合成関数の偏微分

■問題

$3$\displaystyle{z=f $ x,y $ }$ , $3$\displaystyle{x=r\cos {\theta}}$ , $3$\displaystyle{y=r\sin {\theta}\text{\hspace{1} }}$ （平面の極座標変換）ならば

$3$\displaystyle{{ $ \frac{\hspace{2} dz \hspace{2}}{\hspace{2} dx \hspace{2}} $ }^{2}+{ $ \frac{\hspace{2} dz \hspace{2}}{\hspace{2} dy \hspace{2}} $ }^{2}}$ $3$\displaystyle{={ $ \frac{\hspace{2} dz \hspace{2}}{\hspace{2} dr \hspace{2}} $ }^{2}+\frac{\hspace{2}1\hspace{2}}{\hspace{2} {r}^{2} \hspace{2}}{ $ \frac{\hspace{2} dz \hspace{2}}{\hspace{2} d{\theta} \hspace{2}} $ }^{2}}$

となることを示せ．

■ヒント

合成関数の偏導関数を参照せよ．

■解説

$3$z$ を $3$r$ で偏微分する．

$3$\displaystyle{\frac{\hspace{2} {\partial}z \hspace{2}}{\hspace{2} {\partial}r \hspace{2}}=\hspace{1}{f}_{x}\frac{\hspace{2} {\partial}x \hspace{2}}{\hspace{2} {\partial}r \hspace{2}}+\hspace{1}{f}_{y}\frac{\hspace{2} {\partial}y \hspace{2}}{\hspace{2} {\partial}r \hspace{2}}}$

$3$\displaystyle{=\hspace{1}{f}_{x}\text{\hspace{1} }\cdot \text{\hspace{1} }\cos {\theta}+\hspace{1}{f}_{y}\text{\hspace{1} }\cdot \text{\hspace{1} }\sin {\theta}}$ 　

$3$z$ を $3$\displaystyle{{\theta}}$ で偏微分する．

$3$\displaystyle{\frac{\hspace{2} {\partial}z \hspace{2}}{\hspace{2} {\partial}{\theta} \hspace{2}}=\hspace{1}{f}_{x}\frac{\hspace{2} {\partial}x \hspace{2}}{\hspace{2} {\partial}{\theta} \hspace{2}}+\hspace{1}{f}_{y}\frac{\hspace{2} {\partial}y \hspace{2}}{\hspace{2} {\partial}{\theta} \hspace{2}}}$

$3$\displaystyle{=\hspace{1}{f}_{x}\text{\hspace{1} }\cdot \text{\hspace{1} } $ - r\sin {\theta} $ +\hspace{1}{f}_{y}\text{\hspace{1} }\cdot \text{\hspace{1} }r\cos {\theta}}$

$3$\displaystyle{=r $ - \hspace{1}{f}_{x}\sin {\theta}+\hspace{1}{f}_{y}\cos {\theta} $ }$

以上より

$3$\displaystyle{{ $ \frac{\hspace{2} dz \hspace{2}}{\hspace{2} dr \hspace{2}} $ }^{2}+\frac{\hspace{2}1\hspace{2}}{\hspace{2} {r}^{2} \hspace{2}}{ $ \frac{\hspace{2} dz \hspace{2}}{\hspace{2} d{\theta} \hspace{2}} $ }^{2}}$

$3$\displaystyle{={ $ \hspace{1}{f}_{x}\text{\hspace{1} }\cdot \text{\hspace{1} }\cos {\theta}+\hspace{1}{f}_{y}\cdot \sin {\theta} $ }^{2}}$ $3$\displaystyle{+\frac{\hspace{2}1\hspace{2}}{\hspace{2} {r}^{2} \hspace{2}}{ \{ r $ - \hspace{1}{f}_{x}\sin {\theta}+\hspace{1}{f}_{y}\cos {\theta} $ \} }^{2}}$

$3$\displaystyle{= $ f_{x}^{2}{ \cos }^{2}{\theta}+2\hspace{1}{f}_{x}\hspace{1}{f}_{y}\cos {\theta}\sin {\theta}+f_{y}^{2}{ \sin }^{2}{\theta} $ }$

$3$\displaystyle{\text{\hspace{1} \hspace{1} \hspace{1} \hspace{1} }+\frac{\hspace{2}1\hspace{2}}{\hspace{2} {r}^{2} \hspace{2}} \{ {r}^{2} $ f_{x}^{2}{ \sin }^{2}{\theta}- 2\hspace{1}{f}_{x}\hspace{1}{f}_{y}\cos {\theta}\sin {\theta}+f_{y}^{2}{ \cos }^{2}{\theta} $ \} }$

$3$\displaystyle{= $ f_{x}^{2}{ \cos }^{2}{\theta}+2\hspace{1}{f}_{x}\hspace{1}{f}_{y}\cos {\theta}\sin {\theta}+f_{y}^{2}{ \sin }^{2}{\theta} $ }$

$3$\displaystyle{\text{\hspace{1} \hspace{1} \hspace{1} \hspace{1} }+ $ f_{x}^{2}{ \sin }^{2}{\theta}- 2\hspace{1}{f}_{x}\hspace{1}{f}_{y}\cos {\theta}\sin {\theta}+f_{y}^{2}{ \cos }^{2}{\theta} $ }$

$3$\displaystyle{=f_{x}^{2}{\cos }^{2}{\theta}+f_{x}^{2}{\sin }^{2}{\theta}+f_{y}^{2}{\sin }^{2}{\theta}+f_{y}^{2}{\cos }^{2}{\theta}}$

$3$\displaystyle{=f_{x}^{2} $ { \cos }^{2}{\theta}+{ \sin }^{2}{\theta} $ +f_{y}^{2} $ { \sin }^{2}{\theta}+{ \cos }^{2}{\theta} $ }$

$3$\displaystyle{=f_{x}^{2}+f_{y}^{2}}$

$3$\displaystyle{={ $ \frac{\hspace{2} {\partial}z \hspace{2}}{\hspace{2} {\partial}x \hspace{2}} $ }^{2}+{ $ \frac{\hspace{2} {\partial}z \hspace{2}}{\hspace{2} {\partial}y \hspace{2}} $ }^{2}}$

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最終更新日： 2023年8月29日