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Directional Derivative and unit vectors

  1. Jul 23, 2006 #1
    What happens if a unit vector is not used in calculating the directional derivative. From when I worked it out the directional derivative is multiplied by a scalar if a unit vector is not used. So I gather that the directional derivative must be calculated by a unit vector. Because it is still calculating a rate of change and a rate of change multiplied by a number is not the same answer. I was just curious if my answer was on the right track or way off. Thanks.
     
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  3. Jul 23, 2006 #2

    quasar987

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    Here's a more useful interpretation IMO.

    In the same way as [itex] f(x+a) \approx f(x) + af'(x)[/itex] (sometimes known as Euler's approximation) in single variable calculus,

    [tex]f(\vec{r}+\vec{A}) \approx f(\vec{r}) + \nabla f (\vec{r}) \cdot \vec{A}[/tex]

    gives a first order approximation of the value of f(r+A).
     
    Last edited: Jul 23, 2006
  4. Jul 23, 2006 #3
    A directional derivative is just the projection of a function's gradient along some direction. To get the projection of a vector in a particular direction you take the dot product of the vector with a unit vector in the direction you're looking at.

    If you don't use a unit vector, then your directional derivative will be multiplied by the length of the vector that you do use.

    quasar, I think you need to fix a few errors in your post :yuck:

    [tex]f(x+a) \approx f(x) + af^\prime (x)[/tex]

    [tex]f(\vec{r}+\vec{A}) \approx f(\vec{r})+\vec{A} \cdot \nabla f(\vec{r})[/tex]

    (note, this is basically just taking a Taylor expansion up to first-order)
     
    Last edited: Jul 23, 2006
  5. Jul 24, 2006 #4

    HallsofIvy

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    So, in other words, if your original vector wasn't a unit vector, just divide your answer by its length:
    [tex]D_{\vec{v}}f= \frac{\vec{v}\cdot\del f}{|\vec{v}|}[/itex]

    Of course, since [itex]\frac{\vec{v}}{|\vec{v}|}[/itex] is the unit vector in the direction of [itex]\vec{v}[/itex], that is exactly the same as reducing to a unit vector in the first place.
     
  6. Jul 24, 2006 #5

    quasar987

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    Strangely, \del doesn't do anything in Latex afaik. [itex]\nabla[/itex] is \nabla and [itex]\partial[/itex] is \partial. I'm sure everyone knew, but here it is anyway,

    [tex]D_{\vec{v}}f= \frac{\vec{v}\cdot\nabla f}{|\vec{v}|}[/tex]
     
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