Answer to Question #147584 in Differential Geometry | Topology for anjali

Question

Calculate the normal and the geodesic curvatures of the following curves on
the given surfaces:
(a) The circle γ(t) = (cost, sin t, 1) on the paraboloid σ(u, v) = (u, v, u^2 + v^2).

Answer 1

The normal curvature, k_n, is the curvature of the curve projected onto the plane containing the curve's tangent T and the surface normal u; the geodesic curvature, k_g, is the curvature of the curve projected onto the surface's tangent plane

and normal curvature, $K_n$

The unit normal vector is given by (http://mathonline.wikidot.com/the-frenet-serret-formulas): $N(s)=\frac{T'(s)}{||T'(s)||}$ , where $T(s)=\frac{\vec{r}'(s)}{||\vec{r}'(s)||}$

where $r(s)$ is an arc-length parametrization of $\vec{r}(t)$

$\vec{r}(t)$.

(a) We point out that one has an arc-length parametrization in this case. The length of the circle is $2\pi$ and it corresponds to the interval $[0,2\pi)$ for possible values of $t$. Thus, t=s. We have:

$r'(t)=(-sin(t),cos(t),0)\Rightarrow ||\vec{r}'(t)||=1.$

$T(t)=(-sin(t),cos(t),0)$; $T'(t)=(-cos(t),-sin(t),0)$;

$N(t)=(-cos(t),sin(t),0)$.

The geodesic curvature is (https://encyclopediaofmath.org/wiki/Geodesic_curvature):

$k_g=\frac{(r',r'',n)}{|r'|^3}$ , where $n$ is a normal vector for the surface. The normal vector is:

$n=(2u,2v,-1)$ (see https://mathworld.wolfram.com/NormalVector.html).

After substitution of the curve equation, we get that $n=(2cos\,t,2sin\,t,-1)$ on the points of the curve. Thus,

$k_g=\begin{vmatrix} -sin\,t & cos\,t &0 \\ -cos\,t & -sin\,t &0\\ 2cos\,t & 2sin\,t &-1 \end{vmatrix}=-1$

The normal curvature is related to the second fundamental form, and an expression for it is

$$k_n=L \dot{u}^2+2M\dot{u}\dot{v}+N\dot{v}^2$$

Where there is the surface on the paraboloid $\sigma (u,v)=(u, v, u^2+v^2)$ then

$L=\sigma_{uu} \cdot \vec{N},\ M=\sigma_{uv} \cdot \vec{N}, L=\sigma_{vv} \cdot \vec{N},$ where $\vec{N}$ is the normal vector to the surface.

But on the paraboloid, $z=0$. So, $\sigma_{uu}=\sigma_{uv}=\sigma_{vv}=0$ and also $L=M=N=0$ thus $k_n=0$

Hence the solution for $K_g$ and $K_n$