OK. So I can start. So we were still talking about tangent space, singular and non-singular points. And we had introduced first the tangent space of a fine algebraic set, so the x, the n, the fine algebraic set. Then we had that the NP is a point in x. We had introduced the tangent space at p to x to be the zero set of all the differentials at p for all f, where f lies in the idea of x. We'd seen this can also be, and this will by itself be some, this dpf is a linear polynomial. So this will be a linear subspace of a n. And in fact, the vector space, because it goes to zero. We had seen that one, and this can also be written as the zero set of, so if the ideal of x is generated by some elements f1 to fr, this can also be written as the zero set of the differentials. And maybe we saw that this is also the same as taking the Jacobian of f1 to fr at p, where the Jacobian is the matrix of partial derivatives. So Jane, put? Yeah, yeah, that's what I meant, obviously. That's what I should have written. So this map given by these differentials is the Jacobian, and then the kernel. OK, one could, just so that you have some more intuition, one could also see it in a different way if p is the origin. Then one can also say it is following the differential to f at p to at zero is just you take, so f is a polynomial. So you just take the polynomial f, and you take its homogeneous part of degree one. That's the same, OK? So that's maybe a bit simpler than, I mean, it's equivalent to that formula. But it somehow gives you more feeling that you really have some kind of linear approximation to the zero set. OK, and I wanted to, OK, so we had to introduce now another version of the tangent space, and I wanted to prove that it's equivalent. So if x is just a variety, p and x, then we can define another definition of the tangent space, the more general one, is just we take this to be, we take the maximal ideal at p, o p, o x p is a local ring, and m p maximal ideal. We take m p, model m p squared, the dual vector space of this. So this would be the same as the maps v from m p to k with the property that if I take v and restrict it to m p squared, this is equal to 0. And now we want to see that if we have in a fine variety, this definition coincides in a suitable sense with the other one. So at least as vector spaces, these two versions of the tangent space are isomorphic. And so I have to say, so I have to write down some kind of isomorphism. It's slightly clumsy. But first maybe I'll also write something else. We had for a morphism of varieties, we also had a map of tangent spaces, which we had defined in terms of, I mean, for a fine varieties. We had defined the map of tangent spaces in terms of the differentials of the components of the map. And now there is also a version of this here in this more abstract setting. So if phi from x to y is a morphism of varieties and phi of p is equal to q, then we have the differential of phi at p is the map, which is somehow given by pushing forward. So we take dp phi is a map, should be a map from tpx to tq. Why? And how can we do this? Now we want to have a map from mp. So if you have a map from mp, what mp squared to k, we want to make a corresponding map here. So if we have a, so we assume we have such a map v from mp to k, we stick to mp squared at 0, we can associate to it the map v composed with phi. We first take the element with phi star, I expect. So if we have a phi star goes from O xp from O yq to O xp, namely by composition with phi. And so we have this. So one could say if you know that phi star is just composition with phi. So if we have for an element f here, we compose it with phi, and then we apply v to it. OK, so this is the kind of most natural map that one can have. But anyway, we are not going to use this very much, or maybe not even almost at all. I just wanted to have it for completeness. Yeah, yeah. That is usually when you want to do the computation, then you will need to most of the time to go to the fine case. Because for concrete computations, this is a little bit abstract. So we don't know precisely most of the time what mp and mp squared is and so on. So if you have a variety, which is a hypersurface in some affine space or something, it's much simpler to work with the other definition. And so you can go to the fine case. But if you want to prove theorems, this is much more useful. So now we want to prove that both definitions of the tangent space are isomorphic vector spaces. The question is, where do we have this? Yeah. And this is a little bit, maybe a little bit long, and afterwards we're not deferred to it. But at least one can see how it goes. So I will write for the moment, write so now let x in a n in a fine variety. And we write, say, small tpx for the old definition of the tangent space. So p and x. So the one which is given as a zero set of the differentials, that is right now with small t. I'm going to show that the two t's are isomorphic twitcher. So first I want to, so I have to define an isomorphism between these two vector spaces. So first, so I make some definition. So if f is an element in the coordinate ring of x, then I say I want to and a is an element in this tangent space. I want to say what the evaluation of the differential of f at a is. So we say define dpf of a to be dpf of a, where f, so where the class of f is equal to f. So f is a polynomial. So what do I mean by this of a? So if I write a equal a1 to an, it's just a vector in an. So dpf of a is just sum i equals 1 to n df by dx i at p times ai. We just evaluate this polynomial of dp1 at a. And we define it like this. And note that this makes sense because this is a reasonable definition. If you have any other representatives, so we know that any two representatives of this element f in ax differ by an element in the ideal of x. And the elements a in the tangent space are precisely laws which lie in the kernel of the differential. So this will give the same. So this is already fine. So what I mean, only on the tangent space. And in the same way, so then we can also go to the maximal ideal. So if, so let, so p was a point in x after all. So now if h equal to f divided by g is in the maximal ideal at p. So it means f and g are elements in ax. And f of p is equal to 0. And g of p is different from 0. Then we define dph of a. a is again an element in the tangent space to be what? To be df dpf of a divided by g of p. And check that this is well defined. So if we have another, if you write f as another, h another way as a quotient of such polynomials. And you know what the equivalence relation is. Then you should get the same result for all a. And you can also check if h is an element in mp squared. Then dph of a is equal to 0 for a. This is basically because you can then write it as a sum of products of two such. And then you use the Latinx rule. And you find that you get 0. OK? What? Well, so. This is because I've assumed here that I'm in mp. So in principle, you would expect. So what? No, no, h is certainly not a unit. No, no, no. But I mean, this formula is a special case of a formula that you know, which looks quite different. And maybe then, as he asked, I can tell you. So if h is an element in o xp, I can define dph of a. I don't need it. But you know, it's confusing that this formula looks very different from what we're used to in calculus. Then I can define this. And this is where I define by the usual formula, which I hope I remember. So dpf of a times g of p minus f of p times dpg of a divided by g of p squared. And now we use that f of p is equal to 0. So we forget this term. And we divide by this. Yeah, the value is 0. That is the point of being the maximal ideal. So this is precisely the elements which at the point are 0. So that is a bit slow. So this is this. So this defines me this thing. So therefore, we have a map, which I call dA. So thus, for A in the old tangent space, so A, an element in the old tangent space, we have, say, an element dA, mp, if you want mp, mp squared to k, which is therefore an element in the new tangent space of x. And so we define a linear map, which I would be called delta, from tpA, tpx to tpx, which sends tangent vector A to kind of this d delta A. And the claim is that this is an isomorphism. And therefore, you just basically, so the map is by evaluating the differential of the function in mp mod mp squared on the tangent vector. So this is where this duality comes from. So the theorem is, so this map, delta, is an isomorphism. And the second statement, which we will not also not prove. And it's not so important, but for completeness. So using delta to identify tpx and tpx, we find that the two definitions of the differential of amorphism are identified. So for phi from x to y, amorphism are identified. So if you, I think you understand. So basically you have the, so we have this old definition of tp phi. And we have the new definition. This is the new definition of tp phi. And here we have this diagram commutes. OK? That's what I mean by these words. OK, I will only prove the first part. Second is if you want an exercise, it's just a little bit. One has to kind of go through the details. It's not so exciting. Anyway, so fun. So we have to show that this map, which I unfortunately just wiped out, but you have it crystal clear in your head, is an isomorphism. So we have to show it's injective. It's obviously a linear map. We have to show it's injective and surjective. So we prove the injectivity first. That's the easy part. So we have, remember we have x in a n. And p was a point in x. And so write ti for the class of xi. So xi are the coordinates. x1 to xn are the coordinates on a n. Minus the value of xi at p. So this is still a polynomial. And then the class of that in the local ring at p actually lies in mp, because it's 0 at the point p. So we have some elements in mp. We can evaluate our wonderful. We can see what happens if we look at this. So now, so assume we have our tangent vector a in tpx. So then we ask ourselves if we take delta of a. So remember that we had this map delta. Would send a tangent vector a to da. So if we take da and apply it to xi, what is it? So we are supposed to take the differential of ti. So we are supposed to take the differential of ti and apply it to a. So the differential of ti is just the class of xi. You apply it to a, you get ai. So thus, if a is in the kernel of this delta, that's da is equal to 0, then it follows that all ai are equal to 0. Simple as that. Now, so subjectivity is not much more difficult. We want to show that we, so we have seen that we can pin down what the tangent vector a was by looking what it does on these coordinate functions. Now, in order to prove the subjectivity, we have to see that if it does it, the correct thing for the coordinate functions, then it is the correct tangent vector. So therefore, we will, we want to show that somehow the tangent space, the mp mod mp squared, is generated as a vector space by these ti. Then it follows to show subjectivity. It is enough to show that ti, so t1 to tn generate mp mod mp squared as a vector space. Then, but I should maybe a little bit more say about it. So yeah, OK, I come back to this in a moment. And then I, so what I mean is, so if I take an element delta in tpx, so a tangent space in the new sense, so that means that delta is a map from mp mod mp squared to k. So then, we want to get a vector here. And we know how to get the vector. Namely, its coordinates will be what happens if I apply it to the ti, if the map is a. So therefore, we let ai equal to delta of ti. And we put a equal to a1 to an. And the claim is that delta is equal to da.   So in other words, that v is equal to delta of a. So the map is subjective. OK, so it's clear that if I take, by definition, precisely, we see that v of ti is equal to delta a of ti. So if it is true that these generate this as a vector space, therefore, the two are equal. Because they are equal on a space which generates. And so this is just then, one just has to see that this happens. It's quite simple. So let us look at an element. If I have f equal to g mod h is an element in mp, then I can write f, if I write f, minus g divided by h of p. Now, this is just a. So as usual, when I write it like this, it means that g of p is equal to 0, and h of p is not equal to 0. And here you see, I've just taken, this now is just an element in ax, because this is just a number. If I take this, I can write this as g h of p minus h divided by h times h of p. Just put it on the common denominator. But what you see here is, this thing is 0 at p, and this thing is also 0 at p. So this is an element in mp squared. So thus, if I replace, so in every equivalence class in mp mod mp squared is an action element in ax. So thus, mp mod mp squared is generated by classes of elements, say, a of x. So I don't have to look at quotients of polynomials. So but it's clear by definition that ax, the xi are just the coordinates. And if I make the shift here, these are still coordinates of an, however, just shifted. So kx1 to xn is also generated by these shifted ones. And therefore ax is also generated as a ring by the ti. So I have that ax can be written as kt1 to tn. It's given by all polynomials in the ti. I mean, I don't say it's a polynomial ring. There are some relations between them. But it can be written like this. It's a ring generated by these. And a polynomial of degree bigger equal to 2 in the ti. So a homogenous polynomial, well, say, a monomial of degree bigger equal to 2 in the ti will lie in mp squared. Because I know the ti lie in mp. It's in the maximal idea. So if I multiply two or more together, we lie in mp squared. So therefore, in mp model mp squared, it's generated as a vector space by the ti. So this proves that it's maybe not so extremely illuminating. But at any rate, we find that these two definitions of the tangent space are equal. And as I said, this old definition somehow has the advantages quite concrete. You can also view it as saying you just take a linear approximation of your variety, just taking the zero sets of the first order part of your polynomial. And so that's kind of very concrete. On the other side here, this will turn out to be quite useful also because it's very similar to how you view things in differential geometry. It is very reasonable to view a tangent vector as something which acts on functions. So you could, for instance, so one classical way of, I mean, more elementary way of looking at the different geometries, you look at all curves to the point. And you look at all the tangent vectors to the curves. So the derivatives along the direction of the curve. I don't know which one Retzor uses. But anyway, there are several versions of defining tangent vectors. And many of them involve somehow. So to say it one last time, a tangent vector is in some sense something is the same as the operation as taking the derivative along that vector. And so here we have the tangent vector in the old definition as a kind of vector in some vector space in which the variety is a subspace. And this new definition translates it into taking somehow the derivative along the direction of regular functions along the direction of the tangent vector. And these are equivalent viewpoints on the same thing. OK, so maybe that's enough now. We want to just do a few more. So first, now we want to prove to generalize this result that we had in the beginning. We had proven that if we have a hyper surface, then the non-singular points form an open dense subset. And now I want to prove that this is true for any variety. So the non-singular points form an open dense subset. So that every variety is almost everywhere smooth. So this is the following theorem. So let the x be a variety. Then we have that the regular of x, so the set of all non-singular points, is an open dense subset of x. So then along the way we prove another statement, namely for all points p and x, we have that the dimension of the tangent space of x at p is bigger equal to the dimension of x. So you recall that in the definition of non-singularity, we had that a variety, so a point is a non-singular point at a variety if the dimension of the tangent space is equal to the dimension of the tangent space. Now if you have a singular point, it's the other point where it's not equal. But the statement here is that if it is not equal, it must be bigger. So the dimension of the tangent space can never be too small. It can only be too large. So well, actually it's not one. It's always we kind of mix up the two proofs. So x has an open cover. We will reduce to the fine case and then use the old definition of the tangent space. So it has an open cover by a fine varieties. And it's clear the theorem is true if it is true for each open set in the cover. Because here we say that the regular locus of x is an open dense subset of x. If we have an open cover of x, and for each open set in the cover, the intersection of that open set with the regular locus is dense in that open set, so then x-ray is open in x, open and dense in x. Because you can check such things by checking it for an open cover. And this statement is also true. Here I'm just saying something about every point in x. The dimension of the tangent space of x is the same as the dimension of the tangent space at p of an open subset of x. Because it only depends on the maximum ideal at p. So we can, in order to check the theorem, it's enough to check it for each open set in an open cover. And we take an open fine cover. So therefore, we can assume that x in an is a closed sub-variety. Just we replace the elements in the open cover. In the open and fine cover will be isomorphic to closed sub-varieties. The statement is obviously invariant and isomorphism, so we can as well assume this. So we have immediately gotten rid of the general setting. We are, again, just talking about the fine varieties. So let's assume, let's say, f1, the ideal of x is equal to the ideal generated by some elements f1 to fr, where the fi are some polynomials. Then we know that the dimension of the tangent space of p of x, when I just write it down first, is equal to n minus the rank of the Jacobian of f1 to fr. If you recall, the tangent space of p of x is the kernel of the Jacobian as a map from k to the n to k to the r. And so we get the dimension of the kernel is this. So this is of the Jacobian at p. So this is a matrix of polynomials. We evaluate p. So the locus where this matrix has a certain, has a rank smaller equal to something is a zero set of certain minus of this matrix. So of the determinants of a certain, of all the determinants of a certain size. So therefore, we know, so thus, thus, so if you look at the, so thus for all d, xt is a set of all p and x, which I defined to be a set of all p and x. That's the dimension of tpx is bigger equal to d, is the size, is the zero set of certain minus of this matrix. So it's closed in x for all d. So every d, I get a closed subset xd. And now we choose. So and obviously it's clear that, by the way, this is written that xd contains xd plus 1, contains xd plus 2, and so on.  They might all be equal, but at least, or I mean, but at least, you know, the bigger the d is, the more conditions you have. So we choose the largest d such that xd is still equal to x. For instance, if we put d equal to 0, the dimension of the tangent space will be at least 0. So xd is certainly equal to x. If we put d larger than the dimension of the, than the n, the dimension of the an, as the tangent space is a subspace of an, then xd will be the empty set. So there will be some last time when xd is equal to x.  This is, by definition, not equal to x. And it's an, this will be an open, dense, no. So it's not, by definition, it's not the empty set. And it's open because xd is closed. So this is an open, there's an open subset of x, because a complement of this closed subset. And it's dense because it's a non-empty open subset. And x is irreducible. So then we know, by our definitions, the dimension of the tangent space of x at p is bigger equal to d for all p in x. Because we have chosen d to be the largest one such that xd is still equal to x. So for all p in x, we have the dimension of the tangent space bigger equal to d. And the dimension of the tangent space of p of x is equal to d for all p in x0, which is an open, dense subset. So this means we prove the theorem, both part 1 and 2, if we prove that d is equal to the dimension of x. Because then it precisely says, for all points in x, the dimension of tpx is bigger than, bigger equal to the dimension of x, which is part 2. And there's an open, dense subsets where the dimension of the tangent space is equal to the dimension of x, which means there's an open, dense subsets where x is smooth, which is part 1. So we only have to show this if I have it. Now we come back to this theorem that we proved. The other time to prove the statement that two different definitions of dimension were equal, namely that every variety is birational to a hypersurface in a to the dimension of x plus 1. So x is birational to a hypersurface in x. And they find space of dimension of x plus 1. Now this is something we proved. So it's birational. So there are open subsets which are isomorphic. So let u in x be a non-empty open subset, which is isomorphic under this birational map to the hypersurface. So maybe hypersurface, I call it y, which is isomorphic, say, to an open subset of, say, the regular locus of y. So you have to remember that x is birational to a hypersurface y in x plus 1. It means there's a non-empty open subset u tilde of x, which is isomorphic to an open subset of y. But we know if we have a hypersurface, we have proven that it contains an open subset, that the regular locus of y is an open subset. So we intersect this open subset of u, which is of y, which is birational to the open subset here with the regular locus. We still have an open subset. And we can find an open subset of x, which is isomorphic to it. But now we know that y has dimension, the same dimension of x. So if I have a point p in y, the dimension of the tension space is the dimension of x. So then the dimension of tpx is equal to the dimension of x for all points p and u. But now we have, so u is an open subset of x, a non-empty open subset of x, and x0 is a non-empty open subset of x. So x is irreducible. Any two non-empty open subsets intersect. So we have u intersected x0 is non-empty. So but on the other hand, the dimension of the tension space for every point of u is the dimension of x. And for every dimension of tpx, on the one hand side, it's equal because we are in u to the dimension of x. So p is in u. And on the other side, so let p in element in u intersect at x0, as p is in x0, it is equal to d. So obviously, it follows that d is equal to the dimension of x. And that proves it. So it's not so, you know, but OK. So in some sense, if you think of it, the proof is not so very complicated.  OK, so we know now that every variety has an open-dense subset, which is non-singular. So I want to make one last statement about this. This is about how one can a slightly more simple way check the non-singularity of a variety by, I mean, I will maybe just leave it brief because it's not so exceedingly exciting. So if x in an is in a fine variety, and ix is generated by f1 to fr. We find that p in x is non-singular, if and only if the rank of this Jacobian at p is bigger equal to n minus the dimension of x. OK, and this is clear because, I mean, by definition, the dimension of the tangent space is n minus this rank. So by definition, it's non-singular, if and only if here we have equality. But we know, by what we just proven, that the other inequality is true. We know that the rank of this is smaller equal to n minus dmx. So if we have this, then they're equal. And the second statement is the same in the projective case. So first, I have to write it. So if we have some polynomials in kx0 to xr xn homogeneous, then p in x is non-singular if and only if I write down the same. So I'm sorry. So I first have to define the Jacobian of this. In this case, I just take the same Jacobian, but depending on all n plus 1 variables, no? Let 1 by dx0 until df1 by dxn. And here goes on, dfr by dx0 until dfr by dxn. Then p in x is non-singular if and only if I can't forgot. So I have, again, I assume the homogeneous ideal of x is generated by these elements f1 to fr. Then p in x is non-singular if and only if. The dimension of the rank of the Jacobian is bigger equal to n minus dimension of x. So the same statement. It's a slightly more complicated if you think of it, because here we actually, I mean, we are in the projective case. We have to somehow reduce to the affine case. So you can assume that our point p lies in the open subset u0. So the first coordinate is 1. And then you can express this Jacobian in terms of the Jacobian of these. But it's still a little bit tricky, because you actually have now one more. This matrix is slightly larger than what you would think. So you could worry that the rank of the matrix might be too large. But it, so let me, I mean, as I said so much, I have to maybe explain properly. So we can assume that p is of the form 1 a1 to an. And this is the point in an. And we put fi, x1 to xn, to be the large fi of 1, x1 to xn. So we de-homogenize. And then we know by definition that p in x is non-singular, if and only if it is non-singular as a point of the intersection of x with the n. So that means if and only if a equal a1 to an is a non-singular point of the 0 set of this, which is just equal to x intersected with the n. I mean, identifying this thing with the 1 with the vector in an given by a1 to n. And if you look at it, if we take dfi by dxj for j from 1 to n at the point 1 a1 to an, this is the same as dfi by dxj at the point a1 to an. So somehow if you want to compute it, it's the same thing. Actually, it's just in a very strong sense the same thing. So that means if we write down this here, j of f1 to fr at our point p, this is equal to fdf1 by dx0 until dfr by dx0. And here at p. And the rest is just j of f1 to fr at a. Matrix looks like that. So by definition, we have that our point a is a non-singular point of this thing even only if the rank of j of f1 to fr of a is bigger equal to n minus the dimension of x. Now, and you want to claim that the same is true for this bigger matrix. So we have to see that the matrix does not, the rank of the matrix does not get larger when we add this row, the first column I think it's called. Well, and so that means we want to show that the first column is a linear combination of the other columns. And this follows from Euler's formula. I actually didn't want to explain it, but as I started, I cannot give it anything like that. So we have the fi, a homogeneous of some degree, which I may be called di. And so there's Euler's, there's the well-known Euler formula, which says that if you take sum j equals 0 to n, xj dfi by dxj, this is equal to di times fi. This is a formula which is some trivial calculation that this will be true. If you make this derivative, you can just see that the factor di, when you make this partial derivative, the factor di comes down and you lose 1xj. I mean, it's not, so that anyway is the statement. And so therefore, if you take dfi by dx0, this can be expressed in terms, say, at the point p. So remember that p is equal to 1 a1 to an. So I put p here, the coordinates of p for the xj. And I bring this one to the other side. So then this will be equal to sum i, if I want to i, j equals 1 to n ai dfi by dxj. Yeah, the point is that at the point p, p is a point in x, these fi are all 0. It looks like that, except there's a minus. I have that the sum of all of these is equal to 0. And I bring this on the other side. So that means, in other words, that the first column is a linear combination of the other columns. OK, so this was just this criterion to check some points non-singular. Maybe just check with, I forgot something. Yes? The Euler formula, well, it's precisely this formula. So if you have a homogeneous polynomial of degree d, then if you take this sum of the partial derivatives multiplied by the variable, you get the degree times the function. That's what I call the Euler formula. And if you have some doubts about it, it's an exercise. It's something that one usually does as an exercise in the second year of analysis. Or even in the first year, it's some formal manipulation. Oh, there are many Euler formulas. Euler is maybe the most, I think, he might be the mathematician who published the most. And he was very much into formulas. So I mean, in principle, if you say Euler formula, I can think of, well, at least a handful of things which I would call Euler formula. This is maybe the easiest thing which has that name.  I'm not even sure whether everything, oh, well. I mean, it's one thing, it's all computations. So it's not that you have much of it is computations. So it's not that you have to sit down for a few months thinking about some deep property. But often it's really just calculations. And he was in Saint Peter's book in this place.  So he would just write his notes. And whenever the pile of his notes was high enough, he would take the notes and give them to the printer. It's maybe, and obviously he could work. He was very brilliant, could work very fast, and was very good with formulas. No, it's not, I mean, if I had things like that, it wouldn't. OK. So now we come to the next paragraph, which is let me first see whether I want to say something more. No, no, I think that's fine. So now I want to talk about non-singular points of curves. So the special case of curves. And somehow, see that if you have a non-singular point of curve, the local ring has a special property. It's a so-called discrete-valuation ring, which basically means that for every function, first for every regular function on the curve, you can say to which order it vanishes at this point. And this is the number that you can associate to the function. But this turns out to be useful. Finally, you can do it for every rational function. So it's like in analysis. If you do complex analysis, you have this thing. You look at meromorphic functions on the plane, and they have some order of pole and some order of 0, depending on whether they are homomorphic at the point. And the same kind of thing with the order, you also have four rational functions on curves. And this turns out to be quite useful. And actually, in the last chapter of my notes, which obviously I didn't really plan to do in the course, because there's only time enough if people ask me to give some extra lectures. And I do not ask you to ask. So in the last chapter, I somehow give some consequences of that.  But I don't think I think you have quite enough exams now so that you're maybe not too eager to have this now. But I will first just set this up, how it starts, what is the local ring of curve looks like, and some first consequences of this. So this is about non-singular curves. So now in this section, we want to talk, so we want to say a curve is supposed to mean just a variety of dimension 1. And the non-singular curve would be a non-singular variety. And we want to study non-singular points of curves. C. And as I said, the special fact that you will find is that the local ring at this point is the so-called discrete valuation ring. So first, I have to again talk a little bit about modules. So recall the definition of module over ring R, maybe A. I'm not sure whether I mean it's very, as you remember, the actions of a module over ring are precise the same as the actions for vector space over field, only that we replace the field by ring. So that means you have a sort of activity for the, so I can just at least say it means you have A is a ring, M is in a B in group, and you have an action from A times M to M, which satisfies the usual actions, which I maybe refuel right down again. So if you have, so if something is A, it's an A. And if something is with the M, it's an M. So if I have A1, A2 times M, this is equal to A1 times A2 times M. So as a strategy, then we have two distributive laws.  So this particularity in M, and we also have A1 plus A2 times M is equal to A1M plus A2M. And finally, we have the obvious thing that the one element of the ring should act as the identity on M. So these were these actions. And we talk, so then in this case, so M is called an A module. If we have this for all A1, A2, and A and A, and all M, M1, M2, and M, as usual. And so we also talked about modules, about module generated by some elements. So if S is a subset in M, then the A module generated by S is a set of all linear combinations of elements in S with coefficients in A. So is the set which I call S, which is the set of all A1S1 plus ANSN for some N. And the AI are elements in A, and the SI are elements in S. Linear combinations are always finite, but S could be an infinite set. But we have just like four vector spaces. And we call M is called the finitely generated module that we had before if it can be written as generated by a finite set. We had also recalled that we know that if we have an ideal, so if I in A is an ideal, then I is by definition an A module in the obvious way by multiplication in A. And its generators as an ideal are the same as its generators as a module. Finally, piece of notation. So if M is a module and I in A is an ideal, we can write IM to be the set of all say B, M, where B is in I and M is in M. And it follows from the fact that I is an ideal, that this is a sub-module of M, A module. And it's obviously IM. It's contained in M. Now I want to at least state the first result that we need, which is the lemma of Nakayama, so somehow the most well known result about modules over local rings, which one uses in many contexts. So let A be a local ring. So if I do it in two parts, first I do the general set up. Let A be a local ring with maximal ideal M in A, maximal ideal. Let large M be a finitely generated A module. So the first statement is, I mean, you have said that if I take these products by definition, if I multiply with all elements of A, I will lend still in M. So I know that IM is contained in M because I is a subset of A. But one could imagine that it happens that IM is equal to M, that it doesn't get small. So if I do this for the maximal ideal, so if it is true that M is equal to the maximal ideal times M, so can this happen? The answer is no. If this is true, then M is 0. And the second statement is somehow corollary. If you write, so A is a local ring. So if you take A divided by M, it's a field. It's M is the maximal ideal. So write K equal to A modulo M. So assume we have some elements, F1 to FR in M. Be some elements. So assume we have such elements, F1 to FR in M, such that the classes F1 to FR generate this quotient M, modulo MM as a K vector space. So if I take the quotient of these two modules, this will be a K vector space because I can multiply with elements of A. But if the element of A lies in the maximal ideal, then I will lie in MM. So we have an action of A modulo M, which is K. So it's a K vector space. So assume that, then F1 to FR generate M as a module. So that means in order to check that we have a set of generators of this module, we only have to check somehow to first order. We only have to check on M modulo MM, which is much easier. OK. But I have to warn you, the assumption that M is a finitely generated AME module comes before. If I have a module, if I have an A module, where A is a local ring, of which I don't know that it's finitely generated, even if I find finitely many elements such that their classes generate this quotient, I do not know that they generate it as an AME module. It might not be finitely generated. I have to assume before that it's finitely generated. Maybe I can still prove the first part, which is relatively simple. So we do it indirect. So assume that M is not the zero module, not. So we know it's finitely generated. So let U1 to UR be a minimal set of generators of M. SA module. So we know there is a finite set. There must be a set with the smallest possible number of elements. And we will want to bring this to a contradiction. We want to find a set with one less generators, a set of generators, which is one smaller. So we note that the last one, UR, is an element of M, which is equal to M times M, according to our assumption. So that is, we can write UR as in a combination of the UI with coefficients in M. So sum I equals 1 to R MI UI. Because you can write any element as in a combination with coefficients in A. And now we multiply with elements in M. And so we then in M. Because M is an ideal in the maximal ideal. So I can bring this to the other side.  That's simple enough. But the point now is that this element is a unit. Why is that? I mean, either it's a unit or it's not. Or it lies in the maximal ideal. We have a local ring. So if not, we have that 1 minus MR lies in the maximal ideal. Because we know that all the nonunits in a local ring lie in the maximal ideal. So we also know, thus, if we take 1, this we can write as 1 minus MR plus MR. So now both lie in the maximal ideal. This will also lie in the maximal ideal, which is obviously contradiction. 1 is certainly a unit. So this is a unit. And so we can bring it on the other. We can divide by it.  Or if I want, I can write it like this. So this is some. So now we see we have written UR as a linear combination of the other UIs. So thus, already U1, so UR minus 1, generate the ideal, generate M, which is a contradiction. We have started with the minimal set of generators. We found one which contains one less element. And so this is impossible. And this proves the first part. Second part we will do next time. And then we'll see why this should help us with our business. I think we meet again on Monday. I also hope that I can correct your exercises finally. I was a bit busy recently with other things. OK.


This morning at approximately five minutes from seven, members of the second district received a run to respond to the 49 and the block control street. Upon their arrival in the scene, they discovered four bodies, three of them in seats. One individual was transported to the hospital. Obviously, this is an ongoing investigation at this time, we're not going to release for particulars as it relates to the ages and the names of the individuals attending the location of their families. I would ask that we all keep those involved, those family members that are going to be facing this horrible tragedy and how it's... What time did you say? It was about five minutes to seven this morning. Six o'clock. How did the call come in? It was members of the second district received a call from communications to respond to the 49 and the block control street.


Let me use a word, prescient. But so does that kind of word, you blur, you know, prescient, I'm not exactly, I think it's prescient? Recent searches on Google. Are New York bagels better because of the water? And this was a disagreement that we all got into. My feeling was no. The short answer is it's mostly not the water. It's mostly how they're prepared, they're boiled. But the water does have something to do with it, something to do with something, something, something around there, some molecules. I wasn't as right as I thought I was, but I was pretty right. East first West Coast bagels, East Coast first West Coast baseball, pizza, baseball.  In fact, they don't care. But I think New York certainly has a handle on that, although LA seems to have had a renaissance as of late. Really truly, what do I know? You guys win bagels and you win pizza. No one's gonna argue with that. Although we win out. So they're gonna get less interesting from there. Truly, I apologize for how boring they're gonna be, but perhaps you can pull something interesting out of them. I look up a lot of words, prosaic. The word prosaic is so beautiful. It's such a beautiful, flowery word. You know, you would think it means beautiful language and a lyrical when in fact, it means the opposite. So a bit of a head fake. Could I even use it in a sentence? Um, didactic. I'm flirting with using it in real life. Pious is good. But you know, you don't wanna be a snob either. And I love language, but there's a bounce, you know?  I just kind of absorbed it and out it goes and sometimes it's right, sometimes it's not, but it's kind of a thrill. Lastly, I looked up a recipe. I was making some soul, the dover, the dover that is. Garlic, caper, lemony, sauce, lightly fried dover sauce. I came across a recipe. I don't know if it's the recipe's fault or my fault. I kind of winged it regardless of bad. It was buttery, just too buttery. The flavor profile was just butter and it smelled like butter and it looked like butter. Butter's good and a ton of butter is delicious in the right thing, but for some reason, this didn't work. I also don't weirdly love the smell of butter.


It's the mat work. The brevity and all that, like I often how I mostly deal with that myself, whether it's with the videos or being productive and of course songwriting is I always have to tell myself like in school, they used to be like, look, if you are a B plus student consistently just by proxy, you usually end up with an A in the class because a lot of times it's just like when I'm writing that B plus paper, and I kind of know it's a B plus paper. I forget that there's that 10% in the class back in school, like 10% of the classes attendance. We forget that and I and that always hurt me. That's why I was never eight plus because I just with that 10% gone, so I'm at I'm at a 90 right now. But if I'm just writing B plus papers consistent consistently, but the other 10% is just attendance, then I'm an A plus student. And from the creative aspect, if I'm dropping B plus videos, every single day or five times a week, I'm an A plus YouTuber. If I'm giving my B plus effort with my coaching, I'm an A plus coach because I'm there, right? Now I tend as a create as a songwriter. I think the song that I just wrote is a B plus at best. But I released the song, but I put the $1,000 in promotion in it, then I'm an A plus. You know what I mean? Where I always mostly have to remind myself of that, right? This interview right now, we're flowing. It's good. It's beautiful. Like, am I sitting here like it? It's definitely the 4.0 level. You know what I'm saying? It's too tight. I don't know. I'm really trying, but I know we're here and I know we're pushing it. And I think that a lot of artists, that would help them along as well. You know what I'm saying? Just from a creative standpoint, I'm like, look, man, finish the song. Try your best to get it to like a B plus level. Release the song. For me, that's an A plus because it will come. The doors will open. You know? Yeah. I think that's it, man. Like half of the game is showing up really. I know, I've heard too many people, right? The whole hard work beats talent when talent doesn't work hard. I know too many people, whether just in sports or not, that that consistency has really made it a thing. The thing is though, too many artists, right, get caught up in the consistency of creating and putting out without considering the consistency of marketing. You have to create consistent awareness. And then you can expect elevation over time. The Russ example is the perfect thing because people don't understand, okay, yeah, Russ dropped the song a week. Perfect. And Gary Vee says, drop like 365 songs a year. Perfect. I can do that. And then it's just going to happen. Wrong. Like what people miss when Russ got on SoundCloud, it was a completely different landscape. And these things evolve quickly. Trust me, coming from somebody in tech, man, like these things evolve super quickly. The incentive, the way these platforms work, right? The incentive for them to give you organic traffic is different because they're trying to attract people onto the platform, right? Once they hit a certain amount, right? And I got, you know, Drew on here, I got Brandon Sean on here. I got all these other billion channels on here and we're good. We don't need people to convince themselves. People almost have to be there just a part of their business or a part of their life, you know, to stay in socially. I don't need to convince people and show them necessarily the best type of content and give them these great organic reaches. They're going to be here regardless. So now my incentive or my best interest is really just monetize this intention, right? Squeeze it for as much money as it's worth. Maybe that kills the business in the long, long term.  I can't expect to post and get this organic traffic. So let's look at SoundCloud. When he was posting, it was a period of time where you post a song and you're going to get a certain amount of organic traffic. It doesn't mean necessarily like how some people would post on Instagram early on and get their very first post. Not trying, not knowing anything, get like 60,000 likes, right? And in 5,000 followers just off of that, it doesn't mean you're necessarily, he was necessarily getting that. But sometimes it'll be like, oh, this gets showed to 5,000 people. This song got showed to 10,000 people. This song blew up to like 100K or this song just got an extra thousand or whatever. It's random, but the point is the algorithm makes every single song get shown to new people, particularly based on the tags and all that stuff in the beginning. Once they get a certain amount of people, again, that's not happening for you. You post a song and if you don't promote it, you got zero views for life, right? So you can't, that's the problem when people follow these strategies as well. Certain things, a lot of times we don't try stuff until somebody else shows that it works. And most people are in a position where if it worked for that other person, so many other people who also flood into that market when they see it works, you're basically putting yourself in a losing position because now you're going to be too late and you're going to be behind the curve. You have to figure out what works for you and sometimes it's going to be doing those things that aren't recommended yet, right?  Yeah, it reminds me of Morehouse, actually, with the Morehouse style back in the day. That's how I know the A. They used to say, if you're on time, you're late. You know what I'm saying? Like if you're early on time, that's the most Morehouse thing to say. I never thought about that because I always thought that way about this in particular. Now that it finally makes sense. I've never had a real lifetime that that actually made sense to me. Now I got it. Yeah, man. This is that A culture, man. This is the network.


This is Carol Barrett and Dean Troyer with another episode of the Newton Design series. Dean, why don't you go ahead and start out and introduce yourself to the crowd. My name is Dean Troyer. I'm the PTL of the OpenStack Client Project and I've been around OpenStack for longer than I care to admit. We're glad to have you here today. Can you tell us a little bit about the OpenStack Client Project? OpenStack Client is an attempt to give OpenStack a consistent command line interface across all of the projects. Great. And in Austin recently, what were some of the hot topics that your team was working through? We had a we had a number of specific things but the thing that kept coming up over and over was support for specific for specific projects and Neutron is the big one that's the one we've lacked for a long time and our next real our last release just before the summit included a lot of Neutron commands. The next release hopefully this week end of May will include the bulk of the rest of the core Neutron commands. Oh that's great. So you were able to go ahead and work through what were the priorities on the support for Neutron and get resolution with the team all way in Austin. Right right and in fact and we got together with the Neutron guys a little bit and some of it was there trying to sort out what they what belonged in OpenStack Client directly and what they were going to implement as plugins and once they got that sorted out we could kind of decide where we needed to go next. Great. And can you say a few more things about what your top priorities are as you come into this cycle? One of the things that got a lot of talk in Austin also was the the load time. It turns out that we've discovered that OpenStack Client is responsible for between one to five minutes depending upon the job of the time it takes to run DevStack in our gate which is a significant amount of time and so we spent a lot of time looking at what we can do to to decrease the Python overhead that's involved in just getting every command started. The support of other APIs is big and one of the things that we're going to do is pull a chunk of OpenStack Client out into a standalone library so that it's more formally available for plugins to use. Right now plugins are using part of our code kind of informally and we want to make that a little bit more formal and the biggest thing that's going to enable is for standalone projects to be able to have standalone commands. Example of that would be ironic if you have an ironic installation that doesn't have much of the rest of OpenStack or any of it you want just the ironic commands and we'll be able to support them in creating a command that looks like OpenStack Client but just supports ironic. Well that's interesting I can see more and more conversation on the email list lately about standalone projects so it seems like an important feature to bring in around now. Yeah we've been hearing more about that. Are you familiar with the themes that we've used in the past to characterize the different capabilities that projects within the OpenStack community are putting in for the releases? Yes, yes I am. Oh great. Are there some key themes that you think will be important as you go through this next cycle for the OpenStack Client? There are a few. We don't necessarily fit in since we're a client side not a server side project with all of them but the one that stands out to me on top is user experience. That's the entire point of the project and in fact we've done a number of user experience studies. We did one in April, early April, and we did one at the summit in Austin with operators. People knowledgeable already about OpenStack to get feedback to find out what we're doing right, what we're doing wrong, that sort of thing. So that's really at the top of our list as far as the themes goes and that probably always will be at the top of our list. The other things that we focus on that fits into that pretty well is the modularity. OpenStack Client is expandable via plugins and a lot of the projects, we only support five projects in the repo directly. So everything else, for example, Trove and Ironic and everything else that's not one of the original, basically the original five, is supported via plugins and so those are the things that we're going to continue to make sure is better. Like I said, the library is going to help with the plugin support. Gotcha. And then as you look on to the Makata timeframe, it sounds like UX and modularity would continue to be high priorities? Yeah, I don't see much changing. Again, our release cycle is independent of the integrated releases. We do throw a stake in the sand saying, okay, we released Makata Metaka, excuse me, and this is what we had current at the time, but we aim to be able to support every supported release. Right now, we still support Kilo. So the things that we're looking at implementing will follow things that are new in a release. So some of the new stuff in Newton we should have ready to go, I hope, by the time we get to that release, but looking out past that, there's nothing specific to the client that we have on the radar yet. It's going to a lot be driven by the features that get implemented that we need to support. That makes sense. Alrighty. Well, thanks, Dean. I think that gives us a good sense of what you're going to focus on during the Newton and some leading edge as we start to think about Makata. So I appreciate you joining us today. You're welcome. Bye-bye.


ing Officer we have been burned in Scotland by O&S reclassification in the past so we should be ultra cautious as we proceed we should take every single bit of advice that we possibly can and if there is any doubt whatsoever those clauses must be removed from the bill and it is incumbent on the Scottish Government to take that advice to publish that advice and to remove any doubt whatsoever because the consequences of not doing so could be dramatic on our universities and it could take many many years to fix Thank you Mr Brown that concludes the debate on Scotland's universities we now move to the next side of business which is Considered of business motion number 14620 in the name of Delford's Patrick on behalf of the Parliamentary Bureau setting out a business programme any member wishes to speak against the motion should press a request speaker now and I call on Delford's Patrick to move motion number 14620 moved No member has asked to speak against the motion therefore I now put the question of the chamber the question is that motion number 14620 in the name of Delford's Patrick be agreed to the motion is therefore agreed to the next side of business is consideration of three business motions I would ask Delford's Patrick on behalf of the Parliamentary Bureau to move motions number 14617 14618 and 14619 setting out timetables for various bills I propose to ask a single question to these motions if any member objects a single question being put please say so now No member has objected to a single question being put therefore I now put the question to the chamber and that is that motion number 14617 14618 and 14619 in the name of Delford's Patrick be agreed to the motions are there for agreed to the next item of business is consideration of a Parliamentary Bureau motion I would ask Delford's Patrick to move the motion the motion will be put a decision time to which we now come there are four questions to be put as a result of today's business can I remind members that in relation to debate on Scotland's universities if the amendment in the name of Angela Constance is agreed the amendment in the name of Ian Gray falls the first question then is amendment number 14596.2 in the name of Angela Constance 14596 in the name of Liz Smith on Scotland's universities be agreed to are we all agreed Parliament's not agreed we move to vote members should cast votes now the result of the vote amendment number 14596.2 in the name of Angela Constance is as follows yes 64 no 54 there were no abstentions the amendment is there for agreed to and the amendment in the name of Angela Constance 14596 in the name of Liz Smith on Scotland's universities be agreed to are we all agreed Parliament's not agreed we move to vote members should cast a vote now the result of the vote on motion number 14596 in the name of Liz Smith as amended is as follows yes 64 no 54 there were no abstentions the motion as amended is there for agreed to the next question is at motion number 14460 in the name of Dolfitt's Patrick on approval for SSI be agreed to are we all agreed the motion is there for agreed to that concludes the decision time we now move to members business members should leave the chamber should do so quickly and quietly


Night combat offers exceptional opportunities for success because of the possibility of achieving surprise and capitalizing on adverse psychological effects on the enemy during darkness. Basically, the fundamentals involved in night attacks are the same as in daylight attacks. However, in a night attack, the problem of control and coordination is much greater. Dispersion is less and movement is considerably slower. You will now see an example of a tank company team conducting a night attack. This is the general situation. The 1st Brigade, 25th Armored Division has been exploiting north with a mission of seizing objective red. Task Force 2-11 has been leading the advance. At dusk, Team Alpha, the Task Force advance guard, is stopped by accurate anti-tank, small arms and artillery fires from hills 501 and 420 south of objective red. The commander of Task Force 2-11, Lieutenant Colonel Hart, has decided to eliminate the enemy positions because he is unable to bypass. He is now issuing his attack order. Since the enemy's major security forces have been driven in by Team Alpha, he feels that surprise can be gained by starting his attack that night without illumination and illuminating when detected by the enemy. He plans to employ teams Charlie and Bravo to conduct the attack from a new direction. When Team Charlie has seized hill 501, designated objective green, and Team Bravo has seized hill 420, designated objective yellow, Team Alpha will lead out in a continuation of the advance to objective red. Team Charlie, commanded by Captain Horton, consists of three tank platoons and one mechanized rifle platoon. Team Bravo, the mechanized rifle company team, is under the command of Captain O'Hare. Team Alpha is commanded by Captain Krebs. At the conclusion of the Task Force Commander's order, the team commanders and the representatives of the support elements will formulate their plans for the operation. Our attention in this film will be focused on Captain Horton and Team Charlie as they prepare for the coming attack. Captain Horton is now back at his own command post. Prior to returning, he issued a warning order to his unit. Now he briefs his subordinates on his tentative plan. This is the plan. Team Charlie will move in column formation from its present location to the attack position, then to the line of departure on these platoon routes. The reason the line of departure is along this main road is because in night operations, control is facilitated by use of readily identifiable terrain features such as highways, ridge lines and stream lines, wood lines and so forth. Captain Horton will move his platoons across the line of departure in platoon columns. Phase line white assigned by the Task Force Commander will be utilized to assess the team commander in controlling the movement of his unit. When Team Charlie reaches the probable line of deployment established by the Task Force Commander, it will take up attack formations and assault the objective. However, if it is detected prior to reaching the probable line of deployment, it would immediately deploy. Beyond and to the flanks of the objective is a limit of advance established by the Task Force Commander to prevent Team Charlie from coming under friendly protective fires. Team Charlie will not cross beyond this line unless ordered to do so by the Task Force Commander. The Task Force Commander has also assigned boundaries to designate the lateral limits of the team's zone of action. At the conclusion of his briefing, Captain Horton and his key personnel will make a visual reconnaissance of the team's routes of movement and zone of action. During his reconnaissance, the team commander selects an attack position. The position must be large enough to accommodate his entire force in the formation it will use in crossing the line of departure. In night operations, the attack position may be closer to the line of departure than in day operations. Captain Horton chooses routes for the movement of his platoons from the attack position to the line of departure. From a vantage point, he selects platoon points of departure from the line of departure. He also selects the direction of attack for each platoon from the line of departure to objective green. In addition to a daylight reconnaissance, if time permits, each platoon leader will make a night reconnaissance to determine the degree of visibility of his platoon sector, his assigned routes, and his platoon points of departure. When Captain Horton and his subordinates are finished with their reconnaissance, Captain Horton will formulate his plan of attack, then issue his order. Attack orders should be complete, covering all details of the operation. They should include the situation, the mission, execution, administration and logistics, and command and signal matters. Briefly, this is Captain Horton's order. The first tank platoon under the command of Lieutenant Turk will attack on the left and seize and secure the west portion of objective green. The second tank platoon under the command of Lieutenant Robinson will attack in the center and seize and secure the north portion of the objective. The third tank platoon under the command of Lieutenant Bushnell will attack on the right and seize and secure the east portion of the objective. Once the attack is detected, or upon reaching the probable line of deployment, all tank platoons will move into line formation and all means of illumination and supporting fires will be employed. Extreme caution must be exercised to coordinate fires and movement on the objective. The mechanized rifle platoon under the command of Lieutenant Reardon will follow the second tank platoon and be prepared to dismount on order. Sergeant Moran, scout squad leader from the armored cavalry platoon, will place guides on the line of departure to assist the platoons in arriving at and crossing the platoon points of departure. Sergeant King, the 4.2-inch mortar observer, will provide fires on call. Lieutenant Condon, the artillery forward observer, will continue the normal harassing fires until the team is detected. At that time, close support fires will be placed on the objective and shifted beyond it on call. He will also be prepared to provide illumination in the zone of action from the time the team crosses the line of departure until the objective is taken and to fire illumination and high explosive rounds beyond the objective. Light aircraft will be used to assist in the illumination of the objective by dropping flares. When required, tanks from Team Alpha will illuminate the team's boundaries. Team Alpha's tanks are used to prevent dissipation of the combat power of the attacking teams. Team Charlie will cross the line of departure in one hour. At the conclusion of Captain Horton's order, the platoon leaders will carry out their troop leading procedures, then return to their platoons and issue detailed orders to their personnel. Thus, every man on the team will know exactly what his job is. Equipment is checked and infrared periscopes are installed. The scout squad now moves out to the line of departure to act as platoon guides. The ground surveillance radar section will assist in securing the line of departure. Ten minutes later, Team Charlie moves out to the attack position. The team moves toward the attack position in column formation. First platoon, command group, second platoon, mechanized rifle platoon, third platoon. In night operations, commanders maintain maximum control of their units by keeping them in column formation as long as possible. At the attack position, the team changes formation from company column to platoon columns and continues forward to the platoon points of departure on the line of departure. At the line of departure, the guides from the scout squad assist the platoons in crossing at their designated points of departure. In night combat, men and vehicles are identified through pre-established means. In this case, guides are identified by luminous armbands. Officers are identified by vertical luminous strips on the rear of their helmets. Non-commissioned officers are identified by horizontal luminous strips, tanks by luminous markers on the turret, and armored personnel carriers by luminous markers or panels. After crossing the line of departure, Team Charlie's tank platoons move out along their directions of attack toward Hill 501 and are now approaching the probable line of deployment. On Hill 501, the enemy forces are alerted to Team Charlie's advance and commence firing defensive fires. Receiving the enemy fire, Team Charlie is forced to deploy into line formation. Denied rifle platoon deploys behind the second tank platoon, ready for commitment by the team commander.


Here we go velocity acceleration and displacement. We spent a lot of time in physics 11 Dealing with that in physics 12. We're gonna assume for the most part that you know it I'm gonna spend one day on it We're gonna be using it over and over and over all year long So the first thing acceleration we defined this last year as a change in velocity Over a time interval and those you that had me last year when I ask you what's changing anything you'll say to me. Oh What is change in anything? Nicole my angel for a candy Final minus initial you see in math notation we get this acceleration equals Change in velocity over time And since change in anything is final minus initial final minus initial do you remember that or do you just read it right there? Mmm We can also write it as this a equals Vf minus vi Over T for what it's worth Emily technically it should be change in velocity over change in time There should be a delta there to remember that from last year But time is always a change in and so we get a bit sloppy and we almost always in physics drop the delta in front of the T because it's just implied This wasn't how you saw it introduced to you last year we rewrote it as this we got the v final by itself Except we're gonna be fussy this year vector vector vector Direction is gonna matter this year Vf equals vi plus at On your formula sheet, but I'll be honest if you're a Good student you'll probably just end up memorizing this one because you use it so often this assumes Uniform or unchanging acceleration if you have changing acceleration you need calculus the math gets ugly Where is there changing acceleration? Brett space shuttle launches as the fuel burns What happens to the overall mass of the spacecraft as the fuel burns? Gets lighter even though the engines are still putting out the same amount of thrust They're having to move less mass the space spacecraft actually accelerate faster and faster and faster and it's beautiful yucky math If we have uniform Acceleration we also can find average speed which I'll sometimes write as the average But if you want the fancy math term and this is what is on your formula sheet the horizontal bar means average How do you find the average how do you find the average let's suppose Brianna has 10 bucks and Emily has 15 bucks What are the average between them? How'd you get 1250 Emily turns out the average is Add V final and V initial and divide by 2 That also implies something it implies Cara that if here is v initial and here is v final and you graph them The average will occur exactly halfway between them if your acceleration is uniform So if you accelerated a constant speed and you start out at 10 meters per second and You hit final 50 meters per second 10 plus 50 is 60 by 2 you hit 30 meters per second exactly halfway through Then we have distance in Physics in physics 10 and science 10 you define distance as average speed times time D equals vt This is before we did acceleration. This was science 10 However, I just told you that the average is This so I plugged it right into there. Is that okay so far? Oh one more thing though V final is v initial plus at so you know what I can replace this v final with right here The initial plus at we can write it like this D equals the initial plus at plus the initial All over to time t. Oh Here's what I see Zay One v initial plus one v initial. I think those are like terms. They're not the cleanest like terms But those are like terms Zay. Do you know what one v initial plus one v initial is I? Can rewrite this as two v initials Plus a t all over to Times t so far so good Mitchell So far so good. I'm gonna multiply the t on to everything Tancy word distribution Front door bomber if you had mr. Rock upper math never understood that one, but whatever it works D equals I'll get to the it Plus a t squared and this is over two and this is over two The whole thing's over two, but instead Kara. I just broke it up. I got a very good reason for doing this Kara look at this first fraction. What's two divided by two? Kayla I was going Kara. I'm sorry Kara there Kayla Kayla What's this first fraction two divided by two? What happens to the twos? And Over here. I have divided by two, but instead of dividing by two I'm gonna write that as multiplying by a half because this is where it comes from Vi t Plus one half a t squared D equals Vi t plus a half a t squared Kayla that's where it came from you don't need to know that I won't ask you to you know redo that on a test But usually I'll try and show you where stuff comes from because I think that helps you both remember and understand the physics so Remember that one from last year Trevor D equals Vi t plus a half a t squared almost all of you memorized it because you use it so often so Let's put that in box number three here on the next page D equals Vi t Plus one half a t squared Oh, except now Sean I need to be vector fussy. That's a vector. That's a vector and that's a vector How do you think I was gonna read that like?  We're going to use this unit is Awkward to derive right now. It actually pops out of work and energy in about two lines really really easily Comes from the fact that kinetic energy is a half and B squared So I'm not going to derive it right now. I'm just going to give it to you and it's this Vf squared equals Vi squared plus two a D and those are all vectors Although here the vectors end up not being quite as important because if you're squaring a negative often the negative vanishes Anyhow, but Do some examples now in physics 11 and in physics 12? I have two acronyms that I always use dope and defect Dope who had me last year. What does dope stand for do you remember? draw little Picture when in doubt dope You have word problem actually to do draw a picture and then defect was another acronym It was my approach and it stood for hey list your data Write down what you know find a formula that has everything And what you're looking for in it, but nothing else insert the numbers and Then crunch the answer. Well, I always say crunch the answer calculate I think was what my teacher used to say if I'm not sure what to do if I come to a new question my dolphin ID Usually it gets me through most of it So we're brushing off cobwebs from physics 11 example 1 a dragster has an acceleration of 18 meters per second squared Who remembers last year? What was 1g of acceleration? how big 9.8 this is not quite 2g's actually look this up dragsters have an acceleration of about this much I Know it's still at least from the internet. So I assume it's actually It begins from rest I Would almost always underline the word from rest because that tells me something very important. What does that tell me? Vi is zero how fast will the dragster be going after 2.1 seconds? So here's my question. What are they asking me to find in this question? first of all, I want to be careful Speed we're doing vectors and if you ever say velocity, I'm always going to say to you which one But this person said it correctly the first time I heard it back there Fine we're being asked to find the finals. So I write down that The reason I always write down what we're trying to find is I've learned over the years my question is no longer blank Brett I'm more relaxed. I'll do better. I've learned that with students over the years writing anything down always helps Now I'm going to defect I'm going to list my data So the first number as I read through here I see the number 18. Hopefully you clued in last year the value of Memorizing what units go with what because I don't actually look at the 18 I look at the meters per second squared and right away. What do I know that is Connor my friend 18 is what? You can write the units when you're deficking or not. I usually don't because I'm in a rush and on a test. I probably would it begins from rest Connor. What did this tell me? I'll write that down and Then I read through part a and what's this 2.1 Connor? Okay? So looking for an equation that has v final a v initial and t in it now in your homework When you get good at this Trevor I don't have a big issue if you jump straight to plugging in numbers on a test or on a quiz I will give you marks for writing out the correct formula Even if everything else is garbage so I'm going to build that habit into you in your notes And on a test and on a quiz. I'm going to always write out the formula. So I heard somebody mutter under their breath I'm pretty sure it's the final equals the initial plus a t which is going to be zero plus 18 times 2.1 18 times 2 is 36 18 times point one is 1.8 36 and 1.8 37 point eight someone want to check me Yes, I can do that in my head I grew up with other counts. Okay units meters per second. All right B How far will it have traveled after 2.1 seconds? Oh? by the way If you're looking for a rough guide as to how big your answer is If you double meters per second, it's roughly miles an hour And I'm assuming you've traveled to the states enough to know a little bit about miles per hour 60 miles an hour is 100 kilometers there. So doubling this It's about 80 miles an hour About 130 140 clicks give or take does that seem reasonable? Sure If I'd gotten 378 meters per second, and I doubled that to 600 miles an hour I'd be saying I don't think you go from zero to 600 in 2.1 seconds. So you can get a rough deal Brett, what's this asking me to find in part B? Mr. Redmond, huh? Got to be really fussy. It's not It is technically the displacement now having said that my big Sin my big mistake. I always mix up the words distance and displacement So I won't jump on you for doing that, but I'll put the vector here and go. Okay So now I have all my defect data listed right above me. I'm not gonna relist it I'm looking for an equation that has D and then some or all of these in it Don't all come up with the answer at once or anything Sorry, no, no, I want the equation. Let's try this again. Okay, right at the answer I'm looking for an equation that has D and some of these other stuff in it I have to assume you guys have forgotten everything Katie. I apologize if you haven't but I have I've learned over the years Go ahead You could use Vf squared equals Vi squared plus 2 a D. I would not you know why Because in using Vf, I'm using my previous answer to find the new answer I'll only use a Calculated answer to find the second answer if I have to because if I got a wrong, I'm getting B wrong I think there is also an equation that has D in it that doesn't use anything calculated And I think it has the D by itself already which one Darn right we can just straight plug-and-shuck D equals Vi T plus a half a T squared I would not take marks off if you use Vf squared equals Vi squared plus 2 a D I would say be a better test writer than that. What did we say Vi was? Whoo God So this is gonna be point five Times eighteen times two point one squared Katie, what's the answer? 39.7 is that two or three sig figs that I'm good with that units Yep Is that right by the way? Yes turn the page How see how long will it take the dragster to reach a velocity of 126 meters per second? Okay, what's this bad boy asking me to find Connor once you done yawning long isn't a variable on my formula sheet I Agree with that So I'll do that What else have they told oh this time they've given me this Connor. What's this? Which one I Agree That's not enough to solve anything. I have to use some other data from the previous page I think we're still assuming that the initial is zero. I think we're still assuming that the acceleration is 18 Is there an equation that has D VF Vi and a in it Connor say yes Okay Watch one and do me a favor once you find it in your head Rearrange it to get the T by itself if you can if you can't that's fine, but we're gonna push you this year Yeah, I don't have them. I don't they're not numbered It's just the first one we did the actual VF equals VF is 80. Is that one? Rewrite it get the T by itself T equals VF minus Vi over a I am totally good if you go straight to this line in fact, I strongly encourage you to go straight to that line and And VF is a hundred and twenty six minus zero divided by 18 126 divided by 18 while I can divide by top by two that's gonna be sixty three. Does it work out evenly? Is it seven point zero even? Yeah This would be wrong by the way How many sig figs? That's one. We said from now on this year two or three. So I'm gonna go seven point zero T time seconds How far will the dragster have traveled once it hits a hundred and fifty three meters per second? I think they're asking me to find a displacement. I think they've given me a hundred to three. That's V final 153 I think the initial is still zero Acceleration is 18 Brett this time. Do I know the time? This time I have to use an equation that has D VFA and I in it This is where I would use the VF squared equals VI squared plus 2 a D equation Except can you do me a favor Brett? Can you get the D by itself in your head and just rattle it off to me? D equals Something squared blah blah blah You do it. I know I'm asking you to do some math in your head first week of second week of school I think isn't it VF squared minus? VI square you'd minus the VI squared over and then you would divide by I Totally agree VF 153 squared minus VI zero squared all over two times 18 more than one number on the top and on the bottom better go with brackets 153 Squared minus zero square. I said I need to type that divided by two times 18 yet 650 Is that two or three sig figs that I'm good with it? Meters units this particular dragster has a top speed of 162 meters per second and then Dragster driver opens his parachute deploys the parachute It takes 12 seconds for the dragster to come to a stop What's its average acceleration reason? I wrote Brianna average acceleration. I don't think it's constant I think initially the faster you're going the more the parachute pulls But when you're nearly slowed down the parachute often I've watched on TV is length and not hardly pulling at all But let's assume it's constant average acceleration otherwise yucky man What's this question asking me to find actually two things? It's got kind of a part a and a part b. So I'm gonna go a little one for part one Andrew, what's the first part asking me to find a Equals question mark. All right, let's work our way back through here. What's this 162? Sorry, I Disagree I don't think it's vf. I Think it's v initial. You know how I know I know what v final is.  Are we speeding up or slowing down? I did Andrew. Are we speeding up or slowing down Andrew? So should our acceleration be positive or negative if we do this, right? I'm expecting a negative answer here. If not, I'll be a little concerned. Oh, I see another number 12. What's that? I Agree Andrew do we have an equation that has acceleration vf vi and t in it? Yeah, can you rewrite that equation getting the a by itself and go a equals? Yep. Yep. I totally agree on the provincial and therefore on my tests You would get a mark just for writing down the correct equation So it's a great habit to get into and it's organized takes one second Andrew, what did you say vf was? I? Agree minus I agree Divided by 12 and oh good. That's gonna give me a negative answer. I have no idea what Bracket zero minus one sixty-two close bracket divided by 12 negative 13.5 Meters per second squared. Yeah. Yeah. Good good turn page. Let's talk about your tests Yo For a second half. Ah, wait a minute. It's a part two. How long do I have to travel? Thank you Thank you Andrew for pointing out my Negligence how long does it have to travel? I think now it's saying Now this is where English is bad because how long can refer to a time or it can refer to a displacement Do you think they want me to find the time here? Say no You know why? Because they told me the time here takes 12 seconds. I don't think that's the ant I don't think they want me to just write 12 seconds or we're done If I gave this to you on a test, I would rephrase this I would say how far does it have to travel before it comes to a stop and I may retype this I keep forgetting to and I do this every year note to sell So I think they want me to find a displacement Do I have all my data listed here, Nicole? Say yes, then I'm not gonna read defect. I got it all there All right, what equation hit? Sorry What equation can I use here? Better have a d in it. So I don't think I'm using vf equals vi plus at because that was like a d in it I don't think I'm using the average equals vi plus vf over two. I think I would use d equals vi T plus a half at square It's got the d by itself now the only issue here bread is that means I'm gonna use the a that I just calculated But I'm pretty sure I'm right and I maybe go back on a test to just double check I really want to make sure having said that on a test This is how I mark Brett Supposing you couldn't find the acceleration Supposing you couldn't do the first part, but you know how to do a second part if you wrote a note to me I can't find part a and you made up a reasonable acceleration you wrote assume a equals 19 meters per second squared and you solved it completely I would work out your answer to see if you did it right and I would give you full marks for part two I will always do that. Okay, the message is Never never ever leave a question blank Always write so so Brandon. We're gonna go like this D equals Vi t plus a half a t squared Vi is 162 Times 12 t plus a half a negative 13.5 12 squared Am I expecting a positive answer or a negative answer here? Which way are we traveling? Forward I think it's we're gonna get a positive answer. We're not traveling backwards yet. Jake. What'd you get? Oh? I thought you looked up cuz you were done Now what'd you get? Sorry, I don't think it's seconds cuz I don't think I'm finding T What am I finding? So what would the unit speed? Is it 972? if we're naughty D equals 972 meters No, ah 162 times 12 plus 0.5 times negative 13.5 Times 12 squared. There is some dissension. I'm getting 972 So if you got something do you get a bigger answer or a smaller answer? Smaller You put a negative there But no negative anywhere else right or you forgot the squared the two most common mistakes back at least once this year after a test You're gonna hear me say squared because about ten people wrote all the right stuff Forgot to do the squared and all got the same wrong answer and I'll notice the pattern when I'm marking it I'll come everyone keeps getting 17.1. That's not. Oh, don't tell me they forgot to hit the squared button on their calculators So don't do that Or Trevor I'll snap. I'll lose it. Can I turn the page now? Whoo-hoo. Let's turn the page You ready for your first shock? I Have based my tests in physics 12 on the used to be provincial that no longer exists And on the used to be provincial there was always listen up an essay question Emily's face just what? sort of the last question of every provincial was a Using principles of physics right to explain question. I Will be doing a bunch of those during the year with you Although I said essay you're gonna find I answer them with almost no English in fact often I'll do a good free body diagram or a couple of lines of work or My last line of defense is I'll make up reasonable numbers and just show that it works But on every one of your tests pretty much there is going to be a using principles of physics right to explain kind of a question Here is your first example of such a question Example three an object starts at rest and it accelerates east at a constant rate Accelerates in which direction Kara? I Have done this stupidly wrong so often as soon as they mention a compass direction I always in the margin draw a compass rose because the number of times I've done stuff wrong drives me crazy Takes one second. So there's my compass rose When it has traveled 10 meters it is moving at speed V check How fast will it be moving when it has traveled 20 meters? Select the best answer V More than 2v Exactly 2v less than 2v but more than V There's how we're going to do this You're going to vote Here's how you're going to vote if you're positive of the answer your hand goes up as high as you can In fact, if you're absolutely positive, you can even stand on your chair if you're fairly sure Not sure Only voting because if I don't vote mr. Duicle notice that I never move my hand and he'll make fun of me and I will Okay, positive all the way to And really haven't got a clue So read the question and you're going to pick a B C or D ready Who says a is the correct answer? Who says B is the correct answer? Z jumps on the bandwagon with Matt Who says C is the correct answer many Who says D is the correct answer? Mitch did you vote? So Everybody look at Mitch Mitch. Do you think a is the correct answer? You haven't voted yet Are you sure or only partly? Yes positive? Maybe not so sure. Okay? I'll catch if you don't vote usually. All right So now here's the here's the rule You can either attack someone else's answer and convince me that they're wrong or you can defend your own answer and convince Me that you're right attack or defend. What do you got? Someone convince me that they're right or that someone else is wrong day What did you pick? You pick D now. Oh you left Matt behind you think D is correct convince me so you wrote this down you said well Vf squared equals vi squared plus 2 a D. Okay Okay, so what Zay is saying I started to make up reasonable numbers, which is a valid approach. He said I'm gonna pretend speed V Which is the initial was what you say. Oh, okay, so you started out here and you said if we're Vi is zero did you get that? A by itself or what did you do now? Let's make up some reasonable numbers Let's say that at 10 meters The speed it was traveling at 10 meters was pick a nice number Emily five So at 10 meters Five meters per second. What was its acceleration? I Don't care. I Could make the acceleration whatever I wanted to by changing the time So I'm going to pretend that it was accelerating at a lovely two meters per second squared And I'm just going to now crunch some numbers and see what happens. Okay, so how fast was it going? after 10 meters five How fast would it be going? After 20 meters well, it would be going vi squared Plus two times what did I say the acceleration was just trying to make up a number to see what would happen? To and here's the key. How far has it traveled? Starting at 10 ending at 20. How far has it traveled? Traveled 10 meters from vi to via that. Okay type in the right-hand side on your calculator and then don't forget to square root if C is right You should get exactly 10. What do you get? Sorry You're jumping on Zay's bandwagon. What did they what did you go with D? Let's see squared plus two times two times ten equals square root Are we faster at 20 meters? Yes. Are we twice as fast? No Says explain your answer using appropriate principles of physics. I wrote this here. I Wrote this here, so I would write assume V at ten is five a is five meters per second squared V 20 equals what did we say using those numbers? 8.06 meters per second Nicole if we'd use different numbers we'd get a different answer, but I don't think we get exactly twice as big What a three dots in a triangle like that mean? It's an abbreviation for therefore feel free to use it.  It can't be the same speed says accelerating That's actually a fairly tough one Sometimes Kara I'll plug in reasonable numbers like we just and that's a great and a lot approach and a lot of you Won't get beyond that. I will try and do some of these algebraically because I'm a nerd as well example for Two cars are at rest facing each other 300 meters apart, but we don't want anybody to die So we're going to assume they're in separate lanes car one accelerates at six meters per second squared up to a maximum speed of 30 meters per second car two accelerates at five meters second squared up to a maximum speed of 35 meters second How long to take the cars to meet? Assumed they are in different lanes, so there is no crash This is actually a fairly tough question You did this last year calling it relative velocity But the only way you did it last year was we had both had the cars already at their top speed the fact that they're accelerating at different rates makes us a fairly yucky question because You'd have to figure out how long it takes this car to accelerate how much time passes How much time passes for this car to accelerate? How far this car is traveled how far this car is traveled? I've talked a long time and I can tell you guys are kind of zoning out So I'm actually going to say don't worry about this one. This was kind of like an advanced bonus one The nerd within me likes it, but that's okay. What's your homework on a separate piece of paper preferably you can do numero, huh? three Four five is good Eight because I've been on a slide before Okay, and you will notice the answers are in brackets I will do that almost all the time or they'll be attached at the end But I will almost never give you homework without the answers let you check your answers I have one more thing for you