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Finding Curl(F) of a vector-valued function

  1. Jun 24, 2014 #1

    JJBladester

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    Gold Member

    1. The problem statement, all variables and given/known data

    Determine the curl of the vector function below.

    [tex]\boldsymbol{F}\left ( x,y,z \right )=3x^2\boldsymbol{i}+7e^xy\boldsymbol{j}[/tex]

    2. Relevant equations

    curl[itex]\mathbf{F}=\mathbf{\nabla}\times \mathbf{F}[/itex]
    [tex]=\begin{vmatrix}
    \mathbf{i}& \mathbf{j}& \mathbf{k}\\
    \frac{\partial}{\partial x}& \frac{\partial}{\partial y}& \frac{\partial}{\partial z}\\
    P(x,y,z)& Q(x,y,z)& R(x,y,z)
    \end{vmatrix}[/tex]

    3. The attempt at a solution

    This problem is solved by my FE review as below. I understand how to solve for the determinant of a 3x3 matrix by rewriting the first two columns to the right of the matrix and obtaining three "+" terms and three "-" terms. I think my partials for x and y are off, perhaps.

    [tex]\mathbf{i}\left ( \frac{\partial }{\partial y}0-\frac{\partial }{\partial z}7e^xy \right )-\mathbf{j}\left ( \frac{\partial }{\partial x}0-\frac{\partial }{\partial z}3x^2 \right )+\mathbf{k}\left ( \frac{\partial }{\partial x}7e^xy-\frac{\partial }{\partial y}3x^2 \right )[/tex]

    [tex]=\mathbf{i}(0-0)-\mathbf{j}(0-0)-\mathbf{k}\left ( 7e^xy-0 \right )=7e^xy\mathbf{k}[/tex]

    The expressions I calculated for [itex]\mathbf{i} and \mathbf{j}[/itex] match what the book has. However, my expression for [itex]\mathbf{k}[/itex] seems to be incorrect. Here is what I calculated for the values of the matrix:

    [tex]\frac{\partial }{\partial x}=6x[/tex]
    [tex]\frac{\partial }{\partial y}=7e^x[/tex]
    [tex]\frac{\partial }{\partial z}=0[/tex]
    [tex]P(x,y,z)=3x^2[/tex]
    [tex]Q(x,y,z)=7e^xy[/tex]
    [tex]R(x,y,z)=0[/tex]

    So my expression for [itex]\mathbf{k}[/itex] was:

    [tex]\left [(6x)7e^xy-\left ( 7e^x \right )3x^2 \right ]\mathbf{k}[/tex]

    I think I went wrong with my calculation of [itex] \frac{\partial }{\partial x}[/itex].
     
  2. jcsd
  3. Jun 24, 2014 #2

    lurflurf

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    Homework Helper

    It is unclear what exactly you have done.
    My guess is that you found
    $$\dfrac{\partial}{\partial x}3\, x^2=6\, x$$
    when you should have found
    $$\dfrac{\partial}{\partial x}7 \, e^x \, y=7 \, e^x \, y$$
    also you are multiplying functions together for some reason

    Look at the solution, of the six terms 5 are clearly zero so only the sixth remains.
     
  4. Jun 24, 2014 #3

    JJBladester

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    Gold Member

    This is why I love physicsforums. You answered me and helped me understand the problem within 20 minutes of my post.

    You're right, I was multiplying the functions when I should not have been. Normally when working with matrices containing only constants, you *do* just multiply elements together. So that seems not to be the case when you have partial derivatives as elements.
     
  5. Jun 24, 2014 #4

    SteamKing

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    Staff Emeritus
    Science Advisor
    Homework Helper

    For the curl, it's helpful to think of the partial derivatives in the second row as operators to be applied to the functions P, Q, and R when the determinant is expanded. Expressing the curl in terms of a determinant is just a handy way to remember how to calculate its various components.

    Taking the i-component of the curl, for example, expanding the determinant gives:

    Ry - Qz, where the subscripts y and z indicate the variable w.r.t. which the partial derivative is taken.
     
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