Directional Derivative

**HAL9000** · January 25th 2013, 01:39 AM

Hello!

I've run into a case, where one should calculate the directional derivative of a function $f$ of several variables at a point on a sphere of radius $R$ in the direction of a normal vector to this sphere. It is known that the function itself can be written only in terms of the length of the argument x, i.e. $|x|$ .

Why is it actually allowed to write $\frac{\partial f}{\partial n}|_{|x|=R}=\pm \frac{\partial f}{\partial |x|}|_{|x|=R}$ , with $\pm$ for the outer/inner normal vector?

**vincisonfire** · January 25th 2013, 09:56 AM

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

**HAL9000** · January 26th 2013, 09:42 AM

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

**Hartlw** · January 26th 2013, 11:01 AM

n = x/Ri + y/Rj + z/Rk
delf = df/dxi + df/dyj + df/dzk = df/dxi
df/dn = delf.n = (df/dx)x/R
IxI = R only at a point where sphere crosses x axis, in which case
df/dn = df/dx, for x pos

**HAL9000** · January 27th 2013, 01:36 AM

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

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

so that

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

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

**ILikeSerena** · January 27th 2013, 03:20 AM

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:

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

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

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.

**HAL9000** · January 27th 2013, 06:45 AM

OK, that was easier than I had thought!

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

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

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

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

Oh my! Simple as that!

**ILikeSerena** · January 27th 2013, 07:02 AM

That looks about right.

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