Main content
Course: Multivariable calculus > Unit 4
Lesson 3: Line integrals in vector fields- 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
© 2024 Khan AcademyTerms of usePrivacy PolicyCookie Notice
Parametrization of a reverse path
Understanding how to parametrize a reverse path for the same curve. Created by Sal Khan.
Want to join the conversation?
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:
http://classicalrealanalysis.info/com/FREE-PDF-DOWNLOADS.php
Silly question:
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!(10 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)
- 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?(3 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)
- 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.(3 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)
- 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) 
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)
Also why can't you just do b<=t<=a for the second one?(1 vote)- Because
a < b, so you can't have both inequalities at the same time. It would be like asking, "Give me all the numbers that are greater than 4 AND lesser than 1".(1 vote)
- I understand that x(a+b-t) works, but how would one get that if he/she didn't know that in advance? Is that just something that's "memorized" (ugh @ that word lol), because I feel like it'd be tricky to "know that"?(1 vote)
- You can start with the guess x(-t), knowing that the minus sign changes the direction. Then you plug in the endpoints. x(b) becomes x(-b) and x(a) becomes x(-a). Since a<b implies, -b<-a you note that the direction is indeed reversed, but the endpoints don't quite match up. Maybe if we added some constant, like x(c-t), so that when you plug in t=b you get x(a).
At this point you might guess the solution c = a + b or solve the equation c-b = a for c.
To summarize, you guess that your solution will look like x(-t) to reverse the direction and shift it to make the endpoints match up.(1 vote)
4:10Video transcript
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.