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## Line integrals in vector fields (videos)

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# Path independence for lineÂ integrals

## Video transcript

What I want to do in this video
is establish a reasonably powerful condition in which we
can establish that at vector field, or that a line 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 d r, and let's say my path looks like this. That's my x and y axis, and
let's say my path looks something like this: I start
there and I go over there to point c. My end point, the
curve here is c. And so I would evaluate this
line integral, this victor field along this path. This would be a path
independent vector field, or we call that a conservative vector
field, if this thing is equal to 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 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. What this is telling me is that
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 for this integral
to be path independent. So before I prove or I show you
the conditions, let's build up our tool kit a little bit. And so you may or may not
have already seen the multivariable chain rule. And I'm not going to 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
want to 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 a third variable, t, so f
of x of t and y of t --that the derivative of f with respect
to t is multivariable. I have two variables
here in x and y. This is going to be equal
to the partial of f with respect to x. How fast does f change as x
changes times the derivative of x with respect to t-- This is a
single variable function right here, so you to take a
regular derivative. So times how fast x changes
with respect to t. This is a standard derivative,
this is a 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 times the derivative
y with respected t. So d y d t. And I'm not going to prove
it, but I think it makes pretty good intuition. This is saying, as I move
a little bit d t, how much of a d f do I get? Or how fast this 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're 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 d x, and this partial y with this d y, you could kind
of imagine the partial of f with respect to t on the x side
of things, and then plus the partial of f with respect
to t in the y dimension. And then that'll give you
the total change of f with respected to t. Kind of a hand-wavy argument
there, but at least to me this is a pretty intuitive formula. So that's our tool kit right
there; the multivariable chain rule. We're going to put that
aside for a second. Now let's say I have some
vector field f-- and it's 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. I'll call that capital F. And this is gradient which
means that capital F is also function of x and y-- so I
don't want to 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, and all that means is that the vector field f of xy--
lower-case f of xy, is equal to the partial derivative of
upper-case F with respect to x times the i-unit vector plus
the partial of upper-case F with respect to y times
the j-unit vector. This is the definition of
the gradient right here. And if you imagine that
upper-case F is some type of surface-- so this is uppercase
F of xy --the gradient F of xy is going to be a vector field
that tells you the direction of steepest descent at any point. So it'll be defined
the xy plane. So on the xy 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 xy plane, it'll tell you the direction
you need to travel to go into the deepest descent. And for this gradient
field it's going to be something like this. And maybe over here it starts
going in that direction because you would descend towards
this little minimum point right here. Anyway, I don't want
to get too involved. And the whole point of this
isn't to really get the intuition behind gradients;
there are other videos on this. The point of this is to get
other 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 going to prove it now --if f is the gradient
of some scalar field, then f is 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, that lower-case f can be written as the gradient
of some upper-case F. So in that case our integral--
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 4t going between a and b. You've seen this multiple
times; this is a definition of pretty much any path
in two dimensions. And then we're going to say f
of xy 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 --times i plus
the partial of upper-case F with respect to y times j. Now, given this what is
lower-case f dot dr going to equal over this
path right here? This path is defined by this
position function right there. Well, it's going to be equal
to, we need to figure out what dr is, and we've done
that in multiple videos. I'll do that on the
right over here. dr, we've seen it
multiple times. Actually, I'll solve
it out again. dr over dt by definition was
equal to dx over dt times i plus-- I don't know why it
got all fat like that --dy over dt times j. That's what dr over dt is. So if we want to figure out
what dr is, the differential of dr, if we want to play with
differentials in this way, multiply both sides times dt. And actually I'm going to
treat dt, I'll multiply it; I'll distribute it. It's dx over dt times dti
plus dy over dt times dtj. So if we're taking the dot
product of f with dr, what are we going to get? So this is going to be the
integral over the curve from-- I'll write the c right there;
we could write in terms of the end points as t once we feel
good that we have everything in terms of t --but it's going to
be equal to this dot that, which is equal to-- I'll try to
stay color consistent --the partial of upper-case F with
respect to x times that, times dx over d t-- I'm going to
write this st in a different color --times dt plus the
partial of upper-case F with respect to y times--
we're multiplying the j components, right? When you take the dot product,
multiply the i components, and then add that to what you get
from the product of the j components --so this j
component's partial of upper-case F with respect to y,
and then we have times-- switch to a yellow --dy over dt times
that dt right over there. And then we can
factor out the dt. Or actually, 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 b, and so this is going to be equal to-- I'll
write it in blue --the partial of upper case F with respect to
x times dx over dt plus-- I'm distributing this dt out --plus
the partial of uppercase F with respect to y. dy over 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 upper-case F with respect to t. Look at this; let me let me
copy and paste that just so you appreciate 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. The driven of any function with
respect t is the partial of that function with respect to x
times dx over dt plus the partial of that function
with respect to dy over dt. I have the partial of
upper-case F with respect to x times dx over dt plus the
partial upper-case F with respect to y. This and this are identical
if you just replace this lower-case f with
an upper-case F. So this in blue right here,
this whole expression is equal to the integral from t is equal
to a to t is equal to b of-- in blue here --the derivative
of f with respect to dt. And how do you evaluate--
let me just the dt in green --how do you evaluate
something like this? I just want to make a point:
this is just this from the multivariable chain rule. And how do you evaluate a
definite integral like this? Well, you take the
antiderivative of the inside with respect to dt. So what is this going
to be equal to? You take the antiderivative of
the inside, that's just f. So this is equal to f of t. And let me be clear. We wrote before that
f is a function. So our upper-case F is a
function of x and y, which could also be written, since
each of these are functions of t, 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 or t. These are all equivalent,
depending on whether you want to include the x's or the
y's only, or the t's only, or them both. Because 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 its antiderivative,
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 home stretch --this is equal to f
of b minus f of a. And if you want to think about
it in these terms, this is the same thing. This is equal to f of x of b
over y of b-- let me make sure I got all the parentheses
--minus f of x of a over y of a. These are equivalent. You give me any point on the
xy 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. This associates a value with
every point on the xy plane. But 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, which is the gradient of the capital F--
remember 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, whatever you want to 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 at our starting point, t is equal to a, this is
the point x of a, y of a, and the ending point, 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 those two points; I didn't care about
the curve in between. So this shows that if F is
equal to the gradient-- this is often called a potential
function of capital F, although they're usually the negative
each other, but it's the same idea --if the vector field f is
the gradient of some scale or field upper-case F, then we can
say that f is conservative or that 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 point are the same. Hopefully found that useful. And we'll some
examples with that. And actually in the next video
I'll prove another interesting outcome based on this one.