 So we've looked at parametric equations. Now we're going to turn to polar coordinates, just changing to polar coordinates. So there's going to be a bunch of exercise problems in this video and we're going to look at polar equations. So rectangular and polar equations. Problem 1.1 derive a conversion from polar to rectangle coordinates. So in the Cartesian plane we have rectangular coordinates and we have any point P as we can see x comma y. Of course we can drop that down to the x-axis which makes this length here x that makes that length y and we can make the hypotenuse just r which would just be the square root of x squared plus y squared. And then this line the hypotenuse makes an angle of theta that goes counterclockwise from the positive x-axis and that's important because that's going to be our convention going counterclockwise from the positive x-axis. So if we draw in these lines it becomes very clear that we can just use some trigonometry because I can now write the cosine of theta and the cosine of theta is going to be adjacent divided by hypotenuse. So that's going to be x divided by r and from through simple algebra I can write x equals r times the cosine of theta. So x equals the r times the cosine of theta if I have the sine of theta that's going to be the opposite divided by the hypotenuse so simple algebras tells me that y equals r times the sine of theta. So if I'm given any point I'm going to say a polar coordinate point polar coordinate point and that point is r comma theta I can convert that to rectangular coordinates so I can get these rectangular coordinates from that so the rectangular coordinates are going to be some point p x comma y and that is going to be simply this point if I'm given r and theta that becomes r times the cosine of theta and r times the sine of theta. So if I need to convert any points given being given in polar coordinates r and theta I can convert that to x and y with this equation. So in problem 1.2 we are asked to derive a conversion from rectangular coordinates to polar coordinates so again I've got this point p and on the Cartesian plane x and y and I've got express this in term of r comma theta and because we have a right angle triangle here for this point just you know filling it in with a triangle I have the fact that r squared equals x squared plus y squared the hypotenuse so I can just write an equation for r that's going to be the square root of x squared plus y squared and we'll see you know that that's plus minus that we're going to see how that works in a later problem so that's that and I also note that that the tangent of theta just seeing that I know what x and y is I'm given the point so the tangent of theta that's going to be y over x and that means I have this equation for theta that's going to be the inverse tangent or the arc tangent of y divided by x so given any x and y value I can convert it to something that contains r and theta and as much as I can now write this point I can write the square root of x squared plus y squared that's going to be my r and I can write the inverse tangent function of y over x as far as my theta value is concerned problem 1.3 convert the point p2,3 so that's in the plane so that's x and y coordinates to polar coordinates so I need to write down these polar coordinates so let's just change color there because we always put our equations that we have to remember in green so I've got to convert this I've given x and y and I've got to convert this into r and theta so remember this is going to be the square root of x squared plus y squared comma the arc tangent or the inverse tangent function of y over x so this is closed down that square root symbol there so that is how I've got to convert this I'm given this point x is 2 and y is 3 and that's how we're going to convert it so let's do that so our point is going to be the square root of 2 squared plus 3 squared and then we need the inverse tangent function of y which is 3 divided by 2 and I can simplify this a little bit it's 2 squared is 4 3 squared is 9 plus 4 is 13 so I'm going to be left with 13 as far as my length is concerned and then the arc tangent the arc tangent of 3 over 2 now let's just try and well let's just look here at the warframe language so again this will be available in the link down below and what I've done is I've created I've created two at least of my own functions here so first of all you give the function a name and because this is designed by us we're going to start it with a lower case but we've got this camel case notation so every word starts with an uppercase letter so that you can distinguish the different words but the first one starts with a lower case to show that it's not an inbuilt function but one that we created so polar to rectangular and then I've got to give it the parameters on which it can work so when I call this function I have to pass r and I have to pass theta because I have polar coordinates and I want to take it to rectangular coordinates and we put these little underscores after some name that we give it and since we're working with radius and angle let's just keep it r and theta but you've got to put these underscores to show that these are parameters then I put colon equals that's just the delay assignment because we don't want anything to be executed now we're just creating this function and then we're going to create this little list so we want to return a list so we put it inside of a set of curly braces and then some element comma some other element and so that will be x and y and what is x well it's r cosine of theta so remember to make a multiplication symbol we just leave a little space there you can see between the r and uppercase c there's a space and theta to get that little theta symbol you can press the escape key on your keyboard type th and then escape again and that's going to give you that theta let me show you if I hit the escape key I get these three little dots I type in th and I hit escape again and I change to a theta comma r sine of theta so there's a little space there as well between the r and uppercase s and theta and I'm going to create another one called rectangular to polar so there I'm giving x and y so remember the underscores and because it's x and y why don't give them the parameter names x and y and that's going to be the square root so I'm using the sqrt function here so in this notebook I'm not going to use the fancy notation in the desktop version so that you can see exactly what it looks like if you want to use the the web version so it's sqrt for square root and that x squared and that's this shift six on my keyboard the carrot symbol so x squared plus y squared and remember there's an order of arithmetical operations so squares are going to come before plus so the two squares are going to happen before the two before the addition so if we want to we can really just put parentheses around this just to force that order of execution but because there is this order of execution that's not going to be a problem comma now the inverse tangent function in the warframe language is arc tan with the uppercase a and uppercase t and then just y divided by x so I can create these two functions so if I now call rectangular to polar as we had in problem 1.3 we've given these values for x and y so if I just pass that I get square root of 13 and then the arc tangent of 3 over 2 now what is the arc tangent of 3 over 2 so we can express that as a numerical approximation with the in function so let's just write the in function and then we pass to that arc tan of 3 over 2 so let's do 3 divided by 2 and let's see what that looks like it's 0.98 so almost 0.98 and that's in radians so almost almost one radian and we know pi over 2 is right at the top that's the y-axis goes up there because I want to show you something else I want you to think about something so it's square root of 13 which is the length of this hypotenuse and then it makes an angle of almost one pi so let's just go here and just think about this for a little bit and I'm going to draw some coordinate axis here so there's my coordinate axis and remember if I go up here that'll be pi over 2 radians up there this is 0 that's pi radians so pi over 2 is 3.14 divided by 2 is about one and a half so one and a half would be there so maybe here let's just make a thinner point here so maybe somewhere in this direction somewhere in this direction I'm going to find this angle here which is this arc tangent let's just write that the arc tangent of 3 over 2 imagine that's there so this is the direction we're going so we go from the positive x-axis we're going from the positive x-axis to this line here in other words we're moving in this direction so from the origin in this direction and we're going square root of 13 units along so the point might be somewhere out there but what if I were to add something to this what if I took this angle arc tangent of 3 over 2 that's an angle remember and I add to that pi radians well if I add to it pi radians I'm actually going in this direction this is the arc tangent of 3 over 2 plus theta that's that direction if I'm going in this direction but I go negative the square root of 13 in that direction so we're going down here but instead of going down here we go negative the square root of 13 that means we're going up here and we end up at the same spot so we can actually write this point also as negative the square root of 13 comma the arc tangent of 3 over 2 plus pi so if we were to do that that would be exactly the same point so remember this angle tells us the direction we're moving in and this going counterclockwise from the positive x direction so if we go up this almost one radian that means we're looking at it from the origin to down that line but if we add theta a pi radians to that we all the way the other way and there's something else going on here because I might as well go around 2 pi and then if I go around 2 pi I'm in the same direction and I can go around 3 pi times so what we should actually have written here is square root of 13 comma the arc tangent of 3 over 2 or that 9.8 plus 2 pi oops 2 pi n where n is this element of the natural numbers starting at zero so because I can go round and around and around and around and I'll still face that way and this one's going to be p negative the square root of 13 comma the arc tangent of 3 over 2 plus pi and I can also go around 2 pi n times there as well with n still being an element of the natural numbers starting out counting at zero so that would be all the points because I can just go with polar coordinates I can just go round and round and round the circle and I'm still going to face the same direction more importantly though I mean we're not going to work with those going around and round so we're not going to work with these but this is definitely very important so just to add pi radians to that meaning I'm facing the opposite direction and then my radius is going to be a negative radius so I put the negative out then that's why I told you it's important to remember that we had plus minus the square root of x squared plus y squared because we can certainly go in the other direction if for our angle we just add pi radians to that so in this problem we're going from polar to rectangular coordinates so we're given this and this is exactly what we had before so let me just change to blue there so r is in this instance negative the square root of 13 and we have theta equals the arc tangent inverse tangent of 3 over 2 that's an angle remember plus pi and remember our point what was our point our point was 2 comma 3 was our point so this is remember always so how do we convert if we're given r and theta so what is this going to be this is going to be our cosine of theta our sine of theta so always going to write that down so let's see if we get to that very same point we had before given this previous problem so let's do that so the point p it's going to be while we're going to have negative the square root of 13 negative the square root of 13 that's r times the cosine and now of this angle and this angle is the inverse tangent of 3 over 2 plus pi so that's going to be our x coordinate and then we've got a look at our y coordinate that's again r so it's negative 13 and we're going to take let's just do that and then we're going to multiply that by the sine of the angle and the angle was the inverse tangent of 3 over 2 plus pi and that is going to be our y value so we can might as well use a calculator I mean I'm hoping that you'll have access to that in exam because to work out these values in pen and paper that's a bit difficult so let's grab the warframe language and there we have problem 1.4 so polar to rectangular so those are my values negative the square root of 13 and then the arc tangent of 3 over 2 plus pi and before we use the pi symbol but you can just use the inbuilt keyword p i with the uppercase p so if we do this we get back to the point 2 comma 3 just as we just as we suspected and we showed in the previous problem so there we go p is 2 comma 3 so we've had those two points we went from 2 comma 3 we went to the square root of 13 and then the arc tangent of 3 over 2 but we also now showing that really that other point the negative square root of 13 and then the arc tangent of 3 over 2 plus an extra pi gives us the exact same point problem 1.5 we asked here to convert rectangular coordinates to polar coordinates find the rectangular coordinate of the polar coordinates so polar to rectangular so let's just write in green always what we have to do what we have to to sort of memorize so we given these r and theta values and we have to go to x and y values so x is going to be r times the cosine of that angle and y is going to be r times the sine of that angle so let's switch to blue so we're going to have p times so r in this instance is 2 times the cosine of pi over 6 pi over 6 there we go and we're going to have 2 times the sine of pi over 6 now we just have to remember what that angle what those angles are and if you can't remember what they are just draw yourself a tiny little neat little triangle let's let's just do that so I've got our little angle triangle here and if we make it so that we have pi over 6 remember pi over 6 is a small little angle so let's draw it out there if we do that pi over 6 pi over 6 then we're going to have 1 2 square root of 3 and if we now look at the cosine of pi over 6 let's go back to blue if we look at the cosine of pi over 6 that is adjacent divided by a hypotenuse so that's going to be 2 times so it's square root of 3 over 2 and then we're going to have 2 times the sine is opposite divided by the hypotenuse so that's 1 over 2 so our point p is going to be square root of 3 comma 1 1.6 find the rectangular coordinates of the polar coordinate point p negative 4 and pi over 3 so it's always we're going to remember how do I get x and y well that's the r times the cosine of an angle and r times the sine of that angle so always write that down and if you write it down often enough it's going to stick in your mind so we're going to have negative 4 times the cosine of pi over 3 and we have negative 4 times the sine of pi over 3 and if you can't remember what those values are remember our little triangle so this instance we have a bigger angle there pi over 3 so let's just draw that out here so that was pi over 3 so this obviously not to scale that's our angle pi over 3 remember that's 1 2 square root of 3 and this is so yeah that is a right angle triangle and the square root of 1 squared plus 3 squared well that's the square root of 4 and the square root of 4 is 2 so our pot news is that so that makes it be easy to remember what these are angle should be so here we're going to have p times that's negative 4 times the cosine of pi over 3 so remember the cosine of of an angle is adjacent divided by a pot new so that's going to be a half and then negative 4 times the sign is opposite divided by a pot new so that's square root of 3 over 2 and if we simplify this a little bit that's going to be a negative 2 there and that's going to cancel out so we have negative 2 square root of 3 so let's just look at the Wolfram language we created those functions in the one of the in the first problem so let's have a look at that and there we go 1 1.6 there so it's polar to rectangular we've got negative 4 and then pi over 3 and if we execute that we see negative 2 comma negative 2 square root of 3 this is we got problem 1.7 again find the rectangular coordinates to the polar coordinates write it down always write it down when you do your exercises so what does x well that's r times the cosine of theta and r times the sine of theta for y and we have this point 3 comma so this polar coordinate the radius is 3 and an angle is three quarters three times pi over 4 so let's get our blue pen out and so we're going to have p times well we've got 3 is our length there and that's the cosine of 3 pi over 4 and that's going to be 3 times the sine of 3 pi over 4 so three quarters of pi so let's go to the Wolfram language and just use our little function that we created there and there we go polar to rectangular 3 comma 3 times pi divided by 4 and we see our solution there we have negative 3 over square root of 2 and 3 over square root of 2 p is negative 3 over square root of 2 and 3 over square root of 2 and remember you can rewrite this as this negative 3 square root of 2 over 2 and 3 square root of 2 over 2 this is you divide multiply the numerator and denominator by square root of 2 over square root of 2 sometimes that just looks needed to have it to have it in that form now the next set of problems here section 2 so this is problem 2.1 actually quite a bit of fun and I hope you get to to learn about these and have to do them so the first one is describe the curve r equals 2 so we've given no information about theta but whatever theta is and that is the crux of the matter whatever theta is that means theta can really be anything so let's write that down theta is going to be on this interval half open interval from zero because remember we always start counting at zero but then it can go to infinity so we can really go around and around and around and around you know forever but whatever we do we have to have this radius of 2 so think about it going around but at a fixed radius so I think you can already know where this is going so if we've got our plane here and no matter what the angle is as we go round round round as long as we are two units away from origin so if we were to do that we just get a circle so what is described here is a circle and the radius of that circle at any point is the radius of 2 and yeah just going around round going around twice is 2 pi but you you know going around this 2 pi but remember we can always add 2 pi into this in as much as for these values n being natural number starting at zero if n is zero that's this 2 pi going around one this one's is 2 pi plus 2 pi you get the point it's quite easy so if we think about the equation for a circle that would be x squared plus y squared equals some radius squared so in this instance the radius is 2 so that's 2 squared and we can solve for y that's going to be 2 squared minus x squared and y equals plus or minus the square root of 2 squared minus x squared and that's going to give us the same circle but here in as a a curve here we can just describe that whole thing is this r equals 2 so in an opposite problem here 2.2 we're not given r r can be anything so what we really are saying here r is an element of this open interval from negative infinity to infinity r can be anything but theta it's got to be fixed so if we have this here at theta equals pi over 4 now that's the angle there and r can be anything as long as that angle is fixed so what do we have we have a straight line a straight line through the origin a straight line through the origin pi equals 4 once again you know well I suppose y equals x is what we could have written there y is just x and the polar form of that just simple pi equals theta at least equals pi over 4 so interesting problem to describe this curve this polar curve r equals twice the sine of theta so theta goes from 0 can go all the way around to 2 pi and around around again but whatever that value is if I plug in the value for theta take the sine of that multiplied by 2 it has to be at that angle a certain radius so the radius is going to change so you know what do we have here well if we look at the right hand side here it looks almost familiar we remember that y equals r times the sine of theta I don't have sine r times the sine of theta I have 2 times the sine of theta but I can multiply both sides by r so on the left hand side I get r times r and on the right hand side I get 2 times r sine of theta so now I have r sine of theta there and I remember as we've written there y equals r sine of theta so I just have y on that side but I also remember on the left hand side I have r squared and r squared is x squared plus y squared when we do these conversions so now on the left hand side I've got x squared plus y squared that equals and on the right hand side I've got 2y so that's very interesting so let's just bring that all on to the one side x squared plus y squared minus 2y and I'm going to leave a little space and equals zero I'll just leave that zero out because what we can do here is just to complete the square if I put plus 1 there and plus 1 remember that's this 2 that we have here divided by 2 squared 2 divided by 2 is 1 so what we have now is x squared plus and now I have here y minus 1 squared if you do that square you're going to end up with y squared minus 2y plus 1 and on the other side I just have 1 so what we have here indeed as well well we have a circle and that's a circle and it has a radius of 1 so a radius of 1 but it's not centered at the origin so it's centered at so on x it's going to be at zero but y we've got to move over one to the positive side because if we subtract one from that we back to the origin at zero so it's a circle of radius radius of 1 and that is centered at the point 0 comma 1 right a pollock equation for the y-axis you know the y-axis goes from negative infinity to positive infinity so from the origin we know that r is just going to be equal to you know on this interval at least from negative infinity to positive infinity but that that doesn't tell us much because if we just wrote that I mean that is radius in any direction what is fixed though is the angle and if we go from the positive x-axis kind of clockwise we're going up theta equals pi over 4 radians if we do that we are on the y-axis so irrespective of what r is as long as we have that angle fixed that is the y-axis so the polo equation for the y-axis is theta equals pi over 4 we're mixing up the problems a bit here's 2.5 back to a kind of problem we've seen before describe the curve that the radius from the origin will always be 4 times the cosine of whatever angle we are at and if we look at the right hand side we remember that x equals r times the cosine of theta so that's very close and we can certainly get it in that form by multiplying both sides by r so on the left hand side I have r times r and on the right hand side it would be 4r cosine of theta and we also remember as with a problem that we saw before that x squared plus y squared remember that's r squared in other words we can just do all these substitutions on the left hand side we have r squared which is x squared plus y squared and on the right hand side we have 4 times x r times the cosine of theta is x so I'm going to bring it to the left hand side I have x squared minus 4x and leave a little space there plus y squared equals the right hand side which is 0 at the moment and again here I can complete I can complete the square so if I were to add a little something there let's add 4 which is just plus 2 squared and that means it's 2 squared I've got to add to the other side as well what I do on the left hand side of the equations sign I've got to do on the right hand side and that gives us x minus 2 squared and if you square that you get to x squared minus 4x plus 4 plus y squared equals 2 squared so again here we have a circle it has a radius equal to 2 and it is centered at where's it going to be centered well we have x minus 2 at the moment so the center has got to be in the opposite direction so that's going to be 2 comma and the the y is is going to be zero so it is this circle with radius 2 centered at the point 2 comma zero so this one always always gets me it's very exciting this one you look at that and you see this this polar curve r equals the tangent of the angle theta and the secant of that angle kind of clockwise from the positive x axis I hope I haven't said clockwise some other time sometimes I just say clockwise when I mean anticlockwise remember these are always anticlockwise the angle so look at that I mean can you figure out in your head what that looks like and we've got to somehow change this and when you see this just change it to stuff that we are working with we're working with signs and cosines so let's write that r equals the sine of theta divided by the cosine of theta and on this side we have one over the cosine of theta so we have r we I suppose we can bring the cosine squared of theta over to the left hand side it's in the denominator there cosine theta times cosine theta and that equals the sine of theta and I think from here it should be pretty obvious what we're going to do we're going to multiply both sides by r if we multiplied both sides by r you know we've got something going on we've got r squared cosine squared of theta and on the right hand side r sine of theta so what's r sine of theta yeah that's just y and what is r squared cosine squared of theta well just think of it think about it for a minute x equals r times the cosine of theta and then if I were to go and just square both sides x squared is r squared cosine squared and that's what we have there r squared cosine squared so it's x squared so y equals x squared the first parabola that you ever that you were ever taught and look at it as a polar curve the tangent of an angle times the secant of that angle given that that is the radius we have a parabola that's a lovely problem problem 2.7 described this curve r equals 6 over the square root of 9 minus 5 times the sine squared of theta if you can just look at that and know what this is hey power to you okay now how shall we tackle this problem we've seen one or two we've seen one or two little tricks we multiply both sides by r but if we multiply both sides by r that's not really going to help us here at all we see a sine squared there and we'd love to get an r squared in there but just looking at this algebraic equation here that's not going to work I think the first thing that you should realize here is get rid of the square root because there's nothing we've dealt with when it comes to these things that that deals with square roots so the great rid of the square root there we can't multiply both sides by r but we can square both sides which in effect does mean the left hand side is going to be r squared but yeah it's not multiplying both sides by r it is squaring both sides so in the numerator we're going to get 36 wait did my little ruler go here we go and that's going to be over and now at least I am rid of this square root symbol now we can start getting somewhere sine squared of theta and if I bring this over to the left hand side I have r squared and I have 9 minus 5 sine squared of theta and that's going to equal 36 and this starts working for me because now I have 9 r squared and I have minus 5 r squared sine squared of theta and that still equals 36 and look at what we have here if we do remember that y equals r times the sine of theta that's what we have always with our conversions and if we now square both sides y squared equals r squared sine squared of theta and we have r squared sine squared of theta so that's just going to be y squared and r squared remember where's my green pen r squared that's just x squared plus nice x squared plus y squared we remember those those are just ones you should of course absolutely know so here we're going to have nine and instead of r squared we have x squared plus y squared remember where that comes from it's just the right angle triangle and that's just the Pythagorean theorem as far as the hypotenuse is concerned that's where we get that minus five times and what is r squared sine squared of theta well that's y squared so we have all x's and y's because you know that's the only why are we doing all of these problems I mean I should have said so with the previous problems but we're trying to get these two to rectangular coordinates because those are the the ones we're familiar with we can describe I mean just looking at a polar curve remember the previous problem the tangent of an angle times the secant of that angle who knew you know who knew so we convert these to rectangular coordinates and it helps us and it makes sense so what we have here is nine x squared plus nine y squared and I have minus five y squared and that's 36 can you see what kind of curve this is going to be I have nine x squared and I have positive four x squared and I have 36 what am I going to do I'm going to divide both sides by 36 and if I divide both sides by 36 let's do that divide by 36 divided by 36 what I have nine over 36 and nine goes into 36 four times and so that's x squared over four which is two squared and I have four goes into 36 nine times so that's going to be let's just fix that tiny little error there where did that x pop in it's an y of course so four goes into 36 nine times so that leaves us with y squared over three squared and 36 divided by 36 is one and what do we have here we have an ellipse who knew from starting here that we can end up with an ellipse problem 2.8 this one you've got to be quicker than me so we see the cos theta on the right hand side so we would love an r in there because we do remember x equals r times the cosine of theta and we do remember that r squared equals x squared plus y squared so now by now you should be familiar with these so we're going to multiply the left side by r so if r times r and on the right hand side we're going to have negative 10 times r times the cosine of theta r times r r squared well that's this x squared plus y squared remember we're taking this into rectangular coordinates because those are the kind of curves we were brought up with imagine at school we were taught polar coordinates before we were taught rectangular coordinates then I suppose we would recognize these things you know as easy as we do now with the rectangular coordinates anyway we have negative 10 there and what is r times the cosine of theta well that's this x so we can bring that over to the other side plus 10x and we still have the y squared and that's going to equal zero that's going to equal zero so I have this idea of just having to complete the square so if I took x squared plus 10x if what can we do there what can we do to complete the square well if I just were to add let's add five squared there if we add five there squared and I still have my y squared that's going to equal five squared over on that side so here if I have x plus five all squared plus y squared equals five squared so whatever I have again I have a circle I have a circle it's going to have a radius of five and where's it going to be centered where's it going to be centered well it's going to be centered at minus five negative five and on the y axis zero and it's negative five because we've got to go in the opposite direction to what we have there to center our circle so again negative 10 times the cosine of an angle so we're starting to develop at least with these easy ones it should be you know rather simple to see that we are at least start to see that we are dealing with circles when we see this kind of polar polar equation problem 2.9 you should be able by now just to look at this and know know intuitively what you have to do what we're looking for is x equals r times the cosine of theta we want y equals the r sine of theta and we want r squared remember that would be x squared plus y squared so can we achieve this if you just look at that the polar curve that we've been given there r equals six times the addition of sine of theta plus the cosine of theta I remember I mean just make it simple for yourself it's six times the sine of theta plus six times the cosine of theta and all we need in there is an r so what do we do to the left hand side multiply by r so r times r and on the right hand side we also have that so it's six r sine of theta plus six times r times the cosine of theta and now r times r is r squared so that just becomes x squared plus y squared and on the left hand side on the right hand side that is six r sine theta is y plus six times our cosine where cosine of theta is x and now we just have to bring it all to the left hand side so I'm going to have x squared minus six x and let's leave a little space because you know what we're going to do y squared minus six y and that's going to equal the right hand side and we'll have to do you know we'll have to do something there so six we divide that six by two we square that so that's going to give us a positive nine there another positive nine there nine and nine is 18 now what we've done on the left hand side we've done to the right hand side competing the square we've done so that will be x then we take the minus there and nine is three squared remember so that's going to be three we square that and think about it x minus three all squared that's x squared minus six x plus nine so no problems there yeah we have y minus three all squared and that's just 18 and 18 is nine times two nine times two and that equals three squared times two and if we if we if we are thinking about squaring that as far as what is the the radius going to be so we need something that we can just something that we can just square so we're going to take the square root of that and square it so we take the square root of that what's three times is nine it's 18 so what we want is the square root of 18 and we want to square that because the radius will then be square root of 18 but what is that well we can bring that three out so that's three square root of two all squared so what we have here again a circle it's going to have a radius a radius of three square root of two and where is it going to be centered well this time we move we move both the x and the y axis so we're going to be centered at three comma three so that's a lovely lovely problem and again we're starting to notice the sort of pattern and I think if you do enough of these you'll start to know what what the curve is going to look like just by looking instead of the rectangular coordinates looking at the the polar equation there so describe the curve r equals 10 times the secant of theta well we know we can't work with secant and dealing with rectangular polar coordinates we deal with sine and cosine and fortunately we can easily rewrite this that's 10 over the cosine of theta and we can simplify this we multiply both sides by the cosine of theta so we have r times the cosine of theta that equals to 10 but we know what r times the cosine of theta is that's x x equals 10 just this vertical line to 11 describe the curve r equals negative three times the cosecant of theta very similar of course to the previous problem we're going to have if I remember just use my blue pen there we go so we can have r equals negative three over the sine of theta multiply both sides by the sine of theta I have r sine of theta and that's going to equal negative three and what is r times the sine of theta that's y so y equals negative three we have this horizontal line going through y equals negative three so now in section three we're going to go the other way so convert the rectangle equation to a polar equation so we see x squared plus y squared equals 25 so that's going to be a circle it's going to be centered at the origin it's going to have a radius of five so how do we convert this to polar coordinates to a polar equation well let's just write these things down just always to remember that x equals r times the cosine of the angle and y equals r times the sine of the angle and we have r squared equals x squared plus y squared then we showed in the first problem you know where this all comes from so let's do that seems to be the only course of action we can take so if we have x equals r times the cosine of theta we've just got a square all of that so we're going to have r squared cosine squared of theta plus here we're going to have r squared sine squared of theta and that looks easy enough to solve and on the right hand side what do we have well we have five squared nothing we can do there so we're going to take out r squared as a common factor there and I'm left with cosine squared of an angle plus the sine squared of an angle and we certainly know that trigonometric identity and that's going to equal five squared so here cosine squared of an angle plus sine squared of an angle that's just one r squared times one is just one so we have r squared equals five squared so r is just going to equal five so that was that's just going to be a circle centered at the origin with the radius of five but it's a square root that we're taking so in strict strict fashion we should actually say negative five as well because if I'm pointing one way for a certain angle I might as well add to add pi to that so 180 degrees to that and point in the opposite direction so that still leaves me at that same point so we have both articles five and articles negative five as an equation for this circle problem 3.2 convert the rectangular equation x squared minus y squared equals one two polar equations just going to write these down all the time so it sticks in your head x equals r times the cosine of an angle and we have y equals r times the sine of an angle make sure you know why this is so and x squared plus y squared equals r squared or r squared equals x squared plus y squared so what do we write for x squared well that will be r squared cosine squared of theta and we have minus now r squared sine squared of theta and that's just going to equal one squared r squared can come out as a common factor there so what are we left with cosine squared of an angle minus sine squared of an angle and that's a trigonometric identity which we should know that equals one squared perhaps it's going to be given to you perhaps not but that is just the cosine of twice the angle that's the cosine of twice the angle and that's going to be one squared so we can divide both sides by this cosine of two times the angle so that's going to be one over cosine of twice that angle and if we I suppose we can really neaten this up we can say r squared equals the secant of two theta and we have r equals well strictly plus minus the square root of the secant of two theta that is our or an easy conversion of x squared minus y squared equals one into a polar equation interesting one this hyperbola x y equals x times y equals four because remember if we multiply both sides by one over x we get y equals four over x hyperbola so as always write these things down so you can remember them in your sleep that's a horrible thing to say oh no just write three just because I want to be so horrible r cosine of theta y equals r times the sine of theta and r squared equals x squared plus y squared I should write that the other way around to make and keep things consistent but anyway x squared plus y squared equals r squared so what are we going to write for x well that's just r times the cosine of an angle and we multiply that by r times the sine of an angle and that has got to equal four so if r times r that's r squared and that's going to be equal to well that's four over we have the cosine of the angle and we have the sine of the angle there and what is that equal to well that'll be four times the secant of that angle and the cosecant the cosecant of that angle but hang on just a little bit we can really do something about this before before we go sort of before we go that far because let's just think about it let's be a slightly more clever about this so we're going to have r squared there and then I'm going to have cosine of an angle and I'm going to have sine of an angle this is showing you sort of something else here and if I were to just multiply both sides by two look at that r squared and I have two times let's just rewrite this as sine of theta cosine of theta and I have that as equal to I've got to multiply both sides by two so that gives me eight on the right hand side but what is this trigonometric identity the twice the sine of an angle times the cosine of an angle well that's r squared and that is sine of twice that angle and that's going to equal eight so I have r squared that equals eight over sine of two times that angle so that's eight times the cosecant of that angle and now we just set with the fact that we have r squared so you're going to take the square root of both sides so just think about that a little bit it might look neater of course this looks a bit neater than that and I've got to put two there it's twice the angle cosecant of twice that angle problem 3.4 convert the rectangular equation x equals 3 to polar to a polar equation so that's just the vertical line it goes through x equals 3 up and down and y goes from positive to negative infinity and we've got to convert this to polar to polar to a polar curve and we know what x is x equals r times the cosine of theta don't get those mixed up ever like I do x equals r times the cosine of theta so now as you think about it I mean it's actually beautiful as the angle changes now we're going from the origin to the positive x axis we go counter clockwise as that angle changes our changes and to such an extent that we have this straight line this vertical line it's actually it's actually quite quite beautiful if you think about it and so I have the fact let's just change our way in blue yes so we have instead of x we write r times the cosine of an angle that's going to equal 3 so r is going to equal 3 over the cosine of that angle and that means if we just needn't things up that's three times the secant of that angle so if you take three times the secant of an angle you're going to land up with this vertical line going through x equals 3 3.5 convert the rectangular equation x plus 2 y equals 3 so that means 2 y equals negative x plus 3 and divide both sides by 2 so that's just the straight line and we've got to convert this to a polar equation I urge you to keep on writing this so that you don't forget x equals r times the cosine of theta and also remember why this is so y equals r times the sine of theta it's all to do just with the right angle triangle and we have x squared plus y squared that equals r squared that's better this time I wrote it in the instead of in the right order so what are we going to plug in for x well we're going to plug in r times the cosine of an angle and we're going to have twice r times the sine of that angle and that's going to equal 3 so you know where do we sort of where do we take this one well at least we can take out r as a common factor so let's do that and I have cosine of an angle plus twice the sine of that angle and that equals 3 and we just take that to the other side 3 over cosine of theta plus twice times the sine of theta and for this one I think it's okay okay it's good enough if you leave it and this you might you know you might need to you know do something else about it but Sydney that is this polar equation being converted from a rectangular equation so here's a lovely problem so we're going to go the other way now when we describe the curves we kind of did this but now we just want to explicitly you know write this down theta equals 3 so it's this angle counterclockwise going from the positive x axis and it's almost all the way around because to the opposite side to the 180 degree because that'll be pi 3.14 so yeah theta equals 3 so it's going to be this line but let's write down an explicit equation for sort of for this line and if you think about what do we have for theta well we actually if you think about the things that we've done let's write that we have that the tangent of theta that's going to equal y over x that's kind of all we have but here we just we have theta we don't have the tangent of theta but we can certainly take the tangent of both sides there's nothing wrong with that so let's say that the tangent of theta well that's going to be the tangent of 3 and that is going to equal y over x and now we simply solve for y so y is going to equal tangent of 3 times x and what's the tangent of 3 well it's about negative 0.14 somewhere there and times x and indeed that is a fixed slope because tangent 3 is just a constant the constant times x so it is just going to be the straight line going through the origin and it has that slope that negative slope so convert the polar equation r equals the tangent of theta to rectangular equations there's a route that you can go that leads you nowhere because we do remember that we have something for the tangent of theta well that's just y over x but i can't write r equals y over x because if i solve then y equals rx that doesn't help me because i still have some a polar form in there so that's not going to help me any but as we've done before we can square both sides r squared equals the tangent squared of theta so i've just squared the right hand side and the left hand side let's just write that down always r squared equals x squared plus y squared so this time around you know remember to write those two equations your two green equations the ones that you should know so r squared i now have x squared plus y squared and the tangent of that angle that's just going to be y squared over x squared so let's simplify this a little bit let's just write it in a neat way so i'm going to have y squared that is x squared and i have x squared plus y squared there so i've just bought the x squared over to the other side and just flipped it around so i have y squared equals x to the power four plus x squared y squared i'm going to bring that over to this side subtracting from both sides so that's going to be minus x squared y squared that equals x to the power four i can take out y squared as a common factor so i'm left with one minus x squared so that equals x over four so i've got y squared equals x to the power four over one minus x squared and i can take the square root of that so plus minus this x to the power four and then over one minus x squared so certainly we can take out that x to the power four as just an x squared so we'll have one over one minus x squared and one minus x squared remember that's one minus six one plus x but nothing we can do about that so we've got our y on the one side and that's what we were aiming for so r equals the tangent of theta we have a rectangular equation for 3.8 these are beautiful problems if you do these it really shows that you understand the work so show that the let's neaten that up that looks horrible show that the polar coordinate point p three comma three quarters pi that that that point lies on the polar equation r equals three times the sine of twice the angle so let's see what happens here on the left hand side so this is r that's theta so we're going to have three on the net on the left hand side on the right hand side we're going to have three times the sine of twice that angle so twice times three pi over four so if we do that that's three over two pi and that's negative one so that's negative one times three that's negative three and three is most definitely not equal to negative three so it's this point really on the line well that kind of is if you remember that i can go in a direction so if we go three quarters pi counterclockwise from the positive x axis that's where we go and then from the origin on that angle we go in the positive direction three what we could also go is another pi radians further so another half circle around so we still at this line with as far as this angle is concerned we just have to go radius in the negative direction now so this point is exactly the same point as negative three negative three comma so i have three pi over four and i've just got to add to that another pi so that's four pi over four that's still the same thing so i'm now left with seven pi over four so if i do that seven pi over four from this angle this point this point so three quarters pi plus another pi that leaves of me with seven pi over four so that's still this exact same point so i still have my radius there and i still have my angle there now let's do the left hand side which is now going to be negative three and is that equal to so i've got to have three times the sine of twice this angle twice this seven pi seven pi over four so if i do that that's still negative one so i have the fact that negative three equals negative three this point and this point is exactly the same point you've got to visualize that in your head so i'm going to a certain angle i'm going in the positive three units along that angle but if i add a half a circle to that i'm going in the opposite direction now and if in that direction i go in sort of in the negative three units i'm still ending up at the same point so this point is indeed on this polar equation so that was such a beautiful problem we just got to do another one and i kid you're not going to be exactly the same but just repeat these again a stick in your head when you see that one in an exam or whatever no problems whatsoever show that the g was weird that the show that the polar coordinate point p three comma three pi over two lies on the polar equation r squared equals nine times the sine of the angle you plug that in the left hand side's going to be positive the right hand side's going to be negative so no doesn't look like that point is on that polar equation that the curve of that polar equation so we've got to think like we did before that we can express this point in another way so we can go in the negative that units for radius but we've just got to take this three pi over two and we add another pi to that now we're going sort of pointing in the opposite direction now the radius this has to be also in the opposite direction so to make this simple we put it over the same denominator there so we have five pi over two five pi over two so negative three five pi over two that is the exact same point those two points are exactly the same point and now if we plug it in on the left hand side we're going to get negative three negative three squared i hope i didn't say it was going to be certain signs on the left hand side forgive me if i did that but negative negative three times negative three that's going to be a positive nine and on this side i have nine times the sign of this angle five pi over two what do you think the sign of five pi over two is it is one and nine equals nine so yes this point is indeed on the curve of this polar equation so two more problems to go and i know i told you use this wolfram language to sketch these things out you know i see very little sort of point in doing these but come on these two are these two are fun so at least let me just show you i'm still going to use the wolfram language but let's let's do this sketch the polar equation r equals one plus cosine of theta so i have a polar equation i've got to sketch this curve and what you see in there i've drawn somewhat to scale what's going on here and what i'm going to do is just plug in values for theta there such that i go up in these increments remember going from the positive x-axis counterclockwise in pi over eight units i'm going to go pi over eight pi over four three pi over eight pi over two so i'm just going all the way around sort of those so that i can make this little table so that i plug all these in and i get values for r you see why i don't like doing this because that's really you know you learn that you know make a little table and draw something that's really school stuff i i don't know i've got mixed feelings about this i don't like like to do it but this one at least the end result is what i like and as much as i like you know these kind of curves um they're quite fun to use so let's go to the wolfram language and what do we have here so what i'm going to do is just create all these values zero pi over eight two pi as a two pi over eight three pi over eight four pi over eight which is pi over two etc i'm just going to create all of them and we use the table function for that so i'm saying eight times pi over eight and a is going to go from zero to sixteen so zero times pi over eight that's this zero so we're starting on the zero line then a goes to one so we have one pi over eight and you'll see now what the end result is so what the table function does is give it a little expression there to work out and then over what values of a because i've used a there but you can use anything else you want so it's just going to iterate over those a equals zero then one then two then three all the way to sixteen and it's going to return all of those values as a list object and i'm just going to assign that to a computer variable called theta and there we have we have zero pi over eight and then two pi over eight three pi over eight four pi over eight five pi over eight six pi over eight seven pi over eight and this is simplifying those and now i'm going to use the association thread function so i'm going to thread these two things together i'm going to pass theta to this equation one plus the cosine of the angle so you see there the one plus the cosine of angle i'm wrapping that as a parameter to the n function so that i get these numerical values back so threading means take my values that i have here theta so it's all those values zero pi of eight two pi of eight three pi of eight and pass them to this function and the n just means express this in a numerical approximation terms for me so we can see all of them so zero is going to end up at two so if i plug in pi equals zero r is going to end up at two if i plug in pi over eight i'm going to be going to 1.923 if i plug in pi over four into one plus the cosine of the angle so one plus the cosine of pi over four that's going to give me 1.7 so i have all these values so i'm just going to drag that to my other screen so that we can just start plugging these in so you've seen how to do that of course this is going to be available on kit up as well so you can just save this to your free version on the wolfram cloud and you can just you know sort of look at that so let's just do this laborious task and again i have mixed feelings of this so zero is going to be let's just choose a different color so zero is going to be at two so we're there so pi over eight is going to be 1.9 so if i go to pi over eight the radius is going to be sort of 1.9 so this becomes a bit on this scale a bit difficult to do so it's going to be there so if i go to pi over four i'm at 1.7 so that's a bit there and if i go three pi over eight that's 1.3 so it's even less and if i go to pi over two i'm at exactly one so it's sort of this arch up and then we're going to come down five pi over eight i'm at 0.6 i should have drawn this bigger shouldn't i this is getting ridiculous and three quarters pi that's 0.3 so it's even closer and if i get to seven pi that's even closer and then if i get to pi that's at zero and then i'm going to go a little bit out again a little bit further out and if i get to three pi over two where are we three pi over two that's going to be at one so we're going to be up there 13 pi over two we're going to be at slightly more and then we're going to be at 1.7 slightly more and slightly more so there we go so let me just show you it's this car deoid pattern so all the way around all the way around and then in isn't that neat so of course do it in the warframe language and have some a proper graph but there we have a car deoid pattern so the next problem is going to be exactly the same let's tackle that one last problem we're going to sketch this almost same exact polar curve of this polar equation but instead of one plus cosine of theta we're going to have one plus twice the cosine of theta let's see how this affects what we're doing here so let's bring up the warframe language and i've done exactly the same by the way you know the previous problem let's just let's spot that so you can see what it should look like it's done beautifully the car deoid pattern there so let's do the same thing so we're still doing this association thread so all these values of of theta that we have zero pi over eight two pi over eight three pi over eight all the way around and i'm passing all those values to this function one plus two times the cosine of the angle this expression i should say and i'm getting a numerical approximation value of that so if we do that they they're all the values so when theta zero one plus two times the cosine of that is going to be three and in one plus two times the cosine of pi over eight is going to leave us at two point eight and we go and we go and we go well instead of doing that because we're you know my graphs look so ugly when drawn by hand let me show you what it looks like so there it's the polar plot is the function that we're using we're saying one plus twice the cosine of theta and we're going to let theta go from zero to two pi so we're going to let it go all the way around kind of clockwise and lo and behold that's what we're going to get so we're going to go out to three units this time when pi is zero all the way around so it's almost like a cardoid but it actually it comes down and makes a little loop there and comes back out again so do that by hand and and i'm sure in the vast majority of courses you are going to be asked to draw this by hand it's making a little table as if you're a little kitty now i'm being very terrible now i'm being terrible really but anyway just come back to the Wolfram language check it out use the polar plot function play around with that table function table function is a very powerful function in the Wolfram language and check it out so that's all the problems for this problem it was actually quite long but they are fun and and just to convert to polar coordinates you're going to see in calculus that helps you out so much it's just such a useful thing to be able to do so do these homework problems do some more leave me a comment down below if you've got some problems that you want perhaps we can make some more videos you make a video let me know about it and you know i'll put it down in the description below as well do do what you want just enjoy these because they really are a bunch of fun and it's a very powerful skill to have to do these things in polar coordinates