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Math Help - Tricky question involving exact differentials

  1. #1
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    Tricky question involving exact differentials

    I've been asking around on this question for a bit, and though I have an answer for the first half of it, I'm a little bit iffy on the solidarity of the answer I have. Primarily, however, I could use a hand with the second half of this, since it isn't worded that well.

    Show that F=\left(\frac{x}{x^2+y^2}, \frac{y}{x^2+y^2}\right) and G=\left(\frac{y}{x^2+y^2}, \frac{x}{x^2+y^2}\right) are such that one of F.dP, G.dP is exact but the other is not (dP=(dx,dy)).

    Let C be the shorter arc of x^2+y^1=1 from (1/\sqrt{2}, -1/\sqrt{2}) to (1/\sqrt{2}, 1/\sqrt{2}).
    Evaluate \int_C F.dP and \int_C G.dP.

    As you can see, this one is actually quite tough. If you could provide any assistance on it.
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  2. #2
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    Quote Originally Posted by Runty View Post
    I've been asking around on this question for a bit, and though I have an answer for the first half of it, I'm a little bit iffy on the solidarity of the answer I have. Primarily, however, I could use a hand with the second half of this, since it isn't worded that well.

    Show that F=\left(\frac{x}{x^2+y^2}, \frac{y}{x^2+y^2}\right) and G=\left(\frac{y}{x^2+y^2}, \frac{x}{x^2+y^2}\right) are such that one of F.dP, G.dP is exact but the other is not (dP=(dx,dy)).

    Let C be the shorter arc of x^2+y^1=1 from (1/\sqrt{2}, -1/\sqrt{2}) to (1/\sqrt{2}, 1/\sqrt{2}).
    Evaluate \int_C F.dP and \int_C G.dP.

    As you can see, this one is actually quite tough. If you could provide any assistance on it.
    An equation is exact if it has a potential function e.g there is a scalar function f(x,y) such that with r=x\vec{i}+y\vec{j}

    \displaystyle \nabla f \cdot d\vec{r}=\frac{\partial f}{\partial x}dx+\frac{\partial f}{\partial y}dy

    Now for F this implies

    \displaystyle \frac{\partial f}{\partial x}=\frac{x}{x^2+y^2}

    Taking the partial integral of both sides with respect to x gives

    \displaystyle f(x,y)=\frac{1}{2} \ln (x^2+y^2)+g(y)

    Since we know what the partial with respect to y is we can determine the function g(y)

    \displaystyle \frac{\partial f}{\partial y}=\frac{y}{x^2+y^2}+g'(y)

    This implies that g'(y)=0 \implies g(y)=c

    So the potential function is

    \displaystyle f(x,y)=\frac{1}{2}\ln(x^2+y^2)

    For Part B:
    Since F has a potential function the fundamental theorem of lines integrals applies and the integral is

    \displaystyle \int_{C}\vec{F}\cdot d\vec{P}=\int_{C}\nabla f \cdot d\vec{P}=f\left( \frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}}\right) - f\left( \frac{1}{\sqrt{2}},-\frac{1}{\sqrt{2}}\right)=0

    Now for G if you try to mimic the process you will find that you cannot find a potential function (the arbitary function of integration will is a function of one variable, but you will find that it is not!

    Because of this the linear integral must be evaluated directly. Since it is a circular arc parametrize it with polar coordinates.
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  3. #3
    Member Abu-Khalil's Avatar
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    Also if \vec F\in\mathrm C^1 and its domain is simply conected you may check if

    \displaystyle\frac{\partial Q}{\partial x}=\frac{\partial P}{\partial y}

    and then conclude that there exists a potential. (but in this case doesn't apply 'cause \mathbb{R}^2\setminus\{\vec 0\} is not simply connected)

    On the other hand, if \vec F=\vec \nabla f and f\in\mathrm C^2 then \displaystyle\frac{\partial Q}{\partial x}=\frac{\partial P}{\partial y} so u may use the contrapositive for \vec G.
    Last edited by Abu-Khalil; February 3rd 2011 at 08:45 PM.
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