In this video, we're going to talk about how to graph polar equations. These include circles, lemmisons, rose curves, and lemniscates. So let's start with a circle. The first equation you may see is r is equal to a cosine theta. Now if a is positive, this is going to be a circle.
directed towards the right. Now, granted, my circle's not perfect, so bear with me. A is basically the diameter of the circle, and this is going to be the center of the circle, so if you go up to find this point here, it's half of A. So if A is greater than 0, if A is positive, it's going to open towards the right. And if A is negative, you're going to get a circle.
Let me draw a good-looking circle this time. you can get a circle that is directed on the left. So keep in mind this is going to be a and this is half of a.
So let's try some examples. Let's say if we have the graph r is equal to 4 cosine theta. If you want to feel free to pause the video and try it yourself. So we're going to have a circle on the right side. Now a is 4, and half of a, well 4 divided by 2 is 2, so that's half of a.
So what we're going to do is travel 4 units to the right, and then up 2 units, and down 2 units. So the circle is going to start at the origin, and it ends at 4 on the x-axis. From the center, which is at 2, we need to go up 2 units, and down 2 units.
and then simply just connect it. So that's how you can graph r equals 4 cosine theta. Let's try another one. Try this one. Let's say that r is equal to negative 6 sine theta.
I mean, not sine, but cosine theta. We can get to sine later. Now, because a is negative, the circle is going to be on the left side. the x-axis so let's travel six units to the right since a is negative six one half of a is negative three now don't worry about the negative sign too much negative sign just tells you if the circle opens to the left now the center is going to be at negative three which is here so we need to go up three units and down three units. So the graph is going to be at the origin, at negative 6, at negative 3, 3, and at negative 3, negative 3. And so that's how you can plot.
the circle. So keep in mind this is equal to a, that distance, and this is also equal to a as well, which means this part is one half of a. So half of a is basically the radius of the circle. So if for some reason you need to find the area of the circle, you can use this equation pi r squared.
The radius is 3, so it's pi times 3 squared. which is 9 pi. So now, the next form we need to know is r equals a sine theta. Cosine is associated with the x values.
So as you can see, the circle was associated with the x axis. Sine is associated with the y values. And so the circle is going to be centered on the y axis.
So let's say if a is positive, then we're going to have a circle that goes above the x-axis centered on the y-axis. So once again, the diameter will still be equal to a, and this portion, the radius, is half of a. So that's when a is positive, or when a is greater than 0. Now in the other case, if a is negative, Or, if a is less than 0, the circle is still going to be centered around the y-axis, but it's going to open in a negative y direction.
So it's going to be below the x-axis. And so the radius, as you mentioned before, is just 1 half of a. And the diameter is equal to a.
So let's try some examples. Let's say if r is 2 sine theta. Go ahead and graph that.
So first we need to travel up two units. A is 2. Half of A, which is the radius, is 1. So we're going to have these two points. Now let's travel one unit to the right and one unit to the left.
So the green dot is the center of the circle. So we're going to travel one unit to the right and one unit to the left from it. And so this is going to be the graph.
Try this one. Let's say r is negative 8 sine theta. Go ahead and work on that example.
Now, the majority of the graph will be below the x-axis, so I'm going to focus on that. So let's travel 8 units down. And then half of a, or 4 units to the right, and 4 units to the left. So the point is going to be at the origin and 8 units down.
The center is 4 units down. So if A is 8, The radius is 1 half of A, which is 4. So we've got to travel 4 units to the right from the center, and 4 to the left. And so this is going to be the graph. So now you know how to graph circles when you're given a polar equation. Now the next type of graph that we need to go over is the Lima Song.
And the equation is r is equal to a plus or minus b sine theta. Now, if you have positive sine, it opens towards the positive y-axis. That is in the upward direction. Negative sine opens in the downward direction, in the negative y direction. You could also have a plus or minus.
B cosine theta. So if cosine is positive, it's going to open towards the right in the positive x-axis direction. And if cosine is negative, it's going to open towards the left. So let's draw the general shape if it opens towards the right. So this is the limousine with the inner loop.
And you get this particular shape if A divided by B is less than 1. Now both A and B represent positive numbers. a and b are both greater than 0 so if you get the graph 3 minus 4 sine theta b is not negative 4 b is positive 4 and a is positive 3 so let's say if it was 3 plus 4 cosine theta both a and b would still be 3 and 4 positive 3 and positive 4 so a and b are not negative Now the next shape that we have if a divided by b is equal to one is the heart shape limousine, also known as the cardiort. And here's the generic shape for it.
So it has like this dimple. So it looks something like that. Maybe I could draw that better.
So that's the cardioid. Now the next one is the dimpled limousine with no inner loop. So that occurs if A over B is between 1 and 2. So let's start with the x-axis.
It's a small dimple. Sometimes it's hard to notice. So that's the dimpled limousine with no inner loop.
The next we need to know is if A divided by B is equal to or greater than 2. So this limousin looks almost like a circle but it's not. There's no dimple and there's no inner loop. So I'm going to start from the left, I'm going to draw it straight up, and then it looks like this. But it's not exactly a circle because, as you can see, the right side is bigger than the left, but it almost looks like a circle.
So that's the limousine without a dimple or an inner loop. So those are the four shapes you need to be familiar with. Let's graph this equation.
Let's say r is equal to 3 plus 5 cosine. What do you think we need to do here? We know this is a type of lemurs salt.
It's in the form a plus or minus b cosine theta. So first, we need to identify a and b. a is equal to 3, and b is equal to 5. Now, we need to see if a over b, if it's less than 1, if it's between 1 and 2. if it's equal to 1 or greater than or equal to 2. So a over b that's 3 over 5 and 3 over 5 as a decimal is 0.6 which is less than 1. Now because it's less than 1 we know we have the lemurson with the inner loop. Now there's four types.
The first type is if it's positive cosine. This graph will open towards the right. The next type is if we have negative cosine, and in that case this graph would open towards the left.
If it's positive sine, then it's going to open in the positive y-direction. And if we have negative sine, it's going to open towards the negative y direction. So it's going to look something like that. So just keep that in mind. That's the first thing that you look for.
So we have positive cosine, which means it should open towards the right side. Now, when graphing this type of limousine, you want to make sure you get four points. 2x intercepts and 2y intercepts.
So let's draw a rough sketch of this graph. This point is actually positive a. It's a units relative to the center. And this other y-intercept is negative a units from the center. The first x-intercept, which is associated with the inner loop, it's the difference between a minus b.
So it's the absolute value difference of a minus b. Or you could just say it's b minus a because b is going to be bigger. Now, the second intercept is the sum of A and B.
And that's all you need to get a good, decent graph. If you can plot those four intercepts, then you should be fine. So let's go ahead and do that. So in this case, we can see that A is equal to 3. So we need to go up 3 units and down 3 units.
so those are the y intercepts now b minus a that's going to give us the first intercept that's 5 minus 3 that's 2 so here's the first intercept and then a plus b that will give us the second intercept that's 3 plus 5 which is 8 so that's how you can find the 2x intercepts Now let's go ahead and graph it. So first, let's start with the inner loop, and then let's go towards the first y-intercept, and then the second x-intercept, and then towards the other y-intercept. So that's a rough sketch of this graph. So the points that you need is 3 and negative 3 on the y-axis, and 2 and 8 on the x-axis. Let's try another example.
So let's say r is equal to 2 minus 5 sine theta. So try this one. The first thing I would keep in mind is what type of, in what direction will it open.
We know that a over b, which is 2 over 5, it's less than 1, so this is a lemur song with an inner loop. But notice that we have a negative sign, so therefore it has to open in a negative y direction. So we can see that a is 2, that's going to give us the y-intercepts, and b is 5. So let's go ahead and graph it.
Well, in this case, because it opens downward, A is actually going to be associated with the X intercepts this time, instead of the Y intercepts. So it's going to switch roles. So we need to travel 2 units to the right, and 2 to the left. If we're dealing with cosine, then a would be associated with the y-intercepts.
But because we're dealing with sine, the rules are reversed. Now, a plus b, that's going to be 2 plus 5, that's 7. And b minus a, 5 minus 2, is 3. So we're going to travel 3 units down, and also 7 units. So we're going to have 2 y-intercepts.
So let's start with the inner loop. And then let's... Let me do that again. And then let's get the X intercept.
And that's it. So make sure you get these two x-intercepts, negative 2 and 2, and the y-intercepts, negative 3, negative 7. And as we mentioned before, because we have negative sign, it has to open in a negative y direction. Now let's try this problem. Let's say that r is 3 minus 7 cosine theta. So go ahead and pause the video and work on that example.
So let's find the value of a over b. a is 3, b is 7. And 3 over 7 is less than 1, because 7 over 7 is 1. So what we have is the inner loop limousine. Now, we're dealing with negative cosine, which means it's going to open towards the left.
So the majority of the graph is going to be on the left side. Now, a is associated with the y-intercepts when dealing with cosine. When dealing with sine, as we saw in the last example, a is associated with the x-intercepts. So we're going to travel 3 units up and 3 units down to get the y-intercepts when dealing with cosine. For sine, you need to know that a is associated with the x-intercepts.
So next, let's find the first x-intercept, which is going to be b minus a, or 7 minus 3, and that's 4. And then a plus b, 3 plus 7, is 10. So because we're going towards the left, we need to... travel four units to the left. That's going to give us the first x-intercept. And then ten units to the left relative to the origin. So that's the second x-intercept.
Now let's go ahead and graph it. So let's start with the origin and let's draw the first inner loop. And then let's focus on the outer loop.
And so that's it. That's how you can graph it. My graph is not perfect, but at least that's the general shape.
You get the picture. Now let's try this one. Let's say that R is 3 plus 3 cosine theta.
What do we need to do here? Well, first, we need to determine what type of limousine we have. A and B are the same.
When A and B are the same, A over B is equal to 1. And in that situation, we have the heart-shaped lemursome, also known as the cardiord. And because cosine is positive, it's going to open towards the right. Now I'm just going to draw the general shape of the cardioid, which it looks like this.
So there is no inner loop. Now when dealing with cosine, the x and y intercepts are going to be a again, a and negative a. Now, the first x-intercept is just going to be the origin. And so we don't have to do anything, it's just going to start from the origin.
Now the second x-intercept is a plus b. We don't have the inner loop, which was a minus b for the inner loop lemur song. So we don't have to worry about a minus b or b minus a. So now let's go ahead and graph it.
So in this example, a is 3. So we're going to go up 3 units and down 3 units. And a plus b, that's 3 plus 3, so that's equal to 6. So the x-intercept is going to be 6, 0. And the y-intercepts are 0, 3 and 0, negative 3. So now that's, we're going to have to graph like this. And that's it.
That's how you can graph the heart-shaped limousine. Try this one. Let's say that R is equal to 2 plus 2 sine theta.
So anytime these numbers are the same, you have the cardiord, or the heart-shaped lumosome. So A over B is going to be equal to 1. Now this time, it's going to open in the positive y direction, since we have positive sine. And A, which is 2, is going to give us the x-intercepts this time.
And A plus B, which is 2 plus 2, or 4, that's going to give us the second y-intercept. So we know the first one is the origin. So those are the points that we're going to have.
And then you just got to graph it. Let's see if I can draw my graph a little bit better. Try this one.
Let's say that R is 4 minus 4 cosine theta. So this time it's going to open towards the left side. So, a plus b, 4 plus 4, that's 8. And then we need to travel a units up, and 4 units down.
So, these are the four points of interest. And then just graph it. And that's it.
Here's the next problem. Let's say that r is equal to 3 plus 2 cosine theta. So in the first example, b was larger than a.
That's when we have the inner loop limbic song and for the cardioid with a heart-shaped limbic saw a was equal to be now in this case notice that a is larger than the so whenever a is larger than B it's either the dimpled with no inner loop or the no dimple limousine with no inner loop so let's figure out what a divided by B is a divided by B is 3 over 2 and 3 over 2 is 1.5 1.5 is between 1 and 2 remember if it's greater than 2 we have to lima song with no dimple and no inner loop but in this case we have the lima song with the dimple but with no inner loop since it's between 1 and 2. Positive cosine opens towards the right, and so the general shape of the graph will look something like this. What you need to know is that this is a and negative a. That's the y-intercept when dealing with cosine. On the right, this is going to be the sum of a and b.
On the left, the difference of a and b. So let's go ahead and graph it. So because a is 3, we're going to have a y-intercept of positive and negative 3. Next, we have a plus b, which is 5. And so that's going to be one of the x-intercepts.
And then a minus b. So that's 3 minus 2, that's 1. So that's going to give us the other x-intercept, which is over there. And now we can graph it. So it's going to look something like this. And so that's the dimpled lemur song with no inner loop.
Here's another example. Let's say that r is 5 minus 4 sine theta. So pause the video and work on that example.
So a divided by b, that's 5 over 4. 5 divided by 4 is 1.25. So a over b is between 1 and 2. So we have the dimple limousine with no inner loop, and it's going to open in a negative y direction. So now this time, A is going to be associated with the x-intercepts because we're dealing with sine.
And since A is 5, we need to travel 5 units to the right and 5 units to the left. Now, a minus b, that's 5 minus 4, that's 1. So that's going to be the first x-intercept. And then a plus b, that's 5 plus 4, so that's equal to 9. And that will give us the other y-intercept. So we're going to get a graph that looks like this.
At least that's a rough sketch, but you get the picture. Now, let's try this one. Let's say that r is equal to 4 minus 2 cosine theta. So this time, A is 4 and B is 2, so this is equal to 2. When A over B is equal to or greater than 2, we have the lemursaw that doesn't have a dimple and it doesn't have an inner loop. But it's very similar to the last example.
Now, we're dealing with negative cosines, so it's going to open towards the left. Now, A is going to give us the x-intercepts, I mean the y-intercepts, when dealing with cosine. But when dealing with sine, A gave us the x-intercepts. So A is 4. That means we have to travel 4 units up and 4 units down to get the y-intercepts when dealing with cosine. Next, we've got to find out the value of a plus b.
So that's 4 plus 2, so that's 6. And it opens towards the left, so that's going to be the bigger side. And then a minus b, that's 4 minus 2, which is 2. So now we can go ahead and graph it. So it's going to look something like that.
Let's try one more example with lemurs songs. So let's say r is 5 plus 2 sine theta. Let's graph this one.
So a divided by b is equal to 5 divided by 2, which is 2.5. So a over b is greater than 2. So once again, we have the lemurs song that has no dimple and no inner loop. And because we're dealing with positive sine, it's going to open towards the positive y direction.
Now when dealing with sine, keep in mind A is associated with the x-intercepts. So we need to travel 5 units to the right, and 5 units to the left. a plus b, that's 5 plus 2, that's 7. So that's going to be the longer side, which faces in the positive y direction.
So we've got to go up 7 units. And the difference between a and b, that's 5 minus 2, that's 3. So we have to travel down 3 units. So we have an x-intercept at 5 and negative 5, and the y-intercepts are negative 3 and 7. So now let's go ahead and graph it. So it should look something like that.
I want to try that one more time. That graph is somewhat circular. So technically it should be more like this. It's not a pure circle, but it's somewhat circular.
Now let's talk about rose curves. The general formula that you need to be familiar with is r is equal to a sine n theta, or a cosine. Now, if n, if it's an even number, then the number of petals is going to be equal to 2n. If n is odd, then the number of petals is simply equal to n.
I will go over some examples. Now, typically you'll see that with sine, sometimes the petals are on the y-axis, but they're not on the x-axis for sine, and sometimes they're on neither. With cosine, sometimes the petals are on the x-axis only, and sometimes they're on both the x-and the y-axis. Now let's focus on the sine graph first. So let's start with this one.
Let's say r is equal to 2 sine 1 theta. If n is 1, we basically have a circle, which you can think of it as one petal. If n is odd, then the number of petals is going to be equal to n, which in this case is 1. So this graph is basically a circle with a diameter of 2. We covered this early in the video.
But you think of it as having one petal. And notice that this petal is centered on the positive y-axis. When n is odd, one of the petals tends to be centered on the y-axis. But let's say if instead of having 1 theta, it's 3 theta. Now, how is it going to look in this case?
So n is odd, n is 3, so the number of petals is equal to n, so we're going to have 3 petals. This time, the petal won't be on the positive y-axis, it's going to be on the negative y-axis. At least one of them will be on it. So if sine, it tends to switch. So just because you have a positive a value, doesn't mean the petal will always be in a positive y direction.
It tends to alternate, so keep that in mind. Now, A is 2, so the length of the petal is going to be 2 units. So, we should have a graph that looks something like this.
Those are the three petals. Now, if we need to find the angles, because we have three petals, it's 360 divided by 3. So, they're 120 degrees apart. So this petal we know it's in a negative y-axis.
If this is 0, this is 90, this is 180, and this is 270 degrees. So this petal is at 270. Now if you take away 120 from 270, 270 minus 120 that's 150, that will give us the pedal in quadrant 2. So this one is at an angle of 150. And if you need to find the first one, it's 150 minus 120, so that's at 30. Some teachers may require you to find the angles. If you need to do that, this is what you can do. Now, it might be helpful to draw circles when graphing the petals. So I'm going to do that for this example.
Because n is 2, we only need to draw two circles. The first one is going to have a radius of 1. The second is going to have a radius of 2. So when drawing the petals, because a is 2, it has to end on the second circle. So this will help you to draw the petal with the correct length if you draw the circle.
So that's how you can graph this particular function. So when n is 1, the petal will be facing towards the positive y direction. at least one of the petals will be.
When n is 3, one of the petals will be facing in the negative y direction. When n is 5, one of the petals will be facing the positive y direction, and it alternates. Now let's say if we wanted to graph... negative 2 sine 3 theta as opposed to positive 2. Now everything will be the same. All you need to do is flip it.
So this petal which is on the negative y-axis is going to flip and be on the positive y-axis. So in this case the graph is going to look something like this. It's not drawn to scale, but that's the general shape.
It's simply going to flip. Now let's try this one. Let's say r is equal to 3 sine 5 theta.
So n is still odd, which means that the number of petals is equal to n and not 2n. So we're going to have 5 petals. Now, when n was 3, and when a was positive, the petal was in a negative y direction.
Now, when n is 5 and a is still positive, it's going to flip in a positive y direction. So let's go ahead and graph it. So let's use circles to do that this time.
So the first petal is going to be over here. And then we're going to have another petal in that direction. Another one here.
and another one there and here's the fifth one. So that's how you can graph this particular function. Now let's find the angles. Let's take 360 and let's divide it by the number of petals, which is 5. 360 divided by 5 is 72 degrees.
So, each petal is 72 degrees apart. Now, we know where this one is located, because it's in a positive y direction, and so that's 90. So, let's start with that petal. If we add 72 to 90, it will give us the angle of the next petal.
72 plus 90 is 162. So this petal is 162 degrees relative to the positive x-axis. Now to get the next petal, add 72 to 162. So this one is going to be at 234 degrees, and that's in quadrant 3. Then if we add 72 again, the next one is going to be 306. Now... Now let's add 72 again. 306 plus 72, that's 378. 378 corresponds to this point, but we want an angle between 0 and 90. So we've got to find a cool terminal angle.
So we can subtract this by 360, and that will give us 18. Which means this is at 18. And 18 plus 72 will give us the one at 90. So keep in mind, you can also start with 90 and subtract it by 72. to get the one at 18. So there's different ways in which you can get the answer but what I would recommend doing is take 360 divided by the number of petals and find one petal in which you know for sure what the angle is. Either a petal on the x-axis or on the y-axis. And then from that petal either add or subtract by this angle to find the angles of the other petals. Now what if n is even?
Let's say if we have r, which is equal to 3 sine 2 theta. What should we do in this case? So we can see that n is 2, and whenever n is even, the number of petals is 2n as opposed to n.
So in this example, we're going to have 4 petals. So let's go ahead and graph it. So here's the first circle. Here's the second one. And here's the third one.
So we're going to have four petals. And when n is even, the petals will not be on the x or on the y axis. They're going to be in between.
So if they're in between the x and y axis, then you know that the angles have to be 45, which is between 0 and 90, and it has to be 135, 225, and 315. So now we can draw the four petals. And they have to be in a third circle because the radius is 3. And that's how you graph it. Now, if you took 360...
and divided by the number of petals, which is 4, you would get 90. And 90 will still give you the angle in between the petals. The difference between 135 and 45 is 90, and the difference between 135 and 225 is also 90. So that will still work, but you don't need to do that for this particular example. Now let's say if the equation was negative 3 sine 2 theta. When sine had an odd n value, when it was like 3, when we added the negative value, the graph, it flipped from being this shape into a shape that looked like this.
Now, because the number of petals is even, and we have symmetry in this graph, in the x and the y direction, the graph won't change. Because we have an even number of petals, Negative 3 sine 2 theta and positive 3 sine 2 theta correspond to the same graph. So whenever n is even, you're going to see that if you add a negative sign, it's not going to change the shape of the graph.
But when n is odd for the sine function, if you add the negative sign, it's going to cause the graph to flip. So for this graph, these two equations correspond to the same graph. Now let's say that r... is equal to 3 sin 4 theta. So because n is even, the number of petals is equal to 2n, or 2 times 4, which is 8. So r, at its max, is 3. Let's draw three circles.
Now we have eight petals. Where should we place the petals? When n was 2, the first angle was at 45. It was 90 divided by 2. So now n is 4. If we take 90 divided by 4, the first angle is going to be at 22.5. If we take 360 and divide it by 8, this will give us the angle between petals.
360 divided by 8 is 45. So if we add 45 to 22.5, the next petal will be at 67.5. Now if we add 45 to 67.5, that will give us an angle of 112.5. And if we add 45 again...
the next one is going to be 157.5 and then after that it's 202.5 and then 247.5 and then 292.5, and if we add 45 again, the last one is at 337.5. So you know where the pedals should be by performing these calculations. So we have 8 pedals. Each of which is going to have an r value of 3. So if you plug in these angles, you should get 3 in the equation. For example, if I take 22.5 and plug it into the equation, 4 times 22.5 is 90. Sine 90 is 1. 1 times 3 is 3. So the petal is going to have a length of 3. And if I plug in 67.5, 67.5 times 4 is 270. Sine of 270 is negative 1. And negative 1 times 3 has a length of 3. Now, keep in mind, when plotting polar coordinates, if r is negative, you need to go 180 in the other direction.
So, for example, even though I plug in 67.5, because I got a negative r value, I won't plot it here. I have to plot it here, which will take me to this point. 67.5 plus 180 will give me the angle 247.5. So, I will still get these angles. The graph won't change.
I just want to show you that if you plug in any of these angles, r will equal 3 or negative 3. But now let's go ahead and graph it. So that's the first petal. Here's the second one. And here's the rest.
So whenever n is even... None of the petals will be directly on the x or the y axis. They're in between for the sine graph. Let's move on to cosine. So in this example, we have an odd n value.
So the number of petals is simply equal to n, which in this case is 3. Now we have positive 2 cosine 3 theta. Whenever n is odd, there's going to be a petal in the positive x-axis if a is positive. Now all we need is two circles because a is equal to 2. The first petal is going to be here, and that's at an angle of 0. If we take 360 and divide it by the number of petals, this is going to be 120. So the next petal is at an angle of 120. And if we add 120 to 120, the next petal is at an angle of 240. So 120 should be somewhere in this region, and here's 240. So that's how you can graph 2 cosine 3 theta.
Now let's try this one. 3 cosine 5 theta. So n is odd, so the number of petals is equal to n once again.
So we have 5 petals in this graph. Now this graph is not like sine, where it's going to switch. The first petal will still be on a positive x-axis, which is good.
But this time, we need three circles, as opposed to two, since a is equal to three. So here's the first petal. It's at an angle of 0 degrees. Now, we know we have 5 petals, so let's take 360 and divide it by 5. And we know it's 72 degrees.
So the next petal is going to be at 72. And then 72 plus 72. That should be 144. So that's the location of the next one, which is closer to the negative x-axis. And then 144 plus 72, that's 216, plus 72 again. The next one is going to be 288. So that's how you define the location of the five petals.
So that's how you can graph 3 cosine 5 theta. Now what about graphing negative 3 cosine 5 theta? So everything is going to reverse. So this petal here will now be on the negative x-axis.
And we're going to start with that. So I'm going to rewrite the equation. And we're still going to have 5 petals, so p is 5. So the first petal is going to be in this location. So that's an angle of negative 180. And 360 divided by 5 is still 72. Well, this is positive 180, actually.
This is 0, 90, and this is 270. So to get the next petal... We need to add 72 plus 180. So that's going to be at 252. And then if we add 72 to that answer, the next one is going to be at 324. And then, 396 is the next one if we add 72. But since 396 is greater than 360, take away 360 from it to get the coterminal angle of 36. And then, 36 plus 72 is 108. And if you add 108 plus 72... you should get what you started with, 180. So now let's go ahead and plot the five petals. And that's it. Now what if we have an even value for n, and we're dealing with the cosine equation?
It turns out that this type of equation is actually easier than the other ones. So n is even, therefore the number of petals is 2n, or 2 times 2, which is 4. Now the good thing about the cosine graph when it's even is that you're going to have petals on the x and the y axis. For this particular graph, there's going to be a petal on the positive x axis, the negative x axis, and the positive and the negative y axis. R is 3, so we need to graph 3 units in each direction.
For this particular problem, I don't need this circle, because I know the petals are on the x and y axis. So here's the first petal. Here's the second one, here's the third one, and here's the fourth. So this is very easy to graph.
And it's easy to find the angles because they're on the axes. So this is 0, 90, 180, and 270. So that's the graph of 3 cosine 2 theta. And because n is even, it's also the graph of negative 3 cosine 2 theta.
So if you put a negative in front, it's not going to change the shape of the graph. Let's try this one. 2 cosine 4 theta. So the number of petals is 2n, or 2 times 4. This time we're going to have 8 petals.
And since we're dealing with the cosine equation, we're going to have some petals on the x and the y axis. But this time I'm going to use the circles. Let's make it 3 cosine instead of 2 cosine. Let's modify this equation a bit. So here's the first circle.
Here's the second one. And here is the third one. So we're going to have a petal in the positive and negative axis. And the same for the y-axis too.
There's the first one here's the second Here's the third and here's the fourth now we need four more So we know these angles are 0 90 180 and 270 And 360 divided by 8 is 45. So each petal is 45 degrees apart. Which means the next one has to be between 0 and 90. That's 45. And then we're going to have one at 135, 225, and 315. And so that's how you can graph it. So that's it for the Rose curves, now you know how to plot all of them.
The next type of graph we need to talk about are the Lemnus gates. Let's start with the cosine graph. So we have r squared is equal to a squared cosine 2 theta. The general shape for this type of graph looks like this, where this is a and negative a. So for example, let's say if we have the graph r squared is equal to 9 cosine 2 theta.
So you need to realize that 9 is equal to a squared. So if a squared is 9, that means a is 3. So we just have to travel 3 units to the right and 3 units to the left. And so we're going to get a graph that looks like that.
Let's try another example. Let's say that r squared is equal to 16 cosine 2 theta. So we can see that a squared is 16, which means a is 4. So we have to travel 4 units to the right.
and 4 to the left. And so that's how you can graph this particular Lemniscate. Now let's talk about the sine version.
So we have r squared is equal to a squared sine 2 theta. And the shape for this graph looks like this. So the symmetry is with respect to the pole and for the other one with where was a squared cosine 2 theta it's symmetric with respect to the polar axis the line theta equals pi over 2 and the pole in case you have to know that for the test. I'm going to focus on just graphing these equations so this is equal to a So let's say that r squared is 4 sine 2 theta.
a squared is equal to 4, and a is equal to 2. Now, for this type of graph... Because it's between the x and the y axis, even though when graphing it, the coordinates are really r and theta, I'm going to use the circles to graph it. So here's the first circle with a radius of 1, and here is the second circle with a radius of 2. So this shows that a is equal to 2, which is... the distance between the origin and this point.
And so that's how you can graph this particular function. Let's try one more example. Let's say that r squared is 9 sine 2 theta.
So you can see that a is equal to 3. So what we need to do is plot three circles. And the points will be the same. This is going to be at a 45 degree angle, and the next one is going to be at a 225 degree angle.
And that's how you can plot this particular Lemnis game. I've changed my mind. Let's talk about the symmetry. So going back to this equation, r squared is equal to a squared sine 2 theta. We said that it's symmetric about the pole.
So what exactly is the pole? It turns out the pole is just the origin. So it's symmetric about the origin.
You can see that this side is the same as that side. Now for the equation r squared is equal to a squared cosine 2 theta, it's symmetric about three things. So first, it's symmetric about the pole. As you can see, it's symmetric about the origin.
The right side and the left side are the same. Next, it's symmetric about the polar axis. The polar axis is basically equivalent to the x-axis. Notice that above and below the x-axis is the same. So this portion of the curve is symmetric about the x-axis.
This side is the same as that side. So in that sense, it's symmetric about the pole axis. And it's also symmetric about the line theta equals pi over 2. So here's pi over 2. That's basically the y-axis.
So it's symmetric about the y-axis, or the line theta equals pi over 2. As you can see, the right side is the same as the left side. So for your test, just know that for this equation, it's symmetric about the polar axis, which is the x-axis. It's symmetric about the pole, which is the origin.
And it's symmetric about the line theta equals pi over 2, which is basically the y-axis. And then for the other equation, r squared is equal to a squared sine 2 theta. It's simply symmetric about the pole or the origin.