Menu Close

Question-59655




Question Number 59655 by aliesam last updated on 13/May/19
Commented by Mr X pcx last updated on 13/May/19
r=ξ(√(x^2  +y^2  )) ⇒(∂r/∂x) =((ξx)/( (√(x^2  +y^2 )))) =((ξx)/r)  (∂θ/∂y) =ξ(y/( (√(x^2 +y^2 )))) =((ξy)/r) ⇒  ((∂r/∂x))^2  +((∂r/∂y))^2  =(x^2 /r^2 ) +(y^2 /r^2 ) =(r^2 /r^2 ) =1  we have θ =arctan((y/x)) ⇒  (∂θ/∂x) =−(y/(x^2 (1+(y^2 /x^2 )))) =−(y/(x^2  +y^2 ))  (∂θ/∂y) =(1/(x(1+(y^2 /x^2 )))) =(1/(x+(y^2 /x))) =(x/(x^2  +y^2 )) ⇒  r^2 { ((∂θ/∂x))^2  +((∂θ/∂y))^2 } =r^2 { (y^2 /r^4 ) +(x^2 /r^4 )}  =(1/r^2 )(r^2 ) =1 ⇒the result is proved
$${r}=\xi\sqrt{{x}^{\mathrm{2}} \:+{y}^{\mathrm{2}} \:}\:\Rightarrow\frac{\partial{r}}{\partial{x}}\:=\frac{\xi{x}}{\:\sqrt{{x}^{\mathrm{2}} \:+{y}^{\mathrm{2}} }}\:=\frac{\xi{x}}{{r}} \\ $$$$\frac{\partial\theta}{\partial{y}}\:=\xi\frac{{y}}{\:\sqrt{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} }}\:=\frac{\xi{y}}{{r}}\:\Rightarrow \\ $$$$\left(\frac{\partial{r}}{\partial{x}}\right)^{\mathrm{2}} \:+\left(\frac{\partial{r}}{\partial{y}}\right)^{\mathrm{2}} \:=\frac{{x}^{\mathrm{2}} }{{r}^{\mathrm{2}} }\:+\frac{{y}^{\mathrm{2}} }{{r}^{\mathrm{2}} }\:=\frac{{r}^{\mathrm{2}} }{{r}^{\mathrm{2}} }\:=\mathrm{1} \\ $$$${we}\:{have}\:\theta\:={arctan}\left(\frac{{y}}{{x}}\right)\:\Rightarrow \\ $$$$\frac{\partial\theta}{\partial{x}}\:=−\frac{{y}}{{x}^{\mathrm{2}} \left(\mathrm{1}+\frac{{y}^{\mathrm{2}} }{{x}^{\mathrm{2}} }\right)}\:=−\frac{{y}}{{x}^{\mathrm{2}} \:+{y}^{\mathrm{2}} } \\ $$$$\frac{\partial\theta}{\partial{y}}\:=\frac{\mathrm{1}}{{x}\left(\mathrm{1}+\frac{{y}^{\mathrm{2}} }{{x}^{\mathrm{2}} }\right)}\:=\frac{\mathrm{1}}{{x}+\frac{{y}^{\mathrm{2}} }{{x}}}\:=\frac{{x}}{{x}^{\mathrm{2}} \:+{y}^{\mathrm{2}} }\:\Rightarrow \\ $$$${r}^{\mathrm{2}} \left\{\:\left(\frac{\partial\theta}{\partial{x}}\right)^{\mathrm{2}} \:+\left(\frac{\partial\theta}{\partial{y}}\right)^{\mathrm{2}} \right\}\:={r}^{\mathrm{2}} \left\{\:\frac{{y}^{\mathrm{2}} }{{r}^{\mathrm{4}} }\:+\frac{{x}^{\mathrm{2}} }{{r}^{\mathrm{4}} }\right\} \\ $$$$=\frac{\mathrm{1}}{{r}^{\mathrm{2}} }\left({r}^{\mathrm{2}} \right)\:=\mathrm{1}\:\Rightarrow{the}\:{result}\:{is}\:{proved} \\ $$$$ \\ $$
Commented by aliesam last updated on 13/May/19
excellent..
$${excellent}.. \\ $$
Commented by Mr X pcx last updated on 13/May/19
ξ=+^− 1
$$\xi=\overset{−} {+}\mathrm{1} \\ $$
Answered by tanmay last updated on 13/May/19
x=rcosθ  y=rsinθ  r^2 =x^2 +y^2   2r×(∂r/∂x)=2x  (∂r/∂x)=(x/r)=cosθ  ((∂r/∂x))^2 +((∂r/∂y))^2 =cos^2 θ+sin^2 θ=1  tanθ=(y/x)  sec^2 θ×(∂θ/∂x)=(∂/∂x)((y/x))  sec^2 θ×(∂θ/∂x)=((−y)/x^2 )  (∂θ/∂x)=((−rsinθ)/(r^2 cos^2 θ))×(1/(sec^2 θ))=((−sinθ)/r)  r((∂θ/∂x))=−sinθ  sec^2 θ×(∂θ/∂y)=(1/x)=(1/(rcosθ))  r((∂θ/∂y))=cosθ  so r^2 [((∂θ/∂x))^2 +((∂θ/∂y))^2 ]  =sin^2 θ+cos^2 θ  =1
$${x}={rcos}\theta \\ $$$${y}={rsin}\theta \\ $$$${r}^{\mathrm{2}} ={x}^{\mathrm{2}} +{y}^{\mathrm{2}} \\ $$$$\mathrm{2}{r}×\frac{\partial{r}}{\partial{x}}=\mathrm{2}{x} \\ $$$$\frac{\partial{r}}{\partial{x}}=\frac{{x}}{{r}}={cos}\theta \\ $$$$\left(\frac{\partial{r}}{\partial{x}}\right)^{\mathrm{2}} +\left(\frac{\partial{r}}{\partial{y}}\right)^{\mathrm{2}} ={cos}^{\mathrm{2}} \theta+{sin}^{\mathrm{2}} \theta=\mathrm{1} \\ $$$${tan}\theta=\frac{{y}}{{x}} \\ $$$${sec}^{\mathrm{2}} \theta×\frac{\partial\theta}{\partial{x}}=\frac{\partial}{\partial{x}}\left(\frac{{y}}{{x}}\right) \\ $$$${sec}^{\mathrm{2}} \theta×\frac{\partial\theta}{\partial{x}}=\frac{−{y}}{{x}^{\mathrm{2}} } \\ $$$$\frac{\partial\theta}{\partial{x}}=\frac{−{rsin}\theta}{{r}^{\mathrm{2}} {cos}^{\mathrm{2}} \theta}×\frac{\mathrm{1}}{{sec}^{\mathrm{2}} \theta}=\frac{−{sin}\theta}{{r}} \\ $$$${r}\left(\frac{\partial\theta}{\partial{x}}\right)=−{sin}\theta \\ $$$${sec}^{\mathrm{2}} \theta×\frac{\partial\theta}{\partial{y}}=\frac{\mathrm{1}}{{x}}=\frac{\mathrm{1}}{{rcos}\theta} \\ $$$${r}\left(\frac{\partial\theta}{\partial{y}}\right)={cos}\theta \\ $$$${so}\:{r}^{\mathrm{2}} \left[\left(\frac{\partial\theta}{\partial{x}}\right)^{\mathrm{2}} +\left(\frac{\partial\theta}{\partial{y}}\right)^{\mathrm{2}} \right] \\ $$$$={sin}^{\mathrm{2}} \theta+{cos}^{\mathrm{2}} \theta \\ $$$$=\mathrm{1} \\ $$
Commented by aliesam last updated on 13/May/19
thank you so much sir
$$\mathrm{thank}\:\mathrm{you}\:\mathrm{so}\:\mathrm{much}\:\mathrm{sir} \\ $$
Commented by tanmay last updated on 13/May/19
most welcome...yours questions are unique..
$${most}\:{welcome}…{yours}\:{questions}\:{are}\:{unique}.. \\ $$

Leave a Reply

Your email address will not be published. Required fields are marked *