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OK, so today we have a new topic.
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And we're going to start to learn about vector fields
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and line integrals.
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Last week we've been doing double integrals for today,
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we just forget all of that, but don't actually forget it.
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You know, put it away in a corner of your mind.
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It's going to come back next week.
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But what we do today, we're going to do line integrals,
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and these are completely different things.
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So it helps, actually, if you don't think of double integrals
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at all while doing line integrals.
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So anyway, let's start with vector fields.
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So what's a vector field?
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Well, a vector field is something that's of a form.
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Well, it's a vector that where m and n, the components,
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actually depend on x and y, on the point where you are.
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So there are functions of x and y.
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So what that means, concretely, is that at every point
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in the plane, you have a vector.
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So in a corn field, everywhere you have corn,
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in a vector field, everywhere you have a vector.
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That's how it works.
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So a good example of a vector field,
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I don't know if you've seen these maps that show the wind.
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But so here's some cool images done by NASA.
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So actually, that's a picture of the wind patterns
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of the coast of California with Santa Ana winds.
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In case you're wondering what's been going on recently.
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So you have all of these vectors that show you
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the velocity of the air, basically, at every point.
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I mean, of course, you don't draw it at every point,
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because if you draw a vector at absolutely all
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the points of a plane, then you just fill up everything,
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and you wouldn't see anything.
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So you just choose points, and you
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draw the vectors at those points.
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So here's another cooling matrix of something done.
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That's a hurricane of the coast of Mexico
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with the winds spiraling around the hurricane.
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So anyway, it's kind of hard to see.
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You don't really see all the vectors.
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Actually, I guess the autofocus is having trouble with it.
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Yeah, I can't really do it.
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So I guess I'll go back to the previous one.
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But OK.
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So anyway, a vector field is something
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where at each point, in the plane,
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we have a vector f that depends on x and y.
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So this occurs in real life when
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you look at velocity fields in a fluid.
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For example, the wind, that's what these pictures show.
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At every point, you have a velocity of a fluid that's moving.
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And the example is force fields.
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So now, force fields are not something out of star walls.
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If you look at gravitational attraction,
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you know that if you have a mass somewhere,
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well, it will be attracted to fall down
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because of a gravity field of the Earth, which
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means that at every point, you have a vector that's
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pointing down.
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And same thing in space, you have the gravitational field
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of planets, stars, and so on.
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So that's also an example of a vector field
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because wherever you go, you would have that vector.
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And what it is depends on where you are.
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So the examples from the real world
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are things like velocity in a fluid
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or force fields where you have a force that depends
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on the point where you are.
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So we're going to try to study vector fields mathematically.
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So we won't really care what they are most of the time.
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But as we'll explore with them, define quantities, and so on,
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we'll very often use these motivations
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to justify why we would care about certain quantities.
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So the first thing we have to figure out
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is how do we draw a vector field?
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How do you generate a plot like that?
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So let's practice drawing a few vector fields.
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Well, let's start with, let's say, our very first vector field
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will be just 2i plus j.
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So it's kind of a silly vector field
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because it doesn't actually depend on x and y.
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So that means it's the same vector everywhere.
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So I take a picture of a plane, and I take the vector 2,1.
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I guess it points in that direction.
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It's two units to the right, 1 up.
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And I just put that vector everywhere.
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So you just put it at a few points all over the place.
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And when you think you have enough so that you understand
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what's going on, then you stop.
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So here probably we don't need that many.
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I mean, here I think we get the picture.
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So everywhere we have a vector 2,1.
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Now let's try to look at slightly more interesting examples.
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Let's say I give you a vector field x times i hat.
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So there's no j component.
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So how would you draw that?
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Maybe I don't need that much.
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Well, so first of all, we know that this guy is only
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in the i direction.
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So it's always horizontal.
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It doesn't have a j component.
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So everywhere it would be a horizontal vector.
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Now the question is how long is it?
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Well, how long it depends on x?
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So for example, if x is 0, then this will be actually
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the 0 vector.
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So x is 0, but here, see on the y axis,
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maybe I should take a different color.
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So if I'm on the y axis, I actually have the 0 vector.
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Now if x becomes positive small, then
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I will have actually a small positive multiple of i.
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So I will be going a little bit to the right.
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And then if I increase x, then this guy becomes larger.
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So I get longer vector to the right.
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If x is negative, then my vector field
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points to the left instead.
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It looks something like that.
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Any questions about that picture?
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No.
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OK, so usually we're not going to try to have very accurate.
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We want actually take time to plot a vector field very
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carefully.
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If we need to, computers can do it for us.
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But it's useful to have an idea of what the vector field does
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roughly, whether it's getting larger and larger in what
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direction and pointing what are the general features.
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So just to do a couple more, I actually
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you'll see very quickly that the examples I take in lecture
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are pretty much always the same ones.
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So you know, we'll be playing a lot
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with these particular vector fields just
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because they're good examples.
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So let's say I give you x i plus y j.
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So that one has an interesting geometric signification,
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the significance.
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If I take a point x y, you know, so there,
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I want to take the vector x y.
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How do I do that?
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Well, it's the same as the vector from the origin to this point.
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So I take this vector and I copy it so that it starts
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at my given point.
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So it looks like that.
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And you know, same thing at every point.
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So it's a vector field that's pointing
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radially away from the origin.
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And its magnitude increases with distance from the origin.
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So you don't have to draw, you know, as many as me.
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But so the idea is this vector field everywhere points
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away from the origin.
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And its magnitude is equal to the distance from the origin.
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So you know, if this were, for example, velocity fields,
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well, that would tell you, you'd
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see visually what's happening to your fluid.
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And like, here maybe you have a source at the origin
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that's pureing fluid out and it's flowing all the way
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away from that.
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OK, let's do just a last one.
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Let's say I give you minus y comma x.
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What does that look like?
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That's an interesting one, actually.
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It's a nice one.
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So let's see.
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So let's say that I have a point xy here.
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OK, so this vector here is x comma y.
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But the vector I want is negative y comma x.
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So what does that look like compared to?
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It's perpendicular to the position to this vector.
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OK, if I rotate this vector, so let me maybe draw a picture
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on the side, if I take the vector xy, then the vector with
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components negative y and x is going to be like this,
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is the vector that I get by rotating by 90 degrees counter
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clockwise.
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And of course, I don't want to put that vector at the origin.
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I want to put it at the point xy.
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So in fact, what I will draw is something like this.
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And similarly here, it would be like that.
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Like that.
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And if I'm closer to the origin, then it looks a bit the same,
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but it's shorter.
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And at the origin, it's 0.
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And when I'm further away, it becomes even larger.
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So see, this vector field, if it was the motion of a fluid,
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it would correspond to a fluid that's just going around
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the origin in circles, rotating at uniform speed.
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So this is actually the velocity field for uniform rotation.
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And if you figure out how long does it take for a particular fluid
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to go all the way around, well, that will be actually 2 pi,
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because the length of the circle is 2 pi times the radius.
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So that's actually at unit angular velocity.
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One radiant per second, or per unit time.
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So that's why this guy comes up quite a lot in real life.
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OK, and you can imagine lots of variations on these.
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Of course, you can also imagine vector fields given
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by much more complicated formulas, and then you'll
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have a hard time drawing them.
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So maybe you'll use a computer, or maybe you'll just give up,
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and just do whatever calculation you have to do
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without trying to visualize the vector field.
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But if you have a nice simple one, then it's worth doing it,
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because sometimes it will give you insight about what we're
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going to compute next.
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OK, any questions first about these pictures?
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No?
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OK, oh, yes.
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Sorry?
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Why?
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You're asking if it should be y and so y, or negative x,
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I think would be the other way around.
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See, for example, if I'm at this point, then y is positive
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and x is 0.
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So if I take y, negative x, I guess a positive first component
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and 0 for the second one.
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So y, negative x would be a rotation at unit speed
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in the opposite direction.
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So yeah, and there's a lot of tweaks you can do to it.
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So if you flip the sides, you get rotation in the other direction.
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Yes?
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How do I know?
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How do I know that it's a unit angular velocity?
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Well, that's because if my angular velocity is 1,
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then that means that the actual speed is equal to the distance
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from the origin.
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Because the arc length on a circle of a certain radius
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is equal to the radius times the angle.
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So if the angle varies at rate 1, then I travel at speed
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equal to the radius.
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And that's what I do here.
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The length of this vector is equal to the distance from the origin.
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I mean, it's not obvious in the picture,
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but really the vector that I put here is the same as this vector
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rotated, so it has the same length.
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So that's why the angular velocity is 1.
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It doesn't really matter much anyway, but...
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Okay.
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So now what are we going to do with vector fields?
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Well, we're going to do a lot of things, but let's...
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You know...
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Let's start somewhere.
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So one thing you might want to do with a vector field is...
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So I'm going to think for now of the situation
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where we have a force.
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So if you have a force exerted on a particle,
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and that particle moves on some trajectory,
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then probably you've seen in physics that the work done by the force
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corresponds to the force dot product with the displacement vector.
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How much you have moved your particle.
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And of course, if you do just a straight line trajectory
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or if the force is constant that works well,
257
00:16:48.160 --> 00:16:52.160
but if you're moving on a complicated trajectory and the force keeps changing,
258
00:16:52.160 --> 00:16:55.160
then actually you want to integrate that over time.
259
00:16:55.160 --> 00:17:00.160
So...
260
00:17:00.160 --> 00:17:02.160
Okay, so we're going to...
261
00:17:02.160 --> 00:17:09.160
The first thing we'll do is learn how to compute the work done by a vector field.
262
00:17:09.160 --> 00:17:19.160
And mathematically, that's called a line integral.
263
00:17:19.160 --> 00:17:27.160
Okay, so physically, remember, the work done by a force is the force times the distance.
264
00:17:27.160 --> 00:17:35.160
And more precisely, it's actually the dot product between the force as a vector.
265
00:17:35.160 --> 00:17:41.160
And the displacement vector for a small motion.
266
00:17:41.160 --> 00:17:45.160
So, say that, you know, your point is moving from here to here,
267
00:17:45.160 --> 00:17:51.160
you have the displacement delta R that's just the change in the position vector.
268
00:17:51.160 --> 00:17:54.160
If it's the vector from the old position to the new position.
269
00:17:54.160 --> 00:17:58.160
And then you have your force that's being exerted,
270
00:17:58.160 --> 00:18:01.160
and you do the dot product between them.
271
00:18:01.160 --> 00:18:06.160
And that will give you the work of the force during this motion.
272
00:18:06.160 --> 00:18:13.160
And the physical significance of this, well, the work tells you basically how much energy you have to provide
273
00:18:13.160 --> 00:18:15.160
and actually perform this motion.
274
00:18:15.160 --> 00:18:18.160
Okay, just in case you haven't seen this in 801 yet.
275
00:18:18.160 --> 00:18:22.160
I'm hoping all of you have heard about work somewhere.
276
00:18:22.160 --> 00:18:26.160
But in case it's completely mysterious, that's the amount of energy you have to provide.
277
00:18:26.160 --> 00:18:30.160
Sorry, actually, no, that's the amount of energy provided by the force.
278
00:18:30.160 --> 00:18:34.160
So if the force goes along the motion, it actually pushes the particle.
279
00:18:34.160 --> 00:18:37.160
It provides energy to it to do that motion.
280
00:18:37.160 --> 00:18:40.160
And conversely, if you're trying to go against the force,
281
00:18:40.160 --> 00:18:45.160
then you have to provide energy to the particle to be able to do that.
282
00:18:45.160 --> 00:18:49.160
So in particular, if this is the only force that's taking place,
283
00:18:49.160 --> 00:18:56.160
then the work will be the variation in kinetic energy of the particle along the motion.
284
00:18:56.160 --> 00:19:00.160
So now that's a good description for a small motion.
285
00:19:00.160 --> 00:19:06.160
But let's say that my particle is not just doing that, but it's doing something complicated.
286
00:19:06.160 --> 00:19:14.160
And my force keeps changing, so somehow maybe I have a different force at every point.
287
00:19:14.160 --> 00:19:18.160
So then I want to find the total work done along the motion.
288
00:19:18.160 --> 00:19:24.160
Well, what I have to do is I have to cut my trajectory into these little pieces.
289
00:19:24.160 --> 00:19:30.160
And for each of them, I have a vector along the trajectory.
290
00:19:30.160 --> 00:19:35.160
I have the force. I do the dot product. And I sum them together.
291
00:19:35.160 --> 00:19:40.160
And of course, the actual answer, well, to get the actual answer,
292
00:19:40.160 --> 00:19:46.160
I should actually cut into smaller and smaller pieces and sum all of the small contributions to work.
293
00:19:46.160 --> 00:19:53.160
So in fact, it's going to be an integral.
294
00:19:53.160 --> 00:20:04.160
So along some trajectory, let's call c the trajectory for curve.
295
00:20:04.160 --> 00:20:19.160
So here, the work adds up to an integral,
296
00:20:19.160 --> 00:20:34.160
which, well, we've write this using the notation integral along c of f dot dr.
297
00:20:34.160 --> 00:20:38.160
So we have to decode this notation.
298
00:20:38.160 --> 00:21:06.160
So one way to decode this is to say it's a limit as we cut into smaller and smaller pieces of the sum of each piece of the trajectory of the force at the given point dot product with that small vector along the trajectory.
299
00:21:06.160 --> 00:21:15.160
And well, that's not how we're going to compute it.
300
00:21:15.160 --> 00:21:17.160
You know, to compute it, we do things differently.
301
00:21:17.160 --> 00:21:31.160
So how can we actually compute it? Well, what we can do is say that actually we are cutting things into small time intervals.
302
00:21:31.160 --> 00:21:37.160
So the way that we split our trajectory is we're just taking a picture every, say, millisecond.
303
00:21:37.160 --> 00:21:41.160
So every millisecond, we have a new position.
304
00:21:41.160 --> 00:21:53.160
And the motion, you know, the amount that we should have moved during each small time interval is basically the velocity vector times the amount of time.
305
00:21:53.160 --> 00:22:03.160
So, in fact, let me just provide this as you know, you do the dot product between the force.
306
00:22:03.160 --> 00:22:09.160
And how much you have moved.
307
00:22:09.160 --> 00:22:13.160
Well, if I just provide it this way, nothing's happened.
308
00:22:13.160 --> 00:22:27.160
So what this thing is actually is the velocity vector, the RdT.
309
00:22:27.160 --> 00:22:42.160
So what I'm trying to say is that I can actually compute my integral by integrating f dot product with the RdT over time.
310
00:22:42.160 --> 00:22:56.160
From time, you know, whatever the initial time to whatever the final time is, I integrate f dot product velocity dT.
311
00:22:56.160 --> 00:23:04.160
And of course, I mean, here when I say f, I mean f at the point on the trajectory at time T.
312
00:23:04.160 --> 00:23:11.160
This guy depends on x and y, therefore it depends on T.
313
00:23:11.160 --> 00:23:13.160
Okay, I see a lot of confused faces.
314
00:23:13.160 --> 00:23:22.160
So let's do an example. And then I will ask you.
315
00:23:22.160 --> 00:23:25.160
So now, yes, that would be probably.
316
00:23:25.160 --> 00:23:30.160
So here I need to put a limit as delta T tends to zero.
317
00:23:30.160 --> 00:23:34.160
So let's get my trajectory into smaller and smaller time intervals.
318
00:23:34.160 --> 00:23:40.160
For each time interval, I have a small motion, which is essentially velocity times delta T.
319
00:23:40.160 --> 00:23:46.160
And then I dot that with the force. And I sum them.
320
00:23:46.160 --> 00:23:50.160
Okay, so let's do an example.
321
00:23:50.160 --> 00:23:53.160
Let's say that we do.
322
00:23:53.160 --> 00:23:58.160
We want to find the work of this force.
323
00:23:58.160 --> 00:24:01.160
So I guess that's the last example that we had.
324
00:24:01.160 --> 00:24:05.160
So it's a force field that tries to make everything in the rotate.
325
00:24:05.160 --> 00:24:09.160
Somehow your first points along these circles.
326
00:24:09.160 --> 00:24:18.160
And let's say that our trajectory, our particle is moving along the parametric curve.
327
00:24:18.160 --> 00:24:24.160
x equals T, y equals T squared for T going from zero to one.
328
00:24:24.160 --> 00:24:31.160
So what that looks like, what maybe I should draw in your picture.
329
00:24:31.160 --> 00:24:39.160
So our vector field.
330
00:24:39.160 --> 00:24:41.160
Well, whatever.
331
00:24:41.160 --> 00:24:47.160
And our trajectory, if you try to plot this, when you see y is actually x squared.
332
00:24:47.160 --> 00:24:54.160
So it's a piece of parabola that goes from the origin to one one.
333
00:24:54.160 --> 00:24:58.160
Okay, that's what our curve looks like.
334
00:24:58.160 --> 00:25:04.160
Okay, so we're trying to find the work done by our force along this trajectory.
335
00:25:04.160 --> 00:25:08.160
I should point out, I mean, you know, if you're asking me, how did you get this?
336
00:25:08.160 --> 00:25:10.160
That's actually the wrong question.
337
00:25:10.160 --> 00:25:13.160
I didn't, you know, this is all part of the data.
338
00:25:13.160 --> 00:25:18.160
So I'm going to draw a force and I have a trajectory and I want to find what's the work done along that trajectory.
339
00:25:18.160 --> 00:25:24.160
These two guys I can choose completely independently of each other.
340
00:25:24.160 --> 00:25:42.160
So the integral along C of f dot dr will be, well, it's the integral from time zero to time one of f dot velocity vector dr dt.
341
00:25:42.160 --> 00:25:49.160
So that will be the integral from zero to one of, okay, let's try to figure it out.
342
00:25:49.160 --> 00:25:55.160
So what is f?
343
00:25:55.160 --> 00:26:01.160
f at a point xy is minus y comma x.
344
00:26:01.160 --> 00:26:07.160
But if I take the point where I am at time t, then x is t and y is t squared.
345
00:26:07.160 --> 00:26:16.160
So here I plug x equals t, y equals t squared, but will give me negative t squared and t.
346
00:26:16.160 --> 00:26:21.160
Okay, so here we'll put negative t squared t dot product.
347
00:26:21.160 --> 00:26:24.160
What is the velocity vector?
348
00:26:24.160 --> 00:26:29.160
Well, the x dt is just one.
349
00:26:29.160 --> 00:26:43.160
The y dt is 2t, so the velocity vector is 1 and 2t dt.
350
00:26:43.160 --> 00:26:47.160
Okay, so now we have to continue calculation.
351
00:26:47.160 --> 00:26:52.160
We get integral from zero to one of what is this dot product?
352
00:26:52.160 --> 00:26:55.160
Well, it's negative t squared plus 2t squared.
353
00:26:55.160 --> 00:27:02.160
I get t squared. Well, maybe I'll write it negative t squared plus 2t squared dt.
354
00:27:02.160 --> 00:27:12.160
But ends up being integral from zero to one of t squared dt, which you all know how to integrate and get 1 third.
355
00:27:12.160 --> 00:27:15.160
Okay.
356
00:27:15.160 --> 00:27:19.160
So that's the work done by the force along this curve.
357
00:27:19.160 --> 00:27:23.160
Yes?
358
00:27:23.160 --> 00:27:30.160
Well, I got it by just taking the dot product between the force and the velocity.
359
00:27:30.160 --> 00:27:43.160
Sorry, that's in case you're wondering, things go like this.
360
00:27:43.160 --> 00:27:50.160
Okay, any questions on how we did this calculation?
361
00:27:50.160 --> 00:27:59.160
Yes?
362
00:27:59.160 --> 00:28:01.160
Ah, why can't you just do f dot d r?
363
00:28:01.160 --> 00:28:03.160
Well, soon we'll be able to.
364
00:28:03.160 --> 00:28:08.160
We don't know yet what d r means or how to use it as a symbol.
365
00:28:08.160 --> 00:28:11.160
Okay, because, you know, we haven't said yet.
366
00:28:11.160 --> 00:28:17.160
I mean, see, this is a d vector r that's kind of a strange thing to have.
367
00:28:17.160 --> 00:28:21.160
And certainly r is not a usual variable.
368
00:28:21.160 --> 00:28:23.160
When we have to be careful about, you know, what are the rules?
369
00:28:23.160 --> 00:28:25.160
What does this symbol mean?
370
00:28:25.160 --> 00:28:28.160
So, I mean, we're going to see that right now.
371
00:28:28.160 --> 00:28:33.160
And then we can do it actually in a slightly more efficient way.
372
00:28:33.160 --> 00:28:35.160
But you can't just use, I mean, r is not a scalar quantity.
373
00:28:35.160 --> 00:28:37.160
R is a position vector.
374
00:28:37.160 --> 00:28:38.160
Right?
375
00:28:38.160 --> 00:28:40.160
So you can't integrate f with respect to r.
376
00:28:40.160 --> 00:28:44.160
We don't know how to do that.
377
00:28:44.160 --> 00:28:50.160
Okay.
378
00:28:50.160 --> 00:28:52.160
So, oh, yes, sorry.
379
00:28:52.160 --> 00:28:57.160
I can't see people over there because of the spotlights, but, yes.
380
00:28:57.160 --> 00:29:01.160
If I were to do.
381
00:29:01.160 --> 00:29:02.160
Sorry.
382
00:29:02.160 --> 00:29:09.160
I still can't hear you, sorry.
383
00:29:09.160 --> 00:29:13.160
Okay, so, question is, if I took a different trajectory from the origin to that point,
384
00:29:13.160 --> 00:29:15.160
1, 1, what would happen?
385
00:29:15.160 --> 00:29:18.160
Well, the answer is I would get something different.
386
00:29:18.160 --> 00:29:21.160
Okay. So, for example, let me try to convince you of that.
387
00:29:21.160 --> 00:29:31.160
For example, say that I chose to instead go like this and then around like that.
388
00:29:31.160 --> 00:29:36.160
Then, you know, first, I wouldn't do any work because here the force is perpendicular to my motion.
389
00:29:36.160 --> 00:29:40.160
And then I would be going against the force all the way around.
390
00:29:40.160 --> 00:29:43.160
So, I should get something that's negative.
391
00:29:43.160 --> 00:29:47.160
Okay. Even if you don't see that, just, you know, accept that phase value,
392
00:29:47.160 --> 00:29:48.160
what I'm going to say now.
393
00:29:48.160 --> 00:29:54.160
The value of a line integral in general depends on how we got from point A to point B.
394
00:29:54.160 --> 00:29:59.160
And that's why we have to compute it by using the parametric equation for the curve.
395
00:29:59.160 --> 00:30:03.160
It really depends on what curve you choose.
396
00:30:03.160 --> 00:30:15.160
Okay. Any other questions?
397
00:30:15.160 --> 00:30:21.160
Oh, yes.
398
00:30:21.160 --> 00:30:25.160
So, yes. What happens when the force inflakes the trajectory?
399
00:30:25.160 --> 00:30:28.160
Well, then, actually, you'd have to solve a different equation,
400
00:30:28.160 --> 00:30:33.160
telling you how the particle moves to find what the trajectory is.
401
00:30:33.160 --> 00:30:37.160
So, that's something that would, you know, be a very useful topic.
402
00:30:37.160 --> 00:30:39.160
And that's probably more like what you will do in 1803,
403
00:30:39.160 --> 00:30:43.160
or maybe you actually know how to do it in this case.
404
00:30:43.160 --> 00:30:48.160
So, what we're trying to develop here is a method to figure out if we know what the trajectory is,
405
00:30:48.160 --> 00:30:50.160
what the work will be.
406
00:30:50.160 --> 00:30:53.160
It doesn't tell us what the trajectory will be.
407
00:30:53.160 --> 00:30:56.160
But, of course, we could also find that.
408
00:30:56.160 --> 00:31:01.160
But here, see, I'm not assuming, for example, that the particle is moving just based on that force.
409
00:31:01.160 --> 00:31:06.160
Maybe, actually, I'm here to, you know, hold it in my hand and force it to go where it's going.
410
00:31:06.160 --> 00:31:10.160
Or maybe there's some rail that's taking it on that trajectory or whatever.
411
00:31:10.160 --> 00:31:14.160
So, I can really do it along any trajectory.
412
00:31:14.160 --> 00:31:20.160
And if I wanted to, if I knew that it's the case, then I could try to find the trajectory based on what the force is.
413
00:31:20.160 --> 00:31:25.160
But that's not what we are doing here.
414
00:31:25.160 --> 00:31:27.160
Okay.
415
00:31:27.160 --> 00:31:35.160
So, next.
416
00:31:35.160 --> 00:31:37.160
Okay. So, let's try to make sense.
417
00:31:37.160 --> 00:31:44.160
I mean, you asked a few minutes ago what can we do directly with dr?
418
00:31:44.160 --> 00:31:47.160
So, dr, see, becomes somehow a vector.
419
00:31:47.160 --> 00:31:54.160
I mean, when I replace it by dr dt times dt, it becomes something that's a vector, but with a dt next to it.
420
00:31:54.160 --> 00:32:12.160
So, in fact, well, it's not really new, but let's see, another way to do it.
421
00:32:12.160 --> 00:32:18.160
Let's say that our force has components m and n.
422
00:32:18.160 --> 00:32:29.160
I claim that we can write symbolically vector dr stands for, it's a vector whose components are dx and dy.
423
00:32:29.160 --> 00:32:33.160
So, now that's a strange kind of vector. I mean, it's not a real vector, of course.
424
00:32:33.160 --> 00:32:36.160
But as a notation, it's a pretty good notation.
425
00:32:36.160 --> 00:32:48.160
Because it tells us that f dot dr is m dx plus n dy.
426
00:32:48.160 --> 00:32:57.160
And so, in fact, well, very often write instead of f dot dr, n integral, n c, will write the line integral,
427
00:32:57.160 --> 00:33:06.160
and then, along c of m dx plus n dy. And so, in this language, of course, what we're integrating now,
428
00:33:06.160 --> 00:33:09.160
rather than a vector field, becomes a differential.
429
00:33:09.160 --> 00:33:12.160
But you should think of the two as being pretty much the same thing, right?
430
00:33:12.160 --> 00:33:21.160
It's like when you compare the gradient of a function, and it's differential, they're different notations, but they have the same content.
431
00:33:21.160 --> 00:33:27.160
Okay, so now, that still remains the question, how do we compute this kind of integral?
432
00:33:27.160 --> 00:33:30.160
Because it's more subtle of an indentation suggests, right?
433
00:33:30.160 --> 00:33:34.160
Because m and n both depend on x and y.
434
00:33:34.160 --> 00:33:41.160
And if you just integrate it with respect to x, you'd be left with y's in there, and you don't want to be left with y's.
435
00:33:41.160 --> 00:33:48.160
You want a number at the end. And see, the catch is along the curve, x and y are actually related to each other.
436
00:33:48.160 --> 00:33:58.160
So, whenever we write this, we have two variables x and y, but in fact, along the curve c, we have only one parameter.
437
00:33:58.160 --> 00:34:02.160
Could be x, could be y, could be time, whatever you want.
438
00:34:02.160 --> 00:34:10.160
But we have to actually express everything in terms of that one parameter, and then we get a usual single variable integral.
439
00:34:10.160 --> 00:34:22.160
Okay, so how do we evaluate things in this language? Well, we do it by substituting a parameter into everything.
440
00:34:40.160 --> 00:35:04.160
Okay, so the method to evaluate is to express x and y in terms of a single variable.
441
00:35:04.160 --> 00:35:18.160
And then substitute that variable.
442
00:35:18.160 --> 00:35:25.160
Okay, so let's, for example, redo the one we had up there just using these new notations.
443
00:35:25.160 --> 00:35:29.160
You'll see it's the same calculation that with different notations.
444
00:35:29.160 --> 00:35:43.160
In that example that we had, our vector field f was negative y, comma x.
445
00:35:43.160 --> 00:35:53.160
So, what we're integrating is negative y dx plus x dy.
446
00:35:53.160 --> 00:36:03.160
And see if we have just this, we don't know how to integrate that. I mean, what you could try to come up with like negative xy or something like that, but that actually doesn't make sense.
447
00:36:03.160 --> 00:36:19.160
It doesn't work. So, what we'll do is we'll actually have to express everything in terms of a same variable, because it's a single integral.
448
00:36:19.160 --> 00:36:29.160
We should have only one variable. And what that variable will be, well, if we just do it the same way, that could just be t.
449
00:36:29.160 --> 00:36:36.160
So, how do we express everything in terms of t? Well, we use the parametric equation.
450
00:36:36.160 --> 00:36:43.160
We know that x is t and y is t squared.
451
00:36:43.160 --> 00:36:50.160
So, we know what to do with these two guys. What about dx and dy? Well, it's easy, we just differentiate.
452
00:36:50.160 --> 00:37:00.160
dx becomes dt. dy becomes 2t dt.
453
00:37:00.160 --> 00:37:11.160
I'm just saying in a different language, what I said of other with dx dt equals 1, dy dt equals 2t. It's the same thing, but written slightly differently.
454
00:37:11.160 --> 00:37:20.160
So, now, so I'm going to do it again. I'm going to switch from one board to the next one.
455
00:37:20.160 --> 00:37:39.160
My integral becomes the integral over c of negative y is minus t squared dt plus x is t times dy is 2t dt.
456
00:37:39.160 --> 00:37:50.160
And now that I have only t left, it's fine to say, oh, I have a usual single variable integral of a variable t that goes from 0 to 1.
457
00:37:50.160 --> 00:38:04.160
So, now I can say, yes, this is the integral from 0 to 1 of that stuff, where I can simplify a bit, becomes just t squared dt, and I can compute it.
458
00:38:04.160 --> 00:38:11.160
I have negative t squared, and then I have plus 2t squared, so I end up with positive t squared.
459
00:38:11.160 --> 00:38:21.160
It's the same as up there. Any questions? Yes?
460
00:38:21.160 --> 00:38:37.160
So, when you do dy, it's the derivative of y, it's the differential of y. Y is t squared, so I get 2t dt.
461
00:38:37.160 --> 00:38:45.160
So, I plug dt for dx, I plug 2t dt for dy, and so on.
462
00:38:45.160 --> 00:38:57.160
And that's the general method. So, if you are given a curve, then you first have to figure out how do you express x and y in terms of the same thing.
463
00:38:57.160 --> 00:39:13.160
And you get to choose in general what parameter we use. You choose to parameterize your curve in whatever way you want.
464
00:39:13.160 --> 00:39:40.160
The note that I want to make is that this line integral depends on the trajectory, c, but not on the parameterization.
465
00:39:40.160 --> 00:40:09.160
So, you can choose whichever variable you want. So, for example, what you could do is, when you know that you have that trajectory, you could also choose to parameterize it as x equals sine theta, and y equals sine squared theta, because y is x squared, where theta goes from 0 to pi over 2.
466
00:40:09.160 --> 00:40:19.160
And then you could get dx and dy in terms of d theta, and you'd be able to do it with a lot of trig, and you would get the same answer.
467
00:40:19.160 --> 00:40:30.160
So, that would be a harder way to get the same thing. So, what you should do in practice is use the most reasonable way to parameterize your curve.
468
00:40:30.160 --> 00:40:40.160
So, you know that here you have a piece of parabola y equals x squared, there's no way you will put sine and sine squared. You could set, I mean, x equals t y equals t squared is very reasonable.
469
00:40:40.160 --> 00:40:59.160
You can even take a small shortcut and say that your variable will be just x. So, that means x, you just keep it, you know, as it is. And then when you have y, you set y equals x squared, dy equals 2x dx, and then you have an integral of x. That works.
470
00:40:59.160 --> 00:41:11.160
So, this one is not practical, but you know, you get to choose.
471
00:41:29.160 --> 00:41:44.160
Okay.
472
00:41:44.160 --> 00:41:49.160
So, now let me tell you a bit more about the geometry.
473
00:41:49.160 --> 00:42:08.160
So, we've said, you know, here's how we compute it in general, and that's the general method for computing a line integral for work. You can always do this, try to find a parameter, the simplest one, express everything in terms of that variable, and then you have an integral to compute.
474
00:42:08.160 --> 00:42:26.160
But sometimes you can actually save a lot of work by just thinking geometrically about what this all does. Okay. So, let me tell you about the geometric approach.
475
00:42:26.160 --> 00:42:47.160
So, one thing I want to remind you of first is what is this vector dr? Well, what is vector delta r? You know, if I take a very small piece of a trajectory, then my vector delta r will be tangent to the trajectory.
476
00:42:47.160 --> 00:43:02.160
So, it will be going in the same direction as the unit tangent vector t, and what is its length? Well, its length is the arc length along the trajectory, which is what we called delta s.
477
00:43:02.160 --> 00:43:06.160
Remember s was distance along the trajectory.
478
00:43:06.160 --> 00:43:22.160
So, we can write vector dr, so we say it's dx, dy, but that's also t times ds.
479
00:43:22.160 --> 00:43:41.160
It's a vector whose direction is tangent to the curve, and whose length element is actually the arc length element. I mean, if you don't like this notation, think about divide everything by dt.
480
00:43:41.160 --> 00:44:05.160
And what we're saying is dr dt, that's a velocity vector. Well, it coordinates the velocity vector is dx dt, dy dt. But more geometrically, the direction of a velocity vector is tangent to the trajectory, and its magnitude is speed ds dt.
481
00:44:05.160 --> 00:44:12.160
So, that's really the same thing.
482
00:44:12.160 --> 00:44:28.160
So, if I say this, that means that my line integral after dr, well, I say that I can write it as integral of mdx plus mdy.
483
00:44:28.160 --> 00:44:43.160
That's what I will do if I want to actually compute it by computing the integral. But if instead I want to think about it geometrically, I could rewrite it as f dot t ds.
484
00:44:43.160 --> 00:45:00.160
So, now you can think of this, f dot t is a scalar quantity. It's the tangent component of my force. So, I take my force, and I project it to the tangent direction to the trajectory. And then I integrate that along the curve.
485
00:45:00.160 --> 00:45:07.160
It's still the same thing. And sometimes it's easier to do it this way.
486
00:45:07.160 --> 00:45:20.160
So, here's an example. Let's say, so, you know, this is bound to be easier only when the field and the curve are relatively simple and have a geometric relation to each other.
487
00:45:20.160 --> 00:45:29.160
If I give you an evil formula with, you know, x cubed plus y to the fifth or whatever, there's very little chance that you will be able to simplify it that way.
488
00:45:29.160 --> 00:45:49.160
But let's say that I'm doing, say, my trajectory is just a circle of radius a centred at the origin. Let's say I'm doing that counterclockwise.
489
00:45:49.160 --> 00:46:07.160
And let's say that my vector field is x i plus y g. So, what does that look like? Well, so my trajectory is just this circle.
490
00:46:07.160 --> 00:46:22.160
So, my vector field, remember x i plus y g. That's the one that's pointing, radially, from the origin. So, hopefully, if you have good physics intuition here, you already know what the work is going to be.
491
00:46:22.160 --> 00:46:27.160
It's going to be zero, because the force is perpendicular to the motion.
492
00:46:27.160 --> 00:46:41.160
Well, now we can say it directly by saying, well, if we take any point of the circle, sorry.
493
00:46:41.160 --> 00:46:51.160
So, if you have any point of the circle, then the tangent vector to the circle will be tangent, well, it's tangent to the circle.
494
00:46:51.160 --> 00:47:07.160
So, that means it's perpendicular to the radial direction, while the force is pointing in the radial direction. So, you have a right angle between them.
495
00:47:07.160 --> 00:47:22.160
So, f is perpendicular to t. So, f dot t is zero, and the line integral of f dot t ds is just zero.
496
00:47:22.160 --> 00:47:36.160
See, that's much easier than writing, well, this is integral of x d x plus y d y. What do we do? Well, we probably set x equals a cosine theta y equals a sine theta. We get a bunch of trig things. Oh, it cancels out to zero.
497
00:47:36.160 --> 00:47:45.160
It's not much harder, but we saved time by not even thinking about how to parameterize things.
498
00:47:45.160 --> 00:47:56.160
Let's do a last one. That was the first one. Let's say now that I take the same curve, see.
499
00:47:56.160 --> 00:48:04.160
But now my vector field is the one that rotates. Negative y i plus x j.
500
00:48:04.160 --> 00:48:26.160
So, that means along my circle, the tangent vector goes like this. And my vector field is also going around.
501
00:48:26.160 --> 00:48:35.160
So, in fact, at this point, the vector field will also be going in the same direction.
502
00:48:35.160 --> 00:48:52.160
So, now, f is actually parallel to the tangent direction. So, that means that the dot product f dot t, remember, that's the component of f in this direction.
503
00:48:52.160 --> 00:49:02.160
That will be the same as the length of f. Right? But what's the length of f on this circle if this radius is a?
504
00:49:02.160 --> 00:49:07.160
It's just going to be a. That's what we said earlier about this vector field.
505
00:49:07.160 --> 00:49:25.160
So, at every point, this dot product is a. Now, we know how to integrate that quite quickly.
506
00:49:25.160 --> 00:49:31.160
Because that becomes the integral of a ds, but a is a constant. So, we can take it out.
507
00:49:31.160 --> 00:49:38.160
And now, what do we get when we integrate ds along c? Well, we should get the total length of a curve.
508
00:49:38.160 --> 00:49:48.160
If we sum all the little pieces of our length, but we know that the length of a circle of radius a is 2 pi a.
509
00:49:48.160 --> 00:49:59.160
So, we get 2 pi a squared. If we were to compute that by hand,
510
00:49:59.160 --> 00:50:07.160
well, what would we do with sets? So, we would be computing integral of minus y dx plus x dy.
511
00:50:07.160 --> 00:50:18.160
We'd probably set since we're on the circle, x equals a cos theta, y equals a sine theta for theta between 0 and 2 pi.
512
00:50:18.160 --> 00:50:31.160
So, then we would get dx and dy out of this. So, y is a sine theta. dx is negative a sine theta d theta.
513
00:50:31.160 --> 00:50:39.160
If you differentiate a cos sine plus a cos theta, a cos theta d theta.
514
00:50:39.160 --> 00:50:51.160
Well, you'll just end up with integral from 0 to 2 pi of a squared times sine square theta plus cos sine square theta d theta.
515
00:50:51.160 --> 00:50:58.160
That becomes just 1. And you get the same answer. It took about the same amount of time because I did this one rushing very quickly.
516
00:50:58.160 --> 00:51:09.160
But normally, it takes about at least twice the amount of time to do it with the calculation. So, that tells you sometimes it's worth thinking geometrically.