I feel like you’re one of my classmate

Can't be more true !

Sameee

So true. I have an exam on mathematical methods tomorrow morning and this helped me summarize three months of lecture in fifteen minutes 😭

you're lucky because our prof never taught this in school 😂

he really should make a correction or risk students getting lost & wasting time trying to verify their answer against the answer of an εxpεrt.

im glad i caught that and that you commented about it. small but crucial mistake

It's a shame I've never tried them, they're much more helpful than Brilliant.org

in the same boat lol, gl with your 4th year

lol same but mine's tmrw :/

yes i think so too

Yes I got that too

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@@KK-xb1zj Jesus is the Word, and the Word was with God and the Word was God! John 1

No. What will make either of these concepts non-continuous, is if there are non-differentiable points or paths in the original function. You will see this visually as a kink or a cusp if you make a graph of the original component of the vector field. As an example, consider the vector field: F = Along the line x=0, the x-component of the vector field is a non-differentiable function. The divergence and curl along this line, is undefined. There will be jump-discontinuities, when you take derivatives of the x-component of the vector field to calculate divergence and curl.

@@larry23100 yes I got that too

You understood this without learning linear algebra!? You must be a GENIUS!!!

@Prodigious147 You must have at least studied vector calculus, right? If not then I have every right to suppose that the age of geniuses is near.

@Prodigious147 Hmm... then all I can do is nothing but wish you best of luck for your mathematical journey. PS: You can start by watching professor Leonard's lectures on calc. He has some of the best lectures around on the entire YT.

Good question, short answer is yes; kzfaq.info/get/bejne/qKhoZqehp7C1o3k.html for anyone else wondering.

Well it’s just that my videos tend to accumulate views very slowly over several years, they don’t get much right away like normal channels. But yes more support is good, please tell your friends to watch and subscribe!

@@ProfessorDaveExplains i seldom share ur videos via my linkedin account

@@ProfessorDaveExplains Alright do some chemistry videos and me and my friends will watch. Like AP chemistry things

buddy I have over 100 general chemistry tutorials already! check out my general chemistry playlist and general chemistry practice problems playlist.

what's that?

Professor Dave Explains abstract algrebra, a set under + and .

oh, i haven't gotten there yet! i need a new point person to write the math scripts as i've gone past what i can handle on my own

Professor Dave Explains oh okay lol, yea ive just started my second year and uni these videos have been a huge help linear algebra was a breeze, combintorics and analysis not so much

The curl is fully able to exist outside of 3D (which should be obvious since reality is 4D). It just can't be represented as a vector field, but rather some other quantity. One way to generalize the curl to arbitrary dimensions is with the exterior or "wedge" product, which returns an oriented plane segment parallel to the two vector inputs rather than an oriented line segment orthogonal to them.

those are all coming!

@@ProfessorDaveExplains i forgot to mention that of surface integral

yes

...and at 8:53, should be -j[d/dz(x^2y-d/dx(xyz)]

They are placeholders so we don't need to write in the contents of the vector field's component functions. You could use any letters you want, but it is common for literature to use the P/Q/R trio in this context.

Dot product and divergence work no matter how many dimensions you have. Dot product means multiply corresponding components, and add up the results. Divergence is the differential operator that is analogous ot a dot product. Cross product and subsequently curl, are calculations that only work in 3 dimensions. Since we live in a 3-d universe, there are plenty of applications of these concepts to physical principles that govern our lives. You can take a curl of a 2-dimensional vector field, and the result will be exclusively in the third direction, perpendicular to both of the original dimensions of the vector field.

The wedge product is indeed a viable alternative to the cross product. It returns an object usually called a "bivector," which acts as an oriented plane segment/area. In adding the dot product to the wedge product, I see you've discovered the geometric product, which between two vectors effectively gives an object that acts like a complex number in 2D and like a quaternion in 3D. (It does _not_ act like an octonion in 4D.) With this product, the divergence and curl of a vector field can be combined into a single complex-like object that I've seen called the "vector derivative." It also gives the shortest version of Maxwell's equation(s) that I've seen: ∇F = J The change in the electromagnetic field is equal to the source density.

Integrate the x-component of the gradient. Call the arbitrary constant of integration C(y, z) Integrate the y-component of the gradient. Cancel terms that are already common in the previous integral. Add terms that didn't exist in the previous integral, in place of C(y, z). Call the arbitrary constant of integration, D(x, z). Repeat for the z-component, and call the arbitrary constant of integration E(x, y). Add up the three results, cancelling terms in common as you do. Terms that are not in common, are terms that are part of the partial constant of integration functions, C(y,z), D(x,z), and E(x,y). When you get to the end of it, call the arbitrary constant of integration K, that is now no longer a function of x, y, or z. K can be any single number, that doesn't depend on any of our function inputs. If the field is conservative, there will be plenty of terms that are common among each integral result. If the field is non-conservative, you will end up with contradictory terms. As an example, suppose our scalar function is: f(x, y, z) = x^2 + x*y*z + z*y^2 + z Find its gradient, and call it F: F = grad f(x, y, z) F = Integrate F's x-component: int y*z = x^2 + x*y*z + C(z, y) Integrate F's y-component int x*z + 2*y*z = x*y*z + z*y^2 + D(x, z) Notice that x*y*z appears in both of the above functions, which means we can cancel it in one of them, and add the two. f = x^2 + x*y*z + z*y^2 + D(z) Now integrate F's z-component int x*y + y^2 + 1 = x*y*z + z*y^2 + z + E(x, y) Combine the terms from all of the above, : f = x^2 + x*y*z + z*y^2 + z + K And you see we now have our original function, with the only difference being the arbitrary constant of integration K. There are an infinite number of potential functions for any given vector field, that all have an identical shape. This is why we have to define a datum of potential energy in physics, where potential energy is by definition zero, for it to be meaningful. Pay close attention to the wording of the problem. If the problem simply says, "find *a* function f(x,y,z), such that grad f(x,y,z) = vector field", then it is OK to omit the arbitrary +K on the end. You can keep it there as a matter of principle, but you are technically correct if you omit it, or make up your own number to take its place. Because you found one function of the infinitely many possible answers. By contrast, if it says "find *the* potential function", then you need to include the +K on the end. The key difference be the article "a" vs "the", in the problem statement wording. Different books or classes may have different conventions for naming this constant. I learned to use K. Most of the time when you use the potential function, you'll end up cancelling this K anyway. But there are some applications where it is of interest to keep it around, and solve for it via boundary conditions.

No difference. Just an adjective to emphasize that it isn't a vector field.

yeah you're correct

It depends on what the vector field represents. Let's assign an arbitrary unit of u, to the quantity represented by the vector field. Assume x, y, and z are all spatial dimensions measured in meters. The units of divergence would therefore be u/m, and likewise for the unit of curl. The unit of second-order derivatives of the vector field, like the Laplacian, would be u/m^2

The curl can be taken in any number of dimensions as long as you use an alternative to the cross product that generalizes nicely. Technically it's possible to take the curl in 1D, but it would always be 0. Curl is ultimately a rotational measure, which looks like a scalar in 2D and like a vector in 3D, but behaves noticeably differently. One generalization of the curl gives its 4D version 6 components, which is notably different from the size of a vector in the same vector space.

A vector field in 2 dimensions in general, consists of two functions of both spatial coordinates, x and y. So F = . Alternatively, F = P(x,y) * i-hat + Q(x,y) * j-hat Both P and Q are functions of both spatial coordinates, and could contain either x, y, or a mixture of both in their definitions. In his example, he is defining P(x, y) to equal y, and Q(x, y) to equal x. Thus, F = , or F = y * i-hat + x * j-hat. It is just a coincidence that P doesn't contain x, and that Q doesn't contain y.