# Geodesic curvature and projection onto the tangent plane

Here is exercise 10 from section 4-4 out of M. 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 tip for this exercise, the autor writes:

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

First, I don'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.) – Ted Shifrin May 10 '17 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.