# Thread: FTC And Path Independence

1. ## FTC And Path Independence

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

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

2. Remember that

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

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

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

So in your case we get

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

Now recall that tangent vector is given by

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

Now to normalize it we get

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

Combining all of this gives

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