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Steps of Customizing the Form Dos 0999

here we're continuing our path down a.very elementary approach to differential.forms and just recall where we came from.so last time we determined the.difference between RN and T PRN in other.words the tangent space based at P of RN.and the idea was that RN were these.points in n space and then the elements.from the tangent space at P of R in.other in other words T PR in those were.n dimensional vectors based at a point P.in RN so I've drawn a little picture.here so R 2 vs T PR 2 so R 2 is just.this plane notice I've got the x-axis.and the y-axis so generally we'll take x.and y to be these coordinate functions.that project to these axes and then what.we have here is this point P which I've.put in yellow in r2 so that point is.just sitting there and then I pick a new.coordinate system at that point P and.that new coordinate system has axes that.are orthogonal to each other DX and dy.and then this vector VP is inside TP r2.and so notice it's a vector that is.based at P and kind of pointing up in.this direction ok so just to reiterate.here here we're thinking about this.point P as being a point in the plane so.in r2 but then we're thinking about this.vector VP as being in this tangent space.of the plane based at r2.ok so the next thing that we want to do.is look at the notion of a one form and.let's just look at the definition so a.one form is a linear function from TPR.in in other words from this tangent.space up to just the real numbers so in.other words the inputs of a1 form our.vectors n dimensional vectors depending.on what.dimension your working end and that.outputs are just scalars and so there's.a bigger notion from this from linear.algebra known as the dual space and.that's what I have right here in other.words Omega is in the dual space of this.tangent space so let's maybe look at.that look at let's make me write down.that so this is the dual space and for.any vector space you have this notion of.a dual space and that's all linear maps.from the vector space up to the base.field so I'll let you guys look into.that if you want to okay so now let's.look at a basic example here so let's.look at our example of r2 and then TP of.r2 in other words the tangent space of.r2 based at P so if we want oh may go to.go from TP are 2 up to R and be linear.we that restricts it to a very special.set up and so that means that Omega has.the following form so if Omega takes an.element from TP R cube we'll just call.that dxdy kind of like we did in the.last video then the fact that this is.linear means you just get a times DX.plus B times dy so already this is.probably looking familiar door something.from vector calculus notice a line.integral over a vector field would have.something like that so a line integral.over a vector field more generally is an.integral of a 1/4 and that's what we're.going is that this type of.generalization ok so just to reiterate.since Omega is a linear function from.this tangent space up to the real.numbers we know that it has to have this.form so it has to take this element from.the tangent space and turn it into the.just this number so here we're thinking.DX and dy are just numbers in this case.now the next thing that I want to notice.is that this is really just exactly a.comma B dotted with DX come.d why so that's pretty easy to see that.that is just the dot product of those.two vectors but we can write this in a.really important way and that is this is.the same thing as the magnitude of our.vector a B and then times this scalar.projection so I'll just write scalar.projection of DX dy onto a B so in other.words the intuition here is that a one.form is a multiple of the scalar.projection onto some line great so in.some ways this one form is defined by.this vector and what's happening is.you're just projecting on to that line.ok now that we have an intuitive feel of.what this one form is in terms of a.scale or projection let's go ahead and.push this up to a one form on T P RN.using the fact that it still needs to be.a linear function from this tangent.space up to R so let's just say we've.got Omega and it goes from T P R n all.the way up to R and so the fact that.it's linear means that when we take.Omega and attack DX 1 up to DX in in.other words that's an arbitrary shape of.an element from T P RN that's just going.to give us this sum a 1 DX 1 all the way.up to a n DX n so you can think about.that in the same way as this intuition.right here you're projecting onto a line.in n-dimensional space and then you're.scaling that projection okay I'm going.to go ahead and clean up the board and.we're going to look at a couple of.simple examples using this intuition so.now we want to do some examples built.off of that intuition that we had and so.let's first of all define this one form.as Omega attacking the vector D X comma.D Y is equal to.three DX plus two dy and the question is.what line does omega project this vector.on to and we can actually figure that.out pretty easily because we know that.the line is in a certain direction and.that direction is 3 comma 2 and we can.see that pretty easily because we have.Omega evaluated at dxdy is going to be.equal to 3 comma 2 dot DX dy by what we.had seen before.and so it's going to project it into.that direction and so notice that 3.comma 2 is parallel to the line 1 comma.2 over 3 so that's pretty easy to see we.just take a scalar multiple of that but.now if we look at that line on the dxdy.axis so if we have this DX dy plane and.then if we go here 1 then that means we.need to go here like 2/3 so that's going.to be right about there so that would be.the line in this case in other words.this line has the equation dy equals 2/3.DX ok so that is the line that this.thing is being projected on to and.recall that that line is happening kind.of in this coordinate system which is.based at P and kind of all living inside.a bigger copy of our two okay good so.I'm going to clean this up and then.we're gonna look at one more for our.next example we kind of want to work.backwards so let's suppose that we know.that a1 form Omega scalar projects.vectors onto the line dy equals 2 DX.with a length of 3 and our goal is to.find like a formula for Omega.and we can do that using the fact that.we know that Omega evaluated at dxdy.needs to be equal to a comma b dot DX.comma D Y in other words it's a DX plus.B dy and in order to scaler project onto.this line we need a comma B to be.parallel to 1 comma 2 and Y 1 comma 2.because this is the vector in the.direction of the line D y equals to DX.and that's pretty easy to see because.this thing has a slope of 2 which means.like the Y component is changing twice.as fast as the X component and that's.exactly what we've got going on here so.that tells us that this a comma B is.really of the form a comma 2 a great now.we need to use the fact that this needs.to have a length of 2 sorry a length of.3 and so in other words we need to take.the length of this thing a comma 2 a and.set it equal to 3 ok great so now let's.just recall that the length of a vector.is going to be the square root of the.sum of the components squared so we have.the square root of a squared and then.for a squared equals 3 so that's that.length condition but that's going to be.the same thing as a times the square.root of 5 equals 3 because notice we've.got 4 plus 1 but that tells us that a.equals 3 over the square root of 5 so.not so nice but that's okay and so what.that tells us is that B is equal to 6.over the square root of 5 and so we.found a and B but those completely.determine our one form and now that.means that we can write Omega in the.following way.dxdy instead of writing as a comma B dot.DX dy I'll just write it as a DX plus b.dy using those values of a and B so here.I'm gonna have 3 over root 5 times DX.plus 6 over root 5 times dy and this.would be this one form that satisfies.these conditions in other words it's the.scalar product projection onto this line.dy equals 2 DX and it multiplies that.scalar projection by a length of 3 okay.so that's a good place to stop.

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Form Dos 0999 FAQs

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How do I fill out the form of DU CIC? I couldn't find the link to fill out the form.

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