2018年度関西大学理系数学第3問 - ある学生の理系科目勉強の部屋

f:id:luv_zawazawa:20180305113240j:image

(1)

$\displaystyle\frac{dx}{d\theta}=\sin{\theta}\cos{\theta}-(1-\cos{\theta})\sin{\theta}$

$=2\sin{\theta}\cos{\theta}-\sin{\theta}=\sin{2\theta}-\sin{\theta}$

(2)

$\displaystyle\frac{dy}{d\theta}=\sin^{2}{\theta}+(1-\cos{\theta})\cos{\theta}$

$=\sin^{2}{\theta}-\cos^{2}{\theta}+\cos{\theta}=-\cos{2\theta}+\cos{\theta}$

$=-2\cos^{2}{\theta}+\cos{\theta}+1=-(2\cos{\theta}+1)(\cos{\theta}-1)$

$\displaystyle\frac{dy}{d\theta}=0$ とすると、

$(2\cos{\theta}+1)(\cos{\theta}-1)=0$

$\displaystyle\theta=\frac{2}{3}\pi,~\frac{4}{3}\pi$

増減表は、以下の通りである。

$\theta$	0		$\frac{2}{3}\pi$		$\frac{4}{3}\pi$		$2\pi$
$\frac{dy}{d\theta}$	×	$+$	0	$-$	0	$+$	×
$y$	×	$\nearrow$	極大値	$\searrow$	極小値	$\nearrow$	×

したがって、

$\displaystyle\theta=\frac{2}{3}\pi$ のとき、極大値 $\displaystyle\left(1+\frac{1}{2}\right)\times\frac{\sqrt{3}}{2}=\frac{3\sqrt{3}}{4}$

$\displaystyle\theta=\frac{4}{3}\pi$ のとき、極小値 $\displaystyle\left(1+\frac{1}{2}\right)\times\left(-\frac{\sqrt{3}}{2}\right)=-\frac{3\sqrt{3}}{4}$

(3) 曲線の長さ

$\displaystyle dL=\sqrt{dx^{2}+dy^{2}}=\sqrt{\left(\frac{dx}{d\theta}\right)^{2}+\left(\frac{dy}{d\theta}\right)^{2}}d\theta$ より

$\displaystyle L=\int_{0}^{2\pi}\sqrt{\left(\frac{dx}{d\theta}\right)^{2}+\left(\frac{dy}{d\theta}\right)^{2}}d\theta$

$\displaystyle=\int_{0}^{2\pi}\sqrt{\left(\sin{2\theta}-\sin{\theta}\right)^{2}+\left(-\cos{2\theta}+\cos{\theta}\right)^{2}}d\theta$

$\displaystyle=\int_{0}^{2\pi}\sqrt{\sin^{2}{2\theta}+\cos^{2}{2\theta}+\sin^{2}{\theta}+\cos^{2}{\theta}-2\sin{2\theta}\sin{\theta}-2\cos{2\theta}\cos{\theta}}d\theta$

$\displaystyle=\int_{0}^{2\pi}\sqrt{2-2\left(\sin{2\theta}\sin{\theta}+\cos{2\theta}\cos{\theta}\right)}d\theta$

$\displaystyle=\int_{0}^{2\pi}\sqrt{2-2\cos{(2\theta-\theta)}}d\theta$

$\displaystyle=\sqrt{2}\int_{0}^{2\pi}\sqrt{1-\cos{\theta}}d\theta$

$\displaystyle=\sqrt{2}\int_{0}^{2\pi}\sqrt{2\sin^{2}\frac{\theta}{2}}d\theta$

$\displaystyle=2\int_{0}^{2\pi}\sin\frac{\theta}{2}d\theta$

$\displaystyle=2\left[-2\cos{\frac{\theta}{2}}\right]_{0}^{2\pi}$

$=-4(-1-1)=8$