we
are back to pi over eight. So not only is 17pi/8 bigger than 2pi, we don’t need an angle
that big…we don’t need to go beyond 2pi, it is also coterminal to pi/8. It is in effect
a repeat angle. Well, now our n value here is even. So, to find the end points of our
petals, n is even and that means that you have twice as many petals, that means you
have eight petals instead of four. But, I only have four angle measures where we have
a maximum value of four. That is because we need to set this equation equal to positive
four and solve, which we just did, and you need to set it equal to negative four and
repeat. So, I am just going to step off, reveal that solution. Maybe you want to pause the
video just to practice it on your own. Then I am going to clean all this up and get ready
to actually draw the graph on the polar grid. nananananan…. So my eight endpoints are
(4,pi/8), (4,5pi/8), (4,9pi/8), (4,13pi/8), and with setting up that equation equal to
negative four we have the other angles of (-4,3pi/8), (-4,7pi/8), (-4,11pi/8), (-4,15pi/8).
Now if you look at this staggered, pi/8, 3pi/8, 5pi/8, 7pi/8, it is going every 2pi/8. I have
a maximum value every 2pi/8 right. Well 2pi/8 reduced to pi/4, and pi over four means that
for every 45 degrees we are getting the end of one of those petals. Now the first petal
is at (4,pi/8). Now this is marked off with standard unit circle angles. Maybe you have
some graph paper that includes basically a line at every 15 degrees. Let’s see here.
Pi/8 is half of pi/4, so pi/8 is really 22.5 degrees. So we have, I am just going to try
and eyeball this, if this is 45 degrees and this is drawn as accurately as i can sketch
it, then there is an end point there. Then every 22.5 degrees and then every 45 degrees,
so there and about there, somewhere around here. Ok, we have (4,pi/8). We go from zero
to four pi over eight. Then, let’s see here. How about it is going to look something like
this. If keep doing that I am just going to mess it up. Now at 3pi/8 here we are actually
at negative four. It is like we are swinging out from zero, we went out to that positive
four, we came back to zero, and then at 3pi/8 which is that line. If that is 1pi/8 that
is 3pi/8 because because we are counting every 2pi/8. With that negative radius we are back
here and it is going to look something like this, only we are cycling around this way,
right? Then at 5pi/8 which is here I believe, one, three, five yes, then we are back to
positive four. Then at 7pi/8 which is here at negative four. That is the 45 degree mark
there. I am trying to draw this the best that I can. Then at 9pi/8 which is here we are
back to positive four. Oops. 11pi/8 we are at negative four. 11pi/8 would be here, but
that is a negative four radius. I hope all of my petals are coming out to be sort of
the same size. And just keeping this pattern on, I am just going to swing this around.
I just heard the static in the mic, I am sorry about that. Where is my 45 degree mark? There
it is. Ok, and that is the best that I can do to sketch the eight petal rose with just
identifying where each of those petals is supposed to be. That is a pretty darn good
sketch. I really don’t think you need to make a list of a million points to find out exactly
if that is there or there. Is that point supposed to be there, or there. etc. I know where the
endpoints are. They all look pretty equal, and that is our sketch of an eight petal rose.
Now I was going to do another example where n was odd. Sorry for the break. I was going
to.. I am cutting a part of the lesson out where I said I was going to skip the next
example where we had a negative a value and an odd value of n just because the length
of the lesson. Then I thought I have the example prepped. I am just going to reveal the example,
reveal the solution one step at a time, and really just give you the opportunity to pause
the lesson and try it on your own. I am not going to go over it because the process is
just like this one. But you will see the next example, you will see the solution revealed,
again pause the video and try it on your own if you like. I think it is a good idea to
make sure that you have the process down. Then we will move on to the next type of polar
graph. Alright then. I said that I was going to just reveal the solution and not even talk
about it because it was so similar to the previous example. But, I realized as I was
working this out, you can see that I added a little bit of extra information about plotting
points with a table. Think… Oh, the sun is out, a different shirt, it is a different
day. I think I may have misspoke a little bit in the previous scene of this lesson.
That rule was when creating tables to just plot a bunch of points to help you graph this
rose curve, I said let theta between zero and pi if n is odd and let theta between zero
and 2pi if n is even. That was a true statement, but I think I kind of emphasized that little
bit I was setting up the equation to find the theta measures where the radius was at
its maximum value. That was incorrect. When setting up an equation to locate the petals,
you want to solve out for theta until you get to 2pi. So, 5pi/3 is the smallest answer
that is less than 2pi. Continuing that pattern on, we got 7pi/3 as our next answer but that
is bigger than 2pi and when you subtract 2pi from 7pi/3 you get pi/3. So these three angle
measures are where you are going to again find those maximum radiuses of two, or the
end points of those three petals. Ok, but what I have done here, is I have made a table
of values to plot the rose out with a bunch of points. You will see that after we pass
pi the points start to repeat. So at zero radians we are at negative two, at pi/6 radians
we are at zero already. At pi/4 we are at 1.4. At pi/3 radians we have a radius of 2.
At pi/2 radians we have a radius of zero. At 2pi/3 we have a radius of negative two.
At 3pi/4 we have a radius of negative 1.4. At 5pi/6 the radius is zero again. At pi we
are at two. Well I already have that point plotted. At pi/6 we are back at zero, but
we hit zero radiuses a lot so no worries there. At 5pi/4 we have a radius of negative 1.4
but I already have that point plotted. At 4pi/3 we have a radius of negative two. So,
you see that after the table of values has passed that theta measure of pi, the points
start to repeat. Now, I also did not change the interval at which I was going count these
theta measurements like I gave you as a tip in the previous scene as well. You will notice
that I don’t have very many points. I just wanted to show you that this could be an issue
when tying to make a big table of values with plotting rose curves. So this is an increment
of pi/6 radians, then we go into an increments of pi/12 radians, and another increment of
pi/112 radians, another increment of pi/6 radians, or 30 degrees, 15 degrees, 15 degrees,
and then 30 degrees. So, I went back through the table of my calculator and I started back
at zero and I filled in some of those points that I skipped, basically counted by every
15 degrees or pi/12 radians. Now to connect the dots in order that they come out of this
table, we get a rose curve that looks this I hope. Now they start to repeat. So that
is our three petal rose for r is equal to negative two cosine of three theta. Now we
are moving on to, let’s see if I can say this right, lemniscates. nanananana… And the
last special type of graph in these polar graphs that we are going to be taking a look
at is something called a lemniscate. They are going to be in the forms of r squared
is equal to a squared times the sine of two theta and r squared is equal to a squared
times the cosine of two times theta. A is of course not equal to zero, because then
our petals would have no length. Now in these forms, these are the only general forms that
we have set equal to r squared, so they are going to be pretty equal to recognize. If
it is not set equal to r squared, then…. I guess you could write it at r is equal to
plus or minus and then have a square root over the whole right hand side of the equation.
These lemniscates that have the sine function in them, they are only going to have symmetry
to the pole. If you have cosine, they are going to be symmetric to the polar axis, the
horizontal axis, the line theta is equal to pi over two, and then finally symmetry to
the pole. Alright, now our last example… I am not going to include an example like
these, like this special type of graph. I want my last example for our lesson to be
a non-special type shape that you can memorize and just know what it looks like before you
even start. We are going to be checking for symmetry and graphing r times the sine of
theta is equal to four. Now to put this into r in terms of theta of course, you are are
going to need to divide both sides by sine of theta. So, we are really just going to
graph r is equal to four over sine theta. This would be a nice format for your equation
to be written, especially if your graphing calculator only has a sine, cosine, or tangent
button. But, if you are able to directly enter cosecant, you might want to write that as
r is equal to four times the cosecant of theta. Now the process of this is going to be just
like all of the other three examples we have done in the lesson. We will check for symmetry,
build up a table of values if need because there are no petal here so we can’t just find
the endpoints of the petals, and then finally plot the points on a polar grid. Polar graphs
can be very unique, very cool looking, very interesting and have a lot of movement. The
roses have the most movement, so I think if you do this in increments of pi/12 or every
15 degrees, if you are doing this in degrees…radians are generally better, you are probably going
to get a decent picture of what the graph looks like. So let’s see. I am going to reveal
the solution one step at a time, if you want to pause the lesson and try it on your own
first. This is the last example. Make sure that you get the process down. Then I will
come back and talk about any issues that we may have and finish up or close our lesson.
nananana.. Alright. We have let’s see here, r is equal to four cosecant of theta. Since
cosecant is based on sine, sine is an odd function so thus our cosecant function is
going to be odd as well. What does that mean. Well, when we check for symmetry for the pole,
we replace theta with negative theta. That means that r is now going to be equal to negative
four times the cosecant of theta. This is because with odd functions if you plug in
opposite numbers, you get out opposite answers. So symmetry to the pole is inconclusive. But
when we replace r and theta with negative r and negative theta everything simplifies
back to the original function, so we are guaranteed to have symmetry to the line of theta is equal
to pi over two. By replacing with r with negative r, our symmetry test for symmetry to the pole
again, it does not fail, but it is inconclusive. So we drew our polar grid. We went from zero
to 3pi/4, just working along with our calculator or the table function of our calculator to
get these values. We see that r is equal to four over the sine of theta is undefined at
zero radians, it will be undefined at pi as well. As we continued through our table of
values and plotting the points as we went, well they are just forming a straight line.
So, that is the end of our last example. I am Mr. Tarrou BAM!!! Go Do Your Homework.

### 53 thoughts on “Graphing Polar Equations, Test for Symmetry & 4 Examples Corrected”

• November 6, 2014 at 7:58 am

First one to comment and was fascinated by the curves. Just a few hours after you uploaded this and already a liked video!

• November 7, 2014 at 3:10 pm

dear sir,
At first, I am grateful for your efforts in teaching Mathematics energetically and I do appreciate that. The only thing that I need from you is to introduce or show us a website for the homework and examples and samples to practice thank youÂ

• November 8, 2014 at 10:58 am

I've been watching your Calculus videos for some time and I like them a lot as you explain them pretty well. I was wondering if you're going to make Discrete Math videos. If possible, please tell me about that. Thank you a lot for everything!!!

• November 12, 2014 at 10:38 pm

Great video as always, Thank you so much

• November 14, 2014 at 3:20 am

• November 23, 2014 at 9:38 pm

thanks

• November 25, 2014 at 5:51 pm

thank you very much, I do appreciateÂ

• February 5, 2015 at 6:35 am

This videos are amazing for me since I'm in trig. ðŸ™‚ life savior

• April 20, 2015 at 12:38 am

Your videos are awesome !!! Great for revision before quizzes !!!
Thanks !!

• April 29, 2015 at 10:21 am

10,000 words of Closed Captioning. Â BAM!!!

• April 30, 2015 at 12:30 am

• May 8, 2015 at 6:00 am

Great video! You have the most beautiful handwriting lol

• August 3, 2015 at 4:14 am

Best online lessons ever. Thank you so much, without your videos i would not be passing Pre-cal! I have subscribed to you and will continue to tell everybody in my class about your awesomeness! BAM!

• November 9, 2015 at 7:28 pm

lol wow man i am in precalc now and your still killing it…BAM

• November 20, 2015 at 6:56 am

I swear I was just thinking to myself "i just want to see a rose curve example, I hope I can find it"… and within the first 30 seconds you introduced the link below…. for a rose curve example. lol you are awesome man, I have yet to come across a math teacher as interesting as you. thank you!!! please continue to share your great talent of teaching!

• December 14, 2015 at 4:00 am

In the limocon, is the convex a perfect circle?

• April 18, 2016 at 2:21 am

Why isn't r=cos(theta/2) symmetric to the polar axis?

• May 2, 2016 at 1:29 am

does the line of symmetry for sine extend all the way down from pi/2 to 3pi/2?

• May 5, 2016 at 3:33 am

ProfRobBob, what do I need to generate a rose curve on my computer? Would like to recreate one that I found online. Any insights will be much appreciated.

• May 13, 2016 at 3:56 am

this is sooooooooooooo good. Its really good to see these full lectures. Its like a class but you have a better way to teach

• June 1, 2016 at 12:31 am

When you randomly grew a mustache at 34:18 XD

• June 1, 2016 at 12:49 pm

THANK YOU SO MUCH ! ðŸ™‚

• July 7, 2016 at 9:33 pm

Does this guy teach at St. Pete High?

• July 11, 2016 at 11:52 pm

hello! for the notes on 18:23, what if (a/b) was to equal 2? would it be a dimpled or convex?

• July 17, 2016 at 4:50 am

this really helped with me learning polar systems in a day taking a year long course over the summer is a pain >)>

• August 17, 2016 at 6:49 pm

• October 13, 2016 at 7:06 am

Haha this video is so long that you grew facial hair at around 38 minutes and then had to go shave it off.

• October 22, 2016 at 2:14 pm

It is very useful to find all kind of polar graphs sketched in one video with great explanation…. #loved it…

• October 25, 2016 at 6:12 am

Ù…Ø¯Ø±Ø³ Ø§Ù„Ø±ÙŠØ§Ø¶ÙŠØ§Øª ÙŠØ¬Ø¨ Ø£Ù† ÙŠÙƒÙˆÙ† Ù…ÙˆÙ‡ÙˆØ¨ Ù…Ø«Ù„Ùƒ Ø¨Ø§Ù„Ø®Ø· ÙˆØ§Ù„Ø±Ø³Ù…ØŒ Ø£Ù†Øª ØªØ¬Ø¹Ù„ Ø§Ù„Ø±ÙŠØ§Ø¶ÙŠØ§Øª Ù…Ù…ØªØ¹Ø©.

• February 15, 2017 at 9:04 am

you really are a saving grace, thank you:)

• February 19, 2017 at 4:26 pm

Thank so much! The video really helped clear things up.

• February 21, 2017 at 6:38 pm

how did u get the 3pi 5 pi 7pi at example 3 in rose curves pls answer.

• February 27, 2017 at 4:03 pm

i wish my teacher taught the way you do! My teacher only obfuscates things!

• March 29, 2017 at 2:50 am

• April 10, 2017 at 4:27 pm

you should be the official math teacher of youtube

• April 29, 2017 at 10:27 am

• May 28, 2017 at 1:48 am

You are saving humanities. Â I don't understand why there are people like you, do great things for free.

• June 1, 2017 at 9:53 pm

THANK YOU SO MUCH PROF ROB BOB, YOUR VIDEOS ARE SO HELPFUL IN FACT YOU ARE A VERY GOOD PROFESSOR.

• June 10, 2017 at 4:04 am

How do you know the width of the pedals?

• October 4, 2017 at 9:46 am

This lesson was very helpful in the exam
Please make some more videos based on bachelor of mathematics in circle,tangent , circular cone , circular cylinder and nature of surface
It would be very helpful to me as my exam is on this Friday so please â˜ºðŸ¤—

• October 25, 2017 at 3:35 am

Why couldn't I have you as my professor in high school!? I would've been like Aristotle by now! nanananana

• November 16, 2017 at 5:35 am

Trying to pass a test tomorrow. Thanks for the crash course!

• February 2, 2018 at 11:09 am

thank u sir.

• March 13, 2018 at 3:07 pm

Thanks sir, great video!

• May 7, 2018 at 11:10 pm

Loving the facial hair. You should keep it like that!

• July 17, 2018 at 11:41 pm

Thank you for the free and high quality lessons, your students should feel lucky!

• July 18, 2018 at 6:16 am

A full hour of Math. Yay!

• September 7, 2018 at 4:20 pm

Mr. Turreau, I love your videos but this one was very convoluted. You do your best when you are solving example problems and teaching as you are solving them.

• September 28, 2018 at 10:08 pm

Watching your videos gives me a good sense of nostalgia of being back in 6th grade math class. When getting taught math was very enjoyable and teachers cared about their students. My high school teachers were meaner than the students. Your students are lucky.

• November 16, 2018 at 12:50 am

how do u made the table starting from 0 , Pi over 6 let me know I have H.W and test to do

• December 10, 2018 at 10:29 am

Wonderful lessons. The effort you put into making these lessons is very evident, and the lectures are really helpful. Thank you so much; keep making more lessons. God bless