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# Vector field line integrals dependent on path direction

Showing that, unlike line integrals of scalar fields, line integrals over vector fields are path direction dependent. Created by Sal Khan.

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• what's the the difference beteen this video and last 2...why different result?
• In the last video, we took the line integral of a scalar field. To think about this, for every value of x and y in the domain of a scalar field, there exists a point f(x,y) which is a number. Just a number, which we represent as the "height" or 'z' coordinate. Taking enough points on the domain x,y, we will get enough heights or 'z' points to map out a surface. When we take the line integral of a scalar field, we are essentially finding the area under the curtain that is formed by the function z=f(x,y) along the path we choose to take. Here direction does not matter because the area of the curtain is the same, no matter if we go 'forward' or 'reverse'.
This video deals with a totally different animal. We are now taking the line integral of a Vector field. That is to say for every position in space r(t), there exists a vector at that point in space. Not just a number any more as was the case with a scalar field. Now each point in space has a vector at that point determined by the vector field. Now when we take the line integral, we are not calculating the "area under a curtain" because this holds no real meaning when dealing with a vector field. When calculating a line integral along a vector field, we are integrating the component of the vector field that is tangent to the path that we take from one point to another. You may want to separate these two "types" of line integrals in your head, because really they are quite different, with different properties, outcomes, and methods of solving.
• what is the difference between ds and dr?
• ds is a vector length, and dr is a vector.
• Is it plausible to think of dependence of path direction as swimming with (positive) or against (negative) a current? What do these values really mean?
• I think it would be perfectly acceptable to think of it that way. The rivers' current would be your vector field and you swimming through the current, either against it or with it, would be your position function. I would just refrain from using an actual person in the current and stick to a particle. People come with hydrodynamics, particles, not so much.
• I don't understand why line integrals over scalar fields are path direction independent, even though ones for vector fields are direction dependent. Isn't a scalar simply a vector with a single dimension?
• This is actually a really good question. The main point here is that there is a difference in definition between scalars and vectors. They should be viewed as two different objects. A scalar is just a magnitude (only one value to describe it), there is no direction associated with it. A vector is defined as a magnitude with a direction (you need more than one value to describe it). At the risk of over-simplifying, the idea of direction is not necessary to describe objects in one dimension. You only need magnitude. So since direction never comes into play for scalars, this is why it doesn't matter for line integrals over scalar fields. I would go back and review the definitions (on wikipedia or something) of both scalars and vectors. I'm not sure if this helps or not...

Edit: Maybe the best way to say it is that the 1D case is a very special case. What this video was showing, was the general case. So in general, direction matters very much, but there are special cases where it doesn't (e.g. scalar fields and conservative vector fields).
• So if direction doesn't matter for scalar fields, then how come basic calculus of 1 variable gives negative areas when integrating from b-->a versus a-->b? Basic single variable calculus just has scalars too...
• If you look at single variable integrals, you'll see that when we get a positive area, it is because the curve we are integrating is above (or mostly above) the x-axis, and we are calculating the area under the curve with the x-axis being our base (for lack of a better word). As Sal says in the video, we get negative values in single variable calculus as well when we reverse the direction because our change in x (the width of those little rectangles) becomes negative when we go from b to a because they represent the change in x, and here the change is negative. In multivariable calculus, it is the same idea. We are calculating the area under a surface, with the curve as our base (hence the idea of a curtain). In these examples we are staying in the positive x,y,z planes and thus are finding positive area. Also, we are calculating the change in both x and y, and the formula is that of the modulus (absolute value) of the derivative of the two values, thus we don't get a negative change in x or y.

For vector fields the idea is different as we are dealing with forces acting upon others with direction. When taking the line integral with a vector field, we are not dealing with areas, but motion and direction. Some other people have explained it better than I can.

Hope this helps!
• 1. Regarding Vector field line integrals, what are their sigificance in physics? Is that kind of the work done by a vector field on something following a curve? I.E., following curve from top to bottom, the work is positive; on the other hand, following curve from bottom to top, the work is equavilent by negative.
2. The substitution by u = a + b - t from f(a+b-t) to f(u) changed the direction of the path, but didn't change the sign of the integral except du = -dt. It is magic.
• why id the limit of integral for -c curve the same as the original curve. Since it is starting from the opposite direction as that of the original curve, shouldn't the limit of the integral be from b to a?
• what's the difference between positon vector valued functions and vector fields?
(1 vote)
• A vector field function will take in a point and output a vector which originates from that point. So maybe this represents the force vector a marble would experience if it were placed at a certain x/y location on a hill.

Position vector functions take in a one-dimensional parameter ( t ) and output a vector pointing to some location. This vector originates from the origin (0,0). They're basically a different way of talking about a location. Instead of referring to the point (2,5) you could talk about a vector that starts at the origin and ends at the point (2,5).

Hope that helps :)
• I'm confused little bit...you look like taking the drivitive of (a+b-t) two times ?
one when they multiplied by x and the 2nd when substituted by u?
• But if the vector field is the gradient of the scalar field wouldn't it be path independent too?
(1 vote)
• Hm. OK. If you have a surface (aka scalar field) which is flat but with some slope, then you will have a surface gradient of parallel vectors pointing in the direction of its slope.
If you complete a circular path through in that vector field, the work done will be 0.
If you start on some point and complete a half-circle clockwise (path A) and, going back to that same point, complete a half-circle counterclockwise (path B), the work done by the vector field for both paths will be equal.
You know this because they must go together to be 0.

Now, when you examine the line integral of that same circle and the sloped flat surface (or scalar field) one half of the circle will not be equal to the other half, and the total line integral will not be 0.

The work done by the gradient of a scalar field and the line integral of a scalar field are not equivelant. One being path independent does not make the other path independent.