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Change of variables | MIT 18.02SC Multivariable Calculus, Fall 2010
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- čas přidán 2. 01. 2011
- Change of variables
Instructor: Christine Breiner
View the complete course: ocw.mit.edu/18-02SCF10
License: Creative Commons BY-NC-SA
More information at ocw.mit.edu/terms
More courses at ocw.mit.edu
10 min vid gave me a better. understanding that a week of doing this in UCLA undergrad. good job MIT
czcams.com/video/vFDMaHQ4kW8/video.html 💐.
math 32b in about week 6 hits different :')
My calculus book decided to write this chapter in the most confusing way possible. I know this shouldn't be that hard but my book makes this out to be rocket science. Thanks for this.
check out Thomas' Calculus, you can buy the cheaper 11th edition, just a few bucks from Amazon.
No seriously though my book was ok until this chapter for no reason
I remember studying this and struggling with it sooo much. But after taking a step back after the class it’s so much easier :)
Especially with these explanations!!
thank you ! this is the most clear graphing ive seen for transforming between coordinates!
Dr. Breiner, great work, and I hope that you are doing well these days in your math career. :)
Amazing! Excellent Job! God bless you and your teaching. Thanks so much!
she slayed this topic.
This video is amazing. Great explanations!
Awesome explanation.
Tiny area can be expressed by 2 tiny small distance times together. This idea is excellent! The jacobin determinate tells you the scaling factor when transform dx and dy into du and dv about the tiny small area which you can figure out from the picture of the transform. For the limit of u , I was wondering if it can be written as (0, -2). Thank you ! A great job done! Many lectures for calculus in my college did not clear the actions taken by the transformation and omit what it means.
coming from MIT OCW to give this video a thumbs up and appreciate the good explanation 🎩
@lesprit1
BCZ Jdxdy=dudv J=determinant=4
or you can look at it this way : the area of the bounded region in X-Y domain is four times the area bounded in the U-V domain
To instructor, thank you.
great lecture !
This helped out so much! :)
great explanation
Lg
Amazing video ! Welldone prepared Lession ! Best regards
Thanks for this
thanks so much
thanks a lot
Perfect!!
Integrate(y^3*(2x-Y)*E^(2x-y)^2,{x,0,2},{y,y/2,(y+4)/2)
Como posso resover este caso
For those who want the answer it's 128/25
Anonymous I got 256/25
Ye its 128/25
I was so confused on Jacobians until I saw this. I was spinning my wheels for two days trying to figure this out.
hi mate, am from the past. how's life? did you finish school? married now? hope you are happy. congrats you made it into the future!.
Good lecture ma'am
what is difference change of variable versus change of coordinates
Why Jacobian determinant 4 is placed in denominator? Shouldn't be it in numerator? That is what textbook shows. Can someone please explain?
@MotoPayton haha... thank you. I was confused too.
If only the triangle uv is as big as 4dx dy we will not be confused as Jacobian stretch is made in the diagram
I am still confused.
Did you forget the Jacobian in the last example?
Hi,
Shouldn't the lower bound of the dv integral be equal to -u instead of 0? (For the same reason the lower bound of the integral du is equal to -v)
If not, please explain why.
I had the same doubt.
No, the outer dv integral doesn't have access to a particular value of 'u' yet.
It may be helpful to think of the outer dv integrand as a function of v, which maps v to some expression that can use v.
∫(v ↦ (...))dv
= ∫(v ↦ (∫(u ↦ (...)) du))dv
= ∫(v ↦ (∫(u ↦ (uv)^4/u) du))dv
= ∫(v ↦ (∫₋ᵥ⁰(u ↦ (uv)^4/u) du))dv
= ∫₀²(v ↦ (∫₋ᵥ⁰(u ↦ (uv)^4/u) du))dv
The inner du integral can use the "v" term in its bounds, because the outer integrand is a function of "v", which you are integrating over at the outermost layer.
You can nest integrals indefinitely and use whatever variables you have access to, and the integrands can themselves contain integrals.
I just wonder whether it's just a global phenomenon or rather convention, that we typically choose to express most calculus problems using x & y, then u & v (when we change variables). DNA to Vitamins. Or it's just to keep instruction universally simple?
Sorry... hope this Isn’t an obstruction to the otherwise impeccable lecture dear Prof. 👋😍
Hi, i am a bit unsure as to where the first graph came from.... how do you know the boundaries of the 3 lines in the very first given graph...
its given, this question asks you to evaluate the integral over region, so she transforms the region to a nicer, equivalent shape to integrate over.
Just a little question to confirm something inmy head:integrating under the region R gives us the area of the region right?
Correct!
Beasting beasting beasting !!!
and just like that i finally get the true meaning of the jacobian
Umm can you help me and tell me how to we compute Jacobian for from X-Y coordinates to polar coordinates
El jacobiano es un vector cierto, usa la misma transformación que usarias para cualquier verctor que quieres convertir de axiales a esfericas
@@davidmendizabal9892 ah sorry man i dont understand spanish and i cant google translate either as one cant copy youtube comments and put it in Google translate
At least give the final answer, even if you don't go through the computation of the integral! (It is 128/25)
isnt it 1/2 or 2?
@@Ftbl_AK They compute a different integral at the end , I was going crazy thinking this for about a minute
Could you please say where we might use change of variables in real physics or engineering scenarios?
she said the main point was manipulating the region, so logically any time you need to perform calculus on a difficult region in real life you would use this method. If youre wondering about specifics it is really dependent on your application as it would apply to so many fields of both subjects. But also many times real life does not produce equations we can manipulate anyways lol
Walking away was awkward, I saw that smirk
good
why is the jacobian 1/4 and 4
dudv = 4dxdy
(1/4)dudv = dxdy
So, she replaces dxdy with (1/4)dudv
If you're still interested watching these few videos will give you a intuitive sense of exactly what is going on. Basically the determinant of the Jacobian is giving you a scaling if you were to apply x(u, v) and y(u, v). So in this example an area of 1 unit would become an area of 4 units.
www.khanacademy.org/math/multivariable-calculus/multivariable-derivatives/jacobian/v/jacobian-prerequisite-knowledge
Good video. Wonder why she walked out like that lol.
hellow mam kaise ho,
0:05 whats a recitation in MIT?
why is the bottom bound for u is -v and not -2?
If you were to use the bounds -2/0 and 0/2, you would get the area of the full rectangle! Using u=-v (the equation of the line) makes it so that it is only the area inside the triangle.
I can add that you need to find the centre of that line and it is where v=-u.
I'd like to know what this girl is up to now!
no reply
That’s not how you do jacobian
Your Jacobian is wrong. Row 1, column 2 (dx/dv) should be 2. And Row 2, column 1 (dy/du) should be -1. thus, your determinant gives you four.
+Shayan Nejadian Its not wrong, Recheck it.
It's not wrong.
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Nyc
for those looking for the answer for verification, it's 0.01
@@sasmitvaidya3594 He died last year.. too late
@@sasmitvaidya3594 He died by covid
Awesome mam.you are very cute and adorable
My calculus book decided to write this chapter in the most confusing way possible. I know this shouldn't be that hard but my book makes this out to be rocket science. Thanks for this.
yeah, the cheap authors and publishers do that. i also faced that with 9th to 12th standard Maths/Physics/Chemistry govenment books. Hated them. I knew books were the problem, and not the subjects.
hellow mam kaise ho,