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I really wish the rest of this playkist was out my midterm is on Wednesday and this is so much more helpful AND ENTERTAINING than the explanations I've seen so far. SO MANY PEOPLE MAKE MATH BORING, you do not, thank you
Can you please make a video explanation of surface integrals? ⊙∀⊙ Your math is really great! Also exterior derivatives, gradients, curl and divergence, I really like you, wish you good luck
For the proof at the end once you expand the dot product you can use chain rule for partial derivatives properties to get a single derivative Then use fundamental theorem of calculus to get the right hand side of the original theorem (For my workings I used r'(t) = (dx/dt, dy/dt, dz/dt))
just a minor nit pick at 8:00, the projection is divided by the norm squared. *b* dot *a* over norm *a* is only the component of *b* in *a* 's direction. In other words, how much the unit vector in *a* direction is scaled. Hence, the extra division by norm of *a*
the way i semi-formally proved the theorem was sort of like this: (click read more if you want to see it, don't click if you want to figure out yourself) (i'll use ellipses to indicate that a function or vector could continue to have any amount of variables/components) first off, dr = r'(t)dt = = ∇f = ∇f•dr = ∇f•r'(t)dt= (∂f/∂x)dx + (∂f/∂y)dy + … a function's complete differential is equal to the sum of its partial derivatives, each multiplied by their corresponding differentials (e.g. (∂f/∂x)dx or anything similar) and the value of this dot product comes out to be in just the right form of this total differential so ∇f•dr = df now, before we move on, recall that we want to take the values of f at certain points plugging in t_0 or t_f alone won't do the job since we're talking multivariable here, but if we can find f in terms of t, that'll work so let's find f(r(t)) this will change f(x, y, … ) into f(x(t), y(t), … ) so now we can evaluate f at the t values now let's get back to the integral: we already know that by the FTC (or by the Generalized Stokes' Theorem) that the integral over an interval of the derivative of a function is equal to the original function at the bounds ∫_C ∇f•dr = ∫_{t_0} ^{t_f} df = f(r(t)) | _{t_0} ^{t_f} = f(r(t_f)) - f(r(t_0)) (the _{} ^{} indicates the bounds) and there we have it ∫_C ∇f•dr = f(r(t_f)) - f(r(t_0))
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Thanks brother, for explaining concepts in simple words, please keep uploading more videos on Multivariable Calculus
Another great video - you have a gift for making murky topics so clear. Thank you!
FYI - this video is not currently part of your math playlist?
so underrated, this needs more views
I love Calculus III so much.
I mean there is nothing wrong with your video, but as a physicist I shiver at the way you write things 😆Great Video!
What do you mean?
Yet another awesome video!
Very good!, Probably having a polarizer for the camera would be good investment.
We love you as math community, keep good work up!
I really wish the rest of this playkist was out my midterm is on Wednesday and this is so much more helpful AND ENTERTAINING than the explanations I've seen so far. SO MANY PEOPLE MAKE MATH BORING, you do not, thank you
please continue doing these videos. they help a lot
Perfect video, and fantastic explanation
Great - Easy for a non-mathematically-inclined listener to follow
Dayum bro, thank you! Excellent presentation, I finally start to understand! Please keep the videos coming!
You did not need to grade this in HDR, but I’m glad you dud
Are you reading my mind? I was just looking for a video like this!
Hey man your videos are great I'm starting to understand something , just wanted to tell you in 2:03 the Subtitle says "two girls" instead of integral
This is extremely good
It wouldn’t be a calculus video without someone bringing up +c
Edit: at 15:32 technically a constant would have been deleted by the derivative
Can you please make a video explanation of surface integrals? ⊙∀⊙ Your math is really great! Also exterior derivatives, gradients, curl and divergence, I really like you, wish you good luck
Coming up soon 🫡
Thank you
Thank you very much for this video❤
For the proof at the end once you expand the dot product you can use chain rule for partial derivatives properties to get a single derivative
Then use fundamental theorem of calculus to get the right hand side of the original theorem
(For my workings I used r'(t) = (dx/dt, dy/dt, dz/dt))
What tablet are you using?
An iPad!
Isn't dr/ds the unit tangent vector, whilst dr/dt the actual tangent vector ?
SHIT , GOD DAMN EFFICIENT
11:46 WHAT ARE YOU DOING??? THIS IS ILLEGAL
bruh it's just the chain rule
so im not the only one writing like this
just a minor nit pick at 8:00, the projection is divided by the norm squared. *b* dot *a* over norm *a* is only the component of *b* in *a* 's direction. In other words, how much the unit vector in *a* direction is scaled. Hence, the extra division by norm of *a*
Please make another example. It seems everyone is using a circle…
the way i semi-formally proved the theorem was sort of like this:
(click read more if you want to see it, don't click if you want to figure out yourself)
(i'll use ellipses to indicate that a function or vector could continue to have any amount of variables/components)
first off, dr = r'(t)dt =
=
∇f =
∇f•dr = ∇f•r'(t)dt= (∂f/∂x)dx + (∂f/∂y)dy + …
a function's complete differential is equal to the sum of its partial derivatives, each multiplied by their corresponding differentials (e.g. (∂f/∂x)dx or anything similar)
and the value of this dot product comes out to be in just the right form of this total differential
so ∇f•dr = df
now, before we move on, recall that we want to take the values of f at certain points
plugging in t_0 or t_f alone won't do the job since we're talking multivariable here, but if we can find f in terms of t, that'll work
so let's find f(r(t))
this will change f(x, y, … ) into f(x(t), y(t), … )
so now we can evaluate f at the t values
now let's get back to the integral:
we already know that by the FTC (or by the Generalized Stokes' Theorem) that the integral over an interval of the derivative of a function is equal to the original function at the bounds
∫_C ∇f•dr = ∫_{t_0} ^{t_f} df = f(r(t)) | _{t_0} ^{t_f} = f(r(t_f)) - f(r(t_0))
(the _{} ^{} indicates the bounds)
and there we have it
∫_C ∇f•dr = f(r(t_f)) - f(r(t_0))