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

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

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

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

  • $$f(x,y,z)= \sqrt{x^2+y^2+z^2}$$


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.