第6章 6.11

(1)

右図のようにA,P,Hを取ると

$AH = r\sin\theta$

$PH = r - r\cos\theta = r(1-\cos\theta)$

$\overset{\frown}{AP} = r\theta$

(2)

$\lim_{\theta\to 0}\frac{PH}{AH^2} = \lim_{\theta\to 0}\frac{r(1-\cos\theta)}{r^2\sin^2\theta} = \lim_{\theta\to 0}\frac{r(1-\cos\theta)}{r^2(1-\cos^2\theta)} = \lim_{\theta\to 0}\frac{r(1-\cos\theta)}{r^2(1-\cos\theta)(1+\cos\theta)} = \lim_{\theta\to 0}\frac{1}{r(1+\cos\theta)} = \frac{1}{2r}$

$\lim_{\theta\to 0}\frac{PH}{\overset{\frown}{AP}^2} = \lim_{\theta\to 0}\frac{r(1-\cos\theta)}{r^2\theta^2}$

ここで$\theta \to 0$のとき$\sin\theta = \theta$と近似できるのでこれから

$= \lim_{\theta\to 0}\frac{r(1-\cos\theta)}{r^2\sin^2\theta} = \frac{1}{2r}$