Question #147584
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).
1
Expert's answer
2020-12-02T18:32:01-0500

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

and normal curvature, KnK_n


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

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

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 2π2\pi and it corresponds to the interval [0,2π)[0,2\pi) for possible values of tt. Thus, t=s. We have:

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

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

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

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

kg=(r,r,n)r3k_g=\frac{(r',r'',n)}{|r'|^3} , where nn is a normal vector for the surface. The normal vector is:

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

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

kg=sintcost0costsint02cost2sint1=1k_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


kn=Lu˙2+2Mu˙v˙+Nv˙2k_n=L \dot{u}^2+2M\dot{u}\dot{v}+N\dot{v}^2

Where there is the surface on the paraboloid σ(u,v)=(u,v,u2+v2)\sigma (u,v)=(u, v, u^2+v^2) then

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

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

Hence the solution for KgK_g and KnK_n


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