WEBVTT
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To graph polar equations, it's first helpful to be able to identify the type of graph that will be formed by
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looking at the structure of the polar equation. The first group of graphs that we're going to examine are
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polar equations that graph lines. These will be written in one of four ways depending on what type of
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line it graphs. The first example that we're going to look at is a vertical line. In polar form an equation
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in the form r cosine theta equals some constant a will be the graph of a vertical line. So for example, this
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equation might be r cosine theta equals 9. Thinking back to r cosine theta being equivalent to x, then this makes sense
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because in rectangular form this equation would be x equals 9. Which we also know graphs a vertical line.
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An equation in the form r sine theta equals some constant a will be the graph of a horizontal line. So for
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example, this graph might have an equation r sine theta equals 5. Knowing that r sine theta is equivalent to y,
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in rectangular form this equation would be y equals 5. Which again we know graphs a horizontal line on a
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rectangular plane. If we have a line that is not vertical or horizontal and does not pass through the pole,
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which on a polar plane means the point (0, 0), we have a line that is not vertical or horizontal and does
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not pass through this point (0, 0). Then the equation will be in the form a r cosine theta plus b r sine theta
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equals c, where a, b, and c are some constants, and a and b are neither one equal to 0. So this equation might
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be in the form 2r cosine theta minus 3r sine theta equals 6. In rectangular form, that equation could be
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written as 2x minus 3y equals 6. Again, we would be able to recognize that as a line that is not vertical
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or horizontal and does not pass through the pole. The fourth type of line that we need to consider is a line
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that is not vertical or horizontal but does pass through the pole and forms an angle alpha with the polar axis.
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On a polar plane, the polar axis is what we generally think of as the x-axis when we're talking about
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a rectangular plane. So that's our polar axis, and this line forms some angle alpha. In this case, we can just
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write the equation in the form theta equals whatever the angle measurement of alpha is. So for example,
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the equation of this line might be theta equals 5 pi over 6. That would not only represent the points along this
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ray of 5 pi over 6, but would also represent the points on the other side of the ray that forms the line because
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recall that if we start at 0 and we have negative values for r, those would move along the other side of
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the 5 pi over 6 ray and form the entire line. So these are the four forms of polar equations that graph lines.
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You should be able to recognize them as lines, and further be able to describe them as vertical, horizontal,
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or lines that are not vertical or horizontal that do or do not pass through the pole.
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Next we will look at polar equations that graph circles. There are three different forms of a polar equation
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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
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circle that is centered at the pole which means its center is at the point (0,0) and has a radius that is the
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length of the absolute value of a. In this case the equation will be in the form r equals a where
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a is some constant. So for example, the equation of this circle might be r equals negative 3.
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That would tell me that this is a circle that has a center at the pole and that has a radius of 3 units.
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The second type of equation that we'll look at is an equation in the form r equals a sine theta. An equation
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in this form will also graph a circle. The circle will be centered on the line theta equals pi over 2, which
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would be on the y-axis if we were talking about a rectangular coordinate system. The center will be the
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absolute value of a over 2 units from the pole and the length of the radius will be the absolute value of
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a over 2 units. So for example, the equation of this circle might be r equals negative 4 sine theta. In this case
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I would know that this is a circle that's centered on the line theta equals pi over 2. The center will be the
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absolute value of a which is negative 4 over 2 units from the pole. The center will be 2 units from the pole,
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which means that the center would be at the point (0, negative 2). The radius will be a length of absolute value of
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a over 2 units, so the radius is 2. The reason that the center was on (0, negative 2) instead of (0, 2)
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is because this coefficient in front of sine theta was a negative 4. Had it been a positive 4, we would have gone
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through the same procedure except that the center would be 2 units above the pole instead of 2 units below.
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The third form of a circle that we're going to examine would be an equation in the form r equals a cosine theta.
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In this case the circle will be centered on the polar axis which recall that means the x-axis when we're
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talking about a rectangular plane. Again the center will be the absolute value of a over 2 units from the pole,
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and the radius will have a length of absolute value of a over 2. So an equation for this circle might be
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something like r equals 2 cosine theta. r equals 2 cosine theta would be a circle that's centered on the
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polar axis. Its center would be the absolute value of a over 2 units away from the pole, so its center would
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be 1 unit from the pole, which means that the center would be at the point (1, 0), and its radius would be 1.
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So given an equation in any of these three forms, you should be able to recognize that it's a circle
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and be able to determine the center and the radius of the circle.
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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
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two forms: r equals a plus b sine theta or r equals a plus b cosine theta. Once you've identified that the
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graph will be a limacon, you need to consider one of four cases by looking at the absolute value of
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a divided by b. The absolute value of this ratio will tell you a little bit more about the shape of the limacon.
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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
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an inner loop and this larger outer loop. As this ratio increases towards the value of 1, the inner loop will
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become smaller and smaller. If the ratio of a over b has an absolute value that is exactly equal to 1, then the
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inner loop disappears, and the graph forms what is called a cardioid. Instead of having an inner loop,
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it has this sharp turn or cusp in the graph. If the ratio increases beyond 1 to any value between 1 and 2,
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then this sharp turn or cusp begins to disappear or become less pronounced, and as the absolute value of
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a over b increases from 1 to 2 the graph will be a limacon that will have this dimple. So it still has this
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turning point, it's just not as sharp or pronounced. As the ratio approaches 2, this dimple will become
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smaller and smaller or less pronounced. Once the absolute value of a over b is exactly equal to 2
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that dimple then disappears, and the graph looks almost like a circle. You can see on my drawing that it looks
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very much like a circle. You have to look closely to see that there's this part that looks almost vertical. It's
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not completely vertical. It does not completely flatten out, but it has this part that is very steep, and so this
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graph is not exactly a circle, but it does look very similar to a circle. So once the absolute value of a over b
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reaches a value of 2 or greater, you will have a limacon, but it will have no inner loop or dimple.
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So that ratio can tell you quite a bit about the graph. Whether the equation is in terms of sine or cosine can
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also give you a little bit more information about the symmetry of the graph. If the graph is in terms of cosine,
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in the form r equals a plus b cosine theta, then the graph will be symmetric along the polar axis.
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You see that all four of the graphs I have drawn here are symmetric with respect to the polar axis.
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All four of these graphs would have equations in the form r equals a plus b cosine theta. If the equation is
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instead in terms of sine theta, then the graph will be symmetric with respect to theta equals pi over 2.
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We'll look at an example in a minute.
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In this example, I see that the equation is in the form r equals a plus b sine theta, which tells me that the
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graph is going to be a limacon. Looking at the absolute value of the ratio a over b, it's equal to the
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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
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limacon with an inner loop. Because the equation is in terms of sine, I know that the graph will be symmetric
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about the line theta equals pi over 2. Now that I know a little bit about what the graph is going to look like,
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I want to select some values to substitute in for theta to get some points to plot. The values that I'm going to
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use for theta are the quadrantal angles because they are easy to figure out and often represent the important
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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
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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
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equation been in terms of cosine, I would have used the quadrantal angles and the angles in the pi over 3
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family. While any angles that you choose will work, these are the best angles to choose in this case.
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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,
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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
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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,
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r will end up equaling 1, and I'll have the ordered pair (1, 0), r, theta. Substituting in pi over 6,
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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).
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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.
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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.
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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
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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
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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.
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7 pi over 6 is in quadrant three where sine is negative, so the sine of 7 pi over 6 is negative 1/2.
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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
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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,
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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,
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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
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what the graph is going to look like. Plotting the first point at (1, 0). I have a point here.
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(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.
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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
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one unit in the opposite direction from pi over 2. So it would be there.
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(0, 5 pi over 6). Here's 5 pi over 6, but again the point is at the pole. (1, pi) would be here.
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(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.
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(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).
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So, the graph would start here at this first point, and then the points would be connected in the order of our
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angles of rotation. I would go from here, to the point on the pole, or the point at the pole, rotate down to form
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the inner loop, back up through the pole, and then around to form the outer loop. As we said from the beginning,
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we end up with a limacon that has an inner loop and is symmetric about the line theta equals pi over 2.
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If the graph happens to be a different type of limacon, such as a cardioid, or a limacon that has a dimple,
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or one that has neither an inner loop nor a dimple, the process of graphing is exactly the same.
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So once you've identified the type of limacon and the type of symmetry, then it's just a matter of
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finding enough points to plot to draw the graph.
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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
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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.
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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.
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After that, if you continued finding points from pi to 2 pi, the graph would simply trace over the petals
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that you've already located. So in this instance, when n is odd, the graph has n petals and you need points
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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,
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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
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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
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the entire graph. In either instance, the length of the petals will be equal to the absolute value of a, so this
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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.
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Looking at this example, I can see that this is going to graph a rose because the equation is in the form
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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
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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
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to the distance from the pole to the outermost endpoint of the petal. So you can see that what we're about
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to do is find the endpoints of the petals. To do that we need to know that the length from the pole to the
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endpoint has to be 3. To find the endpoints of the petals, we need to find the values of theta where r is
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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,
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and I'm going to solve this equation for theta. The other possibility is that r could have a value of negative 3,
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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,
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those will tell me the values of theta where the endpoints of the petals occur. To solve this equation for theta,
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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
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this first step of setting the equation equal to the largest and smallest possible values of r, and if after a few
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examples you want to just jump ahead to this step, that's fine. Alright so now I'm solving this trig equation,
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which means my next step is to undo the sine function and write the general formula that's equal to 5 theta.
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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
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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
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theta is equal to pi over 10 plus 2 pi over 5 times k. Now I begin substituting in values of k, integer values
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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.
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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,
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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
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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.
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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
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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,
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and there were n petals, 5 petals, I only need to find values on an interval from 0 to pi. Therefore, I can
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stop substituting in values here, because if I substitute in 3 for k, I would get a value that exceeds pi.
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If you're not sure, you can continue substituting in values but once you get a value that exceeds pi, you stop.
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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
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this equation for theta, using the same process. Dividing both sides by 3, I get sine 5 theta equals negative 1.
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Now we need to think about the angle that has a sine value of negative 1, and again write our general
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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.
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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
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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.
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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
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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.
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Going back and writing these as ordered pairs, you would have an r-value of 3 and a theta value of pi over 10,
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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
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theta, going in ascending order. So my first ordered pair would be (3, pi over 10). Then my second ordered
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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
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would be (3, 9 pi over 10). Now let's find the values of theta where the graph crosses through the pole.
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To do this we're going to take our equation and replace r with 0. So we're finding the values of theta
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where the r-value is 0. To do this we're going to solve this trig equation, 3 sine of 5 theta equals 0.
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Dividing both sides by 3, I get sine of 5 theta equals 0. Now again I need to undo the sine function and
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set up the general formula that's equal to 5 theta. Here you need to think about the fact that there are
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two places where sine equals 0, and they are separated by pi. Sine equals 0 at 0 and pi, so we can write this
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using just a single general formula. It would be 5 theta equals 0 plus pi k. That takes care of 0, and then pi,
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2 pi, 3 pi, etcetera. Now finish solving for theta by dividing both sides by 5. Zero divided by 5 is still 0, so
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I could write this general formula as just pi over 5 times k. Now again, I would begin substituting in values
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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
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would just give me pi over 5. Substituting in 2, would give me 2 pi over 5. Substituting in 3, would give me
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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
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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.