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ОтветитьBut WHY is the directional derivative given by that dot product?
Ответитьmany thanks!!
ОтветитьNew epsode at MathFlix! 😀
ОтветитьKeep it up. Awesome series
ОтветитьNow when I need to write gravity I can turn it into del mass instead.
ОтветитьCan I use the gradient to model the gradient of sand density on Arrakis?
ОтветитьThank you
ОтветитьThank you, Sir
I am eagerly waiting for the coming lectures.
Coming to learn new knowledge!
ОтветитьAlways wonderful to learn from you, i tried the example of gravitational potential, quite a neat example. Thanks for the video :)
ОтветитьDear Professor Brunton, I heard mathematicians treat the output of the gradient operator as a row vector (for reasons supposedly to do with differential geometry), while engineers and statisticians typically treat it as a column vector. I'm curious if there is a preference for either convention in the optimization / machine learning literature, as well as your own preference.
ОтветитьThank you Steve. Enjoy your sabatical 💗
ОтветитьSir, please can u explain augmented Lagrangian (penalty function)
Ответитьlove this series
ОтветитьI have a question about calculating a gradient of gravitational potetnital:
V = -(GmM)/r
I can decompose r into a form where I keep each axis separate using pythagorean theorem:
r = (x^2 + y^2 + z^2)^(1/2)
now V equals:
V = -(GmM)*(x^2 + y^2 + z^2)^(-1/2)
Now I can calculate a derivative in each direction (x, y, z):
dV/dx = -(GmM)*(-1/2)*(x^2 + y^2 + z^2)^(-3/2)*2x = x(GmM)*(x^2 + y^2 + z^2)^(-3/2) = x(GmM)/(r^3)
y and z can be calculated correspondingly. Now, what's the next step? How do I go from this vector:
∇V=[dV/dx, dV/dy, dV/dz]
to:
F = -r(mMG)/(r^3)
Another nice graphic application of Grad is that it provides the normal direction of the surface T(x,y)=constant. This idea can be very useful for the geometrical understanding of our math problems. For any surface I can write as f(x,y)=constant, I can obtain the normal of this surface at any surface point just obtaining the Grad vector (whatever the complexity of this surface is). This directly applies to plasticity theory with respect to the yield surface and the normality rule.
ОтветитьReally great lecture. Thank you very much!
ОтветитьWorst Earth ever.
ОтветитьHi. So F is minus the directional derivative of V ? Excellent video.
Ответитьso glad to be seeing your lecture. thank you
ОтветитьHi Steve, do you have a video for "Field"? like what is field and what is scalar or vector field? Thank you very much!
ОтветитьHow are you so good at writting backwards?
ОтветитьBest series I found for catching up on uni maths I forgot. Goes into enough detail whilst being short enough to fit in at work
ОтветитьIs the concept of grad similar to maxima and minima?
ОтветитьGreat explanation!
ОтветитьMay I ask where is the I,j,k in nebla (delta ) why you don't put them in the column vector like [d/dx i, d/dy j, d/dz k]
ОтветитьIf there are two sources of heat and the bee is unlucky, it could just stay at a non optimal point
ОтветитьGreat video! Thanx! 😊
ОтветитьThank you!
ОтветитьGreat explanations for easy intuitive understanding of PDEs. I however was wondering whether we can add time in these equations, and how?
Ответитьthanks for your great explaination
ОтветитьThink you sir,,,,,,love from Bangladesh 🇧🇩🇧🇩
ОтветитьSome hints to write V = -mMG/r into cartesian coord (x,y,z) ? I have no clue where to start thx
ОтветитьUC Santa Cruz - Physics 1987
ОтветитьIs he writingevery thing mirrored???
ОтветитьI have an exam on this on monday. ≈50% of students fail the course, where I study. Thank you for the great explanations
ОтветитьGreat!
ОтветитьI didn’t understand how the gravitational force is equal to mMG/r^3 , Isn’t suppoed to be 1/r^2
Ответитьthank you sir
ОтветитьThis is what is good about the internet for education. A good succinct lecture really gets the information across. If you want to learn we are in a golden age. I am very excited for it all.
ОтветитьF = -GMm/r2 (NOT r cube)
ОтветитьSome ants do fly.
ОтветитьThank you so much
ОтветитьGreat lecture! one thing, shouldn't the field have only "one M"? i.e., it should be the gravitational force per unit of mass
ОтветитьTeachings efficient and therapeutic!
Do some more on ODE Sir
Sir, you are the best I have ever known you have done this difficult subject as easy as possible for me and for my friends so you deserve a great thank 🎉🎉❤❤
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