Vector field

**Apprentice123** · April 26th 2009, 09:10 AM

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 ?

**Danny** · April 26th 2009, 10:33 AM

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 ${\bf F} = \nabla f$ for some $f.$
2) If $\nabla \times {\bf F} = 0$
3) Set $\nabla f = {\bf F}$ and solve for $f$ .

**Apprentice123** · April 26th 2009, 10:42 AM

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

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

**Danny** · April 26th 2009, 10:54 AM

Originally Posted by Apprentice123

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

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

Since you only have a two dimensional vector field, say

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

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

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

**Apprentice123** · April 26th 2009, 11:03 AM

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

field conservative = field gradient ?

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

**Danny** · April 26th 2009, 11:13 AM

Originally Posted by Apprentice123

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

field conservative = field gradient ?

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

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

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

**Apprentice123** · April 26th 2009, 11:20 AM

Thank you I understand.

And the statements are correct?

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

and

field conservative = field gradient ?

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