Field Gradient

**Apprentice123** · April 26th 2009, 07:54 AM

Determine if the field is gradient
$F(x,y) = x^2yi + 5xy^2j$

How I can determine

**Apprentice123** · April 26th 2009, 10:30 AM

for gradient $\frac{di}{dy} - \frac{dj}{dx} = 0$ ?

$\frac{di}{dy} = x^2$
$\frac{dj}{dx} = 5y^2$

different of 0, then not is gradient ???

**Calculus26** · April 26th 2009, 11:37 AM

you are correct
F is not a gradient field
there is no scalar funtion f such that F =gradf

**Apprentice123** · April 26th 2009, 11:42 AM

Thank you, the condition $\frac{di}{dy} - \frac{dj}{dx} = 0$ is correct ?

**Jhevon** · April 26th 2009, 11:44 AM

Originally Posted by Apprentice123

Thank you, the condition $\frac{di}{dy} - \frac{dj}{dx} = 0$ is correct ?

it shouldn't be i and j there, but you have the right idea, if $\bold{F} = P(x,y)i + Q(x,y)j$ , then we can find a potential function if $\frac {\partial P}{\partial y} = \frac {\partial Q}{\partial x}$

**Calculus26** · April 26th 2009, 11:45 AM

I know what you are thinking
but
If F =f i +g j

Then df/dy =dg/x is the condition not di/dx = dj/dx

**Apprentice123** · April 26th 2009, 11:58 AM

ok thanks.

To confirm learning
$F(x,y) = (x^2y)i + (5xy^2)j$

Field not is gradient (conservative) because $\frac{d(x^2y)}{dy} \not= \frac{d(5xy^2)}{dx}$

**Jhevon** · April 26th 2009, 12:00 PM

Originally Posted by Apprentice123

ok thanks.

To confirm learning
$F(x,y) = (x^2y)i + (5xy^2)j$

Field not is gradient (conservative) because $\frac{d(x^2y)}{dy} \not= \frac{d(5xy^2)}{dx}$

it should be partial derivatives: $\frac{\partial (x^2y)}{\partial y} \not= \frac{\partial (5xy^2)}{\partial x}$ , so it is not conservative

**Apprentice123** · April 26th 2009, 12:04 PM

ok thanks.

I have one problem with the same F but the question is: Field is gradient ?

It is the same as: field is conservative?

**Jhevon** · April 26th 2009, 12:06 PM

Originally Posted by Apprentice123

ok thanks.

I have one problem with the same F but the question is: Field is gradient ?

It is the same as: field is conservative?

i think you are describing the same thing, but it probably would not be stated like that. saying the field is "conservative" is saying it is the "gradient function of some vector field".

**Apprentice123** · April 26th 2009, 12:10 PM

ok Thank you

**Apprentice123** · April 26th 2009, 12:15 PM

If i have $F(x,y,z) = Pi + Qj + Rk$
The condition is rotational (curl) = 0 ?

**Jhevon** · April 26th 2009, 12:16 PM

Originally Posted by Apprentice123

If i have $F(x,y,z) = Pi + Qj + Rk$
The condition is rotational (curl)F = 0 ?

for F to be conservative, yes. which is what we've said in another one of your posts. don't ask again here

Thread: Field Gradient

LinkBack

Thread Tools

Display

Field Gradient

Similar Math Help Forum Discussions

gradient field

Condition for F to be a gradient field

Using gradient vector field to find value of f

Gradient Vector Field?

gradient field

Search Tags