Path independence for line integrals | Multivariable Calculus | Khan Academy

[@AQ-ej4vv]

it is really amazing how this lesson is still beneficial after 11 years of recording it. thank you khan academy..

[@davidlei6354]

and 3 years after tha comment its still great

[@ieranik1301]

An intuitive explanation is as follows. Think of the graph of F as a mountain range. You want to go from the point (x,y,F(x,y)) to (x',y',F(x',y')). Lets denote the two points by A and B respectively. The line integral is just the work done by you against the gravitational force while you go from A to B. The amount of work you do against the gravity depends on the difference of the heights of A and B, which is F(x.y)-F(x',y'). In other words, the change in your gravitational potential energy is independent of the path you traveled, rather depends on the difference of height between your starting point and ending point. If you take some time to think about it, everything falls into place in this explanation. For example, let the gradient of F be f, which is the direction of the steepest ascent. If you are walking along the direction of f, you are walking upwards in the mountain range. In this case, the dot product of f and dr is positive and your are doing positive work against gravity. Otherwise if your are walking downwards, you are doing negative work. Thus the total amount of signed work remains fixed whatever path you choose.

[@ieranik1301]

and the higher the magnitude of f, the steeper that direction is, and following that direction results in more work.

[@jisyang8781]

5:50 Sal misspoke - gradient points to the direction of steepest ASCENT, not descent.

[@animeshpathak3921]

It can be both?

[@apolloniuspergus9295]

@@animeshpathak3921 No.

[@nandhannatarajan2127]

@@animeshpathak3921 No. steepest descent would be the gradient multiplied by minus 1.

[@animeshpathak3921]

@@nandhannatarajan2127 well yes the direction would be opposite , i thought that was obvious.

[@animeshpathak3921]

ascending up the stairs or descending down doesnt change numbers of stairs present

[@gullisreisen]

Visually you can think of F as a mountain and f therefore is the height-change of this mountain at every Point x,y. Starting from Point a you want to go to Point b. The Integral over any path from a to b over f (the height-change of the mountain) will give you the total height difference between Point A and Point B. This height difference is the same as F(b) - F(a), where F ist just the function describing this mountain. Futhermore it makes sense that it doesn't matter which way you walk up the mountain. The height difference is the same if you go directly where the gradient points (steepest ascent) or walk upwards in a random curvy line. So it's path independant. :)

[@shawon265]

For those who likes Grant's procedure of visualizing, here's how ∇F is always a conservative field: 1. Imagine a 3d graph of F(x,y). Height in z axis at poimt (x,y) corresponds to value of F(x,y) at that point. 2. ∇F. *v* is just directional derivative in direction of *v* . So ∇F.d *r* gives us how much value of F changes if you give a tiny nudge along the curved path. 3. So, integrating from point (x₁,y₁) to (x₂,y₂) along the path will add up all these tiny changes, giving us the total change: F(x₂,y₂)-F(x₁,y₁). 4. Clearly this is not dependent on the path, rather depends on starting and ending points. So any path you take, will give you the same value. Aka, conservative field.

[@rikthecuber]

Thanks that was helpful

[@PriontySaha-cm8zh]

Thanks a lott.

[@Bobo-gj6xy]

Can't believe my textbook did allllll of this in two lines and expects "me" to understand wtf happened. smh! Tnx Sal

[@Cashman9111]

I don't quite get it, if line integral is the area between the curve and function f, then taking different curve we should have pretty much different area ?

[@MrHan-is1ko]

same. my brain hurted after thinking about how they did came up with that in the book.

[@yuhangcao1171]

gradient is the direction that increasing fastest.

[@ptyptypty3]

And just like the regular derivative, the gradient points in the direction of greatest increase

[@superdupe8]

Don't gradients point towards higher f(x, y) values, not smaller ones?

[@HakaTech]

Hi I saw your comment being posted over 2 years ago. Do you have an answer for you question now that you are older or maybe you are studying something completely different? cheers,

[@zaid-ajaj]

+superdupe No, gradients point in the direction of other points of higher steepness, i.e. |slope| is big, like running down a hill (to a minimum)

[@AngeloYeo]

I agree with superdude. Please see : en.wikipedia.org/wiki/Gradient

[@joshua5229]

Yeah, that’s right

[@tejusramakrishna2965]

yes gradients points towards higher f(x,y) values watch this video by Khan Academy: kzbin.info/www/bejne/i2irgpJ9lLNprNE

[@yli5531]

5:26 Gosh I really wished he could use vectors instead of ijs.

[@upasanakalra2206]

This was really helpful .

[@jarikosonen4079]

Then how to calculate the work in closed path case? Earlier example case work was -2*pi for circular path around origo for f(x,y)=x*i-y*j, where a=b and t is from 0 to 2*pi. Maybe because f(x,y)=x*i-y*j can not be result of differentiation of any function at x>0, y=0 (or any 1 point around the origo). But what is easiest way to show the antidifferentiation of f(x,y) would not be continuous at least in 1 point (at t=0- to 0)? Path independence of 'vector' type of the line integrals where vector field can be obtained by gradient of scalar field. Cause the scalar type line integral would produce the area of the wall, what would depend on the path taken. Derivation looks correct though in the context he wants and goes into a lot of detail and this all seems solid proof. This would be results for the work without any type of friction involved. Hope to get also summary of the types of the line integrals, their properties how they are used.

[@kyleberlin6990]

Hey I think your videos are great and extremely helpful because you explain so much. My only suggestion is that you could do less with the indecisiveness.

[@joluju2375]

This demonstrates that if f is the gradient of some F, it implies that f is conservative. But it doesn't demonstrate that if f is conservative, it has to be the gradient of some F. Am I right ? If so, are there conservative fields which are not a gradient of some function ?

[@Person514cs]

Very well explained.

[@ramonstuder1977]

You made my day!

[@DeadliestEvil]

great video i just wish he would add an example on top of the proof

[@Shmiwi]

@Taowhr It's a portmanteau of "Regardless" and "Irrespective", not a double root negation.

[@cziegl3r]

Who's the man? Sal's the man.

[@Cashman9111]

I don't quite get it, if line integral is the area between the curve and function f, then taking different curve we should have pretty much different area ?

[@ajeet3259]

Area is same only if function is conservative in nature

[@sophiawang1209]

Thank you!

[@tomascizmarik1788]

Thank you! very helpful :)

[@nikolaypopov3671]

I don't quite get where did we take that F(x,y) function from, and how come that there is a surface involved?

[@asiergonzalezgarcia54]

There are various methods for finding the potential function from a vectorial function. One is simply doing the integral and then checking the constants for the other variables. The point in this video was not to find it, they have some videos abaut that on the multivariable calculus course. The surface, in this case, may not be a surface. I mean, it's just an standar scalar function, takes some inputs and spits out a single value. That's the potential function. Hope I've been helpful, one year later.

[@ahmedsmart1000]

Watch all KA vids at 2x speed :)

[@HitomiAyumu]

So do I. He talks waaay too slowly.

[@ashar08]

Exactly. Doing that since internet arrived.

[@Boss70305]

what does the Î stand for?

[@jimb4549]

it's sometimes called "i hat" and it denotes the x-component of the vector, "j hat" is the y-component and "k-hat" is the z-component

[@alekseyklintsevich4601]

The proof is beautiful!

[@vkv392]

I REALLY MISS GRANT .... :(

[@erikmoe4767]

Does anyone actually like using ijk over ?

[@leoncio91]

thank you sal, you the man

[@aasimkhalid3754]

Sal i really don't know how to repay you, THANK YOU

[@marknisnisan1282]

conzzzervative. Good video but its hard to see what you're pointing at when you say "this" or "that"

[@Headrum]

So much intuition.

[@TheMightySponge]

Thanks:)

[@MisterTutor2010]

Looks kind of like the proof for the product rule.

[@Phi1618033]

1:43 "irregardless" :/

[@sam_sprague]

Tal Moore Irregardless is in the dictionary, so it is actually a word. It's just not proper.

[@williamgarvey4088]

The definition of whilst is while, another example of a word that is basically a duplicate. Get off your high horse - he's teaching all this stuff because he believes that education should be freely accessible and you guys look down on him because he made a common word mistake? You do realize dictionaries should reflect language, not nice versa? Take the language spanish, they use double negatives sometimes and it's considered correct, so although you may think it's "RIDICULOUS" to say irregardless many languages use double negatives all the time. It's just CONVENTION, and convention changes. You can acknowledge his "mistake" without being a dick about it.

[@williamgarvey4088]

Let me say that I am not arguing that irregardless is proper or standard. If I saw "irregardless" in a formal paper I would definitely point it out and correct it. But I would also correct things like "don't" to "do not." That may not be the best example but the point is that if someone chooses to speak in an improper or nonstandard way it does not AUTOMATICALLY mean what they are saying is incorrect and that they should never use it - hence "don't." Now, we could argue about whether not this video is formal or not and how speech formalities may differ from written, etc. but I'm not really interested in that. What I am trying to say is sure, you can point out that irregardless is generally considered nonstandard or incorrect because it probably will help people understand him in the future (although I have never heard of someone misunderstanding irregardless) however it doesn't require reactions like "Irregardless is not a word." or "irregardless ;/". I think when people have reactions like these they are misunderstanding language and they only see it from the eyes of English class. You should really look up descriptivism vs prescriptivism and the two sides where one side believes language should be standardized ("don't use irregardless!") which has advantages and the other side believes that we should describe language ("people say regardless but irregardless too") and how it is used to understand it which also have advantages. You can say that language when consistently written obeying the rules is better but the fact is many of our rules in the language have extreme amounts of exceptions, and that people change language all the time (selfie, conversate, "I seen it") and when they changes become ingrained they may eventually become a rule. This does not mean I'm arguing against a standard, only that non-standards are okay sometimes, like everyday conversations. Also what I meant by convention was that the term regardless is convention.

[@williamgarvey4088]

I thank you for not resulting to name calling and agreeing to disagree.

[@EDUARDO12348]

lol at 9:10

[@thedivinehostess]

Thanks. Great video

[@norwayte]

Master.

[@luffymaxair]

thanks form MECH 2407

[@ericterry3006]

"irregardless" is not a real word.

[@LogicalPhallusy]

@Taowhr heh heh, I was about to mention that, but I noticed you did already.

[@sarahjassim4511]

اريد مترجم بالعربي ماافهم

[@MoguMogu818]

I'm still confused

[@kyriepotter5984]

It's me, I'm Kathy. I've come home.

[@kam1am]

man his videos are good, but really really slow sometimes

[@FliegendeHollaender]

that'll be 50$

[@maracachucho8701]

9:12 SHE HAS A GLANDULAR PROBLEM

[@khobujkhobuj8670]

it would be clear if u use black board

[@mustafahmed9101]

I'm woke

[@luisRG17]

Thank you!