6.曲线 ({x)^2+(y)^2) y=4 .

6.曲线,在点(2,4,5)处的切线与直线之间的夹角为( ).

参考答案与解析:

相关试题

球面^2+(y)^2+(z)^2+4x+6y+2z+10=0的球心坐标为A.^2+(y)^2+(z)^2+4x+6y+2z+10=0B.^2+(y)^2+(z)^2+4x+6y+2z+10=0C.^2

球面^2+(y)^2+(z)^2+4x+6y+2z+10=0的球心坐标为A.^2+(y)^2+(z)^2+4x+6y+2z+10=0B.^2+(y)^2+(z)

  • 查看答案
  • 6.讨论下列函数的连续性:-|||-(1) (x,y)=dfrac ({x)^2-(y)^2}({x)^2+(y)^2}-|||-(2) (x,y)=dfrac (x-y)(x+y);-|||-(3)

    6.讨论下列函数的连续性:-|||-(1) (x,y)=dfrac ({x)^2-(y)^2}({x)^2+(y)^2}-|||-(2) (x,y)=dfrac

  • 查看答案
  • 设:(x)^2+(y)^2=(a)^2 是正向闭曲线,计算曲线积分 :(x)^2+(y)^2=(a)^2

    设:(x)^2+(y)^2=(a)^2 是正向闭曲线,计算曲线积分 :(x)^2+(y)^2=(a)^2设是正向闭曲线,计算曲线积分

  • 查看答案
  • 设函数y=f(x)由方程(y)^2+(y)^2ln x-4=0所确定,则(y)^2+(y)^2ln x-4=0= ( )(y)^2+(y)^2ln x-4=0(y)^2+(y)^2ln x-4=0(y

    设函数y=f(x)由方程(y)^2+(y)^2ln x-4=0所确定,则(y)^2+(y)^2ln x-4=0= ( )(y)^2+(y)^2ln x-4=0(

  • 查看答案
  • 6.设 (x,y)=dfrac (x-{y)^2+(y)^3}(2x+{y)^2}, 则,lim h(x,y)等于 ()-|||-(A) dfrac (1)(2) (B)1-|||-(C) -1 (D

    6.设 (x,y)=dfrac (x-{y)^2+(y)^3}(2x+{y)^2}, 则,lim h(x,y)等于 ()-|||-(A) dfrac (1)(2

  • 查看答案
  • ( A ) = (x,y,z)|{x)^2+(y)^2+(z)^2=(a)^2,zgeqslant 0} ( B ) = (x,y,z)|{x)^2+(y)^2+(z)^2=(a)^2,zgeqsl

    ( A ) = (x,y,z)|{x)^2+(y)^2+(z)^2=(a)^2,zgeqslant 0} ( B ) = (x,y,z)|{x)^2+(y)^

  • 查看答案
  • 求旋转曲面=(x)^2+(y)^2在点=(x)^2+(y)^2处的法线方程A.=(x)^2+(y)^2B.=(x)^2+(y)^2C.=(x)^2+(y)^2D.=(x)^2+(y)^2

    求旋转曲面=(x)^2+(y)^2在点=(x)^2+(y)^2处的法线方程A.=(x)^2+(y)^2B.=(x)^2+(y)^2C.=(x)^2+(y)^2D

  • 查看答案
  • 6.用适当的变换计算下列二重积分:-|||-(1) iint sqrt ({x)^2+(y)^2}dxdy = (x,y)|{x)^2+(y)^2leqslant 2x} ;-|||-2)x^2 dx

    6.用适当的变换计算下列二重积分:-|||-(1) iint sqrt ({x)^2+(y)^2}dxdy = (x,y)|{x)^2+(y)^2leqslan

  • 查看答案
  • 设f(x,y)= ^4+{y)^2},(x)^2+(y)^2neq 0 0,(x)^2+(y)^2=0 .处是否连续?

    设f(x,y)= ^4+{y)^2},(x)^2+(y)^2neq 0 0,(x)^2+(y)^2=0 .处是否连续?设,试问在点处是否连续?

  • 查看答案
  • 设函数 y=y(x) 由方程 ^3+x(y)^2+(x)^2y+6=0 确定,设函数 y=y(x) 由方程 ^3+x(y)^2+(x)^2y+6=0 确定,

    设函数 y=y(x) 由方程 ^3+x(y)^2+(x)^2y+6=0 确定,设函数 y=y(x) 由方程 ^3+x(y)^2+(x)^2y+6=0 确定,

  • 查看答案