Convert Equations Between Polar and Rectangular Forms

9. Polar Equations

Convert Equations Between Polar and Rectangular Forms - Video Tutorials & Practice Problems

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1

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Convert Equations from Rectangular to Polar

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Hey, everyone, you now know how to convert points from rectangular to polar. But you'll actually be asked to do the same thing for equations, take equations in their rectangular form containing XS and Ys and convert them into their polar form containing RS and theta. Now this isn't quite as complicated as it seems because we already know that X and Y are equal to R cosine theta and R sine theta. So all we have to do is replace our XS and Ys using those formulas and then do some algebra. So with that in mind, we're going to work through some examples together here and soon you won't just be an expert at converting points from rectangular to polar, but also equations. So let's go ahead and dive right in. Now, looking at our first example here, we have the equation Y equals five and we of course want to convert this equation into its polar form. Now earlier, I mentioned that X is equal to R cosine theta and Y is equal to R sine theta as we know. So in this equation, I can simply replace my Y value with R sine theta. Now doing that here, I end up with R sine theta is equal to five. Now that I've replaced that Y all I have to do here is solve for R. Now solving for R in this equation, I can just divide both sides by the sine of the allowing that to cancel out on that left side, leaving me with just R is equal to five over the sine of data. Now I can simple this further here recognizing that one over the sin of data is simply the cos of data. So I end up with my final equation as R equals five times the cosecant of data. And this is my equation in its polar form. Now let's look at our second example here we have Y is equal to X plus one. Now here I have an X and A Y so I need to replace both of them using my formulas here. So doing that, I end up with our sign data is equal to R cosine theta plus one. Now, from here, we want to solve for R. Now, since I have R in two different places here, I want to go ahead and get those on the same side which I can do by subtracting our cosine data from both sides. Now, in doing that, I will cancel on my right side, leaving me on the left with our sine, the minus R cosine data is equal to that one. Now I can go ahead and factor that R out since that's what I'm solving for and I have R times the sine of the minus the cosine data is still equal to one on that right side. Now, from here to fully isolate R I can divide both sides by sine theta minus cosine theta. Doing that. Of course, on that right side as well, sine theta minus cosine data canceling on the left side leaving me with my final equation here R is equal to one over the sine of data minus the cosine data. Now I have my final equation here in its polar form. Now let's look at one final example here. Here we have X squared plus Y squared is equal to 25. Now, since I have an X and A Y, I could go ahead and replace X and Y using R cosine data and R sine data. But there's actually an easier way to do this because we also know that X squared plus Y squared is simply equal to R squared. So I can simply replace X squared plus Y squared here with R squared giving me R squared is equal to 25. Now, solving for R here, I can just take the square root of both sides leaving me with R is equal to five. Now, this makes sense here because I recognize my original equation in its rectangular form was a circle of radius five. So if I end up with R equals five, that would be the same thing in polar form. So when converting equations from their rectangular form to their polar form, remember that X is equal to R cosine theta and Y is equal to R sine theta. But also remember that X squared plus Y squared is equal to R squared. Now, let's keep all of this in mind as we continue to practice. Thanks for watching and I'll see you in the next one.

2

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Convert Equations from Rectangular to Polar Example 1

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Hey, everyone in this problem, we're asked to convert each equation into its polar form. And remember that we can do that by simply replacing XY or X squared plus Y squared with its polar equivalent. So let's go ahead and jump into example a here Y equals X squared. Now I know that Y is our sine theta, X is our cosine theta. So I'm just going to plug those values in where X and Y are respect. So I have our sine theta and that's equal to our cosine theta and we're going to make sure that we're squaring that entire term. Now, when I do square that entire term, that means that my R will be squared as well as that cosine of the. So I end up with R sine theta is equal to R squared cosine squared. The now from here we want to solve for R. Now, on one side, I have R the other, I have R squared. So I can cancel one of those RS on either side, then I have sine theta is equal to R times the cosine squared of theta, then if I do divide by the cosine squared of theta on both sides, I have successfully isolated R. I'm going to go ahead and switch what side it's on because I want R to be on that left side so that R is equal to the sine of the over the cosine squared of data. And we are done here. So let's move on to our second example here. Now, in our problem statement, we're specifically asked to solve for R squared here rather than R. So let's keep that in mind as we solve this problem here, we have four XY is equal to two. Again, I have an X and A Y that I'm going to replace using R cosine theta and R sine theta. So this ends up giving me four times R cosine theta, times R sine theta is equal to two. Now, from here, we want to solve for R. Now, I actually can go ahead and collapse some of this stuff because I have two R's on this side, which means that I'm dealing with an R squared. So this is really four R squared cosine theta sine theta and that's equal to two. Now, from here, I'm gonna go ahead and divide both sides by two since that four and two are both divisible by two. So on that side, I'm left with a one on the right and then I'm left with two R squared cosine theta times sine theta. Now, something that we do want to keep in mind whenever we're converting to polar form is that sometimes we're going to come across equation that end up giving us a trig identity and we need to be able to recognize those trig identities. So as we're simplifying, be sure that you're thinking about trig identities that might be familiar to you. Now, one thing that I see here is this cosine the sine theta which is part of a trig identity. And also I have this too. So if I squish those two things together, put that R squared on the outside, this ends up giving me R squared times two cosine theta s the which using our double angle trig identity is really just the sign of two theta. So I can kind of condense that a little bit using that trig identity. So I have R squared times sine of two theta is equal to one. Now, I want to go ahead and solve for R squared. Remember that's what we were asked to do in this specific problem. So I'm going to divide both sides by the sine of two theta. Now that will cancel on the left and I am left with R squared is equal to one over the sign of two. The, but we can simplify this even further because I know that one over the sign is just the cos so this is really R squared is equal to the cos of two theta. And we are done here having converted our original equation in rectangular form to its polar form. Let me know if you have any questions here and thanks for watching.

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Problem

Problem

Convert each equation to its polar form.

$y-x=6$

A

$6r=\sin\theta-\cos\theta$

B

$r=\frac{\sin\theta-\cos\theta}{6}$

C

$r=6\left(\sin\theta-\cos\theta\right)$

D

$r=\frac{6}{\sin\theta-\cos\theta}$

4

Problem

Problem

Convert each equation to its polar form.

$3y-5x=2$

A

$r=\frac{2}{2\sin\theta-5\cos\theta}$

B

$r=\frac{2}{\sin\theta-\cos\theta}$

C

$r=3\sin\theta-5\cos\theta$

D

$r=\frac35\sin\theta$

5

Problem

Problem

Convert each equation to its polar form.

$x^2+y^2=2y$

A

$r=\sqrt{2}$

B

$r=2\sin\theta$

C

$r=4\sin\theta$

D

$r=2\cos\theta$

6

Problem

Problem

Convert each equation to its polar form.

$x^2+(y-2)^2=4$

A

$r^2=4\sin\theta$

B

$r=4\sin\theta$

C

$r=2\sin\theta$

D

$r=\cos\theta-\sin\theta$

7

concept

Convert Equations from Polar to Rectangular

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6m

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Hey, everyone in converting both points and equations between their polar and rectangular forms, we have just one thing left to learn, taking an equation in its polar form containing R and theta and converting it into its rectangular form containing X and Y. Now this can seem a bit complicated at first specifically because there's not just one single operation that we can do in order to convert these equations every single time. But when we converted equations from their rectangular form into their polar form, we were able to take XY and X squared plus Y squared and replace them with our cosine theta R sine, theta and R squared. So here we're just going to do the opposite. The problem is that our equation won't always initially contain one of these terms. So we're going to have to manipulate our equation in order to get them. Now, here, I'm going to walk you through some strategies to help you manipulate your equation and get to that rectangular form that you want. So let's go ahead and get started. Now looking at our first equation here in its polar form, R equals four, we of course want to convert this equation into its rectangular form. And we also want to go ahead and identify the shape of its graph. Now, even though there isn't one single operation that we can do here, we can still go about converting this equation in a structured way. So let's take a look at our steps here. Now, in our first step, this is where we'll be manipulating our equation. We want it to contain our cosine theta R sine theta or R squared. So here we have some strategies to help us there. Now, our original equation R equals four does not contain any of these terms. So we need to look to these strategies. One of which is to square both sides. Now in squaring both sides of this equation, I end up getting R squared is equal to 16. Now I have one of those terms R squared. So I can move on to my second step and replace that R squared with X squared plus Y squared that I know that it is. So X squared plus Y squared is equal to 16. Now, from here, my equation contains Xs and Ys no, R's no theta. So I can go ahead and rewrite my equation in its standard form. So what does this mean? How do I know that I need to rewrite my equation or don't need to rewrite my equation? Well, it can be helpful at this point to stop and go ahead and try to identify the shape of your graph. Now, looking at this equation X squared plus Y squared equals 16, I know that this is the equation of a circle. So this equation is already in its sort of standard recognizable form of a circle. So I don't need to rewrite any further. Now, let's move on to our next example here. Here we have the equation R equals the C of theta. Now, from here again, we want our equation to contain our cosine theta R sine theta or R squared of which it contains none. So what do we do here? Well, another one of our strategies is to go ahead and rewrite any trick functions in terms of sine and cosine. Now here I have the CC of the, which I can read right as one over the cosine of data giving me R equals one over cosine data. Now, from here, I still don't have any of these terms. So what else can I do? Well, I have a fraction here and whenever I have a fraction, I wanna go ahead and eliminate that fraction by multiplying both sides by my denominator. Now here my denominator is the cosine of the. So multiplying both sides by that cosine that will cancel on the left leaving me with our cosine theta equals one. Now, from here, I do have R cosine data so I can go ahead and replace that R cosine data with X. Now, in doing that I get X equals one, which I immediately recognize as being the equation of a basic line. And since this is already in a recognizable standard form, I don't have to worry about rewriting any further. Now, let's move on to our final example. Here, we're given the equation R equals six sine data. Now, starting from the beginning, step one, we want this equation to contain one of these terms and it currently doesn't have any of them. Now, I do have a sine theta but not an R sine theta. And one of my strategies is actually to go ahead and multiply both sides by R. So let's see what happens when I do that in multiplying both sides by R, I get R squared on that left side is equal to six R sine theta. Now I actually end up with two of those terms R squared and R sine theta. So I can go ahead and replace R squared and R sine theta with Y and X squared plus Y squared. Now, in doing that, I end up getting X squared plus Y squared is equal to six Y. And now my equation again only contains Xs and Ys no R's and theta. So I can go ahead and go about rewriting this equation in its sort of standard form. Now, remember it can be helpful to go ahead and try and identify the shape of your graph at this point. But looking at this equation this is not in a very recognizable standard form in which I can identify the shape of my graph. So I need to think about how I can rewrite my equation. Now, in rewriting your equation, you can consider a couple of different things. For example, if your equation contained a square root, we could eliminate that square root by squaring both sides. Here, we don't have a square root. So we consider another strategy here. If our equation is in the form X squared plus Y squared equals some number times X or some number times Y, we want to go ahead and complete the square. Now, looking at my equation here, I have X squared plus Y squared equals six some number times Y. So I do want to go ahead and complete the square here. So let's go ahead and get set up to do that. Now, the first thing I'm going to do is put all of my terms on the same side by moving this six Y over. So this gives me X squared plus Y squared minus six Y equals zero. Now, we can proceed with completing the square. If you need a refresher on completing the square, feel free to go back to some of our previous videos. But let's continue here. I have this Y squared minus six Y. Now, I know that I can make this into a perfect square by adding some constant. So what perfect square do I want Well, I know that this is Y squared minus six Y. So I want this to become the perfect square Y minus three squared because of that value negative six. So how can I get that perfect square? Well, if I add nine, that does become my perfect square Y minus three squared. But if I add nine on that left side, I also need to add nine on the right side. So let's go ahead and simplify this further. Having completed the square. Now in rewriting this, I end up with X squared plus that perfect square Y minus three squared and all of that is equal to nine. Now, looking at this equation, if I think about what shape this equation would form, I know that this is the equation of a circle that's simply been shifted. So now this equation is in a standard recognizable form. And I'm done, I have converted this equation from its polar form into its rectangular form. Now that we know how to convert all equations and points between polar and rectangular forms. Let's continue practicing. Thanks for watching and let me know if you have questions.

8

example

Convert Equations from Polar to Rectangular Example 2

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3m

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Everyone in this problem, we're asked to convert the equation into its rectangular form and then identify the shape of this graph. And the equation that we're given here is R equals 2/1 plus the cosine of data. So let's jump right into our steps here and get started in taking this into its rectangular form. Now remember that we want to get our cosine theta R sine theta or R squared. So in order to do that here, since I have a fraction, I'm gonna go ahead and eliminate that fraction by multiplying both sides by my denominator. And this one plus the cosine of data. Now multiplying that on both sides, it will cancel on that right side. And I'm going to go ahead and distribute that R to give me R plus R cosine theta and all of this is equal to two. Now, from here, I see that I have this R cosine theta which is great, but I also have this R, which is not something that we typically want to have. Usually we want an R squared and R sine theta or an R cosine theta. Now, this is a sort of unique instance. And in which we actually want to keep that R as is because we already have our cosine theta. So anything else that we do to this equation would affect that? So instead we're going to go ahead and move on to step two and just replace what we have. But how are we going to replace that? R? Well, we know that X squared plus Y squared is equal to R squared. So what if I just take the square root on both sides? Then I know that R is equal to the square root of X squared plus Y squared. So here, that's what we're going to replace our R with the square root of X squared plus Y squared. Then I have that R cosine theta which I can replace with just X. So I have the square root of X squared plus Y squared plus X is equal to two. So I've replaced everything that I need to and I can go ahead and move on to step number three where we're going to rewrite its equation in its sort of standard form. Remember that there's not always just one variable that we're going to be solving for when taking equations from polar to rectangular form. And this can be sort of tricky. But let's go ahead and look at our strategies here. Now, if we see a square root, we should eliminate that square root, which we do have a square root, right. Here. But since I have this term here as well, I'm going to go ahead and bring that to the other side to make squaring a much easier. Then I have the square root of X squared plus Y squared and that's equal to two minus X. Now, from here, I can go ahead and square both sides which will allow me to cancel that radical out. Now, I'm just left with X squared plus Y squared on the left and two minus X squared on the right. I'm gonna go ahead and multiply that out which will give me four minus four X plus X squared. So where can we go from here? Well, I see that I have an X squared and an X squared on both sides. Letting me cancel that out. Leaving me with Y squared is equal to four minus four X. Now, this might not immediately look familiar because it's not a super common equation that you'll see. But this is actually the equation of a horizontal parabola. So we have taken it down to its sort of standard form where we can recognize what type of equation that it is seeing that we have. This Y squared is equal to four minus four X. This is a horizontal parabola. Let me know if you have any questions here. Thanks for watching.

9

Problem

Problem

Convert each equation to its rectangular form.

$r=-4\cos\theta$

A

$r=-4x$

B

$x^2+y^2=4$

C

$(x+2)^2+y^2=2$

D

$(x+2)^2+y^2=4$

10

Problem

Problem

Convert each equation to its rectangular form.

$r=\frac{2}{1-\sin\theta}$

A

$y^2=4-4x$

B

$x^2+y^2=2y$

C

$y=\frac14x^2-1$

D

$x^2-1=y$

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