合成関数の偏微分

■問題

$z=f\left(x,y\right)$ , $x=rcos\theta$ , $y=rsin\theta \text{\hspace{0.17em}}$平面の極座標変換） ならば

${\left(\frac{dz}{dx}\right)}^{2}+{\left(\frac{dz}{dy}\right)}^{2}$ $={\left(\frac{dz}{dr}\right)}^{2}+\frac{1}{{r}^{2}}{\left(\frac{dz}{d\theta }\right)}^{2}$

となることを示せ．

■解説

$z$$r$で偏微分する．

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

$={f}_{x}\text{\hspace{0.17em}}·\text{\hspace{0.17em}}cos\theta +{f}_{y}\text{\hspace{0.17em}}·\text{\hspace{0.17em}}sin\theta$

$z$$\theta$ で偏微分する．

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

$={f}_{x}\text{\hspace{0.17em}}·\text{\hspace{0.17em}}\left(-rsin\theta \right)+{f}_{y}\text{\hspace{0.17em}}·\text{\hspace{0.17em}}rcos\theta$

$=r\left(-{f}_{x}sin\theta +{f}_{y}cos\theta \right)$

${\left(\frac{dz}{dr}\right)}^{2}+\frac{1}{{r}^{2}}{\left(\frac{dz}{d\theta }\right)}^{2}$

$={\left({f}_{x}\text{\hspace{0.17em}}·\text{\hspace{0.17em}}cos\theta +{f}_{y}·sin\theta \right)}^{2}$$+\frac{1}{{r}^{2}}{\left\{r\left(-{f}_{x}sin\theta +{f}_{y}cos\theta \right)\right\}}^{2}$

$=\left({f}_{x}^{2}{cos}^{2}\theta +2{f}_{x}{f}_{y}cos\theta sin\theta +{f}_{y}^{2}{sin}^{2}\theta \right)$

$\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}+\frac{1}{{r}^{2}}\left\{{r}^{2}\left({f}_{x}^{2}{sin}^{2}\theta -2{f}_{x}{f}_{y}cos\theta sin\theta +{f}_{y}^{2}{cos}^{2}\theta \right)\right\}$

$=\left({f}_{x}^{2}{cos}^{2}\theta +2{f}_{x}{f}_{y}cos\theta sin\theta +{f}_{y}^{2}{sin}^{2}\theta \right)$

$\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}+\left({f}_{x}^{2}{sin}^{2}\theta -2{f}_{x}{f}_{y}cos\theta sin\theta +{f}_{y}^{2}{cos}^{2}\theta \right)$

$={f}_{x}^{2}{cos}^{2}\theta +{f}_{x}^{2}{sin}^{2}\theta +{f}_{y}^{2}{sin}^{2}\theta +{f}_{y}^{2}{cos}^{2}\theta$

$={f}_{x}^{2}\left({cos}^{2}\theta +{sin}^{2}\theta \right)+{f}_{y}^{2}\left({sin}^{2}\theta +{cos}^{2}\theta \right)$

$={f}_{x}^{2}+{f}_{y}^{2}$

$={\left(\frac{\partial z}{\partial x}\right)}^{2}+{\left(\frac{\partial z}{\partial y}\right)}^{2}$

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