Convert Equations Between Polar and Rectangular Forms - Online Tutor, Practice Problems & Exam Prep

To convert equations from rectangular to polar form, replace \(x\) and \(y\) with \(r \cos \theta\) and \(r \sin \theta\) respectively. For example, \(y = 5\) becomes \(r \sin \theta = 5\), leading to \(r = \frac{5}{\sin \theta}\). Recognize that \(x^2 + y^2 = r^2\) simplifies conversions. Conversely, to convert polar to rectangular, manipulate equations to include \(r \cos \theta\), \(r \sin \theta\), or \(r^2\). Completing the square may be necessary for standard forms, such as transforming \(r = 6 \sin \theta\) into a circle equation.

Convert Equations from Rectangular to Polar

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 x's and y's and convert them into their polar form containing r's and thetas. Now this isn't quite as complicated as it seems because we already know that x and y are equal to rcosθ and rsinθ. So, all we have to do is replace our x's and y's using those formulas and then do some algebra. 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=5, and we, of course, want to convert this equation into its polar form. Earlier, I mentioned that x=rcosθ and y=rsinθ. In this equation, I can simply replace my y value with rsinθ. Doing that, I end up with rsinθ=5. Now that I've replaced the y, all I have to do here is solve for r. Solving for r in this equation, I can just divide both sides by the sine of theta, giving me r=5sinθ. Recognizing that one over the sine of theta is simply the cosecant of theta, I end up with my final equation as r=5cscθ, and this is my equation in its polar form.

Now let's look at our second example. We have y=x+1. Here I have an x and a y, so I need to replace both using my formulas. Doing that, I end up with rsinθ=rcosθ+1. From here, we want to solve for r. Since I have r in two different places, I get those on the same side by subtracting rcosθ from both sides. This gives r(sinθ-cosθ)=1. To fully isolate r, I can divide both sides by sinθ-cosθ, giving r=1sinθ-cosθ. Now I have my final equation in its polar form.

Lastly, Let's look at one final example where x2+y2=25. Since I have an x and a y, I could replace them using rcosθ and rsinθ, but there's actually an easier way to do this because we also know that x squared plus y squared is equal to r squared. So, I can simply replace x2+y2 with r2, giving me r2=25. Solving for r, I take the square root of both sides, leaving me with r=5. Recognizing that my original equation in its rectangular form was a circle of radius 5, if I end up with r equals 5, 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 rcosθ and y is equal to rsinθ, but also remember that x squared plus y squared is equal to r squared. 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.

Convert Equations from Rectangular to Polar Example 1

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 x, y, or x squared plus y squared with its polar equivalent. So let's go ahead and jump into example a here, y = x^2. Now I know that y is our \sin(\theta), x is our \cos(\theta), so I'm just going to plug those values in where x and y are, respectively. So I have r \sin(\theta), and that's equal to r \cos(\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 theta. So I end up with r \sin(\theta) = r^2 \cos^2(\theta). Now from here, we want to solve for r. Now on one side, I have r, the other I have r^2, so I can cancel one of those rs on either side. Then I have \sin(\theta) = r \cos^2(\theta). Then if I divide by the \cos^2(\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 = \frac{\sin(\theta)}{\cos^2(\theta)}, 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^2 here rather than r. So let's keep that in mind as we solve this problem. Here we have 4xy = 2. Again, I have an x and a y that I'm going to replace using our \cos(\theta) and r \sin(\theta). So this ends up giving me 4 times r \cos(\theta) \times r \sin(\theta) = 2. Now from here, we want to solve for r. Now I actually can go ahead and collapse some of this stuff because I have 2 rs on this side, which means that I'm dealing with an r^2. So this is really 4 r^2 \cos(\theta) \sin(\theta) = 2. Now from here, I'm going to go ahead and divide both sides by 2 since that 4 and 2 are both divisible by 2. So on that side, I'm left with a 1 on the right, and then I'm left with 2 r^2 \cos(\theta) \times \sin(\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 equations 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 \cos(\theta) \sin(\theta), which is part of a trig identity. And also I have this too. So if I squish those 2 things together, put that r^2 on the outside, this ends up giving me r^2 \times 2 \cos(\theta) \sin(\theta), which using our double-angle trig identity is really just the \sin(2\theta). So I can kind of condense that a little bit using that trig identity. So I have r^2 \times \sin(2\theta) = 1. Now I want to go ahead and solve for r^2. Remember, that's what we were asked to do in this specific problem. So I'm going to divide both sides by the \sin(2\theta). Now that will cancel on the left and I am left with r^2 = \frac{1}{\sin(2\theta)}. But we can simplify this even further because I know that one over the sine is just the cosecant. So this is really r^2 = \csc(2\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.

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

Convert each equation to its polar form.

$3y-5x=2$

• A. $r=\frac{2}{3\sin\theta-5\cos\theta}$
• B. $r=\frac{2}{\sin\theta-\cos\theta}$
• C. $r=3\sin\theta-5\cos\theta$
• D. $r=\frac35\sin\theta$

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. $r=2\cos\theta$

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

Convert Equations from Polar to Rectangular

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 \( x \), \( y \), and \( x^2 + y^2 \) and replace them with \( r \cos(\theta) \), \( r \sin(\theta) \), and \( r^2 \). 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, \( r = 4 \), 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 \( r \cos(\theta) \), \( r \sin(\theta) \), or \( r^2 \). So here we have some strategies to help us get there. Now our original equation \( r = 4 \) 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 \( r^2 = 16 \). Now I have one of those terms, \( r^2 \). So I can move on to my second step and replace that \( r^2 \) with \( x^2 + y^2 \) that I know that it is. So \( x^2 + y^2 = 16 \).

Now from here, my equation contains \( x \)'s and \( y \)'s, no \( r \)'s, no \( \theta \)'s. 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, \( x^2 + y^2 = 16 \), I know that this is the equation of a circle. So this equation is already in its sort of standard or 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, \( r = \sec(\theta) \). Now from here again, we want our equation to contain \( r \cos(\theta) \), \( r \sin(\theta) \), or \( r^2 \), of which it contains none. So what do we do here? Well, another one of our strategies is to go ahead and rewrite any trigonometric functions in terms of sine and cosine. Now here I have the secant of theta, which I can rewrite as \( 1 / \cos(\theta) \), giving me \( r = 1 / \cos(\theta) \). 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 want to go ahead and eliminate that fraction by multiplying both sides by my denominator. Now here, my denominator is the cosine of theta. So multiplying both sides by that cosine, that will cancel on the left, leaving me with \( r \cos(\theta) = 1 \). Now from here, I do have \( r \cos(\theta) \). So I can go ahead and replace that \( r \cos(\theta) \) with \( x \). Now in doing that, I get \( x = 1 \), which I immediately recognize as being the equation of a basic line. And since this is already in 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 \( r = 6 \sin(\theta) \). Now starting from the beginning, step 1, we want this equation to contain one of these terms, and it currently doesn't have any of them. Now I do have a \( \sin(\theta) \), but not an \( r \sin(\theta) \). 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 \( r^2 \) on that left side is equal to \( 6 r \sin(\theta) \). Now I actually end up with 2 of those terms, \( r^2 \) and \( r \sin(\theta) \). So I can go ahead and replace \( r^2 \) and \( r \sin(\theta) \) with \( x^2 + y^2 \) and \( y \). Now in doing that, I end up getting \( x^2 + y^2 = 6y \). And now my equation again only contains \( x \)'s and \( y \)'s, no \( r \)'s and \( \theta \)'s. 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 \( x^2 + y^2 = \) 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 \( x^2 + y^2 = 6 \), 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 \( 6y \) over. So this gives me \( x^2 + y^2 - 6y = 0 \). 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 \( y^2 - 6y \). 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 \( y^2 - 6y \), so I want this to become the perfect square \( (y - 3)^2 \) because of that value negative 6. So how can I get that perfect square? Well, if I add 9, that does become my perfect square, \( (y - 3)^2 \). But if I add 9 on that left side, I also need to add 9 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 \( x^2 + (y - 3)^2 = 9 \). 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've 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.

Convert Equations from Polar to Rectangular Example 2

Hey everyone. In this problem, we're asked to convert the equation into its rectangular form and then identify the shape of its graph. Now the equation that we're given here is r equals 2 over 1 plus the cosine of theta. 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 going to go ahead and eliminate that fraction by multiplying both sides by my denominator. In this case, 1 plus the cosine of theta. 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 2.

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, an r sine theta, or an r cosine theta. This is a sort of unique instance in which we actually want to keep that r as is because we already have r 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 2 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 2. So I replace everything that I need to and I can go ahead and move on to step number 3 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. 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 much easier. Then I have the square root of x squared plus y squared, and that's equal to 2 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 (2 - x) squared on the right. I'm going to go ahead and multiply that out, which will give me 4 - 4x + 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 4 - 4x. 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 y² = 4 - 4x. This is a horizontal parabola. Let me know if you have any questions here. Thanks for watching.

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

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. $x^2-1=y$