$x = e^u \cos v$, $y = e^u \sin v$, $z = uv$ のとき、$\frac{\partial z}{\partial x}$ と $\frac{\partial z}{\partial y}$ を求めよ。

1. 問題の内容

$x = e^u \cos v$, $y = e^u \sin v$, $z = uv$ のとき、$\frac{\partial z}{\partial x}$ と $\frac{\partial z}{\partial y}$ を求めよ。

2. 解き方の手順

$\frac{\partial z}{\partial x}$ と $\frac{\partial z}{\partial y}$ を求めるために、連鎖律を使う。

まず、$\frac{\partial z}{\partial u}$, $\frac{\partial z}{\partial v}$, $\frac{\partial x}{\partial u}$, $\frac{\partial x}{\partial v}$, $\frac{\partial y}{\partial u}$, $\frac{\partial y}{\partial v}$ を求める。

$z = uv$ より、

$\frac{\partial z}{\partial u} = v$

$\frac{\partial z}{\partial v} = u$

$x = e^u \cos v$ より、

$\frac{\partial x}{\partial u} = e^u \cos v$

$\frac{\partial x}{\partial v} = -e^u \sin v$

$y = e^u \sin v$ より、

$\frac{\partial y}{\partial u} = e^u \sin v$

$\frac{\partial y}{\partial v} = e^u \cos v$

次に、$\frac{\partial u}{\partial x}$, $\frac{\partial v}{\partial x}$, $\frac{\partial u}{\partial y}$, $\frac{\partial v}{\partial y}$ を求める。

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

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

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

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

ここで、$J = \frac{\partial x}{\partial u}\frac{\partial y}{\partial v} - \frac{\partial x}{\partial v}\frac{\partial y}{\partial u}$ はヤコビアンである。

$J = (e^u \cos v)(e^u \cos v) - (-e^u \sin v)(e^u \sin v) = e^{2u} \cos^2 v + e^{2u} \sin^2 v = e^{2u}$

よって、

$\frac{\partial u}{\partial x} = \frac{e^u \cos v}{e^{2u}} = e^{-u} \cos v$

$\frac{\partial v}{\partial x} = -\frac{e^u \sin v}{e^{2u}} = -e^{-u} \sin v$

$\frac{\partial u}{\partial y} = -\frac{-e^u \sin v}{e^{2u}} = e^{-u} \sin v$

$\frac{\partial v}{\partial y} = \frac{e^u \cos v}{e^{2u}} = e^{-u} \cos v$

最後に、$\frac{\partial z}{\partial x}$ と $\frac{\partial z}{\partial y}$ を計算する。

$\frac{\partial z}{\partial x} = \frac{\partial z}{\partial u}\frac{\partial u}{\partial x} + \frac{\partial z}{\partial v}\frac{\partial v}{\partial x} = v(e^{-u} \cos v) + u(-e^{-u} \sin v) = e^{-u}(v \cos v - u \sin v)$

$\frac{\partial z}{\partial y} = \frac{\partial z}{\partial u}\frac{\partial u}{\partial y} + \frac{\partial z}{\partial v}\frac{\partial v}{\partial y} = v(e^{-u} \sin v) + u(e^{-u} \cos v) = e^{-u}(v \sin v + u \cos v)$

3. 最終的な答え

$\frac{\partial z}{\partial x} = e^{-u}(v \cos v - u \sin v)$

$\frac{\partial z}{\partial y} = e^{-u}(v \sin v + u \cos v)$