Hello wonderful mathematics people. This is Anna Cox from Kellogg Community College. Symmetry test for polar graphs. First symmetry about the pole axis or the X axis in the positive direction. If the point R Theta lies on the graph, then the point R negative Theta or negative R π minus Theta. So if we want symmetry about this pole axis, we need the point R Theta to be reflected. So we need R negative Theta here, negative direction coming out Theta. Or we need to extend the line through the pole, figuring out what that angle is. So that angle would be π minus Theta and then instead of coming out R direction, we would go negative R. So negative R π minus Theta symmetry about Theta equal π halves or the Y axis in the positive direction. If the point R Theta lies on the graph, then the point negative R negative Theta. So if we have R Theta here, if we think about negative Theta coming this way, we need to go back through the pole. So we're going to go in negative R, so negative R negative Theta, or we want the positive angle. So in this case π minus Theta and going out of positive R. So one of these two points symmetry about the pole or the origin. If the point R Theta lies on the graph, then the point negative R Theta. So if R Theta is here negative R Theta negative R telling us we're going through the pole, or we figure out the new angle this π plus Theta rural route, Theta plus π. If two symmetries exist on a polar graph, then the third is also true. Now we have a caution here. If a polar equation passes a symmetry test, then it's graph definitely exhibits that symmetry. However, if a polar equation fails a symmetry test, then it's graph may or may not have that kind of symmetry. We don't know at this point. So if we look at some examples, R equal to -2 cosine Theta, if we put in the point R negative Theta, whenever we see a cosine Theta, because of our even identity, we know that cosine Theta and cosine negative Theta are equal. So if I put in R negative Theta, I get R equaling 2 - 2 cosine negative Theta. Cosine negative Theta is positive because it's an even function. So we can see that this is true, that this point here is on the original graph. Thus we have symmetry about the pole axis because if R negative Theta is on it, then we have that symmetry if we look at a different one like R π minus Theta or negative R negative Theta. If we put in π minus Theta here and use our expansion formula, we get 2 - 2 cosine π cosine Theta plus sine π sine Theta. We know cosine π is -1. We know sine π is 0, so we get R equal 2 + 2 cosine Theta. Hence no information is given. We know that it that particular symmetry didn't exist, but we don't know that it doesn't really still have that symmetry. So let's look at this other point. Negative R negative Theta. If I use negative R negative Theta, here's negative R, here's negative Theta. So now I get negative R equaling 2 -, 2 cosine Theta. That's not exactly the same as the original either, so it gives us out no information. If two symmetries hold, then the third must also be true. Remember that. So we know that one symmetry held. We know that we had symmetry about the pole, but we currently don't have any information about our, π minus Theta or negative R negative Theta. So we don't know anything about Theta equaling Pi halves or the Y axis. So let's look at negative R, Theta. If I put a negative R and Theta, this is also no information given. So then if we look at R Theta plus Pi halves sticking in R, Theta plus Pi halves expanding out, cosine Theta plus Pi halves, cosine Theta, cosine π minus sine, Theta, sine π, our cosine π is -1. Our sine π is 0. So R equal 2 + 2 cosine Theta. It's not exactly the same as the original, so no information is given there either. So all we really know at this point is that we definitely have symmetry about the pole axis. So what we're going to do to graph this is we're going to put in some angle values, zero π thirds, Pi halves, 2π thirds π. When I put in zero for Theta 2 -, 2 cosine 0 is going to give me zero. If I put in π thirds, 2 -, 2 cosine π thirds is going to give me out one. If I put in Pi halves, I get out two, 2π thirds, we get out three, Pi, we get out four. Now we expect symmetry around the pole axis. So I'm not going to do π to 2π because I'm going to expect that symmetry to occur. So if we look at this graph 00, the angle π thirds, we're going to get a radius of one Pi, halves we're going to get a radius of two 2π thirds, we're going to get a radius of three Pi, we're going to get a radius of four somewhere way out there. And if I could draw, the symmetry would make it look the same on the top and the bottom a sideways heart. If we look at another example, R-squared equal negative cosine Theta. If we think about negative R Theta, negative R-squared equaling negative cosine Theta, well negative R-squared is just R-squared, so we have symmetry around the origin. If we thought about R-squared equaling negative cosine of negative Theta, cosine of a negative angle is just the positive because it's an even function. So R-squared equal negative cosine Theta, hence symmetry around the X axis. So if we have two, we know we have to have the 3rd. So here we're going to put in some values. Let's do 0 Pi, 6 Pi fours. Oh, those are all going to be undefined because zero Pi 6 Pi fours is going to give me an R-squared equaling a negative value. So let's do Pi halves 2π thirds 3 Pi fours π halves would give me zero 2π thirds is going to give me root 2 / 2 for my R But the formula said R-squared. So we know that that would be a half three Pi fours. If I put in three Pi fours, we're going to see that the R is positive or negative sqrt 2 / 2 square root of that because R-squared was squared of 2 / 2. So when we square root the R-squared we get positive negative. So doing some symmetry here Pi halves gave me 0 at this point. Here 2π thirds is going to give me out root 2 / 2, but also negative root 2 / 2. Three Π fourths is going to give me out sqrt sqrt 2 / 2 and negative sqrt 2, negative sqrt 2 / 2. If we draw in a few more points, we're going to get a sketch that looks similar to this. When we have these first few points, these points because we have symmetry around the pole. So we have these points here and these points here. We have symmetry around the pole, we have symmetry around the X axis, and we have symmetry around the Y axis or the pole axis and Theta equal π halves. So that's how I know to drawing the rest of the graph. The slope of a polar curve R equal F Theta is given by dy DX not by R prime equaling DF D Theta. Recall that X = r cosine Theta and Y equal R sine Theta. A different way to write that is X equal F of Theta cosine Theta and Y equal F of Theta sine Theta. So if F is a differentiable function of Theta, then so are X&Y when DXD Theta doesn't equal 0. So we're going to have dy DX equaling dyd Theta divided by DXD Theta. This is similar to parametric equations. So dyd Theta we're going to take DD Theta of F Theta sine Theta and we're going to use the product row. So the derivative of F Theta is DFD Theta times sine Theta plus the derivative of sine Theta was cosine Theta times the F Theta. Doing the same thing for the denominator derivative of F Theta is DFD Theta times the cosine Theta. Derivative of cosine Theta is negative sine Theta F Theta times the F Theta by the product rule. So DYDX is really going to just simplify into F prime Theta sine Theta plus F Theta cosine Theta divided by F prime Theta cosine Theta minus F Theta sine Theta. Now if we think about F prime, Theta is really just R prime and F Theta is really just R We could think of this as R prime sine Theta plus R cosine Theta over R prime cosine Theta minus R sine Theta. R Theta is usually a given point, but it doesn't have to be. We could still have just a generic formula there. So if the polar graph RF Theta passes through the origin, thus the F Theta sub zero is 0, then the slope of this is really just R prime sine Theta over R prime cosine Theta. The R primes are going to cancel out, leaving us tangent Theta because if the R is 0, we're going to have these terms here at the end cancel out. R is 0 and we get the R prime sine Theta over R prime cosine Theta. So if we have R equaling -1 plus sine Theta and we want to figure out the slope when Theta is 0 and when Theta is π, our prime is just going to be cosine Theta. So dy DX at -1 zero. If we put in zero here for sine, we get the point -1 for our radius 0 for our Theta. So R prime sine Theta plus R cosine Theta over R prime cosine Theta minus R sine Theta. So in this case my R prime was cosine Theta times sine Theta plus our R is -1 oh, our R is -1 plus sine Theta times cosine Theta over R prime again cosine Theta, cosine Theta minus -1 plus sine Theta times the sine Theta. So now when we actually stick in points -1 zero, this is our generic here. At this point when we stick in negative 10R is -1 Theta 0. So cosine of 0 is 1, sine of 00, negative one plus sine of 0 is 0, cosine of 01 over cosine 0 times cosine 0 minus quantity -1 plus sine 0 times sine 0. When we simplify all this up, we get -1 in the numerator over one in the denominator, or just -1. Doing the same thing for DYDX. When the Theta is π, we know that sine of π is 0, so our radius is going to be -1 when our Theta is π so -1 Pi. So sticking that back into this generic formula up here, cosine π sine π plus the quantity -1 plus sine π times cosine π over cosine π, cosine π minus the quantity -1 plus sine π times sine π. Sticking those in and evaluating those gives us out -1 * 0 + -1 + 0 negative one, IE the slope is one. If we take some time to actually graph the original, the R equal -1 plus sine Theta, it's going to be a graph similar to this. And so then when we think about what's happening at the point -1 zero, here's 0 angle -1 back here. So we expect a slope at this tangent point to be a negative. And it was, it was -1 if we think about Pi -1, here's Pi -1 goes through the pole this way we have our slope here being a +1, which is what we would expect based off of this graph. Thank you and have a wonderful day.