Gradient of a scalar field, divergence and rotational of a vector field

Gradient of a scalar field

Let $f : U \subseteq R^{3} ⟶ R$ be a scalar field and let $\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}, \frac{\partial f}{\partial z}$ be the partial derivatives of $f$ (that is, the derivative with respect to one variable maintaining the others as constants). Then, the gradient of $f$ is: $g r a d (f) = (\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}, \frac{\partial f}{\partial z})$

Let's observe that the gradient of $f$ is a vector, although $f$ is a scalar field. It is necessary to bear in mind that:

The gradient points in the direction in which the directional derivative of the function $f$ is maximum, and its module at a given point is the value of this directional derivative at this point.
It is zero in the inflection points of the function $f$ .
The gradient converts a scalar field into a vector field.

Example

$f (x, y, z) = x^{2} \cdot y - z^{3} \cdot x$

$g r a d (f) = (2 \cdot x \cdot y - z^{3}, x^{2}, - 3 \cdot z^{2} \cdot x)$
$f (x, y, z) = x \cdot \sin (y) \cdot e^{5 \cdot z}$

$g r a d (f) = (\sin y \cdot e^{5 \cdot z}, x \cdot \cos y \cdot e^{5 \cdot z}, x \cdot \sin y \cdot 5 \cdot e^{5 \cdot z})$
$f (x, y, z) = \sqrt{x^{2} + y^{2} + z^{2}}$

$g r a d (f) = (\frac{x}{\sqrt{x^{2} + y^{2} + z^{2}}}, \frac{y}{\sqrt{x^{2} + y^{2} + z^{2}}}, \frac{z}{\sqrt{x^{2} + y^{2} + z^{2}}})$

Divergence of a vector field

Let $F : U \subseteq R^{3} ⟶ R^{3}, F = (F_{1}, F_{2}, F_{3})$ be a vector field. Then, the divergence of $F$ is: $d i v (F) = \frac{\partial}{\partial x} F_{1} + \frac{\partial}{\partial y} F_{2} + \frac{\partial}{\partial z} F_{3}$

Example

$F (x, y, z) = (x^{3} \cdot y, 2 \cdot z \cdot \sin x, \cos z)$

$d i v (F) = \frac{\partial}{\partial x} (x^{3} \cdot y) + \frac{\partial}{\partial y} (2 \cdot z \cdot \sin x) + \frac{\partial}{\partial z} (\cos z) = 3 \cdot x^{2} \cdot u + 0 - \sin z$
$F (x, y, z) = (- 2 \cdot x \cdot y, y \cdot \sin z + y^{2} + z, \cos z)$

$d i v (F) = \frac{\partial}{\partial x} (- 2 \cdot x \cdot y) + \frac{\partial}{\partial y} (y \cdot \sin z + y^{2} + z) + \frac{\partial}{\partial z} (\cos z) =$

$= - 2 \cdot y + \sin z + 2 \cdot y - \sin z$

The divergence converts a vector vectorial into a scalar field.

Rotational of a vectorial field

Let $F : U \subseteq R^{3} ⟶ R^{3}, F = (F_{1}, F_{2}, F_{3})$ be a vector field. Then, the rotational of $F$ is: $r o t (F) = (\frac{\partial F_{3}}{\partial y} - \frac{\partial F_{2}}{\partial z}, \frac{\partial F_{1}}{\partial z} - \frac{\partial F_{3}}{\partial x}, \frac{\partial F_{2}}{\partial x} - \frac{\partial F_{1}}{\partial y})$ or it is also possible to calculate as the following determinant, (bearing in mind that $i, j, k$ are the coordinates to which they correspond): $| \begin{array}{ccc} i & j & k \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ F_{1} & F_{2} & F_{3} \end{array} |$

Example

$F (x, y, z) = (4 \cdot x \cdot e^{y}, x \cdot \ln z, y)$ $r o t (F) = (\frac{\partial (y)}{\partial y} - \frac{\partial (x \cdot \ln z)}{\partial z}, \frac{\partial (4 \cdot x \cdot e^{y})}{\partial z} - \frac{\partial (y)}{\partial x}, \frac{\partial (x \cdot \ln z)}{\partial x} - \frac{\partial (4 \cdot x \cdot e^{y})}{\partial y})$ $= (1 - \frac{x}{z}, 0 - 0, \ln z - 4 \cdot x \cdot e^{y})$

Properties of the gradient, divergence and rotational

If $f$ is a scalar field and $F$ a vector field, then it is always true that

$r o t (g r a d (f)) = 0$
$d i v (r o t (F)) = 0$
$r o t (f \cdot F) = g r a d (f) \times F + f \cdot r o t (f)$
$d i v (f \cdot F) = f \cdot d i v (F) + g r a d (f) \cdot F$

where $\cdot$ is the scalar product and $\times$ the vector product.