Question

find radius of curvature at point (s,psi) of the curve s=a log tan (pi/4+psi/4).

Answer 1

Question #15309 Find radius of curvature at point $(s,\psi)$ of the curve $s = a\log \tan (\pi /4 + \psi /4)$.

Solution. It is impossible to calculate the radius of curvature at any point $(s,\psi)$, since tan could be negative. The formula to calculate radius of curvature (in polar coordinates)

$$R = \frac {(r ^ {2} + r _ {\psi} ^ {2}) ^ {3 / 2}}{| r ^ {2} + 2 r _ {\psi} ^ {2} - r r _ {\psi \psi} |}$$

Here $r_{\psi} = r_{\psi}'$. $r_{\psi} = a\frac{1}{\tan(\pi/4 + \psi/4)}\frac{1}{\cos^2(\pi/4 + \psi)}$, $r'' = a/16(\sec^2(\pi/4 + \psi/4) - \csc^2(\pi/4 + \psi/4))$. If you want to evaluate at some point. First evaluate $r, r_{\psi}, r_{\psi,\psi}$ and put it to the original equation. We took, for instance, $\psi = \pi/3$ and got 2 $\sqrt{\frac{a^2 + 3}{9\left|\frac{a}{(a^2 + 3)^{3/2}}\right|^2 + 3\left|\frac{a^2}{(a^2 + 3)^{3/2}}\right|^2}}$.