## anonymous 3 years ago I guess F is not a conservative vector field. $F(x,y)=e^xsiny\hat{i}+e^xcosy\hat{j}$ $\frac{\partial{P}}{\partial{y}}=\frac{\partial}{\partial{y}}|e^xsiny|=cosy$ $\frac{\partial{Q}}{\partial{x}}=\frac{\partial}{\partial{x}}|e^xcosy|=e^x$ $cosy\neq e^x$

1. 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?

2. anonymous

$\frac{\partial{P}}{\partial{y}}=\frac{\partial}{\partial{y}}|e^xsiny|=e^x cosy$

3. anonymous

treat the respective values as constants for partial derivatives.

4. anonymous

exactly @91

5. anonymous

$\frac{\partial{Q}}{\partial{x}}=\frac{\partial}{\partial{x}}|e^xcosy|=e^x cos y$ so there are conservative

6. anonymous

it is a conservative vector field.

7. anonymous

constant multplied to variable don't go away

8. 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

9. anonymous

only if they are by themselve

10. anonymous

Makes sense. Thank you both!