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

Marks : 6 M

Year : Dec 2014

x $=uv \; \; u= \dfrac{x}{v} \; \; \; \ldots (i) \\ \; \\ \; \\ y=\dfrac{u}{v} \; \; \; \therefore u\;=\; yv \; \; \ldots (ii) \\ \; \\ \; \\ $

From (i)and (ii), $ \dfrac{x}{v}=yv \; \; \therefore v^2=\dfrac{x}{y} \; \; v=\sqrt{\dfrac{x}{y}} \; \; \therefore \; u=yv=y\sqrt{\dfrac{x}{y}} = \sqrt{xy} \\ \; \\ \; \\ Now, \; J = \dfrac{\partial (x,y)}{\partial (u,v)} \;=\; \left| \begin{array}{cc} \frac{\partial x }{\partial u} & \frac{\partial x }{\partial v} \\ \frac{\partial y }{\partial u} & \frac{\partial y }{\partial v} \end{array} \right| \;=\; \left| \begin{array}{cc} v & u \\ 1/v & -u/v^2 \end{array} \right| \;=\; (v) (\dfrac{-u}{v^2}) - u(\dfrac{1}{v}) \;=\; \dfrac{-u}{v} - \dfrac{u}{v} \\ \; \\ \; \\ = \dfrac{-2u}{v} \; \; \; \ldots (iii) \\ \; \\ \; \\ \; \\ \; \\ $

$ J^{-1} = \dfrac{\partial (u,v)} {\partial (x,y)} \;=\; \left| \begin{array}{cc} \frac{\partial u }{\partial x} & \frac{\partial u }{\partial y} \\ \frac{\partial v }{\partial x} & \frac{\partial v }{\partial y} \end{array} \right| \;=\; \left| \begin{array}{cc} \dfrac{\sqrt{y}}{2\sqrt{x}} & \dfrac{\sqrt{x}}{2\sqrt{y}} \\ \dfrac{1}{2\sqrt{xy}} & \dfrac{-\sqrt{x}}{2y\sqrt{y}} \end{array} \right| \;=\; = \bigg( \dfrac{\sqrt{y}}{2\sqrt{x}} \bigg) \bigg( \dfrac{-\sqrt{x}}{2y\sqrt{y}} \bigg) \;-\; \bigg( \dfrac{\sqrt{x}}{2\sqrt{y}} \bigg) \bigg( \dfrac{1}{2\sqrt{xy}} \bigg) \; \; \; \; \; \ldots (iv) \\ \; \\ \; \\ \therefore J^{-1} \;=\; \dfrac{-1}{4y} - \dfrac{1}{4y} \;=\; \dfrac{-1}{2y} \;=\; \dfrac{-1}{2 \dfrac{u}{v}} \;=\; \dfrac{-v}{2u} \; \; \; \; From \; (ii) \\ \; \\ \; \\ \; \\ \therefore J . J^{-1} \;=\; \dfrac{-2u}{v} \cdot \dfrac{-v}{2u} \;=\; 1 \; \; \; \ldots ( From \; (iii) \; and \; (iv)) \\ \; \\ $

Hence Proved.