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Math Help - Function being path independent?

  1. #1
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    Function being path independent?

    I have a question about this problem. I know that I have to parametrize the circle using polar coordinates but after that I am completely lost. My book says that being path independent also means the same as being a conservative field correct? So does that mean that I have to take the partials of those terms? What am I missing here? Thanks!
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  2. #2
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    Quote Originally Posted by cad2blender View Post
    I have a question about this problem. I know that I have to parametrize the circle using polar coordinates but after that I am completely lost.

    My book says that being path independent also means the same as being a conservative field correct? Mr F says: Correct.

    So does that mean that I have to take the partials of those terms? What am I missing here? Thanks!
    If a vector field is conservative then \nabla \times F = 0. Does this happen here?
    Last edited by mr fantastic; December 5th 2009 at 03:22 PM. Reason: Corrected a typo
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  3. #3
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    Quote Originally Posted by cad2blender View Post
    I have a question about this problem. I know that I have to parametrize the circle using polar coordinates but after that I am completely lost. My book says that being path independent also means the same as being a conservative field correct? So does that mean that I have to take the partials of those terms? What am I missing here? Thanks!

    So the parametric equation for the circle is

    \vec{r}(t)=<4\cos(t),4\sin(t)> \implies \vec{r'}(t)=<-4\sin(t),4\cos(t)>dt

    Then the line integral is

    \int_{0}^{2\pi}\vec{F}(r(t))\cdot r'(t)dt=
    \int_{0}^{2\pi}<3(4\cos(t))^2,(4\sin(t))>\cdot <-4\sin(t),4\cos(t)>dt=
    \int_{0}^{2\pi}-192\sin(t)\cos^2(t)+16\sin(t)\cos(t)dt

    =64\cos^3(t)+8\sin^2(t) \bigg|_{0}^{2\pi}=0

    You can see the vector field is conservative.

    \frac{\partial f}{\partial x}=3x^2 \implies f(x,y)=x^3+g(y)

    Now taking the partial with respect to y gives

    \frac{\partial f}{\partial y}=g'(y)=y \implies g(y)=\frac{1}{2}y^2

    So the potential function is

    f(x,y)=x^3-\frac{1}{2}y^2 and notice that

    \nabla f =\vec{F}

    So the vector field is conservative.
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    hmmm ok then, so the question was simply asking me if it was conservative just in different wording? That's what threw me off. Thanks a ton guys
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