大师作文网这几道微积分谁会啊,
来源：学生作业帮 编辑：大师作文网作业帮 分类：综合作业 时间：2024/11/12 23:16:27
这几道微积分谁会啊,
4.dz=z*(2xy)dx+z*(x²+3y²)dy
=2xyzdx+z(x²+3y²)dy.
5.∵x/z=ln(z/y) ==> (-x/z²)(δz/δy)=[y(δz/δy)-z]/(yz)
==> (xy+yz)(δz/δy)=z²
∴δz/δy=z²/(xy+yz).
6.∵令f′x(1,-1)=5+a+=0,
令f′y(1,-1)=-2+b=0,
∴ a=-5,b=2.
∵f′′xx(1,-1)=4>0,
f′′yy(1,-1)=2>0,
f′′xy(1,-1)=-2,
又[f′′xy(1,-1)]²-f′′xx(1,-1)*f′′yy(1,-1)=-4 d(x²y)-d((x+1)²/2)=0
==> x²y-(x+1)²/2=C (C是积分常数).
∵y(1)=0,
∴C=-2,则x²y-(x+1)²/2=-2.
故2x²y-(x+1)²+4=0.
=2xyzdx+z(x²+3y²)dy.
5.∵x/z=ln(z/y) ==> (-x/z²)(δz/δy)=[y(δz/δy)-z]/(yz)
==> (xy+yz)(δz/δy)=z²
∴δz/δy=z²/(xy+yz).
6.∵令f′x(1,-1)=5+a+=0,
令f′y(1,-1)=-2+b=0,
∴ a=-5,b=2.
∵f′′xx(1,-1)=4>0,
f′′yy(1,-1)=2>0,
f′′xy(1,-1)=-2,
又[f′′xy(1,-1)]²-f′′xx(1,-1)*f′′yy(1,-1)=-4 d(x²y)-d((x+1)²/2)=0
==> x²y-(x+1)²/2=C (C是积分常数).
∵y(1)=0,
∴C=-2,则x²y-(x+1)²/2=-2.
故2x²y-(x+1)²+4=0.