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Show F is conservative

  1. Dec 11, 2013 #1
    1. The problem statement, all variables and given/known data
    Show [itex]F=<3x^2y-y^2,x^3-2xy>[/itex] is conservative. Find a scalar potential f. Evaluate [itex]∫FdR[/itex] where C connects (0,0) to (2,1).

    2. Relevant equations
    Conservative if [itex]P_y=Q_x[/itex]

    3. The attempt at a solution
    So it is conservative, but I don't know where to go from here. Thanks
     
  2. jcsd
  3. Dec 11, 2013 #2

    vela

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    What's a scalar potential?
     
  4. Dec 11, 2013 #3
    Where the gradient of f is equal to F, right? I just have no clue how to do this.
     
  5. Dec 11, 2013 #4

    vela

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    I'm sure there are examples you could consult in your textbook. In any case, you're right. If f is the scalar potential, then ##\nabla f = F##, so that means
    \begin{align*}
    \frac{\partial f}{\partial x} &= P \\
    \frac{\partial f}{\partial y} &= Q
    \end{align*} Plug in the function you have for P for this problem, and integrate the first equation. What do you get?
     
  6. Dec 11, 2013 #5
    [itex]x^3y-xy^2+c[/itex]

    My book is so awful
     
  7. Dec 11, 2013 #6
    Never mind I think I got it. Sometimes when I post it jogs my memory. I remember my teacher doing something with a function of y in the place of c. Thanks anyways vela!!
     
  8. Dec 11, 2013 #7

    D H

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    That's correct. You integrated 3x2y-y2 with respect to x to yield x3y-xy2+c. That constant c can be any function of y because what you integrated was a partial derivative with respect to x.

    Consider ##\vec F = 2x\hat x + 2y\hat y##. This is obviously conservative, but now when you do the integrations you get ##U(x,y)=x^2+c## on one hand versus ##U(x,y)=y^2+c## on the other. The way around this apparent problem to realize that the first c is a function of y, the second a function of x. With that, ##U(x,y)=x^2+f(y)=y^2+g(x)##, so ##U(x,y)=x^2+y^2+c##.
     
  9. Dec 11, 2013 #8

    vanhees71

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    The only question left is the sign. Usually one defines the potential such that it adds positively to the total energy, which implies that
    [tex]\vec{F}(\vec{x})=-\vec{\nabla} V(\vec{x}).[/tex]
     
  10. Dec 11, 2013 #9

    D H

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    That's the convention used in physics. Mathematicians often use the opposite convention. This question was asked in the mathematics section, so it's probably more apropos to use the non-negated convention.
     
  11. Dec 11, 2013 #10

    vanhees71

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    Oh dear, that's very confusing for a physicist :rolleyes:
     
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