# Normal curvature along a line of curvature

I have come across the following exercise (the context is curves and surfaces in $\mathbb{R}^3$ and the Gauss map):

If $C=\alpha(I)$ is a line of curvature, and $k$ is its curvature at $p$, then $$k = \mid k_n k_N \mid$$ where $k_n$ is the normal curvature at $p$ along the tagent line of $C$, and $k_N$ is the curvature of the spherical image $N(C) \subset S^2$ at $N(p)$.

I am not sure I understand the question, though... because it seems to me that if $C$ is a line of curvature, the normal curvature should be identical to the curvature of $C$ itself, hence $k = k_n$. Am I mistaken?

$C$ being a line of curvature means that the tangent vector of $C$ at every point is a principal direction, not that its normal curvature is identical to its curvature.

To solve your problem, you will need a formula for curvature that doesn't use parameterization by arc-length since the gauss map doesn't always give a curve parameterized by arc length. Let's use

$k(t) = \frac{|\alpha' \times \alpha''|}{|\alpha'|^3}$ (exercise 12 in section 1-5 of Do Carmo, which you are probably using?)

$C$ being a line of curvature means that the Gauss map $N$ is such that

$N'(t) = \lambda(t) \alpha'(t)$ where $-\lambda(t)$ is the curvature in the direction of $\alpha'(t)$

If you compute the curvature $k_N$ using these two formulas you should get the result after rearranging. Comment if you have more questions :)

• Thanks, that helped me to solve the exercise! Though I am still having a hard time visualizing this curve... Commented Jul 25, 2011 at 23:03
• I tried to find a way to "see" the result too but was unsuccessful so far... Commented Jul 26, 2011 at 3:06
• @Vhailor my computations yield $k=k_N |\lambda(t)|$
– user422112
Commented Apr 4, 2019 at 16:03
• I only get $k=k_n k_2 /k_1$... Commented Apr 19, 2019 at 7:58
• @Vhailor Thank you. I can sleep at night again. :-) Commented May 27, 2020 at 18:49

Let $$\beta=N\circ\alpha$$. Without loss of generality, let's say that $$\beta^{\prime}(t)=dN_{p}(\alpha^{\prime}(t))=-k_{1}\alpha^{\prime}(t)$$, then $$\beta^{\prime\prime}(t)=-k_{1}\alpha^{\prime\prime}(t)$$. Since we don't know if $$\beta$$ is parameterized by arc length, we will use the formula for curvature.

$$\begin{eqnarray*} \left|k_{n}K_{N}\right| & = & \left|k_{1}\right|\frac{\left|\beta^{\prime}\land\beta^{\prime\prime}\right|}{\left|\beta^{\prime}\right|^{3}}\ & = & \left|k_{1}\right|\frac{\left|k_{1}\alpha^{\prime}\land k_{1}\alpha^{\prime\prime}\right|}{\left|k_{1}\alpha^{\prime}\right|^{3}}\ & = & \left|k_{1}\right|\frac{\left|k_{1}k_{1}k\right|}{\left|k_{1}\right|^{3}}\ & = & k. \end{eqnarray*}$$