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Math Help - Curl...

  1. #1
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    Curl...

    Show that the component of ∇φ(x, y) in the i direction of the surface φ(x, y) = exp−(x^2 + y^2) is:

    −(√2)/e

    at the point r = 1/(√2) (i + j).


    Hi. Im having some trouble with the question above. Not really sure where to go with this, but as a starter I know that the curl = dφ/dx i + dφ/dy j. I got these to be:

    dφ/dx = -2x exp-(x^2 + y^2)
    dφ/dy = -2y exp-(x^2 + y^2)

    Thanks
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  2. #2
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    Quote Originally Posted by Croc View Post
    Show that the component of ∇φ(x, y) in the i direction of the surface φ(x, y) = exp−(x^2 + y^2) is:

    −(√2)/e

    at the point r = 1/(√2) (i + j).


    Hi. Im having some trouble with the question above. Not really sure where to go with this, but as a starter I know that the curl = dφ/dx i + dφ/dy j. I got these to be:

    dφ/dx = -2x exp-(x^2 + y^2)
    dφ/dy = -2y exp-(x^2 + y^2)

    Thanks

    Why in "abstract algebra", for hollie mollie's sake?!! . This belongs in the calculus section.
    Besides this some symbols didn't show correctly...the first one was nabla = the gradient of \phi , or what?

    Tonio
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  3. #3
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    Quote Originally Posted by Croc View Post
    Show that the component of ∇φ(x, y) in the i direction of the surface φ(x, y) = exp−(x^2 + y^2) is:

    −(√2)/e

    at the point r = 1/(√2) (i + j).


    Hi. Im having some trouble with the question above. Not really sure where to go with this, but as a starter I know that the curl = dφ/dx i + dφ/dy j. I got these to be:

    dφ/dx = -2x exp-(x^2 + y^2)
    dφ/dy = -2y exp-(x^2 + y^2)

    Thanks
    As Tonio said, this does not belong in the "Linear Algebra and Abstract Algebra" section but in the Calculus section- but I'm not going to bang my head against the wall!

    I hope you do not know "that the curl= dφ/dx i + dφ/dy j"! That's the gradient, \nabla \phi or grad \phi. "Curl" is an operator on vector valued functions: \nabla\times\vec{f}.

    You should also know that D_{\vec{v}} f(x,y)= \nabla f\cdot \vec{v} as long as \vec{v} is a unit vector- and \vec{i} is, of course, a unit vector. That is, the answer to your problem is just \nabla \phi\cdot\vec{i}= \frac{\partial \phi}{\partial x}, evaluated, of course, at x= \frac{1}{\sqrt{2}}, y= \frac{1}{\sqrt{2}}.
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