Does anyone know some links (exclude wiki) that explain the "Directional derivative of a vector field". I've been googling but found none, and i have a task that deals with it but have no literature.

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- Feb 1st 2008, 06:36 AMPinskyDirectional derivative of a vector field
Does anyone know some links (exclude wiki) that explain the "Directional derivative of a vector field". I've been googling but found none, and i have a task that deals with it but have no literature.

- Feb 1st 2008, 07:31 AMtopsquark
- Feb 1st 2008, 07:52 AMPlato
These may help you. They are from a text by Keisler (it is free to download from his website at the Unuversity of Wisconsin).

- Feb 1st 2008, 11:54 AMPinsky
Determine the derivate of a vector field $\displaystyle \vec{a}=yz\vec{i}+xz\vec{j}+xy\vec{k}$

in the direction of vector $\displaystyle \vec{p}=(1,1,1)$

Any chance somebody can solve this with a step-by-step explanation what is done and why? - Feb 1st 2008, 12:41 PMPlato
It seems to me as you may have confused two different concepts.

First, you began this thread by asking about directional derivatives. They apply to functionals: functions from a vector space to the reals.

Then you now appear to be asking vector value functions or vector fields.

There is the concept of the*divergence*in the field. But we don’t find divergence in a particular direction.

Rather divergence is direction at a particular point in the field.

So what exactly are you asking? As put the current question is meaningless.

If $\displaystyle \vec{a}=yz\vec{i}+xz\vec{j}+xy\vec{k}$ is a vector field, do you mean to ask for the divergence at**the point**$\displaystyle (1,1,1)$ - Feb 1st 2008, 03:17 PMPinsky
I have the solution of the task, but i don't know what it means. Perhaps it's divergence and i messed things up. Here it is, maybe you can figure out something out of it.

The text goes:

Find the derivate of a vector field $\displaystyle \vec{a}=yz\vec{i}+xz\vec{j}+xy\vec{k}$ in direction of the vector $\displaystyle \vec{p}=(1,1,1)$ in the point $\displaystyle T(1,1,1)$.

Solution:

$\displaystyle \frac{\partial \vec{a}}{\partial \vec{p}}=(\frac{\partial \vec{a_1}}{\partial \vec{p}},\frac{\partial \vec{a_2}}{\partial \vec{p}},\frac{\partial \vec{a_3}}{\partial \vec{p}})$

$\displaystyle \frac{\partial \vec{a_1}}{\partial \vec{p}}=\nabla a_1 \frac{\vec{p}}{p}=(0,z,y)\frac{(1,1,1)}{\sqrt{3}}= (0+z+y)\frac {1}{\sqrt{3}}=\frac{z+y}{\sqrt{3}}$

$\displaystyle \frac{\partial \vec{a_2}}{\partial \vec{p}}=\nabla a_2 \frac{\vec{p}}{p}=\frac{z+x}{\sqrt{3}}$

$\displaystyle \frac{\partial \vec{a_3}}{\partial \vec{p}}=\nabla a_3 \frac{\vec{p}}{p}=\frac{y+x}{\sqrt{3}}$

$\displaystyle |\frac{\partial \vec{a}}{\partial \vec{p}}|=|\frac{2}{\sqrt{3}},\frac{2}{\sqrt{3}},\ frac{2}{\sqrt{3}}|$ - Feb 1st 2008, 05:20 PMtopsquark
Ah! If I'm right it's something akin to what I have come to call a "vector gradient." (With all due apologies, I don't know what the actual name of it is.)

When the gradient of a scalar function $\displaystyle \vec{\nabla}s$ is taken it returns a vector. But you can define a version of this that acts on a vector as operating on each component of the vector individually. $\displaystyle \vec{\nabla} \vec{v} = \left < \vec{\nabla}v_x , \vec{\nabla}v_y , \vec{\nabla}v_z \right >$.

I've seen it mentioned in my Electrodynamics text, but have never actually used the thing.

-Dan