1. ## Field Gradient

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

How I can determine

2. 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 ???

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

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

5. 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 $\frac {\partial P}{\partial y} = \frac {\partial Q}{\partial x}$

6. 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

7. ok thanks.

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

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

8. 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

9. 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?

10. 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".

11. ok Thank you

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

13. 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