 It's early morning here in the farmland and I just before we start the homework I thought I'd just show you what it looks like. It's always good to come for early morning run So today's homework is all about taking derivatives of parametric equations And it is as easy as taking derivatives of a single variable function So in the description down below you'll find the homework set So you can just download it and try these for yourself first And then we're gonna go through each and every one of those problems In problem 1.1 we are to derive the equation for the first derivative of the parametric equation x comma y equals Some function x which is a function of t and y which is also a function of t So for any value t remember I'm going to be somewhere on this crazy curve here on the right hand side And what we're interested in really we want to take the derivative and we're gonna take the derivative at some point some constant point So let's stick to this little interval here and on this little interval here We can say that y is just a function of x so we're not writing it as a parametric equation here It's just going to be some function of x in the end and We write uppercase f there just to show that Make a difference from y equals f of x with a lowercase f Just to distinguish from what's going on here because in some books you might see that y equals some function of a function of t So we have to remember that this x that we see here it it is a function of t as we have up here It is a function of t. So we just have to remember that or in some books You might have the fact that we have x is x of t So what are we after we are after an equation dy dx So we want the first derivative of a parametric Equation it's as it's written here on the top right So let's do that here on the left hand side We're going to take the derivative with respect to t because at the moment we have functions of t of Y here on the left hand side and on the right hand side We take the first derivative with respect to time of this function f which is a function of x But we remember that x the x here is actually a function of t So if we wanted to really do this dy dt on the left hand side We're going to have to take the derivative of This outside function times the derivative of the inside function if I can use those words outside and inside For referring to the fact that this is nothing other than the chain rule that we have here So we've used just the prime notation there from Newton So let's just stick to Leibniz notation here. So f prime of x, but what is f prime of x? Well, there's f of x The prime and this is with respect to t. Well, that's going to be dy dx So why do y dx? Because at the at the moment we've just written it as it is here Why is some function of x and if you look at this little short segment here Just think back to normal algebra. Why is a function of x? That is what the curve is all about and then we parameterize the curve But on that little section it is so If prime of x, that's dy dx and x prime of t Well, that's dx dt because we're taking the derivative with respect to t of both sides And then by simple algebra and I'm going to put it in green because it's something we should always remember So that's our definition here of the first derivative So I have the fact that dy dx, which is what we were after. That's dy dt over dx dt And that's our final solution. That's our derivation of the first derivative This equation for the first derivative of a parametric equation So very simply we're going to look here on the right hand side and we're just going to take The derivative with respect to t of y and we're going to put that over the derivative of x with respect to t in our problem sets So problem number two we've got the first derivative So let's just write that down. We're going to do green here So we have the y dx and that's equal to dy dt over dx dt Very simple derivation there and now let's be clear about what we are asked here We are asked for the second derivative d squared y over dx squared. That's what we asked for So we have this on the right hand side our definition here in green So let's just take the d dx of both sides So I'm going to take d dx of dy dx And on the right hand side I'm also taking the d dx of dy dt over dx dt So that's what we have there. So let's be clear about this. Let's do something about this and I'm just going to multiply Both the numerator and denominator by the same thing And this thing that I'm going to multiply it by is 1 over dt divided by 1 over dt So that's it and I'm doing that to this but here. So where are we left with that? Well, that's going to equal the d dt over dx dt So that's all I've done and I still have dy dt over dx dt So nothing has changed there On the left hand side, I have what I desire d squared y over dx squared, the second derivative And now, well, let's look at the numerator and denominator of what we have here And we see in the denominator. I have dx dt and another dx dt Let's bring one of them to the numerator. So what I'm going to have is d dt I'm going to have a dy dt And I'm going to have a dt dx And in the denominator, let me get something to make a straighter line here So here So straight I suppose and I still have 1 dx dt in the denominator And what we see here is we can cancel some things out Choose hello there that dt and that dt can go So what we left left with is very simply the d dt Of well, what remains is dy dx over dx dt and there we have it this the this equation for the second derivative Now if you think about it to this dy dt that we have here this dy dx That is a function of t when we take the first derivative of This parametric equation we're going to be left with a function of t and we're taking d dt of a function of of a function of t and in the denominator dx dt Well, we know what x is and it is a function of t there so we can take that as well So let's put this in green because this is something that we should remember d squared y over dx squared and That's going to be the d dt the d dt of The first derivative in our first derivative is dy dx and we just going to divide that by dx dt and Once we see a problem or two it is actually it's it's a matter of just taking two derivatives So we asked to find the first derivative of this parametric equation x comma y that is our cosine of t and our sine of t Let's put it in green because we always put our definitions in green and Remember we have dy dx. That's our first derivative That is going to be equal dy dt first derivative of y though respect to t and first derivative of x with respect to t And that is as far as these parametric equations are concerned So let's just do that. Let's just do dy dx And we have to look at what dy dt is while we have Why they are sine of t and that's it's going to be r times the cosine of t and The x dt and with x dt being our cosine of t So we're going to have negative r Sine of t and we can simplify that by getting rid of the r both in the numerator and denominator so we have this minus the cosine of t divided by the sine of t and That is the first derivative although we can do slightly better. Let's just draw. Let's just draw a little Triangle here on the left-hand side. Let's just do this if I have if I have this that That and I remember that this is my x this is my y and then that means that is going to be r the The radio the the hypotenuse they are so if I'm looking at the cosine of t If this is my angle with respect to what I'm doing, so if I take the cosine Remember that is adjacent of a hypotenuse So that's going to be x over r and remember we have a minus there and the sine is Opposite divided by j synth opposite divided by hypotenuse y over r and that's just going to be minus x over y and That is my first derivative if you want to express it in a slightly different form just remembering that when we do Work with cosine and sine when it comes to these functions that we do we are talking about functions of some angle and We might as well tear remember we always we say tears time And that's usually the problems that we're dealing with but it's not necessarily time This is the parameter and in our instance here. It is this angle here And if we have the two sides and hypotenuse well, we can express it this way Problem number four one point four finally second derivative of the in the same equation so we know what the first derivative was and Let's just jot it down. So we just remember what the first derivative was We had the fact that the ydx and I'm just going to stick with a trigonometric Solution that we had here and remember that was the cosine of t Over the sine of t and there was a negative there So as I said the first derivative is a function of t because now if we remember the equations for the second derivative So let's just write it here d squared y over dx squared That's going to be the d dt Do dt there we go. Let's just get rid of that little mark there There we go. So that's going to be the d dt of dy dx And I've got to divide that by dx dt that is the equation that we derive So we have d squared y over dx squared. That's what we're looking for. So we've got to take the d dt of This function here. So the d dt so d dt of Negative the cosine of t divided by the sine of t So we have to take the derivative here of a of a quotient and So we have to remember little rules there if we wanted to do that We've got to choose one as u and one as v and remember that's going to be will not choose u as the Numerator and v is the denominator. So we're going to have this u prime v minus u v prime over v squared So let's just do that on the side here. So u where shall we do it down here u is going to equal the Negative the cosine is put the negative up there t and v is going to be the sine of t So I'm just doing it the long way, of course, you can just go ahead and do it and much quicker way So cosine of t its derivative is negative of the sine of t But we have the negative there So that's just going to be the sine of t and v prime is going to be the cosine of t and v squared is going to equal sine squared of t So if we were to do this derivative But before we forget remember that we have a denominator here as well And that is going to be the first derivative of the x dt and the x The x of t is this r cosine of t. So remember here we're going to have negative r sine of t in the denominator So d squared y over dx squared that is going to equal so we need u prime v minus u v prime So u prime the sine of t so you know sine of t multiplied by u prime v That's another sine of t and we're going to have minus now. We have u. That's negative the cosine of t times V prime that's the cosine of t There we go, and that's going to be divided by that has to be divided by This v squared and v squared is sine squared of t, but we already have that in the denominator So we're going to have the negative r there and we already have a sine of t there now another sine of t Well, let's try to write it out. So we don't get confused and then we have a not equals multiplied by Sine squared of t so we have d squared y over dx squared So we have sine squared of t sine squared of t and Negative times negative positive plus cosine squared of t and we know that's a trigonometric identity And that's just going to give us one and that's divided by negative r times sine cubed of t So what we're going to have here is negative one over sine cubed of t and Let's just do our little drawing again just to remember that we can express this in terms of x y and r Let's do that and you can see that I Just go back there and just remember that there still has to be an r there So let's not forget about that one and here we have x here. We have y here We have r and here's our angle So if I'm looking at the sine of t The sine of t remember that is opposite divided by our partner news. So that's y over r So if I were to rewrite this, let's go on d squared y over dx squared We have the negative one there We have the r there and then we have One over let's just do this r cubed. So that's going to be y cubed over r cubed And if I were to do that, let's use a different color So let's again that r is going to go and this one is going to be 2 Let's go back to the gray. There we go. So what we have here is still a negative and Because this is in the denominator if I bring it to the numerator That's going to be r squared over y cubed if you wanted to If you wanted to look at it like this negative r squared over y cubed the second derivative So it's actually a nice problem because you've got to remember how to do this derivative here You've got to remember how to do that derivative and then it's just like simple I mean one other thing we had to do is just remember the trigonometric identity that we have here Well, I should say here we have the trigonometric identity and the fact that We can just express this what we have here just in terms of this little and this little a triangle Right triangle that we have there so we can express because remember our problem initially Was given had an r in there so we can't express it this way But it's I think it's quite okay for me if you leave it in that in that format the answer Question 1.5 find the first derivative of another function parametric function tq plus t and t to the power of 7 Plus t plus 1 so this is going to be fairly simple And what this really is is this just reminding us how to do how to do derivatives So let's kick off If you start with these always try and remember your equation So we have the fact that d y dx that is going to be equal dy dt if you write this often enough It becomes second nature and that's what you want. You want unconscious competence So there we are dy dx is dy dt dy dt over dx dt And we have a function x of t and we have a function y of t So this is not a problem the first derivatives should never be a problem So we have d y dx and if we take the first derivative of y with respect to t We're going to be left with 7 t to the power 6 Plus 1 so that's in the numerator and if we look at the denominator the first derivative of tq plus t That's going to be 3 t squared plus 1 and there we have it I mean that's really simple the first derivative of this parametric equation So in problem 1.6, we've still got the same parametric equations there and we asked us to get the second derivative So let's just remind ourselves as always. How do we get the second derivative? Well, that's going to be a d squared y over dx squared. That's what we after we've derived this that's the d dt of dy dx and remember dy dx is now a function of t and Let me get my little ruler out on the screen. There we go. And that's going to be dx dt So that's what we after and we have to remember of course what our first derivative was because we need the first derivative right there So we can remember from the previous problem dy dx. That was simply going to be 7 t to the power 6 plus 1 over 3 t squared plus 1 So again, this is just a good exercise just to remember how to do this derivative because you have to take the derivative of this with respect to t So we're going to have you equals 7 t to the power 6 by all means Just do this in your head if you're comfortable with that. I'm just going to write it out here 3 t squared plus 1 So you prime that's going to be 42 to t to the power 5 and v prime is going to be 6 t and v squared Well, that's going to be 3 t squared plus 1 all squared And we remember we have to have u prime v minus u v prime over v squared That is how we do the quotient rule. So let's just apply that and so what we now want is D squared y over dx squared. So remember we have to use this Definition that we have in green there. So it's d dt of what we have Over here. So we just take the d dt of this first derivative of ours. So u prime is 42 t to the power 5 Times v and there we have v. That's the 3 t squared plus 1 minus u u is 7 t to the power 6 plus 1 and V prime is 6 t and we have to divide this by All the way by 3 t squared plus 1 All squared, but remember all of this is over dx dt So we still have to multiply all of this or divide all of this by dx dt. So well, let's not do that. Let's Stick to convention here. So let's just get rid of that. There we go. So all of this It's still going to be over the x dt and the x dt remember is just 3 t squared plus 1 So where are we d squared y over dx squared dx squared and now it becomes this laborious task of just simplification and Well, first we just have to multiply out and then we can expand. So let's just do that on the right-hand side Yeah, so so you 42 We've got to take 42 and multiply that by 3 and that gives us a hundred and twenty six So that's a hundred and twenty six t to the power seven If we look at 42 times 3 and t to the power 5 times t squared Plus 42 t to the power 5 and then we have a negative there 7 times 6 That's 42 t to the power 7 and then a negative again because we have the little negative there So it's negative 6 t 6 t and now let's just write that out all over the same denominator Because as you can see there, we will have 3 t squared plus 1 we've squared that and we've got another one So now we have it cubed So if we simplify this a hundred and twenty six T to the power we are with t to the power 7 and we have to subtract from that 42 and that leaves us with 84. So that's 84 t to the power 7 and We also have then a plus 42 T to the power 5 and we have a negative 6 t negative 6 t and we're going to put that all over And we can put that all over 3 t squared plus 1 Cubed now, suppose we can simplify that because we can take a 60 out at the in the numerator At least we can take 60 out and then we left with what is 84 divided by 6 that gives us 14 So that's 14 t to the power 6 plus 7 t to the power 4 minus 1 and that's still all over 3 t squared plus 1 or cubed so you see these are nice little exercises Little mental exercises see You know if you can simplify it any further, I suppose this is a this is a an appropriate way to leave to leave the solution So very simple there other than we have to remember how to do the quotient rule And don't forget the DXDT in the denominator here because it's very simple Just to do the numerator DDT of the first derivative And then forget that you still have the denominator that you just have to add to that So do a couple of these problems. I mean you can just come up with your own just write two polynomials there With respect to the parameter as far as x and y is concerned to create these these equations that play around and do a couple of them Just so that you can get back into the swing because usually you know We haven't done this these kind of derivatives for for a while It's just easy to try to remember, you know how to do them by just doing a bunch of bunch of Exercises so that was it the second derivative of this equation So this is a very interesting problem and as much as it's related to physics Remember in physics we take a derivative with respect to time and for for many of the problems in when it comes to motion and We use dot notation which just means with respect take the derivative with respect to time So we also get the first derivative of the parametric equation And we see that this is the generic function x of t and y of t and just to remind ourselves What is dy dx dy dx? That's going to be dy dt Over dx dt So we just keep that in mind So what we have here in essence is this y prime over x prime And I've just done I suppose I've just done the solution here because if we take the first derivative of this Generic function parametric function that is going to be dy dt over dx dt and in dot notation that is x dot Which what am I doing here? It is going to be there we go y dot over x dot Because just a different a different notation that we use So now we have to find the second derivative in problem 1.8 We've just had the first derivative this as always we can to remind ourselves that dy x dy dx equals dy dt over dx dt What an ugly d there we go dy dx and we've had our first Solution here or as far as the first derivative is concerned. That's going to be y dot over x dot There we go now we have to find d squared y over dx squared and We have the quotient rule again and remember. Let's just put the quotient rule out here again That's u prime v minus u v prime over And we have v squared We just have to remember that that's just the derivative of the numerator We still have to do dx dt and the denominator. So let's just never forget to do that so let's just have this upstairs and Upstairs meaning the numerator. So if we have let's just write it out And so we're all clear about it. So we're gonna have u equals y dot and we're going to have v equals x dot So u prime is going to be y double dot. I'm taking the second derivative of there Let's make those dots a bit clear and we're gonna have v prime equals x double dot as well And we have v squared equals x dot squared So now that we have all of those so remember this is just for the numerator You prime v so that's going to be y double dot You prime v and v is x dot Minus u v prime. So that's y dot x double dot and that we've got to divide that by x dot squared So that is divided by x dot squared. But remember that is just the derivative That's just this part. We still have to do this part. So all of that goes over and The x dt. Well, that's just x dot and now we can just clean things up because it looks a bit Untidy and so it's going to be y double dot x dot minus y dot x double dot Double dot there and we're just going to have as the den as the denominator x dot Cubed three of them and that's the solution and remember the dot just means with respect to t So that's the second derivative of y with respect to t times the first derivative of x with respect to t Minus so they the minus there and then y with respect to t Second derivative of x with respect to t all divided by the first derivative of x with respect to t all cubed So we can do a lot of these problems as polynomials and transcendental functions take the first and second derivative but we've got to do something a bit more interesting and This is more the kind of thing that you're probably going to be asked to know about So find the angle at which the cycloid expressed as parametric equations x comma y equals a time a t So let's just make this a little bit more legible a t minus a times the sine of t comma a minus a times the code cosine of t where it meets the x-axis at the origin So if you think back about with the previous Problem said that we have when we just introduced parametric equations We had this idea of the cycloid and what happened with the cycloid is right here when when we asked The orphan language just to use the word from language just to draw these for us a dip down here We're at the origin so we've got to find out what angle there. Well, definitely We have a problem here because the function of continuous at the origin and we'll have to look at it from one side or the other And the angle at which it come in something that can help us there, of course is what is the slope at That point what is the slope at that point? So let's have a look at it now as always we've got to remember that d y dx always our definitions in green here That's d y dt over dx dt and Let's just do that so d y dx and As far as the function y is concerned we have a which is just a constant So the derivative of constant zero We have a negative a cosine of t the derivative of cosine of t with respect to t Well, that's negative the sine of t, but we have a negative there. So that's going to be positive We have a times the sine of t in the numerator Let's have a look at the denominator, which is x dx dt so a t there so we're going to be left with an a and Minus a times the cosine of t first derivative of the sine of t with respect to t that's the cosine of t and We can see that we can get rid of all these a's that we have in as much as if we take the denominator and we just take a out as a Common factor there that's going to be one minus the cosine of t And then those can go and we set with the sine of t over one minus the cosine of t so that is a function Single variable function. It's not a parametric equation anymore dy dx equals This function sine of t minus one over the cosine of t that is our slope at and now we just need to find out that slope And we can look at it and I remember if we if we remember back when t is zero That's exactly well. That's exactly this point here So let's just do that Let's take the limit to find out the limit as t goes to zero But we'll have to come at it from the one side because there's no continuity there. So let's just do the limit as t goes to zero from from the top side of the sine of t over One minus the cosine of t So just finding out what the value is going to be of the slope as we approach that point and of course Sine of zero zero and cosine of zero is one one minus one zero and we have indeterminate form So we just check for the use of this check the the assumptions for the use of L'Hopital's rule So let's just do that. So we're going to use L'Hopital's rule There we go. So that becomes the limit as t goes to Zero from the positive side. So let's do the derivative of sine of t. Well, that's just the cosine of t and If we do this one is a constant So we have negative the derivative of negative cosine of t with respect to t. Well, that's just the sine of t And if we do that as t goes to zero And so the cosine of that that's going to be one over and we approaching zero from the positive side And if we do that, well, we know that's infinity So if the slope as we approaching t equals zero from the right-hand side as we approach that point Slope becomes infinite of that line and an infinite slope. Remember, that's a vertical line So we can say that the cycloid at least as far as this origin is concerned It meets the x-axis at the origin Vertically this is vertically down when we get there we get there if you see if you go from that side as well It's going to be the same. It's going to be Negative infinity, but that's still a straight line. So it meets the cycloid meets that origin there at Vertically at least so although we back to just this These transcendental functions here at least there's a bit of a twisted We asked to get the slope of this parametric curve at a specific value for the parameter at t equals one so as always remember dy dx that's going to equal dy dt Over dx dt. So we always keep that on the right-hand side just to remind ourselves and let's solve the problem So we're going to have dy dx dy dx and Let's take the first derivative of y with respect to t So that's going to be one plus e to the power t and in the denominator Something slightly more exotic. I suppose that's going to be five t to the power four and We have sine. So that's going to be Something the inside function remember that's it's going to be a two pi as far as that's concerned But we still have the cosine of Two pi t There we go and all we need to do now is take this derivative when t equals one. So the y dx When t equals one and we just have to substitute So that's going to be one plus e as far as the numerator is concerned and we have a five there Well, when we plug in t equals one and we're looking at the cosine we're looking at the cosine function of of Two pi t so t is one. So it's the cosine of we're looking for is two pi and If if you think about that it just brings us back to one because No, it has a period of two pi. So that's just going to be one So we just left with two pi and there we go. We have an actual value We have an actual real number there for the slope of this curve at the point Where the parameter equals one So we asked to find the slope problem on 11 find the slope of the curve We have the parametric equations there, but this time not at a specific value for the parameter But at a point so we'll have to find out what t is at that point. So let's just start off Let's just start off by doing that. So This remember is a function. This is x and this is y and at this point x is one and y is one So we can write that t squared plus e to the power t that's going to equal one And we also have t plus e to the power t is equal to one So we see both right-hand sides are one so we can just set the two left-hand sides equal to each other so we have t squared plus e to the power t that equals t plus e to the power t and Very simply those two can go and I have the fact of that Where's my blue? We have t squared We have t squared minus t equals zero and I can take out t and t minus one equals zero So from that we have t equals zero or t equals one Now just think about it We had a square in there So we just have to be quite careful if we plug in t equals one Into x and y we don't get the point one comma one do we? We get one plus e to the power t comma one plus e to the power t. That's not what we want So we definitely want when t equals zero So we know that t equals zero because we plug in zero we get one comma one as the point So as always what is dy dx dy dx that is even equal dy dt if you write it down so many times You're never going to forget it dx dt now, that's not too pretty but you know what I'm You know by now what I'm talking about so let's do that dy dx So we dy dx that's going to be one plus e to the power t divided by Let's just do that's two t plus e to the power t But what do you want the y dx when t equals zero so we just substitute that in One plus e to the power zero. That's this one two times zero zero e to the power zero is one So that's just two so the slope at that point is going to equal two So we're going to get towards finding these horizontal and tangent and vertical tangent lines But let's have a look at this problem Find the coordinates of the highest point on the parametric curve and we see the 96t and then 96t minus 16 squared. Let's have a look at what the plot looks like And there we go 96t and then 96 times t minus 16t squared And it goes up and with the negative 16t squared there we could really think that this is going to be a parabola of pointing down And here about 280, 290 somewhere there On the x-axis we can see the maximum And about 140 somewhere out there we can see where that maximum is going to go as t goes to in this instance I'm letting t go from one to five so Let's remind ourselves because the more you write this down the better it gets So we have dy dx and that's going to equal dy dt Over dx dt So let's have that Let's have that dy dx So the slope we're looking for the slope here and when it comes to y we're going to have 96 minus 32t And in the denominator dx dt we're just going to have 96 and if we simplify this a little bit that's going to be one minus a third t Now at a maximum remember we want the slope to be zero So we can certainly set that one minus a third t equals zero our slope is zero At a low at a local maximum or local minimum So We can solve for t here So we get t equals Three that's going to happen at t equals three and if we look at the point there That's going to be 96 times three And we're going to have 96 times three minus 16 And three squared is nine And I think if you look at that point that's going to be 288 comma 144 just check on the arithmetic there And that's really when we saw the graph that was round about that spot Now we can see from the graph that it's going to be a maximum and the question I asked for the highest point of that graph and with a negative 16t squared the negative there We know it's a parabola facing down and but really to to know that it is Maximum you've got to do the second derivative as well But seeing we we've plotted the graph we know that that is going to be the point the highest point on this parametric curve So the next problem and these become a bit more interesting once again We're going to find the point or points on the curve and we see the parametric curve there Two cosine of t plus sine of 2t and two sine of t plus cosine of 2t Where there is or there are horizontal tangent lines so Let's uh Let's have a look at that curve. So I've Drawn it here. Let me draw drag it on the screen And there we can see parametric curve in the first video We looked at how to do this in the Wolfram language. So I'm just going to drag our t here And we see what's happening We see what's happening there and if we look at this curve here Definitely here where We x is zero and y is one there we find that there is a We find that there is a horizontal tangent line And if we open this up, it looks like it's round about point about 1.5 1.6 there about Definitely 1.6 there about that is just pi over 2 It's around about pi over 2 there. So let's just see how far we get with the problem So there's two things that we have to know about horizontal and about Vertical tangent lines to these parametric curves. So let's start with horizontal horizontal tangent line so we know from just Single variable calculus that it horizontal. So when the slope is zero We with the slope is zero. We have horizontal Tangent line And if we're looking at a fraction that means the numerator The numerator that must be zero while the denominator denominator Is not equal to zero So definitely That is where we're going to do this because it's going to be zero divided by non zero value and that's zero So our tangent line is horizontal or the slope of that tangent line is zero So with that being said, let's look at the derivative So that's dy dx and i'm not going to write the equation for That anymore. We know it's dy dt over dx dt. So let's get dy dt So that is we two sine of t we have there So that's going to be two times the cosine of t And then we're going to have minus two times the sine of two t That is when we have the derivative of y there with respect to t and then if we do If we do x here with respect to t we're going to get negative two times the sine of t And we're going to get plus two times the cosine of two times t So we want to know where this numerator is going to be zero So when is two cosine t minus two times the sine of two t when is that zero? So we can set that equal to zero And let's do that if we have two cosine Two times the cosine of t minus two times the sine of two t equals zero Well, we can just take two out as a common factor And we left with the cosine of t must equal the sine of two t While the denominator cannot be zero and if we just simplify the denominator That means the sine of t the sine of t cannot equal the cosine of two t So that's what we that's what we need to do So we're going to find these values and it's not obviously we can't solve for t there So I'm just going to show you again with the Wolfram language. We're going to do the solve function there So I'm going to say solve when is cosine of t equal equal So you see the double equal sign there. That's a Boolean It's a Boolean logic there. You say is this true or false. So when is this true? And we're going to solve comma for t and we see there all these solutions It's when t equals negative pi over two plus two pi times some constant So that whole series Pi over six plus two pi times some constant and these constants constants must be elements of the integers A pi over two plus two pi and five times pi over six And what I've done here, I've created a computer variable. I've called a numerator And I've done the union of all of these four these four series And the way Sequences I should say The way that we've done that is I've used the table function and I say negative pi over two remember When you use this on the Wolfram cloud just use the ford slash Pi ford slash two to do this division. I'm using the desktop version here. So I can do a bit of fancy mathematical notation plus two pi in and n must be zero one two three four So I'm just going to cycle of all of those Then the next table pi over six plus two pi in and remember to get multiplication I put little spaces behind these values So there's a space between two and pi in between pi and n And you can see I do all of that and eventually I get all these values for which we will get zero in the numerator But I also want to see where the denominator zero because we can't have that So if I go to denominator zero, that's my computer variable And it's the union of all of these and I've done this solve function there as well I see that there's all the values at least the first bit of the values I shouldn't call this sequence or sets See of course just call them a set So this is just the part of an infinite set So the the important part is that I've got to have values for which I'll have a zero in the numerator But I can't have the zero in the denominator. So we're only interested in pi from or t going from zero upwards So we see that pi over six Will give me the numerator of zero, but it'll also give me a denominator of zero. So we can't have that So we move on there's pi over two That'll give me zero in the numerator Pi over two is not in the denominator there So definitely as we saw on the graph above when t is pi over two We are going to get to that horizontal line at zero comma one So definitely pi over two is something that we can choose. So let's do that And we're going to say t equals pi over two We say t is pi over two What is going to happen to dy dx? So dy dx At t equals pi over two If we were to solve for that the numerator is going to be zero The denominator is going to be Some constant And that constant is not equal to zero Therefore my slope is equal to zero So my slope is there is equal to zero And we just have to find on the line where this is and we saw on the graph that it's at zero comma comma one So let's just do that though. So we're going to say x comma y So that's going to be at two cosine of pi over two plus the sine of Two times pi over two. This is pi and then two times the sine of pi over two And plus the cosine of pi And if you solve for that, you're going to get zero comma one So zero comma one we are going to have a horizontal tangent line. So let's just bring up The graph again, and that's our graph at the top and right there and we see it's about You know 1.5 there about which is which is a half pi We get to this horizontal line Now in the next question, we're going to be asked to get Where there's a vertical tangent line at a vertical tangent line. Remember that means the slope is Infinite it's straight up infinite and it certainly looks like here at Zero comma negative three That it is going to meet this y-axis at a tangent at a tangent line that has that is infinite or vertical slope But because we don't have a derivative there, we must be careful because there's a cusp there So we'll have to approach it and let's approach it from the left hand side Um, and let's just see where that value is It's round about there at about 4.7 there abouts And what would that be? Well, we'll see in the next problem. So here we are Problem number 1.14 find the point or points on the curve x comma y. It's the same one as before Where there is a vertical tangent line So if we bring up the off-ramp language again And we saw the dry day about 4.7 there abouts Something might be happening there and that's 3 pi over 2 if you take 3 pi over 2 Land roughly over there and if you put substitute 3 times pi over 2 into our parametric equations, you'll see we get to this point 0 comma negative 3 So certainly something is happening at that value. So let's have a look. So what we want here is first of all is let's just do Our equation there for the derivative dy dx. Remember that was going to be Uh, two times the cosine of t and that's going to be minus two times the cosine of 2t Where did my ruler go? There we go. And that's all over dx dt And dx dt is going to be negative two times the sine of t And then we're going to get twice the cosine of 2t So we want here For vertical asymptotes. So let's look at vertical I should say vertical tangent lines So for vertical tangent lines, we remember from single variable calculus, the slope must be infinite So the first derivative is infinite. We have a vertical tangent line and we're going to get that when the numerator numerator If we have a fraction for the slope that cannot be zero, but the denominator That must Uh, approach zero. Let's put it that. So if I have any non-zero value and divide it in a limit As the limit approaches zero In the denominator there, then of course we get infinity. So that's what we've got to do. So let's bring the warfram language back So this time round we want values where The denominator which we've got here where the denominator Is got to be have b zero But the numerator cannot and we can run through all of these i've done the union there as well negative pi over two Certainly it's up there pi over six. It's up there five times where five times Pi over six five times pi over six is there Three times pi over two. It's there 13 pi over six is there. They're all there And 17 is there seven over two is there So you've got a bit of a problem But we saw in the graph that something special happens at three pi over two So even though three pi over two is here and it is there. So i'm going to get zero over zero But remember if we if we have this limit And we have zero over zero that's in determinate form and we can use l'apital's rule So let's have a look. Let's just have a look at what happens at this point t equals Three pi over two and we saw that that was zero comma negative three. So let's look in the limit And we're going to let t approach three pi over two from the left hand side of this Two times the cosine of t because remember this is the slope First derivative is the slope minus two times cosine of two t divided by negative two times the sine of t plus two times the cosine of two t Well, we saw that that was going to be zero over zero if we just the direct substitution but You know the denominator certainly We were thinking there while it's approaching zero from the left hand side All we now need is for the numerator not to be zero. So we have zero of zero. That's Indeterminate form so we can use l'apital's rule There we go. Let's use his rule And so we're going to take the derivative of the numerator and the derivative of the denominator And there are some assumptions for l'apital's rule. Remember those assumptions from single variable calculus. Don't forget them So we're going to have the limit as t approaches three pi over two from the negative side And let's take the derivative in the numerator And so that's going to be negative two times the sine of t And that's going to be negative four times the cosine of two t over And what do we have? We have a negative two cosine of t And we're going to have negative four times the sine of two t And what is that going to be when t approaches? Three pi over two. So once again Wolfram language just to make things easier And here we have it. We have now just simplified I've taken negative two out as a common factor divided both numerator and denominator by that so that I'm just left with this Sine of three pi over two plus two times cosine of three pi over two times two And that gives us negative three in the numerator now and zero in the denominator So what we have here is negative three over zero, but we're approaching zero from the negative side And that's a vertical slope So definitely here at t equals three pi over two We get a vertical slope and all that's left to do is we must work out this point x comma y x comma y Let's do that x comma y where Where t equals three pi over two so if we can do that And if you substitute that in there and in there indeed, you're going to get the point You're going to get the point zero comma minus three If you do that So that's beautiful And it only only worked really because we had some idea of what this parameter curve looked like And of course when you're in the exam situation, this is going to be a difficult problem Because you'll have to sketch it out a little bit and that's going to take a lot of time and then realize what's going on The crux of the matter here though is to remember if you're looking for horizontal Tangent lines as far as the derivative is concerned. Remember we have dy dt over dx dt So we're always going to have this or at least even if it's over one We are going to have the fact that we have this fraction and we want the numerator to be zero and the denominator not to be zero And if we're looking for vertical tangent lines, we want the numerator to be Nonzero and the denominator to at least approach zero in the limit So that we can have this divided by zero. It's not quite zero It's a tiny tiny amount and in the limit that means that the slope goes to infinity and that's exactly what We want So in part two of this problem said we're just going to look at parametric curves as far as the area Concerned with these curves integration. So the problem 2.1 there reads Consider a single variable function f continues on an interval That's a less than equal to x less than equal to b. So this closed interval from a to b Just a single variable function y equals f of x Now derive an equation for the area enclosed by the parametric curve of this function Given that when we parametrize it the parameter goes from c to d And then with no intersections of this curve on this interval Because you know, we've seen some funny looking curves So we're not allowed any intersection of the curve so curve crossing over itself except perhaps for a Incidence start an endpoint and a clockwise traversal of that interval. So clockwise It's got to be clockwise and you'll see with when we do the problems What happens when we get to going counterclockwise? So i've just written in a shorter driving equation for the area enclosed by parametric curve And if you think we were told we we're just considering at the moment y equals the f of x So it's just some function y equals the f of x And x is on this closed interval going from a to b As simple as that and if you think what the area under the curve is When we come to integration So if we just have the area under the curve that's going to go from an integral Different integral going from a to b of f of x dx That's as simple as that And remember that f of x that's just y so we can probably write it something like this y dx Now we know when we parametrize these functions we get that if x of y that's going to be some Function of t and another function of t when we parametrize this x of t y of t Now let me look at what we have here. We have a dx there in our integral so if we take if we take for instance x of t And we take the derivative of that So we say dx dt dx dt Or remember just with newtonian Notation there. That's this x prime of t And I can say dx equals x prime of t dt But x prime of t if you wanted if your textbook does that differently Well, that's just dx dt dt nothing nothing other than that And we've parametrized this now So what we have in the end we have the different integral going from c to d We still have y there We still have y but y is a function of t remember And uh dx dt dt And there we have it So the integral going from c to d different integral of y dx dt dt And that will give us the area enclosed by a parametric curve In problem two we asked to find the area enclosed by the parametrized curve x comma y that's two cosine of t plus sine of two t comma two sine of t plus cosine of two t And t our parameter there goes from zero to two pi so that closed interval So what I want to do here and what you're going to find with these in some courses Um, it might be slightly easier than than these because they just want you to review your single variable calculus Can you do Can you do these derivatives and integrals when it comes to these trigonometric functions for instance or Other transcendental functions polynomials, of course much easier And can you just remember how to do these some courses? They're going to be quite Difficult to do and you might be allowed to use a computer algebra system Which is exactly what we're going to use what we're going to use here So by all means if you want to practice your single variable calculus Your differentiation integration, please go ahead and do that Yeah, though, we're going to use the Wolfram language Just to do these Motivate integrals for us just to save some time What we want to practice here is the fact though that we have to remember that we need to go Clockwise and then then the area will be given by this definite integral and going from c to d So we have our c and d there of y and then dx dt So that's y of t dx to t dt So if we look at it where we stand at the moment There we have x of t and there's our y of t and what we need to do is get the dx dt So let's just do dx dt at least dx dt And that's a simple enough one for us that won't waste a lot of time So that's going to be negative two times the sine of t Plus and we have twice the cosine of two t So I've just integrated this with respect to t And I've got to multiply that by the y of t So what we're going to have here is not minus the sine Twice the sine of t plus twice the cosine of two t And I'm going to multiply it by y of t. We have the x dt there and dy and just y of t So that's two times the sine of t plus the cosine of two t And then we're going to take the integral of all of that So it's going to be the integral from and there we have c and d there We are told so that goes the definite integral goes from 0 to 2 pi And we're going to have this negative two times the sine of t plus twice the cosine of two t Multiplied by twice the sine of t plus the cosine of two t and all of that dt So what I'm going to do I'm going to bring up the Wolfram language here So we're both going to just look at this definite integral But we're also going to look at this curve Because we are told here that this area is enclosed by this curve on this interval for our parameter Let's have a look at that So here we have the Wolfram language room. Remember I showed you before you can just open a free account on wolframcloud.com I'm using a desktop version here for which I'm paying pretty little money every month And what we can see here Again using the manipulate function because I want to manipulate The the graph to appear as the parameter changes from zero or just over zero We can see they're starting just over zero. So I have a starting point until two pi And there I have the The integral So let's do that and let's just have a look at what it looks like And it goes Like this as we go from zero to two pi So first of all, this is rewind. So if you think about it, we are going Counterclockwise here. So we have a problem as far as our function is concerned And what we do under these circumstances? Well, there are two things you can do the best thing to do is just to put a minus in front of your integral So if you're going to go counterclockwise on that interval your parameters interval just put a minus sign out in front of Of of your definite integral and here we have that definite integral and see I've put the negative sign up front there So I've just typed it in I've multiplied it out, but Just put it in and we see the solution Is two pi so by all means if you want to do If you want to do these integrals by hand go ahead But we're just going to save some time And remember there the negative sign and that's going to end up as two pi The definite integral and going from zero to two pi and closing that Special shape that we saw there. So for the next couple of problems We're going to do I'm still going to use the warframe language just to save a bit of time If you do do these in the warframe language, you'll see here what I have here. I've got an integral sign there I've got the squares up there And there's a writing assistant in the desktop version which allows me to write it like this and that looks very neat And again a very special reason to use the warframe language because you can make your work look look like a hundred notes And it's still going to perform that for you. Remember if you do this if you are going to do this on the web When you do these multiplications, remember just to put a little space there see at the end there cosine of two t There's a little space there a space indicates multiplication instead of the Square symbol there remember you can also just use You can also just use the carrot symbol shift six and then two that would also mean square But of course as I said here, we can do that We can do all of this with a special notation in the desktop version So go ahead. I challenge you to look up the code That will do the integration for you because I've used this special integration symbol here but of course in the In the cloud you'll have to write a function to do that one of the Keyword arguments inside of the warframe language and I challenge you to To find it and see if you also get two pi if you can write out this This integral in this problem We have to find the area enclosed by the ellipse x squared of a squared plus y squared over b squared equals one So we do recognize that as an ellipse and a couple of things We have to remember and the first problem said remember we saw that we can use trigonometric substitution here By using a trigonometric identity and I'm going to put it here So sine squared of t or theta or whatever Plus cosine squared of t equals one So that means we can let's put another color blue We can equate this to that and we can equate this to that and the one equals one In other words, if we were to do this x squared over a squared Equals sine squared of t We saw that was nothing other than x equals a times the sine of t because we can take Square root of both sides and take the a over to the other side So that's what we're going to have here. And of course for the same argument y is going to equal b times the cosine of t So what do we need to enclose this area? Remember we need some parameter So it's going to be the integral going from c to d of the y of t and then dx dt dt So another little thing that we would need here is just dx dt So let's have a look at that dx dt Given that x of t equals a sine of t. So that's going to be a times the cosine of t And what we'll have to do is a times the cosine of t And y of t is b b times the b times the cosine of t And so that's going to be a b cosine squared of t Now let's have a look in the Wolfram language if we just were to plot this function this parameter curve What it looks like so we can get some idea of what the interval is as far as the parameter is concerned So what we've done here is we've just put in a substitute a value for a and b So I made a three and b and b two So so that we can control this ellipse and what you've also noticed there is that my parameter t is going from zero to omega and i'm sitting omega going from zero point zero one To two to two pi. So in essence t is going to go from zero to two pi So let's have a look at what that looks like And again as we increase our parameter first of all noting that we are going clockwise. So that's good And if I open up that little plus sign there, we can see the actual value. So we're going round round round We're going round round round And we're ending up at about six point two eight there about that means two pi So really we're going from zero going from zero to two pi and we're going clockwise So that makes life very easy for us and as much as We're just going to do the integral and going from zero to two pi And what we have here is a b cosine square of t dt a b is they are constant so they can go outside So that's just the cosine square of t dt And if you remember your single variable calculus, that's going to be we still have the a b there And then we're going to have t over two Plus a quarter sign of two t And we've got to go from zero to two pi From zero to two pi and if you think you know, you can just get that easily Of course just type that into the word from language and it's going to give you the solution So what are we going to have is a b And then two pi we're going to have two pi over two Plus a quarter sign of four pi And we have to subtract From this so let's just do that. We're going to subtract Do that and subtract from that just putting zeros in but the sign of zero that's zero And the zero over two that's also zero So nothing is going to happen there. So I'm left with a b and yeah, we just have pi plus And if we take the sign of four pi, that's the zero So that's going to be a zero and we left with the equation that we know for the surface area of an ellipse That's a b times pi And there you go. It works out to what we know Or at least at school. That's what we we were we memorized as far as the area of an ellipse is concerned And now you can see using Parametric curve in the equation for this area enclosed By this curve on this an interval for the parameter that we do get to the same the same result So there we go problem 2.4 Find the area enclosed by this parametric curve So it's a half cosine of t minus a quarter cosine of two t And a half sine of t minus a quarter sine of two t doesn't give us the interval. So, you know, you'll have to figure that one out Usually these things are from zero to two pi But let's have a look at it. Let's have a look at what this curve looks like once again I'm going to go to the warframe language And there we go And we have the problem 2.4. There's my parametric curve there And there enclosed inside of a set of curly braces with a comma in between Parametric plot is the function and I'm going to go from t from zero to two pi But instead of going straight to two pi so it draws the whole thing at once I'm going to wrap this all inside of the manipulate function And so this end Zero to this omega this omega is going to go from 0.01 to two pi So I'm going to go all the way from zero to two pi And then of course I'm just putting my plot range there so that I have this nice little window here So let's open up there and just start drawing and we see what happens Oops, what's happening is we're going counterclockwise. So we have to look at that And all the way around all the way around and it gets to two pi And and we there and that looks like a card wait button So that's quite interesting And now we can just look at the area enclosed by this by this parameter curve So first of all Let's put in the fact that we've already sorted out now that we need zero is less than equal to t is less than equal to Two pi we know that We know that this is the x of t. We know this is the y of t And we know that we have to get dx dt dx dt So that's going to be negative a half sine of t And then negative times negative is positive an eighth sine of two t And we have to multiply that by y of t there And we also remember Be importantly that we're going to go from zero to two pi And that's not what I usually put in green. Remember green is for just for our Definitions and theorems, etc. The things that we have to know about So the area is going to be this it's y of t dx dt Now i'm going to use this opportunity just to show you what happens When we have these counterclockwise examples So that what we're going to do with a counterclockwise example is we're just going to say it's negative Going from a to b or we remember we had c to d But then y of t dx dt. Oh, there must be a dt there dx dt dt So let's do that. Let's clean up all these little mistakes And so that's got to be a dt and just to stick to what we've been using Let's just make sure that we use c and d that looks much better So that would be one option But what you could also do is you could also say a equals going from c to d And then we can do have x of t and we have dy dt dt So we can also have that And once we have that we we can at least also use that as far as As these curves that go counterclockwise are concerned. So let's have a look at this one And we've got it in the warframe language And we can do that multiplication and then do the integral as I said If you're happy to do that just as some exercise for single variable calculus go ahead and do that But here in the warframe language Remember these files will be uploaded for you so you can upload them straight to your free account online And you can have a look at that and I've just drawn it out all the way By doing remember once we have that we have these see see three separate expressions that Are added to each other so we can do the integral of them separately which I've done all the way down there And if we were to run this last one we see that the answer is three pi over eight so by all means go ahead and Three pi over eight see that you get to three pi over eight as well either by doing this multiplication So we're going to multiply this by this as I said, we're going to end up with a couple of expressions and we can take The the definite integral separately for all of those going from t equals zero to t equals two pi Or use your computer algebra system. It'll do this multiplication for you It'll even do this derivative for you and then do the integral for you as well as long as you understand That this this is what it's all about. I think you need to waste time and just rehashing Some all the stuff you've learned before you know about integration. You know about differentiation So let's just get on and understand what's happening here with these parametric curves Let's have a look problem 2.5 calculate the area enclosed by this parametric curve cosine of t comma sine of two t And we see weirdly there our parameter is on this closed interval pi over two to three pi over two So we'll frame language to the rescue. Let's just draw this And there we go. I've written it for us there and look at this curve goes out up So first of all notice that we are going clockwise. So that's sort of problem for us and it's enclosed there So nothing special here. I just want for you to develop appreciation for these wonderful curves that we can create with with parametric curves Much much more interesting. I think that's just when we did single variable calculus. We just had This independent variable x and a dependent variable y and we had this function y equals f of x Great stuff So again, this is this going to be a little bit of mild exercise for us What shall we do? This is Recognize that this is x of t. This is y of t. We need the dx dt And dx dt is going to be negative the sine of t So in other words, we're going to have the negative the sine of t And we're going to multiply that by the sine of two t And when we remember the area in this instance, we're going clockwise. So it's going to go from c to d And we have y of t dx dt Dt So if we just substitute there, we know that this is c. This is d So a area enclosed by this parametric curve is going to go from pi over two All the way to three pi over two and we're going to have negative the sine of t multiplied by The sine of two t dt Once again, we're from language to the rescue But by all means, please do this integral by yourself For challenge, but you can see I've written it out there Shift enter shift return and we get our quick solution four over three So the area enclosed there is four thirds Simple as that. So our last problem here in this problem set We're just going to again find the enclosed area, but this time we have these two polynomials as far as our parametric equations are concerned So reminder, as always, this is the x of t. This is the y of t and Do we need the x dt or dy dt? Let's have a look at the curve So we see it here in problem 2.6. I think you know the code by now And let's just start running And we're going from negative two and we see we're going clockwise. So that's not a problem. Here we go And again, we're going to from negative two to two this time But we see the nice enclosed area there So this one's going to be simple and we are reminded of the fact that area is going to be this definite integral We're going from c to d of the y of t and then dx dt of dt So Let's do that So the area is going to go from negative two to two as far as our definite integral is concerned We have t to two to the power four minus one and we're going to multiply that by three t squared minus four dt And again long-winded multiply that out. This is at least much simpler than some trigonometric identity using some Substitution there as far as trigonometric functions are concerned. There we go We see how it's multiplied out there four minus three t squared minus four t to the power four plus three t to the power six And then that's our expression with respect to t and we're going from negative two to two for this different integral and we get this big fraction 2048 over 35 2048 2048 over 35 There we go do it by hand Good exercise, but probably a bit boring when you watch this on video do it yourself There's the solution and as you can see here, we've done quite a few different problems now I think whatever's thrown at you in the in the tesla exams you'll be able to handle these the area enclosed by a parametric curve