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Math Help - Field Gradient

  1. #1
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    Field Gradient

    Determine if the field is gradient
    F(x,y) = x^2yi + 5xy^2j

    How I can determine

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  2. #2
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    for gradient \frac{di}{dy} - \frac{dj}{dx} = 0 ?

    \frac{di}{dy} = x^2
    \frac{dj}{dx} = 5y^2

    different of 0, then not is gradient ???
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  3. #3
    MHF Contributor Calculus26's Avatar
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    you are correct
    F is not a gradient field
    there is no scalar funtion f such that F =gradf
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  4. #4
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    Thank you, the condition \frac{di}{dy} - \frac{dj}{dx} = 0 is correct ?
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Apprentice123 View Post
    Thank you, the condition \frac{di}{dy} - \frac{dj}{dx} = 0 is correct ?
    it shouldn't be i and j there, but you have the right idea, if \bold{F} = P(x,y)i + Q(x,y)j, then we can find a potential function if \frac {\partial P}{\partial y} = \frac {\partial Q}{\partial x}
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  6. #6
    MHF Contributor Calculus26's Avatar
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    I know what you are thinking
    but
    If F =f i +g j

    Then df/dy =dg/x is the condition not di/dx = dj/dx
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  7. #7
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    ok thanks.

    To confirm learning
    F(x,y) = (x^2y)i + (5xy^2)j

    Field not is gradient (conservative) because \frac{d(x^2y)}{dy} \not= \frac{d(5xy^2)}{dx}
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Apprentice123 View Post
    ok thanks.

    To confirm learning
    F(x,y) = (x^2y)i + (5xy^2)j

    Field not is gradient (conservative) because \frac{d(x^2y)}{dy} \not= \frac{d(5xy^2)}{dx}
    it should be partial derivatives: \frac{\partial (x^2y)}{\partial y} \not= \frac{\partial (5xy^2)}{\partial x}, so it is not conservative
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  9. #9
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    ok thanks.

    I have one problem with the same F but the question is: Field is gradient ?
    It is the same as: field is conservative?
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  10. #10
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Apprentice123 View Post
    ok thanks.

    I have one problem with the same F but the question is: Field is gradient ?
    It is the same as: field is conservative?
    i think you are describing the same thing, but it probably would not be stated like that. saying the field is "conservative" is saying it is the "gradient function of some vector field".
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  11. #11
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    ok Thank you
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  12. #12
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    If i have F(x,y,z) = Pi + Qj + Rk
    The condition is rotational (curl) = 0 ?
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  13. #13
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Apprentice123 View Post
    If i have F(x,y,z) = Pi + Qj + Rk
    The condition is rotational (curl)F = 0 ?
    for F to be conservative, yes. which is what we've said in another one of your posts. don't ask again here
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