The Jacobian 2
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- Опубликовано: 9 фев 2025
- This is the second part of the Jacobian-extravaganza! Here I do an inverse change of variables using the Jacobian, and again I want to show how similar it is to the substitution rule in single-variable calculus.
I love how you thank me before I even get done watching the video. Your content is amazing; you don't have to thank me. I should thank you. So thanks.
Team Jacob!!!
You should show me your chalk magic next time.
Lol. Good thing we have Blu Ray now.
Also, how did you finish that mean ice cream monster?
Thank you. I am learning a lot.
Javier Gómez Morales Hahaha, I couldn't finish it, it was too big 😂😂😂😭😭😭
Do you guys know how they make holy matricies? They differentiate the hell out of it!
You can write it as (1-x^2)/sqrt(1-x^2) than break the fraction into two than 1/sqrt(1-x^2) is a know integral,and than do by parts -x^2/sqrt(1-x^2) and it's easyer than doing with polar coordinates
I think I'll use this when I teach transformations in multiple integral (for reals) in my Calc IV [we're on a quarter/trimester system] course (if it's all right with you); very nice idea, using a trig substitution as a parallelism
Great great thanks a lot 👍🏽
trig substitutions are harder than parts in single variable,but when it's a double integral yeah use polar
I am familiar with replacing dx dy by r dr dtheta, but I can't seem to get there by simply multiplying the expressions for dx and dtheta (without using the Jacobian):
dx = cos(theta) dr - r sin(theta) dtheta
dy = sin(theta) dr + r cos(theta) dtheta
dx dy = r dr dtheta( cos^2(theta) - sin^2(theta) ) + sin(theta) cos(theta) ( (dr)^2 - (r dtheta)^2 )
Even ignoring terms which have squared differentials:
dx dy = r dr dtheta( cos^2(theta) - sin^2(theta) ) = cos(2 theta) r dr dtheta
Also, is it ok to ignore ( r dtheta)^2 at large values of r? Come to think of it, how can r dtheta still be a differential length at large values of r?
That’s correct, (dr)^2 is reaaaaally small, so you can basically ignore it, although squares of differentials aren’t rigorously defined!
Thanks Dr. Payam. Your videos teach me new and interesting things about calculus, and (maybe more important) take me back along old paths but with new questions.
@@stepsvideos I'm glad you asked, I have the same problem.
We all use dx=cosθdr-rsinθdθ and dy=sinθdr+rcosθdθ
We all agree (dr)² and (dθ)² can be ignored
We both get dx dy = (cos²θ - sin²θ) r dr dθ
Dr Peyam gets dx dy = (cos²θ + sin²θ) r dr dθ
I can't find the sign error. Did you ?
@@drpeyam Can you help please ? It drives me crazy you get cos²θ+sin²θ with the jacobian where we get cos²θ-sin²θ without the jacobian. There is something to understand here.
@@joluju2375
The problem, I think, is that you can't simply implicitly differentiate like that when you have a multivariate function. You have to take the partials.
Taking the partials means you end up with four expressions, not two, so a matrix is your only option.
Yahoo!
If S is a disc with radius 1, wouldn't the area be Pi instead of 2/3 of it?
Yes, the area of a disk is pi, but here the integral is actually calculating half of the volume of a ball, which is 1/2 4/3 pi, so 2/3 pi. It’s the base that’s a disk, but the object is a half-ball!
Dr. Peyam's Show oh thanks for explaining, silly me. So the square and disc was the domain all along and not the function themselves. I get it now :D
Yosh i got that result
But why the matrix?
Maybe this helps: Determinants and Volumes ruclips.net/video/MIxTvKXG1jY/видео.html
Proove why the determinant of me is equal to dxdy/drdtheta
You see after I came up with the jacobian I couldn’t proove it so that’s why I’m asking because I made it because my name is Jacob you see
Hahaha
دمت گرم
Lol with the onevariable integral you made an error, which is forgiven in this case! Sqrt(sin²(theta))=|sin(theta)|!=sin(theta). You were just lucky, that sin(theta)>0 for theta in (pi/3,pi/2).
Leonard Romano I mean it's not really a mistake :P I'm aware I was working on an interval where sin is positive, so I didn't put the absolute value on purpose :)
Listen to what he says at 8:38
12:35 dvd u!
we ARE integrating a disk, right?
Payem :)
Looks as if I am not the only one .................................... having watched Walter Lewin. 😂👍