Graphing Other Common Polar Equations - Video Tutorials & Practice Problems

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1

concept

Introduction to Common Polar Equations

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Hey, everyone, when working with polar equations, there are four shapes that you'll encounter rather frequently. Now you're going to have to be able to recognize the equations for each of these in order to classify them and then also be able to graph them. Now, here, I'm going to break down for you the equations and the appearance of each of these commonly occurring shapes. And we'll work through classifying some equations together, laying the foundation to eventually be able to graph those equations as well. So let's go ahead and dive right in. Now looking at the equations for all of these shapes, they all look right, similar, they all contain an R A theta and a trig function. But let's actually dive into their differences here, looking at our first two shapes. Here, we have the cardioid named for its sort of resemblance to a heart shape and the limos. Now looking at the equations for these two shapes, you might notice that they're actually the exact same, they have identical equations. They're going to be of the form R equals A plus or minus B cosine theta or R equals A plus or minus B sine theta. Now, the difference here is going to be with our A and B values. Now, for the cardioid A and B must both be greater than zero. But here A is actually going to be equal to B, they're going to be the exact same value in order to get this cardioid shape. Now, for the limos oe A and B are still going to be greater than zero, but A can either be greater than B or less than B which affects the shape of our graph. If A is greater than B, then the graph of our limos is just going to have a dimple at the top rather than the inner loop that C in our graph here. But if A is less than B, that's when we get the inner loop that appears on our graph. Now the equations for cardioid and Limos Songs are the only equations that are going to contain addition or subtraction. Now let's move on to our third shape here. The rose that looks like a sort of flower with a particular number of petals. Now, the equation of a rose is going to be of the form R equals a times the cosine of N theta or R equals A times the sine of N theta. And for this equation, A can not be equal to zero and N is simply an integer that is greater than, or equal to two, which will actually end up telling us the number of petals that our rose has. Now let's move on to our final shape here, Alumni Gate, which looks like a sort of infinity symbol. Now, looking at our alumni gate here, the equation of Alumni gate is going to be of the form R squared equals plus or minus a squared cosine of two, the or R squared equals plus or minus A squared sine of two. The now here again, a cannot be equal to zero. And the equation of a lemna gate is actually going to be the only one that contains an R squared, making it really easy to recognize these equations when they're going to have that lemniscate shape. Now that we've seen all of these equations for these commonly occurring shapes, let's go ahead and work through classifying some of them. Now looking at our first example here, we're given the equation R equals one plus the cosine of theta. Now here, since I have addition, I know that this equation is either that of a cardioid or a limo. Now, let's look at those A and B values. So here I have a one plus cosine of the, which has this sort of invisible one in front of it, telling me that A and B are both equal to one. Now here, since A is equal to B, that means that this equation must be that of a cardioid. Now, let's move on to our next equation here. Here we have R equals four times the sine of two theta. Now, the first thing that I notice here is that I have this two theta in my argument, which is the same as that of a Lenis gate. But the other thing that I notice here is that R is just R, it's not R squared, meaning that this cannot be the equation of a limit gate. So it has to be that of a rose of the form R equals a sine of N theta where N is equal to two. So this is the equation of a rose. Now being able to classify these equations will allow us to immediately tell what the shape of the graph of that equation is going to be. So let's go ahead and keep that in mind as we continue to practice. Thanks for watching and I'll see you in the next one.

2

Problem

Problem

Identify whether the given equation is that of a cardioid, limaçon, rose, or lemniscate.

$r=4\sin2\theta$

A

Cardioid

B

Limacon

C

Rose

D

Lemniscate

3

Problem

Problem

Identify whether the given equation is that of a cardioid, limaçon, rose, or lemniscate.

$r=3+2\cos\theta$

A

Cardioid

B

Limacon

C

Rose

D

Lemniscate

4

Problem

Problem

Identify whether the given equation is that of a cardioid, limaçon, rose, or lemniscate.

$r=-25\cos2\theta$

A

Cardioid

B

Limacon

C

Rose

D

Lemniscate

5

Problem

Problem

Identify whether the given equation is that of a cardioid, limaçon, rose, or lemniscate.

$r=1-\sin\theta$

A

Cardioid

B

Limacon

C

Rose

D

Lemniscate

6

concept

Cardioids

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Hey, everyone, we're now familiar with the commonly occurring shapes that we encounter when graphing different polar equations. And now it's time to actually graph those shapes starting here with the cardioid. Now, the general shape of the graph of our cardioid will remain the same with the orientation and positioning of it just changing based on the equation that we're actually graphing. Now remember that the equation of a cardioid will always be of the form R equals A plus or minus B cosine theta or R equals A plus or minus B sine. The where A is specifically equal to B. Now, in order to actually graph a cardioid, we have only two things to determine. First, we want to determine that the symmetry and then we simply want to plot points at our quadrant angles. That's all. So let's go ahead and not waste any time here and just jump right into graphing this cardioid. Now, here we're tasked with graphing the equation R equals one plus the cosine of theta. And I can tell that this is the graph of a cardioid because I have addition in this equation and this a value of one is equal to my B value of one because I have this sort of invisible one multiplying this cosine of data. So let's go ahead and jump into our first step here in graphing this cardioid which is to determine at the symmetry. Now, in order to determine the symmetry of a cardioid, all we need to do is check whether our equation contains cosine sine. Since our equation here is a one plus the cosine of theta that tells me that my graph will be symmetric about the polar axis, which is something that I should keep in mind as I can continue into step two where we actually want to find and plot points at our quadrant angle. Specifically remember that our quadrant angles are zero pi over two pi and three pi over two. So let's start here by plugging in theta equals zero in order to find our R value for this angle. Now doing this, I get one plus the cosine of zero and the cosine of zero is one. So this ends up being two. So I can plot my first point at 20, which will end up being a right here on my graph. Now for my second quadrant angle pi over two plugging that into my equation one plus the cosine of pi over two, the cosine of pi over two is zero. So this R value is one and I can plot my second point at one of pi over two which will end up being right here. Then for my third quadrant angle pi plugging in pi to my equation one plus the cosine of pi, the cosine of pi is negative one. So this ends up being zero. Now I can plot this point at zero pi which will just end up being a right at that pole because my R value is zero. Then for my final quadrant angle three pi over two, I want to think back to my symmetry here because I know that my graph is symmetric about this pole axis. I can just take this point and reflect it right over that polar axis without having to calculate that value. So that final point is going to be located at 13 pi over two. Now we have all of these points and we can move on to our final step here which is going to be to connect everything with a smooth and continuous curve. Now, we already know what the shape of our graph is. We know the general shape of a cardio. And here we just want to apply that shape and connect all of these points here. Remember that a cardioid has sort of a bump at the top and then extends out. So our graph is going to end up looking like this. Now remember that you can always calculate some more points if you want to get a bit more precise. But here we have fully graphed this cardioid. Now that we know how to graph cardioid, let's continue getting practice with this. Thanks for watching and I'll see you in the next one.

7

example

Cardioids Example 1

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In this problem, we're asked to graph R equals two plus two sine theta. Now I know that this is the graph of a cardioid because I have addition happening and both by A and B values are equal to each other. So I already know the general shape of this graph and it remains to just determine the orientation and sort of the exact positioning of this on my graph. So let's go ahead and get started with step one first looking at symmetry. Now, since our equation contains a sine function that tells me that my graph is going to be symmetric about the line theta equals pi over two, which we can keep in mind while graphing our points here. Now we want to go ahead and find and plot some points and we're specifically going to use our quadri anal angles starting of course with zero. So plugging in zero for theta, I get two plus two sign zero, which the sign of zero is just zero. So this gives me two. So I can plot that first point at 20, which will end up being a right here. Now pi over two plugging that in for I have two plus two times the sine of pi over two. Now, the sine of pi over two is one. So this ends up being two plus two or four. So I can plot my second point at four pi over two. Now that will end up being a right here at the top of that axis. Now, I want to take a moment to remember my symmetry here because since I'm symmetric about this line, theta equals pi over two, I can go ahead and just reflect this point over onto my other quadrant angle. So I have this other at two pi, then I have just one final point to find at three pi over two. So two plus 2 s3 pi over two, the sine of three pi over two is negative one. So this ends up being two minus two, which is just zero. So here, my final point is actually going to be located at my pole. Now, from here, we want to connect this with a smooth continuous curve. Now, we remember the shape of a cardioid, but remember that the orientation of that can change your furthest point from your will always be the sort of bottom of your cardioid. That's very round. Then right here, we're going to have the sort of dimple heart shape that happens with our cardioid and you want to make sure and around this around. Now it's OK if you don't get super precise and you can always choose to plot some more points here if you want to get a bit more exact or if your professor asks you to get more exact. But this is the general shape and positioning of my cardioid for R equals two plus two science. The thanks for watching and let me know if you have questions.

8

Problem

Problem

Graph $r=2-2\cos\theta$

A

B

C

D

9

concept

Limacons

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Hey, everyone, we can graph A limos just like we graphed a cardioid by determining the symmetry of our graph and finding and plotting points at our quadrant angles. We're just first going to take one additional step and determine whether our limos has a dimple or an inner loop, which we can do easily based on our A and B values given in our equation. So let's jump right into graphing our limos here. Remember that the equation of a limos will always be of the form equal to A plus or minus B cosine theta or A plus or minus B sine theta just like a cardioid except here A is either greater than B or less than B. Now here, the equation that we're tasked with graphing is R equals three minus two sine theta. Now, I know that this is going to be the graph of a limos because I have subtraction happening here and I have an A value of three and A B value of two. So A is greater than B. Now, let's look at our first step here and we just said that A is greater than B. Now, this tells us that our limos will have a dimple rather than an inner loop. So that tells me that my graph will be shaped something like this. And it's just up to us to get more precise with these other steps. Now, in determining whether our graph has a dimple or inner loop, we also learn some more information about our graph whether or not it has a zero. Since here, our graph only has a dimple, it will not have a zero, meaning that our graph will not pass through the. So let's continue on with step two and determine the symmetry of our graph. Now here our equation contains a sine function. So that tells us that our graph will be symmetric about the line theta equals pi over two, which again is something that I want to keep in mind as I move on to my next step where we're actually going to find and plot points at our quadrant angles. Now here at our quadrant angles, I know that I'm going to plug in these values of theta into my equation. So he for that first value of theta zero, I'm going to take three minus two s of zero and the sign of zero is simply zero. So this will end up giving me a value of three. So I can plot that first point at 30, which will be right here then plugging in pi over two to my equation three minus two times the sign of pi over two will give me a value of one. So I can plot this second point at one pi over two right here. Then looking at this next angle pi since I know that my graph is symmetric about this line, theta equals pi over two, I can go ahead and just reflect this point over that line theta equals pi over two in order to get this third point right here at three pi. Now, for my final point plugging in three pi over two, this gives me three minus two times the sign of three pi over two, giving me a value of five. So this final point here is going to be located at 53 pi over two, which will be right here. Now, all that's left to do here is connect all of these points with a smooth and continuous curve and we know the general shape of the graph of a limos. And we know of, of course that this limos has a dimple. So in connecting these points, I want to make sure that I reflect those. Now here, I'm going to go ahead and connect these points with a dimple at the top of my graph here and my graph of this limos will end up looking something like this. Now remember if you're asked to get more precise or if you just want to get more precise, you can always plot more points and that's totally fine. But now that we have fully graphed this limos. Let's continue getting some practice. Thanks for watching and I'll see you in the next one.

10

example

Limacons Example 2

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Hey, everyone in this problem, we're asked to graph the equation R equals one plus two cosine of the. Now I know that this is a graph of a limos because I have addition happening and my A and B values are not equal to each other. Therefore, meaning that this cannot be a cardioid. Now, in keeping in mind that this is the graph of a limos. Let's go ahead and dive into our steps here. Now, for our first step, we want to look at our A and B values I have an A value of one A B value of two seeing that A is less than B that tells me that my limos will have an inner loop rather than a dimple. Now, if our limos has an inner loop, it also means that it will have a zero, which we can go ahead and plot on our graph here right at our pole. Now we can move on to step number two and determine our symmetry. Now, because our equation contains a cosine function that tells us that our graph will be symmetric about the polar axis. Now, this is something we want to keep in mind as we proceed to step three and actually start plotting points. Now we want to plot our points using our quadrant angles. So doing that here, I'm gonna go ahead and plug in zero for theta in my original equation, which is one plus two times the cosine theta in this case zero. Now the cosine of zero is one. So this ends up being a one plus two or three. So I'm going to plot my first point at 30, which is right here on that polar axis. Now plugging in pi over two, I get one plus two times the cosine of pi over two. Now the cosine of pi over two is zero. So this ends up just being one and I can go ahead and plot that point at one pi over two, which will be right here. Now, I remember my symmetry here means that I can just reflect this point over the polar axis giving me another point at my quadrant angle of three pi over two with an R value of one. Now we just need to plug in that pi to get one final point here to get the complete picture of our graph. Now when I plug in pi here I get one plus two times the cosine of pi, the cosine of pi is negative one. So this will end up giving me one minus two, which will give me negative one. Now remember when we have a negative R value. That means instead of counting out towards our angle, we want to count in the opposite direction. So that ends up giving me a point right here again, on that polar axis. Now, from here, we want to connect this with a smooth and continuous curve. Now, remembering back in step one, we said that this limos has an inner loop. So what exactly does that look like? Well, it looks almost identical to the other limos, but now there's an extra little circle or loop in the middle. Now, this point shows us what our inner loop is going to be the point closest to the pole. So I have that inner loop right there. And then my outer loop kind of the rounded bottom of my Lemos is going to be out here at the furthest point away from the pole still having a curved line rounding around to that bottom point. So this is the graph of my limo for the equation R equals one plus two times the cosine of data. Thanks for watching and let me know if you have questions.

11

Problem

Problem

Graph $r=1+2\sin\theta$

A

B

C

D

12

concept

Roses

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3m

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Hey, everyone, we know that the general shape of the graph of a rose has these sort of petals. So it makes sense that in order to actually graph a rose, we need to be able to figure out how many petals our rose has and where exactly those petals are located. Now, we can do this easily by just finding one single petal and then determining the spacing of our pedals, which we can do based on the equation that we're given. Now, this is exactly what we're going to do here. So let's go ahead and jump right in. Now remember that the equation of a rose will always be of the form R equals a times the cosine of N theta or a times the sine of N theta where A is not equal to zero and N is an integer that is greater than or equal to two. Now, the, the specific equation that we're tasked with graphing here is R equals four times the cosine of two theta, which I know is the graph of a rose because it's of this form R equals a times the cosine of N theta. Now let's go ahead and get started with our very first step where we're going to look at our value of N. Now, here N is equal to two and two is an even number. Now, whenever N is even, that tells us that we're going to have two N pedals. So if I take two here and multiply it by value of N, which is also two that tells me that this rose is going to have four petals. Now, with this in mind, let's move on to step two and figure out where that first petal is. Now in order to figure out where our first petal is, we need to look at our value of a. Now in my equation here, I see that A is equal to four. So that tells me that my R value for my pedal is going to be four and all of my petals will be at the same length. So this will actually be the R value for every petal. Now, we need to figure out what theta is for this first petal, which is to based on whether our equation contains cosine or sine. Now here our equation contains a cosine. So that tells us that the here is going to be equal to zero. So I can go ahead and plot that very first pedal at 40, which will end up being right here. Now, we can move on to step number three because now that we have our first pedal we can determine where our other three petals are. Now that very first pedal we know is located at 40. And we also know that all of those petals will be the exact same length. So I can go ahead and fill in four for all of those are values. Now, I just need to figure out how these petals are spaced, which I can do based on the number of petals that we found in step one. Now, we found that our rows will have four petals. So here, if I take two pi and divide it by four, that's how my petals will be spaced. Now, this can simplify down to pi over two. So here if I take this zero of my first petal and add pi over two, that tells me that my second pedal, I can go ahead and plot at four pi over two, which is going to be right here. Then for my next pedal, I'm gonna add another pie over two to get me at pie. So I can plot my third pedal here at four pie which will end up being a right out here. Then for my final pedal here adding another pie over two. For that spacing. I end up at three pie over two and I can plot that very last pedal at 43 pie over two. Now I have the positioning of all four of my petals. So all that's left to do is connect them with a smooth and continuous curve. Now, I know that all of these petals will stem out from that pole because they're all coming kind of out of there to look like a flower. So my graph is going to end up looking something like this. Now again, remember with all of these graphs, if you are asked to get more precise or if you want to get more precise, you can always calculate some more points and that's totally fine. But now that we have fully graphed this rose, let's continue practicing graphing roses together. Thanks for watching and I'll see you in the next one.

13

example

Roses Example 3

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Everyone in this problem, we're asked to graph the equation R equals two times the sine of three theta. Now this is the equation of a rows of the form R equals a sine and theta. So let's go ahead and jump right into our steps here. Now, for our first step, we're going to be looking at our value of N which in this case is three. Remember N is a number that comes before theta in your argument. Now three is an odd number. So that tells us we're going to have N petals in this case N is three. So we're going to have three petals. Now, moving on to step two, we want to actually plot our first pedal. We know that we'll have three of them when we end up with our final graph. But we want to start by plotting that first one. Now, I'm plotting our first pedal. We need to look at our value of A, which looking back at my equation I know is two A is the number that's at the front of your rows equation. So I have an A value of two and I want to figure out where theta is going to be. Now, because I have a sine function of my equation theta is going to be equal to pi over two times N. So we have to do a little calculation here taking pi and dividing it by two times N which we know is three from step one. Now, this leaves me with an angle of pi over six. So my first pedal is going to be graph at two pi over six. So coming up to my angle pi over six and going out to two, I know that my first pedal will be right here. Now, we can find our other petals by figuring out how far they are spaced out from each other and plotting all of them. Now, I know that my first pedal is located at two pi over six, but I know that I have two other petals. So I need to figure out where those are. Now in order to figure out how far your petals are spaced apart, you're going to take two pi divided by the number of your petals, which in this case is three. So these petals will be separated by two pi over three radiance. So in order to determine where our next petal is, I need to take pi over six and add two pi over three. Now, when I do that, I'm going to end up getting five pi over six. Now, all of my pedals are the exact same link. So I can go ahead and fill in all of these R values as that a value of two. So I can plot my second pedal at 25 pi over six, which will end up being right out here on my graph. Then I have one other pedal to account for. So I'm going to add another two pi over three to that five pi over six. And if you do this calculation, you're going to end up getting three pi over two. So I can plot my last pedal at 23 pi over two, which will be right here. Now, we want to connect this with smooth continuous curve. Now remember that for roses, you're always going to go through the center, that's where your petals are kind of stemming out from. So we can go ahead and connect these with smooth continuous curves that look like petals because this is a rose and we're going to try our best to make these curves smooth. But if you don't get them exactly perfect. That's OK. This is the general shape of our graph for the rows R equals two times the s of three theta. Let me know if you have any questions.

14

Problem

Problem

Graph $r=3\cos4\theta$

A

B

C

D

15

concept

Lemniscates

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2m

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Hey, everyone, we know that the general shape of a Leni gate is this sort of infinity symbol or propeller shape with two petals. So in order to graph a le gate, if we just figured out where those two petals were, we'd be good to go. Now, we can actually do this rather easily by just figuring out where our first petal is the same way we did for roses and then simply reflecting that petal over the pole. So let's not waste any time here and jump right into graphing our leda gate. Now remember that the equation of a lambda gate is always going to be at the form R squared is equal to plus or minus a squared times the cosine of two theta or a squared times the sine of two theta. Now here, the equation that we're specifically tasked with graphing is R squared is equal to four times the sine of two theta. Now it's easy to tell that this is the equation of alumni gate because it's the only polar equation that we're working with here that has a square shared value of R. So let's go ahead and jump into step number one. And figure out where that very first pedal is. Now, the first thing that we want to do here is look at our value of a. Now, looking at my equation here, I have R squared is equal to four times the sine of two theta. And something that's really important to remember here is that this four is not a, it's actually a squared. So in order to get a, we need to take the square root of that four in order to get two. So that tells us that our R value for that first pedal is going to be two. Now, in order to determine theta, we need to look at the sine of our equation and also whether we have a cosine or sine function. Now, here we have positive four sine of two theta. So that tells us we're dealing with a positive A squared sine theta, which means that theta will be equal to pi over four for this first pedal. Now I can go and graph this first pedal at two pi over four, which will end up being right here. Now, all that's left to do is reflect this pedal over the pole. Now, remember my pole is what we think of as our origin and rectangular coordinates. So if I reflect this point over that pole, I'm going to end up down here for that second pedal. Now, all we have to do is connect these with a smooth and continuous curve. Remembering that the shape of our lemniscate is a sort of infinity simple. So let's go ahead and connect our points here. Now, coming out from the poll, I'm going to go out to that first point and then out to that second point. And here I have the graph of my lemna skate. Now as for all of these graphs, remember that if you're asked to get more precise or if you just want to get more precise, you can always calculate more points by plugging in values of theta to your equation. But now that we know how to graph a lemniscate. Let's continue practicing together. Thanks for watching and I'll see you in the next one.

16

example

Lemniscates Example 4

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1m

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Hey, everyone in this problem, we're asked to graph the equation R squared equals 25 times the cosine of two theta. Now here, this is the equation for a Lemna scape because I have this R squared, which I know is only something that happens when dealing with equations of lemna gates. Now lemna gates are arguably our easiest shape that we know how to graph. So let's dive right in here. Now, looking at our first step, we want to plot our first pedal here which is going to located at A for R value. And then we need to determine what theta is based on the trig function that we're dealing with. Now remember that in our equation, this value at the beginning is actually a squared. So we need to make sure that we take the square root of that in order to get a, now the square root of 25 is equal to five. So this gives me my value of A, that I will be plotting for that first pedal. Now, we need to determine what theta is. Now here have 25 times the cosine of two theta. So I'm dealing with a positive A squared and a cosine. So that tells me that theta will be located at zero. So I can plot my very first pedal at 50, which will end up being right here on my polar axis. Now, all that's left to do is reflect this pedal over the pole. So doing that here, reflecting this over to the other side. Landing me right here. I have my second pedal and I can move on to my final step and just connect these with a smooth and continuous curve. Now, we know that the general shape of our Lemus gate is a sort of infinity symbol or a propeller or whatever you want to think about this as. But remember that we're always going to go to that hole in there with these sort of petals or propellers or whatever you wanna call these. Now, we have graphed this Lemus gate at R squared equals 25 times the cosine of two. The, let me know if you have any questions.