8.3 解答

8.3

1. 演算子 $\nabla = (\frac{\partial}{\partial x}, \frac{\partial}{\partial y} , \frac{\partial}{\partial z})$を用いると

$${\rm grad} f = \nabla f, {\rm div} {\bf F} = \nabla \cdot {\bf F}, {\rm curl} {\bf F} = \nabla \times {\bf F}$$

(a) $f(x,y,z) = 3x^2 - yz$より

$${\rm grad} f(1,-1,1) = \nabla f(1,-1,1) = (6x,-z,-y)\mid_{(1,-1,1)} =(6,-1,1)$$

(b) ${\bf F} = (3xyz^2,2xy^3,-x^2yz)$より

$${\rm div}{\bf F}(1,-1,1) = 3yz^2 + 6xy^2 - x^2y \mid_{(1,-1,1)} = -3 + 6 +1 = 4$$

(c) ${\bf F} = (3xyz^2,2xy^3,-x^2yz)$より

$${\rm curl} {\bf F} = \nabla \times {\bf F} = \left\vert\begin{array}{ccc} {\hat{\bf i}... ...& \frac{\partial}{\partial z}\\ 3xyz^2 & 2xy^3 & -x^2yz \end{array}\right\vert$$

$$= (-x^2z, 9xyz^2 + 2xyz, 2y^3 - 3xz^2)\mid_{(1,1-1)}$$

$$= (-1,-11,-5)$$

(d) ${\bf F} = (3xyz^2,2xy^3,-x^2yz)$より

$${\rm div}{\bf F} = (3xyz^2,2xy^3,-x^2yz)$$

よって

$${\rm grad}({\rm div}{\bf F}) = (6y^2 - 2xy,3z^2 + 12xy 0x^2, 6yz)\mid_{(1,-1,1)} = (8,-9,-6)$$

(e) $f(x,y,z) = 3x^2 - yz$より

$${\rm grad} f = \nabla f = (6x,-z,-y)$$

よって

$${\rm div}({\rm grad}{f}) = \nabla \cdot (6x,-z,-y) = 6$$

(f)

$${\rm curl}({\rm curl}{\bf F}) = \left\vert\begin{array}{ccc} {\hat{\bf i}} & {\hat{\bf j}} & {\ha... ...ial}{\partial z}\\ -x^2z & 9xyz^2 + 2xyz & 2y^3 - 3xz^2 \end{array}\right\vert$$

$$= (6y^2 - 18xyz - 2xy, -x^2 + 6xz, 9yz^2 + 2yz)\mid_{(1,-1,1)}$$

$$= (-6+18+2,5,-9-2) = (14,5,-11)$$

(g)

$$\nabla \times {f \bf F} = \left\vert\begin{array}{ccc} {\hat{\bf i}} & {\hat{\bf j}} & {\ha... ...-yz)(3xyz^2) & (3x^2 - yz)(2xy^3) & (3x^2 - yz)(-x^2 yz) \end{array}\right\vert$$

$$= (-z(-x^2yz) + 3x^2 - yz)(-2xyz)$$

$$, -y(3xyz^2) + (3x^2 - yz)(6xyz)$$

$$, 6x(2xy^3) + (3x^2 - yz)(2y^3))\mid_{(1,-1,1)}$$

$$= (-1 + 8 - 2, -3 - 24 -14, -12 - 8 -15) = (5, -41, -35)$$

(h)

$$\nabla \times {\nabla f} = \left\vert\begin{array}{ccc} {\hat{\bf i}} & {\hat{\bf j}} & {\ha... ...partial y} & \frac{\partial}{\partial z}\\ 6x & -z & -y \end{array}\right\vert$$

$$= (-1 + 1, 0 , 0) = (0,0,0)$$