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Math Help - How i fond curl F at point (2, 3, 2)

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    Question How i fond curl F at point (2, 3, 2)

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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by king imran View Post
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    Question 2 was asked by another user and was answered here

    personally, i would have used Soroban's method to do it...but CaptainBlack's method was so cool! and it seems to have much wider applications. I don't know what Wallis' formula is though, so you should be able to make a better choice than i can
    Last edited by Jhevon; June 16th 2007 at 11:46 AM.
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by king imran View Post
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    Here's question 1

    Recall that for a vector field, \bold{F} = P \bold {i} + Q \bold {j} + R \bold {k}, \mbox{curl} \bold{F} is given by:

    \mbox{curl} \bold{F} = \nabla \times \bold{F}

    where \times means the cross-product, and \nabla = \frac {\partial}{\partial x} + \frac {\partial}{\partial y} + \frac {\partial}{\partial z}

    That is, \mbox {curl} \bold{F} = \left| \begin{array}{ccc} \bold{i}&\bold{j}&\bold{k} \\ \partial / \partial x & \partial / \partial y & \partial / \partial z \\ P & Q & R \end{array} \right| = \left( \frac{\partial R}{\partial y} - \frac{\partial Q}{\partial z}\right)\bold{i}+ \left( \frac{\partial P}{\partial z} - \frac{\partial R}{\partial x} \right) \bold{j} + \left( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right) \bold{k}



    Here we have \bold {F} = z e^{xy} \bold{i} + 2x^2 z \sin y \bold{j} + \left( 2x^2 + 3y^2 \right) \bold{k}

    and we want the curl at (2,3,2)


    So, \mbox{curl} \bold {F} = \left( \frac{\partial \left( 2x^2 + 3y^2 \right)}{\partial y} - \frac{\partial \left( 2x^2 z \sin y \right)}{\partial z}\right)\bold{i}+ \left( \frac{\partial \left(z e^{xy} \right)}{\partial z} - \frac{\partial \left( 2x^2 + 3y^2 \right)}{\partial x} \right) \bold{j} + \left( \frac{\partial \left( 2x^2 z \sin y \right)}{\partial x} - \frac{\partial \left(  z e^{xy}\right)}{\partial y} \right) \bold{k}

    .............. = \left( 6y - 2x^2 \sin y \right) \bold{i} + \left( e^{xy} - 4x \right) \bold{j} + \left( 4x z \sin y - xz e^{xy} \right) \bold {k}


    Since we want the curl at (2,3,2), we simply plug in x = 2 \mbox { , } y = 3 \mbox { , and } z = 2 . I leave that to you
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    Quote Originally Posted by Jhevon View Post
    Here's question 1 ....
    I hope you learned that from my incomplete tutorial.
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by ThePerfectHacker View Post
    I hope you learned that from my incomplete tutorial.
    i did actually
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