# Geodesic curvature and projection onto the tangent plane

Here is exercise 10 from section 4-4 outin do Carmos's Differential Geometry of Curves and Surfaces:

Show that the geodesic curvature of an oriented curve $$C\subset S$$ at a point $$p\in C$$ is equal to the curvature of the plane curve obtained by projecting $$C$$ onto the tangent plane $$T_p(S)$$ along the normal to the surface at $$p$$.

As a hint, he writes:

Apply the relation $$k_g^2 + k_n^2 = k^2$$ and the Meusnier theorem to the projecting cylinder.

I can't see how the Meusnier theorem applies here, since it deals with curves that are on the same surface. Besides, what cylinder is he talking about?

• There are, of course, other ways to do this exercise (e.g., directly), but he's suggesting you apply Meusnier in one case, but compute the curvature of the curve in the tangent plane directly. (There is a question is signs, by the way. Geodesic curvature has a sign, whereas usually curvature of a curve is taken to be nonnegative.) Commented May 10, 2017 at 21:26

Let $\alpha$ be a unit-speed parametrization of $C$ with $\alpha(0) = p$. The projection of $\alpha$ in $T_pS$ along $N(p)$ is given by $$\beta(s) = \alpha(s)+\langle p-\alpha(s),N(p)\rangle N(p),$$whence $\beta'(s) = \alpha'(s) - \langle \alpha'(s),N(p)\rangle N(p)$ and $\beta''(s) = \alpha''(s) - k_n(\alpha'(s)) N(p)$. For $s=0$ we obtain $$\beta'(0) = \alpha'(0) \quad\mbox{and}\quad \beta''(0) = \frac{D\alpha'}{{\rm d}s}(0).$$Computing the curvature of $\beta$ at $0$ we have $$\kappa_\beta(0) = \frac{\|\beta'(0) \times \beta''(0)\|}{\|\beta'(0)\|^3} = \left\|\frac{D\alpha'}{{\rm d}s}(0)\right\| \sin \theta_1,$$where $\theta_1 = \angle(\alpha'(0), (D\alpha'/{\rm d}s)(0))$. On the other hand, the geodesic curvature of $\alpha$ at $p$ is given by: $$k_g(p) = \left\langle \frac{D\alpha'}{{\rm d}s}(0), N(p) \times \alpha'(0)\right\rangle = \left\|\frac{D\alpha'}{{\rm d}s}(0)\right\|\| N(p)\times \alpha'(0)\|\cos \theta_2 = \left\|\frac{D\alpha'}{{\rm d}s}(0)\right\|\cos \theta_2$$since $N(p)$ and $\alpha'(0)$ are orthogonal unit vectors, where $\theta_2 = \angle((D\alpha'/{\rm d}s)(0),N(p)\times \alpha'(0))$. Since $\theta_1+\theta_2 = \pi/2$, the result follows.
• I know that is an old answer, but I would appreciate if you helped me. I didn't understand why $\beta''(0)=\frac{D\alpha '}{ds}(0)$. $\beta''(0)=\alpha''(0)-\kappa_{n}(\alpha'(0))N(p)$. Why is this the projection of $\alpha''(0)$ on the plane $T_{p}S?$ Commented Nov 5, 2019 at 5:23
• the general formula for projecting $v$ onto a plane with normal vector $N$ is $v-\langle v, N\rangle N$. Now, we have $\langle \alpha '',N\rangle= \langle kn,N \rangle= k\langle n,N\rangle=k_n$. And you can see the equation for $\beta ''$ as the projection of $\alpha ''$ onto the tangent plane, i.e., $\frac{D\alpha '}{ds}$ Commented Nov 29, 2019 at 5:00
• also, you can simplify things a little bit... $k_{\beta}(0)=\|\beta''(0)\|=\|\frac{\alpha '}{ds}(0)\|=|\lambda|\|N(p) \times \alpha '(0)\|$ and because $\alpha '(0)$ and $N(p)$ are orthogonal follows that $k_{\beta}= \left[ \frac{D\alpha '}{ds}(0) \right]=k_g(p)$ Commented Nov 29, 2019 at 5:17
• What does the notation $\frac{Da'}{ds}$ mean? Commented Apr 24, 2021 at 1:53
• It's the tangent projection of $\alpha''$. Commented Apr 24, 2021 at 1:54