• 2022-06-30
    已知\(z = {x^2} + {y^2}\),则在\(\left( {2,1} \right)\)处的全微分dz = ( )
    A: 8
    B: 6
    C: \(4dx + 2dy\)
    D: \(8dxdy\)
  • A

    举一反三

    内容

    • 0

      设X与Y相互独立,且DX=4,DY=2,随机变量Z=3X-2Y,则DZ=(). A: 8 B: 16 C: 28 D: 44

    • 1

      以点\( (2, - 1,2) \) 为球心,3为半径的球面方程为( ) A: \( {\left( {x + 2} \right)^2} + {(y - 1)^2} + {(z + 2)^2} = 9 \) B: \( {\left( {x + 2} \right)^2} + {(y - 1)^2} + {(z + 2)^2} = 3 \) C: \( {\left( {x - 2} \right)^2} + {(y + 1)^2} + {(z - 2)^2} = 9 \) D: \( {\left( {x - 2} \right)^2} + {(y + 1)^2} + {(z - 2)^2} = 3 \)

    • 2

      设\(f\left( {x,y,z} \right) = x{y^2} + y{z^2} + z{x^2}\),则\({f_{yz}}\left( {0,-1,0} \right) = \)( ) A: 1 B: 0 C: -1 D: 2

    • 3

      曲线\( \left\{ {\matrix{ { { x^2} + {y^2} = {z^2}} \cr { { z^2} = y} \cr } } \right. \)在坐标面\( yoz \) 上的投影曲线方程为( ) A: \( \left\{ {\matrix{ { { x^2} + { { \left( {y - {1 \over 2}} \right)}^2} = {1 \over 4}} \cr {z = 0} \cr } } \right. \) B: \( \left\{ {\matrix{ { { z^2} = y} \cr {x = 0} \cr } } \right. \) C: \( \left\{ {\matrix{ {z = {y^2}} \cr {x = 0} \cr } } \right. \) D: \( \left\{ {\matrix{ { { y^2} + { { \left( {x - {1 \over 2}} \right)}^2} = {1 \over 4}} \cr {z = 0} \cr } } \right. \)

    • 4

      已知\( y = {x^2} + 4x \),则\( dy \)为( ). A: \( (2x + 4)dx \) B: \( 2xdx \) C: \( ({x^2} + 4)dx \) D: \( ({x^2} + 4x)dx \)