## Loser66 one year ago If$$F (x,y) =<e^{2y},1+2xe^{2y}>$$ and the curve $$r(t) = <te^t, 1+t>~~~0\leq t\leq 1$$ 1) Find f such that $$F=\bigtriangledown f$$ 2) find $$F \bullet dr$$ along the given curve 3) Use $$\int_C F\bullet dr = \int_r F(r(t) r'(t)dt$$ to evalulate $$\int_C F\bullet dr$$ along the given curve Please, help

1. Loser66

The net is so bad so that I am on and off. Please, leave the guidance, I will be back right after I can access to the net.

2. IrishBoy123

for the first part, it's worth checking the field is [strictly, might be] conservative by checking out $$curl \ \vec F = 0$$, which it is then, as you know that $$\large f_x = e^{2y}$$ and $$\large f_y = 1+2xe^{2y}$$, you solve the DE's remembering that the constant of integration for the first one will be some function of y, and vice versa for the second

3. anonymous

$\frac{d \vec r}{dt}=\frac{dx}{dt}\hat i+\frac{dy}{dt}\hat j$ $\implies d \vec r=dx \hat i+dy \hat j$ $\therefore \vec F. d \vec r=F_{x}dx+F_{y}dy$

4. IrishBoy123

or for 1) as this is conservative, you can integrate along a convenient path, eg (0,0) to (0,y) to (x,y). |dw:1439405196704:dw|