0
$\begingroup$

Let $F:\mathbb{R}^3\to\mathbb{R}^3$ be of class $C^2$ and let $a, b\in\mathbb{R}^3$ be two nonnull vectors. Let $F(x) = F(x_1, x_2, x_3)$.

Consider the vector given by $$\left(\sum_{i=1}^3 \frac{\partial F_1}{\partial x_i} a_i, \sum_{i=1}^3 \frac{\partial F_2}{\partial x_i} a_i, \sum_{i=1}^3 \frac{\partial F_3}{\partial x_i} a_i \right).$$

Denoting by $\widehat{b}$ the unit vector in the direction of the vector $b$, I am trying to understand what $$\left(\sum_{i=1}^3 \frac{\partial F_1}{\partial x_i} a_i, \sum_{i=1}^3 \frac{\partial F_2}{\partial x_i} a_i, \sum_{i=1}^3 \frac{\partial F_3}{\partial x_i} a_i \right)\cdot\widehat{b}$$ does represent.

If, e.g. $b$ is a vector in the direction of the $y$-axis, I would say that $$\left(\sum_{i=1}^3 \frac{\partial F_1}{\partial x_i} a_i, \sum_{i=1}^3 \frac{\partial F_2}{\partial x_i} a_i, \sum_{i=1}^3 \frac{\partial F_3}{\partial x_i} a_i \right)\cdot\widehat{b} = \sum_{i=1}^3 \frac{\partial F_2}{\partial x_i} a_i,$$ but I am not sure what does that scalar product represent in general (I mean, $b$ could be in an "easy" direction, but also in the $x+y$ direction and others more complicated).

Could someone please help me in understanding that? Is that related with directional derivative?

Thank you in advance.

$\endgroup$

1 Answer 1

1
$\begingroup$

You can take $\hat b$ inside the derivative:

$$ \left(\sum_{i=1}^3 \frac{\partial F_1}{\partial x_i} a_i, \sum_{i=1}^3 \frac{\partial F_2}{\partial x_i} a_i, \sum_{i=1}^3 \frac{\partial F_3}{\partial x_i} a_i \right)\cdot\widehat{b}=\sum_{i=1}^3\frac{\partial(F\cdot\hat b)}{\partial x_i}a_i\;. $$

So this is the directional derivative along $a$ of the scalar function $F\cdot\hat b$, the component of $F$ along $\hat b$.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .