at 6:16 shouldnt (v dot nabla)v yield a vector which is multiplied component wise by the divergence of the vector?
@nashs.42067 жыл бұрын
How is it possible that dx/dt = u, if u = u(x,y,z,t)? We are using d, not ∂, which means that x only has a dependence on time (i.e. x is a univariate function). So it's like saying d/dt (f(x)) = 2x*z*t*y. Where/how do the other variables (y,z,t) come into play if we're using the total derivative and not partial derivative?
@mdkashif25603 жыл бұрын
There is some problem in his derivation. Though his results are correct!
@williamjuniorolivaburgos79063 жыл бұрын
I think that if you consider x y and z also as functions of t you can derive the equation although I am not so sure why would that work because of the eulerian focus of the thing
@yousefzeehk2 жыл бұрын
it's because these equations were driven from a lagrangian point of view of the particle so u(t) is only a function of t. you're basically applying it along the pathline not across multiple pathlines or another fluid particles where u(y,z,x,t) would be a function of x,y,z,t .
@remojanja93262 жыл бұрын
Dude u Is the x component So i think you can Just take the derivative of the position with respect to time
@remojanja93262 жыл бұрын
@@yousefzeehkyess
@KakaG2478 жыл бұрын
how does that acceleration equatio come out to be. which laws of partial derivatives are you using. that is what i want to know
@Peter_19867 жыл бұрын
You can figure out the terms for the material derivative by drawing a tree diagram. In this case the arbitrary scalar or vector (let's call it "f") is a function of x, y, z and t, and then x, y and z are in turn functions of t. If you draw four branches from f and write x, y, z and t at the end of those branches, and then draw branches from x, y and z and have all of those branches end with t, then you can easily create terms which all end with a time derivative. For example, the branch from f to x will give you ∂f/∂x (it's a partial derivative because the function f depends on x, y, z and t, and you are taking a derivative of only x), and then the branch from x to t gives you dx/dt --- now, multiply those factors together to get (∂f/∂x)⋅(dx/dt). Then do the same thing for y and z to get (∂f/∂y)⋅(dy/dt) and (∂f/∂z)⋅(dz/dt) and then use the same method for the branch from f to t to get ∂f/∂t, and then finally add up all these four terms to get Df/Dt = (∂f/∂x)⋅(dx/dt) + (∂f/∂y)⋅(dy/dt) + (∂f/∂z)⋅(dz/dt) + ∂f/∂t.
@cooperjacob45483 жыл бұрын
@Yusuf Malcolm Definitely, I've been watching on KaldroStream for since december myself :)
@devinjason68413 жыл бұрын
@Yusuf Malcolm Definitely, I've been watching on kaldrostream for since december myself :)
@izaiahtyler33853 жыл бұрын
@Yusuf Malcolm yup, have been watching on kaldroStream for years myself :)
@PeterTargonski Жыл бұрын
can anybody give me the intro song name????
@ayeshaabdullah22676 жыл бұрын
what is local accelaration and convevtive accelration,,,you shuold telll about it too
@xtheslipknotmaggotx8 жыл бұрын
i dont really get why after applying the chain law we say that it is equal to the divergence of velocity multiplied by it
@chaosui31698 жыл бұрын
the first V is the u,v,w in the expression, the later V is in the dv/dx,dv/dy,dv/dz. Hope this helps
@A.Hisham8620 күн бұрын
it's still doesn't make sense mathematically!! how can the gradient operator work on the vector field? the gradient works only on the scalar field, the divergent does.
@zdravkaivanova56525 жыл бұрын
Thank you!
@beoptimistic58534 жыл бұрын
kzbin.info/www/bejne/joGmmHqKbqefqLM 💐💐💐💐
@md.akiduzzamanabir38152 жыл бұрын
thank you so much sir
@cristhiantenorio165410 ай бұрын
Saludos a la profesora Iris Domínguez de Mecánica de Fluidos. Acá, estudiando para el parcial de mañana. Qué nervios