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Partial Differentiation Troubles

  1. Jun 18, 2009 #1
    1. The problem statement, all variables and given/known data

    Okay, I know that I must be overlooking the obvious here, but here goes.

    Take some velocity function of time and space V(x,y,z,t) and we want to find its derivative, the acceleration vector a(x,y,z,t)

    If we have

    [tex]\vec{V}=u\hat{i}+v\hat{j}+w\hat{k}[/tex]

    Then by chain rule:

    [tex]\vec{a}=\frac{\partial{\vec{V}}}{\partial{t}}+u\frac{\partial{\vec{V}}}{\partial{x}}+v\frac{\partial{\vec{V}}}{\partial{y}}+w\frac{\partial{\vec{V}}}{\partial{z}}[/tex]

    (did I mess something up in the last step? Where did i, j and k go? Sorry...Engineer :blushing:)

    Now isn't it true that (ignoring the first term involving time)

    [tex]a_x=u\frac{\partial{\vec{V}}}{\partial{x}}[/tex] ?


    Because I am confused as to why my book says that


    [tex]a_x=u\frac{\partial{u}}{\partial{x}}+v\frac{\partial{u}}{\partial{y}}[/tex]

    I am not seeing how the two are equivalent?

    Where am I screwing this up? Because I know that it is me :smile:

    Also: feel free to address any notational issues you see with my math. I would like to be able to communicate this stuff clearly and properly
     
  2. jcsd
  3. Jun 18, 2009 #2

    Mark44

    Staff: Mentor

    V is a vector-valued function of x, y, and z, each of which is a function of t alone. So ultimately, V and a are functions of t alone, and it makes sense to talk about dV/dt. At least that's what I think you're talking about. To find a, calculate dV/dt, which means that you calculate dx/dt, dy/dt, and dz/dt.

    For example, if V(t) = (cos t, sin t, t), then a(t) = (-sin t, cos t, 1). You can also write these in the unit vector notation as
    V(t) = cost i + sin t j + t k, and a(t) = -sin t i + cos t j + k.
     
  4. Jun 18, 2009 #3
    This is also known as the convective derivative. You can write it compactly as:

    [tex]\frac{D \vec{V}}{Dt} = \frac{\partial \vec{V}}{\partial t} + \vec{V}\cdot \nabla \vec{V}[/tex]

    Or in terms of components:

    [tex]\frac{D V_{i}}{Dt} = \frac{\partial V_{i}}{\partial t} + \sum_{k}V_{k}\frac{\partial V_{i}}{\partial x_{k}} [/tex]
     
  5. Jun 19, 2009 #4

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    I think you are confusing different problems. The "convective derivative" that Count Iblis mentions applies to the rate of change of a some property of a "fluid" as you move along with the fluid. If you are given the velocity vector [tex]\vec{V}=u\hat{i}+v\hat{j}+w\hat{k}[/tex] giving the velocity vector at some point in the x,y,z coordinate system, the acceleration vector at that point is just [tex]\vec{a}= \frac{d\vec{V}}{dt}=\frac{du}[dt}\hat{i}+\frac{dv}{dt}\hat{j}+\frac{dw}{dt}\hat{k}[/tex]
    What count Iblis gives is the rate of change of velocity if you measured it from a boat moving with the flow.
     
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