Càlcul de la matriu Jacobiana d’una funció vectorial

Per determinar la matriu Jacobiana de la funció vectorial $\mathbf{f}(x, y, z) = \left( z \tan(x^2 – y^2), xy \ln\left(\frac{z}{2}\right) \right)$ en el punt $(2, -2, 2)$, seguim els passos següents. La matriu Jacobiana d’una funció vectorial $\mathbf{f}: \mathbb{R}^3 \to \mathbb{R}^2$ es construeix a partir de les derivades parcials de les seves components respecte a les variables d’entrada.

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Definició de la matriu Jacobiana

La funció donada és:

$$\mathbf{f}(x, y, z) = \left( f_1(x, y, z), f_2(x, y, z) \right),$$

on:

$$f_1(x, y, z) = z \tan(x^2 – y^2),$$

$$f_2(x, y, z) = xy \ln\left(\frac{z}{2}\right).$$

La matriu Jacobiana $J_{\mathbf{f}}$ d’una funció $\mathbf{f}: \mathbb{R}^3 \to \mathbb{R}^2$ és una matriu $2 \times 3$ les files de la qual són els gradients de les components $f_1$ i $f_2$:

$$J_{\mathbf{f}} = \begin{bmatrix}
\frac{\partial f_1}{\partial x} & \frac{\partial f_1}{\partial y} & \frac{\partial f_1}{\partial z} \\
\frac{\partial f_2}{\partial x} & \frac{\partial f_2}{\partial y} & \frac{\partial f_2}{\partial z}
\end{bmatrix}.$$

Calculem cada derivada parcial i les avaluem en el punt $(x, y, z) = (2, -2, 2)$.

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Pas 1: Calcular les derivades parcials de $f_1$

La primera component és:

$$f_1(x, y, z) = z \tan(x^2 – y^2).$$

• Derivada parcial respecte a $x$:

$$\frac{\partial f_1}{\partial x} = \frac{\partial}{\partial x} \left[ z \tan(x^2 – y^2) \right].$$

Com que $z$ i $y$ són constants respecte a $x$, apliquem la regla del producte i la regla de la cadena:

$$\frac{\partial f_1}{\partial x} = z \cdot \frac{\partial}{\partial x} \left[ \tan(x^2 – y^2) \right].$$

La derivada de $\tan(u)$ és $\sec^2(u) \cdot \frac{\partial u}{\partial x}$, amb $u = x^2 – y^2$. Llavors:

$$\frac{\partial}{\partial x} \left[ \tan(x^2 – y^2) \right] = \sec^2(x^2 – y^2) \cdot \frac{\partial}{\partial x} (x^2 – y^2).$$

$$\frac{\partial}{\partial x} (x^2 – y^2) = 2x.$$

Per tant:

$$\frac{\partial f_1}{\partial x} = z \sec^2(x^2 – y^2) \cdot 2x = 2xz \sec^2(x^2 – y^2).$$

• Derivada parcial respecte a $y$:

$$\frac{\partial f_1}{\partial y} = \frac{\partial}{\partial y} \left[ z \tan(x^2 – y^2) \right].$$

Ara $x$ i $z$ són constants:

$$\frac{\partial f_1}{\partial y} = z \cdot \frac{\partial}{\partial y} \left[ \tan(x^2 – y^2) \right].$$

$$\frac{\partial}{\partial y} \left[ \tan(x^2 – y^2) \right] = \sec^2(x^2 – y^2) \cdot \frac{\partial}{\partial y} (x^2 – y^2).$$

$$\frac{\partial}{\partial y} (x^2 – y^2) = -2y.$$

Per tant:

$$\frac{\partial f_1}{\partial y} = z \sec^2(x^2 – y^2) \cdot (-2y) = -2yz \sec^2(x^2 – y^2).$$

• Derivada parcial respecte a $z$:

$$\frac{\partial f_1}{\partial z} = \frac{\partial}{\partial z} \left[ z \tan(x^2 – y^2) \right].$$

Com que $\tan(x^2 – y^2)$ no depèn de $z$:

$$\frac{\partial f_1}{\partial z} = \tan(x^2 – y^2) \cdot 1 = \tan(x^2 – y^2).$$

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Pas 2: Calcular les derivades parcials de $f_2$

La segona component és:

$$f_2(x, y, z) = xy \ln\left(\frac{z}{2}\right).$$

• Derivada parcial respecte a $x$:

$$\frac{\partial f_2}{\partial x} = \frac{\partial}{\partial x} \left[ xy \ln\left(\frac{z}{2}\right) \right].$$

Com que ( y ) i ( \ln\left(\frac{z}{2}\right) ) són constants respecte a $x$:

$$\frac{\partial f_2}{\partial x} = y \ln\left(\frac{z}{2}\right) \cdot 1 = y \ln\left(\frac{z}{2}\right).$$

• Derivada parcial respecte a $y$:

$$\frac{\partial f_2}{\partial y} = \frac{\partial}{\partial y} \left[ xy \ln\left(\frac{z}{2}\right) \right].$$

Ara $x$ i $\ln\left(\frac{z}{2}\right)$ són constants:

$$\frac{\partial f_2}{\partial y} = x \ln\left(\frac{z}{2}\right) \cdot 1 = x \ln\left(\frac{z}{2}\right).$$

• Derivada parcial respecte a $z$:

$$\frac{\partial f_2}{\partial z} = \frac{\partial}{\partial z} \left[ xy \ln\left(\frac{z}{2}\right) \right].$$

Com que $xy$ és constant respecte a $z$:

$$\frac{\partial f_2}{\partial z} = xy \cdot \frac{\partial}{\partial z} \left[ \ln\left(\frac{z}{2}\right) \right].$$

Sabem que:

$$\ln\left(\frac{z}{2}\right) = \ln(z) – \ln(2).$$

Llavors:

$$\frac{\partial}{\partial z} \left[ \ln\left(\frac{z}{2}\right) \right] = \frac{\partial}{\partial z} \left[ \ln(z) – \ln(2) \right] = \frac{1}{z}.$$

Per tant:

$$\frac{\partial f_2}{\partial z} = xy \cdot \frac{1}{z} = \frac{xy}{z}.$$

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Pas 3: Construir la matriu Jacobiana

La matriu Jacobiana és:

$$J_{\mathbf{f}} = \begin{bmatrix}
2xz \sec^2(x^2 – y^2) & -2yz \sec^2(x^2 – y^2) & \tan(x^2 – y^2) \\
y \ln\left(\frac{z}{2}\right) & x \ln\left(\frac{z}{2}\right) & \frac{xy}{z}
\end{bmatrix}.$$

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Pas 4: Avaluar la matriu en el punt $(2, -2, 2)$

Substituïm $x = 2$, $y = -2$, $z = 2$.

Primera fila ($f_1$)

• Component $\frac{\partial f_1}{\partial x}$:

$$\frac{\partial f_1}{\partial x} = 2xz \sec^2(x^2 – y^2).$$

Calculem $x^2 – y^2$:

$$x^2 – y^2 = 2^2 – (-2)^2 = 4 – 4 = 0.$$

Llavors:

$$\sec^2(x^2 – y^2) = \sec^2(0) = \frac{1}{\cos^2(0)} = \frac{1}{1} = 1.$$

Substituïm:

$$\frac{\partial f_1}{\partial x} = 2 \cdot 2 \cdot 2 \cdot 1 = 8.$$

• Component $\frac{\partial f_1}{\partial y}$:

$$\frac{\partial f_1}{\partial y} = -2yz \sec^2(x^2 – y^2).$$

Ja tenim $\sec^2(0) = 1$. Substituïm:

$$\frac{\partial f_1}{\partial y} = -2 \cdot (-2) \cdot 2 \cdot 1 = 8.$$

• Component $\frac{\partial f_1}{\partial z}$:

$$\frac{\partial f_1}{\partial z} = \tan(x^2 – y^2).$$

$$\tan(x^2 – y^2) = \tan(0) = 0.$$

Segona fila ($f_2$)

• Component $\frac{\partial f_2}{\partial x}$:

$$\frac{\partial f_2}{\partial x} = y \ln\left(\frac{z}{2}\right).$$

$$\frac{z}{2} = \frac{2}{2} = 1, \quad \ln(1) = 0.$$

$$\frac{\partial f_2}{\partial x} = (-2) \cdot 0 = 0.$$

• Component $\frac{\partial f_2}{\partial y}$:

$$\frac{\partial f_2}{\partial y} = x \ln\left(\frac{z}{2}\right).$$

$$\frac{\partial f_2}{\partial y} = 2 \cdot 0 = 0.$$

• Component $\frac{\partial f_2}{\partial z}$:

$$\frac{\partial f_2}{\partial z} = \frac{xy}{z}.$$

$$\frac{\partial f_2}{\partial z} = \frac{2 \cdot (-2)}{2} = \frac{-4}{2} = -2.$$

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Pas 5: Escriure la matriu Jacobiana avaluada

La matriu Jacobiana en $(2, -2, 2)$ és:

$$J_{\mathbf{f}}(2, -2, 2) = \begin{bmatrix}
8 & 8 & 0 \\
0 & 0 & -2
\end{bmatrix}.$$

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Resposta final

La matriu Jacobiana de la funció $\mathbf{f}(x, y, z) = \left( z \tan(x^2 – y^2), xy \ln\left(\frac{z}{2}\right) \right)$ en el punt $(2, -2, 2)$ és:

$$\boxed{\begin{bmatrix}
8 & 8 & 0 \\
0 & 0 & -2
\end{bmatrix}}$$