Question

高数急需谢谢

Answer 1

x=e^u + usinv (1)
y= e^u - usinv (2)

(1)+(2)
x+y= 2e^u (3)
1 = 2e^u .∂u/∂x
∂u/∂x= (1/2) e^(-u)

from (3)
x+y= 2e^u
1 = 2e^u .∂u/∂y
∂u/∂y= (1/2) e^(-u)

from (1)
x=e^u + usinv
1=e^u.(∂u/∂x) + ucosv.(∂v/∂x) + sinv. (∂u/∂x)
1=e^u.[(1/2) e^(-u)] + ucosv.(∂v/∂x) + sinv. [(1/2) e^(-u)]
∂v/∂x = (1/2) [ 1- e^(-u). sinv ] /(ucosv)

from (2)
y= e^u - usinv
1=e^u.(∂u/∂y) + ucosv.(∂v/∂y) + sinv. (∂u/∂y)
1=e^u.[(1/2) e^(-u)] + ucosv.(∂v/∂y) + sinv. [(1/2) e^(-u)]
∂v/∂y = (1/2) [ 1- e^(-u). sinv ] /(ucosv)