ベクトル場 $\mathbf{A} = \cos(xy) \mathbf{e}_x + (3xy - 2x^2) \mathbf{e}_y - (3x + 2y) \mathbf{e}_z$ が与えられたとき、以下の偏微分を求める。 1. $\frac{\partial \mathbf{A}}{\partial x}$ 2. $\frac{\partial \mathbf{A}}{\partial y}$ 3. $\frac{\partial^2 \mathbf{A}}{\partial x^2}$ 4. $\frac{\partial^2 \mathbf{A}}{\partial y^2}$ 5. $\frac{\partial^2 \mathbf{A}}{\partial x \partial y}$ 6. $\frac{\partial^2 \mathbf{A}}{\partial y \partial x}$

## 問題の解答

以下に、問題 [3] と [4] の解答を示します。

### 問題 [3]

1. 問題の内容

ベクトル場 $\mathbf{A} = \cos(xy) \mathbf{e}_x + (3xy - 2x^2) \mathbf{e}_y - (3x + 2y) \mathbf{e}_z$ が与えられたとき、以下の偏微分を求める。

1. $\frac{\partial \mathbf{A}}{\partial x}$

2. $\frac{\partial \mathbf{A}}{\partial y}$

3. $\frac{\partial^2 \mathbf{A}}{\partial x^2}$

4. $\frac{\partial^2 \mathbf{A}}{\partial y^2}$

5. $\frac{\partial^2 \mathbf{A}}{\partial x \partial y}$

6. $\frac{\partial^2 \mathbf{A}}{\partial y \partial x}$

2. 解き方の手順

各偏微分を、それぞれの成分ごとに行う。

1. $\frac{\partial \mathbf{A}}{\partial x} = \frac{\partial}{\partial x} (\cos(xy) \mathbf{e}_x + (3xy - 2x^2) \mathbf{e}_y - (3x + 2y) \mathbf{e}_z) = -y \sin(xy) \mathbf{e}_x + (3y - 4x) \mathbf{e}_y - 3 \mathbf{e}_z$

2. $\frac{\partial \mathbf{A}}{\partial y} = \frac{\partial}{\partial y} (\cos(xy) \mathbf{e}_x + (3xy - 2x^2) \mathbf{e}_y - (3x + 2y) \mathbf{e}_z) = -x \sin(xy) \mathbf{e}_x + 3x \mathbf{e}_y - 2 \mathbf{e}_z$

3. $\frac{\partial^2 \mathbf{A}}{\partial x^2} = \frac{\partial}{\partial x} (\frac{\partial \mathbf{A}}{\partial x}) = \frac{\partial}{\partial x} (-y \sin(xy) \mathbf{e}_x + (3y - 4x) \mathbf{e}_y - 3 \mathbf{e}_z) = -y^2 \cos(xy) \mathbf{e}_x - 4 \mathbf{e}_y$

4. $\frac{\partial^2 \mathbf{A}}{\partial y^2} = \frac{\partial}{\partial y} (\frac{\partial \mathbf{A}}{\partial y}) = \frac{\partial}{\partial y} (-x \sin(xy) \mathbf{e}_x + 3x \mathbf{e}_y - 2 \mathbf{e}_z) = -x^2 \cos(xy) \mathbf{e}_x$

5. $\frac{\partial^2 \mathbf{A}}{\partial x \partial y} = \frac{\partial}{\partial x} (\frac{\partial \mathbf{A}}{\partial y}) = \frac{\partial}{\partial x} (-x \sin(xy) \mathbf{e}_x + 3x \mathbf{e}_y - 2 \mathbf{e}_z) = (-\sin(xy) - xy \cos(xy)) \mathbf{e}_x + 3 \mathbf{e}_y$

6. $\frac{\partial^2 \mathbf{A}}{\partial y \partial x} = \frac{\partial}{\partial y} (\frac{\partial \mathbf{A}}{\partial x}) = \frac{\partial}{\partial y} (-y \sin(xy) \mathbf{e}_x + (3y - 4x) \mathbf{e}_y - 3 \mathbf{e}_z) = (-\sin(xy) - xy \cos(xy)) \mathbf{e}_x + 3 \mathbf{e}_y$

3. 最終的な答え

1. $\frac{\partial \mathbf{A}}{\partial x} = -y \sin(xy) \mathbf{e}_x + (3y - 4x) \mathbf{e}_y - 3 \mathbf{e}_z$

2. $\frac{\partial \mathbf{A}}{\partial y} = -x \sin(xy) \mathbf{e}_x + 3x \mathbf{e}_y - 2 \mathbf{e}_z$

3. $\frac{\partial^2 \mathbf{A}}{\partial x^2} = -y^2 \cos(xy) \mathbf{e}_x - 4 \mathbf{e}_y$

4. $\frac{\partial^2 \mathbf{A}}{\partial y^2} = -x^2 \cos(xy) \mathbf{e}_x$

5. $\frac{\partial^2 \mathbf{A}}{\partial x \partial y} = (-\sin(xy) - xy \cos(xy)) \mathbf{e}_x + 3 \mathbf{e}_y$

6. $\frac{\partial^2 \mathbf{A}}{\partial y \partial x} = (-\sin(xy) - xy \cos(xy)) \mathbf{e}_x + 3 \mathbf{e}_y$

### 問題 [4]

1. 問題の内容

次の関数の全微分を求める。

1. $z = f(x, y) = xy \sin(x + y)$

2. $z = f(r, \theta) = r (\cos \theta \mathbf{e}_x + \sin \theta \mathbf{e}_y)$

2. 解き方の手順

全微分の公式を用いる。

$dz = \frac{\partial z}{\partial x} dx + \frac{\partial z}{\partial y} dy$ (1)の場合

$dz = \frac{\partial z}{\partial r} dr + \frac{\partial z}{\partial \theta} d\theta$ (2)の場合

1. $z = f(x, y) = xy \sin(x + y)$

$\frac{\partial z}{\partial x} = y \sin(x + y) + xy \cos(x + y)$

$\frac{\partial z}{\partial y} = x \sin(x + y) + xy \cos(x + y)$

$dz = (y \sin(x + y) + xy \cos(x + y)) dx + (x \sin(x + y) + xy \cos(x + y)) dy$

2. $z = f(r, \theta) = r (\cos \theta \mathbf{e}_x + \sin \theta \mathbf{e}_y) = r \cos \theta \mathbf{e}_x + r \sin \theta \mathbf{e}_y$

$\frac{\partial z}{\partial r} = \cos \theta \mathbf{e}_x + \sin \theta \mathbf{e}_y$

$\frac{\partial z}{\partial \theta} = -r \sin \theta \mathbf{e}_x + r \cos \theta \mathbf{e}_y$

$dz = (\cos \theta \mathbf{e}_x + \sin \theta \mathbf{e}_y) dr + (-r \sin \theta \mathbf{e}_x + r \cos \theta \mathbf{e}_y) d\theta$

3. 最終的な答え

1. $dz = (y \sin(x + y) + xy \cos(x + y)) dx + (x \sin(x + y) + xy \cos(x + y)) dy$

2. $dz = (\cos \theta \mathbf{e}_x + \sin \theta \mathbf{e}_y) dr + (-r \sin \theta \mathbf{e}_x + r \cos \theta \mathbf{e}_y) d\theta$