- [Instructor] What I wanna do in this video is establish
a reasonably powerful condition in which we can establish
that a vector field or that align integral
of a vector field is path independent.
And when I say that, I mean that let's say,
I were to take this line integral
along the path c of f dot dr,
and let's say my path looks like this.
Let's say my path looks like this.
that's my x and y axis, y and x,
and let's say my path looks something like this.
I start there and I go over there to point c just like
or to my end point, the curve here is.
And so I would evaluate this line integral,
this vector field along this path.
This would be a path independent vector field
or we call that a conservative vector field if,
if this thing is equal to the same integral,
the same integral over a different path,
that has the same end point.
So let's call this c1 so this is c1, and this is c2.
This vector field is conservative if,
if I start at the same point, but I take a different path,
let's say I go something like that.
If I take a different path and this is my c2,
I still get the same value.
these integrals, what's this telling me,
is it all it cares about to evaluate these integrals
is my starting point and my ending point.
It doesn't care what I do in between.
It doesn't care how I get from my starting point
to my end point.
These two integrals have the same start point
and same end point, so irregardless of their actual path,
they're going to be the same.
that's what it means for f to be a conservative field
or what it means to be for this integral
to be path independent.
So before I prove or I show you the conditions,
let's build up our toolkit a little bit.
And so you may or may not have already seen
the multivariable chain rule, multivariable chain rule.
And I'm not gonna prove it in this video,
but I think it'll be pretty intuitive for you,
so maybe it doesn't need to have a proof
or I'll prove it eventually,
but I really just wanna give you the intuition.
And all that says is that if I have some function,
let's say I have f of x and y but x and y are then functions
of, let's say 1/3 variable t so f of x of t
and y of t, that the derivative of f with respect to t,
the derivative of f with respect to t, is multivariable.
I have two variables here an x and y.
this is going to be equal
to the partial of f with respect to x,
how fast does f change as exchanges,
times the derivative of x with respect to t.
This is a single variable function right here.
So you just can take a regular derivative.
So times how fast x changes with respect to t.
This is a standard derivative, this is partial derivative,
because at that level we're dealing with two variables
and we're not done plus how fast f changes
with respect to y.
How fast the partial of f with respect to y times
the derivative of y with respect to t.
So dy, dt.
I'm not gonna prove it,
but I think it makes pretty good intuition.
This is saying, as I move a little bit dt,
how much of a df do I get
or how fast does f change with respect to t?
It says, well, there's two ways that f can change.
It can change with respect to x,
and it can change with respect to y.
So why don't I add those two things together
as they are both changing with respect to t?
That's all it's saying and if you kind of imagined
that you could cancel out this partial x with this dx
and this partial y with this dy,
you could have kind of imagined
the partial of f with respect to t on the x side of things,
and that plus the partial of f with respect to t
on the y in the y dimension,
and then that'll give you the total change of f
with respect to t.
Kind of a hand wavy argument there, but at least to me,
this is a pretty intuitive formula.
So that's our toolkit right there,
the multivariable chain rule.
We're gonna put that aside for a second.
Now, let's say I have some vector field f
and this is a different than this f,
so I'll do it in a different color, magenta.
I have some vector field f that is a function of x and y,
and let's say that it happens to be
the gradient of some scalar field.
Let's say it equals the gradient of some scalar field
I'll call that capital F.
And this is gradient which means,
and this is capital f is also a function of x and y.
So I could write f is also, let me write it,
I don't wanna write it on a new line,
I could also write up here.
Capital F is also a function of x and y
and the gradient, all that means is that f,
the vector field f of x, y lowercase f of x, y
is equal to the partial derivative of uppercase F
with respect to x times the eye unit vector,
plus the partial of uppercase F with respect
to y times the j unit vector.
This is the definition of a gradient right here.
This is the definition of a gradient,
and if you imagine that uppercase F is some type of surface,
so if uppercase F is some type of surface like that,
just trying to do my, so this is uppercase F of x, y
the gradient of f of x, y is going to be a vector field
that tells you the direction
of steepest ascent at any point.
So it'll be defined on the x, y plane.
So on the x, y plane, it'll tell you, so let me draw.
That's the vertical axis, maybe that's the x axis,
that's the y axis.
So the gradient of it,
if you take any point on the x, y plane
it'll tell you the direction you need to travel to go
into this deepest descent and for this gradient field
it's going to be something like this.
It's going to be something like this
and maybe over here it starts going in that direction
'cause you would descend towards
this little minimum point right here,
anyway I don't wanna get too involved in that.
And the whole point of this isn't to really you get
the intuition behind ingredients
there are other videos on this.
The point of this is to get a a test
to see whether something is path independent
whether a vector field is path independent,
whether it's conservative and it turns out
that if this exists and I'm gonna prove it now.
If this exists, if f is the gradient of some scalar field
if f is equal to the gradient of some scalar field
then, then f is conservative, conservative
or you could say it doesn't matter what path we follow
when we take a line integral over f,
it just matters about our starting point
and our ending point.
Now, let me see if I can prove that to you.
So let's start with the assumption
that f can be written this way
as the gradient of that lowercase f can be written
as the gradient of some uppercase F.
So in that case, in that case,
our integral, let's see well, let's define our path first.
So our position vector function, we always need one
of those to do a line integral or a vector line integral,
r of t is going to be equal to x of t times i
plus y of t times J for t going between a and b,
we've seen this multiple times.
This is the definition of
this is just a very general definition
of pretty much any path in two dimensions.
And then we're gonna say f, f is going to be f of x, y
is going to be equal to this
it's going to be the partial derivative
of uppercase F with respect to x so we're assuming
that this exists, that this is true
except with times i plus the partial of uppercase F
with respect to y times j.
Now, given this what is f lowercase f lowercase dot dr
going to equal over this path right here
this path is divined by this position function right there.
Well, it's going to be equal to what we need to get
we need to figure out what dr is
and we've done that in multiple videos
so I'll do that on the right over here.
Dr, we've seen it multiple times I'll just
actually, I'll solve it out again.
DR, dt by definition was equal
to dx, dt is equal to dx, dt times i plus dy, d,
I don't know why it got all fat like that.
Dy, dt times j, that's what dr, dt is.
So if we wanna figure out what dr is, the differential dr,
if we wanna play with differentials in this way,
multiply both sides times dt,
and actually I'm gonna treat dt as,
well, I'll multiply it, I'll distribute it.
It's dx, dt times dt, i plus dy, dt times dt, j.
So if we're taking the dot product
of f of f with dr, what are we gonna get?
What are we gonna get?
We are going to get, and this is, let me,
so we're gonna take,
so this is going to be the integral over the curve,
over the curve from, all right, the c right there.
We can write it in terms of the end points of t,
once we feel good that we have everything in terms of t,
but it's gonna be equal to this dot that
which is equal to the partial,
I'll try to stay color consistent,
the partial of uppercase F with respect
to x times that times dx, dt.
I'm gonna write this dt in a different color, times dt plus,
plus the partial of uppercase F with respect
to y times this, we're multiplying the j components, right?
We can take the dot product, multiply the i components,
and then add that to, what you get from the product
of the j component.
So this j component is partial of uppercase F
with respect to y.
And then we have times, let me switch to a yellow,
dy, dt times that dt right over there
times that dt over there, and then we can factor out the dt.
We could factor out the dt
or actually it's just so I don't have
to even write it again,
right now, I wrote it without, well, let me write it again.
So this is equal to the integral,
and let's say we have it in terms of t.
We've written everything in terms of t,
so t goes from a to a.
And so this is going to be equal to, I'll write it in blue,
the partial of uppercase F with respect to x times dx, dt
plus, I'm distributing this dt out,
plus the partial of uppercase F with respect to y, dy, dt,
all of that times dt.
This is equivalent to that.
Now you might realize why
I talked about the multivariable chain rule.
What is this right here?
What is that right there?
You can do some pattern matching.
That is the same thing as the derivative of uppercase F
with respect to t.
Look at this, let me, let me copy and paste this
just so you appreciate it.
Let me copy and paste that.
You copy it, and then let me paste it.
So this is our definition or this is our,
I won't say definition one can actually prove it,
you don't have to start from there,
but this is our multivariable chain rule right here, right?
The derivative of any function with respect to t
is the partial of that function with respect
to x times dx, dt plus the partial of that function
with respect to y, dy, dt.
I have the partial of uppercase F with respect to x, dx, dt
plus the partial uppercase F with respect to y.
And this are identical if you just replace this lowercase f
with an uppercase F.
So this in blue right here,
so this whole expression is equal to,
the integral from a to b, t is equal a to t is equal to b
of, of, in blue here,
the derivative, regular derivative of f
with respect to t, dt.
And how do you evaluate?
Let me read the dt in green.
How do you evaluate something like this?
And I just wanna make a point.
This is just this, from the multivariable chain rule.
And how do you evaluate a definite integral like this?
Will you take the anti-derivative of the inside
with respect to dt?
So what is this going to be equal to?
You take the anti-derivative of the inside,
the anti-derivative of the inside, that's just df
oh, sorry, that's just f.
So this is equal to f of t.
And let me be clear, let me be clear here.
We wrote before that f is a function,
so our uppercase F is a function of x and y
which could also be written since each of these
are functions of t of, it could be written as f of x of t
of y of t, I'm just rewriting it in different ways.
And this could be just written as f of t.
This is the same thing as f of t.
These are all equivalent,
depending on whether you wanna include
the x's and the y's only or the t's only
or them both cause both of the x's and y's
are functions of t.
So this is the derivative of f with respect to t.
If this was just in terms of t,
this is the derivative of that with respect to t.
We take it's anti-derivative, we're left just with f,
and we have to evaluate it
from t is equal to a to t is equal to b.
And so this is equal to, and this is the homestretch,
this is equal to f of b minus f of a.
And if you wanna think about it in these terms,
this is the same thing.
This is equal to f of x of b, y of b,
let me re-check out all the parentheses,
minus f of x of a, y of a, these are equivalent.
You give me any point on the x, y plane, an x and a y,
and it tells me where I am.
This is my capital F, it gives me a height just like that.
But what this tells me this,
this associates a value with every point on the x, y plane.
And what this whole exercise,
remember this is the same thing as that.
This is our whole thing that we were trying to prove.
That is equal to f dot dr.
F dot dr, our vector field for which is the gradient
of the capital F, remember F,
F was equal to the gradient of f.
We assume that it's the gradient
of some function, capital F.
If that is the case,
then we just did a little bit of calculus or algebra
or whatever you wanna call it.
And we found that we can evaluate this integral
by evaluating capital F at t is equal to b,
and then subtracting from that capital f at t is equal to a.
But what that tells you is that this integral,
the value of this integral is only dependent,
is only dependent at our starting point, t is equal to a.
This is the point x of a, y of a, and the ending point
t of b, x of, or t is equal to b which is x of b, y of b.
That integral is only dependent on these two values.
How do I know that?
Because to solve it,
because I'm saying that this thing exists,
I just had to evaluate that thing at both those two points.
I didn't care about the curve in between.
So this shows that if, if f is equal to the gradient,
this is often called a potential function
of capital F, although they're usually
the negative of each other, but it's the same idea.
If the vector field f is the gradient
of some scalar field, uppercase F, then,
then we can say that f is conservative.
F is conservative or that the integral,
the line integral of f dot dr is path independent.
It doesn't matter what path we go on
as long as our starting and ending points are the same.
Hopefully, you found that useful.
I will do some examples with that.
And actually in the next video,
I'll prove another interesting outcome based on this one.