5.3 解答

5.3

1.

1. 位置ベクトル ${\bf r}(t)$に対して， ${\bf v}(t) = {\bf r}'(t)$は速度ベクトルを意味する．幾何学的には運動している物体の接線方向のベクトルである．また， ${\bf a}(t) = {\bf v}'(t) = {\bf r}''(t)$は加速度ベクトルで，幾何学的には，接線方向の加速度ベクトル ${\bf a}_{\hat{\bf t}}$と法線方向の加速度ベクトル ${\bf a}_{\hat{\bf n}}$の和で表わされる．つまり

$${\bf a} = {\bf a}_{\hat{\bf t}} + {\bf a}_{\hat{\bf n}}$$

ここで， $\hat{\bf t}$は接線単位ベクトル， $\hat{\bf n}$は法線単位ベクトルを表わす． 言い換えると，

$$\hat{\bf t} = \frac{{\bf v}(t)}{\Vert{\bf v}(t)\Vert}$$

(a)

$${\bf v}(t) = {\bf r}'(t) = (2 b t - a \pi \sin{\pi t}, \pi a \cos{\pi t} -2 b t )$$

$${\bf a}(t) = {\bf v}'(t) = (-\pi^2 a \cos{\pi t} + 2b, - \pi^2 a \sin{\pi t} - 2b)$$

より$t = 1$のとき

$${\bf v}(1) = (2b,\pi a - 2b), {\bf a}(1) = (\pi^2 a + 2b, -2b)$$

となるので，

$$v = \Vert{\bf v}(1)\Vert = \sqrt{4b^2 + (\pi a - 2b)^2}$$

$$\hat{{\bf t}} = \frac{{\bf v}(1)}{v} = \frac{(2b,\pi a - 2b)}{\sqrt{4b^2 + (\pi a - 2b)^2}}$$

(b) ${\bf r}(t) = (t^3, t^2)$．

$${\bf v}(t) = {\bf r}'(t) = (3t^2,1 )$$

$${\bf a}(t) = {\bf v}'(t) = (6t,0)$$

より$t = 1$のとき

$${\bf v}(1) = (3,1), {\bf a}(1) = (6,0)$$

となるので，

$$v = \Vert{\bf v}(1)\Vert = \sqrt{10}$$

$$\hat{{\bf t}} = \frac{{\bf v}(1)}{v} = \frac{(3,1)}{\sqrt{10}}$$

最後に， $\hat{\bf n}$を求める．

$${\bf a}_{\hat{\bf t}} = ({\bf a}\cdot \hat{\bf t})\hat{\bf t} = ((6,0)\cdot \frac{(3,1)}{\sqrt{10}})\frac{(3,1)}{\sqrt{10}}$$

$$= \frac{(27,9)}{5}$$

より，

$${\bf a}_{\hat{\bf n}} = {\bf a} - {\bf a}_{\hat{\bf t}}$$

$$= (6,0) - \frac{(27,9)}{5} = \frac{(3,-9)}{5}$$

ここで， ${\bf a}_{\hat{\bf n}} = (x,y)$とおくと， ${\bf a}_{\hat{\bf n}} = ({\bf a}\cdot \hat{\bf n})\hat{\bf n}$より

$$(6,0)\cdot(x,y) (x,y) = \frac{(3,-9)}{5}$$

これを$x,y$について解くと

$$6x(x,y) = \frac{(3,-9)}{5}$$

より $x = \frac{1}{\sqrt{10}}, y = -\frac{3}{\sqrt{10}}$

(c)

$${\bf v}(t) = {\bf r}'(t) = ( 0,2t,2(t-1))$$

$${\bf a}(t) = {\bf v}'(t) = (0,2,2)$$

より$t = 1$のとき

$$v = 2, \hat{{\bf t}} = (0,1,0)$$

最後に， $\hat{\bf n}$を求める．

$$\hat{\bf t}(t) = \frac{{\bf v}(t)}{\Vert{\bf v}(t)\Vert} = \frac{(0,2t,2(t-1))}{\sqrt{4t^2 + 4t^2 - 8t + 4}}$$

$$= \frac{(0,t,t-1)}{\sqrt{2t^2 - 2t + 1}}$$

$$\hat{\bf t}'(t) = \frac{-(4t-2)}{(2t^2 - 2t +1)^{3/2}}(0,t,t-1) + \frac{1}{\sqrt{2t^2 - 2t + 1}}(0,1,1)$$

より

$$\hat{\bf t}'(1) = -2(0,1,0) + (0,1,1) = (0,-1,1)$$

したがって，

$$\hat{\bf n}(1) = \frac{\hat{\bf t}'(1)}{\Vert\hat{\bf t}'(1)\Vert} = \frac{(0,-1,1)}{\sqrt{2}}$$

2.

(a) ${\bf r}(t) = (t,t^2,\frac{2}{3}t^3)$

$${\bf v}(t) = {\bf r}'(t) = ( 1,2t,2t^2)$$

$${\bf a}(t) = {\bf v}'(t) = (0,2,4t)$$

より

$${\hat{\bf t}} = \frac{{\bf v}(t)}{\Vert{\bf v}(t)\Vert} = \left(\frac{1}{1 + 2t^2},\frac{2t}{1 + 2t^2},\frac{2t^2}{1 + 2t^2}\right)$$

$$\kappa = \frac{2}{(1 - 2t^2)}, \hat{{\bf n}} = \left(\frac{-2t}{1 + 2t^2},\frac{1 - 2t^2}{1 + 2t^2},\frac{2t}{1 + 2t^2}\right)$$

$$\hat{\bf b} = \left(\frac{2t^2}{1 + 2t^2},\frac{-2t}{1 + 2t^2},\frac{1}{1 + 2t^2}\right),\tau = \frac{2}{(1 + 2t^2)}$$

(b)

$$\hat{\bf t} = \frac{1}{2}\left(-\sin{t},\cos{t},1\right)$$

$$\kappa = \frac{1}{2},\hat{{\bf n}} = \left(-\cos{t},-\sin{t},0\right)$$

$$\hat{\bf b} = \frac{1}{2}\left(\sin{t},-\cos{t},1\right)$$

$$\tau = \frac{1}{2}$$