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 :).