幾何學作業@悟-2013.09.25

3
$ k(t)=\dfrac{|\alpha'\times\alpha |}{|\alpha'|^3} =\dfrac{|ab|}{(a^2\sin^2t+b^2\cos^2t)^{ \frac{3}{2} } }$ ,

$ k'(t)=-3ab\dfrac{(a^2-b^2)\sin t\cos t}{|\alpha'|^5},k'(t)=0$ when $k=0,\frac{\pi}{2},\pi,\frac{3\pi}{2}$ , so it has four vertex.

b
$A(\beta)=\frac{1}{2}\int\limits_0^l\beta\times\beta'ds=\int\limits_0^l(\alpha-rn(s))\times(\alpha'-rn'(s))ds

=\frac{1}{2}\int\limits_0^l(\alpha\times\alpha'-r(\alpha\times n'(s)+n\times\alpha')+r^2n\times n')ds

=A(\alpha)+\frac{1}{2}(2rl+2\pi r^2)$

15
Prove the Poincare inequality (say by Fourier series) and use it to prove the isoperimetric inequality:\\