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Potential function

  1. Jul 18, 2006 #1
    Find the potential function if it exists: v (x1 , x2) = 2 (x1)^2 + 2(x2) + 2, 2(x2)+2x(1)


    may i know how to do this??

    pls help
     
  2. jcsd
  3. Jul 18, 2006 #2

    Office_Shredder

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    The potential function is the function whose gradient = v(x1 x2) (I'll call them x and y for aesthetic purposes). So if Grad(P(x, y)) = v(x, y), that means dP/dx = 2x^2 + 2y + 2, and dP/dy = 2y + 2x.

    Try going from there
     
  4. Jul 18, 2006 #3
    so the answer is v(x,y)=(4x , 2) rite??
     
  5. Jul 18, 2006 #4

    Office_Shredder

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    No..... the potential function of v is going to be a scalar function P, such that P is (essentially) the integral of v. So you're looking for a scalar function whose partial derivatives are the two parts of v
     
  6. Jul 18, 2006 #5

    HallsofIvy

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    No, a gradient is not a vector!

    Find a single, real valued, function of 2 variables, (and x,y is much better than x1, x2!), P(x,y) such that
    [tex]\frac{\partial P}{\partial x}= 2x^2+ 2y+ 2[/tex]
    [tex]\frac{\partial P}{\partial y}= 2y+ 2x.

    What is the anti-derivative of 2x2+ 2y+ 2 (with respect to x, treating y as a constant)? The "constant of integration" may be a function of y since you are treating y as a constant. Now, what is the derivative of that with respect to y?
     
  7. Jul 18, 2006 #6
    so if i change the question to find the conservative vector field,
    than my answer v(x,y)=(4x , 2) is correct rite??
    just do partial derivative rite??
     
  8. Jul 18, 2006 #7

    Office_Shredder

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    teng, (4x, 2) is a vector field.

    You need to supply a scalar field as the answer. You're looking for a scalar field whose gradient is v(x,y).

    For example, if v(x,y) was (3x^2, 3y^2) then the potential P(x,y)=x^3 + y^3 + C

    Because dP/dx = 3x^2, and dP/dy = 3y^2
     
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