8.1 解答

8.1

1. $f(x,y,z) = 0$ の法ベクトル ${\bf N}$ は

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

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，

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

(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

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

(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

$\displaystyle \nabla f(1,2,\frac{3}{2})$	$\displaystyle =$	$\displaystyle (3x^2 - 3yz, 3y^2 - 3xz, -3xy)\mid_{(1,2,\frac{3}{2})}$
	$\displaystyle =$	$\displaystyle (3-9,12-\frac{9}{2},-\frac{9}{2}) = (-6,\frac{15}{2},-\frac{9}{2})$

これより求める平面の方程式は

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

または，

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

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

これより求める平面の方程式は

$\displaystyle z = 0$

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

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

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

$\displaystyle dS$	$\displaystyle =$	$\displaystyle \Vert\frac{\partial{\bf r}}{\partial x} \times \frac{\partial {\bf r}}{\partial y}\Vert du dv$
	$\displaystyle =$	$\displaystyle \Vert\left\vert\begin{array}{ccc} {\hat {\bf i}} & {\hat {\bf j}}... ...u\\ 0 & 1 & 2v \end{array}\right\vert\Vert du dv = \Vert(-2u,-2v,1)\Vert du dv$
	$\displaystyle =$	$\displaystyle \sqrt{4u^2 + 4v^2 + 1}du dv$

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

$\displaystyle dS$	$\displaystyle =$	$\displaystyle \Vert\frac{\partial{\bf r}}{\partial u} \times \frac{\partial {\bf r}}{\partial v}\Vert du dv$
	$\displaystyle =$	$\displaystyle \Vert\left\vert\begin{array}{ccc} {\hat {\bf i}} & {\hat {\bf j}}... ... & 2 \end{array}\right\vert\Vert du dv = \Vert(2\sin{v},-2\cos{v},u)\Vert du dv$
	$\displaystyle =$	$\displaystyle \sqrt{4\sin^{2}{v} + 4\cos^{2}{v} + u^2}du dv = \sqrt{u^2 + 4}du dv$

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

よって

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

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

これより

$\displaystyle dS$	$\displaystyle =$	$\displaystyle \Vert\frac{\partial{\bf r}}{\partial x} \times \frac{\partial {\bf r}}{\partial y}\Vert dx dy$
	$\displaystyle =$	$\displaystyle \Vert\left\vert\begin{array}{ccc} {\hat {\bf i}} & {\hat {\bf j}}... ...1 & f_{y} \end{array}\right\vert\Vert dx dy = \Vert(-f_{x},-f_{y},1)\Vert dx dy$
	$\displaystyle =$	$\displaystyle \sqrt{f_{x}^2 + f_{y}^2 + 1}dx dy = \sqrt{\frac{x^2}{1 - x^2 - y^2} + \frac{y^2}{1 - x^2 - y^2} + 1}$
	$\displaystyle =$	$\displaystyle \frac{1}{\sqrt{1 - x^2 - y^2}}dxd dy$