第1章 1.23

(1)

$\begin{cases} x = \dfrac{1}{t+1} \\ y = \dfrac{1}{t-1} \end{cases} \quad (t = 2)$

$\dfrac{dx}{dt} = -\dfrac{1}{(t+1)^2} \qquad t = 2 \text{ のとき} \qquad \dfrac{dx}{dt} = -\dfrac{1}{9}$

$\dfrac{dy}{dt} = -\dfrac{1}{(t-1)^2} \qquad t = 2 \text{ のとき} \qquad \dfrac{dy}{dt} = -1$

$\dfrac{d^2x}{dt^2} = \dfrac{2}{(t+1)^3} \qquad t = 2 \text{ のとき} \qquad \dfrac{d^2x}{dt^2} = \dfrac{2}{27}$

$\dfrac{d^2y}{dt^2} = \dfrac{2}{(t-1)^3} \qquad t = 2 \text{ のとき} \qquad \dfrac{d^2y}{dt^2} = 2$

これから $\vec{v} = (-\dfrac{1}{9}, -1)$ $\vec{a} = (\dfrac{2}{27}, 2)$

(2)

$\begin{cases} x = \dfrac{1}{(t+1)^2} \\ y = t^2 + 3 \end{cases} \quad (t = 0)$

$\dfrac{dx}{dt} = \dfrac{-2}{(t+1)^3} \qquad t = 0 \text{ のとき} \qquad \dfrac{dx}{dt} = -2$

$\dfrac{dy}{dt} = 2t \qquad t = 0 \text{ のとき} \qquad \dfrac{dy}{dt} = 0$

$\dfrac{d^2x}{dt^2} = \dfrac{6}{(t+1)^4} \qquad t = 0 \text{ のとき} \qquad \dfrac{d^2x}{dt^2} = 6$

$\dfrac{d^2y}{dt^2} = 2$

これから $\vec{v} = (-2, 0)$ $\vec{a} = (6, 2)$

(3)

$\begin{cases} x = \sqrt{t+2} \\ y = t^2 - 3 \end{cases} \quad (t=2)$

$\frac{dx}{dt} = \frac{1}{2\sqrt{t+2}} \qquad t=2\text{のとき} \qquad \frac{dx}{dt} = \frac{1}{4}$

$\frac{dy}{dt} = 2t \qquad t=2\text{のとき} \qquad \frac{dy}{dt} = 4$

$\frac{d^2x}{dt^2} = -\frac{1}{4\sqrt{(t+2)^3}} \qquad t=2\text{のとき} \qquad \frac{d^2x}{dt^2} = -\frac{1}{32}$

$\frac{d^2y}{dt^2} = 2$

これから $\vec{v} = (\dfrac{1}{4}, 4)$ $\vec{a} = (-\dfrac{1}{32}, 2)$

(4)

$\begin{cases} x = \cos t + t\sin t \\ y = \sin t - t\cos t \end{cases} \quad \left(t = \dfrac{\pi}{2}\right)$

$\frac{dx}{dt} = -\sin t + \sin t + t\cos t = t\cos t \qquad t=\dfrac{\pi}{2}\text{のとき} \qquad \frac{dx}{dt} = 0$

$\frac{dy}{dt} = \cos t - \cos t + t\sin t = t\sin t \qquad t=\dfrac{\pi}{2}\text{のとき} \qquad \frac{dy}{dt} = \frac{\pi}{2}$

$\frac{d^2x}{dt^2} = \cos t - t\sin t \qquad t=\dfrac{\pi}{2}\text{のとき} \qquad \frac{d^2x}{dt^2} = -\frac{\pi}{2}$

$\frac{d^2y}{dt^2} = \sin t - t\cos t \qquad t=\dfrac{\pi}{2}\text{のとき} \qquad \frac{d^2x}{dt^2} = 1$

これから $\vec{v} = (0, \dfrac{\pi}{2})$ $\vec{a} = (-\dfrac{\pi}{2}, 1)$