Using nabla with partial derivatives and the Laplace operation $\partial_x^2+\partial_y^2+\partial_z^2$

Source of the problem p.812 here. Suppose

$$\bar{F}(x,y,z)=(xy-z^2)\bar{i}+(xyz)\bar{j}+(x-y^2-z^2)\bar{k}.$$

I am concerned where I need to nabla an unit vector for example with

$$\triangledown \times F.$$

I am particularly uncertain with $\partial_{x}\bar{k}$ (where $\bar{k}$ is an unit vector) -- is it zero or $\bar{k}\partial_{x}$?

This scan reuses the same logic and if the latter is wrong, it will have many errors (particularly in the middle section)

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Needs more prose. – Henning Makholm Mar 20 '12 at 15:19
Why do you think $\partial_x\bar i$ shouldn't be zero? It's the derivative of a constant vector, just like $\partial_x\bar j$; derivatives of constants are zero. – joriki Mar 20 '12 at 15:33
@joriki: I mean $x||\hat{i}$ so $\partial_{x}\bar{i}:=\partial_{x}\hat{i}=1$. Because $\hat{j}\not ||\bar{x}$ and $\hat{j}||\bar{y}$, $\partial_{y}\hat{i}=0$. – hhh Mar 20 '12 at 15:35
@hhh: The vectors $\bf i,j,k$ do not depend on $x$ do they? I.e. are constant? It doesn't matter what direction a constant vector is in relative to anything else; the derivative of a constant is always zero. – anon Mar 20 '12 at 15:37

In line three, you should have $x-2$ in the third component rather than $-2$.

In line four, you computed the curl of your answer in line three (which is $\rm \nabla(\nabla\cdot F)$); instead you should have been computing the curl of the original function $\rm F$, as in

$$\nabla\times \mathrm{F}=\begin{vmatrix} \color{Red}{\mathbf{i}} & \color{Red}{\mathbf{j}} & \color{Red}{\mathbf{k}} \\ \color{Blue}{\partial_x} & \color{Blue}{\partial_y} & \color{Blue}{\partial_z} \\ \color{Green}{xy-z^2} & \color{Green}{xyz} & \color{Green}{x-y^2-z^2} \end{vmatrix}$$

$$=\big(\color{Blue}{\partial_y}(\color{Green}{x-y^2-z^2})-\color{Blue}{\partial_z}(\color{Green}{xyz})\big)\color{Red}{\mathbf{i}}-\big(\color{Blue}{\partial_x}(\color{Green}{x-y^2-z^2})-\color{Blue}{\partial_z}(\color{Green}{xy-z^2})\big)\color{Red}{\mathbf{j}}+\big(\color{Blue}{\partial_x}(\color{Green}{xyz})-\color{Blue}{\partial_y}(\color{Green}{xy-z^2})\big)\color{Red}{\mathbf{k}}$$

$$=-(x+2)y\mathbf{i}-(1+2z)\mathbf{j}+(yz-x)\mathbf{k}.$$

As a general note, since $\partial_x$ denotes partial differentiation with respect to $x$, and $\mathbf k$ is a constant vector, if we encountered $\partial_x\mathbf{k}$ it would be $(\frac{\partial}{\partial x}1)\mathbf{k}=\mathbf{0}.$

The Rule of Sarrus is invalid when you are mixing operators and vectors and scalar components together in a matrix. For each term in the expansion, you must put the differentiation operator to the left of the component so that it can act on it. Above this is depicted as blue-left-of-green.

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...what about if there was a line $z\partial_z\left( \bar{k} \right)$? Is it $0$? – hhh Mar 20 '12 at 15:45
$\bar{k}:=\hat{k}$, $|\hat{k}|=1$ and $\hat{k}||\bar{z}$, it is the 1-length vector in the basis along the $z$ -axis. – hhh Mar 20 '12 at 15:51
@hhh: The Rule of Sarrus (as I see it on WP) is invalid when you mix operators and vectors and scalars together. You must put the operators left of what they act on; just imagine doing the rule but taking all of the lines top-down instead (since $\mathbf{i,j,k}$ are constants they can be put inside the products in any order). – anon Mar 20 '12 at 16:02

The question is not whether $\bar k$ is a unit vector but whether it depends on $x$. From your scanned page, it seems that it's one of the canonical basis vectors and thus doesn't depend on $x$; thus its derivative with respect to $x$ vanishes.

The main problem in your calculation seems to be that you write $\nabla\times \overline F$ but then go on to calculate not $\nabla\times \overline F$ but $\nabla\times (\nabla(\nabla\cdot\overline F)).$

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