Calculating grad/curl/div of a vector?

Question

I'm trying to do some practice for an electromagnetism course, and am trying to calculate the grad, curl and div of a vector $A = (2xy, 3zx, yx^2)$

I know:

Div = $\nabla . A$

Curl = $\nabla \times A$

Grad = $\nabla A$

I have worked out the first two, but I seem to be having a mind blank and I'm getting myself all confused with the Grad component if someone could clarify this?

Also if someone could explain when I might need to use the Laplace Operator? What is it used for?

You can only use the gradient on scalar functions, not on vector fields. — dinosaur, Feb 21 '14 at 12:49
Okay that's why I'm getting confused then! Any chance you could give me an example? — Sarah Jayne, Feb 21 '14 at 12:50
@DavidH yes, but I don't think that this is what Sarah had in mind here. — dinosaur, Feb 21 '14 at 14:06
@SarahJayne for example consider $f(x,y)=x^2+y^2$. This is a scalar function and you can calculcate $\nabla f(x,y)=\left(\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}\right) = (2x,2y)$. — dinosaur, Feb 21 '14 at 14:07
Oh okay, so you apply it to a function! Great thank you so much for clearing that up for me :-) — Sarah Jayne, Feb 21 '14 at 15:05

score 1 · Answer 1 · edited Apr 13 '17 at 12:20

This might help.

You don't take the gradient of a vector field, so $\nabla A$ makes no sense. Instead, the gradient is an operator that acts on scalar functions in $\mathbb{R}^n$ and produces a vector field on $\mathbb R^n$.

Now if $f(x,y)$ is a scalar function, then $\nabla f:=\langle f_x,f_y\rangle$.

As for the Laplacian operator, it is given by $\Delta=\nabla\cdot \nabla$, i.e. the divergence of the gradient. So for example, with the $f(x,y)$ above we would have $$\Delta f:=\nabla \cdot \nabla f=\nabla \cdot \langle f_x,f_y\rangle=f_{xx}+f_{yy},$$ and similarly for higher dimensions.

The Laplacian operator is an extremely important and ubiquitous operator throughout pure and applied mathematics, certainly in areas of PDE and physics.

A simple application of why the Laplacian operator is so central is because many important problems in applied mathematics and physics can be distilled down to this:

Suppose $\mathbf F$ is a given conservative vector field (meaning $\mathbf F=\nabla f$ for some scalar function $f$) which is also divergence-free (meaning $\nabla\cdot\mathbf F=0$). How do we find the $f$ (called a potential function) associated with this vector field $\mathbf F$?

The answer is quickly revealed: $$\mathbf F=\nabla f \implies \nabla \cdot\mathbf F=\nabla\cdot \nabla f\implies 0=\Delta f.$$ The potential $f$ is a solution of Laplace's equation! (In fact, this is why some books refer to it as the potential equation.

Calculating grad/curl/div of a vector?

1 Answers1