$z = x^2 - 2xy + y^2$ であり、$x = r \cos\theta$ および $y = r \sin\theta$ のとき、$\frac{\partial z}{\partial r}$ と $\frac{\partial z}{\partial \theta}$ を求めよ。

## 問題3

1. 問題の内容

$z = x^2 - 2xy + y^2$ であり、$x = r \cos\theta$ および $y = r \sin\theta$ のとき、$\frac{\partial z}{\partial r}$ と $\frac{\partial z}{\partial \theta}$ を求めよ。

2. 解き方の手順

まず、$z$ を $r$ と $\theta$ の関数として表します。

$z = x^2 - 2xy + y^2 = (x-y)^2$なので、$x$と$y$に$x = r \cos\theta$ および $y = r \sin\theta$を代入して、

$z = (r\cos\theta - r\sin\theta)^2 = r^2 (\cos\theta - \sin\theta)^2$

となります。

次に、$\frac{\partial z}{\partial r}$ を求めます。

$\frac{\partial z}{\partial r} = \frac{\partial}{\partial r} [r^2 (\cos\theta - \sin\theta)^2] = 2r (\cos\theta - \sin\theta)^2$

次に、$\frac{\partial z}{\partial \theta}$ を求めます。

$\frac{\partial z}{\partial \theta} = \frac{\partial}{\partial \theta} [r^2 (\cos\theta - \sin\theta)^2]$

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

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

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

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

3. 最終的な答え

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

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

## 問題4

1. 問題の内容

$f(x, y) = \log\sqrt{x^2 + y^2}$ について、$\frac{\partial^2 f}{\partial x^2} + \frac{\partial^2 f}{\partial y^2}$ を求めよ。

2. 解き方の手順

まず、$f(x, y)$ を簡単にします。

$f(x, y) = \log\sqrt{x^2 + y^2} = \log(x^2 + y^2)^{1/2} = \frac{1}{2} \log(x^2 + y^2)$

次に、$\frac{\partial f}{\partial x}$ を求めます。

$\frac{\partial f}{\partial x} = \frac{1}{2} \cdot \frac{1}{x^2 + y^2} \cdot 2x = \frac{x}{x^2 + y^2}$

次に、$\frac{\partial^2 f}{\partial x^2}$ を求めます。

$\frac{\partial^2 f}{\partial x^2} = \frac{\partial}{\partial x} \left(\frac{x}{x^2 + y^2}\right) = \frac{(x^2 + y^2) \cdot 1 - x \cdot 2x}{(x^2 + y^2)^2} = \frac{x^2 + y^2 - 2x^2}{(x^2 + y^2)^2} = \frac{y^2 - x^2}{(x^2 + y^2)^2}$

次に、$\frac{\partial f}{\partial y}$ を求めます。

$\frac{\partial f}{\partial y} = \frac{1}{2} \cdot \frac{1}{x^2 + y^2} \cdot 2y = \frac{y}{x^2 + y^2}$

次に、$\frac{\partial^2 f}{\partial y^2}$ を求めます。

$\frac{\partial^2 f}{\partial y^2} = \frac{\partial}{\partial y} \left(\frac{y}{x^2 + y^2}\right) = \frac{(x^2 + y^2) \cdot 1 - y \cdot 2y}{(x^2 + y^2)^2} = \frac{x^2 + y^2 - 2y^2}{(x^2 + y^2)^2} = \frac{x^2 - y^2}{(x^2 + y^2)^2}$

最後に、$\frac{\partial^2 f}{\partial x^2} + \frac{\partial^2 f}{\partial y^2}$ を求めます。

$\frac{\partial^2 f}{\partial x^2} + \frac{\partial^2 f}{\partial y^2} = \frac{y^2 - x^2}{(x^2 + y^2)^2} + \frac{x^2 - y^2}{(x^2 + y^2)^2} = \frac{y^2 - x^2 + x^2 - y^2}{(x^2 + y^2)^2} = 0$

3. 最終的な答え

$\frac{\partial^2 f}{\partial x^2} + \frac{\partial^2 f}{\partial y^2} = 0$

## 問題5

1. 問題の内容

$z = f(x, y)$, $x = r\cos\theta$, $y = r\sin\theta$ のとき、次の等式が成り立つことを証明せよ。

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

2. 解き方の手順

まず、$\frac{\partial z}{\partial r}$ と $\frac{\partial z}{\partial \theta}$ を $x$ と $y$ の偏微分で表します。連鎖律より、

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

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

次に、右辺を計算します。

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

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

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

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

したがって、$(\frac{\partial z}{\partial x})^2 + (\frac{\partial z}{\partial y})^2 = (\frac{\partial z}{\partial r})^2 + \frac{1}{r^2}(\frac{\partial z}{\partial \theta})^2$ が成り立つ。

3. 最終的な答え

$(\frac{\partial z}{\partial x})^2 + (\frac{\partial z}{\partial y})^2 = (\frac{\partial z}{\partial r})^2 + \frac{1}{r^2}(\frac{\partial z}{\partial \theta})^2$ が成り立つことが証明された。