I'm a prof myself. Great content and delivery! Didn't find you nervous at all, more like wired up a bit but that helps with the material. I would really really love a whiteboard and markers though, easier to find the info on the screen. Keep rocking!
@amiramaz4 жыл бұрын
Love your channel! Keep it up 👌😁
@omargaber31224 жыл бұрын
You seem a little nervous person, but you are a very skilled teacher, I wish the blackboard was more black (very black), I wish you luck always
@josephmartos4 жыл бұрын
Stacking a bit while talking doesnt make somebody a nervous person, we all do it sometimes and its totally normal, even more when you have lots of things in your head.
@Isaac-nw5hv3 жыл бұрын
Thank you very much, this was so useful for my final exam!
@duckymomo79354 жыл бұрын
isnt the example not conservative on a non deleted domain where the domain is not simply connected?
@NutziHD3 жыл бұрын
Yep, if the domain contains a path that isnt contractible to a single point in R^2-{0} (aka not nullhomotopic) the vector field F over said region is not the gradient of any function. This vector field is not everywhere pathindependant: if I memorize correct everytime you go around the origin the integral increases or decreases by 2pi (depending on clockwise or anticlockwise direction). This is the classic example of a non-conservative vector field that has no rotation (this means dF_i/dx_j = dF_j/dx_i for all i,j). The reason for this is that R^2-{0} is not simply connected and you intuitivly speaking hide the rotation of the field in that singularity. I hope that is helpful:)
@GregShyBoy3 жыл бұрын
How about the case when C goes through (0, 0)?
@aminzahedim.75482 жыл бұрын
Very intriguing and a subtle question: I’m not a 100% sure, rather speculating that we first need to “define” what the original integral means in such a case; i.e., a plausible definition-perhaps via a limit of sorts-of the integral F.dr over such a contour (namely, one going through a “pole” of the integrand). If one should define the integral as the limit of the same function over a newly devised contour C_a, index by a positive real number a>0 serving as the radius of a semicircle running past (0,0) instead of going precisely through it, then one might potentially take the value of the original integral as a limit of this replacement integral when the radius approaches zero (a->0). It seems that the major complication arises when one considers the fact that you can run past any point (here the origin) from either left OR right which makes the (uncountably infinite) family of indexed contours {C_a} contain instances of both inclusion AND exclusion of that particular point. In the (presumably rare) incident that the integral happens to be indifferent as to whether the pole is within or without the simply closed curve, everything is fine and well-defined. Otherwise-which I presume happens quite more often-one might have to consider a “principal value” or P.V. for the desired integral and THEN proceed on to the calculation, as the integral in its original form is ill-behaved. Again, I must emphasize that, although I DID take (rudimentary) multi-variable calculus, complex analysis, and mathematical physics courses as an undergraduate, I am by no means an expert in this area and never came across such a situation (to the best of my recollection). If you did consult an actual expert and found the answer, I would be more than grateful if you chose to share it here with everyone as well as me!
@bilalabbad7954 Жыл бұрын
Great channel I learned a lot from you Thanks
@Lklibertad3 жыл бұрын
what about when the field has more than one singular point? for example a circumference i meant for example {-y/(x^2+y^2-a^2) , x/(x^2+y^2-a^2)} has a circumference of radius a ? im supposed to lock that points too ? how?