$ \dfrac{u}{v} \bigg( \dfrac{ \partial x}{ \partial u} \bigg)_v \;+\;
\dfrac{v}{y} \bigg( \dfrac{ \partial y}{ \partial v} \bigg)_u \;=\; 0 \ \; \ $

Mumbai University > First Year Engineering > Sem 1 > Applied Maths 1

Marks : 6 M

Year : Dec 2013

u$x+vy=0 \; \; \; \; \therefore x\;=\; \dfrac{-vy}{u} \\ \; \\ \; \\ \dfrac{u}{x} + \dfrac{v}{y} \;=\; 1 \\ \; \\ \; \\ \dfrac{u}{\dfrac{-vy}{u}} + \dfrac{v}{y} \;=\; 1 \; \; \; \; \; or \; \; \dfrac{u^2}{-vy} + \dfrac{v}{y} \;=\; 1 \\ \; \\ \; \\ \therefore \dfrac{1}{y} \bigg( v - \dfrac{u^2}{v} \bigg) \;=\; 1 \\ \; \\ \; \\ \therefore y \;=\; \dfrac{v^2-u^2}{v} \\ \; \\ \; \\ \therefore \bigg( \dfrac{\partial y}{\partial v} \bigg)_u \;=\; \dfrac{ v(2v)-(v^2-u^2)(1) }{v^2} \;=\; \dfrac{ 2v^2-v^2+u^2 }{ v^2 } \;=\; \dfrac{v^2+u^2}{v^2} \\ \; \\ \; \\ \therefore \dfrac{v}{y} \bigg( \dfrac{\partial y}{\partial v} \bigg)_u \;=\; \dfrac{v}{y} \bigg( \dfrac{v^2+u^2}{v^2} \bigg) \;=\; \dfrac{v^2+u^2}{yv} \; \; \; \; \ldots (i) \\ \; \\ \; \\ \; \\ ux+vy=0 \; \; \; \; \therefore y\;=\; \dfrac{-ux}{v} \\ \; \\ \; \\ \dfrac{u}{x} + \dfrac{v}{y} \;=\; 1 \\ \; \\ \; \\ \therefore \dfrac{u}{x} + \dfrac{v}{\dfrac{-ux}{v}} \;=\; 1 \\ \; \\ \; \\ \therefore \dfrac{u}{x} + \dfrac{v^2}{-ux} \;=\; 1 \\ \; \\ \; \\ \therefore \dfrac{1}{x} \bigg( u - \dfrac{v^2}{u} \bigg) \;=\; 1 \\ \; \\ \; \\ \therefore x \;=\; \dfrac{u^2-v^2}{u} \\ \; \\ \; \\ \therefore \bigg( \dfrac{\partial x}{\partial u} \bigg)_v \;=\; \dfrac{ u(2u)-(u^2-v^2)(1) }{u^2} \;=\; \dfrac{ 2u^2-u^2+v^2 }{ u^2 } \;=\; \dfrac{u^2+v^2}{u^2} \\ \; \\ \; \\ \therefore \dfrac{u}{x} \bigg( \dfrac{\partial x}{\partial u} \bigg)_v \;=\; \dfrac{u}{x} \bigg( \dfrac{v^2+u^2}{u^2} \bigg) \;=\; \dfrac{v^2+u^2}{xu} \; \; \; \; \ldots (ii) \\ \; \\ \; \\ \; \\ $

From (i) and (ii)

$ \therefore \dfrac{u}{x} \bigg( \dfrac{\partial x}{\partial u} \bigg)_v + \dfrac{v}{y} \bigg( \dfrac{\partial y}{\partial v} \bigg)_u \;=\; \dfrac{v^2+u^2}{xu} + \dfrac{v^2+u^2}{xu} \\ \; \\ \; \\ = (u^2+v^2) \bigg[ \dfrac{1}{xu} + \dfrac{1}{xu} \bigg] \;=\; = (u^2+v^2) \bigg[ \dfrac{ux+vy}{xuyv} \bigg] \;=\; (u^2+v^2) (0) \\ \{ \because ux+vy=0 \} \\ \; \\ \; \\ \; \\ \; \\ \; \\ \dfrac{u}{v} \bigg( \dfrac{ \partial x}{ \partial u} \bigg)_v \;+\; \dfrac{v}{y} \bigg( \dfrac{ \partial y}{ \partial v} \bigg)_u \;=\; 0 \; \; \; \; \; Hence Proved. $