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Homework Help: Conservative force

  1. Apr 22, 2010 #1
    1. The problem statement, all variables and given/known data
    show that the force [tex]F(x,y) = (x^{2}+3y+11)\widehat{x} + (3x +5y^{3}+11)\widehat{y}[/tex] is conservative


    2. Relevant equations
    it's conservative if [tex]\nabla X F = 0[/tex]


    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. jcsd
  3. Apr 22, 2010 #2

    gabbagabbahey

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    [itex]\mathbf{\nabla}\times\textbf{F}[/itex] represents the curl of [itex]\textbf{F}[/itex], 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.
     
  4. Apr 22, 2010 #3
    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
     
  5. Apr 22, 2010 #4

    gabbagabbahey

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    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.
     
  6. Apr 22, 2010 #5
    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
    [tex]\widehat{x}, \widehat{y}[/tex]
    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.
     
  7. Apr 22, 2010 #6

    gabbagabbahey

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    No, it isn't. The gradient of a vector, like [itex]\textbf{F}[/itex], 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 [itex]\textbf{F}[/itex] is the cross product of the vector differential operator [itex]\mathbf{\nabla}=\hat{\mathbf{x}}\frac{\partial}{\partial x}+\hat{\mathbf{y}}\frac{\partial}{\partial y}+\hat{\mathbf{z}}\frac{\partial}{\partial z}[/itex] (often called the "Del operator" or "nabla operator") with [itex]\textbf{F}[/itex]. It can be represented by the following determinant:

    [tex]\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}}[/tex]


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