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Thread: Work field

  1. #1
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    Work field

    Find the work realized by field
    $\displaystyle F(x,y) = (2xy, x^2 + cosy)$

    to move a particle from the initial point to end point of the curve $\displaystyle r(t) = ti + costj$ where $\displaystyle 0 \leq t \leq \pi$

    How is the integral ???
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  2. #2
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    Can I use green's theorem?

    $\displaystyle w= \int \int_S [\frac{ \partial (x^2+cosy)}{ \partial x} - \frac{ \partial (2xy)}{ \partial y}]dxdy$

    $\displaystyle 0 \int \int_S dxdy = 0$


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  3. #3
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    Quote Originally Posted by Apprentice123 View Post
    Can I use green's theorem?






    $\displaystyle w= \int \int_S [\frac{ \partial (x^2+cosy)}{ \partial x} - \frac{ \partial (2xy)}{ \partial y}]dxdy$

    $\displaystyle 0 \int \int_S dxdy = 0$

    Greens theorem is only valid for piecewise simple closed curves, and this path is not closed.

    $\displaystyle \vec r(t)=t\vec i +\cos(t) \vec j$

    $\displaystyle \vec r'(t) = \vec i + -\sin(t) \vec j$

    $\displaystyle F(x,y,)=(2xy,x^2+\cos(y))$

    $\displaystyle \int_{0}^{\pi} (2t\cos(t),t^2+\cos(\cos(t)))\cdot (1,-\sin(t))dt$

    $\displaystyle \int_{0}^{\pi}2t\cos(t)-t^2\sin(t)-\cos(\cos(t))\sin(t) dt$

    You should be able to finish from here
    Last edited by TheEmptySet; May 1st 2009 at 05:35 PM. Reason: wrote the wrong equation down
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  4. #4
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    Quote Originally Posted by TheEmptySet View Post
    Greens theorem is only valid for piecewise simple closed curves, and this path is not closed.

    $\displaystyle \vec r(t)=t\vec i +\cos(t) \vec j$

    $\displaystyle \vec r'(t) = \vec i + -\sin(t) \vec j$

    $\displaystyle F(x,y,)=(2xy,y+\cos(y))$

    $\displaystyle \int_{0}^{\pi} (2t\cos(t),\cos(t)+\cos(\cos(t)))\cdot (1,-\sin(t))dt$

    $\displaystyle \int_{0}^{\pi}2t\cos(t)-\sin(t)\cos(t)-\cos(\cos(t))\sin(t) dt$


    You should be able to finish from here
    Not is $\displaystyle \int_0^{\pi} (2tcost, t^2+cos(cost)).(1,-sint)dt$ ??
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  5. #5
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    Quote Originally Posted by Apprentice123 View Post
    Not is $\displaystyle \int_0^{\pi} (2tcost, t^2+cos(cost)).(1,-sint)dt$ ??

    I was a bit careless with notation

    These are Vectors and the dot means the dot product
    $\displaystyle \int_{0}^{\pi} (2t\cos(t),t^2+\cos(\cos(t)))\cdot (1,-\sin(t))dt$

    $\displaystyle
    \int_{0}^{\pi} 2t\cos(t)\vec i+ [t^2+\cos(\cos(t)]\vec j)\cdot (\vec i,-\sin(t)\vec j)dt
    $

    Then this gives the bottome integral in the post before
    Last edited by TheEmptySet; May 1st 2009 at 05:35 PM.
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  6. #6
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    Quote Originally Posted by TheEmptySet View Post
    I was a bit careless with notation

    These are Vectors and the dot means the dot product
    $\displaystyle \int_{0}^{\pi} (2t\cos(t),\cos(t)+\cos(\cos(t)))\cdot (1,-\sin(t))dt$

    $\displaystyle
    \int_{0}^{\pi} 2t\cos(t)\vec i+ [\cos(t)+\cos(\cos(t)]\vec j)\cdot (\vec i,-\sin(t)\vec j)dt
    $

    Then this gives the bottome integral in the post before

    What I said was: $\displaystyle F(x,y) = (2xy)i + (x^2 + cosy)j => (2tcost)i + (t^2 + cos(cost))j$ What did you do with $\displaystyle x^2$ ?
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  7. #7
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    Quote Originally Posted by Apprentice123 View Post
    What I said was: $\displaystyle F(x,y) = (2xy)i + (x^2 + cosy)j => (2tcost)i + (t^2 + cos(cost))j$ What did you do with $\displaystyle x^2$ ?
    Ahhh. I see why you are confused. Sorry

    I have edited the above posts, I wrote down the wrong equation and had y instead of x^2.
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  8. #8
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    Quote Originally Posted by TheEmptySet View Post
    Ahhh. I see why you are confused. Sorry

    I have edited the above posts, I wrote down the wrong equation and had y instead of x^2.
    Ok. Thank you
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