Change of variables | MIT 18.02SC Multivariable Calculus, Fall 2010
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- Опубліковано 7 вер 2024
- Change of variables
Instructor: Christine Breiner
View the complete course: ocw.mit.edu/18-...
License: Creative Commons BY-NC-SA
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I watch all of your videos, Christine, and in each one I marvel at your brilliance and your teaching ability. Many thanks for all of your dedicated work.
The clarity in the exposition of this lecturer is absolutely stunning. She's really awesome.
Great Explanation !! Helped me to clear my concepts on change of variables.. :)
For time 10:40 The easier method is to find the intersection of u =1 and u= 1/v - v, giving the same value v= (Sqrt(5) - 1)/2
at 10:00 just substitute (u=1) in [ u=(1/v)-v ] to get the answer for v
Thank you so much for your work and videos Christine!
( 1 + 5^0.5)/2 is not just any number, it is the golden number
Christine, Newton and Leibnitz would be so proud of you. Absolutely magnificent. And you are so personable.
in case someone is wondering about golden ratio's inverse and how come it is (-1 + sqrt(5))/2 , try multiplying that by golden ratio ypu will get 4/4 = 1
Thank you very much of these lectures. I am a Filipino civil engineer and I am learning more of calculus by watching lectures in youtube.
Forgot to take the absolute value of the Jacobian, otherwise, completely correct. Turns out you get the same answer, but this is not immediately clear, namely:
The ratio between elements that you get is dxdy = abs(1/(2(1-v^2)))dudv. A priori, we don't know if (1-v^2) is positive or negative.
If we look at our initial region (in the xy-plane), we see that our region R lies entirely below the line y = x (can also show this algebraically). This implies y < x, implies y/x < 1 (working in the first quadrant so can divide by x since x positive), implies v < 1, implies v^2 < 1, implies -v^2 > 1, implies 1 - v^2 > 2, implies 1 - v^2 > 0, so indeed, this quantity is always positive, and taking the absolute value does nothing. Still worth noting though.
12:00
Excuse me!
u=(1/v)-v and u=1 then 1=(1/v)-v
So v=(-1+sqrt(5))/2.
I wish to join this awesome MIT!
A great video Professor THANK YOUUUUU !!!.
What happens to u(you) is you are a really good teacher and v(we) learn a lot
okay.i'lll try my hardest.
15:16 Me anytime I'm doing a long math problem. You get side tracked on solving one bit and then you're like "why am I doing this again?"
Thanks 🤍❤️
why did we divide by the jacobian near the end? I thought we multiplied? also why couldn't we just set 1 = 1/v - v instead of all that mess in the middle?
She technically multiplied the Jacobian. for the left side where dA was. So she found (|J|dA)=dvdu. It took me a while to notice she wrote ∂(u,v)/∂(x,y) instead of the the reciprocal. ∂(x,y)/∂(v,u). The Jacobian goes on the side of the transformation that has the variables you differentiated by. She then used substitution to put it into the terms that she wanted to use and divided by both sides in order to make it so it canceled out nicely.
To instructor, thank you
Amazing this solution mem
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Super variables define number comulti equal single Aleph.
this exercise is not easy to know as i am a student in Cambodia.
thx christine
But v should'nt go from 0 to a, it should go from 0 to whatever value some function v(u) has. I would be thankful if someone explained me why they put that in the video.
Dont know if you will read this, since your question is from so long ago, but i had problems figureing that part out too. If you look at it in terms of adding up the area of horizontal slices under the u=1/v - v curve, what youre doing is adding up rectangles with width u-1 and height dv. so if you want to add all of them together, you have to start at v=0 and add them up v=1 to get all of them.
Actually, she seems to have done the inverse:made u go to u(v), so she'd go with v from 0 to a
Thanks :d
Shouldn't the answer be infinite? If we ignore the second curve for a moment and just take the value of the integral 1/x from 2 to infinity, its value is infinite. The value of the integral which is given as the answer is about 0.412. Can someone explain this?
totally lost around 12:00
I got really frustrated that I couldn't even find the bounds...Maybe math is not for me, I suck
MAGNÍFICA
To the right of the region in the UV plan it is un bounded so how can we get the finite area
Why didn't she just put the value u = 1 to find v...
its reallly hard :S
Very professional video. But why were u and v defined as u= x^2-y^2 and v=y/x? Were these chosen to make the bounds of the integration area more manageable? And if so, how did you choose them?
YEs, they were chosen to make it more manageable. That is the whole purpose behind changing the variables in the first place. As for how they were chosen...I got nothing.
It's an exercice. I think the person who wrote the exercice started with the variable changes he wanted to use and the integrale, and only then he chosed y=1/x and x²-y-=1 to make the Integrale and the change of variables match themselves. Most exercice in mathematics are build "from the end to the beginning".
My brain hurts :(
Does someone have the link of the whole playlist of multivariable Calculus?😊
I assume the result is undefined since the first ray approaches infinite length?
Remember just because we are integrating along an infinite bound it doesn't mean the answer is undefined. We're integrating the function 1/x^2 as x approaches infinity which the function approaches 0 as x-> infinity. We're adding up the function's output over that interval so even though the function 1/x^2 approaches 0, there are values to add up as it does approach.
which year of their degree would students be learning this at MIT?
screwed up around 11:58 rubbing the - from +/- then when going to v ended up using the negative anyway, without batting an eye. sorry to all those bringing a nice shiny apple for teacher.
big deal
How so U get the chance of variable equations ?
The golden ratio strikes again
first of all the presenter is not organized and nervous, maybe it's because of the camera maybe she is like this but it makes the explanation hard to follow, second that last substitution 1-v came out of nowhere and as a while the explanation of the substitutions is not very clear, you have to watch it a couple of times for it to make sense, that's ok and just like the presentation is something that is not a big problem what i have a problem is the fact that the region of integration is plain f***ing wrong. the way it's drawn looks like on part goes to infinity(1/x are close to the axis but never touch it) there's no constraints and she basicly gic=ves the same thing twice 1(not 0!) to a(that crazy point with the quadratic equation) and 1/v-v both give the width of the( she says it's turned) function but not the height, this is wrong don't set up your integrals like that
i think you missed the region which is in the 3rd quadrant.
+Deepak Kumar is am i correct
How
Freshman year.
you are doing your duty very well,do much more work hard so that your identity is identify by universe, we leave only our karma in the world nothing else
Christine mam ,tell name of book of vector calculus and integral calculus having good collection of questions of both of these topics
Math is hard.
these MIT TA's are just lol
Sorry Chistine, you could have prepare much better
Jesus
If would speak like you are on methamphetamine it would be easier to understand what you are saying
you're so fine
she made too mistakes 👎👎 wasting time
Thank you so much for your work and videos Christine!