5.1 解答

5.1

1. ベクトル関数の微分，積分はそれぞれの成分の微分，積分で求められる．

(a)

$\displaystyle {\bf F}'(t)$	$\displaystyle =$	$\displaystyle ((\sin{t})',(\cos^{2}{t})', (t^2)')$
	$\displaystyle =$	$\displaystyle (\cos{t}, -2\sin{t}\cos{t}, 2t)$

(b) $\sum \frac{n}{3^{n}}$ について考える．

$\displaystyle {\bf F}'(t)$	$\displaystyle =$	$\displaystyle (\int \frac{1}{\sqrt{1 + t^2}}dt ,\int \frac{1}{1+t^2}dt, \int \tan{t}dt)$
	$\displaystyle =$	$\displaystyle (\log\vert 1 + \sqrt{1 + t^2}\vert,\tan^{-1}{t}, \log\vert sec{t})$

$\displaystyle ({\bf F}\cdot {\bf G})(t) = f_{1}(t)g_{1}(t) + f_{2}(t)g_{2}(t) + f_{3}(t)g_{3}(t)$

より

$\displaystyle ({\bf F}(t)\cdot {\bf G}(t))'$	$\displaystyle =$	$\displaystyle f_{1}'(t)g_{1}(t) + f_{1}(t)g_{1}'(t) + f_{2}'(t)g_{2}(t) + f_{2}(t)g_{2}'(t) + f_{3}'(t)g_{3}(t) + f_{3}(t)g_{3}'(t)$
	$\displaystyle =$	$\displaystyle (f_{1}(t),f_{2}(t),f_{3}(t)) \cdot (g_{1}'(t),g_{2}'(t), g_{3}'(t))$
	$\displaystyle +$	$\displaystyle (f_{1}'(t),f_{2}'(t),f_{3}'(t)) \cdot (g_{1}(t),g_{2}(t), g_{3}(t))$
	$\displaystyle =$	$\displaystyle F'(t)\cdot G(t) + F(t) \cdot G'(t)$

(d) ${\bf F}(t) = (f_{1}(t),f_{2}(t), f_{3}(t)), G(t) = (g_{1}(t),g_{2}(t),g_{3}(t))$ とおくと

$\displaystyle ({\bf F} \times {\bf G})(t) = \left\vert\begin{array}{ccc} \vec{... ... f_{2}(t) & f_{3}(t)\\ g_{1}(t) & g_{2}(t) & g_{3}(t) \end{array}\right\vert$

より

$\displaystyle ({\bf F} \times {\bf G})'(t)$	$\displaystyle =$	$\displaystyle \left\vert\begin{array}{ccc} \vec{i} & \vec{j} & \vec{k}\\ f_{1}... ...f_{2}(t) & f_{3}(t)\\ g_{1}'(t) & g_{2}'(t) & g_{3}'(t) \end{array}\right\vert$
	$\displaystyle =$	$\displaystyle {\bf F}'(t) \times {\bf G}(t) + {\bf F}(t) \times {\bf G}'(t)$