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).
(b) The right circular helix γ(θ) = (a cos θ, a sin θ, bθ) on the right circular cylinder
σ(u, v) = (a cos u, a sin u, v), where a, b > 0 are constants.

Answer 1

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 $\vec{r}(t)$.

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

(b) We calculate the length of the curve between $[0,2\pi)$ and get:

$I=\int_{0}^{2\pi}\sqrt{(x'(\theta))^2+(y'(\theta))^2+(z'(\theta))^2}d\theta=\int_{0}^{2\pi}\sqrt{a^2sin^2(\theta)+a^2cos^2(\theta)+b^2}d\theta=2\pi\sqrt{a^2+b^2}$

We make a change of variables: $s=\theta\sqrt{a^2+b^2}$ . Then the curve takes the form:

$x(s)=a\,cos(\frac{s}{\sqrt{a^2+b^2}}); y(s)=a\,sin(\frac{s}{\sqrt{a^2+b^2}});\,\,z(s)=b\frac{s}{\sqrt{a^2+b^2}}$ . In case we consider an interval $s\in[0,2\pi\sqrt{a^2+b^2})$ we receive an arc-length parametrization.

$\vec{r}'(s)=(-\frac{a}{\sqrt{a^2+b^2}}\,sin(\frac{s}{\sqrt{a^2+b^2}}),\frac{a}{\sqrt{a^2+b^2}}\,cos(\frac{s}{\sqrt{a^2+b^2}}),\frac{b}{\sqrt{a^2+b^2}})$ ; $||\vec{r}'(s)||=1$.

$T(s)=\vec{r}'(s)$ ; $T'(s)=(-\frac{a}{{a^2+b^2}}\,sin(\frac{s}{\sqrt{a^2+b^2}}),\frac{a}{{a^2+b^2}}\,cos(\frac{s}{{a^2+b^2}}),0)$ ;

$N'(s)=\frac{T'(s)}{||T'(s)||}=\frac{a^2+b^2}{|a|}(-\frac{a}{{a^2+b^2}}\,sin(\frac{s}{\sqrt{a^2+b^2}}),\frac{a}{{a^2+b^2}}\,cos(\frac{s}{{a^2+b^2}}),0)$

The geodesic curvature is: $k_g=\frac{(r',r'',n)}{|r'|^3}$ .The normal vector for the parametric surface is: $n=r_u\times r_v=\begin{vmatrix} i & j& k \\ -a\,sin\,u & a\,cos\,u&0\\ 0 & 0&1 \end{vmatrix}=(a\,cos\,u,a\,sin\,u,0)$ . We set $u=\theta$ and get the following curvature:

$k_g=\frac{(r',r'',n)}{|r'|^3}=\frac{\begin{vmatrix} -a\,sin\,\theta & a\,cos\,\theta\,&b \\ -a\,cos\,\theta & -a\,sin\,\theta&0\\ a\,cos\,\theta & a\,sin\,\theta&0 \end{vmatrix}}{(a^2+b^2)^{3/2}}=\frac{1}{(a^2+b^2)^{3/2}}b\,\,0=0$