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Thread: 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 $\displaystyle {\bf F} = \nabla f$ for some $\displaystyle f.$
    2) If $\displaystyle \nabla \times {\bf F} = 0$
    3) Set $\displaystyle \nabla f = {\bf F} $ and solve for $\displaystyle f$.
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  3. #3
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    if I have $\displaystyle F(x,y) = (e^xcosy)i + (-e^xseny)j$

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


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

    $\displaystyle
    {\bf F} = <P,Q>
    $

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

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

    field conservative = field gradient ?


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



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









    field conservative = field gradient ?


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


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

    If $\displaystyle f_x = e^x \cos y$ then integrating wrt to $\displaystyle x$ gives $\displaystyle f = e^x \cos y + g(y)$. Now subs. into $\displaystyle f_y = - e^x\sin y$ which shows that $\displaystyle - e^x\sin y + g'(y) = - e^x\sin y$ which gives $\displaystyle g'(y) = 0$ which gives $\displaystyle g(y)=c$, thus our $\displaystyle f.$
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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 => $\displaystyle \frac {di}{dy} - \frac{dj}{dx} = 0$

    and
    field conservative = field gradient ?
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