$z = f(x, y)$ であり、$x = r\cos\theta$、$y = r\sin\theta$のとき、次の等式を証明します。 $(\frac{\partial z}{\partial x})^2 + (\frac{\partial z}{\partial y})^2 = (\frac{\partial z}{\partial r})^2 + \frac{1}{r^2}(\frac{\partial z}{\partial \theta})^2$

1. 問題の内容

$z = f(x, y)$ であり、$x = r\cos\theta$、$y = r\sin\theta$のとき、次の等式を証明します。

$(\frac{\partial z}{\partial x})^2 + (\frac{\partial z}{\partial y})^2 = (\frac{\partial z}{\partial r})^2 + \frac{1}{r^2}(\frac{\partial z}{\partial \theta})^2$

2. 解き方の手順

まず、$\frac{\partial z}{\partial r}$ と $\frac{\partial z}{\partial \theta}$ を、$x$ と $y$ による偏微分で表現します。

連鎖律（chain rule）を用いると、以下のようになります。

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

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

次に、$\frac{\partial x}{\partial r}$、$\frac{\partial y}{\partial r}$、$\frac{\partial x}{\partial \theta}$、$\frac{\partial y}{\partial \theta}$ を計算します。

$\frac{\partial x}{\partial r} = \cos\theta$

$\frac{\partial y}{\partial r} = \sin\theta$

$\frac{\partial x}{\partial \theta} = -r\sin\theta$

$\frac{\partial y}{\partial \theta} = r\cos\theta$

これらの結果を上記の式に代入すると、

$\frac{\partial z}{\partial r} = \frac{\partial z}{\partial x}\cos\theta + \frac{\partial z}{\partial y}\sin\theta$

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

次に、$(\frac{\partial z}{\partial r})^2$ と $\frac{1}{r^2}(\frac{\partial z}{\partial \theta})^2$ を計算します。

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

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

$\frac{1}{r^2}(\frac{\partial z}{\partial \theta})^2 = (\frac{\partial z}{\partial x})^2\sin^2\theta - 2\frac{\partial z}{\partial x}\frac{\partial z}{\partial y}\sin\theta\cos\theta + (\frac{\partial z}{\partial y})^2\cos^2\theta$

したがって、

$(\frac{\partial z}{\partial r})^2 + \frac{1}{r^2}(\frac{\partial z}{\partial \theta})^2 = (\frac{\partial z}{\partial x})^2(\cos^2\theta + \sin^2\theta) + (\frac{\partial z}{\partial y})^2(\sin^2\theta + \cos^2\theta)$

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

$(\frac{\partial z}{\partial r})^2 + \frac{1}{r^2}(\frac{\partial z}{\partial \theta})^2 = (\frac{\partial z}{\partial x})^2 + (\frac{\partial z}{\partial y})^2$

3. 最終的な答え

$(\frac{\partial z}{\partial x})^2 + (\frac{\partial z}{\partial y})^2 = (\frac{\partial z}{\partial r})^2 + \frac{1}{r^2}(\frac{\partial z}{\partial \theta})^2$ が成立することが証明されました。