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

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Spacelimbus
 2 years ago
Best ResponseYou've already chosen the best response.1The 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?

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

Spacelimbus
 2 years ago
Best ResponseYou've already chosen the best response.1treat the respective values as constants for partial derivatives.

91
 2 years ago
Best 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

Spacelimbus
 2 years ago
Best ResponseYou've already chosen the best response.1it is a conservative vector field.

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

MathSofiya
 2 years ago
Best ResponseYou've already chosen the best response.0Oh 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

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

MathSofiya
 2 years ago
Best ResponseYou've already chosen the best response.0Makes sense. Thank you both!
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