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Conservative fields

  1. Jun 30, 2003 #1

    Dx

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    Determine whether the vector field is conservative. F(x,y)=(e^x sin(y) + tan(y)i + (e^x cos(y) + sec^2(y)j.
    f(x1,y1) = [inte]A to B (F * T) ds = [inte]A to B(e^x sin(y) + tan(y)dx + (e^x cos(y) + sec^2(y)dy = [inte]1 to 0 (e^x sin(y) + tan(y)dx + (e^x cos(y) + sec^2(y)dy

    I am lost from here can anyone help me solve from here plz?
    Dx
     
  2. jcsd
  3. Jul 1, 2003 #2

    HallsofIvy

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    You are given that F=(e^x sin(y) + tan(y))i + (e^x cos(y) + sec^2(y))j.

    The rest of the formula you gave is not necessary. F is the vector field. saying that it is "conservative" (that's really a physics term- the mathematics term is "exact") means that it is the gradient of some scalar function. In other words, is there a function f(x,y) such that F= grad(f)= (df/dx)i+ (df/dy)j.

    If df/dx= e^x sin(y)+ tan(y) then

    df/dxdy= e^x cos(y)+ sec^2(y)

    If df/dy= e^x cos(y)+ sec^2(y)

    df/dydx= e^x cos(y). Notice that df/dxdy is not the same as df/dy/dx: but mixed derivatives have to be equal! F is not the gradient of any function, this vector field is not conservative (exact).

    By the way, if it were F= (e^x sin(y) + tan(y))i + (e^x cos(y) + xsec^2(y))j then it would be conservative. Do you see the difference?
     
  4. Jul 1, 2003 #3
    Or even more directly, if you take the curl of the vector field (something we physicsts do all the time), you'll find conservative field curls are equal to 0. Of course, the curl is little computationally intensive, but the answer is unmistakeable and there is a great deal of satisfaction when you get through the thing :smile:
     
  5. Jul 1, 2003 #4

    HallsofIvy

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    I wouldn't call that "more directly". What taking the curl involves is taking the derivatives as above, then subtracting to see if the result is 0 rather than comparing to see if they are the same.
    I would even point out that the curl requires working in three dimensions while this problem is purely two dimensional.

    You are right that is the same thing. And anyone who talks in terms of "conservative vector fields" rather than "exact differentials" might be more comfortable with "curl" than second derivatives.
     
  6. Jul 1, 2003 #5
    I stand corrected. Sometimes in physics, we have difficulty seeing the forest from the trees. THanx
     
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