## 2018年1月21日日曜日

### 数学 - 解析学 - 多変数の関数 - 微分可能性と勾配ベクトル(極座標、微分鎖律)

1. $\begin{array}{}x=r\mathrm{cos}\theta \\ y=r\mathrm{sin}\theta \end{array}$
$\begin{array}{}\frac{\partial g}{\partial r}\\ =grad\left(f\left(x,y\right)\right)·\left(\mathrm{cos}\theta ,\mathrm{sin}\theta \right)\\ =\left(\frac{\partial f}{\partial x},\frac{\partial f}{\partial y}\right)·\left(\mathrm{cos}\theta ,\mathrm{sin}\theta \right)\\ =\frac{\partial f}{\partial x}\mathrm{cos}\theta +\frac{\partial f}{\partial y}\mathrm{sin}\theta \end{array}$
$\begin{array}{}\frac{\partial g}{\partial \theta }\\ =\left(\frac{\partial f}{\partial x},\frac{\partial f}{\partial y}\right)·\left(-r\mathrm{sin}\theta ,r\mathrm{cos}\theta \right)\\ =r\left(-\frac{\partial f}{\partial x}\mathrm{sin}\theta +\frac{\partial f}{\partial y}\mathrm{cos}\theta \right)\end{array}$
${\left(\frac{\partial g}{\partial r}\right)}^{2}={\left(\frac{\partial f}{\partial x}\right)}^{2}{\mathrm{cos}}^{2}\theta +{\left(\frac{\partial f}{\partial y}\right)}^{2}{\mathrm{sin}}^{2}\theta +2\left(\frac{\partial f}{\partial x}\right)\left(\frac{\partial f}{\partial y}\right)\mathrm{sin}\theta \mathrm{cos}\theta$
$\frac{1}{{r}^{2}}{\left(\frac{\partial g}{\partial \theta }\right)}^{2}={\left(\frac{\partial f}{\partial x}\right)}^{2}{\mathrm{sin}}^{2}\theta +{\left(\frac{\partial f}{\partial y}\right)}^{2}{\mathrm{cos}}^{2}\theta -2\left(\frac{\partial f}{\partial x}\right)\left(\frac{\partial f}{\partial y}\right)\mathrm{sin}\theta \mathrm{cos}\theta$
$\begin{array}{}{\left(\frac{\partial g}{\partial r}\right)}^{2}+\frac{1}{{r}^{2}}{\left(\frac{\partial g}{\partial \theta }\right)}^{2}={\left(\frac{\partial f}{\partial x}\right)}^{2}\left({\mathrm{sin}}^{2}\theta +{\mathrm{cos}}^{2}\theta \right)+{\left(\frac{\partial f}{\partial y}\right)}^{2}\left({\mathrm{sin}}^{2}\theta +{\mathrm{cos}}^{2}\theta \right)\\ {\left(\frac{\partial g}{\partial r}\right)}^{2}+\frac{1}{{r}^{2}}{\left(\frac{\partial g}{\partial \theta }\right)}^{2}={\left(\frac{\partial f}{\partial x}\right)}^{2}+{\left(\frac{\partial f}{\partial y}\right)}^{2}\end{array}$