5.3 点の運動

1.

(a) $ \displaystyle{{\bf v}(t) = (2 b t - a \pi \sin{\pi t}, \pi a \cos{\pi t} -2 b t )}$

$ \displaystyle{{\bf a}(t) = (-\pi^2 a \cos{\pi t} + 2b, - \pi^2 a \sin{\pi t} - 2b)}$

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

(b) $ \displaystyle{{\bf v}(t) = (3t^2,1 )}$ $ \displaystyle{{\bf a}(t) = (6t,0)}$ $ \displaystyle{v = \sqrt{10}}$ $ \displaystyle{\hat{{\bf t}} = \frac{(3,1)}{\sqrt{10}}}$ $ \hat{\bf n} = (1,0)$

(c) $ \displaystyle{{\bf v}(t) = ( 0,2t,2(t-1))}$ $ \displaystyle{{\bf a}(t) = (0,2,2)}$ $ v = 2$ $ \displaystyle{\hat{{\bf t}} = (0,1,0)}$ $ \displaystyle{\hat{\bf n} = \frac{(0,1,1)}{\sqrt{2}}}$

2.

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

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

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

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