The Curl is confusing me, just the determinant

[flyingpig]

Homework Statement

Find the curl of the vector field

\mathbf{F} = <xyz,0,-x^2 y>

The Attempt at a Solution

I am mostly just having problems with computing the determinant. I could just go with crossing the first row and first column. But i noticed that the intermediate step

\begin{vmatrix}<br /> \frac{\partial }{\partial y} &amp; \frac{\partial }{\partial z}\\ <br /> 0&amp; -x^2 y<br /> \end{vmatrix}\hat{i}

Like when we compute the 2 x 2 determinant, we multiply the elements together. But my question is that why we do take granted that multiplying the partial with respect to y with -x²y is equal to taking the partial of -x²y with respect to y?

Also, I tried crossing out from the bottom, where the 0 is to make the determinant easier and I got a different answer. I also tried using the Rule of Sarrus for this 3 x 3 matrix and I also got a different answer.

The one with the Rule of Sarrus is a bit long to type out, so i will only show the one where I crossed with the 0 (I think it's called the co-factor expansion or something)

So crossing the bottom row and first column I get

\begin{vmatrix}<br /> \hat{i} &amp; \hat{j} &amp;\hat{k} \\ <br /> \frac{\partial }{\partial x} &amp;\frac{\partial }{\partial y} &amp; \frac{\partial }{\partial z}\\ <br /> xyz &amp; 0 &amp; -x^2 y<br /> \end{vmatrix} = xyz\begin{vmatrix} \hat{j} &amp; \hat{k}\\\frac{\partial }{\partial y} &amp; \frac{\partial }{\partial z}\end{vmatrix} - x^2 y \begin{vmatrix} \hat{i}&amp;\hat{j}\\\frac{\partial }{\partial x}&amp;\frac{\partial }{\partial y}\end{vmatrix} = xyz\left(\frac{\partial\hat{j} }{\partial z}- \frac{\partial \hat{k}}{\partial y}\right) -x^2 y \left ( \frac{\partial \hat{i}}{\partial y}- \frac{\partial \hat{j}}{\partial x} \right )

The Solution

[PLAIN]http://img641.imageshack.us/img641/1914/unledzw.jpg

 

[olivermsun]

But my question is that why we do take granted that multiplying the partial with respect to y with -x²y is equal to taking the partial of -x²y with respect to y?

It's just a notation -- the operator is applied to the vector components (bottom row) and are collected with the directions i, j, k (first row).

So, keeping that in mind, you can expand along the third row like this:
\begin{vmatrix}<br /> \hat{i} &amp; \hat{j} &amp;\hat{k} \\ <br /> \frac{\partial }{\partial x} &amp;\frac{\partial }{\partial y} &amp; \frac{\partial }{\partial z}\\ <br /> xyz &amp; 0 &amp; -x^2 y<br /> \end{vmatrix} <br /> = \begin{vmatrix} \hat{j} &amp; \hat{k}\\\frac{\partial }{\partial y} &amp; \frac{\partial }{\partial z}\end{vmatrix} xyz<br /> - \begin{vmatrix} \hat{i}&amp;\hat{j}\\\frac{\partial }{\partial x}&amp;\frac{\partial }{\partial y}\end{vmatrix} x^2 y <br /> = \left(\hat{j} \frac{\partial }{\partial z} - \hat{k}\frac{\partial }{\partial y} \right) xyz<br /> - \left(\hat{i} \frac{\partial }{\partial y} - \hat{j}\frac{\partial }{\partial x} \right) x^2 y \\<br /> = -x^2 \hat{i} + 3xy \hat{j} -xz \hat{k}<br />

(Often it's easier to collect by going along the first row just so you get i,j,k terms together).

 

[flyingpig]

I still confused.

1. Is there even a meaning to taking the partial derivative to those "hat" vectors? (at first I thought they were just one and so the derivative is 0, but that was just stupid and I couldn't come up with a good explanation why that is wrong)

2. In \left(\hat{j} \frac{\partial }{\partial z} - \hat{k}\frac{\partial }{\partial y} \right) xyz<br /> - \left(\hat{i} \frac{\partial }{\partial y} - \hat{j}\frac{\partial }{\partial x} \right) x^2 y \\<br /> = -x^2 \hat{i} + 3xy \hat{j} -xz \hat{k}, aren't you still just "distributing" the xyz and x²y into the brackets? Isn't that just multiplying and not really taking the derivative?

 

[SammyS]

It's implied that you take the derivative.

 

[olivermsun]

1. The hat vectors are constant unit vectors which point along the axes: i = (1,0,0), j = (0,1,0), k = (0,0,1). It's just easier to write i, j, k than writing out the vectors. So if you take derivatives they just go outside, e.g., d(2f(x))/dx = 2 df(x)/dx. That still leaves the operator d/dx.

2. The operator notation means "take the derivative." So if I write d/dx f(x) that means take the derivative of f(x) with respect to x, or the same as d/dx f(x) = df(x)/dx.

3. I think you're seeing that "determinant" formula for the curl has some pitfalls. You're right in noticing that, for example, f(x) d/dx should be different from d/dx f(x). So I guess it's best to remember that this is a way to remember the curl in Cartesian coordinates, but it's not properly a determinant. If there's ever confusion about the order, just expand along i, j, k first.

 

[flyingpig]

1)\frac{\partial }{\partial y}\hat{i} = \frac{\partial }{\partial y}&lt;1,0,0&gt; = 0

How does that go "outside"?

 

[Char. Limit]

And this is why I don't like using the determinant method to teach curl.

 

[SammyS]

The derivative of a constant is zero.

The derivative of the product of a constant and a function is the constant times the derivative of the function --- as usual.

 

[olivermsun]

And this is why I don't like using the determinant method to teach curl.

The other ways don't seem any easier to remember, though.

 

[Char. Limit]

The other ways don't seem any easier to remember, though.

This is true. However, they are also not an abuse of notation.

 

[hunt_mat]

The curl is in your question is calculated as:

<br /> \nabla\times\mathbf{F}=\left|<br /> \begin{array}{ccc}<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 /> xyz &amp; 0 &amp; -x^{2}y<br /> \end{array}\right|<br />

This is now expanded in the usual way:

<br /> \nabla\times\mathbf{F} =\left|<br /> \begin{array}{cc}<br /> \frac{\partial}{\partial y} &amp; \frac{\partial}{\partial z} \\<br /> 0 &amp; -x^{2}y <br /> \end{array}\right|\mathbf{i}-\left|<br /> \begin{array}{cc}<br /> \frac{\partial}{\partial x} &amp; \frac{\partial}{\partial z} \\<br /> xyz &amp; -x^{2}y <br /> \end{array}\right|\mathbf{j}+\left|<br /> \begin{array}{cc}<br /> \frac{\partial}{\partial x} &amp; \frac{\partial}{\partial y} \\<br /> xyz &amp; 0<br /> \end{array}\right|\mathbf{k}<br />

The first is just expanded as:

<br /> \left|<br /> \begin{array}{cc}<br /> \frac{\partial}{\partial y} &amp; \frac{\partial}{\partial z} \\<br /> 0 &amp; -x^{2}y <br /> \end{array}\right| =\frac{\partial}{\partial y}(-x^{2}y)-\frac{\partial}{\partial z}0=-x^{2}<br />

 

[hunt_mat]

And this is why I don't like using the determinant method to teach curl.

What method do you use then? I'm interested.

 

[Char. Limit]

What method do you use then? I'm interested.

Setting F = <F_x, F_y, F_z>, the curl of F is defined (in Cartesian coordinates) as:

curl F = \left(\frac{\partial F_z}{\partial y} - \frac{\partial F_y}{\partial z}\right) \hat{i} + \left(\frac{\partial F_x}{\partial z} - \frac{\partial F_z}{\partial x}\right) \hat{j} + \left(\frac{\partial F_y}{\partial x} - \frac{\partial F_x}{\partial y}\right) \hat{k}

It's easy to remember when you look at the order of the first terms in each area. zy xz yx. And then the second term, you just flip the two.