 So this is the first video in the series on vector calculus where we just use homework problems. We're just going to do problems, problems, problems, and by doing these problems we describe what's going on, we learn what's going on and I think that's really the way to tackle it because really in a textbook what are you going to see? You're going to see 1 plus 1 but in the exams you're going to be asked to do some differential equation and that's usually how textbooks work. It's only by going through the problems at the back of each chapter that you really start to understand what's going on. So that's what we're going to do here. So we're just going to do problem after problem and we explain what's going on by doing these problems. So in this first one we're going to look at parametric equations, how to go from a normal function, well I would say normal function in other words the ones that we used to and how to convert them into a set of parametric equations and I'm also going to plot a few. Now we're not going to do that with pen and paper although I suppose in a test and exams it's what you're supposed to do and at least learn how to do that but you know it's tedious it takes a long time and that means you don't do a lot of those and you don't get the chance to experience what's going on. In other words using a computer to do that you can just draw these things very quickly and you can iterate over a bunch of them and see what the differences are by changing some of the values. So it's worthwhile to learn how to do that with a computer so you can do that with Desmos calculator or Geogebra or any of these software packages online but I'm also going to show you in this one at least how to use the Wolfram language so you can just open a free account on the Wolfram cloud I'll show you a little bit about that and we're just going to do some of these parametric curves just using a computer language. So in this first problem we're going to find the parametric equations for the equation y equals x squared. So we're all familiar with y equals x squared if I can just take some time and try and draw it here on the left hand side let's do that and you know it's very very simple but we're going to use this first example just to sort of figure out what is going on here what do we mean by the fact that we say we want parametric equations so we know y equals x squared and we know if x is zero then y is zero remember these are two both real number lines so as an equation or at least as a function remember the function would then be f of x equals x squared and what it does is it takes a value from the set of real numbers and it maps it to another element of the real numbers so if I input one zero zero if I input one it's going to be one if I input negative one it's going to still be one if I do two we're going to have four if I do negative two squared is four and three squared is nine one two three four five six seven eight nine it's going to be way up there and it's going to be way up here so there we have we have sort of this parabola and we all know how difficult these things are to draw there we go that's what we have y equals x squared but now we want to parametrize it we want a parameter and what the parameter is going to do is is this going to allow us to move along this and we use move in quotation marks because we this is all kind of abstract so the first thing that we're really going to do is we've got to find these equations or we've got to find these two parametric equations so in a simple case like this the very first thing would be just to say x equals t and if x equals t that means we have the fact that y equals t squared and there we have it we have the two parametric equations it's it's really as simple as that when it comes to a problem like this so let's see if we if we write down our parametric equation now remember it's y comma x and that's just going to be t comma t squared and remember your x goes to positive infinity and your goes to negative infinity this side so we've got to have the fact that t is on this open interval from negative infinity to positive infinity so what does this all mean how you know how does this this is our final solution remember that's how we write the parametrize two parametric equations here what does it mean well if t is zero then you know if we plug in t equals zero we'll be here by zero comma zero so we still down here as far as as on on this curve is concerned so when t equals one we move all the way up and we're now there when t is two we go all the way up there and when t is three we go all the way up there and so we're only showing integer values here but anything in between as far as this t is concerned it just means we are different spots here we just move we're just moving along this curve and that's what a parametric equation is as simple as that we have now a parameter that allows us to be somewhere on this curve depending on the value of that parameter so here we have problem number two instead of the very simple polynomial we had we just have something slightly more complicated here we see an order two polynomial degree two polynomial y equals x squared plus three x plus two again just a parabola it's a shifted a bit and again it's going to be very very simple as far as what we have to do we have to find values for x and values for y and they both functions of t and again it's very easy just to set x equals t and if x is set to t we just substitute wherever we see x we substitute that we substitute t in there so y is going to be t squared t squared plus three t plus two as simple as that and now for our final solution remember we weren't told specifically as far as x is concerned what the what the interval is or the domain of this function then so we can just write x comma y and that's going to equal t comma t squared plus three t plus two and then negative infinity less than t less than positive infinity and that is the final solution so again it's very simple when it comes to these polynomials all we're going to do is just write some function of x with respect to t and just whatever we chose there we can just put put in there now just want to show you we could just as well do something else let's do that let's say we're going to make x equals t plus one there's nothing wrong with doing that that's going to make my y a bit more complex because every time we see x we have to put in t plus one so that's going to be t plus one squared plus three times t plus one and plus two so that's just a bit more complex we're going to have t squared plus two t plus one and we're going to have plus three t plus three plus another two and where are we left now we've still got a t squared and we have two t and three t that gives us five t we have one plus three plus two and that equals six so this would be another form of parameterizing our curve so that we have x comma y that is going to equal t plus one and then also t squared plus five t plus six and again since our domain is the whole of the real number line we're going to have negative infinity is less than t is less than infinity so there we have another solution and that's also equally valid parameterization of this function of this polynomial y equals x squared plus three x plus two but we want to keep things very simple I think we'd all agree that the parametric equations here in our first solution is a bit more easier to work with than the one in our second solution in this problem problem 1.3 in our problem set there just mind the previous ones where we said this problem one two and three this problem 1.3 so parameterize the equation y equals sign of three x plus cosine of x squared now before we just had polynomials and now we have transcendental function here but it's as simple as before and now let's just take the simplest parametrization and we're just going to let x equals t and then when x is t we're just going to replace that so for y we're going to get by the way these are just my lines I should actually you know just complete this so that you see that it's not minus x in case you get confused there so y equals the sign of three t plus the cosine of t squared as simple as that and then now for my solution my final solution as I say let me do that so you can see I'm just this my different bullet points so in our final solution we're going to have x comma y and that is going to be my parametric equations t and then sine of three times t plus the cosine of t squared and again because we weren't told the domain of x there we're going to just assume it goes from negative infinity is less than t is less than positive infinity and that's our final quick and easy solution so it doesn't matter whether this is a polynomial or transcendental function as we have yet to get a metric function we're just going to do exactly the same by starting with x and just giving it the simplest sort of parametric function and that will just be t so in problem 1.4 we have to parameterize again this very simple polynomial we just back to y equals x squared but now we want to go from one point to another point so we're not just starting at negative infinity going to positive infinity we want this very specific starting point and end point and as we can see there 1 comma 1 so that will be down here and then 3 comma 9 which will be up there so we need values for t so that as t changes we just stick to this little bit we just want to stick to this little bit of the curve between 1.1 and 3.9 so as t changes that's all that's the only place we want to be we want to be nowhere else so what we have to do now if we want to parametrize this is the following so we want usually if we started 1.4 t to be zero at that point because t as far as parameters go when we use when we do apply problems it usually first to time and we want to start at time equals zero so we've got to find a new parametrized function for x such that when t is zero we have the fact that x equals one and just have a look at this think about this if i say x equals 1 plus t now if t is zero i substitute zero and there x is 1 that's beautiful that's exactly where we want to be so what is y going to be well it's x squared so that's one plus t squared and again if t is zero i'm at time equals zero one plus zero is one one squared is one i am at the point one comma one so this function works for me now it's not immediately obvious that it'll always work if we change t that you know it's going to work out properly for us and it will it just depends how t changes now what we want from t equals zero we want to go to t that is more positive we want to move forward in time not backwards in time so you might find that depending on what you do for this crucial step this and put a little block here and a little tick mark that just to show that that is the crucial step now let's have a look at what value must t take given that this is our parametrized function parametric functions what values must t take for us to end up at 3.9 so we know we have this equation x equals 1 plus t so we want x to be 3 in this instance so if we have x equals 3 you know what must t be well at that instance what we have now is we have the fact that 3 equals 1 plus t so 1 plus t that's still my value for x and now we want x to be 3 and that's very simple to go from there and as much as we can just solve for t and we see t equals 2 so when t equals 2 2 plus 1 is 3 x is indeed in position number 3 down here so let's just see when t equals 3 what happens to y so we can have y equals 1 plus what is t it's now 2 squared and indeed that's 3 squared and that's 9 so we are in the right spot 3.9 so we can eventually now write down we can eventually write down our parametric equation now as per usual let me just do this so that there's no ambiguity as to I'm not putting minuses there that's just the bullet points along which we are working is the fact that we have our parametric equations now we can say x and y that's going to be equal to 1 plus t and 1 plus t all squared and now we can also say let's just make that a nice 2 we can also say that 0 is less than or equal to t is less than or equal to 2 because it's only if t is just between those two values that's where that is when we are in this little section here of our curve just between those starting point 1.1 and the n.3.9 so again when you have to solve these problems you have to start with x we'll start with x of course and find some function that includes t such as that when you make t zero that we will be at that point x equals 1 because that's where we have to start there at x equals 1 and then substitute and then you'll have to find out what the value for t is once you get to the other point so we want x to be 3 and that's what we have there we want x to be 3 and now we just say well x is now 3 and our parametric equation for x is 1 plus t so 3 equals 1 plus t we solve for t and then we just see where we land up when t just make sure when we use t then t equals 2 that we are at 9 which of course we will be so we have our final parametric equations there with the domain for our parameter t let's do problem 1.5 and your problem set there find the parametric equations for y equals 2 minus x squared and we can see the little crew drawing there it's a problem of facing downwards and it has an y intercept there of 2 but we want to go from the point 2 comma minus 2 so that's the point right there and then we want to go back up to the point 0 comma 2 so we have to find values for t such that we just move along that section so if we just want to move along that section so how are we going to go about this as always finding parametrized versions of equations of our equation there we've got a set x to be some function of t now remember we would ideally want to start at t equals 0 remember right in the first problems I mean there are many ways to do this but we want to start at t equals 0 so if we want to start at t equals 0 we want a function of x with respect to t that we are at point x equals 2 and then one way to do that would be say x equals 2 and then plus or minus t now I'm going to choose minus t here and you'll see that this is going to work out for us I want you to try positive t and see if that works of course it's going to work but I'll tell you what happens now you'll have to run t backwards in time and as again I use the term time but we have to see this just as a parameter parametric variable and as much as it's not really time but we are using it as time here so we want to move forward in t t goes from 0 up and if you use 2 plus t that's not really going to work for you and please do it and you'll see is this still valid parametric equation for x yes it is because at t equals 0 to minus 0 is 2 and so we are at x equals 2 just as we wanted to be so if that is my equation there my first parametric equation that means y must be 2 minus and then I just substitute for x I substitute 2 minus t squared and let's see what that is that's going to be 2 minus and what do we have here so that's 4 minus 4t plus t squared and that is is this equal to 2 minus 4 that's negative 2 plus 4t minus t squared so I have my parametric equations now and let's see for what value of t because this was for t equals 0 at what value of t are we going to end up at 0 so we have this equation now for for we have this equation here that we can see here as far as x is concerned so now we just want x to be 0 so 0 equals 2 minus t and that just means that's very simple that this means t equals 2 so at t equals 2 when I plug 2 in there 2 minus 2 is 0 we will be at x equals 0 and you see now why I chose 2 minus t if I chose 2 plus t I would have ended up here in my last part here I would have ended up with t equals negative 2 and that's not that's of course not what we want and let's just make sure of course that we are then as far as y is concerned in position 2 so let's just see what y is at t equals 2 so y at t equals 2 let's just write it there t equals 2 and we're going to say that's our equation right up there so that's minus 2 plus 4 times 2 minus I should say they're minus and we have 2 squared so that's going to be minus 2 we're going to have plus 8 and we're going to have minus 4 and that is indeed just 2 so we are in position 0 comma 2 so that's exactly what we want so we eventually have our parametric equations there let me just do my little tick marks here so that you can see that's what they were all about so we're going to have x comma y and that's going to equals 2 minus t and then then this long equation this long polynomial here minus 2 plus 4t why do you say long polynomial that really isn't long t squared and remember now 0 is less than or equal to t is less than or equal to 2 we found t to be 2 when we were up here so there we have it our final solution so this one's just a bit more tricky you'll have to think properly of what your crucial equation is in the crucial one is this very first one x equals 2 minus t and if you said 2 plus t that would still be valid that doesn't say then you would have to run t towards the negative side to get to the second point so just be very careful of that so here we are with problem 1.6 in the problem set find the parametric equation for the line segment from the point 4 comma negative 2 to the point 2 comma negative 1 so if we look at the problem here we see the two points here and we want to go from this bottom point to this upper point that's what we want so that being said we are reminded here and you'll see those in the notes the parametric equations for line segment would just be a plus bt comma c plus dt and then we need to solve find solutions for a and b and c and d and those are just real numbers and t is going to be on this interval from p to q that is how we do it so we when we do these we try to be you know obviously clever about all of this and we're going to start off with with t equals 0 because that's what we want to start at and t2 equals 0 we're going to be right at that point so at t equals 0 if we look at the equation up here we're going to have the following we're going to have x that's going to equal a plus b times 0 t is now 0 so x is just a but we know what x is at that point we're starting at the point 4 so we know a equals 4 similar for y y is going to equal c plus dt but t here is 0 we are t equals 0 so there's nothing there but we know we are at the starting point 4 comma negative 2 so that's going to be negative 2 and we've already solved for a and c now we just have to solve for b and t and unlike the previous problems where we needed to find out what the value of t is going to be here we're not so concerned because it's our choice as to what we're gonna what we're going to do from here on in so let's just do that we can decide and we're just going to say well our next time step or our next parametric step is just going to be t equals 1 as simple as that so where does that leave us well it leaves us with x equals a but we know what a is now it's 4 plus b well t is just 1 and b times 1 is just 1 so that's not a problem and y equals we know c is negative 2 plus d and t again is 1 so that that is not an issue and furthermore we also know for these two points we are now at point 2 comma negative 1 so now that means we can solve for b and we can solve for d that's very simple so b in our instance is just going to be 2 minus 4 that's negative 2 and as far as d is concerned here d if we just take the negative 2 over to the other side that's going to leave us with 1 so we now have we now have these solutions as far as a b c and d are concerned and now all that's left to do let's just draw this line all the way down just keep things nice and neat is that we can write down our parametric equations we have the fact that x comma y that is going to equals we know what a is we know what b is that's 4 minus 2 t and we know it's negative 2 plus t because d is just 1 and we know very simple because we chose this 0 is less than or equal to t is less than or equal to 1 so there we have our final solution so it's very easy actually to do these line segments as long in green i'll always have the definitions in green so we have the parametric equations for line segment a plus bd and c plus dt it's linear it's lines and we just going to go from t equals 0 to t equals 1 and it's very simple at least to solve for the a and c when t is 1 and it's almost as simple to solve for b and d when we make t equals 1 knowing already what a and c are so these are actually very simple problems to do problem 1.7 and your problems that they find the point or points of intersection of the curves and we can see that these curves are parametrically shown here parametric equations here t plus 3 and t squared and the other one is 1 plus s and 2 minus s so we see another parameter variable here s doesn't matter really what we use now think about this where two lines intersect two equations intersect two curves intersect at that point the x values are the same and the y values are the same so if we set this to be x1 and y1 and this one x2 and y2 x1 must equal x2 because at that point of intersection the two x values are the same and similarly for the two y values so we can really very quickly set up two equations and two unknowns because if we set x1 equals x2 so let's say x1 equals x2 if we were to do that we have the fact that t plus 3 t plus 3 equals 1 plus s and when we set y1 equal to y2 we're going to have the t squared equals 2 minus s so if two equations and two unknowns so this is solved for s up here that'll be the easiest so we're going to have s equals we take the negative one over to the other side that is t plus 2 and let's substitute that into our equation here as far as s is concerned so we're going to have t squared equals 2 minus and we have t plus 2 and if we just simplify this it's going to be 2 negative 2 0 and we're just going to have negative t and very simply here if we were just to go down here with t squared equals negative t that means t squared plus t equals 0 we're going to take t out and we have t plus 1 left that equals 0 so we have t equals 0 or t equals negative 1 so for those two values we're going to get our lines of intersection so let's just do this so it's nice and neat when you want to have a look at these let's just keep them nice and neat so what we're going to do now is just to see at t equals 0 so what happens at t equals 0 well at t equals 0 there's a substitute that in as far as these these two points are concerned so if t equals 0 0 plus 3 is 3 so I have 3 and then t squared is 0 squared and that's 0 and when we have t equals negative 1 what point are we going to have then well way up here we can see negative 1 plus 3 well that's just equal to 2 and negative 1 squared is 1 and there's our there we find our two points nothing more to do we can just solve for s and just make make absolutely sure that there was no mistake so let's just do that on the side here but that's our final solution those are two points of intersection so let's just solve let's just solve for s if we solve for s with t when t equals 0 and we plug it in there that means s equals 2 and if we plug in negative 1 we're going to get that s equals 1 and if we plug that those into those two into our second parametric equation there so 2 plus 1 is going to be 3 comma 2 minus 2 is 0 so I get the same point there that's not a problem and when s is 1 1 plus 1 is 2 and 2 minus 1 is 1 and I get exactly the same two points so that's not a problem there so all we did was here because we know what t is we just substitute 2 to t in there so that we could solve for s and we get exactly those two points the crux of the matter here of course is to realize that when curves intersect both of them will have to have the same x value and both will have to have the same y value and we do that we find these two points of intersection because we're setting up two equations in two unknowns so here we have a lovely problem problem 1.8 and the problem said they find the parametric equations for the ellipse x squared over 25 plus y squared over 144 equals 1 now you could solve just y get y on its own and set a parametric equation for x and then get for t but there's a much better way of doing it more eloquent and we give an hint hint actually there and when you see these problems always think of that hint and that's trigonometric identity so we have the most famous one there the easiest one sine squared t plus cosine squared t equals 1 and what we see here if we have a look at let's just do that x squared over 25 well 25 is just 5 squared and if we have y squared 144 well that's just 12 squared and that is 1 no problem there but because both enumerators and denominators are squared I could also have x over 5 all squared plus y over 12 all squared equals 1 and I have something squared plus something squared equals 1 which is what we have up here something squared plus something squared is 1 so what we could do we could just say x over 5 equals the sine of t because if we take the square of both sides we have x squared over 5 and sine squared and that leaves us with the fact that if we just solve for x on its own x equals 5 times the sine of let's put that in parentheses t and we also have that y over 12 well we can set that equal to the cosine of t and that means for us that we have y equals 12 times the cosine of t and you can clearly see that you know of course if we just let if we just make these squares that becomes a square if we make these squares it becomes a square but we just leave you know take the square root of both sides and this is what we left with and now it's very simple for us to write our parametric equation and as much as we're going to have x comma y and that's going to be 5 times the sine of t and we're going to have 12 times the cosine of t and we're not given for x we're not given any domain there so we're just going to have negative infinity is less than t is less than positive infinity and we have our solution there so if you saw the hint there you'll see that it's actually very simple to do and whenever you see problems like this please remember that you could probably use one or two parametric one of a couple of parametric equations to solve this problem this kind of problem another one of these problems find the parametric equations for the hyperbola x squared over a squared minus y squared over b squared equals one and you see the hints there you know that we're going to use trigonometric identities here and if it wasn't for the hints let's us do let's us do the most famous one sine squared of I'm going to use t because we're dealing with parametric equations plus cosine squared of t equals one and that's not what we want though because this is a this is an equation here for hyperbola and so we have minus y squared over b squared so that's not going to help us but it's very easy if you can't remember the one that contains secant and tangent it's just to divide both side by the cosine squared of t so if I divide this by cosine squared of t this by cosine squared of t and this by cosine squared of t what I'm going to be left with sine of a cosine is tangent so that's the tangent squared of t cosine squared of a cosine squared that's this one one over cosine squared is secant secant squared of t and all we all that's left to do is just to take the tangent over to the other side so we're going to have the secant squared of t minus the tangent squared of t equals one and that's exactly what we want because what we have here in essence is that's x over a all squared minus y over b all squared equals one so we have something squared minus something squared is one which is exactly what we have here with our trigonometric identity so very simple here we can say x equals x over a that is that's going to equal the secant of t remember as for the previous problem let's not bother to do the squares there because it's just another step and then y over b that's going to be the tangent of t which means we can very easily solve for x here x is going to be a times the secant of t and here we're going to have y equals b times the tangent of t which means that we have our parametric equations so let's just do this it's all nice and neat so there was step number one another step another step here's our final step and as much as we have our parametric equations x comma y and that's going to equal a times the secant let's make it look neat and then b times the tangent of t and again we weren't given any domain here so we're going to go from negative infinity is less than t is less than positive infinity so where we go in one step we learned we learned how to remember forgot to remember how to get to the secant and tangent squared trigonometric identities and you'll see the second hint there because there's this deeper connection between this hyperbola here and the hyperbolic trigonometric functions so remember that there is another trigonometric identity for hyperbolic trigonometric functions and that says that the hyperbolic cosine squared of a value minus the hyperbolic sine function squared that equals one so in that instance we can we would also have the fact that we have the square so we're going to have that x equals a times the cosh cosine hyperbolic cosine of t and then y was always going to then be b times the hyperbolic sine function of t and then we're going to have our parametric equations x comma y that is going to be a times cosh of t and b times the sinh three awkward pronunciations there and again we're going to have negative infinity is less than t is less than positive infinity so that was going to be the second trigonometric identity that we can use and we can now see start to see this this connection between the hyperbola which we had up here our hyperbola that we have up here and our trigonometric identity as far as hyperbolic trigonometric functions are concerned so there you go two solutions to this problem 1.9 let's go for problem number 10 find the parametric equations for x to the power two thirds plus y to the power two thirds equals eight to the power two thirds you can't read there those are two thirds now we certainly don't know trigonometric identity as far as these things are concerned until we realize that well there's something we can do about this problem and as much as we can rewrite this as x to the power a third all squared and we can write that as y to the power a third and we can write that as all squared and that equals eight to the power a third all squared and remember those are the properties of powers so there's no problem there and we can also divide both sides by the right hand side so we're going to have x to the power a third all squared and we're going to divide that by eight to the power a third all squared and we're going to have y to the power third this is all very laborious and I'm quite sure you can already see where we're going with this and that's eight to the power a third all squared and that equals one and we just back to sine squared of t plus cosine squared of t equals one because we have something squared plus something squared equals one so we have the fact that x to the power a third over let's just write the whole thing out then then it's going to be eight to the power third that is going to equal the sine of t because that's all squared and sine squared of t and from that we know now that x to the power a third x to the power third is going to be eight to the power a third sine of t and we can take that even further by you know taking the exponent to the to the power or taking what we have on the left hand side to the power three and on the right hand side to the power three that this leaves us with x that leaves us with a and that leaves us with a sine to the power three of t and we're going to have exactly the same here when it comes to y to the power a third over eight to the power a third well that's just got to equal cosine of t the cosine of t there that means we can just solve here for y to the power three that's going to be a to the power three cosine of t and then we just take both sides to the power three in other words y is going to equal a cosine to the power three t and we eventually have our parametric equations x comma y and that this is going to equal a sine to the power three of t and we're going to have a cosine to the power three of t and again we weren't given any domain there so we just can say negative infinity is less than t is less than positive infinity and that's it so it was very easy to change what we had way up there as far as our problem was concerned to just to change that into into something that we can use a trigonometric identity and all we had to use was just a very simple trigonometric identity so just remember here we actually just keep a step as much as this step here so let me just write it out actually what we have here is x to the power a third over eight to the power a third and these are both of them are squared so the whole thing is squared and remember that is equal to sine squared of t and we're just taking the square root of both sides and that's where we end up with this here we've just skipped a step and not done all squared I suppose we skipped another step where we can have the square of the numerator and denominator separately and then write it as such but remember all we're doing here is we're using the trigonometric identity sine squared of t plus cosine squared of t equals one as simple as that and that's what we have here summing squared plus summing squared that equals one and we can just simply equate those two equations the left hand side to the left hand side and the right hand side to the right hand side and then each little expression to itself the first one to sine and the second one to cosine very simple problem number 11 that's the last problem in this first set find the parametric equations for x squared plus y squared equals 4 y and of course our minds have got a jump to the fact that we can use a parametric trigonometric identities here but as the problem is stated now that will not be possible so let's just see if we can do something about it x squared plus y squared and let's bring it all to the one side minus 4 y equals 0 and we see the hint there that we can just complete the the square so look at this if we say x squared plus y squared minus 4 y if we do plus 4 and if we do it on this side we have to do it on the other side maintain equality there so if x squared plus and if I have y minus 2 squared that equals 2 squared on the right hand side look y minus 2 squared so that would be y squared minus 4 y plus 4 exactly what we want and now we can say x squared and we can divide both sides by 4 so that's 2 squared plus y minus 2 squared over 2 squared and 2 squared divided by 2 squared is 1 and lo and behold I have another situation where I can use the trigonometric identity sine squared plus cosine squared of t equals 1 because what I have here on the left hand side is something squared plus something squared so I'm going to have the fact that x squared divided by let's just put the 4 there that's going to equal the sine squared sine squared of t and if we were to just solve for that we've got squares on both sides in other words x is going to equal 2 times the sine of t remember if we take the square root of 4 that's 2 and we're also going to have the fact that y minus 2 let's just skip that step of squaring everything that is going to equal the cosine of t in other words if we go there it means y minus 2 equals 2 times the cosine of t and I've got to simplify that even further to get my parametric equations y equals twice the cosine of t plus 2 we've got to bring that 2 over to the other side and eventually I have my two parametric equations and as far as that is 2 times the sine of t and this one's going to be 2 times the cosine of t plus 2 and remember negative infinity is less than t is less than positive infinity so there we go very quick problem just got to realize that there's so many techniques that you've learned about before they are on your hedges just remember and you will remember them and it's just about you know when to pull them out which one to think about and it only comes from doing a lot of these problems so with the hint there that we had at least was just to complete the square and if we complete the square we just sat with a very simple trigonometric identity substitution so those were the 11 problems and the problem set as far as just working with parametric equations are concerned in the second step we're actually going to create curves and very excitingly we're going to use a computer language to create those curves far so that we can sort of develop a better intuition for what's going on and also I suppose as important give us the ability to play around with the values and and see how these curves come out other than doing it by hand so let's answer the next nine questions and that's where we actually have to create these curves we have to to draw them now certainly if you're on class for a test or exam you're going to perhaps be required to do this by hand and that is fun for the first couple of times and it gives you a little bit of intuition as to what's going on but I think it's much more useful to use a computer language to draw these curves and any kind of curve plots parametric curves as we're going to do your parametric equations or curves of parametric equations because you can play around because you just have to change one or two things in a line of code and you get a new curve and you can actually see before your eyes how these things develop and how they change based on the parameters that that or change in in the parametric equations themselves so I think it's just much easier to do a lot of these and if you do a lot of these you can start seeing patterns and and how these things differ so the language that we're going to use is just the Wolfram language I want you to go to wolframcloud.com because you're going to open a browser window in wolframcloud.com and you can just write the code or free of charge just use the wolfram language what I've got here and wolframcloud.com so I've logged in so you can just create a free account but I do have an account mine's called wolfram one and it's a personal account and it costs me a couple of cups of coffee a month that's it you know for what you get this is absolutely phenomenal and then it also gives you the ability to download the desktop version of the mathematical so that you can use the wolfram language on your own desktop so that's absolutely fantastic and so do that and then you can just create a new notebook because a notebook is what we write our code in I just want to show you around a little bit on the top here if you click on documentation you are going to see the wolfram language documentation that's like a massive thick textbook and you can see they've got core language data manipulation analysis visual visualisation and graphics machine learning symbolic and numerical computation high mathematical computation strings graphs images geometry sound and video knowledge representation and natural language time related computation scientific and medical data and computation the list goes on and on and you can just click on this and you know these things open up and you can see what a list is all about for instance all the functions that exist for lists you can click on one of them and it'll tell you how you know how this function works and a little bit of example code so you have a whole in the documentation here you have a whole textbook of how the wolfram language works that's fantastic if we click on language info we get fast intro for programmers elementary introduction book as Stephen Wolfram's book and elementary introduction course you can actually take and if you go for fast introduction for programmers it's a nice little online book that you can go through teach yourself now I've got three massive open online courses on the wolfram language online the first one is from course Sierra all the links will be in the description down below that I've done for the University of Cape Town and you can earn a certificate from the University of Cape Town for the completion of that course and that's all about healthcare well a bit more general but but also healthcare and life science statistics I've got two courses on udemy one specifically for healthcare statistics and using the wolfram language and the other one just a generic introduction to the use of the wolfram language and you can certainly check those out so let's go back and then we get to quick links so quick links just from some useful things there's a one minute video about using notebooks for instance education and training all sorts of resources anyway let's go to the wolfram cloud and what we have here right on the right hand side is the cloud files so there's all my little hard drive you can say in the cloud and I can just click there and I can create a new folder so if you were to create a new folder there's going to be folder one I can click on the little wheel there and say rename it and let's call it my folder and now I have a folder there my folder and I can go into that folder by double clicking on it and here I can create a new notebook or just with that little button or upload files if I have some files to upload like spreadsheet files we want to do some data analysis etc so let's just go home because I've saved everything in YouTube this is my course on vector calculus it's part number one parametric equations and inside of section one parametric equations part one is just this just parametric equations we've just done a bunch of example problems and the notebook I want to show you is parametric curves so if we double click on that that's what's going to open up so there we go this is what a notebook looks like and I've prepared everything for you there you can see this we're busy with problems two so 2.1 through 2.9 and we're going to generate those curves now you can see how beautiful this is a notebook you can see on the right hand side there these cells bigger cells with sub cells and sub sub cells etc and you can see what happens if you hover your cursor in between these it goes horizontal and if you click there a little plus sign opens there so you can enter some code in a text cell there free form input wolfram alpha which you've all heard about plain text or some other style and if you click on some other style you can select between a title sub title section subsection all the way down all sorts of things so this cell is a title cell and you can just start typing then there and this is a section cell subsection cell so you just create in between there you can just in insert little cells that you can use so this is not a course on the use of the wolfram language so please look at all these resources but I think even if you don't do any of those you're going to pick this up very very quickly so the way the wolfram language works it's a functional language so everything is done with a function and they're over 6000 functions I mean you could see here under documentate if we if we were to go here under the documentation there's really there they're very little inside of the stem and beyond you know that's not there higher mathematics we can look at polynomial algebra tensor algebra linear algebra discrete calculus neural networks probability theory random process processes discrete math number theory group theory this is all you can do you can do everything with with the language so what we are doing at the moment is just going to do some plotting so there's a plot function so plot and we have to pass the plot function some information so that you can know what to do those things are called parameters and they go inside of a set of square brackets so you see the opening and closing square brackets there and everything that is passed here they're called parameters as I said and we separate the parameters by a comma okay so the first parameter in other languages we call them arguments is what I want to plot I want to plot plot x squared now in most computer languages the carrot symbol there is is a power notation it's shift six on my keyboard most keyboard so x to the power two so x squared I want to plot x squared but I've got to tell the plot function over what what domain I want this so what x what values must x take so this is called a list because it's inside of curly braces and all the elements are separated by commas so I say for my x axis so my domain I want to go from negative three to positive three and if I hold down the shift key and hit return or enter that cell is going to be executed and I have a beautiful graph here this parabola y equals x squared now I can add some more parameters to this now we're not talking the parameters that we have in parametric equations parameters as far as a computer language are concerned let's do the following we still have x squared we still have this domain from negative three to three but now I want to put a label on on the top of my graph and for that we have an argument or a parameter called plot label and that little sign there is very easy to create you just type minus greater than and it'll be turned into this so let me let's do this by hand so you can just see at least one so we write plot and you see every every function inside of the Wolfram language that starts with plot is going to pop up there and you can just select from them so plot is the one we want anyway if you click that this little chevron sign it'll give you some hints as to what to do there you saw my function and then x going from x min to x max so definitely you can just do that and it sort of helps you out so that you can say x squared and I can hit I can go to my x there that's x and it's going to go from negative three and it's going to go to three but what I want to do I want to add plot label plot label and I'm just going to say minus greater than and immediately it changes into that little arrow and then inside of quotation marks because this is a string just some text I'm just going to use polynomial so let's execute that and now I have my plot but there's also polynomial there by the way there's some some things that open up at the bottom so I can change my background I can add a full I can change the image size and I can as far as the axis is concerned I can turn them off or turn them a different color and all that's going to do it's going to create a new line of code and it's going to add parameters to enact these things that you have chosen so let's not let's add something else and that this time around we're going to add a plot range and we see an outside set of curly braces and then two inside sets of curly braces this one is for the domain this one is for the range or the co-domain let's put it that way so I want the x-axis to go from negative five to five and the y-axis from negative one to ten so instead of this being done automatically I want to force how big this window is and there you go and you can see we have a bit more wiggle room here because we go down to negative one we go up to ten whereas by default you know just went slightly beyond and it went up to about there nine about so you can set this window in which it's plot which it plots and then we can go a step further so you still see my whole plot function here that we had before let me just highlight the whole one so there's the opening and closing you can see the highlighted bit there the opening closing square brackets and the plot function but it all becomes the first parameter in the manipulate function so there's the first parameter and comma and there's the second one so there's this let's just try and highlight it there's the second one so manipulate says I'm going to do everything same plot x squared and x go from negative three till omega and to get that omega is very easy you hit escape escape and it turns into the Greek letter omega they could also use o so I'm going to say on my x axis go from negative three to omega but this manipulate goes right to the end it says take omega and go from negative two to three so what's going to happen my x little window is going to go negative three to two negative three to negative so sorry negative three to negative two negative three to negative one negative three to zero negative three to one negative three to two negative three to three so it's just going to cycle between that start and end value so let me show you what happens when we do this use this manipulate function so I get this little w a slider up here so at the moment at first at the first go w is negative two so that w is now going to take the value negative two so I'm going to say plot x squared from minus three to minus two but as I slide this along as I make w omega I should say more and more I'm plotting more of or I'm inputting at least more of the domain of x and I can even click on that little plus symbol there so I can just go back and forth little steps it's taking steps of 0.5 you can actually set the step size but we won't worry about that now so that's in short just how to plot it's actually quite easy so let's get to these problems so what I'm going to do just increase the screen size a little bit there we go let's look at problem one so a 2.1 then on on your problem said we have a parametric equation which was 5 minus t squared and t over 2 so I'm going to put that inside of my parametric plot function here's my parametric plot function and I'm passing as a list because it's curly braces with a comment between two elements my two parametric equations and then I'm going to say my parameter in the problem the parameter the t goes from 0 to 4 but instead of going from 0 to 4 I'm going to go from 0 to omega and right here at the omega I'm going to go from just beyond 0 0.00 0.01 till 4 and I'm going to show you what's going to happen and then this plot range remember that's just to set my window the window size that I want to plot my domain and my co-domain so if we do this and that's why I say why this is so beautiful because now we can see what happens as as my parameter t changes because I'm going I'm letting t go from 0 to 0.01 so it's just at the beginning there by 5 and then 0.02 so it goes from 0 to 0.02 and eventually we're going to go all the way from 0 to 4 so let's just open this so that I can just click through it and you can see how the function starts growing and that's why we say the parameter allows us to move along the equation that we have so it's a to parameterize the function means I can move along the function and all we're doing here is just to show as as my t value grows from the beginning to the end so we have t going from 0 to 4 as we go from 0 to 4 we can see the this part of the plot for which we have a domain for this parameter the second one is is three cosine of t and three sine of t so you can see them there now in the warframe language if you leave if you leave a space you can see there's a space between the three and the c for cosine now cosine is uppercase c and then oa's uppercase because it's a function one of the 6000 plus functions and all the function starts with uppercase letters there so the space means multiplication so there is a little space there so you can't just say three and directly the c as we would if we write on paper you've got to leave that space alternatively you can wipe out the space and just put a star there shift eight on my keyboard so that's the same as just leaving as just leaving a space once we have a space there it also means multiplication so three times the cosine of t three times the sine of t and we're going from zero to pi once again i'm going from zero and i'm going to set this omega to go from 0.01 to pi so eventually t will go from zero to pi and i've got my little window there so that we can sort of start to see what happens as as we as our parameter goes from its start position to our end position so you can see what three times the cosine of t three times the sine of t so x equals three times cosine of t y equals three times the sine of t and i'm going from zero to pi and that's a half circle so that is the parametric equations for half circle with a radius of three and you can just very quickly change this to four for instance make that a four and make that a four and you'll see that the radius goes by four or you can take t and you can divide it by two and see what the effect is and that's what i mean it can help you very quickly see what these look like let's go to problem number 2.3 we have remember with the problems we had if we have a line segment that's a plus bt and then c plus dt so this is definitely going to be a line segment so if we plot that line segment and let's just go right to the end and you can see that it's going to be a line segment from a starting position to an end position of course we can see where we start and where we end by using the manipulate function problem 2.4 i have t squared minus three and t cubed minus t going from negative two to two as far as the parameter is concerned again i'm going to use the manipulate function so we can see how this develops over time let me just move down so you can see it's really a fascinating parametric parametric equations here look at that that's not a function that you'll just draw y equals x squared for instance now remember the vertical line tests so what we have here you know is not it's not really a function but it is something that we can express with parametric equations beautiful the next one is t cubed minus t and then y is t to the power of four minus two t squared plus two and we're going to go from minus two to two but then i'm starting at negative 1.99 so it's negative two to negative 1.99 and then eventually we work our way up to omega being two so the parameter t goes from negative two to two so let's see what this does as it goes as it goes as it goes as it goes interesting stuff and you can see as the parameter changes what happens to this function 2.6 we have three times the cosine of two t plus sine of five t three times the sine of two t plus cosine of five t we're going from negative four to four so let's have a look at what this complicated function would look like this parametric parametric equations would look like so let's have a look you can see it making a little start there let's go let's go let's go let's go so it's going to be this five pointed one two three four five six seven pointed star there we go it's beautiful in what you can create with these parametric equations just remember that there's a little space there between the three and the c there's a space between the two and the t there between the five and the t just to indicate that those are multiplications so just be aware of that now we wanted to see in problem 2.7 the effect of k and i'm still going to use manipulate but instead of k going from one value to the others i'm going to say k comma and then another little list so k is going to be one two three four five six seven and eight and i'll show you what that does that just gives us this little drop down box here to play with so when k is two that's what happens when k is three that's what happens when k is four this is what happens let's go to six and let's go to eight then you can play to your heart's content to see and this is this is very nice when it comes to working with say with with waves and vibrations for instance we're going to see that these kinds of parametric equations problem 2.8 that's the hyperbolic cosine x equals hyperbolic cosine of t and y equals hyperbolic sine of t and we're going from zero we're going from zero all the way to pi over two so let's have a look at what that looks like and let's go right to the end so we can see this and we can see when you go only go to pi over two it's this upper half of this hyperbola and that's why we have this hyperbolic functions trigonometric functions that's where it comes from now if we use the trigonometric identity that the cosine hyperbolic cosine squared of t minus hyperbolic sine of t both of them squared that's going to equal to one so we can actually solve for this and what we're going to get is the square root so there's the square root function of x squared minus one and negative the square root of x squared minus one so if we were to draw that you can see I'm just using the plot function and because I want to plot two things I'm going to put them inside of a list and separate them by a comma and there you go last problem 2.9 we wanted to look at eight times t minus eight times the sine of t and a minus eight times the cosine of t and we're going to go from negative four pi to positive four pi and we want to know you know what kind of function this is what kind of curve we have here and we can clearly see it's a cycloid and if a increases it's just going to increase the size of these things so that is very definitely a cycloid and in the next section we're going to work out the derivatives of these functions and we're going to ask what is at what slope does this cycloid meet the x-axis every time but we specifically just going to say here at x equals zero at what slope does this cycloid hit so we'll have to take a derivative and we'll have to take the derivative here and we're going to do it from the right hand side as we approach from the right hand side towards x equals zero because you can clearly see we have we don't have a continuous curve here we have this sharp cusp here so certainly we're going to have some problems as far as our derivative is concerned but you'll see I'll let the cut out the box the fact that it meets the x-axis right down in a perpendicular fractional orthogonal and as much as the slope is going to go way up to an infinity right there and that's how it meets the x-axis so that's clearly a cycloid so yes do it by hand but during the course of this course I'm going to show you how easy it is to use the Wolfram language you're probably going to stop doing a bit of parametric equations now just to check out some of these resources on wolframcloud.com just open a free account and start playing I'll put as I said the links in the description down below as to some of the courses and if you really want to take a easy nice introduction to it look at my course on Udemy which is an introduction to Mathematica or the Wolfram language