Finding Curl(F) of a vector-valued function

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SUMMARY

The discussion focuses on calculating the curl of the vector function \(\boldsymbol{F}(x,y,z) = 3x^2\boldsymbol{i} + 7e^xy\boldsymbol{j}\) using the curl formula \(\nabla \times \boldsymbol{F}\). The user initially miscalculated the partial derivatives, particularly for the \(\mathbf{k}\) component, leading to confusion in the determinant expansion. Correctly applying the determinant method reveals that the only non-zero term arises from the sixth term in the expansion, confirming the importance of treating partial derivatives as operators rather than simple multipliers.

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  • Experience with vector-valued functions and their components.
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Homework Statement



Determine the curl of the vector function below.

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

Homework Equations



curl\mathbf{F}=\mathbf{\nabla}\times \mathbf{F}
=\begin{vmatrix}<br /> \mathbf{i}&amp; \mathbf{j}&amp; \mathbf{k}\\ <br /> \frac{\partial}{\partial x}&amp; \frac{\partial}{\partial y}&amp; \frac{\partial}{\partial z}\\ <br /> P(x,y,z)&amp; Q(x,y,z)&amp; R(x,y,z)<br /> \end{vmatrix}

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.

\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 )

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

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

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

So my expression for \mathbf{k} was:

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

I think I went wrong with my calculation of \frac{\partial }{\partial x}.
 
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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.
 
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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.
 
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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