Owen Pierce
01/14/2024 · Senior High School
Ejemplo: Calcule la derivada direccional \( D_{\mathrm{u}} f(1,2) \) si \( f(x, y)=x^{3}-3 x y+4 y^{2} \) y \( \mathbf{u} \) es el vector unitario dado por \( \theta=\frac{\pi}{6} \).
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Para calcular la derivada direccional \( D_{\mathbf{u}} f(1,2) \) de la función \( f(x, y) = x^3 - 3xy + 4y^2 \) en la dirección del vector unitario \( \mathbf{u} \) dado por \( \theta = \frac{\pi}{6} \), seguimos estos pasos:
1. **Calcular el gradiente de \( f \)**:
El gradiente \( \nabla f \) se calcula como:
\[
\nabla f = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} \right)
\]
Primero, calculamos las derivadas parciales:
- Derivada parcial con respecto a \( x \):
\[
\frac{\partial f}{\partial x} = 3x^2 - 3y
\]
- Derivada parcial con respecto a \( y \):
\[
\frac{\partial f}{\partial y} = -3x + 8y
\]
Entonces, el gradiente es:
\[
\nabla f = \left( 3x^2 - 3y, -3x + 8y \right)
\]
2. **Evaluar el gradiente en el punto \( (1, 2) \)**:
Sustituyendo \( x = 1 \) y \( y = 2 \):
\[
\nabla f(1, 2) = \left( 3(1)^2 - 3(2), -3(1) + 8(2) \right) = \left( 3 - 6, -3 + 16 \right) = \left( -3, 13 \right)
\]
3. **Determinar el vector unitario \( \mathbf{u} \)**:
El vector unitario en la dirección de \( \theta = \frac{\pi}{6} \) es:
\[
\mathbf{u} = \left( \cos\left(\frac{\pi}{6}\right), \sin\left(\frac{\pi}{6}\right) \right) = \left( \frac{\sqrt{3}}{2}, \frac{1}{2} \right)
\]
4. **Calcular la derivada direccional**:
La derivada direccional se calcula como el producto punto del gradiente y el vector unitario:
\[
D_{\mathbf{u}} f(1, 2) = \nabla f(1, 2) \cdot \mathbf{u}
\]
\[
D_{\mathbf{u}} f(1, 2) = \left( -3, 13 \right) \cdot \left( \frac{\sqrt{3}}{2}, \frac{1}{2} \right)
\]
\[
= -3 \cdot \frac{\sqrt{3}}{2} + 13 \cdot \frac{1}{2}
\]
\[
= -\frac{3\sqrt{3}}{2} + \frac{13}{2}
\]
\[
= \frac{13 - 3\sqrt{3}}{2}
\]
Por lo tanto, la derivada direccional \( D_{\mathbf{u}} f(1, 2) \) es:
\[
D_{\mathbf{u}} f(1, 2) = \frac{13 - 3\sqrt{3}}{2}
\]
Quick Answer
La derivada direccional \( D_{\mathbf{u}} f(1, 2) \) es \( \frac{13 - 3\sqrt{3}}{2} \).
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