let-w-f-x-y-be-a-differentiable-function-where-x-rcos-and-y-rsin-show-that-f-x-2-f-y-2-w-x-2-1-r-2-w-y-2-help-me-sir-

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

l e t w = f (x, y) b e a d i f f e r e n t i a b l e f u n c t i o n w h e r e x = r c o s θ a n d y = r s i n θ s h o w t h a t {(f_{x})}^{2} + {(f_{y})}^{2} = {(w_{x})}^{2} + 1 / r^{2} {(w_{y})}^{2} ?

h e l p m e s i r

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

w (r, θ) = f (rcos (θ), rsin (θ))

= f (B (r, θ)

\Rightarrow (\frac{\partial w}{\partial r}, \frac{\partial w}{\partial θ}) = (\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}) . (\begin{matrix} \cos (θ) - rsin (θ) \\ \sin (θ) rcos (θ) \end{matrix})

\Rightarrow (\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}) = (\frac{\partial w}{\partial r}, \frac{\partial w}{\partial θ}) . {(\begin{matrix} \cos (θ) - rsin (θ) \\ \sin (θ) rcos (θ) \end{matrix})}^{- 1}

= (\frac{\partial w}{\partial r}, \frac{\partial w}{\partial θ}) . \frac{1}{r} . (\begin{matrix} rcos (θ) rsin (θ) \\ - \sin (θ) \cos (θ) \end{matrix})

\frac{\partial f}{\partial x} = \cos (θ) \frac{\partial w}{\partial r} - \frac{\sin (θ)}{r} . \frac{\partial w}{\partial θ}

\frac{\partial f}{\partial y} = \sin (θ) \frac{\partial w}{\partial r} + \frac{\cos (θ)}{r} . \frac{\partial w}{\partial θ}

\Rightarrow {(\frac{\partial f}{\partial x})}^{2} + {(\frac{\partial f}{\partial y})}^{2} = {(\cos (θ) \frac{\partial w}{\partial r} - \frac{\sin (θ)}{r} \frac{\partial w}{\partial θ})}^{2} + {(\sin (θ) \frac{\partial w}{\partial r} + \frac{\cos (θ)}{r} \frac{\partial w}{\partial θ})}^{2}

= {(\frac{\partial w}{\partial r})}^{2} + \frac{1}{r^{2}} . {(\frac{\partial w}{\partial θ})}^{2}