7.3 Polar Coordinates

[STARTS AT 26:20]

Professor: So we're going to find the symmetry of the rose defined by the equation r equals 3 sine 2 theta and create a graph. 

So let's go ahead and first check for symmetric about the polar axis. Let's evaluate r equals 3 sine 2 theta at negative theta. 3-- don't need parentheses yet-- sine of negative 2 theta, evaluating at negative theta, that is negative 3 sine 2 theta, which is not true. OK, those two were opposites of each other. So that test did not work out. Not that. 

So let's go ahead and try the negative r pi minus theta, since those actually show us the same thing but sometimes one is easier to see than the other. So if I say negative r equals 3 sine and this is of 2 pi minus theta. 

Well, we're going to have 2 pi minus to theta. Now a trigonometric identity actually tells me this is going to be equal to negative 3 sine of 2 theta. And if you notice, this is the same equation we started with-- r equals 3 sine 2 theta. If we divide by negative 1, then we have r equals 3 sine 2 theta. Which means this is symmetric-- symmetric about the polar axis. Symmetric about the polar axis. 

Next, we want to evaluate at pi plus theta. pi plus theta, so 3 sine 2 pi plus theta. That same identity from earlier tells me this is equal to 3 sine of 2 theta, which is the equation I started with. That's great. All right, so we have symmetry. Symmetric about the pole that is (0,0). 

Now we want to test for symmetry about the vertical line, or the line theta equals pi over 2. OK, good. It actually is a vertical line. That is not a typo. Just wanted to check that. 

OK, so if it's symmetrical at that vertical line-- all right so, let's go ahead and plug-in pi minus theta. Which we are going to see is not going to work just like it did not work in part a, because we're actually going to have the same problem. We'll have a negative on one side but not the other. 

But if I take negative r equals negative theta, so this will be 3 sine of negative 2 theta. This is going to be negative 3 sine 2 theta, equals negative r, which is the same as our original equation. So this is a check. Symmetric about theta equals pi over 2. 

So if we were going to graph this, I don't think-- no, I don't have a graph there or a grid. If I were going to graph this, based on this symmetry, we only need to graph one portion of this. So we have (0,0). (0,0), pi over 6 is roughly 2.6. My graph is not going to be the scale. 

pi over 4 we have 3, a radius of 3. Next, we get 2.6 again at pi over 3. And then it pi over 2, we're back at 0. So we have a graph that looks like this. Based on it being symmetric about the pole, we know we're going to have another leaf on this end. 

Based on it being symmetrical at the polar axis, we know we're going to have another leaf here. And then about pi over 2 means we are going to have another leaf over here. So that is the full picture of our graph based on, and this is a rose, but based on that one quadrant. 

[ENDS AT 31:26]