just discovered you last night in frantically studying for my calc test. In a few minutes online, you have made more sense than my prof has all semester. A bit too late to do well in the class, but I will be watching your videos from now on! =)
@Liaomiao When he puts Y1 and Y2, he is saying that the whole Y vector field equation is made up of two separate functions; one is Y1 and one is Y2. It's like drawing two half circles to make a whole circle [for example, in graphing a circle it would be Y1(x)=squareroot(x^2+y^2) and Y2(x)=-squareroot(x^2+y^2) ]. Hopefully that helped :)
At 12:44 I rather preferred to not have heard the word "Volume" as the line integral would have been seen as " an area" The single line integral would be equated to a double integral which is an " area". I guess,
So here the path you drew defines a convex set. What happens if the set is not convex? Taking the max and min points and dividing the path into two will still not get you a function right?
Very good!!! But it was a little difficult to follow the pointer position at the last section of describing what the result equation was, so I repeated that a few times.
I haven't got yet,why have you chosen to pull out a negative sign of the expression? Is this because you wanted to match the results? Or is there some genuine notion behind the same! SOmeone pls help me with the query. Ur response will be gratefully appreciated
Very good, clear example, but a slight criticism for you - at around 7:45, when you're talking about swapping the two functions in the integral, you go to quite exhaustive detail about how how this requires a minus sign on the outside - clarity is always commendable, but given that the subject nature is vector calculus, I think you can assume your watchers have got basic negative multiplying operations covered ;)
You can never be to sure man. That little stuff could get anyone, especially if you're sleep deprived at 1 AM, like I am at the moment. Just because you're learning vector calculus, doesn't mean you're infallible to mistakes and silly misunderstandings.
hmm, dont really see how you can just throw in Y1 and Y2 into the Y part of the respective vector field equation. Sigh, I've been through all the calculus vids and only follow about 95% of it. I think there's still something fundamental about the f(x,y) or f(g(x)) notation that I dont get. I must be slow or something.
6:05 One question. So, you divided the loop (closed path) into two functions. Yes, you divided them into functions and not into two other (open) paths, because the two new functions are just that, functions of _x,_ and they don't have a parameter _t,_ and therefore they don't have an orientation, and so they aren't parametric functions ans thua they don't define an oriented path byt rather just a curve. My question is, is that valid? I don' get it why it's valid. Because you're dividing the original closed path into two *functions* and not *paths,* and therefore you lose the orientation.
Let x=a+t when the curve points to the left, and x=a+b-t when going the other way. More tedious, but you get the same thing in the end. And as another comment said, the bounds from a to b or from b to a reflect the orientation, so it’s fine to not add t as a parameter.
@kylemartin3 oh you should definitely go through the rest of the calc videos. I got a 5 on my BC exam without finishing the precalc class in my school. people were confused
What do you want to know exactly? The path integral of a vector field sums the components of the field along a given path at each point. The components are given by F•dr where dr is the displacement along the path. dr = i dx + j dy means that the change in the radial position vector (the displacement vector) has two components: one horizontal (dx) and one vertical (dy).
P()dx is the line integral, he derives fro that to a double integral which computes the surface area, and it's x-y plane is bounded by (a,b) and (y1, y2).
Simply because if you evaluate the purple definite integral that you see on the right ( which includes the dy) you'd end up with the purple inside-part of The integral on the left.
My opinion: it is still implicitly there, in dr/dt = dx/dt+dy/dt, but we don't need it for this Green's theory derivation, so he cancels dt to have dr=dx+dy.
franco diaz seeing Green's theorem in action can be achieved by looking at examples. A proof has to be all encompassing and abide by mathematical logic. This presentation has neither characteristic. So, presenting it as a proof is misleading. Incomplete knowledge can be more damaging than no knowledge.
when you realize that you were never bad at maths, but the way it was taught to you was too much bs. I am crying right now.
just discovered you last night in frantically studying for my calc test. In a few minutes online, you have made more sense than my prof has all semester. A bit too late to do well in the class, but I will be watching your videos from now on! =)
funny enough i'm also here on my final night.
@@clingyking2774 Good luck! I got through ok, and Khan academy was a really great resource for me back then.
I also have my final tomorrow. I've come across lots of videos, but sal's are the most in depth in this topic.
You alive bro?
@@lukenelissen5722 yeah, that's very satisfying though
@Liaomiao When he puts Y1 and Y2, he is saying that the whole Y vector field equation is made up of two separate functions; one is Y1 and one is Y2. It's like drawing two half circles to make a whole circle [for example, in graphing a circle it would be Y1(x)=squareroot(x^2+y^2) and Y2(x)=-squareroot(x^2+y^2) ]. Hopefully that helped :)
Great video. Unfortunately my current maths course does not have time to prove Greens theorem but this more than make up for it.
Depends on your definition of "understand".
this example was very gud.......I didnt get how we get dx from dr in the previous videos but now its clear......thank Khan.
@rakurai19
He uses smooth draw, if you were wondering...
from 11 months ago :\
pretty good
can someone tell me whether this is the original proof or just the unofficial proof ?
Nice Green's Theorem proof. I find that a lot of professors don't take the time to explain the "why" behind these multifaceted theorems.
At 12:44 I rather preferred to not have heard the word "Volume" as the line integral would have been seen as " an area"
The single line integral would be equated to a double integral which is an " area". I guess,
Thanks
So here the path you drew defines a convex set. What happens if the set is not convex? Taking the max and min points and dividing the path into two will still not get you a function right?
Very good!!! But it was a little difficult to follow the pointer position at the last section of describing what the result equation was, so I repeated that a few times.
I haven't got yet,why have you chosen to pull out a negative sign of the expression?
Is this because you wanted to match the results? Or is there some genuine notion behind the same!
SOmeone pls help me with the query. Ur response will be gratefully appreciated
pulling out a -1 makes it easier to build the inner integral. integral is usually associated with a minus sign rather than a plus sign.
@@pecopeco5907 i mean its still minus sign no?
@rakurai19 he used to use MS paint, like in the older calc and physics videos
tks so much!
Thank you a lot
Will it be different if it's clockwise?
Very good, clear example, but a slight criticism for you - at around 7:45, when you're talking about swapping the two functions in the integral, you go to quite exhaustive detail about how how this requires a minus sign on the outside - clarity is always commendable, but given that the subject nature is vector calculus, I think you can assume your watchers have got basic negative multiplying operations covered ;)
You can never be to sure man. That little stuff could get anyone, especially if you're sleep deprived at 1 AM, like I am at the moment. Just because you're learning vector calculus, doesn't mean you're infallible to mistakes and silly misunderstandings.
Tom
hmm, dont really see how you can just throw in Y1 and Y2 into the Y part of the respective vector field equation. Sigh, I've been through all the calculus vids and only follow about 95% of it. I think there's still something fundamental about the f(x,y) or f(g(x)) notation that I dont get. I must be slow or something.
Good one
6:05 One question. So, you divided the loop (closed path) into two functions. Yes, you divided them into functions and not into two other (open) paths, because the two new functions are just that, functions of _x,_ and they don't have a parameter _t,_ and therefore they don't have an orientation, and so they aren't parametric functions ans thua they don't define an oriented path byt rather just a curve.
My question is, is that valid? I don' get it why it's valid. Because you're dividing the original closed path into two *functions* and not *paths,* and therefore you lose the orientation.
They don’t have an "orientation" but in integrals the order of a and b reflects the direction C is integrated in so it’s valid
Let x=a+t when the curve points to the left, and x=a+b-t when going the other way. More tedious, but you get the same thing in the end.
And as another comment said, the bounds from a to b or from b to a reflect the orientation, so it’s fine to not add t as a parameter.
@kylemartin3 oh you should definitely go through the rest of the calc videos. I got a 5 on my BC exam without finishing the precalc class in my school. people were confused
U feel old yet
What happens if the closed loop line integral goes clockwise? Is there some other double integral that is equivalent to this line integral?
you just have to multiply by -1-
thanks! you rock
At 7:03 why is the need for multiplication and division of -1 ?
I think he made that so that the inner integral can go from y1(x) to y2(x) instead of the opposite
haven't even finished learning derivatives and your understanding vector integral calculus....hmmm find that hard to believe :p
nice job...
splendid!!
Simply delicious!
I'm going crazy trying to find the video where you explain where dr comes from. Can someone help me please?
Not sure this will help, but it is the derivative of a position vector function. I believe it is tangent to the path taken.
What do you want to know exactly?
The path integral of a vector field sums the components of the field along a given path at each point. The components are given by F•dr where dr is the displacement along the path. dr = i dx + j dy means that the change in the radial position vector (the displacement vector) has two components: one horizontal (dx) and one vertical (dy).
How exactly do you pass from integrating over a line to integrating over a region? I know that's correct; I mean the explanation is incomplete.
P()dx is the line integral, he derives fro that to a double integral which computes the surface area, and it's x-y plane is bounded by (a,b) and (y1, y2).
Req in Indonesia min
vvvvvvvvv thanks
@9:54 can anyone explain how it is dy(the whole derivative of y)
Simply because if you evaluate the purple definite integral that you see on the right ( which includes the dy) you'd end up with the purple inside-part of The integral on the left.
Not clear
I get really frustrated when you spend 2 minutes multiplying by -1
I dont get at all why is there no "t" variable which is in the defenition of a line integral.
My opinion: it is still implicitly there, in dr/dt = dx/dt+dy/dt, but we don't need it for this Green's theory derivation, so he cancels dt to have dr=dx+dy.
cool
@kylemartin3 so its not just me..
simply more proof of how fucking awesome sal is.
妙啊
you. are. a. god.
i thought the same thing! haha
IIMPROVEMENT NEEDED!!!
i did not get anything
Haider Ali hahahahahahahahahahhahahahahahahahha
YESSSSSSSSSSSSSSSSSSSSSSSSSSSSSS =D
u oversimplified it son
This over simplified and quite frankly pathetic.
we are not all mathematicians but need an understanding of how Green's theorem works. So this is quite helpful
franco diaz seeing Green's theorem in action can be achieved by looking at examples. A proof has to be all encompassing and abide by mathematical logic. This presentation has neither characteristic. So, presenting it as a proof is misleading. Incomplete knowledge can be more damaging than no knowledge.
@@Dproceeder
Why don't you try and prove it in a better way?