Proving the Properties of Curl

[SithsNGiggles]

Homework Statement

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.

 

[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.

 

[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

 

[lanedance]

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

 

[SithsNGiggles]

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

div(hf) = hdiv(f) + gradh·f,

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

 

[Dick]

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

div(hf) = hdiv(f) + gradh·f,

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

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

 

[SithsNGiggles]

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.

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?

 

[Dick]

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?

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

 

[SithsNGiggles]

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

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

 

[Dick]

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

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.

 

[SithsNGiggles]

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