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Question Number 72886 by mhmd last updated on 04/Nov/19

let w=f(x, y) be a differentiable function where x=rcosθ and y=rsinθ show that (f_x )^2 +(f_y )^2 =(w_x )^2 +1/r^2 (w_y )^2 ? help me sir

$${let}\:{w}={f}\left({x},\:{y}\right)\:{be}\:{a}\:{differentiable}\:{function}\:{where}\:{x}={rcos}\theta\:{and}\:{y}={rsin}\theta\:{show}\:{that}\:\left({f}_{{x}} \right)^{\mathrm{2}} +\left({f}_{{y}} \right)^{\mathrm{2}} =\left({w}_{{x}} \right)^{\mathrm{2}} +\mathrm{1}/{r}^{\mathrm{2}} \left({w}_{{y}} \right)^{\mathrm{2}} ? \\ $$$${help}\:{me}\:{sir}\: \\ $$

Answered by mind is power last updated on 04/Nov/19

w(r,θ)=f(rcos(θ),rsin(θ)) =f(B(r,θ) ⇒((∂w/∂r),(∂w/∂θ))=((∂f/∂x),(∂f/∂y)). (((cos(θ) −rsin(θ))),((sin(θ) rcos(θ))) ) ⇒((∂f/∂x),(∂f/∂y))=((∂w/∂r),(∂w/∂θ)). (((cos(θ) −rsin(θ))),((sin(θ) rcos(θ))) )^(−1) =((∂w/∂r),(∂w/∂θ)).(1/r). (((rcos(θ) rsin(θ))),((−sin(θ) cos(θ) )) ) (∂f/∂x)=cos(θ)(∂w/∂r)−((sin(θ))/r).(∂w/∂θ) (∂f/∂y)=sin(θ)(∂w/∂r)+((cos(θ))/r).(∂w/∂θ) ⇒((∂f/∂x))^2 +((∂f/∂y))^2 =(cos(θ)(∂w/∂r)−((sin(θ))/r)(∂w/∂θ))^2 +(sin(θ)(∂w/∂r)+((cos(θ))/r)(∂w/∂θ))^2 =((∂w/∂r))^2 +(1/r^2 ).((∂w/∂θ))^2

$$\mathrm{w}\left(\mathrm{r},\theta\right)=\mathrm{f}\left(\mathrm{rcos}\left(\theta\right),\mathrm{rsin}\left(\theta\right)\right) \\ $$$$=\mathrm{f}\left(\mathrm{B}\left(\mathrm{r},\theta\right)\right. \\ $$$$\Rightarrow\left(\frac{\partial\mathrm{w}}{\partial\mathrm{r}},\frac{\partial\mathrm{w}}{\partial\theta}\right)=\left(\frac{\partial\mathrm{f}}{\partial\mathrm{x}},\frac{\partial\mathrm{f}}{\partial\mathrm{y}}\right).\begin{pmatrix}{\mathrm{cos}\left(\theta\right)\:\:\:\:\:−\mathrm{rsin}\left(\theta\right)}\\{\mathrm{sin}\left(\theta\right)\:\:\:\:\:\:\:\:\:\mathrm{rcos}\left(\theta\right)}\end{pmatrix} \\ $$$$\Rightarrow\left(\frac{\partial\mathrm{f}}{\partial\mathrm{x}},\frac{\partial\mathrm{f}}{\partial\mathrm{y}}\right)=\left(\frac{\partial\mathrm{w}}{\partial\mathrm{r}},\frac{\partial\mathrm{w}}{\partial\theta}\right).\begin{pmatrix}{\mathrm{cos}\left(\theta\right)\:\:\:\:\:\:−\mathrm{rsin}\left(\theta\right)}\\{\mathrm{sin}\left(\theta\right)\:\:\:\:\:\:\:\:\:\:\:\mathrm{rcos}\left(\theta\right)}\end{pmatrix}^{−\mathrm{1}} \\ $$$$=\left(\frac{\partial\mathrm{w}}{\partial\mathrm{r}},\frac{\partial\mathrm{w}}{\partial\theta}\right).\frac{\mathrm{1}}{\mathrm{r}}.\begin{pmatrix}{\mathrm{rcos}\left(\theta\right)\:\:\:\:\:\:\:\:\:\mathrm{rsin}\left(\theta\right)}\\{−\mathrm{sin}\left(\theta\right)\:\:\:\:\:\:\:\:\:\mathrm{cos}\left(\theta\right)\:\:}\end{pmatrix} \\ $$$$\frac{\partial\mathrm{f}}{\partial\mathrm{x}}=\mathrm{cos}\left(\theta\right)\frac{\partial\mathrm{w}}{\partial\mathrm{r}}−\frac{\mathrm{sin}\left(\theta\right)}{\mathrm{r}}.\frac{\partial\mathrm{w}}{\partial\theta} \\ $$$$\frac{\partial\mathrm{f}}{\partial\mathrm{y}}=\mathrm{sin}\left(\theta\right)\frac{\partial\mathrm{w}}{\partial\mathrm{r}}+\frac{\mathrm{cos}\left(\theta\right)}{\mathrm{r}}.\frac{\partial\mathrm{w}}{\partial\theta} \\ $$$$\Rightarrow\left(\frac{\partial\mathrm{f}}{\partial\mathrm{x}}\right)^{\mathrm{2}} +\left(\frac{\partial\mathrm{f}}{\partial\mathrm{y}}\right)^{\mathrm{2}} =\left(\mathrm{cos}\left(\theta\right)\frac{\partial\mathrm{w}}{\partial\mathrm{r}}−\frac{\mathrm{sin}\left(\theta\right)}{\mathrm{r}}\frac{\partial\mathrm{w}}{\partial\theta}\right)^{\mathrm{2}} +\left(\mathrm{sin}\left(\theta\right)\frac{\partial\mathrm{w}}{\partial\mathrm{r}}+\frac{\mathrm{cos}\left(\theta\right)}{\mathrm{r}}\frac{\partial\mathrm{w}}{\partial\theta}\right)^{\mathrm{2}} \\ $$$$=\left(\frac{\partial\mathrm{w}}{\partial\mathrm{r}}\right)^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{r}^{\mathrm{2}} }.\left(\frac{\partial\mathrm{w}}{\partial\theta}\right)^{\mathrm{2}} \\ $$$$ \\ $$

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