第1章 1.10

(1)

$x = t^3 - t$

$y = t^2 + t \qquad (t = 1)$

$x' = 3t^2 - 1 \qquad t = 1 \text{のとき} \qquad x' = 3 - 1 = 2$

$y' = 2t + 1 \qquad t = 1 \text{のとき} \qquad y' = 2 + 1 = 3$

$x'' = 6t \qquad t = 1 \text{のとき} \qquad x'' = 6$

$y'' = 2$

これから

$\vec{v} = (x', y') = (2, 3) \qquad \vec{a} = (x'', y'') = (6, 2) \qquad v = |\vec{v}| = \sqrt{2^2 + 3^2} = \sqrt{13}$

(2)

$x = t - \sin t$

$y = 1 - \cos t \qquad \left(t = \frac{\pi}{3}\right)$

$x' = 1 - \cos t \qquad t = \frac{\pi}{3} \text{のとき} \qquad x' = 1 - \cos\frac{\pi}{3} = \frac{1}{2}$

$y' = -(-\sin t) = \sin t \qquad t = \frac{\pi}{3} \text{のとき} \qquad y' = \sin\frac{\pi}{3} = \frac{\sqrt{3}}{2}$

$x'' = \sin t \qquad t = \frac{\pi}{3} \text{のとき} \qquad x'' = \sin\frac{\pi}{3} = \frac{\sqrt{3}}{2}$

$y'' = \cos t \qquad t = \frac{\pi}{3} \text{のとき} \qquad y'' = \cos\frac{\pi}{3} = \frac{1}{2}$

これから

$\vec{v} = (x', y') = \left(\frac{1}{2}, \frac{\sqrt{3}}{2}\right) \qquad \vec{a} = (x'', y'') = \left(\frac{\sqrt{3}}{2}, \frac{1}{2}\right) \qquad v = |\vec{v}| = \sqrt{\left(\frac{1}{2}\right)^2 + \left(\frac{\sqrt{3}}{2}\right)^2} = 1$