Question
si \( \mathbf{F}(x, y, z)=(x+2) \mathbf{i}+(y+2) \mathbf{j}+(z+2) \mathbf{k} \) ¿cuál es la divergencia \( \nabla . F \) ?
Ask by Campos Reed. in Mexico
Mar 10,2025
Upstudy AI Solution
Tutor-Verified Answer
Answer
La divergencia de \( \mathbf{F} \) es 3.
Solution
1. La divergencia de un campo vectorial \( \mathbf{F}(x,y,z) = P\mathbf{i} + Q\mathbf{j} + R\mathbf{k} \) se define como:
\[
\nabla \cdot \mathbf{F} = \frac{\partial P}{\partial x} + \frac{\partial Q}{\partial y} + \frac{\partial R}{\partial z}
\]
2. En el campo dado, tenemos:
\[
P(x,y,z) = x + 2,\quad Q(x,y,z) = y + 2,\quad R(x,y,z) = z + 2
\]
3. Calculamos cada derivada parcial:
\[
\frac{\partial (x+2)}{\partial x} = 1,\quad \frac{\partial (y+2)}{\partial y} = 1,\quad \frac{\partial (z+2)}{\partial z} = 1
\]
4. Sumamos las derivadas parciales:
\[
\nabla \cdot \mathbf{F} = 1 + 1 + 1 = 3
\]
La divergencia de \( \mathbf{F} \) es \( \nabla \cdot \mathbf{F} = 3 \).
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Simplify this solution Beyond the Answer
Para calcular la divergencia del campo vectorial \( \mathbf{F}(x, y, z)=(x+2) \mathbf{i}+(y+2) \mathbf{j}+(z+2) \mathbf{k} \), utilizamos la fórmula: \[ \nabla \cdot \mathbf{F} = \frac{\partial F_x}{\partial x} + \frac{\partial F_y}{\partial y} + \frac{\partial F_z}{\partial z} \] Aquí, \( F_x = x + 2 \), \( F_y = y + 2 \), y \( F_z = z + 2 \). Entonces: \[ \frac{\partial F_x}{\partial x} = 1, \quad \frac{\partial F_y}{\partial y} = 1, \quad \frac{\partial F_z}{\partial z} = 1 \] Sumando estas derivadas, tenemos: \[ \nabla \cdot \mathbf{F} = 1 + 1 + 1 = 3 \] Por lo tanto, la divergencia \( \nabla \cdot \mathbf{F} \) es igual a 3.
