Differential Geometry of Curves and Surfaces: Problems with Solutions

Solution:
The position vector of the curve is [tex]\vec{r}(t)=\left(t\cos t,\ t\sin t,\ 3t\right)[/tex], and its tangent vector is the derivative. By the product rule
[tex]\vec{r}\,{}'(t)=\left(\cos t-t\sin t,\ \sin t+t\cos t,\ 3\right)[/tex]
For [tex]t=\frac{3\pi}{2}[/tex] we have [tex]\cos\frac{3\pi}{2}=0[/tex] and [tex]\sin \frac{3\pi}{2}=-1[/tex], so
[tex]\vec{r}\,{}'\left(\frac{3\pi}{2}\right)=\left(0-\frac{3\pi}{2}\times (-1),\ -1+\frac{3\pi}{2}\times 0,\ 3\right)=\left(\frac{3\pi}{2},\ -1,\ 3\right)[/tex]
This vector is non-zero, so the corresponding point [tex]\left(0, -\frac{3\pi}{2}, \frac{9\pi}{2}\right)[/tex] is a regular point of the curve.

Solution:
All four curves are of the form [tex]\frac{x^2}{a^2}+\frac{y^2}{4}=1[/tex] with [tex]a=3[/tex], [tex]2[/tex], [tex]1[/tex], [tex]4[/tex], respectively. With the parametrization [tex]x=a\cos t,\ y=2\sin t[/tex] the point [tex](0, 2)[/tex] corresponds to [tex]t=\frac{\pi}{2}[/tex], and there
[tex]x'=-a\sin \frac{\pi}{2}=-a,\quad y'=2\cos\frac{\pi}{2}=0,\quad x''=-a\cos\frac{\pi}{2}=0,\quad y''=-2\sin \frac{\pi}{2}=-2[/tex]
The curvature of a plane parametric curve and the radius of curvature are
[tex]\kappa=\frac{\left|x'y''-x''y'\right|}{\left(x'^2+y'^2\right)^{3/2}}=\frac{\left|(-a)(-2)-0\right|}{\left(a^2\right)^{3/2}}=\frac{2}{a^2},\qquad R=\frac{1}{\kappa}=\frac{a^2}{2}[/tex]
For [tex]a^2=9[/tex], [tex]4[/tex], [tex]1[/tex], [tex]16[/tex] this gives (the second curve is a circle of radius [tex]2[/tex], as expected)
[tex]R_1=\frac{9}{2},\quad R_2=2,\quad R_3=\frac{1}{2},\quad R_4=8[/tex]
The largest radius, [tex]R=8[/tex], belongs to the curve [tex]\frac{x^2}{16}+\frac{y^2}{4}=1[/tex].

Solution:
First we find the parameters of [tex]M[/tex]: [tex]u=1[/tex], then [tex]u^2-2v=3[/tex] gives [tex]v=-1[/tex], and indeed [tex]u^3-3uv=1+3=4[/tex], so [tex]M=\vec{r}(1,-1)[/tex]. The partial derivatives at [tex](u,v)=(1,-1)[/tex] are
[tex]\vec{r}_u=\left(1,\ 2u,\ 3u^2-3v\right)=(1,\ 2,\ 6),\qquad \vec{r}_v=(0,\ -2,\ -3u)=(0,\ -2,\ -3)[/tex]
A normal vector of the tangent plane is their cross product:
[tex]\vec{N}=\vec{r}_u×\vec{r}_v=\begin{vmatrix}\vec{i}&\vec{j}&\vec{k}\\1&2&6\\0&-2&-3\end{vmatrix}=(6,\ 3,\ -2)\ne\vec{0}[/tex]
Since [tex]\vec{N}\ne\vec{0}[/tex], the point [tex]M[/tex] is regular and the tangent plane is
[tex]6(x-1)+3(y-3)-2(z-4)=0\ \Leftrightarrow\ 6x+3y-2z-7=0[/tex]

Solution:
Let the curve have a regular parametrization [tex]\vec{r}(t)[/tex] of class [tex]C^3[/tex] and lie in the plane through the point with position vector [tex]\vec{r}_0[/tex] and unit normal vector [tex]\vec{m}[/tex]. Then for every [tex]t[/tex]
[tex]\left(\vec{r}(t)-\vec{r}_0\right)\centerdot\vec{m}=0[/tex]
Since [tex]\vec{r}_0[/tex] and [tex]\vec{m}[/tex] are constant, differentiating three times gives
[tex]\vec{r}\,{}'\centerdot\vec{m}=0,\qquad \vec{r}\,{}''\centerdot\vec{m}=0,\qquad \vec{r}\,{}'''\centerdot\vec{m}=0[/tex]
So the vectors [tex]\vec{r}\,{}'[/tex], [tex]\vec{r}\,{}''[/tex], [tex]\vec{r}\,{}'''[/tex] are all perpendicular to [tex]\vec{m}[/tex], i.e. they are parallel to one plane. Hence they are linearly dependent and their scalar triple product is zero:
[tex]\left(\vec{r}\,{}',\vec{r}\,{}'',\vec{r}\,{}'''\right)=0[/tex]
The curvature [tex]\kappa=\frac{\left|\vec{r}\,{}'×\vec{r}\,{}''\right|}{\left|\vec{r}\,{}'\right|^3}[/tex] is non-zero, so [tex]\vec{r}\,{}'×\vec{r}\,{}''\ne\vec{0}[/tex] and the torsion is defined:
[tex]\tau=\frac{\left(\vec{r}\,{}',\vec{r}\,{}'',\vec{r}\,{}'''\right)}{\left|\vec{r}\,{}'×\vec{r}\,{}''\right|^2}=0[/tex]