Hey, everyone. We just learned how to take points given in polar coordinates and convert them to rectangular coordinates, but we also need to be able to do the opposite. Take points given in rectangular coordinates and convert them to polar coordinates. Now looking at this point given in rectangular coordinates here, we see this sort of suspiciously familiar-looking triangle that we've solved time and time again. And that's all this is. It all just comes back down to triangles. So here we're going to keep using everything that we already know, specifically working with the Pythagorean theorem and the tangent function in order to convert points from their rectangular coordinates into polar coordinates. Now here, I'm gonna walk you through how to do that step by step. So let's go ahead and get started.
Now looking at the point that we have here in rectangular coordinates, 3, 4, we're able to form this triangle. And from this rectangular coordinate point, 3, 4, I know that this side length x is equal to 3 and my side length y is equal to 4. Now since I already know x and y, it's just left to find r and theta, which will give me my point in polar coordinates.
Now let's start by finding r. Now looking at my r value here, this is the hypotenuse of my triangle, but I have my other two side lengths already. So I can find that hypotenuse by simply using the Pythagorean theorem, which tells me that c squared is equal to a squared plus b squared, or specifically for this triangle, r squared is equal to x squared plus y squared. Now actually plugging in our values here, I get that r2 is equal to 32 plus 42. Now adding those together gives me 25, so I have r2 is equal to 25, or completely solving that for r, simply r is equal to 5.
Now that I have that r value, let's turn to finding theta. Now looking at theta here, I know that I have my opposite side and my adjacent side, which means that I can set up a tangent function because I know that the tangent of theta is equal to the opposite over the adjacent side. Or specifically for this triangle, for a point given in rectangular coordinates, the tangent of my angle theta is equal to y over x. Now, again, just plugging in the values that I have here, I end up getting that the tangent of theta is equal to that y value of 4 over that x value of 3.
Now in order to actually solve for theta here, I need to take the arctanyx. So here, theta will be equal to the inverse tangent of 4 over 3. Now if we actually plug that inverse tangent into our calculator, we end up getting that theta is equal to about 53 degrees, and I now have both my r and my theta value.
Now I originally started with my point in rectangular coordinates, 3, 4, and now I have my point in polar coordinates, 5, 53 degrees. Now these are the general equations that we're going to use to convert points from rectangular to polar coordinates, but we have to be really careful here because we end up taking an inverse tangent. And the inverse tangent, remember that this function is only defined over the intervals contained in quadrant 1 and quadrant 4. And even though our point here was located in quadrant 1, it won't always be.
So I actually have a step-by-step process for you that will work no matter where your point is located. So let's go ahead and work through these examples together. Now first, we're given this point here, negative 4, 0, and I wanna go ahead and plot this point and then convert it into polar coordinates by following these steps here.
Now starting with step 1, we just want to go ahead and plot this point on our rectangular coordinate system. Now locating this point, negative 4, 0, I end up right here on my x-axis for this first point a. Now with step 1 done, we move on to step 2 in calculating r. Now this is the same equation that we saw above, but just already solved for r having taken the square root on both sides. So here, plugging in my values, I get that r is equal to the square root of negative 4 squared plus 0 squared. Now this ends up being the square root of 16, which is simply equal to 4. So I have an r value here of 4.
Now with r calculated, we move on to finding theta. Now in order to find theta, we need to look at the location of our point and looking at where this point a is on my x-axis, because it's on an axis, I know that theta is going to be one of my quadrantal angles. And thinking specifically about where this point is located, thinking about my angles here going from 0 to pi radians or 180 degrees, I know that my angle here is simply pi. So that gives me my value of theta pi, and I now have my point in polar coordinates 4, pi.
Now let's move on to our next example here. We are specifically given the point negative 1, √(3). Now we're going to restart our steps here starting from step number 1, where we're gonna start out by plotting our point. Now plotting my point here is going to be more useful to know that the square root of 3 is about 1.73 as a decimal. So plotting this point at negative 1, 1.73 ends me up right here in quadrant 2 for point b. Now with that point plotted, we want to go ahead and find r as we have before, taking the square root of x squared plus y squared. Now plugging these values in here, I end up getting the square root of negative 1 squared plus the square root of 3 squared. Now actually adding those together gives me the square root of 4 or simply 2. So I have my r value of 2.
Now all that's left to do is to find theta. Again, we're looking at the location of our point, and I see that my point b is located in quadrant 2. So it's not on an axis. It's not in quadrant 1 or 4 as my original example was, but it is in quadrant 2. So what do we do now? Well, we're gonna start off the same way that we did for our original angle that was in quadrant 1 by taking the inverse tangent of y over x. So let's start there. Theta is equal to the inverse tangent of my y value √(3) over my x value negative 1. Now this simplifies to the inverse tangent of the negative square root of 3, which from my knowledge of the unit circle or by simply typing this into a calculator, I will end up getting a negative pi over 3 for that inverse tangent of negative √(3).
But we're not done yet because let's think about where this point is located or where this angle is located. The negative square root of 3 is here in quadrant 4 and that is not where my point b is located so that wouldn't make any sense. Now in order to make sure that this is located in the right place, because our point is located here in quadrant 2, we need to go ahead and add pi to that angle that we got from the inverse tangent. So this negative pi over 3, I need to add pi to it. This ends up giving me 2π/3, which thinking about where that angle is located, that is located in quadrant 2 as it should be. So here I have my final value of theta, 2π/3, and I have this point now in polar coordinates, 2, 2π/3.
Now when converting points from rectangular to polar coordinates, we're gonna start out the same way. We're going to plot our point on our graph, and we're going to find r by using this equation here. Then when finding theta, we need to pay attention to the location of our point, and then we're good to go. Now that we know how to convert points from rectangular to polar, let's continue practicing. Thanks for watching, and I'll see you in the next one.