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MathSofiya

  • 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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  1. Spacelimbus
    • 2 years ago
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    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?

  2. 91
    • 2 years ago
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    \[\frac{\partial{P}}{\partial{y}}=\frac{\partial}{\partial{y}}|e^xsiny|=e^x cosy\]

  3. Spacelimbus
    • 2 years ago
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    treat the respective values as constants for partial derivatives.

  4. Spacelimbus
    • 2 years ago
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    exactly @91

  5. 91
    • 2 years ago
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    \[\frac{\partial{Q}}{\partial{x}}=\frac{\partial}{\partial{x}}|e^xcosy|=e^x cos y\] so there are conservative

  6. Spacelimbus
    • 2 years ago
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    it is a conservative vector field.

  7. 91
    • 2 years ago
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    constant multplied to variable don't go away

  8. MathSofiya
    • 2 years ago
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    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

  9. 91
    • 2 years ago
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    only if they are by themselve

  10. MathSofiya
    • 2 years ago
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    Makes sense. Thank you both!

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