Vector Derivatives Explained: Definition and Examples

[Jhenrique]

What means:

?

This guy, $\vec{\nabla}_{\hat{\phi}} \hat{r}$, for example, means:

$$\\ \hat{\phi}\cdot\vec{\nabla}\hat{r} = \begin{bmatrix} \phi _1 \\ \phi _2 \\ \end{bmatrix} \begin{bmatrix} \frac{\partial r_1}{\partial x} & \frac{\partial r_1}{\partial y} \\ \frac{\partial r_2}{\partial x} & \frac{\partial r_2}{\partial y} \\ \end{bmatrix}$$
?

 

[jedishrfu]

Its known as the diretional derivative:

http://en.wikipedia.org/wiki/Directional_derivative

and at Mathworld:

http://mathworld.wolfram.com/DirectionalDerivative.html

 

[Jhenrique]

Its known as the diretional derivative:

http://en.wikipedia.org/wiki/Directional_derivative

and at Mathworld:

http://mathworld.wolfram.com/DirectionalDerivative.html

But my ask, you no answered...

 

[Mark44]

Your question was not clear. Were you asking what each of the equations in the first image means? Or were you asking what the following means?
$$\\ \hat{\phi}\cdot\vec{\nabla}\hat{r} = \begin{bmatrix}
\phi _1 \\
\phi _2 \\
\end{bmatrix}
\begin{bmatrix}
\frac{\partial r_1}{\partial x} & \frac{\partial r_1}{\partial y} \\
\frac{\partial r_2}{\partial x} & \frac{\partial r_2}{\partial y} \\
\end{bmatrix}$$
BTW, that matrix product looks flaky to me. On the left side of your equation you have a dot product, which should result in a scalar, but on the right side, you have a 2X1 matrix multiplying a 2 X 2 matrix. That multiplication is not defined.

If you want a clear, concise answer, limit your questions to one per post.

 

[Jhenrique]

Wrong my, the correct would be $$\begin{bmatrix} \phi _1 & \phi _2 \\ \end{bmatrix} \begin{bmatrix} \frac{\partial r_1}{\partial x} & \frac{\partial r_1}{\partial y} \\ \frac{\partial r_2}{\partial x} & \frac{\partial r_2}{\partial y} \\ \end{bmatrix}$$
Anyway, this multiplication is just a tentative of answer my own ask (my unique ask), that is: what means $\vec{\nabla}_\hat{u} \vec{v}$ ?

 

[Mark44]

Wrong my, the correct would be $$\begin{bmatrix} \phi _1 & \phi _2 \\ \end{bmatrix} \begin{bmatrix} \frac{\partial r_1}{\partial x} & \frac{\partial r_1}{\partial y} \\ \frac{\partial r_2}{\partial x} & \frac{\partial r_2}{\partial y} \\ \end{bmatrix}$$

Well, at least the matrix product is now defined. But another problem is that I don't see how this can be equal to the dot product you showed in the OP; namely,
$$\hat{\phi}\cdot\vec{\nabla}\hat{r} $$

Anyway, this multiplication is just a tentative of answer my own ask (my unique ask), that is: what means $\vec{\nabla}_\hat{u} \vec{v}$ ?

 

[Jhenrique]

Well, at least the matrix product is now defined. But another problem is that I don't see how this can be equal to the dot product you showed in the OP; namely,
$$\hat{\phi}\cdot\vec{\nabla}\hat{r} $$

You already understood that I'm lost and that all my equations no pass of speculation. I will not argue anymore about it.

The question is clear:

what means $\vec{\nabla}_\hat{u} \vec{v}$ ?

 

[jedishrfu]

Its a directional derivative of the v vector function in the u direction

 

[Jhenrique]

Its a directional derivative of the v vector function in the u direction

ok, given, u = (u₁, u₂) , v = (v₁, v₂) and ∇ = (∂x, ∂y), how you'll operate u, v and ∇ for get the "directional derivative of the v vector function in the u direction" ?

 

[jedishrfu]

Check this out:

http://tutorial.math.lamar.edu/Classes/CalcIII/DirectionalDeriv.aspx

and here's a concrete example:

http://mathinsight.org/directional_derivative_gradient_examples

 

[Mark44]

This might be what you're looking for - http://en.wikipedia.org/wiki/Normal_derivative#Normal_derivative

Scroll down to this section - Derivatives of vector valued functions of vectors.

 

[Jhenrique]

This might be what you're looking for - http://en.wikipedia.org/wiki/Normal_derivative#Normal_derivative

Scroll down to this section - Derivatives of vector valued functions of vectors.

In other words, $\frac{\partial \vec{f}}{\partial \vec{v}} \cdot \vec{u}$ is the same thing that $$\\ \ \vec{\nabla}\vec{f}\cdot \vec{u} =
\begin{bmatrix}
\frac{\partial f_1}{\partial v_1} & \frac{\partial f_1}{\partial v_2} \\
\frac{\partial f_2}{\partial v_1} & \frac{\partial f_2}{\partial v_2} \\
\end{bmatrix}
\begin{bmatrix}
u_1 \\
u_2 \\
\end{bmatrix} = \vec{\nabla}_{\vec{u}}\vec{f}
$$

(assuming that v = (v₁, v₂) = r = (x, y))

Like I say in the principle... correct?