第1章 1.27

(1)

$\frac{dx}{dt} = \frac{1}{t^2} \qquad \frac{dy}{dt} = -\frac{1}{(t+1)^2} \qquad \frac{dy}{dx} = -\frac{t^2}{(t+1)^2}$

$\frac{d\dfrac{dy}{dx}}{dt} = -\frac{2t(t+1)^2 - t^2(t+1)\cdot 2}{(t+1)^4} = -\frac{(t+1)\{2t(t+1) - 2t^2\}}{(t+1)^4} = -\frac{2t}{(t+1)^3}$

$\frac{d^2y}{dx^2} = \frac{-\dfrac{2t}{(t+1)^3}}{\dfrac{1}{t^2}} = -\frac{2t^3}{(t+1)^3}$

(2)

$\frac{dx}{dt} = \frac{-2t}{(1+t^2)^2} \qquad \frac{dy}{dt} = \frac{4t(1+t^2) - 2t^2 \cdot 2t}{(1+t^2)^2} = \frac{4t}{(1+t^2)^2} \qquad \frac{dy}{dx} = \frac{\dfrac{4t}{(1+t^2)^2}}{\dfrac{-2t}{(1+t^2)^2}} = -2$

$\frac{d\dfrac{dy}{dx}}{dt} = 0 \qquad \frac{d^2y}{dx^2} = 0$

(3)

$\begin{cases} x = \cos^3 t \\ y = \sin^3 t \end{cases}$

$\frac{dx}{dt} = 3\cos^2 t \cdot (-\sin t) \qquad \frac{dy}{dt} = 3\sin^2 t \cdot (\cos t)$

$\frac{dy}{dx} = \frac{3\sin t \cdot \cos t}{-3\cos^2 t \cdot \sin t} = -\frac{\sin t}{\cos t} = -\tan t$

$\frac{d\dfrac{dy}{dx}}{dt} = -\frac{1}{\cos^2 t}$

$\frac{d^2 y}{dx^2} = \frac{-\dfrac{1}{\cos^2 t}}{-3\cos^2 t \cdot \sin t} = \frac{1}{3\cos^4 t \cdot \sin t}$

(4)

$\begin{cases} x = \tan^{-1} t \\ y = \log\left|t + \sqrt{t^2 + 1}\right| \end{cases}$

$\frac{dx}{dt} = \frac{1}{t^2+1} \qquad \frac{dy}{dt} = \frac{1}{\sqrt{t^2+1}} \qquad \frac{dy}{dx} = \frac{t^2+1}{\sqrt{t^2+1}} = \sqrt{t^2+1}$

$\frac{d\dfrac{dy}{dx}}{dt} = \frac{2t}{2\sqrt{t^2+1}} = \frac{t}{\sqrt{t^2+1}}$

$\frac{d^2 y}{dx^2} = \frac{t(t^2+1)}{\sqrt{t^2+1}} = t\sqrt{t^2+1}$