Question

高等数学，第1题的(1)(2)(3)(4)(5)怎么做，需要详细过程，急，求高手

Answer 1

解：
1）
将x,y看成u和v的函数
∂x/∂u=cosv
∂x/∂v=-usinv
∂y/∂u=sinv
∂x/∂v=ucosv
∂z/∂u
=2xy·(∂x/∂u)+x²·(∂y/∂u)-y²·(∂x/∂u)-2xy·(∂y/∂u)
=2xycosv+x²sinv-y²cosv-2xysinv
=(2xy-y²)cosv+(x²-2xy)sinv
∂z/∂v
=2xy·(∂x/∂v)+x²·(∂y/∂v)-y²·(∂x/∂v)-2xy·(∂y/∂v)
=2xy·(-usinv)+x²·(sinv)-y²·(-usinv)-2xy·(sinv)
=(y²-2xy)usinv+(x²-2xy)sinv
2)
求一阶全微分
dz=[e^(x-2y)]dx+(-2)·[e^(x-2y)]dy
dx=costdt
dy=3t²dt
dz=[e^(x-2y)]·(costdt)+(-2)·[e^(x-2y)]·(3t²dt)
因此：
dz/dt
=cost[e^(x-2y)]-6t²[e^(x-2y)]
3）
求一阶微分

du=2xdx+2ydt+2zdz
dx=(e^t)·(cost-sint)dt
dy=(e^t)·(sint+cost)dt
dz=e^tdt
du=2x(e^t)·(cost-sint)dt+2y(e^t)·(sint+cost)dt+2ze^tdt
因此：
du/dt
=2x(e^t)·(cost-sint)+2y(e^t)·(sint+cost)+2ze^t
4)
dz=2uv³wdu+3u²v²wdv+u²v³dw
du=2dx
dv=3x²dx
dw=3dx
dz=4uv³wdx+9u²v²x²wdx+3u²v³dx
dz/dx
=4uv³w+9u²v²x²w+3u²v³
5)
dz=[2x-sin(x+y)]dx+[2y-sin(x+y)]dy

dx=vdu+udv
dy=(e^v)dv
dz=[2x-sin(x+y)]·(vdu+udv)+[2y-sin(x+y)]·[(e^v)dv]
=v[2x-sin(x+y)]du+{u[2x-sin(x+y)]+[2y-sin(x+y)]·[(e^v)]}dv
∂z/∂u
=v[2x-sin(x+y)]
∂z/∂v
=u[2x-sin(x+y)]+[2y-sin(x+y)]·[(e^v)]