굴러가는 원 이동거리

점 $\mathrm{P}$에 대하여 $\mathrm{\vec{OP}}$의 위치벡터를 $\left( x , y \right)$라 하면 $\theta$가 $0$에서 $\displaystyle \frac{\pi}{2}$까지 변할 때, 점 $\mathrm{P}$가 움직인 거리는 $\displaystyle \int _{0} ^{\frac{\pi}{2}} \sqrt{\left( \frac{dx}{d \theta} \right) ^{2} + \left( \frac{dy}{d \theta} \right) ^{2}} \cdot d \theta$이다.

원 $C _{2}$의 중심을 $\mathrm{O}'$이라 하면 $\mathrm{\vec{OO}}' = \left( 5 \cos \theta , 5 \sin \theta \right)$, $\mathrm{\vec{O' P}}$$= \left( \cos ( \pi + 5 \theta ) , \sin ( \pi + 5 \theta ) \right)$$= \left( - \cos 5 \theta , - \sin 5 \theta \right)$ 이므로 $\mathrm{\vec{OP}}$$= \mathrm{\vec{OO'}} + \vec{O}' P$$= \left( 5 \cos \theta - \cos 5 \theta , 5 \sin \theta - \sin 5 \theta \right)$ ∴ $\displaystyle \frac{dx}{d \theta}$$= - 5 \sin \theta + 5 \sin 5 \theta$ $\displaystyle \frac{dy}{d \theta}$$= 5 \cos \theta - 5 \cos 5 \theta$ $\displaystyle \left( \frac{dx}{d \theta} \right) ^{2} + \left( \frac{dy}{d \theta} \right) ^{2}$ $= \left( - 5 \sin \theta + 5 \sin 5 \theta \right) ^{2} + \left( 5 \cos \theta - 5 \cos 5 \theta \right) ^{2}$ $= 50 - 50 \left( \sin \theta \sin 5 \theta + \cos \theta \cos 5 \theta \right)$ $= 50 - 50 \cos 4 \theta$ $= 50 - 50 ( 1 - 2 \sin ^{2} 2 \theta )$ $= 100 \sin ^{2} 2 \theta$ ∴ $\displaystyle \int _{0} ^{\frac{\pi}{2}} \sqrt{\left( \frac{dx}{d \theta} \right) ^{2} + \left( \frac{dy}{d \theta} \right) ^{2}} \cdot d \theta$ $\displaystyle = \int _{0} ^{\frac{\pi}{2}} 10 \sin 2 \theta d \theta$$\displaystyle = \left[ \begin{array}{l} \begin{matrix} \\ \end{matrix} - 5 \cos 2 \theta \end{array} \right] _{0} ^{\frac{\pi}{2}}$$= 10$