Subject:- Applied Mathematics

Marks:- 3

Mumbai Unversity>FE>Sem1>Applied Maths1

$\frac{∂(u,v)}{∂(x,y)}$ = $ \begin{bmatrix} U_x & U_y \ V_x & V_y \end{bmatrix} $ x = uv, y = $\frac{(u+v)}{(u-v)}$ …………(given) we know that JJ’ = 1 ………………..(1) the equation can also be solved by this following method. $\frac{∂(x,y)}{∂(u,v)}$ = $ \begin{bmatrix} \frac{(\partial)x}{(\partial)u} & \frac{(\partial)x}{(\partial)v} \ \frac{(\partial)y}{(\partial)u} & \frac{(\partial)y}{(\partial)v} \end{bmatrix} $ $\frac{∂x}{∂u}$ = ∂(uv)= v. ……………. (2) $\frac{∂x}{∂v}$ = ∂(uv)= u. …………….(3) $\frac{∂y}{∂u}$ = ∂($\frac{((u+v)}{(u-v)}$) = $\frac{((u-v)-(u+v))}{(u-v)^2}$ (u-v-u+v/u-v^2) = $\frac{(-2v)}{(u-v)^2}$ ………………..(4) $\frac{∂y}{∂v}$ = ∂($\frac{(u+v)}{(u-v)}$) = $\frac{((u-v)+(u+v))}{(u-v)^2}$ = $\frac{2u}{(u-v)^2}$ …………………(5) From equation (2), (3), (4), (5) we get, $ \begin{bmatrix} \frac{(\partial)x}{(\partial)u} & \frac{(\partial)x}{(\partial)v} \ \frac{(\partial)y}{(\partial)u} & \frac{(\partial)y}{(\partial)v} \end{bmatrix} $ = $ \begin{bmatrix} V & U \ \frac{-v}{(u-v)^2} & \frac{-u}{(u-v)^2} \end{bmatrix} $ = $\frac{2uv}{(u-v}^2 + \frac{2uv}{(u-v}^2 = \frac{4uv}{(u-v)}^2$ . From (1) we get, JJ’ = 1 J × $\frac{4uv}{(u-v)^2}$ = 1 ………………(let J’ = $\frac{4uv}{(u-v)^2}$ Hence J =$\frac{(u-v)^2}{4uv}$. ∴$e^{2φ}$ = cot$\frac{α}{2}$