Test for concavity.

Question

Prove that $r \cos n\theta = a$ is concave towards the pole . I tried $u= \frac{1}{r}$, and calculate $ u+ \frac{d^2 u}{d\theta^2}$, we get

$\frac{1-n^2}{a}\cos n\theta$, then how to conclude about concavity

Answer 1

If you look here, Robert Israel explained that the curve would be convex if

$$C=r^2 + 2 (r')^2 - r r'' \ge 0$$ Using $$r=a \sec (n t)\qquad r'=a n \tan (n t) \sec (n t)\qquad r''=a n^2 \sec (n t) \left(2 \sec ^2(n t)-1\right)$$ you should arrive to $$C=-a^2 \left(n^2-1\right) \sec ^2(n t)$$ Now, you can use series expansion around $t=\frac{\pi }{2 n}$ and get $$\sec (n t)=-\frac{1}{n \left(t-\frac{\pi }{2 n}\right)}-\frac{1}{6} n \left(t-\frac{\pi }{2 n}\right)+O\left(\left(t-\frac{\pi }{2 n}\right)^3\right)$$ $$\sec^2(n t)=\frac{1}{n^2 \left(t-\frac{\pi }{2 n}\right)^2}+\frac{1}{3}+O\left(\left(t-\frac{\pi }{2 n}\right)^2\right)$$

Just continue.