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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
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

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Loser66
 one year ago
Best ResponseYou've already chosen the best response.0The 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.

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.1for 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

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[\frac{d \vec r}{dt}=\frac{dx}{dt}\hat i+\frac{dy}{dt}\hat j\] \[\implies d \vec r=dx \hat i+dy \hat j\] \[\therefore \vec F. d \vec r=F_{x}dx+F_{y}dy\]

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.1or 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
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