求函数u=x^2+y^2+z^2在椭球面x^2/a^2+y^2/b^2+z^2/c^2=1上点M.(x.,y.,z.)处

1个回答

  • 设F=x^2/a^2+y^2/b^2+z^2/c^2-1

    则其法线方向为:(Fx,Fy,Fz)=(2x/a²,2y/b²,2z/c²),此方向就是外法线方向

    将(2x/a²,2y/b²,2z/c²)化为单位向量得:(x/a²,y/b²,z/c²)/√(x²/a^4+y²/b^4+z²/c^4)

    即cosα=(x/a²)/√(x²/a^4+y²/b^4+z²/c^4)

    cosβ=(y/b²)/√(x²/a^4+y²/b^4+z²/c^4)

    cosγ=(z/c²)/√(x²/a^4+y²/b^4+z²/c^4)

    u=x^2+y^2+z^2的方向导数为:

    du/dx*cosα+du/dy*cosβ+du/dz*cosγ

    =2x*(x/a²)/√(x²/a^4+y²/b^4+z²/c^4)+2y*(y/b²)/√(x²/a^4+y²/b^4+z²/c^4)

    +2z*(z/c²)/√(x²/a^4+y²/b^4+z²/c^4)

    =2(x²/a²+y²/b²+z²/c²)/√(x²/a^4+y²/b^4+z²/c^4)

    由于x^2/a^2+y^2/b^2+z^2/c^2=1

    =2/√(x²/a^4+y²/b^4+z²/c^4)