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independence of path (KristaKingMath)
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- čas přidán 6. 08. 2024
- ► My Vectors course: www.kristakingmath.com/vector...
In this video we'll learn how to prove that a line integral is independent of path. We'll do this by showing that the vector field is conservative, knowing that any line integral inside a conservative vector field will be independent of path.
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Hi, I’m Krista! I make math courses to keep you from banging your head against the wall. ;)
Math class was always so frustrating for me. I’d go to a class, spend hours on homework, and three days later have an “Ah-ha!” moment about how the problems worked that could have slashed my homework time in half. I’d think, “WHY didn’t my teacher just tell me this in the first place?!”
So I started tutoring to keep other people out of the same aggravating, time-sucking cycle. Since then, I’ve recorded tons of videos and written out cheat-sheet style notes and formula sheets to help every math student-from basic middle school classes to advanced college calculus-figure out what’s going on, understand the important concepts, and pass their classes, once and for all. Interested in getting help? Learn more here: www.kristakingmath.com
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Thank you. Your explanation was Krista clear !
Lol, I'm so glad it made sense! :D
Krista, I just love your maths...I think am just in love with u in the Lord. For sure your maths has lifted me high....so high....I understand your videos A to Z. May the good Lord bless u so much. Thanks CZcams, I scored A from my calculus exam in KYAMBOGO UNIVERSITY , UGAND. Thanks tube for your help. Krista, JG, Professor Dave , to me u are fully awesome.... Thanks so much plus, professor Leonard. Thanks so much, great tutors of maths. Thanks so so so much. Muwonge Evans Paul Kyambogo University 2020 year one. Thank CZcams team.
I wasn't understanding a thing about how to go about this problem. But after seeing your video everything is crystal clear. Thank you very much.
Brilliant! 6 hours of class i didn't understand are now clear thanks to this video :)
I have my final for calc 3 in about 4 hours and it wasn't until right now, watching this video that potential functions made sense. Thank you.
+Tim Talbert You're welcome, I'm so glad it finally makes sense! Good luck on your final, I hope it goes great!!
+CalculusExpert.com well it didnt, hah but it was not your fault. i got full points on the potential function problem
this was very very helpful and it's great to watch math videos from female youtubers as a female engineering student myself :)
This video is very helpful and you speak fluently this also makes me feel comfortable when i listen your lecture .thanks a lot 👏
What should be done if there z is also include as third variable to which value it must be compared for equating
Since I started watching you tube videos on maths, my life has never remained the same. Thanks King.
You're welcome, I'm happy to help! :)
@@kristakingmath Given two sets A and B prove that A*B =B*A. How do u attempt such questions please?
Awesome! it helped me to solved a question finally for which i was struggling from 3 days. Thank you so much Krista!
Yay! So glad it could help. :)
You are a life saver...your videos help alot
From 13:20, the integral of gradient function is potential function which is path independent. You should have omitted the integral sign from there onwards.
agree, there should not be an integral sign on the little f. Path integral of F is = little f evaluated on the 2 sets of points. I was confused when she said you don't have to evaluate the integral - the real reason is that there should not be an integral to evaluate at all!
Huge help! But I believed that the last step of fundamental theorem of line integral which is about the plugging in the coordinates need a bit correction.
I've seen all your videos before my calculus 3 exam, you really explained everything clearly, so thank you for these helpful videos
You're very welcome, I hope the exam went great! :)
@@kristakingmath it did, I got an A :D
King, that was so sweet. I have clearly understood....
I'm so glad it all made sense! :)
What happens if both conditions are not met .Does it halt your progress in solving to evaluate it ?
Great explanation! Thanks!
You're welcome!
i dont think i would have passed any math course without your help. if it werent for your awesome channel id be kicked out of college... thank you..
You're still putting in the work! But I'm glad I've been able to help along the way. :)
what if it is a complex line integral. Same rules apply?
What happens when you are dealing with three variables? Showing that the partial derivatives are equal, for the second step, what would you be deriving with respect to? Assuming you have P, Q, R and dx, dy, and dz respectively, what would be the resulting check? Partial of P with respect to what, and so on?
Your just including the component z to the steps. So where she stops at g(y), you would continue the process with z. She showed that the form under the integral is exact in the plane. What you would be trying to find is the form under the integral is exact in the space. Plane is (x,y). Space is (x,y,z).
Thanks!
Thank you
for the last step. should you not have integrated first, then plug in the numbers??
+ash farza nope
Very nice ..thanks a lot
You're welcome, Fatima! :D
Why is it that the function that we solve for is the gradient function and not the potential function?
If this vector field was independent of path wouldn't the line integral equal zero?
only for closed curves, nevermind, i get it
thank you!!!
You're welcome, Anthony!
thanks
nice keep it up!
+Swagger Swagg Thanks!
From 13:20, how come f(x,y) has suddenly become gradient of f(x,y)?
Jiwon Kim yes I saw too (great vid nonetheless)
she explains starting from 12:35. Basically, little "f" is the potential of big "F". Potential is another way of saying gradient. So, F = grad(f). Be sure not to mix up little "f" and big "F".
Yeah that was incorrect, f(x,y) is the potential function whose gradient is equal to the conservative vector field. It is not the gradient, the gradient comes from it. It is the potential function so she shouldn't have put the gradient sign in front of it.
earthisgood yeah but she did gradient of potential function
Thank you
You're welcome, Megha! :)