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Vector Derivative (not del)

  1. May 31, 2008 #1
    What does it mean to operate with [itex]\frac{d}{\vec{v}}[/itex] where v is a vector?

    Say you have another vector q, how do you do [itex]\frac{d\vec{v}} {d\vec{q}}[\itex]?

    What about [itex]\frac{d\vec{v}} {d\vec{v}}[/itex]?
    (Can't remember how to do the proper font sorry)
    Last edited by a moderator: May 31, 2008
  2. jcsd
  3. May 31, 2008 #2


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    Staff Emeritus
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    First you will have to say what kind of vector space [itex]\vec{v}[/itex] is in and what kind of function of v you are talking about. If f: Rn to Rm, that is if the variable, [itex]\vec{v}[/itex] is an n dimensional vector variable and f maps it to an m dimensional vector, the [itex]d\vec{f}/d\vec{v}[/itex], at [itex]\vec{v}_0[/itex] is "the linear transformation from Rn to Rm that best approximates [itex]\vec{f}[/itex] in some region around [itex]\vec{v}_0[/itex]".

    More precisely, a function,[itex]\vec{f}[/itex], from Rn to Rm, is said to be differentiable at [itex]\vec{v}_0[/itex] if and only if there exist a linear transformation, L, from Rn to Rm, and a function [itex]\epsilon(\vec{v})[/itex], from Rn to Rm, such that
    1) [itex]f(\vec{v})= f(\vec{v}_0)+ L(\vec{v}- \vec{v_0})+ \epsilon(\vec{v})[/itex]
    2) [itex]\lim_{\vec{v}\rightarrow \vec{0}}\epsilon(\vec{v})/||\vec{v}-\vec{v}_0||= 0[/itex]

    It can be shown that the linear transformation, L, in (1), is unique and we say that L is the derivative of f at [itex]\vec{v}_0[/itex].

    Notice that, if we reduce this to R1 to R1, we are saying that the derivative is NOT the slope of the tangent line y= mx+ b but, rather, the linear function y= mx.

    If f is from R1 to R3, a "vector valued function of a single real variable", then the L above is linear transformation from R1 to R3 given by
    [tex]Lt= \left<\frac{df_x}{dt},\frac{df_y}{dt},\frac{df_z}{dt}\right>t[/tex]
    which we can think of as being "represented" by the usual derivative vector,
    [tex] \left<\frac{df_x}{dt},\frac{df_y}{dt}, \frac{df_z}{dt}\right>[/tex]

    If f is from R3 to R, a "real valued function of 3 real variables, x, y, z", then the derivative, in the sense above, is the linear transformation from R3 to R given by the dot product
    [tex]\left<\frac{\partial f}{\partial x},\frac{\partial f}{\partial y}, \frac{\partial f}{\partial z}\left>\cdot \vec{v}[/itex]
    which we can think of as represented by the gradient vector,
    [tex]\left<\frac{\partial f}{\partial x},\frac{\partial f}{\partial y}, \frac{\partial f}{\partial z}\left>[/itex]

    More generally, if f is from Rn to Rm, it derivative, at any "point", is the linear transformation which can be represented, in some basis, as the m by n matrix having the partial derivatives of the components of f as elements.

    In any case, of course, [itex]\frac{d\vec{v}}{d\vec{v}}[/itex], where [itex]\vec{v}[/itex] is a vector function from Rn to itself (NOT just a single vector- derivatives are only defined for functions) is the identity transformation on Rn which can be reprsented by the n by n identity matrix.
  4. Jun 1, 2008 #3
    Thanks mate. I didn't expect such a long reply and one quite so theoretical. It'll take me a bit to mull over so cheers!
  5. Oct 1, 2008 #4
    I want to understand exactly how do I derivate vector by vector.
    My problem: I try to solve Euler equations with some method... In this method I use a linearization to the first vector [tex]\hat{Q}[/tex]. I need to write the [tex]\partial[/tex]F/[tex]\partial[/tex]Q.
    Where Q and F are vectors [1,4].
    Can you help me to understand how to build the matrix of linear transformation?

    Thank you, I have found the reference: http://www.met.rdg.ac.uk/~ross/Documents/VectorDeriv.html
    Last edited: Oct 1, 2008
  6. Oct 1, 2008 #5


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    I presume that you mean they are vectors in R2, not just that specific vector!

    Say F= <f(x,y), g(x,y)> and Q= <p(x,y), q(x,y)>. Let r= <x,y>. Then, by the chain rule,
    [tex]\frac{dF}{dQ}= \frac{dF}{dr}/\frac{dQ}{dr}[/itex].

    And those two derivatives are the matrices having the partial derivatives as entries. (Strictly speaking, they are the linear transformations represented by these matrices in the <1, 0>, <0, 1> basis for R2.)
    [tex]\frac{dF}{dr}= \left[\begin{array}{cc}\frac{\partial f}{\partial x} & \frac{\partial f}{\partial y} \\ \frac{\partial g}{\partial x} & \frac{\partial g}{\partial y}\end{array}\right][/tex]

    [tex]\frac{dQ}{dr}= \left[\begin{array}{cc}\frac{\partial p}{\partial x} & \frac{\partial p}{\partial y} \\ \frac{\partial q}{\partial x} & \frac{\partial q}{\partial y}\end{array}\right][/tex]

    So that
    [tex]\frac{dF}{dW}= \left[\begin{array}{cc}\frac{\partial f}{\partial x} & \frac{\partial f}{\partial y} \\ \frac{\partial g}{\partial x} & \frac{\partial g}{\partial y}\end{array}\right]\left[\begin{array}{cc}\frac{\partial p}{\partial x} & \frac{\partial p}{\partial y} \\ \frac{\partial q}{\partial x} & \frac{\partial q}{\partial y}\end{array}\right]^{-1}[/tex]

    The existance of that derivative depends not only on the existance of these partial derivatives but also on dQ/dr being invertible.
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