Question

求高手带做几道高数题，我的邮箱275645313@qq.com

求求求~~

Answer 1

1
2z+x^2y+cos(x-z)=0
2z'x+2xy-z'xsin(x-z)=0 z'x=2xy/[2-sin(x-z)]
2z'y+x^2-z'ysin(x-z)=0 z'y=x^2/[2-sin(x-z)]
dz=z'xdx+z'ydy
2
z=xy+x^2+y^2
z'x=y+2x
z'y=x+2y
z'x=0.z'y=0
x=0,y=0,z极值=0
3
y=x^2,y=1/x交于M(1,1)
y=x^2,x=√y y=1/x,x=1/y
∫D∫xydxdy= ∫[1,3]y∫[1/y,√y]xdx dy
=∫[1,3]y(y/2-1/2y^2)dy=∫[1,3](y^2/2-1/2y)dy=27/6-(1/2)ln3-1/6=14/3-(1/2)ln3

4 x=ρcosθ,y=ρsinθ dxdy=(1/2)ρ^2dρdθ 1<=x^2+y^2<=4 1<= ρ<=2
∫D∫e^√(x^2+y^2)dxdy=4∫[1,2]e^ρ*(1/2)ρ^2∫[0,π/2]dθ dρ
=π∫[1,2]e^ρ*ρ^2dρ
=π(4e^2-e) -2π∫[1,2]ρde^ρ
=π(4e^2-e)-2π*(2e^2-e)+2π∫[1,2]e^ρdρ
=πe+2π*(e^2-e)=2πe^2-πe

5
dx/dy=x/(x^2-y) (x^2-y)dx=xdy x^2dx=xdy+ydx
x^3/3=xy+C
6
dy/dx+y/x=1
xdy+ydx=xdx
xy=x^2/2+C
7
y'=(1/2)+2y/x
xdy=(x/2)dx+2ydx
y=xu,dy=xdu+udx
x^2du+xudx=xdx/2+2xudx
x^2du-xudx=xdx/2
xdu-udx=dx/2
(xdu-udx)/x^2=dx/(2x^2)
d(u/x)=d(-1/2x)
u/x=-1/(2x)+C
u=-1/2+Cx
y=Cx^2-x/2