Question

If ux+vy=a, u/x+v/y=1,show that
\[u/x (∂x/∂u)_v+v/y (∂y/∂v)_u=0\]

Answer 1

my attempt :

Answer 2

\(\large \dfrac {u}{x} (\dfrac{∂x}{∂u})_v+\dfrac{v}{y} (\dfrac{∂y}{∂v})_u=0\)

Answer 3

I should be getting 0 when i add them...not sure if i have done any simple error in any step....or i just need to proceed ahead with that complex algebraic simplification and would eventually be getting 0.....

Answer 4

\(\large \left(\dfrac{\partial x}{\partial u}\right)_v : \)

\[\large\begin{align} ux+vy=a &\implies x + u\left(\dfrac{\partial x}{\partial u}\right)_v + v\left(\dfrac{\partial y}{\partial u}\right)_v = 0 \\~\\u/x+v/y=1 &\implies \dfrac{x - u \left(\dfrac{\partial x}{\partial u}\right)_v }{x^2} -\dfrac{v}{y^2}\left(\dfrac{\partial y}{\partial u}\right)_v = 0\end{align}\]

eliminating dy gives :

\[\large\begin{align} \left(\dfrac{\partial x}{\partial u}\right)_v &= \dfrac{-x(x^2+y^2)}{u(x^2-y^2)} \end{align} \]

Answer 5

similarly we get
\[\large\begin{align} \left(\dfrac{\partial y}{\partial v}\right)_u &= \dfrac{y(x^2+y^2)}{v(x^2-y^2)} \end{align} \]