T[y]=\int^{400km}_{0km}\sqrt{\frac{1+y'^2}{2gy}}dx\\ =\int^{2\pi}_{0}\sqrt{\frac{1+(\frac{dy}{d\theta}/\frac{dx}{d\theta})^2}{2gy(\theta)}}\frac{dx}{d\theta}d\theta\\ =\sqrt{\frac{A}{g}}2\pi
Embed the image to your blog in HTML, Markdown, Textile, BBCode.
+Name +Password or Edit
Recent Referers:
  1. http://d.hatena.ne.jp/ykkpurple/touch/20110608/1307517600
  2. http://d.hatena.ne.jp/ykkpurple/?of=10
  3. http://d.hatena.ne.jp/ykkpurple/20110608/1307517600
  4. http://d.hatena.ne.jp/ykkpurple/?of=5
  5. http://d.hatena.ne.jp/ykkpurple/20110610/1307715598