OK, zato. Zelo smo v izgledanju spasih. Zato, če smo v prvi lečnih lečnih lečnih? Zelo definicije, kaj je zelo prvi lečnih lečnih. Izgledanje. Izgledanje v normalnih lečnih lečnih. Izgledanje v nekaj patologiji, ki sem tukaj vzela, kuspe, nodi. Zelo način. Zelo način. Zelo način. Zelo način. Zelo način. Zelo način. Zelo način. Zelo način. Zelo način. Zelo način. Zelo način. Zelo način. Zelo način. Zelo način. Zelo način. Zelo način. Zelo način. Zelo način. Guys, si Opas. So, this last property induces to ask a key question since for the same trajectory we can have infinitely many reparamet finalization. The natural question is, is there a better one, is there a best parameterization of a space curve? So in order to answer this question, let's see. So, we have a curve from some interval in to 3-space and we fix some fix point in our interval. So, really what we are doing is that we have our interval I. We fix something not a tj. Bolj in tj. Gdje našem ležim... vse. Bila tj. alfa v spas. Zato tj. tj. alfa traja tj. Ilsem se zovem Busheev po kaj glasbe tj. We sem kanalit vonal je s. Zanim s težem je to folija. Vse pogledaj po vnyčitev tj. A možno tj. ki pomožemo. zwečneč so carrišne. Zdaj, imaš na solim Voletiva. Co se vid este. Je tato, da je srečno élite. Šte, da je, da je to. All in, da se je... Tato, da je. Tater. Tater, da je. Tater, da je. Tater, da je. lending. Tater. Tater, da je, da je. Dovrst, danas pa, doj pa jo. Dožel. To je. Tater. Tater. Okay. srednjič, če je to, da je to, da je to. Kaj je to? Kaj je to? Fokon, ki je to, da je počet in taj intervaltuar, če je ta jeznačnača, kaj je to, da je to, da je to, da je to, da je to, da je to, da je to, da je to, da je to. je to je izgledačne vseč, ne znamo do vseč. To je način delovina tega vektora. In je vseč izgledačne. Vseč, je to norma vektora. Zelo vseč, kako je vseč? To je svetovina delovina delovina delovina delovina. Zato, da je svetovina delovina delovina, kaj je to, da imaš vidjevanje deloviti. If the thing inside of the square root goes to zero, you lose the derivatives, in general. You might lose the derivatives. So this is just a continuous function. In general. And since this is the integral of this function, s is the integral of this function, and this is continuous, ide 포zijo sprem. The best thing we can say is that S in general is just C1. So we gain one derivative, but not more than one, by taking the integrals. And of course the first derivative, S prime at T is just the integrant there, evaluated at T. leisure. We cannot make the second derivative because in principle this could not exist, may not exist, ok? So s, in general, the point is that this function does not in general define a referomorphism, ok? So s, in particular s is not a differomorphism ok? Because the thermometer means it's smooth with smooth inverse, so this is smooth to je počka C1. Ale da sem bojila, da so inoziv izgleda. … V kundi, dodao, da ima, da je vse vse s, da je izgleda izgleda? Zato je, da je vse vse vse vse izgleda, da se da se vse vse vse vse vse vse vse vse vse vse vse vse vse vse. Selo, da je svetov poslado. So the only way this is non smooth is when this object becomes the zero vector. Okay. So if the norm then we are okay, then s is indeed a defiomorfism and the only thing I have to be a bit careful is between which intervals, okay, between i which is the domain of the function s. In v enistitju ne vedem ta darga, ta delač, ki je zo vrstvena, ki je SOV.  Kaj je zelo, da ne zelo, da je je vzelo, okej? Tako kurs, tudi je to nekaj nezelo, kurs, ko je učite. Tukaj, tukaj skupaj, tukaj kursi je zelo vziv. Zelo vziv način, če je to, da je to, da je tukaj. Zelo vziv, da se način vziv, če je to, da je to, da je to. V nekaj kaj, vziv, če je to, da je to, da je zelo vziv. needing this Y of I to simplify a little bit the notationsK but then, if it's a definomorpha, so we have that S So between I and J, we have the defiantutz if it's a defiantcco, it has an inverse So i call phi the inverse Vse je z vsem z jah dela po i. Se sem naprej vesim, da vse zeljamo vesim na halj. Vsi daju nekaj, da je zeljamo vesim na halj. Vesim na vsem, da je z našem vsem vsem vsem vsem vsem vsem vsem vsem vsem vsem vsem vsem vsem vsem vsem vsem vsem vsem vsem vsem. Kaj je to? To je z J, z R3. Geometrično je to samo, da je bilo. Tukaj, da smo vse parametričili, je to početno. Tako da je bilo z I, z R3, z J, z R3. A če je to početno? Alfa je bilo vse parametrične vse parametrične vse parametrične. Tukaj da smo vse parametrične vse parametrične vse parametrične vse parametrične. Moj, če je beta z vse parametrične? Beta je alfa kvama napotnevac, da sem tukaj tukaj iz njej. Vse parametručne? Alfa če možite evaluoviti vse parametrične? Vse parametručne vse parametrične? Ne zelo. To je tukaj in tukaj vektor. Zelo sem vzelo v tukaj, pa ne zelo vzelo. Tukaj je tukaj vzelo vzelo in vektor. Kaj je tukaj? Kaj sem vzelo? Faj je vzelo v s. Zelo sem vzelo v s. Zelo sem vzelo v s. S-prime je tukaj. Faj is one over this, evaluated at the right point, of z אז alfa prime of phi s at the point phi s, divided by the norm. Again, if you want it's chain rule, but now it's applied to the composition of s-composed phi. as compos f is by definition the identity, because one is the inverse of the other. So if I take the derivative I get one, but if I use the chain rule I get exactly this formula. Well, this is nice. What is this saying? That the norm, so the norm of the velocity in this new parameterization is equal to one, because I get the same thing for NES. So what is this parameter achieving? This parameter is achieving uniform motion on our trajectory. So parameterized in this way, a regular curve is parameterized in a way that the velocity is constantly equal to one. Of course as a vector it will change, but the norm is constant. We are not saying that the vector beta is constant, just its norm is constant. Now this is a very important class of parameterization, so we should give them a name, so definition. Alpha, a curve in R3. Of course I always assume it's an infinity curve, otherwise I mean all our theory. So this is parameterized with length, and we will not write it again. We will always write p a l for short if it has exactly this property. If the norm is equal to one for infinity. And then what we have just proved essentially in this simple argument is this theorem. Every regular curve can be parameterized with arc length. This is just what we argued, what we proved here. Notes, comments. This parameter is not uniquely defined. Arc length, or S, let me just indicate in the symbol for short. S is not unique. What does it depends on? To define S, I had to use a fixed point, a base point on my curve. So it depends on t0, on this base point that I've chosen right from the beginning. And on the constant of integration. It doesn't matter. In any case, the way two different arc lengths are related is very easy. So we will see it. So examples, one. So for example lines, if you take a straight line, is it? So the way we define the line was as a parameterized. In the parameterized form, yes, the alpha of t, tv plus v0 for some fixed vectors v0 and v. Here the parameter was free to move. So our interval i was the whole real line. So to compute the arc length, we need to fix t0. It doesn't matter. For example, t0 equal to 0, just for example. Question. Is a line a regular curve? Well, what is alpha prime? The question is, does it velocity vector vanishes somewhere? Well, alpha prime is exactly the vector v.  If alpha prime of t is equal to v, to the vector v. So of course, when we write a line in this form, we implicitly assume that v has to be a non-zero vector. Otherwise this straight line is actually the point. So as soon as it really is something geometrically meaningful, it's regular. So we can play the game. And what does the game give us? Well, s of t, by definition, would be the integral I've chosen t0 to be equal to 0. So between 0 and t of the norm of alpha prime. Well, the norm of alpha prime is nothing but the norm of v in du in our notations. But this is nothing but t norm of v. Now, what is this asking me? If I want to write the curve beta, I need to compute phi, the inverse of s. Of course, in this example everything is simple. That's why we are doing it. So what is the inverse? Let me erase everything. So what is phi? Phi will take s. So, in fact, s was taking t to s, by definition. So phi is taking s to t. Now, if this is s of t, what is t of s? So phi of s is equal to s. And this is s. What is t? Is s divided by the norm of v. So now we are done. In the sense we can write immediately beta of s. Beta was what? Was alpha composed phi. Well, but this is alpha. So instead of t, I need to put phi. Beta of s will be s over norm of v times v plus v0. So sometimes I forget, as a notation, this is a vector, plus v0. And now, of course, you see that we have achieved uniform motion. I mean, in some sense it was uniform right from the beginning, because if you parameterize something in this form, the velocity. It is constant. So in this simple example, the only thing we have achieved is really to make this constant equal to 1. Because now if I compute beta prime, of course I get v over norm of v, whose norm is 1. OK. This is clearly the simplest. Now let me give you a slightly less trivial example. These are all excuses also to introduce examples which are useful for later on computations. So let me take from R a planar curve, R2, another famous planar curve, which is called logarithmic spiral. And this is the curve parameterized in the following way.  This is just a convention. Well, the name itself tells you everything, but you can check that the name is not given by a crazy reason. So why this is called the logarithmic spiral? You see, if there wasn't this e to the bt term, if in these components you think this disappear, of course what is a cos t a sin t? It's a circle with center the origin of radius a. Well, and what is the effect of putting, in fact there is in both components there is this, think of this e to the bt to be in front of everything. So what is the effect of this e to the bt? That somehow this is changing as t moves, it's changing the radius. So this particle would like to stay on a circle, but this factor here is continuously changing the radius of the circle that the particle would like to lie on. And how is it changing? As b is negative, as t increases, this radius is going to zero. So then forever. You never reach the origin, of course, but you keep on moving around. Of course it's a terrible picture, but it gives you the idea. Well, let's play our game for this example. What is s of t? In actually computing s of t, we will understand also if it's a regular curve or not. So s of t, again, we have to fix a base point. The interval is the whole real line. Why not taking t not equal to zero? Why not? It's just any point on the domain. Now, s of t would be by definition, the integral between zero is t not. And t of the norm of the tangent vector. OK. Well, let me erase this beautiful picture, because I need space. So what is the, let me indicate first. I leave it as the norm of what, of which vector. I need to take the derivative of this. So it's a, of course there is always, well, doesn't matter, a. So the derivative of the first component, it's b e to the bt cos t minus, no, so this is the derivative of this, minus e to the bt sin t. And this is the derivative of the first component. OK. Derivative of the second, there is always a, and then b e to the bt sin t. And now plus e to the bt cos t. And of course, I'm using not the best notation, because now I'm integrating with respect to t, I left the name t instead of u. OK. You are grown up now. You won't be mistaken by this. OK. So this t and this t don't confuse you. OK. So what is the norm of this vector? The norm of this vector is the square root of this squared plus this squared. You see immediately that here I can take out a e to the bt is everywhere. OK. And a is supposed to be positive, and of course e to bt is positive. So if I take it square and the square root, it remains the same. So this becomes what? The integral between 0 and t of a e to the bt times what? I should take what is left. So now we have taken out e to the bt. So we have this one squared, which is what? b squared cos squared t plus sin t squared, sin squared t. Then there would be minus 2 b cos t sin t. I don't even write it down, because on the other part I see immediately that I have plus 2 the same thing. OK. So the double product will erase. So now here what do I get? I get plus b squared sin squared t plus cos squared t. Everything to the power one-half. OK. Because this was the norm in the t. So how much is this? Well, you see, little miracle, 0 t, a e to the bt. What's left inside this square, this parenthesis? So this is b squared cos squared plus, so this is b squared, and this is 1. OK. So times b squared plus 1 to the power one-half in the t. Oh, you can see this is not, this is a number. And so this is the final result. It's just e to the bt minus 1 over b times a b squared plus 1 to the power one-half. OK. So this is the arc length. Well, actually we should have commented. Is this a regular curve? You see, the norm of the tangent vector is this one. We have computed inside the integral, no? Can this be 0? Is there any value of t for which this is 0? This is a positive number. This is a positive number, and this is a positive number for any t. OK. So this is a regular curve. So our theory applies to it. This is the arc length. So in principle, we could reparametrize our curve, alpha, and construct beta. I'm not going to do it because now there is nothing to learn. But the only thing missing is what? Is writing down the inverse of this function. So this is s of t. I need to write down t of s. OK, which is easy. Is not, OK? Of course, it will be logarithmic. S of t is an exponential. T of s will be a logarithmic. OK. And then I just compose it, and then that's it. Now, OK, but this is just to make a non-trivial example of the computations. Now, what I'm going to say now, I know it's going to sound very mysterious to you. But I would like you to spend a few minutes thinking about this. You should practice a little intellectual game. And in this case, it will be kind of strange, but you will see in going higher dimensions it will become more and more delicate. In the sense that the existence of arc length has an important philosophical implication.  I mean that we are. I'm not trying to offend you. Any of us is a zero-dimensional object. And suppose that our universe is one-dimensional. It's actually, it's a curve. There are particles moving, leaving all our life on a curve. OK. Suppose two of us try to communicate, leaving on two different universe. So what kind of properties you can communicate about your, and one of course is asking the other, how does your universe look like? Well, the only thing you can really communicate is metric properties of your universe. I mean, my universe is made in a way that if I move at distance one, something happens. OK. Now, arc length means that two zero-dimensional beings see the same thing. I would like you to think a bit. It's difficult to, but I mean, since on every universe there is arc length as long as it's a regular curve. So the only thing that they could discover is that one lives in a regular world and the other lives in an irregular world. OK. That's the only thing. As soon as they are both regular curves, they would see the same universe. OK. What it means, there is no one-dimensional intrinsic geometry. OK. Now, this is the first appearance. I propose you this intellectual game to you, because when we will go to surfaces and hopefully to higher-dimensional objects a little bit, this will become more and more interesting. OK. But there is no really geometry, intrinsic geometry of one-dimensional objects. OK. Because of this simple theorem that we just proved. OK. On the other hand, looking from outside, so this happens because your intellectual game has to be imagined to be on a curve and the only thing that you can see is the curve itself. You don't see the world outside by definition, because otherwise that will not be your universe. I mean, the universe is by definition the only thing you see. So the fact that this curve lives in R3 so there is a huge space around, you don't know. That's something you don't know. The only thing you see is your little one-dimensional universe. OK. On the other hand, we are doing geometry, three-dimensional geometry, so we can look from outside and of course we think of two curves. I mean, they can be very different. So this depends not on the intrinsic geometry, but on the way the curve is put in the big space. OK. And that's why we go on. Otherwise everything, otherwise it would be game over. OK. So OK, there is no intrinsic geometry and that's it. Well, there is extrinsic geometry. So the way a curve is put in R3 gives some geometry to the curve.  OK. But on the other hand, looking from outside, I want to detect the difference. So extrinsic geometry must detect the difference between these two things. OK. How do we do it? Well, we associate at every point to a space curve, a special reference. So suppose this is our curve in R3. So this is alpha of i. OK. Now, and here we have our point alpha of t. OK. Suppose this is parameterized with arc length. So from now on, unless alpha is parameterized with arc length. OK. Well, we certainly have one natural vector of R3 associated to a curve. It's tangent vector. OK. At this point. So here we have alpha prime. Well, in fact, it would have been better to call it s. Since it's parameterized with arc length, let me call it directly s. It's this special parameter that we have just defined. OK. So at that point where we know one special vector. It's tangent vector. So let's call it t. So this is by definition alpha prime of s. And now, now t is automatically of norm 1. OK. Because of arc length. OK. This is the key property of arc length. OK. So in particular, of course, the norm is equal to the norm squared equal to 1. So t of s. The scalar product. So this is just to introduce the symbol scalar product. OK. In my notation, t of s, t of s is equal to 1. Now I take the derivative of this. This is an equation, which holds for any s. So if I take derivative, what? How do you make derivative of a scalar product as the standard product? So derivative of the first times the second plus the first times the derivative of the second. OK. So this becomes t prime of s t of s plus t of s t prime of s. So this is from the formula equal to what? Well, equal to 0. OK. And then, of course, I observe the scalar product is symmetric. So this is exactly like t prime t of s is equal to 0. OK. I already canceled the 2 because it's equal to 0, so it doesn't matter. OK. So this means what? This means that the derivative of the tangent vector, so t prime is always orthogonal to t. At every point, the derivative of the tangent vector is orthogonal to the tangent vector itself. OK. This induces a natural definition as usual. So, let me define the function k of s to be the norm of t prime of s. OK. So this function, which takes the norm of this vector, is called the curvature of alpha at s. OK. Again, how regular this function is? Well, this function is the norm of a vector. So it would be smooth. It would be a beautiful smooth function as long as the thing inside the norm stays away from 0. So in general, it's only continuous. So this is only continuous in general. It is smooth if this vector is different from 0 for s. OK. So regularity drops if this vector vanishes. OK. Well, of course, another immediate property of this function is that this function is non-negative because it's the norm of a vector. OK. From now on, so in under computation as we do today, from now on, we assume k is positive. OK. So this vector does not vanish anywhere. If this is the case, the point is that I can divide by k. That's what I gain from this. So if this is the case, I call this definition 1. And then in definition 2, I can define a new vector, a second vector naturally to the curve alpha, which I call n, which will be just t prime of s, divided by its norm. That's why I need to make this assumption. OK. Because otherwise I'm dividing by 0. Of course, I can write it in this by obvious. And this is called the normal vector. OK. So you see, in our picture, well, in my picture, if alpha is moving in this way, as s is increasing, well, I can imagine that alpha prime is changing in this way. So more or less if I want to draw some reasonable picture, my n will be more or less here. And this is the normal vector. But now I am in R3. And I have already two unit vectors, of course, n as norm 1, because it's a vector divided by its norm. So I have two orthogonal vectors of norm 1. So there is only one way up to sine to complete it to an orient to a basis of R3, to an orthonormal basis of R3. OK. So there is only one direction. So as a convention, so my only doubt would be should I take plus or minus the missing direction. So we fix it by this definition here where this symbol for me stands for the vector product. OK. I prefer to call it wedge instead of cross because it introduces the division. So first times the second gives me B. And this is called the B normal vector to alpha at S. So in my picture this would become bit complicated now to draw it properly. But you see now if this is T B of S. Now so if you want we can put everything together in the last piece of the definition. So we have a reference an orthonormal basis at every point of our curve we have constructed an orthonormal basis of R3 which is somehow geometrically remembering how the curve looks like. And this is called frene reference or basis or whatever you want. The only important word there is frene. Three vectors. Is it clear how these three vectors are constructed? It's quite by taking essentially derivatives of alpha and vector products. Well so how the local geometry of our curve should be equivalent to the way these three moves around. So you see you should imagine for any S you should imagine these three vectors moving in time. And the way they move is adapted to the curve. So knowing how they move should be the same thing as knowing the curve. That's now the game we want to play. So what are the relationship between the derivatives of these vectors and geometric properties of these vectors? So we know something by definition. So t prime of S is by definition k times n. This is the definition of k. This is a tautology given the definition. Then we know what another the other key relationship is b is equal to t wedge n. So let's take derivative of this one and let's see what it happens. So what is b prime? I drop of S of S of S of S. So b prime is what? Well how do you take derivatives of a vector product? Have you ever done this exercise? Well this is the moment to do it. Okay? It's the usual rule. Okay? You differentiate the vector products as the standard products. So derivative of the first times a second plus first times the derivative of the second. Well why is that? Exactly. Basically write down if you want explicitly how it is it is really a product of the components of the vectors. So if you write it down in two lines you have a proof. So this is t prime wedge n plus t wedge n prime. Okay? But t prime but t prime and n are proportional. n is by definition the rescaling of norm one of t prime. So what does it happen to a vector product if you take the vector product to proportional vectors? Become zero. So this disappears and we are left with t wedge n prime. Okay? So what we learn out of this? Well we learned that b prime so b prime to t. Because of course out of this for example I extract the information that b prime is orthogonal to t because it's t wedge something. Okay? First bit of information. But then what else I know? Well b prime has to be also orthogonal to what? To b. Why is that? Because it is a family of vectors of constant norm. You see t as norm one for any s. N as norm one for any s. So the vector product these are orthogonal. The vector product as norm one for any s. The norm of b is constantly equal to one. Okay? But then there's always the same argument. If you take the derivative so if b scalar b is equal to one I take the derivative and I get b prime scalar b equal to zero. Exactly as we did for t at the beginning of our story. Now we do it for b. So b prime is orthogonal to b automatically because of this property. And it's also orthogonal to t. But t, n and b are bases of r3. So this vector b prime has to be what? It has to be proportional to n. There is nothing left. So b prime must be b prime of s. So all this together implies what? That b prime of s has to be something that I call tau tau of s, some function because of course for every time there is a proportional factor n. So this is a function. Now this function here tau of s is called torsion of alpha at s. Now again, since I usually make a comment on the regularity of this object remember k was only continuous in general and it's positive so we are in the situation where k is actually a smooth function. How regular is tau? Well, tau is automatically smooth because one way to write down tau being n of norm 1 n is of norm 1. So one way to write down tau is to say that tau is equal to the projection of b prime in the direction of n. These are smooth vectors. So the scalar product is a smooth function. So here you don't have problems of maybe tau equal to 0 because you don't care. It's the projection of a family of smooth vectors on a family of smooth vectors. That's okay. So tau is always smooth. So we see what? The derivative of the vector t has induced us in order also to remember what we are doing. So we have the vector t and looking at the derivative we constructed k. We have the vector b and looking at the derivative we have just constructed tau. So what is left? We have the vector n. So what is n prime giving us? Is it giving us another function or not? Let's go and check. So what do we know about n prime? Again n is a vector of norm 1. So n prime is certainly orthogonal to n by the usual argument. So let's see what else do we know? Of course we know that nt is equal to 0 so nb is equal to 0. If I want to know information about n prime I take these equations and take the derivative. And what do I get? Take the derivative of the first this implies n prime t plus nt prime is equal to 0. The derivative of the second plus n b prime is equal to 0. Ok. Very good. Is there anything here I know already? Yes. And prime t is what I'm looking for. No? But here it's nt prime. What is nt prime? It's here. It's k. And the equation becomes n prime t plus k of s is equal to 0. And the second tells me n prime d plus nb prime nb prime is exactly this one. So plus tau is equal to 0. Ok. So we can make the final summary. This is a theorem which is usually referred to Frenet formulae because it's just a list of remember which are the assumptions. Everything we did all for a curve in R3 parametra is with arc length and don't forget k positive otherwise you have to stop much earlier. And then the theorem tells you just t prime is equal kn, the last one b prime is equal tau n and in between you put what we just discovered and prime is what? Well the component along t is minus k and the component along b is minus tau. Ok. It's an orthonormal frame so if I want to, I mean this is exactly tell me exactly the components do you agree? So this is the proof, I mean there is nothing to prove now, this is done. And actually in some sense two of these are essentially definitions because this one is the definition of k this one is the definition of tau so really the only theorem is here ok very good. Ok, so we have associated to any regular curve with k positive if you want, I mean two functions curvature and torsion and we have this nice picture of how these vectors move around in space as the curve is. So now the obvious question is what are they measuring? I mean what is the geometric meaning what are the geometric meanings of the functions k and tau? Ok. Well the first thing to do is to compute some examples and to start having some feeling and then you can start guessing what are the real meaning any list of example should start from the straight lines remember we want the curve to be parameterized by arc length so when I say line is something like this now to be parameterized by arc length it means that v is automatically a vector of norm one ok so what is the tangent vector what is t? It doesn't matter I'm doing I'm confusing your mind a little bit on purpose so the parameter I'm calling it t and it's arc length ok, don't be confused whatever the name is arc length is that parameter or one of those parameter for which the tangent vector is norm one if it's not called s it doesn't matter ok so what is t? t is the tangent vector so this is t of t is all is equal is constantly v ok so that's good but then what should I do to compute everything else I should take the derivative of this compute its norm and this will give me k and n ok problem t prime is zero constantly equal to zero so straight lines don't follow in this category I have to stop there ok meaning of course I can define k of t k of t is well defined the point is that it's just constantly equal to zero ok and what is wrong is to tell what is n and what is b I don't know and you can also think that it's rightly so ok because if you think of a straight line there is of course a well defined tangent vector but there is no well defined there is not a better normal vector there is certainly a well defined normal plane but why should I pick one normal why there should be one better vector in this plane so it should be it would be wrong if this theory would be identifying one special vector that would be something strange so fortunately it doesn't ok normal vectors to a straight line are all the same and you stop there ok circles well circles we decided that alpha of t, one parameterization one nice parameterization of circles was r cos t over r sin t over r and I don't want to I mean it's a planar curve in some sense I want to remember that they lie in some plane ok for example in the one z equal to zero so I add the third coordinate plus the center vector c which is constant ok well what is alpha prime well alpha prime let's first compute alpha prime because we don't even know if this is regular arc length we have never done this exercise so what is alpha prime of t well alpha prime of t is well of course the derivative of c disappears it's a constant vector so what is left here well this is r times what minus well one over r sin t over r ok one over r cos t over r then zero and now you see why yesterday I played this game this is the moment that you understand I was not completely crazy because r r r ok so what is the advantage now because really it is true t over r is just any number so it's not better than ok so but this is the key consolation ok so this is equal to minus sin t over r cos t over r zero which has the nice property of being a vector of norm one ok for any t this is the nice thing the useful simple thing and what I learned what do I learn from this well I learned that this is a regular curve so I can go on and that t this exactly this t is arc length so I can do exactly the computation in the way I did in the general theory ok so this implies oh sorry this implies that t is arc length let me write it down and t so t is equal to alpha prime now the question is what is t prime t prime just go on taking derivatives so this is the vector minus one over r cos minus one over r sin t over r zero ok what is the curvature the curvature is the norm of this vector ok so k of t how much is the norm of this one over r so it's constant and it's non zero ok so I put in brackets non zero it's important to check because I want to go on with my theory ok so I can decide what is n ok because n would be exactly the unit vector in this direction which is obvious what it is just remove the one over r so it's minus cos t over r minus sin t over r and what is b b by definition is t wedge n ok I don't know which is your favorite way to compute cross products to put them as lines of a matrix and compute the minus ok in any case do it in the way you want and you have to find this ok out of which I find out what this implies so b is a constant vector and this implies that b prime is equal to zero but b prime equal to zero means tau is equal to zero is the zero function ok and that's it I mean there's nothing else you can say ok so circles have constant curvature equal one over the radius and zero torsion ok now we have an exercise the example 3 ok if you take helices in the way I wrote them yesterday you will find out that again I was not cheating you again the parameter was written in a funny form sorry the third component of which vector oh sorry yes of course so for helices the way I wrote them yesterday you will find out again the way I wrote them was a bit strange but it was exactly to achieve the same thing so the parameter I gave you was exactly arc length and then you can play the same game ok and you will find that the curvature is exactly a over a squared plus b squared and tau tau is now non-zero and it's actually minus b over a squared plus b squared ok ok so we can move on to some at least some simple geometric interpretations of these functions so to end this lecture we can give at least so you see we have introduced these functions it's natural to say well suppose that these functions are constant I mean what is this saying to about the curve and in fact among constants if they are actually zero so remember you have to put yourself in the situation where you can actually use the theory of arc length and positive curvature and then for example you can see that alpha is planar if and only if tau is equal to zero we have computed the torsion of the circle and we found zero so now you see that this was not an accident ok so the torsion measures exactly how much you have to stretch so if a curve would like to lie on a plane and instead you are pushing it in r3, in the 3 space I mean there is some kind of effort you have to do and this effort is measured exactly by the function tau ok so it's a very neat interpretation proof first suppose let's first suppose tau is constantly equal to zero why should it be planar what is tau tau remember is the norm of b prime so tau equal to zero means that b is constant because b prime is zero tau equal to zero means b prime is zero is equal to v a constant vector of course this is not particularly important it's always a vector of norm 1 because it's b it's constructed as a unit vector but then if I take the scalar product of alpha of s with this vector v and I take the derivative I'm taking the function alpha of s scalar product this vector v and I take its derivative how much do I get well this is a constant so it doesn't matter the derivative doesn't go here so the only thing I get is t so alpha prime which is equal to t t scalar product v but v was b so t and v must be orthogonal so this is zero but then this means that this function is a constant function it's a function whose derivative is zero at every point so this implies that the function alpha of s scalar v is equal to some number a but what have we written here this means exactly that alpha lies this is the equation of a plane in R3 maybe not passing through the origin ok we never said a is equal to zero ok which is the equation of a plane in R3 ok so this closes this way of the theorem the other way well now you can reverse everything if it's planar it means that there exist a number and a vector for which alpha satisfies this equation and then you go upstairs instead of downstairs if this is true t so the tangent vector satisfies this equation ok and then b has to be this vector here and that's it because if b has to be this vector in particular b is constant and so b prime is zero and so tau is zero so exactly the same stairway of things but red in the opposite direction gives you so the converse is exactly the same thing in the opposite direction ok last thing so this gives a nice interpretation of the torsion what can we say about the curvature well so suppose we have a curve parameterized with arc length such that its curvature is constantly equal to zero in fact k is the zero function I would like to state a theorem like this but if and only if one vote for the straight line somebody else votes for something else I mean winning winning a poll with one vote is ok it is ok if alpha is piece of proof well we have computed the curvature of a line and we have checked so in this way already done ok so the only thing is why should this be true what does it mean that k is equal to zero more or less in the similar spirit so k equal to zero implies what implies that the tangent vector is constant because k is the norm of t prime so that means that t prime is zero so t is equal to a constant vector but t is alpha prime ok and then I integrate so if I know alpha prime what is alpha alpha is of course the integral of alpha prime plus a constant of integration if you want alpha is equal to tv plus v0 and that's it ok probably we this is enough to stop so next time what are we going to do next time you see what we did today was to identify geometric objects naturally associated to a curve the three vectors if it's possible t and b and these two function k and tau now as geometers we ask is this all what does it mean is it all is it all means for example if you give me two functions and you would like them to be the curvature and the torsion of something is it possible well there are some obvious constraints because the function that you want to be the curvature has to be positive especially if there is also a torsion otherwise we do not speak about torsion when tau is when k is 0 so one function certainly has to be positive but is that the only constraint and suppose you can do it in how many curves can have the same curvature and torsion and actually what does it mean how many because of course the space of curves ok so the solution will be exactly yes give me any two functions provided they satisfy nothing the only constraint is what we said and there will be a unique curve but unique up to something and we have to decide everything in mathematics is unique up to something when you classify groups it's unique up to isomorphism when you classify vector spaces it's unique up to linear isomorphism when you classify topological spaces it's unique up to homeomorphism now we are doing another type of geometry so our meaning of unique will be unique up to rigid motions this is the group that it's moving everything around rigid motions of R3 two curves which are obtained one from the other via rigid motion of R3 are totally indistinguishable from our geometric point of view they are the same curve they might actually look quite different but they are the same curve we are ok and still there are infinitely many curves obtained in this way so it's unique only up to so this is the best theorem you can hope and this is the best in fact what it holds and we will prove it next time