Finding Curl(F) of a vector-valued function

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The discussion focuses on calculating the curl of the vector function F(x, y, z) = 3x^2i + 7e^xyj. Participants clarify the use of the determinant method for finding the curl, emphasizing that the partial derivatives should be treated as operators rather than multiplied directly. A common mistake identified is incorrectly calculating the partial derivatives, particularly for the k-component of the curl. The correct approach involves recognizing that many terms will be zero, simplifying the calculation. The conversation highlights the importance of understanding the application of partial derivatives in vector calculus.
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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.
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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