$u = x^2 - y^2$ と $v = xy^3$ という関係式が与えられています。このとき、$(x, y) = (2, 1)$ における $\frac{\partial x}{\partial u}$, $\frac{\partial y}{\partial u}$, $\frac{\partial x}{\partial v}$, $\frac{\partial y}{\partial v}$ の値を求める問題です。

1. 問題の内容

$u = x^2 - y^2$ と $v = xy^3$ という関係式が与えられています。このとき、$(x, y) = (2, 1)$ における $\frac{\partial x}{\partial u}$, $\frac{\partial y}{\partial u}$, $\frac{\partial x}{\partial v}$, $\frac{\partial y}{\partial v}$ の値を求める問題です。

2. 解き方の手順

まず、$\frac{\partial u}{\partial x}, \frac{\partial u}{\partial y}, \frac{\partial v}{\partial x}, \frac{\partial v}{\partial y}$ を計算します。

$\frac{\partial u}{\partial x} = 2x$

$\frac{\partial u}{\partial y} = -2y$

$\frac{\partial v}{\partial x} = y^3$

$\frac{\partial v}{\partial y} = 3xy^2$

次に、ヤコビアン $J$ を計算します。

$J = \frac{\partial(u, v)}{\partial(x, y)} = \begin{vmatrix} \frac{\partial u}{\partial x} & \frac{\partial u}{\partial y} \\ \frac{\partial v}{\partial x} & \frac{\partial v}{\partial y} \end{vmatrix} = \frac{\partial u}{\partial x} \frac{\partial v}{\partial y} - \frac{\partial u}{\partial y} \frac{\partial v}{\partial x}$

$J = (2x)(3xy^2) - (-2y)(y^3) = 6x^2y^2 + 2y^4$

$(x, y) = (2, 1)$ における $J$ の値を計算します。

$J(2, 1) = 6(2^2)(1^2) + 2(1^4) = 6(4)(1) + 2(1) = 24 + 2 = 26$

$\frac{\partial(x, y)}{\partial(u, v)}$ は $\frac{\partial(u, v)}{\partial(x, y)}$ の逆数なので、

$\frac{\partial x}{\partial u} = \frac{1}{J} \frac{\partial v}{\partial y}$

$\frac{\partial y}{\partial u} = -\frac{1}{J} \frac{\partial v}{\partial x}$

$\frac{\partial x}{\partial v} = -\frac{1}{J} \frac{\partial u}{\partial y}$

$\frac{\partial y}{\partial v} = \frac{1}{J} \frac{\partial u}{\partial x}$

$(x, y) = (2, 1)$ におけるそれぞれの偏導関数の値を計算します。

$\frac{\partial u}{\partial x} (2,1) = 2(2) = 4$

$\frac{\partial u}{\partial y} (2,1) = -2(1) = -2$

$\frac{\partial v}{\partial x} (2,1) = (1)^3 = 1$

$\frac{\partial v}{\partial y} (2,1) = 3(2)(1)^2 = 6$

よって、

$\frac{\partial x}{\partial u} (2,1) = \frac{1}{26} (6) = \frac{6}{26} = \frac{3}{13}$

$\frac{\partial y}{\partial u} (2,1) = -\frac{1}{26} (1) = -\frac{1}{26}$

$\frac{\partial x}{\partial v} (2,1) = -\frac{1}{26} (-2) = \frac{2}{26} = \frac{1}{13}$

$\frac{\partial y}{\partial v} (2,1) = \frac{1}{26} (4) = \frac{4}{26} = \frac{2}{13}$

3. 最終的な答え

$\frac{\partial x}{\partial u} = \frac{3}{13}$

$\frac{\partial y}{\partial u} = -\frac{1}{26}$

$\frac{\partial x}{\partial v} = \frac{1}{13}$

$\frac{\partial y}{\partial v} = \frac{2}{13}$