Question Number 72886 by mhmd last updated on 04/Nov/19
$${let}\:{w}={f}\left({x},\:{y}\right)\:{be}\:{a}\:{differentiable}\:{function}\:{where}\:{x}={rcos}\theta\:{and}\:{y}={rsin}\theta\:{show}\:{that}\:\left({f}_{{x}} \right)^{\mathrm{2}} +\left({f}_{{y}} \right)^{\mathrm{2}} =\left({w}_{{x}} \right)^{\mathrm{2}} +\mathrm{1}/{r}^{\mathrm{2}} \left({w}_{{y}} \right)^{\mathrm{2}} ? \\ $$$${help}\:{me}\:{sir}\: \\ $$
Answered by mind is power last updated on 04/Nov/19
$$\mathrm{w}\left(\mathrm{r},\theta\right)=\mathrm{f}\left(\mathrm{rcos}\left(\theta\right),\mathrm{rsin}\left(\theta\right)\right) \\ $$$$=\mathrm{f}\left(\mathrm{B}\left(\mathrm{r},\theta\right)\right. \\ $$$$\Rightarrow\left(\frac{\partial\mathrm{w}}{\partial\mathrm{r}},\frac{\partial\mathrm{w}}{\partial\theta}\right)=\left(\frac{\partial\mathrm{f}}{\partial\mathrm{x}},\frac{\partial\mathrm{f}}{\partial\mathrm{y}}\right).\begin{pmatrix}{\mathrm{cos}\left(\theta\right)\:\:\:\:\:−\mathrm{rsin}\left(\theta\right)}\\{\mathrm{sin}\left(\theta\right)\:\:\:\:\:\:\:\:\:\mathrm{rcos}\left(\theta\right)}\end{pmatrix} \\ $$$$\Rightarrow\left(\frac{\partial\mathrm{f}}{\partial\mathrm{x}},\frac{\partial\mathrm{f}}{\partial\mathrm{y}}\right)=\left(\frac{\partial\mathrm{w}}{\partial\mathrm{r}},\frac{\partial\mathrm{w}}{\partial\theta}\right).\begin{pmatrix}{\mathrm{cos}\left(\theta\right)\:\:\:\:\:\:−\mathrm{rsin}\left(\theta\right)}\\{\mathrm{sin}\left(\theta\right)\:\:\:\:\:\:\:\:\:\:\:\mathrm{rcos}\left(\theta\right)}\end{pmatrix}^{−\mathrm{1}} \\ $$$$=\left(\frac{\partial\mathrm{w}}{\partial\mathrm{r}},\frac{\partial\mathrm{w}}{\partial\theta}\right).\frac{\mathrm{1}}{\mathrm{r}}.\begin{pmatrix}{\mathrm{rcos}\left(\theta\right)\:\:\:\:\:\:\:\:\:\mathrm{rsin}\left(\theta\right)}\\{−\mathrm{sin}\left(\theta\right)\:\:\:\:\:\:\:\:\:\mathrm{cos}\left(\theta\right)\:\:}\end{pmatrix} \\ $$$$\frac{\partial\mathrm{f}}{\partial\mathrm{x}}=\mathrm{cos}\left(\theta\right)\frac{\partial\mathrm{w}}{\partial\mathrm{r}}−\frac{\mathrm{sin}\left(\theta\right)}{\mathrm{r}}.\frac{\partial\mathrm{w}}{\partial\theta} \\ $$$$\frac{\partial\mathrm{f}}{\partial\mathrm{y}}=\mathrm{sin}\left(\theta\right)\frac{\partial\mathrm{w}}{\partial\mathrm{r}}+\frac{\mathrm{cos}\left(\theta\right)}{\mathrm{r}}.\frac{\partial\mathrm{w}}{\partial\theta} \\ $$$$\Rightarrow\left(\frac{\partial\mathrm{f}}{\partial\mathrm{x}}\right)^{\mathrm{2}} +\left(\frac{\partial\mathrm{f}}{\partial\mathrm{y}}\right)^{\mathrm{2}} =\left(\mathrm{cos}\left(\theta\right)\frac{\partial\mathrm{w}}{\partial\mathrm{r}}−\frac{\mathrm{sin}\left(\theta\right)}{\mathrm{r}}\frac{\partial\mathrm{w}}{\partial\theta}\right)^{\mathrm{2}} +\left(\mathrm{sin}\left(\theta\right)\frac{\partial\mathrm{w}}{\partial\mathrm{r}}+\frac{\mathrm{cos}\left(\theta\right)}{\mathrm{r}}\frac{\partial\mathrm{w}}{\partial\theta}\right)^{\mathrm{2}} \\ $$$$=\left(\frac{\partial\mathrm{w}}{\partial\mathrm{r}}\right)^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{r}^{\mathrm{2}} }.\left(\frac{\partial\mathrm{w}}{\partial\theta}\right)^{\mathrm{2}} \\ $$$$ \\ $$