 In this video, we provide the solution to question number 15 from the practice final exam for math 1060, in which case, given this polar graph right here, supposed to find its correct equation. Now we can see this looks like some type of limouson, right? You have this circular-like object with an inner loop here. You could try to use the basic parameters of a limouson and cardioids and things like that to try to do it, but I'm gonna assume we don't know the general formula here. How can we find the correct equation by process of elimination? Given after all, we have six options we have to choose from. So let's look at specific points on the polar graph. So for example, we look at this point right here. The Cartesian point would be two comma zero, but the polar point we're looking at here would be a radius of two and you're gonna get an angle of zero, so it's kind of funny, it's because it's the same point there, right? So we see that your radius should be two at angle zero right there. So which of these equations have that as a possibility? If angle was zero, notice that cosine at zero is equal to one, so it'd be two plus three, which is five, that doesn't work. Sin at zero is zero, so that's a possibility. Again, cosine at zero is one, so you get three plus two, which is five, three plus three, which is six, three plus two, which would be three in that situation, so that doesn't work. And then you're gonna get three plus three, so just by that one single point, we've actually eliminated all the possibilities and we get choice B. Now, let's suppose it didn't work out so conveniently, some other points we could look at would be like this point right here. In this situation, we see that one, two, three, four, five, if the, so the Cartesian coordinates would be zero comma five right here. The polar coordinate would be, and that's the one we really be looking at here, the radius is five, the angle is pi halves, and let's verify here what happens. Sine at pi halves is equal to one, three plus two is equal to five, so that works out. Again, just for further clarification here, if you look at this point, the x and y coordinates would be negative two comma zero, but the polar coordinate would be two comma, in this case, we're gonna get pi as the angle there. So again, what happens at sine? Sine of pi is zero, and so you'll get the radius is again two, so far so good. And then let's just do one more. If you go down this angle right here, take 270 degrees or three pi halves, right? You have the point, it's kind of hard to see because I've drawn over a couple of times, you get this point over here, the Cartesian coordinates would be zero and one, but in terms of polar coordinates, we wanna think of it as negative one comma three pi halves, like so, which again, when you put three pi halves into sine, you'll get negative one, two minus three is equal to negative one. So that does check out with all four of those points right there. So if you're trying to, if you're given a polar graph and you're trying to identify by process elimination, looking at the four quadrental angles, zero pi halves pi and three pi halves will probably sufficient, but like I said in this example, we actually found it just by looking at what happened at a radiant measure of zero.