WEBVTT 00:00:19.602 --> 0:00:26.602 To graph polar equations, it's first helpful to be able to identify the type of graph that will be formed by 00:00:26.605 --> 0:00:32.852 looking at the structure of the polar equation. The first group of graphs that we're going to examine are 00:00:32.855 --> 0:00:39.603 polar equations that graph lines. These will be written in one of four ways depending on what type of 00:00:39.606 --> 0:00:48.103 line it graphs. The first example that we're going to look at is a vertical line. In polar form an equation 00:00:48.106 --> 0:00:56.602 in the form r cosine theta equals some constant a will be the graph of a vertical line. So for example, this 00:00:56.605 --> 0:01:08.851 equation might be r cosine theta equals 9. Thinking back to r cosine theta being equivalent to x, then this makes sense 00:01:08.854 --> 0:01:17.650 because in rectangular form this equation would be x equals 9. Which we also know graphs a vertical line. 00:01:17.653 --> 00:01:25.352 An equation in the form r sine theta equals some constant a will be the graph of a horizontal line. So for 00:01:25.355 --> 00:01:37.611 example, this graph might have an equation r sine theta equals 5. Knowing that r sine theta is equivalent to y, 00:01:37.614 --> 00:01:45.100 in rectangular form this equation would be y equals 5. Which again we know graphs a horizontal line on a 00:01:45.103 --> 00:01:54.352 rectangular plane. If we have a line that is not vertical or horizontal and does not pass through the pole, 00:01:54.355 --> 00:02:02.852 which on a polar plane means the point (0, 0), we have a line that is not vertical or horizontal and does 00:02:02.855 --> 00:02:12.352 not pass through this point (0, 0). Then the equation will be in the form a r cosine theta plus b r sine theta 00:02:12.355 --> 00:02:23.602 equals c, where a, b, and c are some constants, and a and b are neither one equal to 0. So this equation might 00:02:23.605 --> 00:02:38.101 be in the form 2r cosine theta minus 3r sine theta equals 6. In rectangular form, that equation could be 00:02:38.104 --> 00:02:46.851 written as 2x minus 3y equals 6. Again, we would be able to recognize that as a line that is not vertical 00:02:46.854 --> 00:02:54.851 or horizontal and does not pass through the pole. The fourth type of line that we need to consider is a line 00:02:54.854 --> 00:03:03.601 that is not vertical or horizontal but does pass through the pole and forms an angle alpha with the polar axis. 00:03:03.6044 --> 00:03:11.351 On a polar plane, the polar axis is what we generally think of as the x-axis when we're talking about 00:03:11.354 --> 00:03:21.118 a rectangular plane. So that's our polar axis, and this line forms some angle alpha. In this case, we can just 00:03:21.121 --> 00:03:29.850 write the equation in the form theta equals whatever the angle measurement of alpha is. So for example, 00:03:29.853 --> 00:03:42.352 the equation of this line might be theta equals 5 pi over 6. That would not only represent the points along this 00:03:42.355 --> 00:03:50.352 ray of 5 pi over 6, but would also represent the points on the other side of the ray that forms the line because 00:03:50.355 --> 00:03:58.350 recall that if we start at 0 and we have negative values for r, those would move along the other side of 00:03:58.353 --> 00:04:07.852 the 5 pi over 6 ray and form the entire line. So these are the four forms of polar equations that graph lines. 00:04:07.855 --> 00:04:14.351 You should be able to recognize them as lines, and further be able to describe them as vertical, horizontal, 00:04:14.354 --> 00:04:21.852 or lines that are not vertical or horizontal that do or do not pass through the pole. 00:04:24.350 --> 00:04:31.351 Next we will look at polar equations that graph circles. There are three different forms of a polar equation 00:04:31.354 --> 00:04:38.350 that we will examine, all of which form the graph of a circle. The first type of circle that we'll look at is a 00:04:38.353 --> 00:04:45.605 circle that is centered at the pole which means its center is at the point (0,0) and has a radius that is the 00:04:45.608 --> 00:04:52.349 length of the absolute value of a. In this case the equation will be in the form r equals a where 00:04:52.352 --> 00:05:00.600 a is some constant. So for example, the equation of this circle might be r equals negative 3. 00:05:00.603 --> 00:05:20.852 That would tell me that this is a circle that has a center at the pole and that has a radius of 3 units. 00:05:20.855 --> 00:05:28.852 The second type of equation that we'll look at is an equation in the form r equals a sine theta. An equation 00:05:28.855 --> 00:05:37.100 in this form will also graph a circle. The circle will be centered on the line theta equals pi over 2, which 00:05:37.103 --> 00:05:44.848 would be on the y-axis if we were talking about a rectangular coordinate system. The center will be the 00:05:44.851 --> 00:05:51.854 absolute value of a over 2 units from the pole and the length of the radius will be the absolute value of 00:05:51.857 --> 00:06:05.602 a over 2 units. So for example, the equation of this circle might be r equals negative 4 sine theta. In this case 00:06:05.605 --> 00:06:17.348 I would know that this is a circle that's centered on the line theta equals pi over 2. The center will be the 00:06:17.351 --> 00:06:32.601 absolute value of a which is negative 4 over 2 units from the pole. The center will be 2 units from the pole, 00:06:32.604 --> 00:06:44.602 which means that the center would be at the point (0, negative 2). The radius will be a length of absolute value of 00:06:44.605 --> 00:06:58.351 a over 2 units, so the radius is 2. The reason that the center was on (0, negative 2) instead of (0, 2) 00:06:58.354 --> 00:07:05.600 is because this coefficient in front of sine theta was a negative 4. Had it been a positive 4, we would have gone 00:07:05.603 --> 00:07:14.880 through the same procedure except that the center would be 2 units above the pole instead of 2 units below. 00:07:14.883 --> 00:07:23.102 The third form of a circle that we're going to examine would be an equation in the form r equals a cosine theta. 00:07:23.105 --> 00:07:29.850 In this case the circle will be centered on the polar axis which recall that means the x-axis when we're 00:07:29.853 --> 00:07:37.850 talking about a rectangular plane. Again the center will be the absolute value of a over 2 units from the pole, 00:07:37.853 --> 00:07:44.855 and the radius will have a length of absolute value of a over 2. So an equation for this circle might be 00:07:44.858 --> 00:07:55.854 something like r equals 2 cosine theta. r equals 2 cosine theta would be a circle that's centered on the 00:07:55.857 --> 00:08:07.855 polar axis. Its center would be the absolute value of a over 2 units away from the pole, so its center would 00:08:07.858 --> 00:08:21.601 be 1 unit from the pole, which means that the center would be at the point (1, 0), and its radius would be 1. 00:08:21.604 --> 00:08:28.103 So given an equation in any of these three forms, you should be able to recognize that it's a circle 00:08:28.106 --> 00:08:33.855 and be able to determine the center and the radius of the circle. 00:08:35.855 --> 00:08:43.104 The next type of graph that we're going to look at is the graph of a limacon. The equation will be in one of 00:08:43.107 --> 00:08:52.854 two forms: r equals a plus b sine theta or r equals a plus b cosine theta. Once you've identified that the 00:08:52.857 --> 00:09:01.103 graph will be a limacon, you need to consider one of four cases by looking at the absolute value of 00:09:01.106 --> 00:09:09.855 a divided by b. The absolute value of this ratio will tell you a little bit more about the shape of the limacon. 00:09:09.858 --> 00:09:16.852 If the absolute value of a over b is less than 1, then the limacon will have an inner loop. So it will have both 00:09:16.855 --> 00:09:26.351 an inner loop and this larger outer loop. As this ratio increases towards the value of 1, the inner loop will 00:09:26.354 --> 00:09:35.600 become smaller and smaller. If the ratio of a over b has an absolute value that is exactly equal to 1, then the 00:09:35.603 --> 00:09:42.373 inner loop disappears, and the graph forms what is called a cardioid. Instead of having an inner loop, 00:09:42.376 --> 00:09:55.858 it has this sharp turn or cusp in the graph. If the ratio increases beyond 1 to any value between 1 and 2, 00:09:55.861 --> 00:10:02.407 then this sharp turn or cusp begins to disappear or become less pronounced, and as the absolute value of 00:10:02.410 --> 00:10:10.105 a over b increases from 1 to 2 the graph will be a limacon that will have this dimple. So it still has this 00:10:10.108 --> 00:10:20.383 turning point, it's just not as sharp or pronounced. As the ratio approaches 2, this dimple will become 00:10:20.386 --> 00:10:30.123 smaller and smaller or less pronounced. Once the absolute value of a over b is exactly equal to 2 00:10:30.126 --> 00:10:36.854 that dimple then disappears, and the graph looks almost like a circle. You can see on my drawing that it looks 00:10:36.857 --> 00:10:46.354 very much like a circle. You have to look closely to see that there's this part that looks almost vertical. It's 00:10:46.357 --> 00:10:54.107 not completely vertical. It does not completely flatten out, but it has this part that is very steep, and so this 00:10:54.110 --> 00:11:03.104 graph is not exactly a circle, but it does look very similar to a circle. So once the absolute value of a over b 00:11:03.107 --> 00:11:11.355 reaches a value of 2 or greater, you will have a limacon, but it will have no inner loop or dimple. 00:11:11.358 --> 00:11:20.100 So that ratio can tell you quite a bit about the graph. Whether the equation is in terms of sine or cosine can 00:11:20.103 --> 00:11:27.104 also give you a little bit more information about the symmetry of the graph. If the graph is in terms of cosine, 00:11:27.107 --> 00:11:35.584 in the form r equals a plus b cosine theta, then the graph will be symmetric along the polar axis. 00:11:35.587 --> 00:11:42.104 You see that all four of the graphs I have drawn here are symmetric with respect to the polar axis. 00:11:42.107 --> 00:11:50.604 All four of these graphs would have equations in the form r equals a plus b cosine theta. If the equation is 00:11:50.607 --> 00:11:57.601 instead in terms of sine theta, then the graph will be symmetric with respect to theta equals pi over 2. 00:11:57.604 --> 00:12:01.500 We'll look at an example in a minute. 00:12:03.850 --> 00:12:11.101 In this example, I see that the equation is in the form r equals a plus b sine theta, which tells me that the 00:12:11.104 --> 00:12:22.110 graph is going to be a limacon. Looking at the absolute value of the ratio a over b, it's equal to the 00:12:22.113 --> 00:12:32.350 absolute value of 1 over negative 2, which is 1/2. Since 1/2 is less than 1, this graph is going to be a 00:12:32.353 --> 00:12:46.601 limacon with an inner loop. Because the equation is in terms of sine, I know that the graph will be symmetric 00:12:46.604 --> 00:13:01.851 about the line theta equals pi over 2. Now that I know a little bit about what the graph is going to look like, 00:13:01.854 --> 00:13:09.351 I want to select some values to substitute in for theta to get some points to plot. The values that I'm going to 00:13:09.354 --> 00:13:16.351 use for theta are the quadrantal angles because they are easy to figure out and often represent the important 00:13:16.354 --> 00:13:23.851 points on the graph. Then I'm going to use the angles in the pi over 6 family. The reason I'm using pi over 6 00:13:23.854 --> 00:13:31.351 angles is because this equation is in terms of sine and I know that the sine of pi over 6 is 1/2. Had the 00:13:31.354 --> 00:13:37.603 equation been in terms of cosine, I would have used the quadrantal angles and the angles in the pi over 3 00:13:37.606 --> 00:13:45.369 family. While any angles that you choose will work, these are the best angles to choose in this case. 00:13:45.372 --> 00:14:00.852 So, I'm going to use the values of theta, starting with 0, then pi over 6, pi over 2, 5 pi over 6, pi, 7 pi over 6, 00:14:00.855 --> 00:14:11.601 3 pi over 2, and 11 pi over 6. I could finish with 2 pi, but when I substitute in 2 pi, I'll get the same value 00:14:11.604 --> 00:14:25.101 as when I substitute in 0. Starting with 0, r will equal 1 minus 2 times the sine of 0. Since the sine of 0 is 0, 00:14:25.104 --> 00:14:40.351 r will end up equaling 1, and I'll have the ordered pair (1, 0), r, theta. Substituting in pi over 6, 00:14:40.354 --> 00:14:55.851 the sine of pi over 6 is 1/2. 2 times 1/2 is 1. 1 minus 1 gives me 0, so I have the point (0, pi over 6). 00:14:55.854 --> 00:15:15.601 Continuing with pi over 2, the sine of pi over 2 is 1. 2 times 1 is 2. 1 minus 2 gives me a value of negative 1. 00:15:15.604 --> 00:15:24.602 5 pi over 6 is in the pi over 6 family, so I know that the sine will be either positive or negative 1/2. 00:15:24.605 --> 00:15:33.350 Since 5 pi over 6 is in quadrant two, sine of 5 pi over 6 is positive 1/2, which means it will work out to be the 00:15:33.353 --> 00:15:48.350 the same value as substituting in pi over 6. So I get an r-value of 0. Pi lies on the x-axis, so sine is 0, just 00:15:48.353 --> 00:16:02.351 like it was when I substituted in 0, which means I will get an r-value of 1. Now let's substitute in 7 pi over 6. 00:16:02.354 --> 00:16:10.851 7 pi over 6 is in quadrant three where sine is negative, so the sine of 7 pi over 6 is negative 1/2. 00:16:10.854 --> 00:16:32.602 I end up with 1 plus 1 or 2, so my point will be (2, 7 pi over 6). 3 pi over 2 is on the negative y-axis, so sine is 00:16:32.605 --> 00:16:49.854 negative 1. I end up with 1 plus 2, or an r - value of 3. 11 pi over 6 is in quadrant four where sine is also negative, 00:16:49.857 --> 00:17:03.600 so 11 pi over 6 will work out in the same way that 7 pi over 6 did and give me an r - value of 2. Then again, 00:17:03.603 --> 00:17:12.101 if I wanted to finish up with 2 pi, I would get a value of 1 again. Alright, so now let's plot these points and see 00:17:12.104 --> 00:17:23.601 what the graph is going to look like. Plotting the first point at (1, 0). I have a point here. 00:17:23.604 --> 00:17:35.351 (0, pi over 6). My angle of rotation is pi over 6, but my r - value is 0. So the point will be at the pole. 00:17:35.354 --> 00:17:46.350 My third point is at (negative 1, pi over 2). Recall that if I go to pi over 2 and r is negative 1, I plot the point 00:17:46.353 --> 00:17:52.600 one unit in the opposite direction from pi over 2. So it would be there. 00:17:52.603 --> 00:18:08.851 (0, 5 pi over 6). Here's 5 pi over 6, but again the point is at the pole. (1, pi) would be here. 00:18:08.854 --> 00:18:20.851 (2, 7 pi over 6). So here's my 7 pi over 6 angle, and it will lie on the circle with a radius of 2. 00:18:20.854 --> 00:18:39.850 (3, 3 pi over 2) would be here. (2, 11 pi over 6), and then if I circled all the way around I would end up at (1, 2pi). 00:18:39.853 --> 00:18:46.850 So, the graph would start here at this first point, and then the points would be connected in the order of our 00:18:46.853 --> 00:18:59.351 angles of rotation. I would go from here, to the point on the pole, or the point at the pole, rotate down to form 00:18:59.354 --> 00:19:12.353 the inner loop, back up through the pole, and then around to form the outer loop. As we said from the beginning, 00:19:12.356 --> 00:19:23.601 we end up with a limacon that has an inner loop and is symmetric about the line theta equals pi over 2. 00:19:23.604 --> 00:19:30.853 If the graph happens to be a different type of limacon, such as a cardioid, or a limacon that has a dimple, 00:19:30.856 --> 00:19:37.851 or one that has neither an inner loop nor a dimple, the process of graphing is exactly the same. 00:19:37.854 --> 00:19:43.602 So once you've identified the type of limacon and the type of symmetry, then it's just a matter of 00:19:43.605 --> 00:19:47.850 finding enough points to plot to draw the graph. 00:19:50.000 --> 00:19:57.451 The next type of graph that we're going to look at are graphs of roses. The equation for the graph of a rose will 00:19:57.454 --> 00:20:11.353 be in the form r equals a sine n theta or r equals a cosine n theta, where n is a whole number not equal to 1. 00:20:11.356 --> 00:20:35.601 When n is odd, the rose will have n petals. In this case, you will only need to find points in the interval 0 to pi. 00:20:35.604 --> 00:20:41.353 After that, if you continued finding points from pi to 2 pi, the graph would simply trace over the petals 00:20:41.356 --> 00:20:49.349 that you've already located. So in this instance, when n is odd, the graph has n petals and you need points 00:20:49.352 --> 00:20:59.603 on the interval from 0 to pi to complete the graph. If n is even, then the rose will have 2n petals. In that case, 00:20:59.606 --> 00:21:18.009 you will need to find points on the interval 0 to 2 pi in order to complete the graph. So if n is even, the graph 00:21:18.102 --> 00:21:26.853 will have twice n or 2n petals, and you will need to find points from 0 all the way to 2 pi in order to complete 00:21:26.856 --> 00:21:37.099 the entire graph. In either instance, the length of the petals will be equal to the absolute value of a, so this 00:21:37.102 --> 00:21:47.350 will tell you how long each of the petals is in length. Now let's look at an example of how to graph a rose. 00:21:49.800 --> 00:21:54.853 Looking at this example, I can see that this is going to graph a rose because the equation is in the form 00:21:54.856 --> 00:22:08.381 r equals a sine n theta. n in this case is 5 and since n is odd, the graph will have n petals. So this is a rose 00:22:08.384 --> 00:22:29.602 with 5 petals. The length of each petal will be 3, and when I say that the length of the petal is 3, I'm referring 00:22:29.605 --> 00:22:39.102 to the distance from the pole to the outermost endpoint of the petal. So you can see that what we're about 00:22:39.105 --> 00:22:46.602 to do is find the endpoints of the petals. To do that we need to know that the length from the pole to the 00:22:46.605 --> 00:22:58.850 endpoint has to be 3. To find the endpoints of the petals, we need to find the values of theta where r is 00:22:58.853 --> 00:23:12.352 equal to 3, which would be the largest possible value of r. So I can set up this equation 3 sine 5 theta equals 3, 00:23:12.355 --> 00:23:22.352 and I'm going to solve this equation for theta. The other possibility is that r could have a value of negative 3, 00:23:22.355 --> 00:23:37.352 so I also need to set up an equation 3 sine 5 theta equals negative 3. So when r has a value of 3 or negative 3, 00:23:37.355 --> 00:23:46.102 those will tell me the values of theta where the endpoints of the petals occur. To solve this equation for theta, 00:23:46.105 --> 00:23:58.852 I first need to divide both sides by 3. We get sine of 5 theta equals 1. Often you'll see that people might skip 00:23:58.855 --> 00:24:09.352 this first step of setting the equation equal to the largest and smallest possible values of r, and if after a few 00:24:09.355 --> 00:24:16.602 examples you want to just jump ahead to this step, that's fine. Alright so now I'm solving this trig equation, 00:24:16.605 --> 00:24:26.102 which means my next step is to undo the sine function and write the general formula that's equal to 5 theta. 00:24:26.105 --> 00:24:38.352 So I need to think about what angle has a sine value of 1, which is pi over 2. So 5 theta is equal to pi over 2 00:24:38.355 --> 00:24:50.602 plus 2 pi k. That's my general formula. I now need to finish solving for theta by dividing by 5, and I get that 00:24:50.605 --> 00:25:04.851 theta is equal to pi over 10 plus 2 pi over 5 times k. Now I begin substituting in values of k, integer values 00:25:04.854 --> 00:25:17.601 beginning with 0. So starting with a value of 0, if I substitute in 0 for k I get that theta is pi over 10. 00:25:17.604 --> 00:25:29.352 Next I would want to substitute in 1 for k. Substituting in 1, I would have pi over 10 plus 2 pi over 5 times 1, 00:25:29.355 --> 00:25:46.352 so just 2 pi over 5. Rewriting this as 4 pi over 10, I get that theta is equal to 5 pi over 10, which reduces to 00:25:46.355 --> 00:25:59.852 pi over 2. So this is the value I got when I substituted in 0 for k, then 1, and now I'm going to substitute in 2 for k. 00:25:59.855 --> 00:26:15.101 Substituting in 2, I get pi over 10 plus 2 pi over 5 times 2, which would be 4 pi over 5. Changing 4 pi over 5 00:26:15.104 --> 00:26:30.102 to 8 pi over 10, I get a theta value of 9 pi over 10 when k is 2. Remember that because n was odd, 00:26:30.105 --> 00:26:39.352 and there were n petals, 5 petals, I only need to find values on an interval from 0 to pi. Therefore, I can 00:26:39.354 --> 00:26:46.602 stop substituting in values here, because if I substitute in 3 for k, I would get a value that exceeds pi. 00:26:46.605 --> 00:26:53.352 If you're not sure, you can continue substituting in values but once you get a value that exceeds pi, you stop. 00:26:53.355 --> 00:27:04.602 I now have endpoints for 3 petals but remember that we have 5 petals. So to get the other two, I will need to solve 00:27:04.605 --> 00:27:18.850 this equation for theta, using the same process. Dividing both sides by 3, I get sine 5 theta equals negative 1. 00:27:18.853 --> 00:27:25.102 Now we need to think about the angle that has a sine value of negative 1, and again write our general 00:27:25.1055 --> 00:27:39.855 formula that's equal to 5 theta. Sine equals negative 1 on the negative y-axis, so at 3 pi over 2 plus 2 pi k. 00:27:39.858 --> 00:27:53.850 Dividing both sides by 5, I get that theta is equal to 3 pi over 10 plus 2 pi over 5 k. Now we're going to do the 00:27:53.853 --> 00:28:06.386 same thing that we did over here, which is begin by substituting in 0 for k. When k is 0, theta is 3 pi over 10. 00:28:06.389 --> 00:28:16.850 When k is 1, remember that this will become 4 pi over 10, so 3 pi over 10 plus 4 pi over 10 will give me 00:28:16.853 --> 00:28:32.352 7 pi over 10. I can stop there because I know that there are 5 petals, and I have now found 5 values for theta. 00:28:32.355 --> 00:28:44.852 Going back and writing these as ordered pairs, you would have an r-value of 3 and a theta value of pi over 10, 00:28:44.855 --> 00:28:57.101 pi over 2, and 9 pi over 10. So let's list these ordered pairs. As I list them, I'm going to list them in order of 00:28:57.104 --> 00:29:07.603 theta, going in ascending order. So my first ordered pair would be (3, pi over 10). Then my second ordered 00:29:07.606 --> 00:29:32.352 pair would be (negative 3, 3 pi over 10). The next one would be (3, pi over 2), (negative 3, 7 pi over 10), and then the fifth one 00:29:32.355 --> 00:29:48.102 would be (3, 9 pi over 10). Now let's find the values of theta where the graph crosses through the pole. 00:29:48.105 --> 00:29:55.602 To do this we're going to take our equation and replace r with 0. So we're finding the values of theta 00:29:55.605 --> 00:30:11.851 where the r-value is 0. To do this we're going to solve this trig equation, 3 sine of 5 theta equals 0. 00:30:11.854 --> 00:30:26.852 Dividing both sides by 3, I get sine of 5 theta equals 0. Now again I need to undo the sine function and 00:30:26.855 --> 00:30:34.102 set up the general formula that's equal to 5 theta. Here you need to think about the fact that there are 00:30:34.105 --> 00:30:45.352 two places where sine equals 0, and they are separated by pi. Sine equals 0 at 0 and pi, so we can write this 00:30:45.355 --> 00:31:00.352 using just a single general formula. It would be 5 theta equals 0 plus pi k. That takes care of 0, and then pi, 00:31:00.355 --> 00:31:13.099 2 pi, 3 pi, etcetera. Now finish solving for theta by dividing both sides by 5. Zero divided by 5 is still 0, so 00:31:13.102 --> 00:31:23.102 I could write this general formula as just pi over 5 times k. Now again, I would begin substituting in values 00:31:23.105 --> 00:31:35.853 of k beginning at 0. So the first value would be theta equals 0, when k is 0. Then I would substitute in 1, which 00:31:35.856 --> 00:31:51.354 would just give me pi over 5. Substituting in 2, would give me 2 pi over 5. Substituting in 3, would give me 00:31:51.357 --> 00:32:05.602 3 pi over 5, and then 4 would give me 4 pi over 5. If I substitute in 5, I would get pi which would complete 00:32:05.605 --> 00:32:14.101 the graph, but I only need to find points on the interval starting from 0 and up to but not including pi. 00:32:14.104 --> 00:32:18.851 Now that we have the endpoints of the petals and we know the values of theta where the graph crosses 00:32:18.854 --> 00:32:31.602 through the pole, we can construct the graph. So we know that at 0, r is 0. Then we can plot the endpoint of 00:32:31.605 --> 00:32:43.874 the first petal at (3, pi over 10), which would be about right there. These values let me know that the graph 00:32:43.877 --> 00:32:51.101 keeps crossing back through the pole to form each petal. So even though these points won't be visible 00:32:51.104 --> 00:32:56.101 because each one has an r - value of 0, we know that the graph continues to loop back through the pole as we 00:32:56.104 --> 00:33:06.102 draw each petal. So let's plot the endpoints of the other four petals. The second one will be at (negative 3, 3 pi over 10). 00:33:06.105 --> 00:33:15.348 If 3 pi over 10 is about right there, then negative 3 means we're moving 3 units in the opposite direction 00:33:15.351 --> 00:33:25.852 and there's our second endpoint. The third endpoint will be at (3, pi over 2). Then we have an endpoint at 00:33:25.855 --> 00:33:39.351 (negative 3, 7 pi over 10), which will be 3 units in the opposite direction from 7 pi over 10. Then the fifth 00:33:39.354 --> 00:33:53.602 endpoint will be at (3, 9 pi over 10). So we can draw the graph now by starting at 0. The graph will loop 00:33:53.605 --> 00:34:02.863 through the endpoint of the first petal, and back through the pole, down to form the second petal, back up 00:34:02.866 --> 00:34:10.853 through the pole and up to form the third petal, back down through the pole to form petal four, 00:34:10.856 --> 00:34:19.852 and then finally loop back through and form the fifth petal. 00:34:19.855 --> 00:34:27.854 In a case where you have an n-value that is even and you have twice as many petals, the process is exactly 00:34:27.857 --> 00:34:33.852 the same as what you've seen here, as far as finding the endpoints by solving these equations, and finding 00:34:33.855 --> 00:34:38.852 the values of theta where the graph crosses through the pole. The only difference will be that you'll need to 00:34:38.855 --> 00:34:48.851 find points on the interval from 0 all the way to 2 pi, rather than on the interval from 0 to pi like we did here. 00:34:50.850 --> 00:34:57.851 The final type of graph that we're going to look at is the graph of a lemniscate. The equations will be in the 00:34:57.854 --> 00:35:06.350 form r squared equals a squared times sine 2 theta or r squared equals a squared cosine 2 theta. You can 00:35:06.353 --> 00:35:10.851 see that the equations are easy to differentiate from the other ones we've seen today because they look 00:35:10.854 --> 00:35:21.852 quite different. If the equation is in terms of sine, then the graph will look like this. It will be symmetric 00:35:21.855 --> 00:35:34.102 about the pole and the graph will lie on this line, theta equals pi over 4. So it'll be symmetric about this line 00:35:34.105 --> 00:35:45.852 theta equals pi over 4. If the graph is in terms of cosine, then that graph is also symmetric about the pole, 00:35:45.855 --> 00:35:56.050 but then it's symmetric about the polar axis. So if the graph is in terms of cosine your graph will look like this, 00:35:56.053 --> 00:36:05.602 and if the graph is in terms of sine, it will look like this. For example, r squared equals 36 times sine 2 theta. 00:36:05.605 --> 00:36:14.851 By looking at this equation, I would know right away that it was a lemniscate, and that it's symmetric along 00:36:14.854 --> 00:36:27.852 this line theta equals pi over 4. Additionally, I could find the length from the pole to the endpoint. The length 00:36:27.855 --> 00:36:54.102 from the pole to the endpoint of each loop is equal to the value of a. So in this equation, a squared is 36. 00:36:54.105 --> 00:37:11.103 a is 6, and so the length from the endpoint of the loop to the pole is 6 units, and the same would be true here. 00:37:11.106 --> 00:37:17.352 You can see that in the graphs that we've looked at today, some of the graphs could be easily graphed 00:37:17.355 --> 00:37:24.605 using our more familiar rectangular graphing system, the graphs of lines and circles. But the other types 00:37:24.608 --> 00:37:32.140 of graphs we've explored, the limacons, roses, and lemniscates are much easier graphed using the polar 00:37:32.143 --> 00:37:38.100 coordinate system, than if we converted them to rectangular form and tried to graph them in that way. 00:37:38.108 --> 00:37:45.000 So each graphing system is useful depending on what type of graph you're trying to generate.