Directional derivative of a vector field

**Pinsky** · February 1st 2008, 06:36 AM

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.

**topsquark** · February 1st 2008, 07:31 AM

Originally Posted by Pinsky

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.

How about here, or here, or perhaps here though I don't know if this last applies to what you are doing.

(shrugs) Google is working fine for me.

-Dan

**Plato** · February 1st 2008, 07:52 AM

These may help you. They are from a text by Keisler (it is free to download from his website at the Unuversity of Wisconsin).

**Pinsky** · February 1st 2008, 11:54 AM

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

Any chance somebody can solve this with a step-by-step explanation what is done and why?

**Plato** · February 1st 2008, 12:41 PM

Originally Posted by Pinsky

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

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 $\vec{a}=yz\vec{i}+xz\vec{j}+xy\vec{k}$ is a vector field, do you mean to ask for the divergence at the point $(1,1,1)$

**Pinsky** · February 1st 2008, 03:17 PM

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

Solution:

$\frac{\partial \vec{a}}{\partial \vec{p}}=(\frac{\partial \vec{a_1}}{\partial \vec{p}},\frac{\partial \vec{a_2}}{\partial \vec{p}},\frac{\partial \vec{a_3}}{\partial \vec{p}})$
$\frac{\partial \vec{a_1}}{\partial \vec{p}}=\nabla a_1 \frac{\vec{p}}{p}=(0,z,y)\frac{(1,1,1)}{\sqrt{3}}= (0+z+y)\frac {1}{\sqrt{3}}=\frac{z+y}{\sqrt{3}}$
$\frac{\partial \vec{a_2}}{\partial \vec{p}}=\nabla a_2 \frac{\vec{p}}{p}=\frac{z+x}{\sqrt{3}}$
$\frac{\partial \vec{a_3}}{\partial \vec{p}}=\nabla a_3 \frac{\vec{p}}{p}=\frac{y+x}{\sqrt{3}}$

$|\frac{\partial \vec{a}}{\partial \vec{p}}|=|\frac{2}{\sqrt{3}},\frac{2}{\sqrt{3}},\ frac{2}{\sqrt{3}}|$

**topsquark** · February 1st 2008, 05:20 PM

Originally Posted by Pinsky

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

Solution:

$\frac{\partial \vec{a}}{\partial \vec{p}}=(\frac{\partial \vec{a_1}}{\partial \vec{p}},\frac{\partial \vec{a_2}}{\partial \vec{p}},\frac{\partial \vec{a_3}}{\partial \vec{p}})$
$\frac{\partial \vec{a_1}}{\partial \vec{p}}=\nabla a_1 \frac{\vec{p}}{p}=(0,z,y)\frac{(1,1,1)}{\sqrt{3}}= (0+z+y)\frac {1}{\sqrt{3}}=\frac{z+y}{\sqrt{3}}$
$\frac{\partial \vec{a_2}}{\partial \vec{p}}=\nabla a_2 \frac{\vec{p}}{p}=\frac{z+x}{\sqrt{3}}$
$\frac{\partial \vec{a_3}}{\partial \vec{p}}=\nabla a_3 \frac{\vec{p}}{p}=\frac{y+x}{\sqrt{3}}$

$|\frac{\partial \vec{a}}{\partial \vec{p}}|=|\frac{2}{\sqrt{3}},\frac{2}{\sqrt{3}},\ frac{2}{\sqrt{3}}|$

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 $\vec{\nabla}s$ 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. $\vec{\nabla} \vec{v} = \left < \vec{\nabla}v_x , \vec{\nabla}v_y , \vec{\nabla}v_z \right >$ .

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

-Dan

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