Solve Divergence Question: Calculate (B [dot] \nabla)A

[Aristata]

Given two vectors, A and B:

$$A = (x\widehat{x} + 2y\widehat{y} + 3z\widehat{z})$$
$$B = (3y\widehat{x} - 2x\widehat{y})$$

I need to calculate $$(B [dot] \nabla)A$$, as part of a problem. The answer should be:

$$\widehat{x}(3y) + \widehat{y}( -4x)$$


I get:

$$(B [dot] \nabla)A = ((3y) \delta / \delta x - (2x)\delta / \delta y) (x\widehat{x} + 2y\widehat{y} + 3z\widehat{z})$$
$$=(0+0)(x\widehat{x} + 2y\widehat{y} + 3z\widehat{z})$$
$$=0$$

I don't understand what I'm doing wrong. Can someone help me out please? Thanks in advance!

 

[Coto]

Given two vectors, A and B:

I get:

$$(B [dot] \nabla)A = ((3y) \delta / \delta x - (2x)\delta / \delta y) (x\widehat{x} + 2y\widehat{y} + 3z\widehat{z})$$
$$=(0+0)(x\widehat{x} + 2y\widehat{y} + 3z\widehat{z})$$
$$=0$$

I don't understand what I'm doing wrong. Can someone help me out please? Thanks in advance!

Hey Aristata!

First what is $$B \cdot \nabla$$? Above you have written it out slightly wrong (but this may have been just a latex typo.) It should look like:

$$3y \frac{\partial}{\partial x} - 2x\frac{\partial}{\partial y}$$

I would suggest rewriting $$\vec{A}$$ as $$(x,2y,3z)$$ so you would obtain:

$$\left (3y \frac{\partial}{\partial x} - 2x\frac{\partial}{\partial y}\right ) (x,2y,3z)$$.

Now ''multiply'' through the operator as if it were a scalar acting on a vector. Similar to:

$$\lambda \vec{A} = (\lambda a_1, \lambda a_2, \lambda a_3)$$

Since $$\lambda$$ in your case is an operator, you have to perform the action of $$\lambda$$ on each of $$a_1, a_2, a_3$$. In particular take the derivatives of those elements.

Hope this wasn't too vague :).