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Math Help - FTC And Path Independence

  1. #1
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    FTC And Path Independence

    Let F(x,y) = (xy,y^2) and let C be the path y = 2x^2 joining (0,0) to (1,2) in the plane. Evaluate \int_{C} \vec{F} \cdot d\vec{s}

    I know that the it isn't conservative but other than that, I don't know what to do after that.
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  2. #2
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    Remember that

    \int_{C}\vec{F}\cdot d\vec{s}=\int_{C}\vec{F}\cdot \vec{n}ds

    Where \vec{n} is the unit tangent vector. Since you given a function for your path we can parametrize it trivially

    \vec{r}(t)=t\vec{i}+f(t)\vec{j}=<t,f(t)>

    So in your case we get

    \vec{r}(t)=<t,2t^2>,t \in [0,1]

    Now recall that tangent vector is given by

    \displaystyle \vec{T}(t)=\frac{d}{dt}\vec{r}(t)=<1,4t>

    Now to normalize it we get

    \displaystyle \vec{n}=\frac{\vec{T}(t)}{|T(t)|}
    and the arc length
    ds=|\vec{T}(t)|dt=

    Combining all of this gives

    \displaystyle \int_{C}\vec{F}\cdot d\vec{s}=\int_{t_0}^{t_1}F(r(t))\cdot \frac{T(t)}{|T(t)|}|\vec{T}(t)|dt=\int_{t_0}^{t_1}  F(r(t))\cdot T(t)}dt=\int_{0}^{1} <2t^3,4t^4>\cdot <1,4t>dt
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