# Derivatives of vectors

1. Apr 11, 2014

### Jhenrique

What means:

?

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

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

2. Apr 11, 2014

### Staff: Mentor

3. Apr 12, 2014

### Jhenrique

4. Apr 12, 2014

### Staff: Mentor

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.

5. Apr 12, 2014

### 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 $\vec{\nabla}_\hat{u} \vec{v}$ ?

6. Apr 12, 2014

### Staff: Mentor

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,
$$\hat{\phi}\cdot\vec{\nabla}\hat{r}$$

7. Apr 12, 2014

### Jhenrique

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:

8. Apr 12, 2014

### Staff: Mentor

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

9. Apr 13, 2014

### Jhenrique

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" ?

10. Apr 13, 2014

### Staff: Mentor

11. Apr 13, 2014

### Staff: Mentor

12. Apr 13, 2014

### Jhenrique

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?

Last edited: Apr 13, 2014
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