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Math Help - Directional Derivative

  1. #1
    Junior Member HAL9000's Avatar
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    Question Directional Derivative

    Hello!

    I've run into a case, where one should calculate the directional derivative of a function f of several variables at a point on a sphere of radius R in the direction of a normal vector to this sphere. It is known that the function itself can be written only in terms of the length of the argument x, i.e. |x|.

    Why is it actually allowed to write \frac{\partial f}{\partial n}|_{|x|=R}=\pm \frac{\partial f}{\partial |x|}|_{|x|=R}, with \pm for the outer/inner normal vector?
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  2. #2
    Senior Member vincisonfire's Avatar
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    Re: Directional Derivative

    n is the normal vector. It can be pointing in or out of the sphere, whereas |x| points out of the sphere.
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  3. #3
    Junior Member HAL9000's Avatar
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    Re: Directional Derivative

    As far as I know, |x| is NOT a vector.
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  4. #4
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    Re: Directional Derivative

    n = x/Ri + y/Rj + z/Rk
    delf = df/dxi + df/dyj + df/dzk = df/dxi
    df/dn = delf.n = (df/dx)x/R
    IxI = R only at a point where sphere crosses x axis, in which case
    df/dn = df/dx, for x pos
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  5. #5
    Junior Member HAL9000's Avatar
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    Re: Directional Derivative

    I think there is a misunderstanding here. By a function of several variables I mean explicitly

    f: \mathbb{R}^n \rightarrow \mathbb{R}, x \mapsto f(x), x:=(x_1,\dots,x_n),

    so that

    |x|=\left(\sum_{1 \le i \le n} x_i^2\right)^{1/2} is the euclidean length of x.

    I thought this was clear. Anyway, it's my fault. Sorry.
    Last edited by HAL9000; January 27th 2013 at 06:47 AM.
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  6. #6
    Super Member ILikeSerena's Avatar
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    Re: Directional Derivative

    The notation is somewhat arbitrary as long it is clear from the context what it means.

    Since you said it was a directional derivative in the direction normal to the sphere, I would interpret your derivatives to mean:

    {\partial f \over \partial n} = \lim_{h \to 0} {f(\mathbf x + h\mathbf{n}) - f(\mathbf x) \over h}

    {\partial f \over \partial |x|} = \lim_{h \to 0} {f(\mathbf x + h|x|\mathbf{\hat x}) - f(\mathbf x) \over h}

    Although it may also be possible that your vectors need to be normalized, depending on how your book defines a directional derivative.

    See Directional derivative - Wikipedia, the free encyclopedia for more information on how a directional derivative is or can be defined.


    Either way, those 2 forms are the same, except for a possible minus sign.
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  7. #7
    Junior Member HAL9000's Avatar
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    Re: Directional Derivative

    OK, that was easier than I had thought!

    My problem was the normal derivative of a function on \mathbb{R}^n with radial symmetry, i.e.

    f(x)=g(|x|).

    Now it's clear that for n=\pm \frac{x}{|x|} we have

    \langle\nabla{f},n \rangle = \frac{\partial g}{\partial |x|} \left\langle \left( \frac{\partial |x|}{\partial x_1}, \dots, \frac{\partial |x|}{\partial x_n}\right),n\right\rangle =\pm \frac{\partial g}{\partial |x|}

    Oh my! Simple as that!
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  8. #8
    Super Member ILikeSerena's Avatar
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    Re: Directional Derivative

    That looks about right.
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