WEBVTT

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One of the most fundamental
ideas in all of physics

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is the idea of work.

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Now when you first learn work,
you just say, oh, that's

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just force times distance.

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But then later on, when you
learn a little bit about

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vectors, you realize that the
force isn't always going in

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the same direction as
your displacement.

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So you learn that work is
really the magnitude, let me

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write this down, the magnitude
of the force, in the direction,

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or the component of the force
in the direction

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of displacement.

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Displacement is just distance
with some direction.

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Times the magnitude of the
displacement, or you could say,

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times the distance displaced.

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And the classic example.

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Maybe you have an ice cube,
or some type of block.

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I just ice so that there's
not a lot of friction.

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Maybe it's standing on a bigger
lake or ice or something.

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And maybe you're pulling on
that ice cube at an angle.

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Let's say, you're pulling
at an angle like that.

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That is my force, right there.

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Let's say my force is
equal to-- well, that's

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my force vector.

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Let's say the magnitude of
my force vector, let's

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say it's 10 newtons.

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And let's say the direction of
my force vector, right, any

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vector has to have a magnitude
and a direction, and the

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direction, let's say it has a
30 degree angle, let's say a 60

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degree angle, above horizontal.

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So that's the direction
I'm pulling in.

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And let's say I displace it.

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This is all review, hopefully.

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If you're displacing it, let's
say you displace it 5 newtons.

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So let's say the displacement,
that's the displacement vector

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right there, and the magnitude
of it is equal to 5 meters.

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So you've learned from the
definition of work, you can't

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just say, oh, I'm pulling with
10 newtons of force and

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I'm moving it 5 meters.

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You can't just multiply the 10
newtons times the 5 meters.

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You have to find the magnitude
of the component going in the

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same direction as
my displacement.

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So what I essentially need to
do is, the length, if you

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imagine the length of this
vector being 10, that's the

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total force, but you need to
figure out the length of the

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vector, that's the component of
the force, going in the same

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direction as my displacement.

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And a little simple
trigonometry, you know that

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this is 10 times the cosine of
60 degrees, or that's equal to,

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cosine of 60 degrees is 1/2, so
that's just equal to 5.

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So this magnitude, the
magnitude of the force going

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in the same direction of
the displacement in this

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case, is 5 newtons.

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And then you can
figure out the work.

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You could say that the work is
equal to 5 newtons times, I'll

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just write a dot for times.

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I don't want you to think
it's cross product.

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Times 5 meters, which is 25
newton meters, or you could

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even say 25 Joules of
work have been done.

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And this is all review of
somewhat basic physics.

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But just think about
what happened, here.

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What was the work?

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If I write in the abstract.

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The work is equal
to the 5 newtons.

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That was the magnitude of my
force vector, so it's the

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magnitude of my force vector,
times the cosine of this angle.

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So you know, let's
call that theta.

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Let's say it a
little generally.

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So times the cosine
of the angle.

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This is the amount of my force
in the direction of the

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displacement, the cosine of the
angle between them, times the

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magnitude of the displacement.

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So times the magnitude
of the displacement.

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Or if I wanted to rewrite that,
I could just write that as, the

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magnitude of the displacement
times the magnitude of

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the force times the
cosine of theta.

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And I've done multiple videos
of this, in the linear algebra

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playlist, in the physics
playlist, where I talk about

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the dot product and the cross
product and all of that, but

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this is the dot product
of the vectors d and f.

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So in general, if you're trying
to find the work for a constant

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displacement, and you have a
constant force, you just take

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the dot product of
those two vectors.

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And if the dot product is a
completely foreign concept to

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you, might want to watch, I
think I've made multiple, 4

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or 5 videos on the dot
product, and its intuition,

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and how it compares.

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But just to give you a little
bit of that intuition right

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here, the dot product, when
I take f dot d, or d dot f,

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what it's giving me is, I'm
multiplying the magnitude, well

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I could just read this out.

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But the idea of the dot product
is, take how much of this

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vector is going in the same
direction as this vector,

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in this case, this much.

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And then multiply
the two magnitudes.

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And that's what we
did right here.

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So the work is going to be the
force vector, dot, taking the

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dot part of the force vector
with the displacement vector,

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and this, of course,
is a scalar value.

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And we'll work out some
examples in the future where

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you'll see that that's true.

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So this is all review of
fairly elementary physics.

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Now let's take a more
complex example, but it's

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really the same idea.

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Let's define a vector field.

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So let's say that I have a
vector field f, and we're

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going to think about what
this means in a second.

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It's a function of x and y, and
it's equal to some scalar

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function of x and y times the
i-unit vector, or the

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horizontal unit vector, plus
some other function, scalar

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function of x and y, times the
vertical unit vector.

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So what would something
like this be?

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This is a vector field.

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This is a vector field
in 2-dimensional space.

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We're on the x-y plane.

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Or you could even say, on R2.

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Either way, I don't want
to get too much into

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the mathiness of it.

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But what does this do?

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Well, if I were to draw my x-y
plane, so that is my, again,

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having trouble drawing
a straight line.

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All right, there we go.

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That's my y-axis, and
that's my x-axis.

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I'm just drawing the first
quadrant, and but you could

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go negative in either
direction, if you like.

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What does this thing do?

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Well, it's essentially
saying, look.

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You give me any x, any y, you
give any x, y in the x-y plane,

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and these are going to end
up with some numbers, right?

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When you put x, y here, you're
going to get some value, when

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you put x, y here, you're
going to get some value.

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So you're going to get some
combination of the i-

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and j-unit vectors.

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So you're going to
get some vector.

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So what this does, it defines a
vector that's associated with

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every point on x-y plane.

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So you could say, if I take
this point on the x-y plane,

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and I would pop it into this,
I'll get something times i plus

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something times j, and when you
add those 2, maybe I get a

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vector that something
like that.

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And you could do that
on every point.

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I'm just taking random samples.

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Maybe when I go here,
the vector looks

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something like that.

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Maybe when I go here, the
victor looks like this.

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Maybe when I go here, the
vector looks like that.

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And maybe when I go up here,
the vector goes like that.

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I'm just randomly
picking points.

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It defines a vector on all of
the x, y coordinates where

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these scalar functions
are properly defined.

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And that's why it's
called a vector field.

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It defines what a potential,
maybe, force would be,

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or some other type of
force, at any point.

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At any point, if you happen
to have something there.

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Maybe that's what
the function is.

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And I could keep doing
this forever, and

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filling in all the gaps.

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But I think you get the idea.

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It associates a vector with
every point on x-y plane.

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Now, this is called a vector
field, so it probably makes a

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lot of sense that this could
be used to describe

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any type of field.

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It could be a
gravitation field.

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It could be an electric field,
it could be a magnetic field.

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And this could be essentially
telling you how much force

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there would be on some
particle in that field.

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That's exactly what
this would describe.

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Now, let's say that in this
field, I have some particle

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traveling on x-y plane.

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Let's say it starts there, and
by virtue of all of these crazy

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forces that are acting on it,
and maybe it's on some tracks

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or something, so it won't
always move exactly in the

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direction that the field
is trying to move it at.

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Let's say it moves in a path
that moves something like this.

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And let's say that this path,
or this curve, is defined by

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a position vector function.

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So let's say that that's
defined by r of t, which is

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just x of t times i plus y of
t times our unit factor j.

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That's r of t right there.

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Well, in order for this to be
a finite path, this is true

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before t is greater than or
equal to a, and less

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than or equal to b.

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This is the path that the
particle just happens to

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take, due to all of
these wacky forces.

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So when the particle is right
here, maybe the vector field

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acting on it, maybe it's
putting a force like that.

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But since the thing is on some
type of tracks, it moves

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in this direction.

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And then when it's here, maybe
the vector field is like that,

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but it moves in that direction,
because it's on some

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type of tracks.

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Now, everything I've done in
this video is to build up

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to a fundamental question.

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What was the work done on
the particle by the field?

00:10:13.910 --> 00:10:24.960


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To answer that question, we
could zoom in a little bit.

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I'm going to zoom in on
only a little small

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snippet of our path.

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And let's try to figure out
what the work is done in a very

00:10:38.010 --> 00:10:40.470
small part of our path, because
it's constantly changing.

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The field is
changing direction.

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my object is
changing direction.

00:10:43.630 --> 00:10:47.780
So let's say when I'm here,
and let's say I move a

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small amount of my path.

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So let's say I move, this
is an infinitesimally

00:10:55.860 --> 00:10:58.500
small dr. Right?

00:10:58.500 --> 00:11:00.810
I have a differential, it's a
differential vector, infinitely

00:11:00.810 --> 00:11:02.630
small displacement.

00:11:02.630 --> 00:11:06.800
and let's say over the course
of that, the vector field is

00:11:06.800 --> 00:11:08.840
acting in this local
area, let's say it looks

00:11:08.840 --> 00:11:10.480
something like that.

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It's providing a force that
looks something like that.

00:11:13.490 --> 00:11:16.640
So that's the vector field in
that area, or the force

00:11:16.640 --> 00:11:18.750
directed on that particle right
when it's at that point.

00:11:18.750 --> 00:11:18.870
Right?

00:11:18.870 --> 00:11:22.420
It's an infinitesimally small
amount of time in space.

00:11:22.420 --> 00:11:24.440
You could say, OK, over that
little small point, we

00:11:24.440 --> 00:11:26.600
have this constant force.

00:11:26.600 --> 00:11:29.790
What was the work done
over this small period?

00:11:29.790 --> 00:11:32.330
You could say, what's the
small interval of work?

00:11:32.330 --> 00:11:36.120
You could say d work, or
a differential of work.

00:11:36.120 --> 00:11:38.940
Well, by the same exact logic
that we did with the simple

00:11:38.940 --> 00:11:43.810
problem, it's the magnitude of
the force in the direction of

00:11:43.810 --> 00:11:48.550
our displacement times the
magnitude of our displacement.

00:11:48.550 --> 00:11:52.800
And we know what that is, just
from this example up here.

00:11:52.800 --> 00:11:54.810
That's the dot product.

00:11:54.810 --> 00:11:58.340
It's the dot product of the
force and our super-small

00:11:58.340 --> 00:11:59.480
displacement.

00:11:59.480 --> 00:12:07.860
So that's equal to the dot
product of our force and our

00:12:07.860 --> 00:12:09.870
super-small displacement.

00:12:09.870 --> 00:12:13.240
Now, just by doing this, we're
just figuring out the work

00:12:13.240 --> 00:12:16.440
over, maybe like a really
small, super-small dr. But

00:12:16.440 --> 00:12:18.820
what we want to do, is we
want to sum them all up.

00:12:18.820 --> 00:12:21.870
We want to sum up all of the
drs to figure out the total,

00:12:21.870 --> 00:12:25.090
all of the f dot drs to figure
out the total work done.

00:12:25.090 --> 00:12:27.510
And that's where the
integral comes in.

00:12:27.510 --> 00:12:32.570
We will do a line integral
over-- I mean, you could

00:12:32.570 --> 00:12:33.910
think of it two ways.

00:12:33.910 --> 00:12:37.440
You could write just d dot w
there, but we could say, we'll

00:12:37.440 --> 00:12:42.700
do a line integral along this
curve c, could call that c

00:12:42.700 --> 00:12:46.410
or along r, whatever you
want to say it, of dw.

00:12:46.410 --> 00:12:47.800
That'll give us the total work.

00:12:47.800 --> 00:12:49.500
So let's say, work
is equal to that.

00:12:49.500 --> 00:12:54.040
Or we could also write it over
the integral, over the same

00:12:54.040 --> 00:13:00.500
curve of f of f dot dr.

00:13:00.500 --> 00:13:03.580
And this might seem like a
really, you know, gee, this

00:13:03.580 --> 00:13:05.120
is really abstract, Sal.

00:13:05.120 --> 00:13:09.220
How do we actually calculate
something like this?

00:13:09.220 --> 00:13:13.130
Especially because we have
everything parameterized

00:13:13.130 --> 00:13:14.030
in terms of t.

00:13:14.030 --> 00:13:16.130
How do we get this
in terms of t?

00:13:16.130 --> 00:13:19.710
And if you just think about
it, what is f dot r?

00:13:19.710 --> 00:13:21.030
Or what is f dot dr?

00:13:21.030 --> 00:13:23.300
Well, actually, to answer
that, let's remember

00:13:23.300 --> 00:13:25.830
what dr looked like.

00:13:25.830 --> 00:13:36.200
If you remember, dr/dt is equal
to x prime of t, I'm writing it

00:13:36.200 --> 00:13:39.120
like, I could have written dx
dt if I wanted to do, times the

00:13:39.120 --> 00:13:45.180
i-unit vector, plus y prime of
t, times the j-unit vector.

00:13:45.180 --> 00:13:49.320
And if we just wanted to dr, we
could multiply both sides, if

00:13:49.320 --> 00:13:51.850
we're being a little bit more
hand-wavy with the

00:13:51.850 --> 00:13:53.470
differentials, not
too rigorous.

00:13:53.470 --> 00:13:58.480
We'll get dr is equal to x
prime of t dt times the unit

00:13:58.480 --> 00:14:05.070
vector i plus y prime of t
times the differential dt

00:14:05.070 --> 00:14:07.280
times the unit vector j.

00:14:07.280 --> 00:14:09.070
So this is our dr right here.

00:14:09.070 --> 00:14:12.110


00:14:12.110 --> 00:14:16.280
And remember what our
vector field was.

00:14:16.280 --> 00:14:17.440
It was this thing up here.

00:14:17.440 --> 00:14:19.590
Let me copy and paste it.

00:14:19.590 --> 00:14:21.030
And we'll see that
the dot product is

00:14:21.030 --> 00:14:23.360
actually not so crazy.

00:14:23.360 --> 00:14:26.710
So copy, and let me
paste it down here.

00:14:26.710 --> 00:14:31.130


00:14:31.130 --> 00:14:33.820
So what's this integral
going to look like?

00:14:33.820 --> 00:14:37.600
This integral right here, that
gives the total work done by

00:14:37.600 --> 00:14:40.790
the field, on the particle,
as it moves along that path.

00:14:40.790 --> 00:14:44.090
Just super fundamental to
pretty much any serious physics

00:14:44.090 --> 00:14:47.170
that you might eventually
find yourself doing.

00:14:47.170 --> 00:14:48.170
So you could say, well gee.

00:14:48.170 --> 00:14:52.420
It's going to be the integral,
let's just say from t is equal

00:14:52.420 --> 00:14:55.320
to a, to t is equal to b.

00:14:55.320 --> 00:14:58.310
Right? a is where we started
off on the path, t is equal

00:14:58.310 --> 00:14:59.790
to a to t is equal to b.

00:14:59.790 --> 00:15:01.760
You can imagine it as being
timed, as a particle

00:15:01.760 --> 00:15:03.610
moving, as time increases.

00:15:03.610 --> 00:15:07.000
And then what is f dot dr?

00:15:07.000 --> 00:15:10.640
Well, if you remember from just
what the dot product is, you

00:15:10.640 --> 00:15:15.310
can essentially just take the
product of the corresponding

00:15:15.310 --> 00:15:17.740
components of your of
vector, and add them up.

00:15:17.740 --> 00:15:20.070
So this is going to be the
integral from t equals a to t

00:15:20.070 --> 00:15:27.246
equals b, of p of p of x,
really, instead of writing x,

00:15:27.246 --> 00:15:30.740
y, it's x of t, right? x as a
function of t, y as

00:15:30.740 --> 00:15:32.350
a function of t.

00:15:32.350 --> 00:15:33.690
So that's that.

00:15:33.690 --> 00:15:37.600
Times this thing right here,
times this component, right?

00:15:37.600 --> 00:15:39.300
We're multiplying
the i-components.

00:15:39.300 --> 00:15:50.650
So times x prime of t d t, and
then that plus, we're going

00:15:50.650 --> 00:15:52.370
to do the same thing
with the q function.

00:15:52.370 --> 00:15:56.060
So this is q plus, I'll
go to another line.

00:15:56.060 --> 00:15:57.760
Hopefully you realize I could
have just kept writing, but

00:15:57.760 --> 00:15:59.020
I'm running out of space.

00:15:59.020 --> 00:16:09.960
Plus q of x of t, y of t, times
the component of our dr. Times

00:16:09.960 --> 00:16:11.900
the y-component, or
the j-component.

00:16:11.900 --> 00:16:15.530
y prime of t dt.

00:16:15.530 --> 00:16:16.620
And we're done!

00:16:16.620 --> 00:16:17.480
And we're done.

00:16:17.480 --> 00:16:19.300
This might still seem a little
bit abstract, but we're going

00:16:19.300 --> 00:16:23.020
to see in the next video,
everything is now in terms of

00:16:23.020 --> 00:16:25.480
t, so this is just a
straight-up integration,

00:16:25.480 --> 00:16:27.170
with respect to dt.

00:16:27.170 --> 00:16:30.150
If we want, we could take the
dt's outside of the equation,

00:16:30.150 --> 00:16:32.270
and it'll look a little
bit more normal for you.

00:16:32.270 --> 00:16:34.640
But this is essentially
all that we have to do.

00:16:34.640 --> 00:16:38.080
And we're going to see some
concrete examples of taking a

00:16:38.080 --> 00:16:43.230
line integral through a vector
field, or using vector

00:16:43.230 --> 00:16:45.790
functions, in the next video.

00:16:45.790 --> 00:16:46.000