Answer :
g(x,y) = e^xsin(y)
then
f(x,y) = del (g) = d/dx(g) i + d/dy(g) j
it does work out
then
f(x,y) = del (g) = d/dx(g) i + d/dy(g) j
it does work out
In R² vector field F is conservative if:
δM/δy = δN/δx
Where F = M×i + N×j
The solution is
δM/δy = δN/δx ⇒ eˣ×cosy = eˣ×cosy The vector field is conservative
The Potential Function is:
U = - eˣ×cosy +eˣ×siny
The vector field f = eˣ×siny i + eˣ×cosy j
δM/δy = δN/δx ⇒ eˣ×cosy = eˣ×cosy
Then the vector field is conservative
Could be expressed as : f = grd×U
To find the potential function:
δM/δy = eˣ×siny ⇒ ∫dM = ∫eˣ×siny×dy ⇒ M = - eˣ×cosy + g(y)
δN /δx = eˣ×cosy ⇒ g´(y) = eˣ×cosy g(y) = eˣ×siny
U = - eˣ×cosy +eˣ×siny
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