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Math Help - Vector field

  1. #1
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    Vector field

    1) How can I know if the field is gradient?

    2) How can I know if the fiels is conservative ?

    3) How can I find the function potential of vector field ?
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  2. #2
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    Quote Originally Posted by Apprentice123 View Post
    1) How can I know if the field is gradient?

    2) How can I know if the fiels is conservative ?

    3) How can I find the function potential of vector field ?
    1) If {\bf F} = \nabla f for some f.
    2) If  \nabla \times {\bf F} = 0
    3) Set \nabla f = {\bf F} and solve for f.
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  3. #3
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    if I have F(x,y) = (e^xcosy)i + (-e^xseny)j

    what will be: \nabla and f
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  4. #4
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    Quote Originally Posted by Apprentice123 View Post
    if I have F(x,y) = (e^xcosy)i + (-e^xseny)j


    what will be: \nabla and f
    Since you only have a two dimensional vector field, say

     <br />
{\bf F} = <P,Q><br />

    then this would be conservative if P_y = Q_x and if P = e^x \cos y and Q = - e^x \sin y then P_y = -e^x \sin y and Q_x = -e^x \sin y which are identical. So the vector field is conservative and an f exists such that

    f_x = P = e^x \cos y and f_y = Q = - e^x\sin y which gives f = e^x \cos y + c
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  5. #5
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    for the field is conservative \frac {di}{dy} - \frac{dj}{dx} = 0?

    field conservative = field gradient ?


    If f_x = P = e^x \cos y and f_y = Q = - e^x\sin y why f = e^x \cos y + c ???



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  6. #6
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    Quote Originally Posted by Apprentice123 View Post
    for the field is conservative \frac {di}{dy} - \frac{dj}{dx} = 0?









    field conservative = field gradient ?


    If f_x = P = e^x \cos y and f_y = Q = - e^x\sin y why f = e^x \cos y + c ???


    First since the vector field is conservative you know an f exists, i.e. f exists such that f_x = P = e^x \cos y \;{\bf and} \;f_y = Q = - e^x\sin y

    If f_x = e^x \cos y then integrating wrt to x gives f = e^x \cos y + g(y). Now subs. into f_y = - e^x\sin y which shows that - e^x\sin y + g'(y) = - e^x\sin y which gives g'(y) = 0 which gives g(y)=c, thus our f.
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  7. #7
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    Thank you I understand.

    And the statements are correct?

    for that the field is conservative i need have => \frac {di}{dy} - \frac{dj}{dx} = 0

    and
    field conservative = field gradient ?
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