8.1 解答

8.1

1. $f(x,y,z) = 0$の法ベクトル${\bf N}$は

$${\bf N} = \nabla f = (f_{x},f_{y},f_{z})$$

点 $(x_{0},y_{0},z_{0})$を通り，法ベクトル ${\bf N} = (A,B,C)$である平面の方程式は，

$$A(x-x_{0}) + B(y - y_{0}) + C(z - z_{0}) = 0$$

(a) $z = (x^2 + y^2)^2$の法ベクトルは $(4x(x^2 + y^2),4y(x^2 + y^2),-1)$．よって点$(1,1,4)$における法ベクトルは$(8,8,-1)$．これより求める平面の方程式は

$$8(x-1) + 8(y-1) - (z - 4) = 0$$

(b) $f(x,y,z) = x^3 + y^3 - 3xyz$とおくと $f(x,y,z) = 0$の点 $(1,2,\frac{3}{2})$における法ベクトルは

$$\nabla f(1,2,\frac{3}{2}) = (3x^2 - 3yz, 3y^2 - 3xz, -3xy)\mid_{(1,2,\frac{3}{2})}$$

$$= (3-9,12-\frac{9}{2},-\frac{9}{2}) = (-6,\frac{15}{2},-\frac{9}{2})$$

これより求める平面の方程式は

$$-12(x-1) + 15(y-2) - 9(z - \frac{3}{2}) = 0$$

または，

$$8x - 10y + 6z = -9$$

(c) $f(x,y,z) = \sin(x\cos{y}) - z$とおくと $f(x,y,z) = 0$の点 $(0,\frac{\pi}{2},0)$における法ベクトルは

$$\nabla f(0,\frac{\pi}{2},0) = (\cos{y}\cos(x\cos{y}),-x\sin{y}\sin(x\cos{y}),-1)\mid_{(0,\frac{\pi}{2},0)}$$

$$= (0,0,1)$$

これより求める平面の方程式は

$$z = 0$$

2. 曲面 ${\bf r} = {\bf r}(u,v) = (x(u,v),y(u,v),z(u,v))$の面積素$dS$は

$$dS = \Vert\frac{\partial{\bf r}}{\partial u} \times \frac{\partial {\bf r}}{\partial v}\Vert du dv$$

(a) ${\bf r} = (u,v,u^2 + v^2)$

$$dS = \Vert\frac{\partial{\bf r}}{\partial x} \times \frac{\partial {\bf r}}{\partial y}\Vert du dv$$

$$= \Vert\left\vert\begin{array}{ccc} {\hat {\bf i}} & {\hat {\bf j}}... ...u\\ 0 & 1 & 2v \end{array}\right\vert\Vert du dv = \Vert(-2u,-2v,1)\Vert du dv$$

$$= \sqrt{4u^2 + 4v^2 + 1}du dv$$

(b) ${\bf r} = (u\cos{v},u\sin{v},2v)$

$$dS = \Vert\frac{\partial{\bf r}}{\partial u} \times \frac{\partial {\bf r}}{\partial v}\Vert du dv$$

$$= \Vert\left\vert\begin{array}{ccc} {\hat {\bf i}} & {\hat {\bf j}}... ... & 2 \end{array}\right\vert\Vert du dv = \Vert(2\sin{v},-2\cos{v},u)\Vert du dv$$

$$= \sqrt{4\sin^{2}{v} + 4\cos^{2}{v} + u^2}du dv = \sqrt{u^2 + 4}du dv$$

(c) $z = \sqrt{1 - x^2 - y^2}$において $u = x, v = y$とおくと

$${\bf r} = {\bf r}(x,y) = (x,y,f(x,y)) = (x,y,\sqrt{1 - x^2 - y^2})$$

よって

$${\bf r}_{x} = (1,0,f_{x}) = (1,0,-\frac{x}{\sqrt{1 - x^2 - y^2}})$$

$${\bf r}_{y} = (0,1,f_{y}) = (1,0,-\frac{y}{\sqrt{1 - x^2 - y^2}})$$

これより

$$dS = \Vert\frac{\partial{\bf r}}{\partial x} \times \frac{\partial {\bf r}}{\partial y}\Vert dx dy$$

$$= \Vert\left\vert\begin{array}{ccc} {\hat {\bf i}} & {\hat {\bf j}}... ...1 & f_{y} \end{array}\right\vert\Vert dx dy = \Vert(-f_{x},-f_{y},1)\Vert dx dy$$

$$= \sqrt{f_{x}^2 + f_{y}^2 + 1}dx dy = \sqrt{\frac{x^2}{1 - x^2 - y^2} + \frac{y^2}{1 - x^2 - y^2} + 1}$$

$$= \frac{1}{\sqrt{1 - x^2 - y^2}}dxd dy$$

(d) $F(x,y,z) = 0$のとき $\nabla F = (F_{x},F_{y},F_{z})$は曲面 $F(x,y,z) = 0$に直交．$dS$と$dx dy$の面積比は

$$dS \vert\cos{\gamma}\vert = dx dy$$

ここで，$\gamma$は$\nabla F$と ${\hat {\bf k}}$の作る角．これより

$$dS = \vert\frac{1}{\cos{\gamma}}\vert dx dy$$

$$\nabla F \cdot {\hat {\bf k}} = (F_{x},F_{y},F_{z}) \cdot (0,0,1) = F_{z}$$

また，

$$\nabla F \cdot {\hat {\bf k}} = \Vert\nabla F\Vert \cos{\gamma}$$

より

$$\cos{\gamma} = \frac{F_{z}}{\sqrt{F_{x}^2 + F_{y}^2 + F_{z}^2}}$$

したがって，

$$dS = \frac{\sqrt{F_{x}^2 + F_{y}^2 + F_{z}^2}}{\vert F_{z}\vert}dx dy$$