Show that the curvature of the curve vanishes at exactly two points on the curve

Question

Show that the curvature of the curve $$ R(t)=(\sec t, \sec t\tan t),\frac{-\pi}{2}<t<\frac{\pi}{2}$$

vanishes at exactly two points on the curve

how to prove this can any one help ...

Answer 1

For a plane curve given parametrically in Cartesian coordinates as $[x(t),y(t)]$, the curvature is given by $$\kappa = \frac{|x'y''-y'x''|}{\left(x'^2+y'^2\right)^\frac32}$$ Just apply and simplify as much as you can $$x=\sec(t)\implies x'=\tan (t) \sec (t)\implies x''=\sec (t) \left(\tan ^2(t)+\sec ^2(t)\right)$$ $$y=\tan (t) \sec (t)\implies y'=\sec (t) \left(\tan ^2(t)+\sec ^2(t)\right) \implies y''=\tan (t) \sec (t) \left(6 \sec ^2(t)-1\right)$$ Simplify as much as possible and use double angle formulae to end with $$\kappa =\sqrt{2}\frac{ |3 \cos (2 t)-1| \sec ^6(t)}{\left((5-3 \cos (2 t)) \sec ^6(t)\right)^{3/2}}$$ Now, your turn.