Homework Help: Proving properties of curl

1. Jan 25, 2012

SithsNGiggles

1. The problem statement, all variables and given/known data
The curl satisfies

(A) curl(f+g) = curl(f) + curl(g)

(B) if h is real values, then curl(hf) = hcurl(f) + h'·f

(C) if f is C2, then curl(gradf) = 0

Show that (B) holds.

2. The attempt at a solution
I'm not quite sure how to interpret the "h is real valued" part. Does it mean that, as is, h is a scalar quantity and its derivative is a vector?

Any help is appreciated.

- - - -
So far I've been reviewing the definition of curl.
I let f = f(x, y, z) = (f1(x, y, z), f2(x, y, z), f3(x, y, z))

hf = (hf1(x, y, z), hf2(x, y, z), hf3(x, y, z))

curl(hf) = (D2(hf3) - D3(hf2), D3(hf1) - D1(hf3), D1(hf2) - D2(hf1)),

but I don't see where this route is taking me.

Last edited: Jan 25, 2012
2. Jan 25, 2012

DrewD

Probably means h is a real valued function. Something like
$h(x_1,x_2...)\vec{f}(x_1,x_2...)$
so the values of h are scalars, but it is not necessarily constant.

3. Jan 25, 2012

lanedance

yeah I would probably take it as real scalar function of position h = h(x,y,z), multiplying anything else by a vector does not make sense unless it is a vector product

4. Jan 25, 2012

lanedance

also the fact that h' is in a dot product means it is probably the gradient of h

5. Jan 25, 2012

SithsNGiggles

I looked at the problems ahead, and there's a similar one involving divergence, where I have to show that

so I'm not sure if h' is the gradient in the curl question.

6. Jan 25, 2012

Dick

In my reference the h'.f term is the cross product of grad(h) with f.

7. Jan 25, 2012

SithsNGiggles

I ended up continuing with process and was able to simplify the expression to

curl(hf) = h(D2f3 - D3f2, D3f1 - D1f3, D1f2 - D2f1) + ((D2h)f3 - D3h)f2, (D3h)f1 - D1h)f3, (D1h)f2 - D2h)f1)

curl(hf) = h curl(f) + ((D2h)f3 - (D3h)f2, (D3h)f1 - (D1h)f3, (D1h)f2 - (D2h)f1)

Is there any way to simplify the rest of this expression?

8. Jan 25, 2012

Dick

The rest of it looks like grad(h) x f. Where 'x' is the vector cross product.

9. Jan 25, 2012

SithsNGiggles

I suppose there's no way of expressing "grad(h) x f" as "h' f", is there?

10. Jan 25, 2012

Dick

I have no idea what "h' f" is supposed to mean. It's no standard notation I've ever seen. I wouldn't torture yourself over it. You've got the vector formula right.

11. Jan 25, 2012

SithsNGiggles

Good enough, I suppose. Thanks everyone for the help.