$z = f(x, y)$ を全微分可能な関数とし、$x = u \cos\alpha + v \sin\alpha$, $y = -u \sin\alpha + v \cos\alpha$ ($\alpha$ は定数) とするとき、 $(\frac{\partial z}{\partial x})^2 + (\frac{\partial z}{\partial y})^2 = (\frac{\partial z}{\partial u})^2 + (\frac{\partial z}{\partial v})^2$ を示す問題です。

1. 問題の内容

$z = f(x, y)$ を全微分可能な関数とし、$x = u \cos\alpha + v \sin\alpha$, $y = -u \sin\alpha + v \cos\alpha$ ($\alpha$ は定数) とするとき、 $(\frac{\partial z}{\partial x})^2 + (\frac{\partial z}{\partial y})^2 = (\frac{\partial z}{\partial u})^2 + (\frac{\partial z}{\partial v})^2$ を示す問題です。

2. 解き方の手順

まず、合成関数の微分を用いて $\frac{\partial z}{\partial u}$ と $\frac{\partial z}{\partial v}$ を $\frac{\partial z}{\partial x}$ と $\frac{\partial z}{\partial y}$ で表します。

$\frac{\partial z}{\partial u} = \frac{\partial z}{\partial x} \frac{\partial x}{\partial u} + \frac{\partial z}{\partial y} \frac{\partial y}{\partial u}$

$\frac{\partial z}{\partial v} = \frac{\partial z}{\partial x} \frac{\partial x}{\partial v} + \frac{\partial z}{\partial y} \frac{\partial y}{\partial v}$

与えられた $x$ と $y$ の式から、偏微分を計算します。

$\frac{\partial x}{\partial u} = \cos\alpha$, $\frac{\partial x}{\partial v} = \sin\alpha$,

$\frac{\partial y}{\partial u} = -\sin\alpha$, $\frac{\partial y}{\partial v} = \cos\alpha$

これらを代入すると、

$\frac{\partial z}{\partial u} = \frac{\partial z}{\partial x} \cos\alpha + \frac{\partial z}{\partial y} (-\sin\alpha)$

$\frac{\partial z}{\partial v} = \frac{\partial z}{\partial x} \sin\alpha + \frac{\partial z}{\partial y} \cos\alpha$

次に、 $(\frac{\partial z}{\partial u})^2 + (\frac{\partial z}{\partial v})^2$ を計算します。

$(\frac{\partial z}{\partial u})^2 = (\frac{\partial z}{\partial x} \cos\alpha - \frac{\partial z}{\partial y} \sin\alpha)^2 = (\frac{\partial z}{\partial x})^2 \cos^2\alpha - 2 \frac{\partial z}{\partial x} \frac{\partial z}{\partial y} \cos\alpha \sin\alpha + (\frac{\partial z}{\partial y})^2 \sin^2\alpha$

$(\frac{\partial z}{\partial v})^2 = (\frac{\partial z}{\partial x} \sin\alpha + \frac{\partial z}{\partial y} \cos\alpha)^2 = (\frac{\partial z}{\partial x})^2 \sin^2\alpha + 2 \frac{\partial z}{\partial x} \frac{\partial z}{\partial y} \sin\alpha \cos\alpha + (\frac{\partial z}{\partial y})^2 \cos^2\alpha$

これらの和をとると、

$(\frac{\partial z}{\partial u})^2 + (\frac{\partial z}{\partial v})^2 = (\frac{\partial z}{\partial x})^2 (\cos^2\alpha + \sin^2\alpha) + (\frac{\partial z}{\partial y})^2 (\sin^2\alpha + \cos^2\alpha)$

$\sin^2\alpha + \cos^2\alpha = 1$ であるから、

$(\frac{\partial z}{\partial u})^2 + (\frac{\partial z}{\partial v})^2 = (\frac{\partial z}{\partial x})^2 + (\frac{\partial z}{\partial y})^2$

3. 最終的な答え

$(\frac{\partial z}{\partial x})^2 + (\frac{\partial z}{\partial y})^2 = (\frac{\partial z}{\partial u})^2 + (\frac{\partial z}{\partial v})^2$