- Line integrals and vector fields
- Using a line integral to find work
- Line integrals in vector fields
- Parametrization of a reverse path
- Scalar field line integral independent of path direction
- Vector field line integrals dependent on path direction
- Path independence for line integrals
- Closed curve line integrals of conservative vector fields
- Line integrals in conservative vector fields
- Example of closed line integral of conservative field
- Second example of line integral of conservative vector field
- Distinguishing conservative vector fields
- Potential functions
Parametrization of a reverse path
Understanding how to parametrize a reverse path for the same curve. Created by Sal Khan.
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- I'm very lost at3:45, where does he get x(a+b-t) from?(32 votes)
- choosing the quantity (a+b-t) is a nice trick that satisfies
* when t=a, (a+b-t)=(a+b-a)=b
* when t=b, (a+b-t)=(a+b-b)=a
That is, (t) and (a+b-t) reverse values at a and at b(37 votes)
- During4:10. Shouldn't it be b<= t <= a? not a<= t <= b? I mean I understand it is a <= t <= b if you think that the line starts at a, but when I think of it graphically I mean t increases as we move on the positive x-axis direction I think its b<= t <= a? I'm confused which way to think about it, but for now I'll think of it that t strats at the top being t=a and we move down to t=b.(4 votes)
- I think the key thing to realize is that T is what does not change between the two integrals. it's starting value and ending value are the same in both cases.
Therefore, in order to make the integral move in the reverse direction you have to change how the integral interprets the end points (if that makes sense).
Now, if you went and messed with the value of T directly, you couldn't make a direct comparison between the integrals without substituting in and getting back the A+B-T.(2 votes)
- Guys, how do you link all those knowledge together? Is there scheme in khan academy for mathematics? I think I start losing the connection because of my lacking in the English language. I have a weak relationship with it.(2 votes)
- Everything is presented in order.
At the top of the page on the left, beside the Khan Academy logo is a drop down list of subjects. Click on that, then click on math, if you are not already there. The suggested program of study is there.
Now it is possible it could be a language barrier thing, but consider this too:
Math at this level isn't just about doing the calculations, it is more thinking about how these things work the way they do, and that introspective analysis takes time. In a typical university setting, the three courses, pre, differential and integral calculus will take you a year and a half to complete (a year if your university has the trimester system) before getting into multivariable calculus. In that year and a half you are practicing, thinking doing research, usually with a 3 month break to continue reflecting on what you have learned, all over an 18 month period. On average, the typical student cannot internalize all this in less than a year, so it may be that you are rushing a bit. Only you can know the answer.
If you haven't already, please download a calculus textbook and start doing problems, and not just the "differentiate the following" types of problems, but the more advanced ones that force you to think about what you know and apply it - these are the problems that get your mind's internalization machinery working, these are where the connections are forged.
Perhaps take a break from multivariable calculus and go back to single variable calculus, but this time from an analytical point of view, where we talk about the why and how of mathematics. Real Analysis may be what you are looking for. Here are some textbooks made available free by the author. Try the elementary book:
Have you completely finished the pre-calc, differential calc and integral calc sections?
If not, then there could be your answer.
Keep up the great work. I have been following you for a while. You are awesome, RD!(5 votes)
- I don't get what Sal does at3:25.
He says that the curve on the left and the curve on the right are essentially the same, with the difference being only in the case of directions, which is true.
It is true also that they should both yield the same value for x = t and y = t
This means that x = x(t) = x(a + b - t) and y = y(t) = y(a + b - t)
This means that t = a + b - t which implies that t = (a + b)/2 which can only be possible when 'a', 't' and 'b' are in AP, which they surely aren't.
It may work for the points where t = 'a' and t = 'b', but that is obvious, since at 't' = 'a' or 'b', t = (a + b)/2 will be satisfied.
This is just as logical as saying 2 terms are in an AP. And even if they are (which is not possible), there is no progression taking place, since d = 0.
This is the first time I've found it difficult to believe Sal.
Can someone clarify?(2 votes)
- It is true that both curves are generated with the functions x=x(u) and y=y(u). It is not true that x(t) = x(a + b - t). For both curves, c and -c t does go from a to b, but in the first curve, c, the argument goes from a to b with t, in the second curve, -c, the argument goes from b to a. Its true they cover all the same points, but in the opposite order.
Another way of looking at how Sal derived the second parametrization for the reverse path is this:
To follow the same path but in reverse you know you want your argument to go from b to a, but were still assuming that t goes from a to b.
So, to start at b you can plug in (b - (t - a)). That way when t starts at a the t-a term is zero and the argument is b:
(b - (a - a)) = (b - (0)) = b
As t begins increasing the argument starts going down from b, until t reaches b and then our argument evaluates to a:
(b - (b - a)) = (b - b + a) = (0 + a) = a
See? If t goes from a to b, and your argument to your function is (b - (t - a)), then your argument starts at b and goes down to a. And if you distribute out the minus sign there you get:
(b - (t - a)) = (b - t + a) = (a + b - t)
And the last expression is the argument Sal uses.(5 votes)
- What's a scalar field? Maybe I missed the video or part of a video he went over it.(2 votes)
- This might be what you're talking about. https://www.khanacademy.org/math/multivariable-calculus/integrating-multivariable-functions/line-integrals-vectors/v/scalar-field-line-integral-independent-of-path-direction(2 votes)
- It seems to me that the second graph uses a,b in two different contexts. The first is to represent the start and end points on the curve while the second is the actual coordinates of a and b namely (x(a),y(a)) and x(b),y(b)). This is very confusing as it implies that when t=a, it's coordinates are actually x(b),y(b)). Is all of this done simply to prevent a negative t progression? That is, if t progresses positively from a to b then it would have to progress negatively from b to a. Is this the point of the confusion?(2 votes)
- It is done to prevent t from decreasing. If you really wanted to avoid any confusion, you could define a new parameter, s, which is equal to a + b - t. So, when t = a, s = b, and when t = b, s = a. Then you can define the second graph to be a graph of the curve (x(s), y(s)), but still integrate over t.(1 vote)
- Where do you go over parameterizing curves?
I can not find anything about this even though you say in the videos you have?(2 votes)
- The (a + b - t) trick is weakly explained. It seems very important mathematically and the crux of the whole proof here. Can anyone help us all out on this point? I can see many others have stumbled on this part.(2 votes)
- Is the entire path flipped or are we just moving along the same path in reverse? I think it is flipping the curve but I’m not sure because of the symmetry.(1 vote)
- I think it's actually moving the same path in reverse. If you started at point A and ended up at point B by travelling along the curve C between A and B, you could modify the parametric function a little bit so that it has you going along C again but from B to A instead (think of a particle moving as a function of time).(2 votes)
- If we want to find the reverse path of the parameterized function, why can't we directly swap the boundaries? For example, instead of taking the integral from t=a to t=b, we take it from t=b to t=a(1 vote)
What I want to do in the next few videos is try to see what happens to a line integral, either a line integral over a scalar field or a vector field, but what happens that line integral when we change the direction of our path? So let's say, when I say change direction, let's say that I have some curve C that looks something like this. We draw the x- and y- axis. So that's my y-axis, that is my x-axis, and let's say my parameterization starts there, and then as t increases, ends up over there just like that. So it's moving in that direction. And when I say I reverse the path, we could define another curve. Let's call it minus C, that looks something like this. That is my y-axis, that is my x-axis. And it looks exactly the same, but it starts up here, and then as t increases, it goes down to the starting point of the other curve. So it's the exact same shape of a curve, but it goes in the opposite direction. So what I'm going to do in this video is just understand how we can construct a parameterization like this, and hopefully understand it pretty well. And then next two videos after this, we'll try to see what this actually does to the line integral, one for a scalar field, and then one for a vector field. So let's just say, this parameterization right here, let's just define it in the basic way that we've always defined them. Let's say that this is x is equal to x of t, y is equal to y of t, and let's say this is from t is equal, or t, let me write this way. t starts at a, so t is greater than or equal to a, and it goes up to b. So in this example, this was when t is equal to a, and the point right here is the coordinate x of a, y of a. And then when t is equal to b up here, this is really just a review of what we've seen before, really just a review of parameterization, when t is equal to b up here, this is the point x of b, y of b. Nothing new there. Now given these functions, how can we construct another parameterization here that has the same shape, but that starts here? So I want this to be, t is equal to a. Let me switch colors. Let me switch to, maybe, magenta. So I want this to be t is equal to a, and as t increases, I want this to be t equals b. So I want to move in the opposite direction. So when t is equal to a, I want my coordinate to still be x of b, y of b. When t is equal to a, I want a b in each of these functions, and when t is equal to b, I want the coordinate to be x of a, y of a. Right? Notice, they're opposites now. Here t is equal to a, x of a, y of a, here t is equal to b, our endpoint. Now I'm at this coordinate, x of a, y of a. So how do I construct that? Well, if you think about it, when t is equal to a, we want both of these functions to evaluate it at b. So what if we define our x, in this case, for our minus C curve, what if we say x is equal to x of, and when I say x of I'm talking about the same exact function. Actually, maybe I should write it in that same exact color. x of-- but instead of putting t in there, instead of putting a straight-up t in there, what if I put an a plus b minus t in there? What happens? Well, let me do it for the y as well. So then our y, y, is equal to y of a plus b minus t. a plus b minus is t. I'm using slightly different shades of yellow, might be a little disconcerting. Anyway, what happens when we define this? When t is equal to a, when t is equal to a, let's say that this parameterization is also for t starts at a and then goes up to b. So let's just experiment and confirm that this parameterization really is the same thing as this thing, but it goes in an opposite direction. Or at least, confirm in our minds intuitively. So when t is equal to a, when t is equal to a, x will be equal to x of a plus b minus a, right? This is when t is equal to a, so minus t, or minus a, which is equal to what? Well, a minus a, cancel out, that's equal to x of b. Similarly, when t is equal to a, y will be equal to y of a plus b minus a. The a's cancel out, so it's equal to y of b. So that worked. When t is equal to a, my parameterization evaluates to the coordinate x of b, y of b. When t is equal to a, x of b, y of b. Then we can do the exact same thing when t is equal to b. I'll do it over here, because I don't want to lose this. Let me just draw a line here. I'm still dealing with this parameterization over here. Actually, let me scroll over to the right, just so that I don't get confused. When t is equal to b, when t is equal to b, what does x equal? x is equal to x of a plus b minus b, right? a plus b minus b when t is equal to b. So that's equal to x of a. and then when she's able to be why is equal to lie of a plus b minus b, and of course, that's going to be equal to y of a. So the endpoints work, and if you think about it intuitively, as t increases, so when t is at a, this thing is going to be x of b, y of b. We saw that down here. Now as t increases, this value is going to decrease. We started x of b, y of b, and as t increases, this value is going to decrease to a, right? It starts from b, and it goes to a. This one obviously starts at a, and it goes to b. So hopefully, that should give you the intuition why this is the exact same curve as that. It just goes in a completely opposite direction. Now, with that out of the way, if you accept what I've told you, that these are really the same parameterizations, just opposite directions. I shouldn't say same parameterizations. Same curve going in an opposite direction, or same path going in the opposite direction. In the next video, I'm going to see what happens when we evaluate this line integral, f of x ds, versus this line integral. So this is a scalar field, a line integral of a scalar field, using this curve or this path, but what happens if we take a line integral over the same scalar field, but we do it over this reverse path? That's what we're going to do in the next video. And the video after that, we'll do it for vector fields.