Re: Directional Derivative

$\displaystyle n$ is the normal vector. It can be pointing in or out of the sphere, whereas $\displaystyle |x|$ points out of the sphere.

Re: Directional Derivative

As far as I know, |x| is NOT a vector.

Re: Directional Derivative

**n** = x/R**i** + y/R**j** + z/R**k**

**del**f = *d*f/*d*x**i** + *d*f/*d*y**j** + *d*f/*d*z**k** = *d*f/*d*x**i**

df/dn = **del**f.**n** = (*d*f/*d*x)x/R

IxI = R only at a point where sphere crosses x axis, in which case

df/dn = df/dx, for x pos

Re: Directional Derivative

I think there is a misunderstanding here. By a function of several variables I mean explicitly

$\displaystyle f: \mathbb{R}^n \rightarrow \mathbb{R}, x \mapsto f(x), x:=(x_1,\dots,x_n)$,

so that

$\displaystyle |x|=\left(\sum_{1 \le i \le n} x_i^2\right)^{1/2}$ is the euclidean length of $\displaystyle x$.

I thought this was clear. Anyway, it's my fault. Sorry.

Re: Directional Derivative

The notation is somewhat arbitrary as long it is clear from the context what it means.

Since you said it was a directional derivative in the direction normal to the sphere, I would interpret your derivatives to mean:

$\displaystyle {\partial f \over \partial n} = \lim_{h \to 0} {f(\mathbf x + h\mathbf{n}) - f(\mathbf x) \over h}$

$\displaystyle {\partial f \over \partial |x|} = \lim_{h \to 0} {f(\mathbf x + h|x|\mathbf{\hat x}) - f(\mathbf x) \over h}$

Although it may also be possible that your vectors need to be normalized, depending on how your book defines a directional derivative.

See Directional derivative - Wikipedia, the free encyclopedia for more information on how a directional derivative is or can be defined.

Either way, those 2 forms are the same, except for a possible minus sign.

Re: Directional Derivative

OK, that was easier than I had thought!

My problem was the normal derivative of a function on $\displaystyle \mathbb{R}^n$ with radial symmetry, i.e.

$\displaystyle f(x)=g(|x|)$.

Now it's clear that for $\displaystyle n=\pm \frac{x}{|x|}$ we have

$\displaystyle \langle\nabla{f},n \rangle = \frac{\partial g}{\partial |x|} \left\langle \left( \frac{\partial |x|}{\partial x_1}, \dots, \frac{\partial |x|}{\partial x_n}\right),n\right\rangle =\pm \frac{\partial g}{\partial |x|}$

Oh my! Simple as that! (Headbang)

Re: Directional Derivative

That looks about right. :cool: