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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

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  1. Loser66
    • one year ago
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    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
    • one year ago
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    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
    • one year ago
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    \[\frac{d \vec r}{dt}=\frac{dx}{dt}\hat i+\frac{dy}{dt}\hat j\] \[\implies d \vec r=dx \hat i+dy \hat j\] \[\therefore \vec F. d \vec r=F_{x}dx+F_{y}dy\]

  4. IrishBoy123
    • one year ago
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    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|

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