Esercizi sui Campi Vettoriali

Esercizi sui Campi Vettoriali

Esercizi sui Campi Vettoriali

Versione italiana

Esercizi sui Campi Vettoriali

Concetti Chiave

Un campo vettoriale è una funzione che associa a ogni punto dello spazio un vettore. I campi vettoriali sono utilizzati per descrivere fenomeni fisici come il campo elettrico, il campo magnetico e il campo di velocità di un fluido.

Definizione

Un campo vettoriale \mathbf{F}$\mathbf{F}$ in uno spazio tridimensionale può essere rappresentato come:

\mathbf{F}(x, y, z) = (F_x(x, y, z), F_y(x, y, z), F_z(x, y, z)) $\mathbf{F}(x, y, z) = (F_x(x, y, z), F_y(x, y, z), F_z(x, y, z))$

dove F_x, F_y, F_z$F_x, F_y, F_z$ sono le componenti del campo vettoriale.

Tipi di Campi Vettoriali

1. Campi Vettoriali Conservativi: Un campo è conservativo se il lavoro fatto lungo un percorso chiuso è zero. Un campo conservativo può essere espresso come il gradiente di una funzione scalare \phi$\phi$:
   \mathbf{F} = -\nabla \phi $\mathbf{F} = -\nabla \phi$

2. Campi Vettoriali Irrotazionali: Un campo è irrotazionale se il suo rotore è zero:
   \nabla \times \mathbf{F} = 0 $\nabla \times \mathbf{F} = 0$

3. Campi Vettoriali Divergenti: La divergenza di un campo vettoriale misura la tendenza del campo a "divergere" da un punto. È definita come:
   \nabla \cdot \mathbf{F} = \frac{\partial F_x}{\partial x} + \frac{\partial F_y}{\partial y} + \frac{\partial F_z}{\partial z} $\nabla \cdot \mathbf{F} = \frac{\partial F_x}{\partial x} + \frac{\partial F_y}{\partial y} + \frac{\partial F_z}{\partial z}$

Esercizi

Esercizio 1: Calcolo della Divergenza

Calcola la divergenza del campo vettoriale \mathbf{F} = (2x, 3y^2, z^3)$\mathbf{F} = (2x, 3y^2, z^3)$.

Soluzione:

Utilizziamo la formula per la divergenza:

\nabla \cdot \mathbf{F} = \frac{\partial F_x}{\partial x} + \frac{\partial F_y}{\partial y} + \frac{\partial F_z}{\partial z} $\nabla \cdot \mathbf{F} = \frac{\partial F_x}{\partial x} + \frac{\partial F_y}{\partial y} + \frac{\partial F_z}{\partial z}$

Calcoliamo ciascun termine:

1. \frac{\partial F_x}{\partial x} = \frac{\partial (2x)}{\partial x} = 2$\frac{\partial F_x}{\partial x} = \frac{\partial (2x)}{\partial x} = 2$
2. \frac{\partial F_y}{\partial y} = \frac{\partial (3y^2)}{\partial y} = 6y$\frac{\partial F_y}{\partial y} = \frac{\partial (3y^2)}{\partial y} = 6y$
3. \frac{\partial F_z}{\partial z} = \frac{\partial (z^3)}{\partial z} = 3z^2$\frac{\partial F_z}{\partial z} = \frac{\partial (z^3)}{\partial z} = 3z^2$

Quindi, la divergenza è:

\nabla \cdot \mathbf{F} = 2 + 6y + 3z^2 $\nabla \cdot \mathbf{F} = 2 + 6y + 3z^2$

Esercizio 2: Verifica se un Campo è Conservativo

Verifica se il campo vettoriale \mathbf{F} = (y^2, 2xy, 0)$\mathbf{F} = (y^2, 2xy, 0)$ è conservativo.

Soluzione:

Per verificare se un campo è conservativo, calcoliamo il rotore:

\nabla \times \mathbf{F} = \left( \frac{\partial F_z}{\partial y} - \frac{\partial F_y}{\partial z}, \frac{\partial F_x}{\partial z} - \frac{\partial F_z}{\partial x}, \frac{\partial F_y}{\partial x} - \frac{\partial F_x}{\partial y} \right) $\nabla \times \mathbf{F} = \left( \frac{\partial F_z}{\partial y} - \frac{\partial F_y}{\partial z}, \frac{\partial F_x}{\partial z} - \frac{\partial F_z}{\partial x}, \frac{\partial F_y}{\partial x} - \frac{\partial F_x}{\partial y} \right)$

Calcoliamo ciascun componente:

1. Primo componente:
   \frac{\partial F_z}{\partial y} - \frac{\partial F_y}{\partial z} = 0 - 0 = 0 $\frac{\partial F_z}{\partial y} - \frac{\partial F_y}{\partial z} = 0 - 0 = 0$

2. Secondo componente:
   \frac{\partial F_x}{\partial z} - \frac{\partial F_z}{\partial x} = 0 - 0 = 0 $\frac{\partial F_x}{\partial z} - \frac{\partial F_z}{\partial x} = 0 - 0 = 0$

3. Terzo componente:
   \frac{\partial F_y}{\partial x} - \frac{\partial F_x}{\partial y} = 2y - 2y = 0 $\frac{\partial F_y}{\partial x} - \frac{\partial F_x}{\partial y} = 2y - 2y = 0$

Poiché il rotore è zero:

\nabla \times \mathbf{F} = (0, 0, 0) $\nabla \times \mathbf{F} = (0, 0, 0)$

Il campo vettoriale \mathbf{F} = (y^2, 2xy, 0)$\mathbf{F} = (y^2, 2xy, 0)$ è quindi conservativo.

Esercizio 3: Calcolo del Rotore

Calcola il rotore del campo vettoriale \mathbf{F} = (z, x, y)$\mathbf{F} = (z, x, y)$.

Soluzione:

Utilizziamo la formula per il rotore:

\nabla \times \mathbf{F} = \left( \frac{\partial F_z}{\partial y} - \frac{\partial F_y}{\partial z}, \frac{\partial F_x}{\partial z} - \frac{\partial F_z}{\partial x}, \frac{\partial F_y}{\partial x} - \frac{\partial F_x}{\partial y} \right) $\nabla \times \mathbf{F} = \left( \frac{\partial F_z}{\partial y} - \frac{\partial F_y}{\partial z}, \frac{\partial F_x}{\partial z} - \frac{\partial F_z}{\partial x}, \frac{\partial F_y}{\partial x} - \frac{\partial F_x}{\partial y} \right)$

Calcoliamo ciascun componente:

1. Primo componente:
   \frac{\partial F_z}{\partial y} - \frac{\partial F_y}{\partial z} = \frac{\partial z}{\partial y} - \frac{\partial x}{\partial z} = 0 - 0 = 0 $\frac{\partial F_z}{\partial y} - \frac{\partial F_y}{\partial z} = \frac{\partial z}{\partial y} - \frac{\partial x}{\partial z} = 0 - 0 = 0$

2. Secondo componente:
   \frac{\partial F_x}{\partial z} - \frac{\partial F_z}{\partial x} = \frac{\partial z}{\partial z} - \frac{\partial y}{\partial x} = 1 - 0 = 1 $\frac{\partial F_x}{\partial z} - \frac{\partial F_z}{\partial x} = \frac{\partial z}{\partial z} - \frac{\partial y}{\partial x} = 1 - 0 = 1$

3. Terzo componente:
   \frac{\partial F_y}{\partial x} - \frac{\partial F_x}{\partial y} = \frac{\partial x}{\partial x} - \frac{\partial z}{\partial y} = 1 - 0 = 1 $\frac{\partial F_y}{\partial x} - \frac{\partial F_x}{\partial y} = \frac{\partial x}{\partial x} - \frac{\partial z}{\partial y} = 1 - 0 = 1$

Quindi, il rotore del campo vettoriale è:

\nabla \times \mathbf{F} = (0, 1, 1) $\nabla \times \mathbf{F} = (0, 1, 1)$

English version

Vector Field Exercises

Key Concepts

A vector field is a function that associates a vector to each point in space. Vector fields are used to describe physical phenomena such as the electric field, the magnetic field, and the velocity field of a fluid.

Definition

A vector field \mathbf{F}$\mathbf{F}$ in a three-dimensional space can be represented as:

\mathbf{F}(x, y, z) = (F_x(x, y, z), F_y(x, y, z), F_z(x, y, z)) $\mathbf{F}(x, y, z) = (F_x(x, y, z), F_y(x, y, z), F_z(x, y, z))$

where F_x, F_y, F_z$F_x, F_y, F_z$ are the components of the vector field.

Types of Vector Fields

1. Conservative Vector Fields: A field is conservative if the work done along a closed path is zero. A conservative field can be expressed as the gradient of a scalar function \phi$\phi$:
   \mathbf{F} = -\nabla \phi $\mathbf{F} = -\nabla \phi$

2. Irrotational Vector Fields: A field is irrotational if its curl is zero:
   \nabla \times \mathbf{F} = 0 $\nabla \times \mathbf{F} = 0$

3. Divergent Vector Fields:
   The divergence of a vector field measures the tendency of the field to "diverge" from a point. It is defined as: \nabla \cdot \mathbf{F} = \frac{\partial F_x}{\partial x} + \frac{\partial F_y}{\partial y} + \frac{\partial F_z}{\partial z}$\nabla \cdot \mathbf{F} = \frac{\partial F_x}{\partial x} + \frac{\partial F_y}{\partial y} + \frac{\partial F_z}{\partial z}$

Exercises

Exercise 1:

Calculation of Divergence Calculate the divergence of the vector field \mathbf{F} = (2x, 3y^2, z^3)$\mathbf{F} = (2x, 3y^2, z^3)$.

Solution:
We use the divergence formula: \nabla \cdot \mathbf{F} = \frac{\partial F_x}{\partial x} + \frac{\partial F_y}{\partial y} + \frac{\partial F_z}{\partial z}$\nabla \cdot \mathbf{F} = \frac{\partial F_x}{\partial x} + \frac{\partial F_y}{\partial y} + \frac{\partial F_z}{\partial z}$
We calculate each term:

1. \frac{\partial F_x}{\partial x} = \frac{\partial (2x) }{\partial x} = 2$\frac{\partial F_x}{\partial x} = \frac{\partial (2x) }{\partial x} = 2$
2. \frac{\partial F_y}{\partial y} = \frac{\partial (3y^2)}{\partial y} = 6y$\frac{\partial F_y}{\partial y} = \frac{\partial (3y^2)}{\partial y} = 6y$
3. \frac{\partial F_z}{\partial z} = \frac{\partial (z^3)}{\partial z} = 3z^2$\frac{\partial F_z}{\partial z} = \frac{\partial (z^3)}{\partial z} = 3z^2$

So, the divergence is:

\nabla \cdot \mathbf{F} = 2 + 6y + 3z^2 $\nabla \cdot \mathbf{F} = 2 + 6y + 3z^2$

Exercise 2: Check if a Field is Conservative

Check if the vector field \mathbf{F} = (y^2, 2xy, 0)$\mathbf{F} = (y^2, 2xy, 0)$ is conservative.

Solution:
To check whether a field is conservative, we calculate the rotor:
\nabla \times \mathbf{F} = \left( \frac{\partial F_z}{\partial y} - \frac{\partial F_y}{\partial z}, \frac{\partial F_x}{\partial z} - \frac{\partial F_z}{\partial x}, \frac{\partial F_y}{\partial x } - \frac{\partial F_x}{\partial y} \right)$\nabla \times \mathbf{F} = \left( \frac{\partial F_z}{\partial y} - \frac{\partial F_y}{\partial z}, \frac{\partial F_x}{\partial z} - \frac{\partial F_z}{\partial x}, \frac{\partial F_y}{\partial x } - \frac{\partial F_x}{\partial y} \right)$
Let's calculate each component:

1. First component: \frac{\partial F_z}{\partial y} - \frac{\partial F_y}{\partial z} = 0 - 0 = 0$\frac{\partial F_z}{\partial y} - \frac{\partial F_y}{\partial z} = 0 - 0 = 0$
2. Second component:
   \frac{\partial F_x}{\partial z} - \frac{\partial F_z}{\partial x} = 0 - 0 = 0$\frac{\partial F_x}{\partial z} - \frac{\partial F_z}{\partial x} = 0 - 0 = 0$
3. Third component:
   \frac{\partial F_y}{\partial x} - \frac{\partial F_x}{\partial y} = 2y - 2y = 0$\frac{\partial F_y}{\partial x} - \frac{\partial F_x}{\partial y} = 2y - 2y = 0$

Since the rotor is zero: \nabla \times \mathbf{F} = (0, 0, 0)$\nabla \times \mathbf{F} = (0, 0, 0)$ The vector field \mathbf{F} = (y^2, 2xy, 0)$\mathbf{F} = (y^2, 2xy, 0)$ is therefore conservative.

Exercise 3: Calculating the Rotor Calculate the rotor of the vector field \mathbf{F} = (z, x, y)$\mathbf{F} = (z, x, y)$.

Solution:
We use the formula for the rotor:
\nabla \times \mathbf{F} = \left( \frac{\partial F_z}{\partial y} - \frac{\partial F_y}{\partial z}, \frac{\partial F_x}{\partial z} - \frac{\partial F_z}{\partial x}, \frac{\partial F_y}{\partial x} - \frac{\partial F_x}{\partial y} \right) $\nabla \times \mathbf{F} = \left( \frac{\partial F_z}{\partial y} - \frac{\partial F_y}{\partial z}, \frac{\partial F_x}{\partial z} - \frac{\partial F_z}{\partial x}, \frac{\partial F_y}{\partial x} - \frac{\partial F_x}{\partial y} \right)$
Let's calculate each component:

1. First component:
   \frac{\partial F_z}{\partial y} - \frac{\partial F_y}{\partial z} = \frac{\partial z}{\partial y} - \frac{\partial x}{\partial z} = 0 - 0 = 0$\frac{\partial F_z}{\partial y} - \frac{\partial F_y}{\partial z} = \frac{\partial z}{\partial y} - \frac{\partial x}{\partial z} = 0 - 0 = 0$
2. Second component:
   \frac{\partial F_x}{\partial z} - \frac{\partial F_z}{\partial x} = \frac{\partial z}{\partial z} - \frac{\partial y}{\partial x} = 1 - 0 = 1$\frac{\partial F_x}{\partial z} - \frac{\partial F_z}{\partial x} = \frac{\partial z}{\partial z} - \frac{\partial y}{\partial x} = 1 - 0 = 1$
3. Third component:
   \frac{\partial F_y}{ \partial x} - \frac{\partial F_x}{\partial y} = \frac{\partial x}{\partial x} - \frac{\partial z}{\partial y} = 1 - 0 = 1$\frac{\partial F_y}{ \partial x} - \frac{\partial F_x}{\partial y} = \frac{\partial x}{\partial x} - \frac{\partial z}{\partial y} = 1 - 0 = 1$
   So, the rotor of the vector field is: \nabla \times \mathbf{F} = (0, 1, 1)$\nabla \times \mathbf{F} = (0, 1, 1)$