Eulerian vs. Lagrangian description

[coverband]

Homework Statement

A particle moves so that \vec{x} \equiv [x_0 exp(2t^2), y_0 exp(-t^2), z_0 exp(-t^2)]. Find the velocity of the particle in terms of x_0 and t (the LAgrangian description) and show it can be written as \vec{u} \equiv (4xt, -2yt, -2zt), the Eulerian description

Homework Equations

The Attempt at a Solution

\vec {u} = [4tx_0 exp(2t^2), -2ty_0 exp(-t^2), -2tz_0 exp(-t^2)]

The exp's go to 1 !?

Thanks

 

[irycio]

No, i suppose it's just that in your velocity vecotr you can substitute the term x_0exp(2t^2) with x - "first part" from your \vec{x} vector and so on

 

[coverband]

Yeah but why/how can you do that

 

[irycio]

Well, i guess because they're equal. I see no physical meaning of such operation, it's just the mathematical identity that allows you to subsitute it.
Let's consider a single function f(x)=exp(2x). Obviously, \frac{df}{dx}=2exp(2x). And you can write \frac{df}{dx}=2f(x), which is true for all x, but doesn't really mean anything.

E:and so, in your example, you just get x_x=x_0exp(2t^2), u_x=4tx_0exp(2t^2)=4tx

 

[coverband]

Anyone else?