anonymous
  • anonymous
I guess F is not a conservative vector field. \[F(x,y)=e^xsiny\hat{i}+e^xcosy\hat{j}\] \[\frac{\partial{P}}{\partial{y}}=\frac{\partial}{\partial{y}}|e^xsiny|=cosy\] \[\frac{\partial{Q}}{\partial{x}}=\frac{\partial}{\partial{x}}|e^xcosy|=e^x\] \[cosy\neq e^x\]
Mathematics
  • Stacey Warren - Expert brainly.com
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chestercat
  • chestercat
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anonymous
  • anonymous
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?
anonymous
  • anonymous
\[\frac{\partial{P}}{\partial{y}}=\frac{\partial}{\partial{y}}|e^xsiny|=e^x cosy\]
anonymous
  • anonymous
treat the respective values as constants for partial derivatives.

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anonymous
  • anonymous
exactly @91
anonymous
  • anonymous
\[\frac{\partial{Q}}{\partial{x}}=\frac{\partial}{\partial{x}}|e^xcosy|=e^x cos y\] so there are conservative
anonymous
  • anonymous
it is a conservative vector field.
anonymous
  • anonymous
constant multplied to variable don't go away
anonymous
  • anonymous
Oh ok. Thanks in the previous example that I did I got \[\frac{\partial}{\partial{y}}|2x-3y|=-3\] Which lead me to believe that the constant went away
anonymous
  • anonymous
only if they are by themselve
anonymous
  • anonymous
Makes sense. Thank you both!

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