Explanation of implication

Question

Can someone please explain how $$|\sinh \varphi|T_1(s)=T(s)+ \cosh\varphi\cdot u$$ implies $$\sinh^2 \varphi \kappa_1(s)N_1(s)=\kappa(s)N(s),$$ where $\varphi$ and $u$ are constant, $T_1$ and $T$ are tangent vectors of curve $c_1,$ resp. $c,$ $N_1$ and $N$ normal vectors and $\kappa$ its curvature? Both curves are parametrized with arc-length parameter.

Answer 1

Perhaps you better go back to the beginning of the question. As it stands, the problem doesn't make sense, unless we know that $u\cdot T = -\text{sech}\,\phi$.

If I'm right, we're looking at the curve $c_1$ given by $$c_1(s) = c(s) + (\cosh \phi)s u.$$ Then we can have $T_1$ be the unit tangent vector of $c_1$, but $c_1$ is not arclength-parametrized. Indeed, the speed of $c_1$ will be $|\sinh\phi|$ and you need to use the chain rule correction to compute that $T_1'(s) = \upsilon(s)\kappa_1(s) N_1(s)$, where $\upsilon(s)=|\sinh\phi|$ is the speed of $c_1$. This gives you $$|\sinh\phi|\kappa_1 N_1 |\sinh\phi| = \kappa N,$$ as required.

(For the discussion of the chain-rule correction of non-arclength parametrized curves, see pp. 12-14 of my differential geometry text.)