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

Gradient of a scalar field

Let $f: U\subseteq \mathbb{R}^3 \longrightarrow \mathbb{R}$ be a scalar field and let $\dfrac{\partial f}{\partial x}, \dfrac{\partial f}{\partial y}, \dfrac{\partial f}{\partial z}$ 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: $$grad(f)=(\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}, \frac{\partial f}{\partial z})$$

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.

• $f(x,y,z)=x^2 \cdot y- z^3 \cdot x$

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

• $f(x,y,z)=x \cdot \sin(y) \cdot e^{5\cdot z}$

$grad(f)=(\sin y \cdot e^{5\cdot z}, x \cdot \cos y \cdot e^{5\cdot z}, x \cdot \sin y \cdot 5 \cdot e^{5\cdot z})$

• $f(x,y,z)= \sqrt{x^2+y^2+z^2}$

$grad(f)=\Big(\dfrac{x}{\sqrt{x^2+y^2+z^2}},\dfrac{y}{\sqrt{x^2+y^2+z^2}},\dfrac{z}{\sqrt{x^2+y^2+z^2}}\Big)$

Divergence of a vector field

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

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

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

• $F(x,y,z)=(-2 \cdot x \cdot y, y \cdot \sin z+y^2+z, \cos z)$

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

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

The divergence converts a vector vectorial into a scalar field.

Rotational of a vectorial field

Let $F: U \subseteq \mathbb{R}^3 \longrightarrow \mathbb{R}^3 ,F= (F_{1}, F_{2}, F_{3})$ be a vector field. Then, the rotational of $F$ is: $$rot(F)=\Big(\frac{\partial F_{3}}{\partial y}- \frac{\partial F_{2}}{\partial z}, \frac{\partial F_{1}}{\partial z}- \frac{\partial F_{3}}{\partial x},\frac{\partial F_{2}}{\partial x}-\frac{\partial F_{1}}{\partial y} \Big)$$ 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): $$\left| \begin{array}{ccc} i & j & k \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ F_{1} & F_{2} & F_{3} \end{array}\right |$$

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

Properties of the gradient, divergence and rotational

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

1. $rot (grad (f))=0$
2. $div (rot (F))=0$
3. $rot (f \cdot F )=grad (f) \times F + f \cdot rot (f)$
4. $div(f \cdot F) = f \cdot div(F) +grad (f) \cdot F$

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