Introductory Fluid Mechanics L1 p6: Acceleration - Material Derivative

[@stati5tik]

at 6:16 shouldnt (v dot nabla)v yield a vector which is multiplied component wise by the divergence of the vector?

[@nashs.4206]

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?

[@mdkashif2560]

There is some problem in his derivation. Though his results are correct!

[@williamjuniorolivaburgos7906]

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

[@yousefzeehk]

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 .

[@remojanja9326]

Dude u Is the x component So i think you can Just take the derivative of the position with respect to time

[@remojanja9326]

@@yousefzeehkyess

[@KakaG247]

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_1986]

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.

[@cooperjacob4548]

@Yusuf Malcolm Definitely, I've been watching on KaldroStream for since december myself :)

[@devinjason6841]

@Yusuf Malcolm Definitely, I've been watching on kaldrostream for since december myself :)

[@izaiahtyler3385]

@Yusuf Malcolm yup, have been watching on kaldroStream for years myself :)

[@ayeshaabdullah2267]

what is local accelaration and convevtive accelration,,,you shuold telll about it too

[@xtheslipknotmaggotx]

i dont really get why after applying the chain law we say that it is equal to the divergence of velocity multiplied by it

[@chaosui3169]

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

[@PeterTargonski]

can anybody give me the intro song name????

[@cristhiantenorio1654]

Saludos a la profesora Iris Domínguez de Mecánica de Fluidos. Acá, estudiando para el parcial de mañana. Qué nervios

[@zdravkaivanova5652]

Thank you!

[@beoptimistic5853]

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[@md.akiduzzamanabir3815]

thank you so much sir

[@A.Hisham86]

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.

[@bakr6405]

Thank you

[@beoptimistic5853]

kzbin.info/www/bejne/joGmmHqKbqefqLM 💐💐💐💐

[@controlroomq]

You look like Joel from The Last of Us series

[@AdamDvořák-y4n]

No, this is not how it works.

[@Ramchandra-jd8vs]

ewqe