Results 1 to 3 of 3

Thread: Curl...

  1. #1
    Newbie
    Joined
    Apr 2010
    Posts
    11

    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
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Banned
    Joined
    Oct 2009
    Posts
    4,261
    Thanks
    3
    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 $\displaystyle \phi$ , or what?

    Tonio
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor

    Joined
    Apr 2005
    Posts
    19,769
    Thanks
    3027
    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, $\displaystyle \nabla \phi$ or grad $\displaystyle \phi$. "Curl" is an operator on vector valued functions: $\displaystyle \nabla\times\vec{f}$.

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

Similar Math Help Forum Discussions

  1. curl
    Posted in the Calculus Forum
    Replies: 2
    Last Post: Apr 16th 2010, 12:56 PM
  2. Curl Problem
    Posted in the Calculus Forum
    Replies: 1
    Last Post: Jan 20th 2010, 02:48 AM
  3. curl
    Posted in the Calculus Forum
    Replies: 2
    Last Post: Dec 16th 2009, 02:21 AM
  4. Problem with Curl of Curl
    Posted in the Calculus Forum
    Replies: 5
    Last Post: Jul 1st 2008, 02:38 AM
  5. Divergence and Curl
    Posted in the Calculus Forum
    Replies: 1
    Last Post: Mar 29th 2008, 03:22 AM

Search Tags


/mathhelpforum @mathhelpforum