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Dot Product Differentiation question

  1. Aug 6, 2008 #1
    If I differentiate two unit vectors, one with respect to the other, would it just be the dot product between the two vectors (namely the cosine of the angle between them)?

    I don't understand the physical meaning of the result...
     
  2. jcsd
  3. Aug 6, 2008 #2

    HallsofIvy

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    Do you mean, given [itex]\vec{v}[/itex] and [itex]\vec{u}[/itex], differentiate one with respect to the other? That makes no more sense than differentiating one number with respect to another- you have to differentiate a function. If [itex]\vec{f}(\vec{v})[/itex] is a vector valued function of the vector [itex]\vec{v}[/itex] then you could differentiate [itex]\vec{f}[/itex] with respect to [itex]\vec{v}[/itex]. If [itex]\vec{v}[/itex] is in Rn and [itex]\vec{f}(\vec{v})[/itex] is in Rm then the derivative would be a linear transformation from Rn to Rm, representable by an n by m matrix.
     
  4. Aug 6, 2008 #3
    If you treat the first unit vector as just a vector with a given direction and magnitude of 1, then differentiating it with respect to the other unit vector really means how it's magnitude changes if you add 1 but in a different direction. This is equivalent to a so called directional derivative (see http://mathworld.wolfram.com/DirectionalDerivative.html, (6)). In this extreme case you differentiate the first vector by its own unit vector, which becomes 1. then dot product the result with the second unit vector. And yes, it becomes the dot product of the two unit vectors, namely the cosine of their angles.
     
  5. Aug 7, 2008 #4

    HallsofIvy

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    No. "How its magnitude changes if you add 1 but in a different direction" is NOT the directional derivative. The wolfram site you give makes it very clear, as I said before, that you cannot differentiate a "vector", you must differentiate a "vector function".

    It is true that if f(x,y,z) (f is a numerical valued function, not a vector function) and [itex]\vec{v}[/itex] a given vector, then "the derivative of f in the direction of [itex]\vec{v}[/itex]" is
    [tex]\frac{\grad f\cdot\vec{v}}{|\vec{v}|}[/tex]

    But that has nothing to do with "differentiating one vector by another".
     
    Last edited: Aug 8, 2008
  6. Aug 7, 2008 #5
    I'm gonna agree with HallsofIvy here... I think I need to re-look what my question is.
    Thanks.
     
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