promotion image of download ymail app
Promoted
Anonymous
Anonymous asked in 科學數學 · 1 decade ago

工數向量場散度及旋度

麻煩大大幫我求一下

向量場散度及旋度

c向量=[sin(xy),cos(x^2y),e^2x]

1 Answer

Rating
  • 1 decade ago
    Favorite Answer

    Note: 粗體紅字表向量

    向量函數F=(F1, F2, F3)

    散度 div(F)=∇‧F=∂F1/∂x+∂F2/∂y+∂F3/∂z

    旋度 curl(F)=∇xF=

    | i j k |

    |∂x ∂y ∂z |

    | F1 F2 F3 |

    本題C=( sin(xy), cos(x^2y) e^(2x) )

    ∇‧C = ∂/∂x [ sin(xy)] + ∂/∂y [ cos(x^2y) ] + ∂/∂z [ e^(2x)]

    = y cos(xy) - x^2 sin(x^2 y)

    ∇xC=( u, v, w)

    u=∂/∂y [e^(2x)] - ∂/∂z [ cos(x^2 y)]= 0

    v=∂/∂z [ sin(xy) ] - ∂/∂x [ e^(2x) ]= - 2e^(2x)

    w=∂/∂x [ cos(x^2 y) ] - ∂/∂y [sin(xy)]= -2xy sin(x^2 y)- x cos(xy)

    2009-06-21 15:52:44 補充:

    框框表倒三角"∇"

    • Commenter avatarLogin to reply the answers
Still have questions? Get your answers by asking now.