Closed curve line integrals of conservative vector fields Showing that the line integral along closed curves of conservative vector fields is zero
Closed curve line integrals of conservative vector fields
- In the last video, we saw that if a vector field can be
- written as the gradient of a scalar field-- or another way
- we could say it: this would be equal to the partial of our big
- f with respect to x times i plus the partial of big f, our
- scalar field with respect to y times j; and I'm just writing
- it in multiple ways just so you remember what the gradient is
- --but we saw that if our vector field is the gradient of
- a scalar field then we call it conservative.
- So that tells us that f is a conservative vector field.
- And it also tells us, and this was the big take away from the
- last video, that the line integral of f between two
- points-- let me draw two points here; so let me draw my
- coordinates just so we know we're on the xy plane.
- My axes: x-axis, y-axis.
- Let's say I have the point, I have that point and that point,
- and I have two different paths between those two points.
- So I have path 1, that goes something like that, so
- I'll call that c1 and it goes in that direction.
- And then I have, maybe in a different shades of
- green, c2 goes like that.
- They both start here and go to there.
- We learned in the last video that the line integral
- is path independent between any two points.
- So in this case the line integral along c1 of f dot dr
- is going to be equal to the line integral of c2, over the
- path c2, of f dot dr. The line, if we have a potential in
- a region, and we may be everywhere, then the line
- integral between any two points is independent of the path.
- That's the neat thing about a conservative field.
- Now what I want to do in this video is do a little bit of
- an extension of the take away of the last video.
- It's actually a pretty important extension; it might
- already be obvious to you.
- I've already written this here; I could rearrange
- this equation a little bit.
- So let me do it.
- So let me a rearrange this.
- I'll just rewrite this in orange.
- So the line integral on path c1 dot dr minus-- I'll just go
- subtract this from both sides --minus the line integral c2 of
- f dot dr is going to be equal to 0.
- All I did is I took this take away from the last video and
- I subtracted this from both sides.
- Now we learned several videos ago that if we're dealing with
- a line integral of a vector field-- not a scalar field
- --with a vector field, the direction of the
- path is important.
- We learned that the line integral over, say, c2 of f dot
- dr, is equal to the negative of the line integral of minus c2
- of f dot dr where we denoted minus c2 is the same path as
- c2, but just in the opposite direction.
- So for example, minus c2 I would write like this-- so let
- me do it in a different color --so let's say this is minus
- c2, it'd be a path just like c2-- I'm going to call this
- minus c2 --but instead of going in that direction, I'm now
- going to go in that direction.
- So ignore the old c2 arrows.
- We're now starting from there and coming back here.
- So this is minus c2.
- Or we could write, we could put, the minus on the other
- side and we could say that the negative of the c2 line
- integral along the path of c2 of f dot dr is equal to the
- line integral over the reverse path of f dot dr. All I did is
- I switched the negative on the other side; multiplied
- both sides by negative 1.
- So let's replace-- in this equation we have the minus of
- the c2 path; we have that right there, and we have that right
- there --so we could just replace this with
- this right there.
- So let me do that.
- So I'll write this first part first.
- So the integral along the curve c1 of f dot dr, instead of
- minus the line integral along c2, I'm going to say plus the
- integral along minus c2.
- This-- let me switch to the green --this we've established
- is the same thing as this.
- The negative of this curve, or the line integral along this
- path, is the same thing as the line integral, the positive of
- the line integral along the reverse path.
- So we'll say plus the line integral of minus c2 of
- f dot dr is equal to 0.
- Now there's something interesting.
- Let's look at what the combination of the path
- of c1 and minus c2 is.
- c1 starts over here.
- Let me get a nice, vibrant color.
- c1 starts over here at this point.
- It moves from this point along this curve c1 and
- ends up at this point.
- And then we do the minus c2.
- Minus c2 starts at this point and just goes and comes back
- to the original point; it completes a loop.
- So this is a closed line integral.
- So if you combine this, we could rewrite this.
- Remember, this is just a loop.
- By reversing this, instead of having two guys starting here
- and going there, I now can start here, go all the way
- there, and then come all the way back on this
- reverse path of c2.
- So this is equivalent to a closed line integral.
- So that is the same thing as an integral along a closed path.
- I mean, we could call the closed path, maybe, c1 plus
- minus c2, if we wanted to be particular about
- the closed path.
- But this could be, I drew c1 and c2 or minus c2 arbitrarily;
- this could be any closed path where our vector field f has a
- potential, or where it is the gradient of a scalar field,
- or where it is conservative.
- And so this can be written as a closed path of c1 plus the
- reverse of c2 of f dot dr. That's just a rewriting
- of that, and so that's going to be equal to 0.
- And this is our take away for this video.
- This is, you can view it as a corollary.
- It's kind of a low-hanging conclusion that you can make
- after this conclusion.
- So now we know that if we have a vector field that's the
- gradient of a scalar field in some region, or maybe over the
- entire xy plane-- and this is called the potential of f;
- this is a potential function.
- Oftentimes it will be the negative of it, but it's easy
- to mess with negatives --but if we have a vector field that is
- the gradient of a scalar field, we call that vector
- field conservative.
- That tells us that at any point in the region where this is
- valid, the line integral from one point to another is
- independent of the path; that's what we got from
- the last video.
- And because of that, a closed loop line integral, or a closed
- line integral, so if we take some other place, if we take
- any other closed line integral or we take the line integral of
- the vector field on any closed loop, it will become 0 because
- it is path independent.
- So that's the neat take away here, that if you know that
- this is conservative, if you ever see something like this:
- if you see this f dot dr and someone asks you to evaluate
- this given that f is conservative, or given that f
- is the gradient of another function, or given that f is
- path independent, you can now immediately say, that is going
- to be equal to 0, which simplifies the math a good bit.
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