15.8 , 1 a(x,y)Solve the system u = x-y, v= 3x + 3y for x and y in terms of u and v. Then find the value of the Jacobian . Find the image under the transformation of the triangular region with vertices (0,0), (3,3), and (3, -3) in the xy-plane. Sketch the transformed region in the uv-plane.a(u,v)The function for x in terms of u and vis x =G

Jacobian is defined as considering x and y to be two functions with respect to two independent variables u and v then the determinant ∂(x,y)∂(u,v)=∂x∂u∂x∂v∂y∂u∂y∂v is known as jacobian of x and y with respect to u and v.Jacobian is denoted by ∂(x,y)∂(u,v) or J(x,y). If three Function is given with respect to three independent variables then three order determinant exist. The given system is, u=x-y......(1)v=3x+3y...(2) From equation (1), u=x-yx=u+y.....(3) Put x in equation (2), v=3x+3yv=3(u+y)+3yv=3u+3y+3y6y=v-3uy=v-3u6 Put the value of y in equation (3), x=u+v-3u6x=6u+v-3u6x=3u+v6 The value of x and y in terms of u and v is, x=3u+v6,y=v-3u6 Now, ∂(x,y)∂(u,v)=∂x∂u∂x∂v∂y∂u∂y∂v =∂∂u3u+v6∂∂v3u+v6∂∂uv-3u6∂∂vv-3u6 =(3×1)+06(3×0)+160-(3×1)61-(3×0)6 =3616-3616∂(x,y)∂(u,v) =1216-1216 Now expanding determinants, ∂(x,y)∂(u,v) =1216-1216 =12×16--12×16 =112+112 =212∂(x,y)∂(u,v)=16 The image of the vertices (0,0),(3,3) and (3,-3) in xy-plane is, The table for u and v plane, x 0 3 3 y 0 3 -3 u=x-y u=0-0u=0 u=3-3u=0 u=3--3u=3+3u=6 v=3x+3y v=3×0+3×0v=0+0v=0 v=3×3+3×3v=9+9v=18 v=3×3+3×-3v=9-9v=0 (u,v) (0,0) (0,18) (6,0) The graph for uv plane is, The value of x in u and v is, x=3u+v6.