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Math Help - Computing Curl of a function?

  1. #1
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    Computing Curl of a function?

    If r = (x, y, z) and r = norm(r), compute curl [f(r)r], where f is a differentiable function.

    I'm confused at what [f(r)r] means. Would I take each partial derivative of r and dot product it with r?
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  2. #2
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    Quote Originally Posted by hashshashin715 View Post
    If r = (x, y, z) and r = norm(r), compute curl [f(r)r], where f is a differentiable function.

    I'm confused at what [f(r)r] means. Would I take each partial derivative of r and dot product it with r?
    The Key Idea is that f(r)=f(\sqrt{x^2+y^2+z^2}) is a scalar and \mathbf{r} is a vector.

    This gives in Cartesian coordinates

    f(\sqrt{x^2+y^2+z^2})\left( x\mathbf{i}+y\mathbf{j}+z\mathbf{k}\right)= xf(\sqrt{x^2+y^2+z^2})\mathbf{i}+yf(\sqrt{x^2+y^2+  z^2})\mathbf{j}+zf(\sqrt{x^2+y^2+z^2}) \mathbf{k}

    Now we just need to take the curl so we get

    \begin{bmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\\frac{\partial }{\partial x} & \frac{\partial }{\partial y} & \frac{\partial }{\partial  z}   \\xf(\sqrt{x^2+y^2+z^2}) & yf(\sqrt{x^2+y^2+z^2}) & zf(\sqrt{x^2+y^2+z^2}) \\ \end{bmatrix}

    Note that this is using first principles there are product rules for the divergence, gradient, curl ect.
    http://en.wikipedia.org/wiki/Vector_...ifferentiation

    But you should know how to work them this way as well.
    Last edited by TheEmptySet; April 28th 2011 at 01:36 PM. Reason: fix broken link
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