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I guess F is not a conservative vector field.
\[F(x,y)=e^xsiny\hat{i}+e^xcosy\hat{j}\]
\[\frac{\partial{P}}{\partial{y}}=\frac{\partial}{\partial{y}}e^xsiny=cosy\]
\[\frac{\partial{Q}}{\partial{x}}=\frac{\partial}{\partial{x}}e^xcosy=e^x\]
\[cosy\neq e^x\]
 one year ago
 one year ago
I guess F is not a conservative vector field. \[F(x,y)=e^xsiny\hat{i}+e^xcosy\hat{j}\] \[\frac{\partial{P}}{\partial{y}}=\frac{\partial}{\partial{y}}e^xsiny=cosy\] \[\frac{\partial{Q}}{\partial{x}}=\frac{\partial}{\partial{x}}e^xcosy=e^x\] \[cosy\neq e^x\]
 one year ago
 one year ago

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SpacelimbusBest ResponseYou've already chosen the best response.1
The second one, I am not sure what you defined as which, but just to make sure. You are taking the partial derivative of e^x cosy ? with respect to x?
 one year ago

91Best ResponseYou've already chosen the best response.2
\[\frac{\partial{P}}{\partial{y}}=\frac{\partial}{\partial{y}}e^xsiny=e^x cosy\]
 one year ago

SpacelimbusBest ResponseYou've already chosen the best response.1
treat the respective values as constants for partial derivatives.
 one year ago

91Best ResponseYou've already chosen the best response.2
\[\frac{\partial{Q}}{\partial{x}}=\frac{\partial}{\partial{x}}e^xcosy=e^x cos y\] so there are conservative
 one year ago

SpacelimbusBest ResponseYou've already chosen the best response.1
it is a conservative vector field.
 one year ago

91Best ResponseYou've already chosen the best response.2
constant multplied to variable don't go away
 one year ago

MathSofiyaBest ResponseYou've already chosen the best response.0
Oh ok. Thanks in the previous example that I did I got \[\frac{\partial}{\partial{y}}2x3y=3\] Which lead me to believe that the constant went away
 one year ago

91Best ResponseYou've already chosen the best response.2
only if they are by themselve
 one year ago

MathSofiyaBest ResponseYou've already chosen the best response.0
Makes sense. Thank you both!
 one year ago
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