Vector Calculus

**ramdrop** · February 14th 2011, 02:25 AM

This is really starting to irritate me as I can't seem to get anywhere with it:

I need to solve somthing similar to a trigonometric identity but using the definition of grad,div and curl to verify:

$grad(UV) = U grad(V) + Vgrad(U)$

and..

$grad(F.G) = F(curl G) + G(curl F) + (f.grad) G + (G.grad)F$

I have no idea where to begin, I have the definitions but I fail to see how they can help me, can anybody can help me?

**Prove It** · February 14th 2011, 02:38 AM

The first looks to be a simple application of the product rule...

**HallsofIvy** · February 14th 2011, 03:17 AM

If you know the definitions, why not use them to do what is shown on both sides? For example, you should know, from the definition, that $grad(UV)= \frac{\partial UV}{\partial x}\vec{i}+ \frac{\partial UV}{\partial y}\vec{j}+ \frac{\partial UV}{\partial z}\vec{k}$ . Now use the product rule as Prove It suggested.

And be careful how you write the second one. You have one "f" where you mean "F" but more importantly you have written " $F.grad$ " and " $G.grad$ " where I think you really mean $F\cdot\nabla$ and $G\cdot\nabla$ .

**ramdrop** · February 14th 2011, 05:16 AM

Originally Posted by Prove It

The first looks to be a simple application of the product rule...

Maybe I forgot to write in my post, It's obviously going to be the product rule. Yo ucould say that where its normally "f" it's now UV

Originally Posted by HallsofIvy

And be careful how you write the second one. You have one "f" where you mean "F" but more importantly you have written " $F.grad$ " and " $G.grad$ " where I think you really mean $F\cdot\nabla$ and $G\cdot\nabla$ .

That is what the question reads in my tutorial exercises.

**ramdrop** · February 15th 2011, 12:51 AM

Okay, I have done the first question.. finally.

After checking my tutorial exercise sheets, I was wrong and its:

$grad(F.G) = F(curl G) + G(curl F) + (F.grad)G + (G.grad)F$

Still have no idea what to do, I'm thinking I need to prove the identity for the i component and generalize for j and k component?

**Haven** · February 15th 2011, 01:36 AM

Do you mean that
$\nabla(F\cdot G) = F(\nabla \times G) + G(\nabla \times F)+ (F\cdot \nabla G) + (G\cdot \nabla F)$ , where $F$ and $G$ are scalar functions?

if so, we note that for the i component of the LHS is $\frac{\partial}{\partial x}F_iG_i = (\frac{\partial}{\partial x}F_i)G_i + (\frac{\partial}{\partial x}G_i)F_i$ . Where $F_i$ denotes the i component of $F$

Now using the formulas for curl and gradient, we may calculate the RHS and simplify it in order to get it to look like the LHS.

Note the ith component of $\nabla \times F$ is $\frac{\partial}{\partial y}F_k - \frac{\partial}{\partial z}F_j$

**ramdrop** · February 15th 2011, 01:52 AM

Yes I mean that, but that makes absolutely no sense to me.. at the moment I'm plugging in the "definitions" that I know into the RHS.

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