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 \(r \sin \theta\). 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 the equation to include \(r \cos \theta\), \(r \sin \theta\), or \(r^2\). For instance, \(r = 4\) becomes \(x^2 + y^2 = 16\), representing a circle. Understanding these transformations enhances comprehension of polar and rectangular relationships.

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 respectively. So, all we have to do is replace our x's and y's 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=5, and we, of course, want to convert this equation into its polar form. Now earlier, I mentioned that x=rcosθ and y=rsinθ, as we know. So, in this equation, I can simply replace my y value with rsinθ. Now doing that here, I end up with rsinθ=5. 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 θ, allowing that to cancel out on the left side, leaving me with just r is equal to 5sinθ. Now I can simplify this further here, recognizing that sin-1θ is simply the cosecant of θ. So I end up with my final equation as r equals 5cscθ, and this is my equation in its polar form.

Now let's look at our second example here. We have y=x+1. 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 rsinθ=rcosθ+1. 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 rcosθ from both sides. Now in doing that, that will cancel on my right side, leaving me on the left with r(sinθ-cosθ)=1. 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 θ minus the cosine θ is still equal to 1 on that right side. Now from here to fully isolate r, I can divide both sides by sinθ-cosθ, cancelling on the left side, leaving me with my final equation here. R is equal to 1sinθ-cosθ. Now I have my final equation here in its polar form.

Now let's look at one final example here. Here we have x2+y2=25. Since I have an x and a y, I could go ahead and replace x and y using rcosθ and rsinθ, but there's actually an easier way to do this because we also know that x2+y2=r2. So I can simply replace x2+y2 here with r2, giving me r2=25. Now solving for r here, I can just take the square root of both sides, leaving me with r is equal to 5. This makes sense here because I recognize that my original equation in its rectangular form was a circle of radius 5. So 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 x2+y2=r2 is equal to r2. 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 \(x^2 + y^2\) 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 \(r \sin \theta\), \(x\) is our \(r \cos \theta\), 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 gonna 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 \(r\)'s 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 gonna 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 two \(r\)'s on this side, which means that I'm dealing with an \(r^2\). So this is really \(4r^2 \cos \theta \sin \theta = 2\). 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 \(2r^2 \cos \theta \sin \theta\). 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 two. 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 \(r^2 \sin 2 \theta = 1\). 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 θ and converting it into its rectangular form containing x and y. 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 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. 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.

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. 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. 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. 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. 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 θ's. So I can go ahead and rewrite my equation in its standard form. 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 standard or recognizable form of a circle, so I don't need to rewrite it 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 trig functions in terms of sine and cosine. I have the secant of theta, which I can rewrite as 1 / \cos(\theta), giving me r = 1 / \cos(\theta). I still don't have any of these terms. So what else can I do? I have a fraction here, and whenever I have a fraction, I want to eliminate that fraction by multiplying both sides by my denominator. 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. 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). 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. I do have a sine theta, but not an r \sin(\theta). One of my strategies is actually to go ahead and multiply both sides by r. Let's see what happens when I do that. Multiplying both sides by r, I get r^2 on that left side is equal to 6 r \sin(\theta). Now I end up with two of those terms, r^2 and r \sin(\theta). So I can replace r^2 and r \sin(\theta) with x and x^2 + y^2. In doing that, I end up with x^2 + y^2 = 6y. My equation again only contains x's and y's, no r's and θ's. So I can go about rewriting this equation in its standard form. It can be helpful to try and identify the shape of your graph at this point. 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. 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. I have x^2 + y^2 = 6y. So I do want to go ahead and complete the square here. So, by moving this 6y over, I get x^2 + y^2 - 6y = 0. I have this y^2 - 6y. I know that I can make this into a perfect square by adding some constant. I want this to become the perfect square (y - 3)^2 because of that value negative 6. If I add 9, that does become my perfect square, (y - 3)^2. So, in rewriting this, I end up with x^2 + (y - 3)^2 = 9. Now looking at this equation, 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 any 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 = 2 1 + cos ⁡ ( θ ) . 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 cos θ, r sin θ, or r2. 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+cos⁡(θ). 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 + rcos⁡(θ). And all of this is equal to 2. Now from here, I see that I have this rcos⁡(θ), which is great. But I also have this r, which is not typically something that we want to have. Usually, we want an r2, an r sin θ, or rcos θ. Now this is a sort of unique instance in which we actually want to keep that r as is because we already have rcos θ, 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 x2+y2 is equal to r2. So what if I just take the square root on both sides? Then I know that r is equal to the square root of x2 + y2. So here that's what we're going to replace our r with, the square root of x2 + y2. Then I have that rcos θ, which I can replace with just x. So I have the square root of x2 + y2 plus x is equal to 2. So I've replaced everything that I need to, and I can go ahead and move on to step number 3 where we're going to rewrite this 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 much easier. Then I have the square root of x2 + y2, and that's equal to 2 - 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 x2 + y2 on the left, and 4 - 4x + x2 on the right. I'm going to go ahead and multiply that out, which will give me 4 - 4x + x2. So where can we go from here? Well, I see that I have an x2 and an x2 on both sides, letting me cancel that out, leaving me with y2 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 is equal to 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. $x^2-1=y$