dg.differential geometry - mean curvature and polar tangential angle

To elaborate a little on the other answers, here's one approach: If you know the tangential angle $psi$, then you know the angle $pi/2 - psi$ between the point and the outer unit normal, i.e. the Gauss map. Using $theta$, $u(theta)$, and $pi/2 - psi(theta)$, you can write down an explicit formula for the Gauss map. Differentiate it to get the second fundamental form and take its trace (ADDED: You'll need to compute the first fundamental form to do this).

But this appears to require derivatives of both $u$ and $psi$. It's possible that $u'$ disappears from the final formula, but I wouldn't know. (ADDED: I see from a comment that $u'$ is allowed in the formula. In that case, it all definitely works.)

ADDED: Forget what I wrote above. It's easier than that. As Anton points out, the principal curvature directions are the obvious ones. Therefore, it is easy to figure out the principal curvatures in terms of the generating curve (and the circle carved out by each point on the curve). The final answer depends on $psi'$, $u$, and $u'$.

I now see why Anton was so brief. You really should work it all out yourself. It's a straightforward exercise.