Question
f(x,y)=3x2−4y2 . Find the directional derivative at the point (1,−1) in the direction of the vector u=(3,4)
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Answer to a math question f(x,y)=3x2−4y2 . Find the directional derivative at the point (1,−1) in the direction of the vector u=(3,4)
Miles
4.9114 Answers
**
1. **Normalización de\mathbf{u}
El vector\mathbf{u} = (3,4)
\|\mathbf{u}\| = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5
Entonces, el vector unitario es:
\mathbf{\hat{u}} = \left( \frac{3}{5}, \frac{4}{5} \right)
2. **Cálculo del gradiente\nabla f
- Derivada parcial respecto ax
\frac{\partial f}{\partial x} = \frac{\partial}{\partial x}(3x^2 - 4y^2) = 6x
- Derivada parcial respecto ay
\frac{\partial f}{\partial y} = \frac{\partial}{\partial y}(3x^2 - 4y^2) = -8y
Entonces,
\nabla f(x, y) = (6x, -8y)
3. **Evaluación del gradiente en el punto(1, -1)
\nabla f(1, -1) = (6 \cdot 1, -8 \cdot (-1)) = (6, 8)
4. **Cálculo de la derivada direccional**:
La derivada direccional en la dirección de\mathbf{\hat{u}}
D_{\mathbf{\hat{u}}}f(1, -1) = \nabla f(1, -1) \cdot \mathbf{\hat{u}} = (6, 8) \cdot \left( \frac{3}{5}, \frac{4}{5} \right)
= 6 \cdot \frac{3}{5} + 8 \cdot \frac{4}{5}
= \frac{18}{5} + \frac{32}{5}
= \frac{50}{5} = 10
La derivada direccional es10
1. **Normalización de
El vector
Entonces, el vector unitario es:
2. **Cálculo del gradiente
- Derivada parcial respecto a
- Derivada parcial respecto a
Entonces,
3. **Evaluación del gradiente en el punto
4. **Cálculo de la derivada direccional**:
La derivada direccional en la dirección de
La derivada direccional es
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