# Conservative force

1. Apr 22, 2010

### KaiserBrandon

1. The problem statement, all variables and given/known data
show that the force $$F(x,y) = (x^{2}+3y+11)\widehat{x} + (3x +5y^{3}+11)\widehat{y}$$ is conservative

2. Relevant equations
it's conservative if $$\nabla X F = 0$$

3. The attempt at a solution
ok, I know how to take the gradient of a function like F(x,y) = x^2 + 3xy + 3 + y, but I'm not sure how to take the gradient of the function for this question. I've tried many things, including taking the partial derivative of x within the brackets next to the x unit vector, and the partial derivative of y within the brackets next to the y unit vector.

2. Apr 22, 2010

### gabbagabbahey

$\mathbf{\nabla}\times\textbf{F}$ represents the curl of $\textbf{F}$, not the gradient. I'm sure your textbook covers how to calculate a curl, so I suggest you open it up and read that section.

3. Apr 22, 2010

### KaiserBrandon

yes I know it represents the curl. I know how to find the curl if I have the gradient. But I'm stuck on the part where you have to find the gradient

4. Apr 22, 2010

### gabbagabbahey

Gradient and curl are two very different types of derivatives. The gradient takes a scalar function as input and outputs a vector function. The curl takes a vector function as input and outputs a vector function. You do not calculate the curl by first calculating the gradient.

You seem very confused on how to calculate the curl of a vector field, so again, I recommend you open your textbook and read the section on curls.

5. Apr 22, 2010

### KaiserBrandon

the curl is a cross product between the gradient of the force and the force itself. The textbook only tells me how to find the curl given a force in the form of say F(x,y) = x^2 + y^2 + 2, where the finding the gradient is straightforward, and so is using that to find the curl. however, for this question, the force is given in terms of
$$\widehat{x}, \widehat{y}$$
I don't know how to do it using a force given in this way, and my textbook nor the class notes have any information either. And I can't find anything on the internet either.

6. Apr 22, 2010

### gabbagabbahey

No, it isn't. The gradient of a vector, like $\textbf{F}$, would be a second rank tensor (or matrix)...how exactly would you compute the cross product of a tensor/matrix with a vector?

The curl of $\textbf{F}$ is the cross product of the vector differential operator $\mathbf{\nabla}=\hat{\mathbf{x}}\frac{\partial}{\partial x}+\hat{\mathbf{y}}\frac{\partial}{\partial y}+\hat{\mathbf{z}}\frac{\partial}{\partial z}$ (often called the "Del operator" or "nabla operator") with $\textbf{F}$. It can be represented by the following determinant:

$$\mathbf{\nabla}\times\textbf{F}=\begin{vmatrix}\hat{\mathbf{x}} & \hat{\mathbf{y}} & \hat{\mathbf{z}} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ F_x & F_y & F_z\end{vmatrix}=\left( \frac{\partial F_z}{\partial y}- \frac{\partial F_y}{\partial z}\right)\hat{\mathbf{x}}+\left( \frac{\partial F_x}{\partial z}- \frac{\partial F_z}{\partial x}\right)\hat{\mathbf{y}}+\left( \frac{\partial F_y}{\partial x}- \frac{\partial F_x}{\partial y}\right)\hat{\mathbf{z}}$$

I can't believe that. What textbook are you using?