第5章 5.34

(1)

$x = a\cos t \qquad y = a\sin t \qquad z = \frac{ht}{2\pi} \qquad \pi a h = c$

$\left(\frac{dx}{dt}\right)^2 = (-a\sin t)^2 = a^2\sin^2 t \qquad \left(\frac{dy}{dt}\right)^2 = a^2\cos^2 t \qquad \left(\frac{dz}{dt}\right)^2 = \frac{h^2}{4\pi^2}$

$s = \int_0^{2\pi} \sqrt{a^2 + \frac{h^2}{4\pi^2}}\, dt = 2\pi\sqrt{a^2 + \frac{h^2}{4\pi^2}} \qquad \left(h = \frac{c}{a\pi}\ \text{から}\right)$

$= 2\pi\sqrt{a^2 + \frac{c^2}{4a^2\pi^4}}$

$s^2 = 4\pi^2\left(a^2 + \frac{c^2}{4a^2\pi^4}\right)$

$\frac{ds^2}{da} = 4\pi^2\left(2a - \frac{2c^2}{4a^3\pi^4}\right)$

$\dfrac{ds^2}{da} = 0$ のとき $c = 2\pi^2 a^2$ これを $s$ に代入して

$s = 2\pi\sqrt{\frac{c}{2\pi^2} + \frac{c}{2\pi^2}} = 2\sqrt{c}$

(2)

$\pi a h = c$ から

$\pi a h = 2\pi^2 a^2$

$\therefore h = 2\pi a$