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Question Number 59655 by aliesam last updated on 13/May/19

Commented by Mr X pcx last updated on 13/May/19
$r=ξ(√(x^2 +y^2 )) ⇒(∂r/∂x) =((ξx)/(√(x^2 +y^2 ))) =((ξx)/r) (∂θ/∂y) =ξ(y/(√(x^2 +y^2 ))) =((ξy)/r) ⇒ ((∂r/∂x))^2 +((∂r/∂y))^2 =(x^2 /r^2 ) +(y^2 /r^2 ) =(r^2 /r^2 ) =1 we have θ =arctan((y/x)) ⇒ (∂θ/∂x) =−(y/(x^2 (1+(y^2 /x^2 )))) =−(y/(x^2 +y^2 )) (∂θ/∂y) =(1/(x(1+(y^2 /x^2 )))) =(1/(x+(y^2 /x))) =(x/(x^2 +y^2 )) ⇒ r^2 { ((∂θ/∂x))^2 +((∂θ/∂y))^2 } =r^2 { (y^2 /r^4 ) +(x^2 /r^4 )} =(1/r^2 )(r^2 ) =1 ⇒the result is proved$
$r = ξ \sqrt{x^{2} + y^{2}} \Rightarrow \frac{\partial r}{\partial x} = \frac{ξ x}{\sqrt{x^{2} + y^{2}}} = \frac{ξ x}{r}$ $\frac{\partial θ}{\partial y} = ξ \frac{y}{\sqrt{x^{2} + y^{2}}} = \frac{ξ y}{r} \Rightarrow$ ${(\frac{\partial r}{\partial x})}^{2} + {(\frac{\partial r}{\partial y})}^{2} = \frac{x^{2}}{r^{2}} + \frac{y^{2}}{r^{2}} = \frac{r^{2}}{r^{2}} = 1$ $w e h a v e θ = a r c t a n (\frac{y}{x}) \Rightarrow$ $\frac{\partial θ}{\partial x} = - \frac{y}{x^{2} (1 + \frac{y^{2}}{x^{2}})} = - \frac{y}{x^{2} + y^{2}}$ $\frac{\partial θ}{\partial y} = \frac{1}{x (1 + \frac{y^{2}}{x^{2}})} = \frac{1}{x + \frac{y^{2}}{x}} = \frac{x}{x^{2} + y^{2}} \Rightarrow$ $r^{2} {{(\frac{\partial θ}{\partial x})}^{2} + {(\frac{\partial θ}{\partial y})}^{2}} = r^{2} {\frac{y^{2}}{r^{4}} + \frac{x^{2}}{r^{4}}}$ $= \frac{1}{r^{2}} (r^{2}) = 1 \Rightarrow t h e r e s u l t i s p r o v e d$
Commented by aliesam last updated on 13/May/19

$e x c e l l e n t . .$
Commented by Mr X pcx last updated on 13/May/19

$ξ = \bar{+} 1$
	Answered by tanmay last updated on 13/May/19

	$x = r c o s θ$ $y = r s i n θ$ $r^{2} = x^{2} + y^{2}$ $2 r \times \frac{\partial r}{\partial x} = 2 x$ $\frac{\partial r}{\partial x} = \frac{x}{r} = c o s θ$ ${(\frac{\partial r}{\partial x})}^{2} + {(\frac{\partial r}{\partial y})}^{2} = {c o s}^{2} θ + {s i n}^{2} θ = 1$ $t a n θ = \frac{y}{x}$ ${s e c}^{2} θ \times \frac{\partial θ}{\partial x} = \frac{\partial}{\partial x} (\frac{y}{x})$ ${s e c}^{2} θ \times \frac{\partial θ}{\partial x} = \frac{- y}{x^{2}}$ $\frac{\partial θ}{\partial x} = \frac{- r s i n θ}{r^{2} {c o s}^{2} θ} \times \frac{1}{{s e c}^{2} θ} = \frac{- s i n θ}{r}$ $r (\frac{\partial θ}{\partial x}) = - s i n θ$ ${s e c}^{2} θ \times \frac{\partial θ}{\partial y} = \frac{1}{x} = \frac{1}{r c o s θ}$ $r (\frac{\partial θ}{\partial y}) = c o s θ$ $s o r^{2} [{(\frac{\partial θ}{\partial x})}^{2} + {(\frac{\partial θ}{\partial y})}^{2}]$ $= {s i n}^{2} θ + {c o s}^{2} θ$ $= 1$
	Commented by aliesam last updated on 13/May/19

	$thank you so much sir$
	Commented by tanmay last updated on 13/May/19

	$m o s t w e l c o m e . . . y o u r s q u e s t i o n s a r e u n i q u e . .$
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