1. ## Vector field

1) How can I know if the field is gradient?

2) How can I know if the fiels is conservative ?

3) How can I find the function potential of vector field ?

2. Originally Posted by Apprentice123
1) How can I know if the field is gradient?

2) How can I know if the fiels is conservative ?

3) How can I find the function potential of vector field ?
1) If $\displaystyle {\bf F} = \nabla f$ for some $\displaystyle f.$
2) If $\displaystyle \nabla \times {\bf F} = 0$
3) Set $\displaystyle \nabla f = {\bf F}$ and solve for $\displaystyle f$.

3. if I have $\displaystyle F(x,y) = (e^xcosy)i + (-e^xseny)j$

what will be: $\displaystyle \nabla$ and $\displaystyle f$

4. Originally Posted by Apprentice123
if I have $\displaystyle F(x,y) = (e^xcosy)i + (-e^xseny)j$

what will be: $\displaystyle \nabla$ and $\displaystyle f$
Since you only have a two dimensional vector field, say

$\displaystyle {\bf F} = <P,Q>$

then this would be conservative if $\displaystyle P_y = Q_x$ and if $\displaystyle P = e^x \cos y$ and $\displaystyle Q = - e^x \sin y$ then $\displaystyle P_y = -e^x \sin y$ and $\displaystyle Q_x = -e^x \sin y$ which are identical. So the vector field is conservative and an $\displaystyle f$exists such that

$\displaystyle f_x = P = e^x \cos y$ and $\displaystyle f_y = Q = - e^x\sin y$ which gives $\displaystyle f = e^x \cos y + c$

5. for the field is conservative $\displaystyle \frac {di}{dy} - \frac{dj}{dx} = 0$?

field conservative = field gradient ?

If$\displaystyle f_x = P = e^x \cos y$ and $\displaystyle f_y = Q = - e^x\sin y$ why $\displaystyle f = e^x \cos y + c$ ???

6. Originally Posted by Apprentice123
for the field is conservative $\displaystyle \frac {di}{dy} - \frac{dj}{dx} = 0$?

field conservative = field gradient ?

If $\displaystyle f_x = P = e^x \cos y$ and $\displaystyle f_y = Q = - e^x\sin y$ why $\displaystyle f = e^x \cos y + c$ ???

First since the vector field is conservative you know an $\displaystyle f$ exists, i.e. $\displaystyle f$ exists such that $\displaystyle f_x = P = e^x \cos y \;{\bf and} \;f_y = Q = - e^x\sin y$

If $\displaystyle f_x = e^x \cos y$ then integrating wrt to $\displaystyle x$ gives $\displaystyle f = e^x \cos y + g(y)$. Now subs. into $\displaystyle f_y = - e^x\sin y$ which shows that $\displaystyle - e^x\sin y + g'(y) = - e^x\sin y$ which gives $\displaystyle g'(y) = 0$ which gives $\displaystyle g(y)=c$, thus our $\displaystyle f.$

7. Thank you I understand.

And the statements are correct?

for that the field is conservative i need have => $\displaystyle \frac {di}{dy} - \frac{dj}{dx} = 0$

and
field conservative = field gradient ?