Path independence for line integrals | Multivariable Calculus | Khan Academy

What I want to do in this video is establish a reasonably powerful condition in which we can establish that a vector field or that a line integral of a vector field is path independent.

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. 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 endpoint. The curve here is C, and so I would evaluate this line integral of this vector 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 endpoints. So let's call this C1. 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 of these integrals.

What's this 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 endpoint. These two integrals have the same start point and the same endpoint. 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 toolkit a little bit. You may have or may not have already seen the multivariable chain rule. 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, the derivative of F with respect to t, that 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 can take a regular derivative, times how fast x changes with respect to t. This is a standard derivative; this is the 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 going to 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.

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 kind of imagined 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 on the y thing in the y dimension. Then that will 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 going to put that aside for a second. Now let's say I have some vector field F, and this is 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, you know, capital F is also a function of x and y. So I could write, you know, I could write F is also, let me write it. 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. 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 i unit vector plus the partial of uppercase F with respect to y times the j unit vector.

This is the definition of the gradient right here. 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 will be defined on the x y plane.

So on the x y plane, it will 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 the steepest descent.

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 because you would descend towards this little minimum point right here. Anyway, I don't want to get too involved in that, 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 a test to see whether something is path independent, whether a vector field is path independent, whether it's conservative. It turns out that if this exists, and I'm going to 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 F is conservative.

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. 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, 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; this is just a very general definition of pretty much any path in two dimensions.

Then we're going to 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 saying we're assuming that this exists, that this is true with respect times i plus the partial of uppercase F with respect to y times j. Now given this, given this, what is lowercase f dot dr going to equal over this path right here? This path is divided by this position function right there.

Well, it's going to be equal to... well, we need to get... 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. 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/dt times j.

That's what dr/dt is. So if we want to figure out what dr is, the differential dr, if we want to play with differentials in this way, multiply both sides times dt, and actually I'm going to treat dt as... well, I'll multiply... 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 with dr, what are we going to get? What are we going to get? We are going to get, and this is, let me... so we're going to take... so this is going to be the integral over the curve, from I'll write the c right there. We could write it in terms of the endpoints of 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 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 going to write this dt in a different color times dt, plus, plus the partial of uppercase F with respect to y times this times... we're multiplying the j components right? We take the dot product, multiply the i components, and then add that to the 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 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... 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 where t goes from a to b, and so this is going to be equal to, I write 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 copy and paste this just so you appreciate it.

So let me copy and paste that, 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 of uppercase F with respect to y dy/dt. Do you see this? This is 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 to a to t is equal to b of... in blue here, the derivative, regular derivative of F with respect to t dt.

And how do you evaluate something like this? I just want to make a point. This is just from the multivariable chain rule. How do you evaluate a definite integral? Right, this will... 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, the antiderivative 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 are... which could also be written since each of these are functions of t, could be written as F of x of t, 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 want to include the x's and the y's only or the t's only or them both. They're... they're 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 this term, these terms, this is the same thing. This is equal to F of x of b, y of b. Let me make sure I got 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 value, this associates a value with every point on the x y plane.

But 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, 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 on 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, which is x of b, y of b. This integral is only dependent on these two values. How do I know that? Because to solve it, I just... 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 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 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, and we'll do some examples with that.

Actually, in the next video, I'll prove another interesting outcome based on this one.